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2018-04-24T16:37:16.583 | <p>I’ve some small U groove rollers I’m having made, 60mm diameter, 60mm wide, 28mm diameter at ceter of U profile. </p>
<p><a href="https://i.stack.imgur.com/G4OoA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G4OoA.jpg" alt="enter image description here"></a></p>
<p>The rollers need to be silent and have some friction on the rails. Polyurethane seems the most suitable material as it’s the highest load capacity to friction I can find but I’m open to suggestion if anyone can think of something else? I want to maximise how much weight they can take.</p>
<p>I can not harden the polyurethane much as they need to be soft for the friction and to run silent. Also I can’t thicken the diameter of the rollers though I can make the hub smaller to thicken the PU a little. </p>
<p>How do polyurethane wheels fail under pressure? I’m presuming that the PU is displaced and shears of the bonded surface eventually splitting in the centre as in drawing A.</p>
<p>First I though of putting support on ether side as in picture B.</p>
<p>Then I changed the profile of the U groove to have a dip in the centre. The groove should redistribute the pressure away from the centre and into the sides, acting against the side support and on the thickest part of the PU.</p>
<p><a href="https://i.stack.imgur.com/1l0mF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1l0mF.jpg" alt="enter image description here"></a></p>
<p>I was also looking a whether different shaped larger hubs and thinner PU or smaller hubs and thicker PU would be better under strain?</p>
<p><a href="https://i.stack.imgur.com/l9ysw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l9ysw.jpg" alt="enter image description here"></a></p>
<p>Finally I was looking at the possibility of support within the PU to increase it’s load capacity and make it less likely to deform, something like a 3D printed Nylon mesh reinforcement. I’m going to talk to the company who makes the wheels tomorrow but I suspect that making a PU mould with a mesh inside will have problems with getting rid of bubbles and cavities.</p>
<p><a href="https://i.stack.imgur.com/k1St9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k1St9.jpg" alt="enter image description here"></a></p>
<p>Is there is anyone out there with any experience or opinion on how PU behaves under stress, any advise how these wheels fail? Are there any other methods of increasing the load capacity of PU that you may know of? Or is there another material that may be more suitable?</p>
<p>Thanks for taking the time Chris, I can’t see a button to answer you apart from the commet one which has a limit to the letters? So I did it here.</p>
<p>The weight capacity is ok but it’s right there on the edge of ok. Unfortunately I can’t add rollers or increase the outside turning diameter.</p>
<p>I need no slippage, so I need some friction and I need a little give so as not to translate tiny imperfections up the shaft and into the mechanism. 85A hardness PU would do the job, maybe 90A, I don’t know, I’m waiting on harder samples to try them out. </p>
<p>The issue of the wedging effect you brought up is something that has been on my mind, I’ll have a look at the conical idea though I’m hoping that by having the PU bonded to the two side supports will go a long way towards mitigating any wedging effect.</p>
<p>PU can move and flex but can it be compressed? I’m unsure but I suspect that it can not or of it can its only by a tiny amount. So the failure point of the roller is down to the displacement only.</p>
<p>By bonding the PU to the hub and also the two side supports I hoped to have the weight acting on the PU where it has the two bonded surfaces to reduce the displacement that can happen. </p>
<p>My natural reaction has been to increase the thickness of the PU to increase it’s load capacity but the more PU there is and the further it is from the bonded surface surely that means the more it can be displaced? </p>
<p>With this in mind I was thinking that it may be better to make the hub the same shape as the roller and have only a thin layer of PU as in the 3rd drawing down. This would increases the diameter of the hub from 16mm to 25mm at the thinnest point. Do you think the thick hub with the thin layer of PU would be better or the thinner hub with as much PU as possible?</p>
<p>Is there another materiality that would be more suitable? I was trying to think of something like layers of leather or rubber or PU coated fabric or something like this but couldn't think of anything.</p>
| |mechanical-engineering| | <p>In of the first issues you will get is that the area in contact with the rail will flatten, increasing the rolling resistance. In practice this is what tends to limit that capacity and they will become unusable deformed before they actually fail as such. So its really the Youngs Modulus rather than the tensile strength which is the issue. </p>
<p>The most obvious way to improve the capacity is to increase the diameter. If you look at industrial castors the main difference between different capacities is the wheel diameter. </p>
<p>If that isn't possible then the next solution is just to use more rollers. </p>
<p>There is also an issue that concave rollers on rails don't actually work that well, especially under moderate load as the shoulders tend to rub on the rails and wedge you are probably much better off with a conic section like train wheels or using a set of guide wheels like a roller-coaster. </p>
| 21456 | Looking for advice on design of some rollers |
2018-04-25T05:07:12.603 | <p>Given a car configuration with a length of L and width of W. If you can give the top left wheel an angle of A to turn A degree how do you derive the angle of top right wheel for turning . I know that the bottom left angle will just be complimentary of the top left wheel and bottom right wheel is complimentary of top right wheel. This is a swerve drive system so i can control each wheel angle. </p>
| |mechanical-engineering|control-engineering|design|mechanisms|car| | <p>Most road vehicle steering is based on <a href="https://en.wikipedia.org/wiki/Ackermann_steering_geometry" rel="nofollow noreferrer">Ackerman geometry</a> </p>
<p>Put simply the plane of each steered wheel should be perpendicular to a straight line from the centre of rotation the wheel to the centre of the tuning circle. </p>
<p>Normally this is achieved by a mechanical linkage where the steering arms point along a line from the wheel pivot point to the centre of the rear axle. With independent steered wheels you need to translate that geometry to whatever control system you are using. </p>
<p>In practice there are several factors which might make you deviate from pure Ackerman steering, suspension often doesn't travel in a perfectly vertical line and you may want non-zero camper, castor and toe-in in the suspension geometry for various reasons. </p>
<p>Also pneumatic tyres tend to operate with a small slip angle on turning as a result of the finite stiffness of the side wall of the tyre, which is how the turning force is generated. </p>
<p>Modeling and analysing this in detail get very complex very quickly and there is no single 'correct' solution as it will depend on a whole variety if interrelated factors and will often end up being some compromise. </p>
| 21459 | Wheel angle for turning |
2018-04-25T15:16:48.157 | <p>Consider a spring like this (this one CoC hand gripper):</p>
<p><a href="https://i.stack.imgur.com/aWDfK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aWDfK.jpg" alt="CoC hand gripper"></a></p>
<p>They claim on their web site, that the spring will not degrade during time:</p>
<blockquote>
<p><strong>Do your grippers get weaker with time?</strong>
No, so don’t worry if your Captains of Crush No. 1 gripper is feeling easier to you now—it’s because you’re getting stronger! What is mistakenly called “seasoning” by some people is actually a weakening with use, and it’s a reflection of an under-designed gripper with a spring that is bending, which is why it’s not uncommon for low-quality grippers to get narrower and easier as they are used. On the other hand, Captains of Crush grippers hold their line for a lifetime of steady training.</p>
</blockquote>
<p>There is also a lot of discussion on this topic, referencing some re-measurements, but none of that seems really trustable.</p>
<p><strong>My questions, from engineering point of view:</strong></p>
<ol>
<li><p>Is that possible/probable? I mean, can the degradation be small enough (for the given brand) so the user should not care? How likely it is true?</p></li>
<li><p>Is possible what they say about the "low-quality grippers" that they degrade over time significantly. How probable is that?</p></li>
</ol>
<p>I can imagine that both options (degradable and also stable springs) are possible. But how probable it is. Note that such a product cost usally $10-25.</p>
<p>I know this is kind of opinion based question, however the experiment would take few years, so any theory and/or experience based answer can help.</p>
| |steel|springs| | <p>Springs do degrade in length and load. How much and how fast depends on various factors. One very obvious factor is the design, which causes for the same displacement at the grip a different strain in the material. The higher the strain, the higher the wear. </p>
<p>Imagine the spring wire a bending beam. You can generate the same second moment of inertia (which is one factor of generating spring stiffness) with a flat and wide cross section and a high and narrow one. The last one however will generate more strain in the material.
Or, compare a two thinner wires with one thicker one. Or if you have a thicker, but longer wire (more coils). </p>
<p>That said, I also do not think, that any quality manufacturer would deliver a design, that shows an effect of degradation.
But I woulnd't be surprised if you find them. For anything that people have manufactured, you can find examples of ultra low quality. Besides design, also material can be just too crappy. Just, using this to sell something may only show how much this manufacturer just escaped this ultra low quality business.</p>
| 21466 | Spring degradation in hand grip trainer |
2018-04-26T05:03:22.883 | <p>If I have a 4 inch PVC pipe connected to a rotary union, that I want to spin at 200-300 RPM -how do I do it? It appears like the rotary unions themselves have some lateral give -- So, I would need bearings? </p>
<p>Are there rotary unions with built-in motors? Are there motors that can have hollow, 4 inch cores? Does it make sense to get a custom gear machined for this purpose?</p>
| |mechanical-engineering| | <p>You can isolate the other side of the rotary union, and use a belt, or magnetic gears. Traditional gears will put lateral stresses on the rotary joint, which may result in buckling. </p>
| 21479 | How can you spin a pipe connected to a rotary union? |
2018-04-26T08:31:43.450 | <p>Given that we have mechanisms of performing photolithography at 10nm, why are metal filters often meshes? Wouldn't it make a lot more sense to buy a single chunk of metal, and use the UV lasers to poke lots of N micron holes in it?</p>
| |lasers| | <p>There are MANY ways to make filters:</p>
<p>A metal mesh is one. Works well from gravel down to fine sand.</p>
<p>In your coffee pot you use a paper filter. Will take out sand, but will pass suspended clay.</p>
<p>Landscape fabric is often needle punched -- a 50 to 100 micro hole every 5 mm. As a filter it's not very efficient. But it only has to let through a foot or two of water a year.</p>
<p>Another not so common filter are a stack of rough disks. The liquid passes btween the disks. This has fairly low back pressure.</p>
<p>You can make filters out of spun plastic that is heat fused. This is basically what tyvek housewrap is. </p>
<p>The waterproofing in Goretex garments is basically teflon that is stretched so that liquid water had too much surface tension to squeeze through the holes, but water vapour passes thorugh. There is heavy magic in getting the right plastic cross linking and the right stretching.</p>
<p>One type of oil filter consisted of a frame that would hold a roll of toilet paper. Oil pass through the roll lengthwise. </p>
| 21480 | Why are fine metal filters made of meshes? |
2018-04-27T10:07:18.447 | <p>I have recently come across the concept of <a href="https://en.wikipedia.org/wiki/Compact_linear_Fresnel_reflector" rel="nofollow noreferrer">CLFRs</a>, and I want to build one myself, as a private experiment.</p>
<p>I have a rudimentary knowledge of optics (which I'm sure I'll have to refresh on) - any useful links would be appreciated on bringing me up to speed.</p>
<p>My understanding of the technology is as follows (I stand to be corrected if this knowledge is incorrect).</p>
<ol>
<li><p>A linear Fresnel reflector is constructed by taking equal width "strips" out of a parabolic mirror and "translating" the "strips" on to a horizontal surface in such a way that all the strips have bases with the same Y coordinate.</p></li>
<li><p>A compact linear fresnel reflector superimposes two parabolic mirrors (facing oposing directions) and decomposes them into two sets of alternating, equal width "strips" lying on the same horizontal line.</p></li>
</ol>
<p>Is my understanding of the construction of CLFR correct? If yes, is there an algorithm I can use to "slice" a parabola into slices of width <em>w</em>, whilst preserving the reflecting angle of each "slice"?</p>
<p>I can think of a way of generating an algorithm, using geometry and elementary calculus - BUT, I don't profess to be an expert in this area, so I want to know what specialists in this field suggest.</p>
<p>A schematic of a CLFR is shown below:</p>
<p><a href="https://i.stack.imgur.com/GEWN2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GEWN2.jpg" alt="enter image description here"></a></p>
<p>In summary, how may I build a CLFR by "decomposing" two parabolas into two alternating rows of strips of width <em>w</em> ?</p>
| |design|optics| | <p>If the idea of a flat surface cut into a spiral qualifies for your objective, you may be able to make something work based on <a href="https://omnivorenz.wordpress.com/2012/06/09/spiral-fresnel-reflector-a-simple-solar-concentrator/" rel="nofollow noreferrer">Rick Steenblik's design</a>. </p>
<p>From the 1982 Popular Science magazine, the article described how he cut the shape and aligned marks placed on the spiral to attach the reflective material to the cross members. As each segment was "pushed" into position, the outside edge lifted at the angle necessary to provide a common focus.</p>
<p>Allowing for some error, the continuous strip of reflective foil bent progressively greater as the spiral reached the edge.</p>
<p><a href="https://i.stack.imgur.com/zPprL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zPprL.png" alt="flat spiral reflector"></a></p>
<p>I was a young adult when I purchased the template for this item and did not have the resources to build it. I've long since given up on trying to find the template, although if I did, I'd be building one in a flash. I also recall attempting to contact Mr. Steekblik a number of years ago and having the envelope returned to me, probably using the address shown in the article.</p>
<p>Another "dead-end" link can be found at a <a href="http://solarcooking.wikia.com/wiki/Georgia_Tech_Spiral_Solar_Cooker" rel="nofollow noreferrer">solar cooking wiki page</a>. One of the photographs on that site shows clearly the spiral as well as the pairs of mounting holes necessary to accomplish the bend.</p>
<p>It would be great if someone could perform the analysis and create a parametric file for creating this reflector in varying sizes and widths.</p>
| 21494 | How to construct a compact linear Fresnel reflector |
2018-04-27T23:19:59.597 | <p>I'm looking for something like this air valve:</p>
<p><a href="https://i.stack.imgur.com/cIjJK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cIjJK.png" alt="enter image description here"></a>
(as discussed in <a href="https://engineering.stackexchange.com/a/12404/12007">https://engineering.stackexchange.com/a/12404/12007</a>)</p>
<p>But I need to tube to be wider (3/4 in), the tube to be shorter (the shorter the better), and I would like it if the mechanism for opening and closing the valve wasn't so bulky. This is a picture of an air valve at a much larger scale:</p>
<p><a href="https://i.stack.imgur.com/y3OUC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y3OUC.png" alt="enter image description here"></a>
<a href="https://www.homedepot.com/p/Suncourt-12-in-Normally-Closed-Automated-Damper-ZC112/204268743?cm_mmc=Shopping%7cVF%7cG%7c0%7cG-VF-PLA%7c&gclid=CjwKCAjwlIvXBRBjEiwATWAQIp6R6KVXc7l3PrZILy2NjFr2LsexUXmn6eohfsIsHXPDVXo3r9prZBoC_QgQAvD_BwE&dclid=CIXgsZqh2toCFQN6fgodAd8ElA" rel="nofollow noreferrer">https://www.homedepot.com/p/Suncourt-12-in-Normally-Closed-Automated-Damper-ZC112/204268743?cm_mmc=Shopping%7cVF%7cG%7c0%7cG-VF-PLA%7c&gclid=CjwKCAjwlIvXBRBjEiwATWAQIp6R6KVXc7l3PrZILy2NjFr2LsexUXmn6eohfsIsHXPDVXo3r9prZBoC_QgQAvD_BwE&dclid=CIXgsZqh2toCFQN6fgodAd8ElA</a></p>
<p>I'm hoping to find something like that at a 3/4 in scale, where the mechanism is proportionally smaller also.</p>
<p>I only need the valve to be able to withstand the pressure of a person exhaling, which I think is like 3 psi max. So I'm hoping something less bulky than the solenoid in the first link could be used.</p>
| |airflow|valves|electrical| | <p>You are looking for a 3/4" <strong>magnetic</strong> valve or <strong>solenoid</strong> valve. Google these terms, then talk to a supplier. As you want the mechanism inside the duct (or so it appears), you could also look for a magnetic or electrical ball valve. However I beleive at this small size, the electrical componentes outside the duct will be about the same size as the physical body of the valve.</p>
| 21501 | where to find / what to call a 3/4 inch air valve that can be opened/closed electrically? |
2018-04-28T05:35:47.683 | <p>How do I get a rough estimate of how many litres per minute a motor can push?</p>
| |motors|hydraulics| | <p>First order approximation: Do it as a work problem. </p>
<p>Power = energy/time </p>
<p>so Energy = power * time.</p>
<p>Energy = Work = force * distance = pressure * volume. (Do the dimensional analysis...)</p>
<p>So power * time = pressure * volume</p>
<p>rearranging again</p>
<p>power/pressure = volume/time</p>
<p>The rest is picking a compatible set of units so you don't end up with something weird like board-feet per fortnight.</p>
<p>So given a 5 horse motor, and a 100 foot head, how much water can I move?</p>
<p>1 hp = 746 watts
5 hp = 3730 watts</p>
<p>100 foot head = 43 psi.</p>
<p>If you leave it in psi you will end up with a bizarre unit. I think it would be kg<em>m/lb</em>ft*s which gives me a headache trying to visualize.</p>
<p>Easier to conver 43 psi to about 280 kPa</p>
<p>So 3730 W/280 kPa =.013 cubic meters per second or about 13 liters per second.</p>
<p>Add 5% losses to the pump. 10% pump loss. 10% hydraulic losses and you get about 10 liters per second.</p>
| 21505 | Motors, Horsepower, RPM and Litres per minute |
2018-04-28T16:20:22.397 | <p>I just don't get the concept of Boolean Geometry... Can you help me understand how to solve these homework problems?
Thanks!</p>
<p><a href="https://i.stack.imgur.com/2v5BN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2v5BN.jpg" alt="HomeworkProblems"></a></p>
| |mechanical-engineering|geometry| | <p>"Boolean" just means logically true or false. For this exercise you have to figure out which edges exist when the parts intersect or cut.</p>
<p><a href="https://i.stack.imgur.com/xCr5G.png" rel="noreferrer"><img src="https://i.stack.imgur.com/xCr5G.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Partial completion of exercise 1.</em></p>
<p>Exercise 1 requires you to create the combined solid consisting of A and B but with a cutout C. This means that you have the box and wedge combined and a hole cut through with shape C.</p>
<p>I've drawn in a few edges for you to get you started. Note that the top edge of the box has a gap in it where the wedge juts through. If you were to run your finger across the left face you would feel no edge there.</p>
<p>Can you continue from there?</p>
<p>Note that the <strong>∩</strong> symbol is the 'intersect' operation as used in set theory. The result of that boolean operation will be the parts that are common to both objects when merged.</p>
<p>It might be nice to use a different colour to darken the 'hidden' edges - those you can't see from the chosen viewing angle.</p>
| 21509 | Boolean Geometry Help |
2018-04-28T16:49:52.940 | <p>Or alternately what is the name of the device I'm looking for?</p>
<p>My first thought is to make a narrow V shaped chamber by gluing 1/4" acrylic together with 2 face plates and a mid plate with a triangular hole removed. The liquid comes in at the bottom and lifts a slightly heavier than water bead. Faster flow has to get further into the V chamber before it spreads far enough to match the net gravitational down on the bead. </p>
<p>At 10 liters per hour, flow through a 1/4" pipe is about 9 cm/s. </p>
<p>This seems like such a simple way to do it, I'm sure there are off the shelf ones available for a few bucks. But often searching depends on knowing what it's called. </p>
<p><a href="https://i.stack.imgur.com/88pux.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/88pux.jpg" alt="flowmeter"></a> </p>
<p>This is one that I found on ebay, but intended for gasses, not liquids.</p>
<p>this sort of design with 2-4" plumbing is used for pneumatic seed sorters. However in those, you are indifferent to turbulence. For this application to have something reasonably repeatable, I think I need non-turbulent divergence.</p>
<hr>
<p>Purpose: I'm trying to make a fertilizer injector with minimal pressure drop. Overall the notion is to put a slight constriction in a 3/4" pipe creating a small (few oz) pressure drop. Two small diameter hoses connect to the bottom and top of a chunk of PVC pipe with the fertilizer inside it. Since the pipe is pressurized to the line pressure, the differential across the constriction will move water through the device. </p>
<p>A small valve on the 1/4" line can be used to adjust the flow to put the bead at about the same place each time.</p>
<p>Another thought would be to use something like the IV drip chamber. This is a comparable flow rate. For long duration, I don't know if you could maintain air in the chamber. </p>
<p>Are there other effects I could use?</p>
| |fluid-mechanics| | <p>As you're trying to measure small volumes would a volume marked supply vessel feeding your supply valve in conjunction with a timer (watch,phone or stopwatch) give you the answer you require? The only cost would be your time setting valve to required flow, which you'd have to do anyway with an expensive meter.</p>
| 21511 | How can I make an uncalibrated flow meter for water flows of a few liters per hour |
2018-04-30T00:46:12.460 | <p>I have a motor thats rated for 220V single phase. My house has 240V single phase. Im trying to run this motor but it doesn't budge and I don't even hear a hum. I know power is getting to it from checking with a multimeter. Could it possibly have burned from too high of voltage? </p>
<p>Secondly, if wired improperly, say, ground was accidentally switched with one of the hot wires running to the motor, could this somehow burn the motor?</p>
<p>Not sure whats going on with this thing, and I don't see any reset buttons either or fuses either.</p>
<p>Thanks.</p>
| |mechanical-engineering|electrical-engineering|motors| | <p>If it an efficiently designed motor, there will be a 'magnetisation current' component to the input current, which lags almost 90 deg., and a load current component. The magnetisation current (that provides the magnetic field) will increase, but this will give it more torque for the same load current, so the load current (which is greater) will decrease. This means, on full load, the motor will probably take a bit less current.</p>
| 21521 | Running 220V single phase motor on 240V, and did I burn it? |
2018-04-30T20:34:48.330 | <p>I'm trying to build my own weatherstation for my home, I want to measure temperature humidity and wind strenght and direction. For the wind I have a question. I want to measure the wind with a ball on top of a stick (schematic), and measure the bending of the stick with some strain gauges. Is this possible or just a crappy idea. And what material would I need for the stick that it bends for low wind forces too?</p>
<p><a href="https://i.stack.imgur.com/MO7dO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MO7dO.png" alt="enter image description here"></a></p>
| |measurements|wind-power| | <p>Here's another approach: <a href="https://i.stack.imgur.com/GQ4WH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GQ4WH.jpg" alt="enter image description here"></a></p>
<p>While just a bit over the top, using something like this with somewhat lower density materials may be effective. Doesn't turn in to electronic measurements readily.</p>
<p>I have seen simple wind guages that had a flap pivoted on one edge, and the angle from vertical was calibrated with wind speed. </p>
<hr>
<p>A comment above is certainly correct: A round pole will shed vortices and will require some post measurement tricks to get a reasonable number out of it.</p>
<p>You might be able to derive wind speed from the oscillations set up by the vortex shedding. Another method that is fraught with calibration issues.</p>
<p>You also may find that at certain wind speeds you get resonances that turn your flexible pole into so much debris, or give wild answers.</p>
<hr>
<p>If your goal is to make a wind measuring device without moving parts, you might be able to rig something with timing ultrasonic pulses on a round trip cross wind and up/down wind. I bet you could get within a few percent with two sets at 45 degrees to each other.</p>
| 21538 | Wind measure with strain gauge |
2018-04-30T20:59:51.150 | <p>I am designing a lifting device for practical use which is powered by a dc motor with worm gear. The lifting device should be able to hold the load in place at a certain height.however, i read that wormgears are self locking only in situations where shock and vibration are not present, in other words, if the lifting device is not perturbed in any way, the self-locking feature will probably perform as intended. This could present a problem for security reasons and functionality if the lifting device is intended for use in varied conditions. My question is, should I consider others options, like different types of gears or persist with the worm gear</p>
| |mechanical-engineering|gears| | <p>I've seen from simple observation that many lifting devices using winch mechanisms use worm gears. One advantage of using this mechanism is the high gear ratio (reduction) created with few components.</p>
<p>In my search to validate this reference, I found the answer to a <a href="https://engineering.stackexchange.com/questions/7230/designing-non-self-locking-worm-gear?rq=1">post in the Engineering Stack Exchange</a> in which the replying party suggests that a worm gear design will be self-locking even if such a feature is not desired.</p>
<p>The answer also suggests that the angle on the worm gear (lead angle) will reduce or increase the self-locking capacity. This makes sense in that a nearly perpendicular gear engagement will have substantial friction resisting back-drive, while a more oblique angle will allow the teeth to slide more easily.</p>
<p>Even with some vibration and shock, a perpendicular lead angle may not have movement beyond a degree or so of rotation. If more mechanically complex construction is not a problem, one worm gear driving another, driving the winch would increase the self-locking capacity. I attempted to find an image of this type of assembly and failed. It may be excessive and impractical. </p>
<p><a href="https://i.stack.imgur.com/X1RcC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X1RcC.jpg" alt="worm gears x 3"></a>
Pick two!</p>
<p>In the process, I discovered a worm gear being driven by a conventional reduction gear set, which increases the self-locking by requiring that the worm gear and the gear set be back-driven. Using a multi-level planetary gear set as a reduction gear set would fall into the overkill category, I believe, as planetary gear sets of multiple rings have substantial resistance to being back-driven.</p>
| 21539 | Worm gear for a lifting device |
2018-05-01T13:59:06.233 | <p>I am attempting some convection analysis for research I am doing. I was wondering someone could kindly confirm whether I am on the right track. The calculation is based up air being passed over a flat surface and the heat exchange taking place between the 'air' and ' surface'. </p>
<p>I have the following issues...</p>
<p>I have read books that say that hc for air by forced convection is normally within 10 - 10^3 range, however all my calculations seem to result in around 4-5 is this correct as per my example below?. </p>
<p>The example i have based this upon is a 1m2 surface 1m x 1m dimensions and I want to know how long it would take to change the temperature of the surface material. My approach using the example below shows that if I had to add 460 kJ to change the temperature and then calculate the time, would it be the following to determine time to add the energy..... </p>
<p>460000 J / 66.42 j.s = 6925 secs = 1.92 hours, is this correct?</p>
<p><a href="https://i.stack.imgur.com/kHhfI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kHhfI.png" alt="enter image description here"></a></p>
| |mechanical|convection| | <p>You have the right general approach, and the numbers look reasonable. Your airspeed isn't very fast, so I'm not surprised that the convection coefficient you calculate is less than 10 W m<sup>-2</sup> K<sup>-1</sup>.</p>
<p>Note, however, that you're applying the heat flux you calculated at a plate temperature of 288 K for all other temperatures as well. In reality, the convective flux will decrease as the plate heats up to the airflow temperature. Try using the <a href="https://en.wikipedia.org/wiki/Lumped_element_model#Solution_in_terms_of_object_heat_capacity" rel="nofollow noreferrer">lumped capacitance model</a>. You end up getting a time constant (not a total time) of 6925 s; that is, every 6925 s, the temperature different is reduced by (1-1/e) = 73%. Because the heat flux magnitude affects the plate temperature but also depends on the plate temperature, this temperature approaches the airflow temperature asymptotically. People often assume that the process is essentially complete after 2-3 time constants have passed. Does this make sense?</p>
| 21548 | convection query |
2018-05-02T09:18:21.583 | <p>I am building a machine, and one of it's components is a mechanism that will convert rotational energy to a reciprocating horizontal movement.</p>
<p>I am aware of various mechanisms such as the Scotch Yoke shown below:</p>
<p><a href="https://i.stack.imgur.com/ukWZe.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ukWZe.gif" alt="enter image description here"></a></p>
<p>Which perform this translation. In fact, I would have used the Scotch yoke since it is mechanically simpler - however, the problem with the Scotch Yoke is that the <em>extent of the lateral motion is constrained to the diameter of the wheel</em>.</p>
<p>What I want is to be able to create a lateral reciprocating movement of extent <em>NxD</em> where <em>N</em> is a positive real number, and <em>D</em> is the diameter of the wheel.</p>
<p>Is there an existing mechanism that generates lateral displacements that are several times larger than the diameter of the rotating circle/wheel?</p>
<p><a href="https://i.stack.imgur.com/ZoYlT.jpg" rel="nofollow noreferrer">2</a><a href="https://i.stack.imgur.com/ZoYlT.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/ZoYlT.jpg</a></p>
| |mechanical-engineering|mechanisms| | <p>For those viewing this thread four+ years after it was originally posted, Jay Gould has an excellent series of videos on circular to linear motion. He shows many variations on the Scotch Yoke and the subtleties of their differences.
The first video link is here
<a href="https://www.youtube.com/watch?v=QZ2XFTblrC8" rel="nofollow noreferrer">https://www.youtube.com/watch?v=QZ2XFTblrC8</a></p>
| 21562 | Mechanism to translate rotation into reciprocating horizontal movement |
2018-05-02T15:18:51.520 | <p>I'm doing a study on surfaces and I would like to describe all of the surfaces that I expect to encounter in the field and determine which ones will be the most difficult to detect. To do this, I need to determine what attributes in a surface I need to consider. I have no experience in optics so I'm asking this question here in hopes that someone could save me from overlooking important factors.</p>
| |sensors|optics| | <p>Your question lacks important parameters: in addition to the qualities of the surface or material that Pranav mentions, you need to be able to define what "detect" means, what sort of light sensors you think you want to use, what the interfering noise sources are, and more.<br>
If your target surface is not active (like a lightbulb, generating its own output) then you need to know as a minimum what the illuminating sources are. Then you can calculate, based on surface qualities such as <a href="https://en.wikipedia.org/wiki/Bidirectional_reflectance_distribution_function" rel="nofollow noreferrer">BRDF</a> the expected signal magnitude at your sensor location. But there's a huge difference between saying "Look out! you're going to hit that thing!" and saying "Hey, there's a chunk of rock with 20% silicates and 1% manganese at about 240 Kelvins over there"</p>
| 21564 | What attributes of a surface are relevant to light sensors? |
2018-05-03T07:56:01.880 | <p>Is there any scientific comparison between linear and nonlinear systems?</p>
<p>I often hear that</p>
<blockquote>
<p>Nonlinear control is more sluggish than linear control.</p>
</blockquote>
<p>which makes sense. But is there any research or any claim based on practical experience which supports this claim?</p>
| |control-engineering|control-theory|linear-control|nonlinear-control| | <p>"Linear" imposes a set of restrictions. "Non-linear" simply means there are no restrictions.</p>
<p>Many non-linear control schemes can be faster than linear ones. Linear control schemes are restricted to "smoothly" transitioning. Non-linear control can be implemented by suddenly slamming a digital value, for example.</p>
<p>A good example of a fast-responding non-linear control scheme is a typical thermostat. When the room is below the set point, the heater is turned on fully. When above the set point, the heater is turned off fully.</p>
<p>This is obviously the fastest way to get to the desired temperature. Such a system may be less "good" than a linear controller in other metrics, like overshoot, but that's not what you asked about.</p>
<p>The reason linear control schemes are sometimes chosen is not for their speed or other control property, but because they can be mathematically analyzed more easily. There has been much theory and techniques developed around linear systems that don't apply to non-linear systems. Consider Laplace transforms and S-parameter analysis, for example.</p>
| 21578 | Is nonlinear control slower than linear control? |
2018-05-03T10:20:08.133 | <p>Assume we have a cylindrical fin which has the effective length of <strong>L</strong> and its efficiency is given by the equation: $$η=exp(-0.32mL)$$ where $$m=\sqrt{\frac{hP}{kA}}$$ where <strong>P</strong> is circumference and <strong>A</strong> is the cross sectional area of the fin.</p>
<p><strong>If the volume of the fin remains constant, which of the following statements is true?</strong></p>
<p><strong><em>By increasing the length of the fin ...</em></strong></p>
<ol>
<li><p>Heat transfer increases.</p></li>
<li><p>Heat transfer decreases.</p></li>
<li><p>Heat transfer increases then decreases.</p></li>
<li><p>Heat transfer remains constant because the volume is constant.</p></li>
</ol>
| |mechanical-engineering|heat-transfer| | <p>The math for this problem goes like this:
The fin has a circular cross section and a constant volume. We can write a relationship between volume, length and radius: $V = L*pi*r^2$</p>
<p>Solving that for radius: $r = \sqrt{\frac{V}{L \pi}}$</p>
<p>Efficiency is the ratio of the effectiveness of a fin divided by how effective that fin would be if 100% of its area were at the temperature of the base. Total heat transfer is proportional to efficiency times area.</p>
<p>$Q = hA\eta = hL\pi r = hL\pi\sqrt{\frac{V}{L\pi}} = h\sqrt{L\pi V} \eta$ </p>
<p>We can do similar substitutions for efficiency.
$\eta = exp(-0.32\sqrt{\frac{hP}{kA}}) = exp(-0.32\sqrt{\frac{h\pi r}{k\pi r^2}}) = exp(-0.32\sqrt{\frac{h}{k r}}) = exp(-0.32\sqrt{\frac{h}{k \sqrt{\frac{V}{L \pi}}}}) = exp(-0.32\sqrt{\frac{h}{k \sqrt{\frac{V}{\pi}}}}L^{1/4})$</p>
<p>When we put these two terms together we can ignore everything that isn't L for the purpose of this question:</p>
<p>$Q = things*L^{1/2}*exp(-things*L^{1/4})$</p>
<p>The first term with L increases with L, the second decreases and definitely not at the same rate. Put in some dummy values for "things" and you can get this graph with L on the horizontal and Q on the vertical. It clearly goes up and then down.</p>
<p><a href="https://i.stack.imgur.com/d9K8i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d9K8i.png" alt="L vs Q"></a> </p>
<p>All of this math was not really necessary; Common sense could suffice. The fin equation does not account for heat transfer from the fin tip. You need an additional term for that. If we have a very short fin, it will be $\sqrt{very}$ big around, and have very little area (very > $\sqrt{very}$). Even though it will have 100% efficiency, there will be very little heat transfer. If we have a very long fin, the efficiency will be terrible because most of the fin is very far from the heat source and is cold. At some point in the middle there is an optimum.</p>
| 21580 | Heat transfer in a fin with constant volume and given efficiency |
2018-05-03T13:39:52.643 | <p>I have no experience with actual CAD drawings and CNC machines so please forgive me if this is a really stupid question...</p>
<p>Anyway I have a part which I have 3D printed and it contains some internal screw thread for an ISO metric screw. I modelled the threading myself so it will get 3D printed which took some trial and error. Now since I need it in high quality metal I want to CNC it. But I am wondering whether I have to let the screw thread stay in my model or indicate somehow this is actually ISO metric screw thread of this spec? Since I imagine there is special machinery/settings to make the threading.</p>
<p>Any help is appreciated</p>
| |cad|threads| | <p>It’s not a hard no, but do not model threads. They certainly look a lot prettier, but they are mostly unnecessary. They can consume more memory and require more stress on the computer to render them. This is a big factor with larger models with many parts. </p>
<p>Anyway, as @Chris Johns stated, simply indicate the standard thread on the drawing. The machine shop should do the rest. </p>
<p>Furthermore, if you have specific requirement, you could specify the way the threads are to be manufactured(cut or formed)</p>
| 21584 | How to indicate standard screw thread in drawing |
2018-05-04T13:32:03.660 | <p>We all know the typical streamline shape which looks like a tear. <strong>But what is the optimum shape (for speeds 0-200 km/h)?</strong></p>
<p>Is the front a half sphere or is it an ovoid? How are the sides and the end formed? Can it be described in a formula or in a bezier curve? What is the drag coefficient compared to standard tear drops? Image with explanations how the shape must be built to minimize the drag are perfect.</p>
<p>And no, I don't accept flat plates and perfectly laminar airflow :). The shape must be built around a pipe with diameter R and must not be longer than 10R.</p>
| |aerodynamics| | <p>The general rule of thumb for laminar flow is that the tail needs to be about 4 times as long as the nose because it is more difficult to slow down a flow than to speed it up. Equally you want to avoid any geometric discontinuities so a sensible way to do this is treat it as circular arcs at a tangent to each other. </p>
<p><a href="https://i.stack.imgur.com/ZeQVH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZeQVH.jpg" alt="enter image description here"></a></p>
<p>The grey cube is 1x1 aligned with the major axes to give a sense of orientation.</p>
<p>Of course there are a few caveats to this:</p>
<p>Firstly, scale matters in fluid flow and you get to a point where the onset of turbulence is inevitable at a given scale and speed whatever you do. </p>
<p>Equally: when you talk about optimisation this, by definition, implies that you are seeking a useful balance between conflicting requirements. Reducing drag is only one requirement. In reality and even in in fluid dynamics the body under consideration has some purpose. </p>
<p>For example: Aircraft need to generate lift, thrust, control and carry some payload, so in practice a lot of aerodynamics is managing drag and turbulence rather than attempting to eliminate it entirely. </p>
| 21597 | What is the optimal streamlined shape? |
2018-05-04T14:08:33.783 | <p><a href="https://i.stack.imgur.com/nsZJ0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nsZJ0.jpg" alt="enter image description here"></a></p>
<p><sup>Source: <a href="http://www.cbc.ca/news/canada/new-brunswick/friday-flood-new-brunswick-2018-1.4647979" rel="nofollow noreferrer">http://www.cbc.ca/news/canada/new-brunswick/friday-flood-new-brunswick-2018-1.4647979</a></sup></p>
<p>Is there a way to calculate how much weight a vessel can carry (in fresh water) before it submerges?</p>
<p>Assumptions:</p>
<ul>
<li>The water is not disturbed (no waves or wind) and the cargo does not move</li>
<li>The vessel weighs 200 lbs when empty</li>
<li>The vessel's volume is 60 cubic ft</li>
</ul>
| |fluid-mechanics| | <p>A problem is that volume is not exactly displacement. Typically a boat will be a little lower in the middle and / or more weighted in the rear. </p>
<p>Pretend this is a perfect boat and volume = displacement. </p>
<p>Buoyancy is exactly displaced water * water density. </p>
<p>Density of water is nominally 62 lbs per cubic foot. </p>
<p>60 ft^3 * 62 lb / ft^3 = 200 lbs + x lbs<br>
3720 lb - 200 lbs = x lbs<br>
x = 3520 lb</p>
| 21599 | How much weight can a floating vessel carry before submerging? |
2018-05-05T19:02:36.307 | <p>Are there any simplified formulas that could be used to generate plausible airplane, spaceplane/SSTO, and spacecraft performance specs for a game given some basic vehicle and contextual attributes? By plausible, I mean within 25% of the value a full analysis might achieve. For example, could I roughly determine maximum acceleration and top speed from the following attributes:</p>
<pre><code>VEHICLE
thrust (in real units)
mass (in real units)
aerodynamics (any intuitive scale: e.g. 0.0 = horrible, 1.0 = ideal, Boeing
747 set arbitrarily at 0.50)
CONTEXT
atmosphere (density? pressure? friction? How best to measure simply?)
gravity (real units)
angle of travel (0 = parallel to planet, 90 = perpendicular to planet)
</code></pre>
| |modeling|simulation| | <p>The maximum acceleration can be calculated simply by comparing your vehicle's mass and thrust values:</p>
<p>$$F=m*a$$
$$acceleration = \frac{thrust}{mass}$$</p>
<p>The terminal velocity can be estimated by balancing your thrust values and the drag equation. When drag balances thrust, you cannot travel any faster</p>
<p>$$F_D=\frac{1}{2}\rho u^2C_DA$$</p>
<p>$$terminal\ velocity=\sqrt{\frac{2*thrust}{air \ density* drag\ coefficient*frontal\ area}}$$</p>
<p>You have also included 'angle of travel' as a factor - this will affect the total driving force in the direction of travel, and consequently the terminal velocity. If you are flying vertically upwards, the force due to gravity ($F=mg$) must be subtracted from the trust. If you are flying downwards at an angle of $20°$, then the component in the direction of travel ($F=mg\sin(\theta)$) should be added to the thrust.</p>
| 21618 | simplified acceleration and max speed calculations for a game? |
2018-05-06T18:02:19.410 | <p>Do homes increase in temperature due to solar gain with curtains open if it's overcast?</p>
<p>In other words, Does having curtains closed reduce the temperature of the home on an overcast day?</p>
<p>In other words, Does an overcast day prevent the warming effect of the sun hitting a homes window and raising it's temperature?</p>
<p>In other words, Closing the curtains on a sunny clearly keeps a room cooler. Does closing the curtains on an overcast day have the same effect? Why?</p>
<p>Thank you for your time.</p>
| |thermodynamics|temperature| | <p>Closing the curtains will help even on a cloudy day.</p>
<p>Assuming there is minimal heat transfer from the curtain to the room due to infrared radiation of the curtain and assuming minimal convection of hot air trapped between the curtain and the glass, overflowing the curtain rail into the room, the curtain will help by reflecting some of the sun's rays that have penetrated through the clouds and also some of the infrared radiation caused by the clouds. </p>
<p>The difference between a cloudy day and a sunny day is some of the solar radiation energy reflects back into space by the clouds, but not all of it. It depends on reflection index of the cloud. The part of radiation which is not reflected partially heats up the cloud causing infrared radiation to earth and partially reaches the surface of the air. </p>
<p>So having the shades closed still helps even during cloudy days.</p>
| 21625 | Is a Home's Temperature Affected by Solar Gain Through Windows on Cloudy Days? |
2018-05-07T11:40:02.103 | <p><img src="https://i.stack.imgur.com/o6Fs1.jpg" alt="enter image description here">
What is wrong with totally replacing the inlet of a piston engine with a surgical mask?As surgical masks get rid of about 80% of air impurities and dusts.</p>
| |engines| | <p>Air impurities are no issue for an ICE, only dust and other bigger particle matter are. The material of a surgical mask has little flow capacity, so a bigger filter is needed than normal, without good reason. You could use them for an ICE, but as is mentioned before, their flow capacity is way too low. You'd need to glue lots of them together, and fold them into a saw pattern to efficiently package them into a filter box. Then they could properly function as an airfilter. </p>
<p>But it won't weigh less, so it won't contribute in your search to lower the weight of your engine. If you don't live in a dusty area, rather remove the entire filter if it's really important for the engine to weigh less. The engine will wear faster though.</p>
| 21642 | Inlet of an ICE |
2018-05-08T16:02:14.327 | <p>In structural mechanics, we often approximate a partial differential equation by a system of ordinary differential equations in matrix form</p>
<p>$$\boldsymbol{M}\ddot{\boldsymbol{q}}(t)+\boldsymbol{B}\dot{\boldsymbol{q}}(t)+\boldsymbol{K}{\boldsymbol{q}}(t)=\boldsymbol{f}(t),$$</p>
<p>in which $\boldsymbol{M}$ is the mass matrix, $\boldsymbol{B}$ is the damping matrix and $\boldsymbol{K}$ is the stiffness matrix. The vector $\boldsymbol{q}(t)$ is the displacement vector which contains the displacements at specific positions and the force vector $\boldsymbol{f}(t)$ does specify the forces applied at the same positions at which the displacements are specified.</p>
<p>It is possible to deduce the mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$ by introducing finite elements. The damping matrix $\boldsymbol{B}$ is not as easy to determine.</p>
<p>A possible model is Rayleigh’s damping model. It is given by</p>
<p>$$\boldsymbol{B}=\alpha \boldsymbol{M} +\beta \boldsymbol{K}.$$ </p>
<p>Now what I do not understand is the problem of choosing $\alpha$ and $\beta$. Are they obtained by system identification, while testing measured data with simulated data?</p>
| |structural-analysis|applied-mechanics|vibration| | <p>If you know the damping ratios ($\zeta$) at two frequencies, those 4 values determine $\alpha$ and $\beta$ by</p>
<p>$$
\zeta = {1 \over 2}{({\alpha \over \omega} + \beta \omega)}
$$</p>
<p>In this graph, the blue curve is $\zeta$, and it's the sum of the other two curves which correspond to the two terms in brackets in the formula. If you know two points on the blue curve, that uniquely defines its parameters $\alpha$ and $\beta$.</p>
<p>In the special case where damping ratio is proportional to frequency, $\alpha = 0$, the mass matrix $\boldsymbol M$ is ignored, and the formula is exact. Otherwise, it's usually an approximation that's only correct at the two frequencies used to fit it.</p>
<p><a href="https://i.stack.imgur.com/iequQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iequQ.png" alt="Damping ratio vs angular frequency"></a></p>
| 21661 | Rayleigh damping in structural mechanics |
2018-05-08T20:43:08.560 | <p>I grow plum trees from pits using <a href="http://www.canadiantire.ca/en/pdp/mckenzie-21-cell-seed-starting-tray-3-in-0591680p.html#srp" rel="nofollow noreferrer">seed starting trays</a>:</p>
<p><a href="https://i.stack.imgur.com/TPCsF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TPCsF.jpg" alt="enter image description here"></a></p>
<hr>
<p>Unfortunately, I have a problem where the roots of the seedlings grow in circles around the bottom of the cavities of the trays. The circling roots cause problems when the trees grow to maturity (they strangle each other out). </p>
<p>Instead of the roots growing in circles, I need to train them downwards towards the hole (where they will be air-pruned).</p>
<p><a href="https://i.stack.imgur.com/Ka5MO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ka5MO.jpg" alt="enter image description here"></a></p>
<hr>
<p>The problem is normally solved by using <a href="http://www.amaplas.com/ViewProduct.aspx?GroupID=0&InventoryID=5053&CategoryID=52" rel="nofollow noreferrer">seed starting trays that have specialized vertical fins</a> that prevent the root from growing in circles:</p>
<p><a href="https://i.stack.imgur.com/ep4Ab.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ep4Ab.jpg" alt="enter image description here"></a></p>
<p><strong>Question:</strong></p>
<p>I would like to <strong>retrofit</strong> my existing trays (I have several dozen) so that they function like the anti-root circling trays. I think this might be achieved by adding some sort of vertical fin in each cavity that would act as a physical barrier/re-router to the roots (but I'm not sure how to do this).</p>
<p>How can I retrofit my existing trays to prevent root circling?</p>
| |manufacturing-engineering|environmental-engineering|prototyping| | <p>Coffee (stirring) sticks might work.</p>
<p><a href="https://i.stack.imgur.com/ScpEK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ScpEK.jpg" alt="enter image description here"></a></p>
<p>After potting is complete these could be inserted vertically with the narrow edge against the side of the pot in multiple positions around the edge. Seeing as your pots are conical at the bottom you may insert one stick vertically down to the start of the cone and another one on the other side at an angle to form a breaker in the conical section.</p>
| 21663 | How to retrofit seed starting trays to prevent root circling |
2018-05-09T08:00:59.633 | <p>I am bit confused in considering the methods to apply for designing.</p>
<p>Consider a machine, which as several parts mounted on the frame (frame is of any sectional shape for exampe: ´I´ section). Let the length and width of the frame is 2m and 0.6m respectively. Here the Distribution of the masses of the different components on the frame are eccentric in nature. And the machine supports at the four cornes which is similar to the support of the chair.</p>
<p>My question is, how do I calculate the deflection, bending moment, shear stress etc of the Frame or whole machine due to the Forces acting on it.</p>
<p>Since the Forces acting here are eccentric and not exactly on the beam, I think it is not possible to solve it by using Beam theory.</p>
<p><a href="https://i.stack.imgur.com/HKsL9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HKsL9.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/YPrg0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YPrg0.png" alt="enter image description here"></a></p>
<p>So, could anyone suggest me how to carry out this problem to solve it analytically. Or any advices on the study of any related topics are appreciated.</p>
<p>Thank you</p>
| |mechanical-engineering|structural-engineering|structural-analysis| | <p>Lets handle fore the sake of illustration just the horizontal cylinder being supported by four rollers and later we can add the stresses caused by other parts and superpose all of them.</p>
<p>Lets assume the cylinder weighs 400lbs and is touching the bearings at 45 degrees angle, and is not imposing an dynamic loading, or else we need to refer to its manual for dynamic load factors.</p>
<p>The bearings horizontal and vertical reaction is each $$ R_h = 100\sqrt(2) \space = 141lbs \space , \space and \space R_v =100lbs $$</p>
<p>Say the arm of roller bearing is 6inch high and 2 inch in. So we have $$ 6\times 141 = 846 \space lb.in \space - 2\times 100 =646lb.in$$
total clockwise torque to the right beam at each bearing and symmetrical torque at the left beam supports. This causes torsional stress at 2 points on the beam, which we can calculate, given the beam's cross section.</p>
<p>As for the vertical loads which create moment and shear we have for point loads of 100lbs each and again we can calculate the bending moment and shear.</p>
<p>Now we go back to vertical cylinder on the back of diagram. That part can slide back and forth so we have to consider the influence line momentum and shear. But in this case for simplicity we consider the bending moment when the part is at most critical position at mid span and shear when it is at either support.</p>
<p>So now we can apply all the loads at their proper offset from the start of the beam and calculate bending moment and shear and torsion for a two span continuous beam.</p>
<p>I usually use factor of safety of 2.8 for these situations in the absence of a code load factor, but as we said this was just an illustration. Most of the compressors and moving machines have recommended design charts for foundation and support.</p>
| 21671 | How to analyse a frame of a machine? |
2018-05-09T09:00:16.777 | <p>I've been trying to solve this problem for a while with no success. I think I figured out that $m_1$ is $420 kg/hr$ and that $n_{3,A} = 3 kmol/h$ using the mass balance equation:</p>
<p>$input-output+generation-consumption=accumulation$ </p>
<p>where steady state conditions mean there will be no accumulation and the lack of a reaction means that there will be no generation or consumption.</p>
<p>Everything else I've tried to solve for the remaining values has led me to a dead end (420=420, 16=16, etc.) Any help would be greatly appreciated!</p>
<p><img src="https://i.stack.imgur.com/W2Oxv.png" alt="Question"></p>
| |materials|energy| | <p>First, applying your equation to your case, $n_1-(n_2+n_3) + 0 - 0 = 0$ or $n_1 = n_2+n_3$</p>
<p>$$n_{1,a}=n_{2,a}+n_{3,a} \space (1) \\
n_{1,b}=n_{2,b}+n_{3,b} \space (2)\\
n_{1,c}=n_{2,c}+n_{3,c} \space (3)
$$</p>
<p>and since there's nothing that separates/filters the components in different proportions, the output will contain exactly the same fraction as input.</p>
<p>$$ f n_{2,a} + (1-f) n_{3,a} = n_{1,a} \space (4)\\
f n_{2,b} + (1-f) n_{3,b} = n_{1,b} \space (5)\\
f n_{2,c} + (1-f) n_{3,c} = n_{1,c} \space (6)\\
$$</p>
<p>where $f$ is the ratio at which the streams are split into output - the same for all components.</p>
<p>Looking at the table, we have $n_{1,A}$ and $n_{2,a}$. From (1) we can quickly find $n_{3,a}$ : 4-1=3. </p>
<p>With these 3 we can find $f$ from (4): 0.25</p>
<p>Remainder of $n$ is trivial as we have column $1$ - multiply the $1$ column value by 0.25 for column $2$ and 0.75 for column $3$, e.g. $n_{2,b} = 10\cdot0.25 = 2.5$</p>
<p>For totals in kg/h, simply take the units given to find the factor by which to multiply:</p>
<p>$$total_{c} = \sum_i^{a,b,c}{MW_i \cdot n_{c,i}} $$</p>
<p>(remembering the 1000 factor of $kmol$ vs $mol$ - multiply $n$ by 1000 first): For column $1$</p>
<p>(4*10 + 10*20 + 6*30)*1,000 = 40+200+180 = 420,000 kg/h.</p>
| 21673 | How do I solve for the mole and mass flow rates in a system? |
2018-05-09T11:17:02.243 | <p>I am trying to figure out how I can assemble multiple finite elements to obtain the system mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$.</p>
<p>The situation is depicted in a picture (see below, <strong>assume the left side is not fixed</strong>). Now, I thought about splitting the body into a cantilever beam and a disc. The beam has four degrees of freedom in a planar analysis. I will refer to $q_1$ as the vertical displacement at the left side of the beam and $q_2$ is the deflection angle at the left side of the beam. At the right side of the beam, the vertical displace is denoted by $q_3$ and the deflection is denoted by $q_4$.</p>
<p>For the disc, I assume the degrees of freedom as depicted in the picture below. Hence, $q_{\text{S},1}$ and $q_{\text{S},3}$ for displacements and $q_{\text{S},2}$ and $q_{\text{S},4}$ for the angles of deflections (double headed arrow). It is obvious that $q_3=q_{\text{S},3}$ and $q_4 =q_{\text{S},4}$. </p>
<p>If I set up the mass matrix $\boldsymbol{M}_\text{B}$ for the beam I see that it has the format $4\times 4$. The same goes for the stiffness matrix $\boldsymbol{K}_\text{B}$. Assume that the mass matrix of the disc is given by $\boldsymbol{M}_\text{D}$ and the stiffness matrix is given by $\boldsymbol{K}_\text{D}$. How can I assemble the system mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$ from the previous matrices? </p>
<p>Feel free to write down the procedure on a sheet of paper, that would be good enough :). I also appreciate a resource that is explaining this procedure.
<a href="https://i.stack.imgur.com/Ts1Sx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ts1Sx.png" alt="whole system"></a></p>
| |structural-analysis|applied-mechanics|finite-element-method| | <p>The procedure for assembling the system mass and stiffness matrices are the same.</p>
<p>Repeat steps 1 and 2 below for each element matrix, $K_D$:</p>
<p>1) Expand the element matrix to have the same structure as the system matrix, putting zeros where there's no corresponding value in the element matrix. Do this by inserting each element matrix element into the expanded element matrix row with the same DOF that it came from and the column with the same DOF that it came from. For example:</p>
<p>If the DOFs in the system matrix $K$ are in this order:</p>
<p>$$
qs2, q1, q2, q3, q4, qs1
$$</p>
<p>and the DOFs in the element matrix $K_D$ are in this order:</p>
<p>$$
qs1, qs2, q3, q4
$$</p>
<p>and</p>
<p>$$
K_D =\begin{bmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p\end{bmatrix}
$$</p>
<p>then the expanded element matrix would be</p>
<p>$$
\begin{bmatrix}f & 0 & 0 & g & h & e \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ j & 0 & 0 & k & l & i \\ n & 0 & 0 & o & p & m \\ b & 0 & 0 & c & d & a \end{bmatrix}
$$</p>
<p>2) Add (matrix addition) the expanded element matrix to the system matrix.</p>
<p>In real life, you wouldn't actually build the expanded matrix explicitly but just add the non-zero elements.</p>
| 21676 | Cantilever beam with disc at end (structural mechanics) |
2018-05-09T18:30:43.907 | <p>I took this photo behind my place; there was this group of workers for about an hour.</p>
<p>They have a small hose, from the black spool, they were pushing in and out of the manhole and a large hose that didn't seem to move at all.</p>
<p>The truck's engine was quite noisy, so I'm assuming they have a large pump. I'm curious what kind of truck it is and what they could be doing with it.</p>
<p><a href="https://i.stack.imgur.com/dNvtI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dNvtI.jpg" alt=""></a></p>
| |civil-engineering| | <p>The small black hose is a high pressure water hose, with a nozzle attached to its end. Something like that:</p>
<p><a href="https://i.stack.imgur.com/5cCYD.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/5cCYD.jpg" alt="enter image description here"></a></p>
<p>It is used to flush down sand and other sediments
The nozzle usually features some backwards facing jets, so the hose gets pushed inside a pipe. The hose can be pulled back with the windlass.
There's a variety of differenz nozzles, as you can see in the pictures, I think nozzle 2 and 3 have a rotating bit and the chains on nozzle 4 are probably to knock free solid sediments like chalk.</p>
<p>The large hose is a suction tube to either empty the manhole or pump out the sediments flushed down by the pressure hose.
Usually the tanks of those types of vehicles have a bulkhead to separate waste water from fresh water, and a methode to return the sediments inside, while returning the sediment-free waste water to the pipes.</p>
| 21686 | What is this kind of sewer maintenance work? |
2018-05-10T06:26:13.020 | <p>I need to find heat loss from the greenhouse. I know how to find heat loss through the walls and roof , but what about soil ? how to find amount heat conductivity loss to the soil ? I would be gratefull for the any reference/method/ article or etc. </p>
| |thermodynamics|heat-transfer| | <p><strong>Easy answer:</strong> Net heat loss averaged over day/night is zero because the heat that's transmitted into the soil when the greenhouse is hot is stored in the soil and returned to the greenhouse when the greenhouse becomes colder than the soil.</p>
<p><strong>Hard answer suggestion:</strong> This problem is called "One-dimensional transient heat conduction in semi-infinite body". Semi-infinite because the ground is infinitely deep in the downwards direction but not in the upwards direction. It's time dependent. The rate of heat loss will go down as it progresses because the ground becomes hotter. You can write and solve the DEs by hand if you wanted a hard exercise in calculus. Here's somebody's working:
<a href="https://www.thermalfluidscentral.org/encyclopedia/index.php/One-dimensional_transient_heat_conduction_in_semi-infinite_body" rel="nofollow noreferrer">https://www.thermalfluidscentral.org/encyclopedia/index.php/One-dimensional_transient_heat_conduction_in_semi-infinite_body</a></p>
<p>Whiteboard lecture video: <a href="https://www.youtube.com/watch?v=X8ofOO-iHhk" rel="nofollow noreferrer">https://www.youtube.com/watch?v=X8ofOO-iHhk</a></p>
<p><strong>Practical answer:</strong> I would use Finite Element Analysis (FEA) which does all that math for you. Download some FEA software and use time dependent or transient heat transfer analysis. This will also allow you to model 2D and 3D effects like heat flow around the edges. The two 1D suggestions above are not strictly correct because of lateral heat flow under the ground but the bigger the greenhouse, the more accurate the 1D approximation is.</p>
| 21695 | Heat loss from the greenhouse to the soil |
2018-05-10T21:40:07.687 | <p>My strength of materials book provides the following equation to determine the x-component of the normal stress acting on all possible planes of a two dimensional element.</p>
<p>$$
\sigma_{x^{\prime}} = \frac{1}{2}(\sigma_x + \sigma_y ) + \frac{1}{2}(\sigma_x - \sigma_y
)\cos(2\theta) + \tau_{xy}\sin(2\theta)$$</p>
<p>My question is -- in what orientation are the non-primed variables assumed? My instinct is that the initial orientation that the non-primed quantities are given in is irrelevant, only the numerical value of the components matter and the angle from which you measure the new components (primed quantities).</p>
<p>For example, if you are given a 2-D element that is initially inclined at 60 degrees and you have the values of $\sigma_x$, $\sigma_y$, and $\tau_{xy}$ in that orientation and you want to find $\sigma_{x^{\prime}}$ at the 0 degree orientation you would compute the equation above using a value of $\theta = -60$ because you have to measure <em>relative</em> to where you "start".</p>
<p>Is that correct or do I have this completely confused? </p>
| |stresses| | <p>The transformation relation for 2D stresses in matrix form is
$$
\boldsymbol{\sigma}' = \mathbf{Q}^T\, \boldsymbol{\sigma}\, \mathbf{Q}
$$
where
$$
\boldsymbol{\sigma} = \begin{bmatrix} \sigma_{xx} & \tau_{xy} \\ \tau_{xy} & \sigma_{yy} \end{bmatrix} ~,~~ \mathbf{Q} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}
$$
In writing the stresses in matrix form, we assume that the stress components have been expressed in terms of a coordinate system with base vectors ($\mathbf{e}_x, \mathbf{e}_y$) and that the angle $\theta$ is measured counter-clockwise with respect to the $\mathbf{e}_x$ vector. </p>
<p>Therefore, the angle $\theta$ is always zero when you are looking along the $\mathbf{e}_x$ vector, by definition.</p>
<p>If you work out the algebra, you will find that (assuming I've made no mistakes)
$$
\begin{bmatrix} \sigma_{xx}' \\ \sigma_{yy}' \\ \tau_{xy}' \end{bmatrix}
= \begin{bmatrix} \cos^2\theta & \sin^2\theta & \sin2\theta \\
\sin^2\theta & \cos^2\theta & -\sin2\theta \\
-\tfrac{1}{2}\sin 2\theta & \tfrac{1}{2}\sin 2\theta & \cos 2\theta \\
\end{bmatrix}
\begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \tau_{xy} \end{bmatrix}
$$</p>
<p>The inverse relation is
$$
\mathbf{Q} \,\boldsymbol{\sigma}' \, \mathbf{Q}^T= \boldsymbol{\sigma}
$$
In expanded form,
$$
\begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \tau_{xy} \end{bmatrix}
= \begin{bmatrix} \cos^2\theta & -\sin^2\theta & -\sin2\theta \\
\sin^2\theta & \cos^2\theta & \sin2\theta \\
\tfrac{1}{2}\sin 2\theta & -\tfrac{1}{2}\sin 2\theta & \cos 2\theta \\
\end{bmatrix}
\begin{bmatrix} \sigma_{xx}' \\ \sigma_{yy}' \\ \tau_{xy}' \end{bmatrix}
$$
If the initial stress components are with respect to a set of initial axes that are not aligned with ($\mathbf{e}_x, \mathbf{e}_y$) and you want to compute the components along $\mathbf{e}_x$ etc., you will have to use the inverse relationship.</p>
| 21700 | 2-D Plane Stress Transformations |
2018-05-11T00:36:45.613 | <p>How can I determine horizontal force reactions in a fixed on both ends beam like this one?
<a href="https://i.stack.imgur.com/ItzGY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ItzGY.png" alt="Fixed on both ends beam"></a></p>
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis|statics| | <p>The reaction at either end is simply equal and opposite to the axial load in the beam adjacent to it. So what you need to work out is the axial force each side of where <strong>F</strong> is applied.</p>
<p>To work this out you need the plea formula:</p>
<blockquote>
<p>d = PL/EA</p>
</blockquote>
<p>where d is extension, P is axial force, L is the original length, E is Young's modulus and A is cross-sectional area.</p>
<p>Using subscript 1 for the left hand side and 2 for the right hand side, we then get two equations:</p>
<blockquote>
<p>d<sub>1</sub> = P<sub>1</sub>a/EA</p>
<p>d<sub>2</sub> = P<sub>2</sub>b/EA</p>
</blockquote>
<p>We also know that for equilibrium:</p>
<blockquote>
<p>P<sub>2</sub> + P<sub>1</sub> = F</p>
<p>d<sub>2</sub> = d<sub>1</sub></p>
</blockquote>
<p>We can then solve all of these simultaneous equations (I'll leave that step to you), and we'll find:</p>
<blockquote>
<p>P<sub>1</sub> = F * b/(a+b)</p>
<p>P<sub>2</sub> = F * a/(a+b)</p>
</blockquote>
<hr />
<p><em>NB The plea formula works equally well in tension and compression (assuming no buckling). A tensile force leads to elongation, a compressive force leads to shortening. All my workings are on absolute values, if you want you can make P<sub>1</sub> and d<sub>1</sub> negative; this is technically more correct but it adds a layer of complexity that I don't feel is necessary.</em></p>
| 21702 | How can I determine horizontal force reactions in a fixed on both ends beam |
2018-05-11T09:10:09.683 | <p>I've been stuck on this for an hour. (Probem is below)</p>
<p>The problem is with part b. I understand its the basic mass flow in - mass flow out with air. The equation being $$\rho_{air}A_4V_4 = \rho_{air}A_0V_0$$ where we are trying to solve for $V_4$ or the speed of air coming in through pipe 4.</p>
<p>We rearrange to $$V_1=\frac{\rho_{air}A_0V_0}{\rho_{air}A_4}$$ </p>
<p>Everything should cancel to</p>
<p>$$V_1=\frac{A_0V_0}{A_4}$$ </p>
<p>The issue, what on earth is $V_0$???! </p>
<p>In part b, $\frac{dh}{dt}=0.1910\;\frac{ft}{s}$ and is the value I ASSUMED would be used for $V_0$ in this case but no, it gives a value of $27.5 \: \frac{ft}{s}$ which is not correct.</p>
<p>The value supposedly used for $V_0$ is 0.1484</p>
<p>What is this value?
Where does it come from?</p>
<p>The only thing I can think of is that it has something to do with 'average velocity' as stated in the part b question.</p>
<p>Yet I'm drawing a complete blank. And have been for more time than I care to admit. Especially considering the solution is right in front of me.</p>
<p>Help is VERY appreciated at this point.
<a href="https://i.stack.imgur.com/yir7q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yir7q.png" alt="enter image description here"></a></p>
| |fluid-mechanics| | <p>In my opinion, this is a typo and your calculation is correct. In (a), the velocity at the boundary water-air is denoted $dh/dt$. Therefore, when considering only the air volume, and considering the air as incompressible,
$$
\rho_{air}\,A_0\,\frac{dh}{dt}=\rho_{air}\,A_4\,v
$$
as you wrote (though I would use $v$ instead of $V$ for the velocity, since $V$ is often associated with volume).</p>
<p>Using $dh/dt\approx 0.1910\,\mathrm{ft/s}$ yields $v\approx 27.5\, \mathrm{ft/s}$. Using instead the 0.1484 (ft/s) for $dh/dt$ that given in the solution for (b), $v$ calculates to 21.4 ft/s.</p>
<p>PS: Please, unlike the author of the problem, use units in your calculations.</p>
<p>PPS: You might check if there is a newer version of the textbook available, or if there is an errata.</p>
| 21707 | Unknown value in solution of continuity problem? |
2018-05-14T00:59:17.813 | <p>Currently, I am thinking that since electromagnetic waves loose energy over distance, you might be able to figure out the relative distance to each of three radio "beacons" and use this information to triangulate the position of a robot with a radio sensor on-board.</p>
<p>What information each radio signal could I use to measure the distance to correlating beacon? Amplitude, perhaps? The entire system would be running indoors in a relatively small area, so it needs to be relatively precise.</p>
<p>Is this even a reasonable idea, or are there better methods I could use?</p>
<p><strong><em>Note:</strong> I asked a <a href="https://robotics.stackexchange.com/questions/15680/radio-triangulation-with-raspberry-pi">similar question</a> about radio triangulation on the Robotics SE, where someone asked me to look into the Engineering SE instead.</em></p>
| |positions-measurement|radio| | <p>I believe there has been some research done on robot localization based on Wifi signal strength. For example: <a href="http://robotics.usc.edu/~ahoward/projects_wifi.php" rel="nofollow noreferrer">http://robotics.usc.edu/~ahoward/projects_wifi.php</a> and <a href="http://www.cs.cmu.edu/~mmv/papers/10icra-joydeep.pdf" rel="nofollow noreferrer">http://www.cs.cmu.edu/~mmv/papers/10icra-joydeep.pdf</a>. But I believe accuracy is very rough (maybe meters of error).</p>
<p>For more precise radio localization (cm level), you can use ultrawideband (UWB) radios which can measure the time of flight between endpoints. Yes, there are multipath issues indoors, but there are ways to mitigate this. There are two main players here: <a href="https://www.decawave.com/" rel="nofollow noreferrer">https://www.decawave.com/</a> and <a href="http://bespoon.com/" rel="nofollow noreferrer">http://bespoon.com/</a>.</p>
<p>Also FYI, triangulation is when you know the angle to the fixed beacons, trilateration is when you know the range. See: <a href="https://en.wikipedia.org/wiki/Trilateration" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Trilateration</a>. Your proposed application would be trilateration.</p>
| 21732 | Radio triangulation with Raspberry Pi |
2018-05-14T02:16:42.763 | <p>What is the definition of anti-windup? How does it impose the constraints?</p>
<p>What are the advantages of MPC and anti-windup over each other?</p>
<p>Does anti-windup guarantee the constraints or does it just try respecting them?</p>
| |control-engineering|control-theory|pid-control|optimal-control| | <p>Anti-windup is a concept for feedback controllers with integral terms, e.g. PID, to keep the integral term from „overcharging“ when regulating a large set point error. It basically saturates the integral term to keep the system from overshooting the set point.</p>
<p>The <em>classic</em> form of anti-windup, as described above, does not actually ensure satisfaction of input or state constraints of the system, it just enforces a somewhat „auxiliary constraint“ on the integral term. </p>
<p>EDIT: Anti-windup can also be used to ensure satisfaction of state and input constraints by exploiting the unwanted mechanic:</p>
<p>The classic saturation circuit gets extended by a predictor, that predicts the state that corresponds to the control input. A possible violation of constraints can then be detected with saturation. In that case a feasible control input, i.e. a control input such that the resulting state will not violate constraints, is calculated. The details of the approach are explained in <a href="https://www.sciencedirect.com/science/article/pii/S1474667015395100" rel="nofollow noreferrer">this</a> paper.</p>
<p>I am not familiar enough with the above approach to directly compare Anti-windup constraint satisfaction and MPC constraint satisfaction. I read once in some lecture slides about MPC (I'll link to them when I find them again)</p>
<blockquote>
<p>If PID does the job, take PID, otherwise take MPC</p>
</blockquote>
<p>so now that you know how to ensure constraint satisfaction with the former, just take the one that does the job.</p>
| 21733 | Advantage of anti-windup |
2018-05-14T20:26:17.153 | <p>I have a temperature sensor in a -20C freezer. The probe is going into a bottle filled with propylene glycol solution. I want to seal the hole where the probe is inserted into the bottle.</p>
<p>What's a good sealant for this?</p>
| |materials|temperature|seals|adhesive| | <p>In the end, I decided to use a bottle filled with glass beads. This provided the necessary thermal stabilization and it has the added benefit of working below -50C. </p>
| 21738 | How would you seal a hole in an HDPE container that's in -20C environment |
2018-05-15T08:13:49.810 | <p>I have a large boom of a radio antenna weighing $3,500 kg$ which is rotating at $0.5 RPM$. The antenna is a T shape and the boom radius is $14m$. The power to the motor is cut off suddenly and the motor acts as a rigid brake. The boom is not rigid and it continues to move/bend in the same direction for 1 second (at which points it bounces back and fourth until the forces damper out). I am trying to work out the max torque this would apply to the gearbox shaft, assuming there is no friction. </p>
<p>For angular movement, I have designated the centroid of each boom radius as a point load (so a point load at a radius of 7 m on both ends of the boom). </p>
<p>I can work out the angular moment which is $m \cdot r \cdot v$, which works out to $1750 kg \cdot 7m \cdot 0.367m/s$ or $4489.86kg\frac{m^2}{s}$ </p>
<p>This is where I am not sure about though. I have read that torque is simply the change in angular momentum over time. So would that be $4489.86kg\cdot m $ $(44.1kN \cdot m)$?</p>
<p>Would this be an accurate way to determine the torque in the shaft? This is for a real scenario so If I have missed any big considerations please let me know.</p>
| |torque| | <p>Assuming your antenna's boom is solid then its moment of inertia along the long axis is$$ I = 1/12ml^2 $$ approximately which is less than you attributing the mass to the two ends.
The torque this boom causes while stopping and wobbling is
$$ \tau = I\alpha $$</p>
<p>In this case is $$ 1/2-0 = 1/2 \space = \pi $$
So your torque is $$1/12\times 3500\times28^2\times \pi = \pi/12\times 28^2\times 3500 \space Nm $$</p>
<p>But We should consider the vibration of the boom and its dynamic whiplash effect two, which is dependant on the stiffness of the boom and its detailed construction such as bolted together or welded together or what not.</p>
<p>codes in similar situations have a dynamic load factor of 2.8.</p>
| 21745 | How to calculate torque needed to stop a rotating beam over x seconds |
2018-05-15T09:09:02.717 | <p>I am interested on simulating the converse Piezoelectric effect of materials and I have given the mechanical (Young's modulus and Poisson's ratio) and electric material properties (Electric Permittivity, Piezoelectric coupling matrix etc). However from the literature I got double index notation, I am struggling to convert double index notation to triple index notation.</p>
<p>The double index notation from literature is as follows:</p>
<p>Double index notation</p>
<p>$$\begin{bmatrix}
d_{11} & d_{12} & d_{13} & d_{14} & d_{15} & d_{16} \\
d_{21} & d_{22} & d_{23} & d_{24} & d_{25} & d_{26} \\
d_{31} & d_{32} & d_{33} & d_{34} & d_{35} & d_{36}
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 21.3 & 0 \\
0 & 0 & 0 & 21.3 & 0 & 0 \\
-2.6 & -2.69 & 3.65 & 0 & 0 & 0
\end{bmatrix}$$</p>
<p>I would like to convert them to Triple index notation used by some commercial software providers like Abaqus and MSC Marc etc and is written as follows:</p>
<p>$$\begin{bmatrix}
d_{11-1} & d_{11-2} & d_{11-3} \\
d_{22-1} & d_{22-2} & d_{22-3} \\
d_{33-2} & d_{33-2} & d_{33-3} \\
d_{12-2} & d_{12-2} & d_{12-3} \\
d_{23-2} & d_{23-2} & d_{23-3} \\
d_{31-2} & d_{31-2} & d_{31-3}
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & -2.6 \\
0 & 0 & -2.6 \\
0 & 0 & 3.65 \\
0 & 21.3 & 0 \\
21.3 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$$</p>
<p>Am I doing the conversion correctly? Are the values placed in correct positions?
Double Index = Triple Index = Coupling Matrix Value</p>
<p>$$\begin{alignat}{2}
d_{11} &= d_{11-1} &&= 0 \\
d_{21} &= d_{11-2} &&= 0 \\
d_{31} &= d_{11-3} &&= -2.6 \\
d_{12} &= d_{12-1} &&= 0 \\
d_{22} &= d_{22-2} &&= 0 \\
d_{32} &= d_{22-3} &&= -2.6 \\
\vdots \\
d_{16} &= d_{31-1} &&= 0 \\
d_{26} &= d_{31-2} &&= 0 \\
d_{36} &= d_{31-3} &&= 0 \\
\end{alignat}$$</p>
| |electrical-engineering| | <p><strong>Caveat</strong>: I don't know the convention used in Marc. The following is based on what I can recall about Abaqus and Ansys. </p>
<h1>Constitutive relations</h1>
<p>The constitutive relations for <em>linear</em> piezoelectric materials are typically expressed in two forms:</p>
<ol>
<li>the strain and electric field are the independent variables (called the stress-charge or e-form), and</li>
<li>the stress and electric field are the independent variables (called the strain-charge or d-form).</li>
</ol>
<h2>Case 1 : stress-charge or e-form</h2>
<p>The coupled equations in this case are
$$
\begin{align}
\sigma_{ij} &= C_{ijkl} \varepsilon_{kl} - e_{mij} E_m \\
q_i &= e_{ijk} \varepsilon_{jk} + D_{ij} E_j
\end{align}
$$</p>
<p><strong>Abaqus uses this form</strong> because solid mechanics finite element models are more easily solved in strain-driven form.</p>
<h2>Case 2 : strain-charge or d-form</h2>
<p>The above equations can also be written in the form
$$
\begin{align}
\varepsilon_{ij} &= S_{ijkl} \sigma_{kl} + d_{mij} E_m \\
q_i &= d_{ijk} \sigma_{jk} + D_{ij} E_j
\end{align}
$$
<strong>Abaqus allows input of the coupling tensor $d$</strong> because these numbers are easily converted into the $e$ tensor using the relation
$$
e_{mij} = C_{ijkl} d_{mkl}
$$</p>
<h1>Converse piezoelectric effect</h1>
<p>The converse piezoelectric effect is modeled by the first of the d-form equations while the direct effect is modeled by the second. In particular, in the absence of stresses and external forces, the converse effect is written as
$$
\varepsilon_{ij} = d_{mij} E_m
$$</p>
<h1>Conventions</h1>
<p>The symmetry of the strain tensor leads to various ways of representing the components in matrix form. For example, the standard solid mechanics convention is
$$
[\boldsymbol{\varepsilon}] = \begin{bmatrix} \varepsilon_{11} & \varepsilon_{22} & \varepsilon_{33} & \varepsilon_{23} & \varepsilon_{31} & \varepsilon_{12} \end{bmatrix}^T \rightarrow \begin{bmatrix} \varepsilon_{1} & \varepsilon_{2} & \varepsilon_{3} & \varepsilon_{4} & \varepsilon_{5} & \varepsilon_{6} \end{bmatrix}^T
$$
The <strong>Abaqus convention</strong> is
$$
[\boldsymbol{\varepsilon}] = \begin{bmatrix} \varepsilon_{11} & \varepsilon_{22} & \varepsilon_{33} & \varepsilon_{12} & \varepsilon_{31} & \varepsilon_{23} \end{bmatrix}^T \rightarrow \begin{bmatrix} \varepsilon_{1} & \varepsilon_{2} & \varepsilon_{3} & \varepsilon_{4} & \varepsilon_{5} & \varepsilon_{6} \end{bmatrix}^T
$$</p>
<h1>Tensor d in matrix form</h1>
<p>To express the coupling tensor in matrix form, let us expand out each term:
$$
\begin{align}
\varepsilon_{11} & = E_1 d_{111} + E_2 d_{211} + E_3 d_{311} \\
\varepsilon_{22} & = E_1 d_{122} + E_2 d_{222} + E_3 d_{322} \\
\varepsilon_{33} & = E_1 d_{133} + E_2 d_{233} + E_3 d_{333} \\
\varepsilon_{12} & = E_1 d_{112} + E_2 d_{212} + E_3 d_{312} \\
\varepsilon_{13} & = E_1 d_{113} + E_2 d_{213} + E_3 d_{313} \\
\varepsilon_{23} & = E_1 d_{123} + E_2 d_{223} + E_3 d_{323}
\end{align}
$$
Using the <strong>Abaqus convention</strong>, we can write the above as
$$
\begin{align}
\varepsilon_{1} & = E_1 d_{11} + E_2 d_{21} + E_3 d_{31} \\
\varepsilon_{2} & = E_1 d_{12} + E_2 d_{22} + E_3 d_{32} \\
\varepsilon_{3} & = E_1 d_{13} + E_2 d_{23} + E_3 d_{33} \\
\varepsilon_{4} & = E_1 d_{14} + E_2 d_{24} + E_3 d_{34} \\
\varepsilon_{5} & = E_1 d_{15} + E_2 d_{25} + E_3 d_{35} \\
\varepsilon_{6} & = E_1 d_{16} + E_2 d_{26} + E_3 d_{36}
\end{align}
$$
In matrix form,
$$
\begin{bmatrix} \varepsilon_{1} \\ \varepsilon_{2} \\ \varepsilon_{3} \\ \varepsilon_{4} \\ \varepsilon_{5} \\ \varepsilon_{6} \end{bmatrix}
= \begin{bmatrix} d_{11} & d_{21} & d_{31} \\ d_{12} & d_{22} & d_{32} \\ d_{13} & d_{23} & d_{33} \\ d_{14} & d_{24} & d_{34} \\ d_{15} & d_{25} & d_{35} \\ d_{16} & d_{26} & d_{36} \end{bmatrix} \begin{bmatrix} E_1 \\ E_2 \\ E_3 \end{bmatrix}
$$
or
$$
[\boldsymbol{\varepsilon}] = [\mathbf{d}]^T [\mathbf{E}]
$$
where
$$
[\mathbf{d}] = \begin{bmatrix} d_{11} & d_{12} & d_{13} & d_{14} & d_{15} & d_{16} \\ d_{21} & d_{22} & d_{23} & d_{24} & d_{25} & d_{26} \\d_{31} & d_{32} & d_{33} & d_{34} & d_{35} & d_{36} \end{bmatrix}
$$</p>
<h1>Double to triple index map</h1>
<p>From the above we see that, <strong>in Abaqus</strong>, the matrix $[\mathbf{d}]$ is
$$
\begin{align}
[\mathbf{d}] &= \begin{bmatrix} d_{111} & d_{122} & d_{133} & d_{112} & d_{113} & d_{123} \\ d_{211} & d_{222} & d_{233} & d_{212} & d_{213} & d_{223} \\d_{311} & d_{322} & d_{333} & d_{312} & d_{313} & d_{323} \end{bmatrix} \\
& = \begin{bmatrix} d_{11} & d_{12} & d_{13} & d_{14} & d_{15} & d_{16} \\ d_{21} & d_{22} & d_{23} & d_{24} & d_{25} & d_{26} \\d_{31} & d_{32} & d_{33} & d_{34} & d_{35} & d_{36} \end{bmatrix}
\end{align}
$$
Ansys (and possibly Marc) uses a different convention and the numbers will have to be changed accordingly.</p>
<p>If the constants in your initial matrix are arranged according to the <strong>standard mechanics convention</strong>, the above map changes to
$$
\begin{align}
[\mathbf{d}] &= \begin{bmatrix} d_{111} & d_{122} & d_{133} & d_{123} & d_{113} & d_{112} \\ d_{211} & d_{222} & d_{233} & d_{223} & d_{213} & d_{212} \\d_{311} & d_{322} & d_{333} & d_{323} & d_{313} & d_{312} \end{bmatrix} \\
& = \begin{bmatrix} d_{11} & d_{12} & d_{13} & d_{14} & d_{15} & d_{16} \\ d_{21} & d_{22} & d_{23} & d_{24} & d_{25} & d_{26} \\d_{31} & d_{32} & d_{33} & d_{34} & d_{35} & d_{36} \end{bmatrix}
\end{align}
$$
You have to make sure that the convention used in the input 2-index matrix is known before you can write down the correct mapping.</p>
| 21746 | Conversion of Piezoelectric coefficients |
2018-05-15T12:19:29.187 | <p><a href="https://i.stack.imgur.com/J0MkO.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J0MkO.gif" alt="enter image description here"></a></p>
<p>this is deblistering machine by Stripfoil. I wounder to know the mechanism of blades movement ( equally ) with <code>one?</code> screw. when the operator rotates the handle, all blades move with the same distance.</p>
| |mechanical-engineering|mechanisms| | <p>One way to achieve this is by a series of 4 bar linkages </p>
<p><a href="https://i.stack.imgur.com/f4gT4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f4gT4.jpg" alt="enter image description here"></a></p>
<p>The above is used for marking out equally spaced points on a line of arbitrary length. </p>
<p>Pretty simple and elegant. By virtue of the geometry you can choose any two points as the start and end and the distance between them will always be equal fractions of the total. </p>
<p>Obviously in this particular context you would need some extra parts to implement it properly and maybe make the whole thing a bit stiffer, but within reasonable range of adjustment it should do exactly what you are looking for. </p>
| 21748 | What is the mechanism of the blades movement |
2018-05-15T16:45:09.290 | <p>I just purchased an aluminum heatsink that I am using to "wick" away heat from a Peltier; however, I was wondering how I can pull heat from the hot side of the peltier quicker.</p>
<p>I have purchased large, small, medium sized aluminum heatsinks and have not been able to explain why one works better than the other. This is the root of the problem: what determines how fast a heatsink can wick heat away from the hotside of the Peltier?
<a href="https://i.stack.imgur.com/ymYtP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ymYtP.jpg" alt="enter image description here"></a></p>
<p>I was thinking adding copper heat pipes, such as the ones on a laptop heatsink-fan combinations; however, I would need to thermally attach the copper heatpipes to the same heatsink to cool them. I am afraid this might kill the efficiency of the heatsink. </p>
<p>Thank you, and I am looking forward to hearing your suggestions</p>
| |heat-transfer| | <p>Conductivity, surface area, and air flow are the major factors. </p>
<p>Some have a plastic funnel so the fan can be larger than the sink.</p>
<p>I assume (hope) you get better with larger heat sinks and fans. </p>
<p>A case fan also helps. </p>
<p>That looks like a relatively small fan. I get full size cases for better cooling.</p>
| 21755 | How to Pull Heat Quicker from a Thermo-Electric Cooler using an Aluminum Heatsink |
2018-05-15T19:21:22.697 | <p>I have a question regarding the efficiency of lean burn gas engines. Maybe it is just a confusion concerning the terminology of "thermal efficiency", "brake mean pressure", "brake thermal efficiency" etc. but let's see.</p>
<p>Stationary gas engines in combined heat and power plants are usually run at a very high air-fuel-ratio in order maximize efficiency and reduce NOx-emission at the same time. With the introduction of more air, thermal efficiency is increased due to a higher heat capacity ratio of the gas mixture. At the same time this also results in reduced temperatures in the combustion chamber which reduceds the oxidation of nitrogen and therefore lowers nitrogenoxide emissions.</p>
<p>However, in literature a tradeoff between low NOx-emissions and a high engine efficiency is often mentioned.</p>
<p>Also, I assumed that higher temperatures in a combustion chamber were indicative of an efficient combustion, which is also not in accordance with the first paragraph.</p>
<p>Other papers that I have read also indicate a maximum of brake thermal efficiency at the same air-fuel-ratio as the maximum NOx-emissions, which is also not in accordance with the first paragraph.</p>
<p>Does anyone here have a good understanding of lean burn engine mechanisms and can help me understand this? I can also provide more details regarding my own misunderstandings.</p>
<p><strong>Follow up questions:</strong></p>
<p>Once again, thank you for your answer, Mark. Here are some further thoughts including the graphs that confused me initially. Most notably the graph "otto cycle efficiency vs. compression ratio". </p>
<p>Regarding the beforementioned graph: With the increase of lambda, the isentropic exponent gamma is also increased, resulting in a higher thermal efficiency. Does this increase in efficiency also imply an increase of the available power at the crank shaft? My assumption is that the increase in efficiency also increases the available work in the p-V-diagramm of an Otto cycle (area between the 4 points of the different steps of an otto cycle). One thought experiment would be increasing lambda to infinity, therefore maximizing thermal efficiency as the isentropic exponent approaches that of air. However, this does not make much sense now as there will not be any power at the crank shaft.
Or to put this whole paragraph briefly: What do I gain by an increase in thermal efficiency? Where does the statement "lean burn engines are efficient and reduce NOx-emission" originate from, when the minimum of the brake specific fuel consumption is at the same lambda as maximum NOx-emissions?</p>
<p>The argument of the efficiency of a lean burn engine is often used in literature, however the lowest brake specific fuel consumption occurs at a lambda only slightly higher than the stochiometric operation of the engine and quickly rises then. This is very confusing as the previos investigation of gamma would lead to the assumption that I can simply "increase lambda to get more efficiency". What is the motivation behind an increase in thermal efficiency when it cannot be used as mechanical power at the crank shaft? </p>
<p>Please excuse that my questions seem a little naive.</p>
<p>And one last very important question: When investigating lambda, is it correct that with an increase of lambda the mass of the actual fuel is reduced? Or is the mass of fuel usually held constant in such experiments and lambda is increased by adding more oxygen, effectively increasing the total mass of gas in the combustion chamber? (To make a long question short: Is the convention that total mass in the combustion chamber is held constant when varying lambda or is the fuel mass held constant and lambda is varied by adding more oxygen)</p>
<p><a href="https://i.stack.imgur.com/FZB3S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FZB3S.png" alt="Efficiency Increase due to increase in gamma"></a>
<a href="https://i.stack.imgur.com/8nKBv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8nKBv.png" alt="Spark Ignition Engine Fuel Consumption, Power and NOx"></a></p>
<p><a href="https://i.stack.imgur.com/jdkSn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jdkSn.png" alt="Brake Thermal Efficiency and NOx emission at different Air-Fuel-Ratios"></a></p>
| |automotive-engineering|energy-efficiency|biogas| | <p>This is a good, complex question, and the short answer is that it has entirely to do with terminology confusion. To answer it, I have to explain the physics behind the equations, because a lot of single experiments are going into each of these terms, but the net result of the entire system fosters other terminology.</p>
<p>To start with the <a href="https://www.thermalfluidscentral.org/encyclopedia/index.php/Heat_of_Combustion" rel="nofollow noreferrer">heat of combustion</a> is determined in a laboratory where they have a fixed quantity of fuel in a closed chamber that is insulated so heat cannot escape. In this case, we obtain that the reaction would release some amount of heat (in a unit called Joules, which is the amount of energy in 1 second of a 1 Watt power source) for every amount of fuel. Typically this is done on a mass basis (J/kg), but other versions, such as a volume or per-molecule basis (J/L or J/mol) are used. The heat released (in Joules), is a measure of energy. To determine how much energy was released, we actually need to do some backwards calculations. In an ideal world, where the experiment has the stoichiometric (chemically exact amount needed for the reaction - no more, no less) amount of pure oxygen and fuel mixed together, we would run the calculations as follows:</p>
<ol>
<li>Measure the temperature before the reaction</li>
<li>Measure the temperature after the reaction</li>
<li>Note the <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" rel="nofollow noreferrer">specific heat</a>s of the reaction products (Steam and Carbon Dioxide)</li>
<li>Account for the phase change of water if this was done at room temperature.</li>
</ol>
<p>Together, we see a temperature rise, and using properties taken in an earlier experiment (how the temperature of some substances changes when a specific amount of electrical heat was applied), we can find the amount of energy generated during the reaction. Note that <em>your assumption</em> in this experiment is completely accurate - a high temperature rise would indicate the reaction went to completion. If it was low temperature, we would unfortunately see a lot of side reactions which resulted in poor combustion (such as NOx formation). We would call the highest temperature rise possible a value of 1, and anything that got to 90% of this value would be considered 90% efficient. This is a form of thermal efficiency known as <a href="https://en.wikipedia.org/wiki/Fuel_efficiency" rel="nofollow noreferrer">fuel efficiency</a></p>
<p>Another way the temperature in this thought experiment would be lowered is if it was full of an inert gas, such as Helium, that did not participate in the reaction. The limited amount of fuel in the chamber would raise the temperature of the fuel and oxygen, but it would need to also heat up the helium. Such a temperature rise would use some of the available energy. These are both ways we can have low temperature - through poor combustion, and through dilution. One still has a high thermal efficiency, the other is simply a lower temperature.</p>
<p>A final thought experiment, if we had an ideal engine in-taking air from a hot oven, it would be a lot less efficient. The engine would need to be cooled a lot more, and it would have to run very slow. This is partly due to the components over-heating, and partly due to the fuel knocking, but mostly due to the fact that it would be a lot more work to compress that hot air then it would be to compress colder air. As a result, the engine is a lot less efficient. In the early 1820s, it was proven that the most energy you could get using heat was a <a href="https://en.wikipedia.org/wiki/Thermodynamic_cycle" rel="nofollow noreferrer">thermodynamic cycle</a> called the Carnot cycle was the most efficient, and the efficiency depended entirely on how hot you could get the gas, and on how cold the gas started. This Carnot Efficiency, between the two temperatures, could be called 1, and anything that is 90% of this power extraction would be a form of thermal efficiency we could call a <a href="https://en.wikipedia.org/wiki/Carnot%27s_theorem_(thermodynamics)" rel="nofollow noreferrer">brake thermal efficiency</a>, because by physics no engine can be more efficient than that value. It is called a brake efficiency or brake power because you measure the power that you are using at the brake to stop the engine, not the power (in terms of chemical energy flowing through the engine per second).</p>
<p>Note that brake thermal efficiency is different than thermal efficiency in this context - thermal efficiency of the fuel is how much of the fuel was converted from fuel to it's final end products, but the brake thermal efficiency simply refers to how warm the gas becomes. If the gas does not get warm in the combustion, it can not efficiently extract power. As a result, going to the though experiment with the helium, if we flood the engine with air, most of the air will remain unreacted, and act as helium. However, the excess air will ensure all of the fuel would combust. We would have excellent fuel efficiency, but terrible Carnot efficiency, since the gas wouldn't get warm enough for good Carnot efficiency. This low temperature prevents the formation of NOx, but overall the system will not produce as much power as it could have produced if the excess air was not introduced into the system.</p>
| 21759 | Efficiency and NOx emission of lean burn gas engines |
2018-05-16T08:21:18.667 | <p>If you look at the screenshot you will see a composite auxetic material. How would you calculate the E-modulus for it? Is it as simple as the rule of mixtures, <code>E = V1 * E1 + V2 * E2</code>? Or is it more complex because of its structure?</p>
<p><a href="https://i.stack.imgur.com/bTy05.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bTy05.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/BMjgy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BMjgy.png" alt="enter image description here"></a></p>
| |materials|elastic-modulus|composite| | <p>Related: <a href="https://engineering.stackexchange.com/questions/4182/how-do-i-calculate-an-estimate-for-the-properties-of-a-composite-material">How do I calculate an estimate for the properties of a composite material?</a></p>
<p>In this case, we actually have an unusual case of the fourth method. We could look at each individual unit cell of the material and apply a unit force on the structure in the x, y, and z directions. Each application of unit force would correspond to a certain deflection. The net deflection divided by the bounding box dimensions of the unit cell would result in a strain, and the unit force divided by a stress would result in a stress. </p>
<p>However, this simple approach does not appreciate the 3D complexity of the problem. Since we are calculating the response to a force, we are finding the individual relations of the <a href="https://en.wikiversity.org/wiki/Introduction_to_Elasticity/Constitutive_relations" rel="nofollow noreferrer">stiffness tensor</a>. Note that for general anisotropic materials such as this one, we need to find 21 different relationships. However, afterwards, the stiffness tensor is not usable, we need to invert the matrix to find the compliance tensor. The elastic modulus would then be the inverse of the first term of the stiffness tensor, but the usable format would utilize the compliance tensor.</p>
| 21768 | Young's modulus for composite auxetic material |
2018-05-17T08:38:40.787 | <p>In CAD, there are multiple file formats only for the scope of saving polygonized models in three dimensions. </p>
<p>For example, there is the <a href="http://segeval.cs.princeton.edu/public/off_format.html" rel="nofollow noreferrer" title="Object FIle Format">Object File Format</a> and furthermore, there is the much more famous <a href="http://www.fabbers.com/tech/STL_Format" rel="nofollow noreferrer" title="STL">STL</a> format. </p>
<p>What are the different (most important) use cases and reasons to use especially one of them?</p>
| |cad| | <p>The formats have different functions. While it is true that some of the formats duplicate functionality it would be a mistake to think all the formats are interchageable and serve same purpose.</p>
<p>For example stl is the file format intended for 3D printing. All it does is store a bunch of sepaarate triangles. It is meant to be extremely easy to dump. But it does not contain any other reliable info. The fileformat forces you to rebuild all semantic info that may have been in the mesh. The format is nearly unusable for anything other it was intended for.</p>
<p>Usually you then have one or more formats meant for rendering like the obj format which resembles OFF you linked but has info on rudimentary material properties, uv maps normals, connectivity and so on. This makes the format much more usable. But alas unlike STL there is no binary version so these can become really big for highly tesselated meshes.</p>
<p>Then you usually have the intermediate format needed to make 3D PDF files, you get it for free if you intend to support 3D PDF, its nearly useless outside the context.</p>
| 21781 | Why are there multiple ways to save polygonized 3D model data? |
2018-05-17T09:18:33.580 | <p>I have non-rigid beam attached to a gearbox and motor assembly. The gearbox outputs $1000 N \cdot m$ at the shaft. The beam is rotated at $3000 RPM$ around it's center. If the motor is switched off immediately and acts as a brake, is it possible that the torque generated from the momentum of the beam could be greater than the $1000N \cdot m$ that the motor originally produced, and if so how?</p>
<p>I am trying to work out the max torque a beam could create at the gearbox mount when the power is cut suddenly, however we don't know any of the beam characteristics such as where the point loads are, how long it takes to stop etc. All I have is the gearbox and motor specifications.</p>
<p>Edit: This question is different from a similar one I asked <a href="https://engineering.stackexchange.com/questions/21745/how-to-calculate-torque-needed-to-stop-a-rotating-beam-over-x-seconds">here</a>. In the other question I mentioned that the beam was uniformly distributed, however in this example I have no beam information. All I have to go off is the motor and gearbox specifications. </p>
| |dynamics| | <blockquote>
<p>is it possible that the torque generated from the momentum of the beam</p>
</blockquote>
<p>I'm going to stop you right there. Momentum doesn't generate torque. Momentum is moment of inertia times speed. Torque is moment of inertia times acceleration. Just like speed and acceleration aren't the same thing, momentum and force aren't the same thing. </p>
<p>If you apply brakes, the torque that results is whatever the braking torque is. The load doesn't dictate what the braking torque is. The load will determine how quickly it decelerates for a given braking torque, but the brake is what sets the torque. </p>
<p>Again, momentum is $p = mv$, or $L = I\omega$. Torque affects acceleration. It changes the speed, so it changes momentum $\tau = dL/dt$, but that's just because again $L = I\omega$, so really what that's saying is that $\tau = (I)d\omega/dt$, or the more common expression for torque, $\tau = I\alpha$. </p>
<p>So again, to reiterate, the load doesn't set the torque. The brakes set the torque. The load reacts to torque. </p>
<blockquote>
<p>I am trying to work out the max torque a beam could create at the gearbox mount when the power is cut suddenly, however we don't know any of the beam characteristics such as where the point loads are, how long it takes to stop etc.</p>
</blockquote>
<p>If you want to calculate the load, but you don't know anything about the load, you're pretty well out of luck.</p>
| 21783 | How to calculate the max torque due to angular momentum of a beam given motor specs |
2018-05-17T22:55:13.543 | <p>For a plant G(s), the transfer function is given by:
$$G(s) = \frac{(10s+50)}{s(s^2+20s+2500)}$$</p>
<p>Using Ziegler-Nicholas method to get the parameter: $k_p = \dfrac{-1000}{3}$ and $w =-28.868j$.</p>
<p>Is it correct? what should I do in case of negative $k_p$ and how do I calculate the PID parameters?</p>
| |electrical-engineering|control-engineering|pid-control| | <p>The Ziegler-Nichols cycling method is not appropriate for this system. Because of the zero located in -5, the closed loop system is always stable and will not oscillate for proportional compensator (poles will never be conjugated on top of the $j\omega$ axis), as can be evidenced by the root locus plot:</p>
<p><a href="https://i.stack.imgur.com/LME2P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LME2P.png" alt="enter image description here"></a></p>
<p>If you still wish to use the Ziegler-Nichols table for this system you should use a different approach, like the step response method.</p>
| 21795 | Negative gain in Ziegler-Nicholas Oscillation Method? |
2018-05-18T20:59:37.927 | <p>How would one work out the force required to compress flat, using a flat bar on top pressing down, a thick walled (15mm OD 10mm ID) piece of flexible plastic (PVC) tubing i.e. so reducing the x-sectional area inside the tube to 0?</p>
<p>Just looking for any way to approximate the real situation without having to use an FEA simulation.<a href="https://i.stack.imgur.com/1Q8H2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Q8H2.png" alt="enter image description here"></a></p>
<p>Many thanks for your help!</p>
| |compressors|elastic-modulus| | <p>At the crushed configuration we can assume we are dealing with a flat beam.
Width= L of pipe, span = 15pi.</p>
<p>Height = 5mm.</p>
<p>for simplicity we assume three plastic hinges ( we will refine this by assuming more hinges, such as one at either edge of pressure block if need be), one at either crushed edge of the pipe, one at the center of horizontal flatly deflected top of the pipe right under the middle of back block.</p>
<p>Why do we assume this ultimate failure configuration? because we think this is the lowest energy needed, of course we later can try other collapse models.</p>
<p>The two plastic hinges on the side need to be enough to counter the resistance of these hinges M_p.</p>
<p>$$M_p = L5^2/4 \times \sigma_{plastic yield} $$</p>
<p>And the force needed is $$M_p/7.5\times \pi$$</p>
<p>We need to add to above the force needed to crush the pipe at the center which is equal to previous force/2: effective length is double.</p>
<p>This is the first collapse mechanism. now we can assume other modes and compare the results, in a few iterations we get close to required precision.</p>
| 21807 | Force to compress flexible elastic tubing |
2018-05-18T21:41:16.000 | <p>How might one calculate or approximate the negative pressure applied to the outer surface of a cylinder required to expand it a certain distance, when held open rigidly at each end?</p>
<p>The cylinder ID = 20*the wall thickness and the material is very elastic.</p>
<p><a href="https://i.stack.imgur.com/LTr3w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LTr3w.png" alt=""></a></p>
<p>Many thanks for your help!!</p>
<p>Oli</p>
| |pressure|elastic-modulus| | <p>It would be indetrminant structure if we release the axial constraints. In a very crude first estimate one could compare it to a syspended bridge with flexible end supports, or a balloon with built-in geometric constraints. I guess if you don't want to use FEM then, you would want to guess a curve for the expanded shape and solve by energy methods. It would be interesting to start from trig functions for the cross section deformation an see where it goes.</p>
| 21809 | Calculate or approximate force required to expand a thin-walled elastic cylinder held rigidly at both ends? |
2018-05-19T16:44:08.327 | <p>A sealed ball is not connected to anything externally, but can have any complex mechanism inside (mechanical, chemical, or any other). Can a wireless trigger from a remote or any other wireless mechanism make the ball's mass heavier? Take into account that the ball is sealed and therefore only internal changes can occur, no external objects should increase the mass.</p>
<p>PS: Someone suggested that the ball should create a Higgs field inside it which is impractical but theoretically correct.</p>
| |applied-mechanics| | <p>Within the realms of practical engineering (the limits of this SE), <strong>No.</strong></p>
<p>The <a href="https://en.wikipedia.org/wiki/Conservation_of_mass" rel="nofollow noreferrer">Law of Conservation of Mass</a> applies, and as such, if the ball is truly sealed, it cannot become heavier.</p>
<p>Even trying to cheat by using physics, <strong>No.</strong></p>
<p><a href="https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence" rel="nofollow noreferrer">Mass-energy equivalence</a> ($E=mc^2$) shows that an object that a bodies relativistic mass is greater than its rest-mass, that is to say, if a body is moving, it will be heavier than if the same body were stationary. Adding energy to an object adds mass, but, without approaching the speed of light (difficult, to say the least), you would struggle to measure the change.</p>
<p>But... The ball would have to be accelerated by an external energy source for this to work. Any mechanism inside the ball used to make it roll or spin (e.g. a motor) would require its own energy source, and, by the same equation shown above, a <a href="https://physics.stackexchange.com/questions/34421/does-the-mass-of-a-battery-change-when-charged-discharged">full battery weighs more than an empty battery</a>.</p>
<p>Can I make the ball lighter, then? <strong>Yes, but you won't be able to measure it</strong></p>
<p>If you allow the ball to radiate heat to it's surroundings, therefore, loosing energy, you could make the ball <em>lighter</em> (e.g. by running a heater off some batteries inside) without loosing any matter/atoms, but there is no way you could measure a change this small (see the excellent answer on the weight of batteries that I linked above for discussion on this).</p>
<p>One other thought:
If an electromagnet were sealed inside, and the ball were near some iron filings, then the ball could <em>pick up</em> mass from the outside world, without breaking its seal, but, that seems like cheating your rules, rather than a solution.</p>
| 21815 | A sealed ball is not connected to anything externally, but can have any complex mechanism inside. Can the mass change? |
2018-05-19T19:25:22.610 | <p>The shaft is the most important element in my machine, so i rather prefer forging for production technology, the circumscribed circle of the shaft is about 140 mm and the length is about 1400 mm. </p>
<p>What kind of dimensions should i specify on my drawing to help the operator to manufacture it ? The diameter of a circumscribed circle here does't help. </p>
| |product-engineering|forging| | <p><a href="https://i.stack.imgur.com/Zt3hZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zt3hZ.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Hex AF (across-flats) dimensioning.</em></p>
<p>That's how I'd do it (but I'm an electrical engineer so what would I know!).</p>
| 21818 | Production technology for hexagonal shaft ? |
2018-05-20T00:56:40.900 | <p>I am working on this question on my statics assignment:<br/>
Approximately at point B there are two forces, 20kN and 50kN.So far, I have only calculated global equilibrium and got the following values:
<br/>
$D_y=(\frac{80}{7})kN$<br/>
$D_x=(\frac{20}{\frac{\sqrt{13}}{2}})kN$
<br/>
$A_y\approx89.094 kN$
<br/>
My question here is, when I'm drawing the shear force diagram do I sum up the 20kN force and the 50kN force into one resultant and graph it? Or, do I draw them separately. I need help.
<a href="https://i.stack.imgur.com/GwI3d.png" rel="noreferrer"><img src="https://i.stack.imgur.com/GwI3d.png" alt="enter image description here"></a></p>
| |statics| | <p>To answer your question, you just need to break the inclined force into its component X and Y forces:</p>
<p>$$\begin{alignat}{4}
f_x &= \dfrac{20}{\sqrt{1^2+1.5^2}}&&=11.09\text{ kN} \\
f_y &= -1.5f_x&&=-16.64\text{ kN} \\
\end{alignat}$$</p>
<p>and then add the vertical component to the 50 kN force.</p>
<p>That being said, your reactions (other than $D_x$) are incorrect:</p>
<p>$$\begin{align}
\sum M_A &= -66.64\times2 - 6\times4\times\left(2+\dfrac{4}{2}\right) + 200 + 9D_y - 2\times2\times\left(9 + \dfrac{2}{2}\right) - 200 = 0 \\
\therefore D_y &= 29.92\text{ kN}\\
\sum F_y &= A_y + 29.92 - 66.64 - 6\times4 - 2\times2 = 0 \\
\therefore A_y &= 64.72\text{ kN}
\end{align}$$</p>
<p>You can then draw the diagram as normal, remembering to add the inclined force's vertical component to the vertical force at B:</p>
<p><a href="https://i.stack.imgur.com/9JCs7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9JCs7.png" alt="enter image description here"></a></p>
<hr>
<p><sup>Diagram obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis program.</sup></p>
| 21822 | How to draw a shear force diagram when two forces are acting at the same point |
2018-05-21T16:27:36.507 | <p>On a datasheet, sensitivity temperature drift (TCs) is specified in %/K.</p>
<p>I read this as "the percent change per kelvin".</p>
<p>Is this is simple as this sounds, or is there a reason for using kelvin over celsius?</p>
<p>I.e if TCs=0.01 and the temperature changes from 20C to 30C, or from 293K to 303K, then the total percentage change is 0.1%?</p>
| |electrical-engineering|temperature|nomenclature| | <p>Yes, this is the percentage change per kelvin.</p>
<p>Most engineers work in kelvin as some formulae need the absolute temperature such as when calculating the total internal energy, but we sometimes don’t bother such as when dealing with temperature difference, but it was good practice when we were taught it so we continue to use K...</p>
| 21842 | What does "+- %/K" mean? |
2018-05-22T08:51:37.170 | <p>I'm working on the statics problem below:
<br/><a href="https://i.stack.imgur.com/FYh6q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FYh6q.png" alt="enter image description here"></a>
I'm tasked to draw the shear and bending moment diagram for the frame. I'm trying to work out the reactions and I'm taking the moment about pin E.Here's my working so far:<br/>
$$\Sigma M_E=0$$
$$(10)(5)-100-(2*10)(\frac{1}{2}*10+5)+(10)(15)-100-(\frac{30}{\sqrt{2}})+(0.5*5*4)(\frac{1}{3}*5)+A_y(20)=0$$
My question here is, for the triangular UDL, is the distance one-third or two thirds? My lecturer did a similar problem and he said that the distance is two thirds but I don't quite get why it is two thirds. Someone help.</p>
| |statics| | <p>As for the title of your question, any statics textbook should show you the centroid of a right triangle. <a href="https://en.m.wikipedia.org/wiki/List_of_centroids" rel="nofollow noreferrer">This wikipedia article </a>gives the centroids of common shapes.</p>
<p>When replacing a distributed load with an equivalent point load, that point load will act through the centroid of the distributed load. In this case, the equivalent point load for the triangular distributed load will act (1/3)*5 up from Point A.</p>
<p>When summing the moments about a point it is important to indicate your sign convention (clockwise positive or counterclockwise positive). Either convention is acceptable so long as you are consistent throughout. Looking at the work you showed, there appear to be some inconsistencies. For example P1 and M1 (see sketch below for notation) should both produce clockwise rotation about Point E, however your work shows them producing moments with opposite signs.</p>
<p>I won’t write out the whole equation for summing moments about Point E because I don’t want to do the homework for you. You’re on the right track though!</p>
<p><a href="https://i.stack.imgur.com/1osH6l.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1osH6l.jpg" alt="Free Body Diagram"></a></p>
| 21849 | What is the distance at which the triangular UDL is acting on in this problem? |
2018-05-23T18:22:32.907 | <p>I have the following system (at P and Q there is a ball joint; the length of L beam is l; the PB and PA forces are applied at l/2 and are known; Fz is not zero and is known):</p>
<p><a href="https://i.stack.imgur.com/pykcv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pykcv.jpg" alt="enter image description here"></a></p>
<p>The system has 0 degrees of freedom because each ball joint removes three degrees of freedom (6-2*3=0).
I replaced each ball joint with three unknown forces.</p>
<p><a href="https://i.stack.imgur.com/n72HL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n72HL.jpg" alt="enter image description here"></a></p>
<p>The problem is that the second cardinal equation along x gives:</p>
<p>$$-F_z \, l = 0$$</p>
<p>How can I solve the problem?</p>
<p>Thank you very much for your time.</p>
| |structural-engineering|civil-engineering|structural-analysis| | <p>The system is still statically indeterminate, but not due to lack of DOF. You have a singularity. Having correct DOFs only assures you have a chance of being statically determinate, but it does not guarantee this condition. In this case, the singularity is because the beam is allowed to rotate about the x axis unless $F_z$ is an unknown reaction.</p>
<p>As such, in this case, $F_zl = 0$ means $F_z = 0$. Here are your other equations, in no particular order:</p>
<p>$$\begin{alignat}{4}
\sum& F_x &&= A + PB_x + D + PA_x &&= 0 \\
\sum& F_y &&= B + PB_y + E + PA_y &&= 0 \\
\sum& F_z &&= C + G + (F_z = 0) &&= 0 \\
\sum& M_{y@Q} &&= Cl &&= 0 \\
\sum& M_{x@Q} &&= F_zl &&= 0 \\
\sum& M_{z@Q} &&= -Bl + PB_yl/2 - PA_xl/2 &&= 0
\end{alignat}$$</p>
<p>We can immediately see $C = F_z = 0$. Because of $\sum F_z$, $G = 0$. We now have four unknowns and three equations. We can solve for $B$ immediately using:</p>
<p>$$\sum M_{z@Q} = -Bl + PB_yl/2 - PA_xl/2 = 0$$</p>
<p>Rearranging, we can use this result for $B$ to solve for $E$ using:</p>
<p>$$\sum F_y = B + PB_y + E + PA_y = 0$$</p>
<p>But the final equation is singular:</p>
<p>$$\sum F_x = A + PB_x + D + PA_x = 0$$</p>
<p>This cannot be resolved using statics, but require static indeterminate methods.</p>
| 21869 | L beam - Strange result |
2018-05-24T10:51:32.800 | <p>Hardening leads to a material being more brittle.However I can't understand what is strain hardening or strain hardening exponent.Please explain why hardening leads to brittleness also in this answer itself.I would be happy to get a detailed answer of this.</p>
| |mechanical-engineering|materials|metallurgy|metals| | <p>Some metals like steel go through a large range of hardening, while still ductile and under going plastic yield, before getting to the rupture point.</p>
<p>If you consider the entire area under the strain/stress curve up to fracture point it's called toughness module and in steell it is much larger the resilience (elastic) module.
<a href="https://i.stack.imgur.com/Sw7Jn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sw7Jn.jpg" alt="mild steel stress/ strain graph"></a></p>
<p><a href="https://www.cmrp.com/blog/faq/analysis-design/exploring-stress-strain-curve-mild-steel.html" rel="nofollow noreferrer">Source of graph here.</a></p>
| 21880 | What is strain hardening exponent? |
2018-05-24T11:21:02.247 | <p>Please bear with me - I'm a lapsed mathematician and I'm self-studying these concepts.</p>
<p>A question asks the following:</p>
<blockquote>
<p>Water flows in a pipe of diameter 5 m at a velocity of 10 m/s. It then
flows down into a smaller pipe of diameter 2 m. The height between the
centre of pipe sections is 5 m. The density is assumed to be uniform
over the cross sections. The gauge pressure at Boundary 1 is 120 kPa.
Calculate the velocity at the smaller pipe section.</p>
</blockquote>
<p>There is no reason I can see to assume mass flow continuity doesn't apply, and using $v_1A_1 = v_2A_2$ one obtains $v_2 = 62.5$ m/s. However, using the Bernoulli equation while assuming atmospheric pressure at the smaller section, one gets
$$
\frac{p_1}{\rho} + \frac{1}{2}v_1^2 + + gz_1 = \frac{p_2}{\rho}+\frac{1}{2}v_2^2 + gz_2
$$
$$
120+ 50 + 5g = 0 + \frac{1}{2}v_2^2 + 0
$$
$$
v_2 = 20.93,
$$
and in fact this is what the textbook answer gives. I am quite confused as to why mass continuity applies in other situations, even with changes in pressure, but doesn't seem to apply here.</p>
<hr>
<p>My question is: <strong>is this textbook question then ill-posed?</strong> I feel as though by providing too much information about the pipe section without checking the calculations, the question is bound to create a contradiction. The 5m/2m diameters don't actually make it into the final answer.</p>
| |mechanical-engineering|fluid-mechanics| | <p>I think @sam is correct. To provide some intuition here, imagine that we have a pipe that does what is described in the problem but that had constant diameter throughout. In this case, the pressure at point 2 is 169 kPa gauge. If we increase the diameter of the pipe at point 2, the pressure increases. You can think of this as some of the velocity head of the fluid at point 1 becoming pressure head at point 2. If the diameter decreases, the pressure must go down. That is, some of the head of the fluid at point 1 gets converted to velocity head at point 2. As the problem is posed, there is not enough head at point 1 to achieve the velocity at point 2. </p>
<p>To fix this we would either have to increase the pressure at point 1 or increase the height difference. </p>
| 21882 | Can the continuity equation and Bernoulli contradict each other? |
2018-05-24T16:51:03.310 | <p>I recently had a machine shop machine some aluminium panels (3mm thickness) which I then sent to an anodizing shop where the panels were bead blasted and anodized. When they came back they were seriously bent (dimensions of the panel are about 310x130mm). </p>
<p>I did specify 6061-T6 for the alloy but I suspect the machine shop might have used a cheaper non-tempered aluminium. </p>
<p>Is my thinking correct or does this problem come from the anodisation process? </p>
| |mechanical-engineering|machining|aluminum|deformation| | <p>Yeah, this sounds like mishandling. If it wasn't aged, the 6061 is T4 out of the extrusion press, still plenty hard enough with a yield strength about 1/2 of T6. Maybe they sold you O-temper; it is the softest temper, with a yield strength max limit that is at the min limit of T4. Anodizing is really just an accelerated build-up of aluminum oxide, that can be colored; there isn't much heat. There's nothing in that process, except handling, that could warp or bow your parts, especially at a thickness over 0.1". I'm a quality mgr for an extruder, believe me, some folks treat aluminum like it's steel. Our public enemy #1 is surface damage, including dents, scratches, etc.</p>
| 21891 | deformation / bending problem with anodized aluminium panels |
2018-05-25T01:16:50.927 | <p>If I do a port scan from the internet, do I hit a bundle of wires that are physical ports into the motherboard (like a PCE port)?</p>
<p>Or do I actually go into the operating system and interact with the virtual or 'logical' ports of the OS? </p>
<p>I am trying to understand what the difference is between these ports, and why, if they are logical, I would be able to scan <em>any</em> of them from the internet, because it implies my bytes are passing over hardware to reach these 'logical' ports inside the CPU and cache, rather than on the peripheral hardware (before they are rejected)… This feels wrong...</p>
<p>But if they are actually physical, then where are they? And, if I have a daemon listen on a port that is "opened", does this mean that the program is listening to bytes that pass over a physical component on the motherboard, exterior to the cache-ram-cpu area?</p>
| |computer-engineering|computer-hardware| | <p>The ports referred to in a "port scan" are logical.</p>
<p>Consider the fact that all network communication to a computer typically is over a single physical connection, usually Ethernet. There are multiple simultaneous conversations happening routinely. Think about how the network traffic for fetching email in one window, a web browser in another window, various apps accessing servers elsewhere, and all those viruses sending your passwords and bank account details over to my servers all work independently without getting each other's traffic.</p>
<p>At the TCP and UDP level, this is done with virtual ports. Servers are waiting on "well known ports" to accept connections. For example, port 25 is usually reserved for requesting new connections of a SMTP server. Once a connection is established, it is assigned a free port. Subsequent traffic then references that port to identify the connection.</p>
<p>A port scan essentially (this is over-simplified, but that's all that's needed for the basic concept) sends a connection request to every port (they are in a 16 bit namespace) to see which ones the server machine reacts to. Some OS services of particular operating systems and versions are known to have servers on particular ports. Some of there are know to have vulnerabilities. One purpose of a port scan is to find out what ports your machine reacts to, and if there are any you might know how to exploit in doing unintended things with the machine.</p>
| 21898 | Computer "Ports", port scans and daemons |
2018-05-25T03:22:26.117 | <p>I have a crane boom that is rotating and when it is stopped suddenly, the boom flexes back and fourth several times. I have measured the angles between each deflection and have $3.16 ^\circ$, $2.42 ^\circ$, $1.77 ^\circ$ and $1.31 ^\circ$, which if plotted in Excel forms a logarithmic curve equal to $$y = -1.33ln(x) + 3.2221$$ The goal now is to find how to plot this as a sin curve with the period of 3.5 seconds so that it it looks like a deflection vs time chart (or damping curve), similar to the one in red below. Can someone please explain how I can do this?</p>
<p><a href="https://i.stack.imgur.com/JJcG2.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JJcG2.gif" alt="enter image description here"></a></p>
| |mathematics|vibration| | <h1>The model</h1>
<p>The equation for a damped periodic oscillation is
$$
y(t) = A e^{-\lambda t} \cos(\omega t + \phi)
$$
where $A$ is the amplitude, $\lambda$ is the decay constant, $\omega = 2\pi/\tau$ where $\tau$ is the period, and $\phi$ is the phase angle at $t=0$. In your figure, $\phi = -\pi/2$, and therefore your damping curve has the form
$$
y(t) = A e^{-\lambda t} \sin(\omega t)
$$</p>
<h1>Your mistake</h1>
<p>Instead of using
$$
y(t) = A e^{-\lambda t} \quad \quad \text{(Correct)}
$$
you have fit the portion of the curve where $\sin(\omega t) = 1$ using the relation
$$
y(t) = B \ln(t) + C \quad \quad \text{(Wrong)}\,.
$$
This model predicts an infinite displacement $y$ at time $t=0$.</p>
<h1>Data needed</h1>
<p>You have provided angles but not displacements. We have to guess what those are from your fit.</p>
<p>If we fit the correct model to data extracted from your log model, we get the following curves (note that we can calculate the full deflection-time curve after fitting $A$ and $\lambda$ because $\omega$ can be calculated from the period):
<a href="https://i.stack.imgur.com/VUDF3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VUDF3.png" alt="enter image description here"></a></p>
| 21899 | How to work out the damping or deflection curve from deflection angles? |
2018-05-25T06:53:29.187 | <p>How do we calculate the deflection of a cantilever with two or three cross-sections, and different forces acting at the end of each section?
<a href="https://i.stack.imgur.com/x27LV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x27LV.png" alt="image of three-sectioned cantilever"></a></p>
| |structural-engineering|beam|aircraft-design| | <p>Castigliano's method can simplify your calculations to an extent - instead of computing the deflection at every point on the beam, you can calculate the deflection at the points you care about only, and ignore the rest.</p>
<p>First we need to write a moment equation in terms of x. This is fairly straightforward. Defining:</p>
<p><span class="math-container">$$ <x> = \left\{\begin{matrix}
0&\text{if}&x < 0 \\
x&\text{if}&x>= 0
\end{matrix}\right.$$</span></p>
<p>then we have the moment equation for the entire beam as simply:</p>
<p><span class="math-container">$$M(x) = A<l_1-x> + B<l_2+l_1-x> +C<l_3+l_2+l_1-x>$$</span></p>
<p>We also can identify the moment of inertia of each section as <span class="math-container">$I_1,I_2,I_3$</span> respectively. Using the <a href="http://mathworld.wolfram.com/HeavisideStepFunction.html" rel="nofollow noreferrer">Heaviside Step function</a> (which acts as the basis for the <span class="math-container">$<x>$</span> convention - i.e. this convention is simply shorthand for <span class="math-container">$(x)*H(x)$</span>) the Section modulus of each section becomes simply:</p>
<p><span class="math-container">$$EI(x) = EI_1*H(l_1-x)+EI_2*H(x-l_1)*H(l_1+l_2-x)+EI_3*H(x-l_1+l_2)$$</span></p>
<p>Now we differentiate the moment equation with respect to the forces (A,B,C); for example:</p>
<p><span class="math-container">$$\frac{\partial M}{\partial A} = <l_1-x>$$</span>
<span class="math-container">$$\frac{\partial M}{\partial B} = <l_1+l_2-x>$$</span>
<span class="math-container">$$\frac{\partial M}{\partial C} = <l_1+l_2+l_3-x>$$</span></p>
<p>Finally, we have everything we need. To find the deflection at A, we simply take Castigliano's theorem regarding the energy:</p>
<p><span class="math-container">$$U = \int^L_0 \frac{M(x)^2}{2EI(x)}dx$$</span></p>
<p>And therefore the deflection is equal to <span class="math-container">$\frac{\partial U}{\partial P}$</span>, where P is the load at the point you want to test. If there isn't a load, you can just add it into the moment equation, and run through the integral. In this case, you asked for the deflection at the three points, so:</p>
<p><span class="math-container">$$\delta_j = \frac{\partial U}{\partial P_j} = \int^L_0 \frac{M(x)}{EI(x)}\frac{\partial M(x)}{\partial P_j}dx $$</span></p>
<p>(Where <span class="math-container">$P_j = A, B, C$</span> respectively, and <span class="math-container">$\delta_j = x_1, x_2, x_3$</span>, the points under the loads A, B, and C respectively.)</p>
<p>Substitute the different partial derivatives of the moment equation, plug into the integral. If the Heaviside step integral becomes difficult, break the integral into the different lengths. For example, running with the first deflection, since <span class="math-container">$\frac{\partial M}{\partial A} = <l_1-x>$</span>, we only need to compute the integral from <span class="math-container">$0$</span> to <span class="math-container">$l_1$</span>. This simplifies the result immensely:</p>
<p><span class="math-container">$$\delta_1 = \int^{l_1}_0 \frac{A*(l_1-x) + B*(l_2+l_1-x) +C*(l_3+l_2+l_1-x)}{EI_1}(l_1-x) = \frac{l_1^2*(2Al_1+2Bl_1+3Bl_2+2Cl_1+3Cl_2+3Cl_3)}{6EI_1}$$</span></p>
<p>Similar applications utilizing the properties of the <span class="math-container">$<x>$</span> convention can reduce the problem to computing a few additional integrals, which are easily resolved with most computer algebra systems. For a quick double check, the second deflection is:</p>
<p><span class="math-container">$$\delta_2 = \frac{A(2l_1^3+3l_2^2)}{6EI_1}+\frac{B(2l_1^3+6l_1^2l_2+6l_1l_2^2)}{6EI_1}-\frac{B(2l_1^3-2l_2^3)}{6EI_2}+\frac{C(2l_1^3+6l_1^2l_2+3l_3l_1^2+6l_1l_2^2+6l_1l_2l_3)}{6EI_1}-\frac{C(l_1^3+3l_1^2l_3-2l_2^3-3l_2^2l_3)}{6EI_2}$$</span></p>
| 21902 | Multiple Varying cross section beam deflection |
2018-05-26T02:20:58.393 | <p>If a bridge has a parapet that is nota fully solid block and resembles railings with posts, as the following image</p>
<p><img src="https://i.stack.imgur.com/BdN3C.png" alt="bridge parapet"></p>
<ul>
<li>What is this type of parapet called?</li>
<li>What are the top, bottom, post, and hole of the railing-like structure called?</li>
</ul>
<p>Thanks!</p>
| |bridges| | <p>I do a lot of bridge inspections in Ontario, Canada following OSIM (Ontario Structural Inspection Manual). Based on your photo, I would hazard to guess this is an older style barrier system. I would break what you are looking at the side of the bridge into 3 separate components.</p>
<p>At the bottom you have a barrier or safety curb. That approximately 150-200 mm square protrusion at the base of the wall.</p>
<p>The wall itself I would call a parapet wall. </p>
<p>The part at the top I would call a post and railing or hand rail.</p>
<p>I would avoid calling it a post an open railing system as the wall is really solid and looks to be a solid connection with the deck. Meaning in a collision the load would be transferred down and directly to the deck. In a post and railing system the load gets transferred to the posts and the posts transfer it to the deck.</p>
| 21914 | Bridge terminology - railing-like parapet |
2018-05-26T06:37:21.110 | <p>What is the difference between a blade and a vane in a turbine? When constructing velocity triangles, there are two angles at inlet and outlet (α & θ, β & φ), respectively. My textbook defines α & β as blade angles and θ & φ as vane angles. What is the difference between them?<a href="https://i.stack.imgur.com/SU8Cf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SU8Cf.jpg" alt="Velocity triangle of a curved vane."></a></p>
| |mechanical-engineering|fluid-mechanics| | <p>I don't understand why the textbook uses the terms "blade angle" and "vane angle" here, but the point is that you need to consider two sets of angles. </p>
<p>One set ($\alpha$ and $\beta$) describe the fluid flow direction relative to the fixed turbine structure. The other set ($\theta$ and $\phi$) describe the flow direction relative to the rotating blades. The vectors $u1$ and $u2$ in the velocity triangles come from the velocity $u$ of the blade as it rotates.</p>
| 21915 | Difference between 'vanes' and 'blades' |
2018-05-27T07:32:59.990 | <p>I'm a engineering student currently taking Thermodynamics and we have this project for our finals which is a sort of reeling /towing device, the assignment consists in a reeling motor that has to tow a small car up along a slope, loaded with a certain weight, the team that is able to reel in the most weight in the shortest time will win the competition and have the maximum grade and extra credits for the semester.</p>
<p>The rules are:</p>
<p>You can only use 1 Peltier thermoelectric generator module that will deliver 12 volts to the motor of my choosing. No capacitors or energy storage devices. I could use a voltage regulator to rise the voltage but speaking to one of my professors he said that due to the fact the power is just a relationship between current and voltage, rising the voltage wouldn't really make a difference as the current would drop, so my best shot was to shop for a motor with good torque that would surely work under 12 volts in case the peltier module does not deliver its full capacity.</p>
<p>I had 2 options, and because I need to move the biggest load in the shortest amount of time, I had to look for a motor with a good speed/torque relationship that would work under 12 volts in the worst case scenario, but I read somewhere you can make a speed reduction arrangement to help torque like the one below:</p>
<p><a href="https://www.youtube.com/watch?v=LHPduoql93A" rel="nofollow noreferrer">Reducing Engine</a></p>
<p>So I went and bought two different motors,</p>
<p>One with high speed and comparably low torque (to see if I the reducing arrangement can help me get a better torque in the end.</p>
<p><a href="https://rads.stackoverflow.com/amzn/click/B01DVHAW6A" rel="nofollow noreferrer">BEMONOC Small DC Motor 12V High Speed 12000 RPM High Torque PMDC Motor for DIY Parts (0.135 N.m)</a></p>
<p>And one with low speed and high torque, to test both.</p>
<p><a href="https://rads.stackoverflow.com/amzn/click/B071GTTSV3" rel="nofollow noreferrer">Greartisan DC 12V 200RPM Gear Motor High Torque Electric Micro Speed Reduction Geared Motor Centric Output Shaft 37mm Diameter Gearbox (0.2157463 N.m)</a></p>
<p>But I was wondering if you have any suggestions as to how to improve this torque/speed relationship and get the best out of it, I really want to win this competition.</p>
<p>I would truly appreciate any information you can share with me in contribution to my project.</p>
<p>Thank you in advance and sorry for my grammar, English is not my first language. </p>
| |mechanical-engineering|motors| | <p>With such a limited power supply, the key metric that you should be concerned with, is efficiency, i.e. how much of the energy supplied to the motor can <em>actually be used</em> for lifting the weight. For a typical Permanent Magnet DC Motor, peak efficiency occurs at at approximately 90% of the motor's 'no load speed', and 10% of its 'stall torque'.</p>
<p><a href="https://i.stack.imgur.com/QVXTD.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QVXTD.gif" alt="Typical PMDC Motor Performance Graph"></a></p>
<p>Therefore, you will find that using a "low speed and high torque motor" which is able to reel in the load without 'struggling' at all (i.e. the motor itself, before the gearbox, is able to turn at approximately 90% of it's no-load speed), then your system will run more efficiently.</p>
<p>You say that the winners will be able to lift "the most weight in the shortest time" - this is an ambiguous scoring system. Is it:</p>
<ol>
<li>the most weight lifted over a certain distance wins, but if there are two teams lifting the same weight, then the faster team wins?</li>
<li>Teams get a score base on the weight they lifted and the time it took - something like $\frac{Weight}{Time}$?</li>
<li>Something else?</li>
</ol>
<p>Without that information, you can't make an informed decision regarding which motor is likely to be best.</p>
| 21925 | What is your best recommendation to optimize the torque/speed relationship of a motor? |
2018-05-28T14:21:19.753 | <p>I'm currently trying to design a bulk-handling system which minimizes the distance from the bulk material pile(s) where the material is loaded (the volume) to the unloading platform(s) (the point), but I have many possible combinations on the volume shape and the number of unloading point. I would ideally like to iterate through various designs in order to find the most feasible one, and a useful parameter to get from each design would be to get this average travel distance from the pile to the platform. </p>
<p>I have toyed around with the idea of using the centroid as the average distance to a point, but the problem is that the directional component of the centroid makes it such that it doesn't truly represent the average distance. For example, a disk with a radius of 10 meters might have the same centroid as a disk with a radius of 100 meters, but the average distance from the center would be different. </p>
<p>$$\int_{V}r\text{ d}V$$</p>
<p>A slightly better representation would be the moment of inertia, since the squared component better represents average distance (so that opposite directions don't cancel out).</p>
<p>$$\int_V r^2\text{ d}V$$</p>
<p>However, to truly get the average distance, I would need something that looks more like this:</p>
<p>$$\int_V \lvert r\rvert\text{ d}V$$</p>
<p>Is there any simple way of getting this value from a CAD software package, or is it derivable from any other existing mathematical definitions like the centroid or the moment of inertia? Similarly, if this average distance is obtained only relative to the centroid, how could I extrapolate it to another arbitrary point?</p>
| |design|mathematics|computer-aided-design| | <p>Based off of ChP and Fred's answers, I ended up coding the 2-D solution for the problem. I might eventually want to extrapolate it to 3-D, but it shouldn't be too complicated. </p>
<p>I used a "clipped" Voronoi diagram, code examples of which are readily available online, which I used to separate an arbitrary polygon (representing the pile) into different sections, each representing the area closest to one of the given points (representing the unloading/loading points). Then, I placed a uniform point grid inside of the polygon, and solved for the distance of each of these with respect to the closest loading point. After getting an array of these values, I averaged them to get the average distance. It's rudimentary, but it's accurate enough for my particular purpose.</p>
| 21933 | Is there any defined mathematical parameter that can give me the average distance of a volume to a point? |
2018-05-28T21:25:39.233 | <p>If the output of a diaphragm pump is impeded in any way, or meant to be at higher pressure than the input of the diaphragm pump, will it work, or will it stall out and fail to pump? </p>
<p>In addition to this, are there reversible diaphragm pumps?</p>
| |pumps| | <p>Yes, a diaphragm pump is able to pump against a pressure gradient. Typically, positive displacement pumps are able to generate a higher pressure difference, but at a lower flow rate than other technologies, such as centrifugal pumps.</p>
<p>There are not reversible diaphragm pumps, since they rely on one-way-valves for their intrinsic operation.</p>
| 21937 | Can the same diaphragm pump be used to pressurize a container? |
2018-05-31T18:27:54.340 | <p>How would I find the second moment of area of a parallelogram, as shown?</p>
<p><a href="https://i.stack.imgur.com/Dp2Pt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dp2Pt.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|statics|inertia-matrix| | <p>The best way is to create a rectangle by cutting the triangular portion from the left to the right side where it is missing and calculate from the original zero coordinate which is the base width of the triangular portion.</p>
<p>The logic is this the centroid of both triangles is calculated from
<span class="math-container">$$[\frac {bc}{2}\times \frac {2}{3}b]_{left} + [\frac {bc}{2} \times \frac {1}{3}b]_{right}$$</span>
By this statement we realise that the 2 triangular portions are equivalent to a single rectangle added to the rectangular portion of the p'gram but critically we cannot change the origin point to begin at the origin+b or the answer will be wrong.</p>
| 21982 | How would I find the Second Moment of Area of a Parallelogram? |
2018-06-01T09:01:38.070 | <p>What is error in action? And why is it called error "in action".</p>
<p><a href="https://i.stack.imgur.com/Ij38O.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ij38O.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|gears| | <p>From Manual of Gear Design: Spur and Internal gears, volume II:</p>
<blockquote>
<p>The load carrying ability of a pair of gears may be limited by either the beam strength of the gear teeth, or the surface endurance limit of the material. The lower of these two values will be used to establish the load carrying ability of any pair, and must be greater than the maximum dynamic load.</p>
<p>Dynamic Load - whenever there is an error in action between a pair of gears, one of
two reactions must take place: the connected masses must be
accelerated and decelerated to compensate for this error, or the
system must be elastically deformed the amount of the error. In
practice a combination of the two will occur. <strong>Even with perfect gears,
there will be an error in action due to the deformations under load.</strong>
The load caused by these reactions is called the dynamic load.</p>
</blockquote>
<p>Source: <a href="https://www.globalspec.com/reference/68707/203279/gear-tooth-loads" rel="nofollow noreferrer">https://www.globalspec.com/reference/68707/203279/gear-tooth-loads</a></p>
<p>It's called "in action" because the error is not present when the gears are not under load.</p>
| 21989 | What is "error in action" in gears? |
2018-06-01T18:23:53.010 | <p>I have a question on the Bauschinger effect. Let's say that we have a material with yield stress $400 \,\mathrm{MPa}$, and let's say that the actual applied stress is tensile for starters. If we load the material to the point where the stress is, for example, $500 \,\mathrm{MPa}$, then that means that the loading has surpassed the material's yield stress by $100 \,\mathrm{MPa}$ and that the material has entered the inelastic region.</p>
<p>If the direction of stress is now reversed, from tensile to compressive, the Bauschinger effect tells us that the new compressive yield stress will be lower than the initial one and will be $-400+100=-300 \,\mathrm{MPa}$. </p>
<p>What I want to know is: What happens if in my initial loading of the material I don't surpass the yield strength? In other words, if yielding occurs at $400 \,\mathrm{MPa}$, I apply less than that ($350 \,\mathrm{MPa}$, for example). If I now unload and start loading with compressive stress, does the Bauschinger effect apply? Or is the yield stress for compressive loading still $-400 \,\mathrm{MPa}$?</p>
<p>Thanks in advance.
Note: I don't know what tags to add, so if you know of more suitable ones, please add them.</p>
| |mechanical-engineering|materials| | <p>Bauschinger effect is based on plastic strain , so there would be no effect for stresses in the elastic strain ranges. I learned this from Drs N.H.Polakowski and S Mostovoy before there were computers so no tags. However; I am sure they have references in the literature. </p>
| 21997 | The Bauschinger effect |
2018-06-01T19:23:54.797 | <p>I'm using two stepper motors (datasheet below) to pan and tilt a platform. The bottom motor spins fine without the breadboard glued to the top platform. However, it behaves unpredictably and skips steps when the breadboard is glued in place (shown below). My question is why does this happen? Is it related to rotor inertia? </p>
<p><a href="https://i.stack.imgur.com/3ejdB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3ejdB.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/2kI5v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2kI5v.png" alt="enter image description here"></a></p>
| |motors|torque|stepper-motor| | <p>I assume the bottom stepper shaft is affixed directly to the clear plastic platform that supports the top motor and the breadboard, is this correct?</p>
<p>If so, there are two possible causes of your problem. first, the rotary inertia of the assembly consisting of the top motor and the PC board on its lever arm may be too much for the bottom stepper to rotate at the speed with which you are sending it pulses. try slowing down the pulse rate.</p>
<p>second, the combination of board + support arm + top motor may have a torsional resonance close to the step signal frequency. trying to rotate the assembly with a string of discrete pulses may then excite the resonance and kick the torque impulses back to the bottom motor, causing it to jump between cog positions and lose steps. try hitting the bottom motor with one, two, three, ... pulses in a string and watch the assembly closely to see if it has been set into vibration by the end of the pulse string. </p>
| 21999 | Stepper motor skipping steps |
2018-06-01T22:19:38.663 | <p>I have been designing a prototype for a keychain that sprays perfume. I have a metal keychain piece which has a female thread and the plastic sprayer screws right into it and has a male thread.</p>
<p>I want to get these manufacturers ASAP but I need to be ensured it wont leak. I think I will switch my straight thread to a tapered thread, will that help? </p>
<p>What can I do to ensure that it will not leak? </p>
<p>Thanks!!</p>
| |threads| | <p>tapered threads are intended to be leakproof, but the degree to which they resist leaks depends on how tightly they are screwed together. This generally requires the use of a tool of some sort, as it is usually not possible to tighten a tapered-thread joint enough by hand to make it stop leaking. </p>
<p>For this reason you should instead use straight threads and a gasket like a soft O-ring which is trapped between the two pieces when they are screwed together. Unlike the tapered thread joint, these can be tightened enough by hand to be leak-proof. </p>
| 22001 | making a leak proof thread for my keychain idea |
2018-06-02T05:52:07.553 | <p>What is symmetric?Please explain geometrically and in terms of displacements. I have encountered this while looking up for engineering strain and true strain values have totally different compression and expansion parameters for their respective strains.</p>
| |mechanical-engineering|structures|stresses| | <p>My guess would be that talking about symmetry in the context of material science, the meaning of "symmetry" is probably describing direction-dependant properties of a material.</p>
<p>Steel for example is an isotropic material (excluding drawn/stretched steel), meaning that its properties are not dependent on the direction in which you measure them.<br></p>
<p><em>Note, this does <strong>not</strong> mean that compressive and tensile value must be the same. It basically means that if you have for material sample, you will measure the same properties, whether you measure them in $x$, $y$ or $z$ direction.</em></p>
<p>On the other hand, wood, reinforced concrete and other composite materials are usually not isotropic. Wood for example has a much higher strength when loaded in axial direction, compared to a radial or tangential load.<br>
This makes wood an material with orthotropic symmetry, because its properties mainly vary along the three orthogonal directions (ax/rad/tang).<br></p>
<p>There's a formal explanation on the whole subject on <a href="https://en.wikipedia.org/wiki/Orthotropic_material#Condition_for_material_symmetry" rel="nofollow noreferrer">Wikipedia</a>, but it requires some understanding of linear algebra.</p>
| 22007 | What is symmetric? |
2018-06-02T10:23:26.343 | <p>What are the advantages and disadvantages (technical, financial etc.) of having battery swapping system over charging stations for electric cars? As per my reading, battery swapping seems feasible for electric buses and 2-wheelers but not for cars. Is that really the case?</p>
<p>Also, is it possible for the battery to be housed under the bonnet instead of having it under the floor of the car so as to do away with underground swapping (like the one showcased by Tesla)? I know that the battery is placed there so as to lower the center of gravity. However, if there aren't any other significant benefits, can the aforementioned be done and battery swapping be made easier?</p>
| |mechanical-engineering|electrical-engineering|battery|electric-vehicles| | <p>Advantages:</p>
<ul>
<li>Minimal turnover time</li>
<li>Simple, standardized connection</li>
</ul>
<p>Disadvantages:</p>
<ul>
<li>Varying "remaining life" in batteries</li>
<li>Overhead of disposing of spent batteries</li>
<li>Needs standardization (or better, automation) of battery replacement</li>
</ul>
<p>For example regarding varying age/use and recycling, I could start a business "recycling" batteries from BE (British Electricity) by putting them in my car and driving to the Cell station for replacement. I think that to minimize that kind of fraud, you'd have to "subscribe" to a particular vendor, and let them worry about swapping batteries with their competitors (kind of like roaming on phones, a premium service at first but eventually becoming an expected free service).</p>
| 22010 | Battery Swapping in Electric Cars |
2018-06-02T22:21:05.890 | <blockquote>
<p>Dear engineers,</p>
</blockquote>
<p>I have to make an engineering project which is a portable solar phone charger. One thing I need to do in it is to somehow charge the phone (android) from the battery (which is charged by solar panels) and simultaneously send data about the battery level to the phone through the same usb connection. Is this even possible to do? If so, could you please advise a way to do it? Would arduino be helpful?</p>
<p>I know that usb has separate lines for data and power, but I don’t know how a phone can receive this data and whether it will work while charging. I was thinking about using arduino, but it doesn’t seem to be well suited for charging a phone (maximum current output is rather low). Would it work if I made separate circuits, one for charging a phone and another for reading the data from voltmeter and sending it to the phone through the other pins on usb?</p>
<p>I’m sorry for such a tremendous question, I just don’t have much experience with electronics (not even a year at college), and I must somehow do all this stuff. </p>
| |electrical-engineering|power-electronics|circuits|battery|circuit-design| | <p>It is assume that the PV panel is charging a battery such as a LiPo or a Li-Ion battery Cell. A battery Gas gauge circuit such a <a href="http://www.st.com/content/ccc/resource/technical/document/datasheet/ea/b5/01/6e/dd/f0/49/3e/DM00105047.pdf/files/DM00105047.pdf/jcr:content/translations/en.DM00105047.pdf" rel="nofollow noreferrer">STC3117 (Gas gauge IC with battery charger control for handheld applications)</a> and micro-controller. To use a STC3117, micro controller with I2C such a Arduino DIY platform is suitable. </p>
<p><a href="https://i.stack.imgur.com/1DJ11.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1DJ11.jpg" alt="STC3117 I2C"></a></p>
<p><a href="https://i.stack.imgur.com/cFJrb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cFJrb.jpg" alt="Arduino I2C pins"></a></p>
<p>You are correct USB has a data lines which can be used to communicate the status of the portable solar charger. Upon connecting the phone an application on the phone could read the status of the mobile solar charger display the status. This link on <a href="https://www.allaboutcircuits.com/projects/communicate-with-your-arduino-through-android/" rel="nofollow noreferrer">Communicate with Your Arduino Through Android</a> should provide a starting point to communicate to a phone. </p>
<p>Hope this will help you get started.</p>
| 22020 | Simultaneously charging and transmitting data to an android phone using usb |
2018-06-03T11:48:56.277 | <p>Based on the following publication:</p>
<blockquote>
<p>Kouro, S., Perez, M.A., Rodriguez, J., Llor, A.M. and Young, H.A., 2015. Model predictive control: MPC's role in the evolution of power electronics. IEEE Industrial Electronics Magazine, 9(4), pp.8-21.</p>
</blockquote>
<p>Available <a href="https://www.researchgate.net/publication/287796890_Model_Predictive_Control_MPC%27s_Role_in_the_Evolution_of_Power_Electronics?enrichId=rgreq-963aef38d7d14b76effad0c8f661a507-XXX&enrichSource=Y292ZXJQYWdlOzI4Nzc5Njg5MDtBUzozNzgyOTIwNzYwNzI5NjBAMTQ2NzIwMzI2NzM4MA%3D%3D&el=1_x_3&_esc=publicationCoverPdf" rel="nofollow noreferrer">here</a>, it is said that:</p>
<blockquote>
<p>MPC can be also used with <strong>nonminimum</strong> phase plants, which usually appear in rectifiers connected to the grid, avoiding <strong>linear</strong> controllers, which, in these cases, have a very restricted performance [76], [79], [83].</p>
</blockquote>
<p>Dose it mean that linear controllers are not adequate for nonminimum phase systems? If yes, why?</p>
| |control-engineering|control-theory|optimal-control| | <p>A LTI system is nonminimum phase if it has one or more zeros in the right half plane. When using a LTI controller for feedback also then the bandwidth of this controller will be limited to roughly to smallest magnitude of zeros in the right half plane.</p>
<p>This is because a right half plane zero gives a phase drop and slope increase. But near the bandwidth a slope of minus one (time 20 dB per decade) with a phase of -90° is desired. A minimum phase zero, so located in the left hand plane, could be counteracted by a left hand plane pole in order to ensure the desired asymptote at the bandwidth. If this could also be done for a nonminimum phase zero, in which case a right half plane pole would need to be added. According to the <a href="https://en.wikipedia.org/wiki/Nyquist_stability_criterion#The_Nyquist_criterion" rel="nofollow noreferrer">Nyquist criterion</a> this would then require a counterclockwise encirclement of the minus one point of the openloop Nyquist diagram. The magnitude and phase of such counterclockwise encirclement look very similar to a that of a right half plane pole and zero, however when encircling the minus one point you already cross the zero dB line and therefore also give an upper bound for the bandwidth.</p>
<p>So no, a nonminimum phase LTI system can be controller using a LTI controller for feedback, but the bandwidth will be limited.</p>
| 22028 | Are linear controllers inadequate for nonminimum phase systems? |
2018-06-04T00:23:45.437 | <p>I recently installed a VFD to reduce the speed of my pump, however, occasionally the configuration will stop pumping liquid through the system. The pump is a three phase 60 Hz and this issue has happened when running the system at 13 Hz and 40 Hz. Any thoughts?</p>
| |mechanical-engineering|electrical-engineering|pumps| | <p>may be its not getting the enough rotating flux , as you know V/F ratio needs to be maintained constant for motor to run at low speeds... may be your VFD is not providing the appropriate ratio of Votlage to Freq so that its gets right rotating flux... </p>
| 22034 | Why does my VFD and pump configuration stop pumping? |
2018-06-04T01:14:43.557 | <p>I've tried to find some rules of thumb about this without any luck. I'm designing a part with a blind cartridge bearing bore. It's a 624 bearing (13x4x5) and the fit size guides point me to a tolerance of 13.0-13.026mm for the bore for my application, so slip fit on the housing. FWIW the shaft is the rotating part of the design and will be pressed into the inner race. Do I need to create some kind of lip or shelf inside the bore so that the inner race doesn't come into contact with the housing and create friction? It seems like this would obviously be a concern if both the inner race and outer race are the same thickness.<a href="https://i.stack.imgur.com/0XXda.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0XXda.png" alt="housing"></a></p>
| |machining|bearings| | <p>You should probably consider a way to get bearing out of the block as well. Consider putting 3 holes tangential to the bearing to tap the bearing out with a drift. Thin cylinders tend to get cockeyed and gall up the surface, then you would have difficulty inserting a new replacement bearing, more applicable for steel on steel, than steel bearing on a dissimilar metal(or plastic). In tool and die work, it is very common to have tight press fits between inserts and the die block, the insert might be a hard tool steel like D2, while the die might also be D2 or a cheaper steel like O2. In that situation I used to use a anti-galling anti-seize compound, typically copper powder in a oil binder, but molybdenum anti-seize compounds also exist.</p>
| 22035 | Blind bearing bore design |
2018-06-04T09:02:02.290 | <p>How much could this motor lift and how fast? If I connect a belt to its axis about 1cm away or even 10cm away from its tip.</p>
<p>Torque: 1.5Nm
Peak Torque: 10Nm
Power: 9-30A, to 1000W.</p>
<p><a href="https://www.alibaba.com/product-detail/86BLF40-24v-48v-1000W-big-brushless_60644013087.html?spm=a2700.7724838.2017115.56.f5f862211YuM1C" rel="nofollow noreferrer">https://www.alibaba.com/product-detail/86BLF40-24v-48v-1000W-big-brushless_60644013087.html?spm=a2700.7724838.2017115.56.f5f862211YuM1C</a></p>
<p>What kind of motor would I need to be able to lift 50kg about 1m high, in say 4 seconds? And to control finely the exact position within the 1m, sometimes it should lift to 20cm sometimes to 30cm. Is a stepper or servo motor capable of this task? Would a linear actuator be better, but I read those have low duty cycle, and they are slow so they would lift it slowly and wait for next time?</p>
| |motors|robotics|stepper-motor| | <p>The height has nothing to do with the selection of the motor.
Whichever motor type you choose (servo or step motors will allow you control the exact position while simple induction motor just cannot) - you should verify it has sufficient torque capabilities for the desired lifting velocity.</p>
<p>The needed torque in your case is calculated as follows:
1. for a constant speed: the load mass * g * the distance of the weight from the motor axis center (in meters)
2. for accelerating lifting: the load mass * the linear acceleration (in m/t^2) * the distance of the weight from the motor axis center (in meters)</p>
<p>the results for both cases are in Nm units.</p>
<p>Next, you need to convert the desired lift speed to the motor angular velocity. Basically, all you need to do is to divide to linear speed (m/sec) by pulley radius (m) to have it in rad/sec.</p>
<p>Each motor has a specific torque-speed curve. You should make sure your motor can supply the needed torque for the desired speed.
Furthermore, you should relate to the motor peak torque only when this torque is applied for a short period of time. Otherwise, the motor will heat-up and wouldn't be able to supply enough torque.</p>
| 22043 | How much could this motor lift and how fast? |
2018-06-04T16:07:34.237 | <p>I have been doing some research on valve systems for sample analyzers. In many of the valve systems' product documentation they refer to a flow path as 'fully swept'. From the context I can conclude that it increases the rate of purging a previous gas or liquid from the system. My best guess is that it's a path that minimizes 'dead zones' in the flow, where gas or liquid may get caught. Is it simply marketing jargon, or is there a fluid mechanics definition? </p>
<p><a href="https://www.swagelok.com/downloads/webcatalogs/en/ms-02-326.pdf" rel="noreferrer">An example usage of the term (page 7)</a></p>
| |mechanical-engineering|fluid-mechanics|chemical-engineering| | <p>You are correct; as described in product literature, a "fully swept" path "<a href="https://www.designworldonline.com/diaphragm-valve-withstands-rugged-semiconductor-environment/" rel="nofollow noreferrer">minimize[s] entrapment areas, facilitate[s] purging and maximize[s] flow capacity</a>" (SwageLok) and ensures "<a href="http://ppl.saint-gobain.com/uploadedFiles/SGppl-global/Documents/High%20Purity%20Systems.pdf" rel="nofollow noreferrer">low shear</a>" with "<a href="http://ppl.saint-gobain.com/uploadedFiles/SGppl-global/Documents/High%20Purity%20Systems.pdf" rel="nofollow noreferrer">no hold-up volume</a>" (Saint Gobain) and "<a href="http://www.indev-cheaper.it/Brochure/28_trasduttori_ds-pr-solidsense_ii_atex-pt-eng0.pdf" rel="nofollow noreferrer">with no hidden recesses or dead spaces</a>" (SolidSense). </p>
<p>Whether this capability is valuable to you depends on your application. It's certainly industry-specific jargon. Since it's rigorously defined (there either are dead spaces or there aren't) and can be easily verified, on the spectrum of marketing fluff to an objective industry term, it's firmly on the latter side.</p>
<p>My own research, for example, involved sending live biological cells in solution through PEEK tubing and a variety of valves. Increasing the degree of sweeping in the fluid path meant the difference between a quick post-experiment clean with 10% bleach followed by deionized water vs. the possibility of bacterial contamination flourishing in cracks filled with dead cells and the possibility of residual bleach leaking out of crevices and killing my live cells.</p>
| 22054 | What is a 'swept path' in a valve system? |
2018-06-04T19:18:09.227 | <p>I got a mini-project: I want to automatically get rid of the water produced by an indoor air conditioner unit. Since the unit is underground, the water needs to climb ~2m to get to the outside. </p>
<p>I found a cheap pump online that has a max head of 3m and flow rate of 240l/h, the pipes on it have 6mm on the inside diameter. I also have some tubing left over from a gardening project but its 4mm on the inside.</p>
<p>I can figure out how to fit the two together mechanically but since the tubing is tad too small, wouldn't that increase the pressure requirement and thus decrease the max head? </p>
<p>I really don't mind if it affects the flow rate, the amount of water I need to pump out is ~2L/day max so I'm adding a microcontroller to turn it on for just a few minutes out of a day anyways.</p>
| |pumps| | <p>Related: <a href="https://engineering.stackexchange.com/questions/16481/system-curve-for-use-in-determining-pump-operating-point">System curve for use in determining pump operating point</a></p>
<p>The smaller lines will increase the resistance to flow, which will decrease the flow rate. The use able head is reduced, but this wouldn't be a problem with such a low flow rate. However, the lines will still flow, just at a reduced flow rate.</p>
| 22059 | Would too small tubing reduce the max head of a pump? |
2018-06-04T20:56:38.313 | <p>We have an open container sitting at 1000mm off the ground and a receiver tank sitting one metre above it. Container holds 400l of water and is connected to the receiver by 80mm id pipe. I was planning on using extraction fans sitting in the open end of the receiver to create a vacuum to remove the water.</p>
<p>What is a formula I could use to work out how inHg needed to be produced by the fans to remove water in x amount of time?</p>
| |vacuum|vacuum-pumps| | <p>So if your fan doesn't produce a minimum of 1m of water (which is about 2.9 inHg), there would be no flow. Afterwards, it depends upon your fan curve. You would find that the system curve would likely equal 2.9 inHg + K*Q^2, where Q is the flowrate and K is a factor that depends upon how you move the flow - note with smooth curves K is small, and with sharp turns and disruptive contours, K is large.</p>
<p>Wherever your pump curve intersects this system curve is the flow rate. Plain and simple. Once you know the flowrate, divide the volume to move by the flowrate to gain the time estimate.</p>
<p>Edit:</p>
<p>With a wet/dry vacuum style compressor, you're most like to see a curve similar to this one:</p>
<p><a href="https://i.stack.imgur.com/5BmmQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5BmmQ.jpg" alt="enter image description here"></a></p>
<p>At the 40" of water (approx. 1 meter), the vacuum would remove 300 cfm of air (and therefore water) a minute. At less than 100 cfm removal, the vacuum would stall. So, you should expect removal between 100-300 cfm, or between 3-9 seconds.</p>
| 22060 | What is the formula to work out time to remove water from an open vessel with vacuum? |
2018-06-05T09:22:34.877 | <p><a href="https://i.stack.imgur.com/SQPmL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SQPmL.jpg" alt="triangular load single support"></a></p>
<p>From equilibrium: $$C_y = \frac{w_0L}{2}$$</p>
<p>And the load is given by: $$w(x) = \frac{w_0}{L}x$$</p>
<p>Thus the expressions for shear and moment are given by: $$V(x) =\int{-w(x)dx}=-\frac{w_0}{2L}x^2 + C_1\\M(x) = \int{V(x)} = -\frac{w_0}{6L}x^3 + C_1x + C_2$$</p>
<p>How are $C_1$ and $C_2$ determined?.</p>
<p>Cheers</p>
| |mechanical-engineering|structural-engineering|structural-analysis| | <p><a href="https://engineering.stackexchange.com/a/22068/1832">JohnHoltz's answer</a> is correct in that you determine $C_1$ and $C_2$ by looking at the boundary conditions. Looking at the left-hand side of the beam, we know that the cantilever has zero shear and bending moment, which let's us determine (note the use of the auxiliary variable $w$ which describes the load's variation per unit length):</p>
<p>$$\begin{align}
w &= \dfrac{w_0}{L} \\
Q &= \int q\text{ d}x \\
&= \int wx\text{ d}x\\
&=\dfrac{1}{2}w x^2 + C_1 \\
Q(0) &= C_1 = 0 \\
\therefore Q &=\dfrac{1}{2}w x^2 \\
M &= \int Q\text{ d}x \\
&= \int \dfrac{1}{2}w x^2\text{ d}x \\
&= \dfrac{1}{6}w x^3 + C_2 \\
M(0) &= C_2 = 0 \\
\therefore M &= \dfrac{1}{6}w x^3 \\
\end{align}$$</p>
<p>As mentioned by <a href="https://engineering.stackexchange.com/a/22076/1832">Mark's answer</a>, however, this only describes the beam from the left cantilever until the support. The discontinuity caused by the support means the equations are no longer valid and must be recalculated.</p>
<p>You can use Mark's heaviside function, or you can just repeat the process above for the other side as well.</p>
<p>Let's start by pretending that we only have the right-hand side, so that $x=0$ is at point C, and therefore define a new length $\ell$ which describes this cantilever, and $w_c$ as the value of the distributed load at point C.</p>
<p>$$\begin{align}
\ell &= \dfrac{L}{3} \\
w_c &= \dfrac{2}{3}w_0 \\
Q &= \int q\text{ d}x \\
&= \int \left(wx + w_c\right)\text{ d}x\\
&=\dfrac{1}{2}wx^2 + w_c x + C_1 \\
Q(\ell) &= \dfrac{1}{2}w\ell^2 + w_c \ell + C_1 = 0 \\
\therefore C_1 &= -\dfrac{1}{2}w\ell^2 - w_c \ell \\
&= -\dfrac{1}{18}w_0 L - \dfrac{2}{9}w_0 L \\
&= -\dfrac{5}{18}w_0 L \\
\therefore Q &=\dfrac{1}{2}wx^2 + \dfrac{2}{3}w_0 x -\dfrac{5}{18}w_0 L \\
M &= \int Q\text{ d}x \\
&= \int \left(\dfrac{1}{2}wx^2 + \dfrac{2}{3}w_0 x -\dfrac{5}{18}w_0 L\right)\text{ d}x \\
&= \dfrac{1}{6}wx^3 + \dfrac{1}{3}w_0 x^2 -\dfrac{5}{18}w_0 Lx + C_2 \\
M(\ell) &= \dfrac{1}{6}w\ell^3 + \dfrac{1}{3}w_0 \ell^2 - \dfrac{5}{18}w_0 L\ell + C_2 = 0 \\
\therefore C_2 &= -\dfrac{1}{6}w\ell^3 - \dfrac{1}{3}w_0 \ell^2 + \dfrac{5}{18}w_0 L\ell \\
&= \dfrac{1}{162}w_0 L^2 + \dfrac{1}{27}w_0 L^2 -\dfrac{5}{54}w_0 L^2 \\
&= -\dfrac{4}{81}w_0 L^2 \\
\therefore M &= \dfrac{1}{6}wx^3 + \dfrac{1}{3}w_0 x^2 -\dfrac{5}{18}w_0 Lx - \dfrac{4}{81}w_0 L^2 \\
\end{align}$$</p>
<p>Now that we have these values, we just need to shift the origin back to $x=0$ at point A to keep these results compatible with what we got for the left cantilever, so we get:</p>
<p>$$\begin{align}
Q &=\dfrac{1}{2}w\left(x-\dfrac{2}{3}L\right)^2 + \dfrac{2}{3}w_0\left(x-\dfrac{2}{3}L\right) -\dfrac{5}{18}w_0 L \\
M &= \dfrac{1}{6}w\left(x-\dfrac{2}{3}L\right)^3 + \dfrac{1}{3}w_0 \left(x-\dfrac{2}{3}L\right)^2 -\dfrac{5}{18}w_0 L\left(x-\dfrac{2}{3}L\right) - \dfrac{4}{81}w_0 L^2
\end{align}$$</p>
| 22063 | Shear force expression for singly supported triangular load |
2018-06-05T10:57:37.123 | <p>The discs must remain in flat contact to the surface at all times but you can assume no friction while pushing them about.</p>
<p>No using magnets from the other side of the surface, for all intensive purposes assume it is infinitely thick on the other side.</p>
<p>You must be able to move the discs "freely" meaning that normal robotic arms won't cut it due to getting tangled after moving each disc in a circle around the centre maybe once at most.</p>
<hr>
<p>I imagine the answer could be very useful if there is one, I couldn't find one after scouring the internet for a few hours but a solution would open a few doors, especially when it comes to "omnidirectional treadmills", if you replace each disc with a vertical cylinder you could crudely mimic any terrain.</p>
<p>I hope this isn't too theoretical, apologies if it is, I do really think there are numerous practical applications though and for that reason any working answer at all would be useful.</p>
<hr>
<p>Edit: Since it was apparently ambiguous, let me clarify, no using magnets, obviously if a solution uses a machine that has a magnet internally for some reason required for that machine to function then that's fine.</p>
<p>The surface is arbitrary, it could be any material within reason and cannot be changed, assume the area above the surface cannot be altered in any way.</p>
<hr>
<p>Edit 2: My mistake, I completely forgot to mention the surface has a finite size, although the shape could be practically anything as long as no corridors thinner than the sum of both discs' radii are within it, assume it is convex if you cannot find a solution for concave, or just assume it is a rectangle!</p>
| |mechanical-engineering| | <p>Given your multi-edited question, this seems as reasonable an answer as any other: mount each disk on top of a blimp, thus providing the necessary anti-gravity force to keep the disk on the surface. Use the blimp's guidance system (propellors, ailerons, etc) to move wherever you please.</p>
<p>Yes, this is absurd, but your scenario is unrealistic. </p>
| 22066 | Move two discs freely and quickly on the underside of a surface? |
2018-06-05T15:33:03.213 | <p>While browsing a local news site, I stumbled upon this picture of a broken bus door. The glass was fractured due to a rock impact. As far as i know there are no heating elements on the window section in the picture. Now, what interests me is the apparent periodicity and structure of the fracture pattern.</p>
<p>Are periodic fractures something commonly encountered in tempered glass?
Is the mechanism that generates such fractures understood?
Is it a consequence of the manufacturing process?
Perhaps the way the glass is affixed to the door?</p>
<p>Thanks in advance</p>
<p><a href="https://i.stack.imgur.com/fYYK9.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/fYYK9.jpg" alt="Glass fracture"></a></p>
| |glass| | <p>It's an artifact of production - in particular, the construction of the conveyor belt of the machine performing the quenching process.</p>
<p>The pattern can be observed through a polarizing filter in the undamaged glass:</p>
<p><a href="https://i.stack.imgur.com/FRmQy.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/FRmQy.jpg" alt="enter image description here"></a>
<a href="https://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/CarWindowPolarization.jpg/800px-CarWindowPolarization.jpg" rel="noreferrer">source</a></p>
<p>and it's a result of the structure of the conveyor belt through which the surface of the glass is cooled to generate the stress that gives it the special properties:</p>
<p><a href="https://i.stack.imgur.com/iCdZZ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/iCdZZ.png" alt="enter image description here"></a>
<a href="https://www.youtube.com/watch?v=GCnQVVfjabk" rel="noreferrer">source</a></p>
<p>The contact area of the glass with the conveyor is small enough that it doesn't negatively impact the process - all throughout the surface the stress is introduced in sufficient amount, but the amount of stress applied differs by the "openness" of given area, creating this pattern.</p>
| 22072 | Fracture patterns in tempered glass |
2018-06-06T08:51:49.497 | <p>My AC outdoor unit gets direct sunlight (no shading whatsoever) from morning to noon. Temperature normally soars up to 40 degree Celsius (I am from India). I have two questions:</p>
<ol>
<li><p>Is this bad for the AC's refrigerant, does it lose efficiency over time?</p></li>
<li><p>Does an AC compressor located in full sun work harder than one located in the shade?</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/0ZgMq.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/0ZgMq.jpg" alt="enter image description here"></a> </p>
| |mechanical-engineering|thermodynamics|heat-transfer|refrigeration| | <p>In this type, the rated power is determined in such a way that the outdoor unit is not in any way protected by shadows or shields; That is, the outer plate casing essentially casts a shadow on the condenser, i.e. the heat sink. If the system is completely enclosed, ie well-installed and well-fitted, it will not leak for many decades. This, of course, depends on the quality of the copper tube used, the quality of the flange on the tubes. Of course, it increases the lifetime of the climate if it is in a place where it is more protected. Generally speaking, if the outdoor unit is in a sheltered place, it will dust sooner. With a hand-held sprayer, it is advisable to wash more frequently with clean water from the lamellae of the outdoor unit. This keeps the factory performance. If you are in shade and in a sheltered place, less energy is invested in the operation, but of course this is very small overall. Regular cleaning is more important. I noticed that the pipes running outdoors do not have any protection. In Europe, it is standard that these pipes should be routed in an outdoor plastic air channel, because their outdoor insulation will eventually dissipate from the sun's UV radiation. If this happens, it is advisable to put in a new tube. Furthermore, the pipe is more aesthetic when running in a plastic channel. Be careful to drop the pipeline at all points towards the outdoor unit. Many people forget, but the refrigerants in them are no longer as perfectly blended as the banned ozone damaging R22. (Mixing of oil and refrigerant). For this reason, not every point of the system is lubricated (low oil consumption) or even an oil plug may develop. Therefore, they are instructed by thermal engineering engineers to make the piping slightly sloping (ie not a rough slope). Of course there is also a horizontal section if there is no other solution. However, very little climb in the tube is forbidden, especially for inverters with low power. If you pay attention to this, you will be in good friendship with this device for long years. Durable, high quality device. The indoor unit, however, often needs to be cleaned with a fungicidal composition, such as alcohol-containing hospital and solarium disinfectants. Use of chlorine-containing agents is strictly prohibited! Especially, the blower of the indoor unit needs to be cleaned with this alcoholic substance, because in this type of black fungus is highly prone to settling.Greetings from Hungary.</p>
| 22087 | Should we protect our AC's outdoor unit from sunlight?Does it increases the durability of an AC and its refrigerant if it is well shaded? |
2018-06-06T19:38:58.423 | <p>Note, as I have my degree in chemistry, I denote "=>" as the next step in the equation. Also, sorry for any obvious mistakes that I made - as I mentioned, my degree is not in engineering. Thanks!</p>
<p>I am considering a design for a PET spherical toy ball. The ball will have a radius of 0.07 m that will have pressurized air in it at a pressure of 5.81 atm (588698.3 Pa). Also, the ball will include a 35 nm coating of Reduced Graphene Oxide (RGO) as a durable coating on the outside of the ball.</p>
<p>To get the thickness of the walls of the ball, I know that the ultimate tensile strength of PET is 55 MPa. Because I want a 15% safety factor for the ball, I do the following calculation:
$$(1-15\%)\cdot55 = 46.75\text{ MPa} = 46750000\text{ Pa}$$</p>
<p>I then use the reduced ultimate tensile strength to find the thickness of the wall using the formula for a thin-walled sphere where $p$ is pressure, $t$ is thickness, $r$ is radius, and $\sigma$ is stress:</p>
<p>$$\begin{gather}
\sigma = \dfrac{pr}{2t} \\
46750000 = \dfrac{588698.3 \cdot 0.07}{2t} \\
\therefore t = 4.41\times 10^{-4}\text{ m}
\end{gather}$$</p>
<p>I can calculate the burst pressure of the sphere $P$ for pressure (gauge) inside sphere, $FS$ for factor of safety, $\sigma$ for allowable stress and additionally $R_i$ for inner radius, $R_o$ for outer radius:</p>
<p>$$P = \dfrac{(R_o^2-R_i^2)\sigma}{(R_i^2)FS}$$</p>
<p>For calculation of bursting pressure, we take $\sigma$ as ultimate stress for a given material and put $FS=1$.</p>
<p>Note here that I chose the express the ultimate tensile strength of PET as $5.5 \times 10^7$:</p>
<p>$$P = \dfrac{((0.07 + 4.41 \times 10^{-4})^2 - 0.07^2)\cdot 5.5 \times 10^7}{0.07^2 \times 1} = 695183.67\text{ Pa} = 6.86\text{ atm}$$</p>
<p>I now want to calculate the volume of PET needed to create a sphere of the desired dimensions:</p>
<p>$$\begin{align}
V &= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\
&= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3 - \left(\dfrac{0.140}{2}\right)^3\right) \\
&= 1.36 \times 10^{-5}\text{ m}^3
\end{align}$$</p>
<p>I now want to calculate the volume of Reduced Graphene Oxide (RGO) to apply to the surface of the sphere:</p>
<p>$$\begin{align}
V &= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\
&= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4} + 35 \times 10^{-9}}{2}\right)^3 - \left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3\right) \\
&= 1.084366 \times 10^{-9}\text{ m}^3
\end{align}$$</p>
<p>Can you see any mistakes in my calculations?</p>
<p>Please do not ask why I am using PET as a material or other design decisions like that. There are good reasons for them.</p>
| |mechanical-engineering|pressure|structural-analysis| | <p>I don't understand what you're trying to achieve, whether a 15% factor of safety is correct, whether having a factor of safety only on the material strength (instead of a combination of strength-minimizing and load-maximizing factors) or any such thing is adequate.</p>
<p>I also didn't bother checking your calculations (as in, punching numbers into a calculator). But just looking at your reasoning, it seems fine enough.</p>
<p>The only mistake I spotted was in the volume calculations, where you calculate the volume using diameters, but then add the wall thicknesses only once. If you're using diameters, you should add the thicknesses twice, once for each side of the ball.</p>
<p>Or just simplify your life and use radii instead (as you did for the rest of the question):</p>
<p>\begin{align}
V &= \dfrac{4}{3}\pi \left(r_o^3 - r_i^3\right) \\
&= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4}\right)^3 - 0.07^3\right) \\
&= 2.73\times 10^{-5}\text{ m}^3\text{ of PET} \\
V &= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4} + 35\times 10^{−9} \right)^3 - \left(0.07 + 4.41 \times 10^{-4}\right)^3\right) \\
&= 2.18\times 10^{-9}\text{ m}^3\text{ of RGO}
\end{align}</p>
<p>Note that I did not check that your calculations for the values needed for the above (such as 4.41x10<sup>-4</sup> m needed for the wall thickness) were correct.</p>
| 22094 | Difficult Toy Ball Analysis |
2018-06-06T20:51:44.573 | <p>I am wanting to produce custom mass plates for a project, very similar to those found in gyms. Here's a rough idea of what each one should look like:</p>
<p><a href="https://i.stack.imgur.com/jf8aW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jf8aW.jpg" alt="Rough concept of custom mass plate"></a></p>
<p>This is essentially <em>identical</em> to mass plates one can buy online, however I am needing to produce my own, with custom specifications, sizes, etc. Most notably each plate will need a thread through the center hole.</p>
<p>Where would I even start looking for someone who could do this? Looking online, I get confused by all the metallurgical terms. What type of professional am I looking for? I feel clueless.</p>
<p>Thanks in advance.</p>
| |mechanical-engineering|metallurgy|project-management| | <p>At higher volumes, these weight plates are generally cast iron. Here's a link to some more info on that: <a href="http://www.metals-china.com/cast-iron-weight-plate-iron-casting-foundry.html" rel="nofollow noreferrer">http://www.metals-china.com/cast-iron-weight-plate-iron-casting-foundry.html</a> . You'll want to search for "cast iron manufacturer." Adding the country to the end of that search term will help you find someone based on location. Note: Generally, it may be less expensive to manufacture it in Asia, however due to the weight, shipping it back may make it more expensive than making it in the US. Get quotes from both markets to compare.</p>
<p>Now however, casting isn't ideal for having threads. In this case, after casting, the weight plates need secondary processing via machining to get the threads. Afterwards, a third aditional process would be added to it.</p>
<p>From Wikipedia:</p>
<blockquote>
<p>Most plates are coated with enamel paint or hammertone to resist corrosion; more expensive varieties may be coated with chrome, rubber, or plastic </p>
</blockquote>
<p>PS. I'd recommend adding some fillets to your sharp edges for comfort and to prevent cuts to the user of your product.</p>
| 22098 | Where does one go to produce custom metal objects? |
2018-06-06T22:03:55.797 | <p>Cement heats up like crazy. I live in the top floor and the heat is unbearable. Also it has a high carbon footprint. </p>
<p>Why still cement is vastly used? When there are so many better alternatives? </p>
| |mechanical-engineering|civil-engineering|heat-transfer| | <p>as pointed out by others here, it is hard to beat concrete for inexpensive strength and ease of construction. In the dorm building I inhabited in 1970 at the University of California at Davis (a hot place!), the structure was entirely concrete but the roof of the top floor (where I lived) was covered with several inches of insulating foam topped with tar-covered fabric that was painted a light color. The insulation and light color kept the heat out while the concrete underneath it provided the necessary structural strength. </p>
<p>In this way, the advantages of both materials were exploited in combination to produce an inexpensive and strong building that did not get roasted by the sun. </p>
| 22100 | Why don't we use other substitute of cement while building a roof for thermal proofing? |
2018-06-07T20:10:51.213 | <p>If I had an aircraft that used propellers for movement and took the air from the front running it through a tube that gets very small towards the end, would it have any impact on the speed?</p>
<p>Would the aircraft get faster because the ejected air would be faster then when it got in, similar to a waterpipe that is half closed?</p>
<p>Would it be slower because the compressed air would act like a barrier for the new air that comes in?</p>
<p>Or would the aircraft have the same speed regardless of whether we run the air thrusted by the propeller through our cone or not?</p>
<p>Here is an image with what I mean:
<a href="https://i.stack.imgur.com/D1xBX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D1xBX.png" alt="enter image description here"></a></p>
| |airflow|aerodynamics|aircraft-design|propulsion|thrust| | <p>This is done in a jet engine, and partly, by the cowl around the front fan on a jet airliner engine, but on a large propeller, you could only do it slightly without slowing the airflow through the propeller. The extra weight would outweigh the benefits. </p>
| 22121 | Compressing the air thrusted by an aircraft's propeller through a cone like tube |
2018-06-07T21:08:06.660 | <p>Would an aircraft that uses propellers for movement but with a turbo jet like structure where the ignition chamber only heats the air a bit, similar to a hair dryer instead of releasing fuel and combusting it have any impact on the speed of the craft?</p>
<p>Would it just act like a normal propeller or will the aircraft get a little faster?
<a href="https://i.stack.imgur.com/F8bLv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F8bLv.png" alt="enter image description here"></a></p>
| |airflow|aerodynamics|aircraft-design|propulsion|thrust| | <p>As a reference, consider the <a href="https://en.wikipedia.org/wiki/Meredith_effect" rel="nofollow noreferrer">Meredith effect</a>, which applies (applied) in the construction of the P-51 Mustang radiator.</p>
<p>From the linked Wiki site:</p>
<blockquote>
<p>The Meredith effect occurs when air flowing through a duct is heated
by a heat-exchanger or radiator containing a hot working fluid such as
ethylene glycol. Typically the fluid is a coolant carrying waste heat
from an internal combustion engine.<a href="https://en.wikipedia.org/wiki/Meredith_effect" rel="nofollow noreferrer">1</a></p>
<p>For the effect to occur, the duct must be travelling at a significant
speed with respect to the air. Air flowing into the duct meets drag
resistance from the radiator surface and is compressed due to the ram
air effect. As the air flows through the radiator it is heated,
raising its temperature slightly and further increasing its volume.
The hot, pressurised air then exits through the exhaust duct which is
shaped to be convergent, i.e. to narrow towards the rear. This
accelerates the air backwards and the reaction of this acceleration
against the installation provides a small forward thrust.[2] The air
expands and decreases temperature as it passes along the duct, before
emerging to join the external air flow. Thus, the three processes of
an open Brayton cycle are achieved: compression, heat addition at
constant pressure and expansion. The thrust obtainable depends upon
the pressure ratio between the inside and outside of the duct and the
temperature of the coolant.<a href="https://en.wikipedia.org/wiki/Meredith_effect" rel="nofollow noreferrer">1</a> The higher boiling point of ethylene
glycol compared to water allows the air to attain a higher temperature
increasing the specific thrust.</p>
<p>If the generated thrust is less than the aerodynamic drag of the
ducting and radiator, then the arrangement serves to reduce the net
aerodynamic drag of the radiator installation. If the generated thrust
exceeds the aerodynamic drag of the installation, then the entire
assemblage contributes a net forward thrust to the vehicle.</p>
</blockquote>
<p>It's important to note that the primary purpose of this construction was to counteract the drag of the exposed radiator. It's certainly unlikely to generate sufficient thrust to power an aircraft.</p>
| 22122 | Heating the air inside a turbojet instead of combusting it |
2018-06-08T02:13:37.670 | <p>I need a high voltage power supply for the homemade CO2 laser I am building. I understand that a 15kV 60mA transformer is a thing, but its very big and heavy and its out of phase meaning each lug is only 7,500V and the difference is 15,000V. This would work for most applications but not one where i need 15kV positive and GROUND. I have seen simple arc lighter things that generate a supposed 15kV from a small battery, but the amps must be low? How can i design a simple 15kV high amperage transformer that can be smaller than a neon sign transformer?</p>
| |power-electronics|electrical| | <p>Voltage and Current multiply to give power. 15kV times 60mA is the same as 60Volt times 15 Amps (=900 Watts) You would not expect that to be a tiny thing, would you? And that power is without transformation losses.</p>
<p>Keep in mind that 900 watts a common size for an electric space heater. This is not trivial power you are dealing with.</p>
| 22126 | why do 15kV 60mA transformers have to be so big? |
2018-06-08T19:19:23.907 | <p>This mainly would concern Europe, since part of the infrastructure is already in place (trolleys overhead cables). What <strong>economic</strong> reasons prevent electric autos sans <em>large</em> batteries from being feasible in European cities? Lithium is considered by most environmental unfriendly, both extraction, usage(flammable), and disposal. I would see Europe being forefront of electric autos without batteries <em>if</em> it were feasible.</p>
| |electric-vehicles| | <p>One thing about using overhead trolley wires to power vehicles is traffic congestion. </p>
<p>I can remember riding on trolley buses when I was very young & if a trolley bus had to pull over at a bus stop to drop off or pick up passengers it blocked the vehicles behind it. All vehicles behind it had to wait until the bus moved again.</p>
<p>Any vehicle connected to an overhead trolley wire, whether it be bus or car, has restricted movement. It cannot deviate significantly from the line of the overhead trolley wire.</p>
<p>Currently, if something is blocking the path of a self powered vehicle whether it be a combustion engine of a battery the vehicle can go around the obstacle. But vehicles powered by an overhead trolley wire cannot move around. They have to stay where they are until the obstacle is removed.</p>
| 22144 | Overhead cable vs batteries for vehicles |
2018-06-09T13:10:36.710 | <p>in my aquarium Ive got a little pump for bubbling air through the water so it is substained with enough oxygen. Internaly the pump works fairly similar to old door bells, where the mains is flowing through a coil thus creating a magnetic field which moves an armature, which again presses air thorugh a hose and underneath the water's surface. (As from the German Wikipedia, <a href="https://de.wikipedia.org/wiki/Schwingankerpumpe" rel="nofollow noreferrer">https://de.wikipedia.org/wiki/Schwingankerpumpe</a>)</p>
<p>Now one day I looked at my pump as I found out that its housing is very tight so that air cant come to the inside. I wrapped it up completely in plastic foil, but the air output didnt reduce. I followed the hose with my eyes, but it (obviously) didnt show any holes in it. </p>
<p>Still, I can use the pump (without any possible air input) to fill up a balloon.</p>
<p>How is that possible?</p>
<p>Greetings,</p>
<p>~Cr0n</p>
| |pressure|airflow|pumps| | <p>It may by creating small vacuum bobbled by just creating turbulence of enough intensity to create foam bubbles.</p>
<p>Agitating water can produce vacuum cavities and foam. In some cases it is taken advantage of, such as submarine torpedoes. They have been designed to create a shroud of small bubbles to reduce the skin friction drastically and move so fast as to hit the target before it can defend itself.</p>
| 22155 | How does a Vibratory Pump ("Schwingankerpumpe") work? |
2018-06-09T13:11:18.590 | <p>When we can manually measure the discharge flow rate,what for have we developed the concept of venturi meter or orifice metre?</p>
| |fluid-mechanics| | <p>Flow meters can also be used for high-precision applications. Imagine wanting to measure flow of gas, on the order of 0.1 sccm through a very small tube. </p>
| 22156 | Why do we need flow rate measurement devices? |
2018-06-10T14:01:35.580 | <p>Scooters and skateboards are obviously meant for slower speeds, below 5 MPH. Typically if you go faster than that, it becomes harder and harder to control the scooter, and you will fall off. An example of this is the speed wobble, examples easily found on youtube.</p>
<p>What's the deal? Why is it that comparatively, bicycles can handle higher speeds with more stability and comfort?</p>
| |wheels|geometry|bicycles| | <p>Wheels surface area is very, very small, compared to say a bike, tricycle or other wheeled vehicle. That's fine on hard smooth surfaces like say a roller rink or a skateboard ramp. But terrible on pavement and gritty surfaces.
Rubber wheels slightly deform and compress to meet the surface contours they ride on. Hard wheels like skateboards and rollerblades do not. So the weight energy applied to the wheels doesn't cause it to grip the surface very well.
The more surface area you have to adhere to a surface, he safer it is at high speeds.
Case in point look at a standard automobile tire's thickness compared to a race tire.</p>
| 22183 | What makes scooters unstable at high speeds? |
2018-06-11T21:22:04.160 | <p>Consider a dynamical system as bellow that is subjected to an impulse input:</p>
<p>$\sum_{n=0}^N {a_n x^{(n)}}= f_0 \delta(t-t_0)$</p>
<p>the initial values are non-resting (all non-zero):</p>
<p>$x(t_0^-)=x_0\hspace{0.2cm} ;\hspace{0.2cm} x'(t_0^-)=x_1\hspace{0.2cm} ;\hspace{0.2cm} ... \hspace{0.2cm} ;\hspace{0.2cm} x^{(n-1)}(t_0^-)=x_{(n-1)} $.</p>
<p>How can I write the correct initial values for $t=t_0^+$? This is ambiguous since integrating the differential equation above, yields:</p>
<p>$ \sum_{n=1}^N a_n \big[x^{(n-1)} (t_0^+)-x^{(n-1)} (t_0^-)\big] = \sum_{n=1}^N a_n c_{n-1} = f_0$ </p>
<p>where $c_i$s are the jumps in initial conditions. In literature, it is said that only the highest order will undergo a discontinuity, yet it's not clear for me why.</p>
| |control-engineering|control-theory|mathematics| | <p>I asked the question <a href="https://physics.stackexchange.com/questions/411213/non-resting-initial-value-problem-with-impulsive-input">here</a> and I found the answer myself. This is really a simple one really. Here is my answer for this question:</p>
<blockquote>
<p>[...] I revisited this problem with another point of view which gives the answer right away. Since the <span class="math-container">$x$</span> is at least <span class="math-container">$N-1$</span> times differentiable, it's derivatives (up to order <span class="math-container">$N-2$</span>) and itself must be continuous. So no shift until the last derivative. Using the my question's notation,</p>
<p><span class="math-container">$c_n =0\hspace{1cm} \forall n<N-1$</span></p>
<p><span class="math-container">$c_{N-1} =f_0/a_N$</span></p>
</blockquote>
| 22215 | impulse input and discontinuity in initial condition after |
2018-06-13T05:45:20.577 | <p>In the following <a href="http://fumblog.um.ac.ir/gallery/839/mpc_ug.pdf" rel="nofollow noreferrer">book</a>,</p>
<pre><code>Model Predictive Control ToolboxTM (User's Guide)
Alberto Bemporad
Manfred Morari
N. Lawrence Ricker
</code></pre>
<p>At page <a href="http://fumblog.um.ac.ir/gallery/839/mpc_ug.pdf#page=18" rel="nofollow noreferrer">1-6</a>, it has been mentioned that</p>
<blockquote>
<p>If the plant is open-loop unstable, the maximum possible p is the number of control intervals required for the plant’s open-loop step response to become infinite. </p>
</blockquote>
<p>Here, <code>p</code> means the prediction horizon.</p>
<p>This claim does not look reasonable to me.
I have not understood the reason for this claim.
What will happen if I choose the prediction horizon longer?</p>
| |control-engineering|control-theory|optimal-control| | <p>I think the book is using "infinite" to mean "a number that is too big for the computer software to represent", not in the strict mathematical sense. See near the top of page 1-6:</p>
<blockquote>
<p>Open-loop unstable plants: if p*Ts is too large, such that the plant
step responses become infinite during this amount of time, key
parameters needed for MPC calculations become undefined, generating an
error message.</p>
</blockquote>
<p>So, this is practical limitation of how the software works, not a theoretical limitation of the MPC method.</p>
| 22240 | Maximum MPC prediction horizon for an unstable plant |
2018-06-13T18:00:42.117 | <p>I am modelling a plate as 3D deformable solid. My plate has a spatially dependent isotropic thermal expansion coefficient. </p>
<p>I have defined this spatial distribution as a Discrete Field in a tabular form (for the mesh elements). My question is how can I link this distribution to my material definition? Or is it not possible to have spatially dependent thermal expansion in Abaqus/CAE?</p>
<p>Thanks!</p>
| |finite-element-method|abaqus|thermal-expansion| | <p>It turns out that defining predefined field variables in CAE is a feature that was added in the 2018 version. The steps that one needs to follow when creating a field-dependent material property are the following:</p>
<p>1) In the Property module, make your Material dependent on a Field Variable(s) by specifying the number of field variables. In the table you will have to map the material property to the field variable.</p>
<p>2) Create the Field Variable: In the Load module, go to "Predefined fields"->Create->Other->Field and specify the desired distribution, which can be one of several options, including an Analytical Field. This analytical field in my case is the spatial distribution of the expansion coefficient - a function of X,Y,Z.</p>
<p>In case you have an earlier version of Abaqus/CAE, apparently you could use a temperature definition for the Field, and modify your input file to reroute to the field variable. I haven't tried this approach, but it should work just as well. </p>
<p>A couple people suggested parts of this and then I finally figured the rest out through some trial and error. Thanks!</p>
| 22251 | Spatially dependent material property in Abaqus/CAE |
2018-06-13T19:33:18.607 | <p>I have some coefficient values i.e. <code>74.9</code>, <code>81.3</code>, and <code>80.5</code> at frequency centers <code>63</code>, <code>160</code> and <code>400</code> Hz. The values are <code>1/3</code> octave band values. I would like to convert them to the <code>1/1</code> octave band values. Can someone suggest how to do that? </p>
| |acoustics| | <p>The short answer is, you cannot calculate octave-band levels from single 1/3-octave band levels only. For example, in order to calculate the 63-Hz octave band level, you must know the the 50, 63 and 80-Hz 1/3-octave levels, and add them according to the formula for decibel summation as above.</p>
| 22252 | How do I convert coefficient values from 1/3 octave band to 1/1 octave band? |
2018-06-13T23:53:25.043 | <p>Most undergraduate majors require 10-12 courses in the major, and maybe up to three courses from other departments as "co-requisites." But most engineering majors require 15-18 courses in the department itself, plus 10 -15 courses in math, chemistry, physics and computer science as "co-requisites." </p>
<p>In one engineering program I looked at, a "minor" consisted of six engineering courses, plus 14 courses in "co-requisites." I "get" that much, particularly the co-requisites (basically enough for another minor). But the major consisted of 18 engineering courses, three times the "six" for the minor, whereas other majors, require twelve courses (double the six for the minor) as well as co-requisites.</p>
<p>That's 25-33 courses, that is two thirds to four fifths of the whole curriculum, versus about one-third for other majors. Why is that?</p>
| |education| | <p>I went with my daughter's class to the science museum and while we were looking at things, I commented on why it worked that way and gave occasional interpretation of the displays. Impressed, my daughter's friend said, “Gee, Are you some sort of scientist of something?” The answer I had ready was “No, something better. I am an engineer. Scientists only report the news, engineers make it.”</p>
<p>The definition I got for an engineer when I was in high school was “Engineers are science and technology problem solvers.” You can’t solve problems without knowledge of the underlying science, along with the specific technology you are dealing with. When I was in school (BSME, 1981 (I am an old man)), they told us that they were teaching us everything they could fit in four years, it was not enough but it should be sufficient to allow us to function in the real world. As I recall, I had a few “free” electives (I took more math), three (not very free) humanities electives and a few mechanical engineering electives. Everything else was listed out in the requirements when I started. The school knows that much of what they teach is going to be obsolete in a few years so they teach as much of the foundations as they can and how to learn and solve problems. The problem is if you keep squeezing things in, there is not enough room for much else and you wind up with a "course intensive" engineering curriculum </p>
| 22256 | Why are engineering programs so course intensive? |
2018-06-14T06:12:34.083 | <p>I want to use a DC 12/24V motor to rotate an assembly of glass tubes(image attached) through a hollow shaft of 25mm OD and 22mm ID, the shaft is supported on bearings on both the sides(assume minimum friction), two 3mm thick circular plates hold the four tubes together at both the end. The shaft is attached to the center of the disk.</p>
<p>Data:</p>
<ul>
<li>Glass tube: 58 mm OD, 45 mm ID 1.8 m long</li>
<li>Disk: 180 mm OD, 3 mm thick attached at 25 mm from both the edges of the glass tube. </li>
</ul>
<p>Even a general procedure/idea of calculating torque requirement is appreciated</p>
<p><a href="https://i.stack.imgur.com/l3XKY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l3XKY.jpg" alt="Close up of the assembly"></a></p>
<p><a href="https://i.stack.imgur.com/Nrls4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nrls4.jpg" alt="Complete assembly"></a></p>
<p><a href="https://i.stack.imgur.com/giCp3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/giCp3.jpg" alt="Front view of the assembly"></a></p>
| |motors|applied-mechanics|torque| | <p>The torque required during the acceleration phase can be calculated using the formula:</p>
<p>$$T=Ia$$</p>
<p>where: $T = \text{Torque (Nm)}\\ I = \text{moment of inertia (kg.m²)}\\ a = \text{angular acceleration (rad/s²)}$</p>
<p>The moment of inertia of a solid disc is $I=\frac{1}{2}MR^2$, where $M=\text{mass (kg)}$, and $R=\text{radius (m)}$.</p>
<p>In this case, the moment of inertia will be lower than that of a solid disk, since the arrangement of the glass tubes concentrate the mass closer to the axis. Therefore, it is a safe estimate to use for calculation, and will ensure that the selected motor has sufficient power. Using the numbers provided in your question, the radius of the disc is found to be $0.09\text{ m}$</p>
<p>$$I=\frac{1}{2}(23)(0.09)^2=0.0932\text{ (kg.m²)}$$</p>
<p>Using the numbers that you provided in your comment, the angular acceleration can be calculated by converting the steady-state speed $100\text{RPM}=10.47\text{rad/s}$, and dividing by the 'spin up time' in seconds</p>
<p>$$a=\frac{10.47}{2.5}=4.19\text{ (rad/s²)}$$</p>
<p>Combining these two calculated values in the original equation gives the peak torque during acceleration:</p>
<p>$$T=(0.0932)(4.19)=0.39\text{ Nm}$$</p>
<p>It's important to note, however, that the torque generated by any motor is not constant in relation to angular velocity. </p>
<p><a href="https://i.stack.imgur.com/fndYPm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fndYPm.jpg" alt="Torque Speed Curves for different Motor Windings"></a></p>
<p>If you specify a motor which is advertised as being capable of sustaining $0.4\text{ Nm}$ at $100\text{ RPM}$, then you should be OK, since the available torque will be higher at lower speeds. This corresponds to a power of $\approx 4.2\text{ W}$, which is certainly achievable with a small DC motor.</p>
<p>If you want to limit the spin-up-speed to take $2.5\text{ s}$ in any case, then this should be done using e.g. PWM control of the motor.</p>
| 22260 | Calculating the torque requirements of the motor for a specific case |
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