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2018-02-21T10:46:49.140 | <p>I have been struggling to find which node and constraint suffering from 'nodes are used more than once as a slave node' in abaqus error message.</p>
<p>Please help me resolve this problem
Thank you </p>
| |mechanical-engineering|structural-engineering|finite-element-method|abaqus| | <p>in many cases where this type error is detected during job submission a node set is created to identify the offending nodes.</p>
<p>open the odb in the visualization module, go into display group manager, look for node sets with names like "ERROR...." or "WARNING...".</p>
| 19534 | How to know which constraints suffer from 'nodes are used more than once as slave node' in abaqus? |
2018-02-21T20:19:01.007 | <p>I came up with this concept for a partial rotation, lubricant free, low friction bearing for a robotics application. Two complementary sets of tensioned tendon-like cables (or tapes) serve both to constrain the degrees of freedom of the joint as well as the rolling surface. Here's a render of a model I made to illustrate the idea:</p>
<p><a href="https://i.stack.imgur.com/AGUpe.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/AGUpe.jpg" alt=""></a>
<a href="https://i.stack.imgur.com/B9zLU.png" rel="noreferrer"><img src="https://i.stack.imgur.com/B9zLU.png" alt=""></a></p>
<p>I'm sure that I'm <em>far</em> from the first person to have had this idea, but having no clue about its proper name nor about how to describe it, I'm unable to find anything on the net. I guess it's best described as a kind of <a href="https://en.wikipedia.org/wiki/Flexure" rel="noreferrer">flexure</a>, but what exactly? Has this mechanism found any real world applications? </p>
| |mechanical-engineering|mechanisms|bearings|flexures| | <p>I think it is similar to what it is called by <a href="http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs%20Book%20pdf/FUNdaMENTALs%20Topic%2010.PDF" rel="nofollow noreferrer">MIT Fundamentals</a>: <strong>Flexural Rolling Bearing</strong> or <strong>Rolling Contact Flexure</strong> in section 10.14.</p>
<blockquote>
<p>As is often the case, a hybrid system can combine the best of both
worlds. So it is with rolling contact flexural bearings. These bearings allow
for compact rolling element joints to be designed. These joints can sustain
large normal compressive forces due to the rolling contact, and respectable
shear and tensile forces due to the flexible bands. This also make them inherently
preloaded. Most importantly, they allow for rolling motion with a minimal
number of elements and no lubrication which makes them particularly
well-suited for precision instruments medical device applications.</p>
</blockquote>
<p><a href="https://www.youtube.com/watch?v=XIA04EasAFM" rel="nofollow noreferrer">Youtube link for a similar bearing</a> which is called <strong>Compliant Rolling-Contact Element</strong></p>
| 19547 | What is this kind of bearing called |
2018-02-22T04:28:57.117 | <p>This is a practice question that I am trying. I am a student learning about material balance-</p>
<p><a href="https://i.stack.imgur.com/ajacZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ajacZ.jpg" alt="enter image description here"></a></p>
<p>I am trying to find the mass composition of stream D.
The one written in pencil is the values that I have found in advanced- </p>
<p>My attempt- </p>
<p>I have made a material balance around separator 2 ignoring the recycle stream,</p>
<p><a href="https://i.stack.imgur.com/mZv3h.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mZv3h.jpg" alt="enter image description here"></a></p>
<p>I done up a component X balance- </p>
<p>$1250 X + (120 \times 0.3) = (379.08 \times 0.374) + (90.92 \times 0.222) $ </p>
<p>I solve for x $= 0.10076$ </p>
<p>Similarly,I done up a component Y balance-</p>
<p>$125 Y + (120 \times 0.7) = (379.08\times0.626) + (90.92\times 0.778)$</p>
<p>I solve for Y $= 0.1792$ </p>
<p>I know this is wrong as the composition of X and Y total does not add up to 1. Where did I go wrong or what must I do ?</p>
| |materials|chemical-engineering| | <p>How do you get D = 1250?</p>
<p>Should it be A + B - C ie 450 + 130 - 230 = 350 based on separator 1 only, ie what leaves separator 1.</p>
<p>Then when you include separator 2 the +900 will give 1250 arriving at separator 2.</p>
| 19552 | Material balance without chemical reactions |
2018-02-22T14:36:52.747 | <p>Why is hardening and brittle nature so related?What is work hardening or strain hardening as far as true stress strain diagram is concerned?</p>
| |mechanical-engineering|materials|mechanical-failure|failure-analysis| | <p>Brittleness describes the inability of a material to absorb a considerable amount of plastic deformation. Thus, the material fails without a lot of deformation.
Hardening, on the other hand is the process of increasing a material's yield strength $\sigma_y$</p>
<p>The principle used for work/strain hardening has to do with residual strain.
<img src="https://i.stack.imgur.com/RJejY.png"></p>
<p>If you stretch e.g. steel beyond its elastic limit, i.e. beyond $\epsilon_{el}$, or $\sigma_y$, it starts to yield, any further work done on the material will result in plastic deformation and is therefore irreversible. Only the work done up to the elastic limit is reversible.
For example, if you stretch the material in the $\sigma$-$\epsilon$-diagram (~steel) to the point $\epsilon_{pl}$ and then remove the load, a residual strain $\epsilon_r$ will remain. </p>
<p>If you reset your testing apparatus and measure the material's stress-strain diagram again, it will now follow the green line, as the material has undergone strain-hardening. It has now an elastic limit $\epsilon_{el,2}\approx\epsilon_{pl}-\epsilon_r$</p>
<p>The elastic modulus $E$ remains more or less the same. ($E$ is rather affected by the alloy components). Ultimately, its yield strength $\sigma_y$ has increased.</p>
<p>The "green" material will now undergo less deformation $\epsilon$ up to its failure than the initial material (blue), thus it can be considered more brittle.</p>
<p>In general, if you consider <em>one</em> material, you can make it harder, but that will also leave it more brittle.</p>
| 19560 | Hardening and Brittleness |
2018-02-22T16:10:51.800 | <p>I have purchased these copper heat-pipes; however, I am wondering how to bend them without kinking them or cutting them open due to pressure on weak points/bends.</p>
<p><a href="https://i.stack.imgur.com/G4ACs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G4ACs.jpg" alt="enter image description here"></a></p>
<p>These are flat copper heat pipes, so I can't use a pipe bender; however, <strong>these are sold specifically to be bent, according to descriptions</strong>, so I was wondering if anyone here has any experience or tips for bending these flat copper heat pipes.</p>
<p>To be a little more specific, I want to make a 90 degree bend in the middle of the copper heat pipe, without making kinks (which degrade performance).</p>
<p><strong>Furthermore,</strong> would wrapping copper foil around the heat pipes allow for greater thermal dissipation? Apparently some computer modders have seen success; however, I thought it would be appropriate to ask opinions here as I will try to wrap copper foil set in place by thermal paste to the heat pipes.</p>
| |mechanical-engineering|metallurgy|industrial-engineering|home-improvement| | <p>Copper is notoriouos for hardening and breaking under bending or hammering it, or working it any other way. To keep it soft and flexible, you have to anneal it often while working on it. It also bends easier and breaks less soon when you bend it while it's hot. You can heat it with the torch, but you're right about exploding it, getting it too hot may get the pressure too high, and the pipe may pop. Especially since it's filled with some kind of chemical, that may be dangerous. So at least wear goggles and gloves, but rather a full face mask and clothes that cover your arms. Since the pipe is so small, and probably filled with little fluid and mostly gas, i think it will rather 'pop' than explode, but still it can be dangerous.</p>
<p>To bend it you can cut three disks from eg. aluminium or any metal. The middle one will have the radius of the inner radius of the pipe bending, the two outer disks will have that radius, plus the width of the pipe. They will keep it from twisting while you bend it. To minimise the risk of breaking the pipe, keep the radius of your bend as large as possible. Heat the pipe and start making a small bend of a few degrees, reheat it again and bend it a few degrees further, repeat until bent enough.</p>
<p>You're right about your foil thing. Heat pipes are the most effective thing in <strong>transporting heat, not dissipating</strong>. They just relocate the heat from the source to somewhere else. You can't really get rid of any heat with just heat pipes. That's what you need a heatsink for. A heatsink is meant to dissipate heat. So it has to have a large surface, preferably close to the source. Heat pipes are used in laptops since it's not possible to put the heatsink onto the processor. Thus, a heat pipe is used to transport the heat to the side of the laptop, where the heatsink is located. Thermal paste is used for a proper heat conduction between heatsink and heatpipe or source. Keep the paste layer as thin as possible.</p>
<p>Copper is the best material for conducting heat, but aluminium is often used since it's cheaper. I'd rather buy a good heatsink than try and craft it yourself. It probably won't be as effective. When searching for a heatsink, look for the right C/W rating. It basically means how hot it will get when dissipating a certain amount of heat.</p>
<p>5C/W means it will get 5C hotter for every Watt of power it has to dissipate.
When you know the temperature you allow, and the power you have to dissipate, you know how much C/W you need for a heatsink. So lower is better. You can put a fan on the heatsink to make it dramatically more effective. C/W rating is mostly for still air.</p>
| 19564 | How to bend copper heat pipes? |
2018-02-23T14:23:09.017 | <p>Specifically, a 100W CO2 laser. Are there any considerations other than the maximum power for a fiber?</p>
| |optics|lasers|fiber-optics| | <p>CO2 lasers cannot be coupled into standard silica-based fibres, because these fibres absorb light longer than 2.1um. You could use hollow silica fibres, but they are quite recently developed, so might be expensive. Otherwise I‘d just go with the standard approach of free propagation over mirrors OR use a solid state laser which can be coupled into standard fibres easily</p>
<p>Addition: For any laser, whether CO2 or solid state, the laser needs a minimum beam quality (minimum meaning the $M^2$ should be sufficiently small) so that you can guide it with a given fibre. The numerical aperture NA and the core diameter $d_k$ of the fibre are the important values, giving you the following condition for the maximum beam parameter product your laser beam is allowed to have if it should be coupled into a given fibre.</p>
<p>$NA \cdot \frac{d_k}{2} \geq \frac{d_f \cdot \Theta}{4}$</p>
| 19575 | Can I use an optical fiber to direct a laser cutter? |
2018-02-23T14:41:23.753 | <p>I'm currently designing scissors, and I have to describe the tolerance and fits of the parts. What kind of tolerance must be used for the pin holding the two blades together, though allowing for rotations? (i.e. clearance, transition, or interference).</p>
<p>Thank you very much for your insights!</p>
| |manufacturing-engineering|tolerance| | <p>Just looked at my kitchen scissors; clearance approximately of +.001"/+.003".</p>
| 19576 | Tolerance of pin in Scissors |
2018-02-23T15:51:43.537 | <p>i'm trying to connect a D-cut shaft to a plate for transmission of torque</p>
<p><a href="https://i.stack.imgur.com/GEBES.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GEBES.png" alt="enter image description here"></a></p>
<p>This is the motor that i am using with a 'diameter' of 4.5+/-.1mm</p>
<p>My first question is whether I can use 5mm circular universal mounting hub like this to do so? <a href="https://i.stack.imgur.com/lcMzJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lcMzJ.png" alt="enter image description here"></a></p>
<p>Secondly, i have access to a a laser printer. Would you recommend making a mounting hub similar to the one shown above but with a D-Cut hole? It has a precision of .2mm, so along with the uncertainty in the shaft of 0.1mm, I guess I would have to make the 'hole' 4.8mm. Would this potential difference in size mean that the torque may not be transmitted efficiently?</p>
| |motors|torque|power-transmission| | <p>In answer to your questions:</p>
<ol>
<li><p>Yes, you can (and should) use a standard 5mm hub, such as the one shown in your question. The flat on the motor shaft is only there to provide somewhere for a grub screw to bite in (better than having a perfectly rounded shaft).</p></li>
<li><p>No, I would stick with a standard hub with a circular hole, for the reasons stated above. I don't know what the tolerance is on your hub, but it should be a tight fit around the shaft (5mm +0 -0.012). Ignore the 4.5mm dimension, that's just the flat for the grub screw, it's irrelevant.</p></li>
</ol>
| 19578 | How do I connect a plate to a D-cut shaft? |
2018-02-23T17:44:47.567 | <p>Apparently I have a stupid question.</p>
<p>I got curious and crunched some numbers to determine approximately how much air an engine consumes in total to burn a full tank of gas. Something went horribly wrong and I feel stupid.</p>
<p>So theoretically, I can get this number by multiplying the total amount of fuel in the tank, let's say 15 gallons (56.78 liters), by the air part of the air-fuel ratio, which I read is 15:1. So the engine would consume about 225 gallons (851.7 liters) of air per tank of gas?</p>
<p>But that number makes no sense if you take into account the other way to calculate this: engine size. Let's say we have a four cylinder, two liter engine. Since it's four stroke, the air it will consume should be equivalent to its rpm. So at 1,000 rpm it will consume 1,000 liters of air per minute. Right? That's a very long way from a total of 850 liters for the whole tank of gas.</p>
<p>So what did I do wrong?</p>
| |mechanical-engineering|automotive-engineering|fuel-economy| | <p>You probably want to look at stochiometric ratios by mass. </p>
<p>However this is a difficult problem to solve from first principals as for an IC engine air is essentially free and monitoring its consumption in bulk is not a major priority. </p>
<p>I suspect at the mistake you have made is confusing mass ratios with volume ratios. As air is a gas and as such has very low density this make a very big difference. </p>
<p>If we assume that 15:1 air:fuel ratio by mass then we will need 15kg of air for every kg of fuel burned. </p>
<p>Air has a density of about 1.2 kg per cubic metre so that's 18 cubic metes of air per kg of fuel (at atmospheric pressure) </p>
<p>An alternative approach is to consider the swept volume per cylinder and compare with rpm and fuel economy </p>
| 19581 | How much air will an engine use per tank of gas? |
2018-02-25T01:18:27.017 | <p>I am trying to CAD up a drawing for a plate for this motor to mount to. I figured it would be easy since they provided an mechanical drawing, but I have never seen one like this before. Can you please help me read this? It is from China and I don't even have any idea what units it is in. I have read these before but I am completely stuck on this one. I am trying to figure out the spacing of the mounting holes on the front of the motor. Here is the drawing</p>
<p><a href="https://i.stack.imgur.com/Xmxhz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xmxhz.png" alt="enter image description here"></a>, </p>
<p>and here is a picture of the actual motor if that helps at all.</p>
<p><a href="https://i.stack.imgur.com/x4Glp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x4Glp.jpg" alt="enter image description here"></a> </p>
<p>From what I can tell the mounting holes are on the bottom (or left if you rotate it). Any help is GREATLY appreciated. I can't continue my project until I get this figured out.</p>
| |mechanical-engineering|motors|cad|technical-drawing| | <blockquote>
<p>"I am trying to CAD up a drawing for a plate for this motor to mount to.".</p>
</blockquote>
<p>I read the drawing and TinEye.com searched the photo. </p>
<p>For the "Star" bracket you'll need a 114.3 mm recess, 3.2 mm deep to allow for the raised inner section, with a greater than 22.23 mm through hole to allow for the shaft. The bolt hole circles are 106.4 mm and 149.2 mm diameter. Use 5/16" x 18 TPI bolts and 3/8" x 16 TPI bolts; the "Star" bracket uses all 8 bolts, the "L" bracket configuration uses only the 3/8" bolts. Of course you can combine designs and come up with your own.</p>
<hr>
<p>Metric Conversion (rounded to nearest imperial measurement):</p>
<ul>
<li><p>114.3 millimetres = 4 1/2 inches</p></li>
<li><p>3.2 millimetres = 1/8 inch</p></li>
<li><p>22.23 millimetres = 7/8 inch</p></li>
<li><p>106.4 millimeters = 4 3/16 inches</p></li>
<li><p>149.2 millimeters = 5 7/8 inches</p></li>
</ul>
<hr>
<p>Here are example mounting plates for that motor:</p>
<p>Source: <a href="http://www.electricmotorsport.com/nema-143t-c-face-1-2-inch-aluminum-l-bracket-motor-mount.html" rel="nofollow noreferrer">http://www.electricmotorsport.com/nema-143t-c-face-1-2-inch-aluminum-l-bracket-motor-mount.html</a></p>
<p><a href="https://i.stack.imgur.com/KndTl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KndTl.jpg" alt="Motor Mount - L Bracket"></a></p>
<p>Source: <a href="https://m.ebay.com/itm/Manta-10-hp-electric-motor-MOUNT-Etek-Briggs-Motenergy-Permanent-Magnet-STAR-NC-/391984038365?nav=SEARCH" rel="nofollow noreferrer">https://m.ebay.com/itm/Manta-10-hp-electric-motor-MOUNT-Etek-Briggs-Motenergy-Permanent-Magnet-STAR-NC-/391984038365?nav=SEARCH</a></p>
<p><a href="https://i.stack.imgur.com/0MO2k.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0MO2k.jpg" alt="Motor Mount - Star"></a></p>
<p><a href="https://i.stack.imgur.com/UxtKs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UxtKs.jpg" alt="Motor Mount - Star mounted on motor"></a></p>
<p>If you are considering purchasing one of those brackets you can probably find the "L" bracket for a lower price since it looks easy to make and is priced fairly expensive at the place given in that URL. The "Star" bracket is more complicated and CNC machined, for a custom item it's price is more inline (and you may find it difficult to get someone to make you one for less).</p>
<p>Which of the two, or another design entirely, really depends upon what you intend to attach it to.</p>
<p>Those photos ought to give you a hint about how you want to draw it up and the appropriate dimensions to use. The place you bought it from probably has brackets, a drawing of the bracket might help you more than one of the motor.</p>
| 19601 | Please Help me Read this Engineering Drawing |
2018-02-25T18:11:19.630 | <p>Actually, I've asked this question in "English Language & Usage" but someone told me ask it here!</p>
<p>What does it mean by "loops and troughs" in:
"Source has acquired a pattern of loops and troughs."</p>
<p>Here's some context including this phrase: <a href="https://books.google.com/books?id=a6LzCAAAQBAJ&pg=PA482&lpg=PA482&dq=%22loops+and+troughs%22&source=bl&ots=0M9eK7xYHc&sig=afrE7rVYxX8Zxda7_L7NDGYWzYA&hl=en&sa=X&ved=0ahUKEwjhk4Cqz8HZAhUKZ1AKHSFnBsEQ6AEIKzAB#v=onepage&q=%22loops%20and%20troughs%22&f=false" rel="nofollow noreferrer">https://books.google.com/books?id=a6LzCAAAQBAJ&pg=PA482&lpg=PA482&dq=%22loops+and+troughs%22&source=bl&ots=0M9eK7xYHc&sig=afrE7rVYxX8Zxda7_L7NDGYWzYA&hl=en&sa=X&ved=0ahUKEwjhk4Cqz8HZAhUKZ1AKHSFnBsEQ6AEIKzAB#v=onepage&q=%22loops%20and%20troughs%22&f=false</a></p>
| |metallurgy|chemistry| | <p>Have you ever looked at finger prints? Many finger prints have patterns of loops & troughs - though for finger prints the troughs may be called arches. It's just a description of a pattern.</p>
<p>Get a rubber band and change its shape any way you like and it forms all forms of loops. Break the rubber band to create a rubber string and change its shape and you will get different forms of troughs and arches.</p>
| 19614 | What does it mean by "loops and troughs"? |
2018-02-26T18:03:48.060 | <p>I'm building a LED ceiling lamp. The idea is to mix different tones of white to get desired tone.<br>
It will initially have 120W LEDs and maybe more later<br>
It will probably never run at full power but I want to be able to.<br></p>
<p>It consists of two parts. Upper part where psu and electronics are, and lower part where LEDs are.<br>
The parts are separated by sheet metal.<br>
The dimensions are 1020x620x150<br>
The upper parts height is 50mm<br>
The upper part is sealed from top (by ceiling) and from bottom (by metal sheet)<br>
The upper part will have holes in it.</p>
<p>I want to make the lamp as shallow as possible and for that reason I've come up with this idea for airflow generation:</p>
<ul>
<li>Use large fans to lower noise generation.</li>
<li>Turn them on their side.</li>
<li>To force air to move in one direction block top on one side of fans, and bottom on other side of fans (images <a href="https://i.stack.imgur.com/KWQ0g.jpg" rel="nofollow noreferrer">2</a>, <a href="https://i.stack.imgur.com/qVmuo.jpg" rel="nofollow noreferrer">4</a>)</li>
</ul>
<p>Would this kind of configuration work or will I have to use small fans at high rpm?</p>
<p><a href="https://i.stack.imgur.com/7Sbsy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Sbsy.jpg" alt="3D overview of ceiling lamp"></a>
<a href="https://i.stack.imgur.com/KWQ0g.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KWQ0g.jpg" alt="3D close up of fans"></a>
<a href="https://i.stack.imgur.com/XiDMB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XiDMB.jpg" alt="2D schema of ceiling lamp"></a>
<a href="https://i.stack.imgur.com/qVmuo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qVmuo.jpg" alt="2D schema of fans"></a></p>
| |airflow|cooling| | <p>No, this will not work well (as I read your plans). You should configure your fans (of any size) to draw air from inside the chassis and expel it outside the chassis. This will require vents on the perimeter of the chassis to allow air in.</p>
<p>No 'blocking' is necessary except that the outflow face of the fan should be sealed around the outflow vent perforations to ensure all air entering the fan is drawn from the chassis.</p>
| 19626 | Cooling with fans in narrow space |
2018-02-27T07:33:31.790 | <p>How might the "shell" in this picture be manufactured? What process could be used to achieve this geometry? It is made from plastic.</p>
<p><a href="https://i.stack.imgur.com/yCpZdm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yCpZdm.png" alt="render of component in question"></a></p>
| |plastic| | <p>It might also have been made by thermoforming, in which a flat sheet of plastic is heated until it softens and then pressed against a replica of the finished part and allowed to cool and harden. If the process of conforming to the replica is aided by applying a vacuum in the space between the sheet and the replica, the process is called vacuum forming. this is popular for making large objects with thin walls especially if the finished part must be inexpensive. </p>
| 19637 | What manufacturing process could be used for this part? |
2018-02-27T17:44:34.170 | <p>I'm using an engineering software based on <strong>Autodesk AutoCAD 2018 OEM</strong> and I need to assign <strong>keyboard shortcuts to NumPad keys</strong> directly, without any modifier (SHIFT, CTRL, ALT).</p>
<p>On older version of Autocad this was possible, but now even though it's possible to assign NumPad keys directly, they do not work. According to <a href="https://knowledge.autodesk.com/support/autocad-lt/learn-explore/caas/CloudHelp/cloudhelp/2018/ENU/AutoCAD-LT/files/GUID-9970E91B-A4F5-405C-99C1-46AD804A8858-htm.html" rel="nofollow noreferrer">this page</a> on the AutoCAD LT knowledge base: <em>"Valid modifier and key combinations include Number Pad (NUMPADn) keys containing no modifiers"</em> and I hope the same is true for the "standard" AutoCAD (i.e. no "LT").</p>
<p>In fact I'm able to assign a shortcut to - for example - "Scale" function to NUMPAD(8), as in the screenshot below:</p>
<p><a href="https://i.stack.imgur.com/AMTXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AMTXt.png" alt="enter image description here"></a></p>
<p>But then I'm not able to use it because if I press the NumPad 8 with <strong>Num Lock ON</strong> the AutoCAD console reads an 8:</p>
<p><a href="https://i.stack.imgur.com/Qt4cA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qt4cA.png" alt="enter image description here"></a></p>
<p>... and if I keep the <strong>Num Lock OFF</strong> then the console picks the previous command (as it's the same that pushing the <kbd>↑</kbd>).</p>
<p>So if this feature is documented and the <strong>cui</strong> AutoCAD command allows me to choose that keys, how are they suppose to work? How can I assign a key to speed up my work without using a modifier?</p>
| |autocad| | <p>I was able to reproduce the error: basically you cannot assign any NumPad key to AutoCAD 2015 or newer (2018 included) and in any flavour (Full, LT, OEM, etc.) when using a keyboard different from English (mine was Italian).</p>
<p><a href="https://i.stack.imgur.com/ZBq6p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZBq6p.png" alt="enter image description here"></a></p>
<p>So, to <strong>workaround</strong> this bug, change/add an <strong>English keyboard BEFORE starting your AutoCAD product</strong> and then you'll be able to assign the correct Numpad key in the <strong>Customize User Interface</strong> dialog. It will appear as <strong>NUMPAD8</strong> and it will work even after deleting the English keyboard (if not used).</p>
<p><a href="https://i.stack.imgur.com/3GCAu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3GCAu.png" alt="enter image description here"></a></p>
| 19642 | NumPad shortcut keys in AutoCAD 2018 |
2018-03-01T00:42:30.667 | <p>Taking a 1750RPM 110VAC appliance motor down to about 60RPM, maybe a little less. It's a home project on the cheap, so dropping a few hundred dollars on a ready-made gear box is to be avoided, and I'm working with the kids as a school project and so creating by hand as much as possible.</p>
<p>1/2" plywood for an initial prototype/proof of concept, with a planned move to 3D printed gears afterward. Assuming a tooth pitch of 18-25mm, and that we play with the tooth count between the gears to spread out the wear and tear, which would be better for durability, chatter, noise? Fewer gears with larger reduction ratio at each step, or more gears with a smaller ratio? </p>
| |gears|mechanisms| | <p>Why use 3d printed gears ? </p>
<p>3D printing is a terrible way to make gears and you can just buy standard ones with proper tooth profiles off the shelf. </p>
<p>For that sort of high reduction ratio a planetary gearbox is generally preferable to long compound gear trains. </p>
<p>However it sounds like you might be better of with a belt drive. Vee belt pulleys are fairly cheap, can achieve the required ratio without too much compounding and are reasonably quiet and importantly lend themselves well to low-tech fabrication as they are reasonably tolerant of misalignment and have a lot of inherent scope for adjustment. Unlike gearboxes which require pretty good manufacturing tolerances. </p>
<p>For all of these reasons belt drives tend to be the preferred option for stationary tools driven by AC motors. </p>
| 19665 | ~121:1 speed reduction; are more gears better? |
2018-03-01T17:02:45.513 | <p>I am deciding between two design ideas for a project in which I have to build a balsa wood tower that can maximize load weight and minimize structural weight (maximize efficiency). I've included a picture with the two designs I am deciding between. </p>
<p><a href="https://i.stack.imgur.com/U20AL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U20AL.jpg" alt="(Left, tower with height 40 cm, width 5 cm. 10 cross-braces with height 4 cm each. Right, height and width 40 cm and 5 cm. 5 cross-braces with height 8 cm each. Both towers are supported by pin supports on the right and left sides."></a></p>
<p>I am wondering which design would be able to hold greater weight, whether increasing the number of braces changes the load capacity, and in which way it changes (linear correlation, exponential correlation, etc.).</p>
<p>More specifically, I'm considering a load weight of <strong>145 N</strong>, and I am looking for the values of the internal member forces, especially those before buckling (assuming a static system). </p>
<p>My understanding of this subject only extends to truss calculations, considering diagonal bracing (e.g. Howe Truss), which is why I am posting on this forum. Further, if you'd be willing to give a brief explanation of these diagrams from <a href="https://www.aisc.org/globalassets/aisc/research-library/bracing-for-stability.pdf" rel="nofollow noreferrer"><em>Bracing for Stability</em></a>, I'd really appreciate it. <a href="https://i.stack.imgur.com/erzUE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/erzUE.png" alt="Design recommendation for discrete bracing systems for columns, pg. 6"></a><a href="https://i.stack.imgur.com/hzMAO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hzMAO.png" alt="Figure 6 on pg. 6, the accompanying graph"></a></p>
| |structural-engineering|design|statics| | <p>Elements under compression such as the vertical columns in your tower can collapse in two very different ways.</p>
<p>The first is via simple crushing of the member. This happens when the applied load generates an internal stress in the member which is higher than the member's strength.</p>
<p>The second is via buckling. In this case, infinitesimal imperfections in the structure make it "easier" for the member to bow away from the load. Get a plastic straw or a piece of paper and try to place either under compression. You'll notice they just "jump" to the side. However, once the compression is removed, the straw or piece of paper will "jump" right back into its original shape as if nothing'd happened. The theoretical equation for the buckling load (often called Euler buckling) is
$$P_e = \dfrac{\pi^2EI}{(kL)^2}$$
where $E$ is the member's modulus of elasticity, $I$ is its second moment of area (aka, moment of inertia), and $kL$ is the member's unbraced length ($k$ is a coefficient which depends on the member's boundary conditions; in your case, you could conservatively assume $k=1$). Obviously, the real-world buckling load is much lower than $P_e$, since real-world imperfections are actually quite significant, not infinitesimal.</p>
<p>Braces aren't meant to carry any of the applied load. Their purpose is only to guarantee that the principal members (in this case, the vertical columns) do not buckle under compression. So a properly braced structure will instead collapse due to crushing or global buckling (where the entire structure buckles as if it were one member, which the braces can't help against).</p>
<p>To find the unbraced length for your columns, you could rework the Euler's buckling load equation to give you $L$ given the other values (which you can find online). However, given how that equation is highly theoretical (and not used directly in actual engineering), it is probably best to merely create prototypes to see at which length the parts start to buckle and then adopt that.</p>
| 19676 | How does the number of braces in a balsa wood tower affect the load capacity? |
2018-03-01T18:00:08.887 | <p>Modern cars use cogwheel to transfer power from the engine to the wheels.
Steam locomotives <a href="https://i.imgur.com/ou5yWsJ.gifv" rel="noreferrer">used some kind of bars</a> (sorry, I'm not a native speaker) to transfer the power to the wheels.</p>
<p>Why did the engineers not use cogwheels?
Would steam locomotives had been faster if they had used cogwheels?</p>
| |steam| | <h2>Steam locomotives use steam <em>pistons</em>, not steam <em>turbines</em>.</h2>
<p>Gears/cogs would be pointless since there is no rotary source of power on steam locomotives. They use steam pistons, which go back and forth. </p>
<p>As the physics worked out, direct-drive worked out really well with achievable values of piston diameter, stroke/eccentric, and wheel size. <em>Until it didn't.</em> And what got them was the curves.</p>
<h2>Mainline pullers stuck with rods: way too big for gears</h2>
<p>As fully superheated boilers became very powerful, fast passenger locomotives used this power at higher speeds. For them, the side-rod design was perfect. But slow lugging freight locomotives needed more weight on the rail to transfer the power at low speeds. This required more driving axles to spread the weight. That made a single rigid group of driving axles too long for curves. So they split into two (rarely, three) groups of driving axles. Power transfer was done with an engine on each group, usually simple, sometimes compound. Union Pacific's Big Boy had 8 drive axles in two groups (each with a simple engine, still avoiding gears), handling curves like a 4-drive-axle locomotive. </p>
<p><a href="https://i.stack.imgur.com/T6oa4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T6oa4.jpg" alt="enter image description here"></a>
<a href="https://commons.wikimedia.org/wiki/File:28884.jpg" rel="nofollow noreferrer">src</a></p>
<p><em>Taken to absurdity. The Virginian Railway finally gave up and electrified.</em> </p>
<p>At these power levels, 4000-6000 horsepower, gear drive was out of the question: it was an order of magnitude too much power for gears. Even the electric GG1 of the era used <em>twelve</em> massive pinions to transfer a similar amount of power to six axles. </p>
<h2>Much smaller engines could be geared</h2>
<p>Mountain railroads used low power, lightweight locomotives which had to slither up fairly tight curves. Even a very modest side-rod steam engine was too stiff for the curves. They also wasted a lot of precious weight on non-drive wheels, e.g. the pilot truck and the tender. Ephraim Shay solved this problem with, indeed, geared locomotives. Keep in mind these are small locomotives: the largest, Western Maryland #6, has boiler pressure of 200 psi and a top speed of 23 mph. </p>
<p>Ephraim Shay put a drive shaft along one side of the locomotive, gearing to each wheel. The pistons directly cranked the drive shaft. Note the elaborate telescoping drive shafts, most especially important due to its off-center location. </p>
<p><a href="https://i.stack.imgur.com/qGmqA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qGmqA.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/SgqZm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SgqZm.png" alt="enter image description here"></a>
Note the gears. <a href="https://en.wikipedia.org/wiki/Geared_steam_locomotive" rel="nofollow noreferrer">sources</a></p>
<p>Charles Heisler put the drive shaft down the locomotive centerline, and used a "vee-twin" piston arrangement. Note the side rods: that means only one of the two axles is geared to the drive shaft, the side rods transfer power to the other axle. Side rods like that imply perhaps 100 horsepower per axle. </p>
<p><a href="https://i.stack.imgur.com/wpxBk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wpxBk.png" alt="enter image description here"></a></p>
<p>The Climax Manufacturing Co. took Heisler's centerline-shaft arrangement and added a cross-shaft and more gearing to put the steam pistons in an <em>almost</em> conventional location. </p>
<p><a href="https://i.stack.imgur.com/UwmLF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UwmLF.png" alt="enter image description here"></a></p>
<p>Having seen these geared-locomotive arrangements, you can see where they would not "scale" up to multithousand horsepower outputs.</p>
| 19677 | Why did steam locomotives not transfer power by cogwheels? |
2018-03-02T15:01:47.640 | <p>In the answer to this question:</p>
<p><a href="https://engineering.stackexchange.com/questions/10341/difference-between-saturated-liquid-and-saturated-steam">Difference between Saturated Liquid and Saturated Steam</a></p>
<p>How can energy be added with constant temperature? Isn't the energy input itself a change of temperature? </p>
<p>Sorry for the repeat, but I don't have enough reputation to post a comment. </p>
| |thermodynamics|heat-transfer|steam| | <p>Temperature is a measure <strong>proportional</strong> to the average translational energy (bouncing around) of molecules in a substance, but not energy itself. It is an important measure because energy always transfers from high temperature to low temperature bodies. Heat is the energy change during the transfer. </p>
<hr>
<p>For Example: When we heat a fluid at a constant pressure. </p>
<p>$ Q $ is our heat added
$ \Delta t $ is our temperature change, $m$ is our mass, $ C_p $ is our constant pressure specific heat capacity.</p>
<p>$$ Q = mC_p\Delta t $$</p>
<p>Note that a body with a higher $m$ and $C_p$ can be heated by the same amount, but with a lower temperature increase. </p>
<hr>
<p>Between the saturated liquid and saturated gas points, adding or losing heat does affect the temperature at all, but changes the relative proportions of liquid and gas in the mixture. </p>
| 19686 | How can energy be added at constant temprature? |
2018-03-05T02:16:18.867 | <p>Suppose I have a 5kg framed picture that I want to hang on hooks, each of which can hold 2kg. Is there some arrangement of hooks that will make this safe? How many hooks would I need? I thought that I could arrange the hooks as something like a catenary to distribute the load, but I'm having trouble calculating the force on each hook.</p>
<p><a href="https://i.stack.imgur.com/AWMnv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AWMnv.png" alt="enter image description here"></a></p>
<p>What are the techniques I need to use to solve this problem?</p>
| |mechanical-engineering| | <p>I couldn't resist, so: First, you should note:</p>
<ul>
<li>The tension in the string will be the same along its entire length (assuming frictionless conditions)</li>
<li>To use the hooks optimally, you should arrange them that they carry equal load</li>
</ul>
<p>Taking this into consideration, the angle that the string bends at each hook should be the same (the wall hooks should be spaced equally on the arc of a circle between the picture hooks, not a catenary as noted).</p>
<p>The picture below illustrates:</p>
<p>$$ø = \frac{2θ}{3}$$</p>
<p>Or more generally:
$$ø = \frac{2θ}{n}$$
where n = the number of hooks</p>
<p><a href="https://i.stack.imgur.com/d8Zv6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d8Zv6.png" alt="enter image description here"></a></p>
<p>Below is the FBD at one of the frame hooks:</p>
<ul>
<li>$F_t$ is the tension in the string.</li>
<li>$F_p$ is the vertical force on the
frame hook.</li>
</ul>
<p><a href="https://i.stack.imgur.com/5qnCP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5qnCP.png" alt="enter image description here"></a></p>
<p>From this, the tension in the string can be calculated as:</p>
<p>$$F_t = \frac{F_p}{sin(θ)}$$</p>
<p>The FBD at one of the hooks is as follows:</p>
<p>With $F_h$ as the resultant force on the hook.</p>
<p><a href="https://i.stack.imgur.com/IsiJD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IsiJD.png" alt="enter image description here"></a></p>
<p>$F_h$ can be calculated as:</p>
<p>$$F_h = 2F_tsin(0.5ø)$$</p>
<p>This can be generalised as:</p>
<p>$$F_h = \frac{2F_p}{sin(θ)}*sin(\frac{θ}{n})$$</p>
<p>From this you can see that as $θ$ approaches $0$, $F_h$ approaches $2F_p/n$, but $F_t$ approaches infinity. Thus you should consider the tensile strength of the string and the forces that the frame and frame hooks can handle as well.</p>
<p>You can hang your picture with 3 hooks with $θ = 62.11$ degrees, giving you $F_t = 28.285N$ and $F_h = 19.999N$. This will be "safe" assuming that the gravitational constant is 10, wihch it isn't, but that's just nit picking.</p>
| 19715 | Hang heavy picture with multiple hooks |
2018-03-05T02:21:19.327 | <p>This question goes right back to the first principles of mechanics, so to speak. I'm confused as to the nature of forces applied to rigid bodies. Mathematically, a force acts on a particle of a known mass, but I don't see the mathematical connection that allows a rigid body of the same mass to behave similarly under the influence of the same force.</p>
<p>I understand that a rigid body can be obtained by extending the idea of a system with only a finite number of particles to a system with its number of particles tending towards infinity. Thus each particle in a rigid body will have a differential mass. </p>
<p>Therefore, what I don't understand is how a force applied at a particular location on a rigid body doesn't lead to an infinite acceleration due to the infinitesimal mass of the point at which the force is applied. Obviously this is not the case in real life, however this is what occurs when I try to take the definitions that surround systems of particles and try to build up rigid body mechanics around it. </p>
<p>So, does a force applied to a rigid body simply imply that this force is spread over the body? This would mean that the definition of a force on a rigid body is different to that of a force on a system of particles (since if it is acting on the rigid body as a whole it is not acting at a single point), right? What am I missing? </p>
| |mechanical-engineering|mathematics| | <p>Suppose you have body with mass M experiencing a force F. If you model the force as being applied to the whole body, then this gives an acceleration F/M. Now suppose you cut this body into n equal parts. The part that the force is directly applied to will experience force F over mass M/n. So if you model the force as being applied just to one part, that appears to give an acceleration of nF/M. Contradiction? No. We're modeling this as a rigid body. So if one part accelerates, it must take the rest of the body with it. The rest of the body has mass M(n-1)/n. If this mass is accelerating at F/M, then it is experiencing a force of (F/M)(M(n-1)/n) = F(n-1)/n. So if we're modeling the force as applying only to one part, the rest of the body isn't being accelerated by this force, at least not directly. Instead, it must be being accelerated by the part that has the force applied to it. So that part is applying a force of F(n-1)/n to the rest of the body, and by Newton's Third Law, the rest of the body is applying a force of F(n-1)/n to that part. So the total resultant force for the part with an external force is the external force F minus the force F(n-1)/n from the rest of the body, which gives F/n. A force of F/n applied to a mass of M/n gives an acceleration of F/M.</p>
<p>If you want, you can break this analysis down even further. Suppose you have a object of n sections, and you apply a force to a section. Then that section will experience the exterior force plus forces from the adjacent sections. Those sections will experience a force from that original section, plus forces <em>their</em> adjacent sections. And so on. You can then set up a system of equations involving all these forces, relate the acceleration of each part to the net force it experiences, apply the constraint that the acceleration on each part is equal to the acceleration of each other part, and solve that system of equations. Intuitively, we should find that the acceleration for each part is F/M, regardless of how we divide the body into "parts". </p>
<p>Or we can simply recognize that we are dealing with a model to begin with, and different aspects of the model apply to different situations. We can treat the body as being made up of differential masses for such purposes as calculating moments of inertia, but a single mass for calculating acceleration from force. We can treat the body as being a point mass for purposes of calculating what gravitational force it creates, and yet treat forces as being displaced from the center of mass for purposes of calculating torque.</p>
<blockquote>
<p>This would mean that the definition of a force on a rigid body is different to that of a force on a system of particles (since if it is acting on the rigid body as a whole it is not acting at a single point)</p>
</blockquote>
<p>The definition of force is the same. What <em>model</em> we use depends on the context. The term "rigid body" does not refer to any actual physical object. Forces are transmitted through an object at the speed of sound in that object, which cannot be higher than c. No object is truly "rigid". When we talk about a "rigid object", that is simply shorthand for "an object that, in the scenario currently under consideration, deviates so little from the theoretical ideal of a rigid body that it can be modeled as a rigid body to the degree of accuracy required". This is similar to how we talk about applying the ideal gas law to gases, even though there is no such thing as an ideal gas in the real world. So when we model something as a rigid body, we treat all forces applied to it as being applied to its whole mass. When we model something as a system of particles, we model each force as being applied to a specific particle. The actual physical laws are the same in both cases, we're simply applying different <em>models</em>.</p>
<p>Note that treating a force as being applied to a particular point is itself a simplification used in particular models. A full quantum mechanical analysis would have the particle not localized to a point, so the force would not be localized to a point. In fact, in quantum mechanics, forces are themselves carried by particles, and those particles have some positional spread.</p>
| 19716 | Mathematical connection between systems of particles and rigid bodies |
2018-03-05T15:13:25.240 | <p>I need to design a pivot arm which is pulled up by a float (buoyancy force). In the upper positions of the lever arm, there's a mass trying to pull down.
The reason that I'm presenting this problem is to consider all the variables (most significant) in order to build a low-cost prototype.</p>
<p><a href="https://i.stack.imgur.com/APHbe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/APHbe.png" alt="enter image description here"></a></p>
<p>The variables that I consider are:
1.- Volume of the float, the material is polypropylene. I need a low density but high volume float in order to generate more buoyancy force, is that correct?
2.- The distance between the force applied and the pivot "d1". If I generate more torque, more will be the force generated by the buoyancy, right?
3.- The angle of the arm A. </p>
<p>This is my first attempt presenting this problem, so any suggestion will work. The goal is to achieve a bigger force than the 95 N generates by the "object x" in order to reach the 'sealed position' denoted by "d2".</p>
| |mechanical-engineering|fluid-mechanics|torque|experimental-physics| | <p><strong>Edit</strong>: <em>Added the requested drawing + corrected formulae for the 4 bar setup</em></p>
<br/>
<p>As Carl Witthoft mentioned, the buoyancy force <span class="math-container">$F_1$</span> from the block is proportional to the submerged/displaced volume (<span class="math-container">$V_{sub}$</span>).</p>
<p><span class="math-container">$$ F_1=\rho gV_{sub} $$</span></p>
<p>Let's the distance between the Arm A pivot and the intersection of Arms A and B be <span class="math-container">$d_3$</span></p>
<p>Summing the moments around Pivot A (for the 4 bar setup, see below):</p>
<p><span class="math-container">$$F_1 sin( \theta ) \cdot d_1 + F_{objectx}sin(\theta)\cdot d_3=0$$</span></p>
<p><span class="math-container">$$ F_{objectx}= \dfrac{-\rho g V_{sub}\cdot d_1}{d_3}$$</span></p>
<p>(I have negated the weight of the bars here)</p>
<p>So you can move <em>object x</em>, reducing the length <span class="math-container">$d_3$</span>, or to increase the length <span class="math-container">$d_1$</span>.</p>
<p>Other than that, you can only increase <span class="math-container">$V_{sub}$</span>, which will increase anyway as the water rises. But you will have to adjust your geometry to make sure you reach the sealing force at your desired water height.</p>
<br/>
<hr />
<p>Adjusting for the desired water height:</p>
<ol>
<li><p>Calculate the required <span class="math-container">$V_{sub}$</span>, where <span class="math-container">$F_{objectx} $</span> is the force that we need to both suspend the weight of <em>object x</em> (95.6N?) and seal the container.</p>
</li>
<li><p>Make sure that your block volume > <span class="math-container">$V_{sub} $</span>.</p>
</li>
<li><p>Decide how high you want the water to go in order to create a seal (<span class="math-container">$h_{wmax}$</span>). Adjust the geometry to ensure that when <em>object x</em> is in it's sealed position and the water is at <span class="math-container">$h_{wmax}$</span>, <span class="math-container">$V_{sub} $</span> is as calculated.</p>
</li>
</ol>
<br/>
<hr />
<p>I would suggest using a 4 bar linkage instead of your current set-up, so that <em>object x</em> does not rotate, and only transfers force vertically, assuming this was your design intention.</p>
<ol>
<li><p>Add an ARM C parallel to ARM A. It should be connected to the wall on a pivot above point A, and connected to ARM B with a pin, above ARM A.</p>
</li>
<li><p>Change the connection between ARM A and ARM B to be a pin.</p>
</li>
</ol>
<br/>
<p><a href="https://i.stack.imgur.com/8zwEh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8zwEh.png" alt="enter image description here" /></a></p>
| 19726 | Buoyance+torque generated force |
2018-03-05T22:59:30.893 | <p>What kind of motor is suitable for the below requirements?</p>
<ol>
<li>2700 rpm, needs to be accurate, not +- few rpms. I think the best approach is to use a hall effect sensor and adjust current/voltage on the fly? Or a better idea?</li>
<li>Be as noiseless as possible. The noise 12V square fan motors make is perfect.</li>
<li>Torque is not much of an issue as even though it is rotating a 100g load on top of it the load is supported by 4 bearings.</li>
</ol>
<p>What motor am I looking for?</p>
| |motors| | <p>A brushless motor controlled by an electronic speed controller can produce an accurate and relatively powerful system. </p>
<p>Most BLDC motors used for hobby r/c equipment would be suitable and these are capable of supporting a significant axial and tangential load. </p>
<p>When operating at low RPMs (2700 RPM is slow) good quality motors are virtually silent.</p>
<p>The speed controller (ESC) can be controlled by pwm or a coded signal such as sBus which would also provide an interface for a feedback system via a microcontroller.</p>
<p>BLDC motors are specified by the can size and the revs per volt rating. For example a 2216-900kv motor would be approximately 22mm dia, with 16mm magnets, rotating at 900 rpm per volt unloaded. Small motors operate on 2S-4S LiPo batteries or 7v to 14v.</p>
| 19732 | a small noiseless motor with accurate RPM control? |
2018-03-06T14:34:56.807 | <p>Simple one. Some extension springs are wound with an initial tension which acts to pull the coils together, even when not under external loading. </p>
<p>When calculating the preload in the spring, should the initial tension be added to the spring force (e.g. initial length x spring rate)? I have seen a few examples and some people include it, others don't. Which is correct? </p>
| |springs| | <p>After being pointed in the correct direction by an acquaintance I think I have the answer now: </p>
<p><a href="https://i.stack.imgur.com/8JZFF.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8JZFF.gif" alt="enter image description here"></a></p>
<p>It appears that initial tension should be included when calculating load in an extension spring, as shown in the formulae above. However, in some applications the initial tension may be insignificant compared with the load at a given extension. People either ignore it due to it being negligible, or in error. </p>
| 19739 | Calculating pretension of a spring with an initial tension |
2018-03-06T20:29:39.503 | <p>I'm not entirely sure if this is the right place to ask this, so apologies in advance if it doesn't belong.</p>
<p>I have a curved surface with many small, irregular holes in it that I would like to measure. The surface is in a location that makes it awkward to work with, and is effectively immobile. Are there any technologies or techniques that could let me take a 1:1 imprint or copy that is more portable? Being able to 'flatten' the copy would be convenient but not required.</p>
<p>The surface is thin and the reverse is accessible (though again with some difficulty) if at all applicable.</p>
<p>Edit: Specifically, i'm trying to gauge the accuracy of another measuring process, so computer vision techniques are undesirable.</p>
| |measurements|data| | <p>If the holes are small enough and you don't care about spacing measurements, you could use what are called pin gauges. These will obviously not work on non-circular holes (irregular?)</p>
<p><a href="https://i.stack.imgur.com/QV5Ty.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QV5Ty.jpg" alt="pin gauges"></a></p>
<p>These are available in sets of various ranges. The precision is quite good, often measured in tenths of a thousandths of an inch.</p>
<p>If that is excessive in terms of precision and especially in terms of cost, there's a much less elaborate method you can consider.</p>
<p>I initially planned to suggest modeling clay, but it has limits with respect to handling. You would flatten a sheet of the stuff and press it against the surface with sufficient force to create bumps as it enters the holes.</p>
<p>Once peeled off, you should be able to measure the bumps to get the hole sizes.</p>
<p>I think it would be difficult to handle, however, without breaking it.</p>
<p>As an alternative, especially as you have access to the inside/reverse, is to use a silicone molding compound of some high Shore value (hardness). The clay might be useful to create an inner dam as well as a container of sorts for the silicone on the outside.</p>
<p>Silicone molding material, especially the high hardness stuff, is quite durable. This should allow the portability and handling you require. Some compounds have minimal shrinkage, allowing for reasonable precision. It may be necessary to test sample holes to determine exact shrinkage levels, but it would be consistent across a specific range of diameters.</p>
<p>Expanding from the above suggestion would be to seal off the holes with adhesive tape rather than modeling clay or other dam material.</p>
<p>The ideal situation would be that your curved surface has a horizontal orientation, minimizing the damming requirement and the amount of silicone used.</p>
| 19742 | Measurement of immobile curved surface |
2018-03-07T03:10:22.963 | <p>I am making a tea nitrogen laser based off the rimstarorg youtube channel. I have a perfect setup as seen below
<a href="https://i.stack.imgur.com/h4Ypp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h4Ypp.jpg" alt="enter image description here"></a>
I use 11, 1kv diodes to convert my 7,500 volt lug on my neon sign transformer to DC. My dielectric, well, i tried everything from transparency paper, to mylar, to acrylic and i tried multiple layers for each one.
<a href="https://i.stack.imgur.com/8dYyW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8dYyW.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/fUHLg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fUHLg.jpg" alt="enter image description here"></a>
Yeah thats what happened. So I eventually just tried setting up a piece of aluminum foil on the bottom, my dielectric, and then one aluminum sheet with my spark gap, and for some reasonthe spark jumps between the bottom of my 2 angle pieces before it jumps between the TWO CLOSEST POINTS AKA THE SCREW THATS 1mm AWAY!!
<a href="https://i.stack.imgur.com/KoyiC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KoyiC.jpg" alt="enter image description here"></a>
I should also mention when i used just one piece of transparency paper this happened:
<a href="https://i.stack.imgur.com/NSQJm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NSQJm.jpg" alt="enter image description here"></a>
Please help me on where to even go from here, i don't even know what I'm doing wrong at this point</p>
| |lasers| | <p>I finally fixed it, to people who are confused please listen. In some videos you see that there is a spark at the spark gap every 2 seconds or so. This is not gonna be true all the time, in fact there was hardly anything wrong with my laser at all! I just had to leave it plugged in. Yes its loud as hell and it looks like its just sparking with no let up but if you look left at your piece of paper to see the beam, you will be surprised to see that everything has worked!
Check out my laser working
<a href="https://youtu.be/lU-eeZtHVeE" rel="nofollow noreferrer">https://youtu.be/lU-eeZtHVeE</a></p>
| 19752 | Tea nitrogen laser debugging |
2018-03-07T11:16:01.540 | <p>I see tractors with really high wheels and with the carriage placed really high. Similar to this one:</p>
<p><a href="https://i.stack.imgur.com/YE4gN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YE4gN.jpg" alt="enter image description here"></a></p>
<p>How are the wheels powered? Is there is an electric motor for every wheel powered from diesel generator somewhere in tractor?</p>
| |mechanical-engineering| | <p>The wheels are powered by hydraulic motors, one for each wheel.</p>
<p>The pipes for supply and return can be seen on the left side wheels.</p>
<p>Edit:
It could be possible to provide an "ugly" (term as used in the comment below) solution with driveshafts, but, since driveshafts either with Hardy-Spicer joints or Constant Velocity joints have a maximum working angle then they would eat into the clearance available under the machine which is necessary for the crop to pass under without damage.</p>
<p>If one did use the driveshaft solution this would increase the height of the machine and, if the wheelbase stays the same, would increase the chances of it falling over when used on sloping terrain - not all countries are blessed with large flat fields...</p>
<p>Why don't they use electric motors? Well, a good guess is that most agricultural mechanics are already well-versed in hydraulics (all the ones I have met and worked with) and are more than capable of sorting most, if not all, hydraulic problems : spool valve blocks, oneway valves, pumps, motors etc. Going to electric would give them a completely new ball game and they would probably avoid purchasing machines and stick with what they know...</p>
| 19756 | How do tractors with a highly placed carriage power their wheels? |
2018-03-07T12:05:36.320 | <p>I found this <a href="https://engineering.stackexchange.com/a/295/15211">answer</a> to a related question. The part of the answer that's confusing me is: </p>
<blockquote>
<p>Transmitting DC power over a long distance is inefficient. Thus AC supply is a far more efficient to transmit power.</p>
</blockquote>
<p>According to Siemens it's quite the <a href="https://www.energy.siemens.com/hq/en/power-transmission/hvdc/applications-benefits/configurations/long-distance-power-transmission.htm" rel="noreferrer">opposite</a>: </p>
<blockquote>
<p>Whenever power has to be transmitted over long distances, DC transmission is the most economical solution compared to high-voltage AC.</p>
</blockquote>
<p>Also, from <a href="https://en.wikipedia.org/wiki/High-voltage_direct_current#Advantages" rel="noreferrer">Wikipedia</a> </p>
<blockquote>
<p>HVDC transmission losses are quoted as less than 3% per 1,000 km, which are 30 to 40% less than with AC lines, at the same voltage levels.</p>
</blockquote>
<p>Is the posted <a href="https://engineering.stackexchange.com/a/295/15211">answer</a> correct?</p>
<p><strong>- - EDIT - -</strong></p>
<p><strong>Chris H</strong> made a very important observation (see his comment below): The context of the post I mentioned was of <strong>low voltage</strong> whereas I was blindly thinking of high voltage. Indeed I learnt loads by the answers and comments. Thanks. </p>
| |power-transmission| | <p>All else being equal DC transmission is more efficient than AC transmission at the same nominal voltage due to the elimination of reactive losses.</p>
<p>However all else is rarely equal.</p>
<ol>
<li>At a given voltage DC is far more prone to sustaining arcs than AC.</li>
<li>It is only relatively recently that we have developed the ability to convert between DC voltages with reasonable cost and efficiency. At high power levels it is still more expensive and less efficient than transformers.</li>
</ol>
<p>The result is that DC systems have tended to operate at lower voltages than AC systems and this is what has got DC it's reputation for being inefficient.</p>
<p>Voltage has a massive effect on cost and/or efficiency of transmission. If you halve the voltage then to maintain the same level of resistive losses you have to quadruple the size of the conductors. Alternatively you have four-times the losses for the same size of conductors. So the ability to easilly convert voltage allowed efficient and economical AC transmission systems to be built. Even today AC remains the default option for most transmission and distribution needs.</p>
<p>The exception to this is very high power point to point power transmission over long distances, over undersea cables or between non-synchronised grids. In these cases the costs and hazards involved with converting the AC used on the grid to high voltage DC become more justified.</p>
| 19758 | Transmitting power over long distances what is better AC or DC? |
2018-03-07T14:27:55.693 | <p>this is my first project that i am working on. I want to make a IR sensor mount using stepper motors.</p>
<p><a href="https://i.stack.imgur.com/SrDyc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SrDyc.png" alt="enter image description here"></a></p>
<p>I seen this on youtube and would like to make a similar design.</p>
<p><a href="https://i.stack.imgur.com/Wn8H1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wn8H1.png" alt="enter image description here"></a></p>
<p>I plan on using aluminium mounting hubs to transmit the torque from the shafts to the aluminium frame as shown above. My question is how is the torque transmitted from the motors to this camera mount? Is there a bearing and if so, what kind? </p>
| |motors|torque|friction|bearings| | <p>I don't think there's any bearings in there. The mounting hub held rigidly onto the motor shaft with one or two grub screw(s), and the hub itself is rigidly screwed onto the frame with the 4 threaded holes you can see in the photo. As long as the motor can provide enough torque, when the shaft turns, the rest (hub & frame) will follow. Of course, backlash is quite likely to be present, and the motor control is of paramount importance if you want a smooth, jerk-free, motion of the frame, but in essence it's pretty simple, mechanically at least.</p>
| 19762 | Pan and tilt mechanism |
2018-03-07T17:11:54.707 | <p>Today while driving I had to make a quick shift to a lower gear that was not as smooth as it should've been. I've been taught that while shifting to a lower gear you have to be more careful when releasing the clutch than while shifting a gear up. Why is this? I understand the basic mechanism of transmission and clutch. When shifting up, the car can "bump" slightly if the rotational speeds of the engine and wheels are far apart. But why does this effect seem more noticeable when shifting <strong>down</strong>?</p>
| |automotive-engineering|engines|transmission| | <p>That's because when you upshift, you select a lower ratio, so your clutch speed drops. Since you took your foot of the gas, the engine rpm also dropped, and now the clutch and engine speed are close to synced, and hence little force is felt when you engage the clutch. The best way to upshift is to relieve the throttle a little(not fully), so the rpm will drop a bit as soon as you declutch. Meanwhile, shift up and clutch. After practice, no force will be felt. Every gearshift will in/-decrease rpm by about 20-25%. So aim with the throttle at a rpm 20-25% higher or lower when resp. down- or upshifting.</p>
<p>When you downshift, you select a higher ratio, so the clutch speed increases. You should apply a little throttle so the engine spins up and matches the increased clutch speed. Again, you'll feel no force when engaging the clutch.</p>
<p>If you don't do this, the clutch will have to make up for the difference in speed from the engine and the clutch. That costs time and gives you the annoying feedback.</p>
<p>The best way to downshift, is to keep your foot the same on the throttle as you were while cruising. As soon as you disengage the clutch, the engine rpm will increase. Now quickly downshift and re-engage the clutch. While you briefly disengaged the clutch, the rpm rose just enough to match the clutch in the lower gearing, if you did it right. But you have to shift and clutch quickly. It will require some practice, but you'll have supersmooth shifts and impress your passengers.. :p On top of this, you'll save on some wear on the clutch.</p>
| 19771 | Manual transmission: Why downshifting is less smooth than shifting up? |
2018-03-08T00:04:10.663 | <p>Recently came across the following 2-minute YouTube video that claims <a href="https://www.youtube.com/watch?v=KhHrJ45bSVE" rel="nofollow noreferrer">A Balanced Rig Saves Fuel</a></p>
<p>While nicely done, it is a marketing video. Does anyone know of any unbiased studies that support the claim?</p>
| |applied-mechanics|fuel-economy| | <p>Consider to simplify the model to two wheels in line as in a bicycle. For nice round figures, the tires are inflated to 100 pounds per square inch.</p>
<p>The load is two hundred pounds and is placed over the rear tire in such a way as to fully load the rear axle and completely unload the front.</p>
<p>The footprint of the rear tire is now two square inches.</p>
<p>In order to accomplish that footprint, the tire has to deflect along the sidewalls in such a manner as to substantially distort the rubber. This distortion absorbs energy that would otherwise be used to propel the vehicle.</p>
<p>A similar situation is in place when a tire is underinflated. I have a cart similar to that shown in the video. When it is heavily loaded and the tires are at 45 psi, the cart rolls easily. One can allow that the tire footprint matches more appropriately the engineering of the tire and that sidewall flex is at a minimum.</p>
<p>When the tire is underinflated (often!) at say, 20 psi, the footprint is greater, the flex is greater and it sure is much more difficult to pull when loaded.</p>
<p>For a vehicle such as a semitrailer rig, heavy loads over one axle would not necessarily cause visible sidewall flex, but it could be mechanically significant.</p>
<p>The above assessment is a simplification, but appears in the video to be substantiated by the rolling cart test process. When you see someone pulling an airliner with his teeth, you can bet the load is evenly distributed and the tires are either properly inflated or are over-inflated for the purpose of the demonstration.</p>
| 19783 | does payload weight distribution affect rolling resistance? |
2018-03-08T03:22:34.097 | <p><a href="https://i.stack.imgur.com/HdEkH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HdEkH.jpg" alt="enter image description here"></a></p>
<p>Does anyone know what these things on the ceiling of this building are called?</p>
<p>I am referring to those objects with semicircle shape that act as a ceiling on the
top of the building.</p>
| |civil-engineering|concrete| | <p>This is Kuwait National assembly building designed by Jorn Utzon, Danish architect. In your image the building is under construction.</p>
<p>The objects are permanent formwork for roof concrete slab. They create the grooves on top that is going to be populated with reinforcement and mold the stem of T-beam Roof slab.</p>
<p>At the same time, they are nicely arched on the bottom and will play as the dramatic voluted ceiling of the building.</p>
<p>He utilizes this type of architectural feature in many of his designs.</p>
| 19787 | What are those objects of the top of this building called? |
2018-03-08T18:11:32.127 | <p>I have a Structures question about determining strain energy that I am unable to solve. <strong>I have attempted it and I have the answer, I would appreciate if if someone can just tell me where I am going wrong.</strong></p>
<p>Q: A solid conical bar of circular cross-section is suspended vertically. The length of the bar is $L$, the diameter at the base is $D$ and the weight per unit volume is $\gamma$ (equivalent to density x gravity). Determine the strain energy of the bar due to its own weight.</p>
<p>$$\text{Answer} = U = \dfrac{\pi D^2 \gamma^2 L^3}{360E}$$</p>
<p>My work is as follows:</p>
<p>$$\begin{align}
U &= \dfrac{\gamma((1/3)\pi r^2 L)^2)L}{2E\pi r^2} \\
U &= \dfrac{\gamma(\pi^2r^4(1/9)L^2)L}{2E\pi r^2} \\
U &= \dfrac{\gamma\pi r^2L^3}{9\cdot2E} \\
U &= \dfrac{\gamma\pi(d^2/4)L^3}{18E} \\
U &= \dfrac{\gamma\pi d^2L^3}{18\cdot4E} \\
U &= \dfrac{\pi d^2 \gamma L^3}{72E} \\
\end{align}$$</p>
<p>I am incorrect by a factor of 1/5. Where in my working/logic did I go wrong?</p>
<p>I know:</p>
<p>$$U = \dfrac{(\rho gAL)^2L}{2EA}$$</p>
<p>I assume:</p>
<p>$$AL = \text{Volume (of a cone)} = (1/3)\pi r^2 h$$</p>
<p>I am confused as to whether the A in the denominator is the base of the
cone or the volume divided by the height.</p>
<p>I can derive the correct numerator, however I am not sure how my
lecturer arrived at 360E for the denominator.</p>
| |mechanical-engineering|structural-engineering|civil-engineering|structures|energy| | <p>We assume that the wide end of the cone is up, of course—otherwise the stress would go to infinity as the full weight of the cone is applied to a cross-sectional area that tapers to zero. (I incorrectly specified the opposite configuration in my comment above.)</p>
<p>Measure the coordinate $z$ from the bottom (i.e., from the tip of the cone). Then, the diameter at any location $z$ is $$D(z)=\frac{Dz}{L}.$$ Therefore, the corresponding area is $$A(z)=\frac{\pi D(z)^2}{4}.$$ The volume hanging underneath location $z$ is $$V(z)=\frac{\pi D(z)^2z}{12},$$ with corresponding weight $W=\gamma V(z)$ applied across the cross section. Therefore, the stress on any horizontal infinitesimal slice of thickness $dz$ at location $z$ is $$\sigma(z)=\frac{W}{A(z)},$$ and the volumetric strain energy within that slice is thus $$dU=\left(\frac{\sigma(z)^2}{2E}\right)A(z)dz,$$ where I've applied the linear elasticity result that the differential strain energy <em>per unit volume</em> can always be expressed as $du=\sigma\,d\epsilon=\frac{\sigma}{E}d\sigma$ by Hooke's Law, or, integrated, $u=\frac{\sigma^2}{2E}$. Now, integrate $dU$ from $z=1$ to $L$.</p>
| 19799 | Determine the strain energy of a bar |
2018-03-08T21:17:49.567 | <p>I try to design a shaft with yet unknown diameter, i know the distribution of forces and stresses (shear, torsion and normal) in/ along the shaft. I need just one more parameter nl. the tensile strength of martial. I use Rolloff-Matek machine elements table book. In the table i choose a material with tensile stress say $1000 Mpa$, but all the values in the table are normalised for a rod with diameter $d_N=16 mm$.
May i use the tensile stress as mentioned in the tabel ? or i should find a correction factor somewhere else ? <a href="https://i.stack.imgur.com/gdK0T.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gdK0T.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|materials|design|stresses|strength| | <p>There are correction factors in academic table books.
A very detailed explanation is in the "FKM Richtlinie" chapter 1.2 material characteristic.</p>
<p>But you mentioned that you have the book Roloff / Matek: </p>
<ul>
<li>Look at formula 3.7: <span class="math-container">$R_m=K_t*R_{mN}$</span> and <span class="math-container">$R_p=K_t*R_{pN}$</span></li>
<li><span class="math-container">$K_t$</span> is the technological size influence factor – not sure if this is the right translation</li>
<li>And <span class="math-container">$K_t$</span> you can find in your table book. Table 3.11 a) and b) depending what kind of metal you are using</li>
<li>The values of <span class="math-container">$R_{mN}$</span> and <span class="math-container">$R_{pN}$</span> are given in your already shown table. I guess you picked 41Cr4, then <span class="math-container">$R_{mN}$</span> is 1000 MPa and <span class="math-container">$R_{pN}$</span> is 800 MPa</li>
<li>41Cr4 is a metal out of the group “heat treatable steel” – so you have to take table 3.11 a) graph 3</li>
<li>Now you take your desired diameter, say 30 mm, and then you get <span class="math-container">$K_t=0.93$</span> or calculated with the formula <span class="math-container">$K_t=1-0.26*lg(30/16)=0.929$</span></li>
<li>Put <span class="math-container">$K_t$</span> in the formulas above you get <span class="math-container">$R_m=930 MPa$</span> and <span class="math-container">$R_p=744 MPa$</span></li>
</ul>
<p>An additional hint also concerning the comments: the <span class="math-container">$N$</span> in <span class="math-container">$d_N$</span> has nothing to do with the normalization of metal (the heat treatment)! <span class="math-container">$N$</span> stands for standard part size. In your table (nearly) all metals are tested with <span class="math-container">$d_N=16 mm$</span> to have comparability.</p>
| 19801 | How to determine tensile strength while the normalised tensile strength is known? |
2018-03-08T22:37:05.570 | <p>I'm currently taking classical control theory and have some questions regarding basic concepts related to closed loop systems. </p>
<p>Let's say I have a simple servo motor controlled by an electric current. Using a PID controller, my block diagram is as follows:</p>
<p><a href="https://i.stack.imgur.com/CmdCa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CmdCa.png" alt="enter image description here"></a></p>
<p>Now, I'm still struggling to understand some things:</p>
<ol>
<li>At what point is my loop actually producing a signal that my servo can use? And how would I extract it from my control loop?</li>
<li>Why can I simply take my PID'ed error signal and plug it into a plant requiring current as an input? If I had a set current I could calculate the torque on my motor, from which I could then determine the accerleration and thus get the position by integrating that value twice. How do I do such a calculation if I only have a value that tells me to either increase or decrease my current by some factor?</li>
<li>Why is the positional output calculated by my plant reconnected to my sensor in the feedback loop? Why is there an output in the first place? Shouldn't the sensor simply deliver continous data and the loop respond according to my error term?</li>
</ol>
<p>I would be very glad if somebody could answer these basics questions for me, I just can't quiet wrap my head around the basic mechanics of a control loop like this.</p>
<p>Thanks in advance!</p>
| |control-engineering|control-theory|pid-control| | <h2>Your system:</h2>
<p>You have a PID connected to a servo. So you have a closed system connected to a closed system. A servomotor already makes a control loop, that's what being a servo means. So more than likely the servomotor already has a PID inside it (or some other controller), so you don't really need a PID to connect to the servo. you can just connect the <span class="math-container">$P_C$</span> to the servo if you know how it is mapped by the controller in the servo and it is driving by the value you want.</p>
<p>However, it is possible you want to drive some other property than the servo drives. For example lets suppose the servo drives position instead of speed. Or the motor system after the servo is somewhat unknown/hard/problematic and you don't want to drive the value by a measurement at the engine then you want to have a second controller to dive your servo.</p>
<p>Or your system is highly nonlinear and unstable and you want to stabilize it based on 2 parameters. Like say a inverse pendulum.</p>
<p>Or just because you happened to have a servo available.</p>
<h2>Ok for your questions:</h2>
<p>Lets start with the simple one first.</p>
<blockquote>
<ol start="2">
<li>Why can I simply take my PID'ed error signal and plug it into a plant requiring current as an input?</li>
</ol>
</blockquote>
<p>You can, what you effectively then have is a P-Controller with no adjustable gain. It may work, but it can be too slow to react, or too fast. Also it does not have D and I terms so it can not react to sudden and persistent errors. See a P controller rarely reaches the target, it reaches just under or just above the target as the equilibrium of the computation is not spot on target. The I factor would take care of this. Most likely you dont need the D factor.</p>
<blockquote>
<p>1a. At what point is my loop actually producing a signal that my servo can use?</p>
</blockquote>
<p>After the summing of the PID terms. The control signal inside the servo may be a bit harder to obtain without dismantling the servo.</p>
<blockquote>
<p>1b. And how would I extract it from my control loop?</p>
</blockquote>
<p>Connecting a wire to the the node and measuring voltage?</p>
<blockquote>
<p>3a. Why is the positional output calculated by my plant reconnected to my sensor in the feedback loop?</p>
</blockquote>
<p>Its not (atleast not ususally). The sensor is connected to the real plant output. If your doing simulation than the calculation there to work as a stand in for the real deal so that you can test your understanding, the real plant may or may not work as expected.</p>
<blockquote>
<p>3b. Why is there an output in the first place?</p>
</blockquote>
<p>Presumably your plant does something with the output.</p>
| 19803 | How does a closed loop control system actually work? |
2018-03-09T04:51:48.610 | <p><img src="https://i.stack.imgur.com/di8nb.jpg" alt="enter image description here"><img src="https://i.stack.imgur.com/MAYKd.jpg" alt="enter image here"><img src="https://i.stack.imgur.com/axuin.jpg" alt="enter image description here"></p>
<p>In question 3-37,we have
P-p(ATM)=density * g* h and we get h as 8.3 m
But the answer is 1.53 m.How and why?What is wrong in my approach?</p>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>Pressure due to mercury is:
Pm = 13600*9.81*0.12
= 16010</p>
<p>So, that is the pressure created by the blood and it will then rise:</p>
<p>hb = 16010/(1040*9.81)
= 1.57m</p>
<p>Obviously, I have used constants I am aware of - subtle differences will give 1.53m...</p>
<p>Edit the density of mercury is 13560kg/m3 so that is where my error comes in...</p>
| 19805 | Pressure measurement |
2018-03-09T07:29:40.143 | <p>I have a structures problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track. Any assistance would be highly appreciated.</p>
<p><a href="https://i.stack.imgur.com/7yhgw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7yhgw.png" alt="Instructions"></a></p>
<p>My work is as follows:</p>
<p>At point B:</p>
<p>$T = Fd$, where $d = 300\text{ mm}$ and $F = 100\text{ N}$, therefore $T = 30,000\text{ Nmm}$</p>
<p>From torsion, shear stress $\tau = Tr/J$</p>
<p>$$J = \dfrac{\pi}{2}\cdot(r_1^4 - r_2^4)$$</p>
<p>Therefore $\tau = \dfrac{30,000 \cdot 30 \cdot 2}{\pi(30^4 - 25^4)} = 1.37\text{ MPa}$</p>
<p>This is the first answer listed, <strong>which leads me to believe my lecturer listed the answers for (B,A) instead of (A,B)?</strong></p>
<p>However then for point A:</p>
<p>$T = Fd$, I believe $d$ now changes to 330 mm as you have to add the outer
radius of the bar so $T = 33,000\text{ Nmm}$</p>
<p>Following the same process used for B:</p>
<p>$$t = \dfrac{33,000 \cdot 30 \cdot 2}{\pi(30^4 - 25^4)} = 1.50\text{ MPa}$$</p>
<p><strong>This is clearly not the answer my lecturer gave. I am a confused as to
where I went wrong.</strong> I am aware that I did not use the value 250mm but I
am unsure where it applies. <strong>What makes points A and B different, aside
from their distance from the force?</strong> I would really appreciate any assistance in the matter, thank you so much for reading.</p>
| |mechanical-engineering|structural-engineering|structures|steel|stresses| | <p>The reason is, that the pipe is also subject to transverse shear.
<a href="https://i.stack.imgur.com/15FeQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/15FeQ.jpg" alt="enter image description here"></a></p>
<p>This is due to the fact, that this lever doesn't apply a "pure" torque moment, but an eccentric force, which results in torque and shear.
Now you get two different shear stresses when you consider A and B.
As this pipe is subject to the shear force $V_z=100N$, the transverse shear "flows" from B over A towards the opposite side of B.</p>
<p><em>(note: There would be a bending moment $M_y=100N\cdot L_{pipe}$ the base of the pipe as well, but it doesn't result in transverse shear.)</em></p>
<p>You can determine the transverse shear stress, using the following formula
$$ \tau=\frac{V_z\cdot Q_y}{I_y\cdot t} $$
Take a look at <a href="https://engineering.stackexchange.com/questions/15006/shear-stress-formula/15015#15015">this</a> question for further explanation of transverse shear.</p>
<p><strong>further explanation</strong> <em>[edited]</em></p>
<p><a href="https://i.stack.imgur.com/f7xIE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f7xIE.png" alt="![enter image description here"></a></p>
<p>Take a look at this graphic, especially at the transformed coordinated systems. (By convention, a dot in a circle depicts a force/direction facing away from you, and a cross in a circle is a force/direction facing towards you)</p>
<p>In sketch $II$ you can see, how the $100N$ force does create a bending moment, the orange line is a sketch of the deformation curve, that said force would cause.</p>
<p>On the other hand, in sketch $III$ you can see two opposing eccentric forces, resulting in the same torque $T=2\cdot(50N\cdot 0.3m)=30Nm$, which would <em>not</em> result in a bending moment, as they cancel each other out. $M_y=50N\cdot L_{pipe}-50N\cdot L_{pipe}=0$. Furthermore, there would be no transverse shear stress in the pipe as well.</p>
| 19810 | Caulculating maximum shear stress at certain points |
2018-03-11T21:01:48.703 | <p>How do I analyse for, or what do I need to consider, when comparing gravity-opposed vs gravity assisted flow in a tube containing a porous bed that partially shrinks (through partial mass loss) over a fixed time period? </p>
| |chemical-engineering|porous-medium| | <p>With gravity assisted flow, you have a higher potential of channel flow through the media, so sections of your plant packing won't be wetted by the solvent. Making sure you have the optimal flow pattern is one of the primary factors in packed column performance. You indicated a 10 to 15% volume drop in the plant matter, so this poses an issue that over time the solvent flow in will squish the plant matter against one end of the container; the magnitude of this problem will depend on when the extraction of the molecules you want takes place. It's also possible that you could fluidize this plant material after the volume starts to decrease. Fluidizing could be beneficial to get high surface area but you have to control it to avoid the bed compaction effect mentioned above.</p>
<p>The main thing to measure is the pressure drop through the column. Too low a pressure drop means you have channel flow and too high a pressure drop can indicate plant matter compaction. If you can use a clear container to hold the plant matter, you can tune your setup to keep the flow pattern you want - provided you can control inlet pressure of your solvent.</p>
| 19838 | Up flow vs down flow through a porous fixed bed |
2018-03-11T21:17:07.213 | <p>What I've known in my robotics course is that control resolution is given by this equation:
$Control Resolution = Total Range/2^n$
I'm actually confused whether $2^n$ is the number of bits or $n$ is the number of bits, and if the first case is the right one, what does $n$ stand for then, the same for the 2nd case, what does $2^n$ stand for? Thanks in advance.</p>
| |robotics| | <p>$2^n$ is the number of increments and $n$ is the number of bits</p>
| 19839 | Robots Control Resolution |
2018-03-12T03:31:54.563 | <p>As the title suggests, why not use high-powered air pumps instead of ceiling fans to inject hot air from the ceiling down to the cold, bottom floor? </p>
<p>I understand it may seem silly (especially since it may require a lot of energy), but wouldn't it work better than a series of fans in a large, tall warehouse for example? </p>
<p>Wouldn't the pressure differential in air pumps be sufficient to force the hot air at the ceiling down to the floor further and faster than ceiling fans?</p>
| |mechanical-engineering|design|heat-transfer|hvac|building-design| | <p>Mainly cost difference.</p>
<p>And also, a fan is sufficient for the job. The problem being solved is thermal stratification. That's caused by a lack of vertical mixing; the warm air rises to the top, the cold air sinks.</p>
<p>Normally, there's quite a bit of movement in the first two metres or so from the ground - thermal stratification starts when the ceiling gets to around 3 metres or more.</p>
<p>Fans introduce enough vertical mixing, that when combined with general activities at ground-level, the thermal stratification gets sufficiently disrupted, for pretty low capital and operating costs, and with a device that's very easy to fit and to maintain, has a long life, doesn't have filters that need cleaning, and is generally quite quiet in operation.</p>
| 19842 | Why is it more common to use ceiling fans instead of air pumps? |
2018-03-12T21:28:31.083 | <p>I want to convert my son's pedal car to electric. The car is meant for 3-6 year old children.</p>
<p>So far I got the following:</p>
<ul>
<li>50W, 3300 RPM <a href="https://www.dunkermotoren.com/en/products/brushed-dc-motors/detail/8844201221/" rel="nofollow noreferrer">electric motor</a> with 8mm round shaft (no keyway or shaft groove)</li>
<li>10mm threaded rod made of some fairly soft metal to use as a shaft</li>
</ul>
<p>I was thinking to connect the motor to the shaft with XL timing pulleys. With a 10 teeth pulley connected to the motor and a 100T one on the shaft, I would get a 10X RPM reduction and with 10-inch wheels, I calculate that the max speed of the car will be ~15KM/h. Probably still a bit to high but I could do further electronic limiting.</p>
<p>Now I have a few questions:</p>
<ol>
<li>Would there be issues if I combine the smallest pully in the range with the largest one? It seems XL-10T is the smallest and XL-100T is the largest.<a href="https://i.stack.imgur.com/4P7mh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4P7mh.jpg" alt="Timing pulleys"></a></li>
<li>Considering these easily available pulleys are almost always mounted with set screws would the grip be sufficient for this application?</li>
<li>What is the required belt length if the distance between shafts is ~90mm?</li>
<li>What is a practical way to tension the belt for something like this?</li>
<li>Is 10mm thick enough for this belt?</li>
</ol>
| |mechanical-engineering|pulleys|mechanical| | <p>Not being an engineer, I'd run whatever setup fits the confines of the car and wire in a pulse width modulator. You can tune the output of the electric motor to fit your application. Not very pretty, but it will work and PWMs are about <span class="math-container">$10-$</span>15 on ebay.</p>
| 19851 | timing pulleys setup for children's pedal car |
2018-03-13T20:29:32.190 | <p>How can I find a region of attraction for given system? $k<0$</p>
<p>$$\begin{align}
x' =-kx+x^3\\
\end{align}
$$</p>
| |electrical-engineering|control-engineering|control-theory| | <p>If $k < 0$ and the system is $x' = -kx+x^3$ it can be rewritten to $x'=kx+x^3$ with $k > 0$.</p>
<p>Since the equilibrium is $x = 0$, there are two interesting regions, $x < 0$ and $x > 0$.</p>
<p>Region 1, $x < 0$:
$$x'=\underbrace{kx}_{<0} + \underbrace{x^3}_{<0} < 0.$$
Region 2, $x > 0$:
$$x'=\underbrace{kx}_{>0} + \underbrace{x^3}_{>0} > 0.$$</p>
<p>Summarizing the results in a "phase portrait''</p>
<p><a href="https://i.stack.imgur.com/tF8g6m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tF8g6m.png" alt="Phase portrait 1"></a></p>
<p>Here it can be seen that the equilibrium point $x = 0$ is repulsive (unstable), therefore there is no region of attraction (i.e. the region of attraction is an empty set).</p>
<p>However, the question makes more sense for $k > 0$ and $x'=-kx+x^3$.
This system has three equilibria, $x = \pm \sqrt {k}$ and $x = 0$, therefore there are 4 interesting regions:</p>
<ol>
<li>$x < -\sqrt{k}$</li>
<li>$-\sqrt{k} < x < 0$</li>
<li>$0 < x < \sqrt{k}$</li>
<li>$x > \sqrt{k}$.</li>
</ol>
<p>Region 1, $ x < -\sqrt{k} $:
$$ x'=-kx+x^3 < 0 $$
Region 2, $ -\sqrt{k} < x < 0 $:
$$ x'=-kx+x^3 > 0 $$
Region 3, $ 0 < x < \sqrt{k} $:
$$ x'=-kx+x^3 < 0 $$
Region 4, $ x > \sqrt{k} $:
$$ x'=-kx+x^3 > 0 $$</p>
<p>Summarizing the results in a "phase portrait''</p>
<p><a href="https://i.stack.imgur.com/4etqRm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4etqRm.png" alt="Phase portrait 2"></a></p>
<p>Here it can be seen that the equilibrium point $x = 0$ is attractive (stable), while the equilibria $x = \pm \sqrt {k}$ are repulsive (unstable).
Hence the region of attraction is $-\sqrt{k} < x < \sqrt{k}$.</p>
| 19863 | A region of attraction analysis using invariant sets |
2018-03-14T17:27:30.277 | <p>I am looking for a replacement bearing for an old lathe. The lathe spindle assemble features a collet-like bronze bearing with a nut for tightening the inner diameter. </p>
<p><a href="https://i.stack.imgur.com/umA5G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/umA5G.jpg" alt="enter image description here"></a></p>
<p>Unfortunately, I cannot find this style bearing at the required size. Is it possible to use a bearing adapter sleeve inside a plain bronze bearing? </p>
<p><a href="https://i.stack.imgur.com/WFCiw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WFCiw.jpg" alt="enter image description here"></a></p>
<p>If not, are there any suitable alternatives that would allow me to reduce shaft runout?</p>
| |mechanical-engineering|bearings| | <p>I would check McMaster and Misumi to see if a a stock bearing or flange bearing would fit. If not get one sizer smaller and (hand) ream to inside diameter needed.</p>
<p><a href="https://www.mcmaster.com/#bronze-bearings/=1c3onde" rel="nofollow noreferrer">https://www.mcmaster.com/#bronze-bearings/=1c3onde</a>
<a href="https://us.misumi-ec.com/vona2/mech/M0100000000/M0107000000/" rel="nofollow noreferrer">https://us.misumi-ec.com/vona2/mech/M0100000000/M0107000000/</a></p>
| 19873 | Can a sleeve bearing adapter be used with a plain bronze bearing? |
2018-03-15T07:31:05.310 | <p>I am having trouble conceptualizing this problem. I have attempted it but am confused about a few things, I would highly appreciate any insight or advice for this.</p>
<p><a href="https://i.stack.imgur.com/ImAX0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ImAX0.png" alt="image question"></a></p>
<p>I know you need to resolve P into its Y and Z components and that you can move them to the end of the beam.</p>
<p>I understand Py causes bending in the XZ plane and Pz causes bending in the YX plane.</p>
<p>I thought that Pz contributed to σzz and Py contributed to σyy.</p>
<p>But, my lecturer said that both Py and Pz contribute to σxx.</p>
<p><strong>I'm a little confused as to how this is possible when they are orthogonal to each other.</strong> </p>
<p>Both cause bending on the beam (Py causes torsion) but do they not bend the beam in different planes therefore <strong>shouldn't they contribute to different stresses?</strong></p>
| |mechanical-engineering|structural-engineering|materials|structures|stresses| | <p>The confusion seems to arise from a misconception about how $\sigma$ is defined. For example, $\sigma_{yy}$ corresponds to a force in the $y$ direction applied to a surface facing the $y$ direction. Both $\sigma_{yy}$ and $\sigma_{zz}$ are zero essentially everywhere in the shaft (idealizing it as long and narrow) because there's no distributed region where a $y$-or $z$-direction force is internally or externally applied to a $y$- or $z$-direction surface, respectively. </p>
<p>(I'm not including the point loads at either end, which are localized. It's implicit in these problems that stress concentrations can be ignored and that failure doesn't arise directly in the attachment, though it may occur within the shaft <em>adjacent</em> to the attachment or point of load application.)</p>
<p>Instead, $\sigma_{xx}$ and $\tau$ are the key parameters (and the only nonzero stress components in this idealized problem), and the first varies strongly with the location on the shaft. For example, the $y$-direction component of $P$ causes a location-dependent tensile stress $\sigma_{xx}$ (i.e., locations with positive $\sigma_{xx}$) on the top of the shaft as a result of the bending moment). The $z$-direction component causes a location-dependent compressive stress $\sigma_{xx}$ on the left side of the shaft, also arising from the bending moment. The torque causes a shear $\tau$ that's maximized on the surface of the shaft. </p>
<p><strong>The key point is that a single stress component is not necessarily uniform in an object</strong>; here, for example, $\sigma_{xx}$ varies across the surface and interior of the shaft.</p>
<p>This is a very important point, so I suggest reviewing a good mechanics of materials text (e.g., Beer and Johnston) until it's familiar.</p>
| 19876 | Determining σxx on a 3d system with resolved forces (stress, strain, shear strain energy) |
2018-03-15T17:45:01.977 | <p>I need to use this to find a motor that will be able to pull at most 2250 kg at a 1.5 m/s pace. At peak acceleration, I would like this to accelerate at 0.5 m/s<sup>2</sup>. The area that the load will be pulled is a concrete sidewalk with, at its most extreme, a 20° incline.</p>
| |motors|applied-mechanics|torque| | <p>Ignoring friction, because you have not provided any details regarding this in your question, and assuming that the object starts from rest on a $20°$ incline, and accelerates at a peak rate of $0.5m/s^2$ on it's way to a max speed of $1.5m/s$</p>
<p>The force required to stop a static object from rolling down a hill is the same as that required to keep it at a constant speed on that hill, $F_1=mg\sin\theta$.</p>
<p>The additional resultant force required to provide an acceleration of $0.5m/s^2$ can then simply be added using $F=ma$, therefore,</p>
<p>$$F=m(g\sin\theta+a)$$</p>
<p>Which equates to a peak force in your case of:</p>
<p>$$F=2250(9.81\sin20+0.5)=8674=8.7kN$$</p>
| 19880 | How much torque will I need to pull a 2250Kg load? |
2018-03-16T00:22:42.397 | <p>Is there any benefit to using triaxial weave pattern instead of common twill?</p>
| |carbon-fiber| | <p>No, it is not a gimmick. Triaxial weaving has been around for a long time, and can be seen in baskets, hats, etc. etc. It often exhibits superior structural properties such as greater resistance to shear forces, tearing, bursting etc. The sheet will also be generally more isotropic, so if you don't know which direction your forces will be coming from, it may be a better choice.</p>
<p>Whether there is any <em>benefit</em>, however, depends entirely on the specific application. You may not need any of the above properties, making the increased cost not worth it. Further, if you are exclusively loading the fibres in one orientation, then a square weave might be a better choice. What has prompted this question?</p>
| 19889 | Is triaxial carbon fiber gimmick? |
2018-03-16T04:28:12.683 | <p>This is a pretty simple problem, but I'm just not sure how to go about it. I want to find the moment of inertia of a discontinuous hollow circle. But since the shape is so irregular, I'm not sure how to go about solving for it. I've posted the shape in mind, so hopefully that'll clear up any confusion. </p>
<p><a href="https://i.stack.imgur.com/122mM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/122mM.jpg" alt="discontinuous hollow circle"></a></p>
<p>Would an equivalence be needed to calculate the moment of inertia? For example, would I assume that the "new" diameter to calculate with for the outer circle is (0.625 - 0.375)? </p>
<p>Note: All these numbers have inches as units.</p>
| |statics|moments|geometry| | <p>very close to 1E-3 in^4</p>
<p>More precisely: 997E-6 in^4</p>
<p>I wouldn't (didn't) use either of the other algebraic proposals. It's relatively easy to write an equation for the width of cut face at a defined offset from the centre:</p>
<p>R.o = outer radius = 0.3125"</p>
<p>R.i = inner radius = 0.25"</p>
<p>G = gap = 0.375"</p>
<p>width across the outer circle (whether it's actually there or not) at y from centre = 2*sqrt(R.o^2 - y^2) by pythagoras</p>
<p>width across the inner circle = 2*sqrt(R.i^2 - y^2)</p>
<p>cut width = max(0, width across outer-max(G, width across inner)).</p>
<p>That's trivial (5 mins) on excel with a row calcuating the width of each of (say) 500 slices from -0.3125" to 0.3125".</p>
<p>I actually did it in a numerical computer algebra system to get 997E-6, when I do it on Excel and 500 slices I get 996E-6.</p>
<p>Hand check - by eye, it's similar to two rectangles each 0.0625" wide and 0.5" deep, which would have I 1.3E-3, so I probably trust my answer.</p>
| 19891 | Calculating moment of inertia for a hollow discontinuous circle? |
2018-03-16T12:39:49.210 | <p>I recently got interested in an electronic paper.
One design that I liked was the one from Mirasol, <a href="https://en.wikipedia.org/wiki/Interferometric_modulator_display" rel="nofollow noreferrer">Interferometric modulator display.</a> </p>
<p><a href="https://i.stack.imgur.com/5EF4Q.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5EF4Q.jpg" alt="enter image description here"></a></p>
<p>However, what I don't understand, is the following:
<strong>Their subpixels are either ON (color) or OFF (black).
Therefore, I am wondering how they can produce lighter and darker colors?</strong> </p>
<p>By lighter and darker colors, I mean that the color intensity is tuned. For instance for red: </p>
<p><a href="https://i.stack.imgur.com/2zN4G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2zN4G.png" alt="enter image description here"></a></p>
| |electrical-engineering|experimental-physics| | <p>If I understand that picture correctly, the brightness is a function of how many subpixels are turned ON in a given image frame. Your eye integrates all the photons it receives, so if all 16 red subpixels are ON, you'll perceive "red" but also perceive "very bright." If only one is ON, you still perceive "red" but will see "very dark."</p>
<h2>edit to add:</h2>
<p>It occurs to me that the perceived color could in theory be adjusted by toggling the pixels' states at, e.g., 240Hz to be ON for only part of a video frame. However, this has not been achieved in actual devices, and in any case would defeat one of the desired features of this technology, which is to maintain a color image without drawing additional power (the pixels hold state until addressed).</p>
<p>For those who don't know much about the retina, some info:</p>
<p>There are color cones, typically 3 types, each of which is sensitive to a different band of frequencies. The combination of signals from these cones is what the brain processes to sense a color. Along with these are rods, which are more sensitive detectors but are monochromatic (only report total intensity). Combine the intensity signal with the color signal to get teh final perceived shade.</p>
| 19901 | Different Color Intenensities from simple ON/OFF RGB subpixels |
2018-03-17T03:30:26.780 | <p>I want to find a suitable AC or DC gear motor for concrete paste mixing. The application details are given below</p>
<p>Material type - concrete paste
RPM - 30-200rpm
Qty of material - 30Liters
Mixing style - Planetary
To understand the type of mixer, you may consider bakery mixer. </p>
<p>Can somebody help me how to calculate the capacity of gear motor that I have to select for this application?</p>
| |motors|gears| | <p>Concrete mixers usually have 3-4 blades attached to the walls of the mixer and rotating with the mixer tilted 60-70 degrees. Let's assume the diameter of the mixer 0.5 meters</p>
<p>Assuming the blades lifting up carry half of the weight of the mix,
$$w=30/2 (2.5)\space with 2.5 = density\space of\space mix. $$
w = 37.5 kg. And the work mixer has to do at 100rpm is </p>
<p>$$W = (37.5)9.8.100( 0.5)/60seconds/mins= 306.25 watts$$
This is a very rough approximation. One would need to consider the clumping of the concrete, potentially clutching of the mixer, etc. So for a 200rpm mixer, you'd need a 612-watt motor, say one horsepower.</p>
| 19912 | How to find required size of AC gear motor for planetary mixing application? |
2018-03-17T15:39:56.610 | <p>I always thought 20 has to be used for voltage or field strength and 10 for electric power. Now I found this formula which uses power of antennas and they use the factor 20 instead of 10, why?</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Friis_transmission_equation#cite_note-AEH-1" rel="nofollow noreferrer">Friis transmission equation</a></li>
</ul>
| |electrical-engineering| | <p>Look carefully: the equation in linear units has $(\frac{\lambda}{4\pi d})^2$ .
Take the log (as part of the "dB = 10log(x)"), and notice that $log(x^2) = 2log(x)$ </p>
| 20917 | Decibel calculations, when to use 20 and when 10? |
2018-03-17T17:17:46.193 | <p>I am trying to model water gas shift reactor on Aspen using an Rgibbs reactor. Unfortunately whenever i pass the stream into the reactor(HTS) the amount of CO and H2 both are increasing and C in the system is not balanced. Aspen is not prompting an error regarding elemental balance not being intact. I am confused what should I do</p>
| |chemical-engineering|simulation| | <p>Without knowing too much about Aspen (I've been looking for a chemical engineering modeling software, so thank you), the documentation data says <strong>RGibbs</strong> <em>doesn't pay attention to stoichiometry</em>:</p>
<blockquote>
<p>The (Rgibbs Reactor) provides reaction calculations without the need for detailed stoichiometry or yield. The calculations are based on minimizing the Gibbs energy for the system as discussed in page 98"</p>
</blockquote>
<p>This reactor is intended for systems where the exact reactant composition is unknown but the energy is, such as "crude oil" or "natural gas". As such, this reactor won't notice a change in C concentration. It is probably adding the additional thermal energy to change the concentration of the higher thermal energy products, CO and H2, as it is ignoring the Redox mechanism that would force the consumption of CO to generate CO2 to form H2. A better reactor would be the <strong>REquil</strong>:</p>
<blockquote>
<p>When one or more reactions involved are equilibrium reaction, the REquil block can be used. The block requires knowledge of the reaction stoichiometry, and performs chemical and phase equilibrium reactions.</p>
</blockquote>
<p>Using this method would be a better model as the equilibrium constant's dependence on temperature is well documented:</p>
<p>$$K_{eq} = exp(-2.4198 +0.0003855T + \frac{2180.6}{T})$$</p>
<p>Where $T$ is in Kelvins, from the range of 600K - 2000K.</p>
| 20918 | Problems regarding aspen simulation |
2018-03-18T09:17:48.050 | <p>I am looking at a Flood Risk Assessment that uses HEC-RAS to model the predicted levels of flooding on a piece of land. As shown in the attached image, the report lists a number of variables with abbreviated names and I would like to know briefly what each one means and which one of those values shows the predicted height of flood water.</p>
<p><a href="https://i.stack.imgur.com/g5ndW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g5ndW.png" alt="My HEC-RAS data"></a></p>
<p>I found this document: <a href="https://www.novaregion.org/DocumentCenter/Home/View/286" rel="nofollow noreferrer">HEC-RAS Output Variables</a> which gives the full names of the abbreviated variable names. But the actual significance of each variable still make little sense to me.</p>
<p>Thank you in advance for any help you can give me.</p>
| |fluid-mechanics|modeling| | <p>HEC-RAS stands for Hydrologic Engineering Center - River Analysis System. It's an Army Corps of Engineers program for characterizing flow in river or other large open channel systems.</p>
<p>First of all, the output for your question is the channel data at one particular cross section in an open channel. In order from left to right:</p>
<p>Q = total channel flow in cubic meters per second</p>
<p>Min Ch El = elevation of the lowest point of the channel at that point (section)</p>
<p>W.S. Elev. = the elevation in feet of the water surface at that section</p>
<p>Crit W.S. = this is the elevation of the water surface at critical flow;
higher than this and the flow is subcritical, lower than this and the flow is supercritical. Think of a slow meandering river (subcritical) and a rapid, shooting river (supercritical). Compare this elevation to the actual WS Elev to determine which type of flow it is.</p>
<p>E.G. Elev = The is the elevation of the Energy Grade Line, which is the sum of the actual water surface elevation and the additional head derived from the flow velocity. This number represents the elevation if the flow velocity were cleanly directed upwards at this point in the channel. It's important to determine factors of safety of nearby facilities if the flow should become obstructed at this point.</p>
<p>E.G. Slope = The is just the slope of the above Grade Line. It's related to the slope of the channel bottom and the velocity together.</p>
<p>Vel Chnl = The average flow velocity for the channel in meters per second. This is an <strong>average</strong> just to get a feel for the overall flow.</p>
<p>Flow Area = Looking at the cross section of the channel, the area of the flow.</p>
<p>Top Width = the width of the flow section measured at the top free surface.</p>
<p>Froude Number = This number characterizes the flow as subcritical or supercritical. See above for the meanings. If this number is greater than 1, the flow is supercritical, and less than 1 means subcritical. If it's very near or equal to one, the flow can be unstable, varying (surging) back and forth between sub and super - critical. In this case it's well below 1 so the flow will be well behaved, subcritical, meaning essentially gentle flowing.</p>
<p>For a given channel, there will be several of these cross sections with outputs for each section. Put together, they form a picture of the overall flow characteristics for the entire length of the channel being evaluated. Good results come from good inputs, and it takes some experience to know where to take your sections for these programs to yield good results. </p>
| 20924 | Explain each of the values output by HEC-RAS |
2018-03-18T21:41:14.747 | <p>I am trying to calculate the static yield stress of a specimen using the
0.2% offset line on a stress strain curve. I found the static yield stress of the specimen to be the <strong>number in bold (104.1239)</strong>. Given the shape of the
graph and how close the static yield stress appears to be to the UTS, I was wondering if this value actually looked correct? Or could I have accidentally calculated the dynamic yield stress instead? </p>
<p>The material I used in this test was a "low grade aluminum"... which is throwing my thought process off.</p>
<p>I would really appreciate any insight on this.</p>
<p><a href="https://i.stack.imgur.com/Jj0mK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jj0mK.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|materials|structures|stresses| | <p>The mechanical properties of an aluminum alloy definitely vary with the various series designations. As I am not totally sure which series you are working with and how you are doing your calculations, I cannot say whether your actual strength values are correct. </p>
<p>However, it is possible that your yield strength and ultimate tensile strength are not very far apart. For example, in aluminum 1100, the difference between yield strength and ultimate tensile strength is only 5 MPa at room temperature. Your alloy may be similar in nature.</p>
| 20929 | Stress yield of a specimen, result confusion |
2018-03-19T14:57:02.650 | <p>I'm building a mount for a small IR sensor which will look very similar to this, minus the large camera.</p>
<p><a href="https://i.stack.imgur.com/J91Gk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J91Gk.png" alt="enter image description here"></a></p>
<p>I am wondering whether or not it would be necessary to do calculations to ensure that the structure is safe and will not collapse. </p>
<p>The potential problems I see are the bending moment due to the weight of the top motor. Would I calculate this and compare to the yield stresses in the bolts of the bracket? Potential buckling? Would it be worthwhile to use a software such as Ansys to run a stress simulation? The weight of the camera is very small and would not effect any calculations. I am using aluminium. Thanks</p>
| |mechanical-engineering|stresses|aluminum| | <p>In this sport of situation requirement for stiffness tend to dominate so for any halfway acceptable design actual yield probably won't be an issue as a design which is acceptably stiff will be fairly lightly stressed. </p>
<p>FE analysis may help you to optimise a design but its not foolproof and you still need a reasonable degree of knowledge and experience to set up the model properly and interpret the results. </p>
<p>Just by eyeballing the example image I would make a couple of points. </p>
<ul>
<li>The corner joints will tend to be the limiting factor, you're probably better off using short lengths of equal angle extrusion a L shaped brackets than the small blocks shown. </li>
<li>With aluminium is it generally better to bolt through with a nut and bolt then to drill and tap the aluminium itself. </li>
<li>Putting four bolts per side rather than two will greatly improve stiffness, you want to avoid putting any bending load on bolts. </li>
<li>Adjust the dimensions of the structure to keep cantilever lengths to a minimum, for a beam supported at one end only the maximum deflection is proportional to the <em>cube</em> of the unsupported length. </li>
</ul>
| 20932 | Stresses in a camera mount |
2018-03-20T11:43:06.153 | <p>(I'm not sure if this question belongs to this stack exchange, but it seems appropriate to me)</p>
<p>I have a very practical situation and our knowledge is lacking to be able to solve our debate.</p>
<p><strong>The situation</strong></p>
<p>In our country, and I'm guessing everywhere in the world, people seem to always fill their drinking glasses with about one or even half a page of newspaper. I'm talking about regular 20 or 30 cl drinking glasses or mugs.</p>
<p>When asked why they do this, it is to "protect the glass". The most scienty of them told me "the pressure given by the paper inside the glass helps hold its integrity in case the glass undergoing opposite pressure from the outside (which it most likely is considering it's packed in the same box with many other glasses).</p>
<p>My argument is that the glass being a solid and pretty much not flexible at all, you'd have to put a huge amount of paper in each glass to actually make a difference. </p>
<p>Ideally, not adding too much pressure from the inside to not break it that way, but filling it so that there is as little air as possible seems to me that this is the safest option. But that is by far the worst in terms of using paper and time consumption.</p>
<p>What I told them to do is focus on the outside of the glass when packing. Adding a few layers of very scrunched paper to act as some kind of "spring" all outside the glass, therefore spreading the pressure coming from the outside into the paper, instead of helping the glass sustain said pressure by adding pressure from the inside.</p>
<p><strong>The questions :</strong></p>
<ul>
<li>What solution works best to protect the glass ?</li>
<li>What solution is actually viable, not costing 20 minutes per glass and thousands of pages of newspapers.</li>
<li>Is there anything else to this?</li>
</ul>
<p>Note that I am not a native english speaker and there might be false friends between our languages.</p>
<p>By "pressure" I mean the act of applying a force onto a surface.</p>
<p>"Scrunched" was google translated, it should mean "the extreme opposite of flat, like when you make a ball with paper and then expand it again"</p>
<p>Thanks for reading and for your time :)</p>
| |structural-engineering|design|pressure|structural-analysis| | <p>I think of a few benefits of adding the paper inside:</p>
<ul>
<li>I think the glass is weaker from the inside than the outside, so any hard object somehow falling towards the inside of the glass would have a cushion.</li>
<li>In case the glass breaks, the paper inside might also be a better (or at least an extra) barrier for shards, possibly avoiding a chain reaction.</li>
</ul>
| 20941 | Is it useful to stuff my glasses with newspaper to protect them? |
2018-03-20T14:28:16.080 | <p>For a given 2D print, is it recommended to number section views and details views with the same alphabetical list, i.e. Section views A,B and C followed by detail views D, E and F, </p>
<p>or go with different alphabetical list, i.e. Section views, A, B and C, followed by detail views, A, B and C?</p>
| |technical-drawing| | <p>Your first suggestion is correct - there should not be two different views (of any type) with the same letter on any one drawing.</p>
<p>See the quote below from ASME Y14.3-2003</p>
<blockquote>
<p><strong>1.7.3 Identifying Removed Views.</strong> To relate the viewing
plane or cutting plane to its removed view, capital
letters such as A, B, C, etc., are placed near each arrowhead.
The corresponding removed views are identified
as VIEW A-A, VIEW B-B, VIEW C-C, etc. View letters
should be used in alphabetical order excluding I, O, Q,
S, X, and Z. When the alphabet is exhausted, additional
removed views shall be identified by double letters in
alphabetical order, as in AA-AA, AB-AB, AC-AC, etc.</p>
</blockquote>
<p>This applies to <em>all</em> "Removed Views", of any type.
I, O, Q, S, X, and Z are all excluded due to their visual similarity to 1, 0, 0, 5, [blank that needs filling in], and 2, respectively.</p>
| 20943 | How to number section and detail view for a same drawing |
2018-03-21T09:08:23.667 | <p>My pulleys are going to be rotated with a 1000-2000 RPM motor, with a constant RPM controlled by a hall effect sensor, there is going to be almost no torque as one of the pulleys just has a 100 gram optical element attached to it and all the pulleys are on bearings.</p>
<p>Any suggestions on choosing v belt or circular belt? Price is pretty similar and v seems to reduce risk of slipping and has better grip. Cant find a disadvantage of v belts.</p>
<p>I just assume slippage might become in issue with round belts at these RPMs if there isn't enough grip, no other reason to not choose them over v belts, really.</p>
<p>I'd imagine tooted belts wouldn't provide much advantage here besides any backlash or slippage when the system begins rotating but I don't see how it would be more precise after the speed is reached and the speed and load remain constant during operation.</p>
| |mechanical-engineering|motors|pulleys| | <p>Depends on many things. For example I have sometimes used round belts for these applications simply because i happened to have suitably sized O- rings which i could use as belts and didnt have V- belts around. This means i could have a working version today, instead of in a day or two. *</p>
<p>The groove for a round belt may be minimally easier to manufacture with a manual lathe. Also the round belt is slightly easier to install if you do not use a tensioner mechanism as its possible to roll it into place. </p>
<p>Remember that your feedback should come from the final axis thisway it hardly matters if your belt slips. Slippage may be minimal anyway, but if you are really concerned use a toothed belt instead.</p>
<p>* dont underestimate this, as you may end up doing some higher priority work in future. But then if you do not need to concern yourself with assembly and manufacturing choose what you think is best.</p>
| 20957 | choosing between v belt and round belt for fast RPM, low torque application |
2018-03-21T10:17:29.213 | <p>I have a piece of mahinery with a grease fitting in an awkward to reach place. The fitting is of this shape ...</p>
<p><a href="https://i.stack.imgur.com/m860K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m860K.png" alt="enter image description here"></a></p>
<p>... but not screwed into place but pressed. So we can't have it removed and replaced with a pipe leading to a more accessible place.
My question is: Can I mount a tube to the fitting in a way that it stays there for the foreseeable future (a few years)? Grease will be applied twice a year. I have doubts that simply pushing a tube over the fitting and applying a clamp will suffice, as greasing will put some pressure on the thing. Any ideas?</p>
| |mechanical-engineering| | <p>Did you also consider a 90 or 60 degree grease fitting. It would be a cheaper solution to piping to a remote manifold.</p>
<p><a href="https://www.mcmaster.com/#grease-fittings/=1c3ojei" rel="nofollow noreferrer">https://www.mcmaster.com/#grease-fittings/=1c3ojei</a></p>
| 20961 | How to mount a tube to a grease fitting? |
2018-03-21T14:26:35.027 | <p><a href="https://i.stack.imgur.com/GIJTr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GIJTr.jpg" alt="enter image description here"></a></p>
<p>My suggestion was that they are for Tamper-evident band, but there isn't one beneath the cap.<a href="https://i.stack.imgur.com/fIVMQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fIVMQ.jpg" alt="enter image description here"></a></p>
| |plastic| | <p>that part was injection molded in an unscrewing mold which allows fully-formed threads to be molded into a part. Ordinarily, the resulting undercuts would prevent the part from being extracted from the mold but in an unscrewing mold, a motorized shaft inside the mold unscrews the finished part from the mold after the plastic has solidified and the two halves of the mold are being separated. those features in the cap are engagement splines that mate into the rotating shaft that unscrews the finished part out of the threaded half of the mold cavity. </p>
| 20971 | Why there are grooves at inner side of plastic nasal spray cap? |
2018-03-22T10:50:33.193 | <p>I'm trying to fit a shaft into a hole in an acrylic piece that is 5mm thick to cause an interference fit. The shaft has a max. diameter of ⌀5mm and a min. diameter of ⌀4.988mm. I have access to a laser cutter with a precision of 0.2mm. This basically means that i have to cut the hole in the acrylic to a diameter of less than ⌀4.788mm, to ensure an interference fit right? Lets say I cut the hole in the acrylic, using a diameter of ⌀4.788mm, the diameter could be as low as ⌀4.588mm. Would the shaft of ⌀5mm even fit in this hole? Given the material and uncertainty in the laser cutter, what diameter do you think is suitable? Thanks</p>
| |mechanical-engineering|materials|tolerance| | <p>There are a couple of issues. Firstly laser cutting may well not provide good enough tolerances or surface finish to reliably produce the fit you need and you woudl probably be better off drilling the holes, possibly using the laser to create a reference mark or pilot hole. </p>
<p>Secondly poly-carbonate and acrylic don't really like this sort of interference fit and the advice is generally to give holes decent clearance as the stresses involved tend to cause cracks, especially if the holes aren't polished and chamfered. </p>
<p>You would probably be better off drilling a hole with adequate clearance and using an adhesive to secure the shaft. </p>
| 20990 | Interference fit in acrylic/plexiglass (PMMA) |
2018-03-23T03:11:30.910 | <p>I have a friend who isn't properly visualizing energy loss in a moving car.</p>
<p><strong>His idea:</strong> Cars that can run forever without being recharged - - > while the engine rotates the front wheels the move the car forward, there is a device at the back wheels using the back wheels' rotation to generate and store power. The engine draws this restored power to move the car forward and the cycle continues. Now, this is a "perfect" scenario where there's zero energy loss through friction, heat, sound, etc.</p>
<p>I first told him that something similar was already being done: regenerative braking. He said his idea was to generate energy **while* the car was moving.</p>
<p><strong>My counter to why that was rubbish:</strong>
I reminded him the amount of energy in a given system remains constant. Nothing created or destroyed, just transferred form one form to another. Thus he could not "recover" energy that was being spent to move the car forward. He moved on to elaborate that there would be two engines, one rotating the front wheels with an attached dynamo to recover energy, and another rotating the back wheels. I tried to explain he had added another energy consumer, with no way to recover this extra energy being spent...and that's how it would go. You add an extra engine to overcome the dynamo's resistance and move the car, <em>you are spending more than you are recovering</em>, and will eventually have to charge up. To help him visualize, I made a sketch, trying to be more realistic: <a href="https://i.stack.imgur.com/IQsSF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IQsSF.jpg" alt="How I imagine movement/expenditure of energy in vehicle"></a> (please ignore my use of battery capacity, it's only there for the concept, not to be realistic)</p>
<p>I also mentioned to him that this scenario was even ignoring the other loads of problems and dangers that present themselves with having wheels rotating at different speeds. Plus the wear on the slower rotating ties which are effectively being dragged against the road. For some never-to-be-discovered reason, he simply could not understand that to get the slower (front) tires to NOT be a problem, you have to increase the front engine power output for to match the wheel rotation speed of the rear wheels. Again, only recovering a fraction of that energy (where did the rest of it go? He can't understand and I can't figure out how to explain it to him). To counter my logic, he proceeded to mention that trucks don't have an engine attached to all their wheels. Hence, some energy can be captured from those "free-spinning" wheels. <em>Oh my God</em>.
Newton's first law of motion applies here: A car with the engine rotating the wheels towards a given direction goes nowhere if say, a dynamo is offering equivalent or greater resistance in the opposite direction to capture that energy (again, assuming a "perfect" scenario with no energy loss). I tried to elaborate that to move the vehicle, you first exert a force (engine power output) great enough to move the vehicle from its static position (resistance? gravity? I don't understand physics enough to know what to call this), and anything like a dynamo (friction, resistance) serves to make that more difficult, so you can't simultaneously "recover" energy you are spending. I proceeded to explain that regenerative braking worked because the generated energy was being recovered from the moving vehicle's momentum through offering resistance, hence why the car slows down and why you can't sustain-ably use it while the car is in motion. </p>
<p>This guy is super stubborn to being corrected once a very dim bulb lights up in his mind and he believes it a genius idea and can't be dis-proven. I am horrendous at teaching constructively, and would appreciate someone breaking down why his idea is a victim of basic laws of physics. </p>
<p>I even stated if that were possible, it would already have popular applications in industry like electric vehicles.</p>
| |automotive-engineering|energy|regenerative-braking| | <h1>If it worked he would not need the engine to maintain speed</h1>
<p>Ask your friend: </p>
<blockquote>
<p>If this is "perfect" — as you say — with no friction, no losses... then why do you need the engine to maintain cruising speed?</p>
</blockquote>
<p>Once you reach cruising (i.e. constant) speed, the job of an engine is to give energy to the rest of the car for all the energy it loses to air friction and road friction. </p>
<p>If your friend did not have air or road friction — say by being in a <a href="https://en.wikipedia.org/wiki/Hyperloop" rel="nofollow noreferrer">Hyperloop</a> — then all they would need is a shove to get up to speed, and <strong>then they would turn the engine off</strong> and just keep coasting. </p>
<p>So their own premise proves why it will not work, because if they <strong>need</strong> an engine to maintain speed, then they are having losses, and the system is not "perfect" as postulated.</p>
| 21002 | Why can't energy be entirely recovered from a moving car's rotating wheels? |
2018-03-26T07:23:39.533 | <p>I have written a piece of code solving the linear elasticity problem on transversely isotropic materials (before that, the code only supported isotropic materials).</p>
<p>I would like to validate it before using it for other applications, have you got any idea how to do it ? I would like to be sure that the 5 independents coefficients : E1, E3, nu12, nu13, G are well integrated in the code.</p>
| |mechanical-engineering|finite-element-method|elastic-modulus| | <h1>Definitions</h1>
<p>Before we can answer your question, let us look at two standard definitions of terms (from <a href="https://www.asme.org/products/codes.../v-v-10-2006-guide-verification-validation" rel="nofollow noreferrer">ASME Guide for Verification and Validation in Computational Solid Mechanics</a>) :</p>
<p>1) <strong>Verification</strong>: The process of determining that a computational model accurately represents the underlying mathematical model and its solution. </p>
<p>2) <strong>Validation</strong>: The process of determining the degree to which a model is an accurate representation of the real world from the perspective of the intended uses of the model. </p>
<p>You probably mean "verification" rather than "validation" in your question.</p>
<h1>Verification of anisotropic elasticity</h1>
<p>Assuming you want to verify your code for three-dimensional problems, there is a sequence of steps that's needed.</p>
<p>1) <strong>Frame indifference</strong>: Rotate your elements and confirm that stresses are not developed due to pure rotation.</p>
<p>2) <strong>Exact solutions</strong>: Search for exact solutions for 3D anisotropic elasticity in the literature and confirm that your code can reproduce those. (see, e.g., <a href="http://www-personal.umich.edu/~jbarber/Ting.pdf" rel="nofollow noreferrer">http://www-personal.umich.edu/~jbarber/Ting.pdf</a>)</p>
<p>3) <strong>Manufactured solutions</strong>: Create exact manufactured solutions and verify that your code can match those. (see, e.g., <a href="http://www.eng.utah.edu/~banerjee/Notes/MMS.pdf" rel="nofollow noreferrer">www.eng.utah.edu/~banerjee/Notes/MMS.pdf</a>)</p>
<p>If you can pass these tests for a wide range of loading paths, you can be reasonable sure that your implementation is OK. However, I don't think one can prove correctness of an implementation in a mathematical sense.</p>
| 21053 | How to properly validate transverse isotropic elasticity Finite Element Code |
2018-03-26T07:34:00.590 | <p>I am trying to design a lifting mechanism and I would love some insight/opinions as to how I can shed some weight off it. </p>
<p>The current weight I have for my mechanism is 187.5kg, however, this is supposed to be something that is "easily transportable"</p>
<p>I am using stainless steel. Originally we had 20.3m of steel with a cross section of 40mm x 40mm and 10mm thick. Seeing as 187.5kg is quite high I am looking to reduce the weight. I will admit the material and cross section were chosen relatively arbitrarily, but we wanted to be sure that our structure could hold any load without actually worrying about the calculations for that.</p>
<p>The main stress that will be exerted on the beam is bending, as the structure is built like a table, and will have a weight suspended from the middle. I looked up that the yield stress for stainless steel is 193 MPa (I took the lowest value I could find for safety reasons) and I have applied various different cross sections for the steel.</p>
<p>For my original dimensions 40 by 40 by 10mm thick:</p>
<p>$$\sigma\text{ (bending)} = \dfrac{My}{EI}$$</p>
<p>Assuming my table top is completely made of a hollow rectangle of stainless steel:</p>
<p>$$\begin{gather}
M = 200\cdot9.81\cdot500\text{ (lifting 200kgs 50cm away from it)} = 98100\text{ Nmm} \\
I = \dfrac{40^4 - 20^4}{3} = 800000\text{ mm}^4 \\
y = 20\text{ mm} \\
E = 193\text{ MPa} \\
\therefore \sigma_b = 0.0127\text{ MPa}
\end{gather}$$</p>
<p>I worked in mm because 1 kg/(mm*s<sup>2</sup>) = 1 MPa</p>
<p>A reduced, solid cross section of 20mm gave a bending stress of:</p>
<p>$$\begin{align}
M &= 98100\text{ Nmm} \\
I &= \dfrac{20^4}{3} = 53333\text{ mm}^4\\
y &= 10\text{ mm}\\
E &= 193\text{ MPa}\\
\therefore \sigma_b &= 0.0953\text{ MPa}
\end{align}$$</p>
<p><strong>I know stainless steel is a very strong metal, which is why I leaned towards using it originally. I would just really like to know if the logic behind all of my calculations make sense? Am I on the right track for this or have I done something horribly wrong that I am unaware of?</strong> I am aware I did not include self weight, but did not see the need for it. </p>
<p>Thank you for reading, I would appreciate any thoughts on this.</p>
<p>Edit: this is for a university project, I am mainly looking for pointers on my conceptual design and work.</p>
| |mechanical-engineering|structural-engineering|materials|structural-analysis|structures| | <p>Structural shapes are your friend with this. Look at I-beams, channels, and struts...</p>
<p><a href="https://i.stack.imgur.com/3IbZx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3IbZx.png" alt="enter image description here"></a></p>
<p>You won't believe how much more weight this piece can carry than a table that is much heavier.</p>
<p>Stainless will add cost to your project with no structural advantage over carbon steel. The additional elements in stainless prevent corrosion, which you haven't mentioned as a design issue.</p>
<p>Aluminum comes in a zillion flavors, but a stronger aluminum alloy is pound-for-pound much stronger than typical stainless or carbon steel. Aluminum resists corrosion like stainless steel, too... it won't rust outside.</p>
<p>A long horizontal span of a table increases the stress in the top without increasing the capacity. Is there any way to reduce the horizontal span? Also, if you adopt a truss or A-frame design, rather than a cantilever beam, you will achieve much greater strength per weight.</p>
<p>Casters! Put wheels on the thing and let it roll around.</p>
<p>Quick assembly/disassembly. Can you use toggle clamps or quick-release fasteners, like the wheels on a bike, to disassemble it? Then you transport it in pieces rather than the whole.</p>
<p>Don't rule out wood. Use 4 X 4 or 4 X 6 douglas fir or redwood... wood is really light compared with metal for many applications.</p>
<p>Do the statics on your design by making a free-body diagram, and remember the basic truss premise: a truss shape is light and strong.</p>
<p><a href="https://i.stack.imgur.com/J7k5Z.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J7k5Z.gif" alt="Truss"></a></p>
| 21054 | Trying to shed some weight on my design, any thoughts? |
2018-03-26T18:20:01.603 | <p>I submitted the edited inp. file in abaqus. I edited the thickness of shell elements
then after I run the edited inp. file, I am suffering from this error.</p>
<p>Actually, there is another error:
"There are fewer than 2 items on a data line for which a generation sequence (start id, end id, optional increment number) is expected. The keyword in question is related to *ELSET. Please check input data"</p>
<p>does anyone know how to solve these error?</p>
| |finite-element-method|abaqus| | <p>I have solved this problem.
I have made typing mistakes when editing the inp. file.
so it is highly recommended to write with high accuracy, otherwise, it can cause a lot of errors.</p>
| 21059 | in keyword *SURFACE, file "FSECase1cor.inp", line 79025: Unknown assembly element set_M_SURF-1016_E3 (What does this error mean?) |
2018-03-30T17:53:23.303 | <p>This may sound rather vague on its own so I will explain with examples. </p>
<p>The solid booster rockets used by NASA space shuttle used a central bore design because it had a certain characteristic on burn rate and thrust.</p>
<p><a href="https://i.stack.imgur.com/M8rMT.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/M8rMT.jpg" alt="enter image description here"></a></p>
<p>If this data was to be acquired only form experimentation like burning test rockets and not deriving any mathematical relation between bore design and thrust then would it be considered sound engineering?</p>
<p>Are any designs based solely on data from trial and error used in critical mainstream engineering? </p>
<p>Like we use it just because it works and do not investigate why it works like that because it's very complex to understand but still easier to implement.</p>
| |aerospace-engineering|mathematics|manufacturing-engineering|process-engineering| | <p>For complex and expensive systems you usually want to have an understanding of its behavior, cast in models. To model a system (or part of it), a <strong>complete theoretical understanding</strong> is usually not required. </p>
<p>You should, however, have a <strong>complete knowledge</strong> of the phenomena that act on your system. Then you identify the key phenomena and, if it does not yet exist, a theoretical understanding of those phenomena to a degree required to describe your system accurately enough.</p>
<p>The actual work is in finding out which phenomena are important and which are not, what effects to include, what effects to neglect, what depth of understading is actually required, which effects need to be understood, which effects are too expensive to understand now but can be compensated by overly conservative design decisions, etc. Those are some of the tradeoffs that @DanielKiracofe mentions in his answer.</p>
| 21109 | Does every engineering design needs to have a complete theoretical backing or may experimental data suffice? |
2018-03-30T18:28:28.070 | <p>I have a 3d printed (ABS or PLA) case which houses small single board computer. I want to glue it to glass (car's windshield from the inside) so that it sticks nicely and does not vibrate. However, there should be a chance to remove the case from the glass later if wanted without damaging the glass in any way. Will someone recommend suitable adhesive and removal method (heat etc.)?</p>
| |plastic|adhesive|glass| | <p>Because the surface is a windshield and because you wish to have it somewhat removable, consider Solar Mike's suggestion of double sided tape, but add in a parameter for outdoor use. <a href="https://rads.stackoverflow.com/amzn/click/B00004Z4BV" rel="nofollow noreferrer">3M makes a great high density foam tape</a> that I've been using for many years to attach miscellaneous items to the windshield. </p>
<p>The surfaces have to be clean, of course, and the bond between the tape and the plastic should be secure. Because you're considering the plastic to be 3D printed, you may want to sand the surface with fine sandpaper, even to the point of getting it glossy for the best bond. If you construct the part with ABS, performing acetone-vapor smoothing will give you a good surface.</p>
<p>I would suggest to attach the tape to the plastic surface and allow it to bond for a day, then bond the assembly to the windshield. You can see how well the tape is bonding by looking at the glass, pushing against the box to work any visible bubbles out.</p>
<p>I've recently noted that Eclipse Sunshades switched adhesive pads on their product to this 3M product (grey foam adhesive) perhaps because the other verions previously used would release. I'd clean them off, put this 3M stuff in place and be assured of a solid attachment.</p>
<p>Removal means careful slicing with a knife or razor, then additional mechanical removal (fingers scraping) followed by any convenient solvent (acetone - use carefully) or alcohol, although I've not used that, and windows cleaner.</p>
<p><a href="https://i.stack.imgur.com/OxKNx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OxKNx.jpg" alt="mounting tape"></a></p>
| 21111 | Glue plastic to glass with chance for easy removal later |
2018-03-31T12:56:53.623 | <p>I have a solids problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track, (perhaps there was an issue setting up my deflection equations?). Any assistance would be highly appreciated.</p>
<p><strong>Question:</strong></p>
<p><a href="https://i.stack.imgur.com/RDqX2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RDqX2.png" alt="enter image description here"></a></p>
<p><strong>My work is as follows:</strong></p>
<p>Castigliano's theorem: deflection = partial derivative of strain energy U wrt a force F</p>
<p><strong>I split the beam into two parts, A - B and C - B (So I don't have to
deal with the reaction forces at C)</strong></p>
<p><em>Edit:</em> </p>
<p><em>X is the distance from the free end for Mab and Uab</em></p>
<p><em>And it is the distance from F for Mbc and Ubc</em></p>
<p><strong>My first moment:</strong></p>
<p>Mab = Fx</p>
<p><strong>2nd moment:</strong></p>
<p>Mbc = F(x-b)</p>
<p>U = integral from o to x of (M^2/(2EI))</p>
<p>Uab = integral from 0 to b of ((f^2*x^2)/2EI) = (f^2/2EI)*[x^3/3] (0-b)</p>
<p>Uab = (f^2/2EI)*(b^3/3) = ((F^2*b^3)/6EI)</p>
<p>Ubc = integral from 0 to b of ((F^2*(x-b)^2)/2EI) = (F^2/2EI)*[(x -
b)^3/3] (L-b)</p>
<p>Ubc = ((F^2*(L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)</p>
<p><strong>Therefore</strong></p>
<p>Utot = ((F^2*b^3)/6EI) + ((F^2*(L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)</p>
<p>Utot = ((F^2*(b^3+L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)</p>
<p>Utot = ((F^2*(L^3 - (3b(L^2))+(3L(b^2))))/6EI)</p>
<p>Deflection = dUtot/dF = ((2*F*(L^3 - (3b(L^2))+(3L(b^2))))/6EI)</p>
<p><strong>Deflection = ((F*(L^3 - (3b(L^2))+(3L(b^2))))/3EI)</strong></p>
<p><strong>Clearly not the answer,</strong> I am not sure but maybe I went wrong somewhere in the first
couple of steps, perhaps either setting up my moment equations or setting them
up in the integral. But I can't figure it out. Can someone please help ID my error and let me know what I did wrong?</p>
| |mechanical-engineering|structural-engineering|materials|structures|stresses| | <p>Castigliano's theorem says that we can apply a dummy load $P$ at any point of interest; the deflection in that location is $$y=\int_L\frac{M}{EI}\frac{\partial M}{\partial P}\,dx$$ letting $P\to 0$.</p>
<p>Apply the dummy load at the free end of the beam (call this $x=0$). I'm only going to consider the region of the beam to the right of the applied force $F$, since the other part stores no real strain energy. The bending moment in this region is $M=Px+F(x-b)$; thus, $\frac{\partial M}{\partial P}=x$. So we have $$y=\int_b^L\frac{[Px+F(x-b)]x}{EI}\,dx$$ which can be verified for $P\to 0$ to give $y=\frac{F}{6EI}(2L^3-3L^2b+b^3)$, which matches the answer given in the problem.</p>
| 21119 | Using strain energy to calculate deflection |
2018-04-01T22:49:17.270 | <p>I'm trying to figure out how to construct a CNC XY table, ala this instructable: </p>
<p><a href="https://www.instructables.com/id/Low-Cost-Hobby-Servo-XY-Table/#discuss" rel="nofollow noreferrer">https://www.instructables.com/id/Low-Cost-Hobby-Servo-XY-Table/#discuss</a></p>
<p>I'm a noob with mechanical design/mechanical parts, so I don't understand a lot of what this guy is saying. One of those things though, is that you can use (paraphrasing here, with limited understanding) a 10 turn potentiometer with a servo to ... make a motor? </p>
<p>What I'd like to know is, what is a 10-turn potentiometer? Is that somehow different from regular potentiometers? How do you modify a servo with a potentiometer, and how does that allow you to turn something more than the 180 or 90 degrees that a servo allows?</p>
| |motors|servo| | <p>it's not unusual to find a single turn potentiomenter (aka pot) that provides a change in resistance over the range of the turning component. The benefit of a single turn pot is that you can get a large change in resistance with minimal movement. Volume controls are a good example.</p>
<p>A multi-turn potentiometer is just the opposite. It provides a similar range of resistance, but requires that you perform multiple rotations of the control to "travel" from one end of the range to the other.</p>
<p>As an example, a volume control may provide 10K resistance from one point to the other. You can select a 5K resistance (5000 ohms) by approximately turning to the halfway point. Due to the large range of resistance over such a small movement, it is difficult to pinpoint the 5000 ohm figure. For something as objective as volume control, such a coarse level of resistance is acceptable.</p>
<p>On the other hand, with a ten turn pot, as in your example, of the same 10K value, you can easily guess at five turns to get close to the 5000 ohm level, but more important, you can move a quarter turn and have substantially less change than with the single turn pot and be in a better position to get the figure you require.</p>
<p>This allows one to provide for fine adjustment of various devices, such as the servo mentioned in the article you've linked.</p>
<p>If you had a circuit to control a servo, using a single turn pot, you may get 180° movement over the full turn of the pot, but with the ten turn pot, each turn would possibly provide only 10° of movement.</p>
<p>This is a generalization that I'm hopeful will assist your understanding.</p>
<p>You ask how to modify a servo with a potentiometer. A servo is a small motor with a potentiomenter within the housing. The potentiomenter sends a feedback signal to the controller to inform the controller of the location of the servo. What modification do you require of this internal pot?</p>
<p>Some servos are engineered to turn more than 90° or 180° and some can be modified to do so. This would be considered outside the scope of this answer and also can be found by using search terms such as "modify servo rotation" or similar. You may also discover such a reference in the world of <a href="https://www.instructables.com/id/How-to-Modify-Micro-Servo-for-360-Rotation/" rel="nofollow noreferrer">Instructables</a> such as this provided link.</p>
| 21136 | What is a 10 turn potentiometer |
2018-04-02T12:51:16.207 | <p>Consider, while building a new machine which has different elements in it. For instance, a machine having a driving assembly, main body, chassis, crucible etc.
Then, how would I calculate the center of gravity and how would I know the calculated value is an exact value or Cg value gives a balanced system?</p>
<p>Should I calculate a center of gravity of each element or the whole system?</p>
<p>Suppose I need to calculate the center of gravity of each element, how would I calculate? What are the factors to be considered other than weight and distance from the reference plane?</p>
<p>Any suggestions would be appreciated.</p>
<p>Thank you in advance</p>
| |structural-engineering| | <p>You have to know the mass and CG of each individual part.</p>
<p>Then once assembled you set your machine against a corner wall which you will use as xy axis. measure the Dx and Dy, the distance from CG of parts to corresponding X and Y walls.</p>
<p>Then you calculate CG_x as:</p>
<p>$$ CG_x = (\Sigma Mn. Dxn )/W $$ and CG_y as:$$ CG_y = (\Sigma Mn.Dyn)/ W $$ Then you can transfer this X and Y to a more convenient axis, such as one parallel with a prominent axis on your machine.</p>
<p>There are many CAD programs that can calculate CG by the specs and wire-mesh of the machine.</p>
| 21143 | How to find the center of gravity(Cg) of machine? |
2018-04-02T13:02:03.607 | <p>I'm designing a shaft, and i have to choose a material.</p>
<p>I deal with cyclic stress in my machine, so i set up an expression for all the possible stresses and combined it with von Mises criteria for plastic deformation, it gives me the so called <strong>comparison stress</strong>. </p>
<p>This comparison stress should be lower than the endurance limit of the material. I don't have the endurance limit, but instead the tensile strength.
Is there any connection between endurance limit and the tensile strength ? </p>
| |mechanical-engineering|stresses|fatigue| | <p>Steels are relatively unique among metals with a very reliable endurance limit equal to 1/2 of tensile strength. Cast irons definitely do not have this feature. Nickel and cobalt alloys might have it but there is limited data. Fatigue test coupons are smooth with no stress concentration ; not to say that they never contain an inclusion. So fatigue tests give the best possible value; Real world components normally have various stress concentrators. It is the duty of the builder to determine the severity of these stress concentrators. By "endurance limit" I mean the number of stress cycles approach infinity.</p>
| 21144 | Is there any link between tensile strength and endurance limit for steel ? |
2018-04-04T04:33:16.817 | <p>We have steel hollow tubes of 80 mm length and 0.5 mm cross section, that we need to cut in smaller pieces of 10.1 and 10.5 mm length. We have been using utility knifes with very poor results.</p>
<p>I am looking for a tool that can do the job. Something like scissors or pliers, to be handled manually. </p>
<p>Every answer or suggestions is welcomed. </p>
| |mechanical-engineering|applied-mechanics|cutting|tools| | <p>Probably a skilled metalworker will come along and lough at me for this, but I don't think you can cut to 0.1 mm accuracy with many hand tools. I see these ways:</p>
<ul>
<li><p>using a small saw (your tubes are fairly small) and then file down to the desired length. </p></li>
<li><p>I think pliers or bolt cutters could cut through your tube, but will likely chinch it shut</p></li>
<li><p>Or get a tube cutting device - rollers opposite a blade, with a thread to push the blade into the tube. <a href="https://www.homedepot.com/b/Plumbing-Plumbing-Tools-Pipe-Tube-Cutting/N-5yc1vZc4ft" rel="nofollow noreferrer">Look here for these look like</a>. Maybe you can get sub-mm accuracy with these.</p></li>
<li><p>Whatever tool you use, if at all possible use internal support (since your dimensions indicate a very small tube this may be hard to implement), a wood dowel or plug could work (hattip to @fred_dot_u)</p></li>
</ul>
| 21169 | Cutting tool for hollow tubes |
2018-04-04T14:51:54.037 | <p>Yesterday in class this question came up by one of the students in the class. Does anyone know why there is a minus sign in front of the W2 in the figure?<a href="https://i.stack.imgur.com/AJ1gC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AJ1gC.jpg" alt="enter image description here"></a></p>
| |control-engineering| | <p>If you work out the transfer functions from $d$ to $z_1$ and $z_2$ it can be shown that you get</p>
<p>$$
z_1 = W_1 \frac{1}{1+K\,G} d,
$$</p>
<p>$$
z_2 = -W_2 \frac{-K}{1+K\,G} d.
$$</p>
<p>The part of the transfer function of $z_1$ without $W_1$ is often called the sensitivity function. For $H_\infty$ you often choose $W_1$ to be a semi-inverse of a desired sensitivity function.</p>
<p>The part of the transfer function of $z_2$, after cancelling the two minuses, without $W_2$ is often called the control sensitivity function (however <a href="https://www.google.nl/url?sa=t&source=web&rct=j&url=https://www.cds.caltech.edu/~murray/courses/cds101/fa02/caltech/astrom-ch5.pdf&ved=2ahUKEwi3q-iC56HaAhUpsaQKHUd-DE8QFjAAegQIBxAB&usg=AOvVaw3qbmrbZGHBAw5tYwyn1kXI" rel="nofollow noreferrer">here</a> they also call it the noise sensitivity function). For $H_\infty$ you again often choose $W_2$ to be semi-inverse of a desired control sensitivity function.</p>
<p>So in order for $W_2$ to have a more meaningful meaning they added a minus sign into its definition, such that the control sensitivity function is weighted by $W_2$ instead of $-W_2$.</p>
<p>PS: by semi-inverse I am roughly referring to a transfer function which is causal/proper, stable (all poles have a nonzero negative real part) and is an approximation of the inverse of a given transfer function.</p>
| 21179 | Skogestad & Postlethwaite, Multivariable Feedback Control: Analysis & Design, Wiley, 2005 Figure 9.10 Why a negative sign on W2? |
2018-04-04T15:00:17.417 | <p>in most of car cabin filters (for example MANN filter) I see that I need to place it according the "Air flow" label?
But why I need to do that?
What will be wrong if I will put it in opposite side of "Air flow" label? </p>
| |car| | <p>Usually because the strengthening in the filter is optimised for one direction and the layers may also be optimised so the oncoming air is met with a coarse layer first, followed by successively finer layers.</p>
| 21180 | Why do I need to place car cabin filter according Air flow label |
2018-04-04T15:53:15.760 | <p>(please excuse my bad English. I don't speak English)</p>
<p>I have to deal with an old hydraulic machine, recovered from a deposit. I have no knowledge on mechanical engineering. Only civil engineering.</p>
<p>The machine is for doing traction tests on steel bars (for example ribbed bars). The machine is old, from the 70’s or 80’s I guess.</p>
<p>It has a system for plotting the stress-strain/deformation chart on paper, but also has digital instrumentation added, which is being read by an old PC still running Windows XP, and it only provides the raw data from the instruments (as a text file).</p>
<p>The digital system seems to be added, I guess, in the 90’s, and comes with a handbook with some instructions (as for example, how to calibrate the machine), but no explanation on the internal workings.</p>
<p>I need to algorithmically process the raw data produced by the tests, to extract various parameters.</p>
<p>As I had been taught theoretically, the type of chart I should get, must look like this:</p>
<p><a href="https://i.stack.imgur.com/vqyN0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/vqyN0.png" alt="Theorical result"></a></p>
<p>But off course, this is a real world physical test, so I get this kind of chart</p>
<p><a href="https://i.stack.imgur.com/c0FzM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/c0FzM.png" alt="Real data"></a></p>
<p>I had been told that it starts with an accommodation phase which I should ignore, and I need to extract various information from the rest of the chart.</p>
<p>Now, here is a zoom of the “elastic phase”</p>
<p><a href="https://i.stack.imgur.com/td7tH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/td7tH.png" alt="enter image description here"></a></p>
<p>I have no knowledge on why it behaves that way, but I conjecture (guess) that the traction force is increased in discrete steps, and between each force increase, the machine waits some amount of time for the steel bar slowly elongating until it reaches a stress-strain/deformation static equilibrium.</p>
<p><a href="https://i.stack.imgur.com/Ts8cP.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Ts8cP.png" alt="enter image description here"></a></p>
<p>If that conjecture is correct, then I should isolate only the “static equilibrium” points to do calculations, and discard the other, because the other points are transient states, and the theoretical applications of the test are only for static behaviors (civil engineering applications).
If the conjecture is correct, I should discard the transient states and retain only the static data for purposes of static mechanic applications of the traction test, which are the only theoretical applications of the test in civil engineering.</p>
<p><a href="https://i.stack.imgur.com/3HZED.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3HZED.png" alt="enter image description here"></a></p>
<p>But off course, I have no knowledge on dynamic mechanics, and I may be making some huge mistake. I’m not a mechanical engineer.</p>
<p>So, the question is:
Is the conjecture correct?
Should I make some additional considerations?</p>
<p>[Edit] I would welcome if you can give me a book title, for reference, on the dynamical behavior of this type of machine.</p>
| |steel|elastic-modulus|traction| | <p>Edit after AgentP's comment:</p>
<p>As AgentP pointed out, you have swapped your stress-strain axes in the zoomed image. This fact alone refutes your conjecture... Your graph should look like this:</p>
<p><a href="https://i.stack.imgur.com/n0FnY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n0FnY.png" alt="enter image description here"></a></p>
<p>You see vertical increases in stress, and sudden jumps in strain.</p>
<p>A reason the steps might occur:</p>
<p>The steps in your data occur at very specific displacement/strain intervals, this says to me that the displacement read-out has a too low resolution compared to the force read-out.</p>
<p>What to do with the data:</p>
<p>First of all, this data and machine won't cut it if you would like to certify any materials.</p>
<p>If the resolution of the displacement read-out is too low, you would be better off using the points indicating a change in displacement. Judging by the spacing of the force read-out points, their resolution seems much higher, therefor, if you read a change in the displacement read-out, it is more likely that that point will be the most representative of the stress-strain relationship of the specimen at that point. Below is a picture illustrating:</p>
<p><a href="https://i.stack.imgur.com/y7DvX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y7DvX.png" alt="enter image description here"></a>
<em>(Pardon my Paint skills, I'm more of a CAD guy)</em></p>
<p>As for the accommodation zone, as niels nielsen suggests, extrapolate the elastic zone to the x-axis and disregard the data to the left of that line. Below is an image illustrating:</p>
<p><a href="https://i.stack.imgur.com/6kXtb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6kXtb.png" alt="enter image description here"></a></p>
<hr>
<p>P.S. Why your conjecture would be wrong even if you hypothetically did not swap your axes:</p>
<p>The machine wouldn't apply a force on the bar and wait for the bar to elongate to "catch up".</p>
<p>Firstly, you would have to apply the force <strong>very</strong> quickly to see any significant effect, I don't think it would be possible in the elastic zone. And if that was even possible, your graph won't have steps with a horizontal plateau, but it will be a series of bumps, going up in force and elongation, and when the force is stopped being applied, you will see a decrease in stress as the bar elongates to catch up.</p>
<p>Properly built traction testers are built to be strain-controlled, i.e. the machine would have a PID controller or something similar that controls the force applied to ensure that the strain increases at a linear rate w.r.t. time. This reduces the effects of potentially catastrophic failure, especially when testing brittle materials and prevents the jacks from "running away" once the specimen yielded.</p>
| 21183 | Should I discard part of the data (steel traction test) |
2018-04-04T17:52:37.083 | <p>I want to torque a nut to 320 ft lbs, but my torque wrench measures only to 250 ft lbs. Since I know I can get it to 250 ft lbs, is it possible to use that value to calculate the additional degree of rotation it would take to get it to approximately 320 ft lbs? For example... torque to 250, switch to another ratchet to avoid damaging the torque wrench, then rotate the nut an additional 45 degrees.</p>
| |mechanical-engineering| | <p>The key to solving your problem is to understand what the units of torque are and how to apply them to your particular situation. Torque is defined as force per unit length. In every torque problem there is an arm that rotates around a fixed point and force that acts on the end of that arm. The equation that describes the resultant torque is</p>
<p>$\tau = r * F * cos\theta$</p>
<p>Here is a graphic that discribes the equation.</p>
<p><a href="https://i.stack.imgur.com/EXgDD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EXgDD.png" alt="enter image description here"></a></p>
<p>This general equation takes into account that the force acting on the arm may not be perpendicular to it.</p>
<p>Now to asnwer your question. </p>
<blockquote>
<p>is it possible to use that value to calculate the additional degree of rotation it would take to get it to approximately 320 ft lbs?</p>
</blockquote>
<p>No. It is much easier to calculate torque directly. You can do this by placing your wrench on the nut so that is parallel to the ground and then placing a weight on the end. The resultant torque is the length($r$) of the arm times the weight on the end($F$). Since the arm is parallel to the earth and the force of gravity is perpendicular to the earth, $\theta$ is at $90\unicode{xb0}$ and thus $cos\theta$ equals one and does not change the outcome of the equation. </p>
<p>In summary, divide 320 by the length of your bar in feet and place that many pounds of force on the end of your bar at a 90 degree angle and the resultant torque will be 320 foot pounds.</p>
| 21186 | Torquing a nut beyond a known value using degree of rotation? |
2018-04-05T09:46:07.947 | <p><a href="https://i.stack.imgur.com/c0BQ2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c0BQ2.png" alt="enter image description here"></a>Say I have a ring with the center of mass G. Also I rotate the ring around a point P on the ring, and I have a "anything" moving along the ring at the same time with velocity v.</p>
<p>How would one explain the motion of the "anything"? Would one say that it's absolute velocity is the velocity of G + the relative velocity (velocity measured from P) + the rotational velocity of the entire system? Or is some of these terms zero?</p>
<p>Thank you.</p>
| |mechanical-engineering| | <p>The Anything path is a circle with G at its center. So in that reference frame</p>
<p>$$x= \sqrt{1-y^2} $$
But G is turning with its $$ X_p = \sqrt{1-Y^2_p} $$ </p>
<p>you have to transfer the X1 and Y1 to new frame by substituting X1, with (X1-Xp) , and same for y, Y1 substitute with Y1-Yp.</p>
<p>The final path will look like a Lotus pattern of a round flower with oval petals. The radius of this pattern will be twice the radius of your ring and the number of petals depends on the ratio of the to rotations.</p>
| 21202 | Rotating ring with "thing" moving on it |
2018-04-05T19:25:51.420 | <p>I am sorry if this is a strange question. Sometimes I like to visit parliaments from around the world and compare their buildings. </p>
<p><a href="https://i.stack.imgur.com/qP0JV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qP0JV.jpg" alt="enter image description here"></a></p>
<p>After looking at the British house of commons chamber, I noticed that there was sun light coming out of ceiling of the chamber. At first I thought that the ceiling was made of glass and it was just sun light. However after checking the building from outside there seems to be at least another floor on top of the chamber. So it can't be sun light. Then I thought that it must be just some electric light. </p>
<p><a href="https://i.stack.imgur.com/lsFQK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lsFQK.jpg" alt="enter image description here"></a></p>
<p>However, after checking some old images of the chamber(before electricity was used) there is still light coming from ceiling. </p>
<p>Does anyone know how this happens?</p>
<p><a href="https://i.stack.imgur.com/6KsKL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6KsKL.jpg" alt="enter image description here"></a></p>
| |civil-engineering| | <p><strong>It is not sunlight. The lights in the roof are artificial</strong>. Your 'old photo before electricity' is presumably from before the war, when the roof <em>did</em> have real windows, in 'direct contact with the sky'. See this aerial view from 1919 for reference:</p>
<p><a href="https://i.stack.imgur.com/Pzhvdm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Pzhvdm.png" alt="Aerial view of Commons Chamber 1919"></a></p>
<p>There is recorded evidence of resistance to the building of offices etc. above the chamber, as members worried about the lack of daylight.</p>
<blockquote>
<p>Before the Debate goes any further I would like to express my disapproval of the suggestion of the hon. Member for West Walthamstow (Mr. McEntee) and certain other hon. Members that when we come to rebuilding the House of Commons special accommodation should be made available for other purposes above it. If that were done, we would have no daylight whatever in the new Chamber, and having sat, like many others, for two years in this artificial illumination, I certainly look forward to the day when we can have a Chamber very much better lit than the former House of Commons used to be.</p>
<p><a href="https://api.parliament.uk/historic-hansard/commons/1943/oct/28/house-of-commons-rebuilding#S5CV0393P0_19431028_HOC_307" rel="noreferrer">Sir Alfred Beit (St. Pancras, South-East), 28/Oct/1943</a></p>
</blockquote>
<p>The <a href="https://www.parliament.uk/about/living-heritage/building/palace/architecture/palace-s-interiors/commons-chamber/" rel="noreferrer">parliment.uk website</a> states that</p>
<blockquote>
<p>When the Chamber was rebuilt after 1945 at the cost of £2 million, Sir Giles Gilbert Scott designed a steel-framed building of five floors (two taken by the Chamber), with offices both above and below.</p>
</blockquote>
<p>"The Construction of the House of Commons" by Oscar Faber notes further that</p>
<blockquote>
<p>The elaborate wall and ceiling linings to the frame replicate the original timber detailing by Pugin. The frame steps back along its length and addtional height above the Chamber has been allowed for a two-tier roof over the whole rebuilt House of Commons.</p>
</blockquote>
<p>and</p>
<blockquote>
<p>An elaborate network of branched ductwork sent the conditioned air to numerous outlets set into the ornate detail of the Chamber’s gallery structure, and at the upper level of the spring point of the replica roof trusses. Extract ducts were positioned at regular centres in the apex of the roof lining of the Chamber.</p>
</blockquote>
<p>i.e. the new chamber internal roof was designed to look like the old one, but also had extraction ducts etc. added, in keeping with there being in-use office space above.</p>
<p>The phrase "two tier roof" does not refer to the internal roof of the commons, since that is a ceiling, not a roof. Rather, the new roof of the commons simply has two tiers, but this is not accurately picked up by Google Maps. See here:</p>
<p><a href="https://i.stack.imgur.com/5118U.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/5118U.jpg" alt="Close up of Commons Roof showing two-tiers above chamber"></a></p>
<p>Furthermore, a friend of mine who is currently at work inside the House of Commons <em>right now</em> has confirmed both that the look/feel of the roof-lights is artificial, and that the offices on the floor above do not follow a long corridor around the void which would be required to direct daylight down to the commons chamber roof two floors below.</p>
| 21211 | Where does the light in the ceiling of the British house of commons chamber come from? |
2018-04-07T03:49:47.067 | <p>I read the paper "Bilateral Control of Teleoperators with Time Delay",which is written by Robert J. Anderson and Mark W. Spong. In that paper, they defined passivity of an n-port flow as<br>
$\int^\infty_0 F^T (t) v(t) dt \geqslant 0$<br>
$F$ means effort and $v$ means flow.<br>
And they said if a two-port system is passive, the system is stable. They didn't said that directly, but they said an system can be unstable by being nonpassive. They said that without proof, so I want to know why an passive system is stable.</p>
| |control-engineering|control-theory| | <p>I didn't read the entire paper but only checked the statement you mentioned.</p>
<blockquote>
<p>The two-port communication circuit for this system is nonpassive, and is the cause of the instability.</p>
</blockquote>
<p>And later they write again</p>
<blockquote>
<p>[...] it was shown that the instability [...] is due to a nonpassive communication block.</p>
</blockquote>
<p>It looks to me like you mixed up sufficient and necessary conditions here.
They write in the paper, as you already mentioned in your question, that <em>nonpassive</em> implies <em>unstable</em>.</p>
<p>This makes passivity a <strong>necessary</strong> condition for stability in this case.</p>
<p>They did <strong>not</strong> say that it is a <strong>sufficient</strong> condition, which would be how you understood it, that <em>passive</em> implies <em>stable</em>.</p>
<p><a href="https://en.wikipedia.org/wiki/Necessity_and_sufficiency#Sufficiency" rel="nofollow noreferrer">Wikipedia</a> explains it a bit more compact than me.</p>
<blockquote>
<p>If P is sufficient for Q, then knowing P to be true is adequate grounds to conclude that Q is true; however, knowing P to be false does not meet a minimal need to conclude that Q is false.</p>
</blockquote>
<p>I hope I could explain it somewhat understandable, don't hesitate to comment if not.</p>
| 21231 | Passivity and stability |
2018-04-08T02:45:46.347 | <p>I'm not sure if this is an appropriate place to ask this, but I figure the worst that's going to happen is my question will just get closed, so here we go.</p>
<p>I just got my first home defense handgun, a Glock 19 gen 4, and I was wanting to store it within my combination safe. I timed myself at roughly 25 seconds to open the safe and retrieve the gun. This is fast, however, I was wondering if I could make it faster.</p>
<p>Normally you would need to turn the dial like this(Please note i don't count the first 4 turns because that should be muscle memory when opening a safe anyway):</p>
<pre><code>4 times to the left, and end on the first number
3 times to the right, and end on the second number
2 times to the left and end on the third number
1 time to the right and end on 0. (This last step seems to be universal)
</code></pre>
<p>What I would like to do is reconfigure my safe to do something like:</p>
<pre><code>4 times to the left, and end on the first number
1 time to the right, and end on the second number
1 time to the left, and end on the third number
1 time to the right and end on 0.
</code></pre>
<p>I'm essentially cutting out the extra turns of the tumbler. The concept seems solid, and I can indeed make it to the back of the safe tumbler to see the internals.</p>
<p>My question is, is it possible to change not only the combination, but the way the combination is entered, and skip additional turns? The only downside I can see to this is by overriding the functionality of the safe, I may be decreasing the number of permutations for the combination. I can live with that if it means cutting down the retrieval time of my gun to around 5 seconds, which my concept will likely do.</p>
<p>Is this possible?</p>
<p><a href="https://i.stack.imgur.com/GSG2v.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GSG2v.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/85rR3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/85rR3.jpg" alt="enter image description here"></a></p>
| |mechanisms| | <p>You cannot do this with this mechanism. I will now explain in detail how mechanical rotary knob combination lock mechanisms actually work rather than just tell you that you can't.</p>
<p>These kinds of locks are extremely simple, using a series of stacked disks, with a protruding rod on both sides of each disk, and a notch at some random location on each disk.</p>
<p>When you rotate the knob 360 degrees in one direction, the first disk will at some point have its protruding rod make contact with the rod of the second disk. The two disks now move together as a group.</p>
<p>When you rotate the knob a second 360 degrees in the same direction, the knob of the second disk will make contact with the knob of the third disk. The three disks now move together as a group. </p>
<p>This process continues until all disks are moving together as a group. The number at which you are to stop rotating the dial after four turns, will align the notch in the fourth disk with the release lever. </p>
<p>You now need to turn the dial in the opposite direction, so that the protruding pins make contact on the opposite side of each disk, until you are moving all but the fourth disk. You now carefully move the dial so that the notch of the third disk is aligned with the previous fourth disk.</p>
<p>You back out of the number of turns because you are gradually aligning the slots of each disk with the release mechanism, without disturbing the previously positioned disks in the stack.</p>
<p>There is no other way to position the disks other than to gradually back out 4 turns, 3 turns, 2 turns, 1 turn.</p>
<p> </p>
<p>Also due to how these disk combination locks work, there are actually two ways to position the disks that will open the lock, depending on if you start by turning the knob to the left or to the right. Though you will normally only be provided instructions for one of these methods, and only by starting the turning in one direction.</p>
| 21242 | Reconfiguring a Safe to only need Four turns |
2018-04-08T04:25:20.443 | <p>I have a schematic for some gas lines in a wafer processing unit. One of the symbols which occurs is a circle with an x in it that has the left and right quarters filled in to form two black triangles. (I've attached a drawing.)</p>
<p>From context, i can tell that these represent valves to allow or block flowing of the gas. However i would like to find this symbol in some guide to give a more exact interpretation of the meaning, and just to have an authority that i can cite to.</p>
<p>I have not found the symbol in the context of a gas line schematic in any online reference.</p>
<p>Where, and how, can i look?</p>
<p><a href="https://i.stack.imgur.com/5TrFA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5TrFA.png" alt="circle with two black sectors"></a></p>
| |gas|pipelines| | <p>From my experience in bridges, I would expect that to represent a borehole: <a href="https://www.google.co.uk/search?biw=770&bih=742&tbm=isch&sa=1&ei=KSTLWoCyDcvNwAKV3LKYDA&q=borehole%20symbol&oq=borehole%20symbol&gs_l=psy-ab.3..0l2j0i8i30k1.20959.22743.0.23128.9.9.0.0.0.0.79.479.9.9.0....0...1c.1.64.psy-ab..0.9.478...0i10k1j0i13k1j0i7i30k1j0i7i5i10i30k1j0i8i13i30k1j0i13i5i10i30k1.0.wx_0zbo6uy0#imgrc=Lqp9DkRjKgQ8wM:" rel="nofollow noreferrer">google image result</a>.</p>
<p>Perhaps this isn't a pipeline-specific symbol, but a general engineering symbol. I can well believe that boreholes would have been taken to sample the soil/strata before a pipeline is constructed.</p>
<p>In my googled image, it appears with the letters BH next to it to, to signify BoreHole. In my experience there are often also numbers to correlate it with a table of borehole results, e.g. it might say "BH 10436". Are there any letters/numbers near your symbol which could be part of it?</p>
| 21244 | What is a reference that i can use to find the interpretation of a symbol in a gas line schematic? |
2018-04-08T08:42:40.753 | <p>in the following picture, we should find the viscosity of fluid and we have to have the Area surface of cylinder that is in touch with the fluid.
the area surface depends on the bottom of the shape.
but i have no idea what's the shape of this cylinder. what is that curve on the bottom of cylinder? does it mean that cylinder is empty inside or what?
<a href="https://i.stack.imgur.com/WdNA9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WdNA9.png" alt="enter image description here"></a></p>
<p>and the solution of this problem:</p>
<p><a href="https://i.stack.imgur.com/m1Yn7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m1Yn7.png" alt="enter image description here"></a></p>
| |fluid-mechanics|fluid| | <p>There is no info about the shape of that curve in your, ie the book's, question, also, the torque shearing stress is calculated for the gap between the wall and cylinder which is small and much smaller than the gap between the base and the end of the cylinder.</p>
<p>Edit:
After a re-read, the question does also say neglect end effects.</p>
| 21245 | shape of cylinder in the fluid mechanics |
2018-04-08T19:58:54.757 | <p>I am looking at Phase Change Materials as a building fabric option and from reading up on the topic theres something I don’t understand. Hopefully somebody can help explain.</p>
<p>I read that when a substance melts it stores latent energy, otherwise charging the material. When the material freezes again it releases this ‘stored’ energy and this is what is utilised for heating or cooling. However, I’ve also read that some substances can store energy during the freezing process and this is where it loses me...</p>
<p>Am I misunderstanding the process or is this specific to certain materials?</p>
<p>Thanks in advance</p>
<p>Kind Regards</p>
<p>L</p>
| |mechanical-engineering|materials|thermodynamics|chemical-engineering| | <p>Let's take water as an example. to freeze water into ice, you must extract from it a certain amount of heat energy per unit of mass, which is called the specific heat of fusion. This same amount of heat must be added to the resulting ice in order to melt it back into water again. In this sense, the work it took to draw the heat of fusion out of the liquid water and freeze it is "stored" in the resulting chunk of ice, and is available to perform refrigeration work for you at a later time. </p>
<p>Alternatively, the heat that the chunk of ice pulled out of its surroundings as it melted is similarly "stored" in the resulting body of melt water. A device called a heat pump can then be used to pull that absorbed heat back out of the liquid water and exhaust it into a space where heating is needed, thereby turning the water back into ice. </p>
| 21251 | Question on Phase Change Materials |
2018-04-08T21:41:33.387 | <p>How does one determine the equilibrium height/altitude of an enclosure of helium, given payload, weight of enclosure, and net lift? What is the formula that will help calculate roughly at what altitude the enclosure will reach equilibrium with the surrounding atmosphere?</p>
| |fluid-mechanics| | <p>At balance altitude the weight of helium and its container is equal to the weight of displaced air at that level.</p>
<p>Let's assume you let free a semi-deflated balloon, which can potentially get inflated by diminished surrounding air pressure to 100 cubic meters, containing 10 cubic meters of helium weighing at sea level 1.786 kg and your balloon is 4 kg thin plastic, total of 5.786kg. </p>
<p>We need to find out at what altitude the 100 cubic meters of air weighs 5.786 kg. </p>
<p>Hence we need to find at what altitude the atmosphere density is $$5.786/100 = 0.05786$$</p>
<p>By checking International Standard Atmospheric table in Wikipedia <a href="https://en.wikipedia.org/wiki/International_Standard_Atmosphere" rel="nofollow noreferrer">The link</a> and very rough curve fitting, we find the balance altitude at, 21000 meters MSL, Stratosphere.</p>
<p>Obviously, the reality is more complicated and requires refined calculations and considering the event's weather forecast. </p>
| 21253 | How do you determine equilibrium altitude of helium enclosure? |
2018-04-09T19:19:16.840 | <p>I'm attaching a metal pipe to a metal tube as shown in the below diagram. The bolt is threaded through one wall of the pipe and does not pass through the opposite wall of the pipe, but is tightened to the point of bottoming out against it. The bolt passes through unthreaded holes in the tube. It needs to support a load in the direction of the red arrow. I need a bit of a gap between the pipe and tube, so I'll be relying on the ability of the bolt to resist bending under this force. Assuming that nothing else will give out first, what metric would I use to estimate how much weight it can support before flexing the bolts? Is it <a href="https://en.wikipedia.org/wiki/Bending_stiffness" rel="nofollow noreferrer">"bending stiffness"</a> rather than <a href="https://en.wikipedia.org/wiki/Shear_strength" rel="nofollow noreferrer">"shear strength"</a>? Bolts don't typically advertise bending stiffness, so how would I estimate that? I know distance between tube and pipe is a factor because it increases the leverage the force has to bend the bolt.</p>
<p><a href="https://i.stack.imgur.com/PQjkl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PQjkl.jpg" alt="enter image description here"></a></p>
<p>I'd think that filling the void between tube and pipe with nuts/washers would strengthen this arrangement when two flat surfaces are being attached, but since the pipe is round, is there any point? I don't expect a precise answer I just need a way to get a ballpark estimate without bending the bolts before selecting them. Hardware stores don't care for that.</p>
<p>It's for an improvised rail if anyone's wondering. I need upwards of 80% of the circumference of the pipe for the slider to grip it, which is why I'm attaching it in such a dumb way. </p>
<p>Something like this (not my image) will be rolling along it.</p>
<p><a href="https://i.stack.imgur.com/YqdMQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YqdMQ.jpg" alt="enter image description here"></a></p>
<p>I haven't found a way to manufacture a proper cradle for the pipe. (EDIT, a U channel would probably work if I can find one the right size) I know that it would be much stronger if done vertically, but this has some advantages given my situation.</p>
| |bolting|shear|stiffness| | <blockquote>
<p>Assuming that nothing else will give out first, what metric would I use to estimate how much weight it can support before flexing the bolts? Is it "bending stiffness" rather than "shear strength"?</p>
</blockquote>
<p>The metric that you should be interested in is "bending stiffness" instead of "shear strength". Shear strength has to do with opposing forces that are very close to each other. This is the principle that scissors work on. Instead of cutting the paper like a knife the two blades cause the material to exceed its shear strength and you get a nice clean cut.</p>
<p>If you are worried about the bolts bending then you should consider the area of elastic bodies that deals with deflection in beams. In your particular case you have a cantilever beams with a single force acting on it. This is similar to "bending stiffness".</p>
<blockquote>
<p>Bolts don't typically advertise bending stiffness, so how would I estimate that?</p>
</blockquote>
<p>You can estimate the deflection of a cantilever beam with a single point load with this equation.</p>
<p>$\sigma = (Fa^3/(3EI))(1 +3b/2a)$</p>
<p>Where</p>
<p>$\sigma = $ deflection at the end of the beam</p>
<p>$F = $ force</p>
<p>$E = $ the modulous of elasticity</p>
<p>$I = $ the moment of inertia</p>
<p>$a = $ distance from the beginning of the beam to where the force is applied</p>
<p>$b = $ the distance from the point where the force is applied to the end of the beam</p>
<p>See <a href="https://www.engineeringtoolbox.com/cantilever-beams-d_1848.html" rel="nofollow noreferrer">this link</a> for more info.</p>
<p>This will let you determine how much you bolt or "beam" will deflect for a certain force applied. However I would guess that in you situation you do not have easy access to accurate figures for the modulus of elasticity or the moment of inertia for your bolts.</p>
<blockquote>
<p>I'd think that filling the void between tube and pipe with nuts/washers would strengthen this arrangement when two flat surfaces are being attached, but since the pipe is round, is there any point? I don't expect a precise answer I just need a way to get a ballpark estimate without bending the bolts before selecting them.</p>
</blockquote>
<p>There are two ways of reducing the the deflection at the end of your bolt while holding the force and geometry the same. By looking at the above equation we see that we can only change $E$ and $I$. In fact one of your suggestions falls under the moment of inertia category. By installing a larger bolt or adding nuts to a smaller bolt you are increasing the moment of inertia which causes the deflection to grow smaller. You worry that this won't do anything because the pipe is round but it will cause a part of the bolt to flex less which means less deflection at the roller.</p>
<p>The second option you have is to use materials with a high modulus of elasticity. By using a bolt made out of a stiffer material the roller will deflect less. By maximizing these two values you can significantly strengthen the bolts in this system.</p>
<p>Now for a quick note of warning. Per your request of "Assuming that nothing else will give out first" this answer does not take into account many other variables that could lead to structural failure. For example, stiffer bolts are more prone to sudden failure and by adding nuts you can possibly add higher stress to the area in between the nuts and the pipe. Please use common sense when you finalize your design and test it beyond what it would normally handle to ensure it is safe to use.</p>
| 21264 | Determining when bolt will bend |
2018-04-10T21:53:43.927 | <p>I have a backplane.</p>
<p>A backplane is a PCB which more or less serves as a hub for other PCBs. The other PCBs connect to it through its "backplane connectors" and it routes the numerous quantity of signals between those connected PCBs.</p>
<p>On my backplane, the connector ports (backplane headers) were designed too close for testing.</p>
<p>Is anyone aware of backplane-connector right angle adapters such that I could plug into the original header connector, adapt at a right angle, and then
lay a board on its side to plug in at that right angle?</p>
| |mechanical-engineering|electrical-engineering|product-testing| | <p>After more searching and asking,<br>
I think the answer is that<br>
this does not exist.</p>
<p>It makes sense,<br>
since the connection isn't particularly firm.</p>
| 21282 | Backplane connector adapter |
2018-04-10T22:28:58.707 | <p>I changed one laptop keyboard on the new one and while changing it I noticed that it has 30 line bus.
<a href="https://i.stack.imgur.com/bwxrM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bwxrM.jpg" alt="enter image description here"></a></p>
<p>But from my VLSI practice, I know that keyboard can use only 1 data line and 1 clock line (PS/2) or 2 data lines on (USB).</p>
<p><a href="https://i.stack.imgur.com/k1CZ4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k1CZ4.jpg" alt="enter image description here"></a></p>
<p>The question is why it has such number of lines? Is it because it doesn't have a microcontroller in it and it is presumably located on the motherboard? Please explain the structure in detail if you can.</p>
| |electrical-engineering|computer-engineering|computer-hardware|data|computer| | <p>The keyboard contains many switches (100 in my keyboard at this moment) and the simplest solution for a keyboard assembly that contains no decoding circuitry would be for each switch to have its own line going to the motherboard, plus one common ground. This is inefficient. </p>
<p>The simplest scheme to minimize the burden of all those wires in the interconnect would be for the key wires to be arranged into a square array of vertical and horizontal wires and a switch positioned wherever any two wires cross over one another. Pressing a key then connects a specific vertical line with a specific horizontal line and when the motherboard "sees" that connection occurring, it interprets it as a specific key press. This is called <em>square matrix</em> encoding and it allows a reduction in the number of address lines from N+1 to 2*(sqrt(N)) or, in this case, 20 lines. </p>
<p>However, a problem occurs if two keys are pressed simultaneously because then the motherboard "sees" four redundant possibilities instead of two and cannot tell which are real. This is called the <em>phantom key</em> problem and requires a more complex encoding scheme than a square matrix, which means more lines are required to resolve redundancies. </p>
<p>There are various schemes to eliminate redundancies like this and I do not know which was used by the manufacturer of your keyboard, but the result will be more than 20 lines. </p>
| 21283 | Why laptop keyboard has so many bus lines? |
2018-04-12T13:57:52.367 | <p>This should be a pretty straight forward question but I can't seem to find the type of material that is used..</p>
<p>So I am making my own heating coil out of Nickel Chromium wire. I plan on using two bolts in a frame that hold the entire thing together. Obviously the bolts themselves are conductive so my plan is to wrap the bolts in the "heat resistant cardboard" that I see inside of hair dryers and other heating devices that use the same wire. Does anyone know which type of material this is, as I am sure normal cardboard would simply ignite...</p>
| |materials| | <p>The "Heat resistant cardboard" that you are describing is likely '<a href="https://en.wikipedia.org/wiki/Mica#Built-up_mica" rel="nofollow noreferrer">Built-up Mica</a>'.</p>
<p>Clearly this is not "cardboard" (as specified in the original question - now edited to say "material"!), since it is not paper-based, but, I hope it can be used for your application.</p>
| 21303 | Heat resistant material for Nickel Chromium Wire? |
2018-04-13T03:03:56.780 | <p>If we are able to see an object in a certain color,that means it reflects the color.A blackbody is one which has zero reflectivity.Then how come it is not black always?</p>
| |heat-transfer|radiation| | <p>There is no Perfect black body (including black hole see hawkings radiation for details) that is why you can see a small light coming up from every black body and hence no body have emissivity of 1. This means true black is not yet discovered. </p>
| 21307 | Blackbody radiation |
2018-04-13T07:10:54.547 | <p>In this question I’m comparing non-plug-in hybrid fuel/electric vehicles with fuel-only vehicles. Also, I’m not disputing that hybrid cars are more fuel efficient, I’m just trying to understand how.</p>
<p>If I remember two things from high school physics, it’s:</p>
<ol>
<li>Creating energy from nothing is against the law.</li>
<li>Converting energy from kinetic to stored energy and back again is always less that 100% efficient - you end up with less kinetic energy than you put in, the rest being “lost” to the environment in the form of heat or whatever.</li>
</ol>
<p>Both a fuel-only car and a hybrid derive all their locomotive energy from the fuel (ok, and gravity, if you’re going downhill, but let’s ignore that). But the hybrid involves an extra conversion to/from stored energy on the way, i.e. charging and discharging the battery, which according to the second principle above should make it less efficient.</p>
<p>Furthermore, the battery entails additional mass, which, if it were dead weight, would reduce the power/weight ratio of the car, reducing efficiency.</p>
<p>So to achieve greater efficiency, the hybrid engine must somehow make use of energy that would ordinarily be lost in the engine of the fuel-only car. In order for the hybrid to be more fuel efficient than the fuel-only engine, it needs to do this in a way that overcomes the above inefficiencies and then some.</p>
<p>So my question is, what broad physical/mechanical techniques does a hybrid engine use to achieve this, and (in rough numbers) how much of the efficiency gains come from the different techniques?</p>
<p>I know there’s regenerative braking, but how efficient is this in practice? I.e. how much of a car’s kinetic energy that would otherwise be lost through braking is actually captured and re-used as locomotive energy? And is this basically it, or are there other techniques that significantly contribute to the efficiency of a hybrid?</p>
| |automotive-engineering|car| | <p>Paparazzo pretty much nailed it, but I'd like to add a few things. </p>
<p>Regenerative braking is certainly a large part of it, <a href="https://www.geotab.com/blog/why-hybrid-vehicles-are-more-efficient/" rel="noreferrer">but there are other synergies when utilizing multiple engines in an automobile, too.</a></p>
<ul>
<li>The ICE can be smaller and sized closer to what is needed for normal operation instead of peak power since the electric motor helps the ICE during the less common acceleration events</li>
<li>The electric motor moves the vehicle at lower speeds where the ICE is less efficient, specifically when accelerating from a standstill</li>
<li>The electric motor helps move the vehicle at speeds at which the ICE would have to operate further away from its most optimal point, helping to keep the ICE within its optimal range</li>
<li>The ICE can be turned off during idling. This is called stop start and is not limited to hybrids</li>
<li>The ICE can be designed to be more efficient within its narrower operating range since it does not have to be a master of all trades</li>
<li>The batteries and the electric drive allow for regenerative braking which helps recoup some of the energy lost during normal operation</li>
</ul>
<p>In addition to this list from geotab, here are a few more insights to energy balance around an automobile:</p>
<ul>
<li>in certain hybrid setups (unlike through-the-road setups) the motor can charge the batteries during times of idle, utilizing the best load/speed for the motor</li>
<li>being able to undersize the ICE is inherently beneficial to thermodynamic efficiency (similar to geotab point #1). ICE's operate best under high load. Ergo, if you can undersize the motor and force it to constantly be under high load, then kick the electric motor in when the peak load exceeds the capabilities of the ICE
<a href="https://i.stack.imgur.com/Jl231.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/Jl231.gif" alt="enter image description here"></a></li>
</ul>
<p>Other considerations:
I don't have any data for this and at the moment do not have time to research it, but I think that there could easily be a human factor to consider as well. I've driven with many individuals, and I don't think that any one will argue that different people have different driving styles. Considering that some people accelerate slowly, while other accelerate hard and brake hard. Considering the synergies of the hybrid electric-ICE power train, it's possible that the hybrid power train can more robustly serve the two different drivers under equally efficient conditions. Consider that the aggressive driver would lose substantial energy in the braking system - and to get the acceleration they desire would require a larger ICE (think big V8s). they would spend more time NOT under peak load, which is what an ICE needs to be efficient. The hybrid drive train would deliver their desired acceleration, while maximizing thermal efficiency AND recouping the energy spent at the above quoted 60% efficiency.</p>
<p><a href="https://i.stack.imgur.com/eRSRd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/eRSRd.png" alt="Power curves of ICE and electric drive trains"></a></p>
<p>Then imagine that slow driving individual gets in the same vehicle; they would likewise be penalized for their slow driving in a large ICE. Considering the hybrid drivetrain, this individual would no longer be penalized for their tame driving habits. In essence, the hybrid drive train could better service larger extremes in driving style, and is over all more robust in terms of energy efficiency. </p>
<p>Just a thought :) </p>
| 21308 | How do hybrid cars achieve greater fuel efficiency? |
2018-04-13T08:20:35.677 | <p>I've been attempting this fundamental shear force diagram problem for several days, but can't seem to get the correct result. I'm trying to calculate the shear force diagram in terms of $x$, but I'm unsure about the intensity $w(x)$ of the triangular load distribution between $0m \le x \lt 3m$. I am able to calculate the correct result for the latter section $3m \lt x \le 6m$, so I'm a little confused as to what the correct intensity of the triangular load distribution is and how to calculate the correct shear force using the correct intensity $w(x)$?</p>
<p>Below I've attached the problem and calculated the support reactions, which are $A_y=15kN$ and $B_y=15kN$.</p>
<p><a href="https://i.stack.imgur.com/s64b9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s64b9.jpg" alt="Problem and Free Body Diagram with support reaction solutions"></a> </p>
<p>Now, I've attached my free body diagram of the first section between $0m \le x\lt 3m $ and indicated the positive sign convention for this beam. </p>
<p><a href="https://i.stack.imgur.com/Rbpa4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rbpa4.jpg" alt="enter image description here"></a></p>
<p>I then proceeded to find the shear force in terms of $x$ as follows:</p>
<p>$\sum F_y=0:$ </p>
<p>$$15-w(x)·x·\frac12 - v_1 = 0 \quad (eq\ 1)$$</p>
<p>Where $w(x)·x·\frac12$ is the area of the triangular load distribution.</p>
<p>This is where I get confused. My understanding of triangular load distribution in terms of the intensity $w(x)$ is that:</p>
<p>$$w(x)=\frac{w_0x}{L}$$ </p>
<p>Where $w_0 = 10$ and $L=3$ for this problem. </p>
<p>But substituting these values into the intensity $w(x)$ and back into $(eq\ 1)$ gets me the wrong result of:
$$v_1=15-\frac53 x^2$$</p>
<p>After reading multiple textbooks and watching several videos, I finally found out that if the maximum load of a triangular load distribution is at the initial point $x=0$ then the following formula should be applied: </p>
<p>$$w(x)=\frac{w_0x}{L}-w_0$$ </p>
<p>I now understand this a bit, but I am wondering where I could get a good explanation as to why? </p>
<p>I'm struggling to find a good explanation as almost every example I've found in textbooks/videos use triangular load distributions that increase from the initial point and not decrease. </p>
<p>However, after utilising this formula, I still get the wrong solution. My working out is as follows:</p>
<p>$\sum F_y=0:$ </p>
<p>$$15-\Bigl(w(x)·x·\frac12 \Bigr) - v_1 = 0$$
$$15-\biggl(\Bigl(\frac{10x}{3}-10\Bigr)·x·\frac12 \biggr)- v_1 = 0$$
$$15-\biggl(\Bigl(\frac{10x}{6}-\frac{10}{2}\Bigr)·x\biggr) - v_1 = 0$$
$$15-\biggl(\Bigl(\frac{5x}{3}-5\Bigr)·x \biggr)- v_1 = 0$$
$$15-\frac{5x^2}{3}+5x - v_1 = 0$$
$$\Rightarrow v_1=15-\frac{5x^2}{3}+5x$$</p>
<p>The actual solution is:
$$v_1=15+\frac{5x^2}{3}-10x$$ </p>
<p>So I'm not sure whether I'm using the correct intensity $w(x)$ and/or whether the triangle area has been correctly calculated using this intensity $w(x)$.</p>
<p>For the second section $3m\le x\lt6m$ I am able to calculate the correct shear force in terms of $x$, this solution is: </p>
<p>$$v_2=-15-\frac{5x^2}{3}+10x$$</p>
<p>Plotting a diagram of the correct shear forces $v_1$ and $v_2$ in terms of $x$ looks the following: </p>
<p><a href="https://i.stack.imgur.com/ljlOB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ljlOB.jpg" alt="enter image description here"></a></p>
<p>For your reference, this problem (F11.6) can be found in chapter 11 of Statics and Mechanics of Materials (4th Ed. SI edition) by Hibbeler.</p>
<p>I'd appreciate if someone could explain intensity loads for situations similar to above and where I went wrong in my calculations.</p>
<p>Thank you.</p>
<p><strong>Edit:</strong> </p>
<p>After reading a few examples, I found that if I calculate the shear force from the left end I am able to get the correct shear force using my initial intensity $w(x)=\frac{w_0x}{L}$ and not the latter intensity $w(x)=\frac{w_0x}{L}-w_0$. </p>
<p>However I'm unsure why I can't calculate this from the right end? Does it have something to do with the left support $A_y=15kN$ creating a discontinuity? If I calculate from the left end am I correct in changing the section's range to $0m \lt x \le 3m$ to not include the left support $A_y$? </p>
<p>My working out is as follows: </p>
<p>$\sum F_y=0:$ </p>
<p>$$-\Bigl(w(x)·x·\frac12 \Bigr) + v_1 = 0$$
$$-\biggl(\Bigl(\frac{10}{3}(3-x)\Bigr)·(3-x)·\frac12 \biggr)+ v_1 = 0$$
$$-\biggl(\Bigl(10-\frac{10x}{3}\Bigr)·(3-x)·\frac12 \biggr)+ v_1 = 0$$
$$-\biggl(\bigl(30-10x-10x+\frac{10x^2}{3}\bigr)·\frac12 \biggr)+ v_1 = 0$$
$$-\biggl(\bigl(30-20x+\frac{10x^2}{3}\bigr)·\frac12 \biggr)+ v_1 = 0$$
$$-\bigl(15-10x+\frac{10x^2}{6}\bigr)+ v_1 = 0$$
$$-15+10x-\frac{5x^2}{3}+ v_1 = 0$$</p>
<p>$$\Rightarrow v_1=15+\frac{5x^2}{3}-10x$$</p>
<p>This is the correct solution. </p>
| |statics| | <p>Your procedure is correct, but you have made a mistake with the sign convention.
Apparently you are using the same convention as I do, where a vertical load/force is negative, and an upward force is positive.</p>
<p>Consequently, as you've written correctly
$$ w(x)=\frac{w_0}{L}x-w_0 $$</p>
<p>Your mistake happens as you formulate the force equilibrium equation.
With this definition of $w(x)$ you already comply to the sign convention.
If you now formulate the shear force equation and write
$$ V_1=15-\int w(x)dx $$
you basically reverse the sign convention again.
To formulate the force equilibrium equation you have to <strong>sum</strong> all forces, not <strong>subtract</strong> them, thus
$$ V_1=15+\int w(x)dx $$
which leads to
$$ V_1=15+\frac{5}{3}x^2-10x $$
which is the correct result.</p>
<p><strong>Sign convention</strong> <em>[edited]</em></p>
<p>Take a look at your $w(x)$. It's a force pointing downwards, so it should be negative.
You have written it as
$$ w(x)=\frac{w_0}{L}x-w_0 \qquad \mbox{for}\qquad x=\{0...3\}$$
Thus $w(0)=-w_0$, which means your load $w(x)$ is already defined in the coordinate system you specified. If you now sum or integrate and add a minus sign in front of $w(x)$ you basically turn the downward load into an upwards facing load.</p>
<p>Consider, e.g. $w(x)=const.=-w_0$, i.e. an evenly distributed downwards load, thus a negative value. It's easy to figure out, that the reaction at the bearing is $A_y=\frac{1}{2} w_0L $.
Now to find the shear force distribution, if we use your method, we'd write:
$$ V_1=A_y-\int_0^L (-w_0)dx=\frac{1}{2} w_0L - (-w_0)L=\frac{3}{2}w_0L $$
This is the wrong result, because for the shear force distribution calculation we suddenly switch sign convention (subtracting during integration instead of adding).</p>
<p>In the first part you said $w(x)$ was $w(x)=-w_0+\frac{x}{L}w_0$, which goes linearly from $-w_0$ to $0$, which in your example would be $w(x)=\frac{10}{3}(x-3)$. During your second calculation you managed to get the right result because you changed the sign convention of your $w(x)$, as you inserted $w(x)=\frac{10}{3}(3-x)$ (which is an upwards load), but then you turned it downwards again by adding a negative sign in front of the whole term.</p>
<p>I really recommend sticking to <em>one</em> sign convention. Either you say you <strong>sum</strong> all forces, and defined downward loads as negative, or you <strong>subtract</strong> downward loads from upward reactions.</p>
| 21309 | Shear Force Diagram of a Simply Supported Beam with triangular load distribution |
2018-04-16T17:27:46.923 | <p>What is the maximum annual capacity factor that can be expected of a CCGT plant? I see in US in 2015 the <em>average</em> was about <a href="https://www.eia.gov/todayinenergy/detail.php?id=25652" rel="nofollow noreferrer">0.56</a>, but can it be higher? I guess it needs some time for maintenance but I don't know how much. Any guidance on where I can get such information?</p>
| |power|gas| | <p>It's on the link you provided. Look at the middle graph, titled "Distribution of annual capacity factors for natural gas combined-cycle plants (2005, 2015)" and look at the Y axis, ranging from 0-10% up to 90-100%. You can see that there were about 20 plants with a CF greater than 90% in 2015, and about 70 plants with a CF less than 10% that same year.</p>
| 21341 | maximum capacity factors of combined cycle gas power plants |
2018-04-17T19:50:13.077 | <p>I want to find the Poisson's ratio of a 3D model, not the material but the model. How can I do this in ANSYS workbench?</p>
<p>I am using this formula, but I can't get it to work with different models. I am not sure what strain is the correct to use, can someone point me in the correct direction?</p>
<p><span class="math-container">$$\nu=-\frac{\text{Strain in direction of load}}{\text{Strain at right angle of load}}$$</span></p>
<p><span class="math-container">$$\nu=-\frac{\varepsilon_{lateral}}{\varepsilon_{axial}}$$</span></p>
| |ansys|ansys-workbench| | <p>Compute the wave velocity in a 3.05-m-diameter steel penstock having a
wall thickness of 25 mm if it:
i. is embedded in a concrete dam;
ii. is anchored at the upstream end; and
iii. has expansion joints throughout its length.</p>
| 21359 | Finding Poisson's ratio of 3D model in ANSYS |
2018-04-18T15:54:10.357 | <p>We do inspections using a robotic camera in the pipeline. It’s tractor type camera. (Ref. photo)</p>
<p><a href="https://i.stack.imgur.com/gobo1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gobo1.jpg" alt="Pipeline Inspection camera"></a></p>
<p><a href="https://i.stack.imgur.com/QnjGE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QnjGE.jpg" alt="camera inspection video"></a></p>
<p>While driving the camera through the pipeline it gives us the camera tilt degree output. </p>
<p>The camera System gives us output as below,
<a href="https://i.stack.imgur.com/g4mEv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g4mEv.jpg" alt="camera tilt output degree against mar"></a></p>
<p><strong>could please help me to calculate the elevation/altitude of each meter. Using camera tilt degree value</strong>. </p>
<p>As below.
<a href="https://i.stack.imgur.com/9urX0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9urX0.jpg" alt="example elevation graph"></a></p>
<p>Awaiting for your answers. If you need any further info. Let me know. </p>
| |civil-engineering|pipelines| | <p>Assuming you know the original starting position, you can use trigonometry to work out the profile of the pipeline.</p>
<p>If you take coordinates $(x_n,y_n)$, and with a measured 'inside pipe travel distance' of $d_n$ meters inside the pipe, down a slope of $\theta_n°$, then that coordinate can be caluclated:</p>
<p>$$x_n=x_{n-1}+\left(d_n-d_{n-1}\right)\sin\left(\theta_n\right)$$
$$y_n=y_{n-1}-\left(d_n-d_{n-1}\right)\cos\left(\theta_n\right)$$</p>
<p>It is important to note, that while the negative incline of the pipe is increasing, this approach will slightly over-estimate the altitude of each point, and slightly under-estimate as it is levelling out. Over a long run of pipe, this should even out, and the total change in altitude from end to end will be fairly accurate.</p>
<p>Here is an image of the graph that I was able to generate using the data that you supplied above:</p>
<p><a href="https://i.stack.imgur.com/Wkcw0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wkcw0.png" alt="Graph generated from supplied data"></a></p>
| 21370 | Elevation / pipeline Elevation calculation |
2018-04-19T07:40:43.100 | <p>I need to simulate an external pressure load that varies sinusoidally along a circular arc. The sinusoidal variation is with arc length.</p>
<p>I know how to create a uniform pressure load in Patran but do not know how to create a non uniform pressure load. I can't seem to find any guides, tutorials or articles on it either.</p>
<p>If anyone knows how to do this or has a useful guide they could link, your help will be greatly appreciated.</p>
<p>I am using MSC Nastran/Patran student version 2017.0.1</p>
| |structural-engineering|finite-element-method|nastran| | <p>The best way to do it is with fields. In this specific case I had to set up two additional coordinate systems. These were cylindrical coordinate systems the were located at the centre of the two circular arcs which defined the surfaces to which the pressure was applied.
The two fields were then defined for in their respective coordinate frame using a PCL function.</p>
<p>The pressure is then applied like a normal constant pressure but the input is the field and not some static value.</p>
<p>I'm sorry I don't have any screenshots to better illustrate this but I hope it helps someone.</p>
| 21378 | How to create a non uniform pressure load in MSC Patran |
2018-04-20T07:52:54.503 | <p>I want to run a coupled simulation in Ansys - RF heating and deformation of waveguide. Pictures of project layout and settings are <a href="https://i.stack.imgur.com/rAmdl.jpg" rel="nofollow noreferrer">here</a>.</p>
<p>I calculated RF fields and surface losses in HFSS; imported them to steady-state thermal, calculated temperature of the waveguide; calculated structure deformation in static structural. Now what is needed is to run this problem until it converges. But for some reason HSFF does not import calculated deformed mesh from static structural. What else should I do to make it import mesh? Can I then use feedback iterator to automatically recalculate this problem?</p>
| |ansys|ansys-workbench|radio| | <p>Problem was with workbench setup. Static structural node should be placed <strong>over</strong> thermal node, not near it. More detailed description with pictures is avaliable <a href="https://i.stack.imgur.com/RuHHZ.jpg" rel="nofollow noreferrer">here</a>.</p>
| 21397 | Ansys multiphysics import deformed mesh to HFSS |
2018-04-20T13:09:16.787 | <p>How do you do double interpolation for a large set of data in Excel?</p>
<p>The user is required to input an x value and y value then the program has to find the points and do interpolation if necessary. For example, if the x is 3.5 and y value is 4.5 then I'll have 2 x value and 2 y values so I need to do the double interpolation then final interpolation between the remaining values.</p>
<p>I have used a <em>vlookup</em> function and <em>forecast</em> function but it doesn't seem to be working out. I have attached a picture of the data set. </p>
<p><img src="https://i.stack.imgur.com/dXeJO.png" alt="enter image description here"></p>
| |statistics| | <p>It appears that your dataset has integer row and column header values starting at 0. You can try the following code.</p>
<ul>
<li>Turn on Excel's <em>Developer</em> ribbon if it's not already on.</li>
<li>Click the <em>Visual Basic</em> button.</li>
<li>Insert | Module.</li>
<li>Paste the code below into the module.</li>
<li>Adjust the rowOffset value to suit.</li>
<li>Enter the formula =Interpol2D(y, x) into your spreadsheet where you want the interpolation result displayed.</li>
<li>Save as .xlsm (with macro).</li>
<li>Every time you change one of the referenced values the calculation will be run again.</li>
<li>You can monitor the calculated values in the Visual Basic <em>Immediate</em> window.</li>
</ul>
<p>Code:</p>
<pre><code>Function Interpol2D(r, c) 'Row and column.
'By Transistor.
'https://engineering.stackexchange.com/questions/21400/double-interpolation-for-a-large-set-of-data-in-excel/21414#21414
rowOffset = 10 'The row number for the header line of the table.
r1 = Int(r + rowOffset) 'The table row number for the first parameter
r2 = Int(r) + rowOffset + 1 'The next row.
c1 = Int(c) + 2 'The table column number for the second parameter
c2 = Int(c) + 3 'The next column.
'Layout of the four adjacent cells and the interpolated result.
' (r1, c1) *---p-------* (r1, c2)
' |
' |
' r (result)
' |
' (r2, c1) *---q-------* (r2, c2)
r1c1 = ActiveSheet.Cells(r1, c1)
r2c1 = ActiveSheet.Cells(r2, c1)
r1c2 = ActiveSheet.Cells(r1, c2)
r2c2 = ActiveSheet.Cells(r2, c2)
'Interpolate by multiplying the difference between the two cells by the fractional
'part of the parameter and then add in the first cell value.
p = (r1c2 - r1c1) * (c - Int(c)) + r1c1 'The interpolation along the first horizontal.
q = (r2c2 - r2c1) * (c - Int(c)) + r2c1 'The interpolation along the second horizontal.
r = (q - p) * (r - Int(r)) + p 'The interpolation along the vertical.
'Results will be printed out in the Immediate window.
Debug.Print "r1c1:", r1c1, p, r1c2, ":r1c2"
Debug.Print "r: ", , r
Debug.Print "r2c1:", r2c1, q, r2c2, ":r2c2"
Debug.Print "================================================================================"
Interpol2D = r 'Return the result.
End Function
</code></pre>
<p><a href="https://i.stack.imgur.com/q9n2A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q9n2A.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Test data.</em></p>
<p>*Table 1. Debug window result.</p>
<pre><code>r1c1: 12 15 18 :r1c2
r: 16.5
r2c1: 14 17.5 21 :r2c2
================================================================================
</code></pre>
| 21400 | Double interpolation for a large set of data in Excel |
2018-04-20T17:25:53.933 | <p><a href="https://i.stack.imgur.com/PXBOv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PXBOv.jpg" alt="I don’t get how to get the answer to part e. Should we assume reactions at A and B in all three figures are same? If so, why? "></a></p>
<p>I don’t get how to get the answer to part e. Should we assume reactions at A and B in all three figures are same? If so, why?</p>
| |mechanical-engineering|structural-engineering|civil-engineering|statics|beam| | <p>No, they are not the same. I suppose you should calculate distances b and c for same condition as you calculated a in question d) and compare results.</p>
| 21404 | Engineering mechanics(statics) |
2018-04-21T20:39:41.530 | <p>I choose an AC motor (IEC) with frame number 71. The key on the shaft of the motor is $5mm,$, $4.23 mm$ and $6 mm$ in diameter. At other hand i cant't find a safety coupling (torque limiter) with exactly the same dimensions as the key way on the motor. They are all a bit smaller or bigger but they have the same diameter as the motor. for instance $4 mm$, $2.23 mm$ and $6 mm$ in diameter. </p>
<p>what can i do ? Is it possible to use a slightly greater safety coupling or the dimensions of the both, should exactly be the same ? </p>
<p><a href="https://i.stack.imgur.com/o8DrO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o8DrO.png" alt="enter image description here"></a></p>
| |motors|torque| | <p>Once you find a coupling that fits the shaft diameter and the other constraints such as size etc, then all you need to do is to get a machine shop to cut a matching keyway to match the one in the motor shaft.</p>
<p>They will also machine a key to match f necessary. This is a small easy job...</p>
| 21419 | Motor coupling connection |
2018-04-22T21:50:33.260 | <p>Back in the early 60s, Chrysler built a <a href="https://en.wikipedia.org/wiki/Chrysler_Turbine_Car" rel="nofollow noreferrer">turbine car</a> powered by a <a href="https://en.wikipedia.org/wiki/Chrysler_turbine_engines" rel="nofollow noreferrer">turbine engine</a> it had under development.</p>
<p>An interesting feature of the engine was a regenerator system to recover heat from the gases leaving the power turbine, using this heat to preheat the air on its way from the compressor to the combustor. The benefits of this were to improve fuel efficiency and reduce the final exit temperature of the exhaust. The regenerators involved some sort of rotating disks which somehow transferred the heat as they rotated. There are diagrams/animations depicting the gas flow through the engine and (generally) heat path through the regenerators, but in some ways they don't seem to make much sense.</p>
<p><a href="https://i.stack.imgur.com/LBwz6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LBwz6.jpg" alt="enter image description here"></a></p>
<p>My question is: how did the regenerator setup actually work? How was the compressor output air (60 psi?) put into contact with a rotating(?) hot surface without the pressure leaking air away to the exhaust? Why would a regenerator need moving parts at all? Could it not just be a system of tubes within tubes?</p>
| |energy-efficiency|turbomachinery| | <p>The main advantage of a rotary regenerator is that you can pack a lot more surface area into a given volume than with a shell and tube type device important for gas-gas heat exchange where heat capacities and heat transfer coefficients tend to be low compared to liquid-liquid. This obviously has advantages for packaging a turbine in a car. </p>
<p>Essentially they have a fine matrix of eg wire mesh which gets heated as it passed through the hot stream and cooled as it passes through the colds stream so you are basically heating up one part and moving that hot part to the cold side. A good analogy is if you have a bucket of hit water and a bucket of cold water you stick a copper rod in the hot bucket untill it gets hot and then take it out and put in in the cold bucket and repeat. A rotary regenerator is the same principal but continuous. </p>
<p>You do get some cross contamination but in a turbine where you are just mixing intake with exhaust this doesn't matter much and isn't that significant in terms of pressure loss and there is no direct path from one side to the other just the inherent limitations of rotary seals </p>
| 21427 | Why regenerator "wheels" in a turbine engine? |
2018-04-23T20:21:04.713 | <p>My machine design textbook states that the fulcrum pin of a band brake during braking is subjected to a double shear. How do you say it is subjected to a double shear and not to a single one.
By double shear the textbook means:<br>
Double shear stress = 2 * shear stress</p>
| |mechanical-engineering|stresses|mechanical|shear| | <p>From the image you should see that the pin on the left has two "lines" of shear stress while the one on the right only one, bending is of course an issue.</p>
<p><a href="https://i.stack.imgur.com/KJ2nG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KJ2nG.png" alt="enter image description here"></a></p>
| 21440 | Why does the fulcrum pin of a band brake experience "double" shear stress |
2018-04-24T08:59:45.503 | <p>When analyzing machines, how to tell if a point that was attached to another part will have only an $x$ or $y$ component or both?</p>
<p><a href="https://i.stack.imgur.com/PCrDy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PCrDy.png" alt="here"></a> </p>
<p>why wouldn't point $B$, $C$ and $A$ have an $x$ component? </p>
| |structural-analysis| | <p>Whether a reaction has both x and y components is going to depend on the geometry of the machine components and the orientation of the applied load. In the machine example you provided it appears all the components are pin connected, which means they <em>potentially</em> have both x and y components. With experience, you may be able to determine by inspection if one of these components is zero. In the meantime, I find it easiest to simply assume nonzero components for both the x and y direction. Solving the equations of equilibrium will generally reveal if one of the components is zero.</p>
<p>As an example, looking at the machine you provided, we can determine by inspection that there is no x reaction at A or D because for link A-D to be in equilibrium there cannot be forces in the x direction. F_Ax and F_Dx would need to be equal and opposite to satisfy horizontal force equilibrium and that would produce a moment on the link (not allowed by equilibrium).</p>
<p>With no x reaction at D, then for equilibrium of the bottom jaw, there can be no x reaction at E.</p>
<p>For the lower handle, with no x reaction at E and only vertical load applied, there can be no x reaction at C.</p>
<p>For the upper handle, with no x reaction at C and only vertical load applied, there can be no x reaction at B.</p>
<p>In equilibrium we trust.</p>
<hr>
<p>The image below shows an example of y-equilibrium. (I didn’t actually work out magnitudes and directions...just the principle of how forces might flow.)</p>
<p><a href="https://i.stack.imgur.com/of6sRl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/of6sRl.png" alt="Assuming y-reactions only"></a></p>
<p>Looking just at the upper handle it’s possible to imagine that component remaining in equilibrium if we assumed x reactions at B and C (equal and opposite). However, when we follow those reactions throughout the machine it leads to a link AD that is not in equilibrium.</p>
<p><a href="https://i.stack.imgur.com/S4HMGl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S4HMGl.png" alt="Assuming x-reactions"></a></p>
<p>So, we have to assume no x reactions at link AD. When we follow that assumption throughout the machine we end up with no x reactions at all.</p>
| 21449 | Machines components? |
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