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2015-11-29T22:19:11.990

<p>Given this circuit</p> <p><a href="https://i.stack.imgur.com/aQ9eD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aQ9eD.jpg" alt="Question"></a></p> <p>I want to deduce the relationship between the output voltage <span class="math-container">$U_A(t)$</span> an d the input voltage <span class="math-container">$U_E(t)$</span> using Kirchhoff's current law. We're allowed to assume that <span class="math-container">$R$</span>, <span class="math-container">$L $</span>and <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span> are given to us.</p> <p>My solution so far: <span class="math-container">$$ U_E = U_R + U_{C1} $$</span> <span class="math-container">$$ U_{C1}=U_L+U_{C2} $$</span> <span class="math-container">$$U_A = U_{C2} $$</span> <span class="math-container">$$ \rightarrow U_E = U_R+U_L+U_A$$</span> At point <span class="math-container">$K$</span>, i want to use Kirchhoff's current law: <span class="math-container">$$I_R=I_{C1}+ I_L= I_{C1}+ I_{C2} $$</span> Is my last step correct? How to continue?</p>

|electrical-engineering|control-engineering|modeling|

<p>You can write two ODEs describing the relationship, and that would be in the time-domain. Alternatively you could describe things in the frequency domain and get the relationship as a transfer function. I will show the latter. </p> <p>Using Kirchhoff's current law at the two nodes <em>K</em> and <em>A</em>, we get:</p> <p>$$ \frac{U_E-U_K}{R}=\frac{U_K-U_A}{L s}+C_1 s U_K$$ $$\frac{U_K-U_A}{L s}=C_2 s U_A$$</p> <p>These two equations can be solved for the two unknowns $U_A$ and $U_K$. We are only interested in the former.</p> <p>$$\frac{U_A}{U_E}=\frac{1}{C_1 C_2 L R s^3+C_2 L s^2+C_1 R s+C_2 R s+1}$$</p>

6350

connectedness between the output voltage and input voltage

2015-11-30T16:06:43.817

<p>All the electrical vehicles I've ever ride make that humming sound (either cars or trains).</p> <p>What exactly causes that sound? Is it drag force in the gears? Is it a fan?</p> <p>Or is it because of the coils (see <a href="https://en.wikipedia.org/wiki/Coil_noise">coil noise</a>)?</p>

|mechanical-engineering|electrical-engineering|automotive-engineering|power-electronics|

<p>Correct, coil noise is generated in the windings of the motor because of Pulse Width Modulation (<a href="https://en.wikipedia.org/wiki/Pulse-width_modulation">PWM</a>) driving current. Different frequencies of modulation cause corresponding sound frequencies (may create half or double frequency oscillations as well).</p> <p>To change the speed of an electric motor, PWM is required to regulate voltage in DC motors and generate variable frequency wave forms in Variable Frequency Drives (<a href="https://en.wikipedia.org/wiki/Variable-frequency_drive">VFD</a>) for AC motors.</p> <p>A <a href="https://en.wikipedia.org/wiki/Brushless_DC_electric_motor">brushless DC motor</a> is essentially a 3 phase AC motor coupled with an application specific VFD. These are common in electric remote controlled aircraft and often play tones for a power up self check with nothing but the motor acting as a speaker. There are many examples online of people trying this DIY; for example, <a href="https://www.youtube.com/watch?v=kyV6eopQBRM">Brushless motor music (YouTube)</a>.</p>

6361

What causes the sound coming from electrical vehicles?

2015-12-01T23:07:27.200

<p><a href="https://i.stack.imgur.com/bIFBO.png"><img src="https://i.stack.imgur.com/bIFBO.png" alt="enter image description here"></a></p> <p>I have a cantilever shaft 6inches in length supported on one end using a ball bearing. I am having problems of shaft wobbling in the bearing. I cannot have support on the other end.</p> <p>How can I help this situation?</p> <p>The shaft is made of aluminum, rotating at around 120 rpm. </p>

|mechanical-engineering|bearings|

<p>If the dimensions are fixed, a careful selection of alloy or hardness that has a higher inherent damping would also reduce deflection while the shaft is rotating. </p>

6375

How can the wobble of a rotating cantilever shaft be reduced?

2015-12-02T00:51:38.413

<p>I am designing a box and lid (for a home-made micro computer). Using Autodesk Inventor (Student Version), I've created my box which looks something like this:</p> <p><strong>I plan on 3D printing (ABS Plastic) the box and lid in the end.</strong></p> <p><a href="https://i.stack.imgur.com/CdyHb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CdyHb.png" alt="enter image description here"></a></p> <p>And I am going to create a rectangular lid to encapsulate the contents. I know accuracy will be important to the printing, so I want to make sure I have the sizing right. If it helps you answer my question, the dimensions of the top of the box are 3in by 4in and I want my lid and its outer lip to be around 0.1in thick. I would like the seal to be pretty tight (won't fall off easily) but if it's too small I risk breaking the ABS Plastic when putting on or taking off the lid.</p> <p>My question, however, is what should the difference in size between the outer rim of the box's top and the inner rim of the lid be?</p>

|design|plastic|3d-printing|

<p>You seem to want an a slight interference fit so the top will snugly seat into the housing. Depending on your wall thickness and size you could probably oversize the "male" portion of the cap like .002-.003" compared to the "female" dimension. </p> <p>That being said, and understanding that 3D printing generally cannot hold those tolerances I would try an alternative fastening method. Unless you have some time and material to play with to keep tweaking the 3D file for shrinkage and reprint until it fits how you want. </p> <p>Possibly consider a snap fit like so(upper right and lower left examples): <a href="https://i.stack.imgur.com/whOTe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/whOTe.jpg" alt="snap fit examples"></a></p> <p>You could probably get away with only one "tapered beam" and have the top rotate downward along the edge opposite of the snap fit downward and snap into the case. Think tv remote battery cover. You would have better allowances for fit and would be a nice detail for your project. Make sure you have a detail to be able to remove the top like a little slit for a screwdriver or coin to pry it open. </p> <p>You could always use screws as well</p>

6379

What should the size difference be between a lid and a container?

2015-12-02T02:29:42.313

<p>In the book, Design of Liquid-Propellant Rocket Engines, there is a practice problem written as such.</p> <p>"Determine the design values of $(C_*), (C_f), and (I_s)_t$ for the engine thrust chambers of the stages of the hypothetical Alpha vehicle, with the following assumed design parameters."</p> <p>I am given the fuel, the fuel mixture ratio, the nozzle stagnation pressure $ (P_c)_n $, the specific heat ratio, and the nozzle expansion ratio.</p> <p>I can calculate $(C_*)$ fairly easy, the problem is calculating $(C_f)$. The book says that you can derive a theoretical vacuum $(C_f)$ using a figure. However when I look at the figure its listing, It does not give me a clear path to derive the equation.</p> <p>What are the equations I need to calculate $(C_f)$ or the theoretical vacuum $(C_f)$ ( I have the equation used to convert the theoretical vacuum into the actual thrust coefficient. )</p> <p>Here are the pages I'm talking about.</p> <p><a href="https://i.stack.imgur.com/yCgRJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yCgRJ.jpg" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/KEMZg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KEMZg.jpg" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/8Y4vM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8Y4vM.jpg" alt="Fig 1.11 is the reference figure"></a></p> <p>Fig 1.17 is the referenced figure, however it does not show the equation it uses to graph its chart. What would be the equation needed to graph the chart used in fig 1.17</p> <p><a href="https://i.stack.imgur.com/KAhv0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KAhv0.jpg" alt="enter image description here"></a></p>

|aerospace-engineering|thrust|

<p>So I figured it out, I first took this equation.</p> <p>$(C_F) = (CT_F) - (E)(P_a)/(P_C)_N$</p> <p>Then I isolated $(CT_f)$ and substituted $(C_F)$ equation resulting in</p> <p>$C = \sqrt{\frac{2y^2}{y-1)}*\frac{2}{y+1}^{y+1/y-1}*(1-\frac{P_e}{P_C}^{\frac{y-1}{y}})+e(\frac{P_e - P_a}{P_C})} + e\frac{P_e}{P_C}$</p> <p>Then I solved for P_e using this equation</p> <p>$E = \frac{(\frac{2}{y+1})^\frac{1}{y-1}*(\frac{(P_C)_n}{P_e})^\frac{1}{y}}{\sqrt{\frac{y+1}{y-1}*(1-\frac{P_e}{(P_C)_N})^(\frac{y-1}{y})}}$</p> <p>The value $(P_a)$ is ambient pressure, so I will plugin in 14.6959488 in the place of $(P_a)$. 14.6959488 is the pressure in psi at ground level ( I hate imperial but the book's units are imperial ). </p> <p>This will give me a value for $(CT_F)$ which I can then plug back into the original equation</p> <p>$(C_F) = (CT_F) - (E)(P_a)/(P_C)_N$</p> <p>Which will give me the value of $(C_f)$</p> <p>only took me two weeks.</p>

6381

What is the equation to calculate the Thrust coefficient within a set of parameters?

2015-12-02T09:22:45.243

<p>What determines the maximum torque of a DC motor? How is this related to the equations</p> <p>$$E=k\omega$$ $$T=kI$$</p> <p>where</p> <ul> <li>$E$ is electromagnetic force / voltage (V)</li> <li>$\omega$ is rotational speed (rad/s)</li> <li>$T$ is torque (Nm)</li> <li>$I$ is current (A)</li> <li>$\phi$ is flux (Wb)</li> <li>$k$ is the motor constant (V/(rad/s)=Nm/A) such that $k=K\phi$</li> </ul>

|electrical-engineering|motors|torque|

<p>Torque of a motor is directly proportional to current. Therefore, the maximum torque a motor can produce is limited by the maximum current it can handle.</p> <p>The maximum steady state current a motor can handle is limited by the heat it can safely dissipate, which is proportional to the square of the current. This means you have to be careful with that spec. Just 41% more current causes twice the heating, and twice internal temperature rise above ambient, which is likely to cause damage.</p> <p>It is possible to exceed the maximum allowed steady current for short periods of time. Some motors may actually come with such specifications. To really push a motor to its limits without damage requires tracking the assumed temperature. You usually model the temperature as slowly decaying to ambient with a time constant you usually have to guess at or measure. This could be multiple minutes. You can get some idea of this by knowing the maximum allowed steady state current and making a assumption about internal temperature above ambient.</p> <p>This kind of system lets you do some short pulses of somewhat higher current, which you then pay for in a lower allowed maximum current for some time afterwards.</p> <p>Generally you don't want to exceed the maximum allowed steady current by 2x, especially for a brushed motor where there are other limits on the current. Even for a brushless DC motor with good temperature modeling, I'd consider 3-4x a upper limit on short bursts of current intended for extra torque.</p>

6384

What determines the maximum torque of a DC motor?

2015-12-02T10:26:17.377

<p>Assume you have a system model that is controlled by a PID controller (feedback). You want to add feedforward to improve the characteristics of the controlled system. How can you design the transfer function of the feedforward? (Cf. <a href="http://engineeronadisk.com/book_modeling/images/feedback4.gif" rel="nofollow">this image</a>)</p> <p>On slide 18 of <a href="http://research.che.tamu.edu/groups/Seminario/ProcessControl-Fall2012/Handouts/Chap_15_Marlin_2002.pdf" rel="nofollow">this presentation</a>, it is suggested to use a gain and a lead-lag filter:</p> <blockquote> <p>K*(T1*s+1)/(T2*s+1).</p> </blockquote>

|control-engineering|feedback-loop|

<p>The idea of using a feedforward component in a control system is to provide near <strong>instant action</strong> of the plant output to the input command. So a suitable choice of transfer function which connects the command to the plant input directly is an inverse model of the plant. The feedback control can then serve to 'clean up' any error in the feedforward component perhaps due to modeling error or drift in the parameters. </p> <p>Usually the feedforward DC gain used is less than or equal to 1.0.</p>

6385

How to improve a PID control loop by adding feedforward?

2015-12-02T15:33:39.007

<p>I am looking to test the <a href="https://www.tekscan.com/products-solutions/force-sensors/a401" rel="nofollow noreferrer">FlexiForce</a> sensor in measure force, however for accurate readings the load needs to be disturbed evenly over the sensing area. As such I am having "discs" made to cover the sensing area. For my application the thinner the material of the disc the better (1 mm or less in thickness). Which of the following materials will be most suitable to manufacture the disc for the stiffest/strongest possible part?</p> <ol> <li>Brass &amp; Copper</li> <li>Aluminium</li> <li>Stainless Steel</li> <li>Mild/Carbon Steel</li> </ol> <p>The disc will be 25.4 mm in diameter with a thickness of 1 mm.</p> <p>Here's an illustration of what I mean:</p> <p><a href="https://i.stack.imgur.com/83hze.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/83hze.png" alt=""></a></p>

|materials|metals|manufacturing-engineering|deformation|

<p>1) I've used several brands of FSRs, including FlexiForce and you're very unlikely to get results usable in trade; their responses to pressure vary with temperature (because, I suspect, the stiffness of the polymers in the sensor itself vary w/temp.)</p> <p><strong>2) The application guides I've read strongly suggest NOT using extremely stiff materials or sharp radii in direct contact with the sensor that can cause stress concentrations.</strong></p> <p>3) The stiffness and shape of both what's on top of the sensor and what's underneath both matter.</p> <p>4) all of the materials you're considering are orders of magnitude stiffer than the sensor, so they'll likely give you equally wrong and expensive results.</p> <p>5) I suggest you first try either thin disks or very-slightly rounded domes made with something like Nylon or HDPE (High-density polyethylene, AKA cutting-board plastic.) I suspect you can back those with disks of a more rigid material; steel would be fine.</p> <p>6) The disk has to be same-size or slightly smaller than the sensor's active area, or else some of your load will bypass the active area, leading to systemic errors. <strong>And you'd better radius the edges, lest you create a stress concentration right there.</strong></p>

6388

Material selection for weight scale

2015-12-02T16:42:57.263

<p>In commercial fire sprinkler systems or large farm irrigation systems, is the thrust produced by the water a major design consideration? At what stage in the design process is this taken into account?</p>

|systems-design|thrust|

<p>Product design often involves juggling many parameters at the same time, prioritizing them, then testing a design. In sprinklers themselves, the forces associated with the thrust are much lower that the forces the sprinkler experiences during assembly or from internal pressure; so typically they are not a design priority.</p> <p>For irrigation sprinklers the thrust is not a big consideration in the structural design, because the flow, and resultant thrust is comparatively small. It is however important for <a href="https://en.wikipedia.org/wiki/Impact_sprinkler" rel="nofollow">impact sprinklers</a>, as it is their method of operation.</p> <p>Fire sprinklers themselves don't generate a lot of force, but there is a lot of dynamic force on the <a href="http://wermac.org/steel/pipesupports.html" rel="nofollow">pipe supports</a> on large diameter pipes that turn corners to deliver water to the sprinklers. Fluids can exert very large forces without being a free jet. Here is a big nasty document on <a href="http://www.wermac.org/pdf/piping_engineering.pdf" rel="nofollow">pipe engineering</a> that might make your thermo fluids homework look nicer ;-)</p>

6392

Thrust consideration in sprinkler and irrigation systems

2015-12-03T07:04:30.133

<p>I am modelling my structure using FEM, and my structure contains line elements (1D) such as beams and columns and area elements (2D) such as walls and columns.</p> <p>My question is that, how can I proceed to do modelling, and how can I interpret the results for stress and strain, especially at the places where 1D elements meet the 2D elements? Take for example, </p> <ol> <li>A beam's Center of Gravity might not be located within the slab, but in reality the beam and slab are still touching each other. In other words the beam is offset from the slab</li> <li>A wall's center of gravity might not be located within the slab, but in reality the wall and slab are still touching each other. In other words the wall is offset from the slab</li> <li>A beam and a column are supposed to be connected together, except that now the beam is offset in left and column in right. If going by line element, they are no longer touching, although in reality, they still are.</li> </ol> <p>Is this a solved problem? <strong>Any research literature that I can refer to, or any software packages that you can recommend</strong> that handle this gracefully?</p>

|structural-engineering|finite-element-method|structural-analysis|

<p>Say we have a slab supported on a beam.</p> <p><a href="https://i.stack.imgur.com/lVpmp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lVpmp.png" alt="Beam composite with slab"></a></p> <p>The centroid of the slab and the centroid of the beam are not coincident. Fortunately, in FEM software packages the geometric centroid of the element can be offset from the nodes that define the element.</p> <p>The sketch below shows a case where the shells have been offset such that the nodes are at the bottom face and the beams have been offset such that the nodes are at the top face. This way, the section properties accurately reflect the true geometry but the shell and beam elements are able to share nodes (which is essential).</p> <p><a href="https://i.stack.imgur.com/GcpNx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GcpNx.png" alt="Extruded 3D view of beam with slab"></a></p> <p>Now for implementation. Most FEM packages should be able to handle this. I'm familiar with LARSA so my notes below are specific to that software package, but SAP, RISA, MIDAS, LUSAS, ABAQUS, etc. etc. etc. should have equivalent capabilities. Furthermore, there's likely quite a few different ways of solving this problem. What follows is just one of them -- just because it seems like some pictures might help.</p> <p>Here's a wireframe and extruded view of a quick beam-slab model I put together. The plate elements and beam elements are shown at their centroids. Note that the plate and beam elements share nodes but their centroids are offset from one another.</p> <p><a href="https://i.stack.imgur.com/IoY2e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IoY2e.png" alt="Wireframe and Extruded Views"></a></p> <p>In this case, this offset was accomplished in two steps. The beam itself was defined using Section Composer with a geometry that places the node at the top face of the beam. Then, member end offsets were used to offset the top of the beam half the plate thickness downward. The plate elements are defined using nodes at the plate centroid so these two steps shift the beam geometry down so that the top of the beam is 'in contact' with the bottom of the plate.</p> <p><a href="https://i.stack.imgur.com/4PcPy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4PcPy.png" alt="Member End Offsets"></a></p> <p>Basically, look for something like "member end offsets" in your FEM package.</p>

6399

How should one model a structural model with both line elements and area elements?

2015-12-03T17:13:29.183

<p>While this question pertains to smartphone , it is being asked from purely an electrical engineering view (My electrical knowledge is very basic)</p> <p>Trying to understand if charging a cell phone and using it is recommended or not, I came across this <a href="http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries" rel="nofollow noreferrer">Charging Li Ion battery</a></p> <p>Quoting extracts (emphasis mine)</p> <blockquote> <p>portable devices sit in a charge cradle in the on position. The current drawn through the device is called the <strong>parasitic load</strong> and can distort the charge cycle. Battery manufacturers advise against parasitic loads while charging because it induces mini-cycles, but this cannot always be avoided...."</p> <p>A portable device should be turned off during charge. This allows the battery to reach the set voltage threshold and current saturation point unhindered. A <strong>parasitic load confuses the charger</strong> by depressing the battery voltage and preventing the current in the saturation stage to drop low by drawing a leakage current. A battery may be fully charged, but the prevailing conditions will prompt a continued charge, causing stress."</p> </blockquote> <p>I wanted to verify this by testing to the extent possible with limited means. I ran two tests.</p> <p>For measurement of voltage/Current used an app <a href="https://play.google.com/store/apps/details?id=ccc71.at.free" rel="nofollow noreferrer">3C Toolbox</a>, which gives measurement at every 1% change in both tabular and graph form</p> <p>First test was normal charging in air plane mode. Had to keep the phone on to run the app</p> <p>Second test, was to charge the phone while simulating usage. Instead of actual usage, kept screen display on at 50% brightness to simulate usage load, since this would give a steady load and had read elsewhere that this would approximate a 2G data download load</p> <p>I wanted to compare with the graph quoted in the linked article pasted below <a href="https://i.stack.imgur.com/imosa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/imosa.jpg" alt="Battery University Graph"></a></p> <p>Test results for voltage and current are pasted below. The first graph on the left side is of first test and next is of second test (pale gold line at bottom shows screen on status)</p> <p><strong>Charging Time</strong>. Identical in both cases -121 minutes</p> <p><strong>Voltage</strong></p> <p><a href="https://i.stack.imgur.com/Bvodj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bvodj.jpg" alt="Voltage Compared"></a></p> <p><strong>Current</strong> <a href="https://i.stack.imgur.com/k7CpY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k7CpY.jpg" alt="current compared"></a></p> <p><strong>Temperature</strong></p> <p><a href="https://i.stack.imgur.com/2Dkc7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Dkc7.jpg" alt="temperature compared"></a></p> <p><strong>Edit Notes</strong></p> <p>I have edited the question, making it more focused to <strong>validate</strong> my interpretations of the results of test. In this context :</p> <p>a) I fully agree that this is not a rigorous scientific experiment </p> <p>b). Battery University talks of effects observed while charging with device on. Here the comparison from my tests, is not between charging while device switched "off" and "on". It is a comparison between charging while the device is on with " less " load and " more" load (device on with screen "off " and device on with screen "on at 50% brightness")</p> <p>Having made these points, I believe that these tests offer <strong>indicative</strong> trends and is reasonable to compare with Battery University and draw conclusions</p> <p><strong>Inferences</strong></p> <ol> <li><strong>Voltage</strong>. This offers the most striking contrast.</li> </ol> <p>a). While the time taken to reach maximum voltage and voltage levels are nearly same, the voltage graph shows steep ramp up compared to the first. This means that the topping up of charge is gradual in the first case, since more time is spent at various voltages. This translates to longer <strong>battery life per cycle</strong> (assuming equal discharge conditions applied to both cases).</p> <p>b). Battery University says </p> <blockquote> <p>A parasitic load confuses the charger by depressing the battery voltage...., </p> </blockquote> <p>this can be observed in the second case <strong>more prominently</strong> and confirms the behaviour</p> <ol start="2"> <li><strong>Current</strong>. </li> </ol> <p>a). The current graph in the second graph, exhibits a pronounced <strong>saw tooth</strong> shape before it drops off. This corresponds to <strong>mini cycles</strong> caused by <strong>parasitic load</strong> as mentioned by Battery University</p> <blockquote> <p>The current drawn through the device is called the parasitic load and can distort the charge cycle. Battery manufacturers advise against parasitic loads while charging because it induces mini-cycles, ....</p> </blockquote> <p>b). Further, Battery University statement below cannot be established, since in both cases, the battery is being charged while in " on" condition and measured drop is nearly the same when you compare the maximum and minimum</p> <blockquote> <p>A parasitic load confuses the charger by depressing the battery voltage and preventing the current in the saturation stage to drop low by drawing a leakage current. </p> </blockquote> <ol start="3"> <li><p><strong>Temperature</strong>. The battery temperature in second case is marginally higher by 2°C and while it is not significant, it can be argued that it is not <strong>desirable for battery</strong>, since increased internal temperature while charging leads to more internal stress.</p></li> <li><p><strong>Charging Time</strong>. Same in both cases and the charger having a limited capability to supply voltage or current, Would imply that there has to be a trade-off between quality of charging with more load, as is evident from the graphs above</p></li> </ol> <p><strong>Conclusion</strong></p> <p>Inferences above are in line with Battery University and consequently <strong>one can conclude that charging battery, while device is switched "on "or "in use" is NOT a good practice</strong>. This is further supplemented by <a href="http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf" rel="nofollow noreferrer"> best practices</a> shared and amplified by @ericnutsch (thanks) <a href="https://engineering.stackexchange.com/questions/6406/charging-a-smartphone-battery-while-using-it#comment11477_6409">here</a>.</p> <p><strong>Question</strong></p> <p><strong>Seeking anwers for confirmation or negation of conclusion highlighted above (backed by technical sources and citations). Additional inputs on charging methodology implemented by major OEM cell manufacturers to cater for "charge while in use", would be appreciated</strong></p> <p>Please ask if any other information can be provided apart from above and cell phone is Honor 6, with 3100 mAh battery.OEM wall charger of 2A used</p>

|electrical-engineering|battery|

<p>Based on the answers here it shows that interpretation of my results was wrong.</p> <p>This was independently validated and supported by Bruce Huang of Battery University. Main points from him were</p> <ol> <li>Parasitic Load would not delay current drop indefinitely, it does do momentarily.</li> <li>Keeping the screen on at 50% load does not create a significant load to create noticeable parasitic load</li> <li>Charging time may vary depending on the load</li> </ol> <p>Full text of his reply and my answer based on his and inputs obtained from <a href="https://android.stackexchange.com/a/131169/131553">this answer</a>.</p> <p>I would still hope to obtain some inputs on OEM implementation of charging</p>

6406

Charging a smartphone battery while using it

2015-12-04T02:24:26.047

<p>I'm trying to measure water elevation as a function of distance in a water tank, viewed through the glass wall, as shown here:</p> <p><a href="https://i.stack.imgur.com/jWsei.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jWsei.jpg" alt="tank"></a></p> <p>The height variations that's of interest to me is very small. Would anyone know of a lens that can be placed in front of a camera to stretch the image vertically? This will help enhance the vertical resolution of the photo. Other suggestions are also welcome.</p>

|measurements|optics|

<p>How about this?</p> <p><a href="https://i.stack.imgur.com/boZoJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/boZoJ.jpg" alt="enter image description here"></a></p> <p><em>Photo 1. Inclined tank level scale.</em></p> <p>The inclined scale means that graduations can be spread more widely for ease of reading. For example, if 1 mm resolution is required and the line rotated by 84 degrees the 1 mm marks stretch out to 10 mm making it much more easy to read and interpolate. cos(84°) = 0.1. This effectively stretches the vertical resolution as required in the question but without the use of a lens.</p> <p>It's not obvious from the question whether an automated vision system is to be used but either way this should make reading the level somewhat easier.</p>

6410

Lens to "stretch" or enhance vertical resolution in a photo

2015-12-04T06:56:49.583

<p>I need a 8 m x 4 m x 0.6 m RC plinth (mechanical requirement) which needs to be built onto loose gravel. Obviously I am going to over excavate, back fill with selected material and bed my plinth on that. </p> <p>The plinth accommodates 2 x 65 m³ water tanks, placed symmetrically. The tanks are $\phi$3 m, 9 m high and weigh 796 kN when full. The wind load on each creates a 206 kNm moment and horizontal force of 38 kN.</p> <p>I need to design the plinth for bending and check the soil for shear failure. I assume the max. bearing capacity of the soil to be 150 kPa.</p> <p>I want to simulate this using FEM. How do I address the support conditions. I understand that I should use springs as supports, but how would I get the equivalent spring constants if my spring is soil?</p>

|structural-engineering|finite-element-method|foundations|

<p>Before jumping into elastic support conditions, you have to realise that it's <strong>not</strong> a ground beam. For three reasons:</p> <ol> <li><p>With a minor axis width to depth ratio of 6ish, it's not far away from a square. If you stick a wide support somewhere near the ends of the section, the load will simply arch over the supports in pure compression. You'd almost be able to build it in brickwork!</p></li> <li><p>Ground beams by definition span between supports. Your plinth is continuously supported by the fill. It's a raft if anything.</p></li> <li><p>Also, how will you apply the moment? Water does't transfer bending moment.</p></li> </ol> <p>The approximations suggested by Wasabi of the ground elasticity will totally negate your efforts at an accurate assessment of stresses within both the supports and the plinth.</p> <p>If you're going to use an FEA analysis, you'll be looking at modelling the plinth as several layers of concrete cubes, sat on several more layers of cubes of fill, all within a surrounding matrix of cubes representing the gravel. If you're designing a skyscraper have at it, otherwise the soil investigations and modelling effort required would make this one of the most over designed structures I've ever seen. Do the following:</p> <ol> <li><p>Treat the plinth as infinitely stiff</p></li> <li><p>Check that the allowable bearing pressure isn't exceeded by either the dead weigh with a uniform ground pressure, or a triangular ground pressure due to the moment when the tanks are empty, or both.</p></li> <li><p>Design the plinth for bending and shear due to the same load patterns</p></li> <li><p>Ignore the lateral 38. The plinth won't slide due to self weight and friction.</p></li> </ol> <p>You could do this in an afternoon after getting back from the pub. Flippancy aside, I've done similar (your approach) for a multimillion pound project. Don't. It's a pig.</p>

6414

How to model elastic support in FEM?

2015-12-04T15:12:14.390

<p>Below is a picture of two supporting posts in my basement. The posts are directly underneath the kitchen/dining room. Would it be feasible to remove one or both of the posts? What would the cost roughly be to make this change? </p> <p>Note: I know that I should have a structural engineer take a look before actually making any changes. Also, I live in the mid-west. </p> <p><a href="https://i.stack.imgur.com/p9iqJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p9iqJ.jpg" alt="enter image description here"></a></p>

|structural-engineering|beam|

<p>Whenever one speaks about removing a structural element (column or beam), the initial hypothesis must always be:</p> <blockquote> <p>NO! GOD NO! WHAT ARE YOU DOING!</p> </blockquote> <p>The second hypothesis must always be:</p> <blockquote> <p>No. No. No. No.</p> </blockquote> <p>That being said, it can be done. But it is not trivial and requires a serious investigation by a structural engineer. This is a "last resort" issue.</p> <p>Those columns are transferring a load from the beam which can clearly be seen on the ceiling to the foundation. If you remove the columns, then the beam will have to be able to take that load and transfer it across the room to the columns in the walls. The forces the beam will have to withstand will be multiplied many times (possibly 45 times, according to a simplified model).</p> <p>So, with all certainty, if you simply remove the columns, the beam will collapse and probably take the rest of your house (this is the basement after all) and you with it.</p> <p>So, you'd need to effectively redesign the structural layout of your basement. Without columns, you'll need to replace the current beam. That's basically a given. Then you'll have two choices:</p> <ul> <li>Place a much stronger beam in its place, capable of withstanding the increased forces. This almost certainly implies in an increase in beam height (and therefore reduction of usable height under the beam) since beam strength is proportional to the cube of the height.</li> <li>Place a grid of multiple beams (parallel and perpendicular to the current beam) on the ceiling, so that each beam absorbs a very small load and is capable of transferring it across the room. This has the additional issue that you'd need to know where the columns (or structural walls) in your wall are: if you don't have many places to rest your beams, your grid won't be efficient.</li> </ul> <p>There are then a bunch of issues of how to make the beams and the ceiling slab work in tandem to improve the beams' efficiency and to certify that the slab isn't "floating" over the beam, how to tie your new beams to the columns (if the beams are of concrete), limiting deflections and vibrations so that your dinning room doesn't sink or feel like a rock and roll concert when your cat walks over it, how to safely remove the existing columns and beam prior to doing your new layout, etc...</p> <p>So yes, you can do that. But are you really, really, <em>REALLY</em>, <strong><em>REALLY</em></strong> sure you want to?</p>

6424

Can I remove one or both of these support posts in my basement?

2015-12-06T03:14:43.013

<p>I know that digging tunnels is always <em>much</em> more costly than building ways or train above ground.</p> <p>Why doesn't the Channel Tunnel start around the coastline? Why does it have an around 10 km long portion under land on the British side?</p>

|civil-engineering|geotechnical-engineering|rail|tunnels|

<p>The picture, below, of the exaggerated long section of the Channel Tunnel was taken from <a href="https://en.wikipedia.org/wiki/Channel_Tunnel" rel="nofollow noreferrer">Wikipedia</a>.</p> <p><a href="https://i.stack.imgur.com/MFNuz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MFNuz.png" alt="enter image description here"></a></p> <p>_{Full-sized image <a href="https://commons.wikimedia.org/wiki/File:Channel_Tunnel_geological_profile_1.svg" rel="nofollow noreferrer">here</a>.}</p> <p>Some of the limiting factors for the Channel Tunnel are:</p> <ul> <li>Railways don't like steep gradients</li> <li>The tunnels comprising the Channel Tunnel were excavated using tunnel boring machines (TBMs). Like railways, they cannot tolerate steep gradients.</li> <li>The tunnel was excavated in chalk marl (green coloured material in the picture). This was due to its depth (not being too shallow and not being too deep) and its ability to be easily dug but also it would cause major support issues for the tunnels.</li> </ul> <p>If you look at the long section of the tunnel there is some high ground on both coasts. The width of the high ground on the French side is about 2 to 3 km, whereas on the English side the width about 7 to 8 km.</p> <p>For reasons of maintaining a comfortable gradient for the TBMs and the rail line and to position the tunnels in the lower part of the chalk marl combined with the width of the high ground on the coast and where the chalk marl is located within the high ground and because the chalk marl is inclined on the English side of the tunnel, the portal for the tunnel on the English side had to be positioned some 9 to 10 km from the coast.</p>

6435

Why does the Channel Tunnel enter the ground around 10 km from the coast?

2015-12-07T03:09:39.203

<p>I'm designing a little tester for a project which uses a pneumatic cylinder. The pneumatic cylinder's purpose is to apply an impact on a hard surface to test for deflection of the surface. When I looked around catalogs, the pneumatic cylinders have a theoretical force output, for example 3,016 N at 0.6 MPa with an extension speed of 0.8 m/s and a stroke length of 300 mm (preliminary selection although I'm not too sure).</p> <ol> <li><p>How do I calculate the impact energy upon collision with the hard surface, with this information I currently have? There is going to be an amount of impact energy that I need to achieve.</p></li> <li><p>I understand that when it comes to energy, the formula to use is $\frac{1}{2} mv^2$. What should the mass be? Is it the mass of the cylinder or whatever is placed on the piston rod end (I intend to put some kind of bumper at the end)?</p></li> <li><p>Are "impulse forces" something I should be looking at?</p></li> </ol> <p>I can't move on to select and appropriately sized pneumatic cylinder for my tester until I understand how to calculate the force output and impact energy.</p>

|mechanical-engineering|applied-mechanics|

<p>In general pneumatics are quite precise when it comes to force but modeling energy in this context brings in a lot of additional unknowns which make be difficult to measure. </p> <p>The most common method of impact testing is to use an impactor fixed to a pendulum as it's fairly easy to determine the kinetic energy at the lowest point in the swing directly from the initial gravitational potential energy and this gives a very repeatable test. </p> <p>It's also important to bear in mind that, even when you measure kinetic energy that doesn't really give you a general material property which you can extrapolate from in the same way as something like tensile strength as strain rate is usually a very important factor in impact behavior so the priority is usually to simulate the real world conditions you are interested in as closely as possible. </p> <p>If you need to use a pneumatic system for some operational reason then the most useful approach may be to treat it as an air spring. </p>

6441

Pneumatic cylinder force output and impact energy

2015-12-07T12:02:59.790

<p>The high water has some energy E = mgh. In the basic school, we were told that hydropower plants extract the energy of "falling water". Now, <a href="https://physics.stackexchange.com/questions/221243">that banned principle</a> says that if your water falls upon still turbines, you will loose energy. The water will turn into a foam, vapor and heat up until it slows down to the speed of rotating turbine. At this point, half of the "falling energy" is lost. This means that speed of turbines must match the speed of the water stream. </p> <p>Now, the waterfall has the highest speed at the basement and, in order to capture all the energy, you need to stop the water completely because water escaping at speed v, carries energy $E = mv^2/2$, where mass m grows linearly with time (your losses accumulate over time). These losses are proportional to water density m and speed v. Which means we loose more energy with each time unit if outbound stream speed v is higher. The only way to achieve 100% efficiency is to stop the water completely, to v=0. However, this implies no flow and no electricity produced.</p> <p>All these considerations visited me when I did some switching power supplies. Switching power supply achieves 100% efficiency by operating in one of 2 modes: full power, when water input is 100% open to accelerate the turbine and itself, and zero power, when fresh water input is closed and previously accelerated water altogether with massive turbine slow down by the energy consumer (assume a spring that we charge against its force -- it will slow down the stream with turbine). In electronics, the role of moving water and rotor is played by constant inductor, whose inductance is constant, unlike the water, which would arrive all the time and accumulate at v whenever you open the switch. I therefore would stop the water completely in the lowland, before releasing it there, achieving 100% extraction. But, it seems that in real hydro plants and wind mills, turbine rotates constantly, without any switching.</p> <p><strong>So, what theoretical efficiencies can wind and hydro turbines achieve?</strong></p>

|energy-efficiency|renewable-energy|power-engineering|wind-power|

<p>The theoretical maximum efficiency of a wind turbine is the Betz Limit, which is $\frac{16}{27}$ ( ~ 59.3%).</p> <p>The result <a href="http://www.reuk.co.uk/Betz-Limit.htm" rel="nofollow">dates back to 1919</a>.</p> <p>The Betz limit is, as you note, speed invariant. To get an intuition as to why, consider the system speeded up or slowed down. This changes the fluid's velocity, but does not change any of the issues around the dispersion of the fluid after the rotor. This means that the limit is driven not by absolutely velocity, but rather the ratio of the speed after the rotor, to the speed before it. That ratio is 1:3 - the fluid loses two-thirds of its speed in the perfect rotor.</p> <p>It's different in hydro where you've got a vertical drop. There, the turbine can take almost all of the speed out of the water. That's because there's a reservoir both before and after the turbine, and a vertical drop after the turbine. That means that gravity can do the work of dispersing the water after a turbine, and there's enough volume of space immediately behind the rotor for the water to travel into slowly. And as water is much denser, and much less compressible, than air, it displaces it easily. So we get real-world hydro efficiencies of ~90%.</p>

6444

What theoretical efficiencies can wind / hydro turbines achieve?

2015-12-07T14:26:56.197

<p>I'm studying the design of storm sewers and I had two questions relating to air gaps in storm pipes:</p> <ol> <li><p>According to sources I've come across, flow through storm pipes require an air gap so as to maintain open channel flow. Do I need to maintain a specific amount of air gap? Or will any amount of air gap do?</p></li> <li><p>Also, does it <em>really</em> matter whether or not I have an air gap? Do we maintain an air gap so as to prevent a partial vacuum and self siphonage within the pipes? In which case why would it matter? Furthermore, since storm sewer manholes (inlet pits) are open to the air, wouldn't that alleviate any pressure differentials?</p></li> </ol>

|civil-engineering|hydrology|

<p>There are additional features of the air gap besides pressure relief (Especially in small domestic pipes).</p> <ul> <li>Maintenance clean out access</li> <li>View port for viewing if flow is blocked</li> <li>Alleviating a small amount of BOD from system</li> <li>Reducing odor problems with adequate air flow</li> <li>Preventing stagnant water formation while not storming.</li> </ul> <p>All of these should be considered in addition to modeling - otherwise you might eliminate some of the standard "features" normally seen.</p>

6449

How much air gap is required in a storm pipe?

2015-12-08T04:51:30.537

<p>I have two columns of water at different (but constant) temperatures. If I calibrate a pressure transducer using the first column at differing heights, when I measure the second column how do I account for the change in temperature (and water density)?</p> <p>I want to measure pressure in megapascals (MPa) so for the calibration I currently plot pressure transducer mV output at different column heights against the calculated megapascal for the corresponding height and temperature of water using the equation:</p> <p>pressure = height (m) x water density (kg/m³) x gravitational constant (m/s²)</p> <p>But the density of the second column is different so this value needs to be adjusted.</p>

|fluid-mechanics|pressure|

<p>The pressure will be the same regardless of the temperature. So, the two columns will have the correct pressure recording. However, the height of the column will vary. </p> <p>i.e. - Say you have your pressure transmitter reading for calibration a 10C column (999.7 kg/m^3). The results are:</p> <p>0.5m - 4.9 kPa - 7 mV; 1m - 9.8 kPa - 10 mV; 2m - 19.6 kPa - 13 mV;</p> <p>Then you measure the second column at 70C (977.8 kg/m^3), and it reads 7mV. Then the column will still be 4.9kPa at the base, but the new height will be 51.1 cm, not the original 50 cm. The pressure will still be 4.9kPa though.</p>

6460

Calibrating and measuring pressure in Megapascals using a pressure transducer

2015-12-08T14:52:14.787

<p>Normal sealings and gaskets are made from NBR for use in conjunction with aqueous solutions of ethylene glycol which are common coolants for various purposes. </p> <p>However to keep sulphur out of a cooling system and the surrounding area I want to change NBR sealings against silicone sealings. I've read some vague information about reduced stability of these sealings in glycol containg coolants. But what is the detailed behaviour? Will it intumesce, brittle or deteriorate otherwise?</p>

|materials|cooling|

<p>Based on the available data it looks like it will perform well: </p> <ul> <li><a href="http://www.efunda.com/designstandards/oring/oring_chemical.cfm?SM=none&amp;SC=Ethylene%20Glycol" rel="nofollow">efunda.com - (4) Good, both for static and dynamic seals</a><br></li> <li><a href="http://www.pspglobal.com/fluid-compatibility/chemicals-e.html" rel="nofollow">pspglobal.com - 1 = Recommended</a></li> </ul> <p>Temperature and additives may change this, however.</p>

6463

What happens to silicone gaskets (siloxane rubber) in presence of ethylene glycol?

2015-12-09T14:35:09.580

<p>I want to contain a very small PCB (roughly a square centimetre) in a casing. For demonstration purposes I want to 3D print a couple of those casings as well.</p> <p>The most obvious method available would be using ABS for the printed casings. However, nylon appears to be smoother and tougher. Those traits are often a plus during demonstrations. However, since the PCB contains a S band antenna I'm slightly worried about the RF properties of the material used.</p> <p>Are there any downsides to switching from ABS to nylon that I'm not seeing? I'm especially interested in the RF characteristics in 2-4 GHz range. Cost is not an issue.</p>

|electrical-engineering|materials|rf-electronics|

<p>I am not sure if this will help, but when I have used any plastic in antenna designs, or RF power amplifiers, I would put them in a microwave oven and see if they heated up. Most would stay cold, but every now and then some would get warm, and sometimes I could hear some audible arcing. My only experience is at Ham Radio frequencies but I think it may still be a valid test for what you are doing. After all, a microwave oven works at 2.450 GHz. This has always worked for me to see if any insulating material would react to RF.</p>

6481

Are there any differences between using nylon or ABS for RF electronic cases?

2015-12-10T00:35:23.007

<p>How far must the surface temperature of a UL-listed combustion appliance rise to prohibit the manufacturer from specifying it for a zero clearance to combustibles installation?</p> <p>(I looked in UL 127 but was unable to find a straightforward answer there.)</p>

|combustion|heating-systems|ul|

<p>I <em>finally</em> found the answer to this in a BIA (Brick Industry Association) <a href="http://mha-net.org/docs/BIA01.PDF" rel="nofollow noreferrer">test protocol</a> for standardized brick fireplace designs. To quote what they're saying about UL-127:</p> <blockquote> <p>In general, after temperature equilibrium has been reached (or after 12 hours of fueling), UL-127 temperature limits on combustible surfaces are:</p> <ul> <li><p>90&deg;F over ambient room temperature for unexposed (ie, in contact and covered) surfaces and</p></li> <li><p>117&deg;F over ambient room temperature for surfaces exposed to ambient room air</p></li> </ul> </blockquote>

6486

What surface temperature (rise) is required to (dis)qualify a combustion appliance from obtaining a "zero clearance to combustibles" rating?

2015-12-10T17:40:55.017

<p>I have to calculate the uncertainty of measurement for an LVDT Linear displacement transducer used for measuring length. Uncertainty calculations are ruled by ISO 17025 but there are no (at least haven't found) specific guidelines for the mathematical procedure of uncertainty for LVDT sensors. There are many guides for more common measuring instruments like head micrometers and calipers though.</p> <p>Is there any official document to which I can refer to for the requirements and maths for calculating the uncertainty of measurement for this kind of sensors?</p>

|sensors|measurements|mathematics|standards|metrology|

<p>For the Maths of calculating uncertainty the standard document is the <a href="http://www.bipm.org/en/publications/guides/gum.html" rel="nofollow">GUM</a>. Which describes all the maths but can be somewhat unclear if you don't already have some idea how it is supposed to work.</p> <p>Depending on your current level of expertise there are several good introductions to uncertainty calculations. I would recommend <a href="http://www.npl.co.uk/publications/guides/a-beginners-guide-to-uncertainty-of-measurement" rel="nofollow">A Beginner's Guide to Uncertainty of Measurement</a>, which is freely downloadable.</p> <p>For any uncertainty calculation you need a model of your output given the inputs. For your case a very simple model would be purely linear, with distance proportional to voltage measured.</p> <p>$$L=\alpha V$$</p> <p>In this simple case, for fixed $\alpha$, the uncertainty in your distance is $\alpha u_v$ but more generally for a function $f$ with various inputs $x_i$. $$u^2=\sum_i (\frac{df}{dx_i})^2 u_{x_i}^2 $$</p> <p>Your uncertainties in V (or your general $x_i$) can be determined either using prior knowledge or by repeated measurement (at a given distance or similar).</p> <p>Calibration of your system is necessary to determine any coefficients in your model, such as $\alpha$, and to help inform the choice of model itself. The necessary calibration will vary in a case by case basis, but for your setup I guess you will want to measure a series of known distances and plot a calibration curve of distance against voltage.</p> <p>It is important to note that as you have uncertainty in your voltage and in the distance of your standard lengths used for calibration, as well as other non-linearities no there will inevitably be some uncertainty in your calculation of $\alpha$, unlike in my simple example. This will add some additional uncertainty proportional to the voltage to your measurement.</p>

6494

Uncertainty Calculation of LVDT displacement transducers for length measurement

2015-12-11T00:37:31.507

<p>I have a 5" x 4" x 3/8" plastic panel (wall thickness of 1/8", ribbed). I need to slide the panel over a 12" plastic track that is U-shaped (like a skateboard ramp). This motion could be repeated 2 million+ times. </p> <p><a href="https://i.stack.imgur.com/CkuPE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CkuPE.png" alt="enter image description here"></a></p> <p>The panel is rigid and maintains 2 contact edges when traversing the curved surfaces.</p> <p>My question: Should I be worried about wearing-out the plastic components at this scale? What options are there to protect this mechanism?</p> <p>I'm considering:</p> <ul> <li>plain plastic on plastic rails / tracks / sliding pegs</li> <li><a href="http://www.ebay.com/itm/10PCS-MR74-open-4-7-2-mm-Miniature-Bearings-ball-Mini-bearing-4X7X2mm-/400958903666?hash=item5d5b035572:g:yBQAAOSwHnFVrLig" rel="nofollow noreferrer">mini ball bearings</a></li> </ul> <p>Any thoughts on using cheap bearings for this?</p>

|mechanical-engineering|materials|mechanical-failure|bearings|

<p>Without knowing additional details, my attempt at a catch-all answer:</p> <p><strong>Summary:</strong> attempt to minimize wear. Millions of cycles is a lot for even low-load contact surfaces. Use components with high hardness and smooth materials. PTFE or UHMWPE backed by metal for full-contact bearings, and AISI 52100 or AISI 440C for ball bearings. There may be an issue with contact angles at the curves.</p> <h1>Theory</h1> <p>The primary issue with millions of cycles of amply supported contacting surfaces is almost certainly going to be wear. A simple model for wear is given by <a href="http://www.tribology-abc.com/abc/wear.htm" rel="nofollow">the Archard equation</a> - (<a href="https://en.wikipedia.org/wiki/Archard_equation" rel="nofollow">Wiki</a>), which holds that</p> <p>$$ Q = k\:p\:s $$</p> <p>where $Q$ is the wear volume, $K$ is a dimensionless constant, $W$ the normal load, $L$ the sliding distance, and $H$ the hardness of the softest contacting material. The Wiki link separates hardness $H$ from $k$. The idea behind the model is that contacting surfaces don't meet uniformly, but instead at randomly distributed, microscopic height differences called asperities. As a result, the actual contact area is much smaller than the apparent macroscopic contact area, and consequently the contact loads are orders of magnitude larger than the apparent macroscopic load. Thus, the size, distribution, ductility, and hardness of the asperities plays a role, as do lubrication, mean height difference (i.e. smoothness), and the presence of external detritus. The constant $k$ depends on all of these parameters, and is typically thus typically empirically determined. However, knowing the proportionality relationships is useful in making first-passes at material selection. Specifically: increasing hardness reduces asperity formation rate, reducing wear; and decreasing surface roughness decreases number of asperities, reducing wear.</p> <p>Therefore to minimize wear, a smooth, hard material, with an appropriate lubricant should be used. It would also be helpful to minimize external detritus (e.g. dirt, dust) entering the system. You can possibly go either route you describe in the question, depending on the specifics of your setup. I am <strong>assuming</strong> you require motion along a specific path with minimal deviation, like a train on rails.</p> <h1>First-pass Material Selection</h1> <p>For plastic-on-plastic with no balls, it would probably help to use parts formed from a hard metal or ceramic and coated with either polyfluorocarbons (PTFE, e.g Teflon) or ultra-high-molecular-weight polyethylene (UHMWPE). The trick here is that your setup requires changing direction smoothly. Getting custom made, non-linear polymer rails and bearings may be more costly than linear alternatives. Additionally, there is the problem of changing contact angles: the bearings would have to be shaped like the inner surface of a toroid, or like a saddle. If using a more typical cylindrical bearing, when the bearing meets the curve, only the leading and trailing edges will be in contact, which will cause damage.</p> <p>For ball-bearings, there would need to be a special track that solves the issues of leading/trailing edge contact. Ball bearings may be made of metal or ceramic, though ceramics like silicon nitride ($\textrm{Si}_{3} \textrm{N}_{4}$) may be overkill for a low-temperature application as they are more expensive than their metal counterparts. Typically chrome steel (usually near AISI 52100) or stainless steel (usually near AISI 440C) are used as they readily form martensite (a hard, metastable steel phase) during processing and can thus be made quite hard and wear resistant.</p> <p>If the weight truly is so low, adding any bearings at all will dramatically increase the weight. Instead of having bearings, is it possible to shape the panel to fit into a suitable-width groove dug into the track surface? Then make both out of teflon or UHMWPE, and back the track with brass or steel. A thin layer of dense lubricant could even make the panel float slightly, which would reduce wear to virtually zero.</p> <p>If the edges are in contact, as noted in the comments, then a track could still work, but would require rounded panel contact edges. The lubricant "float trick" would be messy and probably wouldn't cause flotation when rounding corners.</p>

6499

Are Rolling-Element Bearings Practical for Light Loads?

2015-12-11T06:58:38.283

<p>What equations are needed to describe the forces and properties of the kind of mechanisms used in a machine that harnesses the gravitational potential energy of a suspended weight to rotate the shaft of a permanent magnet DC generator (i.e. the <a href="http://gravitylight.org" rel="nofollow">gravity light</a>)?</p> <p>Basically, these machines aim to slow the descent of a suspended weight and convert that slow downward motion into high rpm rotary motion that is used to do work. As far as I can tell, this slowdown can be achieved using some kind of governor (in the case of a gravity powered clock air resistance from some spinning baffles is utilized), or by the torque required to turn the shaft on the generator.</p> <p>How can the relationship between the mass of the weight, the torque required to turn the generator, and/or any gearing that might be helpful be described mathematically?</p>

|mechanical-engineering|gears|energy-efficiency|

<p>Taking the force exerted by the mass and converting it to rotational force or torque will involve the following equations:</p> <p>F = g*(mass) where g = 9.81 and<br> T = r*F where r is the radius of the pulley from which the suspended mass is hanging. This is the torque that turns the generator.</p> <p>When other factors such as gears are included the gear ratios need to be multiplied with the torque. Example equations can be seen below:</p> <p>gear 1 radius is 20<br> gear 2 radius is 10<br> if the Torque in the shaft of gear 1 is 5 N.m then the torque in gear shaft 2 is:<br> T2 = 10/20*T1 = 2.5 N.m.<br> Gear calculations can become more complicated with modulus and gear teeth etc being included but this is the basics of it.</p>

6502

Equations relevant to gravity powered clocks and machines such as the gravity light

2015-12-11T20:44:47.790

<p>I was searching for information about pressure-volume processes when I read, in a heat engineering book, "The process is polytropic with $PV^n = \text{constant}$." </p> <p>How can I know that a real life process follows this model? Are all processes related to this heat engineering, pressure-volume, etc., in fact polytropic?</p> <p>And if they aren't then how can I tell when that model is valid?</p>

|thermodynamics|pressure|

<p>No, all thermodynamic processes are not polytropic. As an example, consider a real engine cycle. Any undergraduate engineering thermodynamics text should have a plot showing the true P-V diagram. You also can see <a href="http://4mechtech.blogspot.com/2013/12/indicator-diagram-or-p-v-diagram-actual_9.html" rel="noreferrer">examples</a> online. Clearly parts of this cycle can not be modeled with a polytropic process.</p> <p>A polytropic process is an approximation, much like a linear regression. It is correct to the extent by which the real and approximate curves are similar, however you define "similar".</p> <p>For ideal gases, sometimes with constant specific heats, you can derive certain polytropic relations, e.g. for an <a href="https://en.wikipedia.org/wiki/Isothermal_process" rel="noreferrer">isothermal process</a> of an ideal gas you get $P V =$ constant, for an adiabatic process with no entropy generation you get the <a href="https://en.wikipedia.org/wiki/Isentropic_process" rel="noreferrer">isentropic process</a> relations, etc. These are polytropic processes, and their use often follows the validity of their assumptions. That is, if the gas is ideal and the temperature is constant, the (polytropic) isothermal process is a good model. It should be clear that for other equations of state you will not necessarily get a polytropic process keeping the other assumptions the same.</p>

6508

Are all pressure-volume processes polytropic?

2015-12-11T22:02:13.497

<p>I have a quadrotor aircraft (4 motors/propellers around a center housing) with some electronics in an electronics housing. Looking to cool the electronics with propeller downwash. </p> <p><a href="https://i.stack.imgur.com/RDKZW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RDKZW.png" alt="enter image description here"></a></p> <p>I have a vent on the side of the electronics housing and one on the bottom. I assume the pressure on bottom P0 is atmospheric. Trying to determine which way the air will flow. </p> <p>Would pressure P1 be higher or lower than atmospheric? </p> <p>The propellers do work on the fluid and increase the dynamic pressure but there is also air at high velocities flowing past the side vents. Having trouble consolidating propeller dynamics with Bernoulli's principle.</p>

|fluid-mechanics|heat-transfer|aircraft-design|

<p>It depends on the velocity across P1 and the pressure differential across your blades. It would be hard to determine without testing or using computation fluid dynamics. That is probably overkill and its much easier to just change the geometry to do what you need. Hazzey mentioned a scoop in the comments; that would work well but it would also send rain and dust through. You can also do it the opposite way by putting down facing outlet to generate negative pressure. This would allow you to have a large area on the bottom vent and be able to filter the intake air.</p>

6509

Using propeller downwash for electronics cooling

2015-12-12T17:04:15.657

<p>I have a book that claims:</p> <blockquote> <p>A metal will have more of a tendency to dissolve another metal of higher valency than one of a lower valency.</p> </blockquote> <p>I interpret this as saying that a metal with higher valency will dissolve more easily into a host metal than it would if it had a lower valency. For example, $\textrm{Fe}^{3+}$ as impurity atoms and $\textrm{Na}^{+}$ as host will work better than vice versa.</p> <p>But then in the summary of the chapter it says:</p> <blockquote> <p>A valence that is the same as or less than the host material will have more appreciable solubility.</p> </blockquote> <p>Which of the two statements is correct?</p>

|materials|metallurgy|

<h2>Summary</h2> <p>Your book is correct. The same valence (+3/+3 solvent/solute) will be most likely to have large solubility, while higher valences (+3/+4) will be more likely to have high solubility than lower valences (+3/+2).</p> <p>Putting them in order as an example:</p> <pre><code>Solvent / Solute +3 / +3 Highest Likelihood of Solubility +3 / +4 +3 / +2 Lowest Likelihood of Solubility </code></pre> <h2>Explanation</h2> <p>The Hume-Rothery rules for solubility are necessary, but not sufficient, conditions for complete solid-solution solubility. That means if any of the rules are violated, complete solubility does not occur. However, even if all of the rules are met, complete solubility is not guaranteed. Mixing <em>always</em> occurs due to entropy, though possibly in infinitesimally small ratios.</p> <p>For posterity, the rules are as follows:</p> <ol> <li>Difference of less than 15% in the ratio of the radii of the solute and the solvent, so that</li> </ol> <p>$$ \left( \frac{\left|r_{solute} - r_{solvent}\right|}{r_{solvent}} \right)\leq 0.15 $$</p> <ol start="2"> <li><p>There must be the same or similar crystal structures between the solute and solvent.</p></li> <li><p>Complete solubility occurs when the solvent and solute have the same valency. <strong>A metal will be more likely to dissolve a solute of higher valency than a solute of lower valency.</strong></p></li> <li><p>The solvent and solute should have similar electronegativity. Intermetallics tend to form when the difference is large.</p></li> </ol> <p>The specific part you are interested in is in <strong>bold</strong>.</p> <p>These rules, more or less stated the same way, can be found from several sources, including <a href="https://en.wikipedia.org/wiki/Hume-Rothery_rules" rel="nofollow">The Wiki</a> and <a href="http://www.msm.cam.ac.uk/phase-trans/2004/titanium/hume.rothery.html" rel="nofollow">The University of Cambridge Department of Material Science and Metallurgy</a>. Note that the Cambridge link states the <em>opposite</em> of the bolded portion of rule 3. The Cambridge link is possibly incorrect, but it is difficult to tell based on the literature whether or not rule 3 is even useful.</p> <p>A thorough review of the literature surrounding the rules can be found <a href="http://eprints.soton.ac.uk/172201/" rel="nofollow">at this link</a>. The paper is unfortunately downloadable in Word Document form, but seems to be clean. A brief summary of the relevant part is that the relative valence factor rule (rule 3), specifically the bolded part, seems to only be valid when monovalent Cu, Ag, and Au are alloyed with B-subgroup elements of higher valence (using the Old IUPAC nomenclature, so the right-hand-block of the periodic table). The explanation involves Fermi surface and Brillouin zone interactions in the B-subgroup elements. What I take away from the review is that this particular part of rule 3 is not terribly useful in practice.</p>

6511

How does the relative valency of the impurity and host metals affect solubility in a substitutional solution?

2015-12-13T09:07:18.393

<p>A wooden police baton has to be hung on the back of an officer and released only when s/he wants it to be released. I thought about "push to stow, push to release". This is a thought experiment, I am not a manufacturer...</p> <p>The baton is smooth without grooves so alternative solution of releasing by rotation (like BNC plugs, or push/push pen) won't work. </p> <p>Where can I find a simple design of such contraption or what terms would I use to describe this?</p> <hr> <p>This is a rough model of a four bar (ABCFD) which push point E toward the object to hold (LMNOQ)</p> <p>The object, represent here as rectangle LMNOQ, moves on a fixed bar (HL). When pushed up (represented by piston A1 expanding), point E moves towards the edge NQ. The angle FDB (in the locked position) is > 180 , so it should be locked, as with locking pliers.</p> <p>In the open position, LMNOQ is lower, and point E is far from it. (this model is only for showing the idea. I am sure it should be modified to work in the real world)</p> <p><strong>HOW can a SECOND push unlock point D?</strong></p> <p>[<img src="https://i.stack.imgur.com/7iQf6.png" alt="partial clamping mechanism">] [<img src="https://i.stack.imgur.com/d8aoC.jpg" alt="baton is held"> <a href="https://i.stack.imgur.com/GAp3W.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GAp3W.jpg" alt="baton released"></a></p>

|mechanical-engineering|

<p>The mechanism you're looking for is named <strong>push-push latch</strong>. They are used in a lot of places and there are various patents for that. You can check <a href="https://www.google.com/patents/EP1596030A2?cl=un" rel="nofollow noreferrer">this patent</a> granted at 2005. It is for a small holder but it can be modified for bigger projects. <a href="https://i.stack.imgur.com/yWADU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yWADU.png" alt="Figure from Patent EP1596030 A2"></a></p> <p>However if I were you I'd desing a plastic holder such as the stress holder at <a href="http://kinetxbatons.com/product/" rel="nofollow noreferrer">Kinetxbatons</a> picture given below. Moving parts and mechanisms are prone to errors and dirt is a real problem. You can check snap-fit design books I've linked in another <a href="https://engineering.stackexchange.com/questions/2933/snap-fit-design/3341#3341">answer</a>.</p> <p><a href="https://i.stack.imgur.com/lboZ2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lboZ2.png" alt="Kinetxbatons stress holder"></a></p>

6516

Mechanism for a mechanical latch/clamp

2015-12-14T13:24:30.367

<p>For simplicity we're going to ignore aerodynamics and vehicle mass for a moment...</p> <p>Are small (around 1 litre) car engines designed to tolerate a higher duty cycle than large (2+ litre) car engines? </p> <p>Cars are usually all travelling at roughly the same speed which means the engines will be putting out proportionally similar amount of power, however a small engine would be operating at a greater proportion of its output than a large one to maintain the same speed. </p> <p>For some background to this, I have a 1.1 l car and I spend quite a lot of time with my foot completely flat on the floor, especially going uphill on a high speed road (60 or 70 mph limit) to keep up with the traffic. I think my record's somewhere around 2 minutes on full throttle.</p>

|automotive-engineering|engines|

<p>Most (modern) small and large car engines are designed for 100% duty cycle. This means that at 100% rated power(gas pedal all the way down) the engine can run continuously. Heat dissipation is the limiting factor like Dave Tweed stated. Cars that are not designed to continuously dissipate 100% of the heat generated at max power require the driver to watch the temperature gauge to limit the power use.</p> <p>Modern engines do not have this problem because the engine is <a href="https://en.wikipedia.org/wiki/Governor_(device)" rel="noreferrer">governed</a> (speed regulated) below the cooling capacity of the radiator. Most modern engines use electric fans on the radiators that are independent of engine rpm; greatly increasing continuous cooling capacity.</p> <p>Older cars and "high performance" cars may have power that exceeds the cooling capacity. Any engine that has had the maximum engine speed regulation removed or any engine that can be "<a href="https://en.wikipedia.org/wiki/Redline" rel="noreferrer">red lined</a>" can also overheat. An engine boosting system such as <a href="https://en.wikipedia.org/wiki/Nitrous_oxide_engine" rel="noreferrer">nitrous oxide</a> also exceeds the cooling capacity and thus must be used intermittently.</p> <p>You will often see both large and small cars pulled over for overheating along a steep hill on a hot day. In this instance the "duty cycle" under these operating conditions was not continuous (100%). However, duty cycle is typically not used to describe this behavior, because it is a design expectation that it can operate continuously. The engine was simply operating outside of its designed range.</p> <p>Duty cycle is not influenced by the size of the engine, but rather, duty cycle is a design parameter when designing an engine system. Most cars would be designed for continuous duty, while race cars would be designed for intermittent.</p>

6526

Are small car engines designed to tolerate a higher duty cycle than large car engines?

2015-12-16T08:02:52.703

<p>I am working on a simple program that would help welder to count how much ferrite, austenite etc. there is in weld. I am focusing on stainless steel.</p> <p>I have decided to use 3 graphs: Schaeffler, De Long and WRC. What I don't know is for which materials can I use each of these graphs, to get reliable results. </p> <p>I believe, that in each of those charts I can use only Cr-Ni steels. Am I right? Also, standard in my country (PL) that concerns stainless steels splits them into groups like: ferritic steel, austenitic steel, ferritic-austenitic steel, martensitic steel and heatproof steel (hope I translated it right). So which group of steels can I use in three diagrams I mentioned?</p> <hr> <p><strong>TL;DR</strong> <strong>Which groups of steel can be used in Schaeffler, De Long, WRC diagram to get phase composition of weld?</strong></p>

|steel|metallurgy|welding|

<p><strong>You are not exactly right. The purpose of Cr and Ni in stainless steel, besides the stainless part, is to tailor the microstructure. Cr promotes ferrite, Ni promotes austenite. Other elements have similar effects and must be taken into consideration. Beware of carbide formation changing properties and reducing weldability.</strong></p> <p><strong>The three different graphs plot three different sets of equivalents. Schaeffler accounts for Ni/C/Mn vs Cr/Mo/Si/Nb. De Long accounts for nitrogen as an additional Ni equivalent. The WRC diagram accounts for Ni/C/N/Cu, and Cr/Mo/Nb. The effectiveness ratios are different among the different graphs. What happens <em>in between</em> these different equivalents is <em>not</em> clear, so be careful modeling combinations that don't meet the requirements of any one graph individually. If you have both Cu and Si, for example, none of the graphs can identify what microstructures will occur. Your program can make interpolations, but you should warn any users you have done so and the results may not be correct.</strong></p> <h2>Microstructure Stabilizing Elements</h2> <p>Each of the diagrams has two axes, and points in the diagram area correspond to expected microstructures. Each axis is associated with either <strong>chromium equivalent</strong> or <strong>nickel equivalent</strong>, typically with the former on the horizontal axis and the latter on the vertical axis. In steels, chromium is a <strong>ferrite stabilizer</strong>, meaning that steels rich in chromium will tend to form ferritic microstructures. In contrast, nickel is an <strong>austenite stabilizer</strong>, and steels rich in nickel will tend to form austenite. Naturally, all three diagrams reflect this, with upper-left regions having high Ni equivalent, low Cr equivalent, and high austenite, and bottom-right regions having the opposite. See the Schaeffler diagram below (the other two have similar overall appearance).</p> <p><a href="https://i.stack.imgur.com/l3Ndz.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l3Ndz.gif" alt="Schaeffler Diagram"></a></p> <p>The idea behind an equivalent is that it provides an estimate of the effect of different alloying elements on stabilization of ferrite or austenite. Some elements are "better" at stabilizing their respective microstructure than others. For example, carbon stabilizes austenite 30 times more effectively than nickel, while manganese is about half as effective as nickel. Of course if you want austenite, you can't simply load up the steel with carbon or it will become extremely brittle and hard to work with. The fact that different elements have different efficiencies is what allows engineers to tailor steel properties.</p> <p>As noted on the axes in this particular diagram, the Cr and Ni equivalents include more than just those two elements. Specifically, carbon and manganese are austenite stabilizers, while molybdenum, silicon, and niobium are ferrite stabilizers. A more complete (but probably not exhaustive) list of each type:</p> <ul> <li>Ferrite Stabilizers: Cr, Si, Mo, W, Al, Ti, Nb</li> <li>Austenite Stabilizers: Ni, C, Mn, N</li> </ul> <p>In addition to the ferrite and austenite stabilizers, some elements also form carbides and nitrides, specifically</p> <ul> <li>Carbide Formers: Cr, W, Mo, V, Ti, Nb, Ta, Zr</li> </ul> <p>which can increase strength, but reduce ductility. They can also reduce fatigue properties by providing crack nucleation sites, and can alter creep properties. Carbide formers can also reduce weldability as they tend to form at microstructure boundaries when solidified rapidly and then allowed to cool slowly. Brittle microstructure boundaries, as with carbides, tend to cause excessive brittleness in the weld and greatly reduce impact toughness.</p> <h2>Why Do Stabilizing Elements Work?</h2> <p>Low-carbon steel goes through three phases as it solidifies and cools, starting with ferrite $\delta$-iron, austenite $\gamma$-iron, and then ferrite $\alpha$-iron. In the Fe-Cr phase diagram shown below, note that the upper and lower ferrite regions at the pure iron vertical axis eventually merge with increasing Cr, moving right on the diagram. This indicates that Cr <em>stabilizes</em>, in a thermodynamic sense, the ferrite phase: it is more energetically favorable with high Cr content. The reverse effect occurs with increasing Ni, so that the austenite region spreads outward, eliminating both ferrite regions, as seen in the Fe-Ni phase diagram. Ni makes austenite more energetically favorable.</p> <p><a href="https://i.stack.imgur.com/lJnbK.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lJnbK.gif" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/K5pok.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K5pok.gif" alt="enter image description here"></a></p> <p>See <a href="http://steel.keytometals.com/articles/art50.htm" rel="nofollow noreferrer">this link</a> for a good starting point for more information.</p> <h2>How Should I Use the Diagrams?</h2> <p>The Schaeffler diagram takes into account the following equivalents:</p> <ul> <li>$\textrm{Ni eq}=\textrm{Ni}+30\times \textrm{C}+0.5\times\textrm{Mn}$</li> <li>$\textrm{Cr eq}=\textrm{Cr}+\textrm{Mo}+1.5\times \textrm{Si}+0.5\times \textrm{Nb}$</li> </ul> <p>The De Long diagram has:</p> <ul> <li>$\textrm{Ni eq}=\textrm{Ni}+30\times \textrm{C}+0.5\times\textrm{Mn}+30\times\textrm{N}$</li> <li>$\textrm{Cr eq}=\textrm{Cr}+\textrm{Mo}+1.5\times \textrm{Si}+0.5\times \textrm{Nb}$</li> </ul> <p>The WRC-1992 stainless steel diagram has:</p> <ul> <li>$\textrm{Ni eq}=\textrm{Ni}+35\times\textrm{C}+30\times\textrm{N}+0.25\times\textrm{Cu}$</li> <li>$\textrm{Cr eq}=\textrm{Cr}+\textrm{Mo}+0.7\times\textrm{Nb}$</li> </ul> <p>Sources for <a href="http://www.bssa.org.uk/topics.php?article=121" rel="nofollow noreferrer">Schaeffler and De Long</a> and <a href="http://www.migweld.de/english/service/welding-stainless-steels/wrc-diagram-for-standard-analysis.html" rel="nofollow noreferrer">WRC-1992</a>. The diagrams are only strictly valid when only the relevant equivalent stabilizing elements are present in significant quantities, and no other stabilizers are present. If only Cr and Ni are present, any of the diagrams should be accurate, though you may want to verify that they indicate the same microstructures given the same values of Cr and Ni. In contrast, if both Cu and Si are present in a material, none of the diagrams should be expected to produce accurate results. It is possible to interpolate between the diagrams, but such a model would be a new model, and should be validated experimentally before being asserted as useful or accurate. At the very least, users should be warned that such interpolation is occurring.</p>

6545

Schaeffler, De Long, and WRC welding diagrams - which steels can be used?

2015-12-17T08:04:17.747

<p>For the last hour I have been trying to understand what $\dfrac{\partial p}{\partial s}ds$ is in Euler's equation but I have a problem. You can see this image:</p> <p><a href="https://i.stack.imgur.com/7scvT.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/7scvT.jpg" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/5zEOd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/5zEOd.png" alt="enter image description here"></a></p> <p>I know that $p\ dA = \text{Force}$</p> <p>$γ ds dA = mg$</p> <p>but what is $\left(p + \dfrac{\partial p}{\partial s} dA\right)$?</p>

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

<p>Whenever a fluid flows in a pipe, it flows due to difference in pressure. Due to this difference in pressure, there is a pressure gradient which is $\dfrac{\partial p}{\partial s}$. $\dfrac{\partial p}{\partial s}\ \text{d}s$ is the change in pressure over the length $\text{d}s$. $p+\dfrac{\partial p}{\partial s}\ \text{d}s$ is the pressure at the exit in terms of pressure function $p$ at inlet + change in pressure over length $\text{d}s$. So, $\left(p+\dfrac{\partial p}{\partial s}*ds\right)dA$ is also the force like $p\cdot \text{d}A$ with pressure $p$ at the exit equals $p+\dfrac{\partial p}{\partial s}\ \text{d}s$.Pressure variation in fluid is given by Tailor's series expansion with respect to length $s$ about $p$. $p(s + \Delta s) = p(s) + \dfrac{\partial p}{\partial s}\ \Delta s + O(\Delta s2)$. $O(\Delta s2)$ is the higher order term which is neglected.</p>

6558

∂p / ∂s ds Component in Euler's Fluids Equation

2015-12-18T05:08:20.240

<p>A simple thermostat will turn on at one temperature and off at a higher temperature. This keeps the thermostat from cycling on and off too quickly. The difference between these values is sometimes called <a href="https://en.wikipedia.org/wiki/Deadband#Thermostats" rel="nofollow noreferrer">dead band</a> and sometimes called <a href="https://en.wikipedia.org/wiki/Hysteresis#Control_systems" rel="nofollow noreferrer">hysteresis</a>. Both words apply to more complex processes like hydraulic valves and PID controllers, but what I am curious about is which one is better used to describe a simple thermostat.</p> <p>What term(s) do you use in this context? Please cite sources if you can.</p> <p>For some background, this question came up as a result of this <a href="https://engineering.stackexchange.com/questions/6547/is-it-a-problem-to-have-two-heat-sources-controlled-for-different-temperatures-o">Engineering SE question</a></p>

|control-engineering|terminology|

<p>The difference between the cut in and cut out of a temperature or a pressure control is the 'differential'. Simple controls have set differentials. I would use 'hysteresis' where there is a loss of movement in a mechanical gear and/or lever train when the train reverses direction. This is sometimes termed the 'lash'.</p>

6576

What do you call the difference between the on and off temperatures in a simple thermostat?

2015-12-20T19:25:05.857

<p>I have calculated the shear flows and center of an L shaped profile. I assume they are correctly calculated (I followed the same steps as during the lectures for an I shaped profile). However, I have some problems with the interpretation of the results. The results are shown in my simple sketch.</p> <p>The height of the profile is 1.5 times the base (but this actually doesn't matter) and thickness is very small compared to other dimensions. Given this, I calculated that the shear center is about 0.49xH to the left of the center of gravity (c.g.) to counter act the torsion of the shear flow. <a href="https://i.stack.imgur.com/1jmFQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/1jmFQ.png" alt="enter image description here" /></a></p> <h1>My confusion regarding the results:</h1> <p>The net result of the vertical shear flow is equal to the vertical force V. The net result of torsion of the shear flows is equal to the torsion moment of the vertical force (0.49xHxV). <strong>But there is no horizontal force to counter act the horizontal shear flows. How is this possible?</strong></p> <p>I am confused because in the lectures we have only seen shear flows in symmetric profiles (like C an I).</p>

|mechanical-engineering|stresses|structural-analysis|

<p>Firstly, the shear center is the point at which an applied load produces no torsion on the section. For a singly or doubly symmetric section, the shear center will lie on the axis of symmetry. For the unequal leg angle shown here, we can say by inspection that the shear center must be at the intersection of the two legs because the shear flow will produce no moment about this point.</p> <p><a href="https://i.stack.imgur.com/AMuf5.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AMuf5.png" alt="Shear Center of Unequal Leg Angle"></a></p> <p>The general equation for shear flow in an unsymmetric section is:</p> <p>$$q=\frac{I_xV_x - I_{xy}V_y}{I_xI_y - I_{xy}^2}\sum xA - \frac{I_yV_y - I_{xy}V_x}{I_xI_y - I_{xy}^2}\sum yA$$</p> <p>If the section is symmetric, the $I_{xy}$ term is equal to zero. For a symmetric section subjected only to vertical shear $V_y$ this equation reduces to:</p> <p>$$q = \frac{V_y}{I_x}\sum yA = \frac{VQ}{I}$$</p> <p>But for an unsymmetric section we need the full generalized equation. The result is something like this:</p> <p><a href="https://i.stack.imgur.com/WvK54.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WvK54.png" alt="Shear Flow Diagrams for Unequal Leg Angle"></a></p> <p>Now you can see that if we integrate the shear force over the length of the horizontal leg, the result will be zero. Integrating over the vertical leg will equal the applied shear force $V_y$</p> <p>There is a fairly good discussion/example of this <a href="http://www.ae.msstate.edu/tupas/exp/A14.8_ex1.html" rel="noreferrer">here</a> and <a href="https://books.google.com/books?id=D2JtIxV-WQMC&amp;pg=PA471&amp;lpg=PA471&amp;dq=shear%20flow%20in%20unsymmetric&amp;source=bl&amp;ots=yoJj8yy1Yo&amp;sig=cOa6atR4Kb7Yfa5EMZTpbNdrXvI&amp;hl=en&amp;sa=X&amp;ved=0ahUKEwjQoKSvv-3JAhWHlh4KHRqtCYs4FBDoAQgjMAI#v=onepage&amp;q=shear%20flow%20in%20unsymmetric&amp;f=false" rel="noreferrer">here (page 474)</a></p>

6601

Interpretation of shear flow/shear center of L-profile

2015-12-22T08:54:11.327

<p>For torsional constant, analytical formulas are available for some standard steel sections as shown in this <a href="https://en.wikipedia.org/wiki/Torsion_constant" rel="nofollow noreferrer">wiki page</a>.</p> <p>But what about arbitrary shapes? How can I obtain the formulas for torsional constants? If no such formulas are available, are there any first principles that I can use to deduce the approximate formulas or FEM calculation for torsional constant?</p> <p>Note: a related question about the FEM formulation of torsional constant is being asked <a href="https://engineering.stackexchange.com/questions/6622/how-do-i-use-fem-to-derive-the-torsional-constant-of-an-arbitrarily-shape">here</a>.</p>

|materials|structural-engineering|steel|

<p>For arbitrary thick-walled shapes, there are no generic formulas available. You have to calculate the torsion constant using numerical methods, e.g. FEM or BEM.</p>

6613

First principle derivation for torsional constant

2015-12-22T22:58:38.233

<p>Could someone please explain me why the velocity of point P and P' aren't the same? Appearently it is because the angular velocity of point A and B aren't the same.</p> <p><a href="https://i.stack.imgur.com/GH4r6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GH4r6.png" alt="enter image description here"></a></p> <p>We are only given that the angular velocity of A is 10rad/s clockwise.</p> <p>Why? I would think that, as these two elements are connected , they would all rotate with the same angular velocity. If A rotates with a speed of 10rad/s it would be logical for me that point B just follows and rotates with the same angular velocity.</p>

|kinematics|

<p>The velocity of P' is perpendicular to AD, and the velocity of P is perpendicular to BP. So clearly the directions are not the same. Thus there is relative velocity between P and P' as shown in the diagram. Now, if gyroscopes are mounted at D, P', or any point on AD, except A, they will all measure the same angular velocity. A gyroscope mounted at P will measure the same value as one mounted anywhere between B and P, excluding B.</p>

6621

Why do 2 points on a mechanism not have the same angular velocities?

2015-12-22T23:53:56.660

<p>In <a href="https://engineering.stackexchange.com/q/6613/3353">this question</a> I ask about how to perform a first-principle derivation of the torsional constant of a section. It appears that there is no such analytical derivation for torsional constant, so my question therefore becomes: what about FEM derivation for torsional constant of arbitrary shapes section? </p> <p>Note that I am not interesting in just using a FEM package without actually understanding the basic principle. I want to be able to derive the FEM formulation from first principles.</p>

|modeling|finite-element-method|

<p>Make a long shaft with your chosen section. Fix one end by 3dof restraints at each node. Put a closing plate across the other end and apply a torque, hence back calculate your torsional constant. Double the length of the shaft and check you get the same answer. St Venant's principle is the reason you want a long shaft to do this test.</p>

6622

How do I use FEM to derive the torsional constant of an arbitrary shape?

2015-12-23T09:17:42.740

<p>I'd like to compute the response to a step function of a electrical/thermal system. Generally I can "easily" compute the transfer function $H$:</p> <p>$$H(\omega) = \frac{V_{out}(\omega)}{V_{in}(\omega)}$$</p> <p>Since the Fourier transform ($\mathcal{F}$) of the Heaviside function is (computed with WA):</p> <p>$$\mathcal{F}(\theta(t)) = V_{in}(\omega) = \sqrt{\frac{\pi}{2}}\delta(\omega)+\frac{i}{\sqrt{2\pi}\omega}$$</p> <p>Hence, noting $\mathcal{IF}$ the Inverse Fourier transform:</p> <p>$$V_{out}(t) = \mathcal{IF} \left\{ \left( \sqrt{\frac{\pi}{2}}\delta(\omega)+\frac{i}{\sqrt{2\pi}\omega} \right) H(\omega) \right\}$$</p> <p>To check my math I tried to compute the response for a simple RC system:</p> <p><a href="https://i.stack.imgur.com/6H6lP.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/6H6lP.gif" alt="enter image description here"></a></p> <p>I should get the well known charge of the capacitor. The transfer function:</p> <p>$$H(\omega) = \frac{1}{1+i\omega R C}$$</p> <p>Computing the Inverse Fourier transform ($\mathcal{IF}$) with WA ($R=C=1$) I get:</p> <p><a href="https://i.stack.imgur.com/CKnbV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/CKnbV.png" alt="enter image description here"></a></p> <p>This would be correct if we were going backward in time :/. So the question is... What am I doing wrong?</p> <p>I did the same using Laplace Transforms and everything works fine... But I don't understand why.</p> <p>P.S. I don't want another method, I just want to understand what's wrong in my approach.</p> <p>P.S. the reason why I am using WA is that for my more complicated system I need to compute the Fourier transforms using WA.</p>

|electrical-engineering|mathematics|signal|transfer-function|signal-processing|

<p>The main reason is likely due to Wolfram Alpha applying the inverse Fourier transform as a second Fourier transform. In fact, doing so "flips time" - as can <a href="https://en.wikipedia.org/wiki/Fourier_transform#Invertibility_and_periodicity" rel="nofollow noreferrer">be shown mathematically</a>: </p> <p>Defining the '''flip-time operator''' $\mathcal{P}$ that inverts time, $\mathcal{P}[f(t)] ↦ f(−t)$ $$\begin{align} \mathcal{F}^0 &amp;= \mathrm{Id}, \quad \mathcal{F}^1 = \mathcal{F}, \\ \mathcal{F}^2 &amp;= \mathcal{P}, \quad \mathcal{F}^4 = \mathrm{Id}, \\ \mathcal{F}^3 &amp;= \mathcal{F}^{-1} = \mathcal{P} \circ \mathcal{F} = \mathcal{F} \circ \mathcal{P} \end{align}$$</p> <p>Applying the fourier transform 3 times to the system will get you the version in normal time. Since waves are time consistent, it normally does not matter.</p>

6624

Response of a system to a step function (Heaviside function)

2015-12-23T10:15:03.643

<p>I ve tried to solve this problem in so many ways but still didn't manage to do it...</p> <p><a href="https://i.stack.imgur.com/y9G91.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y9G91m.jpg" alt="enter image description here"></a></p> <p>What would be the correct way to solve it please?</p> <p>This arm of this mechanism has a length of 0,2m. The piston has an angular velocity of 2000 tours/min. What would be the velocity of point D for an angle theta of 60 degrees?</p> <p>I think that what I am missing is the angle formed by the arm and the line, which is 50mm long. Example like here (different exercise):</p> <p><a href="https://i.stack.imgur.com/l0kNH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l0kNHm.png" alt="enter image description here"></a></p> <p>I am trying to look for this angle beta which could help me solve the problem</p> <p>expected answer:2,88m/s</p>

|mechanical-engineering|kinematics|

<ul> <li>First you build a kinematic chain between <em>A</em>, <em>B</em> and <em>D</em> with radius $r$ between <em>AB</em> and the length $\ell$ between <em>BD</em>. The orientation angles from vertical are $\theta$ and $\phi$ respectively. <ol> <li>$(x_B,y_B) = (x_A,y_A) + (-r \sin\theta,r \cos \theta)$</li> <li>$(x_D,y_D) = (x_B,y_B) + (-\ell \sin\phi, - \ell \cos \phi)$</li> </ol></li> <li>You find the angle $\phi$ from the constraint that $x_A-x_D = r \sin\theta + \ell \sin \phi = x_{AD}$ $$\sin \phi = \frac{x_{AD}}{\ell}-\frac{r}{\ell} \sin \theta $$</li> <li>Next you differentiate with respect to time using the chain rule to get the velocities <ol> <li>$(\dot{x}_B,\dot{y}_B) = (-r \dot{\theta} \cos\theta, -r \dot{\theta}\sin\theta)$</li> <li>$(\dot{x}_D,\dot{y}_D) = (\dot{x}_B,\dot{y}_B) + (-\ell \dot{\phi} \cos\phi, \ell \dot{\phi} \sin \phi)$</li> </ol></li> <li>You find the rotational velocity of the connecting rod from the constraint $\dot{x}_D=-r\dot{\theta}\cos\theta -\ell \dot{\phi} \cos \phi=0$ $$ \dot{\phi} = - \frac{r \dot{\theta} \cos\theta}{\ell \cos{\phi}}$$</li> <li>The collar speed is</li> </ul> <p>$$ v = \dot{y}_D = \ell \dot{\phi} \sin \phi - r \dot{\theta} \sin\theta = -\frac{r \sin(\theta+\phi)}{\cos\phi} \dot{\theta} $$ where $\dot{\theta}$ is the rotation of the disk in radians per second.</p>

6625

How to calculate the velocity of a piston-like linkage?

2015-12-24T21:48:10.713

<p>I am in the process of selecting a linear rail system for a 5ft x 10ft cnc laser cutter. <a href="http://openbuildspartstore.com/delrin-mini-v-wheel-kit/" rel="nofollow">Delrin polymer rollers</a> on 2 inch square extruded aluminum tube is an economically attractive solution. I have had success with them on a 3d printer, but I am not sure how well they will handle higher speeds (2m/s) and 12hr per day operation.</p> <p>The environment is reasonably dust free, but not a clean room. I suspect the erosion of the Delrin over time may cause tolerance issues.</p> <p>What are some successful implementations of delrin or other plastic rollers in an commercial/industrial setting? Will the plastic rollers typically outlive the bearings?</p>

|mechanical-engineering|failure-analysis|

<p>Given that you're using extruded aluminium rails polymer rollers are likely to be the best solution as anything harder will likely make the wear rate of the aluminium more of an issue.</p> <p>To be honest using extruded aluminium for a machine tool is always going to be a cheapest rather than best solution compared to ground steel or cast iron.</p>

6639

What is the longevity of Delrin rollers on aluminum linear rail?

2015-12-25T09:07:29.317

<p>I would like to design a long, uniform magnetic field using Hemholtz coils but am not sure how far I can separate them before the field lines diverge and become two loops that go around each coil rather than one which goes through both.</p> <p>I have formulae for finding the magnetic flux at different points but I am not sure how to use them to determine this distance I am trying to calculate.</p> <p>Does anyone have experience in this area who could offer some advice?</p>

|mechanical-engineering|electrical-engineering|magnets|

<p>Short answer: Not very far.</p> <p>I had this question one or two years ago. At that time I did a simple calculation of $B$ along the symmetry line for Helmholtz coils with non-standard distance, i.e. the position of the two coils being different from $+R/2$ and $-R/2$ ($R$ being the radius of the two coils). The homogeneity of $B$ decreases quite quickly when deviating from the standard.</p> <p>Wikipedia gives you the equation. For a single conductor loop, according to Biot-Savart, $$ B(x)=\frac{\mu_0 I}{2}\cdot \frac{R^2}{(R^2+x^2)^{\frac 32}} $$ with $B$ being the magnetic flux density (more precisely its vector component along the symmetry axis of the loop), $\mu_0$ the vacuum permeability, $I$ the current along the conductor loop, $R$ the radius of the conductor loop and $x$ the distance to the center of the loop along the symmetry axis.</p> <p>The following plot shows $B(x-d)+B(x+d)$ for some $d \geq R/2$ for $\frac{\mu_0 I}{2} = R = 1$.</p> <p><a href="https://i.stack.imgur.com/PjU4W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PjU4W.png" alt="enter image description here"></a></p> <p>As you can see, $B$ drops significantly in the center between the two coils once deviating more than 20% from $d=R/2$. You can calculate for yourself how much deviation you wish to accept.</p> <p>Be reminded that this is just an evaluation of the magnetic flux density along the central symmetry line of a coil pair. If you are interested in the full 2D field, things become more complicated. But this simple evaluation gives you some idea what happens when you move the two coils apart.</p> <p>For a longer homogeneous field you need more coils.</p>

6642

How to calculate the maximum spacing between Hemholtz coils?

2015-12-27T01:57:30.460

<p>I would like to essentially flatten or smooth the surface of a viscous liquid after dropping an object into it, without actually touching the liquid. For example, if I drop a rock into a tub of oil, I need the oil to move into the void and return to having a smooth surface as fast as possible.</p> <p>It seems that I need to create some sort of force that acts against the liquid surface from above, or a force from the sides that acts as a kind of 'wiper' to smooth it out. </p> <p>I have looked into using ferrofluids and electromagnets on the sides or bottom of the liquid vat/container, however many demonstrations I have seen result in a 'spiky' surfaces when magnets are brought near the liquid due to non-uniform field lines.</p> <p>I have also looked into trying to suspend diamagnetic nanomaterials in the liquid which could be repelled by a magnetic field from above the vat, however I'm not sure about whether diamagnetic materials exist which would could be used like ferrofluids in the sense of staying suspended in the liquid.</p> <p>My job depends on being able to do this, so if anyone has any ideas I'd really appreciate hearing them.</p>

|fluid-mechanics|magnets|

<p>Some quick dimensional analysis for a Newtonian fluid suggests that the settling time scales as follows:</p> <p>$$\tau = \sqrt{\frac{L}{g}} f\left(\frac{\nu}{\sqrt{g L^3}}\right)$$</p> <p>where $\tau$ is the time it takes for the liquid surface to settle, $L$ is some characteristic length of the dropped object (say, the diameter of a ball), $\nu$ is the kinematic viscosity, and $g$ is the acceleration due to gravity. I expect the function $f$ to increase as the parameter inside it increases, which makes sense. The higher the viscosity, the slower the settling, and the stronger the gravity, the faster the settling.</p> <p>So, influencing these parameters in some way could help. Unfortunately, you have no control over these parameters aside from the viscosity, which you might influence via heating.</p> <p>An oscillation applied to the container could help smooth things out. The speed by which this occurs surely is strongly dependent on the viscosity for a Newtonian fluid. For a non-Newtonian fluid (say, peanut butter), I'm not sure this will necessarily work.</p> <p>I also imagine an air jet applied parallel to the surface could help, depending on the viscosity, especially if its direction is varied. Targeted jets striking at an angle to the surface might help too, but this could require a complicated control system or manual human intervention to do the targeting.</p> <p>Suction applied directly above the hole may help, again, depending on the viscosity.</p> <p>(After writing this, I noticed Chuck said some similar things in the comments. He also recommended a centrifuge, which I didn't think of.)</p>

6659

How to 'smooth/flatten' the surface of a disturbed liquid without contacting it?

2015-12-27T15:29:02.230

<p>I am familiar with the basic concepts of fits and tolerances, and I am aware of the various online calculators for determining them. However, I don't understand how to apply those numbers to my actual CAD design for the parts that I want to fit together. None of the calculators seem to account for the actual manufacturing tolerances of the shop I'm working with (in this case +/- 0.1 mm).</p> <p>So, to be more specific, let's say I have a shaft and hole and I want to work with a nominal size of 6 mm. I want a "locational clearance" fit between them. The calculators tell me the maximum and minimum hole sizes, etc., but I don't know how to gather from this information what I should make the nominal sizes of the hole and shaft in my CAD model. And how can a calculator tell me that anyway, if it doesn't know what size my manufacturing tolerance is?</p> <p>Can anybody elucidate?</p>

|mechanical-engineering|machining|

<p>I think you may be getting a little confused here. There is a fit tolerance, which determines how easy it is to insert and rotate a pin in a joint, and then there is machining tolerance, which is how accurately a shop can match your nominal values. </p> <p>If your shop can't meet your printed fit tolerances, you're using the wrong shop.</p> <p>That said, if you must go with a particular vendor (3d printing, for example), and you have no other choice, then make your nominal value the mean of your MIN and MAX values and then change your tolerance values. </p> <p>So, for an easy example, say your hole should be 6mm +/- 0.05mm. Find the MIN (6-0.05 = 5.95mm) and the MAX (6+0.05 = 6.05mm), and average them - 6mm. Tolerances remain unchanged and this was a trivial problem. </p> <p>Consider though 6mm +0.1mm -0mm. Now do the same, MIN=6mm, MAX=6.1mm, <em>new nominal value of 6.05mm</em>, and new tolerance of +/- 0.05mm.</p> <p>It's functionally the same spec, but the "new" format would get you some dirty looks at a machine shop. However, now you can compare your needs: +/-0.05mm, to the shop capability: +/-0.1mm, and can see that they cannot meet your tolerancing needs and are thus unlikely to produce a satisfactory product. </p>

6667

How do you apply fits and tolerances when designing hole and shaft?

2015-12-28T00:20:40.777

<p>I noticed many kitchen utensils or stand mixer attachments are made with injected aluminum. Aluminum (~\$20/kg) is usually more expensive than stainless steel (~\$10/kg) and is not compatible with dishwashers. </p> <p>For instance, I have a peeler made with anodized aluminum and a KitchenAid attachment with some aluminum parts that I cannot put in my dishwasher. I don't understand why engineers are still making such utensils in this material. </p> <p>Is there any good reason for this?</p>

|steel|aluminum|

<p>The aluminium has different temper, for the utensil usually use 1 series aluminum alloy which is much softer, so it's easier to cast. And aluminum is light, also aluminum can be recycled which is environmentally friendly. Most of all, for business, profits are very important. </p>

6670

Why do manufacturers still make aluminum kitchen utensils?

2015-12-28T18:59:59.827

<p>I have a pump that I want to use to pump a solvent. It is hard black plastic so I'm pretty confident it is either ABS or polypropylene (it wasn't expensive enough to be some more exotic plastic).</p> <p>My solvent will rapidly dissolve ABS but it will work great if its PP. Anyone have a strategy to tell which it is without destroying it if it happens to be ABS?</p>

|plastic|

<p>ABS and Polypropylene have very different chemical resistances, with Polypropylene very resistance and ABS no so much. Acetone, Methyl Ethyl Ketone (MEK), 2-Butanone, and Methylene Chloride will dissolve ABS but not Polypropylene. Thus put a small drop of the solvent on the surface, and mush it around a little with the tip of a knife. If you see some plastic softening and diluting, it's ABS.</p>

6682

How to determine whether plastic is ABS or Polypropylene?

2015-12-29T13:04:38.090

<p>How can I control a Four-Bar Infinity Coupler such that it does not accidentally become a Paralellogram Linkage? It seems to me that, when the bars are all horizontal, gravity could change the movement of the green driven linkage.</p> <p><a href="https://www.youtube.com/watch?v=NZZtXUcRZVs" rel="noreferrer">https://www.youtube.com/watch?v=NZZtXUcRZVs</a></p> <p><a href="https://i.stack.imgur.com/eJLCW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/eJLCW.png" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/WhECH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WhECH.png" alt="enter image description here"></a></p>

|mechanical-engineering|design|linkage|

<p>Gearing is complicated by the fact that the driver bar moves at a constant speed while the driven bar changes speed.</p> <p>Could this gearing Solution work? The gear should only engage when the center bar is fully extended to the right or fully extended to the left.</p> <p><a href="https://i.stack.imgur.com/99uRy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/99uRy.png" alt="enter image description here"></a></p> <p><strong>Update:</strong> </p> <ul> <li><a href="https://www.youtube.com/watch?v=I4V3NqwZG0o" rel="nofollow noreferrer">https://www.youtube.com/watch?v=I4V3NqwZG0o</a></li> <li><a href="https://www.youtube.com/watch?v=aPUcdGnf2uk" rel="nofollow noreferrer">https://www.youtube.com/watch?v=aPUcdGnf2uk</a></li> <li><a href="https://www.youtube.com/watch?v=c9jSFZZX6uA&amp;feature=youtu.be" rel="nofollow noreferrer">https://www.youtube.com/watch?v=c9jSFZZX6uA&amp;feature=youtu.be</a></li> <li><a href="https://www.youtube.com/watch?v=cw0Wco_O600" rel="nofollow noreferrer">https://www.youtube.com/watch?v=cw0Wco_O600</a></li> </ul>

6696

Design of a Four-Bar Infinity Coupler Linkage

2015-12-30T02:35:30.890

<p>For me, it's a bit hard to imagine where shear center can be, and what purpose it serves. <a href="http://www.civilengineeringterms.com/mechanics-of-solids-2/shear-center/">The definition</a> I found online is very vague:</p> <blockquote> <p>Shear center is defined as the point on the beam section where load is applied and no twisting is produced.</p> </blockquote> <p>Given the general profile of a section, is there a proper mathematical derivation/formula that shows how the shear center is defined/derived?</p>

|mechanical-engineering|structural-engineering|beam|

<p>I hope answering this after all this time is not a huge faux pass, but this question comes up on search engines when looking for shear center calculations.</p> <p>First, some preliminaries.</p> <ul> <li>This applies to <a href="https://en.wikipedia.org/wiki/Simply_connected_space" rel="nofollow noreferrer">simply connected</a> cross sections, like channels, I-shapes, L-shapes, etc., but not things like pipes or HSS boxes. -The cross section is defined by the <span class="math-container">$y$</span> and <span class="math-container">$z$</span> coordinates, with the <span class="math-container">$x$</span> coordinate being the axis of the element.</li> </ul> <p>With this out of the way, the following picture illustrates the set-up for this. <a href="https://i.stack.imgur.com/BZDm5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BZDm5.png" alt="Cross section centered on centroid" /></a></p> <p>In the following I refer to the cross sectional domain as <span class="math-container">$\Omega$</span> and its boundary as <span class="math-container">$\partial\Omega$</span>. The first step is to solve for the auxiliary warping function <span class="math-container">$\omega_{0}$</span>, by solving</p> <p><span class="math-container">$$ \begin{aligned} div(grad(\omega_{0}))&amp;=0&amp; \qquad &amp;\textrm{in } \Omega \\ grad(\omega_{0})\cdot n&amp;=\frac{1}{2} \frac{d}{ds}(y^2+z^2)&amp; \qquad &amp;\textrm{on } \partial\Omega \end{aligned} $$</span></p> <p>This is a Neumann problem, so it's only solvable if <span class="math-container">$$ \oint_{\partial\Omega} \frac{d}{ds}(y^2+z^2)\,ds=0, $$</span> luckily this is automatic in our case, as this expression reduces to the expression inside the derivative evaluated at the starting point and endpoint of the path. However, since the path is a closed loop this is automatically 0. All Neumann problems are solved uniquely up to a constant, so fix any single point to an arbitrary value (usually a corner to 0) and you are golden.</p> <p>Now that you have <span class="math-container">$\omega_{0}$</span> the coordinates of the shear center are given by <span class="math-container">$$ y_s=-\frac{I_zI_{y\omega}-I_{yz}I_{z\omega}}{I_zI_y-I^2_{yz}} \qquad \qquad z_s=\frac{I_yI_{z\omega}-I_{yz}I_{z\omega}}{I_zI_y-I^2_{yz}} $$</span> where <span class="math-container">$I_y$</span>, <span class="math-container">$I_z$</span> and <span class="math-container">$I_{yz}$</span> are the second moments of area (usually called moments of inertia), given by <span class="math-container">$$ I_y=\int_{\Omega}z^2\,dydz \quad I_z=\int_{\Omega}y^2\,dydz \quad I_{yz}=\int_{\Omega}yz\,dydz, $$</span> and <span class="math-container">$$ I_{y\omega}=\int_{\Omega}z\omega_{0}\,dydz \quad I_{z\omega}=\int_{\Omega}y\omega_{0}\,dydz $$</span></p> <p>I derived this from an old book I had, but you can find an in-depth explanation and derivation in <a href="https://rc.library.uta.edu/uta-ir/bitstream/handle/10106/1144/umi-uta-2271.pdf?sequence=1&amp;isAllowed=y" rel="nofollow noreferrer">this thesis</a>, the shear center expressions are eqs (2.98) and (2.99) but the whole chapter 2 is worth a read.</p> <p>As an illustration, some examples. The following pictures show the values of <span class="math-container">$\omega_0$</span> (in the colorbar, units are not important) and the <span class="math-container">$y-z$</span> dimensions (in mm) for different simply connected shapes.</p> <p><a href="https://i.stack.imgur.com/TGu0O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TGu0O.png" alt="Centroid and shear center for I beam" /></a> Here's the solution for an I-beam. As expected, the centroid (circle) and shear center (cross) are coincident.</p> <p><a href="https://i.stack.imgur.com/GF6kz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GF6kz.png" alt="Centroid and shear center for an L-shape" /></a> Next we have an uneven L-shape cross section. The centroid is somewhere outside the cross-section, while the shear center, as you'll usually find in the literature, is at the intersection of both legs.</p> <p><a href="https://i.stack.imgur.com/gh2wn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gh2wn.png" alt="Centroid and shear center for a channel" /></a> Finally we have a channel (or C shape or U shape). As expected, the centroid is somewhere to the right of the web (between the flanges) while the shear center is to the left of the web (away from the flanges).</p> <p>As a bonus of doing all this, <span class="math-container">$\omega_{0}$</span> lets you get the torsional properties of the cross section for almost no extra work. First you construct the warping function from the auxiliary warping function by doing <span class="math-container">$$ \omega=\omega_0-z_sy+y_sz, $$</span> then the warping constant <span class="math-container">$C_{\omega}$</span> is given by <span class="math-container">$$ C_{\omega}=\int_{\Omega}\omega^2\,dydz $$</span></p> <p>And the torsional constant of the cross section, <span class="math-container">$J$</span> is given by <span class="math-container">$$ J=I_s-W_0 \\ I_s=\int_{\Omega}(y-y_s)^2+(z-z_s)^2\, dydz \qquad W_0=\int_{\Omega}\left(\frac{\partial\omega}{\partial y}\right)^2+\left(\frac{\partial\omega}{\partial z}\right)^2 dydz $$</span></p> <p>PS: Any tips on how to better format images are welcome.</p>

6705

What is the mathematical derivation for shear center of a beam?

2015-12-30T17:06:34.867

<p>Are there any established (compact and simple) solutions for turning two 5mm rods/shafts side-by-side together by driving a singe motor? The torque is about 5kg-cm. The one I'm thinking of is the coupling rod:</p> <p><a href="https://i.stack.imgur.com/JFAVT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JFAVT.png" alt="enter image description here"></a></p> <p>Are there disadvantages to this solution compared to others, e.g., gearing?</p> <p>If the coupling rod is a good solution, where could I get my hands on those "tabs" where the coupling rod connects? Is the coupling rod supposed to be connected using some small shoulder bolts and nuts?</p>

|mechanical-engineering|motors|

<p>This <a href="https://engineering.stackexchange.com/questions/6696/design-of-a-four-bar-infinity-coupler-linkage/6697#6697">Engineering SE Question</a> explains why a single parallel link will not work. When the link is at <a href="https://en.wikipedia.org/wiki/Dead_centre_(engineering)" rel="nofollow noreferrer">top dead center</a> there is nothing keeping the wheels in time. The mechanism can switch to a lemniscate mode which will make the wheels rotate in opposite directions. To maintain a parallel mode with just links you have to use a second link 90 degrees out of phase.</p> <p>In the following diagram there are two rotating shafts connected by a blue link and a red link. The two links are 90 degrees out of phase. The red link in its current position can not transmit any torque; this is called <a href="https://en.wikipedia.org/wiki/Dead_centre_(engineering)" rel="nofollow noreferrer">top dead center</a>. If only the red link was here; the wheels would be permitted to go in opposite directions. With the addition of the blue link however, the shafts must continue to rotate in the same direction. Correspondingly, when the shafts rotate 90 degrees, the red link transmits the load while the blue link passes top dead center.</p> <p><a href="https://i.stack.imgur.com/ZCHGJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZCHGJ.png" alt="enter image description here"></a></p> <p>Steam engines use links 90 degrees out of phase to couple the two wheels. It can not be seen from one side of the engine though. The blue link is on one side of the engine, and the red is on the other. You can see this in Dave Tweed's answer by looking at the counterweight positions on the engine at the right (there are two separate steam engines in the photo).</p>

6712

Rotating two shafts together - coupling rod?

2016-01-01T05:57:11.507

<p>I am studying the performance of volumetric solar receivers (specifically ceramic foams) versus the change of many parameters: solar flux, surface temperature, porosity, pore diameter, flow pressure and flow velocity. I have seen two ways in the literature to model porous media:</p> <ol> <li>Drawing a complex 3D structure that is analogous to the porous zone.</li> </ol> <p>While this might be a very-problem-specific solution, but turbulence modelling in that case is not a problem.</p> <ol start="2"> <li>Averaging the governing partial differential equations to account for the nature of porous media and introducing new terms to the equations (porosity $\epsilon$ and superficial velocity).</li> </ol> <p>And this is the approach I am taking which is used in porous models in many CFD solvers. </p> <p>However, while reading about the treatment of turbulence in porous media in the <a href="http://aerojet.engr.ucdavis.edu/fluenthelp/html/ug/node274.htm" rel="nofollow">FLUENT user guide</a> I found that turbulence is not exactly modelled:</p> <blockquote> <p>turbulence in the medium is treated as though the solid medium has no effect on the turbulence generation or dissipation rates. This assumption may be reasonable if the medium's permeability is quite large and the geometric scale of the medium does not interact with the scale of the turbulent eddies. In other instances, however, you may want to suppress the effect of turbulence in the medium.</p> </blockquote> <p>I find <em>"medium's permeability is quite large"</em> quite ambiguous (What values are considered <em>large</em>?) and don't know whether my model will be badly affected by this treatment or not and to what extent? </p> <p>So has anyone came through this before? or am I overcomplicating things for a parametric study?</p>

|cfd|turbulence|porous-medium|

<p>Short answer: The FLUENT approach is trivial.</p> <p>Many information are lost due to the space-averaging process - over a representative elementary volume - of the governing equations (which is the essence of every porous model that tries to avoid the complexity of the real geometry of the porous media), So any turbulence model is not to <em>"reproduce the fine structure dynamics of the flow but to take into account information embedded in smaller scale for large scale modelization."</em>[1]. The FLUENT approach simply ignores this fact or even assumes that there is no turbulence energy dissipation or generation due to the porous zone (similar approach was taken by Antohe and Lage to develop a new turbulence model that lead to trivial solutions ($k = 0$ or $\epsilon = 0$)).</p> <p>I found that STAR-CCM+ code follows a similar way in turbulence treatment:</p> <blockquote> <p>The effect of a porous region on turbulent flow depends on the internal structure of the porous medium. Where turbulence is present, the turbulence scales are determined from the geometric structure of the porous medium. As it is not possible for STAR-CCM+ to predict the turbulence scales directly, you specify the appropriate values on the porous region. Turbulence quantities in fluid leaving the porous region are constructed from the user-defined values; <strong>they are not transported from the upstream side of the porous region.</strong></p> </blockquote> <p>So, I think if you are pragmatic enough to have a model that can sacrifice the accuracy of results of the turbulence model you use, these assumptions are the best you can get.</p> <p>[1] $k–\epsilon$ Macro-scale modeling of turbulence based on a two scale analysis in porous media - Francois Pinson, Olivier Gregoire, and Olivier Simonin.</p> <p>[2] A new turbulence model for porous media flows. Part I: Constitutive equations and model closure - Federico E. Teruel and Rizwan-uddin.</p>

6723

How safe is this assumption for turbulence modelling in porous media?

2016-01-01T09:35:52.363

<p>I have a setup in which a bolt threaded on two ends with an unthreaded shank in the middle will be rotated by the unthreaded shank area. That is, the bolt's axis of rotation is roughly the red cross:</p> <p><a href="https://i.stack.imgur.com/Pu9wQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pu9wQ.png" alt="enter image description here"></a></p> <p>I wonder if there's a small part/connector that I could fit over the unthreaded shank so that a shaft could be connected to the connector such that the bolt can be rotated. Perhaps, something like below where the top is a hex/square hole and the bottom is a threaded hole?</p> <p><a href="https://i.stack.imgur.com/qgatq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qgatq.png" alt="enter image description here"></a></p> <p>Alternatively, are there low-profile setups where a bolt can be rotated around the centre?</p>

|mechanical-engineering|bolting|

<p>depends on the application but a TCB Stud may be a possibility - this has a left hand thread on the end opposite the spline and a right hand thread on the end nearest the spline. The left hand thread is tightened by the installation tool into a left-handed threaded nut and then the right hand threaded nut is tightened against the washer against the steel. </p>

6724

Rotating a bolt around the centre

2016-01-02T03:27:18.340

<p>Suppose there is an accelerating mass along a straight line, and you would want to convert that linear motion into rotary motion. Is it possible to convert that accelerating linear motion into constant speed rotary motion? Could you suggest mechanisms that I can use?</p>

|mechanical-engineering|applied-mechanics|

<p>Do you really mean<br> ACCELERATING linear to CONSTANT rotary motion? </p> <ol> <li><p>A variable ratio drive as in the DAF cone drive now used in a number of Japanese gearboxes would work.</p></li> <li><p>Provided the accelerating linear motion is repeatable and predictable, below one turn this can be seen as a "cam".<br> The linear motion is coupled to the cam surface where cam radius from centre decreases in inverse ratio to the linear velocity.<br> For more than one turn the cam becomes a conic surface with again a radius proportional to inverse of velocity as the linear drive accelerates. This becomes impractical for large variations of linear velocity.</p></li> <li><p>Add electronics in the middle and it becomes an alternator-motor set.</p></li> </ol>

6735

Converting accelerating linear motion into constant rotational motion

2016-01-03T01:19:18.717

<p>Recently I watched <a href="http://www.imdb.com/title/tt3659388/" rel="nofollow">The Martian</a> and the space vehicle <em>Hermes</em> in the movie had a rotating platform. I wonder if that would significantly cause the vehicle to be inclined in direction? Does a spinning platform/living space on a space vehicle, used to provide artificial gravity, create momentum that inclines the trajectory of that vehicle?</p>

|mechanical-engineering|aerospace-engineering|

<p><a href="https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum" rel="nofollow">Conservation of angular momentum</a> creates a <a href="https://en.wikipedia.org/wiki/Gyroscope" rel="nofollow">gyroscopic force</a> when the angle of the space craft is changed.</p> <p>The trajectory however is independent of orientation. You could change the velocity of your spacecraft (speed and direction) without changing the orientation (angle relative to the universe). This could be done with multiple rocket engines firing at different amplitudes to get the desired thrust vector or with a rocket engine that could be re-positioned about the crafts center of gravity.</p> <p>If changing the orientation of the craft is important for engine positioning, solar, or other; 2 living spaces could be used rotating in opposite directions. The gyroscopic force would still put a large load on bearings however, but slow changes in orientation would be fine.</p> <p>Another option would be to have large control gyroscopes that have rotational inertia at the same order of magnitude as the rotating living space. These could be rotated to counteract the living space angular momentum and control the angle of the craft.</p>

6744

Does a rotating platform on a space vehicle affect the vehicle's trajectory?

2016-01-04T01:28:09.690

<p>I've been designing an electric skateboard which is driven by only one of the rear wheels (for the moment). As I was picking out motors for the board, I soon realized I would need to find a balance between torque needed to climb hills, and RPM to get a good top speed.</p> <p>So I'm searching for the best KV (RPM/v) motor/gearing ratio combination.</p> <p>For all calculations I've assumed the following:</p> <ul> <li>Total mass (board + rider) = 85 kg</li> <li>Voltage = 12 Cells * 3.2V (LiFePo4) = 38.4 V</li> <li>Max Speed = 40 km/h = 11.11 m/s</li> <li>Hill Grade = 15%</li> <li>Coefficient of friction between rubber and asphalt: 0.65</li> <li>Diameter of wheel = 80 mm</li> </ul> <p>I tried assuming maximum kinetic energy as being equal to maximum gravitational potential energy at the top of the hill.</p> <p>$$E_{k,max} = \frac{1}{2}mv_{max}^2 = E_g = mgh_{max}$$</p> <p>Then solving for $h$.</p> <p>I'd take that and use $y/x = 0.2$, solve for $x$, then use some trigonometry to setup a Newton's Second Law problem and solve for force needed to overcome friction and gravity (Equilibrium problem).</p> <p>Then using $T = F \times r$, find torque needed at the wheels, assume some gearing ratio that keeps that and the $V_{max}$. Then solve for RPM and divide by total voltage for motor KV.</p> <p>My mechanics are rather rusty and I could use some help.</p>

|electrical-engineering|motors|applied-mechanics|energy|kinematics|

<p>The answer provided by Dave is correct but it covers only a very specific case of no-friction and constant speed. There are three forces acting on a vechicle which must be accelerated from 0 to v speed over a slope:</p> <p><a href="https://i.stack.imgur.com/mxvZb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mxvZb.png" alt="enter image description here"></a></p> <p>Cr is the rolling friction coefficient, around 0.01 for cars on asphalt, so probably something less for bikes. Don't know for skateboards, but they have forul large&amp;flat wheels, so maybe it's more similar to car coefficient than bike coefficient.</p> <p>Being it a skateboard over a slope, it has a very slow speed, so we can neglect air drag, but for a generic vehicle you must also take it into account:</p> <p><span class="math-container">$$ F_D = \frac 1 2 \rho C_D A v^2 $$</span></p> <p><span class="math-container">$$ \rho = air density = 1.22 kg/m3 $$</span> <span class="math-container">$$ C_D = \text{drag coefficient (around 0.8 for bike/human, 0.2-0.3 for cars)}$$</span> <span class="math-container">$$ A = \text{Section area or Frontal area, m2}$$</span> <span class="math-container">$$v = \text{speed, m/s}$$</span></p> <p>Inertia force resists to acceleration.</p> <p>I found a calculator: it's designed for robots, so it disregards both wheel and air drags:</p> <p><a href="https://www.robotshop.com/community/blog/show/drive-motor-sizing-tool" rel="nofollow noreferrer">https://www.robotshop.com/community/blog/show/drive-motor-sizing-tool</a></p> <p>This is a much more complex&amp;complete simulator, but I don't understand how to input custom data for motor:</p> <p><a href="https://vesc-project.com/calculators" rel="nofollow noreferrer">https://vesc-project.com/calculators</a></p>

6756

Calculating torque needed to climb a graded hill

2016-01-04T18:18:02.730

<p>We are testing a new ultrasound product using a Texas Instruments PGA450 automotive transducer.</p> <p>Three of us are sharing a lab and we need some kind of ultrasound-blocking partition walls around each product to prevent interference. Think cubicle walls that absorb ultrasound.</p>

|mechanical-engineering|electrical-engineering|automotive-engineering|

<p>Typical towels do a pretty good job of not echoing ultrasound around too much, and also attenuate ultrasound going thru them. Ordinary sheetrock walls are really bad because they are great mirrors for ultrasound.</p> <p>There are special materials for absorbing sounds at various frequencies, but hanging a bunch of large beach towels from the ceiling around each work area will likely be cheaper and just as effective. Two or three towels an inch or so apart should help a lot. You will still either have to put sound absorbent material on the ceiling, or make sure the towels extend to the ceiling.</p> <p>The remaining path will then be by bouncing off the floor. That's not so simple. Some "area rugs" should help, but some of them reflect ultrasound much better than others. Ideally you want something that has roughness extending over &frac12; wavelength or so, but that will be hard to find. Beach towels on the floor will work to attenuate ultrasound, but won't last long as they aren't designed to take that sort of abuse.</p>

6776

How can I block ultrasound from an automotive transducer at about 8'?

2016-01-04T20:39:11.923

<p>I have a DC motor with a shaft with a diameter of x. It would produce some torque T because torque is proportional to radius. If I stuck a pulley on the shaft of radius 2x, then I would produce twice as much torque, right? And then if I attach this to a gear with a ratio of 2:1, I would again double the torque.</p> <p>Here is an image to clarify:</p> <p><img src="https://i.stack.imgur.com/ho88X.jpg" alt=""></p> <p>This bugs me because it means the larger a pulley I can fit on a shaft the more torque I get. What am I missing? Are the RPMs reduced?</p>

|applied-mechanics|gears|torque|

<p>It is more useful to say that torque is proportional to the ratio of the radii of a pair of pulleys or gears. Also RPM is reduced or increased in the same ratio so a small diameter gear driving a larger one increases torque but reduces RPM at the output shaft. </p> <p>E.g., if you had a motor producing 100 Nm of torque at 100 RPM connected to a pulley of diameter 100 mm to an output shaft via a pulley with diameter 200 mm, the torque on the output shaft would be 200 Nm but the RPM would be 50 RPM. </p> <p>To put it another way, if you ignore friction, gearing systems allow you to have more torque and less RPM or less torque and more RPM but you can't gain any power (in fact you always lose a bit).</p>

6780

What is the relationship between drive shaft diameter and torque output?

2016-01-04T21:27:18.527

<p>I'm having trouble identifying the diameter of the reinforced steel bars in the structural steel design for a slab and a beam of a building complex. You can see it <a href="https://www.dropbox.com/s/xszs9wurf8gmk65/POS%20300%2C400%2C500%2C600%2C700%2C800%20%281%29.pdf?dl=0" rel="nofollow">here (PDF)</a>.</p> <p>I'm told there's column design in this file, but I only see beam design. Is the diameter number of the steel before or after the diameter sign? Why are there numbers like 12/20 after the diameter sign? Is there any column design here at all?</p>

|civil-engineering|structural-engineering|

<p>The column design schematics are "POS S1" and "POS S2", on the bottom center-right of the page.</p> <p>The nomenclature here has some peculiarities I'm unsure about, so someone else might come up with a better answer, but here goes... You'll see that there are basically two ways to describe the rebar:</p> <ul> <li>Longitudinal bars tend to be in the form "&#10112; $4\ \phi 16\ l=391\ (2)$": This translates to 4 rebars identified by &#10112; elsewhere which have a 16 mm diameter and are each 391 cm long. I'm not sure what the (2) means.</li> <li>Transversal bars are described similarly (when their dimensions are shown), but they are then often repeated in this drawing in a manner I don't quite understand: "&#10116; $18\ \phi 8/10/20$". I don't understand what the $/10/20$ is supposed to mean. That being said, when their positions along the column are shown as &#10116; $14\ \phi 8/20$, that just means that there are 14 cages made of 8 mm rebar distributed with a 20-cm spacing.</li> </ul>

6783

Identifying the diameter of reinforced steel bars in a column design schematic

2016-01-05T04:38:02.783

<p>The interior girder moment formula for one lane loaded for the AASHTO LRFD method is:</p> <p>$$\begin{align} mg^{SI}_{moment}&amp;=\left(1.75+\frac{S}{3.6}\right)\left(\frac{1}{L}\right)^{0.35}\left(\frac{1}{N_c}\right)^{0.45} \\ &amp;= \left(1.75+\frac{13}{3.6}\right)\left(\frac{1}{100}\right)^{0.35}\left(\frac{1}{3}\right)^{0.45} \\ &amp;= 0.65\ \mathrm{lane/web} \end{align} $$</p> <p>How is this formula derived? I have not been able to find the original research paper.</p>

|civil-engineering|bridges|aashto|

<p>The live load distribution formulas in the AASHTO LRFD Bridge Design Specifications cannot be derived. As I understand it, they are based on calibration to extensive finite element modeling.</p> <p><a href="http://www.trb.org/Main/Public/Blurbs/159050.aspx">This is probably the NCHRP report you're looking for.</a></p>

6788

Derivation of the AASHTO formula of interior girder moment

2016-01-05T09:11:59.517

<p>During the '80s and the '90s, a lot of media cases were made of thermoformed PVC. I believe this trend was created by Disney and Blackbourn with their VHS cases. Then with DVDs, industrial standards for media casing have changed in favor of injection moulded PP.</p> <p>For example, the latest Neo Geo AES cases are made of PP plastic but early Neo Geo AES cases were completely different. Internal trays were clearly made through thermoforming, as they have strong light reflection properties, but I'm not sure how external cases were made. They give the impression of being made of leather but are certainly some kind of plastic imitation. How were the external cases processed? Is it also PVC?</p> <p><a href="https://i.stack.imgur.com/JRj0A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JRj0A.jpg" alt="Alpha Mission complete set: case, manual and cartridge"></a></p>

|materials|process-engineering|plastic|

<p>For injection molding this is typically done by etching or burning(EDM) a pattern onto the surfaces of the injection mold.</p> <p>check out this PDF for more detailed information on this: <a href="http://eschmanntextures.com/fileadmin/user_upload/flyer_neu/texturesguide_eng.pdf" rel="nofollow">http://eschmanntextures.com/fileadmin/user_upload/flyer_neu/texturesguide_eng.pdf</a></p> <p>Search around for injection mold surface texture guide</p> <p>There are plenty of resources out there. </p> <p>re: Is it PVC? Probably not, but you can do a burn test or send it to a lab for material testing.</p>

6790

How can I process plastic to give the impression of a leather finish?

2016-01-07T05:50:11.737

<p>I work in a lab, and I've been trying to replace a lost AC-DC power adapter. I got one recently that has the exact specs that I need, which are apparently pretty rare (15V, 400 mA). However, when I test the power adapter with a voltmeter, it's actually outputting ~22V, despite what it says on the plug. </p> <p>Now, when I plug in the power adapter, the camera "turns on" because the green LED on it comes on. However, when I try to run the camera software, it behaves exactly as it does if I have the camera unplugged or turned off. So the software is not "seeing" it, even though it's plugged in, and plugged into the firewire port on the PC (yes, it uses firewire... ugh). </p> <p>So now I don't know why the camera is not actually working. My hypothesis -- is it possible that the camera is not really working because the power adapter is bad, and that the specs are too far off from what it wants? Even though the LED light turns on, maybe the voltage/current is close enough to make something work, but not close enough for it to actually function properly?? Is this stupid? </p> <p>I'm just not sure if this is a software or hardware problem... </p>

|electrical-engineering|consumer-electronics|

<p>What Russell said.</p> <p>In addition, the reason the LED lights but the device no longer appears to work may be because the LED is just wired to indicate power, and the rest of the device is now blown out due to the much higher than specified voltage you applied to it.</p> <p>Power LEDs are sometimes just a LED and resistor connected in series across the input power. These LEDs don't need to be particularly bright, so are often run at less than their maximum current rating. Many common indicator LEDs are rated for 20 mA, but are still plenty bright enough at 10 mA to tell you the power is on. Figuring the green LED drops 2.1 V and is just connected to the supply with a resistor, your 22 V as apposed to 15 V would result in 54% more current thru the LED. If it was only run at half its maximum current normally, this wouldn't even be pushing it. Even if the LED was set to run at its maximum rating normally, 54% overcurrent for a typical indicator LED won't blow it out. It will be brighter than normal, and it will have a shorter lifetime than rated, but would most likely work fine for days.</p> <p>The circuitry in the rest of the device apparently wasn't so forgiving, and now it's just a hunk of junk.</p> <p>Next time get a <i>regulated</i> supply of the right voltage.</p>

6810

Is it possible that an electronic device "half-works" when the power adapter supplies too much voltage?

2016-01-07T07:18:38.747

<p>I recently bought my first lathe (A circa 50s PREMO, Australian built with tapered roller bearings) and the spindle has a little bit of wiggle parallel to the axis of the spindle. I decided to pull the spindle out to take a look at 'er, but I ran into a spot of bother with the gear that mates with the reverse tumbler. (And just to be clear, I'm absolutely certain the gear needs to come off before the spindle will come out, it's bigger than the ID of the cone pulley)</p> <p>This may seem silly, but I can't work out how that thing is supposed to come off. The gear sits on top of a threaded portion of the spindle (And indeed before getting to the gear I had already removed two nuts from that end of the spindle), so that was my first thing to try. Grabbed it with a chain wrench, gave it a fairly good torque, nothing.</p> <p>Next thing I noticed was that there was a keyway cut in the gear, so I took a look in there. Sure enough, the threads extend under the gear, but curiously there doesn't seem to be a thread cut on the inside of the gear. I would have expected to see the tell-tale profile of the internal thread, but nothing; smooth. My next thought was that it was simple pressed on with a key, but I couldn't see a key, and in fact I'm able to push an allen key down the keyway the full thickness of the gear, there's nothing there. In addition to the lack of a key, there doesn't seem to be a matching keyway on the spindle for the gear to key into.</p> <p>What troubles me is that if there's a key and I try to unscrew it, I'll break something, and if it's threaded and I try to press it off, I'll break something. Where should I go from here?</p> <p><a href="https://i.stack.imgur.com/oLlwN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oLlwN.jpg" alt=""></a> <a href="https://i.stack.imgur.com/DkFdS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DkFdS.jpg" alt=""></a></p>

|gears|

<p>I decided that because of the lack of threads, it was more likely to be pressed on. So I whipped up a little shopmade gear puller, gave a little tug, and off it came. As it turns out, the gear and the spacer that I'd thought to be two pieces was actually one piece, and the key was under the spacer rather than the gear. Mystery solved.</p> <p><a href="https://i.stack.imgur.com/lisxR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lisxR.jpg" alt=""></a> <a href="https://i.stack.imgur.com/Pzrln.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pzrln.jpg" alt=""></a></p>

6811

How can I tell if a gear is pressed or threaded?

2016-01-07T10:15:23.173

<p>Suppose a 'D' shaped shaft with a 4 mm diameter like this type</p> <p><a href="https://i.stack.imgur.com/nOVs5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nOVs5.jpg" alt="enter image description here"></a></p> <p>is fitted inside a module 1 (plastic) gear with a round hole with the same diameter. If the gear has a set screw, will this still be a good fit? Will all the force be transferred to the set screw or will it be distributed onto the plastic surface as well?</p> <p>Should only round shafts fit inside gears with a round hole? Would connecting D shaft to a round hole gear be less preferable than connecting a D shaft to a D hole - even with a set screw? Yes, there's a set screw, but the flat part of the shaft will have a smaller contact surface, right?</p>

|mechanical-engineering|gears|

<h1>Quick Answer</h1> <ul> <li>You can fit D shaped shaft into round hole if diameters are OK.</li> <li>D shaft to D hole without a setscrew will be preferable to D shaft to round hole with setscrew if there is no axial movement. Best is D shaft to D hole with setscrew(s). However you can't always find D holes because they're more costly to manufacture.</li> <li>Shear forces act on setscrew if holding torque isn't exceeded. You can check holding torque below.</li> </ul> <h2>Setscrew holding power</h2> <p>Setscrews' holding power is proportional to its diameter you can check <strong>Marks’ Standard Handbook for Mechanical Engineers, Eleventh Edition, Table 8.2.18</strong>. gives holding power for Cup-Point Setscrews. Maximum load you can get from a setscrew from that table is 7000lb axial holding power for 1 in setscrew. The torque is</p> <p>$\frac{F_{a}}{2}D_{s}$</p> <p>where $F_{a}$ is holding power and $D_{s}$ is shaft diameter in inches. Sorry I don't have metric at hand. </p> <h2>D shaft</h2> <p>Advantage of D shaft is it prevents shaft from spinning inside the gear hub. So it is preffered in gears. </p> <h2>An example</h2> <p>If you choose No0 screw (1.524mm diameter), axial holding power is 50 lb (222N) so your holding torque is 444 Nmm. If you exceed this torque setscrew will fail but because of D shaft it will hold till material failure (on plastic shaft probably).</p>

6812

Using a 'D' shaped shaft with a set screw

2016-01-07T15:28:19.507

<p>I am trying to find a method of calculating the minimum shear area of the following rectangular thread form. It is a 4.125" square thread, 8 T.P.I., straight, right-hand thread. The thread is loaded in the axial direction, thus I would like to calculate the minimum thread shear strength.</p> <p><a href="https://i.stack.imgur.com/28GtV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/28GtV.jpg" alt="Thread"></a></p>

|mechanical-engineering|threads|

<p>The area equation of the thread is PI * D * nut length * .060/(.060*.065) PI = 3.14, D is 4.125-2*.050, nut length not given, probably 1" to 4", .060/(.060*.065) ratio of thread to machined out area</p>

6822

How to find the minimum shear area of a rectangular thread

2016-01-07T21:28:34.023

<p>I am trying to calculate the power distribution of a planetary gear system, specifically a Wolfrom system like this one:</p> <p><a href="https://i.stack.imgur.com/8kRDA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/8kRDA.png" alt=""></a></p> <p>In order to find the output power, I need to find the torque for the rings 5 and 6. Torque and angular velocity of the sun 1 is known. Also I have all the other angular velocities. I am trying to figure out a methodology to design the free body diagram of this system.</p> <p><a href="https://i.stack.imgur.com/NVzkC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/NVzkC.png" alt=""></a></p> <p>If the positive sign is clockwise, how can I add the torque T5 and T6 to this free body diagram?</p>

|mechanical-engineering|gears|

<p>In this answer, I'm going to use the sign convention that torques and angular velocities are positive if their direction points to the right (left would be an equally valid alternative for this answer too) using the Right Hand Rule, i.e. seen as clockwise if looking from the left hand side. Any gears spinning in the opposite direction will have a negative angular velocity.</p> <p>EQUILIBRIUM APPROACH</p> <p>One approach to determining the torques on each gear would be to draw the free body diagram of an individual gear with all the forces (meshing gears, shaft reaction) and external torques (shaft torque) exerted onto that gear. Then, by neglecting inertial forces, use equilibrium to write an expression relating the unknown teeth forces and external torques. An example of this is done for the gear marked with a 1:</p> <p><img src="https://i.stack.imgur.com/jAAgs.jpg" alt="enter image description here"></p> <p>Note the directions of the forces are completely arbitrary, while the input torque respects the sign convention.</p> <p>Applying equilibrium, and eliminating the shaft force (only interested in gear forces and external torques), we get:</p> <p>$$F_{II} = T_{in}/r_1$$</p> <p>where $r_1$ is the gear pitch radius (approximately) or the distance from the gear centre to the pressure line (accurate, but not needed if ultimately interested in external torques). Repeating this for all gears, it should then be possible to relate the external torques to each other, and determine the ratios of each relative to the input torque, the sign of which ratio indicating the direction of the torque.</p> <p>For some gear bodies, there will only be forces acting on it. The following example of a gear train should highlight this:</p> <p><img src="https://i.stack.imgur.com/xQAyl.jpg" alt="enter image description here"></p> <p>Free Body Diagrams: <img src="https://i.stack.imgur.com/EFIPi.jpg" alt="enter image description here"></p> <p>Note how the middle gear has no external torque acting on it.</p> <p>VIRTUAL POWER APPROACH</p> <p>The second approach can be simpler to use, especially if the angular velocities are already known. First of all, a single algebraic equation relating the angular velocities of gears with external torques should be obtained. For simple gear trains, this equation has only two angular velocities, and takes the form: $$\omega_{out} = G \omega_{in}$$ where $G$ is the gear ratio to be determined. For most planetary systems, there can be three angular velocities. An example I will use is the speed rule for a simple <a href="https://en.m.wikipedia.org/wiki/Epicyclic_gearing" rel="nofollow noreferrer">planetary gearbox</a>:</p> <p><img src="https://i.stack.imgur.com/m2A2y.jpg" alt="enter image description here"></p> <p>$$\omega_s = (1+R)\omega_c - R\omega_a$$</p> <p>where subscripts 's','c' and 'a' refer to the sun, carrier and annulus gears. $R$ is the ratio of teeth of the annulus gear to that of the sun gear.</p> <p>Now, we need a power balanced equation: power in = power out. This is where the sign convention comes in handy: since the conventional directions of torque and angular velocity are the same, multiplying the external torque acting on a gear with its angular velocity will give you the value of power in. Therefore, the power balance equation will be the sum of all the power-in products, equal to zero. i.e. In general:</p> <p>$$\sum_i T_i \omega_i = 0$$</p> <p>For my example, this is:</p> <p>$$T_c \omega_c + T_s \omega_s + T_a \omega_a=0$$</p> <p>For a simple gear train, it would now be very simple to substitute the angular velocity equation into the power balance equation, but not immediately so for angular velocity equations with more than two velocities. To solve such equations, you need to set one of the angular velocities to zero, as if holding that gear still, e.g.:</p> <p>Set $\omega_a=0$:</p> <p>Ang. Vel. Equation: $\omega_s = (1+R)\omega_c$</p> <p>Power Balance: $T_c \omega_c + T_s \omega_s=0$</p> <p>Substitute Ang. Vel. Equation into power balance: $T_c/T_s = -(1+R)$</p> <p>Repeat for setting another velocity to zero.</p> <p>This seems like cheating by forcibly setting angular velocities to zero, but it is still valid as the equilibrium solution (i.e. The external torque values) never depended on the angular velocities to begin with.</p> <p>Finally, whichever method was used, a quick check can be made to see if all the torques satisfy global equilibrium. Thanks to the sign convention, global equilibrium is achieved if:</p> <p>$$\sum_i T_i = 0$$</p>

6832

Determine the sign of torques for a planetary gear system

2016-01-09T14:55:44.470

<p>I was studying the design of joints and came to knuckle joints. Generally compressive strength is greater than tensile strength, but <strong>considering knuckle joints compressive strength and tensile strength are considered equal</strong> (as mentioned below). What are the reasons for assuming equal strengths?</p> <p><a href="https://i.stack.imgur.com/BJ2Ek.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BJ2Ek.jpg" alt="enter image description here"></a></p> <p><a href="https://books.google.co.in/books?id=YaYQAgAAQBAJ&amp;lpg=PA126&amp;ots=0xUWYjKLSK&amp;dq=WHY%20KNUCKLE%20joint%20crushing%20stress%20is%20equal%20to%20its%20tensile%20stress&amp;pg=PA126#v=onepage&amp;q=WHY%20KNUCKLE%20joint%20crushing%20stress%20is%20equal%20to%20its%20tensile%20stress&amp;f=true" rel="nofollow noreferrer">Here is a link to the page, shown above, in Google Books.</a></p>

|design|

<p>Assuming they're equal, it allows you to use the more commonly available tensile strength data, while also getting a conservative solution. If it turns out they're not equal, then your design will still work correctly but might not make optimal use of the materials.</p> <p>Assuming uniaxial tensile and compressive strengths are equal is a common practice in engineering and is built into the popular von Mises and Tresca yield criteria which are used for all kinds of structures. If you want to take advantage of higher compressive strength, you can use less common failure criteria like Mohr-Coulomb.</p>

6849

Knuckle Joint: compressive strength equal to tensile strength

2016-01-10T00:57:39.170

<p>I see that POM gears are often not preferred in high torque settings - but how about 0.4Nm? Is that amount of torque too much for module 1 POM bevel gears? Say they are 16-tooth.</p> <p>Is 0.4Nm considered moderately high torque?</p>

|mechanical-engineering|gears|

<p>module 1 16T bevel POM (also called acetal or Delrin) gears are rated at 0.2Nm at <a href="http://www.huco.com/products.asp?p=true&amp;cat=285" rel="nofollow">http://www.huco.com/products.asp?p=true&amp;cat=285</a></p> <p>The webpage also list larger module gears torque ratings.</p> <p>0.4 Nm is considered low torque in industrial machinery, but could be considered high torque in other fields.</p>

6855

Nylon gears for 0.4Nm torque

2016-01-11T00:56:35.047

<p>I came across this question of &quot;can an isothermal process also be adiabatic?&quot;, and at first thought that yes, why not? If the temperature is constant and there is no heat loss to the environment, it seems perfectly possible.</p> <p>Now, I don't have too much experience in this field, but I came across a lot of vague statements that this is not possible at all.</p> <p>Could someone with a bit more expertise give me a clear answer?</p>

|thermodynamics|heat-transfer|

<p><strong>No</strong></p> <p>An <a href="https://en.wikipedia.org/wiki/Isothermal_process" rel="nofollow noreferrer">isothermal process</a> is, almost by definition, a process where the fluid beeing worked upon can keep its temperature constant by exchanging energy with an external reservor, while an adiabatic process is defined by that <em>not</em> happening.</p> <p>All these processes are special cases of <a href="https://en.wikipedia.org/wiki/Polytropic_process" rel="nofollow noreferrer">polytropic processes</a>. An isothermal process as the poytropic coefficient <span class="math-container">$n=1$</span>, an isentropic process - the reversible case of an adiabatic process - has <span class="math-container">$n=\kappa$</span> with <span class="math-container">$\kappa=\frac{C_p}{C_v}$</span>. Looking at the <a href="https://en.wikipedia.org/wiki/Relations_between_heat_capacities" rel="nofollow noreferrer">relation between these heat capacties</a>, this is impossible since they can't be equal.</p> <p><strong>Phase changes?</strong></p> <p><a href="https://engineering.stackexchange.com/a/6863/61">Carlton</a> writes:</p> <blockquote> <p>Another source of storing/releasing energy I can think of is a phase change, i.e. steam-water-ice. Two of those phases can exist simultaneously at a constant temperature but across a range of pressures. Thus there is some capacity to store/release energy without changing temperature.</p> </blockquote> <p>This appears to be wrong: Take a steam/water mix at equlibrium in a pressure cylinder. The volume is reduced by external force, the pressure and temperature of the gas phase rise (adiabatic process). Due to the new pressure, a new equilibrium between the vapor and liquid phase will establish by condensation. The enthalpy of evaporation is then released as heat. For the phase change to store energy without rising temperature, the condensation would have to absorb heat, not release it.</p> <p><strong>Chemical processes?</strong></p> <p>Carlton also suggests:</p> <blockquote> <p>Consider a chemically-reactive gas at equilibrium in an insulated piston-cylinder setup. As the piston is raised by the gas pressure, the temperature and pressure will drop and thus the chemical equilibrium is disturbed. The gas reacts (chemically), releasing energy in the process until a new equilibrium is established at the original temperature. Thus, the whole process would be both adiabatic and isothermal. I don't know of such a reaction, but it is certainly possible.</p> </blockquote> <p>I'm not convinced such a reaction exists and I'd have to (re)learn a lot of chemical physics to begin to answer the question if it is theoretically possible. In effect we are looking for a reaction where a change into the gas phase is exothermal somehow, enough so to offset enthalpy of evaporation. Alternativly we could look for a reaction like this:</p> <p><span class="math-container">$$A_g+B_g {\rightleftharpoons}AB_g$$</span></p> <p>with the synthesis reaction endothermal.</p> <p>Now, I can't prove that this is impossible but I have a strong hunch it is: There would be a tremendous application in energy storage. A huge problem with <a href="https://en.wikipedia.org/wiki/Compressed-air_energy_storage" rel="nofollow noreferrer">compressed air energy storage (CAES)</a> is the heat generated in adiabatic compression, that is also required for adiabatic expansion.</p> <p>Traditionally, the expansion heat is supplied by burning natural gas with the expanded air (so only part of the energy supplied comes from the compressed air), a new proposal was the (canceld) ADELE plant that would have used huge packed bed thermal storage units.</p> <p>If someone had found a chemical reaction that allows <em>de-facto</em> adiabatic-isothermal processes, the killer application for this already here and we would have likely seen it in action.</p>

6862

Can an isothermal process also be adiabatic?

2016-01-11T13:53:43.720

<p>When preparing load calculations for a slab, the quick method is to define the tributary areas of a slab for the beams/columns below. These are generally taken as lines at mid-span of the slab panels and, depending on the support conditions, at the edges at angles of 30 &amp; 60 or 45 &amp; 45 degrees.</p> <p>Where do the 60/30 and 45/45 angle ratios come from? Example distributions are shown in the image below.</p> <p><a href="https://i.stack.imgur.com/lK6UC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lK6UC.png" alt="enter image description here"></a></p> <p>I know that in reality the ratios depend on the stiffness and end conditions of the various structural members, I also take it that the tributary areas are somewhat affected by the load they are carrying. Would excessive loading on one panel, compared to the adjacent panel affect the areas?</p>

|structural-engineering|

<p>The angles of those lines are determined by yield line theory. They are the axes of bending, should the slab decide to fail, which will require the least internal work to yield.</p> <p><a href="https://theconstructor.org/structural-engg/yield-line-theory/6839/" rel="nofollow noreferrer">Here</a> is a very simple, watered down explanation, but if you Google it, you'll find some white papers (or whatever colour you decide to print on :P) with more technical explanations and all the math.</p>

6866

Where do the tributary angles/ratios for tributary areas come from?

2016-01-12T01:04:44.097

<p>I'd like to design some things with bolts, fan belts, levers, motors, etc. So I'd add a bolt here, drill a hole there and maybe take it apart and rearrange. I'd like to do it in 3D and found SketchUp. </p> <p>Can it be used to design, create and print out dimensioned components for manufacture and assembly? I'm not knocking it, but is Sketchup anything other than a really good conceptual visualization tool? </p>

|computer-aided-design|

<p>I've used Sketchup for the last 6 years as my go-to 3D engineering program. When combined in the Pro version with Layout, it can be fully dimensioned, rendered, and turned into full scale 2D production drawings that the shop can build from. In the trial version, without the Layout program, it is only conceptual.</p> <p>That being said - Sketchup is intended for architects. It's built with a bunch of tools for them in mind - including an absolute up and down. As such, I'd only use Sketchup for items on the scale of houses. In my line of work, we build giant ducting systems and process vessels - Sketchup is perfect. </p>

6872

Can Sketchup produce drawings that are adequate to manufacture components?

2016-01-13T01:35:41.500

<p>For a 3D printing application, I need to have a spinning disc(that I will rotate via geared stepper motors) that will wobble as little as possible. By wobble, I mean that it will only rotate around it's axis and not shift side to side. </p> <p>The goal is to build a 3d printer that uses polar coordinates rather than traditional Cartesian coordinates. Instead of using leadscrews, I want to instead rotate two discs. Disc 1 is mounted to disc 2 which is mounted to an immobile surface. If I can precisely rotate both, I should be able to 3d print. So the question is how can I mount them such that they don't require much torque to spin and also spin accurately around their axis?</p> <p>I have ordered some lazy Susan type bearings from McMasterCar, but they wobble too much under use. I've heard that there are axial and thrust bearings and that the ones I ordered are thrust. Should I use a combination of the two (and how would you suggest mounting it?) or should I use the diagonal type that has both combined in one assembly? </p> <p>The disc is 8 inches in diameter and I would like to keep the assembly thin. The disc is spinning parallel to the ground; if I were to mount the bearing underneath, it would be in compression.</p> <p>Here is a sketch showing one of the axis:</p> <p><a href="https://i.stack.imgur.com/vm9RQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vm9RQm.jpg" alt="close up"></a></p> <p>This sketch shows an overview of the whole system:<br> <a href="https://i.stack.imgur.com/HIQ87.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HIQ87m.jpg" alt="system overview"></a></p>

|mechanical-engineering|3d-printing|bearings|

<p>The bearings required will depend on the loads and rotational speeds involved. </p> <p>In the situation you describe you will probably have a combination of axial and thrust loads ie the thrust load is the weight of the disk and axial loads come from torque on the drive shaft from the drive torque and unevenly distributed loads on the disk. </p> <p>You also need to consider how accurately your assembly will be aligned, different types of bearings have different tolerances for misalignment. For a driven shaft you generally need a pair of bearings per shaft so that it is supported at two points to minimise bending loads on the bearings. </p> <p>For this sort of project often the most convenient solution is to use 'pillow block' type bearings which are an assembly which includes the bearing and a housing with mounting holes as a complete assembly and you can just mount the whole thing to a suitable point in your structure without having to design and build a housing for the bearings from scratch. Again you need to make some sort of estimate of the radial and axial loads involved and choose a bearing to suit from the manufacturers specifications. </p> <p><strong>Additional thought</strong> : the situation you describe is broadly similar to the hub on a rear wheel drive car so it might help to look at the way that hubs are designed. It may even be that an off-the-shelf hub unit from a lightweight vehicle/robot (or perhaps an industrial conveyor) would be the easiest solution rather than designing and making something from scratch </p>

6888

How to secure a rotating disc so that it has no wobble

2016-01-14T14:05:17.647

<p>We are designing a diagnostic electronic tool for a vibrating table.</p> <p>In our PCB there are a gyroscope, an accelerometer and a bluetooth module. Our PCB is attached to a vibrating table that works at 12 Hz with 10 mm peak-to-peak amplitude. It has to work for at least three years "always on".</p> <p>When we are choosing the power connector, what are the technical elements we have to check before buying one? Are there special connectors for that specific use? "Vibrating-environment connectors"?</p> <p>We are going to pot the pcb.</p> <p>Our fear is that with that amplitude and with that frequency a normal soldered connector would fail and break itself.</p>

|mechanical-engineering|power-electronics|

<p>Anderson PowerPole connectors are used in high vibration critical systems to make reliable connections. At the board use a ring terminal and screw with a Nylock locknut. </p> <p>I always solder the cables (no crimp) and use Silicone high strand count wires, the type used by radio control model battery power systems. </p> <p>Strain relief may be needed if there is a large differential movement between parts - otherwise leave the cables over-length and allow them to flex naturally.</p> <p>Arbutus</p>

6909

How to choose an electronic connector/cable that works in a vibrating environment?

2016-01-14T15:32:34.497

<p>I am having hard time to fully undestand what those formula's actually mean. I know the meaning of each element:</p> <ul> <li>I : moment of inertia</li> <li>H : angular momentum </li> <li>$\alpha$: angular acceleration</li> <li>$\omega$: angular velocity</li> </ul> <p>$I_{G}\alpha = \dot{\vec{H}}$</p> <p>$I_{G} \dot{\vec{\omega}}= \dot{\vec{H}}$</p> <p>Could someone explain the meaning and/or give and example/context to clarify things?</p>

|mechanical-engineering|dynamics|kinematics|

<p>============<strong>Short answer</strong>===================</p> <p>If you are familiar and understand Newton's second law:</p> <p><span class="math-container">$$F = m\cdot a = m\cdot \dot{v}$$</span></p> <p>Then you should know that one way to interpret it, is that:</p> <blockquote> <p>the acceleration <span class="math-container">$a$</span> of a body with mass <span class="math-container">$m$</span> is proportional to the sum of forces acting on the body.</p> </blockquote> <p>Similarly, keeping in mind that it can be proven that <span class="math-container">$\dot{H} = M$</span>, the equation you describe can be written is a similar fashion:</p> <p><span class="math-container">$$ M = I_G \cdot \alpha= I_G \cdot \dot{\omega}$$</span></p> <p>Where:</p> <ul> <li><span class="math-container">$M$</span> is the resultant moments acting on the body</li> <li><span class="math-container">$I_G$</span> the moment of inertia</li> <li><span class="math-container">$H$</span> the angular momentum</li> <li><span class="math-container">$\omega$</span> the angular velocity [rad/s]</li> <li><span class="math-container">$\alpha$</span> the angular acceleration <span class="math-container">$[rad/s^2]$</span></li> </ul> <p>Essentially, what this equation says is that :</p> <blockquote> <p>the angular acceleration <span class="math-container">$\alpha$</span> of a body with moment of inertia <span class="math-container">$I_G$</span> is proportional to the sum of moments acting on the body.</p> </blockquote> <p>Observe that there is a direct correspondence between:</p> <ul> <li>forces F and moments M</li> <li>mass <span class="math-container">$m$</span> and mass moment of inertia <span class="math-container">$I_G$</span></li> <li>acceleration <span class="math-container">$a$</span> and angular acceleration <span class="math-container">$\alpha$</span></li> </ul> <p><strong>===================Long Answer==========================</strong></p> <p>I won't bother with the long answer, because it would be impossible to give the generic case (see rotating systems of reference) within the space of this answer.</p> <p>You might want to read Dynamics textbooks (Meriam and Kraige , Beer etc)</p>

6911

How to interpete the formula: $I_{G}\alpha = \dot{\vec{H}}$?

2016-01-15T02:13:14.547

<p>I have a setup where I want to move a knife like blade through a body of liquid inside a container and I want to control the blade from outside the container. Apart from using magnets to move the blade, is there any way to form a dynamic seal around the blade as it moves along the container length? </p> <p>I was thinking of using two long lengths of rubber that sandwiched the 1mm blade, but I am not sure how this would hold once some pressure was added to the container, especially if the blade had an oddly shaped cross section (like sharp curves that would be hard to seal).</p> <p>There is a fused silica layer at the top of the container, which is in contact with the liquid.</p> <p><a href="https://i.stack.imgur.com/tPT0U.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tPT0U.jpg" alt="Vat with scraper blade"></a></p> <p>So I am wondering if there are any seals anyone knows of that could be used in this scenario? I'm thinking of possibly using ferrofluid as a seal at this stage.</p>

|mechanical-engineering|fluid-mechanics|magnets|

<p>One option is not to try to design a leak-proof seal, but instead just limit the leaks to acceptable levels and make maintenance possible.</p> <p>My suggestion:</p> <ol> <li>Use rubber seals like you planned to.</li> <li>Add a trough that will catch any leaking resin. Preferrably made of a material such as teflon or silicone that the resin will not stick to.</li> <li>Arrange so that the resin in the trough will eventually harden, making it easy to clean off.</li> </ol>

6914

How to form a seal around a sharp moving blade?

2016-01-15T04:57:22.833

<p>I have an application where I need two motors to drive one shaft via the use of belt drives but I need help on how to arrange such a system. The system constraints are:</p> <ol> <li>motor type - induction motors $(48\ \textrm{V}, 750\ \textrm{W})$</li> <li>both motors will have pulleys of same dimensions</li> <li>both motors will drive single o/p shaft with bigger pulley attached to it<a href="https://i.stack.imgur.com/U0uyA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U0uyA.png" alt="enter image description here"></a></li> </ol> <p>Please see the image of my current work, and please let me know if with this arrangement I can have more output speed and torque. If you have any other ideas, please let me know.</p>

|mechanical-engineering|

<p>Instead of these arrangements have 2 motors with equal qty of v belts and double the qty of v grooves on driven pulley with both motors having takeup device will work perfectly.</p>

6916

Using two motors and belt drives to drive one shaft

2016-01-15T18:24:20.900

<p>According to this <a href="http://home.mit.bme.hu/~simon/publications/Sensys_2004_FloodingTSP.pdf" rel="nofollow">paper</a>, which deals with time synchronization in wireless networks, this kind of delay occurs, however I do not understand its explanation. The paper says the reciever (radio) has to determine the bit offset of the message from a known synchronization byte. </p> <p>The definition from the paper is:</p> <blockquote> <p><em>Byte Alignment Time</em> —the delay incurred because of the different byte alignment of the sender and receiver. This time is deterministic and can be computed on the receiver side from the bit offset and the speed of the radio.</p> </blockquote> <p>If somebody familiar with this phenomenon, can you please explain it to me?</p>

|electrical-engineering|terminology|wireless-communication|rf-electronics|digital-communication|

<p>This happens on a bit level of the receiver electronics. </p> <p>For every byte you have to receive 8 bits. At this point in communication you only decoded the signal (eg. converted the electrical power or voltage) levels to bits. You receive bits in the buffer. </p> <p>You are constantly filling the buffer (on a bit level), so you can check the buffer to see if you received anything. First you receive a lot of 000000, but then at one point message arrives, and bits in the buffer start to have different values. </p> <p>Now if you are lucky first bit of the message arrives on let's say 8th or 16th place in the buffer (or 0th). This are all the bits that also start a logical representation of a byte. </p> <p>If you are not that lucky, you just pushed in a 0 on 8th place of a buffer, and you would start filling the message in on the 9th place, which would in mean first seven bits from the original first byte would be received as first byte. And a second byte would be received as a last bit of the original first byte and first 7 bits of the original second byte and so on. This would mess things up good. </p> <p>You have to be able to slide in some zeros in to the buffer in order to check up with the start filling the message in on the "start of byte position" of the buffer. But since you know how many zeros you had to fill in, you know how much delay this added for certain message (if you know the frequency of transmission). </p> <p>EDIT: Just to make things more exact: you push the preamble in the buffer at random bytes, this is how you know you are getting something. sync is then used to sync the trasmission (eg. fill in the zerros so you start on right bit position).</p>

6925

What is Byte Alignment Time in radio communication?

2016-01-16T13:46:38.757

<p>I have interested in walking bicycle. For the concept is just bicycle that you don't cycling pedal but stepping or striding to drive wheel</p> <p>Which leads me to think that is it better if we use linear or ellipse motion of leg to transmit to bicycle</p> <p>The closest mechanic I could think is piston crankshaft like a car. It convert linear motion to rotation</p> <p>How disadvantage of the mechanic compare to cycling motion?</p>

|power|pistons|bicycles|

<p>The biggest problem with a "two cylinder" crankshaft is that there is a dead spot in the torque curve whenever either "piston" is at TDC or BDC.</p> <p>When you have your feet directly on the pedals of a conventional circular crank, you actually have the ability to apply force over greater than a 180&deg; arc &mdash; by a combination of flexing your ankles and pushing forward/backward with your thigh muscles &mdash; which eliminates the dead spot.</p> <p>If you try to use a mechanical connecting rod with a crankshaft, you need some other way to eliminate the dead spot. This typically involves using more than two pistons, or by incorporating a large flywheel into the mechanism. None of this is practical or desirable on a bicycle.</p> <p>I have seen toy vehicles for small children that use such a mechanism. but invariably, they need to put their feet down on the ground from time to time to push themselves out of the dead spot.</p>

6933

Why there is no piston style pedal bicycle?

2016-01-16T17:39:12.263

<p>How to find out experimentally to measure the vibrational frequency of some type of spring without electronic or electrical instrument?</p> <p>I am just wondering about the advance stage of the discovery of the theory of vibrations it became in the 19th century</p>

|mechanical-engineering|experimental-physics|

<p>If the frequency is high enough, its sound can be compared to a tuning fork, ref <a href="https://en.wikipedia.org/wiki/Tuning_fork" rel="nofollow">https://en.wikipedia.org/wiki/Tuning_fork</a></p>

6935

How to find out experimentally without electronic or electrical instrument to measure the vibrational frequency of some type of spring

2016-01-16T18:59:40.263

<p>How could I design a paper spool or drum that is moving at constant speed without electrical system? What mechanisms are available?</p> <p>I know that some paper tracing system was being used in 19th century to track vibration frequency.</p>

|mechanical-engineering|applied-mechanics|

<p>If you are trying to simulate 19th century technology, based on your other question, you could achieve this to a pretty good precision with a calibrated metronome and a skilled operator turning a crank. That could easily provide enough precision for your vibration measurement experiments.</p> <p>Things don't need to be any more complicated than they have to.</p>

6937

What mechanisms can be used to move a drum at constant speed without an electrical system?

2016-01-17T15:52:00.113

<p>What options do I have for creating a resistance to a mechanical movement apart from simple friction?</p> <p>Assume I have a lever, which will result in either circular or linear motion. I want to be able to apply a resistance to this motion, so that the user will have to apply extra effort to move the lever in either direction. I could use a spring, which would resist motion in one direction, but this would then result in a restoring force when the initial displacement force was removed. I want the extra force to act in both directions, so that the lever is "heavy" to move, but will stay in its new position when displaced.</p> <p>I want the force to be able to be calibrated and remain constant, so I don't want to have something like a brake pad, which will wear, or just a nut to do the joint up tight, as this will be too difficult to achieve a calibrated force.</p> <p>Is there something that could be done with a spring or pneumatic air muscle or by using hydraulics?</p>

|mechanical-engineering|hydraulics|

<p>Your question immediately brings dampers into mind, but I think you might still be happier with friction. It depends on whether you want the resistance to increase with speed of movement (damper) or to remain constant (friction).</p> <p>To keep the friction steady over the lifespan of the device, you can use a spring to push a brake pad against the rotating surface. If they are both made of e.g. steel, wear should be quite slow. As the surface wears, the spring will keep the force constant even if the dimensions change a millimeter or two.</p>

6951

How do I add a resistance to mechanical movement in both directions, apart from friction?

2016-01-17T17:35:35.417

<p>I have an issue with simulink, basically it's to do with a second order system, well two first order systems in series. As I understand it as you increase the damping factor (above 1), the system should respond slower and be more sluggish. The damping factor = (tau1+tau2)/(2root(tau1*tau2)). So, looking at this system with damping factor 1(with it's response): <a href="https://i.stack.imgur.com/1LL8e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1LL8e.png" alt="enter image description here"></a></p> <p>and this system with damping factor 1.19 and it's response: <a href="https://i.stack.imgur.com/u9eZZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u9eZZ.png" alt="enter image description here"></a></p> <p>What gives?! The system with the damping factor 1 hasn't even reached steady state by 250 secs while the system with damping factor 1.19 has had a faster response, why is this happening?</p> <p>Thanks for reading.</p>

|electrical-engineering|control-engineering|chemical-engineering|process-engineering|simulink|

<p><strong>Time Constant and systems</strong></p> <p>A second order LTI system in Laplace domain:</p> <p><span class="math-container">$\hspace{2.5em}$</span> <span class="math-container">$H(s) = \frac{{\omega_{n}}^{2}}{s^{2}+\zeta\omega s+{\omega_{n}}^{2}}$</span></p> <p>The solution is:</p> <p><span class="math-container">$\hspace{2.5em}$</span> <span class="math-container">$h(t) = \frac{{\omega_{n}}}{\sqrt{1-\zeta^{2}}}e^{-\zeta {\omega_{n}} t}sin({\omega_{n}} \sqrt{1-\zeta^{2}}t)$</span></p> <p>Note that the time constant depends on the product of the damping and the frequency!</p> <p>The denominator is called chacateristic equation:</p> <p><span class="math-container">$\hspace{2.5em}$</span> <span class="math-container">$s^{2}+\zeta{\omega_{n}} s+{{\omega_{n}}}^{2}$</span></p> <p><span class="math-container">$\hspace{2.5em}$</span> <span class="math-container">$r_{1,2} = \frac{-\zeta{\omega_{n}}\pm \omega_{n}\sqrt{1-\zeta^{2}}}{2}$</span></p> <p>We have three forms for the solution:</p> <p><a href="https://i.stack.imgur.com/MnHt3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MnHt3.jpg" alt="enter image description here" /></a></p> <blockquote> <p>Overdamped: <span class="math-container">$r_{1} \neq r_{2}$</span> <span class="math-container">$\in$</span> <span class="math-container">$\Re$</span></p> <p>Critically damped: <span class="math-container">$r_{1} = r_{2}$</span> <span class="math-container">$\in$</span> <span class="math-container">$\Re$</span></p> <p>Underdamped: <span class="math-container">$ r_{1} = {r_{2}}^{*}$</span>. Where <span class="math-container">${r_{2}}^{*}$</span> is the complex conjugate of <span class="math-container">$r_{1}$</span></p> </blockquote> <p>In the s plane, should look like this:</p> <p><a href="https://i.stack.imgur.com/AzznH.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AzznH.gif" alt="enter image description here" /></a></p> <p>The figure above show us a complex response (the conjugate is implicated). Note that <span class="math-container">$-\zeta\omega_{n}$</span> its in the real part of the solution! So, is responsible for the time response and, the imaginary part, <span class="math-container">$\omega_{n}\sqrt{1-\zeta^{2}}$</span> is responsible for the oscilation.</p> <p>Note that the real part is the exponential term in the solution!</p>

6953

Problem with process controls/simulink?

2016-01-18T10:30:20.017

<p>First of all, I would like to know about the service life of a <a href="https://en.wikipedia.org/wiki/Flashback_arrestor" rel="nofollow">flashback arrestor</a>. In my laboratory, I got three flashback arrestors {WITTgas RF53N(1/4”NPT-F), Brass} from the ex-students who used those items in 2012 and already finished their graduation. I joined the laboratory in 2015. In this circumstance, I want to know if there is any possible way to check those flashback arrestor. All the flashback arrestors they used are for the premixed combustion in a Constant Volume Combustion Chamber. My purpose is also same. But I want to know first whether those flashback arrestors are dead or alive. </p>

|piping|safety|gas|valves|

<p>The <a href="http://2.wittgas.com/filepool/Datenblaetter/usa/armaturen/flashback_arrestor_rf53n_dn_nsk_nu_u_usa.pdf" rel="nofollow">manufacturer says this</a> about maintanence:</p> <blockquote> <ul> <li><p>annual testing of the non-return valve, body leak tightness and flow capacity is recommended </p></li> <li><p>WITT is happy to supply special test equipment</p></li> <li><p>Flashback Arrestors are only to be serviced by the manufacturer; the dirt filter may be replaced by competent staff</p></li> </ul> </blockquote> <p>Given that you inherit this equipment and don't know how it was treated I would do the recommended testing, or have it done. I would likewise replace the dirt filter.<br> Contact Witt (or their sales rep/service partner) about these steps to get an idea if you can do this yourself, have it done by a technician in house or need external help. </p>

6964

Service Life of a Flashback Arrestor

2016-01-18T14:28:21.247

<p>Lets say you have a measurement device of which you do not know the accuracy, and a reference measurement device. Both measure a variable $x$. The range of interest is $x_0&lt;x &lt; x_1$. How do you determine the accuracy of the unknown device in this range?</p> <p>My course of action would be to gather values for both devices from $x_0$ to $x_1$ and building a distribution of <a href="https://en.wikipedia.org/wiki/Errors_and_residuals">errors</a>. The <a href="https://en.wikipedia.org/wiki/Accuracy_and_precision#ISO_Definition_.28ISO_5725.29">accuracy</a> could then be the error span, $\pm3\sigma$ or something similar - is this correct?</p> <p>Assumptions:</p> <ul> <li>The reference measurement device is calibrated and has virtually no error</li> </ul>

|measurements|statistics|metrology|

<p>I was on a team of quality engineers (but not one of the experts), and they had a visual where they used a 2d plot where X axis was first measurement and Y was second measurement of the same observable feature.</p> <p>They would repeat the measure/remeasure and create what they called a "sausage chart". They would eliminate the outlying 2% of samples and draw a "sausage" around the rest.</p> <p>You could visually see the quality of the measurement system by observing how close the data points fell to the 45deg angle line.</p>

6969

How do you determine the accuracy of a measurement device?

2016-01-19T10:11:22.373

<p>What kind of furnace is used to create alloys for aluminum die casting?</p> <p>I have only been able to find information about how to create alloys or how to process them afterwards in the die casting process.</p> <p>Any insight on which kind of furnace you use to create the initial alloy would be highly appreciated. </p>

|aluminum|alloys|

<blockquote> <p>Aluminum is a metal that can be used in die casting to make parts easily. Many die casting techniques like permanent mold casting and sand casting are used for casting aluminum. Permanent mold casting is highly repeatable and helps to decrease the spin balancing processes and secondary machining operations. On the other hand, sand casting uses temporary molds that are made from wood or metal, making the tooling investment very low. However, per part prices are higher in the case of sand casting than permanent mold castings.</p> <p>In the case of permanent mold casting techniques, the cycle times are short and this will decrease the per part price. However, tooling costs can be very high in this method. When compared to sand castings, permanent mold castings cool faster, offering a finer and uniform microstructure to the parts. This will help to boost the mechanical properties of the product by about twenty percent.</p> <p>Gravity fed permanent metal molds is used these days to manufacture near net shaped parts from different alloys of aluminum. However, it is the duty of the die cast designer to ensure that it is profitable and possible to use permanent molds to manufacture the part. The die cast company should know the limits of the casting method to take full advantage of this die casting technique.</p> </blockquote> <p>Source : <a href="http://pacdiecast.com/aluminum-die-casting/permanent-mold-casting-for-casting-aluminum-alloys/" rel="nofollow noreferrer">http://pacdiecast.com/aluminum-die-casting/permanent-mold-casting-for-casting-aluminum-alloys/</a></p>

6981

Furnace for aluminum alloy creation

2016-01-19T13:54:09.730

<p>I'm working on an application in which sparks (specifically from static discharge) present a severe safety hazard. One component in question is a piece of PVC water pipe. It has been proposed to "ground" the pipe to minimize the risk of static shocks between humans and the pipe, and between other components and the pipe. I am skeptical as to whether this component needs to be grounded at all; it's not conductive, so my intuition is that it isn't able to cause a shock. Normally I would just go along with this, however the proposed methods of grounding involve expensive adhesives and coatings which also have a long lead time for ordering.</p> <p><strong>Can a PVC pipe cause a static discharge? If so, what industry practices are out there to mitigate this phenomena?</strong></p> <p>The pipe is being used in a test fixture for testing energetic materials, i.e. combustible solids. The energetic material is loaded into a length of the pipe, along with some sensors and other components. The material is ignited, after which the pipe is destroyed. The static spark concern is only during the loading process, when the energetic material could be ignited prematurely. The pipe is not used in its normal manner, like to transfer particles or fluids.</p>

|electrical-engineering|safety|

<p>As pointed out, non-conductive plastics can build up a charge. In certain applications industry code call for all piping in ex-hazard areas (or where the inside is an ex-hazard area!) to be conductive material, and grounded. To my knowledge,there's no conducting PVC. Hoever both PE and PP can be bought in a coducting variety (they mix carbon into the plastic matrix.</p> <p>The beaty of PVC ist that the pipes can be glued, with PP or PE you need to weld the pipes or use plug sleeve. I don't know the prices of conductive PP, PE-el is 3-4x as pricey as ordinary PE (and I think with PVC the difference is even bigger).</p> <p>I would seriously consider using (stainless) steel pipes, maybe with press fittings. </p>

6987

Will PVC build up an electric charge and cause a spark?

2016-01-19T17:44:37.480

<p>I am trying to solve the following problem. If someone can just point me in the right direction it would be appreciated. I don't have the textbook and online resources are proving to be absolutely worthless.</p> <blockquote> <p>A piston-cylinder device containing a fluid is fitted with a paddle wheel stirring device operated by the fall of an external weight of mass 47 kg . As the mass drops by a height of 6.1 m , the paddle wheel makes 10800 revolutions. Meanwhile the free moving piston (frictionless and weightless) of 0.59 m diameter moves out by a distance of 0.70 m</p> <p>What is the net work for the system if the outside pressure is 101 kPa?</p> </blockquote>

|thermodynamics|

<p>The work that the system does on the atmosphere by pushing the piston outward is just $$W=\int{PdV}$$ From your problem statement, it looks like you have a constant pressure of 101 kPa and a change in volume of $\Delta V=\frac{\pi}{4}(0.59\ \mathrm{m})^2(0.7\ \mathrm{m})=0.19\ \mathrm{m^3}$. The work done by the system is then $W=(101\,\mathrm{kPa})(0.19\ \mathrm{m^3})=19.3\,\mathrm{kJ}$.</p> <p>On a side note, the energy added to the system by the falling weight is only $\Delta E=mg \Delta h=(47\ \mathrm{kg})(9.81\ \mathrm{\frac{m}{s^2}})(6.1\ \mathrm{m})=2.8\ \mathrm{kJ}$, so either the numbers are wrong or there was some other energy input besides the falling weight.</p> <p><a href="http://www.chem.purdue.edu/gchelp/howtosolveit/Thermodynamics/ExpansionWork.html" rel="nofollow">This link on expansion work pretty much sums it up.</a></p>

6992

Find net work on the system

2016-01-20T16:47:56.673

<p>While I've designed tons of plastics for corrosive service, the fasteners to hold the plastics together are inevitably metallic. There have been a lot of ways to protect them from rusting (due to exposure to things like <em>33% HCl, 50% H2SO4, 50% NaOH, and 15% NaOCl</em>, for example) that I've seen, but here is a new one I haven't heard of until today - <strong>Alloy 20.</strong> It's a nickel alloy apparently and seems to hold up pretty well.</p> <p>One of my concerns about Alloy 20 is whether the parts are easy to obtain and available across the world? How does it compare with Grade 5 Titanium, Hastelloy C-276, and 316 SST as a fastener in plastics?</p>

|mechanical-engineering|materials|chemical-engineering|corrosion|fasteners|

<p>This link specifies that Alloy 20, specifically CARPENTER® alloy 20Cb-3, has an amazing resistance to chemicals containing chlorides and sulfuric, phosphoric, and nitric acids. It also resists pitting, crevice corrosion, SCC, and intergranular attack. It can be used for pickling tanks, gas scrubbers, piping, heat exchangers, pumps, shafts and valves. Synthetic rubber, pharmaceutical production and other process equipment.</p> <p>While Ti 6-4 (Grade 5) is significantly stronger than commercially pure titanium while having the same stiffness and thermal properties ( excluding thermal conductivity, which is about 60% lower in Grade 5 than in CP Ti). Among its many advantages, it is heat treatable. This grade is an excellent combination of strength, corrosion resistance, weld and fabric-ability.</p> <p>Comparing it to alloy C-276, specifically HASTELLOY® alloy C-276, possesses an outstanding corrosion resistance in reducing and oxidizing environments. It maintains corrosion resistance in welded condition, and is excellent resistance to pitting and stress-corrosion cracking (SCC). It is widely used in severest environments in chemical processing, pollution control, pulp and paper.</p> <p>Not too sure about the comparison with alloy 316, but the latter possesses an improved resistance to pitting by solutions containing chlorides and other halides. In addition, it provides excellent elevated temperature tensile, creep and stress - rupture strengths.</p> <p>To answer your last concern, obtaining and availability of Alloy 20 that I am not too sure about.</p> <p><a href="http://www.hpalloy.com/Alloys/corrosionResistant.html" rel="nofollow noreferrer">Corrosion Resistant Materials</a></p>

7015

How does Alloy 20 compare with Ti 6-4, C-276 and 316 stainless for corrosive service?

2016-01-20T22:11:39.683

<p>I am practicing engineering drafting and sometimes I am near on finishing my work but suddenly I make a mistake. Is it possible to remove the ink? If yes, how? Where can I buy such tools? I am using a <a href="http://rads.stackoverflow.com/amzn/click/B001U3RCGQ" rel="nofollow">Staedtler's Mars Matic 0.5mm</a>.</p>

|drafting|

<p>Yes, it can be removed but it is a delicate task requiring some experience...I first did this over 50 years ago. You should be drawing on some heavy velum that can withstand the erasing. You will need an erasing shield and electric eraser with the coarse dark grey eraser to gently sand a layer of paper off with the ink. It's easy to erase a hole. If you're going to do a lot of ink drafting you may want to look into mylar! Goid luck drafting ....computers suck!</p>

7019

Is it possible to remove Staedtler ink from a paper?

2016-01-20T22:25:29.150

<p>I'm curious about pumping a fluid through two different tubes (coming from two different pumps) into a Y connector. The third side of the connector runs into a collector without any resistance (other than that of the length of tubing).</p> <ol> <li><p>If each pump runs at a different rate, say one is twice as strong as the other, do I run any risk of any backing up into the weaker pump? What if one is three times stronger than the other, etc.?</p></li> <li><p>Does the third side of my Y connector (the side the fluid leaves through) now pump with the equivalent of pump a + pump b total power? Are they averaged together? Is it diminished? </p></li> <li><p>Does the type of fluid matter for parts 1 and 2 above?</p></li> </ol> <p>This question assumes that all tubing is the same thickness, but answers regarding different thicknesses could also be interesting (and worth extra credit!)</p>

|fluid-mechanics|pumps|

<h2>Pumps are complicated</h2> <p>Many questions regarding fluid machinery start with these two assumptions:</p> <ol> <li>A pump or gravity is needed to make the fluid go faster</li> <li>High pressure means faster flow</li> </ol> <p>All of these are correct in principle, but none of them are accurate. The first is inaccurate. Due to <a href="https://en.wikipedia.org/wiki/Conservation_of_mass" rel="nofollow noreferrer">conservation of mass</a>, fluid flowing into a pipe on one end will be equal to the flow out the other end. While the pump will make the fluid flow <em>faster</em>, the reality is the pump imparts energy on the flow, until the added pressure loss from the higher flow rate is equal to the added energy.</p> <p>Because of how complicated the energy loss in a pipe can become (<a href="https://en.wikipedia.org/wiki/Reynolds_number" rel="nofollow noreferrer">Reynolds numbers</a>, <a href="https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae" rel="nofollow noreferrer">Darcy friction factor</a>, <a href="https://en.wikipedia.org/wiki/Hydraulic_diameter" rel="nofollow noreferrer">hydraulic diameter</a>, etc.), since most engineering problems deal with water and piping, <a href="https://en.wikipedia.org/wiki/Hazen%E2%80%93Williams_equation" rel="nofollow noreferrer">Hazen Williams equations</a> are typically used. In addition to simplifying the pipe, we simplify the pump. Pumps have a <a href="https://www.youtube.com/watch?v=fiJeOLkLV5I" rel="nofollow noreferrer">curve</a>, which is always convex. By taking the max flow rate (at no pressure), and the maximum pressure (at no flow), a <a href="https://en.wikipedia.org/wiki/Linear_equation" rel="nofollow noreferrer">linear equation</a> can be developed - just use the <a href="https://en.wikipedia.org/wiki/Linear_equation#Intercept_form" rel="nofollow noreferrer">intercept form</a>. This simplifies the pump. Now, a set of equations can be developed called the <a href="https://en.wikipedia.org/wiki/Hardy_Cross_method" rel="nofollow noreferrer">Hardy Cross Method</a>. My <a href="https://rads.stackoverflow.com/amzn/click/com/0135259738" rel="nofollow noreferrer" rel="nofollow noreferrer">textbook</a> shows a typical setup:</p> <p><a href="https://i.stack.imgur.com/EXLHh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EXLHh.jpg" alt="Hardy Cross Setup" /></a></p> <p>A source of unknown pressure is feeding the system 3 cfs of liquid - via valves, pumps, or who knows what. The exits have 1 and 2 cfs - matching the conservation of mass. Again, pressure at these points is unknown, but not unlike pressure loss across a resistor, the energy difference between each point means that energy is conserved across the loop. This is why Hardy-Cross Method can be seen to be the fluid version of <a href="https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws" rel="nofollow noreferrer">Kirchoff's Voltage Law</a></p> <p>As can be seen, the top line (only 6&quot; and a full 3000' + 4000') will get minimal flow, while the bottom line, which is shorter and has larger diameter pipes, should be expected to hold most of the flow. When adding various pumps, it can be seen that even with sufficiently large pumps, circulation starts to become inevitable:</p> <p><a href="https://i.stack.imgur.com/ruN7Z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ruN7Z.jpg" alt="HardyCross Examples" /></a></p> <h3>The Y-Connector</h3> <p>Your Y-Connector method is interesting because you now have an unknown flow rate - but a known pressure. In this case, you need to use the air in the atmosphere to connect the pressures in a loop (See for example, <a href="http://www.innovyze.com/bookstore/cwbsa/Chap5.pdf" rel="nofollow noreferrer">page 15</a> of this pdf, showing how this is done. Also, don't forget to add in all of the <a href="http://www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html" rel="nofollow noreferrer">minor loss coefficients</a></p> <p><strong>However, all of this is a model.</strong> To prevent backflow, <em><strong>always add a check valve to your pumps.</strong></em> Because you never know what could happen, and backflow can be very expensive.</p>

7021

How do pumped flows combine in a Y connector?

2016-01-21T00:01:33.367

<p>I accidentally dropped my <a href="http://rads.stackoverflow.com/amzn/click/B001U3RCGQ" rel="nofollow">Mars Matic 0.5mm</a> and noticed that maybe I broke the needle. I was having a hard time writing with it so I tried to clean it, and while cleaning I accidentally bent the needle part. After all the trouble that I experienced I manage to straighten the needle very carefully.</p> <p>Right now it's working fine but I need to shake it rigorously. I am planning on buying a new one because of the hassle of using it but there is also a side of me that says I should not buy a new one.</p> <p>Is there any other way that I can fix my pen? It's slightly bent now but by shaking it rigorously it can still write.</p>

|drafting|

<p>You can buy a replacement nib. The following picture is from eBay. I just wanted a picture to show you that it is possible to remove the existing nib and replace it. </p> <p>I would suggest you make enquires with pen supplier, or search the internet for a replacement nib.</p> <p><a href="https://i.stack.imgur.com/peN6c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/peN6c.jpg" alt="enter image description here"></a></p>

7023

How do I fix a technical drawing pen with a bent needle?

2016-01-21T17:45:38.973

<p>I am interesting in replacing a solid strut with tubing, but I do not want to destructively test the strut to determine its yield point. I do not want to make assumptions about the yield point based on material, because heat treatment and other factors can affect the yield point.</p> <p>Can I cantilever suspend the strut and record its deflection due to a weight placed on the end, to infer the strength of the strut?</p> <p>In other words lets say the strut deflects 5 inches when it is cantilevered and a weight is placed on its end. The tubing under test is placed the same way and only deflects 4 inches. Can I assume that the tubing will be an adequate replacement for the strut (ignoring factors like fatique, age microcracking, etc etc)?</p> <p>I can do math that would suggest the replacement tubing will be significant lighter and stronger than the bar strut, but I am trying to find a way to actually physically test it.</p>

|mechanical-engineering|structural-engineering|stresses|

<p>The other answers say that this isn't possible in the way that you want to do it, but they don't provide the equations that show why.</p> <h3>Deflection of a Cantilever</h3> <p>The <a href="https://en.wikipedia.org/wiki/Deflection_%28engineering%29#End-loaded_cantilever_beams" rel="nofollow">equation</a> for the deflection of a cantilever beam:</p> <p>$$ \delta = \frac{FL^3}{3EI} $$</p> <p>Where: $F=Force \\ L= \text{Length of cantilever} \\ E=\text{Modulus of Elasticity} \\ I=\text{Moment of Inertial}$</p> <p>As you can see, none of these components have anything to do with strength. The only variable that is affected by the material is $\text{E}$. This is basically constant for steels.</p> <h3>Column Strength</h3> <p>You didn't necessarily ask about the material strength of your tube though. You asked about replacing a strut. A strut is a column. The compressive capacity of a column is ultimately controlled by <a href="https://en.wikipedia.org/wiki/Buckling#Columns" rel="nofollow">Euler Buckling</a> (hint: it rhymes with "oiler").</p> <p>$$F=\frac{\pi^2EI}{(KL)^2}$$</p> <p>Where: $\text{K is dependent on the end connections} \\ \text{(well, see above, everything else is the same)}$</p> <p>This also shows that buckling capacity isn't dependent on material strength. Assuming that the original solid bar is the same material as your replacement tube, there is only one parameter to compare $\text{I}$, (<a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia#Moments_of_inertia" rel="nofollow">Moment of Inertia</a>).</p> <p>Take the diameter of your original solid bar and use that to determine the the required diameter and thickness of the wall so that the moments of inertia match.</p> <h3>What if the materials are not the same?</h3> <p>It the materials are not the same, then the <strong>E</strong> value will change. Everything else in the equations stays the same, so <strong>EI</strong> will need to stay the same. The moment of inertia may go up or down depending on how the modulus of elasticity changes. </p> <p>This ends up meaning that a stiffer strut will have a greater <strong>EI</strong>. This also contributes to a higher buckling force. If buckling is the controlling failure mode, then a strut that deflects less will be stronger. </p> <p><em>Note: Insert standard warnings about there being lots of other factors that could affect this. Don't use this for things related to human life or anything that you care about.</em></p>

7038

Can the strength of a strut be determined by deflections?

2016-01-22T00:51:53.827

<p>I'm interested in learning what I might call "advanced statics" but am not sure what concepts this would include in engineering parlance.</p> <p>Would it actually be called "matrix structural analysis"?</p>

|civil-engineering|statics|

<p>The field of statics is based on analyzing finitely many (solid) members and discrete forces acting in equilibrium. It is a purely algebraic method of analysis (no calculus is necessary). To move to topics more advanced than this while still assuming equilibrium, you have to start eroding away at the "discreteness" assumption. In other words, you have you start considering infinitesimal contributions summed over an area or a volume (i.e. you really need vector calculus to do this for generalized 3d problems). Continuum Solid Mechanics, particularly steady state linear elasticity, is pretty much the next step up from statics if you're analyzing solids in equilibrium.</p> <p>However, there is somewhat of a middle ground between the discrete nature of statics and continuum solid mechanics. The Finite Element Method makes additional assumptions to interpolate a continuous problem into discretely many pieces and resolves the degrees of freedom as a system of equations. But I would consider it more as a method to solve partial differential equations than an "advanced form of statics", because FEM is more general and applies not only to solid mechanics, but also fluid mechanics, heat transfer, electromagentics, and pretty much anything that can be described by a PDE. </p>

7043

What engineering topic builds upon the concepts of basic statics?

2016-01-23T20:34:36.600

<p>I'm well aware of Helmut Hoeltzer's "Mischgerat" analog computer, and I read that it not only served as a kind of "fly by wire" computer on board of the missile, but also as a simulator on the ground that calculated V2 trajectories.</p> <p>But, I didn't find any clue about how it did it - how Hoeltzer calculated optimal trajectory to hit the target (cities). I think the trajectory had to calculated accurately because only by following an accurate trajectory a ballistic missile could hit a city sized target.</p>

|electrical-engineering|

<p>My pardons for addressing this question six years on.</p> <p>The ballistic trajectory for a particular target was calculated by hand by a computing group within the V2/A4 project. If you've seen the movie &quot;Hidden Figures&quot;, just replace African American women with Nordic men.</p> <p>Based on these calculations: 1) the heading was set by orienting the rocket with one of the fins towards the target, which properly aligned the gyroscopes; 2) the range was set by a clockwork mechanism which caused the missile to tilt a set angle at a set time during powered flight; 3) a timer to shut down the engine if needed. At that point Newton took the wheel.</p> <p>The Mischgerät electronic analog guidance computer controlled pitch, roll, and yaw, reacting to inputs from the gyroscopes and optionally from a radio receiver tuned to a heading beam transmitted from the ground. When the gyroscopes' platform was tilted by the clockwork, the Mischgerät commanded the rocket exhaust deflectors and the fin trim vanes, steering the rocket so as to put the gyroscopes level again, and thus tilt the rocket towards the target.</p> <p>Hoelzer developed the Mischgerät and from it his more full featured analog simulation computer. The simulation computer allowed him to validate the Mischgerat's function in two degrees of freedom prior to flight test. He built a couple of examples at Peenemüende, one of which followed him to the US.</p> <p>Sources: Dornburger, Walter (1954), &quot;V2&quot;</p> <p>Ernst, Wolfgang (2008), &quot;Der Analogcomputer als Medium der Zeitmanipulation&quot;, Humboldt-Universität zu Berlin, Seminar für Medienwissenschaft, SoSe 2008 <a href="https://www.cdvandt.org/Busjahn_Analogcomputer.pdf" rel="nofollow noreferrer">https://www.cdvandt.org/Busjahn_Analogcomputer.pdf</a></p> <p>Tomayko, James E. (July 1985). &quot;Helmut Hoelzer's Fully Electronic Analog Computer&quot;. Annals of the History of Computing. 7 (3): 227–240. doi:10.1109/mahc.1985.10025. S2CID 15986944</p> <p>Ulmann, Bernd (Oct 26, 2008), &quot;Von der Raketensteuerung zum Analogrechner Helmut Hoelzers Peenemünder Arbeiten und ihr Erbe&quot;. <a href="http://www.analogmuseum.org/library/hamburg_hoelzer.pdf" rel="nofollow noreferrer">http://www.analogmuseum.org/library/hamburg_hoelzer.pdf</a></p> <p>Photos: Mischgerät, opened - <a href="https://www.cdvandt.org/Mischger3-0.jpg" rel="nofollow noreferrer">https://www.cdvandt.org/Mischger3-0.jpg</a></p> <p>Hoelzer Simulation computer - <a href="https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8789827" rel="nofollow noreferrer">https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=8789827</a> (page 2)</p>

7059

How was the trajectory of the V2 ballistic missile (during boost phase) calculated?

2016-01-23T21:36:50.767

<p>recently I bought 3 solar panels rated at 5V 200 mA each. I want to use them to charge a 5V battery bank to charge a phone. Thinking about the proper way to put them, I thought i can connect all in parallel to get maximum current, but realized that if the sun light was a little weak it will no generate full 5v thus preventing charging. So I decided to put 2 in parallel to give the equivalent of one 5V solar panel, connected in series with the 3rd panel to give the equivalent of 10v. sacrificing a little current to get higher voltage, to allow the charging to happen on a wider range of sun light power. The following picture shows the wiring and the schematic I intend to replicate. </p> <p><a href="https://i.stack.imgur.com/4oFHJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4oFHJt.jpg" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/lHrv9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lHrv9t.jpg" alt="enter image description here"></a></p> <p>Now that I have an equivalent of 10v, 400mA solar panel. I used a 7805 voltage regulator to cut down the excess to 5v. </p> <p><a href="https://i.stack.imgur.com/IbnN3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IbnN3t.jpg" alt="enter image description here"></a></p> <p>Final step, I added a standard diode to prevent the panels from leaking the battery in the shade.</p> <p>and now measuring:</p> <p><a href="https://i.stack.imgur.com/A0RIw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A0RIwt.jpg" alt="enter image description here"></a></p> <p>Questions:</p> <ol> <li>I didn't think about this before putting the diode, but is it okay to put the blocking diode on the ground wire? because I know some applications do not use the ground except for safety (i.e. 3 phase system). it would be helpful also to avoid the 0.7v drop across the diode before the regulator.</li> <li>according to previous calculations, I'm supposed to get a maximum of 10v output before regulation, and considering that the sun was pretty shinny today, why was the reading I got not more then 6v? I have measured across each panel seperatly and got around 5.5V, are the connections right?</li> </ol>

|electrical-engineering|photovoltaics|solar-energy|

<p>A panel &quot;rated&quot; at 5V 200mA will normally put out 5V 200mA at it's maximum power point, in full sunlight.</p> <p>In lower light, it will put out less power.</p> <p>But the voltage depends on the battery, not on the light. When you have more light, you have more current, and the battery charges faster.</p> <p>When you have less light, you have less current, and the battery charges slower.</p> <p>If the battery voltage is higher, the voltage on the panel is above the maximum power point, and current is lost inside the panel, and battery charging is slower.</p> <p>If the battery voltage is lower, the voltage on the panel is below the maximum power point, and voltage is lost inside the panel, but there is no more charging current than at the maximum power point.</p> <p>If the panel is matched to the battery, this is a &quot;self regulating&quot; system, as the battery charges, it's voltage goes up above the maximum power point, and charging slows.</p> <p>When you are charging a lead-acid battery, a self-regulating system is OK. After you overcharge, you just add water. With other kinds of cell, going over-voltage will permanently damage the cell, and self-regulating panels are not used.</p> <p>When there is low light, it does not mean that the panel voltage goes down. It means that the charging current goes down.</p>

7060

Did I wire these solar panels correctly?

2016-01-24T17:16:11.273

<p>It sounds logical untill I tried some thought experiments. So I must be doing something wrong. Or I am missing a fundamental aspect.</p> <p>What I thought of was: </p> <p>-lets say you have an orthogonal parallelepiped which is pulled in tension by a stress-strain test. (in the long axis or z-direction).</p> <ul> <li>the lateral sides are 5 mm and 4 mm and the axial length is 7 mm.</li> </ul> <p>-It has been elongated by the test untill the specimen had an axial length of 11 mm and hypothetically the 4 mm lateral side did not change.</p> <p>So the original volume was 4x5x7= 140 mm^3 and since there is no net volume change the final length of the lateral side that did change has to be: 140= a x 11 x 4 -> a= 35/11 mm</p> <p>now if you calculate the poisson's ratio: v= - Ex/Ez</p> <p>-> Ex = (35/11 -5)/5= -4/11 and Ez= (11-7)/7= 4/7</p> <p>v= 4/11 : 4/7 = 7/11 = 0.64 > 0.5</p> <p>How I understood it: There is no volume change, so when you elongate one direction the other two will compensate this by compressing. And this poisson ratio of 1/2 is in the extreme case that one direction does not compress and so the other side has to take all the compression. Like in the situation that you elongate the axial side to twice it's orginal length, then in order to have same volume the two other directions have to become each 1/4 of their orginal length or one direction has to become 1/2 it's orginal length. This will indeed then give poisson ratio of 1/4 and 1/2. So I don't get why with other numerical examples I don't get this same 1/4 and 1/2 ratio.</p>

|mechanical-engineering|materials|metallurgy|

<p>First poisons ratio for isotropic materials applies equally between each pair of dimensions. That means that if the axial length is stretched $1\%$ both of the lateral dimensions would reduce by $\nu 1\%$. I don't know where you're getting the 0.25 number from.</p> <p>Second, this approximation only works when the strain is small. Lets say we have an orthogonal parallelepiped with dimensions $x$, $y$, and $z$. If we apply a stress $\sigma_x$ streching the x direction and let the y and z dimensions be stress free. Then we'll have:</p> <p>$$\epsilon_x=\frac{\sigma_x}{E}$$ $$\epsilon_y=-\nu\frac{\sigma_x}{E}$$ $$\epsilon_z=-\nu\frac{\sigma_x}{E}$$</p> <p>So the new dimensions would be:</p> <p>$$x'=x(1+\epsilon_x)$$ $$y'=y(1-\nu\epsilon_x)$$ $$z'=z(1-\nu\epsilon_x)$$</p> <p>And the new volume would be:</p> <p>$$V'=x'\,y'\,z'= x\,y\,z\, (1+\epsilon_x)(1-\nu\epsilon_x)^2$$</p> <p>So the ratio of the new volume to the old volume would be:</p> <p>$$\frac{V'}{V}=(1+\epsilon_x)(1-\nu\epsilon_x)^2=1+(1-2\nu)\epsilon_x+(\nu^2-2\nu)\epsilon_x^2+\nu^2\epsilon_x^3$$</p> <p>To get the increase in volume, you can subtract the 1. Then there are three more terms, but since $\epsilon_x$ is assumed to be small then the square and cube terms should be dominated by the linear term. Thus those terms are usually neglected and you're left with $(1-2\nu)\epsilon_x$. If you have an incompressible solid then this value should always be zero, so then $\nu$ must equal $0.5$</p> <p>Let us look at a more general example:</p> <p>$$V=V'$$</p> <p>Is really saying:</p> <p>$$\begin{split}1=(1+\epsilon_x)(1+\epsilon_y)(1+\epsilon_z)=&amp;1+ \\ &amp; \epsilon_x+\epsilon_y+\epsilon_z + \\ &amp; \epsilon_x \epsilon_y + \epsilon_y \epsilon_z + \epsilon_x \epsilon_z + \\ &amp; \epsilon_x \epsilon_y \epsilon_z \end{split}$$</p> <p>but again with a assumption that the strains are small, the product of multiple strains should be really small, so this equation simplifies to:</p> <p>$$0=\epsilon_x+\epsilon_y+\epsilon_z$$</p> <p>Now the general equations for stain in isotropic materials are:</p> <p>$$\epsilon_x=\frac1{E}(\sigma_x-\nu(\sigma_y+\sigma_z))$$ $$\epsilon_y=\frac1{E}(\sigma_y-\nu(\sigma_x+\sigma_z))$$ $$\epsilon_z=\frac1{E}(\sigma_z-\nu(\sigma_x+\sigma_y))$$</p> <p>We can sum these all together to get:</p> <p>$$0=\epsilon_x+\epsilon_y+\epsilon_z=\frac1{E}(1-2\nu)(\sigma_x+\sigma_y+\sigma_z)$$</p> <p>So we see that one again to conserve volume $\nu=\frac12$</p> <p>Now why didn't this work in your example?</p> <p>When you applied the stretching both other directions should have contracted. In order for the second direction to not contract, that direction would also have have to be pulled. This pulling on the second direction would further contract the third direction, but it would also contract the first direction, so in order for that to stay at it's original displacement you'd have to pull even harder.</p> <p>Then you specified that $\nu=-\frac{\epsilon_z}{\epsilon_x}$ but this is only true if there is only stress in the x direction which is not the case for your scenario (because there must have been a stress applied to keep the 4cm side at 4cm).</p> <p>Finally the strain you applied was too large. Even if you let the two lateral dimensions strain freely, you'd get a Poisson's ratio not equal to .5 just because the small strain approximation would no longer be valid:</p> <p>$$\epsilon_x=\frac{11}{7}-1\approx 0.57$$</p> <p>to conserve volume:</p> <p>$$\epsilon_y=\epsilon_z=\sqrt{\frac{7}{11}}-1 \approx -0.20$$</p> <p>Which is quite a bit less than half.</p>

7066

Why is the Poisson's ratio for isotropic 1/4 and the maximum value 1/2 (when there is no net volume change)?

2016-01-25T05:59:33.450

<p>I would like to create beam models having various cross sections on ProE tool.</p> <p>Can I create a beam model like below by only extrusion?</p> <p><a href="https://i.stack.imgur.com/NEwmF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NEwmF.png" alt="enter image description here"></a></p> <p>Can anyone explain what is the best way to model various cross sections of beams?</p>

|mechanical-engineering|structural-engineering|modeling|

<p>If pro-e is anything like solidworks then you need to draw a 2D version of the extrusion pattern. In your case, one large square with the 4 smaller squares within it. </p> <p>Then select the area you wish to extrude and extrude it. </p> <p>I recommend looking up some basic tutorials for ProE as I'm sure this would be covered in the early stages. </p>

7071

beam cross section modeling by extrusion on proe

2016-01-25T06:25:35.537

<p>I've heard that materials that need to slide against each other should be of different materials. Why is that?</p>

|materials|friction|

<p>For a sleeve bushing, you want a soft slippery material with sufficient yield strength to withstand any deformation of the hole surface during operation. For the shaft, you want a hard smooth material with sufficent yield strength to withstand being deformed/bent in order to ensure that the surfaces of the shaft spin true to the axis of rotation at all times during operation. You want as close a possible fit as you can get, you want it to feel slightly snug with a small amount of friction. Spin the parts while not under any load and the rotating surface of the shaft will lap the bushing until there is zero friction and it helps to move the parts in and out along the axis slightly to eliminate as much material as you can from the bushing surface until it rotates freely. If you do all these steps and tolerance your parts correctly and use the right materials, you will have a perfect rotating bushing joint. Make sure to factor in the loads it will be operating under, and make sure to follow the guidelines above, and these kinds of joints can withstand a few hundreds of pounds with very low friction comparable to a ball bearing and gets smoother and looser as time goes on and the bushing continues to wear. You just want to prevent deformation and reduce friction.</p>

7072

Why use bushings of different material than what they contact?

2016-01-25T14:32:16.590

<h1>Background</h1> <p>In the world of <a href="https://www.youtube.com/watch?v=kklBFX62F8U" rel="noreferrer" title="2012 Olympics - Elite Men&#39;s Main Event">bicycle motocross, also known as BMX racing</a>, gearing is a hotly-debated topic.</p> <p>Since the bikes are all single-speed, gear ratio is a fixed number defined as <code>chainwheel / cog</code> (front gear divided by rear gear). Altering your gear ratio is understood as an immediately-noticeable tradeoff between acceleration and top-end speed.</p> <p>Here is a series of common gear ratios:</p> <pre><code>╔════════════╦═════╦════════╗ ║ Chainwheel ║ Cog ║ Ratio ║ ╠════════════╬═════╬════════╣ ║ 43 ║ 16 ║ 2.6875 ║ ║ 41 ║ 15 ║ 2.7333 ║ ║ 44 ║ 16 ║ 2.75 ║ ╚════════════╩═════╩════════╝ </code></pre> <p>In 2012, a company called <a href="http://rennendesigngroup.com" rel="noreferrer">Rennen Design Group</a> created a supposed breakthrough innovation called <a href="http://rennendesigngroup.com/decimal.html" rel="noreferrer">"decimal gearing"</a>. The claim is that through manipulations of tooth profile and ring diameter, in-between gear ratios can be created - for example:</p> <pre><code>╔════════════╦═════╦════════╗ ║ Chainwheel ║ Cog ║ Ratio ║ ╠════════════╬═════╬════════╣ ║ 43 ║ 16 ║ 2.6875 ║ ║ 45.7 ║ 17 ║ 2.6882 ║ ║ 37.7 ║ 14 ║ 2.6929 ║ ║ 43.1 ║ 16 ║ 2.6938 ║ ║ 41 ║ 15 ║ 2.7333 ║ ║ 41.1 ║ 15 ║ 2.74 ║ ║ 52.2 ║ 19 ║ 2.7473 ║ ║ 44 ║ 16 ║ 2.75 ║ ║ 44.2 ║ 16 ║ 2.7625 ║ ╚════════════╩═════╩════════╝ </code></pre> <p><em>Note: Table is not exhaustive.</em></p> <p><strong>For example - a 44.2 tooth gear actually only has 44 teeth, but the tooth spacing, tooth profile, and chainwheel diameter is supposed to have been manipulated to create a larger gear.</strong></p> <p>In the world of BMX racing, the existence of in-between gear ratios like this is a Really Big Deal. Since the man behind Rennen has a Master's from MIT - and since most BMXers would rather hit jumps than do math or measure things - nobody has really ever checked up on whether or not this is valid. Some questions were asked a long time ago in the dusty corners of a BMX forum, but the testing methods didn't properly control for all variables and the thread descended into a bunch of name-calling and ad-hominem attacks.</p> <h1>The Actual Question</h1> <p><strong>Is this physically possible?</strong></p> <p>I understand "gear ratio" to be defined as:</p> <blockquote> <p>For a given gear ratio <code>x / y</code>, one rotation of the gear with <code>x</code> teeth will result in <code>x / y</code> rotations of the gear with <code>y</code> teeth.</p> </blockquote> <p>For a gear ratio of 44/16, one full rotation of the 44 tooth gear (chainwheel) should result in 2.75 rotations of the 16 tooth gear (cog).</p> <p>So for a "decimal ratio" of 44.2/16, one full rotation of the 44.2 tooth gear (which again - only has 44 teeth) is supposed to result in 2.7625 rotations of the 16 tooth gear.</p> <p><strong>My biggest reservation is the fact that a chain-driven drivetrain is a TIMED DRIVETRAIN.</strong> No matter how big or small you make the teeth on the chainwheel, if they fit the chain, they're only going to push as many links through per rotation as the chainwheel has teeth.</p> <p>For a true 44.2 tooth chainwheel, one would expect that 442 links get pushed through over 10 full rotations of the chainwheel - but that's not the case. Only 440 links will ever get pushed through to the cog because only 44 links get pushed through per full rotation of the chainwheel. I actually spent my whole afternoon yesterday taking video and counting links and measuring.</p> <p>But I'm not a scientist. My <a href="http://www.gbcscrusaders.com/main-page.html" rel="noreferrer" title="my graduating class had ten people in it">high school</a> didn't even offer a physics course. I'm just a racer that trains really hard and knows how to do basic math.</p> <p>If this were a belt-driven system, I would completely understand how a manipulation of the chainwheel diameter would change the effective ratio - but it's not. It's a timed drivetrain, limited by the physical dimensions of the chain.</p> <p>I have several hundred dollars and months of training and metrics invested in these stupid chainwheels. If someone could confirm or deny my theories, I would really appreciate it. I just want some closure.</p> <p>Here's a photo of the teeth from a 41 tooth chainwheel on top of the teeth from a 41.2 tooth chainwheel - both are Rennen gears:</p> <p><a href="https://i.stack.imgur.com/j4TBD.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/j4TBDt.jpg" alt="41t and 41.2t teeth"></a></p> <p>Here's a 41t on top of a 41.2t:</p> <p><a href="https://i.stack.imgur.com/UE790.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/UE790t.jpg" alt="enter image description here"></a></p> <p>Here's the 41.2t on top of the 41t, from behind:</p> <p><a href="https://i.stack.imgur.com/q1LRl.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/q1LRlt.jpg" alt="enter image description here"></a></p>

|mechanical-engineering|gears|measurements|

<blockquote> <p>A "44.2 tooth" gear still can only push 44 links of chain per revolution, which pushes a 16 tooth cog around 2.75 times. There's no getting around that.</p> </blockquote> <p>I think that is the question that has everyone's head smoking. So I made a visualization.</p> <p>Here we have small gear with 40px diameter and 4 cogs and a large gear with 80px diameter and 8 cogs, respectively. The chain is indicated by red dots and fits perfectly onto the gears. As one would expect, chain links are picked up at the top of gear and released at the bottom.<br> <a href="https://i.stack.imgur.com/BIoqS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BIoqS.png" alt="4-to-8 ratio"></a></p> <p>In the second picture, I increased the diameter of the larger gear to 90px, but it still has only 8 cogs (I call it the 8.9999/4 ratio). As you can see, I had to move the gears closer together to adjust for the larger circumference (with a less significant increase, as from 41 to 41.2, the chain might only sit slightly tighter).<br> <a href="https://i.stack.imgur.com/XXaec.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XXaec.png" alt="4-to-8.999 ratio"></a></p> <p>So, does increased radius mean that the smaller gear moves faster? I'd say: No.<br> As you can see in the second picture, the chain links only grip at the very bottom of the gear. The cogs move faster than the chain links, which means the gear subtly skids under the chain.</p> <p>Some other answers suggested that this setup might lead to a higher torque. However, I don't think so.</p> <p>Firstly, increasing the size of the larger gear should decrease the torque. Secondly, according to <a href="https://en.wikipedia.org/wiki/Work_%28physics%29#Torque_and_rotation" rel="nofollow noreferrer">Wikipedia</a> work is $W = \tau\phi$. Thus, as long as the angular speed of the chain ($\phi$) does not change, you will get the same torque($\tau$) for your work.</p> <p><strong>Edit:</strong></p> <blockquote> <p>In this setup, the chain only grips at the top of the chainwheel (larger gear); I've visually verified this. The other rollers float between the teeth as you continue on down the gear.</p> </blockquote> <p>If we look closely at your pictures, we can see that the cogs are very close together so I suspect the chain links fit perfectly in-between. Towards the top, the cogs become slimmer. This might cause the tight chain links to slide upwards and become more loose.</p> <p>Here is another illustration (again, I greatly exaggerated the effect):</p> <p><a href="https://i.stack.imgur.com/tI4FY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tI4FY.png" alt="More Sprocketes"></a></p>

7080

Is it possible for a bicycle chainwheel to have a fractional number of teeth?

2016-01-26T13:47:00.400

<p>Does the material with which I make the mold for carbon fiber parts affect the finish the part will get? If it does, I would like to understand how a given material may affect the finish.</p> <p>Another thing that is confusing me is the exact definition of <code>finish</code>. Does it refer only to the smoothness of the surface of the final part, or does it also refer to gloss and polish?</p>

|materials|composite|

<p>The finish of the mould will certainly have an effect on the finished part. How much of an effect depends on the exact process, for example if you use a gel coat then that will pick up the surface texture of the mould pretty closely. Without a gelcoat the ratio of fibre to resin will determine how much the weave of the fabric comes through. </p> <p>In this case it's not so much the mould material as such as the finish of the mould surface and of course any release agent used will have an effect as well. It's certainly possible to get a glossy part straight out of the mould as long as t is prepared properly. In practice you would balance the time and expense you put into maintaining the mould with the amount of time spent on finishing the part after it comes out of the mould. eg if you are making dozens of parts it might well be worth getting the mould as close to perfect as you can to save time spent finishing each part. If you're just making a one-off you might not worry so much about the mould finish and sort out the finish on the part itself.</p> <p>Another consideration is that for high performance laminated parts you may find that the optimum resin to fibre ratio is just enough to wet the fibres and you get the weave of the mat showing through regardless of the mould finish. </p> <p>In terms of definitions finish can refer to overall surface smoothness as well as gloss as these are really just two aspects of the same thing. This is a significantly different thing from the geometric accuracy of the surface. </p> <p>Similarity low volume moulding processes like silicone, plaster etc tend to be more likely to pick up small surface defects than an CNC machined steel mould. </p>

7101

Impact of mold material in carbon fiber part finish

2016-01-26T20:39:10.750

<p>I have heard that error correcting codes are used to store and retrieve data in hard-drives but are they used in the processors in common laptops? Is the information encoded before it is processed?</p>

|electrical-engineering|

<p>Error correction means extra bits are stored along with the data. Think of it as a advanced checksum. A checksum can detect when the data is corrupt. A error correcting "checksum" allows recovering some limited numbers of errors, usually a single bit error in the chunk of data being protected by the checksum.</p> <p>Since this isn't free, it is done only when the benefits are important. In some cases it's cheaper to allow for a higher native bit error rate, but add bits to get the effective overall bit error rate back down via error correcting codes. This is often done with dynamic RAM modules that plug into ordinary PC motherboards. Whether laptops use them or not is best determined by reading the documentation for individual laptops. Magnetic disks is another common application. This time it's because their native bit error rate is unacceptably low for the densities and read/write speed people demand.</p> <p>The error correcting bits are usually determined and checked with dedicated combinatorial logic running in parallel with fetches and stores. That way this process largely doesn't get in the way of memory operations, and is largely transparent to the entity that is reading/writing data from/to the memory.</p>

7107

Error correcting codes in processors

2016-01-27T02:25:29.153

<p>When a fluid, like air, is moving at some velocity and "hits" or collides with a surface: </p> <ul> <li>If the surface is at an angle, the fluid will tend to be propelled in a similar direction (anything else?). </li> <li>If the surface is perpendicular, the fluid will tend to kind of splash out in all directions (anything else?). </li> </ul> <p>For generally turbulent, laminar, or mixed flow conditions? <em>The surface has mass and is relatively stationary for the purposes of this question.</em></p> <p>In each of those scenarios, what are the (qualitative) local effects of physical properties like pressure, density, etc. for both turbulent or laminar flow conditions? Does the air-like fluid compress into a denser form and the local pressure increase where the fluid is coming in contact with the surface?</p>

|fluid-mechanics|

<p>When the velocity of the air is less than about Mach 0.3, there is hardly any change in pressure or density as it impacts the surface. For flows greater than $M=0.3$, you can calculate the stagnation pressure, density, and temperature with the following equations:</p> <p>$$ \frac{p_0}{p}=\left( 1+ \frac{\gamma - 1}{2}\cdot M^2 \right)^\frac{\gamma}{\gamma -1} $$ $$ \frac{\rho_0}{\rho}=\left( 1+ \frac{\gamma - 1}{2}\cdot M^2 \right)^\frac{1}{\gamma -1} $$ $$ \frac{T_0}{T}=1+ \frac{\gamma - 1}{2}\cdot M^2 $$</p> <p>Where $M$ is the mach number and $\gamma$ is the heat capacity ratio. The subscript $0$ is the stagnation value and no subscript is the free-stream value. The fluid "stagnates" at the point of impact with the surface, at which point its kinetic energy transforms into thermal and pressure energy, hence the increase in pressure, density, temperature, etc.</p> <p>Beyond the stagnation zone, its not so easy to calculate the flow field. I ran a test case in ANSYS Fluent with an 80mm pipe shooting air at a solid wall 100mm away at a velocity of 10 m/s. The Reynolds number is about 28,000 for these conditions. I used the Reynolds stress model to calculate turbulence. The pressure and density fields are mostly constant as should be expected for this velocity. The velocity field is plotted in the image below. You can see the air make several vortexes along the wall.</p> <p><a href="https://i.stack.imgur.com/MoLCT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MoLCT.png" alt="enter image description here"></a></p> <p>Hope this gives you some things to think about for your problem.</p>

7111

Air colliding with surface, local effects?

2016-01-27T16:36:41.783

<p>Let's say I have a pump pushing fluid through a 250 ft pipe, 1.5'' diameter, at a rate of 20 gallons per minute. I would like to calculate the delta in energy required to pump this fluid with and without a restriction at the end of the 250 ft line. The restriction results in 0.01 psi pressure drop.</p> <p>What equation is needed to calculate this energy?</p> <p>Update:</p> <p>Based on comments below, I was able to figure this out. First, I need to convert my $\Delta P$ to pressure head using the following formula:</p> <p>$\Delta P = \rho_{fluid} * g * h$</p> <p>where </p> <p>$g =\text{ acceleration due to gravity}$</p> <p>$h =\text{ pressure head}$</p> <p>This formula itself is a simplified form of what can be found <a href="http://www.engineeringtoolbox.com/static-pressure-head-d_610.html" rel="nofollow">here</a></p> <p>$P_2 - P_1 = \gamma (h_2 - h_1)$</p> <p>where</p> <p>$\gamma = \rho_{fluid} * g$</p> <p>This gives us all the ammunition we need with the help of the pump power equation found <a href="http://www.engineeringtoolbox.com/pumps-power-d_505.html" rel="nofollow">here</a></p> <p>$P_{\text{pump_power}} = q * \rho_{fluid} * g * h$</p> <p>where</p> <p>$q =\text{ fluid flow, volume per time}$</p> <p>If you plug in our $\Delta P$ formula from above (you have to rearrange to get the h on it's own), you get the following:</p> <p>$P_{\text{pump_power}} = q*\rho_{fluid}*g*\frac{\Delta P}{\rho_{fluid} * g}$</p> <p>which, with handy cancellations, leads to:</p> <p>$P_{\text{pump_power}} = q * \Delta P$</p> <p>Nice! Dimensionnally, if your flow is in gallon per minute (which it was for me) you'll want to convert to cubic inches per second - this way, if your $\Delta P$ is in psi (which again, mine was) some of your inches will cancel out. You'll then have to convert BACK to feet from inches, but then you can easily go from $\frac{\text{ft-lb}}{s}$ to horsepower or kilowatts or whatever you want.</p>

|energy|

<p>The pump power equation can be used to calculate the energy required in both scenarios. Convert the additional $\Delta P$ to head, then calculate the difference in required hydraulic pump power. Both of these equations are on Engineering Toolbox. </p> <p><a href="http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html" rel="nofollow">Converting head (ft or m) to pressure (psi or bar, kg/cm2) and vice versa</a></p> <p><a href="http://www.engineeringtoolbox.com/pumps-power-d_505.html" rel="nofollow">Calculate pump hydraulic and shaft power</a></p> <p>For the question in your comment about the energy balance equation, Bernoulli's Equation is an energy balance for fluids. </p>

7119

How do I calculate the additional energy required to pump fluid through a restriction?