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2022-07-06T09:33:30.023 | <p>The [depth of groundwater is around 15 to 90 meters][1], while the maximum suction head that can be [created is limited to 10 meters][2] than how this 15 to 90 meters suction head is generated?</p>
<p>[1]: <a href="https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.nationalgeographic.org/encyclopedia/water-tables-and-aquifers/%23:%7E:text%3DAlthough%2520the%2520water%2520table%2520varies,feet" rel="nofollow noreferrer">https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.nationalgeographic.org/encyclopedia/water-tables-and-aquifers/%23:~:text%3DAlthough%2520the%2520water%2520table%2520varies,feet</a>)%2520below%2520the%2520land%2520surface.&ved=2ahUKEwjnmp2A4eP4AhW3R2wGHWuWBEcQFnoECA4QBQ&usg=AOvVaw2aYcIRoTs5PGHslc2XGZPw
[2]: <a href="https://www.tec-science.com/mechanics/gases-and-liquids/how-does-a-drinking-straw-work-principle-maximum-suction-lift/" rel="nofollow noreferrer">https://www.tec-science.com/mechanics/gases-and-liquids/how-does-a-drinking-straw-work-principle-maximum-suction-lift/</a></p>
| |mechanical-engineering|fluid-mechanics|civil-engineering| | <p>So if you put the pump(s) at the bottom of the hole then the suction head is no longer, because you now need the pump(s) to overcome pressure head.</p>
<p>Of course you can use multiple pumps to do 10 meters to fill intermediate reservoirs until you reach the top.</p>
<p>Much like long pipelines that have intermediate pumping stations.</p>
| 51620 | How ground water is brought to surface if maximum suction head is limited to 10 meter? |
2022-07-06T14:49:26.043 | <p>I'm working through a paper to better understand Differential Drive Mobile robots and I'm a bit stuck(fairly early in the paper) on how one of the constraints is derived.</p>
<p>The paper is here, <a href="https://www.hilarispublisher.com/open-access/dynamic-modelling-of-differentialdrive-mobile-robots-using-lagrange-and-newtoneuler-methodologies-a-unified-framework-2168-9695.1000107.pdf" rel="nofollow noreferrer">https://www.hilarispublisher.com/open-access/dynamic-modelling-of-differentialdrive-mobile-robots-using-lagrange-and-newtoneuler-methodologies-a-unified-framework-2168-9695.1000107.pdf</a>.</p>
<p>My question is how is equation 6 derived. Equation 5 defines a no slip lateral slip condition in the ydot robot frame, then states "Using the orthogonal rot matrix R(theta), the velocity in the in inertial frame give -xdot_a * sin(theta) + ydot_a * cos(theta) = 0.</p>
<p>Hmm, now that I think about it if I take the inverse of the Rotation matrix and multiply it by the inertial frame I get xdot_i * -s(theta) + ydot_i * c(theta) = ydot_a which equal 0. This is then defining that no slip condition in the robots y direction. That still doesn't explain why the paper has body frame and I have inertial frame.</p>
<p>Could anyone elaborate?</p>
<p>Thanks!</p>
| |robotics|linear-motion| | <p>Without thinking to much of the correctness of the method, the argument is something like this:<br />
The velocity <span class="math-container">$\dot{y}_a^r=0$</span> is the <span class="math-container">$y$</span>-component of the velocity of the point <span class="math-container">$A$</span> in the body-fixed frame. The same velocity represented in the inertial frame is <span class="math-container">$$\begin{bmatrix}\dot{x}_a \\ \dot{y}_a\end{bmatrix}=R(\theta)\begin{bmatrix}\dot{x}_a^r \\ \dot{y}_a^r\end{bmatrix}=R(\theta)\begin{bmatrix}\dot{x}_a^r\\0\end{bmatrix}$$</span>
We can use this to write
<span class="math-container">$$\begin{bmatrix}\dot{x}_a^r\\0\end{bmatrix}=R(\theta)^T\begin{bmatrix}\dot{x}_a\\\dot{y}_a\end{bmatrix} $$</span> which gives two equations where one is of the form<span class="math-container">$$0 = -\dot{x}_a\sin(\theta)+\dot{y}_a\cos(\theta)$$</span></p>
| 51627 | Differential Drive Kinematic Constraint Determination |
2022-07-06T15:53:23.513 | <p>I understand that this may be incredibly vague, but based on current container ships, does anyone have an idea of how much power a 50,000 TEU (<a href="https://en.wikipedia.org/wiki/Twenty-foot_equivalent_unit" rel="nofollow noreferrer">TwentyFoot Equivalent Unit</a>) container ship might require?</p>
<ol>
<li>For propulsion (assuming a single screw)</li>
<li>For auxiliary power</li>
</ol>
<p>And how much fuel (assuming petroleum)would have to be carried onboard?</p>
<p><a href="https://maritime-executive.com/editorials/50000-teu-the-future-or-not" rel="nofollow noreferrer">https://maritime-executive.com/editorials/50000-teu-the-future-or-not</a></p>
<p>The <em>Algeciras</em>-class container ships have just under 24,000 TEU of capacity, and are powered by MAN 11G95ME engines rated at 101,300 SHP. I'm not sure if this 50,000 TEU ship's power would simply be double that of the <em>Algeciras</em> ships.</p>
<p>n.b.</p>
<blockquote>
<p>While the TEU is not itself a measure of mass, some conclusions can be drawn about the maximum mass that a TEU can represent. The maximum gross mass for a 20-foot (6.1 m) dry cargo container is 24,000 kilograms (53,000 lb). Subtracting the tare mass of the container itself, the maximum amount of cargo per TEU is reduced to approximately 21,600 kilograms (47,600 lb).</p>
</blockquote>
<hr />
<p>UPDATE: Conceptually, the main engine would be a MAN "14G108ME-C" with a 1080 mm bore and 4800 mm stroke. At 70 RPM, total output is 10.8 MW per cylinder with 151.2 MW at the shaft, plus a 10 MW shaft generator, and mean piston speed is 12 m/s.</p>
<p>Emissions reduction is accomplished with direct water/methanol injection, low-pressure SCR, and a sodium hydroxide wet scrubber.</p>
<p>Assuming a 5300mm centre-to-centre distance for the connecting rods, peak-to-peak amplitudes for the secondary imbalance are 574.537mm overall, consisting of 574.301mm second-order, 8.2246mm fourth-order, 0.235618mm sixth order, and 8.43819μm eighth-order.</p>
| |marine-engineering|ships|naval-engineering| | <p><em>Final answer: 150 MW (202,000 hp) for 19 knot propulsion and 30.6MW for auxiliaries.</em></p>
<hr />
<p>EMMA MÆRSK IMO 9321483 (2006) (from <a href="https://en.wikipedia.org/wiki/Emma_M%C3%A6rsk" rel="nofollow noreferrer">wikipedia</a> + <a href="https://www.equasis.org/" rel="nofollow noreferrer">equasis.org</a>)</p>
<ul>
<li>Tonnage 156,907 DWT</li>
<li>Length 397 m</li>
<li>Beam 56 m</li>
<li>Draught 16.02 m</li>
<li>Depth 30 m</li>
<li>Propulsion 81 MW (109,000 hp) Wärtsilä 14RT-Flex96c plus 2 9 MW electric motors</li>
<li>Thrusters 2 Bow (1.75MW) 2 Stern (1.75MW)</li>
<li>Auxiliaries 40 MW 5 Caterpillar 8M32 generators + 8.5 MW TG (Steam powered turbo generator - exhaust heat recovery)</li>
<li>Speed 25.5 knots</li>
<li>Capacity 14,770+ TEU 1,000 TEU (reefers)</li>
</ul>
<p><span class="math-container">$$ \frac{81 MW}{14,770 TEU} = 5.48 kW/TEU$$</span></p>
<p>MADRID MÆRSK - Triple E class - 2nd generation - IMO 9778791 (2017) (from <a href="https://en.wikipedia.org/wiki/Triple_E-class_container_ship" rel="nofollow noreferrer">wikipedia</a> + <a href="https://www.equasis.org/" rel="nofollow noreferrer">equasis.org</a>)</p>
<ul>
<li>Tonnage 210,019 DWT</li>
<li>Length 399.2 m</li>
<li>Beam 58.6 m</li>
<li>Draft 17 m</li>
<li>Depth 33.20 m</li>
<li>Propulsion 2 31 MW (42,000 hp) Twin MAN + 2 2 MW shaft generators/motors</li>
<li>Auxiliaries 13.44 MW = 2 3.84 MW + 2 2.88 MW Hyundai generators + 4.6 MW TG + 2 2 MW shaft generators/motors</li>
<li>Thrusters 2 Bow (2.5 MW)</li>
<li>Speed 19 knots (23 knots)</li>
<li>Capacity 20,568 TEU 1,600 TEU (reefers)</li>
</ul>
<p><span class="math-container">$$ \frac{62 MW}{20,568 TEU} = 3.01 kW/TEU$$</span></p>
<hr />
<p>A quick scan makes it seems that sailing with the Triple-E class produces a better energy usage than the E class. As vessels get larger, energy usage/TEU goes down, but this number is a little misleading since IMO have introduced energy efficiency (EEDI) on 2005+ new designs. Container ships have gotten around improvements in energy efficiency by sailing slower (19 knots for 2017 vs 25.5 knots for 2006) with larger propellors and larger engines. Either way, it will not be above 3.01 kW/TEU.</p>
<p>Now it is 2 engines vs 1 (as required by OP), but using 3.01 kW/TEU, we get propulsion of 150 MW (202,000 hp). This would be the high level for estimation. The state-of-the-art is the Wärtsilä RT-flex96C 84.42 MW (114,800 bhp) (length 27 m, height 13.5 m, weight 2,300 tonnes) used on container ships, so not really sure how it would be fitted as dual engines.</p>
<p>Auxiliaries are a tad harder. They depend on reefer (refrigerated containers) capacity and thrusters. Typically 10% of TEU capacity is reefers. Specifically 1,000 on E class and 1,600 on Triple-E. TEU Reefers can draw up from 5.5kW (frozen) to 9kW (fresh fruit).</p>
<p><a href="https://i.stack.imgur.com/c2aHh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c2aHh.png" alt="enter image description here" /></a></p>
<p><a href="https://www.containerhandbuch.de/chb_e/wild/index.html" rel="nofollow noreferrer">Container Handbook</a> recommends 3.6kW/TEU reefer.</p>
<p>The Triple-E class produces.</p>
<p><span class="math-container">$$ \frac{13.44 MW - 2 \times 2.5 MW}{1,600 TEU} = 5.275 kW/TEU$$</span></p>
<p>Unsure how shaft generators/motors factor in to manouvering.</p>
<p>1,600 of 20,568 = 7.8% and 7.8% of 50,000 is 3,900 reefers. Let's double bow thrusters.</p>
<p><span class="math-container">$$kW = 2 \times 5 MW + 3,900 reefers \times 5.275 kW/TEU = 30.6MW$$</span></p>
<hr />
<p><a href="https://www.emerald.com/insight/content/doi/10.1108/MABR-01-2017-0001/full/html" rel="nofollow noreferrer">Saase (2017/2018)</a> estimates dimensions for a 50,000 TEU to be:</p>
<ul>
<li>Length 521.14 m</li>
<li>Beam 78.20 m</li>
<li>Draught 21.56 m</li>
<li>Air draught 86.08 m</li>
<li>Rows 30</li>
<li>Tiers 27</li>
<li>Bays 32</li>
</ul>
<p>Ports would have to be dredged. Vessel would take up two present bays. Existing loading facilities would not be able to load it. No Suez canal (unless massive dredging, possibly widening). Bridge view issues, so vessel would probably be autonomous.</p>
<p>By 2050, IMO will have a 50% reduction in GHG in place, so heavy crude oil may not be a good choice.</p>
<p><a href="https://newatlas.com/marine/maersk-battery-container-ship/" rel="nofollow noreferrer">Mӕrsk</a> are using a 600 kWh battery in two containers (one control) to act as a buffer to not bring a generator online and meet peak demand loads. Battery charges from waste heat TG.</p>
| 51630 | Power requirements for a 50,000 TEU container ship |
2022-07-10T12:46:11.977 | <p>I need to attach this pressure gauge - <a href="https://www.amazon.co.uk/gp/product/B07CXM3ZGT/ref=ask_ql_qh_dp_hza" rel="nofollow noreferrer">https://www.amazon.co.uk/gp/product/B07CXM3ZGT/ref=ask_ql_qh_dp_hza</a> - to a schrader valve with a pipe to measure the pressure.</p>
<p>The valve has a 1/4 NPT thread. I made the mistake of buying this pressure gauge before checking how compatible it was with a schrader valve and after lots of searching I'm finding it difficult to figure out how to connect them while being airtight.</p>
<p>Could someone recommend an adapter and/or pipe to help me connect them? I'm new to engineering and don't know the best place to look.
I'm questioning if something like this <a href="https://rads.stackoverflow.com/amzn/click/com/B08GJT8B8X" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/dp/B08GJT8B8X/ref=aod_recs_desktop_prsubs_0</a> would work, however I also require at least a foot long pipe to connect them.</p>
<p>Any help would be appreciated.</p>
| |pressure|airflow|pumps|valves| | <p>There are adapters that have a little button in them to depress the schrader pin. But they don't usually have a swivel. So you need the adapter connected to a valve, then to the pipe. Try an A/C shop if you cant find them at your autoparts store. They are used by refrigeration techs quite a lot.</p>
<p><a href="https://mcmaster.com/adapters/thread-type%7Eschrader/" rel="nofollow noreferrer">McMaster</a>
<a href="https://i.stack.imgur.com/sVdnj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sVdnj.png" alt="enter image description here" /></a></p>
| 51659 | How to connect a 1/4 NPT pressure gauge to a schrader valve while being airtight |
2022-07-10T19:08:58.137 | <p>What is the most efficient method to boil water without thermal energy. eg. burning gas. Reducing pressure reduces the energy required but is there something on a molecular level that is the most efficient?</p>
<p>Are microwaves the most efficient or does generating microwaves take more energy that boiling the water??</p>
| |thermodynamics|energy|steam| | <p>This is not a very well-posed question but I will try to take a swing at it.</p>
<p>What you want is a lossless method of delivering heat energy to a mass of water. This would be a simple electric immersion heater dipped into the water, where all the heat generated in the heater gets into the water without any leakage, at 100% efficiency. However, the electricity had to come from somewhere, and if we exclude burning fuel to generate the electricity then this method, while "lossless" in itself, doesn't count. Note also that in this case you are forced to include transmission losses for the electricity.</p>
<p>Microwaves are efficient too, as the efficiency of a magnetron tube can be as high as 70%, but similar comments apply here regarding where the electricity came from.</p>
<p>Using a heat pump to move thermal energy into a tank of water is far more efficient than an immersion heater because it's easier to move heat than it is to create it, and all the efficiency losses of the pump itself are given up to the water too, but here again it takes electricity to run a heat pump so we are once again stuck.</p>
<p>Focusing sunbeams onto a body of water in a container with black walls doesn't require a flame, but the sun itself consists of a scorchingly hot plasma so this technique, strictly speaking, is not "flameless" either.</p>
<p>At this juncture we can propose using a solar cell array to power the heat pump, but with the sun involved we still start out with hot flames!</p>
| 51663 | Boiling water efficiency |
2022-07-11T09:46:20.687 | <p>There are many infrared heaters on the market, that I would put in 3 categories, filament in quartz tube, filament in ceramic, and carbon film (for flat panel heaters).
They all claim to be "infrared heaters" however, as far as I can tell, they all just create heat from electricity and radiate that heat, therefore, they just have a regular thermal radiation spectrum, a blackbody spectrum.</p>
<p>The blackbody spectrum is rather spread out, especially in lower temperatures...So how can these heaters possibly have a narrower band to enable them to claim "near IR radiation" or "far ir radiation" type ? Is it all marketing ?</p>
| |heating-systems|electromagnetism|thermal-radiation| | <p>Each type of heater has a different <em>surface area</em> from which it radiates the rated heat output. This means that to develop the rated power <em>with that surface area</em> requires a different surface <em>temperature</em> for each different radiating area.</p>
<p>So a radiant quartz tube with a tungsten wire inside rated at 500 watts will be typically glowing red or orange (500 watts through the small surface area of the tungsten wire) whereas the carbon film panel heater will not be glowing (500 watts through a large surface area of many square inches).</p>
<p>This means each different heat source in this case will be operating at a different temperature, and each will be producing a blackbody spectrum with a peak somewhere in the IR range- but with a different spectral peak corresponding to that particular temperature.</p>
| 51671 | Do infrared heaters have a special blackbody radiation specturm? |
2022-07-12T12:52:03.860 | <p>So I use accelerometer to get acceleration data. However, I have trouble understanding data. I have been reading manual, but it does not help me:</p>
<pre><code>The measured acceleration data are sent to the OUTL_X, OUTH_X .....
The complete acceleration data for the X channel is given by the concatenation OUTL_X & OUTH_X
and is expressed as a 2’s complement number.
Acceleration data are represented as 12-bit numbers (left justified).
</code></pre>
<p>I understand the concept of 2s complement and left justified number spread in this case, however, I am unable to understand how to arrive at appropriate acceleration values.</p>
<p>The table with couple examples is also provided:</p>
<pre><code>acc value OUTL_X OUTH_X
0 g 0x00 0x00
343 mg 0xE0 0x15
1004 mg 0x00 0x40
-343 mg 0x20 0xEA
-1004 mg 0x00 0xC0
</code></pre>
<p>It would be really helpful if someone helped to figure out how do manufacturers of accelerometer arrive at these values. This is little endian.</p>
| |sensors|acceleration| | <p>the <code>L</code> and <code>H</code> likely represents <code>LOWER</code> and <code>HIGHER</code>. So the Numbers need to be read as</p>
<pre><code> 0 --> 0x0000 --> 0
343 --> 0x15e0 --> 5600
1004 --> 0x4000 --> 16384
-343 --> 0xea20 --> 59936 = 2^16 - 5600
-1004 --> 0xc000 --> 49152 = 2^16 - 16384
</code></pre>
<p>Left justified just means that all measured values have the <em>lower nibble</em> as <code>0</code>; i.e. last Hex digit will always be 0.</p>
<p>Two complement means that to find the magnitude of a negative number, you need to do two's complement once more (or do 2^16 - number as shown above).</p>
<p>From the conversion for 343 mg and 1004 mg, it appears that the conversion formula from the measured number to <code>mg</code> is <code>measured number/16.32</code>.</p>
| 51686 | Understanding data received from sensor (Twos complement) |
2022-07-13T04:40:44.390 | <p>We all know the formula for volume flow rate is, <strong>Q=V<sub>avg</sub>·A</strong>, and that it is constant through a pipe. Now, when a fluid flows in a pipe, a pressure drop will happen thus velocity will decrease and, again to the formula, even if the area is constant the velocity will decrease because of the friction and thus affecting the flow rate.</p>
| |fluid-mechanics| | <p>Hold on, the velocity is not decreasing through the pipe. Who told you that?</p>
<p>Actually the answer is in your question, when the friction occurs, a pressure drop takes place reducing the energy content of the fluid flow.</p>
<p>You actually ignored the fact that the fluid's energy contain its flow energy, which is basically its pressure energy. This pressure will decrease in order to keep the velocity constant in the pipe.</p>
<p>See this photo, you can see that the velocity which is the gap between the hydraulic grade line and energy grade line is constant, while the pressure is not.</p>
<p>When dealing with fluids you should take in consideration that there is a kinetic energy + pressure energy. It is not a rigid body to own only a kinetic energy. The friction has another way to steal energy from the fluid in form of pressure energy.</p>
<p><a href="https://i.stack.imgur.com/CRGqG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CRGqG.png" alt="enter image description here" /></a></p>
| 51695 | Volume flow rate associated with pressure loss in a pipe |
2022-07-13T09:02:55.437 | <p>I am planning on getting an off-grid/hybrid power system for my house.
We have a shortage of inverters in South Africa currently.</p>
<p>Is it possible to have 2 inverters in parallel: one of them 48 V (5 kVA) as primary and another one a 24 V (3 kVA) as secondary?</p>
| |solar-energy| | <p>No. If your battery system is 48V the 24V one will break, and if your battery system is 24V the 48V one won't work.</p>
<p>If you just want a spare one, in case the first one breaks in an emergency - then having the wrong voltage <em>might</em> be okay, but it will be very inconvenient because you'll have to rewire your batteries to get a different voltage. That might or might not be possible, depending on how many batteries you have and which type.</p>
<p>Off-grid inverters will break if you connect the outputs in parallel. If not for the voltage problem, you could put the inputs in parallel, but you'd still have to keep the outputs separate. You couldn't wire them both to the whole house - you could wire one to one part of the house, and the other one to a different part.</p>
| 51697 | Solar power system, hybrid, 48 V & 24 V inverters in parallel |
2022-07-14T11:14:17.740 | <p>I'm working on a robotics project. In this project we need to control the angle position of the wheels. We are going to use an incremental encoder (we didn't find an absolute encoder that was able to to fit into the structure of the system). My question is, is there a method that we can use to store the position of the encoder in case of a power shutdown? I was thinking about ways to save the last position in a non volatile memory (internal microcontroller memory or a sd carf, for example).</p>
| |applied-mechanics|robotics|wheels| | <p>What if someone pushes the wheel while the power is off? The memory won't be correct.</p>
<p>A typical solution to this problem is to use a "home switch". When the robot starts up, it can rotate the wheel in one direction until it bumps into a switch which you have installed in a certain position. Then, the software knows the wheel position, and can track it from then on.</p>
<p>You indicated in the comments that this solution works for you.</p>
| 51703 | Incremental encoder with memory |
2022-07-14T12:41:32.807 | <p>Could anyone tell me what the spring part of <a href="https://uk.rs-online.com/web/p/products/8180599/" rel="nofollow noreferrer">this</a> component is called please so I can source something similar?</p>
<p><a href="https://i.stack.imgur.com/XsYGl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XsYGl.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|mechanisms|mechanical|springs| | <p>The torsion spring with bent ends seems like a good fit for my application. Thanks for helping identify this :)</p>
| 51706 | What is this spring known as? |
2022-07-14T21:20:41.903 | <p>Here a ship that is tilted due an external force . the buoyancy and weight force will create an overturning couple that will rotate the ship in the clockwise direction. Now, the moment equation was defined as Fb*d : where Fb is the buoyant force and d is the distance between the centre of gravity and buoyancy force. I am wondering about which point is this ship rotating? Some refer it to the centre of floatation and some refer it to the centre of gravity . But what is the actual fixed or let's say the pivot point at which the ship is rotating around. How can we know that? It is really mysterious and there is a lack of detailed informations on websites and books. Also if there was a body that is totally immersed (consider it a ball). the same principle would happen. The buyont force will apply a turning moment at a distance from the centre of gravity. but this time the body is fully immersed. Now, will the force act about the centre of gravity also? Because the immersed body has no centre of floatation? I can't understand generally at what point does an immersed or floating body rotate from this buyont force moment.</p>
<p><a href="https://i.stack.imgur.com/JE1I7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JE1I7.png" alt="enter preformatted text here" /></a></p>
| |fluid-mechanics|civil-engineering|ships|water-pressure| | <p>There is a good analogy with balance boards. When the radius of the board bottom is greater than the height of your center of mass, all you have to do to keep the balance is to stay still (metacenter is above the center of mass). In the opposite case, you have to constantly shift your weight to keep the balance.</p>
<p><a href="https://i.stack.imgur.com/vhVSIm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vhVSIm.png" alt="enter image description here" /></a></p>
<p>Buoyancy behaves similarly at least around the starting position. With larger rotations, the buoyancy center may go up or down, which would be similar to a balancing board with general surface on the bottom.</p>
| 51717 | Ship Stability And Overturning Moment |
2022-07-15T04:16:16.247 | <p>At my local gym there is this leg extension machine and I'm trying to figure out the mechanical advantage.<a href="https://i.stack.imgur.com/gjJDS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gjJDS.jpg" alt="leg extension machine" /></a> The pulley attached to the disk (marked by red arrow) confuses me because the disk moves when the force is applied at the pad (marked by green arrow). The pulley moves along with the disk. The rest of the parts on this machine are just cords and fixed pulleys that change the direction of force.<a href="https://i.stack.imgur.com/hgIYA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hgIYA.jpg" alt="leg extension simplified" /></a></p>
| |mechanical-engineering|pulleys| | <p>The porbably easiest way to figure out the mechanical advantage would be to:</p>
<ul>
<li>measure the length <span class="math-container">$r$</span> between the pad that you push against and the yellow axis of rotation</li>
<li>figure out the angle of one full movement - should be about 135°</li>
<li>from this derive the path <span class="math-container">$p$</span> the black pad travels (<span class="math-container">$p=2r\pi\frac{135°}{360°}$</span> or so)</li>
<li>measure the length the weights travel over one full rep. maybe you can bring a chalk and mark the low and the high point during a rep, no need for max load</li>
</ul>
<p>Then the mechanical advantage is the ratio of these two.</p>
<p>My guess is that the marked pully just serves as an attachment point. If you are really really sure the mechanics are as simple as you describe, the advantage is length between axis of rotation to pad divided by length of axis of rotation to outter radius of the marked pulley (to where the rope or band is attached)</p>
| 51723 | What's the mechanical advantage of this leg extension machine? |
2022-07-16T16:48:05.650 | <p>This problem is related to my previous question on the <a href="https://engineering.stackexchange.com/questions/51677/the-generalized-lamis-theorem">generalized Lami's theorem</a>. I would like to see how you solve this problem and compare with my solution. My motivation for this problem is that I have not seen <em><strong>A SINGLE</strong></em> problem of this type on the internet that considers a 4-force system in static equilibrium. All the problems that I have seen consider 3 forces and those that consider 4 never ask for three unknowns, but offer more information in a way that can be solved by vector components. How do you solve this problem using vector components? I apologize for the ugly problem.</p>
<p><a href="https://i.stack.imgur.com/Lx9xF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lx9xF.png" alt="enter image description here" /></a></p>
<p><strong>Note</strong>: The cable for T2 only hangs from the vertical line, <strong>NOT</strong> the horizontal.</p>
| |statics| | <p>The graphic diagram below shows there are infinite solutions (non-unique) to this problem.</p>
<p>Steps:</p>
<ol>
<li><p>Draw the gravity load to scale and mark the ends "a" and "b".</p>
</li>
<li><p>Draw a construction line parallel to the vector <span class="math-container">$T_1$</span> from point "a".</p>
</li>
<li><p>Draw a construction line parallel to the vector <span class="math-container">$T_3$</span> from point "b".</p>
</li>
<li><p>Now make a line parallel to the vector <span class="math-container">$T_2$</span>, but, what is the unique line length required to close the vector loop???</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/jbCke.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jbCke.png" alt="enter image description here" /></a></p>
<p><strong>Let's try another sequence to draw the vector loop.</strong></p>
<ol>
<li><p>Draw a line (3-3) parallel to <span class="math-container">$T_3$</span>.</p>
</li>
<li><p>Draw a line (1-1) parallel to <span class="math-container">$T_1$</span> and let it intercept the line 1-1.</p>
</li>
<li><p>Set the scaled vector 6.21 on line 3-3 at 2 locations, and call the upper points "a" and "b" respectively.</p>
</li>
<li><p>Draw two lines parallel to <span class="math-container">$T_2$</span> and let the lines pass the points "a" and "b", now we get two sets of solutions, which can be more.</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/EBKog.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EBKog.png" alt="enter image description here" /></a></p>
| 51744 | Equilibrium problem with 3 unknown forces | Hanging mass |
2022-07-16T17:18:13.663 | <p>I know about the "whitworth" 3 plate method for making a flat surface but even if you used the side of another of the same material there's no assurance that the two sides would be parallel. This is a property I expect out of gauge blocks for instance, I don't just expect their surfaces to be flat, I expect them to be parallel so that they have a consistent distance from each other too.</p>
<p>Assuming you have a flat surface and nothing else, how does one go about making that? Is there a technique to go from there to make a part with parallel flat surfaces?</p>
<p>I suppose a related question is how you get flat 90-degree surfaces?</p>
| |measurements|machining|metrology| | <p>It's less about how to make them and more about how to measure it.</p>
<p>It actually is a valid method to freehand something and repeatedly check it against gauges until it matches. Since that's so time-consuming and skill-intensive you don't want to do that except to make things in the initial stages, and usually only to make gauges or tools that will obviate the need to repeat similar processes in the future.</p>
<p>So the trick isn't actually how to make "the thing". The trick is how to make the first gauge when you have no gauges to verify it with, and how to verify that all subsequent gauges are good.</p>
<p>The reason that 3-plate method is so important is that it lets you make the first gauge without needing another gauge to verify it. From there the process is mostly using that first gauge (and any other gauges you may have accumulated) to verify if your new gauge is good or not.</p>
<p>Do you know how to measure if something is parallel?</p>
<p>You place it on a surface plate and run a probe along the top face at a fixed height and see if it is actually the same height everywhere. Nowadays, you probably use an indicator mounted to a transfer stand (which will measure the variance in height):</p>
<p><a href="https://i.stack.imgur.com/iAIs7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iAIs7.png" alt="enter image description here" /></a></p>
<p><a href="http://www.tool-precision.com/kpt-62.htm" rel="nofollow noreferrer">http://www.tool-precision.com/kpt-62.htm</a></p>
<p>Or even a coordinate measuring machine (which will measure the actual height):</p>
<p><a href="https://i.stack.imgur.com/pwiyH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pwiyH.png" alt="enter image description here" /></a></p>
<p><a href="https://www.sinowon.com/micromea10128-moving-bridge-scanning-coordinate-measuring-machine" rel="nofollow noreferrer">https://www.sinowon.com/micromea10128-moving-bridge-scanning-coordinate-measuring-machine</a></p>
<p>However, these bely what you actually need: <strong>You do not need to <em>measure</em> (i.e. quantify) the height to do this.</strong> You do not even need to measure the variance in height.</p>
<p>In the most primitive approach, your hands can feel the resistance decreasing, increasing, or staying due to the height decreasing, increasing, or staying the same as you run it along the surface.
Therefore, you need is a base, a vertical beam, and a sharp horizontal point that can be mounted on the vertical beam at any height. A so-called surface gauge:</p>
<p><a href="https://i.stack.imgur.com/2RIKD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2RIKD.png" alt="enter image description here" /></a></p>
<p><a href="https://www.starrett.com/category/111601" rel="nofollow noreferrer">https://www.starrett.com/category/111601</a></p>
<p>Now, you could then freehand and continuously check it until it is parallel. But from what I said above you can probably extrapolate that a much easier method in this case is to run a cutting tool at a fixed height similar to the probe. But if you so desired, you could freehand it to shape it. What you can't do is freehand the measurement part.</p>
<p>You use this to build your parallels. The longer you make your parallels (which also requires making a larger surface plate) higher the resolution you can check for parallelism.</p>
<hr />
<p>Once you have your parallels you can start working on squares. For initial checks, you choose one arm of the square as a reference edge and align it to the parallel and trace the vertical edge. You then flip it and the more square it is the more closely the mirrored vertical edge will overlap/be parallel to the first traced vertical edge. Again, the longer you make the arms the more resolution you will have to check for squareness.
<a href="https://i.stack.imgur.com/NDxjv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NDxjv.png" alt="enter image description here" /></a></p>
<p><a href="https://www.wikiwand.com/en/Square_%28tool%29" rel="nofollow noreferrer">https://www.wikiwand.com/en/Square_%28tool%29</a></p>
<p>Later on you can also do things like mount parallels vertically on a surface plate and run a height gauge on the "unfinished ends" until they read a constant height above the surface plate. At that point your parallels now have ends that are square to the length.</p>
<p>From there you can then mount a horizontal parallel (that does not necessarily need square ends) on top of a vertical parallel (that does have square ends) and run a probe along the length of the horizontal parallel to check if it's height above the surface plate is constant.</p>
<p><a href="https://i.stack.imgur.com/vCinx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vCinx.png" alt="enter image description here" /></a></p>
<p><a href="https://www.hobby-machinist.com/threads/magnetic-cylinder-square.74800/" rel="nofollow noreferrer">https://www.hobby-machinist.com/threads/magnetic-cylinder-square.74800/</a></p>
<p>Also, from having parallels with square ends, you can now now make devices sit on the surface plate and run up and down vertical from the surface plate to more quickly check for squareness without needing to do something like mount two parallels together. But this is not strictly necessary. It's just really convenient.</p>
<p><a href="https://i.stack.imgur.com/dUrEX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dUrEX.png" alt="enter image description here" /></a></p>
<p><a href="http://www.tool-precision.com/kpt-sixty-two-a.html" rel="nofollow noreferrer">http://www.tool-precision.com/kpt-sixty-two-a.html</a></p>
<hr />
<p>At that point you have squares and parallels with square ends. Only then do you start worrying about making a gage block which is parallel, square, and of a particular dimension. Because dimensions are arbitrary. Parallelism and squareness (and by extension cubes), are not.</p>
<p>You would check the that the dimensions of a cube are all equal the same way you checked for parallelism by checking for constant height above a surface plate. And if it was the beginning you could just make any cube, as long as it was a cube and declare that as your unit of length. Because dimensions are arbitrary. Parallelism and squareness (and by extension cubes), are not.</p>
<p>As a result, dimensions tend to be least important thing. Geometry tends to be far more important (parallelism, squareness, roundness, flatness). A gauge block where the dimensions are a bit off but where the ends are parallel parallel is still useful, but a gauge block where the ends are not parallel is useless.</p>
| 51745 | How are flat parallel surfaces made? For example gauge blocks? |
2022-07-17T12:57:59.820 | <p>I recently stumbled across the water tunnel videos by this guy:</p>
<p><a href="https://i.stack.imgur.com/EHXco.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EHXco.jpg" alt="enter image description here" /></a></p>
<p><a href="https://youtu.be/quDLzxmJl5I?t=838" rel="nofollow noreferrer">https://youtu.be/quDLzxmJl5I?t=838</a></p>
<p>He states that for his "high flow" experiments the water speed is at 0.668 m/s. The car model is about 18 cm long. From what I understand about Reynolds numbers this equates to <a href="https://www.wolframalpha.com/input?i=reynolds%20number%20in%20water&assumption=%7B%22F%22%2C%20%22ReynoldsNumber%22%2C%20%22l%22%7D%20-%3E%2218%20cm%20%22&assumption=%7B%22FS%22%7D%20-%3E%20%7B%7B%22ReynoldsNumber%22%2C%20%22Re%22%7D%7D&assumption=%7B%22F%22%2C%20%22ReynoldsNumber%22%2C%20%22v%22%7D%20-%3E%220.668%20m%2Fs%22&assumption=%7B%22F%22%2C%20%22ReynoldsNumber%22%2C%20%22eta%22%7D%20-%3E%220.00089%20Pa%20s%22" rel="nofollow noreferrer">Re = 134702</a>:</p>
<p><span class="math-container">${{0.997 g/cm³ * 0.668 m/s * 18 cm} \over {8.9e-4 Pa s}} = 134702$</span></p>
<p>For roughly the same Re in air and the full car size of 4.5 m in air I come up with a <a href="https://www.wolframalpha.com/input?i=reynolds%20number%20in%20air&assumption=%7B%22F%22%2C%20%22ReynoldsNumber%22%2C%20%22l%22%7D%20-%3E%224.5%20m%22&assumption=%7B%22FS%22%7D%20-%3E%20%7B%7B%22ReynoldsNumber%22%2C%20%22Re%22%7D%7D&assumption=%7B%22F%22%2C%20%22ReynoldsNumber%22%2C%20%22v%22%7D%20-%3E%220.45%20m%2Fs%22&assumption=%7B%22F%22%2C%20%22ReynoldsNumber%22%2C%20%22eta%22%7D%20-%3E%220.0000184481%20Pa%20s%22" rel="nofollow noreferrer">speed of 0.433 m/s (1.56 km/h)</a>:</p>
<p><span class="math-container">${{134702 * 1.8448e-5 Pa s} \over {0.001275 g/cm³ * 4.5 m}} = 0.433 m/s$</span></p>
<p>I like to believe that my thinking here is wrong, as the flow lines in his experiments look "to scale". But from my calculations above it seems that his 0.668 m/s seem to low to represent typical driving speeds at scale?</p>
<p>So to make this a question:</p>
<p>(edited) How are water tunnel experiments equivalent to wind tunnel experiments if, from my napkin calculations above, it seems that to simulate air flow at x km/h you need about the same flow velocity in water?</p>
| |airflow|aerodynamics|hydrodynamics| | <p><strong>TL;DR</strong> Your calculations are correct. As stated by [1], <em>"In flow situations that are insensitive to Reynolds number, or where a test Reynolds number is close to that of a full-size vehicle, water tunnels should be regarded as the preferred option for experimental aerodynamics"</em>. Water tunnels are regarded as a valid approach, but the applicability need to be assessed on a case-by-case basis, mostly due to the discrepency often seen in the Reynolds number.
<em>Since the obective of the video is to give a basic understanding of the flow, I consider the simplifications valid.</em></p>
<hr />
<p>Since the topic at hand is way to large to be satisfactory explained in a SE answer, I highly recommend those interested to have a look at [1]. They greatly compare and discuss the differences in using wind and water tunnels for areodynamic problems. Rather than motor vehicles, their study is focused around air vehicles, which also suits your question.</p>
<p>In the YouTube video you are referencing, from about 5:00-9:00, GraysGarage briefly describes some methods for experimental fluid dynamics, stating that the current objective is to investigate the flow pattern. By not aiming to calculate e.g. drag and lift forces, some scaling errors w.r.t. to the Reynolds number may be neglified.</p>
<p>I suspect the main reason for using a water tunnel in this case is the decreased free-stream velocity obtained due to the high viscosity of water compared to air. This enables easier visualization of complex flow patterns, and consequently a better understanding of the flow dynamics.</p>
<p>However, for the flow pattern obtained in a water tunnel to be representative for a real-life scaled model, some criterias must be satisfied. As is given in Table 1, <em>"the Reynolds number for tests on a model in a water tunnel is at least three orders of magnitude less than that for tests on a full-size vehicle"</em> [1]. For cases where Reynolds numbers are of high importance, such as flow around a circular cylinder, this difference is crucial when interpreting the results. An example of this is seen in the drag coefficent as a function of Reynolds number in Figure 1. [1] states that "<em>Loads measured and flow patterns captured in water tunnels for models with rounded leading edges do not scale well to full-size vehicles, due to different types of boundary layers and separation locations for the two cases</em>". Interestingly, it should also be noted how the use of a wind tunnel still has a difference in Reynolds number when investigating air vehicles, suggesting that care should also be taken when extrapolating such results.</p>
<p>Since the flow patterns of the cars investigated by GraysGarage are dominated by mostly sharp edges in the geometry, I personally believe the flow patterns obtained are representative for use in obtaining a basic understanding of the flow. As GraysGarage also points out, more detailed methods must be applied to achieve the full flow pattern. Then, as is somewhat discussed in the comments on his YouTube video, one should also consider the error obtained by the stationary wheels, which in reality will have large effects on the flow. However, such an analysis was never the intent of GraysGarage, so I therefore consider it off-topic.</p>
<p><a href="https://i.stack.imgur.com/2O7Nn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2O7Nn.png" alt="table1" /></a>
<strong>Table 1</strong> <em>Table from [1, p. 4]</em></p>
<p><a href="https://i.stack.imgur.com/Du8sP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Du8sP.png" alt="figure1" /></a>
<strong>Figure 1</strong> <em>Figure from [1, p. 10]</em></p>
<p>Lastly, [2] is attached to give some further insight to the benefits and drawbacks of flow visualization using <em>Hydrogen Bubbles</em>. GraysGarage also touches on the topic, describing how the use of e.g. dye is disregarded due to the contamination of the water and the following purification needed to start over again. Another benefit is the possibility to create pulses of bubbles travelling downstream, as is applied in the video. This gives a unique possibility to watch how each "particle" behaves inside the flow.</p>
<hr />
<p>[1] Erm, Lincoln P. & OL, Michael V. (2012). <em>An Assessment of the Usefulness of Water Tunnels for Aerodynamic Investigations</em>. URL: <a href="https://apps.dtic.mil/sti/pdfs/ADA582450.pdf" rel="nofollow noreferrer">https://apps.dtic.mil/sti/pdfs/ADA582450.pdf</a></p>
<p>[2] Smith, Charles & Sabatino, Daniel & Praisner, Thomas & Seal, Charles. (2012). <em>Hydrogen Bubble Flow Visualization</em>. URL: <a href="https://www.researchgate.net/publication/303784988_Hydrogen_Bubble_Flow_Visualization" rel="nofollow noreferrer">https://www.researchgate.net/publication/303784988_Hydrogen_Bubble_Flow_Visualization</a></p>
| 51751 | How do water tunnel experiments compare to wind tunnel experiments? |
2022-07-18T20:26:49.880 | <p>I dont know if the title or question makes any sense but when i close my deep freezer i find it difficult to open immediately after closing until after about 2 minutes have passed. How is this done and why? It works all the same with no power source</p>
<p>Does it have something to do with the door seal? I dont get it</p>
| |mechanical-engineering|electrical-engineering|machine-design| | <p>The soft seal on your freezer door allows the door to spring up slightly after the door is closed. But only slightly, because the soft seal (with magnetic closure) prevents the entry of air into the chamber, so the door is held closed by air pressure.</p>
<p>The seal is not perfect, and over a couple of seconds (or minutes), the air pressure equalizes, allowing the door to open easily again.</p>
<p>Before the invention of soft magnetic seals, freezer doors did not 'stick' when closed, and used heavy clamps to keep the door sealed.</p>
| 51774 | What technology is behind a deep freezer door |
2022-07-19T10:32:27.317 | <p>I'm new to mechanical engineering and have a question regarding something that I design using fusion 360.</p>
<p>Here I have 2 wooden rollers that are mounted on 2 side plates thru holes. I'm planning to use sprockets and chain connected to these rods/shafts outside the side plates and a crank handle to power them.
My questions are</p>
<ol>
<li><p>And how do I make sure that these rods wouldn't slide horizontally when the crank handle is rotated that would misalign the sprockets and chain.</p>
</li>
<li><p>Since the thickness of the plate is only 5mm, I don't know if I should use a bearing to hold these rods or the hole is enough or any other thing(I'm not familiar with names of these parts to be used).</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/mcGax.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mcGax.png" alt="design" /></a></p>
| |mechanical-engineering|gears| | <blockquote>
<p>Since the thickness of the plate is only 5mm, I don't know if I should use a bearing to hold these rods or the hole is enough or any other thing(I'm not familiar with names of these parts to be used).</p>
</blockquote>
<p>Yes, absolutely use a bearings or bushings. You want those bearings or bushings. Thickness of the material it is riding in is irrelevant as long as it sits square.</p>
<p>You do not want the outer race of the bearing to rotate relative to whatever it is mounted in, nor do you want the shaft to rotate against the inner race of the ball bearing. That defeats the purpose of the bearing. The goal is to have relative motion <em>only</em> occur between the balls and races of the bearing.</p>
<p>There are at least a few ways to do this:</p>
<ol>
<li>Press fit (works for both inner race and outer race)</li>
<li>Slip fit with set screws (only works for the shaft and inner race)</li>
<li>Slip fit with adhesive such as green threadlocker (works for both inner race and outer race)</li>
</ol>
<p>You can also get bearings and bushings with flanges or shoulders which can help if you are using a slip fit mounting for the outer race.</p>
<p>It is obviously okay for the shaft to rotate against the inner surface of a bushing since that is how a bushing works. Therefore you obviously do not want press fits or adhesives or set screws to prevent the shaft from rotating in the bushing. However, the outside of the bushing should not rotate relative to its mount just like the outer race of a bearing.</p>
<blockquote>
<p>And how do I make sure that these rods wouldn't slide horizontally when the crank handle is rotated that would misalign the sprockets and chain.</p>
</blockquote>
<p>There at at least a few common ways to stop the shaft from sliding lengthwise in the inner race of the bearing, in order of increasing security and elegance, but increasing complexity, cost, and convenience of assembly:</p>
<ol>
<li>shaft collars</li>
<li>grooves with retaining rings in them, but there are also retaining rings designed to dig into the shaft so do not need grooves</li>
<li>put shoulders on the shaft</li>
</ol>
<p>I suppose in some cases you might be able to get away with using adhesive between the shaft and inner race of the bearing, but this is not a proper method to prevent lengthwise sliding.</p>
| 51786 | How to prevent rod/shaft from sliding |
2022-07-19T12:55:57.630 | <p>I am building an imaging platform that requires the application of two pressure sensors shaped into a "hollow" semi-circle, pretty much C-shaped.</p>
<p>To do this, I thought of buying the following <a href="https://store.technimeasure.co.uk/product/flexiforce-a401-force-sensor-25lbf-111n-range/" rel="nofollow noreferrer">sensors</a>, which can be <a href="https://www.tekscan.com/support/faqs/can-i-trim-flexiforce-sensor" rel="nofollow noreferrer">trimmed</a> into a desired shape.</p>
<p>These are relatively expensive and in the end I will have to buy more than two, so I was hoping to buy something more affordable, like this <a href="https://uk.rs-online.com/web/p/strain-gauges/1895556" rel="nofollow noreferrer">second sensor</a>, <strong>but I am not certain this can be cut into the C-shaped sensors I want.</strong></p>
<p>I have tried to contact the website company, but they said they will try to get back to me in some time, as this is a <strong>I.E.E. product</strong> and they are simply the distributors. I have tried to find the company <strong>I.E.E.</strong> online to no avail.</p>
<p><strong>Does anyone know if these pressure sensors can be cut?</strong></p>
<p><strong>Does anyone know where I can find the contact for the I.E.E company?</strong></p>
| |pressure|sensors|consumer-electronics| | <p>No, the sensors described in your link are clearly described as strain gauges. They use a serpentine conductor pattern, as shown on the linked page, and you can not cut that conductor pattern and have a functional strain gauge. Furthermore, a strain gauge doesn't respond to <strong>pressure</strong> applied to it, it responds to <strong>bending</strong> in a direction perpendicular to the serpentine pattern. I don't think any strain gauge will do what you want.</p>
| 51791 | How do I know if I can cut pressure sensors into different shapes? |
2022-07-20T07:26:06.013 | <p>I would like to understand the thermodynamic relation that exist on a big fridge. If you have any links to help me to more understand the physics relation, it would be nice :)</p>
<p>So here is my problem:</p>
<p>There is an electric charge into a big fridge which dissipates X watts into the fridge. The fridge is said to maintained the temperature at T. with T inferior to the ambient temperature.</p>
<p>I also know the electric power consumed by the fridge, which is equal to E. I do not know the transfer ratio between electric power and "calorific power". If you need to add some value, do not hesitate but the less would be the better :)</p>
<p>And finally, I know what is the input (T_input_liquid) and output liquid temperature T_output_liquid , the throughput F of the calorific liquid.</p>
<p>At the end, we may know what would be the COP needed for stabilizing the temperature T into the fridge for X watts.</p>
| |thermodynamics| | <blockquote>
<p>There is an electric charge into a big fridge which dissipates X watts into the fridge.</p>
</blockquote>
<p>We'd normally say "electrical energy" into the fridge. It doesn't dissipate heat into the fridge (which would <em>raise</em> the temperature - it uses the energy to run a compressor which is located outside the cold compartment.</p>
<blockquote>
<p>I also know the electric power consumed by the fridge, which is equal to E. I do not know the transfer ratio between electric power and "calorific power". If you need to add some value, do not hesitate but the less would be the better :)</p>
</blockquote>
<p>I would expect a CoP (coefficient of performance) of about 3. For every 1 watt into the compressor up to 3 watts of cooling (or heat pumping) could be achieved.</p>
<blockquote>
<p>And finally, I know what is the input (T_input_liquid) and output liquid temperature T_output_liquid , the throughput F of the calorific liquid.</p>
</blockquote>
<p>It's not important for the calculations we are doing.</p>
<blockquote>
<p>At the end, we may know what would be the COP needed for stabilizing the temperature T into the fridge for X watts.</p>
</blockquote>
<p>Temperature will stabilise when cooling power matches the rate of heat leakage <em>into</em> the fridge. In refrigeration systems the compressor is oversized and a thermostat is used to switch off the compressor when adequate cooling has been achieved. For your calculations you would need to measure the duty cycle (% on time) of the compressor.</p>
| 51803 | How to know if a fridge would be able to maintain its temperature |
2022-07-24T11:53:42.810 | <p><strong><strong>Updated image attached based on @Translators suggestion comment</strong></strong>
Hi I am looking to make a movable wood cutout via a pulley system on the back my gate/fence (a monkey that tips a top hat when gate is opened and puts top hat back on head when gate is closed). The only part that would move via the pulley is the arm holding the top hat. What kind of pulley system/setup would I need to make this work?
<a href="https://i.stack.imgur.com/2YucO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2YucO.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/vUYO3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vUYO3.jpg" alt="enter image description here" /></a></p>
| |design|pulleys| | <p>I think you're most of the way there.</p>
<p><a href="https://i.stack.imgur.com/u286O.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u286O.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. Rear view of gate, fence and arm.</em></p>
<p>The main tricks here are:</p>
<ul>
<li>Set up the arm so that it always falls under gravity. Add weight at the elbow if required.</li>
<li>Figure out how much motion you want at the point where the string is tied to the arm.</li>
<li>If the gate is opened 90° the horizontal string length will be 70% of its value when the gate is closed. (The gate, fence and string will form a right angled triangle and Pythagoras says that if the two sides are 1 unit long then the hypotenuse (the string) will be <span class="math-container">$ \sqrt {1^2 + 1^2} = \sqrt 2 = 0.7 $</span>.) That means that the string vertical will move 30% of the horizontal length when the gate is closed.</li>
</ul>
<p>e.g. We want 200 mm movement on the end of the arm:</p>
<ul>
<li>200 mm = 30% of horizontal string.</li>
<li>String length = 200 × 100 / 30 = 666 mm.</li>
<li>Distance from hinge to tie point and pulley point = 333 mm.</li>
</ul>
<p>I think a simple eye-hook might suffice rather than a pulley but that depends on the weight of the arm.</p>
<p>If you want to limit the motion when the gate opens then tie a stopper to the horizontal string so that at the required point it bumps into the pulley and won't let the arm drop any further. The horizontal string will then sag.</p>
<p><a href="https://i.stack.imgur.com/0EXM1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0EXM1.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 2. Adding a ring in the position circled will cause the mechanism to fail as there will be no change in length between gate closed and gate open.</em></p>
<p>I can see that my addition of the hinge (drawn before the string) may have caused some confusion as the string passes over the top of the pin.</p>
| 51831 | Engineering a pulley system to make an object move when gate is opened |
2022-07-24T14:33:57.410 | <p>I am trying to compute the deflection of a hollow tube of mild steel, and my computation seems to give radically different results than a couple of online calculators. Apparently one is in error.</p>
<p>The tube is 1.5" nominal Schedule 40, 100" long, with a point load of 100lb at the center. The formula for maximum deflection of a simple supported beam with point load at center is P*L^/48EI. For mild steel, E = 2.9e7. The tube has 1.9" OD and 1.61" ID, which gives its area moment of inertia as 0.310 in^4. Thus the maximum deflection is 0.23".</p>
<p>However, both these calculators:</p>
<p><a href="https://www.easycalculation.com/engineering/mechanical/deflection-round-tube-beams.php" rel="nofollow noreferrer">https://www.easycalculation.com/engineering/mechanical/deflection-round-tube-beams.php</a></p>
<p><a href="https://www.meracalculator.com/engineering/deflection-round-tube-beams.php" rel="nofollow noreferrer">https://www.meracalculator.com/engineering/deflection-round-tube-beams.php</a></p>
<p>give similar and much higher results of about 3.5". They appear to be using a factor of 3, instead of 48, in the denominator in the formula. This is correct for a cantilever beam with a point load at the end. But neither site suggests it's for a cantilever beam, making me wonder if there's something special about round tube beams that I do not understand.</p>
| |mechanical-engineering|beam|deflection| | <p>Your suspicion is correct. The deflection formula was noted below the first cal page. The second should be similar.</p>
<p><a href="https://i.stack.imgur.com/KQ9pM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KQ9pM.png" alt="enter image description here" /></a></p>
<p>Similarly, below the second calculator:</p>
<p><strong>Formula</strong></p>
<p>MI for Solid Round Beams = (pi * (OD^4 - ID^4))/64</p>
<p><strong>Deflection = (length^3 * force)/(3 * E * MI)</strong></p>
<p>Bending Stress = (force * length) * (0.5 * height)/MI</p>
| 51834 | deflection of mild steel hollow tube beam |
2022-07-24T15:08:04.043 | <p>I bought four of these motors, brand new and they all have the same issue. <a href="https://www.amazon.de/RS-550-Various-Cordless-Electric-Drills/dp/B07CXMCNQQ?pd_rd_w=QNEPE&content-id=amzn1.sym.49b0691e-305e-42dc-92a7-9a9f892940e8&pf_rd_p=49b0691e-305e-42dc-92a7-9a9f892940e8&pf_rd_r=WTF7KWRY7W3HBF43QP8G&pd_rd_wg=UPIyO&pd_rd_r=0d8aba03-77cf-4b32-9427-5a637bfa6387&pd_rd_i=B07CXMCNQQ&psc=1&ref_=pd_bap_d_rp_1_t" rel="nofollow noreferrer">Amazon RS-550</a> , <a href="https://rads.stackoverflow.com/amzn/click/com/B07VHN99ST" rel="nofollow noreferrer" rel="nofollow noreferrer">Amazon RS-550</a></p>
<p>I am a bit confused by the Horsepower: 100 watts claim. The motors require 12V from various specifications I found it requires between 0.8A and 1.3A as high peek. I have two variable power supplies:</p>
<p>One which has the following selection:
12V 4.3A 50W all the way to 24V 2.5A 60W</p>
<p>One which has:
3V 1.5A all the way to 12V 1.5A 18W</p>
<p>When connected on any configuration the motors just spin in bursts they spin up and slow down and spin up and slow down with no continous movement.</p>
<p>What am I missing here? Do I have an inadequate power supply? But looking by voltage and amps it should be enough to at least spin the motor under no load.</p>
<p>Looking at this video the power supply should be able to power the motor: <a href="https://www.youtube.com/watch?v=dfiIY1I_MY0" rel="nofollow noreferrer">https://www.youtube.com/watch?v=dfiIY1I_MY0</a></p>
| |motors| | <blockquote>
<p>I am a bit confused by the Horsepower: 100 watts claim.</p>
</blockquote>
<p>You're right. That's the first clue of a dodgy specification. "Horsepower" should be "power".</p>
<blockquote>
<p>The motors require 12V from various specifications I found it requires between 0.8A and 1.3A as high peek. I have two variable power supplies:</p>
</blockquote>
<p>You need to measure the motor DC resistance. From that you can calculate the startup current from I = V/R.</p>
<p>Your power supply will need to reliably supply that much current until the motor gets going. Once running the back-EMF will cause the current drawn to reduce.</p>
| 51835 | RS-550 motor only spinning in short bursts |
2022-07-25T08:20:05.260 | <p>I have two parts: a washer and a rectangular prism.</p>
<p><a href="https://i.stack.imgur.com/AH2AL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AH2AL.png" alt="enter image description here" /></a></p>
<p>I need to connect them side by side. But the joined sides have different curvature.</p>
<p><a href="https://i.stack.imgur.com/w5GrW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w5GrW.png" alt="enter image description here" /></a></p>
<p>How can I merge the sides so that the details look something like this?</p>
<p><a href="https://i.stack.imgur.com/uxwDk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uxwDk.png" alt="enter image description here" /></a></p>
| |solidworks| | <p>you can make the rectangular portion larger,</p>
<p><a href="https://i.stack.imgur.com/zeYQK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zeYQK.png" alt="enter image description here" /></a></p>
<p>and then apply a fillet at the intersection points.</p>
| 51841 | Connecting parts with sides with different curvature in Solidworks |
2022-07-25T15:31:11.473 | <p>I am using Andys APDL.
I want to create Beam Element by entering Node numbers.</p>
<p>Nodes numbers selection is through option “by location“</p>
<p>Then I don’t know how use them in element creation command i.e. E,node#,node#.</p>
<p>Probably the selected nodes need be declared as variable and then variables need be used in element creation command. But I don’t know the syntax.</p>
<pre><code>/NOPR ! Suppress printing of UNDO process
/PMACRO ! Echo following commands to log
FINISH ! Make sure we are at BEGIN level
/CLEAR,NOSTART ! Clear model since no SAVE found
/REPLOT,RESIZE
/PREP7
!*
ET,1,SOLID185
ET,2,BEAM188
SECTYPE, 1, BEAM, CSOLID, , 0
SECOFFSET, CENT
SECDATA,0.25,0,0,0,0,0,0,0,0,0,0,0 ! SMA Wire dia 0.5
!*
!*
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDATA,EX,1,,140e3
MPDATA,EY,1,,10e3
MPDATA,EZ,1,,10e3
MPDATA,PRXY,1,,.3
MPDATA,PRYZ,1,,.3
MPDATA,PRXZ,1,,.3
MPDATA,GXY,1,,5e3
MPDATA,GYZ,1,,5e3
MPDATA,GXZ,1,,5e3
MPTEMP,,,,,,,,
MPTEMP,1,0
UIMP,1,REFT,,,
MPDATA,ALPX,1,,0
MPDATA,ALPY,1,,0
MPDATA,ALPZ,1,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDATA,EX,2,,140e3
MPDATA,EY,2,,10e3
MPDATA,EZ,2,,10e3
MPDATA,PRXY,2,,.3
MPDATA,PRYZ,2,,.3
MPDATA,PRXZ,2,,.3
MPDATA,GXY,2,,5e3
MPDATA,GYZ,2,,5e3
MPDATA,GXZ,2,,5e3
MPTEMP,,,,,,,,
MPTEMP,1,0
UIMP,2,REFT,,,
MPDATA,ALPX,2,,1e-4
MPDATA,ALPY,2,,1e-4
MPDATA,ALPZ,2,,1e-4
RECTNG,,70,,250,
FLST,5,4,4,ORDE,2
FITEM,5,1
FITEM,5,-4
CM,_Y,LINE
LSEL, , , ,P51X
CM,_Y1,LINE
CMSEL,,_Y
!Element edge length = 1
LESIZE,_Y1,1, , , , , , ,1
!*
! Total thickness 1.2
VOFFST,1,1.2, ,
! Active coordinate system is WP
CSYS,4
wpoff,0,0,1.2
NUMCMP,ALL
wpoff,0,0,-.3
VSBW,all, ,DELETE
NUMCMP,ALL
wpoff,0,0,-.3
VSBW,all, ,DELETE
NUMCMP,ALL
wpoff,0,0,-.3
VSBW,all, ,DELETE
NUMCMP,ALL
allsel
CSYS,0
wpro,90.000000,,
CSWPLA,11,0,1,1,
VSEL,S, , ,1
VATT, 2, , 1, 0
VSEL,S, , ,2
VATT, 1, , 1, 11
VSEL,S, , ,3
VATT, 1, , 1, 0
VSEL,S, , ,4
VATT, 1, , 1, 11
/USER, 1
/VIEW, 1, -0.428272868107 , -0.866717917167 , 0.255699828911
/ANG, 1, 67.7726610751
/REPLO
/ZOOM,1,SCRN,0.529172,-0.207734,0.623620,-0.294604
WPCSYS,-1,0
/REPLOT
allsel
VSWEEP,1
VSWEEP,2
VSWEEP,4
VSWEEP,3
allsel
!Apply displacement on thickness areas and Uniform temperature with Pre-stress ON
/SOL
!*
PSTRES,1
FLST,2,4,5,ORDE,4
FITEM,2,4
FITEM,2,9
FITEM,2,13
FITEM,2,18
!*
/GO
DA,P51X,ALL,
TUNIF,1000,
tol=.01
x1=10
x2=20
x3=30
x4=40
x5=50
x6=60
allsel
CSYS,4
NSEL,s,LOC,x,x1-tol,x1+tol
NSEL,a,LOC,x,x2-tol,x2+tol
NSEL,a,LOC,x,x3-tol,x3+tol
NSEL,a,LOC,x,x4-tol,x4+tol
NSEL,a,LOC,x,x5-tol,x5+tol
NSEL,a,LOC,x,x6-tol,x6+tol
NSEL,r,LOC,z,.3-tol,.3+tol
CM,SMAN,NODE ! Group SMA nodes as Component
CMSEL,S,SMAN
nplot
/PREP7
TYPE, 2
MAT, 2
REAL,
ESYS, 0
SECNUM, 1
/ZOOM,1,SCRN,0.170269,-0.876259,0.234494,-0.940468
/AUTO,1
/REP,FAST
/VIEW,1,,,1
/ANG,1
/REP,FAST
! Create Beam Elements for SMA using nodes selected in Component "SMAN"
CMSEL,S,SMAN
nplot
</code></pre>
| |beam|ansys| | <p>You can save a node number in different ways:</p>
<ol>
<li><p>Define the number directly:</p>
<pre><code>N, NODE, X, Y, Z, THXY, THYZ, THZX
</code></pre>
</li>
</ol>
<p>From the docs:</p>
<blockquote>
<p>NODE
Node number to be assigned. A previously defined node of the same
number will be redefined. Defaults to the maximum node number used +1.</p>
</blockquote>
<p>That means you can define a node like this:</p>
<pre><code>NODENUMBER = 999
N, NODENUMBER , 0, 0, 0
</code></pre>
<ol start="2">
<li><p>Get the last created node number:</p>
<pre><code>*GET, NODENUMBER, NODE, 0, NUM, MAX
</code></pre>
</li>
<li><p>Get node number by location:</p>
<pre><code> NODENUMBER = NODE(x,y,z)
</code></pre>
</li>
</ol>
<blockquote>
<p>NODE(x,y,z) returns the number of the selected node nearest the x,y,z
location (in the active coordinate system, lowest number for
coincident nodes)</p>
</blockquote>
<p>I adapted the vm1 example from the docs. When you know the nodal positions it looks like this:</p>
<pre><code>/PREP7
ANTYPE,STATIC ! STATIC ANALYSIS
ET,1,LINK180
SECTYPE,1,LINK
SECDATA,1 ! CROSS SECTIONAL AREA (ARBITRARY) = 1
MP,EX,1,30E6
N, ,0,0,0
N, ,0,4,0
N, ,0,7,0
N, ,0,10,0
NODE1 = NODE(0,0,0)
NODE2 = NODE(0,4,0)
NODE3 = NODE(0,7,0)
NODE4 = NODE(0,10,0)
E,NODE1 ,NODE2 ! DEFINE ELEMENTS
EGEN,3,1,1 ! Generates elements from an existing pattern
D,NODE1,ALL,,,NODE4,NODE3 ! BOUNDARY CONDITIONS AND LOADING
F,NODE2,FY,-500
F,NODE3,FY,-1000
FINISH
/SOLU
OUTPR,BASIC,1
OUTPR,NLOAD,1
SOLVE
FINISH
</code></pre>
<ol start="4">
<li><p>If you don't know the exact position of the nodes, you can loop through a selected set:</p>
<pre><code> /PREP7
ANTYPE,STATIC ! STATIC ANALYSIS
ET,1,LINK180
SECTYPE,1,LINK
SECDATA,1 ! CROSS SECTIONAL AREA (ARBITRARY) = 1
MP,EX,1,30E6
N, ,0,0,0
N, ,0,4,0
N, ,0,7,0
N, ,0,10,0
NSEL,S,LOC,X,0 ! SELECT ALL NODES WITH X-COMPONENT 0. YOU HAVE TO SPECIFY THE RANGE WERE THE NODES ARE LOCATED. I STRONGLY RECOMMEND READING UP ON NSEL
*GET,CURRN,NODE,0,NUM,MIN ! GET MINIMUM NODE NUM IN VARIABLE CURRN
*GET,NUM_OF_NODES,NODE,0,COUNT ! GET NUM OF NODES
NODE1 = CURRN
IND = 1
NUM_OF_NODES_1 = NUM_OF_NODES - 1
*DO,J,1,NUM_OF_NODES_1
IND = IND + 1
*GET,CURRN,NODE,CURRN,NXTH ! GET THE NODE NUMBER IN THE SET HIGHER THAN CURRN
NODE%IND% = CURRN
*ENDDO
E,NODE1 ,NODE2 ! DEFINE ELEMENTS
EGEN,3,1,1 ! Generates elements from an existing pattern
D,NODE1,ALL,,,NODE4,NODE3 ! BOUNDARY CONDITIONS AND LOADING
F,NODE2,FY,-500
F,NODE3,FY,-1000
FINISH
/SOLU
OUTPR,BASIC,1
OUTPR,NLOAD,1
SOLVE
FINISH
</code></pre>
</li>
</ol>
| 51846 | Get Node Numbers to Variable |
2022-07-27T00:38:02.290 | <p>All of the >100,000 tonnes US aircraft carriers are nuclear-powered.</p>
<p>On the other hand, all Chinese and British aircraft carriers are conventional powered.</p>
<p>Which type of aircraft carrier is more cost-effective and reliable in the long run? And, why?</p>
| |ships|naval-engineering| | <h3>It is just a choice.</h3>
<p>Reliability is not at issue here. Both power plants can be built to run whn they need to.</p>
<p>The cost of a nuclear powered ship far outweighs the conventional one. Both building and maintaining the nuclear ship will cost likely twice what the conventional one does. In addition, you have massive training costs for the nuclear sailors and officers, and you need more of them to run a nuclear ship. The training is widely acknowledged as the most difficult academic courses in the US military.</p>
<p>Cost effective is something else. The US wants nuclear carriers because replenishing fuel oil takes away from aircraft launch & retrieval operations. Carriers still need to replenish jet fuel and stores, but avoiding the time to bring aboard ship's propulsion fuel leaves more time for aircraft operations. Current carriers are also very fast, but this is more a design decision than an advantage of nuclear.</p>
<p>The US can afford to deploy these expensive assets, so it does. It can also afford to have enough of these ships to account for the 4+ year refueling times for the ships. We probably shouldn't discount the prestige aspect and the power of the internal nuclear power community, but nuclear ships do what the US Navy wants them to.</p>
| 51860 | Which type of aircraft carrier is more cost-effective and more reliable in the long run? |
2022-07-28T12:56:44.870 | <p>We know that a fluid in reality is not continuous. It has spaces and voids between atoms and molecules.</p>
<p>Continuum approximation is a famous approximation that is taken in any fluid mechanics textbook. It says that even though the fluid has spaces and voids it can be assumed to behave as a continuous media.</p>
<p><strong>Why do we need to assume that a fluid is a continuous media? That is, what was the problem that we were facing when it was not continuous?</strong></p>
| |mechanical-engineering|fluid-mechanics|thermodynamics|fluid| | <p>A continuum fluid model allows us to calculate things using <em>average properties</em> of the fluid at any point throughout its volume.</p>
<blockquote>
<p>For the special situations where the molecular distance does become important, which is in the order of a billionth of a meter, the continuum model does not apply, and requires statistical techniques to study fluid flow</p>
</blockquote>
<p><em>R C Hibbeler, Fluid Mechanics 2nd ed, 2018, Pearson, NY.</em></p>
<p>I believe this answers your question. Fluid on engineering have always been assumed to be continuum fluid until recently. Obviously turbulent flow with a gas mixed with liquid is different because that is a mixture of fluids with different properties.</p>
<p>The question is, when was fluid not treated as continuum model... which would be at molecular level study.</p>
| 51881 | Why do we need continuum approximation in fluid mechanics? |
2022-07-28T18:40:17.880 | <p>Sorry but I'm a little rusty. When solving a beam of any type of section, must the density be included in the moment of inertia of the section?</p>
<p>For example, if I see Navier's formula, I would say that it is not needed, but it seems strange to me</p>
<p>For example, the inertia of a filled circle is:</p>
<p><span class="math-container">$J = \frac{\pi D^4}{64}$</span></p>
<p>and the density does not appear.</p>
| |mechanical-engineering|design|applied-mechanics|beam| | <p>I think you are confused with the "(mass) moment of inertia" and the "second moment of area" as both have the same terms, moment and inertia, in the name, and both are denoted the same - "I".</p>
<ul>
<li><p>The "(mass) moment of inertia", also known as the "rotational inertia", and sometimes as the "angular mass", measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass (which determines an object's resistance to linear acceleration). <a href="https://engineering.stackexchange.com/questions/51885/has-the-inertia-of-a-beam-section-to-incluse-the-material-density">Here</a> is a list of the equations for various situations.</p>
</li>
<li><p>The "second moment of area", also known as the "area moment of inertia", is a geometrical property of an area (cross-sectional area) of a given shape, which reflects how its points are distributed with regard to an arbitrary axis. In "Structural Engineering", it is an important property used in the calculation of the structural elements' deflection and the calculation of stress caused by a moment applied to the element. <a href="https://en.wikipedia.org/wiki/List_of_second_moments_of_area" rel="nofollow noreferrer">Here</a> is a list of equations of various shapes.</p>
</li>
</ul>
<p>Note that the equation you stated is the "second moment of area of a circle". You should avoid denoting it as "J", which generally represents the torsional constant of an area.</p>
| 51885 | Does the inertia of a beam section include the material density? |
2022-07-28T19:16:30.787 | <p>I recently heard a propeller plane, and then very shortly after a jet like noise. After I did some probability calculations, I conjectured that probably these two events are timely correlated.</p>
<p>I also have a hypothesis on why: When the propeller plane is near, the "damping factor" due to the medium (air) not being perfectly "frictionless" is approximately 1, so the experienced loudness of the different frequencies ~ log(amplitude) is approximately described by some formula like the free space path loss formula, so especially the dependance upon the distance is the same for all frequencies, all loose log(d²) DB compared to if you were standing in 1m proximity to the aircraft. Therefore you hear the classical propeller noise.</p>
<p>When the plane distances itself enough, and flies low enough, i suppose the following three effects may change the noise into a white noise jet like noise:</p>
<p>(1) the frequency dependant damping now plays a major role, the air damps the different frequencies differently, damping the classical "propeller frequencies" more than some ohter frequencies that then make up that white noise.</p>
<p>(2) the higher frequencies of the noise are heavily affected by obstacles (the plane flies low and is far away now, so there are obstacles in the way) while the lower frequencies "bend around them" (due to diffraction).</p>
<p>(3) it may also be that with increasing distance and therefore increasing travel time in a real medium like air, sharp frequency distributions naturally get smoothed out over time.</p>
<p>Can/Does the noise of propeller planes change like this? If yes, why does that happen?</p>
| |aerospace-engineering|vibration| | <p>As noted in the comments, what you hear is a combination of engine noise as well as the sound of air over the aircraft and propeller. This will slightly complicate the circumstances, but only slightly. The movement of the aircraft results in a doppler effect.</p>
<p>Sound is compression and rarefaction of air and travels in waves. There is a <a href="https://www.youtube.com/watch?v=h4OnBYrbCjY" rel="nofollow noreferrer">YouTube video</a> explaining that the waves will be subjectively compressed when the source is traveling towards you and will be expanded/rarefied when traveling away from you. This change in frequency means a change in pitch.</p>
<p>The complication arises from the multiple sources of the sound, unlike the automobile horn sound in the video.</p>
<p>One typically experiences a more "pure" example when at a railway crossing and the train is activating the horn during approach and departure. The observer hears a higher pitch at first, an "accurate" pitch abeam of the observer and a lower pitch as the train engine (and horn) departs the crossing.</p>
<p>As a related anecdote, I attended a sailplane competition many stars ago. The racing portion resulted in the sailplanes returning at very high speed and low altitude, as they were carrying water ballast to improve penetration (see <em>sailplane wing loading</em>) and would burn off the speed as well as dump the ballast.</p>
<p>The sound of the sailplane was impressively intense and created the same doppler effect as a jet aircraft at high speed, but without engine noise. Although sailplanes are aerodynamically efficient and aerodynamically "clean," there's still plenty of noise/sound generated at high speed.</p>
| 51886 | (Why) does the sound of propeller planes change with increasing distance? |
2022-07-31T12:47:22.300 | <p>I have this assembly of a cylinder and a block, where the cylinder can rotate inside the block. <a href="https://i.stack.imgur.com/jHQcX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jHQcX.png" alt="enter image description here" /></a></p>
<p>How can I connect the cylinder and the block in a way that is detachable, so that while the connection is close, the cylinder can not be moved, and while it is open, it can be moved? The cylinder is supposed to be able to be set in every rotary position.</p>
<p>One suggestion that I had was to connect them using a screw, as shown in the picture.
<a href="https://i.stack.imgur.com/g6IeZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g6IeZ.png" alt="enter image description here" /></a></p>
<p>However, in this way, the area where they make contact is very small, so I thought maybe there was a better solution? Are there maybe any ressources where you can read about problems like this?</p>
<p>Thanks for any replies and ideas!</p>
| |mechanical-engineering| | <p><a href="https://i.stack.imgur.com/2TYaW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2TYaW.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. A slot and clamp screw provide an even clamping friction around the surface of the mating surfaces and doesn't leave grub-screw marks.</em></p>
<p><a href="https://i.stack.imgur.com/hfFHc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hfFHc.png" alt="enter image description here" /></a></p>
<p>If forced to use a grub screw then a groove can be cut in the cylinder to allow the grub screw to bite on an area that isn't a mating surface between the block and cylinder. This will prevent problems with burrs binding on the mating surfaces.</p>
| 51918 | How can I connect a cylinder and block in a way that can be detached multiple times? |
2022-07-31T14:31:47.057 | <p>I am trying to draw a fin for my final year project, but I could not fully understand the technical drawing explanations that I marked with red markings.</p>
<p>especially R1.281 DEFINES TANGENCY OF R1.312 WITH FIN SURFACE, I couldn't understand the part.
My goal is to make a "bubble fin holder" like in the last picture.
I'm also wondering how we can draw this part of fin .
high resolution image is here
<a href="https://i.imgur.com/OmEKV7J.jpeg" rel="nofollow noreferrer">https://i.imgur.com/OmEKV7J.jpeg</a>
<a href="https://i.stack.imgur.com/X5Whl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X5Whl.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Tjwyc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tjwyc.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/WIBOO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WIBOO.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/hvGbH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hvGbH.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/VcOcz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VcOcz.jpg" alt="enter image description here" /></a>,</p>
<p><a href="https://i.stack.imgur.com/uIX6w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uIX6w.png" alt="enter image description here" /></a></p>
<p><a href="https://i.imgur.com/OmEKV7J.jpeg" rel="nofollow noreferrer">https://i.imgur.com/OmEKV7J.jpeg</a></p>
| |mechanical-engineering|solidworks|technical-drawing|rocketry|airfoils| | <p>You can see in the rightmost view of the drawing that the R1.32 does not meet a spherical surface as suggested by Transistor, but in fact meets the vertical extension of the cylindrical surface beneath.</p>
<p><a href="https://i.stack.imgur.com/O4pZ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O4pZ9.png" alt="Difference between" /></a></p>
<p>The Quickest way to get closer to the drawn geometry (likely within the generous tolerance banding) is to use "Delete and Patch" to simply remove the dome face after you have applied the fillet</p>
<p><a href="https://i.stack.imgur.com/Knaw2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Knaw2.png" alt="Delete Face" /></a></p>
| 51920 | Fin airfoil holder design and technical drawings not understood |
2022-08-02T02:22:56.110 | <p>I'm trying to build a hydraulic power unit with a 5 horsepower electric motor, gear pump and tank. The flow requirement for the system is 1.7 gallons per minute. The pressure requirement is 3000 psi. I'm having some trouble finding gear pumps that are rated for high pressure and low speed. Typically they're at about 20 gallons per minute and some are around 2gpm.</p>
<p>I'm wondering if I get a gear pump that's capable of 2 gallons per minute and run it at a lower speed, will it still be capable of the same pressure that it's rated at at 2 gallons per minute?</p>
<p>Does this have diminishing returns with even lower speeds? For example if I get a pump rated at 4 gallons per minute and run it at less than half the rated speed, will it deliver the same pressure that it's rated for?</p>
| |pumps|hydraulics| | <p>I think 2 GPM with lowering speed to 1.7 GPM will be OK. In fact, engineer always select next larger capacity when (in most of the case) the equipment is not readily available.</p>
<p>For pressure, actually the pressure will depends on back pressure. But as long as the pump at least can push 3000 psi, gear pump should be better choice in term of losing pressure at lower speed (compare to progressive cavity pump, vane pump, and other non rigid sealing mechanism) - maybe second best maintaining pressure at lower speed to Reciprocating Pump only.</p>
<p>So, go for it!</p>
| 51948 | will a hydraulic gear pump be capable of the same pressure at all speeds? |
2022-08-02T17:54:15.380 | <p>I'm discussing safety with a team doing endangered plant conservation work in a small very narrow gulch with walls hundreds of feet high. We see evidence that rocks fall often, and would like to assess the hazard to our once-a-year visiting team who visit briefly to survey the plants. Is it plausible that drought conditions are causing the rockfall frequency to increase? Who is the type of scientist who could analyze the soil type or geo formations type and answer this kind of question?</p>
| |geotechnical-engineering| | <p>You need to consult a <a href="https://en.wikipedia.org/wiki/Geomechanics" rel="nofollow noreferrer">geomechanical</a> engineer. Sometimes they can also be referred to as <a href="https://www.careerexplorer.com/careers/geotechnical-engineer/" rel="nofollow noreferrer">geotechnical engineers</a>.</p>
<p>A geologist can tell you overall geology of the region of interest, but to determine ground stability issues and likelihood of failure and failure modes that is the specialization of geomechanical engineers.</p>
<p><a href="https://australiangeomechanics.org/" rel="nofollow noreferrer">Geomechanical engineers</a> are usually engaged in civil engineering projects or mining.</p>
<blockquote>
<p>Geomechanics is the application of engineering and geological principles to the behaviour of the ground and ground water and the use of these principles in civil, mining, offshore and environmental engineering in the widest sense.</p>
</blockquote>
| 51959 | what kind of expert can analyze drought effect on rock fall risk? |
2022-08-03T15:19:31.933 | <p>Let's say I have two cylindrical objects of the same metal (so they have the same Young's modulus), but object A has a greater diameter than object B.</p>
<p>If both object A and object B are under a tensile strength test, shouldn't object A have a greater yield stress due to a greater cross sectional area?</p>
<p>I understand that both of these objects have the same Young's modulus so their properties would be similar, but engineering stress is defined as force/initial area so the dimensions of the object should impact its ability to withstand permanent deformation.</p>
<p>But this is not the case as my textbook stated that two objects made from the same material will have the same yield stress, and I don't understand why.</p>
| |material-science| | <p>Let's imagine that instead of the two different rods, you have two identical ones. Shouldn't the tensile test require twice the force for the two same rods in paralel than for just a one? If yes, it is twice the force divided by twice the area, which leads to the same stress.</p>
| 51973 | Why doesn't cross sectional area impact yield stress for a given metal? |
2022-08-03T20:07:47.490 | <p>I'm trying to draw a rocket nose cone. Notice the red marked at the bottom left of technical drawing. I couldn't figure out what the diameters 26.870 AND 59.732 mean. I drew endless lines vertically to these points and tried to draw circles at these values, but I couldn't make sense of it.Any help?
high resolution techincal drawing in this link
<a href="https://i.imgur.com/b483Iqd.jpeg" rel="nofollow noreferrer">https://i.imgur.com/b483Iqd.jpeg</a></p>
<p><a href="https://i.stack.imgur.com/VCfuA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VCfuA.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/4Kkl3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Kkl3.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jJmp3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jJmp3.png" alt="enter image description here" /></a></p>
| |solidworks|technical-drawing|rocketry| | <p>It can be drawn if the segments are not tangential to each other.</p>
<p><a href="https://i.stack.imgur.com/a1nAP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a1nAP.png" alt="enter image description here" /></a></p>
<p>If you want tangential transition, you can define it just between the spherical cap and second segment if the cap is not hemisphere (seems nicer), but the second transition between second and third segment cannot be tangential, so the transition points have to be defined by diameters at distances from the table.
<a href="https://i.stack.imgur.com/EhvkY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EhvkY.png" alt="enter image description here" /></a></p>
| 51979 | Rocket Nose Cone Design ,Ogives,technical drawings are not understood |
2022-08-04T04:36:26.873 | <p><a href="https://i.stack.imgur.com/eiUlK.png" rel="nofollow noreferrer">https://i.stack.imgur.com/eiUlK.png</a></p>
<p>Dear all, the gradient of log log graph of equation highlighted in pink is 2. The question is since the scale is log-log, we can take 2 points highlighted in green as gradient that is (1-0.01)/(100-10)=0.011 and why is not 2. What mistake or misconception I have made?</p>
<p>Thank you and have a nice day !</p>
| |mechanical-engineering|electrical-engineering|structural-engineering| | <p>With log log plots, the distance on the graph is the log, not the value. If you calculate out the slope for your numbers, you get:
<span class="math-container">$$
\text{slope}
=
\frac{
\log_{10}{\left(1\right)}
-
\log_{10}{\left(0.01\right)}
}{
\log_{10}{\left(100\right)}
-
\log_{10}{\left(10\right)}
}
= \frac{0 - \left(-2\right)}{2 - 1}
= 2.0
\,.
$$</span></p>
| 51982 | Using 2 points to find gradient in log-log plot |
2022-08-06T06:27:24.913 | <p>Hexagons are considered the strongest grid pattern, I believe, and that's why bees use them? They're an efficient use of space, with maximal structural integrity. I was wondering what a 3-dimensional equivalent would be, for a shape that fits together into a tessellation that's an efficient use of 3D space while having the best structural support.</p>
<p>The sort of application I was thinking of was if you were digging chambers (rooms) for an underground mansion. For this application, you probably want a lot of smallish rooms of a practical shape that'll hold up large amounts of dirt and rock above you.</p>
<p>I was thinking something like a truncated octahedron might work best?</p>
| |structural-engineering|structures|architecture|geometry|mining-engineering| | <p>Your top concern is strength in the vertical direction, correct? Are you sure you aren't tunnel visioned on exotic tessellations? Because I think the actual shape that meets your requirements is rather boring....you are just talking about an underground skyscraper which is a stack of prisms.</p>
<p>In a skyscraper, all the vertical force is taken up by a vertical steel beams (or rather multiple vertical steel beams riveted or welded to each other in practice) that run the length of the skyscraper from top to bottom. This is the strongest ideal configuration for that load. Then to deal with non-idealities you cross brace in an X-shape between vertical beams on the faces to prevent buckling.</p>
<p>If you are concerned with lateral rigidity as your second priority, then you can use a triangular floor plan for maximum lateral strength and rigidity against sideways forces.</p>
<p>Or if you need to balance lateral strength against usable floor space then use a hexagonal or octagonal floor plan. Or just go with square or rectangular floor plans. Honestly it doesn't matter what 2D tessellating polygon you use for your floor plan because if your walls are a vertical planar trusses it will accommodate them all. You just have a bunch of triangular, hexagon, octagonal, or square skyscrapers adjacent to each other.</p>
<p><strong>EDIT:</strong> Except for the ceiling of the topmost floor...that should be arched, domed, or elliptical. Like @Fred mentioned. That way the weight of the ground in the middle of the ceiling gets efficiently transmitted to the vertical truss walls.</p>
| 52007 | What is the most structurally supportive 3D Tessellation? The sort of thing you'd use to build chambers for an underground mansion |
2022-08-06T08:18:32.817 | <p>I am new to the office and I wanted to ask about linear guides like shown in the picture:<a href="https://i.stack.imgur.com/Nys3k.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nys3k.jpg" alt="enter image description here" /></a></p>
<p>(source: <a href="https://www.nskeurope.de/de/linear-motion-control/products/linear-guides.html" rel="nofollow noreferrer">https://www.nskeurope.de/de/linear-motion-control/products/linear-guides.html</a>)</p>
<p>Are the carriages able to move on the bar if you put them on there? And is there a way to keep them fixed in one position? How do you usually attach something to the slider? And are there maybe any ressources where you can look how these linear guides are used for example? Because I was looking online but I did not find any typical uses of them. Thanks for a response!</p>
| |linear-motion| | <blockquote>
<p>How do linear guides work?</p>
</blockquote>
<p><a href="https://www.anaheimautomation.com/manuals/forms/linear-guide.php" rel="nofollow noreferrer">https://www.anaheimautomation.com/manuals/forms/linear-guide.php</a></p>
<p>Linear guides serve three purposes;</p>
<ol>
<li>reduce lateral/radial motion to some minimum slack
<ul>
<li>a wide variety of options from many axial bearings for maximum radial support to a simple rotary axel bearing with internal balls for lower cost just as in drawer slides or sleeve bearing for the lowest cost.</li>
</ul>
</li>
<li>reduce axial friction to some minimum requirement
<ul>
<li>depends on material wear, contaminants, and lubricants</li>
</ul>
</li>
<li>support a moving object with mounting holes on a block.
<ul>
<li>threaded mounting holes exist often metric.</li>
</ul>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/Hq6KO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hq6KO.png" alt="enter image description here" /></a></p>
| 52009 | How do linear guides work? |
2022-08-06T15:53:00.617 | <p>I was interested to find whether it would be more efficient to climb a hill on a lower gear, so I did some calculations.</p>
<p>I have the following data of a custom vehicle:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Gear</th>
<th>Drive force (N)</th>
<th>Max speed (km/h)</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>591</td>
<td>4.87</td>
</tr>
<tr>
<td>2</td>
<td>284</td>
<td>10.12</td>
</tr>
</tbody>
</table>
</div>
<p>I have chosen to calculate for a slope of 8 degrees, so opposing force will be equal to</p>
<p><span class="math-container">$$ F_s = m * g * sin(\alpha) $$</span></p>
<p>Where m is the mass of the vehicle (200kg) and g is gravity, so</p>
<p><span class="math-container">$$ F_s = 200 * 9.81 * sin(8) = 273 N $$</span></p>
<p>I'm ignoring other resistive forces to keep it simple as we can consider them constants at such low speeds.</p>
<p>The drive force is provided from an electric motor with rated power of 1kW, with 80% efficiency, so 800W of usable mechanical power.</p>
<p>I'm calculating the time needed for climbing the distance of 1km with a constant 8 degree slope, assuming we're already at the top speed for the respective gear:</p>
<p><span class="math-container">$$ t = 60 / V $$</span></p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Gear</th>
<th>Time (minutes)</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>12.32</td>
</tr>
<tr>
<td>2</td>
<td>5.91</td>
</tr>
</tbody>
</table>
</div>
<p>Here comes the part I'm not very certain about. If already at maximum speed for the respective gear, the electric motor should only be loaded with the resistive force, thus to calculate the amount of power that will be drawn for both gears I came up with this formula:</p>
<p><span class="math-container">$$ P = P_m * F_s / F_d $$</span></p>
<p>Where <span class="math-container">$P_m$</span> is the motor's power, <span class="math-container">$F_s$</span> is the resistive force from the slope and <span class="math-container">$F_d$</span> is the drive force.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Gear</th>
<th>Power (W)</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>369</td>
</tr>
<tr>
<td>2</td>
<td>769</td>
</tr>
</tbody>
</table>
</div>
<p>And with the resulting powers I also calculate the total amount of energy that will be consumed like this:</p>
<p><span class="math-container">$$ E = P * t $$</span></p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Gear</th>
<th>Energy (Wmin)</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>4544</td>
</tr>
<tr>
<td>2</td>
<td>4546</td>
</tr>
</tbody>
</table>
</div>
<p>I'm not entirely sure about the unit on the energy there but it's not really relevant as I only need to compare the two values and it is identical for both.</p>
<p>Looking at the results it turns out that in both cases the same amount of energy will be spent, which seems a little counter intuitive to me, as if we imagine the same situation on a bike, we'll clearly be more exhausted climbing the hill on a higher gear.</p>
<p>I'm suspicious the method I came up with for calculating the amount of power drawn from the motor on the respective gears might be incorrect.</p>
| |automotive-engineering|power|energy|forces| | <p>You found that the total energy needed to climb the hill is the same despite using two different gear ratios. This is in fact correct because of conservation of energy.</p>
<p>Look at is from another point of view. To get to the top of the hill energy is needed because we are working against gravity. The amount of gravitational potential energy a object contains is given by: <span class="math-container">$$E=mgh$$</span> With <span class="math-container">$h$</span> being the elevation with respect to some datum. Notice that power is not involved and neither is any sort of applied force. For a car climbing a hill <span class="math-container">$m$</span>, <span class="math-container">$g$</span>, and <span class="math-container">$h$</span> will not change whether it climbs fast or slow. Increasing power will only mean you get to the top of the hill faster. The energy is still the same because, for example, doubling the power will halve the time meaning energy will be unchanged. Gears don't change energy, they just trade between force and velocity. This equation might help: <span class="math-container">$$F*V=P=E/t$$</span> In you analogy with the bike, how exhausted you would be is actually not totally correct. While it is definitely easier to climb on a lower gear because of the lower force, you will need to pedal much more to reach the top. If you climb a steep hill on a bike with even a very low gear you will find yourself quite tired by the time you reach the top, even though the pedals are still quite easy to turn.</p>
<p>Just to be sure, imagine that it did take less energy to climb a distance in a lower gear. Well if we climbed in a low gear but then descended in a high gear driving a generator, we would expect that we generated more energy going down than we used going up. This is impossible though because energy must be conserved.</p>
<p>All this, of course, only applies in an idealized situation with no friction, wind resistance, and an assumption that our car is equally efficient at all speeds. In real life the actual motor efficiency curve would determine the most efficient speed with wind resistance skewing the answer slightly slower.</p>
| 52011 | Calculate uphill efficiency on different gears |
2022-08-06T20:19:08.360 | <p>As broadly proven now before the academia, a vehicle including a boat or cart (probably a glider too) can move faster than the speed of wind up to about 2.5-2.8 times with a propeller generating lift on one side of its blades which propels either wheels of the car or the screw of the boat.</p>
<p>(The public debate as broadly settled as of recently: <a href="https://m.youtube.com/watch?v=yCsgoLc_fzI&t=16s;" rel="nofollow noreferrer">https://m.youtube.com/watch?v=yCsgoLc_fzI&t=16s;</a></p>
<p><a href="https://en.m.wikipedia.org/wiki/Blackbird_(wind-powered_vehicle)" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Blackbird_(wind-powered_vehicle)</a>)</p>
<p>Would it be possible to use, say, a 10 times smaller propeller by area one with a propeller diameter than the latest human-sized one with a propeller of ca. 3.5 meters in diameter with a gear of 10:1 gear ration to achieve the same outcome as with the models above? If not, why not?</p>
<p>If yes, would it be possible to set it up with a continuous variable transmission allowing gear ratios to go from 100:1 (100 propeller revolutions to 1 wheel revolution) to 1:1 so as to bring a 10 times heavier body in motion as it continuously changing gear from 100:1 towards 1:1, and achieve substantially the same outcome as seen on the models? If not, why not?</p>
<p>This is for a school project, thank you for any help.</p>
<p>EDIT:</p>
<p>Wind speed would be +15 mph; the total payload plus vehicle weight would be 2 tons, propeller 70 centimeters with a CVT.</p>
| |gears|applied-mechanics|propulsion| | <p>You have to start by looking at the true wind speed that you want to operate in, the speed ratio that you are trying to achieve, the course conditions, and the payload. The rest is all about optimization and what you can actually build. If you are trying for best VMG upwind or downwind, you will need a big propeller. If you just want to try to go faster than the true wind speed in any direction, then that is a lot easier to do, and you don't need as big a propeller.</p>
<p>Driveline efficiency is a huge problem for these machines. You can't afford to loose 5% to a tranny. You need to be able to change gears to suit conditions and target speed ratio, but that usually means changing wheel size or changing a sheave. The actual range of gearing is pretty small - maybe 15% covers all possible operating conditions if you are capable of a speed ratio of 2.0</p>
| 52013 | Could a downwind cart, or “push-me, pull-me” boat of a substantially heavier size be propelled with the use of a CVT, or continuous gear? |
2022-08-08T17:59:40.833 | <p>In general, a kitchen refrigerator is essentially a heat pump. The vapor compression cycle creates a temperature difference, and the insulated compartment stays cold. But how can I determine the heat output of such a heat pump or its COP?</p>
<p>The user manuals only give the average energy consumption per day, but because of the insulation the compressor doesn't have to run all the time, it just turns on when the temperature inside exceeds some threshold, and stops when it reaches another threshold.</p>
<p>As a side question, is a kitchen refrigerator a viable option for low energy cooling? For example, if I have a device with a TDP of less than 300 watts and it needs sub-ambient temperatures, can I just put it in the refrigerator?</p>
| |thermodynamics|heat-transfer|cooling| | <p>I like the answer by @Transistor better but if you don't have the option of turning off the refrigerator, you could get an idea of the COP by measuring the power consumption of the device under somewhat controlled conditions and measuring again after putting in a known thermal load.</p>
<p>Since COP is Q<sub>cooling</sub>/P<sub>in</sub></p>
<p>Say, you ran the device for 24 hours and used 1500 watt-hours.
If you put in a 10 watt light bulb and run it for another 24 hours and now used 1590 watt-hours, your COP would be:</p>
<p><span class="math-container">$$
COP
=
\frac{Q_c}{P_{in}}
=
\frac{Q_c}{1500 Wh}
=
\frac{(Q_c + 24h*10W)}{1590 Wh}
$$</span></p>
<p>Assuming my algebra and arithmetic are correct,
<span class="math-container">$$
Q_c=4000 Wh
$$</span>
so</p>
<p><span class="math-container">$$
COP=2.66
$$</span></p>
<p>As for whether it is economical to run device that puts out 300 watts in your refrigerator, you can work that out the same way.</p>
<p><span class="math-container">$$
COP=2.66 =\frac{300W*24h}{x Wh}
$$</span>
solving for X, you will use about an extra 2700 Watt-Hours each 24 hour period.</p>
<p>At $0.10 per kWh, that's</p>
<p><span class="math-container">$$
cost=
(0.300 kW)(24 hours)($0.10/kWh)
= $0.72
$$</span>
to keep the device running and another:
<span class="math-container">$$
cost=
(2.7 kWh)($0.10/kWh)
= $0.27
$$</span>
to keep it cool.</p>
<p>That's assuming your refrigerator can keep up. If you exceed the cooling capacity of your device, it will run continuously, keeping it as cool as it can until it dies.</p>
| 52041 | Kitchen fridge thermal power and COP |
2022-08-08T19:24:49.287 | <p><a href="https://i.stack.imgur.com/dF4G8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dF4G8.jpg" alt="enter image description here" /></a>I have a light hanging from an arm that is attached to the wall. I need to make sure that the fastener at the wall end is able to hold the weight. How do I calculate this?</p>
| |design|stresses|fasteners|forces| | <p>Say the arm is 2 inches vertically at the base where it is hung by the fastener and the distance between the fastener and the point of pressure is 1.5 inches.</p>
<p><span class="math-container">$$F_{fastener}=\frac{8lbs*27in}{1.5in}=144lbs$$</span></p>
<p>This means your fastener is being pulled out by a nominal force of 144# and the wall on the 1.5 in below must resist a nominal compression force of 144#.</p>
<p>Both the tension and compression should be multiplied by a safety factor of say 1.8.</p>
| 52042 | I am designing a light that is hung from the end of an arm. How do I calculate the amount of force on the fastener where the arm attaches to the wall |
2022-08-09T01:53:30.123 | <p>I don't understand why solid stresses (s11, s22, s33, s12, s13, s23) are reported for EACH joint of a solid (an eight-node brick element). I suppose it is reported per solid, and the stresses correspond to its 6 positive/negative faces.</p>
<p>If we look at joint forces, each joint has 3 translational DOF, instead of 6 values. This "contradiction" confuses me.</p>
<p>Please let me know if any my of understanding is wrong. Thank you.</p>
| |structural-analysis|solid-mechanics| | <p>Your questions relate to the fundamentals of the Finite Element Method. I'll give a brief overview here of what generally happens and then discuss your specific case.</p>
<p>Typically, each element type (line, shell, solid) is associated with a specific mathematical formulation referred to as an isoparametric formulation. Basically what's happening behind the scenes is that each element type is transformed into it's isoparametric type, regardless of its geometry. (if you want to visualize this, one example is that a parallelogram shell element gets transformed into a perfect rectangle in this formulation).</p>
<p>During this process it also creates new specific points where your resultants (forces and moments for line types, force per unit for shells, stresses for solids) can be accurately calculated. It's worth noting that these resultants are NOT exact anywhere else on the element other than at these points (theoretically anyways. Practically we can interpolate them with a high degree of accuracy).</p>
<p>Here's the crux of your issue - there will be as many of these points (referred to as Gauss points) as there are nodes in an element. So a 4-node quad shell will have 4 gauss points, an 8 node solid will have 8 points. Now most software will translate the gauss point results over to the actual node positions (there are reasons for this but they're not very relevant to this discussion).</p>
<p>So there you have it. SAP2000 is showing you stresses at nodes because FEA only allows it to accurately calculate results at the gauss points and then interpolate them over to the nodes.</p>
<p>With regard to why solids only have 3 dofs at each node instead of 6, I'll refer you to Daryl Logan's excellent book on 'First course in FEA'. The simple reason is - the math checks out.</p>
| 52048 | Trying to Understand Solid Stress Output from SAP2000 |
2022-08-09T03:27:44.990 | <p>I've learned that the yield stress of a polymer can be marked on the tip of the first curve on the stress-strain graph of a polymer undergoing a tensile strength test, like so:
<a href="https://i.stack.imgur.com/jrysj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jrysj.png" alt="enter image description here" /></a></p>
<p>As the yield stress - which marks the start of plastic deformation, and necking (which marks the beginning of localized plastic deformation) begin at the same strain, does this mean that polymers don't undergo any uniform plastic deformation and that they only undergo localized plastic deformation?</p>
<p>If so, why?</p>
<p>Help would be much appreciated, thanks.</p>
| |material-science| | <p>The short answer is Yes, some polymer can undergo uniform plastic deformation.</p>
<p>Your question is about necking and <a href="https://en.wikipedia.org/wiki/Necking_(engineering)#Neck_stability" rel="nofollow noreferrer">neck stability</a>. The criteria for the onset of necking is only related to the slope of the true stress <span class="math-container">$\sigma$</span> - true strain <span class="math-container">$\varepsilon$</span> curve,
<span class="math-container">$$ \frac{\mathrm{d}\sigma}{\mathrm{d}\varepsilon} > \sigma $$</span>
You can find this equation in nearly every text book of material mechanics. It applies to all isotropic materials, not only to polymer, but also to steel, aluminum and etc.</p>
<p>The left hand side <span class="math-container">$ {\mathrm{d}\sigma}/{\mathrm{d}\varepsilon}$</span> represents the material's work (strain) hardening characteristics.</p>
<p>As explained in figure 2 of <a href="https://polymerinnovationblog.com/characterization-thermosets-part-21-tensile-testing-polymers-molecular-interpretation/" rel="nofollow noreferrer">this blog</a>, polymer undergoes different stages of hardening during deformation:</p>
<blockquote>
<p>As the semi-crystalline polymer begins to deform, the stress causes the chains in the amorphous region to elongate and uncoil (b and c). At higher strains, the stress is transferred from the amorphous chains to the lamellae causing the crystalline lamellae to become oriented in the direction of the applied stress (d). Parts of the lamellae can break apart forming smaller fold blocks. With further levels of large elongations, the fold blocks continue to deform in the direction of the applied stress and the amorphous regions become highly elongated resulting in a fibrillar type structure.</p>
</blockquote>
<p>This <a href="https://www.researchgate.net/publication/250320450_A_Method_to_Calculate_the_True_Stress_and_True_Strain_for_Tensile_Test_of_Plastic" rel="nofollow noreferrer">journal article</a> explained the three stages of tensile tests of polymer:</p>
<ol>
<li>Shortly after yielding, the polymer chains simply uncoil and, thus, the work hardening is too low for stable deformation, so necking initiates.</li>
<li>Necking evolution: the cross-section of the necking area shrinks.</li>
<li>Necking diffusion: As the lamellae become oriented, the work hardening increases rapidly and, to a certain degree, can stop the necking evolution. Then, the deformation becomes stable and the cross-sectional area at the location of necking does not reduce further. The deformation extends along the
axial direction to the entire length of the testing region of the specimen. The specimen deforms uniformly in this stage.</li>
</ol>
| 52050 | Do polymers undergo uniform plastic deformation? |
2022-08-10T08:04:06.053 | <p>I have a test rig whereby a shaft (an M4 threaded bar) is being pulled ~11.5mm by a mechanism (rather not detail that) in about 20ms. I want to plot the movement of this shaft vs the mechanism pulling it.</p>
<p>I made up my own sensor using an IR switch with a portion of a hair comb passing through it. Of course a custom 3D printed solution would have been ideal but I don't have that.</p>
<p>I got the finest comb I could find and measured the distance between first and last tooth and divided the length by the number of teeth (minus 1) which gives me a 1.54mm pitch, or 0.77mm between tooth and gap.</p>
<p>I made up an L shaped bracket to hold the comb and passed through the IR switch and put a scope on that.</p>
<p>So, I was expecting to see 11.5/1.54 = 7.5 full on/offs on the scope or 15 high/low transitions.</p>
<p>In practice I got much more than I expected. I figured my shaft is bouncing when it hits the stops so tried to damp it by creating a deformable structure with masking tape around the lock nuts that limit travel. This improved things greatly as my last pic shows, but this was the best of several tests. I need data I can trust.</p>
<p>I'm pretty sure the teeth are not hitting the switch, and I shortened the teeth to make them less flexible.</p>
<p>Unfortunately both the comb and switch are black, but hopefully you can see what's going on. For scale, the screws holding the comb are M3.</p>
<p>Any thoughts/suggestions please?</p>
<p>TIA</p>
<p><strong>EDIT 10th Aug</strong></p>
<p>As per the suggestions in the comments, I tried it by hand and was consistently getting the expected results, 7 full waves which is 10.8mm, within 1 step of what I should get and I'm happy that it's repeatable.</p>
<p><strong>EDIT 19 Aug</strong></p>
<p>Sorry for the delay in this update, I've been working on and off at it. I think I'm getting pretty good results now. I re-made the bracket that holds the comb (could not figure a way to attach directly to the bar) and used glue instead of screws.</p>
<p>For damping, I used those felt like things you put on furniture etc to prevent scratches. And, I soaked them in oil to improve damping. Finally, I added a small spring that comes into play near the end of the stoke. See last pic.</p>
<p><a href="https://i.stack.imgur.com/9we6S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9we6S.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/TNgqE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TNgqE.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/Zz4NT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zz4NT.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/K4swB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K4swB.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/Kx0Zi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kx0Zi.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/4pJbF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4pJbF.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/3vsqB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3vsqB.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|measurements|sensors|instrumentation| | <p>The placement of your sensor on that long arm is very bad. It's going to vibrate like a tuning fork.</p>
<p>The arm is long which gives it a lot of flex. And those overly long bolts make it heavy which increases the inertia at the comb. These combine to give a large vibrational amplitude. There is also very little damping, as others have mentioned.</p>
<p>The best solution would be to attach the comb rigidly to the main shaft.</p>
<p>You may see some improvement by lightening the end of the sensor arm as much as possible (perhaps switching to plastic screws or at least shorter ones). Shortening and stiffening the arm, and adding damping (maybe a thin plastic truss).</p>
| 52059 | DIY linear position sensor IR switch - bad results |
2022-08-11T15:29:30.870 | <p>I have the following problem, which I already know the general steps into solving.</p>
<p><a href="https://i.stack.imgur.com/skAmR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/skAmR.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Mr0Jr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mr0Jr.png" alt="enter image description here" /></a></p>
<p>My general approach is: calculate I cracked as we are already told that the section is cracked, and then finally use <span class="math-container">$My/I$</span> to calculate the stress.</p>
<p>But in order to do that I need to calculate the area of steel, so what does<span class="math-container">$ Ast = N16 @ 175mm $</span> centres means? My lecturer drew the following diagram but I still don't get it.</p>
<p><a href="https://i.stack.imgur.com/6PWcV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6PWcV.png" alt="enter image description here" /></a></p>
<p>And then he did this to calculate the area:</p>
<p><a href="https://i.stack.imgur.com/cO2Gb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cO2Gb.png" alt="enter image description here" /></a></p>
<p>Why do you divide to find the area of the steel? How do I interpret his diagram and his calculations? Thank you.</p>
<p>UPDATE:</p>
<p>Why do we do <span class="math-container">$n \times A_{st}$</span>?
<a href="https://i.stack.imgur.com/Z0cIU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z0cIU.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|concrete|reinforced-concrete| | <p>Let's calculate the equivalent reinforcing steel area (<span class="math-container">$Ast)$</span> in a <span class="math-container">$1 m$</span> strip from the given reinforcing configuration (<span class="math-container">$1-N16@175 mm$</span> spaced center to center) using the concept of "consistent reinforcing ratio".</p>
<p>The reinforcing ratio of the <span class="math-container">$1 m$</span> strip is, <span class="math-container">$\rho_1 = \dfrac{Ast}{1 m*d}$</span>, and the reinforcing ratio for <span class="math-container">$1 - N16$</span> at <span class="math-container">$175 mm$</span> spacing is, <span class="math-container">$\rho_{act} = \dfrac{200 mm^2}{0.175 m*d}$</span>, and, <strong>since <span class="math-container">$\rho_1 = \rho_{act}$</span></strong>,</p>
<ul>
<li><span class="math-container">$\dfrac{Ast}{1 m*d} = \dfrac{200 mm^2}{0.175 m*d}$</span></li>
</ul>
<p>With "<span class="math-container">$d$</span>" cancels out, the equivalent reinforcement in <span class="math-container">$1 m$</span> strip, therefore, is,</p>
<ul>
<li><span class="math-container">$\dfrac{Ast}{m} = \dfrac{200 mm^2}{0.175 m} = 1142 mm^2/m$</span></li>
</ul>
<p><a href="https://i.stack.imgur.com/S89JF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S89JF.png" alt="enter image description here" /></a></p>
<p><strong>ADD:</strong> For add'l question in comment</p>
<p><a href="https://i.stack.imgur.com/nmLZ6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nmLZ6.png" alt="enter image description here" /></a></p>
| 52078 | What does$ Ast = N16 @ 175mm $centres means in designing reinforced concrete? |
2022-08-11T18:33:28.927 | <p>According to the following diagram (this is for a p-type semiconductor) :
<a href="https://i.stack.imgur.com/Dg2OV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dg2OV.png" alt="enter image description here" /></a></p>
<p>The acceptor level - which is the hole, has a higher energy than the valence electrons as it is above the valence band. But why does the hole have greater energy than the valence electrons? Shouldn't it have the same energy as an electron as it is simply the broken bond that an electron leaves behind (when the electron gets promoted)?</p>
<p>The text I was studying from also stated that a hole is a place where an electron can be promoted to, but why and how?</p>
| |material-science| | <p>Holes energy is measured down from valence band top because of their positive charge.</p>
<p>Remember that in real world there are no holes. This is quasi-particle introduced in order to describe movement of valence electrons, which is very hard due to high concentration of them. Huge amount of valence electrons is thus replaced by moderate amount of holes.</p>
<p>My advice is to read classic solid state books, for example, <em>Kittel</em> or (and) <em>Aschcroft and Mermin</em>, they step-by-step explain everything from the origins.</p>
<p>P.S. Semiconductor questions should go to electrical engineering.</p>
| 52082 | Why does a hole have greater energy than valence electrons? |
2022-08-11T22:08:15.837 | <p>To expand on my question - Why is the donor level in a an n-type extrinsic semi-conductor called a level instead of a band?</p>
<p>Shouldn't it be referred to as a band instead, as the "donated electrons" are being introduced to the semiconductor from multiple dopant atoms, not a single one?</p>
| |material-science| | <p>Because "width" of <em>dopant level</em> is very small, especially relative to band "width". Due to relatively low concentration of dopant atoms, thus, big distance between them, there is nearly no interaction between dopant atoms' electrons and no electron levels splitting, thus electrons of all atoms have nearly same energy and band is not formed. This can be imagined as dot line on energy diagram.</p>
<p>At high dopant concentration, impurity atoms became close to each other, their electron interaction becomes significant, electron levels split and form a band. Such semiconductors are called <em>degenerated</em>. Such high-doped regions usually form passive parts of devices, like ohmic contacts or emitters.</p>
<p><a href="https://en.wikipedia.org/wiki/Degenerate_semiconductor" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Degenerate_semiconductor</a></p>
| 52089 | Why is the donor level in a semiconductor called a "level"? |
2022-08-13T09:56:34.123 | <p>I have the following questions:</p>
<p><a href="https://i.stack.imgur.com/kwKqM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kwKqM.png" alt="enter image description here" /></a></p>
<p>Would the capacity be the area under the curve? Or can I use this formula: <span class="math-container">$q_c = \frac{k_j \times v_f}{4}$</span> from the greenshields model, where the capacity flow is related to the free flow speed and jam density. If so then <span class="math-container">$q_c = 6600$</span></p>
<p>The following question is :
<a href="https://i.stack.imgur.com/quBQw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/quBQw.png" alt="enter image description here" /></a></p>
<p>I am really not sure how to approach this question. AADT = Annual average daily traffic.</p>
| |transportation| | <p>This is one of those vocabulary problems. If you know what they mean, it should be a matter of applying unit conversions.</p>
<p>I'll make some sense of the words. And then I will do my utmost to forget it, because it's pretty useless to learn some model when you don't even know if reality fits it...</p>
<p>Capacity is the maximum possible veh/hr possible.</p>
<p>The formula you have works if the curve is linear (which we can believe from the graph).</p>
<p>More generally it is <code>max(speed*density)</code>. And here's a derivation of that qc formula.</p>
<pre><code>speed = 110 km/h*(1-density/(120 veh/km)).
speed = vf*(1-density/kj)
speed*density=vf*density-vf*density^2/kj
max of such a parabola occurs where derivative is 0 so:
0=vf-2*vf/kj*density
density = kj/2
speed=vf/2
max(speed*density) = qc = kj*vf/4
</code></pre>
<p>Or 3300 veh/hr (which is probably per direction with an implicit 2 directions based on your mention of 6600)</p>
<p>Next question is irksome in how useless it is. If someone knows 12% of AADT occurs... Then that person could have counted what AADT was when taking measurements to obtain that 12%.</p>
<p>Anyway, it mentions a 55/45 directional split, so 3300 corresponds to the 55%. Word trap is whether that's 55% split of veh/hr or veh/km. Traffic in the measurement sense (rather than "stuck in traffic" sense where it means a section that actually lacks traffic per units of time) is something per time so it is the former (veh/hr). That means the other direction contributes <code>3300/55*45=2700</code> So both dirs it's <code>3300+2700=6000</code> veh in that hr. Divide by .12 aka 12% and you get 50000.</p>
| 52103 | speed-density relationship transportation question |
2022-08-14T16:26:29.943 | <p><a href="https://i.stack.imgur.com/J3mMj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J3mMj.png" alt="My Custom Piece (Measures are in mm) " /></a>
<em><h6>My Custom Piece (Measures are in mm)</h6></em></p>
<p>I'm designing a custom mechanism that needs this piece to be able to function, but when i started thinking about how to manufacture it, I started getting confused about how to make it.</p>
<p>what I know is:</p>
<ul>
<li>how to make a material block into a cylinder using a "Lathe" (For Example).</li>
<li>How to make the hole in the middle of the cylinder using the "Lathe" or the "Milling Machine" (For Example).</li>
</ul>
<p>but when i get to making the saw shape every 60°, I don't know what to use.</p>
<ul>
<li>should I use a CNC Machine? But is it gonna be able to make that shape, since there isn't much space for the <strike>"Drill Bit"</strike> "<strong>Endmill</strong>/<strong>Cutter</strong>"?</li>
</ul>
<p>I know that a similar piece is also used in some one way rotation mechanisms so there should be multiple ways to get this shape.</p>
<p>So if you know some ways to make a saw shaped cylinder,<br />
if you have any advice or suggestion on practically making this piece, or if you even have have some corrections or other ways to get the cylinder or the hole.</p>
<p>it would be appreciated.</p>
<p><strong>EDIT</strong>:
some remarks:</p>
<ul>
<li>This drawing is measured in mm</li>
<li>The purpose of this part is to stop the rotation of a similar part sitting on top of it</li>
<li>So the sharpness of the edges is tolerable, as long as it blocks a similar part's rotation</li>
<li>I like @Jonathan.R.Swift comment about separating the inner cylinder from the outer cylinder.</li>
<li>yes i didn't know at first what <strong>Endmill</strong> is called in English.</li>
</ul>
| |mechanical-engineering|materials|design|manufacturing-engineering|solid-mechanics| | <p>To add to DKNguyen's answer, it depends on the required strength of the part. If the part is highly-stressed, you would <em>hot forge</em> it out of <em>steel alloy</em> in a <em>press</em>.</p>
<p>If the stresses are significant but not excessively high, you would <em>machine</em> it from <em>steel or aluminum</em> in a <em>CNC machine</em>. All the slopes and angles shown in your drawing are easily cut on a 3-axis CNC milling machine.</p>
<p>If the stresses are relatively low and minimizing cost is important, you would <em>die cast</em> the part in <em>copper-aluminum die casting compound</em> in a <em>closed-mold molding machine</em>.</p>
<p>If the stresses are minimal and production volumes are very large and the part will not be subjected to hot conditions during use, you would <em>injection mold</em> the part in <em>glass-filled polymer</em> in an <em>injection molding machine</em>.</p>
| 52120 | What are the practical solutions to manufacturing this mechanical piece? |
2022-08-14T18:15:42.340 | <p>I'm trying to modify my bike's harness.
Only trafficator functionality is supported by means of SPDT switch. I tried to add blinker functionality:</p>
<p><a href="https://i.stack.imgur.com/U8dA2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U8dA2.png" alt="enter image description here" /></a></p>
<p>Kindly excuse the symbols for both Relay and Switch.
This should work all right, but I wanted the trafficator function to override the blinker function. I'm planning to add a switch for the DPDT Relay, which would be my blinker switch.
How should I go about it? A pure electrical solution would be nice with minimal component count and less connections. Thanks!</p>
| |automotive-engineering|electrical| | <p>Here's what I came up with, it's still impractical, and involves lots of components:
<a href="https://i.stack.imgur.com/IQlrf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IQlrf.png" alt="enter image description here" /></a></p>
<p>Sorry, couldn't post in the morning: got the flu, didn't expect downvote.
PS: After attempt at solution:
<a href="https://i.stack.imgur.com/RgtGo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RgtGo.png" alt="enter image description here" /></a></p>
| 52122 | Modifying bike wiring harness to add hazard lights |
2022-08-14T19:18:55.407 | <p>The <a href="https://en.wikipedia.org/wiki/MIT_Daedalus" rel="nofollow noreferrer">MIT Daedalus</a>, a human-powered craft, weighed about 31 kg. This was significantly lighter than its pilot during its record setting trip, though I can't confirm the pilot's exact weight. Still, this raises the question: If the pilot had been half the weight, would it have been possible to safely reduce the Daedalus weight by more than half (aka: exponentially instead of linearly)?</p>
<p>Since smaller birds seem to have an advantage over larger ones, I would assume it'd be easier to achieve flight with a smaller pilot.</p>
<p>Note that the pilot is not half weight due to being malnourished or lacking musculature. The half weight was meant as an easy example. If someone has an example for a 20% lighter pilot, that should be useful.</p>
| |structural-engineering|structural-analysis|structures|aerospace-engineering|aerodynamics| | <p>There is no simple answer because there is an awful lot going on with planes such as <em>Daedalus</em>. That project needed to cover a specific course distance in ground effect over open water. And it needed an allowance for weather. If you go smaller, will you go faster, slower, or the same speed? Doesn't being half the size make the course seem twice and big? How will the aspect ratio of the wings change (spar weight is governed in large part by bending moment, which scales as the forth power). What are the Reynolds number effects on the lift and drag curves? Since you can't scale the weather, how does the perturbation and recovery energy factor in, and how do you scale the control surfaces and control power requirements? How much ground effect are you counting on and how does that work if the scale changes but the sea state remains the same? How does turbomachinery such as a propeller scale? How does the vehicle's stiffness matrix scale. How do the vehicle's stability derivatives scale?</p>
<p>Many of the above questions can be answered with Drela's <em>Flight Vehicle Aerodynamics</em> book. Then you have to figure out how to engineer and build the thing.</p>
<p>The way you approach this in practice is to rank order these effects as best you can, then run sequential optimizations where you add in one more constraint with each run. You may have to analyze and re-linearize the model each go. You can use Lagranian optimization to handle the constraint system. It gets real messy real fast. You end up with a lot of heuristics to speed things up once you think you have the big questions answered.</p>
<p>Much of the design space of interest is fairly flat, and you need really accurate test data to get a good linearized model at each step. That's the real problem here - data on the real performance of built HPV wingsets is not all that robust. Particularly with regard to rigidity requirements.</p>
<p>So the short answer is that every single design variable is going to scale a little differently because they all influence each other. Many are inherently nonlinear, so you have make a guess and linearize around it and see what happens.</p>
<p>Here is the most common scaling list for sailboats, which have a lot of in common as far as problem set-up goes. The list comes from Barkla originally, I cribbed it from <em>Principals of yacht design</em> by Larsson and Eliasson.</p>
<blockquote>
<p><strong>Assumed</strong>: L=Length<br />
Sail area: L^1.85<br />
Beam, depth, freeboard: L^0.70<br />
Keel, rudder - span, cord, thickness: L^0.70</p>
<p><strong>Derived</strong><br />
section areas: L^1.4<br />
wetted hull area: L^1.70<br />
wetted keel, rudder: L^1.40<br />
hull volume:L^2.40<br />
Keel volume:L^2.10<br />
length/displacment ratio: L^0.20<br />
sail area/displacement ratio:L^0.25<br />
2nd moment area lateral:L^3.10<br />
2nd moment area longitudinal: L^3.70</p>
<p><strong>Secondary</strong> There is a whole raft of more ratios that are typical for normal boat, but depend somewhat on the particular design being
scaled.</p>
</blockquote>
<p>In the above list, there was no attempt to optimize the scaling, they just used some basic knowledge from millions of boats and crunched the math for the derived figures. With miniDaedalus, you don't have millions of craft to use as a basis. And the scaling is only considered good for about 0.9 to 1.1 L.</p>
| 52126 | Does the weight of a pilot exponentially affect the necessary weight of human-powered aircraft? |
2022-08-15T12:08:41.803 | <p>I wonder if such a valve exist - a cheap solenoid valve that when open does not restrict air flow?
I've only found solenoid valves for compressed air, although the opening may very (4-12mm) there is a small hole inside no matter what.</p>
<p>This is the one I got but it's not what I need
<a href="https://i.stack.imgur.com/hvV4L.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hvV4L.jpg" alt="enter image description here" /></a></p>
<p>*Motorized ball valves allow air to pass freely but I didn't found a small one and a cheap one.</p>
<p>I'm working on a simple project of a 5015 blower fan that I want to connect to a valve and sometimes allow air to pass, so I can maintain a small pressure difference between the inside and the outside.</p>
<p>I've also tried the plastic one-way valves but they also block most of the air.</p>
<p>Thanks</p>
| |valves| | <p>If DIY is within your requirements, you can construct a device using a non-motorized ball valve. Attach a pneumatic cylinder (or long travel solenoid plunger) to the handle of the ball valve. Your signal to open/close will be the same as with the restrictive pneumatic valve but will have the desired effect of a ball valve, because it is a ball valve.</p>
<p>Cheap is subjective, and difficult to pin down without specifics.</p>
<p><a href="https://i.stack.imgur.com/pyyZl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pyyZl.png" alt="ball valve actuator sketch" /></a></p>
<p>Image created by me. No scale, no proportions provided. Obviously, it will have to be modified to cover parameters such as length of solenoid stroke, force required to move the lever, etc. The solenoid in the diagram is placed at 45° in order to keep the forces balanced. If placed at a 90° orientation, forces would reach excessive levels to ensure closing or opening.</p>
| 52136 | Solenoid valve that does not restrict air flow when open |
2022-08-16T18:59:04.207 | <p>I am working on calculations for an LCA of an operation. The plant operator has provided me a report from an engineering firm who has measured the emissions from their stack. I am trying to estimate the amount of CO<sub>2</sub> produced per year.</p>
<p>The total volume of gas emitted was 2.08Rm3 (25 degrees C, and 101.3 kPa) every two hours. The gas contains 1.30% CO<sub>2</sub>. and the plant runs 1450 hours/year.</p>
<p>I tried to find the density of CO<sub>2</sub>, which is 1.78g/L and multiplied volume (1,508,000 L/year) by (0.013) but I am unsure if this is correct.</p>
<p>The plant is using wood chips in a suspension burner and undergoing complete combustion. But in their report there is 10.40000 kg of carbon monoxide being produced.</p>
| |thermodynamics|chemical-engineering| | <p>As a proofing method you can use the input of fuel to estimate the mass of CO2 produced.</p>
<p>First get the percentage mass of carbon in woodchips (around 50% according to a quick bit of searching)</p>
<p>Then multiply the mass of woodchips burned by that percentage and by</p>
<p><span class="math-container">$$\frac{molar\ mass\ CO2}{molar\ mass\ C} = \frac{12+16+16}{12} = 3.66 $$</span></p>
<p>If there is full combustion then this should come to the same number.</p>
| 52156 | Calculating the mass of CO2 from stack emissions |
2022-08-18T01:30:08.163 | <p>(Question copied largely verbatim from <a href="https://aviation.stackexchange.com/questions/59328/why-do-carburetors-tend-to-produce-richer-mixture-at-higher-altitude">https://aviation.stackexchange.com/questions/59328/why-do-carburetors-tend-to-produce-richer-mixture-at-higher-altitude</a>. I have the identical question and it's really eating at me!)</p>
<p>Simple first-principles analysis of fluid dynamics suggest that the pressure driving fuel into a carburetor venturi should change linearly with air density.</p>
<p>The pressure drop across the venturi is proportional to air density and the fuel is at ambient pressure in the float chamber, so I would expect the fuel flow to reduce proportionally with density, and that response to preserve the fuel-air ratio over changing altitude.</p>
<p>But in practice that does not seem to be the case. Proper response to altitude requires additional modification that most airplane carburetors don't have, so the pilot usually has to manually lean out the engine during the climb. What am I missing here?</p>
<p>More specifically, I would expect that at the same RPM, the volume flow rate will be the same—because the engine pulls in its displacement per revolution. Now velocity in the venturi <span class="math-container">$v$</span> is just</p>
<p><span class="math-container">$$ v = \frac{\dot V}{A} $$</span></p>
<p>Where <span class="math-container">$\dot V$</span> is the volume flow rate and <span class="math-container">$A$</span> is the cross-section of the venturi. So it will also be the same independent of altitude. Since dynamic pressure</p>
<p><span class="math-container">$$ P_d = \frac 1 2 \rho v^2 $$</span></p>
<p>And that is also the pressure that pulls in the fuel (when the float chamber is open to ambient pressure). Substituting mass flow</p>
<p><span class="math-container">$$ \dot m = \rho v $$</span>
<span class="math-container">$$ P_d = \dot m \frac 1 2 v $$</span></p>
<p>and as long as <span class="math-container">$v$</span> is mostly constant,</p>
<p><span class="math-container">$$ P_d \sim \dot m $$</span></p>
<p>So in other words, the pressure which drives the fuel through the jets and into the venturi varies somewhat linearly with the mass of air going through the carburetor. This would suggest that the fuel/air ratio will stay relatively constant... which of course is the opposite of observed behavior.</p>
<p>So this leaves open the dependence of fuel flow on <span class="math-container">$P_d$</span>. If the relation is reasonably close to linear, it should mean the venturi mixes properly by mass. I can see a reason why higher pressure should cause less than linear increase in fuel flow, but not much why it should cause more than linear increase in fuel flow—but that is what the actual behavior would need.</p>
| |carburetor| | <p>TLDR; It's complex, I don't have a simple physical description. It basically comes down to the fuel-air ratio being a function of the square root of pressure whereas air mass is simply proportional to pressure. Maybe someone can add an intuitive explanation in the comments?</p>
<hr />
<h2>Preface</h2>
<p>I figured it out. The answer is subtle and requires a decent dive into the math behind fluid flow.</p>
<h2>Simple carburetor</h2>
<p>Let's consider the constant throttle setting of the below carb:</p>
<p><a href="https://i.stack.imgur.com/GRRHM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GRRHM.png" alt="Ideal flow system of a simple float carburetor" /></a></p>
<p>For simplicity, we will assume both engine RPM and ambient temperature are constant.</p>
<p>We will also assume that the system has no <a href="https://engineeringlibrary.org/reference/head-loss-fluid-flow-doe-handbook" rel="nofollow noreferrer">head losses</a>.</p>
<h2>Fluid dynamics</h2>
<h3>Liquid</h3>
<p>An ideal liquid has no viscosity and its density is constant. Through the Bernoulli Equation, the ideal flow of such a liquid through a passage can be written as:</p>
<p><span class="math-container">$$ \dot{m_l} = A_l \rho_l u_l = A_l \sqrt{2 \rho_l \Delta P_l} $$</span></p>
<p>where</p>
<ul>
<li><span class="math-container">$m_l$</span> is the liquid mass flow</li>
<li><span class="math-container">$\Delta P_l = P_{01,l} - P_{2,l}$</span></li>
<li><span class="math-container">$\rho_l$</span> is the fluid density</li>
<li><span class="math-container">$u_l$</span> is the fluid velocity</li>
<li><span class="math-container">$A_l$</span> is the flow area (in this case, the carburetor jet)</li>
</ul>
<h3>Gas</h3>
<p>Using Bernouilli, the equation for an ideal gas is almost the same:</p>
<p><span class="math-container">$$ \dot{m_g} = A_g \rho_g u_l \phi_g = A_g \phi_g \sqrt{2 \rho_g \Delta P_g} $$</span></p>
<p>where subscripts have changed from <code>l</code> to <code>g</code> and:</p>
<ul>
<li><span class="math-container">$\phi_g$</span> is the gas flow compressibility parameter</li>
</ul>
<h2>Fuel/air ratio</h2>
<h3>Definition</h3>
<p>We'll assume air is an ideal gas, and that the liquid is gasoline. (I apologize in advance for the fact that <em>gas</em> can mean both a phase of matter and a shortened version of "gasoline". Throughout this answer, I will use <em>gas</em> to mean only a phase of matter and <em>gasoline</em> only for petroleum fuel.)</p>
<p><span class="math-container">$$ F_M = \frac{\dot{m_l}}{\dot{m_g}} = \frac{A_l \sqrt{2 \rho_l \Delta P_l}}{A_g \phi_g \sqrt{2 \rho_g \Delta P_g}} $$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$F_M$</span> is the fuel/air ratio</li>
</ul>
<h3>Simplification</h3>
<p>Simplifying this expression by eliminating variables, abstracting out the constant coefficients <span class="math-container">$A_l$</span>, <span class="math-container">$A_g$</span>, and <span class="math-container">$\rho_l$</span>, and-- most importantly-- noting that the pressure differential <span class="math-container">$\Delta P_l$</span> and <span class="math-container">$\Delta P_g$</span> are identical (because in the case of the ideal carburetor both go from the air inlet to the venturi body:</p>
<p><span class="math-container">$$ F_M \propto \frac{1}{\phi_g \sqrt{\rho_g}} $$</span></p>
<h3>Analysis at different pressures, i.e altitudes</h3>
<p>Let's look at thee <span class="math-container">$F_M$</span> at two different pressures, <span class="math-container">$P_0$</span> and <span class="math-container">$P_1$</span>:</p>
<ul>
<li><span class="math-container">$ F_{M,0} \propto \frac{1}{\phi_{g,0} \sqrt{\rho_{g,0}}} $</span></li>
<li><span class="math-container">$ F_{M,1} \propto \frac{1}{\phi_{g,1} \sqrt{\rho_{g,1}}} $</span></li>
</ul>
<p>Dividing these two, and noting that the coefficients eliminated in the <code>Simplification</code> step now fully cancel out, resulting in equality:</p>
<p><span class="math-container">$$ \frac{ F_{M,0}}{F_{M,1}} = \frac{\frac{1}{\phi_{g,0} \sqrt{\rho_{g,0}}}}{\frac{1}{\phi_{g,1} \sqrt{\rho_{g,1}}}} = \frac{\phi_{g,1} \sqrt{\rho_{g,1}}}{\phi_{g,0} \sqrt{\rho_{g,0}}}$$</span></p>
<p>I won't go into the derivation of <span class="math-container">$\phi$</span>, but suffice to say that for the conditions given in the <code>Simple carburetor</code> section, it is constant at all altitude. This results in the very simple equation:</p>
<p><span class="math-container">$$ \frac{ F_{M,0}}{F_{M,1}} = \sqrt{\frac{\rho_{g,1}}{\rho_{g,0}}}$$</span></p>
<h2>Conclusion</h2>
<p>Assuming that condition 0 is at altitude, condition 1 is the sea level reference, and that density, <span class="math-container">$\rho$</span>, is a linear function of pressure, the ratio of the fuel mass flow at seal level to the fuel mass flow at altitude is:</p>
<p><span class="math-container">$$ \frac{ F_{M,alt}}{F_{M,S.L}} = \sqrt{\frac{P_{S.L}}{P_{alt}}}$$</span></p>
<p>Compare this to the ratio of air mass flow at altitude and note the lack of square roots:</p>
<p><span class="math-container">$$ \frac{ \dot{m_{g,alt}}}{\dot{m_{g,S.L}}} = \frac{P_{alt}}{P_{S.L}}$$</span></p>
<p>This is why the ratios get out of whack as altitude changes.</p>
<h3>Worked example</h3>
<p>What is the richening percentage going from sea level on an STP day to 5000m (assuming perfect dry adiabatic lapse)?</p>
<p>At 5000m, according to <a href="https://www.mide.com/air-pressure-at-altitude-calculator" rel="nofollow noreferrer">https://www.mide.com/air-pressure-at-altitude-calculator</a> this is 54kPa, or ~0.5atm.</p>
<p><span class="math-container">$$ \sqrt{\frac{P_{S.L.}}{P_{alt}}} = \sqrt{\frac{101}{54}} = 1.37 $$</span></p>
<p>So the carburetor is pushing a <code>37%</code> richer mixture than at sea level.</p>
<p><strong>NOTE</strong> that this is <em>not</em> the <code>50%</code> which would happen if the carb simply pumped in the same amount of fuel for every engine stroke, no matter the air density.</p>
| 52173 | Why do simple carburetors richen the mixture as altitude increases? |
2022-08-19T17:36:33.677 | <p>I need to submit a lab report.
I've measured:</p>
<ul>
<li>Differential pressure [Volts]</li>
<li>Flow rate [Hz]</li>
<li>Flow Rate [m^3/s]</li>
</ul>
<p>In my report I need to get the Pressure Differential in Pascal but how can I convert Volts to Pascal?(?!?!?!?!?!)</p>
<p>How do the measuring devices can measure pressure in volts and flow rate in Hz in the first place?</p>
<p>Thank you!</p>
<p>Image for demonstration:
<a href="https://i.stack.imgur.com/lwNI6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lwNI6.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|pressure|aerospace-engineering|aerodynamics| | <p>The device will contain a pressure transducer.</p>
<p><strong>transducer</strong><br />
<em>noun</em></p>
<blockquote>
<p>a device that converts variations in a physical quantity, such as pressure or brightness, into an electrical signal, or vice versa.</p>
</blockquote>
<p>In your case it is converting pressure into a voltage.</p>
<blockquote>
<p>... but how can I convert Volts to Pascal?</p>
</blockquote>
<p>You read the manual for the device and find the conversion factor which will be in volts/pascal or pascal/volt. (Note lowercase for SI units named after a person. The symbols are capitalised.)</p>
| 52196 | Converting differential pressure |
2022-08-22T19:16:55.560 | <p>I am trying to figure out the name of a gear type, I think it is a scissor gear, however when looking it up I dont find anything by that name. The gear is simple with a center screw and two nuts. The nuts have links attached to them that then attach to a ring. When the center screw is rotated clockwise the bottom nut is screwed down while the top nut goes up, the system as a whole pulls in (blue). When the center screw is rotated counter-clockwise the bottom nut is screwed up and the top nut is screwed down, the system as a whole pushed up (red). The photo below shows a photo example.</p>
<p>I am curious about the name so I can look more into the gear, and possible alternatives that can withstand bending if one side is pushed (green).
<a href="https://i.stack.imgur.com/8Ntrh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8Ntrh.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|gears| | <p><strong>Scissor Jack Lift</strong> is the correct term and is used in the vertical mode to lift cars.</p>
<p>You would have to brace it in other ways to prevent lateral forces of the green arrow.</p>
<p>This one has a swivel top.</p>
<p><a href="https://i.stack.imgur.com/UmJwn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UmJwn.png" alt="enter image description here" /></a></p>
<p>This is an old Audi jack.</p>
<p>Auto wreckers should have many of these.<br />
<a href="https://i.stack.imgur.com/rp5Qw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rp5Qw.png" alt="enter image description here" /></a></p>
<p>This is an old BMW jack.
Newer ones are single-sided with worm gear & handle in motion axis.</p>
| 52211 | Trying to find the name of a gear system |
2022-08-22T23:33:42.817 | <p>I spent a full day searching for and trying to understand why in this "proof" of virtual work makes sense. I've come to understand the idea from D'Alambert's principle, but I want to understand why this method is so "obvious"</p>
<p>Dark triangles represent internal and external energy from a virtual displacement/deformation. The light triangles represent the internal and external energy from displacement/deformation. Given that the dark triangles are equal in area to each other and the light triangles are equal in area to each other, somehow the white rectangles under eachother are equal to each other, but I don't understand why.</p>
<p>The relevant slide is given below, but I also found this on youtube where it is asserted. <a href="https://youtu.be/qR8NlrP7_VM?t=455" rel="nofollow noreferrer">Here</a>
<a href="https://i.stack.imgur.com/4SHKr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4SHKr.png" alt="visual equality of virtual work" /></a></p>
| |energy|deflection| | <p>So there are a couple of ways to think about this, a sort of intuitive way (Method 1) and a more formal way (Method 2).</p>
<hr />
<h2>Method 1</h2>
<p>You already stated that the dark triangles are equal in area and the light triangles are equal in area. That also means that the the ratio between light and dark is the same for both graphs, that is <span class="math-container">$\frac{W_P}{W_Q}=\frac{U_P}{U_Q}$</span>.</p>
<p>We can also determine that all the triangles from one graph are similar because they have the same angle. That means we can just scale any of the triangles by some factor and it will match another in the same graph. This includes the whole graph, which is one big triangle.</p>
<p>If we take <span class="math-container">$W_Q$</span> and scale it up by <span class="math-container">$S_1=\frac{P+Q}{Q}=\frac{P}{Q}+1$</span>, we will have the whole graph. We can do the same with the other graph with <span class="math-container">$S_2=\frac{F_P+F_Q}{F_Q}=\frac{F_P}{F_Q}+1$</span> to get the whole graph. We know though that <span class="math-container">$\frac{P}{Q}=\frac{F_P}{F_Q}$</span> because the ratio of dark to light is the same (the first equation). That means <span class="math-container">$S_1=S_2$</span>, so the scaling from the dark triangles to the whole graph is the same for each graph. And since we know that the dark triangles are the same area that means that each graph has the same area as a whole. Now it's easy, since we know the light and dark triangles are the same area, that just leaves us with the white rectangles which must also be equal.</p>
<hr />
<h2>Method 2</h2>
<p>This approach is based more on equations.
We know:</p>
<p><span class="math-container">$$\delta_Q*Q=\Delta L_Q*F_Q\tag{1}$$</span>
<span class="math-container">$$\delta_P*P=\Delta L_P*F_P\tag{2}$$</span></p>
<p>We also know that:</p>
<p><span class="math-container">$$W_P\propto W_Q\tag{3}$$</span>
<span class="math-container">$$U_P\propto U_Q\tag{4}$$</span></p>
<p>And thus:</p>
<p><span class="math-container">$$\frac{\delta_Q}{Q}=\frac{\delta_P}{P}\to \delta_Q*P=\delta_P*Q\tag{5}$$</span>
<span class="math-container">$$\frac{\Delta L_Q}{F_Q}=\frac{\Delta L_P}{F_P}\to \Delta L_Q*F_P=\Delta L_P*F_Q\tag{6}$$</span></p>
<p>Now if we take <span class="math-container">$(5)$</span> and multiply each side by <span class="math-container">$\delta_P*Q$</span>, we get:</p>
<p><span class="math-container">$$(\delta_Q*P)*(\delta_P*Q)=(\delta_P*Q)^2\tag{7}$$</span></p>
<p>Rearranged:</p>
<p><span class="math-container">$$(\delta_Q*Q)*(\delta_P*P)=(\delta_P*Q)^2\tag{8}$$</span></p>
<p>Insert <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>:</p>
<p><span class="math-container">$$(\Delta L_Q*F_Q)*(\Delta L_P*F_P)=(\delta_P*Q)^2\tag{9}$$</span></p>
<p>Rearrange again:</p>
<p><span class="math-container">$$(\Delta L_Q*F_P)*(\Delta L_P*F_Q)=(\delta_P*Q)^2\tag{10}$$</span></p>
<p>Insert <span class="math-container">$(6)$</span>:</p>
<p><span class="math-container">$$(\Delta L_P*F_Q)*(\Delta L_P*F_Q)=(\delta_P*Q)^2\tag{11}$$</span></p>
<p>And thus:</p>
<p><span class="math-container">$$(\Delta L_P*F_Q)^2=(\delta_P*Q)^2$$</span>
<span class="math-container">$$\Delta L_P*F_Q=\delta_P*Q\tag{12}$$</span>
<span class="math-container">$$U'_Q=\Delta L_P*F_Q=\delta_P*Q=W'_Q$$</span>
<span class="math-container">$$U'_Q=W'_Q\tag{13}$$</span></p>
<p>And there we have it.</p>
| 52217 | Principle of virtual work "geometric proof" |
2022-08-24T20:45:17.493 | <p>I am looking at part a of the following question,
<a href="https://i.stack.imgur.com/U8qLL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U8qLL.png" alt="enter image description here" /></a></p>
<p>and I can not figure out where this support reaction N is coming from. The solution is as follows.
<a href="https://i.stack.imgur.com/Q052U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q052U.png" alt="enter image description here" /></a></p>
<p>The question and diagram make no mention of N. My understanding of the question is that there are three forces acting on the object, gravity and the two tensile forces from the cable. Where does N come from?</p>
| |statics| | <p>N is just the force of the particle resting on the smooth surface (the shaded thing in Figure 3). The pulleys aren't directly above the particle so it's leaning on the surface. This is also stated in part (c) of the question.</p>
| 52237 | Where does this support reaction come from? |
2022-08-24T21:11:34.157 | <p>If I decrease the weight from top portion of the body keeping all dimensions same, the center of mass shifts downwards, but does it makes it more stable against tipping from perpendicular forces ?</p>
<p>Background: I have a nice sturdy easel style wooden stand which has iron frame for VESA mounting TVs. Till now I've mounted a 55 pound TV on it and it seems pretty stable (w.r.t a general impact from back or front).</p>
<p>I am about to change to a exact same size TV new model which is 40% lighter & is 33 pounds.</p>
<p><a href="https://i.stack.imgur.com/ZAYzE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZAYzE.png" alt="enter image description here" /></a></p>
<p>Per my rough calculation the center of mass is shifting about 2-3 inches downwards but is that factor sufficient ? As per generic family opinion, if I switch to a lighter TV, my stand will be 'more 'prone to tipping. Any advice?</p>
<p>P.S: Dont need any calculations here.</p>
| |mechanical-engineering|structural-engineering|building-physics|homework|centre-of-gravity| | <p>Even though You ask for no calculations a simple equation is helpful here.</p>
<p>A mass, m at a C.G. height, H, on a stand with base, B needs a force, F to topple.</p>
<p><span class="math-container">$$F\geq \frac{m*B}{2H}$$</span>
So if the quantity of <span class="math-container">$m/H$</span> decreases the stand is less stable if it increases it is more stable.</p>
<p>You can plug in your numbers and see what you get!</p>
| 52238 | Impact of decreasing weight from a part of body w.r.t tipping |
2022-08-28T12:40:54.470 | <p>I would like to build a simple homemade <a href="https://en.wikipedia.org/wiki/Gravity_battery" rel="nofollow noreferrer">gravity battery</a> (a weight is pulled to a high height and is then dropped to turn a motor and generate electricity.) My plan is to have a 5kg weight pulled to 2 meters, testing this example and then increasing the weight to 10kg and height to 4 meters.</p>
<p>I was originally thinking of using a gearbox of some sort, in order to reduce the height required to drop the weight, and increasing the RPM at which the motor-generator rotates (due to many motors I’ve found needing between 100-2000 rpm.)
After reading a related <a href="https://engineering.stackexchange.com/questions/44831/using-a-manual-pulley-system-to-repeatedly-lift-a-rock-and-gather-its-falling-en">post on this site</a> about using pulleys to lift a weight, though, it has made me consider whether pulleys would be a better fit.</p>
<p>TL;DR - Would using pulleys, a gearbox, or a combination of the two be beneficial in decreasing the rope used when lifting/lowering a weight and increasing the RPM of the motor-generator?</p>
| |gears|pulleys|energy-storage| | <p>The most efficient power transmission is chain and sprockets. These work very well at a nominal cost.</p>
<p>Gears are probably next, but at a higher cost due to gear fabrication. The closed nature of a gear box should help longevity. Be sure you have gears that run both ways if you're planning on lifting with the gear set - some don't, like worm gears.</p>
<p>Belted pulleys would be next, just a little less efficient. This seems like it would be hard to manage, belts are generally continuous, though I guess they can be bought in length. Pulleys with ropes will be last due to rope friction, but are probably the cheapest option here.</p>
<p>All the above will be affected by the bearings used, simple bushings will be less efficient than good quality roller bearings.</p>
| 52276 | Pulleys vs gears for a gravity battery |
2022-08-29T14:46:33.203 | <p>I'd like to refresh and potentially update my knowledge.</p>
<p>Is there anything like CHAdeMO charging cable to Type 2 car plug connector? or any other solution that would allow me to charge my Type 2 car from CHAdeMO cable / charging station? Or no such thing exists.</p>
<p>My knowledge gained so far says that there isn't anything like that and that designing such thing isn't possible due to wide differences in CHAdeMO and Type 2 interfaces up to the level of different request-answer timeouts, voltages, logic and software behind.</p>
<p>I understand that CHAdeMO technology / protocol / standard is being phased-out worldwide, but here were I live (Poland) there is a serious number of (roughly ever used) public EV chargers with CHAdeMO cable. And I'd like to have an additional option of charging my Type 2 car with such cable in situation when other solutions (like using Type 2 cable / plug or CCS cable) are not available (had such situation only yesterday twice).</p>
| |electric-vehicles| | <p>According to various sources, <a href="https://en.wikipedia.org/wiki/CHAdeMO" rel="nofollow noreferrer">wiki included, CHAdeMO is a DC resource</a> of varying voltage depending on version number. Type 2, also known as L2 or Level 2 is an AC power source and your onboard charger is designed to rectify that alternating current into the DC your battery requires. Also part of the L2 specifications is the ability to communicate between the OBC and the EVSE to provide appropriate power levels to the OBC.</p>
<p>The range of power available at some commercial EVSE units can be from 3kW to 10kW. The above wiki suggests that home units can be purchased to provide power of 22kW, but I've not seen that rating. Our home EVSE is a 10kW unit and the OBC of our Rav4EV is rated to 11kW, providing for a good match.</p>
<p>I suspect it's going to be difficult to impossible to locate an adapter to convert high current/high DC voltage to lower current/lower AC voltage along with the negotiation circuitry which would be required to make such a device work safely.</p>
<p>Your third paragraph is supported by information available.</p>
| 52283 | CHAdeMO to Type 2 connector for charging EV cars |
2022-08-29T20:38:52.280 | <p>I am researching the design of engines and I noticed that most pistons have piston rings. My understanding is that the piston ring helps create a gas-tight seal.</p>
<p>Is the piston diameter machined to exactly match the cylinder diameter? Or is the piston diameter undersized, and then the rings make up the slack and exactly match the cylinder diameter?</p>
<p>I am asking because I would like to try machining a crude piston/cylinder arrangement. My idea would be to machine some round stock steel and purchase piston rings off the internet.</p>
| |mechanical-engineering|pistons| | <p>The piston rings perform two functions:</p>
<ol>
<li><p>they seal the piston tightly in the cylinder bore, so the piston can compress the inlet gases and also develop power without leakage (called <em>blow-by</em>) when the gases burn, and</p>
</li>
<li><p>they scrape the splashed oil off the cylinder walls on the crankcase side of the piston so the oil doesn't get up past the piston and into the combustion chamber, where it will burn with a lot of smoke and foul up the spark plugs.</p>
</li>
</ol>
<p>The <em>compression rings</em> are near the top of the piston and the <em>oil rings</em> are beneath them, near the bottom.</p>
| 52286 | Purpose of the piston ring in an engine |
2022-08-30T05:49:33.990 | <p>I want to design a turned part at the office and I wanted to give it a hexagon socket. I know that screws can have hexagon sockets, but I think it is due to them being produced in batch production. Is it also possible to create a hexagon socket on an individual turned part that is going to be produced by turning and milling (on a cnc lathe) and if so, how could it be done? Thanks in advance for an answer!</p>
| |mechanical-engineering| | <p>The easy way is to not use a hex socket but instead begin not with round stock chucked in the lathe but <em>hexagonal stock</em>. You then turn the part while leaving a convenient length with hex flats intact and use those for rotating the resulting "bolt".</p>
| 52290 | Can you produce a hexagon socket on a turned part? |
2022-08-30T19:57:08.477 | <p>A very similar question to <a href="https://engineering.stackexchange.com/q/52283/39087">this one</a>. Is there any solution available to use CHAdeMO charging cable to power EV car equipped with CCS socket?</p>
<p>My knowledge gained so far says that there isn't anything like that because there are too wide differences in CHAdeMO and CCS interfaces (up to the level of different request-answer timeouts, voltages, logic, software behind, etc.).</p>
<p>I understand that CHAdeMO technology / protocol / standard is being phased-out worldwide, but here were I live (Poland) there is a vast number of public EV chargers with (roughly never used) CHAdeMO cables.</p>
<p>I'd like to have an additional option of charging my CCS car with CHAdeMO cable in situation when options (Type 2 cable or plug or CCS cable) are not available.</p>
<p>I don't know much of anything about Tesla, but I saw Tesla driver powering its car from CHAdeMO (or maybe it was CSS?) cable using some kind of converter or inverter plugged between EV charging station's cable and Tesla's socket. I'd like to know, if similar option exists, if we have CCS instead of Tesla socket?</p>
| |electric-vehicles| | <p>I am not even shure every car will accept DC over the type2 connector of the CCS socket , since as far as I understand it the type2 part is only for the handshake and the DC goes over the 2 big pins under it. Do you know of any cars accepting Dc over the type2 socket ?</p>
<p>regards bart</p>
<p>[email protected]</p>
| 52300 | A converter from CHAdeMO cable to CCS connector for charging EV cars |
2022-09-02T20:08:00.487 | <p>My issue is that downstream, I have a boiler making pressure at 10.54 bar and upstream I have a system that must NOT be heated at any point beyond 130 °C. My plan is to make sure the system pressure upstream stays 3 bar at max and doesn't fall significantly below that as long as the pressure downstream is greater than 3 bar but I can't use any relief valves or anything as they bleed away too much steam. Is there a check valve or any other valve systems that will make sure that the pressure stays less than 3 bar at all times and preferable, more than 2 bar as well? Please don't recommend any electronic systems.</p>
| |pressure|valves| | <p>It's called a pressure regulator or pressure reducing valve.</p>
<p>They work by mechanically actuating a valve based on the outlet pressure (the one you want to keep at 3 bar), choking off the flow when it rises.</p>
| 52327 | Is there a pressure check valve that doesn't bleed the excess pressure? |
2022-09-03T15:50:16.030 | <p>So in this video(<a href="https://www.youtube.com/watch?v=v1z829NF9PA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=v1z829NF9PA</a>) at around 2:35 it says that the alluminium ingots for alluminium foil, after being heated, are cooled down with a mixture of 95% water, 5% oil. Why not simply use 100% water? Because I looked it up online and it seems most oils have a lower specific heat than water, so would be worse for cooling. Is it that oil is somehow cheaper, or some other reason? This episode was in 2001 btw, on Jan 6(so before 9/11)</p>
<p>Thanks</p>
| |heat-transfer|manufacturing-engineering|metallurgy|metals|cooling| | <p>That mixture concentration sounds the same as what is used for CNC coolant to me.</p>
<p>When machining metal we want to use water as a coolant because of its high specific heat capacity but water causes rust, and so it is mixed with a water soluble oil at a concentration of about 5%.</p>
<p>You might then ask "Rust? But it's aluminum." Guess what: aluminum corrodes as well. The oxide layer is not as easily visible, and doesn't lead to as deep a failure as with iron, but aluminum oxide tends to fall off as a power. Not good in clean environments. All the equipment handling that aluminum is also made of steel.</p>
| 52332 | Why 95% Water is used Instead of 100% |
2022-09-05T11:21:15.793 | <p><a href="https://i.stack.imgur.com/7Uzna.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Uzna.png" alt="enter image description here" /></a></p>
<p>Consider the uniform indeterminate beam as shown above. This beam is to be subjected to a single, 80N point load (acting down ) somewhere along its length.</p>
<p>What is the procedure (i.e steps and logic) to calculate the location where application of this load will cause the largest deflection (in mm), i.e. where the "worst case scenario" occurs? My initial intuition tells to assume the maximum sagging will occur if we apply 80N force in the middle, but checking with the online beam calculators proved that the point is closer to the roller rather than equally distanced from both supports.</p>
| |civil-engineering|stresses|beam|bending| | <p><strong>1) - Use table</strong></p>
<p><a href="https://i.stack.imgur.com/TOXjs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TOXjs.png" alt="enter image description here" /></a></p>
<p><strong>2) - Intergrating the second order differential equation</strong> <span class="math-container">$EI\dfrac{d^2y}{dx^2} = M$</span></p>
<p><span class="math-container">$\theta(x) = \dfrac{1}{EI} \int M(x)dx$</span></p>
<p><span class="math-container">$\delta(x) = \int \theta(x)dx$</span></p>
<p>Note, that for a beam supported on both ends, the maximum deflection occurs at a location where the slope of rotation angle (<span class="math-container">$\theta$</span>) is zero.</p>
| 52360 | How to find the location on the 2-support beam where the deflection will be biggest? |
2022-09-06T15:43:55.767 | <p>Refrigerants from CFC series are banned due to environmental concerns,but why not focus on preparing a standard for registration system which minimize or eliminate leakage?
It must be difficult to manufacture such system.but why?</p>
| |thermodynamics|manufacturing-engineering|environmental-engineering|refrigeration| | <h3>Because you can control refrigerant at the source</h3>
<p>If a refrigerant can't be manufactured, it will go away. As prices go up, keeping systems from leaking becomes more important. No one can control procedures in the field. Experience says that any system that can leak will leak.</p>
<p>The bans of CFC's and now HCFC's have been extremely effective at what they were designed to do. That said, <a href="https://research.noaa.gov/article/ArtMID/587/ArticleID/2713/Emissions-of-a-banned-ozone-depleting-gas-are-back-on-the-decline" rel="nofollow noreferrer">crooks in China appear to be making more R-11, contrary to the ban.</a></p>
| 52372 | Why regulations on refrigerant and not on efforts in preparing standard for leak proof refrigeration system? |
2022-09-07T18:41:49.827 | <p>Consider a bar of uniform circular cross-section. We are interested in knowing the elongation of this bar due to its self-weight. This is a common problem in an undergraduate course on Mechanics of Solids.</p>
<p>This is how I have seen it solved in books - <a href="https://www.ques10.com/p/8788/derive-an-expression-for-the-elongation-due-to-s-1/" rel="nofollow noreferrer">Elongation of a bar due to its self-weight</a></p>
<p>Wherever I have seen such a problem, it's stated pretty vaguely. It just says to determine the elongation of the bar and gives the relevant parameters, but no info on how the bar is set up. Which I feel is the main cause of my doubt.</p>
<hr />
<p>Doubt:</p>
<p>To find the elongation, the <a href="https://www.ques10.com/p/8788/derive-an-expression-for-the-elongation-due-to-s-1/" rel="nofollow noreferrer">link</a> has considered an element of thickness dy at y and determined the elongation produced in that element. The force acting on this element is taken as the weight of the part of the bar below it, which is <span class="math-container">$P_y =\rho Ayg$</span>. To find the elongation in this element, it has then used the <span class="math-container">$\frac{PL}{AE}$</span> formula,</p>
<p><span class="math-container">$$\delta (\Delta L)= \frac{P_y dx}{AE}$$</span></p>
<p>The link has then put the value of <span class="math-container">$P_y$</span> and integrated from 0 to L to get the total change in length.</p>
<p>The formula <span class="math-container">$\frac{PL}{AE}$</span> is valid only when the load P applied is gradual (that is it is a gradually applied load that increases from 0 to P). <strong>How did we know in the above case that the load <span class="math-container">$P_y =\rho Ayg$</span> was gradually acting on the element?</strong> Because only when Py acts gradually can we use this formula - <span class="math-container">$\frac{PL}{AE}$</span>.</p>
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis| | <p>The "problem" is that you are solving a problem in the stationary regime, where dynamic effects are neglected.</p>
<p>You are thinking on the transient regime. And in any case, if you consider even the most basic damping for the set-up, it will very quickly converge to the stationary answer.</p>
| 52384 | Elongation due to self weight - why the load acting on an element is gradually applied load? |
2022-09-08T06:58:50.407 | <p>To simplify the problem, I liken the problem to a ship approaching the shore and a device wants to collect the rope in such a way that the rope is neither stretched too much nor released to the extent that it falls into the water.</p>
<p>For this purpose, I thought that a motor with a ratchet should collect the rope at a high speed, but if the rope resists too much, the device should not put too much pressure on it. Because our device is not going to pull the ship to the shore! We just want to collect the rope.</p>
<p>The problem is apparently similar to the function of the clutch, but I am looking for a more geometric and mechanical solution because <strong>the main product is a very delicate device</strong> (not really for a ship but in a fine robotic system) and the implementation of the clutch function in these dimensions requires high technology in terms of materials and precision settings.</p>
| |mechanisms|friction|pulleys| | <p>This is similar to a low-tension rewinder application used for filaments and films / webs.</p>
<p>The simplest solution would be to run a rewinder motor in constant torque mode. This could be achieved with a DC motor by driving it with constant current.</p>
<p>Other methods would feed the rewinder via a roller on a "dancer" arm. If the dancer arm starts to fall below mid-position the motor speeds up. If it goes above mid-position it would run in reverse (until it reaches mid-position).</p>
<p>The complexity of the control system really depends on how complex the motion of the material is. Steady and smooth feed can be rewound with relative ease.</p>
| 52388 | Collecting a released rope with a proper speed |
2022-09-08T14:43:38.247 | <p><a href="https://i.stack.imgur.com/yvGj8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yvGj8.png" alt="enter image description here" /></a></p>
<p>What should I do here? Drying the aggregates prior to mixing make sense to me but would it be costly?</p>
| |concrete| | <p>From my summer student job days in the mid-70's , it was always to control the slump test to obtain the correct water ratio.</p>
<p>Of course aggregate moisture content can be easily measured with an added chemical gas pressure and weighed sample, to predict the reduction of water required for an estimate.</p>
<p>That was my 2nd summer working for RKL in Winnipeg while the 1st yr was lab work doing modified Proctors and sieve analysis.</p>
| 52396 | Question about commercial mix concrete design |
2022-09-08T16:13:40.393 | <p>I am trying to solve a problem where there is a fixed-fixed (clamped-clamped) beam with a point load in the middle (at L/2). The L/H is less then the 10 required for normal beam bending analysis and then results in, what I believe requires, Timoshenko beam analysis.</p>
<p>I went a head and attempted to derive the total displacement in a similar fashion to what would be done in a normal beam bending problem. I start at the moment equation (2nd derivative of displacement, which I will be using the symbol of <span class="math-container">$\delta$</span> for.):</p>
<p><span class="math-container">$$M(x) = \frac{d^2\delta}{dx^2}=\frac{d^2\delta_{bending}}{dx^2}+\frac{d^2\delta_{shear}}{dx^2}=\frac{Fx}{4EI}-\frac{M'}{EI}+\frac{K}{GA}\frac{dv}{dx}$$</span>
We integrate:
<span class="math-container">$$\theta(x)=\frac{d\delta}{dx}=\frac{d\delta_{bending}}{dx}+\frac{d\delta_{shear}}{dx}=\frac{Fx^2}{8EI}-\frac{M'x}{EI}+\frac{KV}{GA}+C_1$$</span>
We know that the shear <span class="math-container">$V$</span> is <span class="math-container">$=\frac{F}{2}$</span><br />
We integrate:</p>
<p><span class="math-container">$$\delta(x)=\delta_{bending}+\delta_{shear}=\frac{Fx^3}{24EI}-\frac{M'x^2}{2EI}+\frac{KFx}{2GA}+C_1x+C2$$</span></p>
<p>We can apply some boundary conditions:</p>
<p><span class="math-container">$$\delta(0)=0=C_2$$</span></p>
<p>We apply another boundary condition:
<span class="math-container">$$\theta(0)=0=\frac{KF}{2GA}+C_1 $$</span></p>
<p>This is the first spot where I am uncertain. I believe that <span class="math-container">$C_1 = \frac{KF}{2GA}$</span>, not <span class="math-container">$C_1 = \frac{-KF}{2GA}$</span> as if it is the 2nd case all of the shear displacement goes to 0. So I continued assuming <span class="math-container">$C_1 = \frac{KF}{2GA}$</span></p>
<p><span class="math-container">$$\delta(L)=0=\frac{FL^3}{24EI}-\frac{M'L^2}{2EI}+\frac{KFL}{2GA}+C_1L$$</span>
<span class="math-container">$$\delta(L)=0=\frac{FL^3}{24EI}-\frac{M'L^2}{2EI}+\frac{KFL}{2GA}+\frac{KFL}{2GA}$$</span>
<span class="math-container">$$\delta(L)=0=\frac{FL^3}{24EI}-\frac{M'L^2}{2EI}+\frac{KFL}{GA}$$</span></p>
<p>Now we solve for M'</p>
<p><span class="math-container">$$M'=\frac{2EI}{L^2}(\frac{FL^3}{24EI}+\frac{KFL}{GA})$$</span>
<span class="math-container">$$M'=\frac{FL}{12}+\frac{2KFEI}{GAL}$$</span></p>
<p>We plug this back into the <span class="math-container">$\delta(x)$</span></p>
<p><span class="math-container">$$\delta(x)=\frac{Fx^3}{24EI}-\frac{\frac{FL}{12}+\frac{2KFEI}{GAL}}{2EI}x^2+\frac{KFx}{2GA}+C_1x$$</span>
<span class="math-container">$$\delta(x)=\frac{Fx^3}{24EI}-\frac{\frac{FL}{12}+\frac{2KFEI}{GAL}}{2EI}x^2+\frac{KFx}{2GA}+\frac{KFx}{2GA}$$</span>
<span class="math-container">$$\delta(x)=\frac{Fx^3}{24EI}-(\frac{FK}{AGL}+\frac{FL}{24EI})x^2+\frac{KFx}{GA}$$</span></p>
<p><span class="math-container">$\delta$</span> max should be at <span class="math-container">$\frac{L}{2}$</span>. So we plug it all back in.</p>
<p><span class="math-container">$$\delta(\frac{L}{2})=\frac{F(\frac{L}{2})^3}{24EI}-(\frac{FK}{AGL}+\frac{FL}{24EI})(\frac{L}{2})^2+\frac{KF(\frac{L}{2})}{GA}$$</span></p>
<p><span class="math-container">$$\delta(\frac{L}{2})=\frac{FL^3}{192EI}-(\frac{FK}{AGL}+\frac{FL}{24EI})(\frac{L^2}{4})+\frac{KFL}{2GA}$$</span></p>
<p>I then run a simple simulation to check the validity of the equation.
<span class="math-container">$L=30mm,H=8mm,T=2mm,$</span>6061 <span class="math-container">$(E=69GPa)$</span>,<span class="math-container">$F=50N,K=\frac{5}{6},v=0.33$</span></p>
<p>I get an anticipated displacement of <span class="math-container">$-0.815\mu m$</span>, which seems inherently wrong as normally a positive number indicates a displacement in the direction of the force vector. In addition when I ran the simulation I got a displacement of <span class="math-container">$1.98\mu m$</span> in the direction of the force vector.</p>
<p>So I am not confident in my approach on this. I believe there are a few locations I might have made mistakes but I am unsure:</p>
<ol>
<li>I am not confident in the <span class="math-container">$C_1$</span> value.</li>
<li>I am not confident in that I took a whole indefinite integral of the whole "equation", I am unsure if the operation needs to be more like: <span class="math-container">$\int (\frac{Fx}{4Ei}-\frac{M'}{EI})+\int \frac{K}{GA}\frac{dv}{dx}$</span> which would result in 2 constants per integration.</li>
<li>I am not confident that I have applied the Timoshenko beam analysis properly on this problem.</li>
</ol>
<p>Any insight would be much appreciated!</p>
<p><a href="https://i.stack.imgur.com/XPgTa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XPgTa.png" alt="Fixed Fixed Beam With Point Load At L/2" /></a>
<a href="https://i.stack.imgur.com/Pichw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pichw.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|beam|mathematics| | <p>I was able to get a hold of Roark's Formulas for Stress and Strain (9th Edition), and there is a section for (8.10) Beams of Relatively Great Depth.</p>
<p>For this problem that is end supported, center load (I have changed variables to match my problem variables):</p>
<p><span class="math-container">$\delta_{shear} = \frac{FLK}{4AG}$</span></p>
<p>So the total deflection is</p>
<p><span class="math-container">$\delta_{max}=\delta_{bending}+\delta_{shear}=\frac{FL^3}{192EI}+\frac{FLK}{4AG}$</span></p>
<p>I ran a handful of simple simulations and I was getting less then 2-3% difference between the hand calculations vs simulations. I played around a bit trying to understand why <span class="math-container">$\delta_{shear}$</span> has a <span class="math-container">$\frac{1}{4}$</span> in it. I believe it is because the max moment of the shear is at the middle and is <span class="math-container">$\frac{F}{2}*\frac{L}{2}$</span>. However I have not dug into that deeply. I hope this answer is able to help some one!</p>
| 52399 | Timoshenko Fixed Fixed Total Beam Deflection |
2022-09-08T16:32:36.737 | <p>I want to build a robot arm (see <a href="https://youtu.be/F0ZvF-FbCr0" rel="nofollow noreferrer">https://youtu.be/F0ZvF-FbCr0</a>) to move a dumbbell of 2 kg (roughly 2.2 lbs).</p>
<p>However, since I have no background in engineering, I have no idea how to calculate / find out if the engines used (MG996r) have enough power to lift the object. How do I solve it? What is the conclusion?</p>
<p>If you come to the conclusion it has not enough power, can you recommend an Arduino-enabled engine that has sufficient power?</p>
<p>I am looking forward to your answer and thank you in advance. :)</p>
| |mechanical-engineering|control-engineering|power|energy|electrical| | <p>In order to determine if the robot arm can lift the object, you would need to determine the torque you need from the motor.
You should read up on how torque exactly works online. There is a lot of information online, here is a good start: <a href="https://theengineeringmindset.com/what-is-torque/" rel="nofollow noreferrer">https://theengineeringmindset.com/what-is-torque/</a>
As explained by the website, gears could make a motor with less torque still able to lift the dumbbell, but this comes at the cost of speed.</p>
<p>Be careful when doing research though, there are also many bad websites explaining torque: If someone explains torque as "force/length" or "force per length" as opposed to "force * length" or "force times length", they have NO IDEA what they are talking about. (And you should find another website.) <strong>The correct unit is N * m (or kg * m or lbs * ft) NOT N/m (or kg/m or lbs/ft).</strong> Again, close the website if they're talking about N/m or something similar.</p>
<p>You're talking about power of the motor, but that is not necessarily what you're looking for here: power would determine how quickly the robot arm could lift the dumbbell, torque determines IF the motor could lift the dumbbell.</p>
| 52400 | mg996r - maximum weight for robot arm |
2022-09-08T23:21:50.087 | <p>Looking at the following question:
<a href="https://i.stack.imgur.com/CPDJL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CPDJL.png" alt="Question" /></a>
The part of the solution I am interested in is as follows
<a href="https://i.stack.imgur.com/zFp3U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zFp3U.png" alt="enter image description here" /></a></p>
<p>Since the tangential velocities are the same, we could use equation 3 to solve for the angular acceleration of gear B. What I'm wondering is, why are the tangential accelerations of the gears at point P the same? I can't seem to wrap my head around this part since they are both obviously spinning at different rates.</p>
| |mechanical-engineering|dynamics| | <p>As you said, both te gears rotate at different rates. At the same time it is important to understand the gears rotate with different "ANGULAR RATES" i.e., both of them rotate with different angular speeds. Since it is often confusing to think in terms of angular rates(angular speed) and linear rates(tangentaial speed/acceleration), think about it in a different way,</p>
<ol>
<li><p>Consider the point 'P' in the image where both the gear mesh with each other. It is clear that the teeth of the driving gear (Gear A) pushes the teeth of the griven gear (gear B). This means one teeth from the driving gear pushes only one teeth in the driven gear. So you can say that the number of teeth which is moving across the point 'P' per unit time is same for both gears.</p>
</li>
<li><p>If you are aware of 'Universal law of gearing' you'll be knowing that in order for two gears to be in mesh, both the gears must have the same module (same teeth size).</p>
</li>
</ol>
<p>Logically you can deduce that if the number of gears per unit time and the gear size is same for both the gears, the tangential acceleration (hence, tangential velocity) must be same for both the gears at the periphery (hence, at the point of meshing 'P' ).</p>
<p>Hope this helps.</p>
| 52409 | Why is the tangential acceleration the same for two gears at the contact point? |
2022-09-10T20:55:33.783 | <p>I have a air dehumidifier in my basement, and at work I am using a coalescing filter for our air compressor. As I see it, my air dehumidifier uses quite a lot of electricity compared to the completely passive coalescing filter.</p>
<p>Why cant we get a low power dehumidifier from a fan and a (presumably slightly larger than average) coalescing filter?</p>
| |dehumidification| | <p>A coalescing filter separates liquids from gasses. You need a mist or droplets for it to work. There are dehumidifiers that work on compression, coalescing, expansion (some with energy recovery), but they aren't really suitable for the home. You need an aftercooler to condense the vapor for removal by the coalescing filter. Efficiency seems to be about 15% thermal.</p>
| 52425 | Why don't air dehumidifiers use coalescing filters? |
2022-09-11T18:48:21.010 | <p>Consider a spring fixed from end A and pulled from the other end B. The displacement of the end B of the spring is <span class="math-container">$x$</span>.</p>
<p><a href="https://i.stack.imgur.com/Kj4mg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kj4mg.jpg" alt="enter image description here" /></a></p>
<p>If I take a point at a distance <span class="math-container">$l_1$</span> from end A, how can I determine the displacement of this point corresponding to the displacement of end B, which is <span class="math-container">$x$</span>?</p>
<p>The free (undeformed) length of the spring is <span class="math-container">$L$</span>.</p>
| |mechanical-engineering|structural-engineering|civil-engineering|applied-mechanics| | <p>You have to know (or make assumptions about) how the spring is constructed.</p>
<p>If the spring is a nice uniform spring of the sort found in textbook illustrations, then you can figure that each section of the spring elongates by the same amount as any other. So if the total spring length is <span class="math-container">$l_t$</span>, and it elongates by <span class="math-container">$x$</span> overall, than any section will elongate by a factor of <span class="math-container">$\frac{l_t} x$</span>.</p>
<p>You can check this for yourself by breaking your textbook spring with spring rate <span class="math-container">$k$</span> into <span class="math-container">$N$</span> shorter springs of equal length, each with spring rate <span class="math-container">$\frac k N$</span>, and doing the math on how much extension you get from pulling on the whole assembly in series, and how much any set of <span class="math-container">$n$</span> springs will extend at the same time. Then for any arbitrary fraction of the spring length, find the limit as <span class="math-container">$N \to \infty$</span>.</p>
<p>If the spring <em>isn't</em> nice an uniform, then the stiffer sections will have less proportional elongation than the sections with more compliance; in that case life gets difficult, but with a knowledge of the incremental spring rate (or, better, compliance) along the length of the spring you could determine this elongation by integrating the spring compliance along the length of the section of interest.</p>
| 52434 | How to find the displacement of an arbitrary point on a spring? |
2022-09-11T22:14:24.930 | <p>Logarithmic cams are used in rock climbing, but involute cams are used to clamp work pieces when machining. Why are cams with two different profiles preferred for these two applications, which seem to have the same goal?</p>
| |mechanical-engineering|design|mechanisms|machine-design|machine-elements| | <ul>
<li>if an involute of a circle is turned uniformly around its center, then the tangents to the involute move in a uniform translation motion.</li>
<li>the mechanical advantage is greater when the gap is just below the maximum range with small deviation</li>
<li>the offset is optimal for repeated gap sizes for a consistent angle of handle and thus not suitable for a random wide range of gaps unlike the logarithmic cams</li>
</ul>
<p><a href="http://www.mozimtec.de/a1.html" rel="nofollow noreferrer">next</a>
<a href="https://i.stack.imgur.com/wrP7X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wrP7X.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/R3nD4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R3nD4.png" alt="enter image description here" /></a> <a href="https://www.pinterest.ca/pin/778278379355520082/" rel="nofollow noreferrer">from</a></p>
<p><a href="http://www.vainokodas.com/climbing/cams.html" rel="nofollow noreferrer">next</a></p>
<p><a href="https://i.stack.imgur.com/AxsXe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AxsXe.png" alt="enter image description here" /></a></p>
<p>The dual logarithmic cam has two advantages for climbers</p>
<ul>
<li>the engagement serves over a wide ratio of gaps.</li>
<li>the mean contact point below horizontal pin axis is constant for any gap and depends on the log factor and thus a constant gear ratio
r = eµø
a = tan-1(µ)</li>
</ul>
| 52443 | Cam-operated clamps: When are logarithmic cams preferred and when are involute cams preferred? |
2022-09-12T15:49:49.760 | <p>I'm quite new to Mechanical Engineering and I'm currently writing a report for a project that is to make a pedestrian truss bridge. I decided on the materials and dimensions of the deck, truss, and safety rails then I proceeded to calculate the dead load of the structure.</p>
<p>However, I'm stuck in calculating the dead weight of the Truss because I'm not sure about the dimensions of a chord in order to calculate its volume. Is there a way to determine cost-effective truss chord dimensions? Or else is there a way to assume the Truss's weight regardless of its volume?</p>
<p>I would really appreciate the help! Thanks.</p>
| |mechanical-engineering|bridges| | <p>This is a civil engineering problem early solved using simple analytics. A bridge is designed according to to its loading design. You need to take account of materials dead weight, live loadings and variable factors. These are set out in Eurocodes. Depending on structural material chosen, it will either be steel, reinforced concrete, or wood. Wood is seldom used as it is subject to highly variable continuum.
Steel is the most often used material for pedestrian bridges and rc the most common.</p>
<p>To get an idea of what is involved, I highly recommend getting your hands on a book that deals with this.</p>
<p>BRIDGE DESIGN IS NOT STRAIGHT FORWARD.</p>
<blockquote>
<p>Design of Structural Elements (3RD ED) by Chanakya Arya. ISBN 9780415467209</p>
</blockquote>
| 52462 | Bridge Truss chord dimensions |
2022-09-13T15:37:26.200 | <p>I was looking in a few different hydraulic power textbooks for symbol information and I noticed that on of them referenced a check valve with this symbol:</p>
<p><a href="https://i.stack.imgur.com/EB97Cm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EB97Cm.png" alt="Full logical symbol of a "Non-Return Valve"" /></a></p>
<p>Is this "full" block symbol ever actually used in industrial diagrams or is this simplified version usually used? Which should I use for documentation for a university report?</p>
<p><a href="https://i.stack.imgur.com/p76ufm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p76ufm.png" alt="Simplified Non-Return Valve" /></a></p>
| |hydraulics|fluid| | <p>The bottom one is a simple check valve.</p>
<p>The top one looks like back pressure from the top will push the spool into the closed position. Forward pressure from below will work with the spring to open the valve for forward feed.</p>
<p>It appears to me that if the top pressure falls that the spring should cause the valve to open in which case the system can drain. (I'm an EE so this isn't really my area but I have worked on a few hydraulic systems.) I can only think that it allows the system to drain or relax after pressure is lost.</p>
| 52479 | Correct Hydraulic Symbol for Check Valve |
2022-09-13T20:32:46.840 | <p>19th Century technology sometimes involves large-scale bearings. I am curious about the "chariot"-style bearings used in lighthouse optics, and the similar bearings used to support rotating telescope domes.</p>
<p>See the illustration of the interior of an observatory dome here: <a href="https://www.dunsink.dias.ie/grubbtelescope/" rel="nofollow noreferrer">https://www.dunsink.dias.ie/grubbtelescope/</a>
The rotating dome sits on a circle of wheels.</p>
<p>The chariot-style lighthouse lens bearing is shown in this video:
<a href="https://www.youtube.com/watch?v=6yCZlbND32o" rel="nofollow noreferrer">https://www.youtube.com/watch?v=6yCZlbND32o</a></p>
<p>The rotating structure that supports the wheels, in both cases, is a bit like the cage in a ball bearing. But I'm not sure whether there's any mechanical advantage to having it rotate freely. Why not fix the wheels to the object that is being supported? Is this something to do with rolling resistance, or is there some other advantage to mounting the wheels on a separate structure that rotates independently?</p>
| |bearings|machine-design| | <p><a href="https://i.stack.imgur.com/tPBMb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tPBMb.jpg" alt="enter image description here" /></a></p>
<p>The big advantage is that there is very little load on the axles (other than the carrier ring) and therefore there is almost no wheel-axle friction.</p>
<blockquote>
<p>Why not fix the wheels to the object that is being supported?</p>
</blockquote>
<p>If you do then you need bearings between the wheels and the axles. As it is, the wheels are acting as roller bearings with a much larger diameter than axle bearings would have so the surface stresses are much lower, wear is lower and a very heavy load can be reliably rotated with a moderately sized gravity motor.</p>
| 52481 | What is the advantage of "carriage"-style bearings for telescope domes, lighthouse lenses etc.? |
2022-09-14T03:31:21.143 | <p><a href="https://i.stack.imgur.com/wag0b.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wag0b.jpg" alt="enter image description here" /></a></p>
<p>Here in this image, you can see the water flow onto the beach to a certain distance. After a while, the water retracts back. If I put a V-shaped collector (like a vertical standing open book) facing the ocean, would I be able to accumulate water in the center of the v shape? I was thinking that instead of the water reaching and spreading out to a certain distance onto the beach, that flow would stop and accumulate vertically upwards. However, I'm not sure if this is correct or partially correct. Would the water retain the same height and just go around the collector? What if the v-shape expands 1/2 mile into the ocean. Would that reduce seepage and cause the water to accumulate vertically in the center?</p>
<p>Any insight would greatly help!</p>
| |mechanical-engineering|fluid-mechanics|civil-engineering|renewable-energy|waves| | <p>Yes water will go higher than the average height of the wave.</p>
<p>Imagine an infintisimally small droplet hit the wall with an angle <span class="math-container">$\theta$</span>. it reflects with an angle <span class="math-container">$2\theta$</span> with respect to the main wave keeping coming straight. so it will spill water over the center flow and raise its height.</p>
<p>The same phenomenon happens to the side of the bow of a boat or ship. there is the great noise of parting of the water and sending it climbing up the sides of the bow and later falling in a white foamy cascade.</p>
| 52484 | Can ocean flows after a crashing wave be accumulated in one point to increase the water height of that location? |
2022-09-15T20:55:14.380 | <p>I have been trying to make a surface loft of a pipe that goes around a U-turn and gradually changes diameter.</p>
<p>I was finally able to complete the surface loft, but something is off in the result. My profile and guide curves are shown below. As you can see, the <strong>profiles</strong> are gradually spaced circles coincident to the <strong>guide</strong> U-curve. It is hard to see in this image, but when I zoom in on the larger diameter pipe, you can see that it is no longer straight like the guide curve.</p>
<p>I've highlighted the portions with low curvature, and while this is a problem, there is a portion on the opposite side with very high curvature. How do I ensure this surface loft in the straight portions of the <strong>guide</strong> is also straight?</p>
<p>I'm using Solidworks v.22.</p>
<p><a href="https://i.stack.imgur.com/wmw2I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wmw2I.png" alt="curves" /></a>
<a href="https://i.stack.imgur.com/a9SnA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a9SnA.png" alt="problem_loft" /></a></p>
| |solidworks| | <p>For the shape that you are attempting to create - it would be more effective to model the tapered tube in its unbent state using a revolve, and then use the Flex feature to bend it around.</p>
<p><a href="https://i.stack.imgur.com/STtiB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/STtiB.png" alt="Finished Item" /></a></p>
| 52502 | Loft doesn't follow guide curve as intended in Solidworks v22 |
2022-09-16T00:05:47.800 | <p>I am seeking help for a textbook question that I am stuck on.
<a href="https://i.stack.imgur.com/mYJdG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mYJdG.png" alt="enter image description here" /></a></p>
<p>My work is attached below. I broke the velocity vector into two components, one that is perpendicular to the wedge, causing the rod to rotate, and the other is the force acting along the surface of the wedge. I am unsure if this is the correct approach, and my answer is in terms of φ, not θ, and I can't find the relationship between the two.
<a href="https://i.stack.imgur.com/3HGF9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3HGF9.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|dynamics| | <p><a href="https://i.stack.imgur.com/99BKi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/99BKi.jpg" alt="enter image description here" /></a></p>
<p>Let the pivot be the point <span class="math-container">$(0,0)$</span> , and let's show the coordinates of the other end of the rod with <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Note that <span class="math-container">$$x=Lcos\theta \quad,\quad y=Lsin\theta$$</span>
and therefore
<span class="math-container">$$dx=-Lsin\theta d\theta \quad,\quad dy=Lcos\theta d\theta \quad (1)$$</span><br />
In time <span class="math-container">$dt$</span> , the rod and the wedge move as shown above. Solid lines and dashed lines correspond to beginning and end of this time interval, respectively. Suppose the sliding end of the rod moves in this time interval from <span class="math-container">$(x_1,y_1)$</span> to <span class="math-container">$(x_2,y_2)$</span> . The difference between the <span class="math-container">$x$</span> coordinates of these points is <span class="math-container">$dx$</span> , and the difference between their <span class="math-container">$y$</span> coordinates is <span class="math-container">$dy$</span> .</p>
<p>The horizontal displacement of the wedge in time <span class="math-container">$dt$</span> is <span class="math-container">$vdt$</span>. Let the oblique edge of the wedge before and after this displacement be represented by <span class="math-container">$L_1$</span> and <span class="math-container">$L_2$</span>, respectively. The equations of these two lines are:
<span class="math-container">$$L_1 : \quad x = y cot \phi + b $$</span>
<span class="math-container">$$L_2 : \quad x = y cot \phi + b + vdt $$</span>
Note that <span class="math-container">$(x_1,y_1)$</span> is on <span class="math-container">$L_1$</span> and <span class="math-container">$(x_2,y_2)$</span> is on <span class="math-container">$L_2$</span>. So:
<span class="math-container">$$b = x_1 - y_1 cot \phi $$</span>
<span class="math-container">$$x_2 = y_2 cot \phi + (x_1 - y_1 cot \phi) + vdt $$</span>
By rearranging and noting that <span class="math-container">$x_2 - x_1 = dx$</span> and <span class="math-container">$y_2 - y_1 = dy$</span> we have:
<span class="math-container">$$dx - cot \phi dy = vdt $$</span>
And by substitution from (1) we have:
<span class="math-container">$$-L(sin\theta+cot\phi cos\theta)d\theta = vdt $$</span>
Or:
<span class="math-container">$$\frac{d\theta}{dt} = \frac{-v}{L(sin\theta+cot\phi cos\theta)}$$</span>
<span class="math-container">$$= \frac{-vsin\phi}{Lcos(\theta -\phi)} $$</span>
Finally, note that in this particular example, where the wedge is moving towards left, <span class="math-container">$v$</span> is negative. Therefore <span class="math-container">$\frac{d\theta}{dt}$</span> is positive.</p>
| 52503 | Finding angular velocity through absolute motion analysis |
2022-09-18T00:07:41.827 | <p>Why do inorganic LED bulbs have a limited lifespan? For example, a manufacturer may quote that a bulb or monitor powered by an organic light emitting diode(s) has a lifespan of 50,000 hours, while it may quote the lifespan of that powered by the inorganic LED(s) to be 70,000 hours.</p>
<p>For OLEDs, it is understandable why they wear out. That is because in chemistry, I learned that polymers are made up of hugely complex, varying molecules that are not inert and are only weakly held between each other by weak van der Waals forces. If a bond breaks within a molecule for example, the original molecule will almost certainly not reform. That is because there are so many possible different atoms to form new bonds with, so the original bond will almost certainly not be reinstated due to probability. Because they are unstable, they self-react, which causes them to self-destruct over time. That is greatly accelerated by external high-energy sources such as ultraviolet rays. Since OLEDs are made out of polymers, that is why they break down over time.</p>
<p>However, inorganic LEDs are made out of glass or ceramic. In chemistry, I learned that glasses and ceramics have a very simple chemical composition with their structure tightly held together by strong atomic bonds. Glasses and ceramics are distinguished between each other because the former has a disordered structure, known as amorphous, while the latter has an ordered structure, known as crystalline. Because of how highly stable they are, they do not change at all under there is an super high-energy event from an external source that causes the bond to break, such as a macroscopic force cracking the material, high temperatures melting the material, or super corrosive chemicals. Similarly, a few days ago, I did an Internet search confirming my suspicions that solid-state ceramic and glass capacitors last forever under normal operation because they do not age at all within design limits.</p>
<p>So, shouldn't inorganic LEDs last forever (geological timescales) when its design parameters are never exceeded under operation, such as never having a power surge? Electrical conductivity is a property of electronic components (of which diode is a specific type, more specifically LED), just like visible light conductivity is a property of glass. Passing light through glass never wears out glass given that it does not overheat, so shouldn't passing electric current of the correct polarity through an inorganic LED never wear out the LED given that it never overheats?</p>
<p>Are the lifespans of inorganic LED bulbs just limited by the driver (required because the LEDs themselves only work with DC in one direction), which presumably typically contains an electrolytic capacitor? Under the absense of freak events, do they fail just because the electrolytic capacitor fails or the solder in the driver oxidizes? If the driver were only made out of materials that are corrosion-proof under normal atmospheric conposition and only contained ceramic/glass capacitors, would that mean the inorganic LED bulb would last geological timescales?</p>
| |electrical-engineering|power-electronics|electrical|electromagnetism|consumer-electronics| | <p>This is not a rigorous answer, but even a piece of steel will change if you just let it sit there for long enough. And I'm not talking about rust; I'm talking about the additives dissolved in the steel. Depending on what you need the piece of steel to do, it might very well be good enough to use even after sitting there for geological time scales. However, sometimes even these changes are too much and the steel needs to be stabilized.</p>
<p>Then consider that semiconductors make use of very finely tuned material compositions. The dopants and additives in there are able to slowly diffuse out. It's not completely stable.</p>
<p>Similarly, if your glass or ceramic plate capacitors don't need to rely on such finely tuned material composition to perform satisfactorily, you can probably tolerate a lot more changes due to the aging of the material. But for a semiconductor that relies on a much more specific material composition, you obviously will tolerate a lot less.</p>
<p>I don't think the material requirements for glass or ceramic plate capacitors is fair comparison to semiconductors. Perhaps more fair would be to look at the aging effects of optical coloured glass filters, though I have no knowledge about their aging effects. However, they are composed of glass with additives and are required to have properties that can deviate much less than capacitors.</p>
<p>Oh, and as for glass being stable. Perhaps you should know that coloured glass filters for optics, particularly UV-pass filters can actually oxidize due to humidity in the air. Furthermore, they actually degrade in the presence of sufficiently strong UV light. So although regular glass may be stable on geological time scales you cannot extend that to glass with additives. Not all glass is the same.</p>
| 52519 | Why do inorganic LEDs wear out? |
2022-09-19T19:44:43.363 | <p>Because of the war in Ukraine I am interested in the question if a nuclear reactor is designed to withstand a missile attack.</p>
<p>In university, back in the eighties, we were assured that a nuclear reactor would withstand an airplane crash, such as a Boeing 747 Jumbo Jet. But perhaps a Russian war missile has much more destructive power? Or is 1 meter of reinforced concrete invulnerable for a stupid missile?</p>
<p>I understand that my question is a bit vague, especially because as an electrical engineer I don't have any expertise on concrete, but these days this question is pretty relevant.</p>
| |safety|reinforced-concrete| | <p>The primary problems would be loss of instrumentation and control.</p>
| 52544 | Can a nuclear reactor withstand a missile attack? |
2022-09-21T20:04:49.650 | <p>This question isn't about how the tire holds weight up against the ground based on its contact patch, I understand that. It's about what's happening inside the tire such that the wheel (the metal hub in the middle, which serves as the inner wall of the pressure chamber) doesn't fall down to rest against the ground (like it does when the tire is flat.)</p>
<p>(This is a rephrasing of a question I asked on Physics.SE which was closed as off-topic and more suitable for Engineering.)</p>
<p>The intuitive answer is "air pressure" but my confusion stems from the fact that, since pressure is roughly equal throughout the tire, the net pressure force on the wheel should cancel out. It feels the same pressure radially inwards all around its circumference, it's just being squeezed in place.</p>
<p>A commenter on the now-closed Physics questions mentioned that the air pressure pre-stresses the tire rubber, which I assume gives it some form of rigidity that allows it to resist the tendency of the heavy wheel to deform the rubber? I.e. the asymmetric force opposing gravity would be applied by the tire rubber pressed against the wheel, not the air. But I don't understand enough about pre-stressing to know if this is accurate/how it works, and it was just a comment, not a full answer. I'd like to know more about the forces at work.</p>
<p>Included is a diagram of my understanding of the forces involved in a tire vs. a simple airbag holding up a mass. Obviously something is missing in the tire diagram, because it doesn't balance.</p>
<p><a href="https://i.stack.imgur.com/m15nj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m15nj.png" alt="A free body diagram of an air bag holding up a mass, and an incomplete free body diagram of a tire with a rigid wheel/hub." /></a></p>
<p>Also included is a diagram of my current understanding of the pre-stress explanation. What I don't understand is why Fr=(Fr2-Fr1) is larger in the presence of air pressure. I understanding why Fr2>Fr1, the tire is more deformed at the bottom and is acting like a spring, I just don't get why the air pressure so drastically increases the force with which it resists deformation.</p>
<p><a href="https://i.stack.imgur.com/5Mywm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Mywm.png" alt="Free body diagram of a tire seen in cross section." /></a></p>
| |automotive-engineering|pressure| | <p>The tire has maximum volume inside when the hub is perfectly centered. This also means that in this configuration, the pressure inside the tire is lowest possible for a given amount of air (given the condition that this amount of air can still exceed atmospheric pressure in this configuration) which, not surprisingly, is also when there is no external forces on the tire other than atmospheric pressure.</p>
<p>The configuration where the hub is sitting on the ground is much lower which means that for the same mass of air inside, the pressure is much higher. Thus the total force this pressure can exert is increased.</p>
<p>If there is enough air in the tire such that there is enough pressure to produce a collective force enough to overcome the weight of the hub and the car connected to it, it will lift up the hub and car and move towards a configuration where there is more volume inside the tire, and more volume means reduced pressure.</p>
<p>The hub will continue to be raised towards center while the volume inside increases as a result until equilibrium is reached which is when the pressure inside produces a force equal to the weight of the car.</p>
<p>In other words, it's no difference than a balloon being spherical and maximum volume versus deforming it to be anything other than a sphere. Except that with a car wheel you're distracted by the hub forming part of the inner wall, and the arrangement lower and higher volume configurations are not as obvious. But nothing has changed from an airbag or balloon.</p>
| 52583 | How exactly does a tire hold up its internal metal wheel against gravity? |
2022-09-24T03:04:26.260 | <p>Is there any way to measure the torque of an engine without reducing the power output of a motor? A ton of dynamometer seem to apply a load on the engine, which works well for static tests, but how can the torque be measured on a running car or boat without affecting the output too much?</p>
| |mechanical-engineering|automotive-engineering|torque| | <p>If the motor is driving something via a conrod, you could try a strain gauge on the surface of the conrod, oriented to measure axial strain. If you know the Young's modulus and cross-sectional area of the conrod, you can covert axial strain to axial load, then the product of axial load in the conrod with distance from the axis of the motor to the point of attachment of the conrod is the torque.</p>
| 52603 | Torquemeter without Changing the Power Output of a Motor? |
2022-09-24T16:37:06.783 | <p>I'm looking to manufacture a coffee table. I would like the top panel of the table to be split into two parts in order to open the top of the table and get access to a cabinet. Each part of the top panel will have a vertical rotation axis and therefore the two rotation axis will be parallel. The rotation angles should be equal in opposite directions.<a href="https://i.stack.imgur.com/eZalL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eZalL.png" alt="enter image description here" /></a> The joint should be hidden below the coffee table. It should be quite flat: maximum 5 to 10 cm. A joint with 3 parallel gears, with one gear on each axis and the third in the middle would work... but the gears would need to have large diameters as the axis would be distant by around 50 - 100 cm.</p>
<p>Do you have better ideas? Maybe with an endless screw in the middle?</p>
| |kinematics| | <p>Depending on whether you care about accurate angles displacement you could use chains or belts.</p>
<h2>Configuration 1 - Belt Not accurate</h2>
<p>If then rotation angles <strong>don't</strong> have to be the same you can get away with using an cross belt drive. See below for the difference between open and cross belt drives and how the cross belt ensure the opposite direction.</p>
<p><a href="https://i.stack.imgur.com/x7Aiu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x7Aiu.png" alt="enter image description here" /></a></p>
<h2><strong>Configuration 2: belt or chain</strong></h2>
<p>another configuration is the following Red and blue are the shafts that you are interested to rotate.</p>
<p><a href="https://i.stack.imgur.com/H0sD6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H0sD6.png" alt="enter image description here" /></a></p>
<p>You could use the configuration above with chains (although you'd need to be able to adjust the tension on one of the yellow pulleys). If you used chain, then you'd be able to get exactly opposite angular rotations.</p>
<h2>Configuration 3: combination gear and chain/belt</h2>
<p>If you need the axles to rotate exactly the same, one way the following:</p>
<p><a href="https://i.stack.imgur.com/S7oKl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S7oKl.png" alt="enter image description here" /></a></p>
<ul>
<li>axle that carries B and C is the motor.</li>
<li>A and B are gears</li>
<li>C and D can be belt or chain (depending on the angular displacement requirements.</li>
</ul>
| 52611 | Kinematic join for parallel axis rotation with equal inverse angles |
2022-09-25T13:57:18.387 | <p>I'm planning a shelving system for my workshop and will have a span a little over 12' that I need to support. I'm not really sure what weight the shelves will need to support in the end but want them to be strong enough that I never have to think about it so probably at least 2000lbs.</p>
<p>In my current design I'm using two 2x8s to cover this span. Will that be enough? If not should I double them up to make a total of four 2x8s or move up to two 2x10s, 2x12s?</p>
<p>The beams are attached to 4x4 vertical posts on both ends and will be attached with joist hangers.</p>
<p><a href="https://i.stack.imgur.com/k7kQK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k7kQK.jpg" alt="" /></a></p>
<p><a href="https://i.stack.imgur.com/SPeVp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SPeVp.jpg" alt="" /></a></p>
| |structural-engineering| | <p>I will Begin with my assumptions.</p>
<ol>
<li>You wish to select the size, species, and grade of a 12-foot-long timber shelf beam with the strength to support a 2000-pound load uniformly distributed over the beam length.</li>
</ol>
<p>a. Possible alternative loading assumptions.</p>
<pre><code> i. 2000 lbs concentrated at center of beam.
ii.2000 lbs is distributed over a specified length of shelf.
</code></pre>
<ol start="2">
<li><p>The lumber specifications are unknown. I have assumed visually
graded Douglass Fir No.1</p>
</li>
<li><p>Tabulated unfactored working stresses based on normal load duration:
Fb = 1000 psi & Fv = 180 psi (Size Factor CF taken as 1.0)</p>
</li>
</ol>
<p>Internal forces acting on member cross section.</p>
<ol>
<li>Bending: M = wL*2/8 = (2000/12/12)[12(12)]^2/8 = 36,000 in-lb</li>
<li>Shear: V = wL/2 = 2000/2 = 1,000 lb</li>
</ol>
<p>Internal stresses acting on a nominal 2x8 cross section. (Actual dimensions = 1.5x7.25)</p>
<ol>
<li>Bending: fb = M/S, S = 1.5(7.25^2)/6 = 13.14 in^3, fb = 36000/13.14 = 2739.6 psi</li>
<li>Horizontal shear: fvh = 1.5(V/A) = 1.5(1000)/(1.5*7.25) = 137.93 psi</li>
</ol>
<p>Stress ratios.</p>
<ol>
<li>Bending: fb/Fb = 2739.6/1000 = 2.74 > 1 NG</li>
<li>Horizontal shear: fvh/Fvh = 137.93/180 = 0.77 < 1 OK</li>
</ol>
<p>As we can see the 2x8 is overstressed in bending by a factor of 2.74. This means you would need three of them based on bending and my assumptions. The best choice is to go deeper if you have the headroom.</p>
<p>The actual depth required for a single 2x board is sqrt[(7.25^2)2.74] = 12 in.</p>
<p>The actual depth required for a <em>double</em> 2x board is sqrt[(7.25^2)2.74/2] = 8.48 in.</p>
<p>Based on this analysis, if bending controls the design, the required size is: <em>Double 2x10</em>.</p>
<p>Check bearing perpendicular to the grain: fc = (2000/12/12)/3 = 4.63 psi</p>
<p>Allowable bearing: Fc = 625 psi > 4.63 psi OK</p>
<p>In addition, the connection to the posts will need to be considered. There may be joist hangars for your situation that will work. Your hardware supplier should be able to tell you. If my assumption regarding the loading, lumber grade & species and anticipated cumulative load duration is wrong, that will change the outcome. These assumptions will need to be either confirmed or corrected before the solution can be known.</p>
| 52620 | Is a 2x8 adequate for this span? |
2022-09-26T03:23:54.440 | <p><strong>This is just a though experiment that I had.</strong></p>
<p>Two hooks (the hooks are just extrusions of the sketches shown in the pictures below, with both hooks being extruded by the same amount).
If we use the flexure formula to calculate the stress at point A in hook 1 and hook 2 then we get the same bending stress despite the two hooks have different shapes. This is because they have the same Moment applied to them, the same MMOI for the cross-section and are the same distance away from their respective neutral axes.
How do I include the effect of the curved shape on hook 1 in my calculation for the bending stress?
<a href="https://i.stack.imgur.com/SiKXY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SiKXY.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/TIjPG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TIjPG.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|solid-mechanics| | <p>It may seem counterintuitive but --barring plasticity effects-- the stresses on point A should be the same in both cases. (this is probably the most difficult concept to grasp in statics 101).</p>
<p>I.e. Unless you have deformations that range into the plasticity region of the material (and other non linear effects) the result should remain the same.</p>
<hr />
<p>If you are interested in the maximum stresses in general (e.g. for a cross-section of the hook, then obviously the shape matters.</p>
| 52622 | Flexure formula insufficient for model |
2022-09-28T08:29:37.287 | <p>When designing shaft keys we calculate compressive stress in hub/shaft/key and compare it with allowable stress. The usual formula is</p>
<p><span class="math-container">$$p = \cfrac{F}{A} = \cfrac{\cfrac{M}{R}}{L_e \cdot t}$$</span></p>
<p>where <span class="math-container">$p$</span> is compressive stress, <span class="math-container">$M$</span> is applied torque, <span class="math-container">$R$</span> is shaft radius, <span class="math-container">$L_e$</span> is effective key length and <span class="math-container">$t$</span> is keyway depth. The <span class="math-container">$t$</span> can be sometimes replaced by <span class="math-container">$\frac{h}{2}$</span>, where <span class="math-container">$h$</span> is key height.</p>
<p>I would like to try and calculate the force <span class="math-container">$F$</span> more precisely although it is not necessary because of other uncertainties and simplifications in the calculations making this still an approximation. It is just an exercise for fun and out of curiosity.</p>
<p><a href="https://i.stack.imgur.com/Dj2C5l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dj2C5l.png" alt="enter image description here" /></a></p>
<p>Using following formulas <span class="math-container">$F = \cfrac{M}{r}$</span>, <span class="math-container">$\cfrac{F_x}{F} = \cfrac{R-y}{r}$</span>, <span class="math-container">$r = \sqrt{(R-y)^2 + (0.5 \cdot b)^2}$</span> I derived formula for normal force <span class="math-container">$F_x$</span> acting on key's surface as a function of parameter <span class="math-container">$y$</span>, where <span class="math-container">$y$</span> is distance from top of the shaft.</p>
<p><span class="math-container">$$F_x(y) = \cfrac{M \cdot (R-y)}{(R-y)^2 + 0.25 \cdot b^2}$$</span></p>
<p><span class="math-container">$$[F_x] = \cfrac{\textrm{N} \cdot \textrm{mm} \cdot (\textrm{mm} - \textrm{mm})}{(\textrm{mm} - \textrm{mm})^2 + 1 \cdot \textrm{mm}^2} = \cfrac{\textrm{N} \cdot \textrm{mm}^2}{\textrm{mm}^2} = \textrm{N}$$</span></p>
<p>Now I would like to calculate the compressive stress <span class="math-container">$p$</span> with my newly defined normal force <span class="math-container">$F_x(y)$</span> but I do not know how to sum the force or what value of parameter <span class="math-container">$y$</span> I should choose.</p>
<p>The questions are:</p>
<ol>
<li>Is my approach correct?</li>
<li>Can I get a single force value, something like <span class="math-container">$F_{total}$</span> with its position <span class="math-container">$y_{final}$</span>? Can I use it to calculate <span class="math-container">$p$</span>?</li>
<li>What value of parameter <span class="math-container">$y$</span> I should choose for the compressive stress calculation?</li>
</ol>
<p>I feel like questions 2 and 3 are dependent on each other. I was thinking about somehow summing the force from <span class="math-container">$y_1$</span> to <span class="math-container">$y_2$</span> with some form of integral. I know I can calculate total force when given a distributed load <span class="math-container">$q(x)$</span> by using formula <span class="math-container">$F = \int_{a}^{b}{q(x)dx}$</span>. The position of said force would be in the centroid of the distribution graph. The problem is that <span class="math-container">$F_x(y)$</span> is not distributed load just by looking at the units (<span class="math-container">$\textrm{N}$</span> instead of <span class="math-container">$\textrm{N/mm}$</span>). Maybe I could go straight for the compressive stress <span class="math-container">$p$</span> and avoid my current force problem or maybe I am misunderstanding the whole problem?</p>
<hr />
<h2>EDIT 1:</h2>
<p>I did the calculations with some numbers using the usual simple formula and NMech's approach and I got different results (<span class="math-container">$\approx$</span> one order of magnitude). Symbolic and numerical integration was done using online tool <a href="https://www.integral-calculator.com" rel="nofollow noreferrer">Integral Calculator</a>. Given <span class="math-container">$M = 10^5 \mathrm{\,Nmm}$</span>, <span class="math-container">$R = 17.5 \mathrm{\,mm}$</span>, <span class="math-container">$b = 10 \mathrm{\,mm}$</span>, <span class="math-container">$L_e = 15 \mathrm{\,mm}$</span>, <span class="math-container">$y_2 = t = 4.7 \mathrm{\,mm}$</span>.</p>
<p><span class="math-container">$$y_1 = R - \sqrt{R^2 - (0.5\,b)^2} = 17.5 - \sqrt{17.5^2 - (0.5 \cdot 10)^2} = 0.73 \mathrm{\,mm}$$</span></p>
<p><strong>Approach 1:</strong> Using the usual simple formula.</p>
<p><span class="math-container">$$F = F_x = \frac{M}{R} = \frac{10^5}{17.5} = \mathbf{5\,714.286} \mathrm{\,N}$$</span></p>
<p><strong>Approach 2:</strong> Using <span class="math-container">$I_p$</span> with outer shaft diameter <span class="math-container">$r = R$</span>, integrating <span class="math-container">$dF(y)$</span>.</p>
<p><span class="math-container">$$I_p = \frac{\pi R^4}{2} = \frac{\pi \cdot 17.5^4}{2} = 147\,324 \mathrm{\,mm}^4$$</span></p>
<p><a href="https://i.stack.imgur.com/rBPh3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rBPh3.png" alt="enter image description here" /></a></p>
<p>Here I don't know what to do with the expression <span class="math-container">$\ln(\mathrm{mm}^2)$</span>. If it was equal to <span class="math-container">$1$</span>, then the final unit would be <span class="math-container">$\mathrm{N}$</span>, which would make sense. From my quick Google search it seems that we should only take logarithm of dimensionless quantity <a href="https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/01%3A_Units_and_Measurement/1.05%3A_Dimensional_Analysis#:%7E:text=The%20arguments%20of%20any%20of%20the%20standard%20mathematical%20functions%20such%20as%20trigonometric%20functions%20(such%20as%20sine%20and%20cosine)%2C%20logarithms%2C%20or%20exponential%20functions%20that%20appear%20in%20the%20equation%20must%20be%20dimensionless.%20These%20functions%20require%20pure%20numbers%20as%20inputs%20and%20give%20pure%20numbers%20as%20outputs." rel="nofollow noreferrer">(1)</a>, <a href="https://math.ucr.edu/home/baez/physics/General/logs.html#:%7E:text=a%20dimensioned%20quantity%3F-,No%2C%20you%20can%27t.,-This%20question%20has" rel="nofollow noreferrer">(2)</a>.</p>
<p><span class="math-container">$$%
F =
%
\int_{0.73}^{4.7}{\frac{10^5 \cdot 15}{147\,324} \sqrt{(17.5 - y)^2 + (0.5 \cdot 10)^2} dy} \approx
%
\mathbf{631.051} \mathrm{\,N}
$$</span></p>
<p><strong>Approach 3:</strong> Substituting <span class="math-container">$I_p$</span> with <span class="math-container">$\frac{\pi \cdot r^4}{2}$</span> and then <span class="math-container">$r$</span> with <span class="math-container">$\sqrt{(R-y)^2 + (0.5 \cdot b)^2}$</span>, integrating <span class="math-container">$dF(y)$</span>.</p>
<p><span class="math-container">$$
F(y) =
%
\int{\frac{2 \, L_e M}{{\pi}} \cdot \left(\left(R - y\right)^2 + (0.5 \, b)^2\right)^{-\frac{3}{2}}dy} =
%
\frac{16 \, L_e M \cdot \left(y-R\right)}{\pi b^3 \sqrt{\frac{4 \, \left(y - R\right)^2}{b^2}+1}} + C$$</span></p>
<p><span class="math-container">$$[F(y)] =
%
\frac{1 \cdot \mathrm{mm} \cdot \mathrm{N} \cdot \mathrm{mm} \cdot\left(\mathrm{mm} - \mathrm{mm}\right)}{1 \cdot \mathrm{mm}^3\sqrt{\frac{1 \cdot \left(\mathrm{mm} - \mathrm{mm}\right)^2}{\mathrm{mm}^2}+1}} =
%
\frac{\mathrm{N} \cdot \mathrm{mm}^3}{\mathrm{mm}^3} =
%
\mathrm{N}
$$</span></p>
<p><span class="math-container">$$F = \int_{0.73}^{4.7}{\frac{2 \cdot 15 \cdot 10^5}{{\pi}\cdot\left(\left(17.5-y\right)^2+\frac{10^2}{4}\right)^\frac{3}{2}}} \approx \mathbf{1\,025.790} \mathrm{\,N}$$</span></p>
<p><strong>Approach 4:</strong> Similar to #2, integrating <span class="math-container">$dF_x(y) = dF(y) \frac{R - y}{r(y)}$</span>.</p>
<p><span class="math-container">$$
\newcommand{\mm}{\mathrm{mm}}
\newcommand{\N}{\mathrm{N}}
F_x(y) =
%
\int{\frac{L_e M \cdot \left(R - y\right)}{I_p}} =
%
\frac{L_e M y \cdot \left(2 \, R - y\right)}{2 \, I_p} + C
$$</span></p>
<p><span class="math-container">$$
[F_x(y)] =
%
\frac{\mm \cdot \N \cdot \mm \cdot \mm \cdot \left(1 \cdot \mm - \mm\right)}{1 \cdot \mm^4} =
%
\frac{\N \cdot \mm^4}{\mm^4} =
%
\N
$$</span></p>
<p><span class="math-container">$$
F_x =
%
\int_{0.73}^{4.7}{\frac{15 \cdot 10^5 \cdot \left(17.5 - y\right)}{147\,324}} \approx
%
\mathbf{597.626} \mathrm{\,N}
$$</span></p>
<p><strong>Approach 5:</strong> Similar to #3, integrating <span class="math-container">$dF_x(y) = dF(y) \frac{R - y}{r(y)}$</span>.</p>
<p><span class="math-container">$$
F_x(y) =
%
\int{\frac{2 \, L_e M \cdot \left(R - y\right)}{{\pi} \cdot \left(\left(R - y\right)^2 + (0.5 \, b)^2\right)^2}} =
%
\frac{L_e M}{{\pi} \cdot \left(\left(R - y\right)^2 + (0.5 \, b)^2\right)} + C
$$</span></p>
<p><span class="math-container">$$
[F_x(y)] =
%
\frac{\mm \cdot \N \cdot \mm}{1 \cdot \left(\left(\mm - \mm\right)^2 + (1 \cdot \mm)^2\right)} =
%
\frac{\N \cdot \mm^2}{\mm^2} =
%
\N
$$</span></p>
<p><span class="math-container">$$
F_x =
%
\int_{0.73}^{4.7}{\frac{2 \cdot 15 \cdot 10^5 \cdot \left(17.5 - y\right)}{{\pi} \cdot \left(\left(17.5 - y\right)^2 + (0.5 \cdot 10)^2\right)^2}} \approx
%
\mathbf{969.253} \mathrm{\,N}
$$</span></p>
<p>Result from approach #4 is smaller than result from #2, same with #5 and #3. This is to be expected because <span class="math-container">$F_x$</span> is a projection of the total force <span class="math-container">$F$</span> onto the x-axis. What staggers me is the huge difference between #1 and other approaches. I though the final force could be somewhere in the interval <span class="math-container">$F \in \langle4000, 8000\rangle \mathrm{\,N}$</span>. Should I have expected these results or is there something wrong?</p>
| |mechanical-engineering|stresses|torque|moments|forces| | <p>The simple answer is
<span class="math-container">$$
F=M/r
$$</span>
and that is acceptable for practical purposes. As OP correctly notes, that does not distinguish between <span class="math-container">$F$</span> and <span class="math-container">$F_x$</span>. That distinction is not considered in typical design practice.</p>
<p>The complicated answer starts with noting that the stress distribution in keys is not uniform. In the question, it's assumed that the compressive stress is uniform but contact with the shaft and hub at the midplane of the key leads to singular stress concentrations. Finite element modeling is probably the best tool for answering your question with more sophistication than is offered by textbooks.</p>
| 52668 | Calculate compressive stress in shaft/key surface from torque |
2022-09-29T09:08:21.480 | <p>Given the nominal size of a bolt/nut, e.g M10, can one construct all other dimensions from it? Do these dimensions differ between manufacturers or are these standardized. If so, where can I find tables with these dimensions. My goal is to make a CAD design library with all these DIN/ISO bolts. For this I need all exact dimensions.</p>
| |cad|fasteners|standards|bolting| | <p>I remember in high school drawing these by hand. The basic design is an hexagonal head chamfered at 45° to the diameter of the inscribed circle to the points of the hexagon.</p>
<p>M10 means 1mm across the flats or specifically the diameter of the inscribed circle.</p>
<p>An M32 would have a 32mm diameter circle.</p>
<p>The height of the head is proportional to the diameter at 50%. So an M10 is over designed compared to an M32 due to material strength properties and tools used.</p>
| 52678 | Are bolt and nut dimensions standardized? |
2022-09-29T09:49:33.043 | <p>I am thinking about the building house heating system like this:
Outside my house I put 12V 1000W wind turbine, for example, like <a href="https://www.aliexpress.com/item/1005003208265431.html" rel="nofollow noreferrer">this</a>.
Inside my house I connect directly without any converter/adapter/controller/battery 12V heater(s) with total power 1000W, for example, like <a href="https://www.aliexpress.com/item/1005001615392228.html" rel="nofollow noreferrer">this</a>.
The idea is that when there is a wind, it will be converted to heat inside my house and my house will be like a "heat battery".</p>
<p>Will it work at all and what can be the showstoppers in this configuration?</p>
<p>I heard and opinion that if I connect 1000W load to 1000W wind turbine it will work only if wind is very strong. In case if wind is not strong enough and wind turbine with small wind gives only 500W, turbine will not start at all or will stop soon after starting due to electric brake effect (kind of <a href="https://en.wikipedia.org/wiki/Regenerative_braking" rel="nofollow noreferrer">Regenerative braking</a>). I will need some kind of smart controller, which will limit the power which goes to the heater to avoid this electric brake effect. My goal is to convert all (or maximum possible) electric energy, which I can get from wind generator to the thermal energy.</p>
<p>It is not only applicable for wind generator but to any electric generator. If I connect electrical generator to bike pedals and start pedaling to get electricity, then I will be fine if I connect 50W light bulb to this generator. If I connect 2000W kettle, then I probably will not be able to turn the pedals for long time, because human body cannot generate so much power. Top athletes can make only 340-360 Watts on bike in long term.</p>
| |wind-power|regenerative-braking|power-generation| | <p>To get the maximum power out of the wind turbine you will need some kind of power converter, or (and this is probably easier) an adjustable load resistance.</p>
<p>Running the turbine with too much or too little load is like running a car in too high or too low a gear. It will work a little bit, but the power output will be reduced.</p>
<p>Matching the turbine to the load is called Maximum Power Point Tracking, or MPPT for short. The same process is used on solar panels.</p>
<p>Heaters are cheap, so a simple way to do this might be to wire five or so in parallel with a switch for each one. As the turbine output increases you can turn on more heaters to keep the load in the optimal range.You will need some math or testing to determine the optimal load for given wind conditions.</p>
| 52679 | Wind generator and heater (high power load) connected directly |
2022-09-30T19:01:33.583 | <p>This is of course inspired by the Nordstream gas pipelines in the Baltic Sea. When it was decided that Nordstream 2 would not start operating after construction finished they still filled it with natural gas at full pressure (a little above 100 bars iirc). When Nordstream 1 was shut down it was also kept full of natural gas at full pressure.</p>
<p>To a lay person this sounds like a reasonable thing to do, after all these pipes are meant to hold and transport natural gas. Then I read in a <a href="https://thelawdogfiles.com/2022/09/nordstream.html" rel="nofollow noreferrer">blog</a> that this apparently is a terrible idea because it will lead to hydrate plugs. Cleaning these out is very complicated and if you do it the wrong way or sloppy there is a high risk of an explosion. The sensible way to shut down a pipeline for a few months is to fully empty it of natural gas and then instead fill it with some inert gas like nitrogen. However I've never seen any talk about this in general news media (so far) and all the news about the Nordstream explosions went from unknown cause to clearly external sabotage within the last two days.</p>
<p>So when the Nordstream gas pipes were shut down, was filling them with natural gas a sensible thing to do or is this something where experts on operating gas pipes should have screamed warnings?</p>
| |gas|pipelines|compressed-gases|maintenance|natural-gas| | <p>The blog is wrong . It requires higher pressures and/or lower temperatures to make hydrates. Hydrates can be a problem at high pressure gas production wells; such as above 5000 psi. That high pressure must be reduced in one or more chokes to deliver less than 2000 psi gas for pipeline transport. There is substantial cooling produced by the expanding gas ; This can cause hydrates at the chokes. So the chokes are heated ( I don't know how ,never worked with it). The gas hydrate can remain solid up to about 70F with high pressure.( I am sure Wikipedia has more info for anyone interested). Nat gas would be the practical material to fill the pipe with because it is available. That quantity of nitrogen would be a logistics problem. Footnote ; there are unimaginable amounts of hydrate in the deeper areas ( high pressure ) of the oceans.</p>
| 52695 | If you shut down a gas pipeline for a few months would you fill it with gas at full pressure? |
2022-10-01T11:26:45.690 | <p>Engineering textbooks have graphs like these for finding the stress concentrations for plates with holes, fillets and what not.</p>
<p><a href="https://i.stack.imgur.com/J7BZB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J7BZB.png" alt="enter image description here" /></a></p>
<p>What should I do when I get a shape that does not have its stress concentrations listed in those graphs? Take for example this shape.</p>
<p><a href="https://i.stack.imgur.com/W3Rmi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W3Rmi.png" alt="enter image description here" /></a></p>
<p>It is so simple yet I cannot find any stress concentrations for a shape like it.</p>
| |stresses|solid-mechanics| | <p>While Tomas Letal answer is probably the best way, I would like to add, that another option if you are trying to do some hand calculations and you are not infront of a workstation (very unusual in the trade nowadays), is to use a worst case scenario.</p>
<p>E.g. for the object that you have in the image, you could use the maximum of the stress concentration of a circle with diameter equal to :</p>
<ul>
<li>the vertical width of the slot. Or,</li>
<li>twice the radius of the corner</li>
</ul>
<p>However, the best and easiest way is to use FEA, just make sure to check the mesh quality around the edges (usually you'd need to perform some sort of benchmarking with known cases).</p>
| 52699 | What should I do if there is no stress concentration for my shape? |
2022-10-02T04:46:28.917 | <p>I'm a software engineer working on theoretical things in my daily work. Lately I wanted to connect a pressure sensor, like <a href="https://rads.stackoverflow.com/amzn/click/com/B07WCPNTT9" rel="nofollow noreferrer" rel="nofollow noreferrer">this one</a>, to a digital device that I can program, so that I can constantly read pressure sensors output and then do stuff depending on what pressure data I read in.</p>
<p>I was thinking about going for the Arduino for that, but feel free to suggest alternatives, as I don't have prior experience with Arduino so far.</p>
<p>In the link above it only says "Output signal: analog". I have unfortunately no education in electrical engineering so my question is if can I connect such a sensor to Arduino (or similar devices)? Do I need to do some soldering?
I guess that there must be some kind of protocol that the output signal must adhere to (I have read Arduino support GPIO)?</p>
<p>Sensor part number in case link dies: RP-S40 Thin Film Pressure Sensor</p>
| |sensors|software|signal-processing| | <p>Yes, you can connect that device to an arduino. Try the left circuit in the second picture first (simple voltage divider). It's the simplest and should work ok unless you need high accuracy.</p>
<p>The device acts electrically as a resistor, and the output signal is the resistance. Some ballpark specs for this are provided, but if you want a lot of detail you need to buy legit parts from suppliers like digikey, not no-name stuff from amazon.</p>
<p>You should expect to do some soldering. It's almost always required when working with small electronics. I won't detail how to get started with that here, just search it, it's cheap, and not very hard.</p>
<p>An arduino is a good choice for getting started with microcontrollers. There is a lot of support available, and the ide, although simplistic works ok without a lot of toolchain setup.</p>
| 52709 | Process this output of an off-the-shelf pressure sensor |
2022-10-03T13:59:33.133 | <p>Given the following kinematic problem, how would one calculate the velocity <span class="math-container">$v_C$</span> of point C given a certain horizontal velocity <span class="math-container">$v_{E}$</span> in point E? Given is that the disk does not slip and cannot move in the vertical direction. Point E can also only move horizontally.</p>
<p><a href="https://i.stack.imgur.com/YStKN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YStKN.png" alt="Kinematic problem" /></a></p>
<p>My train of thought is as follows:
Calculate <span class="math-container">$v_D=v_E+\omega_{DE} \times r_{D/E}$</span>. Then use <span class="math-container">$v_D=v_C+\omega_{BD}\times r_{D/B}$</span>. With these two equations I could find <span class="math-container">$v_C$</span>. However, I cannot find the angular velocities. What am I missing here?</p>
| |dynamics|kinematics| | <p>You need one more equation <span class="math-container">$v_B=v_C+\omega_{BD}\times r_{C/B}$</span> and the fact that <span class="math-container">$\omega_{DE}=\dot{\theta }$</span>.</p>
<p>Now we have <span class="math-container">$v_B=v_C+\omega_{BD}\ \hat{k} \times R_3\ \hat{j}=v_C-\omega_{BD}\ R_3\hat{i}$</span> and since <span class="math-container">$v_B=0$</span>, we get <span class="math-container">$v_C=\omega_{BD}\ R_3\hat{i}$</span>.</p>
<p>For D we have, <span class="math-container">$v_D=v_C-\omega_{BD}\ \hat{k} \times R_3\ \hat{j}=v_C+\omega_{BD}\ R_3\hat{i} = 2\ \omega_{BD}\ R_3\hat{i}$</span></p>
<p>Thus <span class="math-container">$$v_C=v_D/2$$</span></p>
<p>Using <span class="math-container">$v_D=v_E+\omega_{DE} \times r_{D/E}$</span>, we get <span class="math-container">$v_D=v_E\hat{i}-\dot{\theta} \hat{k} \times \{-L_{DE} \cos\theta\hat{i}+L_{DE} \sin\theta\hat{j}\} = v_E\hat{i}+\dot{\theta}L_{DE}\cos\theta\hat{j}+\dot{\theta}L_{DE}\sin\theta\hat{i}$</span></p>
<p>Thus <span class="math-container">$$v_C=\frac{1}{2}(v_E\hat{i}+\dot{\theta}L_{DE}\sin\theta\hat{i}+\dot{\theta}L_{DE}\cos\theta\hat{j})$$</span></p>
<p>(For the points C and D to remain at the constant height with no vertical motion, either <span class="math-container">$\theta=90 {}^{\circ}$</span> or <span class="math-container">$\dot{\theta}=0$</span>.)</p>
| 52720 | Kinematics of actuated disk with no slip |
2022-10-04T04:51:48.313 | <p>What are the design constraints to the diameter of a mine shaft in bedrock, considering the pressure of the ambient rock, its compression strength, the width of the liner, and its compression strength? The ambient rock of the Kola borehole was under such great pressure and strained so easily because it was so hot that it just spontaneously sealed itself. How could one calculate whether that will happen?</p>
| |stresses|geotechnical-engineering|mining-engineering| | <p>It depends on the rock and bore size. If I remember right chalk is a pain for flowing into the bore ,also salt. Any serious hole will use steel casing , likely a few concentric strings. Steel casing strength to 125 ksi is standard and 150 ksi is optional. Welded line pipe is an option for diameters over about 3ft. Then strength will be up to about 80 Ksi . Wall thickness will depend on the manufactures capability. I presume cost is no concern. Collapse strength of steel casing is calculated using API bul. 5C3. There are also other techniques like two concentric casing strings with cement filling the annulus.</p>
| 52729 | What is the largest diameter a vertical mine shaft may have in bedrock? |
2022-10-04T05:59:40.100 | <p>I am going through a PID in which the line size was increased from 2 inch to 4 inch for temperature measurement and then decreased to 2 inch using two reduces. What is the need for reducers before temperature measurement?</p>
| |temperature|piping|pi-diagram| | <h3>You do not need larger pipes to measure fluid temperatures.</h3>
<p>The linked article from your question comments wants a larger diameter to fit "standard" attachment methods. You can't drill a 1 inch hole into a 1 inch pipe, but a T would work fine. You can get into whether or not a thermo well is required. The learning here is don't just trust randomly sourced information about engineering on the internet. I promise you can measure temperature in a 2 inch pipe without making the pipe bigger, and that you can increase the diameter of a pipe with or without multiple reducers.</p>
<p>Maybe the article was written by a pipefitter instead of a welder so they only think in terms of reducers. Maybe your site had a supply of thermo wells and needed to use them, or maybe the fluid requires it. Maybe your original installer and designer was working cost-plus. Certainly why engineers did things the way they did in the past has baffled most of us at one time or another. Looking at the pipe instead of the diagram might help (remember that the map is not the territory).</p>
<p><a href="https://www.bapihvac.com/application_note/using-thermowells-in-small-pipes/" rel="nofollow noreferrer">https://www.bapihvac.com/application_note/using-thermowells-in-small-pipes/</a></p>
| 52730 | Temperature measurement in a process pipe |
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