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2016-08-12T12:02:18.813 | <p>Someone told me to avoid over-sizing when using DC Motors. Why? Is this because of sparking and higher wear (brushes) while the motor is running at a lower power rating?</p>
<p>Can someone explain?</p>
| |electrical-engineering|motors| | <p>This answer assumes that you meant to ask about <em>over-sizing</em> a DC motor rather than <em>over-dimensioning</em> it. over-dimensioning is when you provide too many dimensions such that they might be in conflict with each other based on drafting errors, rounding, or tolerances. For example if you draw a 3 story building and indicate that each floor is to be 10 feet tall, but you also dimension that the overall building must be 31 feet tall, nobody would know where to put the extra foot.</p>
<p>Over-sizing is when you pick an item that is bigger than you theoretically need. For example if structural analysis indicates that you need a 12" tall beam, but you decide to specify a 14" beam to be safe.</p>
<p>Over-sizing motors is common - it is typical in the mechanical world to specify a motor 25 - 50% larger than your calculations show you need as a rule of thumb. Over-sizing motors is useful because there could be some variation in the load the motor supports. A pump may have to run a little faster than it was designed to, or a bearing could be improperly lubricated, increasing friction, for example. In this case, it is desirable for the motor to be able to function even though conditions are not exactly as designed. It is also usually costly to replace a motor with a larger motor, so the relatively small cost of buying a larger motor in the first place is usually worth it.</p>
<p>There are also some down sides to over-sizing motors though, which is why it's unusual to see a motor 100% or 200% bigger than it theoretically needs to be. One obvious issue is cost. Bigger motors are more expensive and buying a bigger motor than you ever need means spending money that didn't have to be spent. There's also an issue with efficiency. Most decently sized motors are more efficient in the top half of their speed range. If they are used at very low speeds, they will waste more energy as heat. This is compounded by the fact that there is more rotating mass in the motor, so it takes more energy to start and stop the machine. In addition, some larger fans have cooling fans attached to the motor shaft to blow cool air across the motor, preventing overheating. At very low speeds, this fan does not move quickly enough to provide cooling air, so the motor is more likely to overheat which will damage the motor.</p>
<p>One other wrinkle is that variable speed controllers have limited resolution, and if your useful range for the motor is all in the lowest 10% of its speed range, your controller might not be able to regulate speed as carefully as you need it to. This is a constraint that can be designed around, but adds complexity and limitations to your system.</p>
<p>So motor size is always a trade-off between the risks of having a motor that is too small to move the load and a motor that is too big and wastes power or overheats. Most industries have settled in the 25 - 50% range but there are certainly valid exceptions where efficiency or reliability are strongly controlling factors.</p>
<p><a href="http://www.farnell.com/datasheets/62656.pdf" rel="nofollow">This Farnell datasheet</a> has a good overview of these principles, and most motor vendors also have tools or guides that will help you with motor sizing. Some even publish efficiency curves for their motors.</p>
| 11059 | Why to avoid over sizing a DC Motor? |
2016-08-14T21:12:11.847 | <p>Suppose I have three 2D drawings prepared in AutoCAD LT.</p>
<p>That 2D drawings are supposed to represent <em>top</em>, <em>left</em> and <em>front</em> projections of the 3D solid body.</p>
<p>How can I view that 3D solid body?</p>
| |drafting|autocad| | <p>I suggest you to watch these videos first:</p>
<p><a href="https://www.youtube.com/watch?v=ZFSV47I7SBg" rel="nofollow">AutoCAD 2D to 3D Conversion Trick</a></p>
<p><a href="https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Core/files/GUID-ECFB3220-6484-4D0B-BB7E-B06AD9F4E856-htm.html" rel="nofollow">Learn how to take 2D drawing designs and ideas and turn them into 3D objects.</a></p>
<pre><code>http://acronymonline.org/2d-flat-geometry-3d-autocad/
</code></pre>
<hr>
<p><strong><em>Also:</em></strong></p>
<p><strong>How to Convert a 2D Model to 3D Using AutoCAD</strong></p>
<blockquote>
<p>by Darrin Koltow, studioD, <code>http://smallbusiness.chron.com/convert-2d-model-3d-using-autocad-31833.html</code></p>
</blockquote>
<p>Use AutoCAD's "Extrude" and "Revolve" commands to turn 2-D designs into 3-D models. "Extrude" is how 3-D modelers to refer to the technique of stretching a 2-D shape into 3-D space. AutoCAD performs this expansion by first extending a new axis at right angles to the 2-D axes on which your 2-D design sits. It then makes a copy of the 2-D shapes you are expanding at a higher location on the axis, while keeping the original shapes at the base of the axis. After converting your 2-D design to 3-D, display it with realistic lighting and shadows using the "Render" command on the Render tab.</p>
<p><strong>Step 1</strong></p>
<p>Click the "File" menu and select the "Open" command from the context menu. Navigate to an AutoCAD file with a 2-D model that you want to convert to 3-D and double-click on it. AutoCAD will load the file for you to convert.</p>
<p><strong>Step 2</strong></p>
<p>Type "Perspective 1" to indicate that you want to view your design with perspective, which means that parallel lines will appear to converge as they do in the physical world. This option is more realistic than parallel projection.</p>
<p><strong>Step 3</strong></p>
<p>Click the “cube” icon at the top right of the canvas and then drag the mouse until the top, right and front sides of the cube are visible. This changes the viewpoint from 2-D to 3-D, allowing you to see the three-dimensionality of the 3-D form that you'll create from the 2-D model.</p>
<p><strong>Step 4</strong></p>
<p>Click the drop-down list at the top of the application window, then click the "3-D modeling" item to change the workspace to one that displays tools for creating and editing 3-D objects. Click the "Home" tab to access the "Modeling" panel, which displays the "Extrude" tool for converting rectangular 2-D shapes into block-like 3-D forms.</p>
<p><strong>Step 5</strong></p>
<p>Click the "Extrude" button; then click a shape of your 2-D model that you want to turn into a block form, as opposed to a cylindrical or spherical form. Press "Enter." The shape you selected will expand into 3-D space and the top of the box formed from the shape will stick to your cursor. Drag the cursor until the box reaches the height you want, and then click the mouse to end the extrusion. Use the "Extrude" command on the remaining parts of the 2-D model that you want to expand into blocks.</p>
<p><strong>Step 6</strong></p>
<p>Click the "Surface" tab and go to the "Create" panel to find the "Revolve" command, which makes cylindrical, spherical and other round forms from 2-D shapes. Click this command, and then click a shape in your 2-D model that you want to convert to a cylinder or other round form. Type the axis around which you want to revolve the shape, which can be "x," "y," or "z." Press "Enter" to perform the revolution, then revolve the other parts of your design that you want to have round 3-D forms to complete the conversion of your 2-D design to a 3-D model.</p>
| 11083 | Convert 2D projections to 3D View in AutoCAD |
2016-08-15T13:47:29.620 | <p>I realise that "bomb" can refer to all sorts explosives, but I am mainly interested in air-dropped bombs. Ever since reading about the <a href="https://en.wikipedia.org/wiki/1967_USS_Forrestal_fire">USS Forrestal fire</a>, I wondered about the safety of bombs when stored or transported in general- but in plain English: suppose I were able to hold a bomb in my hands- say, the B61 Nuclear bomb or the GBU-10 Paveway II- would it be okay to drop it on the ground? How often could I drop it before it damages the bomb? Is there a minimum speed/impact required that could be reached without using an aircraft?</p>
| |mechanical-engineering|safety| | <p>As mentioned already the bulk of the actual explosives used in munitions ect is pretty stable in itself. Clearly bombs need to be stored, handles and transported, often in difficult conditions so large quantities of sensitive explosives would be as much of a hazard to the user as the enemy. </p>
<p>The precise method for detonating a bomb will vary according to its intended use and may include impact fuses (sometimes with a time delay), timed fuses, altitude fuses, proximity detection (especially for air to air missiles) etc etc. For example bombs intended to create large craters may have an impact fuse with a short delay to enable them to penetrate the ground for some distance before detonating. Conversely anti-personnel weapons may be most effective detonated some distance above the ground. </p>
<p>Typically the detonator and associated mechanism won't be fitted untill it is required for use and there will be safety procedures to ensure that this is only done at the correct time, including visual indicators that a detonator is fitted. </p>
<p>In addition there may be other safeguards, for example a safety pin which prevents the detonator from actuating unless it is removed. For impact fuses this could be as simple as a cap over the fuse. </p>
<p>There may also be additional systems to ensure that a bomb cannot detonate untill it has been properly released from the aircraft. A simple way to do this is to have a small turbine at the nose of the bomb which is operated by the airflow over is as it falls and must turn a certain number of times before the bomb is armed. </p>
<p>This problem is somewhat more complex for artillery shells which need to be sensitive enough to detonate when required but not be set off by the forces involved in firing them. </p>
<p>So in summary</p>
<ul>
<li>A bomb will usually require a detonator to be fitted before it is capable of exploding at all. </li>
<li>There will be a variety of additional safety features intended to ensure that the bomb isn't detonated untill the desired time, varying depending on the intended use and sophistication of the design. </li>
<li>The type of fuse used will depend on where the bomb is intended to be detonated relative to its target. </li>
</ul>
| 11089 | Is it safe to drop a bomb? |
2016-08-15T18:57:04.163 | <p>I know the stack exchange does not help on hw problems, but please understand that I am just self learning these topics, I'm not in a class, and I'm using my dads old engineering mechanics book... Now my book says I'm completely wrong on a VERY simple vector problem (<strong>Question 2.6</strong>). This bothers me and want to see if you guys come up with their answer of: 18.95°.</p>
<p>My Answer: 52.02°.</p>
<p><a href="https://i.stack.imgur.com/83O0Q.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/83O0Q.jpg" alt="enter image description here"></a></p>
| |statics| | <p>What steps did you take to try to solve the problem? </p>
<p>b-b is the only "output" for vertical components in force F. I'll call the force in b-b $F_{b-b}$ and the force in F $F_F$. This means that the vertical component of force in $F_{b-b}$ must equal the vertical component of force in $F_F$. That is,</p>
<p>$$
F_F \sin{\alpha} = F_{b-b} \sin{60^{\circ}} \\
$$
Then, just solve for $\alpha$:</p>
<p>$$
\sin{\alpha} = \frac{F_{b-b}}{F_F} \sin{60^{\circ}} \\
\alpha = \mbox{asin}\left({\frac{F_{b-b}}{F_F} \sin{60^{\circ}}}\right) \\
$$</p>
<p>Now, plug in the numbers:</p>
<p>$$
\alpha = \mbox{asin}\left(\frac{150}{400}\sin{60^{\circ}}\right) \\
\alpha = \mbox{asin}\left(0.375*0.866\right) \\
\alpha = \mbox{asin}(0.3248) \\
\boxed{\alpha = 0.3308 \mbox{ rad} \\
\alpha = 18.95^{\circ}}
$$</p>
| 11094 | Simple statics vector analysis problem |
2016-08-16T12:48:42.397 | <p>I'm looking to rivet two sheets of aluminium together with the pop type. I've researched that 3.2mm rivets should go into 3.3mm holes. (Who has a 3.3mm drill bit?) That's not a lot of clearance, which I understand is part of the concept to allow adequate development of shear capacity.</p>
<p>However, how do I practically do this for a row of 20 rivets? I don't think that I can pre drill 20 holes to a tolerance of +/- 0.05mm. Certainly not on my cheap 250W pillar drill. The "How to Rivet" demonstrations on line only show the basic use of a rivet gun on one rivet. How would I join two sheets with a row of pop rivets?</p>
| |mechanical-engineering|rivets| | <p>For large production runs, the base metal may be punched, stamped, laser cut, water jet, etc but for a DIY project, drilling is going to be your best bet. 'Match-drilling' is best, where you drill both layers at the same time ensuring that the two holes line up. Depending on the material and thickness, you could clamp, tack weld, or tape the two layers together before drilling the holes. If the geometry of your joint is conducive, you may be able to do a couple of rivets and then drill the rest of the holes on the already-assembled sandwich, assuring the layers line up.</p>
<p>Pop rivets will work in slightly larger holes too, if necessary, but with reduced holding power. If you have to oversize a hole, it's better to do so on the front side then on the back (blind) side. If you have access to the back side, rivet washers are also available for this situation.</p>
<p>If you match drill and you're still having trouble with tolerances, it may be that your drill press setup is not rigid enough. If that's the case you may want to check out <a href="https://engineering.stackexchange.com/questions/160/how-do-i-keep-my-bit-from-wandering-on-a-drill-press/161#161">this question</a> for some strategies.</p>
<p>Drill bits are generally available in every .1mm increment up to 10mm without being too exotic.</p>
| 11101 | How to install multiple pop rivets |
2016-08-16T16:27:58.430 | <p>I'm hoping this is a relatively simple question to answer, but I just couldn't find it on Google.</p>
<p>With regard to air conditioning (or any cooling system that relies on compressing/evaporating), how do you actually control the 'cooling' temperature? After compressing/pressurizing the refrigerant, I understand that it becomes cool due to releasing heat, however, how do you control the level of coolness? </p>
<p>Is temperature directly related the amount of pressure being applied to the refrigerant? I.e, more pressure leads to lower temperature, less pressure leads to higher temperature?</p>
<p>Thank you!</p>
| |cooling|compressed-air| | <p>The temperature of evaporated refrigerant gas is constant as a property of refrigerant material. And usually it is considerably below the thermostat setting in the controlled space. The design of the system relies on the circulation of cooled air and its gradual warming up to deliver the cool air at desired temperature so that it is still cool enough to be effective and not too cool to be uncomfortable. Thermostats cut off the power to the compressor pump which is the starting point of refrigeration four phase cycle.<br>
Thermostats have built in delays as to switch the system back on or off with enough laps to allow the system get back to equilibrium and prevent freezing of valves or undue heat at the start up of the pump. </p>
<p>Along this path of delivery of the cool air there are incidental materials such as partitions, ducting, furniture, any material with high thermal index, that act as a moderator of temperature by absorbing heat from ambient air and delivering it back when the air coming from the unit it too cold. So the system's sudden switch on and off cycles are tempered into gradients of temperature that are hard to notice to occupants! </p>
| 11105 | How do you actually control the cooling level of AC? |
2016-08-17T13:02:39.060 | <p>I am relatively new (not in terms of registration but in terms of activity) on this site. I don't know if this question is suited here.</p>
<p>I have seen two rail locomotives connected one after another and those two (assuming) hauling the rakes. While this is true, I didn't see a loco-pilot (In India, the engineer who drives a train locomotive is called a Loco-Pilot) in the second locomotive.</p>
<p>If we assume that second locomotive is on neutral gear and not throttling at all, then it makes sense.</p>
<p>If its a case of two locomotive hauling the rakes together, is it a valid assumption that the two have synced to the same speed?</p>
<p>EDIT: The real question is in regards with, how the amount of power required to haul the rakes is divided among the number of locomotives connected?</p>
| |rail| | <h2>Multiple Unit control (or at least how it's typically done)</h2>
<p>If you are wondering <em>how</em> you can get two locomotives to do the same thing when there's nobody in the second locomotive, this is done using a technique called Multiple Unit (MU) control in the US and Multiple Working in British practice.</p>
<p>While older locomotives and self-propelled carsets (especially outside the US, as the 27-pin AAR-standard MU connector was developed from the electrical MU system originally implemented by EMD in their wildly successful E and F series units) used a variety of incompatible systems for this, most modern locomotive designs worldwide have settled on the AAR system. This is in essence a discrete-electrical control system for all basic electrical functions on the locomotive -- the locomotive air braking system is handled through auxiliary air hoses that are connected along with the 27-pin MU jumper cables, while the MU connectors on newer locomotives also carry an AAR-standardized communications bus for advanced, computerized functions (such as detailed alarms).</p>
<p>As a result of this jumpering, each locomotive is effectively synchronized to the lead locomotive's commands -- the crew of the lead locomotive sets up the controls the way they want, and the MU cables transmit that setup to the rest of the consist. The "reverse" and "forward" lines are also transposed at certain points (in each jumper, and also in the connectors on the <em>front</em> of a locomotive) to allow locomotives to run correctly when MUed in while facing "backwards" -- in fact, it is rare (at least in North American freight practice) to see locomotive consists where there isn't at least one backwards facing locomotive.</p>
<p>From the operator's perspective -- they are sitting in the cab of one jumbo locomotive with the combined capabilities of all the locomotives MUed together behind them. Furthermore, the MU wires can be trainlined down passenger coaches to a cab car for push-pull passenger operations, or connected to a receiver box for remote control of the locomotive via radio -- this last one is used for Distributed Power for heavy trains or mountain grades as well as for improvising a Remote Controlled Locomotive out of a regular locomotive for yard use.</p>
<h2>Sometimes, this isn't always the case, though</h2>
<p>However, not every consist (rake over where the OP is) of locomotives is MUed throughout. Some older or smaller locomotives, as well as many engines dedicated to switching duty, may have incompatible (the Brits are plagued by this) or no MU controls, and will wind up "dead in train" if they need to be hauled a long distance. A locomotive can also end up "dead in train" if it needs to be moved somewhere else, but its motive power simply isn't needed to get the train from point A to point B -- too much motive power on the head-end of a train will cause it to derail or break-in-two from excessive drawbar forces. Finally, a locomotive may have broken down somewhere on the line; this means that the locomotive will be hauled "dead" to a shop so that it can be repaired.</p>
| 11111 | Two railway locomotives connected one after another |
2016-08-17T15:19:28.057 | <p>I have been doing research into blood pressure monitors, and I was wondering when the cuff tightens, how do the transducers actually measure the systolic/diastolic pressure in the arteries.</p>
| |electrical-engineering|sensors|biomedical-engineering| | <p>This is actually a really neat question, and it involves a few engineering and anatomical concepts.</p>
<p>Here's what happens when an automatic blood pressure cuff measures your blood pressure:</p>
<ol>
<li><p>The cuff inflates until it reaches a pressure just above systolic pressure (the maximum pressure your blood reaches during a heartbeat). When it reaches this pressure, the main artery in your arm gets closed off because its pressure is not enough to resist the compression of the cuff.</p></li>
<li><p>The cuff then starts to deflate slowly. There is a pressure sensor that measures pressure fluctuations in the cuff. Once the cuff reaches your systolic pressure, the artery opens up just enough to let some blood through. Since the artery is mostly closed, this causes tiny vibrations as the blood squeezes through (this is because the blood flow is "turbulent"). These vibrations are transmitted through your skin and into the air in the cuff. The pressure sensor is sensitive enough detect these tiny vibrations in the air of the cuff once they start, and it records the current pressure of the cuff as the "systolic pressure".</p></li>
<li><p>The pressure in the cuff keeps decreasing, until the tiny vibrations stop (the blood flow becomes "laminar" instead of turbulent). This is the "diastolic pressure", i.e. the minimum pressure during your heartbeat.</p></li>
</ol>
<p>It turns out that doctors do the exact same thing with a manual cuff and a stethoscope: they pump up the cuff until they can't hear any gurgling noises in the stethoscope, then they slowly release until they hear the first gurgling noise (systolic pressure reached). They keep releasing pressure until they can't hear the noise any more, at which point they've found the diastolic pressure.</p>
<p>I've actually performed these measurements myself in a Biomechanical Engineering Measurements course. Cool stuff!</p>
<p>Source: <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1121444/" rel="nofollow">Oscillatory Blood Pressure Monitoring Devices</a> </p>
| 11113 | Transducers in cuff-style blood pressure monitors |
2016-08-17T16:16:11.830 | <p>One night, raining, crossing a street, I noticed strange thing about one of the stopped cars: In front of that car's headlamps, it rained <em>dashed lines</em>. I assumed that the car used LED for lights, running on AC, all with the same phase. (I couldn't recognize the model of the car.)</p>
<p>But what benefit do the AC headlights give? Most car lights run on battery so you can use it even when the engine is off. What powers AC car lamps when the engine is off?</p>
| |electrical-engineering|automotive-engineering| | <p>The headlight LEDs are using DC, but the voltage regulation on all modern power supplies is PWM or <a href="https://en.wikipedia.org/wiki/Pulse-width_modulation" rel="nofollow">Pulse Width Modulation</a>. The electronics switch to full on quickly and full off quickly so there is very little loss in the switching components. This is used in lots of electronics, but in the application you saw it is called a <a href="https://en.wikipedia.org/wiki/DC-to-DC_converter" rel="nofollow">DC DC converter</a>, and permits the car to go from a variable battery voltage to a constant led voltage (or <a href="https://en.wikipedia.org/wiki/Constant_current#Constant_current_power_supplies" rel="nofollow">constant current</a> for higher quality LED power supplies).</p>
<p>Usually a high enough frequency is chosen that it is not visible or displeasing to the eye. If you can estimate the velocity of the rain drop and length of the illuminated path you can calculate the frequency they were using. Velocity in m/s divided by the length in meters for example.</p>
| 11114 | Why do these car headlights run on AC? |
2016-08-18T00:26:27.083 | <p>Say I have a simple reinforced concrete frame (3as shown below) with all beam ends pinned:</p>
<p><a href="https://i.stack.imgur.com/d5TLN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d5TLN.png" alt="enter image description here"></a>
<strong>( the keyplan)</strong> </p>
<p>This is the detailing output:</p>
<p><a href="https://i.stack.imgur.com/C3iMx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C3iMx.png" alt="enter image description here"></a>
<strong>The beam elevation detail ( left), beam cross section detail ( right)</strong></p>
<p><a href="https://i.stack.imgur.com/9agh9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9agh9.png" alt="enter image description here"></a>
<strong>Column cross section</strong></p>
<p>I understand that due to the pins, all beam end moments will be 0. However, at the column joint, will there be any moment? I think no.</p>
<p>But one of my engineering friend think that because this is a <em>reinforced concrete</em> beam, and we design our beam reinforcement bars for the designed moment, this designed moment should somehow be transferred to column (even though the frame analysis shows that there is no moment at this beam joint), and thus the column joint will have moment. </p>
<p><strong>In other words, the presence of reinforcement bars in the beam means that we need to design for moment in the column at the joint, as the reinforcement will create a moment in the column, even though in the analysis this "reinforcement induced" moment is simply not there ( and the analysis shows us that column moment at joint is a 0). Is this a common practice ( as stipulated by convention or demanded by code of practice), or is this something completely unheard of?</strong> </p>
<p>Could he possibly be right?</p>
| |structural-engineering|reinforced-concrete| | <p>You seem confused between how a structure behaves in real life and how it behaves in a structural analysis model. They key point you seem to be missing is that <em>one must reflect the other</em>.</p>
<p>If, in real life, there will be reinforcement continuity between the beam and the column, <em>then the analysis model should not use pin-ended beams</em>; it should use fully-connected beams.</p>
<p>If, in the design, the analysis model is based on pin-ended beams, <em>then the detailing should be done to avoid moment-fixity</em>.</p>
| 11120 | Reinforced concrete design: will there be extra nominal moment on column in a frame structure? |
2016-08-18T00:37:09.913 | <p><em>I am asking this because some of my engineer friends said that they won't want to analyze beam as continuous beam for RC building structures, because it is "unsafe" to do so, whether it has anything to do with less deflection or no, I don't know.</em> </p>
<p>His logic is that <a href="https://engineering.stackexchange.com/questions/9066/why-continuous-beam-has-less-deflection-than-beam-with-support">continuous beam predicts less deflection</a> ( and hence gives more economical design), is his claim valid? </p>
<p>I wonder how does practicing engineers view the modelling and analyzing/designing beam spans as continuous beams ( instead of beams with many supports) when it comes to <strong>reinforced concrete building structures</strong>? Do they generally feel that it is OK to design beam spans as continuous beam ( and hence more economical because the reinforcement bars used will be less), or do they share my friend's opinion?</p>
<p><strong><em>What does the building code of practice ( or common design practice) across the world say about continuous beam design, do they discourage or encourage it? Any reason not to design beam spans as continuous beam?</em></strong></p>
| |structural-engineering|reinforced-concrete| | <p>Continuous beams are safer than simply supported ones in general. They have greater degree of static undetermination. If one section fails, it will form a plastic hinge and will redistribute to the other sections. Then it will form another hinge and another until the whole beam fails. For simply supported beams, we need just one section to fail and it is done. This is in case you design both as needed but something unpredicted happens later.</p>
<p>On the other hand, you can make any beam safer than any other, regardless the static scheme, by simply increasing the size and reinforcement.</p>
| 11121 | Is continuous beam design deemed "unsafe" by any code of practice/design practice for reinforced concrete building structures? |
2016-08-18T12:35:51.037 | <p>I'm my engineering drawing course, do I need to mention and show every dimension in the diagram if I've drawn everything to scale and also written the scale below?</p>
| |drafting| | <p>One should never ever measure things from a drawing. There are several reasons for this:</p>
<ul>
<li>The printer was misconfigured, and the output is no longer in scale. This happens a lot. Even if not the printer has a limited resolution so very small details may be off anyway.</li>
<li>The measurement may be inaccurate maybe the dimension in question is 10.1mm but the machinist measured from the drawing and interpreted it as 10mm. Bad! Also for longer spans the measurement may be a bit skewed etc.</li>
<li><p>The drawing is an acceptance document, what you're documenting is the acceptance criteria. You should thus have all the dimensions that matter to you marked out. Otherwise you may need to pay for all manufacturing defects.</p>
<p>You should not mark out all dimensions just so much that the dimension loop is complete. So to not introduce conflicting information in the drawing. If you just add those dimensions put them in parentheses. This may be useful for the machinist if the acceptance criteria is not really useful for machining for some reason (this can be bad for very accurate parts though).</p></li>
<li><p>If you totally leave out a dimension loop then you are leaving things to the machinists judgement. While this may not be bad as such, you're letting somebody else do the thinking in a way that produces something spectacularly different from your needs. Most shops would not accept such jobs but your local machinist might.</p></li>
</ul>
<p>But no you should try to add as few dimensions as possible that captures what your needs are. And suplement with as much data as the manufacturer needs to get things done.</p>
| 11130 | Do I need to mention every dimension on a to-scale drawing? |
2016-08-19T17:23:33.173 | <p>I'd like to mock up a quick prototype of a chassis assembly for a type of cart, which, in production, would be composed of CNC bent aluminum extrusions (hollow, approximately 5/8" circular profile), and connected using plastic joints. I'm wondering if anyone could recommend methods for prototyping the tubular components.</p>
<p>I'm considering using PVC, but I would like to use something I could bend and shape angles into relatively easily.</p>
| |prototyping| | <p><em>EDIT - I've solved this issue after a bit of searching for similar application requirements, so I'll post my solution:</em></p>
<p>Applying a bit of heat using a heat gun to sand-filled PVC pipe allowed me to slowly bend the pipe into my desired shape. The sand prevents the PVC from collapsing inward as it is bent.</p>
| 11155 | Prototyping tubular components |
2016-08-21T14:18:38.430 | <p><a href="https://i.stack.imgur.com/OSkwJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OSkwJ.png" alt="enter image description here"></a></p>
<p>How do I convert this into a mechanical network? like a circuit network.</p>
<p>my attempted work is this:</p>
<p><a href="https://i.stack.imgur.com/EfGBF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EfGBF.png" alt="enter image description here"></a></p>
<p>is this schematic correct?</p>
| |control-engineering|control-theory|systems-engineering| | <p>I think what you have is correct, provided the friction and spring constants are inverses.</p>
<p>I am more comfortable with the force-voltage analogy and would create the equivalent one as follows.</p>
<p><a href="https://i.stack.imgur.com/6npCG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6npCG.png" alt="enter image description here"></a> </p>
| 11174 | How do I convert this mechanical system into a mechanical network? |
2016-08-22T06:07:00.957 | <p>I think this is very fundamental question.</p>
<p>I am wondering why 0.00259 is used as weight-mass conversion ratio, when inch/pound unit is used in Nastran.</p>
<p>Could anyone explain the reasons?</p>
| |simulation|nastran| | <p>It's the inverse of the <a href="https://en.wikipedia.org/wiki/Gravitational_acceleration">acceleration of gravity</a> g in imperial units.
$$
1 g = 386.1 in/s^2
$$</p>
<p>$$
1/g = 0.00259 s^2/in
$$</p>
<p>You need that since:</p>
<p>$$
weight = mass * g
$$</p>
| 11182 | Where does weight mass conversion ratio '0.00259' come from in Nastran? |
2016-08-23T21:14:00.753 | <p>Is the mass flow rate constant in a thermodynamic cycle? More specifically a Rankine or Carnot cycle.</p>
<p>For example if the mass flow rate of the turbine is 100 kg/s, is the mass flow rate for the boiler / condenser / compressor or pump also 100 kg/s?</p>
| |thermodynamics| | <p><strong>Is the mass flow rate constant in a thermodynamic cycle?</strong></p>
<p>For <strong>general</strong>:</p>
<blockquote>
<p>Depends.</p>
</blockquote>
<p>Here is how;</p>
<p>If one of your components effect the pumping/boiling etc. of course your mass flow rate will change, it may change periodicly or it may not. Also there can be leakage or something that interacts with your fluid, of course that will cause to decrease in your mass flow rate with time.</p>
<p>But mostly, in general thermodynamic cycles, you don't need to worry. Because they build with simple components (boiler/pump/turbine/condenser) and they are not a factor that mass flow rate depends. </p>
<hr>
<p>If we be <strong>specific</strong> about Rankine and Carnot cycles; your assumptions is actually <strong>TRUE!</strong></p>
<p>About the <strong>Rankine Cycle</strong>;</p>
<p><a href="https://i.stack.imgur.com/g2zfs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g2zfs.png" alt="Rankine"></a></p>
<blockquote>
<p>The working fluid in a Rankine cycle follows a closed loop and is
reused constantly. (<em>Wikipedia</em>)</p>
</blockquote>
<p>That "<em>reused constantly</em>" means the total mass of the fluid is constant in cycle, therefore you have to use the same fluid in the same system. Unless there is a malfunction at one of your parts, you can be sure that you mass flow rate is constant.</p>
<p>About the <strong>Carnot Cycle</strong>;</p>
<p><a href="https://i.stack.imgur.com/KsTKT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KsTKT.png" alt="Carnot"></a></p>
<blockquote>
<p>The Carnot cycle is a <strong>theoretical thermodynamic cycle</strong> proposed by Nicolas Léonard Sadi Carnot in 1824 and expanded upon by others in the 1830s and 1840s. It provides an <strong>upper limit on the efficiency</strong> that any classical thermodynamic engine can achieve during the conversion of heat into work, or conversely, the efficiency of a refrigeration system in creating a temperature difference (e.g. refrigeration) by the application of work to the system. It is not an actual thermodynamic cycle but is a theoretical construct. (<em>Wikipedia</em>)</p>
</blockquote>
<p>So we know it is <em>theoretical</em>. But again, if we speaking about "<strong>a simple closed system</strong> <em>(control mass analysis)</em>", yes, the Carnot Cycle will be running with constant mass flow rate because of the same reason of Rankine Cycle.</p>
<hr>
<p>Don't forget to read <em>Mark</em>'s comment about the real life application issues that happens while trying to hold mass flow rate constant.</p>
| 11203 | Is the mass flow rate constant in a thermodynamic cycle? |
2016-08-23T22:33:55.477 | <p>I'm trying to power and agitator (alternating circularly back and fourth) with a crank. I've seen a bunch of complicated mechanisms that do this, but I'm looking for the mechanically simplest method.</p>
| |gears|pistons|mechanisms| | <p>One solution is to use a system with a ring gear that only has teeth on half of its inner circumference, like conceptually shown in the following image:</p>
<p><a href="https://i.stack.imgur.com/f4KOq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f4KOq.jpg" alt="enter image description here"></a></p>
<p>There are three gears:</p>
<ul>
<li>Ring gear (red)</li>
<li>Alternating gear (blue)</li>
<li>Output gear (orange)</li>
</ul>
<p>The alternating gear and output gears can be identical, and the sum of the numbers of teeth of the alternating and output gears should equal that to the ring gear if it had all its teeth instead of half. Otherwise the two inner gears will not properly fit inside the ring gear.</p>
<p>This mechanism works by rotating the ring gear, and the desired alternating motion will be obtained from the output gear. At any one time, the ring gear meshes will one of the inner gears. The inner gear that meshed will the ring gear will rotate in the same direction as the ring gear, causing the other inner gear to rotate in the opposite direction.</p>
<p>For example, let's say we rotate the ring gear CW. If the ring gear meshes with the output gear, the output gear will also turn CW. CW in, CW out. Now, if the ring gear meshes with the alternating gear, the alternating gear turns CW, forcing the output to be ACW. CW in, ACW out: the output shaft has alternated in direction.</p>
<p>The meshing of the ring gear switches between one inner gear to the other every half turn of the ring gear. It is worth noting that the output gear's speed does not change except when the meshing alternates, at which point the speed reverses.</p>
<p>One issue with this concept as-is is that one revolution of the ring gear brings about one full turn of the output gear one way, followed by one full turn the other way: this rotation may be too much for your agitation mechanism. To resolve this, a gear reduction stage should be added to reduce the output rotation:</p>
<p><a href="https://i.stack.imgur.com/lMTtQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lMTtQ.jpg" alt="enter image description here"></a></p>
<p>A gear has been fitted onto the side of the output gear (also orange to indicate its attachment), and this gear meshes with a larger gear (green). The green gear will turn slower and less far than the old orange output gear, and therefore the green gear becomes the new output. Therefore, the amount of rotation of the final output (green) that occurs before the direction alternates, in degrees, $\theta$, can be calculated using the following formula:</p>
<p>$$\theta=360° \cdot N_1/N_2$$</p>
<p>Where $N_1$ is the number of teeth on the gear attached to the side of the old orange output gear, and $N_2$ is the number of teeth on the new green output gear. In the image just above, $N_1=12$ and $N_2=36$, so the rotation angle before direction alternation is 120°. Note that if $N_1 \gt N_2$, you can get more than one output revolution before alternation (although this might not be suitable if the output is subject to high forces resisting motion).</p>
<p>Finally, it should be said that it is possible that, as the meshing transitions from one inner gear to the other, there may be a point where both inner gears are in mesh with the ring gear, jamming the gearset and making mesh alternation impossible. To resolve this, finer gear teeth should be used, as well as removing/filing down one or two of the end teeth on the ring gear to eliminate this dual meshing region.</p>
<p>ALTERNATIVE SETUP (Credit to joojaa's comment)</p>
<p>(@joojaa Please let me know if I've misinterpreted your comment)</p>
<p>An alternative configuration that does away with the ring gear would be like as shown:</p>
<p><a href="https://i.stack.imgur.com/dRKwq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dRKwq.jpg" alt="enter image description here"></a></p>
<p>Where we have the follow gears:</p>
<ul>
<li>Output gear (orange)</li>
<li>2x Large gears (dark red and dark blue)</li>
<li>2x Small gears (red and blue)</li>
</ul>
<p>The small gears should have equal radii, as should the large gears. Either one of the red or blue gear sets should be driven by the crank. The large gears are in constant mesh, so this ensures that the red and blue gear sets rotate in opposite directions. The small gears have half the teeth removed and aligned such that only one of the small gears are in mesh with the output. Like before, as the mesh alternates, so does the direction of rotation of the output.</p>
<p>Note that the rotation of the output gear before alternation is given by:</p>
<p>$$\theta=180° \cdot N_s/N_o$$</p>
<p>Where $N_s$ and $N_o$ are the numbers of teeth on one of the small gears and the output gear respectively. Also, having selected the numbers of teeth for the small and output gears, the big gears must have the following number of teeth (so that the gears fit together):</p>
<p>$$N_b = N_s + N_o$$</p>
| 11204 | Is there a mechanism that can turn circular motion into alternating circular motion? |
2016-08-24T01:01:17.020 | <p>Approximately how many types of stress are known/defined? Ex. Piola stress, Cauchy stress ...</p>
| |mechanical-engineering| | <p>Your examples aren't really "types" of stress, but different ways to <em>measure</em> stress. The real world structure just responds the way it responds - it doesn't worry about what sort of math you used when you tried to model its behaviour. </p>
<p><a href="https://en.wikipedia.org/wiki/Stress_measures" rel="nofollow">https://en.wikipedia.org/wiki/Stress_measures</a> lists a handful different stress measures that have acquired "names". Cauchy stress ("true stress") and Second Piola-Kirchhoff stress are probably the most commonly used.</p>
<p>It doesn't really make sense to talk about a measure of stress in isolation, without also considering how to measure strain, and how to model the constitutive equations that give the relationship between stress and strain. </p>
<p>For example 2-PK stress works nicely with Green strain, which is a convenient way to define the behavior of a body which has "small" elastic deformations superimposed on arbitrary large rigid body motions - in particular, large rigid body rotations, where approximations like $\sin \theta \approx \theta$ are not appropriate. </p>
| 11205 | Approximately how many types of stress are there? |
2016-08-24T08:11:11.353 | <p>A 2 m long beam fixed in one side with a 2000 kg concentrated load at free end. Profile of the beam is HEB100 material S235.</p>
<p>I would like to know if the beam will break under this load. If yes, what is the maximum load it can handle and how do you calculate it?</p>
| |structural-engineering|civil-engineering|steel|structures|beam| | <p>Bending moment at the fixed end for a cantilever with point load at free end is given by: $PL$ where $P$ is the concentrated load and $L$ is the beam's length.</p>
<p>$$M = 2000\ \text{kg} \cdot \dfrac{0.01\ \text{kN}}{1\ \text{kg}} \cdot 2\ \text{m} = 40\ \text{kNm}$$</p>
<p>The maximum stress is $\sigma = \dfrac{M}{Z}$, where $Z$ is the section modulus ($Z = \dfrac{I}{y}$).
$$\sigma = \frac{40\ \text{kNm}\times10^6}{89.9\ \text{cm}^3\times10^3} = 444\ \text{N/mm}^2$$</p>
<p>This stress is well above the the yield of the beam and it will deform plastically.</p>
<p>In fact, it is above the plastic yield too, so it will break.</p>
<p>This doesn't account for lateral torsional buckling because the further calculation is only required if the beam passes this simple test.</p>
| 11211 | Maximum deflection of a beam, fixed in one end and concetrated load at free end |
2016-08-26T00:18:41.877 | <p>I am curious about calculating stress at various points along a barrel as a bullet is being fired through it. The interference fit (press/friction fit) equations immediately come to mind, but I am wondering if these equations are appropriate to use because of:</p>
<p>(i) The non-linear pressure curve of the gas behind a fired bullet</p>
<p>(ii) The bullet is moving through the barrel (dynamic system)</p>
<p>Here is a link to some interfernce fit equations I'm refering to,
<a href="http://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-equations.htm" rel="nofollow">http://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-equations.htm</a></p>
<p>Any pointers?</p>
| |stresses|dynamics|friction| | <p>Indeed, you can use the same equations for calculating the stresses, as the barrel essentially is a thick walled pipe with internal pressure.</p>
<p>Looking at the equations you linked:
(7) "Radial Stress Casued by axial force" (8) "Circumferential Stress Caused by Axial force" are the right ones. (there seems to be a mistake though: the correct term for (8) would be "Circumferential Stress Caused by <strong>internal pressure</strong>"</p>
<p>The nonlinearity of the pressure: the pressure will be the highest at the chamber. Assuming an even wall thickness, check the stresses here and you are ok. In case of changing barrel outer diameter, you shuld look for benchmark pressure curves, and check the stresses at multiple locations along the barrel.</p>
<p>Dynamics: in my opinion this concerns the fatigue life of the barrel. To calculate the lifetime in terms of number of shots fired and survival probability, you will need the Haigh-diagram of the chosen material. But I believe the wear will be the limiting factor.</p>
<p>edit: this paper seems to be dealing with the same issue:
<a href="http://www.slideshare.net/JoshuaRicci/design-of-a-rifle-barrel" rel="nofollow">http://www.slideshare.net/JoshuaRicci/design-of-a-rifle-barrel</a></p>
| 11235 | Are "interference fit" equations appropriate for calculating barrel stress? |
2016-08-26T06:37:02.433 | <p>I am working on chess board using <code>SolidWorks 2016</code> and I've created two rectangles, outer one and inner one with <code>8x8</code> grid using <code>Linear Sketch Pattern</code>:
<a href="https://i.stack.imgur.com/7WF3O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7WF3O.png" alt="enter image description here"></a>Why is my sketch <strong>under defined</strong>?</p>
<p><strong>ADDENDUM</strong>:<br>
It seems I've managed to manage relations of inner rectangle - <strong>I've added dimensions between lines and added <code>Perpendicular</code> relation to base two lines in centre of coordinate system</strong> - so I think now it is fully defined, however, whole sketch is still <strong>under defined</strong>:
<a href="https://i.stack.imgur.com/2ulLR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2ulLR.png" alt="enter image description here"></a>
Why?</p>
| |solidworks| | <p>I've managed to solve the problem. I've deleted all relations and then selected <code>Display/Delete Relations->Fully define sketch</code>.</p>
| 11239 | Under defined sketch - inner rectangle for chessboard |
2016-08-26T11:33:52.650 | <p>Why we study continuous time systems if in the real life we control them using micro-controller which is discrete time .. ?</p>
| |control-engineering|control-theory| | <p>A controller is built around a <em>physical</em> system. What is the open-loop behavior of the system? </p>
<p>You have to sample the physical (continuous-time!) system in order to provide feedback to your discrete controller. What sampling rate should you be using for that microcontroller? <a href="http://www.wescottdesign.com/articles/Sampling/sampling.pdf">Are you sure</a>?</p>
<p>Once you have the physical system modeled and the sampling mechanism designed, how do you develop the controller? Well, with Laplace transforms you could do PID control or state feedback control, but all of those use continuous time differentiators ($s$) or integrators ($1/s$). </p>
<p>So, how do you perform differentiation or integration in a discrete control system? Easy - the Z transform! But, <a href="https://en.wikipedia.org/wiki/Z-transform">which Z-transform</a>? Do you use the <a href="https://en.wikipedia.org/wiki/Bilinear_transform">Tustin Bilinear Z-transform</a>? Well, that's great but it has some <a href="https://en.wikipedia.org/wiki/Bilinear_transform#Frequency_warping">frequency warping</a>. You could maybe avoid that by using a <a href="https://en.wikipedia.org/wiki/Matched_Z-transform_method">Z-transform that maps your poles correctly</a>, or you could use a <a href="https://en.wikipedia.org/wiki/Impulse_invariance">Z-transform that maps the impulse response</a>, or you could do something <a href="https://en.wikipedia.org/wiki/Advanced_Z-transform">more advanced</a>. </p>
<p>Point being that, like all of engineering, there are trade-offs. Every method of converting a continuous time controller to a discrete time <strong>representation</strong> of that controller introduces unwanted aspects. <em>It's up to</em> you <em>to understand the various drawbacks and apply the correct approach.</em></p>
<p>But, how can you understand a drawback unless you have a "perfect" system by which to compare all the others? This is a similar argument for studying ideal/frictionless systems in physics - you need to understand what the <em>best possible outcome</em> is in order to understand your limitations.</p>
<p>Nevermind the fact that you could also implement an analog controller. </p>
| 11242 | Continuous time and Discrete time systems |
2016-08-28T22:18:27.297 | <p>In short I'm designing a 'Humane fly swatter' for my University application into bachelor of Industrial Design. </p>
<p>The idea is to utilise the energy of swatting a fly to capture air through an intake, compress it and ejecting it at a higher velocity in turn drawing more air through the intake creating a slight vacuum, enough to draw the fly into a holding container.</p>
<p>To reiterate I'm not looking to create a big suction from the intake itself. Think of the swatter more as a butterfly net. You are capturing the fly into the funnel and from within the funnel the air velocity would have increased enough to suck the fly the remainder of the way into the holding container.</p>
<p>Simple visualisation of air being compressed and ejected through a formed funnel.
<a href="https://i.stack.imgur.com/liLqT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/liLqT.jpg" alt="Essentially this effect"></a></p>
<p>So I'm wondering if anyone could at the bare minimum provide me with some keywords and concepts that I can use to help me research ways of achieving this and to use this information to inform my concept designs. My biggest issues isn't understanding the concepts but not having to vocabulary to be able to search for what I need.</p>
<p>Thanks</p>
| |vacuum|aerodynamics|compressible-flow|drag| | <p>Why not just make a butterfly net? Or do you mean that the collection vessel is solid? If so, hate to burst your bubble but <a href="https://www.youtube.com/watch?v=u4CQCNfe-KY&t=35s" rel="nofollow">it's not going to work</a>.</p>
<p>The name of the effect you're looking for is the <a href="https://en.wikipedia.org/wiki/Venturi_effect" rel="nofollow">Venturi effect</a>, but it doesn't work quite how you seem to want it to from your description.</p>
<p>The mass of air that flows through the smaller diameter of your funnel has to go somewhere - where exactly is that? If it's a bottle, see the video I linked above - it's not going to work. If it's a mesh, then again, you've just made a more complicated butterfly net. </p>
<p>I hate to shoot an idea down without providing alternatives, but I don't see a way to make workable what you have described. I was motivated to get college degrees in engineering in part so I could critique my own designs, so hopefully you'll let this motivate you (to try something different and/or take some engineering classes) instead of getting discouraged. </p>
| 11262 | What are some concepts I can look into to aide in creating a 'compression funnel'(?) without the aide of a motor? |
2016-08-30T00:09:45.777 | <p>I think I understand how heat pipes work, and thus the following one has me flummoxed:-</p>
<p><a href="https://i.stack.imgur.com/kpb5X.png" rel="noreferrer"><img src="https://i.stack.imgur.com/kpb5X.png" alt="Chinese heat pipe" /></a></p>
<p>It's from a typical bulk order Chinese web site, but I've seem many similarly contorted arrangements. They are very common in the over clocking /modding community. There is never discussion as to which way the heat sink must be oriented for efficient operation.</p>
<p>My example seems ridiculous. With an U bend in the middle, I don't see how heat sinks like this can evaporate at the hot end and condense at the cold end. Surely the condensate will just pool in the U bend? Even with condensate wicking, it must be easier to wick downwards with gravity than upwards fighting gravity.</p>
<p>Are these types of heat sinks just a con? Overclocking /modding heat sinks are never spec'd with a deg. C /W rating or recommended orientation. This would not do in the engineering world. Can it work equally effectively in any orientation?</p>
| |heat-transfer|cooling| | <p>Orientation effects the Heatpipe's Qmax, the maximum watt that particular pipe can carry.</p>
<p>So, a well design Heatsink + Heatpipe system will stay below that. If you are below that in a well designed system, orientation will not affect it.</p>
| 11285 | Do heat pipes work in any orientation? |
2016-08-30T06:32:56.180 | <p>This morning, as with every morning, I had my coffee. However, today it was burnt, because I slightly overfilled the water. I use an Italian Percolator on a gas top.</p>
<p><a href="https://i.stack.imgur.com/CHetv.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CHetv.gif" alt="Diagram of the Italian Percolator"></a></p>
<p><a href="https://i.stack.imgur.com/ZELnx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZELnx.jpg" alt="Picture of the Italian Percolator"></a></p>
<p>Normally, you put it on to boil, and once you hear the water bubbling, the coffee is ready, and chamber C will be full of coffee.
As you can see, you fill chamber A with water, loosely fill B with coffee, and the water then rises through B, up through the red pipe, and flows over into C. However, as the picture shows, there is a small valve on A. If you fill the water above this valve, you get horrible coffee. When you hear it bubbling, very little coffee has risen into chamber C, and you have to wait a long time for it to do this, while water also escapes from between chambers B and C (which screw together). </p>
<p>Why does this happen? I cannot think of why there should be a valve to release what is presumably steam. Surely the machine should work regardless? Why does the steam have to be released for the mechanism to work properly - is it linked to the pressure of the system?</p>
| |heat-transfer|pressure|heating-systems| | <p>I have never used one, but I read about it on the <a href="https://en.wikipedia.org/wiki/Moka_pot" rel="nofollow">Italian Percolator (Moka Pot) Wikipedia Page</a>.</p>
<p>The valve is a pressure relief just like on a pressure cooker; it is for safety and probably not involved with the burnt result. You may want to clean it with some vinegar to be sure it is in good operating condition, but it is probably fine since other pots have been turning out well.</p>
<p>Further down the wiki page it says:</p>
<blockquote>
<p>When the lower chamber is almost empty, bubbles of steam mix with the upstreaming water, producing a characteristic gurgling noise. This "strombolian phase" allows a mixture of superheated steam and water to pass through the coffee, which leads to undesirable results, and therefore brewing should be stopped as soon as this stage is reached.</p>
</blockquote>
<p>Perhaps filling the water higher changes your timing, and higher temperature steam is reaching your coffee grounds.</p>
<p>If I had to guess, without your anecdote, I would say that for the same amount of time, an overfilled pot would result in a more dilute, lower temperature brew, and an underfilled pot would result in dark burnt brew.</p>
<p>You may have to time yourself, sacrifice a few cups, and/or get the <a href="http://rads.stackoverflow.com/amzn/click/B018QHQSB8" rel="nofollow">thermocouple meter</a> out ;-) Good Luck!</p>
| 11298 | How does the Italian Percolator work? |
2016-08-30T13:32:33.817 | <p>To maximize the piezoelectric transducer reception/transmission power we must choose the correct backing/matching layer materials, I'm working on a project to measure a steel tank liquid level (non invasive from the bottom, sensor must be glued) and we have the electronic board finished and the <a href="https://www.steminc.com/PZT/en/piezo-electric-disc-50x21mm-r-1-mhz" rel="nofollow noreferrer">ceramic piezoelectric disk</a>, but we are lacking of knowledge of choosing this material that I mentioned.</p>
<p>The tank that we'll measure is a 3 meters tall, 8mm steel bottom wall.</p>
<p>As far as I read the material selection must take in consideration the acoustic impedance, am I right? Also someone recommend using closed cell foam as backing material and stainless steel as matching layer, but we're not sure about the thickness.</p>
<p>Do anyone have recommendations regarding the correct materials/layer thickness for this application?</p>
<p><a href="https://i.stack.imgur.com/yFq68.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yFq68.png" alt="enter image description here"></a></p>
| |mechanical-engineering|electrical-engineering|materials|acoustics| | <p>I am assuming that you are using the <a href="http://www.ti.com/product/TDC1000-Q1" rel="nofollow noreferrer">TDC1000 Analog Front End (AFE)</a>. This is because you are using an image snippet from the the image below. </p>
<p><a href="https://i.stack.imgur.com/hUq5N.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hUq5N.jpg" alt="enter image description here"></a></p>
<p>If so I suggest that you use <a href="https://en.wikipedia.org/wiki/Cyanoacrylate" rel="nofollow noreferrer">cyanoacrylate</a> (CYA) glue also known as crazy glue. It is important that the transducer has good mechanical coupling to the stainless steel material. I have tried a material similar to closed cell foam and did not have much success. I believe this is because the foam material attenuates the acoustic signal. You also might want to consider roughen the attachment area with a sandpaper. Below are some images to help you. </p>
<p><a href="https://i.stack.imgur.com/a2EVa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a2EVa.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/cwGpt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cwGpt.jpg" alt="enter image description here"></a></p>
<p>Also consider adding some hot glue around the transducer. </p>
<p><a href="https://i.stack.imgur.com/Otafs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Otafs.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/thqDI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/thqDI.jpg" alt="enter image description here"></a></p>
<p>Completed assembly</p>
<p>Also you might need a High Voltage boost circuit to measure 3 meters of liquid level. Note density might also be a factor. </p>
<p>One other factor to consider in improving piezoelectric transducer reception/transmission power is to create a cavity as shown in the below images</p>
<p><a href="https://i.stack.imgur.com/n9Fhm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n9Fhm.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/7eEHq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7eEHq.jpg" alt="enter image description here"></a></p>
<p><strong>References</strong></p>
<ul>
<li><a href="http://www.ti.com/lit/an/snaa266a/snaa266a.pdf" rel="nofollow noreferrer">How to Select and Mount Transducers in Ultrasonic Sensing for Level Sensing and Fluid ID?</a></li>
<li><a href="https://engineering.stackexchange.com/a/2697/110">Can I use a ultrasound sensor to measure water level?</a></li>
<li><a href="https://engineering.stackexchange.com/q/1883/110">What is Radial and Thickness Mode Vibration in a Piezo Electric Ceramic Disc Transducer?</a></li>
</ul>
| 11306 | Piezoelectric matching/backing layer materials |
2016-08-31T17:30:34.840 | <p>I was watching a video on the ti products out currently for wireless microcontrollers, and they mentioned a "multi-standard" CC2650 mcu here <a href="http://bcove.me/uekqgi7q" rel="nofollow">http://bcove.me/uekqgi7q</a>. What makes a microcontroller "multi-standard"?</p>
<p>Thanks!</p>
| |electrical-engineering|embedded-systems|wireless-communication| | <p>I think "multi-standard" for CC2650 that the microcontroller support, BLE, Zigbee, 6lowpan, and RF4CE. It is the same SOC, and you load different software stacks support different wireless protocols. </p>
| 11317 | What is a multi-standard mcu? |
2016-08-31T19:34:28.647 | <p>If I have a small gear that I am turning a large gear with, from what I know that gives me the ability to move heavier things. Like a winch of sorts. Why does turning that little gear allow that mechanical advantage. I saw a video that said gears were really lots of levers. I can’t visualize the lever.</p>
<p>In a car, for example, why is first gear more powerful that fifth for initial movement? I don't know how to visualize it. In order for me to lift a car with a lever, I need a fulcrum close to the car and an effort arm fairly long (I guess). So where's the fulcrum and effort arm in the gear moving the car here?</p>
<p>Is my question worded correctly for what I ask in the body?</p>
<p>edit: Am I correct that gears in a car transmission are mostly for ratio aspect of things? if I look at two gears together, I can see the ratio and one turning twice as much as another. But what is missing is the longer arm that eased the effort. Is that "effort arm" replaced with a strong source of power (an engine) to make up for the missing long effort arm?</p>
<p>edit: This is the video I saw that made me think about it, <a href="https://youtu.be/odpsm3ybPsA?t=3m5s" rel="nofollow">https://youtu.be/odpsm3ybPsA?t=3m5s</a> I started it at the point I'm asking about. It's pretty quick for the part I'm talking about, maybe 2 minutes. You see that guy turning the crank to move that little gear? See the long crank, the "effort arm"? That looks like a lever to me.</p>
<p>Then I saw images somewhere, that had similar ideas and it was showing the crank, like on a well crank, being the "circle" or wheel moving the axle which turns the small gear.</p>
<p>In a car though you don't have that big wheel to turn first. You have the small drive shaft. Is there a bigger wheel or gear I am missing? One that is rotating before the first small gear that turns the big one? edit: Is this the flywheel?</p>
| |mechanical-engineering|gears| | <p>The work-energy principle says that the change in energy in a system is equal to the work done to the system. Conservation of energy means that the sum of inputs must equal the sum of outputs.</p>
<p>Work, in turn, is a force times a distance. That is, $W = Fd$. </p>
<p>So, if the sum of inputs and sum of outputs have to be equal and there's no friction, then:</p>
<p>$$(Fd)_{\mbox{in}} = (Fd)_{\mbox{out}}$$</p>
<p>So, let's talk about the lever first. Assuming both ends of the lever are rigid, motion of certain distance on the input <em>requires</em> motion of a certain distance on the output. How far the output moves for a particular input motion is determined by the location of the fulcrum. FYI, the <em>study</em> of how physical constraints determine input and output motions is called "kinematics."</p>
<p>So, if the input and output distances are fixed by physical constraints, and the input force is a given (you supply a force of $X$), then the only thing in the equation that can change to "balance" the input work and the output work is the output force. </p>
<p>That is, the output force varies as required to keep $(Fd)_{\mbox{out}}$ equal to $(Fd)_{\mbox{in}}$.</p>
<p>Hopefully this all makes sense so far. </p>
<p>A lever doesn't actually move strictly up-and-down, though. It <em>rotates</em> about the fulcrum. The actual distance the input traverses is $L_1\theta$, and the output moves $L_2\theta$, where $L_1$ is the length of the lever from the input side to the fulcrum, $L_2$ is the length of the lever from the output side to the fulcrum, and $\theta$ is the angle of how much the lever rotated. </p>
<p>Define the <em>arc length</em>, or distance actually traveled by the input or output end of the lever to be $s$. The input moves:</p>
<p>$$
s_1 = L_1\theta \\
$$
The output moves:
$$
s_2 = L_2\theta \\
$$</p>
<p>If you divide the output by the input, you can see that:</p>
<p>$$
\frac{s_2}{s_1} = \frac{L_2\theta}{L_1\theta} \\
$$</p>
<p>The thetas cancel, and you're left with:</p>
<p>$$
\frac{s_2}{s_1} = \frac{L_2}{L_1} \\
$$</p>
<p>which can be restated as:</p>
<p>$$
\boxed{s_2 = \left(\frac{L_2}{L_1}\right)s_1} \\
$$</p>
<p>The output distance traveled is equal to the input distance times the <em>ratio</em> of lever arm lengths. You can plug this back into the work equation:</p>
<p>$$
F_1 s_1 = F_2 s_2 \\
F_1 s_1 = F_2 \left(\frac{L_2}{L_1}\right)s_1 \\
$$
Cancel the $s_1$:
$$
F_1 = \left(\frac{L_2}{L_1}\right)F_2 \\
\boxed{F_2 = \left(\frac{L_1}{L_2}\right)F_1} \\
$$</p>
<p>So, these two boxed equations show that the output <em>distance</em> changes by $L_2/L_1$, but the output <em>force</em> changes by $L_1/L_2$. The output force can <em>only</em> go up if the output distance goes down, and vice-versa. This ability to "exchange" force for distance is referred to as "mechanical advantage."</p>
<p>Now, considering this, where a lever can't flip "around-the-world" because it would hit the ground or fall off the fulcrum, a pulley or gear <em>can</em> rotate continuously.</p>
<p>Where before, for the lever, the amount of output motion was dependent on the lengths of the lever arms, here the "levers" are actually gears, and their "lengths" are their radii. </p>
<p>That is, just like before:</p>
<p>$$
s_2 = \left(\frac{r_2}{r_1}\right) s_1 \\
$$</p>
<p>The only difference here is the change from lengths to radii. The total angular distance traversed, $s$, still keeps the same form.</p>
<p>So, if the output gear's radius is very large and the input gear's radius is very small, you get:</p>
<p>$$
s_2 = \left(\frac{\mbox{big}}{\mbox{small}}\right)s_1 \\
s_2 = \left(\mbox{really big}\right) s_1 \\
$$</p>
<p>So now, revisiting the work equation:</p>
<p>$$
W = Fd \\
$$</p>
<p>but, as discussed, the distance traveled isn't quite linear, it's the <em>arc</em> the lever takes about the fulcrum, because the lever <em>rotates</em> about the fulcrum. So you could say instead, that:</p>
<p>$$
W = Fs \\
$$</p>
<p>But, from the definition of arc length:</p>
<p>$$
s = L\theta \\
$$</p>
<p>so, you could substitute:</p>
<p>$$
W = FL\theta \\
$$</p>
<p>You can view or group this two ways - the first is as I did the substitutions here:</p>
<p>$$
W = F(L\theta) \\
$$</p>
<p>but, you could also group that to read:</p>
<p>$$
W = (FL)\theta \\
$$</p>
<p>What is a force times a lever arm? A torque. So you can rewrite the equation as:</p>
<p>$$
W = \tau \theta \\
$$</p>
<p>This is the analogy between linear systems and rotational systems - a force is akin to a torque, and a distance is akin to an angular span. Important to note here that $\theta$ isn't the particular angle the system is currently at, it's the angle <em>through which</em> the system traversed. That is, $\theta = \theta_2 - \theta_1$. </p>
<p>Anyways, hopefully the explanation of how the linear (lever) and rotational (pulley or gear) frames are related makes sense. </p>
<p>I think, more to your question, the "force multiplier" effect that you get with a lever arm, or gear, etc. is a <em>tradeoff</em> between applied force and applied <em>distance</em>. </p>
<p>You do the same amount of work to lift a 1000 pound rock up 1 foot as you do to lift a 1 pound rock up 1000 feet. If you don't have the strength to lift the 1000 pound rock directly, you can use <em>mechanical advantage</em> to <em>trade</em> the 1000 pounds for the 1000 feet.</p>
<h2>:EDIT:</h2>
<p>I drew a picture that hopefully illustrates the relationship between gears and levers. A lever has a fulcrum, which provides reaction forces, and a length on either side of the fulcrum. </p>
<p>A gear (pulley, etc.) is like a lever with equal lengths on either side of the fulcrum that is lashed/tied/bound to another lever. </p>
<p><a href="https://i.stack.imgur.com/hoRs3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hoRs3.png" alt="Gear Transmission"></a></p>
<p>The lashing that "ties" the two levers together is referred to as the gear mesh. Two teeth come into contact with one another, and that physical contact causes one "lever" (gear) to <em>push</em> the other. </p>
<p>I'll add a little more information, in the hopes that more detail will help cement the analogy rather than confuse you. Just like the example I've mentioned - two levers lashed together, if you lash them together too tightly then they're not actually able to move at all. The same can happen with a gear mesh - if the gears are too close together, the mesh is too tight and the assembly won't spin. </p>
<p>Conversely, if the lashing that binds the levers is too loose, then when you change direction there will be some dead band where the lashing is sagging. The input is able to rotate freely before the lashing snaps taught again, at which point the output starts to move. Again, the same thing happens in real gear systems - if the gears are too far apart, or the teeth are too narrow, then there is a void between one pair of teeth and the next. This is referred to as <a href="https://en.wikipedia.org/wiki/Backlash_(engineering)" rel="nofollow noreferrer">backlash</a>. </p>
<p>Again, this might be too much information, but my hope is that you can understand that two gears are like two levers that have been tied together. If the binding (backlash) is too tight then the gears can't move, but if it's too lose then the output tends to get jerked around a lot as the "lashing" goes through the slack-taught-jerk-slack cycle, like a series of flicks instead of a continuous push.</p>
| 11320 | Why do small gears used to turn big gears allow more force with less effort? |
2016-08-31T19:38:51.583 | <p>What shape of object will roll down the fastest through the decline piece of wood if travelling with a straight line assuming there are frictionless ramp to stop them from going to the side given the same amount of material with their smoothness being the same? Could it be sphere or a partial sphere with cross section trimmed off? Please also including the distribution of the mass. Is the mass focused in the center given much faster velocity to roll down? What about the 3D shape generate from other curve for example, elliptical or hyperbola, etc...</p>
<p>Please show your math formulation...</p>
<p>In addition, I know that a hollow cylinder will roll down slower a solid-centered cylinder.</p>
<p>interesting video: <a href="https://www.youtube.com/watch?v=cB8GNQuyMPc" rel="nofollow">https://www.youtube.com/watch?v=cB8GNQuyMPc</a></p>
<p>Interesting thing to learn from this video is that the density doesn't matter, bigger cylinder VS smaller cylinder are the same, even the weight doesn't matter in his experiments... </p>
| |mechanical-engineering| | <p>A solid cylinder with a protrusion like a cam if positioned such that the rolling down starts from the inclined tip of the cam will go down faster. The cam should be formed sharp so as not to change the rotational inertia, J, of the mass too much. The cylinder should be placed in an initial tilted imbalance, ready to fall down.<br>
The length of the ramp should be long enough to let the cylinder roll several times. This will average out the wobbling up and down of CG of the roller.
This geometry will promote rapid start of rotation and faster roll down. </p>
| 11321 | What shape of object will roll down the fastest through the decline piece of wood on Earth? |
2016-09-02T14:56:54.387 | <p>According to this figure whats the difference between leading edge flap and slat and why the slat come with plain wing and TE flap come alone?
thanks.</p>
<p><a href="https://i.stack.imgur.com/Webiz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Webiz.jpg" alt="enter image description here"></a></p>
| |aerospace-engineering|aerodynamics| | <p>This comes down to the fact that flaps are fundamentally changing the geometry of the wing so with flaps deployed the wing is a different shape with correspondingly different characteristics. So when a flap is deployed it is essentially a modification of the main wing. </p>
<p>On the other hand the primary role of slats (distinct form LE flaps) is to control the airflow over the wing, specifically to delay flow separation, and as such don't directly contribute lift themselves but allow the main wing to operate at a higher angle of attack before it stalls. </p>
<p>Some aerofoils also have slots which allow airflow between low and high pressure sides of the wing with the intention of either reducing drag or keeping the airflow attached or both. It's not uncommon to see flaps with slot gaps between the training edged of the main wing and the leading edge of the flap itself. </p>
| 11348 | Whats the difference between leading edge and slat? |
2016-09-04T00:30:48.187 | <p>I am working on a project where 1L of cold water (20C) has to be heated (80C) very quickly (30 seconds). </p>
<p>There were a few designs I was considering</p>
<ol>
<li>Heater cartridge (similar to what you would find in a 3D printer) - I can get up to 800W</li>
<li>Peltier - 10 <a href="http://rads.stackoverflow.com/amzn/click/B00IKDL22O" rel="nofollow">peltiers</a> would add up to about 700W, but would require a power supply that could handle 60A, which would be difficult to find for a reasonable price</li>
<li>Nichrome wire - The challenge with nichrome is how to transfer the heat to water. I was thinking of getting insulated nichrome and rapping that around a metal container that would hold the water. </li>
<li>Blow torch or other flame - If this is the best method, I would need help calculating the 'wattage'. </li>
</ol>
<p>I don't think any of these methods would work because the wattage is too low in any manner. I know water heating is typically measured in kW, but on 120V AC, the circuit would break at 3.6kW (30A circuit breaker). </p>
<p>So my question is what is the best method of heating the water with the given parameters (not limited to the ideas I had)?</p>
| |thermodynamics|heat-transfer| | <p>The specific heat of water is 4.186 J/g°C, so heating 1000 grams of water by 60 °C requires about 250 kJ of energy. Applied over a 30-s period, this represents almost 8.4 kW of power — assuming that none of the power is lost elsewhere along the way.</p>
<p>One approach would be to pre-store the heat that you need for the water in some sort of thermal reservoir. For example, you could have a large block of metal (aluminum is cheap and has a high specific heat of 0.900 J/g°C) that you keep at 80 °C (or somewhat higher) using a heater and a thermostat. The total thermal capacity of the reservoir needs to be significantly higher than that of the water by about 10:1, so you'd need about 50 kg of aluminum. Pass the water through channels in or over the surface of the aluminum, and it will heat up in short order.</p>
<p>You could also use a reservoir of hot water (10 l or so) and pass the water you want to heat through a heat exchanger (e.g., coiled tubing) that's submerged in the reservoir.</p>
| 11353 | How do you quickly heat cold water? |
2016-09-04T18:40:39.393 | <p>I am creating a 24 hour blood pressure monitor that is designed to be sleek and comfortable, I have decided on The Honey Well Series NBP AN transducer (see more info here: <a href="http://www.mouser.com/ds/2/187/honeywell-sensing-basic-board-mount-pressure-senso-740338.pdf" rel="nofollow">Basic Board Mount Pressure Sensors</a>, p13) I want to connect that to a board so that the data can be transmitted wirelessly, or by cable to a phone or computer, how would I go about doing that? </p>
| |electrical-engineering|pressure| | <p>Since your sensor has an analog output. If your comfortable with prototype level work, get an Arduino that has analog inputs then connect a bluetooth module such as the HC-05. Then, you can transmit that data using some code that I wrote to send multiple sensors in columns. You can modify this code to your needs.</p>
<p><a href="http://www.jmaturner.com/sensors.html" rel="nofollow">sending sensor data - serial over bluetooth</a></p>
<p>Your next task would be to write software, but you can view this data on a PC or phone using a terminal program like Teraterm.</p>
| 11358 | Trasmitting data from pressure transducer to phone/web server |
2016-09-05T16:24:06.867 | <p>A translator <a href="http://www.multitran.ru/c/m.exe?a=4&MessNum=336813&l1=1&l2=2" rel="nofollow noreferrer">posted</a> this diagram of a weld joint on a translator's forum (the figure below). What does the abbreviation "B.C." mean on the diagram?</p>
<p><a href="https://i.stack.imgur.com/GQS5n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GQS5n.png" alt=""></a></p>
<p><a href="https://i.stack.imgur.com/ImM77.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ImM77.jpg" alt=""></a></p>
<p><a href="https://i.stack.imgur.com/mMdu0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mMdu0.jpg" alt=""></a></p>
| |terminology|welds| | <p>I'm not aware of BC as standard English nomenclature for welding, but I have a guess for what they mean. The fact that there is a period after the B but not the C makes me wonder if something was cut off or whited out in the original.</p>
<p>I think they mean backgouge (B<strong>G</strong>) which would be a common process in a double sided vee groove weld it it needed to be a complete joint penetration (CJP) weld. This is sometimes indicated with a dotted line like that, although strictly of they backgouge as shown they should have drawn a u groove weld instead of a vee groove on the top side.</p>
<p>There welder will bevel the bottom side using a grinder or plate beveler. Then they will weld the bottom side, but the root of the bottom weld will be contaminated by air from the other side of the weld since there is no backer or backing gas and thus no shielding on the far side. On the top side, using a grinder or a carbon electrode they will remove all of the contaminated root metal and then begin the seconds side weld. This weld if protected on the back side by the metal of the first weld, yielding a sound weld through the whole depth of the joint.</p>
| 11367 | What does the abbreviation "B.C." mean on a weld joint diagram? |
2016-09-06T07:15:13.193 | <p>I'm doing plant design and one of the reactors is a furnace that uses methane to burn. I want to calculate the amount of methane that is required. I have the energy needed and the calorific value of methane but now I need a average thermal efficiency. Just to be clear I want the ratio of energy produced over how much was put in. Thanks</p>
| |chemical-engineering| | <p>The efficiency depends on how well-insulated the reactor is, and how complete the burn is. But broadly speaking, something between 80% and 100%. It also depends on whether you're recovering heat by condensing the water vapour - if not, you want to use the LHV (rather than the HHV) of methane (or equivalently, de-rate the HHV by 10%)</p>
| 11374 | Thermal efficiency of a furnace reactor? |
2016-09-07T02:37:13.727 | <p>Much power is used to get rocket off the platform on the initial launch? Could electric engines used like booster rockets work similar in the way they launch planes of an aircraft carrier? The electric turbines would disconnect at the top of the launch platform that could also be made taller. Could it save fuel by creating lift for the first 1000 ft or so?</p>
<p><a href="https://i.stack.imgur.com/UYZFc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UYZFc.jpg" alt="enter image description here"></a> supplement </p>
| |electrical-engineering|aerospace-engineering|turbines|experimental-physics| | <h1>Electric turbine engines</h1>
<p>First off, "electric turbine engines" don't really exist. What you're probably thinking of is an electric version of the <a href="https://en.wikipedia.org/wiki/Jet_engine#Turbine_powered" rel="noreferrer">jet turbine engine</a>, which is a type of combustion engine (it burns fuel). Air is sucked in the front, compressed by a compressor, then mixed with fuel and burned in a combustion section. A turbine extracts some of the energy from the air and uses it to power the compressor, while the rest of the air shoots out the back of the engine to provide thrust. The important thing to understand is that the turbine is extracting energy from the air, not generating thrust. The burning fuel adds energy to the air which is converted into thrust by the structure of the jet engine. The turbine is just a way of making the jet engine more compact, so it doesn't need a separate power source for the compressor.</p>
<p>The closest electric engine equivalent that I could find would be something like the <a href="https://en.wikipedia.org/wiki/Airbus_E-Fan" rel="noreferrer">Airbus E-Fan</a> (see also the <a href="http://www.airbusgroup.com/int/en/corporate-social-responsibility/airbus-e-fan-the-future-of-electric-aircraft/technology-tutorial/E-Thrust.html" rel="noreferrer">Airbus website</a>). It uses an electric motor to spin a fan, providing thrust. It does not use a turbine, it's more like a fancy propeller.</p>
<p>Apparently <a href="http://www.scientificamerican.com/article/impossible-electric-airplane-takes-flight/" rel="noreferrer">Airbus is hoping to use the E-Fan on commercial passenger jets in the future</a>, albeit using a gas-powered generator to provide electricity, so it might be able to provide enough thrust to launch a small rocket.</p>
<h1>Will this work for your application? (Edit: Space Shuttle booster replacement)</h1>
<p>There are two problems with replacing the solid fuel booster rockets of the space shuttle with electric fans:</p>
<ol>
<li><p>The booster rockets generate <a href="https://en.wikipedia.org/wiki/Space_Shuttle_Solid_Rocket_Booster" rel="noreferrer">13,800 kN of thrust <em>each</em> at liftoff</a>, for a total of 27,600 kN. By comparison, the E-Fan only produces 750 N, therefore you would need 37,000 electric engines to get the same thrust.</p>
</li>
<li><p>The booster rockets burn for 2 minutes and are ejected when the shuttle reaches an <a href="https://en.wikipedia.org/wiki/Space_Shuttle#Solid_rocket_boosters" rel="noreferrer">altitude of 46 km</a>. At that height the atmosphere is <a href="https://en.wikipedia.org/wiki/Atmosphere_of_Earth#/media/File:Comparison_US_standard_atmosphere_1962.svg" rel="noreferrer">so thin</a> that an E-Fan wouldn't be able to produce any thrust. In fact, since the E-Fan works by "pushing" air through its duct, its performance would decrease as the atmosphere thinned, whereas the solid boosters do not need air to operate.</p>
</li>
</ol>
<h1>Conclusion</h1>
<p>You can't replace the space shuttle solid rocket boosters with electric fans/engines, even if you could solve the battery/fuel problem. The electric fans simply do not produce enough thrust. Even if they did they still require an atmosphere to work properly, but the space shuttle needs its boosters to take it most of the way out of the atmosphere. Neat idea, but I'm afraid it just won't work.</p>
| 11390 | What prevents the use of electric engines to assist with spacecraft launches? (Space Shuttle Booster Supplement) |
2016-09-07T17:50:28.137 | <p>Defined the <strong>Gear Ratio</strong> ($i$) as the ratio between the number of teeth on the pinion (driving gear) ($N_{1}$) and the number of teeth on the gear (driven gear) ($N_{2}$). </p>
<p>Why is this better, if better at all, having non-integer values of $i$? Could anyone please illustrate using arguments exploring the fabrication and performance points of view?</p>
| |mechanical-engineering|design|applied-mechanics| | <p>Non-integer gear ratio reduces gear noise and improves service life. Micromo has done a nice job of explainining why [<a href="https://www.micromo.com/media/wysiwyg/Technical-library/Gearheads/25_gearhead_construction_use.pdf" rel="nofollow noreferrer">1</a>]. In short: Assume a gear tooth on gear A is faulty. Let's say the gear ratio is 3:1 (See picture 1). Then, this faulty tooth engages with three locations on the larger gear (gear B) at every rotation of the gear. Imagine this in the long run and it will cause wear only on those three locations. Now imagine the gear ratio was 3.444:1 (See picture 2). then the engaging locations on the larger gear (gear B) keeps shifting at every rotation (see the picture where the numbers on the wheel shows engagement at that number of rotation of gear A), hence wear is distributed among all gear B teeth and each tooth takes less wear in the long run thus more service life.</p>
<p><a href="https://i.stack.imgur.com/Rm7cr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rm7cr.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/dChTq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dChTq.png" alt="enter image description here"></a></p>
| 11394 | Why are spur gear ratios not always whole numbers in practice? |
2016-09-07T23:53:09.057 | <p>I am not sure if this is the right place to ask it, but this is a question that I thought today, and it gave me some curiosity to understand. Imagine that a car will curve, we can say the turn is a bit tight , what are the factors that can help it to flip? I was wondering about some aspects:</p>
<p>if the car has mass, it has inertia, so while it is curving it tends to keep the motion in the same direction that it was instants before the turn. right? So, if the car has more mass, it has more inertia, and since there is friction, one heavier car would flip easier then one lighter, considering that all other possible variables were equal.</p>
<p>Center of gravity, a car with an higher center of gravity would flip easier. The whole inertia of the car distributed to higher heights would be further of the tires(where friction acts), creating angular momentum.</p>
<p>The car being thin because it has less surface contact with the ground;</p>
<p>The car being lighter. This opposes what I've said in "a)", but a heavier car is more difficult to get off the road. A lighter car has more instability.</p>
<p>Am I wrong? In what I'm wrong? What do you think?</p>
<p>Thanks for helping. :)</p>
| |mechanical-engineering| | <p>When a car makes a turn with a radius of R it is subject to an acceleration force, vector, called centripetal force $$ F=m V^2/R $$<br>
this force is applied to the center of gravity of the car and is directed horizontally opposite of center of turn.<br>
This Vector adds to the vector of weight of the car which is directing vertically down. </p>
<p>The sum of these two vectors is a vector drifting out and down diagonally, call it: R fore roll .<br>
If R falls outside of the rectangular area between four wheels the car will start to roll. But it takes a few seconds for it to start from just lifting the inside tires off the road to pass the critical angle of no return. Experienced drivers could maneuver the car back to balance during those early moments. </p>
<p>A car with lower CG and wider wheel-base needs much larger centripetal force to go off balance.
A car with stronger roll control rods will be more stable but will have a hard ride. Fancy cars have computer controlled roll prevention!<br>
having worn out tires on the rear promotes skidding and roll. Road condition on a tight turn such as bumps or puddles promotes roll. Finally sudden jerky braking to avoid a pot-hole is dangerous too.
Some race cars have suspension designed to lower the CG and open the wheel-base in a tight roll!</p>
| 11401 | What aspects can help a car flip? |
2016-09-08T09:43:13.013 | <p>I've recently been on <a href="https://en.wikipedia.org/wiki/R%C3%BCgen" rel="noreferrer">Rügen</a> and I have seen this:</p>
<p><a href="https://i.stack.imgur.com/DkHq6.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/DkHq6.jpg" alt="enter image description here"></a></p>
<p>Does anybody have an idea what this is?</p>
<p>(To give Google a chance: A ship with four "towers" made out of steel on it. They looks a bit like high-voltage electrical towers.)</p>
| |ships| | <p>It's a <strong>heavy-lift vessel</strong>, carrying four identical structures as cargo.</p>
<p>It's very likely they are <a href="http://www.4coffshore.com/windfarms/jacket-or-lattice-structures-aid5.html" rel="noreferrer">jacket foundations for offshore wind turbines</a>. The excellent <a href="http://www.4coffshore.com/windfarms/vessels.aspx?catId=3" rel="noreferrer">4C Offshore has a database of such vessels used on offshore wind farms</a>. It also has a list of all offshore windfarms under construction by country, and tells you which vessels are working on it, so given that you know where the photo was taken, you may able to work out which farm it's supplying.</p>
<p>I think they're going to <strong><a href="http://www.4coffshore.com/windfarms/wikinger-germany-de47.html" rel="noreferrer">Wikinger</a> offshore wind farm</strong>. That's using jacket foundations, is being built as I write this post, and is <a href="http://www.4coffshore.com/offshorewind/index.html?lat=54.834&lon=14.068&wfid=DE47" rel="noreferrer">about 20 km north-east of Rügen</a>.</p>
<p>Here's a jacket foundation <em>in situ</em> at <a href="http://www.4coffshore.com/windfarms/alpha-ventus-germany-de01.html" rel="noreferrer">alpha ventus windfarm</a>, courtesy of <a href="http://eandt.theiet.org/magazine/2014/05/windfarm-foundations.cfm" rel="noreferrer">the IET</a>:
<a href="https://i.stack.imgur.com/t8aoz.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/t8aoz.jpg" alt="enter image description here"></a></p>
| 11409 | What kind of ship is this? |
2016-09-08T20:30:42.273 | <p>I am a fairly recent graduate (I graduated December of 2015 with a Bachelor's of Science in Computational Science) and since graduating I have had an increasing desire to get involved in engineering. </p>
<p>What areas of engineering employ aspects of computational science?</p>
| |mechanical-engineering| | <p>There are several places Computational Science and Mechanical Engineering overlap. Computation is a tool that many engineers use on a daily basis to make design decisions.</p>
<ol>
<li><a href="https://en.wikipedia.org/wiki/Computational_fluid_dynamics" rel="nofollow">Computational Fluid Dynamics (CFD)</a>- Learn <a href="http://www.openfoam.com/" rel="nofollow">OpenFOAM</a> and do some contracting.</li>
<li><a href="https://en.wikipedia.org/wiki/Finite_element_method" rel="nofollow">Finite Element Analysis (FEA)</a>- Learn OpenFoam or <a href="http://gmsh.info/" rel="nofollow">GMSH</a> and do some contracting.</li>
<li>Computation is the only way to solve many ugly differential equations, and a much easier route to solving many of the solvable ones. Engineers know math, but we don't necessarily like math and are not awesome at it. Large engineering firms benefit from in-house math, computational, and data handling expertise.</li>
<li>Find a research firm or supercomputing contractor that hires lots of engineers and computational scientists.</li>
<li>Any other <a href="https://en.wikipedia.org/wiki/Computational_science#Related_fields" rel="nofollow">computational science related field</a></li>
</ol>
| 11420 | What are some overlaps between Computational Scientist and Mechanical Engineering? |
2016-09-08T21:01:05.640 | <p>I am unsure of how to interpret the uncertainty given for a Multimeter I have been using. Everything is listed as follows ±(0.8%+10). I have never seen uncertainty listed this way, with a +"x" after the percent uncertainty. What range does this correspond to exactly, -10.8% to 10.8% or is it -0.18% to 0.18%? Here is a link to the datasheet: <a href="http://www.smart-prototyping.com/image/data/3_equipment/1_electronic/multimeters/100409%20-%20Victor%20VC830L%20Manual.pdf" rel="nofollow">Digital Multimeter Datasheet</a>. Thanks!</p>
| |electrical-engineering| | <p>It means that there are two parts to the error. One part is proportional to the reading (0.8%), and the other part is a absolute error (10).</p>
<p>For example, if a reading is 174, then there is 174 x 0.8% = 1.4 proportional error, plus the 10 fixed error for a total error of 11.4. A reading of 174 therefore means the actual value is in the range of 162.6 to 185.4.</p>
| 11421 | How to interpret this Uncertainty? |
2016-09-09T23:09:23.537 | <p>So I'm trying to create a system that automatically runs evaporative coolers to keep a warehouse at a desired temperature, given the internal and external temperature and humidity. I can get the coolers to run automatically and read the data I want quite easily.</p>
<p>Here's what I need to know: is there a way to calculate how much a cooler cools down a room? I can find out the volume of the room and the airflow of the coolers, so an equation that requires those will be fine. Is there an equation that describes how much a cooler cools a room down?</p>
| |airflow|cooling| | <p>The capacity of an evaporative cooler depends on the relative humidity. The lowest theoretical temperature that the cooler can reach is the wet bulb temperature.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Evaporative_cooler" rel="nofollow">wikipedia article</a> has some discussion on the design and shows how it works with psychometric charts.</p>
<p>For the design you'd look at a 'typical' hot day, compare wet bulb to dry bulb for that day and assume a certain efficiency (e.g. it reaches 80% from dry bulb to wet bulb). The energy in the supply air stream would be:
$$
Q = \dot{m}\Delta_Tc_p\eta
$$
i.e.:</p>
<p>Energy = massflow x temperature drop x specific_heat x efficiency</p>
<p>$\Delta_T$ being the temperature difference before and after the evap cooler using your assumptions.</p>
<p>$Q$ would be your cooling capacity</p>
<p>HTH</p>
| 11442 | How much does an evaporative cooler cool down a room |
2016-09-10T04:25:08.780 | <p>The core question is "What Ingress Protection (IP) rating is required for an item to be dishwasher safe?" Is the answer different for residential versus commercial dishwashers? Is the practical residential answer different than the regulated answer (e.g., for food prep or medical settings). Does the use of dishwashing detergent invalidate the rating in practice? </p>
<p>Note, IP ratings are not cumulative from IPX6 (jets) or less to IPX7 (immersion) or more, so an item might technically need a dual rating, but I've rarely seen that in my very limited research. Are both needed? </p>
<p>The difficulty of performing some tests appears to vary (e.g., 1 meter immersion is pretty simple compared to water jet set at a specific flow rate, angle, and turning rate). Might this lead to some certifications not being sought?</p>
<p>My most immediate interest is shopping for a cooking thermometer, but it seems applicable to a variety of other equipment like small speakers (picnic disaster) and phones (baby vomit) and keyboards. I know at least some hospital/medical keyboards are specified as IP66 or IP67 and also washable and dishwasher safe (and presumably "coffee-safe" and "Coke-safe"). Some flexible membrane keyboards are described similarly but are not rated.</p>
<p>Simple queries such as "What IP rating is dishwasher safe" or "Is IP66 dishwasher safe" were not directly fruitful. Several manufactures describe some of their products as dishwasher safe that are rated variously at IP66 or IP67 or IP57 (but not other comparably rated products even from the same manufacturer).</p>
| |materials| | <p>What you've found is, pretty much, a good guide to the situation.</p>
<p>IP ratings are not designed to measure whether something is dishwasher safe or not. That's not their purpose. And so they just don't map onto whether or not something is dishwasher-safe or not. That is to say, something might be IP66K and not be dishwasher-safe; something else that is IP55 might be dishwasher-safe.</p>
<p>So I can see how one might expect "what IP-rating is dishwasher-safe?" to be meaningful question - but it's not. It's a different kind of rating.</p>
| 11446 | IP Rating Required to be Dishwasher Safe (Commercial and Residential) |
2016-09-10T14:37:04.407 | <p>Consider I have a pole stuck in the ground as illustrated below:</p>
<p><a href="https://i.stack.imgur.com/kTPSN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kTPSN.png" alt="enter image description here"></a></p>
<p>When a force F1 is exerted on the pole, its bottom end will push against the ground it's stuck in; the ground then exerts a resistance force. Intuitively, if F1 is very strong then the pole's upper end will move in F1's direction and if F1 is very weak, the ground resistance Fr keeps the pole in place.</p>
<p>Given some measures of the pole, specifically, how deep it is (L2) and what radius r it has (it is a cylinder), I would like to be able to determine the magnitude required by F1 to move the pole's end.</p>
<p>I have asked a similar question, with more physical details on another stack exchange site, too. <a href="https://physics.stackexchange.com/questions/279345/compute-forces-exerted-on-a-pole-in-ground">Here</a> is the link.</p>
<p>Update:</p>
<p>Assume that the ground is made of only one material, with uniform density, disregarding any "dirty" effects, such as moisture, or temperature. The length of the pole, its radius (it is a cylinder), and its depth L2 are all constant.</p>
<p>Let's view the pole as a lever with its pivot where air, ground, and pole meet. Then we have F1 * L1 = F2 * L2 when the lever is in balance. Now, I would like to find out at what force F1 the pole starts moving; the pole is moving when F1 * L1 > Fr* L2. </p>
<p>The problem would be solved if we find a means to compute Fr. My idea was that as the pole (in the ground)pushes towards the right at every height 0 > L >= L2 (L = 0 on dotted area where air and ground meet), there is a resistance force Fr(L) at each height (view illustration <a href="https://physics.stackexchange.com/questions/279345/compute-forces-exerted-on-a-pole-in-ground">here</a>). For big L this force is small while for small L the force grows. I assume Fr(L) must grow in a linear fashion as it does with levers (F1 * L1 = F2 * L2). Assuming that, Fr could be computed by F2 = 0∫L2 Fr(L) dL.</p>
<p>The solution need not be that exact, the force need only be computed approximately: the force will be used by game simulation the physics need only appear real. All that is required is a means to compute what resistance force Fr the pulling force F1 must overcome to move the pole.</p>
| |mechanical-engineering|simulation| | <p>Depending on the kind of engineer you ask you will get many different options to approach this - your question is pretty open-ended. As a Geotechnical Engineer I'd say your pole can be looked at as a retaining wall. Now, if the ground is modelled as real "dirt" it will have some give. What would happen is that the pole will rotate about some point in the ground (again - you can go into this in more depth) and also bend (or not if rigid). </p>
<p>Now this rotation causes a the rod to displace the soil, which exerts "passive" or "active" lateral pressure, depending on whether it is stressed further of allowed to relax (i.e. rod moving into soil region or moving away from it). These pressures are given by the coefficients of passive and active earth pressure ($K_{p}$ and $K_{a}$) and the vertical effective stress ($\sigma_{v}' = hg\rho$). The coefficients are found from the angle of friction of the soil ($\phi'$) as:</p>
<p>$$
K_{a} = \dfrac{1-\sin \phi'}{1+\sin \phi'}
$$
and
$$
K_{p} = \dfrac{1+\sin \phi'}{1-\sin \phi'}
$$
So passive pressure would be $P_{p} = K_{p}\sigma'$ and so on. A rigid rod will then rotate around the point where all the forces on it are balanced.
You can assume some value of $\phi'$ and be on your way. These pressure will grow linearly with depth, so the resulting forces from the different active/passive wedges will act at 1/3 from the bottom of each wedge. Everything can now be expressed in terms of the distance between the pivot and the ground level. Equilibrium around the pivot will give you its location and then the sizes of the forces can be found.</p>
<p><a href="https://i.stack.imgur.com/OG7ln.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OG7ln.png" alt="Moments with passive pressure ONLY"></a></p>
<p>The picture above is how I would approach this (neglecting the active soil wedges, if included they mirror the passive ones with a limiting pressure as given above). From equilibrium about the orange point you can find its depth and then the values of the pressures (with appropriate Fr and soil parameter $\phi'$ assumed). Or you can assume a value of x and find the Fr that causes movement. Many options possible.</p>
| 11454 | How much force can a pole in ground take before falling? |
2016-09-11T05:17:37.653 | <p>I am supposed to be re-drawing this image to scale
<a href="https://i.stack.imgur.com/vhBjZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vhBjZ.png" alt="Original Drawing"></a> However when I Try to connect the arc R11 with arc R26 using the radius R22 I get a negative # in my calculations. i.e. R26-R22=(-4)</p>
<p><a href="https://i.stack.imgur.com/JQARV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JQARV.png" alt="Using this formula"></a></p>
<p>Is there a different way to draw this? I have tried nearly everything and have completed the rest of the drawing except this one point, Am I missing something here?</p>
| |mechanical-engineering|design|drafting| | <p>Arc 22 does not connect with arc 26 there is a straight horizontal line in between. The hint to this fact is in the width dimension of 44 that would not exist otherwise. There are further hints in the image itself, the small construction line segment shows that its a round of a corner. With a trained eye you can also see in the image the the curvature ends before connect. I have drawn a clarifying image below.</p>
<p><a href="https://i.stack.imgur.com/ySbBz.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ySbBz.png" alt="enter image description here"></a></p>
<p><strong>Image 1</strong>: Showing construction piece by piece and type (color).</p>
| 11456 | Drawing the Connection of 2 Arcs with Radius |
2016-09-11T15:25:49.287 | <p>Why is it not recommend to have pump shut off head lower than its operating head?
Is the answer related to dropping curve by any chance?
Also why is the discharge valve closed when starting the pump operation?
Any help on this appreciated</p>
| |mechanical-engineering|fluid-mechanics|chemical-engineering|pumps| | <p>To address the first point without further clarification from your side:</p>
<p>I think you mean unstable pump curves with that. Otherwise I wouldn't know how you could achieve a higher operating head than shut off head.</p>
<p><a href="https://i.stack.imgur.com/yA7iR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yA7iRm.png" alt="enter image description here"></a></p>
<p>The result would be an oscillating behavior which is not desired.</p>
<p>For the second part of your question the main reasons I know are</p>
<ul>
<li>lower load on the driver during start up</li>
<li>avoiding recirculation if the operating pressure in the pipe is high</li>
<li>maintaining a constant pressure in the pipe, i.e. the valve is beeing opened when the discharge pressure is equal or exceeds the pipe pressure</li>
<li>avoiding a dry pump, especially if the flow on the suction can be interrupted</li>
</ul>
<p>In general there are a lot of information on the two questions and a good book or two could prove helpful if you would like to achieve in depth knowledge. Although there should be a lot of solid information in the web too.</p>
| 11460 | Centrifugal pumps working |
2016-09-13T14:59:59.553 | <p>I am working on project involving designing a recoil spring coil of diameter 10.2 mm and wire diameter 1.25 mm. I tried using a solid core wire but it failed due inadequate tensile strength. Dimensions can only be altered by 5 mm, which is not sufficient to make the solid core wire work.</p>
<p>My professor suggested looking into braided wire rather than a solid core wire. I considered 6x7 wire rope but they are too big. After some googling, I found out there are recoil springs with braided wires with 3 strands. I know it is going to be somewhat less than 3x the strength individual wires. I searched handbooks, reference books, etc. but I could not find any established relations/formulae between strands and tensile strength. How can I determine the tensile strength of a braided steel wire based on the number of strands?</p>
| |mechanical-engineering|materials|steel|springs| | <p>This is because the strands are continually curved and thus the stress distribution inside of them is higher near the centerline (higher curvature) than on the outside.</p>
<p>In addition, the contact forces add stresses in the off-axis directions which raise the internal stress level beyond $\sigma = \frac{T}{A}$.</p>
<p>The math is quite complex, but not unwieldy. Stranded cables offer superior strength for their weight compared to a solid wire of the same overall diameter.</p>
<p>PS. The curvature of a wire whose centerline prescribes a circle of $r$ and has a helix with pitch (distance/angle) of $p$ is $$\rho = \frac{r^2+p^2}{r}$$</p>
| 11490 | How to calculate tensile strength of braided steel wire? |
2016-09-13T16:27:57.263 | <p>So let's say I have some heat generating machine be it a combustion engine or a fridge cooled by a fluid. I don't have a single radiator that is big enough for the amount of heat being generated but I do have a couple of smaller ones so I guess I could connect them together.
Should I put the radiators in series (one connected to the next one and so on) or in parallel (having the coolant intake pipe split into all of the radiators) and why? What would be the most efficient set up?</p>
| |thermodynamics|heat-transfer|cooling|heat-exchanger| | <p>i think series combition of radiatiors would be good because water has cooled two times in one round.</p>
| 11493 | Radiators in series or parallel? |
2016-09-14T16:05:53.570 | <p>I've seen some of the robotics competitions use pneumatic actuators for brute force actions ( like a hammer or lift ), but nothing precise. Is it possible to use pneumatics in a more precise scenario?, I'm assuming this is how pneumatic robotic arms operate, but I'm a little unclear on how they operate so precisely. </p>
| |control-engineering|robotics|pneumatic| | <p>While pneumatic systems can be controlled in a non-discrete manner as GisMofx mentioned, that is generally not the norm because the control system to achieve that movement is more expensive that comparative electric actuators. The result is still less precise and less responsive. This is the reason that all CNC machines are electric. Pneumatic systems are also much less energy efficient than electrical or even hydraulic power systems.</p>
<p>Pneumatic actuators are very hard to precisely position because air is compressible. The flow has to change with inertia and a varying load adding dependencies to the control logic. Pushing a constant load cart back and forth on a rail is hard enough; dynamically loaded systems would be very difficult and the reaction speed would be much lower than hydraulic or electric actuators. So dollar for dollar, an electric actuator will always be more precise than a pneumatic actuator.</p>
<p>The low-cost pneumatic cylinder is offset by the higher cost associated with more complex programming and the expensive bi-directional proportional valve to control the airflow. Hydraulics also use an expensive bi-directional proportional valve, but are much easier to motion control because the load does not greatly affect the flow.</p>
<p>As a general rule:</p>
<ul>
<li>Electronic for highest precision (discrete or motion control)</li>
<li>Hydraulic for high force (discrete or motion control)</li>
<li>Pneumatic for cheap (discrete)</li>
</ul>
| 11504 | Are precision pneumatics possible? |
2016-09-13T12:57:52.313 | <p>I have engineering data emerging from more or less complex constructions.</p>
<p>On a regular base I need outlines from construction data to perform geometric calculations on them. The construction data comes from different CAD packages (e.g. Inventor or SolidWorks). I basically need certain closed polygons stored in a way I can read it with primitive methods like sloppy designed PERL programs.</p>
<p>Edit: The polygons are planar. I.e. the data I want to process is 2D. Basically it often is an outline of a 3D object, e.g. a housing.</p>
<p>So my favourite input format for such extracted polygon data is a CSV-file or plain text with whatever separators are available.</p>
<p>As most CAD-SW can import 3D-Models saved in STEP, I practically can choose which software I use to save the polygon data. The outcome is, I will probably end up having DXF or DWG files, which are the most primitive CAD-files I can get from that. </p>
<p>This leaves me with getting my polygons out of that goddamn DXF into a CSV file. Does anyone have an (preferably) easy way to do that?</p>
<p>Addendum: For increased clarificaiton, I describe my task. I have to design PCBs which fit nicely into oddly shaped housings. While my EDA-sw exhibits a fairly usable routine for importing DXF-data, my self programmed autoplacer does not. So in my EDA file, I may have a nice outline, but this does not help me with placing the components inside that polygon. Hence I need the polygon data separately in my autoplacer.</p>
| |cad| | <p>It seems you are interested in a flat slice of your CAD model. While you could use a 3D file and slice it yourself that seems like a bit overkill as the CAD application is perfectly capable of doing the slices for you.</p>
<h1>quick and dirty</h1>
<p>Ok, so each CAD has a 2D drawing mode, you can save that drawing out as <code>dxf</code> or <code>pdf</code> both are easy to parse. If you don't happen to find a good tool for this turn the <code>pdf</code> into <code>svg</code> that can be easier to parse. This approach can also be done quick and dirty by leveraging such tools as Inkscape or Illustrator. Lets do an example because its easy to do:</p>
<p><a href="https://i.stack.imgur.com/uVHLf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uVHLf.png" alt="enter image description here" /></a></p>
<p><strong>Image 1</strong>: Quick and dirty export drawing as <code>pdf</code>/<code>svg</code> then isolate and read points from that file I <a href="https://graphicdesign.stackexchange.com/questions/73629/create-2-morphable-svgs/73636#73636">used modified version this tool</a> to dump coordinates from <code>pdf</code>. You should be able to do that in Perl easily. <a href="http://pastebin.com/wm8fTdCX" rel="nofollow noreferrer">data available here</a></p>
<h1>Proper Method</h1>
<p>It is possible to access both SolidWorks and Inventor trough a COM bridge so you can access the CAD applications data model directly from your Perl code.This has several benefits but mostly not needing to parse intermediate files. You could select the relevant edges and just traverse them directly from the CAD. Now I only have access to SolidWorks at work but similar approach works in inventor as i have done it.</p>
<p>I had some extra time at work to do some quick <code>VBA</code> code for SolidWorks. The code takes all the lines of a closed sketch, sorts them into polygon (with a naive N^2 algorithm) order and prints them in the <code>VBA</code> debug console.</p>
<pre><code>Option Explicit
Sub main()
Dim swApp As SldWorks.SldWorks
Dim swModel As SldWorks.ModelDoc2
Dim swPart As SldWorks.PartDoc
Dim swSelMgr As SldWorks.SelectionMgr
Dim swFeat As SldWorks.Feature
Dim swSketch As SldWorks.Sketch
Dim numLines As Long
Dim vLines As Variant
Dim dict As New Collection
Dim i As Variant
Set swApp = CreateObject("SldWorks.Application")
Set swModel = swApp.ActiveDoc
Set swPart = swModel
Set swSelMgr = swModel.SelectionManager
Set swFeat = swSelMgr.GetSelectedObject5(1)
Set swSketch = swFeat.GetSpecificFeature2
numLines = swSketch.GetLineCount2(1) 'Exclude crosshatch lines
vLines = swSketch.GetLines2(1) 'Exclude crosshatch lines
Dim startP, endP, line As Variant
For i = 1 To numLines - 1
line = Array(Array(vLines(12 * i + 6) * 1000, _
vLines(12 * i + 7) * 1000), _
Array(vLines(12 * i + 9) * 1000, _
vLines(12 * i + 10) * 1000))
dict.Add (line)
Next i
startP = Array(vLines(6) * 1000, _
vLines(7) * 1000)
endP = Array(vLines(9) * 1000, _
vLines(10) * 1000)
pp startP
pp endP
For i = 1 To dict.Count - 1
endP = NextPoint(dict, endP)
pp endP
Next i
End Sub
Sub pp(point As Variant)
Debug.Print " " & Str(point(0)) & ", " & Str(point(1))
End Sub
Function NextPoint(dict As Collection, point As Variant) As Variant
Dim i As Variant
For i = 1 To dict.Count
Dim data, startRP, endRP As Variant
data = dict.Item(i)
startRP = data(0)
endRP = data(1)
If endRP(0) = point(0) And endRP(1) = point(1) Then
dict.Remove (i)
NextPoint = startRP
Exit Function
End If
If startRP(0) = point(0) And startRP(1) = point(1) Then
dict.Remove (i)
NextPoint = endRP
Exit Function
End If
Next i
End Function
</code></pre>
<p>Since vba is calling COM you ca code you can use nearly any language for example perl implements Win32::OLE that can do the job.</p>
<p><a href="https://i.stack.imgur.com/riuLR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/riuLR.png" alt="enter image description here" /></a></p>
<p><strong>Image 2</strong>: Example part with simple one loop sketch results in <a href="http://pastebin.com/raw/uEETZNjd" rel="nofollow noreferrer">this output</a></p>
<h1>Epilogue</h1>
<p>If you really want to export 3D polygon data and do the slicing manually then i would export either OBJ or STL. But this would be way down on my list of approaches mainly because all other approaches are simpler.</p>
| 11524 | Method to extract closed curve from 3D CAD into CSV or TXT |
2016-09-17T09:20:06.580 | <p>I've recently been reading through introductory texts on the mechanical properties of materials such as <a href="http://link.springer.com/book/10.1007%2F978-94-007-4342-7" rel="nofollow">this</a>. Most start with a description of various material constants such as Young's modulus, Poisson's ratio and a zoo of other properties, and try to determine these properties through a study of dislocations and interatomic interactions.</p>
<p>However, when I read about the very specific test conditions for these kinds of properties, it's not immediately obvious how well these kinds of coefficients and measurements translate to the real world. </p>
<p>For example, I can find coefficients describing hardness, toughness, etc. from laboratory experiments under specific experimental conditions, but it's not obvious that these coefficients can give a more than qualitative description description of how well my nail can hold together a desk, how well my steel truss can hold up a roof, etc., which are ultimately the questions I might be interested in. These coefficients seem to describe very specific geometries, and it seems that I can only get answers about more general geometries using FEA, leaving these kinds of measurements as useful for 1) qualitative descriptions and 2) feeding parameters into FEA.</p>
<p>I think some of the confusion stems from the seemingly arbitrary nature of all these parameters. It appears that someone designed a test and came up with a new parameter every time they were interested in a behavior under different conditions, and there's no minimal degrees-of-freedom type argument that prevents me from coming up with my own test and modulus. </p>
<p>Put another way, what is the argument that the mechanical properties covered by the usual texts (Young's, Poisson's, hardness, toughness, etc.) can give me a complete mechanical description of a system under arbitrary geometries and loads? </p>
<p>Thinking about the parameter space more abstractly, each test can be represented as a material-dependent function that accepts a given material geometry and load geometry and maps it onto strain (throwing out hysteresis and time dependent phenomena for simplification). If we represent our inputs in R^n and our outputs in, say, R^m, what is the minimum number of moduli and constants needed to define this function? What's the argument I won't open up a textbook and find that it defines a new and non-redundant mechanical parameter.</p>
<p>And, if such a well defined relationship exists, where might I find out more about how to go from the parameters derived from well defined experiments (stretching a cylinder, squeezing a flat plate, etc.) and more general material and stress geometries?</p>
| |materials| | <blockquote>
<p>Put another way, what is the argument that the mechanical properties covered by the usual texts (Young's, Poisson's, hardness, toughness, etc.) can give me a complete mechanical description of a system under arbitrary geometries and loads?</p>
</blockquote>
<p>There is no such argument. All models are wrong. Some models are useful. In other words, there is absolutely nothing that can give you a <em>complete</em> mechanical description of any system. But the usual properties are quite useful in making predictions under many (but not all) practical situations. </p>
| 11525 | Value of Mechanical Constants in Real-World Applications |
2016-09-18T11:57:12.440 | <p>I have read a few ISO standards for performing the so-called 'burst pressure' test in which a filter or an element of one is tested to failure. I am curious whether such a failure can occur under normal fuel line pressures (e.g. after filter clogging). To break down:</p>
<ul>
<li>What is a typical fuel line pressure?</li>
<li>Can a blocked/clogged filter 'burst' under such pressure?</li>
<li>If not, what is the worst that could happen?</li>
</ul>
| |pressure|diesel| | <p>Generally speaking the filter is on the low pressure side of the system. That low pressure is either supplied by an in tank lift pump or a low pressure lift pump (usually integrated into the high pressure pump) sucks it from the tank. In 15 years working with an assortment of vehicle, generator, marine diesel engines I've never seen one burst a fuel filter. On automotive use the lift pumps operate at around 5 - 10 psi and flow around a gallon per minute. Worst that could happen in the event of a blocked filter would be the engine stopping, possibly some fuel leaks and you might get the lift pump hot or even cause it to fail if left long enough pumping against a blockage. On some designs the fuel return line from the high pressure pump sometimes goes back into the fuel filter head (to heat the incoming fuel by bleeding a little of the returning hot high pressure fuel back into the cold incoming fuel) before going back to the tank. I'd guess that an internal fault on the high pressure pump fuel return might cause a fuel filter to burst...? Never seen it personally though. </p>
| 11539 | Is a filter burst possible under fuel line pressure? |
2016-09-19T13:33:29.537 | <p>I have the following model:
$$ x[k+1] = Ax[k] + Bu[k]$$
$$ y[k] = x[k] $$
where $x \in \mathbb{R} $ is the state, $A \in \mathbb{R}$, $u \in \mathbb{R}^4$ are the inputs and $B \in \mathbb{R}^4$ and $y[k]$ is the output.</p>
<p>From theory this can be solved using:
$$ y[k] = A^k x[0] + \sum_{i=0}^{k-1} A^{k-1-i}Bu[i]$$</p>
<p>How can I find the time instance $k$ when $y[k]$ has a specific value?</p>
| |electrical-engineering|control-engineering|control-theory| | <p>The answer depends on your initial value and your input. </p>
<p>For specific initial values and inputs there are easy solutions, but in general I think the fastest way is to let the computer solve this for you.
So crank up your matlab / python / mathematica / excel / other tool of choice
and let the brute force of computation serve you.</p>
| 11550 | Find the instance when discrete time model has a specific value |
2016-09-20T03:23:10.430 | <p>From what I've read it seems that hydro turbines can theoretically achieve 100% energy conversion from kinetic to electrical energy. However, I just wanted to double check if this was true.</p>
| |turbines|energy-efficiency|renewable-energy| | <p>Turbine stacking anyone? My truck has dual turbo chargers. The math says that if u convert for liquid instead of gas & increase the slack space to accommodate the liquid pressure variance behind the 1st turbine using Pelton wheels, it allows for the second turbine to have enough head pressure to account for the rotational momentum & friction losses in both turbines. Diameter & length of the slack tube to provide enough head pressure to the second Pelton wheel is crucial, requiring the correct vertical length. This may also require using a reducer nozzle to increase the ram pressure to maintain the second wheels correct rpm. Adjust the flow rate to the 1st wheel to find the slack tube length & diameter to generate the correct flow rate & pressure to the second wheel. I recommend working through the engineering math first.</p>
| 11554 | What is the theoretical maximum energy efficiency of hydro turbines? |
2016-09-20T16:40:57.600 | <p>I'm looking at a motor datasheet where the moment of inertia is specified in units of "oz in sec^2". I would think that inertia should have dimensions of mass * length^2. </p>
<p>I did some further Googling and found that some automatic unit conversion calculators (<a href="http://www.numberfactory.com/nf%20inertia.htm" rel="nofollow">example</a>) let you convert between seemingly contradictory units - the site I've linked lists both kg-m^2 to kg-m-s^2 as acceptable units for moments of inertia. What am I missing here?</p>
| |applied-mechanics| | <p>I spent some time searching on this one, and surprise surprise, Imperial units strike again!</p>
<p>The good old <a href="https://en.wikipedia.org/wiki/Pound_(force)" rel="noreferrer">"pounds-force pounds-mass issue"</a> masked with ounces. This is why any calculation more involved than sizing a pulley should be done in metric, lol.</p>
<p><a href="http://www.orientalmotor.com/technology/articles/motor-sizing-calculations.html" rel="noreferrer">Oriental Motor- Basics of motor control</a>, explains the following:</p>
<blockquote>
<p>Units of Measure for Moment Inertia</p>
<p>The units of inertia are commonly used in two ways, oz-in<sup>2</sup> and oz-in-sec<sup>2</sup>. The former includes gravity, the latter only mass.</p>
<p>Theoretically, inertia is factor of mass so it should not include gravity, however, practically we can not easily measure mass on the earth.</p>
<p>Oriental Motor commonly provides inertia in oz-in<sup>2</sup>. Then, when we calculate the Acceleration Torque in Torque Calculation we divide the total the total inertia by the gravity from.</p>
<p>Gravity = 386 in/sec<sup>2</sup></p>
<ul>
<li>oz-in<sup>2</sup> = Inertia based on weight</li>
<li>oz-in-sec<sup>2</sup> = inertia based on mass</li>
</ul>
<p>Calculation for oz-in<sup>2</sup> to oz-in-sec<sup>2</sup></p>
<p><a href="https://i.stack.imgur.com/Ir2Tn.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Ir2Tn.jpg" alt="enter image description here" /></a></p>
</blockquote>
<p>So an oz-in-sec<sup>2</sup> <strong>does</strong> exist because it's an ounce-force not an ounce-mass.</p>
<p>A kg-m-s<sup>2</sup> <strong>does not</strong> exist, because kg is mass not force. You could convert oz-in-sec<sup>2</sup> with Newtons, but I wouldn't even go there, lol.</p>
<p>The best course of action is to follow Oriental Motor's guide to get to oz-in<sup>2</sup>. Then convert to metric kg-m<sup>2</sup> and don't look back ;-)</p>
| 11561 | Contradictory units for moments of inertia |
2016-09-20T20:56:06.317 | <p>First time poster here. First of all please excuse my english mistakes but most of all my terminology and ignorance on the subject. </p>
<p>Now to the question! A friend of mine asked me if I could help him convert his ceramic 3d printer extruder from compressed air to a piston system (driven by a leadscrew linear actuator). As of now, he told me, he uses 4 bar of pressure to make the clay extrude at the desired rate and that's all I know plus the nozzle and cartridge internal diameter (respectively 1 and 65 mm). What I'm looking for is a stepper motor/lead screw combination that can exert enough linear force to move the clay through the nozzle (the rate of extrusion would be given by the stepping rate of the motor). </p>
<p>Here is what I've done as "homework" but I'm a firmware developer so I'm hoping to get some guidance in solving this problem. Since the nature of this problem is not only practical but also not needing particular precision, what I've done are essentially ballpark, or back of the envelope, calculations.</p>
<ol>
<li><p>converted bar to N/mm<sup>2</sup> (4 bar -> 0.4 N/mm<sup>2</sup>)</p></li>
<li><p>multiplied N/mm<sup>2</sup> by the cartridge cross sectional area ($0.4\text{ N/mm}^2 \cdot 32.5^2\pi\text{ mm}^2$) to get the total force exerted on the clay of ~1320 N (I'll round this to 1400 N to have some margin)</p></li>
</ol>
<p>Knowing the force needed I picked a lead screw from a catalog to get some specs for the next calculations, what I chose is a screw of 10 mm diameter and 2 mm/rev lead</p>
<ol start="3">
<li>using <a href="https://en.wikipedia.org/wiki/Leadscrew" rel="nofollow">this Wikipedia article</a>, I calculated the needed torque to raise a load of 1400 N, assuming that's the clay's resistance according to step #2. (I used the raise formula with the stainless steel coefficient of friction to purposely oversize the system a bit to have some overhead). So the terms were:</li>
</ol>
<p>$$\begin{alignat}{2}
F &= \text{load on the screw} &&= 1400\text{ N} \\
d_m &= \text{mean diameter} &&= 0.01\text{ m} \\
\mu &= \text{coefficient of friction} &&= 0.2 \\
l &= \text{lead} &&= 0.002\text{ m}
\end{alignat}$$</p>
<p>$$
\text{Torque} = \frac{Fd_{m}}{2} \left( \frac{l+ \pi \mu d_{m}}{ \left( \pi d_{m}- \mu l \right)} \right)
$$</p>
<p>As a side note I also used the formula with the lead angle and angle of friction to be sure of my results for which I used:</p>
<p>$$\begin{alignat}{2}
\phi &= \text{angle of friction} &&= 11.3° \\
\lambda &= \text{lead angle} &&= 3.6433°
\end{alignat}$$</p>
<p>$$\text{Torque} = \frac{Fd_{m}}{2} tan(\Phi + \lambda)$$</p>
<p>for a total result of 1.86 Nm of torque (from both formulae!)</p>
<p>Well that's all nice and gives an answer that I like (a torque of 1.9 Nm is quite common for steppers).</p>
<p>But the fact that I like it doesn't mean that it's correct!</p>
<p>What makes me suspicious is that 1400 N of force is more or less what's needed to hold up a 140 kg man (please bear with me on that... I'm trying to use some "everyday experience" to get to a correct answer) and I can extrude the clay from the cartridge with my bare hand... I'm sure I'm not able to put 140 kg on that! </p>
<p>So what I'm asking is essentially if I followed the right path in tackling this problem. </p>
| |mechanical-engineering|pressure|motors|torque| | <p>I'm unable to post a comment as I don't yet have 50 reputation!</p>
<p>It looks to me like you've gone about this problem in a sensible and logical manner. I have a few questions, however.</p>
<ol>
<li><p>Can you confirm if the pressure that your friend uses is really "4 bar" (absoute), or 4 barg (gauge, or above atmostpheric)? You have calculated for 4 barg.</p></li>
<li><p>Your suspicion seems to stem from a mismatch between your human strength and the numbers. This is an exceptionally good way to 'sanity check' your answers! How does the rate at which you are able to push the clay through 'by hand' compare to the 4 bar(g) compressed air system? If the air can push it through at double the rate, then it will require approximately double the force. Does this clear up the discrepancy?</p></li>
</ol>
| 11562 | Calculate proper stepper torque to convert clay extruder from compressed air to leadscrew actuator. Is this the correct way? |
2016-09-21T04:02:29.673 | <h1>Introduction</h1>
<p>I have a file with .rpt extension from ABAQUS which is output of a model. This .rpt contains stresses of elements and displacements of nodes in plane text format. There is an item with this file which i was not able to find in ABAQUS manual fluently. I wanted to find out stresses in this . this is original model and it is completely with STRI3 element.
<a href="https://i.stack.imgur.com/kNTNy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kNTNy.jpg" alt="Original Abaqus Model"></a></p>
<p>This is a section of output file:</p>
<p><a href="https://i.stack.imgur.com/Xkgpp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xkgpp.png" alt=".rpt file content"></a></p>
<p>This does give S11, S22 (normals) and S12 (shear) stresses in defined location.
I believe each element does have 3 points which for integration are used. These points are also known as "Integration Points". In the text file above it seems each element does have 3 integration points. </p>
<h1>Actual Question</h1>
<p>What are exact location of integration points of an STRI3 element, if this STRI3 element does have 3 nodes: N1, N2, N3 with known coordinates (X1,Y1,Z1), (X2,Y2,Z2), (X3,Y3,Z3).</p>
| |finite-element-method|abaqus| | <p>The STRI3 element is a shell element has 3 integration points according to the ABAQUS manual.</p>
<p><a href="https://i.stack.imgur.com/g654T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g654T.png" alt="STRI3 Element in ABAQUS"></a></p>
<p>Forget 3D coordinates (global X,Y,Z) for the moment. Just think of the shell's local 2D coordinate system in x,y. Take the isoparametric representation of any triangle as having coordinates (0,0), (1,0) and (0,1) as shown below:</p>
<p><a href="https://i.stack.imgur.com/iMIim.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iMIim.png" alt="IsoP Triangle Element"></a></p>
<p>The integration points in ($\xi$, $\eta$) coordinates for this element are:
$$A = (\frac{1}{6}, \frac{1}{6})$$
$$B = (\frac{2}{3}, \frac{1}{6})$$
$$C = (\frac{1}{6}, \frac{2}{3})$$</p>
<p>In your I-beam example you have provided it is probably fairly simple as they all look like right-triangles and match the same shape as isoparametric representation of the triangle. The integration points are 1/6 or 2/3 along the base/height depending on which integration point you're looking at. So you probably wouldn't need to worry about shape functions.</p>
<hr>
<p>If your triangles were not simple right-triangles and lied anywhere in space then you would need to do the following...</p>
<p>We can relate the ($\xi$, $\eta$) coordinate system to the ($x$, $y$) coordinate system with these shape functions:
$$N_1 = 1 - \xi -\eta$$
$$N_2 = \xi$$
$$N_3 = \eta$$</p>
<p>At integration point A we have $\xi = 1/6$ and $\eta = 1/6$. Therefore:
$$N_1 = 2/3$$
$$N_2 = 1/6$$
$$N_3 = 1/6$$</p>
<p>So if our triangle had ($x$, $y$) vertices of $(0,0)$, $(3,2)$ and $(1,3)$ then the coordinates of A are:
$$ x_A = \sum N_ix_i = N_1x_1 + N_2x_2 + N_3x_3 = \frac{2}{3}(0) + \frac{1}{6}(3) + \frac{1}{6}(1) = \frac{2}{3}$$
$$ y_A = \sum N_iy_i = N_1y_1 + N_2y_2 + N_3y_3 = \frac{2}{3}(0) + \frac{1}{6}(2) + \frac{1}{6}(3) = \frac{5}{6}$$</p>
<p><a href="https://i.stack.imgur.com/Divj5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Divj5.png" alt="enter image description here"></a></p>
<p>I have very cruedly drawn this example above and you can see that the point coordinates $(2/3,5/6)$ match the position for A in the diagram quite well.</p>
<p>You can repeat this for B and C. Then you would have the coordinates of A, B and C, which are the integration points in the shell's local coordinate system. Then you would have to use a transformation matrix based on the shell's vertices to convert those local x,y coordinates to the global 3D coordinate system in X,Y,Z.</p>
| 11568 | What are exact location of Integration Points in STRI3 element in ABAQUS? |
2016-09-21T08:48:32.060 | <p>Concerning this subject, I consulted the following Wikipedia page: <a href="https://en.wikipedia.org/wiki/Universal_joint#Equation_of_motion" rel="nofollow">Universal joint</a></p>
<p>where the elementary kinematics equation of the U-joint is stated.</p>
<p>But, a little bit further in the article <a href="https://en.wikipedia.org/wiki/Universal_joint#Double_Cardan_Shaft" rel="nofollow">Double Cardan Shaft</a></p>
<p>where the configuration described consists of 2 U-joints, the previously mentioned equation is reversed, i.e:</p>
<p>$\tan \gamma _{1}=\cos \beta \tan \gamma _{2}$,</p>
<p>$\tan \gamma _{2}=\cos \beta \,\tan \gamma _{1}\qquad \tan \gamma _{4}=\cos \beta \,\tan \gamma _{3}$</p>
<p>I do not intend to doubt the validity of the info on the article, but this hit me as a pretty weird thing. </p>
<p>Are these relationships correct?</p>
| |automotive-engineering|kinematics| | <p>The first equation seems correct, and when it is used again it seems to be incorrectly reversed as you point out. But, it did not matter in the end because the mistake was made twice. With the correct equation we still get the same final result.</p>
<p>The equations
$$ \tan \left(\gamma _1\right)=\cos (\beta ) \tan \left(\gamma _2\right) $$
$$ \tan \left(\gamma _3\right)=\cos (\beta ) \tan \left(\gamma _4\right) $$</p>
<p>together with $\gamma _3=\gamma _2+\frac{\pi }{2}$ and $\tan \left(\gamma +\frac{\pi }{2}\right)=\frac{1}{\tan (\gamma )}$, gives</p>
<p>$$\tan \left(\gamma _4\right) = \frac{1}{\cos (\beta
)} \tan \left(\gamma _3\right) =
\frac{\tan \left(\gamma _2\right)}{\tan \left(\gamma _1\right)}\frac{1}{\tan \left(\gamma _2\right)}=
\frac{1}{\tan \left(\gamma _1\right)}=
\tan \left(\gamma _1+\frac{\pi }{2}\right) $$</p>
| 11571 | Universal joint vs Double Universal Joint (CV configuration) kinematics |
2016-09-21T19:01:15.747 | <p>I am trying to get a better understanding of measurements and instrumentation before starting an online class next week. I have been reading a text book and came across a concept question I am stuck on a little bit. It was this: </p>
<p><strong>Design a measurement system that could read temperature, output it to a and regulate the temperature of a room. You have a sensor, space heater, and a computer.</strong> </p>
<p>Another similar one was: </p>
<p><strong>How would you get the temperature of an oven to a digital output.</strong> </p>
<p>I know some of the parts. The sensor would be a sort of thermocouple. You would filter the input and sample it. Do some sort of analog to digital conversion... There are obviously details and parts I'm missing. Putting all of this together is kind of killing my brain though as I just don't get it yet. I would love if someone had the know-how to break it down a little and help me grasp this a little better. </p>
| |mechanical-engineering|measurements|signal-processing| | <p>Those are extremely vague questions, so you could go to varying levels of detail. </p>
<ol>
<li>This is a feedback control question. The short answer: connect temperature sensor and space heater to computer in a room Develop a control method/program on computer to read temperature and switch on/off heater as control method determines necessary. There are many control methods that can be used. Check out this basic Wikipedia article: <a href="https://en.m.wikipedia.org/wiki/Temperature_control" rel="nofollow noreferrer">Temperature control</a> I suggests you read about P, PI, and PID control. </li>
</ol>
<p>Here's another interesting and light article: <a href="https://www.instrumart.com/pages/283/temperature-controller-basics-handbook" rel="nofollow noreferrer">Temperature Controller Basics Handbook</a></p>
<ol start="2">
<li>Find a suitable thermocouple that can withstand the temperatures of the oven. I like to over-rate. Probably a K type with 800F minimum rating. Either get an analog or digital thermocouple amplifier chip. For example, I suggest a max31855k digital output. Connect that and some type of display to a microcontroller an read the sensor once every second and write the value to the display. The Internet has plenty of examples and sample code..especially for Arduino. </li>
</ol>
<p>I suggest you get an Arduino and some thermocouple kits and lcd kits... which you can probably get it all for around $30(or less!) and you can make a temperature controlled thing. See the code and how it works. Experiment. It will give you a much better understanding of how these things interact. </p>
| 11577 | Designing a simple measurement system |
2016-09-22T16:11:26.437 | <p>I'm interested in bringing in a GHT thread compatible product, but my sample's diameter was slightly smaller. GHT has a large (pitch to pitch) diameter of 1.0625 inch. The sample I received has a diameter of 1.0415 inch on the male side, and it doesn't leak and screws in perfectly. </p>
<p>Is this diameter difference meaningful for the long term reliability of my product?</p>
| |mechanical-engineering| | <p>GHT relies on a gasket to seal, not a tight fit between the threads. This makes it more resistant to dirt and also cheaper to manufacture. I would guess that your sizes will work fine and you shouldn't be concerned.</p>
<p>From the sizes you measured, the radial clearance is only .0105 (ten thou.) That might be a little loose for precision threads, but anything smaller than about 2 thou clearance becomes very difficult to assemble.</p>
<p>All that said, the Machinery's Handbook has information on hose threads and gives some tolerances. It's on page 1873 (at least in the 28th edition.) Depending on the material and processes, they list a minimum major diameter for the male thread as low as 1.035. <a href="http://machiningproducts.com/html/NPSH-NH-NHR-Thread-Dimensions.html" rel="nofollow">This</a> website also lists the appropriate dimensional information in case you don't have the handbook. The official standard is <a href="https://www.asme.org/products/codes-standards/b1207-1991-hose-coupling-screw-threads-inch" rel="nofollow">ANSI/ASME B1.20.7</a>. If you are specifying compatible threads for a production run, you should probably consider buying it or at least referencing it on your drawing.</p>
<p>On another type of thread, like NPTF where the threads are the sealing component, tolerances will be tighter. Fittings that have a tapered or flared metal-to-metal surface to seal usually have thread tolerances somewhere in the middle of these two.</p>
| 11585 | Plumbing / Thread Compatibility |
2016-09-22T16:46:15.580 | <p>I am currently designing a motorized zipline trolley that will carry a person on a horizontal cable, so I am trying to calculate the friction on the wheels to determine what kind of motors I need. I initially approached the problem using a static friction model using the coefficient of static friction between two steels. However, my professor told me to replace my model with a rolling friction model, so I took the coefficient of rolling friction between two steels. This greatly reduced the calculated value of friction; however, I feel like there should be more friction because the weight of the person will cause deflection in the cable, so a portion of the wheels' circumference will be in contact with the cable. I want to assume no slipping for the wheels, so I don't know how to account for the static friction while keeping the rolling friction model. How can I calculate the additional frictional force that I am not modeling?</p>
| |friction|wheels|pulleys| | <p>There are a two items I think you are wondering about, the effect of static friction and the extended length of the contact area between the rope and the pulley. </p>
<p>Start up torque on a motor can be greater than the design torque of the motor, so your professor may be suggesting to ignore the rolling friction because of that. I would say it's a valid assumption to assume the motor can overcome the static friction if it's overcoming the dynamic friction. Check out motor-torque curves on engineering toolbox for example. I can't remember the complete explanation and terms for this, but when you start a motor there is not a strong magnetic field produced by the rotor, because it is not moving yet, and so the magnetic field that is being produced by the AC current in the stator gets stronger and produces more torque than when the rotor is spinning at its nominal rating. That's why you have high inrush current when starting a motor, no matter what (assuming you have no starter controls on it). </p>
<p>Secondly, the contact area of the rope on the pulley is irrelevant. You have assumed no slip for starters, so the total frictional force between the rope and the pulley that would be altered by contact length is not going to change. Draw a free body diagram of the pulley. You will have to have tension in the rope between the two pulleys, and you can theoretically make that tension to whatever you want. The tension in the rope (due to the tension you set as well as the mass of the trolley load) will act as the torque on the pulley of the motor. That is the torque the motor has to overcome. </p>
<p>There will be some deflection in the rope due to the load, which you can solve for. Then it's just the added tension in the rope from the load that the motor has to account for since the rope tension is always perpendicular to the pulley. </p>
<p>Break that out into components to and balance then solve for the new tension. I'm on my phone right now and can't enter the equation. But I will try once I have access. </p>
| 11587 | How do I model the friction on a rolling pulley wheel, which holds weight, on a horizontal cable? |
2016-09-22T17:08:29.327 | <p>If you have a saturated steam (quality=1.0) volume that is compressed adiabtically will the steam condense to follow the saturation curve or superheat?</p>
<p>I found in an old textbook: <a href="https://books.google.com/books?id=o5pBAAAAIAAJ&pg=PA444&lpg=PA444&dq=compressing+saturated+steam&source=bl&ots=62g5CGqi45&sig=Ox5v-8gszNyoHnRrnU73gjVJhk4&hl=en&sa=X&ved=0ahUKEwiC3qzmvqPPAhUUImMKHVEHD00Q6AEIUzAH#v=onepage&q=compressing%20saturated%20steam&f=false" rel="nofollow">The Principles of Thermodynamics (1899)</a> on Google the following passage:</p>
<blockquote>
<p>For compression, adiabatic:</p>
<ol>
<li>If we start with pure saturated steam, without admixture of water, it will be superheated by compression.</li>
<li>If the initial steam weight is greater than that of the water, steam is generated by the compression.</li>
<li>If there is more water than steam, steam is condensed during compression.</li>
</ol>
</blockquote>
<p>I am still at a bit of a loss for this makes physical sense and how this is properly captured in an energy balance. (i.e., $\frac{\text{d}U}{\text{d}t} = -\frac{p\text{d}V}{\text{d}t}$)</p>
| |thermodynamics| | <p>An ideal adiabatic compression of steam is a reversible process. This means that on the state of the fluid will follow lines of constant entropy. An alternative view is this is the opposite of an ideal turbine where the initial assumption is ds = 0 (adiabatic expansion in reverse).</p>
<p>For the T-s diagram constant entropy will put you at steam or liquid depending on the condition of the steam-water.</p>
<ol>
<li>If x > 0.5 then compression will lead to steam</li>
<li>If x < 0.5 then compression will lead to liquid</li>
<li>If x ~ 0.5 then compression will stay at roughly 50/50 mixture until supercritical water conditions are met (the maximum of the saturation curve).</li>
</ol>
<p>Or for the Mollier P-h diagram. If you start at "B", compressing the fluid will follow the red line and go to superheated steam.</p>
<p><a href="https://i.stack.imgur.com/Qvzu6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qvzu6.jpg" alt="Temperature-Entropy Diagram for Water" /></a>
<a href="https://i.stack.imgur.com/LZjil.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LZjil.jpg" alt="Mollier Pressure-Enthalpy Diagram of Water" /></a></p>
| 11588 | Compressing saturated steam |
2016-09-24T00:59:01.567 | <p>The theoretical formula for my beam deflection is:</p>
<p>$ v(x) =
\begin{cases}
-\frac{Px}{48EI}(3L^2-4x^2), & 0\le x \le {L\over2}\\
-\frac{P(x-L)}{48EI}(L^2-8Lx+4x^2), & {L\over2}\lt x \le L
\end{cases}$</p>
<p>I need to derive the formulae for the slope $v'$, curvature $v''$ and $v'''$</p>
<p>Is this just a simply case of taking the first, second, and third derivative of the original equation? I know absolutely nothing about structural beams...</p>
| |mechanical-engineering|structural-engineering|mathematics| | <p>Yes.</p>
<p>The first derivative of the deflection is equal to the tangent of the deflection, which for small deflections can be approximated as equal to the angle of rotation of the beam at each point.</p>
<p>The second derivative (times $EI$) is the bending moment along the beam.</p>
<p>The third derivative (times $EI$) is the shear force along the beam.</p>
<p>The fourth derivative (times $EI$) is the distributed load along the beam.</p>
| 11602 | Help an electrical engineering student with a beam deflection question |
2016-09-24T21:41:43.437 | <p>I recently <a href="http://m.gloucestershirelive.co.uk/elmbridge-court-burger-plans-on-track/story-29745204-detail/story.html" rel="nofollow noreferrer">read</a> about a newly proposed 'burger' lane across a roundabout.</p>
<p><a href="https://i.stack.imgur.com/r2ybd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r2ybd.png" alt="burger lane" /></a></p>
<p>From what I can see, it bypasses the main point of a roundabout, a continuous circle of traffic with well defined rights of way. The only reason I can imagine one is to stop a tributary to the roundabout clogging the other entrances,but I don't understand how that would be achieved.</p>
<p><strong>What's the purpose of a 'burger' lane in a roundabout?</strong></p>
| |civil-engineering|transportation|traffic-intersections| | <p>Big vehicles like trucks are often given a priority in these so called lanes so they don't risk flipping. It's also emergency vehicles bypass so they don't have to go around in an emergency. Emergency vehicles zoom past traffic and In america the vehicles like Ambulances and fire engines are often big and have high centers of gravity. Being fast in a straight line helps avoid rollovers.
By contrast, European Ambulance
<a href="https://i.stack.imgur.com/HLpQ1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HLpQ1.jpg" alt="enter image description here" /></a>
Vs. American Counterpart
<a href="https://i.stack.imgur.com/bN7Lj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bN7Lj.png" alt="enter image description here" /></a></p>
| 11611 | What's the purpose of a 'burger' lane in a roundabout? |
2016-09-25T01:02:12.057 | <p>It's common in submarine movies that when suffering some kind of shock (like a near miss by a torpedo or falling deep into a trench), water will start bursting into the cabin and the crew will run around turning little wheels or cranks to stop the leaks.</p>
<p>Is this realistic? What are those wheels <em>doing</em>? Is it even plausible for a submarine to leak without a catastrophic failure?</p>
| |mechanical-engineering|naval-engineering|submarines| | <p>I think that the "cranks" you're talking about are the vent valves that allow air circulation between compartments inside the submarine, which are normally left open in order to allow the air purification equipment to provide clean air to each compartment and remove the stale air.</p>
<p>If a leak from the outside occurs in one compartment, these valves would allow it to flood the entire submarine. Therefore, it is best if the valves for that compartment can be closed, which would allow the rest of the compartments to continue to operate normally. Second best is to close the valves of all of the other compartments, which will allow the air system to flood in addition to the affected compartment and prevent all compartments from getting new air. In this case, it is necessary to surface ASAP.</p>
<p>However, in deep-water events, all of this is for naught, because the bulkheads between compartments are not nearly as strong as the pressure hull, and they will fail regardless of the state of the valves. The only way a crew can survive for any significant length of time is if the pressure hull is still intact. Anything to the contrary you see in movies is mostly "dramatic license".</p>
| 11614 | Do submarines really have cranks for closing leaks? |
2016-09-25T20:31:42.800 | <p>I have designed a product and I found that there is a patent on that specific product. There also seems to be many patents and for the smallest or most obvious use of items. I am wondering if anyone has had experience creating a product and finding there are patents on it and then how do you counter with that? How can you be sure you 100% violate it?</p>
| |product-engineering| | <p>The patent of interest is currently in application stage. Review the marking in the below image. </p>
<p><a href="https://i.stack.imgur.com/WSpuO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WSpuO.jpg" alt="Patent Application - Example"></a></p>
<p>In general from an engineering product design stand point the first to investigate is if the patent is active. Two scenario to look for is as follows. </p>
<ol>
<li>Is the patent currently maintained by the author, if the author is not maintaining the patent, then you will not be infringing on patent, and you should not have to worry about.</li>
<li>Is the patent active. All utility patents are valid for only 17 years from the date the patent was granted. If the patent has expired you should not have to worry about infringing on the patent.</li>
</ol>
<p>If the patents are active then you need to carefully investigate are the claims. Is your engineering design infringing on any of the claims? Looks like their is 20 claims listed in the referenced patent. Below is claim 1 (primary claim) for the patent of interested.</p>
<p><a href="https://i.stack.imgur.com/rZVEb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rZVEb.jpg" alt="Patent Claim"></a></p>
<p>If you believe your product is infringing on the patent application, you will need to hire a patent lawyer to investigate work around. </p>
<p><strong>References:</strong></p>
<ul>
<li><a href="https://patents.stackexchange.com/questions/13539/how-to-apply-for-a-patent-in-the-us-eu/13540#13540">How to apply for a patent in the US/EU?</a></li>
<li><a href="http://www.beemlaw.com/video-gallery/the-three-types-of-patents#.V-sjN_krLIW" rel="nofollow noreferrer">The Three Types of Patents</a></li>
</ul>
| 11624 | Patent for product I designed |
2016-09-26T08:21:10.333 | <p>I am looking about things will make phone emit more radiation and I found some case</p>
<ul>
<li>Bad mobile network signal</li>
<li>Turn on Bluetooth and Wifi</li>
<li>Moving fast</li>
<li>When start dial</li>
<li>When start send message</li>
<li>Battery (not affect)</li>
</ul>
<p>But I do not know enough or not? Anything else can affect phone emit radiation?</p>
<p>Any help would be great appreciated.</p>
| |electrical-engineering|radiation| | <p>Generally the term emissions or RF energy is used rather than radiation. While radiation or even non-ionizing radiation can be used it causes excessive worry and panicking in people who don't understand the differences between types of radiation. No one claims that a lightbulb is flooding the room with radiation but it is, we normally call this radiation light. And the light from a bulb is both more energy that a phone emits and a lot closer in frequency to the dangerous types of radiation than anything you'll get from a phone. In other words if a phone is bad for your health then you need to spend your life in a dark room because the sun or any artificial light will clearly kill you.</p>
<p>Things can be split into two groups - Intentional radiation and unintentional.</p>
<p>Intentional radiation is when the phone is using a radio transmitter. So any extra radio activity will generate more emissions.</p>
<p>When in standby the phone will exchange data with the network at a fairly low duty cycle, just an occasional I'm still here message. Anything that requires more communication will increase the radio activity. So any phone calls, texts, data etc... will increase the amount of information the phone needs to send and so increase the emissions.
There is also a hand off every time you move between cells which generates a small amount of data.</p>
<p>Independent of how busy the phones transmitter is the the phone adjusts its transmit power based on the signal strength so low signal strength will cause it to transmit at a far higher signal power.</p>
<p>So lots of data in an area with weak signal will put out a lot of energy (and drain your battery)</p>
<p>Other radios such as WiFi and bluetooth will also transmit however they are lower power. WiFi will reduce the amount of the phone needs to use the cell network for data and uses a lower power than the phone data link. All things being equal wifi can result in a net decrease in transmitted RF energy however people normally send more data when on wifi. Bluetooth is very low power, unless you are using something that requires constant data, e.g. using a bluetooth headset, the power it radiates is negligible in comparison to the other radios.
As with the cell phone radio, total emissions are very dependent on how busy the radio links are. These are battery powered systems, when not in use they are designed to be as low power as possible. Radio transmitters use a lot of power and so are kept powered down as much of the time as possible.</p>
<p>Finally there are unintended emissions, these are far far lower power than the intentional transmissions but can be measured. All electronic systems emit small amounts of RF energy, the busier they are the more they emit which means anything that uses the phones CPU is causing it to emit a small amount of RF radiation.</p>
<p>Edit -
One minor addition: Connected cables can have an impact (headphones, chargers etc...). While they don't cause any extra emissions directly they can end up acting as antennas and so increasing the efficiency with which the already existing signals are transmitted. This is primarily going to impact the unintentional transmissions but it could also have minor impacts on the intentional transmissions.</p>
| 11629 | What thing will make phone emit more radiation? |
2016-09-26T08:28:20.920 | <p>We have standby diesel generators that are only run a few times a year. How long can we reasonably keep the diesel stored on site before using or disposing of it?</p>
<p>It is accepted that there is a finite shelf life particularly after the UK has started using B5 (5% biodiesel) but I have been unable to find any guidance or papers indicating how long it is recommended to store it.</p>
<p>I know that it depends on how well it is stored (contaminants etc.) but any best practice guidance / documents / journal articles that anyone could point me to would be much appreciated!</p>
| |diesel| | <p>The internet has a range of values from 3 months to a year, depending largely on the political/economic interest of the site in question. Here's a couple quotes that you might want to follow up on:</p>
<p>From <a href="http://www.springboardbiodiesel.com/storing-biodiesel-fuel">springboardbiodiesel</a>,</p>
<blockquote>
<p>The truth is that all fuels will degrade over time. In fact, the EPA
reports that ULSD diesel has a shelf life of between 3-6 months.
Biodiesel, too, has a shelf life that can vary significantly, but with
the proper fuel management, biodiesel's shelf life can be extended
dramatically.</p>
<p>The biggest factors that affect biodiesel storage life include:</p>
<p>Microbial contamination Chemical contamination Exposure to light
Temperature Exposure to air The type of feedstock And the additives</p>
</blockquote>
<p>From <a href="https://www.bellperformance.com/bell-performs-blog/biodiesel-storage-life">BellPerformance</a>,</p>
<blockquote>
<p>All fuels degrade over time. The old school diesel fuels from the
fifties could be kept relatively fresh for 1-2 years. The advent of
ultra low sulfur diesel cut this storage to 3-6 months. When you blend
biodiesel into the mix, the figure changes again. So there's no one
stock answer to the question. A fuel's storage life is going to depend
on a host of factors related to storage conditions.</p>
<p>The influencers of storage life for both conventional diesel and
biodiesel would be contamination with microbes and/or chemicals,
light, storage temperature, oxygen exposure, and the type of biodiesel
feedstock that the fuel was made from.</p>
</blockquote>
<p>From <a href="http://www.associatedpetroleum.com/pdf/AE-BioDiesel-MythsFacts.pdf">AssociatedPetroleum</a>, </p>
<blockquote>
<p>Most fuel today is used up long before six months, and many petroleum
companies do not recommend storing petroleum diesel for more than six
months. The current industry recommendation is that biodiesel be used
within six months, or reanalyzed after six months to ensure the fuel
meets ASTM specifications (D-6751). A longer shelf life is possible
depending on the fuel composition and the use of storage enhancing
additives.</p>
</blockquote>
<p>At the risk of getting booted off this forum, <strong>YMMV</strong> :-)</p>
| 11632 | How long can UK diesel be stored? |
2016-09-26T09:44:37.810 | <p>I'm trying to model a non-linear, static SISO system with</p>
<ul>
<li>input variable $u$</li>
<li>disturbance variable $z$ (can be controlled for identification purposes but not during actual operation)</li>
<li>output variable $x$</li>
<li>system behaviour $N\{x,u,z\}$ is hysteresis afflicted for both $u$ and $z$</li>
<li>$u$ and $z$ influence the hysteresis characteristics of each other </li>
</ul>
<p>I'm searching for any literature that deals with this kind of problem.
It should contain approaches on how to model such a system and if possible a way to invert it, because ultimately I'm trying to linearize the overall behaviour $N^{-1}\{N\} = 1$. </p>
| |control-engineering|modeling| | <p>I found 2 possible approaches:</p>
<ol>
<li>Mayergoyz describes in <a href="http://www.sciencedirect.com/science/article/pii/B9780124808737500037" rel="nofollow">chapter 2 of his book</a> "Mathematical Models of Hysteresis and Their Applications" a Preisach model of magnetostricitve material. He considers a system with two cross-coupled inputs, which suits the problem prefectly.</li>
<li>If the hysteresis of the disturbance value $z$ is negligible, one might try the <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.463.7198&rep=rep1&type=pdf" rel="nofollow">modified Prandtl-Ishlinksi</a> approach, in which the influence of $z$ on the $x$-$u$ hysteresis is modelled by an unambiguous characteristic. Furthermore this approach is highly suited for compensation purposes, since the model can be analytically inverted.</li>
</ol>
| 11634 | Modelling a system with multiple hysteresis behaviours and mutual interference |
2016-09-27T03:46:39.173 | <p>I need to get a motor from an electric motorbike to drive a single-speed BMX Bike chain, and I haven't worked with bikes/sprockets before so I don't know where to start (I was only supposed to be doing the throttle and electronics, not sure how I ended up with the chain drive).</p>
<p>I know I need a sprocket with as few teeth as possible (~10-14 or less) and can be mounted onto a 9.5mm shaft double flattened to 8.4mm, and 4.5mm long, and have the teeth fit a standard single-speed BMX bike chain.(measurements are from a set of calipers, motor doesn't have any specs :P) I have a sprocket from the motor, but it isn't the correct tooth size/spacing for a bike chain.</p>
<p><strong>Is this something that exists or can be made on a part-by-part basis, or do I need to weld something up?</strong></p>
<hr>
<p><img src="https://pbs.twimg.com/media/CtVKT0PVYAEhSyA.jpg" alt="Picture of the motor shaft measurements, I guess twitter isn't a good way to host images"></p>
| |mechanical-engineering|motors|bicycles| | <p>There are quite a few online websites you can find that specialize in the design and fabrication of gears/sprockets.</p>
<p>For example, a really nice website for generating custom sprockets and gears is <a href="http://www.rushgears.com/tech-tools/part-search/build-custom-gears" rel="nofollow">Rush gears</a> because they have a nice sprocket/gear online model generator. You can download the models of gears in CAD format and even order the gear to be made by them in a bunch of materials. Just click the Click to build CAD models and you can adjust a bunch of different specifications.</p>
| 11640 | Designing and fabricating custom gears and sprockets |
2016-09-27T06:16:59.847 | <p>I'm searching for the calculation of the quality of the estimated model output compared to the measured data. I'm working with the System Identification Toolbox. You can see an example here <a href="http://de.mathworks.com/help/ident/ref/compare.html" rel="nofollow">http://de.mathworks.com/help/ident/ref/compare.html</a> at the first graph. Is the percentage calculated with an integral or any other method?</p>
| |electrical-engineering|control-engineering|matlab| | <p>I found the answer. It is at the same page at Output Arguments --> Fit.
The fit is calculated (in percentage) using:
$$
fit=100*(1-\frac{||y-\hat{y}||}{||y-mean(y)||})
$$
where $\hat{y}$ is the estimated model and $y$ the measured data</p>
| 11643 | System Identification Toolbox - How is the quality of the model output calculated? |
2016-09-28T03:18:27.297 | <p>It seems to me that <em>abrasive saw</em> is a term mainly used in the US (the <a href="https://en.wikipedia.org/wiki/Abrasive_saw" rel="nofollow">Wikipedia article on it</a> has no links to pages in any other languages), whereas <a href="https://en.wikipedia.org/wiki/Angle_grinder" rel="nofollow">angle grinder</a> has some 20+ translations. So, can someone clarify whether discs for abrasive saws are interchangeable with (thin/cutting) discs for angle grinders and what other differences there might be between these two tools (even if just in the US)?</p>
| |cutting|tools|grinding| | <p>Generally an abrasive saw is a bench-top tool. It looks like a miter saw but has a grinding disc instead of a toothed blade. The dist will be on the order of 12-16 inches in diameter for a typical saw. Angle grinders are handheld tools with much smaller discs - usually between 4 and 8 inches in diameter. While there are come very large angle grinders and some very small abrasive saws, usually the wheels will be different.</p>
<p>For each, a range of wheels are available. All abrasive saw wheels are designed for cutting on only the outside edge, and will be thick enough to support them selves over their diameter. They will be made of slightly different abrasives, optimized for the metal that you are cutting.</p>
<p>Some wheels for angle grinders are designed to cut along the outside edge as well. These are usually called 'cutting wheels' or 'cutoff wheels.' Because of the smaller diameter, they can bu much thinner than abrasive saw wheels without shattering too easily. Being thin allows them to cut faster, reduces waste from kerfs, and makes them lighter. For angle grinders, other kinds of blades are also available for surface grinding. These wheels are meant to cut on one flat face for blending welds, cleaning metal, contouring, and other surface work. These wheels will be thicker so that they can handle the loads applied. There are a bunch of other types of wheels from flap discs and soft pad grinders to wire wheels for other specific purposes.</p>
<p>So in general, event though they might sometimes be the same diameter in a few cases, discs for abrasive saws and cutoff grinders will not be interchangeable. In any case, be sure to check that the rated rotational speed (RPM) for the abrasive blade matches or exceeds the spindle speed of your tool.</p>
| 11656 | Difference between an angle grinder and an abrasive saw |
2016-09-28T18:42:01.667 | <p>Yesterday I asked this questions on physics part of stackexchange, but I was redirected to engineering side of the force ;-). We all know about the rapidly developing technology of 3D printing. However, this technology is still very very VERY SLOW. For printing objects larger than a matchbox you have to wait days. I wonder if technique and physics will ever allow us to create objects directly from the atoms, by manipulating the setting, structure and connectivity of atoms of a given element. If possible, how can we control the atoms so as to create a fast 3D printer?</p>
| |mechanical-engineering|structural-engineering|3d-printing| | <p>A replicator is a very long way from 3D printing. Although 3D printing is often endowed in the media with quasi magical properties all it really is is multi axis CNC extrusion and not fundamentally different from any other CNC manufacturing process. </p>
<p>The key thing about the replicators in star Trek is that they can produce any material on demand from energy/elementary particles including very complex organic compounds (eg Capt Picard is often seen ordering 'Earl grey Tea : hot'). </p>
<p>In this sort of context fabricating the shape of the object is fairly trivial compared to synthesizing the matter it is composed of. </p>
<p>There have been experiments in fabricating atomic scale objects using electron microscopes but this is still a long way away form making up complex molecules in bulk on demand. </p>
| 11665 | Is the construction of a high-speed 3D printer (replicator from Star Trek) possible? |
2016-09-29T19:26:51.813 | <p>sorry for the odd title. We're working on a project to find a cheaper and safer way to install high voltage power lines in the Scottish highlands. One of my group mates had a really cool idea to instead of actually digging underground, you could just lay the cables and cover them with dirt/some other material which we haven't thought about at the moment. </p>
<p>I am asking because I assume the industry would currently be using the best available technology to them, and they don't use this method; they lay them underground. I'm trying to figure out the potential drawbacks to this proposed method.</p>
<p>1) Initially i thought it could be regulations, i.e. minimum depths etc, however, as far as I can tell from official documentation, there are only guidelines. In general, cables are laid between 0.45m -1m below ground, but not always. For example from a cable map I found, there was an 11kV line just 35mm undeground in Edinburgh. Official documentation seems to imply depths are situational, and as long as a risk assessment is performed, shallower or deeper depths can be allowed.</p>
<p>2) In terms of cost, digging through the rocky terrain of the Scottish highlands would pose significant challenge and cost. Getting heavy vehicles out onto that sort of terrain isn't easy. Using some sort of packing material, such as dirt, would surely be a cheaper option? There would of course be costs incurred due to rough terrain for transport trucks etc, but I don't see how it could cost more than digging through the terrain.</p>
<p>Potential drawbacks I can think of:</p>
<p>1) Using a mound structure would pose issues such as erosion. However, maintenance (which must be carried out on cables anyway) shouldn't be too difficult, especially since faults in the structure would be easy to spot since it is overground. It may require more regular maintenance due to weathering, but that would be offset by the fact that each maintenance trip wouldn't involve unearthing the ground?</p>
<p>2) Potential damage to cables. I don't even know if this is a real drawback (in comparison to the existing system since cable damage isn't very uncommon). I read some case studies from one of the regulatory bodies, and it seems damage came mostly when projects were undertaken without consulting the relevant authorities for cable maps. I assume the mound layout would be very similar, and probably wouldn't encounter a greater frequency of damage to the cables, but this is pure conjecture.</p>
<p>3) It would be an easier target for malicious activity since it would be easier to get to without the need for heavy machinery. However, cable maps aren't very difficult to get, and someone could just as easily target shallow cables using the current system.</p>
<p>Ultimately, does anyone have any criticisms of the potential solution? Or does anyone know of this idea having been implemented but not worked, or was determined to not be feasible? Thank you very much!</p>
| |power-engineering|electrical-grid| | <p>Aside from safety and maintainability concerns, there are also costs with insulation and capacitive losses to consider. See for example: <a href="https://www.quora.com/Why-arent-underground-cables-used-for-power-transmission-in-India" rel="nofollow">https://www.quora.com/Why-arent-underground-cables-used-for-power-transmission-in-India</a></p>
| 11684 | Why shouldn't we put high voltage cables physically underground? |
2016-10-01T21:44:39.463 | <p>I have a Nema 17 stepper motor that does 200 steps per revolution, or 400 800 and 1600 micro-steps if in micro-stepping mode for convenience sake I would like to somehow translate the 200 steps into 360 steps. What gear ratio / micro stepper configuration do I need to convert 200 steps to 360 steps.</p>
<p>Update</p>
<p>If anyone wants an easy way to create an involute spur gear, I found this page, converted the to DXF in illustrator then extruded in Rhino!
<a href="http://geargenerator.com/" rel="nofollow">http://geargenerator.com/</a></p>
| |mechanical-engineering|gears|stepper-motor| | <p>You want a transmission with a ratio R such that $ R \cdot 200 / 360 $ gives an integer number of steps per degree. Then in controlling software you can program it to take e.g. 5 steps to move one degree.</p>
<p>As Brian Drummond mentioned in a comment, 36:20 is one possible ratio, giving 1 step per degree.</p>
<p>Some other options:</p>
<ul>
<li>9:5, equal to 36:20 except smaller gears.</li>
<li>9:1, gives 5 steps per degree, so greater torque but slower speed</li>
</ul>
<p>As to how to implement the transmission, you can do it with either gears or with belts. Some of the ratios above are very basic, I'm sure you could even find a premade 9:1 part.</p>
| 11702 | How do I mechanically convert 200 steps into 360 discreet degrees? |
2016-10-02T02:38:10.793 | <p>A 200×250 mm panel of mass 20 kg is supported by hinges along edge AB. Cable CDE is attached to the panel at C, passes over a small pulley at D, and supports a cylinder of mass m. Neglect the effect of friction.</p>
<p>For those who cannot see the picture. The axis are oriented as follows, x+ is pointing out and to the right, y+ is pointing up and z+ is out and to the left.
<span class="math-container">$A=[0,0,.25]$</span> <span class="math-container">$ B=[0,0,0]$</span> <span class="math-container">$ C=[.2sin(\Theta),-.2cos(\theta),.125]$</span> <span class="math-container">$ D=[.2,.1,0]$</span></p>
<p>I have <span class="math-container">$\sum F_x=0=A_x+B_x+\frac {mg}{\lVert CD\rVert}(.2-.2sin(\theta))$</span></p>
<p><span class="math-container">$\sum F_y =0=A_y+B_y-192.2+\frac {mg}{\lVert CD\rVert}(.1+.2cos(\theta))$</span></p>
<p><span class="math-container">$\sum F_z=0=A_z+B_z+\frac {mg}{\lVert CD\rVert}(-.125)$</span></p>
<p>I am having trouble deciding where I want to make my moment in order to create a system I can solve for m in terms of <span class="math-container">$\theta$</span>.</p>
<p><a href="https://i.stack.imgur.com/vH61D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vH61D.png" alt="Static System in quesiton" /></a></p>
| |statics| | <p>Since you are only interested in the mass $m$, all you have to do is to set the moment about the hinge AB to zero for equilibrium.</p>
<p>This means you have to consider only the two forces $M g$ (panel weight) and $m g$, since only they contribute to the relevant moment.</p>
<p>(If you wish, I can work out the complete answer, but probably this is all you need to arrive at the answer yourself.)</p>
<p>For reference, my calculations give me the result for $m$ as $$ \frac{5 M \sin (\theta )}{\sin (\theta )+2 \cos (\theta )}$$</p>
<p>The plot shows that $m$ increases with $M$ and $\theta$ as expected.</p>
<p><a href="https://i.stack.imgur.com/vrzui.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vrzui.png" alt="enter image description here"></a></p>
<p>An interesting confirmation is that the current configuration will not work for $\theta \gtrsim 2$. </p>
<p><a href="https://i.stack.imgur.com/0o1rz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0o1rz.png" alt="enter image description here"></a></p>
| 11706 | How/where to take moments to solve a static equilibrium equation |
2016-10-03T14:24:49.003 | <p>I want to know if it is possible to control the gearbox of a car with a manual transmission, specifically one with the shift stick in the steering wheel column.</p>
<p><a href="https://i.stack.imgur.com/bZH21.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bZH21.jpg" alt="steer column"></a> </p>
<p>The idea is using linear actuators or hydraulic cylinders to control the gas / clutch pedals and "some kind of mechanism" to control the shifting stick directly or removing the gear stick and installing the mechanism on the gearbox directly. </p>
<p><strong>Is this idea feasible or has it been done before? Or do all autonomous cars use automatic or "tiptronik" gearbox?</strong></p>
<p><em>NOTE: I have taken a look at this post before <a href="https://engineering.stackexchange.com/questions/5838/remote-start-on-a-car-with-manual-transmission">Remote start on a car with manual transmission</a>, but did not find a satisfying answer</em></p>
| |control-engineering|automotive-engineering|gears| | <p>What you've described is not that far from a computer-controlled automatic transmission. In fact, so-called <a href="https://en.wikipedia.org/wiki/Semi-automatic_transmission" rel="nofollow">semi-automatic transmission</a> seems like a quite good match.</p>
<p>Biggest difference is usually the use of a torque converter instead of a clutch. A friction clutch requires a certain amount of "feel" for the car, to know how fast to release it for a smooth start or gear change. A torque converter is more forgiving of timing, which makes it easier to control automatically.</p>
<hr>
<p>And as for retrofitting such a mechanism to an existing car: it surely has been done, for example in remote control cars for movies. But in general there would be little point to do it that way, when automatic and computer controlled gearboxes are available off-the-shelf.</p>
| 11719 | Remote control of a manual transmission |
2016-10-03T19:03:37.673 | <p>For example, if your transfer function is defined as :</p>
<p>$$\frac{H(s)}{Q(s)} = \frac{1}{5s}$$</p>
<p>What would the gain and time constant be in this case (since it's usually in the form of $\dfrac{k}{Ts+1}$ where $k$ and $T$ are the time constant and gain, respectively)?</p>
<p>(This is the transfer function for a tank being filled, but the exit stream is a pump so it is constant. Where $H$ and $Q$ are deviation variables for output and input, respectively)</p>
| |chemical-engineering|process-engineering| | <p>$G(s) = \frac{1}{s}$ is called an integrator. It continuously integrates your signal and therefore it has no time constant, hence $\tau = 0$.</p>
<p>Your system $\frac{H(s)} {Q(s)} = \frac{1}{5s}$ can be written as an integrator multiplied with a gain of $\frac{1}{5}$. </p>
<p>Therefore the pole-zero gain is $k_{pz} = \frac{1}{5}$.</p>
<p>The static gain of your function $k_s = \lim_{s\to 0} \frac{1}{5s} = \infty$ or you might say there is no static gain.</p>
| 11726 | What is the gain and time constant of a system with a constant output? |
2016-10-05T01:54:33.990 | <p>I'm trying to design a jewellery box, with a mirror in the lid. When open, the lid should rest at an angle of about 135 degrees from original closed position.</p>
<p>I've been learning about 4 bar mechanisms, they sound appropriate (frame = box, coupler = lid) BUT in my case, the crank won't go through a full 360 degrees. </p>
<p>I've been using a <a href="http://www.mekanizmalar.com/fourbar.html" rel="nofollow">simulator</a> with crank=150, coupler=160, rocker=frame=225 but I need a way to model the start / finish positions to ensure everything lines up right.</p>
<p>Is there an app, or even a page full of math, covering the use of 4 bar mechanisms as a hinge? (I'm new to the mechanical engineering field, any help appreciated)</p>
<p><strong>EDIT:</strong> The techniques of designing joint positions are apparently what I'm looking for. <strong>Mechanism synthesis</strong> (2- or 3-position synthesis) is the term to look for. It would seem I'm also going to need a full-blown CAD program, as compared to simple web-based apps I've been using</p>
| |mechanisms| | <p>Reading your question leads me to belive that you are not really after a simulator <em>at all!</em> Instead you are looking for a <strong>mechanism synthesis</strong> method, which allows you to design the mechanism. A simulator only shows how the thing would work given measurements but does not help you into placing the joints that you need.</p>
<p>In your case one of the simpler synthesis methods is probably going to suffice, although there are many I would suggest that you use either <strong>2 position</strong> or <strong>3 position synthesis</strong> which are easy to make on your own all you need is a ruler and a compass, or a suitable drawing application/CAD.</p>
<p>The 3 position synthesis is a special case of the 2 position synthesis. So by teaching the 3 position method i will in fact teach you both. I do suggest you draw this is some CAD application as it makes it easy for you to parametrise the values as you may want to fiddle with the joint positions a bit to get a better result (any MCAD can do this). Let get started*:</p>
<ol>
<li><p>Draw a linear span of what connect the joints on the lid. (You may want to draw the lid also so you can see how the entire thing behaves). This will be your first position, i suggest you start with lid closed. Name the joints so you know whet connects where (I will name the joints B1 and C1 for first position. I start with B so they the designed joints will get the name be A and D)</p>
<p><a href="https://i.stack.imgur.com/uC3an.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uC3an.png" alt="enter image description here"></a></p>
<p><strong>Image 1:</strong> Draw the first position (lid closed). Name the joints and position.</p></li>
<li><p>Draw the second position, say where the lid rests in open position. </p>
<p><a href="https://i.stack.imgur.com/IItLv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IItLv.png" alt="enter image description here"></a></p>
<p><strong>Image 2:</strong> Draw the second position (lid open)</p></li>
<li><p>Connect the joints with lines. B1 to B2, C1 to C2. Find the midpoint on the connecting line and draw a perpendicular line at the midpoint. The joint A now need to be on any point at the perpendicular line that hits midpoint B1 and B2 while the joint D needs to be on the other line.</p>
<p><a href="https://i.stack.imgur.com/eoeMT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eoeMT.png" alt="enter image description here"></a></p>
<p><strong>Image 3:</strong> Joints should be anywhere on line A or line D, you pick.</p></li>
<li><p>Due to the set being over constrained and the fact that i can not control the descent direction. I can take a 3rd position and repeat the above method. This will give me an answer that is fully defined. Since i want the lid to approach quite vertically it would be best to position it close to B1. Due to practical considerations on where the joint can be I'm going to place this into a virtual position that happens before open. <strong>Repeat steps (2-3)</strong></p>
<p><a href="https://i.stack.imgur.com/B0nGd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B0nGd.png" alt="enter image description here"></a></p>
<p><strong>Image 4</strong>: The second constraint forces the joints A and B to be in the position where the 2 constraints cross.</p></li>
</ol>
<p>You still need to check that you havent got a degenerate solution. That's it. After doing this a few times i'm sure you have the ultimate positions. Go search your own solution. There are more sophisticated methods, but ultimately there is no perfect way to choose. It is always a bit open to human intuition, there are more solutions possible (but that's good this makes them patenttable). But 3 points synthesis should suffice for you. Like i said this is the place where you crack open your CAD as it makes it easy for you to explore the space </p>
<p>* I am drawing in illustrator so i can not easily remake the positions but if you use a CAD then it would be easy to just drag things until your position is good enough.</p>
| 11749 | 4 bar mechanism simulator |
2016-10-05T04:49:31.110 | <p>Is there a specific way to figure out the gear ratio I would need for a wind up mechanism that could last around one year? Would I start out with a really high tooth and step down too a small enough ratio for a complete cycle to last a year? </p>
<p>Do I need to devise a polynomial function, then use calculus to solve for the gear ratios, number of teeth, and time?</p>
| |mechanical-engineering|gears| | <p>Starting with the gear ratio is the wrong way to look at this problem. Start by determining the total energy you need for one year. No amount of gearing can fix it if the fully wound coil doesn't hold the energy you need.</p>
<p>Once you know the energy you need, you look for a spring coil that can store at least a little more than that. Only then are you ready to consider gear ratio. The spring spec will give you the number of turns over the coil's useful discharge life. Only you can tell how many turns of some shaft you need to run your device for one year. The overall gear ratio is then the latter divided by the former.</p>
| 11750 | Time duration for a mechanical wind up mechanism? |
2016-10-05T13:36:14.820 | <p>I have the formula for the translational stiffness which is :
<img src="https://latex.codecogs.com/gif.latex?k=%5Cfrac%7BA%5Ccdot&space;E%7D%7Bl%7D" title="k=\frac{A\cdot E}{l}" /></p>
<p>However, what I would like to find is the rotational stiffness. Is there any formulas to convert from one to an other knowing the pitch of the screw and the diameter ?</p>
| |mechanical-engineering|bolting|stiffness| | <p>The translational stiffness is written as
$$ k_l = \frac{A\,E}{\ell},$$
where the stiffness $k_l$ is in $\left[\frac{N}{m}\right]$, the area $A$ is in $\left[m^2\right]$, the young's modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$.</p>
<p>$$\,$$</p>
<p>The rotational stiffness is written as
$$ k_r = \frac{G\,J}{\ell},$$
where the stiffness $k_r$ is in $\left[\frac{Nm}{rad}\right]$, the second moment of area (or torsion constant) $J$ is in $\left[m^4\right]$, the rigidity modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$.</p>
<p>$$\,$$</p>
<p>Assuming linear elasticity the following relation between the rigidity modulus and the young modulus holds
$$ G = \frac{E}{2(1+v)},$$
where $v$ is the poison ratio.</p>
<p>$$\,$$</p>
<p>The second moment of area (or torsion constant) for a circle is
$$ \iint\limits_{R} r^2\,dA =
\int_0^{2\pi}\int_0^r r^2\left(r\,dr\,d\theta\right) =
\int_0^{2\pi}\int_0^r r^3\,dr\,d\theta =
\int_0^{2\pi} \frac{r^4}{4}\,d\theta = \frac{\pi}{2}r^4 =
\frac{A\,r^2}{2}.
$$</p>
<p>$$\,$$</p>
<p>Hence the rotational stiffness of a cylinder can be rewritten as function of $k_l$
$$ k_r = \frac{\frac{E}{2(1+v)} \frac{A\,r^2}{2}}{\ell} =
\frac{A\,E}{\ell} \frac{r^2}{4(1+v)} =
k_l \, \frac{r^2}{4(1+v)}
$$</p>
<p>This equation does not hold for any other shape! </p>
<p>When looking at a screw, there is a thread at the outside which has a significant effect on the rotational stiffness.</p>
<p>For example, leaving half the diameter open in the cylinder, the stiffness will only decrease $0.0625\%$ which is insignificant. Increasing the outer radius will increase the stiffness significant.</p>
| 11755 | Convert translational stiffness to rotational stiffness of a screw |
2016-10-05T16:06:04.300 | <p>I understand that the inside of a carbonated aluminium beverage can is pressurised. When the tab is pulled up,a rivet is lifted. Does this cause the can to depressurise? So then the can can be pierced with less force.</p>
<p>Thanks.</p>
| |mechanical-engineering|pressure| | <p>Yes. You are correct. The can is first depressurized as the rivet is lifted allowing the lever action of the tab to pop open the scored can opening</p>
<p>Watch this short video showing this is detail: <a href="https://youtu.be/ekv0kprA3AY" rel="nofollow">https://youtu.be/ekv0kprA3AY</a></p>
| 11757 | When beverage can stay-on tabs are lifted do they depressurise the can by lifting the rivet? |
2016-10-06T00:31:28.227 | <p>I know the ratio at which the cross sectional area changes to length is Poisson's ratio but is this why necking is more apparent in some materials than others or is the necking effect only dependent on the material's ductility? </p>
| |mechanical-engineering|civil-engineering| | <p>It is important to think about the underlining physics of a tensile test. What material are you testing? What is the geometry you are testing? How much global strain is being applied? With these questions in mind, I think it is necessary to think about this problem both in terms of elastic and plastic deformations. </p>
<p>Consider the stress-strain behavior for two materials, one with no plasticity (traditionally rubbers) and one that exhibits plasticity (traditionally metals). First let's focus on the elastic material. It is possible to imagine a situation were differences in Poisson's ratio manifests as differences in necking behavior. One might consider there to be necking in a simple rectangular tensile specimen. In this situation the boundary conditions are such that displacements at the ends of the specimen are zero. Therefore, during deformation a curved edge profile might develop. For perfectly elastic deformations this is where it is clear that differences in Poisson's ratio might affect the relative necking between two materials. If the Poisson's ratio is zero, then we would not expect there to be any geometric changes in the sample--the rectangular specimen just gets longer. However, on the other end of the spectrum, if the material is incompressible, $\nu = 0.5$, the initial and final configurations must have the same volume. Geometrically this means, given the fixed zero displacement boundary conditions of our test, there needs to be some distribution of deformation, necking, within the specimen. </p>
<p>In the context of plasticity the picture gets a bit more complicated. The short answer is that the final geometry of a tensile specimen will be dependent on its material properties. This includes both the plastic and elastic behavior (including the Poisson's ratio) of the material. Materials that are more ductile have a higher strain to failure, and this feature, when comparing elastically similar yet plastically dissimilar materials, predominately drives differences in necking behavior. If you have two materials, one which can withstand 2% strain and one 20% (assuming $\epsilon_{yield} = 1\%$), each increment of strain the second material can withstand relative to the first will contribute to differences in the final geometries of the specimens. In this example the failure of the material is driving the final configuration of the neck.</p>
<p>Finally, it is important to realize that one necking starts to occur things are no longer as simple as they appear. Once the specimen starts to neck the stress state with in the gage is, if it ever really was, no longer uniform. This means that at the smallest section of the neck stresses increase (dropping crosssectional area) and new components of stress start popping up. </p>
| 11767 | Does Poisson's effect explain why the necking effect is more apparent in some materials during a tensile test? |
2016-10-06T05:06:49.520 | <p>I am new to thermal management.</p>
<p>I want to lift about 250 W of heat from hot end of thermoelectric cooler/peltier cooler (TEC). </p>
<p>The temperature of hot side of TEC is 30°C. </p>
<p>I want to use some kind of heat exchanger/chiller to accomplish this task.</p>
<p>I am using equation</p>
<p>$$ P = h S (T_s-T_f) $$
where </p>
<p>$P$ = heat to be removed = 250 W</p>
<p>$h$ = heat transfer coefficient </p>
<p>$S$ = area of contact between hot side of TEC & fluid (air/water) used
for convective heat transfer</p>
<p>$T_s$ = temperature of hot end</p>
<p>$T_f$ = temperature of fluid (air/water)</p>
<p>I do not know how to calculate $h$. If I calculate that, it will give me temperature of fluid. </p>
<p>How do I compute $h$?</p>
<p>And is my approach correct?</p>
| |mechanical-engineering|thermodynamics|heat-transfer|heating-systems|heat-exchanger| | <p>This is a convective heat transfer problem you're asking about. Calculating the coefficient isn't normally possible.</p>
<p>What you can do is read up on stuff like fin geometry on heat sinks, and CFD. The long-term solution here is to derive an equation for your system, something like P = f(X) where X is (for example) the length of a fin. Then, perform a CFD simulation for X = 1 to get P for that particular geometry. Then you can scale your fin until you get the right value for P.</p>
<p>It's get a bit more complicated because you probably shouldn't use dimensioned parameters for something like this. Instead, consider something like the Reynold's number of the system.</p>
<p>Some more reading to help you get started:</p>
<ul>
<li><a href="http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node128.html" rel="nofollow noreferrer">Heat Transfer From a Fin</a></li>
<li><a href="http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html" rel="nofollow noreferrer">Convective Heat Transfer</a></li>
</ul>
| 11769 | Lifting 250 watts of heat using some heat exchanger/heat sink |
2016-10-08T12:25:25.363 | <p>Tension and compression re-bars in a concrete beam are connected by stirrups vertically tied together. If the space in between (this corresponds to web of an I-Beam) carrying shear load predominantly why are stirrups not placed along $\pm 45^0$ principal stress direction for better material economy ?</p>
<p>Such a design practice could be in aerospace engineering for lightweight construction but not found in the originating civil/structural construction practice ? </p>
| |structural-engineering|stresses| | <p>In days of old (not sure how old) stirrups did get installed at the ends of beams in an inclined fashion. I believe CHBDC currently still has clauses for it, but I never see it used for new design. I have used it however in evaluating old designs. </p>
<p>Additionally to inclined stirrups, it was also common practice at one point to bend some of the bottom longitudinal bars up at the end of the beams where moment was decreasing and they were no longer required. However since the shear was increasing some of this bent bar was also used for shear reinforcing. </p>
| 11791 | Stirrups are not shear bearing diagonals but just spacers? |
2016-10-10T00:34:28.137 | <p><strong>How to deal with thermal expansion in a city wide heating system?</strong></p>
<p><a href="https://i.stack.imgur.com/fptiK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fptiK.jpg" alt="Picture of the pipes in street"></a></p>
<p>Background:</p>
<p>As part of the program to combat pollution in and around Beijing, my adopted <a href="https://en.wikipedia.org/wiki/Zhuozhou" rel="nofollow noreferrer">city of Zhuozhou</a> (population, over half million, about 90 minutes Southwest of central Beijing, inhabited for at least 5,000 years) is installing a hot water heating system. In the last 3 weeks, they have torn up every major street in town, and they are in the process of installing pipes. </p>
<p>There are two pipes because the water will be conserved; according to my father in law, the hot water will flow through one pipe, they enter nearby buildings, flow through the radiators in the building, then return back to the central heating facility.</p>
<p>Of course, everyone is happy as this will stop a lot of coal burning, and vastly improve the air quality. Nobody seems too bothered that every street in town has been dug up at once. While traffic is horrendous, it seems like they want centralized heat.</p>
<p>What I have seen thus far is that they put the pipes close to each other, then weld them together.</p>
<p>I think when the heat is running, the length and size of these pipes will change drastically. <strong>How do you make sure the street above the pipes won't buckle? When the pipes get longer, how to prevent them from leaking? Won't the welds break? Also, are they a special type of pipe?</strong></p>
<p><a href="https://i.stack.imgur.com/M8anh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M8anh.png" alt="enter image description here"></a></p>
| |civil-engineering|temperature|piping|heating-systems| | <p>Thermal expansion axially in pipes will be far less than you expect and the calculations for it demonstrate it. Steel is quite strong and elastic in both tensile and compressive states compared to concrete or stone.</p>
<p>In these pipes I shouldn't expect them to expand much beyond 1mm in every 200m and steel can take enormous compressive stress loads and generally its capacity is described by Young's modulus as <span class="math-container">$210kN/mm^2$</span>.</p>
<p>It is only above 800°C that steel becomes more ductile and faces structural creep issues of any concern.</p>
<p>Buried S bend joints can absorb expansion stresses every 500m and remain within the elastic range of the material so, no need to worry much beyond the completion date of the project and traffic woes.</p>
<p>What is far more interesting to me is the design lifespan of the project as there is less than 50 years of oil left and systems like this will become white elephants of a bygone energy age.</p>
| 11804 | How to deal with thermal expansion in a city wide heating system? |
2016-10-10T03:25:58.230 | <p>I'm a Junior computer engineer working on a project involving a 3d printer. It is a mobile printer with the print head attached to wheels which move along the x,y axis of the surface it is printing on. The wheel movements may be off by less than a millimeter as the printer is moving. This causes a serious problem after printing a few layers as each layer does not land on top of the other. </p>
<p>I am in the process of implementing a sensor to detect the location of the print head with respect to some arbitrary reference point. My task is to implement a feedback control loop to correct the printer's movements and have the object printed properly. </p>
<p>I have an idea of how I want to solve this; using the sensor to detect the current location, then read the code to check what is the desired location, I then implement code to move the difference. </p>
<p>My question is where does the concept of PID control even fit in? The following equations were taken from wikipedia for time/frequency domains. </p>
<p><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/7e13a0cb7bc543e507ae58384edcebcdcace5d27" alt="equation"></p>
<p><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/842e237127cf16ee7e961c84116df3946c887b99" alt="equation"></p>
<p>I am confused about the values of t and s in the above equations. I have numeric values for input, output, error and the K constants. None of the values are functions so I don't understand what happens with the derivative and integration involved. Sorry if this seems like a dumb question but I am a complete novice to control theory and have been learning from online articles.</p>
| |control-theory|3d-printing| | <p>e(t) and u(t) are functions.</p>
<p>e(t) shows the difference between required location and current location.
u(t) is the voltage input to the motor.</p>
<p>That means let suppose your printer head is at 4cm w.r.t reference point at t=1s but it must be at 5cm w.r.t reference point so the difference will be 1cm at t=1s. So the voltage input the motor will be u(1)=Kp(1)+Ki(1+e(t-1))+Kd(1-e(t-1))/1. So when t=1.1s e(1.1) will decrease due to previous effort on the motor by u(1). And soon error will reduce to zero. Hence u(t) and e(t) are varying with time they are functions. </p>
| 11807 | Implementing PID feedback control |
2016-10-10T20:16:38.460 | <p>I'm looking for a control method for a production process with the following characteristics:</p>
<ul>
<li>1 control variable</li>
<li>many process parameters (+-50)</li>
<li>1 resulting variable</li>
<li>continuous measurement of resulting variable with 30-90 second delay</li>
<li>complex physics that govern the process and numerous factors that determine the (non-linear) relation between control variable and resulting variable</li>
</ul>
<p>With constant process/input parameters, the distribution of the resulting variable is normal.</p>
<p>The control method should ensure that the resulting variable should be above a set minimum and that the average of the resulting value should be as close to the limit as possible. Being above the lower limit makes the product more expensive, being below the lower limit however is more expensive: the product is scrapped.</p>
<p>Now I want to try the following: I have want to make a statistical model using some process parameters as input predicting the control variable that leads to the desired value for the resulting variable. Next to the value of the control variable, I also want to determine the 'confidence' or distribution of that value. Then I want to state a portion of the time the control variable must be above the lower limit (i.e. 3 sigma). Consecutively I'd like to set the setpoint for the control variable at the such a value that I achieve the given portion above the lower limit. To illustrate what I envision I made the illustration below.</p>
<p><a href="https://i.stack.imgur.com/E4uRa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E4uRa.png" alt="Feed forward Control Optimization based on prediction confidence"></a></p>
<p>Do you think this setup makes any sense? Am I missing any obvious alternative solution that is much simpler? Is this a known approach? </p>
| |control-engineering|steel|process-engineering|optimal-control| | <p>In a totally different field, I've approached this by building a model that predicts a distribution rather than a single value. Some examples I've used:</p>
<ul>
<li>Quantile regression (Neural Network, using a different (pinball) loss function for each quantile)</li>
<li>Predicting both Mean and Standard deviation</li>
</ul>
| 11818 | How to optimise feed-forward control of a process based on a prediction using the prediction confidence? |
2016-10-11T05:56:17.237 | <p><a href="https://i.stack.imgur.com/XUpGx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XUpGx.png" alt="enter image description here"></a>
For the connection above, I think that once the beam is loaded, their is no room for rotation of the beam.</p>
<p>But as per British Standard this is a pinned connection. So I have been trying to determine how a connection is defined as Pinned or fixed. After reviewing American and British codes, I have come to the conclusion that steel structure designers need to follow their Standard connections for design (for which they have done experiments) or conduct experiments on connections to determine to evaluate whether its a pinned or connections based on certain criteria. Is my understanding correct?</p>
| |steel|structures| | <p>Pinned connection in structural context doesn't mean free rotation/zero moment. Pinned connection means much bigger rotations and much lower strength compared to the connected members. According to Eurocode 3 (EN1993-1-8:2005 section:5.2), if the moment strength is below 0.25 of the connected beam and its stiffeness is below a limit (relative to the beam) then it should be considered nominally pinned, provided adequate rotational capacity is available (for example through ductile critical mechanism of failure). </p>
<p>Normally one <strong>should find the moment strength and stiffness of the connection and classify it against the limits</strong>. A connection can be classified either as full strength/rigid or pinned or semirigid. </p>
<p>For the connection in question, it is obvious that no transfer of force occurs through the beam flanges. The bending moment finds its way through the flanges mostly. So this connection is ineffective to transfer enough of the beam end moment. Also take into account that the connection is bolted to the web of the column which is quite flexible. </p>
<p>Finally, i want to emphasize that the following statement is not accurate:</p>
<blockquote>
<p>once the beam is loaded, their is no room for rotation of the beam.</p>
</blockquote>
<p>Don't assume that based on the undeformed geometry!</p>
| 11824 | Steel Structure connection pin or fixed |
2016-10-12T00:25:06.803 | <p>I am currently tinkering with a railgun experiment, and I was thinking of using a liquid nitrogen loop to cool the rails down to very low resistance, and inject some nitrogen into the barrel between the rails, to remove some oxygen and prevent corrosion, and excessive arcing. I would run the liquid nitrogen through a hole down the length of the rail, and into the barrel through much smaller holes connecting to the main line in the rail. The nitrogen would go from the tank through a pump, through one rail, into a tubing to the other rail and back to the tank.</p>
<p>Now I know nitrogen is magnetically transparent, and being a mostly inert gas doesn't run the chance of reacting unexpectedly. Some things I am unsure about:</p>
<ol>
<li><p>I have looked around but it seems there are not a lot of places that sell tubing and fittings for liquid nitrogen. Is there some place I can browse to look for standard size fittings and tubing, and a small pump to carry liquid nitrogen through my system? The other difficulty is that I need to keep fittings and tubings to magnetically transparent non conductive materials, otherwise the magnetic field while firing (let's say 10 kJ) will mangle them. I am guessing there will be some custom fabrication involved, but still an idea of the basic sizes will help me make some design decisions.</p></li>
<li><p>At what pressure is it safe to carry liquid nitrogen through a system like this? Obviously the pressure will not be very high, but since I need to bleed some gas into the barrel, I need to set a target main line pressure to calculate injector channels size.</p></li>
<li><p>Is there an inherent danger I am overlooking with running an inert gas through a rail that will conduct a 1 ms pulse of 10 kJ? Obviously it won't ignite or explode, but is there a chance it will ionize at those relatively low energies?</p></li>
</ol>
<p><strong>Edit:</strong></p>
<p>After some more looking, it seems PTFE (Teflon) is the right material for the job, but its stability under a high magnetic field is unclear. Is there a better material for this?</p>
<p><strong>Edit 2:</strong></p>
<p>This is a crude cross section of the proposed arrangement. Orange is the copper rails, dark grey is PTFE, light grey is payload.</p>
<p><a href="https://i.stack.imgur.com/E2OSG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E2OSG.png" alt="enter image description here"></a></p>
<p><strong>Edit 3:</strong></p>
<p>There has been concern about the heat dissipation in the system, and I should mention that the system is fed power through some large inductors that should dissipate the bulk of the heat. They are in place to stretch out the pulse in the first place. The dissipation in the rails should be very, very low, considering oxygen-less copper at 77K has an abysmally low resistance. The coils will probably also get the nitrogen treatment.</p>
| |pressure|power-electronics|hydraulics|cooling|electromagnetism| | <p>Regarding questions 2 and 3, I think you need to look deeper in the chemistry, reactivity, and heat transfer characteristics of Nitrogen.</p>
<p>1) It is not a noble gas. It is reactive, and you benefit from this reactivity each time you eat a plant protein.</p>
<p>2) I have not designed a railgun and I don't have much sense of the thermal inefficiencies here. But I think it would be a good idea to consider what kind of heat your launch pulse is going to transfer to the liquid nitrogen in the system. Depending on the temperature of the nitrogen (presumably just at -196 C, i.e. boiling point) then (some?) nitrogen in the system is likely to "flash" into its gaseous form during the launch, and pressure will go up considerably. (Assuming atmospheric pressure and ideal gas behaviour, it will take up ~700x the volume as soon as it gasifies.) Obviously you'd want to make sure the sudden gas pressure doesn't immediately blow up all the tubing, or the barrel. </p>
<p>Some aspects of this system will be similar to a water-cooling system in a foundry or a power plant, and there will be much more engineering experience with that coolant. You're just dealing with something 300 C colder. </p>
<p>Possibly you've already modelled out this thermal stuff but if so, mention it in the question to prevent us from worrying. :) </p>
<p>Your project sounds like fun, good luck! </p>
| 11833 | Some questions about liquid nitrogen cooling |
2016-10-12T06:16:28.417 | <p>Is there a word for the tendency of some metals to make crayon-like marks on other metals? For example, a lead bullet makes faint marks on some steels, but not on others with a similar surface finish.</p>
| |metallurgy| | <p>"Galling" is probably the word you want here. It's the tendency of a (usually) soft metal to break down under pressure and adhere to the harder material.</p>
<p>It's common with soft metals like aluminium, so that's why I think this is what you've noticed on lead too. However harder materials like stainless steel can gall too in certain circumstances, so it's not a hard and fast rule.</p>
<p>(The authoritative place to look would be to read up on Tribology by the way - that's the field of surface wear and other effects.)</p>
| 11839 | Word for the tendency of one metal to rub off on another? |
2016-10-12T07:58:38.240 | <p>Here's the setup, going through those four valves will be some fluid, probably water, but I need some fishing line going through this cylinder to be sealed. The line is anchored at two points, the top and bottom, and will be used to draw two points together. This apparatus is to heat and cool wound up fishing line, as that can be used as a muscle fiber. When hot water is applied, it contracts, when cool water is applied, it expands. This does mean that the line will be moving some percentage of the length, which may also mean that the diameter will change slightly.</p>
<ul>
<li><a href="http://spectrum.ieee.org/tech-talk/robotics/robotics-hardware/fishing-line-makes-superhuman-artificial-muscles" rel="nofollow noreferrer">Fishing Line Makes for Superhuman Artificial Muscles</a></li>
</ul>
<p><a href="https://i.stack.imgur.com/UOUsd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UOUsdm.jpg" alt="enter image description here"></a></p>
| |fluid-mechanics|heat-transfer| | <p>A high-quality seal that diameter would be very difficult to fabricate. You always want to avoid seals in designs when you can help it. For example, you could fix one end inside the cylinder so that you only have one moving end to worry about.</p>
<p>If you allow for some leakage and run the water at low pressure (less than 12" of head) you could probably get away with melting a small cup of wax around the line. Have the wax be as tall as the diameter of your primary tube. The longer distance will reduce water leakage. Pull the line taught while it cools so the path is straight. It should pull free with very little force, and with little movement during operation, it should provide a fair amount of operating life.</p>
<p>A more robust solution would be to put a diaphragm (as mentioned in the comments) on the end cap instead of a seal. The diaphragm will act like a soft spring, and can be compensated for with other spring components if necessary.</p>
<p>A design that you may or may not have considered; would be to use a single tube as your tension member and run hot and cold fluid through this tube. The design would be simpler, the volume of fluid per tension force would be lower, and the heat transfer would be quicker. Something to think about.</p>
| 11842 | Can I make a fishing line going through a small hole waterproof? |
2016-10-12T19:38:37.030 | <p>I am working with a peristaltic pump to move liquid to/from a reservoir through a closed loop of tubing. The motor can rotate clockwise or counter-clockwise when set with either of the directional buttons below. </p>
<p><a href="https://i.stack.imgur.com/Azc9e.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Azc9e.jpg" alt="enter image description here"></a></p>
<p>The direction of rotation can also be controlled remotely using contacts #1/2 in the green barrier strip below:</p>
<p><a href="https://i.stack.imgur.com/cHAli.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cHAli.jpg" alt="enter image description here"></a></p>
<p>I have little experience with this kind of work. Would this be a simple enough task to connect, program and use a controller to alternate the direction of the motor for some time interval? </p>
<p>Even if someone could describe the process for me in simple terms, it might be helpful in my understanding of how to approach the challenge. </p>
| |control-engineering|pumps| | <ol>
<li>Find the manufacturers spec sheet if possible. This will detail the voltages, pin configuration, and more control details. </li>
<li>If the spec sheet specifies serial commands like 706Astor mentioned, you may be able to use a usb-to-serial converter and a serial console like <a href="http://www.chiark.greenend.org.uk/~sgtatham/putty/download.html" rel="nofollow">putty</a> for testing it. You could then send serial commands via <a href="https://pythonhosted.org/pyserial/" rel="nofollow">python</a> or other programming language.</li>
<li>If serial is not an option, test the inputs with an <a href="http://www.apogeekits.com/images/alligator_leads_red_black.jpg" rel="nofollow">alligator clip</a> or other testing leads. I assume based on the diagram that connecting 1 and 2 makes it run counter-clockwise, leaving them disconnected is clockwise. Connecting 3 and 4 makes it run, disconnecting makes it stop. Connect a 1.5v AA or AAA battery up to 5(+side) and 6(-side) to test its speed at 1.5/5 = 30% speed.</li>
<li>If your tests work out and you now understand the operation, and all you want to do is change the direction at some interval, you might be able to use a <a href="http://rads.stackoverflow.com/amzn/click/B012FSL2GK" rel="nofollow">cheap dc timer relay</a> to connect contacts 1 and 2.</li>
<li>If you need more control or more adjustment on the timing, you will want to get an <a href="https://www.arduino.cc/en/Main/ArduinoBoardUno" rel="nofollow">Arduino UNO</a>. They are very popular with lots of tutorials and internet community support. There are also some good tutorials on <a href="https://learn.adafruit.com/category/learn-arduino" rel="nofollow">Adafruit</a>.</li>
</ol>
| 11850 | Using a controller to remotely alternate the direction of a peristaltic pump |
2016-10-13T03:53:42.183 | <p>I was wondering, in laymen terms, what is the exact definition of renewable energy and what are its various forms, which ones are better and in what situations can they / can they not be used.</p>
<p>How do you weigh the cost of setting up this technology with its payoff, and how do you set these benefits apart from those derived by investing in other areas such add healthcare, research, and development in other areas such add food and disease fighting?</p>
<p>Sends like there are manpower-based investments and research based investments.</p>
<p>I realize this is a very broad question, but tried Wikipedia and could not find a laymen answer to my broad question.</p>
<p>NOTE: If my question is to broad, then please just pinpoint this very basic question: how do you define renewable energy?</p>
| |renewable-energy| | <h3>overview:</h3>
<p>Renewable energy is one for which the input energy is renewed on human timescales.</p>
<p>Remember, energy is always conserved. It doesn't come from nowhere. Pretty much all the energy we use on Earth comes from the stars - and almost all of it comes from one star in particular, the Sun.</p>
<h3>not renewables:</h3>
<p>Uranium (for <strong>nuclear</strong> power) is in the earth after being manufactured in generations of stars. This takes billions of years, and is thus not renewable. <strong>Coal</strong>, <strong>gas</strong> and <strong>oil</strong> are in the earth after having been living organisms (plants, animals, etc) millions of year ago, and are thus not renewable.</p>
<h3>renewables:</h3>
<p><strong>Wind</strong>, <strong>solar</strong>, <strong>hydro</strong>, <strong>biomass</strong> and <strong>waves</strong> are all powered by incoming solar energy every second of the day, and so are constantly replenished at human timescales. The <strong>tides</strong> are renewed every day by the relative motion of the Earth and the Moon (and to a much lesser extent, the sun too). <strong>Geothermal</strong> energy is renewed every second of the day by slow nuclear reactions deep in the ground, and in some cases by being replenished constantly by the heat emitted from magma.</p>
<h3>endnote:</h3>
<p>You've asked a very broad range of questions. Many of them, <em>when refined</em>, would be best handled on our Sustainability Stack, where we already have quite a few <a href="https://sustainability.stackexchange.com/questions/tagged/renewables">questions on renewable energy</a> - the economics, the engineering, and so on. You'll also find questions on the <a href="https://physics.stackexchange.com/questions/tagged/renewable-energy">physics concepts behind renewable energy, on our Physics site</a>.</p>
| 11856 | What is the exact definition of renewable energy |
2016-10-15T10:41:08.093 | <p>I am a computer engineering student and currently having my undergraduate thesis. My thesis is about a chicken egg incubator and i need to control the humidity inside the incubator.</p>
<p>Right now, i dont know how to calculate the transfer function of the humidity control system. Please someone point me to the right direction.</p>
<p>Is it possible to get the transfer function if i can give a step input (turn on the humidifier) and then get the s-domain of the output, then divide the output over the input because transfer function = output/input? or is this method difficult considering im only an undergraduate computer engineering student?</p>
| |thermodynamics|control-engineering|chemical-engineering| | <p>If a system can be approximated as a (stable) linear time invariant system, than any bounded input which contains all frequencies (for digital signals limited to the Nyquist frequency) can be used to identify the system.</p>
<p>The Laplace transform of a step function is $s^{-1}$, so high frequency behavior might be more prone to be hidden below a noise floor. So you would need more measurements in order to average away this noise, if you are interested in those high frequencies.</p>
<p>Once you have multiple measurements of input data, which contain "all frequencies", and output/system-response data, then you can approximate the frequency response function of the system, denoted with $P(s)$, using,</p>
<p>$$
P(s) \approx \frac{\sum Y_i(s)\,\bar{U}_i(s)}{\sum U_i(s)\,\bar{U}_i(s)},
$$</p>
<p>where $Y_i(s)$ is the <a href="https://nl.wikipedia.org/wiki/Fast_Fourier_transform" rel="nofollow">fast fourier transform</a> (FFT) of the $i$th measurement of the output, $U_i(s)$ the FFT of the $i$th input data, and $\bar{U}_i(s)$ means taking the <a href="https://en.wikipedia.org/wiki/Complex_conjugate" rel="nofollow">complex conjugate</a> of $U_i(s)$ (which will help reduce the amount of noise in $P(s)$ the more measurements you use). Usually you would also apply a <a href="https://en.wikipedia.org/wiki/Window_function" rel="nofollow">window function</a> to each time domain signal, before calculating the FFT. A common choice is the <a href="https://en.wikipedia.org/wiki/Window_function#Hann_.28Hanning.29_window" rel="nofollow">Hanning window</a>.</p>
| 11877 | How to model a humidity control system inside a box? |
2016-10-17T12:05:58.090 | <p>I am trying to derive the isentropic flow equations for a compressible gas by myself and at the end, I have different formulation than the one in the literature. Can you please tell me what am I doing wrong?</p>
<p>So we have a nozzle:</p>
<p><a href="https://i.stack.imgur.com/YE0vB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YE0vB.png" alt="enter image description here"></a></p>
<p>If we do an energy balance and consider that the kinetic energy and the potential energy in point 1 is negligible, one ends up with this relation:
$h_1=h_2+\dfrac{v_2^2}{2}$</p>
<p>By using the ideal gas relation $h = CpT$, and if we then divide the equation by Cp and then by $T_2$, we end up with something like this:</p>
<p>$\dfrac{T_1}{T_2}=1 + \dfrac{v_2^2}{2T_2Cp}$</p>
<p>And finally using $Cp = Cv + R$, denoting $Cp/Cv = k$, using the formula for the mach number $M = v/c$ and the speed of sound in gas $c = \sqrt{(T_2Rk)}$, we end up with this relation:</p>
<p>$\dfrac{T_1}{T_2}=1 + \dfrac{k-1}{2}M^2$</p>
<p>But from the <a href="https://www.grc.nasa.gov/www/k-12/airplane/isentrop.html" rel="nofollow noreferrer">literature</a> one can find that this formula is written like this:</p>
<p>$\dfrac{T_t}{T}=1 + \dfrac{k-1}{2}M^2$</p>
<p>The question is... Is the total temperature $T_t$ in their equation the same as $T_1$ in our case? And if the nozzle discharges to the ambient, is $T$ in their formulation the same as $T_2$ in ours? I'm a bit confused with the symbols and the meanings and I want to learn how this works.</p>
<p>P.S. This is a copy of a question from <a href="https://physics.stackexchange.com/questions/286940/isentropic-flow-equation-derivation-vs-literature">enter link description here</a>. I have been suggested to try and ask here.</p>
| |fluid-mechanics|thermodynamics|compressed-gases|compressible-flow| | <blockquote>
<p>Is the total temperature $T_t$ in their equation the same as $T_1$ in our case? </p>
</blockquote>
<p>$T_t$ is called the stagnation temperature, which is the temperature of a flow when it's brought to rest ($v=0$), the $T_1$ you defined in the first equation is called the stagnation temperature since you ignored kinetic energy.</p>
<p>$$C_pT_t = C_pT + \frac{v^2}{2}$$
$$T_t = T + \frac{v^2}{2C_p}$$</p>
<blockquote>
<p>And if the nozzle discharges to the ambient, is $T$ in their formulation the same as $T_2$ in ours?</p>
</blockquote>
<p>Yes, the arbitary $T$ in the equation is for any location in the nozzle including the discharge.</p>
| 11894 | Isentropic flow equation derivation |
2016-10-17T21:45:08.487 | <p>Consider a simple structure such as a beam, plate, ... Assume that you know how to determine the different modes of vibration of the structure. You are given an external input, such as a force or moment at a certain location. How can you then determine which modes will be excited how much? (Just a general question, I am not considering a specific case.)</p>
| |structural-analysis|modal-analysis| | <p>They way to answer this question is transform the dynamics model into modal coordinates, and see what happens to the force term.</p>
<p>Suppose we can describe the stiffness and mass properties of the structure as matrices $\mathbf K $ and $\mathbf M$, and its displacement as a vector $\mathbf x$, in a physical coordinate system, and we apply a vector of forces that vary sinusoidally in time, $\mathbf F e^{i\omega t}$ to the structure.</p>
<p>The equation of motion of the system is then $$(\mathbf K - \omega^2 \mathbf M) \mathbf x e^{i\omega t} = \mathbf Fe^{i\omega t}$$</p>
<p>We can cancel the $e^{i\omega t}$ terms and reduce this to $$(\mathbf K - \omega^2 \mathbf M) \mathbf x = \mathbf F$$
(Note, I'm ignoring damping, for the sake of making things a bit simpler - doing that doesn't affect the final conclusion).</p>
<p>We can find the normal modes of the system and write the eigenvectors as a matrix $\mathbf \Phi$.</p>
<p>We can write the displacements $\mathbf X$ as a linear combination of the eigenvalues, i.e. $\mathbf X = \mathbf \Phi \xi$ where $\xi$ is a vector.</p>
<p>Substitute that in the equation of motion, and pre-multiply both sides by $\mathbf \Phi^T$ and we get $$\mathbf \Phi^T(\mathbf K - \omega^2 \mathbf M) \mathbf \Phi \xi = \mathbf \Phi^T \mathbf F$$
or $$(\mathbf \Phi^T\mathbf K \mathbf \Phi - \omega^2 \mathbf \Phi^T\mathbf M \mathbf \Phi) \xi = \mathbf \Phi^T \mathbf F$$</p>
<p>Now, if we use <em>mass-normalized</em> eigenvectors, we know that $\mathbf \Phi^T\mathbf M \mathbf \Phi$ is a unit matrix, and $\mathbf \Phi^T\mathbf K \mathbf \Phi$ is a diagonal matrix of the eigenvalues squared. So the matrix equation becomes a set of scalar equations, and for the $i$th mode we have
$$(\omega_i^2 - \omega^2)\mathbf \xi_i = \mathbf \Phi_i^T \mathbf F$$</p>
<p>Translating that equation back into words answers the OP's question: For each mode, you take the scalar product of each eigenvalue with the applied forces to find the "modal component of the force" (i.e. right hand side term of the final equation), and from that you can find the relative amplitude of that mode (i.e. $\xi_i$).</p>
<p>In many cases we only apply a force to a single degree of freedom of the structure. Then, the result is fairly intuitive - there are two effects which are relevant, when they are taken together: </p>
<ol>
<li><p>Looking at the right hand side of the final equation, <em>The modes that will be excited most are those with the biggest displacements at the point where the force is applied</em>.</p></li>
<li><p>Looking at the left hand side of the same equation, <em>The modes that will be excited most are those whose natural frequencies are close to the forcing frequency</em>. </p></li>
</ol>
| 11903 | How to know which modes are excited by a given input in modal analysis? |
2016-10-18T01:32:23.593 | <p>I'm trying to reverse engineer a set of spur gears that I have on a model RC car, but they are tiny, so there's a limit to what I can measure with my calipers. My plan is to create a new gear design with a few important changes that will mesh with these existing gears, then have it professionally 3d printed in nylon.</p>
<p>I've used this image as a reference for what data I might want to obtain:</p>
<p><img src="https://www.unf.edu/~aschonni/Research/design%20analys%20fabr%20verific/gear%20tooth/Gear_design_analysis_files/image005.jpg" alt=""></p>
<p>And here's an image of the gears in question:</p>
<p><img src="https://i.stack.imgur.com/ViFFm.jpg" alt=""></p>
<p>So far I have these measurements for each gear:</p>
<p><strong>Gear 1:</strong></p>
<ul>
<li><p>Teeth: 10</p></li>
<li><p>Pitch Diameter: 5mm</p></li>
<li><p>Outside Diameter: 6mm</p></li>
<li><p>Root Diameter: 4mm</p></li>
</ul>
<p><strong>Gear 2:</strong></p>
<ul>
<li><p>Teeth: 22</p></li>
<li><p>Pitch Diameter: 11mm</p></li>
<li><p>Outside Diameter: 12mm</p></li>
<li><p>Root Diameter: 9.5mm</p></li>
</ul>
<p>Note: I don't know how to obtain the base diameter, only the root diameter, since the base diameter doesn't seem to be directly measurable. Also, the pitch diameters are a guess, but I'm assuming for simplicity the people who designed them probably picked 5mm and 11mm to go with 10 and 22 teeth respectively. It's the sort of thing I'd probably do. Also it matches with the centre-to-centre measurement of the meshed gears being 8mm, which is the same as (5mm + 11mm)/2.</p>
<p>I'm thinking of using this OpenSCAD script to replicate the gear designs:</p>
<p><a href="https://github.com/openscad/MCAD/blob/dev/gears/involute_gears.scad" rel="nofollow noreferrer">https://github.com/openscad/MCAD/blob/dev/gears/involute_gears.scad</a></p>
<p>Do I have enough data to make a replica of these gears using that script? If not, what else do I need? The script's comments seem to indicate that I need to know the pressure angle, but looking at the diagram I linked, I can only figure that out if I have the base diameters.</p>
| |mechanical-engineering|design|gears| | <p>Given these measurements I think these are standard Module 0.5mm (Mod 0.5) gears, and that alone plus a little web search should let you design for them to mesh correctly, and <a href="http://www.technobotsonline.com/mod-0.5-gear-range.html">buy</a>, <a href="http://www.thingiverse.com/thing:683030">design</a> or <a href="http://www.chronos.ltd.uk/acatalog/Gear_Cutters_and_Arbors.html">manufacture</a> any further gears that will mesh correctly with them. </p>
<p>Easiest to measure:<br>
Outside diameter / (Tooth count + 2) = 0.5mm. </p>
<p>Useful cross check:<br>
Pitch diameter / (Tooth count) = 0.5mm. </p>
<p>Technically you may have to consider the possibility of incompatible Pressure Angles, but for modern (< 50 year old) injection moulded gears, I suspect that's such a remote possibility as to be non-existent. </p>
<p>I deal with 14.5 degree pressure angle Imperial gears on pre-WW2 lathes but as far as I know everything recent uses 20 degree pressure angles.</p>
| 11905 | Reverse-engineering spur gears - do I have enough data to make a gear that will mesh correctly? |
2016-10-19T10:35:33.580 | <p>I want to do some experiments on the dynamic behavior of the beam, plate and so on. Four boundary conditions I want to simulate: </p>
<ul>
<li>Free</li>
<li>Clamped</li>
<li>Simply supported</li>
</ul>
<p>For free boundary condition, I can use a string to tie the beam or plates up and hang them on some place. For clamped boundary, I can use some bolts to fix the beam or plate on a base. But how can I simulate the simply supported boundary conditions?</p>
| |mechanical-engineering|applied-mechanics|dynamics| | <p>For steel beams, these are relatively simple.</p>
<p>A simply supported beam requires two different supports: one to withstand vertical and horizontal forces (pinned support) and one to withstand only vertical forces (roller support).</p>
<p>Here are examples of each (click the links to see more examples and different types):</p>
<p><a href="https://www.google.com/search?q=pinned+support&tbm=isch" rel="nofollow noreferrer">Pinned supports</a></p>
<p><a href="https://i.stack.imgur.com/Gj9We.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gj9We.gif" alt=""></a><br>
<sub>(source: <a href="http://web.mit.edu/4.441/1_lectures/1_lecture13/pinned_1.gif" rel="nofollow noreferrer">mit.edu</a>)</sub> </p>
<p><a href="https://www.google.com/search?q=roller+support&tbm=isch" rel="nofollow noreferrer">Roller supports</a></p>
<p><a href="https://i.stack.imgur.com/brHTV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/brHTV.jpg" alt=""></a><br>
<sub>(source: <a href="http://ceephotos.karcor.com/wordpress/wp-content/uploads/20110923-kjans-Bridge-Roller-Support-Manistee-MI.jpg" rel="nofollow noreferrer">karcor.com</a>)</sub> </p>
| 11922 | How to simulate the simply supported boundary condition in experiments? |
2016-10-19T15:05:04.573 | <p>I am trying to make like a "periscope" that i can use with a mobile camera.
I have made a very quick prototype with two mirrors and it "works" but you can see the interior of the walls so it is not a good solution. It currently looks like this <a href="https://i.stack.imgur.com/MKPRT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MKPRT.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/vzUuv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vzUuv.jpg" alt="enter image description here"></a></p>
<p>So as I said the problem is that you can see the "walls" on the interior of the periscope when i put the camera in there.
Is there anyway I can adjust this or do I need to look at an optional solution? </p>
<p>What i want to achieve is to make the backcamera of the phone to be like a frontcamera and have it under the camera that is vertical. (just like my current model). Can I somehow adjust the periscope (build it better, add more mirrors, maybe add a magnifying glass?) or do I need to build something else completely in order to make it work flawlessly without interior walls etc being in the way making the quality bad.</p>
<p>Any tips, tricks, links is very much appreciated!</p>
| |optics| | <p>Ok let me draw some pictures you right now have the following situation:</p>
<p><a href="https://i.stack.imgur.com/0o3Pr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0o3Pr.png" alt="enter image description here"></a></p>
<p>If you add a convex lens to this setup you get:</p>
<p><a href="https://i.stack.imgur.com/EoJNf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EoJNf.png" alt="enter image description here"></a></p>
<p>Which does work. However you may want to add a concave lens after the convex lens so that you can more readily adjust the beam focus and shape giving you more freedom to shape your periscope.</p>
| 11926 | Periscope camera with a mobile phone, how to make walls not visible on the camera/alternative solution? |
2016-10-19T17:55:39.077 | <p>The problem given is as follows:</p>
<p>A Carnot cycle has a heat engine fluid efficiency of 30%. The heat transfer to the fluid in the boiler <em>Qh</em> happens at 270 centigrade. At 270 centigrade, the entropy of saturated vapour and liquid water are 6.001KJ/(kgK) and 3.067KJ/(kgK), respectively. Determine the net work produced by the Carnot engine.</p>
<p>I am pretty sure to solve it you need to use the Carnot efficiency equation so 0.30 = <em>W</em> / <em>Qh</em> but I am not too sure how to get <em>Qh</em> from the information provided. Please explain or put me on the right track if you can!</p>
<p>Thank you in advance!</p>
| |thermodynamics| | <p>Assuming reversible heat addition process:
$$q = \int_{s_1}^{s_2} T \,ds = T \int_{s_1}^{s_2}ds=T\Delta s $$
$$ w = \eta_{carnot} \,q$$</p>
| 11927 | How do you calculate the work of a Carnot Heat Engine using entropy? |