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2019-09-11T07:06:11.373 | <p>Recently I noticed a concrete <a href="https://en.wikipedia.org/wiki/Speed_bump" rel="nofollow noreferrer">speed bumper</a> created near my parking. On enquiring the purpose, the committee who took the decision told me that the purpose of the bumper is to safeguard an existing pipeline below the bumper from heavy vehicles. Some significant points:</p>
<ol>
<li>Existing road in the complex is made up of concrete and is plain. The road has not suffered any damage from any heavy vehicle in past.</li>
<li>The pipe is made up of polypropylene or some similar material. The diameter of pipe is about 6 inch. It is there to direct the water collected from roofs to rainwater harvesting reservoir.</li>
<li>Speed bumper is solid and made up of concrete.</li>
<li>Heavy vehicle here is a water tanker truck with capacity in multiples of kilolitres (Usually 5KL) filled to the top with water.</li>
</ol>
<p><strong>Question</strong><BR> Is it necessary to raise such a speed bump to prevent any harm to pipeline below concrete plain road? Is it helpful, has no effect or worse in avoiding impact on the pipeline below?</p>
| |civil-engineering|structural-analysis|structures| | <p>A speed bump not only does not help to protect a submerged pipe, but it may also even damage it more because of the impact of heavy truck axle hitting and falling down from the bump after it just has crossed it.</p>
<p>An axel impact on hitting a flat landing area after dropping the face of a speed bump can be several times the weight of the axel. While an axel just passing on, will push the road just barely more than its weight depending on its speed and tire air pressure and some other small factors. We all experience the discomforting jerk on the suspension when we dismount a bump. Some people as a reflex brake partially after the tire rolls down the bump which makes things worse because of the car balking. </p>
<p>The only way that something looking like a speed bump can help is when it spans the pipe, not installed before it and has been actually engineered to work as a bridge such as a bilaterally tapered concrete slab designed with foundations saddling the pipe and adequate rebars. </p>
| 31093 | Can a speedbump safeguard a pipeline located below? |
2019-09-11T12:47:10.833 | <p>How is the allowable number of axle-load to failure equation actually used in the software? Is there any documentation regarding the calculation of the tensile strains?</p>
| |pavement| | <p>The allowable number of axle-load repetitions to failure, <span class="math-container">$N_f$</span> is used to calculate the incremental fatigue damage using Miner's law for each analysis interval (which is equal to one month, or two 15-day intervals if there is a freezing cycle in the month). ME pavement design procedure considers the following inputs in the damage calculation process:</p>
<ul>
<li>Truck traffic: Different FHWA truck classes, different axle loads, tire configurations (single, tandem, tridem and quad axles), hourly and monthly adjustment factors, etc. </li>
<li>Temperature variation within the pavement structure as a function of time. This is used to calculate asphalt mixture stiffness <span class="math-container">$E_{HMA}$</span> which is used to calculate <span class="math-container">$N_f$</span>.</li>
</ul>
<p>Pavement ME calculates the horizontal tensile strain at the bottom of AC layer (for bottom-up fatigue cracking cracking) and at the AC surface (for top-down fatigue or longitudinal cracking) using JULEA (linear elastic analysis procedure). Within each damage computation interval, the strains are calculated for each <span class="math-container">$i^{th}$</span> group of axle loads (axle passes, <span class="math-container">$n_i$</span>) at critical locations that are defined separately for different axle types. These strains are then used to calculate allowable load repetitions <span class="math-container">$N_{f,i}$</span> for the <span class="math-container">$i^{th}$</span> load group, and damage <span class="math-container">$D_i = n_i / N_{f,i}$</span>. Cumulative damage due to all traffic is the sum of incremental damage caused by each load group in all analysis intervals.</p>
<p><strong>References</strong></p>
<ol>
<li>National Cooperative Highway Research Program, “Guide for Mechanistic-Empirical Design of New and Rehabilitated Pavement Structures, Part 3, Chapter 4 - Design of New & Reconstructed Flexible Pavements.” NCHRP Project 1-37A Final Report, Washington, D.C. (2004). [Contains definition of critical strain computation locations]</li>
<li>American Association of State Highway and Transportation Officials, “Mechanistic-Empirical Pavement Design Guide - A Manual of Practice”, 2nd edition, Washington D.C. (2015)</li>
</ol>
| 31097 | What is the "allowable number of axle-load to failure equation" used in Pavement ME Design? |
2019-09-11T13:08:25.853 | <p>There's basically two ways train wheels can operate, the flanges could be either "inner" or "outer".</p>
<p><a href="https://i.stack.imgur.com/8MhrA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8MhrA.png" alt="Inner- and outer-style wheels"></a></p>
<p>Switches can be made in equivalent ways for inner and outer flanges.</p>
<p><a href="https://i.stack.imgur.com/oPBa0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oPBa0.png" alt="enter image description here"></a></p>
<p>We could expect that, just like some countries, as well as different railway companies within countries, picked up different rail gauges, they could have also varied between inner- and outer-flanges railways.</p>
<p>We could imagine that, just like today some countries have left-hand and right-hand traffic, there would be inner and outer style rail wheels. Except this is not the case, the "inner" style is almost universal. Almost, because the "outer" style indeed did exist in railways' debuts.</p>
<p><a href="https://i.stack.imgur.com/ALZf7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ALZf7.jpg" alt="Richard Trevithick railway (England)"></a></p>
<p>Is there a technical reason that makes the "inner" style preferable ?</p>
| |mechanical-engineering|rail|transportation| | <p>If you leave off the flanges and just use cones that are solidly connected to an axle, then cones that taper outward (get <em>smaller</em> as they go out) tend to be self-centering but undamped (or underdamped, I'm not sure which).</p>
<p>This is because if the axle is offset, the wheel that's to the outside has a larger diameter, and tends to drive the axle to turn toward the center. The "outer flange" design you should would do the opposite.</p>
<p>So the "inner flange" picture that you present is mostly not relying on the flanges to keep the axle centered -- it's the taper of the wheels that does that. The flange is there to damp oscillations, and for when the system is overwhelmed by circumstances.</p>
<p>(There's a nice video of this on YouTube, but I'm too lazy right now to find it -- try searching on "train tire design", or maybe those words with the word "stability" tossed in).</p>
| 31098 | Is there any reason why "inner-flange" style trains are almost universal? |
2019-09-12T13:36:28.513 | <p>I come from Maths background, but recently I encountered problem where a channel is described as "flat fading channel". I would like to make some conclusions based on that fact. I suspect that for radio engineers this is self explanatory, but not for me. Do we prefer a channel to be flat fading vs selective fading? Can we even make the distinction here? If no what are the pros and cons of each?</p>
| |electrical-engineering| | <p>The names stem from the fact that communications engineers tend to think in the frequency domain.</p>
<p>A flat fading channel is one where the entire signal is scaled by the fading, i.e. the signal is multiplied by a gain <span class="math-container">$a(t)$</span> which (usually) varies more slowly than the bandwidth of the channel.</p>
<p>A selectively fading channel is one where some time-varying, frequency-selective process is at work. As an example of a baseband channel, this would be as if you're talking on the phone, and the frequencies from 1000Hz to 1200Hz were being randomly notched out and added back in.</p>
<p>Flat fading is easier to deal with, because the whole signal is unmolested other than being changed in amplitude.</p>
<p>Frequency selective fading is much harder to deal with (although using OFDM would make it easier), because the impulse response of the channel is constantly changing, so not only is the signal distorted, but the equalizer that you'd need to reassemble it is constantly changing.</p>
| 31111 | Flat Fading vs Selective fading, which one is better? |
2019-09-12T19:43:28.410 | <p>What are the likely causes of a stepped crack up and away from the edge of a lintel? See the diagram below.</p>
<p><a href="https://i.stack.imgur.com/EbDhn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EbDhn.jpg" alt="enter image description here"></a></p>
| |civil-engineering|masonry| | <blockquote>
<p>What are the likely causes of a <strong>stepped</strong> crack <strong>up and away</strong> from the edge of a lintel?</p>
</blockquote>
<p>Shearing causes stepped cracks and bending causes vertical cracks.</p>
<p><a href="https://i.stack.imgur.com/2hb6Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2hb6Y.jpg" alt="Shear vs. Bending Cracking"></a></p>
<p><a href="https://i.stack.imgur.com/SNei3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SNei3.jpg" alt="Shear Cracking Remediation"></a></p>
<p>See: "<a href="https://gharpedia.com/diagonal-cracks-brick-walls/" rel="nofollow noreferrer">Diagonal Cracks in Brick Walls and Its Measures</a>" or "<a href="https://www.helifix.co.uk/products/remedial-products/helibar-remedial/" rel="nofollow noreferrer">HeliBar Remedial</a>" where the causes and methods for remediation are explained. Two different causes and means to prevent worsening of the problem.</p>
| 31119 | Cause of stepped crack up and away from lintel |
2019-09-13T00:40:59.980 | <p>On a recent trip to Cape Elizabeth, Maine, I came across this fog signal:</p>
<p><a href="https://i.stack.imgur.com/utCFA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/utCFA.jpg" alt="photo of Cape Elizabeth fog signal"></a></p>
<p>It appears to have an electromagnetic diaphragm in the middle, surrounded by concrete and some very curiously shaped steel pieces. Why is the steel surrounding the horn shaped the way it is?</p>
<p>I've found some material on fog signal design, but nothing looks remotely close to what I see here.</p>
<h2>Appendix:</h2>
<ul>
<li><a href="https://uslhs.org/light_lists/lighthouse_list.php?id=43" rel="nofollow noreferrer">US Lighthouse Society - Cape Elizabeth (last updated 2004)</a></li>
<li><a href="https://www.us-lighthouses.com/images/archive/c/cape-elizabeth-lighthouse-and-fog-signal.jpg" rel="nofollow noreferrer">https://www.us-lighthouses.com/images/archive/c/cape-elizabeth-lighthouse-and-fog-signal.jpg</a></li>
<li><a href="https://en.wikipedia.org/wiki/Cape_Elizabeth_Lights" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cape_Elizabeth_Lights</a></li>
</ul>
| |acoustics| | <p>The steel structure will be designed as a diffraction grating to create a directional sound beam.</p>
<p>To make a directional sound source, the area covered by the source needs to have linear dimensions several times bigger than the wavelength of the sound, and at low frequencies that means tens of meters. It is impractical to make the actual sound producing device as large as that, but the steel structure "guides" the sound to make the complete structure equivalent to a single loudspeaker cone the same size as the whole frontal area of the device.</p>
<p>The same principle is used on a smaller scale when creating loudspeaker "stacks" at rock concerts, etc. A tall narrow "stack" of speakers generates a sound field which covers a wide area horizontally (i.e. a wide enough angle to cover the whole of the audience) but a focussed beam in the vertical direction, so you are not wasting energy on sound radiating up into the air, or down into the ground close to the speaker location, but radiating all the sound "horizontally" into the audience area.</p>
<p>Incidentally this is also why high-frequency loudspeakers are <em>small</em> (only a few mm in diameter,) so that they do <em>not</em> produce a directional beam of sound that can only be heard if you are positioned directly in front of the speaker.</p>
| 31124 | Why is the Cape Elizabeth Fog Signal shaped like this? |
2019-09-13T19:02:51.580 | <p>In v2.3 of the software, we noticed that in the rehabilitation flexible calibration tab, there are no standard deviation inputs for thermal cracking. They do appear in the new flexible rehabilitation tab. Was standard deviation for thermal cracking removed for flexible rehab designs? If so, when and why?</p>
| |pavement| | <p>The thermal cracking model in Pavement ME Design prior to v2.3 only considered the predicted transverse cracks for the overlay layer and not the transverse cracks that reflected through the overlay layer that originated in the existing layer. The standard deviation equations which correspond to the transverse cracking model only, were removed from the rehabilitation design calibration file when the new reflection cracking model was implemented (v2.3). </p>
<p>Currently, the software calculates the total transverse cracking which a function of the reflected transverse cracking from the existing pavement and the thermal/transverse cracking from the new overlay layer. Therefore, the standard deviation equation in the software is now based on the total thermal/transverse cracking value instead. </p>
| 31135 | Pavement Mechanistic-Empirical Design application (v2.3) - Rehabilitation Flexible Calibration Factors |
2019-09-14T06:00:53.090 | <ol>
<li><p>When I start the car, why do I hear the engine working when I'm not pressing the gas pedal?</p></li>
<li><p>How exactly does the torque and power of the gears work?
<br>
Torque = Radius * Force
<br>
Power = Torque * Speed
<br>
Do the different gears have different torque? They should, because they have difference radiuses, right? I'm a bit confused on how torque and power relates to the torque and what happens when we switch gears.</p></li>
<li><p>How automatic transmission cars detect what torque to put into the wheels and how do they put it?</p></li>
<li><p>How electric cars detect what torque to put into the wheels and how do they put it?</p></li>
</ol>
| |mechanical-engineering|gears|torque|car| | <p>The noise you hear is the engine running at the idle speed. To provide minumu energy for lubrication, charging the battery, air conditiong, water pump etc. In newer cars it has been eliminated for duration of intermittent stops to save fuel and pollute less. The computer will start the engine the moment you release the brakes.</p>
<p>The gear box, manual and automatic, function is to extend the range of engine torque. Because gasoline engines normal RPM is not spread enough to cover all the torque demand of the car.</p>
<p>The gears torque has to do with their arrangement in the gear box and as you said to the ratio of their RPM to that of the engine crankshaft. A gear turning 1/2 times faster provides 2 times more torque.</p>
<p>The automatic transmissions used to be controlled mechanically by the engine RPM, carburetor vacuum and rear axel RPM which has to do with the speed of the car.</p>
<p>But modern transmissions past 90s are controlled by a computer with many sensors, such as the air and engine temperatures, road condition, humidity, altitude, etc. </p>
<p>Electric cars, Tesla namely, don't have a transmission. They have all the torque range needed because they have wide range of RPM up to 20kRPM.</p>
<p>They have a computer which converts the direct current of the battery to alternative 3 phase current whit variable frequency, read variable RPM, via an inverter. Completely controlled by the master computer and in many cases even override the driver to correct his or her errors. </p>
| 31137 | A few questions about cars |
2019-09-14T09:43:48.783 | <p>The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs 6000 lb.</p>
<p><a href="https://i.stack.imgur.com/YiQXT.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YiQXT.gif" alt="enter image description here"></a></p>
<p>When using the equilibrium of moment equation at point C, The Tension in the string is found to be 2957.13 lb.</p>
<pre><code>6000(5) = 5(T) + 10(T)*(3/(34)^(1/2))
</code></pre>
<p>I tried to use the equation of equilibrium for Vertical Forces. </p>
<pre><code> 6000 = T + T(3/(34)^(1/2))
T = 3961.714 lb
</code></pre>
<p><a href="https://i.stack.imgur.com/GJrUC.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GJrUC.gif" alt="enter image description here"></a></p>
<p>Is there any mistake in applying the equilibrium vertical condition? I'm having trouble finding any faults. The diagram for forces is given below</p>
| |stresses| | <p>You're on the right track but as noted by commenter Sam, the free body diagram for beam AC should include the support reactions from the pin at point C.</p>
<p><a href="https://i.stack.imgur.com/05nKbl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/05nKbl.jpg" alt="Free Body Diagram of Beam AC"></a></p>
<p>Then, the general procedure using the equations of equilibrium will be:</p>
<ol>
<li>Sum moments about point C to solve for cable tension T
<span class="math-container">$$\Sigma M_C = -T(5m) - T \frac{3}{\sqrt{34}}(10m) + W(5m) = 0$$</span></li>
<li>Sum forces in vertical direction to solve for the vertical reaction at point C, (Cy)
<span class="math-container">$$\Sigma F_y = T + T\frac{3}{\sqrt{34}} - W + C_y = 0$$</span></li>
<li>Sum forces in the horizontal direction to solve for the horizontal reaction at point C, (Cx)
<span class="math-container">$$\Sigma F_x = T\frac{5}{\sqrt{34}} - Cx = 0$$</span></li>
</ol>
<p>Note that I just assumed directions for Cx and Cy in the sketch. If the equilibrium equations produce a negative value for either Cx or Cy it simply means the force acts in the opposite direction to the one I assumed.</p>
| 31140 | Calculating Forces using equations of equilibrium |
2019-09-14T13:07:52.433 | <p>I was studying a book on control systems where I read that a block diagram is a specialized high-level type of flow chart.</p>
<p>What is the difference between a normal flow chart and a block diagram?</p>
| |control-engineering|control-theory|process-engineering| | <p>The ISO EN 10628-1 has this example for a basic flow chart (which block diagram may also refer to):</p>
<p><a href="https://i.stack.imgur.com/qIypT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qIypT.png" alt="block diagram"></a></p>
<p>The blocks can be unit operations, processes, or parts of a plant, the lines can be material or energy flows.</p>
<p>An example process flow chart looks like this:</p>
<p><a href="https://i.stack.imgur.com/u5OMf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u5OMf.png" alt="enter image description here"></a></p>
<p>The "block" are no longer blocks but symbols for specific devices, amounts/flow rates should be specified at the lines. Important valves etc. may be drawn.</p>
<p>And the next level of detail would be a P&ID:</p>
<p><a href="https://i.stack.imgur.com/DBmM7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DBmM7.png" alt="enter image description here"></a></p>
<p>Individual devices inlcuding actutators, names/numbers for individual components, sensors, pipe class of piping etc.</p>
<p>At the end of the day, you have to ask yourself what you want to say with your drawing and what level of detail is required, either by your contract or by further work you do based on the diagram. The above hopefully helps with navigating this.</p>
| 31141 | Block diagram vs flow chart? |
2019-09-15T06:39:47.037 | <p>I am trying to figure out the lifting mechanism of a manual Crank Adjustable Height Standing Desk. Please see the image below:</p>
<p><a href="https://i.stack.imgur.com/MQBol.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MQBol.jpg" alt="enter image description here"></a></p>
<p>My question is what it is inside the metal frame (check the red marked area)
that is causing the entire system go up and down when I move the crank? </p>
| |mechanical-engineering|hydraulics|mechanisms|actuator| | <p>the crank handle is connected to the little black box, inside of which there are gears that rotate the metal shaft you can see running through the box and across the back of the table support. the right hand end of that shaft enters the support leg on the right which has another gearbox in it that rotates a shaft leading down the length of the support leg. That shaft has threads cut into it which engage a nut fixed to the lower portion of the support leg. rotating the crank handle then rotates the threaded shaft inside the leg which causes that shaft to push up or pull down on the upper half half of the support leg via the bearings inside the gearbox that is inside the top of the leg. This raises or lowers the table. </p>
| 31150 | Mechanism of a manual Crank Adjustable Height Standing Desk |
2019-09-15T20:53:08.457 | <p>I was trying to look for an L/D ratio published everywhere on the web and I can't seem to find it (for a project). Since I couldn't find it, I wanted to try calculating myself. Can't seem to find <span class="math-container">$C_l$</span> or <span class="math-container">$C_d$</span> though without some sort of paywall on research papers.</p>
<p>The equations are:</p>
<p>L = <span class="math-container">$\frac{1}{2} C_{l} \cdot \rho v^2 A$</span></p>
<p>D = <span class="math-container">$\frac{1}{2} C_{d} \cdot \rho v^2 A$</span></p>
<p>So L/D would just be:</p>
<p><span class="math-container">$\frac{L}{D} = \frac{C_l}{C_d}$</span></p>
<p>or would the effective area be different for L and D, where L is the area of the top/bottom view of the plane, and D is the area of the front/back view?</p>
<p>The only quantities I have are:</p>
<p>v = Mach 0.92 = 315.62 <span class="math-container">$\frac{m}{s}$</span>..............................top speed</p>
<p><span class="math-container">$\rho$</span> = 0.003996 <span class="math-container">$\frac{kg}{m^3}$</span>....................density of air at 40,000 ft (assuming it's flying at its maximum altitude)</p>
<p>I couldn't really find any area of the plane; only wingspan and vertical length, but that would give an inaccurate area of the plane. How would I determine this? If <span class="math-container">$\frac{L}{D}$</span> is just <span class="math-container">$\frac{C_l}{C_d}$</span> does anyone happen to know <span class="math-container">$C_l$</span> and <span class="math-container">$C_d$</span>? Thank you in advance.</p>
| |aerospace-engineering|aircraft-design|drag| | <p>Glide ratio, or glide angle meaning when you trim the plane to glide level with the engines off, is numerically equal to L/D ratio. </p>
<p>As a start guess I would say with a wing load of 330kg/m2 and very small wing aspect ratio it shouldn't be more than 12- 14.</p>
<p>This site has more info on F-117 Nighthawk. <a href="https://jackryan.fandom.com/wiki/F-117_Nighthawk" rel="nofollow noreferrer">Jackryan fandom</a></p>
| 31158 | How do you determine the L/D ratio of F-117 Nighthawk? |
2019-09-15T08:37:29.470 | <p>I made a gimbal at home with PVC. Used some cement filled PVC pieces as hanging weight to make the phone steady. But when I shoot with it, I feel some tilting movements in the video due to the momentum of these weights. How can I reduce this?</p>
<p>Is it really possible to make a gimbal at home which can produce some decently stabilised videos? <a href="https://i.stack.imgur.com/IBNjH.jpg" rel="nofollow noreferrer">image of gimbal</a></p>
| |mechanical-engineering| | <p>I made one for my son's camera.</p>
<p>The issue is usually with the bearings. I used 2 pairs of bearings taken from those "spinners" and put them on a threaded rod.</p>
<p>Used double nuts to "just" load the bearings to provide some damping and he had good results when he was taking videos while skiing...</p>
| 31159 | How to reduce tilting in self-made gimbal |
2019-09-16T15:15:22.330 | <p><a href="https://i.stack.imgur.com/8eMKJm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8eMKJm.png" alt="enter image description here"></a></p>
<p>A pinned system comprising bars of the same material and cross-sectional area is subjected to a vertical load P.</p>
<p>I am required to form a relationship between F<sub>AB</sub> and P.</p>
<p><strong>The first part of the solution is as such,</strong></p>
<p><a href="https://i.stack.imgur.com/detT3m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/detT3m.png" alt="enter image description here"></a></p>
<p><span class="math-container">$cos\theta=\frac{\delta_{AB}}{\delta_{AC}}$</span></p>
<p><strong>But how is the angle <span class="math-container">$\theta$</span> the same after deflection? Shouldn't it be smaller?</strong>
And I can't seem to form a relationship without assuming the theta as the same.</p>
| |mechanical-engineering|structural-analysis|structures|beam|deflection| | <p>This is not a statically determinate system. To find how the load splits between the members, start by finding their stiffness.</p>
<p>Since the materials and cross-section areas are the same, the stiffness is inversely proportional to the length.</p>
<p>Length of AC = <span class="math-container">$L\cos\theta$</span>.
If the stiffness of AB is <span class="math-container">$K$</span>, the stiffness of AC is <span class="math-container">$K/\cos\theta$</span>.</p>
<p>Now consider a small downward displacement <span class="math-container">$x$</span> at point A. The length of AB changes from <span class="math-container">$L$</span> to <span class="math-container">$$\sqrt{L^2\sin^2\theta+ (L\cos\theta + x)^2} \approx \sqrt{L^2 + 2Lx\cos\theta}\\ \approx L + x\cos\theta$$</span> to first order in <span class="math-container">$x$</span>, using the Binomial theorem .</p>
<p>So the tension in AB is <span class="math-container">$Kx\cos\theta$</span> and the downward component of the tension is <span class="math-container">$Kx \cos^2\theta$</span>.</p>
<p>The tension in AC is <span class="math-container">$Kx/\cos\theta$</span>.</p>
<p>So we have <span class="math-container">$P = 2Kx\cos^2\theta + Kx/\cos\theta$</span> and <span class="math-container">$F_{\text{AC}} = Kx/\cos\theta$</span>.</p>
<p><span class="math-container">$P = F_{\text{AC}}(2cos^3\theta + 1)$</span>.</p>
<p>From the diagram <span class="math-container">$\cos\theta = \sqrt{3}/2$</span>, so <span class="math-container">$P = F_{\text{AC}}(3\sqrt{3}/4 + 1)$</span>. </p>
<p>i.e. <span class="math-container">$F_{AC} = 0.435\,P$</span>.</p>
| 31169 | Pin joint system displacement question |
2019-09-16T19:33:41.157 | <p>What kind of automatic transmission would be the best to connect an electric engine with the rear axle of a RWD car?</p>
<p><a href="https://i.stack.imgur.com/NFGZ1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NFGZ1.png" alt="enter image description here"></a></p>
<p>Specifications:</p>
<ul>
<li>200HP, 3000rpm electric motor</li>
<li>top speed 220km/h</li>
</ul>
<p>The automatic transmission should have higher output RPM and be able to bring the car to top speed of 200km/h.</p>
| |gears|electric-vehicles|transmission| | <p>The essential question here is why or why not use a transmission in an electric car. Here are the underlying issues. </p>
<p>It is commonly asserted that a DC electric motor is a constant-torque device, developing the same torque at standstill (0 RPM) that it does at its rated RPM (in this case, 3000). Advocates of the "no-transmission" position cite this as a reason why electric motors in cars do not need transmissions, but instead can be used in direct-drive mode.</p>
<p>But the flaw in this reasoning is that power- the capacity to perform work at a certain rate- is the product of torque <em>and RPM</em>, and it is this <em>product</em> which accelerates the car. </p>
<p>This means that in the case of a motor which is starting from zero RPM to accelerate a car, its <em>power output</em> at near-zero RPM is <em>near zero</em>, and the motor does not develop its rated power output <em>until it is running at its rated RPM</em>. </p>
<p>This means that to maximize the power output of the motor and thereby maximize the car's acceleration, we need to get the motor all the way up to 3000RPM as quickly as possible, which means <em>gearing the motor down</em> when starting off from zero speed. </p>
<p>Then, when we reach 3000RPM, we <em>upshift</em> the transmission to a "taller" gear ratio, and shift again when the motor comes up to 3000RPM, and repeat until the drag force on the car body is equal to the force applied to the pavement by the rear wheels. this process is designed to keep the motor running at or near its maximum power point for as long as possible. </p>
<p>In fact, the fastest acceleration will be had if the transmission is capable of holding the motor right at 3000RPM throughout the car's acceleration from zero wheel speed to whatever its top speed is. This means that to <em>maximize acceleration</em>, a transmission is necessary, and the best one will be a continuously-variable one that locks the engine speed at 3000RPM at all times.</p>
| 31172 | Electric engine and transmission |
2019-09-17T10:28:28.023 | <p>The web of a H-section column has been completely corroded at the air-moisture interface with the concrete encasement leaving a hole in the web. The proposal is to break out the existing concrete and weld plates onto the web that will bridge over the hole in the web and will be welded on all four sides to good steel.</p>
<p>The plates will be on both sides of the web. If the web is 8 mm what should the thickness of the plates be? At a minimum should they be 4 mm to reinstate the loss of section?</p>
| |structural-engineering|civil-engineering|steel|welding|corrosion| | <p>You need to shore the structure first. Welding heat will lower the strength of the steel.</p>
<p>then you should use needle gun and grinder to bring the corroded material to bright metal.</p>
<p>The patch plates should be thicker than 1/2 of the web thickness. H.S bolts are preferred method of fastening, but if welding is used it should be intermittent and done in alternate runs to avoid heat strains.</p>
<p>The job must be primed and sealed before newly removed concrete is replaced with high strength repair concrete with proper acrylic admixture for strength and curing. </p>
| 31185 | Plate thickness for repair of completely corroded web |
2019-09-17T11:46:39.047 | <p>Should I consider them as a cylinder to find their inertia? Any tips will be helpful.</p>
| |gears| | <p>The thread's mass is half of the mass of a cylinder with a thickness of the difference between inner and outer. It's CG is 1/3 out from inner circle. 1/3*1/2= 1/6. </p>
<p>If you calculate the I of a cylinder with a radius R = r inner + 1/6 thread depth, you're fine.</p>
| 31186 | How to calculate the moment of inertia of a worm gear? |
2019-09-18T12:46:38.240 | <p>A client wishes to use an existing steel beam to lift a 1 tonne load using a sling. The 3 m long beam is measured as UB 203x133 visually observed at both ends to be secure with end-plates with all four bolts present.</p>
<p><strong>calculation</strong></p>
<p>For conservatism assume the lowest credible strength steel i.e. S275 steel and the lightest section i.e. UB 203 x 133 x 25. From the <a href="https://www.steelforlifebluebook.co.uk/ub/ec3-ukna/buckling-resistance-s275/" rel="nofollow noreferrer">Blue Book</a> for <span class="math-container">$L_{ef}$</span> = 3 m, the design moment capacity <span class="math-container">$M_{Rd}$</span> = 54.6 kNm (<span class="math-container">$c_1$</span> = 1.00).</p>
<p>There is no dead load on the beam. Use partial safety factor <span class="math-container">$\gamma$</span> = 1.6 for imposed load.</p>
<p>Design load <span class="math-container">$F_{Ed}$</span> = 10 kN x 1.6 = 16</p>
<p>By inspection beam is OK in shear. All 4 bolts present in each end plate so connection is OK. Check beam in bending.</p>
<p>Treat the beam as simply supported and assuming sling midspan:</p>
<p><span class="math-container">$M_{Ed} = \frac{PL}{4} = \frac{16 \times 3}{4} = 12$</span> kNm < 54.6 kNm <span class="math-container">$\therefore$</span> lift OK</p>
<p>Please provide feedback on this assessment with a criticism or suggestions for improvement.</p>
| |structural-engineering|civil-engineering| | <p>If you use the sling with a winch you have to consult the winch's manufacturer instructions usually printed on a tag or side of the winch.</p>
<p>Usually, they recommend a dynamic load factor of 3 to 4 which will make your factored load moment 40kN but still less than allowable 54.6kN.</p>
<p>A ton is 9.81kN, but rounding up is ok.</p>
<p>the rest of your work seems to be ok.</p>
| 31201 | Assessment of steel beam for lifting to Eurocode 3 |
2019-09-18T19:03:34.297 | <p>I want to set up steel truck fenders production line, but have no idea how to do that. Something like this picture.</p>
<p>Edit1:
I have two plans to form final curved shape , using a hydraulic press or using a customized rolling machine .
Final products must have no riples .</p>
<p><img src="https://i.stack.imgur.com/DD7aB.jpg" alt="enter image description here"></p>
| |mechanical-engineering|structural-engineering|metal-folding| | <p>Your raw material will come in the form of a large spool of sheet metal containing a strip of steel hundreds of meters long. You will need an <em>unspooler</em> to feed steel off the spool and a <em>straightener</em> to take the curvature out of the steel strip. Then you need either a <em>punch press</em> or a <em>shear</em> to cut the sheet metal blanks to size and trim their corners. To put ripples or corrugations into the cut blanks, you will need a <em>roller die</em> or a <em>rolling mill</em>. To put a folded edge onto the sides of the blank you will need either a <em>bending brake</em> or a set of <em>progressive dies</em>. To bend the fenders into their final curved shape you will need a <em>sheet metal press</em> and to smooth out any resulting ripples you will need an <em>ironing press</em>. </p>
<p>To attach the mounting brackets you will need an <em>electric spot welder</em>. Finally, you will need either an <em>electrostatic spray booth</em> to paint them or a <em>powder coating rig</em>. </p>
| 31203 | How to form steel truck fenders |
2019-09-19T07:34:28.933 | <p>I have an <code>.inp</code> file of Ansys that is basically a mesh, that could be imported into Ansys APDL and converting to <code>.cdb</code> file.</p>
<p>But I also have another <code>.IGES</code> file that I should import in the same model and assemble on the first one. How do I do that ?</p>
| |finite-element-method|ansys|ansys-workbench|ansys-apdl| | <p>You could import the <code>.inp</code> file containing your mesh into <em>Ansys Mechanical APDL</em>, then import your <code>.iges</code> as well (through <code>File/Import/IGES</code> or sometimes <code>Archive Model/Read</code> in <code>/prep7</code> works smoother), give your attributes to the different areas, mesh them, and export the whole model in <code>.cdb</code>.</p>
| 31207 | How do I import an .inp and an IGES together into Ansys workbench? |
2019-09-19T07:58:30.570 | <p>I found this bouncy ball in my garage. It is too deflated to use and too inflated to store, so I would like to inflate it or deflate it. It doesn't have a standard valve for a bike pump: the only break in the plastic shell is this solid white plug that I was unable to remove.</p>
<p><a href="https://i.stack.imgur.com/rZ8HO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rZ8HO.jpg" alt="Bouncy ball"></a></p>
<p><a href="https://i.stack.imgur.com/eLk5Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eLk5Y.jpg" alt="White plug on the bouncy ball"></a></p>
<p>What is this plug, and can I inflate or deflate the ball?</p>
| |pumps|valves| | <p>Yes, normally you can pull those plugs out - takes a strong grip or « folding » the ball to be able to push on the inner end of the plug...</p>
<p>Here is a photo with such a plug taken out:</p>
<p><a href="https://i.stack.imgur.com/w9wXX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w9wXX.jpg" alt="Photo of plug out of the ball"></a></p>
| 31208 | How to inflate or deflate a bouncy ball that lacks a standard valve? |
2019-09-20T02:05:05.583 | <p>I'm new to the FEA world and trying to learn the ropes through resources available online. I've imported a model into LISA FEA that I would like to analyze, however, when I press solve to conduct my analysis I'm getting an error message that states the following:</p>
<blockquote>
<p>Error: Element 1 is collapsed. More than one local node share the same local node.
Warning: Elements 395, 400, ... overlap each other.
Failed</p>
</blockquote>
<p>Being so new to this, I do not have any idea as to how to interpret this. M question is: how can I tweak my model to get rid of these error messages. Namely, how do I "uncollapse" an element? How do I stop elements from overlapping?</p>
<p>Although this is a LISA specific question, and I would appreciate advice as to how to overcome this directly through LISA, I am also open to general advice applicable to other types of FEA. If anyone could give me a few pointers as to how this issue is dealt with in FEA software, I would be greatly appreciative.</p>
| |finite-element-method|software| | <p>I don't use LISA, but the most likely cause of both problems is that the geometry of the element mesh is invalid.</p>
<p>The "collapsed element" message is saying that two (or more) nodes in element 1 have the same node numbers, or they are at the same position in space. For example if you have a 4-node "rectangular" element, you can't squish it down into a triangle shape by putting two "corners" at the same place, or you can't have a weird shape where two "opposite" sides of the element cross over each other, etc.</p>
<p>The "overlapping element" message is saying that two (or more) elements cover the same area or volume. The elements have to fit together like building blocks, or tiles. They can't overlap in an arbitrary way.</p>
<p>To see exactly what the problem is, display the model and highlight the element numbers in the error messages, and then figure out how to make a mesh that doesn't have those problems.</p>
| 31220 | Issue with collapsed element in LISA Finite Element Analysis |
2019-09-20T11:52:24.597 | <p>This is probably very well explained but I'm lacking the terminology knowledge and that's keeping me from finding the information I need.</p>
<p>I have a frame to which I need to attach a weight carrying arm. The arm will need to rotate vertically (1 degree of freedom) but will not be doing complete turns thus a bearing is overkill. The arm has a rectangular profile which is hollow inside, so I was planning to weld 2 small pieces of sheet metal to the frame that will form a bracket inside of which the arm will fit then drill a hole and put a bolt through but I'm afraid that due to the vertical rocking motion the bolt will quickly come loose.</p>
<p>What would be the proper solution in this case?</p>
| |automotive-engineering|metallurgy|joining| | <p>Long story short : In any moving part you need to use a bearing, however there are many, many types of bearings. The main two types are : <a href="https://en.wikipedia.org/wiki/Plain_bearing" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Plain_bearing</a> <a href="https://en.wikipedia.org/wiki/Ball_bearing" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Ball_bearing</a></p>
<p>Also important is the direction of the bearing :
<a href="https://en.wikipedia.org/wiki/Thrust_bearing" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Thrust_bearing</a></p>
<p>If you google on 'buy bearing' you can find them in all shapes and sizes on any technical supplier website. </p>
<p>Anyway if you insist on using the good old 'this is a bolt and can be used as a bearing' technique just use two nuts.
One nut as obvious the other as a counter nut tightened to the first one. </p>
<p>Good luck! </p>
| 31224 | Fastening rotating surfaces |
2019-09-20T16:45:27.553 | <p>The exact reason why adding less than optimal level water but just enough water to drench and dive the whole sasuage in during boiling in a pot isn't the optimal quicking cooking method? My intiuion is that adding a swimming pool of water or a few drip of water won't cook the sasuage.</p>
| |heat-transfer| | <p>As long as the sausage is completely immersed, you have optimized heat transfer between the water and the sausage, which will yield the shortest cooking time. Furthermore, by putting in no more water than necessary to accomplish this, you have minimized the heat-up time for the pot and water. </p>
| 31227 | Why adding less than optimal level water but just enough water to drench the sausage in general isn't the quickest way to cook? |
2019-09-20T16:46:42.840 | <p><a href="https://i.stack.imgur.com/vrXQk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vrXQk.png" alt="enter image description here"></a></p>
<p>From <a href="https://en.wikipedia.org/wiki/Pressure_measurement#Bourdon_gauge" rel="nofollow noreferrer">Wikipedia:</a></p>
<blockquote>
<p>Bourdon tubes measure gauge pressure, relative to ambient atmospheric pressure, as opposed to absolute pressure;</p>
</blockquote>
<p>But looking at pictures such as the one above, it seems there is only one connection into the tube, the one where the gas whose pressure is being measured enters. It would seem that the deformation of the tube is only dependent on the pressure of the gas. So why does the Bourdon tube measure gauge pressure relative to atmosphere instead of absolute pressure?</p>
| |pressure|measurements| | <p>The atmosphere is acting on the outside surface of the tube.</p>
<p>So any internal pressure is acting against the atmospheric pressure, which means if you need absolute pressure readings you need to accurately determine the atmospheric pressure at that height and compensate to sea level as necessary.</p>
| 31228 | Why does Bourdon tube measure relative pressure? |
2019-09-23T06:21:38.583 | <p>I have some homework question that i require some help understanding.</p>
<p>1) For the first question highlighted in red (referring to diagram viii), I know for sure that reactions at the supports is dependent on the length a because it determines the magnitude of moment it creates about the support and that it is not dependent on H because it is in the direction of the force(weight P) itself. </p>
<p>However, what I am uncertain about is the cable CE. Is this considered a two-force member in which tension at both ends will cancel each other out, hence varying the length L will not cause the reaction forces at A and B to change?</p>
<p>2) For internal loadings, I am guessing that it will depend on the length L, because for internal loading, we are viewing the forces at the particular point and hence varying the angle of elevation will change the Rx and Ry component of the forces that the point and that it does not cancel out.</p>
<p>And also, is it true that for a two-force member that is in equilibirum, the net translational forces and rotational moment will be 0 throughout the entire segment?</p>
<p>Thanks in advance.</p>
<p><a href="https://i.stack.imgur.com/jM0av.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jM0av.png" alt=""></a></p>
| |structural-engineering|structural-analysis| | <p>I don't want to take away the opportunity to work out the exact details for yourself, but I'll talk about the thought process I would use to approach these questions, and hopefully that will help clarify things.</p>
<p>Firstly, while I understand that the problem statement says to complete (a) before (b), I think at the point that you're getting stuck, it makes sense to sketch a few free body diagrams. It's part of the learning process, and as you do more and more of them you'll start developing a sort of intuition. My tendency is to run a couple quick 'thought experiments' either in my head or on paper, investigating two extremes. That's some of what I'll be doing below. Bear in mind that there are often multiple valid approaches to arrive at a solution - this is just one possibility.</p>
<p><strong>Idea 1: Dependency of external reactions on structure geometry</strong></p>
<p><strong>1.1 Do external reactions depend on length 'a'?</strong></p>
<p>Perhaps the quickest mental check of this one is to think about...what if 'a' were suddenly set to '2a'...would it change the support reactions? Given that the applied load is unchanged, we certainly can't say the support reactions would also be multiplied by 2. This suggests the support reactions aren't dependent on the numerical value of 'a' but on something else.</p>
<p>Here's a thought experiment that will lead us toward what's really going on. Structure viii is a simply supported beam, so let's first imagine a more straightforward situation. What if we had a simply supported beam with single vertical point load at midspan.</p>
<p><a href="https://i.stack.imgur.com/ZUQHlm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZUQHlm.jpg" alt="simply supported beam with midspan point load"></a></p>
<p>Do we even need to know the quantitative beam length to determine the support reactions? No, we don't. We simply need to know the load is applied at midspan. That is, we need to know the <em>ratio</em> of the beam lengths to the left and right of the load. This is enough to tell us that half the applied load will go to each support. The same principle holds true for instances where the vertical load is offset from midspan. The length ratio (not the lengths themselves) is what determines the simply supported beam reactions. You'll see this when you complete part (b) of the assignment and go through the exercise of solving for the support reactions.</p>
<p><strong>1.2 Do external reactions depend on the vertical position 'H' of the applied load?</strong></p>
<p>You're on the right track with this one. Mathematically I guess it relates to how we can slide a force along its line of action. My first thought is usually the physical intuition that whether I'm holding, say, a can of paint on a 6" string or a 12" string, I've got to support the same vertical load from the can of paint.</p>
<p><a href="https://i.stack.imgur.com/EnnV3m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EnnV3m.jpg" alt="cans of paint"></a></p>
<p><strong>1.3 Do external reactions depend on the height 'L' of the vertical member?</strong></p>
<p>At first glance, this one seems like a harder question, but the important thing to remember is that with support reactions we're investigating <em>external</em> equilibrium. For this simply supported beam, we don't need to know what's happening inside the structure to find the external reactions. It's like how for a statically determinate truss, we can find the external reactions without investigating individual truss member forces or worrying about the specifics of truss geometry. So, for structure viii, we can solve for our support reactions using our equations of equilibrium without knowing 'L' at all. As with idea 1.1, this will probably become more apparent when you work through part (b) of the assignment.</p>
<p><strong>Idea 2: Dependency of internal loading on height 'L' of vertical member</strong></p>
<p>Now we're getting into internal equilibrium, so you're correct that the cable geometry matters. If we assume that the cable is continuous and that at the point where the cable connects to the beam, the cable is passing around some sort of frictionless pin - then we can say that the tension, T, at every point in the cable must be the same (and equal to the applied load). Changing height L of the vertical member change the angle of the cable. As you noted, when we hold T constant and vary the cable angle, the x- and y- components of T vary. This means the internal forces in beam segment CD and vertical member DE are also changing.</p>
<p><strong>Idea 3: Two-force members and equilibrium</strong></p>
<p>For any member in static equilibrium (two-force or otherwise) the equations of statics will be satisfied at <em>every</em> point in the structure (including sections cut through a member). That is, <span class="math-container">$\Sigma F_x = 0$</span>, <span class="math-container">$\Sigma F_y = 0$</span>, and <span class="math-container">$\Sigma M = 0$</span>. Satisfying these equations is one way we can determine internal forces (shear, axial, moment) at points along a member.</p>
| 31260 | Concept on two-force member |
2019-09-23T19:16:06.037 | <p>Suppose I have the system:</p>
<p><span class="math-container">$\dot{x} = Ax+Bu$</span></p>
<p><span class="math-container">$y=Cx+Du$</span></p>
<p>and the following Hamiltonian matrix:</p>
<p><span class="math-container">$H=\begin{pmatrix}
A & \frac{1}{2}B^TB\\
-CC^T&-A
\end{pmatrix}$</span></p>
<p>I want to find the value of <span class="math-container">$\gamma$</span> which is the bound of the <span class="math-container">$H_{\infty}$</span> norm, so it is the value such that <span class="math-container">$\left |T(j\omega) \right |_{\infty }<\gamma$</span>.</p>
<p>I know that for the bounded real lemma, if the eigenvalues of <span class="math-container">$A$</span> have negative real part, and <span class="math-container">$I\gamma^2-DD^T>0$</span>, then the Hamiltonian have no eigenvalues on the imaginary axis. I also know that if the eigenvalues of <span class="math-container">$A$</span> have negative real part, then <span class="math-container">$\left |T(j\omega) \right |_{\infty }<\gamma$</span> holds.</p>
<p>But my question is : how do I find the value of <span class="math-container">$\gamma$</span>?</p>
<p>I have been told the result is <span class="math-container">$\gamma=0.5$</span> but I really can't get to this result. I have tried using the Shur's complement to see if this matrix is negative definite (so before doing that i switched sign to the Hamiltonian). In this way i thiught that if the A matrix is negative definite, it has all eigenvalues with negative real part, so the resulting value of <span class="math-container">$\gamma$</span> form the computation would have been the searched value. But I don't find the desired result. Maybe I am missing a point and doing something wrong, or maybe I am complitely on the wrong path.</p>
<p>Can somebody please help me? Thank's in advance.</p>
| |electrical-engineering|control-engineering|control-theory| | <p>I believe you should take a look of the real bounded lemma. Because the hamiltonian matrix has another shape. I quote the book from Zhou - Essentials of Robust Control p.238.
<span class="math-container">$$ H=\begin{bmatrix}
A+BR^{-1}D^{\top}C & BR^{-1}B^{\top} \\
-C^{\top}(I+DR^{-1}D^{\top})C & -(A+BR^{-1}D^{\top}C)^{\top} \end{bmatrix} \
$$</span>
where <span class="math-container">$R=\gamma^2I-D^{\top}D$</span>. </p>
<p>I would compare both hamiltionian matrixes and try to find <span class="math-container">$\gamma$</span>.</p>
| 31268 | value of $\gamma$ in the H-infinity norm |
2019-09-24T14:19:44.123 | <p><a href="https://i.stack.imgur.com/bQlsz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bQlsz.png" alt="Modified Sine"></a></p>
<p>I'm trying to create an excel sheet that will use the modified sine equation above to output a 360° displacement chart when a user types in the acceleration, velocity, total distance traveled and start and stop angles. From what I've been told previously this is the equation I have to use, but my background is more on the excel side of things so I'm having trouble understanding exactly how to break this equation down into something I can use.</p>
<p>Could someone please explain how to solve this equation using the variables that will be available that I described above? Does this equation require any more or any less variables than what will be available?</p>
| |mechanical-engineering| | <p>You can define the <code>h</code> and <code>beta</code> constants, either in a cell or as a <a href="https://www.ablebits.com/office-addins-blog/2017/07/11/excel-name-named-range-define-use/" rel="nofollow noreferrer">defined name</a>.
After that, you can create a column with the <code>theta</code> parameter and another column with the calculated <code>y</code> parameter.</p>
<p>The formula for the <code>y</code> should use the <a href="https://support.office.com/en-us/article/if-function-69aed7c9-4e8a-4755-a9bc-aa8bbff73be2" rel="nofollow noreferrer">IF</a> and <a href="https://support.office.com/en-us/article/and-function-5f19b2e8-e1df-4408-897a-ce285a19e9d9" rel="nofollow noreferrer">AND</a> functions. Suppose the <code>theta</code> parameters are on column A. The formula should be something like</p>
<pre><code>=IF(AND(0<=A1,A1<beta/8),h*(PI()/(4+PI()))*A1/beta - 1/4/(4+PI())*SIN(4*PI()*A1/beta),IF(AND(beta/8<=A1,A1<7*beta/8),h*(2*/(4+PI()))+PI()/(4+PI()))*A1/beta - 9/4/(4+PI())*SIN(4*PI()*A1/3/beta),h*(4/(4+PI()))+PI()/(4+PI()))*A1/beta - 1/4/(4+PI())*SIN(4*PI()*A1/beta) ) )
</code></pre>
<p>To break down the formula, the IFs statements define which equation to use.
The criteria is accessed with the AND function, since you have a range on beta. The equation itself is straightforward.</p>
| 31272 | Modified Sine Equation |
2019-09-25T12:09:51.713 | <p>We know that <a href="https://en.wikipedia.org/wiki/Blaise_Pascal" rel="nofollow noreferrer">Blaise Pascal</a> built a <a href="https://en.wikipedia.org/wiki/Pascal%27s_calculator#Carry_mechanism" rel="nofollow noreferrer">mechanical calculator</a> which had a carry mechanism called a sautoir. </p>
<p><a href="https://i.stack.imgur.com/PIr68.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PIr68.jpg" alt="enter image description here"></a></p>
<p>You can see a video of it working <a href="https://youtu.be/OpTAODDU1pg" rel="nofollow noreferrer">here</a>, and <a href="https://youtu.be/-li5VWnqhmI" rel="nofollow noreferrer">here</a>. </p>
<p>We <a href="https://books.google.com.au/books?id=piNNAAAAMAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false" rel="nofollow noreferrer">know that</a> <a href="https://en.wikipedia.org/wiki/Charles_Babbage" rel="nofollow noreferrer">Charles Babbage</a> wrote a book of <a href="https://en.wikipedia.org/wiki/Mathematical_table#Tables_of_logarithms" rel="nofollow noreferrer">log tables</a>. </p>
<p>In doing this, Babbage was inspired to create Log Tables mechanically, leading to his design of the <a href="https://en.wikipedia.org/wiki/Difference_engine#targetText=A%20difference%20engine%2C%20first%20created,small%20set%20of%20polynomial%20coefficients." rel="nofollow noreferrer">Difference Engine</a>. </p>
<p>My question is: <strong>Did Charles Babbage's Difference Engine use the sautoir carry mechanism from Pascal's calculator?</strong></p>
| |gears|mechanisms| | <p>Short answer is no. Longer answer is that the Babbage Difference engine has taken the <em>concept of a sautoir</em> and implemented something much more complicated. </p>
<p>Here we have two wheels representing values in the calculation, and the gear in the middle that adds.</p>
<p><a href="https://i.stack.imgur.com/cAatZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cAatZ.png" alt="enter image description here"></a></p>
<p>Then to carry numbers up - there was an external device:</p>
<p><a href="https://i.stack.imgur.com/eSPVv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eSPVv.png" alt="enter image description here"></a></p>
<p>Keep in mind that a single number is represented vertically on a column, so the values and the carrying device were replicated vertically.</p>
<p><a href="https://i.stack.imgur.com/cMkrp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cMkrp.jpg" alt="enter image description here"></a></p>
<p>See also this <a href="http://www.meccano.us/counter/index.html" rel="nofollow noreferrer">Counter</a> built by a person who built a <a href="http://www.meccano.us/difference_engines/rde_2/index.html" rel="nofollow noreferrer">Difference Engine</a> built out of <a href="https://en.wikipedia.org/wiki/Meccano" rel="nofollow noreferrer">Meccano</a> - which implemented its own carry mechanism:
<a href="https://i.stack.imgur.com/P77LV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P77LV.png" alt="enter image description here"></a></p>
<blockquote>
<p>This arm will engage between two of the double brackets on the detent assembly of the adjacent dial and cause it to advance by one unit.”</p>
</blockquote>
<p><strong>Bibliography</strong></p>
<ul>
<li><a href="https://www.youtube.com/watch?v=vdra5Ms__9s" rel="nofollow noreferrer">Babbage's Difference Engine No. 2, Part 2: The Calculation Section</a></li>
<li><a href="http://www.meccano.us/" rel="nofollow noreferrer">Tim Robinson's Meccano Computing Machinery web site</a></li>
</ul>
| 31280 | Did Charles Babbage's Difference Engine use the sautoir carry mechanism from Pascal's calculator? |
2019-09-26T13:21:22.177 | <p>I'm learning about how dog clutches work particularly in sequential transmissions.</p>
<p>I'm having a hard time understanding the principle which ensures dogs will always mesh correctly. I can easily imagine a situation in which dogs will collide and switching gears will not work such as when the speed of the input shaft matches the speed of the intermediary shaft and dogs are aligned together or when the engine is off and the vehicle is not moving then dogs can again collide.</p>
<p>I'm designing my own clutch-less transmission using dog gears, because I thought they are most appropriate, so I'm trying to understand the principle, not so much the example cases I mentioned above.</p>
| |mechanical-engineering|automotive-engineering|gears|transmission| | <p>I appear to have found an answer but I can not find the source to give credits.</p>
<p>As the shafts are rotating there is a chance the dogs will collide but assuming there is lubrication, they will just slide on each other until they eventually fall into place and engage the gear.</p>
| 31283 | How don't dog clutches collide? |
2019-09-26T21:03:19.063 | <p><a href="https://i.stack.imgur.com/4mfLr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4mfLr.jpg" alt="enter image description here"></a></p>
<p>We're building this thing in AutoDesk Inventor and SolidWorks, yet I don't know what it is. Can someone please tell me? Thanks.</p>
| |mechanisms| | <p>It is a Z bracket.</p>
<p><a href="https://www.google.com/search?q=z%20bracket&rlz=1C1GCEA_enCA851CA851&sxsrf=ACYBGNQyVIUw3sjDhx4SgP4k2X11wypl4A:1569607940286&source=lnms&tbm=isch&sa=X&ved=0ahUKEwj8qN6dzfHkAhXJjFkKHaKuAFkQ_AUIEigB&biw=1536&bih=722" rel="nofollow noreferrer">Google Search</a></p>
| 31289 | How could I identify the object in the question body? |
2019-09-27T02:12:37.167 | <p>This is a general question about root locus plots. I notice that there is a source that says:</p>
<p>"The root locus begins at the finite and infinite poles of G(s)H(s)". </p>
<p>Note the part that says 'finite and INFINITE poles'. </p>
<p>Should the part that says 'infinite' be omitted? The root locus only begins at the FINITE open loop pole locations, right?</p>
<p>If the statement is indeed correct - about cases where a root locus may start at 'infinite' pole locations, then are there any examples of such cases?</p>
<p>Thanks all!</p>
| |control-engineering| | <p>You can do a root locus plot of a system with more zeros than poles, even though such a thing is physically impossible. If you do, you find poles coming in towards your zeros, just as a system with an excess of poles has them running off to infinity.</p>
| 31293 | root locus question - root locus 'begins' at finite open loop pole locations |
2019-09-27T06:24:35.003 | <p>In industrial manufacturing lines, there is a a limit on the number of discarded parts per million.I need to have a sense about this, e.g. How different is 4 ppm from 1 ppm and how much does it indicate the improve in production quality?</p>
<p>To me considering the price of a part as 1€, 1 ppm and 4 ppm seem so trivial</p>
| |manufacturing-engineering|industrial-engineering|statistics| | <p>If we talk about quality and 6-sigma (which roughly means 1 ppm defects), the 1 ppm refers to test escapes, not test rejects. </p>
<p>Even if a part only costs \$1, if testing fails to catch a bad part and we accidentally send it to a customer, where it fails in the field, the eventual cost (in failure analysis, corrective actions, documentation, time spent communicating with the customer, etc) can be much, much more than \$1.</p>
<p>There is also a benefit to demanding 6-sigma (1 ppm defects) or better quality from your vendors.</p>
<p>If you buy a \$1 part, and assemble it into a \$100 (or \$1000) product, where it causes the whole product to fail, then again the cost of that defect is much higher than the cost of the faulty part itself.</p>
| 31295 | Talking in terms of 6 sigma, why 4 ppm would be very costlier than 1 ppm (part per milion)? |
2019-09-29T15:37:11.437 | <p>Around where I live there's these soapbox/gravity kart races (google "carros de rolamentos") in which the wheels are built simply from ball bearings. As I was building my own, I began wondering how I could easily mount my large bearings onto my kart, and it occurred to me that an easy way to do it would be to get some smaller bearings inside the ones previously mentioned and mount the resulting pair onto the frame, like this:</p>
<p><a href="https://i.stack.imgur.com/mvLi4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mvLi4.png" alt="enter image description here"></a></p>
<p>More importantly, I was under the impression that this would improve the performance of the kart in terms of speed. I am not completely sure though, hence why I am asking this question. </p>
<ul>
<li><p>On one hand, I find that my older kart reaches some sort of terminal velocity when the inner and outer races of the bearings can only move so fast in relation to each other. So, the inner bearing would sort of take some load out of the larger one by rotating in the same direction, allowing the outer-outer race to rotate faster in relation to the inner-inner race (or, in the worst case scenario, I would get no gain but also no loss, unless the smaller bearing spins backwards). </p></li>
<li><p>On the other hand, I asked this to a student of mechanical engineering and he said he doubted I would get any gain, because there would be more overall friction, which does make some sense, but does not convince me much.</p></li>
</ul>
<p>Also:</p>
<ul>
<li><p>We can assume that they will be well coupled together (I can file them if needed and will probably weld them afterwards).</p></li>
<li><p>I can't run any practical tests at the moment because I haven't bought the smaller bearings yet (I don't know if it's worth it)</p></li>
<li><p>From my research, this kind of configuration has only been mentioned in some videos about fidget spinners, and sure enough, it seems to rotate fairly well.</p></li>
</ul>
| |mechanical-engineering|applied-mechanics|bearings| | <p>Model airplane motors run at 50000 RPM with ball bearings; I don't think that your limit is due to speed.</p>
<p>Basically, the assembly you picture will have the friction of, and be as rugged as, just using the inner bearing. The outer will just be going along for the ride (and adding weight and expense to your assembly).</p>
| 31321 | Cascaded ball bearings |
2019-09-30T09:52:55.927 | <p>I am writing a finite element code for my final year project of BS Mechanical Engineering. The geometry is an integration of several parts composed of different materials. I don't have exact values of material properties. However, I do have ranges of them. For example, for Steel1020, Young's modulus varies from 205-215 GPa. Should I use lower limit of Young's modulus or higher limit of Young's modulus, or average (geometric / arithematic)?</p>
| |mechanical-engineering|structural-engineering|structural-analysis| | <p>As <a href="https://engineering.stackexchange.com/a/31331/1832">recommended by @Solar Mike</a>, the most "professional" solution is to let the end-user decide. Give them some input mechanism (a textbox, slider or dropdown, for example) so they can define which value to adopt. This is what you'll see in basically any modeling software you'll find in the wild.</p>
<p>However, it's also very important to note the observations made in <a href="https://engineering.stackexchange.com/questions/31330/determination-of-youngs-modulus-for-use-in-analysis/31331#comment56301_31330">@alephzero's comment</a>: it almost doesn't matter. The difference between 205 and 215 MPa is 4.5%. Given the errors involved in basically every step of an engineering design calculation*, that's peanuts.</p>
<p>Now, that being said, if you really want to take a strict approach to this, you need to look at the worst case. This will often be to simply adopt the lowest value. But not always.</p>
<p>And in this case, it might very well not be. If you are sandwiching different materials, one will be stronger than the other(s). In this case, you'll probably want to adopt the highest stiffness for the material with the lowest yield strength, and visa-versa.</p>
<p>This way, if these elements need to work together, the strong material will transfer the most force to the weaker material, increasing the likelihood that the weaker material yields first.</p>
<p>But then again, maybe the strong material would yield even earlier if it were stiffer (if it were right on the edge of yielding and the weaker material's increased stiffness is what's holding off the "straw that'd break the camel's back").</p>
<p>So this is a non-trivial question which depends on the specific case being analyzed. So, if you want to be strict about this, you'd either need to look at the specific case to determine the worst case or (if your program doesn't take ages to run) just run the model with the different combinations of moduli for each material and select the worst case.</p>
<hr>
<p>* Every step of the process is an approximation: materials aren't perfectly elastic, they'll likely have different properties along a beam, supports and connections aren't perfectly fixed/pinned/springy, beam dimensions will be different, the welder will come to work drunk that day, etc.</p>
| 31330 | Determination of Young's Modulus for use in Analysis |
2019-09-30T15:31:44.037 | <p>I've watched a few videos explaining how actuators work and it seems to me that a regular, non-blocking, actuator would freely move if not powered. But theory and practice are rarely the same thing.
I've never owned an actuator, and nobody on the entire internet have tried to push on an actuator that is not powered, hence this question.</p>
<p>To give a bit more clarity, I'll describe my application and what I want to use the actuator for. There's this door that I often forget to close. I want the door to close after some time, if open.</p>
<p>I don't need the actuator to ever open the door, just to close it, if open for more than, say 5 minutes. Edit: The closing mechanism has to be on the "pull to close" side of the door.
I want to be able to manually open and close the door, as that will still be the primary mode of operating the door.
The logic part will be done using an arduino and some sort of sensor that figures whether the door is open or not.</p>
<p>An actuator seems to fix my problem, but only if the shaft can freely move when not powered. So, can a cheap actuator <a href="https://www.aliexpress.com/item/32984863011.html" rel="nofollow noreferrer">like this</a> be moved freely when not powered? If not, what would my options be?</p>
| |actuator|linear-motors| | <p>Your primary question, if a linear actuator can be moved with the power off, will be no, unless you find a mechanism specifically created for that purpose. </p>
<p>These actuators are effectively a pinion gear design with a very high reduction ratio. You'll note in the link you provided that heavy lift mechanisms such as a motorcycle lift and a medical bed are presented for this type of use. They work well because you can position the bed/lift top where you wish and stop without additional locking mechanisms, although you'll find most have a safety latch of some sort.</p>
<p>For your application, closing a door, this type of linear actuator will be poorly suited.</p>
<p>I can envision a more mechanically complex arrangement which would use a linear actuator. Consider the manual release for a garage door opener. The chain and carrier for the opener can be considered the linear actuator.</p>
<p>If I disengage the manual release, then activate the button or remote, the door will remain closed. I can open and close manually the garage door without interference from the motorized mechanism, as long as I do not engage the latch system that is part of this garage door device.</p>
<p>If the linear actuator in fully-open mode has a pusher block attached, and remains in open mode, your door will operate normally. A sensor tells your system that the door is closed after the appropriate period.</p>
<p>If not, the device actuates the actuator and pushes the door closed, until the sensor informs the device that it is closed, at which point, the actuator returns to the original location, ready to push again on command.</p>
<p>There may be other approaches using clutches, cables, etc., but this is the one that popped into my alleged mind as an option.</p>
<p>EDIT: similar method to my suggestion, based on new information regarding the placement.</p>
<p>Consider that the actuator is on the inside of the door, as specified. When idle (door closed), the actuator is retracted, to a point where the end of the ram is not in contact with the door.</p>
<p>At the end of the ram is a magnet of sufficient strength, matching placement on the door of a metal panel.</p>
<p>After the door is ajar for the specified time, the ram/actuator will travel to the limit, "collecting" the door in the process. If that aspect is undesirable, a limit switch could be placed on the actuator, triggering motor reversal. Simple microswitches such as those used in 3D printers are inexpensive and easily found online.</p>
<p>When the actuator contacts the door, either at full travel or with the switch activated, the magnet will pull the door closed as the ram retracts. When the door is closed, the additional space between the ram and the door will disengage the door, returning it to normal service.</p>
<p>If magnets are not desirable, a more complex design using the limit switch concept to trigger a latch and socket mechanism could be implemented. Additionally, a low-to-medium force detent mechanism might work, but that would require full extension to engage. Think of a horizontal plate on the door with a large diameter hole in the surface. The end of the ram would have a sphere that has to deflect or be deflected in order to engage the hole. Once engaged, reversal will pull the door closed until the retraction force pops the sphere free from the hole. Less reliable, less fussy to build, but still an option.</p>
| 31334 | Can linear actuators be moved manually? |
2019-10-01T09:22:44.897 | <p>When analysing and then designing a rigid jointed, statically indeterminate structure, I can solve for all the forces and moments using the Hardy Cross (Moment Distribution) Method.</p>
<p>However, I am unsure about how to solve the deflections on the beams and columns. I need these values in order to design based on serviceability limit states and verify deflection checks. </p>
<p>Can you please demonstrate (or point out) a method to quickly and effectively find the displacements in a multi-storey, statically indeterminate structure. </p>
| |structural-engineering|civil-engineering|structural-analysis|structures| | <p>To rigorously solve for deflections of indeterminate structures, the most feasible method is the <a href="https://en.wikipedia.org/wiki/Direct_stiffness_method" rel="nofollow noreferrer">Direct Stiffness Method</a> (aka matrix methods). This is generally the approach finite element method software packages (FTool, SAP2000, RISA, etc.) are implementing when dealing with line elements.</p>
<p>For structures with a relatively small number of degrees of freedom it is possible to implement the direct stiffness method and arrive at a solution (somewhat) by hand. In my matrix methods class at university we generally solved the matrices using a calculator or by implementing a routine coded in MATLAB. So really, we were doing exactly what the FEM software does but more slowly and/or transparently.</p>
<p>Prior to the advent of computers (including programmable calculators) engineers would have used approximate methods of analysis to ballpark deflections. One very simple example of this might be that if we know the load on a beam we can determine an upper-bound deflection by considering it to be simply supported. When doing a hand check of computer calculations, engineers today will still take this approach. That is, we make simplifying assumptions and get a back-of-the-envelope feel for whether the computer solution is appropriate.</p>
<p>Computer-based analyses are fairly essential for the design of today's structures by today's design codes. The goal of hand calculations is rarely to exactly duplicate the computer solution. Rather, the goal is to use hand methods and engineering judgement to validate the computer solution.</p>
| 31345 | Deflections Of Statically Indeterminate Structures |
2019-10-05T19:53:58.540 | <p>The concrete slab (?) supporting the swimming pool (≈ 10 x 10m x 1m) of my building has clearly a 'belly':</p>
<p><img src="https://i.stack.imgur.com/Uw3iK.jpg" alt="concrete slab belly"></p>
<p>I'm not sure whether this was built this way or evolved over time, however, my best guess is that after 25 years the heavy load of the swimming pool deformed the structure and even the pipes that goes right under it.</p>
<p> </p>
<p><strong>Questions</strong> </p>
<ul>
<li>What is the technical term for this? </li>
<li>Is this a problem or not? Why? </li>
<li>Any action should be taken in order to prevent an accident?</li>
</ul>
| |civil-engineering|concrete|building-design| | <ul>
<li>What is the technical term for this?</li>
</ul>
<p>Bending downwards is called <strong>sagging</strong>. (Technically "sagging" isn't moving downwards; it's bending in a shape that looks like the mouth on a smiley face).</p>
<p>One thing you'll see with concrete structures is <strong>creep</strong>. This is where under constant load, the deflection increases over time.</p>
<ul>
<li>Is this a problem or not? Why?</li>
</ul>
<p>Sagging deflection is inevitable. If you put a load on something it <strong>will</strong> move. Whether this movement is big enough to be visible or not it another question, but it <strong>will</strong> move.</p>
<p>When designing something we often check for two different failures: ultimate strength and deflection. <em>(Note: There are plenty of other things we check for, this is a simplification.)</em> Ultimate strength check = is it going to fall down. Deflection check = is it going to deflect so much that it scares people.</p>
<p>When assessing (i.e. checking) an existing structure, we generally only do ultimate strength checks; we don't worry so much about whether people are worried or not...</p>
<p>So, your structure has clearly failed the deflection limits, as it's scared you! But that's no indication as to whether or not it has exceeded its ultimate strength limits, so its no indication of whether there's actually a problem or not.</p>
<ul>
<li>Any actions should be done in order to prevent an accident?</li>
</ul>
<p>Get a structural engineer to assess it. You've said in comments that the structure has been inspected due to this deformation - as long as that inspection has been carried out by a suitably experienced/qualified engineer, then you have no need to worry. I have no idea of how you satisfy yourself of their qualifications in your country.</p>
| 31396 | Concrete Slab Deformation |
2019-10-06T21:24:55.493 | <p>If creep behavior can begin at 0.6T(melting) but, hot working for a pure metal starts at 0.4Tm, what is the explanation for this? </p>
| |materials|metals|creep| | <p>Hot work involves macroscopic deformation of the metal; no time is allowed for creep to take place so the concept of "creep" as such doesn't exist during hot work. </p>
<p>The objective of hot work is to deform the workpiece to its desired shape at a temperature at which it happens to be easily malleable and will not strain-harden in the process. This means that the temperature you choose depends on exactly how soft you want it to get and how hot you need to get it so that solid-state diffusion through the lattice is more or less fully activated. </p>
| 31401 | Creep behavior in metals |
2019-10-08T10:36:31.737 | <p>I would like to try out a component that operates like a combination of the two pictures provided - spring loaded pull pin, with a pin finish like a tubular latch (chamfered/filleted finish).</p>
<p>Do these exist?</p>
<p>If somebody can tell me their correct name or better still, provide me with a link to a manufacturer's site that would be great.</p>
<p>Thank you</p>
<p><a href="https://i.stack.imgur.com/6HXXg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6HXXg.jpg" alt="Spring Loaded Pull Pin"></a>
<a href="https://i.stack.imgur.com/yuprc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yuprc.jpg" alt="Chamfered/Filleted Pin Finish"></a></p>
| |springs| | <p>Thank you GisMofx and M.Bauer for the advice, those website will come in handy for future research.</p>
<p>Norelem actually have one, which can be found <a href="https://www.norelem.com/us/en/Products/Product-overview/Flexible-standard-component-system/03000-Spring-plungers-Indexing-plungers-Stops-Centring-positioning-components-Ball-lock-pins-T-slot-nuts/Indexing-plungers/03089-10-Indexing-plungers-with-rotation-lock-and-lead-in-chamfer.html" rel="nofollow noreferrer">here</a>.</p>
<p>I also found a spring-loaded slide bolt latch from McMaster-Carr which will also suit what I need (<a href="https://www.mcmaster.com/2206a21" rel="nofollow noreferrer">Link</a>).</p>
| 31414 | Spring Loaded Pin with Chamfered Pin Finish? |
2019-10-08T15:07:55.150 | <p>I'm designing a circular shaft that's going to carry a spread radial load of about 200kg and also transmit a 160Nm torque. I'm not very familiar with materials but what I'm guessing I'll be able to get at a reasonable price locally is some lower grade steel.</p>
<p>Replacing the spread load with a single force in the middle of the shaft I guess will look something like this</p>
<p><a href="https://i.stack.imgur.com/pUCcc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pUCcc.png" alt="enter image description here"></a></p>
<p>Both points A and B are symmetrical so I'm taking in consideration only half of the shaft and calculating for that.</p>
<p><span class="math-container">$$T = F\ell = 200 \cdot 0.5 = 100\text{ kNm}$$</span></p>
<p>Where <span class="math-container">$T$</span> is the torque at point <span class="math-container">$A$</span> created by the force <span class="math-container">$F$</span>.</p>
<p>The tensile stress on the shaft will then equal</p>
<p><span class="math-container">$$t = \dfrac{T}{W}$$</span></p>
<p>Where <span class="math-container">$t$</span> is tensile stress and <span class="math-container">$W$</span> is section modulus.</p>
<p>For a solid circular face the section modulus equals</p>
<p><span class="math-container">$$W = \dfrac{\pi d^3}{32}$$</span></p>
<p>Now I have to assume either <span class="math-container">$t$</span> or <span class="math-container">$d$</span> in order to solve. I looked it up in <a href="http://www.matweb.com/search/datasheet.aspx?bassnum=MS0001&ckck=1" rel="nofollow noreferrer">this</a> table and it looks like the tensile yield strength of steel is 350 kPa so I'm taking a conservative estimate of 300 kPa for <span class="math-container">$t$</span>.</p>
<p><span class="math-container">$$\begin{align}
d^3 &= \dfrac{32(T / t)}{\pi} \\
&= \dfrac{32 \cdot (100e3 / 300e6}{\pi} \\
&= 0.003397028 \\
\therefore d &= 0.150\text{ m}
\end{align}$$</span></p>
<p>Obviously there is something wrong either in my logic or calculations because I'm getting unrealistic results.</p>
<p>I went ahead and calculated the torsion stress as well with <a href="https://engineering.stackexchange.com/a/17202/20333">this formula</a>. I found that <a href="https://www.quora.com/What-is-the-shear-strength-of-mild-steel" rel="nofollow noreferrer">shear stress for steel is about 200-300 kPa</a> so I'm again taking a conservative 200.</p>
<p><span class="math-container">$$\begin{align}
d &= 1.72 \cdot \left(\dfrac{M}{\tau_{max}}\right)^{1/3} \\
&= 1.72 \cdot \left(\dfrac{160}{200e6}\right)^{1/3} \\
&= 0.0159 ~ 16mm
\end{align}$$</span></p>
<p>which seems about right.</p>
<p>I was thinking of getting a 20 mm diameter profile but I would rather play it safe and ask for some help instead of just assume it's going to hold.</p>
| |mechanical-engineering|materials|steel|strength| | <p>@php_nub_qq, Even if its a bit late, my main comment is that you probably have messed up with the engineering units.</p>
<p>The tensile stress you are referring to, is probably 3 order of magnitude lower than the actual UTS even of a common grade steel (300MPa instead of 300 kPa). That allow would mess up and produce significant differences in the loads.</p>
<p>Also I concur with @Amit in that your method you describe is valid only for the bending moment imposed from the 200kN load. In your calculation you do not account for the shear stress due to the torque. Due to my late answer I will only point you to the following link <a href="https://www.engineeringtoolbox.com/torsion-shafts-d_947.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/torsion-shafts-d_947.html</a>, so that you can calculate <span class="math-container">$\tau_{xz}$</span></p>
<p>After that, you could use the maximum stress theory or Von Mises (it will be much more straightforward in your case and also differences in most cases are negligible).</p>
<p><span class="math-container">$$\sigma_{VM} = \sqrt{\sigma_{xx}^2 + 3 \cdot \tau_{xz}^2}$$</span></p>
<p>For justification about that, you best do a bit of reading at Norton or Shigley's books, and look at design of shafts under combined loading.</p>
<p>EDIT 1: I performed a rough analysis (linear) and the estimated normal stresses (due to bending) compared to the shear stresses (induced by torsion) are approximately 50 times larger in the scenario you are describing.</p>
<p>E.g. assuming a maximum tensile stress of 200 MPa, the required diameter is approximately 4.7[cm]. For that diameter the torsionally induced shear stress is about 8[MPa].</p>
<p>Bottom line, from an engineering point of view you are ok just considering the bending stresses only.</p>
| 31415 | Calculating shaft diameter |
2019-10-08T19:25:44.787 | <p>Let'say the steel tank would be submerged deep in the ocean with 136 atm ambient pressure then the other tank with the same material would just be used to contain 136 atm pressure. How would the thickness of the wall differ? </p>
<p>Also, if the tank is designed to be sunk deep in the ocean, would it be cheaper to design a tank that could contain 136 atm pressure then pressurize it at that pressure so that it would have zero forces acting on the tank's walls if the ambient pressure is also 136 atm?</p>
| |pressure|steel| | <p>when the tank is pressurized with 185atm it is very simple to calculate the thickness needed to hold the pressure.
Say the radius of the spherical tank is R. the max pressure is at a great circle passing through the center and is equal to <span class="math-container">$ \pi*R^2*186 $</span>.</p>
<p>This pressure has to be supported by the tension stress on the thickness of the peremeter of the tank, t, at that circle.</p>
<p><span class="math-container">$t = \frac{\pi R^2*186}{2 \pi R* \text{allowable tension stress of steel}}$</span> </p>
<p>Which is a reasonably small number. </p>
<p>But if the tank will be exposed to 185atm external pressure it is susceptible to catastrophic buckling and explosive crushing. so it has to be designed for moment resistance in case of buckling, which will turn out to be mostly following empirical methods which includes thick wall vessel design and possible use of rib stiffeners or use of thick hardened steel or special materials.</p>
| 31418 | What are the design difference of a steel tank that would withstand 136atm inside and 1atm ambient compared to 1atm inside and 136atm ambient? |
2019-10-09T01:11:28.620 | <p>I was given a challenge question from my teacher that I need some assistance in solving. The problem is to merely identify the two-force members. I understand the ideas of how to determine a two-force member (colinear forces, equal magnitude, etc.), but in this problem, they just aren't clicking for me. Thanks in advance for the help!</p>
<p><a href="https://i.stack.imgur.com/WA7q8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WA7q8.png" alt="enter image description here"></a></p>
| |statics| | <h2>None of these elements are two-force members</h2>
<p>Before anything, it's worth remembering what two-force members are: bars which only have two equal-and-opposite colinear forces applied, one at each extremity. Given these conditions, we also know the forces must be purely axial, with no shear component. Given they have no shear component, they'll also suffer no bending moment.</p>
<p>Now, it's worth mentioning that there are actually two ways of looking at that definition. One is strictly based on the structure itself, while the other considers the applied loading. The simple case which exemplifies the difference is a cantilever beam with only an axial force applied.</p>
<p>Is that beam a two-force member? If you only look at the structure, then <strong><em>no</em></strong>. It's obvious that under basically any other loading condition, this beam would present shear and bending moment. But if you only think of the currently applied forces, then <strong><em>yes</em></strong>: it only suffers equal-and-opposite colinear axial forces.</p>
<p>I'm of the opinion that the best way to think about it is by only looking at the structure: if you can come up with a classic loading pattern which generates bending moment, it's not a two-force member<sup>1</sup>.</p>
<hr>
<p>So let's look at the structure ignoring its loads:</p>
<p>The bottom beam AF is basically a simply-supported beam with a cantilever. If you put a vertical load anywhere along that span, the whole thing'll have shear and bending moments, so not a two-force member.</p>
<p>The bend which connects AF to the rest of the structure is obviously not a two-force member: axial loads aren't collinear. And even if you consider that as two separate members, their shared node is fixed. Therefore, any deflections will cause relative rotations and therefore bending moments.</p>
<p>As for the "box" at the top, well, that's also a bunch of fixed beams. Any loading in that region will lead to bending moments all over the place.</p>
<hr>
<p>For the record, for those who prefer to consider the given loading, the same conclusion applies:</p>
<p><a href="https://i.stack.imgur.com/oO5LI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oO5LI.png" alt="enter image description here"></a></p>
<p>That's the bending moment diagram (ignore the actual values, I got lazy with the loading inputs). As we can see, all relevant members have bending moment and the box is unloaded (which in my book doesn't count as being a two-force member).</p>
<hr>
<p><sup>1</sup> For the pedants thinking of truss members with uniform loads applied, just remember the "classic loading pattern" for truss members is for uniform loads to be replaced by nodal forces. And nobody likes a pedant.</p>
| 31422 | Statics Two Force Member Problem |
2019-10-09T15:44:54.707 | <p>Hello I am a simple rock climber, and would like to figure out how much force is applied to The lobe of a cam/the rock when the cam is loaded. So the cam lobe is circle the ramp or cam goes from 7/16”-11/16” I’m not sure that matters or if just the point of contact matters. If I ignore friction and those kind of nussance variables what is my formula if I 180lbs simply hang on the cam. (Is it a simple lever calculation)!</p>
<p><img src="https://i.stack.imgur.com/2tKWf.jpg" alt="enter image description here"></p>
| |mechanical-engineering| | <p>The curve of a spring loaded cam is certain curve called exponential spiral with the unique property that the angle it touches the wall of the crack and a horizon remains the same (13-14 degrees) no matter haw much the cam expands.</p>
<p>This small angle means the force applied by the cam to the crack is always <span class="math-container">$F_{horiz}= W_{weight}*cotan(13)= W*4.3$</span> </p>
<p>Which after many tests has proven to be the safest and most practical.</p>
<p>here is the detail calculations. <a href="http://www.valleygiant.com/cam_math.html" rel="nofollow noreferrer">cam math.</a></p>
| 31433 | Force applied to a cam follower |
2019-10-09T22:54:25.110 | <p>Lately I've been watching some videos on machining and hydraulics as it always interested me.</p>
<p>I was toying with the idea of building a budget CNC mill with threaded rod for the axes, but the issue of backlash was bugging me.</p>
<p>Then I thought about using hydraulic pistons instead. If used in conjunction with a positive displacement pump and using a large hydraulic transmission ratio the issue of precision could be sorted out, and the incompressibility of the fluid would help to keep backlash to a minimum.</p>
<p>I searched but found little info on hydraulic actuators for precision positioning, so I believe there's something I must be missing</p>
| |mechanical-engineering|fluid-mechanics|hydraulics|machining| | <p>beware ethan48 answer and these corporate smears and myths.</p>
<p>At this level you can use water hydraulics or use oil hydraulics with filtered vegetable oil instead. you can create a hydraulic actuator in a lathe <a href="https://www.youtube.com/watch?v=-a0-rPURjeI" rel="nofollow noreferrer">https://www.youtube.com/watch?v=-a0-rPURjeI</a> <a href="https://www.youtube.com/watch?v=i-auVa-T180" rel="nofollow noreferrer">https://www.youtube.com/watch?v=i-auVa-T180</a> however there's the possibility of using hydraulic stepper motors with 4 radial pistons or hydraulic motors with appropriate gear reduction like Maho hydraulic cnc machines. hydraulics makes possible not only the emancipation from petrol oil but from electric energy grid by the use of mechanical wind power, connecting a pump to a VAWT wind turbine, or use the same turbine attached to an air compressor filling an air tank to use the compressor as motor and move the hydraulic pump when there's no wind.</p>
| 31438 | Hydraulic linear actuators as CNC axis instead of screws |
2019-10-09T23:06:42.867 | <p>I observed the rebuilding of a 12KW, 120V gas-powered generator. I was amazed at how small the generator portion was (roughly the size of an electric motor you might find in a major home appliance). The windings were pretty thin magnet wire, like you would find in an appliance motor.</p>
<p>Getting the 100A generator output to the breaker box requires heavy-duty wire. That same current originates in the generator windings. How is the thin wire in the windings able to carry 100A (which far exceeds what wire that size is rated for)? </p>
| |generator|current|wire| | <p>My question was based on an appearance that the thin winding wire seemed to violate the laws of physics, so there might be some other factors at play. Neil_UK answers this in <a href="https://electronics.stackexchange.com/a/325358/115681">generator coils: max power given wire gauge</a> on the Electrical Engineering site. The bottom line is no. It's mainly a matter of using multiple coils in parallel to manage the current handling requirements per coil. Other factors, like the duty cycle and dissipating heat also come into play.</p>
| 31439 | How can thin wire in generator windings carry the high output current? |
2019-10-10T13:21:59.343 | <p>For purposes of usage in home/auto and light industrial circumstances, what are the differences? What are the pros and cons, and suited uses, of each? </p>
<p>Thanks!</p>
| |materials| | <p>Silicone grease and paste are frequently meant the same thing. They are used for lubrication and because they wrap the area of application they protect the parts from rust and elements. When they are used in applications such as plumbing orings they have to be rated food grade, or for electrical uses they have to be rated.</p>
<p>Silicone spray is mostly used as thin lubricant and locked tread breaker.</p>
<p>Sometimes difference in consistency is just to solve access issues. There are sprays of silicone with different properties for different applications which are hard to reach.</p>
| 31449 | Silicone paste vs. silicone grease vs. silicone spray lube |
2019-10-10T13:56:35.083 | <p>Isochoric means constant volume (In SI engine: piston should not move when the spark is ignited). In reality piston always keeps displacing up and down means volume changes. So, how does this make it a Isochoric Heat Addition in power stroke.</p>
| |mechanical-engineering|pistons| | <p>the <em>rate</em> at which heat is added is so fast that the piston has no opportunity to move during the process. Furthermore, when the piston is at the top of its stroke (which is where the heat is added) the crank-and-connecting rod mechanism yields a finite residence time of the piston in that position for a span of crank rotation, which means that even though the crank never stops <em>rotating</em>, the piston spends a finite amount of time in an <em>almost motionless</em> state at TDC. </p>
| 31450 | What practically is constant volume heat addition in SI piston engine? |
2019-10-10T19:13:28.220 | <p>I'm trying to reduce the rpm of a combustion motor. It operates at 2000 rpm with 3HP, and I need it to work at around 12 rpm (no matter the power). I've made the math with sprocket and chains, but the biggest sprocket is of 472 mm of diameter, which is too big.
Is there any cheap and easier way to achieve this?
Thanks!</p>
| |motors| | <p>2000/12 is about 167, so you need to gear it down a lot.</p>
<p>To a sensible 1st-order approximation, you don't want to gear down by more than 10:1 in any one stage (I'm not a mechanical engineer; if one of them speaks up feel free to believe them).</p>
<p>If I were throwing this together from surplus parts, or even building a one-off machine or a short production run, I think I'd run belt drive to a gearbox with my desired total reduction. So maybe 4:1 belt reduction to a 40:1 box, or 8:1 belt reduction to a 20:1 box.</p>
<p>One stage of reduction on a belt drive to one more stage of reduction with a gear drive wouldn't be horribly impractical. Because of the large torque multiplication, you'd need an enormous belt drive on the final stage which is why I'm suggesting gears or chains.</p>
<p>If you're building 10,000 a year, then you can afford to get an integrated motor/gearbox -- I suspect that's another story.</p>
<p>Or I'd go with 200:1 reduction because it's a nice round number, and run the motor at 1200 RPM.</p>
<p>Keep in mind that if the motor can do 3HP then the potential torque output at 12 RPM is around 1300 foot-pounds*, which is ginormous. Size your shafts accordingly or put in torque-limiting clutches, and don't get in its way.</p>
<p>* For those who noted my earlier math error -- I did the torque calculation from the RPM and power, without regard to gear ratio (and I used a calculator). So it should be correct.</p>
| 31454 | Trying to reduce RPM of a combustion motor |
2019-10-10T19:48:31.497 | <p>If you're double-bagging some groceries, and one of the bags has tears in it while the other is intact, which bag should go on the inside?</p>
| |structural-engineering| | <p>Put the torn bag inside, just to let it absorb the possible moisture of veggies and other freshly washed fruits.</p>
<p>It is interesting to note that some tears at critical locations where they could concentrate stress to outer bag can lead to tearing of the outer bag, while left alone by itself it could safely contain it's load. Same idea as the shallow scorelines the glazing guys put on a plane of glass to cut it.</p>
| 31455 | Double-bagging problem: which bag on the inside? |
2019-10-10T23:40:53.170 | <p>I have a statics problem that I am running into some difficulties with. What is being asked here is to determine the support reaction at the points A and C. This would normally be no problem, but when I set up my equations (sum of the forces in the x-direction, sum of the forces in the y-direction, and total moment) I realize that there are too many unknowns for the problem to be solved. </p>
<p>Is there a way to assume something in order to eliminate one or two of the unknowns? (I am guessing it will be in the x-direction, but that's just a hypothesis.) Or should the problem be solved in a completely different way?
Thanks for the help!</p>
<p><a href="https://i.stack.imgur.com/PWZk2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PWZk2.png" alt="enter image description here"></a></p>
| |mechanical-engineering|civil-engineering|statics| | <p>In this kind of questions, I believe that most of the attention should be focused on the general approach and the philosophy of attacking. Afterwards, each individual may proceed by writing its own equations and solving them in its prefer way.</p>
<p>In our particular case, I guess you examined the whole structure, realized that there are 4 reactions (Ax, Ay, Cx and Cy) and thought to yourself - OK, here we have a statically indeterminate system (3 equations and 4 unknowns), so we must apply a unique technique in order to solve it.</p>
<p>However, the trick here is hidden in the middle joint. The structure is divided into two parts (AB and BC) which are connected through a pin joint. This joint brings another two reactions to the table (Bx and By). You may first be tempted to think that "hi, these are internal forces, they cannot add any useful information. I could alternatively create an artificial cut in any other place along the structure and look for the internal reactions there".</p>
<p>However, this would be a wrong way of thinking. If we cut our structure in any other zone, we would have to deal with 3 reactions in the cross section (two forces and a bending moment). The pin joint actually eliminates the bending moment which in turn allows us the solve the problem.
You may look at each part separately now. Part AB has 4 reactions and 3 equilibrium equations and the same is true for part BC. Since two of these reactions are the same (Bx and By) we end up with 6 equilibrium equations and 6 unknown (Ax, Ay, Bx, By, Cx and Cy). At this point we realize that we can solve the problem without applying sophisticated methods (using stiffness and calculating deflections and etc.) </p>
<p>@kamran showed you a good way of solving it. Just pay attention to the nice approach he used when writing the equations of part BC - Since there are no external forces exerted on it, the total reaction of joint C must be directed towards joint B. Otherwise, there would be a total moment applied on Joint B. The same is true, of course, to point C. By finding a relationship between Cy and Cx he was able to come up with the 4th equation you just look for.</p>
| 31461 | Determining Support Reactions |
2019-10-11T14:04:21.733 | <p>In a paper in the context of controller design for DEAP actuators I came upon the following sentence:</p>
<blockquote>
<p>Furthermore, the nonlinear behavior of the
control plant requires either an adaptive or a robust design in
order to ensure a precise and accurate control.</p>
</blockquote>
<p>What is the relationship between a robust/adaptive controller and the nonlinearity of the control plant?
Isn't it somehow always necessary to design a robust controller no matter whether the control plant is linear or not?
Or are nonlinear plants in general more sensitive to parameter variations, e.g. nonlinear systems sensitive to initial conditions (chaotic behaviour)?</p>
| |control-engineering|control-theory| | <p>The authors are assuming that you're going to make a linear or mostly linear controller after finding a linearized model of the plant.</p>
<p>It's not that nonlinear plants are <em>sensitive</em> to model variations, it's that for a certain class of nonlinear plant you can treat the plant as a linear time-varying system that has time-varying disturbances.</p>
<p>So in that sort of a design paradigm, you use some design process that works with time-varying plants: either you make an adaptive controller that's constantly discovering the up-to-date plant model and controlling it in some optimal way, or you make a robust controller that will control a linear time-varying plant that varies in the way your linearization does over the range of operating points of the real, nonlinear plant.</p>
| 31467 | Nonlinear behaviour of control plant and adaptive or robust design |
2019-10-11T16:47:29.320 | <p>Since the voltage required to bridge the gap between the central electrode and the ground electrode of a spark plug can reach up to 40 000V and even more, my question is, how does the battery not die since the spark plug is grounded to it. Is all that energy lost in the ionization of the gas and thus the voltage reduced to ~12V?</p>
| |electrical-engineering|battery|car| | <ul>
<li>All current flows in a closed loop circuit.</li>
<li>The ignition coil primary current flows from the battery through the coil and returns to the battery through the chassis circuit.</li>
<li>The ignition coil secondary current flows from the ignition HT terminal through the spark plug, across the gap and returns to the ignition coil through the chassis circuit.</li>
</ul>
| 31471 | What is the voltage at the ground electrode of a spark plug? |
2019-10-12T20:37:57.487 | <p>I need a shaft that can be stretched along its length while also being resistant to torque enough to be a functional shaft (see pictures). I have a rotating sphere within another rotating sphere, hence the use of universal joints. But due to the universal joints I can't use a shaft that is long enough to slide up and down. You can see in my images where there's a shaft missing. I don't know of the existence of any rubber that is sturdy enough to withstand the torque applied while also being stretchy enough to span that gap. Perhaps there's other ways to achieve what I am trying to achieve without the use of universal joints (I am open to suggestions).</p>
<p><a href="https://i.stack.imgur.com/hMhLk.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hMhLk.gif" alt="enter image description here"></a> </p>
<p><a href="https://i.stack.imgur.com/vj1cb.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vj1cb.gif" alt="enter image description here"></a></p>
<p>Edit:</p>
<p>Perhaps I could 3D print a sort of origami shaft that unfolds?</p>
<p><a href="https://i.stack.imgur.com/N6jan.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N6jan.jpg" alt="enter image description here"></a></p>
<p>This is getting to be a bit ridiculous lol what I'm trying to make shouldn't be rocket science</p>
| |materials| | <p>There are sliding shafts used for agricultural machinery commonly called a "lemon shaft" and they are available in several sizes.</p>
| 31485 | Is there such a material that is resistant to torque while easily stretching when under torsion? |
2019-10-13T10:12:42.903 | <p>When analysing a structure using traditional methods (e.g. Hardy Cross, Slope Deflection, Force Method, etc.), it is a first-order elastic analysis. My question is, how would one analyse a statically indeterminate frame by hand but taking into account the second-order effects. This is more from an academic perspective and that is why I am asking for hand calculations using traditional methods or even direct stiffness method. </p>
| |structural-engineering|structural-analysis| | <p>After completing some further research, I have come across a few approximate methods which were described in the book: Stability Design of Steel Frames. These include the two-cycle iterative method and modified slope deflection equations which can both analyse statically indeterminate structures while taking into account second-order effects. Thanks for all the answers. </p>
| 31493 | Second Order Elastic Analysis |
2019-10-13T23:43:46.280 | <p>How can I calculate the power and torque required for the motor on a wheeled robot or vehicle if a particular acceleration or movement up an incline is required?</p>
| |motors|torque|wheels| | <p>The required tractive torque of the motor shall be equal to the tractive torque at wheel <span class="math-container">$$T_w = F_t \cdot r_w,$$</span> (i.e. Tractive Torque = Tractive Force x mean wheel radius).
Where,</p>
<ul>
<li><span class="math-container">$r_w$</span> is the mean wheel effective radius</li>
<li><span class="math-container">$F_t = F_r + F_g + F_d + F_{ie}$</span>
with,</li>
<li><span class="math-container">$F_r$</span> the tyre rolling resistance (can be in the form of <span class="math-container">$C_{rN}$</span> - simplified and treated as independent of velocity)</li>
<li><span class="math-container">$F_g$</span> the forces due to gradient (depending on slope angle, can be positive of negative)</li>
<li><span class="math-container">$F_d$</span> the aerodynamic drag (as a function of air density, drag coefficient, vehicle cross sectional area, and squared of vehicle velocity)</li>
<li><span class="math-container">$F_{ie}$</span> = equivalent inertial force (during acceleration) - (including linear and rotational inertias, due to vehicle mass and rotating component of gear train and wheels).</li>
</ul>
<p>So you must determine the acceleration required to reach you designed speed.
And must also include overall transmission efficiency in order to get the right size of the motor.</p>
<p>Just to visualize for 1000 kg car and everything else constant:
on flat road @ 25 km/h requires 1.0 kW power;
on 5<span class="math-container">$^\circ$</span> gradient climbing @ 25 km/h requires 6.9 kW power;
on flat road accelerating @ 1.4 m/s<span class="math-container">$^2$</span> (i.e. 0 - 25 km/h in 5 sec) requires 10.6 kW power.</p>
<p>So you really need to determine you design specifications in order to size the right motor...</p>
| 31501 | How can I calculate the power and torque required for the motor on a wheeled robot/vehicle? |
2019-10-14T14:38:14.007 | <p>I'm totally ignorant in this area, and probably didn't know how to properly google my question - sorry..</p>
<p>I found this specification here: (<a href="https://si.shimano.com/pdfs/ev/EV-RH-IM10-2045B.pdf" rel="nofollow noreferrer">https://si.shimano.com/pdfs/ev/EV-RH-IM10-2045B.pdf</a>). My guess is that M10 is the about the metric thread specification (<a href="https://en.wikipedia.org/wiki/ISO_metric_screw_thread" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/ISO_metric_screw_thread</a>). But what is the meaning of <code>B.C. 3/8"</code> then?</p>
<p>Another question about this spec: (<a href="https://si.shimano.com/pdfs/ev/EV-FH-2200-2202B.pdf" rel="nofollow noreferrer">https://si.shimano.com/pdfs/ev/EV-FH-2200-2202B.pdf</a>) - what would be the meaning of <code>6-11/32"</code> - it can't be a range, or can it?</p>
| |mechanical-engineering| | <p>It looks like the hub axle has two different threads on it: M10 probably in the middle and 3/8" either side of the middle section. Not sure what the "B.C." refers to.</p>
<p><a href="https://i.stack.imgur.com/F3XZ6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F3XZ6.png" alt="enter image description here"></a></p>
<p>This is confirmed by the fact that the hub nuts (1), which go on the extremities are 3/8", whereas the lock nuts (18), which go on the middle section, are M10. </p>
<p>I think <code>6-11/32"</code>means 6 inches and (or plus if you prefer) 11/32 of an inch. Indeed, if you do:</p>
<p><code>(6 + 11/32) * 25.4 = 161.77mm</code></p>
<p>and the spec says 161mm. You can apply the same logic for 16mm and 6-17/32:</p>
<p><a href="https://i.stack.imgur.com/8lN9S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8lN9S.png" alt="enter image description here"></a></p>
| 31509 | What are the meanings of `B.C. 3/8"` and `6-11/32"`? |
2019-10-15T02:47:38.490 | <p>This is a prework for a lab, I haven't seen things in the picture before and have no idea on how to transfer this to an equation.</p>
<p><a href="https://i.stack.imgur.com/pExay.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pExay.png" alt="enter image description here"></a></p>
| |signal| | <p>The wavy symbols are integrators, which means that:</p>
<p><span class="math-container">$$\frac{dy}{dt} = x_1$$</span>
<span class="math-container">$$\frac{dx_1}{dt} = x_2$$</span></p>
<p>Then the "+" sign gives you (the triangles are "gains", which means multiply):</p>
<p><span class="math-container">$$x_2 = x(t) - a_0 y(t) - a_1 x_1$$</span></p>
<p>If you combine all this together, you get your differential equation(s).</p>
| 31526 | How to read the block diagram for the 2nd order integrator feedback system? |
2019-10-15T10:12:17.393 | <p>What is the difference between perpendicular and transverse loads? I saw this on a forum for 3d printing, but I couldn't find any online explanations on what the difference between both measures of loading was.</p>
<p><a href="https://i.stack.imgur.com/NDysz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NDysz.png" alt="enter image description here"></a></p>
| |mechanical-engineering|3d-printing| | <p>3D Printed objects are made of layers, and are intrinsically weaker in one direction, as you can 'pull the layers apart'. In this case 'transverse' means 'in the vertical direction as built', and 'perpendicular' means 'in any of the horizontal directions as built'.</p>
<p>The graph in question is with reference to infill patterns - these are generated in the slicer and can either be 2D (eg hexagonal tubes, the cross section is the same from one layer to the next) or 3D (eg cubes on their corners). This non-uniformity of the 2D patterns is why the patterns fail in compression (what the graph shows) differently according to the orientation. Imagine crushing a toilet roll tube from the top vs from the side.</p>
| 31531 | Difference between perpendicular and transverse loads? |
2019-10-15T20:49:59.717 | <p>My question essentially is how much water to power green house LED lights. Specifics bellow :</p>
<p>Say I have a warehouse with a flat roof with a bunch of water tanks or the whole thing being a water tank. Using say 30 feet High and 20'Lx20'W. How much power could I generate utilizing the roof to store water?</p>
<p>One lighting fixture could run 400 watts. </p>
<p>I was pondering these last few days on this, I've googled a couple of equations but not sure I used the right ones. Thanks for your input in advance.</p>
| |electrical-engineering|electrical| | <p>Not very much.</p>
<p>One liter of water has a mass of about 1kg. The acceleration due to gravity is 9.8m/s^2. One Joule is one Newton exerted over one meter. So one liter of water falling one meter loses 9.8 Joules of potential energy.</p>
<p>Your roof is 9 meters high, and has an area of 36 meters (6m * 6m). If you cover that with a meter of water that's 36 cubic meters, or <span class="math-container">$36 \cdot 10^3$</span> liters. The average drop is 9.5 meters, so you get <span class="math-container">$w = (9.8 \mathrm{m/s^2})(9.5 \mathrm{m})(36 \cdot 10^3 \mathrm{l})(1 \mathrm{kg/l}) = 3.35 \mathrm{M J}$</span>.</p>
<p>Which sounds pretty impressive, until you do just a bit more math and discover that's 0.93 kilowatt-hours, or enough power for your 400W panel for about two hours, twenty minutes -- and that's assuming 100% energy conversion, where less than 50% is probably more realistic.</p>
<p>There's some <span class="math-container">$x^2$</span> terms in there, but you're talking about needing a water tank that's taller than your greenhouse, and enough rain to keep it filled.</p>
| 31537 | How much rain water run off do I need to power an LED greenhouse |
2019-10-16T17:57:16.913 | <p><img src="https://i.stack.imgur.com/PtbYw.jpg" alt="enter image description here"></p>
<p>I don't know how this is made and I don't know the name of the method to search for it to learn more.</p>
| |manufacturing-engineering|aluminum| | <p>If the extruder has a die in the exact shape you're wanting, it might be worth asking them. If you can get past the extruder's quote filter (our's is at hundreds of thousands of pounds per year), if they don't have a die for the part you're wanting, you'll be buying an extrusion die (think <span class="math-container">$3k - $</span>6k). Do you want close tolerancing, that'll be extra. Depending on the cross-sectional area of the part, you could wind up with 1,000's of parts just from one billet, which you're paying for as well. With a solid shape, very little is wasted, except for maybe the charge-welds and an inch or two of the billet butt-discard (where the impurities, oxides, etc are, you don't want those). If you're planning on purchasing a lot of parts or parts that the extruder already does, you can get it done, but they aren't interrupting their schedule for a billet or two of a one-off. To explain the process the easiest, we always tell our customers to imagine a very large Play-Doh fun factory, except the Play-Doh is near 1,000f, and instead of your hand providing the horsepower, there's probably 4 huge hydraulic pumps supplying 5,000 tons of it.</p>
| 31543 | What is the manufacturing method used to make aluminium C channel |
2019-10-17T02:42:57.120 | <p>In my class I'm in, our lecturer presented frequency response methods such as the nyquist plot and bode diagrams. Then when designing a lead compensator later, he said one of the specifications should be a steady state error to a step response below a certain value. Why do we consider step responses? I have been reading ogata's book and the first sentence in the frequency response chapter says that frequency response is specifically for the case of sinusoidal inputs. </p>
<p>It doesn't make sense to me to analyse the frequency response over a large range of <span class="math-container">$\omega$</span> either since a non repeating signal has an infinite frequency anyway? </p>
<p>If someone could explain this it would be helpful, I can't post my lecture slides here, but I hope I was clear enough.</p>
| |control-engineering| | <p>In linear analysis, any real input can be decomposed into a set of step inputs, and the response is just the superposition of the step responses.</p>
| 31548 | Why do we analyse frequency response for step inputs? |
2019-10-17T15:37:26.020 | <p>Why don't engineers build highways with corners such those on NASCAR speedways?</p>
| |highway-engineering|roadway| | <p>Public roads must be safe to handle slow, heavy traffic and be safe to pull over a stalled car under a snow storm.</p>
<p>They should be flat enough to make it easy for emergency services such as ambulances, firemen to set up a triage and evacuate or help injured people out.</p>
<p>There are standards for road curves and turns and maximum slopes both uphill and sideways, and signage.</p>
<p>Provision for safety at night and black ice conditions and safe visibility around the turns at night should be put in place.</p>
<p>They should be sloped gently to allow for surface runoff water to drain.</p>
<p>Highway design is a specialty in civil engineering. A safe highway will save hundreds of lives over time.</p>
| 31559 | NASCAR corners on public roads |
2019-10-17T21:11:13.317 | <p>What kind of professional/architect/engineer builds agricultural systems, specifically automated hydroponic systems? </p>
<p>Things such as growing, cultivation, lighting, plumbing, harvesting, processing and packaging systems, all in one facility, much like a factory.</p>
<p>For example, if I wanted to setup such a facility, who would I have to consult?</p>
| |mechanical-engineering|structural-engineering|control-engineering| | <ul>
<li><p>Someone with a background in agriculture, biology or biotechnology can design the basic process and determine the principle mass and energy flows - how much area, how much artificial lighting if any, how much water, air changes if aneclosed building, nutrient or fertilizer dosing etc.</p></li>
<li><p>The basic design stage should allready consider the logistics of your plant, but also things like workplace health and safety</p></li>
<li><p>Civil and structural engineers should design the main vessels and the general plant layout</p></li>
<li><p>There are a number of works you specified that would be their own mechanical engineering speciality: HVAC but mostly processing and packaging equipment. Others, like plumbing/pumping can probalby be done by most MEs with some experience in plant design</p></li>
<li><p>Electrical and control systems are on or two additional engineers you need.</p></li>
</ul>
<p>In theory, the process designer designs the process, then the civil engineers size and place the vessels in their drawings, then the ME figure out the plumbing and at last the MEs show up to design the electrical systems for the machinery. In actuality, you will iterate through this process or some steps of it a few times.</p>
<p>This is about a dozen specialities you need. However, sometimes you can buy expertise along with equipment, this migh be the case with processing or packaging equipment or dosing stations for nutrients. You still need someone on your team with a basic understating to talk to them, however!</p>
<blockquote>
<p>For example, if I wanted to setup such a facility, who would I have to consult?</p>
</blockquote>
<p>You either find a consulting engineering firm with a lot of experience in designing and planning (and building!) such a plant. Or you find anexperienced hydroponics expert (the first bullet point) and a consulting firm with experience in a relevant field like wastewater.</p>
<p>Alternativly, you find a company specialized in building and selling hydroponics farms. This will give you less choice over the final plant - the company will want to work with their preferred components and concepts, and crucially they will want to reuse as much egnineering work from previous projects as they can.</p>
| 31563 | What type of engineer/professional designs hydroponic farms? |
2019-10-18T06:28:36.360 | <p>A friend of mine asked me this question, but I don't know if it is possible or not. If it is possible, please show me how to construct it. If it is not possible, please prove it. Thanks.</p>
| |technical-drawing| | <p><a href="https://i.stack.imgur.com/oNZYu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oNZYu.png" alt="object"></a></p>
<p>Created in Fusion 360, this is a 3D object with a cross section of a circle when viewed from above, a triangle when viewed from one side and a square when viewed from another side 90 degrees from the first.</p>
<p>ortho views below:</p>
<p><a href="https://i.stack.imgur.com/59Puj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/59Puj.png" alt="circle cross section"></a></p>
<p><a href="https://i.stack.imgur.com/TXeoH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXeoH.png" alt="square cross section"></a></p>
<p><a href="https://i.stack.imgur.com/U9BT8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U9BT8.png" alt="triangle cross section"></a></p>
<p>The process is thus: </p>
<ol>
<li>sketch and extrude a square, creating a cube. select a face/plane,</li>
<li>sketch and extrude a triangle, boolean intersect the two solids</li>
<li>select a face/plane, sketch and extrude a circle, boolean intersect.</li>
</ol>
<p>This is the sequence in creating an ambigram, a three dimensional object in which the appearance changes when viewed from different (orthogonal) angles. </p>
| 31567 | Is is possible to form a 3D object from the front, side, top views with a triangle, a square and a circle respectively? |
2019-10-18T13:13:01.887 | <p>I'd like to know the tensile strain beneath the asphalt concrete (AC) layer in my design structure. Usually, the strain is used to calculate the number of axles (Nf) that the pavement structure can take before failure. Do you know how/where I can find that?</p>
| |pavement| | <p>The tensile strain beneath the AC layer is not an output in Pavement ME Design due to the vast number of calculations performed within the Asphalt Pavement Design System (APADS) analysis module. </p>
<p>The size of the output file is too large to output (3 – 10gb or more). The reason for the large file size is due to the analysis being performed for each month, sub-month, sub-seasons, vertical locations, all axle configurations, axle load spectra’s for each axle configuration, and all traffic wander locations.</p>
<p>The “.fat” intermediate output file summarizes the monthly fatigue damage for the vertical depth and wander locations. The maximum fatigue damage value between all wander locations for the last vertical depth location is used for the bottom-up fatigue cracking calculations.</p>
| 31573 | Does Pavement ME Design output the tensile strain beneath an AC layer? |
2019-10-18T22:35:52.463 | <p>Does the quality of a microphone affect the frequency and intensity(dB) of the sound it can pick up?</p>
<p>Do all microphones pick up sound as it is in real life? </p>
| |structural-engineering| | <p>TimWescott is right: not all microphones are alike in their output level and frequency response. Microphones with good frequency response are expensive in rough proportion to their bandwidth. In the recording industry, it is common to choose a specific microphone brand and type for recording a specific instrument or voice because its response matches the output of the source. </p>
| 31581 | Does the quality of a microphone affect the frequency and intensity(dB) of the sound it can pick up? |
2019-10-19T05:08:26.607 | <p>LMTD of a counter flow heat exchanger is 20°C and cold fluid enters at 20°c & hot fluid at 100°C, mass flow rate of cold fluid is twice that of mass flow rate of hot fluid. <span class="math-container">$C_p$</span> of hot fluid is twice that of cold fluid's. How to find the exit temp. of Cold fluid?</p>
<p>I did it in this way.</p>
<p>Its 40°C</p>
<p><span class="math-container">$$\begin{align}
LMTD &= \dfrac{\Delta T_1 - \Delta T_2}{\ln\left(\dfrac{\Delta T_1}{\Delta T_2}\right)} \\
\Delta T_2 &= \Delta T_1 \\
\therefore LMTD &= \dfrac{0}{0}
\end{align}$$</span></p>
<p>Let <span class="math-container">$\Delta T_2 = x\cdot \Delta T_1$</span></p>
<p><span class="math-container">$$\begin{align}
LMTD &= \dfrac{\Delta T_1 - x\cdot \Delta T_1}{\ln\left(\dfrac{\Delta T_1}{x\cdot \Delta T_1}\right)} \\
&= \dfrac{(1-x)\Delta T_1}{\ln\left(\dfrac{1}{x}\right)} \\
&= \dfrac{(1-x)\Delta T_1}{-\ln x}
\end{align}$$</span></p>
<p>Applying L'Hopital Rule:</p>
<p><span class="math-container">$$\begin{align}
LMTD &= \lim_{x\rightarrow1} x \cdot \Delta T_1 \\
&= \Delta T_1 = \Delta T_2
\end{align}$$</span></p>
<p>20 = Cold outlet Temp - (Cold inlet Temp = 20)</p>
<p>Cold outlet Temp = 40</p>
<p>Similarly, The Output Temp. of Hot Fluid is 80°C</p>
<p>Is this approach correct?</p>
| |heat-transfer|heat-exchanger| | <p>Your approach is one method to solve <a href="https://www.chegg.com/homework-help/questions-and-answers/balanced-heat-exchanger-one-capacitances-fluids-equal-e-effectiveness-affected-show-effect-q1665483" rel="nofollow noreferrer">a balanced heat exchanger</a>. I might instead have simply divided the equation by <span class="math-container">$\Delta T_2$</span> to obtain</p>
<p><span class="math-container">$$ \Delta T_{LMTD} = \Delta T_2 \left[\frac{R - 1}{\ln(R)} \right]$$</span></p>
<p>Finding the limit as <span class="math-container">$R \rightarrow 1$</span> gives the same answer.</p>
<p>An alternative approach with the above is to graph the numerator <span class="math-container">$R - 1$</span> and denominator <span class="math-container">$\ln(R)$</span>.</p>
<p><a href="https://i.stack.imgur.com/8dP6l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8dP6l.png" alt="Graph of numerator (red) and denominator (blue) of function"></a></p>
<p>The graph shows in the limit as <span class="math-container">$R \rightarrow 1$</span> that the ratio of numerator to denominator approaches unity.</p>
| 31585 | Heat Exchangers |
2019-10-20T21:38:33.920 | <p>I am a graduate student trying to sharpen the direct stiffness method skills. However, I am stuck at the beginning.</p>
<p>The problem is given in the image.</p>
<p>Can someone help me to write down all the nodal and member forces with equations ?<a href="https://i.stack.imgur.com/T1Tro.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T1Tro.jpg" alt="enter image description here"></a></p>
| |civil-engineering|structural-analysis|finite-element-method|homework| | <p>Because the specific value are not given, I will give a general overview of the method. When applying the direct stiffness method for trusses, the following steps must be used:</p>
<p>Find the local stiffness matrix for each individual element as following:</p>
<p><span class="math-container">$\begin{bmatrix}AE/L & 0 & -AE/L & 0 \\0 & 0 & 0 & 0\\-AE/L & 0 & AE/L & 0 \\0 & 0 &0 &0\end{bmatrix}$</span></p>
<p>You also need a beta matrix in order to convert the local coordinate system into global coordinates.</p>
<p><span class="math-container">$\beta = \begin{bmatrix} cos\theta & sin\theta & 0 & 0 \\-sin\theta & cos\theta & 0 & 0\\0 & 0 & cos\theta & sin\theta \\0 & 0 &-sin\theta & cos\theta\end{bmatrix}$</span></p>
<p>Then, <span class="math-container">$[\beta]^T[k_l][\beta] = [k_g]$</span></p>
<p>Complete the same process for all matrices and then assemble global matrices to create the system matrix [K]. Apply boundary conditions for the problem.</p>
<p>Then solve for the nodal displacements by using the equation, <span class="math-container">$[K]^{-1}[F] = [U]$</span></p>
<p>Where F is the forces at the joints and U is the nodal displacements.</p>
<p>Expand [U] into elemental displacements <span class="math-container">$[U_{1,2,3}]$</span></p>
<p>Multiply each by their own original beta matrices in order to produce <span class="math-container">$[\delta_{1,2,3}]$</span></p>
<p>Multiply each original stiffness matrices by their delta values like so:
<span class="math-container">$[k_l][\delta_{1,2,3}] = [F_{e 1,2,3,}]$</span></p>
<p>This will produce the elemental forces. </p>
| 31602 | 2D- Truss Member Analysis |
2019-10-22T12:33:02.960 | <p>I have a bar which I want to move as if it was pivoted on a point beyond its end.</p>
<p>e.g.</p>
<pre><code>-------- *
/
/
/
/
</code></pre>
<p>Is there a linkage, or other mechanism, which can let a bar move as if it were pivoted at a point beyond its end (the asterisk), yet not extend as far as the (virtual) pivot point?</p>
<p>The mechanism can extend to the left, above and below the bar, but cannot extend towards the (virtual) pivot point.</p>
<p>(I have already thought of using two arcs to constrain the ends of the bar, but that seems prone to jamming.)</p>
| |mechanical-engineering|mechanisms| | <p>Many of the car hoods or trunks do exactly what you are about to do. They open using a trapezoid link and a spring loaded shock absorber. They do this to give more space than a simple hinge, so the mechanic can work under the hood.</p>
<p>Some garage doors use the same mechanism and you can buy the parts from hardware stores.</p>
| 31618 | Is there a linkage or other mechanism which can simulate a pivot, without extending that far? |
2019-10-22T16:52:48.690 | <p>I've been doing some reading into spinning wheels and old-fashioned sewing machines and so far as I can tell, they use a system similar to a piston in an engine. The main difference being, the oscillation is coming from a treadle. It got me wondering about something though. With the way the linkage works, rotation to oscillation should work fine at any speed. That being said, couldn't oscillation to rotation be messed up if the oscillations are happening too slowly?</p>
<p>If the connecting rod doesn't pass the middle point (where the rod is angled at 0°), another oscillation might cause the connection rod to push backwards, reversing the rotation. Also, if the rod is angled at 0° (almost exactly), the linkage might jam because the oscillation isn't being directed one way or another? If that is the case, the crank/wheel would need to have enough momentum to pass that spot when the next oscillation happens?</p>
| |mechanical-engineering|mechanisms| | <p>I've actually sat at a treadle sewing machine and made seams.</p>
<p>Yes, if they're going too slow they stop. They have flywheels that lower that speed, but it's still there.</p>
<p>Note that with a treadle machine you rock the treadle, which can both push and pull on the connecting rod. With some pedal-operated machines (such as the spinning wheels I've seen) you only push on the pedal, and the connecting rod only pulls. That works on a spinning wheel because you generally want the wheel going at a constant and fairly rapid rate. With a sewing machine, you need finer control of the speed, and you want to be able to start at (almost) all positions and to go slower for delicate work.</p>
<p>With a treadle sewing machine, when you're ready to sew you use the hand-wheel on the machine to get it started (and over the top), then use the treadle to continue. Starting it by hand means that you can feel which way you need to rock the treadle to continue.</p>
<p>At least for the electric sewing machines that were in use through the 1970's, you did the same thing: the motors are generally poor at going slowly, and if you're doing delicate work you want direct control over the thing anyway -- so you use the wheel to get things started right, then the motor once you're sure you're ready to go.</p>
| 31620 | Do crank-sliders require some minimum amount of momentum to work in linear-to-rotary convertions? |
2019-10-23T07:05:52.390 | <p>I am watching this video(<a href="https://www.youtube.com/watch?v=gVLhrLTF878" rel="nofollow noreferrer">https://www.youtube.com/watch?v=gVLhrLTF878</a>) and beginning at around the 2:20 mark, it explains the significance of an expansion valve with the following ideas:</p>
<ol>
<li>Restricts refrigerant flow to lower refrigerant pressure.</li>
<li>Decreasing surrounding pressure <span class="math-container">$\lt$</span> liquid(refrigerant) pressure implies boiling liquid.</li>
<li>Decreasing surrounding pressure around a liquid allows the liquid to evaporate. The evaporation takes some of the kinetic energy from the liquid which consequently lowers the temperature of the liquid.</li>
</ol>
<p>I have listed the ideas as to how I have understood it, but number 2 and 3 seem contradictory to me. Maybe temperature and pressure do not have a linear relationship? Please, anyone, clarify any misunderstandings I have.</p>
| |pressure|compressed-air|refrigeration| | <p>Both boiling and evaporation involve changing a liquid to a gaseous state. Both are driven by the liquid/vapor system not being in equilibrium as far as the actual vapor pressure vs. The vapor pressure associated with the saturated state.</p>
<p>Boiling means the conversion process is happening throughout the bulk of the liquid and bubbles are forming. Evaporation is the same process, but it is happening on the surface of the fluid.</p>
<p>The refrigerant compressor pumps vapor. Let's assume it operates on a constant volume basis. The pumping rate is proportional to vapor density at the pump inlet. Cooling the vapor at constant pressure increases density and flow rate. Increasing the pressure at constant temperature increases density and flow rate.</p>
<p>At the expansion valve, fluid is being pulled through because the compressor is removing vapor from the exit side of the evaporator coil. The fluid at the expansion valve inlet is liquid. By design, you don't want vapor there. The exit of the expansion valve is mixed phase. There is a big pressure drop across the valve. As soon as the fluid enters the low pressure side, some of it flashes to vapor. The refrigerant will be in saturated equilibrium at the low-side pressure. Most of it is still liquid, but it is bubbly. The remaining liquid gets vaporized in the evaporator and cools the room.</p>
<p>A TXV valve like in the video has two inputs. There is a knob that adjusts the pressure drop at a given sense temperature (only to be adjusted by a qualified tech), and a temp bulb that senses the evaporator vapor temperature and changes the size of the passage in the valve. The result is a complicated balancing act.</p>
<p>If the valve closes a bit (because the bulb senses too low a temp in the vapor line), the pressure difference increases. After a bit, the low side of the valve (bubbly liquid line) will be cooler and lower pressure. And the flow rate will be less as well. The density of the vapor on the low side will be less. This is the key result. By design, closing the valve a little results in a lower vapor density at the compressor inlet, reducing flow rate, and ensuring that only vapor is getting sent to the compressor. The lower flow rate means the vapor line temp rises, providing the system's negative feedback. You can't show this by just looking at the valve - you have to look at how the entire system responds to the valve, and it is designed so that it responds this way when working properly and properly charged. It may not work this way if there are problems in the system. Lots of these valves get replaced when they aren't bad, but the tech didn't find the real problem.</p>
<p>The TXV valve effects a change to the refrigerant flow rate. It senses the vapor line temperature, and changes the flow rate to keep the vapor line temperature above the saturation temperature. Most are pressure compensating so that in effect, they do a decent job of maintaining a fixed superheat in the low side vapor line.</p>
<p>Compare with AXV, automatic expansion valve - <a href="https://neilorme.com/AEV.shtml" rel="nofollow noreferrer">https://neilorme.com/AEV.shtml</a></p>
| 31627 | What is the idea behind an expansion valve? |
2019-10-23T23:30:54.810 | <p><a href="https://i.stack.imgur.com/mnUcu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mnUcu.png" alt="diagram"></a>
To calculate the moment of inertia at o, I tried to apply (1/3mq^2) but I am not sure if it is correct. Any help would be appreciated.</p>
| |dynamics| | <p>It looks like the question assumes the rod to be of negligible mass. </p>
<p>Then the rotational moment of inertia is just <span class="math-container">$I=mq^2 $</span></p>
<p>You may have confused the moment of inertia of a rod of mass m with length q. </p>
| 31634 | Moment of Inertia |
2019-10-24T13:29:34.157 | <p>I've been looking at ways to re-concentrate a formic acid-water binary mixture with the VLE diagram below.</p>
<p><a href="https://i.stack.imgur.com/vaJE8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vaJE8.png" alt="enter image description here"></a></p>
<p>Am I correct in saying that to concentrate a solution of say , 40% formic acid by weight I could (excluding all recycle streams for simplicity) implement a distillation column at 1 bar to produce a distillate at the azeotrope (78%) and then pass this distillate into a second column operating at 0.1 bar to distill further to 95%?</p>
<p>Aditionally, would operating the 1bar column for a distillate of say only 70% (past the 0.1 bar azeotrope) then performing the remaining concentration in the low pressure column result in any significant energy savings due to less steam required at the lower pressure?</p>
<p>Lastly, if I was to START with a solution of 85% formic acid, already beyond both azeotropes, I could just perform simple distillation in a single column at either pressure point to obtain the 95% product?</p>
| |chemical-engineering| | <p>OP Quote:</p>
<blockquote>
<p>Am I correct in saying that to concentrate a solution of say , 40% formic acid by weight I could (excluding all recycle streams for simplicity) implement a distillation column at 1 bar to produce a distillate at the azeotrope (78%) and then pass this distillate into a second column operating at 0.1 bar to distill further to 95%?</p>
</blockquote>
<p>Answer: Yes</p>
<p>OP Quote:</p>
<blockquote>
<p>Additionally, would operating the 1bar column for a distillate of say only 70% (past the 0.1 bar azeotrope) then performing the remaining concentration in the low pressure column result in any significant energy savings due to less steam required at the lower pressure?</p>
</blockquote>
<p>Answer: Significant is relative. But if the feed solution for the 1 bar column is 70 wt% as compared to 40 wt% in the first paragraph, you will save energy in operating at the 70 wt% specification because you will have less stages to get to the azeotrope and likely less vapor to generate to meet formic acid production rate targets. The lower pressure column will work and use the same amount of energy whether the feed to the first column is 40 wt% or 70 wt% because the second column always has azeotrope condition feed.</p>
<p>OP Quote:</p>
<blockquote>
<p>Lastly, if I was to START with a solution of 85% formic acid, already beyond both azeotropes, I could just perform simple distillation in a single column at either pressure point to obtain the 95% product?</p>
</blockquote>
<p>Answer: Yes</p>
| 31647 | Azeotropic Distillation - Am I interpreting this VLE diagram correctly? |
2019-10-24T22:06:22.650 | <p>Why is it that a total condenser and partial reboiler are used in a distillation column and it is not the other way around?</p>
| |chemical-engineering| | <p>Partial reboiler is related to that fact that most times, there is material left in the bottom of the column. This can be because you need to maintain a level of liquid because the bottom take off is a purified stream you want (as product or waste treatment input) or there is not enough area to maintain the proper amount of boil up.</p>
<p>The total condenser is related to the fact that most times you want to condense the purified upper stream as that is the product (and since liquids have lower specific volumes than gases, it makes storage and transport more efficient) or you want a liquid stream into a second column or other downstream unit operation.</p>
<p>It is not a law of the universe that the partial reboiler and total condenser are the only way distillation can be done, but it is very common in current industries and so it is often discussed.</p>
| 31656 | On partial reboiler and total condenser in distillation column |
2019-10-25T04:54:17.300 | <p>Most electrical transformers that I am aware of have either copper or aluminum cable wound around an iron core.</p>
<p>Can the iron core be replaced with a copper core? Would there be any performance differences?</p>
| |electrical-engineering| | <p>It would not be a good idea at all - a transformer is effectively a magnetic circuit, so you need a low reluctance path for the magnetic flux to flow through. Copper has a high magnetic reluctance, so would not be a good choice.</p>
<p><a href="https://en.wikipedia.org/wiki/Magnetic_reluctance" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Magnetic_reluctance</a></p>
<p><a href="https://en.wikipedia.org/wiki/Permeability_(electromagnetism)#Values_for_some_common_materials" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Permeability_(electromagnetism)#Values_for_some_common_materials</a></p>
| 31659 | Can the core of electrical transformers be made solely from copper? |
2019-10-25T16:12:07.693 | <p>I am confused whether a lead screw and a worm screw are the same exact thing. In my engineering workshop lab, they come up as separate things. For example, a worm screw is there in the indexing head of the milling machine (as part of the worm drive), while the lead screw is part of the feed mechanism in the milling machine and the lathe machine. They both look like the exact same thing. Can someone tell me what's the difference?
<a href="https://en.wikipedia.org/wiki/Leadscrew" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Leadscrew</a>
<a href="https://en.wikipedia.org/wiki/Worm_drive" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Worm_drive</a></p>
| |mechanical-engineering|terminology| | <p>One way to think of it is this:</p>
<p>A Worm, it's a short "threaded cylinder" with relatively few teeth. This rotates and in doing so, moves a worm wheel with much more teeth than it. Typically this results in a rotating motion as @Eirc Shain suggested, but it could also be used to drive a long Rack (see 'Rack and Pinion') in a linear way. If the rack has more teeth, and it's engaged "from the side" and not wrapped around the threaded cylinder, then that cylinder is a worm. While it may look like a thread form, it's actually more correct to think of them as elongated teeth, with the number of thread starts being effectively the number of teeth.</p>
<p>On the other hand, a lead screw is much longer and with more thread turns than the item that engages with it (usually a nut or a half nut), and that engagement wraps around the cylinder. This always outputs a linear motion.</p>
<p>Does this make things any clearer?</p>
| 31666 | Difference between lead screw and worm screw? |
2019-10-25T22:55:48.497 | <p>For an application of displacement measurement under 1 kHz, I have 3 options to choose from. The application is automotive. Unfortunately I don't have deep electronics knowledge and the catalogues contain so many abbreviated expressions in the names of their connectors, etc.</p>
<p>What table of comparison in parameters would you suggest to classify them as best and to also justify them with my needs in this frequency range?</p>
<ul>
<li>The temperature range is up to 1000°C.</li>
<li>The movement range is ±50mm</li>
<li>Frequency : automotive range but let's say 1 kHz..
bolting can be the solution due to the temperatures.</li>
</ul>
<p>I also have seen a table like this, but I would like to widen it to more parameters :</p>
<p><a href="https://i.stack.imgur.com/AQZMg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AQZMg.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|automotive-engineering|vibration| | <p><strong>Edited</strong> based on new information provided</p>
<p>The most important thing to start with is: what resolution DAQ do you need? 16 bit is fairly standard, cheaper ones will be 12 or 14 bit. more expensive ones will be 20 or 24 bit. </p>
<p>Based on the specs you provided, Firstmark series 173-176, the conversion is 165 mm = supply voltage. Max supply voltage is 35VDC, but you could use anything. You didn't provide the min frequency you are interested, so let's just assume you need all the way down to 0 Hz, i.e. needs to be DC coupled. Most basic DAQ systems would be able to read -10V to 10V. You could get something that would read up 35V, you'll need something more expensive and complicated, so just assume you are keeping it simple and use 10V supply voltage. </p>
<p>A 16 bit ADC would set +32768 = 10V, and -32767 = -10V (note 2^16=65536, and we are using half that range for positive voltages and half for negative, hence 65536/2=32768... note that as I'm assuming DC coupled above, half the voltage range is unused... i.e. the DAQ system is setup to use -10 to +10, but you'll only be using 0 to 10V. If you can assume AC coupled you can use the whole range, but in that case the minimum frequency you can read would be a few hertz, won't get DC). So the smallest voltage you could read is 0.0003V. </p>
<p>With the assumption that 10V=165mm, this means the smallest displacement you could read is 0.005 mm. That would appear to meet your specifications. So a 16 bit DAQ should meet your needs, you don't need to spring for anything more expensive than that. On the other hand, a lot of cheaper systems might be 12 bit. Those will be 16 times worse resolution, i.e. minimum value you could read would be 0.08 mm. Probably would not work for you. </p>
<p>you were also asking about "40 kS/s", that's 40,000 Samples/second. That is not the maximum frequency that you can record however. General rule of thumb would be divide sample rate by 2.5 to get the maximum frequency you can record, which is called the "bandwidth". In this case 40k/2.5 = 16 kHz. You are referencing 7.2 kHz in the comments, which is less that I would have expected for a 40k sample rate. But since you said that you only care about signals at 300 Hz or maybe 1kHz anyway, this will be plenty for you. </p>
<p>while I'm thinking about it, another thing you should be looking at is whether the DAQ has built in anti-aliasing filters. Virtually all of the good ones do, but some cheaper ones do not. Unless you are 100% sure you know what you are doing, make sure you get one with an anti-aliasing filter. </p>
<p>I'll also say that most of the big DAQ manufacturers have pretty helpful sales engineers. Just call the company up and tell them what you are trying to do. They can probably point you in the right direction. </p>
| 31674 | How to compare multiple data acquisition systems? |
2019-10-28T15:53:00.417 | <p>I am attempting to solve ANSYS mechanical (via workbench) using GPU acceleration, but I keep getting the following error: </p>
<blockquote>
<p><strong>* WARNING *</strong> CP = 79.594 TIME= 15:40:08
The GPU accelerator capability is not valid when using the memory<br>
saving option (MSAVE command) for the PCG solver. The GPU accelerator<br>
capability is disabled for this solution. </p>
</blockquote>
<p>How can I turn off the MSAVE option? It seems to be default for a model this size and I cannot find the way to turn it off.</p>
<p>I am using V19.2.</p>
| |ansys|ansys-workbench|ansys-apdl| | <p>I found the answer in the end with a lucky guess: </p>
<p><a href="https://i.stack.imgur.com/jl8OA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jl8OA.png" alt="enter image description here"></a></p>
<p>Simply enter "MSAVE, off" into the command line arguments in the advanced solver settings. </p>
| 31689 | Using GPU Acceleration in ANSYS Mechanical |
2019-10-29T00:35:06.963 | <p>I'm a software engineer that works on custom embedded devices. I program these devices through a typical RJ-45 ethernet cable. In most cases, I can connect my computer to these devices through a typical network switch and update multiple devices at the same time.</p>
<p>However, some devices are not compatible with others, and will cause issues if connected. For example:</p>
<p><a href="https://i.stack.imgur.com/u28Js.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u28Js.png" alt="Isolated Network Design"></a></p>
<p>I cannot have traffic from Device A reach Device B, or vice versa. Typically, I only need to be connected to one at once, so my current solution is just moving an ethernet cable that is connected to my computer to Device A or Device B, but I'd like to improve that.</p>
<p>I'm looking to either build or buy a togglable network switch, where the system can be in one of two states: Computer to Device A (and vice versa) or Computer to Device B (and vice versa). Basically, I minimally need to toggle ports on a network switch on and off.</p>
<p>I haven't been able to find something like this online, so I was wondering if I can rig a switch up by splicing some network cables and toggling them in some way, but I'm worried about interference and potential byte corruption.</p>
<p>Any initial suggestions?</p>
| |embedded-systems|computer-hardware|networking| | <p>I ended up finding the ethernet toggle switch on Amazon that I was looking for. This works perfectly for what I need:</p>
<p><a href="https://smile.amazon.com/gp/product/B077NYH15H/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1" rel="nofollow noreferrer">https://smile.amazon.com/gp/product/B077NYH15H/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1</a></p>
<p>I think the term I was originally looking for was an "Ethernet Selector".</p>
| 31694 | Network Switch with Togglable Ports |
2019-10-29T08:02:44.783 | <p>I have been drifting RWD (rear wheel drive) cars for some years now. I tend to think about every engineering solution that other drivers use and how does it work before applying it to my car.</p>
<p>What has been baffling me from some time now is: <strong>Why using LSD (limited slip differential) results in bigger speeds when drifting?</strong></p>
<p>Food for thought:</p>
<ul>
<li><p>Welded differentials (100% lock) are very common in this motorsport since you are dealing with kinetic friction on rear tires anyway - which is constant regardless of their rotation speed (unless you match tire speed to ground and grip using static friction).</p></li>
<li><p>Open differentials (0% lock) on the other hand practically make drifting (powersliding) unusable as almost all of the engine torque is routed to unloaded wheel resulting in slowing down.</p></li>
<li><p>Slow motion drifting video: <a href="https://www.youtube.com/watch?v=OG0cyjqDJCw" rel="nofollow noreferrer">https://www.youtube.com/watch?v=OG0cyjqDJCw</a></p></li>
</ul>
| |friction|wheels|power-transmission|traction| | <p>To facilitate my explanation I'll follow the same scenario and the following definitions:<br>
I'll consider all explanation considering the car doing a left curve.<br>
<span class="math-container">$W_{rL} =$</span> For Left Rear Wheel<br>
<span class="math-container">$W_{rR} =$</span> For Right Rear Wheel<br>
<span class="math-container">$WD = $</span> for Welded differential<br>
<span class="math-container">$LD = $</span> for Limited-slip differential (btw Torsen is a type of <span class="math-container">$LD$</span>)<br>
<span class="math-container">$OD = $</span> for Open differential<br>
<span class="math-container">$r_v = $</span> for rotation speed on the wheel<br>
<span class="math-container">$L_F = $</span> for Lateral force (byproduct of the mass of the vehicle) </p>
<p>When you start cornering what makes the car slip (drifting) is the resultant force from the sum of the force from rotation of the wheel and the lateral force resultant from the mass of the vehicle.</p>
<p>In the normal cornering (without drift) the <span class="math-container">$W_{rL}$</span> will rotate slower than the <span class="math-container">$W_{rR}$</span>. And the lateral force will be concentrated on the right side of the car.<br>
The <span class="math-container">$W_{rR}$</span> will be subjected to more lateral force than <span class="math-container">$W_{rL}$</span>. </p>
<p>With <span class="math-container">$OD$</span> the car will first accelerate <span class="math-container">$W_{rR}$</span> until the wheel start slipping. On this moment almost all the power of the car goes to <span class="math-container">$W_{rR}$</span> and your car will desacelerate until the slipping stops (and if you continue accelerating) there will be a loop without never slipping on <span class="math-container">$W_{rL}$</span>, and so <em>no driftting here</em>.</p>
<p>With <span class="math-container">$WD$</span> the car will start the drifting with the same speed on both wheels. In this case the minimum speed to get a true drift will need to surpass the minimum threshold from <span class="math-container">$W_{rL}$</span>. But the <span class="math-container">$W_{rR}$</span> will rotate with the same speed from <span class="math-container">$W_{rL}$</span> (<span class="math-container">$r_v$</span> is distributed 50% on <span class="math-container">$W_{rL}$</span> and 50% on <span class="math-container">$W_{rR}$</span>). But our lateral force is not the same with let's say <span class="math-container">$L_F$</span>=20% on <span class="math-container">$W_{rL}$</span> and <span class="math-container">$L_F$</span>=80% on <span class="math-container">$W_{rR}$</span> (the correct <span class="math-container">$L_F$</span> will vary acording with entry speed on curve and curve angle).<br>
The result in this case is that with <span class="math-container">$WD$</span> the tire grip on <span class="math-container">$W_{rL}$</span> is better than the grip on <span class="math-container">$W_{rR}$</span> because of the inbalance of the sum of the forces.</p>
<p>With <span class="math-container">$LD$</span> the car will start drifting first on <span class="math-container">$W_{rR}$</span> like with <span class="math-container">$OD$</span> <strong>but</strong> our <span class="math-container">$LD$</span> will garantee that some power remains in <span class="math-container">$W_{rL}$</span> and the speed keeps increasing until <span class="math-container">$W_{rL}$</span> starts slipping too.<br>
The moment the drifting start, the <span class="math-container">$LD$</span> will distribute the <span class="math-container">$r_v$</span> on both wheels to get a balance in the resultant force. So the values now will be something close to:<br>
- on <span class="math-container">$W_{rL}$</span>: <span class="math-container">$L_F$</span>=20% + <span class="math-container">$r_v$</span>=80%<br>
- on <span class="math-container">$W_{rR}$</span>: <span class="math-container">$L_F$</span>=80% + <span class="math-container">$r_v$</span>=20%<br>
With this combination the grip on the wheels will be the same and no lost power like with <span class="math-container">$W_{rR}$</span> like with the <span class="math-container">$WD$</span>.</p>
<p>Sorry by the lack of images to help the understanding.<br>
I'm new with stackexchange and I need to fiddle more with the commands to learn how to create some helping image and put here.</p>
| 31698 | Faster car drifting (motorsport) when using limited slip vs welded differential |
2019-10-29T17:01:31.347 | <p>At my current role we have a hell of a time trying to get NPT threads to seal. We avoid them where we can but sometimes its just faster and cheaper to spec and buy parts with NPT threads.</p>
<p>We are using pipe dope and I fully trust in my techs skills. Is there any other methods or products out there that can help ensure a good seal ?</p>
| |piping| | <p>With good quality threads, 3 wraps of PTFE tape can seal NPT connections above 1,000 psi.
SAE J1926 straight thread o-ring ports will seal a lot easier though. I've also used this company in the past: <a href="https://www.highpressure.com/" rel="nofollow noreferrer">https://www.highpressure.com/</a></p>
| 31707 | Leaky NPT fixes |
2019-10-29T17:57:15.923 | <p>A standard practice in many academic labs, which I'm told is "bad practice", is to use RG58 cables (i.e. standard BNC cables) for power distribution. </p>
<p><strong>Question 1:</strong> Why is this considered bad practice?</p>
<p><strong>Question 2:</strong> Is there a different sort of cable which is better for power distribution and compatible with BNC connectors?</p>
| |electrical-engineering|power-electronics|power-transmission| | <p>Because it is not a power cable (Q1) using power connectors (Q2).</p>
<p>BNC cable has a solid conductor and a ground braid to electrically shield and mechanically protect the conductor.</p>
<p><strong>Electrical reason:</strong> Some ground braids are made of steel, which is not as good a conductor as copper, so voltage losses to the wire would be higher than copper wire. Basically, wasted power.</p>
<hr>
<p>Edit....</p>
<p>Current creates heat (<span class="math-container">$I^2 R$</span> losses). The ampacity rating of a power cable is determined by the maximum temperature the cable's insulation can handle. The BNC wire was never meant to carry power, so the current flowing may cause the insulation to break down and cause a fire.</p>
<hr>
<p><strong>Safety reason:</strong> The main reason BNC cable should not be used to send power is because of the connectors. The outside of the metal BNC connector is uninsulated, so you have a shock risk, if you disconnect the load terminal while power is being applied. </p>
<p><strong>Common sense reason:</strong> You may know what you have done so are not at risk, but an unsuspecting person is <strong>NOT</strong> expecting power on a signal cable.</p>
<p>Even if you found an insulated BNC connector and added a warning sign, if it was disconnected a user may connect a normal connector negating your bad practice. </p>
<p>From: <a href="https://yaoota.com/en-ng/product/generic-bnc-male-connector-uhf-walkie-talkie-antenna-28-inch-price-from-jumia-nigeria" rel="nofollow noreferrer">Generic BNC Male Connector UHF Walkie Talkie Antenna 2.8 Inch For Kenwood TK-388</a></p>
<p><a href="https://i.stack.imgur.com/Sw7g2m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sw7g2m.png" alt="Insulated BNC Connector"></a></p>
<p>No one expects power over a BNC cable. So power distribution over BNC cables is bad practice.</p>
| 31708 | Why is it bad to run power through BNC (RG58) cables? |
2019-10-30T08:17:26.797 | <p>For any testing that a part is suppose to pass a test, let's say the leakage test of a seal flange. </p>
<p>How should I individualize the cause? </p>
<p>I know there is root cause analysis and for different cases I probably need Design of Experiment (DOE) analysis for considering all cases of test to find the root cause?</p>
<p>I have not seen any papers that talks about doe analysis and root cause analysis at the same time. </p>
| |mechanical-engineering|mechanical-failure|product-testing|failure-analysis| | <p>I think you might be mixing up the intentions of the two analysis tools you you mentioned - Root Cause Analysis and Design of Experiments. Consider thinking of it this way:</p>
<p>Root Cause Analysis is something you do in order to help identify, well, the "Root Cause" of some issue or problem. There are multiple ways to go about doing a root cause analysis in the same way there are multiple way to clean your house. Over time, industry has developed a number of analysis "tools" that can be used to <em>help</em> in figuring out a root cause. Short of explaining each tool, for now I suggest just reading up a bit on some of the following techniques:</p>
<ul>
<li>5-Whys: Basically keep asking "why" it happened until you get to an answer that you can control</li>
<li>Fishbone Diagram: visual tool to help map out the different sources that can contribute to the failure - machine related, material related, people related, environment related, etc.</li>
<li>Pareto Charts, Scatter Charts, or other graphs: help to identify trends or interactions that might signal when or how a problem started. </li>
<li>Failure Modes and Effects Analysis (FMEA): A very methodical tool for evaluating and how various root causes/issues drive different failure modes and their effects which also considering their severity and a number of other factors. This is what's considered a "bottom-up" approach to root cause analysis. </li>
<li>Fault Tree Analysis (FTA): The "top-down" brother of an FMEA where a fault is traced down through all the possible chains of events that could lead to it. </li>
<li>Anticipatory Failure Determination (AFD): A creative approach where you try to figure out HOW to actually MAKE the issue happen. In your case, if you <em>wanted</em> to MAKE seal to leak, how would you do it? </li>
</ul>
<p>These are all tools that are regularly used in risk based industries (think auto manufacturers, medical devices manufacturers, aerospace, etc.) but certainly in lots of other places as well. </p>
<p>Now, the second topic you mentioned - Design of Experiments - while also an option to help with root cause determination, it more suited to helping test a theory, or develop a design or specification. It's commonly used in determining how variation in different inputs can impact the output. Consider something as simple as boiling an egg. Your inputs could be a number of things: </p>
<ol>
<li>Amount of Water</li>
<li>Temperature of Water</li>
<li>Time spent boiling</li>
<li>Amount of salt added to water</li>
<li>Age of the egg</li>
<li>Size of the pot</li>
</ol>
<p>A DOE is performed to analyze the compound interactions between this. In a very high level summary the team would perform the following steps:</p>
<ul>
<li>First decide what they wanted to consider a "correctly" boiled or "good" version of a hard boiled egg WAS. What are our specifications for what we WANT. </li>
<li><p>Then there are some ways to eliminate inputs or "factors" which don't really influence the outcome. The more factors we have to evaluate the more complicated the DOE. In this case, we can probably reasonably eliminate:</p></li>
<li><p>The size of the pot (unlikely to impact the finished egg - assuming of course its big enough to fit the egg and hold water)</p></li>
<li>Age of the egg maybe? Do we feel that this strongly influences the final product?</li>
<li>Amount of water - as long as it's boiling (and we don't break the egg) then this probably doesn't matter. </li>
</ul>
<p>This leaves us with 3 "factors" that we now want to evaluate (and which also seem reasonable):</p>
<ol>
<li>Temperature of the Water</li>
<li>Amount of salt added</li>
<li>Time spent boiling. </li>
</ol>
<p>Once a DOE is set up, we would define a number of different experiments where we boiled an egg while varying those parameters in known amounts. Eventually we'd end up with some data that we could statistically analyze to help us understand what input factors we want to control. We'd end up finding that the temperature of the water is somewhat influenced by the amount of salt added. We'd also learn that the amount of time we needed to boil was correlated with the temperature. Eventually, through our analysis, we'd likely come to the conclusion that while temperature is important, the amount we raise it by adding an acceptable amount of salt, doesn't impact the way our egg end up nearly as much as does the time we actually spend boiling it! Not long enough and we end up with raw egg. Too long and we turn it to dried rubber. Our DOE has helped us identify that, when we now go to boil eggs, controlling the time we boil is the best way to guarantee us an egg made exactly the way we want it!</p>
<p>Now, while this may seem a super trivial example, hopefully you can envision how the concept of a DOE as a "tool" could be extremely useful. And just as there were multiple different tools for performing RCA, the same exits for design analysis!</p>
<p>Hope this helps to better understand how these tools relate. </p>
| 31718 | DOE analysis or root cause analysis? |
2019-10-30T15:59:40.550 | <p>I want to calculate the voltage at node <span class="math-container">$A$</span> respective to the ground.</p>
<p><a href="https://i.stack.imgur.com/WbzrW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WbzrW.jpg" alt="enter image description here"></a></p>
<p><span class="math-container">$R_1=1.8\text{k}\Omega, R_2=3.8\text{k}\Omega, R_3=1.8\text{k}\Omega, R_4=5.8\text{k}\Omega, R_5=2.4\text{k}\Omega$</span> and <span class="math-container">$V1=4.3V, V2=2.3V$</span></p>
<p>I know that voltage <span class="math-container">$V=-V2+V1=2V$</span> then <span class="math-container">$R_1$</span> and <span class="math-container">$R_2$</span> in series so <span class="math-container">$R_{12}=R_1+R_2=5.6\text{k}\Omega$</span> then the current is <span class="math-container">$I=\frac{V}{R_{12}}=0.356\,\text{mA}$</span></p>
<p>But how do I calculate the voltage at node <span class="math-container">$A$</span> respective to the ground? Is it over <span class="math-container">$R_{45}$</span> or <span class="math-container">$R_{3}$</span>? </p>
| |electrical-engineering|electrical|circuits|circuit-design| | <p><span class="math-container">$R_4 \parallel R_5 $</span> and <span class="math-container">$R_3$</span> form a voltage divider:</p>
<p><span class="math-container">$$V_A = \frac {R_4 \parallel R_5}{R_4 \parallel R_5 + R_3} \times (V_1 - V_2)$$</span></p>
<p>Or alternatively using series/parallel rules and Ohm's Law:</p>
<p><span class="math-container">$$I_A = \frac {V_1 - V_2}{R_4 \parallel R_5 + R_3}$$</span>
<span class="math-container">$$V_A = I_A \ (R_4 \parallel R_5)$$</span></p>
| 31721 | Voltage at node A respective to the ground |
2019-10-30T17:36:24.593 | <p>Using the sylvac UL 4
Each time I invoke the (send data) function the sylvac LED panel reports '<strong>no.data</strong>'. To mention the obvious but nothing is received at the connected USB device. </p>
<p>My question is. . <strong>How do force the sylvac device to (have data) to send?</strong> Can anyone explain what this 'no.data' actually means from the calipers perspective?</p>
<p><a href="https://i.stack.imgur.com/HU3U2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HU3U2.png" alt="enter image description here"></a></p>
| |measurements| | <p>No.data: Value could not be sent via Bluetooth® Technology. Check the Bluetooth® connection on instrument and master device (on/off). Check if you are within the max. transmission range 5-15m.</p>
<p><a href="https://www.sylvac.ch/faq?view=topic&id=2" rel="nofollow noreferrer">https://www.sylvac.ch/faq?view=topic&id=2</a></p>
<p>I have now resolved this issue in my case: the caliper's BT connection needs to be set up using the app that Sylvac has made for this. Note, if you've used the phone's BT procedure, it's the wrong way.</p>
<p><a href="https://play.google.com/store/apps/details?id=ch.sylvac.Anywhere" rel="nofollow noreferrer">https://play.google.com/store/apps/details?id=ch.sylvac.Anywhere</a></p>
| 31723 | Sylvac UL 4 - USB - (Send data) reports 'no.data'? |
2019-10-31T09:09:23.303 | <p>While my research about the untreated surface of moulded cast I read that the properties such as tensile strength, 0,2% proof stress and so on are different (much smaller) from the properties you can find in the tables. That’s because the samples have a treated surface to evaluate the material properties.</p>
<p><strong>Now my question is</strong> how can I determine the thickness of the surface where my properties are much smaller?</p>
<p>The background is that I have a bore near the untreated surface loaded with hydraulic pressure and for safety I need a specific thickness of material around the bore. And because the material properties of the untreated surface are much smaller, I will not take this area into account to my calculations. If it is needed: the regarded cast is GJS 400.</p>
<p>Many thanks in advance!</p>
| |mechanical-engineering|materials|metallurgy|casting|molding| | <p>There is no significant loss of cast surface strength in ductile cast iron . In fact the ASM Handbook ( 9th ed.) shows higher strength near the surface , the amount of strengthening depends on grade and thickness of casting. Basically caused by the cooling rate; faster cooling = more strengthening.</p>
| 31730 | (moulded) cast properties of the untreated surface |
2019-10-31T15:44:21.897 | <p><a href="https://i.stack.imgur.com/Hymba.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hymba.png" alt="enter image description here"></a>I am studying derivation of bernoulli equation from FM White's fluid mechanics and in that pressure distribution is as shown in the image 1. I am unable to understand force calculation as shown in image .</p>
| |mechanical-engineering|fluid-mechanics| | <p>There seem to me to be three main things that appear initially confusing about the derivation. I believe they both have to do with the <span class="math-container">$\frac{1}{2} dpdA$</span> term. I think the <span class="math-container">$-dp(A+dA)$</span> term is quite self explanatory.</p>
<p>I think the three main points are:</p>
<p>(1) Why is the term of the form <span class="math-container">$\frac{1}{2} dpdA$</span></p>
<p>(2) Why doesn't the <span class="math-container">$dA$</span> term take into account the geometry of our streamtube (i.e. why doesn't the area use the formula for the side surface area of a frustum instead of using <span class="math-container">$dA$</span>). </p>
<p>(3) Why doesn't the resultant force take into account the angle of deviation of the side of the streamtube when calculating the force, (i.e. why is the term <span class="math-container">$\frac{1}{2} dpdA$</span> instead of <span class="math-container">$\frac{1}{2} dpdA \theta$</span> where <span class="math-container">$\theta$</span> is the angle of deviation from the axis of symmetry of the streamtube.)</p>
<p>The answer to part (1) is that when we remove the constant pressure <span class="math-container">$P$</span> acting on all surfaces, we notice that we are left with a linearly increasing pressure, from the 'left' of streamtube, to the 'right'. If you are familiar with distributed loads from structural mechanics, the general formula for calculating the average force across a section with varying force, is <span class="math-container">$\frac{1}{2}Fl$</span>, where <span class="math-container">$F$</span> is the force distribution and <span class="math-container">$l$</span> is the length of our section. This is analogous to how we treat the pressure, i.e. think of it as a 'pressure distribution' of sorts. The average pressure is then just <span class="math-container">$\frac{1}{2}dp$</span>. Note here that we merely assume a linear increase of the pressure across the volume. (I do not know the answer for why this is a valid assumption for full disclosure.)</p>
<p>To answer part (2), This is simply because we are dealing with such small changes in area the actual geometry doesn't figure into the derivation. An extremely tiny change in surface area of an extremely tiny cube is exactly analogous to an extremely tiny change in area of a frustum for example.</p>
<p>The answer to part (3) is simply that pressure is isotropic. Thus regardless of the angle, we know the force in the direction of <span class="math-container">$S$</span> will still be <span class="math-container">$\boldsymbol{n}p\partial S$</span> where <span class="math-container">$\boldsymbol{n}$</span> is the unit vector in the specified direction, <span class="math-container">$p$</span> is the pressure, and <span class="math-container">$\partial S$</span> is the surface element. Have a look at this <a href="http://www.atm.damtp.cam.ac.uk/people/mem/FLUIDS-IB/dyn.pdf" rel="nofollow noreferrer">http://www.atm.damtp.cam.ac.uk/people/mem/FLUIDS-IB/dyn.pdf</a> which will give a much better (and detailed) explanation than I have here.</p>
<p>Hopefully this covers the more odd circumstances (they were at least the most peculiar aspects of the formula to me ). Note I assumed the other term <span class="math-container">$-dp(A+dA)$</span> is straightforward; it is simply an expression of force acting on the 'right' side surface of the element.</p>
<p>Hope this helps somewhat.</p>
| 31734 | Can anyone explain calculation of net force due to pressure in a streamtube of fixed control volume and infinitesmal size |
2019-10-31T18:15:59.117 | <p>Suppose i have 4 empty syringes. And by empty i mean they were having the air sucked out of them, and they <strong>"contain vacuum"</strong>. Assume the head of the syringe is blocked so vacuum can be formed.</p>
<p>I want to understand how to choose between this 4 different syringes.</p>
<p>My goal is to have the <strong>piston reach the highest speed possible</strong>. I don't know how to calculate what is better.. should i increase the diameter? - should i keep the diameter the same but make the syringe longer.. what is better?</p>
<p>When i release the piston - the atmospheric pressure will make the piston move pretty fast - and blow up the head of the syringe. </p>
<p>This is not a real world example. Just something for me to learn. I don't know what the piston friction with the walls is - but is the same for all 4 syringes so does not matter when comparing them.</p>
<p>Also i don't know the mass of the piston either.. i mean we can use a gut number like 0.25 kg. And the bigger piston is 1kg.</p>
<p>Now how to calculate all this?
<a href="https://i.stack.imgur.com/DMfJu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DMfJu.png" alt="how to calculate the speed and force of the piston in a vacuum syringe?"></a></p>
<p>Also i'm interested in forming an intuition of what to expect given changes to the design of this syringes rather then exact numbers. What happens if i increase the diameter 3 times.. is it better or worse .. that kind of thing. </p>
<p>Does it matter how long the syringe is? Can it happen that if i make it too long then the piston is actually slowed down? Is this piston constantly accelerating while is in the syringe and there is still vacuum in front of it .. or not actually? Can air to air turbulence slow things down? Where is the highest acceleration at? Maybe in the middle of the syringe ? </p>
<p>I say this because if we go down to 1 mm tube, then capillarity is so high that you need to push on it like crazy to make things move. I'm thinking about what force is needed to squeeze that fluid in a medical equipment (needle + tube). So when it comes to tubes like this, maybe having a higher diameter is better?! Maybe this applies to air also?</p>
<p>I can put the numbers in a calculator and get the result, but i'm interested more on how to think about this, not just the formula. How do you break it down in your mind - and not be wrong?</p>
<p>But i do want to lean how calculate the exact numbers as well. Thanks :)</p>
| |pistons| | <p>I think you're dismissing friction far too easily.</p>
<h2>Frictionless</h2>
<p>Since you asked for it here it is:</p>
<p>The forces on the piston are pressure (from the atmosphere on the back of the piston) and friction. So we have:</p>
<p><span class="math-container">$$P\,A=F=m\,a$$</span>
Where <span class="math-container">$P$</span> is pressure, <span class="math-container">$A$</span> is area, <span class="math-container">$m$</span> is mass, and <span class="math-container">$a$</span> is acceleration.</p>
<p>Since the area, pressure, and mass are constant, the accerations must be as well. Thus we can use the equations of constant acceleration:</p>
<p><span class="math-container">$$d=\frac12 a t^2$$</span>
<span class="math-container">$$\Delta V=a\,t$$</span></p>
<p>Combining these two can give you final velocity in terms of distance and acceleration:</p>
<p><span class="math-container">$$V=\sqrt{2\,d\,a}$$</span></p>
<p>Now we can do some more interesting things if we make some assumptions about how the piston mass scales with different dimension syringes. For example, we could assume that the density of the syringe was constant, and thus the mass was proportional to the length, and proportional to area:</p>
<p><span class="math-container">$$m=\rho\,d\,A$$</span></p>
<p>if we plug that in we get:</p>
<p><span class="math-container">$$P\,A=\rho\,d\,A\,a$$</span></p>
<p>The areas cancel. So the cross section of the picton wouldn't matter at all if our assumptions were correct.</p>
<p><span class="math-container">$$a=\frac{P}{\rho\,d}$$</span></p>
<p><span class="math-container">$$V=\sqrt{2\,d\,\frac{P}{\rho\,d}}$$</span></p>
<p>The length/ distance cancels out too!
<span class="math-container">$$V=\sqrt{2\frac{P}{\rho}}$$</span></p>
<p>Now, from your images it actually looks like your piston is the same length regardless of the length of your syringe. This would result in the mass only being proportional to area:</p>
<p><span class="math-container">$$m=\rho_2\,A$$</span></p>
<p><span class="math-container">$$V=\sqrt{2\,d\,\frac{P}{\rho_2}}$$</span></p>
<p>So in this case: the longer the faster.</p>
<p>However, speed might not be the factor you really care about. If you care about how hard the piston is going to hit, you might care more about impulse or kinetic energy. Ballistics usually measure impact energy when concerned with how strong an impact will be.</p>
<p><span class="math-container">$$KE = \frac12 m\,v^2 = d\, P \, A$$</span></p>
<p>The energy that you impart into the piston is always equal to the volume of your vacuum multiplied by atmospheric pressure. This is true regardless of the mass of your piston.</p>
<h3>Friction</h3>
<p>It's likely that friction plays a significant role in the final velocity.</p>
| 31738 | How to calculate the speed and force of the piston in a vacuum syringe? |
2019-11-01T15:11:50.163 | <p>I need a heated surface for a project and was thinking of using one of <a href="https://www.amazon.co.uk/dp/B0758CQN6M/ref=cm_sw_r_cp_apa_i_5OeVDb9FW7WFR" rel="nofollow noreferrer">these</a> PCB heater beds. But I'd need to drill a few holes in it - can I do this without stopping it working? Thanks</p>
| |heating-systems|3d-printing| | <p>If it is constructed like <a href="https://reprap.org/wiki/PCB_Heatbed" rel="nofollow noreferrer">this one</a> then at best you'll have dead stripes where you've drilled through conductors; at worst you'll short some adjacent stripes together and you'll (probably) get hot spots.</p>
<p>PCB layout software is cheap; you may want to see if you can get the part made as specified in the Wiki I link to (or use a service that uses gold flash rather than tin plate -- gold flash is much thinner, so it won't disturb the resistivity of the copper as much).</p>
<p>If you can, design a board with holes. Run an annular ring of copper around each hole, with stripes going into and out of the rings. Current will conduct around the holes, then go back into stripes where it'll do it's job of heating up the board. Without a lot of painful calculation the heating won't be perfectly uniform, but it should be pretty good -- or you can do a lot of painful calculation, and get it spot on.</p>
| 31747 | Drilling PCB 3d printer bed |
2019-11-03T09:02:58.810 | <p>I am a recent electronics engineering graduate, currently legally working as an embedded software engineer in a company.</p>
<p>In the future, I would like to become a freelancer, like signing up on a freelancing website and getting hired to do real projects.</p>
<p>Currently, as a side/weekend gig, I help engineering students with their diploma projects. It's obviously illegal but it's a very common practice around here. But frankly, it's a great opportunity for me to earn good money and learn my trade with real hands-on experience.</p>
<p>In the future, when looking for real customers, they will likely ask for a portfolio of my work. What am I going to show them? Some diploma projects of random people that I supposedly made? I'm afraid that would look very unprofessional.</p>
| |electrical-engineering|ethics| | <p>If you in the USA there are couple of good ways to start building a profile. </p>
<ul>
<li>Volunteer at your local <a href="https://www.firstinspires.org/robotics/frc" rel="nofollow noreferrer">FIRST robotics</a> team. This is good place to get started learning new skills and using current skills</li>
<li>Be an active member of the communities such as <a href="https://engineering.stackexchange.com/questions">engineering stackexchange</a> where you can start building an online profile of you skills. You can asked embedded engineering type of question as well as answer engineering questions. </li>
<li>You can contribute to open source projects, through platforms such as github. </li>
<li>Solving a problem such as <a href="https://engineering.stackexchange.com/questions/31770/which-kind-of-accelerometer-vibration-sensor-are-suitable-for-sensing-vibration">Which kind of accelerometer/vibration sensor are suitable for sensing vibration of building floor caused by human foot strike?</a> in open source platform help build a technical profile.</li>
<li>Sharing you solutions via platform such as youtube, or other social media platform can help build a technical profile. </li>
<li><a href="https://perscholas.org/" rel="nofollow noreferrer">Par Scholas</a> is great place to volunteer to share your technical skills, learn new skills and a build a technical project portfolio </li>
<li><a href="https://turing.io/" rel="nofollow noreferrer">Turing School of Software and Design</a> is another good place to start building a technical profile. </li>
</ul>
<p>References:</p>
<ul>
<li><a href="https://insights.dice.com/2017/01/11/tech-pros-volunteer-fun-ways/" rel="nofollow noreferrer">How Tech Pros Can Volunteer in Fun Ways</a></li>
<li><a href="https://www.yearup.org/about-us/vision-mission-values/?location=national-us/" rel="nofollow noreferrer">Year up</a></li>
<li><a href="https://chicktech.org" rel="nofollow noreferrer">Chick Tech</a></li>
<li><a href="https://techbridgegirls.org/" rel="nofollow noreferrer">Tech Bridge Girls</a></li>
<li><a href="https://www.bizible.com/blog/volunteer-opportunities-for-tech-professionals" rel="nofollow noreferrer">10 Volunteer Opportunities For Tech Professionals To Transform Their Communities</a></li>
</ul>
| 31771 | How do I transition to legit freelancing, coming from a background of unethical projects? |
2019-11-04T14:43:03.233 | <p>I need to devise a way to ensure only NG passes through a pipe, and no condensates or other liquids. It is ok to shut down the pipeline if liquids would be detected. The dimensions involved are small and space limited.</p>
<p>Is there a valve which shuts if liquid starts passing through but stays open for gas? Aside from a involved solution with a trap with a sensor attached, is there any other way to solve this?</p>
| |fluid-mechanics|gas|multiphase-flow| | <p>What are you searching for is called: </p>
<p><strong>Thermostatic expansion valve or (TXV).</strong> </p>
<p>You have one of those in your air conditioner. Because the compressor of AC's are designed to handel only vapour. </p>
| 31794 | Valve/Filter allowing through gas only, no liquid |
2019-11-07T14:27:20.260 | <p>In relation to <a href="https://3dprinting.stackexchange.com/questions/5850">this question about 3D printing</a>, I wonder how much a 200 mm square glass plate heated only in a central 15 cm square region would bow upwards or downwards.</p>
<p>According to the thermal expansion of glass, the linear expansion for 15 cm and 100°C deltaT would be 0.05 mm.</p>
<p>Using the formulas provided in <a href="https://planetcalc.com/1421/" rel="nofollow noreferrer">https://planetcalc.com/1421/</a> I input 200 mm as chord length and 200.05 as arc length. The result is 2 mm height which seems a lot to me. Using 150 mm and 150.05 mm I get 1.7 mm which is still a lot.</p>
| |glass|thermal-expansion| | <p>You are right that your numbers are over-estimates. They would be "correct" for a plate with no resistance to bending, which was only supported along two edges and heated across its full width between the unsupported edges.</p>
<p>This is a hard problem to produce a simple "formula", because in the real situation the bending will be reduced by the stiffness of the glass, and because the deformed shape is not part of a circle but a bulge in the center of a square. In real life there will be some transition between the temperature of the heated part and the surrounding "frame" as well, and not a sudden temperature change.</p>
<p>The best way to get a theoretical answer would be to model this with finite element analysis. Alternatively, learn from other people's practical experience, which is quicker and cheaper than learning from your own practical experience!</p>
<p>FWIW the simpler case of a circular plate was apparently studied as part of a PhD back in 1960, with this report as a spin-off: <a href="http://naca.central.cranfield.ac.uk/reports/arc/rm/3245.pdf" rel="nofollow noreferrer">http://naca.central.cranfield.ac.uk/reports/arc/rm/3245.pdf</a></p>
| 31842 | How much an unevenly heated glass plate bows? |
2019-11-08T07:46:59.427 | <p>Suppose I have a heat generator with a connected heat storage tank. If the heat generator remains the same and the heat storage tank is dimensioned larger, the running time of the heat generator increases due to the higher storage losses.
Why is this so?</p>
<p>Normally, the surface with exponent 2 and the volume increases with exponent 3 - should the heat storage tank not have fewer losses? </p>
| |thermodynamics|heat-transfer|heating-systems|energy-storage| | <p>Heat loss due to conductivity of the walls of your container assuming the ambient outside temperature,
<span class="math-container">$$ \frac{\Delta Q}{\Delta t*A}=-k \frac{\Delta T}{\Delta x} =C_{constant}$$</span></p>
<p>In the above equation, the left part is heat loss per unit of area per second and the right side is the thermal conductivity, K multiplied by the temperature gradient which is constant. x is the coordinate perpendicular to the temperature gradient of the tank wall. </p>
<p>So every unit of the surface of the tanks loses heat at the same rate.</p>
<p>Meaning the larger tank loses more heat because it has more surface area.</p>
| 31853 | Storage losses of a heat storage tank |
2019-11-10T15:05:08.103 | <p>I'm not an engineer, just a novice hobbyist, so I'm not sure my terminology "compression clamp" is correct, but I'm attaching a drawing showing what I'm trying to accomplish. </p>
<p>This plate will be placed over two rods and then "screwed (partly) shut" so that the plate will "squeeze" the rods and not budge under normal operational stresses. The machine screw is inserted into the hole the pink arrow points at and threads into the hole the yellow arrow points at.</p>
<p>My question: It seems to me that the threads will align properly if the gap between those two sections is cut in advance of the threading operation. Is that right? Or should the pink-arrow hole be left smooth-walled and only the yellow-arrow hole be threaded?</p>
<p><a href="https://i.stack.imgur.com/EfiRO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EfiRO.png" alt="clamping plate"></a></p>
| |threads| | <p>The pink Arrow hole should be left smooth.</p>
<p>If both holes have threads when you turn the screw into the pink hole as soon as it passes the gap and reaches the yellow hole, it will keep the gap space constant.</p>
<p>Turning the screw more will just drive it into both holes while keeping the gap at the same space till the head of the screw jams onto the lower surface of the pink hole.</p>
<p>Any further attempts at tightening the screw will either damage the treads or the screw.</p>
<p>The pink hole should be left smooth and a right size washer used to allow the screw to tighten the bracket safely.</p>
<p>Or alternately you can use a half thread screw similar to this photo.</p>
<p><a href="https://i.stack.imgur.com/6BWnI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/6BWnI.png" alt="enter image description here"></a></p>
| 31872 | threading a compression clamp |
2019-11-11T02:25:53.557 | <p><a href="https://i.stack.imgur.com/DWgGa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DWgGa.jpg" alt="Air conditioner device installed near the ceiling"></a></p>
<p>Almost all air conditioners (AC) are installed on top. For house small office use, will be like the picture. For office with central air conditioner, the blowing come from the ceiling through duct. But however, still there are few occasion where the air conditioner is installed near the floor, like once I saw in a historical cathedral church, due to esthetic's consideration.</p>
<p>My question, is any thermodynamic reason so almost all air conditioners are installed on top? If any, what is that?</p>
| |thermodynamics| | <p>The cold air has a higher density so it always settles down, by installing the AC on top, the cold air can interact faster with hotter air, so the cooling process is much faster. </p>
<p>If you install the AC on the floor, the cold air, doesn't naturally travel upward, and even if it does, thanks to the density difference and the air flow of the AC, it then, would be less desirable for human being think about cold feet and hot upper body! </p>
<p>Usually the heat is coming from the outside not the floor, windows for example, they pass the radiation, and simplify the conduction so the heat rate transfer is higher where the windows are installed. In a ten feet high room, an AC should be installed at 8 feet high, slightly above the windows, so the cold air can interacte faster with the hotter air. </p>
<p>There is also a technical issue here, duo to condensation you have to accommodate an extra duct to redirect the water to somewhere else, if you level the AC with floor then you need a pump to let the water flow, however by installing the AC at the recommended height, the gravity does the trick. </p>
| 31877 | What is the thermodynamic consideration so almost all air conditioners are installed on top? |
2019-11-11T09:48:41.973 | <p>I want to replicate the kind of interface you see on inboard boats, pumps, or food blenders. Basically, I want put a motor in a sealed vessel, and have it turn a prop outside the vessel, in water. </p>
<p>Is this usually done just with a layer of grease over a bearing? Are there any alternatives?</p>
| |mechanical-engineering|waterproofing|transmission| | <p>The seal you are looking for is commonly called an oil rotary seal ring.</p>
<p>I found many brands and sizes with a quick search.</p>
<p>the range of prices for a 1-inch shaft is around $5.</p>
<p><a href="https://i.stack.imgur.com/saCes.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/saCes.jpg" alt="oil rotary seal"></a></p>
| 31882 | How are water-tight transmissions made, like the kind you see on inboard boats, pumps or food blenders? |
2019-11-12T09:04:54.453 | <p>70% of the Earth's surface is covered in water, why can't a hydroelectric power plant be set up on each and every river? Is it that its too expensive?</p>
<p>Why is electricity still a problem in countries that have abundant number of rivers? Why can't each and every river be associated with a hydro power plant through dams? What makes a river unsuitable?</p>
<p>My basic question is that Earth has got abundant amount of water. Then why is (hydro) electricity still scarce in some places? Is it that its too expensive?</p>
<p>My research shows that it might be expensive to set up but it is quite cheap in the long run compared to other power plants besides itself being low-maintenance.</p>
| |water-resources|renewable-energy| | <p>Hydroelectric power is (indirectly) solar and geothermal power:
Heat causes evaporation, temperature gradients create flow of air and water vapor to higher elevations, clouds form, rain falls, and hydroelectric power simply harvests the gravitational potential energy originally imparted by heat to the raindrops.</p>
<p>Therefore, you can harvest hydroelectric power (a) anywhere it rains, and (b) anywhere between where it rains and a lower elevation so long as there is a continuous and preferred canal to supply the drainage from the high-rainfall areas.</p>
| 31896 | Why is hydro-electric power still scarce in some places? |
2019-11-12T09:31:56.220 | <p>A tangent ogive body has a specific radius for a given length, so how would you determine length if only given the radius? </p>
| |aircraft-design| | <p>Looking at the the image below:</p>
<p><span class="math-container">$L=\sqrt{r^2-x^2}$</span></p>
<p><span class="math-container">$x=r-\left(\frac{D}{2}\right)$</span></p>
<p><span class="math-container">$L=\sqrt{r^2-\left(r-\left(\frac{D}{2}\right)\right)^2}=\frac{1}{2}\sqrt{-D (D - 4 r)}$</span></p>
<p><a href="https://i.stack.imgur.com/z9R0k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z9R0k.png" alt="Ogive Diagram"></a></p>
| 31897 | For a tangent ogive, given a radius, can you determine it's length? |
2019-11-13T11:54:51.100 | <p>I have a simulation model of a cylindrical hot water storage tank. The heat loss is determined by the surface area and the heat transfer coefficient. </p>
<p>A configuration of the storage tank with 900 liters was validated with a real scenario. </p>
<p>Now I would like to test how the scenario changes when the storage tank gets bigger. Thereby my surface becomes bigger, but I don't know if the heat transfer coefficient changes as well.</p>
<p>Does a larger storage tank usually improve the insulation so that the heat transfer coefficient changes?</p>
| |thermodynamics|heat-transfer|heating-systems|energy-storage| | <p>In a tank with no fluid movement and no heat source with steady ambient temperature, the temperature of fluid are layered strata roughly in onion layers but favoring the top of the tank, meaning the heat is gradually decreasing from the top in ellipsoid layers roughly following the geometry of the cylinder in you question, imagine inverte flames getting colder as they get near the tank walls and gently sliding up till they reach the top and flare open turn back down in a perpetuous circulation.</p>
<p>So if you increase the volume of your tank, it means you are providing new insulation for the hot core.
So in the larger tank the surface temperature will be less than the smaller tank while keeping similar distribution topography.</p>
<p>So the ratio of heat loss will be smaller than jast smaller ratio of surface to volume of fluid.</p>
| 31915 | Heat transfer coefficient of a cylindrical hot water storage tank depending on its size |
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