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2016-12-07T14:27:28.150 | <p>I am looking for making a <strong>chip</strong> or a <strong>microchip</strong>, <em>which can be traced with the help of the satellites</em>. Means it is merely a chip and when I tune my satellite to a particular frequency I would get to know that this particular chip is present here and so. But it should be a chip only means no battery and all i.e. <em>no other circuitry</em>. I am thinking so if it is possible now and if not please suggest something similar to it so that I can use that in the similar way.
Thanks</p>
| |electrical-engineering|telecommunication|wireless-communication|communication| | <p>Firstly you seem to be under the common misunderstanding that the GPS satellites somehow calculate where a GPS receiver is. They don't. The satellites are transmit only systems (ignoring the command and control radio links with their control stations) and a GPS receiver is a receive only system.</p>
<p>As user114749 indicated RFID is the closest to what you want and even that doesn't quite fit the requirements, it still needs an antenna.</p>
<p>A 900MHz RFID tag doesn't actually transmit anything, it's not picking up anywhere near enough power to run a transmitter, it works by modulating the reflections of the readers signal bouncing off its antenna.<br>
A good tag can have a read range of ~20 meters when running the transmitter at 1W with a moderately directional antenna. Radio follows the inverse square rule, if you double the range you need 4 times the power (a 6 dB increase). But with RFID since the signal has to reflect back to the transmitter, you lose 6 dB getting to the tag and another 6 dB getting back to the transmitter meaning 12 dB of losses for every doubling of the range.<br>
The lowest possible practical orbit is around 350km (the fuel cost to maintain a stable orbit at this altitude is very high but it's possible). So when your satellite is directly over the tag you are trying to read we have log(350km/20m) / log(2) doublings of our 20 meter range. That works out as just over 14.<br>
Which means we need 14 x 12 dB = 168 dB more power needed at the transmitter over our 1 W RFID reader.<br>
So in order to pick up a tag that is directly under it we need our RFID reading satellite to broadcast with around 63,000 TW of power, about 2.5 times the total worlds power output.</p>
<p>To locate a tag you'd need a range of several times this (add an extra 0 or two to the power number), you can't fly directly over every point on the earth, and a fleet of satellites to give reasonable coverage. But those are relatively small issues in comparison to getting it to work with a single satellite to start with. </p>
<p>These numbers may be off by a little bit. In fact they could well be off by a lot. But unless I've gotten it wrong by a factor of a few million it's not even close to being practical.</p>
| 12720 | How can I make a GPS traceable microchip? |
2016-12-07T17:24:50.653 | <p>According to ASME B31.8-2008 to calculate minimum thickness if a pipe under internal pressure you use:
$$t=\dfrac{PD}{2(SEW+PY)}$$</p>
<p>I have seen this formula re-written in terms of pressure of the fluid within the pipe as follows:</p>
<p>$$p(psig)=\dfrac{2t \cdot S \cdot E}{D-2 \cdot t \cdot Y}$$</p>
<p>Assuming that the reduction factor $W=1$, how do they get to the rest of the formula?</p>
| |mechanical-engineering| | <p>Getting from one form of the equation to the other involves a little bit of algebraic manipulation and factorisation:</p>
<p>First, assuming $W=1$, multiply both sides of the equation by $2(SE + PY)$:</p>
<p>$$2tSE + 2tPY = PD$$</p>
<p>Then, bring terms that have $P$ in them to one side and factorise:</p>
<p>$$P(D-2tY) = 2tSE$$</p>
<p>Finally, divide by $(D-2tY)$ on both sides:</p>
<p>$$P = \frac{2tSE}{D-2tSE}$$</p>
<p>So, for a given pressure, a minimum thickness is specified. So intuitively, for a given thickness, a maximum pressure should be specified. Therefore, minimum thickness $t$ can be substituted with "given thickness", and simultaneously the given pressure $P$ can be substituted with "maximum pressure", hence giving you the second equation.</p>
<p>A very similar approach is used to obtain the first equation from the second.</p>
| 12723 | Max. Allowable pressure in process pipe |
2016-12-09T12:54:29.687 | <p>In regulation, if we represent a system with a state space model, we end with three matrices A, B and C (without taking into account feed-forward part) that define a specific part of the following equation:</p>
<p>$\dot{x}=Ax+Bu$</p>
<p>$y=Cx$</p>
<p>Is it also possible to define the transfer function $G=\frac{N}{D}=C(sI-A)^{-1}B$?</p>
<p>If we want to find the poles, we search when the denominator ($D$) is equal 0 and for the zeroes when the numerator ($N$) is 0.</p>
<p>My question is, if we want to define which part of the state space model influence whether the poles or the zeroes. I would say by looking at the previous equation that $A$ is the only one to influence the poles and $B,C$ act on the zeroes.</p>
<p>Unfortunately, the corrections made to my exam gave a different answer: $C$ influences the poles and all the matrices the zeroes.</p>
<p>I don't understand, why is that the case?</p>
| |control-engineering|transfer-function| | <p>Half of your statement is correct: If we are considering the following system:</p>
<p><span class="math-container">$$G : \begin{aligned} \dot{x} &= Ax+Bu \\ y &= Cx \end{aligned}$$</span></p>
<p>then <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, and <span class="math-container">$C$</span> all influence the zeros of <span class="math-container">$G$</span>, but <span class="math-container">$C$</span> certainly does not affect the poles. The poles <span class="math-container">$G$</span> are the eigenvalues of <span class="math-container">$A$</span>.</p>
<p>This makes sense from a purely common-sense consideration: the matrix <span class="math-container">$C$</span> is the output matrix, it is only capable of scaling and adding the states to result in an output. It has no effect on the states themselves, therefore it won't affect things like stability of <span class="math-container">$G$</span> (which is related to the poles).</p>
<p>You can easily see this if you look at the transfer function equation you provided, with the inverse expressed as the determinant and the adjunct (or adjugate):</p>
<p><span class="math-container">$G(s) = C(sI-A)^{-1}B = \frac{C \text{adj}(sI-A) B}{\text{det}(sI-A)}$</span></p>
<p><span class="math-container">$C$</span>, <span class="math-container">$B$</span>, and the adjunct of <span class="math-container">$A$</span> all appear in the numerator, but only the determinant of <span class="math-container">$A$</span> appears in the denominator.</p>
<h1>However...</h1>
<p>Here's where your problem is likely coming from: If we consider <span class="math-container">$G$</span> to be a system plant being controlled through some feedback element with a transfer function <span class="math-container">$H(s)$</span> (like the following picture from <em>Modern Control Engineering</em>, 5th ed. by Ogata):</p>
<p><a href="https://i.stack.imgur.com/rOUoE.png" rel="noreferrer"><img src="https://i.stack.imgur.com/rOUoE.png" alt="Image of a feedback system" /></a></p>
<p>Then the overall transfer function for the entire system, from reference input <span class="math-container">$r$</span> to output <span class="math-container">$c$</span> is</p>
<p><span class="math-container">$$\frac{C(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)}$$</span></p>
<p>Now, since <span class="math-container">$G(s)$</span> is in both the numerator and the denominator, you can see that <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, and <span class="math-container">$C$</span> affect both the poles and zeros of the <strong>closed-loop</strong> transfer function.</p>
<p>In conclusion:</p>
<ul>
<li>Only <span class="math-container">$A$</span> affects the poles of the system plant <span class="math-container">$G$</span></li>
<li><span class="math-container">$A$</span>, <span class="math-container">$B$</span>, and <span class="math-container">$C$</span> affect the zeroes of the system plant <span class="math-container">$G$</span></li>
<li><span class="math-container">$A$</span>, <span class="math-container">$B$</span>, and <span class="math-container">$C$</span> affect both the poles <strong>and</strong> zeroes of the closed loop system</li>
</ul>
<p>Reference: <em>Modern Control Engineering</em>, 5th ed. (2009) by Katsuhiko Ogata</p>
| 12755 | Which equations of a state space model influence the poles and zeroes? |
2016-12-09T14:28:38.143 | <p>GPS receivers are able to operate at signal level of -150 dbm or less, but mobile phones need far more than -115 dbm level signal. Why GPS receivers are so much more sensitive than GSM or WiFi receivers? See these receivers </p>
<ol>
<li><a href="http://www.quectel.com/product/prodetail.aspx?id=77" rel="nofollow noreferrer">GNSS L86</a></li>
<li><a href="http://www.quectel.com/product/prodetail.aspx?id=93" rel="nofollow noreferrer">Wi-Fi FC10</a></li>
</ol>
| |electrical-engineering|wireless-communication| | <p>There are a few factors at play.</p>
<p>Firstly as Olin indicated the signal bandwidth and data rates are different.</p>
<p>It is a fundamental rule of information theory that the maximum amount of data you can pass through a data channel depends upon the channel bandwidth and the signal to noise ratio. See <a href="https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem" rel="nofollow noreferrer">the Shannon-Hartley theory</a> for details. </p>
<p>This means that if you have a lower data rate or a wider channel you can cope with a higher signal to noise ratio.</p>
<p>The GPS navigation message sends data at 50 bits per second over a channel that is around 20MHz wide. WiFi 802.11n tries to get 72Mb/s or around 1 million times more data over a channel that is 20MHz wide.</p>
<p>This would indicate that GPS can cope with significantly lower signal levels than WiFi</p>
<p>But that only tells you what is theoretically possible, not how to achieve it.</p>
<p>A GPS signal is very predictable. Each satellite is broadcasting a pesudo random data sequence which repeats every millisecond, the data is then added on top of that every 20ms. The GPS reciever knows what this pattern is and looks for it in the background noise. It does this by making a guess at the timing of the signal and then moving this expected timing around until the correlation between the expected and recieved signals is maximised. By doing this the GPS reciever can effectivly find a signal that is weaker than the random background noise.</p>
<p>This has the added advantage that since the signal is random it looks like noise unless you are looking for it. This means that all the satellites can broadcast on the same frequency at the same time without interfering with each other.</p>
<p>Another advantage of this system is that the primary information of interest to a GPS reciever is the exact timing of the recieved signals. By maximising the correlation between the expected and actual signals you get a very robust and accurate way of determining the exact signal timing.</p>
| 12756 | Why GPS receivers are more sensitive than GSM or WiFi receivers? |
2016-12-09T17:06:08.103 | <p>I just finished my first controls course (on linear control systems) and we learned about 3 types of controllers: PID, State Space, and Frequency Response/Loopshaping. From what I understand, PID has the advantages of being really easy and it can be applied without knowing the dynamics of the system. State space provides really good responses. But loopshaping seemed very guess and check when we were doing it and kind of arbitrary. In addition, we got to actually implement the first two on a multirotor for our labs and didn't have time to do the last beyond homework. So when would it be advantageous to use loopshaping?</p>
| |control-engineering|control-theory| | <p>When designing a controller the first priority should always be stability and then performance/robustness.</p>
<p>I do not know what method you used for tuning a PID controller, but I would suspect that it might not be easy to proof stability. Increasing performance would then probably involve increasing the gains. And it is probably hard to make a PID controller robust.</p>
<p>For state space controller design you can look at the closed loop dynamics (matrix) and check whether it is stable (Hurwitz for continues time and Schur for discrete time). For performance you can try to place the closed loop poles as far as possible into the left half plane. I haven't looked at it, but there will probably be methods for robustness in state space. For disturbance rejection you can find a state estimator using a Kalman filter (by solving a Riccati equation).</p>
<p>For loopshaping you can proof stability using a <a href="https://en.wikipedia.org/wiki/Nyquist_plot" rel="nofollow noreferrer">Nyquist plot</a> using the <a href="https://en.wikipedia.org/wiki/Nyquist_stability_criterion" rel="nofollow noreferrer">Nyquist stability criterion</a>. For performance you can just increase your bandwidth (sometimes defined at the point where the open loop crosses the 0 dB). Gain and phase margin can be used to see how stable/robust your closed loop system is. Closely related to this (but not often mentioned) is the modulus margin, which is basically the shortest distance to the minus one point in the Nyquist plot. This inversely relates to the highest peak of the sensitivity function of the closed loop system, which is a measure of the worst disturbance rejection. There are also other methods for robustness, such as H infinity.</p>
<p>So a PID controller might be the easiest, but it is much less advanced. The advantage of state space control is that it can also be applied to MIMO systems, which is harder to do for loopshaping. But the advantage of loopshaping is that you can just measure a frequency response function and design a controller for that directly, while for state space you first have to fit a model onto it first. Also if your system is continues, but also has delays it is harder to capture that into a state space model, but for loopshaping this does not change anything.</p>
| 12762 | What benefits does loopshaping have over other control design methods? |
2016-12-10T17:40:33.553 | <p>Picture shows the electrical system. Analytically determine and draw system response $u_2(t)$ for the step input shown (use Laplace transform) (tp>>0).
<a href="https://i.stack.imgur.com/fWTQe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fWTQe.jpg" alt=""></a></p>
<p>So, I know to solve RC system transfer function and get G(s)=1/(RCs+1), but I don't understand how to do this problem, we have never done it in class and professor often asks this question. If anyone could help me out, thank you!</p>
| |electrical-engineering|control-engineering|control-theory| | <p>Because this is a linear and time-invariant system, you can use the principle of superposition. </p>
<p>If the input to $u_i(t)$ is $y_i(t)$, then </p>
<ul>
<li>The response to $\alpha u_i(t) $ is $\alpha y_i(t) $</li>
<li>The response of $u_i(t)+u_j(t)$ is $y_i(t)+y_j(t)$</li>
<li>The response of $u(t-t_p)$ is $y(t-t_p)$</li>
</ul>
<p>The response of the system to a unit step input $\theta (t)$ is $1-e^{-\frac{t}{R C}}$. (I assume you have no problem with this.)</p>
<p>The input to the system can be split into two. </p>
<ul>
<li>$5 \theta (t)$ when $0\leq t\leq t_p$. The response of the system to this input is $5(1-e^{-\frac{t}{R C}})$.</li>
<li>$5 \theta (t)-\theta(t-t_p)$ when $t_p < t$. The response is $5(1-e^{-\frac{t}{R C}})-(1-e^{-\frac{t-\text{tp}}{R C}})$.</li>
</ul>
<p>The final response is a sum of these two piecewise functions.</p>
<p>I have plotted the result for $R=1$, $C=1$ and various values of $t_p$. The dashed line shows the response if the second step was not applied. </p>
<p><a href="https://i.stack.imgur.com/0E7rT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0E7rT.png" alt="enter image description here"></a></p>
| 12769 | Analytically determine and draw the system response to a piecewise step input for an electrical system |
2016-12-11T01:06:45.827 | <p>It seems that most bridge are composed of <a href="http://www.nvfnorden.org/library/Files/Utskott-och-tema/Broar/Conference-2014/05_NVF%20Iceland%202014.pdf" rel="nofollow noreferrer">concrete</a>, e.g in Latvia</p>
<p><a href="https://i.stack.imgur.com/nabm8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nabm8.png" alt="enter image description here"></a></p>
<p>However, concrete encompases a lot of bridge types:</p>
<blockquote>
<p>Most concretes used are lime-based concretes such as Portland cement
concrete or concretes made with other hydraulic cements, such as
ciment fondu. However, asphalt concrete, which is frequently used for
road surfaces, is also a type of concrete, where the cement material
is bitumen, and polymer concretes are sometimes used where the
cementing material is a polymer.</p>
</blockquote>
<p>Which concrete material do modern civil engineers prefer when designing bridges over motrways less than 30 meters above the underpass?</p>
| |civil-engineering|bridges|concrete| | <p>In structural situations, <a href="https://en.wikipedia.org/wiki/Concrete" rel="nofollow noreferrer">portland cement concrete</a> is what is used.</p>
<p>The properties of <a href="https://en.wikipedia.org/wiki/Asphalt_concrete" rel="nofollow noreferrer">asphalt</a> change with temperature and it never really "sets". It is a viscous liquid that cannot support long-term loads.</p>
<p>There really shouldn't be any confusion as to what is being described in that chart as "concrete".</p>
| 12773 | What is the preferred composition for modern concrete bridges? |
2016-12-11T17:17:20.580 | <p>I'll soon get my engineering diploma (~MSc, from a "renowned-but-not-in-the-top-3" so-called french <em>grande école</em>). The degree itself is not specialized, but I mainly studied mechanical and industrial engineering.<br/>
I aim to work in the industry – or, in other words, I don't want to make an academic career... even if I am still not sure, what position I will look for (R&D, project management, or maybe even entrepreneurship). I'd be ok to work for some years (4-12 years) abroad (Europe of further), but I think that I'd like to come back to France at some point.</p>
<p>So I'm now wondering if doing a PhD in France (i.e. during 3 years) would be worth it, regarding my estimated career plan.<br/>
Question: What would then be the <strong>pros and cons I should consider</strong> before choosing to do (or not) a PhD?</p>
<hr>
<p><sub>This question have been asked on Academia.SE. According to comments/advices there, I do ask it again here.<br/> I wanted to add these tags, but I don't have enough rep on this site to do it: <code>job</code>, <code>career</code>, <code>phd</code>.</sub></p>
| |mechanical-engineering|industrial-engineering| | <p>A PhD is specifically geared towards academic research and so if you are working in a more 'pure' field like maths or physics it is certainly a good thing. However engineering is not by any means a pure academic field and so there is a reasonable argument that industrial experience is at least as valuable. </p>
<p>Equally PhD research tends to be quiet niche and you may be limited in what you can research by the facilities of your chosen institution and the interests of the available academic supervisors. </p>
<p>If you are unsure it may not be bad advice to look at working industry for a year or so and then evaluating your options. </p>
<p>One thing about engineering is that it is always very useful to have some experience of getting your hands dirty at actually making things. Ultimately engineering is about pragmatic problem solving, something which academic qualifications are not necessarily all that great at teaching. </p>
<p>There is also a potential problem that if you have a PhD you end up overqualified for junior positions but don't have the specific knowledge and experience required for more senior ones, especially if you don't have a very strong interest in a particular field. </p>
<p>There is also an argument that a year spent working in a foundry, fabrication or machine shop will give you an insight into the real world practicalities of engineering which you will not get elsewhere. </p>
| 12788 | Young engineer wanting to work in the industry: is PhD worth it? |
2016-12-11T18:28:36.543 | <p>Can I inject plastic in a RTV mould?
Will the interior of the mould get deformed by pressure?</p>
| |pressure|plastic|deformation| | <p>RVT silicone is commonly rated by Shore Hardness. some grades are hard and heat resistant enough to be used for centrifugal casting of low temperature metal alloys (aka white metal). </p>
<p>However in practice for plastic casting they are best suited to 'cold cast' resins such as Polyester, Epoxy and especially Polyurethane. </p>
<p>Generally injection moulding is specified for high volume production using rigid steel dies and for low to medium volume production filled resins are more practical and can provide equivalent or better mechanical properties. </p>
<p>If you can provide more details of your application we can give you more specific advice.</p>
<p>You can certainly do very good quality castings, in terms of both detail and mechanical strength with gravity casting of PU resins in RTV rubber moulds in moderate quantity without needing injection moulding. </p>
| 12791 | Can I inject plastic in a RTV mould? |
2016-12-12T06:15:27.237 | <p>I am wondering how to determine screw size for a system.</p>
<p>Here is the attached pictures of an example system.
<a href="https://i.stack.imgur.com/xmNSz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xmNSz.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/M3hC2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M3hC2.png" alt="enter image description here"></a></p>
<p>Lets say the mass of object, m = 10kg
<br>distance, x = 0.5m, x2=0.5m</p>
<p>with moment = 0,
<br>0 = F*1.0m - 100N*0.5m
<br>F = 50N as the screw clamping force</p>
<p>is the calculation to determine required clamping force correct?
<br>If its correct, how to determine the required size and number of screw?
(FYI, socket head cap screw is used) </p>
| |mechanical-engineering|structural-engineering|civil-engineering| | <p>I believe your equillibrium is very wrong. This is like a rigid body and we have to consider all forces applied to it. The base shouldn't be part of the rigid body, since the screw force would be then an internal one, with no way to find it. So, an apporopriate equillibrium would be as shown in the figure below:</p>
<p><a href="https://i.stack.imgur.com/p5Xa4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p5Xa4.png" alt="rigid body equiilibrium"></a></p>
<p>The forces are:</p>
<ul>
<li>Weight F</li>
<li>Reaction from base at rotation point R</li>
<li>Screw tension T</li>
</ul>
<p>Assuming small displacements, the equillibrium equations should be:
$$ F + T - R = 0 \\ F\cdot 0.5b - T \cdot 0.5d = 0 $$</p>
<p>From the second one we get:
$$ T = F\frac{b}{d}$$</p>
<p>If b/d is let's say 5, then T = 5F !!!
That means the screws are loaded with 5x the weight F. It can only get worse with a more slender post.</p>
<h2>How to improve it</h2>
<p>Add some stiffeners between the base and the posts. The rotation point is moved away from the screws, offering a larger lever arm for T, and simultaneously reducing the lever arm of the weight F.</p>
| 12800 | How to determine which size of screw needed? |
2016-12-13T16:49:44.763 | <p>This is more of a general physics question to help me understand how to choose sports headsets in the future, however it is too specific to a certain use case (bluetooth headset) to belong in the Physics StackExchange.</p>
<h3>Problem:</h3>
<p>3 of my 4 sports bluetooth headsets suffer from choppy sound quality (signal interference?) the minute I step outside. It seems to be especially bad under a clear sky no matter the time of day.</p>
<p>The 4th and the most expensive headset does not suffer from this, but it has a different (more compact) design. The worst performance was by the cheapest sports headset (from ebay) and it was unusable outdoors.</p>
<h3>Observations:</h3>
<ul>
<li>In all cases my phone is located in my trouser pocket, but moving it closer to my head (the headset) seems to lessen signal interference, but not remedy the issue completely.</li>
<li>In all tests the headsets have been fully charged.</li>
<li>The speed of movement also seems to play a factor, but I am not able to reproduce that affect every time.</li>
<li>I can say that clothes are not a factor, as the same clothing indoors produces no interference.</li>
</ul>
<h3>Theories:</h3>
<ul>
<li>background EM radiation coming from outer space?</li>
<li>EM radiation coming from our sun? Although I have not observed if the moon reflecting the EM from the Sun plays a part or not.</li>
</ul>
| |electrical-engineering|telecommunication|wireless-communication| | <p>As you might have already realized there is a good reason that some of Bluetooth headset are inexpensive. Poor design. High quality, well designed Bluetooth headset tend to be expensive. Also not all expensive Bluetooth headsets are high quality.</p>
<p>While it is difficult to root cause failures without physically analyzing the hardware, I will provide some possibilities. Since the device is a handset the device is using Bluetooth classic. Either way some understanding of Bluetooth technology might serve well.</p>
<p>Bluetooth operates in the 2.4 MHz frequency band, more specifically between 2402 and 2480 MHz frequency. If the 2 MHz bottom and 3.5 MHz upper guard band are included then the total bandwidth is 2400 to 2483.5 MHz. Bluetooth classic uses 79 channels, of which each channel has a 1 MHz bandwidth. More importantly Bluetooth uses a radio technology called <strong>frequency hopping spread spectrum</strong>, and if adaptive frequency hopping (AFH) is enabled it usually performs 800 hops a second. Bluetooth divides data into packets and each packet is transmitted on one of the 79 channels.</p>
<p><a href="https://i.stack.imgur.com/LWdgo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LWdgo.png" alt="Bluetooth Classic 80 channels"></a></p>
<p>When transmitting each of the data is subjected to <a href="https://en.wikipedia.org/wiki/Path_loss" rel="nofollow noreferrer">path loss</a> (or path attenuation). That is when the propagation of electromagnetic waves is subjected to a reduction in power density. In your case this path loss is influence by propagation medium such as dry or moist air as well as distance between the transmitter and receiver.</p>
<p>In your situation propagation losses would include absorption loss or penetration loss which a when the signal passes through a medium not transparent to electromagnetic waves or diffraction losses, which is when the radio signal is obstructed by an opaque obstacle.</p>
<p>The signal transmitted by your device also takes multiple paths before arriving at the receiver. The signal received by the receiver might vary in strength, and propagation time etc.</p>
<p>A well designed system will account for all of the above and many more factors. In your case the inexpensive headset might have low <strong>Receive Signal Level</strong> (RSL), or might not have accounted for <strong>fade margins</strong>. A good design will include about 20 to 25 dbm fade margin. So if your headset has a RSL of 40 dbm, and a receive threshold is about 60 dbm. The design has accounted for 20 dbm of fade margin.</p>
<p>Also note that different environmental factors cause different level of signal attenuation. Example a water drop can cause about 5 dbm of attenuation. Not all environmental factors are present at all times. Depending on the application designer choose applicable fade margins.</p>
<p>I hope I have given you a good over view of Bluetooth, Radio Frequency and electromagnetic propagations help you make an informed technical decision on your next purchase. </p>
<p><strong>Reference:</strong></p>
<ul>
<li><a href="http://www.aowireless.com/blog/bid/34066/Does-Weather-Effect-Wireless-The-5-Misconceptions-Part-1" rel="nofollow noreferrer">Does Weather Effect Wireless? The 5 Misconceptions - Part 1</a> </li>
<li><a href="http://smallbusiness.chron.com/signal-interference-issues-bluetooth-technology-58273.html" rel="nofollow noreferrer">What Are Some Signal Interference Issues for the Bluetooth Technology?</a></li>
<li><a href="http://support.trangosys.com/hc/en-us/articles/216202223-Receive-Signal-Level-RSL-or-RSSI-is-Too-Low" rel="nofollow noreferrer">Receive Signal Level (RSL or RSSI) is Too Low</a></li>
<li><a href="https://engineering.stackexchange.com/a/13292/110">What is the difference between Bluetooth Low Energy and Bluetooth BR/EDR in Park mode?</a></li>
</ul>
| 12822 | Why do bluetooth headsets get interference (choppy sound quality) outdoors? |
2016-12-14T17:29:42.797 | <p>As I see it, if you're going to calculate a twin-spool turbojet engine, you</p>
<ul>
<li>Determine expected max aircraft speed, and engine thrust at that speed and when stationary.</li>
<li>Calculate peak mass of air flow</li>
<li>Calculate energy needed to propel this mass of air</li>
<li>Calculate bypass ratio</li>
<li>Calculate compressor properties</li>
</ul>
<p>And from here I have no idea where to move on. Some books I've read just go through the engine from compressor to exhaust nozzle, but none explains which parameter of the engine is primary and where do calculations start and how they're performed.</p>
<p>Would you give me a clue?
Thank you!</p>
| |turbines|turbomachinery|propulsion| | <blockquote>
<p>Determine expected max aircraft speed, and engine thrust at that speed and when stationary.</p>
</blockquote>
<p>Well, sort of. At the very beginning, you may know the <em>desired</em> thrust at a given condition, but you have to work all the way through the calculations to get the <em>actual</em> thrust at that speed. If the actual thrust and desired thrust don't match, then you would need to change the design assumptions (e.g. change fan diameter, or compressor pressure ratio or something) in order to meet that.</p>
<p>What I'm trying to say is, I think you may be combining the concept of how do I design a jet engine to meet a certain performance goal, with how do I evaluate the performance of a particular jet engine design.</p>
<p>To do the later, you start with upstream pressure, upstream temperature, and velocity, and the known design of your engine, and walk from front to back. At the end of that process you will know thrust, mass flow, fuel consumption, etc for that given design and operating conditions. </p>
<p>To do the former, you start with a set of assumption, calculate the performance, and iterate until you have the desired design. For a large commercial turbofan, this process would typically be done to optimize the performance at cruise (since that is where the plane spends the majority of them), and then you would verify that other conditions (e.g. sea level static takeoff) are acceptable. </p>
| 12844 | In what order turbojet engine parameters are calculated? |
2016-12-14T21:16:27.537 | <p>It's written that at constant pressure & temperature for Otto & Diesel cycles
the efficiency of diesel is more than efficiency of Otto.<br />
Efficiency = Work / Heat added = 1- (Heat rejected / Heat added )</p>
<p>from fig :<br />
-(4->1) Heat rejected is same<br />
-(2'->3) Heat added for diesel is greater than (2->3) )heat added for Otto</p>
<p>so for diesel the term (Heat rejected / Heat added ) is less</p>
<p><strong>so efficiency of diesel is more.</strong></p>
<p>What make it not clear for me is that<br />
work ( Cv(T4-T3) ) (3->4) is same for both & heat added for diesel is more than that of Otto.<br />
So if we use eff= Work/(heat added)<br />
<strong>Eff of otto will be more since less heat will be added and same work is done</strong>,<br />
What makes it wrong?</p>
<p><a href="https://i.stack.imgur.com/G3cb8.jpg" rel="nofollow noreferrer" title="P-V For same pressure and temperature"><img src="https://i.stack.imgur.com/G3cb8.jpg" alt="P-V For same pressure and temperature" title="P-V For same pressure and temperature" /></a></p>
<p>source : <a href="http://slideplayer.com/slide/10748942/" rel="nofollow noreferrer">http://slideplayer.com/slide/10748942/</a> slide 30</p>
| |mechanical-engineering|thermodynamics|energy-efficiency| | <p>As @Brian Drummond said ,i overlooked the work done between 2' and 3</p>
| 12849 | cycle efficiency for (otto -diesel -dual) at constant Pr & Temp |
2016-12-15T01:21:19.937 | <p>I try to understand how is possible to turn a photo receptive sensor into an emitter of light.
<br>
<em>Settings, tools, modulation</em>..<br> </p>
| |sensors|modulation| | <p>In the general case, you can't. Not all processes are reversible. For example, a Cadmium-sulfide photoresistor won't emit light when you apply a voltage to it.</p>
<p>There really shouldn't be the expectation that this would work in general since there are so many every day examples. Your car doesn't put gas back into the tank when going down a hill. You don't get less hungry by walking down a mountain. Etc, etc, etc.</p>
| 12851 | How to turn a photo receptive sensor into an emitter of light? |
2016-12-15T02:11:15.073 | <p>Consider the two scenarios in this picture: <img src="https://qph.ec.quoracdn.net/main-qimg-28763a636910916c63557f31d77183d8?convert_to_webp=true" alt="Built Up Member Constructions"></p>
<p>On the left, calculating Q=A'y[bar]' is easy. A is the area of the top board since the cross sectional area of that board corresponds to the area on the opposite side of the shear plane to the centroid of the member. </p>
<p>I am confused about how to choose Q on the right hand side scenario. The two shear planes are perpendicular to the plane that separates A' from the rest of the member, meaning the definition I used for the first case on the left simply doesn't apply. Based on other sources I have consulted, the area I should choose is the area of the board in the center of the two upright members. I have no good justification for this choice.</p>
<p>Edit: More information: Before I was introduced to the concept of shear flow, I was taught how to make shear stress calculations at any given height above or below the centroid of a beam cross section. To calculate this shear stress, a first moment of area (Q) needed to be calculated. This Q value was explained to me to be the area of the cross section of the beam on the opposite side of the desired horizontal shear plane.
<img src="https://www.codecogs.com/users/23287/Shear-Stress-004.png" alt="CSA diagram"></p>
<p>Take the example above. In that diagram, the shear plane is a plane running along the longitudinal axis of the beam at a height $y$ above the neutral axis. The $Q$ value = [Area of the hatched region (which is all of the area on the opposite side of the shear plane from the neutral axis)]$\times$[$\bar{y}'$(height of the centroid of this area above the neutral axis)].</p>
<p>That is how I understand Q. Now let's apply this understanding of Q to figure b in the first picture. The two shear planes are vertical planes that run along the longitudinal axis of the beam. Thus, a horizontal plane cannot be defined co-planer to the 2 vertical shear planes to denote that I want to calculate the shear flow along this horizontal plane. In short, my definition of Q has broken down spectacularly. How is Q defined for case (b)? </p>
| |mechanical-engineering| | <p>As preamble, I note that the equation for this shear stress / force / flow is something that seems to suffer from almost everyone adopting a different nomenclature. I learnt it as longitudinal shear = Q A y[bar] / I. In that form, the longitudinal shear is a force per unit length along the section, and Q is the shear force in the section.</p>
<p>However, I will adopt the nomenclature that the question is adopting. In this answer, I'm adopting:</p>
<p>S = longitudinal shear force per unit length across a defined boundary of interest, that boundary being perpendicular to a plane section. This boundary is normally a single plane on which you could slice off a part of the section, but it doesn't have to be.</p>
<p>V = vertical shear force in the plane section.</p>
<p>A = an area (see below).</p>
<p>y = a distance (see below).</p>
<p>I = second moment of area of the whole plane section.</p>
<p>The equation in question is then S = V A y / I.</p>
<p>(The question has an intermediate value Q = A y, but I'm going to ignore that - you can substitute it back in if you like, having calculated A and y, at any point in this discussion.)</p>
<p>The value of A is the area of a 'chunk' of the section. The value of y is the distance from the neutral axis of the whole section to the centroid of the sliced off part. It doesn't actually matter if the chunk is the result of a single planar slice, or an arbitrary portion of the section.</p>
<p>In case b there's a chunk that's the top plank. It's a rectangle d wide and b high. The centroid of that part is exactly central, b/2 from the top and bottom faces of that plank. The y value will be the distance from the centre of that plank to the neutral axis of the whole section.</p>
<p>For sake of discussion, suppose the neutral axis of the whole section is 0.75xc from the bottom edge. Suppose also that a=d+2xb and e=b+c (ie, as it appears to be drawn).</p>
<p>In case a):</p>
<p>A = a * b</p>
<p>y = 0.25 * c + 0.5 * b</p>
<p>in case b):</p>
<p>A = b * d</p>
<p>y = exactly the same as case a), because the centroid of area A in case a) is at the same place as the centroid of area A in case b)</p>
<p>The second half of the question contains an error about what y is. The value y in the equation is not the distance from the centroid of the whole section to the cut plane. That is, y is not what is labelled as y in the diagram, it should be the distance from the whole section neutral axis at NA to the centroid of the hatched area.</p>
<p>It's worth noting that the S = V A y / I equation is only valid in the case of a prismatic section where the chunk is arranged such that A y and I do not vary along the length of the member. Some thinking about the derivation of the equation is relevant (and explains the assertions made above):</p>
<p>The axial bending fibre stress at any point on a section is calculated from stress = M y / I, where M is bending moment, y is now the distance from the neutral axis to the point where the stress is calculated and I is the second moment of area of the whole section.</p>
<p>If you consider any arbitrary chunk of the cross section, the axial force (F) on that whole chunk is the integration across the area of the chunk of (M y / I). That is, F = M A y / I, if A is the area of the chunk and y is back to being the distance from neutral axis of the whole section to centroid of the chunk.</p>
<p>Now, consider repeating that an increment along the member. There's now a different axial force on the chunk, and the way that force got there was by longitudinal shear across the perimeter of the chunk. Or, to put it another way, if x is position along the beam, the longitudinal shear flow across the chunk boundary = dF/dx = d(M A y / I)/dx. However in that differentiation, A y and I are all constant with varying x, so dF/dx = dM/dx A y / I. However, we know dM/dx (rate of change of bending moment along the beam) as something else - it's the vertical shear force in the section: dM/dx = V.</p>
<p>Hence longitudinal shear across the chunk boundary S = dF/dx = V A y / I</p>
<p>The important thing to note is that this derivation depends absolutely upon A y and I not varying with x - if they do vary, you can't simply recalculate S = V A y / I with different A values along the section, because with an A that varies with x (for example), d(M A y / I)/dx does NOT equal dM/dx A y / I. In that case you need to apply the differentiation product rule, so d(M A y / I)/dx = (dM/dx A + M dA/dx) y / I. Actually, of course, if A is varying with x it's almost certain that y and I are likewise varying with x, and you'll need to multiply out all four terms as varying-with-x.</p>
| 12852 | Determining the first moment of area (Q) when calculating the shear flow of a built up member |
2016-12-15T21:24:16.553 | <p>I've been looking at using Aluminium 6082-T6 structures in impact energy absorption mechanisms. I planned to do some material testing, made specimens for testing then was distracted with another project for about a year. When I finally tested these specimens they broke in a surprisingly brittle manner with a much higher than expected proof stress.</p>
<p>I'm wondering whether 6082-T6 is an alloy that age hardens at room temperature as one of the technicians suggested this as the cause. If so, does anyone know of any studies into this phenomenon?</p>
<p>This is would be rather important for impact energy absorption as increasing proof stress can result in higher risk of fatality. </p>
| |materials|aluminum| | <p>That alloy/temper should be stable when it leaves the manufacturer. I have a piece of 6061 T6 that I've been hardness testing (Rockwell) over the last 10 years, I don't believe the hardness has ever changed; if it has, it wasn't enough to affect its performance.</p>
<p>You didn't say if you had tensile tested them before you got distracted. It may be that the extruder or heat-treater, left them in the age oven too long. Depending on the temperature they were running, they could've blown the grains, the result would be brittle material. If you have the means, do grain size analysis. Our 6082 is usually between a 4-6 ASTM. If it's brittle, the grain size might be down below 3 ASTM. We had a customer leave a part in a braze furnace too long, and then complained our tube leaked. The grain size was measured at a -2 ASTM :-O Sorry, not our fault.</p>
| 12863 | Natural age hardening of aluminium 6082-t6 |
2016-12-16T15:34:49.250 | <p>I made a calculation for a pipe that needs to handle 1 MPa pressure. I based my calculation on <code>ASME VIII div1</code></p>
<p>My calculation says that my pipe needs to be 3 mm thick. But do I also need to account for the tolerances pipe manufactures have on the thickness?</p>
<p>This should be 12,5% according to a colleague of mine (sorry I don't have the source) but I am quite confident he is correct about this.</p>
<p>I can't find a note in the <code>ASME VIII div1</code> about this maybe I over-read this.</p>
<p>So the question is: do I need to add the pipe wall thickness tolerance to my calculated thickness to be theoretically correct according the <code>ASME VIII div1</code>? </p>
| |mechanical-engineering|pipelines| | <p>The ASME calculation gives you a minimum allowable wall thickness. You have to ensure that the actual thickness is equal to, or greater than the minimum allowable. Therefore you have to account for the manufacturing tolerance by specifying a pipe schedule which is certain to be at least the minimum calculated value. So yes, you have to add the tolerance to the calculated value.</p>
| 12878 | Pipe wall thickness and production tolerances |
2016-12-16T21:13:10.593 | <p>One of the cellular vendors wants to install a 180 foot monopole antenna in our neighborhood. They claim that there is significant growth in LTE consumers. So in order to better serve the customers they need to install a new monopole. Following are two reasons that they provided recently.</p>
<ul>
<li>The current towers are over capacity thus cannot reliably serve the
consumers</li>
<li>The cellular signal is weak therefore consumers cannot make phone calls</li>
</ul>
<p><strong>Question:</strong> What are the alternate ways to solve this problem beside installing a new monopole? Are there new LTE/Cellular technology that can help reduce cell tower capacity as well as permit the consumer to make phone calls reliably using cells phones? What are any future technology that might address this issue?</p>
| |electrical-engineering|wireless-communication| | <p>US base Cellular vendors (Such as Verizon, Sprint, AT&T) in combination with cell phone manufactures are starting to introduce a technology call Carrier-branded wifi calling. This is similar to Skype, Google Hangouts, Facebook Messenger and WhatsApp, except the phone dialer is integrated to the smartphone's wifi radio. So the cellular connection is weak, then the phone will automatically switch over to Wifi-Calling. </p>
<p>Note as I understand the LTE uses VoIP (VoLTE) to make the call. Skype like services use VoIP. This is an integration of both wifi and cellular technologies. Depending of the wifi connection the consumer is likely to experience better voice quality. Currently a VoIP call will requires about 1MB/minute of data for a voice call. The typical video call will take about 6-8 MB/minute of data for a video call. </p>
<p>As consumers move over to Wifi calling the need for cellular towers will decrease thus will be used for true mobile connectivity. Thus the residential consumer will less depend on cell phone towers. </p>
<p><strong>References</strong></p>
<ul>
<li><a href="https://www.verizonwireless.com/support/wifi-calling-faqs/" rel="nofollow noreferrer">Wi-Fi Calling FAQs</a></li>
<li><a href="https://www.cnet.com/news/what-you-need-to-know-about-wifi-calling/" rel="nofollow noreferrer">Everything you need to know about Wi-Fi calling</a></li>
<li><a href="http://www.gsma.com/network2020/technology/volte/" rel="nofollow noreferrer">VoLTE Realising revenue potential with Voice over LTE</a></li>
</ul>
| 12881 | What are alternative to installing LTE / Cell Tower to address capacity issue? |
2016-12-16T21:19:06.343 | <p>I am currently designing a small weather station that will be suspended from two fiber optic cables and a power wire. You can imagine for now a three point attachment to each wire with a wind vane suspended below, like <a href="https://www.dropbox.com/s/r0jf7pxd119rt79/20161218_095830.jpg?dl=0" rel="nofollow noreferrer">so</a>. I need some help figuring out the best way to support it while allowing rotations.</p>
<p>Basically, I would like to be able to have the device adjust its pitch so that if one fiber gets pulled vertically, the wind vane will stay level. The yaw adjusted by the force from wind. I am not sure how important roll will be to add in, so for now I have not considered it too much.
The reason for controlling pitch is that as the fibers are pulled up in the air, one may be </p>
<p>My basic idea now is to have a rod go through a small hole in the top part of the device, where the device can rotate, adjusting the pitch. You can see the 3 holes for inserting that rod; <a href="https://youtu.be/sNi7g8Lp--c" rel="nofollow noreferrer">here</a> is a video of the device. I would choose one nearest the center of mass (to be determined). For a sense of scale, those holes are about 1.5 mm in diameter. The whole device is small, and should be about 100-200g total.</p>
<p>I plan to have a clevis structure attach to that rod; what my fingers do in the video. After that it will have to go to an assembly above that will allow yaw rotation (not designed yet).</p>
<p>The problem I find with this design for pitch is there is no damping, so the device oscillates upon disturbances instead of finding an equilibrium point. Also, I have had some trouble finding rods and other parts at this size.</p>
<p>Any suggestions on how to best design this and parts that may be useful would be great. </p>
<p>Thanks!</p>
<p>(P.S. I tried putting the third image, but stack exchange limits me to 2 links until my reputation score is at least 10.)</p>
| |mechanical-engineering|bearings| | <p>First, the functional name of the asked mechanism is spherical bearing. Please have a look at <a href="https://en.m.wikipedia.org/wiki/Spherical_bearing" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Spherical_bearing</a></p>
<p>You can use a spherical bearing to allow continuous rotation (in yaw) while also allowing for some pitch.</p>
<p>Damping can be enhanced by inserting some sort of friction to the assembly.</p>
| 12882 | Weather Station Gimbal Support |
2016-12-17T14:46:56.897 | <p>I am working on product where we need pogo pins, as copper is better conductor of electricity I was searching for the copper pogo pins. After about a month of searching, it is come to my notice that most of the manufacturers are dealing with pogo pins not of copper but of brass. </p>
<p>Even when I inquired about the pogo pin that suited voltage specification of 250V 15A. The manufacturers insisted on using Brass than copper.</p>
<p>Apart from this I also seen many smart watches which were using pogo pins were using Brass instead of copper?</p>
<p>Is there any basic issue with using copper? What are the scenarios when we should chose Brass over the copper while building electrical connectors?</p>
| |mechanical-engineering|electrical-engineering|materials| | <p>For pogo pins to work correctly, there must be a good electrical connection across the sliding joint between the tip and base parts of the pin. If the joint develops a high resistance and current flows through the spring inside the pin, that introduces a random and potentially variable amount of inductance into the connection which can affect high frequency signals.</p>
<p>Copper is a very "soft" material after it has been annealed, but any mechanical strain causes significant work hardening in the material which would lead to poorly-defined geometric tolerance and mechanical stiffness across the sliding joint. Brass doesn't have those disadvantages. </p>
<p>To create the required mechanical stiffness, copper pins would need to be thicker than brass, but thin brass pins already have good enough electrical conductivity for the application, so the mechanical disadvantage of using copper are more important than its theoretical electrical advantage.</p>
| 12891 | Why manufacturers prefer Brass over Copper in electrical conductors? |
2016-12-18T21:40:30.803 | <p>I noticed that blades of HP compressor are shorter in axial dimension next to its LP counterpart. </p>
<p>Why?</p>
<p>I understand why it is smaller in radial dimension (this is because compressed air has smaller volume, and whole air passage is converging), but it seems to me that longer (again, in axial dimension) blades can transfer more energy to the air being compressed.</p>
<p>But it is obvious to be some good reason to make them small.</p>
<p>So why?</p>
| |turbines|turbomachinery|propulsion| | <p>There is a practical limit to the amount of compression that can be achieved in a single stage of an axial compressor. As a general principle, the rotor blades increase the velocity of the gas without much increase in pressure, and the stator blades increase the pressure while reducing the velocity.</p>
<p>The changes in the gas velocity are mainly in the <em>circumferential</em> direction. The maximum circumferential exit velocity of the gas that can be produced is related to the circumferential velocity of the rotor blades themselves (i.e. radius times angular velocity). </p>
<p>If you can design a blade with a short chord length which can do that (as in real-life compressors) there is no value in making the chord length of a single compressor stage longer. Even if you could get a higher compression ratio when the machine was running at one particular design condition of flow rate and gas input temperature and pressure, there would be a greater loss of performance and efficiency away from that design point.</p>
<p>The way you get a "longer chord length," to give a higher compression ratio, is by making a multi-stage compressor with alternating rotors and stators. Jet engine compressors may have 10 or more stages in total, on two or three different rotors running at different speeds.</p>
| 12903 | turbojet axial compressor: blades of HP compressor |
2016-12-20T15:49:12.553 | <p>I need to conduct nano-indentation within a glovebox of a split-test cell for battery testing. I know that vibration may be a potential issue, and I should probably buy an isolation table. However, I do not know how to select the proper isolation table for this application. Does anyone have any advice? Thanks!</p>
| |mechanical-engineering|vibration|sound-isolation| | <p>It's not possible to give an exact answer, but here are a few things you may want to consider:</p>
<ol>
<li><p>Do you want to place the isolation table inside the glovebox or outside it? On the outside, it will isolate building vibrations, but soundwaves hitting the glovebox surface and anyone touching it will still cause movements. On the inside, however, you are heavily limited by the size of the box.</p></li>
<li><p>How much space and weight do you have available? Very sturdy isolation tables may need reinforced floor and will be difficult to move into the room. However, making the table heavier is the easiest way to improve the vibration isolation.</p></li>
<li><p>How much isolation do you need? This is the hardest part to estimate, and isolation table manufacturers use several ways to report attenuation performance. One way is just to try it out: if you need more isolation, add another isolation device on top of the previous one.</p></li>
</ol>
| 12911 | Selection of an isolation table |
2016-12-22T11:18:12.403 | <p>Metal oxides are hard in nature, that's what my professor says. Technically Iron oxide is a metal oxide and must be hard. </p>
<p>Then <strong>is it used in industries for surface finishing operations like buffing?</strong> If not, why so?</p>
| |industrial-engineering| | <p><a href="http://www.cooksongold.com/Jewellery-Tools/Jewellers-Rouge-Powder-120g4oz-prcode-998-193" rel="nofollow noreferrer">Jeweller's rouge</a> is fine Iron III oxide. It is generally used for fine polishing of relatively soft metals like gold and silver as well as some plastics and minerals. </p>
<p>It isn't aggressive enough to be used on harder metals like steel except perhaps for very fine specialist polishing operations. Consider that aluminium oxide typically forms a hard, stable layer on metal whereas rust has very friable structure. </p>
<p>Aluminium oxide is one of the most widely used general purpose abrasives (usually colour coded brown), zirconium oxide is used for more aggressive stock removal at larger grit sizes (usually colour coded blue) and other high performance 'ceramic' abrasives (colour coded red) are sometimes used for stock removal on hard to grind materials. </p>
<p>As well as their hardness the performance of different abrasives depends on how the particle wear for example aluminium oxide tend to become smoother as it is used which can be good for surface finishing whereas zirconium tends to fracture and so retains an aggressive cut for longer making it more effective for stock removal especially at larger grit sizes but at grits higher than about 120 the is no significant advantage over aluminium oxide. </p>
<p>Diamond grit may be used for some applications, especially lapping and honing of hard metals either as a paste or embedded in a metal substrate.</p>
<p>There are also various proprietary compounds, typically an abrasive powder blended with grease or wax used for specific buffing operations on specific metals. </p>
| 12930 | Use of iron oxide as surface cleaning agent |
2016-12-22T16:21:29.747 | <p>I apologize for the confusing question title. I work with area rugs, and have almost never done measurements before. I am tasked with converting these measurements into the right unit.</p>
<p>I have four measurements:</p>
<p>6</p>
<p>19</p>
<p>39</p>
<p>81</p>
<p>These are all in Square Inches I believe. On the spreadsheet it shows "Square Footage (in.)"</p>
<p>I need to convert them to Square FT. I'm unfamiliar with measurements, but is it any different than regular inches? Would 6 square inches be 0.5 square ft?</p>
<p>Thank you.</p>
<p>EDIT: Some were saying the question is confusing, I agree. If it helps, here is the product width for each of these measurements. </p>
<p>6 = 24 inches wide</p>
<p>19 = 31 inches wide</p>
<p>39 = 63 inches wide</p>
<p>81 = 94 inches wide</p>
<p>I really wish I could provide more information, but I am just as confused on the entire thing as anyone else.</p>
| |measurements| | <p>Let's think about how we work out the area in inches. What we are simply doing is multiplying the lengths (in inches) of the shape (let's say rectangle for simplicity) to get the square footage. ie. </p>
<p>$$a * b = area_{inches}$$ where $a$ and $b$ are sides of the rectange (measured in inches).</p>
<p>Instead of using the length of the sides in feet, we could convert it into inches first. Given that there are 12 inches in a foot (or $1/12$ feet in an inch):</p>
<p>$$\frac a {12} * \frac b {12} = area_{feet}$$
$$ { a * b \over 144} = area_{feet}$$</p>
<hr>
<p>This can also be derived from the units, we can treat the units as an equation themselves.</p>
<p>$$ feet = {{inches} \over 12} $$
$$ feet^2 = ({{inches} \over 12})^2 $$
$$ feet^2 = {{inches}^2 \over 12^2} $$</p>
<p>This sort of logic can be used for any unit conversations. Given the relationship of the base units (in this case inches and feet), more complicated relationships can be worked out.</p>
<hr>
<p>For example, using your 81 square inch measurement. If we replace the ${inches}^2$ with $81$, we get 81 inches squared in square feet</p>
<p>$$ feet^2 = {81 \over 12^2} $$
$$ feet^2 = 0.5625 $$</p>
| 12940 | How to convert Square FT (in.) to Square FT (ft)? |
2016-12-22T16:36:18.877 | <p>I want to understand how a laser printers laser writes on a photosensitive drum. I am trying to understand this process.</p>
<blockquote>
<p>the photosensitive cylinder is illuminated by a laser beam that projects above the negative of the image.</p>
</blockquote>
<ol>
<li>After you illuminate with laser beam what happens exactly on the photosensitive drum? What is this projection?</li>
<li><em>negative image</em>? Where does this come from?</li>
<li>When the drum receives this laser light, the photosensitive drum material generates a current, voltage? How and what is handled this photosensitive drum variation due to the absorption of light?</li>
</ol>
<p>I understand the process <code>drum-toner</code> but I don't understand this step well:</p>
<blockquote>
<p>when you want to perform printing, it is loaded with positive electric energy from the corona wire, an electric current thread that runs through it in all its length.</p>
</blockquote>
| |lasers| | <p>You are apparently asking about the Xerographic process, such as used in original copy machines and now in laser printers.</p>
<p>The process works by charging up the surface of a cylinder (the drum) to a high voltage. The drum then rotates so that any parts of it that are charged pick up black particles, called the <i>toner</i>. Then a piece of paper is pressed against the drum as it rotates further. That causes the black stuff to stick to the paper. The paper is then heated and squished to make the black stuff stick to the paper permanently.</p>
<p>Note that you end up with black on the paper wherever there was electric charge on the drum. As I described things above, the drum was charged everywhere, so the paper ends up all black.</p>
<p>If only we could selectively discharge parts of the drum, then we could end up with white areas on the paper. Fortunately there is a way. Instead of making the drum out of purely insulating material, we make it out of material that insulates unless it gets hit with light. Early drums were made from selenium, for example. If we connect the inside of the drum to low voltage (discharge potential), then all we have to do is shine light on the areas where we ultimately want the paper to be white. The light discharges the high voltage on the surface of the drum, which causes that area to not pick up toner, which won't make the paper black.</p>
<p>So now we have a way of exposing a image onto the drum surface, and ending up with that image on a piece of paper. This is exactly what early copiers did. A strip of bright light was shone on the original to be copied. The image of that was optically focused onto the drum as it rotated and the bright light strip moved along.</p>
<p>Laser printers use the same principle, except that a modulated laser "writes" onto the drum to discharge it, instead of the optical projection of the paper to be copied. Since the modulation of the laser doesn't have to come from scanning a original, you now have a "printer" that can discharge any pattern on the drum, which means it can produce any pattern of black/white on the resulting piece of paper.</p>
<p>Modern copiers are really digital scanners with a laser printer back end. There is no longer a direct optical connection between the original and the drum. This allows for image processing, enhancements, storing the result, or even generating the pattern in a computer without ever scanning any original (printer versus copier).</p>
| 12941 | How does a laser printer write with light on the photosensitive drum? |
2016-12-22T18:28:01.437 | <p>I've written a program to generate gears and write them to dxf files. I thought all was well until I looked at the mesh of two gears.</p>
<p><a href="https://i.stack.imgur.com/hpLE2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hpLE2.jpg" alt="Gear Mesh"></a> </p>
<p>From everything I've read there should be no backlash when the pitch circles touch. So I'm ether reading wrong, or my calculations are skewed. If I drop the pressure angle to like 8, the teeth fit nicely. The image is with the angle set to 14.5 This is the actual code, I derived it from <a href="http://faculty.tamucc.edu/psimionescu/PDFs/ASEE-2008.pdf" rel="nofollow noreferrer">here</a></p>
<pre><code>//angle of involute to outside circle
//see fetch.pdf, equ (6),(7) then from...
//polar_radius = base_radius * sqrt( 1 + t^2 )
//where t is the arc AB
//the limit of polar_radius is the outer_radius
// so solve for t at the limit
// t= ( ( outside_diameter / base_diameter )^2 - 1 )^1/2
//for the end point of the involute arc
double Dop= gp.od / gp.bd; //ouside diameter / base diameter
double t= sqrt( Dop * Dop - 1 ); //angle of tangent to involute arc, at limit
double Ac= t - atan( t ); // angle of involute arc, at limit
</code></pre>
<p>then I create a line string for the involut curve as:</p>
<pre><code>for( size_t step= 0; step <= ITERS; ++step )
{
double arc= t / ITERS * step;
*ls+= bg_point( base_radius * ( cos( arc ) + arc * sin( arc ) ),
base_radius * ( sin( arc ) - arc * cos( arc ) ) );
}
</code></pre>
<p>And thinking about it, how could you not have backlash when the teeth arc from the base circle, the point where the line starts at half a tooth/space distance. But I'm just not sure.</p>
<p>Also, I'm planing to put all the code on GetHub and will do it sooner than later if needed.</p>
<p><strong>UPDATE:</strong>
The fix is in and now it looks like it should. This is with a 20 degree pitch.
<a href="https://i.stack.imgur.com/rRlir.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rRlir.jpg" alt="20 degree pitch meshed"></a></p>
<p>I've uploaded to GetHub using the MIT license. If I think this useful, maybe others will also. The user interface needs a lot of work, gear params are still hard coded. Only these two files have any thing to do with gears.</p>
<p><a href="https://github.com/lakeweb/spur-gear-gen/blob/master/project/gear_gen.h" rel="nofollow noreferrer">gear_gen.h</a></p>
<p><a href="https://github.com/lakeweb/spur-gear-gen/blob/master/project/gear_gen.cpp" rel="nofollow noreferrer">gear_gen.cpp</a></p>
<p>The rest is a CAD drawing interface. Thanks again SprocketsAreNotGears!!</p>
| |design|gears| | <p>First of all, there should indeed ideally be no backlash between the teeth. There is a slight problem with the code, which stems from the misunderstanding that consecutive points of the involute curves that lie on the base circle occur a half-tooth's distance away from each other. If this were the case then, yes, there would be backlash. However, in fact, it is actually the points of the involute curves at the <strong>pitch</strong> circle that are a half-tooth's distance apart. I'll illustrate this below.</p>
<p>What your program is currently doing is as below:</p>
<p><a href="https://i.stack.imgur.com/msM8w.jpg"><img src="https://i.stack.imgur.com/msM8w.jpg" alt="enter image description here"></a></p>
<p>Note that each of the radial lines are one of the $2N$ division lines (for a gear with $N$ teeth) and each are separated by an equal angle of $\frac{\pi}{N}$ radians. Note the black dots are the points of the involute curves that lie on the base circle, and these lie on the equal division lines according to your program. This is not the case for standard gears with no deliberate backlash implemented.</p>
<p>Take note of the following diagram, which is basically a zoom-in of the previous:</p>
<p><a href="https://i.stack.imgur.com/RmyaT.jpg"><img src="https://i.stack.imgur.com/RmyaT.jpg" alt="enter image description here"></a></p>
<p>Note the pitch point on the involute curve, and take note of the angle $\theta_P$ that occurs between the radial lines passing through the base and pitch points. We will need this angle later on.</p>
<p>We want to instead have the pitch points on consecutive involute curves separate by equal angles, and so the following diagram should be obtained: </p>
<p><a href="https://i.stack.imgur.com/QgBOj.jpg"><img src="https://i.stack.imgur.com/QgBOj.jpg" alt="enter image description here"></a></p>
<p>Note that the black dots are the pitch points that are lying on the equal division lines. To get from the previous incorrect case to the correct one, this involves rotating the involute profiles about the centre of the gear by the angle $\theta_P$: the involutes that curl out clockwise from the base circle will be rotated by $\theta_P$ in an anticlockwise direction, and the ones that curl out anticlockwise will be rotated by $\theta_P$ clockwise. Now it is a matter of determining the value of $\theta_P$, and this can be done by looking at the polar parametric equation you were provided for the involute curve.</p>
<p>Let's determine the value of $t$ at the pitch circle, which we will call $t_P$. Using:</p>
<p>$$\rho = r_B \sqrt{1+t^2}$$</p>
<p>And noting that the pitch radius is $r_P = r_B/\cos{\phi}$, where $\phi$ is the pressure angle, then:</p>
<p>$$\frac{r_B}{\cos{\phi}} = r_B \sqrt{1+t_P^2}$$</p>
<p>And, by algebraic manipulation, it can be shown that:</p>
<p>$$t_P = \tan{\phi}$$</p>
<p>Note that $\phi$ is in radians. Now, $\theta_P$ can be determined using the second of the polar formulae:</p>
<p>$$\theta = t - \arctan{t}$$</p>
<p>And so,</p>
<p>$$\theta_P = \tan{\phi} -\phi$$</p>
<p>Rotate your profile curves by this angle in the appropriate directions, and that should solve the issue! (Note how $\theta_P$ drops to zero as the pressure angle decreases, aligning with the observation you made earlier.)</p>
<p>The best of luck with your gear generator! :)</p>
| 12944 | Involute Gear Calculation |
2016-12-24T19:30:50.080 | <p>I find this scheme <a href="http://www.explainthatstuff.com/laserprinters.html" rel="nofollow noreferrer">here</a></p>
<p><a href="https://i.stack.imgur.com/qYuwI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qYuwI.png" alt="printer"></a></p>
<p><strong>1</strong> - is <em>data stream</em> come from computer. This 'data' is an electric signal moudulated by the n°<strong>2</strong><br>
<strong>2</strong> - this is the <strong>electronic circuit</strong> inside printer that <br></p>
<p><strong>a)</strong> activates the corona wire <code>→</code> <strong>b)</strong> the corona wire charges up the photoreceptor drum<br>
<strong>a)</strong> at the same time circuit activates the laser to make it draw the image of the page onto the drum. The laser beam doesn't actually move: it bounces off a moving mirror that scans it over the drum</p>
<p>My question are: </p>
<ul>
<li><p>when data stream come <em>from</em> computer to the printer <em>inside</em> the electronic circuit what happens inside this electronic circuits ?
How mechanism 'tranform' this electric signal of <em>data</em> into instruction for laser diode and moving mirror ?<br>
But this step is possible only if you have correct modulation between data stream and instruction laser otherwise printing should be wrong and electronic circuit may break or not work properly.<br>
What modulation technique is used for laser diode ?</p></li>
<li><p>which components inside electronic printer is needed to drive and control this ?</p></li>
</ul>
| |electrical-engineering|lasers|circuits|modulation| | <p>What you are asking in similar to how the beam in a CRT monitor is modulated to make lighter and darker areas on the screen.</p>
<p>The lasers used in these devices can be electronically controller to vary their light output. This is done in sync with the laser beam being mechanically swept across the drum. As the laser hits each spot on the drum, the electronics modulates it to get the desired amount of light.</p>
<p>A moving mirror sweeps the laser beam across the drum to make one dimension of the picture, and the drum moves orthogonal to that to make the other dimension of the picture. The result is a 2D image "painted" on the drum with laser light. Again, this is much like the electron beam being swept across the face of a CRT and modulated per pixel to make a 2D image on the face of the CRT.</p>
| 12961 | How the electronic circuit control laser diode? |
2016-12-27T15:17:51.353 | <p>I have an electronic device that uses black potting materiel similar the one below to I believe to protect the electronic components from elements.</p>
<p><a href="https://i.stack.imgur.com/y6WYR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y6WYR.jpg" alt="enter image description here"></a></p>
<p>I would like to safely remove this material. This material might be some type of epoxy. There are plenty of suggestions how to remove potting material (<a href="https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=removing%20potting%20electronics" rel="nofollow noreferrer">removing potting electronics</a>). Some indicate that the chemicals in the potting might be harmful.</p>
<p>How can I safely remove the potting material? Through observation can I safely narrow down the potting material? </p>
| |electrical-engineering| | <p>Cold Crack
IPA and crushed dry ice makes a low low cold slush. Apply the super cold slush to circuit board potting compound and chip it away.
Caution
Use heavy gloves this slush is super cold and can freeze off skin or fingers.
Cryo-fluid from dry ice and rubbing alcohol that goes down to -110</p>
| 12982 | How to remove potting material from an electronic package? |
2016-12-27T17:46:39.277 | <p>I'd like to mount a shelf on the wall. I have two shelf supports to do so, like this:</p>
<p><a href="https://i.stack.imgur.com/pluMo.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pluMo.png" alt=""></a></p>
<p>Relative to the shelf, where do I place the (blue) supports to achieve the best distribution of load?</p>
| |mechanical-engineering|statics| | <p>21.6% from the end, assuming the same DEFLECTION at the middle as at the end.
22.5% from the end, assuming the same STRESS at the middle as at the end.</p>
| 12984 | How should I position two shelf supports for the best distribution of load? |
2016-12-27T22:28:00.633 | <p>I am trying to design something that rotates 90 degrees based on a small (10mm or so) linear motion, using <a href="https://www.youtube.com/watch?v=09jZsBDHalE" rel="nofollow noreferrer">this basic setup</a> as an example. </p>
<p>There are a lot of questions about how long the arms need to be, how far they need to anchor from the point of rotation, etc. that i need to know. I am assuming there is some formula that governs the relationship that someone can point me to?</p>
| |mechanical-engineering|design| | <p><strong>There is a mechanism called:</strong> <a href="https://www.google.com.tr/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=slider%20crank%20mechanism" rel="nofollow noreferrer">SLIDER CRANK MECHANISM</a></p>
<p>One can copy one of these mechanism for just converting linear to rotational motion. In most of the cases, this will do the trick. </p>
<p>But for general, if you want to design a mechanism that will do your desired motion, get ready to do some work!</p>
<hr>
<p>First, here some links you will probably need for getting the dimensions for your particular shape:</p>
<ul>
<li><a href="http://saltire.com/atlas/CrankRocker.html" rel="nofollow noreferrer">Web Atlas of Linkages</a></li>
<li><a href="https://www.cedarville.edu/cf/engineering/kinematics/ccapdf/ccpagef.htm" rel="nofollow noreferrer">Four-bar Coupler Curves and Centrodes</a></li>
<li><a href="http://www.saltire.com/LinkageAtlas/Crank%20Rocker.html" rel="nofollow noreferrer">Atlas of Linkages - Crank Rocker</a></li>
<li><a href="http://www.mekanizmalar.com/fourbar.html" rel="nofollow noreferrer">Four-Bar Linkage and Coupler Curve</a> (Especially this one!)</li>
</ul>
<p>Then you must start your motion (PVA) analysis, you can use some schema like that:</p>
<p><a href="https://i.stack.imgur.com/JNUEu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JNUEu.png" alt="by Virtual Labs"></a></p>
<p>I don't think you will need it, if you only want to perform your motion. But if you want to have an analysis over your mechanism (for every position of crank), you can start with inspecting <a href="http://facultad.bayamon.inter.edu/elay/mecn4110/ch04%20Position%20Analysis.pdf" rel="nofollow noreferrer">these slides</a>.</p>
<hr>
<p>Here it comes, there are some example of mine because the process is much or less same.</p>
<h2>Four Bar Mechanism for Straight Line</h2>
<p>This is the desired shape:</p>
<p><a href="https://i.stack.imgur.com/bj2EG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bj2EG.png" alt="Straight Line"></a></p>
<p>These are the calculations for PVA analysis:</p>
<p><a href="https://i.stack.imgur.com/HuKHY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HuKHY.png" alt="Theta"></a></p>
<p><a href="https://i.stack.imgur.com/4UWoU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4UWoU.png" alt="Alpha"></a></p>
<p>and some more of course...</p>
<p>After all, to sum up:</p>
<ol>
<li>Finding the dimensions of linkes for the desired shape</li>
<li>Writing and solving Vector Loop</li>
<li>PVA Analysis</li>
<li>Test final mechanism in MATLAB or Solidworks.</li>
</ol>
<p>Here is MATLAB generated mechanism: (Also <a href="https://www.youtube.com/watch?v=Ax-GNOOiQmc" rel="nofollow noreferrer">here</a> @youtube with Solidworks)</p>
<p><a href="https://i.stack.imgur.com/nswNe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nswNe.png" alt="Four Bar for Straight Line"></a></p>
<p>Hope this helps, I wanted to help as much as I can because I've been experienced your suffer and it will be good to be informed when you are at the start.</p>
| 12988 | Formula for converting linear to rotational motion |
2016-12-28T17:33:45.707 | <p>I'm trying to increase the Helium detection capabilities of a sniffer unit. Could a cyclonic separator, like that used in the oil/gas industry, with enough fluid velocity move the heavier gas molecules in air (O<sub>2</sub>, N<sub>2</sub>, Ar, CO<sub>2</sub>) to the sides of it while keeping the lighter elements (He, H<sub>2</sub>) in the center to be detected by a sniffer?</p>
<p>We're trying to detect leaks of helium from refrigeration systems on an assembly line. we're designing a positive air box to prevent any outside contamination from the filling and recovery stations. The box will have a blower attached to it to pull the volume of the box in 10 seconds. because most of the gases in the box are not our trace gas we have a very low mbar l/s level (allowable is 9e-7) that is on the borderline of the minimum of our sniffer unit we are looking to use. The sniffer unit has the precision of 1e-7 mbar l/s at 300 SCCM and 1e-6 mbar at 3000 SCCM. We are going to have an array of sniffer heads in the center of this conic that will sample the air at the center (with the thought that the lighter elements would be there) leading to a more accurate reading because it is sampling less of the non-trace gas. This process of sampling and detection along with the lifting/closing of doors and shifting of the equipment is done on a conveyor belt within a 40 second time period.</p>
<p>The current system is a person at the assembly line with a hand held sniffer going around the connections, but are getting readings that there are leaks even though there might not be, probably due to the helium charging station only 3 feet away from the leak testing area. we are looking to automate the process for less false readings and quicker turn over time in the process.</p>
| |mechanical-engineering|fluid-mechanics|airflow|gas|cyclones| | <p>Cyclones are designed to quickly and cheaply remove contaminants that would tend to settle out of your process gas (or liquid) if it stopped flowing. They're very good at controlling coarse PM, not so good at controlling fine PM and really quite poor at controlling ultrafine PM. What you're looking for here is so far past ultrafine PM control on this spectrum that you would never achieve a useful result.</p>
<p>Even if you could slightly concentrate your heavier gases around the outside of the cyclone, where are they going to go? In a cyclone, the volumetric flow is the same at the inlet as it is at the outlet. The only thing that gets knocked out is the particulate, when it impacts the wall and settles down into the hopper. This works because that particle would tend to settle anyway, if it weren't entrained in your process flow, and near the wall of the cyclone the flow approaches zero (recall the <a href="https://en.wikipedia.org/wiki/No-slip_condition" rel="nofollow noreferrer">no-slip condition</a> of fluids 101). The same can't be said of a theoretical packet of slightly-heavier-than-average air.</p>
<blockquote>
<p>We are going to have an array of sniffer heads in the center of this conic that will sample the air at the center (with the thought that the lighter elements would be there) leading to a more accurate reading because it is sampling less of the non-trace gas.</p>
</blockquote>
<p>I'm not sure this really ends up being more accurate, given that your sample is no longer representative of the mixture on which your leak detection threshold of 9 x 10<sup>-7</sup> mbar·L/s is based. This isn't a situation where you want to set your leak detection threshold at zero.</p>
<p>I think any clever gas stratification scheme would just over-complicate the process. Using the PPV box might be enough by itself to make your client happy. Otherwise, if you just need a little bit more sensitivity and you can't go with a more sensitive device, I'd keep thinking about optimizing the sampling conditions—can your client afford to spend 90 seconds per unit rather than 40, if it cuts down on false positives? Or maybe your client would be willing to perform some more reliable leak test on a sample, rather than testing every unit, if they can accept a non-zero defect rate.</p>
| 13004 | Can cyclonic separators be used to separate heavier and lighter gases? |
2016-12-29T09:33:51.663 | <p>Consider a 1D point-mass moving along an axis. A force $u$ is applied as control. There is no gravity or other forces involved. The system can be described in state space equations as:</p>
<p>$$\begin{align}
A &= \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{bmatrix} \\
B &= \begin{bmatrix} 0 \\
0 \\
\dfrac{1}{M} \end{bmatrix} \\
C &= \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \\
D &= [0]
\end{align}$$</p>
<p>The shown system is controllable, but not observable. Not even structurally observable and most certainly not fully observable. Thus, it should be impossible to construct an observer for this system.</p>
<p>However, if I know the initial state of the system, I can compute the full state at every time, i.e. by integrating the system's output. How does this fall in line with the concept of observability? How would I incorporate the initial state into the equations? </p>
<p>I can't find the error in my train of thought, but I am certain there is one. Do I misunderstand observability?</p>
| |control-engineering|control-theory|linear-control|linear-systems| | <p>Observability means that you can estimate the complete state using only the output, without knowing the initial state. In other words, you have to figure out where you are without knowing where you were initially. </p>
<p>A more practical reason why this rarely works is that when you are limited by non-perfect sensors and non-zero sampling time, taking the integral of the acceleration will cause growing errors in your estimates of positions and velocity. Thus, even if you know the initial state you will "lose track" of it over time. </p>
| 13013 | Why is it impossible to create an observer for this not fully observable system? |
2016-12-29T22:41:19.893 | <p>I have some high pressure and vacuum lines off of a tool from the 1980s and they are labeled as, for example, "3/8 imperial - eastman," which seems to be different from how normal compression connections are labeled.</p>
<p>Is this the same as regular imperial units, or is there some difference I need to be aware of? Where can I find these parts?</p>
| |pressure|compressed-air|vacuum|tools| | <p>"Imperial Eastman" is the name of the now-defunct company that originally manufactured the tool. It exists today as <a href="http://imperial-tools.com/about-us" rel="nofollow noreferrer">the Imperial Tools brand of Stride Tool</a>. The part you mention is probably just a standard 3/8" fitting or hose but if you're concerned about it, you could take a look at their <a href="http://imperial-tools.com/support/replacement-parts" rel="nofollow noreferrer">replacement parts guides</a>.</p>
<p>If you want to get the parts elsewhere and you're worried about compatibility, you should be able to get equivalent parts from any large supplier. If you can find the actual part number on the fitting (not just the nominal dimension), you may be able to find the equivalent part in a document like <a href="https://www.parker.com/parkerimages/Parker.com/ParkerStore2014/Global%20Assets/PDFs/Product%20Info/4400_B10_Imperial%20Eastman%20to%20Parker.pdf" rel="nofollow noreferrer">this Equivalent Products List for Parker</a> and not even have to get on the phone.</p>
| 13021 | Are "Imperial Eastman" compression connections sized the same as regular imperial connections? |
2016-12-30T02:53:26.033 | <p>I have two basic questions about helicopters. Let's say the z-axis is the vertical axis from ground to sky. I know the tail rotor is there to counteract the angular momentum effects of the main rotor, and keep the total angular momentum about the z-axis at 0.</p>
<ol>
<li><p>Say the tail rotor spins about the x-axis. What keeps the body of the helicopter spinning about the x-axis in the opposite direction? I assume it's a slight tilt of the main rotor forward to balance this effect. Am I right?</p></li>
<li><p>How are the speeds of the two rotors are kept coordinated? I assume there must be a sensor/feedback mechanism by which rotor speeds change to counteract spinning.</p></li>
</ol>
| |aerospace-engineering|aerodynamics| | <p>A conventional helicopter has three main flight controls. </p>
<ul>
<li><p>Torque pedals : these function is a similar way to the rudder on a fixed wing aircraft. They work by adjusting the pitch of the blades of the tail rotor to produce more or less thrust, allowing the pilot to balance the torque from the rotor blades and also control the yaw of the aircraft. </p></li>
<li><p>The cyclic lever : this controls the pitch of individual rotor blades at a specific point in their rotation, controlling pitch and yaw. This is analogous to tilting the rotor disk at about its centre. </p></li>
<li><p>Collective pitch : this controls the pitch of all of the rotor blades simultaneously controlling overall lift. As the pitch is changed the throttle must also be used to adjust engine power to maintain a more or less constant rotational speed as the load on the blades changes. </p></li>
</ul>
<p>It is worth reiterating that it is the pitch of the blades which provides the direct control over thrust and lift and engine power is adjusted to keep their rotational speed withing an optimum range. </p>
<p>In a conventional helicopter the tail rotor and main rotor are directly coupled together via the gearbox and only their blade pitch rather than speed can be adjusted independently ie the tail rotor speed is proportional to the main rotor speed. </p>
<p>There are designs which deviate from the conventional layout though for example there are tail designs which use ducted airflow rather then an external rotor and also various configurations of twin rotor which dispense with tail thrust altogether eg the Chinook which has fore and aft rotors and various
Russian military helicopters which have coaxial rotors. </p>
| 13022 | Questions about helicopter rotors - conservation of angular momentum |
2016-12-30T17:22:29.170 | <p>I am trying to model the strain generated by inserting an electrode device into brain tissue. I've had some success in doing this in ANSYS Explicit Dynamics. <a href="https://i.stack.imgur.com/VTTFB.jpg" rel="nofollow noreferrer">Here's a link to a GIF showing the penetration</a>.</p>
<p>My issue is that I had to do a very fast insertion (10 m/s) in order to have a faster computation time (~18 h). The actual insertion speed is on the order of 0.1 mm/s, which ends up being 5 seconds of simulation. 5 seconds of explicit dynamics simulation is days of computation on my computer! My question is, can I do this sort of slow penetration/ballistics model in ANSYS transient dynamics or some other implicit model? Or am I stuck with Explicit dynamics and the super long computation time?</p>
| |finite-element-method|ansys| | <p>I don't know anything about the material properties of brains, but if you are applying some loads/displacements at 0.1mm/s, I don't understand why you are doing dynamics at all.</p>
<p>I expect the inertia forces caused by slowly deforming the material will be completely negligible compared with the elastic forces. A nonlinear stress analysis (probably with large displacement capability switched on) should work fine. Try it, and compare the results with the dynamics model you already ran.</p>
| 13029 | How can I reduce the computation time required to simulate a brain tissue penetration model? |
2017-01-02T21:10:57.560 | <p>I am working with silicone injection molding as a hobby and I am wanting to make a ring.</p>
<p>I have tried with 100% silicone and the results are a clear ring that is not quite the consistency I want. I would like to experiment with making it more or less elastic if possible, as well as different swirled colors.</p>
<p>How can I change the elasticity of a silicone ring and how can I introduce swirling colors into it as well?</p>
<p>I searched around for other existing things, and this silicone ring is close to what I want to make.</p>
<p><a href="https://groovelife.co/products/silicone_ring_breathable_groove?variant=13528608580&gclid=CjwKEAiAkajDBRCRq8Czmdj-yFgSJADikZggxMT5d91MuDELqOKU6eFGvRvJWkAbHWH1_Gxu9of58RoC9oDw_wcB" rel="nofollow noreferrer">https://groovelife.co/products/silicone_ring_breathable_groove?variant=13528608580&gclid=CjwKEAiAkajDBRCRq8Czmdj-yFgSJADikZggxMT5d91MuDELqOKU6eFGvRvJWkAbHWH1_Gxu9of58RoC9oDw_wcB</a></p>
| |structural-engineering|chemical-engineering|industrial-engineering| | <p>If you are using RTV silicone of the type used for prosthetics and animatronics then you can get a variety of grades with different mechanical properties, commonly rated by Shore Hardness. </p>
<p>You can also get additives known as 'deadeners' which create a more matte surface and reduce the slippery feel. </p>
<p>To create swirl colours you want to make up 2 or more batches of silicone of different colours mix in the catalyst , (pigments are available to add to clear silicone) and either mix them lightly once both area catalysed or inject them into the mould at the same time. Different ratios and mixing methods will produce different results. </p>
| 13063 | silicone injection molding - making a flexible ring |
2017-01-02T23:16:05.640 | <p>I know that according to the thin airfoil theory all airfoils (symmetric or not) have lift curve slope of $2\pi$. I also know that finite wings should have less inclined lift curve because of the downwash. Then I saw this formula used for finite elliptic wings which is fine.
$$
\frac{dC_L}{d\alpha}=a=\frac{a_0}{1+a_0/\pi AR}
$$
The problem is $a_0$ is not $2\pi$ in examples that above formula is used. I am thinking that since non of the downwash effects are present for airfoils (infinite wings) $a_0$ should always be $2\pi$. What am I missing? Is it because not all airfoils are thin?</p>
| |fluid-mechanics|aerospace-engineering|aircraft-design|aerodynamics| | <p>In general, most airfoils only approximately display the $2\pi$ lift slope as predicted by <em>thin</em> airfoil theory. That is because airfoils are not actually infinitely <em>thin</em> in practice, and will deviate from thin airfoil theory by a small amount. $a_0$ is left in the above equation so one can predict the elliptical finite wing lift slope (effects of aspect ratio), $a$, given the best predicted/measured infinite wing lift slope $a_0$. The thin airfoil theory will predict a finite wing lift slope of $2\pi$ rad$^{-1}$ or 0.1097 deg$^{-1}$. It is common to measure airfoil lift slopes near the 0.1097 deg$^{-1}$ ($2\pi$) for the NACA 4 and 5 series airfoils with thickness ratios below 10%. However, once you go above this thickness, the measured lift slope begins approaching 0.1 deg$^{-1}$ or 5.73 rad$^{-1}$ for smooth airfoils with thickness ratios near the upper limit of 24%. Similarly, if roughness effects are taken into consideration, the measured lift slope is even lower and reaches 0.1 deg$^{-1}$ at only a thickness ratio of 14% and can reach as low as 0.085 deg$^{-1}$ or 4.87 rad$^{-1}$ near thickness ratios of 24%. This is all detailed in the plot below from the Theory of Wing Sections (excellent reference text).</p>
<p><a href="https://i.stack.imgur.com/JSnFK.png" rel="noreferrer"><img src="https://i.stack.imgur.com/JSnFK.png" alt="enter image description here"></a></p>
<p>Oddly enough, for the smooth NACA 63-series and 64-series, the lift slope will actually increase with increasing thickness ratios. However, the roughness effects keeps the measured lift slope around 0.11 for thickness ratios up to 24%. This is detailed in the following 2 plots.</p>
<p><a href="https://i.stack.imgur.com/qScNg.png" rel="noreferrer"><img src="https://i.stack.imgur.com/qScNg.png" alt="enter image description here"></a></p>
<p>In summary, the $a_0$ is left as a variable for the designer to be predicted or measured using vortex panel methods or experimental wind tunnel testing, to which the designer can use the above equation to estimate the effects of aspect ratio on the elliptical finite wing lift slope, $a$, given the predicted/measured $a_0$.</p>
| 13066 | Do all airfoils have a lift curve slope of $2\pi$? |
2017-01-04T22:32:35.320 | <p>Any idea on how this type of pneumatic solenoid works? I know that NC is normally closed, NO normally open, but not sure how air flows? Thanks! </p>
<p><a href="https://i.stack.imgur.com/uCcu6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uCcu6.png" alt="enter image description here"></a></p>
| |airflow|compressed-air|pneumatic| | <p>When no voltage is applied to the solenoid, air will flow between port 2 'Common' and port 3 'Normally Open,' but not to port 1 'Normally Closed.' When voltage is applied, air will flow between port 2 and port 1, but not to port 3. We could likely assume that the direction of air flow is intended to be from port 2 to ports 1 & 3, but it isn't a guarantee and it may not matter to the valve. Note that in pneumatic and hydraulic circuits, normally closed and normally open have the opposite definition as in electrical circuits. For a valve, closed means <em>not passing</em> air/oil, whereas for a relay closed means <em>passing</em> electrons.</p>
<p>A more standard diagram for this valve would be #3 in the image below:
<a href="https://i.stack.imgur.com/6fRSa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6fRSa.jpg" alt="Example valve diagrams"></a></p>
<p>Although it's worth noting that the 'T' designation implies a hydraulic valve, and numeric port designations are more common for pneumatic valves.</p>
| 13085 | pneumatic solenoid operation? |
2017-01-04T22:53:24.153 | <p>During the heat treatment process, people use clay to harden the steel in certain areas like the spine. Then they have to go through another process to make the steel tough. I need very hard steel and want to know will the hard steel be brittle and able to withstand large amounts of pressure? I ask because the steel is heavily stressed from the forging process. Also, can I make a thinner piece of steel as hard as a thicker piece with this method?</p>
| |steel|heat-treatment| | <p>In steel hardness is closely related to tensile strength so a hardened steel will have both high hardness and high tensile strength. </p>
<p>Toughness is a bit more difficult to quantify than strength but in general is the ability to resist impact and resistance to brittle fractures. Ductility plays a part in toughness as it tends to help a material to resist crack propagation. </p>
<p>For simple steels heat treating is usually a balance between hardness and toughness. Cooling a hardenable steel from critical temperature (around 850 C) creates a very hard but brittle material, tempering (heating to a lower temperature, around 200-300 C) partially reverses the hardening process and improves toughness and ductility. Controlling the tempering temperature allows the relative hardness/toughness to be controlled to an extent. </p>
<p>The exact properties you get from a given heat treating process will depend on the alloy used and different alloys will have different heat treating requirements. </p>
<p>Prior to hardening residual manufacturing stresses and grain growth can be mitigated by <em>annealing</em> and/or <em>normalising</em> if necessary. </p>
<p>Having said that heat treating steel is a complex subject on which many large and complex books have been written and it's not easy to summarise in a short answer. It is also important that you read up on the precise engineering definitions of the terms involved eg <em>hardness, toughness, tensile strength, ductility</em> etc as they have very specific meanings in this context and talking about 'strength' generally can cause confusion. </p>
| 13086 | Is hardened steel better than tough steel? |
2017-01-05T01:53:25.077 | <p>Being a mechanical engineering novice (I stick closer to solid state electronics), when I look inside of any toy-grade RC car, I see a series of different-sized gears between the motor shaft and the drive axle. </p>
<ul>
<li>Why does a single-speed RC car require a gearbox? </li>
<li>Instead, why can't a gear on the motor shaft connect to a gear on the axle?</li>
</ul>
<p>So far, two of the plausible explanations I have come up with are that</p>
<ol>
<li>The motor turns at a different RPM than desired by the axle, thus
conversion is required.</li>
<li>To put protection between the spinning of the axle and the
spinning of the motor.</li>
</ol>
<p>I ask because I am setting out to make my own Lego car driven by a DC brush motor and have no idea what to do for a single speed gearbox other than hook one gear to the motor shaft and one gear to the axle and let it go.</p>
| |mechanical-engineering|gears| | <p>While it is certainly possible to connect input and output shafts with gears there are a number of reasons why having a separate gearbox might be desirable. </p>
<ul>
<li>For more complex compound gear systems the gearbox casing provides mounting points for shaft bearings for intermediate gears. For larger gear rations compound gears are generally more compact than a single pair of gears with a large difference in diameter. </li>
<li>The gearbox casing allows lubricant (oil or grease) to be retained and helps prevent contamination by dust, debris, water etc from outside.</li>
<li>Having a self contained modular gearbox can make assembly and maintenance easier and can allow for a standard gearbox to be used for a number of different applications. </li>
<li>Gears and their shafts need to be accurately aligned to work effectively, a rigid casing helps with this. If one of the shafts needs to move or has large loads on it, as in the case of a drive shaft in a vehicle with suspension the gears can be separated from the loads on the shaft via constant velocity joints or other articulated couplings. </li>
</ul>
| 13089 | Why do single speed RC cars use a gearbox? |
2017-01-05T03:04:44.633 | <p>Is it due to the PET material that it is made of or is it more the shape and structure of the bottle? If it is due to the shape/structure of the container, is it possible to achieve such pressure resistance using another plastic?</p>
| |materials|pressure| | <blockquote>
<p>Is it due to the PET material that it is made of or is it more the shape and structure of the bottle? </p>
</blockquote>
<p><strong>Short answer</strong>: <em>Partially because of both</em>. With that I mean it would be possible to produce weaker bottles changing the material or changing the usual design of the bottles. </p>
<blockquote>
<p>If it is due to the shape/structure of the container, is it possible to achieve such pressure resistance using another plastic</p>
</blockquote>
<p><strong>Short answer</strong>: It is because of both, being impossible to say what contributes more without a definition of contribution to resistance. <em>Yes</em>, it would be possible to use other polymers. </p>
<p><strong>Additional info</strong></p>
<p>Comment about the material:
The resistance to pressure of polymer containers usually depends on the combination of three tensil properties: </p>
<ul>
<li>ultimate tensile strength (force needed to break it pulling from extremities);</li>
<li>tensile elongation
(how much you can pull it apart until it rips);</li>
<li>tensile modulus ( ratio of stress to elastic strain in tension).</li>
</ul>
<p>The table below illustrates these values for some polymers:</p>
<p><a href="https://i.stack.imgur.com/u0Rqq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u0Rqq.jpg" alt="Tensile property polymers"></a>
<a href="http://www.matweb.com/reference/tensilestrength.aspx" rel="nofollow noreferrer">http://www.matweb.com/reference/tensilestrength.aspx</a></p>
<p>As shown in the table, PET doesn't have the highest values. However, it does withstand reasonable pressure and certainly within the limits of proposed use. </p>
<p><strong>Comment about the shape</strong>:
Intuitively, if standing the pressure was the only requirement, the ideal shape would be a sphere. Although a sphere would allow thinner walls without deforming, I believe it is not usually chosen because of ergonomics and transportation as others have commented ( I have seen one brand of water that has spherical bottles, but I think it was done for aesthetics). The most common designs just prioritise a shape that is easy to hold and without sharp edges. </p>
<p><strong>Why PET was chosen even though it is not the most resistant polymer</strong></p>
<p>Although many other polymers could satisfy the pressure requirement, I believe the reason of why polyethylene terephthalate (PET, material of most soda bottles) is of widespread use is <strong>compatibility</strong>. Even though PET is what we see everyday, polypropylene (PP), Ethylene-Vinyl Alcohol Copolymer (EVOH), Polyethylene Terephthalate (PET),
and other plant derived plastics are used for containers <a href="https://i.stack.imgur.com/u0Rqq.jpg" rel="nofollow noreferrer">1</a>. Although I did not do extensive research on their applications, I believe come of blends do satisfy the pressure requirements of 200~700 KiloPascals inside a PET bottle <a href="https://i.stack.imgur.com/3gHnI.jpg" rel="nofollow noreferrer">2</a>.</p>
<p>( The comments below are my personal justification on the compatibility argument)</p>
<ol>
<li><strong>Machinery</strong>
The process of producing machinery for polymer molding usually is polymer specific, that is, not always a machine that molds one polymer can mold another and therefore it is cheaper for industry as a hole to adopt a convention.</li>
<li><strong>Regulation</strong>
To test and approve a polymer that stays in long and close contact to food is extremely expensive. Also, considering that the approval must happen for the health agency of every country where the product is sold, there is no reason to make a change for another polymer if you know one that attends your health requirements and that has been tested for decades. I believe one evidence for this is that most os the polymers that I cited are used for bottles of liquids (sometimes pressurised) but not for human consumption. </li>
<li><strong>Recycling</strong>
Just as for producing, recycling processes are usually polymer specific, for what they needed to be separated. I don't believe they thought of that by the time the industry adopted the PET as the standard for carbonated drinks, but today it is an extra reason. </li>
</ol>
<p>I believe that because of these reasons and others (please suggest edits) the <strong>transition of PET to anything will take a long time</strong> and would be very bothersome. For example, we still were not able to transition completely from glass. In many countries glass is still used for carbonated drinks. This transition was proposed because PET is lighter, doesn't break, cheaper, resistant to microrganisms (which is not so good today as it takes a long time to degrade in nature) and I believe because it was developed in a time when it was believed that men were unable to consume all petroleum available on Earth. However PET is no perfect polymer, it can't for example hold bear for very long, since it has considerable permeability to oxygen. Below there is a table with some permeability data of common use polymers [3] (unfortunately it doesn't include the other polymers that I said can be used for bottles, but you can find the values online; note that the permeability depends on the gas)</p>
<p><a href="https://i.stack.imgur.com/3gHnI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3gHnI.jpg" alt="enter image description here"></a></p>
<p>At the moment what is being done is research on different blends of PET that can fill the gaps. In the future, however, there might be a new polymer that takes its place, hopefully addressing the environmental concerns. </p>
| 13090 | Why can soda bottles withstand so much pressure? |
2017-01-05T11:17:48.473 | <p>Which is better to use in contructing ceiling frame? RHS or SHS? And why?</p>
| |structural-engineering|structures| | <p>Obviously this depends on the relative size of the sections.</p>
<p>Assuming that the rectangular section's smallest dimension is equal to (or greater than) the square section's side and their thicknesses are equal, then the rectangular section will always be stronger than the square, for any load in any orientation. This is because the rectangular section's geometric properties (namely area and moment of inertia) will always be greater.</p>
<p>For the same reasons, if the rectangular section's largest dimension is equal to (or lesser than) the square section's side, then the square section will always be stronger than the rectangular section for any load in any orientation.</p>
<p>If, however, the rectangular section's smallest dimension is lesser than the square section's side but it's largest dimension is greater, then you need to calculate the section properties and see which is best:</p>
<p>$$\begin{align}
A &= BH - bh \\
I &= \dfrac{BH^3 - bh^3}{12}
\end{align}$$</p>
<p>where $B$ and $H$ are the section's width and height, respectively, and $b$ and $h$ are the hole's (so $b = B - 2t$, where $t$ is the section's thickness). The moment of inertia also depends on the rectangular section's orientation ("standing up" means $H>B$, "sideways" means $B>H$).</p>
<p>It's possible to get dimensions where the rectangular section has a greater moment of inertia but smaller area or vice-versa. In these cases, the best section also depends on the expected loading: moment of inertia is important for bending and torsion, while area is important for normal and shear loading.</p>
<p>Also, there may be case-specific height restrictions or architectural/aesthetic reasons to choose one over the other, but that's obviously not in the scope of this answer.</p>
| 13092 | Rectangular hollow section vs square hollow section |
2017-01-05T12:10:19.563 | <p>i'm a chemical eng. student from Italy and i'm "sizing" ("setting"?) a heat exchanger to cool some octane with toluene.</p>
<p>Is it possible that pressure drops shell side in a 1,2 pull-through floating head heat exchanger are lower than 0.001 atmosphere? I used really wide buffles and buffle cuts opened at 45%. The heat exchange is in range and the pressure drops tube side are "normal" (0.2 atm). I just find it weird that shell side they are so low. Can someone help me?</p>
| |heat-transfer|chemical-engineering|heat-exchanger| | <p>This is a bit of a vague question but I can give you two pieces of information which may help:</p>
<ol>
<li>In theory the pressure drop can be arbitrarily low. That is to say that if the flowrate is small enough relative to the size of the flow path, then the pressure drop could be 0.001 atm.</li>
<li>In practice, for any heat exchanger design you would not have a pressure drop that low. If your pressure drop is that low I suggest the heat exchanger is over-sized and uneconomical to produce.</li>
</ol>
| 13095 | Can pressure drops in a heat exchanger shell side be super low? |
2017-01-05T14:03:34.870 | <p>I'm attempting to find the highest free stream velocity over a flat plat before a sensor placed 'delta' away from it experiences turbulent flow. When I plot this in Matlab it seems that the turbulent boundary layer thickness decreases with velocity. This seems pretty counter intuitive to me, am I thinking of this in the wrong way? Or am I making a silly coding error?</p>
<p>These are the governing equations</p>
<p>$$\begin{align}
\delta &\approx \dfrac{0.37x}{Re_x^{1/5}} \\
Re_x &= \dfrac{\rho u_0x}{\mu}
\end{align}$$
<a href="https://en.wikipedia.org/wiki/Boundary_layer_thickness" rel="nofollow noreferrer">Source</a></p>
<p><strong>EDIT</strong> : Do not include any 'clear' or 'delete' code when posting, because many people do not want their workspace cleared.</p>
<p><em>Matlab Code</em> </p>
<pre><code>% clc
% clear all
rho = 1025; % density of water
mu = 0.00108; % dynamic viscosity
x = 65e-3; % distance from leading plate edge
U = linspace(0,10); % Free Stream Velocity
Re = (rho*U *x)/mu;
turb_lay = (0.382*x)./(Re).^0.2
lam_lay = (5*x)./(Re).^.5
plot(U,lam_lay,U,lam_lay2,U,turb_lay)
xlabel('Velocity (m/s)')
ylabel('Boundary Layer Height (m)')
legend('Laminar Boundary', 'Turbulent Boundary')
</code></pre>
| |fluid-mechanics| | <p>As the velocity increases, the fluid will not be able to form a cohesive boundary layer because the fluid particles will not have the needed residence time near the boundary layer to add to it. </p>
| 13099 | Turbulent Boundary Layer Thickness as Velocity Increases |
2017-01-05T13:32:00.463 | <p>These days I'm designing a 5-DOF parallel manipulator with Autocad. Though I know that some people might say that Autocad is not a proper software for robot designing but I only have access to this software at the moment. So here is my question (and I thank you in advance for your help) :</p>
<p>$\textbf{Parametric constraints:}$ Unfortunately I do not know how to put geometrical constraints on my parallel manipulator though I have searched a lot on net and watched some videos about it on youtube. By geometrical constraint I mean putting some restrictions on for example two 3d objects that they can make some restricted moves with respect to each other. For example I have a spherical joint in my robot which I want it to perform some rotations without actually leaving its socket (as depicted in the following basic picture).
<a href="https://i.stack.imgur.com/3eO8T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3eO8T.png" alt="enter image description here"></a></p>
| |design|autocad| | <p>I can only answer the question you asked about. You asked about a spherical joint, so I'll give advice about the spherical joint.</p>
<p>First, are you using Autocad? That looks like Inventor. They are two separate software packages. </p>
<p>The two ways you could constrain (mate) the spherical joint would be to make a "cup" (spherical cut) on the socket and then constrain the ball to be tangent to the cup (cup is same radius as the ball) or concentric to the cup (not necessarily the same radius). </p>
<p>The other way would be to create a reference point on the cup and a reference point on the ball and constrain the two points to be coincident. That's probably the way I would do it, though a concentric mate would probably be the same thing behind the scenes. Tangent mates would be a lot more taxing on the software and would bog down your simulation. </p>
<p>You can use driven constraints to get a joint to move to a particular parameter's value - <em>those</em> would be parametric constraints. Check out constraint limits; I forget what they're called in Inventor off hand, but you can essentially make "end stops" for actuator min/max strokes, etc.</p>
<p>If you're in academia then you might try to see if you can get an academic-use copy of Matlab. There is the SimMultibody package that lets you do probably a lot of what you want to do anyways, along with the Robotic System toolbox. Don't confuse the latter with the Robotic Toolbox - the Robotic System Toolbox is offered by Mathworks; the Robotic Toolbox was written by a third party and is out of date. </p>
<p>If you have any other <em>specific</em> questions, please ask another. </p>
| 13101 | Designing a Parallel Robot by Autocad |
2017-01-06T10:53:06.257 | <p>I'm having a presentation about dc generators and I wanted to demonstrate how it works. Anyone have an idea on how to build one. I left my electric motors at home because of Christmas break, now I can't find anything.</p>
| |electrical-engineering|motors|generator| | <p>For a presentation it might makes sense to reproduce one of <a href="https://en.wikipedia.org/wiki/Faraday's_law_of_induction" rel="nofollow noreferrer">Faraday's</a> original experiments which give more of a visible insight into what is going on than a modern generator which is essentially just a sealed box. </p>
<p>Something like the experiment in <a href="https://www.youtube.com/watch?v=vwIdZjjd8fo" rel="nofollow noreferrer">this video</a> would be quick and easy to set up. </p>
<p>Obviously making a rotating armature which produces a steady current is a bit more difficult than just moving a magnet in and out of a coil but in some ways the simple version demonstrates the underlying concept better. </p>
| 13115 | How do I build a dc generator at home for my presentaion? |
2017-01-06T12:32:56.383 | <p>why the torque is negative in this case ? the question ask for determine the angle of twist of gear C with respect to B . </p>
<p>We have to 'fix' the B , and determine the twist of C ? So , imagine that we cut off the shaft and gear , now only left with gear B and gear C connected by a shaft . So , when B is fixed , the gear C rotate , with the direction of torques pointing to the left ( by using right hand grip rule ) ,
So , the finger point towards B , so it's negative ? <a href="https://i.stack.imgur.com/Gmlah.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gmlah.png" alt="11"></a></p>
| |mechanical-engineering| | <p>they are not opposing. Their rotation is a ratio regarding the number of teeth n1/n2. If you fix b so that there is torsion I think you're moving in the wrong direction. The torque* down the shaft is constant. The torque is relative to the diameter of the gear i.e. F*r = T. The angle is simply geometric unless it is restricted by an outside torsional force. They both rotate in the same direction but with different torque. </p>
| 13116 | angle of twist of gear relative to another gear |
2017-01-07T03:41:05.567 | <p>In this problem, why did the author use <span class="math-container">${(30×10^3)} {/s} = {VQ} /{ I} $</span> only?
Shouldn't it be <span class="math-container">$4(30×10^3) /s = VQ / I$</span>? </p>
<p>I said so because the board at the centered is connected by 4 bolts , 2 on the left and 2 on the right . The faces on the right also resisted by the same arrangement of nails , right ?
<a href="https://i.stack.imgur.com/pH5tw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pH5tw.png" alt="22"></a>
<a href="https://i.stack.imgur.com/JK9uV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JK9uV.png" alt="226"></a></p>
| |structural-engineering| | <p>The four connected boards are encompassed by the inertia in the shear equation. In the inertial formula they account for the four connections (the multiplier 4). </p>
<p>The shear force on the element in question is defined by the bodies and forces that are acting on it, and inertia is the resistance to rotational movement (simplification to the extreme but I think you get my point), so by using the inertial mass of the connected elements we can deduce the force on the main body by accounting for the connected masses.</p>
<p>Let's go back to what shear force is on a body. In equilibrium you should be able to make a theoretical slice in the body at any point and have a balanced system (axial, shear, rotational), the 4 outside the brackets of the inertia equation (I) accounts for the four masses connected.</p>
| 13127 | shear stress resisted by bolt |
2017-01-07T08:51:38.030 | <p>Our teacher gave us a project before Christmas break, it was to build a miniature house, she didn't state what specific materials should be used, so I built a cardboard house. She didn't accept it because it wasn't firm enough for her. Now I want to know if it is possible to build a miniature concrete house?
Considering I can't find any tutorials online I'm guessing it has a high chance of collapsing, but I don't know, that's why I came here.</p>
| |civil-engineering|structures|architecture| | <p>If you must use concrete, you could consider ShapeCrete -- it's a clay-like moldable concrete, more info could be found on their web site www.shapecrete.com . They also have a projects gallery that you can use to decide whether this material will suite your needs. It is available from Amazon.</p>
| 13132 | How do I build a miniature concrete house? |
2017-01-07T21:16:40.977 | <p>How can the vertical wind direction be measured?</p>
<p>For a project I have to design a wind turbine and I would like the blades to make slight adjustments based on the wind direction in order to achieve the most efficient energy conversion. Whilst I can use a wind vane to get the direction of the wind on a horizontal plane; I have been unable to find a method for finding the direction of the wind on a vertical plane.</p>
<p>I would appreciate any help.</p>
| |mechanical-engineering|aerodynamics| | <p>Given the budget of up to £3m, then lidar scanning seems to be current state of the art for 3-d profiling of wind for active management of wind turbines.</p>
<p>Here's a <a href="http://www.dewi.de/dewi_res/fileadmin/pdf/publications/Magazin_37/08.pdf" rel="nofollow noreferrer">sample business case</a> for one.</p>
<p>The WINDCUBE v2 is a vertical wind profiler that uses doppler pulsed lidar. <a href="http://www.norcowe.no/doc/Documents%20for%20web/SMI%20S%20Presentasjonar/E%20Santiago,%20A.,%20Leosphere.pdf" rel="nofollow noreferrer">Here's a presentation on it</a>.</p>
<p>As far as I know, lidar has mostly been used in offshore wind. There are quite a few academic papers on its use in the wind industry. I don't know what, if anything, has been done about processing vertical-plane wind data, but there's an increasing number of conference= and journal- papers on yield optimisation for wind turbines, so there could be quite a bit of existing literature in the field on this already.</p>
| 13138 | Vertical wind measurements |
2017-01-07T23:40:57.997 | <p><a href="https://i.stack.imgur.com/IhsLg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IhsLg.png" alt="enter image description here"></a></p>
<p>The above is a past exam question from an introductory structural analysis course, one in which although we have studied the Euler buckling load equation, we have just been given parameters for the equation based on the standard end support conditions (fixed/fixed, fixed/pinned, etc.). I don't think the above example fits any of the standard end conditions and am therefore finding it difficult to derive an equation for $P_{cr}$
I have tried to use the bending moment equation by including torque (that opposes bending) from B and the net load (due to spring C) as follows:
$$
T_B = \beta_r \theta
$$
Taking a small deflection x in the vertical direction and y in the horizontal:
$$
tan\theta = \frac{dy}{dx} \approx \theta
$$
$$
\therefore T_B = \beta_r \frac{dy}{dx}
$$
Using the bending moment equation:
$$
M=-EI\frac{d^2y}{dx^2}
$$
$$
\therefore \sum M=-EI\frac{d^2y}{dx^2} + \beta_r \frac{dy}{dx} = Qy
$$
The net load $Q$ can be expressed in term of the force P and the reaction force due to the spring at C as:
$$
Q = P - \beta dx
$$
Leaving the final second order equation:
$$
EI\frac{d^2y}{dx^2} - \beta_r \frac{dy}{dx} + (P-\beta dx) y = 0
$$
<p>
This is beyond the scope of our course, so I have no idea if:<br>
a) The above equation is correct or solvable (and if so how one goes about solving it)<br>
b) There is a different approach using more basic techniques (which is more than likely the case)</p>
| |structural-engineering|structures|buckling| | <p>To answer the specific question: (a) No, that equation is irrelevant and (b) Yes. see below.</p>
<p>The question isn't about Euler buckling. It says the bar is rigid. Euler buckling only applies to a flexible bar - the Euler formulas include Young's modulus, and also the moment of inertia of the bar which is not mentioned in the question.</p>
<p>One way to solve the question is to use the principle of virtual work, and compare the strain energy in the springs with the work done by the external force P.</p>
<p>You might also notice that the condition $\beta_r = 3\beta L^2/2$ is related to the geometry of the triangle ABC.</p>
| 13141 | Critical Buckling Load for a Spring Supported Bar |
2017-01-08T13:14:21.553 | <p>Is there any way to display element coordinate system in ANSYS Mechanical APDL GUI (the same way the global coordinates system is displayed)?</p>
| |ansys|ansys-apdl| | <p>Under the menu <em>PlotCtrls > Symbols</em>, you can find the following window where the option <em>ESYS Element coordinate sys</em> you are looking for is available.</p>
<p><a href="https://i.stack.imgur.com/I2GMv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I2GMv.png" alt="Symbols window"></a></p>
<p>As previously answered, it's also available via the following command:</p>
<pre><code>/psymb,esys,1 ! or 0 to remove it.
</code></pre>
<p>You can look for <code>/help,/psymb</code> for more information.</p>
| 13148 | Display element coordinate system using Ansys Mechcanical APDL |
2017-01-08T14:33:57.323 | <p>My friend and I are currently working on a cloud chamber cooled by the compressor of a deep-freezer. We have built an isolated box that contains the cold pipes of the compressor. The pipes cool down to a whopping -40C. On the top of our box lies an aluminium plate, and on top of that a well isolated chamber. With the aluminium plate 5 or 6 cm away from the compressor pipes, the plate just gets -15C. Connecting the tubes and the plate with a folded fin heatsink gives us -20C. Moving the plate down onto the tubes results in a much better -35C, however this is not a very favorable option as this makes the plate harder to look at. Unfortunately, bending the pipes is not an option either.</p>
<p>We think the air in the box is simply not thermally conductive enough. The heatsink was not doing a very good job either, presumably because the fins where too thin and were in contact with a lot of air. </p>
<p>Our question is: What would be the best way to transport the "coolness" of the pipes towards the plate. Would it be better to fill the box with some sort of liquid that would transport heat more efficient then air? </p>
<p>We are using a computer fan to circulate the air in the box. The fan is pointed upward, so the airflow flows via the pipes onto the plate.</p>
<p>Images:
<a href="https://i.stack.imgur.com/UaXTa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UaXTa.jpg" alt="Top of the box with the aluminium plate"></a>
<a href="https://i.stack.imgur.com/OsQTf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OsQTf.jpg" alt="Top of the box without the aluminium plate, you can see the pipes"></a></p>
| |heat-transfer|temperature| | <p>The goal is to remove heat energy from the aluminum plate to reach a temperature of at least $-15\unicode{xb0}$F = $-26.1\unicode{x2103}$ by increasing heat transfer between the plate and the refrigeration tubing. </p>
<p>One option is to try to increase convection. This would increase the "h" value in the heat transfer equation below.
$$Q = h*A_S*\Delta T$$
$Q$ is the Heat Transferred. $h$ is the heat transfer coefficient. $A_S$ is the surface area. $\Delta T$ is the temperature difference. </p>
<p>Adding liquid might work, but would be heavy and messy.</p>
<p>I would suggest trying to increase conduction between the plate and the tubes. The distance between the plate and the tubes is too much for using heat transfer compound. An inexpensive option could be to lightly crumple up aluminum foil and put it under the plate in such a way where it contacts both the plate and tubing. </p>
| 13149 | Cooling an aluminium plate |
2017-01-08T17:24:06.170 | <p>How I can calculate or find a value for the start-up cost of a natural gas boiler? The boiler will have a capacity of about 700 kW and is intended for heating the water for a greenhouse.</p>
| |boilers|cost-estimation| | <p>For accurate cost estimation you will need to find a vendor and have them supply you a "budgetary quote" for your 700 kW boiler. The vendor can usually get this type of quote to you more quickly than a firm quote because they are not held to that price or terms; though I find that frequently the price is the same when we go to purchase.</p>
<p>When I need an estimate in the next 10 minutes I try to find something on the internet with a price that I can scale. I find a unit closest to the type and capacity I need. Usually economy of scale ensures that I over estimate, but it doesnt hurt to give yourself some safety factor.</p>
<blockquote>
<p><a href="http://www.supplyhouse.com/Lochinvar-KBN701-660100-BTU-Knight-XL-High-Efficiency-Boiler" rel="nofollow noreferrer">574,000 BTU Boiler</a> for 14,000USD<br>
700 kW is 2,390,000 BTU<br>
2,390,000 BTU / 574,000 BTU = 4.16<br>
4.16 * 14,000 USD = 58,240 USD<br>
58,240 USD * 1.2 safety factor = 70,000 USD<br></p>
</blockquote>
<p>Obviously you will have to do this for all your major components, piping, etc.</p>
<p>For installation costs you typically estimate some number of labor hours and multiply by your "shop rate." Accounting should be able to give you this number as it factors in all employee costs like benefits, vacation, training, tool overhead costs, etc.</p>
| 13151 | How to calculate the start-up cost of a natural gas boiler |
2017-01-08T19:51:24.200 | <p>I'm using existing architectural designs and some mechanical designs too, all CAD files in formats *.dwg, *.dgn and the drawings have smart solid geometries including voids. Using CAD software what is a void space and why is it important to create voids as an object?</p>
| |autocad|cad|geometry| | <p>You can add a void geometry in a window object in order to make a hole when you insert the window in a wall. Autodesk Revit works like this.</p>
| 13153 | What is a void in CAD |
2017-01-09T02:06:58.160 | <p>I got this stand from a friend of mine years ago. Its unbranded. I want to use it to hold a 55 gallon fish tank which can weigh about 456 pounds to 622.5 pounds.</p>
<p>How can i calculate a estimate of how much weight this stand will hold ? </p>
<p>The wood is 61 in by 17 in and about 3/4 in thick. The metal underneath it is 2 inches. The metal legs are 28 inches high and 1 inch on each side. </p>
<p>If not, then is there any way i can easily modify the stand for more support? Perhaps stick some 2x4s or cinderblocks under it? </p>
<p><img src="https://i.stack.imgur.com/eRlRo.jpg" alt="a busy cat">
<img src="https://i.stack.imgur.com/sNgeC.jpg" alt="a busy cat">
<img src="https://i.stack.imgur.com/iHvqL.jpg" alt="a busy cat">
<img src="https://i.stack.imgur.com/Mx72H.jpg" alt="a busy cat">
<img src="https://i.stack.imgur.com/eQnJ3.jpg" alt="a busy cat">
<img src="https://i.stack.imgur.com/oYoaS.jpg" alt="a busy cat"></p>
| |structural-engineering|civil-engineering|structures| | <p>If you look carefully at most fish tanks, the bottom glass is actually not touching the table underneath. Disclaimer: I've only dealt with tanks up to about 30 gal. The bottom is held up a small distance because it sits inside the bottom rectangle of the frame.<br>
Assuming your tanks are built similarly, you can easily see that the only thing which matters for the table is the strength of the outer frame, not the top itself. Despite what some comments say, I'd be surprised if those two-inch frame parts could not handle considerable load. The legs are fine, since they only take a compressive load. </p>
<p>Running a test load w/ cinderblocks or pig iron will always outrank theoretical analysis, of course. Alternatively, take a look at aquarium stands designed for such tanks and see what the dimensions of the metal parts are (and look at the L- or I- shape of the beams). </p>
| 13155 | How much weight can this stand hold? |
2017-01-09T14:16:29.247 | <p><a href="https://i.stack.imgur.com/nlFcC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nlFcC.png" alt="enter image description here"></a></p>
<p><a href="https://en.wikipedia.org/wiki/File:DieselElectricLocomotiveSchematic.svg" rel="nofollow noreferrer">Image Source</a></p>
<p>Why not just take the AC from the alternator to the motors? Why go AC to DC to AC again?</p>
| |mechanical-engineering|electrical-engineering|rail|transportation| | <p>To expand on this, controlling the speed of the motors requires feeding it AC and varying the frequency. The generator puts out AC at a fixed frequency. So you need to convert fixed-frequency AC to variable-frequency AC. The most straightforward way of doing that is to rectify the AC to DC, then invert it back to AC again.</p>
<p>A variable-frequency drive is the standard way of accomplishing this. Most take AC input into a rectifier, then have an inverter stage feeding the motor. Occasionally it makes more sense to have a single large rectifier creating a large DC bus, and then feeding several inverters off that single bus.</p>
| 13163 | Why use rectifier / inverter in locomotives? |
2017-01-09T14:21:34.947 | <p>I have a part made from 5mm alloy plate and I need to cut a large cutout in the middle of said part:</p>
<p><a href="https://i.stack.imgur.com/3IW2o.png"><img src="https://i.stack.imgur.com/3IW2o.png" alt="An alloy part requiring a large cutout"></a></p>
<p>As seen above I intend to use a small cutter to cut through the stock (not necessarily plunging as shown here) and cut around the inner diameter of the part instead of roughing out the entire area of the cutout which would take a lot longer. </p>
<p>My concern is that when the cut is complete the large piece of internal cutout could get caught by the cutter and thrown around the machine enclosure, wreaking death and destruction as it goes. Is this something I should be worried about and if so, what's the usual strategy for making this type of feature?</p>
| |machining|safety|cnc| | <p>Some CAM software offers the option of generating <em>tabs</em> or <em>bridges</em> that look like this:<a href="https://i.stack.imgur.com/V4k3Z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V4k3Z.jpg" alt="CAM Tabs"></a>. </p>
<p>The position, qty and size of the tabs are configurable in the CAM software. Here's an example configuration dialog for Fusion 360:</p>
<p><a href="https://i.stack.imgur.com/ZadXm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZadXm.png" alt="Fusion 360 dialog with 'tab' configuration"></a></p>
<p>The tabs can be 'cleaned up' in a secondary operation.</p>
| 13164 | CNC milling of large cutouts |
2017-01-09T23:41:47.927 | <p>I'm wondering what causes the price difference between a <a href="http://www.cablewholesale.com/products/serial-modem-cable/modular-adaptors/product-31d3-37200.php?utm_source=GoogleShopping&utm_medium=cpc&utm_term=31D3-37200&utm_campaign=Modular%20Adapter%2C%20Gray%2C%20DB25%20Male%20to%20RJ45&gclid=Cj0KEQiAhs3DBRDmu-rVkuif0N8BEiQAWuUJr6FoZ_2-R_gjiAJBk-9abyF9pj5E38frxKYkp7pGCTwaAgqD8P8HAQ" rel="nofollow noreferrer">Serial-to-RJ45 adapter</a> and a <a href="https://gridconnect.com/uds1100.html" rel="nofollow noreferrer">Serial-to-Ethernet server</a>. Given a serial-interfaced instrument, can one use either to connect the instrument to a computer? (Of course, the first must be wire correctly first).</p>
| |electrical-engineering| | <p>To answer your first question: the Serial-to-RJ45 adapter is just wires, the Serial-to-Ethernet server is a computer in a box.</p>
<p>I can't answer your second question without knowing more about your equipment, but perhaps this information will help you.</p>
<h2>Serial-to-RJ45 adapter</h2>
<p>This adapter to me looks like it is designed to permit users to use inexpensive ethernet cables in place of relatively rare serial cables. In order for it to work, you need to have a serial port on your computer. The ethernet jack on your PC will not output serial signals and this adapter will not turn them into serial signals.</p>
<h2>Serial-to-Ethernet server</h2>
<p>This is a small computer which connects to your local network and has a serial port which you can connect equipment to. It comes with software that allows you to use it as if your computer had the serial port itself.</p>
<h2>Other notes</h2>
<ol>
<li><p>If your computer doesn't have a serial port, USB-to-serial adapters are available very cheaply</p></li>
<li><p>There are a lot of different serial communication protocols. Generally "serial" on an instrument means RS-232 or RS-485, but could be something different. Make sure you understand the requirements of your equipment before buying something that may not work.</p></li>
</ol>
| 13172 | Difference between Serial-to-RJ45 adapter and Serial-to-Ethernet server |
2017-01-11T04:32:09.900 | <p>For this problem , i found that the internal force AB at point A and point B pointed in the same direction ( as shown) in my working , so , how they cancel off each other ???Since they cant cancel off each other , so they are not stable , right ? The structure is statically indeterminate ? </p>
<p><a href="https://i.stack.imgur.com/szLHe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/szLHe.jpg" alt="22"></a>
<a href="https://i.stack.imgur.com/OuMey.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OuMey.jpg" alt="222"></a></p>
<p><a href="https://i.stack.imgur.com/ve1LE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ve1LE.jpg" alt="2131"></a></p>
| |structural-engineering| | <p>As @AndyT stated, you need to add the horizontal reaction at A.</p>
<p>You start by finding the reactions. Your global equilibrium equations are</p>
<p>$$\begin{align}
\sum F_x &= R_{A,x} + F = 0 \\
\therefore R_{A,x} &= -F = -4\text{ kN} \\
\sum F_y &= R_{A,y} + R_{B,y} = 0 \\
\therefore R_{A,y} &= -R_{B,y} \\
\sum M_A &= 8R_{B,y} - 3F = 0 \\
\therefore R_{B,y} &= \dfrac{3F}{8} = 1.5\text{ kN} \\
\therefore R_{A,y} &= -1.5\text{ kN}
\end{align}$$</p>
<p>Now, looking at the nodal equilibrium diagrams, we have for B:</p>
<p><a href="https://i.stack.imgur.com/7dHZy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7dHZy.png" alt="enter image description here"></a></p>
<p>Before going into the math, it is worth noting that this truss is composed of two <a href="https://en.wikipedia.org/wiki/Special_right_triangle#Common_Pythagorean_triples" rel="nofollow noreferrer">3-4-5 triangles</a>, so we already know $\overline{AC}$ and $\overline{BC}$ are 5 m long.</p>
<p>So $\overline{BC}$ is the only member that can resist the vertical reaction, and it does so by suffering compression equal to</p>
<p>$$\begin{align}
\sum F_{B,y} &= R_{B,y} + \dfrac{3\overline{BC}}{5} = 0 \\
\therefore \overline{BC} &= -\dfrac{5R_{B,y}}{3} = -2.5\text{ kN (compression)} \\
\sum F_{B,x} &= \overline{AB} + \dfrac{4\overline{BC}}{5} = 0 \\
\therefore \overline{AB} &= -\dfrac{4\overline{BC}}{5} = 2\text{ kN (tension)}
\end{align}$$</p>
<p>And then looking at A:</p>
<p><a href="https://i.stack.imgur.com/G3NuB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G3NuB.png" alt="enter image description here"></a></p>
<p>$$\begin{align}
\sum F_{A,y} &= R_{A,y} + \dfrac{3\overline{AC}}{5} = 0 \\
\therefore \overline{AC} &= -\dfrac{5R_{A,y}}{3} = 2.5\text{ kN (tension)} \\
\sum F_{A,x} &= R_{A,x} + \overline{AB} + \dfrac{4\overline{BC}}{5} = 0 \\
\therefore \overline{AB} &= -\dfrac{4\overline{BC}}{5} - R_{A,x} = 2\text{ kN (tension)}
\end{align}$$</p>
<p>Both nodes agree that $\overline{AB}$ is in tension, so it checks out. Also, $\overline{AC}$ and $\overline{BC}$ have equal but opposite results, which is as it should be (and, if you actually do the work for node C, you'll see they are also correct).</p>
| 13188 | internal forces in truss |
2017-01-11T14:05:31.257 | <p>For a beam supported at both ends with a mass in the middle, the flexural modulus is given by:</p>
<p>$$E_b = \frac{F L^3}{48I d}$$</p>
<p>with I as the area moment</p>
<p>$$I = \frac{1}{12}wh^3$$</p>
<p><a href="https://i.stack.imgur.com/87kWQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/87kWQ.png" alt="Flexural modulus measurement"></a></p>
<p>How does the expression for modulus change if we have one fixed end as in the diagram below?</p>
<p><a href="https://i.stack.imgur.com/kjN4R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kjN4R.png" alt="scenario"></a></p>
<p>ja72 states <a href="https://physics.stackexchange.com/a/279972/12568">in this answer</a> that </p>
<p>$$E_b = \frac{F L^3}{3I d}$$</p>
<p>But I'm not clear on the derivation.</p>
| |structural-engineering|structural-analysis|elastic-modulus|moments| | <p>The fundamental beam equation is</p>
<p>$$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI\dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$</p>
<p>Which basically translates to "the fourth derivative of the deflection function is equal to the applied load". In fact</p>
<ul>
<li>the first derivative is the tangent of the deflection, which for small angles is approximately equal to the angle of deflection</li>
<li>the second derivative is the bending moment</li>
<li>the third derivative is the shear force</li>
<li>the fourth derivative (to repeat myself), is the applied load.</li>
</ul>
<p>All deflection results are obtained using this equation.</p>
<p>To simplify this answer, I'm going to start off from the second derivative, the bending moment, since it (and the ones after it) is trivial to find by inspection.</p>
<p>For the simply-supported beam with a concentrated load at midspan, we have:</p>
<p>$$\begin{align}
M &= \begin{cases}
\dfrac{Fx}{2} &\text{ for } x \in [0,\dfrac{L}{2}] \\
\dfrac{F(L-x)}{2} &\text{ for } x \in [\dfrac{L}{2}, L] \\
\end{cases} \\
EI\theta = \int M\text{d}x &= \begin{cases}
\dfrac{Fx^2}{4} + C_1 &\text{ for } x \in [0,\dfrac{L}{2}] \\
\dfrac{FLx}{2} - \dfrac{Fx^2}{4} + C_2 &\text{ for } x \in [\dfrac{L}{2}, L] \\
\end{cases} \\
EI\delta = EI\int \theta\text{d}x &= \begin{cases}
\dfrac{Fx^3}{12} + C_1x + C_3 &\text{ for } x \in [0,\dfrac{L}{2}] \\
\dfrac{FLx^2}{4} - \dfrac{Fx^3}{12} + C_2x + C_4 &\text{ for } x \in [\dfrac{L}{2}, L] \\
\end{cases}
\end{align}$$</p>
<p>We know that $\delta(0) = \delta(L) = 0$ and that $\theta\left(\dfrac{L}{2}^+\right) = \theta\left(\dfrac{L}{2}^-\right)$ and $\delta\left(\dfrac{L}{2}^+\right) = \delta\left(\dfrac{L}{2}^-\right)$ (that is, the deflection and angle is continuous at $\dfrac{L}{2}$).</p>
<p>So you solve it:</p>
<p>$$\begin{gather}
\delta(0) = C_3 = 0 \\
\delta(L) = \dfrac{FL^3}{4} - \dfrac{FL^3}{12} + C_2L + C_4 = 0 \\
\therefore C_4 = -\dfrac{FL^3}{6} - C_2L \\
\theta\left(\dfrac{L}{2}^+\right) = \theta\left(\dfrac{L}{2}^-\right) \\
\therefore \dfrac{FL^2}{16} + C_1 = \dfrac{FL^2}{4} - \dfrac{FL^2}{16} + C_2 \\
\therefore C_1 = \dfrac{FL^2}{8} + C_2 \\
\delta\left(\dfrac{L}{2}^+\right) = \delta\left(\dfrac{L}{2}^-\right) \\
\therefore \dfrac{FL^3}{96} + \dfrac{FL^3}{16} + \dfrac{C_2L}{2} = \dfrac{FL^3}{16} - \dfrac{FL^3}{96} + \dfrac{C_2L}{2} - \dfrac{FL^3}{6} - C_2L \\
\therefore C_2 = -\dfrac{9FL^2}{48} \\
\therefore C_1 = -\dfrac{3FL^2}{48} \\
\therefore C_4 = \dfrac{FL^3}{48}
\end{gather}$$</p>
<p>Now, by inspection we can easily see the deflection is at the midspan, so let's calculate it (doesn't matter which $\delta$ equation you choose).</p>
<p>$$\delta\left(\dfrac{L}{2}\right) = \dfrac{1}{EI}\left(\dfrac{FL^3}{96} - \dfrac{3FL^3}{96}\right) = -\dfrac{FL^3}{48EI}$$</p>
<hr>
<p>The same process can be repeated for a cantilevered beam, only it's much simpler:</p>
<p>$$\begin{align}
M &= FL - Fx \\
EI\theta = \int M\text{d}x &= FLx - \dfrac{Fx^2}{2} + C_1 \\
EI\delta = EI\int \theta\text{d}x &= \dfrac{FLx^2}{2} - \dfrac{Fx^3}{6} + C_1x + C_2 \\
\theta(0) &= C_1 = 0 \\
\delta(0) &= C_2 = 0 \\
\therefore EI\delta &= \dfrac{FLx^2}{2} - \dfrac{Fx^3}{6} \\
\therefore \delta(L) &= \dfrac{FL^3}{3EI}
\end{align}$$</p>
| 13196 | Flexural modulus for a beam fixed at one end |
2017-01-11T17:54:15.843 | <p>Can tensile strength be the same as yield strength? Can Flexural Yield Strength be related to the yield strength of a material in any way?</p>
| |strength|yield-point|unit| | <p>Strength can mean different things in different contexts and technical definitions can vary significantly from what is commonly understood by the word 'strength'. </p>
<p>For structures where actual forces are more relevant you might talk about rated loads, safe working loads or design loads in conjunction with factors of safety but this usually needs to be qualified by specific details of how the load is applied and is much less generally applicable than stress. </p>
<p>In terms of yield it is generally more precise to use the term yield stress. This is a numerical value which applies to a particular material. For example Ultimate Tensile Stress is a common indicator of the resistance of a material to failure. For many ductile materials eg steel UTS is somewhat higher than yield stress, usually by some concistant factor for a given material. For example metric bolt grades are defined by UTS and yield stress as a factor of UTS eg for an 8.8 bolt UTS is 800 MPa and yield stress is 80% of that. </p>
<p>Also crucial is how failure is defined in a specific context and there are quite a few possible modes of failure. </p>
<ul>
<li>Yield : the point at which plastic deformation begins to occur. The material remains deformed after the load is removed but may still be capable of supporting a significant load. </li>
<li>Fracture : the material 'breaks' and is no longer able to support any meaningful load. </li>
<li>Excess strain : the material deflects elastically to the point where it is no longer able to carry out its intended function. This does not necessarily imply permanent damage or deformation to the material. </li>
</ul>
<p>Note also that we tend to use different terms for material properties and structural properties, ie rigidity for a material vs stiffness for a structure. </p>
<p>When discussing flexing it is important to maintain a clear distinction between stiffness/rigidity and strength. Where stiffness is the degree to which a material deforms under a certain load expressed as a ratio (strain) and strength is the load it can resist per unit area without failing. </p>
<p>Equally the stresses experienced by a part are as much a function of its geometry as its material properties. </p>
<hr>
<p>For homogeneous materials (including most engineeriong metal alloys) yield stress is a pretty generally applicable figure, what is more important is the model used to determine stresses as even a simple applied load can result in complex internal induced stresses. </p>
<p>However for certain other materials such as many natural materials, textiles and composites the situation becomes rather more complex as both strength and stiffness can be directional. </p>
<p>Also nominal values for stress are generally based on static loads and apparent material properties can change significantly with strain rate or loads applied over long periods of time. Some materials may also be subject to fatigue failure under cyclic loads well below the static yield stress. </p>
| 13200 | Definitions of strength? |
2017-01-11T19:33:37.457 | <p>Few months ago I opened the boards of an old, scrap radio-cum-tape-recorder (Philips DR183); whose tuning knob (that adjusts the gang); was very peculiar. </p>
<p>( * I did not found its datasheet on web)</p>
<p><a href="https://i.stack.imgur.com/6Xklj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Xklj.png" alt="inside"></a></p>
<p>The channel-indicator is a narrow rod at the tip of red plastic-thing. That indicator moves along a straight-linear segment when I turns the wheel. </p>
<p><a href="https://i.stack.imgur.com/Ytcxu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ytcxu.png" alt="what its work look like"></a></p>
<p>However, it is just a rough sketch. <em>I could not figure out, what mathematical principle/ geometric theorem being used in this mechanic. It is neither like pulley and rope; and nor-like a piston-with-crankshaft. <strong>My question is</strong>, on what-basis they have chosen the angles, distances and ratios to make such movements?
Or, For say if I try to make one with cardboard etc; what angles, lengths, ratios etc I have to choose, and in which logical basis?</em> </p>
<p>Here I've provided some of its allowed conformations. </p>
<p><a href="https://i.stack.imgur.com/0JrEJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0JrEJ.jpg" alt="Steps"></a></p>
<p>Please feel free to move this question if it is more appropriate for any other SE site, such as <a href="https://math.stackexchange.com/">mathematics.SE</a> . </p>
| |mechanisms|geometry| | <p>The mechanism shown in your picture is known as a <a href="https://en.wikipedia.org/wiki/Four-bar_linkage" rel="nofollow noreferrer">four-bar linkage</a>. These mechanisms are used quite often in mechanical engineering to convert rotational motion into complex trajectories, and vice-versa.</p>
<p>The specific mechanism you've shown, as BrianDrummond mentioned in his comment, is <a href="https://en.wikipedia.org/wiki/Chebyshev's_Lambda_Mechanism" rel="nofollow noreferrer">Chebyshev's Lambda Mechanism</a>. This four-bar linkage is often used to produce a linear motion from the rotational motion of a crank, and can be seen in some simple walking robots. There are other ways to do produce linear motion from rotational motion, such as a <a href="https://en.wikipedia.org/wiki/Four-bar_linkage#Slider-crank_linkage" rel="nofollow noreferrer">Slider-crank</a>, another type of four-bar linkage.</p>
<p>As for how to design such mechanisms, the methods are too involved to give an adequate answer on this site, but you can easily find resources online by searching "four-bar linkage synthesis." Keep in mind that there are two different ways of going about the design: graphical methods and analytical methods. With graphical methods you place a series of anchor points and trajectories on a piece of paper, then using some simple geometrical tricks you can easily generate a four-bar that performs that task. Analytical methods involve having a mathematical expression of the trajectory and using that to solve for your four-bar.</p>
| 13203 | How this radio knob converting a rotation into a straight-linear motion? |
2017-01-11T19:36:04.810 | <p>I've been making technical drawings for various parts I've designed. Since these parts were meant to be 3D printed, this is mainly just to have them in my portfolio so I have something to show/talk about at the upcoming career fair (I'm an undergraduate ME student).</p>
<p>I've encountered difficulty with one particular part, which is designed to hold the motor for a perpetually spinning top. Because the motor ran a little hot when turning the rather heavy disc it was attached to, I used the SolidWorks wrap tool to remove portions of the cylindrical casing in order to encourage heat dissipation.</p>
<p><a href="https://i.stack.imgur.com/ko1FR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ko1FR.png" alt="Technical drawing of a motor holder."></a></p>
<p>As you can see in the image, SolidWorks is unable to annotate the resulting dimensions, at least not using the basic tools I'm familiar with. So my question is: how would the dimensions of this part be represented on a technical drawing? Relatedly, since it probably determines the answer: by what process could a part like this be machined, if 3D printing was not an option and the arguably unnecessary complexity could not be removed?</p>
| |solidworks|drawings| | <p>"Try working on a collaborative project with half a dozen large companies all using different CAD software, and the probability that "the solid model" will look the same when they all import it into their own system is just about zero." </p>
<p>I disagree, step file translation and exchange is industry standard works great.. Undimensioned drawings are common practice... Partial dimensioned drawings are also called "Critical Feature Drawings" or "Limited Dimension Drawings".</p>
<p>Google it..</p>
| 13204 | Technical drawing for a part with complex (non-line, non-arc) geometry |
2017-01-11T23:05:47.460 | <p>I need a pump that can remove water from certain volume (~300ml). Water mixed with washing agent (soap) is pumped into this volume using another pump ~5-20L/minute. This volume is not completely sealed - air can get inside (sometimes easily, sometimes in very limited quantities). I need to guarantee that water will not spill out of this volume. In my understanding I need a water transfer pump, but such that can run dry for a while and work as vacuum pump. I looked what market has to offer, but didn't find anything like that.</p>
<p>So the question is - does such pump exist at all? If so, how it is called, so I can search for it online?</p>
| |pumps|vacuum-pumps| | <p>Look at <a href="https://www.dekkervacuum.com/resource-library/knowledge-database/principles-of-operation/two-stage-liquid-ring-vacuum-pump/" rel="nofollow noreferrer">liquid ring vacuum pumps</a> (<a href="http://www.graham-mfg.com/usr/pdf/TechLibVacuum/212.PDF" rel="nofollow noreferrer">sizing tips</a>) (really a version of the vane pumps mentioned by ericnutsch). Any fluid they suck will add to the liquid ring, excess fluid will be discharged. Talk to an applications engineer if the pump will have a problem with soap. These pumps exist at flowrates around 10-20 l/min. </p>
<p>In your situation, however, I would look at diaphragm pumps first and at other solutions if the pulsed flow is a problem.</p>
| 13206 | Transfer pump suitable for air, water and their mixture |
2017-01-12T01:45:53.300 | <p>I imagine this is an easy concept but I do not know much about mechanical engineering design.</p>
<p>I am trying to make a "knob" like a control knob in solidworks that turns and is locked in place after twisting. I am thinking of a twisting knob in a labyrinth game or something like a foosball table knob that is locked in place after turning, or just a general control knob on an oven. I want to figure out how to design the "interface" that lets the knob turn yet stay in place after turning but do not know enough vocabulary to point myself in the right direction. Does the "interface" I am looking for have a name? How does one go about making it?</p>
| |mechanical-engineering|solidworks| | <p>A cogged positioning system uses the concept of a <em>detent</em> - a pin, ball, pawl or bump that fits against a mating cavity. Usually there is some kind of spring to allow the detent to disengage when enough force is applied. This gives the familiar click as a knob is rotated.</p>
<p>So to answer the question the interface is called a detent. Google search <em>detent ball</em> for examples of how a detent works.</p>
<p>Wiki definition here <a href="https://en.wikipedia.org/wiki/Detent">https://en.wikipedia.org/wiki/Detent</a></p>
| 13208 | Make something that turns yet locked in place |
2017-01-13T06:40:21.540 | <p>I have a laptop computer (Dell L701X) which requires an 11.1 V battery with a rating of 8400 mAh (Battery Type R795X, 9 cell, 11.1 V, 90 Wh, 8400 mAh, other battery ID are: R1122J-733 & B052R796-9021).</p>
<p>If I place a new 11.1 V battery with the same physical dimensions in the computer, but with a rating of 5200 mAh the computer doesn't like the battery. On start-up the computer beeps & gives a text warning on the black start up screen. The on-keyboard battery indicator flashes between white & amber when both my old & new battery are in the computer. After boot up, the battery icon in the task bar acknowledges the 5200 mAh battery but the computer will not charge the battery - the battery is new and it only has a charge of 17%.</p>
<p>Does the computer need a 8400 mAh battery or is there a fault with the battery charger in my computer?</p>
| |electrical-engineering|battery| | <p>If dell recommends the use of a 5200mAh battery then it should work. There is something else not in sync with the laptop. I would suggest that you update the BIOS firmware. All thought the computer recognizes the lower capacity battery, there might be some parameters that need to be updated. Below is link that might help you.</p>
<ul>
<li><a href="http://www.dell.com/support/home/us/en/4/Products/?app=drivers" rel="nofollow noreferrer">Dell Drivers & Downloads Identify your product</a></li>
</ul>
<p>Although highly unlikely your power adapter might not match in new battery. If you a getting error similar to the one below you might want to follow the direction. </p>
<p><a href="https://i.stack.imgur.com/i34Oj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i34Oj.jpg" alt="enter image description here"></a></p>
<p>There might be some discrepancies between your laptop BIOS, power adapter and the new battery. Here is an example where there was inconsistency between the power adapter and the BIOS. </p>
<p><strong>Before</strong></p>
<p><a href="https://i.stack.imgur.com/An6Se.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/An6Se.jpg" alt="Batter adapter not recognized"></a></p>
<p>After replacing the power adapter.</p>
<p><a href="https://i.stack.imgur.com/0nmeA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0nmeA.jpg" alt="enter image description here"></a></p>
<p>I have listed the links to help you further perform root cause analysis. </p>
<p><strong>References:</strong></p>
<ul>
<li><a href="http://www.dell.com/support/Article/us/en/19/266588/EN" rel="nofollow noreferrer">AC Adapter Not Powering on the Computer or Charging the Battery</a></li>
<li><a href="http://www.dell.com/support/article/us/en/19/SLN85590#Issue7" rel="nofollow noreferrer">How to improve your laptop battery's performance?</a></li>
<li><a href="http://pcroger.com/dell-laptop-battery-will-not-charge/" rel="nofollow noreferrer">Dell Laptop Battery Will Not Charge</a></li>
<li><a href="http://en.community.dell.com/support-forums/laptop/f/3518/t/19452828" rel="nofollow noreferrer">(XPS L701X) Battery not charging, computer shuts down instantly if unplugged</a></li>
</ul>
| 13231 | Battery mAh ratings |
2017-01-13T17:25:23.947 | <p>Trying to design a 2:1 gear ratio planetary gear, and I'm not sure how to begin. In the past I've had luck using <a href="http://geargenerator.com/" rel="nofollow noreferrer">this</a> site to create an illustrator file that I can extrude in Rhino. This does not generate planetary gears, and I'm also unclear as to how design a planetary gear with a 2:1 ratio.</p>
| |mechanical-engineering|gears|servo| | <p><a href="https://i.stack.imgur.com/xSThp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xSThp.png" alt="enter image description here"></a>
(Image from Wikipedia)</p>
<p>When designing planetary gears, there are three different types of gear present to be designed:</p>
<ul>
<li>Annulus gear, an internal gear surrounding the other gears (shown above in red)</li>
<li>Sun gear, an external gear in the middle of the gear set (shown in yellow).</li>
<li>Planet gears, external gears surrounding the sun gear, usually three or four of them are present (shown in blue).</li>
</ul>
<p>First of all, it's worth noting about internal gears. The sun and planet gears, being external gears (i.e. teeth point outward), are the exact same as any other kind of spur gear. However, the annulus gear is an internal gear, with the teeth pointing inward. An internal gear may look like a circle with an external gear subtracted from inside of the circle, and it almost is; there is just a slight difference in how far the teeth extend from the pitch circle. Not noting this difference in internal gears can lead to gear teeth undesirably colliding before properly meshing, causing the gear set to jam. Therefore, if you are fine with gears being output as .dxf files, then I suggest using <a href="http://hessmer.org/gears/InvoluteSpurGearBuilder.html" rel="nofollow noreferrer" title="Dr. Rainer Hessmer's gear generator">Dr. Rainer Hessmer's gear generator</a>, as it allows you to create both external and internal gears.</p>
<p>When it comes to actually designing these three different types of gears, the following equation, which I'll refer to as the <em>speed rule</em>, should be observed:</p>
<p>$$\omega_C (1+\rho) = \omega_A \rho + \omega_S$$</p>
<p>$\omega_C$ is the rotation speed of the <em>carrier</em>, which is shown in green in the above image. $\omega_A$ is the rotation speed of the annulus gear and $\omega_S$ is the rotation speed of the sun gear. Rotation speeds are all defined to be positive in the same direction of rotation. $\rho$ is the ratio of the number of teeth on the annulus gear to the number of teeth on the sun gear. It is worth noting that $\rho$ cannot be less than or equal to 1.</p>
<p>Often, out of the carrier, the annulus gear and the sun gear, one is selected to be fixed in place ($\omega=0$), one is selected to be the input and one is selected to be the output (it is however possible to have two inputs or two outputs). By fixing one of the rotation speeds, a gear ratio can be determined between the input and output, given as follows:</p>
<p>Fix the carrier:
$\omega_C = 0, \frac{\omega_S}{\omega_A} = -\rho$</p>
<p>Fix the annulus:
$\omega_A = 0, \frac{\omega_S}{\omega_C} = \rho+1$</p>
<p>Fix the sun:
$\omega_S = 0, \frac{\omega_A}{\omega_C} = 1+\frac{1}{\rho}$</p>
<p>It is now a matter of selecting one of the above configurations that give you the appropriate gear ratio. Sometimes it requires a bit of trial and error: stick with a configuration and determine the value of $\rho$. If $\rho$ is less than or very close to one, try another setup.</p>
<p>For your case, with a gear ratio of $2:1$, let's try the following configuration:</p>
<p>Fixed: carrier</p>
<p>Input: sun</p>
<p>Output: annulus</p>
<p>Note that in this configuration, $\frac{\omega_S}{\omega_A} = -\rho$, so the input and output need to be rotation in opposition direction to get a positive value of $\rho$. Let's say the input is spinning in a "negative" direction and the output in a "positive" direction. Therefore:</p>
<p>$$\frac{\omega_S}{\omega_A} = \frac{-2}{1} = -2$$</p>
<p>$$\rho = 2$$</p>
<p>A value of $\rho$ which is greater than 1 is obtained: this configuration seems to work well. Now we enter the second phase of the planetary gear design: assigning the number of teeth.</p>
<p>First of all, the planet gears are generally spaced equally apart from one another. To achieve this, the number of teeth of both the sun gear and the annulus should be a multiple of the number of planets. Otherwise, difficulties may arise when it comes to assembling the gears together. This is a process that may require some trial and error, and it may even required very slightly changing the value of $\rho$ to make it easier to make the teeth number multiples.</p>
<p>In our example let's assume that there are four planet gears. Therefore, it can be seen that a sun gear of 31 (not multiple of 4) teeth is not appropriate, but one of 32 (multiple of 4) teeth is. If the sun gear has 32 teeth, the annulus gear will have $32 \times \rho = 64$ teeth. 64 is also a multiple of 4 and is therefore appropriate.</p>
<p>Once the number of teeth for the sun and annulus gears are selected, the number of teeth for the planet gear is calculated using the following formula:</p>
<p>$$\text{planet teeth} = \frac{\text{annulus teeth} - \text{sun teeth}}{2}$$</p>
<p>Note the division by two: this means that the number of teeth on the annulus and sun gears must both be odd numbers, or both be even numbers. If this turns out not to be the case, doubling the number of teeth on the sun and annulus gears quickly resolves the issue. In our example, the number of teeth on each planet gear is 16.</p>
<p>With the number of teeth selected, now select an appropriate module or diametral pitch, depending on how strong you want the gears, and that is the planetary gear set designed.</p>
<p>So in summary:</p>
<ul>
<li>Use speed rule to select appropriate configuration </li>
<li>Determine ratio of annulus teeth to sun teeth</li>
<li>Determine number of teeth for each gear</li>
<li>Select the module/diametral pitch</li>
</ul>
| 13248 | How to I design a planetary gear with a gear ratio of 2:1 ( doubling the torque) |
2017-01-14T00:51:53.653 | <p>I am not sure if this is the best place to put this, but if I'm in the wrong place please tell me where I can go. So I know a peltier module is not the most efficient, however my goal is to make a hand warmer / cooler with a petlier module. My operating temperatures will be around 60f to 90f, nowhere near the max of most peltiers. My first question is can it be powered by a small battery? Second, would I need a heatsink with this kind of temperatures that I want to produce and if it's required, is there any passive cooling I can use without being thick? Third, is this idea practical?</p>
| |electrical-engineering| | <p>For heating, it would be practical. There are <a href="https://www.milwaukeetool.com/power-tools/cordless/2345" rel="nofollow noreferrer">battery powered coats</a> that are currently on the market. A peltier module pumps heat, plus makes a heat; so you get more heat than electricity you put in. So instead of getting 100% heat from electricity in a resistive heater, you are getting 150% heat (for example) from electricity in a heat pump (peltier module in this case).</p>
<p>A heatsink will increase the efficiency by allowing more heat to be pumped. Lots of variables, but lets say an increase from 150% to 180%. Add a fan to that heatsink and say an increase from 180% to 200%. I would recommend starting with a heatsink at least, and increasing or decreasing the chosen voltage to achieve your desired temperature. If you have a 12v module, start with something around 6 volts. Remember not to drain lipo batteries too low and to watch battery chemistry or you could damage them or start a fire.</p>
<p>Good luck!</p>
| 13250 | Peltier Module to warm hands? |
2017-01-15T04:23:29.840 | <p>In a <a href="https://en.wikipedia.org/wiki/Split-flap_display" rel="nofollow noreferrer">split-flap display</a> there is a barrel which rotates by a small angle N times to get to some next display element. However I fail to understand, how it is possible that the barrel rotates <strong>exactly</strong> by some desired angle? How is the rotation error after each flip is not accumulating? The only explanation I can think of is that it has sensors that sends signal back to the the motor to stop it. </p>
<p>Or is it possible that a motor can make exact rotations without error accumulation? I don't think it is possible unless the motor itself has built-in controller systems which resets the position each time to eliminate the error.</p>
| |mechanical-engineering|electrical-engineering|gears| | <p>The barrel is usually rotated by a ratchet mechanism, not directly by a rotary motor. The ratchet might be driven by such a motor (e.g., a clock) or by a solenoid (e.g., railroad station status board).</p>
<p>In the former case, the display is initialized manually. In the latter, there is usually a microswitch that indicates the index position of the barrel.</p>
| 13263 | How split-flap display rotates by exact angle |
2017-01-15T04:28:03.740 | <p>Should standard SI units be used in this formula or are other units supposed to be used for the formula for stress in a thin-walled spherical pressure vessel?</p>
<p>$$\sigma_{yield}=\frac{P_g r_i}{2t}$$</p>
<p>where $\sigma_{yield}$ is the yield strength of the material, $P_g$ is the gauge pressure, $r_i$ is the inner radius of the sphere, and $t$ is the thickness of the sphere wall.</p>
| |pressure|measurements|unit| | <p>Yield should be in MPa, Pressure Pa, and R in mm t is thickness therefor mm</p>
| 13264 | Units to use in this formula? |
2017-01-15T14:24:06.860 | <p>In the solution , we notice that the 2823.6 is the result of carry over factor from BA to AB . Why we dont need to do the same thing for BC to CB ? <a href="https://i.stack.imgur.com/lS69U.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lS69U.jpg" alt="dd"></a>
<a href="https://i.stack.imgur.com/fSpNN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fSpNN.jpg" alt="dd"></a></p>
| |structural-engineering|structural-analysis| | <p>Because it is an end pin joint, it is assumed that no moment can occur in the joint. </p>
<p>The Moment carried from BA to AB because AB is a fixed joint, and thus incur a moment as a result of the moment at joint B.</p>
<p>Check out the <a href="https://en.wikipedia.org/wiki/Moment_distribution_method" rel="nofollow noreferrer">Moment Distribution Method</a> Wikipedia page and look at their worked example, it should help clear things up. </p>
| 13271 | carry over factor in beam ( moment distribution ) |
2017-01-16T14:50:49.170 | <p>In a laminar flow along an Annular cross section with heating in one side and insulation in the other, what is the value of the Nusselt number in this flow? and is the diameter in the Nusselt expression Nu=h*D/k the hydraulic diameter?</p>
| |fluid-mechanics|thermodynamics| | <p>For</p>
<p>$Re < 2300$</p>
<p>$0.1 \leq Pr \leq \infty$</p>
<p>$0.1 \leq Pr \frac{d}{h} < 10^4$</p>
<p>I found the correlation</p>
<p>$Nu_m = \left( 49.37 + (1.615\frac{Pe \cdot d}{h})^{\frac{1}{3}} -0.7)^3\right)^{\frac{1}{3}} K$</p>
<p>With</p>
<p>d = hydraulic diameter</p>
<p>h = length of pipe</p>
<p>$K = 1$ for gases</p>
<p>$K = \left(\frac{Pr_f}{Pr_w}\right)^{0.11}$ for liquids</p>
<p>$Pr_f$ = Pr for fluid temperature</p>
<p>$Pr_w$ = Pr for wall temperature</p>
<p><a href="http://www.schweizer-fn.de/waerme/waermeuebergang/waerme_uebergang.php" rel="nofollow noreferrer">Source</a></p>
<p>I'm not a native speaker, so someone please feel free to add a source for that correlation in English.</p>
<p>And if the geometry is not circular, you use the hydraulic diameter, as always.</p>
<p>You may derive it from</p>
<p>$d_h = \frac {4 A}{U}$</p>
<p>Where</p>
<p>A denotes the crosssectional area and U the circumference.</p>
| 13283 | What is the value of Nusselt number in laminar flow? |
2017-01-16T15:47:48.397 | <p>My research has found a <a href="https://www.quora.com/What-is-the-difference-between-a-plunger-and-a-piston" rel="nofollow noreferrer">Quora link</a> and another <a href="http://www.drucklufttechnik.de/www/temp/e/drucklfte.nsf/b741591d8029bb7dc1256633006a1729/959ffe76f432a534c125662500774784?OpenDocument" rel="nofollow noreferrer">source</a> giving two different, albeit not contrasting, definitions. Also, as far as I have seen, looking up "plunger compressor" also refers me to pumps.</p>
<p>I am looking to understand the proper definitions, and see some animated GIFs should you have some, that exactly tell these two apart.</p>
| |mechanical-engineering|pumps|compressors| | <p>To the best of my knowledge, the main apparent difference between plunger and piston is that piston is short and plunger is long. You can see it on the picture in the @ericnutsch answer.</p>
<p>Hydraulic engineers employ plungers in different situations. Personally, I am only aware of two.</p>
<p>First is pressure increase, as @GisMofx asked and as @PeerapongKladkrayang responded. Simplifying the computational schematic without loss of generality but emphasizing the considering effect, one can say that pressure produced by a pump equals to the force exerted on a piston/plunger rod divided by the piston/plunger area: <span class="math-container">$$p = \frac{F}{A}$$</span> Let's consider two cylinders of the same crossectional area. Let's place a piston to the first cylinder and a plunger to the second one (as it is shown on the picture in the answer of @ericnutsch). Let's exert the same force, say <span class="math-container">$F = 100N$</span>, on the rods of both of them. Let's say that the area of the cylinder and, therefore, the area of the piston is <span class="math-container">$A_{pist} = 10 m^2$</span>; and the area of the plunger is two times smaller: <span class="math-container">$A_{plun} = 5 m^2$</span>.</p>
<p>Now let's compute what pressure we can obtain with piston and plunger in the same cylinder exerting the same force on them: <span class="math-container">$$p_{pist} = \frac{100N}{10m^2} = 10N/m^2$$</span> <span class="math-container">$$p_{plun} = \frac{100N}{5m^2} = 20N/m^2$$</span> See: in the same cylinder reducing the working crossectional area twice, I gain a double amplification in resulting pressure.</p>
<p>Therefore, plunger allows me to increase working pressure without increasing the size of my cylinder. It may be helpful in some cases when I am restricted in space dedicated for cylinder installation. E.g., I am working on creation of a compact excavator. Everything needs to be as small as possible, providing as much resulting pressure as possible.</p>
<p>The tradeoff here is that I lose a lot of power on pumping liquid between the plunger and the inner cylinder wall. Hence, efficiency of a plunger pump should be smaller. It means that I can pump out less fluid than in a piston pump, and the outer power of plunger pump is smaller (it can be shown through some formulas).</p>
<p>Another application of plungers is in positive displacement pumps. Such as axial piston pumps, for instance. When a plunger moves in the cylinder, a liquid film between the plunger and the inner wall of the cylinder prevents the plunger from getting stuck. It is as if the liquid film supports the plunger just because the plunger is long. In turn, piston is short. And even if it had a supporting liquid film it would've easily gotten stuck, merely because it is short.</p>
<p>Notice, that distinguishing between plunger pump and piston pump is a product of engineering practice. It is not a scientific term. Therefore, there is no a strict general definition of a plunger and a piston. Many years ago, when engineers were designing early hydraulic devices, they figured out that it is a smart idea to reduce a working crossectional area in a hydraulic cylinder. That's it. If you tried and ask a fluid mechanics professor about piston/plunger definition - you would most likely get an answer that one is short and the other is long (as I did in my time).</p>
<p>Also, notice that there is much more to add about the second application. But speaking about relative lengths of pistons and plungers - that's the only thing I can say.</p>
| 13287 | What is the difference between a piston and a plunger compressor/pump? |
2017-01-17T01:09:53.690 | <p>I'm an EE student by trade, but I've started a project that requires some mechanical analysis, and I've run up against a bit of a wall in my calculations.</p>
<p>The system in question that I'm looking at is an actuator, connected via ball joint to a rod, which is then connected with a ball joint to a platform. During operation, the platform will be constrained to movement in either only the x direction, or only the y direction. I'm trying to find out, based on the force of the actuator and the geometry, what the X and Y components of theses forces will be.</p>
<p><a href="https://i.stack.imgur.com/gbX60.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gbX60.png" alt="enter image description here"></a></p>
<p>(Apologies for the cartoonish drawing). The actuator is on the bottom left. For the purposes of calculation, I'm assuming that the rod is rigid, has zero mass, and that the actuator does not need to move to apply any amount of force.</p>
<p>I'm using static equilibrium to calculate the X and Y components of the force on the upper right hand side, which will be the X and Y force that my actuator applies to the moving platform.</p>
<p>The actuator will apply an initial force Fa to the rod. Since both supported ends are ball joints, I know that there can be no moments on either end of the rod, only forces acting in-line with it.</p>
<p>In the case of $\theta_M$ being equal to $\theta_R$, the solution is straightforward: all of the force is being applied in-line with the rod, so we set the sum of forces in X and Y equal to zero, and solve.</p>
<p>However I'm having trouble wrapping my head around what happens if $\theta_M \neq \theta_R$.</p>
<p>My naive approach to solving this would be as shown below:</p>
<p><a href="https://i.stack.imgur.com/dv3v5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dv3v5.png" alt="enter image description here"></a></p>
<p>Break down the initial applied force into components that are parallel with the rod, and components that are perpendicular to the rod. </p>
<p>Here's the part I'm unsure about. Since the actuator can't move, the perpendicular force <em>should</em> be cancelled by an equal and opposite normal force of the actuator pushing back against the rod, correct?</p>
<p>However something in the back of my head is telling me that I'm missing something. </p>
<p>Intuitively, if we were to apply the actuator force not in-line with the axis of the rod, I'd think that there would be some kind of lever action present. An example would explain what I'm trying to say best I think. Can I not analyze a system like this statically? Or am I just completely off base?</p>
<p>EDIT: to make the point I'm trying to get at clear, I was able to find a mechanism that outlines what I'm talking about:</p>
<p><a href="https://i.stack.imgur.com/IC8ae.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IC8ae.jpg" alt="enter image description here"></a></p>
<p>If the wheel is driving the slider, won't the output force of the slider be at its peak value when members 2 and 3 are near parallel to one another? I feel like the same effect should be at play in my system above, but I don't know how to mathematically articulate that. Or my intuition about when the force is at peak value could be totally wrong.</p>
| |statics|dynamics| | <p>Your equations are mostly correct, but you're missing the tangential load on the actuator:</p>
<p>$$\begin{array}{l}
{F_a}_p & = & \text{Force parallel to the actuator arm at the joint} \\
{F_a}_t & = & \text{Force tangent to the actuator arm movement at the joint} \\
F_p & = & \text{Force parallel to the rod} \\
F_t & = & \text{Force perpendicular to the rod}
\end{array} $$</p>
<p>$$F_p={F_a}_{p} \cos(\theta_m - \theta_r) - {F_a}_{t}sin(\theta_m - \theta_r)$$
$$0=F_t={F_a}_{p} \sin(\theta_m - \theta_r) + {F_a}_{t}cos(\theta_m - \theta_r)$$</p>
<p>These equations can be rearranged to give:
$${F_a}_p=F_p \cos(\theta_m - \theta_r)$$
$${F_a}_t=F_p \sin(\theta_m - \theta_r)$$</p>
<p>Or rearranged again:</p>
<p>$$F_p = {F_a}_t \csc(\theta_m - \theta_r)$$</p>
<p>This does indeed give you a multiplying effect as the difference in angles approaches 0.</p>
| 13291 | Equilibrium of a Rod with an Actuator |
2017-01-17T09:39:06.980 | <p>I am studying Gaudí and how does he created some of his work. I found that he used inverted weight models (as the shown in the picture below) to recreate the catenary. In many articles it is pointed as a good practice to simplify and find a better/more sophisticated and solid structure, but I can't see why.</p>
<p><a href="https://i.stack.imgur.com/1467f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1467f.jpg" alt="enter image description here"></a></p>
| |structures|architecture| | <p>A catenary is the shape a cable will hang in under self-weight. Cables only work in tension, not in bending. If you take this shape and flip it upside down, then (due to gravity now operating in the opposite direction relative to the structure) the structure is purely in compression, with no bending.</p>
<p>Clearly this wouldn't work for a cable, as they don't work well under compressive loads. But it does work well for concrete structures, or for steel sections.</p>
<p>Axial force <strong><em>P</em></strong> over an area <strong><em>A</em></strong> causes a stress of <strong><em>P/A</em></strong>. Moment <strong><em>M</em></strong> on a section with an elastic section modulus <strong><em>Z</em></strong> causes a stress <strong><em>+M/Z</em></strong> on one edge, linearly decreasing to 0 at the neutral axis, and further decreasing to <strong><em>-M/Z</em></strong> on the other edge.</p>
<p>So, if you have a section under compression and bending, then stress at one edge is equal to <strong><em>P/A + M/Z</em></strong>, and at the other edge <strong><em>P/A - M/Z</em></strong>. Clearly if you design so that your maximum stress (P/A + M/Z) is equal to your allowable stress, then most of your section is at less than the allowable stress. However, if you only have compression, then your whole section is at P/A, and if you design such that this is equal to your allowable stress then your whole section is fully utilised - much more efficient.</p>
| 13298 | Does the use of the catenary in an inverted weight model imply a more solid structure? |
2017-01-17T23:35:25.920 | <p>I'm trying to build a self-regenerating door lock as a side project in college and i'm having some difficulty accurately calculating the gear trains output (in rpm), the purpose of this gearing system is to operate a AC generator. Any help at all would be much appreciated!</p>
<p>I've estimated that a torque of about 1.12 Nm will be used to turn the door handle down 50 degrees. This torque will be transmitted through the shaft which is attached internally to a one way gear. The one way gear is attached to both a fly wheel and the gear train.</p>
<p>My goal is to get an output of about 100 rpm to operate the generator. My question is what factors must i include in the calculation? do i need to include things such as the weight of the gear train and how do i factor in the fly wheel? Thanks!</p>
| |generator| | <p>Obtain a cheap plastic hand crank generator or several and play.<br>
That will show you much of what you need to know. </p>
<p>If you turn the handle 1/8th turn (45 degrees) in 1/4 second then its<br>
RPM is 60/time_to_turn x fraction of turn =<br>
60s/0.25s x 1/8 turn ~~= 30 RPM.<br>
In engineering 30 ~~~= 100 in many cases :-).<br>
But if you want closer to 100 RPM then a 3 or 4 to 1 step up gear ratio will give around 100 RPM. </p>
<p>SO coupling the shaft to the alternator with a one way clutch and a say 4:1 speed up gearbox gives you about what you want. One stage of the epicyclic gearbox in many battery electric drills may be about right. These may have multiple stages to get higher ratios. </p>
<p>Just turning the chuck on a battery electric drill and measuring motor voltage produced may be worthwhile. Slightly tidier than <a href="http://media.treehugger.com/assets/images/2013/02/macgyver-generator.jpg.650x0_q70_crop-smart.jpg" rel="nofollow noreferrer">this</a> ideally.</p>
<p>A useful rule of thumb is Power is ~~~= RPM x kgm
Here RPM = 100 and torque = torque in / gear ratio = 1.12 Nm/4 = 0.28 Nm ~= 0.028 kg.m<br>
So power = kg.m x RPM = 0.028 x 100 = 2.8 Watts.
This is produced only during 1/4 second of turning so energy<br>
= power x time = 2.8W x 0.25s = 0.9 Watt seconds = 0.9 Joule.</p>
<p>A useful amount of energy.</p>
<p>My experiments with cheap plastic squeeze type hand alternators shows they make about 1 Watt with very enthusiastic cranking so the above calculation is "in the order of right".</p>
<p>Flywheel does not alter energy delivered but can modify user handle-feel and rate of delivery - eg if a flywheel is spun up by a 1 way clutch it may spin down over 1+ seconds and deliver energy at a less peaky rate. </p>
<p><strong>______________________________________</strong> </p>
<p><strong>Bonus:</strong> Obtaining energy from the opening or closing door is liable to be rather more productive if suitable to purpose. The door can eg have a spring closer and drive the alternator on one or both halves of open/close cycle. Energy taken from this source is liable to less noticed and easier to obtain in greater quantity. </p>
<hr>
<p><a href="https://www.google.co.nz/search?tbm=isch&sa=1&q=hand+crank+generator&oq=hand+crank+generator&gs_l=img.3..0l9.989370.994934.0.995430.20.15.0.5.5.0.358.2325.2-9j1.10.0....0...1c.1.64.img..5.15.2338...0i67k1j0i10k1.tJlPO4CMl9o" rel="nofollow noreferrer"><strong>Many</strong> ideas here</a><br>
<a href="https://www.google.co.nz/search?tbm=isch&sa=1&q=hand+squeeze+generator&oq=hand+squeeze+generator&gs_l=img.3...44995.47128.0.47963.8.8.0.0.0.0.226.1697.2-8.8.0....0...1c.1.64.img..0.1.225...0i7i30k1j0i8i7i30k1.dnq3vanXmNo" rel="nofollow noreferrer"><strong>and here</strong></a> - each image links to a related page.</p>
<hr>
<p>Cheap plastic hand crank alternator.
Mechanical efficiency is usually poor.
Note gear ratio is probably higher than apparent as another gear is probably not shown. ie their alternator spins MUCH faster than yours. The plastic Z-ish shaped piece in the flywheel acts as a one way clutch/pawl.<br>
<a href="https://commons.wikimedia.org/wiki/File:%D0%9C%D0%B5%D1%85%D0%B0%D0%BD%D0%B8%D1%87%D0%B5%D1%81%D0%BA%D0%B8%D0%B9_%D1%84%D0%BE%D0%BD%D0%B0%D1%80%D1%8C_inside_(1).jpg" rel="nofollow noreferrer">From Wikipedia</a></p>
<p><a href="https://i.stack.imgur.com/10fwc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/10fwc.jpg" alt="enter image description here"></a></p>
<p><a href="http://patentimages.storage.googleapis.com/US7893551B2/US07893551-20110222-D00000.png" rel="nofollow noreferrer">Patent imahge</a><br>
from here <a href="https://www.google.com/patents/US7893551" rel="nofollow noreferrer">hand squeeze generator</a> </p>
<p>You can buy <a href="http://www.ebay.com/itm/3-Phase-AC-Micro-Mini-Wind-Hand-Brushless-Generator-Teaching-Model-DIY-Motor-24V-/272526120811" rel="nofollow noreferrer">these</a> for far far too much $ on ebay OR pull apart old CD or DVD drives.</p>
<p><a href="https://i.stack.imgur.com/4CMEx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4CMEx.jpg" alt="enter image description here"></a> </p>
<p>Stepper motors also may serve. Old large ones from 5.25" floppy drives (remember them :-) ) are usually better.</p>
| 13310 | How to accurately calculate the rpm from a gear train |
2017-01-19T08:39:15.300 | <p>I need to calculate the plastic section modulus of a rectangular section with rounded corners. First I need to know the formula for first moment area of a quadrant. I can't find it anywhere on the internet. Does anyone know?</p>
| |civil-engineering|mathematics| | <p>I don't have a simple formula but that is the way i would handle it. Assuming the four corners are circular with equal radius r, so that symmetry exists, the first moment of area of <strong>one half</strong> of the section should be:</p>
<p>$$S_{half} = S_{half\,rect}-2S_{corner}$$</p>
<p>where </p>
<ul>
<li>$S_{half\,rect} = b(h/2)(h/4) = \frac{bh^2}{8} $ the 1st moment of
area of the half rectangle (with intact corners) around neutral axis and</li>
<li><p>$S_{corner}$ the 1st moment of area of the removed material from each
corner, around neutral axis.</p>
<p><a href="https://i.stack.imgur.com/GpzzT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GpzzT.png" alt="basic shape"></a></p></li>
</ul>
<h2>How to find $S_{corner}$</h2>
<p>We have to find its area and the distance of its centroid from the section neutral axis. It helps to consider the removed corner as a quarter circle subtracted from an r x r square (see figure below). The area is then simply found, by subtracting form the small r x r square, a circular quarter (see figure below):
$$A_{corner} = r^2 - \frac{πr^2}{4} $$</p>
<p><a href="https://i.stack.imgur.com/DMlOz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DMlOz.png" alt="detail of the removed corner"></a></p>
<p>The centroid of the removed corner area, relative to the top edge, is similarly found considering the centroid of the rxr square (red dot) which is located $y=r/2$ below top edge and the centroid of the quarter circle (blue dot) which is located $y=r-\frac{4r}{3\pi}$ from the top edge. Combining them both, we get the distance of centroid of the removed corner from the top edge:</p>
<p>$$ y_{corner} = \frac{1}{A_{corner}}\left(r^2\cdot\frac{r}{2} - \frac{\pi r^2}{4} \left(r-\frac{4r}{3\pi}\right)\right) $$</p>
<p>Having found the above, the first moment of area of one removed corner around the section neutral axis is:
$$ S_{corner} = A_{corner}\left(\frac{h}{2} - y_{corner}\right)$$</p>
<h2>Finally</h2>
<p>Substituting $ S_{corner}$ to the 1st formula, we get the 1st moment of area of the half section $S_{half}$. Then, the plastic modulus of the total section, taking advantage of symmmetry, is:</p>
<p>$$Z=2S_{half}$$</p>
<p>For verification of the result <a href="http://calcresource.com/cross-section-rectround.html" rel="nofollow noreferrer">a rounded rectangle calculator</a> may prove handy. </p>
| 13326 | What is the first moment area of a rounded rectangle? |
2017-01-19T20:28:27.133 | <p>How can I make an elevator that goes up while twisting like in <a href="https://youtu.be/KCElBEfwLfc" rel="nofollow noreferrer">this video</a>? (The man that made this used an old printer)</p>
| |mechanical-engineering| | <p>In this case it looks like the lift and rotation elements are mechanically separate ie the lift mechanism sits on a rotating turntable. </p>
<p>Here the key element is to be able to provide electrical power to a rotating assembly, I can't see in detail how this is done in this particular case but <a href="https://en.wikipedia.org/wiki/Slip_ring" rel="nofollow noreferrer">slip rings</a> would be the conventional solution. </p>
| 13340 | How to make a mechanical elevator with rotation? |
2017-01-19T23:17:42.650 | <p>Does some material/metal exist that can be stretched and still transfer current? </p>
<p>For example:
Say I have a rectangular item made of some metal or material(unknown to me at the time). On both ends, I have two small sensors attached. These sensors do nothing but want to establish communication between each other. For the purpose of this question, let's refer to them as sensor A and sensor B. Sensor A sends a message to Sensor B. Since the metal is not very stretched, it manages to transfer the message in electrical current to Sensor B in an elapsed time of .1 second.</p>
<p><a href="https://i.stack.imgur.com/yG2xK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yG2xK.png" alt="enter image description here"></a></p>
<p>Now, imagine that we stretch the metal a bit more, where its width is lowered and its length is increased. Since the message now has to travel further, I would expect the delivery to be greater than .1 second. </p>
<p><a href="https://i.stack.imgur.com/ebcB8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ebcB8.png" alt="enter image description here"></a></p>
<p>Note: Numbers used are just an example.</p>
<ul>
<li><p>What sort of material or metal can I use to pass current, that when stretched, would alter the total distance a current would have to travel?</p>
<ul>
<li>Ideally, the two sensors will at the minimum be separated by 3-4 inches, and the metal/material would only stretch up to 1 inch.</li>
</ul></li>
</ul>
| |mechanical-engineering|electrical-engineering|metals| | <p>First of all, with your 4 inch distance, the propagation time wouldn't be 0.1 seconds. It would be 0.000 000 000 339 seconds. Measuring such short time differences is a significant challenge and there are no obvious off-the-shelf solutions capable of it.</p>
<p>But more than that, the propagation time doesn't depend only on the distance. The shape and surroundings of the conductor also affects it, due to <a href="https://en.wikipedia.org/wiki/Transmission_line">transmission line effects</a>. Thus even if you had the propagation time, it would need a more accurate modelling of the situation to arrive at a distance reading.</p>
<p>You may want to ask another question about what kind of distance sensors would be applicable to your problem, but you'll also have to give more details about the actual goal that you want to achieve.</p>
| 13343 | Strechable object that can transfer current |
2017-01-21T01:52:27.167 | <p>I was working on a project where I have to launch a ball. The current design has two wheels that are side by side with the ball going between them. I noticed that in almost every commercial ball launcher, the wheels are stacked on top of one another rather than side by side. </p>
<p>I don't understand what advantage this has over my current system. It seems a lot more accurate and precise, but I don't see why.</p>
| |wheels| | <p>Quite a few of the machines allow to tilt the pair of wheels to arbitrary angle too. Some allow tuning speed of the two motors separately.</p>
<p>It's all about curveballs and <a href="https://en.wikipedia.org/wiki/Magnus_effect" rel="nofollow noreferrer">Magnus Effect</a>. One wheel going faster will give the ball a spin on top of the forward speed - and the ball trajectory will curve in direction perpendicular to rotation axis and current velocity vector. "Stacked" layout allows to pitch the ball a longer distance, by counter-acting gravitational drop, and keeping the trajectory more straight. Sideways placement will tend to curve the ball to a side. Which you want depends on how you want to train.</p>
| 13364 | Why do baseball pitching machines have wheels stacked instead of side by side? |
2017-01-22T00:02:00.843 | <p>I'm a 3d artist and I'm trying to produce a type of scaffolding within a biodome. </p>
<p>For lack of knowledge in technical terms, I'm having a hard time learning about the design pattern/structure that I must follow (I suspect just making a triangle-array would be ignorant).</p>
<p><strong>What is the technical term for the triangular scaffolding often used in future-esque projects</strong>? I've seen it used to hold up giant solar arrays as well as biodomes, etc. </p>
<p>If I only knew what it was called I could go ahead and research it!</p>
| |architecture| | <p>You may be thinking of geodesics, commonly seen in geodesic domes. </p>
<p>"Geodesics" or geodetics refer to lines following a <a href="https://en.wikipedia.org/wiki/Geodesic" rel="nofollow noreferrer">"great circle"</a> route across a surface. They can produce strong structures with no internal reinforcement, maximising usable space (in principle ... in practice, filling the space with cuboid furniture is problematic!)</p>
<p>One early exploitation of this was in late 1920s airship designs, where it is associated with <a href="https://en.wikipedia.org/wiki/Barnes_Wallis" rel="nofollow noreferrer">Barnes Wallis</a> and his successful R100 design.</p>
<p>The structure has a lot of redundancy and can be quite resistant to damage, a property that proved useful in a later Barnes Wallace design, the <a href="https://en.wikipedia.org/wiki/Vickers_Wellington" rel="nofollow noreferrer">Vickers Wellington</a> bomber from early WW2.
<a href="https://i.stack.imgur.com/n6Jpr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n6Jpr.png" alt="enter image description here"></a>
The post-war Vickers Viking airliner followed the geodesic design, one survived a bomb exploding in the toilet but landed successfully.</p>
<p>Post war geodesics are more commonly associated with another genius and dreamer, <a href="https://en.wikipedia.org/wiki/Buckminster_Fuller#Geodesic_domes" rel="nofollow noreferrer">Buckminster Fuller</a>. He famously championed finding peaceful uses for weapons of war.</p>
<blockquote>
<p>It is now highly feasible to take care of everybody on Earth at a higher standard of living than any have ever known. It no longer has to be you or me. Selfishness is unnecessary. War is obsolete. It is a matter of converting the high technology from weaponry to livingry.</p>
</blockquote>
<p>Whether he knew of Wallis's weaponry or independently invented geodesics for "livingry" I don't know.</p>
<p>A common variant is composed of pentagons and hexagons with slightly raised centres. using just two slightly different sizes of triangle, as can be seen here (not quite complete in the photo). This greenhouse, made of lightweight 2x1 timber and polycarbonate survived quite a few hurricanes in the Hebrides.</p>
<p><a href="https://i.stack.imgur.com/63qWB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/63qWB.jpg" alt="enter image description here"></a></p>
<p>Hence the relatively recently discovered C60 atom mentioned in the other answer, which earned the name Buckyball, or Buckminsterfullerene for obvious reasons.</p>
| 13370 | What do they call that elegant triangular scaffolding often used in future-oriented architecture? |
2017-01-23T08:56:46.110 | <p>For the highlighted part , why we need to replace the support at A with a roller guide ? We could see at A , it's a pinned support , right ? </p>
<p>Pinned support will resist horizontal and vertical reaction , why the author said it will not resist vertical force ? Is the author wrong ?
<a href="https://i.stack.imgur.com/gMhaW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gMhaW.png" alt="fsf"></a></p>
| |structural-engineering|structural-analysis| | <p>The answer is in your quote/picture:</p>
<blockquote>
<p>In order to draw the deflected shape properly, the capacity of the beam to resist the applied function must be removed so the beam can deflect when the function is applied.</p>
</blockquote>
<p>The Muller-Breslau principle is that the influence line is equal to: the deflected shape of the relevant force/moment at the relevant location <em>if the resistance to the effect at the location of interest is removed</em>.</p>
<p>In order to find the influence line for a vertical reaction at A, we therefore want to find the deflected shape for a vertical force at A, when there is no vertical support at A.</p>
<p>If we had left the vertical support in at A, then a vertical force applied at A would give us no deflection, and hence the influence line would be zero everywhere. Clearly that's wrong, as we know a force applied just next to A would give us a reaction at A!</p>
| 13388 | qualitative influence line of pinned support |
2017-01-23T13:06:04.727 | <p>The following picture is pretty self-explanatory. I need a connection for a beam that is coming in very close to a concrete wall edge (north/south beam in picture). The embedded plate with nelson studs in already in place. I was thinking a T type welded plate with bolts already fixed to plate (as drawn in purple), but they wont be able to weld both sides of the plate. Maybe with a full penetration backed weld? Any better ideas on how to fix this beam? Required shear for full beam capacity is about 175 kN and factored load is about 50 kN to give a quick idea.
<a href="https://i.stack.imgur.com/W8GPO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W8GPO.png" alt="enter image description here"></a></p>
| |structural-engineering|shear|connections| | <p>Full penetration weld would be ok. I am more concerned about that bolts between the beam and the concrete edge. It could be difficult to tight them and inspection would be problematic too. Also you loose a considerable part of the flanges, at the beam edge, and therefore you cause a severe drop of the moment strength. I don't know if this is acceptable for your design. </p>
<p>A more simple solution i came up, is shown in the figure below. The beam is cut a little shorter in order to be flush with the concrete face and thus avoid flange trimming. A plate (pink color) is fastened with fillet welds at the anchor plate. A filler plate (green) is used to fill the space between the pink plate and the beam web. With this design, access for inspection and repairs remains good. No installation problems too. </p>
<p>Another advantage from placing the pink plate parallel to the edge, is that no bending is transferred to it. As a result, you avoid any additional bending of the anchor plate, tension of the anchors and compression stresses on the concrete. But you may induce more shear to the anchors (not certain if it is worst, because shear is induced by your design too). </p>
<p><a href="https://i.stack.imgur.com/Qa1dX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qa1dX.png" alt="enter image description here"></a></p>
<p><strong>What verifications are needed:</strong></p>
<ul>
<li>Fillet Welds of the pink plate</li>
<li>Anchors due to the additional shear (there will be a torsional component) </li>
<li>Bolts (considering the filler plate)</li>
<li>Pink plate in bending/shear</li>
<li>Beam web</li>
</ul>
| 13395 | Connection for beam very close to concrete wall |
2017-01-23T17:25:55.897 | <p>(Discrete systems for my question are further divided into periodic and event based)</p>
<p>By reading around I have a feeling that analog does not equal continuous and that digital does not equal discrete. </p>
<p>For instance, in the case of a periodic discrete control the measurements may not be digitized, not even quantized and simply fed into a controller. Is this correct? </p>
<p>If yes, does this mean that there can be 6 types of systems? Analog-continuous, analog-periodic, analog-event_based, digital-continuous, digital-periodic, digital-event_based?</p>
| |electrical-engineering|control-theory| | <p>Let's go back to the original fundamental definitions of the terms.</p>
<p>"Analog" mean that there is an <em>analogy</em> between two things. For example, in a mercury thermometer, there is a relationship between the height of the mercury in the thermometer tube and the outside temperature. Or in a car tachometer, there is a relationship between the angle of the pointer on the dial and the speed of the engine. Or in an analog clock there is a relationship between the angle of the two hands and the time. A very common type of device is where there is an electrical voltage and there is a relationship between the voltage and something else. e.g. the voltage is related to the temperature or the engine rpm. Bottom line: analog simply means that one thing is represented by analogy to another thing.</p>
<p>"Digital" means that a thing is not represented by another thing, but directly by a number. So in a digital thermometer, we don't have a level of mercury in a tube, we have a number. In a digital tachometer, we don't have a pointer on a dial, we have a number. In a digital clock, we don't have hands we have a number. So in digital electronics, we don't have a voltage that is proportional to a quantity, we have a number. That number is usually commonly encoded in binary, although it could be something else, like balanced ternary.</p>
<p>Now, "continuous" means that given any two values, there are an infinite number of intermediate value. E.g. like the real numbers. In between 1.0 and 1.1 there are an infinite number of intermediate values.</p>
<p>"Discrete" means that there are some values which do not have any values in between. e.g. like the integers. There is no integer between 1 and 2. </p>
<p>Now, continuous and discrete may refer to either time or the quantity being measured. e.g. you can have continuous time systems, discrete time systems, but you can also have a continuous valued waveform or a discrete quantized waveform. </p>
<p>So, strictly speaking, analog does not equal continuous and digital does not equal discrete. However, in common usage in the context of digital signal processing, analog systems are usually continuous in time and value, and digital systems are usually discrete in time and value. So the terminology does get abused somewhat. </p>
| 13403 | Is analog|digital different from continuous|discrete? |
2017-01-24T22:58:21.500 | <p>I paddle several different types of small craft in the ocean and bays near my home. One phenomenon I've observed is beyond my understanding of drag on a narrow displacement hull. When paddling in water less than maybe 6 feet in depth, there is a very noticeable increase in drag, which increases as the depth decreases. This is true for all the types of boats I paddle, from a large 6-person Hawaiian outrigger canoe (L=40', width=2', draft=0.67') to a one-person racing kayak (L=20', width=1.5', draft=0.33'). I should add that hull speeds are generally in the 6-8 mph range. I've read that rowing shells also encounter this same drag and they are longer, narrower, but with drafts in more or less the same range.</p>
<p>I've read one explanation that the boundary layer on the hull makes contact with the bottom and the bottom increases the drag on the outer part of the boundary layer which in turn is transmitted to the hull. This seems highly unlikely to me, especially when the water is more than a maybe a foot deep. I don't believe the boundary layer from a 20 foot kayak extends that deep. As I said, this is noticeable at water depths up to 6 feet or more. </p>
<p>I've also heard that pressure waves from the hull bounce off the bottom and reflect back up to the hull and cause drag but as an engineer that doesn't really sound very rigorous to me. </p>
<p>It seems to me there might be some interaction between the bow wave of a boat and the bottom, since I know that the dynamics of surface waves do extend approximately as deep as their wavelength, and at these speeds the wavelength of a bow wave is certainly on par with the water depth. Can anyone give me a properly defensible answer to this?</p>
| |hydraulics|marine-engineering|drag| | <p>Note that the mean water depth around the vessel is reduced
when the depth-based Froude number is close to 1.
It is not just that the vessel sinks and trims, which would imply
that the wetted surface area would increase significantly.
Finite width effects further complicate the situation.</p>
| 13430 | Boat Hull Drag in Shallow Water |
2017-01-25T00:29:02.063 | <p>I'm designing something with a vertical linear actuator. It is a quite large actuator lifting a load. I'm anticipating sway from the actuator as it moves up and down so I'm creating a support structure. What is the best way to determine how much this sway will be? I can figure out the weight of the load and the acceleration of the actuator </p>
<p>Thanks for the help!</p>
| |mechanical-engineering|structures|torque|stability|actuator| | <p>A sketch would be helpful here, but you should have an addition linear bearing to stabilize the load and then you decouple any moment loads on the actuator. If space/design allows, you could use your actuator in tension instead of compression which would reduce/remove any buckling conditions</p>
<p>Regarding "sway", you should start with a free body diagram, and then apply some simple statics to get deflections and stresses. </p>
| 13434 | Vertical Linear Actuator Sway Movement |
2017-01-25T08:33:20.517 | <p>It's given that the applied force is 10kN at F and 25kN at E . I have trouble of finding the other real force in the member of the trusses .
So , i start by drawing the FBD at F . I have 10kN point downwards , so the reaction is 10kN upwards , so it's in equilibrium . But , <strong><em>i really have no idea how the author get the reaction 15kN ( in tension) pointing to the left and to the right .</em></strong> Can someone explain about it?</p>
<p><img src="https://i.stack.imgur.com/G3f9x.jpg" alt=""></p>
<p><img src="https://i.stack.imgur.com/roR5v.jpg" alt=""></p>
| |structural-engineering|structural-analysis| | <p><em>Note: a <strong>reaction</strong> is an external force, such as at A and D. The internal forces in members are not reactions; you could refer to them as "axial loads" in this case.</em></p>
<p>Remove unknowns one at a time. The easiest are the external reactions at A and D. Take moments about A to get the reaction at D, take moments about D to get the reaction at A. Resolve vertically to make sure they add up!</p>
<p>Next, draw free body diagrams at nodes with the fewest members; these are likely to allow you to eliminate unknowns. In this example, resolving vertically at A will give you the force in member AB, given that you know the reaction at A. Resolving horizontally at A will give you the force in member AF in terms of the force in member AB.</p>
<p><em>Then</em> you can resolve a free body diagram at B, which will give you that the force in member AF is equal to the force in member FE.</p>
<p>Continue in this manner and you can get the forces in all members.</p>
| 13438 | finding real force in trusses |
2017-01-25T16:08:33.420 | <p>I dont understand how the diagram of M/EI of conjugate beam drawn , can someone explain about it ?
According to conjugate beam theorem , i know that the M/EI represent the w(x) , which is force per unit length of the beam . I dont understand why the author draw in in 2 directions ( upwards and downwards) ...
<a href="https://i.stack.imgur.com/Rh1yj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rh1yj.png" alt="ddd"></a></p>
| |structural-engineering|civil-engineering|structural-analysis| | <p>The purpose of the conjugate beam method is to find deflection and slope of the beam.</p>
<p>This method uses equilibrium equations in conjugate beam, instead of using moment area diagram in real beam.</p>
<p>First, you must know the support of both real and conjugate beams, and then you <a href="http://www.mathalino.com/sites/default/files/reviewer-strength/06-beam-deflections/conjugate_support_real_support.gif" rel="nofollow noreferrer">transform</a> the support of the real beam to the support of the conjugate beam, and draw it.</p>
<p>Second, you plot the M/EI of "real beam" on the conjugate beam.</p>
<p>Finally, you take the equilibrium equations and determine the reactions at the conjugate beams. The reactions of the conjugate beam, V and M, will equal to θ and Δ, respectively, for the real beam.</p>
| 13444 | Conjugate beam (M/EI) |
2017-01-26T17:18:19.023 | <p>When talking about roller supports in a statics problem, we say that the roller prevents only the vertical translation of the beam, but not the rotation or the horizontal translation of it. So there is only one vertical reaction applied on the beam by the roller.</p>
<p>Also, we know a beam is connected to the roller using a pin, and the reaction force is the result of the pin that acts on the beam, as shown in this picture:</p>
<p><a href="https://i.stack.imgur.com/9od1C.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9od1C.gif" alt=""></a></p>
<p>Now, suppose that I have a beam that is connected to a roller and I am going to push the beam in the horizontal direction (rollers don't prevent translation in the horizontal direction). What reactions should I expect?</p>
<p>I think that the pin is going to apply a force in the horizontal direction (due to the contact between the pin and the beam), which means that there is a reaction force in the horizontal direction. Is that true? If it is true, then why don't we consider horizontal reactions when analyzing rollers?</p>
| |civil-engineering|statics| | <p>In a rigid system, the roller transfers lateral forces to the pinned fixings and can only resist perpendicular forces.</p>
| 13457 | Why do we ignore horizontal reactions at roller supports? |
2017-01-26T18:39:42.313 | <p><a href="https://i.stack.imgur.com/FVm5U.png"><img src="https://i.stack.imgur.com/FVm5U.png" alt="enter image description here"></a></p>
<p>For example, for yield point you can say that it is a point after which the material will have plastic (permanent) deformation. I'm wondering if there is any similar meaning to the tensile strength? And is there any importance to it?
The thing is, whenever we calculate tension of anything (in college), the tension in the material must always be lower than it's yield strength. So it makes me wonder why do we care about ultimate tensile strength?</p>
| |mechanical-engineering|materials| | <blockquote>
<p>For example, for yield point you can say that it is a point after which the material will have plastic (permanent) deformation. I'm wondering if there is any similar meaning to the tensile strength?</p>
</blockquote>
<p>Yes: <strong>tensile strength</strong>, or <strong>ultimate tensile strength</strong> (UTS), is the point at which, if the load is not reduced, the material ruptures or fractures and thereby becomes discontinuous. Rupture is commonly viewed as a form of material failure. Note that for brittle materials rupture occurs at the UTS. For ductile materials the UTS often does not coincide with rupture because the material is able to change shape to accommodate the strain. The shape change, or plastic deformation, is limited because the volume of the material is constant, hence why necking occurs.</p>
<p>After the yield stress, metals work harden, which is why required engineering stress increases with increasing strain. While work hardening is occurring the material is also necking, up to the UTS. With strain increasing beyond the UTS, voids and cracks begin to form within the material's microstructure and start to coalesce into successively larger voids, until they bridge and the material fails. Voids and cracks can not support loads, so the effective cross-sectional area of the material decreases, hence the decreasing engineering stress with increasing strain after the UTS. Once the voids have completely bridged the cross-section of the material, rupture has occurred.</p>
<p>In a sense the UTS can be viewed as a "point-of-no-return" for constant-stress applications. Once the UTS has been reached the material will inevitably rupture if the load is not sufficiently reduced in time.</p>
<blockquote>
<p>And is there any importance to it? The thing is, whenever we calculate tension of anything (in college), the tension in the material must always be lower than it's yield strength. So it makes me wonder why do we care about ultimate tensile strength?</p>
</blockquote>
<p>Absolutely, it is quite important for shaping metals. As an example, consider the process of wire drawing. Wire drawing is the process of reducing a wire stock from a larger diameter to a smaller diameter through mechanical force. Wire drawing is a continuous process involving very large spools of material moving very quickly through dies, to produce many miles of wire per hour. Without going into mathematical details, the wire-drawing process requires precise control of stresses to achieve optimal plastic deformation. If the UTS is inadvertently reached anywhere in the wire, the wire will rupture and stop the entire continuous process. Most likely the spool will be ruined, because the wire can not be rejoined, and customers expect a certain continuous length when they make a purchase. Restarting the wire drawing process can be a time-consuming waste of materials and thus costly. Therefore, wire manufacturers rely on knowledge of the UTS of the metal they are working to avoid costly mistakes.</p>
<p>The reason most introductory materials and mechanics courses only involve the elastic regime, below the yield point, is because the calculations are less complicated. Plastic deformation requires a greater understanding of the mathematics behind material flow, as well as incorporation of strain hardening and other fairly involved phenomena. That said, plastic deformation and the UTS are of utmost importance for shaping of ductile materials in industrial processes.</p>
| 13458 | Does the ultimate tensile strength of a material have a physical meaning? |
2017-01-27T12:44:25.847 | <p>I am trying to source the parts for a <a href="http://reprap.org/wiki/Category:Wilson_II" rel="nofollow noreferrer">Wilson II 3D printer</a>, and the recommended struts are made from 2020 aluminium extrusions, like so:</p>
<p><a href="https://i.stack.imgur.com/cKLj5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cKLj5.png" alt="2020 Aluminium extrusion" /></a></p>
<p>However, in Bangkok, I can only find this type:</p>
<p><a href="https://i.stack.imgur.com/y7mat.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y7mat.png" alt="2020 Aluminium extrusion in Bangkok" /></a></p>
<p>Note that the profile/shape of the slot is markedly different from the image above.</p>
<p>Whilst looking about on the web, I have also found other types, like these:</p>
<p><a href="https://i.stack.imgur.com/KfVZg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KfVZg.png" alt="Various types of 2020 Aluminium extrusion" /></a></p>
<p>Again, the profile of the slot varies considerably.</p>
<p>My question is whether standard full size</p>
<p><a href="https://i.stack.imgur.com/gPhHI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gPhHI.png" alt="Full T-nut" /></a></p>
<p>and half size T nuts</p>
<p><a href="https://i.stack.imgur.com/2fMA0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2fMA0.png" alt="Half T-nut" /></a></p>
<p>will still fit all of these different types? Or are there different T-nuts for the different types of extruded 2020 aluminium, depending on the profile/shape of the slot.</p>
<p>Also, why are there the different designs, does each type have an advantage for particular applications (with respect to strength, weight, etc.)?</p>
<hr />
<h3>Update</h3>
<p>For completeness and accuracy, the 2020 extrusion, that I obtained in Bangkok, actually has this profile:</p>
<p><a href="https://i.stack.imgur.com/MoVBE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MoVBE.jpg" alt="2020 Aluminium Extrusion BKK#1" /></a></p>
<p>and here is a more in-focus photo, of four, in a 2 x 2 configuration:</p>
<p><a href="https://i.stack.imgur.com/Vzce3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vzce3.jpg" alt="2020 Aluminium Extrusion BKK#2" /></a></p>
<p>As ericnutsch states, in his answer, only regular hexagonal M5 nuts will fit.</p>
| |aluminum|fasteners| | <p>I don't think that any profile has any major advantage over any other. I use <a href="http://openbuildspartstore.com/v-slot-linear-rail/" rel="nofollow noreferrer">openbuilds v-slot</a> on a lot of my projects (its 2020 designed for perpendicular v rollers) and all the <a href="https://www.aliexpress.com/item/100pcs-lot-M5-Nickel-Plated-Carbon-Steel-Hammer-Nut-Aluminum-Connector-T-Fastener-Sliding-Nut-For/32317449512.html" rel="nofollow noreferrer">aliexpress t-nut hardware</a> I have bought works fine.</p>
<p>Worst case, you can use regular nuts with the aluminum rail you have access to; they are not as convenient but will work fine.</p>
| 13471 | Will standard M5 T-slot nuts fit these different 2020 aluminium extrusions? |
2017-01-28T09:58:59.693 | <p>I'm measuring thicknesses of ~15 $\mu$m nickel thin films electroplated on a silicon substrate. I only have access to a SEM for this task. I need to cut the samples to get a side view of the film. However, the nickel film is plastic and the silicon undergoes a brittle fracture. As a result the nickel film often detaches from the substrate near the fracture point before I have the chance to cut it, making imaging impossible.</p>
<p>How could I cut through both the plastic thin film and the brittle silicon simultaneously? I'm getting access to liquid nitrogen in a week or so with the intention of freezing the sample. Is there a ~room temperature process which would allow me to prepare samples meanwhile? </p>
| |materials|measurements|thin-films| | <p>Liquid nitrogen performed poorly in this case. The change in temperature and tension popped the relatively thick film straight off the substrate!</p>
<p>The best results were achieved with encasing the small square samples in epoxy. It allowed the samples to be cut with a diamond saw without even fracturing the silicon. Tool marks were apparent and the film suffered, but in this case it was enough to estimate the thickness of the film.</p>
| 13477 | Cutting silicon samples with plastic thin films |
2017-01-28T11:49:04.247 | <p>I have an absolute value encoder with 14-bit resolution and want to implement a speed controller. For high speeds, the offset between the old and the new position value is large and so the accuracy of the estimation for velocity is high. If I target a low speed, then the accuracy of my estimation shrinks, since I have only a little difference between two steps of the encoder.</p>
<p>I see two solutions to increase the resolution. First, the time between two measurements can be increased, but then the speed controller begins to fail for very large speeds.</p>
<p>Trajectories for the position controller could be calculated to come up with a velocity controller.</p>
<p>What do you think is the best solution for speed control to achieve a large dynamic of speed and how is this done normally for industrial applications?</p>
| |control-engineering| | <p>It will depend on your application, but I would consider a hall effect sensor with one or more magnets the "default" sensor for rotation speed.</p>
<p>The dynamic range is essentially infinite, since a digital timer can record short (uS) intervals all the way up to months fairly easily.</p>
<p>The only limitation is that you need to measure for some fraction of a rotation, like 1/4 turn if you have 4 magnets.</p>
<p>You can use an encoder in a similar way by counting the time between encoder steps, rather than the number of steps in a fixed interval. This may be suitable for very low speeds (<60 rpm).</p>
| 13478 | Speed estimation with absolute encoder |
2017-01-28T13:33:00.523 | <p>Recently I found it essential to make my room soundproof and for this I have worked out the following way:</p>
<p>The walls of my room are made of cement. I want to add two glass layers (of different thickness) inside them. To make it clear, think of my room for the time being a hollow cement cube. Now I want to add one hollow glass cube inside this cement cube (whose dimensions are intentionally kept less than the cement cube) and then do the same for the second glass cube (whose dimensions are also intentionally kept less than the first glass cube). Now what I want to do is to suck out the air between two consecutive layers by a <a href="https://en.wikipedia.org/wiki/Stopcock" rel="nofollow noreferrer">stopcock</a>.</p>
<p>My questions are</p>
<ol>
<li>Will this procedure make my room soundproof? </li>
<li>Will my room be durable?</li>
</ol>
| |civil-engineering|sound-isolation| | <p>If I am understanding your question correctly you want to surround your room a vacuum so sound waves will not travel through your glass walls. In theory this would work very well but there are some major obstacles to overcome. </p>
<p>Say your room is quite small, $12$ by $12$ by $8$. From the standpoint of a vacuum, earths atmosphere is at $14.7$ psi. $12$ feet times $12$ feet equals $152$ tons of force on your ceiling and floor due to atmospheric pressure. ($12\ \text{ft}\times 12\ \text{ft} = 20726\ \text{in}^2$, $20726\ \text{"}^2 \times 14.7\ \text{lb/in}^2 = 304819\ \text{lb}$ or $152\ \text{tons}$)</p>
<p>Also at some point you are going to have to interface your inside box with your outside box for support and access. Any loud noises would vibrate the inside box which would vibrate your doorframe/supports, which would vibrate your outside box and the air as well. While the noise would probably be minimal it would not be 100% soundproof.</p>
<p>So to answer to your questions </p>
<ol>
<li>Theoretically yes but I doubt you have the time or money.</li>
<li>No it would not be durable without very thick glass or some kind of internal and external support.</li>
</ol>
<p>If you want to soundproof your room I would look into getting some foam padding specifically made to cancel sound waves. These will absorb sound waves and deter sound waves from traveling through your walls.</p>
<p><a href="http://www.soundproofcow.com/soundproofing-101/soundproof-a-room/" rel="nofollow noreferrer">Here</a> is a website that will help you get an idea of how to do what you want based on the reason you want to sound proof your room.</p>
| 13480 | Is my procedure applicable to make a room soundproof? |
2017-01-29T02:55:32.770 | <p>consider following diff equations
$$
\dot{v} = \frac{f_x}{m} = \mu g\\
\dot{\omega} = \frac{T}{J} - \frac{Rf_x}{J}\\
\dot{R} = \eta
$$
where the input is $T$. There is a help equation for slip ratio. The unknown variation is $\dot{R}$.where $\eta$ is a state and variable that I declared. $f_x$, $J$, $m$ and $g$ are known. $R$ is known at the beginning but then it varies.
$$
\lambda = 1-(\frac{R\omega}{v})^n, \:\: n=\left\{-1,1\right\} $$
What I have done is to define some state variable:
$$
x = [x_1,x_2,x_3]^T = [\omega, v, R]^T \\
u = T\\
y = [x_1, x_2]$$
When I build the system I end up with a singular matrix. What kind of observation approach does work on such system to observe $$x_3$$. I really want to use a slide mode observer. Otherwise everything else works.</p>
<p>I would appriate it alot if someone gave me a hint.</p>
| |control-engineering|control-theory|dynamics| | <p>Its not a complete answer. But I found out that the dynamics need redefining such that a transformation of the state is performed. A system can be generically observable if the systes can be expressed by output and input and finite numbers of their derivatives i.e
$$x=X(y, \dot{y},..., y^j , u , \dot{u}, ..., u^i)$$
Where
$$j,i \in \mathbb{N}_+$$
Denote the number of derivation.
When the dynamics are transformed to a simpler and standard form you can take the unkowns as uncertainties and handle the system using desired observer.</p>
<p>More about this can be found in the book Advances and Applications of Nonlinear Control. </p>
| 13489 | Nonlinear observer of an unknown variation state |
2017-01-29T21:51:30.877 | <p>When you buy plastic sheets such as acrylics, polycarbonate sheets etc, they come with a paper/plastic protective film. Like in this image : </p>
<p><a href="https://i.stack.imgur.com/15YQX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/15YQX.jpg" alt="enter image description here"></a></p>
<p><a href="http://qdjnd.en.alibaba.com/productshowimg/50172628-200203123/Protective_Film_for_Acrylic_Sheet_PMMA_Panel.html" rel="nofollow noreferrer">Protective Film for Acrylic Sheet PMMA Panel</a></p>
<p>You notice that there is no adhesive (at least not something tacky) that is present on the sheets, and there is no residue when the sheet is peeled off, I was wondering if anyone can shed some light on how these protective films are applied ?</p>
| |plastic|adhesive| | <p>The material is called "Electrostatic Protective Film".</p>
<p>It is manufactured by letting the plastic film cool in a strong electrostatic field: there are high-voltage electrodes on top and bottom of the plastic. This causes the polar molecules in the plastic to align so that one side of the film has a positive charge and the other side has a negative one.</p>
<p>This charge will persist quite long, and will keep the film adhered to the plastic sheet by electrostatic forces. However, because the electrostatic force is quickly reduced by distance, it is important that no dust gets between the protective film and the base plastic during manufacturing.</p>
| 13497 | How are protective sheets attached to plastics? |
2017-01-30T09:49:14.383 | <p>I have here a BLDC motor that is controllable with FOC and I already implemented a position controller. The next step is to implement a velocity controller, but I actually don't see how to control the velocity at very low speeds. The speed is proportional to the reference torque producing current i_d for FOC and so for low speeds, there is only a very low current and so nearly no power. This immediately can be seen, that the motor becomes bumpy for low speeds. The implemented speed controller looks like this and works fine for speeds around 50rpm or higher. </p>
<p>i_q_ref = i_q_ref_old + K_P<em>err_speed + K_I</em>sum_err_speed</p>
<p>The speed measurement is pretty accurate.</p>
<p>What kind of other control structures can be used to achieve a velocity controller for low speeds with FOC?</p>
| |control-engineering|motors| | <p>If you already have a functional position controller, you can just integrate the target speed to get a target position (which is moving). This should work relatively well for low speeds, but probably not when the motor starts moving a significant fraction of a revolution per control loop iteration.</p>
<p>But I think the underlying problem is probably either latency or inaccuracy in the speed measurement. I assume it comes from some kind of encoder? At low speeds, it can take a long time for the motor to move to the next encoder tick, thus the velocity change is not immediately visible to the controller. And if you don't have any smoothing on the velocity reading, it can render PID D-term quite ineffective; and if you add smoothing, it will increase the latency.</p>
<p>One way to increase the speed of the feedback loop is to form some kind of a feed-forward system. Consider having a Kalman filter with state variables (velocity, inertia, external_torque). Then you can update that Kalman filter by both your velocity measurements when you get them, and more often by torque values that you drive to the motor.</p>
| 13499 | Speed control of a BLDCM at low speed |
2017-01-30T16:54:14.747 | <p><a href="https://i.stack.imgur.com/fUPEo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fUPEo.jpg" alt="enter image description here"></a></p>
<hr>
<p>Does the left red wheel turn in the same direction as the leftmost orange wheel?If so P and S go up? If not, Q and S go up?</p>
<p>My reasoning:</p>
<p>If so</p>
<ol>
<li>Leftmost orange cc</li>
<li>Left red cc (by assumption)</li>
<li>Leftleast orange clock --> P goes up</li>
<li>Gray cc</li>
<li>Right least orange clock</li>
<li>Right most orange clock (similar to assumption)</li>
<li>Right red cc --> s goes up</li>
</ol>
<p>If not</p>
<ol>
<li>Leftmost orange cc</li>
<li>Left red clock</li>
<li>Leftleast orange cc --> Q goes up</li>
<li>Gray clock</li>
<li>Right least orange cc</li>
<li>Right most orange clock (similar to assumption)</li>
<li>Right red cc --> s goes up</li>
</ol>
| |machining|pulleys|turbomachinery| | <p>When two wheels/pulleys/cogs are connected by a belt, when one of the wheels turns the other one turns in the same direction as the first wheel that turns.</p>
<p>When two cogs/gears are positioned so they are touching, when one cog is turned in one direction the other cog turns in the opposite direction.</p>
<p>So for your example:</p>
<ol>
<li>The upper left yellow cog turns counter clockwise (ccw)</li>
<li>Because its connect to the red cog via a belt, the red cog also
rotates ccw and the middle yellow cog turns clockwise (cw) causing
weight P to rise.</li>
<li>Because the middle yellow cog rotates cw the black cog rotates the
opposite direction, ccw. This causes the upper right yellow cog to
turn cw.</li>
<li>The upper right yellow cog is connected to the lower right yellow
cog via a belt, so the lower right yellow cog also turns cw.</li>
<li>The lower right yellow cog touches the lower right red cog which
then turns the opposite direction, ccw.</li>
<li>This results in weight S being raised.</li>
</ol>
| 13503 | How do Cogs and belts work? / Cogs and belts in an ICAS exam |
2017-01-30T23:23:46.437 | <p>I would like to calculate or find the electrical resistance (in ohms) of aluminium oxide at or past its melting point, for the purposes of using it as an electrolyte in a molten state.</p>
<p>My apologies for the broadness I just can't seem to find anything specific to this. Even if you don't have an answer due to the nature, any direction for a conclusion to this question would be greatly appreciated.</p>
| |electrical-engineering|chemical-engineering|metallurgy| | <p><a href="https://www.researchgate.net/publication/231422619_The_Electrical_Conductivity_of_Liquid_Al2O3_Molten_Corundum_and_Ruby" rel="nofollow noreferrer">Here</a> you can find the resistivity of the molten $Al_2 O_3$ at $2400 {}^o C$: $384 ± 5\% \Omega m$. I've found this link with lesser than 10 seconds of googling.</p>
<p>Its melting point is only $2072 {}^o C$ (<a href="http://de.wikipedia.org/wiki/Aluminiumoxid" rel="nofollow noreferrer">source</a>). Its resistivity is probably smaller as the referred data. Checking this isn't very simple, because you can't easily extrapolate from the available experimental data (parameters such this tends to varies greatly around the phase shifts).</p>
<p>In your actual case, it depends also on its shape. For a correct calculation, you could make a numerical simulation depending on its temperature distribution. But, as a fist rule, you can relative easily estimate it by simplifying the calculation to a constant-temperature tube.</p>
| 13505 | Electrical Resistance of molten aluminium oxide |
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