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2017-07-01T17:03:27.483 | <p>Can a Prince Rupert's drop be made of metal instead of glass?</p>
| |materials|pressure|stresses|metals|cooling| | <p>Diving deep, given the correct material/component, I was think if it is possible to use a strong external fast changing EM field (electro-magnetic) to control the shape of melting metal and keep it suspended/floating in space, and cool down using water or a high speed jet of liquid air.</p>
<p>this will result in a material as hard as rupert's drop with no tail (no achilles heel).
this can be used for military and space exploriation.
I was wondering if a similar process has been used (secretly) in military industry already.</p>
<p>----- pioneering towards the Prince Rupert's drop with short or no tail, in glass, and in metail.</p>
| 16122 | Is a metal Prince Rupert's drop possible? |
2017-07-02T22:15:55.710 | <p>Going through some theory about epicyclic gears I came accross a comment that says that the spin velocity of the gear irrelevant if perfectly balanced, I don't understand why this might be true, is it because it would be constant and thus can be measured without variations?</p>
| |mechanical-engineering|gears|dynamics|vibration| | <p>This depends greatly on the context. Here is a counter-example, a situation in which the spin velocity of the gear is relevant to the dynamics of the overall system, even if that velocity is constant. That situation is high speed gears on aircraft (e.g. in the engines). At sufficiently high speed, the gear will exhibit the gyroscopic effect. Then when the plane executes a maneuver angular velocity (e.g. nose pitch up on takeoff), there is a gyroscopic reaction moment on the gear, which change the loading on the bearings, and possibly shift the gear off center a little. Of course, this is probably a rare example. Most gears do not get fast enough to have significant gyroscopic moment, and most gearboxes are not rotated like in an aircraft. </p>
| 16135 | Why is the spin velocity of a gear irrelevant if perfectly balanced? |
2017-07-03T18:10:12.927 | <p>I want to channel light from an LED indicator through some clear, circular acrylic rod. I want a rough-ish idea of the smallest bend radius through which the acrylic can be bent and maintain TIR. I will be experimenting, and would like an equation to guide the design.</p>
<p>The surrounding medium will be air. I don't know if there are different acrylic grades that need to be factored in, hopefully not because retailers don't seem to give further information.</p>
<p>The wavelength will vary, but only visible light is of interest. So I'd like an equation to give minimum bend radius, given a particular wavelength and rod diameter.</p>
| |electrical-engineering| | <p>This is a well-known and multiply-engineered problem, as you might imagine. The simplest fix, at least conceptually, is to coat the exterior with a metallic layer so that you no longer depend on internal reflection & Snell's Law. At this point, your don't need the core (glass) any more, just a shiny tube. </p>
<p>But for situations (glass, or cladded optical fibers) where you do depend on total internal reflection, you want a material with the highest possible index of refraction. Follow the equations in other answers to see why. The usual problem is that high-index materials tend to have a higher absorption coefficient as well. One way or the other you lose some of your signal.</p>
<p>Finally, if you have money, you might be interested in <a href="https://en.wikipedia.org/wiki/Photonic-crystal_fiber" rel="nofollow noreferrer">"holey fibers"</a> </p>
| 16144 | Tightest bend radius of circular acrylic rod light pipe for total internal reflection |
2017-07-03T20:19:06.337 | <p>This is the Jacobian of a robot arm (IBB IRB 120) with already specified joint angles (0,0,0,0,0,0). This Jacobian is singular for this configuration. </p>
<pre><code>J =
[ 0, 17/50, 7/100, 0, 0, 0]
[ 187/500, 0, 0, 0, 0, 0]
[ 0, -187/500, -187/500, 0, -9/125, 0]
[ 0, 0, 0, 1, 0, 1]
[ 0, 1, 1, 0, 1, 0]
[ 1, 0, 0, 0, 0, 0]
</code></pre>
<p>What motions are unachievable with this specific configuration? How to find that out conveniently with MATLAB?</p>
| |robotics|kinematics|matlab| | <p>In MATLAB use ''svd'' to find the singular vectors corresponding to zero singular values of the Jacobian. The corresponding right singular vectors describe the input directions (i.e. movement in the joint space) that result in zero movement in the task space; the corresponding left singular vectors describe the output directions (in the task space) in which no movement is possible through any combination of movements in the joint space.</p>
<p>The Jacobian defines the relation between infinitesimal displacements $dx$ in the task space and infinitesimal displacements $dq$ in the joint space:
$$dx = \mathbf{J} dq$$
One way to calculate unachievable directions of (infinitesimal) motion from the Jacobian is to use it's singular value decomposition:
$$\mathbf{J} = U \Sigma V^T$$
The left singular vector matrix $U$ and the right singular vector matrix $V$ are orthogonal and span the (infinitesimal / local linearized) task and joint spaces respectively; the diagonal matrix of singular values $\Sigma$ determines the scaling between the individual input and output directions.</p>
<p>Unachievable motions are scaled versions of singular vectors corresponding to zero singular value(s). A right singular vector $V$, corresponding to a zero singular value, implies that this input direction (i.e. movement in the joint space) will result in zero movement in the task space. A left singular value $U$, corresponding to a zero singular value, implies that (due to the zero scaling factor) movement in this output direction is not possible.</p>
| 16147 | Unachievable motions because of singular Jacobian |
2017-07-03T22:11:49.710 | <p>I've been puzzling over a question / class of questions I'd like to solve via rough estimate before turning to CFD analysis.</p>
<p>Say I have a pressure vessel with one inlet and one outlet. The pressure vessel is at some initial pressure $P_0$. The outlet is at some fixed pressure $P_{out}$. The inlet has some mass flow rate, $\dot{m}$ (assume it's attached to an infinite upstream reservoir and a mass flow controller).</p>
<p>With a compressible fluid, a known mass flow rate, and a set volume in the vessel, what mechanism would you use to approximate the final steady state (volume-average, we'll say) pressure in the vessel? Assume that the outlet pressure is such that the flow will choke. Assume no heat transfer to the walls and that the gas is ideal. </p>
<p>One solution I have thought of is to perform a dumb iterative solution:</p>
<ul>
<li>Pick some time step, $dt$</li>
<li>Add $\dot{m} * dt$ mass to the tank</li>
<li>Compute the new pressure in the tank via the ideal gas law</li>
<li>Use choked pipe flow equation to determine $\dot{m}$ at exit (knowing exit area)</li>
<li>Subtract $\dot{m}_{exit} * dt$ mass from the tank</li>
<li>Repeat until "convergence"</li>
</ul>
<p>Possibly you could do some Bernoulli-streamline-style analysis, but I'd like to also apply this sort of approximation to a multi-inlet / multi-outlet problem. This feels like some form of the Hardy Cross method might be applicable as well. Interested in anybody's suggestions.</p>
<p>Graphs of what the above technique gets me for a 1 kg/s flow rate into a 5 m^3 volume vessel, initial conditions T = 290K & P = 1 Pa in the vessel:</p>
<p><a href="https://i.stack.imgur.com/dQU0o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQU0o.png" alt="Mass flow rate convergence"></a></p>
<p><a href="https://i.stack.imgur.com/2i4uy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2i4uy.png" alt="Pressure convergence"></a></p>
<p>Edit: Added a few figures from my "estimation."</p>
| |fluid-mechanics|cfd| | <p>We model the system simply for small mass flow rates compared to the size of the vessel simply:</p>
<p><a href="https://i.stack.imgur.com/FlS5O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FlS5O.png" alt="enter image description here"></a></p>
<p>At point 1, the pressure can be near infinite, or the density can be near infinite, or simply the capacity may be near infinite as far as we can tell. At this point, one thing is known for certain - no matter what the mass flow regulator needs, this system has enough energy to feel the need. However the reservoir supplies it, compared to the other points, $h_1 \approx \infty$. However, since the mass flow rate is fairly small compared to the remainder, each removal of mass will keep the temperature in the reservoir fairly constant. It's fair to state that $T_1 = T$, since $V_1 \approx 0$. After all, the tank is so massive compared to the tiny amount that leaves that the tank will warm or cool the fluid compared to the amount that has left so the outside will keep $h_1$ constant.</p>
<p>By the dynamics of the mass flow regulator, we know that $\dot m$ mass leaves the reservoir. We don't know much about point 2, but it's very clear that $h_2 \lll h_1$. If this wasn't the case, the reservoir in question wouldn't be able to supply a consistent energy supply and mass supply through the line in question. The pressure is still completely unknown, but $P_1 \ggg P_2 > P_3 \ggg P_4$, or the fluid wouldn't leave the system at all. However, across the regulator, the tiny mass moving of the fluid compared to the size of the piping calls into question the <a href="http://www.thermopedia.com/content/1145/" rel="nofollow noreferrer">Stanton Number</a> of the fluid. At a high Stanton number, it can be shown that a lot of heat moves from the pipe back into the fluid, reestablishing $T_2 = T$, and bringing $h_2$ up, but less than $h_1$ since $P_2 < P_1$</p>
<p>As established above, the temperature of $T_2$ should be brought back to room temperature, $T$. While the energy is unknown, we are attempting to find $P_3$. Since we enter a reservoir, $V_3 \approx 0$, and the fluid stagnates. The reservoir walls may be warming the fluid inside. Since the mass flow rate is small compared to the mass of the fluid inside the reservoir during steady state situation, it can be seen the reservoir should warm the fluid until it reaches $T$. Thus, the stagnation temperature, $T_{0_3} = T$. With this temperature known, the choked flow equation can be modified to reveal $P_{3_{final}}$, since the fluid, having stagnated:</p>
<p>$$\dot m = \dot m_{choked} = A_e(P_0 = P_3)\sqrt{\frac{k}{R(T_0 = T_3 = T)}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}} $$</p>
<p>$$P_{3_{final}} = \frac{\dot m}{A_e\sqrt{\frac{k}{RT}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}} $$</p>
<p>Indeed, with these assumptions, you've effectively turned the reservoir at point 3 into it's own mass flow regulator, at the price of keeping the pressure and temperature inside of the reservoir 3 constant.</p>
| 16148 | Steady state approximation for compressible fluid flow (manifolds and other pressure vessels) |
2017-07-04T14:59:17.123 | <p>I'm trying to calculate the values of pressure, temp, speed and density of air going through an axial compressor (12-stage) a helicopter. I've read a few bachelors' thesis and papers but wasn't able to find anything particularly helpful for this task. Can anyone offer advice or, ideally, a paper/textbook with relevant information? Thanks in advance.</p>
| |fluid-mechanics|aerospace-engineering|compressors| | <p>This would be a good start :
Engineering Thermodynamics, Work and Heat Transfer (4th Edition). / Rogers, GFC; Mayhew, YR.</p>
<p>Still got mine and use it..</p>
| 16156 | Help me find a paper/textbook on axial compressor in turboshaft engine |
2017-07-04T23:53:24.837 | <p>I'm an engineering coop at a vibrations consultant. I was given a graph plotting a series of applied frequencies against a floor's acceleration response to them. I was asked to find the natural frequency and damping ratio, but I haven't been able to find the damping ratio. I've taken a course in vibrations, but we were always given more about the scenario (mass of the object, spring constant, etc.). I've asked him to at least give me a starting point, but he's refused.</p>
<p>My boss also asked me to find the theory behind how the damping ratio is determined empirically. All the resources I've found online either reiterate what I already learned in my courses, or are so advanced that I can't even begin to understand them.</p>
<p>My questions are: how is the damping ratio determined empirically, and what is the general way to solve this problem?</p>
<p>For context, my vibrations course focused on 1 DOF systems, teaching the basics behind damping ratio, natural frequency, and general vibrations. We did questions about base excitation and rotating imbalances in machinery.</p>
| |mechanical-engineering|vibration| | <p>The <a href="https://en.wikipedia.org/wiki/Q_factor" rel="nofollow noreferrer">Wikipedia page</a> is somewhat electrically focused, but it should be good enough to get you going. </p>
<p>Step 1: determine the frequency at which the response peaks, call it fc</p>
<p>step 2: determine the frequencies at which the power is 1/2 of the peak power (i.e. amplitude is sqrt(2)/2 times the peak). there are 2 such frequencies, one higher than the peak and one lower. call them f2 and f1</p>
<p>step 3: Q=fc / (f2-f1), and damping ratio zeta = 1/(2*Q)</p>
| 16162 | Determining damping ratio empirically? |
2017-07-06T14:47:10.883 | <p>Are these pieces of equipment independent from each other, or one requires also the other one to be in a system (chiller)?</p>
| |thermodynamics| | <p>They are 2 distinct pieces of equipment that both utilize variable speed drives. and one is not a requirement for the other.</p>
<p>The compressor is used to adjust the pressure/flow rate of the working fluid in the system, adding a variable speed drive gives highly efficient control over this. This can make the system more efficient at part load.</p>
<p>At the condenser the variable speed drive controls fans that pass air (not the working fluid) over an atmospheric heat exchanger, effectively allowing you to alter its thermal conductivity. This gives additional control over the temperature at the condenser exit.</p>
| 16180 | Difference between Variable Speed Compressor and Variable Speed Condenser(Fan controlled)? |
2017-07-06T14:48:14.617 | <p>Hot humid east TX; Large subdivision pool is warm ,92F. A plan is being make to cool it with evaportive spray . One data point is : dry bulb= 82.0 , wet bulb = 79.7 F. This is likely typical average of 24 hours. The cost of pump and spray may be $ 3000 ( to purchase). How much cooling could be expected ? My thought is at this high humidity it will not be cost effective but I can't put numbers to it. My bad: by "large subdivision pool ", I mean 10,000 sq ft / 450,000 gal.</p>
| |thermodynamics|cooling|evaporation| | <p>I'm going to take the formal question statement as "Is evaporative cooling for a pool effective?"</p>
<p>There are a handful of factors that need to be considered. First and foremost though, is <strong>wet bulb temperature</strong>. The wet bulb temperature is <strong>theoretically</strong> the coldest temperature that you will ever be able to achieve utilizing a wet cooling system (a system that evaporates water to reject heat). Taking this into consideration, the absolute coldest that you will be able to make the pool given your stated data points and method is the wet bulb temperature of 80 Fahrenheit. Realistically, you will probably only be able to come within 2-5 degrees F of this (82-85 Fahrenheit) due to mass transfer limitations and lack of crossflow packing & drift elimination in a small residential unit. For sake of the argument, assume that 80 F is the best you can do using wet cooling.</p>
<p>If 80 F is suitable and you would like to pursue this, the next step is to determine total <strong>capacity</strong> required. <strong>Capacity</strong> when talking about heat rejection is measured in MegaWatts, kiloWatts, Watts, or BTU/hr. To help understand this concept, think about the total energy coming into the pool that's heating it up; likely the predominant source is sunlight being absorbed into the pool (there may be some heat transfer from the dry bulb temp of the air and the water, but odds are it's negligible, or possibly even cooling the pool off a small amount via some natural evaporation). So if we want to keep the pool at a constant, cooler temperature, we must have enough <strong>capacity</strong> to remove all of the heat that the sun is putting into the pool. Because we have a very small amount of knowledge about your pool, I'm going to make some considerable assumptions <a href="https://www.troublefreepool.com/threads/9205-Water-Absorption-and-Heating-from-Sunlight" rel="nofollow noreferrer">utilizing previously calculated values</a>. So we're going to assume a crude measurement of a peak solar energy absorption that induces a 0.75 F / hr (0.41 C / hr) increase in pool temperature. Note that this is NOT accurate if your pool is considerably different geometry, or has a different level of reflection on the pool floor. Additionally I will use a per-unit-of-water calculation so that you can extrapolate this to your pool size as this was not provided. An example was provided for a pool size of 10,000 US gallons</p>
<p>Heat capacity of water: 4.184 kJ/kg*k;
Q = heat flow rate;
1 US gallon water = 3.785 kg</p>
<p><a href="https://i.stack.imgur.com/2Oebc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Oebc.jpg" alt="enter image description here" /></a></p>
<p>So, what this is saying: for a 10,000 US gallon pool, at the same conditions described in the link, we should expect to need at least 18 kW of heat removal ability to keep the pool cool.</p>
<p>Remember that the coldest we can get is going to be the wet bulb temp; but to get to that wet bulb temp, we have to be able to reject the same amount of heat that's coming into the pool from the sun that we previously calculated (18 kW). To do this, you need to size your evaporative cooler. Unfortunately, evaporative coolers are sized and sold on a "CFM" basis, as their BTU performance is variable depending on inlet conditions (namely temperature and humidity). You can find <a href="http://www.breezair.com/us/help-me-choose/model-specifications" rel="nofollow noreferrer">some manufacturers that will rate their units</a> based on very tight design conditions; but note that a 15 kW (50,000 BTU/hr) unit at 0.3 IWG conditions may not be able to achieve 15 kW regularly. Accurate sizing of the evaporative cooler will require measurement of heating degree days, historic wet bulb temperatures, and then analysis of this against said design criteria. Alternatively, you could just way over shoot your kW target (if you need 18, maybe go with 30 kW worth, or two 50,000 BTU units). Note: if you try to size an air conditioner using BTU's or tons and translate that to a comparable evaporative cooler by using the CFM required to cool the same square footage/volume, you will undersize the evaporative cooler - The cooler is only designed to hit its design wet bulb temperature approach, NOT to match the same BTU output of an air conditioner.</p>
<p>I do not know which model(s) you're looking at, but just looking at some home depot styles and some on the link above, it seems like most coolers are around <span class="math-container">$100-$</span>200 USD/1000 CFM. If we extrapolate <strong>very crudely</strong> based on the Australian design conditions used by breezair, you can achieve ~1.5 kW / 1000 CFM. To hit 18 kW for a 10,000 US gallon pool would require 12,000 CFM, or roughly <span class="math-container">$ 1200 to $</span> 2400 in capital expense. This doesn't include the power to run the fan (~1kW * (12 hrs / day) * <span class="math-container">$.15/kWh =~ $</span>2/day to operate at peak load), or water costs (<span class="math-container">$0.10/gal * ~100 Gal / 12 hrs operation =~ $</span>10/day peak operation)</p>
<p>One other concern with this is cooling water source. It would generally not be recommended to utilize pool water for the evaporative effect, as all the minerals in the water will concentrate up over time as you evaporate water out and recirculate it back into the pool - this will eventually cause scale buildup in the pool water, making it undesirable to swim in. If one discharges the water somewhere else as opposed to back into the pool, then you'd be wasting the treatment chemicals (bleach etc), which would be an additional cost.</p>
<p>Alternatives to utilizing a swamp cooler / evaporative cooler is to simply remove the heat source in the first place. If all the heat is coming from the sun, then building a way to provide shade to the water may provide a substantial benefit.</p>
| 16181 | Is evaporative cooling for a pool effective? |
2017-07-07T07:31:26.470 | <p>Watching Apollo 13 and rereading Lost Moon, Jim Lovell's exciting recollection of the events on the crippled spaceship, I began to wonder. Why does NASA maintain the gasses at the triple point? A pressure vessel designed to handle very high vacuum (which H2 and O2 triple points are near complete vacuum) and low temperatures is thicker and more dense then a pressure vessel designed to handle modest amounts of pressure. Why wouldn't they store these gasses at higher pressures, at or near atmospheric pressure?</p>
| |mechanical-engineering|thermodynamics|pressure|aerospace-engineering| | <p>If you want a few hundred pages of gory detail on gas management, I found an original research report <a href="https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19720016146.pdf" rel="nofollow noreferrer">here</a>. </p>
<p>Looking at excerpts from a <a href="https://books.google.com/books?id=8MHBDAAAQBAJ&pg=PA117&lpg=PA117&dq=Apollo%20mission%20oxygen%20storage%20at%20triple%20point&source=bl&ots=skmrTx4iiz&sig=dPb7-GyxXf7hC1UPlI7cBGQkoY8&hl=en&sa=X&ved=0ahUKEwjbkODVmPfUAhXD6YMKHUUeBwUQ6AEIQzAI#v=onepage&q=Apollo%20mission%20oxygen%20storage%20at%20triple%20point&f=false" rel="nofollow noreferrer">book on cryogenics</a>, one thing that's pointed out is that the triple point provides a stable pressure inside the container even as energy is added or removed, because the energy just changes some of the material from one state to another. By comparison, a can of LOX needs pressure management because a little heat produces a gas head, and so on. </p>
| 16192 | Triple point storage in space travel |
2017-07-07T12:47:09.093 | <p>I recall there was a set of "engineering numbers" that were to be used when deciding how big to make something. So that if, for example, you wanted your screw to be 1.36mm you'd instead round up or down to the nearest of these prescribed numbers, perhaps to 1.2 or 1.5. This way everyone's things were more likely to be the same size because instead of randomly selecting sizes that were particular to a specific situation, we'd all be using the same set of pre-selected numbers instead. Does anybody know what this concept was called? Thanks!</p>
| |engineering-history| | <p>It is also worth adding that preferred numbers are highly contextual as what numbers are actually sensible will depend a lot on manufacturing processes and existing standards.</p>
<p>For example many standard steel sections, even when nominally metric are actually based on inch sizes so you often see fairly odd mm dimensions on rolled steel sections, simply because it is too much trouble to convert to round metric numbers for no real benefit. Especially for things like structural sections and pipe where new stock needs to be compatible with legacy installations. Similarly BSP is still the norm for gas and hydraulic fittings. </p>
<p>Equally with the rise of international trade most things that need to be standardised are so this is much less of an issue that it once was. </p>
<p><em>This probably should be a comment really but is a bit too long</em>. </p>
| 16196 | Pre-Defined Numbers to be used in Sizing Mechanical Parts |
2017-07-07T14:38:34.993 | <p>From time to time I come across bicycles with a particular geometry (or rather a lack of). Most of them have no seat and a single tube which simply connects wheels and has the bottom bracket (if that still makes sense) in the middle of it. </p>
<p>What I'm talking about are bicycles like these:
<a href="https://i.stack.imgur.com/sBLvt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sBLvt.jpg" alt="occam cycle"></a>
<a href="https://i.stack.imgur.com/C5QS8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C5QS8.jpg" alt="micro pedalflow"></a></p>
<p>Shouldn't the frame split in two by the force put on the "bottom bracket" while pedaling? Is the "bottom bracket" stay strengthened in any way? Does the lack of the seat tube imply there is less load on the frame? I assumed that the "diamond" shape of a bicycle offsets the bending and compressing forces that act on a beam structure.</p>
| |mechanical-engineering|structural-engineering|design|structural-analysis|bicycles| | <p>While a single tube like that isn't as structurally efficient as a triangulated frame there is no reason it can't take the required loads as long as it is chunky enough. Note that, especially in the top picture the beam is clearly a lot more substantial in section than the tubes in more standard cycles. Equally you can't see how thick the walls of the tube are. </p>
<p>So the chances are that any weight saving from having less tubes is offset by the fact that the ones you do have need to be more substantial but here I suspect that the main aim is to make the whole thing as compact as possible. </p>
<p>Also, in most cases the limiting factor in bike frame design is stiffness rather than outright strength and most failures are due to either impact damage, manufacturing defects or long term fatigue. </p>
<p>From experience something like 25x50x2mm rectangular mild steel tube will support an adults body weight over this sort of length with no trouble whatsoever. </p>
| 16202 | How are single tube bicycles prevented from collapsing? |
2017-07-07T17:47:43.793 | <p>I'm learning theory of axial compressors for my summer internship and I find some of the equations puzzling, although I do understand the way they were derived.</p>
<p>The equation bellow is for temperature increase on rotor.</p>
<p><a href="https://i.stack.imgur.com/XMuYQ.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XMuYQ.gif" alt="Temperature change equation in Axial compressor"></a></p>
<p>Could someone elaborate on why isn't the $\Delta T$ proportional to other factors like the number of blades? Is it correct to assume that even one blade is considered to be able to accelerate all the fluid going through the compressor to the tangential velocity considered in this equation, given by $U =\omega$r? This would then mean that the number of blades and their geometry, except the main angles, will only play a role in energy losses and distribution of load on individual blades? </p>
| |mechanical-engineering|fluid-mechanics|thermodynamics|aerospace-engineering|compressors| | <p>It is usually a completely different group of people doing compressor aerodynamic design versus overall thermodynamic cycle design. The compressor aero group, using a variety of computer simulation codes (ranging from simple spreadsheets all the way up to 3 dimensional unsteady CFD) are going to generate a <a href="https://en.wikipedia.org/wiki/Compressor_map" rel="nofollow noreferrer">"compressor map"</a>. The compressor map gives the pressure ratio, mass flow through the compressor, and efficiency for given rotor speeds (and perhaps other variables). This allows the thermodynamic cycle guys to treat the entire compressor with just a few simple numbers. All of the details like the number of blades, the number of vanes, the stagger angles of the blades, etc, etc get boiled down to a simple map. That way the thermodynamics guy does not have to know or even care about how many blades there are. He doesn't even care how many stages are in the compressor. He just knows that at 10,000 rpm the compressor is giving him 10 lbm/s, 15:1 pressure ratio and is 90% efficient (or whatever), and at 12,000 rpm it's giving 13 lbm/s at 18:1 ratio, etc. etc. The thermodynamic cycle then takes that compressor map, along with similar input from the turbine aerodynamics guys, the combustor aerodynamics guys, the inlet and exhaust aerodynamics guys, and combines it into an overall picture of the engine performance. There will be iteration between those groups. For example, the cycle guy may determine that he has more than enough thrust but he needs more efficiency. So the compressor aero guy then changes number of blades or something, now maybe it is 9.5 lbm/s at 14.5:1 ratio but 93% efficient at 10,000 rpm. </p>
| 16210 | Axial compressor - change of temperature not proportional to number of blades and their geometrical properties? |
2017-07-09T21:19:50.677 | <p>I'm interested specific on electricity consumption changes on 2 different situations inside an elevator.</p>
<ol>
<li>When the elevator is stopped and people enters the elevator, does it change in any way the electricity consumption? Simple logic tells as that the elevator in this state is stopped by brakes, and brakes pressure is not affected by the amount of people being inside the elevator, so there should be no difference on electricity consumption on this case.</li>
<li>If elevator is going up/down, and a passenger is walking, jumping inside the elevator does this makes any difference on power consumption of the elevator? or regarding the external motor we can see all the cabinet as a unit and the forces that interact inside the cabinet doesn't change anything regarding the external force that is trying to make all the cabinet move ? If it's true how is called this principle?</li>
</ol>
| |energy|electrical| | <ol>
<li><p>When power fails the elevator must remain safe, which means the brakes will be released by applying power.</p></li>
<li><p>The force required to move the cabin will change according to how the passenger is accelerating himself. Similar how a phone can vibrate all on its own by oscillating a mass inside of it. The mean power required to lift the cabin+passenger will remain the same regardless of what he's doing in there.</p></li>
</ol>
| 16227 | Elevator standby + moving passanger , electricity consumption changes |
2017-07-10T13:21:52.353 | <p>I'm trying to calculate how much simply-supported round beams of several materials and diameters deflect under their own weight plus an additional weight in the middle.</p>
<p>I know the material densities so I can calculate weight of the entire beam, and I know the material Young's Modulus so I can calculate Moment of Inertia.</p>
<p>I've got the deflection due to an added weight calculated correctly, but I'm stuck on deflection due to the shaft's own weight.</p>
<p>Several places have given the formula:</p>
<p>$$\delta C = \frac {5 q L^4} {384 E I}$$</p>
<p>eg: <a href="https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams</a></p>
<p>But I am not sure what <code>q</code> represents.</p>
<p>Wikipedia describes that as "Uniform load on the beam (force per unit length)" and I see other sites mention it in Newtons.</p>
<p>Is that gravity's acceleration (<code>9.80665 m/sec^2</code>) times the beam's own weight in kilograms to get Newtons of downward force?</p>
<p>eg: <a href="https://en.wikipedia.org/wiki/Newton_(unit)#Examples" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Newton_(unit)#Examples</a></p>
<p>For example, a 102g beam would exert 1N of downward force, so <code>q</code> would be <code>1</code>?</p>
| |structural-engineering|civil-engineering|beam|deflection| | <p>q is going to be the distributed load along the span of the beam. If you are looking at just the self weight, you need to calculate that weight (based on density and cross sectional area) as a distributed force and apply that as q. </p>
<p>Important stuff to understand and learn, but you might be able to use this <a href="https://skyciv.com/free-beam-calculator/" rel="nofollow noreferrer">beam deflection calculator</a> to check your answer. </p>
| 16232 | Round beam deflection under its own weight |
2017-07-10T20:52:45.583 | <p>I am interested in designing a scale model semi-monocoque wing for a glider. I've done some research on the framework and internal structure. I have found out about the spars, ribs and stringers and I am trying to produce a design with the highest strength to weight ratio. I am planning on making the wing 0.5m long with a chord length of 0.1m along the cross section (so no tapering).</p>
<p>Here is my design so far: <a href="https://i.stack.imgur.com/a2b9w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a2b9w.jpg" alt="design plans of glider wing"></a></p>
<p>If anything is unclear please let me know. In my first diagram, I have placed 2 spars through the wing about 10% of the chord length from the leading edge and 15% from the trailing edge, leaving 75% of the chord length between the two spars.</p>
<p>In the second diagram is a rough plan of how many ribs would be used and where they would be placed: the solid outline is the wing outline, the thick black lines are the spars and the dashed lines are the ribs.</p>
<p>I have used a triangular configuration, as the triangle is the strongest shape and resists shear, torsion, compression and tension better than any other shape. Then I have placed a rib going down from the top corner of each of the 3 triangles to the bottom of each. Then I have placed ribs (or stringers) between each of the bottom corners of the triangles.</p>
<p>My question is, how can I improve or optimise that design? Is there a certain number of triangles I should use and how big should they be relative to the wing? Should they be equilateral triangles or is isosceles stronger? Also, how many ribs should I be considering using?</p>
<p>I'm not looking for anyone to figure out exact quantities. If there is some kind of calculation I need to carry out to find the optimal number or triangles and ribs, I would be happy to do that. Unless it is way beyond A-level maths/physics standard then I probably will not be able to do it. But I would like advice on how I can figure it out. If my question is vague or unclear, please inform me, and I will gladly edit it.</p>
<p>Thank you</p>
| |mechanical-engineering|design|structural-analysis|aerospace-engineering|geometry| | <p>If you want to design something, the first step is to find out what the design needs to do - i.e. what are the forces which will act on the spar.</p>
<p>You can find the wing loading from the weight of the plane plus the maximum "g" force you want it to survive in flight (and in crash landings, while you are learning to fly it!) </p>
<p>You also need to think about the torsion forces on the wings - read about the difference in the chord position of the "center of pressure" and the "neutral axis" of the beam created by your wing spars.</p>
<p>Then, you can find the bending and torsional stresses in the spars, and calculate how big they need to be to carry those loads.</p>
<p>The "nice" thing about beams is that the stresses are <em>statically determinate</em> - i.e. changing the shape of the beam doesn't change the loading pattern (except for the change in weight of the wing, of course). </p>
<p>That should get you started answering your question. The lightest design will probably have spars that are tapered along their length - that may be more important than the exact layout of the "web" connecting the spars, for weight saving.</p>
| 16239 | Advice for designing internal structure of glider wing |
2017-07-11T09:34:48.917 | <p>I am asking here for your help and advice on how to read holes schematic notations.</p>
<p>On the attached picture I've marked with a green color part of notation that I don't know how to read / understand. And will be thankful for any advice on how to interpret this notes.</p>
<p>With the red color I've marked a holes with countersinks. But I'm failing to pick up necessary details from the sketch to understand the size/angle/depth of this countersinks. (I understand that they can be missing, but I tend to think that they might be in the notations that I don't understand)</p>
<p>Thank you for considering this question!</p>
<p>source: the drawing was taken from this youtube video - <a href="https://www.youtube.com/watch?v=2FDeeIlVAtY&feature=youtu.be&list=PL4mNbJmXLK0dn7wdGYOncwFRAQoFIgXiA&t=3219" rel="nofollow noreferrer">link</a></p>
<p><a href="https://i.stack.imgur.com/oW9Dx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oW9Dx.jpg" alt="enter image description here"></a></p>
| |measurements|drilling|drawings| | <p>Just to add.</p>
<p>The green is perpendicularity in respect to datum C. Meaning, the called out feature must be perpendicular to datum feature C within 0.003.</p>
<p>I'm also seeing the condition "45 degrees 0.030 - All holes". The countersink will be an additional 0.030 at 45 degrees from the thru hole values (since the countersink won't be thru). I'm not entirely sure if the larger hole callouts are pointing to the OD or ID of the countersunk hole but you'll simply add or subtract in accordance to the condition: "45 degrees 0.030 - All holes". Hope that helped.</p>
| 16247 | Read holes schematic drawing notation. How to? Particular case |
2017-07-11T10:06:17.423 | <p>Yes, the one where you put lighter close to nozzle of deodorant, light it up and start to spray out the deodorant. I suppose that this is simillar to how gas stoves work, so it shouldn't explode. Even when I stop pushing on spray button, flame gets closer to deodorant and it feels like its your last moment, flame can't get back into the can and cause a fireshow, becasue there's no air in the can. Worst thing that can happen is nozzle meltig down over time and flame getting bigger but not out of control. Right?</p>
<p>Also, when things can go bad? I know that i shouldn't throw deodorant can into flame, but can it overheat just from being used as a flamethrower and boom?</p>
| |fire| | <p>No, the flame can't go upstream back into the can and explode what is in the can. This is no different than the flame in your gas stove can't go back up the pipe and explode the whole distribution network.</p>
<p>Fire is a chemical reaction between fuel and oxygen. There can't be fire in the nozzle or in the can because there is no oxygen there.</p>
| 16248 | Will deodorant flamethrower ever explode? |
2017-07-11T13:58:17.060 | <p>I was wondering whether the hysteresis (in particular, mechanical motion systems) can be considered under the class of friction.</p>
<p>From theory, I know that the frictional forces are direction, and velocity dependent. In the same sense, they also possess memory in the special cases of frictions. So, is it correct (without abusing any definitions) to consider the hysteresis observed in motion systems to be under the class of friction. Because the hysteresis in motion systems also characterises the same characteristics.</p>
<p>Simply put, under what conditions one can classify the hysteresis under the frictional nonlinearity, and if not, under which conditions this fails to do so?</p>
| |mechanical-engineering|friction| | <p>Hysteresis is not the same as friction, but friction is part of hysteresis, it is normally considered as the play or "free movement" in joints etc that cause a system to take a slightly different route "down or return" compared to the "up or outward" travel.</p>
| 16253 | Can the hysteresis phenomenon be considered as a (type of) friction |
2017-07-11T18:25:58.337 | <p>I have a shaft with splines on it that I want to create a drawing for in Solidworks because I may need to have this specific spline replicated on a part in the future. Does anyone have any documentation on how to identify a spline? Or possibly a table of standard splines used in industry based off of shaft size?</p>
| |mechanical-engineering|manufacturing-engineering|computer-aided-design|documentation| | <p>Depending on the type of spline you're defining, there are different characteristics that you'll need to identify the spline to either a machinist or to the customer who will need to design the mating part. Some of them are pretty straightforward and will be needed regardless of the type of spline, such as physical diameters (major and minor), number of teeth, the class of fit and spline shape (involute, flat, etc.) </p>
<p>For the most basic splines I work with, we can define it with only a major and minor diameter, the number of teeth, and the distance between two parallel tooth edges, because it's a flat side spline. However, something like an involute spline requires more information, such as a pitch diameter, pressure angel, effective tooth width, and distance between pins of a given size. This is because the complex shape makes it impossible to directly measure the spline geometry in a meaningful way, so we require both design dimensions and measurable dimensions. The spline table below is an example for an involute spline - the dimensions listed fully define the spline and allow us to measure it for quality purposes. I would include a table like this on any spline machining print so that you have the information readily available.</p>
<p><a href="https://i.stack.imgur.com/jmw8Vm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jmw8Vm.jpg" alt="Spline Table Example"></a></p>
<p>If you are using a standard spline, such as a spline defined by ANSI B92.1, you could refer to it on a customer print as just the ANSI standard spline with only the information necessary to locate it in the standards table (should be pressure angle, class of fit, and either number of teeth or pitch diameter.) Then the standard will provide all other information, or the methods to calculate the information necessary to manufacture the spline.</p>
| 16258 | Shaft Spline Information |
2017-07-12T11:52:23.310 | <p>Imagine some thing put in the presence of loud pure white noise, say at a level that might come from a teenager's HI-FI set to annoying. And the thing is like a champagne glass, sheet metal box or other thin walled vessel. I'm not considering anything like a cannon ball or anvil.</p>
<p>I can foresee two scenarios:-</p>
<ol>
<li>The thing vibrates at it's natural self resonant frequency, but only with limited amplitude as the power content of white noise at a specific frequency is relatively small.</li>
<li>The thing does nothing whatsoever as all the various frequencies present in the white noise induce vibrations that cancel each other out.</li>
</ol>
<p>Which is it, or might there be a third possibility?</p>
| |dynamics|vibration| | <p>The thing will vibrate at all frequencies. It will vibrate more (be amplified) at frequencies that are close to a natural frequency. Effectively, the system will act as a band pass filter (NOT a band notch filter) </p>
<p>I did a very similar experiment in graduate school many times. It wasn't a teenagers hi fi set or a wine class, but it was a random input to a vibratory system. </p>
| 16268 | Can a thing resonate whilst exposed to white noise? |
2017-07-13T07:54:45.803 | <p>I am trying to find reliable information and comparisons about the energy production of renewable energy technologies, specifically, how solar and wind alternatives compare to nuclear fusion.</p>
<p>Why is it generally consider, as I have read, that nuclear fusion is the ultimate renewable energy? Is it really that clean? Would a fusion reactor be safe? Is the infrastructure required cheaper than solar and wind (considering long-term returns)?</p>
<p>If the answer to all or some of the above questions is affirmative, then why countries around the world invest anything at all in solar and wind? Why aren't the efforts directed to obtaining nuclear fusion?</p>
<p>I'm guessing a reasonable answer to these questions could be the inmediate mitigation of ongoing climate change and shortage of non-renewable resources. But then we should say that solar and wind alternatives are a 'transition' towards nuclear fusion. In this case, the main objective for countries worldwide should be obtaining nuclear fusion. But, in the future, when we harness nuclear fusion, wouldn't solar and wind become obsolete?</p>
<p>P.S. I just asked this question in <a href="https://physics.stackexchange.com/">Physics</a> and it wasn't well received by some user because it is about 'politics'. Please, note that I am not interested in a political opinion, but a scientific assessment. The political and economic issues that may be involved are inevitable given the nature of the question. I posted here because now I think it is more appropriate for this site.</p>
| |energy-efficiency|renewable-energy|nuclear-technology|infrastructure| | <p><strong>Fusion is widely extolled by advocates as being safe (but is that collective opinion justified)?</strong></p>
<p>I would prefer to begin my discussion comparing fusion with renewable energy systems with an examination of safety as in my opinion, safety should always come first when we compare energy alternatives.</p>
<p>Some Reasons offered why Fusion should be Safe -
- Only tiny amounts of fusion fuel are inside a fusion reactor at any one time. Continuous operation of a fusion power plant is maintained by continual refuelling with the fuel mixture (deuterium-tritium or deuterium-deuterium), so the fuel inventory in the plasma chamber at any time is gram quantities or less and sufficient only for about one minute of operation. The ITER tokamak will contain only grams of fusion fuel as it runs. A NIF laser fusion capsule contains only about 0.17 milligrams of Deuterium-Tritium fuel (Basic premise – If the fuel is not inside the reactor, it can’t even theoretically explode inside the reactor or become a factor in an accident)</p>
<p>-No greenhouse gasses produced </p>
<ul>
<li><p>No reactor meltdowns or decay heat burn downs (Chernboyl, Three Mile Island, Fukushima incidents cannot happen)</p></li>
<li><p>No mining of fuel (fuel is extracted from water)</p></li>
<li><p>Billions of years of fuel available on earth (versus hundreds for fossil fuels)</p></li>
<li><p>1/1000ths of the radioactive waste of fission power. Nearly all materials become activated to some degree by energetic neutron bombardment. Neutron reactions in DT fusion reactors will inevitably create radioisotopes. The principal radioactive materials present in a DT or DD fusion reactor will therefore be tritium and neutron activated structural materials surrounding the reaction volume in the fusion chamber. </p></li>
<li><p>Radioactive waste, mostly neutron activated fission chamber structural material remains radioactive ~20 years versus hundreds of thousands of years for Minor Actinides in spent fuel produced by fission reactors. DT fusion reactors would however burn and require storage of significant quantities of Tritium, a moderately radioactive nuclear isotope of hydrogen. Tritium gas is hard to secure and small leaks in reactor tanks and plumbing could result in some loss of tritium to the environment. Tritium decays while emitting weak Beta radiation and under most circumstances is not seriously hazardous to humans. The deuterium-deuterium (DD) fusion process, dispensing with tritium as a fuel and the requirement for Tritium storage. DD fusion does however produce Tritium as part of the DD fusion reaction, but this Tritium does not tend to accumulate but is immediately burned in place inside a DD fusion reactor. </p></li>
<li><p>Fuel for energy would no longer be a geopolitical issue (huge benefit for humanity) </p></li>
<li><p>No nuclear weapons from waste products (again, huge benefit for humanity)</p></li>
</ul>
<hr>
<p>To have success in fusion and renewable energy innovation -</p>
<p>you have to get the BIG PICTURE right</p>
<p>The ultimate potential of D-D Fusion is very great and eclipses the potential of all other forms of energy production including nuclear fission and renewable energy systems-</p>
<p><strong>Fusion Energy longer than the earth has existed or the sun will burn</strong></p>
<p>The complete conversion of deuterium nuclear fuel releases an energy content of 250 x 10^15 joules per metric ton of deuterium. The quantity of deuterium in the world’s oceans is estimated at 4.6 x 10^13 metric tons. Deuterium present in seawater will yield around 5 x 10^11 TW-years of energy. In the year 2016 the entire planet consumed around 17 TW-years of energy, which means that the energy content of the deuterium in seawater would be enough for 29.5 billion years of energy supply. </p>
<p>To give all 10 billion people expected to live on the planet in 2050 the level of energy prosperity we in the developed world are used to, a continuous average use of power of 6 kilowatts per person as is typical in Europe, we would need to generate 60 terawatts as a planet—the equivalent of 900 million barrels of oil per day. </p>
<p>The time since the earth first formed = 4.54 billion years.</p>
<p>The time until the sun burns out = 5 billion years. </p>
<p>The deuterium in the sea is capable of completely powering planet earth at a level of 60 Terawatts for 8.33 billion years </p>
<p>(longer than the earth has existed or the sun will burn).</p>
<hr>
<p><strong>D-D fusion of pure Deuterium separated from sea water is a practical fusion reaction and practical fuel for ICF fusion</strong></p>
<p>When talking about near term prospects for fusion, fusion scientists frequently neglect to mention that mankind's first successful terrestrial release of fusion energy (Ivy-Mike experimental test shot of 1952) produced in excess of an exawatt of fusion power from D-D (not D-T) fusion. Use of cheap and ubiquitous Deuterium separated from sea water was the very first fusion reaction to work. Futher, it deserves to be pointed out that Ivy-Mike D-D fusion worked the very first time it was tried, and never failed to work.</p>
<p>Hundreds of Magnetic Confinement Fusion devices that have been built over the course of 60+ years have never once even for milliseconds produced break-even fusion power or fusion with energy gain. ICF fusion of cheap and common Deuterium from sea water worked the first time it was tried (Ivy-Mike 1952) and thereafter never failed to work in hundreds of underground nuclear tests held at NTS and on the Pacific Test Range (Marshall Islands). </p>
<p>Large tokamak magnetic fusion designers often tend to discount the deuterium-deuterium (D-D) fusion reaction because it is not possible to achieve the necessary fusion reaction rates and break-even energy with D-D fusion using tokamaks and stellarator devices such as we currently are able build. The situation is different, however, with Inertial Confinement Fusion which typically employs fusion plasmas that are literally 10^11 times more dense than the typical ion plasma densities used in MCF tokamaks or stellarators. </p>
<p>The power than can be drawn from a plasma at fusion conditions is proportional to the second power of the plasma ion density. Even small improvements in plasma ion density will result in significant improvements in the performance of a fusion experiment.</p>
<p>There is an impressively large Range of plasma density in current and planned fusion experiments
<a href="https://i.stack.imgur.com/bGzJp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bGzJp.png" alt="Plasma ion density of typical current fusion experiments"></a></p>
<p>There is literally a factor of 10^10 to 10^11 difference in plasma ion density between MCF and ICF fusion experiments. This large difference has implications for commercial fusion technology as the denser the plasma, the more power that can be realized from a fusion burn and the smaller the physical fusion reactor can be while achieving fusion energy with energy gain.</p>
<p>The D-D reaction takes place at higher temperatures than can typically be achieved in magnetic confinement tokamaks and typically has a lower fusion cross section leading to lower fusion reaction rates on earth relative to D-T fusion . </p>
<p>D-D fusion is more complex and proceeds with multiple fusion side reactions, but when all of the side-chains run to completion, D-D fusion actually produces more energy than D-T fusion per starting mass of reactants. D-D fusion produces T and He3, which in a secondary reaction burn with D. This was practically demonstrated in early thermonuclear tests like the 15 Megaton Ivy Mike fission triggered deuterium experimental test in 1952. </p>
<p>D-D fusion does normally produce Tritium</p>
<p>D-D Fusion of Deuterium fuel produces energy through four reactions:</p>
<p>D + D -> He-3 + n + 3.268 MeV</p>
<p>D + D -> T + p + 4.03 MeV</p>
<p>(side chains)</p>
<p>D + T -> He-4 + n + 17.588 MeV</p>
<p>He-3 + D -> He-4 + p + 18.34 MeV</p>
<p>The net effect of these four fusion reactions taken together is:</p>
<p>6 D -> 2 He-4 + 2p + 2n + 43.243 MeV</p>
<hr>
<p>Fusion and neutrons -</p>
<p>DT and DD fusion release large amounts of their fusion energy in the form of fast neutrons.</p>
<p>The task of shielding fast neutrons requires</p>
<p>• Fast Neutrons (E >1 MeV) must first be thermalized with high</p>
<p>neutron scattering materials (typically materials that have a high percentage of light atoms like hydrogen)</p>
<p>• These thermal neutrons can then be absorbed or shielded with material having high cross sections for neutron absorption</p>
<p>Absorption of thermalized neutrons can be done by relatively light materials like polyethylene plastic unlike gamma radiation that require lots of high-z material (lead, concrete, water) to absorb the penetrating gamma. LLNL's Dr. Ralph Moir has pioneered several molten salt fusion reactor concepts (HyLife II and PACER Revisited). A thick liquid falling wall of molten salt loaded with a neutron absorber like 50 micron particles of Boron Carbide as a slurry can absorb damaging fast neutrons. Such a thick falling wall of molten salt can effectively protect fusion chamber structure and largely prevent neutron activation of fusion reactor materials. This permits molten salt ICF fusion reactors to have long 30+ year commercial lifetimes and to at reactor end of life produce little or no neutron activated waste.</p>
<hr>
<p><strong>Energy Density</strong></p>
<p>The issue of producing fusion with energy gain while simultaneously achieving a high enough power density are dominant themes of the near future in selection of which energy technology is best suited to produce competitive power plants that communities will want to build to produce safe and dependable power.</p>
<p><strong>Comparison of fusion and “renewable” energy systems from the standpoint of energy density</strong></p>
<p>Source Joules per cubic meter</p>
<p>Solar Radiation = 0.0000015</p>
<p>Geothermal = 0.05</p>
<p>Wind at 10 mph = 7</p>
<p>Tidal water = 0.5-50</p>
<p>deuterium in the form of D2O (heavy water) = 6.88 x 10^10 MJ/cubic meter </p>
<p>note: about 20% by mass of heavy water or D2O is deuterium
Deuterium is a gas at standard temperature and pressure but heavy water D2O is a easily and safely stored as a liquid. D-D fusion is in excess of million times more energy dense than any renewable energy system.</p>
<hr>
<p><strong>Practical fusion will always be 50+ years ago (not 50 years away)</strong></p>
<p>Mankind came into possession of a practical way of generating energy from fusion over 50 years ago with the Ivy-Mike nuclear test that produced fusion energy from pure Deuterium via DD fusion.
Practical fusion will always be 50+ years ago (not 50 years away)
Today, there are smaller pure fusion devices designed to make clean energy (not blast effects) from pure DD fusion of Deuterium separated from sea water. One such design is called mini-Mike, which produces a small predictable controlled energy yield of 250 GJ.</p>
<h2>Note: Deuterium separated from sea water is totally non-radioactive and fusion of this fuel produces only totally non-radioactive nuclear waste</h2>
<p><strong>Inertial Confinement Fusion has already been PROVEN to work and produce fusion ignition and fusion with energy gain</strong></p>
<p>Both LANL and LLNL were successful in producing full ICF fusion ignition during the last years of the cold war era. ICF fusion is experimentally demonstrated and proven to work and this places ICF fusion in a different category than any other form of fusion..</p>
<p>The energy needed to ignite an inertially confined thermonuclear fusion reaction in liquid (or solid) deuterium-tritium (DT) is not that large; it is on the order of not more than 10 MJ or about the same amount of chemical energy stored in about 1.25 cups of automotive gasoline.
The problem is that this energy must be compressed in space (focused down to an area less than a 2 mm) and in time (to less than 3 nanoseconds).</p>
<p>Pure Inertial Confinement Fusion that does not use nuclear fission to produce the conditions for fusion is today driver limited.</p>
<p>It is still not experimentally possible to build a laser (or ion particle accelerator) large enough to produce DT fusion ignition. Still, many people, including Congress, in the late 1980s and early 1990s wanted to know for certain if inertial confinement fusion would ultimately work and actually produce net energy from fusion. To answer this question, in the final few years on underground nuclear testing, both LANL and LLNL designed a series of test shots called Halite-Centurion. Halite-Centurion series shots were fusion related add on shots piggy-backed onto shots already on the schedule. These shots were designed to utilize a small portion of the X-rays produced from the primary of an experimental device through a line of sight to a remote fusion experiment housed some distance away in the underground experimental test canister. Lasers and Ion-beam fusion drivers such as were available at that time could not provide the driver energy required to produce fusion ignition - but X-rays from a remotely ignited fission device could provide the driver energy needed (>10 MJ energy delivered into a spot of about 2 mm in a time of less than 3 nanoseconds) .</p>
<p>Halite-Centurion fusion experiments in the Nevada desert worked and reliably and repeatedly and produced full fusion ignition of small sub-gram samples of DT fuel (in small filled spheres). These experiments were once classified but DOE allowed senior scientist Dr. John Lindl to declassify and reveal about half of the fusion related project information. Halite-Centurion experiments are referred to in a fundamental document for the design of the NIF facility: the 91 pages paper of John Lindl entitled “Development of the indirect-drive approach to inertial confinement and the target physics basis for ignition and gain”, published in 1995 in the AIP/Physics of plasmas. Halite-Centurion success was used by LLNL managers to sell the NIF program to Congress in the mid-1990s.</p>
<p>What is lacking in current ICF fusion experiments to support building of practical ICF fusion power plants is a fusion driver with characteristics close enough to the driver employed by LANL and LLNL in Halite-Centurion tests to permit full fusion ignition with energy gain. A driver capable of delivering greater than 10 MJ into a focused area less than a 2 mm in time of less than 3 nanoseconds. The current NIF driver laser driver is capable of delivering about 2 MJ total energy and is only able to achieve "alpha heating" phase of fusion operation - a stage that immediately precedes fusion ignition. LASNEX and HYDRA computer simulations on which LLNL based the engineering design of the NIF laser were overly optimistic in predicting the required fusion driver characteristics to produce ignition. Failure of the LASNEX and HYDRA simulations resulted in NIF being built too small to achieve the goal of fusion ignition and for NIF to serve as the basis of LIFE ICF fusion power plant technology.</p>
<p>Source documents (with links where possible) -
NY Times article published at the time of Halite-Centurion field tests - Secret Advance in Nuclear Fusion Spurs a Dispute Among Scientists
<a href="http://www.nytimes.com/1988/03/21/us/secret-advance-in-nuclear-fusion-spurs-a-dispute-among-scientists.html?pagewanted=all" rel="nofollow noreferrer">http://www.nytimes.com/1988/03/21/us/secret-advance-in-nuclear-fusion-spurs-a-dispute-among-scientists.html?pagewanted=all</a></p>
<p>The following document contains what John Lindl was permitted to release publicly regarding Halite-Centurion ICF by DOE
"Development of the Indirect‐drive Approach to Inertial Confinement Fusion and the Target Physics Basis for Ignition and Gain." John Lindl. Page: 3937. AIP Physics of Plasma. American Institute of Physics, 14 June 1995.</p>
<p><a href="https://i.stack.imgur.com/sdpaL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sdpaL.png" alt="Light ion driver pure fusion power plant concept"></a></p>
| 16282 | How is the energy production of solar and wind compared with the prospect of nuclear fusion? |
2017-07-14T16:49:05.313 | <p>does anyone have any resources for designing a bulk material handling system for an injection molding outfit? Currently we get material in gaylords and vacuum it to the hopper on each molding machine independently. I would like to start buying pellets by the truckload and stored in a silo to be fed to each machine from a central location. I am unsure of what equipment I would need other than the silo itself to do such a thing. Any leads?</p>
| |mechanical-engineering|manufacturing-engineering|plastic| | <p>There are several companies out there that can assist with silo/hopper design, which would allow easy offloading of truck to silo,, but that can also offer a solution to use vacuum pneumatic conveying to each machine as you currently have.</p>
<p>Of course there are many technical things to consider throughout the system, therefore it will be best to get a company that has delivered similar system to provide a solution. Each of these components are a science in itself and you can spend lots of money on a system that in the end could be useless if not done correctly. </p>
<p>Some of these include:</p>
<p>Silo design:</p>
<ul>
<li>Material flow out of the Silo: typical you would want a mass flow system rather than a funnel flow system (this will ensure you use the older material first and reduce the risk of buildup in the silo).</li>
<li>Do you want level measurement or weight of the silo to know when you are running low and have to order a new truck. You can add load cells in the supports of the silo to its main structure (typical suupliers will be mettler toledo for load cell and Endress Hauser or Vega for level measurement)</li>
<li>Also, the silo must be earthed to ensure you don't have static buildup (typical for plastic pellets)</li>
<li>A bag filter for when you fill the silo (assuming you will use a supplier with pneumatic pump fitted to the truck) - a typical supplier will be Donaldson</li>
<li>an over under valve or bursting disk for the silo to ensure you don't over pressurise it or accidentally get a vacuum generated within the silo that could cause it to implode</li>
<li>Depending on the flow ability of the product, the size of the silo you are considering you will have to consider whether you need devices to assist with the material flow out of the silo. This will be as the material will be under its own pressure and dependent on the shape of the plastic pallets. Typical devices that might get added is an inverted cone to break material pressure and air pads to assist the material to flow down the chute of the silo. alternative materials could also be used to assist with the material flow - stainless steel has better flow than mild steel, ultra-high molecular polyethylene (UHMWPE) sheets can be used to line the chute.</li>
</ul>
<p>Vacuum pump system:</p>
<ul>
<li>What type of system do you want to have - Lean or dense phase (dense phase will be at a slightly higher pressure and slower velocity but will be at lower running cost but with higher upfront costs)</li>
<li>You will need a feeder that slowly feeds the material to your vacuum line (this could be a star feed or a rotofeeder, which will gauge the amount of material that gets fed).</li>
<li>The vacuum pump sized according to the pressure drop of your system and the material flow rate you want to achieve</li>
<li>Air inlet filter with valve at you silo</li>
<li>Piping sized based on your route and pump sizing</li>
<li>Receiving cyclones/hoppers at your units as what you currently have</li>
</ul>
<p>Typical global suppliers for these systems will be Flexicon and Schenck Process, but there are smaller companies that can offer similar solutions in every country.</p>
| 16301 | Designing bulk material handling system for injection molding |
2017-07-15T00:10:44.897 | <p>Hy everyone, new here, I don't know much about how high pressure water pumping works - ( i'm just a programmer ) but i need to know if we can create water pressures <code>above 100 000 psi</code> for a crazy <code>water-jet cutting project</code> i have in mind :)</p>
<p>What's the maximum pressure we can reach inside a single pump / or by connecting pumps in series or parallel? </p>
<p>I <strong>would like to go to pressures of 500 000 psi or even higher if possible</strong>. I have no practical intuition of what that actually means :) </p>
<p>But i know that the higher the pressure the better - my reasoning is - if it takes <code>5 minutes</code> to cut a big thick block of iron at <code>100 000 psi</code>- will take <code>1 minute</code> with <code>5 times that pressure</code> - ( assume that the other parameters will be scaling as well.)</p>
<p>Taken <a href="http://waterjets.org/archive/buyers-guide/general/choosing-a-pump/#crankshaft_vs_intensifier_pump" rel="nofollow noreferrer">from here</a>: </p>
<blockquote>
<p>At pressures of 60,000 PSI (4,100 bar) and higher, metal fatigue
becomes a serious issue with many components. Although pumps that can
reach 100,000 PSI have been around for many years, nobody runs them at
such pressures because of the extreme maintenance issues involved. For
this reason, most manufacturers purposely limit their pumps to below
60,000 PSI (4,100 bar).</p>
</blockquote>
<p>This sounds serious - but i'm not too concerned about efficiency here. </p>
<p>I'm interested to <strong>know more about the theoretical limits</strong> that exist with creating such pressures.. not so much if a pump doing this is available or not for commercial purposes. </p>
<p>What i have in mind is suppose to be a very serious industrial machinery - custom build - all it matters is to cut as fast as possible. All the other concerns like cost or <code>metal-fatigue</code> are secondary.</p>
<p>And also can you please help me picture how big the dimensions of this <code>pump / system of pumps</code> will actually be? What volume it needs to take?</p>
<p>Thanks so much :) Have a nice day!</p>
<p><strong>EDIT 1:</strong> </p>
<p>Form @alephzero and @Solar Mike comments (thank you both) i realized there is an important difference between the pressure and the flow rate. This is an xy question and my question title is bad.. sorry about that.</p>
<p>The bottom line is - i need this machine to cut faster - as faster as it can get. I don't think we reached the limits of what's possible with our water jet cutting technologies - and if we did, i want to understand why?</p>
<p><a href="https://i.stack.imgur.com/XoYeb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XoYeb.jpg" alt="enter image description here"></a></p>
<p><strong>So what are the parameters we need to change - for this cut to happen 5 times faster?</strong> Or 10 times faster - or whatever i pick. When we will reach the maximum theoretical and then the practical limit of that this pumps can do.</p>
<p><strong>It will help if we put pumps in series or parallel?</strong> <strong>It will help if we increase the pressure?</strong> <strong>Or the flow rate?</strong> (but how to increase the flow rate - if the diameter is fixed) </p>
<p>I thought that by increasing the pressure we can make the cut faster. @Solar Mike comment made me realize that i don't need pressure - i need a 5 times higher flow rate - in order to cut 5 times faster. But because the orifice is a certain diameter - i think we can't do it without increasing the pressure.(maybe i'm wrong about this.)</p>
<p>And for now don't worry about how the sand will be mixed with this higher speed jet. Based <a href="https://youtu.be/YBqgofmdKc8?t=91" rel="nofollow noreferrer">on this video</a> it seems you can't mix sand if the water jet is at higher speed/ orifice is smaller - but i think i can get around that by combining more water jets into a single one inside a diamond chamber, like this:</p>
<p><a href="https://i.stack.imgur.com/PdKjR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PdKjR.png" alt="enter image description here"></a>
Not sure if this schematic will mess up the jet shape - but anyway the first problem is to figure out how to get that jet moving faster. Then we worry about how to add sand to it.</p>
<p><strong>EDIT 2:</strong> </p>
<p>Based on @Mark answer (thank you) i understand now that is not even possible to build such a pump - because the the maximum pressure we can possibly have is <code>150 000 psi</code> - if that <code>pump is made out of titanium</code>. </p>
<p>There is no material stronger than that - except diamond, graphene and such - which are not an option. (I <a href="https://www.youtube.com/watch?v=iH_-SIE4zlM" rel="nofollow noreferrer">found a company that makes nozzles out of pure diamond</a>, specially designed for water-jet cutting - but they don't make any pumps :)) - and for sure what @alephzero said about diamond strengths depends on the angle that force acts on it - applies here. )</p>
<p>Just to keep things simple let's stay with 100 000 psi for now. (such a bummer ..)</p>
<p><strong>I still want to cut faster. What can i do?</strong></p>
<p>If cutting faster is related to flow rate - then i need to increase flow rate somehow.</p>
<p>It follows that the only way to do that - is to increase the orifice diameter. (the other way - is to increase the pressure - which is not an option)</p>
<p>If i double the diameter - and increase the power appropriately - that will <code>quadruple the flow rate</code>.</p>
<p>Now because the <code>diameter is double</code> - the <code>water-jet cross section has quadrupled</code> - the <code>amount of material</code> i need to cut out <code>has quadrupled also</code> - the same as with flow rate. (not 100% sure about this - but is my first intuition)</p>
<p>What i'm trying to understand now is: <strong>So what will happen to cutting speed by increasing the diameter - is this a zero-sum game ??</strong></p>
<p>I mean, <code>trying to increase speed</code> -> i <code>increase the flow rate</code> -> by <code>increasing the diameter</code> -> which <code>which increases the jet cross section</code>-> which then <code>increases the amount of material i need to cut</code> -> which <code>lowers the cutting speed all over again</code>. <strong>So i go nowhere with this?</strong></p>
<p><strong>Or can be beneficial to increase the diameter?</strong> My intuition can't confirm anything here- i need someone to help me understand what will happen in the case i increase the diameter.</p>
<p>Is there a way to get out positive form this situation? Or is a hard theoretical limit like the one with pressure? </p>
<p>I rely don't like increasing the diameter - but seems we are out of options here - that's kind of sad..</p>
<p>Thanks everyone, this question got out of hand here - but promise i will not ask more questions :)</p>
| |fluid-mechanics|pumps| | <p>It is theoretically possible to drive a water jet to pressures of, say, 7,000,000 psi using explosives with a high detonation velocity, the shaped charge effect and a single use device. Higher pressures may well be possible with money, effort and optimization, for example swapping the RDX-based explosives in common shaped charges (detonation velocity around 8 km/s) with octanitrocubane (10 km/s).</p>
<p>Conventional shaped charges drive parts of the metal liners to velocities around 7-14 km/s. As the density of these liners (usually copper) is much higher than water, we may expect water's maximum possible explosive-driven velocity to be even higher. However there are other issues such as vaporisation of the water when a shockwave passes through it (reflecting off the free surface as a rarefaction wave), but you're still projecting "water" albeit in the vapor phase. For argument's sake let's start at several km/s. From there you can estimate the dynamic pressure by 0.5 x density x velocity^2.</p>
<p>At these pressures, struck materials behave like fluids and yield may be ignored as a good approximation. In fact - based on my experience simulating exploding and imploding systems driven by high explosives - you get good approximations to reality by treating the impact as fluid or particle-based. You won't need any sand in the water here and the cutting would be almost instantaneous :)</p>
<p>Yes, there are explosively-driven water jets out there, mostly for niche applications like disposal of IEDs by severing the detonation train. It was found that water jets, even those traveling at a few km/s, are not likely to initiate explosives compared to say metal jets traveling at the same velocities.</p>
| 16306 | What's the maximum water jet psi we can create? |
2017-07-15T17:00:41.263 | <p>I am researching a system that appears to have a hydraulic arm with some kind of "handbrake" as pictured below:</p>
<p>System Description:</p>
<ul>
<li>Table that can raise and lock into place</li>
<li>Locking mechanism triggered by user releasing lever</li>
<li>Locking mechanism released by user pulling lever
<a href="https://i.stack.imgur.com/Yo3F8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yo3F8.png" alt="desk"></a></li>
</ul>
<p>I'm new to hydraulics, and I don't know what kind of component this is called. I've searched hydraulic cylinder with handbrake and pneumatic cylinder, but I'm not getting any good information. Can anyone point me in the right direction?</p>
<p>Thanks in advance!</p>
| |hydraulics|pneumatic| | <p>It is a gas spring. They are rated in units of pounds and travel in inches. Typically ranges are 10 to 200 pounds, and 6 to 24 inches. Here is one example :
<a href="http://rads.stackoverflow.com/amzn/click/B002UCAHAE" rel="nofollow noreferrer">https://www.amazon.com/JR-Products-GSNI-5300-60-Gas-Spring/dp/B002UCAHAE/ref=lp_16413861_1_3?s=industrial&ie=UTF8&qid=1500165912&sr=1-3</a> </p>
<p>The particular one being shown in your diagram is a specialized version with a locking valve which keeps the gas spring compressed until the valve button is pressed. They are a less common because they are used in more specialized applications, whereas the plain devices are used anywhere a door, lid, or window might be raised easier with the aid of a spring.</p>
| 16310 | Hydraulic Arm - Handbrake |
2017-07-17T19:56:58.680 | <p>This is part of an electric resin-extruding machine.</p>
<p>What's the part, and if possible, who makes it?</p>
<p>Any additional information regarding what the part is, or what the assembly is called, is greatly appreciated.</p>
<p>Here's the part:</p>
<p><a href="https://i.stack.imgur.com/Kd05d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kd05d.jpg" alt="Connected plate"></a></p>
<p>Here's the back side:</p>
<p><a href="https://i.stack.imgur.com/s79q7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s79q7.jpg" alt="Plate back side"></a></p>
<p>Here's where it connects:</p>
<p><a href="https://i.stack.imgur.com/wt7l0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wt7l0.jpg" alt="Connect to assembly"></a></p>
<p><a href="https://i.stack.imgur.com/UH29Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UH29Y.jpg" alt="Connected example"></a></p>
| |mechanical-engineering|gears|pumps| | <p>After extensive searching, it appears the part is referred to as a "driving flange" by the manufacturer's technical diagram, which I've included below.</p>
<p>Additionally, the pump itself seems to be an SPT gear metering pump made by Slack & Parr.</p>
<p><a href="https://i.stack.imgur.com/JwyZm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JwyZm.jpg" alt="Technical Diagram"></a>
<a href="https://i.stack.imgur.com/1mOX7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1mOX7.jpg" alt="Technical CAD Drawing"></a></p>
| 16322 | Identifying a connected plate on a gear liquid-metering pump |
2017-07-18T02:57:45.107 | <p>I already know solution, I just want to understand something in part b.</p>
<p>A vertical cylinder fitted with a piston contains air initially at 120kPa and 100 celcius. The piston is restrained by a set of stops. The piston diameter is 0.1m, its mass is 5kg, and the ambient pressure and temperature are 100kPa and 20 celcius. The cylinder is then cooled as heat is transferred to the surroundings.</p>
<p>a) At what temperature does the piston fall below the stops?</p>
<p>b) How far down (as a fraction of its initial height) has the piston moved when the contents of the cylinder has cooled to the ambient temperature?</p>
<p>a)
So this is assumed to be a 3-step process:</p>
<p><a href="https://i.stack.imgur.com/5zWNL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5zWNL.png" alt="enter image description here"></a></p>
<p>$ P_2 = P_{external} + \frac{mg}{A} $</p>
<p>Volume is <em>supposedly</em> constant from 1 to 2:</p>
<p>$ \frac{P_1 V_1}{T_1} = m R = \frac{P_2 V_2}{T_2} $</p>
<p>So $ \frac{P_1}{T_1} = \frac{P_2}{T_2}$</p>
<p>$ T_2 = T_1 \cdot \frac{P_2}{P_1} $</p>
<p>b)</p>
<p>Since 2 -> 3 is <em>supposedly</em> a constant pressure process:</p>
<p>$ \frac{P_3 V_3}{T_3} = \frac{P_2 V_2}{T_2} $ or $ \frac{V_3}{V_2} = \frac{T_3}{T_2}$</p>
<p>So $ \frac{V_2 - V_3}{V_2} = \frac{\delta h}{h_2} $</p>
<hr>
<p>So my question is, if the ideal gas equation is:</p>
<p>$ P V = mR \cdot T$</p>
<p>$m$, $R$ are constants</p>
<p>and if the temperature changes, then on the left hand side either pressure of volume can change, how is it known that from process 1 to 2, volume stays constant?</p>
<p>The problem postulates that heat of system will be cooled down by the surrounding temperature (20 C), nothing's said about neither pressure nor volume. So either of them can change... how does one know that volume stays the same from process going from 1 to 2?</p>
| |thermodynamics|gas|pistons| | <p>The solution for this problem relies upon the fact that the process is reversible, that is every state undergone by the system is in equilibrium.
Consider the initial state. Now the pressure inside the cylinder is 120 kPa. If we consider the equilibrium of the piston, 4 forces are acting upon it. They are: gravitational force(weight of piston) acting downward, atmospheric pressure acting downward,reaction force exerted by the stops acting downward andpressure inside cylinder acting upward. By doing the calculations we can see that the net downward force exerted by the atmospheric pressure and the weight of piston is 834 N. But the upward force exerted by the gas in the cylinder is 942 N. Hence a downward force of 108 N is required to be applied by the stops as a reaction force. Since reaction force is a surface force this force cannot be applied unless the piston and the stops remain in contact, ie the volume of the gas inside the cylinder remains constant.
The need for such a reaction force to maintain equilibrium will exist until the pressure inside the cylinder reduces to about 106.24 kPa . This is the process 1-2.
After reaching the state point 2 , the force exerted by the stops will be zero. So further reduction in pressure inside cylinder is impossible since it would disrupt the equilibrium. Hence, if temperature is again reduced the volume will decrease, as seen in process 2-3</p>
| 16330 | Thermodynamic, ideal gas problem |
2017-07-18T04:16:47.473 | <p>I know major oil pipelines do exists either underground or under the seabed. However, these usually move crude/unrefined oil from refinery to refinery. Yet we don't have refined petroleum or gas directly piped from refinery to the gas station? I mean we have done it with electricity, water, sewage, stormwater and piped LPG gas (for cooking), why do we still see petrol takers plying the roads to top up gas stations that you go to top up your car?</p>
<p>NOTE: I understand this is easier to achieve in big cities and Urban areas rather than laying pipework for rural areas. But I so not yet have concrete info this is implemented in long term gas stations located in the cities. </p>
| |pipelines|petroleum-engineering|infrastructure| | <p>The statement part of you questions is mostly incorrect. There are a great
number of product pipelines ( At least in the US) . There are a few stations that have pipelines to refineries or bulk plants but that is unusual. Some of the problems have been mentioned ; certainly the biggest problem would be getting the plethora of permits for each line. Another tremendous problem is the ignorant public ; For existing pipelines the most common cause of significant leaks is "third party" damage. Where a contractor for a city, state, etc, puts a hole in an 18" pipe because he didn't see it or all the warning signs. Can you imagine if there were a million 1" diameter pipes running all over? Corrosion caused by interference in cathodic protection is already a problem with pipelines in urban areas. However , there are many more local supply lines than you can imagine, for example I expect about every major airport is supplied by pipe ( jet and av gas). Most interstate truck stops have pipelines,etc.</p>
| 16331 | Oil refinery pipeline direct to gas stations |
2017-07-19T20:49:31.327 | <p>What are the <strong>properties</strong>, <strong>purposes</strong> and <strong>relationships</strong> between: <strong>Markers</strong>, <strong>Blocks</strong> and <strong>Parts</strong> in MSC Adams?</p>
<p>I don't understand how to use them properly.</p>
| |simulation|cad|software| | <p>Markers are coordinate systems they represent a position and orientation. Parts represents separate rigid bodies and ground. </p>
<p>Each actionable part consists of atleast one marker which represents the center of gravity of a part. Parts and most other modeling elemenrs act as groups to markers.</p>
<p>Markers represent the mathematical unerpinnings of systems, each joint (constraint) has a marker in both parts, forces emanate to and from markers, and measuremenrs are made to markers. For the simulation to work all you need are parts, markers, forces and joints.</p>
<p>Rest are convenience, you do not need geometry for example. For example geometry it does make easier to generate innertia values, softbody stfnesses and interpretting the result animations.</p>
| 16356 | How do markers, blocks, and parts work in MSC Adams? |
2017-07-19T22:25:30.210 | <p>I have a recessed shelf which holds three components that each generate heat:</p>
<ul>
<li>A projector</li>
<li>A computer</li>
<li>An audio power amp</li>
</ul>
<p>The shelf is located at the top of the wall and open at the front but closed on the 5 other sides. Because of this, the airflow created by the component fans is simply not enough to cool the space and things are beginning to overheat.</p>
<p>I'd like to install a fan to cool the space but not sure what the best placement and airflow direction would be to cool the space. One consideration is the direction of the existing projector fans which create a bit of an airflow pattern on their own. I'd also like to keep the heat inside the house since it's beneficial in the winter.</p>
<p>My question is basically, what is the optimal direction and placement of a single fan in order to exchange the air and cool the space?</p>
<p><strong>EDIT:</strong> To clarify, I intend to cut a hole in one of the existing surfaces and install the fan to move air between rooms, <em>and not</em> just to blow air around inside the shelf.</p>
<p><a href="https://i.stack.imgur.com/rkGVF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rkGVF.jpg" alt=""></a></p>
| |airflow|cooling| | <p>You have a couple good options. Don't worry about the fans that cool the pc or projector. They will be of little influence since they are far from the cool air that needs exchanged. Even if they were near a boundary, any tiny benefit in complex considerations like this lost when you upgrade hardware. The key is to look for the source of reliably cool air, existing hvac airflows, and to work with natural convection.</p>
<ol>
<li><strong>Cool in, existing hole.</strong> Place a fan at the bottom of the box near the already open face and blow inward. The projector throw will probalby keep you from going all the way to the front. Closer to the front will be more effective. This requires no new holes, and its flow is not limited by closed doors or other hvac airflows in your house. Simple and reliable.</li>
<li><strong>Hot out, existing hole.</strong> Place the fan at the top of the box near the already open face and blow hot air out. Same benefits and limitations as number 1. The difference between these only comes into play when you are worried about enclosure positive pressure and filtering.</li>
<li><strong>Hallway hole, hot out top or cold in bottom.</strong> Make sure to cut the hole larger and select a larger grating than just the fan. This will reduce performance with some recirculation, but will ensure your hvac system doesn't fight it and closed doors don't seal it off. The down side is that loud audio will leave your theater room through this hole.</li>
<li><strong>Stairway hole, cold in from bottom.</strong> I do not recommend air out the bottom, because you are working against natural convection. This could also have the same noise leaking issues.</li>
</ol>
<p>On a side note, make sure you purchase a <a href="http://rads.stackoverflow.com/amzn/click/B00EON40CS" rel="nofollow noreferrer">"80+" rated powersupply</a> for your pc. They are 80% efficient or greater and will reduce the amount of heat you have to get rid of. In my experience they are also quite a bit more reliable.</p>
<p>When selecting fans get one with a high cfm (cubic feet per minute) rating and a low dB (decibel) noise rating. Something like this <a href="http://rads.stackoverflow.com/amzn/click/B008G3OLRQ" rel="nofollow noreferrer">200mm 700rpm one</a>. If you use more than one fan put them in parallel with both of them on "blow cold in bottom" or "blow hot air out the top". This will effectively double your airflow. Putting them in series with one on hot and one on cold will not increase total air exchange by very much.</p>
| 16359 | Most efficient airflow pattern to cool a recessed shelf |
2017-07-20T10:19:52.037 | <p>From communication prospective why can't higher frequency ranges be reflected from atmospheric layers and what is the effect of parameters (like temperature, pressure, etc) of the atmospheric layer on reflection of waves for sky wave and space wave propagation.</p>
| |telecommunication|wireless-communication| | <p>Different materials reflect, absorb, and scatter different wavelengths of light. The ionized layer at the edge of earth's atmosphere happens to reflect a certain range of wavelength while transmitting others. This is due to the physics of the material.</p>
<p>The reason you can't use the atmosphere to reflect shorter wavelengths for the purposes of communication, is because the atmosphere doesn't reflect those wavelengths in the first place.</p>
<p>The primary parameters affecting the atmosphere's reflectivity is the amount of ionization and the depth of the ionized layer. That largely depends on solar activity. There is a significant difference in this layer between night and day. It is also subject to individual solar phenomena, like flares or coronal mass ejections.</p>
| 16367 | Why only the spectrum range of radio waves can be reflected from atmospheric layers while higher ranges (Infrared-Visible Light-Ultraviolet) can't? |
2017-07-21T21:01:02.353 | <p>This question might look silly. I am not knowing the correct engineering terms to use in this question. I am trying to explain using the words what I know in English. </p>
<p>We have been hired some construction workers for a work in our backyard. I have some doubt on reinforcing beam for slabs. I came through following 2 models </p>
<p>.</p>
<p><strong>1.</strong> Model one</p>
<p><a href="https://i.stack.imgur.com/mFk9z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mFk9z.png" alt="Insert the steel bar for slab through the steel bear created for the beam"></a></p>
<p>.</p>
<p><strong>2.</strong> Model two</p>
<p><a href="https://i.stack.imgur.com/5C5us.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5C5us.png" alt="Put the steel bar for slab above the bar made for beam."></a></p>
<p>According to my perception, I believe the second one is much better in terms of strength. Because the beam is supporting the weight of slab. But my friend suggested me the first one is the right way. But I am not aware why the first one is better than the second. Could any one please help me out in this case to identify the better one.</p>
| |structural-engineering|beam|reinforced-concrete| | <h2>Model 1 is always better, but may need some modifications</h2>
<p>Whenever you have two reinforced concrete elements, you always need to facilitate the transfer of internal stresses between them. This is done by "mixing" their reinforcement.</p>
<p>For instance, model 2 will have no steel between the slab steel and the beam steel. This means that the connection between them will be very fragile and weak to horizontal shear, meaning that there's a risk of a horizontal crack like this:</p>
<p><a href="https://i.stack.imgur.com/KMpB1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KMpB1.png" alt="enter image description here"></a></p>
<p>With Model 1, this isn't a risk because the vertical steel from the beam ("stirrups") will resist these forces.</p>
<p>Model 1 also has an advantage in that it makes the beam itself stronger, since it allows the beam to behave as a much taller beam (including the height of the slab). Indeed, it even allows the beam to behave not as a rectangular beam, but as a stronger T-shape beam, using some of the slab to resist some of the beam's internal forces (how much depends on your country's standards and codes).</p>
<p>If the beam and the slab will be poured simultaneously, then that's it, and you can stop reading this answer.</p>
<p>However, if there's a chance that the beam will be built first and then the slab will be poured over it, then that means that the beam's stirrups and negative reinforcement (the longitudinal bars at the top of the beam) will be outside of the concrete when the beam is initially built but the slab hasn't yet been poured. This means that the beam's weight and that of the freshly-poured slab (before it gains enough stiffness to do anything) will have to be resisted by the "short" beam alone (without the added height from the slab), so some additional steel will be necessary:</p>
<p><a href="https://i.stack.imgur.com/6bPIv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6bPIv.png" alt="enter image description here"></a></p>
<p>You are basically reinforcing the same beam twice: once considering it as the short beam (without the slab), and once as the tall beam (with the slab). To avoid confusion as to how to define how much reinforcement goes into each position, here's a list:</p>
<ul>
<li>positive reinforcement (longitudinal steel at the bottom of the beam): calculate the necessary reinforcement for the short beam to resist the beam's and the slab's self-weight, and then calculate the reinforcement for the tall beam to resist any additional loads (dead loads, live loads, etc). Add these two numbers and you have the total necessary positive reinforcement.</li>
<li>negative reinforcement (longitudinal steel at the top of the beam) and the shear reinforcement (stirrups) for the short beam: calculate the necessary reinforcement for the short beam to resist the beam's and the slab's self-weight. You may need to just put the minimum reinforcement.</li>
<li>negative reinforcement (longitudinal steel at the top of the beam) and the shear reinforcement (stirrups) for the tall beam: calculate the reinforcement for the tall beam to resist any additional loads (dead loads, live loads, etc). You may need to just put the minimum reinforcement.</li>
</ul>
| 16385 | Reinforcing beam and slab with correct steel bar structure |
2017-07-22T13:44:41.550 | <p>Since an evaporator also a heat exchanger, how can the inlet and outlet temperatures of the HEX accepted as the same, i.e. isothermal? In this case, cold stream (inlet and outlet of the evaporator in the refrigeration cycle) does not incur any change in temperature, which requires an infinite size of the heat exchanger.</p>
| |evaporation| | <p>Assuming you mean evaporator as part of a vapour compression cycle, then because the fluid enters as a liquid and leaves as a gas, which means that the temperature can be the same but the internal energy changes due to the latent heat of vapourisation.</p>
| 16396 | Why is it generally assumed that the evaporator works isothermally? |
2017-07-24T05:20:44.503 | <p>I am, in actuality, a network engineer and not a trained mechanical engineer. It was kind of a hobby of mine until recently, when my boss gave me the opportunity to try my hand at mechanical engineering through CAD design (Autodesk Inventor) about two years ago and I took to it like a duck to water. I have been working in that subject since then.</p>
<p>My problem, however, is that I am unfamiliar with most of the best-practices of mechanical engineering and it has been a "trial by fire" sort of situation. Currently my main problem has to do with the alignment of a chain-drive system used for balancing a counterweight and lifting mechanism.</p>
<p>Details:</p>
<ul>
<li>Horizontal alignment is great in my CAD design, however when I actually order and build out the products, I find it extremely difficult to get the alignments properly set and I end up spending hours fidgeting with the axle and chain sprocket alignment... I am currently using aluminum extrusions (via MISUMI) for the frame design.</li>
</ul>
<p>In the images below, you will find an example (several components have been redacted due to proprietary design). The images show the main components for operating the lift mechanism. One side houses the sprockets for the Counterweight, the other for the forklift itself. It is controlled by an AC servo motor which attaches to a custom gearbox, and finally to one of the chain axles (the counterweight).</p>
<p>**The design must use these aluminum extrusions, however this makes it very difficult to properly align the bearing housings for the axles, often making everything off-center or skewed, Even when using an accurate measure and square.</p>
<p>Can anyone suggest some improvement points or alternative types of components for mounting the axle/sprockets?</p>
<p>Side Profile:
<a href="https://i.stack.imgur.com/owZZx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/owZZx.png" alt="Side Profile"></a></p>
<p>Perspective Profile:
<a href="https://i.stack.imgur.com/WkTnE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WkTnE.png" alt="Perspective Profile"></a></p>
| |mechanical-engineering|design|mechanisms| | <p>A very common method for shaft alignment is by using shims and a <a href="https://ae01.alicdn.com/kf/HTB1vXQjMpXXXXaKXXXXq6xXFXXXh/2-IN-1-font-b-Dial-b-font-font-b-Indicator-b-font-0-0-8mm.jpg" rel="nofollow noreferrer">DTI</a> (dial test indicator). Having a <strong>solid bed plate</strong> for the bearing housings is a must-have. Depending on the load, the aluminium extrusions may flex, which makes any fine adjustments to the bearing housings redundant. The alignment process generally involves using a or better yet a <a href="https://i.ytimg.com/vi/PNMfJLaoUqo/maxresdefault.jpg" rel="nofollow noreferrer">laser alignment kit</a>. There are plenty of videos on YouTube that document the process of using a DTI to align shafts with a central coupling, but these can be extrapolated to using it with one shaft across two bearings.</p>
| 16411 | how can I improve my chain-drive lifting system? |
2017-07-24T06:47:23.790 | <p>I'm designing a water proof case for a power bank to be used in an occasional underwater condition. The input and out put port components already have a gasket or sealing ring. Now to close the case I am planning to mill a metal with all walls and floor and then screw the top along with and O-ring. I haven't used an O-Ring before and wanted to know how to select an O-ring for a rounded rectangle groove (Does the circumference being equal to the grove length , enough). Is there any handbook for the same?</p>
| |waterproofing|fluid| | <p>With any O-ring seal, static or dynamic, the key parameter is "O-ring Squeeze" which is defined as the percentage reduction in cross sectional area when compressed from the free state. Since the o-ring acts like a spring, this squeeze is directly related to the sealing force. The greater the squeeze, the greater the sealing force, and the greater resistance to leakage...at least in general. </p>
<p>As with most things, more is not always better, too much squeeze can result in a poor joint since it can cause excess stress in the mating parts, excess friction (in the case of a dynamic seal), and O ring damage since there's a greater risk of the o-ring getting pinched between the two mating surfaces. </p>
<p>In designing O-ring seals in stainless steel with good surface finish, I would try to stick to < 15% squeeze for dynamic seals and 15%-30% for static seals using a standard NBR O-ring from Parker, but it really depends on the type of O-Ring, the materials involved and all of the other specifics of the application. </p>
<p>In your case, you'll want to use an O-ring that has a circumference close to the mean perimeter of your groove. You can, of course buy custom O-rings, but those can be expensive. </p>
<p>There's a lot that goes into the design of a good O-ring joint. The best resource I've found is the <a href="http://promo.parker.com/promotionsite/oring-ehandbook/us/en/ehome" rel="nofollow noreferrer">Parker O-ring Handbook</a>. </p>
| 16413 | O-Ring Selection for a round rectangle groove |
2017-07-24T07:58:38.007 | <p>I have googled for a while and I can't seem to find the correct term to refer to this.</p>
<p>Imagine you have a tunnel in where trains, cars, or anything goes through.</p>
<p>This tunnel might elements along its extension, such as cracks, fissures, weak spots, etc. To visualize this in paper, a 3D figure must be "projected" into 2D. </p>
<p>I'm looking for the word to refer to this kind of map.</p>
<p>The closest I've come to is calling it "A projection map" but is there a more suitable term for this?</p>
<p><a href="https://i.stack.imgur.com/Z9ZND.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z9ZND.jpg" alt="enter image description here"></a></p>
<p>In japanese it can be called a "展開図(てんかいず) Tenkaizu", but a dictionary translation, or a google translation don't fit since that term is also used to describe the diagrams of how to fold paper to build something. (like a cube)</p>
<p>I tried in the "English" site, but I was referred to this site instead.</p>
<ul>
<li><a href="https://english.stackexchange.com/questions/402469/how-do-you-call-a-map-that-shows-the-characteristics-of-a-train-tunnel-3d-in-p?noredirect=1#comment955561_402469">What do you call a 2D map that shows the characteristics of a 3D
train tunnel?</a></li>
</ul>
| |civil-engineering| | <p>Within the tunnelling environment I work in we refer to that sort of image as a 'developed elevation'. </p>
| 16414 | How do you call a map that shows the characteristics of a train tunnel (3D) in paper (2D)? |
2017-07-24T23:27:28.390 | <p>I have a set of equations of motion describing a planetary gear train of 18 DoF (sun, 3 planets, carrier and ring), they have the general form of:</p>
<p>$$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{Kq}=F(t)}$$</p>
<ul>
<li>$K$: stiffness symmetrical matrix</li>
<li>$M$: Mass symmetrical matrix</li>
<li>$G$: Gyroscopic skew symmetrical matrix</li>
<li>$\Omega_c$: angular speed</li>
<li>$F$: excitation</li>
<li>$q$: generalized coordinates</li>
</ul>
<p>While trying to solve the eigenvalue problem with Matlab, I noticed there´s a whole bunch of ways to represent the equations in state space form and I don´t understand what the main difference is, or which one I should use, since when calculating I get different results in the eigenvectors.</p>
<p>But for the free vibrations I have the equation without damping:</p>
<p>$$\mathbf{M{\ddot q}+{Kq}=0}$$
<strong>1.</strong>
and to solve the standard eigenvalue problem of:</p>
<p>$$\mathbf {\omega^2M\phi_i-{K}\phi_i=0}$$
I use the Matlab code:</p>
<pre><code>[v,D]=eig(K,M)
root=sqrt(D)
natural_frq=root/(2*pi)
</code></pre>
<p><strong>which gives me all real natural frequencies and eigenvector per frequency.</strong></p>
<p><strong>2.</strong>
I also tried calculating them in the form of:
$$\mathbf{{\dot x}=Ax+Bu}$$</p>
<p>$$ \begin{bmatrix} \mathbf{\dot q} \\ \mathbf{\ddot q} \end{bmatrix} = \begin{bmatrix} \mathbf{0} & \mathbf{I} \\ \mathbf{-M^{-1}{K}} & \mathbf{-M^{-1}{G}} \end{bmatrix} \begin{bmatrix} \mathbf{ q} \\ \mathbf{\dot q} \end{bmatrix}+ \begin{bmatrix} \mathbf{0} \\ \mathbf{-M^{-1}} \end{bmatrix}u$$</p>
<p>and by using:</p>
<pre><code>[v,D]=eig(A)
</code></pre>
<p>I get similar results but in complex numbers (there´s no need for the square root), so with some extra code I choose only the imaginary parts and select the positive values of the eigenvalues vector, which instead of 18 elements is now 36. <strong>The frequencies I get with this method against the first ones are paractically the same, but the eigenvectors are not</strong>, they are also complex and the results don´t look correct.</p>
<p><strong>3.</strong>
Then there's another state space form:</p>
<p>$$ \begin{bmatrix} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{M} \end{bmatrix} \begin{bmatrix} \mathbf{\dot q} \\ \mathbf{\ddot q} \end{bmatrix} + \begin{bmatrix} \mathbf{0} & \mathbf{-I} \\ \mathbf{K} & \mathbf{G} \end{bmatrix}\begin{bmatrix} \mathbf{ q} \\ \mathbf{\dot q} \end{bmatrix} = 0$$</p>
<p>If I use the code:
[v,D]=eig(-B,A)
I get similar results as the last method, I'm not sure which form of state space this form is, I would guess it's $$A{\dot x} = -Bu$$.</p>
<p>The main question is that I don't know which is the best solution to my problem and why I get complex frequencies in the second method, if my angular speed $$\Omega_c$$ is zero, all damping terms should also go to zero and thus not affect the system results , right?</p>
<p>EDIT: why do I get very different values when calculating the eigenvectors through the standard eigenproblem form $$\mathbf {\omega^2M\phi_i-{K}\phi_i=0}$$ and with the state space?</p>
| |mechanical-engineering|gears|dynamics|frequency-response|eigenvalue-analysis| | <p>You basically want to solve for the homogeneous solutions of the system</p>
<p>$$
\textbf{M}\, \ddot{\textbf{q}} + \textbf{D}\, \dot{\textbf{q}} + \textbf{K}\, \textbf{q} = \textbf{0}.
$$</p>
<p>It can be shown that each solution will be of the form</p>
<p>$$
\textbf{q}_i(t) = \textbf{u}_i\,e^{\mu_i\,t},
$$</p>
<p>where $\textbf{u}_i$ is a vector of the same dimension as $\textbf{q}$ (usually normalized to a length of one) and $\lambda_i$ is a scalar. In general any homogeneous solution can be constructed by taking a linear combination these solution (I will ignore for now that you can also have <a href="http://mathworld.wolfram.com/JordanCanonicalForm.html" rel="nofollow noreferrer">Jordan blocks</a> of bigger then one by one, which will increase the possible solutions).</p>
<p>Substituting the solution into the homogeneous equation yields</p>
<p>$$
\textbf{M}\, \textbf{u}_i\,\mu_i^2\,\,e^{\mu_i\,t} + \textbf{D}\, \textbf{u}_i\,\mu_i\,\,e^{\mu_i\,t} + \textbf{K}\, \textbf{u}_i\,e^{\mu_i\,t} = \left(\textbf{M}\, \mu_i^2 + \textbf{D}\, \mu_i + \textbf{K}\right) \textbf{u}_i\,e^{\mu_i\,t} = \textbf{0}.
$$</p>
<p>Since $e^{\mu_i\,t}$ is always non-zero (for finite time) it can be ignored since the remaining terms on the left hand side of the equation still has equal to zero in order to satisfy the right hand side. Doing so already starts to make the equation look more like a <a href="https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Generalized_eigenvalue_problem" rel="nofollow noreferrer">generalized eigenvalue problem</a></p>
<p>$$
\textbf{A}\, \textbf{v} = \lambda\, \textbf{B}\, \textbf{v},
$$</p>
<p>however $\mu_i$ appears both linearly and quadratically. In the special case when $\textbf{B} = 0$ then the generalized eigenvalue problem can be formulated using $\textbf{A} = \textbf{K}$, $\textbf{B} = \textbf{M}$, $\textbf{v} = \textbf{u}_i$ and $\lambda = -\mu_i^2$. Another special case is when $\textbf{M}$, $\textbf{D}$ and $\textbf{K}$ are all simultaneously diagonalizable by pre and post multiplying them by the same matrices. Namely in that case the differential equation can be written as a system of decoupled second order differential equations, like a single mass-spring-damper system. One way such transformation could be constructed when $\textbf{M}$, $\textbf{D}$ and $\textbf{K}$ are symmetric is</p>
<p>$$
\mathbf{U} = \begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \cdots & \mathbf{u}_n\end{bmatrix},
$$</p>
<p>where $\mathbf{u}_i$ are the eigenvectors of the eigenvalue problem with $\textbf{B} = 0$. The matrices with which you pre and post multiply would then be $\mathbf{U}^\top$ and $\mathbf{U}$ respectively. Often if the system is not simultaneously diagonalizable and the damping is small, then assuming it is can be a good approximation (so setting the non-diagonal elements of $\mathbf{U}^\top \mathbf{B}\, \mathbf{U}$ to zero). A specific form for which this is the case is called Rayleigh damping, when $\mathbf{B} = \alpha\,\mathbf{M} + \beta\,\mathbf{K}$.</p>
<p>The advantage of all the previous mentioned methods is that the generalized eigenvalue problem you have to solve are of the same dimension as $\mathbf{q}$. However if none of cases mentioned previously apply to your system you would have to solve the generalized eigenvalue problem of twice the dimension of $\mathbf{q}$. There are multiple ways of formulating this. A few of them are</p>
<p>$$
\begin{bmatrix}
\mathbf{I} & \mathbf{0} \\
\mathbf{0} & \mathbf{M}
\end{bmatrix}
\begin{bmatrix}
\dot{\mathbf{q}} \\ \ddot{\mathbf{q}}
\end{bmatrix} =
\begin{bmatrix}
\mathbf{0} & \mathbf{I} \\
-\mathbf{K} & -\mathbf{D}
\end{bmatrix}
\begin{bmatrix}
\mathbf{q} \\ \dot{\mathbf{q}}
\end{bmatrix},
$$</p>
<p>$$
\begin{bmatrix}
\mathbf{I} & \mathbf{0} \\
\mathbf{0} & \mathbf{I}
\end{bmatrix}
\begin{bmatrix}
\dot{\mathbf{q}} \\ \ddot{\mathbf{q}}
\end{bmatrix} =
\begin{bmatrix}
\mathbf{0} & \mathbf{I} \\
-\mathbf{M}^{-1}\mathbf{K} & -\mathbf{M}^{-1}\mathbf{D}
\end{bmatrix}
\begin{bmatrix}
\mathbf{q} \\ \dot{\mathbf{q}}
\end{bmatrix},
$$</p>
<p>$$
\begin{bmatrix}
\mathbf{D} & \mathbf{M} \\
\mathbf{M} & \mathbf{0}
\end{bmatrix}
\begin{bmatrix}
\dot{\mathbf{q}} \\ \ddot{\mathbf{q}}
\end{bmatrix} =
\begin{bmatrix}
-\mathbf{K} & \mathbf{0} \\
\mathbf{0} & \mathbf{M}
\end{bmatrix}
\begin{bmatrix}
\mathbf{q} \\ \dot{\mathbf{q}}
\end{bmatrix}.
$$</p>
<p>For this extended state space it can easily be seen that the solutions will be of the form</p>
<p>$$
\begin{bmatrix}
\mathbf{q}_i(t) \\ \dot{\mathbf{q}}_i(t)
\end{bmatrix} =
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix} e^{\mu_i\,t}.
$$</p>
<p>Substituting this into the three equations above and factoring out $e^{\mu_i\,t}$ yields</p>
<p>$$
\begin{bmatrix}
\mathbf{I} & \mathbf{0} \\
\mathbf{0} & \mathbf{M}
\end{bmatrix}
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix} \mu_i =
\begin{bmatrix}
\mathbf{0} & \mathbf{I} \\
-\mathbf{K} & -\mathbf{D}
\end{bmatrix}
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix},
$$</p>
<p>$$
\begin{bmatrix}
\mathbf{I} & \mathbf{0} \\
\mathbf{0} & \mathbf{I}
\end{bmatrix}
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix} \mu_i =
\begin{bmatrix}
\mathbf{0} & \mathbf{I} \\
-\mathbf{M}^{-1}\mathbf{K} & -\mathbf{M}^{-1}\mathbf{D}
\end{bmatrix}
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix},
$$</p>
<p>$$
\begin{bmatrix}
\mathbf{D} & \mathbf{M} \\
\mathbf{M} & \mathbf{0}
\end{bmatrix}
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix} \mu_i =
\begin{bmatrix}
-\mathbf{K} & \mathbf{0} \\
\mathbf{0} & \mathbf{M}
\end{bmatrix}
\begin{bmatrix}
\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i
\end{bmatrix}.
$$</p>
<p>These are all just generalized eigenvalue problems. I am not an expert on numerically solving these problems, however I have few things that can be noted. Namely the second problem formulation requires a matrix inverse, which might not exist, but does reduce it to a normal eigenvalue problem (not sure if this requires any less computation time in general). The third problem formulation will yield symmetric matrices if $\textbf{M}$, $\textbf{D}$ and $\textbf{K}$ are symmetric as well, which might have some attractive properties when doing numerical calculations. So depending on your system and solver a different formulation might give more accurate results.</p>
| 16426 | What's the difference of using different state-space forms for solving eigenvalue problem? |
2017-07-25T20:52:30.507 | <p>I am building a device and I tried to Google an answer to this question but I am an electrical person and lack knowledge in this field. </p>
<p>I need to know what the relative humidity inside an enclosure (of my device) will be for a given ambient temperature. If I know:-</p>
<p>Relative humidity outside the enclosure (95% RH)
The ambient temperature (40 deg C)
The air temperature inside the enclosure (60 deg C)</p>
<p>Since the air inside the enclosure is hotter because of the electronics I know the RH inside will be lower but I do not know how to go about quantifying and justifying it on paper.</p>
| |mechanical-engineering|temperature|environmental-engineering| | <p>Using the +10 to +120°C psychrometric chart with sensible heating from 40°C, 95% RH to 60°C gives approximately 35% RH. If you need a more precise answer I would look up the formulas which can be found in ASHRAE Fundamentals or CIBSE Guide C.</p>
<p>See image below for how to look this up on the chart:
<a href="https://i.stack.imgur.com/KCDCu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KCDCu.png" alt="psychrometric chart with markup"></a></p>
| 16445 | Correlating relative humidity outside and inside an enclosure |
2017-07-26T02:22:28.040 | <p>Pressure pumps are used to apply pressure to a system. What would happen if someone attempted to fill, for example, a bicycle tire (with a check valve) to 2 ATM of pressure using a pressure pump that could only supply 1.5 ATM of pressure? As the gas cannot go backwards due to the check valve during filling, could the pressure be expected to rise in the tire to 2 ATM?</p>
| |pressure|compressors|compressed-air|wheels| | <p>No, because the pump is designed to limits and if you exceed its working or maximum pressure then, for example, its seals fail and it just does not pump any more.</p>
| 16452 | Compression and check valves |
2017-07-26T12:35:28.490 | <p>Instead the pump relied on the diesel generators, why didn't it just use the residual heat to generate electricity?</p>
<p>P.s. The question is very specific to the residual heat generated after inserting the control rods and the cooling pump which failed to cool down the system since it has no electricity when the diesel generators failed.</p>
| |nuclear-engineering| | <p>According to the <a href="https://en.wikipedia.org/wiki/Fukushima_Daiichi_nuclear_disaster#Events" rel="nofollow noreferrer">Wikipedia entry</a>:</p>
<blockquote>
<p>[The earthquake] produced maximum ground g-forces of 0.56, 0.52, 0.56 (5.50, 5.07, and 5.48 m/s2) at units 2, 3, and 5 respectively. This exceeded the earthquake tolerances of 0.45, 0.45, and 0.46 g (4.38, 4.41, and 4.52 m/s2). The shock values were within the design tolerances at units 1, 4, and 6.</p>
<p>When the earthquake struck, units 1, 2, and 3 were operating, but units 4, 5, and 6 had been shut down for a scheduled inspection. Reactors 1, 2, and 3 immediately shut down automatically; this meant the plant stopped generating electricity and could no longer use its own power. One of the two connections to off-site power for units 1–3 also failed, so 13 on-site emergency diesel generators began providing power.</p>
</blockquote>
<hr>
<h2>:EDIT:</h2>
<p><a href="https://engineering.stackexchange.com/questions/16459/why-didnt-the-fukushima-turbine-supply-electricity-to-the-residual-heat-removal/16461?noredirect=1#comment29580_16461">OP asked</a>:</p>
<blockquote>
<p>yes, but isn't "immediately shut down automatically" means inserting Control rods? There is still unstoppable nuclear fission which create the residual heat, I'm just wondering why they don't utilize the residual heat to cool itself</p>
</blockquote>
<p>Well, the "residual heat" (<a href="https://en.wikipedia.org/wiki/Decay_heat#Power_reactors_in_shutdown" rel="nofollow noreferrer">decay heat</a>) <em>is</em> an effect that generates heat, but (speaking from military reactor experience, not civilian) one of the goals after a scram or fast insertion is to recover the plant, and that means not letting the primary cool down too much. This is assisted by automatically closing the main steam valves on the steam generators, which isolates the secondary plant and thus also isolates the turbine generators. </p>
<p>In order to recover the plant, you need to establish <em>why</em> the shutdown occurred, assess any damage, and then start bringing the plant back online. The entire time this is happening though, the secondary plant piping is cooling down. You can get into a scenario where you may need to do an accelerated secondary warmup, which still takes time despite being "accelerated."</p>
<p>To address your question directly, I'll make a couple points that are <em>pure speculation on my part.</em> </p>
<ol>
<li>Biggest point in my mind is that there was an assumption that backup power would be available regardless of current catastrophic conditions. There's no need to cool the plant down too quickly if you have the luxury of time thanks to a bank of diesel generators. </li>
<li>Next biggest point - In the heat of the moment, you need to respond to <em>what is happening</em>. You can't take time to consider the "what ifs" that <em>could</em> happen down the line; you need to address the current casualty. You can say, "The reactor just tripped offline, what if the generators go down?" but you could also say, "The reactor just tripped offline, what if the main steam valves get stuck shut?" or, "..., what if the cooling water pumps won't restart?" The list of hypothetical questions goes on. It was up to the engineers that designed the power plant to consider the likely scenarios, and they codify the actions you should take in a book of <em>emergency procedures.</em> If they were doing their jobs correctly, <strong>the people running the plant were performing the actions dictated in a checklist that was written before the plant was commissioned.</strong> If the checklist said not to use decay heat for backup power (or failed to instruct them to use decay heat for backup power) then that would be why they didn't.</li>
<li>Available decay heat may have been insufficient or borderline. Bear in mind that when you make enough power to "just cool the primary," that implies that you also need to make enough power to run all of the auxiliary machinery for the secondary, too. Depending on the plant, you need to run multiple primary coolant pumps - these are NOT small pumps! Picture at the end with people for scale. But you need a means to return the condensed steam back to the steam generator, so you need condensate pumps running, and booster pumps running, and main feed pumps running, you need to have the lubricating oil system online, so you need to run oil pumps (again, not small for a large commercial turbine), and you need to have cooling water to cool the condenser, so you need to have those pumps online, also. I don't have a number on the percentage of full power that the auxiliary machinery consumes, but it is not insignificant. All this is to say that using decay heat to cool the primary plant down may result in a rapid cooldown that exceeds safety limits. The reactor vessel itself has stainless steel walls that are 8-10+ inches thick (20-25+ cm), and the temperature gradient across the thickness there and in the primary piping could result in stresses that cause <em>permanently bending</em> (plastic deformation). This is generally the concern for rapid cooldowns; heating up too fast could lead to <a href="https://en.wikipedia.org/wiki/Fracture#Brittle_fracture" rel="nofollow noreferrer">brittle fracture</a>, where part of the primary plant could suddenly and completely fail. </li>
</ol>
| 16459 | Why didn't the Fukushima turbine supply electricity to the residual heat removal system? |
2017-07-26T15:54:11.493 | <p>I know how to use thermal resistance circuits to calculate heat transfer in Cartesian and cylindrical coordinates, but I don't know how to deal with real-world insulation values, and I was not able to find much about this online.</p>
<p>For example, I have this hot water tank that has an insulation jacket with 2" R-12.5 insulation. For simplicity, let's assume the insulation is the only thing that separates the hot water from cold outside air (no heat transfer from bottom of tank, top of tank, etc. Just through the insulation jacket). Assume water tank thickness aside from the 2" insulation jacket is negligible.</p>
<p>So it goes hot water -> insulation -> cold air outside. I can look up a free convection coefficient for the tank's hot water and can look up a convection coefficient for air as well. That allows me to use the formula
$$
R=1/(h*2*pi*r*heightofinsulationjacket)
$$
twice (once for water and once for air).</p>
<p>So the problem is how to deal with the insulation jacket. How do I interpret 2" R-12.5 insulation? How do I factor in the thickness of the insulation? The surface area is taken care of in the heat transfer equation.</p>
<p>$$
Q''=(Tf-Ti)/R
$$
or
$$
Q=(Tf-Ti)/R*A
$$</p>
<p>Updated: How do I use an R-value in cylindrical coordinates?</p>
<p>e.g.
$$
R/A=1/(hW*2*pi*r1)+Rins/(whatAreaHere)+1/(hA*2*pi*r2)
$$</p>
<p>I am missing the area for my R-value insulation</p>
| |heat-transfer|energy-storage| | <p>@geekly: agreed. Also, insulation thickness will increase surface area for external convection, as:
$$ Q_{jacket \to air} = 2\pi (l_{heater})(r_{heater} + t_{jacket})h(T_{jacket} - T_{air})$$
This effect prompted the critical radius of insulation for cyclinders:
<a href="https://en.wikipedia.org/wiki/Thermal_insulation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Thermal_insulation</a>:
$$ r_{critical} = \frac{k}{h} = \frac{t_{jacket}}{R_{jacket}h}$$</p>
| 16464 | How do you use an insulation jacket's R-value for calculating heat transfer from a hot water tank? |
2017-07-27T13:50:35.320 | <p>This is a theoretical question.
It is known that in "classic" vibration analysis natural frequencies can be found by solving an eigenvalue problem, from the undampened vibration equation like</p>
<p>$\mathbf{M}\mathbf{\ddot{u}} + \mathbf{K}\mathbf{u} = \mathbf{0}$</p>
<p>where $\mathbf{u}$ is the displacement vector.
But, with reference to this situation
<a href="https://i.stack.imgur.com/MfooU.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MfooU.png" alt="Cantilever beam with cable"></a></p>
<p>I started to wonder if this type of analysis is not correct here because $\mathbf{K}$ (stiffness) of the cable is not constant but $\mathbf{u}$-dependant (equal to zero when $\mathbf{u}$ is a displacement of compression). </p>
<p>So I would like to ask if someone could suggest me another strategy, or show why the eigenvalue analysis could be used also in this case.
Thank you all.</p>
| |vibration|eigenvalue-analysis| | <p>The equation you showed is an ordinary differential equation. It works for solving natural frequencies of discrete systems (ie lumped rigid masses connected by massless springs). Here you have a continuous system. You will need to set up two partial differential equations, one for the cable and one for the beam. The resulting PDE can be discretized into a system of ODEs (eg galerkin method, finite element method). Thr details of how to do this are too complicated to go into here. Suggest that you pick up a textbook on vibrations of continuous systems </p>
| 16478 | Strategies for cable vibration: eigenvalues are suitable? |
2017-07-27T15:20:54.547 | <p>What is meant by a "system curve" that is used to determine the operating point of a pump. I know what a "pump curve" is, but I often hear about a "system curve" being generated so that the pump operating point can be determined by where the pump curve and system curve intersect.</p>
| |mechanical-engineering|fluid-mechanics|pumps| | <p>Imagine if you were pumping through a single pipe. That's all the pump does, is take in water from a source and pumps it along a very long length of pipe. Since this is just plain piping, the friction loss isn't difficult to determine, but you need to know the velocity to calculate the <a href="https://en.wikipedia.org/wiki/Reynolds_number" rel="nofollow noreferrer">Reynolds number</a>. Without it, you can't use a <a href="https://en.wikipedia.org/wiki/Moody_chart" rel="nofollow noreferrer">Moody Chart</a> to calculate a friction factor to find out the loss. But, with a good estimate of the flow rate, the estimated range of the flow rate could be guessed at, and a friction factor can be guessed. With a friction factor estimate, but a flow rate unknown, you finally settle on the pressure loss across this long length of pipe:</p>
<p>$$\Delta P =f_D \frac{\rho V^2}{2}\frac{L}{D} = f_D \frac{8\rho Q^2}{\pi^2}\frac{L}{D^5}$$ </p>
<p>Where P is pressure, Q is the flow rate, $\rho$ is the density, L is pipe length, D is diameter, and $f_D$ is the friction factor. Let's take cast iron 100mm pipe, highly turbulent regime, running for 10km. Plugging all the numbers in this theoretical situation, we come to (hypothetically):</p>
<p>$$\Delta P = 0.304 Q^2 \frac{kPa}{(\frac{L}{s})^2} $$ </p>
<p>Now we have a formula to give an idea for pump sizing! More importantly, we have a curve. Plugging in any arbitrary value of Q (in L/s) would give you the pressure loss across the pipe in kPa. This curve is the system curve, and it naturally looks like a parabola. This is, in general, true for all fluid systems without any type of controls response (acting under natural behavior). You can plot this curve on top of the pump curve, and find out where the system will reach equilibrium. </p>
<p>Note, it isn't hard with such a simple system to get a lot of liquid out. More complex system have a lot more intricate mechanics, but the general parabola rule still applies. Thus, most people work through the complex mechanics of their system to simplify it into a single point. In our case, operating at 25 L/s would mean a pump that has to pump 190 kPa. This is the single operating point. Typically, many engineers will slightly increase these values, so they will always find an operating point that's safe. In this case, going to 30 L/s would mean a 275 kPa. Thus, the only parabola that goes through the point 30 L/s, 275 kPa and the origin (there is only one parabola that does this) would be the system curve.</p>
| 16481 | System curve for use in determining pump operating point |
2017-07-27T01:36:15.730 | <p>I am analysing the forces within a certain frame, and for some reason there do not seem to be sufficient equations of equilibrium to solve for all the internal forces. Is there something about this structure than makes it unsolvable without more information?</p>
<p>There is a frame, ABCDE, shown below. Two equal forces are applied to it (both shown as F). Assume we know F, w, and h (and from the illustration, l = w/2). Let's try to find the forces within member ADC.</p>
<p><a href="https://i.stack.imgur.com/la6mJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/la6mJ.png" alt="Frame illustration"></a></p>
<p>First, we can analyse the frame as a whole. It is straightforward to show that Ay = By = F.</p>
<p>Next, we should be able to analyse member ADC by itself.</p>
<p><a href="https://i.stack.imgur.com/Db9WX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Db9WX.png" alt="Member ADC"></a></p>
<p>If we sum forces in the y-direction, we get:</p>
<pre><code>Ay + Cy - F = 0
</code></pre>
<p>Since we already found that Ay = F when we analysed the whole frame, we know that Cy = 0</p>
<p>If we sum forces in the x-direction, we get:</p>
<pre><code>Ax + Dx + Cx = 0
</code></pre>
<p>And if we sum moments around A, we get:</p>
<pre><code>F*w/4 + Dx*h + Cx*w/2 = 0
</code></pre>
<p>Without going any further, we can see that we have two equations with three unknowns (Ax, Dx, and Cx).</p>
<p>If we look at the other members in the frame, we don't get any new information, I don't think... So what's missing? How can this simple frame be an unsolvable physical conundrum?</p>
| |structural-engineering|statics| | <p>It's statically indeterminate. It's far from unique - most real world structures are.</p>
<p>Trivially / intuitively, the frame 'works' if member DE is omitted (ie, you could take it out and the frame still stands up), so we can say immediately that there's a valid solution with member DE having zero axial force.</p>
<p>Also, we can see that if you put a tension into that member (for example, if it's slightly too short and you stretch it to fit it in place) the frame would still work. Thus, there's a valid solution with member DE having a non-zero axial force.</p>
<p>To analyse it, you need to know more. The normal assumption would be that the members are all unstressed if no loads are imposed (ie, if the members are weightless and F is removed, there's no member actions anywhere). Then, if you know the relative stiffness of the members you can work out to what degree the forces distribute. If DE is very axially compliant and AC and CB are very flexurally stiff, the frame would tend towards acting as if DE wasn't there - DE would have a very low force. If DE is very stiff, and the sides are relatively flexible, then DE would tend to prop the sides apart and pick up lots of load.</p>
| 16489 | Structural Analysis Conundrum |
2017-07-28T15:35:48.117 | <p>How do those mechanisms such as the ones in cars work - when you open a tray or glove-box, and it opens slowly by "easing in".</p>
<p>I would like to build my own compartment with a sliding cover that would slowly open by with the push of a button.</p>
<p>Thanks</p>
| |mechanical-engineering| | <p>One method is they use a piston/ cylinder arrangement that has a small hole to allow the air in / out to slow down the rate of movement caused by the weight or a spring .</p>
<p>As for a name : damper or damping piston comes to mind.</p>
| 16499 | Achieving a "Soft" open mechanism |
2017-07-29T21:48:16.887 | <p>How much would it cost to build an oil rig type platform, that floats like an oil rig, but doesn't have the functions of drilling oil and things like that? Basically what I'm trying to say is, how much would it cost to build a large, floating metal platform to put in the ocean? </p>
| |drilling| | <p>Google found an answer in about 5 minutes for me:
<a href="http://www.offshore-mag.com/articles/print/volume-72/issue-7/rig-report/reviewing-rig-construction-cost-factors.html" rel="nofollow noreferrer">http://www.offshore-mag.com/articles/print/volume-72/issue-7/rig-report/reviewing-rig-construction-cost-factors.html</a></p>
<ul>
<li>Drilling equipment - 10% to 30% of total cost.</li>
<li>Non-drilling-related equipment - similar.</li>
<li>Total equipment cost - 30% to 60% of total.</li>
<li>Cost of cheapest jackup rigs - approx. \$200m</li>
<li>Total cost cheapest semi-submersible or drillship rig - approx. \$500m - \$600m</li>
</ul>
<p>So you are looking at something like \$100m minimum for a bare platform with no equipment.</p>
| 16523 | How much would a regular sized oil rig type platform cost to build? |
2017-07-31T20:08:00.603 | <p>What is the advantage of adding alloy elements in a steel in order to increase the tempering temperature while keeping hardness and ductility the same as before?
If I wasn't clear enough, I want to know what difference makes a higher tempering temperature, if hardness and ductility will be the same.</p>
| |materials|heat-treatment| | <p>Temperature resistance is especially useful for machine tools eg drills, lathe tools ect etc. </p>
<p>Mechanised cutting generates a lot of heat and equally cutting tools need to be tempered fairly hard for obvious reasons and it is very easy for something like a drill to get well into the normal tempering range of carbon steel, thus losing its original temper. </p>
<p><em>High speed steels</em> for example are widely used for general purpose drill bits etc as they allow higher speeds and/or feed rates to be used while remaining in an acceptable temperature range. </p>
| 16549 | Why adding alloy elements in a steel for increasing tempering temperature while keeping the same hardnes and ductility? |
2017-07-31T22:13:03.393 | <p><a href="https://i.stack.imgur.com/T1piV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T1piV.png" alt="Circuit 1, gain of 1"></a></p>
<p><a href="https://i.stack.imgur.com/0o3Xf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0o3Xf.png" alt="Circuit 2, gain of ~30"></a></p>
<p>In the case of the first circuit, The ratio between the input voltage Vin and the voltage at the collector and the ratio between the input voltage Vin and the voltage at the emitter are supposed to be around 1. </p>
<p>In the case of the second circuit, the ratio between the input voltage Vs and the voltage at the output Vout is around 30. </p>
<p>My understanding of the circuit (please correct me if this is incorrect):</p>
<ol>
<li>I know the purpose of the resistors in the center region of circuit 2 is to bias the transistor and make it operate at a specific Q-point. </li>
<li>The purpose of the 10uF capacitors is to filter any DC component of the input and output signal, as well as to prevent changing the biasing.</li>
<li>The capacitor across Re is a bypass capacitor, which is supposed to prevent the AC component from affecting the gain through negative feedback.</li>
<li>The purpose of the emitter resistor is to provide negative feedback.</li>
<li>The purpose of the collector resistor (and I am not sure about this) might be to limit the current from the source.</li>
<li>The purpose of the resistor at the output (I am not 100% sure about this either) might be to work as a load</li>
<li>The resistor next to the input signal Vs is used to step down the voltage of the signal, as to use a small signal model in the circuit instead of having to use the large signal model.</li>
</ol>
<p>Assuming the following: a beta (Ic/Ib) of around 250, R1 of 75k and R2 of 33k for both circuits, and an input voltage Vin of 4 volts peak to peak with a frequency of 1khz for the first circuit, </p>
<p>What controls the voltage gain in circuit 1 and 2 at the collector, and why is the gain in the first circuit only 1?</p>
| |electrical-engineering|circuit-design| | <p>In this circuit:</p>
<p><a href="https://i.stack.imgur.com/NPfnP.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NPfnP.gif" alt="" /></a></p>
<p>The gain from Vin to Vout2 is about 1, and to Vout1 about -1.</p>
<p>That the two gains have the same magnitude with opposite sign can be deduced from the fact that at first approximation, the emitter and collector currents are equal. Since the emitter and collector resistors are the same, the voltage changes across them due to the same current will also be the same.</p>
<p>Actually the emitter and collector currents aren't exactly equal. The emitter current is the collector current plus the base current. However, due to the gain of the transistor, the base current is a small fraction of the collector current. For example, if the gain of the transistor is 100, then the emitter current relative to the collector current is 101 to 100. That can be approximated to 1:1 for many practical purposes.</p>
<p>So now the question comes down to why the magnitude of either gain is 1. This can be answered by looking at the emitter voltage relative to the base voltage. At first approximation, the B-E voltage is fixed, usually around 700 mV for silicon BJTs in their active useful range. Since gain is about how much <i>changes</i> are amplified, the fixed B-E offset doesn't matter. AC-wise, the emitter voltage is the same as the base voltage.</p>
<p>Note that the input capacitor is proof that this circuit is not intended to work at DC. That capacitor specifically blocks DC, and only passes frequencies above some rolloff. Basically, the capacitor forms a high pass filter against the impedance of R1, R2, and the apparent impedance looking into the base.</p>
<p>This circuit is often used with an serial R-C network between the two outputs. The middle point between the R-C added become then a phase shifting network (0 to 180 degrees theoritical)</p>
| 16551 | What controls the gain in a BJT amplifier, and why does this circuit have a gain of unity? |
2017-08-01T02:17:38.923 | <p>I have a project where I need to be able to measure the volume of remaining liquid in a bottle based on weight. The idea is to use a Force Sensing Resistor: </p>
<p>(model FSR402-Short Tail: <a href="http://www.interlinkelectronics.com/FSR402short.php" rel="nofollow noreferrer">http://www.interlinkelectronics.com/FSR402short.php</a>)</p>
<p><a href="https://i.stack.imgur.com/lGXrj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lGXrj.jpg" alt="image"></a><br>
<sub>(source: <a href="https://www.interlinkelectronics.com/assets/images/FSR402_short.jpg" rel="nofollow noreferrer">interlinkelectronics.com</a>)</sub> </p>
<p>My issue is that I can't seem to get an accurate consistent reading on the pressure exerted on the device when any kind of shifting (such as replacing / refilling the bottle, moving the device location, etc) would create a huge variance in the data points. It can vary from 20-150 points easily... which is too large of a margin since 150 points could equate to about 500ml of liquid.</p>
<hr>
<p>The idea was to loosely hang the bottle & connector to the pressure plate (see second image) using two centrally located screws. the pressure plate would have a set of two rubber feet perpendicular to the screw placements to allow the pressure to be focused on the rubber footing with the FSR.</p>
<p>This produced two main issues that I did not foresee with the CAD design:</p>
<ul>
<li>over-tightening/under-tightening of the screws</li>
<li>inaccurate or inconsistent readings of the pressure from very minor shifts or bumps.</li>
</ul>
<p><strong>What can I do to improve my design, shown below?</strong></p>
<p>1) Top perspective view of pivot mechanism.
<a href="https://i.stack.imgur.com/ibAHH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ibAHH.png" alt="Top perspective view of pivot mechanism."></a></p>
<p>2) Side perspective view, translucent with descriptions.
<a href="https://i.stack.imgur.com/WoAnP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WoAnP.png" alt="Side perspective view, translucent with descriptions."></a></p>
| |mechanical-engineering|design|cad|mechanisms|autodesk-inventor| | <p>This is not an appropriate sensor for what you are trying to do. This type of flat pressure sensor is sensitive to localised pressures. For example, from a piece of dirt or an uneven surface. You should use a load cell which is designed for your application. For example, something like: <a href="http://www.omega.co.uk/pptst/LCMKD.html" rel="nofollow noreferrer">Omega LCMKD-20N</a> would be more suitable.</p>
<p>You can also consider this <a href="http://www.omega.co.uk/literature/transactions/volume3/weigh.html" rel="nofollow noreferrer">brief introduction to designing a weighing system</a>.</p>
| 16554 | How can I *more accurately* measure the weight of a hanging bottle? |
2017-08-01T10:25:10.320 | <p>I would like to convert a continuous rotation into two 180° intermittent rotations. A standard Geneva drive layout can not do this, is there any variation of a Geneva drive that can do this?</p>
<p><a href="https://i.stack.imgur.com/ei4Ae.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ei4Ae.png" alt=""></a></p>
<p>Above is an example of a 4 slot Geneva drive capable of four 90° intermittent rotations.</p>
| |mechanical-engineering|gears| | <p>It's simpler to add a 1:2 ratio gear ration on the output of the geneva drive to get your 180° rotation per 90° of the intermittent</p>
| 16558 | How can I produce two 180-degree intermittent rotations with a Geneva drive? |
2017-08-02T01:56:56.997 | <p>If we consider a stationary car, a moderate wind hitting it broadside might tilt it a bit, but not move it otherwise. Now, consider the car moving perpendicular to that wind (e.g., driving straight forward). If the car were a rolling rigid body, it would still move straight ahead so long as the wind is not strong enough to overcome the friction between the tires and the road. However, experience says that it can sometimes require a significant steering input into the wind to keep the car moving straight ahead. What about the dynamics and architecture of a vehicle make it tend to steer itself when subjected to a crosswind?</p>
| |mechanical-engineering|automotive-engineering|applied-mechanics|dynamics| | <p>This is primarily caused by "side slip" or "<a href="https://en.wikipedia.org/wiki/Slip_angle" rel="nofollow noreferrer">slip angle</a>". This occurs because tires are not rigid, but pneumatic bladders of flexible rubber and steel cord. During normal operation (including crosswind and turns) your tires operate with static friction between the road and surface. Exceptions would be skidding to a stop or "peeling out". When the car is moving, it is easier to steer because the tire can flex and recover, flex and recover as it travels around and around. If you have ever tried to turn a steering wheel (without power-steering) when a car is stationary, it is quite difficult because you are dragging the rubber across the surface as opposed to the "flex and recover" mode of operation when it is rolling. This is the reason the car doesn't steer itself or move horizontally when it is parked.</p>
<p>Side slip is caused by any horizontal force on the tire during rotation. The image below shows how the front tires flex in a turn. This is very similar to how the tires would flex during crosswind loading.</p>
<p><a href="https://i.stack.imgur.com/IPOlU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IPOlU.png" alt="enter image description here"></a><br>
<a href="http://technicalf1explained.blogspot.com/2012/10/f1-tirespart-2.html" rel="nofollow noreferrer">Formula 1 image source</a>
<br><br><br>
Here is another image showing side slip with the surface contact patch.<br>
<a href="https://i.stack.imgur.com/uAJMj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uAJMj.jpg" alt="enter image description here"></a><br>
<a href="http://www.mgf.ultimatemg.com/group2/suspension/chassis_and_handling/slip_angle.htm" rel="nofollow noreferrer">Slip angle article and image source</a></p>
<p><br><br>
A secondary cause is that the crosswind force and "<a href="https://en.wikipedia.org/wiki/Caster_angle" rel="nofollow noreferrer">caster</a>" in the front wheels causes the vehicle to want to turn away from the crosswind. Castor essentially means the pivot point is placed ahead of the place the wheel contacts the surface. Some castor in automobile design is desirable because it makes the steering self correcting. This is responsible for the feedback force you feel in the steering wheel during a crosswind. The drivers steering wheel torque input is required to oppose this crosswind force or the car will quickly drift off the road.</p>
<p>This may be easier to visualize on a shopping cart. While you push a shopping cart, a person pushes it to the left causing a moment on the two front caster wheels rotating them in to turn in the direction the force was applied. After the force leaves the wheels will realign with the single force pushing them forward.</p>
<p><a href="https://i.stack.imgur.com/197dv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/197dv.jpg" alt="enter image description here"></a></p>
| 16573 | Why does a vehicle under crosswind need corrective steering? |
2017-08-03T12:53:23.490 | <p>I have to calculate a ratio of a planetary gear set. It's an existing gear set and had been already implemented for years ago(works fine). Somehow it does not fit the general formula of planetary gear set which is</p>
<p>R = 2*P + S (number of teeth)</p>
<p>On my gear set:
R = 62
P = 21
S = 18<br>
62 != 21*2 + 18</p>
<p>So how can i calculate the correct ratio?</p>
| |gears| | <p>While R = 2*P + S is typical, the deviation doesn't stop you from calculating the gear ratio. The carrier is over sized from a tight fit, but it seems to be functioning. A 22 gear planet would help, but these are unusual. So let's just calculate:</p>
<p>The overall gear ratio of a simple planetary gearset can be calculated using the following two equations, representing the sun-planet and planet-ring interactions respectively:</p>
<p>$$S\omega_s + P \omega_p - (S + P)\omega_c = 0 $$
$$R\omega_r - P \omega_p - (R - P)\omega_c = 0 $$</p>
<p>where $\omega_r,\omega_s,\omega_p,\omega_c$ is the angular velocity of the Ring, Sun, Planet and Planet Carrier respectively, and $R,S,P$ are the Number of teeth of the Ring, the Sun gear and each Planet gear respectively. From the above we can deduce that:</p>
<p>$$S\omega_s + R\omega_r = (S + R)\omega_c$$
$$-\frac{R}{S} = \frac{\omega_s - \omega_c}{\omega_r - \omega_c}$$</p>
<p>Assuming $\omega_{r} \neq \omega_c$. If we know which are fixed, we can go further. Calling $n=\frac{R}{S}$</p>
<p>Case 1, Sun fixed: $$\frac{\omega_c}{\omega_r}=\frac{n}{1+n} =\frac{R}{R+S}$$
Case 2, Carrier fixed: $$\frac{\omega_r}{\omega_s}=-\frac{1}{n}=-\frac{S}{R}$$
Case 3, Ring Fixed: $$\frac{\omega_c}{\omega_s}=\frac{1}{1+n} =\frac{S}{R+S}$$</p>
<p>In your example, $n=62/18 = 3.44$, so Case 1 is 31:40, Case 2 is 9:31, and Case 3 is 9:40.</p>
| 16596 | Marginal Planetary Gear Set |
2017-08-03T23:12:17.583 | <p>How does a <a href="https://en.wikipedia.org/wiki/Bicycle_brake#Side-pull_caliper_brakes" rel="nofollow noreferrer">single pivot side-pull caliper bicycle brake</a> work? The brake cable only pulls on one of the arms, so what is causing the force on the other arm? I understand that it is a normal force due to the cable housing, but why is this force developed?</p>
<p><strong>EDIT</strong></p>
<p>I'll elaborate on my questions a bit. Let $A$ be the point where the cable is attached, $B$ be the point where the housing pushes on the other arm, $P$ be the point of pivot, $C$ be the point where the first arm touches the rim and $D$ be the point where the second arm touches the rim. $APC$ is then the arm pulled on by the cable and $BPD$ is the arm potentially pushed by the housing.</p>
<p>Let's say the cable pulls at $A$ with some tension force $T$. A normal force $N$ from the wheel must then push on $C$, in order for the torque with respect to $P$ to be zero. We have $N = (|AP|_x/|CP|_yS) $, if $x$ and $y$ are in the horizontal and vertical direction respectively, and $T$ is vertical and $N$ horizontal.</p>
<p>The force $N$ gives a lateral force on the rim and a torque on the wheel. They could be balanced by a normal force of the same magnitude at the other brakepad. By symmetry, $B$ would then have to be pushed by a force of magnitude $T$. </p>
<p>But couldn't they also be balanced at the hub? Does it have to do with the construction of the wheel, i.e. because it's spoked? If the wheel were in the form of a solid cylinder (which would of course be very impractical!), would the situation be different?</p>
<p><a href="https://i.stack.imgur.com/Ya5kZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ya5kZ.png" alt="enter image description here"></a>
<sup>image source <a href="https://commons.wikimedia.org/wiki/File:Bicycle_caliper_brake_highlighted.jpg" rel="nofollow noreferrer">wikipedia</a></sup></p>
| |mechanical-engineering| | <p>Force diagrams often gloss over complications that are assumed to be common knowledge. Here I will be annotating the full system with multiple step interpolations, but for now without force arrows. A single-pivot side-pull caliper brakes is composed of the following elements, pictured below:</p>
<ul>
<li>brake line</li>
<li>brake line housing (aka noodle)</li>
<li>brake line puller (ie, a handle)</li>
<li>left caliper (green) attached to pivot point</li>
<li>right caliper (red) attached to pivot point</li>
<li>spring, which pushes the calipers apart</li>
<li>wheel/tire</li>
</ul>
<p><a href="https://i.stack.imgur.com/JeyDn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JeyDn.png" alt="full model" /></a></p>
<p>Let's simplify the model by removing the spring and truncating the right caliper (red). We also assume the rigidity of the brake line and the brake line housing is enough to support the left caliper (green), which without the spring would otherwise fall into the tire. When the handle and brake line are pulled, it is intuitive that only the right caliper (red) contracts.</p>
<p><a href="https://i.stack.imgur.com/XxPYp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XxPYp.png" alt="simplified model v1, animated" /></a></p>
<p>Let's now incorporate the complete right caliper (red) so that only the spring is missing. At first and until the tire is contacted, only the right caliper (red) contracts. Just as before. However with the right caliper (red) locked into place and unable to further close, continued pulling of the handle forces the brake line taut, reducing the length of brake line which runs between the <strong>L2-L3</strong> points. Since the brake line housing cannot contract or collapse it forces the left caliper (green) shut, balancing the force diagram.</p>
<p><a href="https://i.stack.imgur.com/OaN6u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OaN6u.png" alt="simplified model v2, animated" /></a></p>
<p>Now really, the left caliper (green) isn't fixed and would collapse into the tire. So we introduce a spring to force the calipers apart when the handle and brake line are no longer being pulled. We would also like the calipers to be balanced - to make contact with the tire at the same time and with equal force. Fortuitously, the spring has just that effect. Recall that the force to compress a spring grows linearly with distance. In our previous model, regardless of the force applied along the brake line, once the brake makes contact with the tire the right caliper (red) will not move. In other words, upon contact there is an equal and opposite force. With the spring installed this force grows linearly rather than instantaneously. As before, tension between the <strong>L2-L3</strong> points is counteracted by the cable housing, forcing the left caliper (green) shut.</p>
<p><a href="https://i.stack.imgur.com/JDdHd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JDdHd.png" alt="full model, animated" /></a></p>
<p>The trick is balance the brakes so the calipers close equally. We have two variables: the spring stiffness and the length of the brake line housing. I've never balanced these brakes but I imagine that the spring cannot be tightened. Which leaves the length of housing; I imagine that properly balancing such brakes is difficult. For finer adjustments and as with many bicycle systems, the length of the housing can be adjusted by screw. Refer to the original image:</p>
<p><a href="https://i.stack.imgur.com/l0ifz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l0ifz.jpg" alt="brake line housing adjustment screw" /></a></p>
| 16603 | How does a single pivot side-pull caliper bicycle brake work? |
2017-08-05T17:05:26.860 | <p>As my first serious mechanical engineering project, I've built a wall-mounted folding table using of a pair of 4-bar mechanisms (WIP, ignore the wall mounts): <a href="https://i.stack.imgur.com/Dw1kf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dw1kf.jpg" alt="folding table with 4-bar mechanism"></a> </p>
<p>I'm using fisheye bar ends with spherical bearings, like this: <a href="https://i.stack.imgur.com/TCPNJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TCPNJ.jpg" alt="fisheye spherical bearing bar-end"></a></p>
<p>The mechanism behaves more or less the way I'd intended, but the spherical bearings have introduced a lot of lateral motion into the mechanism, which I'd like to minimize.</p>
<p>What is a good way to minimize or eliminate this lateral motion? Should I replace the fisheyes with clevis bar ends?</p>
| |mechanical-engineering|stability| | <p>For reference, I'm answering my own question with the solution I used. </p>
<p>I replaced half of the fish-eye bearings with clevis bearings (on the wall end of each of the bars), which has eliminated most of the lateral motion while still allowing enough flexibility to actually mount the thing. </p>
<p>There is still some lateral motion but, because the table has one side flush with the wall in both the fully-open and fully-closed configurations, it doesn't matter in practise.</p>
| 16626 | Eliminating lateral motion in a pair of 4-bar mechanisms |
2017-08-05T21:24:35.643 | <p>If I was designing a simple gabled house with a 21 ft vaulted ceiling as the peak of the roof, and the house was about 32 ft x 40 ft with no load bearing walls (other than the 4 exterior walls)... would I need a beam across the entire peak of the house? If so would it have to be steel cause of the 40 ft span?</p>
| |structural-engineering|beam| | <p>You would not need a steel beam . Although steel would give you the lowest profile. An engineered wood beam would be more typical ; truss type or laminated beam . I guess a wood beam would be about 2 feet high/deep. They will both meet code anywhere but you you will need to know the weight of your roof and determine how you will provide lateral stability. So professional design will be needed.</p>
| 16630 | Structural engineering: Would I need a beam for this? |
2017-08-06T14:18:57.383 | <p>What is difference between the <strong>tailboard</strong> and the <strong>bood lid</strong>? Can you show me where each part is? </p>
<p>Thanks for help. </p>
| |automotive-engineering| | <p>A tailboard would be the name of the opening panel found at the rear on a pick-up type vehicle, while a boot lid is the name for the panel that closes the boot or trunk. An SUV has the rear panel called a door and if the rear window opens it is likely to be called a hatch - also hatch-back.</p>
| 16636 | Difference between the tailboard and boot lid. |
2017-08-06T17:20:14.423 | <p>I'm trying to measure the actuation force of buttons on the <a href="https://en.wikipedia.org/wiki/PlayStation_Vita" rel="nofollow noreferrer">Playstation Vita</a>, in preparation of building an external device that presses these buttons automatically. Specifically, I'm trying to measure the force required to push the buttons on the D-Pad on the left, and the action buttons on the right.</p>
<p>Googling "how to measure actuation force" will give you pages of results of mechanical keyboard enthusiast forums, and they often mention measuring the actuation force by stacking pennies / whatever coins you have until the key is depressed and the computer recognises the keypress. I can't do that normally here, since the buttons are really small (<1cm for the action buttons) and even the smallest coin my currency has cannot be stacked onto it without toppling over.</p>
<p>In an attempt to mitigate the problem, I made the following setup. It still uses coins as weights to push down the button, but I rested the coins on the ruler that in turn rests on another object of somewhat equal height/thickness to the game console, in this photo, the remote control seen in the background. I conducted the experiment by continuously adding coins with a game running on the console until it registers a keypress, and then summing up the weights of the coins.</p>
<p><img src="https://i.stack.imgur.com/CBRPL.jpg" alt="experimental setup"></p>
<p>However, the results are rather inconsistent and questionable. I repeated my experiments several times (by knocking down the coin tower and stacking it up again) as standard practice, and I got values ranging from <b>100g to 180g</b>, which feels extremely high for buttons on a handheld game console, and that large spread seems to be telling me that my experimental technique is just plain wrong.</p>
<p>Some possible causes of inconsistencies from the setup:</p>
<ol>
<li>The coins are not the same - I'm stacking 5 different coins at once</li>
<li>The coins are not stacked in the same order each time I redo the experiment</li>
<li>The center of the coin tower is not aligned with the center of the button on the game console - I don't know how to do this without the coin tower making contact with the neighboring buttons or it toppling over</li>
</ol>
<p>Some steps I have taken to possibly improve accuracy:</p>
<ol>
<li>The ruler used is a stiff metal ruler and does not bend significantly</li>
<li>The ruler is kept horizontal</li>
<li>The only points of contact of the ruler are the remote control seen in the background and a single button on the D-Pad.</li>
</ol>
<p><b>TL;DR:</b> How can I improve the accuracy of my setup, or other better ways to measure actuation force of such small buttons?</p>
<p><b>Update & Final Results</b> I used the weighing scale method, and the results are 80g for the D-Pad and about 120g for the action buttons on the right. Thanks everyone!</p>
| |mechanical-engineering|measurements|torque|experimental-physics| | <p>If you want a quick and somewhat inaccurate hill-billy engineering answer, I would take a digital kitchen scale and place the PSV ontop of it, tare it, then load up a program to test the buttons (game, menu, etc). with the scale device zero'd out, slowly depress the button until it is read by the program. a full actuation can and will differ from the triggering of the circuit. Wait for the program to read the triggering rather than the button's actuation, and add a few extra grams to it.</p>
<p>If you want a precise measurement... I have no idea what you have available for building as far as "home equipment" is concerned.</p>
| 16641 | How can I accurately measure the actuation force of small buttons (<1cm diameter) using home equipment? |
2017-08-07T18:27:29.103 | <p>I want to lift a mass of 1kg so how much force is required ?</p>
| |motors|torque| | <p>Mass is measured in kg and weight measured in N. </p>
| 16652 | Force Required to Lift a WEIGHT of 1Kg |
2017-08-07T22:53:56.067 | <p>I want to measure the distance between the back of a human body, and the chest. This is denoted in the image below, where Point A is the back and Point B is the chest:</p>
<p><a href="https://i.stack.imgur.com/V7GOV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V7GOV.png" alt="enter image description here"></a></p>
<p>I do not wish to measure the diameter of the chest. Rather, the direct path from Point A, to Point B ( Through the Ribcage ). Do any <strong>mobile sensors</strong> exist that can help me capture this data, which can be worn by the applicant at all times? ( During sleep, walking, running, etc.)</p>
| |sensors| | <p>The simplest answer would be "a large pair of calipers". In fact, they make such a tool for exactly this task. See <a href="http://www.seritex.com/gpm-large-spreading-caliper-rounded-ends-108/" rel="nofollow noreferrer">http://www.seritex.com/gpm-large-spreading-caliper-rounded-ends-108/</a></p>
<blockquote>
<p>The GPM Large Spreading Caliper with rounded ends provides accurate
measurement of the distance between two points, where one or both
points may not be on a flat surface. This Large Spreading Caliper can
also be used for the measurement of Anterior-Posterior chest depth.</p>
<p>The ends of the spreading branches are rounded for the patient's
comfort, as opposed to the GPM Large Spreading Caliper model 109,
which has pointed ends.</p>
<p>The GPM Large Spreading Caliper with rounded ends is manufactured from
chrome plated steel. Hinged at the base, the two curved branches with
rounded ends are connected by a pivotal ruler, which is engraved in
millimeter intervals from 0 to 60 cm. The branches can be immobilized
in any position by a screw located on the slider which guides the
ruler. The ruler gives the precise distance between the two points
over which the measurement has been taken.</p>
</blockquote>
<p>Or you could improvise by using something like this: </p>
<p><a href="https://i.stack.imgur.com/fvrmi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fvrmi.jpg" alt="enter image description here"></a></p>
<p>NOTE: The OP changed the question to include the "mobile and wearable" requirements, after this answer was written.</p>
| 16655 | Accurately measure the distance from chest to back |
2017-08-08T08:54:05.730 | <p>I have a Problem with a transformation.</p>
<p>The energy function is the following:</p>
<p>$$I(u_1,u_2,P)=\int_R [\frac{\mu}{2}(u_{i,j} + u_{j,i}) u_{i,j}-Pu_{i,i} dx_1 dx_2 - \int_R f_i u_i dx_1dx_2 +\int _S t_iu_idS$$</p>
<p>where</p>
<p>$u_{i,j}=\frac{du_i}{dx_j}$ and Einstein's summation convention is applied (sum over repeating indices).</p>
<p>The task is to compute the variation $\delta I=0$ </p>
<p>In the result sheet they do following transformation, which I do not understand:</p>
<p>$$\begin{align}
\tag{1}I(u_1,u_2,P)&=\int_R [\frac{\mu}{2}[(\delta u_{i,j} + \delta u_{j,i})u_{i,j}+(u_{i,j} + u_{j,i}) \delta u_{i,j}] dx_1 dx_2 +[...]\\
\tag{2}I(u_1,u_2,P)&=\int_R [\mu(u_{i,j} + u_{j,i})\delta u_{i,j}dx_1 dx_2 +[...]
\end{align}$$</p>
| |fluid-mechanics|finite-element-method|fluid| | <p>I found an answer:</p>
<p>As $I= f(u_1,u_2)$ it follows that $\{i,j\}=1,2$</p>
<p>\begin{align}
\tag{1} u_{ij}\delta u_{ji}+ u_{ji}\delta u_{ij}=& \sum_i \sum_j (u_{ij}\delta u_{ji}+ \delta u_{ij}u_{ji}) \\
\tag{2} =&u_{11}\delta u_{11}+ u_{12}\delta u_{21}+u_{21}\delta u_{12}+u_{22}\delta u_{22}+ \\ &\delta u_{11}u_{11}+ \delta u_{12} u_{21}+\delta u_{21} u_{12}+\delta u_{22} u_{22} \\
=&2u_{11}\delta u_{11}+ 2u_{12}\delta u_{21}+2u_{21}\delta u_{12}+2u_{22}\delta u_{22}\\
=&2u_{ij}\delta u_{ji}
\end{align}</p>
<p>Therefore:</p>
<p>\begin{align}
\tag{1}I(u_1,u_2,P)&=\int_R [\frac{\mu}{2}[(\delta u_{i,j} + \delta u_{j,i})u_{i,j}+(u_{i,j} + u_{j,i}) \delta u_{i,j}] dx_1 dx_2 +[...]
\\ &= \int_R [\frac{\mu}{2}[\delta u_{i,j}u_{i,j} + \delta u_{j,i}u_{i,j}+u_{i,j}\delta u_{i,j} + u_{j,i}\delta u_{i,j}] dx_1 dx_2 +[...]
\\ &= \int_R [\frac{\mu}{2}[2u_{i,j}\delta u_{i,j}+ \delta u_{j,i}u_{i,j}+ u_{j,i}\delta u_{i,j}] dx_1 dx_2 +[...]
\\ &= \int_R [\frac{\mu}{2}[\delta u_{i,j}u_{i,j} + \delta u_{j,i}u_{i,j}+u_{i,j}\delta u_{i,j} + u_{j,i}\delta u_{i,j}] dx_1 dx_2 +[...]
\\ &= \int_R [\frac{\mu}{2}[2u_{i,j}\delta u_{i,j}+ 2u_{j,i}\delta u_{i,j}] dx_1 dx_2 +[...]
\\ &=\int_R [\mu(u_{i,j} + u_{j,i})\delta u_{i,j}dx_1 dx_2 +[...]
\end{align}</p>
| 16662 | Variational Problem of slow steady flow of viscuous fluid in 2D |
2017-08-09T06:04:44.057 | <p>Quite often while calculating angular velocity we come across the quantity radians per second. How do I treat radian dimensionally?</p>
| |mechanical-engineering|fluid-mechanics|applied-mechanics|mathematics|mechanical-failure| | <h2>TL/DR</h2>
<p>As the other answers have stated, radians are dimensionless.</p>
<h2>Cycles and Turns</h2>
<p>This is a question that seems to come up often, and from many different quarters; not only from students and lay people, but I have seen plenty of professionals for whom the concept of radians causes difficulty.</p>
<p>One way that seems to help some people's intuition is to put aside the idea of radians entirely, and instead talk about <em>cycles</em> and <em>turns</em> of a circle.</p>
<h2>As the World Turns</h2>
<p>Consider, for example, the rotation of the Earth around the Sun. We can call each time the Earth completes a trip around the sun as a <em>cycle</em>, or a <em>full turn</em> about its path. Note that it <strong>doesn't matter</strong> that the Earth's trip around the Sun is NOTHING like a circle; the <em>trip itself</em> is the cycle.</p>
<p>Using this terminology, calculating the average angular velocity of the Earth is very simple: <span class="math-container">$1 \frac{cycle}{year}$</span>.</p>
<p>The conversion from per year to per seconds is rather trivial:</p>
<p><span class="math-container">$1 \frac{cycle}{year}\frac{1~year}{365~days}\frac{1~day}{24~hours}\frac{1~hour}{3600~seconds}=1 \frac{cycle}{31.5576E6~seconds}$</span></p>
<p>Furthermore, the unit for <span class="math-container">$1 \frac{cycle}{second}$</span> is <span class="math-container">$hz$</span>. So:</p>
<p><span class="math-container">$1 \frac{cycle}{year}=1 \frac{cycle}{31.5576E6~seconds}=316.9~\text{microhz, or μhz}$</span></p>
<p>Again: <span class="math-container">$1 hz = 1 \frac{cycle}{second}$</span>, or we can also say <span class="math-container">$\frac{turn}{second}$</span>. Remember: <strong>a cycle and a turn are the same thing</strong>.</p>
<h2>Circles as Useful Models</h2>
<p>It turns (<-- funny!) out that <em>circles</em> are a very useful way to <em>model</em> cycles/turns. Even though the path of the Earth <em>not a circle</em>, we can use a circle to <em>map</em> the position of the Earth at any point in time we are interested in against other things we are interested in. We'll use the seasons as an illustration.</p>
<p>To make things easier, let's assume that a solar year- 1 cycle- starts at the Winter solstice: December 21st. Therefore, we can have the following model of a year:</p>
<p><a href="https://i.stack.imgur.com/v96Dp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v96Dp.png" alt="enter image description here" /></a></p>
<p>So now, we can make the following fairly intuitive statements:</p>
<ol>
<li>Winter occurs at 0 turns, 1 turn, 2 turns, etc etc.</li>
<li>Spring occurs at 0.25 turns, 1.25 turns, 2.25 turns, etc etc.</li>
<li>Summer occurs at 0.5 turns, 1.5 turns, 2.5 turns, etc etc.</li>
<li>Spring occurs at 0.75 turns, 1.75 turns, 2.75 turns, etc etc.</li>
</ol>
<p>Because we have chosen a circle as a <em>model</em> of a solar cycle or turn, we now have some very useful tools available for use "for free". These tools can help us to analyze our model in a mathematically rigorous way. To put it another way: we chose a circle precisely because we know all kind of cool things about circles that we can use to help us understand a solar cycle better. I'll provide an example at the end.</p>
<h2>What Is a Radian, Really?</h2>
<p>One of the things we get for free: for circles, we know that the relationship between the <em>radius</em> and the <em>circumference</em> is a <em>constant ratio</em>. In keeping with what we have done so far, we are just going to refer to the <em>circumference</em> of the circle as <em>one turn</em>, or <em>one cycle</em>. That ratio is:</p>
<p><span class="math-container">$\frac{1~\text{turn}}{1~\text{radius}}=2\pi\rightarrow 1~\text{turn}=2\pi~\text{radii}$</span></p>
<p>If we divide both sides of that equation by <span class="math-container">$2\pi~\text{radii}$</span>, we finally arrive at the definition of a radian. Which is the <em>ratio of the circle radius to one full turn of the circle</em>:</p>
<p><span class="math-container">$1~\text{radian}=\frac{1~\text{turn}}{2\pi~\text{radii}}=\frac{1}{2\pi}$</span></p>
<p>But recall that for our <em>circle model</em>, the turn and the circumference are the same thing. Therefore the units for the radii and the turn cancel each other out, and we have a dimensionless radian as expected.</p>
<p><span class="math-container">$\frac{1~\text{circumference}}{2\pi~\text{radii}}=\frac{X\require{cancel} \cancel{Unit}}{Y\require{cancel} \cancel{Unit}}=\frac{X}{Y}=\frac{1}{2\pi}$</span></p>
<h2><a href="https://www.youtube.com/watch?v=ZfYBsuBzulE" rel="nofollow noreferrer">What Does it All Mean, Basil?</a></h2>
<p>It turns out, <em>radians</em> are often a lot more convenient to use than <em>turns</em> and/or <em>cycles</em>. Here's a couple examples:</p>
<ul>
<li><p>Because we understand how radians and <em>turns</em> are related, we can <strong>intuit</strong> much of the rest of geometry without having to memorize formulae. One example is the arc length formula:</p>
<p><span class="math-container">$L_{arc}=\theta\cdot radius$</span>, where <span class="math-container">$\theta$</span> is in radians.</p>
<p>Where does this formula come from? It is simply a result of the <em>definition</em> of a radian. Observe:</p>
<p><span class="math-container">$1~\text{radian}=\frac{1~\text{turn}}{2\pi~\text{radii}}=\frac{1}{2\pi}$</span></p>
<p>The arc length <span class="math-container">$L_{arc}$</span> is some <em>portion</em> of a total turn, or <span class="math-container">$Decimal\cdot\text{turns}$</span>. So:</p>
</li>
</ul>
<p><span class="math-container">\begin{equation}
\begin{split}
& Decimal~\text{radians}=\frac{Decimal\cdot\text{turns}}{radius} \Rightarrow \\
\\
Decimal\cdot\text{turns}= & L_{arc}=Decimal~\text{radians}\cdot radius=\theta\cdot radius
\end{split}
\end{equation}</span></p>
<ul>
<li><p>Another very useful set of tools we get for free by virtue of choosing a circle as our model is the entire field of trigonometry. Trigonometric relationships are almost always expressed using radians, degrees, or sometimes gradians. However, in keeping with the above, you could easily use <em>turns</em> of the circle instead of radians. For example, the equation for <span class="math-container">$\text{sin}$</span> is:</p>
<p><span class="math-container">$\text{sin}~\theta=\frac{opposite}{hypotenuse}$</span></p>
<p>If we make <span class="math-container">$\theta=\frac{1}{4}\text{turn}$</span>, then:</p>
<p><span class="math-container">$\text{sin}~1~quarter~turn=\frac{1~radius}{1~radius}=1.0$</span></p>
<p><a href="https://i.stack.imgur.com/28G0C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/28G0C.png" alt="enter image description here" /></a></p>
<p>But if you want to use a calculator, the calculator doesn't know about <em>turns</em>. It knows about radians and degrees. But from above we know that <em>1 turn</em> is <span class="math-container">$2\pi$</span> radians, so 1 quarter turn is just <span class="math-container">$\frac{2\pi}{4}$</span> radians, or <span class="math-container">$\frac{\pi}{2}$</span> radians.</p>
</li>
</ul>
<h2>Back to Earth</h2>
<p>We started with the Earth revolving around the Sun. Can we do anything interesting relating the circle model, radians, and the seasons together? Indeed we can.</p>
<p>We know that the Earth's rotation is tilted at an axis of about 23.5 degrees from vertical (with respect to the sun), which is:</p>
<p><span class="math-container">$23.5/1~\text{turn}=23.5/360=0.0653~\text{radians}$</span></p>
<p>We also know that this tilt is at maximum at the solstices; the Earth's Southern hemisphere is pointed toward the Sun on December 21st, and the Northern hemisphere pointed toward the Sun on July 21st. The angle with respect to the Sun is zero at the equinoxes: March 20th and September 22nd.</p>
<p>Can we approximate the tilt on some arbitrary date? We can easily do that using our circle model. How about on George Washington's birthday: February 22nd.</p>
<p>We can ascertain using geometry that the relative inclination of the axis varies with the cosine of the angle of the Earth's position around the Sun, starting with position 0 at the Winter solstice:</p>
<ol>
<li>Winter: <span class="math-container">$\theta$</span> = 0 turns <span class="math-container">$\Rightarrow$</span> 23.5 deg tilt <span class="math-container">$\Rightarrow \text{cos}~\theta=1.0$</span></li>
<li>Spring: <span class="math-container">$\theta$</span> = 0.25 turns <span class="math-container">$\Rightarrow$</span> 0 deg tilt <span class="math-container">$\Rightarrow \text{cos}~\theta=0.0$</span></li>
<li>Summer: <span class="math-container">$\theta$</span> = 0.5 turns <span class="math-container">$\Rightarrow$</span> 23.5 deg tilt <span class="math-container">$\Rightarrow \text{cos}~\theta=1.0$</span></li>
<li>Spring: <span class="math-container">$\theta$</span> = 0.75 turns <span class="math-container">$\Rightarrow$</span> 0 deg tilt <span class="math-container">$\Rightarrow \text{cos}~\theta=0.0$</span></li>
</ol>
<p>Starting at the Winter solstice, how many turns are we on February 22nd?</p>
<p><span class="math-container">$(Jan~1-Dec~21)=9~days$</span></p>
<p><span class="math-container">$(Feb~1-Jan~1)=31~days$</span></p>
<p><span class="math-container">$Feb~22=22~days$</span></p>
<p><span class="math-container">$Total=62~days$</span></p>
<p>How many turns is that?</p>
<p><span class="math-container">$62/365.2425=0.16975~\text{turns}$</span></p>
<p>We'll need to convert to radians to use a calculator. So what is the approximate portion of the 1 yr trip that has been traveled from the starting position measured in radians?</p>
<p><span class="math-container">$0.16975~\text{turns}=0.16975\cdot 2\pi~\text{radians}=1.0666~\text{radians}$</span></p>
<p>As shown above, the cosine varies from 1.0 at position 0 (23.5 deg tilt) to 0.0 at 1 quarter turn (0 deg tilt). So:</p>
<p><span class="math-container">$Tilt_{Feb~22}≈\text{cos}(1.0666~\text{radians})\cdot 23.5~\text{deg}=11.35~\text{deg}$</span></p>
<p>This is approximate because we have used a circle to model the change in tilt over time. However, in reality the Earth's speed around the sun is not constant, so this is just an estimation. Our circle model is only an approximation of reality.</p>
| 16680 | Is radians a dimensionless quantity? |
2017-08-09T09:51:44.157 | <p>How do I say my weight? In which SI units? I see people are using Kg as a unit of denoting weight from my childhood. Answer this question "What's your weight?".</p>
| |unit| | <p>Weight is a force and is expressed in Newton (N). Mass is expressed in kilogram (kg).</p>
<p>However, in informal (non-scientific) language, people often express weight in kg, although this is not correct strictly speaking.</p>
<p>The relation between the two is $F=mg$, with $F$ the weight (N), $m$ the mass (kg) and $g$ earth's gravity constant.</p>
<p>See also <a href="https://en.wikipedia.org/wiki/Mass_versus_weight" rel="noreferrer">here</a>.</p>
| 16684 | How do I denote someone weight |
2017-08-10T15:40:38.833 | <p>I have an O-shaped aluminum (7075) profile WxDxH - 40mm x 200mm x 25mm
The inner channel is 20 mm wide and 10 mm deep - runs through the entire length of the bar. </p>
<p>Inside, I have a 20 mm x 10 mm bar (also Al 7075). As you can see from the attached picture, on the opposite face there's a hole in which a flat screw is used to fasten the two pieces in position.</p>
<p><a href="https://i.stack.imgur.com/Pvpxg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pvpxg.png" alt="enter image description here"></a></p>
<p>This telescopic mechanism needs to withstand large loads >50 kg (>110 lbs) on both ends and I'm worried that using only a screw would mean that there will be slippage.</p>
<p>Is there a better way to design a secure clamp for such a telescoping mechanism? I need to be able to do fine positioning and have the confidence that it won't move.</p>
<p>Thanks in advance for your help.</p>
| |mechanical-engineering|aluminum|mechanisms| | <p>If I were building this I would do three things to make the assembly secure, you may think it is over-kill but I will throw this out to you anyway.
Where you have the threaded hole to receive the locking screw, I would have a flangemount locknut threaded insert instead.
Instead of a plain screw I would use a set screw with a pointed tip that would dig in to the aluminum to give you more holding power.
And last but not least , have several of these set screw assemblies spaced along the length of the tube, don't rely on just one screw to hold 100 lbs!</p>
| 16704 | Aluminum sliding mechanism clamping |
2017-08-10T18:06:17.690 | <p>I am creating subwoofer boxes out of MDF or Baltic Birch. My local makerspace has a CNC machine. I am going to make all of the cuts with it. </p>
<p>What software is generally used or will work to get a file for a CNC machine? In the <a href="http://www.sector67.org/blog/faq/#target-toc-12" rel="nofollow noreferrer">makerspace's FAQ</a> they say DXF or STL but it's under the section regarding Laser Cutting. So I'm not sure if that's the needed file format.</p>
| |modeling|cnc| | <p>You'll want to find out what formats the CAM software which they usually use for this machine accepts.</p>
<p>For hobbyist machines, SVG is a frequent option. Initially popularized by PartKam:</p>
<p><a href="https://partkam.com/" rel="nofollow noreferrer">https://partkam.com/</a></p>
<p>it had direct support for CAM in the gcodetools plug-in for Inkscape (until that was deprecated), and many computer aided manufacturing (CAM) programs will import SVGs.</p>
| 16706 | Looking for program and file format for CNC Milling |
2017-08-10T19:54:42.457 | <p>I'm trying to determine the amplitude vs the excitation frequency of a multi dof dynamic system. I'm very confused from reading different methods, although I understand there's a very general one, which is not giving me the results that I think should be when plotting the response.
Considering a system with no damping $\mathbf{M{\ddot x}+{Kx}=F(t)}$, and by assuming the harmonic excitation of the force is a cosine the system becomes $\mathbf{-MX{cos(\omega t)}+{KX}cos(\omega t)=fcos(\omega t)}$. From here the cosines get cancelled out so
only $\mathbf{[K-M{\omega^2}]X=f}$ remains. By solving for the $X$ vector in the end I get $X = \frac{F/M}{|\omega_n^2-\omega^2|}$.</p>
<p>So with Matlab I wrote the following code, with parameters I found on an example from brown.edu, in which the explanation is not complete, my system is actually a 24 DoF system, so I need the code a bit automated:</p>
<pre><code> k1 = 2;
k2 = 1;
k3 = 1;
m1 = 1;
m2 = 1;
M=[m1 0; ...
0 m2];
K=[k1+k2 -k2; ...
-k2 k3+k3];
F=[1;...
1];
[V,D] = eig(K,M);
D = diag(sqrtm(D));
frequencies = (D);
[rows,~] = size(V);
f=linspace(0,0.7,100);
omega=2*pi*f;
Mag = M\F;
Xt = zeros(rows, length(omega));
for i = 1:rows
for j = 1:length(omega)
Xt(i,j) = abs((frequencies(i)^2 - omega(j)^2)\Mag(i)) ;
end
end
figure
for m = 1:length(frequencies)
plot(omega, Xt(m,:));
set(gca,'YScale','log')
hold on
axis tight
end
</code></pre>
<p>Where the main part is the <code>Xt(i,j) = (abs (frequencies(i)^2 - omega(j)^2)\Mag(i) );</code> , which follows the determined formula for the displacements of each degree of freedom.</p>
<p>When plotting what I get is:
<a href="https://i.stack.imgur.com/l8RLs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l8RLs.png" alt="enter image description here"></a></p>
<p>which shows the masses vibrations being at resonance, but from what I understand there should be more peaks and some antiresonance.</p>
<p>What am I missing in the general math?</p>
<hr>
<p><strong>EDIT_1:</strong> I changed the bit of the code to calculate the absolute values of the equation like: $$X(\omega) = \left|\sum_n \frac{F/M}{\omega_n^2 - \omega^2}\right|$$ but I get the same when plotting. I found another solution but I don't understand how its determined or why choose it.
From ${M{\ddot x_i}+{Kx_i}=f(t)sin\omega t}$ the steady state solution is $x_{iss} = x_{is} sin \omega t + x_{ic} cos \omega t$ , where $i = 1,2$ , then
$$ \begin{bmatrix} \mathbf{-(\omega)^2M+{K}} & \mathbf{0} \\ \mathbf{0} & \mathbf{-(\omega)^2M+{K}} \end{bmatrix} \begin{bmatrix} \mathbf{x_{ic}} \\ \mathbf{x_{is}} \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ \mathbf{f} \end{bmatrix} $$
which is then solved to get ${X_{i}} = (x_{ic} + x_{is} )^{1/2}$
First, I don´t understand very well if just $x_{iss}$ is substituted in ${M{\ddot x_i}+{Kx_i}=f(t)sin\omega t}$ for $x_i$. And is this just assuming a sum of $cos$ and $sin$ instead of my first approach of only $cos$? Then, do I need to consider adding the transient response also, which I´m actually not sure how it would be. </p>
<hr>
<p><strong>EDIT_2:</strong> I think I figured out what I was doing wrong. I changed the equation from $$X(\omega) = \left|\sum_n \frac{F/M}{\omega_n^2 - \omega^2}\right|$$ to $$ X = [K-M\omega^2]^{-1}F$$ which mathematically I though they were the same??? The code part that changed is as follows:</p>
<pre><code>for j = 1:length(omega)
Xt(:,j) = abs((K-M*omega(j)^2)\F);
end
figure
for m = 1:length(frequencies)
plot(omega,Xt(m,:))
hold on
axis tight
end
</code></pre>
<p>And the result plot is:
<a href="https://i.stack.imgur.com/49106.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/49106.png" alt="enter image description here"></a></p>
<p>For my extended MDoF system I got:
<a href="https://i.stack.imgur.com/oacKP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oacKP.png" alt="enter image description here"></a></p>
<p>In which I excited all DoF, even when I only excite a single one, it is still very hard to notice what's going on, is there a better way of analysing the response?</p>
| |dynamics|vibration|frequency-response|eigenvalue-analysis| | <p>The code in your OP seems to be for a 2 DOF system not a 24 DOF system. You could check that by printing the "frequencies" variable from your eigensolution.</p>
<p>The reason you are not getting anti-resonances is because you are taking the absolute values in the wrong place. You should be calculating $$X(\omega) = \left|\sum_n \frac{F/M}{\omega_n^2 - \omega^2}\right|$$ not $$X(\omega) = \sum_n \frac{F/M}{\left|\omega_n^2 - \omega^2\right|}$$</p>
<p>The contribution to $X$ from each mode <em>changes its phase</em> by 180 degrees, and therefore <em>changes sign,</em> when the frequency passes through the corresponding $\omega_n$. The antiresonances occur when the contributions from different modes cancel out.</p>
<p>Note, you will probably see the antiresonances better on a plot if you use a log scale for $X$.</p>
<p>EDIT following comments by the OP:</p>
<p>You also need to decompose $F$ into its modal components and apply them to the corresponding terms in the sum. If $$M\ddot x + K x = F.$$ Pre-multiply by the eigenvectors $\Phi$: $$\Phi^T M\ddot x + \Phi^TK x = \Phi^T F$$
But you know $\ddot x = -\omega^2 x$, so $$(-\omega^2 \Phi^T M + \Phi^TK) x = \Phi^T F$$ Also $\Phi^T M \Phi = I$ for mass-normalized eigenvectors, and $\Phi^T K \Phi = \text{diag}\,(\omega_n^2)$, so inserting $\Phi \Phi^{-1}$ into the left hand side gives $$(-\omega^2 + \text{diag}\,(\omega_n^2)) \Phi^{-1}X = \Phi^T F.$$ Decomposing $X$ into its modal components gives $$X = \sum_n \Phi_n\xi_n,$$ so you end up with the decoupled equations $$(-\omega^2 + \omega_n^2) \xi_n = \Phi_n^T F$$.</p>
<p>In other words, if $\mathbf{f}$ is the vector applied forces, the $F$ in each term of your sum should be $\Phi_n^T \mathbf{f}$ where $\Phi_n$ is the $n$th column of the eigenvector matrix.</p>
<p>Finally, you need to recombine the $\xi_n$ to get $X$, and <em>then</em> take its absolute value.</p>
| 16712 | Undamped forced vibration for a multi dof system |
2017-08-11T06:46:34.920 | <p>i'm trying to apply free body diagram to simplified excavator consept.
<a href="https://i.stack.imgur.com/vwTuc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vwTuc.jpg" alt="Excavator Consept"></a> </p>
<p>As you see, earth drill causes moment on the arm. If we know the "L", we can find F = M / L. But let's say ground friction is very very high. The F force cant move the excavator. Then the arm is going to try take excavator down. Because the arm comes from upper not the ground level. </p>
<p>So the problem is how can i find the force and its moment arm which causes taking down?</p>
| |structural-engineering| | <p>I will try to answer your question.</p>
<p>The problem, from what I can understand, goes something like this. </p>
<p><em>How are you going to determine the maximum weight the excavator arm can support without tipping over?</em></p>
<p>Your understanding is quite wrong there. It is not as simple as M = F*d.</p>
<p><a href="https://i.stack.imgur.com/vRITn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vRITn.jpg" alt="See image for the representation of points and forces considered."></a></p>
<p>As you can see, the excavator arm will create a vertical force downward, say F. The arm is hydraulic-system controlled therefore you can safely assume that the arm is free from rotation and can be considered rigid.</p>
<p>In your drawing, the F is located in the point where the arm is connected to the body. This is quite wrong because that point has nothing to do with the overall tipping action. You should consider the <strong>pivot point</strong> wherein your excavator would tip. In that case, it is the front end of the steel track of the excavator (in case that the arm is directly in parallel to the tracks). It is labelled as point <strong>O</strong> in the image.</p>
<p>Summing up moment at point O yields the following equations:</p>
<p>$$\begin{align}
M &= FL && \text{(overturning moment)}\tag1 \\
M &= F_1L_1 && \text{(righting moment)}\tag2
\end{align}$$
where $F$ is the load carried in the arm; $F_1$ is the load (weight) of the counterweights, including the engine, the body, etc; $L$ is the distance between $F$ and $O$, $L_1$ is the distance between $F_1$ and $O$.</p>
<p>Equating EQ-1 to EQ-2 yields:</p>
<p>$$\begin{align}
FL &= F_1 L_1 \\
F &= F_1 L_1 / L
\end{align}$$</p>
<p>This means that the arm capacity is dependent on the weight of the body, the center of gravity of the excavator and its location, and the distance of the excavator arm from the body.</p>
| 16716 | Effect of Moment on Excavator arm |
2017-08-11T11:59:50.277 | <p>I've recently learned about free-flow intersections. In particular, the <a href="https://en.wikipedia.org/wiki/Diverging_diamond_interchange#Double_Crossover_Merging_Interchange" rel="nofollow noreferrer">DCMI</a>, and the <a href="https://en.wikipedia.org/wiki/Stack_interchange" rel="nofollow noreferrer">Stack</a> interchanges.</p>
<p>What are the disadvantages of using them, compared to the "normal" signaled intersections? do they cost significantly more?</p>
| |highway-engineering|traffic-intersections|traffic-light| | <p>Free flow intersections take up a lot more real estate due to the wider curves needed for the higher speeds. Every time traffic crosses the roads need to be put onto a different grade. This means bridges and ramps to lift the traffic to that grade and back down again. Ramps (and the levies they are built on) also take up space and moves interaction with the main roads away from the actual intersection.</p>
<p>If traffic flow is not heavy then a simple priority regulated diamond interchange with just single lane traffic all around is more than enough. With a little forethought adding a third lane on the bridge to allow for left-turning traffic to not block other traffic will help flow for not that much extra cost.</p>
| 16719 | What are the disadvantages of free-flow intersections? |
2017-08-11T13:47:19.647 | <p>I am using Dynamic Implicit to model a rolling problem. But I am not really sure what to choose as Initial increment. The default suggestion by Abaqus is the step time. </p>
| |finite-element-method|abaqus| | <p>By default Abaqus decides time increments adaptively. The initial time increment is typically a guess. If the guess is too large to solve the equations for the first time increment, it guesses smaller and repeats the process. If the guess could be larger and the equations still solve well, it increases the guess. </p>
<p>If your simulations run in a reasonable time frame, don't worry about the time increment. If you have to run a bunch of similar simulations and would like to speed things up a little, open up the monitor while your simulation is running and watch how Abaqus adaptively sets the first increment. Write down what it settles on, and use that as the first increment size for future simulations. This will save time because Abaqus can jump straight to the right answer without a ton of guessing.</p>
<p>If you have to choose a fixed time increment scheme, the proper time increments depend on a lot of factors, including the physics you are trying to simulate and the non-linearity of your problem with respect to time. </p>
| 16720 | How to decide initial size increment in Abaqus Dynamic Implicit? |
2017-08-11T13:47:40.923 | <p>Regarding Solidworks 2014 (SP2)...</p>
<p>In an assembly I have three identical copies of a single transparent acrylic component.</p>
<p>How may I give them different colours? When I change the colour of the first one to green (material...edit) they all become green even though only the first one is selected.</p>
<p>Clearly this change is getting pushed all the way back to the .sldprt file, but I only want this to happen at the assembly:component instance level.</p>
<p>I have tried to do this using a design table in the .sldprt file but I cannot figure out how to configure the columns. Also I couldn't persuade the appearance manager to understand what I want as the "copy/paste Appearance" features seem to overwrite the Appearance rather than adding new ones.</p>
<p>Edit: @SBarry's suggestion worked a treat...
<a href="https://i.stack.imgur.com/Iqhzi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Iqhzi.png" alt="enter image description here"></a></p>
| |solidworks| | <p>Couple of ways you could do it, but this method is pretty easy:</p>
<p>1) Go to the DisplayManager (Appearances) tab - 4th tab in the Feature Manager area (left side of the screen).</p>
<p>2) Search for the appearance that matches your plastic material. It might say "polycarbonate plastic"</p>
<p>3) Right click the appearance and click "Copy Appearance"</p>
<p>4) Go back to the Feature Manager (your tree listing all the parts)</p>
<p>5) Select the individual components you want to be a specific color from your feature tree by CTRL + clicking the name of each component</p>
<p>6) Right click your selection and click "Paste Appearance"</p>
<p>7) Go back to the Appearances tab, right click & edit the new appearance to be the color you want. It should maintain transparency & general material appearance</p>
| 16721 | How to have multiple copies of the same component all with different colours in Solidworks? |
2017-08-11T15:38:49.983 | <p>I have taken old speakers I found lying around and connected them to a 3.5mm auxiliary connection, however when I plug this into my laptop/phone the sound comes through just really quiet and sometimes only the bass. How can I get the speakers working more efficiently/louder? I'm assuming the Auxiliary connection that I took from old broken earphones is causing the problem, can I get something better?</p>
| |materials| | <p>You will need an amplifier of some sort : stereo and after that the choice is huge.</p>
| 16722 | Wiring old speakers |
2017-08-12T11:07:53.443 | <p>I have a pretty basic question. </p>
<p>I need to install a timing pulley on a ball screw shaft (6mm):
<a href="http://uk.rs-online.com/web/p/products/2816419/" rel="nofollow noreferrer">http://uk.rs-online.com/web/p/products/2816419/</a></p>
<p>The processed end (the pointy side) of the shaft is secured by a support mount, I need to place a timing pulley on the other side. I'm not sure how to go about this. E.g. this pulley: <a href="http://uk.rs-online.com/web/p/timing-belt-pulleys/0744508/" rel="nofollow noreferrer">http://uk.rs-online.com/web/p/timing-belt-pulleys/0744508/</a></p>
<p>Would a mounting hub or a pulley with a holding screw work? Thanks.</p>
| |mechanical-engineering|gears|pulleys| | <p>For such a configuration the screw will need to be fixed at one or both ends so it cannot turn but move freely back and forth; unless you want the timing pulley and ACME nut to traverse back and forth then the screw must not rotate or move freely on its axis.</p>
| 16733 | Mounting a timing pulley on a ball screw shaft |
2017-08-13T08:01:29.020 | <p>I'm getting more serious about a combat robot design I've been contemplating for several years now (BattleBots, Robot Wars, etc) and find myself overwhelmed by the sheer variegation of ASTM and AISI grades and alloys. I hope that by detailing the application, those familiar with the nomenclature can steer me in the right direction.</p>
<p>The overall design is a full-body spinner - somewhere around 0.75m in diameter, at least 2500 RPM. (Approximately 205 MPH weapon speed, though this is a minimum target.) The circumference is lined with five or six solid-mount blades approximating the size and shape of forestry mauls, though slightly thinner - while primarily relying upon inertia and brute force, I would like to retain some degree of sharpness/cutting.</p>
<hr>
<p>Structural integrity is the key attribute, remaining intact (if deformed to some degree) in spite of high-speed impacts with fixed hazards, opponents, and/or similar spinning weaponry. Any significant separation would be of serious concern owing to the high-speed rotation.</p>
<p>The blades will be mounted to the outside of the body with high-strength bolts, anchoring roughly half the length and 60% the mass of each individual blade. (I have no intention of ever welding these pieces as even a fractured blade would be swiftly replaced.) </p>
<p>Withstanding impacts and remaining affixed with minimal distortion is highly desirable. Retaining a sharp edge would be nice, but this is mostly wishful thinking - for the sake of other traits, some sacrifice must be made.</p>
<hr>
<p>What little I understand of this sort of metallurgy has me looking at ASTM-S1, which seems to be recommended for high-speed impacts and sudden shock-loading. (Also assuming those terms mean what I think they do.)</p>
<p>Any guidance/recommendations in this department would be greatly appreciated.</p>
| |materials|steel|robotics| | <p>Answer 2: Because you are not building thousands of the Bots, You need to look at materials that are already heat-treated for strength and toughness. Just thinking of the material and not the form : rear axel( of rear wheel drive ), Torsion bars , leaf and coil springs. Construction equipment ; bulldozer blade ( edge if it is separate material from the bulk, bucket teeth ( if replaceable ), bulldozer treads, blades/arms of big chipper shredder ( hammer mill ) used to chop tree trunks ( Asplendunk is one name). The chipper is almost what you are making. These are educated guess ; all likely have more carbon than optimum for toughness. But it gives some ideas where to look. </p>
| 16737 | Which types of steel would be most appropriate for blunt cutting weaponry on a combat robot? |
2017-08-13T09:06:22.510 | <p>The wikipedia page for groundworks (earthworks) does mention contour trenching and cut & fill but not a word about bring tunnels. Is a bored tunnel part of groundworks discipline/field? If not what discipline does it belong to.</p>
| |civil-engineering|tunnels| | <p>In my experience <strong>groundworks</strong> refers to the site activity of shallow work in the ground, such as trenches (for foundations or drainage) and slab foundations.</p>
<p>Tunnel construction comes under the term <strong>tunnelling</strong>. Tunnelling is a specialist activity: you wouldn't want a general groundworks contractor to try boring tunnels, and you wouldn't want a tunnelling contractor to dig shallow foundations.</p>
<p>On the design side they are all part of the discipline of <strong>geotechnical engineering</strong>. Again, tunnelling is specialised on the design side too, and will be done be specialised geotechnical engineers. Foundation design may be carried out by structural engineers or geotechnical engineers, mainly depending on how complicated it is. Small residential building foundations are designed by rules of thumb based on the soil type; skyscraper foundations take much bigger loads and rules of thumb are not sufficient.</p>
<hr>
<h2><strong>Research</strong></h2>
<p>It's surprisingly difficult to find a definition for <strong>Groundworks</strong>.</p>
<p>There are plenty for "groundwork" as:</p>
<blockquote>
<p>the foundation, basis, preliminary or basic works</p>
</blockquote>
<p><sup><a href="https://en.oxforddictionaries.com/definition/groundwork" rel="nofollow noreferrer">oxford dictionaries</a>, <a href="http://www.dictionary.com/browse/groundwork" rel="nofollow noreferrer">dictionary.com</a></sup></p>
<p>Much of the usage of "groundwork" seems to be metaphorical, comparing "preliminary work" for any type of project to laying the foundation for constructing a building.</p>
<p><a href="https://en.wikipedia.org/wiki/Groundworker" rel="nofollow noreferrer">Wikipedia</a> does have a page for "groundworker" as somebody who:</p>
<blockquote>
<p>clears the site, lays a foundation, installs drainage and other pipework, and may build roads if necessary</p>
</blockquote>
<p>A typical <a href="http://www.malcolmgroup.co.uk/construction/civil-engineering-groundworks/" rel="nofollow noreferrer">website</a> for a "groundworks" contractor talks about:</p>
<blockquote>
<ul>
<li>Re-profiling the earthworks to required design levels</li>
<li>Drainage and roads</li>
<li>Foundations and buildings</li>
<li>Ground floor slab</li>
<li>SurfacingFinishing touches</li>
</ul>
</blockquote>
<p>This seems consistent with the definition for "groundworker" - someone who does shallow depth work in the ground, typically for buildings.</p>
<hr>
<p><strong>Geotechnical engineering</strong> is defined as:</p>
<blockquote>
<p>Geotechnical engineering is the branch of engineering concerned with the analysis, design and construction of foundations, slopes, retaining structures, embankments, tunnels, levees, wharves, landfills and other systems that are made of or are supported by soil or rock.</p>
</blockquote>
<p><sup><a href="http://www.ejge.com/Definition.htm" rel="nofollow noreferrer">ejge.com</a></sup></p>
| 16738 | Is tunnel boring considered groundworks? |
2017-08-14T13:53:30.890 | <p>I want to build drawers for my kitchen. On Youtube I found this Channel, where a guy does woodworking. In a video about making kitchen drawers, he says that the bottom surface of the drawer will be stronger if the joint connecting the bottom to the sides is positioned lower. <a href="https://youtu.be/wLMQ_6Y29pY?t=5m15s" rel="nofollow noreferrer">https://youtu.be/wLMQ_6Y29pY?t=5m15s</a> (5:15 to 6:03).</p>
<p>To me, it should make pretty much no difference. Am I wrong? </p>
| |structural-engineering|stresses|wood| | <p>Using a tie-and-strut model makes it easier to explain the difference here. A tie-and-strut model is a simplification of the structure which describes where the structure is under tension, and where it is under compression. The model gets effectively simplified into a truss. Below we have two examples of the axial load on members of two different trusses with identical loading (positive values are tension, negative are compression):</p>
<p><a href="https://i.stack.imgur.com/KGh4y.png" rel="noreferrer"><img src="https://i.stack.imgur.com/KGh4y.png" alt="enter image description here"></a></p>
<p>Looking at the values, you'll see that the results are mirror images of each other: same values, but different signs. Importantly, notice that the first truss has a negative (compressive) force of 634.9 between the supports and the upper chord, while the second one has a positive (tensile) force of 634.7 between the supports and the lower chord.</p>
<p>Now, looking at the video, we basically have two possible layouts, which are converted into the overlaid trusses:</p>
<p><a href="https://i.stack.imgur.com/a7fLe.png" rel="noreferrer"><img src="https://i.stack.imgur.com/a7fLe.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/74YKI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/74YKI.png" alt="enter image description here"></a></p>
<p>Comparing with the truss results shown above, it is clear that the first diagonal emanating from the support will suffer compression in the first format and tension in the second layout.</p>
<p>Now, what will resist the forces represented by that diagonal? The combined plywood layers, of course. However, that force needs to be transmitted from layer to layer. In the case of compression, that transfer is done by contact. However, in tension, that transfer occurs via the glue. If the glue were to fail, then the plywood layers will separate and the structure will progressively collapse:</p>
<p><a href="https://i.stack.imgur.com/IJSuY.png" rel="noreferrer"><img src="https://i.stack.imgur.com/IJSuY.png" alt="enter image description here"></a></p>
<p>Meanwhile, with the other orientation, that force is compressive, so such a failure is impossible (other failure states will certainly happen before the "compressive strut" collapses).</p>
| 16751 | Which drawer design can hold more weight? |
2017-08-14T19:40:03.243 | <p>In order to identify <a href="https://en.wikipedia.org/wiki/Steinhart%E2%80%93Hart_equation" rel="nofollow noreferrer">Steinhart-Hart coefficients</a> for a thermistor, three known temperatures may be used to find three resistances. </p>
<p>What is the best way to choose these temperatures to calibrate from? </p>
<p>Is it more beneficial to use three temperatures within the typical operating temperature of the thermistor, or to take a broad range of temperatures (say 0, 50, and 100°C)?</p>
| |electrical-engineering|sensors|measurements|temperature|calibration| | <p>It depends to some extent. Normally you'd assume you should look at the datasheet, see what it's operating limits are and then use the centre point for your 3 points. If the thermistor is not entirely linear however, you'll be off at certain temperatures across the range. Personally i'd find out what my expected limits are for the application and calibrate for those limits. I'd even take more than 3 points for that reason.</p>
| 16759 | Thermistor - Steinhart-Hart Coefficients |
2017-08-15T14:32:30.230 | <p>Originally I was going to use accelerometers, but they are to big for the disk I'm trying to measure.</p>
<p>Basically I have a set of planetary gears for which I need to measure the vibrations in the x and y components, mainly. I´ll be performing modal testing with an instrumented hammer and a laser single point vibrometer.</p>
<p>Since my gears are straight teeth spur gears I didn't take into account the "z" coordinate or vibrations of the system for my dynamic model, so if I point the laser perpendicularly to the gear´s plane the vibrations being measured by the laser will be those in z, which I'm not interested in.</p>
<p>I read that if I had two Laser Doppler Vibrometers I could measure the in-plane motions I'm looking for, but since I only have one I was wondering if there´s a way of decomposing the projected component by pointing the laser at certain angle?</p>
| |optics|vibration| | <p>There's just no way that you are going to turn a single Z measurement into an X or Y measurement. With multiple Z measurements, you could calculate pitch and yaw angles, but for a spur gear, that's not going to get you want you want as the gear is going to translate, not bend or rotate. </p>
<p>Suggestion 1: Glue something to the spur gears that sticks up far enough out of plane that you can get the laser to reflect off of it in the X and Y direction. e.g. glue a small reflective cube to the disk. Point the laser at it in the X direction, perform the impact test, and then move the laser to the Y direction, and repeat. </p>
<p>Suggestion 2: regarding accelerometers... How big of a disk are we talking here? They do make some very small accelerometers. e.g. <a href="https://buy.endevco.com/22-accelerometer.html" rel="nofollow noreferrer">Endevco Model 22</a> is a mere 0.14 grams. </p>
| 16770 | Is it possible to measure x & y displacement components with a single point (out-of-plane) vibrometer? |
2017-08-16T15:18:20.947 | <p>Why does a ductile rod in torsion fail at an angle perpendicular to its axis? I know the reason for brittle material failing in pure shear but I don't understand the reason for ductile material.</p>
| |mechanical-engineering|applied-mechanics|mechanical-failure| | <p>As a rule of thumb:
When brittle materials are subjected to torsion they fail in the plane, where tension is at its highest, i.e. at a 45° angle.
Ductile materials on the other hand fail in the plane of maximum shear stress.</p>
<p>Take a look at Mohr's Circle for pure shear.</p>
<p><a href="https://i.stack.imgur.com/JGq6F.png" rel="noreferrer"><img src="https://i.stack.imgur.com/JGq6F.png" alt="Mohr's Circle for pure shear"></a></p>
<p><em>Maximum-Shear-Stress Theory</em> states:</p>
<blockquote>
<p>[...] yielding of the [ductile] material begins when the absolute maximum shear stress in the material reaches the shear stress that causes the same material to yield when it is subjected <em>only</em> to axial tension.<br>– <em>R.C. Hibbeler, Mechanics of Materials, p.525</em></p>
</blockquote>
<p>for ductile materials: $$ \tau_{max}=\frac{\sigma_y}{2} $$
As you can see in the graphic, $\tau_{max}$ is reached before $\sigma_y$, thus the plane of failure is perpendicular to the primary axis. </p>
<p>For brittle materials on the other hand, <em>Maximum-Normal-Stress Theory</em> says:</p>
<blockquote>
<p>[...] a brittle material will fail when the maximum tensile stress, $\sigma_1$, in the material reaches a value that is equal to the ultimate normal stress the material can sustain […]<br>– <em>R.C. Hibbeler, Mechanics of Materials, p.528</em></p>
</blockquote>
<p>for brittle materials: $$ \tau_f=\sigma_f $$
So, under pure shear it fails in tension at a 45° angle.</p>
| 16779 | Failure in torsion |
2017-08-18T09:08:20.903 | <p>I am trying to figure out how much torque this piece of metal that I have can deal with before breaking.</p>
<p>Are there any tools online that can help me calculate these items?</p>
<p>The hardness of this iron is 167-229Ha
tensile strength is >=58kg/mm2
elasticity is >= 45%
density is 0.785</p>
<p>Currently the pipe is a hexagaon that's 55mm in diameter, then the ends are cylinders that are 40mm in diameter.</p>
<p>Since this is early in the design phase I'd like to be able to adjust torque and the size of the metal pipe so that it wouldn't break.</p>
<p>Currently I have 1200RPM 15HP motor that I have reduced down to 85RPM which if I calculated correctly should be around 926 ft-lbs of torque.</p>
<p>Are there any tools that will help me convert from ft-lbs to lbs-ft3 or calculate these values?
<a href="http://www.endmemo.com/sconvert/kg_m3lb_ft3.php" rel="nofollow noreferrer">one such online calculator.</a></p>
| |motors|gears| | <p><a href="http://www.engineeringtoolbox.com/torsion-shafts-d_947.html" rel="nofollow noreferrer">This page</a> and <a href="http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect07.d/IAST.Lect07.pdf" rel="nofollow noreferrer">this page</a> would be good places to get you started. </p>
<p>In addition to my comment above, you'll probably want to design to the yield strength and not the ultimate tensile strength. </p>
| 16800 | ft-lbs to ft-lbs3 or newtons as well as strength of metal |
2017-08-18T17:31:42.530 | <p>Most cheap pump can achieve, according to the Amazon page, around 5 PSI and I would need something a way higher that can still be controlled.</p>
<p>I have <a href="https://www.amazon.ca/4inch-Normally-Closed-Electric-Solenoid/dp/B01M0UEISP/ref=sr_1_3?ie=UTF8&qid=1503076215&sr=8-3&keywords=electric%20air%20valve" rel="nofollow noreferrer">this</a> solenoid valve.</p>
<p><a href="https://i.stack.imgur.com/1R5VP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1R5VP.jpg" alt="enter image description here"></a></p>
<p>Can I use pulse-width modulation (PWM) with a longer time period to control the flow of a pressurized liquid? Will that damage the valve? What is the accuracy of the valve?</p>
| |electrical-engineering|pressure|pumps|valves|liquid| | <p>Yes, you can control flow/pressure with an on/off solenoid, using electronic PWM. EPRs such as these from SMC do exactly that. The lifetime of the valves in # of cycles needs to be considered, as well as their switching time.</p>
<p><a href="https://content2.smcetech.com/pdf/ITV.pdf" rel="nofollow noreferrer">https://content2.smcetech.com/pdf/ITV.pdf</a></p>
<p><a href="https://i.stack.imgur.com/L6UZl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L6UZl.png" alt="EPR Working Principle" /></a></p>
<p>However, if you want smoother control, you can use proportional solenoids, which can smoothly open and close to control flow/pressure. You still use PWM control usually to supply a constant current, however the frequency of PWM will be significantly higher with a proportional solenoid.</p>
<p><a href="https://content2.smcetech.com/pdf/PVQ.pdf" rel="nofollow noreferrer">https://content2.smcetech.com/pdf/PVQ.pdf</a>
<a href="https://www.burkert.com/en/content/download/7480/229630/file/Product-Overview-Proportional-Valves.pdf?id=44" rel="nofollow noreferrer">https://www.burkert.com/en/content/download/7480/229630/file/Product-Overview-Proportional-Valves.pdf?id=44</a></p>
<p><a href="https://i.stack.imgur.com/MzuEL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MzuEL.png" alt="Proportional Solenoids" /></a></p>
| 16803 | Can PWM be used to control the flow of a solenoid valve? |
2017-08-19T12:22:44.133 | <p>Currently software and maps represent earth in many different ways, the latest is Point Clouds as this is the most common surveying tools currently available. For the needs of visualisation, construction etc. this seems to suffice for now. Not sure this is an engineering question but I tried in the philosophy forum and been advised this is applied science.</p>
<p>What is the most accurate representation of earth in a digital format (or mathematical equation):</p>
<ul>
<li>3D splines</li>
<li>solids and primitives</li>
<li>breaklines/contours</li>
<li>polygonal mesh, TIN</li>
<li>square 3D pixels</li>
</ul>
<p>?</p>
| |cad|geometry|surveying| | <p>You are confusing methods of organizing data with the accuracy of that data or the accuracy of the interpolation method. The two are orthogonal.</p>
<p>For example, you could describe a surface in Z as a function of X and Y as a rectangular grid of points. At each point, the Z for that X,Y is known explicitly. For X,Y between points, you do a 2D linear interpolation between the 4 nearest points.</p>
<p>Alternatively, you could describe the surface as a set of contour lines, like a topo map. For any point not on a contour, you interpolate between the two nearest countours.</p>
<p>Which one is more accurate? Either can be made arbitrarily accurate. If you put a grid point every mile and compare that to a map with 10 foot countours, the contour map will usually give you far better accuracy. On the other hand, consider a grid point every meter versus 100 foot contours. Most of the time the grid will be much more accurate.</p>
| 16818 | What is the most accurate geometry to represent earth |
2017-08-19T23:59:26.493 | <p>I'm attempting to make the rigid equivelent of a gate tensioner/turnbuckle. I'm doing this for a couple of reasons:</p>
<ol>
<li>I like the idea of not fussing with cables</li>
<li>I'm using this experience to learn about the technology behind the turnbuckle</li>
</ol>
<p>I've mocked up something that <em>almost</em> works. I've connected two hook-bolts to a threaded rod (5/16-18-36") via connectors. </p>
<p>It seemed like it would work in my head; I thought I would just be able to tighten one of the connectors to tension the gate. However, in practice, all I accomplish when tightening a connector is tightening one side, but then loosening the other side. The devise comes apart and fails miserably.</p>
<p><strong>What kind of mechanism would it take to make this work as expected?</strong> In summary, I want to shorten the length of the device/rod by tightening a connector.</p>
<hr>
<p><a href="https://i.stack.imgur.com/Ab3E9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ab3E9.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/KHmzh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KHmzh.jpg" alt="enter image description here"></a></p>
<hr>
<p>For bonus points: What is the principle called in Mechanical Engineering theory that would explain the problem where one side tightens, but the other side loosens?</p>
| |mechanical-engineering|threads| | <p>I managed to find a turnbuckle that had thread that fit standard threaded rod. </p>
<p><em>Note: Turnbuckles seem to often have non-standard thread. Maybe the thread is tighter for the heavy loads they sustain (relative to the length of thread)?</em></p>
<p>I used a turnbuckle and replaced one of the hooks with a threaded rod (bent in a hook shape). All parts have 3/8" thread. </p>
<p>It seems to do the job well enough.</p>
<p><a href="https://i.stack.imgur.com/yHZ31.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yHZ31.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/VH480.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VH480.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/jW3tW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jW3tW.jpg" alt="enter image description here"></a></p>
| 16822 | Rigid tensioner/turnbuckle: One side tightens, but the other side loosens |
2017-08-20T02:50:46.623 | <p>I did some calculations to find out the moment of inertia and physical qualities of some metal.</p>
<p>I am using typical medium carbon steel.</p>
<p>I'll use a density of 7.8 g/m^3 until I get better numbers.
There's a solid circular shaft that's 1000mm long
The diameter of this shaft is 40mm</p>
<p>This gives me: Polar moment of area 251327 mm^4</p>
<p>The next part is the motor.</p>
<p>This will be a shaft connected to a 15HP 1200RPM motor but I'll be adding a gearbox to bring the RPM down to about 85RPM this will give me about 1256 newtons/meter of force.</p>
<p>I am a bit lost at this junction.</p>
<p>Do I need to convert the newtons to MPa or the Polar moment to some other units?</p>
<p>How do I compare the 1256 newtons to the amount of force this cylinder can withstand 251327 mm^4</p>
<p>would this pipe bend or break under the max torque load?</p>
<p>How can I relate 1256N.m to 251327mm^4 to see if the pipe would break?</p>
| |gears|modeling|moments| | <p>Andrew has provided some good information with regard the formula for shear stress.</p>
<p>What I disagree with is the standard yield strength ($\sigma_y$) values for a medium carbon steel. I think Andrew has forwarded UTS values. For this I think you should be working in the $275MPa$ for low strength, to $450MPa$ for high strength range, with most structural steel being $355MPa$.</p>
<p>To add apply this to your query, you have:</p>
<ul>
<li>$T = 1256 N m$ (torsional moment)</li>
<li>$r = 0.02 m$ (radius)</li>
<li>$J = 25.133 cm^4$ (polar moment of area)</li>
<li>$\tau_{allow} = \frac{\sigma_y}{2} $</li>
</ul>
<p>So, if the calculation results in $\tau_{max} > \tau_{allow}$ then you know you are in trouble regarding any deformations / damage to the shaft.</p>
| 16823 | moment of inertia forces and torque |
2017-08-20T15:30:39.723 | <p>Please explain me the difference between $\lim_{x->0}\frac{\partial E}{\partial x}$ and $\lim_{x->0}dE/dx$.In physics I encountered something similar while reading about Newton's Law of Fluids.While in F.M. White its done using partial derivatives.I want to know the physical difference instead of the highly mathematical one. Having said that I am well conversed with the first principle of derivatives.</p>
| |mechanical-engineering|electrical-engineering|mathematics|experimental-physics| | <p>Consider a surface <code>S</code> in 3-space and its gradient </p>
<p>$ {\hat{i} }\frac{\partial{S}}{\partial{x}} + {\hat{j} }\frac{\partial{S}}{\partial{y}} $ . Then each partial represents the slope along the given axis, as though you'd taken a 2-dimensional slice of the surface and looked at the slopeof the line that resulted.<br>
If you then find the direction (vector in x and y) which maximizes the gradient, you'll find the path of steepest descent. </p>
| 16829 | Difference between partial derivatives and derivatives in physics |
2017-08-20T18:08:45.530 | <p>I understand that Hydrostatic pressure and Soil pressure exist concurrently. I'm curious how these separate loads interact with each other.</p>
<p>Full disclaimer: I am not an engineer nor do I pretend to be one. I recognize that these are very broad & elementary questions; I am at the early stages of my understanding. Please be assured I am not, and will not, attempt to engineer actual structures. </p>
<ol>
<li>Does the pore size limit, in any way, the surface area available for
hydrostatic loading on the back of a retaining wall? I assume it
does not. It would stand to reason that water under pressure would
suspend the soil particles and push its way to the point where most
of the surface area on the back of the wall was under a hydrostatic
load?</li>
<li>Soil particules become suspended in water while a hydrostaic load is
present. Does this (or any other water related factor) affect the
overall soil load or does it remain the same?</li>
<li>Is the total load on the wall simply $\frac{H^2}{2}(\gamma_s K_a + \gamma_w)$ (Where $\gamma_s$ and $\gamma_w$ are the soil and water densities, respectively)?</li>
</ol>
| |pressure|geotechnical-engineering|retaining-wall|soil-mechanics| | <ol>
<li><em>Does the pore size limit, in any way, the surface area available for hydrostatic loading on the back of a retaining wall (I assume it does not)?</em></li>
</ol>
<p>Pore size has no bearing on hydrostatic surface area load on a retaining wall. What does change is the structural load-bearing strength of the soil in the water saturated soil. The more water the less strength in the soil. Imagine that where there is saturated soil it behaves exactly like a fluid (water causes mudslides because the water content exceeds the liquid limit of that soil under gravity and explosively collapses like a river down slopes.)</p>
<ol start="2">
<li><em>Soil particles become suspended in water while a hydrostatic load is present: does this (or any other water related factor) affect the overall soil load or does it remain the same?</em></li>
</ol>
<p>Soil particles don't become suspended in an hydrostatic load but become inert structurally proportionally to the liquid limit state of that soil relative to the depth of soil. Think of doing a rock in a bucket of water. The surface of water rises volumetricaly. The problems arise when the rock particles become very small and are classed as clay. It is impossible to remove all the water from clay because it is bonded electronically to the soil itself. The capacity for clay to <strong>adsorb</strong> water is well understood and documented and causes immense damage to buildings worldwide but it also means that the mass of clay behind a retaining wall will be directly influenced by the water content. This water content will also determine the pressure against which a retaining wall must theoretically be designed to withstand in addition to the safety factor relevant to the local building regulations for such structures. </p>
| 16830 | Retaining Wall: Saturated Soil & Hydrostatic Pressure |
2017-08-20T22:02:21.190 | <p>I know that a 230/400V 3phase motor requires 230V on each coil. No more and no less. So on a 3*230V grid it would be wired in Delta, on a 3*400V it would wired in Y.</p>
<p>I went to evening school to learn about this and my course on the subject has several diagrams on this, most resembling these 2:
<a href="https://i.stack.imgur.com/7mXXA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7mXXA.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Wo4xv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wo4xv.png" alt="enter image description here"></a></p>
<p>We also put this in practice and wired the motors and tested errors etc.</p>
<p>But thinking about it now I noticed that we never grounded the center point of these connections. How can a Y-config at 3*400V produce 230V on the coil? </p>
<p>Is there a mathematical way of calculating this? It's confusing since there's 3 voltages that are all averaged at 400V but measured at any given moment they're 3 very different voltages connected to eachother. </p>
<p>I have however also seen images online of a delta/Y connections that do ground the centerpoint:
<a href="https://i.stack.imgur.com/PSCre.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PSCre.jpg" alt="enter image description here"></a>
And here it totally makes sense that the coil get 230V because we know that L-N voltage is sqrt(3) of L-L voltage.</p>
<p>I feel like I'm missing something here. </p>
| |electrical-engineering|motors| | <p>Your understanding so far is correct. If you draw your phasor diagrams to scale you can solve this from basic trigonometry.</p>
<p><a href="https://i.stack.imgur.com/8oA9M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8oA9M.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Phasor diagram for delta and star connections</em></p>
<p>You should, when you have finished, see that the phase to phase voltages are $ \sqrt 3 $ times the phase to star voltages.</p>
<blockquote>
<p>But thinking about it now I noticed that we never grounded the center point of these connections. How can a Y-config at 3*400V produce 230V on the coil?</p>
</blockquote>
<p>Now realise that without a neutral connection the star / wye point will only be in the centre if all three phases are balanced. If one phase is more highly loaded then the star point will be pulled towards that phase.</p>
<p>The reason that we have a neutral star point at all is that the currents into this point all sum to zero at every instant. </p>
<blockquote>
<p>Is there a mathematical way of calculating this? It's confusing since there's 3 voltages that are all averaged at 400V but measured at any given moment they're 3 very different voltages connected to each other.</p>
</blockquote>
<p>I tend to think graphically and my illustration below may help.</p>
<p><a href="https://i.stack.imgur.com/wSkKT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wSkKT.png" alt="enter image description here"></a></p>
<p><em>Figure 2. (1) The red phase current is the geometric sum of the black and blue. (2) When black and blue are equal and opposite the red must be zero. (3) When black falls to zero the other two must sum to zero.</em></p>
<p>I hope that helps.</p>
| 16835 | How does a Y connection in a 3 phase motor create 230V without a neutral/ground |
2017-08-21T03:52:08.810 | <p>(This is probably a simple question, but I've been staring at it so long I'm not sure I know how to do it any more. I have an Excel spreadsheet with like 3 different ways to calculate this, and they all give different results, so I know I've screwed up somewhere.)</p>
<p>I would like to build a do-it-yourself 1700-degree-Celsius (though will settle for 3000F) electric muffle furnace because I like electricity and playing with fire. I am trying to get an idea of the cost, which means I need to know how thick the insulation need to be.</p>
<p>Unfortunately, materials rated for those kinds of temperatures tend to both be expensive and have relatively high thermal conductivity values, so I plan to use enough high-temperature insulation to reduce the chamber temperature to about 1300C/2400F, surrounded by cheaper, lower-conductivity insulation to bring the exterior temperature down to the target. I want the inside of the furnace as large as I can affordably make it, so instead of a rule of thumb I'd like some real equations I can play with the parameters of.</p>
<ul>
<li>The furnace will be of a front-loading box design.</li>
<li>The furnace will run on 120V AC, with a target power dissipation of 1200W.</li>
<li>For convection calculations, I am assuming a 10 W/m^2K heat transfer coefficient as this will be free-standing. I also assume room temperature is 21C and I'd like to target a maximum of 43C on the exterior surface.</li>
<li>My current calculations use a 12" x 12" x 8" furnace chamber, though I'll probably settle for whatever it turns out I can afford.</li>
</ul>
<p>I am considering the following refractory materials as insulation:</p>
<ul>
<li>Kast-O-Lite 30 LI Insulating Castable: Max 3000F/1650C, max ~0.75 W/mK</li>
<li>Alumina Fiberboard: Max 3100F/1700C, max ~0.45 W/mK</li>
<li>NC-28 Insulating Fire Brick: Max 2800F/1540C, max ~0.44 W/mK</li>
<li>NC-26 Insulating Fire Brick: Max 2600F/1430C, max ~0.32 W/mK</li>
<li>NC-23 Insulating Fire Brick: Max 2300F/1260C, max ~0.27 W/mK</li>
<li>8# Ceramic Fiber Wool: Max 2000F/1100C, max ~0.3 W/mK</li>
</ul>
<p>(I'm thinking about using Satanite, ITC-100HT, and ITC-296A as coating materials, but I can't find data about how much that would improve the conductivity ratings.)</p>
<p>I'm also pretty new to this, so any advice at all is appreciated. (Hell, maybe I'm overthinking this, the ratings mean nothing, and I should just put "a bunch" of insulation in an see what happens.) I plan to use MoSi2 elements and a nitrogen atmosphere, and will be sintering various powdered metals, including steel. I'll be slapping a PID controller on it for controlled ramp-soak phases.</p>
| |thermodynamics|thermal-conduction|thermal-insulation| | <p>First off, the design problem is not trivial and it looks like you have done your homework. On high temperature, small vessels like this you need to consider the 3 dimensional transfer of heat (the wall area is increasing as you move out). Or you can be conservative and much more easily calculate it with all the insulation acting at the outside dimensions.</p>
<p>So lets say you want a 6"x6"x6" firebox with 6 inches (0.152m) of insulation surrounding. Your external surface area of the 18"x18"x18" box would be 18*18*6walls = 5184in^2 or 3.4m^2</p>
<p>At a 1700-20 =1680C differential and 0.4 W/mK
<a href="https://en.wikipedia.org/wiki/Thermal_conductivity#Equations" rel="nofollow noreferrer">Thermal Conductivity Equation</a><br></p>
<p>1680K * (3.4m^2 / 0.152m) * 0.4 W/mK = 15kW worst case
(Notice how units cancel)</p>
<p>To check ourselves, this <a href="http://sentrotech.wpengine.com/wp-content/uploads/2015/03/1700C-666-Data-Sheet.pdf" rel="nofollow noreferrer">commercially available 6"x6"x6" unit</a> uses 4kW. So our numbers were probably a bit too conservative. Calculating it at the 3/4 thickness (14x14x14) gives us 0.76m^2 and 3.4kW. This is likely pretty close. The manufacturer will make sure their amount of heating is some amount greater than the heat loss at maximum temperature so the max temperature can be reached in a reasonable amount of time.</p>
<p>Radiation losses will also become significant at 1700C. A stainless or aluminum reflective shield should be used for your outer layer.</p>
<p>Stacking different rated materials is possible, by running a duplicate calculation (furnace in a furnace). Make sure to have <strong>lots</strong> of safety factor on the maximum material temperature, as those conductivity ratings are not always very close. They will often provide you "at least" the value they are publishing. More insulation is better for most customers, but if you are counting on that external oven to only get to a certain temperature; too high of thermal resistance will result in that material over heating and failing.</p>
<p>Again, a good test of what is possible in a DIY type environment is to look at what is <a href="http://www.mtixtl.com/1700CCompactMuffleFurnace4.7x4.7x4.71.7L-KSL-1700X-S.aspx" rel="nofollow noreferrer">commercially available</a>. Look at the watts vs insulation vs firebox size of what vendors are selling. Then remember they do this for a living so your diy design (while much cheaper for you) will at best require 50% more power, 50% more insulation, or achieve say 200F less temperature.</p>
<p>Sounds like a fun and challenging project! Targeting a <a href="https://www.youtube.com/watch?v=M5oKvHktmNU" rel="nofollow noreferrer">1300C furnance</a> might be a better starting point if you have not built any high temperature devices before. Good luck!</p>
| 16840 | How thick do I need to make the insulation in my homemade furnace? |
2017-08-22T14:49:31.697 | <p>I need to sketch some diagrams involving basic symbols like springs and dampers, specifying angles, perpendicularity etc... something like the images below:</p>
<p><a href="https://i.stack.imgur.com/d5Up7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d5Up7.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/q1AkE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q1AkE.png" alt="enter image description here"></a></p>
<p>Obviously my first option was word, then power point, the visio, in which the latter was a bit more productive, but I haven't found any other software that will let me do the diagrams more productively and have an elegant professional look. </p>
| |mechanical-engineering|design|dynamics|diagram|technical-drawing| | <p>This is more of a long comment than an answer as i can not advice any specific software. First, if you intend to do anything professional in print or web productions than yous shouldn't be looking in direction of Microsoft for anything. Much abused does not mean any good.</p>
<p>First tier would be to use direct vector drawing apps. In this category you have:</p>
<ol>
<li>Illustrator</li>
<li>Autocad</li>
<li>Inkscape</li>
<li>Corel draw etc.</li>
</ol>
<p>They are more than capable of doing these.</p>
<p><a href="https://i.stack.imgur.com/kLmTR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kLmTR.png" alt="enter image description here" /></a></p>
<p><strong>Image 1</strong>: 12 minutes of playing with Illustrator, and that's about all I have time for.</p>
<p>You can also use pacakges designed to do programmatic input like Tex and TiKz for example. Also <strong>Mathematica</strong> works quite well for both images, and you can animate in it.</p>
| 16852 | What software is used for drawing dynamic & mechanical diagrams? |
2017-08-22T18:22:00.203 | <p>I"m working on a design and having difficulty calculating possible misalignment.</p>
<p>I have a shaft (diameter D, tolerance t) that is running through two bushings. The two bushings are separated by a distance L. The two bushings are of width W1 and W2 (measured along the axis parallel to the shaft), with internal diameter of D1 and D2 with tolerances of T1 and T2. (see image below)</p>
<p><a href="https://i.stack.imgur.com/DVCJC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DVCJC.png" alt="enter image description here"></a></p>
<p>How do I go about calculating the potential misalignment of the shaft? When I started this I though it was going to be a quick 5 minute challenge but the more I work on it the more difficult it's seeming.</p>
| |mechanical-engineering|tolerance| | <p>I would like to point out that your CAD application can most likely do this for you. Just make a parametric drawing and sweep the values you need. This may be more practical as you go forward since the cad application can effortlessly make changes based on geometric constraints and you don't need to calculate this.</p>
<p><a href="https://i.stack.imgur.com/E1pM9.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E1pM9.gif" alt="enter image description here"></a></p>
<p><strong>Image 1</strong>: Parametric sweep of one value. No harder than typing 3 values. Note the drawing s intentionally exaggerated.</p>
<p>Making the sweep by script is not much harder. This opens up statistical analysis by montecarlo simulation. Doing it this way reduces chance of error made during the mathematical analysis. Of course, this is both a good thing and a bad thing. Analytical analysis can give insights into global minima, but as the complexity of your analysis grows this might not be feasible so a local minima may suffice.</p>
<p>But yeah you can calculate this by hand too, the problem is that one you start adding the things you now consider trivial this may no longer be a trivial extension.</p>
<p><a href="https://i.stack.imgur.com/N19CA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N19CA.png" alt="enter image description here"></a></p>
<p><strong>Image 2</strong>: Vector expression</p>
<p>You can express this as a equation of vector expressions</p>
<p>$$
\vec a + \vec b + \vec c = 0
$$</p>
<p>where $\vec a$ is known, magnitude of $\vec b$ is known and direction of $\vec c$ is linked to vector $\vec b$. So vector $\vec b$ is $D \cdot \{sin(\theta), cos(\theta)\}$ therefore $\vec c$ is $x \cdot\{-cos(\theta), sin(\theta)\}$. Since you have 2 directions and 2 unknowns this is solvable. But i wont take this further.</p>
| 16854 | Bushing and Shaft Alignment |
2017-08-23T10:49:22.653 | <p>I am trying to figure out am I doing this calculation correctly.</p>
<p>To get the maximum shear stress for a solid cylindrical pipe I need two formulas:</p>
<p>Moment of inertia = pi/2*r^4</p>
<p>Not sure what this formula is called but its:</p>
<p>max = T*r/ Moment of inertia</p>
<p>I tried to calculate these numbers for a solid shaft that is 20mm in diameter with a torque of 1255nm.</p>
<p>1255*0.02/((pi/2)*0.02^4)</p>
<p>Pascals = 99869726.7902
MegaPascals = 99.87</p>
<p>The material that I have has a yield strength of 490 MPa.</p>
<p>Is the calculation saying that even at 20mm radius the stress will be below the 490MPa that would cause this pipe fail elastically?</p>
| |mechanical-engineering|applied-mechanics|modeling|moments| | <p>Your calculation is almost correct. The only problem is that you confused diameter with radius in the calculation.</p>
<p>If the diameter is 20[mm] (i.e. radius =10mm) then the torsional stress is 798.95 MPa.</p>
<p>If the diameter is 40[mm] (i.e. radius =20mm) then the torsional stress is 99.87 MPa.</p>
<p>Please notice that although you have unwittingly only doubled the diameter, the stresses decreased 8 fold (<span class="math-container">$2^3$</span>), due to the non linearity of the problem. So, you can easily get it wrong.</p>
| 16857 | Maximum torsion shear stress in cylinder |
2017-08-24T15:30:59.660 | <p>I have a shape that I'd like to calculate the polar moment of inertia for:</p>
<p><a href="https://i.stack.imgur.com/tKlv5m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tKlv5m.png" alt="enter image description here"></a></p>
<p>the diameter is 600mm; the hexagon in the middle sides are 70mm each; the holes around the hexagon are 130mm from the center of the disk and they are 16mm.</p>
<p>$$ \frac{\pi D^4}{32} - \frac{5\sqrt3}{8}s^4$$</p>
<p>How can I remove the 5 holes from the calculation? Is that even something that's done?</p>
<p>[EDIT]
The disk is 30mm thickness.
The mass of the disk is 66.5Kg</p>
<p>The mass of the small holes is total 0.3Kg</p>
<p>The hexagon mass is 3Kg</p>
<p>weight without holes is 66Kg</p>
<p>mass with the circular cutouts is 66.5-0.3 = 66.2kg
mass with the hexagon cutout is 66.5-3 = 63.5Kg
mass with all the cutouts is 63.3Kg</p>
<p>Which percent change in weight would help calculate the polar moment of inertia for this shape?</p>
| |mechanical-engineering|applied-mechanics|modeling|moments| | <p>Without looking up all the formula's for you, the approach to this problem is rather simple. Find the polar moment of a solid cylinder, and subtract off the
polar moment of the holes.</p>
<p>For the off center holes you need to use the parallel axis therem, so you will have</p>
<p>$$I_\text{assembly} = I_\text{solid cylinder} - I_\text{hexagon} - 6 (I_\text{small cylinder} + m r^2 )$$</p>
<p>the expressions you need are all on <a href="https://en.wikipedia.org/wiki/List_of_second_moments_of_area" rel="nofollow noreferrer">Wikipedia</a>.</p>
| 16873 | polar moment of inertia for compound shape |
2017-08-24T17:36:37.983 | <p>I am working with a normalized ball-screw drive where the nut has holes for screws to screw onto the moving part. The catalogue says these holes have diameter 13.5 mm. What is the appropriate screw size (metric) to use on such a part?</p>
| |design| | <p>I would suggest you use the largest diameter bolt you can find that fits through the hole, and that you can be sure you'll be able to install properly. By "install properly" I mean that you leave enough clearance between the bolt and 13.5mm hole to account for misplacement of the holes in whatever you're mounting the nut to. </p>
<p>In my work, when I design a clearance hole for a bolt, I usually go 0.5mm above the nominal size of a bolt, and we feel comfortable going down to 0.2mm above nominal if we need a more snug feet. A 13.5mm hole for a 12mm bolt is a lot of slop, so I would suggest a small washer if you go this route.</p>
<p>If you have imperial sizing available, a 1/2" bolt will have a nominal diameter of 12.7mm, so that would probably actually be your best bet. 13mm bolts do exist, but they're not common. </p>
| 16875 | Hole diameter vs Screw Diameter |
2017-08-27T20:54:57.747 | <p><strong>"Ponder the path of your feet, and let all your ways be established." Proverbs 4:26</strong></p>
<p>Over the last two years I have had a strong will to become an inventor within the field of
computer-technology/robotics,a field where I have many visions and concepts for projects,
and have been persistent and consistent in my ways, so that I may reach a point that I
can assume the title of engineer. </p>
<p>I'm currently working full-time(lot's of over-time as a laborer), yet with all my free time I try to mold myself
into the man that can execute these projects I visualized. I'm currently trying to design and develop my first project, and I wasted a lot of time picking different books
up, not knowing where to start my research and discipline. </p>
<p>Should I have started with programming? Electronics? Mechanics? Eventually I realized that I had a strong
apptitude for mathematics, and decided to order a used college algebra book from online, and I have been making
progress ever since(would be done with the txtbook if it wasn't for the amount of hours I worked every week.)
Electronics was a path that I always felt was on this path that I am becoming aware of, and in different periods
of my life I would find myself picking up Eectronics books and for reasons beyond my understanding I would never get far. </p>
<p>Yet now that I have developed a solid foundation of mathematics, I am beginning to understand and comprehend Electronics
theory, and I am confident that I can now begin to progress to tinkering with components.
I have also decided that maybe it would be a good thing not to try to discipline myself so much, so I
have enrolled into a computer science degree program, and my long term educational goal Is to become a computer engineer.</p>
<p>With all this being stated, what is the path? We all have a journey, we all have a point in space-time that we
have visualized, but how do you arrive at that point? How do I arrive at the point where I have designed and developed
my first robotics project?</p>
<p>I'm seeking feedback from the community so that I may arrive at this point as efficiently as possible. Thank you.</p>
<p>**P.S.: Excuse me for my lack of formatting, this is only my second post on Stack-Exchange, and it is a learning curve for me.</p>
| |robotics|computer-engineering| | <p>Starting from a fairly low level academic background, I think the best way to "develop your first robotics project" is to focus on the practical side, not the theory. </p>
<p>Get a robotics kit like Lego Mindstorms, build some of the tutorial projects, then follow your own ideas. Mindstorms is not just a "kid's toy" - MIT were involved in designing it, and it is used at university level for robotics lab work. It has the advantage of providing ready-made <em>subsystems</em> that do relevant tasks which you can interconnect to make your complete robot, instead of having to design everything yourself from scratch. </p>
<p>Of course you <em>can</em> design your own subsystems from scratch and use them with the rest of the Mindstorms kit, if you need to or want to learn how to do that. But in the real world, engineers don't design stuff from scratch just "because they know how to", if they can buy it ready-made at a reasonable price.</p>
<p>I wouldn't discourage anyone from starting a computer science degree program - so long as you understand that computer science is fundamentally a <em>theoretical</em> subject, not a <em>practical</em> one. Realistically, your math background might not be strong enough to start a CS course right now. You might do better to switch to computer engineering, or mechatronics - they are more focused on "practical engineering" and actually "making stuff" rather than the theory of "why stuff works" - though of course CE and mechatronics will also cover the <em>essential</em> parts of the theory.</p>
<p>Note, "making stuff" in the previous paragraph includes "making software" as well as "making hardware". You probably don't need to know how to design and write your own compiler for a new computer language (which you would learn how to do in a CS course) or design a new type of CPU chip to run your newly invented language efficiently - but you do need to know how to <em>use</em> a "standard" programming language and like C, C++ or Java and "standard" embedded computing hardware like an Arduino, a Raspberry Pi, or a smaller microcontroller system to interface with the hardware of your robot. A CE course is likely to be more relevant for that than CS.</p>
<p>The same applies for "learning electronics". Get some (fairly cheap and simple) kits, build them, then modify them. When (note, "when" not "if"!) your modifications don't work, you can always ask why on Electronics SE! The theory will make more sense if it you can relate it to a practical context. Theory on its own is all very well, but the <em>real</em> learning starts when you try to understand what just caused some of your circuit components to "release their magic smoke" and die!</p>
| 16896 | I Say To myself:"I Am An Engineer!"...How Do I Arrive At That Point? |
2017-08-29T15:23:45.607 | <p>I have a 20 cm long glass tube of outer diameter <code>6.4 mm</code> and inner diameter <code>4.3 mm</code>. Air at <code>Ti = 20 °C</code> is pushed through the glass tube at <code>2 litres per minute</code>. The glass tube has a wire wound resistance heater on it which we can assume is perfectly thermally insulated from the ambient air.
My aim is to find out the power needed to heat the air flowing through the tube to <code>Te=650 °C</code> at the tube exit. </p>
<p>After doing the calculation using </p>
<blockquote>
<p><img src="https://latex.codecogs.com/gif.latex?%5Cdot%7BQ%7D=%5Cdot%7Bm%7Dc_p(T_e-T_i)&space;=%5Crho%5Cdot%7Bv%7Dc_p(T_e-T_i)%5C%5C" title="\dot{Q}=\dot{m}c_p(T_e-T_i) =\rho\dot{v}c_p(T_e-T_i)\\" /></p>
</blockquote>
<p>Where <img src="https://latex.codecogs.com/gif.latex?%5Cdot%7BQ%7D" title="\dot{Q}" /> is the power supplied by the heater (<code>W</code>),</p>
<p><img src="https://latex.codecogs.com/gif.latex?%5Cdot%7Bm%7D" title="\dot{m}" /> is the mass flow rate (<code>kg/s</code>),</p>
<p><img src="https://latex.codecogs.com/gif.latex?%5Crho" title="\rho" /> is the density of the air <code>kg/m^3</code> at the bulk temperature <img src="https://latex.codecogs.com/gif.latex?T_b=(T_i+T_e)/2" title="T_b=(T_i+T_e)/2" />,</p>
<p>I get the power required as being <code>12.5 W</code>. This seems to me to be too low. Unfortunately I do not know where I am going wrong. Does anybody have any idea where I am making a mistake in my calculation?</p>
| |thermodynamics|heat-transfer|heating-systems|glass| | <p>If I perform a back of the envelope calculation (using just the density of air at standard conditions, 1.225 kg/m3):</p>
<pre><code>Converting to SI units of mass flow rate:
2 SLM of air -> 0.04 grams / sec = 0.00004 kg / sec
Cp = 1005 J / kg K
Te = 923.15 K
Ti = 293.15 K
P = 0.00004 * 1005 * (923.15-293.15) = 25.326 W (or J / s)
</code></pre>
<p>Your density is probably different by a factor of two or so: @ 350C, air density = 0.566 kg / m3. Everything looks in order to me.</p>
<p>Yes, the required power may seem low, but look at how little mass is flowing through the section that is being heated every second. 2 liters a minute sounds like a lot until you break it down into grams per second. Electrothermal heating is also a fairly lossless process, so if you perform this experiment, you will likely not need to increase this number by too much to account for inefficiencies.</p>
| 16909 | Problem finding temperature gain of air through heated glass tube |
2017-08-30T04:05:19.460 | <p>I am working on a small project where I, from a relatively confined space in a well that has a diameter of 30 cm, will need to transport the water upwards about 2 meters.</p>
<p>I have been looking at ram pumps but cant really see a way to utilise that as i dont have a "higher" water source. I have also been looking at electric pumps on Alibaba and it seems that i'll land at around 2W, which seems like a lot for an iot device - i really dont want to have to recharge the device every week, esp. since there is little energy to harvest nearby.</p>
<p>The requirements are:</p>
<ul>
<li>Ability to pump water upwards 2 meters</li>
<li>Flow of around 12 liters per hour would be enough</li>
<li>Ideally 12V or lower</li>
</ul>
<p>What are the best options? :) I know that product recommendations are not allowed, but what techniques should I look at overall? Is there a resource somewhere that compares the efficiency of different pumping methods?</p>
<p>[EDIT] RE comments: In terms of effeciency, I mean as low wattage as possible. I wont have easy access to recharging (solar and wind are out, but potentially the energy of the water flowing down into the well could produce a minor charge for the battery). If we can get to a point where the battery only needs to be changed once a month, that would be perfect.</p>
| |pumps| | <p>12 liters per hour and 100% efficiency is:<br>
12l/hr * 1kg/l * 1hr/3600s * 2m * 9.81m/s^2 = <strong>0.0654 Watt</strong></p>
<p>Best case efficiency on a pump this small without a military budget will probably be 25%.</p>
<p>To operate for one day on a 12v battery you would need:<br>
(0.0654w/12v)/0.25 * 24hr = <strong>0.5AmpHr/day</strong></p>
<p>I don't know the details of your setup, but it would probably be more efficient (for the cost) to pump one liter up to a gravity holding chamber (perhaps with a level switch) 12 times a day, rather than run a very tiny pump continuously.</p>
<p>A very small <a href="https://www.aliexpress.com/item/1PCS-used-DC-6-V-submersible-pump-sinking-pump/32690622459.html" rel="nofollow noreferrer">centrifugal pump</a> or <a href="https://www.aliexpress.com/item/2PCS-Micro-Vacuum-Pump-Medical-Mini-Negative-Pressure-Pump-DC-Pressure-Small-Air-Pump-DC3V-6/32791745569.html" rel="nofollow noreferrer">diaphragm pump</a> would probably be your best bet for the low head application. A <a href="https://www.aliexpress.com/item/24V-Brushless-DC-Mini-Water-Pump-Submersible-40H-24110-11M-450LPH-with-DC-Plug-Low-Noise/32606474788.html" rel="nofollow noreferrer">brushless centrifugal pump</a> will generally have higher efficiency if you can find one small enough.</p>
<p>If there is any available light (even indirect), solar may be applicable for this low demand application.</p>
| 16915 | Most efficient/low power consumption way to pump water |
2017-08-31T18:52:22.497 | <p>I see some example of DLX assembly instructions such as:</p>
<p><code>LW Ri, 0 (Rz)</code></p>
<p>but I can't find out why <code>Rz</code> has brackets around it.</p>
| |electrical-engineering|embedded-systems|computer-engineering| | <p>Basically, "Rn" means "the contents of register n" and "(Rn)" means the "data in the memory location whose address is in register Rn". </p>
<p>For memory access, you can also add a constant to (Rn), so 4(R8) means "add 4 to the address in register 8 (but don't change the data that is in R8!), and use the data in that memory location".</p>
<p>See <a href="http://www.cs.utexas.edu/~pingali/CS378/2015sp/lectures/DLX.pdf" rel="nofollow noreferrer">The DLX Instruction Set Architecture</a> for examples (e.g. on Page 8.)</p>
| 16928 | DLX assembly synthax |
2017-09-01T09:05:56.590 | <p>We know that moment of inertia is commonly known as second moment of area. Which is $\int_{}{} y^2 dA$. </p>
<p>However there another moment of inertia called as mass moment of inertia given by $\int_{}{} r^2 dM$ </p>
<p>Also we know that radius of gyration is by definition,</p>
<p>$r_{mass} =\sqrt{ \frac{ \int_{}{} r^2dM }{\int_{}{}dM} }$</p>
<p>Similarly </p>
<p>$r_{area} = \sqrt{ \frac{ \int_{}{} r^2dA }{\int_{}{}dA} }$</p>
<p>So my question is whether this $r_{mass}$ and $r_{area}$ are same for any arbitrary object? Of course the axes about which integral is evaluated are the same. </p>
| |mathematics|moments| | <p>Area moment of inertia and mass moment of inertia are two different things.</p>
<p><strong>A) Mass moment of inertia (or moment of inertia):</strong> </p>
<p>This is the resistance offered by a <em>solid body</em> when subjected to rotation (or application of torque). The mass moment of inertia is given by </p>
<p>$$I = \int r^2 dM $$</p>
<p>The radius of gyration is the distance at which the mass can be considered to be concentrated (with respect to the axis of rotation). For two wheels with the same overall mass but different distribution of the mass, the one with mass concentrated near the axis of rotation has less moment of inertia. This is the 'rotational' equivalent of the 'mass' in translational motion, thus sometime also known as the 'angular mass'. Moment of inertia has units $kg m^2$. <a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia" rel="nofollow noreferrer">Here is a list</a> of moments of inertia some regular geometric shapes. Check out <a href="https://www.youtube.com/watch?v=7PRZ5IIT5Tg" rel="nofollow noreferrer">this video</a> by the YouTube channel 'smarter every day' for a physical sense of the moment of inertia. </p>
<p><strong>B) Area moment of inertia (or second moment of area):</strong></p>
<p>The area moment of inertia is a property of a $2$D section or surface of a body. For a solid body, the area moment of inertia can be defined only for the cross sections. The second moment of area is not to be confused with mass moment of inertia. The area moment of inertia is used to characterize the resistance offered by a body (such as a beam) for deflection under load. Its units are $m^4$. </p>
<p><a href="https://en.wikipedia.org/wiki/List_of_second_moments_of_area" rel="nofollow noreferrer">Here is a list</a> of area moments of inertia of a few regular $2$D shapes. Note that, for a circular shaft, <a href="https://en.wikipedia.org/wiki/Polar_moment_of_inertia" rel="nofollow noreferrer">the polar moment of inertia</a> is more relevant. </p>
<p>Coming to the question, </p>
<blockquote>
<p>So my question is whether this $r_{mass}$ and $r_{area}$ are same for
any arbitrary object? Of course the axes about which integral is
evaluated are the same.</p>
</blockquote>
<p>Provided that a solid body has uniform cross sectional area and uniform second moment of area (both about the axis of rotation), $r_{mass}$ and $r_{area}$ will be same if the material is isotropic (i.e. uniform density). </p>
<p><strong>Proof:</strong></p>
<p>For an object with uniform thickness $h$ unit and constant cross sectional area $dA$ and constant second moment of area $I_{area}$ along the thickness, the total mass will be
$$ M = \int dM = \rho h \int dA $$. </p>
<p>In such as case, </p>
<p>$$r_{mass} = \sqrt{\frac{\int r^2 dM}{\int dM}} = \sqrt{\frac{\rho h \int r^2 dA}{\rho h \int dA}} $$</p>
<p>where, $\rho$ is the constant density of the material. This is same as $r_{area}$. However, it is to be noted that the mass moment of inertia was found out for the entire body whereas the area moment of inertia is only valid for a surface of the body (cross section) thus it needs to be same along the axis, same holds true for area of the cross-section itself.</p>
| 16938 | Is radius of gyration same for both types of moment of inertia? |
2017-09-02T09:19:52.353 | <p>There are Tait-Bryan angular changes $(\dot{\phi}, \dot{\theta}, \dot{\psi})$, which describe the angular change rates of the body-frame within the inertial frame (world frame).</p>
<p>And then there are angular change rates $(p,q,r)$ of the aircraft within its own body frame.</p>
<p>To convert these angular change rates between each other, a rotation matrix is needed.</p>
<p>$\begin{bmatrix} p \\ q \\ r \end{bmatrix} = \begin{bmatrix} 1 & 0 & -sin(\theta) \\ 0 & cos(\phi) & sin(\phi)cos(\theta) \\ 0 & -sin(\phi) & cos(\phi)cos(\theta) \end{bmatrix} \cdot \begin{bmatrix} \dot{\phi} \\ \dot{\theta} \\ \dot{\psi} \end{bmatrix}$</p>
<p>Why is that? Why aren't they exactly the same?</p>
| |aerospace-engineering| | <p>When plane is flying north, roll axis goes from north to south. </p>
<p>When the plane is flying west the roll axis goes from west to east.</p>
<p>So no they are not the same.</p>
| 16945 | Rotation of aircraft's body-frame |
2017-09-02T17:12:18.110 | <p>According to James Cameron, the submarine barely had any space for him to move around etc., and as he approached the bottom of the Mariana Trench, the submarine shrieked down by a few inches (due to pressure).</p>
<p>Why was the challenger deep designed to be just big enough for James Cameron to fit in?</p>
| |pressure| | <p>The design of the submarine to be used in such purpose depends on the following (arranged from least important to most):</p>
<ol>
<li><p>Cost - The amount used to build the submarine. With respect to Solar Mike's and Fred's comments, the larger the submarine, the more money you will use to fund the project.</p></li>
<li><p>Buoyancy - The larger the submarine, the harder it gets to make it remain submerged. Remember that buoyant force is equal to the weight of the fluid displaced by the object, therefore, having bigger submarine requires bigger engines, thus requires more fuel. Fuel efficiency is very important as depth is about 11 kms below MSL.</p></li>
<li><p>Design strength - The walls of the submarine shall be designed to resist 107,910 kPa of pressure (gauge) because the pressure is directly proportional to the depth. Stress = PD / 4t. That means, as the diameter becomes bigger, the thicker the vessel should get to resist pressure. If the submarine becomes too thick, this goes back to item no. 1 (cost efficiency) and no. 2 (submarine now becomes very heavy and will need a lot of power to propel upwards)</p></li>
</ol>
<p>Hope this helps.</p>
| 16951 | Why was the challenger deep so small? |
2017-09-03T13:35:55.863 | <p>In the context of physical asset management (BIM etc) what is a Design Element? I have come across the term several time but never in an official document - can anyone explain the term and provide a definition in case it is an actual professional term</p>
| |cad|construction-management| | <p>Wikipedia explains the term "Design Element" quite well. I don't expect there is a formal "official definition" of the term Design Element, but the relevance to BIM (Building Information Modelling) seems fairly obvious. </p>
<p>See <a href="https://en.wikipedia.org/wiki/Visual_design_elements_and_principles" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Visual_design_elements_and_principles</a>
<a href="https://en.wikipedia.org/wiki/Graphic_design#Experiential_graphic_design" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Graphic_design#Experiential_graphic_design</a></p>
| 16965 | What is Design Element in the context of asset management |
2017-09-06T01:01:18.950 | <p>Does anyone know what this mechanism for linear movement is called? I have a different model of something like this, and the black chain appears rigid, so I assume it is the mechanism moving the laser head. I could be wrong though.</p>
<p><a href="https://i.stack.imgur.com/JvXR2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JvXR2.jpg" alt="The mechanism in question."></a></p>
| |mechanisms| | <p>The head probably moves by rack and pinion or a hidden timing belt.</p>
<p>The cable management is done by a <a href="https://en.wikipedia.org/wiki/Cable_carrier" rel="nofollow noreferrer">cable carrier</a>. It works by having each joint only have a limited rotation so that it can support itself when held horizontally.</p>
| 16993 | What is this linear movement mechanism called? |
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