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engineering.stackexchange.com

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2015-05-10T20:28:06.510

<p>I was always taught that electricity flows through the path of least resistance. </p> <p>The problem that I'm having is that I've seen/done this: Take 3 wires have one wire break off into the 2 other wire (like a fork in the road) and put electricity through the first single wire. </p> <p>Now I take a multi-meter and test the end of each of the wires that forked from the original wire. Both wires show current/energy.</p> <p>How can this be possible, this happens even if I use 2 wires with different conductivity. </p> <p>The ending output might differ between the wires, but both wires are getting energy, even though one of them is more resistant. </p> <p>Can someone please explain this to me?</p>

|electrical-engineering|

<p>Ohm's law states that current through a resistor is voltage over resistance. Resistors in parallel will see an equal voltage. This means that each resistor will get some current through them. The one with lower resistance will get more. But they will both conduct current and the total current through the circuit will be the sum of the current through each.</p> <p>Given that you can work out that formula for resistance of 2 different resistors in parallel is: </p> <p>$$\frac{1}{R_{tot}} = \frac1{R_1}+\frac{1}{R_2}$$</p> <p>Otherwise to get a 5 $\Omega$ resistance from 2 10 $\Omega$ resistors would be much more finicky as under your assumption even the slightest change in resistance would cut off the slightly stronger resistor.</p>

2766

Path of least resistance misunderstanding

2015-05-10T21:57:28.567

<p>This may be a trivial problem, but I'm having a tough time with conceptualization. </p> <p>I'm using a code for modeling flow out of a tank. I'm not experienced in fluid dynamics; I'm more experienced with chemical aspects of the problem. I want to use pressure boundary conditions (a constant pressure at the bottom of the tank and atmospheric pressure at the tank outlet). The code is for incompressible flow and I must input the pressure values in terms of pressure/density (units of m2 s-2).</p> <p>The problem is I don't understand which densities to use for each boundary condition. At the bottom of the tank, I'm guessing that I should use the tank fluid density (or some weighted average if there is more than one kind of fluid layer in the tank). But what should I use at the top (where the boundary condition is the pressure of the atmosphere)? When I divide atmospheric pressure by the density of air, I get a value that is higher than the bottom boundary of the tank and flow goes the wrong way (down, not up to the outlet). When I divide atmospheric pressure by the same density that I use for the bottom boundary (the density of the fluid in the tank), my output shows pressure values that are lower than atmospheric. It seems like I should get this, but I don't. Any help would be appreciated, but unfortunately, I'm not allowed to post the code.</p>

|fluid-mechanics|

<p>The reason you're dividing the pressure's by the density is that in the naiver stokes equations that describe fluid flow have three types of terms: inertia terms that are multiplied by the fluid density, pressure gradient terms, and viscous terms that are multiplied by the fluid viscosity. If you divide the whole equation by the fluid density then you can combine your absolute viscosity and your fluid density to get the kinematic density. Then you only have one fluid property to worry about, which makes solving equations easier, but you have to divide your pressure by your fluid density.</p> <p>This means that the density you want to divide by is the density of the fluid you're simulating flow for. If you're simulating multiple fluid densities in the same model you cannot use this simplification. However, it seems like you aren't simulating the air, so you can safely just use the fluid density.</p> <p>One other thing to note is that it's only pressure gradients that matter in incompressible flow, so often times people will use gauge pressure (absolute pressure minus one atmosphere) as it doesn't make a difference in the flow results and remembering to add the one atmosphere everywhere is annoying. So if the pressure at your inlet is in gauge pressure than the pressure at the top of your tank should be 0 not one 1 atm.</p>

2768

Conceptual problem: how to define pressure boundary conditions for stratified fluid in units of pressure/density?

2015-05-11T07:32:39.047

<p>I've recently discovered a very awesome software tool for the synthesis and analysis of mechanisms. The tool is called GIM, from Faculty of Engineering in Bilbao, Department of Mechanical Engineering, University of the Basque Country (UPV/EHU) "The software is intended for educational purposes, in particular destined to the field of kinematic analysis, motion simulation and synthesis of planar mechanisms" as well as for the "static analysis of mechanical structures." I'm doing some personal dynamic art work with six bar linkages (not for profit) so I think I'm okay here. </p> <ul> <li><a href="http://www.ehu.eus/compmech/software/" rel="nofollow noreferrer">Link to download GIM software</a> </li> <li><a href="https://www.youtube.com/watch?v=0xMHisJR384" rel="nofollow noreferrer">Four Bar Tutorial Video</a></li> <li><a href="https://www.youtube.com/watch?v=2DkrrYgSLMc" rel="nofollow noreferrer">Synthesis Sample Video</a></li> </ul> <p>I will say the interface wasn't totally intuitive.. after a few minutes in the users guide I was able to figure out the double click thing. (Duh! Read the messages on the bottom bar.)</p> <p>I do have one issue that is driving me crazy. I'm actually trying to easily compare different design alternatives. I'd just like to display coupler point output path, but I can't figure out how to do that.</p> <p><img src="https://i.stack.imgur.com/09GIG.jpg" alt="four bar mechanism with motion paths displayed"></p> <p>Has anybody else used this tool? Anybody know the secret to just showing a single output path? </p> <p>Oh, and yes I'm aware of the <a href="http://www.saltire.com/HTML5/Mechanisms/4%20Bar%20Linkage.html" rel="nofollow noreferrer">www.saltire.com software</a> and that would work for displaying output in a nice format, but I won't have the synthesis and full analysis tools available. </p>

|mechanical-engineering|software|linkage|kinematics|

<p>Okay, it took me way too long to figure this out. I'm thinking this is just not a very intuitive interface. Here's the fix for anybody else who makes it here. Click #1, Click #2 (carefully!), then check the box at #3.</p> <p><img src="https://i.stack.imgur.com/sAwHL.jpg" alt="4 bar display output path"></p>

2771

GIM Mechanism Software (Kinematics / Analysis / Synthesis)

2015-05-11T20:59:23.277

<p>I want to affix a cantilever to wall. I will support the other end of the cantilever with a strut made of wood, that attaches to some point on the wall below the cantilever, as shown in this sketch (click for full resolution):</p> <p><a href="https://i.stack.imgur.com/z3Cj2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z3Cj2m.jpg" alt=""></a></p> <p>At what angle will the strut provide the greatest vertical strength/support for the free end of the cantilever?</p>

|structures|statics|stresses|wood|

<p>I had a school assessment and decided to look into this. I believe you have calculated the angle that results in the least stiff strut - not the strongest. I would like to think that a higher moment of inertia (I) is better as it reduces the effect that force has on the rotation of the strut - this contradicts your answer. Yet you are correct in saying that a reduction in E the modulus of elasticity is better for the structures strength. I believe that this demonstrates the fact that this problem requires a better solution somewhere between 54.7 and 0 degrees because we do not just want to minimise EI but find a balance. Here is my reasoning/approach to the problem: the theoretical most optimal angle is 0, as theta approaches 0 and the side lengths extend to infinity and simulate a bridge or 2 fixed points (most optimal). this does not take into account the material weight and hence is unreasonable. for anyone looking into this as-well I would think angles of 45 or less to be optimal. Please (logically) correct me if I am wrong.</p>

2780

What angle - for a strut - provides the greatest vertical strength/support for a cantilever?

2015-05-12T06:54:03.450

<p>If you look at the man-made objects around you, you will see that almost all of them are either rectangular, or parallel and perpendicular. There are a few circles here and there, but the overwhelming trend in basically everything humans make now is the rectangle. This is true from the computer user interface that you're looking at now and the typography in it all the way up to the shape of building blocks in your city, and if your city is completely planned, to the shape of large sections of it as well, and the fields around it. Even some administrative borders are also rectangular, which seems like the only geometric trend with these.</p> <p>The product and graphic designers apparently only follow this trend and make everything rectangular because that's what it looked like when they started to work with a particular medium. They didn't probably set the trend.</p> <p>You could argue that rectangles are easier to pack together, but triangles and hexagons are just as easy to pack, and there are numerous other shapes, polygonic or curved, that pack perfectly well.</p> <p>The concept of rectangles even applies to the third dimension and there most things are box like, our fridges, houses etc. </p> <p>If a list were to be prepared of the primary motivations for such rectangular practise, then what would top that list ?</p> <p>Moreover, the question also applies to the relevance of such motivations in the current age of 3D printing, and modular and adaptive architecture. </p>

|mechanical-engineering|structural-engineering|design|engineering-history|geometry|

<h3>Buildings</h3> <p>The primary reason is gravity. When humans first started building things structures that were perpendicular to the ground stayed up more. Stacking rocks for example, the straighter you could get it the more stable it would be, therefore the more 'square' the rocks you used, the easier it would be to get it straight. This has not changed, while it's possible to build walls at an angle, these need to be stronger and have more support. Actually if you build a long enough 'rectangular' structure, the top wouldn't be flat, it would bend with the curvature of the Earth. However, at scales we deal with on a day-to-day basis, the easiest shape to build with is rectangular.</p> <p>Triangular structures can also be strong, but you get less usable space inside the structure for the same amount of material used.</p> <h3>Human devices</h3> <p>Gravity comes into it again, though indirectly. Also geometry plays a big role in what shape our human devices are.</p> <p>Due to gravity, there are lots of things to see on the surface of Earth, such as trees, rocks, wild animals, food, etc. but not all that much up in the air and downwards there is just the ground. As we evolved, it was more important to see things to the left and right than up and down, such as predators or potential food. So we evolved two eyes, arranged horizontally, giving us a better field of vision in the horizontal plane than the vertical one. Our arms and hands also evolved to allow us to manipulate things horizontally more than vertically. This means if you're looking at something or interacting with it (such as your monitor and keyboard) it's much easier to do so in a horizontal direction.</p> <p>An arc shape (similar to the shape a car windscreen wiper makes) might actually be more ideal than a rectangle for interacting with, but it is more difficult to manufacture, so ease of manufacturing also comes into it. We do have curved interfaces, but generally when we need to pack lots of things into a small yet accessible space, such as a car dashboard, or an air-traffic-controller's desk, where function is more important than simplicity.</p> <p>Rectangles are much less universal in smaller human devices. Circles can also be simple to make, for example mugs/vases/bowls on a potters wheel, table-legs on a lathe. This is where geometry comes in. If you want to enclose something, for example hold tea in a mug, using the same amount of material you can hold more with a circular shape than a square one. However if the purpose is to spread things out, to make them accessible (for example words on a page or items on a table) moving them away from each other at right angles will get them further apart for a shorter move-distance. So the function plays a much bigger role than physical viability for smaller items.</p> <p>Design for aesthetics does of course also affect things, but they need to remain functional otherwise they would become useless, so this limits what can be done in this regard and guides the shape of things around us.</p>

2787

Why are so many things around us rectilinear?

2015-05-12T11:45:01.943

<p>I'm trying to get my bearings (a somewhat heavy-handed expression) on pump design and I'm finding in the various books I've procured on the subject that there is very little consideration given to the output pressure developed by a pump.</p> <p>If there isn't much detail to this question, I apologize; I'm still in the far-below-basic level of understanding. Obviously, head is a key parameter of pump design, but my concerns are with output pressure. The reason being, I'm considering for example rocket engine design where the key parameters of pump requirements are chamber pressure and flow rate. I understand that flow rate varies with head by the characteristic Q-H curves.</p> <p>So, how does pump head vary with or affect the output pressure developed by a pump? Are the two concepts related, or are they independent of one another? If they are related, how would I go about figuring out the required head parameters from the pressure and flow requirements?</p>

|mechanical-engineering|fluid-mechanics|pumps|

<p>Head and pressure are effectively the same thing.</p> <p>It's impossible to look at a pump and say "this pump will provide 10 bar of pressure." One of my favorite aspects of fluids has always been how interconnected an entire fluid system is, and how it stabilizes itself on its own. So the pressure that's actually supplied depends on the rest of the system: the losses in the pipe, the inlet and outlet conditions of the fluid to the system, the fluid itself, and how much power you're supplying to the pump. </p> <p>Hopefully you've come across the Bernoulli equation somewhere in your reading.</p> <p>$$P+{\rho}gz+{\frac12}{\rho}v^2=constant$$</p> <p>(Remember that this can only be used for incompressible flows, but since we're talking about pumps and probably only common liquids at relatively low speeds, the incompressible assumption is fine. It only becomes an issue when $Mach&gt;0.3$.)</p> <p>In the form above, we see that the terms together will output a constant in units of pressure. However, we can rewrite to be in the form</p> <p>$$\frac{P}{\rho g}+z+\frac{v^2}{2g}=constant$$</p> <p>In this case, the constant has units of length (more specifically, units of length of a particular fluid, e.g. mmHg), and it's referred to as the head. But this term is fundamentally no different from the constant we get in the first equation, it's just expressed differently. $H=P_{total}/{\rho g}$, but because this is an imcompressible flow, the density is constant, and we know that $g$ won't change significantly unless we're moving the fluid over a massive height differential. </p>

2792

Output pressure of centrifugal pump

2015-05-12T13:18:28.760

<p>I am working with a very simple 3-DOF damped LTI spring-mass system. As an exercise I am altering two parameters simultaneously (stiffness and damping coefficients connecting mass 1 to mass 2) to see which combination minimizes:</p> <ul> <li>Settling time</li> <li>Overshoot</li> </ul> <p>The objective is to determine which parameter has the greater effect on the performance, either the stiffness or the damping. </p> <p>My question is: How can I determine if changing stiffness or damping is having more of an effect on the aforementioned criteria ? Is there a way to quantify how they contribute to the performance ?</p>

|dynamics|springs|vibration|stiffness|

<h1>Equations of Motion</h1> <p>From what I understand from the question and your comments, you can write the equations of motion of the 3-dof oscillator in the form:</p> <p>$$M \frac{d^{2}x}{dt^{2}} + C \frac{dx}{dt} + K x = f$$</p> <p>where the vector $x$ represents the displacements, $M$, $C$ and $K$ are the mass, damping and stiffness matrix respectively, and $f$ is some forcing vector (e.g. impact, step function, etc.). Therefore, this type of equation may be classified as a Linear Time-Invariant (LTI) system.</p> <h1>Damping models</h1> <p>There are several ways to solve this type of LTI systems, for instance numerical time integration, and as mentioned in the response by @am304, different models to represent the damping matrix $C$.</p> <p>Rayleigh damping is indeed one form:</p> <p>$$C = aM + bK$$</p> <p>where parameters $a$ and $b$ are chosen arbitrarily. You could also use modal damping, which is similar and may be written in the form:</p> <p>$$V' C V= 2 x_iW$$</p> <p>where $W$ is a diagonal matrix containing the linear eigenvalues of your system, $V$ the associated eigenvectors and $x_i$ is the damping coefficient (which may vary from mode to mode but can be considered as a single constant as a first approximation). The main advantage of this method is that the damping carries with it the inherent dynamics of your linear system and is diagonal in the modal domain.</p> <h1>Performance</h1> <p>In terms of performance, with respect to the overshoot and the settling time, they will be both dependent on the damping term as well as the stiffness, since the two quantities are related in the two modeling strategies here proposed. However, for a given stiffness matrix, the solution will reach a steady-state with a specific amplitude, and the overshoot and settling time will solely depend on $x_i$ as shown in the picture below (Last mass displacements for: $x_i=0.2$ (blue), $x_i=0.5$ (red) and $x_i=0.8$ (black)).</p> <p><img src="https://i.stack.imgur.com/nTSen.jpg" alt="Last mass displacements for: xi=0.2 (blue), xi=0.5 (red) and xi=0.8 (black)"></p>

2796

Dominant parameters in the dynamical response of a linear oscillator

2015-05-12T16:00:54.030

<p>I want to fix a wire rope horizontally between two trees with a T-connector in the middle of the wire rope to hang a tire swing from.</p> <p>Something like this:</p> <pre><code>tree-------wire rope------???-----------------------tree I I I I I I I I I I I I I I I I tire swing I I I I I </code></pre> <p>I looked for a T-shaped connector but have not found a solution.</p> <p>What kind of connection can I use to hang the tire swing from the middle of the wire rope? I need something that will not deteriorate during the use of the swing.</p>

|mechanical-engineering|

<p>A guy on youtube does this professionally. <a href="https://www.youtube.com/watch?v=jWVL6EmBaZ4&amp;ab_channel=PatrickBrandt" rel="nofollow noreferrer">https://www.youtube.com/watch?v=jWVL6EmBaZ4&amp;ab_channel=PatrickBrandt</a> IMO rope breaks over time and I prefer extra strength for my kids. He uses EHS 5/16 guy wire [11k breaking strength] straight through the tree. Arborists do it all the time to shore up trees. I am about to do this in my yard. Ordering from here. <a href="https://www.westechrigging.com/" rel="nofollow noreferrer">https://www.westechrigging.com/</a></p>

2798

How can I make a tee connection on a wire rope?

2015-05-12T19:30:03.300

<p>I don't understand how to use the information in a stepper motor datasheet to understand how much torque it can generate. I was going to do a simple board and try to lift a small weight as an example but I realize I need to know how much torque the motor can generate. I'll probably have a little spool out there to hold the string for the weight, which probably also changes how much torque it can deliver.</p> <p>All I see in this <a href="http://www.mercurymotion.com/products/zlbjmd/st-24byj.pdf" rel="noreferrer">datasheet</a> is "in traction torque" and I'm not sure what that means. I don't see any graphs or other information that relate how much input voltage/power will give me how much torque. What happens if I run it at 5 V instead of 12 V, for instance, or if I current-limit it to 1 mA?</p> <p>I'm just trying to learn what to look for and how to read the datasheet properly.</p>

|mechanical-engineering|electrical-engineering|linear-motors|

<p>@StainlessSteelRat linked to a very good resource for <a href="http://www.allaboutcircuits.com/vol_2/chpt_13/5.html" rel="nofollow">stepper motors at All About Circuits</a>, but I fear he didn't address your questions. I'll go through your question line by line.</p> <blockquote> <p>I'll probably have a little spool out there to hold the string for the weight so that probably changes how much torque it can deliver too.</p> </blockquote> <p>First, the sentence above is wrong. The radius of the spool will not change the torque. It will change the weight you can lift, but only because $Fr=T$, where $F$ is the force available for lifting, $T$ is the torque output, and $r$ is the radius of the spool. BTW, this method works fairly well, I have done it myself.</p> <blockquote> <p>Anyway all I see in this datasheet is "in traction torque", and I'm not sure what that means</p> </blockquote> <p>Based on the torque curve from All About Circuits, it is probably almost equal to the holding torque. As you see from that curve, torque is fairly constant at the low end of motor speed. All About Circuits also mentions that in stepper applications, the speed of the stepper motor should be gradually accelerated.</p> <blockquote> <p>What happens if I run it at 5V instead of 12V for instance. Or if I current limited it to 1mA.</p> </blockquote> <p>If you run stepper motors at a higher voltage (within rated limits), the current, and therefore holding torque, will go up. Another way to look at it is that you can run the motor faster for the same torque. I don't have any equations here, but increased current leads to increased force in electromagnets. If you current limit the solenoid, the holding torque will go down.</p> <p>One last note: @am304 makes an important note about half/quarter stepping. The torque will reduce for half/quarter stepping because the magnets in the stepper motor are acting in opposing directions, reducing the net torque.</p> <p>Sorry I don't have any equations, but experiments are fun right?</p>

2802

Reading stepper motor datasheets to get torque and speed

2015-05-12T23:40:40.827

<p>The term Systems Engineer has always interested me as it generally involves many different fields of engineering.</p> <p><strong>Systems Engineer - Interdisciplinary</strong></p> <p>The definition I'm familiar with is usually defined as an interdisciplinary profession of engineering, where the engineer has experience in a number of fields and uses this during the design of a system.</p> <p><strong>Systems Engineer - IT related</strong></p> <p>After searching for jobs here in Australia, I've found many systems engineer positions. However almost all of them are relating to IT jobs, referring to the programming of computer systems at the system level.</p> <p><strong>Question</strong></p> <ol> <li>Why is there a conflict of these two definitions?</li> <li>Which one was first?</li> <li>And is Systems Engineer no longer valid in the interdisciplinary sense?</li> </ol> <p><strong>Notes</strong></p> <p>Please forgive any mistakes and assumptions I have made, I'm new and ready to learn.</p>

|engineering-history|terminology|systems-engineering|

<p>sys·tem /ˈsistəm/ </p> <p>noun: a set of connected things or parts forming a complex whole, in particular.</p> <p>a set of things working together as parts of a mechanism or an interconnecting network.</p> <p>plural noun: systems "the state railroad system"</p> <p>synonyms: structure, organization, arrangement, complex, network; </p> <p>informal: setup "a system of canals" </p> <p>en·gi·neer /ˌenjəˈnir/ </p> <p>noun: engineer; plural noun: engineers</p> <ol> <li>a person who designs, builds, or maintains engines, machines, or public works.</li> </ol> <p>synonyms: originator, deviser, designer, architect, inventor, developer, creator; mastermind </p> <p>"the prime engineer of the approach" </p> <p>verb: engineer; 3rd person present: engineers; past tense: engineered; past participle: engineered; gerund or present participle: engineering</p> <ol> <li>design and build (a machine or structure).</li> </ol> <p>"the men who engineered the tunnel"</p> <p>Thus, a Systems Engineer is a person who designs, builds, or maintains a set of connected things or parts forming a complex whole. Doesn't matter if that complex whole is a rocket ship, an airplane, a computer network, or a waste treatment facility. The Systems Engineer is the guy that makes things work. </p> <p>Think of the classic "glass half full, glass half empty" test for optimism/pessimism. Any engineer can tell you the problem isn't half full or half empty. The problem is that the volume of the glass is twice a much as it needs to be to efficiently contain the liquid. The Systems Engineer is the guy who comes up with a way to successfully reduce the volume of the glass so it works properly, with maximum efficiency. </p> <p>While the liquid is still in it. </p> <p>And while people (customers) continue to drink out of it. </p> <p>Which is why "Freakin' Awesome" should be a perfectly acceptable job title. </p> <p>LEM</p>

2807

Is systems engineer an IT profession or an interdisciplinary field of engineering?

2015-05-13T04:59:24.827

<p>I wish to calculate air leakage in a piston cylinder arrangement. One side is pressurized and the other is at atmospheric pressure. All steel construction. No piston rings or lubrication. The piston oscillates at about 50 to 80 times per sec, amplitude 5 mm, and is used as an indentation marking tool.</p> <p>I know the piston diameter, cylinder diameter, piston length, air pressure and roughness. I need to calculate the leakage in liters per minute to determine: <strong>If the clearance is increased, how much increase in piston length is required for the same leakage?</strong> Less clearance gives more air efficiency, but is difficult to manufacture.</p>

|mechanical-engineering|fluid-mechanics|pistons|

<p>If the piston is in (slight) clearance inside the cylinder, it's a bit like calculating the flow rate through a capillary, with the difference in diameter being the capillary diameter and the contact length between the piston and the cylinder the capillary length. There are various online calculators to help you with the actual calculation, such as <a href="http://www.tlv.com/global/TI/calculator/air-flow-rate-through-orifice.html" rel="nofollow noreferrer">this one</a>. If you're after the theory, it's based on <a href="http://en.wikipedia.org/wiki/Bernoulli%27s_principle" rel="nofollow noreferrer">Bernoulli's principle</a>.</p> <p><strong>EDIT</strong></p> <p>The length doesn't actually come into account in Bernoulli's equation, only the pressure and flow rate, if you neglect the difference in altitude. If you want to take account of the length, then you also need to look at the friction and things start to get messy. The most common approach is generally the <a href="http://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation" rel="nofollow noreferrer">Darcy-Weisbach equation</a>. It's expressed as pressure loss, but you can rearrange the equation to work out the fluid velocity and therefore the flow rate. Careful which friction factor you use, the Darcy factor is 4 times the so-called Fanning friction factor. For laminar flow, it is generally approximated as $\frac{64}{R_e}$ where $R_e$ is the Reynolds number.</p> <p>The air properties are also dependent on pressure and temperature, but assuming dry air at ambient temperature at standard atmospheric pressure (not exactly true, but it won't be far off) and a pressure drop of 1 bar between the piston chamber and the atmosphere, I get a flow rate of ~20.4 mL/min with a 1mm length and ~1.7mL/min with a 12mm length. I checked <em>a posteriori</em> that the assumption of laminar flow was valid.</p> <p><strong>Darcy-Weisbach equation</strong></p> <p>Pressure loss form:</p> <p>$\Delta P = f_D * \frac{L}{D} * \frac{\rho u^2}{2}$</p> <p>Re-arrange to give flow rate, given that $Q = A * u$</p> <p>$Q = A * \sqrt{\frac{2}{\rho} * \frac{\Delta P * D}{f_D * L}}$</p> <p><strong>Friction factor</strong></p> <p>Assume flow is laminar (need to check <em>a posteriori</em>):</p> <p>$f_D = \frac{64}{R_e} = 64 * \frac{\nu A}{QD}$</p> <p>which gives (using $\nu = \mu / \rho$):</p> <p>$Q = \frac{A * \Delta P * D^2}{L} * \frac{1}{32 \mu}$</p> <p><strong>Numerical application</strong></p> <pre><code>D = 0.04e-3; % [m], diameter L = 1e-3; % [m], length dP = 1e5; % [Pa], pressure difference mu = 1.846e-5; % [kg/(m*s)], dynamic viscosity of dry air at 1atm and 300K rho = 1.2922; % [kg/m^3], air density at standard conditions for P and T A = pi*D^2/4; % [m^2], cross area Q = A*dP*D^2/(L*32*mu); % [m^3/s], flow rate Re = Q*D/(A*(mu/rho)); % [-], Reynolds number </code></pre> <p><img src="https://i.stack.imgur.com/PAEi3.png" alt="Flow rate"></p> <p>This gives a flow rate of ~20.4 ml/min and a Reynolds number of ~758, which is less than 2000, so the assumption of laminar flow is valid.</p> <p>Change the length to 12mm, and you get a flow rate of ~1.7 ml/min and a Reynolds number of ~63.</p>

2812

Calculating leakage in an all steel piston cylinder arrangement

2015-05-13T08:59:46.440

<p>I am trying to understand the difference in the definition of &quot;fault tolerance&quot; and &quot;robustness&quot; when it comes to system design.</p> <p>For instance, what would be the main design considerations or what would be the definition of the terms when looking at engineering problems? In particular, how do those terms apply to mechanical or electrical power engineering?</p> <p>I found some definitions but they were centered on computational engineering. The document didn't give much insight in how to define those terms in other engineering disciplines.</p> <blockquote> <p>Robustness: The capability to cope with unknown errors during execution and provide the system services all the same.</p> <p>Fault-tolerance: The capability to tolerate a certain set of errors defined during the development of a system.</p> </blockquote> <p><a href="http://ti.tuwien.ac.at/ecs/teaching/courses/hwswcode_vu_WS_2011/protocols/9-fault-tolerance" rel="noreferrer">Source</a></p> <p><strong>What would be the difference between a fault tolerant and robust design?<br /> Can a hard distinction be made?</strong></p> <p><strong>Which techniques or design principles are applied to ensure a fault tolerant design, and which are applied to ensure a robust design?</strong></p>

|mechanical-engineering|electrical-engineering|mechanical-failure|

<p><strong>Robustness</strong> (of a system): The system keeps its expected performance (not failing) even when the real operation conditions (external to the system) are not the same as the ones assumed in the design. The broader is the range around the expected conditions more robust is the system. As a simple example you can consider a car that was designed to run over relatively smooth asphalt and you happen to use the car in a very poorly maintained road. The more robust is the car design, less will be the impact of the poorly maintained road on the performance (for example comfort of passengers) or expectation of failure (kilometers run before breaking) of the car.</p> <p><strong>Fault tolerant</strong> (system): The system keeps the expected performance (not failing) even when one of its components (which are internal to the system) fail. For this analysis is important to know the criticality of each component and also the likelihood of failure of that component.</p> <p>To make a system fail-tolerant you make each component less critical (one plane with more than one engine and that don't need all engines working in order to fly is fault-tolerant on engine failures). A plane with just one engine is not fault-tolerant on the regard of engine failures.</p> <p>If it is not practical to make the system fail-tolerant on regard of a specific component, you make that component more robust, so the likelihood of that component to fail is reduced (which means that component can cope a wide range of operating conditions when you look at it as a system).</p>

2814

What is the difference between fault tolerance and robustness of a system?

2015-05-13T14:33:36.993

<p>Most modern aircraft engines, such as the one depicted below taken from <a href="http://en.wikipedia.org/wiki/Turbofan" rel="nofollow noreferrer">Wikipedia</a>, are composed of several compressor stages which are driven by a turbine (or several), and a combustion chamber in-between, in order to increase the temperature of the flow.</p> <p><a href="http://commons.wikimedia.org/wiki/File:Turbofan_operation.svg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MRrIB.png" alt="Modern Turbofan Sketch"></a></p> <p>In general, manufacturers and designers focus on increasing compression ratios as well as combustion temperatures for efficiency enhancement.</p> <p>My question is, under simplifying assumptions such as perfect gas, no energy losses or friction, and constant inlet temperature and velocity: How is the efficiency of this thermodynamic cycle evaluated? How can one quantify the efficiency gain from a pressure or temperature increase?</p>

|mechanical-engineering|thermodynamics|aerospace-engineering|aircraft-design|

<p>Gas turbines are modeled using Brayton cycle which in the simplest case consits of:</p> <ol> <li>Isentropic compression (in a compressor) </li> <li>Constant-pressure heat addition (combustion chamber) </li> <li>Isentropic expansion (in a turbine)</li> <li>Constant pressure heat rejection <img src="https://i.stack.imgur.com/KT5G5.png" alt="enter image description here"></li> </ol> <p>And since efficiency is defined as Net output work / Heat input, Efficiency can be easily related to temperature of cycle states as follows:</p> <p><img src="https://i.stack.imgur.com/fAqiQ.png" alt="enter image description here"></p> <p>Processes 1-2 and 3-4 are isentropic, and P2 = P3 and P4 = P1. Thus:</p> <p><img src="https://i.stack.imgur.com/sRRqJ.png" alt="enter image description here"></p> <p>And finally efficiency can then be related to compression ratio as follows:</p> <p><img src="https://i.stack.imgur.com/MEQN6.png" alt="enter image description here"></p> <p>However most gas turbines do not operate on this simple theoretical ideal conditions as isentropic compression&amp; expansion, constant pressure heat addition, single stage compression and single stage expansion. And in such cases the modeling and efficiency analysis are far more complex than the ideal cycle.</p> <h2>Abbreviations:</h2> <ul> <li>$w_{net}$: mechanical net work, Turbine power minus compressor power</li> <li>$q$: heat, addition or rejection. $q_{in}$ = heat addition and $q_{out}$ = heat rejection </li> <li>$k$: specific heat ratio $Cp/Cv$</li> </ul>

2821

Efficiency in a gas turbine or aircraft engine

2015-05-14T02:56:45.873

<p>This is a continuation of me trying to understand torque and stepper motors in my other <a href="https://engineering.stackexchange.com/questions/2802/reading-stepper-motor-datasheets-to-get-torque-and-speed">question</a>. I'm trying to understand the torque a motor would be required to generate to lift a small weight, and the formulas involved.</p> <hr /> <p>The first part of my question is to verify if I am calculating this correctly:</p> <p>Let's say I have a 450 g mass (roughly one pound) then the force of gravity pulling it down is:</p> <p><span class="math-container">$\begin{align} F &amp;= ma \\ &amp;= 0.450 \:\mathrm{kg} * 9.8 \:\mathrm{m}/\mathrm{s}^2 \\ &amp;= 4.41 \:\mathrm{N} \\ \end{align}$</span></p> <p>If I have a stepper motor with a spindle for my string that pulls up my motor with a radius of 5 cm. I think my torque needed would be:</p> <p><span class="math-container">$\begin{align} T &amp;= Fr \\ &amp;= F * 0.05 \\ &amp;= 0.22 \:\mathrm{Nm} \\ \end{align}$</span></p> <p>So now if I want to move that mass I need to find a stepper motor that can output more than 0.22 Nm of torque, right?</p> <hr /> <p>The follow-on to my question is that if I want to see how fast I can move it then I need to look at a Torque speed curve, right?</p> <p>My confusion is this: do I have to ensure that I'm moving slow enough to get the torque I need, or does that curve say if you need this torque you won't be able to go above this speed because the motor won't let you?</p>

|mechanical-engineering|motors|torque|stepper-motor|

<p>You have the right concept, but slipped a decimal point. 5 cm = 0.05 m. The gravitational force on your 450 g mass is 4.4 N as you say, so the torque just to keep up with gravity is (4.4 N)(0.05 m) = 0.22 Nm.</p> <p>However, that is the absolute minimum torque just to keep the system in steady state. It leaves nothing for actually accellerating the mass and for overcoming the inevitable friction.</p> <p>To get the real torque required, you have to specify how fast you want to be able to accellerate this mass upwards. For example, let's say you need at least 3 m/s&sup2;. Solving Newton's law of F = ma:</p> <p>(0.450 kg)(3 m/s&sup2;) = 1.35 N</p> <p>That, in addition to the 4.4 N just to balance gravity means you need 5.8 N upwards force. At 0.05 m radius, that comes out to a torque of 0.28 Nm. There will be some friction and you want some margin, so in this example a 0.5 Nm motor would do it.</p> <p>Note also that torque isn't the only criterion for a motor. Power is another important one. For that you have to decide what's the fastest speed you want to be able to pull the mass upwards at. Let's say 2 m/s for sake of example. From above, we know the highest upwards force is 5.8 N.</p> <p>(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W</p> <p>After accounting for some losses due to friction and leaving a little margin, the motor should be rated for about 15 W minimum.</p>

2826

Understanding required torque for a motor lifting a weight

2015-05-15T02:44:28.287

<p>This is a follow up question to my question <a href="https://engineering.stackexchange.com/questions/2826/understanding-required-torque-for-a-motor-lifting-a-weight/2829#2829">here</a> about torque and stepper motors.</p> <p>There Olin explains:</p> <blockquote> <p>Note also that torque isn't the only criterion for a motor. Power is another important one. For that you have to decide what's the fastest speed you want to be able to pull the mass upwards at. Let's say 2 m/s for sake of example. From above, we know the highest upwards force is 5.8 N.</p> <p>(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W</p> <p>After accounting for some losses due to friction and leaving a little margin, the motor should be rated for about 15 W minimum.</p> </blockquote> <p>I'm unclear about what kind of Watts we're talking about here, is this just Watts in the mechanical sense or in the electrical sense. In other words should I be thinking about needing to deliver 15W as in 15V @ 1A? </p> <p>I'm also wondering about controlling the amount of power delivered to the motor. In steady state I guess it would be Input Voltage divided by motor coil resistance? But to change the position where the current and voltage are changing the current must be the input voltage / (reactance of the coil plus the coil resistance). But the reactance changes with frequency... so do I have to look at the rise time of the input voltage to understand the frequency involved and then try to calculate from there.</p> <p>I'm a little lost on that point, but trying to read more and understand.</p>

|mechanical-engineering|motors|stepper-motor|

<p>In the text you quoted, I was talking about mechanical power out of the motor. This is what the power rating of a motor, given no other qualifications, means. A "15 W motor" therefore can put out 15 W of mechanical power under the right conditions.</p> <p>No motor is 100% efficient, so more electrical power than that will need to be supplied to the motor. For example, if the motor is 85% efficient at a particular operating point where it produces 15 W of mechanical power, it will require (15 W)/85% = 17.7 W of electrical power. The remaining 2.7 W that doesn't come out in the form of mechanical power will be dissipated as heat.</p> <p>Your electrical model of the a motor isn't quite right. Think of it as a resistor in series with a voltage source. Yes, coils are in fact inductive, but the time constant of the inductance with the resistance in the system is usually far less than the mechanical time constant and so isn't relevant.</p> <p>The resistance is fixed and is the actual DC resistance of the coils. The voltage source is a function of the motor's speed, and always <i>opposes</i> the applied voltage when the motor is spinning in the direction that it is being pushed by the applied voltage. This voltage source represents the motor acting as a generator.</p> <p>The torque produced by the motor is proportional to the current thru it.</p> <p>When you first apply a voltage to a stantionary motor, the voltage source is 0 and all the applied voltage is across the resistance. This causes a large current to flow, called the <i>stall current</i>. As the motor speeds up, the voltage source opposes some of the applied voltage. That leaves less votlage across the resitor, which reduces the current, which reduces the torque. Eventually a balance is reached where the current drops to the point where it produces just enough torque to keep the motor going at that speed.</p> <p>For a unloaded motor, little torque is required to keep it going, and the internal voltage source opposes most of the applied voltage. There is little voltage left across the resistor, so the current drawn by the motor is much lower than the stall current. If the motor is mechanically loaded, then a higher torque is required to keep it going, so the steady state speed will be lower, the internal voltage source lower, more voltage across the resistance, and the current higher.</p>

2836

Power and motor current

2015-05-16T00:42:28.253

<p>I'm designing a solenoid valve for an automated irrigation project that I'm working on. After having looked at some designs, I have a question.</p> <p><img src="https://i.ytimg.com/vi/SwqM8zpmAD8/0.jpg" alt="example design"></p> <p>For most of the designs I've seen, the orifice is smaller than the pipe diameter. Wouldn't this reduce the flow rate? If this is the case and if I'm using such a design, do I make that opening bigger?</p>

|fluid-mechanics|pressure|valves|

<p>The total flow resistance is the sum of its parts. Pipes are long, valves are short. Making a pipe wider is relatively inexpensive, and making a valve wider is relatively expensive. As you make the opening wider, everything about the valve has to be beefed up, as the forces involved will scale along with the size of the opening. So on balance, youd always expect valve openings to be smaller relative to pipe dimensions, in a cost optimized system.</p>

2841

Solenoid valve pressure drop

2015-05-17T00:38:37.407

<p><strong>tl;dr:</strong> After an extended conversation with an old-timer, I realized a few things:</p> <ol> <li>The single most valuable measurement for the majority of people will be water-depth-in-well.</li> <li>The second most valuable will be water-flow-from-well.</li> <li>The &quot;bubbler&quot; solution discussed below has another major weakness (in addition to the frailty of air pumps): the introduction of oxygen into the well water will cause oxide formation, leading to mineral encrustation of not only the opening of the tubing, but extending all the way up inside to wherever its normal level would be. He knows because he's had to deal with something almost exactly analogous and it was a major hurdle. Larger size tubing will slow down the process, but eventually the tubing will be blocked.</li> <li>We are reexamining the solution that uses a bladder-in-tank with differential pressure sensor. He had specific ideas about how to do this that sound doable (but there are still some details to be dealt with).</li> <li>Oh, and he solved the tank problem in about 10 seconds. Put a pressure sensor on the pipe from the tank to the pressure pump. Ignore the spikes that happen when the pump kicks in, and we have exactly the pressure reading we want with cheap, well-understood sensors. Sheesh! It was so obvious once he said it I almost kicked myself.</li> </ol> <p>I thank all of you for your ideas and your analysis. If anyone is interesting in seeing how the project unfolds, keep an eye on <a href="http://waterunderground.net/" rel="noreferrer">waterunderground.net</a>. It's pretty empty at the moment, but should have more content in a month or so.</p> <h3>Backstory</h3> <p>I am designing an open-sourced well &amp; water-usage monitoring system for people in Northern California. The goal is to be able to measure water flow from well-to-tank, tank-to-house, and tank-to-irrigation, plus monitor the water depth in the tank and the well. Our current target parts cost is under \$200 for a system including CPU, 3 flow sensors, and 2 pressure sensors, although we think we may be able to get it closer to \$100 after a few design iterations.</p> <p>We appear to have the flow sensor portion solved now that we finally have a supplier of Female G1 =&gt; U.S. 1&quot; slip adapters to integrate cheap Hall effect sensors into a standard U.S. piping environment. The depth measurement solution is not so straightforward.</p> <p>I'm asking for a sanity check on my reasoning here before I go off and start buying stuff that is wrong, either in size, type, or altogether.</p> <h3>Problem statement</h3> <p>I need a <em>low-cost</em> way to measure the depth of 2 columns of water with moderately decent accuracy, say +/- 5%. Although our own property is site Alpha 1, we would like a solution that scales up, or down, for other properties with similar needs.</p> <p>We have:</p> <ol> <li>A 3,000 gal storage tank that is approx. 8.5' of water when full. Other tanks are of similar height +/- 5'.</li> <li>A water well. Our own well is 75' deep w/ 37' of water. Other wells in the area are as shallow as 30' w/ 15' of water, or as deep as 300' w/ 70+' of water.</li> </ol> <p>We have the following criteria:</p> <ol> <li>No more than \$30 for the tank and (hopefully) no more than \$50 for the well. Lower costs would be great.</li> <li>Solution must integrate <em>in some manner</em> (handwave) with an Arduino, BeagleBone Black, or similar low-cost controller.</li> <li>A continuous readout is desirable, but something that triggers every 15, 30, or &lt;whatever&gt; minutes would be acceptable.</li> <li>No electronics/electrical systems <em>in</em> the well or tank.</li> <li>No metal in the well or tank, with the possible exception of material used to weigh down the tubing that goes into the water.</li> <li>The solution should work reasonably well (no pun intended) for wells from 35' deep w/ 15' of water, up to wells 300' deep w/60+' of water.</li> </ol> <p>Amongst several solutions considered so far, our current front-runner is a &quot;bubbler&quot;, as described in <a href="http://www.sensorsmag.com/sensors/leak-level/a-dozen-ways-measure-fluid-level-and-how-they-work-1067" rel="noreferrer">this article</a>:</p> <blockquote> <p>A bubbler-type level sensor is shown in Figure 3. A dip tube having its open end near the vessel bottom carries a purge gas (typically air, although an inert gas such as dry nitrogen may be used when there is danger of contamination of or an oxidative reaction with the process fluid) into the tank. As gas flows down to the dip tube's outlet, the pressure in the tube rises until it overcomes the hydrostatic pressure produced by the liquid level at the outlet. That pressure equals the process fluid's density multiplied by its depth from the end of the dip tube to the surface and is monitored by a pressure transducer connected to the tube.</p> </blockquote> <p>We are planning on using:</p> <ol> <li>A 1/4&quot; to 3/8&quot; open-ended tube weighted down (or better yet, zip-tied to the well's up-pipe) to hang a short distance above the bottom (we can get closer in the tank, but wells tend to silt up so that will be within a couple of feet). The small down-tube is a strong point in favor of this approach because almost nothing is going into the well itself.</li> <li>Some (cheap) source of air pressure sufficient (300+ kPa) to blow all of the water out of the tube in the well. Once the value from the sensor plateaus it means we're blowing bubbles and we can convert pressure to feet of water.</li> <li>At the top we tee the tube into a differential pressure sensor, such as the <a href="http://cache.freescale.com/files/sensors/doc/data_sheet/MPX5500.pdf" rel="noreferrer">Freescale MPX5500DP</a>, which can handle up to 500 kPa, which translates to approx. 160' of water. They have a slightly more accurate one (the 5100 series) for shorter columns, such as in the tank. We selected the differential sensor to allow for varying atmospheric pressure.</li> <li>The specifics of the Arduino turning the air pump on/off have not been decided, but I believe it will be straightforward once we know what kind/size of pump we are trying to control.</li> </ol> <p>Note: although we can easily calibrate the reading from the tank sensor, the well may be more problematic. In our own case we have a way to use a drop-line to directly measure the well depth and water column height, in other cases this may be difficult.</p> <h3>Questions</h3> <ul> <li>Is there anything about this approach that is fundamentally flawed?</li> <li>Will temperature changes (primarily in the tank, not so much in the well) make any real difference here?</li> <li>Other than the volume of air needed for different diameters of tubing, will a pump have to work harder to achieve a given pressure if we use a larger or smaller down-tube?</li> </ul> <h3>Update to answer questions:</h3> <p>User null asked if there was unnecessary redundancy in the system; wouldn't just the depth in the tank be sufficient? Not really. Each of the measurements gives us some information the others do not. Although there is some overlap in what is being measured, I see that as an opportunity for a sanity check on the system.</p> <p>For example, if the measured flow from the well does not have a fairly close correlation (shifted in time because of the tank) with the combined flows to the house and irrigation system, then <em>something</em> is out of whack.</p> <p>Combining the flow-from-well chart with the well-water-depth chart can give critical information about the well's <a href="http://en.wikipedia.org/wiki/Groundwater_recharge" rel="noreferrer"><em>recharge rate</em></a>. If recharge is dropping off, then we have some <em>serious</em> trouble coming toward us.</p> <p>Finally, if our well-water-depth is dropping <em>and we aren't using that much water</em> then it could mean that one of our neighbors, say the the 300 acre vineyard about 1/2 mile up the hill, is over-pumping. Unfortunately, California is the only state without any regulation of below-ground water, so we can't stop them, only get ready to order a 3,500 gallon load of water for $175 a pop.</p>

|civil-engineering|fluid-mechanics|sensors|pumps|water-resources|

<p>RE: for your well:</p> <p>In geotech, piezometers measure water depth in monitoring boreholes.</p> <p>Here's a professional sensor: <a href="https://www.geokon.com/4500-Series" rel="nofollow noreferrer">https://www.geokon.com/4500-Series</a></p> <p><a href="https://i.stack.imgur.com/3FCMK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3FCMK.png" alt="piezometer" /></a></p> <p>Here are possible DIY discussions:</p> <p><a href="https://www.envirodiy.org/topic/monitoring-well-or-piezometer-water-level-sensor/" rel="nofollow noreferrer">https://www.envirodiy.org/topic/monitoring-well-or-piezometer-water-level-sensor/</a> <a href="https://i.stack.imgur.com/iPaal.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iPaal.png" alt="DIY" /></a> <a href="https://www.envirodiy.org/construction-of-water-level-monitoring-sensor-station/" rel="nofollow noreferrer">https://www.envirodiy.org/construction-of-water-level-monitoring-sensor-station/</a> <a href="https://i.stack.imgur.com/tDuHw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tDuHw.png" alt="DIY" /></a></p>

2852

Low cost, moderately accurate water depth measurement

2015-05-18T03:38:48.707

<p>I once meditated on the fact that a car, in downhill, would consume more fuel when in neutral to keep the engine on, rather than in gear. To keep the engine on when in gear, in fact, you mostly need <em>the gravity</em> and enough thrust to keep it rotating, and therefore no fuel (just motor oil).</p> <p>This is what I'd expect cars to do but I'm probably missing something. How's actually the story? And if it's not like this, would/could my idea work somehow? (note: patent pending :P)</p>

|automotive-engineering|fuel-economy|

<p>The answer depends a great deal on the control system around the engine, and how fast you are going down the hill relative to the gear you are in.</p> <p>For a basic engine system, like 1960s or earlier, at idle throttle setting the fuel usage would go up a bit with engine speed. The higher engine speed would make a higher vacuum, which would pull in more fuel-air mixture. If the downhill speed and gearing come out to the same engine speed as idle, then the fuel usage would be the same as idle. If you used a higher gear so that the engine would go slower, then it might use a little less fuel. However, there isn't a lot of engine speed room below idle. You run the risk of the engine bucking, which is not a good idea to subject the power train to.</p> <p>To meet late 1960s pollution limits, some cars were equipped with a "decel valve". This sort of effectively stepped on the gas for you when the engine was undergoing rapid deceleration, or was externally driven (rapid deceleration is being "externally" driven from the momentum of the engine). The reason was that this condition with low fuel input would cause more pollutants to be emitted. The short term expedient was to give it more fuel. In such a case, running the engine faster than idle due to coasting down a hill in a sufficiently low gear would definitely use more fuel than idling in neutral down the hill.</p> <p>With more modern control systems, it's hard to know. Some cars detect this condition and effectively shut down the engine. In that case, you are better off with the engine engaged than in neutral. For example, my Honda Civic hybrid routinely shuts down the engine and re-starts it as the conditions dictate. Going down a hill uses no fuel at all, and actually charges the battery a bit depending on speed and gear. Many modern cars are also capable of shutting off some of the cylinders under light load. In that case it again would take less fuel to keep the engine engaged than to let it idle.</p>

2859

While going downhill, does my car consume less fuel when in neutral or in gear?

2015-05-19T01:51:04.870

<p>If I take a large thin plate and I uniformly drill a very large number of holes (diameter "d") very close together (minimum spacing "s"), what will be the stiffness of this new plate? When I say "stiffness" I mean the derivative of average macroscopic displacement with respect to stress. With no holes it would be Young's modulus.</p> <p><a href="http://en.wikipedia.org/wiki/Circle_packing#Uniform_packings">Here</a> is a way to visualize how the drills are to positioned, i.e. close packed circles with space in between.</p> <p>This is the question I want to answer. Of course, the stiffness will depend on which direction the stress is applied - for example, if I am pulling along the "grain" it will be stiffer than if I turn the plate 45 degrees and pull.</p> <p>I thought about doing some sort of analytic model (found a few papers but they are hard to understand). Or a finite element model. However, I am interested in how the parameters "d" and "s" (see above) change the stiffness for each type of circle packing pattern. I am stuck now, what would you recommend as the best way to approach this problem?</p> <p>Also, if this problem has already been solved (wouldn't be surprised) and someone could point me to a reference that would work too.</p>

|mechanical-engineering|stiffness|elastic-modulus|

<p>A simple first order approach would be to treat the plate like a composite material, with the holes acting as a medium with no modulus. The <a href="https://en.wikipedia.org/wiki/Rule_of_mixtures" rel="nofollow">rule of mixtures</a> , treating the "holes" as fibers with 0 modulus, would yield a modulus of 0. So, the <a href="http://www.doitpoms.ac.uk/tlplib/fibre_composites/stiffness.php" rel="nofollow">Semi-Emperical Halpin Tsai</a> would be better:</p> <p>$$ \eta = \frac{\frac{E_f}{E_m} -1}{\frac{E_f}{E_m}+1} $$</p> <p>$$ E_c = \frac{E_m(1+\eta f)}{(1-\eta f)}$$</p> <p>Since $ E_f = 0 $, we have $ \eta = -1 $, so:</p> <p>$$ E_c = E_m(\frac{1-f}{1+f}) $$</p> <p>In the case of the square packing arrangement, the circles occupy 78.54% of the area. So, the "composite" would be ~12% of the original modulus. Again, this would be a first order approximation to save you from running 50 finite element models. Then, run your FEA to watch out for the stress concentrators for final design.</p>

2865

Elasticity of a drilled plate

2015-05-19T12:02:10.323

<p>If you have a look at the car's wheels, you'll notice that they have holes which can be of different forms (mostly circular or rectangular). </p> <p>Why do they have such holes? Doesn't that reduce the stiffness of the wheels? </p> <p><img src="https://i.stack.imgur.com/XjBCgm.jpg" alt="enter image description here"></p>

|mechanical-engineering|automotive-engineering|wheels|

<p>Many of the answers so far have mentioned that part of the purpose of the holes is weight reduction, but most of them don't express why weight reduction in the wheels is important. There are two major reasons; the first (also mentioned by Steve Ives) is that the suspension systems in vehicles operate better if the 'unsprung' mass is kept as low as possible, and the second (not mentioned so far) is that <strong>shaving weight from the wheels contributes more significantly to performance than shaving weight from the rest of the vehicle</strong>. </p> <p>To see why this is true, consider the energy that the engine must put into the vehicle to get it moving at a speed $v$: $$ E=\frac12 m_tv^2+\frac12 I\omega^2 $$ For the wheels we can express the angular velocity in terms of the linear velocity as $\omega=\frac{v}{r}$ where $r$ is the radius of the wheel. The moment of inertia of the wheel can be expressed as $I=\eta\ m_dr^2$ where $m_d$ is the mass of the wheel. Here $\eta$ is a number between $\frac12$ and 1, but is closer to 1 since a wheel is closer to a <a href="http://en.wikipedia.org/wiki/List_of_moments_of_inertia">thin hoop than a solid disk</a>. Sticking all of this back in gives $$ \begin{align} E&amp;=\frac12 m_tv^2+\frac12\eta\ m_dr^2\frac{v^2}{r^2}\\ &amp;=\frac12 (m_s+m_d)v^2+\frac12\eta\ m_dv^2\\ &amp;=\frac12[m_s+(1+\eta)m_d]v^2, \end{align} $$ where I have used $m_s$ as the non-rotating mass of the vehicle. So, you can see that <strong>shaving mass from the wheels is equivalent to shaving a factor of $1+\eta\simeq 2$ as much mass from the non-rotating parts of the car</strong>.</p> <hr> <p>There is an additional, relatively minor, effect due to angular momentum for which it is advantageous to reduce the weight of the wheels. Due to conservation of angular momentum, the body of the car will tend to roll toward the outside of a turn when the wheels are rotated to initiate the turn. Reducing the moment of inertia of the wheels reduces their angular momentum and thereby reduces the amount of body roll upon steering.</p>

2873

Why do car wheels have holes?

2015-05-20T04:58:27.930

<p>In my engineering class we must make a machine that can sort 15 marbles with 3 different types in less than 2 minutes. I plan on having one marble go over/under a light sensor with a flashlight on the opposite side so it can read its reflection, then have a gear with three cups spin according to what the light sensor value is. </p> <p>Do you think that the light sensor will be able to read the value and spin in less than 1 second? Or should I put something to stop the marble while it is being read?</p>

|electrical-engineering|sensors|optics|robotics|

<p>A sensor's switching frequency will usually be in the order of milliseconds, so it definitely won't be a problem. However the highest delay you will have in the system is turning to the right cup to dispense the last marble that was sensed. Attach a two bit absolute encoder on the spinning gear to always find the shortest path to the right cup. </p> <p>As for the marble feeder you can use a vertical tube stack of the marbles to be sensed. Two gates can be used along the tube feeder, one that will dispense the marbles one by one from first to last to a chamber for them to be sensed, and a second one to dispense it in the corresponding cup.</p> <p>This task should be a fast one to solve using this setup. Hope this helps!</p>

2884

How fast can a VEX robotics light sensor operate?

2015-05-21T09:03:46.213

<p>A neat image <a href="http://en.wikipedia.org/wiki/British_Rail_Mark_3#/media/File:BR_Mk.IIIa_TSO_No.12604_%288074749189%29.jpg" rel="noreferrer">from Wikipedia</a>:</p> <p><img src="https://i.stack.imgur.com/c8JHcl.jpg" alt="railway carriage side looks neat"></p> <p>Here the railway carriage side looks like it is crafted from a single sheet of metal (if we ignore the doors and the windows and some minor parts like that small thing near the carriage number) - no seams, no rivets, no anything just a 20+ meters wide and about 3 meters tall single sheet of metal.</p> <p>Is it indeed possible and reasonable to craft a carriage side of such large single sheet or should I expect that there're neatly hidden seams somewhere?</p>

|materials|rail|metals|

<blockquote> <p>Is it indeed possible</p> </blockquote> <p>Yes</p> <blockquote> <p>and reasonable to craft a carriage side of such large single sheet</p> </blockquote> <p>It can be reasonable, but typically isn't. To form an entire side of the carriage from one sheet of steel would require a forming press that is gargantuan. However, one could instead have three more reasonably sized presses and then weld the sections together. Further, this could be more flexible - changing out the press plates is easier and faster, storing smaller press plates is cheaper and easier, if the machine breaks down you can still use the other presses, whereas if one piece of the gargantuan press breaks down, the entire operation might be waiting for the repair.</p> <p>There are a whole host of issues that follow the same reasoning - if a panel is damaged, it's expensive and time consuming to rework, and affects a much larger part of the carriage than a single damaged panel would. The pieces are much harder to work with, move, and fasten, etc, etc.</p> <p>So it's very unlikely that it's a single seamless sheet of steel or aluminum.</p> <blockquote> <p>or should I expect that there're neatly hidden seams somewhere?</p> </blockquote> <p>You should expect seams, and in fact, as <a href="https://engineering.stackexchange.com/questions/2895/is-this-railway-carriage-side-likely-made-of-single-metal-sheet#comment4955_2895">ratchet freak pointed out</a>:</p> <blockquote> <p>"The bodyshell is [...]of full monocoque construction with an all-welded mild steel stressed skin,"</p> </blockquote> <p>Monocoque is a technique where the skin or surface of the construction serves an integral support purpose. </p> <p>The seams, therefore, may also provide support as ribs, and so you may find that having seams provides an added benefit, or perhaps reduces the need for internal supports.</p> <p>The seams, of course, are not visible for both aerodynamic and visual appeal purposes, and it's relatively easy to hide weld lines with grinding, sanding, and polishing.</p> <p>So without actual proof of manufacturing process, I believe you can safely assume smaller sheets with invisible seams are more likely than single seamless formed sheets.</p>

2895

Is this railway carriage side likely made of single metal sheet?

2015-05-22T11:47:17.743

<p>I have noticed when examining performance of various heating elements in different atmospheres that in a vacuum maximum temperatures are lower than the in air temperatures. Why is this?</p> <p>(Typical heating element alloys are Nickel-Chromium and Iron-Chromium-Aluminum)</p> <hr> <p>As Olin Lathrop says below the evaporation of the element becomes significantly increased in a vacuum. I found this interesting quote in an old book:</p> <blockquote> <p>This idea of working in a vacuum brought with it new difficulties as regards the construction of a furnace. The ordinary platinum resistance furnace proved of little value, as, owing to diminution of pressure, the volatility of the platinum foil, at high temperatures, was increased to such an extent that it was destroyed in a short time.</p> </blockquote>

|thermodynamics|temperature|

<p>One reason is because in vacuum there is no convection to help remove the heat. All heat loss from the element is either by radiation, or via conduction thru the supports and wires. Technically this only means lower maximum heater power, but that may be dumbed down to temperature.</p> <p>Another reason is due to evaporation of the heating element. At high enough temperatures, this can matter. With 1 atm ambient pressure, evaporation is greatly reduced at low partial pressures, like heater element metals have at glowing temperatures. In vacuum, even a little pressure is greater than ambient (because ambient is theoretically 0), so there can be significantly faster evaporation, reducing the life of the heating element.</p>

2904

Why do heating elements have lower rated temperatures in vacuums?

2015-05-22T12:54:26.960

<p>I have just begun studying mechanics of materials and I am struggling to understand intuitively <strong>how to select the area</strong> in first moment of area calculations. I was hoping someone has a relatively easy explanation.</p> <p>The problem arises when calculating the shear stress <span class="math-container">$\tau$</span> at a specific point in a beam due to a specified shear force. For <span class="math-container">$\tau_{xy}$</span> the calculations seem to be the same: <img src="https://i.stack.imgur.com/5eI5E.jpg" alt="Example Problem" /></p> <p><span class="math-container">$\tau_{xy}$</span> due to <span class="math-container">$V(x)$</span> at point A requires the calculation of the first moment of area, <span class="math-container">$Q$</span>, shaded here: <img src="https://i.stack.imgur.com/FxZFBm.jpg" alt="Area 1" /></p> <p>However, then the problem requires me to find <span class="math-container">$\tau_{xz}$</span> due to <span class="math-container">$V(x)$</span> at point B. The shaded area under is the area we have to use, and my question is why?</p> <p><img src="https://i.stack.imgur.com/tAXdZm.jpg" alt="Area 2" /></p> <p>I know this is probably a very banal question for you guys, but I just really want to understand this, and browsing the web has led me nowhere.</p>

|mechanical-engineering|materials|beam|stresses|

<p>You're looking for $\tau_{xz}$, which pertains to horizontal shear flow along the top flange. (Rather than the vertical shear flow considered in $\tau_{xy}$.) As Mark noted, the shear flow starts from the center of the flange and flows outward/down.</p> <p><a href="https://i.stack.imgur.com/2HHJI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2HHJI.png" alt="Shear Flow"></a></p> <p>The formulation for the shear stress calculation remains essentially the same. Only the area changes -- we're simply cutting out a different section, and because of the tube shape we have to make two cuts to remove our section. (Versus and I-shape where only one cut is required.)</p> <p><a href="https://i.stack.imgur.com/wrpdx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wrpdx.png" alt="Section Cuts"></a></p> <p>This <a href="http://kisi.deu.edu.tr/mehmet.aktas/Dersnotlari/6.pdf" rel="nofollow noreferrer">presentation</a> has a pretty good explanation of shear stress.</p>

2905

How do you select the correct area to consiser for first moment of area calculations?

2015-05-23T14:01:24.447

<p>System with compressed air seems more expensive and harder to build and integrate neither using simple electromagnetic drives to open public transport doors.</p>

|transportation|motors|compressed-air|

<p>Do you mean, for example, a bus entry door? It may be as simple as the "service factor" for motors ("electromagnetic devices") as compared to pneumatics. Inexpensive motors with sintered bronze bushings wear out, and better ones with actual bearings must be routinely lubricated. Air-operated cylinders are cheap and last a very long time, with the air typically containing traces of added oil. </p> <p>The two are really not all that different, and this has been a hot-button topic for a long time. Electrical systems require a generator/alternator, regulator, wiring, control device, and actuator. Pneumatic systems require a compressor, tank/oiler, tubing, valve, and piston. Electrical systems (wiring) can short or open; pneumatic systems can plug or leak.</p> <p><a href="http://www.automationworld.com/motion-control-systems/how-decide-between-pneumatic-and-electric-actuators">This article</a> gives some pros and cons between the two, with interesting cost-analysis.</p> <p><strong>Electro-Mechanical</strong>:</p> <ul> <li>Provide precise control and positioning.</li> <li>Help adapt machines to flexible processes.</li> <li>Low operating cost.</li> <li>Most economical when deployed in a moderate scale in processes where their performance advantages can be a benefit and when the electronics are separate from the actuator to segment and minimize replacement costs.</li> <li>Stepper motors are an economical choice for accurate positioning at lower speeds. However, steppers may lose synchronization with the controller when employed open loop without an encoder or if they are undersized for an application. </li> <li>Servos, by definition, are closed loop and provide superior performance at high speeds, albeit at a higher cost.</li> <li>Components of an electric actuator include the mechanical actuator that translates motor rotation to linear speed and thrust, the motor, an electronic driver or amplifier to power the motor, and a controller to control motion. The total cost for these components ranges from 800 USD to 3,000 USD and up.</li> <li>While component costs of electric actuators are high, operating costs are low. High component costs often deter the use of electric actuators because savings in operating costs compared to pneumatics are often not adequately considered or are outright ignored.</li> </ul> <p><strong>Pneumatic Cylinders</strong>:</p> <ul> <li>Provide more force and speed per unit size than any other actuator except hydraulic. </li> <li>Force and speed are easily adjustable and are independent of each other.</li> <li>Are most economical when the scale of deployment matches the capacity of the compressor.</li> <li>Small compressors are efficient and economical when used for a small number of devices.</li> <li>Low component costs.</li> <li>High operating costs (install, replacement cylinders, electricity for compressor (76%))</li> </ul> <p>Added: In addition to the points @ratchet freak has made, mechanical break-away (limit) systems are possible for motors; however they are seldom used. Typically the motor is sized to generate torque below a threshold value as to not cause harm. However, two issues arise: 1. This is a mechanical linkage, and would not be as repeatable, reliable, or self-correcting as a (pressure regulated) pneumatic system. And 2. If the motor failed, the actuator is usually locked due to mechanical gearing (thus relying on break-away linkage.)</p> <p>It is possible to implement a <a href="https://www.parker.com/portal/site/PARKER/menuitem.7100150cebe5bbc2d6806710237ad1ca/?vgnextoid=f5c9b5bbec622110VgnVCM10000032a71dacRCRD&amp;vgnextfmt=EN&amp;vgnextdiv=&amp;vgnextcatid=4491073&amp;vgnextcat=POWERROD+LINEAR+MOTOR+CYLINDER&amp;Wtky=#">cylinder as a linear motor</a>; essentially a solenoid with two or more coils. But it would be heavy, use lots of copper wire, and use an elaborate drive system (especially if to limit maximum thrust.)</p>

2917

Why does public transport uses compressed air to open doors instead of electromotors?

2015-05-23T15:44:27.637

<p><strong>Background</strong></p> <p>For a washing process after a crystallization we want to know how much filter area we would need and if said area will work for our fixed volume flow. </p> <p><strong>Calculation</strong></p> <p>$\dot{V_F}$ has a given value</p> <p>Velocity $u$ should be the superficial velocity and is at a fixed value we found in our mechanical engineering script.</p> <p>With $\frac{\dot{V_F}}{u} = A$ we can calculate our needed filter area. Question is, will that area work for our volume flow. </p> <p>With $\dot{V_F} = \frac{A \cdot \Delta p}{\eta_F(\alpha_w \cdot L )+ f_M}$</p> <p>we can then calculate $L$.</p> <p>$\Delta p = \Delta p_{filter medium} + \Delta p_{filter cake}$</p> <p>$\Delta p_{filter cake} = \frac{(1-\epsilon)^2}{\epsilon^3 \cdot x^2_{32}} \cdot 150 \cdot L \cdot \eta_F \cdot u$</p> <p>Now we insert $L$ in the equation for $\Delta p_{filter cake} $ and we get the exact $\Delta p_{filter cake}$ that we originally assumed when calculating $L$. Actually we expect to get a different value in order to find all variables via iteration.</p> <p>Somewhere we made a mistake in our assumptions I think, but I can't see where right now.</p> <p>I know that $\dot{V_F}$ is a function of $L$ however we can't change $\dot{V_F}$. And we insert it anyway in our equation for $L$ so I think that assumption is valid.</p> <p>English is not my mother tongue, so please excuse any spelling / grammar mistakes. Furthermore this is not a given tasks, but a problem we stumbled upon during our group project (designing a plant).</p>

|fluid-mechanics|chemical-engineering|process-engineering|fluid-filtration|

<p>Your math seems to be sound. It looks like you calculated everything correctly up to where you calculated $L$. At that point you should have all of the variables solved for and there should be no need to refine your numbers. </p>

2920

Filter Area for given flow rate

2015-05-23T19:54:09.733

<p><img src="https://i.stack.imgur.com/iUTmzm.jpg" alt="image from the measurement result"></p> <p>I know the current should be higher than 0.67 mA, because I know phones use current in the order of 1A. Also the charger is able to support up to 1A, it doesn't make sense that the charger would have such a superior current than the phone will draw. </p> <p>I connected the multi-meter in series with the phone charger. That was done by stripping the red wire then cutting it, and then I held the 2 rods of the multi meter at both ends. The phone indicated that it was charging.</p> <p>Since the phone indicated that it was charging and the battery was not full, I would take out the possibility that the phone disconnected the charger automatically. Also, I'm not sure about the idle state of the phone. The phone was turned on, and it has about a 3 inch screen. The battery is 1800 mAh. It still doesn't make sense that even in "idle" state it will draw such a small current while charging. Please let me know if I need to provide any more information about the measuring method, I didn't elaborate much on it because I thought it was simple, but I totally understand the possibility of a small mistake.</p> <p>Here's another picture of the way of measurement, the reading is 0.45mA (from a different phone).</p> <p><img src="https://i.stack.imgur.com/rxM3b.jpg" alt="2nd measurement picture"></p>

|electrical-engineering|measurements|rf-electronics|

<p>It seems like the Monsoon Power Monitor might be the right tool for the job of determining current draw. This would give you very accurate current readings in real time.</p>

2924

How much current is this phone drawing?

2015-05-26T00:51:54.240

<p>I'm working on a project that uses a strong stepper motor to power a wheel that needs to be able to "coast" freely at times. The motor is too stiff when not engaged, and along with my gearing ratio, makes it impossible to get the free-spinning coast** that I want from the wheel when the motor is not powered. I only need the wheel to turn in a single direction, so I've been playing around with a design (see image) where the motor slides just a bit to engage via a small set of gear tracks to pull itself into the other gear. I was thinking that I would simply just reverse direction when I wanted to pull the motor away to enable free-spinning. Unfortunately, it doesn't work so well. The gear does engage, but the little bit of play it has when it slides causes it to make contact with the very last tooth of the tracks at the same time. This causes it to make a grinding noise, and certainly isn't efficient.</p> <p>Does anyone know how I could get this to do what I want? I've been searching online, but haven't found anything and I am not sure of how to technically describe what I'm trying to do in a few search terms. So, I'd be grateful for any links to similar designs as well.</p> <p>** Sorry, I forgot to add an important detail that will explain why I need to find a way to disengage the motor... The motor will turn the wheel forward, but when disengaged, the wheel needs to be able to turn freely in either direction.</p> <p><img src="https://i.stack.imgur.com/35K2j.png" alt="image"></p> <p>So, it's two days later, and this is what I'm going to try next:</p> <p><img src="https://i.stack.imgur.com/RniiF.png" alt="image"></p> <p>It consists of:</p> <p>A drive gear (in red) that turns counterclockwise, with a one-way clutch bearing mounted to it. The clutch bearing turns freely clockwise.</p> <p>There is an arm with an intermediate gear that will make contact with the large gear when engaged.</p> <p>There is a spring attached to the arm. The spring pulls the arm down, to engage the gear. The important thing is that the spring is NOT strong enough to overcome the resistance of the idle stepper motor. So the arm will not move unless assisted by the stepper motor.</p> <p>So here's how I hope this will work, assuming we start with it not engaged:</p> <p>The stepper motor turns on, counterclockwise. The movement, in addition to the spring, engages the intermediate gear. The clutch bearing prevents the arm from dis-engaging the intermediate gear, yet allows the stepper motor to turn the gears freely.</p> <p>To disengage, we simply reverse direction of the stepper motor just a few degrees. The clutch bearing will lock, pulling the arm away.</p> <p>I have yet to prototype this, as my 3D printer is down for a few days. But I would be interested in feedback and whether anyone thinks this will work.</p>

|gears|stepper-motor|

<p>You can use a SPRAG type clutch from Altramotion.com </p> <p>or a backstop that would spin the opposite direction which could act like a engagement gear. </p> <p>But i still believe you can get a all in one unit from them for a SPRAG clutch that will be useful for you. </p>

2934

How can I get a stepper motor to engage with a freely spinning wheel?

2015-05-26T06:40:26.810

<p>I am working on modeling concrete fibers (metallic fibers) as a mathematical model. My work is for my thesis. I am a PhD student of numerical analysis, but I am working on a real tunnel project. </p> <p>I have had issues with the distribution of fibers in the concrete. I am trying to find a way to develop a stochastic differential equation for the fibers.</p> <p>I have the following questions: </p> <ol> <li>Is there any mathematical model for concrete fiber? (not statistical model) </li> <li>Is there any technical information available for how concrete fibers behave?</li> </ol>

|civil-engineering|materials|modeling|concrete|

<p><a href="https://engineering.stackexchange.com/questions/4182/how-do-i-calculate-an-estimate-for-the-properties-of-a-composite-material">Related - how do i calculate an estimate for the properties of a composite material</a></p> <p>The reference to <a href="https://www.library.ucdavis.edu/wp-content/uploads/2017/03/HDBK17-3F.pdf" rel="nofollow noreferrer">Mil Handbook 17F</a>, p. 213 is summarized here:</p> <p><a href="https://i.stack.imgur.com/J7Rbi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J7Rbi.png" alt="enter image description here" /></a></p> <blockquote> <p>Computation of effective elastic moduli is a very difficult problem in elasticity theory and only a few simple models permit exact analysis. One type of model consists of periodic arrays of identical circular fibers, e.g., square periodic arrays or hexagonal periodic arrays ... These models are analyzed by numerical finite difference or finite element procedures. Note that the square array is not a suitable model for the majority of Uni-Directional Composites since it is not transversely isotropic.</p> <p>The composite cylinder assemblage (CCA) model permits exact analytical determination of effective elastic moduli ... Consider a collection of composite cylinders, each with a circular fiber core and a concentric matrix shell. The size of the cylinders may vary but the ratio of core radius to shell radius is held constant. Then...</p> </blockquote> <p><a href="https://i.stack.imgur.com/LOxId.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LOxId.png" alt="enter image description here" /></a></p> <p>(Where <span class="math-container">$V_f$</span> is the volume fraction of fibers to the total amount of material. <span class="math-container">$X_{m}$</span> is a property of the matrix, <span class="math-container">$X_{f}$</span> is a property of the fiber, and <span class="math-container">$E, G, k$</span> are the elastic modulus, shear modulus, and bulk modulus properties. The bulk modulus, k, can be computed for isotropic materials as <span class="math-container">$\frac{E}{2(1-\nu-2\nu^2)}$</span>, where <span class="math-container">$\nu$</span> is the Poisson's ratio. The G without a subscript is a typo, and should be replaced with <span class="math-container">$G_m$</span>)</p> <blockquote> <p>A preferred alternative is to use a method of approximation which has been called the Generalized Self Consistent Scheme (GSCS). According to this method, the stress and strain in any fiber is approximated by embedding a composite cylinder in the effective fiber composite material. The volume fractions of fiber and matrix in the composite cylinder are those of the entire composite. Such an analysis ... results in a quadratic equation for the shear modulus...</p> </blockquote> <p><a href="https://i.stack.imgur.com/kRVB9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kRVB9.png" alt="enter image description here" /></a></p> <p>The net algorithm is to compute the effective bulk modulus <span class="math-container">$k^*$</span>, <em>12</em> poisson's ratio <span class="math-container">$\nu_{12}^*$</span>, and young's modulus <span class="math-container">$E_{1}^*$</span> first, then use the quadratic formula listed to calculate the second shear modulus, <span class="math-container">$G_2^*$</span>. Using <span class="math-container">$G_2^*$</span>, <span class="math-container">$E_2^*$</span>, <span class="math-container">$\nu_{23}^*$</span>, and <span class="math-container">$G_1$</span> can be calculated. These are in the local coordinate system of the fiber. To translate to global coordinates:</p> <p><a href="https://i.stack.imgur.com/AHLti.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AHLti.png" alt="enter image description here" /></a></p> <p>We can then rotate the fiber to find the properties of the uni-directional composite to find the properties in an arbitrary direction:</p> <p><a href="https://i.stack.imgur.com/wkSOI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wkSOI.png" alt="enter image description here" /></a></p> <p>where Qbar is the rotated matrix, and Q is the original inverse matrix. For a stochastic model, the angle of the fiber and the volume fraction can be the inputs, and the outputs would be the resulting properties. Note that for a uniform random distribution, it is possible to integrate the Qbar matrix as theta varies from 0 to <span class="math-container">$2\pi$</span>, then divide by <span class="math-container">$2\pi$</span> to obtain a symmetrical matrix. The results from this method match well with data on random fiber materials in the fiberglass industry.</p> <p>As you asked about a differential equation, we'd need to review the appropriate theory from this point on. For example, the classical plate equation, <span class="math-container">$$\nabla^2\nabla^2 = \frac{q}{D}$$</span>, works partly. We have to include another stoichastic variable, the height of the fiber inside of a block of concrete. The closer the fiber is to the top, the stiffer the block will be against the bending load. The block can be divided into arbitrary segments of uniform thickness, and the volume of the fibers in each segment is added, generating different Qbars. A different distribution would result in different properties of the block:</p> <p><a href="https://i.stack.imgur.com/6fBoD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6fBoD.png" alt="enter image description here" /></a></p> <p>This matrix, called the ABD matrix, would then redefine the plate equation as follows:</p> <p><span class="math-container">$$D_{11}\frac{\partial^4 w}{\partial x^4} + 2(D_{12}+2 D_{66})\frac{\partial^4 w}{\partial x^2 \partial y^2} + D_{22}\frac{\partial^4 w}{\partial y^4} = q(x,y)$$</span></p> <p>for the simplest of cases (B matrix irrelevant, no transverse loading, etc...). The cases go stranger from there, but can be derived from the original derivations, but stopping when the model says to assume the stress is proportional to the stain.</p>

2935

Developing a stochastic differential equation model for concrete fibers

2015-05-26T15:40:08.587

<p>I was trying to find the moment of area about point A on the following figure:</p> <p><img src="https://i.stack.imgur.com/mN8FS.jpg" alt="enter image description here"></p> <p>I prefer to use integration, since it can be extended to different shapes, and did this:</p> <p>$\displaystyle Q_z=\int_{-25}^{25}\int_{-23}^{35}y \, dy \, dz=-17400 \:\mathrm{mm}^3$</p> <p>However, the solution I have been given states that $Q_z=17400 \:\mathrm{mm}^3$</p> <p>This made me wonder about two things:</p> <ol> <li><p>What is the meaning of the sign (if there is any)?</p></li> <li><p>What have I done wrong in my calculation (aka "What am I not understanding here?")</p></li> </ol>

|mechanical-engineering|geometry|

<blockquote> <p>What is the meaning of the sign (if there is any)?</p> </blockquote> <p>There is no physical meaning associated with the negative sign in your result; it just means you got the math wrong.</p> <p>The integral definition of the first moment of area is $\iint_A {r(y,z) \, dA}$, or the sum over the entire cross-sectional area of the product of a differential area element and its perpendicular distance from the axis of interest.* Both area and distance in this sense are conceptually non-negative scalars, regardless of which side of the axis you're on, so your answer must always be non-negative as well.</p> <blockquote> <p>What have I done wrong in my calculation (aka "What am I not understanding here?")</p> </blockquote> <p>The sign of your answer is not the only problem.</p> <p>If you're going to go through the integration process, you must be careful to use the proper distance function $r(y,z)$, which in this case is the <a href="http://en.wikipedia.org/wiki/Euclidean_distance" rel="nofollow">Euclidean distance</a>.</p> <p>In the case where your axis of interest is parallel to one coordinate axis, your distance function is invariant with respect to that axis. If you define your coordinate axes as shown in the figure, with the origin at the centroid of area, then the coordinates of point $A$ are $(-23,20)$ and your distance function is simply the one-dimensional Euclidean distance between the differential area element with height $y$ and the height $y_A$ of the line parallel to the z-axis that contains the point $A$:</p> <p>$$r(y,z) = |y-y_A| = |y+23|$$</p> <p>Your integral is now:</p> <p>$$Q_z = \iint\limits_A |y+23| \,dA$$</p> <p>All that's left, recognizing that $dA=dy\,dz=dz\,dy$, is to determine the limits of integration. With the origin at the centroid, as in the figure, the limits of integration are the bounds of the area:</p> <p>$$Q_z = \int\limits_{-25}^{25} \int\limits_{-35}^{35} |y+23| \,dy\,dz$$</p> <p>It's hard to say <em>exactly</em> what your mistakes were without seeing your step-by-step calculations. Perhaps when you split up the absolute value into its piecewise definition, you forgot the second term? An alternative approach would be to redefine the origin at $A$, in which case your distance function would be nearly correct, but you'd have had to to recalculate the y-limits of integration.</p> <p>The answer will be the same no matter where you put the origin.</p> <hr> <p>_{* The equation I have written here is the general two-dimensional form, where the axis of interest is not necessarily parallel to either coordinate axis.}</p>

2940

First Moment of Area - Meaning of Sign/Sense

2015-05-27T14:29:06.947

<p>What is the minimum number of TET elements you need, in order to fully fill a cube(HEX element)?</p>

|finite-element-method|

<p><img src="https://i.stack.imgur.com/BmlXi.png" alt="5 tet elements"></p> <p><strong>Image 1</strong>: 5 tetrahedral elements can form a cube.</p> <p><a href="https://www.youtube.com/watch?v=XAhgw2Z4T2M" rel="nofollow noreferrer">This video</a> shows how to decompose a cube into five tetrahedra. Not saying it can't be done with less, but I can't figure out how :-)</p>

2949

Minimum number of tetra elements required, to represent a cube?

2015-05-27T16:11:54.837

<p>Each power plant has a load gradient characteristic, describing to what extent it can change the output of its generators in a given timeframe. Reading the literature, I know that, for example, a gas turbine power plant has a vastly higher load gradient than a one that uses coal as its primary energy.</p> <p>However, I cannot find a description which parts of the plant contribute the most to the load gradient. I can imagine that a gas-powered plant can change its output faster because there's no mill involved as in a coal power plant, but where does its gradient exactly come from?</p> <p>More specifically: What constitutes the load gradient of</p> <ul> <li>a gas turbuine power plant,</li> <li>a Lignite (brown coal) power plant</li> <li>a bituminous coal power plant</li> <li>a nuclear power plant?</li> </ul> <p>And why is there a difference between lignite and bituminous coal power plants?</p> <p>I'd very happy to read up on the details if one could name a book. I've read several books that describe how these plants work, but fail to mention what creates or limits their load gradient.</p>

|electrical-engineering|thermodynamics|power-engineering|

<p>A gas turbine plant basically uses jet engines optimized for shaft power instead of exhaust push to drive the generators. These can be controlled quickly because there is little stored energy. You reduce the fuel flow rate, and the turbine pretty much has to slow down quickly, within a few seconds, taking tens of seconds to reach the new steady state. The combustion chamber is small, and the little heat stored in its walls and the turbine blades is small compared to the overall power flowing thru the device.</p> <p>A large coal or oil fired plant has a large boiler with water/steam pipes running thru it. These things take some time to heat up and cool down. Then there is also a lag from less heat into the boiler before the less steam makes its way to the turbine. Pressures and flow rates also have to be adjusted so that the liquid to gas phase change continues to happen in the right region of the pipes in the boiler. Simply shutting off the fuel to the boiler abruptly can possibly cause damage. All this takes more time than for a jet engine to slow down or speed up due to a change in its fuel flow.</p>

2953

What limits a power plant's load gradient?

2015-05-28T21:40:02.537

<p>I am working on a compressed air energy storage system. The size and weight of the system are heavily constrained; nothing should (ideally) exceed a few pounds. For this reason, I would like to store the energy (compress the gas) and extract the energy (make the gas do work) with the same mechanism <strong>in a rotary fashion</strong>. </p> <p>Essentially what I need is a <strong>rotary</strong> air compressor that, when air is forced through in the opposite direction, doubles as a pneumatic motor. I'm working with fairly high pressures (I'm estimating a few hundred psi) but low volume. In my search, I have found a plethora of compact rotary air compressors and rotary pneumatic motors, but there is hardly comment on what systems would work as both. </p> <p>To me, it seems very intuitive that an air compressor could have these properties, but I don't want to jump to any conclusions. I have looked at several compressors, and the most applicable to my situation seem to be:</p> <ul> <li>Centrifugal compressor</li> <li>Axial flow compressor</li> <li>Rotary screw compressor</li> <li>Rotary vane compressor</li> </ul> <p>The centrifugal compressor is ideal, but it seems the least likely in my eyes to be reversible, at least with any efficiency. I also looked at pneumatic motors, of which there were fewer available. Most applicable seemed to be the:</p> <ul> <li>Rotary vane motor</li> </ul> <p>Other systems, such as the pietro motor, were obviously not applicable in my light weight, compact application. The correlation between the rotary vane compressor and the rotary vane motor is promising, but I would like to know about any options I have. </p> <p><strong>What rotary gas compression systems can double as motors powered by the gas they compress?</strong></p> <p><strong>EDIT</strong> The answer most likely lies in the similarity between a radial (centripetal) turbine and a centrifugal compressor. </p>

|mechanical-engineering|motors|compressed-air|energy-storage|compressed-gases|

<p>I'm not an expert on air compressors or motors, but from my limited knowledge I think as you say that the centrifugal compressor would be the best for compression and a Tesla Turbine would be ideal as the motor. I think it should be possible to mount them on the same shaft but in separate airtight chambers, with some valves such that when the turbine is in operation the air is pumped out of the compressor chamber, and vice versa, so as not to cause undue resistance from the other impeller. Alternatively a clutch/grip mechanism that selects which one turns with the shaft.</p> <p>Such a device could be considered a both-ways compressor/motor. To try to do it both ways with an impeller optimised for one of those scenarios seems like you will always be inefficient in the other.</p>

2966

Air compressor that doubles as a pneumatic motor?

2015-05-28T22:46:41.087

<p>What would the adhesive strength of chewing gum be? Would it be enough to support a car bumper? I'm more interested in how much tensile strength chewing gum has. I'm working on a project regarding the used cars movie, and in the movie, the salesman repaired a bumper that was unconnected with chewing gum.</p> <p>The bumper I'm thinking of is the one featured in the beginning of the movie, a '57 Chevy sedan. I want to see if it is even plausible for chewing gum to support a car bumper, so assume a fixed load of just the bumper. Since it was the '50s I would guess the bumper would be 25 pounds at least.</p>

|materials|automotive-engineering|

<h1>Short Answer</h1> <p>Yes it can hold. Because there are 10 brackets on a '57 Chevy front bumper and some more on rear bumper but not all of them are always under load conditions. A broken bracket or lost screw can be compensated with chewing gum.</p> <h1>Long Answer</h1> <p>I haven't seen an actual '57 Chevy Sedan however with some chewing gum can hold bumper for a time if its used for lost screws on side brackets. Because of the design philosophy it seems there are a lot redundancy in bumper brackets. Modern bumpers are expected to break easily but in 50's they were sturdier. </p> <h2>Front Bumper</h2> <p>Below is the assembly image I've found from <a href="http://www.trifive.com/forums/showthread.php?t=28049" rel="nofollow noreferrer">TriFive.com</a> <img src="https://i.stack.imgur.com/8pTd3.png" alt="&#39;57 Chevy Sedan Front Bumper Assembly"> </p> <p>As you see the side bumper screws are under shear only under acceleration. So a chewing gum will hold it.</p> <p>If brackets in the middle are loose some chewing gum can hold it for time again there are 10 brackets. But I'm not sure you can reach them without loosening sides. </p> <h2>Rear Bumper</h2> <p>Rear bumper assembly is from TriFive.com again. <img src="https://i.stack.imgur.com/UlSvp.png" alt="Rear bumper &#39;57 Chevy"></p> <p>If bumper sagged because of a few screws got loose or bracket legs are broken chewing gum can hold for some time. However again it won't hold for long. </p>

2967

Adhesive strength of chewing gum

2015-05-29T01:14:45.657

<p>I'm working on designing a Cold Spray system that uses a convergent-divergent nozzle to achieve a 800 m/s exhaust velocity. I'm assuming 400 psi nitrogen at room temperature is provided at the inlet of the nozzle. The throat area is 0.0859 square inches (tiny, I know). </p> <p>$$ \begin{align} \dot{m} &amp;= \rho\ v\ A \\ \rho &amp;= 31.920\ \frac{\text{kg}}{\text{m}^3} \text{(N}_2\text{ at 400 psi)},\\ v &amp;= 353.575\ \frac{\text{m}}{\text{s}} \text{(speed of sound at 400 psi)},\\ A &amp;= 5.54*10^{-5}\ \text{m}^2,\\ \Rightarrow\qquad\dot{m}&amp;=0.625\ \frac{\text{kg}}{\text{s}} \end{align} $$</p> <p>assuming $M = 1$ in the throat.</p> <p>This figure seems absurdly large. Can someone shed some light on the situation?</p>

|mechanical-engineering|aerospace-engineering|

<p>You're not that far off. In your analysis, you've just neglected the fact that the gas will accelerate as it passes through the converging nozzle. This is basically converting potential energy, in the form of pressure, into kinetic energy, velocity. As a result, the gas will no longer be at 400 psi when it reaches the throat.</p> <p>The compact form to compute mass flow through a choked nozzle is </p> <p>$$ \dot{m} = C_d A \sqrt{\gamma \rho P \left( \frac{2}{1+\gamma} \right)^{\frac{\gamma+1}{\gamma-1}}} $$</p> <p>where $\gamma$ is the ratio of specific heats (1.4 for nitrogen), and $C_d$ is an dimensionless coefficient that depends on the efficiency of the nozzle design (let's guess 0.5 for now).</p> <p>Applying the above, I got $0.18 \, kg\cdot s^{-1}$.</p> <p>As a sanity check I compared this result to manufacturer reported values for flow control nozzles that I use in my lab. When I scaled up their values for a $.0252 \, in.$ nozzle at $100 \, psig$ up by area and pressure, I got $0.15 \, kg\cdot s^{-1}$. Seems reasonable.</p>

2968

Convergent-Divergent Nozzle Design from Exhaust Velocity

2015-05-28T23:54:37.397

<p>I'm planning to build a laser shooting target. Something like this:</p> <p><a href="https://i.stack.imgur.com/RXQlM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RXQlMm.jpg" alt=""></a></p> <p>In the long term I want to build something advanced like a professional grade target, in which the user is able to tell exactly where was the shot, but for now I'd be happy with a basic one like in the image.</p> <p>Basically whenever a laser shot hits the black area it counts as a shot inside. I'm just unsure about how can I do it since I don't want it to easily confuse a shot with regular light, and if possible I would like to use optics side-effects to avoid cross-shooting (whenever one shoots someone else target, while misaligned with it).</p> <p>To begin with, one has to be 10 meters away from the laser target, with a variation of a meter at most. The targets are held at a meter and a half high. The laser of a gun, in conformance with the regulations, has a ~3 mW peak during 15-20 ms of activity. The black area is about 7 cm in diameter.</p> <p>I don't know the official specifications for a laser target in terms of light, but I found a brand with the specifications of <a href="http://www.apeom.cz/products/pentathlon/" rel="nofollow noreferrer">one target</a> (end of the page). It states "Sun resistance: 70k Lux." Does that mean that it is able to distinguish a laser shot from the ambient light at up to 70k lux ambient light? What is a simple way to understand the amount of light that represents?</p> <p>My idea so far its having a photo sensitive transistor in a PCB bolted to the back of the target. An aluminium sheet in the front with a circle in the middle of it, and a smoked acrylic sheet in its back (to fill the circle). And either a fresnel or a biconvex lens right next to the acrylic.</p> <p>Both the biconvex lens and the acrylic converge the frontal light to a point (the point where the photo transistor stays), so that spot becomes mostly light coming exactly from the front of the target, and becomes sort of immune to most ambient light (since it will refract to the points around the transistor).</p> <p>The problems I see so far are:</p> <ul> <li>Which one (fresnel/biconvex) has the shortest light converging point?<br> (I would prefer the target as slim as possible)</li> <li>If one shoots the edge of the valid area, it's not a straight shot. Assuming one is perfectly aligned with the target, and the weapon is at its height, it would already become a ~90.2° incision. Being a little misaligned (half meter) and being half a meter lower than the target while shooting the center would represent a ~90.4° incision. Being misaligned and shooting the edge opposed to the misalignment direction would be an even more tilted shot. Is that little tilt enough to make the lens focus out of the transistor? It seems like a really small deviation.</li> <li>Is it worth it to get a sheet of red cellophane between the acrylic and the transistor? It would filter the non-red light, wouldn't it?</li> <li>To avoid counting lights turning on or a cloud passing by as valid shots, can I can count 30 ms from the moment in which the light peak happens, and see if it has disappeared?</li> </ul> <p>So basically the main question is:</p> <p>Would either of these two lenses solve my problem, by converging the light coming from one direction into one point? Would it be practical? Are there better alternatives (like leaving the transistor alone without focusing; mirrors; etc.)?</p> <p><strong>EDIT:</strong> Floris guess was probably right. The signal seems to be modulated. I managed to have a shot at the spec sheet of one of the guns:<img src="https://i.stack.imgur.com/s8i2C.jpg" alt=""> Those graphs mean that with UIPM 2014 signal it pulses 10 times in 15.6 milisecs? Should I get my target also recognising those pulses? I bought a BH1750FVI light sensor module, can those kind of sensors recognise differences of light in fractions of miliseconds? <strong>EDIT2:</strong> Found the official regulations: <img src="https://i.stack.imgur.com/yyXN1.png" alt="Regulations"> <img src="https://i.stack.imgur.com/JAtLp.png" alt="Regulations"> <img src="https://i.stack.imgur.com/nKmRO.png" alt="Regulations"> <img src="https://i.stack.imgur.com/4aixX.png" alt="Regulations"></p> <p>Together with what was already answered, it should be enough to help anyone with the same question.</p>

|electrical-engineering|optics|photonics|

<p>Briefly:</p> <p>The way to distinguish your signal from any others (and incidentally get a massive gain in signal to noise ratio) is to use modulation. The simplest modulation scheme just turns the laser on and off very quickly (a few MHz) and your detection circuit (synchronized with the switching of the laser) ADDs detected signal when the laser is on, and SUBTRACTS when the laser is off. This is the principle of lock-in amplifiers. You can build a simple one with cheap CMOS switches, and they are incredibly effective. Any signal not at the "detected" frequency will be sometimes added and sometimes subtracted - so averaging over a few 1000 cycles it will be reduced to almost nothing.</p> <p>Filtering light from wavelengths you are not interested in helps as well - this is why your IR remote has a dark red window, for example (on the receiver as well...)</p> <h1>Further thoughts</h1> <p>Having read your question again, and in light of the comments below, here is a bit more detail.</p> <p>In general for a situation like you describe (where you want to detect a "signal" (the laser hit) in the presence of "noise" (the other guns, ambient light, the sun...) you have to look for the ways in which one signal is different from the others. This comes down to three things:</p> <ol> <li>Spatial difference</li> <li>Wavelength difference</li> <li>Temporal (modulation) difference</li> </ol> <p>Let's take each of these in turn. </p> <h2>Spatial difference</h2> <p>You know roughly where the gun is positioned that is pointing at the target. We can use that to eliminate "cross fire". Let's look at this figure:</p> <p><img src="https://i.stack.imgur.com/QcnUm.png" alt="enter image description here"></p> <p>As you can see - if you have an on-axis aperture near the focal plane of your lens, then light that comes in off-axis will be rejected. How big that aperture needs to be follows directly from the geometry of the target - shooter - neighbor configuration. You already said your shooter would be approximately 10 m ± 1 m from the target - and if you are trying to keep the target "thin" (maybe a focal length of 20 cm) the focal plane would be very close to the focusing distance of the lens: $\frac{10\cdot0.2}{10-0.2}=20.4 cm$.</p> <p>The angle subtended by the aperture at the center of the lens will be the acceptance angle - any lasers shooting outside that range will be blocked by the aperture:</p> <p><img src="https://i.stack.imgur.com/3slJU.png" alt="enter image description here"> </p> <p>Now for the choice of lens. A (plastic) Fresnel lens will be thinner, lighter and cheaper than another kind of lens - and sufficient for this application. They don't suffer much from spherical aberration, so it is possible to have a relatively large lens with a short-ish focal length. These may be hard to find, but <a href="http://www.teachersource.com/product/fresnel-lenses/microscopes-magnification" rel="nofollow noreferrer">I found a very cheap 7"x10" lens</a>. You could cut it to size and stack two or three on top of each other to give you the focal length you need (the shorter the focal length, the thinner the target; but otherwise there is not much reason to spend the money, certainly for a proof of principle). If your target area is only 38 mm, there is a cheap 50 mm biconvex lens available on the same site: <a href="http://www.teachersource.com/product/biconvex--biconcave-lens-sets/light-color" rel="nofollow noreferrer">http://www.teachersource.com/product/biconvex--biconcave-lens-sets/light-color</a> . Disclaimer: I am not associated with this site or the products.</p> <p>General note about lenses: fresnel is thinner (and often cheaper) than conventional; little to be gained in this application from plano-convex vs biconvex (although plano would allow the front of the target to be flat).</p> <h2>Wavelength difference</h2> <p>Lasers are "monochromatic", meaning their energy covers a very narrow range in the spectrum. This means that if you have the right filter, you can block a lot of "non-laser" light. The narrower the filter, the better the relative blocking. For example, you can get a <a href="http://www.omegafilters.com/products/filters/bandpass/635bp5.html" rel="nofollow noreferrer">635 nm filter with 5 nm FWHM</a> - that's centered on a "better" red laser pointer. </p> <p><img src="https://i.stack.imgur.com/x0PEs.png" alt="enter image description here"></p> <p>That will cut out an enormous amount of external light - if you imagine (for estimating) that sunlight is uniform between 400 and 800 nm, then cutting out everything that is not within 5 nm of the laser cuts the sunlight contribution by 80 x ($\frac{5}{400}$). It would be expensive - but shows it can be done. How well your "red acetate sheet" will work is something you would have to experiment with...</p> <p>An important reason to consider a filter is <em>saturation</em>: in the next step we will look at modulation, but that can only work if the detector is still linear over the range of intensities that are incident. If the detector (or electronics) are saturated, there will be no change in signal when the input intensity changes - so you would not be able to detect the small signal on top of the large one. By filtering "a good chunk" of the sunlight out, you make the system less sensitive - reduce the need for a massive dynamic range.</p> <h2>Modulation</h2> <p>The final point - and the one I started this post with - concerns modulation. In its simplest form, you can look for a sudden increase in intensity. For example, if you put a high-pass filter (series capacitor, shunt resistor) on the output of your photocell you would only see a pulse if the intensity suddenly increases. This would happen with a laser pulse, but not with sunlight. You could feed that signal into a comparator, and it would detect the pulse.</p> <p>More complex systems could use pulse modulation - you can learn about this, and get kits for very little money, from <a href="https://learn.sparkfun.com/tutorials/ir-communication" rel="nofollow noreferrer">https://learn.sparkfun.com/tutorials/ir-communication</a> . The problem with the transmitters on these systems is that throw a wide beam (so you don't have to "aim" at the TV). Replacing them with a narrow beam transmitter (like a laser diode) you should be able to re-use most of the other components. This would allow you to use several guns (each programmed with their own "code") and tell them apart.</p> <p>But as I mentioned earlier, the really cool system uses a lock-in amplifier scheme; using a cheap quartz oscillator you should be able to detect the signal regardless of phase (derive the quadrature signal from the oscillator and have two detection channels; then add the square of these signals together to have intensity independent of phase). As long as your detector does not saturate, this method is incredibly sensitive: it used to be one of my favorite optics demonstrations to show how you can "see" a small LED light up from across the room when the oscilloscope shows "nothing but noise". There is a nice description in <a href="http://www.phys.utk.edu/labs/modphys/lock-in%20amplifier%20experiment.pdf" rel="nofollow noreferrer">this article - especially part D</a>.</p>

2974

Focusing light from a laser shot into one point

2015-05-31T12:08:08.527

<p>My question is "what is the need of using auxiliary carry?" I mean 'AC' flag is set when D3 generates a carry. Then why not different flags for D2 or D4? </p> <p>Why suddenly we are treating D3 carry in a special way? </p>

|electrical-engineering|embedded-systems|

<p>Auxillary carry flag is used in microprocessors to carry out BCD arithmetic operations.</p> <p>As hexadecimal is used to carry out operations in microprocessor, we need to find a way to convert the answers in decimal( if we want decimal arithmetic). In 8085 microprocessor this process is carried out by commands like DAA and DAS. These need Auxillary Carry flag's status for functioning.</p> <p>For example:</p> <p>If we want to perform BCD addition on 38 and 29 we can use <em><strong>DAA</strong></em> command after <em><strong>ADD</strong></em> instruction.This will check whether an auxillary carry has been generated(or lower nibble is greater than 9) or not. In this case it has as <em>9+8=11</em> in hexadecimal, so reading this it will add 06 to the answer microprocessor has got i.e. 61. Adding 06 to 61 we get 67 which is the correct BCD addition of 38 and 29.</p> <p>DAA work with Accumulator only. Two conditions of DAA are:</p> <ol> <li>Add 06 with A, if Lower Nibble &gt; 9 or Auxillary Carry = 1</li> <li>Add 60 with A, if Upper nibble &gt; 9 or Carry = 1</li> </ol> <p>Similarly, <em><strong>DAS</strong></em> command is used in the same way for BCD subtraction in 8086 microprocessor but it is not available in 8085.</p>

2986

What is the need of having a separate flag as Auxiliary carry in microprocessor 8085?

2015-05-31T17:59:30.210

<p>Some laser applications consist simply of concentrating light into a small spot. Two example applications are laser welding and cutting. In these cases a CO₂ laser is often used which needs a regulated power supply, a water cooling system, and a supply of CO₂ gas. </p> <p>Why do these applications use a laser instead of a simpler (i.e. incoherent) light source such as an AC powered arc lamp?</p>

|optics|lasers|

<p>This is just an addition to the answer of Chris Mueller.</p> <p>When you think about lasers, you always think about an aperture with lot of mirrors, lenses and optics in general. Lets say that you manage to create a focused (wide spectrum) beam at one point, now you want to bring it to the point of application. With a wide spectrum it won't work well, as the beam will lose focus for the different wavelengths. while passing through the optics. </p> <p>To give you a picture of the optical phenomena look at this famous <a href="http://en.wikipedia.org/wiki/The_Dark_Side_of_the_Moon" rel="nofollow">album cover</a>. You will notice that the exiting light is spread out(along the spectrum) and lost focus.</p> <p>You could refocus it, but that is impractical. For other considerations refer to the answer of Chris Mueller.</p>

2991

Why are lasers used for concentrated light applications instead of incoherent light sources?

2015-05-31T19:05:56.117

<p>Lasers have a wide range of applications, but can be very costly to produce at high power/intensity. Since light is nothing more than a narrow band of the electromagnetic spectrum, it makes me wonder if the same high power/intensity applications can be achieved at a lower cost by concentrating other wavelengths (e.g. Infrared, radio, etc.) </p> <p>A cursory google search revealed nothing of this sort, which makes me suspect that i this is not possible. But i just want a clear answer as to whether other electromagnetic waves (besides light) can be concentrated and propagated as a beam invthe same way as a laser. If possible, i'd also like to understand if/why larger wavelengths hinder our ability to concentrate these waves into high power/intensity applications.</p>

|electrical-engineering|lasers|photonics|

<p>I would just like to add, in addition to the other answers regarding the EM spectrum, that anything that follows the <a href="https://en.wikipedia.org/wiki/Wave_equation" rel="nofollow">wave equation</a> can be focused because all of the principles of constructive and deconstructive superposition apply. </p> <p>Things outside the EM spectrum that follow the wave formula include pressure waves (when focused, commonly referred to as <a href="https://en.wikipedia.org/wiki/Sonar" rel="nofollow">sonar</a>), and even matter waves, such as focused ocean waves, called <a href="https://en.wikipedia.org/wiki/Rogue_wave" rel="nofollow">rogue waves</a>.</p>

2992

Can all electromagnetic spectra be focused like a laser?

2015-06-01T13:07:21.207

<p>I'm interested in the usage of a self-Locking hex nut with a nylon insert (ISO 7040 or DIN 985). How is it used correctly? Is there a prescribed usage, or a norm that describes how it should be used? </p>

|civil-engineering|bolting|

<p>The biggest consideration is that the thread must fully engage the nylon ring. If your bolts are too short, the nylon won't do it's job by deforming to the shape of the threads. Depending on how critical your application is, in many industries they are considered single-use as the nylon gets less effective the more times the nut is installed.</p> <p>For very long studs, power driving the nut can generate a lot of heat, melting the nylon insert. Because of this some people insist on only installing nylocs using hand wrenches, although this is only a practical concern in very rare circumstances.</p> <p>As far as the application, nylon insert nuts don't perform well in high temperature environments, or in the presence of chemicals that attack the specific plastic used. It's also worth noting that they are not the most robust of thread-locking options, so in very high vibration applications, or applications where the bolt forms a pivot, they may not be advisable. All-steel locknuts, or pinned nuts may be a a better option in such cases.</p>

3000

How is a self-locking hex nut (ISO 7040) used?

2015-06-01T19:52:11.580

<p>I'm using a 10/100/1000 BASE-T fiber transceiver. </p> <p>What does this mean?</p>

|lasers|

<p>The 10/100/1000 refers to the transmission rate in megabits per second<br> = millions of bits per second.<br> Also may be shown as Mbps or Mb/s.</p> <p>1000 Mb/s is a data rate typically achieved by modern LAN interfaces.</p> <p>10 and 100 mb/s are the maximum rates achieved by older standards. A 10/100/1000 label is indicating that the device operates at the 1000 Mb/s rate where it is interfacing with equally capable equipment, but is also able to work at the 10 or 100 Mb/s rate and standards when the equipment it is interworking with is limited to those rates.</p> <p>As hazzey notes, Base-T does not apply to fibre based equipment. It is a designation indicating that twisted pair is used as the transmission medium.<br> The original <a href="http://en.wikipedia.org/wiki/Ethernet" rel="nofollow"><strong>"Ethernet" LAN interface</strong></a> (now part of IEEE 802.3-xxx) used coaxial cable and operated at 10 Mb/s. When twisted pair interfaced alternatives were introduced, the name BASE-T was introduced for twisted pair rather than coaxial cable interconnected equipment. 10 Base-T was used to designate 10 Mb/s twisted pair equipment.<br> "Just to confuse" designations 10Base5 and 10Base2 were designations relating to thin-coax 10 Mb/s Ethernet in the 1980s. </p> <p><a href="http://en.wikipedia.org/wiki/Gigabit_Ethernet" rel="nofollow">Wikipedia - Gigabit Ethernet</a> - nominally re 1000 Mb/s systems but covers the historical development.</p> <p>Ethernet <a href="https://www.perle.com/products/10-100-1000-media-converter.shtml" rel="nofollow">Fibre and twisted pair interfaces</a> exist but that is not what you are dealing with.</p> <hr> <p>Related</p> <p><a href="http://en.wikipedia.org/wiki/Ethernet_over_twisted_pair" rel="nofollow">Ethernet over twisted pairs</a></p> <p><a href="http://en.wikipedia.org/wiki/10BASE2" rel="nofollow">Wikipedia - 10BASE2</a></p> <p><a href="http://en.wikipedia.org/wiki/10BASE5" rel="nofollow">Wikipedia - 10BASE2</a></p> <p><a href="https://www.google.co.nz/search?q=ethernet&amp;biw=1280&amp;bih=789&amp;tbm=isch&amp;tbo=u&amp;source=univ&amp;sa=X&amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;ved=0CFAQsAQ#tbm=isch&amp;q=ethernet+base-10" rel="nofollow">10BASE2 &amp; 10BASE5 eternet</a> and<a href="https://www.google.co.nz/search?q=ethernet&amp;biw=1280&amp;bih=789&amp;tbm=isch&amp;tbo=u&amp;source=univ&amp;sa=X&amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;ved=0CFAQsAQ#tbm=isch&amp;q=ethernet+10base5" rel="nofollow">here</a> and <a href="https://www.google.co.nz/search?q=ethernet&amp;biw=1280&amp;bih=789&amp;tbm=isch&amp;tbo=u&amp;source=univ&amp;sa=X&amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;ved=0CFAQsAQ#tbm=isch&amp;q=ethernet+10base2" rel="nofollow">here</a></p> <p><a href="https://www.google.co.nz/search?q=ethernet&amp;biw=1280&amp;bih=789&amp;tbm=isch&amp;tbo=u&amp;source=univ&amp;sa=X&amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;ved=0CFAQsAQ" rel="nofollow">Ethernet</a></p>

3005

What is the "10/100/1000 BASE-T" laser/transceiver rate unit signifying?

2015-06-02T17:38:46.190

<p>Lift produced by an airplane wing is related to airspeed - this much is clear; a plane moving too slowly will stall. But what is that relation? Linear? Quadratic? Exponential? I don't need the exact equation, which is surely quite complex, just the character of the relation.</p>

|fluid-mechanics|aerospace-engineering|

<p>It's all based in air pressure. Lifting requires velocity because it needs to create pressure gradients between the wing profile, so the air that is lower in pressure in the bottom part of the wing tend to equalize with the pressure in the upper part. When you stall, that means that you are no longer creating pressure between the upper and lower part of the wing profile. </p> <p>That's why you have different wing profile geometries, so that you can make different pressure ranges in different velocity scales.</p>

3018

How is lift related to airspeed?

2015-06-02T23:21:04.027

<h3>Summary</h3> <p>A short way to phrase this question is:<br> <em>Shear</em> describes the translation of one part of a body relative to another given an applied force.<br> What is the equivalent rotational term? That is,</p> <p><strong><em>What</em> describes the rotation of one part of a body relative to another given an applied moment of force?</strong></p> <h3>In more detail:</h3> <p>I'm trying to find the right word for a particular type of distortion. I have an assembly of items connected to be approximately solid. I mount one flat to a wall, and put some force or moment onto the opposite flat side so that the assembly distorts. In particular, I want to correctly describe the relative rotation of the two flats (in any direction), and I want this term to describe only the rotation, and exclude any translation (so that if it only shears without twisting I'd like the quantity to be zero).</p> <p><img src="https://i.stack.imgur.com/hL8dm.png" alt="" /></p> <p>I thought of <strong>torsion</strong>, but definitions of torsion always seem to refer to a bar or an elongated axis of some type. Is a bar required for &quot;torsion&quot; to be meaningful? I thought of <strong>twist</strong>, but that doesn't necessarily imply a dynamic quantity, so, for example, one could freeze in a twist, and something can be twisted without forces applied. I don't think <strong>bend</strong> is correct because that could just refer to translation. Is there a better word that I'm not thinking of, or even a concise phrase?</p> <p>If there is a word that matches my needs, a reference that I could review would be very helpful.</p> <hr> <h3>Why &quot;bending&quot; is not the word I need</h3> <p>Below is a sketch of two objects that are bending along the same curve (both by intuition and the equations), yet the rotation of the ends is different in each. I'm interested in a word or phrase for describing the rotation of the ends (or a tendency for the ends to rotate or not, etc).</p> <p><img src="https://i.stack.imgur.com/LS3iD.png" alt="" /></p> <p>Also, according to wikipedia, <a href="http://en.wikipedia.org/wiki/Bending" rel="nofollow noreferrer">bending</a> &quot;characterizes the behavior of a slender structural element&quot; under load, but I'm not interested in slender elements (and only drew the above to illustrate bending). The first is the common way a beam would bend, but not everything bends like a beam, and I am hoping to find a word that characterized the particular difference illustrated here.</p> <h3>Things that don't bend like normal beams</h3> <p>A reasonable model is a highly anisotropic material, rigid in one axis but less so in the others. For example, a bundle of parallel fiber optic cables.</p> <p>I think another would be a tube filled with water at high pressure, like a fire hose. (I don't know how this bends, but I doubt it bends like a beam.)</p> <p>Here's a sketch of a structure that doesn't bend like a beam, in that the side parallel to the wall stays parallel, even though the rest of the structure bends in a normal way. The black dots are meant to represent rotating joints. Even if this item were long and slender I think the ends would stay parallel. The point of the third picture is that you don't necessarily get to see what's on the inside (but it could still be important to describe that the two walls don't rotate with respect to each other in response to an applied force).</p> <p><img src="https://i.stack.imgur.com/JnHsI.png" alt="" /></p>

|mechanical-engineering|statics|terminology|

<p>What you're describing is a subset of the general concept of rigidity (or stiffness). I think the measurement you're looking for is the inverse of rotational stiffness (<a href="http://en.wikipedia.org/wiki/Stiffness#Rotational_stiffness" rel="nofollow noreferrer">as defined by Wikipedia</a>):</p> <blockquote> <p>A body may also have a rotational stiffness, <em>k</em>, given by $$k=\frac {M} {\theta}$$ where</p> <p><em>M</em> is the applied <a href="http://en.wikipedia.org/wiki/Moment_(physics)" rel="nofollow noreferrer">moment</a><br> <em>θ</em> is the rotation</p> </blockquote> <p>You could say something like, <strong>"The rotational stiffness of the assembly is very high, such that the rotation of the free end with an applied bending moment is nearly zero."</strong></p> <p>If you need to come up with a term for the inverse of rotational stiffness, you might get some mileage out of "skew," as in, "The end of the assembly opposite and parallel to the wall will not tend to skew with respect to the wall when any force is applied."</p> <p>Informally, this use of "skew" may just make sense to the reader; formally, it's a bit tricky to justify, and an audience of engineers may second-guess it. You could argue that there are more parallel lines between parallel planes than there are between intersecting planes, because all parallel lines between intersecting planes must be parallel with the line of intersection, therefore intersecting planes are more "skewed" because there are fewer ways to pair lines between them that are <em>not</em> skew. <a href="http://en.wikipedia.org/wiki/Skew_lines" rel="nofollow noreferrer">You'd be on shaky ground with the mathematicians, though.</a></p> <p>That's my direct answer to your vocabulary/documentation question. The rest of this will be a bit more free-form, starting with a response to <a href="https://engineering.stackexchange.com/questions/3024/twisting-or-torsion-of-a-non-bar-vocabulary#comment5274_3040">your comment on AndyT's answer</a>:</p> <blockquote> <p>my assembly doesn't really have clear surfaces (think of two spheres stuck together). So instead of, say, "torsional rigidity", I'd need something like "given any two virtual surfaces, one close to the binding surface and one close to the free end, when a force is applied to the free end, the two virtual surfaces will experience only a small relative rotation".</p> </blockquote> <p>You'll need to at least define these virtual surfaces initially. One of them should be easy, since it's a plane defined by a wall. If the wall is curved, you might use the plane having a normal vector that averages the axes of the fasteners used to mount the assembly, which passes through some "centroid of mounting" (call this the "virtual wall" if you like). A more complex geometry will demand a more complex explanation; efforts to reduce the complexity of that explanation are unreasonable if and when they impair the effectiveness of the explanation.</p> <p>Beams and shafts aside, the difference between rotational stiffness and <a href="http://en.wikipedia.org/wiki/Torsion_(mechanics)" rel="nofollow noreferrer">torsional rigidity</a> has to do with the nature of the forces applied. Rotational rigidity corresponds to bending moments, while torsional rigidity corresponds to twisting moments; in a torsional context, your virtual plane(s) would be <a href="http://en.wikipedia.org/wiki/Plane_of_rotation#Plane_of_rotation" rel="nofollow noreferrer">planes of rotation</a>. The same would not be true in a bending context.</p> <p>For an analogy, consider the motion of a <a href="http://en.wikipedia.org/wiki/Rubik%27s_Cube" rel="nofollow noreferrer">Rubik's Cube</a>. When you twist one of the layers of the cube, the angle through which the layer moves is within the plane of the cube. Now, imagine twisting a softball-sized lump of rubber in the same way that you would twist a Rubik's Cube. Because the rubber is solid, the half in one hand is constrained relative to the half in the other hand, and the moment you apply deforms it. How much it deforms depends on its <em>torsional stiffness</em>; the internal stresses are shear stresses.</p> <p>Likewise, consider the motion of an accordion. The two ends of the accordion are initially parallel but, when you hold the accordion in your two hands, you can manipulate it so that the ends are not parallel (because of its flexible diaphragm). Now recall that lump of rubber. Holding it in the same grip as before, you can apply a bending moment to deform the rubber in a different way, corresponding to bending rather than twisting. How much it deforms now depends on its <em>rotational stiffness</em>; the internal stresses are normal stresses.</p> <p>If we have a slender beam or shaft, torsion is only practically relevant with respect to a single axis, and bending is only practically relevant with respect to the other two (i.e., the major and minor axes of a beam). If we have a lump of rubber, bending and torsion could occur <em>and be significant</em> in any direction. We don't have the advantage of simple equations developed for classical beams and shafts (or even <a href="http://en.wikipedia.org/wiki/Timoshenko_beam_theory" rel="nofollow noreferrer">less-simple methods</a> used for, e.g., short beams), but "torsion" and "bending" are still meaningful.</p> <p>If you define virtual planes, the angle between the two planes quantifies the relative rotation you're after. <a href="https://engineering.stackexchange.com/questions/3024/twisting-or-torsion-of-a-non-bar-vocabulary#comment5322_3024">Ethan48's suggestion of using the term "dihedral angle"</a> is technically correct, but probably more abstract/general than you need—identify the planes and say the angle between them is $\theta$. If you know that angle to be much smaller than $M$, don't even go through the trouble of naming the inverse of rotational stiffness; just claim that $\theta \approx 0°$, justify it with "high rotational stiffness" and be done with it.</p> <p>Remember, if you go through the trouble to formally define virtual planes, that points fixed to the plane when the assembly is unloaded must remain collinear so that the virtual plane doesn't warp into a curved surface under loading. (Do you want to measure the angle between curved surfaces? I don't!)</p>

3024

How to describe the rotation of opposite ends of a body due to an applied moment

2015-06-03T09:26:54.690

<p>I have the utilization rates of several machines for each week over a year. These differ per week because of the occurrence of machine failures and changes of orders. Meaning that one week a machine could be a bottleneck and the other week not. I want to rank these machines based on the utilization and so determine their criticality. However if I calculate the mean utilization, information gets lost. The same is when I count the events the utilization is bigger than a threshold. For example; </p> <ul> <li>machine A has a utilization of 50% and 14 weeks the utilization > 90%</li> <li>machine B has a utilization of 70% and 7 weeks the utilization > 90% </li> <li>machine C has a utilization of 89% and 6 weeks the utilization > 90%</li> </ul> <p>In my opinion machine A is more critical than machine B. So ranking based on a threshold value seems reasonable. However, when comparing machine B and C this is not that clear. Does anyone know a statistical/mathematical method to get a good comparison?</p>

|statistics|industrial-engineering|

<p>The problem is a multi-criteria decision analysis (MCDA), as I want to rank the machines based on several (available) criteria. With initially the following criteria: downtime costs and utilization. However with utilization there was the following problem: if the events are counted that surpasses a threshold or the average utilization is used, both situations would lead to loss of information. As it is an MCDA, both criteria can be added, thus the criteria are now:</p> <ol> <li>downtime costs per time unit</li> <li>average utilization per time span</li> <li>number of events the utilization that surpasses a threshold (e.g. 95%)</li> </ol> <p>As these factors are not directly commensurable and differ in importance, a weighted method is needed. To select the appropriate method 7 guidelines are used (Guitouni and Martel, 1998);</p> <ul> <li><strong>Guideline G1</strong>: Determine the stakeholders of the decision process. If there are many decision makers (judges), one should think about group decision making methods or group decision support systems (GDSS).</li> <li><strong>Guideline G2</strong>: Consider the Decision Maker (DM) `cognition' (DM way of thinking) when choosing a particular preference elucidation mode. If he is more comfortable with pairwise comparisons, why using tradeoffs and vice versa?</li> <li><strong>Guideline G3</strong>: Determine the decision problematic pursued by the DM. If the DM wants to get an alternatives ranking, then a ranking method is appropriate, and so on.</li> <li><strong>Guideline G4</strong>: Choose the Multi Criteria Analysis Problem (MCAP) that can handle properly the input information available and for which the DM can easily provide the required information; the quality and the quantities of the information are major factors in the choice of the method.</li> <li><strong>Guideline G5</strong>: The compensation degree of the MCAP method is an important aspect to consider and to explain to the DM. If he refuses any compensation, then many MCAP will not be considered.</li> <li><strong>Guideline G6</strong>: The fundamental hypothesis of the method are to be met (verified), otherwise one should choose another method.</li> <li><strong>Guideline G7</strong>: The decision support system coming with the method is an important aspect to be considered when the time comes to choose a MCDA method.</li> </ul> <p>Based on all guidelines except guideline G6, this results in the following suitable methods: Analytic Hierarchy Process (AHP) or Promethee II. However, AHP has the assumption that inner and outer criteria are independent. A correlation test showed that there exists significant correlation between criteria 2 and 3, thus the criteria are not independent. Promethee II is therefore the appropriate method for ranking the machines given my situation. </p> <hr> <p>Guitouni, A., Martel, J.-M., 1998. Tentative guidelines to help choosing an appropriate MCDA method. Eur. J. Oper. Res. 109, 501–521.</p>

3030

How to compare utilization rates?

2015-06-03T15:09:56.840

<p>If you have a prestressed concrete beam or slab and are jacking a cable from both ends, does the order of operations alter the result?</p> <p>For a symmetric curve, the tension diagram along the cable's span after friction losses is always shown as symmetric. </p> <p><img src="https://i.stack.imgur.com/zmWaO.gif" alt="enter image description here"></p> <p>Is this the case regardless of whether the ends were jacked simultaneously or one after the other?</p> <p>If the ends are jacked simultaneously then the symmetric friction loss makes perfect sense. However, if the cable is jacked first from the left and then from the right, I'd expect the tension profile to be asymmetric.</p> <p>When jacking from the left, the cable is pulled and on its left end the cable force is equal to $P_0$, the jacking force. On the right side, the cable force is $P_0 - \Delta P$, where $\Delta P = P_0\left(1-e^{-(\mu\alpha+kL)}\right)$ is the friction losses. The left side is then anchored (disregard the anchorage slip losses for a moment). The right side is then jacked. Effectively, this jacking adds a force of $\Delta P$ to the right end, so that it too is now at $P_0$. However, wouldn't the left end also be affected with an increment equal to $\Delta Pe^{-(\mu\alpha+kL)}$, such that the diagram is asymmetric?</p> <p>What about if one considers the fact that when the left side is anchored, anchorage slip losses occur? So the diagram should go: friction losses from jacking the left side, anchorage slip losses from anchoring the left side, friction losses from jacking the right side and then anchorage slip losses from anchoring the right side.</p> <p><strong>EDIT</strong> After a few honest-to-God minutes freaking out thinking I was the biggest idiot on Earth I realized that what @sanchises commented doesn't rule out my question.</p> <p>The cable in my question does remain in static equilibrium, as can be seen here:</p> <p><img src="https://i.stack.imgur.com/KzP5V.gif" alt="enter image description here"></p> <p>When jacking from the left, the cable suffers at its left extremity a force of $P$. Friction losses along the way ($\Delta P$) cause the right extremity to end up with $P - \Delta P$. The right extremity is then jacked, which effectively applies a force of $\Delta P$. Friction losses also apply here, now with a value of $\Delta\Delta P$, such that the left anchor is additionally jacked with $\Delta P - \Delta\Delta P$.</p> <p>This results in the left anchor with a force of $P + \Delta P - \Delta\Delta P$ and the right anchor with only $P$. These are not, however, the only forces acting on the cable: the friction losses are what keeps it in static equilibrium. Globally, we have to the left $$P + \Delta P - \Delta\Delta P + \Delta\Delta P = P + \Delta P$$ and to the right $$\Delta P + P - \Delta P + \Delta P = P + \Delta P$$ both of which equal the total jacking force (first $P$ on the left and then $\Delta P$ on the right).</p>

|civil-engineering|structural-engineering|beam|concrete|prestressed-concrete|

<p>After the discussion with @sanchises in the comments, the answer is that, yes, the order of operations in jacking does affect the final tension diagram of the cable. A symmetric cable simultaneously jacked will have a symmetric tension diagram, while the same cable jacked from one side and then the other will not. The diagram actually ends up as a line almost parallel to the tension diagram after the jacking of the left side:</p> <p><img src="https://i.stack.imgur.com/cmAUP.gif" alt="enter image description here"></p> <p>The cable in this example is jacked from the left side (grey line) and loses around 15% of its tension at the right anchor ($\Delta P$). It is then jacked from the right side by that same amount, adding to the tension profile according to the yellow line, which results in the blue line, with the left anchor being incremented by $0.15\cdot0.85=13\%$.</p>

3038

Is the order of operations in jacking a cable relevant?

2015-06-03T19:08:43.163

<p>I'm building a pizza oven and, to do so, I bought a whole load of engineering bricks (not <a href="http://www.travisperkins.co.uk/Ibstock-Brick-Class-B-Perforated-Engineering-Red/p/123759">these</a> but similar). They've got five holes vertically through them, each hole is 2x6x6.5 cm (80 ml) and the walls, both end and dividing, are 2 cm thick by 6.5 cm high.</p> <p>When the oven is fired up it will reach ~500 °C. If we use some good solid concrete as masonry for the bricks, which will seal the ends of the holes, I'm worried that as the air heats up, the pressure inside will turn them into a sort of ceramic bomb that will make cooking pizza a rather more exciting experience.</p> <p>Is that likely to happen? How can I calculate the pressure that will be exerted on the bricks by 80 ml of air at 500 °C?</p>

|mechanical-engineering|thermodynamics|pressure|mechanical-failure|masonry|

<p>Engineering bricks (solid) are fired at 1200 degrees in a kiln so don't worry about your complicated mathematics , your pizza oven will never get to that temp.if it did or even if it got much much higher the bricks would just melt together.(They call them clinkers) I am a bricklayer with 47 years experience and a lot of that time was building and repairing kilns. When you repair a kiln you only use normal clay for the crowns (internal)as once the kiln is fired up the clay turns hard just like the bricks you are cooking. You only use fire bricks for the fire holes and they are cemented together with texacrete.</p>

3041

Would engineering bricks explode if used to build an oven?

2015-06-04T02:10:11.593

<p>All these processes sound like structural change caused by heating.</p>

|process-engineering|metallurgy|

<p><strong>Annealing</strong> is a heat-treatment process for stress-relief of a material. Annealing facilitates reduction of internal stresses and total elastic energy stored in inter-atomic bonds within a treated material. The term is used for appropriate heat-treatments of metals, ceramic glasses, and high-performance polymers. An example would be a cold-rolled steel billet annealed so that it can be worked further into final products. The process of annealing is used whenever internal stress are unacceptable, such as prior to a machining step. Machining changes the distribution of internal stress by removal of material, which is rebalanced by the machined part deforming.</p> <p><strong>Curing</strong> is a heat-treatment process for accelerating a chemical reaction, and is generally used in the context of polymeric and polymeric composite materials. Thermoset polymers which do not set in a reasonable time at room temperature are cured at higher temperatures in a curing oven. An example would be a pre-impregnated or prepreg carbon-fiber mat draped over a mold and then cured so that it retains the shape of the mold. The monomers react much more rapidly at the curing temperature, and form a thermoset polymer, hardening the material prepreg mat. Personally, I have never heard curing applied to any other class of material.</p> <p><strong>Sintering</strong> is a heat-treatment process for causing a powdered material to become a monolithic bulk material by diffusion between individual powder particles. Sintering may occur in either or both of the solid and liquid states. Consider a slurry mixture of alumina powder and wax binder pressed into shape. If the binder is melted off, the alumina powder that remains is only mechanically bonded. That green form may then be sintered to bind the alumina particles together into a monolithic bulk in the same shape as the green form. The same process may conceivably be applied to any material that can undergo diffusion at reasonable rates and temperatures, though it is most commonly used in the context of ceramic and metallic materials, especially high-melting-point refractory metals. Personally, I have never heard it applied to polymers, as their melting points (if such exists) are so low they are simply processed in the liquid state. The process of sintering is used where (1) the material's melting temperature is unreasonably high or (2) parts of acceptable quality may be produced more rapidly by sintering than by casting or machining, such as in low-stress, complex-shaped steel gears.</p>

3047

What's the difference between annealing, curing and sintering?

2015-06-04T21:27:52.493

<p>There are multiple algorithms out there for generating synthetic EKG signals (for test and validation of heart rate estimation algorithms). All you plug in is the desired heart rate, and the algorithm outputs an EKG signal with the desired heart rate.</p> <p>I need to create a similar algorithm - but for pulse oximetry. I would like to implement a signal generator which generates a synthetic infrared and red (photodiode) signal based on a user-specified/pre-defined SPO2 rate. I then want to take the synthetic red and infrared signal and let them be processed by an SPO2 estimation algorithm. If the algorithm outputs an SPO2 estimate which is close to the pre-defined (i.e. the true SPO2) SPO2 value, then I know that the algorithm works.</p> <p>How would I go about creating such a signal generator? </p>

|biomedical-engineering|

<p>Short: </p> <p>(1.) Mindless and without understanding method that would work well.</p> <p>This is as quick and easy as is likely to be possible, should work 'well enough' and would be 'easy enough' to implement without any in depth understanding past ensuring that simulated sensor output characteristics (impedance etc) match those of the target sensor.</p> <ul> <li><p>Obtain known good SPO2 meter and sensor.</p></li> <li><p>Monitor data signals from LEDs and record</p></li> <li><p>Connect sensor to "patients" or test samples of choice and adjust SPO2 level. (Patients can easily 'adjust' SPO2 level over a wide range by controlled breathing. Ask me how I know :-) ).</p></li> <li><p>At a range of SPO2 levels log waveforms and store in simulator</p></li> <li><p>'Play back' signals as required.</p></li> </ul> <p>(2.) The path to understanding - much harder. </p> <p>See <strong>"Real World sensor Outputs"</strong> below for a set of raw and processed sensor output signals for IR and visible red LEDs in an SPO2 measurement environment. Using the related evaluation kit (or an equivalent circuitry of your choice) would allow examination of SPO2 sensor waveforms at various SPO2 levels. From these it would be "simple enough" to provide analog levels corresponding to equivalent sensor outputs either at selected SPO2 levels or, potentially, continuously variable ones. Available on ebay. </p> <p>As noted below, the transferability of sensor signals from one system to others from other manufacturers is far from certain.</p> <p>Note: Below I have provided detailed information from a single supplier. I have no involvement with the company or their products. They seem to "know their stuff" and examples show that TI have used their equipment in their development work. </p> <hr> <p><strong>Background:</strong> </p> <p>Simplistically, SPO2 is measured by comparing the attenuation of two wavelengths of red light by a portion of the body which has blood flowing through it. The wavelengths are chosen such that one is attenuated in a manner which is a function of Oxygen content when it passes oxygenated haemoglobin and the attenuation of the other is independent of haemoglobin oxygenation percentage. Oxygenated haemoglobin attenuation is a function of blood vessel optical path length and haemoglobins saturation versus Oxygen partial pressure "transfer function". </p> <p><strong>Haemoglobin saturation versus oxygen partial pressure.</strong> </p> <p><img src="https://i.stack.imgur.com/uFrQS.jpg" alt="enter image description here"></p> <p>The oxygen-unnaffected signal provides a reference attenuation against which the affected signal can be compared to deduce oxygen saturation level, after taking the factors mentioned above into account. </p> <p>For a much more complete and competent explanation see <a href="http://www.apmkr.com/bio-device/transmittance_oximeter.htm" rel="nofollow noreferrer"><strong>here</strong></a></p> <p>Not too surprisingly, it's not that easy. SPO2 assessment is a black art - far less straight forward than one might expect when considering the method used. I base this claim on an interesting and detailed Hewlett Packard report which I read many years ago, which said as much, and then set out to demonstrate how empirical the art is in practice. That this is still the case to some extent is amply demonstrated by the myriad internet pages and ads comparing brand x with the seller's brand Y and showing how unreliable the other brand is under various conditions. For example: </p> <p>Example: Brand A &amp; Band B versus two of "<em>ours</em>"</p> <p><img src="https://i.stack.imgur.com/P1NK7.png" alt="enter image description here"></p> <p>And again ...</p> <p><img src="https://i.stack.imgur.com/FG1gA.png" alt="enter image description here"></p> <p>So - if I was required to build a SPO2 sensor simulator, in the absence of known good sensor signal data, I would take as many reputable SPO2 meters as I could find and observe the signals produced by their sensors under various conditions and displayed SPO2 values. An issue is that sensors and SPO2 meters may be to some extent matched. A sensor pair which works with brand A SPO2 meter may implement assumptions which make it unsuited for use with a brand B meter. Determining to what extent this is true is one of the first steps in evaluating algorithm performance.</p> <p>Fortunately, at least one company seems keen to help you do this ...</p> <hr> <p><strong>Real world sensor outputs:</strong></p> <p>To start - super cheap complete pulse-oximeters on ebay for about $US20 all up. Unbelievable - and performance totally unknown <a href="http://www.ebay.com/itm/CMS50DL-SPO2-Monitor-OXYMETRE-SATUROMETRE-POULS-METRE-ECG-OXYMETER-OM1-/120971428378?pt=LH_DefaultDomain_71&amp;hash=item1c2a75821a" rel="nofollow noreferrer">CMS50DL SPO2 Monitor OXYMETRE SATUROMETRE POULS METRE ECG OXYMETER OM1</a> - the "OM1" in the title is probably a hat-tip / pretend similarity to an APM model. </p> <p>Buy one. Test. IF it works (IF) buy several.<br> Pull apart, bring out LED signals, use, play ... . </p> <p>and - much more "real" </p> <p>APMKOREA <a href="http://www.apmkr.com/" rel="nofollow noreferrer">Home Page</a> - easier to look at than describe - sensors evaluation boards, application notes, waveform photos, ... . </p> <p>APM provide both PPG (essentially pulse waveform) and SPO2 measurement equipment. By the time I realised they were overlapping I had somewhat mixed PPG and SPO2 references as they overlap. Sorting out which is which I leave as "an exercise for the student" as anyone interested enough to follow this up will probably find it all of interest and relevance. </p> <p><strong>Likely of most relevance -</strong> APM's ICOM1 (RS232) and ICOM2 (USB) <a href="http://www.apmkr.com/bio-device/SpO2_module_detail.pdf" rel="nofollow noreferrer">Pulse oximetry modules</a> </p> <p><a href="http://www.apmkr.com/bio-device/reflectance_oximeter4.pdf" rel="nofollow noreferrer">SPO2 and PlethysmoGraphy discussion</a> looks useful. </p> <p><a href="http://www.apmkr.com/bio-device/SpO2_module_detail.pdf" rel="nofollow noreferrer">SPO2 modules</a> </p> <p>APMKOREA <a href="http://www.apmkr.com/spo2_oximeter_blood_sensor.htm" rel="nofollow noreferrer"><strong>" Leading PhotoPlethysmoGraphy Sensor Technology"</strong></a></p> <p>Offer a wide range of emitters, detectors, reflective and transmissive (through tissue) sensors. </p> <p>Plus an <a href="http://www.apmkr.com/bio-device/PPG%20with%20evaluation%20kit.pdf" rel="nofollow noreferrer">sensor evaluation and development board</a> which will work with onboard reflective sensors or external transmissive or reflective ones. </p> <p><img src="https://i.stack.imgur.com/qPvwa.jpg" alt="enter image description here"></p> <p><strong>And finally</strong>, the heart of what looks helpful is these example signals - presumably available from their evaluation systems for SPO2 levels of your choice</p> <p><img src="https://i.stack.imgur.com/Z1GuJ.jpg" alt="enter image description here"></p> <p><a href="https://i.stack.imgur.com/Z1GuJ.jpg" rel="nofollow noreferrer"><strong>Larger version here - MUCH easier to read</strong></a></p> <p>plus</p> <p><a href="http://www.apmkr.com/bio-device/pd_oximeter.htm" rel="nofollow noreferrer">Appears to be very useful application guide</a></p> <hr> <p>Nonin <a href="http://www.nonin.com/Accurate-Pulse-Oximeter" rel="nofollow noreferrer">"Acurate Pulse Oximeter"</a> </p> <p>Ecostore <a href="http://www.echostore.com/wireless-oximeter-cms50e.html" rel="nofollow noreferrer"> Wireless pulse oximeter</a></p> <p>Related:</p> <p><a href="http://www.anesthesiaweb.org/lung-function.php" rel="nofollow noreferrer">Lung function and anaesthesia</a></p>

3058

How to generate a synthetic infrared and red LED signal for SPO2 calculation

2015-06-05T03:34:50.300

<p>The obvious answer is a compass. I have an old (1952) transit that has a compass which seems to be accurate to 15'. However, that doesn't seem too incredibly accurate at all. Furthermore, most modern digital theodolites I see don't appear to have a compass (correct me if I'm wrong). I'm curious how a surveyor actually finds north with high accuracy. Are there digital theodolites or other instruments with compasses? Are north-south lines set out with GPS? I'd appreciate any insight.</p>

|measurements|surveying|

<p>You can use software like Stellarium to know at what exact time the sun will cross the meridian. At that time, you should cast a straight shadow on the terrain using a vertical pole. This line (the shadow) will go true North-South. In fact, you can know the angle with respect to true North of any straight pole shadow at any time of day. With that information you can figure out where is true north to decent accuracy, if you don't trust your compass.</p> <p>Other way of determining true North would be at night, using what's called a polar scope. These have a reticle that shows the aparent orbit of Polaris (Northern hemisphere) and 4 stars of Octans (Southern hemisphere). You just need to point and align the reticle with the stars.</p> <p>If that's not enough, you can use Plate Solving. I'm not too familiar but I remember it involves taking long-exposure photographies of the stars usign a star-tracker mout and adjusting your azimut and altitude until no star trails are present. The longer the exposure, the greater the accuracy of true north. You will need a software to analize the photographies and solve the star positions using a catalogue.</p>

3061

What instrument does a surveyor use to accurately find North?

2015-06-05T15:35:15.687

<p>I'm building a model of a long vehicle with 6 axles and I want to be able to steer it. How do I calculate the amount each axle should steer to minimise dragging of some wheels round sideways as it turns?</p>

|mechanical-engineering|automotive-engineering|

<p>What you are talking about is called <a href="http://en.wikipedia.org/wiki/Ackermann_steering_geometry">Ackerman steering</a> and is pretty easy to get approximately right, but requires non-linear proportional steering control to get fully right.</p> <p>If you put an imaginary focus/center point of steering somewhere to the side of the vehicle, in the plane of the axis of rotation for each wheel, then the axis through the center of each wheel should all meet at that point. The math is simple trigonometry, but building a mechanical steering system that gets this right for many wheels for all degrees of steering is hard. (For a rover robot I built, I ended up driving each wheel using a servo, because math for servo controllers is easy :-)</p>

3066

How to calculate the steering ratios on multiple axle vehicles?

2015-06-06T06:37:45.013

<p>I've heard that the part of a tire which is towards the inner side of a vehicle has material which is more wear resistant than the part of a tire on the outer side of the vehicle. </p> <p>In short, I have difficulty with identifying the inner side of a tire. Is there a difference in the material properties of tires from the inner side to the outside?</p>

|mechanical-engineering|materials|automotive-engineering|

<p>There are <a href="https://www.kaltire.com/whats-difference-directional-asymmetrical-symmetrical-tread-patterns/" rel="nofollow">Tread pattern differences</a>, which allows part of the tire to make good contact with the ground for performance (grip). Part of the tire with wide channels to evacuate water (typically called ribs), and sipes so the tire wears properly, etc. </p> <p>An Asymmetrical tire is often referred to in the tire world as an inside outside tire, where a directional tires typically have a V pattern with the v point toward the hood of the car, and the 2 ends facing the rear. To answer the question of rubber on the surface or thread of the tire being different, this is incorrect it remains the same across the thread, while the rubber used for the sidewall in some cases is different. <a href="http://swampertires.blogspot.com/" rel="nofollow">This describes the different terms typically used.</a> while if you search for Goodyear you can find a tire with Kevlar in it which is stronger side wall. The inner side of the tire leads me to believe you might have a worn suspension part in the front causing premature tire wear, or that the alignment of the vehicle is off, I would post more links however I do not have the rep it appears.</p>

3073

Are tires symmetrical & are their material properties uniform?

2015-06-06T23:35:28.910

<p>My apologies for a likely very simple question. But what is this nut which I have on a micrometer head called, and what is the tool for removing it called?</p> <p><img src="https://i.stack.imgur.com/PjMOA.jpg" alt="micrometerhead"></p>

|mechanical-engineering|nomenclature|

<h1>Short Answer</h1> <p>The nut on it is called a <strong>slotted lock nut</strong> or <strong>locking bearing nut</strong>. You can remove it with a <strong>hook spanner</strong>.</p> <h1>Long Answer</h1> <p>I guess your micrometer is Japanese and thus probably doesn't conform DIN standards but for slotted lock nuts you can check <strong>DIN 981</strong> or <strong>DIN 1804</strong> and for hook spanner you can check <strong>DIN 1810</strong></p> <h2>Why use Lock Nut</h2> <p>The main reason is screw loosening. With time all screws get loose (no pun intended) and on a device such as micrometer that loose fit will cause errors. That nut, locks bearing into place. Below is a similar Lock Nut for bearings from Misumi USA, you can check <a href="http://us.misumi-ec.com/pdf/fa/2012/p1_0917.pdf" rel="nofollow noreferrer">datasheet here</a> <img src="https://i.stack.imgur.com/uee2u.gif" alt="Hard Locking Bearing Nut Misumi USA"></p> <p>Information about lock nuts and hook spanners from SKF but I don't have enough rep to post more links.</p>

3077

What is this nut on a micrometer head and what is the tool for removing it called?

2015-06-07T01:35:19.467

<p>I have some university work to do based on a real world application with five or <a href="http://en.wikipedia.org/wiki/Six-bar_linkage" rel="nofollow">six bar mechanism</a>, but I am unable to find such a device. Are there real applications of such mechanisms? I've considered some, such as the trigger of a gun and a locking plier, but they are all <a href="http://en.wikipedia.org/wiki/Four-bar_linkage" rel="nofollow">four bar mechanisms</a>.</p>

|mechanical-engineering|kinematics|linkage|

<p>To find a ready-made device that utilizes a six-bar mechanism, you can dig through some of the classical resources below. Instead of pointing out a few devices, I think it would be better to point you to these databases.</p> <ol> <li><strong>507 Mechanical Movements: Mechanisms &amp; Devices</strong>. <a href="http://www.amazon.in/507-Mechanical-Movements-Mechanisms-Devices/dp/1603863117" rel="nofollow">(amazon link)</a>. This book is also available for free as an animated database.</li> <li><strong>Mechanisms and Mechanical Devices Sourcebook</strong>. <a href="http://www.amazon.in/Mechanisms-Mechanical-Devices-Sourcebook-Edition/dp/0071704426" rel="nofollow">(amazon link)</a> </li> </ol> <p>I have used most of them frequently while attending a course on mechanisms design and found all of them equally useful.</p>

3078

Applications of Five-Bar Mechanisms

2015-06-07T03:00:57.307

<p>I am affixing a servo motor with a machined enclosure to a base plate with drilled holes. The current design uses eight M-2.5 bolts in a figure U pattern (3-2-3) around the enclosure.</p> <p>Tightening that many bolts/screws takes time, and the small gauge makes them finickier than larger-gauge bolts (missed nut thread starts, etc.)</p> <p>If I were to change this design to use fewer, bigger, bolts, what would be the pros/cons of this? Is the amount of position slop significantly different between, say, three M-6 bolts, and eight M-2.5 bolts? Is there a formula for the affixing force that N bolts of size S will provide?</p>

|bolting|

<p>In the machinery industry the holo-krome bolt selectors are used regularly to get tightening torque and resultant bolt tensions: the green one is for inch and blue is for metric: <a href="https://www.google.com/#q=holo-krome+screw+selector+cards" rel="nofollow">https://www.google.com/#q=holo-krome+screw+selector+cards</a></p> <p>If feasible, the miniature sizes such as 2.5mm are avoided. Simplified, bolt strength is proportional to cross section area, or PI*R^2. In comparing a 6mm to a 2.5mm, the correspond radii are 3 and 1.25, The corresponding strength ratios are approximately 3^2 and 1.25^2, or 9 to 1.56, or a ratio of 5.8. </p> <p>The bigger screws may need blue locktite to keep from becoming loose due to vibration: the length of the screw under tension generally needs to be 4x the diameter to be considered vibration resistant.</p>

3080

Many small bolts, or few large bolts?

2015-06-07T11:08:09.377

<p>I want to create a solar powered kettle but first, I need to know how much power will be required. </p> <p>This means, I think, I'll need to know how much water I want to bring up to a certain temperature. </p> <p>I'm trying to understand how to work this out, and I'm frankly stumped.</p> <p>Is it possible to work this out, if I knew I wanted to bring 0.5litres to 60 degress C, how much power would I need to achieve this within 30 minutes?</p> <p>My end goal here is to find a suitable solar panel powerful enough.</p>

|mechanical-engineering|electrical-engineering|thermodynamics|solar-panel|

<p>The amount of water and how much you want to heat it only tells you the <i>energy</i> required, not the power. Power is energy per time, so it matters how fast you want to get there.</p> <p>You say you want to heat 500 ml of water by 60&deg;C in 30 minutes. Do the math.</p> <p>The specific heat of water is 4.18 J/g&deg;C. The total energy required is therefore:</p> <p>(500 g)(60&deg;C)(4.18 J/g&deg;C) = 125 kJ</p> <p>This energy spread out over 30 minutes (1800 seconds) is:</p> <p>(125 kJ)/(1800 s) = 70 W</p> <p><b>However</b>, that is only the power that has to go into the water. Your 500 ml of water sitting in some container at near boiling temerature is most likely loosing more than that to the environment. Put another way, your heater has to not only provide enough power to raise the temperature of the water, but also to overcome losses of this heat to the enviroment. At only 70 W into the water, the heat loss will be significant.</p> <p>With a lot of care spent on insulation, maybe 150 W is enough. It's really hard to guess since we have no idea what your mechanical system looks like. A few 100 W would be better.</p> <p>Also, consider that 30 minutes is a long time to wait for a cup of boiling water. If the system can just do that in full sun, then it won't work in anything less than ideal circumstances. 500 ml of water is also not a lot of stuff to heat. I think you are grossly underspecifying your heater. You may get 500 ml to boil in 30 minutes, but I expect you won't be happy with the performance of the heater overall.</p> <p>Figure you get about 1 kW per square meter of full sunlight, so a few 100 W doesn't really take a large collector. However, you have to pay attention to the aligment, and focus the rays on a small spot roughly the size of the bottom of the kettle.</p>

3085

Solar powered kettle - how much power for 1/2 litre

2015-06-07T14:56:01.723

<p>From what I understand, a control systems engineering job is almost always a senior level position. What kind of entry level jobs do most engineers have before they are qualified to actually design control systems? Is a PE license necessary for this kind of work?</p> <p>I am currently working on a masters degree in electrical and computer engineering. I am mostly interested in control theory and that is the subject of my thesis.</p>

|control-engineering|control-theory|employment|

<p>I recommend you to learn control theory while implementing what you learn directly. a useful example to do so ,this course for mobile robots using control theory <a href="https://www.coursera.org/learn/mobile-robot/home/info" rel="nofollow">https://www.coursera.org/learn/mobile-robot/home/info</a> and in each lecture you can implement what you learn on a simulator on MATLAB and you can even implement it on a hardware; small mobile robot. -You can also read or use as a reference "Modern Control theory, for William Brogan" i believe it is the best control theory reference i ever seen. and finally GOOD LUCK :)</p>

3087

How does one become a control systems engineer?

2015-06-08T23:54:06.047

<p>So I got myself questioning what could be worse for the driver... a collision of two identical cars at equal speed (frontal crash) or the same car with the same speed crashing through a wall? The first case I see it would double the impact, but also it will absorb the energy into the other car structure, otherwise, in a solid and rigid wall, all the energy would come back to the vehicle.</p> <p>Which situation is worse for the passengers?</p>

|automotive-engineering|safety|

<p>I'm guessing you haven't seen the <a href="https://www.youtube.com/watch?v=r8E5dUnLmh4" rel="nofollow">Mythbusters episode</a> on this. Then, see <a href="http://www.wired.com/2010/05/mythbusters-energy-explanation/" rel="nofollow">this writeup</a> that explains why they are correct, since Mythbuster's explanation leaves something to be desired. </p> <p>In short, it's exactly the same. In the 2 car example, each car has the same amount of energy (as each other, and as the car in the 1 car example) since they are all going the same speed (and have the same mass). In the two car crash, the total energy in the crash is doubled, however, the energy is distributed equally between the two cars. Therefore, the energy for each car is identical to a single car hitting a ridged wall. The tests in Mythbusters illustrates this.</p>

3099

Which is Worse: Car vs. Car or Car vs. Wall?

2015-06-09T01:44:08.630

<p>The dimensions of the container for the water is 6.06m by 2.44m the depth of water is at 1m equating to 14.786m^3 or 14,786 ℓ of water.</p> <p>The sytems is a closed system with a pump circulating the water 150 ℓ/min @ 1100 bar. How long would it take for the water to heat up from 20*C to 30*C?</p> <p>How would the time change if the depth of water increase to 2m (ie. 29.572 ℓ). </p> <p><strong>Q:</strong> What formula(s) do I need to use to find the time required for the water heat up from 20*C to 30*C. Also what other variable is required for me to calculate this time?</p> <p>Any pointers will be highly appreciated.</p>

|thermodynamics|

<p>These two very simple to use formulas will probably allow easy order of magnitude scoping. </p> <p><strong>Time to heat with given heating power</strong> </p> <p>T = V x 1000 cc/l x 4.17 x K / W seconds or T = 4170 V.K/W</p> <p><strong>Power required to heat in time T</strong> </p> <p>Power = W = = V x 1000 cc/l x 4.17 x K / T Watts or W = 4170 V.K/T</p> <p>Where </p> <p>T = seconds<br> V = litres<br> K = degrees K (or C) rise<br> W = Watts heating power<br> 4.17 = Specific heat of water over range 20 C to 30 C adjusted for mean water density so units are J/cc/K</p> <p>This is increased by any losses<br> eg electric kettle uninsulated is 85% - 95% efficient Pumping energy adds partially to heating. </p> <hr> <p>The key factor is the amount of heat stored in water per cc per degree K (or degree C) rise in temperature. Once this is known all else can be calculated.<br> A more useful than many table of <a href="http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html" rel="nofollow"><strong>water thermal properties</strong> may be found <strong>here</strong></a> on the well-worth-bookmarking "Engineering Toolbox" site. Rather than giving just a few figures at a few selected temperatures this gives properties at a wide range of temperatures, so it can be established if temperature has a significant effect on the relevant properties in the context of the current problem. </p> <p>The specific heat of water is an average of about 4.18 Joule per gram* per degree K across the 20 C to 30C temperature range, or 4.17 Joule per cc per degree K. The 0.01 J/unit/K is not going to make much difference to your result. </p> <ul> <li>The per cc and per gram figures differ slightly because the density of water is very slightly less than 1 g/cc across this range. The specific heat will be <strong>about</strong> 4.18 Joule per gram and 4.17 J/cc per degree K. The usual value given is 4.182 J/g/K, often without reference to temperature range.</li> </ul> <p>Other useful (and necessary) facts:</p> <ul> <li><p>One Joule is provided by 1 Watt of heating per second or<br> 1 J = 1W/s</p> <p>1 litre = 1000 cc</p> <p>1 metre^3 = 1000 litre And 1 litre = 10 x 10 x 10 cm or 100 x 100 x 100 mm</p> <p>Total_energy = SH x cc x delta-C = SH x litres x 1000 x delta-C</p> <p>Power = energy /second = Total_energy / seconds.</p> <p>Pressure matters little here except as it may affect water density.<br> Look up compressibility of water to see how much this matters.</p></li> </ul> <p>Doubling water head affects volume (of course) but has an insignificant effect on pressure.</p> <hr> <p>Power input</p> <p>Energy in Volume of V litres = V x 1000 cc/l x SG g/cc x SH x delta_T</p> <p>V = 14,786 (user figure)<br> delta_T = 10 K (user figure)<br> SG = 1 (actual figiure rolled into SH - see text)<br> SH = 4.17 J/cc/K (SG is included in this - see text) </p> <p>Energy = 14,786 x 1000 x 1 x 4.17 x 10 = 616,576,000 J or W.s</p> <p>So you can heat it in 1 second using a 617,000! kW heater.<br> This may be hard.</p> <p>or in one hour with 616,576/3600 seconds = 171,000 Watts of heat (+ losses) </p> <p>Or using a 3kW element as used in domestic hot water heaters.<br> (Max size available is usually 3 kW - larger for special uses). 616576000/3000 =~ 57 hours! </p> <p>To heat it in and '8 hour day' = 616.6 MJ/(8 x 3600) = 21.4 kW.</p> <hr> <p>Heat capacity of tank may be significant. Insulation of tank will be significant.<br> Other losses may occur.<br> But that gives you an idea.</p>

3101

Calculating time to heat up water at flow rate of 150 ℓ/min. @ 1100 bar

2015-06-09T11:47:12.390

<p>I would like to get a simple estimation of the parameters for a humidity chamber system. The humidity chamber has one constant inflow ($k_1$) and and outflow which depends on the humidity of a neighbor humidity chamber (diffusion, modeled as $k_2(RH_{ext} - RH)$. $RH$ is the humidity of the chamber, $RH_{ext}$ is the humidity of the external chamber. I have the following system description (first order): $$ \frac{dRH}{dt} = k_1 + k_2(RH_{ext} - RH) $$</p> <p>Now, I have measurement data for $RH$ (output) and $RH_{ext}$ (input). $k_1$ could be seen as a second (constant) input. I would like to estimate the parameters $k_1$ and $k_2$ in a simple manner (maybe least-square based) using my (discrete) measurement data in Matlab, but I don't know how I could proceed. Use the system identification toolbox? Other ways?</p>

|measurements|modeling|control-theory|systems-engineering|

<p>It sounds like you have <a href="http://uk.mathworks.com/products/matlab/" rel="nofollow">MATLAB</a>, so it's a simple linear fit:</p> <ul> <li>From your vector $RH(t)$, compute $\frac{d RH}{dt}$ by numerical differentiation</li> <li>Create a matrix $A$ with ones on the first column and the values of $RH_{ext}(t) - RH(t)$ on the second column</li> <li>Create a matrix $B$ with the values of $\frac{d RH}{dt}$ (use 0 as the initial value)</li> <li><p>Your result vector is given $A$ \ $B$</p> <p><code>dRH = [0; diff(RH)./diff(t)];</code></p> <p><code>A = [ones(size(RH)) RH_ext-RH];</code></p> <p><code>k = A\dRH; % k(1) is k1, k(2) is k2</code></p></li> </ul>

3106

Parameter Estimation for First Order System

2015-06-09T14:31:28.457

<p>When studying the strength of materials at the introductory level, we often make five simplifying assumptions.</p> <ol> <li>Material is homogeneous and isotropic. </li> <li>Material obeys Hooke's law </li> <li>Body is assumed to be prismatic </li> <li>Effect of self-weight is neglected </li> <li>Load is assumed as static load. </li> </ol> <p>Under what circumstances is assumption 4 valid? When can the effect of self-weight be neglected and when must it be included?</p>

|mechanical-engineering|materials|

<p>(Warning: this answer is skewed towards static applications - dynamic applications like the self-weight of a moving cable or a vehicle are much more complicated.) In actual applications, self weight is usually significant and it's fairly rare that you can neglect it for a real-world system. However, there are a number of ways to address self-weight with varying degrees of accuracy depending on the situation.</p> <p>There are only a few applications where self-weight can be written off completely as insignificant based on engineering judgement. These would be situations where the self weight is a very small portion of the total weight. One example that comes to mind would be an item hung from wire rope. The weight of the wire rope may well be 5 lbs for a load of 5,000 lbs. I would be comfortable ignoring self weight in that scenario.</p> <p>The main reason that in academic exercises, it's tempting to ignore self-weight, is that it adds an iterative step to design. You size a beam for example to meet a loading requirement (and if you're clever, an estimate for the self-weight) but then once you pick the beam and know the true self-weight, you have to re-evaluate the structure. This is a cumbersome process, and especially when working towards a specific right answer as in a textbook problem, it can take many iterations to find the perfect beam.</p> <p>When checking a design, it is necessary to validate it against your acceptance criteria (probably dictated by relevant codes) including the self-weight, but that doesn't mean that you have to follow this iterative process throughout the entire design. One option is to add an estimate of your self-weight to the other downward loads you are designing for, and process the entire design based on that increased loading criteria. Then at the end, you would add in the appropriate gravity loads, and return your other loads to their proper values for final checks. Because self-weight acts on specific members, rather than spread out evenly, it's important that your initial estimate is conservative. This method saves a lot of time when manually calculating things, but only if your self-weight estimate was conservative enough. If you can't afford to be modestly conservative in your design, then it's likely that your design won't pass the final checks.</p> <p>In the era of computer-based structural modeling, of course, each iteration is nearly free, so there is little advantage to adding this extra step. Further complicating the discussion is the fact that when seismic loading governs, it is also influenced by the dead weight, so there is a second-order effect of any approximation.</p> <p>So the short answer to your question is: it is almost never OK to completely ignore self-weight, but there are some tricks to reduce it's impact on the design process.</p>

3110

When can the effect of self-weight be neglected?

2015-06-09T15:43:24.007

<p>I have been <a href="http://www.lab-quip.co.uk/dynamic-cone-penetrometer-dcp" rel="nofollow">reading about DCP</a> and watched <a href="https://www.youtube.com/watch?v=UadgMNG-LWA" rel="nofollow">a video</a> of how it is done (the video is not in English). But there are some things that are still unclear and I would greatly appreciate some help and clarification. I have these questions:</p> <ol> <li>Is there a standard number of times that the hammer must be dropped?</li> <li>What is recorded after an $x$ number of hammer drops? <ul> <li>Lat/long coordinates of where test occurred?</li> <li>How far the metal head penetrated the ground (in cm)?</li> </ul></li> <li>What happens to that data? What does it correlate with to give us the strength of the ground? What is that strength measured in?</li> </ol>

|geotechnical-engineering|

<h3>Standard</h3> <p>The official description of the test can be found in one of the following depending on your location of interest:</p> <ul> <li><a href="http://shop.bsigroup.com/ProductDetail/?pid=000000000030087025" rel="nofollow noreferrer">BS 1377 Part 4</a></li> <li><a href="http://www.astm.org/Standards/D6951.htm" rel="nofollow noreferrer">ASTM D 6951</a> or <a href="http://www.astm.org/Standards/D3441.htm" rel="nofollow noreferrer">ASTM D 3441</a></li> </ul> <p>These should have all of the answers that you could possibly want. I'll answer your questions directly though.</p> <h3>Questions</h3> <ol> <li>Is there a standard number of times that the hammer must be dropped?</li> </ol> <p>Typically the test is run continuously from the surface of the ground to a depth of interest. It could also be run from the bottom of a drilled hole if the soil layers above that point are not of interest.</p> <ol start="2"> <li>What is recorded after an xx number of hammer drops?</li> </ol> <p>The depth of penetration is recorded after each drop. The test is continuous from beginning to end. The penetration depth should be continuously increasing as the test progresses. That is unless refusal is reached for some reason.</p> <ol start="3"> <li>What happens to that data? What does it correlate with to give us the strength of the ground? What is that strength measured in?</li> </ol> <p>Yes, there are various conversion equations to convert the number into other values. This <a href="http://docs.lib.purdue.edu/cgi/viewcontent.cgi?article=1544&amp;context=jtrp" rel="nofollow noreferrer">research report</a> shows some comparisons.</p> <p>Some typical comparisons are California Bearing Ratio (<a href="https://en.wikipedia.org/wiki/California_bearing_ratio" rel="nofollow noreferrer">CBR</a>) and density. These comparisons are based off of best fits of experimental data and are dependent on soil type.</p> <p>In general think of the Dynamic Cone Penetration Test as a cheap and quick check of subgrade strength.</p>

3115

Dynamic Cone Penetrometer (DCP) results and correlation

2015-06-10T12:11:25.607

<p>Yesterday, I tried to find constants $k_1$ and $k_2$ for the following system:</p> <p>$$\frac{\mathrm d RH}{\mathrm dt}=k_1+k_2(RH_{ext}-RH).$$</p> <p>Now, I have some first computations for $k_1$ and $k_2$ and would like to test how good a simulation with these parameters works. I think I could also write this as state space model:</p> <p>$$\dot{x}=-k_2x+\left[\begin{matrix} k_1 &amp; k_2\end{matrix}\right]\left[ \begin{matrix} u_1 \\ u_2\end{matrix} \right],$$</p> <p>$$y = x,$$</p> <p>where $x = RH$, $u_1=1$ ($const.$) and $u_2=RH_{ext}$. I have measurement data for $RH_{ext}$ in Matlab (vector), a time vector $t$ with the same length and would like to do a simulation for $RH$.</p> <p>Any suggestions how I could proceed? Simulink is possible, but just a script-based solution would be even better.</p>

|measurements|modeling|simulation|systems-engineering|

<p>Yes, if you have values for <code>k1</code> and <code>k2</code>, data for <code>RH_ext</code> and <code>t</code>, your suggestion would work although I think you have your matrices wrong. You probably need to define <code>x</code> as <code>RH - k1/k2</code> to get rid of the constant term:</p> <pre><code>A = -k2; B = k2; % x_dot = -k2*x + k2*u C = 1; D = 0; % y = x sys = ss(A,B,C,D); y = lsim(sys,RH_ext,t); RH_sim = y + k1/k2; </code></pre>

3122

Simulation of First Order System (Matlab & Simulink)

2015-06-10T18:10:07.387

<p>I have this differential equation:</p> <p>$\dot{\eta (t)}=\frac{-2\eta(t)}{C(\widehat{R}+\Delta R)}+\frac{K^2 L}{C E^2 (\widehat{R}+\Delta R)}+K$</p> <p>which, my book says, correspond to a LTI system, asymptotically stable with a constant input:</p> <p>$\frac{K^2 L}{C E^2 (\widehat{R}+\Delta R)}+K$</p> <p>It says also that the steady state value of the variable $\eta (t)$ is</p> <p>$\tilde{\eta (t)} = \frac{K C(\widehat{R}+\Delta R)}{2} + \frac{K^2 L}{2E^2}$</p> <p>The book doesn't say how to get this, but I reached the same result solving the differential equation (after a change a variable it can be solved separating variables) and calculating the limit for $t\rightarrow\infty$. My question is: is it really necessary to solve the differential equation or is there any property I can use to avoid it? I'm asking this because looking at the expression of the steady state value you can see that is equivalent to the ratio of the input and the coefficient of $\eta(t)$. Is it random or am I missing something?</p>

|control-theory|mathematics|

<p>Think for a moment about what it means for $\eta(t)$ to have reached its <strong>steady</strong> state. It means precisely that $\dot{\eta}(t)=0$. If you plug that into your first equation and solve for $\eta(t)$ you get your solution as you've already observed. </p>

3130

Steady state value, a possible shortcut?

2015-06-10T19:32:53.503

<p>One antonym word I can think of is "derating". If I say "derate motor for pump X", it can mean "select a motor with lower power output, since the pump doesn't need a more powerful motor".</p> <p>But what about selecting a more powerful motor? How can I say "I want to ____ the motor for this pump" and mean "up" the motor, "bump up" the motor, "choose-a-stronger" motor?</p> <p>I am looking for a phrase or single word, that can be used to speak of this concept clearly. For example when explaining the concept to the sales team or to the customer (user of the engineering application).</p>

|motors|sales|communication|

<p>Quite sure the word used is "oversized". When you say to an engineer that a motor is oversized for that application (for that pump) the engineer will not think the motor dimensions are too big (and will not fit) but he will understand as the motor rated values would exceed the minimum requirements by an unexpected high amount. The context of the conversation will clarify if you mean "oversized" as a good thing (heavy-duty) or as a bad thing (waste of money, energy, space or even security compromise in the case the motor can destroy the pump for example).</p>

3132

Is there a universally accepted way to refer to choosing a more powerful motor for an application?

2015-06-11T14:09:29.923

<p>The four sides of the <em>Great Pyramid of Giza</em> are oriented to the four cardinal directions of the compass (north, west, east, south).</p> <p><strong>Did I.M. Pei do the same with the <em>Louvre Pyramid</em>?</strong></p>

|engineering-history|architecture|

<p>It doesn't appear so:</p> <p><img src="https://i.stack.imgur.com/zGrdW.jpg" alt="enter image description here"> (screen shot <a href="https://www.google.be/maps/@48.860611,2.337644,745m/data=!3m1!1e3" rel="nofollow noreferrer">from google maps</a>.)</p>

3139

Is the Louvre Pyramid oriented to the four cardinal points of the compass?

2015-06-11T16:21:06.690

<p>$\frac{0.25}{s^2+0.5s}$</p> <p>Can I use formulas for 2nd order systems in <a href="http://www.ece.unm.edu/course/ece345/lectures/LectureNotes4.pdf" rel="nofollow">this pdf</a>?</p> <p>If not, how can I understand if this sytem is underdamped, damped, etc?</p>

|control-engineering|

<p>You can initially consider the system to be $$\frac{\frac{0.25}{\epsilon}\epsilon}{s^2+0.5s+\epsilon}$$ </p> <p>This gives $\omega =\sqrt{\epsilon }$ and $\zeta =\frac{0.25}{\sqrt{\epsilon }}$.</p> <p>In the limit $\epsilon \to 0$, $\omega \to 0$ and $\zeta \to \infty$. From the latter we conclude that it is damped to the hilt.</p>

3141

How to calculate of Second-order system Frequency and Time domain values?

2015-06-11T17:20:10.830

<p>I was wondering how the values of thermodynamic properties as enthalpy, entropy, internal energy, specific heat, specific volumes ... etc, of gases and liquids are calculated using programming libraries such as <code>cantera</code> and <code>thermopy</code> as an example.</p> <p>How are steam tables made for instance? How do we get all these values of properties if not experimentally?</p>

|mechanical-engineering|thermodynamics|numerical-methods|

<p>If you assume the gas you are dealing with is a semi-perfect gas (e.g. the heat capacity is not a function of pressure, and the gas follows ideal gas law), then you can calculate all properties (in the order of enthalpy, internal energy, entropy and gibbs free energy) if you can measure the specific heat capacity at constant pressure (cp, sometimes, a fitted polynomial) base on the basic relations. E.g. $dh=cpdT$, $u=h-RT$, $ds=(dh-vdp)/T$, etc</p> <p>The steam table I believe is calculate from EOS as suggested by @sturgman. The vapor liquid saturation line can be deduced (though not very accurately) using equal area rule if the EOS is cubic.</p>

3143

How are thermodynamics properties of gases and liquids calculated numerically?

2015-06-12T03:28:32.003

<p>I bought a germicidal UVC florescent lamp online and i would like to put it in a an 20 inch box for sanitation uses. After going to the hardware store I couldn't find any light that is compatible with the T8 light bulb other then this 15 Watt fixture:</p> <p><a href="http://www.homedepot.com/p/Lithonia-Lighting-1-Light-Flush-Mount-White-Ceiling-Fluorescent-Closet-Light-CUC8-15-120-LP-S1-M4/202193137" rel="nofollow">http://www.homedepot.com/p/Lithonia-Lighting-1-Light-Flush-Mount-White-Ceiling-Fluorescent-Closet-Light-CUC8-15-120-LP-S1-M4/202193137</a></p> <p>The maximum current it can support is 0.27A. The UVC bulb function at 0.6 A. Here is the link for the bulb:</p> <p><a href="https://www.1000bulbs.com/product/7342/GERM-3000008.html" rel="nofollow">https://www.1000bulbs.com/product/7342/GERM-3000008.html</a></p> <p>So the the questions are:</p> <ol> <li>Would their be too much current drawn that the heat my damage the lamp if I install it this way? (even though I would be using it for 15 min. maximum each time)</li> <li>If get a new lamp that has compatible power with the closet light above, would the lamp give the specific light wavelength required to get ultra-violet radiation (250nm)? Because I know that different ballasts would give different light temperatures.</li> </ol>

|electrical-engineering|power-electronics|

<p>Well that's a lot of light (and heat) for fixture rated for a 15W lamp. And of course, UVC light is not good for skin and eyes. Here is some <a href="http://www.americanairandwater.com/lamps.htm" rel="nofollow">more info</a> about UVC lamps.</p> <ol> <li>Actually, no, the opposite. The lamp will attempt to draw more current from this ballast than it can provide, causing the ballast to get warmer than normal. Meanwhile, the lamp will not produce the full quantity of light it is supposed to, since it is not getting the full power it is rated for.</li> </ol> <p>Furthermore, UVC lamps appear to have some <a href="http://www.light-sources.com/?q=germicidal-uvc/products/electronic-ballasts-uvc-germicidal-lamps" rel="nofollow">unique qualities</a> which may make them incompatible with traditional fluorescent fixtures. Namely, <em>"[UVC Lamps] generate harmonic and transient voltages and currents. They exhibit a seemingly strange characteristic – the voltage decreases with increasing current. This process is known as negative AC resistance. Therefore, gas discharge devices, such as germicidal lamps, are inherently unstable; once lit, the current increases without limit unless other circuit elements - ballasts - are used to restore and maintain operation."</em></p> <ol start="2"> <li>As long as a new lamp is a "UVC" type, rated for 15W, it may emit light just fine (but of course, be a "smaller" lamp, and produce less light.) However, if UVC lamp operation is incompatible with the ballast in the fixture (designed for traditional fluorescent lamps) then it might not work at all, or destroy itself, or the ballast.</li> </ol> <p>Note that since UVC is a rather unusual light frequency, some materials may be incompatible with it, such as the plastic cover for a traditional fluorescent lamp. I'm not sure, but UVC light could damage this cover, or not pass through it at all. Note that most <a href="http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&amp;field-keywords=t8+uvc+light+fixture&amp;rh=i%3Aaps%2Ck%3At8+uvc+light+fixture" rel="nofollow">UVC fixtures</a> seem to have no covers over the lamps at all. At 46W per lamp, they are going to get a lot warmer than 15W lamps also, so the added airflow may be necessary simply to cool them.</p> <p>My advice is to get a lamp and fixture designed for the purpose you have in mind. It may be more expensive, but it will be more robust and operate correctly, without fear it may overheat or fail catastrophically.</p>

3149

What will happen if I install a 46 watt fluorescent bulb on a 15 watt light fixture?

2015-06-12T03:52:35.640

<p>I'd like to know which stainless steels are good for heating on an induction stove. Wikipedia <a href="http://en.wikipedia.org/wiki/Induction_cooking#Cookware" rel="nofollow">notes</a> that the surface <s>resistance</s> resistivity is a good figure of merit. However, that article only lists the surface <s>resistance</s> resistivity for two types of stainless steel -- the better being 432. However, I cannot find 432 from my favorite supplier, although they do have 430.</p> <p>Will 430 have a similar surface resistance? Is the a resource where I can find the surface resistance of other stainless steels? I have searched around but only found results for the two stainless steels that wikipedia lists.</p> <p>EDIT: Looking at the <a href="http://dx.doi.org/10.1109/TIA.1973.349892" rel="nofollow">IEEE article</a> that underlies the wikipedia article, I think that the wikipedia article should use "surface resistivity" instead of "surface resistance". I was using the terms in the wikipedia article. Sorry for any confusion. I will edit the wikipedia article if appropriate.</p>

|materials|steel|

<blockquote> <p>... that article only lists the surface resistance for 430 &amp; 432 stainless steel -- the better being 432. However, I cannot find 432 from my favorite supplier, although they do have 430. Will 430 have a similar surface resistance?</p> </blockquote> <p>Yes.<br> 430 and 432 are listed as having almost identical resistivities - see below.</p> <p>Surface resistivity by itself is not an adequate figure of merit.<br> Inductive heating is a function of AC skin depth and resistivity. At lower frequencies skin depth increases and higher resistivity materials are required to produce the desired load. As frequency rises skin depth decreases and total material resistance increases due to less depth x area volume and lower resistivity materials are better.</p> <p>The recommendations to use stainless steels is related to the typical frequencies of current induction cookers. This is mainly limited by the economics of currently available high power electronic drivers. As driver costs fall with time frequencies will be able to be increased, reducing skin depth and allowing lower resistivity metals to be used. </p> <p>430 and 432 are listed as having almost identical resistivities - see below.</p> <p>Care should be taken to distinguish between resistIVITY and resistANCE. </p> <p>ResistIVITY is a property of the material and expresses the resistance between opposite faces of a unit cube of material but does not tell you actual resistance that results when dimensions are other than a unit cube.<br> Resistivities are usually expressed either in microOhm.inch or microOhm.centimetre. I'll call these mOin and mOcm respectively. </p> <ul> <li>The reason for the units of Ohms x length is that, starting with a unit cube resistance increases with increasing thickness and decreases with increasing face area. So R for some other shape = Resistivity x t/A Ohms.<br> To have units of Ohms, resistivity must cancel the t/A, so units of resistivity = Ohms x A/t = Ohms x length.</li> </ul> <p>ResistANCE is the electrical resistance to current flow in Ohms of the actual heating material used in the shape/dimensions used. The pot base offers Ohmic resistance to induced current as a result of material resistance for a slab the shape of the base and a depth controlled by the skin depth at the frequency used.</p> <hr> <p>Resistivities are usually expressed either in microOhm.inch or microOhm.centimetre. I'll call these mOin and mOcm respectively.</p> <p>1 mOin = 2.54 mOcm</p> <p>Links to many Stainless steel data sheets <a href="http://www.aksteel.com/markets_products/stainless.aspx" rel="nofollow">here</a></p> <p>430 - 24 micro-Ohm.Inch (mOi) From <a href="http://www.aksteel.com/pdf/markets_products/stainless/ferritic/430_Stainless_Steel_DS_201503.pdf" rel="nofollow">here</a></p> <p>430 ultraform (added Titanium) - also 24 mOin (so suspect)<br> From <a href="http://www.aksteel.com/pdf/markets_products/stainless/ferritic/AK%20430%20ULTRA%20FORM%C2%AE%20Stainless%20Steel%2020130617.pdf" rel="nofollow">here</a></p> <p>432 - 24.5 mOin<br> 304 - 29 mOin<br> Both from your wikipedia link <a href="http://www.aksteel.com/markets_products/stainless.aspx" rel="nofollow">here</a></p> <p>304 - 69 &amp; 72 mOcm (about 28 mOin)<br> 347 - 72 mOcm (~= 28 mOin)<br> 316 - 75 mOcm (~= 30 mOin)<br> All from <a href="http://hypertextbook.com/facts/2006/UmranUgur.shtml" rel="nofollow">here</a><br> (suspiciously) Identical figures <a href="http://eddy-current.com/conductivity-of-metals-sorted-by-resistivity/" rel="nofollow">here</a> </p> <hr> <p><strong>Skin effect &amp; skin-depth:</strong></p> <p><a href="https://en.wikipedia.org/wiki/Skin_effect" rel="nofollow">Skin effect - Wikipedia</a></p> <p><a href="http://www.rfcafe.com/references/electrical/skin-depth.htm" rel="nofollow">Skin deth formula &amp; notes</a></p> <p><a href="http://chemandy.com/calculators/skin-effect-calculator.htm" rel="nofollow">Skin effect depth calculator</a> - does not have an entry for stainless steel - but see simple associated formula. Skin depth scales with 1/(sqrt(relative permeability))</p> <p>Depth ~= Sqrt(Resistivity/(Pi x Freq x absolute permitivity)) see above reference</p> <p><a href="http://ecee.colorado.edu/~ecen3400/Chapter%2020%20-%20The%20Skin%20Effect.pdf" rel="nofollow">Textbook chapter</a> 7 page pdf.</p>

3150

What stainless steels are good for induction heating?

2015-06-12T04:54:03.063

<p>What is safe and what is acceptable in the context of Engineering.</p> <p>Specifically speaking we are in the process of designing a test rig which will pump water. What is a maximum safe temperature for the water to achieve without infringing employee safety. ie water at ${90}$ $^oC$ will burn so clearly that's not safe. </p> <p><strong>Q:</strong> How can I calculate/find in order to justify the value of the maximum safe water temperature to operate the test rig.</p>

|safety|

<p>A common refrain in these kind of discussions is "there is no such thing as safe, only safer." Nothing we make is perfectly safe in an absolute manner - we don't design buildings to withstand asteroid impacts and we don't design cars to survive falling off a bridge. Instead, when we say safe, we usually mean "safe enough" - the level of risk is acceptable based on an understanding of all the ways we anticipate it could fail.</p> <p>The hard part of this is defining the acceptable level of risk for a given situation. In many fields, a governing body has already defined the acceptable level of risk and written a code/standard/regulation that we get to follow. This saves everyday engineers and designers from having to spend a lot of time deciding what is and isn't acceptable. For people working in a field that doesn't have such standards, or people developing the standards, a risk assessment (however formal) is required to decide what risks can be tolerated. These assessments involve many factors, including:</p> <ul> <li><strong>What are the consequences of failure?</strong> If your water is just warm enough to cause discomfort, you may be able to tolerate that risk. If it is so hot that it could cause permanent injury, your risk tolerance should be much lower.</li> <li><strong>How likely is a failure?</strong> If the water is in an open container that frequently splashes on people, you should keep it at a cooler temperature than if it's in an insulated tank and could only escape in a catastrophic equipment failure.</li> <li><strong>What are the options for mitigating this risk?</strong> If using room-temperature water just means that your experiment will take a little more time, it may be prudent to just not heat the water at all. If your experiment can only be done at high temperatures, then you might try to mitigate the risk by not allowing employees near the hot water, or requiring them to wear special insulating clothing.</li> </ul> <p>So getting to a specific number depends on a variety of other aspects of your design. Certainly one conservative approach would be to follow whatever regulations are in place for tap water in your jurisdiction. Keeping the water in the neighborhood of 38 degrees celsius would be one reasonable option - about what is comfortable for people to touch. Of course if your process requires employees to keep their hands in the water for prolonged periods of time, even this may be too hot. </p> <p>However if the water is piped and insulated and employees will not be exposed to it directly, you could consider following standards established for steam heat or perhaps industrial process piping if they more closely match your application. In broad strokes, these alternate guidelines will allow you to keep your water at higher temperatures in exchange for building more robust piping systems with additional safety features to reduce the frequency of ruptures.</p> <p>Without more specifics of your application, system, and jurisdiction, we can't conclusively offer a specific number, but hopefully this gives you some pointers for finding the right value.</p>

3152

How does one define 'safe' in engineering?

2015-06-12T19:51:41.537

<p>I need to install an actuators to LABVIEW base test system because the old actuators are defective. We cannot purchase the old actuators because they are obsolete. The old actuators communicated using the COM ports 1 and 8. The COM port are hard coded in the LABVIEW program. The new actuators communicate via a USB interface. But COM ports 1 &amp; 8 not available. The systems states that they are "in use".</p> <p>The LABVIEW based test systems is running on a Window 7 machine. Is there a way to point the new actuator USB interfaces to COM port 1 and 8? </p>

|software|labview|

<p>What you might need to do is to remove the "in use" port and then assign the respective COM port. Follow is a suggested set of steps.</p> <ul> <li>Open command prompt (Start > Run) and type: <strong>cmd</strong> </li> <li>At the prompt type: <strong>set devmgr_show_nonpresent_devices=1</strong> </li> <li><p>Followed by C:> devmgmt.msc </p> <p><a href="https://i.stack.imgur.com/t4r14.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5vToz.jpg" alt="Administrator Window"></a><br> _{<em>Click on image for a larger version of the image.</em>}</p></li> <li>Next in the device manager view > Show hidden files </li> </ul> <p><a href="https://i.stack.imgur.com/TU0Jd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BxWIG.jpg" alt="Device Manager-1"></a><br> _{<em>Click on image for a larger version of the image.</em>}</p> <ul> <li>Expand COM by clicking on the "+" and uninstall the COM ports</li> </ul> <p><a href="https://i.stack.imgur.com/kwCG1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A8S48.jpg" alt="Device Manager-2"></a><br> _{<em>Click on image for a larger version of the image.</em>} </p> <ul> <li>From device manager select the COM port that the device is currently attach to and right click </li> </ul> <p><img src="https://i.stack.imgur.com/Okm4J.jpg" alt="COM Port Configuration"> </p> <ul> <li>Select Properties -> Port Setting -> Advance -> Assign the desired COM port from the drop down </li> </ul> <p><a href="https://i.stack.imgur.com/bg4gM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FDT4I.jpg" alt="Advance Setting for COM port"></a><br> _{<em>Click on image for a larger version of the image.</em>}</p> <p>Let me know how it works out for you</p>

3163

Unable to install Actuators to a LABVIEW based Engineering Test System

2015-06-12T20:48:46.163

<p>I have been learning the program embedded systems using Keil µVision 4.7 free version from Mentor graphic. So far all the learning has be using projects that have been already pre-defined. This has worked well so far. I would like to take the next step. Below are some of the several tutorial that I have tried with out much success. </p> <ul> <li><a href="http://www.keil.com/support/man/docs/uv4/uv4_creating_apps.htm" rel="nofollow">Creating Applications using Keil µVision IDE from Mentor Graphic</a></li> <li><a href="https://www.youtube.com/watch?v=Z_BwDGf7FTs" rel="nofollow">Create a new project using Keil uVision4 for STM32 (tutorial)</a></li> <li><a href="http://www.circuitstoday.com/getting-started-with-keil-uvision" rel="nofollow">Getting Started with Keil uVision</a></li> </ul> <p>What the minimum step to create a new an Embedded System project using Kiel µVision (Including defining the micro controller, and even setting up the debugger) so issues describe below are avoided?</p>

|electrical-engineering|embedded-systems|

<p>Here are some steps for a ARM Cortex-M4</p> <ol> <li>Close your current project in keil µVision, menu: projects->close project</li> <li>Select "New uVision Project..." from "Project" menu <img src="https://i.stack.imgur.com/q7ilH.jpg" alt="enter image description here"></li> <li>open an empty new one, menu: projects->create a new µVision project <img src="https://i.stack.imgur.com/ChA09.jpg" alt="enter image description here"></li> <li>Select a Target processor <img src="https://i.stack.imgur.com/qufIE.jpg" alt="enter image description here"></li> <li>You get a popup that asks if you want to copy the startup assembler file into your project startup startup_xxxx.s.<br> <img src="https://i.stack.imgur.com/yNVps.jpg" alt="enter image description here"></li> <li>now you see the empty project with just the startup file included. <img src="https://i.stack.imgur.com/YQ0yB.jpg" alt="enter image description here"></li> <li>Project is ready for embedded software development. It is good idea to rename "Target 1" and "Source Group 1" that describes the project</li> </ol> <p>Right click on Target 1 > Select Manage Project Items -> Update Project Targets, Groups and Files <img src="https://i.stack.imgur.com/n2xdW.jpg" alt="enter image description here"></p> <ol start="8"> <li>In order to startup our micro-controller, we need to add 'system_.c' file to our project. Keil already has this file defined. To add manually, Right-click on our "Startup" group and select "Add Existing Files to Group 'Startup'...":</li> </ol> <p><img src="https://i.stack.imgur.com/Zhp4H.jpg" alt="enter image description here"><br> a. Keil already have common startup files for some microcontrollers. They are stored at /Keil/ARM/Startup folder<br> <img src="https://i.stack.imgur.com/z8RX9.jpg" alt="enter image description here"><br> b. From there, for my LaunchPad I will move to TI folder, and then TM4C123 folder:<br> <img src="https://i.stack.imgur.com/NduA0.jpg" alt="enter image description here"><br> c. And there i can see my system_TM4C123.c file:<br> <img src="https://i.stack.imgur.com/DXjmK.jpg" alt="enter image description here"></p> <ol start="9"> <li>now you need to add the copied file into the project. project->manage->components,.. or simply click the icon with the three coloured boxes add there the startup_TM4C123.c file. you should then see it listed on the left window bar.</li> <li>now create your xxx.c file containing the main procedure. you need to add it to the project again, like you did in the previous step.</li> <li>and now you need to set some registers in the controller. for now you can simply do that at the beginning of the main procedure.</li> </ol> <p>Note: Most the screen shots were created by screen name vasily.sib on piazza forum. I copied these improved them for my purpose.</p>

3165

How to create Embedded System Program from Scratch using Keil µVision?

2015-06-13T22:07:23.317

<p>So i had this experience when driving in a cold night, where my windshield just got all blurry. I knew that on the drivers manual they say you should turn the hot air and soon it will go off, but my friend that was on the passenger seat said the cold air will work too. We turned the AC off and re-experienced the fact. I got the chronometer and and got the time of both situations, defog with hot and cold air and there were really small diference btween them. How can i explain this in terms of thermodynamics? </p>

|thermodynamics|

<p>I just came from a heavy rain in a cold day, and wanted to know if any body had posted about this. My facts are that using only the heater, is more delayed and besides that all other car windows get fogged, There after I left the heater on, and turned on the AC, and the result is amazing, not only the windshield cleared, but all the windows around, to the point that I didn´t need to use the rear defogger.</p>

3174

Which is better: Defog a car windshield with cold or hot air?

2015-06-14T23:45:36.407

<p>I need to find the emission power of a WiFi mobile phone antenna. I have checked the IEEE 802.11 specifications but it says that the power transmission is variable and depends on the mobile phone fabricant and power saving options. </p> <p>I also searched for information about some specific mobile phones but the datasheets available on the Internet seem quite weak and do not say anything about WiFi power transmission.</p> <p>Is there a complete datasheet of mobile phones? How can I determine the transmission power in the WiFi antenna of a specific mobile phone? </p>

|electrical-engineering|power-electronics|wireless-communication|

<p>There is a list of some widely used devices and their transmit power (and many other properties) in an easily digestible table: <a href="https://clients.mikealbano.com/" rel="nofollow noreferrer">https://clients.mikealbano.com/</a> The most smartphones seem to be just around 20 dBm and the maximum lies around 30 dBm.</p>

3183

How can I find the emission power of a WiFi mobile phone antenna?

2015-06-15T09:15:38.480

<p>I have to put together a small solar system for a college project. </p> <p>I have solar cells with an output of 3.65A and 0.6V. Now if I paralel two groups of two cells connected in series I will double the amps and volts. And the output power will be like 8W. </p> <p>The question is, do a 1.2V panel really work? Can I charge a 12V battery? I mean the panel will be 8W and high amperage low voltage. </p> <p>Do I need to have a minimum 12V panel to charge battery? This means I will have to put in series 20 cells to obtain 12V. </p> <p>Are there high voltage low amperage solar cells so I don't need to put together so many cells? </p>

|electrical-engineering|solar-panel|photovoltaics|

<p>If you want to charge a 12 V battery, you need a bit more than 12 V.</p> <p>One way to achieve that is to put enough individual cells in series. Another way is to have the panel produce a lower voltage and use a electronic circuit called a <i>boost converter</i>. These convert low voltage at high current to high voltage at low current.</p> <p>If you're clever, you can build in optimal use of the solar array and battery charge management into the boost converter. However, if the panel can only put out 8 W, which is 670 mA at 12 V, then you can just size the battery so that it can take 670 mA indefinitely without harm. Or, if this is a lead-acid battery, just have the boost converter produce 13.6 V when it can, but never more. A "12V" car batter, for example, can take that indefinitely.</p>

3184

8W Solar panel high amperage low voltage

2015-06-15T10:35:18.380

<p>I have <a href="http://www.allpumpsdirect.co.uk/files/1806869/uploaded/FLJ%20Duplex%202%20Tech%20Data.pdf" rel="nofollow">specifications for a positive displacement pump</a> where, in the data table on the last page, the given flow rate is referred to in the column header as 'Open Flow.'</p> <p>Is this the maximum flow generated by the pump?</p>

|mechanical-engineering|fluid-mechanics|pumps|

<p>The open flow is the flow rate through the pump where no pressure is developed. That is, if you have fluid coming in and out horizontally, and the input and output hoses are open to the outside air (not a closed system developing pressure. This is the most flow the pump will ever create.</p> <p>Usually though, we use pumps to either lift fluids vertically, or to pressurize a closed system. In order to do either of these, you need a higher pressure on the outlet side of the motor than on the inlet side. In these cases, the pump will run slower as it has to exert more pressure. The curves on the second page of your link show you for each bar of pressure you need to develop, how much the flow rate will be reduced. In the case of a pressure piping system, you'll know the pressure rise you need directly. If you're using the pump to push fluid uphill, the pressure will depend on the vertical rise (usually referred to as 'head') and the density of the fluid. There will also be frictional losses which may be worth accounting for depending on the length of your piping system. For water, each meter of head is roughly .1 bar, closer to .09 for oil.</p>

3187

What does 'Open Flow' mean for a pump flow rate?

2015-06-16T11:29:21.090

<p>I have been assigned to build a ramp for a small car. I have only one A2 sheet of paper, scissors and a stapler. What would be the best slope profile, in order to move the car as far as possible after leaving the ramp, using only gravity? Any other ideas on construction? </p> <p>The car is 1:43 size. The weight is 100 g. The car must stay on the track and there is no initial push.</p>

|applied-mechanics|

<p>So there's a number of concerns that you want to think about. Some of these you'll have to determine experimentally, others you'll be able to calculate and then iteratively improve based upon your observations.</p> <p>It should be understood / obvious that the higher up you can get the car, then the faster it will be able to go, and ultimately travel the furthest distance. The challenge though is transferring the vertical potential energy into horizontal motion. So the taller and longer of a ramp you can make, then the furthest your car will end up traveling.</p> <p>Use a fixed, rigid plate to form the initial ramp. Run some experiments to determine the steepest angle the plate can form but still allow the car to roll smoothly off of the ramp. Having the car crash at the end of the ramp doesn't do you any good and wastes any energy that was converted.</p> <p>Once you know your angle of attack for the ramp, you'll want to start experimenting with various ramp designs. This is where the principles of beam deflection start coming into play. </p> <p>I'd look into C and I (or H) beams as they ought to provide optimal stiffness at the edges of the beams while minimizing the amount of material required. An H beam design may allow you to join two beams together and fasten them in areas where the car won't be impacted by the staples. Another advantage of those beams is that they'll provide rails to keep the car on the track. That said, you'll need to fold under the edges of the H beam where the ramp meets the horizontal surface the car will travel along.</p> <p>Likewise, look into triangular beams to create the struts that will support and lift your ramp up into the air.</p> <p>If you'd like to do some of the beam calculations ahead of time, it's pretty easy to search for Young's modulus of elasticity for paper. That parameter can be affected by paper weight and coating, so that's research you'll want to do on your own.</p> <p>Knowing the angle of attack and your preferred beam design will allow you to lay things out on the sheet of A2 paper that you have. This represents your final constraint inasmuch as you have a fixed amount of material you can use.</p>

3206

How to build a paper ramp for a small car?

2015-06-16T11:58:58.760

<p>I have the task to connect 6 input sources of fluid with 9 output (sinks) sources. Each input should be connected to each output. Each input should be able to distribute the fluid to only 1 output at a time. I have straight pipes, Y connections and valves, that I can use. I should use the MINIMUM amount of valves and Y connections as possible. </p> <p><strong>Is there a standard approach to reducing the number of components required for a design like this?</strong></p> <hr> <p>This is the best I could get to so far: <img src="https://i.stack.imgur.com/ZUt7z.jpg" alt="enter image description here"></p>

|fluid-mechanics|pipelines|

<p>I recommend approaching problems like this by starting with a base case and working up from there until you see a pattern.</p> <p>For piping, the base case is a straight case with one input flowing to one output. Depending on the exact specs of the problem, this requires either 0 or 1 valves; it does not require any branching.</p> <pre><code>I------O (or, I--▷◁--O) </code></pre> <p>Add an input to the base case:</p> <pre><code> I--▷◁--Y---O / I--▷◁--+ </code></pre> <p>Add an output to the base case:</p> <pre><code>I---Y--▷◁--O \ +--▷◁--O </code></pre> <p>Add both an input and an output to the base case:</p> <pre><code> I--▷◁--Y---Y--▷◁--O / \ I--▷◁--+ +--▷◁--O </code></pre> <p>Here's a <a href="https://i.stack.imgur.com/qVj3W.png" rel="nofollow noreferrer">nicer-looking diagram</a>, if you prefer.</p> <p>What you can see from looking at these simple cases is that:</p> <ul> <li>A single input or output requires no valves.</li> <li>If there is more than one input or output, each requires one valve.</li> <li>Each input or output beyond the first requires one branch.</li> </ul> <p>Since your problem has multiple inputs and multiple outputs, the minimum number of valves is equal to the total number of inputs and outputs and the minimum number of branches is equal to two fewer than the total number of inputs and outputs.</p>

3207

Distribute 6 inputs to 9 outputs with pipes

2015-06-16T21:30:06.937

<p>I work in consultancy and I am designing a product for a customer. The customer has requested a specific PSU. I have contacted this PSU manufacturer and they request 6 L/min of coolant flow and minimum 3 Bar. I now have to specify a suitable pump to deliver these fluid characteristics to the PSU.</p> <p>I have confirmed this is not a typo for "maximum pressure," but no reasoning for specifying a minimum pressure is forthcoming.</p> <p>My confusion is why the flow rate AND minimum pressure are specified. The manufacturer does not know what pressure differential I require to generate the specified flow around my bespoke circuit - so why specify a minimum pressure AND a flow rate?</p>

|fluid-mechanics|pressure|power-electronics|cooling|computer-hardware|

<p>The pressure requirement could also be to raise the temperature at which the water boils. The boiling point of water depends on the pressure it's at; higher pressure raises the boiling point, as the following image shows.</p> <p><img src="https://i.stack.imgur.com/ZPggh.png" alt="Image taken from www.engineeringtoolbox.com"></p> <p>As you can see, water at atmospheric pressure boils at 100 C, but water at 3 bars boils at about 130 C. When water starts to boil, it becomes less effective at removing heat. So, the PSU internals might operate hotter than 100 C, which requires that the water be under pressure to prevent boiling. See <a href="https://en.wikipedia.org/wiki/Critical_heat_flux" rel="nofollow noreferrer">this Wikipedia article</a> on critical heat flux for a more detailed explanation.</p> <p>On a side note, coolant reaching its critical heat flux was a major factor that doomed reactor #4 at the Chernobyl nuclear power plant.</p>

3215

Why is minimum pressure specified on a PSU water cooling circuit?

2015-06-17T02:28:36.597

<p>I have here a heating element that broke when it was used to ramp a furnace to 750 C, breaking at 740. The element had previously been used for higher temperatures and appears that it may have possibly shorted. The shape of the element is custom, as well as the furnace, and so we had a custom element machined with a specific wind, but this new element is straight. I have few tools and am limited in buying tools; So, whats the best way I can give the coil this shape? Some things to consider:</p> <ul> <li><p>The previous coil broke at what appears to be a weak point. It's possible that it shorted, but its also possible that it just gave out. Therefore, with shaping it, there needs to be as few of strained points as possible.</p></li> <li><p>It needs to remain planar, which is our largest problem right now. I attempted to curl it around a cylinder, and it now spirals upward (we have more coil if we need to start over.) If it doesn't remain planar, shorting will be likely.</p></li> <li><p>When we attempted to spiral around a cylinder, it gives a good general shape, but will generally expand after releasing tension.</p></li> </ul> <p>Below are two photos of what we are dealing with. The poorly coiled lustrous one is the new, and the burnt black piece in the ceramic is the old.</p> <p><img src="https://i.stack.imgur.com/CoRPI.jpg" alt="enter image description here"></p> <p><img src="https://i.stack.imgur.com/3YMvj.jpg" alt="enter image description here"></p>

|mechanical-engineering|

<p><strong>EDIT</strong> Just read the comments, #3 might be the best option, as #1 won't work, and #2 wont shape it, just hold it in place. </p> <p>First thing I am assuming is that this stays in the ceramic as it is heated. </p> <p>A couple things I would try:</p> <ol> <li>Fit the coil in the ceramic like the bottom picture is, but with the new coil, then put something heavy on top and let it sit there for a long while. Hopefully that would work, but no promises. </li> <li>Find a thin, non-conducting material that can withstand at least twice the temperature that it will be subjected to (Possibly even more ceramic), and put a plus shape over the heating element to hold it in place. </li> <li>Bend the coil bit by bit to get it to stay in the right shape. start by bending it to where it should be, then releasing it. If it does not stay, bend it past the point you want it at. Keep bending it farther until it stays in the place you want it. You may need to do very small sections at a time, so this will take a very long time. Unfortunately, this may cause a little weakness in the metal.</li> <li>Carve a wood shape that matches the ceramic and use that to form the spring using method 1 </li> </ol> <p>Good Luck, I hope you can get something to work!</p>

3220

How do I manually shape this heating element?

2015-06-17T08:36:55.010

<p>I recently built the basic construction for a crane out of lightweight material. Basically the head of the crane contains an electric servo motor which spins the head of the crane by pushing itself away from the tower (the rotor is fixed to the axis which is fixed to the tower. The axis isn't fixed to the head in any way).</p> <p>All electronics required to operate the motor and more are inside the head. The only thing I should extend to the rest is a 240 V cable.</p> <p>Since the head is allowed to make multiple rotations (preferably limitless), <strong>how do I avoid snapping my cable by rotating it too often?</strong> I'm aware such solutions exist, but I have no experience in this area so I have no idea what to ask for specifically.</p> <pre><code>/---------------------\ | | | motor | \----------|----------/ | ___________|___________ \ | / \ | / </code></pre>

|electrical-engineering|motors|

<p>The device you are looking for is called a <a href="https://en.wikipedia.org/wiki/Slip_ring" rel="noreferrer">slip ring</a> such as the one shown below. These devices use internal components which maintain contact as they spin (old ones used brushes) allowing them to turn by an unlimited amount in either direction. </p> <p><a href="http://www.mercotac.com/?gclid=CP-b7NnWlsYCFYUXHwod-y0Afw" rel="noreferrer"><img src="https://i.stack.imgur.com/YJyZ6.gif" alt="enter image description here"></a></p>

3221

Avoid snapping an electric cable from too much rotation

2015-06-17T14:34:06.153

<p>How do we start a nuclear chain reaction? For example, if we wanted to do it this weekend, what would we need, and what are the basic steps? I'm not looking for a text book response or highly detailed information. I'm only curious about the most basic components. For example, we need to take A and B and submerge it into C? That's all I'm asking. If you had a daughter, and she asked, "What happens in a nuclear power plant?", what would you tell her?</p>

|nuclear-technology|nuclear-engineering|

<p>Nuclear fission is spontaneous. In fact, if you have a critical mass of fuel assembled, you need to take explicit steps to keep it from going into chain reaction, typically by adding neutron-absorbing material, AKA "control rods" to the assembly. When you remove the control rods, the reactor starts up by itself.</p>

3227

How does a nuclear reactor initiate nuclear fission?

2015-06-17T20:57:46.077

<p>My North American facility is looking at a major shut down this fall. In the area, three other large facilities are also doing either shutdowns or expansions in the fall. </p> <p>A last minute project that I have been assigned requires a couple of weeks of machining custom components. This is an issue as none of the local machine shops are likely to be able to fit the work within their schedule in the required time frame.</p> <p>At this point I think my only option is to start looking beyond the local shops. What should one look for when trying to source non-local machining? Are there any red flags to look for when talking to a shop? </p>

|machining|project-management|

<p>At very highly level the three areas that need to manage are scope, cost and time also known as the triple constraints. </p> <p><strong>Scope:</strong><br> Make sure the sponsor clearly defined the project. It is highly unlikely that the sponsor will document the scope. Therefore you need to document the scope and get some type of agreement from the sponsor. An e-mail might be a good start.</p> <p>Make sure all drawings and other documents are clear with sufficient details. Recently I encountered there was a situation where specification state, use ABS (Acrylonitrile Butadiene Styrene), without any other details. These types of situations create confusion, delay, and unnecessary costs.</p> <p><strong>Cost:</strong><br> Make sure your sponsor, you and the vendor clearly understand the project related cost. It is good to request for quotes from multiple sources, and understand all the deliverable. Considering the current situation expect pay more than the market price. It is good idea to have about a 10% buffer. </p> <p><strong>Timing:</strong><br> Make sure the machine shop clearly understand the project timing. Don’t push the machine shops, they may agree to timing that they are unable to fulfill. Make an attempt to have regular communication with the vendor. This is very important. </p> <p>It might be a good idea to seek recommendations from your current local machine shops. </p> <p><strong>Reference:</strong> </p> <ul> <li><a href="http://www.pmi.org/default.aspx" rel="nofollow noreferrer">Project Management</a> </li> <li><a href="https://pm.stackexchange.com/">Project Management Stack Exchange</a></li> </ul>

3234

Going distant in sourcing machining

2015-06-18T07:10:50.970

<p>Trivially you can prevent rocking with many vehicle by eliminating the suspension. But is it possible to allow a floaty suspension that glides over potholes, without also having to rock back and forth, with some kind of stabilization? I know that it's possible in buildings, but having a fixed platform really helps.</p> <p>I ask because the rocking motion is one of the primary sources of carsickness, and a bus trip this evening left me feeling rather queasy. This seems like something that could be improved, especially in such expensive vehicles.</p>

|mechanical-engineering|applied-mechanics|automotive-engineering|

<p>The <a href="http://en.wikipedia.org/wiki/Citro%C3%ABn_Xantia" rel="nofollow noreferrer">Citroën Xantia Activa</a> was one of the earliest mass production active suspension vehicles I know of.</p> <blockquote> <p><strong>Activa active anti-roll bars</strong></p> <p>In 1994, the Activa technology was introduced, which is an extension to the Hydractive II suspension, where two additional spheres and two hydraulic cylinders are used together with computer control to eliminate body roll completely. This technology is more broadly known as active suspension, and the Xantia Activa has exceptional road holding comparable to true sports cars. It employs active anti-roll bars.</p> <p>In the Swedish magazine Teknikens Värld's moose test the 1999 model of Xantia V6 Activa still holds the record speed through the manoeuvre - faster than the Porsche 996 GT2.</p> <p>UK Models of the Activa came fitted with a XU10 2 litre turbocharged engine also fitted to the Citroën XM 2.0CT and Peugeot 605 SRi. It produced 150 bhp and 171 lb ft of torque and was a 'low-blow' type for smooth power delivery rather than outright bhp.</p> </blockquote> <p>So it's very possible to create a flat platform.</p> <p>Anecdotally, friends as kids used to get car sick more often in Citroën cars with hydraulic suspension than normally sprung cars, my guess is that the ride makes you feel sea sick.</p>

3238

Is it possible to design a bus that doesn't rock?

2015-06-19T00:54:54.150

<p>My project team and I delivered the design of a 14ha urban infrastructure development in Sydney for a Client. The project was to hand over the development at a Tender level of design, which would be tendered upon by a number contractors. The succesful bidder would then take on the rest of the project as a Design and Construct contract, and complete the works.</p> <p>As a part of our deliverables, we issued all engineering plans in pdf and AutoCAD dwg, and all design files including the 12d data terrain models and design strings. We have now had a request from the winning contractor to provide the entire 12d '.project' file, which includes all functions, modifiers, appliers and range files. Providing this would save the contractor a significant amount of time, even considering all of the design strings and models which have already been issued.</p> <p>Our involvement in the project has been completely handed over, and we consider all of our responsibilities departed. There are a number of people in our team who consider it not appropriate to provide the .project file, whilst there are a number of people who don't see it as an issue.</p> <p>Has anyone run into this issue before, or has anyone got any guidance on whether it should be provided? Are there any risks in sending the information? Does anyone have any thought's on where our Client's ownership of the data ends and our Intellectual Property begins?</p>

|civil-engineering|modeling|design|computer-aided-design|

<p>Before I start, let me disclose that I don't know a lot about the details of infrastructure projects or your file format, but I think the general considerations are fairly universal.</p> <p>There are a number of considerations when providing your digital model to a contractor, and in my work I have decided to offer them my model in some situations, but not in others. Here are the main factors:</p> <ul> <li><p><strong>Linear Time.</strong> By providing a model, you can probably save the project significant linear time by reducing the amount of work the contractor needs to in order to prepare their shop drawings, allowing the work to start and finish sooner. This is probably the most compelling reason to provide the model. It's worth noting that in some industries where file formats and workflows aren't well standardized, sometimes this efficiency can't actually be realized as things have to be re-drawn or re-modeled anyway for the target platform.</p></li> <li><p><strong>Risk.</strong> If there is a discrepancy between the model and your drawing, there should be language that instructs the contractor to follow the drawing. However, you have no control over how the contractor uses your model, and they may uncover an error in the model that you never noticed or didn't care about because it is never shown in the drawings. Depending on the contract language, these errors could expose you and your employer to additional liability.</p></li> <li><p><strong>Proprietary information.</strong> By giving the contractor your design as a complete model, rather than just output drawings, you make it much easier for them to copy designs you may have developed over multiple projects and re-appropriate them on other jobs. While a well-written contract will make that clearly illegal, it's very hard to translate that into recovering financial losses. In addition, any custom objects (blocks, toolsets, plug-ins, etc.) you have created for the software may be transmitted. Again, this kind of intellectual property is hard to control once it is out in the world.</p></li> <li><p><strong>Costs.</strong> Unless someone made a major mistake, the contractor should have bid the job assuming they had to work off of your drawings (the contract documents) and do or redo any ancillary work required to develop their shop drawings and fabricate/build the job. Likewise, your contract was just to produce the drawings, and the model is simply a tool you used. You may have created a model that is good enough for your drawings, but making it good enough to build off of would cost you money. For the added quality expectation, and the added risk, it would not be unreasonable to charge a fee for providing the model. Even if your model is perfect, the contractor may have questions about some part of it, or ask you to convert it to another format, or otherwise have modest requests that mean you need to devote time to the project that you didn't bid on.</p></li> </ul> <p>By way of example for the discrepancy issues, when making a drawing for a machined part, I might not draw every feature to scale in my model out of laziness. While this is a bad practice, if I override the dimension and the output drawing clearly conveys the correct information, I have met my contract requirement in the design/engineering role. However, if I then provide the model, and the machine shop doesn't check the written dimension, the piece would be fabricated wrong. Each party will feel confident that this is the other party's fault. If I had never provided two redundant sources of information, blame would have been much easier to assign. This is the same reason that traditionally, we don't dimension segments as well as the whole that they add up to. We try to avoid opportunities for information we provide to be internally inconsistent.</p> <p>It's obviously not directly applicable to your project, but here in the US, the <a href="http://www.aisc.org/WorkArea/showcontent.aspx?id=25256" rel="nofollow">AISC Code of Standard Practice</a> (AISC 303-10) answers contract questions like these for steel projects when they aren't clearly addressed by a specific contract. See section 4.3 for the details, but the general points are:</p> <ul> <li>The contractor can only use the model with permission of the engineer, and only for the one job (not to extrapolate to future jobs)</li> <li>The contractor has to acknowledge that the model is not a contract document (with an exception for projects where the original engineering deliverable was a digital model in lieu of drawings) and is superseded by information on the drawings.</li> <li>The contractor is still responsible for verifying all information</li> <li>The contractor has to remove information not relevant to their scope of work.</li> <li>(In the commentary:) The engineer may choose to charge the contractor a fee for use of the model, without selling the intellectual property rights to the contractor.</li> </ul> <p>In the end, the decision is very job specific. On a fast job with a team that has good existing relationships, it may be worth it to provide the model for free as a good will gesture. On very large projects with lots of locked up risk, it might be wise to only provide what you are contracted to. There may exist a middle ground where providing your model is an option, but only for a price.</p>

3247

Sending out a 12d Civil Design Model

2015-06-19T13:35:22.110

<p><a href="https://i.stack.imgur.com/Fieak.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fieakm.jpg" alt="enter image description here" /></a></p> <p>I have to solve the differential equation for radial disks, and therefore I need the shear force distribution. The disk is rotationally symmetric.</p> <p>I want to use the superpositions principle and set 3 areas.</p> <ol> <li><p><span class="math-container">$0 \le r \le D_i / 2$</span></p> </li> <li><p><span class="math-container">$D_i / 2 &lt; r \le k / 2$</span></p> </li> <li><p><span class="math-container">$k/2 &lt; r \le D / 2$</span></p> </li> </ol> <p><a href="https://i.stack.imgur.com/fGjnv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fGjnvm.jpg" alt="enter image description here" /></a></p> <p>The shear forces are for area 1</p> <p><span class="math-container">$\sum F_z = 0$</span></p> <p><span class="math-container">$F_{rz1} \pi r - \frac{\pi r^2}{2}p = 0$</span></p> <p><span class="math-container">$F_{rz1} = \frac{p r}{2}p$</span></p> <p>Area 2</p> <p><span class="math-container">$0 = -F_{rz2} \pi r + F_s \pi k $</span></p> <p><span class="math-container">$F_{rz2} = F_s \frac{k}{r} $</span></p> <p>Area 3</p> <p><span class="math-container">$0 = F_{rz3} $</span></p> <p>The solution confirms <span class="math-container">$F_{rz1}$</span> and <span class="math-container">$F_{rz3}$</span> for me. However it states that only after <span class="math-container">$F_{rz3}$</span> is known one can solve for <span class="math-container">$F_{rz2}$</span> for the area <span class="math-container">$D_1/2 &lt; r \le D / 2$</span> The solution should be <span class="math-container">$F_{rz2} = F_s \frac{k}{2r}$</span></p> <p>Did I cut correctly?<br /> What am I missing here?</p> <p>I know realized I had to insert <span class="math-container">$k/2$</span> instead of <span class="math-container">$k$</span> and then I get the correct solution for</p> <p><span class="math-container">$0 = -F_{rz2} \pi r + F_s \pi k/2 $</span></p> <p><span class="math-container">$F_{rz2} = F_s \frac{k}{2r} $</span></p> <p>However the question that remains is, why the hint to extend the area? Why is it not possible to calculate <span class="math-container">$F_{rz2}$</span> before <span class="math-container">$F_{rz3}$</span> when it apparantly is possible?</p>

|mechanical-engineering|mathematics|

<p>When working with superposition method, any way of approaching the disc works - so long as you account for the reaction forces properly. In this context, it is important to ensure that the sum of the forces is always 0, so unknown reactions (like your simple support in the middle) can be factored in appropriately.</p>

3252

Shear Force Distribution on disk

2015-06-19T20:43:22.043

<p>When testing a pressurized pipe with off-standard flanges, often times the materials available to fabricate a blind flange of the correct size are limited to what is around.</p> <p>I've made it a habit of referring to Roark's formula's for stresses and strains, and going to the table for large deflection circular plates for these. I use the area inside the bolt circle. They have three boundary conditions available:</p> <ol> <li>Simply supported (neither fixed nor held)</li> <li>Fixed but not held (no edge tension)</li> <li>Fixed and held.</li> </ol> <p>Bolted connections are unusual - especially since I have a full face flanges. I've designed to both simply supported and fixed but not held. The simply supported works, the fixed but not held does not. However, due to the flatness out to the OD of the face, it seems like the rotation of the edge should be 0, so it should be fixed but not held. I'd like to make these thinner, so it would be nice if someone can demonstrate an accurate boundary condition (or perhaps that 80% of the fixed + 20% of the simply supported would be conservative via FEA). I'd like to see better research into the mechanics of design for blind flanges (which is rarely covered in the context of the Boiler and Pressure Vessel Code, or elsewhere).</p> <p>The main question is: what boundary condition for the edge should I use?</p>

|mechanical-engineering|piping|bolting|

<p>Given the clamping force of the bolts, I would think that Fixed and Held would be appropriate.</p> <p>One way to test this is to calculate the edge tension (I don't recall if Roark's has formulations for this) and compare this to the clamping force of the bolts (i.e., the static frictional force between the flanges due to the clamping will need to be greater that the edge tension).</p> <p>I'm not certain that this case really meets the criterion of "large deflections" for the blind flange. The ones I've seen used are pretty beefy in thickness compared to the span and have almost no deflection.</p>

3257

What boundary condition should I use for the edge of a blind flange?

2015-06-20T08:23:30.653

<p>I have machined parts that are made of aluminum and anodized black. The reason that I chose aluminum is because:</p> <ol> <li>It can easily be machined and is low cost and is common, so easy to buy</li> <li>It is light weight compared to steel</li> <li>It can take heat</li> <li>It is easy to finish with oxidation</li> </ol> <p><img src="https://i.stack.imgur.com/8CmIt.jpg" alt="1![2![3"><br> The parts are used on a CNC machine so being lightweight is important so that the machine can move faster.</p> <p>But, I need my part to be significantly lighter weight, so would like to switch materials, but not pay a lot to get a prototype made. </p> <ul> <li>Is there another metal that is a good substitute? </li> <li>Can a plastic laminate be used? For example, plastic parts that I can get machined then add a high temperature tape to reflect heat?</li> </ul>

|mechanical-engineering|aluminum|cnc|

<p>I'm not sure how your part looks, but I can tell you about temperature and plastics.</p> <p>I've blasted <a href="https://en.wikipedia.org/wiki/Phenol_formaldehyde_resin" rel="nofollow">phenolic</a> composites with <strong>600 °F (310 °C)</strong> and haven't had a problem. There are high temperature epoxies and vinyl esters that get to lower temperatures - and those are a lot more commonly available. I wouldn't touch them with anything over <strong>300 °F (150 °C)</strong> though.</p> <p>All composites range from <em>0.05 - 0.07 pci (1.38 g/cc - 1.97 g/cc)</em>, depending on if you use more reinforcement, or more resin. The phenolic can be as strong as an epoxy laminate, in testing. For glass reinforcement, you can guarantee breaking strengths of <strong>25 ksi (200 MPa)</strong>. If you designed your parts with a professional, orienting the fibers, you can get above <strong>500 MPa</strong>. Wikipedia cites breaking strengths going into the GPa range, but there is no guarantee your part will have that - those are lab sample strengths, not actual usable part strengths. Bear in mind fiberglass is fatigue resistant (better than your aluminum!), but only if the part is designed with a factor of safety of 5.</p> <p>Hope that helps you understand the world of high temperature composite substitutes.</p>

3259

Lightweight substitutes for machined aluminum parts

2015-06-21T19:05:01.283

<p>It is possible with a microphones array to isolate a sound coming from a direction. Is it possible with an array of photo diodes to isolate a light signal coming from one direction?</p> <p>For example having 4 photo-diodes oriented at the same direction, spaced by 2 cm, and facing lamps at the ceiling, is it possible to isolate the amplitude of each light source reaching the photo-diodes from different directions (different lamps). What would be the algorithm or the method used for doing such thing?</p>

|electrical-engineering|

<p>This concept in general as it applies to microphones is known as <a href="https://en.wikipedia.org/wiki/Beamforming">beamforming</a>, and is most commonly applied to sound and radio waves. It is mostly aimed at wave forms that carry signals (ie change significantly over time,) but similar concepts could be used in your example.</p> <p>In general terms, beamforming on the receiving side comes in two flavors. Either you know the location of the signal you want to receive, and you phase-align your sensors to be more sensitive to that direction, or you know what your signal 'looks' like, and you can locate it based on phase differences in your array of sensors. </p> <p>You are proposing the second application, so the difficult thing for you is knowing how to distinguish one light source from all of the others. For example, if you had only one light in a dark room, it would be easy to identify your target signal. In your situation with multiple lights, if they all produce the same wavelengths, it will be nearly impossible to distinguish them if they are all on at the same time. If they produce different colors, you could possibly distinguish them, but you would still have a different problem.</p> <p>In order for beamforming based on phase shift to work, you have to be able to identify each received waveform, and index them against each other. There are two problems here. The first, inherent problem is that conventional light bulbs produce essentially the same wave form forever, dimming very slowly, until they burn out. The problem with this is that since the waveform has no distinct changes in it, you can't index it to find time delays. when you see the same signal from two sensors, you might notice that they are 180 degrees out of phase, but you wouldn't be able to tell if they are just half a wavelength different, or 1000.5 wavelengths different. In order to tell how much more the signal has been delayed traveling to one sensor, you need a distinguishable change in the waveform (for example the light getting brighter or changing color.) The second, more pragmatic problem is that I'm not aware of any photodiode that is so responsive that you could actually observe the waveform of light. I'm not an electronics expert though, so they may exist.</p> <p>So we need a strategy to make the waveform of the light source less uniform, and ideally to add a much lower frequency signal which is easier to detect. Assuming you have control over the lights, this should be easy to do. At its simplest, you could turn on one light at a time, and watch for the first light to hit your sensors. Since you know how far apart your sensors are, and you can easily identify the relative delays between the first and last sensor the light hits, you can use <a href="https://en.wikipedia.org/wiki/Multilateration">Multilateration</a> to locate the source. If it's not an option to only have one source on at a time, there are other ways to make the source identifiable such as perhaps turning it from 100% brightness to 80% brightness or changing it from white to blue at intervals. How significant these changes will be, and how robustly your system will detect them will depend on the sensitivity of your instruments and the degree of background noise.</p>

3263

Light signal isolation

2015-06-22T12:26:39.303

<p>Every scientific calculator I've come across has a DRG button that controls whether trigonometric functions use units of degrees, radians or grads yet I've never seen or heard of any system that actually uses grads. When is the grads setting useful?</p>

|measurements|

<p>Grade is often used for inclination, such as looking at how steep a road is. The units are not as "crazy" as they seem, when you think of grade as a percentage. In other words, a hill can go down at anything from 0% (flat) to 100% (45 deg angle). (rise/run) Based on some comments added below, it is not clear if the G on this calculator is actually grade or not.</p>

3268

What are grads used for on a scientific calculator?

2015-06-22T17:37:19.177

<p><a href="https://youtu.be/WOoUVeyaY_8?t=284" rel="nofollow noreferrer">In this video</a>, the video emphasises the very low recoil level of the AA-12 fully automatic shotgun. In particular, at <a href="https://youtu.be/WOoUVeyaY_8?t=300" rel="nofollow noreferrer"><strong>this point in the video</strong></a> the presenter uses one hand to relatively lightly grip the gun while firing 300 rounds per minute. Note that the presenter does not brace the shotgun against his shoulder. </p> <p>The image below is a frame from the video at 5m:03 - it shows the gun in"mid burst" - two cartridge cases can be seen in the air, one having just been ejected and the other is closer to the camera. The users 'stance' appears to be intended to show the effortlessness of holding the gun while firing. If he is in fact using typical 12 gauge shells (which he implies that he is), holding a "normal" shotgun like this when firing would result in it being torn from his hands. </p> <p><img src="https://i.stack.imgur.com/AB2cV.jpg" alt="enter image description here"></p> <p>In the video the presenter claims that the shotgun exhibits, for all intents and purposes, zero recoil. He remarks that there is a spring system in the buttstock that counters the momentum of the recoil. </p> <p>I can think of two ways that the spring system could work:</p> <ol> <li>The first recoil (or something else) compresses the spring, and then the spring extends upon the generation of the second recoil. </li> <li>Each recoil compresses a stiff spring that extends more slowly than the recoils compress it. Assuming a sufficiently quick firing rate over a sufficiently protracted period, complete compression of the springs would obtain; consequently, the user would need to abstain from firing the shotgun until the springs extended, lest s/he absorb the recoil herself. </li> </ol> <p>However, it seems to me that those systems would work only if the user braced the shotgun against something (i.e. his/her shoulder). </p> <p>What is the simplest spring system that could produce recoil absorption comparable to that exhibited in the video?</p>

|mechanical-engineering|springs|firearms|

<p>There is a pervasive misconception that recoil can be "absorbed" by such things as springs and proprietary gas systems. This is not true; eventually all the recoil produced by launching the bullet must be absorbed by the shooter. The only thing that a spring/gas system can do is spread the recoil over a longer period of time, so that the force acting on the shooter's body is reduced.</p> <p>In physics terminology, recoil is analogous to momentum, not energy. Momentum is the integral of force with respect to time, $P=\int{F \cdot dt}$, and momentum is a conserved quantity; i.e. it can't be destroyed. The bullet and escaping propellant gas gain momentum during the launch because they're subjected to a huge force for a very short period of time. As a consequence of Newton's third law of motion, the gun and shooter gain the same amount of momentum as the bullet, but in the opposite direction. To dissipate this momentum, you can either absorb a lot of force in a short period of time, or absorb a lower amount of force over a longer period of time (preferable). For automatic weapons, an ideal recoil system would dissipate the momentum to the shooter in such a way that they would just feel a constant force pushing on them; i.e. no impulses from the shots.</p> <p>To answer the question, the spring system to do this would be a progressive-rate spring combined with a reactive damper. This type of setup is used on some high-end car suspensions, but I imagine it could be adapted for firearms as well. The trick would be to adjust the reaction of the spring and damper so that they reduce the force exerted on the shooter to a minimum.</p>

3273

What is the simplest spring system that could produce recoil absorption comparable to that exhibited by the AA-12 assault shotgun?

2015-06-23T22:02:27.137

<p>We've had unusually dry weather and our well has run so low that we had to call in a water hauling company to refill it a few days ago.</p> <p>The water man told me that even though he dumped in ~ 1000 gallons, a great part of it will seep out of the well to the surrounding (drier) aquifer, the same way groundwater makes its way inside the well in normal conditions. Now a few days later, the water is low again.</p> <p>I'm thinking some kind of drinking-water-safe plastic poly sheet sown or welded together at the seam would create a long and flexible tubing that could be progressively immersed inside of the well (with weight at the end) to effectively waterproof it on the inside. </p> <p>The well could then be refilled to the top and hold all the water that it's containing, instead of slowly seeping it out. Under normal conditions, the sleeve can be removed.</p> <p>The well is 2 feet wide, 15 feet deep. 2 ft diameter, 3 ft long concrete pipe sections stacked on top of each other. </p> <p>There's also another well on the site (very low water as well) that's 4 feet wide, 15 feet deep - a much better candidate for the concept as it could hold twice as much water.</p> <p>Does this sound doable? What are the problems I might face?</p>

|pressure|plastic|water-resources|waterproofing|

<p>Well liners are generally made to order. One of your issues will be to find a company that can make what you want. </p> <p>You will need to ensure the liner is the correct size and that it does not tear when inserted into the well, while being filled with water, or when it is being removed from the well.</p> <p>You mention your wells are cased with concrete pipes. They will mostly be smooth on the inside of the well. However, there may be rough spots or chunks of concrete missing, or the joints may be rough and have sharp edges.</p> <p>Every insertion and removal of the liner risks tearing or weakening the liner.</p> <p>Another option, if the have the space, is to place a water tank on the surface and re-plumb the downpipes from all the roofs to discharge rain water (when it falls) into the tank. During droughts the tank can be fill by water brought in by a tanker.</p>

3300

Can I seal the inside of a water well to use it as water reservoir when the well is dry?

2015-06-23T23:38:25.450

<p>I'm trying to do some processing of optical emission spectra from sputtering plasmas, and am confused by what NIST means by <a href="http://www.nist.gov/pml/data/handbook/index2.cfm">'easily reversed' in their notation</a>. Anyone have a clue?</p>

|optics|

<p>This means that the line in question can be apparent or not depending on whether the spectrum has been absorbed for other reasons. Or as <a href="https://en.wikipedia.org/wiki/Spectral_line#Opacity_broadening" rel="nofollow">Wikipedia states</a>:</p> <blockquote> <p>...the reabsorption near the line center may be so great as to cause a self reversal in which the intensity at the center of the line is less than in the wings.</p> </blockquote> <p>Also, there is a paper that takes this topic more in depth: <a href="http://epsppd.epfl.ch/Praha/98icpp/f089pr.pdf" rel="nofollow">Effect of Self-Reversed Spectral Lines and the Temperature of the Switching Arc Plasma</a></p>

3303

What does 'easily reversed' mean in the NIST spectral database?

2015-06-24T02:06:03.477

<p>Is there a formula to calculate the grip generated by a tyre on a vehicle for a given surface?</p> <p>I have a 13,000 kg machine pulling nearly 200 N of weight. I need to figure out if we need to put any more weight onto the machine so that it has enough traction to start the initial pull. It's going to use pneumatic tyres and the surface is concrete.</p>

|mechanical-engineering|automotive-engineering|

<p>According to the <a href="http://www.engineeringtoolbox.com/friction-coefficients-d_778.html">engineering toolbox</a> the coefficient of static friction between a tire and pavement ranges between $\mu_{wet}=0.2$ for a wet surface and crappy tire and $\mu_{dry}=1$ for a dry surface and high quality tire. You can use this to calculate the total amount of force that the tires will be able to apply with $$ F_{tire}=\mu F_N=\mu Mg\cos\theta $$ where $\theta$ is the slope of the road, with $\theta=0$ being a flat road. The total amount of force required of the tires to keep your load from rolling backward (with slightly more needed to get it moving forward) is given by $$ F_{load}=(W+Mg)\sin\theta $$ where $W$ is the weight of the load. So, the maximum slope that your vehicle will be able to operate on is given by the point at which the backwards force of the load becomes equal to that which the tires can apply; $$ \theta=\tan^{-1}\left(\frac{\mu Mg}{W+Mg}\right)=11.3^\circ, $$ assuming that the pavement is wet and your tires aren't particularly great. This doesn't account for any small undulations in the pavement or bumps in the road, but your tractor to load weight ratio is so high that you shouldn't have any problems. </p>

3306

Calculating the grip of a tyre for a given surface

2015-06-24T13:16:54.330

<p>I've been working with SC magnet quench simulations. The conductor is made from NbTi in a Cu channel with a 5:1 ratio of Cu:NbTi. Opera3d has a quench program with an example that I began working from. In a quench scenario using a solenoid made from the conductor I described above heat propagates at different speeds throughout the bulk of the solenoid. For this case, heat conduction along the conductor (azimuthal) is fastest. Axial and radial propagation are much slower, with radial being the slowest.</p> <p>Opera's documentation documentation has the following description regarding the anisotropic thermal conductivity:</p> <blockquote> <p>Nonlinear anisotropic thermal conductivity properties are defined by three expressions based on the functions Cu_Kappa(T), Bulk_Kappa_r(T), and Bulk_Kappa_z(T).</p> <p>For the azimuthal thermal conductivity of the bulk material,the conductivity of copper, Cu_Kappa(T), will be scaled by the copper factor. It is assumed that this is significantly higher than the conductivity of NbTi and dominates the thermal conduction in this direction. The bulk properties radially and axially are to be taken directly from the <strong>tables of measured values.</strong> <em>(Note that the data used in this example are fictitious, but with similar characteristics to real materials)</em>.</p> </blockquote> <p>So, the values used in the example are claimed to be taken from measured values.</p> <p>My question is this: Can I create the radial and axial thermal conductivities from the thermal conductivities of NbTi and Cu along with their proportions in the conductor?</p> <p>Here is what I've done so far in trying to find the answer:</p> <ol> <li>Google searches. There are many articles and web pages which describe anisotropic <span class="math-container">$\kappa$</span>. But all the information I've found assumes you already know the <span class="math-container">$\kappa$</span> values.</li> <li>Hand calculations. I made the following model and tried to think of the problem in terms of thermal resistivity, which I then assumed I could treat like electrical resistances, i.e. use rules for parallel and series resistance to calculate the effective resistance.</li> </ol> <p><img src="https://i.stack.imgur.com/tR8xv.png" alt="Conductor Unit Cell" /></p> <p>There is one more piece to this, however, and that is the insulation. Around the unit cell there is polyester insulation. It is 0.27 mm and 0.25 mm thick the edges that run in the axial and radial directions, respectively.</p> <p>I am going to try accounting for this by summing the materials in their respective direction, each scaled by a packing factor.</p> <p>Does this sound like a good approach? I've spent a bit of time on this already and have no idea if the answer to that approach will resemble the physical model.</p>

|thermodynamics|magnets|superconduction|

<p>You are not too far off with parallel and series resistors method.</p> <p>The <a href="https://en.wikipedia.org/wiki/Rule_of_mixtures" rel="nofollow">Rule of mixtures</a> for composite design works in the same fashion - by treating each of the <em>loads</em> as either springs in parallel (when loaded in the direction of the fibers), or as springs in series (when loaded opposite to the direction of the fibers). (See this nice <a href="http://www.doitpoms.ac.uk/tlplib/bones/derivation_mixture_rules.php" rel="nofollow">explanation</a> if the Wikipedia explanation seems a bit difficult).</p> <p>The result works for determining a wide variety of composite properties. From Wikipedia:</p> <p>$$(\frac{f}{k_f} + \frac{1-f}{k_m})^{-1} &lt; k_{composite} &lt; fk_f + (1-f)k_m $$</p> <p>can be used to determine bounds on your mixture. I'd really run the analysis with both values, and take the most conservative case.</p> <p>If that doesn't work, I use <a href="http://www.lib.ucdavis.edu/dept/pse/resources/fulltext/HDBK17-3F.pdf" rel="nofollow">MIL HDBK 17-3F</a> for analysis. Page 223 shows the equation, which has a slightly larger lower bound by placing additional emphasis on fiber conduction. The insulation can then be added to the radials, again using the rule of mixtures (it's a pretty useful rule).</p> <p>Unfortunately, I can't help you on Opera3D parameters.</p>

3311

Calculating Anisotropic Thermal Conductivity for two Materials

2015-06-24T17:03:32.877

<p>Our component sourcing organization purchases large amount of electronic components from suppliers. I believe this is through a bid process. Engineering provides the sourcing organization specification, and they purchase these parts for the plant to use in manufacturing. The issue is that recently the plant has been receiving a increasing amount of capacitors which might be defective. This is causing our products to fail. I have checked the few new incoming capacitors using fluke multimeter, and the measurements are with in specification. </p> <p>It is not practical to continuously inspect these capacitors, and I have been tasked to recommend the solution. How does one <strong>technically</strong> address similar issues? </p>

|electrical-engineering|statistics|quality-engineering|

<p>Looking for good and reliable suppliers might be one of best solutions. Vendors such as Digi-key, mouser, Newark always supply high quality electronic components. But understandably they come at a higher cost. If directly purchased from a manufacture it might be good idea to request for test data from suppliers to validate that product is meets specification. Most Suppliers have the test data readily available and if the product meets specification they will gladly share the data almost immediately. If the capacitors are purchased through an intermediary then you could request for sample data. A sample size of 30 pieces is good sample size to perform statistical analysis.</p> <p>Alternatively a small sample set of components can be tested and the measured data can be analyzed using basic statistics as part of incoming inspection. If the capacitors are SMT this might become tricky because randomly sampling across the SMT reel will cause issues for the SMT pick and place machines.</p> <p>Regardless after obtain this data consider performing some basic statistical analysis such as averages, standard deviation, and process capability analysis. This type of simple first order analysis will give a good indication if the incoming material is with in specification.</p> <p>Below is 30 randomly generated data point to illustrate the example basic analysis. The data points are based on 10µF capacitor with +/- 10% tolerance. Some data points are marginally below or above the lower or high tolerance limits. </p> <p>\begin{array}{| c | c | c | c | c | c | c | c | c | c |} \hline 9.40 &amp; 10.00 &amp; 10.10 &amp; 11.30 &amp;10.10&amp; 8.80 &amp;9.60 &amp;9.20 &amp;11.50 &amp;8.90 \\ \hline 8.60&amp; 11.30&amp; 10.00 &amp;11.10&amp; 10.60&amp;11.10 &amp;10.40&amp; 9.10&amp; 11.50&amp; 8.90\\ \hline 8.70&amp; 10.70&amp; 11.20&amp; 10.30&amp; 10.30&amp; 11.50&amp; 10.80&amp; 9.70&amp; 9.50&amp; 10.80 \\ \hline \end{array}</p> <ul> <li><strong>Average</strong> ($\mu$): 10.17 </li> <li><strong>Standard Deviation</strong> ($\sigma$): 0.93 </li> <li><strong>C$_{p}$</strong> = 0.35 </li> <li><strong>C$_{pl}$</strong> = 0.41 </li> <li><strong>C$_{pu}$</strong> = 0.29 </li> <li><strong>C$_{pk}$</strong> = 0.29 </li> </ul> <p><img src="https://i.stack.imgur.com/fxJv2.jpg" alt="Data1"> <img src="https://i.stack.imgur.com/a3QQT.jpg" alt="Data 2"></p> <p>Base on the example data it appears that the capacitor value distribution is very wide. Looking are such data analysis certain characteristics such as the capability of the process and incoming material can be ascertained. This type analysis can help make simple decisions about the capacitors such as whether to accept or reject the material. <a href="https://engineering.stackexchange.com/questions/390/how-can-the-vendor-be-approved-disapproved-based-on-validation-test-data/571#571">How can the vendor be approved/disapproved based on validation test data?</a> post closely related to the topic and some the equations discussed in greater detail. Hope this response helps address the issue.</p>

3315

How to identify low quality electronic components early?

2015-06-24T18:22:12.483

<p>I'm doing a static analysis on a uniformly loaded cantilever beam in ANSYS APDL. I'm using BEAM188 with all the default options, so it uses linear form. for the element behaviour, which I think means the bending moments and tensions are linear within the element.</p> <p>I want to plot the bending moments along the beam with an area contour, for that I'm using:</p> <ul> <li>ETABLE,MI,SMISC,3 </li> <li>ETABLE,MJ,SMISC,16 </li> <li>PLLS,MI,MJ,1,0</li> </ul> <p>This are the axial stresses due to bending, it looks very similar to the bending moment, it shows discontinuities between elements and for each element it shows the same value on both nodes. </p> <p><a href="https://i.stack.imgur.com/53s55.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/53s55m.png" alt="enter image description here"></a></p> <p>Why does MI=MJ? Shouldn't there be a linear variation of the moment in the elements? Does it make sense for it to be discontinuous between elements? Or is this just a bad plot and I should plot the values on the nodes I get with PRESOL,M as opposed the plot each I and J node values for each element.</p> <p>The results overall seem good, but I'm not sure if this is a correct representation of the behaviour of the elements.</p> <p>I'm sorry for such a basic question. I hope I wasn't too confusing.</p>

|mechanical-engineering|finite-element-method|

<p>You might have got the answer if not, come to your direct question for plotting actual bending moment. Just choose the element option key for "quadratic" instead of linear and re run the problem and plot the BMD.</p>

3317

Results for Bending Moment using BEAM188 on ANSYS APDL

2015-06-25T00:47:24.340

<p>If I have an electric vehicle on an incline (m=17,300 kg) I want to figure out how much energy it would generate by regenerative braking. </p> <p>I've used $a= g \cdot \sin({\theta}) $ and $f=m\cdot a$.</p> <pre><code>at 05 deg incline, acceleration is 0.85 m/s/s; this creates a force of 11114.972 N at 10 deg incline, acceleration is 1.70 m/s/s; this creates a force of 22145.352 N </code></pre> <p>So bearing these values in mind if I don't want to accelerate and maintain a set speed I will need to create a braking force of 11114.972 N at 5 deg incline - this in turn will keep the machine at the same speed (ignoring friction, bumpy/uneven surface, etc.). </p> <p><strong>How can I figure out the Watts generated for the regenerative braking force of 11114.972 N?</strong></p> <p>The DC motor (I think it's 3 phase) is a 105 kW motor rated at 350 V and 300 A, max torque is 190 Nm @ 285 V drawing 285 A. (That's all the information I have for the motor.) Efficiency of the motor would be 0.85 (Assumed).</p>

|electrical-engineering|automotive-engineering|regenerative-braking|

<p>The calculation perfectly right, bu the normal electric vehicle weights just 1/10th on the weight that you have considered. Hence the results are scaled in that ratio (10x more than what the real world EV could generate)... Hope that helps!</p>

3321

How can I determine the power generated by a regenerative braking system?

2015-06-25T09:02:53.780

<p>I was reading about <a href="https://en.wikipedia.org/?title=CD-ROM">how CD-ROMs work</a> and I came to this: </p> <blockquote> <p>Although it might seem simplest to use a pit to record a 0 and a land to record a 1, it is more reliable to use a pit/land or land/pit transition for a 1 and its absence as a 0,so this scheme is used.</p> </blockquote> <p>Now, I'd like to know why?</p>

|optics|computer|computer-engineering|

<p>The explanation is incorrect. The laser is never more than half "in the pit". The reflection is land or 1/2 pit plus 1/2 non pit.</p>

3324

Why is it more reliable to use the land/pit transition in a CD-ROM?

2015-06-25T10:42:30.957

<p>In lab we have Struers Tegrapol-21 grinding and polishing machine. I use it for sample preparation. Usually I put 6 samples at once and prepare them with sample holder. However today I need to prepare 3 samples. I checked <a href="http://www.struers.com/resources/elements/12/101288/TegraSystem%20brochure%20English.pdf" rel="nofollow noreferrer">brochure</a> and it's not clear. <a href="http://www.struers.com/resources/elements/12/282889/TegraPol-21%20-25%20-31%20-35%20TegraForce-3%20-5%20TegraDoser-1.pdf" rel="nofollow noreferrer">Manual</a> wasn't helping either. </p> <p>I'm trying to understand why do I use two different type of holders. Ease of loading is an issue however I'm looking to learn is there a difference in grinding/polishing performance. Why and when I should choose one over the other? </p> <p>Here is photo of single holder (left), sample holder(right) and Tegrapol-21.</p> <p><img src="https://i.stack.imgur.com/J26QU.jpg" alt="Tegrapol-21"></p> <h2>Question</h2> <ol> <li>Can I use sample holder for 3 samples?</li> <li>Can I use single holder for 3 samples? </li> <li>If both of them is possible which one is preferred?</li> </ol>

|materials|metals|grinding|

<p>I personally have the tegrapol 11 and I use it with the tegraforce 1</p> <p>Our units sample mover plate holds three specimens simultaneously (concept is the same though). Google tegrapol 11 and tegra force 1 for images if this is unclear.</p> <p>Many times I run my sample mover plate both with only one and only two specimens at a time merely leaving the last hole(s) empty.</p> <ol> <li><p>If I were working with this unit and I were concerned with using the sample mover plate sans all 6 holes being filled: I would distribute the three specimens (skip every other hole).</p></li> <li><p>I would NOT use the single sample mover plate for three specimens.</p></li> <li><p>As with 2. I personally would use the six-holed sample mover plate for three specimens.</p></li> </ol>

3331

How to use sample holder on Metallography specimen preparation machine Struers Tegrapol-21