CreationDate
stringlengths 23
23
| Body
stringlengths 57
10.9k
| Tags
stringlengths 5
114
| Answer
stringlengths 39
16.3k
| Id
stringlengths 1
5
| Title
stringlengths 15
149
|
|---|---|---|---|---|---|
2022-04-27T17:49:33.203 | <p>What is a rough cut figure I could use to determine what the thermal conductivity of 10mil and/or 20mil of air sandwiched between two plates (perfectly conducting)?</p>
<p><a href="https://i.stack.imgur.com/zM2gd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zM2gd.png" alt="enter image description here" /></a></p>
<p>I understand that convection dies at a certain point radiation and conduction takes over since the air is not moving much if at all when there are small gaps. At what point does convection stop with a variable gap?</p>
| |fluid-mechanics|thermal-conduction|thermal-insulation| | <p><a href="https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1919.0002" rel="nofollow noreferrer">This paper</a> reviews the measurements available at the time of the thermal conductivity of air, and comes up with a typical value of <span class="math-container">$2.2\times 10^{-2}\,\mathsf{W}/\left(\,\mathsf{m}\,\mathsf{K}\right)$</span>. From kinetic theory, one would expect the exact value to depend on the flux density of molecules and on the mean free path, i.e. on temperature and pressure.</p>
<p>Since the configuration of interest has two horizontal-plane plates separated vertically, the type of convection at issue is known as <a href="https://en.wikipedia.org/wiki/Rayleigh%E2%80%93B%C3%A9nard_convection" rel="nofollow noreferrer">Rayleigh-B'enard convection</a>, and one would expect the convection to shut down when the <a href="https://en.wikipedia.org/wiki/Rayleigh_number" rel="nofollow noreferrer">Rayleigh number</a> drops below a critical value of <span class="math-container">$1\,708$</span>.</p>
| 50755 | What is the thermal conducivity of Air between two plates? |
2022-04-28T09:29:02.130 | <h1>Background</h1>
<p><a href="https://en.wikipedia.org/wiki/Henry_Maudslay" rel="nofollow noreferrer">Henry Maudslay</a> invented the "three plate method" for producing high-precision <a href="https://en.wikipedia.org/wiki/Surface_plate" rel="nofollow noreferrer">surface plates</a>, apparently required to produce replaceable parts with high precision. For example, <a href="https://industrytoday.com/engineering-surface-plates-the-most-important-tool/" rel="nofollow noreferrer">this website</a> claims that:</p>
<blockquote>
<p>To ensure all parts could be produced identically, Maudsley introduced the requirement of a true plane surface for measurement.</p>
</blockquote>
<p>In another example, <a href="https://youtu.be/sFrVdoOhu1Q?t=390" rel="nofollow noreferrer">in this video at 6:30</a>, the presenter describes a way to get two "coplanar" surfaces by bonding granite surface plates with epoxy, which apparently helps achieve high precision using a machine tool.</p>
<h1>My question</h1>
<p>How does an extremely flat surface help with the following?</p>
<ol>
<li>Producing replaceable parts (like screws, bolts etc.)</li>
<li>Achieving high precision with machine tools</li>
</ol>
| |machining| | <p>There are basically 2 ways you can make parts fit together.</p>
<ol>
<li><p>Make the parts crudely, and then file them down and test the fit until everything works.</p>
</li>
<li><p>Make the parts with precise tolerances, so that the parts fit every time without modification.</p>
</li>
</ol>
<p>#1 is obviously much simpler. You can create very nice mechanisms this way without any precision measuring tools. But it has a serious drawback. All the parts of your product (lets say a musket) are now "unique" and not interchangeable with the same part from another musket. The more industrial things got (wars, large factories, etc), the more of a problem this became, creating a need for products with parts that could be simply removed and replaced without fitting by a craftsman.</p>
| 50763 | How do precision surface plates allow for the manufacturing of replaceable parts, and why are they needed for precision machine tooling? |
2022-04-28T14:49:48.653 | <p>I would like to secure components to an aluminium sheet that is 3mm thick. Ideally I do not want to drill holes all the way through the sheet because it is important that a seal is maintained.</p>
<p>I am therefore wondering whether it might be realistic to fasten my components to the aluminium sheet using partially tapped holes in it. For example, a 3mm diameter tapped hole that goes to a depth of 2mm in the aluminium sheet.</p>
<p>To me, these dimensions seem pretty tight. Could anyone offer any advice or an opinion on this?</p>
| |machining|aluminum|fasteners|threads|cutting| | <p>A <a href="https://www.travers.com/product/nachi-0715228-flat-bottom-high-performance-drill-20-149-108" rel="nofollow noreferrer">1.8mm flat bottom drill bit</a> and a <a href="https://www.travers.com/product/nachi-91002-thread-forming-tap-12-488-440" rel="nofollow noreferrer">2mm bottoming tap</a> will do what you need. You will need to use a drill bit collar or a depth limiter on your drill to prevent going all the way through. You will want to test the process on an non-critical piece of aluminum first. It would also be a good idea to test if the strength of the connection is going to be acceptable in your application.</p>
<p>In addition to the stud welder, welded nut, and expansion shield ideas; you could <a href="https://en.wikipedia.org/wiki/Brazing" rel="nofollow noreferrer">braze</a> on a nut or stud. Brazing aluminum is easier than welding it and you can use a metal other than aluminum for the nut or stud. You will need <a href="https://rads.stackoverflow.com/amzn/click/com/B00012Y0FS" rel="nofollow noreferrer" rel="nofollow noreferrer">Aluminum brazing rods</a> and a <a href="https://www.homedepot.com/p/Bernzomatic-Trigger-Start-Torch-Kit-TS4000KC/203368730" rel="nofollow noreferrer">MAPP gas torch</a>. Again practice on a scrap piece of aluminum first.</p>
| 50769 | Is it possible to tap a hole partially through a 3mm aluminium sheet? |
2022-04-28T22:55:44.860 | <p>Ok, here's the background: I'm a BASE jumper and usually I jump from stationary objects like bridges, cliffs, etc.</p>
<p>But one jump I would love to do at some point involves exiting from the roof of a moving vehicle and clearing a railing of a yet-to-be-named bridge. Knowing that Newton's law of inertia would be at play, I was hoping someone could help me come up with a way to calculate exactly when I should jump from the vehicle so that I'm at the exact halfway point of the bridge's length 1, 2, and 3 seconds after jumping from the vehicle.</p>
<p>These are all just placeholder values so let's say the vehicle is traveling at 15mph and the bridge it's traveling over is 200ft long. Would a headwind of 8mph change anything? Or a tailwind?</p>
| |kinematics|projectiles| | <p>Your speed along the bridge after the jump is the same as the speed of the vehicle minus air friction, which we can ignore for small spans of the jump.</p>
<p><span class="math-container">$$ X_1s =\frac{V_{vehicle*1}}{3600}$$</span></p>
<p>The range of the jump (so you know if you clear the railing) can be estimated like this, again ignoring air friction. Note this time we talk about the speed of your jumping out of the car!</p>
<p><span class="math-container">$$R = V_0 \sqrt{\frac{2h}{g}}$$</span></p>
<ul>
<li>R = rang</li>
<li>V_0 = speed of your jump perpendicular to the motion of the car</li>
<li>h = difference between the car roof height and the railing</li>
<li>g = 9.8m/s^2</li>
</ul>
<p>Mind all the units must conform.</p>
| 50776 | Computing distance traveled jumping from moving vehicle |
2022-04-29T16:02:21.990 | <p>In the circuit below, the switch has been in position 1 indefinitely long and suddenly switches to position 2 at t=0. I'm asked to find the initial voltage of the inductor at t=0.</p>
<p>My professor lists the answer as 15-5(5)=-10 V (not sure where he's getting -5(5)), but I don't see why.</p>
<p>Here's how I'm thinking through the problem:</p>
<p>Since the initial current in the inductor is 5 A at t=0- (from when the switch is at position 1) and then 0 A at t=0+ (when the switch is at position 2, the capacitor prevents any current), di/dt is "infinite" (technically the derivative does not exist) meaning that the voltage should also be "infinitely high." This also makes sense intuitively as the sudden drop in current will produce an arbitrarily large induced EMF (at least in the ideal case).</p>
<p>Is there an error in my reasoning? I'd appreciate any feedback!</p>
<p><a href="https://i.stack.imgur.com/rKNSs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rKNSs.png" alt="enter image description here" /></a></p>
| |electrical-engineering|circuits| | <p>The whole idea of using inductors is the fact that their currents do not change instantly. You assumed infinite di/dt, but that doesn't happen. The inductor generates just as high voltage as needed to allow the current to continue so that it decays gradually with finite di/dt. The voltage is L(di/dt) if it really is an inductor.</p>
<p>Let's assume a moment the switch is a practical one. Even during the flight of the switch there's an arc inside. That's because, as I said, the inductor will not allow current changes to happen instantly. Otherwise it's not an inductor, but something else.</p>
<p>It's just like a permanent item which has non-zero mass doesn't allow instant changes of its velocity, its velocity changes always gradually. A guy named Newton (never heard?) became famous when he published a book (=Principia) which told it among other as popular and practically useful things.</p>
<p>Arc needs energy and if the flight of the switch lasts long enough the magnetic energy is used before the switch reaches position 2. In that case the inductor current is zero when the switch reaches position 2. But this obviously is an ideal switch. Its's flight time is zero, so at t=0 the inductor current is that already said 5A. The situation changes obeying a 2nd degree differential equation, because there's also a capacitor taken along.</p>
<p>The problem is not properly defined before one knows the initial charge of the capacitor at t=0. Of course, one can guess that it might be zero, but nothing already shown tells it.</p>
<p>BTW to avoid an analog problem with capacitors try to learn to start to think that the voltage between the poles of a capacitor never changes instantly. The voltage changes as current charges or discharges the capacitor. In this case the current goes through the inductor and you have already got to know that the initial value of that current is 5A, not zero. It's true that in practical circuits often the currents of the capacitors decay towards zero if there's no sources of permanently changing voltages, but here the initial value is 5A as the inductor declares if it really is an inductor.</p>
<p>In this old case <a href="https://electronics.stackexchange.com/questions/282053/how-does-the-inductor-really-induce-voltage">https://electronics.stackexchange.com/questions/282053/how-does-the-inductor-really-induce-voltage</a> you can find a practical induction self-teaching machine. Do not try it to others.</p>
| 50785 | Inductor Voltage at t=0 |
2022-04-30T10:27:55.173 | <p>I just bought a vacuum chamber which comes with a little square weaved fiberglass mat with some silicone rubber on it's border, and I can't find any information about it's use. Even in the comment people tell the don't know what it's for.</p>
<p>Can anyone tell about it ?
<a href="https://i.stack.imgur.com/dnWPW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dnWPW.jpg" alt="enter image description here" /></a></p>
| |vacuum| | <p>Either for the chamber to sit on, to avoid contact with the support surface for heat or noise transfer or to sit inside the chamber so any object placed inside is not in contact with the base.</p>
<p>Notice the corners have been cut off, so perhaps inside is more likely.</p>
| 50795 | what is the purpose of the fiberglass mat we get with some vacuum chambers? |
2022-04-30T11:54:55.253 | <p>why <strong>DCEP</strong> is generally used in consumable welding process?</p>
<p>As per my understanding more heat is generated where the electrons collide(anode),
so in consumable welding,we want deposition and also want penetration,</p>
<p><strong>DCEN</strong> would give <strong>more heat at the work piece</strong> (as the electrones would be bombarding the workpiece) also some heat at the electrode due to collision of ions.<strong>okay?</strong></p>
<p>So the question is why to use <strong>DCEP</strong> when we can get deeper penetration with the <strong>DCEN</strong>?
(is melting of work piece is necessary or deposition of molten metal from the electrode?)
Or if i am wrong what is the reason for getting more penetration in DCEP?</p>
| |mechanical-engineering|metallurgy|metals|welding|joining| | <blockquote>
<p>"Why is DCEP generally used in consumable welding processes?</p>
<p>As per my understanding more heat is generated where the electrons collide (anode), so in consumable welding, we want deposition and also want penetration,".</p>
</blockquote>
<p>Welding is about three things: the best process (SMAW, GMAW, SMAW or TIG) for what is to be accomplished, a balance of the parameters for the work, and knowledge of the exceptions applicable to the process and work.</p>
<p>It's never one or two things, penetration and deposition, unless an exception applies; for example, for extremely thin metal you wouldn't want penetration and deposition.</p>
<p>Source: <a href="http://www.difference.minaprem.com/joining/difference-between-dcen-polarity-and-dcep-polarity-in-arc-welding/" rel="nofollow noreferrer">"Difference Between DCEN Polarity and DCEP Polarity in Arc Welding"</a>:</p>
<blockquote>
<p>"Direct Current Straight Polarity (DCSP) or Direct Current Electrode Negative (DCEN): It occurs when electrode is connected with the negative terminal of the power source and base metals are connected with the positive terminal.".</p>
<p>"Direct Current Reverse Polarity (DCRP) or Direct Current Electrode Positive (DCEP): Here the base metals are connected with the negative terminal of the power source, while the electrode is connected with the positive terminal.".</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/fBjFG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fBjFG.png" alt="enter image description here" /></a></p>
<p>A consumable welding process is one where the electrode becomes a part of the finished weld, GTAW is an example of a non-consumable process.</p>
<p>DCEP has been used for GTAW welding of magnesium or aluminum sheet metal using larger electrodes and lower currents, DCEP provides the necessary cleaning action.</p>
<ul>
<li>Video - Weld.com YouTube - <a href="https://youtu.be/Wtgxcx9TjxA" rel="nofollow noreferrer">"Can you Weld Magnesium with DCEP? | Weld.com Forum"</a></li>
<li>Video - Weld.com YouTube - <a href="https://youtu.be/CHkx8cca-SQ" rel="nofollow noreferrer">"TIG Welding Aluminum with DCEP | TIG Time"</a></li>
</ul>
<p>See this article on Miller's blog: <a href="https://www.millerwelds.com/resources/article-library/pro-tips-for-understanding-tig-welding-waveforms-and-controls" rel="nofollow noreferrer">"Pro Tips for Understanding TIG Welding Waveforms and Controls"</a>:</p>
<blockquote>
<p>"If you weld aluminum with DC EN, a ‘skin’ of aluminum oxide will be floating on the puddle, hampering the flow of the metal you’re trying to weld.".<br />
[In reference to their <a href="https://www.millerwelds.com/products/dynasty" rel="nofollow noreferrer">Dynasty</a> welder, in asymmetrical squarewave mode they write:]<br />
"For most work, 75% EN is preferred, since 25% EP is usually sufficient for cleaning, and any more than that limits the total heat input, which could decrease the weld penetration.
".</p>
</blockquote>
<p>The relation between the EN ratio setting and welding result are shown in the table below:
| Negative Direction | EN Ratio Setting | Positive Direction |
|:----------:|:----------:|:----------:|
| Decreasing | EN Ratio | Increasing |
| Slow | Wire Melting Speed | Quick |
| Small | Gap Tolerance | Large |</p>
<p><a href="https://i.stack.imgur.com/Lvk3R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lvk3R.png" alt="Waveform diagram, showing the ratio of EN and EP" /></a></p>
<blockquote>
<p>"So the question is why to use DCEP when we can get deeper penetration with the DCEN? (is melting of work piece is necessary or deposition of molten metal from the electrode?) Or if i am wrong what is the reason for getting more penetration in DCEP?".</p>
</blockquote>
<p>DCEP tends to cause the electrode to melt (see exception above) which is necessary for <strong>consumable</strong> welding. Penetration can be obtained by increasing the amperage and/or decreasing the surface area of the electrode (example FCAW).
Increasing the wirespeed (amperage) also increases the deposition.</p>
<p>One exception is using 7018 to weld open root. (Sources: <a href="https://youtu.be/R_8rs1wMnDo" rel="nofollow noreferrer">1</a>, <a href="https://weldingweb.com/vbb/threads/706024-7018-on-DCEN?p=8713503#post8713503" rel="nofollow noreferrer">2</a>). See: Hobart Brothers' article: <a href="https://www.hobartbrothers.com/2013/07/7018-welding-rod-amperage/" rel="nofollow noreferrer">"The Basics of Welding: 7018 Welding Rod Amperage & Beyond"</a>, 7018 is usually DCEP.</p>
| 50797 | Consumable arc welding Polarity Effect on heat input |
2022-05-01T03:50:15.110 | <p>I have a conceptual idea for a thruster for a VTOL aircraft/drone which should produce lift by utilizing a upward-flowing airflow to produce low air pressure over the top surface of an airfoil that is embedded within a vertical pipe.</p>
<p>I would like to be able to calculate what the drop in air pressure will be, but being that I am neither an aerospace or aeronautical engineer, I do not know which aerodynamic equation to use to determine this.</p>
<p>I have made some CAD drawings of this conceptual VTOL aircraft/drone thruster to help illustrate and explain how the thruster should produce lift/thrust. (Note: The embedded airfoil is fastened to the vertical pipe with screws and it is colored orange to make it stand out from the rest of the thruster.)</p>
<p><a href="https://i.stack.imgur.com/9WuHm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9WuHm.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/0DvZ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0DvZ9.png" alt="enter image description here" /></a></p>
<p>I am thinking of using a Drag Equation that I found on a NASA website to calculate the drop in air pressure, but I'm not sure if it is the right one to use. Reference <a href="https://www.grc.nasa.gov/www/k-12/airplane/drageq.html" rel="nofollow noreferrer">https://www.grc.nasa.gov/www/k-12/airplane/drageq.html</a></p>
<p><a href="https://i.stack.imgur.com/crKBj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/crKBj.png" alt="enter image description here" /></a></p>
<p>So, my question is what aerodynamic equation should be used to determine the drop in air pressure over an airfoil embedded within a vertical pipe?</p>
| |mechanical-engineering|fluid-mechanics|pressure|aerospace-engineering|aerodynamics| | <p>The equation you quote isn't a physical law, it's a definition of <span class="math-container">$C_{\textrm{d}}$</span>. You will need it, but for finding the value of <span class="math-container">$C_{\textrm{d}}$</span>, there's no substitute for doing a scale model experiment, geometrically similar to and at the same Reynolds number as (and unless the Mach number is very small, also at the same Mach number as) the situation about which you want to make predictions.</p>
| 50802 | What aerodynamic equation should be used to determine the drop in air pressure over an airfoil embedded within a vertical pipe? |
2022-05-01T12:51:35.243 | <p>Does protuberance affects dimensions like tooth tip/root height, pitch diameter etc?
In my books there is a short sub-chapter about the protuberance but nowhere is said if it affects these basic dimensions for outer-normal gearing:
These are my calculations for two gears of outer normal gearing:</p>
<p><a href="https://i.stack.imgur.com/dLd8q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dLd8q.png" alt="enter image description here" /></a></p>
<p>So if I was to considier protuberance of the gearing to minimize pitting effect how would this protuberance affect these dimensions i.e. would the formulas and coeficients change if so, then how and to what value?
Or am I just simply ignoring the fact that the gearing will be with protuberance and I just simply move on to next step?</p>
| |mechanical-engineering|gears| | <p>The author of this question probably meant profile shift (ISO 21771, 4.2.9). While protuberance is generally a parameter in gear machining, the effect of "moving the involute" might appear similar to profie shift.
Answering the question:
When you design gears with different profile shift coefficient (and/or in combination with pressure angles different from 20°), you may as well choose a tooth tip hight and tooth root height which are off (not h*a = 1). Thus you may gain a lot of freedom designing gears with optimized contact ratio (5.4.7.1), flank- and root-stresses and more leading to better efficiency, lower noise excitation and better power to weight ratio.</p>
<p>However, if you are only just beginning to design gears with different profile shift, you should probably keep the height factors constant to the V-circle. In effect, tip and root diameter should change with profile shift.</p>
| 50806 | Does protuberance affects basic gear dimensions like tooth tip height, tooth root height etc.? |
2022-05-02T00:16:27.680 | <p>This is a capstan for a stringbender for an acoustic guitar. When actuated, the String Wheel turns about 15-18 degrees. It does NOT spin, but simply turns, then is pulled back to "zero" when released.</p>
<p>My problem is that when I tighten the nut (bottom) the entire shaft binds and sticks. I assume that the 2 washers in the center are the problem, but am not sure how else to assemble it. Without those washers, the entire bearing rubs against the surface of the Housing and sticks. The small washers are the diameter of the bearing's center shaft, and so I thought they would release it from sticking to the Housing.</p>
<p>It's a simple setup (I want to keep it that way), but I would like to be able to tighten the shaft properly. I've attached sketches of the parts, and a photo of a finished piece on a guitar. Thanks.</p>
<p>--john</p>
<p><img src="https://flutney.com/wp-content/uploads/2022/05/Acoustic-Capstan-Illustration.jpg" alt="Text" /></p>
<p><img src="https://flutney.com/wp-content/uploads/2022/05/Acoustic-Capstan-Top-View.jpg" alt="Text" /></p>
<p><img src="https://flutney.com/wp-content/uploads/2022/05/Bender-Detail.jpg" alt="Text" /></p>
| |bearings| | <p>You need a collar around the shaft in the middle hole of the housing.</p>
<p>At the moment you are clamping the rotating assembly to the housing between the two middle washers. Fit a collar so the washers are held clear of the housing at that point.</p>
| 50809 | How to Keep Bearings From Binding |
2022-05-02T02:57:20.530 | <p>I am designing a simple centrifuge to separate soil by particle size (as well as decanting water). My goal is to be able to continuously funnel in 500 <span class="math-container">$cm^3$</span> of soil and 1.25 liters of water and have the components of the soil (clay, silt, sand) separated from each other and funneled into individual containers for each component. Water will be easily decanted and I know how to do so.</p>
<p>The problem is that I am having difficulties trying to figure out how to easily and continuously <em>extract</em> the soil layers without remixing them together. It is important that the machine is able to sort the soil layers while the centrifuge is still spinning. This is due to waiting for the centrifuge to stop spinning/restart spinning adding large time wastes over many kilograms of soil processed.</p>
<p>I was thinking about adding opening/closing dividers to attempt to keep the layers separate while funneled out, however, there are two problems with this:</p>
<ol>
<li>The ratio/amount of soil components (again, clay, silt, and sand) is unknown. Installing opening/closing dividers will have to be at fixed positions and as the soil components are unknown, dividing along these fixed positions will cause some remixing of the soil.</li>
<li>I am hesitant to introduce moving parts on a spinning part spinning at 2,500 RPM (the inner drum of the centrifuge). This seems like an unnecessary safety risk that I would like to avoid, if possible.</li>
</ol>
<p>Here is a diagram of a simple example of what the inner drum looks like:
<a href="https://i.stack.imgur.com/HdsMj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HdsMj.png" alt="Simple soil centrifuge" /></a></p>
<p>What is supposed to be happening in the centrifuge is similar to if you grabbed some soil, mixed it with water, and let it set, but exceptionally faster. If I wait for it to settle naturally, I have to wait 4-5 days for a small 500g sample to fully settle, whereas with a continuously flowing/separating centrifuge (and additional mechanisms), I can expect to sort hundreds of kilograms within a day. This, along with it automatically sorting soil components, drastically improves my workflow.</p>
<p>Short explanation of expected sorting behavior:</p>
<ul>
<li>Sand particles (defined as particles with diameters from 0.05 mm to 2 mm) should separate closest to the wall of the centrifuge due to being the densest particles in the soil.</li>
<li>Silt particles (defined as particles with diameters from 0.002 mm to 0.05 mm) should separate and collect in the "middle" of the soil solids due to it being the moderately dense particles in the soil.</li>
<li>Clay particles (defined as particles with diameters less than 0.002 mm) should separate closest to the water layer due to it being the least dense particles in the soil. I also expect organic matter to collect close to this layer, but I know of some ways of eliminating the organic matter.</li>
<li>Water, of course, would be the furthest away from the wall due to it being much less dense than any of the soil solids.</li>
</ul>
<p>It is not required for the sorting to be perfect. For instance, I do not expect perfectly sized particles that are perfectly sorted. I do expect some silt to be mixed with clay, sand with silt, etc. However, I would prefer this to be minimal with a reasonable setup.</p>
<p>I greatly look forward to your suggestions. Additionally, if I have missed any oversights, I appreciate pointing them out as well. Thank you very much.</p>
| |mechanical-engineering| | <p>I would suggest a horizontal classifier - fluid with soil mixed in flows through, solids settle out, where solids settle out will depend on particle size, density and flow regime.</p>
<p><a href="https://i.stack.imgur.com/LmGjU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LmGjU.png" alt="enter image description here" /></a></p>
<p>These are more appropriate for a continuous application than you batch process. I suggest to build a setup where water flowrate is adjustable, but constant throughout one run, and then slowly mix in your slurry upstream of the pump (peristaltic pumps handle solids reasonably well).</p>
<p>an alternative, probably better, would be to simply sieve your slurry (several stacked sieves). Sieves for grain sizes 0.002 mm are available.</p>
| 50810 | Help with Continuously Separating and Sorting Solids out of a Centrifuge |
2022-05-04T23:38:12.377 | <p>Context: I study aerodynamics and want to design and build an RC wing to put the theory I learned into practice. The essence of the theory I learned is: Given some mass of wing building material, how do we build the most efficient wing? To actually do this in practice and apply the theory, I need an estimate of the strength of my wing building material (more specifically the spar).</p>
<p>Question:
I do not know structural mechanics but I need an estimate of the max bending moment a rod can withstand. Is there any convenient "bending strength coefficient" which I can look up for the rod material, then plug into some relatively simple equation to find the max bending moment of a given rod? If not what would be the easiest way to get this estimation without fully diving into structural mechanics.</p>
| |structural-engineering|structural-analysis|beam|moments|bending| | <p>The wing design is the art of designing for strength in the most efficient way to make the plane as light as possible, while not sacrificing the strength.</p>
<p>The part of the wing that carries the bending moment is called spar, basically acting like a cantilever beam, and it is never a simple rod or other simple uniform cross-section.</p>
<p>This is for example the figure of the spar of a Cessna 210, a small single-engine plane. The bigger the plane gets the more complex the spar and its attachments get.</p>
<p>As we see it is a complex assembly of formed sheets of aluminum/ titanium that grows wider and adds layers to get stronger as it nears the haul of the airplane.</p>
<p>It has many openings at places where cutting a hole could save weight but not impede strength.</p>
<p>This same philosophy runs through the rest of the structure of the plane like the fuselage, the rudder, and elevators. The entire structure is light and perforated but strong to take the eventual storm and high G loads.</p>
<h1>Edit</h1>
<p>After OP's comment, I add this link to a book on airframe structure.</p>
<p><a href="https://air.flyingway.com/books/Airframe-Stuctural-Design.pdf" rel="nofollow noreferrer">Airfram Structural Design book.</a></p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/g5wgR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g5wgR.png" alt="C210 spar" /></a></p>
| 50841 | Max bending moment estimate |
2022-05-05T18:04:22.110 | <p>Which will be more efficient in term of fuel price:</p>
<ol>
<li>Transport via roadways, say truck or motor vehicle.</li>
<li>Transport via waterways, say via ship or motor boat.</li>
</ol>
<p>Assumptions:</p>
<p>Same caliber engines (same power) are used in either ways. Fuel is same. The distance is same for both water ways and land ways. The aim is to find which would be needing lesser Fuel, if both are travelling with same speed:</p>
<p>Considering real life scenarios, like considering friction on land and turbulence in water.</p>
| |transportation|fuel-economy|fuel| | <p>From the <a href="https://www.eea.europa.eu//publications/rail-and-waterborne-transport" rel="nofollow noreferrer">European Environment Agency - Rail and Waterborne</a> cited in Jon Arnt's answer:</p>
<p><a href="https://i.stack.imgur.com/LzqqO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LzqqO.png" alt="enter image description here" /></a></p>
<p>Chart shows gm CO2 / tonne km.</p>
<p>For my preceived interpretation of the question. The best way to ship cargo on land would be rail at 24 gm, then Inland WaterWays at 33gm, which are significantly better than heavy goods vehicles at 137gm.</p>
<p>But for shear mass of cargo the best way is via ships. 7gm would correspond to container ships.</p>
<p>From <a href="https://higherlogicdownload.s3.amazonaws.com/SNAME/a09ed13c-b8c0-4897-9e87-eb86f500359b/UploadedImages/SNAME%20EEDI%20finalf.pdf" rel="nofollow noreferrer">EEDI Explained</a>
<a href="https://i.stack.imgur.com/BMpDr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BMpDr.png" alt="enter image description here" /></a></p>
<p>Even generic cargo vessels >400 gt produce less than 30 gm CO2 / tonne nautical mi. 1 nautical mi = 1.852 kms. Larger the vessel the less polution is produced per tonne of cargo.</p>
<p><a href="https://www.cn.ca/repository/popups/ghg/Carbon-Calculator-Emission-Factors#:%7E:text=This%20truck%20fuel%20consumption%20factor,2e%2Ftonne%2Dkm." rel="nofollow noreferrer">Carbon Calculator Emission Factors</a> from CN (Canadian National) railway report: 2.5-6 gm CO2e/tonne-km for bulk carriers; 8.3 gm CO2e/tonne-km for containerships; 12.1 gm CO2e/tonne-km for rail; and 63.4 gm CO2e/tonne-km for trucks.</p>
<p>If there is a water path, cargo transport consumes significantly less fuel than a road path.</p>
| 50851 | Which is fuel efficient, roadways transport vs waterways transport? |
2022-05-05T22:00:59.877 | <p>I've read through the <a href="https://ciechanow.ski/mechanical-watch/" rel="nofollow noreferrer">excellent guide to how mechanical watches work by Ciechanowski</a>, but it leaves some questions unanswered.</p>
<p>The main spring of the watch, when it loses energy, can of course be re-wound using the watch's crown, and the guide above explains in detail how that happens.</p>
<p>But the hairsping of the watch, the one that drives the balance wheel, must lose energy at some point as well through friction, with the balance wheel oscillations dying down as a result. How is energy restored to that spring, i.e. how is it re-wound?</p>
<p>On a related note, the guide above also shows that the balance wheel is stopped, through friction, by a special lever during the process of time adjustment. If the balance wheel is thus stopped at a particularly bad time (i.e. when it's exactly at its midpoint position), wouldn't <em>all</em> of the energy in the hairspring be dissipated when the balance wheel is stopped? How does it re-start once the watch is out of time adjustment mode?</p>
| |mechanical-engineering|mechanisms|springs|coil-spring|watch| | <p>I enjoyed the detailed and beautifully animated diagrams you linked.</p>
<p>The source of energy to the balance wheel is the end of the fork that is pressured by the jewels acting intermittently by the scapement gears.</p>
<p>As soon as the balance wheel is released it will be accelerating by the force of the fork handle, either clockwise or anticlockwise.</p>
| 50853 | How is energy in the hairspring restored in a mechanical watch? |
2022-05-06T03:32:38.020 | <p>I am having a difficult time solving the following problem. The only thing we learned in class was parallel continuous stirred-tank reactor (<a href="https://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor" rel="nofollow noreferrer">CSTR</a>) with equal-sized reactors and equal flow rates for each reactor (which I think also implies the volumetric flow rate is the same across all reactors).</p>
<p>When the 1st order liquid-phase reaction <span class="math-container">$A \rightarrow B$</span> (elementary) is carried out in a single CSTR (initial concentration of species A = <span class="math-container">$C_{A0}$</span>, reactor volume = <span class="math-container">$V_R$</span>, volumetric flow rate = <span class="math-container">$ \nu_0 $</span>), one can achieve 80%
conversion. The same reaction is to be conducted in two CSTRs connected in parallel while keeping the
total reactor volume constant at <span class="math-container">$V_R$</span>. How should we split the volumetric flow rate when the final
conversion (<span class="math-container">$X_{final}$</span>) is 0.786.
<a href="https://i.stack.imgur.com/PjVPk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PjVPk.png" alt="enter image description here" /></a></p>
| |chemical-engineering| | <p><strong>Initial Case</strong> Applying a mole balance to the single CSTR in steady state yields
<span class="math-container">\begin{align}
v_0C_{A0} - v_0C_{A} - kC_{A}V &= 0 \\
C_{A} &= \frac{v_0C_{A0}}{v_0 + kV} \tag{1}
\end{align}</span>
The conversion, combining with Eq. (1), is
<span class="math-container">\begin{align}
X_1 &= \frac{C_{A0} - C_A}{C_{A0}} \\
X_1 &= \frac{C_{A0} - \dfrac{v_0C_{A0}}{v_0 + kV}}{C_{A0}} \\
X_1 &= 1 - \frac{v_0}{v_0 + kV} \\
X_1 &= \frac{kV}{v_0 + kV} \rightarrow kV = \frac{X_1v_0}{1 - X_1} \tag{2}
\end{align}</span>
We will use Eq. (2) later.</p>
<p><strong>Second Case</strong> Now the volumetric flow rate splits in two. Let me call that value in the upper branch <span class="math-container">$ \alpha v_0 $</span>, and the one for the lower branch <span class="math-container">$(1 - \alpha)v_0$</span>. Since we have a bifurcation, the concentration of reactant <span class="math-container">$A$</span> doesn't change, so the value is still <span class="math-container">$C_{A0}$</span> when entering both reactors.</p>
<p>We denote the exit concentration in the reactor at the top <span class="math-container">$C_{A1}$</span>, and at the bottom <span class="math-container">$C_{A2}$</span>. Applying a mole balance to both CSTR's in steady state, with equal volumes but of value <span class="math-container">$V/2$</span>, yields
<span class="math-container">\begin{align}
\alpha v_0C_{A0} - \alpha v_0 C_{A1} &- (k C_{A1})\frac{V}{2} = 0 \\
(1 - \alpha)v_0C_{A0} - (1 - \alpha)v_0C_{A2} &- (kC_{A2})\frac{V}{2} = 0 \\
C_{A1} = \frac{\alpha v_0C_{A0}}{\alpha v_0 + kV/2} &\hspace{0.25 cm}
C_{A2} = \frac{(1 - \alpha)v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{3,4} \\
\end{align}</span>
Now, upon mixing the streams, the volumetric flow rate recovers its value <span class="math-container">$v_0$</span>. However, the concentration changes, and we denote its value by <span class="math-container">$C_{A3}$</span>. This one is obtained by a mole balance at steady state, combining with Eqs. (3) and (4)
<span class="math-container">\begin{align}
\alpha v_0C_{A1} + (1 - \alpha)v_0C_{A2} &= v_0 C_{A3} \\
C_{A3} &= \alpha C_{A1} + (1 - \alpha)C_{A2} \\
C_{A3} &= \frac{\alpha^2 v_0C_{A0}}{\alpha v_0 + kV/2} +
\frac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{5} \\
\end{align}</span>
The conversion, using Eq. (5), yields
<span class="math-container">\begin{align}
X_2 &= \dfrac{C_{A0} - C_{A3}}{C_{A0}} \\
X_2 &= \dfrac{C_{A0} - \dfrac{\alpha^2 v_0C_{A0}}{\alpha v_0 + kV/2} -
\dfrac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2}}{C_{A0}} \\
X_2 &= 1 - \frac{\alpha^2 v_0}{\alpha v_0 + kV/2} -
\frac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{6} \\
\end{align}</span>
Now we combine Eq. (2) with Eq. (6)
<span class="math-container">\begin{align}
X_2 &= 1 - \frac{\alpha^2 v_0}{\alpha v_0 + \dfrac{X_1v_0}{2(1 - X_1)}} -
\frac{(1 - \alpha)^2v_0}{(1 - \alpha)v_0 + \dfrac{X_1v_0}{2(1 - X_1)}} \\
X_2 &= 1 - \frac{\alpha^2}{\alpha + \dfrac{X_1}{2(1 - X_1)}} -
\frac{(1 - \alpha)^2}{(1 - \alpha) + \dfrac{X_1}{2(1 - X_1)}} \tag{7} \\
\end{align}</span>
Eq. (7) is a non-linear equation that we need to solve. Note that if <span class="math-container">$\alpha = 1/2$</span>, the left-hand side yields 0.8, which is reasonable.</p>
<p>It is informed to us that <span class="math-container">$X_2 = 0.786$</span> and that the original conversion was <span class="math-container">$X_1 = 0.8$</span>. Upon solving numerically, we found actually two possible solutions
<span class="math-container">$$ \boxed{\alpha \in (0.33500,0.66500)} $$</span>
However, since both branches are "identical" to each other, it is the same if 33.5% goes up, and 66.5% goes down, or vice versa.</p>
| 50857 | Continuous stirred-tank reactor (CSTR) in parallel problem |
2022-05-06T14:35:19.113 | <p>Which equation can be used to find failure based on ply orientation.
Tension,shear & bearing faliure to validate against fem analysis
<a href="https://i.stack.imgur.com/F7HGB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F7HGB.png" alt="enter image description here" /></a></p>
| |aerospace-engineering|mechanical|machine-design|aircraft-design|composite| | <p>I suggest to familiarize with the "yield line theory" and stress concentration around bolt holes. I don't know what you mean of "equation can be used", as all three are potential failure modes to be investigated over the assumed failure areas with respect to the resulting stress, whether is tensile, shear, bending, or combined stresses.</p>
| 50862 | Failure criteria for CFRP single plate bolted joint |
2022-05-06T15:34:38.967 | <p>I have been practising the superposition method lately and I am confused in terms of drawing the directions when sectioning. Here is the problem:
<a href="https://i.stack.imgur.com/NfqoH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NfqoH.png" alt="enter image description here" /></a></p>
<p>Heres is the process of converting a indeterminate system into a more solvable indeterminate system drawn by my lecturer.</p>
<p>Can someone please explain the direction of shear and moment drawn? For example, Why is it up and clockwise for the sectioning on the left beam? I understand that the opposite side of the beam must have opposite shear/moment.</p>
<p><a href="https://i.stack.imgur.com/OCdzO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OCdzO.png" alt="enter image description here" /></a></p>
<p>According to my textbook:</p>
<p><a href="https://i.stack.imgur.com/Z9HOU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z9HOU.png" alt="enter image description here" /></a>
Shouldn't we assume down and anticlockwise for the left most beam, like this diagram suggests?</p>
| |structural-engineering|structural-analysis| | <p>This is the first step to simplify the analytical model by applying the equivalent load concept as depicted below:</p>
<p><a href="https://i.stack.imgur.com/6MJ8v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6MJ8v.png" alt="enter image description here" /></a></p>
<p>Another way to examine the direction of forces at where the cantilever is cut off.</p>
<p><a href="https://i.stack.imgur.com/7NSZL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7NSZL.png" alt="enter image description here" /></a></p>
| 50864 | Direction confusion for shear and moment when sectioning a beam |
2022-05-06T15:51:30.563 | <p><a href="https://i.stack.imgur.com/ny17y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ny17y.png" alt="enter image description here" /></a></p>
<p>I have done some calculations:</p>
<p><a href="https://i.stack.imgur.com/0Rxv9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Rxv9.png" alt="enter image description here" /></a></p>
<p>My lecturer just said that exit loss <span class="math-container">$\frac{u^2}{2g}$</span> is small for laminar flow so ignore it. But never explains why it is small?</p>
| |fluid-mechanics| | <p>The assumption that the velocity head is small is due to velocities in laminar flow generally being low. This assumption is supported by the observation that laminar flow is more commonly seen in smaller diameter systems where high velocities are typically kept low to keep the frictional losses low. Recall that Reynolds number is directly proportional to velocity and the diameter in pipe flow calculations.</p>
| 50865 | Why is the exit loss $\frac{u^2}{2g}$ is small for laminar flow? |
2022-05-10T19:49:25.580 | <p>Flying gliders can have glide ratios (ratio of horizontal distance vs vertical distance) around 50 or 60. How would that change when the fluid density changes, in particular under water. Could a craft designed for under water gliding achieve a similar glide ratio, or does the increased drag reduce it? Would the reduction be linear to density?</p>
| |fluid-mechanics|submarines| | <p>If all things were somehow magically equal, density does not affect things because although density increases drag it also increases lift by the same proportion. So the glide <em>ratio</em> doesn't change.</p>
<p>But not all things are equal and what density does influence is the Reynolds number and <em>this</em> changes the glide ratio. The Reynolds number is the ratio of inertial forces in the fluid to viscous forces. Higher Reynolds number indicates the fluid behaves with more inertia and less viscosity (like water) and lower Reynolds numbers indicate the fluid behaves with more viscosity and less inertia (like honey).</p>
<p>The Reynolds number also depends on speed. Increasing fluid density has the same effect as moving faster where both increase the Reynolds number. Increasing speed or density helps inertia overpower viscosity.</p>
<p>Another effect that influences Reynolds number is "size" or more specifically the path length along which the fluid is guided and flows across the surface of the object. This is the chord for airfoils.</p>
<p>In fact, water is so dense compared air such that a submarine experienced a higher Reynolds number than an airplane even though it is traveling much more slowly through it's fluid medium.</p>
<p>Similarly, as aircraft and flying animals get smaller, this also reduces the Reynolds number since the chord length of their airfoils decreases and their smaller size tends to translate to slower flight speeds.</p>
<p>For an airplane to experience the same Reynolds number as a submarine it would have to fly impractically slow (almost a standstill) or be impractically small (microbe size). For a submarine to experience the same Reynolds number as an airplane it would need to move impractically fast (torn apart by the drag and friction from travelling at hypersonic speeds or faster through the water while also vapourizing it) or be ENORMOUS (I suspect it would not fit into the ocean).</p>
<p>That's why there are hummingbirds are all small, or birds that glide like albatrosses or soar on thermals like vultures are all huge.</p>
<p>Insects, in particular, experience a Reynolds number in air so low that they are swimming through the air more than they are flying through it. It means they can't glide but the benefit is that hovering is easier.</p>
<p>So for your actual question:</p>
<p>If we're talking about glide ratios, the glide ratio of a submarine I would expect it to be higher than that of an airplane...if submarines were actually built with wings to produce lift capable of supporting its own weight.</p>
| 50928 | Glide Ratio of an unpowered "sub" |
2022-05-11T15:28:15.840 | <p><a href="https://i.stack.imgur.com/BbG0B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BbG0B.png" alt="enter image description here" /></a></p>
<p>I'm not really sure how is the current, represented,
usually it was represented as the positive OX axis and the voltage (potential difference) was between -90 and 90.</p>
<p>Is the current -35 below the voltage that is on the OX axis?</p>
| |electrical-engineering| | <p>Yes. Since the phase of the voltage is not explicitly mentioned, it can be assumed that it is the reference and hence aligned to the OX axis.</p>
<p>The current phase is specified in the question. The question asks for the impedance to be found which requires voltage magnitude and phase also to be known / specified. This justifies the above assumption that the voltage phasor is the reference phasor to which the current is referenced to.</p>
| 50934 | Find the values of resistance R and inductance L in the R-L-C circuit |
2022-05-12T01:16:25.650 | <p>The following is an indeterminate frame consisting of two beams that are rigidly connected to each other at the corner support.</p>
<p><a href="https://i.stack.imgur.com/mHlBP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mHlBP.png" alt="enter image description here" /></a></p>
<p>I want to find the support reactions.</p>
<p><a href="https://i.stack.imgur.com/HLT5j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HLT5j.png" alt="enter image description here" /></a></p>
<p>As far as I can see, this is a symmetric structure with symmetric loading (please do correct me if I am wrong).</p>
<p>And as I understand it, a symmetric structure with symmetric loading has symmetric reactions.</p>
<p>From here, the chain of logic I would follow is:</p>
<ol>
<li>The corner roller support has a vertical reaction (V2) of 0, for symmetry to be possible.</li>
<li>Similarly, H1 = 0, for symmetry.</li>
<li>V1 = - H2, again for symmetry of reactions.</li>
</ol>
<p>Then, from equilibrium, we can say:</p>
<p><a href="https://i.stack.imgur.com/isnl0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/isnl0.png" alt="enter image description here" /></a></p>
<p><strong>However, this is incorrect.</strong> The mark scheme for this question has a different solution.</p>
<p>Could someone please tell me where in my process I have made a mistake?</p>
<p><strong>EDIT</strong>: This is what the mark-scheme says:
<a href="https://i.stack.imgur.com/6Tgfi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Tgfi.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|structures|frame| | <p>The members and loads of the frame are symmetric about joint "B", but the supports are not, which is the source that causes non-conforming deflections when loaded. The sketches below depict each case of the deflection of the frame.</p>
<p><a href="https://i.stack.imgur.com/A1rxj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A1rxj.png" alt="enter image description here" /></a></p>
<p><strong>ADD:</strong></p>
<p>The roller support at "B" is free to move in x-dir. The sketches below show the lateral displacement and corresponding joint rotation due to the respective loads "F" and "H".</p>
<p><a href="https://i.stack.imgur.com/5Vp5w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Vp5w.png" alt="enter image description here" /></a></p>
<p><strong>Simplified Frame Model:</strong></p>
<p><a href="https://i.stack.imgur.com/OETgy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OETgy.png" alt="enter image description here" /></a></p>
| 50937 | Symmetry to determine the support reactions of a statically indeterminate frame |
2022-05-12T10:52:53.757 | <p>I need to complete the 3D stiffness matrix of steel concrete composite columns under all kinds of stresses (axial and rotational as shown in the image below). So because the section is a Square Hollow Section 150 x 150 x 16mm, filled with concrete, I need to transform the concrete into equivalent steel. How can I go about this to obtain the equivalent, Young's Modulus E, Shear Modulus G, Area to use and Section Moment of area I and Polar Moment of area J. Any advice on the procedure to solve this? Images of the section and the matrix are below.</p>
<p><a href="https://i.stack.imgur.com/4Pc5B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Pc5B.png" alt="3D Stiffness Matrix" /></a>
<a href="https://i.stack.imgur.com/HM4Fr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HM4Fr.png" alt="SHS filled with Concrete Composite Column" /></a></p>
| |structural-engineering|structural-analysis|stiffness| | <p>You can simply calculate "equivalent" cross-sectional stiffness values.</p>
<p>Note that all non-zero terms of the stiffness matrix depend on one of these values: <span class="math-container">$EA$</span> or <span class="math-container">$EI$</span>. These are the cross-sectional axial and rotational stiffnesses, respectively.</p>
<p>Obviously, when looking at entire beams, another important parameter is the beam's length <span class="math-container">$L$</span> (to multiple exponents), but here we're just looking at the cross-section, so we only care about <span class="math-container">$EA$</span> and <span class="math-container">$EI$</span>.</p>
<p>Given that the stiffness matrix only cares about <span class="math-container">$EA$</span> and <span class="math-container">$EI$</span>, we can actually replace them entirely with <span class="math-container">$K_a$</span> and <span class="math-container">$K_r$</span> (axial and rotational stiffness).</p>
<p>The wonderful thing about doing this is that it exposes an important aspect of the stiffness matrix: it doesn't care about the beam's elastic modulus, area or moment of inertia. It only cares about its <em><strong>stiffness</strong></em>.</p>
<p>We can therefore do whatever we want with <span class="math-container">$E$</span>, <span class="math-container">$A$</span> and <span class="math-container">$I$</span> as we see fit, so long as we get a reasonable stiffness at the end.</p>
<p>In this case, we're dealing with a composite beam made up of a square concrete section and a hollow square steel section. How do we find the stiffness of this beam? Well, a good first guess is to simply add the stiffness of each of component:</p>
<p><span class="math-container">$$\begin{align}
K_a &= E_c A_c + E_s A_s \\
K_r &= E_c I_c + E_s I_s
\end{align}$$</span></p>
<p>(the math is a bit simple in this case since both concrete and steel have the same centroid; if that weren't the case, then their moments of inertia would have to be calculated relative to the composite centroid, see <a href="https://engineering.stackexchange.com/a/49799/1832">here</a> an example of how to do so)</p>
<p>This is a pretty conservative method: you aren't taking into consideration any interaction between the components. For example, the concrete's true axial stiffness will likely be larger than <span class="math-container">$E_c A_c$</span> due to the lateral constraint imposed by the steel section (as described <a href="https://engineering.stackexchange.com/a/40931/1832">here</a>).</p>
| 50940 | Transforming the Area of a Hollow Square Section filled with concrete into the equivalent area of just steel |
2022-05-12T17:57:00.043 | <p>Looking at guides for building wooden frame gates I almost always find the following design recommendation:</p>
<p><a href="https://i.stack.imgur.com/IHbjr.png" rel="noreferrer"><img src="https://i.stack.imgur.com/IHbjr.png" alt="enter image description here" /></a></p>
<p>The idea is that the cross brace will distribute load from the top outside corner into the bottom of the hinge bracket. I understand this logic and am sure it works but wouldn't it be preferred to use a tension brace rather than a compression brace in this application?</p>
<p>For example virtually every recommendation for a wire rope support shows the opposite:</p>
<p><a href="https://i.stack.imgur.com/k8kFG.png" rel="noreferrer"><img src="https://i.stack.imgur.com/k8kFG.png" alt="enter image description here" /></a></p>
<p>In this case the wire rope supports the far end under tension, but is there a reason a 2x4 couldn't also be used under tension rather than compression? Is the former design more effective for some reason?</p>
<hr />
<p>I realize that using the 2x4 for tension would require stronger joinery but it's certainly possible to secure it more than well enough to take the load.</p>
<p>Additionally a common theme here seems to be that the load on the gate is too trivial to really care either way but I strongly disagree with that statement. Wooden gates almost never hold up to time, and we aren't talking decades here...most wooden gates will sag within 1-2 years. Wood is certainly a troublesome material but plenty of critical structures have been made from wood with proper engineering and have held up much longer than this. So maybe we should consider complicating things more? It's certainly cheaper and easier to use more complicated joinery/a 99 cent steel bracket and 2x4s than it is to fabricate an all steel gate frame.</p>
<p>Are 2x4s just undersized to support a load like this or is it possible the bracing can make a difference?</p>
| |structural-engineering| | <p>I think @r13 nailed it.</p>
<p>FWIW...</p>
<p>My preference for the cable version is based on the following, FWIW:</p>
<ol>
<li>It adds less weight to the gate.</li>
<li>I like the "mechanical assist" look.</li>
</ol>
<p>And most importantly...
3) It allows for adjustability when there are changes to either side post or the frame of the gate itself.</p>
<p>I have a privacy fence gate on the side of my house that has an 8' span using a cable support and after 35 years it still closes perfectly thanks to a handful of adjustments to the tension cable over the decades. YMMV.</p>
| 50946 | Why do wooden gate designs recommend a brace under compression instead of tension? |
2022-05-13T06:06:34.383 | <p>Axial deformations are ignored. This structure is under the specified force:</p>
<p><a href="https://i.stack.imgur.com/TI8Q1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TI8Q1.png" alt="Structure" /></a></p>
<ul>
<li>Is the internal shear force in column BC equal to that of column DE?</li>
<li>If we cut the structure at middle of BC and DE columns:
<ul>
<li>Internal moment at the middle of DE is zero, right?</li>
<li>But the moment at the middle of BC is <em>not</em> zero, right?</li>
<li>So, how it has static equilibrium? I don't get it :(</li>
</ul>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/sdntJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sdntJ.jpg" alt="Cut frame" /></a></p>
| |structural-engineering|civil-engineering|structural-analysis|shear|bending| | <p>For your first question regarding whether the shear in BC equals the shear in DE: yes, they are the same.</p>
<p>An easy way of seeing this is that some fraction of <span class="math-container">$P$</span> is going to be absorbed by BC as shear, which will then be transferred to CD as an axial load, only to then become shear in DE again.</p>
<p>As for your main question: you haven't included all the loads acting on that slice of the frame. You've drawn the moments, but forgot about all the other forces acting on that part of the structure. Specifically, the axial and shear forces at the cut points (mid-BC and mid-DE).</p>
<p>If you were to solve this entire structure, you'd find that BC and DE are under axial loads due to the support reactions (ABC under compression, DE under tension). These opposing loads are a force couple which generates a moment equal and opposite to the moment at mid-BC:</p>
<p><a href="https://i.stack.imgur.com/tcgML.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tcgML.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/GlVM9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GlVM9.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/dDTFd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dDTFd.png" alt="enter image description here" /></a></p>
<p>In the example above, we have that mid-BC has a bending moment of <span class="math-container">$\dfrac{133.2}{2} = 66.65\text{ kNm}$</span>. However, BC and DE are under 13.33 kN compression/tension, a force couple which generates a moment of <span class="math-container">$13.33 \cdot 5 = -66.65\text{ kNm}$</span> (negative because it's clockwise), perfectly cancelling out the bending moment at mid-BC.</p>
<hr />
<p><sub>Diagrams obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis program</sub></p>
| 50959 | Justification of static equilibrium for a cut structure/frame |
2022-05-13T10:39:59.067 | <p>This YouTube video <a href="https://www.youtube.com/watch?v=nlG2dsYRagI" rel="nofollow noreferrer">https://www.youtube.com/watch?v=nlG2dsYRagI</a> shows how an electric engine is made but I don't understand how friction between the stator and rotor is dealt with.</p>
<p>At 3:50, we see the rotor inserted inside the stator and at 4:32 we can see a hand turning the shaft. I don't see any lubrication, how can this be possible ?</p>
<p>At 3:34, we can see a calibration of something and the woman says in french that the rotor will turn around steel bearings, I guess the lubrication will be made there : In the connection between rotor center and the steel bearings but still we can see that there's no space between the rotor and the stator.</p>
| |mechanical-engineering|electrical-engineering|friction|engines| | <blockquote>
<p>... there's no space between the rotor and the stator.</p>
</blockquote>
<p>There is always an air gap between the rotor and stator.</p>
<p>Iron has a permeability of about 1000 times that of air so the trick for maximum efficiency is to keep the flux in iron and minimise the air gap in the magnetic path where losses would occur. Minimumising air gap will require higher tolerances and therefore will be more expensive to manufacture.</p>
| 50962 | How friction between stator and rotor is dealt with in an electric engine/motor |
2022-05-13T14:49:10.383 | <p>in the figure below the reaction torque equation is given by Fg41.x trying to rock the ground plane about the O2 pivot due to gas forces . My question is if the piston with the cylinder whole assembly were shifted down by a distance L as shown in figure , will the horizontal (Fg) force contribute to the reaction torque equation ? Because this gas force was not creating a moment in the first case since its moment arm was passing through the pivot , but now after shifting L distance, a moment vertical arm is in present i.e. (T12 will now become = Fg.L+Fg41.x) , is that true?</p>
<p><a href="https://i.stack.imgur.com/vd4Jl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vd4Jl.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/WrXLn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WrXLn.png" alt="enter image description here" /></a></p>
<hr />
<p><strong>Edit</strong></p>
<p>Here is a figure that may help you understand me.</p>
<p>In this case in the figure with an offset length=L won't the gas force acting on the wall of the cylinder (Fg41x * L) generate a moment about O2 pivot which will try to rock the cylinder ground link 1 around this O2 pivot? and thus the total gas torque would be the torque in the first case (were the Fg41x moment arm was passing through the pivot and thus produce no moment) + the new generated moment from Fg41x?</p>
<p><a href="https://i.stack.imgur.com/o9lTF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o9lTF.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|torque|dynamics|machine-design|engines| | <p>@r13 here is a figure that may help you understand me see here in this case in the figure with an offset length=L won't the gas force acting on the wall of the cylinder (Fg41x * L) generate a moment about O2 pivot which will try to rock the cylinder ground link 1 around this O2 pivot? and thus the total gas torque would be the torque in the first case (were the Fg41x moment arm was passing through the pivot and thus produce no moment) + the new generated moment from Fg41x?</p>
<p><a href="https://i.stack.imgur.com/o9lTF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o9lTF.png" alt="enter image description here" /></a></p>
| 50966 | Question about engine reaction torque from gas forces |
2022-05-13T18:30:42.407 | <p>It seems to me that bicycles have proven themselves to be the fastest human-powered vehicle in common usage. By “human-powered” I mean it has no fuel or power source other than the inputs given by a human, i.e. no engines, no sails, no clever use of gravity. I wonder if this is a function of the design of the bicycle or just a coincidence. In other words, is the bicycle the fastest possible human-powered vehicle, or is just that no-one has built a faster vehicle? Are there faster human-powered vehicles that I don’t know about?</p>
<p>For standards, lets say we are talking about locomotion on a level (no incline) road surface.</p>
<p>If the bicycle is the fastest, can someone give a technical reason why this should theoretically be so? For example, why not a tricycle, unicycle, or something without wheels, perhaps?</p>
| |bicycles| | <p>To understand why a bike wins, consider the needs and the costs of maximizing the power output of a human body and minimizing losses.</p>
<p>Human body cannot move continuously. Joints rotate but need to rotate in the opposite direction after. If one were to move in spurts like this, the drag losses would be much greater. Bike casette helps smooth out the speed.</p>
<p>There is an additional alternating moment to cancel out too. As legs pump, there is need for a twisting moment to maintain orientation. You may see cyclists swing their bikes left and right between their legs sometimes in order to reduce the effects of their pumping legs. Bikes can impart this moment to the ground via their two wheels (two points of contact).</p>
<p>Supposing a person unicycle managed to somehow cancel this while travelling fast, it would involve a lot of movement of mass (flailing arms to cancel pumping legs?) which in turn cause air movements that make it inefficient compared to the bike simply relying on the normal force from the road.</p>
<p>Knowing this, what else might show promise? - Would you care to design a unicycle or hamster wheel incorporating a bike casette, and a mechanism that allows it to be powered via symmetric movements (think person running like a cheetah), sacrificing steering and control for energy efficiency and hence speed?</p>
| 50968 | Is the bicycle the fastest possible human-powered vehicle? |
2022-05-14T08:33:30.570 | <p>I have the following truss and I want to calculate the reactions using the flexibility method
<a href="https://i.stack.imgur.com/89PI6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/89PI6.png" alt="enter image description here" /></a></p>
<p>Using this method means that I need to break this indeterminate system into a determinate system and two redundant systems (with the reintroduced member force) in this case. In other to obtain these subsystems, I need to release members (or reactions).</p>
<p>Say I want to release 2 of the members from the truss, how do I know which members to release?</p>
<p>Do the members that I remove have to be the members that are connected the supports? (ie to use the compatibility equation).</p>
<p>So a follow up question is where does the compatibility equations apply? (Only at the supports?)</p>
<p>I have tried to remove 2 members that are connected to the supports but using a software it tells me that the structure is unstable. (with full hinges applied etc)</p>
<p><a href="https://i.stack.imgur.com/TWne5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TWne5.png" alt="enter image description here" /></a></p>
<p>I then proceed to remove other 2 members, they are stable, but they are not connected to supports. Does the compatibility equation apply here?</p>
<p><a href="https://i.stack.imgur.com/Y1FDR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y1FDR.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis| | <p>Hint - On top of pin supported ends (one more than the equations for global equilibrium conditions), the two diagonal members in the middle panel possess one degree of redundancy. So you should release the horizontal restraint at one of the pin supports, and release one diagonal member addressed above.</p>
<p>For compatibility, apply a unit load at the released support, and a pair of unit loads at the joints connecting the released diagonal member, the unit loads should be in the opposite directions.</p>
<p><strong>ADD:</strong></p>
<p><a href="https://i.stack.imgur.com/9AbKS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9AbKS.png" alt="enter image description here" /></a></p>
<p>The compatibility should be carried out at the joint/member with the red unit load, to close out the horizontal gap at the released support, caused by the releases.</p>
<p><strong>Note:</strong></p>
<p><a href="https://i.stack.imgur.com/wBFNq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wBFNq.png" alt="enter image description here" /></a></p>
| 50974 | Solving an indeterminate truss with static redundancy of 2 confusion |
2022-05-14T08:35:58.683 | <p>That sounds like a stupid question but hear me out.</p>
<p>For something like a rocket where the goal is to fly in a straight line the use of the tail fins is obvious. But air-to-air missiles are designed to be maneuverable. Aim9 sidewinder pictured:</p>
<p><a href="https://i.stack.imgur.com/SQMtD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SQMtD.jpg" alt="aim9 sidewinder" /></a></p>
<p>Wouldn't having large tail fins spoil this maneuverability? If anything, I would expect the fins to be somewhere near the CG, like an airplanes wings. A plane with large wings at the back would fly like a dart, even with canards near the nose.</p>
<p>On some missiles like the Javelin, the fins are placed this way.
<a href="https://i.stack.imgur.com/VFmxW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VFmxW.jpg" alt="javelin missile after launch" /></a></p>
<p>And on some anti aircraft missiles like the stinger there are barely any fins at all. (there are canards which haven't deployed yet in the picture).</p>
<p><a href="https://i.stack.imgur.com/QeIm4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QeIm4.jpg" alt="stinger missile in flight" /></a></p>
<p>Obviously it's convenient to not have large fins, but why can this work for the stinger and not for larger missiles like the sidewinder?</p>
<p>My <em>guess</em> is that the high speed of missiles like the sidewinder gives the tube enough body lift to maneuver the missile, and some additional stability is needed. Whereas the much slower javelin does not have enough "grip" on the air just from the missile body.</p>
<p>Is that true? Or are there other factors at play?</p>
| |aerospace-engineering|aerodynamics| | <p>The basic problem is that for a long skinny anything travelling longways, the aerodynamic center for pitch and yaw moments is 25% from the nose, just like the lift center for a flat plate is 25% from the leading edge. And the aerodynamic moments work to increase the perturbation - forward feedback - assuming the CoG is further aft. Given a tiny perturbation, the missile wants to pivot 90 degrees and go sideways. You have to put the fins behind the CoG so that they provide a corrective moment (negative feedback in pitch and yaw) that is larger than the perturbation's forward feed back from the missile body.</p>
<p>Additionally, the missile has to carry properly on the pilon and launch in a controlled manner from all the planes that carry it. There were a lot of separation problems during launch testing. They still have separation problems during testing today.</p>
<p>Pull the ink out of a cheap Bic pen and drop it from about 10 feet. It will start to turn sideways. Play with scotch tape fins until it falls straight.</p>
<p><a href="https://i.stack.imgur.com/FHuhs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FHuhs.png" alt="enter image description here" /></a></p>
<p>Below is good intro to "finner" missile stability derivatives. The purpose of the document is to derive correction factors for wind tunnels, but it presents the uncorrected stability derivatives as the starting point. See in particular section II - Intro to slender body theory.</p>
<p><a href="https://apps.dtic.mil/sti/pdfs/ADA040670.pdf" rel="nofollow noreferrer">https://apps.dtic.mil/sti/pdfs/ADA040670.pdf</a></p>
| 50975 | Why do air-to-air missiles have tail fins? |
2022-05-15T05:31:04.540 | <p>There was a video of a train that had tvs in the windows so it appeared the rider was moving across the landscape. If the train was stationary the passenger would not believe the ride was real because they would not feel the drag from acceleration. See racing arcade games.</p>
<p>Is there any way to simulate the acceleration effect? As if you were in a race car going from 0 to 60?</p>
<p>I think about this a lot</p>
<p><a href="https://i.stack.imgur.com/dl9u5.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/dl9u5.jpg</a></p>
| |experimental-physics|acceleration|drag| | <p>The rides in Disneyland do a great job of combing the visual cues of the ride with actually accelerating the vehicle or boat or rocket but in a carefully designed short span giving the impression of larger acceleration during a longer time.</p>
<p>The displacement follows a pattern of fast short acceleration amplified with the visuals. then a gentle recoil while visual cues give the impression of coasting.</p>
| 50987 | Is there a way to simulate the drag effect created by acceleration while in a stationary object? |
2022-05-15T20:21:32.267 | <p>R-134a refrigerant has recommended pressure levels to keep a system properly charged:</p>
<p><a href="https://i.stack.imgur.com/aCFtm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aCFtm.jpg" alt="enter image description here" /></a></p>
<p>My question is if you're trying to make an ac system as cold as possible, and you're willing to ignore potential side effects, should you slightly over charge, or under charge the system?</p>
<p>The reason I ask is when an ac system is low on freeon, one of the dangers is the evap coil gets so cold it literally freezes over. So it seems to me like you'd want to "under charge" the system as much as possible to the point it's just about to freeze but does not.</p>
<p>However I've also heard if your ac is slightly undercharged the best way to make it blow colder is to bring it up to 'spec' even if it's not currently freezing.</p>
<p>These 2 things sound contradicting to me, so I'm just wondering what is the right way to interpret this?</p>
| |refrigeration| | <p>Firstly, lets agree on what you mean by blow colder. You want more cooling. This is not the same as having the air come out the vents colder.</p>
<p>The setup that works best is different for every out door temp, indoor temp, and indoor humidity combination. So it isn't a trivial problem at all.</p>
<p>Also, you probably care about this when it is hotter than standard conditions outside. The California Energy rating is more applicable to these conditions, compared to the Federal SEER.</p>
<p>When it is really hot out, the problem often is getting the heat out of the condenser and into the outside air. To improve this, add refrigerant to boost the highside pressure and increase the condensation temp. It is pretty common to have to adjust the refrigerant load in large systems seasonally. The system check that tells you if you have it right is to calculate the supercooling at the condenser exit.</p>
<p>The evap pressure and temp may increase also, but the added density should improve cooling. If you lose air delta T across the evap coil, it isn't working.</p>
<p>Modern efficient systems don't have the adjustability that the older systems with their higher compression ratios did. I used to run 175 psi condenser pressures in South Florida in the summer with R22. Nothing less would dehumidify the hotels.</p>
<p>If you run into obstinate permitting people who simply won't permit the property for an adequately sized unit, look for the brands with the highest compression ratios and highest water extraction numbers.</p>
| 50990 | At what refrigerant pressure does an air conditioner blow coldest? |
2022-05-16T14:38:27.700 | <p>I have a motorcycle frame. It is made form forged aluminum. There is a pressed headset in the frame that I need to remove - it was pressed in there with machined force. Some folks remove the headset (from the opposite end) with a large hammer and punch, taking up to an hour of hitting remove the headset.
I am concerned to create stress fractures either in the weld/join if in the steering column itself with constant hard hitting.
These frames are made for abuse (jumps, etc), but im concerned this constant hitting would create weakness in the frame.
I’m interested to know how strong forged aluminum is and if damage can be done to the structural integrity with this kind of treatment.</p>
| |aluminum| | <p>The proper way is to press it apart; there is always a big enough press for money. Being aluminum , differential thermal expansion is not likely to help. Whatever the alloy and heat-treatment, it can be damaged with a big enough hammer. The stress of use is not close to the amount of stress a hammer can produce. One approach would be to do whatever necessary to get it apart; Then examine it with dye penetrant to see if there are cracks. That is, if good examination shows no cracks, there very unlikely to be any "almost cracks". And aluminum welds are not as strong as any heat-treated component.</p>
| 50998 | What kind of stress can forged aluminum take? |
2022-05-17T09:18:48.767 | <p>Hi I've been stuck on a problem for one of my exam revisions, we were asked to identify the natural frequency of the following structures</p>
<p><a href="https://i.stack.imgur.com/uBAke.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uBAke.png" alt="enter image description here" /></a></p>
<p>I know that the equation for the stiffness is as follows<br>
Left Column : <span class="math-container">$k= 12EI/L^3 $</span> <br>
Right Column : <span class="math-container">$k2=3EI/L^3 $</span> <br>
Equivalent Stiffness : <span class="math-container">$15EI/L^3$</span><br>
Hence the natural frequency of the column will be <span class="math-container">$$w_n=\sqrt{15EI/mL^3} $$</span>
Given <span class="math-container">$EI:8*10^{11} Nmm^2$</span>, <span class="math-container">$L: 2.5m$</span> , <span class="math-container">$m:250kg$</span>
The problem is when i tried to do the problem without converting the EI into SI unit of <span class="math-container">$Nm^2$</span> I will have a different result in terms of radians. i.e
<span class="math-container">$$wn(withoutconverting) :w_n=\sqrt{15(8*10^{11})/250(2500)^3}=1.75 rad/sec $$</span>
<span class="math-container">$$wn(with conversion) :w_n=\sqrt{15(8*10^{11})*10^{-6}/250(2.5)^3}=55.42 rad/sec $$</span>
The second answer seems to make more sense, as we have a very stiff column with a small mass it should have a really high natural frequency, however the answer key provided insisted that the first one is the correct way to approach it. How do i approach this and is my approach of converting it to SI unit correct? Thankyou so much for the help</p>
| |structural-engineering|civil-engineering|structural-analysis|vibration| | <p>The basic unit of the natural frequency is <span class="math-container">$Hz$</span>, <span class="math-container">$cycle/s$</span>, or just <span class="math-container">$1/s$</span>.</p>
<p>Let's check the results by using "<span class="math-container">$m$</span>" and "<span class="math-container">$mm$</span>" as the base unit -</p>
<p><strong>1. Use <span class="math-container">$m$</span> as the base unit:</strong></p>
<ul>
<li><span class="math-container">$\omega^2 = \dfrac{N-mm^2}{kg*m^3} = \dfrac{N*10^{-6} m^2}{kg*m^3} = \dfrac{N}{kg*10^6m} = \dfrac{kg*m}{kg*10^6 m*s^2} = \dfrac{1}{10^6s^2}$</span></li>
</ul>
<p><strong>2. Use <span class="math-container">$mm$</span> as the base unit:</strong></p>
<ul>
<li><span class="math-container">$\omega^2 = \dfrac{N-mm^2}{kg*m^3} = \dfrac{N-mm^2}{kg*10^9mm^3} = \dfrac{N}{kg*10^9mm} = \dfrac{kg*m}{kg*10^9mms^2} = \dfrac{kg*10^3mm}{kg*10^9mms^2} = \dfrac{1}{10^6s^2}$</span></li>
</ul>
<p><strong>Note: The base unit of $N is kgm/s^2, which needs to be converted too.</strong></p>
<p>Your mistake/discrepancy was caused by terminating the conversion too early before getting to the bottom of it.</p>
<p><strong>Check:</strong></p>
<p><span class="math-container">$EI = 8*10^{11} N-mm^2 = \dfrac{8*10^{11}kgm-mm^2}{s^2}*\dfrac{10^3mm}{m} = \dfrac{8*10^{14}kg-mm^3}{s^2}$</span></p>
<p><span class="math-container">$\omega = \sqrt{\dfrac{15*8*10^{14}kg-mm^3}{250kg*(2500mm)^3*s^2}} = 55.42Hz$</span></p>
| 51013 | Natural Frequency of a Shear Frame |
2022-05-17T12:29:39.770 | <p>I am wondering if a car that weighs 2,5kg and a engine power output of 1hp would be just as fast when going in a straight line like a 1250kg car with 500hp (they both would have a weight-to-horsepower ratio of 2,5:1). To make things easier, lets ignore things like efficiency, traction and aerodynamics. Are there other factors I am not aware of?</p>
| |car|acceleration| | <p>Acceleration is determined by power-to-weight ratio. Two vehicles with the same power-to-weight ratio will accelerate at the same rate.</p>
<p>Note however that top speed in level motion does not depend on weight, it depends on another ratio: power-to-drag. Two vehicles with the same power-to-drag ratio will achieve the same top speed- but the heavier one will take more time to get there.</p>
| 51014 | Is the effect of weight-to-horsepower ratio linear? |
2022-05-18T00:36:48.530 | <p>In a AWD Tesla Model 3, why is the front motor (OEM 1120960-00-E) different from the one in the back (OEM 1120980)? Why not just use the same one for both front and back? I'm sure that there was a reason for this and I would like to know what insight it brings forth... It's for an EV project that I am working on.</p>
| |electrical-engineering|motors|automotive-engineering|electromagnetism|electric-vehicles| | <p>As SolarMike suggested, available volume is part of the deal.</p>
<p>The second motor is actually idle most of the time, being called upon for dramatic acceleration or when wheel slippage is detected. As such, it doesn't need to send as much power (or torque) as the main motor does.</p>
<p>BTW, Tesla is planning, for either the cybertruck or the roadster or both, to produce models with a single front motor but separate motors for each rear wheel (tons of torque!), and it's probable that heavy-duty EV trucks will be built with a motor for each wheel. This gets rid of mechanical differentials as a side benefit.</p>
| 51024 | In a AWD Tesla Model 3, why is the front and back motor different? |
2022-05-19T09:52:21.107 | <p>I was posed a very similiar block diagram in my exam from this book (Alan V Oppenheim Ronald W Schafer - Discrete-Time Signal Processing-Pearson Education) but couldn't solve it:</p>
<p><a href="https://i.stack.imgur.com/K2H6K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K2H6K.png" alt="enter image description here" /></a></p>
<p>I want to solve for
<span class="math-container">$y[n]+a_{a}y[n-1]+...=b_{a}x[n]+b_{b}x[n-1]+...$</span>
I tried using w[n] but always failed, since both, w[n] and y[n] depend on w[n-1] always leaving one unknown.
Here's how far I got:
<span class="math-container">$
\\
\begin{aligned}
w[n] &=x[n-1]+y[n-1]+2w[n-1]\\
y[n] &= 2y[n-1]+2w[n-1] -2x[n] +2w[n]
\end{aligned}
$</span></p>
<p>I also know, that the solution to the problem is:
<span class="math-container">$
y[n]-8y[n-1]=-2x[n]+6x[n-1]+2x[n-2]
$</span></p>
<p>Any help is greatly appreciated :)
Thanks in advance for any help :)</p>
| |transfer-function|signal-processing|diagram| | <p>You can label each node and form a matrix equation:
<a href="https://i.stack.imgur.com/Ka1Ry.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ka1Ry.png" alt="enter image description here" /></a>
You get equations:
<span class="math-container">$$\begin{align}x_0&=x[n]\\
x_1&=-x_0+x_3+x_4\\
x_2&=2x_1\\
x_3&=z^{-1}x_0+2x_4\\
x_4&=z^{-1}x_2+z^{-1}x_5\\
x_5&=x_3\end{align}$$</span>
or matrix equation
<span class="math-container">$$\begin{bmatrix}
1&0&0&0&0&0\\1&1&0&-1&-1&0\\0&2&-1&0&0&0\\-z^{-1}&0&0&1&-2&0\\0&0&-z^{-1}&0&1&-z^{-1}\\0&0&0&-1&0&1
\end{bmatrix}\begin{bmatrix}x_0\\x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}
=\begin{bmatrix}x[n]\\0\\0\\0\\0\\0\end{bmatrix}$$</span>
Your output <span class="math-container">$y[n]$</span> is equal to <span class="math-container">$x_2$</span> which takes the solution
<span class="math-container">$$y[n]=-2x[n]\left(\frac{z^{-2}+3z^{-1}-1}{8z^{-1}-1}\right)\hspace{1.5cm}(**)$$</span>
i.e. you get the transfer function
<span class="math-container">$$T(z)=\frac{2z^2-6z-2}{8z-z^2}=\frac{2+6z-2z^2}{z^2-8z}$$</span>
To get the difference equation form we use <span class="math-container">$(**)$</span>
<span class="math-container">$$y[n]\left(8z^{-1}-1\right)=x[n]\left(2-6z^{-1}-2z^{-2}\right)$$</span>
<span class="math-container">$$8y[n-1]-y[n]=2x[n]-6x[n-1]-2x[n-2]$$</span></p>
| 51032 | How to get difference equation from this block diagram |
2022-05-19T19:19:30.670 | <p>If the car's acceleration is higher when the wheel's angular acceleration is higher(tangential acceleration=alpha x radius). Why are higher gears not used to accomplish the acceleration of the car instead of lower gears?</p>
| |mechanical-engineering|automotive-engineering|gears|power-transmission|engines| | <p>Every engine will start to choke when demand torque from the transmission, or better said from the tires, gets close to the engine's output torque at that state, RPM, breathing, engine capacity to shed the heat, and many other factors.</p>
<p>All engines need a burst of torque to accelerate their angular speed and that torque comes from the gap between available and demand torque.</p>
<p>A good driver or a good automatic transmission will keep the demand torque under available torque to let the engine breathe and rev up and deliver the best performance till the RPM nears the maximum engine's design RPM and available torque is so much more than the demand torque that will allow up-shifting for speed without choking.</p>
<p>And then and only then what you mean to say (if I understand correctly) is applied, higher gears are engaged or even overdrive.</p>
| 51034 | Shifting Between Gears To Achieve Higher Acceleration |
2022-05-21T05:44:51.460 | <p>I have the following question:</p>
<p><a href="https://i.stack.imgur.com/HyXXR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HyXXR.png" alt="enter image description here" /></a></p>
<p>I have no trouble drawing the influence line but I have problems with finding the magnitude. Can someone please explain how to get <span class="math-container">$\frac{8}{3}$</span>? How should I use the unit relative rotation to find the height of the triangles?</p>
<p><a href="https://i.stack.imgur.com/xngwJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xngwJ.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/H4SWe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H4SWe.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis| | <p>This question is to investigate the moment at an internal point due to the influence of the unit load "1". The solution is illustrated below.</p>
<p><a href="https://i.stack.imgur.com/ARotE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ARotE.png" alt="enter image description here" /></a></p>
<p>You shall repeat the steps to obtain the coefficient at "E".</p>
| 51049 | Influence line question for calculating maximum bending moment |
2022-05-24T18:11:16.373 | <p><em>Preface: I have no formal experience in design or mechanical engineering. I am simply a backyard mechanic looking to see if a particular assembly exists as an off-the-shelf product before designing and producing a custom one-off part.</em></p>
<p>The part I am looking for is similar to the floor mounted parking brakes that you could push into a locked position, then push again and have the device return via spring. Except more specifically, something that operates in a pull-only configuration (via a pull cable) and would have only 2 major positions, instead of the dozen or so ratchet positions of a parking brake.</p>
<p>Operation is something like this:
You pull a cable, it pulls some mechanical rod a certain distance, and locks the the rod into that position. Then to disengage the mechanism, you would pull the cable again, which would free the rod from the locked position and a spring would return the assembly to the disengaged position.</p>
<p>The application is fairly general, but I am specifically looking to use it as a way to operate a fuel valve via pull-only cable, or a gear shifter fork via a pull-only cable.</p>
<p>I'm having a hard time finding similar assemblies on google. Looking up things like "mechanical toggle actuator", "cylindrical linear ratcheting", or "barrel cam toggle actuator" doesn't bring up any parts that could be used in this manner.</p>
<p>Is what I'm describing an off-the-shelf part anywhere? Or am I out of luck for an existing product.</p>
| |actuator| | <p>Your question indicates an option for push-only, while the body of the question references pull-only. One could argue for a simple means to invert the motion, hence this answer.</p>
<p>Parker Pens invented a ball point retraction system that is in common use today. <a href="https://www.youtube.com/watch?v=MhVw-MHGv4s" rel="nofollow noreferrer">Engineerguy on YouTube</a> provides a very clear explanation of the workings of this push to engage, push to retract mechanism.</p>
<p>Another YouTuber, <a href="https://www.youtube.com/watch?v=3_wPH904a_8" rel="nofollow noreferrer">This Old Tony</a> has a much longer (18 minutes) video of a DIY push-latch mechanism.</p>
<p>With a little bit of exploration and modification, your answer may be found in one of these or similar videos. I'm partial to the drunken cone/triangle toT mentions in his video.</p>
<p>Yet another YouTube creator, <a href="https://www.youtube.com/watch?v=VA7UGVCpcFk" rel="nofollow noreferrer">Coffredom</a> provides for a 3D printed mechanism, which exposes the design and structure for more clarification. The comments section includes an untested link for the 3D printing files. Screen capture below from the linked video.</p>
<p><a href="https://i.stack.imgur.com/ESK4p.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ESK4p.jpg" alt="diy latch video screen capture" /></a></p>
| 51078 | Finding the name of: Mechanical toggle actuator utilizing single direction of force (pull only/push only) |
2022-05-26T02:40:32.410 | <p>What happens if an EV's motor is stalled, i.e. prevented from rotating while attempting to accelerate? Can this result in over-current damage to the batteries, motor, or speed controller circuits etc.</p>
<p>I know that for normal hobby DC motors, this scenario can be extremely harmful and will almost certainly result in severe (often catastrophic) damage to the motor or controller circuits, if the stall condition isn’t promptly resolved.</p>
<p>How is this risk mitigated in an electric vehicle?</p>
| |electric-vehicles| | <p>A stalled motor's cooling fan is not turning, at the same time its windings are dissipating lots of electrical power into heat. The heat ruins the insulation on the wire in the windings, and the windings then get short-circuited. This lets still more current flow through the motor, and the drive circuits themselves then fail due to excessive power dissipation in them.</p>
<p>This <em>locked armature condition</em> can be protected against by monitoring armature RPM as a function of drive power. High drive power + no RPM = locked armature which condition causes the driver circuit to <em>fold back</em> i.e., to reduce the drive power automatically to a value that is known to be survivable by a motor with no fan cooling.</p>
<p>It is also common to put thermistors on the drive circuit transistor heat sinks and on the inside of the motor casing, to generate a power shutdown command if either set of components gets too hot for any reason.</p>
<p>Alternatively, it is easy to implement a <em>current sense</em> circuit on the driver board which automatically limits the drive current to a level which will not overtemp the output transistors.</p>
<p>Note in this context that each of these error condition detectors is also used in audio amplifiers of modern design; in fact, you will almost always find at least two entirely independent power interrupt systems in audio amps to redundantly catch any error condition that could arise during operation. One important difference between the first generation of solid-state audio amps designed in the mid-1960's (which were notoriously prone to blowing up) and all later designs is the presence of more than one error condition detector circuit.</p>
| 51095 | What happens if an EV motor is stalled? |
2022-05-27T17:43:33.840 | <p>I am curious about integration of renewable energy in electricity grids. I am a mechanical engineer and understand energy and power but I have little knowledge about electricity. I will be thankful if you help me understand what can go wrong in a power grid when we add intermittent renewable energy sources, namely solar and wind. I found helpful information about inertia in answers to <a href="https://engineering.stackexchange.com/questions/2245/quantifying-inertia-on-the-electricity-grid">this question</a>. I have also learned that stability in a power grid means stability in its voltage, frequency and phase. Here are my questions:</p>
<ol>
<li><p>Suppose a simple AC circuit containing a source, a resistor and a switch. If the switch is suddenly turned on, electricity flows through the resistor. Is my understanding correct that in this experiment the frequency of the source tends to fall? Does this experiment have any effect on the source's voltage and phase? My reasoning for the decline of frequency is that the resistor dissipates some energy (as heat), which means some energy is taken from the generator, which means the generator is left with less energy to maintain its angular velocity with.</p>
</li>
<li><p>Suppose the switch in the circuit of question 1 is turned on and off repeatedly at irregular intervals. Does this experiment lead to different results than that of question 1?</p>
</li>
<li><p>Suppose a large number of circuits of question 1 connected in parallel. If the switches of the circuits are turned on and off randomly and repeatedly and independently from one-another, can we assume the ensemble of the circuits has the effect of a uniform load on the source?</p>
</li>
<li><p>In a grid based on conventional energy sources, I think of renewables as the switch in the above example. That is, when renewable electricity is available, I consider the switch is <em>off</em>, the renewable source is feeding the circuit, and the main (conventional) AC source is disconnected from the circuit. On the contrary, when renewable power is <em>not</em> available, the switch is turned <em>on</em>, and the main source is connected to the circuit and feeds the resistor. I understand electric loads are not just resistance, and I am considering them as things that consume a certain kWh of energy. Are my assumptions reasonable or am I missing something here? Is it reasonable to say that in such a grid the main challenge is to maintain frequency stability?</p>
</li>
<li><p>In a hypothetical grid based on solar and wind (that is, with little or no conventional sources), I am guessing the main challenge is to maintain voltage stability with minimum storage costs. Is my guess reasonable or am I missing something here? Is it true that voltage stability can be achieved by adding storage to the grid? What can be done for achieving frequency and phase stability?</p>
</li>
</ol>
<p>Thank you for your time.</p>
| |electrical-engineering|power-engineering|stability|ac| | <p>In a traditional grid, the frequency stability is maintained by the rotating mass of the generator set. If the generator suddenly becomes easier to spin, or more difficult to spin, (because you have switched a switch on or off), the steam turbine and generator may gradually get faster or slower -- like a moving car at the top or bottom of a hill. Short term frequency stability just depends on the rotating weight: if you have a couple of slow rotations, you give it some more steam, and at the end of the day you expect (and are required) to have had the correct number of rotations over the last 24 hours. Sometimes the weight wasn't enough, and extra rotating weight is added: flywheel installations.</p>
<p>With modern electronics, it is not necessary to use rotating weight. Electronic devices can react to the slope of the alternating current, adding more energy to make the voltage and current rise and fall faster, or shunting energy away to make the voltage and current rise and fall slower, not having to take the average over a few cycles and then add or reduce steam to accelerate or decelerate the turbine.</p>
<ol>
<li><p>Gensets slow down when you add load, just like any other engine.</p>
</li>
<li><p>Gensets are regulated to maintain constant speed. If you have hills, it may confuse the speed regulation, just as your speed control on your car can be confused. There are specifications and requirements on how fast you can change the load on a grid without the grid falling over: if you operate a load outside those specifications and expectations, you could be going downhill when the speed control is trying to make the car go faster, then be going uphill when the speed control is trying to make the car go slower. This is called 'control theory', and is a characteristic of the 'control system', of which the generators and solar/wind are only a small part.</p>
</li>
<li><p>In normal operation, load made up of lots of houses is just one large load. If something goes wrong, different parts of the load may go out of synchronization with each other or with the supply. You may consider it as a 'loosely coupled' system of multiple weights and springs: you can overdrive part of the system to make it over-oscillate, or stop, or go faster or slower.</p>
</li>
</ol>
<p>4,5) For grid control, frequency stability is not a 'different thing' than voltage stability or current stability. They are all aspects of the same thing. Adding storage to a grid doesn't magically give it more voltage stability or phase stability any more than adding a large gas tank to your car does.</p>
<p>Wind and Solar have less rotating mass than 'conventional' does, which removes one form of stability. But solar can be switched off and started up in with sub ms delay. The control system has to be engineered to correspond to the different characteristics, which may include batteries and electronic frequency regulation.</p>
| 51114 | Basic questions about power grid stability with variable loads and sources |
2022-05-28T00:18:18.387 | <p>I'm trying to retrofit a <a href="https://i.guim.co.uk/img/media/edc7e5b39855f7a486aeb53414e7cb6bbf2ca3e3/0_211_5184_3110/master/5184.jpg?width=300&quality=45&auto=format&fit=max&dpr=2&s=3d53a0037230b597b24cd53e2b653091" rel="nofollow noreferrer">pull cart</a> with electric motor. So, one wheel will have an electric hub motor and it is connected to a rotating shaft to the other non-powered wheel. The driveshaft is connected to the undercarriage of the cart via <a href="https://res.cloudinary.com/rsc/image/upload/b_rgb:FFFFFF,c_pad,dpr_1.0,f_auto,h_843,q_auto,w_1500/c_pad,h_843,w_1500/F7508977-01?pgw=1&pgwact=1" rel="nofollow noreferrer">pillow block bearing</a>.</p>
<p>As the powered wheel rotates, the other wheel follows and rotates, that's the idea.</p>
<p>Steering should be done by the user with their hands/arms, not by the motors.</p>
<p>I wonder if the cart will slowly steer unwantedly to the left and right, since only one motor is attached to only one wheel?</p>
<p>P/S: I can't have two motors at two wheels because if both are not completely in sync (which is hard to achieve), the cart will steer to the left or right ny itself (imagine how a a tank steers). I will also have a switch to turn on/off the motor and PAS sensor. I will figure out the speed control issue in another post.</p>
| |mechanical-engineering|electrical-engineering|motors|bearings|electric-vehicles| | <p>If only 1 wheel is driven, and the other can spin freely, yes it will want to cut aggressively towards the opposite side from the wheel.</p>
| 51117 | Will the cart steer to the left/right by itself in this design? |
2022-05-28T09:22:25.250 | <p>I'm traveling to a mountain hut in Austria every year which is in the middle of nowhere. If you're lucky you can sometimes get one bar and a weak Long-Term Evolution (LTE) or Edge connection for a few minutes if you put your mobile phone at a certain position.<br />
What would be a cheap way to enhance the signal to get a stable connection everywhere around the hut (doesn't even have to be inside)?</p>
| |electrical-engineering|telecommunication|signal-processing| | <p><strong>1.-</strong> Mobile operators are gold mines.</p>
<p>Their business model is that people communicate through <strong>their towers</strong> AND using devices, <strong>ONLY</strong> the devices that the operators have agreed upon with mobile phone manufacturers.</p>
<p>If you work for a mobile operator or contractor, or work for certain companies that have specialised equipment, you may be able to walk up the mountains with a <strong>narrow beam antenna</strong>, with a <strong>good set of amplifiers</strong>, and pick up tower signals that conventional mobile phones can't.</p>
<p>For instance, do you have a <strong>portable spectrum analyzer</strong>?</p>
<p><a href="https://www.electrorent.com/uk/search?search_term=Fieldfox+Handheld+Spectrum+Analyzer" rel="nofollow noreferrer">https://www.electrorent.com/uk/search?search_term=Fieldfox+Handheld+Spectrum+Analyzer</a></p>
<p><strong>Keysight Field fox</strong></p>
<p><a href="https://www.electrorent.com/uk/search?search_term=keysight%20fieldfox&creative=&keyword=keysight%20fieldfox&matchtype=e&network=o&device=c&msclkid=408e60b398de1aafdb740330eba8858c&utm_source=bing&utm_medium=cpc&utm_campaign=Europe%20%7C%20UK%20%7C%20AW%20%7C%20Search%20%7C%20Manufacturers&utm_term=keysight%20fieldfox&utm_content=KEYSIGHT%20%7C%20Fieldfox" rel="nofollow noreferrer">https://www.electrorent.com/uk/search?search_term=keysight+fieldfox&creative=&keyword=keysight%20fieldfox&matchtype=e&network=o&device=c&msclkid=408e60b398de1aafdb740330eba8858c&utm_source=bing&utm_medium=cpc&utm_campaign=Europe%20%7C%20UK%20%7C%20AW%20%7C%20Search%20%7C%20Manufacturers&utm_term=keysight%20fieldfox&utm_content=KEYSIGHT%20%7C%20Fieldfox</a></p>
<p><strong>Rohde&Schwarz FPH or FSH</strong></p>
<p><a href="https://www.electrorent.com/uk/search?search_term=Rohde%20%26%20Schwarz%20Handheld%20Spectrum%20Analyzer&creative=&keyword=%2BRohde%20%2B%26%20%2BSchwarz%20%2BHandheld%20%2BSpectrum%20%2BAnalyzer&matchtype=e&network=o&device=c&msclkid=a834f36db18f1b5d166490eedf762ed9&utm_source=bing&utm_medium=cpc&utm_campaign=Europe%20%7C%20UK%20%7C%20AW%20%7C%20Search%20%7C%20Manufacturers&utm_term=%2BRohde%20%2B%26%20%2BSchwarz%20%2BHandheld%20%2BSpectrum%20%2BAnalyzer&utm_content=R%26S%20%7C%20Handheld%20Spectrum%20Analyzer" rel="nofollow noreferrer">https://www.electrorent.com/uk/search?search_term=Rohde+%26+Schwarz+Handheld+Spectrum+Analyzer&creative=&keyword=%2BRohde%20%2B%26%20%2BSchwarz%20%2BHandheld%20%2BSpectrum%20%2BAnalyzer&matchtype=e&network=o&device=c&msclkid=a834f36db18f1b5d166490eedf762ed9&utm_source=bing&utm_medium=cpc&utm_campaign=Europe%20%7C%20UK%20%7C%20AW%20%7C%20Search%20%7C%20Manufacturers&utm_term=%2BRohde%20%2B%26%20%2BSchwarz%20%2BHandheld%20%2BSpectrum%20%2BAnalyzer&utm_content=R%26S%20%7C%20Handheld%20Spectrum%20Analyzer</a></p>
<p>do you also have a <strong>logperiodic antenna</strong>?</p>
<p>The receiver side is not enough because to transmit you are going to need another amplifier and another antenna, that could be the same antenna, but then you need more compoments.</p>
<p>Instruments, amplifiers .. current .. battery .. more battery.</p>
<p><strong>2.-</strong> If you say you can set up something in place, then a <strong>local mobile or landline operator</strong> may have a product to bring coverage with for instance
a microwave link.</p>
<p>Here there's a one-off installation including a roof antenna, plus indoor equipment usually supplied by the same operator.</p>
<p>It may be the case that the installation and the equipment costs, all are covered by the operator, as by law they sign down to supply coverage for emergencies, and not only for 999. But then again, do they really want to extend their coverage to just one hut 'up there'?</p>
<p><strong>3.-</strong> This is why <strong>Satellite communications</strong> may be of interest.</p>
<p>Satellite operators offer coverage where big mobile operators consider it's not worth or cannot their coverage.</p>
<p><strong>used Globalstar $375</strong></p>
<p><a href="https://www.ebay.co.uk/itm/325269451418?chn=ps&norover=1&mkevt=1&mkrid=710-153316-527457-8&mkcid=2&itemid=325269451418&targetid=4584757337008491&device=c&mktype=&googleloc=&poi=&campaignid=431353847&mkgroupid=1298523655396099&rlsatarget=pla-4584757337008491&abcId=9301942&merchantid=87779&msclkid=63f1557ade751520dfbd36f9af3ace14" rel="nofollow noreferrer">https://www.ebay.co.uk/itm/325269451418?chn=ps&norover=1&mkevt=1&mkrid=710-153316-527457-8&mkcid=2&itemid=325269451418&targetid=4584757337008491&device=c&mktype=&googleloc=&poi=&campaignid=431353847&mkgroupid=1298523655396099&rlsatarget=pla-4584757337008491&abcId=9301942&merchantid=87779&msclkid=63f1557ade751520dfbd36f9af3ace14</a></p>
<p><strong>Iridium 9505A used</strong> (new above £1k) <strong>for £400.-</strong></p>
<p><a href="https://www.ebay.co.uk/itm/155269109306?chn=ps&norover=1&mkevt=1&mkrid=7101533165274578&mkcid=2&itemid=155269109306&targetid=4584826055637462&device=c&mktype=&googleloc=&poi=&campaignid=412354546&mkgroupid=1299623041023876&rlsatarget=pla-4584826055637462&abcId=9300541&merchantid=87779&msclkid=a2f3db409266146f1334b8c3e329b585" rel="nofollow noreferrer">https://www.ebay.co.uk/itm/155269109306?chn=ps&norover=1&mkevt=1&mkrid=7101533165274578&mkcid=2&itemid=155269109306&targetid=4584826055637462&device=c&mktype=&googleloc=&poi=&campaignid=412354546&mkgroupid=1299623041023876&rlsatarget=pla-4584826055637462&abcId=9300541&merchantid=87779&msclkid=a2f3db409266146f1334b8c3e329b585</a></p>
<p><strong>Thuraya XT-Lite £635.-</strong></p>
<p><a href="https://www.amazon.co.uk/Thuraya-NA-XT-LITE-Satellite-Phone/dp/B00RBQYO5I/ref=asc_df_B00RBQYO5I?tag=bingshoppinga-21&linkCode=df0&hvadid=80126967116376&hvnetw=o&hvqmt=e&hvbmt=be&hvdev=c&hvlocint=&hvlocphy=&hvtargid=pla-4583726541606934&psc=1" rel="nofollow noreferrer">https://www.amazon.co.uk/Thuraya-NA-XT-LITE-Satellite-Phone/dp/B00RBQYO5I/ref=asc_df_B00RBQYO5I?tag=bingshoppinga-21&linkCode=df0&hvadid=80126967116376&hvnetw=o&hvqmt=e&hvbmt=be&hvdev=c&hvlocint=&hvlocphy=&hvtargid=pla-4583726541606934&psc=1</a></p>
<p><strong>4.-</strong> If <strong>SMS only</strong> is an option, perhaps this <strong>Zoleo £199.</strong>- , gloal SMS, solves it.</p>
<p><a href="https://www.amazon.co.uk/ZOLEO-ZL1000-Satellite-Communicator/dp/B07X59RH7T/ref=asc_df_B07X59RH7T?tag=bingshoppinga-21&linkCode=df0&hvadid=80814232605788&hvnetw=o&hvqmt=e&hvbmt=be&hvdev=c&hvlocint=&hvlocphy=&hvtargid=pla-4584413755259847&psc=1" rel="nofollow noreferrer">https://www.amazon.co.uk/ZOLEO-ZL1000-Satellite-Communicator/dp/B07X59RH7T/ref=asc_df_B07X59RH7T?tag=bingshoppinga-21&linkCode=df0&hvadid=80814232605788&hvnetw=o&hvqmt=e&hvbmt=be&hvdev=c&hvlocint=&hvlocphy=&hvtargid=pla-4584413755259847&psc=1</a></p>
<p><strong>5.-</strong> If it's about staying updated, <strong>receiving only</strong>, an affordable solution may be a <strong>satellite pager</strong> :</p>
<p><a href="https://www.satphone.co.uk/product/iridium-motorola-9501-kyocera-sp-66k-satellite-pager-used/" rel="nofollow noreferrer">https://www.satphone.co.uk/product/iridium-motorola-9501-kyocera-sp-66k-satellite-pager-used/</a></p>
<p>Thanks for reading.</p>
| 51120 | What is the best way to enhance an existing LTE signal on a mountain hut? |
2022-05-28T17:30:12.910 | <p>For a variable pitch propeller where the angle of attack can be adjusted during flight, are the individual blades similar in design to standard propellers, with a low angle of attack at the edge of the propeller (3-6 degrees), and gradually increases towards the root (14-16 degrees); or are the propeller blades one singular angle of attack throughout the entire length of the blade? I'm presuming the blades have a differing angle of attack at the root and tip, but of a lower magnitude to prevent the root AOA on the root of the blade exceeding the critical angle?</p>
| |aerospace-engineering|aerodynamics|propulsion| | <p>All propellers have to have a gradually increasing twist as we get closer to the hub to maintain the same angle of attack with the relative wind at that particular position.</p>
<p>The reason is the change in the angle of the relative wind as we move from the tip to the root.</p>
<p>The relative wind is the vector sum of the airplane's speed and the speed of the point along the length of the propeller.</p>
<p>Relative wind tilts more toward the airplane's axial speed near the root as compared to near the tip which is more aligned with the plane of the propeller. Just compare the tangential speeds <span class="math-container">$V_t=\omega R$</span> the greater the R is greater Vt.</p>
<p><span class="math-container">$\vec{V_r}= \vec{V_t}+\vec{V_{axial}}$</span></p>
<ul>
<li><span class="math-container">$V_r=$</span> relative wind</li>
<li><span class="math-container">$V_t=$</span> propeller tangential speed</li>
<li><span class="math-container">$V_{axial}= $</span>airplane speed</li>
</ul>
| 51122 | Angle of attack on variable pitch propeller? |
2022-05-29T12:20:36.077 | <p>I'm reading about buckling of timber members from Eurocode 5: Design of timber structures part 1.</p>
<p>Formula 6.30 gives the relative slenderness of a member:</p>
<p><span class="math-container">$$\lambda_{rel.m} = \sqrt{\frac{f_{m,k}}{\sigma_{m,crit}}}$$</span></p>
<blockquote>
<p>where <span class="math-container">$\sigma_{m,crit}$</span> is the critical bending stress calculated according to the classical theory of stability, using 5-percentile stiffness values.</p>
</blockquote>
<p><span class="math-container">$f_{m,k}$</span> is the characteristic bending strength of the timber.</p>
<p>Formula 6.32 gives the critical bending stress for solid rectangular cross-section:</p>
<p><span class="math-container">$$\sigma_{m,crit}=\frac{0,78b^2}{hL_{ef}}E_{0,05}$$</span></p>
<p>I'm curious about this last formula, how does it result as the critical bending stress from classical theory of stability? I'm used to seeing the flexural buckling calculated with the Euler formula:</p>
<p><span class="math-container">$$\sigma_{crit}=\frac{\pi^2EI}{hbL_{eff}^2}$$</span></p>
<p>Simplifying the <span class="math-container">$I$</span>, using the normal formula <span class="math-container">$I=\frac{bh^3}{12}$</span> for rectangular member I get:</p>
<p><span class="math-container">$$\sigma_{crit}=\frac{\pi^2Eh^2}{12L_{eff}^2}$$</span></p>
<p>The result does not look similar to what is provided in Eurocode. Some difference in constant for example would not be so surprising as some corrective factors might be used, but the form is completely different.</p>
<p>So, I wonder if someone has better information on the critical bending stress for a rectangular section? Where is my derivation going wrong?</p>
<p>Thank you!</p>
| |buckling|stability|wood| | <p>You are comparing lateral-torsional buckling from bending with flexural buckling from compression (column buckling). That comparison doesn't make sense.</p>
| 51130 | Stability of solid rectangular wood member |
2022-05-30T08:54:27.390 | <p>I've been told that it does because it's still searching for networks or something, which doesn't really make any sense to me. But what does fit somewhat with that are notices in the android interface that say enhanced location services will allow WiFi networks to be scanned for even when WiFi is off, though this isn't exactly the same thing.</p>
| |electrical-engineering|electromagnetism|consumer-electronics| | <p>All electric and electronic devices emit EM waves as a matter of course because changing current flows will produce waves like a motor, clock, or pocket calculator. But then there are also devices that intentionally produce emissions as part of their function like a radio.</p>
<p>Sort of like the difference between your lawnmower which produces sound incidentally compared to a speaker whose purpose is to produce sound.</p>
| 51137 | Does an iPhone on airplane mode continue to emit electromagnetic radiation field? |
2022-05-31T04:08:02.233 | <p>I would like to understand what's the best optimization in terms of design for a single mold, dual cavity of a two-part box (top and bottom parts) which should be hold together with 4 screws without any clip. For other reasons not drawn, both parts should be roughly the same height - therefore the box is cut in the middle.</p>
<p><a href="https://i.stack.imgur.com/U8tH6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U8tH6.png" alt="enter image description here" /></a></p>
<p>From a manufacturing point of view only, which following injection molding option is better for a single mold with two cavities?</p>
<p>Option A:</p>
<ul>
<li>to have 4 pillars on both bottom and top parts, that are at same height of each part. Basically, the 4 screws will be inside the pillars in the middle of the box.</li>
</ul>
<p><a href="https://i.stack.imgur.com/xAOFu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xAOFu.png" alt="enter image description here" /></a></p>
<p>Option B:</p>
<ul>
<li>to have 4 pillars on one part ONLY, that are at the double height of that part. The 4 screws will be close to the main wall of the other part.</li>
</ul>
<p><a href="https://i.stack.imgur.com/Fnd3u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fnd3u.png" alt="enter image description here" /></a></p>
| |manufacturing-engineering|abs|injection-molding| | <p>The optimal solution is usually thus:</p>
<p>You move each of the cylindrical features into its nearest corner and mold it in flush with the side walls.</p>
<p>You extend all eight of these <em>bosses</em> (that's what those tall cylindrical features are called) flush with the parting line at the center. You make four on one side into hollow tubes where the tube goes all the way through the boss to the outside of the part and the tube has enough diameter to pass a screw without interference.</p>
<p>The bosses and the side walls of the part are then <em>relieved</em> by tapering them by about 1 to 1.5 degrees, so the molded part will free itself more easily from the mold cavity.</p>
<p>On the mating half of the box, you put small shallow holes into the ends of the bosses that are the right diameter to engage the screw threads.</p>
<p>To assemble the box, you fit the two halves together and run a long screw into each of the thru-holes on one half so the threads engage the holes in the tips of the bosses on the other side, and tighten the screws.</p>
<p>Note also that it is standard design practice to <em>pilot</em> the mating surfaces together so that they <em>interlock</em> when pressed together. This will automatically bring the screw features into proper alignment during assembly, so the screw point will "find" the hole it is supposed to screw down into.</p>
<p>To save a few pennies on each assembly, you can step the diameters of the thru-holes so as to pass the <em>head</em> of a screw and then stop the screw head near the bottom of the thru-hole with a short section of smaller diameter hole. This lets you use a shorter (cheaper) screw.</p>
| 51140 | Injection molding: What is best for a dual cavity single mold? |
2022-05-31T09:53:37.710 | <p>When waiting for the underground, I've always been wondering why we could not absorb the kinetic energy of the train to slow it down at the station and then to release this energy to help it accelerating during the starting up. This would also reduce air pollution due to braking at underground stations.</p>
<p>Indeed, in frequently-stopping, trains such as underground trains, most of the energy is used to accelerate, so I am wondering whether the train could be slowed down by a large spring that would be compressed while arriving a stop station, then blocked when the train is stopped for the travelers to get out/in, then released to help the train accelerate to continue its route.</p>
<p>Of course, the train should not go to the opposite direction when releasing the spring, so a system should be designed so that the spring can be released in the correct direction.</p>
<p>This idea looks simple and I'm sure engineers has already thought about it, however, I've not found any discussion on this on the web. I found that on some Spanish lines, the kinetic energy during braking is converted to electricity and used by other trains from the same line (<a href="https://journals.sagepub.com/doi/10.1177/0954409711429411" rel="noreferrer">research paper</a>), but why not keeping the mechanical energy to release it directly with a spring method?</p>
| |springs|energy-efficiency|energy-storage|regenerative-braking|braking| | <p>On the London Underground, many stations use gravity. The line runs uphill into the station, and then departs downhill. This gives a large amount of energy storage, for free, with no additional moving parts requiring maintenance. It's somewhat inflexible, the amount of rise needs to be tied into the train's operating speed.</p>
<p>It obviously can't be applied on surface railways, or for road vehicles, but underground you have that extra degree of freedom.</p>
| 51142 | Could a mainspring be used to generate train acceleration during starting-up from the kinetic energy stored during its slowing down? |
2022-06-01T13:40:52.180 | <p>I'm trying to retrofit an electric motor (such as <a href="https://www.alibaba.com/product-detail/3000W-48V-60V-72V-84V-96V_1600103779994.html?spm=a2700.details.0.0.6777758443el7c" rel="nofollow noreferrer">this</a>) to a pull cart as depicted below. I got most things figured out except for how to slow down the motor to a pedestrian speed.</p>
<p>Could I slow it down by reducing the amperes? Or must I use a lower gear ratio?</p>
<p><a href="https://i.stack.imgur.com/k4R2H.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k4R2H.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|electrical-engineering|motors|gears|bicycles| | <p>The specific motor that you linked is a <a href="https://en.wikipedia.org/wiki/Brushless_DC_electric_motor" rel="nofollow noreferrer">BLDC motor</a> and requires a controller. The multitude of wires coming out is a tell.</p>
<p>This controller will be what will regulate the speed. By turning on and off certain coils in a rythm.</p>
<p>To set the speed to your liking you will need to program the controller to your needs.</p>
| 51152 | How to slow down an electric motor for pull cart? |
2022-06-02T08:03:19.593 | <p>I'm looking for advice on how to depict the HPLC pump on the P&I Diagram. What is the symbol (or combination of symbols) to indicate that it has 4 switchable inputs and one combined output with the built-in high-pressure pump? (+ <em>in-situ</em> flow indicator)</p>
<p>Additionally, I'm looking for the symbol of the position switch valve. It's the thing like this:</p>
<p><a href="https://i.stack.imgur.com/rCNno.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rCNno.jpg" alt="enter image description here" /></a>'</p>
<p>Have several inputs, out of which only one can be selected, and permanently attached the common output.</p>
| |pi-diagram| | <p>FPLC manufacturer Pharmacia (now Cytiva) draws the pumps and injection valves in their AKTA systems like this. FPLC is a low-pressure HPLC. It sounds like you want the symbol for the injection valve, rather than the pump itself?
<a href="https://i.stack.imgur.com/6AYh9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6AYh9.png" alt="enter image description here" /></a></p>
<p><a href="https://btiscience.org/wp-content/uploads/2014/04/FPLC_Optional_Config.pdf" rel="nofollow noreferrer">https://btiscience.org/wp-content/uploads/2014/04/FPLC_Optional_Config.pdf</a></p>
| 51164 | Piping and Instrumentation diagram symbols for Position valves and HPLC pumps |
2022-06-02T15:40:37.757 | <p>I built a (relatively) portable air-blowing device that purifies room-temperature ambient air (HEPA, carbon filter), and then heats and humidifies it to 37 °C and RH85%. I use the machine to do aerobic exercise indoors on a treadmill/stationary bicycle: conditioning the air prevents airway constriction (common for people with disabilities like asthma, COPD, long COVID etc). I now wonder if I can put the machine on a cargo bicycle which would allow me commute to work (the machine uses about 1.5 kW, and there are 2 kWh batteries available these days).</p>
<p>The air-purification and heating part is easy, but the humidifier part does not seem to be. The humidifier is, essentially, a stainless (316) bucket of water (~10L) with a thermostat (an off-the-shelf item -- a sous-vide oven). The air is blown over the water surface(about 2 sq feet). Is there a passive (or a low-cost active) method to keep the bucket balanced and to keep the water horizontal to within 10 mm over 30x30 cm area, to prevent the water spilling and being blown inside of the breathing mask? I looked into multi-tube Nafion humidifiers from Perma-Pure, but these are not rated for humans (rated for hydrogen fuel cells), and are too costly (many Ks).</p>
<p><a href="https://i.stack.imgur.com/zxJmc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zxJmc.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|hvac| | <p>Random idear. Use a gel blanket on top of the heated water reservoir. The gel will acquire water from below and transpire it into the air. The water can be heated normally. Sodium polyacrylate is used in food and hygiene products. Basically, you want a lid made of deposable diapers.</p>
<p>A lid made of thick PVA sponge foam might be good enough, and could be tested for $10.</p>
| 51174 | portable humidifier (balancing a bucket of water on a bicycle) |
2022-06-03T11:47:45.787 | <blockquote>
<p>Calculate the machining time required to reduce the shaft diameter by 6 mm if the maximum depth of cut is 2 mm, and also the parting off (cut off) time for the shaft, knowing that: the shaft diameter = 50 mm, shaft length = 500 mm, longitudinal feed = 0.2mm/rev, cross (traverse) feed = 0.1 mm/rev and cutting velocity (v) 16 m/min.</p>
</blockquote>
<p>This is a problem about parting off then turning what I don't understand is the definition of depth of cut here I initially assumed that the depth of cut is 6mm but however when I read the max. depth of cut =2mm this confused me.</p>
<p>Also the turning part seems to be done on multiple stages longitudinal and transverse but exactly how do I calculate the number of stages?</p>
| |mechanical-engineering|design|machining|lathe|metalwork| | <p>OK so you have a shaft of length 500 mm and diameter 50 mm. You need to reduce the diameter by 6 mm. For your material and setup, your cutting speed (that is, how fast the tool can travel over the shaft) is <span class="math-container">$\dot{c} = 16 m/min$</span>, and your maximum depth of cut is 2 mm. Longitudinal feed is given as 0.2 mm/rev, transverse feed 0.1 mm/rev.</p>
<p>I am going to assume that machine set time, measurement time, tool change time, etc are all zero. I will also assume that there is no facing cut to be made, and that the original metal you would be using is greater than 500 mm, since you are parting it off. I am also going to assume that your machine can be set to ANY rpm.</p>
<p>Since your machine will reduce the <em>radius</em>, this means that your total longitudinal cuts should come out to <span class="math-container">$\Delta = 3mm$</span>. This would be done most efficiently with two cuts, of <span class="math-container">$\delta_1 = 2mm$</span> and <span class="math-container">$\delta_2 = 1 mm$</span>. You would then have a parting cut.</p>
<p><strong>Pass 1:</strong>
Longitudinal pass cutting <span class="math-container">$\delta_1 = 2mm$</span>, across length <span class="math-container">$l=500mm$</span>, original diameter <span class="math-container">$d_0 = 50 mm$</span>. First, find the spindle RPM:
<span class="math-container">$$\nu_1 = \frac{\dot{c}}{\pi d_0} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(50 mm)}=102 rpm$$</span>
Your travel speed is your longitudinal feed <span class="math-container">$f_z = 0.2 mm/rev)$</span> times your spindle speed:
<span class="math-container">$$ \dot{z}_1=f_z \nu_1 = 20.4 mm/min$$</span>
From there, you can get your tie taken:
<span class="math-container">$$ l = \dot{z}_1t_1 \implies t_1=\frac{l}{\dot{z}_1}=\frac{500 mm}{20.4 mm/min} = 24.51 min$$</span>
<strong>Pass 2:</strong> Your second pass should be identical to the first, except that the depth of cut is 1 mm. Normally, reducing your depth of cut means you can adjust your cutting speed by a factor <span class="math-container">$F_d$</span>, but we will disregard that here since you don't have enough information to find it.
Since your new diameter is <span class="math-container">$d_1=46mm$</span>, we will recalculate the spindle RPM:
<span class="math-container">$$\nu_2 = \frac{\dot{c}}{\pi d_1} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(46 mm)}=111 rpm$$</span>
<span class="math-container">$$\dot{z}_2=f_z\nu_2 = 22.2mm/min$$</span>
<span class="math-container">$$t_2=\frac{l}{\dot{z}_2}=\frac{500 mm}{22.2 mm/min} = 22.52 min$$</span></p>
<p><strong>Pass 3:</strong> This is the parting off. New diameter is <span class="math-container">$d_2=44 mm$</span>, and transverse travel feed is <span class="math-container">$f_x=0.1 mm/rev$</span>. Lets get our new spindle speed and travel speed:
<span class="math-container">$$\nu_3 = \frac{\dot{c}}{\pi d_2} = \frac{(16 m/min)(1000 mm/m)}{(3.1416/rev)(44 mm)}=116 rpm$$</span>
<span class="math-container">$$\dot{x}_3=f_x\nu_3=(0.1 mm/rev)(116 rpm)=11.6 mm/min$$</span>
<span class="math-container">$$r = \frac{d}{2} = \dot{x}_3t_3 \implies t_3=\frac{d}{2 \dot{x}_3}=\frac{44 mm}{2(11.6 mm/min)}=1.90 min$$</span></p>
<p><strong>Total time:</strong> Your total time is the sum of these three passes, since we are ignoring everything else:
<span class="math-container">$$t=t_1+t_2+t_3 = 24.51 min + 22.52 min + 1.90 min = 48.93 min$$</span></p>
<p>Based on the assumptions made to solve the problem (and, barring a math error) this should be the machining time.</p>
| 51190 | What is the definition of depth of cut in machining? |
2022-06-03T15:50:13.563 | <p>I know that the minimum number of reinforcement bars in circular concrete section of columns are 6, is it the same for circular tie rod section?</p>
| |civil-engineering|applied-mechanics|reinforced-concrete|eurocodes|cross-section| | <p>The longitudinal reinforcement in a concrete column serves two functions - 1) to provide the required strength to resist the applied forces, and 2) acting as support for the transverse reinforcement (hoops/ties). The reason for the minimum reinforcement requirement is based on the latter, because, in general, a column is a compression member, for which the shear stress and buckling are the main concerns, which can be minimized/eliminated by providing the minimum amount of stress confining reinforcement - longitudinal bars and hoops (for circular columns).</p>
<p>If a column is mainly provided to resist tension, as tension ties/piles, the former reasons for requiring the minimum reinforcement are no longer exist, instead, you shall provide reinforcing with minimizing the potential cracks in mind, for which, there is no guidance exists, but to exercise by the engineering judgment at a case by case base - a single steel rod encased in the concrete protective layer is acceptable in the non-corrosive environment, but more bars are needed otherwise.</p>
| 51194 | what is the minimum number of reinforcement bars in a circular tie rod? |
2022-06-04T03:11:59.243 | <p>I am looking for a good working fluid whose boiling point is in the range of 150 - 180 °F for a waste heat recovery system. The fluid should be minimally flammable, preferably nontoxic, and low in environmental impact. Does anyone know of any working fluids that match this description?</p>
| |fluid-mechanics|fluid| | <p>Use water. It is cheap, plentiful, nontoxic and plentiful- plus it has an very high specific heat. All of these things make it ideal for heat transfer purposes. Just choose the <em>system operating pressure</em> to be enough <em>below</em> ambient to depress the boiling point to the range you desire.</p>
| 51204 | What is a good working fluid in the temperature range of 150 - 180 °F? |
2022-06-04T15:35:04.620 | <p>If two 24in x 6in x 1in thick plates of 6061-T6 aluminum are temporarily bonded together with <a href="https://www.3m.com/3M/en_US/p/d/b40069433/" rel="nofollow noreferrer">3M Super Trim Adhesive</a> (<a href="https://multimedia.3m.com/mws/mediawebserver?mwsId=SSSSSuUn_zu8lZNUM8_xnx2vov70k17zHvu9lxtD7SSSSSS--" rel="nofollow noreferrer">MSDS</a>), would this be a nightmare to separate?</p>
<p>The plate is rated for a max temperature of 572°F (300°C) so I was hoping that a heat gun and thin putty knife would make separating easy without affecting the temper much. But the adhesive information only states "heat resistant." Is this asking for trouble?</p>
<p>Edit: Why I don't want to use clamps - the sandwich is too large to be mill-drilled all at once, so it must be repositioned. The holes must be at (close as possible) right angles to each other, so the pieces <em>cannot shift</em> while moved and re-indexed.</p>
| |machining|adhesive|heat| | <p>Yes, it would be a nightmare to separate. That adhesive is designed for high strength, temperature resistant applications. That max temp is the point at which it looses it's rated strength. It may still be a ginormous pain to get apart even at significantly higher temperatures.</p>
<p>If a spray-on solution is a must, try a weaker option like 3M™ Repositionable 75 Spray Adhesive which is meant to be removed after application.</p>
<p>If a high-strength adhesive is essential, use hot-melt glue which you can melt to release after you're done machining or cyanoacrylate glue which you can remove with acetone (though it may take a while to soak).</p>
<p>My final thought for you is to tack in some bolts, rivets, or welds and then remove them when you're done.</p>
<p>Best of luck!</p>
| 51213 | Temporary spray adhesive to hold two plates together for drilling? |
2022-06-05T05:20:08.677 | <p>I need to design a long (6-7m) pole for plucking fruits from tall trees.</p>
<p>The following are the requirements:</p>
<ol>
<li>The pole material needs to be as light as possible.</li>
<li>The pole should not be flexible meaning it should not swing in air due to its own weight or the weight of the front attachment or the fruits.</li>
</ol>
<p>I need to know the following:</p>
<ol>
<li>What would be the ideal material for the pole?</li>
<li>What mechanical properties ( strength/ stiffness/ cross section etc.) affect the rigidity of the pole?</li>
</ol>
| |mechanical-engineering|structural-engineering|stresses|solid-mechanics|material-science| | <p>You will want a material with a high stiffness, and strength to weight ratio.</p>
<p>If cost is no object, then carbon fiber is probably the best option.</p>
<p>Fiberglass, titanium and aluminum are also good options, with aluminum and fiberglass being the most realistic if you're on a budget.</p>
<p>The next question will be how to structure the "pole" to increase the rigidity. You can look at cranes and other tall/stiff constructs for inspiration.</p>
<p><a href="https://i.stack.imgur.com/34MHW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/34MHW.png" alt="enter image description here" /></a></p>
<p>The simplest option that will still give decent results would be a tube. Using a larger diameter will increase the stiffness, but will increase the tendency to buckle (assuming a fixed qty of material for unit length).</p>
<p>You can decrease the droop by using a cable to support the picker end similar to the crane pictured.</p>
<p>The non-picking end of the pole will be subject to greater bending forces than the end. Therefore a tapered design would be optimal, if it's feasible.</p>
| 51223 | Ideal design for a long fruit plucking pole |
2022-06-05T12:50:49.827 | <p>The goal is to find the optimum values of the diameter and the thickness of the circular cross-section of a cantilevered beam in bending, with the lowest weight (cross-sectional area). The main criterion is to make sure that the internal stress does not exceed a limit anywhere in the cross-section. The length of the beam has already been chosen. The approach so far has been to find the internal loading of the beam from the external loads, then use the following formula. I am using the thin-walled assumption for the tubular cross-section.</p>
<p><span class="math-container">$\sigma = \frac{M_{y}}{I_{yy}} x$</span></p>
<p>However, it appears that I can increase the radius indefinitely, with the same area, and always achieve a better cross-section.</p>
<p><span class="math-container">$\sigma = \frac{M_{y}}{\pi R^{3} t} x = \frac{2 M_{y}}{(2 \pi R^{} t) (R^{2})} x = \frac{2 M_{y}}{(const) (R^{2})} x$</span></p>
<p>Is the limit solely based on manufacturability, or is there something else I should consider? Also, if it is based on manufacturability, could you give some advice on what the limits would be for composites or metals? I am currently considering either an aluminum alloy or carbon fiber. This is for a beam for an eVTOL vehicle that is connected to a propeller, so if you have comments relating to my approach let me know. I am currently considering only the vertical force from the propeller since I think it would be the critical load.</p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam|aerospace-engineering| | <p>The equation for bending stress is <span class="math-container">$\sigma_b = \dfrac{My}{I}$</span></p>
<p>For a <strong>thin-walled</strong> circular shaft, <span class="math-container">$I = \dfrac{\pi d^3t}{8}$</span>, <span class="math-container">$d = 2r$</span> (diameter)</p>
<p>Rewrite the equation, with <span class="math-container">$y = d/2$</span>:</p>
<p><span class="math-container">$\sigma_b = \dfrac{4M}{\pi d^2t} \le \sigma_{(a)llowable}$</span> for the composite or metals.</p>
<p><span class="math-container">$d^2t = \dfrac{4M}{\pi \sigma_a}$</span></p>
<p>In order to satisfy the least weight criteria, we need to find the dimensional parameters "<span class="math-container">$d$</span>" & "<span class="math-container">$t$</span>", and optimize the <span class="math-container">$d/t$</span> ratio.</p>
<ul>
<li>For stability and buckling concerns, the <span class="math-container">$d/t$</span> ratio shall be kept within <span class="math-container">$3300/f_y$</span>.</li>
</ul>
<p>Now, you can set "<span class="math-container">$t$</span>" as a function of "<span class="math-container">$d$</span>", and should be able to find the optimum section that satisfies all criteria.</p>
| 51231 | How to design the thickness and diameter of a cantilever beam in bending for both composites and metals? |
2022-06-05T17:05:26.087 | <p>If a car is moving with a constant velocity this means that resistive loads (like air drag friction) should equal the driving force from the car; but I know that the governors always cancel the effect of these resistive loads acting on the engine by supplying more fuel.So how will the cars move at constant speed if always any change in load due to resistive forces(like friction gravity ..etc) is cancelled from the governor by supplying more fuel?</p>
| |mechanical-engineering|automotive-engineering|torque|aerodynamics|car| | <p>Governers aren't really a thing on road-going vehicles. Governors are used on constant speed engines to maintain an engine speed. Single speed road vehicles aren't very useful.</p>
<p>A governor works by feeding more or less fuel as engine load changes. Lawn mowers typically have a simple governor that maintains engine speed as load (more or less grass) changes. Industrial diesel engines may have a governor to maintian speed for a generator, air compressor, etc. Some governors can be adjusted and some will always have a set speed.</p>
<p>Governers do not always supply "more fuel." They supply enough fuel to maintain the desired engine speed.</p>
<p>I leave out the cateory of Cruise Controls, which I consider just an automated version of a foot on an accelerator pedal.</p>
| 51233 | Governors and acceleration effect on IC engine |
2022-06-06T08:13:13.927 | <p>Our kitchen electric oven has a large glass door approx 530 x 440mm, overall thickness 35mm.</p>
<p>There is a front and back glass pane, no insulation and the air cavity is open at the bottom.</p>
<p>I'm wondering if it would be cost effective to insulate the door by filling the cavity with a suitable high temperature insulating material?</p>
<p>Electricity will soon be nearly 0.50gbp per kWh round here, hence my thoughts.</p>
<p>On the other hand, perhaps the warm dry air in the door cavity is already a reasonable insulator? Certainly you can touch the front of the glass without danger of being burnt.</p>
| |heat-transfer|thermal-insulation| | <p>If the air cavity is so thin as to prevent convection, it will be almost as good an insulator as a vacuum.</p>
<p>For this reason the air cavity in insulating double-pane glass is normally made so thin as to prevent convection.</p>
<p>It's not perfect, but you'd probably get more effect by gluing reflective foil to the inside surface, to prevent radiation loss.</p>
| 51244 | Heat Loss from Domestic Oven Door |
2022-06-08T15:14:44.267 | <p>I have a slide here which says following:
For the control loop</p>
<p><a href="https://i.stack.imgur.com/MdvqO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MdvqO.png" alt="Control loop" /></a></p>
<p>...apply <span class="math-container">$d$</span>, measure <span class="math-container">$u$</span> and <span class="math-container">$y$</span>, calculate</p>
<p><span class="math-container">$\hat{H}(f) = \frac{S_{yu}(f)}{S_{uu}(f)}$</span></p>
<p>but then typically</p>
<p><span class="math-container">$E\{\hat{H}(f)\} \ne H(f)$</span></p>
<p><strong>Can someone explain to me the meaning of the last equation?</strong> Is it correct to read this line as: "The expected value of <span class="math-container">$\hat{H}(f)$</span> is not equal to <span class="math-container">$H(f)$</span>", and if yes, why is it like this?</p>
| |control-engineering|system-identification| | <p>In the context depicted in the diagram, the transfer function estimate can be shown to be biased (wrong in expectation) due to the feedback loop. The input will, due to the feedback, be correlated with the noise <code>n</code> that appears in the output, causing the bias.</p>
<p>See, e.g., "System modeling and identification" by Rolf Johansson, chapter 4, for more details.</p>
<p>In fact, several classical identification methods are biased when operating on closed-loop data. This includes several common subspace-based identification methods. Notably, the prediction-error method (PEM) is unbiased also for closed-loop data due to it explicitly taking causality into account (subspace id typically does not).</p>
| 51287 | Measure a frequency response function |
2022-06-08T19:10:23.160 | <p>This has bothered me for many years, but I'm only now coming around to ask about it.</p>
<p>I don't know if this is unique to Sweden or something, but for whatever reason, when I try to plug things into the standard power strips (see photo), they physically won't go in. If I try another socket on the same strip, it might fit there.</p>
<p>They look visually identical! One will go in and the next won't. For example the standard power connectors for Sega Saturn and PlayStation. Or SNES. Or any console. Or anything. It seems completely random.</p>
<p>I've carefully examined the plugs. They look exactly identical. Nothing special about them that would "block" them. Yet they physically cannot be pushed down, in spite of no plastic parts being in the way, and the metal pins are perfectly aligned to the holes.</p>
<p>Yes, I have tried both sides. Many, many times.</p>
<p>I'm starting to wonder if this is some sort of security mechanism and that the power strip has some sort of mechanical "blocker" when it detects that too many devices are connected or something. But how could that possibly work? There are multiple unused sockets in my power strip right now, and I just cannot connect my PlayStation/Saturn cable into those. I have to unplug an already attached one and use its place instead. Even though they look exactly identical.</p>
<p>I cannot stress this enough: both the plugs and the connectors on the power strip look exactly identical. No special colour coding, or different shape, or extra/lacking pins. Nothing like that whatsoever. They physically won't go into some of the sockets on the power strip.</p>
<p>It would mean a ton for me to finally get this maddening fact of life straightened out. Sometimes I feel like I've gone completely mental and am hallucinating.</p>
<p><a href="https://i.stack.imgur.com/xKjAG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xKjAG.jpg" alt="enter image description here" /></a></p>
| |electrical-engineering|power|power-electronics|electrical|electricity| | <p>Apparently the type of power strip your picture shows is one with something like a child protection. I.e. the green parts (kind of a slider) visible in the picture.</p>
<p>Those are mechanical devices which will block anything entering the holes except they are pushed aside really simultaneously.</p>
<p>Now, there's something happening which prevents certain plugs from being inserted in certain sockets. There are multiple reasons and some might exacerbate the problem when coming together.</p>
<ol>
<li><p>Wear out. The mechanics behind this child protectors can wear out. So they loose precision, get out of balance or show increased friction.</p>
</li>
<li><p>The plugs you want to insert might be bent a little bit. This could result in slightly assymetric buildup of force on the two plastic sliders behind the holes.</p>
</li>
<li><p>The prongs you want to insert might be scratched at the tip increasing friction between the prong and the plastic sliders.</p>
</li>
<li><p>There are mainly two types of plugs in Europe. Plugs with only two contacts and with no protective earth have pins with a smaller diameter. That might cause the pins pressing uneven and off-center in the holes on the sliders.</p>
</li>
</ol>
<p>Some of these four causes might work together that enough asymmetry is gained to block the sliders from giving way.</p>
| 51289 | Why do identical-looking power plugs randomly not fit in some sockets in the same power strip? |
2022-06-09T00:16:43.420 | <p>I apologize if the question is very basic, I admit I'm not the best at math and physics.</p>
<p>I am doing an experiment with a sphere, the measurements are as follows:</p>
<ul>
<li>The sphere has a diameter of 330 mm (13 inches)</li>
<li>D1 measures 75mm</li>
</ul>
<p>I want D2 to be attached to D1 and both ends of D2 to be attached to the sphere. D1 must be attached to one end of the sphere. <strong>I want to know how long D2 should be.</strong> How can I do that calculation?</p>
<p><a href="https://i.stack.imgur.com/XQia3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XQia3.jpg" alt="enter image description here" /></a></p>
<p>Update:
<a href="https://i.stack.imgur.com/bd2H6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bd2H6.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/UwVKT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UwVKT.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|mathematics|experimental-physics| | <p>First, consider a circle with same diameter as the sphere, with the distance <span class="math-container">$d_2$</span> being the projection of <span class="math-container">$D_2$</span> onto the plane and assuming <span class="math-container">$d_1=D_1$</span> relative to your figure:<a href="https://i.stack.imgur.com/Nlkee.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nlkee.png" alt="enter image description here" /></a>
The radius of the circle is equal to the radius of the sphere, i.e. <span class="math-container">$r=165$</span>mm.<br />
We get a planar measure for <span class="math-container">$d_2$</span> that is
<span class="math-container">$$d_2=2\sqrt{r^2-(r-d_1)^2}$$</span>
To get the length <span class="math-container">$D_2$</span> in 3D, we multiply <span class="math-container">$d_2$</span> by <span class="math-container">$\pi/2$</span>
<span class="math-container">$$D_2=\pi\sqrt{r^2-(r-d_1)^2}$$</span></p>
| 51292 | Calculations of a sphere |
2022-06-09T08:38:14.483 | <p>I have a chain drive which is moving a load of 200kg. I have 95Nm on the input shaft which has 15 teeth, which results in 152Nm on the output with 24 teeth.</p>
<p>The shafts are supported on bearings on both sides and the sprockets sit in the middle. I'm trying to calculate the lateral forces that will be generated on the bearings and I'm not really sure how to do that.</p>
<p>Taking for example the output shaft, its sprocket is 0.095m in diameter.</p>
<p><span class="math-container">$$152/0.0475=3200N$$</span></p>
<p>That would mean there is about 300kg of force acting on the sprocket's teeth. This is already a little weird to me as I'd imagine forces wouldn't exceed the total weight of the load, but okay.</p>
<p>I'm having trouble figuring out how much of that will get transferred to the bearings in lateral forces?</p>
| |statics|chain|forces| | <p>Depending on where the torque was measured, the force "<span class="math-container">$F$</span>" equals the torque divided by the "diameter" of the sprocket.</p>
<p><a href="https://i.stack.imgur.com/mVVQR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mVVQR.png" alt="enter image description here" /></a></p>
| 51299 | How to calculate forces on bearings in a chain drive |
2022-06-09T11:48:35.383 | <p>I'm trying to cast an impeller but I have a problem. The melted aluminium doesn't go inside the thin parts of the mould. Here is the mould before casting:</p>
<p><a href="https://i.stack.imgur.com/6X3Lu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6X3Lu.jpg" alt="enter image description here" /></a></p>
<p>After the casting:</p>
<p><a href="https://i.stack.imgur.com/g5uCw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g5uCw.png" alt="enter image description here" /></a></p>
<p>After breaking the mould:</p>
<p><a href="https://i.stack.imgur.com/rSbfp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rSbfp.png" alt="enter image description here" /></a></p>
<p>As you can see the vanes of the impeller are not produced. It should be something like this:</p>
<p><a href="https://i.stack.imgur.com/nuiA0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nuiA0.jpg" alt="enter image description here" /></a></p>
<p>What's the problem? How to figure it out?</p>
| |casting| | <p>Your objective seems to be a cast aluminum turbocompressor wheel. The techniques for casting such thin sections are highly specialized- you just can't achieve the desired result when gravity-casting aluminum in a fused sand mold in your back yard.</p>
<p>There's also a number of good reasons why turbocompressor wheels are not made of cast aluminum in the first place: as-cast aluminum is simply not strong enough to stand up to the stresses in that application. Superalloys high in chrome, nickel, etc. are required.</p>
| 51301 | Problem in casting with aluminium? (misrun?) |
2022-06-09T11:53:34.523 | <blockquote>
<p>I want to preface this by saying I am software engineer so each part of this robot build has been a lesson on its own. From welding, to CAD design, pillow block bearings, etc, etc....</p>
</blockquote>
<p>I am working on a DIY Rover using the following components:</p>
<ul>
<li><a href="https://rads.stackoverflow.com/amzn/click/com/B07KVVL1NM" rel="nofollow noreferrer" rel="nofollow noreferrer">4 DC Motors</a> from electric scooters: 24V, 250W. Connected to axle with #40 roller chains and sprockets.</li>
<li><a href="https://lawnmowertirestore.com/wheels/lawn-mower-wheels/16x6-50-8-turf-tire-and-rim-for-lawn-and-garden-mower-75-bearing-3.html" rel="nofollow noreferrer">4 Lawn Mower Wheels</a>. They are 16x6.50-8 for a 3/4" axle with a keyway</li>
<li><a href="https://www.ezmobilitybattery.com/mk-battery-m34-sld-g-m34sldg-gel.html" rel="nofollow noreferrer">A Pair of 12 Volt - 60Ah GEL Batteries</a> : I pulled them from a discarded electric wheel chair. They are fully charged, and they show <strong>26V</strong> when connected in series</li>
<li>I replaced the stock sprockets on the motors with <a href="https://rads.stackoverflow.com/amzn/click/com/B00VI4S1LC" rel="nofollow noreferrer" rel="nofollow noreferrer">these</a> #40 chain 12 teeth, 1/2" pitch.</li>
<li>I put the same sprockets on the axles</li>
</ul>
<p><a href="https://i.stack.imgur.com/XdiuO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XdiuO.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/kAMyM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kAMyM.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/vn42I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vn42I.jpg" alt="enter image description here" /></a></p>
<p>The problem is that the robot does not begin to move. When I lift the wheel off the ground, they move fine, and spin fast depending on how much throttle I give them.</p>
<p>However on the ground they don't move, which is typically a <strong>torque problem</strong>.</p>
<p>However I am not so convinced because these motors are supposed to be for electric scooters and each one capable of moving an adult person, weighing 175+ lbs.</p>
<p>Here are my diagnostic steps:</p>
<ol>
<li>Using a FLYSKY controller and <a href="https://www.dimensionengineering.com/products/sabertooth2x12rc" rel="nofollow noreferrer">12x2 Sabertooth</a> motor controller. Unfortunately this controller has an over current protection, and those batteries are massive, so it just shuts itself down whenever I move throttle just a little bit. But again, when the wheels are off the ground they are spinning fine, so the connections are fine.</li>
<li>To avoid the over-current protection problem, I purchased this <a href="https://rads.stackoverflow.com/amzn/click/com/B089273KVZ" rel="nofollow noreferrer" rel="nofollow noreferrer">60Amp DC Motor Controller</a>. I connected it to just one wheel so far, and spliced a Multimeter into the connection to measure voltage. I can see the motor will get up to 20V when I turn the Potentiometer dial, and I can see the car begin to try to move.</li>
</ol>
<p></p>
<p>Another question is: what's a good way to measure how many Amps the motor is trying to pull?</p>
<p></p>
<p><strong>My conclusions are:</strong></p>
<ul>
<li>The chains are just too heavy for this motor. It's a #40 chain which is way overkill. The sprocket on the motor shaft and the axle are also too heavy</li>
<li>The tires maybe too heavy. But again, it's 4 x 250W motor that's getting 24V and probably up to 10Amps</li>
</ul>
<p><strong>What is your advice?</strong></p>
<p>What would you do to fix this problem with minimal changes to the design? I.e. just replacing the sprockets? Or are the tires also too heavy?</p>
<p>How would I keep the tires and motors as is? Replace the sprockets and chains? Figure out a way to give a higher initial Amperage?</p>
| |mechanical-engineering|motors|torque|power|robotics| | <p>The motors have a rated speed of 2750rpm. It doesn't look like there's much of a gear ratio, so at the moment the motors spin at the same speed as the wheels. With 16" wheels, 2750rpm is about 130mph (rpm*diameter*pi, then converting inches/minute to mph). I'm guessing you don't want it to go that fast, and that speed wouldn't be possible with a total power of 1kw, or 1.3hp.</p>
<p>So you want to gear the motors down a lot, which increases the wheel torque. If you want a design speed of say 20mph, that's a gear ratio of <span class="math-container">$\frac{130}{20}=6.5$</span>, so the wheel sprocket should have 6.5 times as many teeth as the motor sprocket and be 6.5 times the diameter.</p>
<p>That will also give you 6.5 times as much torque at the wheels</p>
| 51302 | Torque Problems vs Motor Problem? |
2022-06-09T13:45:26.037 | <p>Sites like the Lockheed Helendale RCS or the Tonopah test range are only known to have an <a href="https://www.thedrive.com/the-war-zone/15746/lockheeds-helendale-radar-signature-test-range-looks-right-out-of-science-fiction" rel="nofollow noreferrer">underground pit</a> to support raising or lowering of test articles. But is it technically possible to have a fully underground test range?</p>
<p>For example the Helendale RCS Range has a "runway" like structure sitting in between the test chamber and the antenna array. For an underground facility I propose the runway like path would be enclosed in a horizontal TBM-bored or cut/cover tunnel rather than digging up an entire area. Also noting the pit one and two in the center of the "runway" if the runway has to be fit into an underground tunnel. What are some of the design considerations of the tunnel that would minimize EM interference as opposed to having the entire structure aboveground?</p>
<p>Then there is also the question of how to get a test article, potentially the size of an aircraft to the underground location. But that's a question for later</p>
<p>The Nevada test side has alot of potential to reuse the subterranean structures as test ranges.</p>
<p><a href="https://i.stack.imgur.com/WdaQv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WdaQv.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|electromagnetism|tunnels|radar| | <p>It would probably be designed in the same manner as an EMC test chamber, except WAY bigger.</p>
<p><a href="https://i.stack.imgur.com/T3YDb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T3YDb.png" alt="emc test chamber" /></a></p>
<p>That being said, even with very sophisticated radar absorbing walls I'm not sure you could reduce the reflections enough to not drown out something like a stealth aircraft.</p>
| 51306 | What are design considerations for an RCS test range be relocated fully underground? |
2022-06-09T18:06:01.927 | <p>I have an old laptop which I revived with the addition of a SSD. The problem now with this laptop is that it gets very hot causing thermal shutdowns in the middle of work. I thought of a lot of solutions and finally repasted the heat sink, bought a cooling pad and removed the bottom cover. It is now running fine. Now that I am free with my projects, I am thinking of converting this laptop to a tablet pc. Now my question is that "Can I weld a copper pipe directly to the existing one?"
Will welding it increase the thermal capacity of the PC? Or will there be issues due to other scientific reasons?</p>
| |cooling|computer-engineering|computer-hardware|heat|heatsink| | <p>I think, it is necessary to find a cause of overheat. This may be due to bad thermal contact somewhere:</p>
<ul>
<li>between integrated circuit and metal pad through thermal paste or pad</li>
<li>between metal pad and heat pipe</li>
<li>between heat pipe and heat exchange plates</li>
</ul>
<p>Heat exchange plates could be blocked by dust/dirt.</p>
<p>Heat pipes may be damaged and should be replaced. Typically, all heatsink system of notebook is replaced.</p>
<p>Fan may be not creating enough air flow due to thickened old grease or contamination.</p>
<p>May be there is any software problem which causes CPU overloading.</p>
<p>Continuous and cyclic IC overheat leads to breakage of BGA solder balls which connect IC to board and irreparable breakage of flip-chip joint between IC crystal and flip-chip substrate, so, overheat problem requires solution.</p>
<p>Simple copper pipe has much lower heat conductivity than thermal pipe, except if it is filled with circulating liquid.</p>
| 51308 | Can I weld a copper pipe into a CPU heat sink for better cooling? |
2022-06-09T23:15:33.560 | <p>I don't know much about levers and hydrostatic/hydraulics myself, so I hope I don't make too many misconceptions.</p>
<hr />
<h2><strong>The idea:</strong></h2>
<p><strong>Hypothetically</strong>, if you had a 40km+ long indestructible lever on a indestructible fulcrum, you could lift the Eiffel Tower with only your body weight.</p>
<p>With this in mind, the idea is to use these same principles (hydraulics/hydrostatics and levers) in order to <strong>increase the distribution of force without needing a 40km lever</strong> so an average human being could power this system with the strength of their arms and/or legs.</p>
<p>Yes, I would be "increasing" the "length" where the load is, but I would also increase the "length" where the force is being applied, increasing the distance in which the first moves. It makes me wonder if it would make sense at all, because even though you increase the length of the lever, it still has the same forces in the "same" proportions.</p>
<h2><strong>The simplified system would be like this:</strong></h2>
<ul>
<li>First, a class-1 lever where the load would be a
hydraulic/hydrostatic cylinder, and the force would be the hand/legs
of a person.</li>
<li>Second, I don't know much about these subjects myself: Option 1 would
to make a larger and shorter piston at the end of the lever that
would impel a longer but thinner cylinder.</li>
<li>At the end of this thinner cylinder would be the final load, which,
if wasn't a simple system laying on the ground, would be the weight
of the load, which I would put at worst at
300 kg in total.</li>
</ul>
<hr />
<h2>How much bigger this system and/or the parts of the system would be required to be in order to lift something from 200-300 kg in weight with only the human force of a arm/leg?</h2>
<hr />
| |mechanical-engineering|fluid-mechanics|statics|mechanisms|hydrostatics| | <p>Levers and hydraulic cylinders are not typically used for the distribution of force, but are more frequently used to transmit and otherwise modify applied force.</p>
<p>Per your reference to <a href="https://www.softschools.com/examples/simple_machines/class_one_lever_examples/511/" rel="nofollow noreferrer">class one levers</a>, the fulcrum is located between the load and the force. This also (and obviously) results in a change in direction of the force with respect to the movement of the load. What cannot be overlooked is the mechanical advantage of a level with the fulcrum not located precisely in the center of the distance between the load and the applied force. Image from linked site.</p>
<p><a href="https://i.stack.imgur.com/GJ53n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GJ53n.png" alt="class one lever image" /></a></p>
<p>If the load is located one-third of the total way from the fulcrum, the applied force will be doubled at the load, but the travel of the load will be one-half of the applied force travel. This aspect doesn't appear to be a consideration in your question.</p>
<p>In the world of hydraulics, a similar ratio exists, but is applied to the surface area of the piston to which the force is applied in relation to the piston surface area of the loaded piston.</p>
<p>The <a href="https://www.engineeringtoolbox.com/hydraulic-force-calculator-d_1369.html" rel="nofollow noreferrer">math is a bit more complex</a> than the simple class one lever, however and is left as an exercise to the reader. The image is from the linked site.</p>
<p><a href="https://i.stack.imgur.com/Hf48J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hf48J.png" alt="hydraulic force image" /></a></p>
<p>In simplistic terms, a small diameter force cylinder can create a "force multiplier" on the load cylinder if the load cylinder is a larger diameter. As before, there is a compensation applied in that the travel of the smaller cylinder is much greater than that of the larger cylinder. As it can be impractical, for example, to have a 30 meter long small diameter force cylinder, a series of valves allows repeated / multiple strokes of a class one lever (!) to apply hydraulic fluid to the system, moving the load cylinder appropriately.</p>
<p>In the question, the suggestion to use a larger (diameter) cylinder at the force end is the reverse of the desired objective.</p>
<p>Consider a <a href="https://www.researchgate.net/figure/Principle-of-hydraulic-jack_fig1_309235120" rel="nofollow noreferrer">hydraulic bottle jack</a>. Image from linked site.</p>
<p><a href="https://i.stack.imgur.com/iSulV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iSulV.png" alt="hydraulic bottle jack image" /></a></p>
<p>In the above image, the handle is not a class one lever, but it could be if desired or required. For practical purposes, the displayed handle accomplishes the necessary movement of the force cylinder, regardless of lever class.</p>
<p>The load cylinder is the piston of the jack and presents a cross section of greater diameter than the force cylinder. Missing from the image is the necessary return oil flow valve and passages for the jack to function in the real world.</p>
<p>Hydraulic floor jacks can be rated to lift as much as 10,000 pounds / 4500 kilograms using a reasonable length lever as shown in the above image operated by an ordinary human being. The key factor in this consideration is that many pumps of the lever is required to move that weight an appreciable distance.</p>
<p>If a specific and restricted objective is created (distance of travel, mass of load) the parameters of the lever length, force piston diameter and travel can be adjusted appropriately.</p>
<p>In relation to the comment added to my post, a hydraulic pump in an excavator or piece of similar heavy equipment is effectively a quantity of cylinders or means of imparting force to hydraulic fluid. I have seen swashplate pumps with sixteen very small cylinders. One rotation pumps quite a bit of fluid to the destination cylinder/s but also requires substantial energy, typically from an electric motor or petrol engine.</p>
| 51310 | Could a hybrid system of mechanical levers and hydrostatic/hydraulic cylinders be able to lift 200+ kg with the force of an arm? |
2022-06-10T16:15:59.750 | <p>I have tons of questions in regard to what is, and isn't, in accordance with ISO standards when drawing P&IDs. So, I'm looking for a resource which gives a complete overview of the standards.</p>
<hr />
<p><strong>Meta:</strong></p>
<p>I realize this question may veer towards being off-topic, but user <em>RickSupportsMonica</em>'s answer to <a href="https://engineering.meta.stackexchange.com/questions/10/the-engineering-se-position-on-recommendation-finding-stuff-questions?newreg=7c471965a578494fb830845485bedb53">this meta question</a> argues questions asking for standards ought to be allowed. I agree. This isn't me asking for your opinion about what's the best resource. I'm asking for a resource that satisfies two objective, decidable criteria:</p>
<ol>
<li>The resource contains ISO standards.</li>
<li>The resource is complete (that is, all standards are included).</li>
</ol>
<p>Furthermore, the issue of ephemerality is removed if the resource is a reguarly updated site. And, it definitely isn't something that is only going to help a small number of people, given this is a basic thing that helps anyone drawing P&IDs in accordance with ISO standards. So, the only issue left is the fact that this is answerable by a simple link, though to me, that isn't really a problem. I understand if the majority of users here find it a problem though, and thus decide to close my question.</p>
| |standards|pi-diagram|international| | <p>Well basically the text of the standard is available at:</p>
<p>iso.org</p>
<p>Though ISO has a number of local distributors, that may be more convenient to deal with or even get free acces through university libraries or physically at their library. So for example I use SFS online because thats the distributor of my local standards as well as ISO standards and my organisation has a contract with them.</p>
| 51318 | Where to find a complete overview for ISO P&ID standards? |
2022-06-10T17:28:45.397 | <p>For what I think I know, the mechanical advantage of a lever comes in exchange for the distance the load travels upwards, independent from the type of lever (I think).</p>
<p><a href="https://i.stack.imgur.com/5HXPb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5HXPb.png" alt="enter image description here" /></a></p>
<p>So, let's say, in a (class-1) lever with a long area where the load is applied, but the effort is also applied in an even longer area. Is there a way of "shortening" it while maintaining the mechanical advantage?</p>
<p>I thought of putting a lot of small levers in sequence (with different lengths to compensate for the increase of force), one applying effort until the total of levers would make the same amount of force as the long single lever.</p>
<p><a href="https://i.stack.imgur.com/6ISJC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ISJC.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/RRCDj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RRCDj.png" alt="enter image description here" /></a></p>
<p>Let's say, someone then would organise them in a spiral, so they occupy less space.</p>
<p>Would that even be possible, or it is just a nonsensical question?</p>
<hr />
<p>By the way, the intention with this question is to lift around 200 kg with the force of a leg.</p>
| |mechanical-engineering|statics| | <p>Although you need to move the ends of the lever through the correct distances to get the correct advantage -- work equals force times distance -- you don't need to have the same linear distance to do so. A worm screw does effectively the same, without the long levers.</p>
<p>A simple example is a double lever system where the second lever points back towards you. Two 2:1 levers give you a 4:1 advantage, but the load point is right next to your hand, and the length of the lever system is only 2+1=3, not 4+1=5.</p>
| 51320 | Is there a way of "shortening" a mechanical lever, but keeping its mechanical advantage intact someway? |
2022-06-10T18:43:31.667 | <p>Let's say you have two pumps, one always running and the other in reserve. Maybe they're switched regularly, or maybe the other is only used when the first is broken/down for maintenance.</p>
<p>In a P&ID, drawn after ISO standards, will the reserve equiptment and/or the label of the reserve, equiptment be shown? I see a few different possibilities:</p>
<ol>
<li>Only one pump is shown, perhaps labelled with the label of the pump that was first installed. Nothing else.</li>
<li>Only one pump is shown, but it is labeled "Pump x/x + 1/x + 2/..."</li>
<li>All pumps are shown, with some kind of symbolization shown that the other pump(s) are not operating when one pump is operating. In the case of the reserve pumps only being used when the primary pump is down, this can easily be shown with normally-closed valves on the process lines going from the reserve pumps. In the case of periodic switching however, where there is no main pump, I have no idea how this is to be shown.</li>
</ol>
| |standards|pi-diagram|international| | <p>From <a href="http://www.driso.ir/standards/iso/ISO_10628_1_2014_,_Diagrams_for.pdf" rel="nofollow noreferrer">ISO 10628-1 Diagrams for the chemical and petrochemical industry —
Part 1: Specification of diagrams</a></p>
<blockquote>
<p>4.4 Piping and instrumentation diagrams (P&ID)</p>
</blockquote>
<blockquote>
<p>All equipment, valves, and fittings shall be represented in accordance with ISO 10628-2.</p>
</blockquote>
<blockquote>
<p>4.4.2 Basic information</p>
</blockquote>
<blockquote>
<p>The piping and instrumentation diagram shall contain at least the following information:</p>
</blockquote>
<blockquote>
<p>a) function and type of equipment and machinery, including drives, conveyors, and <strong>installed back-up/reserve equipment</strong>;</p>
</blockquote>
<p>Essentially, P&ID are the most detailed. That document is a good read because it talks about different types of diagrams AND has examples.</p>
| 51322 | ISO standards for drawing reserve equiptment in P&IDs |
2022-06-12T09:09:09.177 | <p>I am building a 4 wheel drive rover robot. I was very lucky to find four good condition wheel chair motors on the side of the road. I had to disassemble the wheel chairs of course but well worth the effort.</p>
<p><a href="https://i.stack.imgur.com/6XiYR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6XiYR.png" alt="enter image description here" /></a></p>
<p>I also have 4 John Deere riding mower tires similar to these:</p>
<p><a href="https://i.stack.imgur.com/IE4WW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IE4WW.png" alt="enter image description here" /></a></p>
<p>they are the same size however the ones I have already have a keyway in them. You'll notice of course that they have no mounting holes in their hubs.</p>
<p>My solution to this problem is as follows:</p>
<ol>
<li>Mount a pillow block bearing on the outside of the body of the robot</li>
<li>Mount another pillow block bearing on the inside, but facing backwards.</li>
<li>Bolt a piece of steel that is the same size as the inside pillow block frame</li>
<li>Bolt the motor's default mounting to the piece of steel.</li>
</ol>
<p>Another option of course is drilling holes into the hub of the tires I have, but I am concerned about ruining the tires while doing so</p>
<p>These are the solutions based on the tools and skills I have. I don't know how to create a CAD drawing and have it 3D printed, or CNC machined online.</p>
<p>The upfront time investment in learning CAD just to design this piece makes it less practical than using the tooling and experience I already have.</p>
<p>My questions are: what are you thoughts on this idea? How would you go about connecting these two for a rover robot.</p>
| |motors|robotics|wheels| | <p>Assuming I'm understanding your solution correctly that sounds decent, and it may be necessary if the wheelchair motors don't have bearings large enough to take the full load of the wheel. You will probably want a spider coupler, or a belt between the wheel shaft and the wheelchair motor to handle any misalignment. Soft mounting the motors might also work.</p>
<p>If the wheelchair motors DO have strong enough bearings, you could make an aluminum adapter that attaches the motor flange to the hub. I would turn it on a lathe or 3D print it (you're probably going to want to learn it sometime). Basically make one side of it conform to the hub shape, and attach it with bolts, (or 1 large bolt through the shaft hole). The other face would have a hole pattern that matches the motor.</p>
| 51331 | Connecting a brushed direct drive motor to a wheel with no holes in the hub |
2022-06-12T20:16:14.983 | <p>Based on this feedback control loop
<a href="https://i.stack.imgur.com/rjY3r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rjY3r.png" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$w$</span> is the desired setpoint, <span class="math-container">$x$</span> the process variable, <span class="math-container">$e$</span> the error value, <span class="math-container">$y$</span> the correcting value and <span class="math-container">$z$</span> the disturbance value.</p>
<p>Did I interpret the components correctly for this context:</p>
<p>A robot has a ultrasonic sensor which measures the distance to the nearest object in centimetres. The robot has two motors, a motor for the left wheels and a motor for the right wheels. The robot should hold 50 centimetres distance to the nearest found object. In order to adjust the current position of the robot, the motor speed for both motors are calculated as follows <span class="math-container">$(currentDistance - 50) * 2$</span>.</p>
<p>In this context, <span class="math-container">$w$</span> is 50, the distance which should be kept to the nearest found object, <span class="math-container">$x$</span> the current measured distance and <span class="math-container">$e=x-w$</span> the error. The controller is a P-controller which multiplies the error by a constant 2.
The output of the controller is therefore <span class="math-container">$y=e*2$</span> which is used to set the new speed of the motors (plant). If this is not correct, please correct me.</p>
<p>The question I have is, what could be the disturbance <span class="math-container">$z$</span> in this context. Is there actual any disturbance here? Could I just say the disturbance is 0?</p>
| |control-engineering|pid-control|feedback-loop| | <p><span class="math-container">$z$</span> could be a change in wind direction, a change in the supply voltage (requiring adjustment of the PWM to the motors to compensate), a change in the slope of the surface that the objects are on due to movement, etc. It's some external interference with the plant.</p>
| 51334 | How to determine the components of a feedback control loop given following context? |
2022-06-13T12:41:38.113 | <p><a href="https://i.stack.imgur.com/AYl0a.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/AYl0a.jpg" alt="enter image description here" /></a></p>
<p>Why is a bridge designed like this?</p>
<p>The depth of the section at pillars is more than the depth at middle.</p>
<p>If I model this as a simply supported beam having load at mid span then the bending moment will be maximized at the middle and the area is also less at the middle. So, this will lead to higher bending stress.</p>
<p>So, why is it designed like that?</p>
| |mechanical-engineering|structural-engineering|civil-engineering|beam|bridges| | <p>The bridge in the photo appears to be a post-tensioned concrete box girder balanced - cantilever. This is a 'continuous' structural form, meaning the spans are not simply supported, but are continuous over the tops of the piers. As pointed out in the digrams in one of the other answers, this will lead to large 'hogging' type moments (of opposite sign to the 'sagging' moments at mid-spans) over the piers.</p>
<p>Depending on how the cantilevers are balanced, the bending moments under dead loads may be designed to be effectively zero at the midspans (but not always designed this way). Under different live load conditions both hogging and sagging type moments may be experienced at the midspan, again depending on the design.</p>
<p>The bridge spans may have a shallow arched shape, but it is certainly not an arch in terms of structural action and there is effectively no 'arching action' - contrary to what some of the other answers have implied. To model or analyse it as an arch would be seriously incorrect.</p>
<p>To develop arching action the supports would require exceptional rigidity against longitudinal movement (even more so in this case since the 'arch' is very shallow). As it is, the leaf piers are relatively flexible and do not have anything like the necessary rigidity to develop arching action. Additionally, box cantilever decks of this form are usually supported on guided bearings at all but one of the piers/abutments. These bearings permit free longitudinal movement to release thermal and creep (and arching) effects.</p>
| 51347 | Why is this bridge thickest above the support pillars instead of the mid-span where the bending stress is highest? |
2022-06-13T16:54:39.540 | <p>I'm reading about buckling of timber members from Eurocode 5: Design of timber structures part 1.</p>
<p>Section 6.3.3 gives instructions to check the stability of beams subject to compression, bending or combination of both.</p>
<p>Formula 6.30 gives the relative slenderness of a member:</p>
<p><span class="math-container">$$\lambda_{rel.m} = \sqrt{\frac{f_{m,k}}{\sigma_{m,crit}}}$$</span></p>
<blockquote>
<p>where <span class="math-container">$\sigma_{m,crit}$</span> is the critical bending stress calculated according to the classical theory of stability, using 5-percentile stiffness values.</p>
</blockquote>
<p><span class="math-container">$f_{m,k}$</span> is the characteristic bending strength of the timber.</p>
<p>Formula 6.31 gives the equation for critical bending moment for lateral torsional buckling:</p>
<p><span class="math-container">$$\sigma_{m,crit}=\frac{\pi \sqrt{E_{0.05}IG_{0.05}I_t}}{WL_{eff}}$$</span></p>
<p>Formula 6.32 gives the critical bending stress for solid rectangular cross-section:</p>
<p><span class="math-container">$$\sigma_{m,crit}=\frac{0,78b^2}{hL_{ef}}E_{0,05}$$</span></p>
<p>I'm curious about this difference between formulae 6.31 and 6.32. The general formula 6.31 is the usual formula for lateral torsional buckling that I'm familiar with. For some reason, a different formula for rectangular cross sections are given. One thing that strikes me is that this formula does not include the shear modulus <span class="math-container">$G$</span>, or the torsional stiffness <span class="math-container">$I_t$</span>.</p>
<p>Why isn't the lateral torsional critical moment of rectangular cross sections dependent on shear modulus <span class="math-container">$G$</span> or torsional stiffness? Are the two formulas even equivalent (does the specific formula for rectangular sections follow from the general one?)</p>
| |structural-engineering|buckling| | <p>Bit late to the party, but there's an explanation to be found in the book 'Structural Timber Design To EC5' section 4.5.1.2:
<a href="https://i.stack.imgur.com/fJGD0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fJGD0.png" alt="enter image description here" /></a></p>
<p>Seems like the removal of the torsional rigidity term is an engineering decision rather than a mathematical one. Equation 4.7(a) in the image will simplify to the EC5 softwood equation if you use the approximation G = E/16.</p>
| 51350 | Why isn't lateral torsional buckling of rectangular timber members dependent on torsional stiffness? |
2022-06-14T07:37:53.730 | <p>Scenario:
A robot has a ultrasonic sensor which measures the distance to the nearest object in centimetres. The robot has two motors, a motor for the left wheels and a motor for the right wheels. The robot should hold 50 centimetres distance to the nearest found object. Therefore the motors have to change the speed of the wheels. The minimum speed is -100 which means that the wheels move backward with maximum speed. The maximum speed is 100 which means the wheels move forward with maximum speed. It can be that the object changes position abruptly for example if the object is a hand.</p>
<p>Which controller would you choose? P,I or PI? I would say P because if the position changes abruptly the robot can react instantly whereas using a PI or just I controller could lead to a collision if the integral is so big that if the object changes position abruptly the integral could still be big enough to move toward the object and may collide. The PI controller could reach the distance faster than the P controller but in this context the PI controller would not be suitable because of the disturbance (moving object) or?
Which controller would you use here and why?</p>
| |control-engineering|control-theory|pid-control| | <p>It all depends on what you want.</p>
<p>The easiest is a pure proportional (P) controller. It will give you an output that is proportional to the difference from the target value.</p>
<p>Pro: It's pretty stable (as long as you're not going too high with the gain)</p>
<p>Con: There will always be an difference left because the closer you get to your target the smaller the control signal gets. If your target is moving you'll have a difference which is proportional to the speed.</p>
<p>To get around this, you introduce an integral part (I). This integrates the error over time and makes sure that the error can get to zero.</p>
<p>Pro: It can get the error to zero</p>
<p>Con: It can be slow or lead to oscillations</p>
<p>To get around this, you introduce a differential part (D). This differentiates the error signal. So if the error suddenly changes, you get an immediate reaction. It makes the reaction fast so you can reduce the gain on the integral and with all three parts together nicely tuned you get a fast and stable control circuit.</p>
<p>So, start with a purely proportional system. If this does what you need -> great.
If you need a more dynamic and precise control add I and D.</p>
| 51357 | Which controller is more suitable in this scenario? P, I or PI controller? |
2022-06-16T06:30:37.123 | <p>Recently I am learning about the MaFaulDa machine failure database. I am wondering about the unit of column zero (tachometer signal) measured by Monarch Instrument MT-190 (as announced by database publisher). How to convert that data to rpm?
Thanks for any feedback.</p>
| |motors|automotive-engineering|vibration| | <p>There seem to be an easier method for determining the speed of the machine (I found it with the help of Bing AI):</p>
<p>from scipy.signal import find_peaks</p>
<h4>detect peaks</h4>
<p>peaks, _ = find_peaks(sig, prominence=3) ##since we want to determine the intervals between any peaks >3</p>
<h4>calculate interval between peaks</h4>
<p>interval = (peaks[1] - peaks[0])*0.00002
speed = 50/interval</p>
<p>print(interval, speed)</p>
<p>where sig = signal.iloc[:,0] since the first column (zeroth column) is the speed sensor's voltage readings. I hope this helps, as I have a very difficult time to transfer learning a model from this signal to CWRU (with a speed of ~1750 CPM)</p>
| 51395 | Understanding the units of columns in MaFaulDa Dataset |
2022-06-17T07:30:11.810 | <p>If a building the shape of pyramids slops downwards, wouldn't it minimize the risk of fall from heights, whether intentionally or by accident (like fire/bomb scare), from a balcony or window opening? That is discounting falling in a lift shaft or stairwell? Perhaps Egyptians were smart in that sense?</p>
| |building-design|building-physics| | <p>If you take the <a href="https://www.wonders-of-the-world.net/Pyramids-of-Egypt/Dimensions-of-the-pyramids-of-Egypt.php" rel="nofollow noreferrer">ancient Egyptian pyramids</a> as an example, the slope angle of the sides is greater than 50 degrees. You can also experiment with pyramid dimensions and walls angles using <a href="https://www.blocklayer.com/pyramid-calculator.aspx" rel="nofollow noreferrer">this website</a>.</p>
<p>People may not "fall" to their deaths from such pyramids, but they will <em>slide</em> to their deaths.</p>
<p>To prevent anyone from sliding to their death from a pyramid the wall angles would need to be less than the angle of sliding, which for most situations would be less than 30 degrees.</p>
| 51405 | Would pyramid shape buildings prevent people from falling to their deaths? |
2022-06-17T12:04:56.103 | <p>Regarding piston acceleration variations due to the geometric nature of crankshaft and connecting rod motion, in relation to connecting rod length and bore to stroke ratio.</p>
<p>Why is <a href="https://2.bp.blogspot.com/-wKbBBNQvwJU/T5_XKRt9BBI/AAAAAAAALr0/Erakyh7Ymk4/s400/Scotch+yoke.gif" rel="nofollow noreferrer">Scotch-Yoke mechanism</a> not a viable choice for reciprocating to rotary motion conversion, applied to internal combustion piston engines? (whatever the purpose and scale of this engine)</p>
| |mechanical-engineering|engines|linkage| | <p>Making an engine using scotch-yoke mechanisms for all pistons on a multi-cylinder engine will increase the cost and add a lot of needless complexity and mass, especially rotating mass that won't improve the output very much.</p>
<p>Then also consider that there are increased friction losses as well.</p>
<p>We did a laboratory experiment where we measured the angular velocities of the piston comparing scotch-yoke to the classic crank & con rod. Plotting the relative velocities etc Probably still have the writ-up somewhere but it's in the attic...</p>
<p>The piston still has the same time from TDC to BDC etc as that is based on the crankshaft rotation.</p>
| 51406 | Why is Scotch-Yoke mechanism not used for internal combustion reciprocating piston engines? |
2022-06-17T12:30:27.213 | <p><a href="https://i.stack.imgur.com/XAsmB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XAsmB.jpg" alt="enter image description here" /></a></p>
<p>What's that visible airflow over the airplane wing and what causes it?</p>
| |mechanical-engineering|fluid-mechanics|aerospace-engineering|fluid| | <p>Specifically, what you are seeing here is a <em>vortex</em> which is basically a sideways tornado. Vortices occur when moving air forms a twisted stream when flowing over an object with a complicated shape. The air pressure in the <em>core</em> of the vortex is lower than ambient and if the air is near the dew point temperature, the moisture condenses out at the core and "decorates" the vortex so it becomes visible.</p>
<p>Vortices commonly form at the sharp corner of an extended flap or aileron and at the trailing edge corner of a wingtip. They slightly reduce the lift generated by the wing in the vicinity of the tip. <em>Tip winglets</em> are designed to minimize this effect by blocking air under the wing from spiraling up and around the tip of the wing and cancelling the low pressure zone on the top of the wingtip.</p>
| 51408 | Airflow visible over the airplane wing |
2022-06-20T10:14:27.923 | <p>I am designing a simple structure using PROKON. The structure is a full 3D space frame. When I run the analysis I get an error called the Paradiso Error. This error only occurs when the structure is modeled in 3D however when a frame is modeled in 2D I do not get this error and the frame analysis can be completed.
I am wondering how to get rid of this error. Thanks
<a href="https://i.stack.imgur.com/iYrol.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iYrol.png" alt="The modeled structure in 3D" /></a></p>
<p><a href="https://i.stack.imgur.com/uPG8i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uPG8i.png" alt="The paradiso error message" /></a></p>
| |structural-engineering|civil-engineering|structural-analysis|software| | <p>A zero-pivot error means your stiffness matrix has a zero element on the diagonal. This in turn is most likely caused by your structure containing (at least) one mechanism.</p>
<p>A good check would be to remove all releases at element joints and looking at the deformed shape. Another good idea is to change your pinned supports to fixed.</p>
<p>After you get the model to run you can re-apply joint releases and original supports a few at a time until you find what is generating your mechanism.</p>
| 51437 | Prokon Paradiso Error |
2022-06-22T06:11:19.953 | <p>Context: I am a hobbyist and I don't have experience with industrial solutions</p>
<p>I am starting to prototype a farming equipment gps guiding system and I was looking fore some feedback regarding my approach.
I was looking into using a ublox-zed-f9p connected to a processing board that is going to spin up some servo motors that should be driving the wheels.
Also to note that the wheel equipped machine would have somewhere close to 8t max.</p>
<p>The main questions that I have regarding this project:</p>
<p>Is the ublox-zed-f9p good as an industrial solution ?</p>
<p>What processing boards would you recommend ? (I am used to raspberries and ESPs)</p>
<p>What makes a component to classify as industrial ?</p>
<p>Thank you in advance, I am new here, so if I did something wrong, please point it out :)</p>
| |mechanical-engineering|electrical-engineering|embedded-systems|servo| | <p>An 8 ton hobby machine? That's quite a hobby! To answer your questions:</p>
<ol>
<li><p>ublox-zed-f9p is probably fine for production, but it's a terrible choice for prototyping because it's got pad, not wire, connections. It will be very challenging for you to solder to that. You can only really install it on a PCB. Many parts come in prototyping (wire) and production (bond pad) variants. Find one of those so you can start with something easy to prototype and then scale up to custom PCBs.</p>
</li>
<li><p>Really depends what you're doing. If you're trying to do AI controls/image recognition/something else computationally expensive you need a chipset that is designed for that. If you want to have an LCD control panel you probably want to run something like Linux or Windows on the chip so you can use a GUI framework to design the UI. Raspberry PI could be enough to prototype with, but you will want something weatherized when you put it on a piece of equipment that actually has to go in the field. If you're going the AI/automation route, I would look at chips designed for self driving cars.</p>
</li>
<li><p>There's no hard line between industrial equipment and other equipment, but some features that industrial users might care about include:</p>
</li>
</ol>
<ul>
<li>Weatherized package that meets certain specifications ensuring it is safe to use outdoors or in a factory where surfaces get washed regularly</li>
<li>Reliable supply chain able to handle user's demand for the equipment</li>
<li>Ability of the supplier to provide technical support to project engineers</li>
<li>Easy integration in automated manufacturing processes (e.g. chipset with pads)</li>
</ul>
| 51456 | Industrial (Farming) high precision gps tracking solution |
2022-06-23T04:49:54.543 | <p>Consider the following composite shaft which consists of 3 different materials with modulus of rigidity of <span class="math-container">$G_1$</span>, <span class="math-container">$G_2$</span> and <span class="math-container">$G_3$</span>. Shaft 1 has length 2L, and shafts 2 and 3, L. Assume they are all 'glued' together.</p>
<p><a href="https://i.stack.imgur.com/BbufH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BbufH.png" alt="enter image description here" /></a></p>
<p>Knowing that the left-half of shaft 1 is connected in parallel to shaft 2 (and one could say something similar to the right-half and shaft 3), I would like to prove that the torsional rigidity <span class="math-container">$(GJ)$</span> of each side of the composite shaft is given by</p>
<p><span class="math-container">$$(GJ)_{\text{left}} = G_1 J_1 + G_2 J_2 \qquad (GJ)_{\text{right}} = G_1 J_1 + G_3 J_3$$</span></p>
<p>I just assume that this is correct based on how we find the equivalent spring constant of wo parallel ones, and that, in this case, GJ is kind of a torsional spring constant.</p>
| |solid-mechanics|strength| | <p>Your expression is correct, the proof is carried out below for the left segment of the composite bars:</p>
<p>From static force equilibrium, we have</p>
<p><span class="math-container">$T = T_1 + T_2$</span> ----- (1)</p>
<p>Furthermore, the angle of twist <span class="math-container">$\phi$</span> must be the same for both parts,</p>
<p><span class="math-container">$\phi = \dfrac{T_1L}{G_1J_1} = \dfrac{T-2L}{G_2J_2}$</span> ----- (2)</p>
<p>From (2),</p>
<p><span class="math-container">$T_1 = T\dfrac{G_1J_1}{G_1J_1 + G_2J_2}$</span> ----- (3a), and</p>
<p><span class="math-container">$T_2 = T\dfrac{G_2J_2}{G_1J_1 + G_2J_2}$</span> ----- (3b)</p>
<p>Substitute eqs (3a) and (3b) into (2), and we get</p>
<p><span class="math-container">$\phi = \dfrac{TL}{G_1J_1 + G_2J_2} = \dfrac{TL}{(GJ)_{composite}}$</span></p>
<p><a href="https://i.stack.imgur.com/i0TRE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i0TRE.png" alt="enter image description here" /></a></p>
| 51471 | Torsional rigidity of a composite shaft in parallel |
2022-06-25T02:33:30.047 | <p>This question stems from the online thermodynamics notes provided by <a href="https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node17.html" rel="nofollow noreferrer">web.mit.edu</a>. I'm having some trouble understanding the example given of air expanding into a vacuum.</p>
<p>Consider the transient, unsteady process of filling a tank, initially evacuated, from a surrounding atmosphere, which is at pressure <span class="math-container">$P_0$</span> and temperature <span class="math-container">$T_0$</span>. The tank is insulated, so the process is adiabatic.</p>
<p><a href="https://i.stack.imgur.com/cAUw3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cAUw3.png" alt="filling an evacuated tank" /></a></p>
<p>The author of the notes states that work of magnitude <span class="math-container">$P_0V_0$</span> is done <em>on</em> the system. However, I thought that because the pressure in the tank is zero, the work done (<span class="math-container">$PdV$</span>) would be zero.</p>
<p>Why is it the case that the work done on the system is <span class="math-container">$P_0V_0$</span> and not zero?</p>
| |thermodynamics|gas|vacuum| | <p>the work done on the system is has caused it to move <span class="math-container">$V_0$</span> volume of gas up to the pressure of <span class="math-container">$P_0$</span></p>
<p>As if you had blown a deflated balloon to that size and that pressure, you worked moving air out of you lung to the balloon by the magnitude of <span class="math-container">$$P_0V_0$$</span></p>
<p>The fact that the tank is vacated give you the chance to do more work on the gas as opposed to what you think, else if there was some pressure left in the tank the work would be <span class="math-container">$$W=(P_0-P_{tank})V_0 < the \ emty\ tank\ case$$</span>
Remeber we are considering the gas getting into the tank and how much work has gone into pumping it. this work is coming from the ambient gas sorrunding the system!</p>
| 51488 | Work done during transient filling of a tank |
2022-06-25T23:46:34.120 | <p>What is the technical name for this kind of road segment?</p>
<p><a href="https://i.stack.imgur.com/aiL7W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aiL7W.png" alt="enter image description here" /></a></p>
<p>I think the purpose of the lane on the right is to bypass the traffic lights at the intersection.</p>
<p>I've heard that kind of lane referred to as a <em>"sliparound"</em> lane. But I doubt that's the proper term.</p>
<p><strong>Edit:</strong> The location is Ontario, Canada.</p>
| |civil-engineering|terminology| | <p>The name is the same as its purpose of service - "Exclusive Right Turn Lane", or "Right Turn Only Lane". Right turn lanes can significantly improve the capacity and level of service of signalized intersections.</p>
| 51495 | Technical name for a road "sliparound" lane |
2022-06-26T18:52:56.830 | <p>How can we find a relation between <strong>welding bead</strong> and <strong>outer diameter</strong> of a pipe? Welding bead is inside of the pipe and I am using M.S. pipe.</p>
<p>Dimensions are listed below:</p>
<p>Pipe OD=25.4mm</p>
<p>Thickness=2mm</p>
<p>Bead Length=1.5mm</p>
<p>Bead Height= 1.5mm</p>
<p>Mandrel OD=19mm</p>
<p><a href="https://i.stack.imgur.com/ls4Uu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ls4Uu.jpg" alt="enter image description here" /></a></p>
<p>I need to bend pipe but <strong>welding bead</strong> is not allowing <strong>mandrel</strong> to bend and all I need is to find a relation between <strong>welding bead</strong> and <strong>outside diameter</strong> of pipe. Do we have <strong>standard</strong> for that?</p>
| |mechanical-engineering|structural-engineering|materials|welding|bending| | <p>A Weld Bead is not a defect, and in fact is necessary to expel all of the Oxide impurities formed during the Resistance Welding Process. The issue is with the Purchase Order and/or Drawing. ID and/or OD Scarfing must be ordered. If you do not order the Tubes that way, then the manufacturer will not add that to the process. It is cheaper and easier for them to run the Tubes with the Weld Beads and no Scarfing.</p>
| 51505 | Relation between welding bead and outer diameter of a pipe? |
2022-06-27T16:43:46.920 | <ul>
<li>Z-Wave operates at <strong>868MHz</strong> (in EU) and has up to ~100-meter range</li>
<li>LoRa operates at <strong>868MHz</strong> (in EU) and has up to ~15 km range</li>
</ul>
<p>My knowledge says that the lower the frequency = the higher the range/penetrability.</p>
<p>However, both these technologies operate at the same frequency, so how come LoRa is able to achieve such a long-distance?</p>
<p>Furthermore, I'm a bit confused, as to how power consumption plays a role in the above situation. I'm assuming a high-powered signal means that the waves themselves have a higher amplitude, thus making it harder for other signals to interfere.</p>
<p>With that assumption, does that mean that LoRa uses a lot of power to reach such a long-distance?</p>
| |waves| | <p>Three factors:</p>
<ol>
<li><p>Transmitter power: LoRa, for example, typically uses about 160x the power of Z-Wave. However, this alone cannot explain a 150x increase in operating range given that signal strength falls off with the square of distance (so you would expect that extra power to get you to perhaps 1250 meters of range).</p>
</li>
<li><p>Beam-forming: Some protocols are able to directionally transmit signals instead of sending them in all directions. This increases the range by concentrating the power.</p>
</li>
<li><p>Error correction: (Over simplifying here because this is a complex topic) Some protocols send the same data many times to ensure it reaches it's destination even if there is a lot of interference (which increases with distance). Others send the data only a few times, which is faster but limits the range/reliability. Many protocols dynamically adapt how many copies of the data they send based on operating conditions.</p>
</li>
</ol>
| 51517 | What causes wavelength technologies to have longer range than other despite having the same frequency? |
2022-06-27T17:15:18.080 | <p>My system is a fluid exposed to light and I want to calculate the temperature change with time at a point, hence there are no spatial effects or convection due to the flow of the fluid.</p>
<p>The heat source is expressed as fluence multiplied by the absorption coefficient [W/cm<sup>3</sup>]. Given the same fluence, I want to see the effect of the absorption coefficient with time at a point in space.</p>
<p>What I have found similar is the lumped capacitance method and the equation is given by</p>
<p><span class="math-container">$-hA (T-T_{inf}) = \rho V c \frac{dT}{dt}$</span></p>
<p>however, there is no internal heat generation and it considers boundary effects.</p>
<p>This leaves me with <span class="math-container">$Q = mc(T-T_{inf})$</span>, and <span class="math-container">$Q$</span> is just energy.</p>
| |mechanical-engineering|heat-transfer|optics|fluid| | <p>If it is a real world system, its temperature should not rise at a constant rate forever. Instead it will try to achieve some kind of an equilibrium. For a fluid with mass <span class="math-container">$m$</span>, capacity <span class="math-container">$c_p$</span>, constant light energy input <span class="math-container">$\alpha F$</span>, environment temperature <span class="math-container">$T_{env}$</span> and heat transfer coefficient with the environment <span class="math-container">$U$</span>, this equation should apply:</p>
<p><span class="math-container">$$m c_p \frac{dT}{dt} = \alpha F + \left(T_{env}-T(t)\right) U$$</span></p>
<p>Assuming only the fluid temperature will change in time, you can separate the variables and integrate both sides using starting fluid temperature <span class="math-container">$T_0$</span>:</p>
<p><span class="math-container">$$\int\limits_{T_0}^{T}\frac{1}{T-T_{env}-\frac{\alpha F}{U}} dT = -\frac{U}{m c_p}\int\limits_0^{t}dt$$</span></p>
<p>Resulting expression for fluid temperature <span class="math-container">$T$</span> as a function of time <span class="math-container">$t$</span> should look something like this:</p>
<p><span class="math-container">$$T(t) = T_{env}+\frac{\alpha F}{U} + \left(T_0-\frac{\alpha F}{U}-T_{env}\right)\exp\left(-\frac{U}{m c_p}t\right)$$</span></p>
<p>From this you can see that at the start <span class="math-container">$t=0$</span>, the temperature <span class="math-container">$T(t=0)=T_0$</span> because the exponential is 1 and other terms cancel out. Equilibrium is reached in infinity, <span class="math-container">$T(t\rightarrow \infty)=T_{env}+\frac{\alpha F}{U}$</span> as the exponential term goes to 0. It is a typical exponential function where you have the highest rate of temperature change at the start and it will decrease in time as the system approaches equilibrium.
<a href="https://i.stack.imgur.com/pPEO6l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pPEO6l.png" alt="enter image description here" /></a></p>
| 51518 | How to solve transient heat conduction equation at a point in space with volumetric heat source? |
2022-06-28T01:11:24.207 | <p><strong>Some background</strong></p>
<p>I am reading through a <a href="https://www.elsevier.com/books/an-introduction-to-atmospheric-gravity-waves/nappo/978-0-12-385223-6" rel="nofollow noreferrer">textbook on atmospheric fluid dynamics</a>. The author is discussing something called the Boussinesq relations, and eventually the following expression arises:</p>
<p><span class="math-container">$$
\frac{1}{\theta}\frac{d\theta}{d z}=\frac{1}{\rho}\left(\frac{1}{c_s^2}\frac{dP}{dz}-\frac{d\rho}{dz}\right)
$$</span></p>
<p>where <span class="math-container">$c_s$</span> is the speed of sound and <span class="math-container">$\theta$</span> is the potential temperature. The <a href="https://en.wikipedia.org/wiki/Potential_temperature" rel="nofollow noreferrer">potential temperature</a> is the temperature a parcel of air would have if it were compressed (or expanded) adiabatically from its initial pressure to a pressure of 1000 mbar. By using the definition of potential temperature and the ideal gas law, you can arrive at the relation above through differentiating with respect to <span class="math-container">$z$</span>.</p>
<p><strong>Question</strong></p>
<p>The author states the following:</p>
<blockquote>
<p>Because the rates of motion of the atmosphere and gravity waves are generally much less than the speed of sound, the term <span class="math-container">$\frac{1}{c_s^2}\frac{dP}{dz}$</span> can be neglected.</p>
</blockquote>
<p>My question is how I can reason about the term <span class="math-container">$\frac{1}{c_s^2}\frac{dP}{dz}$</span> to convince myself that it should be small for most atmospheric phenomena? Or, in other words, how can I show that <span class="math-container">$\frac{dP}{dz}$</span> is small compared to <span class="math-container">$c_s^2$</span>? In what sense is the derivative of pressure comparable to a speed squared?</p>
| |fluid-mechanics|pressure|air| | <p>You might find more satisfying the justification given by Holden and Hakim in <em>An Introduction to Dynamic Meteorology</em>: "For buoyancy wave motions
[<span class="math-container">$\left|\frac{\rho}{\theta}\frac{d\theta}{dz}\right|\gg \left|\frac{1}{c_s^2}\frac{dP}{dz}\right|]$</span>; that is, density fluctuations due to pressure changes are small compared with those
due to temperature changes." (I've used Nappo's notation for the relation in brackets.)</p>
<p>On the other hand, Satoh's <em>Atmospheric Circulation Dynamics and General Circulation Models</em> simply asks us to consider the simplification arising "[i]n the case that the speed of sound <span class="math-container">$c_s$</span> approaches infinity".</p>
| 51522 | Analyzing the term $\frac{1}{c_s^2}\frac{dP}{dz}$ for atmospheric phenomena? |
2022-06-29T01:23:47.867 | <h2>Background</h2>
<p><a href="http://www.digipeer.de/index.php?media=DMA_FA_014_14400&size=2" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JWNps.png" alt="enter image description here" /></a></p>
<p>The part above is from 1944 V2 plans which I am using as real world practice examples to teach myself 3D modeling using Inventor LT 2021. Therefore there is no original designer/engineer I could got to for clarification.</p>
<p>I know how I attempted to model this piece but I am not sure on the proper way to model this.</p>
<h2>What I did</h2>
<p>I started out by drawing up the section on the left in a 2D sketch and then added some cutting shapes.</p>
<p><a href="https://i.stack.imgur.com/7Rml0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Rml0.png" alt="enter image description here" /></a></p>
<p>I then revolved the large rectangle on the right side to generate a section straight pipe. I followed this up by extruding triangles at the top and bottom to generate the angled cuts.</p>
<p><a href="https://i.stack.imgur.com/paxuJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/paxuJ.png" alt="enter image description here" /></a></p>
<p>I then proceeded to create a work plane on the face of the lower cut and proceed to added two shapes. The first to be essentially a plate to be extruded into the remaining pipe. The section a bounding box the extrude and remove material from the lower pipe that does not belong.</p>
<p><a href="https://i.stack.imgur.com/RCjC5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RCjC5.png" alt="enter image description here" /></a></p>
<p>I then applied a 6 mm fillet to the inside edge of the added plate and pipe, followed by an 10 mm fillet to the outside edge.</p>
<p><a href="https://i.stack.imgur.com/QVAID.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QVAID.png" alt="enter image description here" /></a></p>
<h2>Problem</h2>
<p><strong>The problem is the last and final step of applying a 15 degree chamfer to top face of the pipe.</strong></p>
<h2>What I tried</h2>
<p>I tried applying a chamfer using the tool which was probably the closest result I got but it had a problem. The tool requires either fixed distance on both faces or a fixed distance on one face and an angle from that face. The problem is the width of the top face varies ever so slightly due to the angled cut through the piped. Since the 15 degree angle is relative to the horizontal plane, the depth along the side of the pipe also varies. Below is an endview of the problem.</p>
<p><a href="https://i.stack.imgur.com/XsnWe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XsnWe.png" alt="enter image description here" /></a></p>
<p>I also tried creating triangle to cut out the required shape. I initially was going to revolve it, but that does not work as the path it need to follow is not circular. Then I thought about using the sweep tool and selecting the inside edge and tip of the triangle. However the resulting cut from that was not good at all. It was as if the triangle did not remain in its original horizontal position but got twist as it followed the path.</p>
<p>Setting up using the sweep tool</p>
<p><a href="https://i.stack.imgur.com/AckFI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AckFI.png" alt="enter image description here" /></a></p>
<p>Result</p>
<p><a href="https://i.stack.imgur.com/3bzgf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3bzgf.png" alt="enter image description here" /></a></p>
<p>The blue line is the 15 degree from horizontal line from the sketch at the low end. Sweep has some serious issues relative to the desired output.</p>
<p><a href="https://i.stack.imgur.com/I3K3L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I3K3L.png" alt="enter image description here" /></a></p>
<p>So the last thing I tried was loft. I made 4 sketches, one at each quadrant, and selected closed loop. The result were not good. The resulting shape did not follow the radius of the pipe when viewed on end. I even added a rail of the inside edge which improved things but still had problems.</p>
<p>I just forgot, I also tried a helix (coil) cut, but the path of the helix is not the same as the angled cut through the pipe.</p>
<h2>Question</h2>
<p><strong>Is there another way of generating this chamfer or removal?</strong></p>
| |cad|technical-drawing|drafting|drawings|autodesk-inventor| | <p>Chamfer is an angle or distance from a surface. Since your surface varies in angle to the horizontal, you will not get the desired results the way you did it.</p>
<p>Lots of sufficiently fine approximations via sweeps, etc but it seems like people want some exact features, so here's the surface modeling solution. (Sorry no screenshots) Use Thicken/Offset with an Output of Surface to offset the inside of the pipe outward. Distance is likely to be at least 4 mm (can be more than 4 depending on what you want to do). Then use Move Body to shift by (the offset distance from the thicken) * tan(15deg) along the pipe axis. You may have to negate, but thankfully your pipe likely lines up with a lettered axis. Use the Patch feature and select the upper edges of both the pipe interior and this new outer pipe surface. You can do some neat things with the resulting surface patch such as stitching into a solid with surfaces, or Extrude to.</p>
<p>If instead of revolving the solid, you started your pipe as a surface of 36 mm diameter, trimmed it to your 8 mm slant (extruding a line can make a surface to trim to), used thicken by 4 mm, moved the outer surface body, made a couple of patches at top and bottom, and stitched the four into a pipe, the 15 degree "chamfer" surface would already be there prior to cutting the 40 mm feature and filetting.</p>
| 51534 | How to place a chamfer on the end of an angle cut pipe |
2022-06-29T21:25:12.830 | <h1>Problem</h1>
<p>I have a STEP file with a few holes for bolts. I'm hoping to bolt this part into stock material in Solidworks.</p>
<h1>Example</h1>
<p>My first thought is to extend the holes into other connected components in an assembly to ensure both parts now have the same hole profile.</p>
<p>I imported a STEP file into this basic assembly.
<a href="https://i.stack.imgur.com/CxdOX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CxdOX.jpg" alt="The assembly has one internal part and one STEP file." /></a></p>
<p>Then I mated the part to the assembly so it can't move.
<a href="https://i.stack.imgur.com/YSfLI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YSfLI.png" alt="enter image description here" /></a></p>
<p>Then I drew onto the face on the assembly that needed to be cut.
<a href="https://i.stack.imgur.com/9AXaF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9AXaF.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/a1uEE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a1uEE.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/VfQZr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VfQZr.png" alt="enter image description here" /></a></p>
<p>Then I exited the sketch, and Solidworks took me to the "Extruded cut" dialog.
<a href="https://i.stack.imgur.com/M5Z48.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M5Z48.png" alt="enter image description here" /></a></p>
<p>Now holes are cut through the material.
<a href="https://i.stack.imgur.com/NTkIb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NTkIb.png" alt="enter image description here" /></a></p>
<p>Finally, I find and import the STEP models of the bolts they provided to my assembly. Each one is manually mated.
<a href="https://i.stack.imgur.com/S7tKM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S7tKM.png" alt="enter image description here" /></a></p>
<p><strong>There has to be a more standard/convenient way to bolt in COTS parts, I'd hope?</strong></p>
| |mechanical-engineering|solidworks|bolting| | <p>The only other way to do this that I know of is to draw circles and extruded cut as usual to make the holes, then edit the circles in the assembly and create an external reference between your drawing and the holes in the STEP part.</p>
| 51547 | SOLIDWORKS-Bolting a COTS part from a STEP File onto an Assembly |
2022-07-01T02:19:23.447 | <p>The snow load duration factor is 1.15 times the standard 10-year sustained maximum load rating as long as the accumulated time under this load is not more than 2 months. Does this mean that</p>
<ol>
<li><p>a. the snow load being at the duration factor of 1.15 for 2 months continuously (experimental
case) is at the same level of safety as</p>
<p>b. permanently being at the standard load rating (control case)</p>
</li>
</ol>
<p>when assuming that the experimental case is well under the under standard load rating for the rest of the year, or</p>
<ol start="2">
<li>does the experimental case impart additional risk compared to the control case from the long-term perspective?</li>
</ol>
<p>Basically, this is asking whether the standard practice allows the control case to use up the safety margin (factor of safety minus one) of the standard load rating only because its short term means that the probability of other loads happening simultaneously will be negligibly small, or whether the safety margin of the standard load still applies on top of the maximum load at full duration allowed by the duration factor.</p>
<p>For example, if the snow season is especially long one year but not especially intense at any given moment, meaning that the roof at any given time is at the forseeable maximum snow load accounted by the duration factor but that it is at that load for a whopping 4 months rather than the expected maximum of 2 months, does that mean the roof will cave in, probably causing the pancake effect to collapse the entire building too?</p>
<p>As for another example under main question, will having the load at any given time within the timeframe moderately exceeding the snow load duration factor, but not having the duration exceed that, make the roof or even building collapse?</p>
<p>Under the hypothetical case that load duration factors in standard practice already use up all of the safety margins for the standard load ratings, this means that exceeding the load duration factor also exceeds the safety margin. Since this exceeds the safety margin, does this mean that exceeding the load duration factor will guarantee a collapse?</p>
<p>In civil engineering in general, does being at the rated limit mean that there is a negligible chance of failure if everything is done perfectly (too small to be measured, such as 1ppb) and does being at the upper limit of the safety margin practically guarantee a failure if everything else is done perfectly (too large to measure a successful case, such as 99.9999999% failure rate)?</p>
| |structural-engineering|civil-engineering|load-spreading| | <p>The problem is you deciding to push the limit in one area and the owners will have to know that and maintain all the supports for the roof that entire 100 or 1000 years. Not likey, and it isn't collapse that is the outcome, it's FAILURE. One small part failing that isn't visible starts the process and then a portion of the system fails or "collapses", the failure may be at the foundation because of the roof load. I had a house that a stack of timber holding a portion of an upper floor shear a concrete foundation. That ended with removing 13 cubic yards of soil from below grade and replacing 20ft of foundation. It was a concrete foundation over a 100 years old and that why real professionals say "They don't build them like they used to." Pause ......."Thank God". It's a line used on remodeling or updating old homes. You pull something open and say that. Everyone on site knows we have a problem. Everywhere in the world you're only seeing the 1 out of a 100 buildings that didn't fall down or burn.</p>
| 51562 | Would exceeding the snow load duration factor guarantee a collapse? |
2022-07-01T11:28:07.823 | <p>I am working with a variable power laser system that has a wavelength of 0.8μm – 1.1 μm. The laser set up in the lab must be able to weld any gauge of material together for the experiment to take place. Thinner gauge material sheets must be supported in order for a proper weld to take place. Thus, the introduction of a supporting wheel or device was introduced. Currently, the laser beam welds or cuts the test sample but continues on afterwards. Damaging the supporting mechanism below. A stainless steel planish wheel has been experimented with to support the material and resist damage by the laser beam. This only lasts 5 or 6 welds until it needs to be replaced. A proper life cycle for the material should be upwards of 1,000 weld cycles. Is there a proper way to properly diffuse the laser beam reducing damage below the test sample?</p>
| |optics|lasers| | <p>Generally the support structure is designed with a void behind the cutting area so that the beam has space to diverge (rather than "diffuse") enough that the power density has dropped to a harmless (to the machine) level.</p>
<p>Given the problem I would consider a glass support but I have no idea about your setup.</p>
| 51566 | Diffusing a Laser beam |
2022-07-01T14:23:30.607 | <p>A huge number of specimens are built. For each specimen, soil is mixed with cement (and Zeolite) with different percentages. It's intended to test, measure and compare the impact strength of the specimens.</p>
<h1>One test option</h1>
<p>One test approach is to drop a ball on the specimens and count the number of drops before the specimen collapses/breaks. There is a problem with that, some specimens require hundreds of the drops before collapsing/breaking. Dropping the ball for hundreds of times is done <strong>manually</strong> which is a daunting task for the students doing the test.</p>
<p><a href="https://i.stack.imgur.com/m9FTN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m9FTN.png" alt="Drop ball test" /></a></p>
<h1>Question</h1>
<p>I wonder if there is any other approach for testing which as cheap, but less of a manual work?</p>
<p>Students can build simple devices. Is there any design for a simple device which can do the test automatically, i.e. without the need for dropping the ball <strong>manually</strong> hundreds of times?</p>
<p>Thanks =)</p>
<h1>Ball size</h1>
<p>We cannot use a larger ball to reduce the drop count. Since some specimens are breaking with a few ball drops :(</p>
| |civil-engineering|product-testing|soil|impact|soil-mechanics| | <p>It sounds like you could modify the procedure and equipment used for a soil standard penetration test <a href="https://en.wikipedia.org/wiki/Standard_penetration_test" rel="nofollow noreferrer">(ASTM D1586)</a>. In this test, a hammer of specific weight is dropped a specific distant to strike a rod. The test is automated in that the hammer is re-raised mechanically.</p>
| 51568 | Testing specimens by dropping a ball on them and counting the number of drops required before the collapse. A better way? |
2022-07-02T00:34:55.930 | <p><a href="https://i.stack.imgur.com/GlW00.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/GlW00.jpg" alt="a building with dark patches on its brick exterior under every window" /></a></p>
<p>I don't recall seeing the walls like this before, so I don't think it's just accumulated soilage. If it's wetness, what design flaw causes the windows to do that? Today was very hot and muggy so it's likely the AC was running all day.</p>
| |structures|waterproofing| | <p>Cold air is spilling down the wall from the window. This permits condensation on the block. I bet it was a <em>really</em> calm day too.</p>
| 51576 | How did this building's windows get like this? |
2022-07-04T15:04:18.820 | <p>It is common setup to have two lead rods that carry a platform that needs to move on linear bearing. The platform itself is controlled by a separate lead screw connected to a stepper.</p>
<p>I'm looking for very lightweight and small setup and wonder what kind of problems should I expect if instead of two rods + lead screw I would use one rod and a lead screw that would use as second rail for the device.</p>
<p>I understand that the end of the lead screw (maybe start as well) need to be put into some kind of bearing to provide structural support and accuracy.</p>
<p>Question:</p>
<ol>
<li>What kind of problems should I expect if any? In terms of control accuracy stability, etc.</li>
<li>What kind of support recommended for such a lead screw (one bearing, two bearings) and what kind (since standard bearings for example 8mm are actually bigger than 8mm lead screw)?</li>
</ol>
<p>Edit: It is for movement of optical system. I mostly for accurate slow movement - i.e. the speed should be constant and as accurate as possible.</p>
| |mechanical-engineering|stepper-motor|rail| | <p>You may plan something like this:</p>
<p><a href="https://i.stack.imgur.com/AGsOO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AGsOO.jpg" alt="enter image description here" /></a></p>
<p>The green screw rotates and the blue block moves along the purple rod.</p>
<p>You may remember or be able to imagine this very real situation:</p>
<p><a href="https://i.stack.imgur.com/fQNmg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fQNmg.png" alt="enter image description here" /></a></p>
<p>There's no screw, only the rod is in the hole of the blue block. The gravity cannot make the block to slide down, because the weight of the block generates enough torque to make the lower edge of the hole to bite the rod.</p>
<p>The screw can cause the same, but it also tightens the block to a tilted position which surely causes damage.</p>
<p>Long sleeve with rounded edges around the rod and tight tolerances may prevent the unfortunate tighted tilted position to develop, but 2 rods at the opposite sides of the screw would allow less tight tolerances and shorter sleeves.</p>
<p>There exists numerous single rod solutions in actuator mechanisms. Study them.</p>
<p>For more advanced answers insert more known data of the conditions that your system must stand and limitations, too.</p>
| 51598 | Using lead screw with lead rod instead of two rods and lead screws |
2022-07-05T08:40:05.427 | <p>I've been trying to figure this out for days and can't seem to get it. I'm trying to create a DIY winch. The motor can pull up to 5500W. It'll be moving things between 45-90 kg (100-200lb) at a steady speed of 1-2m/s.</p>
<p>The motor has a lot of speed, but I need torque, so I've been looking at planetary gearboxes. However, I can't figure out how much Newton-meters of torque the gearbox should rated for.</p>
<p>I'm looking at 10:1 gearboxes. Basically, I want to know how I can calculated what Nm the gearbox will be subjected to so I can get the right one.</p>
| |gears|torque| | <p>The largest torque applied to the gearbox occurs at the low-speed side. In this case, this is at the load-side of the gearbox, as opposed to the motor-side.</p>
<p>So, all you need to do is compute the torque the load exerts directly to the gearbox. If we know the maximum load to be lifted, and we know the radius of the winch pulley (or more accurately, the horizontal distance between the centre of the gearbox output shaft and the centre of the load), then the torque can be found from the product of the radius and the load in newtons.</p>
| 51607 | How to calculate torque on gearbox? |
2022-07-06T00:23:27.950 | <p>In the figure attached showing the throttle position versus an IC engine produced torque, I didn't understand why does the torque value at a specific rpm vary for different throttle positions. My knowledge for these curves is that the torque actually increases due to more air-fuel mixture entering the chamber. now if we took the 20% throttle position from the graph, or any other percent, we can see that the torque is increasing;but how does that happen when the amount of air-fuel mixture is held fixed (ie : the throttle opening is fixed at (20% for example); and thus the air-fuel mixture entering the combustion chamber is fixed and predetermined by this percentage. Then how would the torque value increase? unless of course we vary the throttle position?</p>
<p><a href="https://i.stack.imgur.com/cuTKm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cuTKm.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|automotive-engineering|torque|engines|combustion| | <p>The Torque/Speed curves (T vs w) are generated for a specific throttle position. IC Engines don't make T very well at low w. It's physics; The pistons need to draw in the fuel-air mixture. (e.g., to hold constant T at w = 0 would mean the valves seal perfectly and theirs no blow-by the rings. Need to get fuel/air in and exhaust out. 4-stroke.) This is why T is increasing with RPM for ICE/Otto Cycle.</p>
<p>So, the basic shape is a T vs w curve that rises to a peak T value, then levels off and/or decreases. (An electric motor curve is down sloping due to back EMF (and other effects depending on control.)</p>
| 51612 | Throttle position as a function of engine torque and speed |
2022-07-06T04:40:08.507 | <p>Why were so many pre-1970 bridges built so low to the water that they do not even allow a single-level stand-up boat (serving as water buses or small fishing/research vessels) to pass underneath without raising or swinging the span? If they're built so low with short spans anyway, why not make them a pontoon bridge by making their light-duty piers floating platforms to save tons of time, money, and difficulty? Examples include the Oregon Trunk Rail Bridge (also known as the Celilo Bridge), the old Biloxi Bay Bridge (1962-2005), Norfolk Southern Lake Ponchartrain Bridge, Lake Ponchartrain Causeway, the concrete arches of the old Seven Mile Bridge (originally for railroad), etc.?</p>
<p>Why not build the non-shipping channel portions as high as a boardwalk pier, so that small single-floor stand-up vessels can pass under with traffic simultaneously passing overhead, just like how all bridges built since the 1970s have [been built at least as high as the Burlington Northern Railroad Bridges 5.1 and 9.6 (also known as the BNSF Railway Bridge 5.1 and BNSF Railway Bridge 9.6, respectively) across the Willamette River and Columbia River (both in Portland, Oregon and the second one also being in Vancouver, Washington), respectively]?</p>
<p>For example, see how the lowest spans of the Chesapeake Bay Bridge-Tunnel are built significantly higher than those of the Lake Pontchartrain Causeway, the lowest spans of the replacement Biloxi Bay Bridge (2007-present) are built way higher than those of the original (1962-2005), and the lowest spans of the replacement Seven Mile Bridge (1982-present) are built significantly higher than the original (1916-1981). Also, this phenomenon isn't limited to just the United States alone or even when also including Canada -- it is a global phenomenon. For example, look at how the lowest spans of the Annai Indira Gandhi Road Bridge (1988-present) are way higher than the those of the Pamban Bridge (1914-present) in India. The video below shows how much higher the replacement bridge (estimated completion around late 2024) for the Hanshin Namba Line over the Yodo River in Osaka, Kansai, Japan is compared to the original (1924). Also, note how the Denpo-Ohashi Bridge (the tied arch bridge, estimated to have been completed in the 1930s) in the near background is significantly lower than the Shin-Denpo-Ohashi Bridge (completed in 1958 as part of National Route 43) right behind it. All of the bridges mentioned here for this video are level.
<a href="https://youtu.be/n9pBhmlQsQA" rel="nofollow noreferrer">https://youtu.be/n9pBhmlQsQA</a></p>
| |civil-engineering|bridges|hydrology|infrastructure| | <h3>Cost</h3>
<p>It costs more to build a 23 mile bridge 20 feet (30 feet, 40 feet?) higher in the air.</p>
<p>When it was built, the Lake Ponchartrain Causeway wasn't blocking anything - there was zero freight traffic from the west shore of the lake. So the extra expense of building it higher would have laughed the engineer out of the room.</p>
<p>You can drive a personal boat under it.</p>
| 51616 | Why were so Many Older Bridges Built so Low to the Water? |
2022-07-06T07:06:16.527 | <p>In my company, we use IR Thermography.
But I was thinking, why IR? Why not microwave or radio wave or ultraviolet even.
Is it because of some unique property of infrared or the cost-effectiveness?
Or is the thermal imaging also done with other methods other than infrared?</p>
<p>I have little understanding of radiation, so please bear with me.</p>
| |heat-transfer|radiation|thermal-radiation| | <p>In theory you can measure temperatures at different points of a solid item surface and present them with colors on a plot. Collect enough temperatures and plot them accurately enough and that's your manually scanned thermal image.</p>
<p>In many applications that method is not acceptable. For ex. to see an intruder before he knows he's detected you cannot ask him stop for making your pointwise temperature measurements. In such cases thermal cameras are more useful because they work with thermal radiation.</p>
<p>Physicist Max Planck became world famous about 120 years ago by inventing the physical laws behind how the thermal radiation (=generated by the random thermal motion in materials) depends on the temperature - before him the phenomenon was known only experimentally, but it was known.</p>
<p>Actually ultraviolet and visible light could be also be used for thermal imaging, but substantial amounts of them need so high temperatures that nobody could live in such conditions. Think about red or white hot iron to get a picture. The radiation is strongest in the infrared range in temperatures where we can live.</p>
<p>Microwave emissions caused by thermal motion in materials occur as well, but detecting them needs complex high cost equipment. But the radiation in common temperatures is strongest at infrared range, strongest usually at wavelengths about 10 micrometers. Microwave emissions of electronic devices would easily cause harm for thermal imaging at microwave range by saturating the imaging devices with overwhelming radiation intensity created by other means than thermal motion.</p>
<p>Microwave thermal imaging is still used for ex. to get some evidence for cosmological theories or to prove them false. The needed equipment for one single image isn't handheld, they can cover square miles.</p>
<p>For better knowledge read these: <a href="https://en.wikipedia.org/wiki/Thermography" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Thermography</a> and <a href="https://en.wikipedia.org/wiki/Black-body_radiation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Black-body_radiation</a></p>
| 51618 | Why is infrared used in thermal imaging and not microwave or radio wave or ultraviolet? |