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2015-02-09T12:45:47.267 | <p>I have used a cheap multimeter to measure voltages in simple DC circuits, but I have seen pictures of them plugged straight into the mains and used to measure various home-built generators.</p>
<p>Why doesn't the higher voltage fry the multimeter, and also in theory could a small cheapo multimeter be safely used to measure very high voltages? If you get the setting wrong on the dial, does this matter?</p>
<p>I'm not planning to plug one in, nor would I recommend anyone who doesn't know what they're doing to do this either, I'm just wondering how it works.</p>
| |electrical-engineering|measurements| | <p><strong>WARNING</strong>: Note that <strong>some cheap meters are not suitable for use with 230 VAC AC mains.</strong> Some meters may have AC voltage ranges able to conceptually measure to well above AC mains voltage BUT have internal componentry not certified, suitable or safe at eg 230 VAC. <strong>Use of such meters to measure such voltages is akin to a safer than usual game of "Russian Roulette" which still may end in death.</strong></p>
<p><strong>_________________</strong></p>
<p>Failures may occur to power dissipation in components or to voltage breakdown even when power dissipation is within bounds.
Voltage ranges are usually less stressed than most other ranges when overrated. </p>
<p>Auto-ranging meters start at the highest range and work down until the reading becomes a certain percentage of full scale on that range. Switching can be via eg MOSFETs used to either short resistors in dividers or to pick voltages from appropriate tapping points.</p>
<p>In cheap meters protection is limited.<br>
Current much over 200 mA on low current ranges will usually blow the internal fuse.<br>
Too high current on the high amps range (10A, 20A, will blow the shunt or a fuse if fitted.<br>
High voltage on lower voltage AC or DC volts or on current ranges will sometimes destroy the meter (Ask me how I know :-) ). </p>
<p>All that said: Over ranged inputs will not necessarily stress other than the input circuitry. Higher than minimum necessary input resistor wattages can protect short term. Zener diodes or other clamps can stop high voltage getting into the circuitry proper.<br>
Very keen manufacturers can provide electronic switches. These can be a simple as a high voltage MOSFET in series with the input) which can be turned off when needed. This will add some error due to voltage drop but this can be controlled and designed for. </p>
<p>So eg an overvoltage condition is applied to a current input, the input resistor starts to dissipate excessive power, the inside end of the resistor is clamped by a zener or TVS (transient voltage suppressor) and a fast MOSFET switch is triggered to disconnect the overload. The higher than necessary power dissipation rating of the input circuitry provided enough time for the protection to act. </p>
<hr>
<p><strong>Real world example:</strong></p>
<p>This Intersil application note -<br>
<a href="http://www.intersil.com/content/dam/Intersil/documents/an04/an046.pdf" rel="nofollow noreferrer"><strong>AN046 Building a Battery Operated Auto Ranging DVM with the ICL7106</strong></a> provides specific examples of the design of autoranging equipment and how the issues involved are addressed.</p>
<p>Here is how the front end looks conceptually:</p>
<p><img src="https://i.stack.imgur.com/CXqXD.jpg" alt="enter image description here"></p>
<p>and here is how it ends up in practice</p>
<p><img src="https://i.stack.imgur.com/0WB5r.jpg" alt="enter image description here"></p>
<p><a href="http://www.intersil.com/content/dam/Intersil/documents/an04/an046.pdf" rel="nofollow noreferrer">http://www.intersil.com/content/dam/Intersil/documents/an04/an046.pdf</a></p>
| 557 | How does a multimeter protect itself from high voltages? |
2015-02-09T22:48:52.813 | <p>The campus where I work has a long covered walkway (~.5 mile) which has several labeled pipes running under the roof (chilled water, fuel oil, air...). All of the pipes run dead straight <em>except</em> for the natural gas lines, which have little loops spaced about every 250ft, as seen in the attached image (the lowermost, yellow line. There's another natural gas line hidden above all the others, which also does the same thing.)</p>
<p>The line isn't branching at these points, and there doesn't seem to be any need to divert the pipe in order to support it. I've looked at some building codes to see if I could find a reason (or even a requirement) to insert these.</p>
<p>Any ideas as to what these are? It's driving me batty!</p>
<p><img src="https://i.stack.imgur.com/VhLi2.jpg" alt="enter image description here"></p>
| |mechanical-engineering|building-physics|pipelines| | <p>If the other pipes are carrying liquids with a greater specific heat coefficient then they would not be expected to experience the same temperature ranges, i.e. the water pipes will be at some average of the outside temp and the water temp.
A low-pressure gas line would be made of lighter material and the gas inside would equalize with the external temp pretty quickly. So the code for a gas line and a liquid line (or a light pipe and a heavy pipe) might be different.</p>
<p>Expansion joints are often used for exposed liquids pipelines, like the Alaska pipeline, that are subject to temperature extremes and cover long distances. Buried pipelines don't experience the same range of temperature and don't normally employ expansion loops.</p>
| 569 | What is the purpose of these diversions in a natural gas line? |
2015-02-10T13:06:50.863 | <p>For Newtonian fluids, the sinking (or rising) speed of a particle is (reasonably) well known. What is a relationship for shear thinning fluids?</p>
| |fluid-dynamics| | <p>Doing a brief review of the literature, for spherical particles at high Reynolds numbers, <a href="http://books.google.com/books?id=ekIrBgAAQBAJ&lpg=PP1&ots=T2jDGPKMHP&lr&pg=PA87#v=onepage&q&f=false" rel="nofollow">it appears that you can use standard drag curves</a> to get the drag coefficient as long as you use the following Reynolds number:</p>
<p>$$\mathrm{Re}_\text{PL} \equiv \frac{\rho V^{2-n} d^n}{m}$$</p>
<p>This Reynolds number assumes the fluid follows a <a href="http://en.wikipedia.org/wiki/Power-law_fluid" rel="nofollow">power-law</a>. $n$ is the flow behavior index, which is less than 1 for a shear thinning fluid. $m$ in the above equation is $K$ on Wikipedia.</p>
<p>So, if you just want the terminal velocity, you can plug this in the <a href="http://en.wikipedia.org/wiki/Terminal_velocity#Physics" rel="nofollow">standard equations for terminal velocity</a>, as all they assume is that you know the drag coefficient at the terminal velocity state. (You can derive them from a steady force balance or from the unsteady case. I recommend trying both just to get a feel for these sorts of problems.)</p>
| 585 | Sinking speed of particle in shear thinning fluid? |
2015-02-10T19:12:02.523 | <p>NIL (Nanoimprint lithography) and hot embossing seem to mean the same thing in many instances. Is hot embossing a subset technique of NIL? Or are they completely different methods?</p>
| |lithography| | <p>I had the same question recently and was doing some research a wile ago. Sorry that I don't have all the literature prepared.</p>
<p>NIL is for very small structures, the mold is created with a lithography process. Depth are usually not very large. Can be very fast implemented in R2R. There are two basic types:</p>
<ol>
<li>Thermal NIL: Mold is places on substrated, mold is heated at first, then cooled down and separated from substrate with finished imprint.</li>
<li>UV NIL: Same as above but instead of a heating source you have a UV source.
[Source: N. Kooy et al. A review of roll-to-roll NIL Nanoscale Research Letters 2014, 9:320]</li>
</ol>
<p>Hot embossing is also used for small structures but also for bigger ones, small structures with great depth can be achieved.
In the process the substrate is preheated and in liquid phase. The mold is pressed on the substrate. Temperature and pressure are controlled. During cooling the mold is removed. Speeds are slower, depending on how big the structure is can be very slow. The mold can also be created by lithography but not necessarily as they often exceed the size possibilities of lithography. The molds are bigger than the imprinted structure depending on the flow characteristics of the material imprinted. Delamination happens during cooling.
[Source: M. Heckle et al. Hot embossing - The molding technique for plastic microstructures - Microsystem Technologies 4 (1998) 122-124]</p>
<p>Differences:
Hot embossing is: slower, for bigger structures, or small but great depth (holes, lenses, deep channels...), mold is bigger than structure, substrate is preheated, temp. and pressure are always controlled, mold isn´t necessary made with lithography. </p>
<p>For sure there are things I missed but I hope it helps someone.</p>
| 589 | What is the difference between NIL and hot embossing? |
2015-02-10T20:22:09.863 | <p>If I know the tensile strength of a pin, how do I compute the total strength of multiple pins? That is, 2 objects are connected by multiple pins, evenly distributed, each with a height of $2mm$ and a cross sectional area of $3.33mm^2$. For some reason the answer of just multiplying the tensile strength of each pin by the number of pins, doesn't make 100% sense to me.</p>
| |materials| | <p>The short answer is yes, it is that simple. </p>
<p>Think about it this way. The pins are connected to end plates of a given area. If all we're worried about is the tensile strength, we can place a one dimensional force on each end plate to put the system in pure tension. Now, split the system into a couple parts and do a force balance. </p>
<p>If we split it right down the middle, so that we're seeing the actual force on the top plate, and the internal force along each of the pins. These two forces have to be equal, otherwise the system would be in motion, meaning the pins have yielded or fractured. </p>
<p>Now, to calculate the stress in the pins, we simply divide the force applied across all the pins by the total area. Because in this case, the pins are all of even cross sectional area, the stress will be divided evenly between them. The implication of this is that none of the pins should fail before the others, and any difference in their tensile strength is due to manufacturing defects or variations, so we can just multiple the tensile strength of a single pin by the number of pins to get the overall tensile strength.</p>
<p>There is an important assumption here that the pins are distributed evenly on the plates. If the pins are lopsided, they're going to end up under some amount of bending, which complicates the stress calculations in them and adds to the stress that they see under the same applied load. </p>
<p>If we make the pins of variable cross-sectional area, then the load begins to concentrate in the larger pins, but it should still end up with the same stress in each pin because the load will distribute proportionally to the area, which is how we calculate stress anyway. </p>
<p>Where this gets more complex is if the pins are made of different materials or treated differently to change the tensile strength. This is essentially creating a composite material, and there we have to take into account volume fractions to calculate new yield and tensile strength numbers, but also consider which material fails first. For further reading I'd suggest chapter 3 in <a href="http://web.mit.edu/course/3/3.225/book.pdf" rel="nofollow">this MIT book</a> (that is a 128 page pdf, beware if you're on mobile). </p>
| 591 | How do I compute combined tensile strength? |
2015-02-11T01:51:06.607 | <p>I'm writing a presentation for people and I need an analogy to convey the following problem:</p>
<blockquote>
<p>Person A wants to sell a liquid, so he fills a tank, measures volume
and sends to Person B. Now Person B receives the liquid and wants to
know if he received the volume he bought from Person A, so since this
liquid is now stored in his tank, Person B measures the liquid volume.
Person B measures won't match exactly Person A measures - due to
losses and imprecisions in both measures.</p>
</blockquote>
<p>These are oil tanks and I need to explain this to really varied audience that contains people unskilled in metrology. Also, Person A and B can be the same person in different places - which is actually the case that I'm studying.</p>
<p>I think an analogy for the tanks would be good, but I can't think anything! How can I explain this problem in a way that a lay audience will understand?</p>
| |education|metrology| | <p>I'm not sure of the specifics of your problem, it seems to be relatively straightforward to me. I have previoulsy considered addressing the subject of measurement errors with schoolchildren in the following manner - it may be of help to you and your audience or change in subject may confuse and distract them from your topic, that's up to you to decide.</p>
<p>Step 1 - give all the students and length of string (considerably longer than a metre) and ask them to put two knots in the string exactly a metre apart, based on their own estimation. The smart ones will use a length they already know as a guide - most common is their own height.</p>
<p>Step 2 - measure the lengths achieved against a ruler and see who got closest. A leaderboard of some kind can make this more fun. This can lead into a discussion of estimation in general. Plotting a scatter plot of the measurements might also be interesting and lead into a discussion about distributions (with enough students I think I'd expect a roughly normal distribution?)</p>
<p>Step 3 - now change the ruler for a tape measure and get everyone to remeasure someone elses string - they will inevitably get slightly different results. Write these down and look at the differences. Perhaps now also draw out a scatter plot of the differences. The change in results should bring up all sorts of questions to prompt your discussions on metrology. Examples could be:</p>
<ul>
<li>Are the metre ruler and the tape measure the same length? Which of these is the most accurate measure of a metre? How can we test them?</li>
<li>Why do two measures of the same thing come out different? Was there an element of judgement by eye on the part of the measurer? Did the string stretch? Did the tape measure stretch? Were the conditions in the room the same? Have the knots tightened or slipped?</li>
<li>Was the level of precision used in recording the measurements the same? What is the right level of precision? Did we agree on what part of the knots we were measuring to?</li>
<li>Is there a general trend in the difference results? Are they dominated by bias or random errors?</li>
</ul>
<p>I've done steps 1 and 2 with students, along with similar experiments on time and mass. (Generally they are roughly ok at estimating a metre, largely underestimate how long a second is, and are wildly inaccurate at estimating a kilogram). Step 3 is an extension that I've planned but not yet gotten round to trying out.</p>
<p>I think you could also carry out similar experiments/demonstrations with your exact problem of liquids in containers as well.</p>
| 594 | Communicating losses and imprecision in liquid measurement by analogy |
2015-02-11T02:28:29.930 | <p>Movable span bridges (<a href="http://en.wikipedia.org/wiki/Bascule_bridge">bascule</a> and <a href="http://en.wikipedia.org/wiki/Vertical-lift_bridge">lift</a>) typically employ a counterweight to help reduce the amount of energy and size of motor required to move the span. Due to variations in the the construction of the bridge and counterweight, the theoretical balance of both the span and the counterweight may be off. </p>
<p>After a movable span is complete, how close to perfectly balanced do the span and counterweight need to be?</p>
| |civil-engineering|bridges| | <p>This <a href="https://heavymovablestructures.org/assets/technical_papers/64.pdf" rel="noreferrer">paper</a> from the <a href="https://heavymovablestructures.org" rel="noreferrer">Heavy Movable Structures</a> group goes into detail about how roadway movable spans are balanced. It references AASHTO (<a href="http://www.transportation.org/Pages/Default.aspx" rel="noreferrer">American Association of State Highway and Transportation Officials</a>) manuals such as those below:</p>
<ul>
<li><a href="https://books.google.com/books?id=wSkIJaDO1ooC&lpg=SA2-PA2&ots=TxgSGZ89r8&dq=AASHTO%20movable%20spans&pg=SA1-PA14#v=onepage&q&f=false" rel="noreferrer">AASHTO LRFD Movable Highway Bridge Design Specifications</a></li>
<li><a href="https://books.google.com/books?id=IKwfVvt0FOAC&lpg=PT252&dq=how%20to%20balance%20a%20movable%20span%20bridge&pg=PT252#v=onepage&q&f=false" rel="noreferrer">AASHTO Maintenance Manual for Roadways and Bridges</a></li>
</ul>
<p>The general guidelines are as follows:</p>
<ul>
<li>The spans are balanced so that the dead load reaction is greater than 1,000 kips per girder when the span is down. This keeps the span down even if the locking mechanism isn't working.</li>
<li>The balance is calculated multiple times by both the contractor and the engineer based on theoretical and fabricated weights.</li>
<li>The counterweight is designed so that the weight can be adjusted from the theoretical. It must accommodate a 5% heavier or 3.5% lighter span.</li>
<li>The counterweight can be adjusted by adding or removing individual 1 cu. ft. blocks. These blocks are typically concrete but can also be steel or lead.</li>
<li>The final check of weight can be done with strain gauges.</li>
</ul>
| 595 | How accurately are moveable span bridges balanced? |
2015-02-11T12:55:57.317 | <p>I've attempted to answer the following question and I need to know if I am on the right track:</p>
<blockquote>
<p><strong>LOGIC GATES</strong></p>
<p>Question 3</p>
<p>A manufacturing plant uses two tanks to store certain liquid chemicals
that are required in a manufacturing process. Each tank has a sensor
that detects when the chemical level drops to 25% of full. The sensors
produce a HIGH level of 5V when the tanks are more than one-quarter
full. When the volume of chemical in a tank drops to one-quarter full,
or less, the sensor puts out a LOW level of 0V. It is required that a
single green light-emitting diode (LED) on an indicator panel shows
when <strong>both</strong> tanks are <strong>less</strong> than one quarter full.</p>
<p>3.1 Show the truth table</p>
<p>3.2 Derive the Boolean expressions.</p>
<p>3.3 Show how 3 NAND gates can be used to implement this</p>
<p>(Consider the wording very carefully in this question)</p>
</blockquote>
<p>It's question 3.1 and 3.2. The last sentence of the main question is a bit tricky. It says "Show when both tanks are less than 1/4 full."</p>
<p>Here is my solution:</p>
<ul>
<li>Both inputs (namely Tank A and Tank B) need to be LOW in order to indicate a LOW output</li>
<li>The only logic circuit to satisfy the above is an OR circuit.</li>
<li>0 + 0 = 0; 0 + 1 = 1; 1 + 0 = 1; 1 + 1 = 1</li>
<li>From the above I believe OR operations is correct, since A and B are both LOW and thus causing the output to be low.</li>
</ul>
<p><strong>I was considering AND Circuit</strong> but then 0 x 1 = 0 and 1 x 0 = 0. These operations conflict with the question. <strong>The output must only show LOW when BOTH inputs are LOW</strong>.</p>
<p>Therefore my solution for 3.1 and 3.2 would be using an OR gate. Is that correct?</p>
| |electrical-engineering| | <p>So let's break this down, piece by piece.</p>
<blockquote>
<p>The output must only show LOW when BOTH inputs are LOW. </p>
</blockquote>
<p>We also know that the sensors read High (5V) when the tank volume is above 25%.</p>
<pre>
| Tank 1 | Tank 2 | Monitor LED |
|--------|--------|-------------|
| High | High | Lit |
| High | Low | Lit |
| Low | High | Lit |
| Low | Low | Dark |
</pre>
<p><em><strong>Please note</strong>: For the sake of this explanation, we're assuming 0 means low and 1 means high. So the logic diagram above is "upside down" from a traditional representation that starts low and iterates to high.</em></p>
<p>And if we look at a typical AND gate, we find that we're close but our logic is backwards.</p>
<p><img src="https://i.stack.imgur.com/3kZKH.gif" alt="AND gate logic"></p>
<p>So let's introduce the NOT gate:</p>
<p><img src="https://i.stack.imgur.com/A7ir9.gif" alt="NOT gate logic"></p>
<p>But if we put the NOT gate after our AND gate, we end up with a NAND gate and that's <strong><em>not</em></strong> the logic diagram we want.</p>
<p><img src="https://i.stack.imgur.com/3WySQ.gif" alt="NAND gate image">
<img src="https://i.stack.imgur.com/bjHl8.gif" alt="NAND gate logic">.</p>
<p>Dang! that brings us back to square one, right? Nope, not quite. What happens if we use two NOT gates and move them to the other side of our AND gate?</p>
<p><img src="https://i.stack.imgur.com/KXosN.png" alt="Inverted NAND"></p>
<p>That seems to do the trick! Our High signals become Low, and our Low signals become High.</p>
<p>I'll let you work out the algebraic expressions for that diagram; it should be trivial to translate now.</p>
<hr>
<p>The fun part is trying to express this using just NAND gates. The trick is to use the NAND gate as an inverter <em>and</em> realize that our logic is backwards from the voltage that's present.</p>
<p>If we layout the three NAND gates like this:</p>
<p><img src="https://i.stack.imgur.com/BATVG.png" alt="3 NAND gates"></p>
<p>And if Tank 1's sensor is fed to both gates of <code>U1</code>, and Tank 2's sensor is fed to both gates of <code>U2</code>, we're using the first tier of NAND gates as inverters. Looking at the logic diagram, we see that our green LED will be lit in the cases we want, and not lit in the case when both tanks are below 25%.</p>
<pre>
| Tank 1 | Tank 2 | T1 V | T2 V | !T1 V | !T2 V | 2nd NAND | LED |
|--------|--------|------|------|-------|-------|----------|-----|
| High | High | 5 | 5 | 0 | 0 | 5 | Lit |
| High | Low | 5 | 0 | 0 | 5 | 5 | Lit |
| Low | High | 0 | 5 | 5 | 0 | 5 | Lit |
| Low | Low | 0 | 0 | 5 | 5 | 0 | |
</pre>
<hr>
<p>You also asked:</p>
<blockquote>
<p>Therefore my solution for 3.1 and 3.2 would be using an OR gate. Would I be correct on this?</p>
</blockquote>
<p>Looking at the truth table, with 0 for Low and 1 for High:</p>
<p><img src="https://i.stack.imgur.com/VinDV.gif" alt="OR truth table"></p>
<p>And that would work too. But using an OR gate doesn't necessarily set you up to easily understand how to use the NAND gate configuration.</p>
<hr>
<p>If nothing else, this exercise should help you understand the duality in being able to express positive and negative boolean logic.</p>
| 600 | Logic gates for an indicator light to show the state of two sensors |
2015-02-11T15:33:08.663 | <p>In <a href="https://engineering.stackexchange.com/questions/591/how-do-i-compute-combined-tensile-strength">this question about combined tensile strength</a>, all of the tensile bars are of the same material. But what if the case that they are made of different materials?</p>
<p>Say we have $200$ bars of $A=10 (\rm mm^2)$. </p>
<ul>
<li>$100$ bars of steel with $E=200 \, (\rm {GPa})$, at tensile limit, $\sigma=200\, (\rm {MPa})$, $\eta=20 (\%)$. </li>
<li>$100$ bars of aluminum with $E=60 \, (\rm {GPa})$, at tensile limit, $\sigma=80\, (\rm {MPa})$, $\eta=10(\%)$. </li>
</ul>
<p>What is total strength (the maximum force capacity) of the bars? Assume that one end is fixed and the other is a rigid body moving in longitudinal direction, so that all bars have the same change in length. </p>
<p>Note: We have to assume that the material is of plasticity, and will not hardening after yield. (The tensile limit is the same as the yield strength.) This may be beyond the text book of elasticity theory.</p>
| |mechanical-engineering|materials| | <p><a href="https://engineering.stackexchange.com/a/592/41">My answer</a> of that question covers this a little bit, and the link fully explains it, but I can go into more detail so you don't have to wade through that giant PDF. I'll link it again here at the bottom, because I'm going to reference equations in it for this answer. </p>
<p>As you might imagine, the total strength of a composite such as the example above is a combination of the two materials. The proportions in which each material contribute to the overall strength are related to the volume fractions of the constituent materials. Because in this case, the lengths of the bars is always equal and the bars are of constant area along their length, we can simplify the volumes to areas. This means that we can compute the elastic modulus using a version of equation 3.1 in the book.</p>
<p>$$E_c=E_{St}A_{St}+E_{Al}A_{Al}$$</p>
<p>Here, St and Al are steel and alumin(i)um, respectively, the c subscript refers to the composite, and A is the area fraction, not the total area. So in the example problem, $A_{St}=0.5$ and $A_{Al}=0.5$. </p>
<p>To calculate the stress in the bars, we simply use the definition of stress, given that we can measure the strain and we know the strain is the same in all bars.</p>
<p>$$\sigma_c=\epsilon E_c=\epsilon E_{St}A_{St}+\epsilon E_{Al}A_{St}=\sigma_{St} A_{St}+\sigma_{Al} A_{Al}$$</p>
<p>At this point, though, it's important to note that we are seeing different stresses in the bars of different material, which is a natural result of them having the same strain but different elastic moduli. Unless the materials break at the same strain, one of them will fail first, at which point, the same load will be applied to the whole system, but only one type of bar will still be intact to carry that load. The total cross sectional area of the system is reduced as well though.</p>
<p>We need to determine which material breaks first. Because we're applying a known strain rate here, we can find the ultimate strain of each material using the stiffness and ultimate strength. Once we know the strain at which the first material breaks, we plug that strain into the second equation above to see the total stress in the composite at this elongation. However, because the same force is being applied over a smaller area, the strength of the composite after the breakage of the first material (in this case, the steel) is given by </p>
<p>$$\sigma_{TS,c}=\sigma_{Al} A_{Al}$$</p>
<p>We can compare this value of tensile strength to the value of tensile strength just before the steel breaks. If the value before steel fracture is greater, then the steel fracture results in the fracture of the entire system. If the value after steel fracture is greater, the system will continue to carry load until it reaches the tensile strength of just the aluminum bars. </p>
<p><a href="http://web.mit.edu/course/3/3.225/book.pdf" rel="nofollow noreferrer">Mechanical Properties of Materials</a>, David Roylance, 2008.</p>
| 609 | How to Compute Total Strength of Composite Tensile Bars |
2015-02-11T23:21:32.310 | <p>I have a uniform load applied to the underside of a plastic sheet which is to be resisted by the combination of a bolt, washer, and nut as depicted below (the other end of the bolt is secured and will not budge). </p>
<p><img src="https://i.stack.imgur.com/B5unK.png" alt="Diagram of nut, bolt, washer, and ASA plastic."></p>
<p>The thickness of the Luran S 777k ASA (Acrylonitrile Styrene Acrylate) plastic is currently unknown, but to get the ball rolling let's just say it is 1/8" thick (EDIT: confirmed). The plastic will bear directly on the square washer. </p>
<p>For modeling purposes, I have decided to assume the plastic is totally fixed/rigid and that the load is being applied to the bolt. In this model, it is the washer that is bearing on the fixed plastic. </p>
<p>I expect that the limiting factor of this system is going to be the pull-through capacity of the plastic as (I assume) the washer will tear it apart way before the washer or the bolt fail. How can I determine the pull-through capacity of this system? Is my assumption that the pull-through capacity of the plastic is the limiting factor valid? </p>
<hr>
<p>My idea is to simply obtain the shear resistance of the plastic material online somewhere, and calculate the pull-through capacity like so:</p>
<ul>
<li>Shear strength of the material is $\tau_{\text{ASA}}=??$</li>
<li>Load-resisting cross-sectional area: $$A=4s_{\text{washer}}\cdot t_{\text{plastic}}$$</li>
<li>Therefore the pull-through capacity of washer bearing on plastic is: $$P_t=\tau_{\text{ASA}}\cdot 4s_{\text{washer}}\cdot t_{\text{plastic}}$$</li>
</ul>
<p>However, I'm concerned there may be other considerations at work here that I need to take into account. Is it really this simple? </p>
<hr>
<p>Thanks to hazzy for the simple but reassuring answer, and the points he made. </p>
<p>It turns out that the limiting consideration for this particular problem seems to be fatigue/cyclic bending of the plastic around the washer, since in actuality the load is being applied at either end of the plastic section (instead of uniformly as I proposed in the model above). </p>
| |mechanical-engineering|materials|applied-mechanics|plastic| | <p>You are KIND OF RIGHT !</p>
<p>IN THIN PLASTIC DEFORMATION ONLY</p>
<p>OD of steel washer = OD</p>
<p>Thickness of Material (Pastic) = Tp</p>
<p>Shear Area = (OD+2.5*Tp)*pi()*Tp</p>
<p>Shear force is Shear Area * MPa = KN force.</p>
<p>The plastic deformation forms a "bulb" effect same goes for concrete piles
in the ground ! the diameter x 2.5 is the effective area for the end of the pile.
This gets adjusted should the depth be less than 2.5 x the diameter.</p>
<p>Garret Krampe</p>
| 616 | Finding the pull-through capacity of a nutted bolt and washer in ASA plastic |
2015-02-11T23:58:32.340 | <p>My understanding is that cavitation occurs in the flow of a liquid when the static pressure drops below the vapor pressure, even intermittently. So even if the <em>time-averaged</em> static pressure (what you might measure) is above the vapor pressure, the pressure fluctuations from turbulence or other unsteadiness could be large enough to cause cavitation locally. So comparing the time-averaged static pressure against the vapor pressure isn't enough; you need to add some extra cushion to account for the pressure fluctuations. (This is my interpretation, not having read too deeply into this.)</p>
<p>So, in various books, websites, and journal articles I have seen two different types of dimensionless numbers for estimating whether the flow through a valve or nozzle cavitates. They are generally called the cavitation index or cavitation number. They take one of two forms:</p>
<p>$$\sigma = \frac{p_\text{in} - p_\text{vapor}}{p_\text{in} - p_\text{out}}$$</p>
<p>or</p>
<p>$$\sigma = \frac{p_\text{in} - p_\text{vapor}}{\tfrac{1}{2} \rho V^2}$$</p>
<p>where $p_\text{in}$ is the inlet pressure, $p_\text{out}$ is the outlet pressure, $p_\text{vapor}$ is the vapor pressure, $\rho$ is the liquid density, and $V$ is some characteristic velocity of the flow (say, in the nozzle case, the velocity at the outlet). Some forms of this number are inversions of the numbers above, but these aren't that different.</p>
<p><em>What is the difference between these parameters?</em> Based on energy conservation you can relate the pressure drop to the flow rate, but typically there is an empirical coefficient added in to account for non-idealities. Is there something else I am missing?</p>
<p><em>Is one form preferred over the other?</em> Best I can tell whether to use one or the other depends on what sort of data you have (so, for flow over a turbine blade, the velocity form is preferred), but I've seen both even for nozzles.</p>
<p><em>Where can I get accurate data to predict cavitation based on these numbers?</em> I've tried using some data on atomizer nozzles from various journal articles but generally they use different forms of the cavitation number. Some of the data suggests the flow through the nozzle will cavitate at the pressures I want, but other data for similar nozzles suggests it won't. I'm not sure what the source of the inconsistency is. My understanding could be faulty, the cavitation number model could be too simplistic, the data could be inaccurate, etc.</p>
| |mechanical-engineering|fluid-dynamics|multiphase-flow| | <p><strong>The difference between the two equations</strong></p>
<p>The cavitation number is the ratio of the static pressure difference to the dynamic pressure difference. So, if you want to use the first equation, you would need to take the pressure using a Pitot tube to measure the total pressure, whereas if you want to use the second equation you will need to measure the freestream velocity, but I would recommend measuring it upstream rather than downstream because of possible effects of acceleration and boundary layer growth. Also, your $V$ should be $V_{in}$ such that it corresponds to the same location where $p_{in}$ is measured, because this equation is derived from Bernoulli's equation which says the energy is conserved along a streamline. </p>
<p><strong>Is one form preferred over the other?</strong> </p>
<p>In all my experience working in cavitation research for many years, we have almost always used the latter equation you mentioned (although I have mainly been working in hydrofoils and propulsion systems). The reason is that we could get more accurate non-intrusive velocity measures using <a href="https://en.wikipedia.org/wiki/Laser_Doppler_velocimetry">Laser Doppler velocimetry (LDV)</a> than by using an intrusive method.</p>
<p><strong>Where can I get accurate data to predict cavitation based on these numbers?</strong></p>
<p>It is difficult to use experimental data to predict the cavitation number because of the differences in things like turbulence intensity and air nuclei content, which are difficult to match in reality with controlled laboratory methods. Traditionally, in my circles, this is done by running some CFD analysis codes on your design. There are two different approaches here: (1) compute the average mean flow using a RANS or LES technique, and (2) using a bubble dynamics code which will model the air nuclei, but requires a flowfield (either from experimental measures or from the from the CFD model). If you use a typical RANS CFD model to compute the flow-field, it should give you the the pressure coefficient which has a very similar definition to the cavitation number:</p>
<p>$$C_P = \frac{P-P_\infty}{\frac{1}{2}V^2}$$</p>
<p>If you are doing some CFD calculation on your nozzle, you should find the location of minimum pressure, and that is the place where cavitation should occur. You can infer the cavitation number from this pressure coefficient as:</p>
<p>$$\sigma = -C_P^{min}$$</p>
<p>where $C_P^{min}$ is the minimum pressure in your nozzle. I explain this in more detail in this <a href="http://www.sname.org/communities1/resources/viewtechnicalpaper/?DocumentKey=7663cdf9-7481-4556-84ae-cbfabf204cb1">paper</a>. However, this will only give you an idea of the time averaged cavitation inception number. Most people don't go to such detail in trying to get such an accurate prediction of cavitation inception, unless it is absolutely critical.</p>
<p>If you want to get a more accurate number, you need to consider that cavitation inception requires three things to happen at the same time: (1) a local area of pressure which is below the vapor pressure of water, (2) an air nuclei which enters into that low pressure region, and (3) the air nuclei must be in the low pressure for a significant enough time that it basically rapidly grows, becomes unstable and hence collapses. The way people have been able to more accurately estimate this is by using a using a Lagrangian method that simulates sending air nuclei through an Eulerian CFD dataset. Some of the real experts in this field are the people at Dynaflow-inc.com. I might suggest taking a look at this paper: </p>
<blockquote>
<p>Chahine, G.L. "Nuclei Effects on Cavitation Inception and Noise", 25th
Symposium on Naval Hydrodynamics, St. John's, NL, Canada, Aug. 8-13,
2004. <a href="http://www.dynaflow-inc.com/publications/pdf_documents/2004/ONR-SNH25-Chahine-1.pdf">PDF here</a></p>
</blockquote>
<p>However, if you don't want to go to all that trouble, I would recommend that you compute an estimate of the pressure fluctuations $p'$ based on the ambient turbulence intensity of your flow, and then subtract this value off of your mean pressure to get a better estimate of cavitation number. You should be able to get this value out of the turbulence model if your using a RANS technique. If you are looking at possible CFD techniques to use, unless you have a lot of money to spend, I might suggest looking into using OpenFOAM.</p>
| 617 | Estimating whether the flow through a valve or nozzle cavitates |
2015-02-12T20:42:33.570 | <p>A friend of mine is a structural engineer and he often complains to me about how little sleep he gets at night worrying over whether or not he accounted for all the little details involved in his design. </p>
<p>For all the double and triple checking one might do, buildings are very complicated structures. Sometimes its impossible to account for every possible situation. Of course, one does to the best of their ability to try to account for everything. But if something goes wrong, say a building either collapses or some other less severe problem causes injury or death, I would assume that the engineers can be held accountable.</p>
<p>Are they covered by some sort of malpractice insurance in a similar fashion to doctors?</p>
| |structural-engineering|liability| | <p>Air's answer is pretty thorough already, and the liability issue is truly complex. In my liability training at work, we've been told of situations where engineers from every firm with any association to the machinery or structure in question are named in suits, not necessarily because they're all at fault, but (1) so that their fault can be determined by a judge and (2) because in a civil case, settling may be cheaper than paying for a lawyer to try and get the company/engineers dismissed from the suit. </p>
<p>What I want to do is point out a good example of a catastrophic failure, the level of research that can go into finding out what happened, and the consequences for the firms and engineers involved. One of my professors favorite case studies was the <a href="http://en.wikipedia.org/wiki/Hyatt_Regency_walkway_collapse">Kansas City Hyatt Regency walkway collapse</a>. </p>
<p>In short, two walkways, at the fourth and second floor levels and directly above/below one another, collapsed during an event the hotel hosted, killing 114 people and injuring. It was the largest loss of life in a structural collapse in the US from the time that it happened in 1981 until the World Trade Center collapsed on Sept 11th. </p>
<p>There was a large investigation by the National Bureau of Standards, a link to which can be found <a href="http://fire.nist.gov/bfrlpubs/build82/art002.html">here</a>. That PDF is 370 pages long, including 95 pages of appendices. </p>
<p>As a result of the investigation, the engineering firm and the engineers who signed off on the final drawings lost their licenses to practice, though it appears they avoided any criminal charges and that most of the civil damages were paid by the owners of the hotel. </p>
<p>Obviously, all cases are different, so while they criminal charges were dismissed here, it's possible they wouldn't be in another case. I think the important thing to take away here is that a lot of the work engineers do can be dangerous if not done properly, and being involved on a project that ultimately injures someway means you are extremely likely to have to answer for your design choices at some point, even if they are sound and you have done nothing wrong. </p>
| 637 | Do structural engineers carry insurance for catastrophic errors? |
2015-02-13T04:50:09.690 | <p><a href="https://en.wikipedia.org/wiki/Maglev">Magnetic levitation</a> systems of transportation such as trains seem to be commonplace in early 2000's science fiction, and while some systems exist currently, they aren't exactly common.</p>
<p>It seems to me that this technology has grown unpopular (or at least has fallen into obscurity) in recent years and I'm curious as to why that may be.</p>
<p>Is it primarily technical factors or perhaps technology issues that prevent wider adoption of mag-lev transportation systems? Or is there something else that's slowed the rate of installation?</p>
| |mechanical-engineering|civil-engineering|transportation| | <p>For a given method of transport to become the most popular, it needs to be the safest, cheapest, most efficient way to get from point A to point B, relative to comparable forms of transportation. In this case, the comparable transportation method is to use normal trains, which run on coal, electricity, etc.</p>
<p>Let's compare the two:</p>
<p><strong>Safety</strong></p>
<p>This is sort of hard to compare because the two methods involve different ways of killing people. There are the old "run-someone-over" and "hit-a-service-vehicle" scenarios, and in fact <a href="https://en.wikipedia.org/wiki/Lathen_train_collision" rel="nofollow">the only major (and fatal) accident on a maglev train</a> occurred about eight and a half years ago. 23 people died when the train hit a service vehicle. However, the crash was attributed to human error, not the maglev technology.</p>
<p>Putting these types of accidents aside, maglev trains aren't any more dangerous that normal trains. People can die if they accidentally touch power lines, but normal electric trains pose the same risks. Fires aren't any more likely to be started, and fuel can't be spilled. So safety isn't an issue.</p>
<p><strong>Cost</strong></p>
<p>Here, there <em>is</em> a difference. In a Congressional report, <em>Report to Congress: Costs and Benefits of Magnetic Levitation</em>, it was estimated that maglev tracks would cost 40-100 million dollars per mile, compared to 10 million dollars for high-speed rail (HSR). The cost can vary given the area the train is passing through. The report estimates that a maglev system would cost 1.92 times as much as an HSR system in rural areas, 1.22 times as much in suburban areas, 1.20 times as much in mountainous areas, and 1.13 times as much in urban areas.</p>
<p>That's very interesting, because it shows that the technology might be accepted in some areas but not others. Maglev subway systems and <a href="https://en.wikipedia.org/wiki/Elevated_railway" rel="nofollow">elevated railways</a> could become more prevalent in cities but not in rural areas. It wouldn't take too much of a change in infrastructure.</p>
<p>Different routes lead to different costs per mile, with the cost of routes in the American Northeast Corridor being one and a half times as much as average. Fortunately, thus setup is an outlier, and is really due to the high population in the region and the replacement of the existing rail lines.</p>
<p>Different methods range in price, too. The Congressional report also compares the programs of different countries. They vary quite a lot, though, and there doesn't seem to be a definite pattern.</p>
<p>One last thing: The report was written in 2005. Since then, the value of the dollar has changed and technology has improved. I think that the cost of maglev trains has gone down quite a lot since then.</p>
<p><strong>Efficiency</strong></p>
<p><a href="http://large.stanford.edu/courses/2010/ph240/ilonidis2/" rel="nofollow">This paper</a> (admittedly short) compares a German <a href="https://en.wikipedia.org/wiki/Transrapid" rel="nofollow">Transrapid</a> maglev train and the <a href="https://en.wikipedia.org/wiki/ICE_3" rel="nofollow">ICE 3</a> high-speed train, also developed in Germany. The energy usage is about the same, though at lower speeds Transrapid has the edge over ICE 3. There isn't data for higher speeds, though, which is odd, because the newest Transrapid models can go much faster than the data explains.</p>
<p><a href="http://www.21stcenturysciencetech.com/articles/Summer03/maglev_3.pdf" rel="nofollow">This article</a> shows that maglev trains are still many times more efficient than airplanes or cars. It's also a bit speculative (using maglev as a launch system for spacecraft?!), but it's fairly comprehensive. But it's a bit dated, like the Congressional report.</p>
| 644 | Could magnetic levitation transportation systems become a commonplace technology? |
2015-02-14T01:27:23.320 | <p>I've recently been learning about FET amplifier circuits and I'm looking for some information that can help explain why the calculated and simulated Gain Values can differ by up to 15% in some cases.</p>
<p>this image is an example of the first FET amp.
<a href="http://www.electronics-tutorials.ws/amplifier/amp14.gif" rel="nofollow">http://www.electronics-tutorials.ws/amplifier/amp14.gif</a>
which gives the following values
calculated Gain = -11.8
simulated Gain = -13.4
difference = 12.698%
any help explaining this difference will be greatly appreciated. </p>
| |electrical-engineering|power-electronics|circuit-design| | <p>If you measure the gain, there is usually some variability in the measurements. This means that you would have to measure enough times to get a statistic distribution for the gain. The computed value should be equal to the mean of the distribution.</p>
<p>What I'm also curious about: is the gain given in Decibels (dB)? </p>
| 650 | Percent (%) error between calculated gain and measured gain of FET amplifiers |
2015-02-14T03:58:12.383 | <p>I've always wondered why refrigerators don't have some of their parts located outdoors like an air conditioner.</p>
<p>In warm weather, it seems like it would make sense to have the condenser outside like an AC unit to avoid heating the room. In cold weather, it seems like it would be much more efficient to have the condenser outside to cool down faster. Why is the condenser still indoors?</p>
<p>First, does this exist in some places? Have I personally just never seen it? Maybe for large commercial, refrigerator/freezers?</p>
<p>If not, would the cost simply outweigh the benefit? Is it cheaper to have the AC cool the room down after the fridge has heated it slightly rather than install another condenser outside? In cold weather, is the fridge acting as an efficient space heater, so putting the condenser outside not really get you anywhere? I'd be interested to see if a cost/benefit analysis has been done.</p>
<p>Finally, is it just convenience? Is it simply much easier to just move the fridge in, plug it in, and be done? </p>
| |mechanical-engineering|thermodynamics|refrigeration| | <p>In the winter, you're paying to run the refrigerator when you shouldn't need to run it at all. However, since we're also paying for heat and the refrigerator is generating heat then our furnace doesn't run as much and it doesn't make much difference.</p>
<p>In the summer, we're paying to cool our homes and the heat from the refrigerator makes that cost more. So this is really the biggest problem but it can be solved without moving the condenser outside. Both options have already been mentioned.</p>
<p>It's much easier to pipe air in and out then it is to pipe a pressurized coolant. If we insulate the back of the fridge and insulate the duct going outside we could exhaust the heat. Of course, you could simply use a fridge in a garage or on enclosed patio perhaps.</p>
<p>Sure the fridge would cost more to make but it's just a bit more plastic and insulation. If the manufacturer can claim it's 32% more efficient or even 25% that a huge selling point these days. And yes the duct work would cost more for sure but we need a duct anyways for the stove vent.</p>
<p>In the end, times changes, what was economic before might one day make total sense. And you could even do it yourself. It's just a matter of how good you want it to look. I'm sure many people could even recess it into a wall or cabinet so you wouldn't even see it. Others could match the paint, bevel the edge and make it look totally normal.</p>
<p>Just my opinion.</p>
| 651 | Why don't refrigerators have parts outdoors? |
2015-02-14T14:39:18.577 |
<p>I was recently reading up on the <a href="http://en.wikipedia.org/wiki/Hyatt_Regency_walkway_collapse" rel="noreferrer">1981 Hyatt Regency walkway collapse</a>. It was a tragic loss of life and haunts many to this day.</p>
<p>The linked article does a better job of summarizing things, but here's a summarized version of what led up to the collapse:</p>
<p>The engineering firm (Gillum & Associates) designed the walkways to be supported on really long, threaded rods. The company supplying the rods (Havens Steel) didn't like the specified rods because they were going to be expensive and likely to be damaged during construction. Havens Steel sent an alternate design to Gillum & Associates which was used for the construction.</p>
<p>The alternate design didn't meet code, catastrophically failed, and led to the loss of many lives.</p>
<p>The Missouri Board of Architects, Professional Engineers, and Land Surveyors found Gillum & Associates guilty of negligence and revoked all of the engineering licenses assigned to Gillum & Associates and its employees.</p>
<p>What I'd like to understand better is <em>why</em> Gillum & Associates was held liable for the disaster. <strong>Wasn't it really Havens Steel's fault for proposing the alternate design to begin with?</strong> If Havens Steel had built according to the original design proposed by Gillum & Associates, then the disaster never would have occurred.</p>
| |licensure|ethics| | <p>At the time , my understanding was that high strength ,quench and tempered rods were specified . The contractor used hardware store threaded rod with less than half the strength without the engineers approval.</p>
| 662 | Why were engineers negligent in the 1981 Hyatt Regency walkway collapse? |
2015-02-14T16:25:28.327 | <p>I am working on an idea that involves a jet engine, but I don't quite know how the fuel system works. In a combustion engine like that in a car, you have nozzles that spray the fuel into the piston chambers (I think, correct me if I am mistaken), and then it's ignited when the piston compresses the chamber (sometimes using a spark plug, sometimes it doesn't). How is fuel mixed with air and ignited inside a jet engine? I know that compressed air comes from the intake system (the fan blades pull in and compress air and direct it into the part where it mixes with fuel), but how does fuel get mixed in?</p>
| |mechanical-engineering|fluid-dynamics|aircraft-design| | <p>The place where air and fuel are mixed is the <strong><a href="https://en.wikipedia.org/wiki/Combustor" rel="nofollow">combustor</a></strong>, also known as the <strong>flame holder</strong>:
<a href="https://upload.wikimedia.org/wikipedia/commons/1/1c/Combustor_diagram_componentsPNG.png" rel="nofollow"><img src="https://upload.wikimedia.org/wikipedia/commons/1/1c/Combustor_diagram_componentsPNG.png" alt="Combustor #1"></a></p>
<p>The <em>diffuser</em> takes in the compressed air and slows it down (remember that aircraft with jet engines are traveling extremely fast). If the air goes too fast, it won't burn enough; if it goes too slow, it won't provide enough thrust.</p>
<p>Air goes in through gaps in the <em>liner</em>. There are holes all along it, called <em>primary holes</em>, <em>intermediate holes</em> and dilution holes*. These further slow the air down, and help it enter the main part of the combustion chamber. However, most of the air goes through the <em>swirler</em>, which mixes the air as it enters. The <em>dome</em> is a related device.</p>
<p>Inside this chamber is an <em>igniter</em>, which creates the sparks needed for the fuel to burn. Next to it is the <em>fuel injector</em>, a pipe which inserts fuel into the chamber.</p>
<p>The air follows a series of complex paths inside the combustor:</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/4/49/Combustor_diagram_airflow.png" rel="nofollow"><img src="https://upload.wikimedia.org/wikipedia/commons/4/49/Combustor_diagram_airflow.png" alt="Combustor #2"></a></p>
<p>This picture should give you a good idea of them. Now you probably understand the names of the holes into the main chamber! They cool the combusted air (entering through the swirler) as it goes out the other end and into the next (and rear) section of the engine.</p>
| 664 | How is fuel mixed with air in a jet engine? |
2015-02-14T18:31:57.687 | <p>I'm trying to construct a stainless steel high pressure containment vessel (pressure reactor) (2000 psi or so). Yes I realize that this is a don't try this at home type of thing, and I plan to put some lexan in front of it before I do any testing. </p>
<p>I'm looking at 2" stainless pipe with a wall thickness of .148"(rated 4000 psi), and I'm trying to figure out the best way to seal the ends. I plan on welding one end cap on, but the other end needs to be removable. I'm not certain how to calculate how much pressure a threaded pipe end would be able tolerate, or if perhaps some sort of high pressure triclamp. I've seen some industrial pressure reactor vessels, but they are rates for something like 13,000 psi, way overkill, they are also super expensive. Looking for a DIY lower cost alternative. Any info would be appreciated.</p>
| |mechanical-engineering|pressure|piping| | <p>Use Flange & Blind Flange ANSI 2"-2500# WNRTJ ASME B16.5 A-182 F304. With a gasket RTJ SS304 and bolting.</p>
| 666 | How to seal a pressure vessel |
2015-02-14T19:23:55.787 | <p>This is inspired by the discussion and controversy surrounding the <a href="https://en.wikipedia.org/wiki/Keystone_Pipeline" rel="nofollow">Keystone Pipeline</a>.</p>
<p>The main part of the Keystone Pipeline system is about 3,400 kilometers long, stretching across a large portion of the United States. The Keystone XL extension would add another long section to it. Altogether, the length of all the segments would be . . . oh, something very big, I imagine (though I acknowledge that none of the oil will travel through all sections).</p>
<p>What is the practical limit for how long an oil pipeline can be?</p>
<p>In the interest of narrowing this down, I'd like to focus on two sub-questions:</p>
<ul>
<li>Is the structural integrity of the pipeline at risk the longer it gets?</li>
<li>Is there any possibility of adverse effects on the oil over longer spans?</li>
</ul>
| |mechanical-engineering|structural-engineering|pipelines|petroleum-engineering| | <p>Same as gas or water,etc, pipelines ; As long as someone is willing to pay for the construction. Presumably because they can profit by delivering the oil/gas/products to a destination. Other comments refer to the "nuts and bolts" of design and construction.</p>
| 667 | What is the maximum length of oil pipelines? |
2015-02-16T05:50:12.400 | <p>I have a chicken coop I plan on modifying to automatically open and close the door at dawn and dusk.</p>
<p>What sort of actuator or mechanism would be appropriate for operating the small, side-hinged door?</p>
<p>Constant power is not available (solar) plus it needs enough holding force to keep foxes out.</p>
| |electrical-engineering| | <p>You could build something out yourself and connect to a lead-acid battery which charged during the day like this one <a href="http://makeitbreakitfixit.com/2016/08/30/diy-home-automation-chicken-enclosure/" rel="nofollow noreferrer">http://makeitbreakitfixit.com/2016/08/30/diy-home-automation-chicken-enclosure/</a></p>
| 686 | Electromechanically open and close a small, hinged door? |
2015-02-16T20:34:29.427 | <p>Suppose I have a sealed enclosure, say <a href="http://www.hoffmanonline.com/product_catalog/catalog_item_detail.aspx?cat_1=34&cat_2=2346&cat_3=78347&catIDs=78347,186&itemIDs=3239,4185&catalog_item=4185" rel="nofollow">this one from Hoffman</a>:</p>
<p>The box is 16"x12"x6", made of galvanized steel with a wall thickness of .0635" or .0785" (gauge 16 or 14). I'm planning to put some electronics in this box, which will generate an arbitrary amount of heat. How can I use this information to estimate how much the air inside the box will heat up above ambient outside the box?</p>
| |mechanical-engineering|thermodynamics|temperature|heat-transfer| | <p><a href="http://www.automationdirect.com/static/specs/encstratusheatexchangers.pdf" rel="nofollow">http://www.automationdirect.com/static/specs/encstratusheatexchangers.pdf</a></p>
<p>According to this document, heat load transfer through the walls of a metal enclosure is estimated in BTU/H as 1.25 x surface area (sq ft) x temperature differential (degrees F).</p>
<p>Converting to W/degC, you would get about .66 times the surface area of the box in square feet, or 7.1 times the surface area in square meters.</p>
<p>Or in degC/W, that's 1.52 ft^2, 218.9 in^2, or .14 m^2. Divide by the surface area to get the thermal resistance of the box.</p>
| 693 | How do I determine the thermal resistance of a metal box? |
2015-02-17T14:40:48.747 | <p>My car has ultrasonic proximity sensors to help me park. I've noticed that when motorcycles whiz past me the proximity alarm goes off. I originally thought that the motorcycles were just too close, but now I have observed that that isn't the case; cars or other road users at a similar distance and speed do not set off the sensors. It doesn't seem to matter what speed I am traveling.</p>
<p>I am not interested in troubleshooting some aspect of the alarm. I'm just trying to understand why this occurs. What special characteristic of a motorcycle or its movement causes it to be detected by the sensors when other road users are not detected?</p>
| |electrical-engineering|sensors|automotive-engineering| | <p>It is likely that the noise of the motorcycle is causing the sensor to activate.</p>
<p>This <a href="http://static.catalogs.rockwellautomation.com/pub-files/13676018925183f2e400f4c/pub/Ultrasonic-Sensing/Of/page.pdf" rel="nofollow">listing</a> of the advantages and disadvantages of ultrasonic sensors lists one of the disadvantages as:</p>
<blockquote>
<p>While ultrasonics exhibit good immunity to background noise, these sensors are still likely to falsely respond to some loud noises, like the hissing sound produced by air hoses and relief
valves.</p>
</blockquote>
<p>A loud motorcycle could fall into that same level of noise (sound pressure).</p>
<p>Other <a href="https://dlnmh9ip6v2uc.cloudfront.net/datasheets/Kits/HRXL-MaxSonar-WR_Datasheet.pdf" rel="nofollow">sensor spec sheets</a> that I have seen have phrases like:</p>
<blockquote>
<p>Many acoustic noise sources will have little to no effect on the reported range...</p>
</blockquote>
<p>This would likely be why the sensor is not going off all the time, but is triggered by loud noises.</p>
<p>The noise from a motorcycle is about <a href="http://www.gcaudio.com/resources/howtos/loudness.html" rel="nofollow">100dB</a>.</p>
| 1705 | How do ultrasonic proximity sensors detect motorcycles differently from cars and trucks? |
2015-02-17T14:47:53.910 | <p>Bridges are designed for the loads that come from the vehicles that are expected to cross them. This includes the weight the vehicle and any dynamic loads that may be introduced from movement of the vehicle. Dynamic loads may be from "bouncing" or from hitting a joint or pothole.</p>
<p>Initially it would seem obvious that more load is applied to a bridge while the vehicles are in motion (weight of vehicle plus dynamic load). The dynamic loads are proportional to the travelling speed of the vehicles, but as vehicles go faster, they typically are spaced farther apart.</p>
<p>When vehicles are stopped, they typically are much more closely spaced than when they are moving.</p>
<ul>
<li>Can there be a situation where more load is on a bridge because of closely spaced <strong>parked</strong> vehicles than from <strong>moving</strong> vehicles that are farther apart?</li>
<li>Are these two situations covered in bridge design?</li>
</ul>
| |civil-engineering|bridges|structural-engineering| | <p>Which is more significant depends upon the bridge in question, certainly its length and the characteristics of the loading applied. For this discussion I am assuming highway traffic loading.</p>
<p>The question refers to dynamic effects, and it is worth noting that this is more than just impact from striking a pothole. If an elastic single span simple beam is instantaneously loaded with a force, the peak deflection that results is twice the deflection under the same force in the static case, and this effect is unrelated to the effect of instantaneous spikes in applied loading due to vehicles hitting surface irregularities (potholes etc).</p>
<p>For most highway bridges, I believe the dynamic effect is more significant. Fast-but-spaced-out traffic is more onerous than slow-and-bunched traffic. However, this conclusion is based upon the observation that there are more short bridges (spans of up to tens of metres) than long bridges - there is not a general principle that gives one universal answer.</p>
<p>It's relatively simple to determine that for a very short bridge whether the traffic is queuing or not is irrelevant - if the bridge is shorter than one vehicle, only one vehicle (or one axle) will be on the bridge, so whether there is a queue or not does not influence the number of vehicles loading the structure or the loading. Conversely it's easy to imagine that a very long bridge (several hundreds of metres long), one vehicle hitting a pothole will have a negligible effect, since even if the load from that one vehicle doubles instantaneously, if there are hundreds of vehicles on the deck it won't have a proportionately great effect.</p>
<p>In UK practice, trunk road bridges are designed and assessed to Highways Agency documents (so-called 'BD's and 'BA's). The vehicular highway loading is in two 'flavours' - HA is 'normal' traffic, and HB is an arbitrary loading used to examine the characteristics of the bridge under abnormal loading.
HA loading for design is defined in <a href="http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol1/section3/bd3701.pdf" rel="nofollow">BD37</a> and the derivation includes allowance for impact - read Appendix A: "the impact effect of an axle on highway bridges can be as high as 80% of the static axle weight and an allowance of this magnitude was made in deriving the HA loading", though the influence of impact reduces as the span becomes greater.</p>
<p>Traffic queues not only give rise to nose-to-tail bunching, they potentially give rise to vehicles squeezing closer side-to-side. In the BDs this is referred to as 'lateral bunching', which is where more vehicles crowd onto the structure.</p>
<p><a href="http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol1/section3/bd3701.pdf" rel="nofollow">BD37</a> allows for both impact and lateral bunching simultaneously - ie, it assumes that you have a tightly-packed traffic jam that's also travelling at high speed. This obviously doesn't happen, but is what the code encapsulates.</p>
<p>When assessing existing structures, however, the UK standards don't apply both effects together. <a href="http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol3/section4/bd2101.pdf" rel="nofollow">BD21</a> is the code for assessing structures. Clause 5.23 specifically addresses this question (UDL and KEL are two component parts of HA loading):</p>
<p>"The HA UDL and KEL have been derived using a lateral bunching factor to take into account the possibility that, in slow moving situations, more lanes of traffic than the marked or notional lanes could use the bridge. Probabilistic analysis shows that maximum impact effects, which occur at high speeds, should not be considered together with maximum lateral bunching. Comparison of the effects of alternative traffic speed and bunching situations have led to the conclusion that high speed high impact effect with no lateral bunching is the most onerous criterion for bridge loading. The HA UDL and KEL are therefore to be adjusted in order to eliminate the lateral bunching factor by dividing by
the following Adjustment Factor (AF)"</p>
<p>The adjustment factor is a relatively large number for loaded lengths up to 20m, but tails off to 1.0 (ie divide by 1.0, so don't change the value) at 40m loaded length. </p>
<p>It's not possible to make definitive statements from this (ie, you can't say "below 20m it's dynamic, above 40m it's traffic queues"), because the loading is to some degree empirical and derived after a probabilistic analysis, and includes guesswork (described as "by estimation" in the standard) with allowance for the findings of sensitivity analysis. Again, BD37<a href="http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol1/section3/bd3701.pdf" rel="nofollow">1</a> Appendix A has discussion of this:</p>
<p>"For long loaded lengths, the main factors affecting the loading are the traffic flow rates, percentage of heavy vehicles in the flows, frequency of occurrence and duration of traffic jams and the spacing of vehicles in a jam. These parameters were determined by studying the traffic patterns at several sites on trunk roads, by load surveys at other sites and, where the required data was unobtainable, by estimation. A statistical approach was adopted to derive characteristic loadings from which nominal loads where obtained. Sensitivity analyses were carried out to test the significance on the loading of some of the assumptions made."</p>
<p>It is worth noting that the above is a bit non-rigorous with respect to spans. The length that matters with respect to load derivation is the 'loaded length', which is not always the same thing as the span. For a single simply-supported span, if you're examining the bending effect, the two are synonymous, but very many bridges are more complex than that (eg, multiple continuous spans, or integral abutments etc). Loaded length is the length over which the element of structure is loaded, and when designing you should choose the length that gives rise to the most onerous effect for the element you are designing for doing the calculations with respect to that particular element. This is often the whole length of a span, but there are cases where (eg) loading a shorter length has a greater effect, particularly in continuous structures, partly because the shorter the loaded length used, the greater the intensity of load to be applied.</p>
| 1706 | Does a roadway bridge experience more load when vehicles are parked or when they are moving? |
2015-02-17T20:05:05.127 | <p>I'm giving SolidWorks a try and am following one of their tutorials on creating a part. When I create circles they are looking a bit "boxy" to me. My graphics card has 512MB of ram and I have checked the rendering settings that I found and everything seems to be OK - but I would have expected something more "circular" and less rough looking, but I'm new to working with CAD - is this normal?</p>
<p><img src="https://i.stack.imgur.com/koy1P.png" alt="Boxy Edges"></p>
<p><img src="https://i.stack.imgur.com/ylBSY.png" alt="Full screen View"></p>
<p><img src="https://i.stack.imgur.com/5umIB.png" alt="Rendering Settings"></p>
| |computer-aided-design|solidworks| | <p>Yes, this is normal. By default, SolidWorks renders curves on the screen using less than the highest possible level of detail that your monitor is capable of displaying. The point of this is to allow the screen to be redrawn quickly as you edit your design and change the view. The more detailed the curve, the longer it takes to redraw the screen; depending on your hardware, this could make the software feel unresponsive or "laggy" and make it harder to use.</p>
<p>You can adjust the Image Options in the Document Properties tab to render curves with more detail; see <a href="http://help.solidworks.com/2014/English/SolidWorks/sldworks/HIDD_OPTIONS_IMAGE_QUALITY.htm?id=b432045cef064e2db19d520f397724c7#Pg0" rel="noreferrer">documentation here</a> and this screenshot from <a href="http://help.solidworks.com/2014/English/api/swconst/DP_ImageQuality.htm" rel="noreferrer">the API help pages</a>:</p>
<p><img src="https://i.stack.imgur.com/GsHJw.gif" alt=""></p>
<p>The slider at the top of the window controls how precisely curves are drawn:</p>
<blockquote>
<ul>
<li><strong>Low - High (slower) and Deviation:</strong> (For assemblies, available only if Apply to all reference part documents (below) is selected.)
The slider controls the image quality resolution, and Deviation is the
maximum chordal deviation in effect. Move the slider or type a value
in Deviation. The slider setting and deviation value are coupled and
are inversely proportional.</li>
</ul>
</blockquote>
<p>If all you're concerned about is confirming that your circle is actually a circle even when it is rendered with corners, you may not want to increase this setting, as it will have a negative impact on performance.</p>
| 1709 | Are boxy edges normal in SolidWorks when working with a drawing? |
2015-02-17T23:44:38.207 | <p>What with all the work done by the FAA recently on regulating drone usage, I thought it would be nice to turn my eyes to the sky for another, less-known type of aerial denizen: the <a href="https://en.wikipedia.org/wiki/Airborne_wind_turbine" rel="noreferrer">aerial wind turbine</a>.</p>
<p>Wikipedia has all the relevant information, so I won't rehash it here. The most important thing here is that aerial wind turbines - and I'm mostly interested in aerostat-based designs - are free-floating, just like certain types of tethered drones, which have given the FAA headaches because they're not like most aircraft.</p>
<p>I did find <a href="http://www.avweb.com/avwebflash/news/aerial_wind_turbine_204758-1.html" rel="noreferrer">this</a> article, which says</p>
<blockquote>
<p>Regulations and technological restrictions suggest it may not happen very soon, or at all, but some researchers believe aerial turbines will be tapping high-altitude winds for power generation sometime in our future and perhaps within the decade.</p>
</blockquote>
<p>and</p>
<blockquote>
<p>That said, companies currently working on the project seem very loosely formed. They are in the development phase and are aware of federal airspace restrictions.</p>
</blockquote>
<p><a href="http://www.faa.gov/uas/civil_operations/" rel="noreferrer">Here's</a> the relevant FAA page.</p>
<p>I suspect that FAA regulatory measures on drones cover aerial wind turbines, but I'm not positive, and I'd like to see some specific mention of them in regulations, if possible. I assume they're treated as (tethered) drones, but I haven't been able to find any direct mention.</p>
<hr>
<p>Getting to the point: Aerial wind turbines are, in general, much smaller than typical wind turbines. There are obviously some factors that make it easier for wind turbines built on the ground to be larger (support, for instance). But larger turbine blades, while a source of trade-offs, can be much more helpful.</p>
<p>What regulations (in the United States), if any, limit the size of aerial wind turbines?</p>
| |aerospace-engineering|renewable-energy|regulations| | <p>I think you may be misinterpreting aerial wind turbines as a type of aircraft. The FAA section of the Code of Federal Regulations defines an aircraft as <a href="http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&node=pt14.1.1&rgn=div5#se14.1.1_11" rel="nofollow">a device that is used or intended to be used for flight in the air.</a> I suppose the real debate comes down to what you define as "flight," which the CFR does not deal with. While an aerial wind turbine certainly is supporting itself in the air through aerodynamics, it's still tethered to the ground and I wouldn't exactly call that "flight" in the same sense as it's being applied to the aircraft being regulated in these codes. </p>
<p>This question is really more complex than anyone can answer right now, because I have a feeling that in the not-so-distant future (<10 years?) a company manufacturing aerial wind turbines will submit some sort of request for a ruling to the FAA on whether or not they're aircraft, and if they disagree, they'll file a suit and go to court over the definition of "flight." But for now, let's assume they aren't aircraft. </p>
<p>Traditional wind turbines are regulated by <a href="http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&node=pt14.2.77&rgn=div5" rel="nofollow">Title 14, Part 77</a> of the CFR, and I think the aerial wind turbines would be regulated with the same code. They're merely an obstruction that is fairly high and has some capacity to move around with the wind, depending on how tightly they're tethered. </p>
<p><a href="http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&node=pt14.2.77&rgn=div5#se14.2.77_117" rel="nofollow">Section 17</a> describes an obstruction as anything higher than 499 feet above ground level, with lower height restrictions near airports, for obvious reasons. However, <a href="http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&node=pt14.2.77&rgn=div5#se14.2.77_19" rel="nofollow">Section 9</a> states that the FAA must be notified of any construction going higher than 200 feet. The key here is that the regulations say absolutely nothing about whether or not you are legally allowed to build something over 200 feet. It simply says that you should notify the FAA, and they will make a determination. </p>
<p>If you are below 500 feet, the code doesn't appear to be explicit on this fact, but it seems you'd find little argument with your request. Over 500 feet, they'll look into it a little more deeply, because that is officially an obstruction. But they're still free to say "We don't think this will be an issue, go ahead." </p>
<p>So do the regulations explicitly limit the size? No. They make no limitations on what you can build, just that you have to get approval once you pass certain restrictions, dependent on your location and proximity to airports. </p>
| 1713 | Do regulations in the United States limit the size of aerial wind turbines? |
2015-02-18T03:57:53.167 | <p>In the US, typical structural steel welding is covered by either AWS D1.1 or AWS D1.5 (bridges). Both of these codes cover a wide range of carbon steels.</p>
<p>AWS has a separate code for welding stainless steels, AWS D1.6. There is some mention of welding to carbon steels, but this is mostly talking about choosing the filler metal and not the testing.</p>
<p>Neither D1.1 nor D1.5 cover stainless steel. </p>
<p>In this specific situation, I am looking at a fillet weld between a carbon steel plate and a stainless steel plate. </p>
<p>What is the procedure for qualifying (PQR and WPS) a weld between the two materials? Does one code control over the other? Would the weld have to be qualified per both specifications?</p>
| |steel|structural-engineering|welding| | <p>According to <a href="http://awo.aws.org/wp-content/uploads/2014/stainlessconference/Campbell-AnUpdateonAWSD16StructuralWeldingCode-Stainless%20Steel.pdf" rel="nofollow">this slideshow</a> it looks like a weld between stainless and carbon or low-alloy steels would be within the scope of D1.6 entirely. It would not be prequalified so it would have to be qualified by testing. Table 4.2 (reproduced in the slide show) covers the test types if you want to qualify with a CJP, but you would need a full copy of D1.6 to see the acceptance criteria and specific geometry for each test. If the fillet weld is all you need to cover, you could use table 4.4 and qualify with a fillet weld only (again, geometry is going to have to come from the full code.)</p>
<p>For what it's worth, this is a relatively young standard, and the third edition is expected out this year, so there's potential for some change.</p>
| 1715 | What is the process for qualifying a weld between structural steel and stainless steel? |
2015-02-18T07:53:38.490 | <p>I've seen it in several data sheets - it is a measure of error of some kind, of course. The problem is I dont know the exact meaning. I've seen it in the context of repeatability, accuracy and linearity.</p>
<p>An example is the following data sheet: <a href="http://www.smcpneumatics.com/pdf/americansmc/ZSE40_ISE40/ZSE40_and_ISE40_Series_Digital_Switches.pdf">smc data sheet</a> (On page 3)</p>
| |nomenclature| | <p>FS = FULL SCALE = maximum reading.</p>
<p>It means that the accuracy is such that the reading is <strong>probably</strong> within + or - 0.5% of the <strong>FULL SCALE</strong> reading.</p>
<p>This is a very important and easily overlooked qualification of the result.<br>
If I have a reading of 1 Volt and the accuracy is +/- 0.5% it means that the actual result should lie in the range 1 - 0.5% x 1 to 1 + 0.5% of 1<br>
= 0.995V to 1.005 V</p>
<p>However - if I measure the result on the 10V range then 0.5% of 10V = 0.5% of Full Scale<br>
= 0.05V.
So 1V +/- 0.5% of FS<br>
= 0.95V to 1.05V.</p>
<p>On the 100V range, 1V +/- 0.5% FS lies in the range<br>
0.5V to 1.50 V. !!!!</p>
<p>The reason for specifying results in this manner is that the error experienced on a given range tends to largely be constant regardless of the actual reading. So, as the input gets smaller the error becomes increasingly large in proportion. </p>
<p>So eg on a 100V range a reading of 0.5V +/- 0.5% FS lies in the range<br>
0 to 1V !</p>
| 1718 | What does "$\pm$ 0.5% F.S." mean? |
2015-02-18T12:02:45.663 | <p>Most introductory books on control theory usually start their <a href="http://en.wikipedia.org/wiki/State_observer">state observation</a> part by introducing the Luenberger observer, and after that they might continue by introducing the <a href="http://en.wikipedia.org/wiki/Kalman_filter">Kalman filter</a>. When reading papers from academic journals I have also come across all kinds of fancy nonlinear estimation methods (sliding-mode observers, passive observers, etc...).</p>
<p>However, from my short experience, it seems like there is only one type of state-estimation method used outside academia: the Kalman filter. </p>
<p>Is my observation correct or are other types also being used? If so, why is the Kalman filter so predominantly used over other state observers? Is it because it's fairly easy to implement?</p>
| |control-engineering|kalman-filters| | <p>I can only speak for the industry I have worked in (heavy machinery). I have only seen Kalman filters used in practice as observers. </p>
<p>Most of the data sources in heavy machinery tend to be quite noisy (pressure or accelerometer sensors). Kalman filters (as compared to simpler Luenberger observers) provide better resilience when faced with high noise levels. I have seen them overall behave more robustly than Luenberger observers. I have seen extended Kalman filters used for sensor fusion in nonlinear systems as well.</p>
<p>The various fancier methods may often have increased computing requirements which make implementation harder in embedded microprocessors. Additionally, the general popularity of Kalman filters means that there is a higher chance managers and engineers outside of controls areas have at least heard of them before. This sort of "brand recognition" can help when selling a solution internally in a large company that isn't focused on controls. Along with this, the support in various libraries or in packages such as Simulink/Matlab is quite old and has been stress tested considerably already.</p>
| 1722 | Types of state observers/estimators actively used in the industry? |
2015-02-18T02:05:22.357 | <p>The <a href="http://en.wikipedia.org/wiki/Tsar_Bomba" rel="nofollow">Tsar Bomba</a> packed a yield of 50 megatons into a package 2.1m in diameter by 8m long. Assuming that the lead tamper modification was not used (which boosts the yield of the Tsar Bomba design to 100 megatons), how much would the yield scale as a multiple-stage thermonuclear warhead (assume that extra stages are added as-needed) was scaled up, say to 3.5m diameter and 20m length?</p>
| |nuclear-technology|nuclear-engineering| | <p>I am also not a nuclear engineer, but here are my thoughts on it:</p>
<p>My impression is that you might have trouble scaling a weapon in an arbitrary dimension but that they should scale basically linearly with mass assuming an acceptable configuration of the weapon.</p>
<p>In practice you reach a maximum size for your h-bomb based on the input energy available and various mechanics that they try to keep secret. However, to detonate an h-bomb you simply need an adequate energy flux, the source of that energy flux is irrelevant. In practice it's an fission bomb because nothing less than this can provide the needed energy. However, there's nothing about the fission part of it that's actually needed--a fusion detonator works even better. (Besides, the fusion stage of an h-bomb actually contains a simplified fission device in it's heart--there is a rod of Pu-239 there, after the energy flux has crushed the sides of the fusion device in then the result gets crushed lengthwise, turning the rod into a critical mass and setting off the crushed lithium deutride.)</p>
<p>Thus you can simply stack up the fusion stages of h-bombs one after another, the energy flux from one sets off the next. You can also arrange multiple fusion devices around the fission trigger. Just watch your configuration that you don't destroy a stage before it's fired.</p>
| 1730 | How does thermonuclear warhead yield scale with size? |
2015-02-19T03:24:30.607 | <p>The situation is a steel baseplate that has to be anchored to concrete to resist lateral loads. </p>
<p><img src="https://i.stack.imgur.com/7yWzS.png" alt="base plate with 8 holes"></p>
<p>The lateral load on the base plate is great enough that more than one or two anchor rods are required. The base plate has room for 8 anchor rods.</p>
<p>The holes in the base plate will be somewhat over-sized per <a href="https://www.aisc.org/DynamicTaxonomyFAQs.aspx?id=1840" rel="nofollow noreferrer">AISC</a> to aid in setting the base plate over over the anchor rods. The anchor rods have already been cast into the concrete. The nuts on the anchor rods will not be tightened enough to consider the connections to be slip-critical. There will be some movement of the base plate required before it engages the anchor rods.</p>
<p>Variations in the placement of the anchor rods and the hole locations will mean that not all of the anchor rods will be engaged at the same time.</p>
<p>In this situation, how do you determine how many anchor rods can be considered to be effective in resisting the load? The number of anchor rods will directly effect the size of each anchor rod.</p>
| |structural-engineering|bolting|equipment-selection| | <p><a href="https://www.aisc.org/uploadedFiles/Research/Research_Reports/Kanvinde%20-%20Shear%20Transfer%20in%20Exposed%20Column%20Base%20Plates.pdf" rel="nofollow">This report (large PDF warning)</a> on base plate design says that the AISC <em>Steel Design Guide I</em> instructs the use of plate washers fillet welded to the top of the base plate, with the inner diameter of the washer being much closer to the rod diameter, to ensure initial engagement of the rod. Specifically, I found that information in Sections 2.3.2 (pp.8-10), 3.3.1.2(p.21), and a visual example in Figure 3.9. </p>
<p>While that doesn't directly answer your question, it seems the answer is "You can't determine that, so we put in other design elements to account for it."</p>
| 1733 | How to determine the number of effective anchors in a base plate? |
2015-02-19T23:37:42.743 | <p>Imagine I have an important axle in the gearbox of a motor. I want to support it with ordinary <a href="https://en.wikipedia.org/wiki/Ball_bearing" rel="nofollow">ball bearings</a> at several places along its length. I need both the outer and inner races to stay at fixed diameters at a fixed distance from each other.</p>
<p>However, I <em>can</em> change the number of balls in between the races. I can choose any number of different bearings (though each one has to have a certain diameter, so there's actually a limit). I could squeeze an arbitrary number in, so long as that number is below the maximum and can still support the inner and outer rings.</p>
<p>What determines the optimal number of balls used?</p>
| |mechanical-engineering|bearings| | <p>The number of balls in a bearing affects a couple of different properties of the bearing:</p>
<ul>
<li>Bearing Speed</li>
<li>Bearing Strength</li>
<li>Bearing Life</li>
<li>Bearing Cost</li>
</ul>
<p>A <strong>Full Complement</strong> bearing means that the bearing has as many ball as can physically fit. Having the maximum number of balls means that there isn't enough space left for a cage. The cage keeps the balls from touching each other. When the balls are allowed to touch and rub, the maximum speed of the bearing is reduced. A full complement of balls increases the load capacity by 30% over a bearing with fewer balls in a cage. (<a href="http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs%20Book%20pdf/FUNdaMENTALs%20Topic%2010.PDF" rel="noreferrer">Reference</a> page 16)</p>
<p>The <strong>Number of Balls</strong> affects the load on each individual ball. The equation for the force per length unit of bearing diameter is: </p>
<p>$\frac{F}{ZD^2}$ </p>
<p>where: $Z=\text{Number of balls}$</p>
<p>The number of balls directly affects the load on each bearing and thence the life of the bearing. (<a href="http://nhbb.com/files/catalog_pages/HiTech_Catalog.pdf" rel="noreferrer">reference</a> page 46)</p>
<p><strong>Bearing Life</strong> is controlled by a number of factors. These factors are:</p>
<ul>
<li>Type</li>
<li>Race material</li>
<li>Ball material</li>
<li>Load</li>
<li>Speed</li>
<li>Lubrication</li>
<li>Cleanliness</li>
</ul>
<p>All of these are factors that affect the number on cause of bearing failure: <strong>Fatigue</strong>. (<a href="http://nhbb.com/files/catalog_pages/HiTech_Catalog.pdf" rel="noreferrer">Reference</a> page 45)</p>
<p>The final consideration is <strong>Cost</strong>. Using standard, off-the-shelf bearings are the cheapest. If one of these bearings can be used in your project, it is quite possible that this the best solution. This is even if the project has to be redesigned slightly to accommodate the bearing. Anything that makes your bearing <em>special</em> is likely to <a href="http://www.nmbtc.com/bearings/white-papers/bearing-designing-to-lower-cost/" rel="noreferrer">cost a lot more</a>.</p>
<p>So in the end, <strong>there may not be a definable optimum number of balls</strong>. The best bearing will likely be the one that meets your <strong>speed</strong>, <strong>load</strong> and <strong>life</strong> requirements while being available for a <strong>reasonable price</strong>.</p>
| 1741 | What determines the optimal number of balls between races in a bearing? |
2015-02-20T15:24:06.147 | <p>It may just be my perception, but it seems like water main breaks (at least in Pittsburgh PA) are more common in the winter during the cold weather. It may just that they are more news worthy in the winter (water+cold=ice > news). </p>
<p>Are water mains more likely to break in the winter? If so what can be done to limit or prevent the occurrence? </p>
| |pipelines|infrastructure| | <p>Water main failures are greater in the winter months than in the spring or summer. As a water utility professional in northern Illinois we have around 120 failures each year with 20 + years of data. Each and every year we have 90 of the 120 between December and March (Fact). The question that indeed is hard to define is why? Here are the facts I see year after year and I base my opinion on.</p>
<ul>
<li>Each year as the frost gains depth and speed 1 foot to 3 feet, between December 15th and January 7th we will have 25 main breaks.</li>
<li>From the 7th of January till the 21st of Feb. frost stays at 3 feet we will have 40 main breaks.</li>
<li>From Feb 21st till March 20th as the frost is coming out quickly 3 feet till 1 foot or less we will get 25 main breaks. </li>
<li>Many of the older pipes are made from cast iron that is very rigged and brittle. its like trying to bend a candy cane. 95 percent of failures are on this type of pipe.</li>
<li>The newer pipe is made from ductile iron pipe that is less rigged, allowing it to flex and bend with the frost speed and depth both in and out.</li>
</ul>
<p>As far as frozen water mains the only two I have known of were both shallow 4 foot or less, a dead end run, and very few service lines on them 1 or two.
I could go on and on about the subject but be it known I have been watching frost, cold weather, age of pipe, and material composition for 24 years and the whole water main failure is still a great mystery many days.
Water Utility Superintendent Northern Illinois. </p>
| 1753 | Why do water mains break in the winter? |
2015-02-20T19:34:43.573 | <p>I have a small <a href="http://en.wikipedia.org/wiki/IP_Code" rel="nofollow">IP66</a> <a href="http://www.hammondmfg.com/1554FLP.htm" rel="nofollow">polycarbonate enclosure</a> with a <a href="http://en.wikipedia.org/wiki/Oxygen_sensor#Zirconia_sensor" rel="nofollow">Zirconia-based oxygen sensor</a> which produces significant amounts of heat (the internal sense element runs at 450C using a 1.5W heating element, sensor surface reaches 85C). The enclosure also contains sensitive analog electronics to read the sensor. The residual heat from the sensor is driving the ambient temperature in the enclosure up past 40C (measured on the outside of the enclosure front panel). The oxygen sensor and another gas sensor are only rated for ambient temperatures of 50C maximum, so I'm worried about keeping those in spec. I'm also concerned about needing temperature compensation for the analog circuits.</p>
<p>How can I remove the excess ambient heat from the enclosure while still maintaining an IP66 rating?</p>
| |mechanical-engineering|electrical-engineering|heat-transfer|cooling| | <p>You could pre-cool the gas entering the enclosure, the heat would then go into re-warming up this gas.</p>
<p>This could be done by passing the gas through a simple finned radiator-pipe in a temperature controlled environment (basically a pipe through a fridge/freezer). This might be a good way to do it if you want to control the temperature that the gas re-enters the chamber as you could control the temperature of the pre-cooler and monitor the temperature that goes back in.</p>
| 1758 | How to remove heat from an IP66 enclosure? |
2015-02-20T20:46:22.240 | <p>I'm doing an investigation about building a bridge over a railway line and its <a href="http://en.wikipedia.org/wiki/Overhead_line#Overhead_catenary" rel="nofollow">catenary system</a>. I'm mainly interested in the electromagnetic environment on the bridge as it is affected by the presence of the catenary.</p>
<p>Is this an issue that I need to be concerned with?</p>
<p>I have no experience in either of these fields and that is exactly why I would like to get insight from people who do work in these particular fields. <strong>To answer some of the questions that have been raised in the comments section:</strong></p>
<blockquote>
<p>What order of distance, what voltage? AC/DC? | South Africa or ...? | I'd expect regulatory requirements for unrelated reasons would place you well outside EM interaction range.</p>
</blockquote>
<p>Ideally I would like the bridge to be of reasonable height such that the pedestrians wouldn't mind taking the stairs. I have freedom to reduce the height of the catenary system for exactly this reason. It is a 25 kV AC catenary system. What is the standard EM interaction range?</p>
<blockquote>
<p>What electromagnetic enviroment? You haven't specified any. Will this bridge parallel a power line? Be near a large radio transmitter? What makes you think there are electromagnetic issues here at all?</p>
</blockquote>
<p>According to <a href="http://en.wikipedia.org/wiki/Railway_electrification_system" rel="nofollow">this Wikipedia article</a>, AC high voltage lines "give off" electromagnetic radiation. The bridge is not close to anything really (no power line or large radio transmitter), it just goes over the train track and its catenary.</p>
| |electrical-engineering|civil-engineering|bridges|rail| | <p>"standard EM interaction range" depends a <strong>lot</strong> on frequency, or its inverse which is wavelength. Your 25 kV catenary probably is 50 Hz or 60 Hz, same as your national grid. That means the wavelength is 6000 kilometers. That's far more than your dimensions.</p>
<p>Hence, you can ignore the electro-magnetic aspects and treat this as an electro-static problem. Arcing is the main risk. Here there are two concerns that need tackling: </p>
<ol>
<li><p>Can you prevent catenary movement? Some bridges have isolators beneath them to keep the catenary away from the bridge. </p></li>
<li><p>if the previous mechanism fails, will it fail safely? If the catenary would hit the structure, there will be a rather spectacular light show and the catenary might weld itself to the underside of your bridge. That will result in some high currents. Will the people on the bridge itself be safe? Metal may not be the best surface choice.</p></li>
</ol>
<p>Also keep in mind that dry air is a much better isolator than wet air/rain. It's not just water forming a direct path, raindrops already are an issue. Your bridge probably should be designed to direct rainfall away from the catenary.</p>
| 1760 | Building a steel footbridge over a railway line and its catenary |
2015-02-20T20:42:10.647 | <p>I am constructing an autonomous, unmanned boat for voyages of several months. Traditional autopilots use a linear actuator to move the tiller but require several adjustments per minute. They will consume a lot of electricity and probably wear out mid-voyage. We are looking at wind-vane designs to steer the boat. Wind-vanes are able to steer for great distances without adjustment. The boat will be approx. 5m in length.</p>
<p>In the image below, the 'Course Setting' is done by manually rotating the wind-vane and dropping a pin to hold this position of the two discs. We need a way to rotate the vane fairly precisely and then lock it into position. We are using 12V DC. After it is locked, it should draw no power.</p>
<p><img src="https://i.stack.imgur.com/jhODM.gif" alt="aux rudder trim tab"></p>
<p>As you can see, the current design (not our drawing) would have any electromechanical mechanism hanging off of the end of the boat and attached to the rod of the trim tab.</p>
<p>Is there anything we can use to turn and then lock the wind vane in place? Preferably something that could be proofed against wind and water.</p>
| |mechanical-engineering| | <p>I believe the simplest modification to the drawing that meets the requirements is to replace the top disk in the coupler with a standard spur gear. Replace the bottom disk with a worm gear engaged with the upper gear. This would allow the top to be rotated to the desired position, while the bottom only would rotate small amounts so you wouldn't have to worry about twisted cables.</p>
| 1763 | Precise rotational movement and locking |
2015-02-21T09:55:56.060 | <p>According to BS5950, a beam section can be classified as plastic, semi-compact, compact or slender. For the same section area, a H-beam can take axial compression (without buckling) better than an I-beam, and as such, uses a different strut curve in the code:</p>
<p><img src="https://i.stack.imgur.com/JQFqH.png" alt="enter image description here"></p>
<p>Now, I understand that a H-beam has a wider flange compared to an I-beam, but at what point, precisely, does this transition from I- to H- occurs? For example, is a 400x300 (depth x width) beam considered a H- or an I-beam?</p>
<hr>
<p><strong>Update:</strong></p>
<p>Extracted from BS5950 guide, the following table shows H-beams (also known as universal columns, some of which with depth greater than width. This is the reason why I don't believe the differentiation is so straight forward.</p>
<p><img src="https://i.stack.imgur.com/tRRXw.png" alt="Section Property Table"></p>
| |civil-engineering|steel|beam| | <p>BS5950-1:2000 Clause 1.3.23 defines an H-section as having "<em>an overall depth not greater than 1.2 times its overall width</em>", and Clause 1.3.25 defines an I section as having "<em>an overall depth greater than 1.2 times its overall width</em>".</p>
<p>Note that at exactly a ratio of 1.2, it would be an H section not an I section.</p>
| 1767 | At what point does an I-beam becomes a H-beam? |
2015-02-21T21:28:27.043 | <p>I'm building a software application that uses GPS for a purpose related to roads. </p>
<p>I'd like to know how many decimal places of GPS data should be stored to provide measurements that are accurate to within a few feet?</p>
| |surveying| | <p>The circumference of the earth is about 25,000 miles <em>[40,075 km]</em>, or 131.5 Mfeet. That divided by 360 is 365 kfeet/degree <em>[111.3 km/degree]</em>. A degree value with 4 decimal places has a implied accuracy of 0.00005 deg, or 18 feet <em>[5.5 m]</em>. That's roughly the accuracy of a typical consumer grade GPS anyway, so there is little point going farther. If you have a special (and expensive) surveying grade GPS, then you can use 5 decimal places and get about 1.8 foot <em>[0.5 m]</em> numerical accuracy. Each extra digits reduces the numerical error by a factor of 10. Degrees with 6 fraction digits would be specifying location to about 2 inches <em>[about 5.5 cm]</em>.</p>
<p>Note that the above was assuming the worst case where the 360 degrees are spread out over the entire circumference of the earth. That is true for longitude at the equator, and latitude everywhere. The numeric accuracy of longitude is what is computed above scaled by the cosine of the latitude.</p>
| 1776 | How many decimal places of GPS should be stored to be accurate within a few feet? |
2015-02-22T19:21:19.357 | <p>Lightning strikes have been known to cause massive amounts of <a href="http://www.lightningmaster.com/White-Papers/Lightning-Damage" rel="noreferrer">damage</a>. The <a href="http://www.lightningsafety.com/nlsi_lhm/lpts.html" rel="noreferrer">stats on a lightning bolt</a> are:</p>
<blockquote>
<p>current levels sometimes in excess of 400 kA, temperatures to 50,000 degrees F., and speeds approaching one third the speed of light</p>
</blockquote>
<p>These are massive numbers, but lightning protection systems are designed to draw the lightning away from the building or structure that they are protecting. Lightning protection systems can be simply thought of as lightning rods connected to the ground via cabling (downconductor).</p>
<p><img src="https://i.stack.imgur.com/kndwb.png" alt="enter image description here"></p>
<p>The NOAA <a href="http://www.nws.noaa.gov/directives/sym/pd03041006curr.pdf" rel="noreferrer">specification for lightning protection</a> requires that lightning rods be at least 0.5in (13mm) in diameter. The downconductor is a similar size copper cable (<a href="https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes" rel="noreferrer">4/0 AWG or 12mm</a>). The allowable amperage for this type of wire is only about <a href="http://www.cerrowire.com/ampacity-charts" rel="noreferrer">250A</a> for constant current. I realize that this is more of a heat limit rather than a instantaneous current capacity limit.</p>
<p>From this <a href="http://www.lightningsafety.com/nlsi_lhm/conventionalLPT.pdf" rel="noreferrer">paper on lightning protection</a> (page 28):</p>
<blockquote>
<p>Positive feedback on the operation of a lightning protection system is seldom documented and most often not even noticed. Only in some rare cases can it be documented that a lightning protection system has been struck if it works properly and there is no damage. There is sometimes evidence at the strike termination point which can be noted during a careful inspection, but it is seldom cost effective for the owner of a lightning protection system to obtain the expertise necessary to conduct such a careful inspection.</p>
</blockquote>
<p>How can a seemingly small 0.5in (13mm) piece of metal handle a lightning strike with little or no visible damage much less without being completely destroyed?</p>
| |electrical-engineering| | <p>The current limit specification for a wire is limited by the heat that current will produce, and how much heat the wire can dissipate before getting too hot. "Too hot" depends on the circumstances. You will see higher current ratings for the same type of wire in chassis wiring applications than the electrical code allows for home installations, for example. This is mostly due to how hot too hot is. The ultimate limit for extreme applications is that the conductor not melt. Temperatures anywhere near that would be unsafe running along wooden supports inside a wall in a house.</p>
<p>As you say, the wire is rated for 250 A <i>continuous</i>. Lightning is decidedly not continuous. 1 ms is "long" for the time of the main lightning stroke. There may be several strokes in one event, but the total time is still short, and the other non-main strokes will have significantly less current.</p>
<p>Do the math. You say the wire is 12 mm in diameter, so has a cross-sectional area of 113 mm² = 113x10<sup>-6</sup>m². The resistivity of copper at 20°C is 1.68x10<sup>-8</sup> Ωm. A 1 meter length of this wire therefore has a resistance of</p>
<p>(1.68x10<sup>-8</sup> Ωm)(1 m)/(113x10<sup>-6</sup> m²) = 149 µΩ</p>
<p>The power with 400 kA thru this resistance is then:</p>
<p>(400 kA)²(149 µΩ) = 23.8 MW </p>
<p>Times the 1 ms time the current is applied yields the energy:</p>
<p>(23.8 MW)(1 ms) = 23.8 kJ</p>
<p>The density of copper is 8.93 g/cm³, and our 1 m length has a volume of 113x10<sup>-6</sup> m³, which is 113 cm³.</p>
<p>(113 cm³)(8.93 g/cm³) = 1010 g total mass of copper</p>
<p>The specific heat of copper is 0.386 J/g°C.</p>
<p>(23.8 kJ) / (0.386 J/g°C)(1010 g) = 61°C</p>
<p>This means putting 400 kA through a 12 mm diameter copper wire for 1 ms will cause a temperature rise of 61°C. That's a rather extreme value for a lightning strike. The main stroke is usually substantially shorter than 1 ms, and the other strokes have substantially less current. However, even with these numbers it shows that while the wire will certainly get toasty for a while, it is well within the wire's ability to handle without any structural failure.</p>
| 1785 | Why doesn't a lightning strike destroy the lightning rod? |
2015-02-22T21:22:16.370 | <p>Measuring things like a martial artist's strike can be challenging because it is a complex motion and common vernacular will use imprecise or incorrect terms. This question is an attempt to identify the applicable measurements (quantification) and the correct units behind those numbers.</p>
<hr>
<p><em>Background:</em><br>
In a <a href="https://engineering.stackexchange.com/q/1742/16">recent question of mine</a>, I asked for sensors that could be used to measure a martial artist's strike, such as a kick or punch. My primary intent was to demonstrate to the student how their technique improved their strike. The answers there are great, and definitely addressed my question.</p>
<p>What led to this question is that I received a follow-up, clarifying comment asking if I wanted to measure force, energy, or power. To which I responded that I simply needed something measurable to show the difference or improvement to the student.</p>
<p>There was a subsequent comment<sup>*</sup> that got me thinking:</p>
<blockquote>
<p>From what you write here, I would like to add that it is neither force nor pressure nor energy that you actually want to measure.</p>
</blockquote>
<p><sup>*</sup><sub>The comment has since been deleted, but I started this question shortly after it was left.</sub></p>
<p>What got me thinking about that second comment is while I've seen many references using units of force, there are other ways of quantifying a martial arts strike.</p>
<p>For example:</p>
<ul>
<li><a href="http://www.washingtoncitypaper.com/articles/39469/straight-dope-the-physics-of-punching-someone-in-the-face" rel="nofollow noreferrer">This reference</a> prefers to use 'force' and condemns using 'pressure'</li>
<li><a href="http://www.livestrong.com/article/1003413-lightweight-vs-heavyweight-punch-power/" rel="nofollow noreferrer">This reference</a> also uses force, but mixes up their terminology when describing acceleration and velocity, so their usage of terminology is suspect.</li>
<li><a href="http://www.sciencebuddies.org/science-fair-projects/project_ideas/Sports_p020.shtml#background" rel="nofollow noreferrer">This reference</a> focuses on velocity, momentum, and energy.</li>
<li><a href="https://www.youtube.com/watch?feature=player_detailpage&v=sfnGkV6qmTw#t=34" rel="nofollow noreferrer">This reference</a> uses $lb_F$ for their measurements. Coincidentally, this is the series that started me down my original line of thought with the other question.</li>
</ul>
<hr>
<p><em>My question:</em><br>
What is the correct terminology for measuring (quantifying) the magnitude of a martial artist's strike versus another strike?</p>
<p>The measurement should be applicable across a variety of strikes (punches, kicks, elbows, etc...) and valid for different practitioners. Likewise, I'm not worried about defining good versus bad; I want to be able to quantify the strike. This ties into basic process improvement - we measure the process we want to improve and compare the measured results. </p>
| |measurements|biomedical-engineering|terminology| | <p>A martial artist's live strikes can be quantified between practitioners using <strong>force applied as pounds per square inch</strong>. The basic premise is (of course) force equals mass times acceleration. The importance is pairing with the <strong>area of contact</strong> used with which it is delivered - <strong>inverse square law</strong>. These two are the main pair.</p>
<p>There are of course wild complications in live striking - like transverse compression waves and [insert just about anything here]. In basis, transferring force can be qualitatively assessed through trained observation and the particular study of <strong>change of momentum, relative position and angle of attack relative to centre-mass</strong>. (Radial and longitudinal effects should be gestured rather than demonstrated).</p>
<p>I believe that there are some factors (not including deformation and shear stress) which cannot be accurately repeated in simulation and do require quantification/demystification - but is much harder to do so.</p>
<p>Ultimately, practitoners seek the obvious quantification of a strike: <strong>timing</strong> and <strong>accuracy</strong>. It should take into account the effectiveness of a technique arising as a result of competing motions. A trained student's <strong>degree of automaticity</strong> should also be assessed. </p>
| 1788 | What was that?! Quantifying the impact of a martial arts strike |
2015-02-23T11:57:48.563 | <p>I am reading a novel set in the future. A planet has been found where there are extensive deposits of diamond. Presumably diamond is mined much like marble is here on earth (there is not a lot of detail on this).</p>
<p>A character has a knife where the blade/shaft is a single piece of diamond. In 2015 here on Earth you can purchase <a href="http://www.fieldandstream.com/blogs/gun-nuts/2013/03/diamond-blade-meridian">diamond blade knives</a>, but they seem to be all steel (no diamonds per comment by @mart). There are some <a href="https://en.wikipedia.org/wiki/Diamond_blade">saw blades that have pieces of diamond attached</a> to a steel blade.</p>
<p>Assuming large quantities of reasonably priced materials, would it be possible to design and manufacture a functional knife made from a single section of diamond (all but the handle)?</p>
| |materials| | <p>These people just arent getting it. A diamondoid knife made from nano structured diamond would have the hardness of diamond but the elastic toughness and plastic deformation of steel. You would use molecular assemblers to make this. These are manmade enzyme like machines that bond carbon atoms into diamond composite. Think fiberglass. Fiber Diamond. These engineers are not thinking along these lines. Aggregated diamond nano rods also called ultra hard fullerite exists and is stronger and harder than diamond. We're talking flexible diamond. They gave made diamond needles that are elastic like rubber but diamond hard.</p>
| 1794 | Could a knife blade & shaft be made of 100% diamond? |
2015-02-24T19:53:24.217 | <p>I used LabView for a lot of my BME undergrad, but the labs focused mainly on things dealing electrical signals.</p>
<p>In the real world I'm working mostly on ME projects, but we lack a good deal of experimentation equipment. </p>
<p>What are the limitations of LabVIEW as it relates to ME experiments?</p>
| |mechanical-engineering|measurements|product-testing|labview| | <p>LABVIEW can be easily be used for ME related experimentation.</p>
<p>One such example would be to use an actuating mechanism to exercise a user interface on a mechanical DUT, which dispense a specific material quantity into a holding container. The weight of the material is measured using a weighing scale which communicates to a computer or controller. Also the test system also includes a digital manometer.</p>
<p><img src="https://i.stack.imgur.com/PDmPN.jpg" alt="Actuator">
<img src="https://i.stack.imgur.com/P6HLB.jpg" alt="RS232 Communication"></p>
<p>LABVIEW commands the actuator via RS232, and the weight of the material is captured using a scale. The weight data is communicated back to LABVIEW via RS232 serial communication. </p>
<p><img src="https://i.stack.imgur.com/XA2iF.jpg" alt="Scale">
<img src="https://i.stack.imgur.com/T9bSv.jpg" alt="Manometer"></p>
<p>This is an example of using basic tools to develop fairly complex ME experiment,both electrical and mechanical tools.</p>
<p><img src="https://i.stack.imgur.com/hk549.jpg" alt="Diagram"></p>
<p><strong>Limitation of LABVIEW</strong></p>
<ul>
<li>Labview is a very capable tool for bench top experimentation either in ME or EE environment. To the most part capability of the tools is limited by users understanding and experience of the LABVIEW software and other tools.</li>
<li>Labview’s graphical programming environment doesn't blend with the traditional structured or OOP programming environment. Therefore maintenance and enhancement is a limitation in a traditional sense. National Instrument will argue against my opinion.</li>
<li>Labview offers basic program structures such as for loops, if then else, and while loops. This is sufficient for typical basic bench top testing software. But Labview has limitation in implementing advance structure like a binary search tree or recursion. This can be done, but not too elegant.</li>
</ul>
<p>With the growth in LABVIEW, it is almost necessary to have fairly modern computer to use LABVIEW software. But with currently a good enough computer can be purchased for a bargain. </p>
<p>In summary, limitation is mostly the skill of the LABVIEW user. </p>
<p><strong>References:</strong></p>
<ul>
<li><a href="http://www.miraiintertech.com/home/downloads/SCN-SCL-Manual2011.pdf" rel="nofollow noreferrer">Mechatronics Cylinder</a></li>
<li><a href="http://cn.mt.com/cn/zh/home/supportive_content/product_documentation/installation_instructions/PS_Comm_Tst/jcr:content/download/file/file.res/Ser_Communication_Test_PS_scale_7-23-09.pdf" rel="nofollow noreferrer">Serial Communication Test (RS232) PS Scale</a></li>
<li><a href="http://tmi.yokogawa.com/files/uploaded/BU7673_00E_015_1.pdf" rel="nofollow noreferrer">Digital Manometers</a></li>
</ul>
| 1824 | LabView or another software for experiments |
2015-02-24T21:45:35.510 | <p>I have a metric thread pitch gauge that came in a tap and die set, it has pitches like $0.75$, $0.8$, $1.25$, etc. which is the distance between each thread. However there are two gauges that say $27$ and $28$ - I thought maybe it means $0.28\text{ mm}$ or possibly $0.28\text{ inches}$ if they threw in some non-metric ones, but it is neither of these (it's about $0.9\text{ mm}$) what are these?</p>
| |threads| | <p>As mentioned these are 'teeth per inch' threads. they find their way onto metric thread gauges because BSP threads are still the standard in many countries for tings like gas fittings. </p>
<p>This is partly because of legacy issues where it would simply be too much hassle to change over millions of fittings on things like industrial gas bottles and the associated regulators and fittings as opposed to nuts and bolts which have a much higher turnover and in many cases will allow a straight swap for the nearest equivalent metric size. </p>
<p>Another advantage is that discourages bodges using standard nuts and bolts for gas fittings which have specific safety requirements. </p>
<p>Also BSP type fittings are generally so specialist that there is little advantage in standardising to metric as they will only ever be used to mate to each other. </p>
| 1829 | What does "28" and "27" mean on my metric thread pitch gauge? |
2015-02-25T02:42:17.167 | <p>I was thinking about fighter aircraft like the <a href="http://en.wikipedia.org/wiki/Hawker_Siddeley_Harrier">Harrier</a> which have <a href="http://en.wikipedia.org/wiki/Thrust_vectoring">thrust vectoring</a> and computer-controlled stabilisation nozzles. Also an <a href="http://en.wikipedia.org/wiki/Dihedral_%28aeronautics%29#Anhedral">anhedral angle</a> on the (small) wings. The aircraft relies so much on the computer stabilisation, and flies like a brick if the engine fails, so why have those wings at all? Couldn't you get as much lift from wings of the same area that run parallel to the fuselage, that is much shorter (left-to-right) but deeper (front-to-back) wings? And wouldn't this create much less drag reducing the thrust requirements?</p>
<p>Having thought about this I realised that for fighters it's probably not a good idea as they would lose a lot of pitch-maneuverability, but for passenger jets, would a roll-stabilised approach with a short wing that runs the full length get the same lift while reducing drag? Is such a wing viable?</p>
| |aircraft-design| | <p>Helicopters are proof you don't need wings at all.</p>
<p>However, short wings (left right), no matter how long (front back), are less efficient than wide wings. You can see this from basic physics without having to understand anything about how wings actually work.</p>
<p>Consider a plane in straight, level, and steady flight. The net force on the air is to push down on it by the weight of the plane. That force is produced by imparting momentum downwards on the air immediately surrounding the plane as the plane flys by. Momentum is mass x velocity. In this sense stubby wings and wide wings are equivalent. You can get the same momentum by pushing a little air a lot (stubby wing), or a lot of air a little (wide wing).</p>
<p>However, consider the power requirement. Power is proportional to the <i>square</i> of velocity times the mass. Therefore, pushing a little air at high speed takes more power than pushing a lot of air at low speed. Stubby wings transfer more power into the air for the same lift. This extra power shows up as higher drag, which ultimately requires more pushing from the engines to overcome.</p>
<p>Wide and thin wings are best for efficiency, but there are structural limits and other trade offs. Note that wings of gliders (where efficiency is very important since the power comes from altitude loss) are very wide, but thin in the other two dimensions. They also can't carry much payload, in part because the wings are too fragile to support it.</p>
<p>Everything is a trade off. Jet fighters have other important criteria, like maneuverability, high top speed, good cockpit visibility, small radar cross section, etc. It is often useful to give up some efficiency in return for these other features. It all depends on what the plane is supposed to do.</p>
<p>Take a look at the F104 as a example of stubby and thin wings. It was fast, but also very tricky to fly, with several pilots lost due to inability to control the plane.</p>
| 1840 | Is a wing that runs the length of an aircraft viable? |
2015-02-25T09:47:35.883 | <p>In submerged conditions galvanic corrosion effects exist between metals with dissimilar nobleness and the presence of seawater as electrolyte.</p>
<p>Does reducing the contact area by using an nonconducting paint <em>on the interface layer between the two metal parts</em> help delay this process? Or is any the solid contact electrical route, e.g. via a bolt or so, enough to make the deterioration speed limited by the electrolytic process c.q. the surface area in contact with the electrolyte?</p>
| |materials|steel| | <p>No solid contact is required at all, as the seawater provides the electrical route. In order to prevent galvanic corrosion, you would need to electrically isolate your two metals. That said, I can see that a solid electrical route would allow faster transfer of electrons, speeding up corrosion, and hence having no solid electrical route would delay corrosion.</p>
| 1843 | Subsea galvanic corrosion protection by paint |
2015-02-25T10:31:13.053 | <p><strong>The setup</strong></p>
<p>Consider a cylinder inside another cylinder (placed with their symmetry axis horizontal).</p>
<p>The inner cylinder is about Ø100 mm in diameter and a centimeter or two in lenght/thickness. On the perimeter there is a thread. This cylinder is <strong>fixed</strong> in rotational motion, and cannot rotate.</p>
<p>The outer cylinder is much longer. It has a thread on the inside through half of its' lenght or so and this thread fits with the inner cylinder thread.</p>
<p><strong>The question</strong></p>
<p>Now, when rotating the outer cylinder, the inner cylinder will move back and forth (the inner cannot rotate but can move translationally).</p>
<p>My issue is that the outer cylinder might rotate many revolutions (it follows a bicycle wheel), and therefor at some point the inner cylinder will move <em>out of</em> the thread of the outer cylinder.</p>
<p><strong>Is there a mechanical and smart method to let such a cylinder leave and <em>reenter</em> the thread of the outer cylinder?</strong> E.g. the outer cylinder would rotate in one direction untill the inner cylinder leaves the thread. Then when the outer cylinder is reversed in rotational direction, I want the inner cylinder to reenter the thread and start moving the opposite way.</p>
| |mechanical-engineering|threads| | <p>The simplest solution would be to use a spring such that it exerts a small force to push the moving cylinder back in. As it 'unscrews' the spring will be tensioned (or compressed) and as the wheel starts spinning the other way, the small force will help the thread to re-enter.</p>
| 1847 | How to let a "screw" leave and reenter a thread |
2015-02-25T15:21:54.267 | <p>As I was browsing <a href="http://www.greenbookee.com/iso-965-1/" rel="nofollow noreferrer">ISO 965-1</a> it defines tolerances on a plenty of variables regarding metric threads, but it gives the pitch of the thread as a simple number from an allowed set ( 3 – 2 – 1,5 – 1– 0,75 – 0,5 – 0,35 – 0,25 – 0,2 ), and no tolerances on these are ever given.</p>
<p><img src="https://i.stack.imgur.com/59SoE.png" alt="enter image description here"></p>
<p>Why is it so? Are they defined in some other standard or is there a specific reason why they aren't defined to a given tolerance? Specifically, when I'm using a threaded rod as a lead-screw actuator, I'd like to know what inaccuracy can I expect coming from imprecision of pitch - how accurately do 1000 turns of a M10x1 threaded rod convert to 1000mm travel distance of the nut (minus backslash etc)? </p>
| |mechanical-engineering|threads| | <p>I think the pitch is implied by the major and pitch diameter.</p>
<p>The tolerance of these $D_{maj}$ and $D_{pitch}$ are given in <a href="http://en.wikipedia.org/wiki/ISO_965" rel="nofollow noreferrer"><code>ISO 965-2</code></a></p>
<p>I'll attempt juggling with the math in the <a href="http://en.wikipedia.org/wiki/ISO_metric_screw_thread" rel="nofollow noreferrer">ISO metric screw wiki</a>, (<em>please correct if wrong</em>):
<img src="https://i.stack.imgur.com/5esRY.png" alt="http://en.wikipedia.org/wiki/ISO_metric_screw_thread"></p>
<p>$(3/8)H=(D_{maj}-D_{pitch})\cdot(1/2)$</p>
<p>$H=P\cos(30^\circ)=P\sqrt{3}/2$</p>
<p>$P = \underbrace{(2/\sqrt{3})\cdot(8/3)\cdot(1/2)}_{1.5396}\cdot (D_{maj}-D_{pitch})$</p>
<p>For an M10x1.25 fine thread (could not find the M10x1.00) this would result to:</p>
<p><code>M10x1.25</code>: $D_{maj}=[9.972...9.76]$ $D_{pitch}=[9.16...9.042]$ --> $P=[0.92..1.43]$</p>
<p>So for a $1000mm$ rod, this would be a nominal $800$ turns with a minimum of $698$ and a max of $1083$.</p>
<p>Quite a range I think! I also found this picture of a mismatch failure <a href="http://www.boltscience.com/pages/screw8.htm" rel="nofollow noreferrer">(src)</a>:
<img src="https://i.stack.imgur.com/TvLEj.png" alt="broken"></p>
| 1853 | Is the tolerance of pitch of a thread standarized? |
2015-02-26T15:06:37.470 | <p>In a program on <a href="http://www.npr.org/2015/02/25/389008046/a-hard-look-at-the-risks-of-transporting-oil-on-rail-tanker-cars" rel="noreferrer">NPR</a> that I was listening to, there was a bit about a bridge that from the description sounded to a layman as unsound and is still in use. The program described it as an old wooden railway (and I'm aware that has its own set of challenges) bridge with rotting timbers.</p>
<p>In the United Sates, if a member of the public sees a bridge (railway, tramway, car/truck, foot, bike, etc...) that is of questionable soundness, what is the process for him or her to determine who us responsible for it, and if they (the responsible parties) should look at it and have it evaluated? </p>
<p>Is there a particular agency that is responsible for regulating bridges and ascertaining their safety?</p>
| |civil-engineering|bridges|regulations| | <p>In the United States, you can often use <a href="http://nationalbridges.com/" rel="nofollow">nationalbridges.com</a> and <a href="http://uglybridges.com/" rel="nofollow">uglybridges.com</a> to ascertain the current status of a bridge, along with some basic information about the owning agency/maintenance agency. Ordinary road/highway bridges are usually inspected on a biennial basis, and given a sufficiency rating out of 100. If a bridge is identified as having structural problems it will be listed as "structurally deficient". Those with substandard geometry, safety railing, or insufficient vehicle capacity will be listed as "functionally obsolete".</p>
| 1864 | How should the public raise questions about unsafe structures in the United States? |
2015-02-27T20:14:32.750 | <p>I am now in charge of a product line my company has been shipping for a decade. One of the previous product engineers was... shall we say, less than conscientious about sustainability and proper documentation. We have shipped hundreds of units, of multiple design variants, under the <em>exact same part number</em>. The manual presently reflects only one variant, meaning many users can't use it. And we've had multiple instances where a user has tried to reorder a unit by part number, only to find that what we ship them does not match what they already had.</p>
<p>Obviously, this is terrible. One does not change the specs or user interface of a product without also changing the part number. We will avoid such things in the future. But my question is about the past.</p>
<p>We have, on paper, documentation indicating what design variant each serial number corresponds to. My thinking is to create a spreadsheet, and name each variant retroactively, so we can at least support users that call in or place reorders. We would then create proper manual(s) so that the user can, based on their serial number, understand the operation of the units they have.</p>
<p>But I'm just making up that solution. It occurs to me that there may be formal, industry-standard methods of dealing with such things. Is there a procedurally-correct way to handle my existing install base?</p>
| |project-management|documentation|product-support| | <p>Tracking the various product versions by serial number is a good idea, but be careful to standardize that as well. Years ago while working for an electronics manufacturer in service and repair, I discovered two units; one was serial number 0024, the other was 00024. They had changed from a four-digit number to a five-digit one and in the process created a duplicate (except for the number of leading zeros). To avoid confusion in the future we relabeled the four-digit one to be 0024A instead. Not the best answer perhaps, but it worked.
You may need to do something similar. The suggestion to document the various versions in one document with subsections and notes is a good one, if done properly.</p>
| 1880 | I have inherited a product with inconsistent designs sharing the same part number. What do I do? |
2015-02-27T23:57:58.973 | <p>The <a href="https://en.wikipedia.org/wiki/Concorde" rel="nofollow">Concorde</a> had a famously high <a href="http://www.boeing.com/commercial/aeromagazine/aero_12/whatisaoa.pdf" rel="nofollow">angle of attack</a> (and <a href="https://www.aviation-history.com/theory/angle_of_attack.htm" rel="nofollow">pitch angle</a>) when landing. This led to its famous <a href="https://en.wikipedia.org/wiki/Droop-nose" rel="nofollow">droop nose</a>:</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/1/11/Concorde_landing_Farnborough_Fitzgerald.jpg" rel="nofollow"><img src="https://upload.wikimedia.org/wikipedia/commons/1/11/Concorde_landing_Farnborough_Fitzgerald.jpg" alt="Concorde"></a></p>
<p>The Concorde's Russia counterpart, the <a href="https://en.wikipedia.org/wiki/Tupolev_Tu-144" rel="nofollow">Tupolev Tu-144</a>, had the same issue.</p>
<p>These are the only two supersonic airliners to have been built, so there is, admittedly, a small sample size. The Tu-144 has also been derisively called "the Concordski", as some claim that its designers stole plans from the Concorde. So there are clear similarities.</p>
<p>However, there are some supersonic airliners on the drawing board. Will they, too, have high angles of attack when landing? As a follow-up question: Will they, too, be fitted with droop noses?</p>
| |aerospace-engineering| | <p>A long trapezoidal wing grows significantly in lift coefficient and area through flaps/slats and Fowler flaps, respectively. Delta wings, especially those using elevons, essentially have no capability to employ high-lift devices. A confluent design driver for delta wings is that they stall only at very high angles of attack; a vortex sheet rolls up in front of both edges and makes the wing have a high effective camber after the rest of the airflow goes over the sheet.</p>
<p>The drawback of using vortex lift is the longer landing gear and very bad lift-to-drag ratio, approximately 3. This additional drag is what made losing two engines on the Concorde unrecoverable at the takeoff speed.</p>
<p>As far as nose-droop on future designs, I would expect to see the use of cameras and monitors instead. Triple redundant, battery-backed electronics are likely lighter weight than a mechanical hinge.</p>
<p>Fun fact: it's impossible to generate lift at supersonic speeds without emitting external shockwaves, some towards the ground. Because we know that any aircraft will generate lift and that the creation of shockwaves is a source of aerodynamic (non-isentropic compression/expansion) drag, a low lift co-efficient given by a large wing implies some aerodynamic savings at cruise.</p>
| 1882 | Do all supersonic airliners have a high angle of attack when landing? |
2015-02-28T00:30:05.980 | <p>I am trying to use a Piezo Electric Ceramic Disc Transducer for ultrasonic sensing measurements for fluid levels. A Piezo Ceramic Disc with a Resonant frequency of 215 KHz or 1 MHz is under consideration. These devices specify a Radial and Thickness mode vibration configuration.</p>
<p><img src="https://i.stack.imgur.com/3f84k.jpg" alt="Piezo Electric Ceramic Disc Transducer" /></p>
<p><strong>Primary Question</strong>: What is Radial and Thickness mode vibration configuration?</p>
<p><strong>Secondary Question</strong>: Is Axial and Thickness mode vibration mean the same mode?</p>
<p><img src="https://i.stack.imgur.com/DlCir.jpg" alt="Transducer1" />
<img src="https://i.stack.imgur.com/IRdp3.jpg" alt="Transducer2" /></p>
<hr />
<h2>Part Specifications</h2>
<p><em>Piezo Ceramic Disc 1 MHz,Thickness mode vibration</em></p>
<ul>
<li>Dimensions: 15mm diameter x 2.1mm thickness</li>
<li>Resonant frequency fr: 1 MHz±3 %</li>
<li>Electromechanical coupling coefficient Kt:≥39%</li>
<li>Resonant impedance Zm: ≤7.6 Ω</li>
<li>Static capacitance Cs: 900pF±15%@1kHz</li>
<li>Test Condition: 23±3 °C 40~70% R.H.</li>
<li>fr, Zm, Kp => Thickness mode vibration</li>
<li>Cs => LCR meter at 1KHz 1Vrms</li>
</ul>
<p><em>Piezo Ceramic Disc R 215 KHz, Radial mode vibration</em></p>
<ul>
<li>Dimensions: 10mm diameter x 2mm thickness</li>
<li>Resonant frequency fr: 215 KHz±5 KHz</li>
<li>Electromechanical coupling coefficient Kp:≥60%</li>
<li>Resonant impedance Zm: ≤6 Ω</li>
<li>Static capacitance Cs: 450pF±15%@1kHz</li>
<li>Test Condition: 23±3 °C 40~70% R.H.</li>
<li>fr, Zm, Kp => Radial mode vibration</li>
<li>Cs => LCR meter at 1KHz 1Vrms</li>
</ul>
<p><a href="https://i.stack.imgur.com/cyzLw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cyzLw.png" alt="Piezoelectric Modes of Vibration" /></a></p>
<hr />
<p><strong>References</strong>:</p>
<ul>
<li><a href="https://www.americanpiezo.com/knowledge-center/piezo-theory/vibration.html" rel="nofollow noreferrer">Modes of vibration for PIEZOELECTRIC elements</a></li>
</ul>
| |mechanical-engineering|electrical-engineering|sensors| | <p>Texas Instruments has extensive material on this Piezo Electric disc for applications of measurements fluid level. You can search their website for <a href="https://www.steminc.com/PZT/en/piezo-disc" rel="nofollow noreferrer">STEMINC</a> and you will see two discs used on Thickness mode. Both discs have frequency 1MHz and are used with a couple of their controllers. The discs are SMD10T2R111WL and <a href="https://www.steminc.com/PZT/en/piezo-ceramic-disc-1-mhz" rel="nofollow noreferrer">SMD15T21R111WL</a>. STEMINC also has these same dimensions discs with Radial frequency but the Texas Instruments Controller TDC100 for liquid level measurement will work with the 1MHz version.</p>
| 1883 | What is Radial and Thickness Mode Vibration in a Piezo Electric Ceramic Disc Transducer? |
2015-02-28T04:25:51.817 | <p>I am involved in an upcoming product launch and we recently had a discussion about serial numbers. We will of course use variant-specific model identifiers and revision numbers, but couldn't see any good reason to include any leading zeros in serial numbers, or otherwise make them fixed-length with padding numbers, prefixes, characters, etc. But this seems to be a common practice.</p>
<p>Is there anything wrong with 1, 2, ... 10, 11, 12, ... 100, 101, 102, ... ? What possible advantages are lost by just starting at 1 and counting up?</p>
| |documentation| | <p>Fixed-width serial numbers are just easier to handle. It also allows you to know where the serial number is within a overall product ID string without special rules to parse the string.</p>
<p>For example, if the serial number is expressed in hexadecimal in the string, it can contain letters A-F in addition to the digits 0-9. If the part ID string might also contain a suffix or prefix with these letters, then it becomes ambiguous with a variable-length serial number and without some delimiter. Consider "07933A". Is the whole thing the HEX serial number, or is 07933 the serial number and "A" a suffix denoting a configuration option, for example.</p>
<p>Fixed-width serial numbers strings are also easier to see in a list, and sort on.</p>
<h2>Serial number generation</h2>
<p>You didn't ask about this, but you have to think carefully about what/how new serial numbers are generated. This is probably more important than how exactly the serial number will be listed on the nameplate. I have seen a number of screwups, usually as a result of giving manufacturing too much leeway, or too much information so that they <i>think</i> they know how to create new serial numbers.</p>
<p>What I usually advise now is a dedicated piece of hardware that is pre-programmed with a range of serial numbers, then gives out those serial numbers sequentially. When the range is used up, it refuses to give out another serial number. The hardware must be returned to a central serial number management place to get updated with a new range, or swapped out entirely. You don't give any documentation about the internals of this device to manufacturing, whether internal or contract, remote or local. This device is a necessary part of the overall production process, usually part of the final test jig.</p>
<p>One advantage of this method is that it's easy to assign serial number ranges to different production lines or even different manufacturers. Engineering will need to create internal prototypes and sometimes samples, so they have one or more of these units too. Any engineer can grab the unit floating around engineering to assign a guaranteed unique serial number to a new prototype.</p>
<p>This dedicated hardware need not be big or expensive. It can be as simple as a USB dongle or small device connected to the USB. I've done this with a USB microcontroller and external EEPROM. Again, production gets no documentation on this device other than how to use it. They can be given multiple devices as spares, which is OK since each has its own unique serial number range. We usually programmed these devices with a range of 65536 (2<sup>16</sup>) serial numbers. To update the serial number range required soldering a temporary jumper between two pads for that purpose, and running a special program on the host that only one person at the home office had.</p>
<h3>Screwup #1</h3>
<p>Apollo Computer made workstations in the 1980s. Each had to have a unique serial number. Among other things, these serial numbers were used to allow each node to make its own globally unique ID numbers used in the file system. The serial numbers were a hexadecimal string. A bunch of years after production started, someone noticed that the serial number of their node was well beyond the number of nodes supposedly produced. It turns out production was assigning serial numbers in decimal, even though each digit was really hexadecimal. This was fixed by having them make new units by filling in the unused serial numbers with digits A-F in them.</p>
<p>In this case no real harm was done, except that it made tracking units by serial number ranges awkward. Below a certain number, ranges were divided into decimal-only and hexadecimal.</p>
<h3>Screwup #2</h3>
<p>Serial numbers were assigned by the production test system, which included a PC. There were multiple production lines, and each PC was carefully assigned a particular range. The contract manufacturer in China was given explicit instructions to notify us in case of any trouble with any of the production test systems.</p>
<p>One day someone noticed that some of the 10s of thousands of little gizmos received from the factory in China had duplicate serial numbers. This was bad, because the larger system these gizmos were part of relied on the serial number to uniquely identify each gizmo, which was in large part the point of the system.</p>
<p>It turns out that the factory in China had done something stupid with one of the production test PCs and messed up the data on it. Instead of telling us like they were supposed to, they knew they could fix the problem on their own and we'd never know. Their fix was to do a complete disk copy from the working system to the broken system. From then on the two systems assigned duplicate serial numbers.</p>
| 1888 | Is there a good reason for long and fixed-length serial numbers? |
2015-02-28T17:51:46.813 | <p>I am specifying a product made in Australia, which is made of "Grade 520 Steel." The manufacturer of the product lists a minimum yield stress of 520 N/mm² (75,400 psi) and a minimum ultimate stress of 650 N/mm² (94,300 psi) with a minimum elongation of 20% and Young's modulus of 205 kN/mm² (29,700 ksi.) They describe the steel as "A fine grain micro alloyed carbon steel which is fully weldable." The raw steel form should be solid round bar.</p>
<p>My problem is that the plan checker wants a reference to a normative documents (ISO, EN, ASTM spec, etc.) for the material grade. I cannot find who (if anyone) defines this grade of steel. What is the standard organization and standard number that would govern this material?</p>
<p>I have, of course contacted the manufacturer, but what's above is all that they've been able to tell me so far. The only material specification '520' I've been able to find is a withdrawn ASTM tube spec. This component is significantly over-designed, so I'm not concerned with the actual properties, just finding the appropriate document. There are other products available made with ASTM materials, but for visual reasons we'd prefer the Australian product.</p>
| |materials|steel|standards| | <p>I saw some material for tension rods that match the material properties that you gave, but these places do not list the steel used. It might be proprietary:</p>
<ul>
<li><a href="http://www.ronstanrigging.com/arch_w/systemrange.asp?RnID=704" rel="nofollow">Ronstan</a></li>
<li><a href="http://www.daversteels.co.uk/bar_systems_load_capacity" rel="nofollow">DaverSteels</a> </li>
</ul>
<p>A third manufacturer <a href="http://www.macalloy.com/tension-rods" rel="nofollow">Macalloy</a>, has a similar component that also doesn't list a material specification, but it does have an <a href="http://www.macalloy.com/system/files/tension-rod-systems.pdf" rel="nofollow">independent certification</a> of its material properties. This leads me to believe that this is a custom steel that is only guaranteed by its given properties and the agreements between the company and the foundry.</p>
| 1895 | Normative Document for 'Grade 520' Steel |
2015-02-28T21:12:00.727 | <p><strong>Downdrag</strong><br>
<a href="https://www.dot.state.oh.us/Divisions/Engineering/Geotechnical/Geotechnical_Documents/2007%20Workshop%20Downdrag%20in%20Foundation%20Design.pdf" rel="nofollow">Downdrag</a> in foundations occurs when soil layers consolidate and settle under additional load. This typically occurs when additional fill is placed on top of the existing soil. The actual downdrag force comes from the friction of the soil against piles as the soil moves down. It is assumed that soil settlement goes along with downdrag.</p>
<p><strong>Retaining Walls</strong><br>
The controlling load for cantilever retaining walls is usually the <a href="https://en.wikipedia.org/wiki/Lateral_earth_pressure" rel="nofollow">lateral earth pressure</a>. Any vertical load on the <a href="https://en.wikipedia.org/wiki/Retaining_wall#Sheet_piling" rel="nofollow">sheet pile</a> or <a href="http://www.haywardbaker.com/WhatWeDo/Techniques/EarthRetention/SoldierPilesAndLagging/default.aspx" rel="nofollow">soldier pile</a> is usually so small that it is ignored.</p>
<p>This retaining wall is designed to retain additional fill over consolidating soil. The piles are embedded below the consolidating soil and will not settle. Would the downdrag or settling soil affect the calculation of the horizontal soil forces?</p>
| |civil-engineering|geotechnical-engineering|retaining-wall| | <p>It depends on the construction sequence and support conditions; however in my experience most walls like these are built by installing the piles and then excavating the open side.</p>
<p>As you've noted, consolidation settlement is a concern for piles because the 'downdrag' force effectively reduces their capacity as a percentage of the vertical resistance is now being used to counter the downdrag.</p>
<p>Walls, however, provide capacity against lateral movement, and therefore loss of vertical capacity is insignificant. The consolidation, however, could effect the lateral pressures (depending on the construction method / support conditions; see note at end).</p>
<p>The lateral force applied to the retaining structure is calculated as:</p>
<p>$$\sigma_h = K \sigma_v$$</p>
<p>Where:</p>
<ul>
<li>$\sigma_h$ is is the resulting lateral earth pressure</li>
<li>$K$ is the lateral earth pressure coefficient</li>
<li>$\sigma_v$ is the applied vertical pressure at the calculation location</li>
</ul>
<p>If the pile is installed in situ prior to excavation then the lateral earth pressure can be based on the 'at rest' condition (see note at end), as the soil is never allowed to displace enough to generate the full 'active' stress.</p>
<p>The at rest lateral earth pressure coefficient for solis is:</p>
<p>$$K_0 = (1 - \sin \phi ' ) \times OCR ^ {( \sin \phi ' )} $$</p>
<p>Where:</p>
<ul>
<li>$K_0$ is the 'at rest' lateral earth pressure coefficient</li>
<li>$\phi '$ is the effective stress friction angle of the soil being retained</li>
<li>$OCR$ is the overconsolidation ratio ( > 1 for overly consolidated soils)</li>
</ul>
<p>As the soil consolidates, therefore, the at rest lateral earth pressure increases; and therefore the lateral loading that the wall must structurally resist and retain increases.</p>
<p><strong>Note</strong></p>
<p>The 'at rest' condition can only be maintained so long as the retaining wall does not move. As the wall displaces, the lateral earth coefficient of the retained soil tends towards the smaller, 'active' coefficient (that is not proprotional to consolidation).</p>
<p>It is therefore a conservative assumption to say that the wall must sustain 'at rest' pressures.</p>
<p><a href="https://books.google.co.uk/books?id=lYoS4VXJPJ4C&pg=PA206&lpg=PA206&dq=embedded%20retaining%20wall%20at%20rest&source=bl&ots=ZeFYLruEeD&sig=zq6c6HdLNYuEjKrVuKj3I52ql5U&hl=en&sa=X&ei=o-DyVIuXIMfa7Abau4HgDg&ved=0CFUQ6AEwCQ#v=onepage&q=embedded%20retaining%20wall%20at%20rest&f=false" rel="nofollow">Tomlinson</a> suggests that $K_0$ should only be used when the movement of the structure is less than $5 \times 10^{-4} $ the retained height in normally consolidated soils; I would expect that limit to be lower for overly consolidated soils, as they require a larger strain to mobilse the 'active' condition.</p>
<p>It should be noted that, to the old British Standards at least, these kinds of walls are typically designed considering only the bending moment determined from the active / passive pressures, with an applied factor of safety (circa 1.4 - 1.5); potentially accounting for this.</p>
| 1900 | Does downdrag affect cantilever pile retaining wall design? |
2015-03-01T17:40:49.340 | <p>What happens to the working of petrol engine when it is emptied and filled with diesel or to a diesel engine emptied and filled with petrol? Will the engine be able to operate and, if not, why not?</p>
| |mechanical-engineering|automotive-engineering|combustion| | <p>IC engines work in cycles, in a four-stroke IC engine there are four cycles</p>
<ol>
<li>Intake stroke</li>
<li>Compression stroke</li>
<li>Combustion stroke</li>
<li>Exhaust stroke</li>
</ol>
<p>During intake stroke the air-fuel mixture is sent into the engine cylinder, later in the compression stroke, it is compressed and then in the ignition stroke fuel gets ignited to take out power. In the exhaust stroke, the residue is sent out of the chamber.</p>
<p>There are two types of IC Engines based on their method of fuel ignition</p>
<ol>
<li>SI (spark ignition: typical gasoline or petrol engines)</li>
<li>CI (Compression Ignition: typical diesel engines) </li>
</ol>
<p>The way they are named like that is because in SI engines the fuel is ignited using a spark plug and in CI engines fuel is compressed to such high pressure that it reaches a temperature near to its autoignition temperature.</p>
<p>*This auto-ignition temperature is an important thing to notice</p>
<p>The autoignition temperature or kindling point of a substance is the lowest temperature at which it spontaneously ignites in normal atmosphere without an external source of ignition, such as a flame or spark.</p>
<p><a href="https://i.stack.imgur.com/GovS3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GovS3.png" alt="table with different fuels' auto-ignition temperatures"></a></p>
<p>As shown in the figure, it is clear that diesel has a lower auto-ignition temperature than gasoline. For diesel, it is 210°C; for gasoline, it is 280°C.</p>
<p>In CI engines the air is taken during the intake stroke and it is compressed to a high pressure where the air will reach a temperature of more than 210°C, then diesel is injected using a fuel injector. As soon as the diesel comes in contact with the air it gets ignited.</p>
<p>Now when you put gasoline inside the diesel engine, the air during compression still will be at a temperature ranging 210°C - 220°C and the Gasoline entered has an auto-ignition temperature of 280°C. This makes fuel impossible to ignite, hence the engine will not start. If it starts it will stop within seconds due to the accumulation of fuel in the cylinder.</p>
<p>Whereas SI Engines use a different method to ignite the fuel, first the fuel is mixed with air in the carburetor. This mixing can take place only when the fuel is in vapour state. For that, the flash point of the fuel must be very low.
(flash point is the lowest temperature at which a liquid can form an ignitable mixture in air near the surface of the liquid. The lower the flash point, the easier it is to ignite the material.)</p>
<p>Now coming to flash point of gasoline it is -44°C at normal atmosphere temperatures it can easily form vapours. But diesel has a flash point of 55°C. When you put diesel into the petrol engine (SI) it will not form an ignitable mixture with air, even when a spark is generated inside the cylinder the diesel inside will be in liquid form which is not the ideal condition for flame generation. Thus the engine will not get started.</p>
<p>That is the reason why the vehicles are not compatible with other fuels than they are designed for.</p>
<p>*NOTE:- This is most raw explanation for this question, more in-depth explanations may be there please refer to them as well.</p>
<p>Fun To Know:- Due to higher compression ratios diesel engines are usually used for high torque applications like load carrying and power generation. Petrol or gasoline engines are used in commercial cars and bikes due to their light weight and faster response. </p>
| 1903 | What happens when you put the wrong type of fuel in an internal combustion engine? |
2015-03-01T18:26:26.137 | <p>I'm thinking of developing a pocket 'power bank' that could provide both 6V and 12V - a charger to charge either a phone (with 6V) or a notebook computer (with 12V). </p>
<p>The electrical diagram is simple enough; switch attachment of the terminals like below.</p>
<p><img src="https://i.stack.imgur.com/CVHVv.png" alt="schematic"></p>
<p>The schematics is not everything though. I'd like the charger to have two sockets, one providing 6V, the other - 12V, and to "sense" what to serve by having a plug in either of the sockets (let's say using both at a time is forbidden; I can place them in such a way that both plugs won't fit at a time, say, a slider that opens either of them but never both).</p>
<p>Best if the "sensing" was done in the simplest, mechanical way (e.g. there are sockets that provide an extra "sensing" contact that shorts to mass if the plug is in), but if that would prove too difficult or mechanically complex, a more complex electronic solution would work too, providing it won't drain the batteries when nothing is plugged in. (and of course the circuitry must accept external charging current - accepting only one of the two, either 6 or 12V is fine). </p>
<p>And of course while just cutting one of batteries off for 6V would be the easiest solution, it would halve the capacity for 6V application and cause parasitic charging issue when switching to 12V after the 6V power has been depleted, so it's not really an acceptable solution.</p>
<p>So, how to build such a circuitry - that provides power from two batteries, in series or in parallel depending on which socket was used?</p>
| |mechanical-engineering|electrical-engineering| | <p>I made a fast drawing with a bit more proper electronic:
<a href="https://i.stack.imgur.com/iOoRr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iOoRr.png" alt="voltage switcher"></a></p>
<p>You cannot put two electrical sources in parallel, this will break your system imediatelly or on the long term.</p>
<p>I am using Not gates because they are the most common, also the simplest to build.</p>
<p>Connect the green with the green ($12V$) and orange with orange ($6V$) at the moment the circuit is on $12V$, press the switch and it will work on $6V$
(you need to work a bit the switch so what is not connected is then connected to the ground)</p>
<p>I also indicated that you can charge your batteries <em>in series please</em></p>
| 1905 | Switching batteries from series to parallel - technical solution |
2015-03-01T19:20:06.230 | <p>As a follow up to <a href="https://engineering.stackexchange.com/questions/1864/how-should-the-public-raise-questions-about-unsafe-structures-in-the-united-stat">this question</a>, I wonder if structural integrity monitoring information can be provided real-time and to a sufficient degree that safety decisions can realistically be made from the information. </p>
<p>To put things in context, in the USA and other parts of the world federal, state, and local governments regularly receive structural integrity reports on infrastructure, including bridges. These reports are the results of assessments that are performed on a scheduled basis, and it can be many months or possibly years between assessments. </p>
<p>Bridges are an important part of federal, state, and local government economy since they provide critical infrastructure. It is also important to protect historical land marks such as the Golden Gate Bridge. </p>
<p><img src="https://i.stack.imgur.com/zrPSW.jpg" alt="Golden Gate Bridge Picture 1"><br>
<sub><em>Golden Gate Bridge, an iconic bridge in the United States</em></sub></p>
<p>And the consequences of insufficient information can be quite severe. For example, in August of 2007 the <a href="http://en.wikipedia.org/wiki/I-35W_Mississippi_River_bridge" rel="nofollow noreferrer">I-35W Mississippi River bridge</a> collapsed, which resulted in the loss of many lives.</p>
<p><img src="https://i.stack.imgur.com/6h2Ab.jpg" alt="bridge collapse 2007"><br>
<sub><em>I-35W after the collapse</em></sub></p>
<hr>
<p><strong>Question:</strong></p>
<ul>
<li>Are there technologies that could provide real-time status such that they would help forewarn federal, state, and local governments of impending bridge failures? For example, is there something that could have predicted the I-35 collapse and notified the public in order to avoid the loss of life? It seems like this <a href="http://www.gizmag.com/wireless-bridge-sensor/19380/" rel="nofollow noreferrer">wireless sensor to monitor structural integrity of bridges</a> would work, but I'm not certain as the domain is outside of my experience.</li>
</ul>
| |civil-engineering|structural-engineering|bridges|sensors| | <p>Real-time monitoring can be done and has already been done. The technology is there and as pointed out by hazzey, is a matter of cost versus the risk involved.</p>
<p>As a practising Engineer, I am/was involved in two deep excavation projects in two different countries where a cut-and-cover tunnel is/was constructed right under a major highway bridge. In both cases, traffic closure or diversion is not an option. The Building Authorities in both projects requested for real-time structural health monitoring as the consequences of a mishap can be catastrophic.</p>
<p>In both projects, an array of optical prisms together with robotic total stations are deployed to monitor bridge pier movements. The total stations are networked to a server where data is being uploaded continuously. The server itself is connected to the internet, where parties involved can download the data and process the information. In addition, some systems can generate SMS alerts in case where a predetermined level is breached.</p>
<p>Before any monitoring is done, an impact/damage assessment has to be carried out to determine:</p>
<ul>
<li>the predicted soil movement (horizontal and vertical) at the foundation of the bridge based on the construction method</li>
<li>the kind of forces that will be induced on the piles</li>
<li>the structural capacity of the existing piles as determined from as-built drawings</li>
<li>the existing load on the piles during service</li>
<li>the reserve capacity of the piles</li>
<li>the allowable lateral movement and settlement of the bridge foundation</li>
</ul>
<p>This is followed by a condition survey to assess the current state of health of the bridge. It usually includes visual inspection for cracks and measurements on bridge bearing movements.</p>
<p>Reinforced concrete and steel bridges, under normal circumstances, do not require real-time monitoring as they do not fail suddenly. Bridges that have been properly designed tend to fail in serviceability before they fail structurally, giving sufficient warning before they collapse. Also, regular inspection and maintenance of the bridges help to ensure that the bridges remain serviceable.</p>
<p>With enough data, real-time monitoring can capture long-term movement trend and smooth out reading errors. In cases of fatal design, human error in interpreting survey records, or human error in assessment of the bridge structural capacity, real-time monitoring would not be of much help to avert a disaster.</p>
| 1906 | Can the structural integrity of bridges be measured in real-time for safety? |
2015-03-01T21:58:32.903 | <p>On a recently launched Russian diesel-electric submarine, the rear propeller has two distinct features. You can see spheres at the base of every propeller blade:</p>
<p><img src="https://i.stack.imgur.com/vDXwO.jpg" alt=""></p>
<p>Also, the trailing edge of the shaft has four adjoining fins in line with the axis of rotation:</p>
<p><img src="https://i.stack.imgur.com/Nb9C4.png" alt=""></p>
<p>Most importantly, what is the purpose and function of the spheres, and what's beneath the veil in the first image? I believe that it may differ from the second image.</p>
<p>It's made of a different material, so possibly a sacrificial element. </p>
<p>Is it to reduce noise? Is it to smooth flow or prevent cavitation?</p>
<p>Notice that in the first image the spheres are at the leading edge and in the second image the spheres are at the trailing edge.</p>
<p>There is a related <a href="https://www.youtube.com/watch?v=hTrg2nhh0-E" rel="noreferrer">youtube video</a>.</p>
| |mechanical-engineering|fluid-mechanics|propulsion|marine-engineering|naval-engineering| | <p>Submarine propulsor designers are least concerned about hub vortex cavitation as they are primarily concerned about cavitation inception, or the onset of cavitation, which first occurs in the tip vortex regions, where the velocities are highest. Once cavitation occurs you are no longer operating in stealth mode. However, the hub vortex does waste a significant amount of energy.</p>
<p>Therefore, the fins looks to me to be a kind of hub stator used for diffusing the hub vortex and thereby increasing propulsor efficiency. Take a look at this <a href="http://articles.maritimepropulsion.com/article/IMPROVED-PROPULSIVE-EFFICIENCY-BY-WEAKING-THE-PROPELLER-HUB-VORTEX84713.aspx" rel="nofollow">article in maritime propulsion</a>. I would guess the covered picture could be some kind of <a href="http://www.mol.co.jp/en/pr/2011/1145.html" rel="nofollow">propeller boss cap fin system</a> which would even be more effective in improving propulsive efficiency.</p>
<p>The spheres you refer to are a bit of a mystery to me. They would create horseshoe vortices and convect momentum downward into the hub vortex region, which would slightly increase the pressure in the in hub vortex wake and help to alleviate flow separation when the hub sharply ends. However, it wouldn't seem to me this was their ultimate purpose, as generating secondary flows are not usually desirable unless there is another advantageous purpose. </p>
<p>Given that there has been quite a bit of research the past 10-15 years in drag reduction by polymer or microbubble injection, I would expect a state-of-the-art submarine to be employ such a system. See for example this <a href="http://www.sciencedirect.com/science/article/pii/S0894177708001386" rel="nofollow">paper by Nouri and Sarreshtehdari</a> specifically about microbubble injection to reduce drag in the hub vortex. They are optimally placed for feeding into the hub vortex.</p>
| 1908 | Purpose of spheres and fins on submarine propeller |
2015-03-01T23:00:33.383 | <p><a href="http://en.wikipedia.org/wiki/Turbo-compound_engine" rel="noreferrer">Turbo compound</a> engines extract some exhaust energy by using it to power a turbine connected to the driveshaft. Why are these engines not more widely used?</p>
| |mechanical-engineering| | <p>I’ve been doing a lot of reading on this and what we haven’t seen is a form of turbo compounding but instead using a small turbine engine capable of running itself when a steady torque is required. If married to a reciprocating engine with variable valve timing, the valves could be held open; the turbine generating the power. Under acceleration the turbine could operate as a turbocharger using bleed air intercooled as boost. </p>
| 1910 | Why are turbo compound engines not more widely used? |
2015-03-02T13:11:14.063 | <p>I have a rf module transmitting from an antenna. As you see in Figure 1, I have a radio mast that sends waves in all directions. I don't want this. My plan for controlling wave direction is Figure 2. </p>
<p>How do I control RF wave direction? </p>
<p><img src="https://i.stack.imgur.com/z5sj6.png" alt="enter image description here"></p>
| |electrical-engineering|telecommunication| | <p>Directional antennas generally use one of two priciples: Reflection or phasing.</p>
<p>Reflection works just like it does with optical mirrors, except that the wavelength is much much longer and therefore the construction of something that reflects may seem unintuitive. Most of the world's really high gain* antennas are parabolic dishes, which focus the power in a relatively narrow beam. For example, large radio telecscopes are like that. Common rooftop UHF TV antennas also usually worked on this principle.</p>
<p>Phasing means multiple elements with careful spacing and being fed with a specific phase shift of the carrier between them. If arranged properly, the multiple elements all add to a wave going one direction, but cancel each other out going other directions.</p>
<p>For example, consider two vertical antennas that each radiate equally in all horizontal directions. These antennas, I'll call them A and B, are spaced 1/4 wavelength apart. B is fed with a signal that lags A by 1/4 wavelength. The signal radiated by A in the direction of B will be in phase with B by the time it gets there. The signals from the two antennas will then add going in the direction from A to B. However, the signal emitted by B in the direction of A will find A 180° out of phase, so the two signals cancel. This is a simple system that has a gain of 3 dB in the A to B direction compared to one of the antennas by itself.</p>
<p>Phased arrays can contain more than just two phased emitters. More emitters allows tighter control over the beam shape. The Aegis radar is a good example of this technique taken much farther. The large number of individual emitters all acting together form a narrow beam, which is narrow enough to focus on a single aircraft. One advantage of this system is that the direction can be steered by electronics very quickly, much faster than a reflecting antenna can be mechanically re-aimed.</p>
<p> </p>
<ul>
<li>"Gain" when applied to antennas doesn't mean the antenna can radiate more RF power than it receives electrically. Ideally, antennas radiate all the electrical power they receive. The difference is how concentrated it is in some directions. Gain is how much a antenna can concentrate the power in its optimal direction relative to a standard emitter. There are two references commonly used for 0 dB gain, a isotropic radiator and a dipole. A dipole has a donut-shaped radiation pattern, so has gain along its plane relative to a isotropic radiator. Both references are commonly used, so if it is not stated the statement is either ambiguous or it is implied by other context.</li>
</ul>
| 1920 | Control RF wave direction or path |
2015-03-02T17:45:28.383 | <p>I thought my electric car charging unit uses 6.6 kW of power. However, I found the label and it actually says 6.6 kVA. When I saw this I thought something along the lines of...</p>
<blockquote>
<p>Well, $ P=VI $, therefore kVA must be the same thing as kW... strange, I wonder why it's not labelled in kW.</p>
</blockquote>
<p>So a quick Google search later, and I found <a href="http://www.dieselserviceandsupply.com/Power_Calculator.aspx">this page</a>, which has a converter that tells me 6.6 kVA is actually just 5.28 kW. The <a href="http://en.wikipedia.org/wiki/Watt">Wikipedia page for watts</a> confirmed what I thought, that a watt is a volt times an ampere.</p>
<p>So what part of all this am I missing, that explains why kVA and kW are not the same?</p>
| |electrical-engineering| | <p>Another way to understand why kVA is different to kW, is the Power triangle.</p>
<p><a href="https://i.stack.imgur.com/H69PY.png" rel="noreferrer"><img src="https://i.stack.imgur.com/H69PY.png" alt="enter image description here" /></a></p>
<p><strong>Figure 1: power Triangle (source: <a href="https://www.electricaltechnology.org/2019/08/difference-between-active-and-reactive-power.html" rel="noreferrer">Electrical technology</a>)</strong></p>
<p>The power triangle shows that the total (<em>apparent</em>) power is the <strong>vector</strong> sum of the active (or real) power and the reactive power.</p>
<ul>
<li><strong>Active Power</strong> <span class="math-container">$P$</span>: this is the actual power that its used/is available (electrical engineers apologies for the poor choice of words here). This is the portion of the apparent power that is available to be converted to work/heat.</li>
<li><strong>Reactive Power</strong> <span class="math-container">$Q$</span>: this is portion of the power that is stored in the system. That power is stored in the capacitive or inductive elements of the systems (every system has those in some small degree). This is why the LED of a power supply takes some time to fade when you pull it off a plug.</li>
<li><strong>Apparent power</strong> <span class="math-container">$S$</span>: this is the quantity measured in kVA. It is what is calculated by measuring the voltage and the current and multiplying them together.</li>
</ul>
<h2 id="why-apparent-and-real-power-are-not-the-same-p19m">Why Apparent and Real power are not the same</h2>
<p>When measuring an AC line, you can see, that the maximum values of voltage and current can be reached at the same time or not. See image below:</p>
<p><a href="https://i.stack.imgur.com/nTMxV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/nTMxV.png" alt="enter image description here" /></a></p>
<p><strong>Figure 2: power Triangle (source: <a href="https://www.electronicshub.org/power-factor-and-its-correction/" rel="noreferrer">Electronics Hub</a>)</strong></p>
<p>When the current and the voltage <strong>reach</strong> the maximum and minimum values at the same time, then the Apparent power (S) and the Real Power (P) are equal. There is no reactive element.</p>
<p>If the current and the voltage <strong>do not reach</strong> the maximum and minimum values at the same time, then the Apparent power (S) is greater than the Real Power (P). In that case there are some times that the product of <span class="math-container">$V*I$</span> is positive and others that it is negative.</p>
<p>In the (ideal) case of sinusoidal waveforms of Voltage and Current, the delay between voltage and current when expressed in the angular quantity <span class="math-container">$\phi$</span>, also determines the Power factor (which is <span class="math-container">$\cos(\phi)$</span>).</p>
| 1926 | Why is kVA not the same as kW? |
2015-03-04T19:50:21.837 | <p>...or in other words, why doesn't the material just stick to the screw, rotating in place, without progressing along its length?</p>
<p>In the simplest case, the answer is obvious: gravity. If it's a granular or liquid material and the screw is tilted to a side, it will just roll/slide/flow along the blade, to remain on the bottom side of the tube. </p>
<p><img src="https://i.stack.imgur.com/KE01J.gif" alt="enter image description here"></p>
<p>But then, <a href="http://en.wikipedia.org/wiki/Archimedes%27_screw#Variants" rel="noreferrer">Wikipedia</a> says:</p>
<blockquote>
<p>A variant of the Archimedes' screw can also be found in some injection molding machines, die casting machines and extrusion of plastics, which employ a screw of decreasing pitch to compress and melt the material. Finally, it is also used in a specific type of positive displacement air compressor: the rotary-screw air compressor. On a much larger scale, Archimedes' screws of decreasing pitch are used for the compaction of waste material.</p>
</blockquote>
<p>In such case the forces - pressure, shearing, viscosity, adhesion, all would outweigh gravity, and probably in some cases friction against the tube walls. For example, in injection molding, what is there to keep the half-molten plastic from forming a clumping mass, sticking to the screw and keeping rotating in place without any progress? When the pressure increases, which force prevents the material from backing up into area of lower pressure?</p>
| |fluid-mechanics| | <p>This is effectively an addition to other answers:</p>
<p><strong>An original Archimedes screw for liquids does not work as most people imagine it does,</strong><br>
and the diagram does NOT show how an original Archimedes screw works for fluids. The diagram is valid - it's just of something else which works in a similar manner and has become synonymous with the original design in peoples' minds. And both versions do not work as most people imagine them to. </p>
<p>In an original Archimedes Screw the outer cylinder is integral with the "screw" and turns with the screw - there is no moving seal. </p>
<p>When a marble or solid object that cannot fit through the sealing gap is used the two screws work the same. When a fluid is used the difference is important. From the reference below:</p>
<ul>
<li>An analysis, using the lifting of marbles instead of water, is used in almost all nineteenth century texts. The lower end of the helical tube dips into a dish of marbles and scoops one up. The helix continues to revolve, and the marble is continually being lifted a short distance up an inclined plane. The frictional forces are small, and the marble keeps rolling down an infinite succession of inclined planes formed by the revolving helix. At the same time the marble resides at the local low spot on the helix, and is carried up the slope by forces perpendicular to its local motion. </li>
</ul>
<p>The key point is that the "payload" sess a downwards ramp the whole way and simply "runs downhill.</p>
<p>A good illustration of the principle is given on <a href="http://physics.kenyon.edu/EarlyApparatus/Fluids/Archimedes_Screw/Archimedes_Screw.html" rel="noreferrer"><strong>this page</strong></a><br>
They use tubing so there is no doubt about the "seal".<br>
This image demonstrates what the fluid or other payload "sees".</p>
<p><img src="https://i.stack.imgur.com/dMHyO.jpg" alt="enter image description here"></p>
<p>Effectively the fluid sits in "buckets at all times and there is no opportunity to escape. </p>
<p>In typical original style screws the whole outer casing is sealed to the "screw" and the outer casing rotates with the screw. Consideratuiomn of a cross section shows that as with the tubing each batch of water sits in a container and NOT in a sloping tube.</p>
<p>Even in the stationary casing systems mentioned below the payload still usually sits in a "bucket" and runs forever downhill down a ramp as it ascends (!). </p>
<p>The crucial point of the original system is that the screw can be turned at any speed and even stopped and there is no leakage (apart from any that results from poor construction). Here is a "modern" real world device that works in this manner. This allows the device to be turned by hand at slow speed, or rotated intermittently, or with pauses, with no loss of fluid. Modern sealing techniques and relatively fast and consistent speeds of rotation allow systems where the casing is stationary.</p>
<p><img src="https://i.stack.imgur.com/03dUo.jpg" alt="enter image description here"></p>
<p><a href="http://www.math.nyu.edu/~crorres/Archimedes/Screw/applications/ScrewFarmer1.jpg" rel="noreferrer">Image from here</a><br>
From <a href="http://www.math.nyu.edu/~crorres/Archimedes/Screw/Applications.html" rel="noreferrer">this very good page</a></p>
<p>The nature of the original design is clearly explained by the <a href="http://en.wikipedia.org/wiki/Archimedes'_screw" rel="noreferrer">Wikipedia - Archimedes Screw</a> page.<br>
While they suggest that whether the inner and outer were sealed in original designs, even a brief consideration of the available facts shows that it was indeed the case. viz</p>
<ul>
<li>"Depictions of Greek and Roman water screws show them being powered by a human treading on the outer casing to turn the entire apparatus as one piece, which would require that the casing be rigidly attached to the screw." - Wikipedia</li>
</ul>
| 1958 | What makes the material travel through Archimedes' Screw? |
2015-03-05T23:16:05.160 | <p>Does the dead weight tester technology provide capability for an accurate, low pressure calibration standard in the range of zero to 150 cm H<sub>2</sub>O, and if so is there a manufacturer that provides such an instrument?</p>
<p>If not, what is the best primary standard for that range of pressure in terms of accuracy?</p>
<p>I suspect for dead weight testers there might be a lower limitation due to the balance of gravitational force and frictional force. I am looking to measure gauge pressure.</p>
| |measurements|pressure|metrology|standards| | <p>In this situation, you might do just as well to use a U-tube manometer. Is some respects, this is a sort of dead-weight tester where the fluid itself is the dead weight. In this device pressure is determined from the difference in height between two connected volumes of fluid. Usually this is just a transparent tube bent into a 'U' shape, placed upright in front of a convenient means of measuring the relative heights of the free surfaces of the liquid.</p>
<p><img src="https://i.stack.imgur.com/u1Szx.gif" alt="U-tube Manometer"></p>
<p>One end of the tube is connected to the pressurized volume, while the other is left open. The pressure difference may be calculated from the height difference.</p>
<p>$$ P = \rho g h $$</p>
<p>Where $\rho$ is the density of the liquid ($\sim 1000 \, kg/m^3$ for water) and $g$ is gravitational acceleration ($9.81 \, m/s$ in most places). Or, if you need your measurements in cm of water, simply use water as the working fluid and the height difference will be exactly the quantity you want. This is basically why "length units of liquid" can be used as a unit of pressure*.</p>
<p>If a standard manometer is not sensitive enough, you can tilt it off of vertical by some known angle. As a result, the liquid will travel a larger, more easily measured, distance along the tubes to achieve the same difference in height.</p>
<p>Despite the seemingly primitive technology involved, measurements of this kind can be quite accurate when performed correctly. There's a nice <a href="http://www.nist.gov/calibrations/upload/pmc-2.pdf" rel="nofollow noreferrer">document from NIST</a> that goes into detail about how to maximize the accuracy of these devices. Also a <a href="https://books.google.com/books?hl=en&lr=&id=FzgXBQAAQBAJ&oi=fnd&pg=PA229&dq=ruthberg+%22pressure+measurements+for+the+range%22&ots=dprESfnvpb&sig=GyEALrIq4xvbPJow9U80JoYFD7Y#v=onepage&q=ruthberg%20%22pressure%20measurements%20for%20the%20range%22&f=false" rel="nofollow noreferrer">review by Ruthberg</a> specifically for low pressure measurements.</p>
<p>*I don't mean to endorse these units of pressure. The pascal ($Pa$ or $N/m^2$) is the correct SI unit of pressure derived from the newton and the meter.</p>
| 1970 | Dead weight pressure tester for very low positive pressures |
2015-03-06T09:02:15.897 | <p>I work in the middle of London, in an area full of large office blocks. Across the road from my office they have started construction of a large building (10 stories plus). Over the last few weeks, diggers have dug a large (and vertical walled) hole. Lorries have taken the resulting dirt and old concrete away, leaving a very neat hole.</p>
<p>In the last day or so, the lorries have returned with new dirt (or the old dirt crushed) and diggers have been putting it back in the hole (and compacting it).</p>
<p>Why put the dirt back? Surely leaving the hole deeper would allow for deeper basement (or digging it shallower would be cheaper)?</p>
<p>I'm not a structural engineer, so this is all lost on me, but I'm fascinated.</p>
| |structural-engineering|geotechnical-engineering|building-design|foundations| | <p>In addition to the other answers, some soils (e.g. clay) expand and contract as their moisture content changes, and wet soil expands as it freezes. This movement is rarely even across site, and differential movement causes a building to crack.</p>
<p>Foundations are dug down to below an expansive soil and below the frost line to avoid movement. They are then built up with crushed rock or concrete, which doesn't expand with moisture or freezing temperatures.</p>
| 1977 | Why dig out and then fill in before building a large structure? |
2015-03-06T13:11:42.143 | <p>Whenever I've come across notation for specifying how much reinforcing steel is in a member, it is typically by giving the area of steel as a percent of the cross-sectional area of the member.</p>
<p><img src="https://i.stack.imgur.com/eO22J.png" alt="enter image description here"></p>
<p>For example in this picture, you could say that the reinforcing steel is <code>0.02</code> or <code>2%</code>. You could also specify the reinforcing steel given a schedule, e.g. <code>#8 @ 12" O.C</code>.</p>
<p>I recently came across a piece of software that was asking for the reinforcing steel of a floor slab in units of <code>mm^2/mm</code>. I cannot figure out how they came across those units. How could I calculate the reinforcement in these units given I know how much steel is in the slab in the other conventions I listed earlier?</p>
| |civil-engineering|structural-engineering|reinforced-concrete| | <p>Your second bar description is what you need to compare it to:</p>
<blockquote>
<p>#8 @ 12" O.C.</p>
</blockquote>
<p>A #8 bar is 0.79 in<sup>2</sup>. This gives us $\frac{0.79in^2}{12in}$ or in<sup>2</sup> over in. This is the similar to your mm<sup>2</sup>/mm. (Obviously the units are not interchangeable!)</p>
<p>This method of describing reinforcing is useful if you don't care about the actual size of rebar used. It could be #8 at 12" or #6 at 6.625".</p>
| 1980 | How do units of mm^2/mm represent the reinforcing steel area in a slab? |
2015-03-06T14:41:51.470 | <p>We are working on some equipment that is going to be attached to existing, 3,000 psi reinforced concrete floor slabs. In some locations, the best way to do this is to through-bolt and 'sandwich' the slab with a piece of threaded rod (5/8" rod in our case.) On the top side of the slab, our equipment has a large steel bearing surface. On the bottom side, we expect we'll need steel plate washers. We were thinking of using square washers cut from 1/2" A36 steel. What I don't know is how big of a square would be 'standard.'</p>
<p>Is there a chart or standard practice for what size plate washer to use on what size fastener? I couldn't find one in AISC, but may have missed it, or maybe it's in an ACI standard? If there's no prescriptive chart, what checks would typically be done to size it by calculation?</p>
| |structural-engineering|standards|fasteners|reinforced-concrete| | <p>There are some standard square washer sizes. These may not specifically be for your configuration, but they may help to decide on a size to start with.</p>
<ul>
<li><p>ASTM F 436 Square Washers are found in the AISC Steel Manual on the table that shows other bolt and washer diameters. These are specifically for beveled washers and the sizes given are only minimums. You can see the sizes <a href="http://www.portlandbolt.com/products/washers/hardened-beveled/" rel="nofollow">here</a>. For a 5/8" rod, the washers are 1.75" square.</p></li>
<li><p>Square washers are also used in timber construction. You can find some typical sizes <a href="http://www.portlandbolt.com/products/washers/square-plate/" rel="nofollow">here</a>. These washers are 2.5" square for 5/8" rod.</p></li>
<li><p>Custom washers might be the way to go. Since none of the washer types listed above are specifically for bolting to concrete, a custom size might be required. The references above give a starting point, but the final size might be controlled by the allowable bearing strength on concrete.</p></li>
</ul>
<p>Making a square washer is a simple machining task, so going the custom route may not add a lot of cost.</p>
| 1983 | Sizing plate washers for through-bolting to concrete |
2015-03-06T17:37:53.700 | <p>I am interested in a microcontroller for a low power application. I have been advise to look at MSP430 or Microchip PIC microcontrollers. I also wonder if <a href="http://www.arm.com/products/processors/cortex-m/cortex-m0.php" rel="nofollow">ARM-Cortex-M0</a> is a good choice too. </p>
<p>At a high level the system will have two analog sensors, few GPIO to control LED, and actuation mechanisms. The systems is intended to be powered with standard batteries.</p>
<p>What are the critical parameters that warrant attention in researching for a suitable low power micro controllers?</p>
| |electrical-engineering|embedded-systems| | <p>When low power consumption is important, you have to look at the whole picture. A microcontroller with low Joules/cycle is only one thing to look at.</p>
<p>A good start would be to pick a battery and micro combination where the micro can run directly from the battery over the useful voltage range of the battery. Many PIC microcontroller, for example, can run from 1.8 or 2.0 V up to 5.5 V. That nicely covers a single lithium ion cell, two or three primary "1.5 V" cells in series, etc.</p>
<p>Another important strategy is to let the micro sleep as much as possible. If you only need to check something once a second, then maybe you can have the micro wake up for 1 ms every second. That means you need to look at sleep current carefully. It also means you want one with a built-in RC oscillator to minimize startup time, assuming a few percent clock rate accuracy is good enough. There are lots of tradeoffs.</p>
<p>You didn't give enough information to have some idea of the computing power required, so we can't say what would be adequate. But in general, take a look at any Microchip PIC with the "XLP" (extra low power) feature. In particular the 16F1xxx series has some impressively low power devices.</p>
| 1987 | How to select a micro controller for a low power application? |
2015-03-06T19:19:02.307 | <p><a href="http://en.wikipedia.org/wiki/Sinkhole" rel="noreferrer">Sinkholes</a> have been known to occur in the <a href="http://www.cnn.com/2014/08/05/world/asia/seoul-skyscraper-sinkholes/" rel="noreferrer">middle of cities</a> or <a href="http://corvettemuseum.org/enews/backups/media/sinkhole.htm" rel="noreferrer">other locations</a> where they affect buildings:</p>
<p><img src="https://i.stack.imgur.com/TxpQYm.jpg" alt="sinkhole"></p>
<p><a href="http://www.dep.state.fl.us/geology/geologictopics/sinkhole.htm" rel="noreferrer">Some areas</a> are more prone to sinkholes than other areas because of the presence of old mines or limestone bedrock. Even in these areas where sinkholes are more common, humans continue to build buildings and even <a href="http://indianapublicmedia.org/news/sinkholes-discovered-monroe-airport-17956/" rel="noreferrer">airports</a>.</p>
<p>How do engineers prevent or mitigate sinkholes? Is it as low-tech as dumping in rock until no more fits?</p>
| |geotechnical-engineering| | <p>Whether it's natural subsidence, like sinkholes or human induced subsidence like the collapse of underground engineered chambers or mining subsidence the two ways of dealing with it are backfill or leaving it alone and the enforcement of an exclusion zone.</p>
<p>Where backfill is used for the remediation of subsidence it's generally loose rock fill, because it's the cheapest form of backfill and excavations aren't going to be established against the backfill, particular in civil situations. In mining, the subsidence backfill can be loose rock, sand or tailings, or cemented rock, sand or tailings; depending on circumstance, what materials are available and how much the company is prepared to pay.</p>
<hr>
<p>Edit: 26 March 2015</p>
<p>I came across this picture of a sinkhole being backfilled with concrete on the <a href="http://www.msn.com/en-au/news/world/giant-sinkholes/ss-AA9Xam4?ocid=mailsignout#image=5" rel="nofollow noreferrer">MSN news website</a>.</p>
<p><img src="https://i.stack.imgur.com/cQYBD.jpg" alt="enter image description here"></p>
<hr>
<p>Additional Information 26 March 2015</p>
<p>One group of <a href="http://www.rembco.com/news/sinkholes-101/" rel="nofollow noreferrer">geotechnical contractors</a> in the US advocates excavation and backfilling of sinkhole where the bedrock in no deeper than 4.5 m (15 ft). For deeper holes it recommends grouting and for very deep holes it recommends cap grouting.</p>
<p>The <a href="http://efotg.sc.egov.usda.gov/references/public/WI/527.pdf" rel="nofollow noreferrer">Karst Sinkhole Treatment</a> document by the Natural Resources Conservation Service recommends establishing of a buffer zone around the hole and backfilling. Depending on circumstances the backfill will include loose rock, concrete and if necessary geotextile.</p>
<p>When floods hit <a href="http://www.sott.net/article/263232-Flood-leaves-70-sinkholes-in-Calgary" rel="nofollow noreferrer">Calgary</a>, in Canada, in June 2013, creating numerous sinkholes, the holes were backfilled.</p>
<p>The <a href="http://www.sinkholeguide.com/sinkhole-faqs.htm" rel="nofollow noreferrer">Sinkhole Guide</a> recommends backfilling sinkholes with,</p>
<blockquote>
<p>“native earth materials or concrete. Broken limestone rip-rap or a
concrete plug in the bottom of the sinkhole often helps create a
stable foundation for the fill. Above that, add clayey sand to form a
barrier that will help to prevent water from seeping downward through
the hole and enlarging it further. Lastly, add sand and top soil, and
landscape to surrounding conditions. Additional fill may be necessary
over time, but most holes eventually stabilize.”</p>
</blockquote>
<p>According to the <a href="http://www.fhwa.dot.gov/engineering/geotech/hazards/mine/workshops/kdot/kansas04.cfm" rel="nofollow noreferrer">US Department of Transportation</a>, fly ash in grout has been used to backfill sinkhole/subsidence holes in abandoned mines.</p>
<p>In North Dakota, the <a href="http://www.psc.nd.gov/docs/amlarticles/structural-monitoring-at-a-pressurized-grout-project.pdf" rel="nofollow noreferrer">Public Service Commission</a> backfills subsidence hole in abandoned mines in that State. </p>
| 1988 | How do engineers deal with sinkholes? |
2015-03-07T11:25:26.880 | <p>Is the smaller gear (pinion) always mounted to the input shaft when meshed with a bigger gear that is mounted on the output shaft? Are there places where the bigger gear drives the smaller gear?</p>
| |mechanical-engineering|gears| | <p>Any mechanical clock or watch relies on the motive power applied to the large gear (the "wheel") which drives the smaller one (the "pinion"). Thus the weight in a longcase clock is suspended by a cord, rope or chain from the "great wheel" (usually making a rotation every 12 hours) and the rate of rotation is geared up to the escape wheel (which often has the seconds hand mounted on it).</p>
<p>Note that the tooth form is usually different when gearing up : friction is critically important in a clock, transmitting high forces is usually less so (and is usually catered for by making the great wheel thicker than the others). So the teeth are usually cycloidal in form, where the deep part of the slot in a tooth is approximately rectangular, which means the base of a pinion tooth is undercut. This is a fundamentally weaker tooth form, especially since pinions may have as few as 6 teeth, but runs freely with little friction and zero pressure angle (see below). </p>
<p>For example
<img src="https://i.stack.imgur.com/4yqOL.jpg" alt="enter image description here"></p>
<p>(from <a href="http://www.ryerson.ca/~v7chan/cncproject/gear/Cycloid%20Gear%20Calculator.html" rel="nofollow noreferrer">this page</a>)</p>
<p>An extreme case is the lantern pinion<br>
<img src="https://i.stack.imgur.com/6KFCB.jpg" alt="enter image description here"><br>
(from <a href="http://www.donaldsonworkshop.com/chapdetail.php?project=clock&chapter=08ClickWheelsPinions.txt" rel="nofollow noreferrer">this page</a>) where the pinion tooth is completely undercut!</p>
<p>You never lubricate the teeth of a clock or watch wheel : that only adds viscosity (i.e. friction), wasting power, and does nothing to eliminate wear. This is because the contact surfaces of the teeth roll over each other, there is no sliding motion involved. (The pivots, unless running in ballraces, do need lubrication. John Harrison employed ballraces for a prototype marine chronometer).</p>
<p>In contrast, while gearing speed down also involves contact surfaces rolling over each other, the purpose is usually to amplify force, and to do that with the least material, a stronger tooth form is required. This is normally an <a href="http://en.wikipedia.org/wiki/Involute_gear" rel="nofollow noreferrer">involute tooth form</a>, where each tooth is wider at the base, like a wedge.
<img src="https://i.stack.imgur.com/puHpt.jpg" alt="enter image description here"></p>
<p>This means the teeth press each other outwards as well as turning each other, at an angle known as the pressure angle (usually 20 degrees in modern gears, formerly 14.5 degrees). Thus the axles are pushed apart, increasing friction on the pivots and requiring a stronger gearbox. (The animation on the Wikipedia page exaggerates the pressure angle). Traditionally, involute pinions are only cut down to 12 teeth, with the 20 degree PA making for more friction but stronger teeth with wider roots.</p>
<p>So : yes, gearing can be used to increase rotational speed, but it usually requires a different tooth form, otherwise it loses a lot of power to friction.</p>
| 1995 | Mechanism of gear rotation |
2015-03-07T19:12:07.967 | <p>Pressure treated lumber is specified for many exterior applications because of its resistance to insect damage and fungal rot. But how does it compare to untreated wood, mechanically?</p>
<p>For example, consider a rim joist supporting the ground floor of a residential structure with a pier and post foundation. If the joist has been damaged by rot in a location that is impractical to fully protect from exposure, I might be tempted to replace the joist with a pressure treated member of the same nominal dimension (in <em>addition</em> to proper flashing) for added protection in that location.</p>
<p>Since this is an existing structure, by far the easiest approach is to use a member with the same dimension to replace the rotted joist. However, this relies on the new member meeting the same load-bearing requirements as the old member.</p>
<p>Building codes should provide enough wiggle room that in this particular example, there's not much of a safety concern for the homeowner. After all, the rotted joist had not failed, and it would definitely have less strength than the member was originally rated for. In practice, treated and untreated members may be manufactured from different wood species with different mechanical properties to begin with; for the purposes of this question, assume the species is constant.</p>
<p>Does pressure treatment result in a member with more or less strength in tension, compression or torsion? Does it affect the durability of the wood* <em>in ways not related to rot or insect damage?</em></p>
<hr>
<p><sup>* Not the fasteners; that's a different issue that's pretty well-covered online. See <a href="http://www.strongtie.com/productuse/ptwoodfaqs.html">this page from Simpson</a>, for example.</sup></p>
| |structural-engineering|wood| | <p>Pressure treatment does have a small, but documented effect on the strength of the member, particularly if it is 'incised' (has slots or holes cut into it as part of the pressure treating process.) If you're working to American codes, according to the <a href="http://www.awc.org/helpoutreach/faq/faqFiles/Pressure_treat_valu.php">American Wood Council</a>, pressure treated wood is limited to a maximum duration factor of 1.6. This wouldn't matter for your example of fixing a house, because the duration factor would already be much smaller. Effectively, this just restricts the use of pressure treated lumber for resisting impact loads.</p>
<p>More importantly, if it is incised which is often the case, the NDS has a derating factor of 20% for the strength of the member except for compression perpendicular to the grain, which is not derated at all. Note that there is also a 'wet service factor' for any member that will be exposed to moisture, regardless of any treatment.</p>
<p>Pressure treated lumber intended for structural applications should have a research report that covers usage, labeling, and special instructions. These often have useful information. For example, here are the research reports for <a href="http://www.icc-es.org/Reports/pdf_files/ESR-1081.pdf">two</a> <a href="http://www.icc-es.org/Reports/pdf_files/ESR-3038.pdf">random</a> brands. (Warning: PDF links.)</p>
| 1998 | How does pressure treatment affect the mechanical properties of lumber? |
2015-03-07T22:40:33.667 | <p>A <a href="https://en.wikipedia.org/wiki/Epicyclic_gearing" rel="noreferrer">planetary gear</a> is a gear that rotates not around an axle but around another gear known as a sun gear, or star gear. I have designed prototype systems in the past that used a single planetary gear rotating around a rotating gear. However since then almost every system I've seen using planetary gears have used multiple gears, most commonly in sets of <a href="http://forums.autodesk.com/autodesk/attachments/autodesk/78/316168/1/sun%20and%20planets.gif?nobounce" rel="noreferrer">3</a>, <a href="https://d2t1xqejof9utc.cloudfront.net/screenshots/pics/2260e59ecf6198bdcdcb266aeae6cf64/medium.jpg" rel="noreferrer">4</a>, <a href="http://www.bncgears.com/file/0_2011_06_24_194353.gif" rel="noreferrer">5</a> and <a href="http://www.intechpower.com/Portals/62516/images/planetary_gear.gif" rel="noreferrer">6</a>.</p>
<p>What is the mechanical advantages and disadvantages of using multiple planetary gears and what is the optimum number of gears?</p>
| |mechanical-engineering|dynamics|statics|gears| | <p><strong>Advantages</strong></p>
<p>I'll start off by quoting from the "Benefits" section of <a href="https://en.wikipedia.org/wiki/Epicyclic_gearing#Benefits" rel="nofollow">the Wikipedia article</a>:</p>
<blockquote>
<p>The load in a planetary gear train is shared among multiple planets, therefore torque capability is greatly increased. The more planets in the system, the greater the load ability and the higher the torque density.</p>
</blockquote>
<p>The more the merrier. The same idea is covered <a href="http://www.gearsolutions.com/article/detail/5837/epicyclic-gearing-a-handbook" rel="nofollow">here</a>:</p>
<blockquote>
<p><strong>Torque Splits</strong> <br>
When considering torque splits one assumes the torque to be divided among the planets equally, but this may not be a valid assumption. Member support and the number of planets determine the torque split represented by an “effective” number of planets. This number in epicyclic sets constructed with two or three planets is in most cases equal to the actual number of planets. When more than three planets are used, however, the effective number of planets is always less than the actual number of planets.</p>
<p>Let’s look at torque splits in terms of fixed support and floating support of the members. With fixed support, all members are supported in bearings. The centers of the sun, ring, and carrier will not be coincident due to manufacturing tolerances. Because of this fewer planets are simultaneously in mesh, resulting in a lower effective number of planets sharing the load. With floating support, one or two members are allowed a small amount of radial freedom or float, which allows the sun, ring, and carrier to seek a position where their centers are coincident. This float could be as little as .001-.002 inches. With floating support three planets will always be in mesh, resulting in a higher effective number of planets sharing the load.</p>
</blockquote>
<p>The single-planet arrangement doesn't change the torque capability, but the multiple-planet arrangement does. You can increase the number of planets without limit and still continue to increase it.</p>
<p>Another advantage of using more planets is less energy loss. Wikipedia claims that there is an energy loss of only about 3% per stage, but it doesn't back that up with an inline citation, nor does it specify the number or arrangement of planets. <a href="http://www.cs.berkeley.edu/~sequin/CS285/2011_REPORTS/CS285%20final%20paper_Eric&Jessie.pdf" rel="nofollow">This</a> gives the same feature as a "typical energy loss."</p>
<p><strong>Disadvantages</strong></p>
<p>The obvious one here is that the more planets you add, the more complex the system becomes. If there's a major failure, it could be amplified by having more planets (alternatively, it could be argued that some arrangements provide redundancy).</p>
<hr>
<p>Is there an optimal number of gears? It really depends on your application. Complexity can be a huge issue, and with complexity comes cost. Repairs, in turn, can be expensive if something goes drastically wrong. If the application involves a great load and torque, then more gears can be helpful.</p>
| 2001 | What is the benefit of additional planetary gears? |
2015-03-07T22:40:44.457 | <p>Chainsaws use oil to lubricate the chain. The usage is quite minor, probably less than 1cc per minute of work. The oil is normally supplied from a refillable compartment to the chain only when the chainsaw is running.</p>
<p>How is the delivery managed? </p>
<p>A pump? The capillary effect?</p>
<p>If it's a pump, I'd be interested in details - I mean, I have a hard time imagining a pump this slow (<1cc/min) supplied from the high-RPM engine... or does it use an extensive gearbox to slow down? A separate motor?</p>
<p>[note: I mean the oil used for chain lubrication, not mixed with fuel for the two-stroke engine of motorsaw]</p>
| |mechanical-engineering|pumps| | <p>I sent this question along to a forester of 20+ years and this was his response:</p>
<blockquote>
<p>So...... back in the day......... the first saws had a manual oiler.
This was a younger lumberjack in training who operated the oil can. He
held this job until one of the more senior lumberjacks cut off
something vital and he was promoted.</p>
<p>The next generation of oilers was still manual. There was a thumb
plunger on the case. It was located near the center of the case and
was operated by the same thumb/hand which operated the throttle. The
theory being that you could give one good squeeze and cut/oil at the
same time. the operator was also able to titrate the oil use to the
dryness or wetness of the wood.</p>
<p>One of the posters was correct in that there have been worm gears used
also. These are very effective in that they are throttle based. they
work as fast as the saw requires and in theory you don't burn up bars
and chains. There is an adjustment screw which has numbers from 1-5
with the preset usually being a 2. I believe that this is the current
style in use. The output is based on a motor speed of 9000 rpm. There
are some other pump styles in use but these are the most reliable.
They are mechanical. They leak on the floor when not in use. Gravity!</p>
</blockquote>
| 2002 | How is oil supplied in chainsaws? |
2015-03-08T00:17:04.670 | <p>In common US usage, the heights of tall things are sometimes converted to a "number of <a href="https://en.wikipedia.org/wiki/Storey" rel="nofollow">stories</a>". The thinking is that people can better compare heights to similar tall buildings that they might have seen. </p>
<p>The US conversion is usually:
$$\text{Number of stories} = \text{Round}\Big(\frac{\text{height in feet}}{10}\Big)$$</p>
<p>I assume that something similar is used in metric countries like:
$$\text{Number of storeys} = \text{Round}\Big(\frac{\text{height in meters}}{3}\Big)$$</p>
<p>I am assuming that like most things that are common usage, this conversion is not correct in practice. </p>
<p>What is a more accurate height of a building story?<br/>
How does this cause confusion when comparing the height of a building in "stories" to the actually number of stories?</p>
| |structures|metrology| | <p>There's a handy-dandy table <a href="http://www.ctbuh.org/TallBuildings/HeightStatistics/HeightCalculator/tabid/1007/language/en-GB/Default.aspx" rel="nofollow">here</a>:</p>
<pre><code> Office | Residential/hotel | Function Unknown
or Mixed-Use
floor-to-floor height (f) 3.9m | 3.1m | 3.5m
Entrance lobby level floor-to-floor height 2.0f = 7.8m | 1.5f = 4.65m | 1.75f = 6.125m
Number of mechanical floors above ground s/20 | s/30 | 2/25
(excluding those on the roof)
Height of mechanical floors 2.0f = 7.8m | 1.5f = 4.65m | 1.75f = 6.125m
Height of roof-level mechanical 2.0f = 7.8m | 2.0f = 6.2m | 2.0f = 7.0m
areas / parapets / screen walls
H = Building height
f = Typical occupied floor-to-floor height
s = Total number of stories
m = meters
</code></pre>
<p>These numbers are just averages.</p>
<p>Using this, it is possible to derive the height of an office building:
$$H_{\text{office}}=3.9s+11.7+3.9(s/20)$$
The same can be done for a residential building:
$$H_{\text{residential}}=3.1s+7.75+1.55(s/30)$$
The page then gives some comparisons of real buildings and the variation from the formula. However, while the "aggregate" variation for residential buildings (for example) comes out to 0.36%, a better indicator of accuracy would be to take the absolute value of the variation.</p>
<p>So, in summary: The average height depends on the function of the building (i.e. office vs. residential). But there are, of course, deviations from these figures.</p>
| 2004 | Is a building story actually 10ft (3m)? |
2015-03-08T08:43:33.363 | <p>I have got a spinning top from a Canadian manufacturer, and I am deeply amazed by how neatly it spins.</p>
<p>My interest in physics has led me to try to find out at what top speed I can spin the top and how that compares to other tops made from other materials and other shapes.</p>
<p>I have glued a little sticker to it and recorded it with my phone's camera but the frame rate is far too slow to be able to count the number of rotations over time.</p>
<p>I don't have access to specialized equipment but I can measure or approximate the volume, mass, density, temperature and probably also the moment of inertia. How do I measure the angular speed of the top? </p>
| |measurements|applied-mechanics| | <p>The answer could be to get hold of a high-speed LED stroboscopic light. How easy/expensive this will be depends on the angular speed you are dealing with. There are even some phone apps which will go to moderate speeds.</p>
<p>You tune the speed of the stroboscope to make the mark on your spinning top appear stationary. But this will be difficult because the speed will not be constant.</p>
| 2006 | Practical setup to measure angular velocity |
2015-03-09T02:03:45.357 | <p>When looking at thread descriptions, one of the <a href="https://en.wikipedia.org/wiki/Screw_thread#Lead.2C_pitch.2C_and_starts" rel="noreferrer">basic properties</a> is always the number of thread starts.</p>
<p><img src="https://i.stack.imgur.com/HmBHi.png" alt="wikipedia thread starts"></p>
<p>As far as I could tell, all of the major standard bolt threads are single-start. This includes:</p>
<ul>
<li>Unified Standard (UNC, etc.)</li>
<li>National Pipe Thread (NPT, NPS)</li>
<li>British Standard</li>
</ul>
<p>I only found one standard thread that can also come in a multiple-starts: <a href="http://www.amesweb.info/Screws/AcmeScrewNutThreadDimensions.aspx" rel="noreferrer">ACME</a>. </p>
<p>What are the reasons why single-start threads are so common and multiple-start threads are rare? I am specifically interested in bolts and other fasteners.</p>
| |threads|fasteners| | <p>Advantages of double lead threading: Starting ease and eliminating cross-threading (i.e. happy customers). </p>
<p>Common usages:</p>
<ul>
<li>Any bottle cap or jaw lid</li>
<li>Catsup bottles</li>
<li>Tooth paste tubes</li>
<li>Split bolts (Kearneys)</li>
<li>Flash light end caps</li>
<li>Any assembly not requiring high tension</li>
</ul>
| 2015 | Why are most standard bolt threads single start? |
2015-03-09T02:39:12.847 | <p>I've heard the term <em>spiral curve</em> used to describe a section of highway that is more aesthetically pleasing to the driver's eye. However, I believe I've driven on enough road to say that I can't definitively tell the difference between any given curve other than how "sharp" it is.</p>
<p>Could anyone explain how one can determine if a section of curve on a highway is classified as a "spiral" curve, and are there other advantages besides "making it look good"?</p>
| |civil-engineering|highway-engineering|surveying|rail| | <p>Q: ...how one can determine if a section of curve on a highway is classified as a "spiral" curve, and are there other advantages besides "making it look good"?</p>
<p>A: The purpose of a spiral or transition curve has nothing to do with aesthetics. They were originally introduced on railways for safety reasons in conjunction with superelevation as explained here: <a href="http://www.abcls.ca/wp-content/uploads/pdfs/Notes_on_Transition_and_Terminal_Curves.pdf" rel="nofollow">Notes on Transitional and Terminal Curves</a>. On high speed highways, they permit a driver to enter/exit a circular curve with gradual increase/decrease in the deflection of the steering wheel. The mathematical form of the spiral may vary from jurisdiction to jurisdiction, and may have changed in recent decades as hand-held calculators and other computing devices replaced the hard-copy field tables once used by surveyors. One common form is the <a href="https://en.wikipedia.org/wiki/Euler_spiral" rel="nofollow">Euler Spiral or clothoid</a>. I have read that in India the usual transition curve is a third-order hyperbola, and I have been told that in Germany autobahns are designed as a continuous series of connected clothoids with no tangential sections or circular curves. </p>
| 2017 | What is a Spiral Curve, and How is it Different from a Normal Curve? |
2015-03-10T05:29:20.740 | <p>I have an 80 g·cm motor with a rotational frequency of 15,000 rpm. I want to lift a weight of 2 kg at a speed of 0.5 m/s. How do I calculate the gear ratio required?</p>
| |mechanical-engineering|gears| | <blockquote>
<p>I have a 80gcm motor with a rpm of 15000.<br>
I want to lift a weight of 2 kg at a speed of 0.5 m/s.<br>
How do I go about calculating the gear ratio required for this?</p>
</blockquote>
<p><strong>Firstly - is it possible?</strong></p>
<p>In particular, is there enough input power available for the desired output power?</p>
<p>To within about 2% a very handy formula applies - it can be derived conventionally and seen that several factors happen to cancel nicely.</p>
<p>Watts = kg x metres x RPM</p>
<p>80 gram∙cm = 0.080 kg x 0.01 m</p>
<p>So for input W = 0.080 kg x 0.01 m x 15000 = 12 Watts.<br>
This is the maximum Wattage you can deliver if properly geared at 100% efficiency<br>
(we should be so lucky).</p>
<p>Desired power = Force x distance per unit time<br>
Watts = Joules/sec = mg∙d/s</p>
<p>= 2 kg x $g$ x 0.5 m/s = 2 x 9.8 x 0.5 = 9.8 Watts</p>
<p>So to work at all overall efficiency needs to be at least 9.8/12
or greater than about 82%.<br>
That's potentially doable but also potentially difficult.</p>
<p><strong>Now to the actual problem.</strong></p>
<p>The following assumes that the output weight or force is taken from the end of a radius of the driven "gear". If output is instead taken from eg a windlass drum at lower diameter to the driven gear the ratios will be scaled based on the relative diameters. Ignore that for now. </p>
<p>Torque_in x RPM_in = Torque_out x RPM_out at 100% efficiency</p>
<p>or RPM_out = Torque_in x RPM_in / Torque_out at 100% efficiency </p>
<p>So:</p>
<p>RPM out = 0.080 kg x 0.01 m x 15000 RPM / (2 kg x 0.5 m) = 12 RPM</p>
<p>So gear ratio = 15000/12 = 1250:1</p>
<p>The specification of not just output Torque but actual force (2 kg x $g$) constrains the actual output pulley size if output is taken off at the pulley radius.</p>
| 2029 | How do I calculate the gear ratio to lift a weight at a constant speed? |
2015-03-11T13:05:31.940 | <p>I have an welded steel mesh, like in the following image.</p>
<p><img src="https://i.stack.imgur.com/qT2lx.jpg" alt="steel mesh">.</p>
<ul>
<li>Material: steel</li>
<li>Diameter of the wires: 4mm</li>
<li>Mesh: 15x15cm</li>
</ul>
<p>I found the following information about the welds:</p>
<ul>
<li><strong>the minimal relative strength of welds in shear: F<sub>ms</sub> = 4.4kN</strong></li>
</ul>
<p>I want cut out an rectangle from this mesh (dimensions are on the following image) and want hang it on the wall, <strong>suspended by <em>two</em> screws</strong>.</p>
<p>Now I want hang to the welding points some weights. (on the image the drop-like things) :), in two scenarios:</p>
<ol>
<li>uniformly attach to <em>each welded point</em> some weight (like the blue "drops")</li>
<li>attach one big weight to the center (the red drop)</li>
</ol>
<p><img src="https://i.stack.imgur.com/8VFs9.png" alt="enter image description here"></p>
<ul>
<li><p>What is the weight I can hang on the mesh (in both scenarios) without significant structural deformations (2% elongation)?</p></li>
<li><p>I don't want have broken welds. Will the welds fracture before the bars yield or vice versa? At what load will the first of these failure modes occur?</p></li>
</ul>
<p>Also, i'm looking for some "guide" how it should be calculated - if it is possible to do in easy (read: "idealized") form, without specialized software. I would be happy to calculate it myself, but need some "how to". Understand than using finite-elemets or such will give most precise result, but is here some "easier/simpler" way?</p>
| |mechanical-engineering|steel|structural-engineering| | <p>The weld points will definitely be the limiting factor (think about a cargo net, where both the strands and joints are flexible, but usually the strands are straight and the joints are at whatever angle they need to be at to let the strands be straight.)</p>
<p>Since the "shear" strength is listed in linear kN I'm going to assume that that is the effective strength of one joint resisting a force that would slide one bar along the other.</p>
<p>I'll further assume that the steel used has a tensile yield strength of 370 MPa.</p>
<p>I think a good assumption is that the weld joints can be approximated as having a circular cross section. We can estimate the diameter of the cross section:</p>
<p>$$F=\frac{\sigma}2\,\pi\,r^2$$
$$d=\sqrt{\frac{8F}{\sigma\,\pi}}=\sqrt{8\frac{4.4\,\mathrm{kN}}{370\,\mathrm{MPa}\,\pi}}=5.5\,\mathrm{mm}$$</p>
<p>Well, that's unreasonable...so either those welds are made of some super alloy, or the maximum shear load of those welds is less than 4.4 kN.</p>
<p><strong>Edit:(</strong> Actually according to <a href="https://engineering.stackexchange.com/a/2058/1122">this answer</a> the welds can be significantly larger than the wires so this might not actually be unreasonable. <strong>)</strong></p>
<p>In fact the maximum <em>tensile</em> load a 4 mm diameter cylinder of 1018 steel could hold before yielding is about 4.4 kN.</p>
<p>So let's proceed as if the diameter of the weld is 1 mm. Each joint could then handle some torque $\tau$ according to:</p>
<p>$$\tau=\frac{\pi\,\sigma_y\,d^3}{24} \approx 50 \,\mathrm{N\cdot mm}$$</p>
<p>There are approximately 250 joints so the maximum total torque is approximately 62 N·m.</p>
<p>The torque produced by the weight be calculated via Virtual Work:</p>
<p>The elongation of a ridged mesh $\epsilon$ is approximately half the radians of deformation $\gamma$.</p>
<p>The gravitational potential energy gained from a uniformly distributed mass by stretching is:</p>
<p>$$E=\frac12 m\,g\,h\,\epsilon= \frac14\,m\,g\,h\,\gamma$$</p>
<p>$$\tau=\frac{dE}{d\gamma}=\frac14\,m\,g\,h$$</p>
<p>Solving for mass gives:</p>
<p>$$m=250\frac{4\,\tau}{g\,h}\approx 2\,\mathrm{kg}$$</p>
<p>This indicates that if the strength of the welds is close to 1018 Steel, and the weld diameter is close to 1 mm then it wouldn't even come close to holding up its own weight (8 kg) when hung at a 45 degree angle. However, if the weld points are 2 mm in diameter then it could hold 15 kg allowing it to support its own weight with a safety factor of 2.</p>
<p>If this were my project I would do destructive testing on one weld joint to see how much torque it takes to yield it. That would allow you to plug in for $\tau$ and get a reasonable estimate for the maximum load.</p>
| 2041 | Hanging a steel mesh without significant deformation |
2015-03-11T16:42:38.097 | <p>I need to replace the fan on my graphics card. The original fan is exactly like <a href="http://www.dhgate.com/product/original-for-sapphire-apistek-graphics-card/190676310.html" rel="nofollow noreferrer">this one</a> but there is another fan that seems to be a <a href="http://www.dhgate.com/product/original-sauter-ga51s2l-12v-0-13a-blade-diameter/230471693.html" rel="nofollow noreferrer">new revision</a> of the same model number from the same manufacturer (NNTG vs. NNTM). The main difference seems to be the shape of the blades:</p>
<p><img src="https://i.stack.imgur.com/uSkiv.png" alt=""></p>
<p>Neither I nor the DHGate seller can find any data about the fans (CFM, noise) so that's why I ask here.</p>
<p>To make up my mind I need an answer to both questions:</p>
<ol>
<li><strong>Which blade design is likely to cool the GPU the most?</strong></li>
<li><strong>Which one is likely to be the quietest?</strong></li>
</ol>
<p>The safe choice would be to get the exact same one (NNTG) but if the new design can be more efficient with the same noise, or as efficient but quieter, I'd prefer the new one.</p>
<p>Note that I don't want to buy an aftermarket fan (even if the prices are similar), I just want a replacement for the old one.</p>
<p>EDIT 2015-04-26 : Finally I bought both and, unlike what I said in the comments, I found a way to quickly switch the fan without taking out the whole heatsink assembly, allowing me to test both fans.</p>
<p>Despite the general consensus that NNTM was supposed to cool the GPU better, temperature with NNTG is actually 5°C cooler than with NNTM (after 10 minutes of running glxgears in full-screen). Maybe the motor doesn't match the power needed for the redesigned blades of NNTM, I don't know.</p>
<p>Anyway, if you need to replace a GA512SL fan for a RadeonHD like I did, buy the NNTG model :-)</p>
| |mechanical-engineering|cooling| | <p>Factors determining fan airflow (heat reduction): </p>
<ol>
<li>Blade pitch.</li>
<li>Blade shape and size.</li>
<li>Motor power and RPM.</li>
</ol>
<p>In your case 3 is the same, 1 and 2 in the "newer version are bigger", so it will cool better.
Main source of noise in fans is coming from the bearing and the electromotor (same in both cases) and from the blades which are slightly larger. Considering that our hearing works on logarithmic scale, I highly doubt that the increase of blades's size in the second model will double the noise, i.e. make it possible to discern from the first. Based on that, draw your conclusion :).</p>
| 2044 | How does the shape of CPU/GPU fan blades affect their performance and loudness? |
2015-03-11T21:51:51.500 | <p>On old bridge plans I have occasionally seen a note to "burr threads after installation". This was done to keep the nuts of bolted connection from loosening all the way and falling off of the bolt.</p>
<p>An example of this practice can be seen in this <a href="https://books.google.com/books?id=bxAkAAAAMAAJ&pg=PA1524&lpg=PA1524&dq=bridge%20burr%20threads&source=bl&ots=B3fbOBL1C5&sig=e_QglH2tkXRRS7bhha4sFVPXtKo&hl=en&sa=X&ei=HrAAVfPsJIKXgwSh44CACQ&ved=0CEoQ6AEwBg#v=onepage&q=bridge%20burr%20threads&f=false">Google Books</a> excerpt from <em>Bridge Engineering, Volume 2</em>:</p>
<blockquote>
<p>Nuts on the floorbeam hangers have a tendency to work loose; this we generally remedy by putting on check nuts where the thread is long enough to permit; if not, we burr the thread after adjusting.</p>
</blockquote>
<p>That book is from 1916, but I have seen this on much more recent bridge plans. Sometimes the note is along the lines of "intentionally damage threads ...". </p>
<p>The practice of damaging the threads is still an option per this <a href="http://www.aisc.org/DynamicQuestion.aspx?Grpid=6&QueId=1760&ste=_Preview&id=2100&type=0">AISC response</a>.</p>
<p>This isn't such a big issue for normal bolted connections where the bolt could be cut (or otherwise destroyed) and replaced with a new one. This is a problem for anchor bolt connections where one end of the anchor bolt is embedded in concrete. These bolts can not be easily replaced.</p>
<p>The intention of damaging the bolt thread is to keep the nut from working loose. The expectation seems to be that with enough torque the nut would effectively re-tap the bolt as it is removed.</p>
<ul>
<li>Is it feasible to back the nut off of the bolt without destroying the bolt? </li>
<li>Would there be any confidence that the anchor bolt could be reusable after this procedure is complete? </li>
</ul>
| |structural-engineering|bridges|bolting|threads| | <p>I asked a friend who is a highly competent EE by day - but who restores super sized steam engines by night and collects olde heavy metal engines etc and has much experience with old large corroded items. His comments:</p>
<hr>
<p>It really depends on the specific situation. </p>
<p>For mild steel bolts in good condition which have not been too excessively deformed, and where the nuts are in reasonable condition, it is often possible to "re-form" the damaged threads by applying sufficient torque to back the nut off. In some case you can encourage this to happen by carefully grinding the deformed end of the bolt to get it back to something close the original nominal thread diameter. You can also increase your chances if you apply a penetrating lubricant (and any lubricant is better than none). Applying heat may also help if that is possible - and if you can apply heat and then apply lubricant at a temperature where it is not broken down the lubricant will tend to be drawn into the thread as further cooling occurs.</p>
<p>If the nut is in poor condition or the shaft of the bolt has corroded and become "waisted" your probability of success will be pretty low. If just the nut is corroded you may be able to split the nut with chisel, but you then have the problem of how to reform the threads. If sufficient bolt end exists you may be able to use a die to cut new threads, but you will then have to use a bush under the new nut to take up the length of the remaining old thread as the new thread is unlikely to run into the old properly.</p>
<p>In some cases you may be able to reduce the diameter of the bolt and cut a new thread at the reduced diameter. If you are willing to make you own nut, you have full freedom as to the diameter you choose - at the expense of a non-standard thread for your successors to deal with when its their turn.</p>
<p>For high tensile bolts your job is much harder and I think the chance of success pretty low - and you will probably need more specialised equipment. A carbide shell type cutter which fits over the bolt thread may allow the nut to be cut away and perhaps a die grinder could be used to clean up the thread so it can be re-cut (but even HSS dies will struggle with high tensile steel).</p>
<p>There are special repair kits available that work a bit like concrete anchors. They have a nut with a segmented sleeve which has sawtooth ridges on the inside and is slightly tapered. You slip it over the (suitably prepared) bolt end and then tighten the nut. As the nut tightens it compresses the sleeve which in turn bites into the bolt. Don't ask me what they are called or who makes them.</p>
<p>There is also some process used in civil engineering that allows a failed steel tie in a blind hole to be repaired in-situ. It must be some kind of welding process but I have no idea how it might work. I believe that it was used to do some repair work on the bridge over the Grafton gully [in Auckland, New Zealand]. I think it is also be used to extend existing concrete bridge structures when the roadway underneath the bridge must be widened.</p>
<p>Source: Ken Mardle.</p>
| 2050 | Can nuts safely be removed from bolts where the threads have been intentionally damaged? |
2015-03-12T16:08:06.983 | <p>I was doing a question involving a T-Beam question wrong since I was selecting the wrong area to use for $Q_{max}$ when trying to maximise the shear stress.</p>
<p>$$\large{\tau}= \frac{VQ}{It}$$</p>
<p>A friend argued that the area both above and below the Neutral Axis should be equal but despite trying multiple times, I have been unable to get the same result. </p>
<p><strong>Is this statement correct?</strong> (and if not, why)</p>
<p>It would seem by argument of equal values both above and below the beam that perhaps it would be correct, even if the beam is unsymmetrical. </p>
<p><img src="https://i.stack.imgur.com/HugpN.png" alt="T-Beam"></p>
| |structural-engineering|beam|statics| | <p>The answer to the headline question is yes, it is equal. It physically has to be equal, since you can't have two different values of a particular directional stress at one point.</p>
<p>The quoted equation is for shear stress, where
V = total shear force at the location in question;
Q = first moment of area 'beyond' the point considered;
t = thickness in the material perpendicular to the shear;
I = Moment of Inertia of the entire cross sectional area.</p>
<p>The problem seems to be the assumption that the cross-sectional area above and below the neutral axis is equal (ie, the embedded question "the area both above and below the Neutral Axis should be equal ... is this statement correct" to which the answer is no)</p>
<p>Neutral axis is the line about which the first moment of area each side is equal, not the line about which the areas each side are equal.</p>
<p>Applying that equation above and below the neutral axis, V an I are properties of the whole section so identical wherever you do the calculation on this cross-section, t is a function of the level at which the calculation is done so identical in both applications, and Q is equal by definition (that's the definition of the neutral axis), so the answer has to be identical since all the numbers substituting into the equation are identical.</p>
<p>Example: for the diagram shown, if T=10, W=60, H=70, then the tee shown has a cross-section of 1200, and both flange and stem each have area 600. However, the neutral axis is NOT at the bottom surface of the flange (ie, is not where there is equal area above and below). In this example, the neutral axis is at 22.5 from the top surface, or 12.5 below the level at which the areas are equal either side. So there's 725 area above the neutral axis, and 475 below the neutral axis.</p>
<p>Calculating for above the neutral axis, there are two rectangular regions,</p>
<pre><code>Q = 10 x 60 x 17.5 + 12.5 x 10 x 6.25 = 11281.25
</code></pre>
<p>Calculating for below the neutral axis,</p>
<pre><code>Q = 47.5 x 10 x 23.75 = 11281.25
</code></pre>
<p>I = 552500 and t = 10 at the neutral axis, so whatever value of V is used, $\tau$ is the same.</p>
| 2059 | Is the Qmax of an T-Beam equal if calculated both above and below the neutral plane? |
2015-03-13T04:00:33.250 | <p>I am building a device that needs to be inside of a microwave oven but somehow needs to communicate with another device outside of the microwave.</p>
<p>How can I reflect microwaves so they don't come into contact with the electronics in the device, while still allowing the device to transmit and receive data?</p>
<p>The frequency of the microwaves in a microwave oven is pretty much the same frequency that is used in <a href="http://en.wikipedia.org/wiki/Bluetooth" rel="nofollow">Bluetooth</a>. I was thinking the only possible way of doing this is to have an antenna that is small enough that it can slip outside of a microwave oven through the door (but small enough so the door is still able to close properly).</p>
<p>Could I use some material to protect the electronics of the device from microwaves? Is there a way to have this device communicate with another device outside of the microwave oven without the microwaves interfering with it?</p>
| |mechanical-engineering|electrical-engineering|materials|telecommunication| | <p><strong>Your best option for wireless communication in this situation is some kind of optical link.</strong></p>
<p><a href="http://en.wikipedia.org/wiki/Microwave_oven#Principles" rel="nofollow noreferrer">Consumer microwaves</a> typically operate at a frequency of 2.45 GHz and <a href="http://en.wikipedia.org/wiki/Bluetooth" rel="nofollow noreferrer">Bluetooth</a> operates from 2.4 GHz to 2.485 GHz, so any shield you design for one will be equally effective for the other. You could design a Faraday cage which gives you enough suppression at 2.45 GHz but still allows enough power at higher frequencies (<a href="http://en.wikipedia.org/wiki/Extremely_high_frequency" rel="nofollow noreferrer">EHF</a> or <a href="http://en.wikipedia.org/wiki/Terahertz_radiation" rel="nofollow noreferrer">THF</a>) to communicate. However, there aren't many devices which operate at those frequencies, and designing one yourself is not a simple feat. An antenna will likely suffer from the same problem. If it transmits at 2.4 GHz it will also receive at that frequency, and the load caused by this will likely also destroy your device. </p>
<p>Instead, you should build a Faraday cage around your device <a href="https://physics.stackexchange.com/a/89090/35024">which suppresses 2.45 GHz but still allows light to pass through</a>. A cage with 0.5 cm holes should give you more than enough suppression since the wavelength of 2.45 GHz is ~12 cm. Building an optical link may sound complicated, but the hardware is fairly cheap and you might be able to steal some of it from a television remote. The <a href="http://en.wikipedia.org/wiki/PSK31" rel="nofollow noreferrer">amateur radio</a> community has some experience building DIY optical communications systems which you can draw on. You can obtain hardware for the encoding and decoding using either the <a href="http://en.wikipedia.org/wiki/PSK31" rel="nofollow noreferrer">PSK31</a> or <a href="http://en.wikipedia.org/wiki/Radioteletype" rel="nofollow noreferrer">RTTY</a> protocols from the same set of devices they use for radio communications.</p>
| 2062 | Protecting device from microwaves in microwave oven.. but allowing Bluetooth? |
2015-03-13T18:26:07.460 | <p>Suppose I want to ship my product via UPS, or whatever other professional carrier service. What vibration and shock forces should I design my product to withstand? And how can I effectively test my product against those forces?</p>
| |vibration|product-testing|shock| | <p>UPS has, or at least had when I was transit testing packaged products, a specific packaging test standard based on <a href="http://www.ista.org/" rel="nofollow">ISTA</a> standards and <a href="http://www.astm.org/Standards/D4169.htm" rel="nofollow">ASTM D4169</a>. There's also a MIL-STD, 810 something, but I don't remember the exact number.</p>
<p>Typically, for a parcel carrier, the test, regardless of standard, consists of environmental conditioning, a "10 pt" drop test and a vibration test. </p>
<p>The environmental conditioning typically represents the extremes of what would be expected. In our case, we conditioned products for 12h @ +150F and -30F with uncontrolled humidity. The referenced standards discuss other conditions such as a high humidity environment. </p>
<p>The height of the drop is determined by the packaged weight. The product is dropped onto a rigid surface without inducing any vertical motion as follows: </p>
<ol>
<li>The bottom vertex of the carton which includes the carton seam. </li>
<li>Each of the three edges radiating from that vertex. </li>
<li>Flat on each of the 6 faces </li>
</ol>
<p>Following that, the product is subjected to random vibration (typically 0.52 gRMS) for 1 hour on each of the three axes with a vibration profile according to <a href="http://www.astm.org/Standards/D4728.htm" rel="nofollow">ASTM D4728</a>, ISTA or a profile developed from data collected by monitoring your particular distribution stream.</p>
<p>As to what forces you design your product to withstand dependent on a lot of things, mainly what's the cost of product breakage? What are the specific stages of getting your product to the customer? How much does it weigh? and so on. How much engineering and money are you willing to put into the packaging to protect the product? Can you make the product itself robust to transportation, or is it delicate and requires significant packaging? Like most engineering undertakings, it's a tradeoff and there's no one good answer that applies to all products. I recommend looking at the referenced standards, and using those as a guide. </p>
<p>If you want to get testing done without buying a test lab of equipment, there are plenty of labs across the country that will help you develop a test plan and perform the testing. I have personal experience working with <a href="http://www.lansmont.com/" rel="nofollow">Lansmont</a>, they make packaging test equipment, perform testing in house and provide consulting on packaging design. (<strong>Note:</strong> My relationship with Lansmont is strictly as a prior customer, I have no personal, financial or other ties to them). They are great to work with. UPS also has a test lab and provides packaging design support.</p>
<p>I could go on and on about this stuff, but this is already long enough. If you have more questions, want further information or clarifications, I'll be glad to answer them.</p>
| 2068 | Are there standard vibration and shock specs for shipment of a product? |
2015-03-13T19:11:53.423 | <p>A common method of repairing old <a href="http://en.wikipedia.org/wiki/Rivet#High-strength_structural_steel_rivets" rel="noreferrer">riveted structures</a> is to <a href="http://www.state.nj.us/turnpike/documents/DCA2011SS-13-PAGES.pdf" rel="noreferrer">replace any damaged rivets with new high strength bolts</a>. </p>
<p>When this is done, the new bolts are typically installed like they would be on a new structure which results in a bolt pretension. My concern is that the one bolt location might be effectively slip-critical while the rest of the existing rivets are something less than slip-critical. This may concentrate stresses on the one bolt.</p>
<p><strong>When new high strength bolts are installed within an array of existing rivets, is there a concern with stress concentrations resulting from the increased clamping force from the bolt?</strong></p>
| |structural-engineering|bolting|rivets| | <p>According to AISC J1.9 rivets can be considered as sharing the load with high strength bolts, specifically in joints designed as slip-critical as per J3. The apply this section to both new work and alterations.</p>
<p>The commentary on this section points out that this is because the ductility of the rivets.</p>
<p>This means that you do not need to be concerned with the distribution of load across the various fasteners as long as your holes are appropriately sized for slip critical connections. What's not clear to me is if the faying surface requirements of RCSC for slip critical connections would apply.These would obviously be hard or impossible to verify when replacing just some fasteners in an existing connection.</p>
<p>The second page of <a href="http://msc.aisc.org/globalassets/modern-steel/steel-interchange/2004/2004v09_si_web.pdf" rel="nofollow">this</a> Modern Steel Construction article suggests that while the riveted connection is not slip critical, simply tightening the new bolts as if it were is enough to consider the bolts and rivets to share the load, but won't make the joint slip-critical as a whole.</p>
<hr>
<p>The <a href="http://www.boltcouncil.org/files/2ndEditionGuide.pdf" rel="nofollow">RCSC Design Guide</a> (PDF link) for bolted and riveted connections has a more detailed explanation in section 14.2.2, which suggests that in connections that are under significant load, the AISC rule may be a simplification.</p>
| 2069 | Are stress concentrations a concern when replacing rivets with high strength bolts? |
2015-03-13T21:26:35.323 | <p>I was reading up on <a href="https://en.wikipedia.org/wiki/Fly-by-wire" rel="noreferrer">fly-by-wire</a> development, and I saw a short section about <a href="http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20070013704.pdf" rel="noreferrer">fly-by-wireless</a> technology. It seems like a great idea, with the potential to lower costs, weight and complexity. I can see a possible scenario where it could be an issue, though:</p>
<ol>
<li>Two aircraft are very close together (e.g. on a runway or flying in formation).</li>
<li>One pilot transmits commands through the aircraft's fly-by-wire system to other parts of the aircraft. </li>
<li>The other aircraft accidentally receives the signal because it is so close.</li>
<li>Things get very bad very quickly.</li>
</ol>
<p>The thing is, I haven't been able to find any technical specifications regarding fly-by-wireless systems, and I have no idea if the transmission would be powerful enough to reach the other craft, nor if it would then be interpreted as actual data sent from <em>that</em> aircraft's pilot.</p>
<p>Is this cross-interference between fly-by-wireless systems possible? If so, how can it be mitigated?</p>
| |electrical-engineering|aerospace-engineering|emc|telecommunication|product-testing| | <p>No, crossed communications would be a total non-issue.</p>
<p>Think of it like a home wireless network. All your devices at home can talk to each other on your WiFi network, and the same is true for your neighbour. But your devices can't talk to your neighbour's devices because they are on a different wireless network.</p>
<p>Such wireless networks in planes would have to be far more robust than your typical home network, as there are other potential issues:</p>
<ul>
<li>Accidental interference from another craft that doesn't cross-communicate between craft, but potentially blocks communication on one or both networks.</li>
<li>Deliberate interference (jamming) of the communications by for example a terrorist on the plane or nearby.</li>
<li>Deliberate interception or spoofing of communications to hi-jack the plane. This would require hacking into the secure network and would therefore be much harder than jamming.</li>
</ul>
<p>Some ideas to mitigate these problems might be:</p>
<ul>
<li>Directional antennae on the wireless equipment. This could help to make jamming or interception much harder because other signals would be blocked so any would-be hacker would need to have their equipment physically located directly between the antennae.</li>
<li>Multi-part encryption keys and certificates in the devices and stored on a card held by the pilot. This would allow the equipment to verify commands were coming from a verified source (the pilot/cockpit) and also ensure that nobody else could listen in or intercept/spoof the commands.</li>
</ul>
| 2071 | Is interference between aircraft an issue for fly-by-wireless technology? |
2015-03-14T01:30:55.137 | <p>I am a programmer facing an electronic question, so I thought here is the place to ask!</p>
<ul>
<li>I have a magnetic field sensor which provides me with magnetic field values (XYZ axis) 250 times a second. </li>
<li>An electronic circuit with a programmable microprocessor controls a coil, which can change the magnetic field close enough to this sensor. Yet, the two devices are not strictly positioned, so I cannot relay on positioning measurement.</li>
<li>I wish to send 2 different types of signals from the circuit to the sensor, which can withstand strong noises on the magnetic field, and be shorter than half a second (Yes, seconds - we are in 250 Hz !)</li>
</ul>
<p>Current solution is to transmit a 17Hz square wave, then 12Hz, then 17Hz for signal A, reversing the order for signal B. But in order to detect these signals it is needed to send long enough waves, which brings the signals to length around 1.5 seconds.</p>
<p>So, my question here is: Is there a way, let's say a pattern, like musical pattern that can be used to signal faster, and still be reliable?</p>
| |electrical-engineering|signal|magnets| | <p>With a sample rate of 250 Hz, you could easily send 100 bits/second or more. In 0.5 second, that would be on the order of 50-60 bits.</p>
<p>The question is, what bit patterns should you select that are readily distinguishable from each other and also from outside noise and interference? A common solution is to use pseudorandom bit sequences, also known as <a href="http://en.wikipedia.org/wiki/Barker_code">Barker Codes</a>, that have the desired low cross-correlation.</p>
<p>A DSP technique known as "<a href="http://en.wikipedia.org/wiki/Matched_filter">matched filter</a>" can be used to detect the codes.</p>
| 2074 | Best way to send a signal in a very low frequency (250Hz)? |
2015-03-14T12:39:34.807 | <p>Trucks and smaller vehicles with diesel engines run on diesel fuel which looks more or less like kerosene - less flammable than gasoline, low viscosity liquid. However larger diesel engines <a href="https://en.wikipedia.org/wiki/W%C3%A4rtsil%C3%A4-Sulzer_RTA96-C" rel="nofollow">like this 120 thousand horsepower marine engine</a> have more or less the same design but use fuel oil which looks very different from diesel fuel - much higher viscosity and I'd guess igniting fuel oil at room temperature is a challenge.</p>
<p>How does it happen that engines of the same design use two different fuels? If one of them is superior to the other then why won't they all stick to that superior fuel?</p>
| |mechanical-engineering|combustion|engines|diesel| | <p>Diesel and heater oil are the same thing , the tiny difference is the additive package ( a fraction of 1 % of the total). Diesel may have a better additive to prevent wax from crystallizing and plugging lines in cold weather , for example. From time to time in various locations there may be different limits on sulfur ; depending on the refinery this may or may not result in different S levels in the products. Heavy marine fuel is a different from diesel because marine engines have heaters to reduce the viscosity of heavy oils like "bunker C".</p>
| 2081 | Why do some diesel engines run on diesel fuel and others on fuel oil? |
2015-03-14T12:58:17.830 | <p>Suppose I have two parts with holes aligned and I have to install a solid rivet into the "combined" hole.</p>
<p>Clearly the rivet diameter must be slightly smaller than the hole diameter, otherwise it simply won't fit. So I heat this slightly smaller rivet, insert it into the hole and then deform its tail so that the rivet now has two heads and sit properly in place.</p>
<p>Assume I follow all the procedures and best practices.</p>
<p>Does the deformation only affect the tail or does it also make the middle of the rivet expand and fill the gap between the hole walls and the rivet?</p>
<p>In other words, if I wait till the rivet cools down and then grind the rivet head off - will it leave its place easily or will I have to force it out?</p>
| |structural-engineering|steel|rivets| | <p>The short answer is: <strong>Yes, as the rivet head is formed, the shank of the rivet is also deformed to fill the hole.</strong></p>
<p>Just like structural bolts, rivets are installed in holes that are 1/16 inch larger in diameter than the rivet. This clearance allows for manufacturing imperfections and clearance for easily placing the rivet in the hole.</p>
<p>A US military specification for riveting (<a href="http://www.engineersedge.com/MIL-R-47196A.pdf" rel="noreferrer">MIL-R-47196A</a>) states in 3.3.3.3:</p>
<blockquote>
<p>The driven rivet shall completely fill the hole...</p>
</blockquote>
<p>The figures in that MIL Spec show that rivets that do not completely fill the hole are one of the ways that a rivet can be unacceptable.</p>
<p>An image of this from a <a href="https://mechanicalinfo.wordpress.com/tag/cold-riveting/" rel="noreferrer">web page on riveting</a> shows this as well:</p>
<p><img src="https://i.stack.imgur.com/8Vu11.jpg" alt="Rivet diagram" /></p>
<h3>Video of Hydraulic Riveting</h3>
<p>You can also see <a href="https://www.youtube.com/watch?v=I9Q5OHYDbvo" rel="noreferrer">video of a hydraulic rivet on YouTube</a> that shows just how much deformation is involved with riveting.</p>
| 2082 | Do solid rivets fully fill the holes when installed properly? |
2015-03-14T13:08:37.043 | <p>AFAIK when you heat up a piece of steel to high temperature and let it cool down slowly (as opposed to shock cooling by say submerging it into water) steel gets softer - that's called tempering. Steel getting softer means it can bear less load without getting deformed.</p>
<p>Suppose we connect parts of steel skyscraper with steel rivets. We heat them up in fire and then put them into holes and deform their tails. While the rivet is inserted inside a rather cool surrounding construction (and being deformed) it cools down which more or less resembles tempering process.</p>
<p>So it looks like installing rivets leaves them tempered and so rather soft and prone to deformation.</p>
<p>How is this process of rivets getting softer and "weaker" accounted for?</p>
| |structural-engineering|steel|rivets| | <p>Structural steel are not usually intended to be heat-treated. </p>
<p>Heat treatment involves heating the steel beyond a critical temperature (typically around 850 °C) and then cooling it rapidly to change its crystal structure. This results in a 'stressed' structure which tends to increase tensile strength and hardness at the expense of toughness and ductility. The tempering process involves reheating hardened steel to a very specific temperature (typically in the rang 180-300 °C) to partially reverse this process with the specific tempering temperature allowing fine control over the balance between hardness and toughness. </p>
<p>However if the steel has not been quench-hardened in the first place then heating and slow-cooling will not have any great effect on its mechanical properties.</p>
<p>To put this another way tempering and annealing reverse the effect of quench hardening but won't soften steel which hasn't been hardened. </p>
<p>Generally structural steels are designed to have good tensile strength and ductility in a non-heat-treated state. </p>
<p>Clearly there is little point in hardening rivets for either hot- or cold-forming so once they cool they just go back to the state they were in before they were heated.</p>
| 2083 | How is steel tempering addressed when installing pre-heat rivets? |
2015-03-14T14:57:35.473 | <p>I was once told that the speed limits on banked highway curves (specifically off-ramps) were determined by assuming zero friction between the car and the road, such that as long as you stayed on the correct path of travel, the banking of the curve would prevent a vehicle from sliding outwards. That is, the angle of the banking would mean that the portion of your gravity vector normal to the road surface provides enough centripetal force without counting on your wheels to provide any 'sideways' friction.</p>
<p>I am not a civil engineer and have no familiarity with the relevant codes, but I was wondering if this is true, or a myth. If it's not true, is there a simple formula used to set these speed limits, or is it a much more complicated procedure/judgment?</p>
| |civil-engineering|highway-engineering|transportation| | <p>This information comes from a <a href="http://www.iowadot.gov/design/dmanual/02a-02.pdf">design document</a> by the Iowa DOT (US). It might not apply everywhere in the world, but the considerations are probably universal.</p>
<p>For clarity, the amount of banking of a turn is typically called <em>super elevation</em>. At least in the US, this is given as a percentage for roadways.</p>
<ul>
<li>The <strong>maximum</strong> super elevation is <strong>8%</strong>.</li>
<li>The typical <strong>high-speed</strong> super elevation is <strong>6%</strong>.</li>
<li>The typical <strong>urban</strong> super elevation is <strong>4%</strong>.</li>
</ul>
<p>The <strong>6%</strong> typical super elevation was chosen for:</p>
<blockquote>
<p>This reduces the risk of slow moving vehicles sliding down a superelevated roadway during winter conditions.</p>
</blockquote>
<p><strong>Speed</strong></p>
<p>Design speed of a highway is not always the same as the posted speed. Drivers have this habit of not following the speed limit. Designers know this and make the roadway safe at these increased speeds as well.</p>
<p>There are only two ways that speed will affect the curve radius. Either:</p>
<ol>
<li>The design speed has been set and the geometry limits are based off of that speed.</li>
<li>The geometry has been set and the design speed is reduced to still provide adequate levels of safety.</li>
</ol>
<p><strong>Curve Radius</strong></p>
<p>Both the super elevation and design speed are taken into consideration to determine the minimum radius of the curve. These also consider the maximum side friction from both the vehicle's tires and the comfort of the driver. These values are listed in tables for easy look up or in a spreadsheet.</p>
<p>The friction of the vehicle's tires on the road is incredibly variable. It depends on the weather, condition of the road, condition of the tires, etc. Because of this, the values used for design are very conservative.</p>
<p>The other factor controlling curve design is driver comfort. It is surprising how ofter roadway design is controlled by human factors. In this case, the typical driver will be concerned about the amount of horizontal acceleration they are feeling and slow down before the vehicle is in danger of sliding sideways.</p>
<p><strong>Result</strong></p>
<p>So in the end, there are limits on the super elevation of curves that are based on low friction. But this is the controlling factor at low speed and not high speed.</p>
<p>These super elevations are no where near the amount that would be required for no side friction. As you can imagine you never know when there will be a traffic jam on a road because of a wreck. You don't want trucks to tip over toward the center of the curve if they are forced to stop!</p>
<p>The minimum curve radius is then determined based on looking up the information in tables. Often there are other concerns that also might after the curve such as sight distance or vertical curves.</p>
<p><strong>How it works in design</strong></p>
<p>Setting the roadway cross section details (of which super elevation is one) is usually done last in the sequence of design. The sequence usually goes like this:</p>
<ol>
<li>Set design details, e.g. design speed, road type, number of lanes, etc.</li>
<li>Set the horizontal alignment. This would be where the curve radius is set. This is also where geometric constraints are found that might require reducing the design speed.</li>
<li>Set vertical alignment. This stage can also reveal more geometric constraints.</li>
<li>Set details of cross section, e.g. super elevation, cross slopes, embankment slopes, ditches, etc.</li>
</ol>
| 2085 | How are maximum speed limits determined for banked curves? |
2015-03-14T21:32:54.417 | <p>Some roadway intersections that are near railroad tracks have signs that light up when a train is approaching. These signs warn that certain turns are not allowed because of the train. One of these signs is shown below from the <a href="http://www.edmonton.ca/transportation/ets/safety_security/lrt-traffic-signs-and-signals.aspx" rel="noreferrer">City of Edmonton</a>.</p>
<p><img src="https://i.stack.imgur.com/WDkw5.jpg" alt="No right turn on train sign"></p>
<p>My understanding is that normal train crossing signals are the responsibility of the railroad and that traffic signals are the responsibility of the highway department. This wouldn't normally seem like a big problem, but these are two completely different systems and organizations.</p>
<p>Obviously there is some way that the two owners coordinate and communicate the train warning information.</p>
<p>How does the train warning signal get passed from the railroad signal to the traffic signal? Is this as simple as a wire run from the train signal that is "high" while lights are on? Is there standard way that this connection is done?</p>
| |rail|traffic-intersections|traffic-light| | <p>The answer is slightly more complex than what the OP proposes -- this is due to two factors:</p>
<ol>
<li><p>The interconnect must be <em>fail safe</em> -- i.e. if the interconnect circuit fails, the system must be able to detect this and report a problem to both highway and railroad maintenance crews. To this end, special supervised circuits are used, using two relays in opposite states to detect a failure of any single relay to operate correctly.</p></li>
<li><p>The interconnect must trigger the traffic light preemption cycle sufficiently before the railroad crossing signals trigger to allow queues to clear off the track before the train arrives. This may require both extended advance warning times (in excess of the 20 second regulatory minimum) and the use of advance warning outputs from the railroad grade crossing predictor -- just about all constant warning time predictors support advance preemption, but older DC or AC/DC three-track-circuit systems and Audio Frequency Overlay detectors may not.</p></li>
</ol>
<p>Furthermore, there are special programming concerns on the highway side as well -- the grade crossing preemption cycle must be set up correctly to provide sufficient track clearance green before it blocks all movements over the grade crossing, either using red arrows or illuminated blank-out signs to prohibit turning movements onto the tracks, while also allowing pedestrians to clear from the railroad crossing (this can be critical in geometries where the railroad crossing bisects or is in close proximity to the intersection).</p>
<p>Preemption reservice (where the traffic signal controller handles a second preemption request that closely follows the first) and extra logic on both sides (highway and rail) may be necessary when multiple tracks or switching movements are involved, and coordinated signalization and preemption are needed when multiple traffic signals are in close proximity to the grade crossing. Diagonal crossings (either cutting multiple approaches or bisecting the intersection) pose even more challenges, sometimes requiring multiple grade crossing predictors to be interconnected to each other as well as to the highway traffic signal controller.</p>
<p>If you want more gory detail on the intricacies of this than you ever need, the FHWA has an <a href="http://safety.fhwa.dot.gov/hsip/xings/com_roaduser/07010/sec04b.cfm#i" rel="nofollow noreferrer">excellent handbook chapter on this topic</a>. In particular, Exhibit 3 in the pre-emption sidebar (reproduced below) provides a sample diagram of a fail-safe interconnect relay circuit.</p>
<p><a href="https://i.stack.imgur.com/LfOSR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LfOSR.jpg" alt="FHWA Highway-Rail Grade Crossing Handbook, Exhibit 3"></a></p>
| 2091 | How are train crossing signals and highway traffic signals coordinated? |
2015-03-15T03:24:33.990 | <p>I'm an artist building a project which one could think of as a "sliding door" – a very narrow panel running on grooved wheels riding in tracks at top and bottom. It will be motorized and driven by a stepper motor, timing belt and pulleys. I will probably end up driving it from top and bottom too so it doesn't rack. </p>
<p>This is the gist of it:</p>
<p><img src="https://i.stack.imgur.com/drVuw.jpg" alt="enter image description here"></p>
<p><strong>My question:</strong> does it matter if the load is on the top set of wheels or the bottom set? </p>
<p>I realize this is a dead simple "more efficient to push or to pull" type question but the answer is not obvious to me.</p>
| |mechanical-engineering| | <p>Driving from the top is better. If you think about it on an intuitive level, when you pull from the top, the weight of the unit 'wants' to stay under the wheels, so your door tends to stay straight up and down. If you pull from the bottom, the weight of the door is tempted to lag behind and tip over, causing the whole mechanism to bind. This is because the center of mass of the door lags slightly behind the driven end as the door starts and stops moving.</p>
<p>If everything were perfectly sized and aligned, this wouldn't be a problem, but in practice, driving from the bottom will be more difficult.</p>
<p>One thing to keep in mind is the risk of over-constraint. If the tracks at the top and the bottom are not very well lined up with each other, the wheels are likely to get stuck. In entertainment, we build mechanisms like this all the time (we call them 'elevator doors' or 'sliders.') We would typically detail this so the top is driven, and supported by two or more wheels on a track that is strong enough to support the whole door. At the bottom, we'll run a tab (probably made of steel) through a slotted block (usually made of plastic - UHMW for example.) The tab at the bottom is just to keep the door from getting pushed in a way that would twist the track at the top, it doesn't actually support any of the weight.</p>
<p>Here is a sketch of what that solution looks like in cross section. It can be scaled to any size, but the dimensions shown are easy to do with minimal tooling. <img src="https://i.stack.imgur.com/yFta6.png" alt="enter image description here"></p>
<p>If you do something similar, you can make the slot bigger than the tab, reducing the risk of binding. One other strategy, if you have enough space above and below the door, is to use a <a href="https://www.mayline.com/sft277/651.pdf" rel="nofollow noreferrer">mayline</a> <a href="https://books.google.com/books?id=A30rBgAAQBAJ&pg=PA402&lpg=PA402&dq=mayline%20cable%20rigging&source=bl&ots=wyptc4dVsW&sig=iYt5n2TLvxDNo5w2eA1TyJvSCkE&hl=en&sa=X&ei=LR4FVf3cMYOuogSl7oA4&ved=0CEIQ6AEwAw#v=onepage&q=mayline%20cable%20rigging&f=false" rel="nofollow noreferrer">mechanism</a> which uses two opposing cables to keep the door square. These are more common for things that move vertically, but can be used horizontally as well.</p>
| 2094 | "Sliding door" - is top weight bearing more efficient? |
2015-03-15T06:35:49.177 | <p>In designing mechanical parts in an assembly, sometimes the mating surfaces are raised faces. I've been told it's to provide a better seal. However, for piping (flanges), it seems to be the consensus that a raised face flange does not provide any extra seal functionality, but has more to do with the manufacturing method (forged vs cast).</p>
<p>My question doesn't necessarily apply just to piping flanges, but to designing parts in general. Is it desirable to use a raised face on mating parts, and when (always, sometimes, etc)?</p>
| |mechanical-engineering|design| | <p>In general you would raise mating surfaces in order to make them easier to machine.</p>
<p>Having the surface raised above the surrounding material means that you don't have to worry so much about tools colliding with edges of the feature and it's fairly intuitive that skimming a raised surface is a simpler operation than machining a recessed pocket and it minimises the amount of material that you need to remove. </p>
<p>Similarly if you are machining a raised surface with a bit of 'meat' to it you can concentrate on surface finish and flatness without too much concern about the precise depth of cut and you have some material to work with if you need to rework a part during original manufacture or in-service maintenance/repair. </p>
<p>A raised surface also means that the overall dimensions of mating parts can be less precise as slightly oversized parts will just overlap rather than interfering with anything. </p>
| 2096 | Why raised face for mating parts? |
2015-03-15T23:34:27.277 | <p>I found the following comments on direct and indirect control in <a href="http://www.elsevierscitech.com/pdfs/JPC_SurveyPrize.pdf">"Survey on iterative learning control, repetitive control and run-to-run control." by Wang, Gao, and Doyle</a>:</p>
<blockquote>
<p>There are two application modes to use the learning-type control. First, the learning-type control method is used to determine the control signal directly, and this kind of learning-type control is called direct learning-type control. Second, there is a local feedback controller in each cycle and the learning-type control is used to update the parameter settings of the local controller, so this kind is called indirect learning-type control. The methods that can be used for designing direct learning-type control and indirect learning-type control will be discussed in Section 4 and in Section 5, respectively.</p>
</blockquote>
<p>What is the difference between direct and indirect learning control? My understanding is that in indirect control you can alter the control parameters and the input signal as opposed to just the input signal in direct control; is this correct? I also do not understand the significance of a "local" controller.</p>
| |mechanical-engineering|control-engineering| | <p>I have only skimmed through the paper. The terminology is put forward by the paper to organize it.</p>
<p>Figure 6 in the paper is the key (bottom of page 1595). That is a diagram of indirect learning control.</p>
<p>Here, the learning controller takes two inputs: one is the system output, and the other is the output of the 'local controller'.</p>
<p>What is the local controller? The paper says: "In principle, any real-time feedback control law can be chosen as the local control." It seems that the local controller is a real-time feedback controller, such as a PID controller (<a href="https://en.wikipedia.org/wiki/PID_controller">proportional-integral-derivative controller</a>), but it could be more complex, or even a simple open loop controller. The feedback is shown in the diagram, where the output of the system is fed into the local controller.</p>
<p>So, what is the other input to the local controller? That is the output of the learning controller. For the purpose of this paper, it is whatever input the designer considers to be useful. The examples in the paper range from setting the duration of a batch process, through setting the references of a PID controller, to updating parameters of advanced controllers such as model-based or neural network controllers.</p>
<p>The paper describes direct learning control as an arrangement without the intervening, 'local', real-time controller. In other words, the learning controller is connected directly to the process or system under control. <strong>Update:</strong> This direct connection will modify the setpoint of a real-time feedback controller.</p>
| 2101 | What is the difference between direct and indirect learning control? |
2015-03-16T04:13:59.720 | <p>In case of Uranium, things are lot easier, because $UF_6$ is gaseous from around 60 $\,^{\circ}\mathrm{C}$.</p>
<p>I know, that in most cases, plutonium isotope separation is not needed, because there is no need to separate its fissile isotopes. But there is an exception: <a href="https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator" rel="nofollow noreferrer">radioisotope thermoelectric generators</a> (used to power remote devices such as spacecraft and unmanned polar equipment) need ${^{238}}Pu$.</p>
<p>I think, Plutonium can be separated from the depleted fuel cells with <a href="https://engineering.stackexchange.com/questions/26/how-does-nuclear-fuel-reprocessing-work">chemical methods</a>, but how to separate ${^{238}}Pu$ from the result?</p>
| |nuclear-technology|chemical-engineering|nuclear-reprocessing| | <p>Pu238 is not usually isotopically separated from spent nuclear fuel for exactly the reason you pointed out, it would be very difficult. Instead most of what we have comes from one of two different processes.</p>
<p>The first is bombardment of Np237 (also made in a nuclear reactor) with neutrons. It will become Np238 and then undergo a beta emission (fancy way of saying it spits out an electron from one of the neutrons in the nucleus) forming your Pu238.</p>
<p>The other option is to bombard Am241 with neutrons to produce Am242, (excluding the metastable nuclei) this in turn will do another beta decay (like our Np238) and turn into Curium-242 ( Cm242 ) . Cm242 is unstable and will decay by alpha emission (fancy way of saying it coughs out a helium nucleus) thus forming our Pu238.</p>
<p>And that is about it, there are other paths to Pu238 but these are the easiest to perform so far. Eventually we will run a thorium fuel cycle, when that happens there will be an abundance of this radioisotope.</p>
<p>Hope this helps!</p>
| 2104 | How to separate isotopes of Plutonium? |
2015-03-16T16:49:13.207 | <p>I have a vibration motor contained in an object. The motor looks similar to this:</p>
<p><img src="https://i.stack.imgur.com/ww4Eq.jpg" alt="Typical vibration motor"></p>
<p>When the motor is running, I want to simulate the movement of the container object that results from the vibration of the motor. I'm able to model the parts/assemblies in SolidWorks. How can I model or simulate the movement of the motor and the container object?</p>
| |mechanical-engineering|vibration|modeling|simulation| | <p>@Narada is correct, but I their answer is a bit short on detail, so I will expand on it.</p>
<h1>Calculating Force</h1>
<p>First, you can calculate the force that the motor exerts on the containing object from the acceleration of the mass at the end of the motor shaft. As the mass rotates, it accelerates toward the motor shaft at a rate $\omega^2 y$, where $\omega$ is the angular velocity and $y$ is the distance from the motor shaft to the <a href="https://en.wikipedia.org/wiki/List_of_centroids" rel="nofollow">centroid</a> of the mass. If we assume that the mass is a perfect semi-circular prism, then $y = \frac{4R}{3\pi}$, where $R$ is the radius of the mass.</p>
<p>Once the acceleration is known, the force can be computed by Newton's Second Law, $F = ma = m\omega^2 y = \frac{4mR\omega^2}{3\pi}$.</p>
<p>The direction of said force is always directed toward the motor axis, so as it spins it alternates between purely horizontal force and purely vertical force (or two different horizontal directions if the motor is mounted standing up). We can decompose the force into vertical and horizontal forces by using sines and cosines, and realizing that the angle of the mass is equal to $\omega t$, where $t$ is the elapsed time.</p>
<p>$$F_{x} = \frac{4mR\omega^2}{3\pi} \cos{(\omega t)}$$
$$F_{y} = \frac{4mR\omega^2}{3\pi} \sin{(\omega t)}$$</p>
<p>Here $F_x$ is the horizontal force and $F_y$ is the vertical force.</p>
<h1>Simulating Vibration</h1>
<p>If you have the Solidworks simulation software, you can apply the $F_x$ and $F_y$ loads at the point of the vibration motor and Solidworks should give you the vibration behavior of the structure.</p>
<p>Alternately, if you just want to know how far the vibration motor will move your object and the vibration motor is mounted close to the center of the object, you can just use Newton's laws of motion and treat your whole object as some mass with an applied force. </p>
<p>Noting that the angular velocity must be the same for the vibration motor and the object (assuming a rigid body with no other vibration motors) one can see that $m_{motor}y_{motor} = m_{obj}y_{obj}$. Since all but $y_{obj}$ are known, you can easily get an estimate of how far the object will move. The object will alternately move to $\pm y_{obj}$ in both directions.</p>
| 2107 | How can I model or simulate motion caused by an internal vibrating motor? |
2015-03-17T02:19:35.227 | <p><strong>Construction Joints</strong></p>
<p>Construction joints in concrete footings or walls are places where the concrete pour can be stopped so that everything doesn't have to be poured at once. Later, once the concrete has hardened, the next part of the concrete can be placed against the previous pour. These locations are sometimes planned for in the design, and special details are added to the plans to show how the construction joint is to be formed.</p>
<p>A typical detail of one of these joints from <a href="http://epg.modot.org/index.php?title=751.34_concrete_pile_cap_non-integral_end_bents" rel="nofollow noreferrer">MODOT</a> is shown below:</p>
<p><img src="https://i.stack.imgur.com/QqZXw.jpg" alt="Typical construction joint details"></p>
<p>These joints have a "key" formed in them so that when the next batch of concrete is placed, there isn't a flat shear plane across the entire wall thickness. The "key" is supposed to help to transfer shear across the joint. In the US, the blockout is typically formed by placing a 2x4 (2 in x 4 in nominally) board in the fresh concrete.</p>
<p>Sometimes a rubber or plastic waterstop is cast into the concrete at these locations to help keep water from flowing through the joint.</p>
<p><strong>Shear key design</strong></p>
<p>While I have seen this detail used many times, I have never seen a calculation done to design it. This exact same detail is used no matter what the thickness of the wall is or the shear force that may be acting on the wall.</p>
<p>How do you design (or check) a shear key like this? </p>
<ul>
<li>Do you only consider shear-friction across the joint, ignoring the key?</li>
<li>Do you only consider shearing of the concrete in the key?</li>
<li>Is it some combination of the two?</li>
</ul>
| |civil-engineering|structural-engineering|reinforced-concrete| | <p>Typically one does not consider the key when checking a slab - footing interface. This is based on ACI318 Section 15.8.1.4, which discusses the transfer of lateral forces to footings and directs the reader to Section 11.6. 11.6 which says this,</p>
<blockquote>
<p>11.6.3 A crack shall be assumed to occur along the shear plane considered.</p>
</blockquote>
<p>Therefore, you do not consider the key to be acting in any way structurally because you are to assume that a crack will form at the plane defining the intersection of the wall and footing.</p>
<p>Section 11.6.4 goes on to discuss how the nominal shear-friction capacity of the cross section is calculated, what coefficient of friction to use, and how much the concrete should be intentionally roughened at the cross section.</p>
<p>In my experience, these keys are typically added (by the request of the architect) to combat water intrusion at the concrete joint.</p>
| 2112 | Design behind typical keyed construction joints in concrete walls |
2015-03-17T16:01:04.487 | <p>We have a simple class 1 lever:</p>
<p>$$\begin {array}{c}
\text {5,000 kg} \\
\downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \\
==================== \\
\hphantom {==} \triangle \hphantom {==============} \\
\vdash \text{1 m} \dashv \vdash \hphantom {======} \text{4 m} \hphantom {======} \dashv \\
\end {array}
$$</p>
<p>The lever ($===$) is 5 m long. The fulcrum ($\triangle$) is 1 m from one end of the lever. The lever has a object sitting uniformly upon it weighing 5,000 kg.</p>
<p>How do I compute the upward force that needs to be exerted at the end of the 1 m side of the lever to keep the lever stationary? $F = (W \times X)/L$ is simple when the weight is applied at the very end of the lever. But what happens if the weight is distributed along the lever?</p>
<p>Our final goal is to tether the free end (on the 1m side) to keep the lever level and we need to know how strong the tether should be.</p>
| |mechanical-engineering|statics| | <p>A uniformly distributed load can be considered to act in its centre. Working in kg and m:</p>
<p>Clockwise moment about the left hand end = 5000 * 2.5 = 12500
Anticlockwise moment about the left hand end = F * 1 (where F is the reaction at the fulcrum)</p>
<p>These must be equal for it to be balanced, giving F=12500kg</p>
<p>Resolving vertically (the total downward force must equal the total upward force), taking T as the reaction on the tether: T + 5000 = 12500, therefore T = 7500kg.</p>
<p>Or converting into N (as you say you want a force, and kg is mass not force) then T = 7500 * 9.81 = 73575N = 73.6kN</p>
| 2114 | How to calculate lever force when lever has uniformed distributed load? |
2015-03-17T20:04:10.807 | <p>I have seen several block diagrams with diagonal inputs, drawn with an arrow through the block.</p>
<p><img src="https://i.stack.imgur.com/tHTpI.png" alt="enter image description here"> <a href="https://www.google.com/search?q=indirect+adaptive+control&biw=1550&bih=720&source=lnms&tbm=isch&sa=X&ei=CYYIVbfkFImegwT41oHwCA&sqi=2&ved=0CAcQ_AUoAg#imgdii=_&imgrc=gULVia1RSqWyPM%253A%3BgD-PX1e3g17aiM%3Bhttp%253A%252F%252Fieeexplore.ieee.org%252Fiel5%252F5%252F20157%252F931488%252F1104583.gif%3Bhttp%253A%252F%252Fieeexplore.ieee.org%252Fiel5%252F5%252F20157%252F931488%252F931488.plain.html%3B444%3B246" rel="noreferrer">and others like that (via Google image search)</a></p>
<p>What is difference between this and a typical signal input? What are the dotted lines signifying?</p>
| |mechanical-engineering|electrical-engineering|control-engineering|systems-design| | <p>The diagonal input generally means that it changes some parameter of the block, such as its gain, without otherwise becoming part of the signal that the block is processing.</p>
| 2120 | What is the significance of a diagonal input in a control diagram? |
2015-03-18T10:44:46.117 | <p>I think, it could lead to, for example, very environment friendly <em>and</em> very efficient urban traffic.</p>
<p>But I also think, it required very precise part construction.</p>
<p>Do such things exist? At least in plan?</p>
<p><em>Extension:</em> Clarification added, waiting for acceptable answer again.</p>
<p><em>Extension #2:</em> Despite the "common sense", diesel engines aren't so simple. Although there is no spark, there is a part which injects the fuel in exactly the perfect moment into the piston. This happens with around 2000atm pressure. Maybe this would be hard in case of gases.</p>
| |mechanical-engineering|combustion|diesel| | <p>If you count gasoline as "gas", many car manufacturers are already testing homogeneous charge compression ignition engines. It is where they use the compression stroke to ignite the fuel mixture in the cylinder. The fuel is not direct injected, it is premixed in the intake charge before it makes it into the cylinder, so diesel type direct injectors are not necessary. </p>
<p>The same technology could be used for other gaseous fuels such as compressed natural gas.
Such as the experimental compression ignition natural gas engine made by toyota, referenced here <a href="http://papers.sae.org/2007-01-0176/" rel="nofollow">http://papers.sae.org/2007-01-0176/</a></p>
<p>So such things do exist, though uncommon.</p>
| 2124 | Does a diesel-like (ignition by compression) engine running on gaseous fuel exist? |
2015-03-18T10:48:17.857 | <p>I am new to mechanical engineering, although I have a scientific background (postgrad in Mathematics), and I (mostly) code for a living.</p>
<p>I have an idea about creating a mechanical device; I envisage it will entail gears, linkages and actuators.</p>
<p>I have a rough idea where things will fit but I would like to be able to test and tweak the design in software before building the actual device. As a point of clarification, when I say "test," I mean view via animation, for example, whether two items will collide when in motion, or if there is sufficient leeway between them as they move past each other.</p>
<p>This allows me to do the design and simulation testing of the parts, before finalising the design and then building the physical system from the design.</p>
<p>The stages are:</p>
<ol>
<li>Build the 3D design in software</li>
<li>Run simulation to see if it "works," if not fix design and iterate</li>
<li>Build physical system from design that "works"</li>
</ol>
<p>I have figured out that the system consists of three subsystems working together.</p>
<p>So, I would like to design and test each sub-component before integrating them into the complete system.</p>
<h3>My question then is this:</h3>
<ul>
<li>Is this how design is done in the real world?</li>
<li>What are the pros and cons of the scheme I have planned?</li>
</ul>
<p>I am intending to use FreeCAD to do the design and testing.</p>
| |mechanical-engineering|systems-design|computer-aided-design| | <blockquote>
<p>I am new to mechanical engineering, although I have a scientific
background (postgrad in Mathematics), and I (mostly) code for a
living.</p>
</blockquote>
<p>My observation is that there are plenty of analogues between how software is designed, tested for, coded (heh, in that order), and distributed - and how mechanical models are described in CAD. Maybe this is because most of what I do is parametric modeling, e.g. describing the computer a recipe on how to produce a model from parameters, that can then be tested.</p>
<p>You certainly sound to be on the right tracks.</p>
<p>As with software tools, you need to pick and know your tools well, to work with them and not against them.</p>
<hr>
<p>Meta note: this question is going against the StackOverflow <a href="https://engineering.stackexchange.com/tour">tour</a>. It advices e.g. against opinionated answers and those ".. with too many possible answers or that would require an extremely long answer". What this might mean is someone closes the entry eventually.</p>
| 2125 | How is a mechanical system designed and tested without building a physical model? |