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bshada

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engineering.stackexchange.com

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2022-12-29T17:13:43.293

<p>I make specialized flapper valves that are similar to O-Rings. The current model uses aluminum sheets as base disks that adhere to the rubber (2 part urethane). The rubber is poured into the waxed mold and then topped with an aluminum piece. Unfortunately, there is currently no way to remove the rubber from the mold without damaging the mold or rubber material.</p> <p>The mixing, degassing, pouring, and subsequent removal of the rubber must be done perfectly in order to create a seal. Since this process is done almost entirely by hand, there is a large margin of error and I end up with a lot of waste. Not only that, but I spend an extensive amount of time cleaning the mold to remove even the specks of dust. Also, I believe the aluminum piece may be preventing the rubber from moving easily making it more difficult to get a good seal with even the slightest imperfection. I should also note that the center hole is what gets worn away the fastest and is my most common issue with age.</p> <p>Previous thoughts have been to weld handles to the aluminum pieces to make removal easier. However, welding would warp the aluminum. Another idea was to put screws through the aluminum but this would result in a rubber piece with holes/tears. A different idea was to remove the aluminum entirely and make the rubber much thicker. The center hole would still need to be reinforced though and this method may reduce the life of the flapper valve. Current implementation has added &quot;ears&quot; to the edge of the aluminum but as these sit flush against the mold, they are still near impossible to remove without damaging the mold.</p> <p>I am not an engineer nor am I familiar with this type of work so pointing me in the right direction, any thoughts, and any comments are appreciated; thank you!</p>

|design|manufacturing-engineering|valves|seals|injection-molding|

<p>This community is not dead, but it is assuredly populated by people (like me) who make a living by being paid to solve problems exactly like yours.</p> <p>Now, it is common for small design and manufacturing firms to design products and processes that they lack the expertise to troubleshoot when things go wrong on the factory floor. It is also common for them to lack the money to hire experienced troubleshooters to solve those problems. This is a fact of life...</p> <p>Anyway, here are some ideas.</p> <p>Review the mold design and be sure to include generous <em>draft angles</em> throughout, of order ~1.5 degrees per side.</p> <p>Consider including either a <em>stripper plate</em>, <em>ejector pins</em> or <em>ejector blades</em> to the mold design to pop the part out of the mold when finished.</p> <p>Machine the white mold base out of teflon.</p> <p>Buy you some industrial-grade <em>mold release compound</em> and spray it onto the mold surfaces before each run.</p> <p>Send out the aluminum lid piece for teflon coating. Most big cities have a teflon shop that resurfaces commercial cookware with fresh teflon. Seek one out.</p> <p>If you do not know what draft angles, mold release, stripper plates, ejector blades or pins are, you really have no business trying to run a process like this, and you should consider farming out the manufacture of the valve membrane to someone like Vernay Laboratories in Ohio.</p>

53702

Improving O-Ring/flapper valve injection mold ->non-engineer

2022-12-29T20:34:09.853

<p>I have one named constraint, R1, and now I want to define another constraint as R1-5. How can this be done? I been reading <a href="https://wiki.freecadweb.org/Expressions" rel="nofollow noreferrer">here</a> and I manage to make it work with product and division but not sum or subtraction:</p> <p><img src="https://i.ibb.co/zFZkqTT/Screenshot-from-2022-12-29-21-16-24.png" alt="Screenshot working" /></p> <p><img src="https://i.ibb.co/bQd2GwT/Screenshot-from-2022-12-29-21-07-16.png" alt="Screenshot fail" /></p> <p>As seen, I get an error when I try to use <code>+</code> or <code>-</code> operators. I am using FreeCAD 0.20.1 in Ubuntu 22.04.</p>

|cad|

<p>All you need to do is add units for the number! i.e. -5mm</p>

53705

How to express one constraint as a function of another with sum in FreeCAD

2022-12-29T20:59:47.790

<p>Is the curtain wall glass facade on the building in the following picture a <a href="https://en.wikipedia.org/wiki/Brise_soleil" rel="nofollow noreferrer">brise soleil</a>?</p> <p><a href="https://i.stack.imgur.com/P2dT7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P2dT7.jpg" alt="enter image description here" /></a></p>

|building-design|architecture|

<p>Referenced Wikipedia page defines Brise Soleil (sun breaker) as &quot;an architectural feature of a building that reduces heat gain within that building by deflecting sunlight&quot;.</p> <p>Image is Block 185, in Austin, Texas (ironically a Google building). <a href="https://earth.google.com/web/search/block+185,+austin/@30.26702921,-97.75225407,218.52148517a,605.31049591d,35y,-1.34290836h,88.02023491t,0r/data=CigiJgokCYU1yo_GRjxAEa0qrYalZi3AGVlSLhob3DvAIfNtDlqhhmHA" rel="nofollow noreferrer">Google Earth</a> shows building oriented to south. Wing shape is facing west-ish. Note building is on top of Shoal Creek, west of the building.</p> <p><a href="https://i.stack.imgur.com/EVo1j.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EVo1j.jpg" alt="enter image description here" /></a></p> <p>From <a href="https://austonia.com/google-tower-austin-skyline" rel="nofollow noreferrer">austonia: Architect involved in Austin's new Google tower outlines considerations that led to unique design:</a></p> <blockquote> <p>The previous land development code for the city required that any projects in that area up against a body of water like a creek, lake or river needed to have a slope down in the form of a 40-degree setback to preserve solar access.</p> </blockquote> <blockquote> <p>The design team also had to consider the Texas heat, which they handled with automatic shades plus solar sunscreens on the curved portion facing west.</p> </blockquote> <p>The shape has more to do with allowing as much sun access to Shoal Creek (to west) and Lady Bird Lake (to south) as possible. The sail is not a sun breaker and requires sunscreens on the sail and shades (with plants) on the south side of the building.</p>

53706

does this building feature a brise soleil?

2023-01-01T11:13:59.887

<p>I'm currently building a music instrument (string instrument), and I am at the step where one applies certain braces to the soundboard. Since I actually don't know how the bracing will affect the sound (or basically anything) I thought it would be a good starting point to model the soundboard and the applied braces with CAD, and then apply an FEM simulation that will show to me the effect of the braces.</p> <p>To my Background, I'm a physicist, but besides a course in technical mechanics 1 (statics), and some time selfstudying continuum-mechanics and the cauchy-stress-tensor, I don't have a background in engineering.</p> <p>The program I'm using can perform a modal analysis on the soundboard, and modal analyses are also is the predominantly used way to model e.g. violin bodys with FEM technology (for example in this publication]<a href="https://asa.scitation.org/doi/10.1121/1.3383977" rel="nofollow noreferrer">1</a>, or in this <a href="https://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=&amp;cad=rja&amp;uact=8&amp;ved=2ahUKEwjKpcHPkqb8AhW6R_EDHboICZ8QtwJ6BAgJEAI&amp;url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DAzo-UYfN4Hg&amp;usg=AOvVaw3exDVzF0c6OdogbE5Za4gP" rel="nofollow noreferrer">youtube video</a>).</p> <p><strong>But of what worth are those simulations actually?</strong> In the end, I roughly want to know how well the soundboard is able to transmit frequencies to the air that have been present in the strings before, that means how well an amplitude in a frequency in the strings will translate to an amplitude in air.</p> <p>A modal analysis will show me in what ways (and with what frequencies) the body will vibrate on its own (without external force applied). But it won't show me the amplitude of the soundboard (because it's abitrary): following (<a href="https://engineering.stackexchange.com/a/2576/15922">this stack exchange answer on modal analysis</a>, the amplitudes of the eigenmodes can have any value, similar to the length of an eigenvector that also can have any value.</p> <p>Granted, a periodic force will excite a vibrational modes amplitude the more if it meets its frequency, but <strong>doesn't it also play a role where this force is applied?</strong> When I apply period forces at a point on the soundboard where a certain mode of that soundboard has a node, I would expect the mode not to be excited at all for example.</p> <p>Additionally, I know from a simple harmonic oscillator that its amplitude, subject to an external driving force, will reach the maximum at its resonance frequency. Is the same true here? <strong>When I apply a driving force to the soundboard, will it be sufficient to check how it affects the eigenmodes of the soundboard, because this will be the strongest excitations anyway?</strong></p> <p>As an addendum, I am aware that I don't model the transfer from soundboard to air at all. For now, I only want to model the transfer from amplitudes in the strings to amplitudes in the soundboards vibrations.</p> <p>So to make the question short and concise: What can (and what can't) modal analysis tell me about the transfer of periodic motion from the bridge (where the strings are attached) to the soundboard? And since this information alone is probably not enough - When modelling the effect of external forces on the soundboard, will it be enough to only look at the frequencies of the previously found out eigenmodes?</p>

|finite-element-method|modeling|vibration|acoustics|modal-analysis|

<blockquote> <p>What can (and what can't) modal analysis tell me about the transfer of periodic motion from the bridge (where the strings are attached) to the soundboard?</p> </blockquote> <p>Very often in engineering, we find ourselves in a situation where getting the exact complete answer is difficult and time consuming. But often there is a simplified analysis method that gets us part of the answer quickly and easily. Often time part of the answer is actually enough to solve the problem, so we don't even need to bother with the complicated analysis.</p> <p>For example, in your problem, you really want to know transmissibility (and/or resonant amplification). A forced response analysis will tell you that, but it can be long and complicated (and often requires inputs that you might not even know, like the damping). A modal analysis, only tells you natural frequency and mode shape. But we know that resonant amplification is related to the frequency ratio between the excitation and the natural frequency. So we can use this to get part of the answer.</p> <p>E.g. let's say you are concerned with an excitation frequency of 440 Hz. If the natural frequency is 40,000 Hz then you don't even need to run a forced response analysis to know that the resonant amplification is basically nothing. Maybe it's 1.0001 and maybe it is 1.0002 but who cares. Further, if brace A gives you a natural frequency of 40,000 Hz and brace B is 41,000 Hz, then they are both who cares conditions, there is no reason to prefer one over the other. You can save yourself the hassle of running a forced response analysis.</p> <p>Now on the other hand, if the excitation frequency is 440 Hz, and the natural frequency is 430 Hz, well now you know that the resonant amplification could be significant. You don't know the exact answer until you run the full forced response analysis, but you know it is definitely more than nothing. And if brace A gives you 430 Hz and brace B gives you 40,000 Hz, well that alone might be enough to tell you to prefer one over the over, even without running forced response.</p> <p>In my work (not musical instruments but very much vibration), I use modal analysis as a first screening criteria. When I come up with a design, I first look at the natural frequencies. If they are not where I want then, then I start tweaking the design to move them around. When they start to get close to where I want them, then I start looking at forced response analysis.</p> <blockquote> <blockquote> <p>When I apply a periodic force at one point, is there a simple way to determine how mutch an eigenmode will be driven by that point? I would guess that the bigger the eigenmode vibrates at that point, the better the force couples to the eigenmode.</p> </blockquote> </blockquote> <p>Yes, that is it exactly. In mathematical terms, the response will be the dot product of an excitation force vector and the mode shape vector.</p> <p>Overall I think you might benefit from an undergraduate textbook in vibration theory. Rao could be good. The most recent edition is typical textbook expensive, but a used copy of the previous edition is reasonable: <a href="https://rads.stackoverflow.com/amzn/click/com/0132128195" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Mechanical-Vibrations-5th-Singiresu-Rao/dp/0132128195</a></p>

53732

Effect of bracing on a music instrument top - is FEM modal analysis enough to predict this?

2023-01-02T06:52:30.610

<p>I'm working on the following problem:</p> <blockquote> <p>Use a stainless-steel pipe with an internal diameter of 0.016m and a length of 2.5 m to conduct a convective heat transfer experiment. During the experiment, the stainless-steel tube is directly heated by direct current with a voltage of 5V and current of 900A, and the inlet temperature of the water is 20°C. The flow velocity is 0.5 m/s, the pipe is insulated externally, and the heat loss is ignored. Find the surface heat transfer coefficient and the temperature difference of the convective heat transfer in the tube.</p> </blockquote> <p>Since there is no width of the pipe, I can't use the specific heat of stainless steel to figure out the temperature of the pipe. I resorted to calculating the added temperature through the water's specific heat, and got 30.7 °C as the outlet temperature, but I'm not confident in this solution at all.</p> <p>How should I go about it? Thanks in advance.</p> <p>Edited to add my calculations: I used the Joule heating formula <img src="https://latex.codecogs.com/svg.image?Q=I%5E%7B2%7D%5Ctimes&space;R%5Ctimes&space;t=900%5E%7B2%7D%5Ctimes&space;0.005(5)%5Ctimes&space;%5Cfrac%7B2.5%7D%7B0.5%7D=22500&space;J" title="https://latex.codecogs.com/svg.image?Q=I^{2}\times R\times t=900^{2}\times 0.005(5)\times \frac{2.5}{0.5}=22500 J" />. Then, I divided Q by the specific heat capacity per the mass of water, so <img src="https://latex.codecogs.com/svg.image?%5Cpi%5Ctimes&space;0.016%5E%7B2%7D%5Ctimes&space;2.5%5Ctimes10%5E%7B3%7D%5Ctimes4186" title="https://latex.codecogs.com/svg.image?\pi\times 0.016^{2}\times 2.5\times10^{3}\times4186" />. The result was 10.69°C, so it would be 30.69°C for the final temperature.</p> <p>I wasn't sure if this was the correct way to go about this, since the problem seems to stress on the stainless-steel bit, so I was wondering if there is another way to do this more accurately.</p>

|mechanical-engineering|thermodynamics|heat-transfer|convection|

<p>I agree with your calculations. The temperature rise should be about 10.7 °C.</p> <p>The stainless steel aspect is mostly irrelevant other than it should ease any worries about rust, etc., affecting the result.</p> <p>Now, what's the answer to the other bits? I'd have to do some more thinking and research before I could answer confidently.</p>

53741

How do I get the temperature of a stainless steel pipe that's being heated by a direct current?

2023-01-02T11:13:50.823

<p>All solar collector and boiler tank systems that I can find on the market use a liquid to circulate the heat between solar collector panels (or PVT panels) and boiler tanks. Is there a reason why they do not use expandable gas and a compressor instead?</p> <p>I was looking for an air conditioning system that would not waste excessive heat during the summer, or at least use it for preheating tap water.</p> <p>This is what I came up with:</p> <p><a href="https://i.stack.imgur.com/tsrvd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tsrvd.png" alt="Combined heating cooling system" /></a></p> <p>Since I'm a software engineer, this idea will evidently be full of &quot;bugs&quot; that need to be ironed out. But let’s start with the first most obvious question: Why don't boilers use gas instead of water (or even worse glycerine where freezing may be an issue)?</p>

|hvac|solar-energy|photovoltaics|boilers|solar|

<p>Heat pumps don't magically generate extra heat from a source, they just move it around, specifically we use them to move heat energy in the direction opposite to where it would normally move (from cold to hot).</p> <p>When used for heating, they rely on the almost unlimited amount of heat available in the air as a source. They concentrate that heat into a small amount of fluid at a higher temperature. The total energy hasn't changed, just that some of it is now inside your house.</p> <p>Solar thermal panels are a limited source of heat energy that's already hot enough. If you added a heat pump, you'd get a smaller amount of fluid at a higher temperature but the same amount of energy.</p>

53744

Why isn't heatpump technology used for solar collector panels and boiler tanks?

2023-01-02T16:46:31.017

<p>Problem: I have a 98 lb box with dimensions of 45&quot;l x 41&quot;w x 27&quot;h that holds a holiday decoration. I need to lift it to a shelf I made that is 103&quot; above the floor. The space I have between the shelf and the ceiling is 32&quot;, so I have 5&quot; of clearance to get the box on the shelf. Second part of the problem is I am an EE trying to be an ME ;)</p> <p>At first, I was going to try a 100 lb bicycle/kayak lift that has two hooks and reinforce holes on the front and back of the box and lift it that way. That method may work, but it needs more than 5&quot; from the ceiling to the top of the box so I would also have to be under the box pushing it up and in.</p> <p>My other idea was this simple pulley system. The plan is that I can lift the box as high as possible up to the edge of the shelf with one person holding the line, and I push the edge of the box onto the edge, then have then loosen the line to push it the rest of the way. I'm looking for the best way to do this.</p> <ol> <li>Where to find the pulleys? I hope I can find a pulley for the ceiling joist that has a plate with holes for one bolt in front and one in back of the wheel to screw into a 2&quot; joist. And I need a pulley with an eye on the bottom to connect to the box.</li> <li>How to connect the box pulley to the box? Maybe a simple dual sling of a two ropes to go under the box? Or spreader bars from 2 x 4s?</li> <li>Other than the ceiling pulley being able to clear the box and have the ceiling eyelet being over the shelf to be able to push the box over the end of the shelf, how should I properly calculate the space between the eyelet and the ceiling pulley.</li> </ol> <p>The shelf is the brown rectangle on the left. The ceiling joist is oriented inline with the rope and pulleys, so the ceiling pulley would have a bolt to the left of the wheel and a bolt to the right of the wheel in the image.</p> <p><a href="https://i.stack.imgur.com/wZyUK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wZyUK.png" alt="enter image description here" /></a></p>

|mechanical-engineering|applied-mechanics|pulleys|

<p>In similar situations I have improvised a ramp. This makes things much safer imho.</p> <p>I did this to get solar thermal panels up about 15 feet. I used 4 scaffold boards and joined them 2 in series. The joint made for a stable and safe halfway point in the pulling process.</p>

53748

Can I use this simple pulley system to lift a large 100lb box?

2023-01-03T12:30:56.327

<p>I want a simple mechanism to rotate something 180 degrees when another thing moves linearly past/through it. The below is what I have, but it can only work with an offset, so the movement is never fully 180 degrees. No gears/belts etc., it must be a simple mechanism. Any ideas? <a href="https://i.stack.imgur.com/VmSao.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VmSao.png" alt="enter image description here" /></a></p>

|mechanical-engineering|mechanisms|linear-motion|

<p>You need two slots in 90 degrees angle on each other for the wheel<a href="https://i.stack.imgur.com/fF21u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fF21u.png" alt="enter image description here" /></a></p>

53762

Simple mechanism to convert linear motion into 180 degrees rotation

2023-01-03T22:19:42.857

<p>I have a set of time-series data that consists of inputs <span class="math-container">$u_k$</span> where <span class="math-container">$ u \in R $</span> and <span class="math-container">$k = 1 ... T$</span>, and outputs <span class="math-container">$ y_k $</span> where <span class="math-container">$ y \in R^2 $</span> and also <span class="math-container">$k = 1 ... T$</span>, from a given system. I believe this system can be modeled in discrete canonical state space form as <span class="math-container">$$ x_{k+1} = Fx_k + Gu_k $$</span> <span class="math-container">$$ y_{k+1} = Cx_{k+1} $$</span> In this case, <span class="math-container">$ y_{k} = x_{k} $</span> so the form becomes <span class="math-container">$$ y_{k+1} = Fy_k + Gu_k $$</span> Given all of the <span class="math-container">$ y_k $</span> and <span class="math-container">$ u_k $</span> I am fairly sure that I should be able to solve for <span class="math-container">$ F $</span> and <span class="math-container">$ G $</span>, but how do I actually do this? Some work on paper got me nowhere and I can't seem to find anything on the internet.</p>

|mathematics|linear-systems|linear-control|

<p>The answer to this turned out to be to frame the problem as a least-squares problem. Specifically I found &amp; used <a href="http://paginapessoal.utfpr.edu.br/avargas/courses-1/identificacao_sistemas/identificacao_sistemas/1_materialApoioStateSpace.pdf" rel="nofollow noreferrer">State Space Model Identification by Least Squares</a>. This allowed me to solve for the rows of <span class="math-container">$F$</span> and <span class="math-container">$G$</span> one-by-one.</p>

53766

How to solve for discrete state space matrices given input and output

2023-01-04T16:18:44.460

<p>I have confusion on MRC (Maximal Ratio Combining) method, which I would like to clarify. An example is where I have 1 transmit antenna and 4 receive antennas. The concept of MRC is to combine all the signals from the 4 receive antennas such that the combined SNR is increased (given the phase is corrected). To achieve this, a weight is computed for each antenna branch, such that it weights the received signal in proportion to the signal strength.</p> <p>Once the MRC process is completed, are we required to equalize the channel effect with some equalizer (for e.g. ZF or MMSE). Most of the online materials refer to MRC as a equalizer, which is causing confusion. With my understanding, we should equalize, as the MRC is only increasing the SNR but not inversing the channel effect.</p> <p>If we require to use an equalizer (ZF / MMSE) after combining, which channel coefficient should we use? Because once the received symbols are combined, we just have a vector (with increased SNR). But the channel coefficients are different between 1 tx antenna and each 4 rx antenna.</p>

|wireless-communication|digital-communication|

<p>Combining signals can improve SNR as you have said. Equalization on the other hand reduces the effect of distortion, which could be deterministic, in which case combining signals would not improve the transmission as the effect would be the same in all the received signals. So yes you are right, you could have a situation where it is beneficial to both equalize and combine signals.</p> <p>For the channel coefficients, you would have to estimate the channel coefficients for a specific application, in the same way as you would estimate them if you weren't doing any combining: for instance this could either be by explicit physical modelling of the transmission medium, or experimentally by applying a test signal and fitting a model.</p> <p>I have borrowed from the answer to this almost identical question: <a href="https://www.dsprelated.com/thread/4790/confusion-with-maximal-ratio-combining-mrc-in-simo-ofdm" rel="nofollow noreferrer">Confusion with Maximal Ratio Combining(MRC) in SIMO OFDM</a>. There's some discussion about the order of combining/equalization too which might be of interest, as this has implications for estimating the channel coefficients.</p>

53772

Should we use Maximal ratio combining and an equalizer together?

2023-01-05T06:39:53.787

<p>What are the limitations of trains that never stop. Can we have shuttles decelerating from the train's speed to 0km/h at the station for alighting passengers, while another shuttle picks up boarding passengers from the station, and accelerates to the train's speed to catch up? (Even if this is only done for a few stops on the whole line, and the main train still has to decelerate slightly, I imagine it will still lead to a fair bit of time saved, especially for metro.)</p> <p>This seems to be a 'vernacular' idea in the minds of some. I was reminded of it when reading a <a href="https://www.carscoops.com/2008/06/train-that-never-stops-at-stations/" rel="nofollow noreferrer">vague pitch by Peng Yu-lun</a>. There must be many reasons this hasn't worked/won't work that I am missing.</p> <p>I'm personally most curious about this in the context of urban metros, though it could be relevant to longer-distance rail and HSRs too.</p> <p>(First time asking a question here, hope it's not off-topic/too unspecific . Personally I wonder about this whenever I'm taking the subway. I'm from Singapore if you'd like to provide additional context.)</p> <p>Edit: sorry for my slow updates. Thanks for the comments. I never managed to find the phrase 'slip coach' when googling, that helps a lot. Transistor's answer/other answers mentioning safety (if there's a crash) and infra/cost (platform) have been enlightening. I promise to read up and mark/accept an answer eventually/soon.</p>

|rail|

<p>I worked several decades ago as a railway signal engineer. We used to say that everyone would like an express train that ran non-stop from their origin to destination but, of course, that doesn't suit everybody else on board. Your question seems to try to address the problem.</p> <p>Using self-powered multiple units coupled together would offer a theoretical solution on normal rails.</p> <pre><code>Trains move from left to right. 1 [d][c][b][a] Approach [stn] 2 [d] [c][b][a] Decouple [stn] 3 [d] [c][b][a] Stop [stn] 4 [h][g][f][e] [d] Next train [stn] d starts off 5 [h] [g][f][e] [d] Transition [stn] 6 [h] [g][f][e][d] Recoupled [stn] e, f &amp; g couple up with d. </code></pre> <p>How it works:</p> <ul> <li>All passengers for the next station move to the last car or set of cars. (This requires connecting corridors.)</li> <li>Corridor doors are closed.</li> <li>At normal braking distance from the station the last set is decoupled. The front section runs on at line speed. The decoupled set brakes to a halt at the station.</li> <li>Passengers disembark and the set sits there (blocking the line for other traffic) until the next express is due.</li> <li>The set accelerates out of the station on time so that it will be at express speed when the express reaches its rear. The express couples up, the passengers don't feel a thing and the journey continues.</li> <li>Meanwhile the last set of the express comes to a halt at the station and the procedure repeats.</li> </ul> <p>There are multiple problems with this scheme.</p> <ul> <li>Any delay or mishap with the shuttle acceleration may result in an emergency stop of the express or a fatal collision.</li> <li>Every station along the express route will have a train set sitting idle for most of the time.</li> <li>Every station along the route will have to have platforms on the express line. This platform will be blocked for most of the time.</li> </ul> <p>There will be many more problems. I don't think it's a runner.</p>

53780

Trains that do "never" have to stop

2023-01-05T11:28:05.777

<p>I have a 5.5 kW motor that I'm using for testing. I tried using it to lift a 15lb weight but it struggled. I reduced the amount of rope on the spool (there was a lot and then it was able to lift it at about 1ft per second. However, it was taking in 100 amps to do so!</p> <p>The setup is drawn in a link below, it's like this:</p> <p>The motor is attached to a spool that has rope, the rope goes about 2-3 meters up into a pulley and comes back down. The rope is then tied to a 15lb dumbbell. The spool diameter is 50mm (25mm radius) and the motor is rated for 10Nm of torque.</p> <p>The rope effectively makes a bigger spool when wrapped, which comes to about 120mm diameter when the motor couldn't lift the weight. I left most of the rope off to retest when it finally lifted it.</p> <p>Now, using the formulas and what I understood from here: <a href="https://engineering.stackexchange.com/questions/2826/understanding-required-torque-for-a-motor-lifting-a-weight">Understanding required torque for a motor lifting a weight</a></p> <p>I calculated that if I wanted to move 15lb (6.8kg) at 2m/s:</p> <p><em><strong>6.8 * 9.8m/<span class="math-container">$s^2$</span> = 66.64N</strong></em></p> <p><em><strong>6.8 * 2m/<span class="math-container">$s^2$</span> = 13.6N</strong></em></p> <p><em><strong>Torque Needed = (66.64 + 13.6) * 0.025 = 80.24N * 0.025 = 2Nm (The 0.025 is the spool radius of 25mm converted to meters)</strong></em></p> <p>The battery output was 44 V and the motor took 100 amps to move the weight, totaling to 4400W. I calculated that moving this weight should take less than 200W. The motor is rated for 10Nm of torque, yet according to the formula it only needs 2Nm! I may hook up a stick to the motor and use a scale to directly measure it's torque output.</p> <p>Is there something wrong in my setup? Do these formulas not apply the way I'm using them? Any insight would be helpful, thanks!</p> <p>The motor is connected to one end of the spool. From end to end, the spool is about 80mm long.</p> <p><strong>After Testing:</strong></p> <p>Just did some testing. With very little rope on the spool, it can match gravity and barely pick it up at around 60amps. As it takes more amperage, it starts accelerating the weight. It can just barely start picking up the weight with all the rope around the spool.</p> <p>However, even with the spool filled to the brim, with 120mm diameter, the math says:</p> <p><em><strong>(66.64 + 13.6) * 0.025 = 80.24N * 0.06 = 4.8Nm</strong></em></p> <p>That's still only half of what the 10Nm specification of the motor!</p> <p>Definitely how much rope is on the spool is affecting the torque needed. However, does the distance the rope is from the motor (whether at the same end or the opposite end of the spool as opposed to the motor) play a role? Is there a way to incorporate the distance of the rope from the motor into the calculations?</p> <p>Would supporting the spool at the other end help with this?</p> <p><a href="https://i.stack.imgur.com/AGmdg.png" rel="nofollow noreferrer">Setup Drawing</a></p>

|motors|torque|

<p>I directly measured the torque of this motor. It seems that while rated for 10Nm, it's only able to output around 4Nm. This nearly perfectly correlates well to the equations.</p> <p>It seems that while the site claims 10Nm, they are also selling 3 different versions of the motor. All 3 versions are given the same spec sheet with the only difference being their KV ratings. I have the highest KV rated motor, meaning the lowest KV rated motor is likely the one able to produce the 10Nm of torque specified. It's very disappointing that they would slap the 10Nm label on all 3 versions when it's clearly not true and are all rated for 5kW.</p> <p>Everyone's help was appreciated. It's good to know that my methodology was not flawed.</p>

53783

Motor capable of 10Nm can barely lift 15lb

2023-01-06T16:27:30.080

<p>We have a claw machine that grabs a bunch ping pong balls that then drops them onto an incline that leads them to a 2 inch in diameter pvc pipe that they must all fall through.</p> <p>The issue is that when dropping several ping pong balls at once instead of falling into the hole multiple balls will block each other from the entrance and they are so light gravity doesn't seem to help.</p> <p>The balls need to be light as they are because they are shot through a leaf blower after falling through the tube.</p> <p><a href="https://i.stack.imgur.com/wbJ0c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wbJ0c.jpg" alt="enter image description here" /></a></p> <p>We've tried various shapes of inclines that funnel to the pipe. For example a bowl with a hole in the middle but that seemed to cause the same problem.</p> <p>We are currently at a convention showing this so a solution that would require just a trip to a nearby cvs would be ideal.</p> <p>Anyway we can make it so the balls don't jam themselves into the entrance?</p>

|pumps|

<p><a href="https://i.stack.imgur.com/L2SDw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L2SDw.png" alt="enter image description here" /></a></p> <p><em>Figure 1. Option 1.</em></p> <p>Try a baffle on the right. This should be a little less than the radius of the ping-pong ball so that balls approaching from the right will roll away from the hole and fall into the gulley some distance from the hole.</p> <p>I suspect that the oversized rectangular opening is not helping. Try making it a little larger than required by the ball.</p> <p><a href="https://i.stack.imgur.com/0Yixx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Yixx.png" alt="enter image description here" /></a></p> <p><em>Figure 2. Option 2.</em></p>

53793

How to prevent ping pong balls from clogging the entrance to a tube they must be able to individually pass through

2023-01-07T03:29:20.750

<p>In this video(<a href="https://youtu.be/QvLdVWS_DsM" rel="nofollow noreferrer">https://youtu.be/QvLdVWS_DsM</a>) it says that a cord lock works like this: there is a cylinder which the string moves around, and there is another cylinder which is rough. When you pull out and let go, the string catches on the rough cylinder. The cylinder and string both go up, and the rough cylinder stops when the groove ends, and pinches the string so it stops.</p> <p><a href="https://i.stack.imgur.com/p6SfT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p6SfT.png" alt="a picture showing the above" /></a></p> <p>But then how it unlocks is unclear to me. It says you have to pull to the center of the window horizontal side, so the rope lets go. But I don't understand why you need to pull to the center.</p> <p>Could someone please explain? Thanks!</p>

|mechanisms|pulleys|product-engineering|

<p>Here is a picture of one in my blinds.</p> <p>The serrated roller is the movable one.</p> <p>The cord is nearly vertical and it touches the serrated roller.</p> <p>The cord will lock if released in this orientation, and has to be held towards the right to prevent locking.</p> <p><a href="https://i.stack.imgur.com/ZdHE8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZdHE8.jpg" alt="enter image description here" /></a></p>

53797

How A Cord Lock in A Window Unlocks

2023-01-07T21:01:41.900

<p>I have a motor with an 8mm shaft diameter. I'm trying to make my own hub for direct mount driving a wheel. What diameter hole size should I drill for the hub's shaft which couples with the motor shaft? Would these be the same standards as clearance hole drillings for screws? For example, an M3 screw has a suggested close fit drill size of 3.2mm and a normal fit drill size of 3.4mm. Should I drill an 8.2mm diameter then?</p>

|mechanical-engineering|motors|machining|drilling|

<p>You probably want a sliding fit or transition. 0.2mm clearance is far too much, H7g6 sounds about right, hole is 0.005-0.029 mm bigger than the shaft. You'll need to use a reamer or a D bit.</p>

53809

What should be the clearance hole size for mounting a hub on a motor shaft?

2023-01-09T21:08:55.313

<p>If a cable is attached to a support on each side, and there is a load acting on it somewhere along its length.</p> <p>Is there going to be a vertical reaction force only if the the cable is angled?</p>

|mechanical-engineering|structural-engineering|structural-analysis|structures|mechanical|

<h3>The reaction force is a vector in the direction (or opposite) of the cable.</h3> <p>Note that the only way to have a horizontal cable is by having the other support higher. But if you do, a horizontal vector will resolve into only a horizontal force. The sin of 0 is zero, so an angle of zero will yield zero vertical force.</p>

53828

What will the reaction forces of a support be in case of an angled cable

2023-01-10T07:32:15.973

<p>I made this problem to better understand how supports react to tension forces.</p> <p>Here we have a weight Mg attached to a cable, the left side of the cable is angled at 10 degrees, while the right side is horizontal. The cable is supported by L2 and L1</p> <p>Now what I am interested in knowing is how the supports L2 and L1 will react. I am assuming there will be no bending moments at either L2 and L1 as this is a cable.</p>

|mechanical-engineering|structural-analysis|

<p>Assuming that the posts are rigid (non deformable) and the rope is weightless, then pole L1 should experience only horizontal forces at the top (which will be converted to shear forces for the pole).</p> <p>Additionally you are right that :</p> <ul> <li>there will be no bending moments transferred (because its a rope)</li> <li>the force on post L2 from the rope will be forming a 10<span class="math-container">$^o$</span> (clockwise) with the horizontal axis, and its vertical component should be equal to <span class="math-container">$Mg$</span>. i.e. <span class="math-container">$$F_{L2} \sin(10^o) = Mg $$</span></li> </ul> <p>therefore: <span class="math-container">$$F_{L2} = \frac{Mg}{\sin(10^o)} $$</span></p> <p>and also for the force on the pole L1 (<span class="math-container">$F_{L1}$</span>) the magnitude will be:</p> <p><span class="math-container">$$F_{L1} =F_{L2}\cos(10^o) \rightarrow F_{L1} =\frac{Mg}{\sin(10^o)}\cos(10^o) \rightarrow \boxed{F_{L1}= \frac{Mg}{\tan(10^o)}}$$</span></p> <p>And the direction will be towards the left.</p>

53835

Tension forces in a cable and reaction at the supports

2023-01-10T14:18:29.400

<p>In a Newton's cradle, just like the one from the picture, where will the load of the spheres be applied on the right-hand beam?</p> <p>Will it be a transverse load? The wire is connected laterally, not at the top or bottom of the beam, which is why I am wondering.</p> <p><a href="https://i.stack.imgur.com/qwrCW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qwrCW.jpg" alt="Newton's cradle" /></a></p>

|mechanical-engineering|dynamics|

<p>For a simple pendulum with a swing angle maximum of <span class="math-container">$\theta_m$</span>, the tension in the string caused by the sphere is the sum of two forces, centripetal and the component of the weight of the sphere along <span class="math-container">$\theta$</span> :</p> <p><span class="math-container">$$ T= rmv^2+mgcosθ $$</span></p> <p>By conservation of energy, we have <span class="math-container">$ \ mg\Delta h=1/2mv^2 :$</span> <span class="math-container">$$\frac{1}{2}mv^2=mg(cos\theta- cos\theta_m)$$</span> <span class="math-container">$$T=r2mg(cos\theta-cos\theta_m)+mgcos\theta$$</span></p> <p>In your case, this tension is shared by the two strings and has two components on the right-hand beam. one on the vertical plane <span class="math-container">$T_v$</span> along angle <span class="math-container">$\theta $</span> and one on the horizontal plane <span class="math-container">$T_h$</span>. Let's call the angle of the string from the vertical plane of the right beam <span class="math-container">$\phi.$</span> <span class="math-container">$$T_v= \frac{T}{2sin\phi}$$</span> And <span class="math-container">$$T_h=\frac{T}{2cos\phi}$$</span></p>

53837

Where will the load be applied on the beam

2023-01-11T17:43:44.247

<p>If i mount a torque flange so that it connects 2 axles. Does it measure the difference between the torque applied to each axle?</p>

|torque|

<p>I assume you refer to something like the following:</p> <p><a href="https://i.stack.imgur.com/cOVX2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cOVX2.png" alt="enter image description here" /></a></p> <p>This is a torque measurement device that measures the torque which is carried by each shaft. I.e. assuming</p> <ul> <li>you have two shafts connected with the torque flange and</li> <li>if one shaft is driven by a motor with <strong>torque M</strong> on one end</li> <li>and the other shaft is driving a generator (or a brake)</li> </ul> <p>Then if the system is not accelerating or decelerating rotationally, the torque sensor should read torque M (i.e. all torque supplied by the motor will &quot;go through&quot; the torque flange -- and the shafts for that matter-- and will be consumed at the generator/brake).</p> <p>If on the other hand the system is increasing or decreasing the rpm, then the torque sensor will only read part of the torque (and with a lot of oscillations) -- the part of the torque that will go through the flange has to do with the acceleration of the rotational masses past the torque flange.</p> <hr /> <p>In general the torque flanges usually use a strain gage principle. I.e. when torque passes through the flange it create a twisting angle. That angle can be measured and because of the calibration of the torque flange sensor it is possible to know what is the torque &quot;passing through&quot; the sensor.</p>

53850

What does a torque flange measure?

2023-01-12T13:56:06.257

<p>With CSTR (continuous stirred-tank reactor), it is in many cases benificial to connect multiple CSTR in series. I wonder if there is a purpose for tubular reactors to be connect in series as well?</p>

|chemical-engineering|

<p>As you may have thought, connecting tubular reactors in series could easily be replaced by a larger reactor. But, sometimes, these reactors are sold in modules and you can add more or less modules according to demand. I could use as an example series of adsorption columns or membrane separators.</p>

53862

Is there a reason to connect tube reactors in series?

2023-01-12T20:18:06.403

<p>I live in a tall, older apartment building with a modest workout area. There is a smith machine loaded with about 500 lbs of weight. The floors are padded with a rubbery material. I have deadlifted in this room, infrequently, and being mindful of how I set the weights down. However, I usually stay between 100-200 lbs as I have not had a conversation with the manager regarding the load that this room can structurally support. What I am trying to say is that, the room seems to have been reinforced, or at least, cleared for the load associated with most gym activities.</p> <p>I would like to hang a ~40 lbs punching bag in this room from the smith machine. However, I don't want to contribute to significant structural aging of the building. I have a really poor understanding of physics, and load. I am wondering whether, the bouncing of a punching bag, could be significantly more load than the smith machine that is already in use. I am thinking a 35-40 lbs punching bag, perhaps, bungeed to a weight on the floor to avoid excessive swing. Obviously, the final word should come from the building manager regarding any kind of consultation that has been done about this room in the building. I am just trying to get a general idea.</p>

|structural-engineering|

<p>If a 40 lb bag has a noticeable effect on the structural stresses in your building, you need to run out of the building right now, cause it will collapse at any moment.</p> <p>In seriousness, in a building that meets US code, it is a non-event.</p>

53866

Structural load from swinging (punching) bag

2023-01-13T21:48:57.250

<p>I have never built anything with gears before, and I find myself needing to do just that. Given that I have very little background, I'm mostly looking to be pointed in the right direction.</p> <h2>The Task</h2> <p>I need to design and build a rotation stage that can both rotate freely in both directions, while simultaneously being able to be elevated through ~0.5 m. Both motions should be driven independently, and I cannot put a motor in the room the platform will live in. The motors need to be about 5 m away, in a different room, around several corners. So, I figured this sounds like a job for a drive shaft (or maybe belts/chains, although I am trying to avoid as much position uncertainty as possible). The actual power requirement is low --- no more than a few kg on the platform as it's being moved. Pardon the potato drawing.</p> <p><a href="https://i.stack.imgur.com/OBvc5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OBvc5.jpg" alt="A rotating, elevating platform" /></a></p> <h2>The Problem</h2> <p>The problem I keep running into is how do I keep the rotational motion system working as the platform is moved through its full linear range. Doing either alone is straight forward --- bevel gears or a worm for the rotation, or a rack and pinion for the translation. My problem is getting them to be able to do both, without the translation uncoupling the rotation system.</p> <h2>Help</h2> <p>I can't imagine that this isn't a solved problem, somewhere. I just don't know where to look for the answer. I've put a few days into Googling and, while I am now much more comfortable with the types of gears out there, I haven't found anything that starts to explain how I should proceed with combining elements. Any resources people know that could help, or places to look at similar designs would be very helpful.</p>

|design|gears|machine-design|power-transmission|

<p>I suspect that your answer is a well-lubricated splined shaft. The splined shaft is the translation shaft, while the rotation is a matching internally-splined gear being turned by the rotation input.</p> <p>The translation input as a rack and pinion will apply the vertical force from below the translation shaft, with an appropriate surface to mate with lubrication or a bearing.</p> <p>Created via Tinkercad and a combination of graphic software packages to create the gearing portion:</p> <p><a href="https://i.stack.imgur.com/JqT7S.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JqT7S.jpg" alt="mockup of lift and rotate" /></a></p>

53873

A rotating, elevating platform driven by drive shafts

2023-01-14T11:10:31.160

<p><a href="https://i.stack.imgur.com/kbYWs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kbYWs.png" alt="Diagram" /></a> I am doing an analysis on a shaft loaded by a gear at point B and supported by three bearings at points A, C and D. Is it possible to solve for reaction forces in the vertical direction for points A, C and D knowing the vertical component of B is given?</p>

|mechanical-engineering|bearings|

<p>It is possible with some simplification, which is however generally the case when modelling a real situation. Provided that the supports are basically pinned (unrestrained rotations) and deformations are small, you can use FEA like approach. Although there may be simpler analytical solution, FEA approach does not require much thinking, you just need to be precise with stiffness coefficients and it is practical to use computer for the system of equations :).</p> <h1>FEA beam element</h1> <p>A beam is governed by set of 4 linear equations (for example from <a href="https://people.duke.edu/%7Ehpgavin/cee421/beam-element.pdf" rel="nofollow noreferrer">here</a>):</p> <p><span class="math-container">$$K\cdot U = F$$</span></p> <p>or</p> <p><span class="math-container">$$EI\left[\begin{matrix} \frac{12}{L^3} &amp; \frac{6}{L^2} &amp; -\frac{12}{L^3} &amp; \frac{6}{L^2} \\ \frac{6}{L^2} &amp; \frac{4}{L} &amp; -\frac{6}{L^2} &amp; \frac{2}{L} \\ -\frac{12}{L^3} &amp; -\frac{6}{L^2} &amp; \frac{12}{L^3} &amp; -\frac{6}{L^2} \\ \frac{6}{L^2} &amp; \frac{2}{L} &amp; -\frac{6}{L^2} &amp; \frac{4}{L} \end{matrix}\right]\cdot \left[\begin{matrix} w_i \\ \varphi_i \\ w_j \\ \varphi_j \end{matrix}\right] = \left[\begin{matrix} F_i \\ M_i \\ F_j \\ M_j \end{matrix}\right]$$</span></p> <p>where <span class="math-container">$w_i$</span>, <span class="math-container">$\varphi_i$</span>, <span class="math-container">$w_j$</span> and <span class="math-container">$\varphi_j$</span> are transverse displacements and rotations at beam endpoints <span class="math-container">$i$</span> and <span class="math-container">$j$</span> with corresponding forces and moments <span class="math-container">$F_i$</span>, <span class="math-container">$M_i$</span>, <span class="math-container">$F_j$</span> and <span class="math-container">$M_j$</span></p> <h1>System of 3 elements</h1> <p>For your situation, you will need 3 interconnected elements, which is also governed by a system of linear equations:</p> <p><span class="math-container">$$K_s\cdot U_s = F_s$$</span></p> <p>With lengths of the beams denoted as <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> and stiffness coefficients (shown here just for <span class="math-container">$a$</span>):</p> <p><span class="math-container">$$k_{12, a} = 12\frac{E_a I_a}{L_a^3}$$</span> <span class="math-container">$$k_{6, a} = 6\frac{E_a I_a}{L_a^2}$$</span> <span class="math-container">$$k_{2, a} = 2\frac{E_a I_a}{L_a}$$</span></p> <p>The system of equations would look like this:</p> <p><span class="math-container">$$\left[\begin{matrix} \color{blue}{k_{12, a}} &amp; \color{blue}{k_{6, a}} &amp; \color{blue}{-k_{12, a}} &amp; \color{blue}{k_{6, a}} &amp; &amp; &amp; &amp; \\ \color{blue}{k_{6, a}} &amp; \color{blue}{2k_{2, a}} &amp; \color{blue}{-k_{6, a}} &amp; \color{blue}{k_{2, a}} &amp; &amp; &amp; \\ \color{blue}{-k_{12, a}} &amp; \color{blue}{-k_{6, a}} &amp; \color{blue}{k_{12, a}}+k_{12, b} &amp; \color{blue}{-k_{6, a}}+k_{6, b} &amp; -k_{12, b} &amp; k_{6, b} \\ \color{blue}{k_{6, a}} &amp; \color{blue}{k_{2, a}} &amp; \color{blue}{-k_{6, a}}+k_{6, b} &amp; \color{blue}{2k_{2, a}}+2k_{2, b} &amp; -k_{6, b} &amp; k_{2, b} &amp; &amp; \\ &amp; &amp; -k_{12, b} &amp; -k_{6, b} &amp; k_{12, b}+\color{green}{k_{12, c}} &amp; -k_{6, b}+\color{green}{k_{6, c}} &amp; \color{green}{-k_{12, c}} &amp; \color{green}{k_{6, c}} \\ &amp; &amp; k_{6, b} &amp; k_{2, b} &amp; -k_{6, b}+\color{green}{k_{6, c}} &amp; 2k_{2, b}+\color{green}{2k_{2, c}} &amp; \color{green}{-k_{6, c}} &amp; \color{green}{k_{2, c}} \\ &amp; &amp; &amp; &amp; \color{green}{-k_{12, c}} &amp; \color{green}{-k_{6, c}} &amp; \color{green}{k_{12, c}} &amp; \color{green}{-k_{6, c}} \\ &amp; &amp; &amp; &amp; \color{green}{k_{6, c}} &amp; \color{green}{k_{2, c}} &amp; \color{green}{-k_{6, c}} &amp; \color{green}{2k_{2, c}} \end{matrix}\right]\cdot \left[\begin{matrix} w_a \\ \varphi_a \\ w_b \\ \varphi_b \\ w_c \\ \varphi_c \\ w_d \\ \varphi_d \end{matrix}\right] = \left[\begin{matrix} F_a \\ M_a \\ F_b \\ M_b \\ F_c \\ M_c \\ F_d \\ M_d \end{matrix}\right]$$</span></p> <h2>Boundary conditions</h2> <p>The system of equations needs to be modified using boundary conditions, so all the unknowns are in the left vector and the right vector is completely known:</p> <p><span class="math-container">$$K_{s,mod} \cdot U_s = F_{s, mod}$$</span></p> <p>There is a known force <span class="math-container">$F_b$</span> at point B, which is reflected in the right vector. There are also known lateral displacements <span class="math-container">$w_a$</span>, <span class="math-container">$w_c$</span> and <span class="math-container">$w_d$</span>, which are all 0 and this is most easily implemented by replacing corresponding 3 equations by <span class="math-container">$1\cdot w_i = 0$</span>.</p> <p><span class="math-container">$$\left[\begin{matrix} 1 &amp; &amp; &amp; &amp; &amp; &amp; &amp; \\ \color{blue}{k_{6, a}} &amp; \color{blue}{2k_{2, a}} &amp; \color{blue}{-k_{6, a}} &amp; \color{blue}{k_{2, a}} &amp; &amp; &amp; \\ \color{blue}{-k_{12, a}} &amp; \color{blue}{-k_{6, a}} &amp; \color{blue}{k_{12, a}}+k_{12, b} &amp; \color{blue}{-k_{6, a}}+k_{6, b} &amp; -k_{12, b} &amp; k_{6, b} \\ \color{blue}{k_{6, a}} &amp; \color{blue}{k_{2, a}} &amp; \color{blue}{-k_{6, a}}+k_{6, b} &amp; \color{blue}{2k_{2, a}}+2k_{2, b} &amp; -k_{6, b} &amp; k_{2, b} &amp; &amp; \\ &amp; &amp; &amp; &amp; 1 &amp; &amp; &amp; \\ &amp; &amp; k_{6, b} &amp; k_{2, b} &amp; -k_{6, b}+\color{green}{k_{6, c}} &amp; 2k_{2, b}+\color{green}{2k_{2, c}} &amp; \color{green}{-k_{6, c}} &amp; \color{green}{k_{2, c}} \\ &amp; &amp; &amp; &amp; &amp; &amp; 1 &amp; \\ &amp; &amp; &amp; &amp; \color{green}{k_{6, c}} &amp; \color{green}{k_{2, c}} &amp; \color{green}{-k_{6, c}} &amp; \color{green}{2k_{2, c}} \end{matrix}\right]\cdot \left[\begin{matrix} w_a \\ \varphi_a \\ w_b \\ \varphi_b \\ w_c \\ \varphi_c \\ w_d \\ \varphi_d \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \\ F_b \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right]$$</span></p> <p>When you solve this modified system of equations, you will get the whole vector <span class="math-container">$U_s$</span>. Finally, you can simply calculate the force and moment vector <span class="math-container">$F_s$</span> just by multiplying system stiffness matrix <span class="math-container">$K_s$</span> by the displacement vector <span class="math-container">$U_s$</span>:</p> <p><span class="math-container">$$F_s = K_s\cdot U_s$$</span></p> <p>For your situation, the resulting forces should be: -12329.700012800344, 22682.5, -57303.35631973247, 46950.556332532804.</p> <p>Although you need shaft cross section for this approach, the resulting forces should be independent of it (provided the cross section is constant for the whole shaft).</p>

53879

Vertical reaction on a shaft with 3 bearing supports

2023-01-14T13:55:26.933

<p>Say we have ten cameras surrounding an object. The object changes its color with a certain frequency (or any information other than color, such as a text on it). The point is if all cameras are triggered simultaneously, they will capture their frame such that the object in all those frames has the same color in it (or the same text or information).</p> <p>For example, if the text on the object changes with a frequency of 1 Hz, it is obviously possible that we trigger the cameras at the same time and see the same text in all frames. However, if the frequency of the change of that text on the object changes, my question is to what realistic extent can we obtain the same information/text from all frames.</p> <p>P.S., An additional question would be if this possiblity would hold if the trigger is made wireless (assuming all the cameras have the same distance from the wireless transmitter).</p> <p>---edit:</p> <p>All cameras look at a set of LEDs (3 or 4 ones) which are very small but their intensity can be controlled by myself so exposure time can be small (should be very small). But the LEDs move/vibrate and my goal is that the frames basically are from the same position instance. My other intuitive example is if a text is changing with high frequency and all frames have the same text if taken simoultaneously.</p>

|optics|signal-processing|frequency-response|metrology|

<p>Regardless of the data interface, many cameras offer hardware triggering (sometimes software configured). Once you've got a hardware trigger you can wire whatever you like up to it, crucially connected in parallel.</p> <p>I've used hardware triggering down to microsecond-scale repeatability and jitter on Basler Ace cameras, for example. I would expect several cameras of the same model to trigger at the same speed, and with some cameras you can calibrate the trigger delay. Some models in the range go down to 1mus exposure times, all or most have hardware triggering. There are rival makes, but I can't recall the other one I used with similar specs, and I'm working on slower cameras these days.</p> <p>I've also built <a href="https://github.com/ChrisHodgesUK/mtrigger" rel="nofollow noreferrer">hardware triggers for DSLRs</a>, but never tested their timing accuracy. I'd expect millisecond scale or quite likely far better, but ~millisecond exposure times are as fast as you'll get with those.</p> <p>For wireless triggering, there may be off-the-shelf solutions, but in a controlled environment I'd look at triggering on a simple infrared pulse with a phototransitor-based receiver (possibly filtered optically) per camera and one transmitter with several LEDs- just be sure not to pick up the trigger pulse on the cameras. Timing accuracy form propagation delays isn't a problem with µs-scale timing as that would be hundred of metres at light speed</p>

53884

How simultaneous can we trigger multiple cameras' frame acquisition

2023-01-14T13:57:40.167

<p>In this video(<a href="https://www.kidzsearch.com/kidztube/watch.php?vid=65fbcb9ca" rel="nofollow noreferrer">https://www.kidzsearch.com/kidztube/watch.php?vid=65fbcb9ca</a>), at timestamp 0:27, it says that in mechanical pencils, the fingers have a taper in them that results in them flying out horizontally, unclamping the graphite.<br /> I am confused about why this happens. Could someone please explain? Thanks!</p>

|mechanical-engineering|product-engineering|forces|

<p>The plastic fingers are moulded in the open position. During assembly and operation they are compressed.</p> <p><a href="https://i.stack.imgur.com/B7C1d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B7C1d.jpg" alt="enter image description here" /></a></p> <p>I imagine that the fingers are moulded so that with the ring in the position shown they are in pre-tension. i.e., They are not fully open so that they still have some opening force in reserve.</p>

53885

Why Do the Fingers in Mechanical Pencils Fly Out?

2023-01-18T02:57:37.393

<p><a href="https://i.stack.imgur.com/QJ0Mx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QJ0Mx.jpg" alt="Curvature line of face and flank of a spur gear" /></a></p> <p>I have difficulty to understand the profile of a spur gear, the part I put in a red box line inside a box, line number 2 in the picture, or the part number 6 and 7. How it is determined? I read or watched many about spur gear. When talking about spur gear, many times there are involute also mentioned. But so far I don't understand what is its relation to gear profile?</p> <p>How is a spur gear profile is determine?</p>

|gears|

<p>You could device other gear shapes, in practice only involutes (of a circle) and cycloidal gears are used. However, there are a lot of shape modifications to gears you can make making the subject complex indeed.</p> <p>The reason why you find the shape hard to understand is because its not one of the commonly used basic shapes (line, circular arc) nor one of the easy graphs of one dimension (paraboloa...) that usually makes up of most designs.</p> <p>Instead the shape is the involute of a circle. Easiest way for you to make it is to tie a pen on a string around a round object and keeping the string tight as you draw the curve. This is the involute for that particular circle. This makes the curve quite simple in cartesian vector form. Its essentially a rotating vector + a perpendicular vector to that that is as long as the distance traveled. *</p> <p><img src="https://i.stack.imgur.com/ppXC6.gif" alt="enter image description here" /></p> <p><strong>Image 1:</strong> Formation of the involute shape.</p> <p>This in mathematical parametric terms forms a function for x and y as for example as follows, you get the following parametric function:</p> <pre><code>x = r * sin(t) - r * t * cos(t) y = r * cos(t) + r * t * sin(t) </code></pre> <p>Where t is the angle in radians and r is the radius of the base circle. Yes that means the involute is different for different circles.</p> <p>But why this shape? Well, you want generally the surfaces to be rolling instead of sliding since rolling has less friction. So you want the contact to be point like in all possible cases. Now the easiest even contact is a line. if you draw a line for the contact you get either an cycloid or a involute depending on wether the contact line is perpendicular or not to the rotating surface.</p> <p>The main benefit of the involute is that if the line starts a bit on a different position but moving in same direction its still the same involute. Which means your gear has some tolerance for the gear center distances.</p> <p>* although you can express the curve as a polar formulation with a one dimensional function. I find that its really hard to understand while the cartesian formulation is easy enough to describe and syhesize from the description especially if your comfortable working with vectors.</p>

53925

How to determine curvature of face and flank of a spur gear? What is the involute in gear design?

2023-01-18T03:26:22.597

<p>So I noticed that some cups are shaped like frustrums or truncated cones. I looked up why and here(<a href="https://www.reddit.com/r/askscience/comments/2v151h/why_are_cups_cone_shaped/" rel="nofollow noreferrer">https://www.reddit.com/r/askscience/comments/2v151h/why_are_cups_cone_shaped/</a>) it says that</p> <ol> <li>They are easier to stack</li> <li>They are easier to hold.</li> </ol> <p>I am confused by the second point. Why would they be easier to hold? Wouldn't the holder need to bend the palm of their hand, making it harder to hold?</p> <p>Thanks</p>

|product-engineering|

<h2>TL;DR: it is due to the effect of a mechanical edge.</h2> <p>When you are holding a cup pressure is exerted by the fingers (essentially). Like the following image. Pressure is always normal to the surface.</p> <p><a href="https://i.stack.imgur.com/WnRss.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WnRss.png" alt="enter image description here" /></a></p> <p>That pressure results in a force (black arrow in image below) that if it is analysed in the vertical and the horizontal you get the following.</p> <div class="s-table-container"> <table class="s-table"> <thead> <tr> <th>Common shape</th> <th>inverted shape</th> </tr> </thead> <tbody> <tr> <td><a href="https://i.stack.imgur.com/1csqB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1csqB.png" alt="enter image description here" /></a></td> <td><a href="https://i.stack.imgur.com/o40tf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o40tf.png" alt="enter image description here" /></a></td> </tr> </tbody> </table> </div> <p>you will notice that the vertical component is opposite in those cases. I.e. when you increase the pressure the cup will be pushed downwards.</p> <hr /> <p>One thing that pop to mind (excuse the pun) wat the ice lollies (a type of frozen icecream in a cone shaped packaging) . If you ever held one of those and tried eat it, you know that you need to squeeze it, to make it pop out.</p> <p><a href="https://i.stack.imgur.com/lmzKq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lmzKq.png" alt="ice lolly " /></a></p> <p>The principle is the same. If you inverted the cone then as you increased the pressure the glass would be pushed downwards (more easily that in the opposite case).</p> <hr /> <p>Final note: <strong>Lab glassware</strong></p> <p>One of the few exceptions, when an inverted cone shape is used is in Lab glassware like the following.</p> <p><a href="https://i.stack.imgur.com/cOLXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cOLXt.png" alt="enter image description here" /></a></p> <p>However you may notice that there is still a part of the glassware with a straight neck and also there is a visible bottle lip at the top (for lab clampers to be able to hold). The main reason that this the preferred shape, is that it secures the liquid in place and it enables easier stirring with little chance of spilling.</p>

53926

Why Cups are Frustrum-Shaped

2023-01-18T22:29:51.923

<p>I'm learning about PID controllers and one thing I don't quite understand from the pseudocode and formulas is the apparent persistence of the &quot;Integral&quot; portion.</p> <p>For example, imagine the position is low and is adjusted with minimal oscillation to reach the set point. At that point however, the &quot;Integral&quot; will be low because of the duration of time spent low, and will forever stay low, thereby forcing the system to eventually compensate by overcorrecting and staying &quot;high&quot; for a period of time.</p> <p>I feel like it would be appropriate to &quot;reset&quot; it when the system stabilizes or have an ongoing dampening process which slowly dampens it to zero.</p> <p>Does that make sense or is there something I have misunderstood?</p>

|pid-control|

<p>You are absolutely correct. In fact this is a common, real world problem, even with industrial PID controllers. It is often referred to as &quot;integral windup&quot;.</p> <p>There are many solutions, but they fall outside of the &quot;pure&quot; PID algorithm. The general idea is that the integral factor is there to compensate for droop when the system is at steady state. So if you're not near steady state you may want to zero out the integral error accumulator. Another crude solution is to limit the maximum accumulated error. The overshoot you described will still occur, but the amplitude will be limited.</p>

53935

Do you need to reset the Integral part of a PID controller?

2023-01-19T05:08:01.523

<p>I'm a high-school engineering student, and for my final project I wanted to make a robotic arm.</p> <p>However, I'm stuck on how to minimize friction between 2 surfaces (both made out of aluminum 6061-T6) when moved by a stepper motor. I can't seem figure out how to implement something like a ball-bearing or other friction-reducing parts into my design to negate issues like heat or the scraping / scratching of the surfaces.</p> <p>Here's a quick drawing of what my design currently looks like: <a href="https://i.stack.imgur.com/yFuq0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yFuq0.jpg" alt="My Design" /></a></p> <p>With this design, would there be any possible way to implement the aforementioned friction-reducing parts / bearings?</p>

|friction|bearings|machine-design|

<p>As drawn, you could nylon or oilite washers as spacers on the shaft between the white walls and grey-U. Second best would just be to use brass or bronze since you don't want aluminum rubbing on aluminum. That will promote a lot of wear and galling. Brass and bronze rubbing on pretty much anything else are much better for withstanding wear and galling. But this is all a bit slipshod since this will allow your setup to withstand stronger horizontal loads than vertical loads but your vertical loads are going to be higher than the horizontal loads. You have excessive strength where you can't use it. As drawn, where the the stepper motor shaft simply passes through a clearance hole in the white wall unsupported, the stepper motor bearings and their mounting bolts will take all the vertical load.</p> <p>For best vertical load capability, your stepper motor shaft should rest on separate ball bearings that are inset into the white walls. Many bearings have the inner race a bit longer and wider than the outer race so it protrudes a bit. With proper positioning, the inner race will protrude beyond the white wall and the grey-U can rest against that to maintain a gap. Even if the bearing doesn't have this feature, all you need to do is get a small tube or thin ring with the same outer and inner diameter as the inner race and slip it over the shaft on that side and it will act as a spacer. In this setup the sidewall load capability is whatever the axial load the ball bearings can withstand and different ball bearing designs can withstand different amounts. Ball bearings by design will withstand higher radial loads (vertical loads in your setup) than axial loads (horizontal loads) but that's probably fine since your vertical loads are going to be higher than your horizontal loads. If you need extra horizontal strength <em>then</em> you seat the the bearings so they sit within the thickness of the white wall (remove any tube or ring spacers on the shaft, reposition the bearing, or select a bearing with no inner race protrusion), and add the aforementioned washers to transfer horizontal loads directly into the white wall rather than having the bearing take them.</p> <p>And even if there are no ball bearings set in the white walls (the stepper motor shaft just passes through a clearance hole in the white wall completely unsupported such that the stepper motor bearings and mounting screws take all the load) you are fixing the grey-U to the shaft somehow anyways. So just fixing it so that the grey-U cannot slide along the axis of the shaft and position it so there is a gap between the white wall and the grey-U.</p>

53936

Where would I place a bearing between 2 surfaces?

2023-01-19T09:18:00.343

<p>Have a look at this water wheel (located at Hama, Syria).</p> <p>Most spoked wheels seem to be designed with spokes that radiate outward from the center of the hub (bicycle wheels being the notable exception). All of the spokes in this wheel are offset from the center of rotation.</p> <p>Is there a good reason for this? it seems to add extra levels of complexity for no added benefit (that I can see). <a href="https://i.stack.imgur.com/rtQHp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rtQHp.jpg" alt="Ancient water wheel" /></a></p>

|water-resources|engineering-history|

<p>Radial lacing &quot;tortures&quot; the spokes of the wheel because when torque needs to be transferred from the hub to the ring when accelerating (or from ther ring to the hub during braking).</p> <div class="s-table-container"> <table class="s-table"> <thead> <tr> <th style="text-align: center;">Radial no load</th> <th style="text-align: center;">Radial with torque</th> </tr> </thead> <tbody> <tr> <td style="text-align: center;"><a href="https://i.stack.imgur.com/TV6hXm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TV6hXm.png" alt="enter image description here" /></a></td> <td style="text-align: center;"><a href="https://i.stack.imgur.com/bORFa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bORFa.png" alt="enter image description here" /></a></td> </tr> </tbody> </table> </div> <p>I.e. if radial lacing is used then the spokes would need to act as beams transferring a bending moment (but their second moment of area is really small so they would deform and result in a pattern similar to cross lacing).</p> <hr /> <p>The Different patterns of lacing (Two cross, 3 cross etc), by having an offset from the center of the wheel, they allow the carrying elements to act more (&quot;more&quot; is the essential word here) as rods (elements carrying only axial forces). (The cross parameter improves the geometric stability of the pattern).</p> <p><a href="https://i.stack.imgur.com/svOy5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/svOy5.png" alt="Different lacing patterns" /></a> <strong>Figure: Different lacing patterns (source: <a href="https://bestwheelset.com/36-spoke-wheel-lacing-patterns/" rel="nofollow noreferrer">best wheel set</a>)</strong></p> <p>That is the reason that probably this watermill has this shape (although most surviving watermills are designed by wood beams with enough cross-section that gives them a second moment of area that doesn't allow much deformation). However, the wooden beams are in stark contrast to the spokes of a bicycle wheel in terms of second moment of area.</p>

53938

Why aren't these spokes radial?

2023-01-19T14:17:02.523

<p>I bought an under-sink electric boiler/hot water storage unit, for instant almost-boiling water. This device has an 'eco' mode, which broadens its hysteresis to 10 degrees C: it will allow the stored water temperature to drop to <span class="math-container">$T_\text{target} - 10$</span> before turning on the element.</p> <p>This made me curious, since naively I would think that it will turn on less often of course, but for longer (or hotter) when it does; consuming just as much energy overall.</p> <p>If we assume:</p> <ol> <li>the element has a fixed resistance (e.g. it does not deliberately vary with <span class="math-container">$T_\text{target} - T_\text{now}$</span>), i.e. a fixed kW consumption <span class="math-container">$P$</span> s.t. <span class="math-container">$E=P\cdot{t_\text{heated}}$</span> and we need only worry about <span class="math-container">$t_\text{heated}$</span>;</li> <li>the unit has a fixed efficiency across all <span class="math-container">$T_\text{now}$</span>;</li> <li>standby power is negligible;</li> </ol> <p>(so as to avoid specifics about the particular device)</p> <p>I know:</p> <ol> <li>the stored unheated water will cool exponentially over time;</li> <li>the heated water will rise in temperature as an inverse exponential, asymptotic to <span class="math-container">$T_\text{element}$</span>;</li> </ol> <p>let:</p> <ol> <li><span class="math-container">$t_{\text{fall},x}$</span> be the time taken for the unheated water to fall from <span class="math-container">$T_\text{target}$</span> to <span class="math-container">$T_\text{target} - x$</span>;</li> <li><span class="math-container">$t_{\text{rise},x}$</span> be the time taken for the heated water to rise from <span class="math-container">$T_\text{target}-x$</span> to <span class="math-container">$T_\text{target}$</span>;</li> </ol> <p>then:</p> <ol> <li>what is the relationship between these exponents, what else are they a function of if it's really the case that <span class="math-container">$$\dfrac{t_{\text{rise},x}}{t_{\text{rise},y}} \ne \dfrac{t_{\text{fall},x}}{t_{\text{fall},y}}$$</span>?</li> <li>if they <em>are</em> only a function of time, <span class="math-container">$T_\text{target}$</span>, and <span class="math-container">$T_\text{ambient}$</span> (room/insulation, or the element, for cooling or heating respectively), then surely 'eco' would be setting an appropriate <span class="math-container">$T_\text{target}$</span>, not the max allowable <span class="math-container">$T_\text{target}-T_\text{now}$</span>?</li> </ol>

|thermodynamics|heat-transfer|heating-systems|fluid|

<p>the heating is not asymptotic. Its roughly linear with time. The reason is that when you are heating you are supplying constant heat flux (ie. heat energy per unit of time). The underlying equation is that:</p> <p><span class="math-container">$$Q = m \cdot c_p \delta T$$</span></p> <p>where:</p> <ul> <li>Q is the heat energy provided</li> <li>m is the mass of the liquid</li> <li><span class="math-container">$C_p$</span> is the heat capacity of the liquid</li> <li><span class="math-container">$\delta T $</span> is the temperature difference.</li> </ul> <p>So if you differentiate with time (keeping constant m and constant <span class="math-container">$c_p$</span> then)</p> <p><span class="math-container">$$\frac{dQ}{dt} = m\cdot c_p \cdot \frac{d T}{dt}$$</span></p> <p>The reason that it takes longer is because the losses to the environment increase however, they are significantly less in the heat energy balance.</p>

53941

Optimal maintenance temperature and allowable drop for energy efficient hot water storage

2023-01-20T20:45:05.540

<p>So, I was searching on google/youtube about <a href="https://en.wikipedia.org/wiki/Minimally_invasive_procedure" rel="nofollow noreferrer">minimally invasive procedures</a> and its tools to reach deep and delicate spaces of the human body, and it came to my mind the possibility of using it for CNC machining.</p> <p>For example, these could reach deeper and produce incredibly complex shapes just like the 3D printing machines do, but without the weaknesses that normally come with the second method.</p> <p>The first problems I can imagine would be the difficulty in automatic control (but that depends on the design of the robot, for example, <a href="https://youtu.be/DpLiFFnifU0" rel="nofollow noreferrer">this one</a> would be easier to control) and how to subtract material in confined spaces (maybe fiber lasers, water jet cutter or wire EDM), but even so, <strong>why aren't continuum robots used for CNC machining?</strong></p> <p>The following picture is a cutaway metal 3d print of an Aerospike Rocket Engine.</p> <p><a href="https://i.stack.imgur.com/pP52y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pP52y.png" alt="Picture of a cutaway metal 3d print of an aerospike rocket engine" /></a></p>

|machining|robotics|machine-design|cnc|

<p>Continuum Robots is a kind of flexible robot, which is composed of flexible materials and many small continuum structures, which can realize highly flexible movements. In contrast, CNC machine tools usually use rigid materials and joint structures during machining, and their movements are strictly controlled to achieve high-precision machining.</p> <p>Although continuum robots have advantages in some application areas such as medical surgery and flexible industrial automation, they are not common in the field of CNC machining for the following reasons:</p> <p>High precision requirements: In CNC machining, it is usually necessary to process and position materials with high precision to achieve the expected geometry and size. Because the structure of the continuum robot is relatively flexible, it is difficult to achieve high-precision processing and positioning like a CNC machine tool.</p> <p>Slower machining speed: Compared with rigid robots, continuum robots usually take longer to complete the same machining task. This is because the movement mode of the continuum robot is realized through the overall deformation, and the movement speed is slower than that of the traditional robot.</p> <p>High cost for industrial applications: Continuum robots usually require the use of high-strength flexible materials and complex sensor and control systems, which lead to their high cost. Therefore, for some industrial applications, their cost may exceed the budget.</p> <p>So while continuum robots excel in some applications, they are less common in CNC machining</p>

53951

Why continuum robots aren't common on CNC machining processes?

2023-01-23T11:32:05.000

<p>What software(s) would you recommend to simulate a water wheel that will generate 3D graphics?</p>

|software|

<p>After reviewing the options, I was surprised to find that Ansys Fluent did not provide the most detailed 3D simulation, but FLOW 3D and SimScale (cloud). Ansys Fluent generates a realistic 2D simulation, but a comparatively poorer 3D one, among the examples I could find.</p> <p><a href="https://i.stack.imgur.com/PenSX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PenSX.png" alt="SOLIDWORKS Flow Simulation" /></a></p> <p>SOLIDWORKS Flow Simulation</p> <p><a href="https://i.stack.imgur.com/kakJ0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kakJ0.png" alt="Ansys Fluent" /></a></p> <p>Ansys Fluent</p> <p><a href="https://i.stack.imgur.com/aOWmv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aOWmv.jpg" alt="FLOW 3D" /></a></p> <p>FLOW 3D</p> <p><a href="https://i.stack.imgur.com/WGEZO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WGEZO.jpg" alt="SimScale" /></a></p> <p>SimScale</p>

53978

What software(s) are suited to simulate a water wheel generating 3D graphics?

2023-01-28T10:15:30.037

<p>I've heard that cars parked on an incline should leave the car in either 1st gear or reverse, because this supposedly provides the most engine resistance should the wheels start turning.</p> <p>But my question is: Wouldnt using a higher gear cause more resistance for the wheels, given that in a high gear each revolution of the wheel causes a greater number of revolutions in the engine than a lower gear would?</p>

|torque|

<blockquote> <p>given that in a high gear each revolution of the wheel causes a greater number of revolutions in the engine than a lower gear would?</p> </blockquote> <p>It is exactly the other way around:</p> <ul> <li><p>In a low gear, the engine makes many revolutions for one turn of the wheel. This provides more torque for acceleration.</p> </li> <li><p>In a high gear, the engine makes few revolutions for one turn of the wheel, this allows you to drive high speeds with relatively low engine RPM.</p> </li> </ul> <p>Hence, when you push a parked car that has engaged 1st gear, you will make the engine do many revolutions for a low speed of the car: The car provides more engine resistance than when it would have been in a higher gear.</p>

54026

Does a parked car provide more engine resistance when left in a low gear or high gear?

2023-01-28T15:09:12.200

<p>what speed must a bullet be fired at to turn into gas when hitting a steel plate.</p> <p>from watching this <a href="https://9gag.com/gag/ajVmzAq" rel="nofollow noreferrer">video</a> on 9gag, i see that the bullet almost turns into liquid, what speed would it need to hit it to turn into a gas?</p>

|experimental-physics|building-physics|

<p>You would have to heat it up to the vaporisation temperature and you have also 2 phase changes, from solid to liquid and then from liquid to gas. If you assume kinetic energy is converted into heat and some loses, the speed <span class="math-container">$v$</span> comes up from equilibrium:</p> <p><span class="math-container">$$\frac{1}{2}\cdot m \cdot v^2 = m\cdot c_p \cdot \Delta T+ m\cdot \Delta H_{fus} + m\cdot \Delta H_{vap} + H_{losses}$$</span></p> <p>where:</p> <ul> <li><span class="math-container">$m$</span> is bullet mass</li> <li><span class="math-container">$c_p$</span> is heat capacity</li> <li><span class="math-container">$\Delta T$</span> is temperature difference between vaporisation and initial temperature</li> <li><span class="math-container">$\Delta H_{fus}$</span> and <span class="math-container">$\Delta H_{vap}$</span> are latent heats of fusion and vaporisation</li> <li><span class="math-container">$H_{losses}$</span> represents losses to air and steel plate</li> </ul> <p>From this, the minimum speed <span class="math-container">$v_{min}$</span> (ignoring losses) will be:</p> <p><span class="math-container">$$v_{min} = \sqrt{2}\cdot \sqrt{c_p \cdot \Delta T+ \Delta H_{fus} + \Delta H_{vap}}$$</span></p> <p>Here are minimum speeds for lead and iron:</p> <div class="s-table-container"> <table class="s-table"> <thead> <tr> <th>substance</th> <th>liquefaction speed m/s (km/h)</th> <th>vaporisation speed m/s (km/h)</th> </tr> </thead> <tbody> <tr> <td>lead</td> <td>161.02 (579.67)</td> <td>678.89 (2444)</td> </tr> <tr> <td>iron</td> <td>322.22 (1160)</td> <td>922.11 (3319.6)</td> </tr> </tbody> </table> </div>

54028

what speed must a bullet be fired at to turn into gas when hitting a steal plate

2023-01-31T13:04:14.853

<p>What is the critical buckling force needed to be applied on a system of made out of two parts?</p> <p>The parts of the system are as depicted in the picture:</p> <ol> <li>incompressible elastic beam - on top</li> <li>compressible support beam - on the bottom</li> </ol> <p>Upon applying enough compression force the top beam will buckle and the bottom will compress. I managed to follow worked examples for calculating the Euler buckling of the top beam leading to <span class="math-container">$P_{cr} = \frac{π^2\kappa}{l^2}$</span> where <span class="math-container">$\kappa$</span> is the bending rigidity and <span class="math-container">$l$</span></p> <p>is the compressed length of the bottom beam and the projection of the top beam on the x-axis. The bottom beam is modeled as a spring using Hooke's law <span class="math-container">$F = -k(l_0-l)$</span></p> <p>I am having problem connecting all the pieces of the system together to be able to find the two behaviors where under the critical load the system is flat and above it it undergoes compression and buckling. I would like to also know the energy of the system but that should be relatively easy to obtain from the force by integration.</p> <p><a href="https://i.stack.imgur.com/JfWkY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JfWkY.png" alt="enter image description here" /></a></p>

|compressors|buckling|

<p>Assuming that the support to which they are attached cannot rotate or deflect other than laterally, the buckling limit of the system equals the buckling limit of the system when the weaker bar buckles.</p> <p>There is no incompressible elastic material. Actually, there is no rigid material, only a convenient way of solving problems!</p> <p>Therefore when the first bar has buckled the entire load <span class="math-container">$F$</span> has to be supported by the remaining member and that member already could not stop the deflection under force <span class="math-container">$F$</span> even with the assistance of the pre-buckling resistance of the failed member.</p> <p>But if the support can deflect or bend, the system has to be assessed as a whole. The share of lower bar in carrying <span class="math-container">$F $</span> is</p> <p><span class="math-container">$$F\_l=F*A_l*E_l/ (A_lE_l+A_{top}E{top})$$</span> So we can calculate back <span class="math-container">$F$</span> from this.</p>

54063

What is the critical buckling force needed to be applied on a system of made out of two parts?

2023-02-01T12:35:15.533

<p>I noticed while reading about compressors that they have various types and are generally made for a specific function. However, are there types of compressors (or pumps) that can compress both water and water vapor?</p>

|mechanical-engineering|fluid|compressors|

<p>Just a remark to your terminology: if the main function is the TRANSPORT of gas, the device is called VENTILATOR. The pressure change is &quot;small&quot; (relative to an arbitrary reference). If the main function is the COMPRESSION of the gas, the device is a COMPRESSOR (high pressure change, eventually small mass flow). The label PUMP is mostly used for liquids. The TRANSPORT of liquids often need high pressure changes because of GRAVITATION. However as SolarMike says, that is not a compression in the sense that the rise in density is neglible in technical applications.</p>

54072

Are there compressors that can compress both water and water vapor?

2023-02-01T18:52:43.207

<p>I am trying to come up with a way to clean the air in my one-car garage/DIY-woodshop, removing fine sawdust particles smaller than 1-micron, and thought that maybe one or two <a href="https://youtu.be/hIuH-2naozI" rel="nofollow noreferrer">Corsi-Rosenthal boxes</a> would do it.</p> <p>Could the fan orientation be reversed so air is blown <em>into</em> the box, with the filters also reversed so the air flows through them out into the room? Sawdust would then collect on the inside of the filters and disposal would be easier.</p> <p>What would happen in that scenario if the filters couldn't keep up with the fan? If, say, 2&quot; thick filters didn't have enough air flow and what was needed were 4&quot; thick filters? Could the force of the air from the fan dislodge the dust from the filters? Would &quot;blowback&quot; result, with dust coming back out of the box through the fan?</p>

|airflow|air-filtration|

<p>In this link is what you are describing. One could add some legs and add a 5th filter opposite the fan so the static pressur is reduced and less risk of dust escaping the fan blade boundary. I hope this helps, and please update on how the reverse CR Box is performing if possible. Thanks.</p> <p><a href="https://www.facebook.com/reel/688198306589401?mibextid=NnVzG8" rel="nofollow noreferrer">https://www.facebook.com/reel/688198306589401?mibextid=NnVzG8</a></p>

54076

What are the downsides of reversing fan and filter direction on a Corsi-Rosenthal box so the fan blows air into the box?

2023-02-02T22:07:08.450

<p>I have a eigenfrequency simulation <span class="math-container">$ M * \ddot{\vec{x}} + K * \vec{x} = \vec{0} $</span> and want to add point masses to certain nodes. The only example I have is a 1D matrix where all of the mass at node 1 is added to the x1/x1 element in the matrix.</p> <p>Since my matrix has 3 main diagonal elements at node 1, x1/x1, y1/y1 and z1/z1, do I add 1/3 at each of these?</p> <p>I first added the entire mass (which makes sense because if I only move in x direction, I still need to move the whole mass) but then the sum of all masses in the matrix is not consistent with the mass of the system. Which way is correct?</p>

|finite-element-method|eigenvalue-analysis|

<p>Full point masses should be on lumped mass matrix diagonal for each direction. Sum of the mass matrix elements does not have to be the full mass, that would make no sense. Full gravity force si pulling you down with the same intensity no matter if you are standing on the ground or accelerating in a horizontal direction.</p>

54085

How to add point mass in 3D mass matrix

2023-02-02T22:20:39.717

<p>I am interested in building an experimental lubricant-less pneumatic tube system in which the canister will be propelled up to a high-velocity, reaching a peak velocity of approximately 100 m/s before the canister is forced to slow down and come to a stop.</p> <p>The canister should attain a high velocity within the pneumatic tube by being propelled by ambient air pressure behind it and there being very low air pressure in front of it inside the tube (similar to the Hyperloop). I plan to create this very low air pressure either through the use of a vacuum pump, or perhaps by sealing one end of the tube with tape and then pulling the canister backward through the entire tube and then releasing it.</p> <p>I have spent time researching on the Internet for what would be an ideal material to use for the O-rings that will wrap around this canister. The challenge I'm facing is trying to find O-rings that will keep the ambient air pressure from leaking around the canister and getting into the tube, and at the same time being slippery enough to glide over the tube's internal surface, whether the tube is made out of metal or PVC.</p> <p>I have the option of putting standard rubber O-rings around the canister, yet this would require me to lubricate the pneumatic tube with some type of grease/oil before each test and I don't wish to lubricate several hundred feet of tubing for each test.</p> <p>I have considered making the felt-covered, rubber O-rings, yet I think a lot of air may push through the felt. I have also considered making O-rings out of tightly wound silk, yet I think silk O-rings will not last very long. So, I am seeking suggestions on what would be an ideal material to use for these special-purpose O-rings.</p> <p>What would be an ideal material to use for O-rings on a canister in a lubricant-less, high-velocity pneumatic tube system?</p>

|mechanical-engineering|materials|design|applied-mechanics|friction|

<p>Precise machining of metals can result in a cylinder/piston fit so tight that it cannot be inserted if the exit hole is blocked. This would be impractical, in my opinion, in your application, as temperature is likely to be a factor over any appreciable distance.</p> <p>Directly related to this, however is the concept of using a slippery material such as UHMWPE (Ultra High Molecular Weight PolyEthylene) as the casing. This will reduce the overall friction of the assembly.</p> <p>For sealing, PTFE O-rings are available. PTFE is related to Teflon™ and is quite slippery as well. It's possible you may find a custom &quot;not-O&quot; ring with a flange that will seal against the tubing (lip seal) when under pressure, akin to a one-way flapper valve, but even a conventional O-ring will be more slippery than rubber compounds.</p>

54086

What would be an ideal material to use for O-rings on a canister in a lubricant-less, high-velocity pneumatic tube system?

2023-02-05T12:06:21.433

<p>I will attach the photos of the questions, it was my quarterly test paper <a href="https://i.stack.imgur.com/9ruif.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ruif.jpg" alt="enter image description here" /></a></p> <p>Consider a continuous line charge distribution along the circumference of a circle of radius 1 unit in the xy plane. With the origin at the center of the circle and using the cylindrical (p. q, z) coordinate system. the charge density y, varies along the circle as y = Ksin^2φ, 0&lt;φ&lt;2pi, K is a constant.</p> <p>(a)Determine the total Charge in the system</p> <p>(b) Determine the electric field vector at the centre of the circle</p> <p>(c) Determine the potential difference between the points A (x=100,y=0,z=0) and B(x=0,y=100,z=0)</p>

|electromagnetism|

<p>I think total charge in the system is just integral around the circle: <span class="math-container">$$Q = K\int\limits_0^{2\pi} \sin^2 \varphi d\varphi = K\cdot \left[\frac{\varphi}{2} - \frac{\sin(2\varphi)}{4}\right]_0^{2\pi} = K\cdot \pi$$</span></p> <p>If you try to draw how the distribution looks like, it would be something like an eclipse in polar coordinates. So from the center point of view, effects of charges cancel out and electric field vector should be 0.</p> <p>Similar reasoning works for the last point, where I think the difference is also 0, due to symmetry.</p>

54133

I have 2 questions i am not able to solve in applied electromagnetics

2023-02-06T01:42:11.680

<p>Conditions: there is a steel plate insulated at the bottom and facing the clear night sky, the ambient air temperature is 2°C, the cosmic microwave background radiation is 3K, and the convective heat transfer coefficient is 10 <img src="https://latex.codecogs.com/svg.image?W/m%5E%7B2%7D%5Ccdot%20K" alt="formula" />. My job is to calculate the temperature of the steel plate at thermal equilibrium. The steel plate doesn't have given dimensions.</p> <p>I've used <img src="https://latex.codecogs.com/svg.image?Q_%7Bc%7D=h%5Ctriangle%20T" alt="formula" /> for the convective heat transfer, and <img src="https://latex.codecogs.com/svg.image?Q_%7Br%7D=%5Csigma%20%5Cepsilon%20T%5E%7B4%7D" alt="formula" /> for the radiative heat transfer. Using 0.8 as the emissivity of steel, <img src="https://latex.codecogs.com/svg.image?Q_%7Bc%7D+Q_%7Br%7D=0" alt="formula" /> yields -22.158°C which is close enough to the textbook's answer of -20.9°C. This textbook has provided wrong answers multiple times in the past, so I don't fully trust it and this answer doesn't seem right to me.</p> <p>Attempts to calculate in Kelvin have resulted in even more unlikely answers. Is -20°C plausible? Or where did I go wrong?</p> <p>Edited to add calculations:</p> <p><img src="https://latex.codecogs.com/gif.latex?Q_%7Bc%7D%3Dh%5Ctriangle%20T%3D10%5Cast%20%282-x%29" alt="formula" /></p> <p><img src="https://latex.codecogs.com/gif.latex?Q_%7Br%7D%3D%5Cepsilon%20%5Csigma%20%5Ctriangle%20T%3D0.8%5Cast5.67%5Cast10%5E%7B-8%7D%5Cast%28x%5E%7B4%7D-%28-270.15%5E%7B4%7D%29%29" alt="formula" /></p> <p><img src="https://latex.codecogs.com/gif.latex?Q_%7Bc%7D%3D-Q_%7Br%7D" alt="formula" /></p> <p>My calculator puts out: <img src="https://latex.codecogs.com/gif.latex?x%3D-22.159%5E%7B%5Ccirc%7DC" alt="formula" /> as the solution.</p>

|mechanical-engineering|thermodynamics|heat-transfer|convection|thermal-radiation|

<p>I get a value closer to -17.40 C = 255.75 K using an iterative solution (manually in Excel).</p> <p><span class="math-container">$Q_C = 10*(2-(-17.40)) = 194 \, W/m^2$</span></p> <p><span class="math-container">$Q_r = 0.8*5.67\times10^{-8}*((-17.40+273.15)^4-(3)^4) = 194.06 \, W/m^2$</span></p> <p>Note that convection can use either C or K but radiation must use K.</p> <p>Since -17.4 C satisfies both equations, the answer in the textbook is wrong.</p>

54136

Can a steel plate reach -20°C due to radiative cooling when the ambient temperature is 2°C?

2023-02-09T13:36:46.030

<p><a href="https://i.stack.imgur.com/ftJ0G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ftJ0G.png" alt="enter image description here" /></a></p> <p>I've tried using various calculators but that has just confused me even more so I'll describe my scenario and hopefully I can get some answers. In the picture above, each item is described below</p> <p><strong>A:</strong> Timber Loft Joist, Width 3.5m, Height 225mm, Depth 45mm. Either end of each joist sits on top of a block wall 100mm thick and 2.5m in height. From this link <a href="https://www.engineeringtoolbox.com/floor-joist-capacity-d_1832.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/floor-joist-capacity-d_1832.html</a>, it says that these joists can support 3kN/m2.</p> <p><strong>B:</strong> OSB board, 18mm thick</p> <p><strong>C:</strong> Mild Steel plate, Width 400mm, Length 400mm, Depth ?mm</p> <p><strong>D:</strong> Load, 120x120mm, Weight 150kg</p> <p>With my very limited engineering knowledge, I'm assuming that I need <strong>C</strong> to spread the 150kg load from <strong>D</strong> so that it doesn't punch a hole through <strong>B</strong>. If this is the correct approach, what would be an adequate thickness for <strong>C</strong> to accomplish this. Any thoughts or suggestions would be greatly appreciated as it'll help me with knowing the viability of my project.</p>

|load-spreading|

<p>This can be handled like a beam resting on two pin supports, with good approximation. A plate can support a bit more than a beam. I suppose that the OSB alone, along the major axis, would be strong enough to support this load but just to be on the safe side.</p> <p>let's pick a 4mm thick plate with:</p> <ul> <li>M = moment kg.mm</li> <li>S = beam section modulus =<span class="math-container">$bh^2/6$</span></li> <li>h = plate thickness, 4mm</li> </ul> <p><span class="math-container">$$ M=Pl/4=150kg∗400mm/4=15000kgmm$$</span> <span class="math-container">$$\sigma= M/S=15000/ \frac{bh^2}{6}=15000/ (\frac{400*4^2}{6})=15000/1066=14.06kg/mm^2&lt;25kg/mm^2 \text{ allow stress of mild steel}$$</span></p> <p>So the safety factor is <span class="math-container">$ 1.75 &gt;1.6 \ $</span>and is okay.</p> <p>In my comment, I had recommended SF of 2.8 but with your adding more info 1.6 is okay.</p> <p>The blocking under the plate between joists is for joists' lateral support and is required by most codes.</p>

54168

Steel plate to distribute load on top of joists

2023-02-10T07:30:40.683

<p>Here I have a system where there is 18 bar of Nitrogen in a valve with a piston. The pressurized nitrogen is below the piston, above the piston, and also in the ventilation bore in the piston. Nitrogen can move freely between these three volumes.</p> <p>Since the upper surface area of the piston is greater than the lower surface area of the piston, I would expect a difference in &quot;pneumatic force&quot; between the top and bottom of the piston, which would drive the piston downwards (as the upwards pneumatic force is lower than the downwards pneumatic force).</p> <p>However the small ventilation bore in the piston makes me feel unsure. I can't think of a reason why this ventilation bore would prevent the piston from functioning this way, as pressure should still act on every surface equally, right?</p> <p>So my question is: will the piston experience a net downwards force proportional to the difference in areas as a result of the difference in surface areas from the top of the piston to the bottom (even if the piston is ventilated)? <a href="https://i.stack.imgur.com/YEsCN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YEsCN.png" alt="Piston with ventilation bore and different surface areas" /></a></p>

|fluid-mechanics|pressure|pressure-vessel|

<p>Yes, even if the ventilation bore has an oval or kidney shape or is significantly larger. The difference between the top and bottom areas times approximately 17 bars (18 bars-1 atm) will be the net force acting on the piston.</p> <p>Despite its size or shape, the 18 bars of pressure acting on the ventilation bore's walls cancel each other out. The remaining forces are the top and bottom forces!</p>

54177

If a piston has a larger upper surface area than lower surface area experience a force downwards even if that piston is ventilated?

2023-02-10T17:06:45.890

<p>Sealed bearings are a common component on modern bicycles.</p> <p>Even if you take off the plastic covers of sealed bearings, the ball bearings inside will not fall out.</p> <p>How is such a part assembled in the first place?</p>

|manufacturing-engineering|bearings|bicycles|

<p>Usually they are pressed together with a hydraulic press. Also, given the ball, count there can be sufficient space for the balls to be placed in one half and then spread around the periphery and held in position with the spacing device - which can be plastic or metal and in one or two pieces which get glued, welded or riveted together.</p> <p>Some bearings are designed, like taper roller bearings, to disassemble easily.</p>

54183

How are sealed bearings assembled?

2023-02-13T07:50:57.187

<p>In nuclear fusion the atoms are in a hot plasma and the temperature is &gt;5000K for them to posses kinetic energy and overcome repulsion to fuse , this process generates some heat(which is equal to the mass defect).But how do we actually extract that given heat if the surroundings are already so hot ?</p>

|heat-transfer|nuclear-engineering|nuclear-technology|

<p>Most of the energy release in a fusion reaction is in the form of the kinetic energy of the reaction products plus a bit more in the form of radiative energy carried off by gamma rays. the kinetic energy of the reaction products is captured as <em>heat</em> in the part of the reactor known as the <em>first wall</em>, when the fast-moving products (helium nuclei) collide with the stuff from which the first wall is made (graphite, for example). The gamma rays also collide with the first wall and through a series of successive <em>scattering events</em> with nuclei inside the first wall they impart a bit more heat energy to the first wall as well. The heat from the first wall is then captured by heat transfer pipes that carry the heat out of the reactor and into a steam generator.</p> <p>The plasma in which the fusion reactions occur must be confined in such a way that the plasma never comes into contact with the first wall, otherwise the fusion process will be immediately <em>quenched</em> and the reactor will shut itself down.</p> <p>Since there is no material known which can stand up to fusion temperatures, the plasma confinement must be done either with magnetic fields or via <em>inertial confinement</em>.</p> <p>Inertial confinement is the process in which the input energy required to trigger the fusion reaction is dumped into the fusion pellet so fast that the inertia of the pellet itself prevents it from escaping the reactor until the fusion process is finished. At that point, the input energy dump is over too which means the inertial confinement ceases and the energy (fast-moving helium nucleus plus gammas) streams freely out towards the first wall.</p>

54215

Heat extraction in Nuclear Fusion

2023-02-13T11:41:24.167

<p>Remember those pull down ceiling lamps from the 80s? A lamp hanging on a steel cable and you could adjust the height by just pulling or pushing the lamp and it would stay there. There was probably a gear and a spring in there, because you could adjust the balance by rotating a knob inside the mechanism. And it would click when moved down. But I cannot find any details.</p> <p>Does anyone know how this mechanism is called and how it works?</p> <p><strong>EDIT</strong>: I found a picture.</p> <p><a href="https://i.stack.imgur.com/654Qwm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/654Qwm.jpg" alt="Ceiling Lamp" /></a></p>

|mechanical-engineering|

<p>According to <a href="https://www.lightingcompany.co.uk/ceiling-lights-c3/rise-and-fall-lights-c10" rel="nofollow noreferrer">lightingcompany.co.uk</a>, this particular type of product is called a rise and fall light. It appears there are three types, one type in the original image uses coiled wiring to contain and release the necessary changes in wire length, and includes a spring loaded winch assembly, akin to retractable dog leashes in common use today.</p> <p>Another version is represented by @Jonathan R Swift's image, using a counterweight and pulleys.</p> <p>The third version encases the winch inside a decorative shell, and the winch wraps the wiring onto itself when the fixture is raised.</p> <p>From the linked site:</p> <blockquote> <p>Up and down over table lights are often referred to as rise and fall lights and The Lighting Company has a great selection of rise and fall ceiling lights including pull up and down pendant lights. When raised up, a rise and fall light will give a wide spread of light and when lowered over a table will create a more intimate dining table light. Adjustable rise and fall pendant lights which pull up and down are a great choice of lighting for over a dining table, as breakfast bar lighting or for lighting over a kitchen island. The Lighting Company's selection includes double and triple rise and fall lights with two, three or more separately adjustable up and down lights on a bar suspension which are ideal for lighting over larger dining tables allowing you to choose the height of each light. Traditional French rise and fall pendant lights and contemporary or retro style industrial style pull up and down ceiling lights are also popular choices.</p> <p>Rise and fall lighting works best above tables and for lighting central kitchen islands, providing light where and when it is needed. Pull down lights also make a good choice of light fitting for a high ceiling as they can be raised or lowered as required. Pull down lighting can also be effective either side of the bed as an alternative to bedside table lights. The simple rise and fall pendants have a curly cable which allows the light to be lowered and raised back up again. The larger bar suspensions are suspended on sets of wires and the more traditional rise and fall lights have a balanced counterweighted mechanism to control the up and down movement of the light. The Lighting Company's selection includes traditional, classic and period rise and fall lights as well as modern rise and fall suspensions. The Lighting Company has a large selection of over table lighting, pendant lights that are also worth considering.</p> <p>Retractable rise and fall pendant lights have a spiralled cord which you can push up and pull down to alter the lighting effect. The other style (which tends to be more traditional), is a winch and pulley style light.</p> </blockquote> <p>The above section covers the &quot;what is it called.&quot; The how it works portion is relatively simple.</p> <p>A drum or winch body includes a spring and ratchets. The flat coil spring (typical) is lightly loaded when the winch body is fully loaded with the wire/cable being retracted.</p> <p>As the cable is pulled out, the spring accumulates tension. When the desired length is reached, a small release of tension on the wire causes the ratchet teeth to engage, holding the cable at that length.</p> <p>Ratchet mechanisms vary, but most require an additional pull to effect the release, which allows the spring to retract the cable to the desired location.</p> <p>Links abound for this information, but my contribution is from direct exposure by dismantling various spring loaded products and examining the works.</p>

54217

How does a pull down ceiling lamp work?

2023-02-13T12:14:29.667

<p>In flat slab system, say that the total length is 48m x 24m, column with drop panel with spacing of 6m... In this case , main bar outermost top bar (T1) and outermost bottom bar (B1) should be parallel to the 24m , right?</p> <p>I came across an example that main bar T1 &amp; B1 is parallel to 48m... Which one is correct ?</p> <p>Or it doesnt matter that whether mian bar is parallel to shorter / longer dimension of the overall flat slab system ?</p>

|structures|

<p>The direction of the main bars (T1 and B1) in a flat slab system depends on various design factors such as the span length, load distribution, structural efficiency, and code requirements.</p> <p>In general, the main bars are usually placed parallel to the shorter dimension of the slab to reduce the slab's overall span length and provide better structural efficiency. This is because longer spans tend to produce higher bending moments, which can increase the size and number of reinforcement bars required to resist the loads.</p> <p>However, there may be cases where the main bars are placed parallel to the longer dimension of the slab due to specific design requirements or to minimize the number of drop panels. In these cases, the design must account for the increased span length and ensure that the slab is adequately reinforced to resist the loads.</p> <p>In summary, the direction of the main bars in a flat slab system depends on the specific design requirements and considerations, and it can be parallel to either the shorter or the longer dimension of the slab.</p>

54218

Flat slab main bar direction longer or shorter

2023-02-14T04:23:34.170

<p>I'm considering repurposing a part of a rice farm into a farm house. I've seen this done elsewhere and after about 10 years, the entire floor just sank. (I saw in 10 years, didn't know the situation before). My question is what special considerations should I keep in mind to ensure that ground is solid. I will make a brick and mortar house.</p>

|civil-engineering|

<p>Two things cause soil subsidence in this context. 1) is the oxidative and also anoxic decay of all the accumulated organic matter that settles into the bottom of the rice pond through the decades, and 2) is soil deformation and slow flow due to the water content of the subsoil (which water leaks laterally out of the remaining ponds nearby).</p> <p>You can't do anything about settling due to decomposition of organics in the soil; note that this process also releases significant amounts of methane and carbon monoxide under the structure, which can then diffuse upwards and accumulate in the crawlspace to hazardous concentration levels.</p> <p>Avoiding soil deformation and flow requires that the structure be positioned far enough away from the ponds that the subsoil remains dry.</p>

54235

Repurpose part of a rice farm for a farm house

2023-02-15T04:45:14.733

<p>What is the metallic alloy used in camera cases?</p> <p>They are manufatured by what process?</p> <p><a href="https://i.stack.imgur.com/6ewCb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ewCb.jpg" alt="enter image description here" /></a></p>

|metallurgy|alloys|

<p>In mass production complex shapes like that are usually cast, then the casting is machined where accuracy and good finish is required. Magnesium is a good candidate for casting and Googling around about cast camera casings it seems many if not most of them use magnesium.</p> <p>This is the Samsung NX1 which is magnesium. <a href="https://i.stack.imgur.com/sx1KK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sx1KK.jpg" alt="enter image description here" /></a> <a href="https://en.wikipedia.org/wiki/Magnesium_alloy" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Magnesium_alloy</a></p> <p>Huh. Apparently magnesium is even lighter than aluminum. I did not know that.</p>

54246

What is the metallic alloy used in camera cases?

2023-02-17T13:13:18.330

<p>I did simple observation when I was heating water using stainless steel container and aluminum container. Indeed they are not exactly same but quite comparable. What I discovered was that, with the same amount of water I put inside them, I found like below:</p> <ul> <li>Using the aluminum container, the water get boiled faster.</li> <li>The aluminum container get cooled faster counted since the boiled water get boiled and the flame is turned off or I lifted the container off from the stove.</li> <li>When the water get boiled and the water is poured out, the aluminum container is get cooled faster compared to the stainless steel with the same treatment.</li> <li>After the water get boiled and poured out, the stainless steel container is hotter than the aluminum (for this, I didn't do scientific measurement with measurement tool, just based on my finger sense).</li> </ul> <p>I can say that <em>stainless steel container is like it has inertia to the thermal</em>. The container and the water inside was slower get heated/boiled but that container also slower get cooled. I am not very sure what kind of stainless steel was it, but probably it was 316. I searched in internet, its thermal conductivity is <strong>16.5W/m.K</strong>, while thermal conductivity of aluminum is <strong>247W/m.K</strong>, 15 times than the stainless steel's. There is also <a href="https://www.hardrok.com.au/thermal-activity-in-stainless-steel-compared-to-other-metals#:%7E:text=Thermal%20Conductivity%20Of%20Stainless%20Steel,watts%20per%20kelvin%20per%20metre." rel="nofollow noreferrer">mentioned</a> that stainless steel has among the lowest thermal conductivity of any metal.</p> <p>What I am asking is, what property the stainless steel has so it slower get heated but also slower get cooled as it has thermal inertia? Is that due to that thermal conductivity?</p>

|thermodynamics|heat-transfer|aluminum|

<p>I'll add another perspective, -- although probably DKNGuyen answer is what you are after ( I upvoted it). The reason I am also providing another answer is because you are also considering the rate (how fast the material changes temperature). From my view both properties a) <strong>heat capacity</strong> <span class="math-container">$c_p$</span> of the material, and b) <strong>heat conductivity</strong> <span class="math-container">$k$</span> of the material play a role.</p> <p>Heat capacity was covered in DKNguyen answer thoroughly, so I'll just say that it basically shows how much heat energy in Joules is required to raise the temperature of the material. The <strong>Specific Heat Capacity</strong> is in <span class="math-container">$\frac{J}{kg\cdot K}$</span> and translates to how much energy energy is required to raise by 1 degree Kelvin 1 kg of the material.</p> <p>However, because you are also looking for the rate of something getting hotter and colder, the thermal conductivity also plays a role. Thermal conductivity shows (simply put) how easy heat flows through a material. If there is no conductivity then there is an energy buildup, which leads to higher temperature at one end and no change at the other. So you could have a material with low specific heat capacity that is a good insulator (low conductivity) that would not allow the liquid inside the container to get hot.</p> <p>E.g. wood has a heat capacity about double of aluminium (1.76 compared to .9 kj/kgK), but its density is almost 1/3 of the aluminium (aluminium has 2700 kg/<span class="math-container">$m^3$</span> while most wood floats so it under 1000 kg/<span class="math-container">$m^3$</span>). So by volume, the heat capacity is less. However, you'd <em>never be able to boil water inside a wooden container</em>, because wood is a poor conductor of heat.</p> <p>So, if you are after boiling the liquid inside then the property of <strong>thermal diffusivity</strong> is probably the most relevant:</p> <p><span class="math-container">$$ D = \frac{k}{c_p\cdot \rho}$$</span></p> <p>where:</p> <ul> <li>D is the thermal diffusivity</li> <li>k is the conductivity coefficient</li> <li><span class="math-container">$c_p$</span> is the specific heat capacity.</li> <li><span class="math-container">$\rho$</span> is the density</li> </ul> <p>Basically, Diffusivity would show how fast would the temperature rise at one end of a bar that is heated from the other end. Higher values would indicate faster temperature rise.</p>

54281

What property cause a stainless steel slower get heated and slower get cooled, compared to aluminum

2023-02-18T08:58:46.083

<p>I am a novice in automatic control, the theoretical basis is not very good. We have a large electric clamping jaw, single degree of freedom, the motor is controlled by the torque output, the motor has a maximum stable clamping stiffness (This parameter is related to the gain margin, the greater the gain margin the greater the clamping stiffness can be). There is a rigid coupling between the motor and the gearbox.</p> <p>We simulated the Porter diagram of the motor with matlab and got the curve shown in the first diagram, Gm is 78.9dB.</p> <p>Then we changed the coupling to a flexible (elastic) coupling and got the second diagram below, Gm became larger, 87.7dB. My question is, when the coupling is changed to flexible, intuitively I feel it will cause resonance. But why is the gain margin bigger instead. Also, the part of the Gm increase appears a little to the right of the peak resonant frequency. Does this resonant frequency have any effect on stability?</p> <p><a href="https://i.stack.imgur.com/AVhUr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AVhUr.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/b2JR5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b2JR5.png" alt="enter image description here" /></a></p> <p>In addition, there is another question. For the above figure, the gain margin of 87.7, when I set the maximum gain to 87.7 near the stability will become worse, or that the frequency of my control signal reaches the resonant frequency when the stability will be affected</p>

|control-engineering|control-theory|pid-control|matlab|linear-systems|

<h2>&quot;Actual&quot; Gain Margin</h2> <p>Let the phase at 10 rad/s to 60 rad/s be -165 deg. i.e, 20 deg away from 180 deg. At those frequencies, this 20 deg distance from -180 deg will vanish if the system (for some reason) has an additional time delay of 35 ms and 6 ms respectively. Is the timing uncertainty in the various sub systems in your control system smaller than these numbers ? (I don't know how well your system is built and <em>modeled</em>).</p> <p>If not, your <em>actual</em> phase cross over frequency could occur between 10 and 60 rad/s and hence your <em>actual</em> gain margin may be as small as 50 dB.</p> <p>If the plot is from theoretical numbers, actual hardware could have phase cross over at 10 rad/s and increasing gain by 78 dB to get additional stiffness would lead to instability (possibly at oscillating at 10 to 60 rad/s rather than the resonant frequency).</p> <p>Marked as <code>A, D</code> below.</p> <p><a href="https://i.stack.imgur.com/epGKx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/epGKx.png" alt="bode plot" /></a></p> <h2>Behavior after increasing gain</h2> <blockquote> <p>For the above figure, the gain margin of 87.7, when I set the maximum gain to 87.7 near the stability will become worse, or that the frequency of my control signal reaches the resonant frequency when the stability will be affected.</p> </blockquote> <p>When you increase the gain by 87.7 dB, the theoretical phase cross over frequency occurs at about 600 rad/s (marked as C in the figure above). At that frequency, 600 <em>microseconds</em> of timing uncertainty can eat away what ever gain (or phase) margin is left. Similar comment about the point marked <code>A</code>.</p> <p>Also, note the points <code>A</code> and <code>C</code>. The new <em>gain cross over frequency</em> is at those points. i.e. you are operating your system at near zero phase margin. you can expect the system to have oscillations which are lightly damped and take several seconds to settle.</p> <h2>Effect of increased gains on Phase margin</h2> <p>Text books tell you that you need to maintain 60 deg phase margin where possible. It is preferred <em>not</em> to go below 60 deg phase margin unless you can tolerate responses with overshoot and slowly converging oscillations.</p> <p>From your figure, it appears that increasing gain by about 15 dB will reduce your phase margin to 60 deg. So that is possibly the limiting factor for increasing gains.</p> <h2>Effect of flexible attachments</h2> <blockquote> <p>My question is, when the coupling is changed to flexible, intuitively I feel it will cause resonance.</p> </blockquote> <p><a href="https://i.stack.imgur.com/VXHXB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VXHXB.png" alt="superimposed bode plots" /></a></p> <p>It appears that the peak response has increased in magnitude when flexible coupling is used. So, I think your intuition is correct to some extent.</p>

54293

What is the effect of the resonant frequency of the system function Porter diagram on the stability of the system and how to analyze it?

2023-02-22T19:03:19.090

<p>I have approximate measurements only, because I don't have physical access to the device. This chain is driving a small conveyor. The link length is about 9.5mm, ~9mm at narrow width and ~11mm at wide width. The chain is stamped with <code>890</code> number.</p> <p><a href="https://i.stack.imgur.com/KcD8j.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KcD8j.jpg" alt="enter image description here" /></a></p> <p>Is this a standard size chain? What is that standard, so I can look up a matching sprocket?</p>

|mechanical-engineering|chain|

<p>Looking at the photo, I believe the marking is not 890 but <strong>06B</strong>, a standard size for roller chains according to DIN ISO 606 with a pitch of 9,525 mm.</p>

54345

What type of chain is this?

2023-02-22T20:39:30.237

<p>So, I have 72 hydraulic McKibben actuators with 30cm in length and 14cm in width under 400 kPa (or 4 bar) that lifts a 5 ton weight that I want to actuate in a third of a second (continuously, actuating and un-actuating).</p> <hr /> <p>The explanation below is about <em>McKibben actuators</em>, <strong>if you have no clue about how these work</strong>, just imagine that these are hydraulic cylinders that work at 4 bar, have 30cm in length and 40cm of diameter and just move by 6cm.</p> <hr /> <p>McKibben muscles normally contract 20% of its length and increase 30-40% of its diameter.</p> <p>So this means that a single one of these muscles have an uncontracted volume of 4.618 liters inside of it.</p> <p>Taking into consideration the change of length in 20% and diameter in 40%, the contracted McKibben muscle would have a volume of 6.244 liters inside of it.</p> <p>The difference between these two would be 1.626 liters, which means that if I wanted to actuate all of the muscles in a third of a second, I would need 578.88 liters per second, so <strong>34732,8 liters per minute</strong>.</p> <hr /> <p>Most commercial hydraulic pumps have a certain limit of RPM, liters per minute and pressure. I don't think that even if the torque of the pump motor was super low, it would be able to pump as much fluid as this.</p> <p>So, regardless of making a custom pump or not, I believe that this is a really inefficient way of actuating these.</p> <p>One would need a lot of energy to pump 34000 liters of hydraulic oil, and it would also take a lot of energy to actuate cylinders that work around 12000 PSI, but need a few ml.</p> <hr /> <p>So, the question:</p> <p>Is there a way of achieving some kind of ideal balance between working pressure to fluid flow? Or I will need to test/calculate option to option until I find a certain balance?</p>

|fluid-mechanics|pumps|hydraulics|

<p>What you are describing is referred to in dynamical systems analysis as an <em>impedance matching problem</em>.</p> <p>For optimum transfer of <em>power</em> (effort x flow), the impedance of the <em>source</em> must equal the impedance of the <em>load</em>.</p> <p>See Karnopp &amp; Rosenberg, <em>System Dynamics: A Unified Approach</em> to learn how to solve for the impedances.</p>

54349

Is there a way of find the ideal balance between pressure and fluid flow in a hydraulic system?

2023-02-23T06:07:26.087

<p>I can't find a 16 teeth sprocket with hub and 18mm bore for 06B roller chain. There are sprockets for ANSI #35 chains with these specifications, though. The pitch is practically the same, the roller has a bit smaller diameter (5.08mm vs. 6.35mm) and is a bit narrower (4.77mm vs. 5.72mm). #35 sprocket will fit 06B chain with some lateral slack. Is this substitution going to work in a light application with about 10lbs chain tension?</p>

|mechanical-engineering|chain|

<p>They are not going to mesh properly because of the different roller diameter. I put a quick sketch together to illustrate it:</p> <p><a href="https://i.stack.imgur.com/niWhT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/niWhT.png" alt="#35-6B-chain" /></a></p> <p>The rounded tooth base is made to match the diameter of the chain roller, and only in the tooth base the pitch is correct. The rollers of the 6B chain are too large to fit into the tooth base of a #35 sprocket, so the rollers need to climb up on the tooth flanks to still fit. In the climbed-up position, the pitch doesn't match anymore, so this chain can never properly mesh with the sprocket.</p> <p>American Standard and British Standard (which the DIN ISO 606 orientates on) chains appear similar, but are not interchangable. You need to either find a 6B sprocket to fit your machine or change the entire chain drive to #35.</p>

54352

06B chain on #35 type sprocket

2023-02-23T10:19:58.387

<p>Very new to thermal conductivity calculations, so I'm aware there are already queries on here that may cover this, but I'm struggling to apply them to this instance.</p> <p>I'm trying to calculate the U-Value of a composite material. The data I have is the thickness of the materials used, and their thermal conductivity in W/mK. How do I use these to calculate the thermal conductivity of the material as a whole?</p> <p>My attempt so far has been to calculate the individual R-Values by dividing the thickness of the material in metres by the W/mK, and then sum those R-Values, and then divide 1 by that sum to get a U-Value. I'm sure I'm off somewhere though, as the U-Value has ended up huge!</p> <p>Appreciate any assistance, fairly sure it's a simple answer for someone familiar with the topic!</p>

|thermal-conduction|thermal-insulation|thermal-radiation|

<p>The answer depends on the configuration of your composite material.</p> <p>If the materials are arranged such that heat must transfer through each and every material in sequence, you would treat those material elements as a circuit in series: the total resistance to thermal conductivity is equal to the sum of the thermal resistances (R total = R1 + R2 +... Ri). In addition, the thermal &quot;current&quot; is your heat flow (Q) and the thermal &quot;voltage&quot; is the temperature difference (T2-T1 = Delta T). Simply solve the equation using the equation I = V / R or Q = Delta T / R total</p> <p>If, however, the materials are arranged such that heat may transfer through one material or another, you would treat those material elements as a circuit in parallel: the inverse of the total resistance to thermal conductivity is equal to the sum of inverses of the thermal resistances (1 / R total = 1 / R1 + 1 / R2 +... 1 / Ri). The thermal &quot;current&quot; and &quot;voltage&quot; are the same as shown above.</p> <p>If you have a combination of the two, you will solve for an R at whatever locations are arranged in parallel and then use them in conjunction with the sections that are in series in the Q = Delta T / R Total equation.</p>

54357

How to calculate the thermal conductivity of a composite material

2023-02-23T19:38:16.637

<p>The <a href="https://en.wikipedia.org/wiki/Lam%C3%A9_parameters" rel="nofollow noreferrer">Wikipedia article</a> has a completely separate table for Lamé parameter conversions in the 2D case. Philosophically, why is there even a different table for 2D if these parameters are meant to express intrinsic material qualities? Practically, how does one derive and use the 2D values?</p>

|elastic-modulus|

<p>Let me answer here with my own findings, since I am currently working on elasticity of membranes (non-linear membrane model, publications are to appear):</p> <p>For any isotropic membrane (2D) material, there are two membrane Lamé parameters <span class="math-container">$(\lambda,\mu)$</span>. We call <span class="math-container">$\lambda$</span>: first membrane Lamé parameter and <span class="math-container">$\mu$</span>: second membrane Lamé parameter of that material. They are defined analogously to the classical three dimensional Lamé parameters, by expansion of the stresses close to the natural state/reference configuration:</p> <p>For any membrane motion (family of time-dependend membrane deformations) starting in the reference configuration, there are two parameters <span class="math-container">$\lambda$</span> and <span class="math-container">$\mu$</span>, such that the derivative of the time-dependend membrane second Piola Kirchhoff stress tensor <span class="math-container">$S(t)$</span> with respect to time at <span class="math-container">$t=0$</span> can be computed from the derivative of the time-dependend membrane Cauchy-Green deformation tensor <span class="math-container">$C(t)$</span> at <span class="math-container">$t=0$</span> as follows:</p> <p><span class="math-container">$$ S'(0)= \frac{\lambda}{2}\mathrm{tr}(C'(0))\mathrm{Id}+\mu C'(0). $$</span></p> <p>They obtain a physical interpretation by choosing specific membrane motions, like pure shear (<span class="math-container">$\mathrm{tr}(C'(0))=0$</span>) and pure dilation (<span class="math-container">$C'(0)=a\cdot\mathrm{Id}$</span>). Then for membranes it turns out that <span class="math-container">$\mu$</span> is the membrane shear modulus <span class="math-container">$g$</span> and <span class="math-container">$\lambda+\mu$</span> is the membrane bulk modulus <span class="math-container">$k$</span>. (Note that a membrane pure dilation would correspond to an equibiaxial extension in three dimensions.)</p> <p>Concerning the relation inbetween the 3D and 2D values I think:</p> <p>Let <span class="math-container">$\Lambda$</span> and <span class="math-container">$M$</span> denote here the classical 3D Lamé parameters of a given isotropic material (Note the difference in the units: <span class="math-container">$M,\Lambda$</span> are in <span class="math-container">$\tfrac{N}{m^2}$</span>, whereas <span class="math-container">$\lambda,\mu$</span> are in <span class="math-container">$\tfrac{N}{m}$</span>). Then the membrane Lamé parameters for an isotropic elastic film with thickness <span class="math-container">$d$</span> in terms of the 3D Lamé parameters of that material are given by <span class="math-container">$$ \mu =d\cdot M $$</span> and <span class="math-container">$$ \lambda =d\cdot\frac{2M\Lambda}{\Lambda+2M} $$</span></p> <p>(Any elastic modulus of an isotropic material has its own membrane version, and these can be computed as in the table in Wikipedia. I would also very much like to know the source for that table. Thanks for your question.)</p>

54359

Why are Lamé parameters different in two dimensions?

2023-02-24T00:25:11.083

<p>I am learning about theoretical DC motors where the rotational motion is produced due to torque on either side of a coil.</p> <p>However, after researching real DC motors it seems to me that the winding pattern of the 3 coils would cause torque on each side to approximately cancel out , since the 2 ends of the coil are not on opposite sides of the circle, but rather wound about 1/3 of the armature (see image). <a href="https://i.stack.imgur.com/6sUjf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6sUjf.png" alt="enter image description here" /></a></p> <p>Instead, the motion is produced due to the 3 sections of the armature being magnetised as N and S poles to produce an attraction and/or repulsion from the stator poles.</p> <p>However, N and S poles are just an abstraction which apply equivalently to a theoretical DC motor, where one side of the coil is a N pole and the other is a S pole (e.g. in the image below the bottom of the coil is a N pole). <a href="https://i.stack.imgur.com/cgnke.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cgnke.png" alt="enter image description here" /></a></p> <p>Therefore, the underlying principles of the two motors must be the same, yet I cannot understand why. How can torque have any effect in the real motor construction? What equations are used to calculate the speed of rotation of a real motor if equations relating to torque on the coils are not applicable?</p>

|motors|electromagnetism|

<p>It's more than just the coils. In the case of a motor whose armature is made of some magnetic material (I'm going to say &quot;iron&quot;, even if it isn't), then the magnetic field produced by the armature is not, necessarily, lined up with the magnetic field that would be produced by the coils in isolation, or if they were wound on some non-magnetic material.</p> <p>So in the top picture, the armature does act pretty much like a bar magnet that has two south poles each 60 degrees off of vertical, and one north pole pointing down. The <em>average</em> magnetic field of the armature is pointing straight down, and since the magnetic field of the stator is across the page, that generates a torque.</p> <p>This is really asking too much from what's supposed to be an analogy, but I'm pretty sure that if you could actually measure the forces on each leg of the armature in the top picture you'd see this: the upper left one would be pulled down and to the left, contributing to torque because it &quot;wants&quot; to be aligned to the stator's north pole; similarly, the upper right one would be pushed up and to the left, trying to get away from the stator's south pole; the bottom leg would be pulled straight to the right, trying to simultaneously get away from the north stator pole and to the south stator pull.</p> <p>The net result would be the torque shown.</p>

54366

Physics behind a realistic and theoretical DC motor

2023-02-24T01:34:26.080

<p>I heard the following statement <a href="https://engineering.stackexchange.com/a/54322/40848">https://engineering.stackexchange.com/a/54322/40848</a></p> <p><strong>if we have a type III system, or one that has three or more low-frequency poles that we're closing around, then we have at least two gain margins: a low-frequency margin that defines the minimum gain for stability, and a high-frequency one that defines the maximum gain for stability.</strong></p> <p>Can anyone explain that or use an image to show that a type III system has two gain margins. I can't quite understand this thing, it seems too abstract.</p>

|control-engineering|control-theory|pid-control|frequency-response|linear-systems|

<p>Consider a plant with a transfer function equal to <span class="math-container">$$H(s) = \frac{250000}{s^2 \left(s^2 + 1000 s + 250000 \right)} \tag 1$$</span></p> <p>For some good reason (when I've done this it's for low-frequency disturbance rejection) we want to wrap it with a full PID controller:</p> <p><span class="math-container">$$G(s) = 250 + 50 \frac{1}{s} + 25 \frac{s}{0.005 s + 1} \tag 2$$</span></p> <p>The closed-loop response of this combination is <a href="https://i.stack.imgur.com/NGDX3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NGDX3.png" alt="enter image description here" /></a></p> <p>So, all-in-all a nice closed-loop response, if your goal was a loop that closes at around 7Hz.</p> <p>The <em>open loop</em> response is:<a href="https://i.stack.imgur.com/EIMG6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EIMG6.png" alt="enter image description here" /></a></p> <p>Note that the phase plot crosses -180 degrees at two frequencies, and two gains: a little over 0.2Hz and a gain of +40dB, and around 30Hz and a gain of maybe -20dB.</p> <p>What this tells you is that you don't <em>just</em> have a unidirectional gain margin. If your gain were to rise by 20dB, then your system would go unstable. But the story doesn't end there -- if your gain were to <em>fall</em> by 40dB, then your system would <em>also</em> go unstable. So -- you have two gain margins, both of which must be observed.</p> <p>(Note that in real life these margins can be worse than this -- I just whipped something up on the fly -- margins of <span class="math-container">$\pm$</span> 6dB aren't unheard of).</p> <p>As a test, I calculated the pole positions for the closed-loop system as given by (1) and (2), the system with the gain reduced by 43dB, and the system with the gain increased by 23dB. You can see that both of the cases where the gain margin is violated have unstable poles.</p> <div class="s-table-container"> <table class="s-table"> <thead> <tr> <th>as designed</th> <th>43dB down</th> <th>23dB up</th> </tr> </thead> <tbody> <tr> <td><span class="math-container">$s = -520 \pm j85$</span></td> <td><span class="math-container">$s = -500 \pm j7.7 $</span></td> <td><span class="math-container">$s = -603 \pm j225$</span></td> </tr> <tr> <td><span class="math-container">$s = -132$</span></td> <td><span class="math-container">$s = -200$</span></td> <td><span class="math-container">$s = -9.6$</span></td> </tr> <tr> <td><span class="math-container">$s = -16 \pm j9.0$</span></td> <td><span class="math-container">$s = 0.0016 \pm j1.33$</span></td> <td><span class="math-container">$s = 7.8 \pm j208$</span></td> </tr> <tr> <td><span class="math-container">$s = -0.20$</span></td> <td><span class="math-container">$s = -0.20$</span></td> <td>$s = -0.20</td> </tr> </tbody> </table> </div> <p>For reference, here's the Python script I used to do this analysis:</p> <pre><code>#! /usr/bin/env python3 import control import matplotlib.pyplot as plt import numpy as np omega_0 = 500 plant = control.TransferFunction([1], [1, 0, 0]) * \ control.TransferFunction([omega_0 ** 2], [1, 2 * omega_0, omega_0 ** 2]) tau = 5e-3 k_i = 50 k_p = 250 k_d = 25 freq = np.geomspace(0.01, 100) controller = k_p + k_i * control.TransferFunction([1], [1, 0]) + \ k_d * control.TransferFunction([1, 0], [tau, 1]) print(f'{controller = }') print(f'{plant = }') open_loop = plant * controller closed_loop = 1 / (1 + 1 / open_loop) print(f'{control.poles(open_loop) = }') print(f'{control.poles(closed_loop) = }') print(f'{control.poles(1 / (1 + 1 / (0.007 * open_loop))) = }') print(f'{control.poles(1 / (1 + 1 / (14 * open_loop))) = }') mag, phase, _ = control.frequency_response(open_loop, omega=2 * np.pi * freq) fig, ax = plt.subplots(nrows=2) fig.suptitle('Open-loop Bode plot') ax[0].semilogx(freq, 20 * np.log10(mag)) ax[0].grid(True) ax[0].set_ylabel('mag, dB') ax[1].semilogx(freq, np.unwrap(phase) * 180 / np.pi - 360) ax[1].hlines(-180, freq[0], freq[-1], 'r') ax[1].grid(True) ax[1].set_ylabel('phase') ax[1].set_xlabel('frequency, Hz') mag, phase, _ = control.frequency_response(closed_loop, omega=2 * np.pi * freq) fig, ax = plt.subplots(nrows=2) fig.suptitle('Closed-loop Bode plot') ax[0].semilogx(freq, 20 * np.log10(mag)) ax[0].grid(True) ax[0].set_ylabel('mag, dB') ax[1].semilogx(freq, np.unwrap(phase) * 180 / np.pi) ax[1].grid(True) ax[1].set_ylabel('phase') ax[1].set_xlabel('frequency, Hz') # plt.figure() # control.root_locus(plant * controller, kvect=np.geomspace(0.01, 1)) plt.show() </code></pre>

54369

why Type III systems has at least two gain margins?

2023-02-25T21:27:56.987

<p>One of the legs of a bed is bent (I believe it buckled). The bent can be corrected to be straight, however under the application of load to the bed, the leg will slowly return to the bent position. My goal is to fix the leg in place and stop it from sliding.</p> <p><strong>My current idea is to increase friction between the bed leg and floor.</strong> As a first pass, I tried putting an old rubber slipper underneath the bent leg after correcting it(best I could come up with). It seemed to help a bit, but on heavier load (eg: a person getting on and off), the bending happens again.</p> <p>I believe the material of the bed leg is some sort of plastic, and that of the floor is granite.</p> <p><a href="https://i.stack.imgur.com/jTj3X.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jTj3X.jpg" alt="enter image description here" /></a></p>

|friction|

<p>Here are a few possible solutions you can try to stop the leg from sliding:</p> <p>Furniture Grippers: You can buy furniture grippers, which are small pads that go underneath the legs of furniture to provide more friction and prevent sliding. These are available in many different sizes and shapes, and can be easily attached to the bottom of the bed leg. Look for grippers that are specifically designed for use on hard floors like granite.</p> <p>Rubber Pads: Another option is to use rubber pads or discs underneath the bed leg. These can also be purchased at most hardware stores or online. The rubber material will help provide more grip and prevent sliding.</p> <p>Anti-Slip Tape: Anti-slip tape is another option that can be used to increase friction between the bed leg and the floor. This tape has a rough surface that provides more traction, and can be easily cut to size and applied to the bottom of the bed leg.</p> <p>Glue or Adhesive: If none of the above solutions work, you can try using a strong adhesive or glue to attach the bed leg to the floor. This should be a last resort, as it may damage the floor if you ever need to remove the bed in the future. Use a strong adhesive that is designed for use on plastic and granite surfaces.</p> <p>Before trying any of these solutions, make sure the bed leg is straightened as much as possible to prevent any further bending. Good luck!</p>

54387

How to stop a leg from sliding?

2023-02-26T18:59:08.507

<p>I am using this little internal turning tool I have gotten from some company long ago as it was too worn out for them to use. By the time i gotten it for me as hobbyist, it was stil good for me to do some simple turning. Now after some time I think it starts to wear out. I would like to buy a new one, but I cannot figure out what kind of internal turning tool this is. Perhaps some internal turning tool to make internal screw thread?</p> <p>Can someone help me to identify what kind of tool this is and where it is used for so perhaps I can find a new one to buy. The thing itself is like 3cm so it is pretty small.</p> <p><a href="https://i.stack.imgur.com/u8u68.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u8u68.png" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/DOAg3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DOAg3.png" alt="enter image description here" /></a></p>

|tools|lathe|

<p>It looks a lot like a <a href="https://eshop.phorn.de/SM-System-105?changeLanguage=38" rel="nofollow noreferrer">PH HORN 105 system</a> boring insert, e.g.</p> <ul> <li><a href="https://www.mscdirect.com/product/details/77421105" rel="nofollow noreferrer">R105105 TN35, Solid Carbide</a> $36</li> </ul> <p>And you can get a holder for it, e.g.</p> <ul> <li><p><a href="https://www.mscdirect.com/product/details/04148474" rel="nofollow noreferrer">BU105 holder (new, 1&quot; OD)</a> $183</p> </li> <li><p><a href="https://www.mscdirect.com/product/details/04148474" rel="nofollow noreferrer">BU105 holder (new, 3/4&quot; OD)</a> $161</p> </li> <li><p><a href="https://gen3industrial.com/ph-horn-bu105-1000-5-01-boring-bar-holder-supermini-1-bu105/" rel="nofollow noreferrer">BU105 holder (used, 1&quot; OD)</a> $98</p> </li> </ul>

54396

Identify small internal turning tool

2023-02-28T13:30:29.610

<p>I'm looking to make something like a paint shaker - but I'm wanting to create perfectly rotational reciprocating motion. The picture below shows what I've been considering. The interface could be a problem, especially after a bit of use. I'm just wondering if anyone had a better way of doing it? <a href="https://i.stack.imgur.com/AgIsa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AgIsa.png" alt="enter image description here" /></a></p> <p>Many thanks,</p>

|mechanical-engineering|mechanisms|

<p>Something like this would work.</p> <p><a href="https://i.stack.imgur.com/vxEID.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vxEID.png" alt="enter image description here" /></a></p>

54409

Mechanism to convert simple rotational motion to reciprocating rotation

2023-02-28T17:19:42.973

<p>Take the simplest example of some uniform bar being stressed by a force N, I understand that we need to adjust an angle to determine the stress in terms of shear and principal stresses on that plane, but what are the physical reasons for this? Is it that at a point there will be different internal forces between particles that are positioned around that point depending on their relative position to the point?</p> <p>Why is shearing experienced at an angle to an axial force? What causes there to be a shearing effect on a plane that is at an angle to the direction of N?</p>

|stresses|solid-mechanics|

<p>What helped me understand this was drawing a square on the side of the bar and then watch what happens when you simply stretch the bar in axial direction.</p> <p><a href="https://i.stack.imgur.com/iUqhf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iUqhf.png" alt="enter image description here" /></a></p> <p>The physical reason is for this is a force equilibrium. Each square can in general experience normal and shear stresses on its edges. Combination of the normal and shear stress on the edge has to result in a force in axial direction in case of the stretched bar.</p>

54410

Why is it that plane stress varies?

2023-03-03T04:48:47.123

<p>I'm thinking about buying a hub motor that seems to be relatively common in the alibaba/ebay world but has a pretty strange shaft coming out of it (see pictures). It has a D profile that doesn't go all the way to the end of the shaft, and it seems to be 24.8mm in diameter, not 25 or 25.4. <a href="https://i.stack.imgur.com/jG7B6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jG7B6.png" alt="hub drawing" /></a><a href="https://i.stack.imgur.com/IUdx6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IUdx6.jpg" alt="hub image" /></a></p> <p>Is this some common size / shape that I'm not familiar with? Is there some specific part that it is supposed to mate with?</p> <p>Barring that, how would you design a simple, cheap mount for this motor? I have some ideas, but I'm interested in seeing what others come up with.</p> <p>I've also asked the supplier these questions but I don't have a lot of faith that I'll receive any feedback.</p>

|mechanical-engineering|motors|electric-vehicles|

<p>The motor design in the image provided is consistent with a hub motor for a bicycle or other electrically powered vehicle. I expected to see a matching flat on the opposite side of the shaft, to accept a typical bicycle fork. This does not exclude the concept that this shaft is designed to be fixed in place while the hub/motor assembly rotates, providing motivation to the vehicle on which it is installed.</p> <p>An appropriate mount would be a flat bottom fork with a &quot;cork&quot; or latch type securing device pushing the axle of the motor into the bottom, preventing rotation. One could also use a radiused-bottom slot with a set-screw of some sort to prevent rotation, but the flat bottom design provides for greater surface contact.</p> <p>The power/control wiring exits from the shaft, confirming that the shaft does not rotate.</p>

54422

How to mount hub motor with strange shaft profile

2023-03-04T06:11:13.710

<p>When I was meshing a helical coil, I've noticed these &quot;cube&quot; symbols located at the midpoint of a the block's edges:</p> <p><a href="https://i.stack.imgur.com/H3PCr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H3PCr.png" alt="Ansys ICEM CFD, Cube symbols on block edges" /></a></p> <p>Blocks with these symbols tend to have a lot of strange/unpredictable behavior. In one instance, I noticed that one of these blocks had bends along the edge lines that I expected to always be straight.</p> <p>This block was created by extruding the face below it along a helical curve.</p> <p>Here is an image of the geometry and blocking:</p> <p><a href="https://i.stack.imgur.com/l8paF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l8paF.jpg" alt="Geometry &amp; Blocking" /></a></p> <p><strong>What do these cube symbols mean, and can I replace these blocks with &quot;regular&quot; ones that always have straight edge lines?</strong></p>

|cad|ansys|cfd|ansys-workbench|meshing|

<p>If your geometry is curved, chances are it was meshed using higher order cells. Check out <a href="https://resources.system-analysis.cadence.com/blog/msa2021-mesh-order-explained-understanding-high-order-mesh-generation" rel="nofollow noreferrer">this article</a> for more information.</p> <p>The &quot;cubes&quot; at the edge centres would be additional nodes to define polynomial distribution of a quantity across a cell. Depending on the type, you can have additional nodes in the middle of edges, additionally in the middle of faces and there can also be one in the cell center.</p>

54429

In Ansys ICEM CFD, what do these "cube" symbols on the midpoint of block edges mean?

2023-03-05T12:07:08.607

<p>I have come across two definitions of kinematic viscosity and I am wondering about their relationship and significance in engine oils. <br> One definition describes kinematic viscosity as momentum diffusivity, while the other defines it as resistance to flow and shear due to gravity (commonly used in lubricants-specific texts). I am curious why there are two different definitions for the same fluid property and whether these definitions are related in any way.</p> <p>Additionally, I am curious why kinematic viscosity is more significant than dynamic viscosity for engine oils. I have noticed that oil companies only mention kinematic viscosity on oil bottles, such as SAE 5W-30, without mentioning dynamic viscosity. Is dynamic viscosity not useful practically?</p>

|mechanical-engineering|fluid-mechanics|automotive-engineering|fluid|

<p>In the daily practice of monitoring the output of a full-scale oil refinery, the process engineers need a simple, repeatable and quick field test for oil viscosity that could be performed right on the factory floor. Many years ago they settled on measuring the time it took for an oil sample to drain out of a standard measuring cup with a calibrated hole drilled in the bottom of it.</p> <p>The guy who invented the test was named Redwood, and the viscosity unit was named the <em>Redwood-second</em>.</p>

54434

Two Definitions of Kinematic Viscosity and it's Importance in Engine Oils

2023-03-06T08:09:20.040

<p>I'm wondering how to accomplish some kind of gear box that would reverse only the direction of one side of the two shafts on it, like in the picture below.</p> <p><a href="https://i.stack.imgur.com/2bLto.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2bLto.png" alt="enter image description here" /></a></p>

|mechanical-engineering|control-engineering|automotive-engineering|gears|

<p>You just gave me a great idea, i think. Check it out.</p> <p><a href="https://i.stack.imgur.com/acabQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/acabQ.png" alt="enter image description here" /></a></p> <p>Whats inside the dotted line 1 moves together with the main shaft, in a way that when the shaft goes down the bevels in 1 are connected to the side shafts making them move together in the same direction and when the shaft goes up bevel 2 connects to the side shafts making them rotate in opposite directions, this will gave me the effect i wanted.</p>

54445

Half bidirectional transmission shaft gear box solve

2023-03-06T21:11:08.593

<p>I have an aluminum cooling plate with a water path inside. The structure is simple- one part with a snake-like cut and a flat cover. To seal this thing I have designed a cut for an o-ring. The problem: design is tied to electronic parts and therefore changes from time to time, so i can't order o-ring of a certain length.</p> <p>I can see certain IP67 connectors that have seals that are not o-rings, but rather look like a compound.</p> <p><a href="https://i.stack.imgur.com/5rp7Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5rp7Q.png" alt="enter image description here" /></a></p> <p>Is there a compound that would form a seal, similar to what oring does - but without me worrying about length and exact profile?</p>

|seals|

<p>There are a couple of ways to accomplish your objective. The messier method would be to use a suitable RTV silicone compound, frequently found in squeeze tubes. The nozzle of the tube can be cut to provide an appropriate diameter for your purpose. Typical RTV silicone compounds require outgassing in order to properly cure, which for some products creates a vinegar-like odor. Newer versions also outgas but do not smell the same. If provision is not made for this release of gas, curing time is extended substantially.</p> <p>One could also use a two-part silicone molding material, available in liquid A/B components as well as in a putty A/B form. Both types cure via chemical reaction and usually involve relatively short curing times, especially compared to RTV types. The putty type can be manipulated by hand to create &quot;ropes&quot; to fit your application.</p> <p>If the cut is created via machining, one can duplicate that bit in a piece of suitable stock, apply the choice of silicone, allow it to cure and remove the now-ready-to-install &quot;O-ring&quot; in the part.</p>

54448

Compound to replace o-ring

2023-03-07T10:08:35.093

<h1>FEA engine</h1> <p>I'm using CCX finite elements analysis engine along with its pre- and post-processor CGX:</p> <p><a href="http://www.dhondt.de/index.html" rel="nofollow noreferrer">http://www.dhondt.de/index.html</a></p> <p><a href="https://github.com/calculix" rel="nofollow noreferrer">https://github.com/calculix</a></p> <h1>Example</h1> <p>I'm looking at one example here that is using <em>steel</em> material:</p> <p><a href="https://github.com/calculix/CalculiX-Examples/blob/a3ef0b86de47ad3d8f7e4bb39a3200c839a55f2c/Elements/Solid/solid.inp#L7" rel="nofollow noreferrer">https://github.com/calculix/CalculiX-Examples/blob/a3ef0b86de47ad3d8f7e4bb39a3200c839a55f2c/Elements/Solid/solid.inp#L7</a></p> <h2>Elastic specifications</h2> <p>In the example above, the declaration is done for Young’s modulus, Poisson’s ratio, and Temperature:</p> <pre><code>*ELASTIC,TYPE=ISO 210000,0.333333333,0 </code></pre> <h2>Density</h2> <p>Mass density is initialized by:</p> <pre><code>*DENSITY 7.85e-9 </code></pre> <h2>Gravity</h2> <p>Gravity, i.e. acceleration vector, is defined like this:</p> <pre><code>*DLOAD Eall,GRAV,9810.,0.,-2,0. </code></pre> <h2>Input values</h2> <p>Therefore, the input values are:</p> <ul> <li>Young’s modulus = <code>210000</code></li> <li>Mass density = <code>7.85e-9</code></li> <li>Gravity acceleration = <code>9810.</code></li> </ul> <h1>Question</h1> <p>What are the <em>units of measurement</em> for the above example?</p> <h2>My guess</h2> <p>I know the gravity acceleration is <code>9.8 m/s2</code>. So, <code>9810.</code> means <code>mm/s2</code>.</p> <p>But I'm confused about Young’s modulus and Mass density. What are their units? They should be consistent with the <code>mm</code>. But I cannot figure them out.</p> <h1>Solved</h1> <p>This table from the latest CalculiX solver <a href="http://www.dhondt.de/ccx_2.20.pdf" rel="nofollow noreferrer">documentation</a> helps with unit conversion. The units of the above example are <code>mm,N,s,K</code>.</p> <p><a href="https://i.stack.imgur.com/qi5xI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qi5xI.png" alt="Table of units conversion" /></a></p>

|materials|structural-analysis|finite-element-method|measurements|steel|

<p><strong>DISCAIMER</strong>: I haven't used calculix.</p> <p>From this <a href="https://web.mit.edu/calculix_v2.7/CalculiX/ccx_2.7/doc/ccx/node4.html" rel="nofollow noreferrer">page</a>, I found out that there are 3 predominant unit systems that are being used</p> <ul> <li>m,kg,s,K : meter, kilogram, second, Kelvin</li> <li>mm,N,s,K: millimeter, N, second, Kelvin</li> <li>cm,g,s,K: centimeter, g, second, Kelvin</li> </ul> <p>You are right in that the length units seems to be consistent with mm, so most likely the second is the one being used. If you notice though the second is the only one that used Force (N) instead of mass (kg, or g).</p> <p>This is consistent with the Modulus is 210000 MPa (Mega Pascal) i.e. <span class="math-container">$\frac{N}{mm^2}$</span> (modulus of steel is 201000 - 210000MPa).</p> <p>However it gets a bit confusing with the density. More specifically:</p> <ul> <li>Mass density of steel should be equal to 7800 <span class="math-container">$\frac{kg}{m^3}$</span>.</li> </ul> <p>In the page mentioned above you can see that density is given by: <span class="math-container">$\frac{N\cdot s^2}{mm^4}$</span> in the 2nd system of measurement. The reason is that because <span class="math-container">$mass = \frac{Force}{\left(\frac{Length}{time^2}\right)}\rightarrow mass = \frac{Force\cdot time^2}{Length}$</span>, and volume is <span class="math-container">$Length^3$</span>, therefore the units for density are:</p> <p><span class="math-container">$$[density] = \left[\frac{Force \cdot time^2}{Length^4}\right]$$</span></p> <p>The actual value of steel density is 7800 <span class="math-container">$\frac{kg}{m^3}\equiv \frac{N\cdot s^2}{m^4}$</span>. So when you convert m to mm you get</p> <p><span class="math-container">$$7800\cdot \frac{N\cdot s^2}{m^4}=7800\cdot \frac{N\cdot s^2}{(1000mm)^4}=7800\cdot \frac{N\cdot s^2}{10^{12}mm^4}=7800\cdot 10^{-12}\frac{N\cdot s^2}{mm^4}$$</span></p> <p>and finally: <span class="math-container">$$7800\cdot 10^{-12}\frac{N\cdot s^2}{mm^4}\Rightarrow 7.8\cdot 10^-9\frac{N\cdot s^2}{mm^4}$$</span></p>

54454

Units of measurement for structural analysis

2023-03-07T18:08:08.787

<p>I'm designing a machine that can be simplified as a log splitter. It will have a long, central beam that will resist both the forces of a force pushing out and the moment that will be caused by the force being resisted a distance away from the centerline of the beam. For simplicity, the two sides will have equal/opposite loads and moments. <a href="https://i.stack.imgur.com/AuKIq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AuKIq.png" alt="enter image description here" /></a></p> <p>It has been a few decades since I was in engineering school, so I want to check my assumptions:</p> <ul> <li>As far as beam deflection, the tensile force pulling the beam apart can be ignored</li> <li>As far as beam deflection, the important load is the torque/moment being applied by the force times the distance from the centerline.</li> <li>To use standard beam deflection formulas, I can model half the beam as a fixed cantilever beam with an end moment, because the other half will act as the &quot;wall&quot; the cantilever beam is attached to.</li> <li>For maximum tensile load, I would use the tensile load from the force as if it was applied to the center-line, and then add the forces caused by the moment.</li> </ul> <p>Assuming those assumptions are correct, I only have one remaining question: if the cross-section of the beam is not symmetrical top to bottom, how do I determine what the centerline of the beam is to determine the moment applied? (Assuming the stand offs are perfectly rigid)</p>

|beam|moments|cross-section|

<p>If by &quot;centerline&quot; you mean neutral axis in case of pure bending, it should simply go through the cross-section centroid.</p>

54459

Deflection of a log splitter beam

2023-03-10T08:06:36.880

<p>I have to calculate the pull forces of Fab and Fac. I have used extended Pythagoras to find the length of AC and AB but I have no idea how to calculate the forces. I also tried to use cartesian coordinates because I found a PFD with the same assignment somewhere on internet with the calculations but I don't understand it (<a href="https://people.utm.my/pauziahmuhamad/files/2018/12/Lecture_15.pdf" rel="nofollow noreferrer">https://people.utm.my/pauziahmuhamad/files/2018/12/Lecture_15.pdf</a>).</p> <p>Could someone help me with a solution preferably without the use of cartesian coordinates but with is ok, thanks in advance.</p> <p><a href="https://i.stack.imgur.com/FAjah.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FAjah.png" alt="" /></a></p>

|mechanical-engineering|mathematics|forces|

<p>Let's call the point vertically above O in the middle of B and C, D , the angle <span class="math-container">$AD$</span> makes with <span class="math-container">$AO$</span>, <span class="math-container">$tetha$</span> , its force <span class="math-container">$F_{1}$</span> then it length<br /> <span class="math-container">$$ L_{A D}=(0.6^2+0.2^2)^(1/2)=0.632m$$</span></p> <p>And the length of AC,</p> <p><span class="math-container">$$L_{AC}=\sqrt{0.633^2+0.2^2}=0.663m$$</span> We get the moment aboutO.</p> <p><span class="math-container">$$\Sigma m_O=0 \quad \rightarrow 375*0.6= 0.3m*cos(F_1)$$</span></p> <p><span class="math-container">$$cos(F_1)=375(0.6/0.3)=750 $$</span></p> <p><span class="math-container">$$F_{AD}=750\frac{0.633}{0.6}=790N$$</span> This force, 790N is divided between the AB and AC. <span class="math-container">$$790/2=395N$$</span></p> <p><span class="math-container">$$F_{AB}=F_{AC}=395\frac{0.663}{0.632}=414.73N$$</span></p> <p>I used your figure to show the details.</p> <p>.</p> <p><a href="https://i.stack.imgur.com/Puk8U.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Puk8U.jpg" alt="trss" /></a></p>

54478

Calculating forces on cables

2023-03-12T17:12:49.680

<p>For example in this link you can see a 2 speed gearbox for a high speed boat <a href="http://www.coanracing.com/custom-block2/transmissions.html/xtreme-marine-transmission.html" rel="nofollow noreferrer">http://www.coanracing.com/custom-block2/transmissions.html/xtreme-marine-transmission.html</a> , what even is the purpose of having different speeds in a boat? In the link it says it allows faster accleration and better manuevering at low speeds but how exactly does a transmission help with those things in a boat or a ship and why dont other ships also use them? This just seems like its only used by high speed boats but i might be wrong.</p>

|gears|transmission|ships|

<p>It's for exactly the same reason that cars have multispeed transmissions, as follows.</p> <p><em>Power</em> is always the product of torque times RPM (times the right conversion constants to account for units). This means an internal combustion engine produces peak power when it is running at the highest RPM it can handle without blowing up.</p> <p>But when initially accelerating from zero velocity, the engine cannot spin its propeller at full speed because its propeller is pitched to match the load presented to the engine when the boat is travelling at its maximum speed. By putting a reduction gear into the driveline, the engine can then spin all the way up to its redline, at which point it is screaming its guts out- and delivering its maximum power output to the prop and hence developing maximum acceleration for the boat.</p> <p>This extra oomph sets the boat in motion as quickly as possible- after which you clutch out the reduction gear, slam it into direct drive and max the throttle until you hit redline again.</p> <p>Note that it is common for boat owners to use one prop for yanking water skiers up and out of the water as quickly as possible, and a second prop for maximum top speed. The pull prop often has four blades for maximum thrust and the speed prop will have just two. A compromise prop will have three blades. So on water ski day, the pull prop is installed and on race day, the speed prop is installed.</p> <p>This process is done in racing cars and motorcycles with as many as 6 or 7 different drive ratios, to keep the engine running up at its max power point no matter what the wheel speed is.</p> <p>In airplanes, this same thing is done by changing the blade pitch instead of shifting gears. You set <em>fine pitch</em> for takeoff to spin the engine up to max power and as your speed builds, you dial in more bite to the blades. In inexpensive planes (just like in boats), you instead choose between installing one of two different propellers: the fine-pitch model is called a <em>climb prop</em> and the deep-pitch model is called a <em>cruise prop</em>.</p>

54498

Why do high speed boats have 2 speed transmission?

2023-03-13T17:03:10.963

<p>this is my first post here. I'm a software engineer by trade, but I have a 3d printer so that makes me a mechanical engineer, right?</p> <p>Clearly not. However I do have some rudimentary skills and have built some basic things. This project is my most ambitious to date. I have a design in mind that I think is relatively simple, and I wanted some feedback on it.</p> <p>First, the concept: an array of 8 laser pointers mounted into a device on the ceiling, which will point to the locations we need to place balls for various games (because he doesn't want to mark up his very nice felt on his very nice table). We play snooker, cowboy pool, and other games that have frequent ball-placement in spots that we have to eyeball, and it would be a lot easier (and cooler) if we had a laser pointer to mark the exact spot for each ball.</p> <p>Well I've been getting into 3d printing and hobby electronics like Arduino etc, so I think this is a doable project, and I have a design already in mind for this device. Like I said, it's very simple (but also, very complex, depending on how you look at it).</p> <p>The design (I'm not great with drawing so I will describe the design carefully, but I can illustrate anything people have trouble visualising):</p> <p>8 laser &quot;modules&quot; that pop into a central device. Each module has 2 bearings, one on the side (perpendicular to the table), and one on the bottom (parallel with the table). The bearings each have a gear attached to them. The bottom gear/bearing turns the whole laser module in a circular motion. The side gear/bearing has an axle which goes through (or attaches to) the center of the laser pointer, and can move the laser pointer linearly.</p> <p>So each laser platform can be turned on an axis perpendicular to the table to do a &quot;circular sweep&quot; by turning the bottom gear, and the laser can be moved in a &quot;linear sweep&quot; using the side-gear (with an axis parallel to the table). Essentially it's just 8 little robots inside of a housing structure. The housing structure is responsible for engaging the circular sweep of each module, so it has 8 gears in it, to one for each laser module.</p> <p>The modules have the linear sweep &quot;built-in&quot; or in other words, controlled separately. So the central unit is responsible for turning the laser modules, but they do the linear sweep independently of the main unit.</p> <p>Here's my first question - the only way I can think of to make this work is to have independent servos on the linear sweep of each laser, that way the module can be moved with the circular sweep without having to worry what direction the linear sweep drive motor is oriented. In other words, I'm planning to have 16 RC servos. 2 for each laser module. Each module has one servo for the circular sweep drive, and one servo for the linear sweep drive.</p> <p>Yes, it's a lot of parts that can fail, but RC servos are really cheap, the idea is that all the parts I use will be cheap parts, and we can just snap in new ones if they break. But maybe there is a simpler design that's within my design skills that I haven't thought of, because I'm not a mechanical engineer. I'm very open to suggestions.</p> <p>The next part of my question is a very naive and kinda technical one. My idea for saving the positions of the lasers for particular games is this:</p> <p>On the other end of each drive axle, I was thinking I will place a potentiometer. The orientation of the potentiometer for a particular game can be saved by the device to storage, SD or something, I'll need to write the code for it, obviously, but the basic idea is to orient the lasers manually once, save the position using software, and then when we want to play Snooker we just hit the Snooker button and the lasers orient themselves in place.</p> <p>My question here is, given the level of accuracy I will need to have in order to save/load a position of the lasers, and re-orient them in software when we change the game, can saving and reloading the locations using a potentiometer measurement be accurate enough? Will I need expensive pots? Or should I consider a different approach from the start? I don't think a pure mathematical solution would work, the amount of skill it would take to mount the device on the ceiling &quot;perfectly&quot; to support a 100% math-driven approach is out of my reach, and besides, stuff moves over time, even pool tables, floors and ceilings. So I think saving and loading of positions using sensors is the way to go.</p> <p>Any and all ideas are welcome. Thanks for your time!</p>

|mechanical-engineering|electrical-engineering|robotics|

<p>It's probably cheaper for you to get a crapload of laser pointers for every position you want rather than to build gearmotors and all that. Definitely easier to engineer but I can see why you might not want to since the end-result is cumbersome.</p> <p>With regards to your proposed gimbaled approach, rather than eight separate laser pointers on gimbals and rails, might I suggest just one laser pointer pointing into a spinning mirror to scan the entire table and you instead turn the laser on whenever it points at a position you want to light up? The design and hardware involved is more complex but there is ultimately less hardware involved. If you have the means to get such a thing machined and the mathematical prowess, you might be able to design a mirror where rotation about one axis scans the laser across the entire table in a spiral. If you hook it up to a motor with encoder it mostly becomes straightforward programming problem.</p> <p>This is the sexiest approach in the most compact package and could be augmented with vision for calbration or position sensing rather than relying on blind motor coordinates. That said, I can understand if you feel it is beyond you</p> <p>But the more conventionally approach would probably be two motors each spinning for a mirror for the X and Y axis respectively with one mirror re-directing the laser into the other.</p> <p>RC servos aren't the most accurate devices. Linearity of input signal vs position isn't exactly the most important for the applications they were designed for and the position vs input signal will also vary from servo to servo. You're going to have to calibrate each servo at minimum due to repeatability between devices, or outright set custom coordinates for each position if the linearity is bad enough. And they may not stay constant with age either. With 16 servos you would likely might be fiddling with the system all the time.</p> <p>There are servos that use encoders which are more precise and should be more repeatable between devices. These look similar to RC servos and often accept things like serial inputs. They are targeted towards the hobbiest robotics market. They also cost more like high-end RC servos. Sometimes a lot more than the highest end RC servos. So if you need 16 of them this is probably not an option.</p> <p>In short, I do not recommend you rely on rotary position encoders as a primary method of positioning your laser.</p> <p>Let's split this into actuators vs position sensors now:</p> <p><strong>ACTUATORS:</strong></p> <p>If you are adding external position sensing anyways, then you might as well go for nicer solution: Brushless gimbals. These are targeted towards camera platforms for hobby UAVs. They will be smoother and quieter than RC servos. They have less torque but that doesn't really matter for your application as long as you design the gimbal so that the load is rotated around the COG so there isn't always a cantilever constantly applying load to the motor. The motor torque is just there to accelerate and decelerate the load into position.</p> <p><strong>POSITIONS SENSORS:</strong></p> <p>Brushless gimbals, by nature of their intended application tend to use gyroscopes and accelerometers on their control board for holding position and motion stabilization. You could surely finagle accelerometers for use as attitude (tilt, such as yaw and roll) sensors but the heading is an issue. You probably don't want to use a compass or light beacon or anything similar for heading. That said, it may be useful to mount accelerometers and gyros on each gimbal to track the motion of the gimbal itself. In your application, you would need accelerometers for pitch and roll and gyros for yaw. Best results would come from gyros and accelerometers for all 3-axis but this would likely be excessive for your application is rather static. However, this would be more motion control and not position sensing.</p> <p>For positionin sensing, consider using machine vision. Hobbiest level devices like CMUCam exist with onboard processing hardware and object detection and analysis libraries which should expedite this process. You use the camera to calibrate your lasers or outright position them. An empty green pool table is rather ideal for this kind of thing. It enables you to use just one camera as a position sensor for all motors and makes mounting accuracy of those motors almost irrelevant. You can try to get the camera to calibrate to the four corners of the table or fix four light beacons to the table or lasers to shine at four points on the table as calibration points for the camera.</p> <p>You could use a camera with the RC servos too, but there's not much point except as maybe a stepping stone.</p> <p>In any case, whatever gimbal you use, make sure it's designed such at the gimbal motor's load is rotating around the COG, not as a cantilever constantly applying load to the motor.</p>

54510

Building a laser-pointer array for my buddy's pool table

2023-03-13T21:03:26.427

<p>So, I was trying to figure out how to figure out how much force I would need to apply to these articulated arms in order to make them extend or &quot;contract&quot;, but I can't figure out.</p> <p><a href="https://i.stack.imgur.com/lVyfn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lVyfn.png" alt="Photo of a torque reaction arm" /></a></p> <p>I tried to search on google, but I couldn't find any information/formulas on the subject, only product lists and ads selling said articulated arm.</p> <hr /> <p>To be honest, what made me interested on knowing the answer to this was the megabots mech that uses a similar mechanism for its legs, but the question isn't about it, I'm just wondering.</p> <p><a href="https://i.stack.imgur.com/MCL0a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MCL0a.png" alt="Photography of a megabot mech" /></a></p>

|mechanical-engineering|mechanisms|linkage|forces|

<p>In each joint, The moments are the weight of the load plus the weight of the lift arm(s) multiplied by the distance to the joint.</p> <p>setting the moments sum about the joint equal to zero will give equilibrium. An increase in the spring tension, F, or hydraulic actuator force will lift the arm or vice versa.</p> <p><span class="math-container">$$\Sigma M_{joint}=0 \quad P*D-sin(\theta)*F_{spring}*C=0$$</span></p> <p>The wider the distance K the less power is needed to lift the load.</p> <p>I used a drafting light but the concept is the same.</p> <p>.</p> <p><a href="https://i.stack.imgur.com/kiXW1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kiXW1.jpg" alt="lamp" /></a></p>

54512

How to calculate the mechanical advantage of torque reaction arms (or balanced articulated arms)?

2023-03-14T03:25:58.683

<p>I am trying to understand how to describe and calculate the shape a length of piano wire makes from its own weight when supported vertically at only one end. Is it a catenary or a parabola? Specifically I am trying to calculate how long a piece of piano wire I would need to have it arc exactly 180 degrees to contact the same plane to which it is attached, and the distance from the attachment point and this unattached contact point.</p> <p>How would this equation differ for a thin strip of spring steel? Simply using different Young's modulus and Poisson ratios for x and y components?</p> <p>BTW, this isn't a student homework problem, I am a machinist making a strange apparatus for a science museum, and trying to push beyond empirical solutions to these kinds of problems. I'm sure there is a very broadly applicable deformation equation which could be applied here, but I don't know what to search for.</p> <p>Thanks in advance!</p>

|mechanical-engineering|springs|buckling|deformation|

<p>Rafe</p> <p>Here is the Delphi code that does all the calculation. It should be straightforward to convert it to C or your favourite language.</p> <p>The wire is modelled as a chain of short straight segments, each one is angled slightly relative to its predecessor according to the bending moment it's seeing.</p> <p>You give it a wire diameter and length and it calculates the coordinates of the segments.</p> <pre><code>procedure TForm1.PaintBox1Paint(Sender: TObject); const n: integer = 100; // number of segments (wire rod is divided into short straight segments) const density = 7850; // kg per m^3 const modulus = 200e9; // pascal = N per m^2 const dia: double = 0.0005; // dia of wire in m const g = 9.81; // acc of gravity var i,j,pass: integer; dL: double; // length of each segment MoI: double; // moment of inertia of each segment moment: double; // moment applied to a segment; mass,c,ang: double; // used in calculating new curvature x,y,curvature: array[0..10000] of double; // position and curvature of each segment begin with Paintbox1,Canvas do begin pen.color:=clgray; moveto(0,400); // 400 is the height of the table on the screen lineto(width,400); pen.color:=clblack; dL:=SpinEdit1.Value/n/1000; // SpinEdit1 sets the total length in mm dia:=FSpinEdit1.Value/1000; // FSpinEdit1 sets the wire dia in mm MoI:=pi*dia*dia*dia*dia/64; // moment of inertia // initialise the curvature of each segment // it's a semicircular arc // curvature = 1/radius for i:=0 to n do curvature[i]:=pi/(n*dl); mass:=dL*pi*dia*dia/4*density; // mass of each segment // simulate the wire springinging into place // 100 iterations for pass:=1 to 100 do begin // calc position and curvature of each segment y[0]:=0; x[0]:=0; ang:=0; for i:=1 to n do begin ang:=ang+dl*curvature[i]; x[i]:=x[i-1]+dl*sin(ang); y[i]:=y[i-1]+dl*cos(ang); end; // calc new curvature of each segment for i:=0 to n do begin moment:=0; for j:=i+1 to n do moment:=moment+(x[j]-x[i]); moment:=moment*mass*g; c:=moment/(MoI*modulus); curvature[i]:=curvature[i]*0.9 + c*0.1; // change in curvature is damped end; end; moveto(round(x[0]*1000),round(400-y[0]*1000)); for i:=0 to n do lineto(round(x[i]*1000),round(400-y[i]*1000)); end; Label3.Caption:= 'dist='+inttostr(round(x[n]*1000)); // horizontal distance to touching place end; </code></pre>

54514

Describing 180 degree arcing/deflection of wire supported vertically at one end

2023-03-17T20:14:31.227

<p>So, I'm trying to calculate how much fuel/energy I would need to input in a homemade monotube steam turbine engine in order to achieve 15hp (11kw), the idea is that the engine will run at this specific power, not that I will take 11kw out of a generator.</p> <p><a href="https://i.stack.imgur.com/kF7Z2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kF7Z2.png" alt="Illustration of the homemade monotube steam turbine engine" /></a></p> <p>(I didn't add a water pump because it is out of space on the image, but just pretend that in the bottom check valve there is a water pump)</p> <p>I didn't knew where to start, so I looked on a <a href="https://jakadofsky.com/index1.php?bereichID=7&amp;lang=en" rel="nofollow noreferrer">website about micro jet turbines</a> and found that the 10kw (13-14hp) turboshaft has a length of 30cm and a diameter of 14cm and can achieve 100,000 RPM. Even though this doesn't necessarily means that the turbine is 14cm in diameter, I took that value to base myself.</p> <p>With that in mind, I used an <a href="https://www.codecalculation.com/htm/calculate/mechanical/engines/steam-turbine-performance/" rel="nofollow noreferrer">online steam generator calculator</a> that I found on the internet, and writing down that I would achieve <strong>10 bar</strong> of pressure on the input, <strong>450-500 ºC</strong> of temperature and <strong>11kw</strong> of output, the calculator gave the conclusion that I would need <strong>0.02797 kg</strong> of mass flow per second.</p> <p>I assumed that this mass flow means flow of steam, I also assumed that I would need to convert 0.2 liters of water into steam per second, which would give 12 liters per minute. However, I asked in other websites, and some people gave me the answer that it should be 0.02 liters of water per second, which would be just around 1.2 liters per minute.</p> <hr /> <p><strong>The question:</strong></p> <p>Taking into consideration that the heat source is 450 ºC, the copper tube has 6.35 mm of outer diameter with 0.79 mm of wall thickness, and the heat transfer of the copper tubing, what should be the length of the copper tube in order to convert 0.2 liters of water per second into 0.02kg of steam?</p>

|fluid-mechanics|thermodynamics|heat-transfer|steam|steamengine|

<p>ASME (USA Clever Engineers) state that copper shall not be used above 100psi... because it loses strength too much above 200deg.C. So do not go to 450psi !!!</p>

54558

How long a spiral copper tube would need to be in a monotube boiler in order to achieve 0.02 kg of steam per second?

2023-03-20T14:21:24.973

<p><a href="https://i.stack.imgur.com/KjJ8p.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KjJ8p.jpg" alt="enter image description here" /></a> I found this in the USS Yorktown aircraft carrier (it’s open for public tours)</p> <p>I know it’s a 3 phase connector of some kind, but what type of plug connects to it? Is it a nationwide standard or something specific for the ship?</p> <p>In addition to the 3 phases, there’s also 3 separate holes at the top of for ground</p>

|electrical-engineering|marine-engineering|

<p><a href="https://i.stack.imgur.com/xlgzO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xlgzO.png" alt="enter image description here" /></a></p> <p>The surface Navy had (has?) a whole set of casualty electrical power systems for use in restoring power. This is part of those systems. I say surface Navy because I sure don't remember this on my submarine.</p>

54583

What is the name of this power connector used in the USS Yorktown?

2023-03-20T17:54:56.347

<p>lets take a component like a fan blade as an example. Is it right to conclude that it will fail even when the stress is within permissible limits but its operating frequency is close (close enough than the resonance separation margin) to first mode natural frequency?</p> <p>Kindly let me know your views. Thanks</p>

|stresses|solid-mechanics|frequency-response|fan|

<p>It depends on what it is made of, and how large the vibratory stresses happen to be. If the stresses are low and the material is steel, it is possible that a <em>fatigue failure</em> due to crack propagation will not occur, so long as the stress is below that steel's <em>fatigue limit</em>. It is more likely to occur with higher stresses and aluminum instead of steel, because aluminum and its alloys do not possess a fatigue limit.</p> <p>More information on metal fatigue can be had from an introductory materials science text like Shakelford or Van Vlack.</p>

54588

Failure of component due to natural frequency vs stress

2023-03-20T21:21:29.080

<p>I will be mounting a sprocket gear on the 2 inch input shaft to my custom piece of industrial machinery I am preparing. The sprocket gears are available in 1 15/16 and 2 inch bore diameter. Would there be an advantage to having the end section of my shaft turned down to 1 15/16 in ? I feel like if the shaft was a bit smaller in that section, it would help keep the sprocket in place, and act as sort of a built in snap ring to hold the lateral position of the gear.</p>

|gears|machining|

<p>The word you're looking for is &quot;shoulder&quot;. It might help but it's kind of pointless during operation if you are using a setscrew. Might help with assembly I guess since there is less thinking involved If it were square or D-shaft it and no setscrew then it would help since you would only need one snap ring or shaft collar.</p>

54591

Advantage to reduced diameter where a gear will mount to shaft?

2023-03-22T02:54:21.547

<p>Why was Trinity Lake's <a href="https://cdec.water.ca.gov/jspplot/jspPlotServlet.jsp?sensor_no=9157&amp;end=&amp;geom=small&amp;interval=30&amp;cookies=cdec01" rel="nofollow noreferrer">control regulating discharge</a> increased from 0 to 1,477 CFS on March 6 .. 20, 2023? (During historic drought in California.) What is &quot;control regulating discharge&quot;?</p> <p><a href="https://i.stack.imgur.com/sJElO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sJElO.png" alt="enter image description here" /></a></p> <p>It seems difficult to justify discharge from either a consumption or flood control standpoint. Currently water consumption is low because of constant rains across the state, combined with strict irrigation limits due to continuing drought. While most large California reservoirs are at seasonal average capacity as of March 20, 2023, Trinity Lake is at only 50% of its average for this time of year. Most consumers have access to lower-altitude sources like Oroville (which is currently overflowing) and Shasta (which is just 70 miles downstream of Trinity and currently at seasonal average capacity). In this situation, wouldn't we expect water managers to <em>retain</em> water in high-altitude lakes, particularly the ones that are taking longer to fill up?</p>

|hydrology|

<p>As Walt's answer points out, water is being dumped for &quot;flow augmentation&quot;. I.e., environmental purposes. The specific program is the <a href="https://www.trrp.net/" rel="nofollow noreferrer">Trinity River Restoration Program</a> which aims “to restore and maintain the Trinity River’s anadromous fishery resources … [by] … rehabilitating the river itself”</p> <p>TRRP is currently <a href="https://krcrtv.com/news/local/high-releases-from-trinity-lake-concern-residents" rel="nofollow noreferrer">under investigation</a> for massive water waste during historic drought.</p>

54602

Trinity Lake Leak

2023-03-22T03:38:17.197

<p>I create Auto CAD Drawing using Primary units decimal (mm). I need to add alternative unit as Architectural. What is the suitable multiplier for all units?</p> <p><a href="https://i.stack.imgur.com/rmqr0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rmqr0.jpg" alt="Modify dimension Style" /></a></p> <p>After adding new value</p> <p><a href="https://i.stack.imgur.com/0P3Np.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0P3Np.jpg" alt="Updated Dims" /></a></p>

|autocad|

<blockquote> <p>In 1930, the British Standards Institution adopted an inch of exactly 25.4 mm. The American Standards Association followed suit in 1933. By 1935, industry in 16 countries had adopted the &quot;industrial inch&quot; as it came to be known, effectively endorsing Johansson's pragmatic choice of conversion ratio.</p> </blockquote> <blockquote> <p>In 1946, the Commonwealth Science Congress recommended a yard of exactly 0.9144 metres for adoption throughout the British Commonwealth. This was adopted by Canada in 1951; the United States on 1 July 1959; Australia in 1961, effective 1 January 1964; and the United Kingdom in 1963, effective on 1 January 1964. The new standards gave an inch of exactly 25.4 mm, 1.7 millionths of an inch longer than the old imperial inch and 2 millionths of an inch shorter than the old US inch.</p> </blockquote> <p><a href="https://en.wikipedia.org/wiki/Inch" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Inch</a></p>

54603

What is the suitable multiplier for all units?

2023-03-25T01:36:49.457

<p>As mass flow rate increases so does the heat coefficient. Is there a way to directly relate this? Is it because of the flow going from laminar to possibly turbulent also help increase heat transfer. This is an example based on a Shell and Tube heat exchanger.</p>

|heat-transfer|

<p>Relation between heat transfer coefficient and mass flow rate could be easily related only in laminar flow, which will basically not occur in correctly designed heat exchanger. Heat transfer is strongly influenced by turbulence but also by other factors, so in the end, the difference in complexity of heat transfer in laminar vs. turbulent flow is greater than between the flows in these regimes.</p> <p>For shell and tube heat exchangers, there are semi-empirical methods like the Kern's or the Bell-Delaware. You could also check out <a href="https://hedhme.com/content_map/?link_id=666&amp;article_id=259" rel="nofollow noreferrer">Flow Stream Analysis</a>.</p>

54625

Importance of mass flow rate on the overall heat transfer coefficient

2023-03-25T10:58:00.747

<p>What mechanisms can be used to apply fast vibration to an object? One way is <a href="https://engineering.stackexchange.com/questions/54409/mechanism-to-convert-simple-rotational-motion-to-reciprocating-rotation?noredirect=1#comment103342_54409">this</a>.</p> <p>What are some other ways to do it?</p> <p>I'm looking to go as fast as possible - 100s of Hz if possible.</p> <p>Thankyou</p>

|mechanical-engineering|mechanisms|vibration|

<p>100 Hz is what we call „bass“ in music. So a well proven device is called „speaker“, which comes in many flavors, and which is excited by electric signals, like sine, square, saw, noise, voice etc.</p> <p>Try picking a suitable system and modify it.</p> <p>Here‘s a short video about <a href="https://www.pond5.com/de/stock-footage/item/22039903-sugar-loud-speaker-cone-vibrating-sound" rel="nofollow noreferrer">sugar on a vibrating speaker</a>.</p>

54627

Mechanism for fast vibration

2023-03-25T19:07:52.647

<p>When producing lime on a lime kiln, there exists the possibility of overburning the original limestone and producing the so-called <strong>dead-burnt lime</strong>, with a much lower reactivity than quicklime.</p> <p>Some informal source states that dead-burnt lime is just sintered quicklime: if this is the case, how does sintering affect reactivity, provided the product is milled to the desired granulometry after production?</p>

|process-engineering|

<p>After calcium carbonate (lime rock) is heated in a kiln to form calcium oxide, it must be <a href="https://en.wikipedia.org/wiki/Slaking" rel="nofollow noreferrer">slaked</a> (mixed with water) to form calcium hydroxide. This may occur right away in a sugar processing plant, or later for premix concrete. Note that the mined lime rock is never perfectly pure and contains impurities such as silica (the primary component of common beach sand). If the kiln temperature gets above 1300C, impurities in the lime rock such as silica <a href="https://en.wikipedia.org/wiki/Vitrification" rel="nofollow noreferrer">sinter or vitrify</a> and become glass. The glass is intermixed in the structure and greatly reduces the rate at which the calcium oxide can dissolve in water and consequently the over heated lime is termed &quot;dead burnt lime&quot; or &quot;overburned lime&quot;. Grinding will certainly help by increasing the surface area and improve the acceptable range of overheating, but the impurities are intermixed at a molecular level and will still reduce the rate of the reaction.</p>

54631

What is exactly dead-burnt lime, and why is less reactive than quicklime?

2023-03-26T11:44:06.103

<p>This is meant for mechanical closed system for hydroponics. Two wheels connected by same shaft so the wheel going down lifts the water back with the other interconnected wheel to a slight incline. I suppose the torque created by falling water should be higher and so there's should be a difference in wheel's cups volume or radius. A high reservoir also could be used to be replenished by gravity ballast lifting pulleys once a while. Any suggestions on feasibility?</p>

|pumps|wheels|hydrostatics|lifting|

<p>The device you are describing is called a <a href="https://en.wikipedia.org/wiki/Pressure_exchanger" rel="nofollow noreferrer">pressure exchanger</a>. They are common on high pressure reverse osmosis systems and use <a href="https://energyrecovery.com/desalination/px-pressure-exchanger/" rel="nofollow noreferrer">positive displacement pump/turbine</a> instead of wheels. The efficiency is not 100%, (no perpetual motion) so additional input is required; such as pumping additional water to keep the system going.</p> <p>I would not recommend it for your application because the cost for such a system is high and the energy recovery at the low pressure would be very low.</p>

54636

shaft connected double water wheel to lift water back in closed loop

2023-03-26T13:20:40.893

<p>In this blueprint, there's the line with a cross with diameter of <code>.115</code> at the far right side. <a href="https://i.stack.imgur.com/UIYTa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UIYTa.png" alt="enter image description here" /></a></p> <p>What does this line means?<br /> Does it means that the section with length <code>.13</code> has an inclination?</p> <p>Also, can I assume that the 3 last sections on the right (rounded, then straight then angled) all have the same length?</p>

|mechanical-engineering|design|technical-drawing|

<blockquote> <p>The symbol or variable for diameter, ⌀, is sometimes used in technical drawings or specifications as a prefix or suffix for a number (e.g. &quot;⌀ 55 mm&quot;), indicating that it represents diameter. For example, photographic filter thread sizes are often denoted in this way.</p> </blockquote> <p>See <a href="https://en.wikipedia.org/wiki/Diameter" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Diameter</a>.</p> <blockquote> <p>The diameter symbol (⌀) (Unicode character U+2300) is similar to the lowercase letter ø, and in some typefaces it even uses the same glyph, although in many others the glyphs are subtly distinguishable (normally, the diameter symbol uses an exact circle and the letter o is somewhat stylized). The diameter symbol is used extensively in engineering drawings, and it is also seen in situations where abbreviating &quot;diameter&quot; is useful, such as on camera lenses. For example, a lens with a diameter of 82 millimeters would be engraved with &quot; ⌀ 82 mm &quot;.</p> </blockquote> <p>From <a href="https://en.wikipedia.org/wiki/%C3%98#:%7E:text=The%20diameter%20symbol%20(%E2%8C%80)%20(,letter%20o%20is%20somewhat%20stylized)." rel="nofollow noreferrer">Wikipedia Talk</a>.</p> <hr /> <p>Q1. What does this line means?<br /> A1. Diameter symbol.</p> <p><a href="https://i.stack.imgur.com/m7fe7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m7fe7.png" alt="enter image description here" /></a></p> <p><em>Figure 1. The ⌀ 0.115 indicates the centres of the radius.</em></p> <p>Q2. Does it means that the section with length .13 has an inclination?<br /> A2. No, it's showing the centres of the radius cut. Because the widest section of the pin has diameter of 0.107 we know that the curve doesn't meet the pin outer diameter at a right angle but at some angle &gt; 90°.</p> <p>Q3. Also, can I assume that the 3 last sections on the right (rounded, then straight then angled) all have the same length?<br /> A3. No. The machinist would probably -</p> <ul> <li>Cut the circular groove.</li> <li>Cut the flat top on the right-hand end.</li> <li>Profile it to 60°.</li> <li>Turn the cylindrical section down to 0.0645.</li> </ul> <p>The length of the flat section would just be whatever it turns out to be. If it's not dimensioned then it isn't important for the function of the pin.</p>

54637

What does this line with cross means in the blueprint?

2023-03-26T14:32:47.420

<p>Hypothetical question: What would happen if I were to cool down a steel cylinder to make it contract, and then slide it all the way into a cylindrical hole in a room-temperature solid steel box? The hole's diameter is very slightly bigger that's the diameter of the cylinder when cold, but meaningfully smaller than the diameter of the cylinder when in room temperature.</p>

|pressure|metals|temperature|

<p>Every material can stretch or compress, even steel. The equations used to predict how tightly the rod would be squeezed in the hole are the same equations that would apply if the hulk stuck his fingers in the hole, stretched it out like a rubber band and placed it around the rod. Or, more realistically, if there was a taper and you hammered the rod in.</p> <p>All of these scenarios are called <a href="https://en.wikipedia.org/wiki/Interference_fit" rel="nofollow noreferrer">interference fits</a>, since if the steel were to relax into it's natrual shape, the parts would interfere with each other.</p>

54640

Steel expanding in a tight space

2023-03-27T00:32:20.657

<p>I was researching opportunities to learn more about manufacturing and machining and found that much of the education is driven through 3yr-4yr apprenticeships. I found several postings for remote opportunities. How is this possible? How can one get &quot;on-the-job&quot; training with CNC lathes and mills etc. when not physically present? Do you just watch on-demand videos for 8hrs/day?</p>

|machining|education|

<p>Some thoughts. However, the comment, to contact said companies and ask, is a very good one.</p> <h2>1. HR-related</h2> <p>It may be that HR on behalf of its company or client tries to attract students via &quot;remote&quot;, who are reluctant due to OR even hindered by Covid-19 infections.</p> <p>Remember, Covid is &quot;ended politically&quot;, not biologically. Cures to complex long-covid are still not found, affecting considerable parts of (previous) employees, students etc., i.e. practically disabling them &quot;forever&quot; at the moment.</p> <h2>2. Technology-related</h2> <p>More and more machines have digital interfaces. I.e. as long as you don't have to touch and repair it, you can obtain all data all day long and control it or change its behavior on-line.</p> <p>You may even be asked to model or simulate wrt to the machine(s), e.g. to solve long lasting problems or to do sth. &quot;innovative&quot;. <em>(to contrast &quot;new to us&quot; - innovative - with inventiveness - &quot;new to the world&quot;)</em></p> <p>And yes, there will certainly be training sessions on-line, be it video, presentation, video-conferencing etc.</p>

54644

How is a Remote Apprentice Machinist Possible?

2023-03-27T13:30:13.747

<p><a href="https://i.stack.imgur.com/M6yCj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M6yCj.png" alt="Stepped Inclined Beam" /></a></p> <p>I'm applying a force to a stepped inclined beam (that's what I'm calling it; don't know what the correct technical term would be) in which it's essentially a cantilever with the first section angled at some angle <span class="math-container">$\theta_1$</span>. After some distance the beam is angled, or bent, at some angle <span class="math-container">$\theta_2$</span>. There is a point load, <span class="math-container">$F$</span>, applied to the end of the beam. I would like to calculate the maximum deflection of the beam (I assume it's at the end), but I am not exactly sure how to approach this problem.</p> <p>Of course I'd need to calculate the reactions at the support. Would I then need to split the beam at <span class="math-container">$x_1$</span>, calculate deflections and then do the same at <span class="math-container">$x_2$</span>? Any help will be greatly appreciated. Thanks guys</p>

|beam|statics|deflection|

<p>The <a href="https://en.wikipedia.org/wiki/Flexibility_method" rel="nofollow noreferrer">flexibility method</a> of <a href="https://en.wikipedia.org/wiki/Virtual_work" rel="nofollow noreferrer">virtual work</a> is an alternate (and potentially easier) way solving such multiple-segment elasticity problems. Instead of applying cantilever deflection theory to each beam, one adds up the strain energies, performs a matrix inversion, and differentiates the result with respect to the end motion. An example for a similar geometry is <a href="http://john.maloney.org/Papers/Maloney_IMECE_2000.pdf" rel="nofollow noreferrer">here</a> (starting with Eq. (15)).</p>

54650

Calculating deflections on an inclined beam

2023-03-27T22:08:49.303

<p>All the &quot;energy-efficient&quot; continuously-variable-speed electric motors I've seen in consumer applications (e.g., HVAC, water pumps – all motors producing on the order of one horsepower) appear to convert incoming AC to DC, which is then modulated by an ECM to achieve the variable speed.</p> <p>The variable speed AC electric motors in this power class that I know of just use multiple windings, and so the number of speeds equals the number windings that can be unenergized – typically 2 out of 3.</p> <p>Are there any electric motor designs that can efficiently modulate AC to directly drive the motor over a wide range of speeds or torque, without converting to DC?</p>

|electrical-engineering|motors|

<p>Are there AC input variable speed drives without a DC conversion (without a DC-link)? That answer is yes: <strong>Cycloconverters</strong>.</p> <p>But that's not the motor. In your first paragraph, you are not describing motors; you are describing variable speed drives. A motor is just a dumb motor of coils and maybe magnets. A modern variable speed drive is a box of electronics that takes DC and modulates it to produce waveforms to properly commutate the motor at various speeds. If the power source is AC it rectifies the AC to DC first. It provides continuously variable speed and speed control.</p> <p>The variable speed motors are variable speed but this does not mean continuously variable. No rectification or modulation occurs. The coils inside have taps so you can choose to energize fewer coils to run the motor as if it had fewer turns which changes the speed. There is no rectifying or modulating going on. It is still just a dumb motor so there is no speed control. So your first and last paragraph is muddling motors and variable speed drives with each other. Motors don't modulate and there are no motors that convert AC to DC.</p>

54656

Do any continuously-variable-speed electric motors use Alternating Current?

2023-03-29T06:28:33.057

<p>After 10^6 cycles a component is considered to have infinite life. What makes 10^6 cycles a deciding factor for infinite life? Why not 10^8?</p>

|structural-engineering|solid-mechanics|machine-design|aircraft-design|fatigue|

<p>No, the only metal that has an indefinite life at some stress level is steel. All other metals have a limit to fatigue life. A rule of thumb for steel is unlimited life with maximum stress cycles less than half of the tensile strength.</p>

54683

Infinite life of a material, Fatigue Strength

2023-03-29T15:12:03.983

<p>I have a high order discrete transfer function model which has the following form. When I use the allmargin function to get the gain margin, I get the following result:</p> <pre><code> G_uncoupled = tf(1,[0.25 0.5 0]); G_uncoupled_d = c2d(G_uncoupled,0.001,'zoh'); D_d = tf([1 -1],[0.001 0],0.001); H_d = tf([(1 - exp(-10*2*pi*0.001)) 0],[1 -(exp(-10*2*pi*0.001))],0.001); delay_d = tf(1,[1 0],0.001); L_uncoupled = (G_uncoupled_d*delay_d^3)/(1 + G_uncoupled_d*D_d*H_d^1*1*delay_d^3); marginInfo = allmargin(L_uncoupled); </code></pre> <pre><code> marginInfo = struct with fields: GainMargin: [336.0257 9.8081e+05] GMFrequency: [37.5848 1.7955e+03] PhaseMargin: [90.0569 84.1254] PMFrequency: [0.0019 0.6617] DelayMargin: [8.2179e+05 2.2191e+03] DMFrequency: [0.0019 0.6617] Stable: 0 </code></pre> <p>When I plot the bode plot of this transfer function, I get the following graph, from the graph I can see two phase margins, but I only see a gain margin 336.0257(50.5dB) at 37.5 rad. For the second frequency 1.7955e+03, I can't see the gain margin at this frequency from the Bode plot, how did MATLAB get this value or what does this value represent?</p> <p><a href="https://i.stack.imgur.com/3Kmzy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Kmzy.jpg" alt="enter image description here" /></a></p>

|control-engineering|control-theory|matlab|simulink|

<ul> <li>Try right clicking and there should be a menu item along the lines of <code>show-&gt;all margins</code> as opposed to <code>show-&gt;minimum margin</code>.</li> <li>Phase plot crosses <em>180 deg</em> at multiple frequencies since any odd multiple of 180 deg is also effectively 180 deg.(180, 540, 900, 1260 etc.). The bode plot shown in the question, the phase plot crosses 540 deg apart from the 180 deg crossing. Visually the 540 deg crossing appears to be around <span class="math-container">$2\cdot 10^3$</span> rad/s.</li> </ul>

54692

MATLAB allmargin() function return multiple GainMargin, but how to identify them on the Bode plot?

2023-03-31T09:06:29.930

<p>I am studying geotechnical engineering and specifically the chapter in B.M. Das book of the same name, 6th edition.</p> <p>On the topic of shear forces, reference to internal friction angle, <span class="math-container">$\phi$</span> is made. My understanding is that the friction angle is the angle in a right triangle with legs of magnitude <span class="math-container">$F_N$</span>,force normal to a specific plane, and <span class="math-container">$\mu F_N$</span>, the friction force parallel with the plane, where <span class="math-container">$tan(\phi)=\frac{F_N \mu }{F_N}=\mu$</span>.</p> <p>My question is if <span class="math-container">$\phi$</span> is constant for all planes in a specific soil, i.e. isotropic, or can it be anisotropic?</p> <p>Further, is the answer the same for rocks and minerals?</p>

|geotechnical-engineering|soil-mechanics|

<p>No, research has shown inherent soil anisotropy, such as naturally deposited sand's bedding angle, has a substantial correlation with the stress-strain response of sand. <a href="https://www.researchgate.net/publication/311412282_Influence_of_soil_inherent_anisotropy_on_the_behavior_of_crushed_sand-steel_interfaces" rel="nofollow noreferrer">research</a></p> <blockquote> <p>For instance, Oda and Koishikawa (1979), Siddiquee et al. (2001), and Azami et al. (2010) showed that the bearing capacity of shallow foundations resting on inherently anisotropic sand decreases with the increase in the inclination of the bedding plan</p> </blockquote> <p>Quoted from the linked document!</p> <p>Rocks of a sedimentary nature, as well, are showing major differences in shear friction angle. Slates is a good example!</p> <p><a href="https://i.stack.imgur.com/kGCrj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kGCrj.png" alt="shear strength and bedding angle" /></a></p>

54717

Is internal friction angle isotropic in soil?

2023-04-03T17:09:37.420

<p>So basically in a video I was taught that, for a submerged cylinder the forces acting on the bottom curved surface is equal to the weight of water directly above + imaginary water weight, if someone could explain me this thing, I would be grateful.<a href="https://i.stack.imgur.com/Rn3E4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rn3E4.jpg" alt="enter image description here" /></a></p>

|mechanical-engineering|fluid-mechanics|civil-engineering|

<p>I think the word &quot;imaginary&quot; is not helping here.</p> <p>The force on the bottom of the cylinder is the weight of the round bottom column water above it, period.</p> <p>And that weight can be calculated by breaking the volume of the round bottom column into a rectangular and a half circle. Exactly as you have sketched.</p> <p>To show what you mean in another way possibly easier to envision, let's assume the cylinder was floating on the surface of the water so its center was level with the surface of the water.</p> <p>Then the force applied to the bottom of the cylinder would be the sum of infinitesimally small columns of water. Let's say the cylinder with radius R starts at the origin.</p> <ul> <li>R radius of the cylinder The force on the bottom is the sum of infinitesimally small columns of water with the width dx of the function</li> </ul> <p><span class="math-container">$$Y=\sqrt {R^2-X^2}$$</span></p> <p><span class="math-container">$$F_{bottom}=\Sigma Y *dx$$</span></p> <p><span class="math-container">$$F_{bottom}= \int_0^{2R} \sqrt(R^2 - x^2) dx $$</span></p> <p>And we already know that integral is equal to <span class="math-container">$ \ \pi*R^2/2$</span>, which is the are of half circle.</p> <p>So if we move the cylinder dawn to the depth of D. the force on the bottom will be</p> <p><span class="math-container">$$F_{bottom}=2RD+ \pi R^2/2$$</span></p> <p>The force on the top again is exactly as you have shown</p> <p><span class="math-container">$$F_{top}=2RD-\pi R^2/2$$</span></p> <p>And buoyancy is <span class="math-container">$$F_{bottom}- F_{top}= \pi R^2 \quad = area \ of \ cylinder$$</span></p> <p>For simplicity, I assumed the axial length of the cylinder as one!</p>

54753

Why forces on bottom curved cylinder surface equal to the weight of water and weight of imaginary water?

2023-04-04T08:16:03.353

<p>I am using a servo motor 130st-M15015LB with the AASD-30A driver. I have to use the acceleration with position control mode. I can see in the documentation that If I use Pn110 to Pn112, I can have a time of acceleration between 5 to 1750ms, but when I try, it doesn't work, (it works until 500ms +-), I have been trying to solve this problem for one week, but I really don't understand (I'm a software developer). If someone can help me, that would be great!</p> <p>Example, I tried:</p> <p>With the combination of Pn110 = 400, Pn111 = 300, Pn112 = 100 It works,</p> <p>but when I put Pn110 = 1500, Pn111 = 1200, Pn112 = 300 It doesn't work, the AASD-30A driver say that I cannot use these parameters,</p> <p>(Sorry for my English)</p>

|mechanical-engineering|mechanical|acceleration|servo|

<p><strong>Update</strong></p> <p>The servo motor and his driver are fake, so some registers didn't work, it's normal.</p>

54766

Acceleration of servo motor with AASD-30A

2023-04-05T05:10:56.470

<p>When compressed air passes through a nozzle, its velocity increases while its temperature decreases due to the pressure drop. Why is this process not utilized as a refrigeration system similar to air conditioner? Could the high exit velocity at the nozzle exit be a potential barrier to this application?</p>

|mechanical-engineering|fluid-mechanics|hvac|refrigeration|

<p>If pressurization of air (or other gas I suppose) were free this might be viable. We also need to dump the heat we make compressing the air. Certainly we could make a nozzle that had whatever pressure drop required. We might struggle figuring out how to get heat into that moving air, it isn't the best way to transfer heat. But air definitely cools when it expands, so the possibility is there.</p> <p>The advantage of the standard refrigeration cycle is that it moves more heat than the energy used. You could never get that multiplier effect with direct expansion.</p>

54773

Use of nozzle as a refrigeration device

2023-04-05T22:01:38.473

<p>I am trying to figure out the inflation/how to calculate the time it takes to fill something to a certain pressure.</p> <p>I have an <a href="https://www.boteboard.com/products/inflatable-dock-7-classic" rel="nofollow noreferrer">inflatable dock</a> I am filling with a <a href="https://outdoormaster.com/products/electric-paddle-board-pump-shark-ii" rel="nofollow noreferrer">portable pump</a>.</p> <p>The dock is 7ft x 7ft x 8in -&gt; which is about 32.75 cubic feet (or 245 gallons or 925 liters) and is supposed to fill it to at least 6 PSIG (20.7 PSIA)</p> <p>The pump has 2 stage - 350 L/min (12.3 CFM) to 1PSIG, and then 70L/min (2.47 CFM) up to 20PSI.</p> <p>So I am trying to map out how long it would take to fill this. The problem is I see varying formulas. A) <a href="https://www.vmacair.com/blog/work-time-fill-type-questions-using-simple-logic" rel="nofollow noreferrer">Here</a> and B)<a href="https://www.about-air-compressors.com/fill-an-air-tank/" rel="nofollow noreferrer">here</a>, but also the problem is this is for tanks and not inflatables, so is there a difference. My assumption is that if I use, the method in Source A example 1, the assumption is it starts at atmosphere right? Is it that it should take 925L/350L/min to get to 0 PSI, and then do the rest?</p>

|fluid-mechanics|pumps|compressed-air|

<p>I would take it to mean -</p> <ul> <li>1 PSIG is really just expanding the flat dock to its inflated size so you're going to need 925 (L) / 350 (L/min) = 3 minutes to get that far.</li> <li>20 PSI = 1.37 bar (above atmospheric) so that means you're going to need to pump in another 925 × 1.37 = 1267 L of free air. This will occur at 70 L/min so will take 1267 / 70 = 18 minutes.</li> <li>Total time = 3 + 18 = 21 minutes.</li> </ul> <p>This all assumes that the pump works to specification but I suspect that it might vary during the high-pressure pumping and you might find that the specifications may have been written by an optimist.</p> <hr /> <p>From the comments:</p> <blockquote> <p>What if I were to add 100 ft of 1inch ID tubing? By my calculations this should only add ~15 L extra to fill, but is there a way to account for pressure drop?</p> </blockquote> <p>Pressure drop will only occur while there is air flow. As the pressure in the inflatable reaches that of the compressor the flow will reduce and so will the pressure drop. You can check the drop by inflating at the compressor with the 100 ft of hose in line and a pressure gauge at the compressor and at the inflatable. (Do a calibration check first by by comparing both gauges with the hose outlet blocked.)</p>

54784

How long it takes to fill an inflatable to a certain pressure?

2023-04-06T00:24:42.840

<p>Let <span class="math-container">$X$</span> be a scalar random variable with pdf <span class="math-container">$f(x)=e^{-2|x|}$</span>. Among the class of symmetric 3 -point scalar quantizers find one which minimizes the MSE in quantizing <span class="math-container">$X$</span>. Compute the MSE in this case.</p> <p>I find that f(x) is the Laplacian distribution with <span class="math-container">$\mu = 0, b = 1/2$</span>. By using the Lloyd-max scalar quantizer, I can find that the threshold is <span class="math-container">$[-1/2, 1/2]$</span> and quantization levels are <span class="math-container">$[-1, 0, 1]$</span>.</p> <p>Now, I need to find <span class="math-container">$d=M S E=E\left[(X-\hat{X})^2\right]$</span>, but I can't compute the mean-squared error during integration.</p> <p>How I can make progress to find the MSE?</p>

|electrical-engineering|signal-processing|

<p><span class="math-container">$$MSE=E\left[(X-\hat{X})^2\right]$$</span></p> <p>From the definition of the expectation operator.</p> <p><span class="math-container">$$ \int_{-\infty}^{\infty}{(x - \hat{X})^2 f_X(x)} dx$$</span></p> <p>Split the integral into the different quantization regions.</p> <p><span class="math-container">$$ \int_{-\infty}^{-\frac{1}{2}}{(x - (-1))^2 e^{2x}} dx + \int_{-\frac{1}{2}}^{\frac{1}{2}}{(x - (0))^2 e^{-2|x|}} dx + \int_{\frac{1}{2}}^{\infty}{(x - (1))^2 e^{-2x}} dx $$</span></p> <p><span class="math-container">$$ \int_{-\infty}^{-\frac{1}{2}}{(x + 1)^2 e^{2x}} dx + 2\int_{0}^{\frac{1}{2}}{x^2 e^{-2x}} dx + \int_{\frac{1}{2}}^{\infty}{(x - 1))^2 e^{-2x}} dx $$</span></p> <p><span class="math-container">$$ \frac{1}{8}e^{-1} + \left(\frac{1}{2} - \frac{5}{4}e^{-1} \right) + \frac{1}{8}e^{-1} $$</span></p> <p><span class="math-container">$$ \frac{1}{2} - e^{-1} \approx 0.132 $$</span></p>

54787

Minimizing the MSE with symmetric scalar quantizers

2023-04-06T14:49:01.697

<p>For the little I know about turbine engines, I know that these run at enormous high temperatures and thus, need nickel based superalloys.</p> <p>From what I know, jet engines use a smart use of air flow throughout the parts of the engine (such as the combustion chamber) in order to make an air layer of protection between the metal walls and the flames.</p> <p>... But you have to keep a balance between air cooling with the energy used to compress the air in order to increase efficiency.</p> <hr /> <p>Well, in the scenario that one wouldn't use such superalloy because of its high price, if someone were to inject water with the combustion jet ejecting the combustion chambers directly to the turbine, would the water avoid the material melting?</p> <p>... Or it would just make unnecessary steam?</p>

|thermodynamics|heat-transfer|cooling|turbines|

<p>Water injection for the sake of cooling down the combustion gas would miss the point. An air engine is designed to have the highest temperature possible at the first stage of the turbine. The higher the temperature the higher the fuel efficiency of the air engine. The temperature is limited by several factors, e.g. emissions or materials.</p> <p>The cooling air used for the blades in the hot section provides a very thin layer of reduced temperature to the surface of the blades. The cooling air acts as a local (!) cushion to keep the hot combustion gas away from the blade surface. It has not the purpose to cool down the overall combustion gas temperature in the hot section, because that would effectively decrease the overall turbine efficiency.</p> <p>If you would inject water upstream of the turbine to cool down the combustion gas, you would cool down the overall temperature of the gas, which you don't want to keep the temperature as high as possible. Not to mention the extra equipment which needs to be carried by the plane.</p> <p>BTW: cavitation wouldn't be the limiting factor. The combustion chamber operates at temperatures far beyond 1000°C, so the water droplets would evaporate before they hit the blades, if the injection is properly designed. In fact there are new concepts of water enhanced air engines, which inject steam to the combustion chamber to increase the efficiency. Though the purpose of the steam is not to cool down the temperature, but to increase the mass flow through the turbine.</p>

54796

Would injecting water in a turboshaft engine avoid turbine material meltdown?

2023-04-07T01:02:23.483

<p>I understand in residential plumbing, that a thermal expansion tank is needed before the hot water tank. From my understanding this is to allow for the water in the hot water tank to expand, instead of creating large amounts of pressure, when the water is being heated while the system is closed.</p> <p>However, I'm most curious about on large scale, industry projects. I am in a power generation course at the moment and we are discussing steam power plants. Is thermal expansion not a factor in this situation? It hasn't been mentioned in class (this course is largely applied thermodynamics, not power plant design). Is this because the fluid is actually flowing and not stationary, or are there expansion tanks that we just haven't discussed in class?</p> <p>It is my understanding that pumps provided the pressurization for the system, and not the steam.</p> <p>Thanks!</p>

|mechanical-engineering|thermodynamics|power-generation|

<p>No need for a thermal expansion tank in the water steam cycle of a steam power plant.</p> <p>A thermal expansion tank is added to residential plumping, because your plumping system only contains water and no steam. As water is approximately incompressible, it generates such a high pressure through thermal expansion that it would burst the whole pipe if there was no thermal expansion tank.</p> <p>The thermal expansion tank technically adds a gas to the plumping system, but keeps it contained to the thermal expansion tank by separating it from the water with a flexible bladder. The gas is compressible. Therefore it can buffer the thermal expansion of the water and keeps the overall pressure of the pipe system within the specification.</p> <p>The water steam cycle in a steam power plant already contains two phases namely water and steam. Any fluctuation of the average water temperature would indeed change the total water volume, but the change is small and buffered by the compressible steam in the system. Nobody needs to worry about it.</p>

54800

Is thermal expansion of water a consideration in steam power plants? If so, how is it managed?