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2015-10-02T08:49:10.083 | <p>How can I choose between Implicit or Explicit Analysis in ABAQUS for a simple rectangular composite structure with 2 layers of unidirectional lamina? I am trying to do a linear-static analysis to capture the tensile strain for the specimen. </p>
| |finite-element-method|abaqus| | <p>Further to existing answers: in a dynamic problem, an explicit method computes time-derivatives for use in updating through a timestep, by inserting into the dynamical equations the values of the fields and their spatial derivatives at the start of the time-step; an implicit method computes time-derivatives for use in updating through a timestep, by inserting into the dynamical equations the values of the fields and their spatial derivatives at the end of the time-step. You may spot that there's a challenge in implementing an implicit method, in that, in advance of updating through the timestep, one doesn't know the values of the fields and their spatial derivatives at the end of the time-step: for linear differential equations, this can be sorted out exactly through a matrix inversion; for nonlinear differential equations, one has to estimate the values of the fields and their spatial derivatives at the end of the time-step iteratively.</p>
| 5627 | What is the difference between implicit and explicit analysis in ABAQUS? |
2015-10-02T09:54:56.790 | <p>From time to time I am tasked with additions or alterations to old buildings or structures. I would need to establish the "as built" structure and from that analysis see how the amended/new structure suits the existing structural components.</p>
<p>The challenge lies in the assumptions one is forced to make. Over the years things change, so do the material properties and standard member sizes. For example, I have come across a bombed bridge that had to be rehabilitated in Angola. The bridge was built by the Portuguese government some time before the Angolan war. The bridge was constructed from large steel I-beams as main bearers. We could not establish the type of steel used for the I-beams, and the sectional properties we established by measuring the beams with a tape measure.</p>
<p><strong>How do engineers determine the material properties of members of old or historic structures?</strong> I am not aware of any document or database that captures all steel (or rebar) manufacturers of the world with their respective product lists and material properties over time, or what design standards were historically applicable at various times in various regions of the world.</p>
| |materials|civil-engineering|structural-engineering| | <p>When it comes to concrete and steel, there were few standards for materials and design prior to about 1920, and you may not find any prior to 1905, at least that is my experience. Reinforced concrete technology was largely proprietary prior to about 1910 - as such the first multi-story reinforced concrete building in the UK (Weaver's Warehouse 1897 to 1906 I believe) was built under license.</p>
<p>The Civil Engineer's Handbook in the UK can be a good guide for what was going on in the early days (UK and Commonwealth), and the Institution of Civil Engineers library has older copies of this. For example, the water to cement ratio was found in the 1923 version of the Handbook in the UK for the first time (known as the Abrams Ratio and was expressed as volume of cement to volume of water rather than today's w/c m/m ratio).</p>
<p>The design of the Weaver's Warehouse indicated that the use of reinforcement at that time resulted in loads being supported by an 'arching' effect rather than the 'ductility' design approach of present. This means that older structures could be at greater risk of brittle failure modes. I suspect that this is because there was little knowledge of embedment length, etc prior to the 1930's, and steel was often 'fixed' at the ends. The proprietary construction process of that project required that the slabs/beams be load tested as the structure was built.</p>
<p>From memory the first codes for steel and cement (eg BS 12) in the UK were released in the early 1900's, around about 1906 or so, I don't have the exact dates. Cements were far more coarsely ground in those days, and chemically very much different. There was little C3A and gypsum added, for example, prior to about 1920, as it was introduced at a later date when manufacturers found it could reduce the energy and cost of production. Furthermore, because little was known about the bonding of reinforcing steel and embedment length, it was often smooth, not deformed, prior to 1920. </p>
<p>On the other side of the Atlantic, the American Concrete Institute was founded around 1904, and the first building regulations were adopted around 1910 with a number of standards in place around 1912. The Portland Cement Assocation, with Duff Abrams later as its first Chief Engineer, was founded around 1917. Abrams was to originate the ACI concrete mix design method, the concept of water to cement ratio and reinforcing steel bonding capacity in the US.</p>
<p>What I have found in the US and Europe on structures dating back to about 1910 (but mostly during the 1920's) is that you can find just about anything, and testing is strongly advised. It is not unusual to find concrete typically designed and produced on a loose volume basis (ie by scoop or shovel 1:2:4 and no water to cement ratio). One project we looked at, an older bridge foundation and deck, had concrete mixture composition that varied from one end of the bridge to the other: the aggregates were a rounded river gravel on one end and the other end had a crushed granite, and they blended together as they reached the centre.</p>
<p>Another significant issue is the foundation design and construction. You can find anything. I have seen structures carrying load successfully for 50 years or more but with timber piles that were located only half under the cap (ie extending beyond the edges). So again there is no substitute for investigation and/or testing.</p>
<p>For concrete there is an international standard BS EN 13791 that includes statistical strength assessment for structures where no history of the structure is known. This can be of use for comparing the concrete to modern structural codes. But if there is to be a change of loading condition then there is no substitute for identifying cover depths and steel/section geometry using non-destructive tests. And of course reinforcing steel corrosion can be a significant factor.</p>
<p>For structural steel a critical issue is often the bolts/welds as well as the sections. The assessment is very much of an art here, and may require radiographic/ultrasonic/torque testing if the situation demands it. Very old steel structures are often riveted which introduces further complexity. As far as I am aware, there aren't any historical guides on bolts and rivets. Previous posts have given some references that could be of use for the sections. If in doubt about a steel structure, you can obtain coupons from the sections and have them tested for tensile capacity and ductility.</p>
| 5629 | Determining the properties of members of historic structures |
2015-10-02T15:13:41.620 | <p>In the context of Bernoulli, does "static pressure" typically refer to $p$, or to $p + \rho gz$?</p>
| |fluid-mechanics| | <p>I think there is no unique technical term to define pressures in different situations.
As a reference states <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html" rel="nofollow noreferrer">here</a>: <span class="math-container">$p_\text{static}=\rho g h$</span></p>
<p>If we consider the data computed from any incompressible CFD solver, the data of pressure term indeed includes the hydrostatic part when gravity <span class="math-container">$g$</span> is available.</p>
<p>I do prefer to think the fluid pressure (incompressible) in <a href="https://nptel.ac.in/courses/112104118/lecture-16/16-1a_hydro_static_pressure.htm" rel="nofollow noreferrer">this way</a> for
<span class="math-container">$p+\frac{\rho v^2}{2}+\rho g z=p_0$</span>：</p>
<ol>
<li><span class="math-container">$\rho g z$</span> is the potential at <span class="math-container">$z_A$</span> not the hydrostatic pressure.
If zero is the reference position, the hydrostatic pressure or pressure due to acceleration should be <span class="math-container">$p_{hs}=\rho g (-z)=-\rho g z$</span>, since gravity has a negative direction.</li>
<li><span class="math-container">$p$</span> raised in the previous equation has two components: hydrodynamic pressure <span class="math-container">$p_{hd}$</span> and hydrostatic pressure <span class="math-container">$p_{hs}$</span>, which is to say <span class="math-container">$p=p_{hd}+p_{hs}$</span>.</li>
<li>Hence, it is quite obvious that static pressure <span class="math-container">$p_{s}=p_{hd}=p_0-\frac{\rho v^2}{2}$</span></li>
<li>Dynamic pressure <span class="math-container">$p_{d}=\frac{\rho v^2}{2}$</span></li>
<li><span class="math-container">$p_s+p_d=p_0$</span>(total pressure)</li>
</ol>
<p>As for the common situation, pressure <span class="math-container">$p$</span> is the static pressure <span class="math-container">$p_s$</span> (hydrodynamic pressure <span class="math-container">$p_{hd}$</span>) plus hydrostatic pressure <span class="math-container">$p_{hs}$</span> (or pressures due to inertial force). Excluding the rest balance, <span class="math-container">$p_s$</span> or <span class="math-container">$p_{hd}$</span> is the real reason to make flow move (to balance dynamic pressure <span class="math-container">$p_d$</span>).</p>
| 5635 | Is gravity included in "static pressure"? |
2015-10-02T17:30:03.270 | <p>I'm designing a crucible / extruder, and right now it has a .6mm hole to extrude from. Will this work? A chart with various metals listed would be nice.</p>
<p>Pictures:<br>
<sub><em>Click to expand</em></sub> </p>
<p><a href="https://i.stack.imgur.com/UPzKX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UPzKXt.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/qdmj3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qdmj3t.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/UYLOJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UYLOJt.png" alt="enter image description here"></a></p>
| |mechanical-engineering|thermodynamics|metallurgy|metals|experimental-physics| | <p>Making a nozzle with that long and thin of a hole is not feasible, but you could make a larger hole behind it and just have the last 3mm be 0.6mm in diameter. This would be feasible in a ceramic, but due to cost, I might make the nozzle into an insert that goes into a larger piece made of cheaper material.</p>
<p>Things to be concerned with:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Surface_tension#The_breakup_of_streams_into_drops" rel="nofollow">Stream breakup due to surface tension.</a></li>
<li>Flow velocity/pressure</li>
<li>Heat transfer</li>
<li>Manufacturability </li>
</ul>
<p>Each of those could be their own question, answerable by people with different expertise. So I'd recommend asking them separately.</p>
| 5640 | How small can you get a stream of molten steel to be at low pressure? |
2015-10-05T01:49:44.163 | <p>I've heard the <a href="https://en.wikipedia.org/wiki/Mark_48_torpedo" rel="nofollow">Mk. 48 torpedo</a> is wire-guided, why not make it wire powered? The range would only be limited by the length of the wire. I'm wondering why this is not used and what problems it would have.</p>
| |mechanical-engineering|electrical-engineering| | <p>The main reason is that the power requirement for running the torpedo is vastly larger than that for merely sending control signals to it. This means <i>much</i> thicker wire would be needed to actually power the torpedo over the wire. Even if cost is not a consideration, a usefully long spool of such thick wire would be very large.</p>
<p>You can reduce the size of the wire by increasing the voltage, which thereby reduces the current at the same power. However, that then increases the demands on insulation, which takes significant space. Thin insulation, like enamel coated, can't stand off much voltage. To make things even worse, this all has to work in the a conductive medium (seawater).</p>
<p>I don't know how much power a torpedo requires, but probably a few kW at least. Let's say 5 kW just to pick something (if anyone knows the real number, please tell us). That could be 10 A at 500 V, 50 A at 100 V, etc. Even considering the heat sinking provided by the seawater, 50 A still requires a substantial wire, and 100 V insulation is going to double the thickness, which quadruples the volume.</p>
<p>I'm going to stop here since figuring out the volume of wire, volume and weight of copper, power loss, and voltage drop requires knowing too many specifics that I can only guess at. However, it should still be obvious that this is going to be a lot worse than the torpedo spooling out double-stranded #30 wire with thin insulation behind it.</p>
| 5663 | Is it possible to power a torpedo via a wire? |
2015-10-05T10:57:10.960 | <p>I am having trouble to calculate the differential equations of a simplified loading bridge.</p>
<p>The system is build as shown in the picture below (just a sketch):</p>
<p><a href="https://i.stack.imgur.com/YEj4q.png" rel="noreferrer"><img src="https://i.stack.imgur.com/YEj4q.png" alt="enter image description here"></a></p>
<p>If I use the Newton approach, I am getting the following equations by neglecting friction, air resistance and changes in the length of rope:</p>
<p>$$
m_k \ddot{x}_{k} = F_{A} + F_{S} \sin(\varphi) \\
m_G \ddot{x}_{G} = -F_{S} \sin(\varphi) \\
m_G \ddot{z}_{G} = m_{G} g - F_{S} \cos(\varphi)
$$</p>
<p>When I look at the kinematic relationships from the gripper (the circle with the weight $m_G$) I get the following equations.</p>
<p>$$
x_{G} = x_{k} + l \sin(\varphi) \\
z_{G} = l \cos(\varphi)\\
\varphi = \omega t = \dot{\varphi} t
$$</p>
<p>I know the weights $m_k$ and $m_G$ and the length $l$ but the values are not important right now.</p>
<p>The goal is to have two differential equations at the end.
One equation shall show the relationship between the driving force $F_A$ and the path of the trolley $x_k$ (with derivations)
The other equation shall show the relationship between driving force $F_A$ and angle of the rope $\varphi_G$.</p>
<p>After that I want to make the transfer functions (Laplace transformation etc.) but that is not the problem.</p>
<p>The problem is that I can not seem to find those equations. My best approach so far looks like this:</p>
<p>$$
m_{k} \ddot{x}_{k} = F_{A} + F_{S} \sin(\varphi)
$$</p>
<p>So that means if</p>
<p>$$
m_G \ddot{x}_{G} = -F_{S} \sin(\varphi) \\
F_{S} \sin(\varphi) = -m_{G} \ddot{x}_{G} \\
$$</p>
<p>I can say:</p>
<p>$$
m_{k} \ddot{x}_{k} = F_{A} - m_{G} \ddot{x}_{G} \\
$$</p>
<p>and if I derive $x_{G}$ like this:</p>
<p>$$
x_{G} = x_{k} + l \sin(\varphi) \\
\dot{x}_{G} = \dot{x}_{k} + l \dot{\varphi} \cos(\varphi) \\
\ddot{x}_{G} = \ddot{x}_{k} + l \left[ \ddot{\varphi} \cos(\varphi) - \dot{\varphi}^{2} \sin(\varphi) \right]
$$</p>
<p>I am actually getting stuck here because I can not find a way to eliminate $\varphi$ from the equations. The addition theorems are not helping me at all (or I'm using them correctly).</p>
<p>Does anyone have an idea of how I should continue at this point?
I hope I don't need a complete solution. I am actually more interested at doing this myself and hope to get a push towards the right direction.</p>
| |mechanical-engineering|control-engineering|modeling|mathematics| | <h2>Kinematics and dynamics</h2>
<p><a href="https://i.stack.imgur.com/94sUM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/94sUM.png" alt="enter image description here"></a></p>
<p>Those are the steps to solve problems of this nature.</p>
<ol>
<li>Analize the kinematics of the system.</li>
</ol>
<p>$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $_{o}\vec{r}_{RP}$</p>
<p>$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $R(\varphi) _{B}\vec{r}_{RP}$</p>
<p>$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big(x_{k}î + 0j + 0k \big)$ + $\big(\sin(\varphi)l î + 0j + \cos(\varphi)lk\big)$</p>
<p>$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big[\big(x_{k}+\sin(\varphi)l\big)î + 0j + \big(\cos(\varphi)l\big)k\big]$</p>
<p>note: $R(\varphi)$ is a <a href="https://en.wikipedia.org/wiki/Rotation_matrix" rel="nofollow noreferrer">rotation matrix</a> and $x_{G} = x_{k}+\sin(\varphi)l$.</p>
<p>Taking the time derivatives:</p>
<p>$\hspace{5.em}$ $\dot{x_{G}}$ = $\dot{x_{k}}+\cos(\varphi)\dot{\varphi}l$</p>
<p>$\hspace{5.em}$ $\ddot{x_{G}}$ = $\ddot{x_{k}} + l\cos(\varphi)\ddot{\varphi} - l\sin(\varphi)\dot{\varphi}^{2}$</p>
<ol start="2">
<li>Use Newton's equation:</li>
</ol>
<p>$\hspace{5.em}$ $m_{k}\ddot{x_{k}} = F_{A} - m_{G}\ddot{x_{G}}$</p>
<p>Substitute $x_{G}$:</p>
<p>$\hspace{5.em}$ $m_{k}\ddot{x_{k}} = F_{A} - m_{G}\big(\ddot{x_{k}} + l\cos(\varphi)\ddot{\varphi} - l\sin(\varphi)\dot{\varphi}^{2}\big)$</p>
<p>$\hspace{5.em}$ $\big(m_{k}+m_{G}\big)\ddot{x_{k}} + m_{G}\big(l\cos(\varphi)\ddot{\varphi}\big) - m_{G}\big(l\sin(\varphi)\dot{\varphi}^{2}\big)= F_{A}$</p>
<p>For the z axis:</p>
<p>$\hspace{5.em}$ $F_{Z}$ = $m_{G}g-l\big(\cos(\varphi)\dot{\varphi}^{2}+\sin(\varphi)\ddot{\varphi}\big)$</p>
<ol start="3">
<li>Use Newton's second law for rotation:</li>
</ol>
<p>$\hspace{5.em}$ $I\ddot{\varphi}$ = $F_{Z}l\sin(\varphi)-\big(m_{G}\ddot{x_{G}}\big)l\cos(\varphi)$</p>
<p>$F_{Z}l\sin(\varphi) = m_{G}gl\sin(\varphi)-l^{2}\big(\cos(\varphi)\sin(\varphi)\dot{\varphi}^{2}+\sin(\varphi)^{2}\ddot{\varphi}\big)$</p>
<p>$\big(m_{G}\ddot{x_{G}}\big)l\cos(\varphi) = m_{G}\big(l^{2}\cos(\varphi)^{2}\ddot{\varphi}\big) - m_{G}\big(l^{2}\cos(\varphi)\sin(\varphi)\dot{\varphi}^{2}\big)+m_{G}\ddot{x_{K}}l\cos(\varphi)$</p>
<p>Using trigonometry identities:</p>
<p>$\hspace{5.em}$ $\big(I+m_{G}l^{2}\big)\ddot{\varphi}$ = $m_{G}gl\sin(\varphi)-m_{k}l\cos(\varphi)\ddot{x_{k}}$</p>
<ol start="4">
<li>Done! Now you can rest...
$\ddot\smile$</li>
</ol>
| 5665 | Differential equations of a (simplified) loading bridge |
2015-10-05T13:26:22.700 | <p>I understand the difference between the two in terms of how they function; one will release all of it's energy in small pulses with the other creates a constant stream of radiation.</p>
<p>My question is for materials processing such as marking or cutting. Why would I use one over the other?</p>
| |materials|optics|machining|lasers|cutting| | <p>Pulsed lasers release their energy in very short pulses which can have incredibly high <strong>peak</strong> powers. A run-of-the-mill nanosecond laser will have a peak power in the multi-kilowatt range while a femtosecond laser can easily reach into the megawatt range. In contrast, CW lasers generally do not reach power levels in excess of a few hundred Watts.</p>
<p>When these pulsed lasers are focused to a small spot size, the intensity is generally high enough to significantly alter the molecular state of the material through non-thermal processes such as ablation. These non-thermal processes are useful for creating very small features because they have a smaller <a href="https://en.wikipedia.org/wiki/Heat-affected_zone" rel="nofollow">'heat affected zone'</a>. The very high peak powers can also make use of nonlinear absorption in materials which would otherwise be transparent (and therefore not able to be processed by that wavelength). </p>
<p>In contrast, the energy from a tightly focused CW laser is deposited over long enough timescales that the material simply heats up and melts. This can be more useful for processes such as welding. </p>
| 5667 | Difference between continuous wave laser and a pulsed laser |
2015-10-07T22:32:50.567 | <p>The Reynolds Number is defined as $Re = \frac{\rho v L}{\mu}$, where $\rho$ is the density, $v$ is the fluid velocity, $L$ is the characteristic length, and $\mu$ is the dynamic viscosity. By definition, inviscid flow implies that $\mu=0$. By this formula, this would make the reynolds number infinite and thus turbulent. </p>
<p>Is it possible to have laminar inviscid flow? Must inviscid flow always (necessarily) be turbulent?</p>
| |fluid-mechanics|turbulence| | <p>No, inviscid flows are not necessarily turbulent. If there is nothing to "trip" the turbulence, then the flow will remain laminar. Features which could trip the turbulence include vibration, small temperature fluctuations, any geometric imperfections, velocity field imperfections, and other similar things.</p>
<p>For example, <a href="https://en.wikipedia.org/wiki/Potential_flow" rel="nofollow">potential flow</a> is a type of inviscid flow. Potential flow solutions are laminar solutions to the Navier-Stokes equations.</p>
<p>As another example, if care is taken to avoid vibration and other flow imperfections, it seems you can generate laminar pipe flows at arbitrarily high Reynolds numbers. <a href="http://www.annualreviews.org/doi/abs/10.1146/annurev-fluid-122109-160652" rel="nofollow">Laminar pipe flows have been obtained at Reynolds numbers of about 100,000 under these conditions</a>, which is far higher than the typical transition Reynolds number of about 2,000. From what I understand, there is no indication that 100,000 is any hard limit; you probably could get higher with better experimental setups.</p>
<p>How I think about it is this: Viscosity helps damp out flow imperfections, allowing laminar flows to occur with more imperfections. If you have a truly inviscid flow, it needs to be perfect to not trigger instabilities which lead to turbulence. If you were able to obtain an inviscid flow, you should expect turbulence due to the imperfections inherent in reality.</p>
<p>To answer a question you put in a comment in the other answer, yes, I do believe using a turbulence model for inviscid flow is prudent. For RANS, the Reynolds stress still will exist if the flow is inviscid, and for LES the same is true for the residual stress.</p>
| 5685 | Is inviscid flow necessarily turbulent? |
2015-10-08T00:25:27.780 | <p>I am going to machine something out of a small 40mm long x 8mm diam tungsten rod. The majority will be hollowed out using a larger bit, but at the very tip, I need a small hole, the smaller the better between .05mm and .6mm. I understand an L/D for that small hole should be less than 60%? Did I get that right?</p>
<p>Can this be done with a standard micro carbide bit? I guess the length of the hole can be adjusted to make this work out.</p>
<p>The purpose of the nozzle will be to pour out accurately micro amounts of melted steel. My parts include the ability to push steel wire at the rate of 1/27 mm^3 / step (of a stepper motor), so as you can see a tiny amount! But if not possible, I can bump that number up some.</p>
| |machining|microfluidics| | <p><strong>Reposting my comment for clarity:</strong></p>
<p>"Before we get into making this: have you confirmed that a wall thickness of ~0.3mm around the hole is going to be thick/strong enough to resist the pressure required to push the molten steel through your nozzle/hole in the first place? That's very thin.... I think you might have to look at a higher ratio than 60%"</p>
<p>Not to mention the difficulty in actually performing a drill operation with that much precision. </p>
<p><strong>My answer:</strong></p>
<p>Just saw your other <a href="https://engineering.stackexchange.com/questions/5640/how-small-can-you-get-a-stream-of-molten-steel-to-be-at-low-pressure" title="question">question</a>. As per the recommendation on the accepted answer a 3mm deep hole is probably much more reasonable than the 0.3mm one that you're proposing (due to the pressure forces that the steel will be at blowing out the tip of your nozzle). </p>
<p>This would give a L/D ratio of 5 (500%) which off intuition sounds hard to machine. I can't comment on whether it would work with a bit sticking 3mm out of the chuck but if it doesn't then you could always drill the hole in 1mm (or smaller) segments. What I mean by this is extend the bit out of the chuck by 1mm and drill into the piece by that amount. Then change the bit to extend by 2mm out of the chuck and drill it again. Repeat.</p>
<p>On the deeper drills the bit would be supported by the surrounding tungsten, and on the earlier passes the bit would extend out only a short amount and thus be unlikely to break due to lateral load. </p>
<p><strong>Realistic</strong> </p>
<p>It would be best to contact the machine shop where you plan to get this made. If you're making it yourself I would have a play around with a small amount of tungsten bar and see what's possible. </p>
<p><strong>edit in response to comments</strong></p>
<p>I had this sort of thing in my mind:
<a href="https://i.stack.imgur.com/b2f07.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b2f07.jpg" alt="Diagram of nozzle"></a></p>
<p>However now I realise that it would be more sensible/easier to just leave the drill profile in the nozzle:
<a href="https://i.stack.imgur.com/EQEjy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EQEjy.jpg" alt="Realistic diagram"></a></p>
<p>Which negates my concerns. </p>
| 5686 | What's the smallest pin-like hole I can machine into a piece of tungsten using a metal lathe and common carbide tools? |
2015-10-08T03:46:57.153 | <p>They've printed jet engines with it that work after polishing. So how is the accuracy measured and how accurate is the DMLS process? I can't find the required info anywhere. </p>
| |machining|manufacturing-engineering|lasers| | <p>Accuracy will vary a lot depending on what you are interested in.</p>
<p>At the simplest level you are limited by the layer thickness, which is in the $20-100~\mu m$ range. This places a limit on the accuracy of the dimensions that can be achieved in the z direction. Theoretically you can achieve very high accuracy in x and y, but in reality you are limited by the roughness of the the part.</p>
<p>Indeed, the real problem with most of these additive technologies is that the resulting surfaces are very rough, hence why polishing is needed to finish the surface. I don't know for DMLS specifically, but for related methods, such as <a href="https://en.wikipedia.org/wiki/Selective_laser_melting" rel="nofollow">SLM</a>, $30~\mu m$ roughness would not be bad. Generally, the roughness is highest where the part is angled with respect to the layers as you then can get additional effects from going over steps.</p>
| 5689 | What is the accuracy of direct metal laser sintering (DMLS)? |
2015-10-08T23:38:25.087 | <p>I have a 10 mm depth x 5 cm length 99% Tungsten rod. How can I machine the described hole so that instead of having a bevel from the tip of the main drill-hole bit, it has a perfect 90 degree corner all around as if the tip of the drill bit were like a flat router bit?</p>
<p>If this is not feasible, I can think about a design that doesn't use this larger hole. For this design, the inside of the large hole will be filled with an AlO3 ceramic tube that snuggly surrounds 1mm D of round stainless steel wire that flows through the center.</p>
<p><a href="https://i.stack.imgur.com/wGS3p.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wGS3p.png" alt=""></a></p>
| |mechanical-engineering|machining|metals| | <p>Wow, tungsten, huh? Whenever I see exotic materials, usually a design mistake is being made. The other answer by starrise is wrong for various reasons. Counterbores are normally only used for making sockets, not boring blind holes.</p>
<p>If you actually want to do this, the right strategy would be to set up the work piece in a four-jaw chuck on a lathe and rough the bore with a carbide jobbing drill mounted in the footstock.</p>
<p>Your next step will be to dress a cylindrical form in Aluminum Oxide J 40-60 grit using a rotary dresser. Mount the form in the tailstock of the lathe and then finish the hole. Clean the lathe carefully, because now it will have grit all over it. If your bore needs to have a good finish (not apparent that it does, given what you have written so far), you can repeat the process with a 200 grit form.</p>
<p>Next you need to get an extension mount for your tailstock and mount a smaller diameter carbide drill to rough the small bore. Once again use a grinding form to finish the small bore. Depending on the tolerance of the small bore you will probably need to redress the form because it will wear down as it grinds out the bore.</p>
<p>Obviously everything needs to be concentric all the time, so you have to be checking runout carefully on every step.</p>
<p>All this work can be also be done in a drill press (or mill) if you make a suitable workholding fixture. Using a drill press or mill your maximum precision will be 0.005" and 0.002" respectively. A lathe will be 0.0002".</p>
| 5701 | How can I finish a round hole so it has 90 degree corners (flat bottom)? |
2015-10-09T14:10:41.717 | <p>A question asked about the <strong>necessary curing pressures for a composite resin in the autoclave</strong>. Looking through the corresponding <a href="http://hexcel.com/Resources/DataSheets/Prepreg-Data-Sheets/F155_us.pdf" rel="nofollow">datasheet</a> (HexPly F155), I'm confused:</p>
<blockquote>
<p><strong>Cure Procedure</strong></p>
<ol>
<li>Apply vacuum of 22 inches (74 kPa) Hg minimum.</li>
<li>Apply 85 + 15 – 0 psig (586 + 103 – 0 kPa) pressure for laminates.</li>
<li>Apply 45 + 15 – 0 psig (310 + 103 – 0 kPa) pressure for sandwich.</li>
<li>Vent vacuum bag to atmosphere when pressure reaches 20 psi (138 kPa).</li>
<li>During cool-down when the part temperature falls below 140°F (60°C), pressure can be released and the test panel removed from the
autoclave and debagged.</li>
</ol>
</blockquote>
<p>The first step is clear, but not the next steps, for instance <code>85 + 15 – 0 psig (586 + 103 – 0 kPa)</code> is causing me a headache and I can't make sense of it.</p>
<p>I'd guess that one is the external positive pressure and the other is negative vacuum pressure under the sheet, but I couldn't for instance have a 103 kPa vacuum.</p>
| |composite|composite-resin| | <blockquote>
<p><strong>Cure Procedure</strong></p>
<ol>
<li>Apply vacuum of 22 inches (74 kPa) Hg minimum.</li>
<li>Apply 85 + 15 – 0 psig (586 + 103 – 0 kPa) pressure for laminates.</li>
<li>Apply 45 + 15 – 0 psig (310 + 103 – 0 kPa) pressure for sandwich.</li>
</ol>
</blockquote>
<p>I believe Steps #2 and #3 are providing you a tolerance for the applied pressure.</p>
<p>So, for Step #2, it's telling you to apply 85 psig. You can go +15 psig higher (for a total of 100 psig) but should not go lower than 85 psig. Normally, this is written as <span class="math-container">$85^{+15}_{-0}$</span> psig or 85 +15/-0 psig.</p>
<p>For Step #3, apply 45 psig, with an upward tolerance of +15 psig (total 60 psig) but go no lower than 45 psig total pressure. Normally, this is written as <span class="math-container">$45^{+15}_{-0}$</span> psig or 45 +15/-0 psig.</p>
| 5706 | How do I interpret the curing pressure of a composite from this data sheet? |
2015-10-09T18:12:35.390 | <p>I'm trying to figure out the theoretical calculations for a Stirling Engine system so I can determine optimal piston size, optimal piston displacement and number of pistons. I want to calculate which configuration has the most amount to torque to turn something else, and the highest amount of RPM's.</p>
<p>What I have managed to achieve:</p>
<p>(1) Internal energy = Thermal difference + Work</p>
<p>$$ΔU = Q + W$$</p>
<p>(2) Work = Rotational Kinetic Energy</p>
<p>$$W = \dfrac{1}{2}Iw^2$$
(3) Assuming that Inertia is standard wheel</p>
<p>$$I = \dfrac{1}{2}Mr^2$$</p>
<p>(4) Equation I have so far</p>
<p>$$\dfrac{1}{4}Mr^2w^2 = ΔU - Q$$</p>
<hr>
<p>I am unsure on how I can relate the equation to the pistons. According to HyperPhysics </p>
<blockquote>
<p>Internal energy is defined as the energy associated with the random,
disordered motion of molecules.</p>
</blockquote>
<p>That sounds like pressure to me, and I know that:</p>
<blockquote>
<p>Pressure = Force / Area</p>
</blockquote>
<p>But I think I need a better way to express the physics of pistons.</p>
<p>The Q component of internal energy is also difficult for me to find, because a heat source (like a candle flame), will be creating a heat differential on the piston. This heat differential in Joules, will be a little difficult to find as well.</p>
| |automotive-engineering|experimental-physics|pistons| | <p>To "optimize" requires a desired outcome. Sterling engines, like most mechanical/thermodynamic systems are highly complex. You could optimize for efficiency, cost, power-to-weight, and many others. For learning the theory before you get to that stage I would recommend the wiki article, then do web searches based on your new vocabulary after that.</p>
<p><a href="https://en.wikipedia.org/wiki/Stirling_engine#Theory" rel="nofollow">https://en.wikipedia.org/wiki/Stirling_engine#Theory</a></p>
| 5709 | Find optimal configuration for Stirling Engine |
2015-10-10T00:39:46.823 | <p>I have a 15 pound weight that I use to crush 8.4 oz red bull cans. A metal barbell with a handle and 2 ends. Not 100% flat ends, but close to flat. About 12 inches from end to end.</p>
<p>I hold the barbell by one end and lower it to crush a standing redbull can. sometimes the can crushes completely, other times it gets crooked. There doesn't seem to be much predictability in regards the various methods of striking it. Only that multiple small strikes allows it to be mended and crushed by hand also.</p>
<p>So, my question is - how can I predictable crush the can with the least effort? I think my strikes are consistent but the cans are in need of some pre-crushing to create fault lines were it will crash like a controlled demolition, or at least that is my current theory.</p>
| |mechanical-engineering| | <p>There was a very slight wobble of the weight as it went from "standing" to "dropping". The standing position had no wobble, but it created a wobble when released. The solution was hold the weight freely and to control the wobble, so when allowed to drop by gravity, no additional forces occur in transition. Can crushes are then relatively predictable. Only it hurts my back a little from the extra forces and time.</p>
| 5715 | Crushing drink cans |
2015-10-10T05:58:34.710 | <p><a href="https://i.stack.imgur.com/s9LVf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s9LVf.png" alt="enter image description here"></a></p>
<p>Hi, I'm having some problems with this truss question that I have in my homework. So far this is my solutions.</p>
<p>I worked out my solutions to the reaction forces but don't know if they're correct.</p>
<p>The next step is the find the internal forces ZB
YB
SU
SV
TV
RT
RS
QS
JL
JK
NP
MP
HO
HM
IO
MO</p>
<p>Is there an easier way to find all these? e.g. are some of them the same?. Wouldn't using either method of joints or sections be too tedious here?.
Thanks. </p>
<p>[![enter image description here][2]][2]</p>
| |civil-engineering|structures|statics| | <p>It may be worth reconsidering your deduced zero force members. But for this question it is not really necessary to determine which are zero force members in advance as this will come out in the analysis.</p>
<p>It is not important if the truss is determinate or not in order to get the member forces your are interested in (highlighted in red). Approach this question using the <a href="http://www.learnengineering.org/2013/08/structural-analysis-method-of-sections.html" rel="nofollow noreferrer">method of sections</a>. I have marked sensible section cuts in green. You can get the member forces of JL and JK by considering the equilibrium of joint J. It is worth taking note that the member forces requested hints at the appropriate solution method.
<a href="https://i.stack.imgur.com/PdcO1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PdcO1.jpg" alt="Figure1"></a></p>
<p>For example, in order to get the member forces RT, RS, and QS make cut 3:
<a href="https://i.stack.imgur.com/6srGl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6srGl.jpg" alt="Figure2"></a></p>
<p>Notice that since you have the reaction forces at A already the member forces RT, RS, QS can be solved by equilibrium.</p>
<p>Getting member forces for NP, MP, and MO is a little bit more tricky, but once you have member force for MH (by making cut 4) you can make the cut marked in blue.</p>
<p>It is a bit tedious using method of sections, but there isn't a faster method that I am aware of. Even modelling this in a structural analysis software package would probably take longer ...</p>
| 5719 | Loaded Truss problem |
2015-10-10T06:47:34.927 | <p><a href="https://i.stack.imgur.com/SkVHG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SkVHG.png" alt="enter image description here"></a></p>
<p>I'm supposed to be finding the reaction forces first but from the FBD of the entire truss where am I supposed to start. If I take the moment at A then I get 3 unknowns at Bx Dy and Dx. If I get the sum of forces in x direction equals to 0 then I get 3 unknowns again. Same thing with the Y direction, I get two unknowns. </p>
| |civil-engineering|structural-engineering|statics| | <p>You seem to be treating all the connections between members as fully fixed but rather they are all hinges with the exception of joint E. The contact at E can be treated like a roller, thus there's a force normal to member DE as shown below which I have called $F_E$:</p>
<p><a href="https://i.stack.imgur.com/O0yFc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O0yFc.png" alt="Member DE"></a></p>
<p>Consider the sum of moments about D:</p>
<p>$\sum M_D = 0: F_E(5) = 650(7.5)$</p>
<p>$\therefore F_E = 975 \text{ N}$</p>
<p>Consider the sum of forces in the direction normal to the member:</p>
<p>$\sum F_n = 0: F_E = 650 + F_D$</p>
<p>$F_D = F_E - 650$</p>
<p>$F_D = 975 - 650 $</p>
<p>$\therefore F_D = 325 \text{ N}$</p>
<p>Now we can split $F_E$ into its x and y components:</p>
<p>$F_{E,x} = F_E\text{cos}(30) = 975\text{cos}(30) = 844.4 \text{ N}$</p>
<p>$F_{E,y} = F_E\text{sin}(30) = 975\text{sin}(30) = 487.5 \text{ N}$</p>
<hr>
<p>Let's look at the rest of the system including the resultant force of $F_E$ at E - remember to switch the directions of $F_E$ as this is how it acts on member AE:</p>
<p><a href="https://i.stack.imgur.com/4FjLw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4FjLw.png" alt="The rest of the system"></a></p>
<p>I won't do the rest but you should be able to do some force balances in the x and y-axes and then a moment balance about A to get the reactions: $A_x, A_y \text{ & } B_x $</p>
<p>You could use <a href="https://skyciv.com/3d-structural-analysis-software" rel="nofollow noreferrer">SkyCiv Structural 3D</a> to check your solution for this as shown below but you need to be careful so that you model the connections properly:</p>
<p><a href="https://i.stack.imgur.com/EG2CL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EG2CL.png" alt="SkyCiv 3D Structural Analysis Software"></a></p>
| 5720 | Frame/truss problem with sliding joint |
2015-10-10T14:00:46.633 | <p>I have an application (a tube with parts in it) which needs a certain static pressure in it in order to contain a sustainable air flow. Unfortunately the fan I wanted to use in a push-configuration is not able to create this pressure. Thus I thought that I could install another fan with a higher cfm at the end of the tube (in pull-configuration), and thereby increasing the static pressure? Would that be useful, or is there another way to increase the static pressure in the tube?
Schematic is:
<a href="https://i.stack.imgur.com/nqQbo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nqQbo.png" alt="enter image description here"></a></p>
<p>Neither F1 or F2 are able to provide the necessary static pressure alone.</p>
| |fluid-mechanics| | <p>In order to get the air through the pipe, the pressure rise of the first fan has to be so high, that the pressure-loss within the tube is compensated.</p>
<p>Since your first fan (F1) is not powerful enough you can add a second fan (F2) to rise the pressure at the end of the pipe to ambient pressure, or use a more powerful fan (F3) instead.</p>
<p><a href="https://i.stack.imgur.com/6o0Wn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6o0Wn.png" alt="setch"></a></p>
<p>The sketch shows the static pressure on the y-axis and the tube at the x-axis</p>
| 5723 | Increase static pressure |
2015-10-11T02:26:05.770 | <p>Assume I put an RF antenna in front of an object (typically the face of someone sitting in a chair, so able to move about +/- 10cm in all directions) at a typical distance of 1m. If I consider the system antenna/face as a resonant cavity, I'd get a fundamental resonant frequency of about 150MHz.</p>
<p>I have a general physics education but no experience at all with RF antennas and resonators, so I have no orders of magnitude in mind. I realize such a 'resonant' cavity would be extremely poor, but how poor? My question is: is it conceivable to have enough resonance with basic antenna and processing electronics to be able to measure the mean distance to the face with a precision of say 1cm (i.e. measure the mean resonance frequency with about 1.5MHz precision), or is it an obvious and complete no go?</p>
<p>If it were conceivable, how 'clean' would the background need to be? Assuming there could be many unconstrained background objects (typically at a distance twice larger) all in a small room, could I just pick up the main resonance frequency around 150MHz for my measurement or would all the reflections / interferences with the other objects + room walls mess the signals completely up?</p>
<p>Note: I'm not expecting a detailed solution to my problem (although I'll happily take it if it comes :) I'm more looking for an "expert's feeling" like "yes, people have done that for decades", "maybe, you'd have to try" or "No way dude, you're completely dreaming". </p>
<p>In any case, if anyone sees any other (as cheap as possible, and most importantly automated and non intrusive) way to get such a measurement (+/- 1cm at 1m distance) please let me know.</p>
| |electrical-engineering|rf-electronics|distance-measurement| | <p>I do not believe this will be an effective scheme. The <a href="https://en.wikipedia.org/wiki/Q_factor" rel="nofollow">$Q$ of a resonator</a>, a number which describes the strength of the resonant peak, <a href="https://www.rp-photonics.com/q_factor.html" rel="nofollow">is given by</a>
$$
Q=\frac{f_0}{\delta f}\approx f_0\frac{2\pi\ t_{RT}}{\ell},
$$
where $f_0$ is the resonant frequency, $\delta f$ is the FWHM of the resonance, $t_{RT}$ is the round-trip time, and $\ell$ is the round-trip loss. The approximation is for $\ell\ll1$, but it actually overestimates the $Q$ value for $\ell\sim1$ so it will be useful for this case.</p>
<p>The biggest problem for your application are the losses. I don't know exactly how much power a human face will reflect back towards the RF antenna, but I would think 5% is a relatively high value. Putting $\ell=0.95$ in the above equation gives $Q\sim3,$ which is a very low value. Good resonators have $Q$ values between 100 and 10,000. Using the other definition of the $Q$ parameter given above you can see that the width of the resonant peak will be
$$
\delta f=\frac{f_0}{Q}\simeq50\ \text{MHz}.
$$<br>
So, the accuracy you can expect in the measurement is on the order of 50 MHz. You may be able to do somewhat better with a really clean measurement, but the accuracy won't be anywhere near 1.5 MHz. </p>
| 5728 | Measuring the distance to an object with RF resonance |
2015-10-08T05:21:18.700 | <p>I'm thinking about using the <a href="https://www.bopla.de/en/enclosure-technology/product/ultramas/ultramas-um-309-l-enclosures/um-32009.html" rel="nofollow">Bopla UM 32009 housing</a> for my product. It has little standoffs for attaching PCBs. These standoffs are made of plastic, and seem to be drilled (or molded?) to a diameter of 2.0 mm.</p>
<p>My questions: What type of screw would be appropriate for this hole diameter? A regular M2 machine screw won't work well. An M2.5 might work, but would probably deform the standoff.</p>
<p>An M2.2 self tapping screw might work I suppose, but I think the hole is a bit large for this (recommended is something like 1.8 mm).</p>
<p>Is there some kind of imperial screw that fits well into a 2.0 mm plastic hole?</p>
| |plastic|fasteners| | <p>Typing "screws" in the search field of the Bopla website finds <a href="https://www.bopla.de/en/service/technical-information/plastic/general-information-on-plastic-enclosures/screws-and-attachments-for-plastic-enclosures.html" rel="nofollow">this link</a>.</p>
<p>It looks like the screw you need is has a shank diameter of 1.7–1.8 mm and a thread diameter of 2.2 mm. The head is rounded.</p>
| 5735 | What screw for 2.0 mm plastic hole in housing |
2015-10-12T15:48:11.197 | <p>I live in Melbourne, Australia and a lot of the creeks here have been diverted through underground storm water pipes. I guessed that this is because land is highly sought after for development, and a lot of these creeks have been built over, however there are also a lot of cases where there is nothing built over the top of them at all. They simply carry the normal water flow through a small concrete pipe, leaving an empty valley on the surface which would only carry flood waters during heavy rainfall. What's the reason for going to the cost and effort of doing this?</p>
<p>Is it a safety concern having creeks in residential areas? Do creeks have some effect on the land which would affect nearby buildings? Is it just for the convenience of being able to cross them without a bridge?</p>
<h3>Example</h3>
<p>Dandenong Creek in Bayswater area @ -37.836524 lat, 145.254465 long (water flow goes from the right to the left of the image):</p>
<p><a href="https://i.stack.imgur.com/BDt0d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BDt0d.jpg" alt="Dandenong Creek"></a></p>
<p>I've also seen a creek which briefly goes both above and below ground. The rushing water of the underground section could be seen through an overflow while the topside creek was mostly just stagnant water. It seemed completely pointless and made no sense to me.</p>
| |civil-engineering|water-resources| | <p><strong>What's the reason for going to the cost and effort of doing this?</strong></p>
<p>In many places, placing a stream into a culvert can be cheaper in the long run. Streams (particularly fast flowing ones) erode the land and may require erosion protection added in the future. </p>
<p>An advantage is reduced land use: Land that does not have a stream running through it can be used for other purposes, although the culvert will need to, obviously, be assessed before plomping (technical term..) a multi-storey car park on top of it! </p>
<p><strong>Is it a safety concern having creeks in residential areas?</strong></p>
<p>Highly dependent on the area in question and the nature of the creek or stream. Naturally, a stream is a hazard to people of all ages, particularly kids though. But so is a culvert if the access to it is not cordoned off. As answered previously, an open grassy pitch is more favourable than a flowing stream. </p>
<p><strong>Do creeks have some effect on the land which would affect nearby buildings?</strong> </p>
<p>Creeks have a tendency to meander in the long term (decades to centuries), and deeply scour the nearby banks in the short term (years). Scoured banks can become steep and rocky, and even collapse. If there are buildings located nearby, then this can cause structural damage. </p>
<p><strong>Is it just for the convenience of being able to cross them without a bridge?</strong></p>
<p>As an engineer with Highway experience you'd be amazed at how many culverts manage water flow run under many of our roads. In Scotland, it's not unusual for there to be a 600mm diameter culvert every few hundred metres or so on many trunk roads and motorways helping ease water through them. </p>
<p>Most of the time, these are never seen or acknowledged as they can be down steep road verges. This is because these roads tend to be built very straight (much like railways), opposed to older roads that simply follow ancient paths and travelling routes. </p>
<p><strong>A final note:</strong> </p>
<p>Culverts can be used to change the velocity of flowing water. For example if a stream has a narrow channel, the water will flow quickly causing it to quickly erode the bank. You can slow the water velocity in an area by passing it through a wide culvert. With a wider area of flow, the water will travel more slowly. This can cause advantages down stream. Similarly, the same effect can be caused upstream by forcing the water through a narrower passage. This will cause the water to back-up up stream reducing it's speed. </p>
| 5749 | Why are creeks sometimes piped underground in urban areas? |
2015-10-12T19:08:39.697 | <p>I'm running a simple tensile test simulation for a single element. Under Results I made a Chart object with Stress vs Strain, which can be manually exported to a csv file by right clicking on the table. The trouble arises from the parametrization. I've parametrized temperature in Workbench so that the same simulation is run once for each temperature. However, the results are not saved for each substep for each design point, and it would be tedious to have to open Mechanical, re-solve, and export the Chart as many times as I have temperatures.</p>
<p>Ideally there would be a macro that runs every time the simulation is solved in Mechanical, exporting the Chart's table to a csv (appending it if possible so that the data for all temperatures is in the same file).</p>
<p>Is there any way to do this? I would imagine it could be achieved with JavaScript in Mechanical (for which I cannot find any documentation). </p>
| |mechanical-engineering|simulation|ansys|finite-element-method| | <p>There are several options for doing this:</p>
<ol>
<li>If there are only a few values to be saved per design point, you could use 'output parameters'. </li>
<li>If there is a lot of data to be saved, the report generator may be useful. Some information can be found here: <a href="https://www.sharcnet.ca/Software/Ansys/17.2/en-us/help/wb2_help/wb2h_projectreports.html" rel="nofollow noreferrer">Working with Project Reports</a> </li>
<li>An APDL snippet can be used in the results* tree. Some information here: <a href="http://www.padtinc.com/blog/tag/design-points" rel="nofollow noreferrer">Saving Mechanical APDL Plots in a Design Study</a> </li>
<li>IronPython may be used. This requires you are an expert in Ansys and know programming well. </li>
</ol>
<p>See also <a href="https://caeai.com/sites/default/files/automated_postprocessing.pdf" rel="nofollow noreferrer">Path Plots and Automated Post Processing in MADPL and Workbench</a> </p>
<p>*Be careful to place the snippet in the results section of the model, otherwise it will run before in stead of after solving. </p>
| 5750 | ANSYS Workbench/Mechanical: Automatically export chart? |
2015-10-12T23:46:41.957 | <p>This is a basic control theory question, since Control Theory is a part of applied mathematics but also of engineering I was unsure whether to ask this here. </p>
<p>The question says:</p>
<p>Given the transfer function of a system is $G(s)=1/(s^2+3s+2)$, consider the design of a PI closed loop control system with unit feedback using proportional gain $k_p$ and integral gain $k_i$, both of which are positive. Determine the range of gain for which the closed loop system is stable.
What I did was this, I went the Routh-Hurwitz way: </p>
<p><a href="https://i.stack.imgur.com/2tb8f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2tb8f.jpg" alt="My solution"></a></p>
<p>After completing the Routh table, I went ahead and reasoned a little about what conditions need to be met in order to avoid sign changes on the main column, however it seems that these conditions are never met! I get that $k_i$ should be less than zero when the problem clearly specifies it will always be positive.
Have I done something wrong? Is my reasoning right? Is the answer "The controller is never stable"?</p>
| |control-engineering|control-theory|transfer-function| | <p>There is a mistake in your expressions. The coefficient of $s$ is $2+k_p$. </p>
<p>The conditions are: $$\frac{1}{3} \left(3 \left(k_p+2\right)-k_i\right)>0$$ $$k_i>0$$ $$k_p>0$$</p>
<p>This simplifies to: $$k_p>0$$ $$0<k_i<3 k_p+6$$</p>
<p><strong>Update:</strong></p>
<p>If the transfer function is $$ \frac{1}{s^2+3 s + c} $$ where c is some positive constant, then the conditions simplify to: $$k_p>0$$ $$0<k_i<3 ( k_p+c)$$</p>
| 5753 | Determine the range of values for a PI controller |
2015-10-12T00:18:41.260 | <p>I'm wondering what the technical name is for this kind of stepper-motorized arrangement (a positioning system? an XY table?)?</p>
<p><a href="https://i.stack.imgur.com/JLtsL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JLtsL.png" alt="enter image description here"></a></p>
<p>What is it typically used for?</p>
<p>It appears to have two steppers as opposed to generic DC motors, so it's presumably for precision work, and I'm trying to see if it would be appropriate for my (experimental biology-related) application.</p>
<p>The unstable kind of two-axis arrangement I suppose would make it too wobbly for something like CNC-machining, but perhaps might work well for scientific experiments.</p>
| |mechanical-engineering|stepper-motor| | <p>cartesian 2 axis robot</p>
<p>2 axis robot</p>
<p>cartesian coordinate robot</p>
<p>xz axis robot</p>
<p>xz gantry</p>
<p>For your purposes, compare the required precision, the load weight, and the control board and display, pc link, if it uses GCode or some other control. </p>
<p>Note that this kind of cartesian robot system has been popularized by 3d printers and there are masses of fan info's for steppers and gantry's and 3d printer related control displays.</p>
| 5755 | What is this kind of X-Y motorized system? |
2015-10-13T14:32:24.623 | <p>I have raw data obtained from EMED barefoot scan containing a matrix of pressure sensors over about 70 frames. This totals 70 matrices that record a snapshot of the pressure over the duration of a person's natural walk over the pressure plate. </p>
<p>I wanted to ask if anyone knows the algorithm that is typically used to determine the Center of Pressure Line (indicated by the path on the heatmap below). One thing I have tried is to consider each row as an array and find the index of the maximum pressure across the row, and create a line by matching up all the points that correspond to these indices. My approach fails to produce a smoothed line (even when using Gaussian smoothing).</p>
<p><a href="https://i.stack.imgur.com/Bnjqi.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bnjqi.gif" alt="center of pressure line across foot heatmap"></a></p>
| |biomechanics| | <p>Actually, scipy has a function that does exactly this. I used it on each of the 70 arrays that make up the time series, and points generate a smooth line.</p>
<p><a href="https://docs.scipy.org/doc/scipy-0.16.1/reference/generated/scipy.ndimage.measurements.center_of_mass.html" rel="nofollow">https://docs.scipy.org/doc/scipy-0.16.1/reference/generated/scipy.ndimage.measurements.center_of_mass.html</a></p>
| 5763 | Center of Pressure Line from Barefoot Scan (EMED) |
2015-10-14T02:35:34.203 | <p>I'm having problems understanding the worked answer for this problem: </p>
<p><a href="https://i.stack.imgur.com/29pZi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/29pZi.png" alt="Problem definition"></a></p>
<p>Here's part of the worked answer:
<a href="https://i.stack.imgur.com/Rxy6q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rxy6q.png" alt="Part of worked answer"></a></p>
<p>I don't understand why $\delta_C = \dfrac{4}{9} \delta_A$ . Can you please explain that part of it?</p>
| |mechanical-engineering|civil-engineering| | <p>Between ABC there is a single member. Hence the angle between AB and BC is always 90 degrees. Hence if AB rotates by theta, then BC also rotates by theta. As the length AB is 90mm, and the length AC is 40mm, then (by similar triangles):</p>
<p>$\delta_C = \dfrac{40}{90} \delta_A$</p>
| 5771 | Virtual Work - Determining Relationships in deflections |
2015-10-14T18:34:18.793 | <p>Turbulent flow involves extremely small scale fluctuations which are usually too small to resolve by using the Navier-Stokes equations directly. To overcome this, we model each transient variable $\Phi$ (e.g. velocity, pressure, etc...) as the sum of a time average quantity $\bar{\Phi}$ and some small fluctuation. By introducing both of these new variables into the navier stokes equations, we obtain more unknowns than equations. In order to obtain a unique solution, we need to introduce a turbulence model which consists of at least one additional equation.</p>
<p>There are a number of different turbulence models developed from the Boussinesq Hypothesis, which roughly says that the Reynolds stresses are proportional to the mean rate of deformation by way of the so-called "turbulent viscosity". The different turbulence models vary by how they estimate the <em>turbulent viscosity</em>. Models of this type include, but are not limited to:</p>
<ol>
<li>$k-\epsilon$ models</li>
<li>$k-\omega$ models</li>
<li>Mixing Length models</li>
<li>Spalart-Allmaras models</li>
<li>Reynolds Stress models</li>
</ol>
<p>With such a wide selection of turbulence models, it can be difficult to assess which model is appropriate for each modeling situation. I understand that with infinitely many potential simulation needs, there is no one-size-fits-all answer. But, are there any general guidelines on how to select an appropriate turbulence model? What factors should I consider when selecting a model? What kinds of problems are known to be effective for the models listed above?</p>
<p>While I'm mostly interested in the time-averaged turbulence models, I also welcome responses for spatially-averaged models (e.g. Large Eddy Simulation models).</p>
| |fluid-mechanics|turbulence| | <p>Disclaimer: As you requested <em>general guidelines</em> here they are. After having a converged solution, you need to investigate it thoroughly and look for clues that the choice of the turbulence model altered the flow in an un-physical way.</p>
<p><strong>My Rules of Thumb for Turbulence Models</strong></p>
<ol>
<li><p>The more additional transport equations you solve:</p>
<ul>
<li>the more possibilities exists to make errors (e.g. wrong assumptions)</li>
<li>the better the capabilities of the simulation are to capture the physics involved</li>
</ul></li>
</ol>
<p>Usually <em>mixing length</em> models have ONE additional transport equation, k−ϵ and k−ω have TWO and Reynolds Stress models have SEVEN. <br/>
The Spalart-Allmaras model is a highly tuned model and has only one additional transport equation. It was tuned for flight regimens (Reynolds and Mach number) of airplanes. The application in other fields might result in bigger errors.</p>
<ol start="2">
<li><blockquote>
<p>k−ϵ models<br></p>
</blockquote></li>
</ol>
<p>The k-ϵ-Model is commonly optimised for external flow (around bodies)</p>
<ol start="3">
<li><blockquote>
<p>k−ω models<br></p>
</blockquote></li>
</ol>
<p>The k−ω-Model is commonly optimised for internal flows (inside pipes, close to walls)</p>
<ol start="4">
<li><blockquote>
<p>Mixing Length models<br></p>
</blockquote></li>
</ol>
<p>Are very very basic and will only produce meaningful results when the turbulence modelling was not necessary in the first place.</p>
<ol start="5">
<li><blockquote>
<p>Spalart-Allmaras models<br></p>
</blockquote></li>
</ol>
<p>Are very robust, fast and widely tested. Additional testing is required when not applied in their actual scope (flow around airplanes)</p>
<ol start="6">
<li><blockquote>
<p>Reynolds Stress models<br></p>
</blockquote></li>
</ol>
<p>Cutting Edge Models need a lot of computational power and are not as robust.</p>
| 5778 | How do I select an appropriate turbulence model? |
2015-10-15T01:53:19.380 | <p>Weathering steel is supposed to form a protective patina when exposed to wetting and drying cycles. This protective patina prevents more corrosion from occurring. This is what differentiates it from regular (carbon) steel.</p>
<p><strong>In a situation where the environment does not allow the patina to form, does weathering steel perform any differently than carbon steel?</strong></p>
<p>I am thinking that there might be situations where weathering steel without a patina might actually corrode faster (perform worse) than carbon steel. The alloy differences may cause the change.</p>
<p>My situation is a steel deck with a layer of aggregate on top. The aggregate will likely keep the top of the deck from ever completely drying or at least greatly slow the process. This will keep a patina from forming. Various people that I have talked with at the client have said that they would rather have carbon steel instead of weathering steel. They seem to think that it will perform better.</p>
<p>I know that the <em>real answer</em> is to provide a layer of protection (waterproofing, paint, etc.) over the steel, but this is not an option. I wanted to provide my exact situation in case someone requests specifics.</p>
| |steel|corrosion| | <blockquote>
<p>In a situation where the environment does not allow the patina to form, does weathering steel perform any differently than carbon steel?</p>
</blockquote>
<p>Weathering steel is specifically designed to form a protective coating (i.e., patina) of rust that prevents the material underneath from corroding.</p>
<p>Referring to the <a href="https://en.wikipedia.org/wiki/Weathering_steel" rel="nofollow noreferrer">Wikipedia article</a> (1) on the topic,</p>
<blockquote>
<p>Using weathering steel in construction presents several challenges. Ensuring that weld-points weather at the same rate as the other materials may require special welding techniques or material. Weathering steel is not rustproof in itself. If water is allowed to accumulate in pockets, those areas will experience higher corrosion rates, so provision for drainage must be made. Weathering steel is sensitive to humid subtropical climates. In such environments, <strong>it is possible that the protective patina may not stabilize but instead continue to corrode</strong>. For example, the former Omni Coliseum, built in 1972 in Atlanta, never stopped rusting, and eventually large holes appeared in the structure. This was a major factor in the decision to demolish it just 25 years after construction. The same thing can happen in environments laden with sea salt. Hawaii's Aloha Stadium, built in 1975, is one example of this.</p>
<p>The rate at which some weathering steels form the desired patina varies strongly with the presence of atmospheric pollutants which catalyze corrosion. While the process is generally successful in large urban centers, the weathering rate is much slower in more rural environments.</p>
</blockquote>
<p>As you say, it would seem that your case, where the steel cannot fully dry, the protective patina cannot form.</p>
<blockquote>
<p>My situation is a steel deck with a layer of aggregate on top. The aggregate will likely keep the top of the deck from ever completely drying or at least greatly slow the process. This will keep a patina from forming. Various people that I have talked with at the client have said that they would rather have carbon steel instead of weathering steel. They seem to think that it will perform better.</p>
</blockquote>
<p>However, given the conditions you've specified, <strong>I cannot see how regular, uncoated carbon steel would perform any better than weathering steel in these conditions</strong>. I would expect that both will corrode approximately at the same rate in these conditions. My assumption is backed up by the information available on <a href="http://www.steelconstruction.info/Weathering_steel" rel="nofollow noreferrer">SteelConstruction.info</a> (2):</p>
<blockquote>
<p>Alternate wet/dry cycles are required for the adherent ‘patina’ to form. Where this cannot occur, due to continuously wet or damp conditions, a corrosion rate similar to that of ordinary structural steel must be expected. Examples include weathering steel elements submerged in water, buried in soil or covered by vegetation. If weathering steel is used in such cases, it should be painted and the paint should extend above the level of the water, soil or vegetation.</p>
</blockquote>
<p>Regular carbon steel will be more cost effective than weathering steel, so I assume this is where your client's concern lies.</p>
<p>Personally, I would go with galvanized steel for the decking, or put some sort of <a href="http://www.sumtercoatings.com/epoxy-primers.html" rel="nofollow noreferrer">epoxy primer</a> over the steel (I assume stainless steel is out of the question).</p>
<hr />
<p><strong>References</strong></p>
<p>1.) <a href="https://en.wikipedia.org/wiki/Weathering_steel" rel="nofollow noreferrer">Wikipedia - Weathering Steel</a></p>
<p>2.) <a href="http://www.steelconstruction.info/Weathering_steel" rel="nofollow noreferrer">SteelConstruction.info - Weathering Steel</a></p>
| 5782 | Does weathering steel corrode the same as carbon steel when it doesn't develop a protective patina? |
2015-10-16T08:26:13.677 | <p>What is the difference between <em>Transport Phenomena</em> first edition (1979) and second edition (2002)?<br>
Which one do you recommend to be bought, taking into account that I am a student and that the first one is cheaper?</p>
| |fluid-mechanics|heat-transfer|chemical-engineering| | <p>Taken from the <a href="http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0470115394.html" rel="nofollow">Wiley</a> page:</p>
<blockquote>
<p>About the Revised 2nd Edition: Since the appearance of the second
edition in 2002, the authors and numerous readers have found a number
of errors--some major and some minor. In the Revised 2nd Edition the
authors have endeavored to correct these errors. A new ISBN has been
assigned to the Revised 2nd Edition in order to more easily identify
the most correct version.</p>
</blockquote>
<p>New content:</p>
<blockquote>
<ul>
<li>Revised to include more extensive reference to applications of material covered and the addition of appendices.</li>
<li>Expanded coverage of: transport properties in 2-phase systems, boundary-layer theory, heat/mass transfer coefficients, dimensional
analysis and scaling</li>
</ul>
</blockquote>
<p>Concerning your issues with the English language used, your comment exhibits a good grasp of the English language and I would argue that a good understanding of Calculus is much more valuable than being a native English speaker for reading this book. Besides, many canonical literature is written in English and it would be good to get some practice in reading and understanding such texts. Good luck!</p>
| 5804 | "Transport Phenomena" (Bird, Stewart, Lightfoot) difference between edition 1 and 2 |
2015-10-16T19:28:13.597 | <p>In Rankine cycle there's the boiler where water gets boiled into overheated steam. On input there's a pump that delivers more water, and on output there's a turbine that picks up the energy of the compressed steam.</p>
<p><a href="https://i.stack.imgur.com/bhQ5I.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bhQ5I.png" alt="enter image description here"></a></p>
<p>The steam pressure is roughly identical against the turbine and the pump; the pressure of the boiler section.</p>
<p>What makes the steam power the turbine instead of backing up and forcing the pump to turn backwards? - well, this one is simple, power delivered to the pump. But then how comes the turbine produces more power than the pump takes? The pump, after all, must overcome the same pressure that propels the turbine and deliver the same amount of water that is being ejected as steam. I'm missing some significant element of the device. What is it?</p>
| |thermodynamics|steam| | <p>The main reason is that it takes much less energy to compress a liquid than a gas by the same pressure difference. The pump takes a little bit of energy to compress the water, but a huge amount of energy is released when the steam expands in the turbine. This is why a phase change is used in the Rankine and related cycles. </p>
<p>Another thing to consider is that the pump doesn't need to over-match the <em>power</em> of the turbine, as you said - it only needs to over-match the <em>pressure</em>.</p>
<p>You can see the difference in energy between gas and liquid compression when looking at an enthalpy-vs-pressure table.</p>
| 5813 | How in Rankine cycle the turbine generates more power than the pump takes? |
2015-10-17T13:06:39.367 | <p>Just something I was wondering about when I've seen <a href="https://i.stack.imgur.com/xVr4T.jpg" rel="noreferrer">this</a> on reddit today. Unlike in firearms where the pressure propels the projectile, in a railgun, wouldn't the pressure ahead/behind of the projectile slow it down instead? If so, would it be by a significant amount?</p>
| |mechanical-engineering|pressure|firearms| | <p>For both conventional guns and rail guns, the resistance from pressure ahead of the projectile is usually negligible. Venting probably won't help much either if the projectile gets near the speed of sound because the air ahead of the projectile won't have time to move out of the way. You could fire in an evacuated chamber to completely avoid the air resistance, but the gains probably wouldn't justify the added complexity.</p>
| 5818 | Does negative pressure inside the barrel affect low-powered railguns? |
2015-10-18T23:23:20.613 | <p>What does trip angle mean in the context of <a href="http://www.pqcontrols.com/m410.php" rel="nofollow">slope / level sensors / inclinometers</a>?</p>
| |sensors|terminology| | <p>From the link that you provided, it would seem that trip angle is a preset angle at which the sensor will "trip" or "switch on". The results of this can depend on purpose of the tilt meter. It can be wired to sound an alarm, turn on a light, etc.</p>
<p>From this <a href="http://www.spectronsensors.com/appsheets/SAN-218-2109.pdf" rel="nofollow">spec sheet</a> on another tile switch, the trip angle can be adjustable. The spec sheet also gives an example circuit. </p>
<p>Another <a href="http://www.pqcontrols.com/m420.php" rel="nofollow">sensor on the site</a> that you linked gives possible uses.</p>
| 5832 | What is the meaning of a "trip angle" in a level sensor? |
2015-10-19T12:54:08.443 | <p>I'm making a remote starter for a car using a Raspberry Pi. The problem is that the car has a manual transmission. The driver needs to remember to place the gear shift in neutral when parking, otherwise all kinds of bad stuff could happen when the car is started in-gear. </p>
<p>What possible solutions do I have for making an interlock that prevents the car from starting in gear? It's a 2001 Honda Accord, if that helps. I have seen transmissions with sensors (switches) on every gear, but this transmission does not have those. I thought of monitoring the speedometer sensor, but I think that info would come too late. I'm thinking I might have to put one or more switches on the shift lever to detect its position, but this sounds like a lot of work.</p>
| |electrical-engineering|automotive-engineering| | <p>They already have remote starters for cars with manual transmission. I know as i used to own one. The way a remote starter in a manual tramission works, or at least the one I owned, relied on the operator essentially arming the remote starter before leaving the car.</p>
<ol>
<li>having the engine running.</li>
<li>place take the manual transmission out of gear and into neutral.</li>
<li>engage the parking parking break</li>
<li>Take your fee off the clutch and brake. This proves to the starter that the car is in neutral. Otherwise it would stall.</li>
<li>Turn off the car.</li>
<li>Everyone get out of the car without touching the brake or clutch. (Drives door may need to be last to close, I cant remember)</li>
<li>Lock the doors</li>
<li>System is now armed as the car is supposedly empty and therefore cant be put into gear. (does not account for convertibles with top down or people going in through open windows)</li>
</ol>
<p>The system disarms under the following circumstances:
Car gets unlocked/Door opens
Break is touched</p>
<p>If the remote starter has been fired/initiated, the car will continue to run when a door is unlocked/opened.</p>
<p>I believe you need to insert the key and turn it to run (one stop before engaging the starter before you put your foot on the break.</p>
<p>To rearm the system if it has been disabled (ie someone opened a door to grab something they forgot, you need to start the car and go through the arming shutdown procedure again.</p>
| 5838 | Remote start on a car with manual transmission |
2015-10-19T14:59:08.427 | <p>What's the mechanism behind disposable syringes jamming? How do you prevent this from happening or reverse the effect? They are a lot cheaper.</p>
<p>The problem is that once the syringe is pressed all the way in, the plunger sticks and can not be pulled back out.</p>
| |mechanical-engineering|medical-devices| | <p>There's a story on them <a href="http://www.pbs.org/newshour/rundown/self-destructing-syringes-force-safer-injection-practices/" rel="noreferrer">at PBS</a>. From the article:</p>
<blockquote>
<p>A nurse injects a patient with a syringe of antibiotics, reloads and moves on to the next patient in line. The syringe isn’t sterilized, the needle is not replaced, and the patient is at risk of contracting a disease from the very shot that is supposed to cure him.</p>
</blockquote>
<p>...</p>
<blockquote>
<p>About 40 percent of all injections are given with unsterilized, reused syringes and needles, reports the World Health Organization. An estimated 1.3 million deaths — and 21.7 million new Hepatitis B infections — occur each year as a result of the unsafe practice.</p>
</blockquote>
<p>...</p>
<blockquote>
<p>The devices, called auto-disable syringes, jam up after one injection and can’t be reused. The casing is engineered to break if someone tries to forcefully refill it.</p>
</blockquote>
<p>...</p>
<blockquote>
<p>Regular disposable syringes can also make their way back into the marketplace after being used. People will take syringes from a health clinic’s trash, wash them and repackage them for resale.</p>
<p>“You can find in a market a perfectly well packed syringe, then you look closely and you can still see some blood in the syringe from the last person,” he said.</p>
</blockquote>
<p>The article continues with all the reasons you can think of for not wanting to re-use a syringe.</p>
<p>The syringe appears to be invented by a company called <a href="http://www.starsyringe.com/product-k1.htm" rel="noreferrer">"Star Syringe"</a>, who has licensed their intellectual property to 13 manufacturers. From their website:</p>
<blockquote>
<p>A small ring etched on the inside of the barrel allows the specifically adapted plunger to move in one direction and not the other. After one complete injection is given the plunger will automatically lock in place and break if forced, thus rendering the syringe useless.</p>
</blockquote>
<p>The following graphic is from <a href="http://www.zdnet.com/article/this-syringe-will-now-self-destruct/" rel="noreferrer">ZDNet</a></p>
<p><a href="https://i.stack.imgur.com/8DJPM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/8DJPM.png" alt="Auto-disable syringe" /></a></p>
<p>Finally, there's a video on YouTube that <a href="https://www.youtube.com/watch?&v=jmsyuPUS-EM" rel="noreferrer">shows the operation of the syringe</a></p>
| 5840 | What is the purpose and cause of disposable syringes being single use only? |
2015-10-20T21:03:42.257 | <p>Here's my situation: </p>
<p><a href="https://i.stack.imgur.com/dMIvO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dMIvO.png" alt="enter image description here"></a></p>
<p>I'm trying to drop a payload via parachute with some accuracy. I'm worried that in the above setup, the line will interfere with parachute deployment or - if not deployment - operation during flight (imagine wind pushing the chute into the line). </p>
<p>I haven't tested this yet but plan to. Does anyone know how the payload to anchor connection might be engineered to ensure full deployment throughout the descent? </p>
| |mechanical-engineering|aerodynamics| | <p>The chute should probably be shaped to pull itself away from the line like a kite.</p>
<p>It may be simpler however to use the line itself as the brake instead of using a parachute.</p>
| 5860 | Guiding a parachute via anchor: Possible? |
2015-10-22T12:34:54.370 | <p>I have a small vessel with a rectangular cross-section which needs to be ventilated at a relatively slow rate, ~$3\ \tfrac{\text{L}}{\text{hr}}$. The connections to the vessel will be tubes with an approximately 0.5" diameter, but the cross-section of the vessel is 4" x 1". The diagram below gives a rough idea of the cross-section (viewed from above) with the height of the chamber being about 1". The part circled in orange is the topic of this question. The flow will be forced in from the left and will exit passively from the right (or vice versa).</p>
<p>How can I couple the 0.5" tube to the chamber such that the flow in the chamber is uniform across the entire cross-section for the entire length of the duct? In addition, how can I couple the two in the smallest amount of space possible?</p>
<p><a href="https://i.stack.imgur.com/JkH0K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JkH0K.png" alt="enter image description here"></a></p>
| |fluid-mechanics|airflow| | <p>Quick and dirty CFD simulation of your problem using ANSYS Fluent 14.5: I used a 2D duct, 8" x 4" with a 45-degree angle going from the inlet pipes to the main chamber. Assuming 3 liters per hour flow through a half-inch diameter pipe gave me an inlet velocity of 6.6 mm/s. Air enters from the left and exits through the right. Inlet and outlet were set to a constant pressure of 101 kPa. I used the realizable $k-\epsilon$ turbulence model with standard wall functions. Computed velocity contours are below:</p>
<p><a href="https://i.stack.imgur.com/Mhs9o.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Mhs9o.png" alt="enter image description here"></a></p>
<p>You can see that the flow becomes very uniform by the time the angled parts meet the main duct. It may be hard to read off the image, but the difference in velocity between points A and B is about 1 mm/s.</p>
<p><strong>Update</strong></p>
<p>I re-ran the problem with the inlet going directly into the main duct - no expander section. Velocity contours are as follows:</p>
<p><a href="https://i.stack.imgur.com/yQcwB.png" rel="noreferrer"><img src="https://i.stack.imgur.com/yQcwB.png" alt="enter image description here"></a></p>
<p>Velocity at point A is nearly zero (a dead-spot). Velocity at B is about 4 mm/s. The flow through the duct becomes uniform after about 1 inch into the duct, where I'm defining uniformity as when the velocity at the center of the duct is within 1 mm/s of the velocity near the wall.</p>
<p>The most interesting thing about the second case is the formation of dead spots in the corners. The flow is fairly uniform, but the dead spots may be bad if you need good ventilation there.</p>
| 5878 | How can I create uniform laminar flow in a short duct? |
2015-10-22T20:53:36.683 | <p>Let's say I have a bar attached to the rim of a rotating wheel, <a href="https://i.stack.imgur.com/Zdg7G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zdg7G.png" alt="sketch"></a>, and it's in a vacuum so we can ignore drag. </p>
<p>If the wheel is rotated, does the bar experience a lateral force (perpendicular to the bar)? What if the rotation speeds up impulsively: will the bar experience a lateral force then?</p>
| |mechanical-engineering| | <p>In a vacuum acceleration is all towards the centre of rotation. Acceleration and force are vectors in the same direction, hence with no lateral acceleration there is no lateral force.</p>
<p>In the real world it's not in a vacuum and hence there is an additional force acting on the bar, caused by drag. This drag is lateral to the bar.</p>
<p>An increase/decrease in rotational speed (assuming the impulse for this change of speed comes from a torque at the centre of rotation) would cause an angular acceleration, i.e. the acceleration would not be towards the centre of rotation. As previously mentioned, acceleration and force are vectors in the same direction, hence the angular acceleration (which is lateral to the bar if taken over an infinitely small time period) would result in a lateral force on the bar.</p>
| 5880 | Does a rotating bar experience lateral force? |
2015-10-22T20:57:29.887 | <p>More specifically, I'd like to know if unmanned rockets need to be specially designed to withstand the heat generated from air friction and shock waves during launch. I found the diagram below on the NASA website which shows some fairly high stagnation temperatures, but the problem is that I don't know how fast a typical rocket travels on its way to, say, low earth orbit. I also don't know what the velocity profile looks like, i.e. Mach number vs. altitude.</p>
<p>Most of the information I've found only talks about orbital velocity and reentry temperatures, neither of which is very useful.</p>
<p><a href="https://i.stack.imgur.com/qBriJ.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/qBriJ.gif" alt="enter image description here"></a></p>
| |aerospace-engineering| | <p>Surprisingly not as hot as expected. At standard temp of 25°C at sea leval launch. Will peak at 400°C, then begin to dissipate. As rocket nose cones designed to dissipate thermal energy by piercing air away from them.</p>
<p><a href="https://i.stack.imgur.com/sv0cV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sv0cV.jpg" alt="enter image description here" /></a></p>
<p>The higher the altitude the colder the air gets. At 50,000 feet, air temp, what little air there is, is -67°F (-55°C)
Temps rise......between 55 - 175 thousand feet, then begin massive decline.</p>
| 5881 | How hot does a rocket nose cone get during launch? |
2015-10-22T23:42:29.920 | <p>Please note, this is not a shopping question; this is a <em>what type of component</em> works best for this scenario question. </p>
<hr>
<p>I have an induction heater circuit and a 5A max. / 24V wall wart, and I'd like to switch off / on the output of the wall wart so that I can control the power output of the induction heater. </p>
<p>About 7 years ago I found something by the name of SSR or SCR (or something) on digikey and that cost a reasonable amount (< 10 USD), and now I can't find it any longer. The SSR's on there that can handle that much power cost around 50 USD, which is too much to design with. The frequency can be low (5-100 Hz) range, since it <em>is</em> a relatively slow heating process.</p>
<p>I plan to switch the relay on/off with a raspberry pi, and will probably need a transistor in between. I don't want a mechanical relay since the thing will probably break before even making real use of the product.</p>
<p>The component I found prior was upright flat-rectangular with 4 thru-hole terminals coming out of the bottom smaller side. I found some that look like that but like I said they were in the 50 USD range. Also, that component if I recall correctly was for switching on/off AC for a different type of heater, so maybe <em>that's</em> why it was cheaper? </p>
<p>What types of alternative components should I be considering based upon these requirements?</p>
| |electrical-engineering|power-electronics|consumer-electronics| | <p>If you want to use an SSR, you'll need to get one designed for switching DC. Like <a href="http://www.aliexpress.com/item/solid-state-relay-SSR-25DD-25A-actually-3-32V-DC-TO-5-110V-DC-SSR-25DD/1154754963.html?spm=2114.031010208.3.41.qum1ti&ws_ab_test=searchweb201556_2_71_72_73_74_75,searchweb201527_4,searchweb201560_9" rel="nofollow">this one</a> Unlike mechanical relays which can switch AC or DC, SSR's are designed for switching either AC or DC loads.</p>
| 5883 | Identifying affordable alternatives for low-frequency high-current DC-DC solid state relays |
2015-10-23T01:59:50.227 | <p>I am in a first-year engineering design group where we are in the process of designing a cooling vest for motorcyclists for use in the city (1-2h length). Water will circulate through the vest, absorbing body heat, and then that heat will be dumped into a heat sink.</p>
<p>We initially decided on using Peltier chips for our heat sink, but when we brought this up with our instructor he was unenthusiastic (they are highly inefficient apparently). Then we decided to look into short-acting (1-2h) endothermic chemical reactions, but here ran into information overload.</p>
<p>What kind of practical options are there for such a heat sink?</p>
| |heat-transfer|design|cooling| | <p>What you first need to establish is your temperatures. Body temp min and max; ambient min and max. Select the worst case. Then take some guesses as to how much heat you have to pump. Is the vest insulated on the outside or does it have to cool the ambient as well? I would blind guess that you will need about 200 watts of cooling by the time you figure your losses.</p>
<p>Peltier modules are inefficient, but have a much nicer form factor than an R-134A compressor (like what is sitting under your fridge). You will have to weigh the befits of each. With the low temperature differential it will likely turn out that Peltier modules are actually more efficient than a prebuilt compressor designed for a high temperature differential. Also keep in mind that if you pay top dollar you can get a lot more efficient peltier modules than the $5 Chinese ones. Here are some nice reasonably priced ones with good specifications; <a href="http://tetech.com/peltier-thermoelectric-cooler-modules/high-performance/" rel="nofollow">tetech.com</a>. From the COP graph in the specifications pdf you can calculate very precisely how efficient a module will be for a given temperature differential.</p>
<p>Im not sure about your application, but remember that endothermic reactions are either not sustainable (throw them away when you are done) or require recharging. Recharging requires cooling; so you are just delaying the need to have a heat pump somewhere else in the process.</p>
| 5885 | What are heat sink options for a circulating-water cooling vest? |
2015-10-24T18:17:12.103 | <p>We are given a standard diagram of a rotating cylinder with the parabolic shape in the rotation.</p>
<p>Given:</p>
<ul>
<li>the distance between the vertex of parabola to the top of the parabola.</li>
<li>the original height of the water. </li>
</ul>
<p>I have a really general question regarding this problem. From where the vertex of the parabola begins, up to the top, the pressure distribution varies with radius? And from the vertex of the parabola to the bottom of the cylinder, it only varies with height, since radius is fixed?</p>
<p>My question is: can I split the two different areas and integrate $P = \dfrac{\rho\omega^2}{2} -\gamma z$ twice?
Once while holding the radius constant but the height a variable, and another while holding the height constant but integrating with respect to the radius. </p>
<p>I'm planning on using $\text{d}F= P\cdot\text{d}A$ when integrating each one.</p>
<p>A link to an image that looks similar to the one I have : <a href="http://demonstrations.wolfram.com/FluidRotatingInACylinder/" rel="nofollow">http://demonstrations.wolfram.com/FluidRotatingInACylinder/</a></p>
| |fluid-mechanics| | <p>I would like to add to Carltons answer and go a little more in depth on how to approach problems like that in general.</p>
<p><strong>Hydrostatics or Hydrodynamics</strong></p>
<p>There are two possible solutions to the problem. You can either choose a reference frame that rotates with the fluids or that stays still and observes the system from the outside. This will determine if hydrostatics are allowed or not.</p>
<p><strong>Rotating reference frame</strong></p>
<p>The origin of the coordinate system (polar coordinates) is at the top of the cylinder, positive z axis direction is upward, against the gravity.</p>
<p>Since there is no movement in our system as we rotate the same way the fluid does, hydrostatics can be used.</p>
<p>$$\nabla p = \varrho \vec{f}$$
Due to the rotation of our reference frame we have to add the centrifugal force as an external force.
$$\vec{f} = \begin{pmatrix}
\Omega^2r\\
0\\
-g
\end{pmatrix}$$</p>
<p>Hence
$$\begin{pmatrix}\frac{\partial p}{\partial r}\\
\frac{1}{r}\frac{\partial p}{\partial \varphi}\\
\frac{\partial p}{\partial z}
\end{pmatrix} = \varrho \begin{pmatrix}
\Omega^2r\\
0\\
-g
\end{pmatrix}$$</p>
<p>Successively Integration yields</p>
<p>$p(r,\varphi,z)=\varrho \Omega^2 \frac{r^2}{2} + C(\varphi,z) \tag{1}$</p>
<p>Inserting in the second row</p>
<p>$\frac{1}{r}\frac{\partial \varrho \Omega^2 \frac{r^2}{2} + C(\varphi,z)}{\partial \varphi} = 0$</p>
<p>$C'(\varphi,z) = 0$</p>
<p>Hence $C(z)$</p>
<p>Inserting in the third row yields</p>
<p>$C'(z) = \frac{\partial C(z)}{\partial z} = - \varrho g$</p>
<p>$C(z) = - \varrho g z + D$</p>
<p>Hence</p>
<p>$$p(r,z)=\varrho \Omega^2 \frac{r^2}{2} - \varrho g z + D$$</p>
<p>With the boundary condition that $p(r=0, z=0)=p_a$ it follows that</p>
<p>$$p(r,z)=p_a - \varrho g z +\varrho \frac{\Omega^2 r^2}{2} \tag{2} $$</p>
<p>This approach is more flexible as you can account for any changes in the problemset without relying on a formula that is just 1 solution for 1 specific problem. </p>
<p><strong>Fixed reference frame</strong></p>
<p>In this case we observe the system from the outside and we can no longer use hydrostatics for our problem. </p>
<p>Here Navier-Stokes comes in handy.</p>
<p>In polar coordinates for an inviscid, stationary flow the equation for the radial direction is</p>
<p>$u\frac{\partial u}{\partial r} + \frac{v}{r}\frac{\partial u}{\partial \varphi} + w\frac{\partial u}{\partial z} - \frac{v^2}{r} = -\frac{1}{\varrho} \frac{\partial p}{\partial r} + f_r$</p>
<p>with $\vec{v} = \begin{pmatrix}
u\\
v\\
w
\end{pmatrix}$</p>
<p>As there are now no external forces acting on the system $f_r=0$ and we omit all zero terms that follow from $u=0$ and $z=0$ as we only have a velocity in $v$ direction, it follows that.</p>
<p>$- \frac{v^2}{r} = -\frac{1}{\varrho} \frac{\partial p}{\partial r}$</p>
<p>with $\Omega = \frac{v}{r}$</p>
<p>$$\Omega^2 r = \frac{1}{\varrho} \frac{\partial p}{\partial r}$$</p>
<p>which equals (1).</p>
<p>Repeat with v and z direction and the same boundary condition you get to (2).</p>
<p><a href="https://i.stack.imgur.com/mOB6H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mOB6Hm.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/uwWU4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uwWU4m.png" alt="enter image description here"></a></p>
<p>I prefer general solutions over specific solutions that why I felt the need to add to this answer.</p>
<p><strong>Calculating force</strong></p>
<p>Now you can integrate over the walls of the cylinder.</p>
<p>$F = \int_A p(r=R,z) R d \varphi dz$</p>
| 5905 | Force exerted on wall of a rotating cylinder filled with fluid |
2015-10-25T09:08:25.237 | <p>Sorry if this is the wrong forum, I'm not an engineer but I'd need a pro to answer my question.</p>
<p><strong>Short version:</strong>
Are there any tests or studies, if a carpet in an apartement can reduce noise from the apartment below. I'm searching for numbers rather than oppinions. The noise comes in all frequencies and shapes, from trampling to yelling, machine noise etc.</p>
<p>If there aren't any tests, please give me your oppinion, if you're an experienced professional in this field.</p>
<p><strong>Long version:</strong>
I need to find a way to reduce noise in my apartment. I'm living in the topmost floor, so only the noise from below is the problem. Apparently our landlord put laminate in all apartements, without any form of insulation. My neighbors aren't the type of people that seem to be (neither could be made) aware, that noise exists.</p>
<p>After browsing the web for months, trying out various earplugs for >300$, hanging thick curtains, rearanging my apartement etc. I'm just searching for something more. If it's 5% less noise - fine. If it doesn't help at all I don't want to spend time and money on it.
I know, that I propably should better find a new place to live or make my neighbors be more silent, but this is not a perfect world, things cannot be done in the best possible way.</p>
<p>Not sure why Mythbusters never tested this stuff.</p>
<p>EDIT:
<a href="https://i.stack.imgur.com/BcsHw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BcsHw.jpg" alt="Soundproof bed"></a>
<strong>RED</strong> = Insulation (Foam, ..., whatever)
<strong>White</strong> = Heavy wood
<strong>Blue</strong> = Inner shell, also heavy wood
What if I layed carpet and built something like this? I set up directional lamps for every wall and the ceiling. Sure, light is not sound, but it bounces too, right? Or will that totaly not work? Maybe I worsen it by that, like I build a trumpet..</p>
| |architecture| | <p>I have a background in acoustics, noise control and vibrations. I haven't done much in the way of residential applications though. </p>
<p>Honestly, your best bet is to lay down heavy carpet with an underlayment designed to maximize sound absorption, like <a href="http://www.soundisolationcompany.com/solutions-products/soundproof-floors/carpet-floors/privacy-premium-carpet-underlay/" rel="nofollow">this one</a>. That doesn't mean that your problem will go away, or even be affected significantly.Since you're hearing sounds of all frequencies, in all likelihood there are "flanking paths" that allow sound from your downstairs neighbors into your apartment. These are typically things like ducts, pipes, walls and other vertical passages inherent in a multistory building. </p>
<p>The reason there aren't a lot of quantifiable measures and material ratings for sound reduction is that the testing required to give good results adds to the product cost. Most people are simply looking for a rug or carpet that will make their room look great. They aren't willing to pay a lot more for a material that's been tested. In any case, too much of a material's effectiveness in a given application depends on other factors that are beyond the control of the manufacturer. Just to give you an idea, consider what happens when sound waves hit a material surface. The energy from the sound can do three things. It can be reflected, it can be transmitted through or it can be absorbed by the material. What happens depends not just on the material, but also a great deal on the frequency of the sound, the angle of incidence and what's behind the material (i.e. the structure of the building). </p>
<p>It's entirely possible to design and construct buildings that minimize noise pollution between rooms/units, but for pre-existing structures, it can be all but impossible</p>
| 5909 | Carpet for noise reduction (tests / studies) |
2015-10-25T22:37:41.507 | <p>In the buckling of columns we know that:</p>
<p>$$P = \dfrac{n^2\pi^2EI}{L^2}$$</p>
<p>The smallest value of P occurs when $n=1$ which gives a simple buckling shape (one wave):</p>
<p>$$P_{cr} = \dfrac{\pi^2EI}{L^2}$$</p>
<p>However for $n > 1$, as shown below the buckling shape is more complex and has many waves:</p>
<p><a href="https://i.stack.imgur.com/x2Mm9.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/x2Mm9.jpg" alt="Buckling shapes"></a></p>
<p>My question is do the buckling mode shapes for $n > 1$ ever occur in reality? If the column begins to buckle as per the shape for $n = 1$ then wouldn't it just continue to buckle like this until failure? How would the other buckling modes ever occur?</p>
| |structural-engineering|beam|columns|buckling| | <p>Whether or not buckling modes with $n>1$ exists depends on how you look at the structure.</p>
<p>As @hazzey notes in his answer, columns with bracing may display buckling modes with $n>1$. These buckling modes, however, are simply equivalent to the $n=1$ modes of the individual segments that compose the column. To be clear, this doesn't mean that the segments behave independently (you'll never have two consecutive unbraced lengths buckling to the same side), only that any $n>1$ mode can be composed by a series of continuous $n=1$ modes for the unbraced lengths.</p>
<p>So, if you have a column with a single bracing which buckles, do you consider that an $n>1$ mode for the entire column or an $n=1$ mode for each of the unbraced lengths? Both? Your call.</p>
<p><a href="https://i.stack.imgur.com/6AJff.png" rel="noreferrer"><img src="https://i.stack.imgur.com/6AJff.png" alt="enter image description here"></a></p>
<p>To paraphrase @starrise's comment on @hazzey's answer, this can be demonstrated by looking at the buckling equation:
\begin{align}
P &= \left(\dfrac{n}{L}\right)^2\pi^2EI \\
P_{column,\,n=2} &= \left(\dfrac{2}{L}\right)^2\pi^2EI \\
P_{segment,\,n=1} &= \left(\dfrac{1}{\frac{L}{2}}\right)^2\pi^2EI = \left(\dfrac{2}{L}\right)^2\pi^2EI \\
\therefore P_{column,\,n=2} &= P_{segment,\,n=1}
\end{align}</p>
| 5910 | Buckling: Do buckling mode shapes of n > 1 occur in reality? |
2015-10-26T15:57:06.453 | <p>I'm trying to determine a level of vibration testing I should perform on my product. I understand electrical stress/life testing and I'm looking for a mechanical equivalent.</p>
<p>I have a DC motor running in my application that generates a certain level of vibration. I'm planning to do some environmental testing to my electronics and one of the tests is vibration, so I plan to subject the electronics to the level of vibration above what the motor produces, my thinking being that a higher level may be equivalent to electrical life testing. My product needs to last ten years so I'd like to know that the vibration I see now won't cause problems in several years time. </p>
<p>For instance, I can put an accelerometer on my electronics and measure ~2 g. To simulate over the product's lifetime should I be applying a higher level of vibration, say 3-5 g?</p>
| |mechanical-engineering|vibration|product-testing| | <p>Q- The aim of this research study is mainly based on the comparison of measured frequency response function results for the threshold with and without damage, and comparing them with the theoretical model to update the modal parameters.
The study will be performed on isotropic beam, to study the vibration characteristics. First Step, the beam will be tested without damage, and the frequency response function will be measured, then the natural frequencies and mode shapes, can be evaluated. To ensure these values, analytical model based on modal analysis approach will be utilized to predict the structural dynamic response of the beam and comparing it with the experimentally FRF. Second Step, cut in the beam, will be induced and getting again the FRF experimentally which will be used as input data to the analytical model to get the modal parameters. Last Step, the results from the previous steps will be treated to update the modal parameters (mass and stiffness matrix).</p>
| 5923 | What level of vibration testing should I perform on my product? |
2015-10-26T16:38:07.030 | <p>I'm not sure how to calculate force $P$ when the moment is a given with the force $P$ at an angle. </p>
<p>I'm also having trouble breaking the moment into the $x$ and $y$ components.</p>
<p><a href="https://i.stack.imgur.com/SXy9z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SXy9z.jpg" alt="force P causes a moment of 500 N-m about point A. Determine force P"></a></p>
| |mechanical-engineering|torque| | <p>$$
\tau = 500 \mbox{Nm} \\
$$</p>
<p>$$
\tau = r \times F \\
\tau = |r| |F| \sin{(\theta)} \\
$$</p>
<p>If the lower left angle is $43^{\circ}$, then the top angle from the bar $r$ to vertical is then $180 - 90 - 43 = \boxed{47^{\circ}}$. This means that $\theta$, the angle between the bar $r$ and the force $P$ is $47^{\circ} + 15^{\circ} = \boxed{\theta = 62^{\circ}}$.</p>
<p>Then,</p>
<p>$$
\tau = |r| |F| \sin{(\theta)} \\
|F| = \frac{\tau}{r \sin{(\theta)}} \\
|F| = \frac{ 500\mbox{Nm} }{(1.8 \mbox{m} ) \sin{(62^{\circ})}} \\
$$</p>
<p>$$
\boxed{F = 314.6 \mbox{N}} \\
$$</p>
| 5927 | Calculate the force when given the moment |
2015-10-28T10:23:35.860 | <p>I want to know if the sensor nodes in any cluster in a wireless sensor network may detect events in a different ratio i.e.some nodes could detect events more others in the same cluster, so is power consumption in these nodes will be higher than other nodes?
i want to know if there is an application in which this assumption or question is possible with take in our concentration that we can applied clustering to increase the lifetime of this network.</p>
| |telecommunication| | <p>The function of a node and how it interacts on the mesh network (assuming a mesh-connected node) are two different things.</p>
<p>Such nodes almost certainly will have a microcontroller in them that performs the node function, like measuring temperature periodically or whatever. This micro then also usually decides when it should attempt to communicate on the mesh network.</p>
<p>Obviously, the micro can take many measurements per mesh connection, or connect many times without taking any measurements. The two are independent.</p>
<p>If the mesh is only powered up at regular intervals, then a node can't report any data no matter what activity it might detect. It has to wait until the next mesh interval. Until then, it has no way to pass on any of its data.</p>
<p>Some nodes might consume more power if whatever they are connected to is more active. Such activity could cause a node to wake up and do something, then go back to low power state until the next event. Other types of nodes just sample something periodically, so wouldn't take more power. It's all up to whatever policies are encoded in the firmware of the micro.</p>
| 5940 | Could a one sensor node in a cluster sense a data more the others? |
2015-10-29T18:18:57.320 | <p>Hooke's Law defines a linear-elastic relationship between stress and strain. </p>
<p>$$ \sigma = E\epsilon $$</p>
<p>Steel behaves very much like a linear-elastic material, following Hooke's Law closely. It does, however, display non-elastic behaviors such as relaxation. Relaxation is the behavior wherein a member under constant strain displays variable (and reducing) stress over time. </p>
<p>My question is: is relaxation plastic? If the relaxed member were released, how would it behave? Would it follow a path defined by its elastic modulus? If this is the case, then it will end with a plastic deformation, no? After all, when stressed, the member will have reached $\left(\sigma_1, \epsilon_1\right)$. After relaxation, it will reach $\left(\sigma_2, \epsilon_1\right)$. Once released, it would have to reach $\sigma =0$, which implies in $\epsilon = \epsilon_1 - \dfrac{\sigma_2}{E}$ and since $\sigma_2 <\sigma_1$, that implies in a nonzero $\epsilon$. </p>
<p>Or is there some other behavior? Does the elastic modulus change to allow for a return without plastic deformations? </p>
| |materials|steel| | <p>In short, yes, relaxation should probably be considered plastic deformation, as plastic strain is defined as <strong>non-recoverable deformation</strong> when applied stresses are removed.</p>
<h1>Definitional Explanation</h1>
<p>Assume you have a sample of some material, in this case steel, and want to apply a load for an extended period of time, sufficiently long for noticeable relaxation to occur. The load is not sufficient to leave the elastic regime. Just after the load is applied but before relaxation begins, the strain in the material due to the load is $\varepsilon_{0}$. If the load is immediately removed, again before relaxation proceeds, all $\varepsilon_{0}$ is recovered and the steel material returns to its original shape.</p>
<p>If instead the material experiences the load sufficiently long so that relaxation proceeds, and the load is removed, then $|\varepsilon_{1}|<|\varepsilon_{0}|$ is recovered. As a result, not all of the strain is recovered. Therefore, there must have been <strong>non-recoverable strain</strong> of $|\varepsilon_{0}-\varepsilon_{1}|>0$ due to relaxation. Therefore, by definition, relaxation is plastic deformation.</p>
<h1>Thermodynamic & Kinetic Explanation</h1>
<p>If the definitional explanation is insufficient, we can also look at this from a thermodynamic and kinetic point of view. Suppose for the moment the steel is instead a single crystal of pure iron. Elastic strain stores energy in the crystal lattice. Because the energy is higher than its rest state, there is free energy available to do work, and thus a driving force for reorganization of atoms in the crystal lattice. There are also point defects in the lattice in the form of vacancies, or missing atoms. Random fluctuations cause neighboring atoms to fill the vacancies, which results in the vacancies moving around the lattice. The vacancies provide a means for reorganization of the atoms.</p>
<p>Note that if the strain is not isotropic (i.e. is not purely hydrostatic), then the lattice strain field makes vacancies slightly larger in tensile-strain directions than in compressive-strain directions. As a result, the energy barrier to moving in the tensile directions will be lower than in the compressive directions. Think of the atoms being squeezed out from between their compressive-direction neighbors along the tensile directions. There will thus be a net flow of atoms in the crystal, with atoms tending to move from directions of high compression to directions of high tension. The overall, long term effect is to extend the crystal in the directions of tension and shorten the crystal in the directions of compression, causing a <strong>non-recoverable deformation</strong>. The same effects occur with multiple grains, except the mechanics are complicated by the presence of grain boundaries and varying crystal orientations. The same effects also occur with the presence of interstitial atoms like carbon, and they probably have negligible effect on vacancy motion as they don't get in the way (though I am not 100% sure of this part, see <strong>note</strong> below).</p>
<p>The above is a most-likely theory based on theories of vacancy flow and grain boundary migration due to thermal stresses (e.g. creep and grain-growth) and from dislocation motion, which have been observed directly. The described behavior for relaxation, however, has not been observed directly to the best of my knowledge (i.e. with a tunneling electron microscope).</p>
<h2>Note</h2>
<p>*Interstitial atoms will have lower energy in interstitial sites aligned with the tensile directions, as those sites are slightly increased in volume. This is related to anelastic strain and martensite formation, but may or may not have an impact on relaxation. However it is worth noting that purely axial strain may induce anisotropic properties in the steel.</p>
| 5962 | Is steel relaxation plastic? |
2015-10-30T01:00:59.637 | <p>By definition, prestressing is a self-balanced load. This is because it is in fact merely an applied state of internal stresses. </p>
<p>This can also be demonstrated mathematically (in this case for a simple parabolic cable) via Lin's load-balancing method. </p>
<p>For a cable which follows the layout given by
$$ y(x) = \dfrac{4e}{L^2}x^2 $$
where $e$ is the vertical distance between the maximum and minimum points of the parabola, $L$ is the span of the cable, and $x=y=0$ is located at the minimum point of the layout, the equivalent load is
$$ q = \dfrac{8Pe}{L^2}$$
where $P$ is the prestress force. </p>
<p>Therefore, the total vertical load given by the distributed load is $\dfrac{8Pe}{L}$. </p>
<p>This force is countered by the vertical forces applied at the anchors. These are equal to $P\sin(\theta) \approx P\theta$ and $y'(x) \approx \theta$, so the forces are each equal to $Py'\left(\frac{L}{2}\right) = \dfrac{4Pe}{L}$ and combined are equal (and opposite) to the total distributed load. So the total global load applied is null. </p>
<p>Now, what if we consider friction losses between the cable and the duct in a post-tensioned beam? Then the combined forces at the extremities remain equal to $\dfrac{8Pe}{L}$, but the total distributed load will be reduced by the losses, implying in a non-balanced load, which is impossible. </p>
<p>This may be adjusted by the anchorage slip losses, but what if we assume a hypothetical anchor which doesn't slip? Is this impossible and slippage is necessary to make the load balanced? </p>
<p>And how are the remaining losses (elastic deformation, creep, shrinkage and relaxation) affected? </p>
| |civil-engineering|structural-engineering|concrete|prestressed-concrete| | <p>Comments by @Mr.P nudged me to realize that there is a redistribution of equivalent loads due to losses which cannot be trivially encompassed by Lin's method.</p>
<p>To demonstrate this, take the following simply-supported beam (ignore all concepts of units or scale here, this is a thought exercise).</p>
<p><a href="https://i.stack.imgur.com/tE6mi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tE6mi.png" alt="enter image description here"></a></p>
<p>The bending moment diagram solely due to prestress for this beam will be the following polygonal diagram (before any losses, assuming <span class="math-container">$P\cos\theta \approx P$</span>):</p>
<p><a href="https://i.stack.imgur.com/0E1RB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0E1RB.png" alt="enter image description here"></a></p>
<p>This is equivalent to the diagram obtained by a concentrated vertical load at midspan equal to <span class="math-container">$F = 2P\sin\theta \approx 40$</span>, which is balanced out by the two vertical loads at the supports, each equal to <span class="math-container">$P\sin\theta \approx 20$</span> (in the opposite direction).</p>
<p>However, let us now consider friction losses. Let's assume they cause a 10% reduction to the stress in the tendon at midspan. This implies that the bending moment at midspan will also be reduced by 10%, and will therefore equal 90. However, the diagram's profile is no longer polygonal.</p>
<p>That can be easily observed by looking at the bending moment at any other point. In an isostatic structure, the bending moment is simply equal to <span class="math-container">$P \cdot e$</span>, where <span class="math-container">$e$</span> is the distance between the cable and the centroid. Looking at quarter-span, the bending moment before losses was equal to <span class="math-container">$100\cdot0.5=50$</span>, or exactly half of the mid-span moment. To calculate after losses, however, we need to calculate the force at this point. Simplifying things considerably, let's assume the loss here is half of that at midspan, so only 5%.<sup>1</sup> In that case, the bending moment at quarter-span will be equal to <span class="math-container">$95\cdot0.5=47.5$</span>, which is <strong><em>not</em></strong> equal to half of the mid-span moment of 90.</p>
<p>Indeed, the bending moment diagram from support to midspan becomes of the form</p>
<p><span class="math-container">$$M = Pe\left(\dfrac{2x}{L}-0.1\left(\dfrac{2x}{L}\right)^2\right)$$</span></p>
<p>where <span class="math-container">$e$</span> is the distance of the cable to the centroid at midspan, <span class="math-container">$L$</span> is the span, and <span class="math-container">$0.1$</span> represents the 10% loss of prestress at midspan.</p>
<p><a href="https://i.stack.imgur.com/yZgr4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yZgr4.png" alt="enter image description here"></a></p>
<p>Getting the first and second derivatives of this equation gives us the equivalent concentrated load at midspan and the uniform load distributed along the entire span, respectively:</p>
<p><span class="math-container">$$\begin{align}
M' = Q &= Pe\left(\dfrac{2}{L} - 0.2\dfrac{4x}{L^2}\right) \\
Q\left(\dfrac{L}{2}\right) &= \dfrac{1.6Pe}{L} \therefore F = 2Q\left(\dfrac{L}{2}\right) = 32 \\
M'' = q &= \dfrac{0.8Pe}{L^2} = 0.8
\end{align}$$</span></p>
<p>Therefore, the equivalent loading which generates the correct (simplified) bending moment diagram after friction losses is the following:</p>
<p><a href="https://i.stack.imgur.com/fhpli.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fhpli.png" alt="enter image description here"></a></p>
<p>This implies in a new uniform load which did not appear before losses, a representation of the load redistribution described by @Mr.P. The equivalent concentrated load at midspan is also no longer equal to <span class="math-container">$F = 2P\sin\theta$</span>, which would have resulted in <span class="math-container">$F \approx 36$</span>. The prestress, however, is balanced, since <span class="math-container">$32 + 0.8\cdot10 = 40$</span>, as expected.</p>
<hr>
<p>The same reasoning can be applied to parabolic cables. For the following beam:</p>
<p><a href="https://i.stack.imgur.com/cwQKa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cwQKa.png" alt="enter image description here"></a></p>
<p>the uniform equivalent load is <span class="math-container">$q = \dfrac{8Pe}{L^2} = 2$</span>, which generates a total upwards force of <span class="math-container">$2\cdot32=64$</span>, to be cancelled out by the concentrated downwards forces of 32 at each support. The bending moment diagram before any losses is:</p>
<p><a href="https://i.stack.imgur.com/Ql0sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ql0sn.png" alt="enter image description here"></a></p>
<p>However, assuming once again a 10% loss due to friction at midspan, and that the friction-loss profile is linear, the bending moment diagram becomes:</p>
<p><a href="https://i.stack.imgur.com/RMmyU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RMmyU.png" alt="enter image description here"></a></p>
<p>which has the following cubic equation (with <span class="math-container">$x=0$</span> at the midspan):</p>
<p><span class="math-container">$$M = 2.00\cdot128\left(1-\left(\dfrac{2x}{L}\right)^2\right)\left(0.9+0.1\dfrac{2x}{L}\right)$$</span></p>
<p>getting the first and second derivatives, I can find the equivalent loading, which in this case is equal to:</p>
<p><a href="https://i.stack.imgur.com/Vx4em.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vx4em.png" alt="enter image description here"></a></p>
<p>Given what we saw with the first example, it comes as no surprise that there's an increase in the distributed load near the supports. It also makes intuitive sense that the distributed load is reduced near the midspan. How that concentrated load at midspan comes to be, however, I have no idea. However, once again the prestress is self-balanced: <span class="math-container">$\dfrac{1.8+2.4}{2}\cdot32-3.2=64$</span>.</p>
<p>This answer therefore answers the question posed of how prestressing with losses is self-balanced: the equivalent load is redistributed, but the total value is not modified. That being said, I cannot explain how to calculate this redistribution in a general case because, well, I don't know how that's done.</p>
<hr>
<p><sup>1</sup> <sub>Though friction losses are usually quite linear (or polygonal), this is a poor assumption. After all, the 10% loss at midspan must include the effect of the concentrated angle change at that point. At quarter-span, the losses are only due to linear friction loss, and will therefore probably be substantially lower than 5%. That being said, we can just state that 10% is at the point immediately before the angle change, where the bending moment approaches 100 but where only linear friction losses have occurred.</sub></p>
<p><sub>All diagrams obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis tool.</sub></p>
| 5967 | How is prestressing with losses self-balanced? |
2015-10-30T07:16:32.320 | <p>for example :</p>
<p><strong>Sender:</strong> A cellular signal tower(Base Station(BS)) sends data(voice, GPRS, VOIP) to the cellphone. The sender will obviously have the carrier frequency and more power(voltage+current) to generate the carrier signal.</p>
<p><strong>Receiver:</strong> A cell phone which receives the data signal (voice, GPRS, VOIP) from the BS. This requires less energy because the cell phone is receiving data.</p>
<p>I want to know, when the receiver sends data(voice,GPRS,VOIP) to the BS, how does it do this?</p>
<p>Does the cell phone generate it's own carrier signal?</p>
<p>If yes, then how can a 3.7 volt battery be available for 8-12 hours?</p>
<p>OR</p>
<p>The receiver's signal goes on to same carrier signal which is generated by the BS.</p>
<p>If yes, How? Does the carrier signal use refraction/reflation properties to send data back to BS?</p>
| |electrical-engineering|telecommunication| | <p>When a cell phone is sending, it generates its own carrier. The power for doing this comes from the cell phone's battery.</p>
<p>Batteries in a cell phone may last 12 hours, but that is not while transmitting. Depending on how "off" the cell phone is, it is only occasionally listening to the cell tower, and even less often sending anything to it. When not moving, it may send nothing at all for long periods of time.</p>
<p>The battery wouldn't last 12 hours with you continuously talking on the cell phone. That requires frequent transmitting, which uses a lot more power than just sitting there, even in standby mode listening for others to call it.</p>
<p>Transmitting power also depends on distance to the cell tower. Actually the phone doesn't measure distance, but the cell tower tells it how well the message was received. The phone cranks up the power as needed achieve whatever it considers a acceptable error rate.</p>
<p>Despite all that, the predominant energy drain on the battery can be illuminating the screen. If you want long life per charge, don't make the phone power up the screen.</p>
| 5973 | How does a carrier signal work from a receiver's perspective? |
2015-10-30T22:36:52.547 | <p>I'm in school for electronics engineering. Would someone lay out the high level steps in becoming a professional engineer (PE) in the state of California?</p>
| |electrical-engineering|licensure| | <p>California has a few notable exceptions - they try to favor graduates from the in-state schools, but are very picky on structural engineers. Here they are, in no particular order:</p>
<ul>
<li>You do not need to have 4 years of experience if you are a resident and graduated from a California School. 2 years is acceptable.</li>
<li>For all licenses except Civil, working in industry (v. studying under a guiding engineer) is considered acceptable experience, so long as the test score is high enough</li>
<li>The ethics examination needs to be completed by hand and submitted. You do not know if you passed until you get your information back.</li>
<li>Feedback isn't done by e-mail, it's done by postcards. Always include 2 self addressed post cards in your packet when you mail it in.</li>
<li>Reviewers are assigned based upon last name. Redoing the packet will mean that it will get reviewed again by the same reviewer.</li>
<li>Electrical, Mechanical, and Civil have some interesting rules on personal recommendations. Civil can only have other Civil. Mechanical can have Civil or Mechanical. Electrical can only have Civil or Electrical. Anything else (besides structural!) can have any recommendation desired.</li>
<li>To gain a structural license, you have to have a Civil license. Then another 4 years of additional experience studying as a licensed Civil engineer under a structural engineer. Then you have to get Structural Engineer recommendations, then take the 2 part structural examination. It's very complicated for structural engineers.</li>
</ul>
| 5982 | What is the sequence of exams to become a Professional Engineer in California? |
2015-10-31T02:53:00.950 | <p>How do I know when to use an upward or a downward curve when I'm drawing a bending moment diagram for a uniformly-distributed load (UDL) and what's the most reliable way of identifying the maximum bending moment?</p>
| |statics| | <p>The direction a bending moment diagram is drawn depends on the sign convention being used. The most common sign convention is:</p>
<p><a href="https://i.stack.imgur.com/ag9He.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ag9He.gif" alt="PositiveSignConvention"></a></p>
<p>It is good practice to indicate the sign convention used whenever you draw a BMD.</p>
<p>So if we are interested in the sign of the bending moment for a UDL we can take a cut such as:
<a href="https://i.stack.imgur.com/hWNCF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hWNCF.jpg" alt="enter image description here"></a></p>
<p>In which case we can see from the sum of moments at the cut:</p>
<p>$$ M = Rx - w\frac{x^2}{2} $$</p>
<p>And whatever the sign found from this is the sign for the BMD. </p>
<p>For simple structures you can obtain the sign of the bending moment by thinking of the deformed shape. If the deformed shape resembles that shown in the sign convention above then the sign of the BMD will be positive (tension on bottom). So for a simply supported beam with a UDL acting downwards the bending moment would be positive. </p>
<p>Be aware that sometimes a BMD is drawn with the positive moment axis pointing downwards, this is called drawing 'on the tension side' and is common in concrete design as it indicates where reinforcement should be placed.</p>
<p>The fastest, most reliable way of finding the maximum bending moment, especially for simple cases, would be to use established beam design formulas such as: <a href="http://www.awc.org/codes-standards/publications/da6" rel="nofollow noreferrer">DA 6 - Beam Design Formulas with Shear and Moment Diagrams</a>. If you knew where the location of maximum bending moment was, you could take a cut at that location and calculate it directly as shown above. Otherwise, you will need to resort to integrating the shear force (either directly, or graphically) to obtain the BMD and then locating the maximum that way.</p>
| 5983 | Drawing Bending Moment Diagrams |
2015-10-31T06:35:28.450 | <p>In addition to power and performance, one of the reasons VW cheated their emissions results was to improve the fuel economy of their cars. Why exactly does this happen?</p>
<p>I'm familiar with cars and how cars work so feel free to get technical.</p>
<p>And a follow-up question: Where is the "sweet spot" between burning more gas per mile but having that burned gas release fewer emissions, and burning less gas per mile but releasing more emissions per gallon burned?</p>
| |mechanical-engineering|automotive-engineering|environmental-engineering| | <p>Because most controls reduce efficiency, since they are aimed at protecting the catalyst - not reducing pollution. An air/fuel ratio of 14.7/1 keeps the catalyst happy. Engines can and do run cleaner and better without the catalyst, but the EPA is not only decades behind chemistry - it is not concerned with the environment.</p>
| 5984 | Why does reducing the amount of emissions control improve the fuel economy of a car? |
2015-10-31T15:49:06.700 | <p><strong>Regarding micro inverters for solar power systems:</strong></p>
<p>Is there technically a difference between an inverter classified as a "micro inverter" and and inverter classified as a "pure sine wave" inverter? Or is it simply a matter of using extra jargon to describe the size of the inverter?</p>
<p><strong>Problem I'm trying to solve:</strong></p>
<p>I'd like to purchase a micro inverter based solar system, where each panel has it's own dedicated inverter. Enphase seems to have coined the term "micro inverter" for this purpose, but the reseller doesn't offer a micro inverter that is big enough for a 315 watt panel.</p>
<p>So I'm wondering if I can purchase a pure sine wave inverter rated at 400 watts, for each 315 watt panel. Would those wire up the same way as the "micro inverter" solution?</p>
| |electrical-engineering| | <p>Stop.</p>
<p>You can't just attach <em>any</em> inverter on a PV panel or array, even if it does have the right power rating.</p>
<p>Your inverter needs to be a PV inverter specificaly. It has to be designed to do max-power-point tracking. That's because the panel doesn't have any specific potential difference: that has to be imposed on it by the rest of the system.</p>
<p>Also, if you're connecting to the grid, the inverter may have to meet specific local regulations (e.g. G83 in Great Britain).</p>
<p>The whole idea of micro-inverters is that you have one per panel. And any micro inverter will be a sine-wave generator. The need for "purity" of that sine wave, and the ability to synchronise phases with the grid, as well as behaving properly upon grid disconnection, is what goes into things like G83 certification.</p>
| 5988 | "Micro Inverters" vs "Pure Sine Wave" for Solar Systems |
2015-11-01T11:30:45.920 | <p>I have the same problem as described in this question:<br>
<a href="https://engineering.stackexchange.com/q/5535/2844">How to calculate the resulting temperature for a stream of air through a hot tube?</a>, but I want to increase the transferred heat by adding fins onto the inner side of the tube, parallel to the flowing air. Is that useful? If yes, how does the formula mentioned in the answer (<a href="https://engineering.stackexchange.com/a/5536/2844">https://engineering.stackexchange.com/a/5536/2844</a>) change?<br>
Furthermore, how do I get from the temperature difference to the transferred heat power?</p>
| |heat-transfer| | <p>Internally finned tubes are not that common because of the high cost of manufacture. There aren't many sources for this kind of heat transfer. </p>
<p><a href="https://www.ruor.uottawa.ca/bitstream/10393/10870/1/EC52380.PDF" rel="nofollow noreferrer">This</a> paper provides a detailed description of numerical methods used to calculate heat transfer in a finned tube with fully developed laminar flow. </p>
<p>It evaluates fins with the following measurement limits: </p>
<p>Number of Fins:$$8\le M\le32$$
Fin Length (ft):$$0.2\le\ell\le0.9$$
Fin angle:$$1.5°\le\beta\le3°$$</p>
<p><a href="https://i.stack.imgur.com/fTZRk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fTZRk.png" alt="enter image description here"></a></p>
<p>If you do have laminar flow, you can use the following chart to estimate the Nusselt number for a certain fin design:</p>
<p><a href="https://i.stack.imgur.com/F7UTE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F7UTE.png" alt="enter image description here"></a></p>
<p>Use the Nusselt number to calculate the heat transfer coefficient, h:</p>
<p>$$Nu= \frac {hd}{k}$$</p>
<p>Optimally, you would use the hydraulic diameter rather than the nominal tube diameter. <a href="http://buet.ac.bd/me/icme/icme2001/cdfiles/Papers/Thermal/29_FINAL_TE-71(159-163).pdf" rel="nofollow noreferrer">This</a> uses a method of determining hydraulic diameter of an internally finned tube using the ratio of cross sectional area of flow to the wetted perimeter. This area would be irritating to calculate for fins and I don't think it would affect the result any more than the other assumptions we have made. </p>
<p>Also of course, use the thermal conductivity, k, value for your fluid. </p>
<p>Lastly use this equation to get your heat transfer:</p>
<p>$$Q=hA_s\theta_{lmtd}=mC_p\Delta{T}$$</p>
<p><a href="http://www.banglajol.info/index.php/JME/article/download/3473/2912" rel="nofollow noreferrer">This</a> paper analyzes T shaped internal fins and gives a good analysis on how fins affect friction factor and a comparison of smooth tube vs finned. </p>
| 5995 | Calculate thermal flow for air flowing through a tube, with fins |
2015-10-31T22:20:01.323 | <p><strong>TL;DR: how can I calculate the disturbance at each joint due to coupling forces in a two-link planar robot manipulator actuated by two independent DC motors?</strong></p>
<hr>
<p>I'm studying control theory and trying to work through a simple example using a two-link planar robot manipulator:</p>
<p><a href="https://i.stack.imgur.com/hfUPe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hfUPe.png" alt="enter image description here"></a></p>
<p>My goal is to simulate PID control of the planar two-link manipulator, where each joint is actuated by an independent DC motor. The input is two continuous signals $$\theta_1(t), \theta_2(t)$$</p>
<p>which represent the desired angle for each joint at time <em>t</em>.</p>
<p>Following this paper: <a href="http://www.scirp.org/journal/PaperInformation.aspx?PaperID=22044" rel="nofollow noreferrer">Modeling a Controller for an Articulated Robotic Arm</a>, I can obtain an expression for the voltage I need to apply to each motor to achieve a desired angle. The paper also describes the PID controller necessary to maintain the desired angle given an error signal $$e(t) = \theta_{desired}(t)-\theta_{actual}(t)$$</p>
<p>Since each motor is controlled independently, coupling effects among joints due to varying configurations during motion are treated as disturbance inputs. <strong>My question is: how can I model this coupling effect in order to "simulate" the error signal <em>e(t)</em> for system under ideal conditions?</strong> By ideal conditions I mean that the disturbance due to coupling among joints accounts for 100% of the error signal.</p>
<p>My current thought is to use an expression for the dynamics of the two-link planar manipulator, as per the following book: <a href="http://www.cds.caltech.edu/~murray/books/MLS/pdf/mls94-complete.pdf" rel="nofollow noreferrer">A Mathematical Introduction to Robotic Manipulation</a></p>
<p><a href="https://i.stack.imgur.com/FPqIJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FPqIJ.png" alt="enter image description here"></a></p>
<p>This way, at time <em>t</em>, we determine the voltage to achieve the desired angles for each independent motor, then plug that into our DC motor model to obtain the generated torques: $$\tau_1, \tau_2$$</p>
<p>We then plug these torques into the dynamics to get the actual angles, taking into account the coupling forces, and then use the actual angles compared to the desired angles in order to generate the error signal that feeds into the PID control loop.</p>
<p>Does this approach make sense? If not, where have I gone wrong and how can I simulate the error signal due to coupling forces?</p>
<p><strong>Edit:</strong> one commenter points out PID control may not be optimal for this problem. If this is the case, what alternative control strategies should I use?</p>
| |control-engineering|control-theory|dynamics|kinematics| | <p><strong>Non-Linear Control Approach</strong></p>
<p>Your approach is correct! I just want to point it out some tips.</p>
<p>$1^{st}$: Try to get the canonical form for the generalized coordinates:</p>
<p>$M(q) \ddot{q}$ + $C(q,\dot{q}) \dot{q}$ + $g(q)$ = $u$ </p>
<p>Were $M(q)$ is the inertia matrix, $C(q,\dot{q})$ is the Coriolis matrix, $g(q)$ is the gravitational vector and, $u$, is your input for the controller.</p>
<p>$2^{nd}$: The goal is to make the resulting dynamics of the system an double integrator:</p>
<p>$a_{q} = \ddot{q}$</p>
<p>Note that the resultant dynamic is decoupled!</p>
<p>$3^{rd}$: Using the set of equations: </p>
<p>$M(q) \ddot{q}$ + $C(q,\dot{q}) \dot{q}$ + $g(q)$ = $u$</p>
<p>$u$ = $M(q) a_{q}$ + $C(q,\dot{q}) \dot{q}$ + $g(q)$</p>
<p>$a_{q} = \ddot{q}$</p>
<p>You just gonna need to choose the variable $a_{q}$ for the controller. (Substitute those $u$'s to verify the double integrator)</p>
<p>$4^{rd}$: Defining the error as $e = q - q_{d}$, were $q_{d}$ is your desired joint position.</p>
<p>$\ddot{e} + K_{1}\dot{e} + K_{0}e$ = 0</p>
<p>The explicit form:</p>
<p>$a_{q} = \ddot{q}_{d} - K_{1} \big( \dot{q}- \dot{q}_{d} \big) + K_{0} \big( {q}-{q}_{d} \big) $</p>
<p>If your matrices $K_{0}$ and $K_{1}$ are symmetric and positive-defined, the error dynamic is asymptotically stable.</p>
<p><a href="https://i.stack.imgur.com/onEz1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/onEz1.jpg" alt="System + COntroller"></a></p>
<p>To see more, I suggest the reading of this <a href="https://books.google.com.br/books/about/Robot_Modeling_and_Control.html?id=wGapQAAACAAJ&source=kp_cover&redir_esc=y" rel="nofollow noreferrer">book</a>. ( The figure is from there ).</p>
| 6006 | Feedback control of two-link planar manipulator |
2015-11-02T09:45:26.133 | <p>For a project using rotating voice coils motor in order to balance a system I need to determine the damping coefficient of these voice coils in order to insert it in a ADAMS model and analyse its performances.</p>
<p>Starting with the oscillation data obtained during a physical test on a mock-up, I used the Matlab tool <a href="http://www.vibrationdata.com/" rel="nofollow noreferrer">vibration data</a> in order to find the damping ratio of these voices coils. To find this value the vibrationdata tool makes a FFT and then a "Half-Power Bandwidth Curve fit" method...
At the end I got : damping ratio = 0.061, as well as the natural frequency $(f_n)$ 3.1Hz and the Q factor 5.8</p>
<p>As I was not sure about either to insert a damping coefficient or damping ratio value in my ADAMS model, I sent an email to MSC and they answered me "In Adams, bushing required the damping coefficient values, in the units as:
- newton-sec/meter
- newton-meter-sec/deg" </p>
<p>So, I have to calculate the damping_coefficient using the damping_ratio, natural frequency...</p>
<p>This is obtained using the expression : $$c = 2\,m\omega_n \times \text{(damping ratio)}$$
with $\omega_n = 2\pi f_n$</p>
<p>I get a certain value but I am totally not sure about the use of the mass $m$ in this expression as I want to get the damping coefficient of a rotating system...
Can I directly replace this mass by the mass inertia coefficient around which the system rotate ($I_{yy}$ for example from the Matrix) ?
Or what else should I do ?
I would like to specify a damping coefficient in Translation in my ADAMS model...</p>
<p><a href="https://i.stack.imgur.com/KLflm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KLflm.png" alt="enter image description here"></a></p>
<p>On this awful drawing I just tried to sketch the two rotating voice coils in blacks with the magnets (right and left) rotating around the axis y... I would like to know what should I consider in term of mass or inertia to find the damping coefficient in the expression above.</p>
| |mechanical-engineering|vibration| | <p>You need to find the effective mass of a rotating rod at the point where the damping force is applied. This equals $$m_{eff} = \frac{I_{pivot}}{\ell^2} $$ where $I_{pivot}$ is the MMOI about the pivot, and $\ell$ is the distance where the force is applied to the pivot.</p>
| 6012 | How to determine the damping coefficient? |
2015-11-02T11:46:32.560 | <p>Most/all nuclear fission reactors require cooling. Due to radioactive decay heat, this is true even when the reactor is switched off. Hence, there are requirements for nuclear plants to have power at all times: for example, in the United States, they must have backup battery power for 4–8 hours (<a href="http://www.pennlive.com/midstate/index.ssf/2011/03/nuclear_power_plants_in_us_vul.html" rel="nofollow">example source</a>), with the assumption that power is subsequently restored.</p>
<p>If, power is not quickly restored to a nuclear reactor and there is a prolonged <a href="https://en.wikipedia.org/wiki/Loss-of-coolant_accident" rel="nofollow">loss of cooling</a> event, what is the risk of large-scale radioactive contamination of the environment?</p>
<p>For the purpose of this question, assume nuclear reactor models that are currently common in the United States or western Europe.</p>
| |safety|nuclear-technology|nuclear-engineering| | <p>I just thought that I would clarify something that Fred said. In an emergency shutting down the nuclear chain reaction is the easy part. The main issue is the decay heat. The decay heat comes from the residual radioactive components in the core.</p>
<p>This decay heat starts off at approximately 7% of the total reactor power. However, it drops off exponentially with time. In about an hour it is down to about 1%. This may not seem like a lot of heat to remove however when considering a 3GWth (~1GWe) that is 30MW that has to be removed for an extended period. Without electricity providing circulation of cooling fluid etc. This is very hard!! </p>
| 6013 | When backup cooling fails in reactors without passive safety, what is the risk of radioactive contamination? |
2015-11-02T20:45:50.640 | <p>I know that the neutral axis occurs where the sum of stresses (bending + axial) is equal to 0. I have found the stresses and drawn a diagram to represent forces. How do I determine the neutral axis?</p>
<p><a href="https://i.stack.imgur.com/wwGxq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wwGxq.png" alt="Question"></a></p>
<p><a href="https://i.stack.imgur.com/bm4Dj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bm4Dj.jpg" alt="What ive done so far"></a></p>
| |mechanical-engineering|structural-engineering|structures|applied-mechanics| | <p>In order to calculate the neutral axis under this definition (see my comment under the OP), you need to figure out the stress distribution for each of the applied loads and then superpose them.</p>
<p>So, for this we need to calculate the properties:
$$\begin{align}
A &= 80\cdot80 - 60\cdot60 = 2800 \\
I &= \dfrac{80\cdot80^3 - 60\cdot60^3}{12} = 2333333
\end{align}$$</p>
<p>So, under the axial force, the stress distribution is uniform and equal to (positive for tension)
$$\sigma = \dfrac{N}{A} = \dfrac{72000}{2800} = +25.71$$</p>
<p>Now, under the moment, the stress distribution is linear, and the maximum stresses are equal to
$$\sigma = \dfrac{My}{I} = \dfrac{(72000\cdot(60+40))\cdot(\pm40)}{2333333} = \pm123.43$$
where the stresses on the left side are negative (compression) and on the right side are positive (tension). The stress profile due to bending can therefore be described by the equation
$$\sigma = -123.43 + \dfrac{2\cdot123.43}{80}y$$
where $y=0$ is on the left side.</p>
<p>Now, adding the effect of the axial load's stress, this equation becomes
$$\sigma = -123.43 +25.71 + \dfrac{2\cdot123.43}{80}y$$</p>
<p>So, to find the neutral axis, you just need to find the zero of this equation:
$$\begin{gather}
\sigma = 0 = -97.72 + \dfrac{2\cdot123.43}{80}y \\
y = \dfrac{97.72}{\left(\dfrac{2\cdot123.43}{80}\right)} = 31.7
\end{gather}$$</p>
<p>So there you have it, the neutral line is 31.7 mm from the left side.</p>
| 6016 | Finding Neutral Axis of a simple shape undergoing axial load |
2015-11-02T23:07:32.820 | <p>I can't find anything on this. Every source I can find on UV cured polymers describe the process and how they're cured using UV radiation but nothing on what the resistance to UV is once it has cured.</p>
<p>Specifically I'm looking at the Stratasys PolyJet (<a href="http://www.stratasys.com/materials/material-safety-data-sheets/polyjet" rel="nofollow">http://www.stratasys.com/materials/material-safety-data-sheets/polyjet</a>) line of resins which are supposed to be very similar to PMMA in mechanical properties but I can't find anything on whether this extends to UV.</p>
<p>If anyone with more knowledge on polymers can help me out that would be greatly appreciated.</p>
| |materials| | <p>I can provide an anecdotal answer. We use UV cured adhesives to bond pieces inside of our UV lasers. These lasers produce intense UV light, some of which is scattered around inside of the laser head. The exposure to those specific UV wavelengths is much more intense than the exposure of a piece of material laying out in the sun. Bond failure, however, is not an issue for us. </p>
| 6017 | UV resistance of UV cured photopolymers |
2015-11-03T02:57:00.033 | <p>Surprisingly this hasn't been asked before, so I must be missing something simple. </p>
<p>We use engineering stress and engineering strain in this eq. Stress = (Young's modulus) × (strain). This eq. is used in analysis of bending beams, twisting shafts and in buckling. So the final equation of bending $(\frac{M}{I} = \frac{\sigma}{y})$ and torsion $(\frac{T}{I} = \frac{\tau}{r})$ will give us value of engineering stress but not the value of stress.</p>
<p><strong>Why are we considering engineering stress instead of true stress while we know it will not give correct value of stress?</strong></p>
<p>Some things I read are:</p>
<ol>
<li>Difficult to measure.</li>
<li>Not that much of a difference and we can just apply a Factor of Safety.</li>
<li>"We don't consider materials to change their cross-sectional area after loading, since we design to have no plastic deformation the elastic region is most important, therefore what happens after the proportional limit is not important"</li>
</ol>
<p>Firstly, 1 and 2 are not real reasons for me. Number 3 seems plausible since we always design in the elastic region, but is this it? <strong>Does engineering strain even give valid information after the proportional limit?</strong></p>
| |materials|structural-engineering|stresses| | <p>Adding to @starrise's answer:</p>
<p>With regards to your dismissal of reasons 1 and 2, you are forgetting to consider the cost-benefit analysis regarding them. As @starrise showed in their answer, the difference is usually not substantial (though other materials will usually have larger differences).</p>
<p>On the other hand, the materials always show intrinsic variation of their values. Steel's modulus of elasticity has a range defined in different articles from $\pm6\%$ <a href="http://www.jcss.byg.dtu.dk/~/media/Subsites/jcss/english/publications/probabilistic_model_code/steelpr.ashx?la=da">[a]</a> to $\pm15\%$ <a href="http://www.assakkaf.com/Papers/Journals/Uncertainties_in_Material_Strength_Geometric_and_Load_Variables.pdf">[b]</a> (for the 95% confidence interval).</p>
<p>So, what's the point of considering the true strain in everyday engineering practice if all the other properties (including yield strength and cross-section dimensions) will have random fluctuations that are all but certain to drown out the "error" due to the use of engineering strain instead of true strain?</p>
| 6020 | Why do we even use engineering stress? |
2015-11-03T23:03:22.640 | <p>Magnetic locks seem to be fairly common on the doors of modern buildings, but I question their reliability when compared to a traditional, mechanical lock. I realize that the strength that it would take to actually pry a door away from a good electromagnetic lock would certainly be enough to damage the door or frame, just as it would be with a mechanical lock.</p>
<p>If you had an electromagnet strong enough, however, wouldn't it be possible to negate a magnetic lock by fastening your oppositely-polarized electromagnet on the outside of the door opposite the magnet on the inside so that, when activated, the outer magnet would overpower the attraction between the inner magnet and the magnetic material on the door, effectively opening the locked door?</p>
<p>If so, I feel this would be a pretty big disadvantage when compared to a traditional mechanical lock. Not that a powerful and portable electromagnet is necessarily a readily available and affordable thing, but...</p>
<p>Ignoring the practicality of actually owning and operating such a powerful electromagnet, is this even a plausible thing to do?</p>
| |mechanical-engineering|electrical-engineering|magnets| | <p>A lot of the responses here are very interesting but the inverse relation of magnetic fields and magnetic strength must be considered. if two attracting magnets are a certain distance apart (say 1 inch) their attraction is for example 2lb. if the two magnets are 1/2 inch apart their attractive force is 4lbs and a quarter inch 16 pounds. all simplified numbers. In a magnetic lock, the two units are touching and therefore very strong. any external magnetic force applied to the outside of a door would be the thickness of the door (plus a little) away from the electromagnet. to overcome a mid-range magnet of 800lbf you would need a much much larger electromagnet (exponential). this seems to be as much of a physics question as an engineering question. If you have an hour, this guy is very engaging and demystifies magnets and electromagnets in one shot. it is filmed 10 years ago at UC Berkley but is meant to be understood by the likes of me so that is saying a lot...</p>
<p><a href="https://www.youtube.com/watch?v=WxitGR-9qGA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=WxitGR-9qGA</a></p>
| 6031 | Can a magnetic lock be negated or overcome by a stronger electromagnet? |
2015-11-04T14:16:09.203 | <p>I look at PID controler and do not understand derivatve term. I know it's used in order to not let the signal to overshoot the set value. BUT when the sigal is arising, its derivative is positive. So when we add the positive derivative term, the signal gets even higher, so its derivative becomes even higher!
I think that if the signal is rising too fast it should be slowed down by derivative term, which means that derivative term should produce NEGATIVE value to DECREASE the signal that controlls the object. How does it work in reality? Is there any good simulation online of how PID works? I've seen some simulations, but they weren't very informative.</p>
<p>One more question: I want to check if I understand the integral term properly. Let's assume our PV is arising and getting closer and closer to the SP. During that time integral part of the PID output is GROWING all the time, until the PV reaches SP, than the integral part has already grown so much that the PV overshoots the SP. So PV rises above SP and the integral part becomes smaller and smaller (but still bigger than zero), and finally it becomes so small, that it gets negative smaller than zero). That casues the PV to start decreasing. It gets smaller and smaller and again, integral part is so big (on the minus part this time) that the PV goes down too much and gets smaller than the SP.
And that's why integral part causes oscillations. Was my example right? Is that how it really works?</p>
| |control-engineering|control-theory| | <p><span class="math-container">$$ e = r - y$$</span> </p>
<ul>
<li><span class="math-container">$e \rightarrow$</span> error</li>
<li><span class="math-container">$r \rightarrow$</span> reference input</li>
<li><span class="math-container">$y \rightarrow$</span> actual output of the plant</li>
</ul>
<p>If <span class="math-container">$e$</span> is increasing (<span class="math-container">$\left[\frac{d e}{d t}\right]>0$</span>) this means that <span class="math-container">$y$</span> is becoming smaller and smaller compared to <span class="math-container">$r$</span>. So the input needs to be increased to increase <span class="math-container">$y$</span>. </p>
<p>I think your analysis of the integral term is right.</p>
| 6035 | How does the DERIVATIVE term slow down the signal? |
2015-11-04T22:44:48.653 | <p>I'm trying to solve the question attached below (please note the handwritten values are ones that I calculated/not part of question) I am assuming that to find the load $q$ I need to get the bending stress (which from my calculation was $45\text{ N/mm}^2$) and substitute into equation of Bending stress $= \dfrac{My}{I}$ to solve for M which should be equated to $\dfrac{ql^2}{4}$ then find $l$, however I think I messed up in the first part of the question because I'm not taking the area of part A into consideration, can you please advise me on how to approach depth-varying beams?</p>
<p><a href="https://i.stack.imgur.com/FrDpv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FrDpv.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|structures|stresses| | <p>As always, the first thing you need to do is calculate your properties.
\begin{align}
A_A &= 150\cdot10 + 6\cdot(200-10) = 2640\text{ mm}^2 \\
\overline{y}_A &= \dfrac{150\cdot10\cdot195 + 6\cdot190\cdot95}{2640} = 151.8\text{ mm} \\
I_A &= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(151.8-195)^2+\dfrac{6\cdot190^3}{12}+6\cdot190\cdot(151.8-95)^2 \\
I_A &=9919274\text{ mm}^4 \\
A_B &= 150\cdot10 + 6\cdot(100-10) = 2040\text{ mm}^2 \\
\overline{y}_B &= \dfrac{150\cdot10\cdot95 + 6\cdot90\cdot45}{2040} = 81.76\text{ mm} \\
I_B &= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(81.76-95)^2+\dfrac{6\cdot90^3}{12}+6\cdot90\cdot(81.76-45)^2 \\
I_B &=1369647\text{ mm}^4
\end{align}</p>
<p>Now, you need to observe the three sources of stress in your structure:</p>
<ol>
<li>The axial load at point B;</li>
<li>The moment due to the eccentricity of the load at point B;</li>
<li>The moment due to the load $q$ to be determined.</li>
</ol>
<p>So, what you need to do is calculate the stresses due to the known sources (#1 and #2) to see what slack you have for the load $q$. Strictly speaking you should check points A and B, but I'll only do A here.</p>
<p>So, the axial load at point B is equal to $70\cdot2040=142800\text{ N}$. This immediately gives us a uniform stress of
$$\sigma_1 = \dfrac{142800}{2640} = 54.09\text{ N/mm}^2$$</p>
<p>However, the resultant force of the load at point B is located at the centroid of the section at point B, which is different from the centroid at point A. This causes a moment, which is equal to
\begin{align}
M &= 142800\left((200-151.8)-(100-81.76)\right) = 4278288\text{ Nmm} \\
\therefore \sigma_2 &= \dfrac{My}{I} = \dfrac{4278288\cdot(200-151.8)}{9919274} =20.79\text{ N/mm}^2
\end{align}
Note the stress is being calculated in the top fiber, since that is where the tensile stresses occur.</p>
<p>So, just these loads give you already $\sigma = 54.09+20.79=74.88\text{ N/mm}^2$, meaning you are allowed an increment of $\sigma_3 = 115-74.88 = 40.12\text{ N/mm}^2$.</p>
<p>You are correct that you need to back-calculate the load via stress, so:
\begin{align}
M &= \dfrac{qL^2}{2} \\
\sigma_3 &= \dfrac{My}{I} \\
\therefore \sigma_3 &= \dfrac{qL^2y}{2I} \\
\therefore q &= \dfrac{2I\sigma_3}{yL^2} = \dfrac{2\cdot9919274\cdot40.12}{(200-151.8)\cdot2500^2} = 2.64\text{ N/mm}
\end{align}</p>
<p>I may have bundled the calculations, but that's the general gist of it.</p>
| 6041 | Calculating stresses of section with varying depth |
2015-11-05T09:04:40.973 | <p>What is the difference between an automatic transmission (AT) and an automated manual transmission (AMT)?</p>
| |mechanical-engineering|automotive-engineering|power-engineering| | <p>In answering this question I'm assuming you already know the difference between an Automatic Transmission (AT) and a Manual Transmission (MT).</p>
<hr>
<h2>Conventional Automatic Transmission (AT)</h2>
<ul>
<li>Minimal to no power interruption</li>
<li>Automatic moving-off</li>
<li>Various gear ratios</li>
<li>Efficiency: 90-95%</li>
</ul>
<hr>
<h2>Automated Manual Transmission (AMT)</h2>
<p>Functionality is the same as a Z-speed manual transmission in constant-mesh or synchromesh but the process of shifting gears and moving off are automated.</p>
<ul>
<li>Power interruption like a manual transmission (MT)</li>
<li>Automatic moving-off</li>
<li>Efficiency: 92-97% (pretty much the same as a MT)</li>
<li>Reputation for poor shift quality and reliability</li>
</ul>
<p>There are more - I wasn't sure if you were looking for functional differences or characteristics. See the reference I've provided for more info - it is a fantastic textbook.</p>
<hr>
<p>Reference:</p>
<p>Automotive Transmissions: Fundamentals, Selection, Design and Application (Naunheimer et al.)</p>
| 6047 | What is the difference between an automatic transmission and an automated manual transmission? |
2015-11-05T09:20:01.100 | <p>Figure 2 in <a href="http://uk.mathworks.com/help/simulink/examples/modeling-an-automatic-transmission-controller.html" rel="nofollow noreferrer">this tutorial</a> represents the shift map/schedule of an automatic transmission. How is this map constructed? I want to know how to design a shift map myself for another transmission.</p>
<p>The shift map is below:</p>
<p><a href="https://i.stack.imgur.com/vFfNqm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vFfNqm.png" alt="enter image description here"></a></p>
| |mechanical-engineering|automotive-engineering|power-engineering| | <p>The crucial piece of information is the torque curve of the engine ie. the function which relates torque to engine speed. </p>
<p>Any gearbox, manual or automatic, has two primary concerns the first is to allow access to the desired level of torque at any given road speed, for example changing down to give extra torque for acceleration eg overtaking on a hill. The second consideration is to have the engine at the most fuel efficient RPM for a particular cruising speed. Many automatic gearboxes will automatically change down if the throttle is fully depressed as full throttle is interpreted as a demand for maximum torque. </p>
<p>There are also some circumstances where slightly different considerations apply. For example when traction is limited eg in the middle of a high speed corner or in wet or icy condition it may be desirable to select a higher gear than normal to limit torque and make the torque output less sensitive to throttle input to better control traction.
Similarly downshifting in advance of a hazard allows engine braking to contribute to overall braking effect but conversely too aggressive a downshift can lock the driving wheels. </p>
<p>The most basic system simply change up or down at set points to keep the RPM in a defined range. More sophisticated systems may also work in concert with other systems such as ABS, stability control or traction control. But in general the goal is to control the relationship between engine RPM and road speed. </p>
<p>It is also important to consider the gear ratios used as these many not necessarily be equal steps between all gears and gear ratios will be selected according to the characteristics of the engine and the expected use of the vehicle. </p>
<p>The most sophisticated systems will aim to emulate the sort of decisions that a driver makes when selecting a particular gear with a manual box. These can also be linked to engine mapping to create different modes related specific road conditions or to achieve a balance between performance and economy. </p>
<p>For example an economy mode might change up earlier to keep RPM in the most efficient rage while a 'sport' mode might attempt to keep RPM in the region of peak torque/power. </p>
<p>In high performance cars these modes may also be related to differential and suspension settings and include modes for standing starts and overtaking. </p>
<p>The specific characteristics of an engine's torque curve are often referred to 'drivability', it is generally desirable that an engine has a predictable response to throttle and a reasonably flat and linear torque curve or at least a broad RPM range where this is the case. </p>
| 6048 | How is the shift map of an automatic transmission constructed? |
2015-11-05T21:51:20.493 | <p>I saw a <a href="http://www.cnn.com/2015/11/05/travel/macau-studio-city-wheel/index.html" rel="nofollow">story on CNN today</a> about a figure-eight ferris wheel being built in Macau.</p>
<p>I'm curious what path the cabins take, and how this occurs. Do they actually follow a figure eight pattern? Do they move from one wheel to the other somehow?</p>
<p>The story has a picture with the caption</p>
<blockquote>
<p>The cabins don't actually cross in the middle. They follow a gourd-shaped track. Each ride takes around 15 minutes.</p>
</blockquote>
<p>I don't understand what that means. A "gourd-shaped" track?</p>
| |mechanical-engineering| | <p>It looks like the cars don't actually move in a figure eight, though it is designed to look like they do from afar. In the image below you can see that the cars travel around an hourglass-shaped (or gourd-shaped) track which is behind the outer yellow steel structure that looks like a figure eight. </p>
<p><a href="http://www.travelandleisure.com/articles/studio-city-macau-ferris-wheel" rel="noreferrer"><img src="https://i.stack.imgur.com/R5fuF.jpg" alt="enter image description here"></a></p>
<p>They describe the track as gourd-shaped because some gourds, such as the one shown below, describe the shape of the track very well even though others do not.</p>
<p><a href="http://blogs.ntm.org/rachel-chapman/2012/07/19/an-old-nahuatl-tale/gourd/" rel="noreferrer"><img src="https://i.stack.imgur.com/h2WNLm.jpg" alt="enter image description here"></a></p>
| 6056 | How does the figure eight ferris wheel work? |
2015-11-07T16:31:29.847 | <p>I would like to know what mechanical parts can be used to convert a rotation movement from an axis to another.</p>
<p>For example: You have the shaft of a motor rotating vertically, but you need to rotate a part horizontally.</p>
<p>I already found the worm drive assembly but I would like to know if an "exhaustive" list exists somewhere with pro/con for each.</p>
| |mechanical-engineering|gears| | <p>The go-to solution is <a href="https://en.wikipedia.org/wiki/Bevel_gear">bevel gears</a>. They are used when the shafts would intersect and allow the same ratios as normal gears.</p>
<p>You can also make one of the gears a <a href="https://en.wikipedia.org/wiki/Crown_gear">crown gear</a> so the other gear can be a normal spur gear.</p>
<p>A <a href="https://en.wikipedia.org/wiki/Universal_joint">universal joint</a> is also an option if the angle isn't too big.</p>
| 6069 | Transform/change/convert rotation axis |
2015-11-10T08:04:23.520 | <p>I'm trying to find a material suitable to be printed and with the properties of whiteboard, easy to clean and reusable. I'm having difficulties to find alternative materials, or providers.</p>
<p>As far as I know <a href="http://www.magicwhiteboard.co.uk/" rel="nofollow">magic whiteboard</a> is an alternative but it isn't reusable, only last two times.</p>
<p><strong>EDIT</strong></p>
<p>When I say printable I mean than I can print on it, lines or other shapes and images like on paper and I want this to be permanent. I didn't mean that it needs to be printed in a 3D printer, sorry.</p>
<p>Maybe I'm looking for simple whiteboard, but I need to print on it and make it lightweight and handy (mobile phone size). In addition I would like to be able to use different and thinner pens on it.</p>
| |materials| | <p>I use polycarbonate film which is used for binding manuals. Polyvinyl chloride sheets are equally good but when using whiteboard markets you must encaustic they are stressed within 24 hours to avoid nasty stains</p>
| 6089 | Printable and reusable whiteboard material |
2015-11-10T17:55:19.880 | <p>If we have a simple adiabatic cylinder with a piston on top and we heat the gas from the inside, the gas starts to expand and move the piston up. This process is isobaric; but if the pressure on the piston doesn't change, how does it move up?</p>
| |thermodynamics| | <p>In reality, such a process is not <em>truly</em> isobaric; a small rise in temperature will cause the gas to build up an infinitesimally small amount of pressure, which moves the piston up an infinitesimally small distance, which restores the pressure to its original value. If you assume that the piston can react quickly to the heat being applied, then it behaves nearly isobarically. Or, in mathematical terms, if $\Delta T$ is the rise in temperature from one state to the next, the process becomes isobaric as ${\Delta T \to 0}$. This is also referred to as a quasi-static process.</p>
<p>Textbooks frequently make simplifications like this because they don't expect students to ask such questions!</p>
| 6097 | How does expanding gas in isobaric process move a piston? |
2015-11-11T09:35:06.280 | <p>I have a school case study to prepare a (business) plan for office building construction with at least one level of underground parking. I have it pretty much done, but I still need to clarify one thing (teacher is not available):</p>
<p>(1) What is implied by "Underground water level varies from 5-6m" for the construction? Currently I ignore it. Does it affect the underground parking if I consider only 1 level? Does it have (any/significant) impact on building costs?</p>
<p>(2) What would be depth of typical base/foundation for an office complex building (3 stories, 50k sqm)? Does the underground water interfere in any way?</p>
<p>Tried googling for relation of underwater level and foundations, but couldn't really find anything. </p>
| |building-design| | <blockquote>
<p>(1) What is implied by "Underground water level varies from 5-6m" for the construction?</p>
</blockquote>
<p>Typically this means that the groundwater table starts between 5m and 6m below grade. In other words, if you dig over 5m into the soil, you will likely hit the water table and have water in your excavation.</p>
<p>Construction sites use water mitigation strategies to deal with groundwater, which can be anything from using sump pumps to putting a slurry wall around the excavation. It all depends on the in situ condition and cost.</p>
<blockquote>
<p>(2) What would be depth of typical base/foundation for an office complex building (3 stories, 50k sqm)?</p>
</blockquote>
<p>Without knowing anything about the subsurface condition and what kind of equipment will be housed in the building, this question is really impossible to answer.</p>
<p>You could have bedrock 3m down. The soil could be cruddy organics or sand, which would require drilling piles down to a more stable soil strata. Your office building could house a heavy piece of equipment, requiring a special foundation.</p>
<p>Also, not knowing the footprint of the building makes it difficult to guess at a foundation design.</p>
<blockquote>
<p>Does the underground water interfere in any way?</p>
</blockquote>
<p>Groundwater pressure can make a building buoyant, which needs to be considered in design of the foundation.</p>
<p>You will likely have to include some kind of foundation drainage system into your building design to shed the groundwater away. Also, a sump pump system in the foundation will likely be needed, similar to what is done in residential basements.</p>
| 6103 | Office Complex Foundations/Base Depth and Undeground Water |
2015-11-12T00:17:49.883 | <p>For example i have an exhaust pipe with very hot air from burning biofuel, and i have a turbine at the end of the exhaust that rotates and generates electricity.</p>
<p>Why is it more efficient to use the heat to boil water then use the steam produced to rotate the turbines? As in why is more electricity generated by using steam to rotate the turbines as opposed to not using steam when burning the same amount of biofuel?</p>
| |electrical-engineering|turbines|steam| | <p>A two phase chemistry that makes pressure using heat is necessary. </p>
<p>A pressure cooker with only air makes a lot less pressure than with a liter of water. </p>
<p>The water is in effect potential pressure stored in a cold state.</p>
<p>Supercritical fluids are actually more efficient than steam but require higher pressure vessels, and a lot of ice CO2. and other exotic substances.</p>
| 6108 | Why is using steam to rotate a turbine more efficient? |
2015-11-12T20:31:05.983 | <p>If biofuel is burnt, for example wood, and the stove I am building is made of metal, what is the most appropriate metal material I should use to avoid excessive corrosion in the exhaust pipe of the stove?</p>
| |materials|combustion|corrosion| | <p>Just to add some unique engineering solutions that are used, depending upon applications:</p>
<ol>
<li>Brick works wonders - it resists the acid in the smoke well and can withstand the heat. Not metal, but well worth it if the application is right.</li>
<li>For power plants, typical material use is acid brick lined steel.</li>
<li>For industry, such as fume exhaust of potential violently reactive chemicals (not just limited to combustion), typically Fiberglass Ducting is utilized, or fluoropolymer coated steel ducting. Both have excellent corrosion resistance, and the flouropolymer coating can withstand 150C.</li>
</ol>
| 6121 | What metal material to use for stove exhaust pipe? |
2015-11-13T15:52:31.983 | <p>Lavoisier's principle of mass conservation states that the mass of a system must remain constant over time. But, as far as I know, there's no similar principle for density.</p>
<p>So, is it possible to develop an alloy as dense as gold but made of constituents less dense than gold?</p>
| |materials|alloys| | <p>Unfortunately, your best bet is abusing the existence of isotopes of tungsten to produce "heavy" tungsten. A potential, but unexplored, alternative is to alloy tungsten with an interstitial atom other than carbon.</p>
<p>There is no way to make tungsten look or behave like gold beyond superficially, i.e. spraypainting it or electroplating it.</p>
<p>No other element or mixture of elements can make it to the density of gold without requiring denser constituents.</p>
<h1>Explanation</h1>
<p>There are only a finite number of elements to play with, unfortunately, so options are very limited. I am going to assume when you say "less dense than gold" you mean without using any elements that have a density higher than gold at standard temperature and pressure, and in their as-typically-processed, pure states. To have an alloy with the same density of gold but be made of constituents that all have lower density than gold would require very special and unusual properties of those materials. Density of single crystals is controlled by four factors: crystal structure, atomic weight, interatomic spacing, and quantity of point defects.</p>
<h2>Crystal Structure</h2>
<p>Changing the crystal structure of an elemental metal can change its density. The change in density is usually very small, on the order of $1\%$, so we would need to start with either tungsten (W, $19.25\ \textrm{g/cc}$) or uranium (U, $19.05\ \textrm{g/cc}$). Other appropriate elements are simply too far below the density of gold.</p>
<p>Unfortunately tungsten occupies only the body-centered cubic (BCC) structure at all temperatures and pressures when solid (<a href="https://upload.wikimedia.org/wikipedia/commons/2/22/Phase_diagram_of_tungsten_(1975).png">phase diagram</a>). Uranium has an orthorhombic structure at room temperature, and changes first to tetragonal around $950\ \textrm{K}$, then to BCC at about $1050\ \textrm{K}$ (<a href="https://upload.wikimedia.org/wikipedia/commons/2/2c/Phase_diagram_of_uranium_(1975).png">phase diagram</a>).</p>
<p>It's worth noting that as the temperature increases, density decreases for two reasons: an increase in vacancy density, and an increase in interatomic spacing due to higher average atomic kinetic energy. It is unlikely uranium will have density $19.3\ \textrm{g/cc}$ at any temperature at atmospheric pressure (we'll get to elevated pressure momentarily). At any rate, you'd have a hard time fooling anyone with a radioactive slab of glowing hot metal. Or at the least you'd probably have folks in black suits knocking at your door.</p>
<h2>Atomic Weight</h2>
<p>We can change atomic weight by adding or removing neutrons from existing elements to create isotopes. How you would go about mass-producing specific isotopes is beyond the scope of this answer. The stable isotopes of Tungsten are <a href="https://en.wikipedia.org/wiki/Isotopes_of_tungsten">listed here</a> and have 182 to 186 nucleons, weighing between $181.95$ and $185.95\ \textrm{g/mol}$.</p>
<p>The atomic weight of Tungsten is $183.84\ \textrm{g/mol}$, and the value is an average of the isotopic content of typical samples of Tungsten containing the various stable isotopes. To increase the density of tungsten to that of gold, an increase of $0.25\%$ is required, so increasing the atomic weight to $184.3\ \textrm{g/mol}$ should be sufficient.</p>
<p>Uranium has no stable isotopes, but it does have two that last for billions of years, specifically 235 and 238. Unfortunately those are practically the only isotopes in existence, and virtually all of it is 238. The next isotope up, 239, only has a half life of 30 minutes, so by the time you made it it would already be decaying. Even still, that isn't enough of an increase in atomic weight to push it up to the density of gold.</p>
<p>It's also worth noting that, even if you went this route, neutron bombardment creates byproducts like hydrogen and helium which create density-reducing defects such as vacancies and voids, so you would have to reprocess your material after you make it to remove the defects. Alternately, you could process via atomic centrifuge set up to process tungsten, instead of uranium, and isolate the isotopes, then recombine them to the appropriate average atomic weight.</p>
<h2>Point Defects</h2>
<p>More vacancies would only reduce the density, which means we would need to start with something heavier than gold, so that won't work. More substitutional atoms would also generally require at least one denser component as density virtually always follows a linear rule of mixtures in alloys, so that won't work either. Because of the linear rule of mixtures issue, virtually all multi-metal alloys are out, and only pure elements have any real chance of approaching the density of gold without using elements that have higher density than gold.</p>
<p>The last point defect to consider is interstitial defects, where atoms small enough to fit between the metal atoms are used as alloying elements. Interstitials generally increase the density, as with carbon in steel. Unfortunately, tungsten does not dissolve carbon at standard temperature and pressure, as shown in the <a href="http://patentimages.storage.googleapis.com/EP1606426B1/imgf0001.png">phase diagram</a>, so that idea is out for carbon. I am unable to find a phase diagram for nitrogen, so it is unclear whether it would be possible to produce a tungsten-interstitial alloy. If it is up to the requisite $0.25\%$ atomic percent, then this method may also work.</p>
<h2>Interatomic Spacing</h2>
<p>Aside changing the magnitude of the electromagnetic force constant, the only way to change interatomic spacing is via the application of stress. In this case, we would want hydrostatic compressive stress. Again, we only need an increase of $0.25\%$ for tungsten, so a decrease in lattice constants of one-third that or $0.083\%$ would suffice. That is a strain of $0.00083$. The <a href="https://en.wikipedia.org/wiki/Tungsten">modulus of tungsten</a> is $411\ \textrm{GPa}$, so the required hydrostatic compressive stress is $341\ \textrm{MPa}$, or a pressure of close to $50,000\ \textrm{psi}$. Fooling anyone by this method would require some interesting apparatus.</p>
| 6128 | Alloys with same density as gold |
2015-11-14T04:49:34.043 | <p>The question has been surfacing periodically in the media and books over the past 100 years or so. I realize we're talking about a lot of material and work hours to begin with; the subsequent maintenance would be a nightmare. There are no power stations, nor radio towers between the continents; there is no viable way (or is there?) to build in vents along the way; the bottom of the ocean is uneven, and there's a mountain ridge in the middle; and who knows what problems would be encountered if the engineers decided to actually drill under the ocean floor all the way. </p>
<p>Pumping the air out might turn out to be extremely expensive. Leaving the air in might create problems for the trains: that's a lot of air in a tight space to push through. </p>
<p>How feasible is it? Or is it even technically possible to have a tunnel under the Atlantic? </p>
| |civil-engineering|tunnels| | <p>I'm assuming you are asking about a tunnel between North America and Europe.</p>
<p>There are two types of tunnel that "might" be possible: excavating a tunnel through the rock beneath the sea bed or sinking prefabricated sections of tunnel to the sea floor, joining them up and sealing them against water inflow.</p>
<p><em><strong>Problem #1 - Depth of the North Atlantic Ocean</strong></em></p>
<p>The depth of the <a href="https://en.wikipedia.org/wiki/Atlantic_Ocean" rel="noreferrer">Atlantic Ocean</a> is between 3700 and 5500 metres. Such depths are going to make it extremely difficult to sink the tunnel sections, assemble them, ensure they can survive the water pressure and be water tight.</p>
<p>Additionally, such deep water will increase the ground stresses in the rock in the sea bed, which will be problematic for an excavated tunnel.</p>
<p><em><strong>Problem #2 - Geological Reconnaissance</strong></em></p>
<p>For an excavated tunnel, the sea bed will need to drilled along the proposed path of the tunnel at closely spaced intervals so core sample can be taken, so that geologists can determine what type are rock or soil (sediments) are there and ascertain material properties for the rock. They will also need to know what geological structural discontinuities are present, such as faults, where they are located, how they are orientated and how adverse they could be.</p>
<p>Drilling in waters 5500 metres presents a whole lot of challenges, one of them being the drill string would be unsupported for the first 5500 metres of its length.</p>
<p>Geological reconnaissance would also be required for a sunk tunnel to ensure the sea bed could support the weight of the tunnel sections. Having sections of tunnel subside into soft sediments is something that would need to be prevented.</p>
<p><em><strong>Problem #3 - Ventilation</strong></em></p>
<p>Ventilation shafts rising though 5500 m of water would be problematic in terms of construction, support and possible impact by whales, ships or submarines. They would also need to be water tight to prevent water inundation of the tunnel.</p>
<p>The only way to ventilate such a tunnel would be via ventilation tunnels parallel to the main tunnel. These could double-up as service tunnels for inspections, maintenance and rescue, should there be a fire in the main tunnel.</p>
<p>For reasons of safety, ventilation tunnels would need to be duplicated, at least.</p>
<p>Ventilation fans would need to be placed at the ends of the vent tunnels. They would need to be large and they would consume a lot of electricity.</p>
<p><em><strong>Problem #4 - Traversing an Active Volcanic Zone</strong></em></p>
<p>There is an entity called the <a href="https://en.wikipedia.org/wiki/Mid-Atlantic_Ridge" rel="noreferrer">Mid Atlantic Ridge</a>. This is an active volcanic zone that is pushing apart the continental plates that separate the Americas from Europe and Africa. This ridge is visible above the water on Iceland.</p>
<p>Putting a tunnel on or through the Mid Atlantic Ridge may not be possible. If it were possible, the tunnel would require regular maintenance to remedy damage done by the Ridge.</p>
<p>Putting ventilation tunnels through the Mid Atlantic Ridge would be give additional ventilation problems.</p>
<p><em><strong>Overall</strong></em></p>
<p>The distance between Canada and Ireland is close enough to 3000 km.</p>
<p>The cost of construction a trans Atlantic tunnel would be hideously expensive!</p>
| 6134 | Is a Transatlantic tunnel technically possible today? How feasible would it be? |
2015-11-14T17:42:32.860 | <p>I'm trying to understand and identify the equations to use in defining the relationship between wind velocity, turbine rotor diameter, and power output for a wind turbine.</p>
<p>To simplify my question, let's use the following assumptions:</p>
<ul>
<li>Average wind speed of 8 m/s. </li>
<li>Rotor length of 40m for an overall diameter of 80m.</li>
<li>Ideal conditions, so we can neglect direction of wind with respect to the positioning of the turbine.</li>
</ul>
| |energy|renewable-energy|power|wind-power| | <p>The power generated depends on the product of two factors:</p>
<ol>
<li>the windmill (wind turbine) power curve</li>
<li>the windspeed curve or histogram, % at each windspeed.</li>
</ol>
<p>Here's a plot of 2 wind turbine types <span class="math-container">$\times$</span> 3 synthetic (Weibull-distributed)
windspeed curves:</p>
<p><a href="https://i.stack.imgur.com/QGuDY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QGuDY.png" alt="enter image description here" /></a></p>
<p>Let's compare this with the theoretical <span class="math-container">$A \, v^3$</span>,
as described e.g. in MacKay,
<a href="https://withouthotair.com/" rel="nofollow noreferrer">Sustainable Energy -- without the Hot Air</a>
p 263 ff.
Consider the effects of <span class="math-container">$A$</span> and <span class="math-container">$v^3$</span> on power separately:</p>
<p><span class="math-container">$A$</span>: the two wind turbines here have rotor diameters 101 and 141m,
so <span class="math-container">$A$</span> ratio ~ 2 but max power ratio ~ 1.4.
This doesn't tell us much: maybe one could build a wind turbine with twice the MW
at <span class="math-container">$2 A$</span>, but it would cost more than it's worth ?</p>
<p><span class="math-container">$v^3$</span>: the power ratios in table form</p>
<pre><code>Ratios of MW at [6 7 8] / 6 m/s
E-101/3050 [100 132 161] %
E-141/4200 [100 131 159] %
</code></pre>
<p>show a factor ~ 1.4, not 2 <span class="math-container">$\sim (7.5 / 6)^3$</span> .</p>
<p>(Fine print: why synthetic windspeed curves ?
Real data is hard to find (worth \$\$\$).
There are of course lots of histograms with average 8 m/s.
<a href="https://en.wikipedia.org/wiki/Weibull_distribution" rel="nofollow noreferrer">Weibull</a>
give a family of curves or histograms with the same average,
and a parameter <span class="math-container">$k$</span> that shifts less wind at lower / more at higher windspeeds.)</p>
<p>See also:
<a href="https://wind-data.ch/tools/powercalc.php" rel="nofollow noreferrer">wind-data.ch</a>,
<a href="https://wind-stats.readthedocs.io" rel="nofollow noreferrer">wind-stats</a>,
<a href="https://globalwindatlas.info" rel="nofollow noreferrer">globalwindatlas.info</a> -- nice gui, legend re-scale</p>
<hr>
Summary: is there a simple model of wind turbine power<span class="math-container">$( A, v, height, cost )$</span>
that fits power actually generated to within say 10 % ? Sounds unlikely.
| 6144 | Relationship between wind velocity, turbine rotor diameter, and power output |
2015-11-14T22:39:47.933 | <p>I've recently designed what is essentially an aluminium box. I've done the entire design process of CAD software SolidWorks and then done some simulations of loading of the box.</p>
<p>Through iterations I've figured out the minimum thickness of the aluminium (AL6061 T6) box material to be quite small - 0.25 mm. This seemed almost too small to me so I looked up kitchen 'tin' foil for comparison. The <a href="https://en.wikipedia.org/wiki/Aluminium_foil" rel="nofollow">Wikipedia page</a> puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box.</p>
<p>I'm aware that tin foil is probably not AL6061 and it's probably not tempered at all but this has thrown me into quite a bit of confusion as to whether or not I can trust my simulation results (I've not given any specific details of my simulation since I don't want to turn this into a homework style question - but I'm happy to share if it helps).</p>
<p>So my question is; can I trust the results at AL6061 T6 is significantly stronger than aluminium foil (it would help if I can find what aluminium the foil is made of)?</p>
| |design|computer-aided-design|aluminum| | <blockquote>
<p>The Wikipedia page puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box.</p>
</blockquote>
<p>Ooook. Let's look at this, from an engineering perspective.</p>
<p>First, scales. The smallest measurement from Wikipedia vs the largest is a 33x factor difference. This is perhaps hard to conceptualize in mm, but imagine a 1m stick. Now imagine 33 of them. This is the difference between those measurements.</p>
<p>This means you need to be <em>very</em> careful looking at those ranges and saying "oh that means my measurement is about tin foil." Understanding orders of magnitude is a very important engineering principle.</p>
<blockquote>
<p>So my question is; can I trust the results at AL6061 T6 is significantly stronger than aluminium foil (it would help if I can find what aluminium the foil is made of)?</p>
</blockquote>
<p>It is good to question unintuitive software results. But be careful to not combine different things.</p>
<p>You may (as in this case) attempt to compare things which are not good to compare. Experience helps with this.</p>
| 6148 | Intuition vs software results |
2015-11-15T00:42:01.457 | <p>I need a small air compressor for a robotics application. I currently have one of those 12V tire inflators (200 psi, 1 CFM or so) that is sized well for this purpose (sample image below).</p>
<p>However, in the future I would like to replace this with a lighter compressor with similar specifications (for weight savings). Most of the weight of these compressors is a large DC brushed motor (the whole big silver cylinder in the photo below), while the actual compression module is relatively small. These motors are roughly 350W peak (30A, 12V from spec sheet). Similar DC brushless motors (by wattage) are <150 grams. Certainly they might be at a higher RPM, but even with gears this would still be a lot smaller.</p>
<p>Is there any reason that these huge (3-4 kg) brushed motors couldn't be replaced by tiny brushless motors? If so, why can't I find such devices anywhere? It wouldn't be cost effective for these cheapo inflators (you need more complex controls for brushless), but it seems like there would be enough demand for lightweight compressors for someone to manufacture a few.</p>
<p><a href="https://i.stack.imgur.com/Sj1eM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sj1eMt.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|compressors| | <blockquote>
<p>If so, why can't I find such devices anywhere?</p>
</blockquote>
<p>In a word - <strong><em>Cost</em></strong>. </p>
<p>Running some quick searches for 300W DC motors, the brushless motors were typically 3x - 5x more expensive than their brushed counterparts. And as you pointed out, brushless motors require additional and more expensive controls than a comparable brushed motor. </p>
<p>It's worth pointing out that small compressors like the one you asked about are commodity grade products. Profit margins on them are slim and purchase decisions are frequently driven by cost alone. It is the rare consumer who would consider paying a significantly higher amount of money for a comparatively smaller benefit of decreased weight.</p>
<p>And I think you're overestimating the demand for lighter weight compressors. Most places that I'm aware of that are using compressors purchase them infrequently, and they tend to be fixed installations where the weight of the compressor doesn't matter that much. And for the cases where weight and portability is a concern, cost becomes an over-riding factor.</p>
<hr>
<p>That said, would you be able to replace the existing motor with a brushless motor? Presuming you have the ability to create the additional controls and gearing to go along with using the brushless motor, I don't see why not. And for what I suspect is going to be a relatively low utilization rate, you're likely going to see an equivalent lifespan of the product.</p>
| 6150 | Lightweight Air Compressor |
2015-11-15T01:56:27.710 | <p>Do we have the technology to illuminate a small region on the dark side of the Earth with an artificial satellite in orbit around the Earth? By illuminate I mean bright enough to light objects for regular humans nearby to see better than they could have otherwise. Assume low Earth orbit and no moon to make it easier.</p>
| |aerospace-engineering| | <p>Sure. Even a cubesat could do this (if only for a few seconds). </p>
<p>Put a relatively powerful laser (maybe 100W?) on a satellite, focus and aim it at a square meter or so of ground, and light it up. 100 km (LEO) is short enough that a good laser can focus down to a pretty small land area, meaning a smallish laser can focus all of its power on a small area. </p>
<p>A 100W laser is about as bright as 10 100W incandescent light bulbs, and only weights a few kilograms. Add some solar panels and batteries, and your satellite can light up your 1 square meter of dirt for a bit each orbit.</p>
| 6152 | Illuminate the Earth with an artificial satellite? |
2015-11-15T12:16:15.840 | <p>Is there any emitter/receiver (passive or active) systems that allow distance estimation in the range of 1~2 meters with an accuracy of few centimetres and don't need to have a direct line of sight as optical systems mostly do?</p>
<p>I was thinking about using such sensors as nodes to put on a suit that a human can wear. As a result, a graph can be obtained where each node represents one of those sensors and each edge describes the distance from this node to the other nodes. Using graph optimization techniques, the graph would converge to the correct joint position of the human. Such a system could replace expensive motion capture methods using calibrated external cameras.</p>
| |electrical-engineering|sensors|distance-measurement| | <p>Ultrasonic sensing is can be used for distance measurement. Ultrasonic proximity sensing technology is commonly used in automotive as part of <a href="https://en.wikipedia.org/wiki/Advanced_driver_assistance_systems" rel="nofollow noreferrer">Advanced Driver Assistance Systems</a>. The basic theory behind this technology is <a href="https://en.wikipedia.org/wiki/Time_of_flight" rel="nofollow noreferrer">Time-of- Flight</a> (TOF). Below few images to help visualize the idea.</p>
<p><a href="https://i.stack.imgur.com/xGM0A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xGM0A.jpg" alt="Park Assist-1"></a>
<a href="https://i.stack.imgur.com/BO8P4.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BO8P4.gif" alt="Park Assist-2"></a></p>
<p>To achieve this you can use <a href="http://www.ti.com/product/pga450-q1" rel="nofollow noreferrer">PGA450 Ultrasonic Analog Front End</a> (AFE) from Texas Instruments. As I understand this sensor is capable of distances up to 7M.</p>
<blockquote>
<p>The PGA450-Q1 device can measure distances ranging from less than 1 meter up to 7 meters, at a resolution of 1 cm depending on the transducer-transformer sensor pair used in the system.</p>
</blockquote>
<p>Below is a diagram describing the concept.</p>
<p><a href="https://i.stack.imgur.com/fc7RS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fc7RS.jpg" alt="PGA450 AFE"></a></p>
<p>Also you might want to check out <a href="http://www.ti.com/product/TDC1000-Q1/datasheet" rel="nofollow noreferrer">TDC1000 Ultrasonic AFE</a>. I am much more familiar with this part. But I have not used this part for proximity sensing. Once again below is a diagram describing the application. </p>
<p><a href="https://i.stack.imgur.com/1rGvC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1rGvC.jpg" alt="TDC1000 AFE"></a></p>
<p>Also TDC1000 has a configuration where the emitter and receiver are seperate that might be interesting.</p>
<p><a href="https://i.stack.imgur.com/5OY8N.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5OY8N.jpg" alt="TDC1000 AFE Separate emitter and receiver"></a></p>
<p>Below are some references that you might find interesting</p>
<hr>
<p><strong>References</strong></p>
<ul>
<li><a href="http://www.freescale.com/applications/automotive/adas:ADAS" rel="nofollow noreferrer">Advanced Driver Assistance Systems (ADAS)</a></li>
<li><a href="https://engineering.stackexchange.com/questions/2176/is-the-reliability-of-automotive-sensor-technology-impeding-the-success-of-self">Is the reliability of automotive sensor technology impeding the success of self driving car?</a></li>
<li><a href="http://www.ti.com/lsds/ti/sensors/ultrasonic-sensing-applications.page#proximity-sensing" rel="nofollow noreferrer">Proximity sensing</a></li>
<li><a href="http://www.ti.com/tool/tida-00151" rel="nofollow noreferrer">Automotive Ultrasonic Sensor Interface for Park Assist or Blind Spot Detection Systems</a></li>
<li><a href="https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&cad=rja&uact=8&ved=0CE0QtwIwCGoVChMIq9Di2K2WyQIVhGs-Ch3QhwPb&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DrjEO4PsTnP4&usg=AFQjCNFccjB5OWF4_n_Ilz4HJjbe_uHfmw&sig2=pJojaKTNTbfsEzLIVa0fww&bvm=bv.107467506,d.cWw" rel="nofollow noreferrer">PGA450-Q1 for Ultrasonic Sensing</a></li>
<li><a href="http://www.engineering.com/Library/ArticlesPage/tabid/85/ArticleID/8307/Ultrasonic-Sensor-SOC-can-Sense-Objects-and-Distance-for-Automotive-and-Industrial-Designs.aspx" rel="nofollow noreferrer">Ultrasonic Sensor SOC can Sense Objects and Distance for Automotive and Industrial Designs</a></li>
<li><a href="https://engineering.stackexchange.com/a/2697/110">Can I use a ultrasound sensor to measure water level?</a></li>
<li><a href="https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CB0QFjAAahUKEwi76_DN0JjJAhXCPT4KHSQkDuw&url=http%3A%2F%2Fwww.ti.com%2Flit%2Fsnaa220&usg=AFQjCNHiIW0c4gybaffO0MlWBYrpGYK47g&sig2=4jIjgrHtE75F1TqW7ReYNw" rel="nofollow noreferrer">Ultrasonic Sensing Basics for Liquid Level Sensing, Flow Sensing, and Fluid Identification Applications</a></li>
</ul>
| 6157 | Short range distance estimation systems |
2015-11-15T12:25:42.243 | <p>I know definition but how would you explain to complete layik ( as me ) what is MEMS and how it is used. </p>
<p>Examples: Silicon nanopore membranes (SNM) have been prototyped from silicon substrates by an innovative process based on MEMS technology.</p>
<ul>
<li><a href="http://whatis.techtarget.com/definition/micro-electromechanical-systems-MEMS" rel="nofollow">micro-electromechanical systems (MEMS)</a></li>
</ul>
| |mechanical-engineering|applied-mechanics|power-electronics|systems-engineering| | <p>One way to think of MEMS is regular mechanical systems, but at a very small scale. Often these tiny systems are fabricated using technology developed for making silicon electronic chips, like nanometer-scale photolithography and etching.</p>
<p>However, describing MEMS as just downscaled regular mechanical systems is doing the concept injustice.</p>
<p>Various physical properties scale differently with linear dimension. For example, simply scaling a ant to the size of a elephant wouldn't work. The mass the legs have to support is proportional to the volume (cube of the linear scale), but the strength of the legs only proportional to their crossectional area (square of linear scale). Such a system can't be scaled up linearly and still function as the original.</p>
<p>As a result of different scaling of different physical properties to the tiny dimensions of MEMS, some things we do routinely at the human scale are impossible, but new things are possible that would be impractical at human scale. For example, rotary electric motors with conventional windings would be extremely difficult in MEMS. However, at such short distances, the electrostatic force can be harnesses much more easily, and MEMS motors using it are quite possible and efficient. At nanometer distances, just a few volts can cause quite significant forces relative to the much much lower mass (remember, mass scales with the cube of linear dimension) of the parts.</p>
<p>The above is just one example, but illustrates how you enter a different world at micrometer or nanometer dimensions.</p>
| 6158 | What is micro-electromechanical systems (MEMS)? |
2015-11-15T15:11:56.637 | <p>I am studying embedded systems and microcontrollers and now I have to face the interrupts. I understood the main concepts around them and how to implement them inside a C code, but I want to know how they work at hardware level. </p>
<p>I found this scheme on the internet:</p>
<p><a href="https://i.stack.imgur.com/kvTLq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kvTLq.png" alt="enter image description here"></a></p>
<p>But I cannot understand it at full. Can you explain me a bit? Don't worry to use technical terms, I think that I will understand them. However I prefer to search some terms that I don't know rather than limiting your knowledge while answering.</p>
<p>Nice engineering at everyone!</p>
| |electrical-engineering|embedded-systems| | <p>A interrupt is when the processor goes and executes some pre-determined code based on some event. This is done so that, with the cooperation of the specially executed code, the processor can return to executing the original code in such a way that its operation is not affected other than some additional time taken.</p>
<p>That's all it is, but there's a lot behind this relatively short definition. For example, the event is usually some hardware completing a task (so that the software now needs to "service" it) or a external event, but can also be caused internally in the CPU. Examples of the latter in some processors are when a unrecognized opcode is encountered for execution, a divide by 0, attempt to reference non-existant memory, or even a condition deliberately set by the software. There is great variation between processors in the details of how interrupts are handled. Saying something as simplistic as the program counter is pushed onto the stack and the interrupt code is executed is missing many important issues and is downright wrong for some processors. Not all processors even have a stack, for example.</p>
<p>Since the interrupt envolves interrupting the existing code to execute something else, and the processor must be able to resume the existing code, some state has to be saved as part of the interrupt mechanism. At the very least, this requires saving the PC (program counter). The mechanism for storing the PC and then later restoring it to resume execution of the original code is usually done similarly to how subroutines are handled on that processor. For example, the PC could be stored in a special place, special register, pushed on a stack, etc. All these and more have been done historically.</p>
<p>Since it must be possible for interrupts to be transparent to the interrupted code (other than a delay between two instructions), this usually means some other state must be saved and restored on entry and exit of the interrupt routine. What state needs to be restored, and whether hardware mechanisms are provided to do this or facilitate this is highly dependent on the processor architecture. For example, old style PICs save the PC on the hardware stack for you, and the rest is up to the interrupt code. Anything it uses must be saved first so that it can be restored later. For example, on these PICs you can't do much without using the W register, so just about any useful interrupt routine would need to save the W register early on and then restore is shortly before exiting the interrupt code.</p>
<p>More high end processors might provide a whole set of "shadow" registers just for use during interrupts. The hardware switches to using these alternate registers automatically upon interrupt and switches back on return from interrupt. In this case, the interrupt code can just use registers normally without worrying about trashing them from the interrupted code's point of view.</p>
<p>There are many many processor architectures out there, and the details of how interrupts are handled can vary significantly between architectures. Some processors have a single location execution goes on interrupt. Others have a different location for each posssible interrupt condition. On processors with virtual memory or protection mechanism to handle hostile processes, it gets more complicated. Multiple threads of execution makes things complicated (which process' context does the interrupt code run with, for example?). Then there are interrupt priorities, or not. You generally don't want the same interrupt condition to be able to interrupt the interrupt code that is trying to handle it.</p>
<p>There are many many details and many many schemes have been devised to implement interrupts, but they all conform to the definition in the first paragraph.</p>
<p>The best way to learn how interrupts work is to understand them on one specific machine. For starters, pick a simple one, like a original PIC 16. Once you understand that, it will be easier to understand the extra wrinkles added by higher end processors.</p>
| 6162 | Hardware behind interrupts |
2015-11-16T02:37:53.643 | <p>I want to convert rotary motion (the orange propeller) onto a square path (the dark gray path). Is this a valid way to go about it? Would it be possible to use a square peg instead of a round one (the red circle) in order to keep the peg level as it travels? If not how could I keep it level? I'm a noob.</p>
<p><a href="https://i.stack.imgur.com/wwoX2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wwoX2.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/3Apv8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Apv8.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/LNMMa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LNMMa.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/GTvkA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GTvkA.jpg" alt="enter image description here"></a></p>
<p><strong>Update</strong> 2015/11/16 - Posted another option as an answer below</p>
<p><a href="https://i.stack.imgur.com/zfV8s.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zfV8s.gif" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>The reuleaux triangle is a great solution but I would prefer something that has fewer parts and is more compact...</p>
<p>Here's another option. The idea is to sandwich a bent peg between offset tracks. So the sandwich goes like this: Bottom Track, peg, Propeller, Top Track (Transparent). And Yes, I would round out the track corners more than depicted.</p>
<p><a href="https://i.stack.imgur.com/SruXM.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SruXM.gif" alt="enter image description here"></a></p>
| 6170 | Is this Square Track Cam valid? |
2015-11-16T12:10:04.283 | <p>I need to design the roof structure for a sports facility. I have no information with regard to HVAC design at this stage. Also, the HVAC may change in future. Thus I don't know what load to apply to the structure. I intend to give the HVAC and electrical engineers a weight envelope in which they can operate.</p>
<p>Is there a standardised list available which would give an indication of what loads I can assume for this?</p>
| |structural-engineering|hvac| | <p>I don't think what you are looking for really exists, in terms of a standardized list of HVAC/services loadings. Building codes don't usually specify dead loads, probably because they are, in principal, known. Typically only imposed (live) loads are specified. From the buildings I have worked on the services loadings were essentially copied from similar buildings/agreed with the HVAC contractor. There are some generic guidelines available. For example:</p>
<p><a href="http://www.amazon.co.uk/Structural-Engineers-Pocket-Book-Eurocodes/dp/0080971210" rel="nofollow">The Structural Engineers Pocketbook</a> uses 0.15 kPa for ceiling and services</p>
<p>The Arup Structural Scheme Design Guide has 0.25 kPa nominal and 0.4 kPa for HVAC (I only have an old version so don't know if that has changed...)</p>
<p>The <a href="http://www.steelconstruction.info/Design_codes_and_standards" rel="nofollow">Steel Construction.info</a> site (which is a pretty good resource) has 0.25 kPa as being a typical services loading.</p>
<p><a href="http://ascelibrary.org/doi/book/10.1061/9780784412916" rel="nofollow">ASCE 7-10</a> in Table C3-1 lists a dead load of 4 psf as a minimum 'mechanical duct allowance'. Which is likely where the 4 psf (0.19 kPa) mentioned in the comments originally came from and is consistent with the above references.</p>
<p>The <a href="http://www.designingbuildings.co.uk/wiki/Structural_systems_for_offices#Standard_allowances_for_a_dead_load" rel="nofollow">Designing Buildings Wiki</a> has 0.85 kPa for Raised floors, ceiling and building services equipment combined. </p>
<p>Hard to say if these would be applicable to your situation. Also
you would need to deal with heavy equipment (condensers, fans, heaters, etc) which are not considered in these loadings. </p>
<p>The HVAC contractors tend to want to put holes all over your structure as well, so it would probably save time and money to consult with them early in the design and sort out what they need.</p>
| 6174 | Dead Load / Self Weight for ceilings & HVAC |
2015-11-17T04:19:21.503 | <p>Does the speed of the water flowing radially outwards from a water jet impacting a sink depend on how far the tap is opened?</p>
<p>At the moment, I understand that the maximum velocity of the water in the vertical jet is constant regardless of how far the tap is open (as it is accelerated by gravity, and disregarding air-resistance), but the volume flow rate is greater the further the tap is opened.</p>
<p>I'm thinking that a higher volume flow rate results in a greater depth around the bottom of the jet which reduces the effect of friction between the water and the sink as the contact surface of the water has reduced, which would decelerate the water at a lower rate than if it were shallower, but am not sure.</p>
<p>Could someone explain precisely <em>how</em> opening a tap affects the radial speed of the water around the bottom of the jet?</p>
<p>EDIT</p>
<p>I should have clarified that, for my question, the area is variable.
What I had in mind was more like a <em>variably sized circular hole beneath a constant-height water tank</em>. I used the word "tap" for simplicity, but I realise now this is a bad substitute, and has caused more confusion than intended.</p>
| |fluid-mechanics| | <p>The speed at which water leaves the tap is not constant, as it has a finite area, which remains roughly constant, so increased flow does indeed increase the velocity. ($\dot m= \rho\, VA$)</p>
<p>However, we can still ask the question of how the flow profile of water flowing outward from a jet is dependant on flow rate when jet velocity is held constant.</p>
<p>First let's calculate the reynolds number:</p>
<p>$$Re=\frac{\rho V D}{\mu} \approx \frac{1000\frac{kg}{m^3}\,2\frac{m}{s}1\,cm}{900 Pa\,\mu s}\approx 20000$$</p>
<p>So the viscous effects are going to be tiny during the transition, which means we can use bernoulli's equation. The pressure will be at atmospheric, before and after the transition, and the height will be the same, so the velocity will be the same.</p>
<p>In conclusion, the spray will just get thicker, but remain at the same velocity when the flow rate is increased. Of course, far from the jet the viscous effects will begin to matter more as the fan thickness decreases, and the reynolds number drops. Then indeed the thicker flow will remain at high speed for longer.</p>
| 6190 | Does opening a tap in a sink affect the radial speed of the water around the jet? |
2015-11-17T19:11:38.180 | <p>I want to know why ceramic or disk capacitors are not rated directly like electrolyte ones? Is it just a convention or some other reason is behind this? Like, the ceramic ones are rated as 762k. then the value of the capacitance is 7600pF. (7-1st significant digit, 6-2nd significant digit and 2=multiplying factor(100))</p>
<p>But an electrolyte one is directly printed like:- 3400uF.</p>
<p><a href="https://i.stack.imgur.com/QEOvY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QEOvY.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Nn8U9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nn8U9.jpg" alt="enter image description here"></a></p>
| |electrical-engineering| | <p>There is not enough space.</p>
<p>But they do manage to find enough space on the bigger ceramic components though.</p>
| 6196 | Why Ceramic Capacitors Are Not Rated Directly Like Electrolytic Ones? |
2015-11-18T04:19:27.050 | <p>I was a doing a placement at a company, and they were designing a new heat pump. I was told the more compact you can make a heat-pump, the better it is for efficiency, and they mentioned something about this was because the power density was increased.</p>
<p>I get there are benefits from using less material and making something lighter which would improve efficiency in that regard, but I don't get the benefit of a higher power density in a heat pump. I understand that in a heat engine like a combustion engine you want a high compression ratio so you'd want to make the thing as compact as possible, but why is a better power density also good for a heat-pump? Is it this just the thing with all types of machinery which have some sort of volume which is compressed? If so Why?</p>
<p>Edit: I didn't really want to say COP but rather efficiency to keep the question more general. These guys seemed pretty certain that having a small high rpm heat pump gave improved performance, whilst I assume they meant COP could be something else they are looking at. Maybe a better way to ask this is: Are there any theoretical benefits to the COP of a heat pump by making it smaller?</p>
| |mechanical-engineering|thermodynamics|heat-transfer| | <p>As far as I know the COP of a heat pump doesnt vary greatly with the size of the unit only with the technology used, quality of components, and the correspondingly price. As a general engineering rule, its easier to make something cheaper/more effective when you have less constraints. So having no footprint constraint would generally make the unit cheaper/higher COP.</p>
<p>I do not work in HVAC, but generally a large building will invest in two large heat pump air handlers integrated into one system. They have two for redundancy purposes and no more than two because maintenance increases with increased number of units.</p>
<p>If the reliability of small units were very high, they could potentially be a better solution because of the elimination of the large ducting required. With current technology however, small units increase facility maintenance. Another aspect is control. One or two large units is easy to control. Many small units, all with different set points and pressure differentials can cause a lot of maintenance headaches.</p>
| 6203 | Are smaller heat-pumps more efficient? |
2015-11-18T13:38:15.003 | <p>I'm an undergraduate student of Electrical Power Engineering currently in 3rd semester, one of the courses we study is Digital Logic Design(dld) and I have found much interest in it but I am unclear whether the subject would have application in our field i.e. power engineering or not....Could anyone guide me that how should I study dld with what things in mind related to future perspective?</p>
| |electrical-engineering| | <p>Having studied electronic and electrical engineering, I covered both the power/analogue and the digital part. There were certain power designs where using a digital microcontroller (to generate a square wave of a certain time period and duty ratio) helped.</p>
<p>There are methods of obtaining oscillators using discrete analogue components (a Wien Bridge oscillator), but incorporating a 555 timer (digital) makes things much simpler. </p>
<p>Digital Electronics are used in systems control, so it still has relevance to eletrical engineering. </p>
<p>Digital Logic Design and VLSI technology were still my favourite topics as it shows how the theoretical boolean logic is implemented using hardware a nano-scale.</p>
| 6212 | Relation of Digital Logic Design to Electrical Power Engineering? |
2015-11-18T20:28:32.723 | <p>I have a problem in understanding of gears (spur gears). I designed some spur gears but I don't understand how the modulus works. I saw that this modulus is some standard thing and it's calculated by diametral pitch. Do I need to make this diametral pitch from standards, or could I make whatever I want? How does this work?</p>
| |gears| | <p>Whenever you choose to design standard gears, you typically have two main parameters: the number of teeth $N$ and the module, or modulus $m$. The module can generally be interpreted as the distance between the gear teeth. In fact, the module is defined as the circular pitch $p$ (the circumferential distance between the same point on two neighbouring teeth) divided by pi.</p>
<p>The reason the parameter 'module' is popular is because it allows you to work with rational numbers. You can calculate the gear's pitch diameter from the circular pitch $d=Np/\pi$, but this means either the pitch diameter or the circular pitch is an irrational number. It is not useful or practical to have either a standard set of values for gears (circular pitch) or the distance between gear shafts (from pitch diameters) be based on irrational numbers.</p>
<p>So instead, the module is preferred, and used in sets of standardised values (i.e. you are more likely find off the shelf gears of module like 4mm or 5mm than something like 4.123mm). You can calculate pitch diameter from the module via $d = Nm$ and all the variables can be rational numbers. For example, a gear of 16 teeth and module of 2.5mm turns out to have a pitch diameter of 40mm. All nice round numbers, easy to work with without having to bring out a calculator too often.</p>
<p>Because module can be interpreted as the distance between gear teeth, it is also important to note that only gears with the same module can mesh together. Also, in general, gears with a large module can withstand higher bending stresses at the root of teeth.</p>
| 6215 | Spur gears design |
2015-11-19T01:07:00.053 | <p>Are there any established solutions for truning a gear (or wheel) half-circle in one direction, back in the opposite direction, and repeating? Mechanism must be driven by a motor and continuous rotary motion. Compact and simple mechanisms preferred. Below are two possible solutions I dreamed up. Which one is better and why? Or what should I do instead?</p>
<p>One idea I had was to use a motorized disk (orange) with a slotted shaft(yellow) to create reciprocating motion. Then by extending the shaft and connecting it to a second disk (Red) I could achieve it, I think.</p>
<p><a href="https://i.stack.imgur.com/EgflS.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EgflS.gif" alt="enter image description here"></a></p>
<p>Another Idea I had was to use a full-square Scotch Yoke (Yellow) with gear teeth on the bottom. Properly sizing the connecting gear would allow for 180 degrees reciprocating motion.</p>
<p><a href="https://i.stack.imgur.com/mDWE7.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mDWE7.gif" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/DelpQ.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DelpQ.gif" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>Selection of a mechanism highly depends on the end application. Here are some ways to evaluate a mechanism vs your application:</p>
<ol>
<li>Input (360 deg rotation in this case)</li>
<li>Output (180 deg rotation motion in this case)</li>
<li>Relationship between input and output (mechanical advantage, velocity, and acceleration for a given position). (linear, parabolic, sinusoidal, other crazy polar math, etc)</li>
<li>Number of moving parts</li>
<li>Number of rolling surfaces (rollers, gears)(expensive)</li>
<li>Number of sliding surfaces (pins, slots)(reduced life)</li>
<li>Number of rotation points (bearing locations)</li>
<li>Can the mechanism be statically and dynamically balanced (for high speed)</li>
<li>What is the footprint of the mechanism</li>
<li>What loads does the mechanism have to endure (material required, fatigue stress, etc)</li>
<li>Backlash (new and worn)</li>
</ol>
<p>You may be able to use <a href="https://en.wikipedia.org/wiki/Four-bar_linkage#Classification" rel="nofollow">crank-rocker four bar linkage</a>. I haven't sketched it out, but 180 degrees is likely its theoretical maximum. A spring at either end of the range would help avoid top dead center, but power transfer at those high angles would be difficult. In either example you could replace the orange pin connection with a four-bar. That would eliminate one pin and slot giving you a longer service life.</p>
<p>Four bar linkages are attractive because they employ only low cost rotation connections that have long service lives. Rack and pinion is well suited for when low backlash linear output is required, but can be expensive. Slots and linear bearings are less desirable because of service life, but sometimes reduce the number of components or provide the desired output.</p>
| 6221 | Rotary Motion to 180 Degrees Reciprocating Mechanism |
2015-11-19T10:51:00.163 | <p>I'm well aware that there are many different factors to be considered, such as ease of installation/maintenance, conversion from AC to DC, environment, distance, speed, frequency of turns, and so forth.</p>
<p>Setting all of those aside for now, suppose we build a 10-mile long demonstration track with some hills and turns along the way and featuring both a third rail and an overhead cable. Suppose further that we have two trains (one equipped with a pantograph, the other with contact shoes). They have the exact same weight and are programmed to accelerate and decelerate at the same rate, and run at the same speed.</p>
<p>Which one will have consumed more electricity at the end of the ten-mile run?</p>
<p>Bonus Question: A third rail is a lot wider and thicker than an overhead wire. Is more electricity wasted that way? </p>
| |electrical-engineering|energy-efficiency|power| | <p>Your "Bonus question" holds the answer here. Assuming that the two trains are perfectly identical, and that the power consumption is identical (which, if you have the same weight, motor, and size, it will be), you are down to transmission loss.</p>
<p>The resistance of a conductor is R=L*Rho/A.
L=length
Rho=resistivity (material property)
A=cross sectional area</p>
<p>Assuming that L is the same for both systems, it comes down to cross-sectional area. The rail is larger, so less power is lost in transferring the power to the train.</p>
<p>P=I^2*R</p>
<p>P=power, I=current, R=resistance (from above)</p>
<p>So, in the act of transferring current to the train, power is lost. And more is lost in the lines/rails than in actually transferring to the train, and because the rails have lower resistance, they waste less power.</p>
<p>Note: Typically, when overhead lines are used to power trains, it is due to safety concerns. If people are around, and could potentially walk on the rails, like an above-ground, street level ("trolley") train, it makes a lot of sense to not electrocute them when they step on the rails.</p>
| 6223 | Third Rail vs Overhead Cable: which is more efficient in terms of electricity consumption? |
2015-11-19T19:15:26.153 | <p>I need to spin a threaded rod to raise a picture on a frame. The weight of the picture + the parts that it hangs on will be around 6 pounds. Two motors are going to be used to lift it via threaded rods (like in 3d printer platforms) but I need them to lift about 8" in less than 4 seconds.</p>
<p>I am unfamiliar with the differences between a DC and stepper motor and my worry is that when the picture is coming back down, a DC motor could overrun or would it not have enough power to strip the rod or nut (even over time)?</p>
<p>With a larger stepper, say like a 128oz/in and an M8x1.25 rod (.31" with 20 threads/in) is it reasonable for me to use steppers or should I go with DC and rely on timing to stop the lowering motion? </p>
| |mechanical-engineering|motors|stepper-motor| | <p>Direct answers to question:</p>
<ul>
<li><p>Whether a motor could overrun depends on motor power, momentum of the load, and resistance of the system. This could be a problem even with switches, but can be overcome with careful design.</p></li>
<li><p>Steppers are overkill if you need to move between two points without stopping between them. See below.</p></li>
</ul>
<p>What you need is a linear actuation system, translating rotational motion from a motor into linear motion.</p>
<p>Threaded rod can be used in principle, but it is not designed for linear actuation. It is designed for static structural applications. Threaded rod has sharp threads which can suffer damage and can wear quickly. The products of wear, and damaged (especially folded-over) threads can cause the actuator to seize during operation. Instead I recommend using something like an acme screw with appropriate nuts. The threads are broader and blunted, and the steels used are typically much harder than those of threaded rod with a smoother surface, all contributing to lower wear rates and less risk of damage, reducing the risk of seizing. Threaded rod also has finer pitch, resulting in less distance per rotation, and is less mechanically efficient than acme, lead, or ball screws.</p>
<p>The choice between a DC brushless motor and a stepper motor should be based on repeatability. A stepper is designed for ensuring that X number of electric power pulses translates into exactly Y distance, every time, within some tolerance, without any feedback mechanism. A brushless motor can not achieve the same repeatability without a tight feedback mechanism. If you don't need repeatability then a brushless motor is cheaper and easier to build working electronic systems for as it doesn't require pulse train generation, and operates with continuous DC power. I am assuming that you intend to raise/lower something between two positions and not stop anywhere in between.</p>
<p>Depending on momentum, braking is also a concern to ensure you don't overrun the limit switches or damage anything when stopping. You have a 6 pound mass, but you don't give a speed. If the speed is significant, then braking could be an issue. Otherwise, the limit switches could swap the DC motor from the power source to a bank of resistors tied back into the motor: a braking loop. The short-circuit from the motor to itself provides braking, and the resistors dump the energy as heat outside the motor.</p>
<p>If I have your scenario correct, then a brushless motor directly connected to an acme screw (or similar) with limit switches at each end, with a braking loop as required, is a simple design.</p>
<p>Proper sizing and quantitative design of each component of the overall system could each be its own question, if you need more information.</p>
| 6228 | High RPM - Stepper vs Brushless Motor |
2015-11-19T23:41:08.030 | <p>A wifi user is in a different room than the router. The computer is having a hard time connecting and receiving the wifi signal.</p>
<p>Can the wifi signal from the router to the computer be improved by opening a door to the room where the computer is?</p>
| |signal|wifi| | <p>I live in a big old 1920s built house. The bedroom is one floor above the router, and one room adjacent. The floors in my house are wooden, the walls are stone, the cielings are high and the doors are solid wood. If I try to stream video on Chromecast in my bedroom with the bedroom door closed, it intermittently buffers. Chromecast even grumbled that the wifi signal was weak and recommended using a different connection. (side note, but my phone picks up full wifi signal). If I open the bedroom door, and the kitchen door where the router lives, it NEVER buffers. I have watched whole films and shows without any issue.</p>
<p>Perhaps it is just a coincidence? I thought. But I have repeated this experiment on several different days, at different times of day, and it always yields the same result. Video can be streaming in HD beautifuly, if I go downstairs and close the kitchen door and then the bedroom door, within 5 seconds the video is buffering. I have even tried it with the doors closed over a good 30 minute period. I get small bursts of video followed by minutes of buffering.</p>
<p>So, the numbers shown above, although probably quite accurate and scientifically measured ....</p>
<p>"Generally, the amount of impedance added by the door is a negligible amount and would not be sufficient to noticeably improve the quality of the signal."
In my case it is not negligible, and makes the difference between watching video, and watching a spinning icon.</p>
<p>Quite often a real world experiment will give different results to the science backing the accepted answer. This happens all the time, isn't engineering great.</p>
| 6230 | Can wifi signal reception be improved by opening a door? |
2015-11-20T15:14:08.613 | <p>I have a rotary pump and want to gain knowledge about its characteristics. I have an uncalibrated flow meter. At best I want to get a flowrate/pressure-diagram. </p>
<p>I understand, I need to analyse the flow sensor in advance.</p>
<p>Things I have:</p>
<ul>
<li>Voltage and current measurement equipment</li>
<li>Counter and/or microcontroller for evaluating rotary sensor pulses</li>
<li>oscilloscope</li>
<li>vast set of tubing and fittings which can be connected to the pump and the flow sensor.</li>
<li>graduated beaker</li>
</ul>
<p>Things I don't have:</p>
<ul>
<li>Pressure sensors</li>
<li>calibrated sensors</li>
<li>calibrated pumps.</li>
</ul>
<p>My questions:</p>
<ul>
<li>Can I gain pump characteristics without usage of pressure sensors?</li>
<li>How can I calibrate the flow sensor without a calibrated source of flow rate?</li>
<li>What is a setup for getting the full Q/P-Diagram without?</li>
</ul>
| |mechanical-engineering|pumps| | <p>I am working under the assumption that we are talking about a small pump with a flow rate appropriate to measure using a graduated cylender. It's a bit tricky with the tools described but I would make a vertical column using your vast set of tubing as a home made pressure sensor. Also you can use a stop watch and your graduated cylender to calibrate your flow sensor. Use the cross sectional area and mark the pressure required to push the fluid to various heights. </p>
<p>Run the pump and see how high it can push the column. This will give the maximum pressure. Turn off the pump, cut the tubing at a known height and measure flow rate using your freshly calibrated flow sensor. Calculate your pressure output using head and friction losses at the known flow rate. Repeat for various heights of column to create the various data points on your Q/P curve.</p>
<p>For large pumps this will be impractical.</p>
| 6238 | Method for determining pump characteristics |
2015-11-20T16:27:47.437 | <p>My father-in-law would like to install a hinged stair system in his garage that allows him comfortable access to his attic space.</p>
<p>He wants the stairs to swing up flush with the ceiling when not in use, and then to easily swing down when he needs it, and plans on using a set of pulley and weights as a counterbalance to make raising and lowering the stairs easy.</p>
<p>I am thinking of the stairs as a "lever", where the lever itself is the "load" (i.e. the load is distributed). If he places the ropes in the middle of the stair system, then obviously the "effort" is in the middle, while the fulcrum will always be at an extreme end.</p>
<p>My confusion is, since the load is distributed evenly, is this a second class lever or a third class lever? Really I don't care what class it is, but it would be nice to understand how to roughly calculate the weight required to lift the stair system. :)</p>
<p>Obviously we can just install it and keep adding weight to see what works in order to solve the problem, but I'd like to rough out some calculations first to see if there are any gotchas.</p>
<p>I found this <a href="https://engineering.stackexchange.com/questions/2114/how-to-calculate-lever-force-when-lever-has-uniformed-distributed-load">link</a> to a related Question, but this situation is just different enough that I can't tell if it solves my problem?</p>
<p>Any hints on what the equation(s) would look like to calculate the required weight (in pounds) to lift the stair system that weighs, say, $L$ pounds, and the ropes are $d$ distance from the hinge?</p>
| |applied-mechanics|statics| | <p>The answer for this will depend on the location of where you attach your counterweight rope, but the generic solution is as follows:</p>
<p>Suppose you have a staircase of weight $W$. It can be assumed that this weight is centered at the midpoint of the staircase, or $\frac{L}{2}$.</p>
<p>Additionally, suppose you have a countering force of $F$ supplied by your counterweight, at a location of $d$ from the hinge.</p>
<p><a href="https://i.stack.imgur.com/zgqIh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zgqIh.png" alt="FBD"></a></p>
<p>This is a simple balance of moments about the hinge of the staircase, where</p>
<p>$$\frac{WL}{2} = Fd$$</p>
<p>You should know everything in the equation except for $F$. Solving for $F$, you find that</p>
<p>$$F = \frac{WL}{2d}.$$</p>
<p>You will see that the closer you attach the counterweight rope to the hinge, the larger the countering force you will need. Just think of a teeter-totter on a playground - the heavier kid always wins.</p>
<hr>
<p>It should be evident, but make sure that you use consistent units when solving this equality (i.e., don't mix feet and inches, etc.). Also, you will want your counterweight slightly heavier than what's required so that it keeps the staircase closed without any outside effort.</p>
<p>If you're looking for a cheap and effective counterweight, drywall screws with sand filling in between works quite well. A concrete block works fine too but requires more preparation and mess.</p>
| 6239 | Calculate effort to raise hinged lever with distributed load |
2015-11-23T01:15:17.043 | <p>I've see the same lids every day but I never really thought about their construction. There is an indent in a "Solo Travel Lid" for coffee cups that is just above the hole that you drink the coffee from.</p>
<p><a href="https://i.stack.imgur.com/cHn26.png" rel="noreferrer"><img src="https://i.stack.imgur.com/cHn26.png" alt="enter image description here" /></a></p>
<p>You can see the crescent shaped indent in the image. What is it for? Does it somehow increase fluid flow? Is it just a good spot for your upper lip? If so, why is there not a matching one for the nose? I tried googling this but I got no answer.</p>
| |design|applied-mechanics| | <p>From the "Solo Traveler" <a href="http://www.google.com/patents/US4589569" rel="nofollow noreferrer">patent</a>:</p>
<blockquote>
<p>A more specific object of the present invention is to provide a lid having an opening formed therethrough to enable drinking, and having a recess formed in the lid adjacent the opening to accommodate the upper lip of one drinking from the cup</p>
</blockquote>
<p>This is also illustrated in the patent diagrams.
<a href="https://i.stack.imgur.com/K5c8M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K5c8M.png" alt="Patent Image"></a></p>
<p>I came across this <a href="http://www.theatlantic.com/technology/archive/2011/05/the-rise-of-the-plastic-disposable-coffee-cup-lid/238573/" rel="nofollow noreferrer">interesting article</a> on the history, use, design, and styles of disposable lids. The Solo Traveler is prominently featured for its excellent aesthetics and design . The article reiterates that:</p>
<blockquote>
<p>the Solo Traveler lid was designed to accommodate the nose and lip of a drinker. In accomplishing this design goal, the necessary height of the lid made it useful for foam-topped gourmet coffees.</p>
</blockquote>
<p>Additionally, when upright, the recess acts as a reservoir (complete with drainhole) to catch drips/spills/splashes that might otherwise run down the cup and create the infamous coffee stain ring.</p>
| 6257 | From an engineering standpoint, what is the purpose of the indent in a coffee lid? |
2015-11-23T18:07:37.427 | <p>When you compress a bicycle pump and seal the opening, the pump will automatically return to it's initial position when you release it. I suppose this is how air suspension works as well?</p>
<p><a href="https://i.stack.imgur.com/durma.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/durma.png" alt="enter image description here"></a></p>
<p>if you have a internal combustion engine without any gas and turn it over - for instance with a kickstarter, until the piston reaches the very top on the compression cycle - why doesn't it return to the bottom by itself by the air compression alone? Is there anything blocking this, or is the air simply not powerful enough to turn the engine over?</p>
<p>See comments from <a href="https://physics.stackexchange.com/questions/220153/turning-an-engine-over-with-engine-compression">Physics StackExchange</a></p>
| |pressure|energy|compressed-air| | <p>There is another reason for this in multi-cylinder engines. While one cylinder is compressing the air, another is decompressing it. The net result is that the torque as a function of position gets smoother as the number of cylinders is increased.</p>
| 6268 | Turning an engine over with engine compression |
2015-11-25T00:16:19.183 | <p>I am trying to bore out a red brass pipe from 3/8" to 10mm using my drill press and a standard twist HSS bit. Running the drill at 600rpm I can't get more than half an inch in before the drill starts binding. I tried running first with cutting oil, and then, thinking maybe the oil wasn't sticky/viscous enough, tried with Permatex silver, which only got me a few mm further before binding.</p>
<p>My guess is that I need some sort of metal-cutting bit that has a narrower shank so that there isn't friction except where it's actively cutting. If so, what are such bits known as?</p>
<p>Is there any other practice I might be missing to make this work?</p>
| |mechanical-engineering|machining|drilling|brass| | <p>Reamers mentioned in an ericnutsch's answer are good solution (remember that these reamers are designed to increase diameter of existing aperture up to 0,5mm), but may be the better solution is to use senker.
But seems your problem is not in using wrong reamer type.
I would suggest you to check the straightness of your bit and drill. May be a drill or a bit have some wobbling due to overheating or overloading in past, also your drill motor must have enough power (rpm is not a major parameter). Lack of power may cause the same effect with using senker or drill bit.</p>
| 6284 | How to drill larger bore in pipe? |
2015-11-25T21:15:02.287 | <p>My goal is to have an array of LEDs arranged in such a way that I can light the view of my action cam. My action cam has a horizontal field of view of 175°.</p>
<p>My first idea was to determine the angle of light emission of a given LED and stack them next to each other, each time being rotated by their angle of light emission around each LEDs vertical axis. So if I have an angle of light emission of 15°, I need equally spaced 175 / 15 = 12 LEDs. To make mounting easier, I could create an element that is shaped in a circular way, so I don't have to calculate each rotation but only watch out for the same distances between the LEDs.</p>
<p>Then I would do the same for the vertical axis.</p>
<p>Is this the best approach or is there a better approach?</p>
| |mechanical-engineering|electrical-engineering|design| | <p>Instead of generating a perfect set of lights, which may be expensive, it should be possible to correct some uneven illumination via software post-processing.</p>
<p>One solution is to generate a static, white-field exposure of the space you wish to image. Essentially, drape the area you wish to image with something as close to perfectly white as possible, and then capture an image of the area with the same illumination setup you will use on whatever you wish to capture. The drape should be white and matte, i.e. not have specular (mirror-like) reflection, but instead have diffuse reflection, like paper or a cloth. Once you've captured the static, white-field image, you have captured all of the uneven-ness of your illumination setup. You can then subtract the uneven-ness from any future images to correct for uneven illumination.</p>
<p>A drawback to this method is that it works best with flat objects, as the illumination uneven-ness may change if a 3D object is inserted into the frame.</p>
<p><a href="http://www.mathworks.com/help/images/examples/correcting-nonuniform-illumination.html" rel="nofollow">This Matlab tutorial</a> gives a general idea of one way of correcting uneven illumination. Their method involves determining background illumination via software (which you can also do). Since you have access to your own hardware, you may also capture the background illumination as discussed above, and follow the remainder of the procedure.</p>
<p>Ideally, non-uniformities in illumination should still be minimized so as to ensure you capture as much information as possible through a wide-spread histogram distribution. If the illumination non-uniformities dominate the overall image lightness, then you aren't capturing as much lightness information from the objects in the frame.</p>
<p>Photography.SE may also be a useful source of information.</p>
| 6304 | Equal illumination of an actioncam's field of view |
2015-11-27T06:01:27.993 | <p>I posted this question on <a href="https://electronics.stackexchange.com/questions/202839/three-heaters-on-single-pcb">electronics SE</a>, but was suggested that I post it here.</p>
<p>I basically having three heaters (serpentine pattern) on a PCB (fiberglass board), as the following image shows:<a href="https://i.stack.imgur.com/OiMs1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OiMs1.jpg" alt="PCB Heater"></a></p>
<p>The ones at the end would be heated to 95°C, the ones next to it at about 50-55°C, while the middle one to about 72°C. They are spaced 0.3mm apart.</p>
<p>The problem is because the 55°C heater is between heaters with higher temperatures, the heat gets transferred onto that area and goes well beyond 60°C, which I'm not able to control.</p>
<p>I tried removing the area between the heaters to see if it would help somehow, but the problem persists. </p>
<p>What can I do to "confine" the temperature to the heater area? </p>
| |electrical-engineering|materials|heat-transfer|temperature|power| | <p>Well, air has <a href="https://en.wikipedia.org/wiki/List_of_thermal_conductivities" rel="nofollow">worse thermal conductivity than practically any solid</a>, so chances are that a hole will decrease the heat transmission. There are however solids with thermal conductivity very close to that of air, so if that board was made (on purpose) from one of those... you won't see much improvement for your hassle.</p>
<p>And since you mentioned fiberglass while I typing this (doesn't quite look a common FR4 I have to say), then you might hope for halving of the conductivity given that Wikipedia give 0.024 Wm<sup>-1</sup>K<sup>-1</sup> for air and 0.045 for fiberglass. I suspect with fiberglass though there can be a lot of variation between between various flavors because it's porous.</p>
<p>According to <a href="http://www.ewh.ieee.org/soc/cpmt/presentations/cpmt0412b.pdf" rel="nofollow">this presentation</a> thermal conductivity of FR4 is a lot better at 0.25 W/m·K (5x) than that of pure fibeglass given by Wikipedia. (And actually that presentation agrees with <a href="https://en.wikipedia.org/wiki/FR-4#Properties" rel="nofollow">https://en.wikipedia.org/wiki/FR-4#Properties</a> but on the through-plane value). In-plane value is even greater there at 0.81 W/m·K to 1.059 W/m·K. I'm guessing the epoxy binder in FR4 contributes a lot to that. So on conducted heat alone you could hope for 10x (or even 40x) improvement in thermal isolation a with hole then. But then there's also radiated heat and depending how the board is oriented convection may contribute too. There are some examples/caculations on the web for radiative heat transfer between some simple geometries, e.g. <a href="http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html#SECTION07541000000000000000" rel="nofollow">here</a> but I couldn't find something for yours.</p>
<p>There is <a href="https://www.mentor.com/pcb/hyperlynx/thermal/" rel="nofollow">software that can do this sort of complex thermal simulation</a>, alas I don't know any free&good such, and it depends of course on having good material data. Actually, of the freebie stuff <a href="http://poweresim.com/newhelp/EN/thermal.jsp" rel="nofollow">http://poweresim.com/newhelp/EN/thermal.jsp</a> might help to some extent, but it seems it can only do some chunky blocks.</p>
<p>Actually, one thing you could do to estimate this is to position a termocouple in air at the same distance form the hot traces as the cold[er] traces are. And then measure what this registers. I suspect this would be a resonable approximation for what a hole could hope to achieve (ignoring he self-heating of the cold[er] traces).</p>
| 6323 | Having three heaters on single PCB |
2015-11-27T13:28:46.553 | <p>This might be a stupid question, but I'm new to mechanics, so please bear with me. Suppose I have a compound gearing system defined in the image below (not to scale). </p>
<p><a href="https://i.stack.imgur.com/ZAqiM.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ZAqiM.jpg" alt="compound gear system"></a></p>
<p>The green gear is the driving gear, and the blue gear is a compound gear. </p>
<p>Suppose I have a driving torque of 10kg.cm, and the ratio is 1:6, this means the output torque is 60kg.cm, correct?</p>
<p>At which point is this torque increase? Is it on the outer edge of the smaller gear (A), or larger gear (B)?</p>
| |mechanical-engineering|gears|applied-mechanics|torque| | <p>I think that the most useful answer to your question is that, in practice we are most concerned about the relative torque between two linked shafts. </p>
<p>To put it another way the forces on gears are only of interest to people who design gears and if you are designing a gearbox it's the ration of torque/RPM which is of primary interest. </p>
<p>In your example shaft Shaft B will see 6x the torque of shaft A and rotate at 1/6 RPM. At this point the diameter of the smaller gear on shaft B is irrelevant until we know what it is coupled to. </p>
<p>Having said that you are correct in saying that the ratio of torque between shaft A and shaft B is 1:6 and the ratio of angular velocity(RPM) is 6:1 ie shaft B has 6 time more torque and rotates at 1/6th of the speed. </p>
<p>For it to be true compound gear train you also need a shaft C coupled t the smaller gear on shaft B. </p>
<p>If the gear on shaft C was 6cm in diameter then you would have a final drive ratio of 1:(6x6) ie 1:36 </p>
| 6326 | Compound gear system ratio |
2015-11-28T23:05:42.363 | <p>Since all it really is is a heating element, what happens if I use a 240v/50Hz clothes iron with a 110v/60Hz supply?</p>
| |electrical-engineering| | <p>In general, anything with a heating element will produce only 1/4 the heat on 1/2 the voltage, since power is proportional to voltage squared.</p>
<p>In other words, you'll probably be disappointed with the results.</p>
| 6343 | What happens if I use a 240v clothes iron in USA? |