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2020-01-15T06:35:39.040 | <p>I had/have a 3D camera back in the day, and I gotten a few pictures. I kinda stopped using it since there wasn't anything I could do with the MPO files. However, the camera might gain a new life if I can turn pictures into a CNC carving by grayscale the MPO files. Does anyone know if this is possible and where it is?</p>
<p>BTW I wasn't sure if this is the right subgroup to ask this to. If it isn't, then please kindly direct me to the sub group I should ask such a question to.</p>
| |3d-printing|cnc| | <p>This was answered at another place. There is 2 programs that I know about now.</p>
<p><a href="https://triaxes.com/articles/from-stereo-pair-to-lenticular-print/" rel="nofollow noreferrer">https://triaxes.com/articles/from-stereo-pair-to-lenticular-print/</a></p>
<p><a href="http://www.depthmask.com/priceen.html" rel="nofollow noreferrer">http://www.depthmask.com/priceen.html</a></p>
<p>none so far is that great when it comes to the details</p>
| 32778 | Is it possible to grayscale a MPO file (3D camera) for a CNC carving |
2020-01-15T15:40:39.370 | <p>I am putting together a basic machine for “pad printing” extremely small and precise parts. The design is very much the same as most homemade drill presses and consists of linear rods, flanged bearings and compression springs, but in place of a drill bit is a 30mm conical silicone rubber pad/stamp. The object being stamped will be positioned underneath on a precision XYR stage. A basic hand lever will be used to press the the rails/block downwards, applying minimal pressure. Scale will be approximately 10-15” tall. The blocks will be made of aluminium at this point and will be mounted to a very large solid base. Screws will he used to fix each component together. My goal, once set up, is for the pad to land in the exact same spot every time and with absolutely no movement or variation. It needs to be very rigid and with no slop.</p>
<p>I would like to know what diameter linear rods I should use, and if it makes much difference. I was originally looking at 40mm but this seems large?</p>
<p>Should I use standard steel rods? I have also considered introducing a ball spline shaft to prevent rotation.</p>
<p>I am not sure what size or type screws would be best suited for holding this thing together.</p>
<p>Are there any other factors I am overlooking that will help out? </p>
<p>I have come up with a few different layouts and I may be overthinking it. Do any of them make a difference to the overall result and accuracy of the press? Is there any particular direction I should be taking?</p>
<p>Thank you <a href="https://i.stack.imgur.com/66HMX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/66HMX.jpg" alt="concepts"></a></p>
| |springs|linear-motion|coil-spring| | <p>it may be necessary to construct more than one design (known as iteration) in order to refine your final objective. "Very rigid and with no slop" is a rather general description. "A tolerance of 0.10 mm" would be more precise.</p>
<p>My understanding of pad printing is that one requires the impression pad to be re-inked from a source located near the destination. This suggests that your design requires a rotating head, which is another location of "slop" in the construction.</p>
<p>If you do not require maximum efficiency, you could get away with having the re-inker in the same location as the parts to be inked, with suitable alignment guides to ensure that the precision is maintained.</p>
<p>If you require the re-inker to be offset, your drawings don't show a precise enough option.</p>
<p>Generally speaking, for the most precision, use rectangular/linear assemblies and avoid cylindrical mounts for anything moving. My mini-milling machine has a rectangular column, while my drill press has a cylindrical column, as the drill press does not offer the precision of the mill. Related to this build question, the milling machine does not rotate about the z-axis and would not be capable of re-inking in the offset manner.</p>
<p>One of your drawings shows a pair of vertical rods. This approximates a rectangular support structure, but allows for torsion of the rods in the rotation of the z-axis. The 40 mm diameter referenced is not going to be "too big" for this type of structure, based on the description of the precision required.</p>
<p>Unrelated to this build, have you considered photo-lithography to accomplish your objective?</p>
<p>There are many details omitted from your design description, but enough of a basic question that this post may give you assistance.</p>
| 32784 | Help with linear rod drill press like design/materials |
2020-01-15T18:43:31.807 | <p>I have a stepper motor linear actuator assembly <a href="https://www.amazon.co.uk/Akozon-Automatic-Actuator-Aluminum-Effective/dp/B07JYGRFFL/ref=asc_df_B07JYGRFFL/?tag=googshopuk-21&linkCode=df0&hvadid=375410926273&hvpos=1o2&hvnetw=g&hvrand=9849643731108924450&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=1006948&hvtargid=pla-633082499162&psc=1&tag=&ref=&adgrpid=77488740915&hvpone=&hvptwo=&hvadid=375410926273&hvpos=1o2&hvnetw=g&hvrand=9849643731108924450&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=1006948&hvtargid=pla-633082499162" rel="nofollow noreferrer">like this for example</a>, but the stepper shaft seems able to move axially back and forth - hence when I change direction it has to do a few rotations, forcing the ballscrew against one end or the other, before the rail platform itself starts moving. Why could this be happening and what can I do to fix this? </p>
<p>If anyone has a diagram showing the internals of a NEMA stepper motor that would be really helpful also. I've seen many on the internet but none showing me enough to understand this problem. Many thanks!</p>
| |mechanical|stepper-motor|actuator|linear-motion| | <p>Constrain the driven shaft and use a coupling to the motor.</p>
| 32789 | How to constrain stepper motor inside case? |
2020-01-16T12:56:25.677 | <p>I'm trying to cut some bendable flaps on aluminium sheets for hooking/clipping onto some fabric. The flaps I have in mind look like this:</p>
<p><a href="https://i.stack.imgur.com/FFklU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FFklU.png" alt="enter image description here"></a></p>
<p>Are these common features? Does anyone know the geometry of the cutout? I wish SolidWorks had them as standard features.</p>
| |design|cad|metals|cutting|metal-folding| | <p>They will need to be sized appropriate to the load that will be applied, which means that they are not “standard”.</p>
<p>You need to make sure the part that bends is sufficiently strong and will have sufficient life when it is flexed often.</p>
| 32795 | Bendable flaps on sheet metal |
2020-01-17T09:59:11.843 | <p>Let's assume I have a cogeneration plant that produces heat with a nominal output. The heat is given into a heat storage tank, which is filled from above. The maximum flow temperature of the cogeneration plant is 85 °C, if this occurs an emergency shutdown takes place. The maximum water temperature in the heat storage tank is 95 °C. </p>
<p><a href="https://i.stack.imgur.com/GcuNy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GcuNy.png" alt="enter image description here"></a></p>
<p>Is it theoretically possible in this case to reach a water temperature of over 85 °C only with the cogeneration plant?</p>
| |electrical-engineering|energy|heating-systems|energy-efficiency|power-engineering| | <p>when the cogeneration plant switches of upon reaching 85°C flow and without another heat source, there's no way to reach 95°C.</p>
<p>Are you sure the switch-off happens at 85°C flow, and not at 75°C return <em>to</em> the cogernetion plant(this is a typical value for Otto-cycle cogeneration plants for the emergency cooler to kick in)? </p>
<p>If you have other heat sources and want to achieve a higher flow temp. you can use a setup with flow like this: cold site of storage -> cogeneration plant -> boiler -> hot side of storage. </p>
| 32806 | Question about flow temperature and storage temperature in a cogeneration plant |
2020-01-18T03:20:08.587 | <p>My prof is the type who makes up his own questions (rather than pulling them from textbooks). This is his latest work... Anyone got any ideas on how to solve this? He assures me that you don't need velocity or chord or additional info (like Mach numbers) to figure it out and I keep ending up with 3 unknowns with only 1 or 2 equations! Here's the question:</p>
<blockquote>
<p>High-performance gliders have very large aspect ratios in order to
soar efficiently (for reasons we will discuss later in the course). A
1:30 scale glider model with an aspect ratio of AR=30 is tested in a
water tunnel, and found to have a lift coefficient of CL=0.8 at a
Reynolds number of Re=2.0*10^6. Based on this result, what is the
maximum mass of a full-scale glider, operating in air at the same
Reynolds number, that can maintain level flight? Give your answer in
kg. </p>
<p>Acceleration due to gravity: g=9.81 m/s^2</p>
<p>For water: ρ=1000 kg/m3 and μ=1.0*10^-5 kg/ms</p>
<p>For air: ρ=1.2 kg/m3 and μ=1.9*10^-5 kg/ms</p>
</blockquote>
| |mechanical-engineering|aerodynamics| | <p>I discovered the solution method through more banging head against a wall:</p>
<p>You have to do a double-substitution of velocity and chord, together, and you only need to focus on the prototype.</p>
<p>Solve Reynolds equation for u<em>c (u</em>c = Re*(mu)/(rho))
Use Aspect ratio to find area in terms of chord only (S = c^2 * (AR))
Plug area into lift force equation
Rearrange to get (u<em>c)^2 in the equation and solve using values for (rho), (u</em>c), C_L, and (AR).</p>
| 32819 | Aerodynamics - How can you solve for mass when velocity & chord are not given? |
2020-01-18T07:07:20.423 | <p>I am currently trying to model the buckling behaviour of a cylinder using sol 400 nonlinear static analysis using Nastran. I am using the arc-length (modified Riks) method as apparently it can increment load and displacement in both positive and negative directions, as opposed to Newtons method which only increases the load or displacement increment. However, my problem is that the analysis always stops after a certain load due to convergence problems. </p>
<p>I have tried changing many of the parameters such as MINALR/MAXALR ( min/max arc-length ratio between increments) and MXINC(max increments), however I still don't get any good results. </p>
<p>I feel like I am missing a parameter that enables the load to decrease after a maximum load has been reached. </p>
<p>I am aware that load control and displacement control is possible, but my understanding is that arc-length method uses neither of these, and is sort of a combination of the two. My model has a force applied without any enforced displacement. Is this maybe the problem?</p>
<p>Any other ideas which might help me?</p>
<p>Thanks in advance.</p>
| |structural-engineering|structural-analysis|aerospace-engineering|buckling|nastran| | <p>If your model is very symmetrical (e.g. a circular cylinder with an axial compressive load, modeled with a regular grid of elements not an automatically generated irregular grid) you can hit problems because when you reach the instability, there are several possible unstable paths and no way for the program to choose between them. (In the circular cylinder example, the cylinder could theoretically buckle in any lateral direction.)</p>
<p>One solution is to make the model unsymmetrical, so it prefers to buckle in one particular direction. For example, apply a small side load in the direction you want it to buckle.</p>
| 32823 | For arc-length method in nastran, how to analyse post-buckling behaviour with load decrease |
2020-01-19T00:28:52.177 | <p>I have a electronics assembly company, and we run two assembly lines. They consist of several machines, with the heaviest being around 4000lbs, on 6 metal casters. We currently have the line on a concrete slab, inside a large converted warehouse.</p>
<p>The concrete slab is badly spalled, and apparently didn't have a vapor barrier when originally poured. If we leave anything heavy on the floor (cardboard box), it will be dripping with water the next day. We maintain a humidity level of 45%, which probably brings more water up also. We are located in Florida, so have a high water table.</p>
<p>The landlord is willing to work with us to fix this issue, and a contractor recommended we do a vapor barrier on the slab, then a grid of 2x4's, then tongue and groove plywood. On the plywood we would install ESD safe vinyl tiles. They said this is common in gyms and dance studio's. They use rubber strips under the 2x4's to help dampen noise and vibration.</p>
<p>It's hard to find information on this method, is this suitable for this amount of weight, in an industrial environment? I really don't see of many other options if this isn't acceptable to do. I would assume we would do the 2x4's either 6" or 12" on center.</p>
| |concrete|wood| | <p>The use in gyms and dance studios is not a good reference. Those floors are designed for dynamic impact loads from people dropping gym equipment and dancers jumping around. They need to be able to absorb the impact energy by deflecting under load, and therefore the plywood is introduced as a "soft" element which is able to bend with relative ease so the dancers don't hurt their knees and ankles when landing. The distance between the 2x4's is not just a maximum distance but also a minimum distance to prevent the floor from becoming too stiff.</p>
<p>This is generally not the best way to design floors for heavy machinery especially if the machinery is able to induce vibrations in the underlying structure. Then it will be dancing around. But assuming the machinery won't be moved around all that much, it is an option to create hardpoints to support it — supports that don't rely on the stiffness of the plywood. And by reducing the distance between the 2x4's and increasing the thickness of the plywood, it is possible to design it for fairly large point loads. So it is possible to do it right, but don't accept a design that has been copy-pasted from a gym or dance studio. Demand that is redesigned for the weight as well as any vibrations of your machinery and it should be okay.</p>
<p>For this type of use, I would normally recommend an epoxy-based coating applied directly on the concrete, but in your case it may be impossible to get the concrete dry enough for the coating to stick properly. So maybe a bitumen-based membrane on top of the concrete, a concrete screed on top of the membrane to protect it and then an epoxy-based coating on top of the screed could be an alternative approach capable of supporting very heavy machinery.</p>
| 32828 | Raised plywood false floor for heavy equipment usages |
2020-01-19T02:20:16.227 | <p>I want to design an artificial leg for disabled people. The idea is design of a Device for patients with problems on one side of the leg from the knee down. This device first detects intent to move through movement of the hip. The knee flexes when it detect force in the ankle foot. Then it decides the type of motion needed for the foot by using force sensors placed on the sole of the healthy foot. These force sensors are placed on three different regions; heel, metatarsal and toe. Using the human gait pattern in relation to force and time. This device can segment the pattern of the person’s gait and analyze the required gait of the disabled ankle-foot so it may assist the patient to move appropriately. This device also assists in telerehabilitation as it records the patients gait data and sends it to a medical professional for efficient follow up and prevention of further injury or fall. </p>
<p>I have little knowledge on where to start and I want to take courses. I know I need mechanical and electrical engineering lessons. However, I don't know which courses to take and where to take them. Any recommendations?</p>
| |mechanical|robotics|electrical| | <p>You will need to do some math and sizing calculations along the way so by all means check out the course suggestions in the other answers.</p>
<p>Most of the time, engineering isn't about inventing something new but finding a solution that also fits to your problem (like a specific type of sensor or bearing). Courses on robotics or prosthetics building should give you many hints about what parts etc. exist. Additionally, read up as much as you can on existing prosthetics how they go about things, try to spend some time at a relevant trade fair and have things shown and explained to you. Spend time with people using prosthetics and try to understand their gripes.</p>
<p>I'm in a completely different field but to me, discussions on why <em>this</em> solution was chosen over <em>that</em> solution are often enlightening.</p>
<p>Lastly, never underestimate manufacturability: Can the thing I design actually be built?</p>
| 32830 | what to learn to design mechatronics device that move |
2020-01-19T22:54:50.993 | <p>There are the "consumer" negative ion detectors which are about 50 USD, reviews appear to indicate that they are not measuring very much at all. Then there are the professional ones which are $800 and up. As I understand it the customary standard measurement is in millions of ions per cubic centimeter; the precision needed is not great, 25% or 20M/cm3 will do, range 10 to 300 would be very helpful. Any ideas?</p>
| |electrical-engineering| | <p>The standard measurement for a genuine, <em>high-density</em> negative ion generator is 1,000,000 negative ions (anions) per cm3 <em>at a distance of 1 meter from the ionizer</em>. That is the level that Dr. Michael Terman of Columbia University used in his well-known study of the treatment of depression using negative ions. I believe he has a patent on that.</p>
<p>If you are looking for an accurate ion counter, look at the <a href="https://www.alphalabinc.com/product/aic/" rel="nofollow noreferrer">Alpha Lab air ion counter</a>. That's what we used in <a href="https://web.archive.org/web/20180826040944/http://negativeiongenerators.com/index.html" rel="nofollow noreferrer">our negative ion generator business</a> (which we might sell because we are retiring).</p>
<p>On the web, you'll find several variations of a two-transistor negative ion detector schematic. It can be quite useful; the LED just starts to glow at ~9000 ions/cm3. (In fact, our IDS-2 ion detector was based on that circuit. Google is your friend.) And the model IG-133* series we manufactured --that generated that ion density-- made its LED just glow <em>at 8' away</em> from the ion emitter on those models. Therefore, it was quite useful in determining whether a particular ionizer was actually a high-density ionizer or not.</p>
<p>You can also make a simple but useful one with an NE-2 neon lamp in parallel with a .001 uF disc ceramic capacitor. Solder a 6" piece of bare wire to one end, and hold the other end between your thumb and forefinger. Unless a static charge has built up on your body, it will slowly blink within 6" from a high-density ionizer. We used to give these (IDN-1) away with every ionizer purchase.
<hr>
This is <strong>not</strong> spam. <strong>We no longer sell any of these products.</strong></p>
| 32836 | Simple negative ion detector circuit? |
2020-01-20T22:45:26.727 | <p>First time asking question here.</p>
<p><strong>I'm not a mechanical engineer!</strong></p>
<p>I have a simple question that I can't explain the answer to it.
I have a double gear (2 concentric gears with different radius) and I don't understand why the torque is the same for both.
I found in the internet that the torque should be the same because they are cocentric, but don't understand why.</p>
<p>Thanks!</p>
<p>Gear Example:</p>
<p><a href="https://i.stack.imgur.com/7V1jF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7V1jF.jpg" alt="enter image description here"></a></p>
| |gears|torque| | <p>Consider, for example, the shaft passing through the two gears has a torque of 10Nm and the diameter of the larger gear is twice as the diameter of small gear, <span class="math-container">$ R_{large}= 2*r_{small} $</span>.</p>
<p>The tangential force on the circumference of the large gear is then <span class="math-container">$\frac{10Nm}{R}=\frac{10Nm}{2r}\quad$</span> but the tangential force on the small gear is twice as much<span class="math-container">$ =\frac{10Nm}{r} $</span></p>
<p>Therefore the torque produced by the small gear is <span class="math-container">$ =\frac{10Nm}{r}*r=10Nm.$</span> </p>
<p>And the torque produced by the big gear is <span class="math-container">$=\frac{10Nm}{2r}{2r}=10Nm. $</span></p>
<p>So regrdless of the diameter of the gears the torque on them is equal. </p>
| 32850 | Torque between concentric gears |
2020-01-21T18:53:00.563 | <p>I'm a computer scientist doing a dissertation in robotics and I'm designing a mechanical arm that mirrors a humans, as such I am wanting to pivot the entire arm from a "shoulder" using a stepper motor and some gearing, I was looking at holding torques for different stepper motors and realised that I'm not actually sure how to calculate the force that is going to be put on the arm and the motor when the arm is straight out perpendicular, is this the same as the holding torque? </p>
<p>I think part of my confusion is I am not to sure what terms to actually search, so sorry if this has already been answered before.</p>
| |mechanical-engineering|gears|robotics|stepper-motor| | <p>When the arm is straight out it has a torque <span class="math-container">$T= mg*L \quad $</span>with L the distance between arm's CG and rotation center.</p>
<p>Your stepper motor should provide this torque to keep the arm in balance.</p>
<p>if you need to rotate the arm up and down you need an additional torque such that if you rotate the arm with angular acceleration, <span class="math-container">$\alpha$</span>
<span class="math-container">$$T_{rotation}=I\alpha$$</span> </p>
<p><span class="math-container">$$ I= 1/3mL^2$$</span></p>
| 32859 | Calculating the torque required to move and hold an arm parallel to the floor |
2020-01-21T18:54:24.773 | <p>I need help designing simple circuits that resonate around specific frequencies. Basically I am looking for recommendations as to what size capacitor, inductor, and resistors I should get. Sorry if this is not the right forum - I am new to all of this!</p>
<p>We will use them to model neuronal resonance in an undergrad course. Each student group would get a breadboard + components, simple audio / function generator (adjusting input frequency) and measure resulting AC voltage with a multimeter. They will estimate the resonant frequency range by noting which input frequency produces the highest voltage response on the multimeter, and then change out components of the circuit to see how their properties affect this output.</p>
<p>I was hoping to create circuits that resonate around 8 Hz, 16 Hz, and anywhere between 30-35 Hz. These don't have to be exact.</p>
<p>Thanks for your help.</p>
| |circuit-design| | <p>Assuming that you have an <span class="math-container">$RLC$</span> circuit, the resonant frequency will be
<span class="math-container">$$ f_0 = \frac{1}{2\pi\sqrt{LC}} $$</span>
so, theoretically, you can pick either <span class="math-container">$L$</span> or <span class="math-container">$C$</span> arbitrarily and then calculate the other for a given resonant frequency.</p>
<p>Unfortunately, you want circuits that are resonant at low frequencies. For reasonable values of <span class="math-container">$C$</span> you will need very large <span class="math-container">$L$</span>. As an example of something similar you could look at the crossover networks used with subwoofer speakers. In the real world, high-value inductors also tend to have a high parasitic resistance, because they are made with many turns of wire. This will tend to make the resonance appear less ideal.</p>
<p>If you have an audio generator or function generator you may not be able to drive the low impedances that you will get at resonance. You will need <span class="math-container">$R$</span> to be at least 50Ω, perhaps several hundred ohms. For students to be able to observe a significant change in behavior at resonance, the total impedance of the <span class="math-container">$RLC$</span> should be significantly greater than <span class="math-container">$R$</span> when close, but not at, resonance. This will also push you to large inductors.</p>
<p>I suggest that you try calculating some values and then do some shopping to see what you can purchase. When I have done these lab exercises I used circuits that were resonant in the tens of kilohertz range, and the components were fairly easy to find and easy for students to work with.</p>
| 32860 | Designing RLC circuits at specific frequencies for teaching use |
2020-01-21T20:11:27.130 | <p>it is known that the stability of a system with negative feedback may be analized through the Nyquist stability criterion, which is based on observing the number of turns around the point (-1;0) in the complex plane.</p>
<p>But, what if we want to apply the same criterion for a positive feedback system? Is it correct or should be modified in some ways?</p>
<p>Can we say also for positive feedback that the system is unstable if and only if there are at least 1 pole with positive real part?</p>
| |control-engineering|control-theory|systems-engineering|stability| | <p>You could just multiply <span class="math-container">$H$</span> with minus one, change the positive feedback to negative feedback and proceed with the standard analysis for the Nyquist stability criterion. Here <span class="math-container">$H$</span> is the transfer function in the feedback path as shown in <a href="https://www.electronicshub.org/wp-content/uploads/2015/07/Negative-feedback-signal1.jpg" rel="nofollow noreferrer">this image</a>.</p>
| 32863 | Nyquist stability criterion for positive feedback |
2020-01-22T17:55:36.657 | <p>On my air tool compressor or scuba gear you can see or hear a diaphragm regulator sloughing off a tiny amount of gas to the atmosphere when you adjust the regulator down. Does a natural gas regulator do the same if there is variation in the supply line pressure?</p>
| |compressed-gases| | <p>According to a valued associate from Washington DC Gas, residential regulators can vent small amounts of raw natural gas to the surrounding environment and cannot be enclosed. Any enclosed regulator say, in a ship, must have active ventilation measures. Interior appliances like water heaters are regulated from the external regulator and do not vent.</p>
| 32873 | gas regulator exhaust gas during diaphragm adjustment? |
2020-01-23T14:22:01.443 | <p>Imagine you are in the mountains and it is very cold. The walls are colder than the air, so air close to the wall will lose heat to the wall, and by losing heat it might lose its humidity (water condenses to the wall). Now I know this is bad for the paint on the wall. However, in some sense, it is heating up the wall, which will lessen the cooling effect of the air. Please tell me if my logic is correct.
Also, if the walls are now wet, could they allow for more condensation? Or it is impossible because all the surface is covered in water?
PS: House without heating nor insulation.</p>
| |thermodynamics|heat-transfer| | <p>Certainly we could calculate the heat gain to the wall due to condensation (a glass of ice water will melt faster in humid climates due to the heat gain from condensation).</p>
<p>In practice, though, this will be insignificant compared to typical winter cooling loads. Also, you will never heat the wall above the dew point of the air (temperature at which water condenses out of the air), which is likely in winter to be quite low.</p>
<p>You will never be able to actually heat the room this way because to heat the room you need the temperature of the wall to be above the temperature of the air in the room (heat always moves from high temperature to low temperature). Once the wall is hotter than the air next to it, no water will ever condense from the air.</p>
| 32879 | Wall Humidity Condensation |
2020-01-23T14:19:16.213 | <p>If I'm in the wrong section of stack exchange, please help me go to the right place.</p>
<p>I am trying to find out if this combination of motor shaft and clamps will work out.</p>
<p>I found these clamps/bushings</p>
<p><a href="https://www.maedler.de/product/1643/1621/spannsaetze-com-b-bohrung-6-bis-100mm" rel="nofollow noreferrer">https://www.maedler.de/product/1643/1621/spannsaetze-com-b-bohrung-6-bis-100mm</a></p>
<p>and I'm curious if they will work out with the kind of shaft on this motor </p>
<p><a href="http://mehatron.rs/step-motor-nema-34-86x155-60a-130nm-sa-duplom-osovinom" rel="nofollow noreferrer">http://mehatron.rs/step-motor-nema-34-86x155-60a-130nm-sa-duplom-osovinom</a></p>
<p>The main problem is that the shaft of that motor is machined and the clamp relies on the full surface of the shaft.</p>
<p>By the way, those clamps work for the motors with a shaft containing removable key ( for example MiGE 130ST-M15015 ).</p>
<p>Aliexpress is full of these shaft couplings, but i do not believe that they are sufficient to hold these kinds of torque levels, for example :</p>
<p><a href="https://www.aliexpress.com/item/32967906911.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-9&btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53" rel="nofollow noreferrer">https://www.aliexpress.com/item/32967906911.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-9&btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53</a></p>
<p>or these :</p>
<p><a href="https://www.aliexpress.com/item/32956726953.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-8&btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53" rel="nofollow noreferrer">https://www.aliexpress.com/item/32956726953.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-8&btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53</a></p>
<p>The good thing about these sahft couplers is that they can be made to fir the shaft perfectly, with 2 screws for the 2 machined parts i was talking about.</p>
<p>The forces are around 10 Nm and I want to avoid any kind of slipping.</p>
| |motors|coupling| | <p>I am going to say that no, your chosen coupling would not work. Based on the K-K cross-section in the drawing, there would be a large portion of the shaft that is not in contact with the coupling.</p>
<p>I believe what you need is a coupling where it is held against a flat in the shaft by a set screw, like the one here (though in your size): <a href="https://www.omc-stepperonline.com/shaft-coupling/5mm-6mm-flexible-coupling-18x25mm-cnc-stepper-motor-shaft-coupler-st-fc02.html" rel="nofollow noreferrer">https://www.omc-stepperonline.com/shaft-coupling/5mm-6mm-flexible-coupling-18x25mm-cnc-stepper-motor-shaft-coupler-st-fc02.html</a></p>
<p>You will need to match your torque requirements to the coupling as well.</p>
| 32880 | Motor shaft mounting with clamps and bushings |
2020-01-24T18:49:45.457 | <p>I am stuck on this problem for a while:</p>
<p>I have a square-shaped camera, with:</p>
<ul>
<li>1000 pixels on vertical side</li>
<li>1000 pixels on horizontal side</li>
</ul>
<p>I want to know the number of meters per pixel depending of the distance, I mean:
If I have a 1 meter long object which is 100 meters far from the camera, how many pixels of the resulting picture will be occupied by the object?</p>
<p>I am not sure if it is useful, but we can estimate the length and size of the camera:</p>
<ul>
<li>0.05 meters vertical</li>
<li>0.05 meters vertical</li>
</ul>
<p>How to calculate the number of meters per pixel as defined above?</p>
<p>EDIT:</p>
<p>From first answers and the given formula, I am wondering:</p>
<ul>
<li>Does the formula depends of how far from the lens is the pixel device?</li>
</ul>
<p>SECOND EDIT:</p>
<p>I have this data: The Optical Form Factor of the camera is: 1/2.5''</p>
<p>Is it related to the lens focal length? I did not find a definition on wikipedia and only <a href="https://blogs.cisco.com/datacenter/qsfp-dd-an-evolutionary-approach-to-400gbe-interconnect" rel="nofollow noreferrer">this blog article</a> that I don't understand</p>
| |mathematics|optics|geometry| | <p>Independent of what you are imaging, the ratio of object dimension to it's image dimension is a function of the lens focal length and the distance to the object. </p>
<ul>
<li><p>Knowing the sensor dimensions allows you to determine if the image will "fit on the sensor". </p></li>
<li><p>Knowing the sensor pixels per distance allows you to express the distance in pixels.</p></li>
</ul>
<p><strong>Image dimension / lens-focal-length = Object dimension / distance to object.</strong> </p>
<p>So eg a 1m long object at 100 metres from the lens, with a 200mm lens, will result in </p>
<p>ID / 0.2m = 1m / 100m or
image dimension = 1 x 0.2 / 100 = 2 mm</p>
<p>If the sensor is a full frame "35mm" sensor of typically 24 x 35 mm size, in landscape mode, and the object is eg a 1m vertical fencepost, then the 2mm image will occupy 2/24th = 8.333%n of the sensor height. </p>
<p>If the sensor is a 4000 x 6000, 24 megapixel sensor then the image of the 1m fence[post will be 4000 pixels x 2/24 = 333 pixels tall on the sensor.<br>
So 1 pixel is 1/333 of a meter at 100m at 200mm focal length = 3 mm at the object,<br>
Which is awesomely impressive. </p>
<p>Each pixel on the sensor is 24mm / 4000 = 6 um.<br>
Which is also awesomely impressive. </p>
<p><a href="https://i.stack.imgur.com/bhpm0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bhpm0.jpg" alt="enter image description here"></a></p>
<p><strong>In the image above, "Height" to base size is the same for both triangles.</strong> </p>
<p><strong>So: Image-height / focal-length = Object height / distance to subject.</strong><br>
So Image-height = focal-length x Object height / distance to subject.** </p>
<hr>
<p>Note:</p>
<p>Edited - I had a factor of 10 error initially.<br>
I KNEW my 200mm lenses were not THAT good, so went back and found the error. </p>
| 32892 | How to calculate meters per pixel for a given camera? |
2020-01-25T03:41:02.103 | <p>So, I have been doing research into the use of Insulated <em>Composite</em> Concrete Forms (ICCFs) for small-scale construction, and that has led me to looking at the Rastra ICCF system, as it has been in production for a long time, is relatively well documented, and is versatile enough to answer a variety of building design needs; in particular, Rastra parts can not only be used to build <em>walls</em>, but <em>horizontal assemblies</em> (floors/ceilings/roofs) as well.</p>
<p>However, the published documentation for Rastra ICCFs stops short of providing span charts for their use in horizontal assembly construction. I contacted Rastra's technical support regarding this omission, and they sent me a PDF file that included the following drawing and determination-table (span chart) for Rastra floor systems using Trigon-type reinforcing girders and supplemental reinforcement laid in a bed of Rastra end elements, then finished with a concrete pour:</p>
<p><a href="https://i.stack.imgur.com/gYoM1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gYoM1.png" alt="drawing of Rastra floor using end elements"></a></p>
<p><a href="https://i.stack.imgur.com/1NEeD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1NEeD.png" alt="Rastra floor span table"></a></p>
<p>Most of this is clear enough (although I'm not sure how to get my paws on prefabricated reinforcing girders of the style they used, given that I'm across the pond from them); however, it does not answer one question for me: are the loads given across the top of the posted table inclusive or exclusive of the floor assembly's unit weight (inherent deadload)?</p>
| |structural-engineering|building-design|reinforced-concrete| | <p>The load values in the table is described as the design load with no further qualifiers and that will generally mean the total load including all dead load and live load as well as safety factors.</p>
<p>It would usually be problematic to specify the capacity exclusive of the inherent dead load because multiple load safety factors need to be checked, that is both a set of factors with high safety on live load combined with lower safety on dead load as well as a high safety on dead load combined with a lower safety on live load, so the critical safety factor on the inherent dead load would depend on the size of the live load and non-inherent dead load.</p>
<p>Since you write that you're across the pond from the manufactorer, I'll add just for good measure that since material safety factors etc. vary from country to country, such a table is only valid for the jurisdiction it was intended to be used in.</p>
| 32898 | Are the design load values in this floor determination table (span chart) inclusive or exclusive of the given floor weight (inherent dead load)? |
2020-01-25T16:10:22.073 | <p>Consider a mass of about <span class="math-container">$130 \pm 10 $</span> kg, falls from height, let say <span class="math-container">$2.5 $</span> m.
In order to dampen the vibration and sound, I use silencer or drop pads. I was wondering if introducing the macroscopic projections, helps to reduce the thickness of the shock absorber in comparison to the classic flat surface shock absorbers.
<a href="https://i.stack.imgur.com/c38BE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c38BE.jpg" alt="enter image description here"></a> </p>
<p><a href="https://i.stack.imgur.com/Wj7hQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wj7hQ.jpg" alt="enter image description here"></a></p>
<p>After reading Kamran's answer:
I think the impact energy results in transversal deformation of the projected surface, that is in contrast to the vertical compression of the flat surface. So there would be less energy transmitted to the floor. </p>
| |mechanical-engineering|materials|dynamics|shock|impact| | <p>No, look at it this way: </p>
<p>you are removing material that would otherwise help break the fall.</p>
<p>All it does is give the mass a chance to penetrate deeper into the drop pad.</p>
<p>It would be ok if you add it to existing pad for a thicker pad if you are looking for less noise and less stress to the floor. </p>
| 32912 | Does the surface roughness play any role in shock absorbing? |
2020-01-26T12:56:27.683 | <p>Consider this image of a loop shaping design:</p>
<p><a href="https://i.stack.imgur.com/ghIlY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ghIlY.png" alt="enter image description here"></a></p>
<p>here which is the bandwidth of the system? I have a doubt because until now I have considered as bandwidth to the system the one of the complementary sensitivity function, but I heard someone sayong that it is not a good measure for the bandwidth, so it is better to consider as bandwidth the one of the sensitivity function.</p>
<p>Can somebody clarify this?</p>
| |control-engineering|control-theory|systems-engineering| | <p>There are many definitions of bandwidth. Typically it is determined from the open loop transfer function and not from the sensitivity or complementary sensitivity functions. </p>
<p>Assuming a SISO feedback loop, with plant <span class="math-container">$P$</span> and controller <span class="math-container">$C$</span>, a common definition of bandwidth is the 0 dB crossover frequency of the open loop transfer function <span class="math-container">$PC$</span>. Assuming that the sensitivity can be written as <span class="math-container">$\frac{1}{1 + PC}$</span> this is also (roughly) the frequency where the sensitivity peaks. </p>
| 32918 | Doubt on bandwidth of a control system |
2020-01-27T10:16:05.220 | <p>I have a transfer function with a RHP zero in <span class="math-container">$s=+1$</span>, and I am trying to show in Matlab thath this limitation is present. In particular, initially I had a MIMO system, which I decoupled and then I started doing this analysis on the two separate SISO system that came out of the decoupling.</p>
<p>My code is the following:</p>
<pre><code>s = tf('s');
G = [2/(s+1) 3/(s+2);1/(s+1) 1/(s+1)];
d_12 = -(G(1,2)/G(1,1));
d_21 = -(G(2,1)/G(2,2));
D_simp = [1 d_12;d_21 1]; %decoupler
F1 = G*D_simp;
k1 = 5;
k2 = 2;
k3 = 50;
loop1 = loopsens(F1(1,1),k1/s);
loop2 = loopsens(F1(1,1),k2/s);
loop3 = loopsens(F1(1,1),k3/s);
figure;
bodemag(loop1.Ti,'r',loop2.Ti,'b',loop3.Ti,'g'),grid
figure;
bodemag(loop1.Si,'r',loop2.Si,'b',loop3.Si,'g'),grid
</code></pre>
<p>where I am trying to increase the bandwidth of the system by increasing the gain. </p>
<p>The expected result is that if I increase the bandwidth , I should have worse performances as the bandwidth increases, but what happen is exaclty the opposite:</p>
<p><a href="https://i.stack.imgur.com/ldMRl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ldMRl.png" alt="enter image description here"></a></p>
<p>can somebody please help me solve this problem? </p>
| |electrical-engineering|control-engineering|control-theory|systems-engineering|signal-processing| | <p>This is because for all three of of your choices for the gain the closed loop system is unstable. Namely, for just an integrator as controller the gain should be below 1.5 in order for the closed loop system to be stable and all of your gains are above that.</p>
| 32932 | Problem with RHP zero in transfer function |
2020-01-27T21:11:11.253 | <p>For brushless DC motors we have hall sensors to keep track of rotor; however, for stepper motor we don't know the actual position of the rotor and therefore while changing pulses with some speed it may happen that the rotor cannot keep up with the rotating magnetic field and lags behind. Then some unexpected behaviors may occur. How do we make sure that this is not going to happen? In practice, does it happen just above some speed value, i.e. below that speed everything is ok and no such problem occurs? Or what are some other aspects to consider for avoiding such scenarios.</p>
<p>Thanks</p>
| |control-engineering|motors|pid-control|stepper-motor| | <p>Although the advertised advantage of a stepper is that it can be driven "open loop" (no position or velocity feedback required for accuracy, instead you just keep count of the steps) this is only true when the speed with which the step signals are being fed to it are below a certain rate threshold. Beyond that threshold, the stepper armature begins to lag behind the steps and position errors then occur. </p>
<p>Furthermore, the "stepper approximation" only holds when the compliance and inertia of the load fall within certain bounds. Outside those bounds, the relationship between the number of steps sent to the stepper and its angular position is broken and position errors again result. </p>
<p>This limits the use of steppers to certain applications where the stepper speed and its load are well-specified to be within the stepper's <em>accuracy envelope</em>, as specified by the manufacturer. </p>
<p>Outside that envelope, the designer must use a DC motor and shaft encoder, plus a closed-loop control system that uses position reports from the encoder to "juice" the motor in such a manner as to minimize or eliminate position errors. </p>
<p>This stepper-versus-DC motor-and-encoder tradeoff is well-known within the design community. </p>
| 32943 | Stepper motor speed control problem |
2020-01-28T14:08:18.730 | <p>A piston pump has a swash plate that pivots back and forth to control the flow rate of the hydraulic fluid.</p>
<p>Is there a version of this pump where the swash plate rotates like a cam instead of the pistons rotating?</p>
| |pumps| | <p>Here are two categories of pumps with non-orbiting pistons: </p>
<ol>
<li><p>Axial piston pumps with rotating swash plate.<br>
Operating principle at 0:48 - <a href="https://www.youtube.com/watch?v=sP3gc4b8Z5k" rel="nofollow noreferrer">https://www.youtube.com/watch?v=sP3gc4b8Z5k</a><br>
Commercial examples:<br>
Dynex Rivett PFXXXX and PVXXXX series<br>
Oilgear PFBX and PFCX series </p></li>
<li><p>Radial piston pumps.<br>
Operating principle at 0:15 - <a href="https://www.youtube.com/watch?v=2T-6jDm_ebI" rel="nofollow noreferrer">https://www.youtube.com/watch?v=2T-6jDm_ebI</a><br>
Commercial examples:<br>
Bosch-Rexroth PR4-XX series<br>
HYDAC HRK series<br>
HAWE R series </p></li>
</ol>
<p>Need more info about inlet/outlet pressure, flow, temperature, working fluid, noise limits, motor speed/power/torque, geometric constraints, etc... to recommend a specific pump. </p>
| 32951 | Rotating Swashplate on Piston Pump |
2020-01-28T17:50:16.930 | <p>I have the following transfer function:</p>
<p><span class="math-container">$G(s)=\frac{-s^3 +s}{s^4+3s^3+2s^2}$</span></p>
<p>and this is a transfer function of a plant with two poles in the origin. So I want to desing a controller with two poles in the origin in order to have zero steady state error. So the controller I want to desing is of the type:</p>
<p><span class="math-container">$C(s)=\frac{k}{s^2}$</span></p>
<p>now, I want to find the k such that the closed loop is stable, but I am having some problems in doing so. I am using the Routh criterion, and I have that the closed loop is:</p>
<p><span class="math-container">$\frac{-k(s-1)}{s^4+2s^3-ks+k}$</span></p>
<p>and the table for the Routh criterion:</p>
<p><span class="math-container">$1$</span>|<span class="math-container">$0$</span>|<span class="math-container">$k$</span></p>
<p><span class="math-container">$2$</span>|<span class="math-container">$-k$</span>|</p>
<p><span class="math-container">$\frac{k}{2}$</span>|<span class="math-container">$k$</span></p>
<p><span class="math-container">$\frac{-k^{2}-2k}{\frac{k}{2}}$</span></p>
<p><span class="math-container">$k$</span></p>
<p>and where I have omitted terms it means that there is a zero and the bar <span class="math-container">$|$</span> has been used to say that these numers are in different columns.</p>
<p>By doing this I obtain that the closed loop is stable if <span class="math-container">$k>0$</span> and <span class="math-container">$k<-4$</span>, but if I use these values I have that the closed loop is unstable.</p>
<p>And also I would like to ask if there is a simpler way to find the value of <span class="math-container">$k$</span> instead of using the Routh Criterion, for example using Matlab.</p>
<p>What am I doing wrong?</p>
| |control-engineering|control-theory|systems-engineering|stability| | <p>To solve this problem I would:</p>
<ol>
<li><p>Sketch the block diagram of the control network.</p></li>
<li><p>Derive the required closed loop transfer function in terms of <span class="math-container">$G(s)$</span> and <span class="math-container">$C(s)$</span>. </p></li>
<li><p>Substitute the expressions for <span class="math-container">$G(s)$</span> and <span class="math-container">$C(s)$</span> to obtain the closed loop transfer function in terms of <span class="math-container">$s$</span>. </p></li>
<li><p>Calculate the poles of the closed loop transfer function using your preferred method. Alternatively, you could use the Routh criterion to evaluate stability.</p></li>
</ol>
<p>It should not matter which closed loop transfer function you select (e.g. sensitivity, complementary sensitivity, etc.), they should all have the same closed loop poles.</p>
<p>To do this in Matlab use <code>tf</code> to define your plant and controller transfer functions; then use <code>feedback</code> to calculate the closed loop transfer function (or better use the expression for the closed loop transfer function in terms of <span class="math-container">$G(s)$</span> and <span class="math-container">$C(s)$</span> derived above). You can then calculate the closed loop poles from this expression using the <code>pole</code> command.</p>
<p>To get an expression in symbolic form you will likely have to use the symbolic toolbox. </p>
| 32954 | Problem finding value of the gain such that the closed loop is stable |
2020-01-28T23:08:51.370 | <p>In a PE (polyethylene) foam factory there have been some incidents with static electricity that have created minor fires. Fire is a concern in such a factory, as PE is highly flammable, and even more as PE foam uses propane or butane gas in its manufacturing process (that's what creates the bubbles in the foam).</p>
<p>The issues we've been having are mostly related to the thinnest of the foams we produce, which is 1 mm thick. It's formed by extrusion into a flat fabric of about 1m width and we form rolls of 350 linear meters. The machine's speed is about 78 meters / minute.</p>
<p>The foam is rolled over cardboard cores in a torque machine (the shaft is chrome-coated steel). The plant is located in a very dry, cold place (humidity is often around 40-60% and temp ranges between 10-28°C).</p>
<p>I've gathered the following observations while analysing this problem:</p>
<ul>
<li>While the roll is below 45cm in diameter, the static electricity is negligible (I don't have an instrument to measure it, I just get my arm near the roll and see if the hair is drawn towards the foam).</li>
<li>Once the diameter gets beyond the 45cm diameter the static is noticeable, to the point of firing little sparks (when it reaches the max diameter).</li>
<li>The static builds up first near the sides of the roll, even though it can be felt in the center too. All of the incidents result from a spark igniting the roll <em>in any of the sides</em>, never in the center.</li>
<li>The foam goes through a series of rolls (puller, thkcness measurment) before being formed into a roll. Most of those rolls are metallic (steel) and in direct contact with both the fabric and the chasis of the machine (grounded). I can't notice static in the fabric as it leaves this series of puller rolls.</li>
<li>There's static buildup no matter if the foam is in contact with the shaft or not (sometimes it moves sideways, leaving the cardboard core and "touching" the shaft).</li>
<li>I suspect some slipping between each layer of foam while the roll is formed (due to the shaft rotating slightly faster than the foam is fed), but I can't confirm (the mechanism is synchronized by a PLC).</li>
</ul>
<p>My hypothesis is that the static buildup is being caused by friction of the foam with the surrounding air, at a rate that is faster than its ability to release it through the shaft (its only contact with ground).</p>
<p><a href="https://i.stack.imgur.com/EvgFN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EvgFN.png" alt="PE foam roll"></a></p>
<p><strong>Questions</strong>:</p>
<ol>
<li><p>Shall I discard the idea that the static buildup is being caused by friction between layers of the foam in the roll (as they're the same material, theoretically they're not apart in the triboelectric series)?</p></li>
<li><p>Other than spraying the area (or even the foam) with any watery solution to increase the conductivity of the air, what would you suggest to keep the sparks from jolting and causing fires?</p></li>
</ol>
<p>2A. If a watery solution spray is the best solution there is, what makes a good mixture?</p>
<p>2B. Would a metallic side plate in contact with the shaft and the side of the roll be a good, safe solution? If so, what material would you use? </p>
<ol start="3">
<li>How could I measure the static electricity built up into the roll to ascertain a "safe limit"?</li>
</ol>
| |electrical|fire|polyethylene| | <p>Since asking the question I went to study the subject a little bit more. I want to share my conclusions, hoping that they could be of help to someone else.</p>
<p>Static buildup can't be caused by friction (or contact) between layers of the same material. They don't make a triboelectric pair. Most likely, the static is being generated by the foam slipping in the surrounding air.</p>
<p>This brings us to the second point: try to humidify the air. That could be accomplished by spraying the area, but more technically by installing humidifiers.</p>
<p>In addition to installing the side plate you mentioned (a good conductor, such as copper is the obvious pick, as it doesn't require any mechanical properties), it's a good idea to ensure that the chassis of the rolling machine is properly grounded (as well as those structures in which the material slips).</p>
| 32956 | Static electricity in PE foam rolling |
2020-01-29T01:29:20.173 | <p><a href="https://www.thespruce.com/wonderboard-backer-board-1822682" rel="nofollow noreferrer">WonderBoard</a> is a cement-based backer board mainly used for tiling in wet places. Specifically, its components are Portland cement, expanded polystyrene (EPS) beads, aggregates, and fiberglass reinforcement.</p>
<p>It looks like WonderBoard is also a trademark/</p>
<p><a href="https://i.stack.imgur.com/RkJnm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RkJnm.jpg" alt="enter image description here"></a></p>
<p>Is there a way to find who holds the trademark?</p>
| |home-improvement| | <p>The (r) symbol indicates this is a registered trademark, and as such has registered with the US Patent and Trademark Office. The USPTO maintains a public searchable <a href="http://tmsearch.uspto.gov/bin/gate.exe?f=tess&state=4807:luw4a3.1.1" rel="nofollow noreferrer">database</a> of all registered marks. A little trial and error reveals the pertinent search term here, is "Wonder Board" including the space. This <a href="http://tmsearch.uspto.gov/bin/showfield?f=doc&state=4807:luw4a3.2.2" rel="nofollow noreferrer">mark</a> is registered to Modulars, Inc. This same corporation appears to hold the mark "wonder-board" as well, dating from the 70s and 80s.</p>
| 32959 | Is there a way to find who holds the trademark? |
2020-01-29T15:46:23.433 | <p>Consider we have the third order plant described by the equation:</p>
<p><span class="math-container">$$ y = G(s)u $$</span></p>
<p>where <span class="math-container">$\ G(s) = \frac{b_{2}s^{2}+b_{1}s+b_{0}}{s^{3}+a_2s^2+a_1s+a_0} $</span></p>
<p>If we assume that all parameters <span class="math-container">$\ a_0,a_1,a_2,b_0,b_1,b_2 $</span> are unknown then we can develos a linear parametric model of the form:</p>
<p><span class="math-container">$$ z = θ^{*Τ}φ $$</span></p>
<p>where</p>
<p><span class="math-container">$\ θ^*=[a_2\ a_1\ a_0\ b_2\ b_1\ b_0]^T $</span>, <span class="math-container">$\ φ = [-\frac{s^2}{Λ(s)}y \ \ \ -\frac{s}{Λ(s)}y \ \ \ -\frac{1}{Λ(s)}y \ \ \ -\frac{s^2}{Λ(s)}u \ \ \ -\frac{s}{Λ(s)}u \ \ \ -\frac{1}{Λ(s)}u]^T $</span> ,</p>
<p><span class="math-container">$\ z = \frac{s^3}{Λ(s)}y $</span></p>
<p>Now, let's suppose that the parameters <span class="math-container">$\ a_2,a_1,a_0 $</span> are known and their values are:</p>
<p><span class="math-container">$\ a_2 = 3 $</span></p>
<p><span class="math-container">$\ a_1 = 1 $</span></p>
<p><span class="math-container">$\ a_0 = 2 $</span></p>
<p>How can we develop a linear parametric model of the plant considering these parameters known and as a result the vector <span class="math-container">$\ θ $</span> should be <span class="math-container">$\ θ^* = [b_2 \ \ \ b_1 \ \ \ b_0]^T $</span> ? </p>
| |mechanical-engineering|control-engineering|control-theory|transfer-function| | <p>As <code>fibonatic</code> commented the proper linear parameterization in terms of the unknown <span class="math-container">$\ b_2,b_1,b_0 $</span> parameters is as follows: </p>
<p><span class="math-container">$\ y = G(s)u $</span> is in terms of a differential equation: </p>
<p><span class="math-container">$\ \dddot{y} + a_2\ddot{y}+a_1\dot{y}+a_0y = b_2\ddot{u}+b_1\dot{u}+b_0u $</span></p>
<p>Due to the fact that we only have measurements of the input <span class="math-container">$\ u $</span> and output <span class="math-container">$\ y $</span> of the system, we can't use their derivatives. As a result, we filter each term with a third order stable filter (poles of filter need to be negative) <span class="math-container">$\ Λ(s) = s^3 + λ_1s^2 + λ_2s + λ_3 $</span> where the coefficients <span class="math-container">$\ λ_1,λ_2,λ_3 $</span> are chosen though the poles of the filter. Filtering the above differential equation results in no use of differentiators and produces the following equation: </p>
<p><span class="math-container">$\ \frac{s^3}{Λ(s)}y \ + a_2\frac{s^2}{Λ(s)}y \ + a_1\frac{s}{Λ(s)}y + a_0\frac{1}{Λ(s)}y \ = b_2\frac{s^2}{Λ(s)}u \ +b_1\frac{s}{Λ(s)}u \ + b_0\frac{1}{Λ(s)}u $</span></p>
<p>Let's define <span class="math-container">$\ z = \frac{s^3}{Λ(s)}y \ + a_2\frac{s^2}{Λ(s)}y \ + a_1\frac{s}{Λ(s)}y + a_0\frac{1}{Λ(s)}y \ \ $</span> since <span class="math-container">$\ Λ(s),a_2,a_1,a_0,y $</span> are known. Our equation now becomes:</p>
<p><span class="math-container">$\ z = [b_2 \ \ b_1 \ \ b_0] [\frac{s^2}{Λ(s)}u \ \ \frac{s}{Λ(s)}u \ \ \frac{1}{Λ(s)}u]^T $</span> which is in the form: <span class="math-container">$\ z = θ^{*Τ}φ $</span> with <span class="math-container">$\ θ^{*} = [b_2 \ \ b_1 \ \ b_0]^T $</span> and <span class="math-container">$\ φ = [\frac{s^2}{Λ(s)}u \ \ \frac{s}{Λ(s)}u \ \ \frac{1}{Λ(s)}u]^T $</span>. So, now we have the linear parametric model in terms of the unknown parameters <span class="math-container">$\ b_2, b_1,b_0 $</span>. </p>
<p>Correspondingly we can come up with the linear parametric model in terms of the parameters <span class="math-container">$\ a_2,a_1,a_0 $</span> being unknown and the parameters <span class="math-container">$\ b_2,b_1,b_0 $</span> being known. Following the same procedure, the linear parametric model is: </p>
<p><span class="math-container">$\ z = θ^{*Τ}φ $</span> where <span class="math-container">$\ z = \frac{s^3}{Λ(s)}y \ - b_2\frac{s^2}{Λ(s)}u \ - b_1\frac{s}{Λ(s)}u - b_0\frac{1}{Λ(s)}u \ \ $</span> since <span class="math-container">$\ Λ(s),b_2,b_1,b_0,y $</span> are known, <span class="math-container">$\ θ^{*} = [a_2 \ \ a_1 \ \ a_0]^T $</span> and <span class="math-container">$\ φ = [-\frac{s^2}{Λ(s)}y \ \ -\frac{s}{Λ(s)}y \ \ -\frac{1}{Λ(s)}y]^T $</span>. So, now we have the linear parametric model in terms of the unknown parameters <span class="math-container">$\ a_2, a_1,a_0 $</span>.</p>
| 32965 | Develop linear parametric model of a third order plant |
2020-01-29T16:01:34.237 | <p>I have a network of stainless steel pipes that are distributing water throughout a building. </p>
<p>I know the flow rate of water entering the system is 600 IGPM, but am unsure of how this water will be distributed. I have attached a simplified schematic of what the pipe network looks like. <a href="https://i.stack.imgur.com/jB4P0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jB4P0.png" alt="enter image description here"></a></p>
<p>Is it safe to assume the flow rate throughout this network will be the same, even though the diameters change, based on the continuity equation?</p>
<pre><code>A1*V1 = A2*V2
</code></pre>
| |mechanical-engineering|hydraulics|fluid| | <p>The flow rate through the network will most definitely not change only with the area of the pipes.</p>
<p>As water flows through a system, its pressure (often referred to as head) is reduced by friction along the way and changes in elevation.</p>
<p>Pipe friction is a function of pipe diameter, pipe length, fluid characteristics, and pipe wall smoothness. Every bend in a pipe also introduces additional head loss. The head loss due to this directly affects the flow rate through the pipe.</p>
<p>But the key thing that will affect flow through the network is the usage by each end user. I doubt that these pipes just empty into rooms. So the flows in each branch pipe will be determined by the outlet of the pipes. In a non-industrial use we can probably assume that water requirements fluctuate greatly. For industrial use we can determine flows by direct measurement (buckets, etc), or flow measurement devices in the pipes. If we had static pressures throughout the system we might be able to back in to flows, given that we know the total. I feel like this might have to be an iterative process.</p>
| 32967 | Distribution of water through a pipe network |
2020-01-30T13:26:05.187 | <p>I've heard this term used a lot in literature but I don't understand what the technique is they are referring to. Please help me out here. </p>
| |mechanical-engineering|manufacturing-engineering|casting|production-technology| | <p>According to p.33 of <a href="http://www.iitg.ac.in/engfac/ganu/public_html/Metal%20casting%20processes_1.pdf" rel="nofollow noreferrer">this presentation</a> 'bedding in' is a process of packing the molding sand by ramming the sand around and under the pattern until the sand is tightly packed and even with the parting line. There is also a glossary with foundry and casting terms defined <a href="http://www.atlasfdry.com/glossaryb.htm" rel="nofollow noreferrer">here</a>. This is used when the parts to be cast are quite large, often as a step in pit molding or floor molding. A description of pit molding is found at <em><a href="https://www.coursehero.com/file/p1nqcj7/1219-CLASSIFICATION-OF-MOLDING-PROCESSES-Molding-processes-can-be-classified-in/" rel="nofollow noreferrer">coursehero.com</a></em>: </p>
<blockquote>
<p>12.20.3 Pit Molding Usually large castings are made in pits instead of drag flasks because of their huge size. In pit molding, the sand under the pattern is rammed by bedding-in process. The walls and the bottom of the pit are usually reinforced with concrete and a layer of coke is laid on the bottom of the pit to enable easy escape of gas. The coke bed is connected to atmosphere through vent pipes which provide an outlet to the gases. One box is generally required to complete the mold, runner, sprue, pouring basin and gates are cut in it.</p>
</blockquote>
<p>Floor bedding is discussed in <em>MANUFACTURING PROCESSES</em> By J. P. KAUSHISH. An excerpt from Google Books (also found <a href="https://books.google.com/books?id=FDIfTrE3BjUC&pg=PA205&lpg=PA205&dq=bedding%20in%20pit%20molding&source=bl&ots=dx6k6G1tcL&sig=ACfU3U1_uNOYHfTENFOufxmGjQORg9ZhJw&hl=en&ppis=_e&sa=X&ved=2ahUKEwim3JvNvqznAhVulHIEHXyACyUQ6AEwFHoECAsQAQ#v=onepage&q=bedding%20in%20pit%20molding&f=false" rel="nofollow noreferrer">here</a>)</p>
<p><a href="https://i.stack.imgur.com/nv1UA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nv1UA.png" alt="enter image description here"></a></p>
| 32980 | What does "bedding-in" mean in casting technology? |
2020-01-30T16:00:11.560 | <p>I have a doubt concerning the closed loop frequency responce of a control system. </p>
<p>Suppose I have the following situation:</p>
<p><a href="https://i.stack.imgur.com/zu0iP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zu0iP.png" alt="enter image description here"></a></p>
<p>in which is plotted the frequency response of a system with different badwidths. In this case I know that in generale the performances are good as the bandwidth of the closed loop increases and I know that the price to pay to increase the bandwidth is a high control effort, but is there some other drawbacks in increasing the bandwidth of a system? </p>
<p>I was thinking abouth the noises, which are at high frequencies, this if the bandwidth is at high frequencies, and the system does not roll off fast, the noises enter into the system. </p>
<p>But I am not sure about what I said.</p>
<p>I know that as the badwidth increases, also the reference tracking gets better, but I also see that the roll off is not smooth, I so I was interpreting also this as a fact that there are worse performances with respect to the noise rejection.</p>
<p>Also, I know that there are some limitation of increasing the bandwidth of a system. So, considering a stable and minimum phase system, I know that the limit in bandwidth ig given by the hardware. </p>
<p>But if I increase the bandwidth is there some other limitation? By doing some experimentaition I have see that the disturbance rejection gets worse after a certain frequency, is it due to noise?</p>
<p>Can somebody please help me?</p>
| |electrical-engineering|control-engineering|control-theory|systems-engineering| | <p>The three main drawbacks for a high bandwidth are:</p>
<ol>
<li><p><strong>Noise amplification</strong>.
The higher the bandwidth, the more you are going to act on high frequent stuff in your error signal.
Noise typically dominates at the higher frequency and therefore the system will start reacting to it.
Therefore, the noise will not be rejected / ignored but amplified.</p></li>
<li><p><strong>Robustness</strong>.
In general, a high closed loop bandwidth will reduce the robustness margins (e.g. phase margin, modulus margin and gain margin).
If you have tight margins, you can increase them with some advanced filtering (e.g. notch filter).
However, the accuracy of how 'good' you can tune these filters depends on the model accuracy.
Thus, if there is a mismatch between your model and the setup, or the setup changes over time, this can have a negative effect on your robustness margins and can even cause instability.</p></li>
<li><p><strong>Large control signals</strong>.
In general, a high bandwidth requires a high gain.
When the error is relatively large, this results in large control signals.
Then the input signal may be saturated, resulting in nonlinear plant behavior, which may be undesired.</p></li>
</ol>
| 32982 | What are the drawbacks of inceasing the bandwidth of the system? |
2020-01-31T05:59:43.840 | <p>I ask this because I want to get into robotics(mainly the building of it), so I wanted to go to college to learn mechanical engineering but now that I have found out so many different types of engineering(other than just mechanical and electrical, which are what I thought where the only engineering degrees) I don't know for sure which one would be best for me.</p>
| |robotics| | <p>Mechatronics is the relatively new engineering discipline that combines mechanical and electrical engineering, with robotics right at the forefront.</p>
<p><a href="https://en.wikipedia.org/wiki/Mechatronics" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Mechatronics</a></p>
| 32990 | Would mechanical engineering work robotics and if not what field does? |
2020-01-31T21:16:12.027 | <p>I am an engineering student. I want to get into microscopy research and bio-science. So I want to buy a fair enough compound microscope. I searched for them and there are so many variants from different manufacturers. So, I couldn't select one.</p>
<p>My budget is $1000. I know that sounds way too overkill for entry level, but just say that this is total amount I have to spend for a good microscope. I am currently in India and associated price for variants of major manufacturer is not mentioned here so if anyone can mention price(as in India), would be appreciated.</p>
<p>Which manufacturer and which variant should you recommend specifically?</p>
| |optics| | <p>Olympus scopes are well-known for their excellent optics. Used examples are plentiful and modest in cost. They are easily customized with things like binocular eyepieces, mechanized stages, and camera output ports and both their eyepieces and objective lenses are available in a variety of magnification powers. They are also available as <em>metallographic</em> or <em>confocal</em> microscopes which operate on reflected light instead of transmitted light; these saw wide use as wafer inspection microscopes in wafer fabs in the US.</p>
| 33008 | Compound Microscope recommendation |
2020-01-31T22:08:41.157 | <p>I'm trying to help a student with a project. The task is to generate CNC G-Code from a Solidworks hole table export (CSV) using VBA.</p>
<p>The problem is the export format is going to be a difficult to parse. Samples include the following:</p>
<pre><code>TAG,X LOC,Y LOC,SIZE
A1,15,50,<MOD-DIAM> 10 <HOLE-DEPTH> 10
B1,30,80,<MOD-DIAM> 8 THRU ALL
M10 - 6H THRU ALL
C2,134.50,18,<MOD-DIAM> 8.50 THRU ALL
<HOLE-SPOT><MOD-DIAM> 23.79 <HOLE-DEPTH> 7.50
D2,75,24.50,<MOD-DIAM> 13.50 THRU ALL
</code></pre>
<p>Each tool size seems to be assigned a letter and a sequence number. That's fair enough and leads to the hope that we could sort by first letter. The problem is that the metric tap sizes follow the same format. I'm sure there are other problems the more we dig.</p>
<p>Are we missing a trick here on the export or has anyone any ideas? I can't find any Solidworks documentation on the matter.</p>
<hr>
<p><strong>Solution</strong></p>
<p>Based on Jim S's answer below it became clear that complex holes (tapped and counterbore, for example) are split by CR-CR-LF. This is easy enough to filter out in VBA.</p>
<p><a href="https://i.stack.imgur.com/jNcT5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jNcT5.png" alt="enter image description here"></a></p>
<p><em>Figure 1. A screen grab of the code as displayed by Notepad++ with View | Show Symbol | Show end of line turned on. These three lines should all be one.</em></p>
<p><em>Code snippet 1</em></p>
<pre><code>Private Sub CommandButton1_Click()
'From http://codevba.com/office/read_text_file_into_string_variable.htm#.XkHkHM77R1I
Dim strFilename As String: strFilename = Application.ActiveWorkbook.Path & "\HoleTablecsv"
Dim strFileContent As String
Dim iFile As Integer: iFile = FreeFile
Open strFilename For Input As #iFile
strFileContent = Input(LOF(iFile), iFile)
Close #iFile
a = Replace(strFileContent, vbCr & vbCrLf, " ")
Debug.Print a
End Sub
</code></pre>
| |solidworks|cnc| | <p>The problem is that the CSV export is including the new line characters that are part of the hole callout. So this:</p>
<pre><code>A1,2.07,9.78,<MOD-DIAM> .257 THRU
5/16-18 UNC THRU
</code></pre>
<p>should really look like this:</p>
<pre><code>A1,2.07,9.78,<MOD-DIAM> .257 THRU 5/16-18 UNC THRU
</code></pre>
<p>You can perform a string replace of any pair or new line characters with some delimiter of your own (say, {new-line}) to make parsing easier, and you can get data like this:</p>
<pre><code>TAG,X LOC,Y LOC,SIZE
A1,2.07,9.78,<MOD-DIAM> .257 THRU{new-line}5/16-18 UNC THRU
A2,2.53,6.28,<MOD-DIAM> .257 THRU{new-line}5/16-18 UNC THRU
B1,4.72,5.03,<MOD-DIAM> .197 THRU{new-line}M6X1.0 - 6H THRU
B2,5.88,8.49,<MOD-DIAM> .197 THRU{new-line}M6X1.0 - 6H THRU
B3,7.74,6.57,<MOD-DIAM> .197 THRU{new-line}M6X1.0 - 6H THRU
C1,6.05,3.38,<MOD-DIAM> .266 THRU{new-line}<HOLE-SINK><MOD-DIAM> .531 X 82°
C2,7.09,9.74,<MOD-DIAM> .266 THRU{new-line}<HOLE-SINK><MOD-DIAM> .531 X 82°
C3,7.69,2.61,<MOD-DIAM> .266 THRU{new-line}<HOLE-SINK><MOD-DIAM> .531 X 82°
C4,9.63,8.89,<MOD-DIAM> .266 THRU{new-line}<HOLE-SINK><MOD-DIAM> .531 X 82°
</code></pre>
| 33011 | Solidworks hole table export data format |
2020-02-01T05:16:49.160 | <p>The other day, some restauraunts in the union were closed "due to steam outage." I had seen the venting of course, but had thought little about why the school has a steam system. As best I can find, steam is usually used for heat distribution from a central heat plant; more efficient, etc. But the average yearly low temperature here is 57 degrees. What purpose does the system serve?</p>
| |civil-engineering|heating-systems| | <p>Steam networks are used for heating (there are still heaters in use on the gulf coast), creating hot water, and any other energy loads a university might need. I suspect that the restaurants in question didn't have hot water.</p>
| 33017 | Why does a Gulf Coast campus need an active steam network? |
2020-02-01T19:45:07.737 | <p>Good afternoon, I have recently ordered a normally closed relay to fix on my car's daylight-running lights. When the main lights come on the DLR should switch-off. Unfortunately amazon has delivered instead of a normally closed 2 normally open relays (!!?). Is there a clever way to emulate the normally closed relays functionality with two normally open relays?
Thanks in advance.</p>
<p>regards
perry</p>
| |electrical-engineering|power-electronics|car|electronics| | <p>You have a chance. This might work as a temporary solution.</p>
<p><a href="https://i.stack.imgur.com/A1oUW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A1oUW.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Switching DLRs off with an NO when headlights are on.</em></p>
<p>How it works:</p>
<ul>
<li>When the ignition is on and the headlights are off the headlamps will provide a low enough resistance to allow the RLY to pick up. The normally open (NC) contact will close and the DLRs will turn on.</li>
<li>When the headlamps are on there will be 12 V on both sides of the relay so no current will flow and the relay will drop out switching off the DLRs.</li>
</ul>
<p><a href="https://i.stack.imgur.com/uDuC9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uDuC9.png" alt="enter image description here"></a></p>
<p>*Figure 2. Modification to detect both dip and high beam status.</p>
<ul>
<li>The DLRs will turn on only if both DIP and HIGH are off.</li>
</ul>
<p>Note that in both circuits I've powered the relay from a line switched by the ignition. This will avoid draining the battery when the ignition is turned off.</p>
| 33023 | Emulate normally closed relay with 2 normally open relays |
2020-02-02T05:37:40.827 | <p>Could a Stirling Engine replace the cooling system of an engine?</p>
<p>Instead of having a clutch and fan attached to the main drive shaft to cool an engine, could a Stirling Engine and clutch share that drive shaft to cool and add torque? What problems may arise?</p>
| |mechanical-engineering|fluid-mechanics|thermodynamics|design|heat-transfer| | <p>Hypothetically yes. However the existing engine must run at a peak torque/rpm to run the engine to act as a heat pump.
Much like turbochargers, some companies are installing a small battery to run the turbos even when the engine is idle or off (start-stop systems), so when the engine reactivates its turbo lag is eliminated.</p>
<p>MSI developed a stirling cooling system for CPU gaming rigs <a href="https://i.stack.imgur.com/SWKo6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SWKo6.jpg" alt="enter image description here" /></a></p>
<p>But the thermal threshold and temperature is what makes and ideal engine run.</p>
<p>GM's Mild Hybrid system uses a oversized startup motor to run the engine when it's off or in low power mode, saving fuel to reduce winding startup. Stirling engine as a heat pump can be run on an additional battery<a href="https://i.stack.imgur.com/ATs88.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ATs88.jpg" alt="enter image description here" /></a></p>
| 33026 | Stirling Engine Cooled Engine |
2020-02-02T23:17:47.333 | <p>I am building a desoldering gun. While the pump I am using provides decent suction, I would like a quick powerful vacuum burst to prevent the solder from solidifying in the nozzle rather than in the chamber.</p>
<p>I figure there must be the equivalent of a capacitor where a vacuum is held in reserve and released when needed. Does such a device exist? If so what is it called. Finally, are they available in small sizes, say the size of a tennis ball?</p>
| |airflow|vacuum|pneumatic| | <p>The device is called a "receiver" in pneumatics - this is basically the tank you see on any garage compressor, or much larger and more elaborate setups in industrial settings. Its function can be modelled almost exactly like a capacitor. It stores energy by accumulating air inside of it (the integral of the incoming flow, just as charge is the integral of current), and will maintain a pressure (analogous to voltage) in proportion to its volumetric capacity after the external pressure source is removed. It even acts as a low-pass filter, essentially smoothing out the inconsistent input from the compressor, in fact this is the reason why garage compressors almost always have a tank attached. </p>
<p>In a lot of cases in the industrial world, receivers are seen at the point of usage as well, for example a valve that must close quickly, a pneumatic conveying line, or an air cannon. In those cases it's analogous to a capacitor used to store energy for a very high current but short discharge, such as a camera flash.</p>
| 33039 | What is the equivalent of a capacitor in pneumatics? |
2020-02-03T02:12:02.213 | <p>an <a href="https://engineering.stackexchange.com/q/33041/24455">acoustic anechoic chamber</a> has mainly 3 features</p>
<p>stopping sound getting in;
stopping sound getting out;
reduce echoey.</p>
<p>Is there a term to describe a room only used to stop sound getting in?</p>
| |terminology| | <p>Acoustic test booth </p>
<blockquote>
<p>However, acoustic tests require special spaces. They should be conducted in a place where the sounds can be isolated and any unwanted noises are kept outside. This is where acoustic test booths come in.</p>
</blockquote>
<p><a href="https://www.enoisecontrol.com/acoustic-test-booth-chamber/" rel="nofollow noreferrer">https://www.enoisecontrol.com/acoustic-test-booth-chamber/</a></p>
<p>Specifically, a hemi-anechoic chamber has some sound-reflective inside surfaces, typically the floor, in order to simulate operating conditions.</p>
| 33044 | Is there a term to describe a room only used to stop sound getting in? |
2020-02-03T13:54:23.843 | <p>I need a source of air pressure for an air bearing (3D printed cylinder that expands air upwards like an air hockey table). I would like to be able to control the flow of air. I don't have a lab air supply or advanced equipment, this will basically be a homemade build.</p>
<p>An off the shelf electric pump might be a good supply source. What do I need to regulate the flow?</p>
| |airflow|pumps|compressed-air| | <p>As was mentioned in the comments, you can get off-shelf air compressors, which are quite good at regulating pressure. Now if you want to regulate the flow rate, that becomes a bit more tricky.</p>
<p>Your first (cheap ~ £10) option is to go for something like this (see <a href="https://www.smc.eu/en-eu/products/as-standard-w-built-in-one-touch-fitting~134135~nav" rel="nofollow noreferrer">https://www.smc.eu/en-eu/products/as-standard-w-built-in-one-touch-fitting~134135~nav</a> for details):</p>
<p><a href="https://i.stack.imgur.com/brIl7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/brIl7.png" alt="enter image description here"></a></p>
<p>This will allow to set your flow rate to a constant value <strong>for a given pressure</strong>, but you'll have to measure the flow rate in some other way. Also, if the pressure changes, your flow rate will also change, even if the setting on the speed controller hasn't changed.</p>
<p>Your second and more expensive option, is to go for a mass flow controller, such as this one (see <a href="https://www.bronkhorst.co.uk/en/products/gas_flow_meters_and_controllers/elflow_select/" rel="nofollow noreferrer">https://www.bronkhorst.co.uk/en/products/gas_flow_meters_and_controllers/elflow_select/</a> for more details):</p>
<p><a href="https://i.stack.imgur.com/StnjH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/StnjH.png" alt="enter image description here"></a></p>
<p>Now this will allow you to measure and control the flow rate to whatever value you desire and maintain that flow rate regardless of inlet pressure variations. This is controlled by a software running on your PC and communicating with the instrument via a serial cable (you'll need a serial to USB converter). Now this will set you off ~£1,200 so not cheap, but it's pretty fool-proof.</p>
| 33051 | How can I build a regulated pressurised air supply? |
2020-02-04T12:17:59.833 | <p>I have a very long beam which is subjected to a torque (I have the moments and shear forces with respect to X along the beam). I can calculate the critical load when the beam would bend if it was loaded axially, however I don't know how to calculate the critical torque on the beam before it "buckles" by twisting. How would I go about calculating this critical load/torque?</p>
| |stresses|beam|buckling| | <p>There is no buckling under torsion in a beam, there is lateral web buckling under vertical loads under certain circumstances.</p>
<p>A beam under pure torque will gradually twist due to St Venant and warping torsion until it yields with no sudden loss of strength as in buckling under axial load, however, under combined axial stress and torque it will buckle and needs to be analyzed. </p>
<p>this is the section torque moment under pure torsion.</p>
<p><span class="math-container">$$T=G*I_t \phi^ \prime -EI\phi^ { \prime \prime \prime }$$</span></p>
<p>where</p>
<p>T is the torsional moment at the cross-section.</p>
<p><span class="math-container">$\phi^ \prime \ and\ \phi^ { \prime \prime \prime } $</span> are first and third derivateves WRT X.</p>
<p><span class="math-container">$I_w $</span> is the warping constant</p>
<p><span class="math-container">$ I_t $</span> is ST Venant's torsional constant.</p>
| 33061 | Twist "buckling" critical load |
2020-02-04T16:13:24.507 | <p><a href="https://royalsocietypublishing.org/doi/10.1098/rspa.2013.0689" rel="nofollow noreferrer">https://royalsocietypublishing.org/doi/10.1098/rspa.2013.0689</a></p>
<p>A chain of some sort goes upwards, turns in a half circle (according to the drawn diagram), and goes back down.
<a href="https://i.stack.imgur.com/aSDuW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aSDuW.png" alt="enter image description here"></a></p>
<p>Focusing on the half circle, the equation given by the paper to account for its centripetal force is
<a href="https://i.stack.imgur.com/sdtLr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sdtLr.png" alt="enter image description here"></a></p>
<p>"In our initial analysis, we assume the curved region at the apex of the fountain is small enough, and has tight enough curvature, that the centripetal acceleration is much larger than the gravitational acceleration (v2/r≫g). The centripetal acceleration is then provided by the tension in the chain. If the (local) radius of curvature is r and the tension in the curved region is Tc" Also the lambda is mass per unit length</p>
<p>Why is Tc divided by r in the left half of the equation? If it was force per unit length then I would expect Tc to be divided by (pi)r instead of just r.</p>
| |mechanical-engineering|applied-mechanics|mechanical|pulleys| | <p>Consider a small element of chain with length <span class="math-container">$dl$</span> at the top of the arc.</p>
<p>Its mass is <span class="math-container">$\lambda\,dl$</span>.</p>
<p>It is curved through an angle <span class="math-container">$dl/2r$</span> on each side of the vertical.</p>
<p>So the downward force on it is <span class="math-container">$2T_c\sin(dl/2r) \approx T_c\,dl/r$</span> (the first 2 is because there are two free ends of the element)</p>
<p>The acceleration downwards is <span class="math-container">$v^2/r$</span> (ignoring gravity).</p>
<p>Newton's second law gives the equation in the paper.</p>
| 33064 | calculation of tension in curved chain for chain fountain (Mould effect) |
2020-02-05T12:34:27.193 | <p>Will epoxy resin dissolve a silicone spatula, cake icing spatula? Will it work well enough to get the job done without contaminating the resin?</p>
| |composite-resin| | <p>Silicone molds are often used for epoxy resin casting, so I don't see any problems.</p>
| 33082 | will silicone spatula be damaged by epoxy resin? |
2020-02-06T04:20:56.150 | <p>When I happened to see pictures of a highway in the desert, the question arose in my mind. The highway will probably be buried if a sandstorm rages. So is there any solution to this problem?
<a href="https://www.sciencenews.org/blog/wild-things/algal-blooms-created-ancient-whale-graveyard" rel="nofollow noreferrer">https://www.sciencenews.org/blog/wild-things/algal-blooms-created-ancient-whale-graveyard</a></p>
<p><img src="https://www.sciencenews.org/wp-content/uploads/2014/02/wt_LaFamilia_South_CerroBallena_free.jpg" alt="enter image description here"></p>
| |civil-engineering| | <p>The usual solution is to build a fence with lots of slats in it about 20 feet away from the edge of the road. The fence slows down the airflow and thereby causes the sand in it to fall out and accumulate right behind the fence, thereby keeping most of it off the road. </p>
<p>Since most of the sand carried by the wind scoots along close the surface of the ground, the fence does not have to be more than 4 or 5 feet tall to catch most of the sand. </p>
<p>Once the fence is about half-buried in sand, its effectiveness diminishes and the sand around it has to be plowed away. </p>
| 33101 | How is a highway prevented from being buried by sand in the desert? |
2020-02-06T18:25:38.863 | <p>I was wondering why motorcycle brakes have so much more stopping power than bicycle hydraulic brakes (Than can be found on mountain bikes for example).</p>
<p>The rotor has a bigger size which gives better leverage for the brake pads and thus more stopping power. </p>
<p><strong>But I don't understand the following:</strong></p>
<p>If the moment (force times lever length) exerted on the brake lever (by the rider's hand) is the same (is this a correct assumption) and the total amount of lever travel is the same... the amount of clamping force the hydraulic brake's caliper's pistons exert on the pads must be equal (I think)(since the same amount of energy is used at the lever.</p>
<p>Or is the free travel (the distance the pads have to travel before they hit the disc) much smaller on a motorcycle disc brake resulting in a shorter (free throw) of the lever which means there is more percentage of the lever travel that can be used to 'build pressure' meaning a smaller piston can be used which would give more clamping force at the pads?</p>
<p>The brake pads are much larger on motorcycles but then again because of the increased surface area one would have to exert more force on the bigger brake pads to get the same pressure/area. </p>
<p><strong>Are my conclusions incorrect?</strong></p>
<ul>
<li>Is the pad material a lot softer on the motorcycle brake pads?</li>
<li>What design changes would enable higher clamping force with the same lever throw/force exerted on lever (if any design changes could accomplish this)?</li>
</ul>
<p>Another example (light motorcycle vs heavy motorcycle):
my motorcycle (Suzuki gn250, 140 ish kg wet weight) has a hydraulic front (disc) brake. A Honda Goldwing (very heavy motorcycle , 380 kg wet weight) also has a hydraulic front (disc) brake. <strong>What would be the main differences that enable the Goldwing's front brake to have so much more power?</strong></p>
<p>Another question:
The Goldwing has 2 discs/2 calipers in the front. There is more pad material indeed but the same thought (more surface area of pads means more force needs to be exerted in total on the brake pads in order to get the same force/area on the contact patch between pads and disc surface).
<strong>Is even softer material used on the brake pads for 2 rotor front brake systems?</strong>
Or am I missing something?</p>
<p>Some disc brakes have 2 piston calipers, some have 4 piston calipers. It is suggested that the 4 piston caliper variants have more stopping power..
Is it true that the 4 piston calipers have more stopping power in general? If so, what is the reason behind this?</p>
<p>Thank you!</p>
| |hydraulics|mechanical|friction| | <p>Swept area INCLUDES all material of the rotor, not just the pad surface area. The entire rotor itself acts like a heat sink.</p>
| 33114 | Hydraulic Brakes - Why motorcycle brakes more powerful than bicycle brakes? |
2020-02-06T19:21:42.787 | <p>I've got a flashlight with a 3.7 V 18650 cell that charges with a 3.5 mm barrel plug. They provided adapter that says 4.2 V +/- 0.5. Can I use a 5 V USB to barrel plug adapter to charge it?</p>
<p>UPDATE
So the adapter clearly says +- .5 v. Is this an obvious misprint. It must be .05 right?</p>
| |pid-control| | <p>I think maybe, definitely NOT do this. Measurement shows the provided adapter/charge controller ranges 4.0 to 4.3 v and current at 200-220 mA.</p>
<p>Plugging into an un charge controlled 5 v 1A power supply immediately drives the current to 1.1 A.</p>
<p>Certainly leading to overheating, possibly venting and leaking, venting leaking and lithium fire, rapid venting, explosion, and then fire. Pick your poison.</p>
<p>UPDATE</p>
<p>Although the adapter is rated at 500 mA; when connected to a 1 A USB load tester it supplies only 250ma.</p>
<p>So, someone tell me, where is the charge controller, in the flashlight or in the wal wart?</p>
<p>FINAL HOPEFUL UPDATE</p>
<p>In a disturbing development after charging at 4 v for 180 mAH the voltage across it dropped to 3.0 and it started getting hot, AND the flashlight is only intermittently shining any light at all.</p>
| 33116 | Can I charge 18650 Li-ion cell from 5 V USB? |
2020-02-07T14:34:21.220 | <p>For comparison: in Euler's formula, we calculate the critical force such that when the column is subjected to it, it buckles. As far as I understand, the yield strenght of the material doesn't really matter.</p>
<p>However, when facing eccentric load, secant formula is used and a maximum strain is calculated, instead of a critical force. Does it makes sense then to use yield strength as a maximum admissible strength? </p>
<p>Edit: also, when derivating the formula for the maximum strain, one already considers the compressible strain and the bending combined. Thus, when calculating the maximum strain, I don't need to further make any considerations about their interations, correct?</p>
| |mechanical-engineering|structural-engineering| | <p>We need to check the column for both the maximum stress with the secant formula and critical buckling load if the eccentricity is small and the column is slender and tall. As we know critical buckling load in medium and short columns gives wrong, over the yield strength results, so we always check for P/A< (allowable stress) too.</p>
<p>The secant formula already calculates the maximum stress under the combination of the axial load and moment stress due to the bending of the column.</p>
<p><a href="https://i.stack.imgur.com/7t15W.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7t15W.gif" alt="secant formula"></a></p>
| 33129 | What's the usefulness of maximum stress calculated on Secant's formula for buckling? |
2020-02-07T23:34:00.790 | <p>I am making a project using 10mm aluminum extrusions. I am trying to connect all three extrusions like the picture below shows and I am going to make my own bracket (in red) out of aluminum since the variety of choices is very scarce for this 10mm size, but I will include a picture of something similar so you can get an idea of what I'm making. I am wondering which arrangement is stronger and more rigid- the left one, or right one? Or is there an even better way to do it? <a href="https://i.stack.imgur.com/vdQv3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vdQv3.png" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/yKgvV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yKgvV.jpg" alt="enter image description here"></a></p>
| |structural-engineering|beam|building-design| | <p>I would say the configuration on the right would be the best if you cut the top (vertical) extrusion at a 45 degree angle at the bottom where it meets the "45 degree strut". This way all the struts will be flush mounted together. Then you can add side straps and joining plates have the same geometry as the strut-intersection area. </p>
| 33141 | What is the strongest way to fasten 8020 extrusions in this arrangement? |
2020-02-11T00:38:56.727 | <p>Consider the system:</p>
<p><span class="math-container">$$ y(s)=b\frac{e^{−sτ}}{s+a}u(s) $$</span></p>
<p>where a,b are unknown, constant parameters and <span class="math-container">$\ τ>0 $</span> is an unknown delay constant. How can I obtain a plant parametric model for this system in the following form ? </p>
<p><span class="math-container">$$ z = \theta^{*T}\phi \ + η $$</span></p>
<p>which satisfies the following condition:</p>
<ul>
<li><span class="math-container">$\tau = 0 \Rightarrow η = 0 $</span></li>
</ul>
| |control-engineering|control-theory|modeling|mathematics|transfer-function| | <p>Solution to the linear parameterization of the given system:</p>
<p><span class="math-container">$\ y = \frac{b}{s+a}e^{-sτ}u = \frac{b}{s+a}e^{-sτ}u \ + \ \frac{b}{s+a}u \ - \ \frac{b}{s+a}u = \frac{b}{s+a}u \ + \ \frac{b}{s+a}(e^{-s\tau} \ - \ 1)u $</span> </p>
<p><span class="math-container">$$ sy = -ay+bu+b(e^{-s\tau}-1)u $$</span></p>
<p>We filter both sides of the equation with a stable first order filter </p>
<p><span class="math-container">$$ Λ(s) = s + l_1, \ l_1 >0 $$</span></p>
<p>and the final parametric model is now:</p>
<p><span class="math-container">$\ \frac{s}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u$</span></p>
<p>And the last equation can be written in the form: </p>
<p><span class="math-container">$$ z = \theta^{*T}\phi + η $$</span></p>
<p>where <span class="math-container">$\ z = \frac{s}{Λ(s)}y, \ \theta^* = [a \ b], \ \phi = [-\frac{1}{Λ(s)}y \ \ \frac{1}{Λ(s)}u]^T $</span> and <span class="math-container">$\ η = \frac{b}{Λ(s)}(e^{-sτ}-1)u $</span>. </p>
<p>It is obvious that for <span class="math-container">$\ \tau = 0 \to η=0. $</span> </p>
<p>The new input vector of the parametric system <span class="math-container">$\ z $</span>, can be derived directly from the input of the initial system <span class="math-container">$\ y. $</span></p>
<p><strong>EDIT: Slightly different approach</strong></p>
<p><span class="math-container">$$ \frac{s}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u $$</span></p>
<p>From this point, we can come up with a linear parametric model which contains the initial input vector <span class="math-container">$\ y $</span> following the procedure below:</p>
<p><span class="math-container">$$ Λ(s) = s + l_1 \Rightarrow s = Λ(s) - l_1, \ l_1>0$$</span></p>
<p><span class="math-container">$$ \frac{Λ(s) - l_1}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u \Rightarrow $$</span></p>
<p><span class="math-container">$$ \frac{Λ(s)}{Λ(s)}y - \frac{l_1}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u \Rightarrow $$</span></p>
<p><span class="math-container">$$ y = -(a-l_1)\frac{1}{Λ(s)}y +b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u $$</span></p>
<p>and the last equation can be written in the form:</p>
<p><span class="math-container">$$ y = \theta^{*T}\phi + η $$</span></p>
<p>where <span class="math-container">$\ \theta^* = [(a-l_1) \ \ b], \ \phi = [-\frac{1}{Λ(s)}y \ \ \frac{1}{Λ(s)}u]^T $</span> and <span class="math-container">$\ η = \frac{b}{Λ(s)}(e^{-sτ}-1)u $</span>. </p>
<p>It is obvious that for <span class="math-container">$\ \tau = 0 \Rightarrow η=0. $</span> </p>
| 33176 | How to obtain plant parametric model of given system? |
2020-02-11T16:06:54.137 | <p>I'm writing a story set in NYC and trying to describe certain industrial objects. I'm having trouble with specific technical terms for the items that I've marked w/ arrows. E.g. I don't want to use the generic term "support" or "wall" if there's a more specific word like viaduct, truss, pier, girder, silo, etc. Also specific materials (are they steel or iron, or a specific kind of design,style, type). </p>
<p>After extensive research, I'm still not confident which technical terms apply to these specific things. Hoping a civil or other engineer might be able to shed some light. p.s. for the Smith-9th elevated train, what's this black material and when/why was it put on if anyone knows).</p>
<p>Thanks in advance!</p>
<p>R</p>
<p><a href="https://i.stack.imgur.com/4TfPt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4TfPt.jpg" alt="278W in Brooklyn"></a>
<a href="https://i.stack.imgur.com/soDUy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/soDUy.jpg" alt="278W in Brooklyn"></a>
<a href="https://i.stack.imgur.com/yDuil.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yDuil.jpg" alt="Brooklyn Bridge"></a>
<a href="https://i.stack.imgur.com/Rz6qf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rz6qf.jpg" alt="Smith-9th Street Station, Brooklyn"></a>
<a href="https://i.stack.imgur.com/uuANQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uuANQ.jpg" alt="Smith-9th Street Station, Brooklyn"></a>
<a href="https://i.stack.imgur.com/nakq4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nakq4.jpg" alt="Smith-9th Street Station, Brooklyn (also, what's this black material and when/why was it put on if anyone knows)"></a>
<a href="https://i.stack.imgur.com/rwXSd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rwXSd.jpg" alt="Baysite Fuel Depot"></a>
<a href="https://i.stack.imgur.com/uWoXc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uWoXc.jpg" alt="Oil depot and cement company"></a></p>
| |structural-engineering|civil-engineering|structures| | <p>There are a whole lot of terms in your photos.</p>
<h3>Photo 1:</h3>
<p>That vertical structure is referred to as a <strong>PIER</strong>. Another style of pier which as far as I know is a series of piles and a pier cap is referred more specifically as a <strong>BENT</strong>. Piers perform the job of supporting the bridge between abutments by carrying the vertical (and horizontal loads) to ground. The vertical element of the pier the arrow is pointed at I would classify as a <strong>PIER SHAFT</strong> or <strong>PIER COLUMN</strong>. The material being steel plate. The horizontal piece at the top attached to the leg, is referred to as the <strong>PIER CAP</strong>. It just is to support the girders and transfer their loads to the shafts. The shaft carry the load down to its footings, normally below ground, which are normally spread footings or piles.</p>
<p>Also of note in this photo are the <strong>GIRDERS</strong>. There are 4 of them These are the main longitudinal structural shapes supporting the bridge. Girders are normally either steel, concrete, or wood. They can also come in a variety of shapes, I, box, and trapezoidal being some of the more common ones. In this case you have steel plate I girder. Their job is to carry the vertical load of the deck to its supports (abutment or pier normally). This particular bridge is a bit of an odd ball as it has additional smaller longitudinal members running parallel with the girders. I would call these smaller members <strong>STRINGERS</strong>. They are transferring the vertical load from the deck to the Diaphragms.</p>
<p>Above the pier cap you will see a transverse steel plate member running between girders and it tappers as it extends out past the exterior girder. This member is referred to as <strong>DIAPHRAGM</strong>. The diaphragm performs several tasks. The main being to help redistribute the load between girders so the girders deflect more evenly, or behave as a group. The other job they do is help prevent twisting of the girders. Not that big of an issue on short straight bridges, but can be more of an issue on curved and longer spans. At the abutments and piers, the diaphragms also tend to be a bit beefier. This allows them to be used as jacking points. This allows the bridge to be lifted slightly to allow bearings to be replace and minor repairs if any being performed around the bearings. The diaphragms will also help transfer some of the horizontal loads the bridge deck may experience. Due to the stringers that are being supported by the diaphragms, one could argue that the diaphragms in this case are actually <strong>FLOORBEAMS</strong>. I would personally hesitate to call them that though as that is a term more often used with truss style bridge as opposed to a slab on girder bridge which is what I was assuming was shown in photo 1.</p>
<p>Beneath the girders you will notice some small steel members in an X pattern. This is commonly called <strong>CROSS BRACING</strong> These are normally L shaped sections, but can also be flat plate or rods to name a couple of other options. Their designed normally so one diagonal will be in tension and the other diagonal is in compression. The compression is assumed to overload the capacity of the diagonal and it will deflect/buckle out of shape and therefore not carry significant structural load. However the members will work in tension quite nicely. Since tension resistance is pretty much based on pure cross sectional area, it does not really matter if the member is a rod, L-section, plate or a cable in theory (never seen one as a cable though).</p>
<p>The portion outside of the outer most girder (exterior girder) is referred to as the <strong>CANTILEVER</strong>, bridge deck cantilever, deck cantilever, etc. It gets the cantilever name by the way the forces from it are transferred back to the exterior girder.</p>
<h3>Photo 2:</h3>
<p>The arrow pointing up is pointing at a portion of what appears to me to be a cantilevered bridge deck. If you are referring more to the under side of the deck as opposed to the extension of the deck from the longitudinal support then the term I would use is <strong>SOFFIT</strong>. Soffit means the under side of the deck or ceiling.</p>
<p>The arrow pointed to the left I would simply call a wall. It appears very solid. I am not aware of a special term for it and will add that it is not common for a bridge as it adds significant weight. Bridge design tends to try and keep bridges light.</p>
<p>The arrow pointed to the right I would refer to as a <strong>BARRIER</strong>. There are 3 common forms of barriers. Parapet wall which looks very rectangular and is usually made of solid concrete. It can be topped with a pedestrian handrail if appropriate. Barrier wall which has beveled traffic face with a toe portion. Its similar to the parapet wall except its design is such that the car tire will impact first normally and it helps redirect the car slightly before the rest of significant impact occurs. The third type is an Open railing system. In this case you have an open railing system and more specifically a post and railing system</p>
<h3>Photo 3:</h3>
<p>The left arrow is pointed at the end of a <strong>CROSS BEAM</strong>. This is one element that makes up a truss style bridge. Its job is to transfer horizontal load to the adjacent truss. It also helps improve the torsional stiffness of the bridge so the truss does not twist and collapse over long spans. Truss bridges where you have the tops of the truss connected and the traffic passes through the middle of the truss are referred to as Through Trusses. There is a shorter style of truss bridge where there are no transverse members attached at the top. It has a couple of names I am aware of: Pony Truss and Saddle Truss.</p>
<p>The right arrow is pointed at a member in a truss system call a <strong>VERTICAL</strong>. As pointed out by others, The vertical plan of a truss is made up of 4 main types of elements. Verticals which as the name suggests go up and down. Diagonals that go from top to bottom on some angle. Top chord which is the upper usually horizontal members. Bottom chord which is the lower usually horizontal member. Where the chords connect with with vertical or diagonals is usually referred to as a Node, though that comes up more for structural analysis purposes, and may just use connection in the field. Normally the top chord is in compression and the bottom chord is in tension.</p>
<p>Also of interest in this photo is that the transverse member instead of being an I beam or Channel, or Angle (L-section), is what I would call a built up member. Without taking too close a look I would guess it is a pair of Channels (C-Sections) back to back and stitched together with baton plates. <a href="https://engineering.stackexchange.com/questions/15140/is-there-a-specific-name-for-diagonally-braced-truss-members-like-those-on-the-q">This question</a> has a more detailed description of this member make up.</p>
<p>Also note that small roughly 45 degree short member at the end of the transverse beam that connects to the vertical. This is commonly referred to as a knee brace. Its job is to help stiffen the connection of the transverse member to the vertical. This is what helps with the overall torsional stiffness of the bridge.</p>
<h3>Photo 4:</h3>
<p>This is a <strong>DIAGONAL</strong> as discussed in an earlier photo. If this is the last member of the truss, meaning it's lower end is resting on a pier or an abutment, then you may also refer to this as the end diagonal. It will be the diagonal carry the largest axial load of all the bridge diagonals. As a result the end diagonal will usually be a larger member than interior diagonals.</p>
<h3>Photo 5:</h3>
<p>This is an <strong>END DIAGONAL</strong>. See description of photo 4.</p>
<h3>Photo 6:</h3>
<p>The lower arrow is simply a truss or frame being used as a pier to support the bridge. If the pier trusses were all interconnected, you could refer to this as trestle. Though trestles are more commonly made out of wood and more often than not rail bridge. There are definitely exceptions to this though.</p>
<p>The upper arrow looks like its some sort of wall. I would suspect non-structural and more like a debris fence. Really hard to say though with out getting different angles and closer look.</p>
<p>The black material looks like some sort of wrap used to either catch debris and keep it from falling on people and property below. If if was only on the truss members I might guess it would be netting to prevent birds from nesting and crapping on the bridge. Bird crap can be quite corrosive for steel. Oh I just thought of this, if that structure is slated for demolition, the black covering could be a blast net / mat intended to contain the shrapnel / flying debris caused by the use of explosives in explosive demolition.</p>
<h3>Photo 7:</h3>
<p>The upper arrow seems to be pointed at some red pipe. I would assume this is steel pipe and it is painted red to make it stand out so people know its part of the fire suppression system as opposed to the plumbing for the roof drain. It is not a structural member.</p>
<p>The lower arrow is pointed at a beam. I would probably refer to this as a cross beam. I may also call it a brace as it is connecting the two columns.</p>
<h3>Photo 8:</h3>
<p>The arrow on the left is pointed at what I would call storage tanks. If its oil or gasoline I would refer to them as fuel tanks. notice the wall around the tanks. This is a safety precaution. Those walls make a second tank or bath tub if you will. odd are the volume inside that external wall is close to 1.25-1.5 times the volume of the inner tank. The outer wall should contain the contents of the tank in the event of a failure/leak of the inner tank.</p>
<p>The arrow on the right pointed at the round metal structure I would call those structures hoppers or silos. This is really depended on what is being stored inside them. Dry contents like sand or grain I would definitely use one of these two terms.</p>
<p>The other arrow pointed at the rather square structure I am not sure what I would call it from this angle other than a building or structure.</p>
<h3>Materials</h3>
<p>For the most part, metal used in bridges in north america will all be structural steel. The strength of the steel has changed significantly over the years, but it is all steel to the best of my knowledge.</p>
| 33189 | Vocabulary: NYC Elevated Highway & Trains, Industrial Park |
2020-02-12T21:21:22.293 | <p>I am having trouble understanding how to approximate the solution to this problem using the Ritz method and the weak form:</p>
<p><span class="math-container">$$\frac{d^2u}{dx^2} - u=0; \ \ x \in [0,1]$$</span>
<span class="math-container">$$u(x=0)=0; \ \ \frac{du}{dx} \bigg|_{x=1} = 20$$</span></p>
<p>We multiply the strong form of the equation by a weight function <span class="math-container">$w$</span> and integrate over the domain by parts to get</p>
<p><span class="math-container">$$ \int_0^1w\left[\frac{d^2u}{dx^2} - u \right] dx = \int_0^1 w \frac{d^2 u}{dx^2} dx - \int_0^1wu \ dx$$</span></p>
<p><span class="math-container">$$\int_0^1 \frac{du}{dx} \frac{dw}{dx} dx - \int_0^1 wu \ dx - 20 w(1) = 0$$</span></p>
<p>which is the weak form. Now, I am confused because when I try to change this into a matrix equation, I practically try to solve</p>
<p><span class="math-container">$$\int_0^1 \frac{du}{dx} \frac{dw}{dx} dx = \int_0^1 wu \ dx + 20 w(1) $$</span></p>
<p>If I try to use, say <span class="math-container">$$u\approx\hat{u}=a_1 \phi_1 + a_2 \phi_2$$</span></p>
<p>I should be able to form a linear problem <span class="math-container">$$\mathbf{K} \vec{a} = \vec{b}$$</span></p>
<p>where <span class="math-container">$\mathbf{K}$</span> is a <span class="math-container">$2 \times 2$</span> matrix obtained from the left hand side of the equation where <span class="math-container">$w \to \phi_i$</span> and <span class="math-container">$u \to \phi_j$</span> to get the entries <span class="math-container">$K_{ij}$</span>, but this substitution would not work to obtain the components of <span class="math-container">$\vec{b}$</span> since I this substitution gives me terms of <span class="math-container">$\phi_i$</span> and <span class="math-container">$\phi_j$</span> and this makes no sense since <span class="math-container">$\vec{b}$</span> is indexed only by one subindex. I know I am not understanding something, but this was my professor's explanation and I am confused.</p>
<p>Thank you so much for your help!</p>
| |finite-element-method|numerical-methods| | <p>Everything in your OP looks correct.</p>
<p>I think you have just got into a muddle with the notation, and what is known and what is unknown.</p>
<p><span class="math-container">$\phi_1$</span> and <span class="math-container">$\phi_2$</span> are known functions, i.e. you <em>choose</em> two sensible functions for the problem you want to solve, to give a "good" approximation to the solution you expect.</p>
<p>You can differentiate them with respect to <span class="math-container">$x$</span>. Let <span class="math-container">$\dfrac{d\phi_i}{dx} = \phi'_i$</span>. </p>
<p>If you let <span class="math-container">$w = \phi_1$</span>, your integral equation becomes <span class="math-container">$$\int_0^1 (a_1 \phi'_1 + a_2 \phi'_2)\phi'_1\,dx - \int_0^1(a_1\phi_1 + a_2\phi_2)\phi_1\,dx = 20 \phi_1(1).$$</span>
You can evaluate the integrals <span class="math-container">$\int\phi_1^2$</span>, <span class="math-container">$\int \phi_2\phi_1$</span>, <span class="math-container">$\int {\phi'_1}^2$</span>, and <span class="math-container">$\int \phi'_2\phi'_1$</span> (either analytically or numerically) since the functions <span class="math-container">$\phi_i$</span> and <span class="math-container">$\phi'_i$</span> are known.</p>
<p>You then get an equation of the form <span class="math-container">$$K_{11} a_1 + K_{12} a_2 = b_1$$</span> where you know the <span class="math-container">$K$</span> and <span class="math-container">$b$</span> terms.</p>
<p>Similarly letting <span class="math-container">$w = \phi_2$</span> gives you the equation <span class="math-container">$$K_{21} a_1 + K_{22} a_2 = b_2.$$</span></p>
<p>These are the two rows of the matrix equation <span class="math-container">$\mathbf{K}\vec a = \vec b$</span>.</p>
| 33207 | How to use the Ritz method with the weak form to approximate solution of differential equation |
2020-02-13T10:00:42.013 | <p><strong>What i am looking for:</strong></p>
<p>I am looking for a formular to calculate the water output in <span class="math-container">$l/s$</span>.
My goal is to match a pump (pumping the water back up) with the following setup, to keep it going with as few corrections as possible.</p>
<p><strong>Where i am at the moment:</strong></p>
<p>Currently i have Torricelli's law for an ideal system.</p>
<p>But with this i am not sure if i have to take <span class="math-container">$h = h_2 - h_3$</span> or <span class="math-container">$h = h_2 - h_3 + h_1$</span>?</p>
<p><strong>Restrictions:</strong></p>
<p>The pipe is flexible.</p>
<p>I don't know:</p>
<ul>
<li>The friction of the pipe.</li>
<li>The angles.</li>
</ul>
<p><strong>My setup:</strong>
<a href="https://i.stack.imgur.com/Rfcey.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rfcey.jpg" alt="Sketch of the setup."></a></p>
<ol>
<li>An upper water tank with a water level <span class="math-container">$h_2$</span> mm (which is consistant!).</li>
<li>A lower water tank.</li>
<li>A pipe with it's intake <span class="math-container">$h_3$</span> mm above the upper Tank bottom. And it's outlet <span class="math-container">$h_1$</span> mm below the upper tank bottom.</li>
<li>The pipes diameter is <span class="math-container">$d$</span> mm.</li>
<li>The pipes length <span class="math-container">$l_1$</span> mm.</li>
<li>The water flow is started by sucking (like stealing fuel).</li>
</ol>
| |fluid-mechanics|pipelines|fluid| | <blockquote>
<p>Currently i have Torricelli's law for an ideal system.</p>
</blockquote>
<p>You could have started out with Bernoullis Equation, but it doesn't really matter here.</p>
<blockquote>
<p>But with this i am not sure if i have to take h=h2−h3 or h=h2−h3+h1?</p>
</blockquote>
<p>Neither, if the free jet exiting the pipe is not continous, then I'd say you would use <span class="math-container">$\Delta h = h_1+h_2$</span>. I don't see any reason to place the pipe so far above the surface level of the lower container that this would be a problem, so I will answer that the height is:</p>
<p><a href="https://i.stack.imgur.com/706J8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/706J8m.png" alt="enter image description here"></a></p>
<p>It can even dip into the lower container. The height difference is always the difference between the two water surfaces, it does not matter where exactly you place the pipe.</p>
<p>I studied the so called steam filament in fluid dynamics, so we image a streamline (green) from point 0 to 3 and follow it. (hence I recommended Bernoulli as a starting point)</p>
<p><a href="https://i.stack.imgur.com/tf3Rx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tf3Rxm.png" alt="enter image description here"></a></p>
<p>Now I'd like to imagine the parts that cancel each other out, e.g. the inverted U shaped top section, the fluid needs to go up a little and go down the same distance. Overall there is no height difference then.</p>
<p>So I marked all parts that cancel out each other and I get:
<a href="https://i.stack.imgur.com/nAxl7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nAxl7m.png" alt="enter image description here"></a></p>
<p>Which is consistent with what I said earlier.</p>
<p><em>I know this is a very over simplyfied way to regard the problem, however I thought the visualization may help here. You can calculate everything with bernoulli and get the same result.</em></p>
<blockquote>
<p>The velocity of the surface.</p>
</blockquote>
<p>What surface? You said they are constant, i.e. the containers are rather large (one of the assumptions of Toricelli btw). If not you need to modify your calculations bearing in mind that:</p>
<p><span class="math-container">$v = \frac{\partial h}{\partial t}$</span></p>
<p>and need to use the version of <a href="https://ocw.mit.edu/courses/mechanical-engineering/2-25-advanced-fluid-mechanics-fall-2013/inviscid-flow-and-bernoulli/MIT2_25F13_Unstea_Bernou.pdf" rel="nofollow noreferrer">bernoulli for instationary flow</a>.</p>
<blockquote>
<p>The angles.</p>
</blockquote>
<p>They don't matter, see above.</p>
<p><strong>edit</strong> in regards to Scott Dunnington's comment, I need to clarify that my approach of course neglects friction in the pipe (as you did with your assumptions).
However you can extend the Bernoulli equation with a term that accounts for these additional pressure drops.
It probably won't matter, but just for the sake of complecity.</p>
| 33211 | Calculating the flow rate of a siphon system |
2020-02-15T08:23:37.153 | <p>I am trying to find the exact tool which i assume is a modified t ruler to purchase but i cannot see any exactly like it. Any help would be appreciated:</p>
<p>Picture attached:</p>
<p><a href="https://i.stack.imgur.com/6yE38.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6yE38.png" alt="enter image description here"></a></p>
<p>It is adjustable and has a fixed right angle piece on top so it is square with the material you are marking. All the t rulers i find only are essentially a standard ruler with an adjustable moving piece. I need the T ruler with the t at the top and then the adjustable moving part.</p>
| |mechanical-engineering|tools| | <p>I'd consider this to be an identification question related to engineering, rather than a product recommendation question.</p>
<p>A search using "scriber and depth gauge" presents a large number of inappropriate devices and a smaller number of what appear to be the right device:</p>
<p><a href="https://i.stack.imgur.com/jI6Hk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jI6Hk.png" alt="scriber depth gauge"></a></p>
<p>As one might note, this is not of the same scale, but appears to be an accurate match.</p>
<p><a href="https://www.youtube.com/watch?v=dcO-WNlVnh8" rel="nofollow noreferrer">This video</a> presents one method to construct a scriber and depth gauge or the same smaller scale. One could expect it is thus possible to build at a larger scale.</p>
<p>Searching for larger versions was fruitless, but only a short period of time was used for the search.</p>
| 33236 | Trying to find this exact type of T Ruler? |
2020-02-16T21:11:44.870 | <p>I want to build a shelf that can I can place near a wall and still be able to rotate it around 180 degrees. I therefore would want to mount the shelf on a rotating pole and to reduce the needed distance to the wall would have a sliding mechanism that lets me slide the shelf off-center to one side (see the following sketch).</p>
<p><a href="https://i.stack.imgur.com/Bes9S.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bes9S.jpg" alt="Sketch of the sliding mechanics"></a></p>
<p>I built the following prototype out or cardboard to verify the idea. </p>
<p><a href="https://i.stack.imgur.com/Aqnex.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Aqnex.jpg" alt="Cardboard Prototype"></a></p>
<p>So given I can get the center pole to be stable enough (for example by attaching it to the wall) is it feasible to have such a sliding mechanism that can hold a weight of around 40kg in the off-center position and still slide? Am I missing any other critical problems here? Any hints for similar existing parts or mechanisms would be greatly appreciated.</p>
| |structural-engineering|stresses|wood| | <p>What you are looking for is already available as 70" TV mount, pull down over fireplace, in many Internet shopping sites for prices ranging between $50-100.</p>
<p>I add a photo just as an example.</p>
<p><a href="https://i.stack.imgur.com/1i10b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1i10b.png" alt="enter image description here"></a></p>
| 33254 | Is it feasible to mount a 40kg shelf on a rotating and sliding pole? |
2020-02-17T11:34:04.770 | <p>We've been requested to find the resonant frequencies of a 3rd order plant in class and were presented to the following method to do so:
<a href="https://i.stack.imgur.com/FCFC5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FCFC5.png" alt="Example included"></a></p>
<p>My question is, when using the <span class="math-container">$\ \omega_r $</span> formula, to find both of the resonant freq's do I need to place <span class="math-container">$\ \omega_{n1} $</span> with <span class="math-container">$\ ζ_{1} $</span> to get <span class="math-container">$\ \omega_{r1} $</span> (and <span class="math-container">$\ -\omega_{n2} $</span> with <span class="math-container">$\ ζ_{2} $</span> to get <span class="math-container">$\ \omega_{r2} $</span>),
or I am supposed to use the same <span class="math-container">$\ ζ_{2} $</span> (of the Pole) for both, like this:</p>
<p><span class="math-container">$$ \omega_{r1} = f(\omega_{n1}, ζ_{2}) $$</span></p>
<p><span class="math-container">$$ \omega_{r2} = f(\omega_{n2}, ζ_2) $$</span></p>
<p>I hope the question is clear enough, thanks!</p>
| |control-engineering|control-theory| | <p>The resonance is caused by the poles, therefore <span class="math-container">$\omega_r = \omega_{n_2} \sqrt{1-2\zeta_2^2} \approx 2.9462$</span>.</p>
<p>The zeros do not cause a resonance, but an antiresonance.
Thus, the antiresonance frequency is located at
<span class="math-container">$\omega_{ar} = \omega_{n_1} \sqrt{1-2\zeta_1^2}\approx 0.9592$</span>.</p>
<p>You can verify this by drawing the Bode Diagram<a href="https://i.stack.imgur.com/VHeeA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VHeeA.png" alt="Bode Diagram"></a>
Note that there is a small deviation between the values obtained using the formula and the Bode Diagram created with MATLAB.
The small error is caused by the poles and zeros being close to each other, while the used formula only holds for a pure second order system.
The further away the poles are from each other, the smaller the effect and the better the formula will estimate the (anti)resonance frequency.</p>
| 33261 | Control System; Finding resonant frequency of 3rd order |
2020-02-18T09:37:17.323 | <p>DB9 female port often have female slots for two screws to secure the male plug onto it. What is the type of this screw?</p>
<p>Word 'screw' is mentioned 7 times in Wikipedia DB-9 article but not the type.</p>
<p>Same port is a COM port on computer. Old school mouse port too.</p>
<p>At least Farnell, parts distributor have these screws categorized under: d-sub-jack-screws, and many items say M3 that being the type of the thread and the size both, 3 millimeters.</p>
<p>However, I have an unknown screw that fits into the mentioned female slot, and I have a legit M3 type screw, and the threads are way denser on the standard M3 screw. They have the same thickness tho.</p>
<p>Two leftmost ones fit, and are of the type of which I want to know the designation, the one on the right is M3</p>
<p><a href="https://i.stack.imgur.com/PA4R1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PA4R1.png" alt="enter image description here"></a></p>
| |computer-engineering| | <p>At least 3 sizes of threads seem to be in common use on DE-9/DB9 connectors:</p>
<p>#4-40 UNC</p>
<p>M2.6 x 0.45</p>
<p>M3 x 0.5</p>
<p>And, the screws that go into the attachment blocks are about 5.7mm long, but it is a specialty screw that isn't threaded all the way up to its head.</p>
<p><a href="https://omronfs.omron.com/en_US/ecb/products/pdf/en-xm3_l_xm2_l_xm4k_xm4l.pdf" rel="nofollow noreferrer">https://omronfs.omron.com/en_US/ecb/products/pdf/en-xm3_l_xm2_l_xm4k_xm4l.pdf</a></p>
<p><a href="https://www.datasheets.com/en/part-details/jey-9s-1a2b-j-s-t-mfg-co--ltd-34474395" rel="nofollow noreferrer">https://www.datasheets.com/en/part-details/jey-9s-1a2b-j-s-t-mfg-co--ltd-34474395</a></p>
| 33274 | Screw type associated with COM port/DB-9 |
2020-02-18T15:20:14.383 | <p>I'm looking to determine the difference between the two yield strengths on a Matweb material property table for <a href="http://www.matweb.com/search/datasheet.aspx?matguid=77b8eca6a7324775827b768f95637200" rel="nofollow noreferrer">DuPont Performance Polymers Zytel® 105F BK010 PA66</a>. The top is for "50% RH; ISO 527-1/-2" per the comments and the bottom is "DAM;ISO 527-1/-2". I need to know what 50% RH and DAM mean so I can make an informed decision. Any help would be appreciated. </p>
<p><a href="https://i.stack.imgur.com/x7AiG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x7AiG.png" alt="Image from Matweb Table"></a></p>
| |materials|plastic| | <p><a href="https://www.nfsa.gov.au/preservation/preservation-glossary/relative-humidity-rh" rel="nofollow noreferrer">The relative humidity</a> (RH) is the ratio of the actual water vapour pressure to the saturation water vapour pressure at the prevailing temperature. For example – if a cubic metre can hold 100ml of water at 20 degrees centigrade (273 K) and it does contain 100ml then it is said to be 100% RH. If the same cubic metre of air at the same temperature only contains 50mls of water then it is described as 50% RH.</p>
<p><span class="math-container">$\rm{RH = \frac{p}{p_s}}$</span></p>
<p><a href="https://acronym24.com/dam-meaning/" rel="nofollow noreferrer">DAM Meaning</a> is: Dry As Molded.</p>
| 33280 | What does 50% RH mean? |
2020-02-19T10:09:33.057 | <p>Looking at a model of the HMS Ark Royal I found myself intrigued by the prominent 'poles' lining the flight deck. Some investigation on Google reveals that they can apparently be lowered to lay horizontally, presumably to keep them out of the way of any aircraft landing or taking off.</p>
<p><a href="https://i.stack.imgur.com/T8Wna.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T8Wna.jpg" alt="enter image description here"></a></p>
<p>(Source: <a href="https://www.hydro-international.com/content/article/discovering-h-m-s-ark-royal" rel="nofollow noreferrer">https://www.hydro-international.com/content/article/discovering-h-m-s-ark-royal</a>)</p>
<p>The only hint of a purpose that I found was an old Google Group where someone mentions they are radio antenna. Unfortunately, I could not find a more authorative source.</p>
<p>So my question remains, what are these?</p>
| |mechanisms|marine-engineering|ships| | <p>They are HF radio masts that are hinged down for flying operations.</p>
<p>My initial thoughts they were radio direction finders or airframe recovery booms, but my research does not support it. </p>
<p>In port they are up. With planes in the air, they are down. Four port, four starboard. Two forward, six stern. They are connected by cables.</p>
<p>From <a href="https://www.the-blueprints.com/blueprints/ships/carriers-uk/4038/view/hms_ark_royal_1939/" rel="nofollow noreferrer">the-blueprints.com</a>.</p>
<p><a href="https://i.stack.imgur.com/GuuxP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GuuxP.png" alt="enter image description here"></a></p>
<p><a href="https://external-preview.redd.it/WuW3Z-PY_RxXHWT7gWhAtiixqGqIOehyCg0WzSCtfMQ.jpg?auto=webp&s=d9091df063954c286b5f544bd8e6c12b0de09b87" rel="nofollow noreferrer">Deployed as vessel was sinking (Higher quality image).</a> You can clearly see end curves down towards water. Deployed they are below flight deck. Clearly seen from YouTube video <a href="https://youtu.be/Rn77MzyWs9M?t=39" rel="nofollow noreferrer">With The 'ark Royal' (1940)</a> as a plane takes off. </p>
<p><a href="https://i.stack.imgur.com/VF00p.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VF00p.jpg" alt="enter image description here"></a></p>
<p>They are substancial as seen from YouTube video <a href="https://youtu.be/oTsR8IMpQzU?t=307" rel="nofollow noreferrer">HMS Ark Royal - 14 de Noviembre de 1941</a>.</p>
<p><a href="https://i.stack.imgur.com/vxdIX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vxdIX.jpg" alt="enter image description here"></a></p>
<p>The timing is around when Radio Direction Finding was introduced (which is why I answered the question) and RDF was used by the HMS Rodney supporting the Arc Royal. The Arc Royal was built before the Illustrious.</p>
<p>From: <a href="https://ethw.org/Radio_Location_Goes_to_Sea_-_Chapter_2_of_Radar_and_the_Fighter_Directors" rel="nofollow noreferrer">Radio Location Goes to Sea - Chapter 2 of Radar and the Fighter Directors</a></p>
<blockquote>
<p>The new-construction carrier HMS Illustrious was commissioned in April 1940, and would be the first RN carrier to have RDF. Later, while exercising with her air squadrons off Bermuda, Illustrious tried using RDF to guide fighter intercepts. HMS Formidable would be the second RN carrier to have RDF, commissioning in October 1940.</p>
</blockquote>
<p>These antennas are common to carriers. The next Arc Royal and the Illustrious have them too. All carriers of the period and later have them. </p>
<p><a href="https://i.stack.imgur.com/SX2CY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SX2CY.png" alt="enter image description here"></a></p>
<p>From <a href="https://books.google.co.uk/books?id=ROE7BAAAQBAJ&pg=PA64&lpg=PA64&dq=uss%20essex%20anatomy&source=bl&ots=RizN2wUWBm&sig=suZbWrpKwN4_kT4xHdvxzhkEssE&hl=en&sa=X&ved=0CF4Q6AEwDGoVChMIipvOydv0xgIVpCnbCh1O3Qyj#v=onepage&q=radio&f=false" rel="nofollow noreferrer">Essex Class Aircraft Carriers of the Second World War</a></p>
<blockquote>
<p>Lexington (CV-16) ... SBD Dauntless dive-bombers are marshalled at either end of the flight deck, and the <strong>radio masts</strong> are hinged down for flying operations.</p>
</blockquote>
<p><a href="https://forums.appleinsider.com/discussion/11888/" rel="nofollow noreferrer">anyone serve on an aircraft carrier or in the know about them?</a></p>
<blockquote>
<p>I served as an electronics tech onboard the USS Nimitz for three years and those were my babies. Those are two piece 35 foot HF whip antenna. They are in the down position during flight operation (and generally when underway) to keep jet wings safe. They are manually lifted when in close confines. They have nothing to do with HFDF or fall arrest.</p>
</blockquote>
<p>And finally, you can trust the <a href="https://i.pinimg.com/originals/33/8e/00/338e004f4bf9ad8e3d62b639907e5a43.jpg" rel="nofollow noreferrer">model world</a> because they really do do their research.</p>
<p><a href="https://i.stack.imgur.com/WfzWY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WfzWY.jpg" alt="enter image description here"></a></p>
<p><a href="https://sketchfab.com/3d-models/ark-royal-3ce8cf93dc8b4779bc373849a4d32388" rel="nofollow noreferrer">Sketchfab 3d Model of Arc Royal.</a></p>
| 33295 | What are the 'poles' off the side off this aircraft carrier's flight deck? |
2020-02-19T10:11:08.080 | <p>I’m working on a DIY project and need some assistance. I have a single motor that can lift a load of 500 pounds.</p>
<p>Is it possible to use 4 smaller motors to lift the same load as long as the 4 smaller motors equal the same torque as the single motor?</p>
<p>This isn’t a question about gearing or placing the 4 motors on a single shaft....each of the smaller motors would work independently to lift the load.</p>
<p>My thought is this... You have a 3’ lever on a fulcrum situated at a given distance (let’s say the fulcrum is exactly in the center of the 3’ lever. 100 foot pounds is placed on the lever, but the load does not lift. Is it possible to place another lever against the load with the exact same conditions as the 1st lever. Then, both levers are given 100 foot pounds....will the load lift with the added mechanism.</p>
<p>This is hypothetical of course.... I would love to understand this.</p>
| |torque| | <p>Ignoring losses, the total force on the object is all that matters. So 4 motors each supplying 25% of the torque would be exactly equivalent to 1 motor at 100% torque.</p>
<p>This applies everywhere. Two people each exerting 50 pounds can lift a 100 pound object; two levers double the force applied; two locomotives double the pulling power, etc.</p>
| 33296 | Lifting a Mass with one or multiple motors |
2020-02-19T12:22:05.653 | <p>If I have a uniform rod <code>--</code></p>
<p>And then I push on each end toward the center <code>>--<</code></p>
<p>So it bends <code>>^<</code></p>
<p>What shape does it make? Is it a parabola, a catenary, or something else?</p>
<p>Assume constant stiffness along the rod. I'm using the rod to draw an aerodynamic shape, and I'd like to know what shape I'm drawing. I've read about parabolas and catenaries, but I don't know the answer to this question.</p>
<p>Cheers!</p>
| |aerodynamics|drafting| | <p>For a slender column, pinned at both ends, the typical shape would be:</p>
<p><span class="math-container">$$y(x)=\delta_{max}\sin\left(\frac{\pi x}{L}\right)$$</span></p>
<p>where <span class="math-container">$\delta_{max}$</span> is the maximum deflection.</p>
<p>see <a href="https://www.continuummechanics.org/columnbuckling.html" rel="nofollow noreferrer">https://www.continuummechanics.org/columnbuckling.html</a> for more details.</p>
| 33298 | What shape does a rod bend into when you push the ends together? |
2020-02-19T17:59:18.140 | <p>When you compare single wall tent to double wall tent usually you will hear short summary "double wall tent prevents condensation from building up".</p>
<p>So in most fair comparison it should work like this -- I have imaginary single wall tent (i.e. tent with only one layer, the rainfly) and using it will cause condensation to build up. OK.</p>
<p>So now let's add to this tent inner tent (mesh layer) and this prevents condensation.</p>
<p>Hmm... I don't see it. I mean, condensation occurs when the warm air is cooled down to the point it no longer can hold steam. I guess the mesh barrier can slow down the process (i.e. you would have warm space inside, middle space between layers, and the cold air outside), the warm air from inside not so soon reaches the outer layer. But eventually it will reach it.</p>
<p>Second guess would be that inner mesh gets a little wet, and by this it helps the air to hold the rest of the amount of steam. But this effect I would say is a condensation anyway, just not visible because of the nature of the mesh.</p>
<p>So is there any other effect that contributes to the notion "inner tent prevents condensation"?</p>
| |steam|air| | <p>They don't prevent condensation. They just reduce the chances of condensation getting onto your kit down to practically nothing, if you do it right.</p>
<p>The condensation stays on the inside of the outer layer. Your kit is on the inside of the inner layer. The condensation doesn't cross the gap between the inner and outer layer. It runs down the inside of the outer layer. And then it can't pond inside the inner layer.</p>
| 33303 | The physics behind the condensation in tents? |
2020-02-21T07:47:18.163 | <p>I wanted to modify my kneading machine. In fact it can work for at maximum 5 minutes, and then I have to let it cool for 20 minutes (it is a very cheap model), and it is a problem for me.</p>
<p>So I was thinking of adding a sort of cooling system, for instance a fan (like that used in motherboards). </p>
<p><a href="https://i.stack.imgur.com/lX6jN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lX6jN.png" alt="enter image description here"></a></p>
<p>I was thinking of removing the red "hat" of the machine and replacing it with a bigger hat, so that there will be space for the cooling system.</p>
<p>But I have the following questons:</p>
<p>1) Do you agree with me with the position of "cooling ports" from with the air will enter and exit (IN, OUT)?</p>
<p>2) How can I realize the bigger hat? Which material? </p>
<p>3) How can I connect the bigger hat to the machine?</p>
<p>4) I was thinking of doing this modification because it is a cheap model and because I think that the only problem of using it for more than 5 minutes is the excessive heat. So if we solve it with a proper cooling system, I can use it for instance for 20 minutes. Do you agree with me about this last statement?</p>
<p>No problems about the electrical supply of the fan because I will build an external supply circuit for it.</p>
| |mechanical-engineering|materials|manufacturing-engineering|cooling| | <p>The motors used in inexpensive, occasional-use appliances often do not have cooling fans in them because they are not designed for continuous operation. This means that cutting holes in the top case of your mixer might not help get the heat out of the motor windings- unless you added a fan for this purpose. That would be a time-intensive proposition, requiring several hours. You then have to ask yourself the question of what is an hour of your time worth?</p>
<p>In general, if your appliance hits its thermal cutout after 5 minutes of operation, it is the wrong tool for the purpose and should be replaced by one of higher capacity. </p>
<p>An alternative solution is to buy a second mixer of the same type, and let the one cool down while using the other.</p>
| 33338 | Building a cooling system for a kneading machine? |
2020-02-23T02:32:44.017 | <p>I'm trying to create a place for these PCBS to mount onto:</p>
<p><a href="https://rads.stackoverflow.com/amzn/click/com/B072Z7Y19F" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/ELEGOO-Prototype-Soldering-Compatible-Arduino/dp/B072Z7Y19F/ref=pd_cart_vw_crc_1_2/133-3052301-7665733?_encoding=UTF8&pd_rd_i=B072Z7Y19F&pd_rd_r=75e160a5-3830-464f-8ad1-cb79cd1c127b&pd_rd_w=rUo6r&pd_rd_wg=K9U0z&pf_rd_p=3ffc9231-b582-4361-9816-b39fdbe2f421&pf_rd_r=12SMA97QX1HJGFXAQERM&psc=1&refRID=12SMA97QX1HJGFXAQERM</a></p>
<p>but I don't know where to position the mounting holes for the PCB so that it can be
screwed onto a standoff because the documentation for them only states how long each side of it is. </p>
<p>I already bought the PCBs and have a circuit soldered to one of them so I can't buy better documented ones. I've currently tried printing screw holes at varying distances to test for the ideal distance but the plastic bends too much to be reliably compared to the ideal unbending mounting board, so I would need to waste dozens of grams of plastic per failed hole distance test to make thick prints to accurately test.</p>
<p>Anyone here have a solution to this?</p>
| |cad|3d-printing|prototyping| | <p>Measure them using a caliper.
The best method will be to measure the inner and outer distance between two holes. The center-to-center distance is the average of the two measurements.</p>
| 33359 | How do you create a mounting device for a PCB if the PCB's mounting hole positions aren't documented? |
2020-02-23T02:38:00.050 | <p>I am curious how taking the most basic display such as this OLED one that works with the ardunio is driven:
<a href="https://i.stack.imgur.com/PqHTp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PqHTp.png" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/deitO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/deitO.png" alt="enter image description here"></a></p>
<p>i.e. We are using just a clock and data line to drive a 128x64 pixel display. The pixels containing the OLEDs with TFTs are arranged in a cross-hatch design, but what is the link between them? There are probably shift-registers to cut down on the number of pins, correct? Any guidance is helpful, thanks. </p>
| |electrical-engineering|computer-engineering| | <p>Pretty much every display is multiplexed. Typically each column is turned on (pulled to positive supply, for example) in turn and the rows for the pixels to be turned on in that column are switched to the opposite polarity (pulled to negative supply in this example). Usually the sequence happens at high speed so that there is no noticeable flicker.</p>
<p>For very small displays the rows and columns can be driven by the microcontroller. For larger displays this would require too many pins on the controller so the data to be displayed is somehow shifted in serially and shift-registers and latches hold the data within the display. This can have an advantage for displays that change relatively infrequently because the microcontroller can send the data to the display and then forget about it while it gets on with more important stuff. Meanwhile the display can continuously refresh based on the last data it received.</p>
<blockquote>
<p>We are using just a clock and data line to drive a 128 × 64 pixel display. </p>
</blockquote>
<p>The simplest implementation of this display would require 128 + 64 pins to drive it. Obviously this is not possible on an Arduino.</p>
<blockquote>
<p>The pixels containing the OLEDs with TFTs are arranged in a cross-hatch design, but what is the link between them? There are probably shift-registers to cut down on the number of pins, correct?</p>
</blockquote>
<p>Correct.</p>
| 33360 | How is the most basic display driven? |
2020-02-23T17:45:17.913 | <p>I read that wine bottle produces sound when u blow sideways onto its rim because
some frequencies of the white noise produced in your mouth gets amplified when they match the natural frequency of the bottle. But is the tight neck of the bottle necessary for this to happen? If I blow into a closed end tube with uniform width, can't I produce the same effect? Why do I need the small opening?</p>
| |fluid-mechanics|airflow|compressed-air| | <p>This phenomenon is known as flow-induced resonance. We learned about this in the submarine Navy, as water flowing over an open hole in the deck will produce a tone, and making tones is a bad thing for subs. The only links I could find for this were to scientific papers about the effect.</p>
<p>You personally need the small opening because you can't produce enough flow across the bottle if the opening is large. But a large enough flow would work on any size hole/tube/bottle combination, although the resulting resonance may be outside a human's perceivable frequency range (20 Hz to 20 kHz for young healthy individuals).</p>
| 33365 | How do cavity resonators differ from tubes of uniform width? |
2020-02-24T16:43:13.277 | <p>Why do we keep speed of turbofan of turbofan jet engine, low.</p>
<p>And what is the maximum thrust a turbofan jet engine., how much can thrust a turbofan jet engine produce with just compressed air.</p>
| |mechanical-engineering|fluid-mechanics|applied-mechanics|aerospace-engineering|jet| | <p>The fan speed is kept low because the blades in it must not exceed the speed of sound during operation, otherwise the efficiency of the fan will go down and noise generation will go up. </p>
<p>I am not sure your second question is meaningful; all jet engines generate thrust by discharging a large mass flow rate of compressed air from their tailpipes. They compress that air by burning fuel and capturing some of the heat energy thus produced in a turbine which then spins the compressor.</p>
| 33384 | Effect of speed of turbo fan in bypass jet engine |
2020-02-25T15:17:40.660 | <p>I used minutely sampled wind speed data for one year to calculate the yield power from specific wind turbines to be used in electric vehicles charging station.</p>
<p>I am supposed to calculate the number of electric vehicles that can be charged within each day given the total constant energy from wind turbine.</p>
<p>My question is:</p>
<p>How many batteries can I charge using a 100 kW supercharger if I have energy of 45 MWh for one day? Battery capacity is 50 kWh, it takes 21 min to fully charged (or up to 80% for example) using 100 kW supercharger.</p>
| |wind-power|electric-vehicles| | <p>It would be very unusual to have a wind turbine power a charging station without having a buffering device for energy storage such as an electrolyser or a battery of its own. But if it's a textbook calculation and these things are not there, then one approach is given below.</p>
<p>The limiting factor for battery charging is the current. If it takes 21 min at the supercharger to charge 80%*50 = 40 kWh, then the current is 40 kWh/(100 kw * 21/60) = 1.14 Amps. This is the max charging current to the battery. So one way to solve the problem is to look at the minute by minute amperage from the wind turbine and use the minimum of the current from the WT and the max charging current.</p>
| 33393 | number of electric vehicles that can be charged with specific amount of energy |
2020-02-25T15:18:14.373 | <p>According to Wikipedia (I know) a parallel manipulator is a single end effector driven by several serial chains. Does this include a simple scissor lift mechanism? And is there a better definition of what a parallel mechanism actually is?</p>
| |mechanical-engineering|mechanisms| | <p>A parallel mechanism is one where multiple "chains" connect the moved part to the ground. The parallel portion references that each chain (really a set of actuators and linkages) can move separately at the same time, and each is required to move in separately to achieve the desired platform orientation. </p>
<p>Your two legs are a parallel mechanism. For a dog, its 4 legs are a parallel mechanism.</p>
<p>A serial mechanism is a single leg - it extends, retracts, and twists to provide the needed input to the body. The position of each joint is dependent on the supporting joint.</p>
<p>A scissor lift is really just a simple mechanical advantage device that lifts a platform up and down only. We could define it as a limited version of a parallel mechanism where the 2 serial mechanisms are linked, but I don't think the definition fits well.</p>
| 33394 | Is a scissor mechanism considered a parallel manipulator |
2020-02-27T13:39:36.913 | <p>I'm stuck half way through construction of parametric conical helix antenna on a dieletric. I have created the intersection line but I'm struggling to make it a planar surface on top of the conical object as shown in the picture attached. I also have attached the Solidworks file.</p>
<p><a href="https://i.stack.imgur.com/Jnmtp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jnmtp.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>See below for how I set up my Global Variables, and for all equations that I used. The only formulae that actually needed calculations are the taper angle for the helix (calculated using basic trigonometry from the Min/Max diameters of the cone, and the height) and the height of the helix (the height of the final object, plus 2x the pitch. This allows it to be cut to height and have a smooth runout - I hope this is correct.</p>
<p><a href="https://i.stack.imgur.com/NEQ7W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NEQ7W.png" alt="Equation Editor View"></a></p>
<p>The steps that you need to take to model this item are as follows:</p>
<ol>
<li>Set up your global variables <em>before</em> you start modelling.</li>
<li>I renamed 'Top Plane' as 'Base Plane', since this will be the base of the cone. I then inserted a plane called "Max Height", set to an offset from the Base Plane, defined by the global variable "Height". Always define your planes or axes as early in the history tree as possible.</li>
<li><p>Set up your sketch for the cone. I started with an infinite length centreline - this allows us to use diameter dimensions, rather than needing to divide the global variables by 2. The cone is defined by two horizontal construction lines, one coincident with the origin (and base plane), and the other with the 'Max Height' plane. Note that the "Height" is not dimensioned here, it's driven by the location of the plane. I have added an extension past the Base and Max Height planes equal to the "Pitch" global variable.</p>
<p><a href="https://i.stack.imgur.com/vSxjym.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vSxjym.png" alt="cone sketch"></a></p></li>
<li><p>Revolve the Cone - It should automatically select the infinite centreline as the axis of rotation.</p></li>
<li><p>Generate your Helix. I started a sketch on the base face of the cone, and used convert entities to make a circle coincident with the bottom edge. You can't type equations directly into the helix tool (like you can with regular dimensions), so you need to just type a starting point, before moving to the Equation editor to add values as shown in my first image. I set it to be defined by Pitch and Height, with taper enabled. I put my Cone profile on the right plane so that a start angle of 0 deg will be coincident with the Cone Profile Sketch. You can see in the image below how to select the helix definition - if height and revolution is more appropriate, you can select it here, and adjust your global variables as required.</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/RzuMR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RzuMR.png" alt="how to select helix defintion"></a></p>
<ol start="6">
<li>Set up your Antenna profile sketch. This should be coincident with the Cone profile, and the start of the Helix. I have set the width and thickness equal to their respective global variables. Note how the angle of this is defined by the Cone Profile sketch, which is in turn driven by the Max/Min diameters.</li>
</ol>
<p><a href="https://i.stack.imgur.com/ayndR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ayndR.png" alt="antenna profile"></a></p>
<ol start="7">
<li>Sweep the Antenna along the Helix. Uncheck 'Merge result' in the options to keep this as a separate body from the cone.</li>
<li>Cut off the bottom of both parts - using Cut with Surface and the Base Plane</li>
<li>Cut off the top of both parts - using Cut with Surface and the Max Height Plane</li>
<li>Sit back and admire your handiwork</li>
</ol>
<p><a href="https://i.stack.imgur.com/gUQSB.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gUQSB.gif" alt="parametric helix antenna"></a></p>
| 33424 | Solidworks - parametric conical helix antenna on a dielectric |
2020-02-28T20:06:14.290 | <p>When I open Paint, I change the size to 500x500. It appears more as a rectangle than a square. Has anyone else had this happen? Is there an explanation? Not sure if this is a thing that differs on different computers.</p>
| |computer-engineering|computer| | <p>Looks like a square to me - are you sure you selected the pixel checkmark, and also unchecked the "Maintain aspect ratio" box before typing in your second '500'?</p>
<p><a href="https://i.stack.imgur.com/QlHZlm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QlHZlm.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/LkJmo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LkJmo.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/K4Lzs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K4Lzs.jpg" alt="enter image description here"></a></p>
| 33430 | Are pixels not squares and more like rectangles? |
2020-02-28T23:44:13.677 | <p>I want to keep the pressure inside a balloon (effectively - a rubber vessel) constant, as the balloon is compressed. I'm wondering if a regulator like <a href="https://in.rsdelivers.com/product/norgren/r72g-2gk-rmn/norgren-pneumatic-regulator-g-1-4/2390670" rel="nofollow noreferrer">this</a> (<a href="https://en.wikipedia.org/wiki/Pressure_regulator?wprov=sfla1" rel="nofollow noreferrer">wiki page here</a> showing mechanism - I believe) can do it? They're designed to regulate a higher relatively constant pressure (eg large volume at 30bar) down to <1bar in a smaller vessel to keep that constant. I think these are one way tho - if the pressure starts increasing in the outlet the inlet will close off, rather than dissipate that increase back to the source, whereas I want to effectively use it to do exactly that. Would this work? Or is there a better way to do this? </p>
<p>Thanks for all help given!</p>
| |pressure|pressure-vessel|pneumatic| | <p>Have a second separate balloon that expands as the first is compressed.</p>
<p>This will also keep the pressure constant as the volume is increased if that is part of your requirement.</p>
| 33433 | Regulating pressure down inside vessel as it increases |
2020-03-01T11:48:24.177 | <p>The vehicle running at Hill station.. Same time vehicle moves back ward direction in manual transmission system. So avoid the problem.. I think ratchet pinion mechanism suitable! or not?</p>
| |mechanical-engineering|automotive-engineering| | <p>There is a system called a "hill holder" in cars which engages the parking brake on a hill while stopped. When the clutch is engaged to drive up the hill, the parking brake is slowly released by the mechanism. Manual transmission Subarus had this feature years ago. </p>
| 33450 | Problem for backwards moving vehicle in hills |
2020-03-01T17:46:06.523 | <p>Distinguish between dead center and live center in a center lathe tool when this terms are used in the context of work holding in a lathe </p>
| |mechanical-engineering| | <p>A live centre is driven by the spindle. A dead centre is on the non drive side. A revolving dead centre is also on the non drive side but has bearings so that it can revolve with the workpiece.</p>
| 33452 | Distinguish between dead center and live center |
2020-03-02T07:14:18.060 | <p>I have a technical drawing, but I can't understand the "h2". The "h2" points to a hole, and I don't understand it. Can you please give me an answer? Thanks!</p>
| |technical-drawing| | <p>H2 refers to the tolerance of the hole - whether it’s an upper or lower case letter shows if it’s a hole or shaft tolerance. In this case, as you mention it’s on a hole, I would expect it to be “H6”, which means for an 6.00mm hole, it would be acceptable if manufactured in the range 6.00-6.03mm in diameter.</p>
<p><img src="https://i.ytimg.com/vi/c8TKftViusQ/maxresdefault.jpg" alt="enter image description here"></p>
| 33458 | What does "h2" mean on technical drawing? |
2020-03-02T13:19:24.423 | <p>I have been struggling with the following problem for a few days now and would greatly appreciate some assistance.</p>
<p>I'm new to control systems and have very little experience with PD controllers and never heard of the pole placement method so please correct me where I'm wrong.</p>
<p>The main problem I seem to be having is that there are three poles <span class="math-container">$\ s(s+2)(s+5) $</span> all other examples i have done have had only two <span class="math-container">$\ (s+2)(s+5) $</span>.</p>
<p>So I don't see how I can you the standard formulas I have been using:</p>
<p><span class="math-container">$$ s^2+2ζω_ns+ω_n^2 $$</span></p>
<p><span class="math-container">$$ s^2+(K_d-1)s+(K_p+1) $$</span></p>
<p>Like I said I'm very new to this so please forgive me if anything I've said is completely wrong.</p>
<p>Thanks in advance for any assistance.</p>
<p><a href="https://i.stack.imgur.com/mOk4g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mOk4g.png" alt="Control systems Question"></a></p>
| |control-engineering|control-theory| | <p>First of all let's consider the aspect of dominant poles. You have to control the plant:</p>
<p><span class="math-container">$$ G(s) = \frac{1}{s(s+2)(s+5)} $$</span></p>
<p>The poles of this plant are located at : </p>
<ul>
<li><span class="math-container">$p_1 = 0$</span></li>
<li><span class="math-container">$p_2 = -2$</span></li>
<li><span class="math-container">$p_3 = -5$</span></li>
</ul>
<p>Dominant poles in general are the poles which are closer to the imaginary axis and define the response (behaviour) of the system. In this particular system the poles that are closer to the imaginary axis are: <span class="math-container">$\ -2,0 $</span>, so we can say that they dominate the dynamic behaviour of the system. It is always good if possible to model a system by a second order transfer (while obtaining the steady state behaviour) due to the simplicity of the math involved in the deisgn process. In order to reduce the system‘s order and simultaneously maintain the steady state response we can bring this system up to the system: </p>
<p><span class="math-container">$$ G(s) = \frac{\frac{1}{5}}{s(s+2)} $$</span></p>
<p>We can use MATLAB to check the steady state responses of the two systems. These are shown at the graph below:</p>
<p><a href="https://i.stack.imgur.com/UiLOU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UiLOU.jpg" alt="enter image description here"></a></p>
<p>It is obvious that both systems have the same steady state responses. Now, we continue with the design process.</p>
<p><strong>First Step:</strong> Find the closed loop transfer function. </p>
<p><span class="math-container">$$ G(s) = \frac{1/5}{s(s+2)} \rightarrow Plant $$</span>
<span class="math-container">$$ G_c(s) = K_p+K_ds \rightarrow PD \ Controller $$</span></p>
<p>and the closed loop transfer with unity feedback is:</p>
<p><span class="math-container">$$ H(s) = \frac{0.2(K_p+K_ds)}{s^2+s(2+0.2K_d)+0.2K_p} $$</span></p>
<p>The characteristic polynomial (denominator of the closed-loop transfer function) is:</p>
<p><span class="math-container">$$ s^2+(2+0.2K_d)s+0.2K_p = 0 $$</span></p>
<p><strong>Second Step:</strong> Pick the gains of the controller with respect to the system's design specifications.</p>
<p><span class="math-container">$$ ζ = 0.707 $$</span>
<span class="math-container">$$ t_s = \frac{4}{ζω_n} \Rightarrow 2=\frac{4}{ζω_n} \Rightarrow ω_n = 2.8289 \ rad/sec $$</span></p>
<p><strong>Third Step:</strong> Obtain the desired characteristic polynomial with respect to the desgin specifications:</p>
<p><span class="math-container">$$ s^2+2ζω_ns+ω_n^2 = 0 $$</span></p>
<p><span class="math-container">$\ ζ: $</span> Damping Ratio </p>
<p><span class="math-container">$\ ω_n: $</span> Natural Frequency</p>
<p>which by plugging in the vaues for <span class="math-container">$\ ζ $</span> and <span class="math-container">$\ ω_n $</span> becomes:</p>
<p><span class="math-container">$$ s^2 + 4.0001s + 8.0027 = 0 $$</span></p>
<p><strong>Fifth Step:</strong> Match the coefficients of the system's characteristic polynomial with the corresponding ones of the desired characteristic polynomial. By doing so we come up with the values of <span class="math-container">$\ K_p $</span> and <span class="math-container">$\ K_d $</span>.</p>
<p><span class="math-container">$$ K_p = 40.0135 $$</span>
<span class="math-container">$$ K_d = 10.0005 $$</span></p>
<p><strong>Sixth Step:</strong> Check the step response to verify the requirements. By plugging in the values for the gains <span class="math-container">$\ K_p $</span> and <span class="math-container">$\ K_d $</span> to the closed loop transfer function we obtain the following step response of the system:</p>
<p><a href="https://i.stack.imgur.com/lsqnz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lsqnz.png" alt="enter image description here"></a></p>
<p>It is shown that the settling time is: <span class="math-container">$\ t_s=1.86sec $</span></p>
<p>However, if you put these particular gains and simulate the initial third order system, you will find out that the settling time is larger than the settling time of the equivalent second order system. Below is the graph of simulation of the initial system (blue line) and the equivalent second order system (red line).</p>
<p><a href="https://i.stack.imgur.com/XbgNt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XbgNt.png" alt="enter image description here"></a></p>
<p>This happens because the third pole of the initial system <span class="math-container">$\p_3 = -5 $</span> is not that far away so we can ignore it. It is near the dominant poles which means that it also affects considerably the dynamic behaviour of the system. However, it is important to notice that at steady state the two systems reach the same amplitude which means that the equivalent system we obtained is correct. In general, we can approach a third order by a second order system when the third pole of the third order system is sufficiently further away than the dominant poles. For example consider the following system: </p>
<p><span class="math-container">$$ H(s) = \frac{1}{s(s+2)(s+30)} $$</span></p>
<p>The poles of this system are: <span class="math-container">$\ 0, -2, -30 $</span>. The third pole of this system is sufficiently bigger than the two dominant poles. This system is approached by the second order system:</p>
<p><span class="math-container">$$ G(s) = \frac{\frac{1}{30}}{s(s+2)} $$</span></p>
<p>The step responses of these systems, which are similar, are:</p>
<p><a href="https://i.stack.imgur.com/BSq3f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BSq3f.jpg" alt="enter image description here"></a></p>
<p>If we follow the same design process, we obtain the following PD controller: </p>
<p><span class="math-container">$$ G_c(s)=240.3213+60.0631s $$</span></p>
<p>Now, if we simulate the initial third order system and the equivalent second order system in series with this PD compensator (and unity feedback), we obtain the following graph:</p>
<p><a href="https://i.stack.imgur.com/Y8fCI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y8fCI.jpg" alt="enter image description here"></a></p>
<p>It is obvious that the step responses of the two systems are almost identical. The difference in settling time is <span class="math-container">$\ 0.03 $</span> seconds, which is neglectible. In conclusion, the aspect of dominant poles is one hundred percent correct when there is a pole which is enough bigger than the poles which are close to the imaginary axis. </p>
<p>Below, I place some MATLAB code which produces the above results.</p>
<pre><code>s = tf('s');
Kp = 240.3213; % Or any gain
Kd = 60.0631% Or any gain
initial = tf([0 0 0 1],[1 32 60 0]); % Or any system
equivalent = tf([0 0 1/30],[1 2 0]); % Or any equivalent
pd_compensator = Kp+Kd*s;
initial_comp = series(pd_compensator,initial);
equivalent_comp = series(pd_compensator,equivalent);
closed_initial = feedback(initial_comp, 1);
closed_equivalent = feedback(equivalent_comp, 1);
figure
step(closed_initial)
hold on
step(closed_equivalent)
hold off
</code></pre>
| 33462 | Proportional-Derivative compensator design |
2020-03-02T16:38:58.443 | <p>I have a question regarding inertia and friction between gears.
I have a gear which is connected to the rod at one end and an electric motor which is connected to the other end. I know the torque of the rod and I want to calculate the torque of the motor I need.
I have been told that I can't lay on transmission ration only and I also have to include the calculations of the inertia and friction between gears.</p>
<p>Can please someone explain me how should I do it (every example will appreciated).</p>
| |mechanical-engineering|gears|friction|transmission| | <p>To turn your gear from rest, you'll need to overcome the inertia of your entire system (rod and gear).</p>
<p>Torque = inertia x angular acceleration. You'll need to figure out what angular (rotational) acceleration you need/can get from your motor. You can approximate inertia with I = (1/2) x mass x radius^2.</p>
<p>You can approximate gears being 95% efficient if you're using spur gears (straight teeth). This means 5% of the energy put in is being lost as friction.
So your required motor torque = (operating torque + acceleration torque)/ 0.95 efficiency [friction].</p>
| 33465 | Inertia and friction between gears |
2020-03-03T07:18:20.350 | <p>I am thinking of building a indirect Peltier AC to cool a room of approx size 30m<sup>3</sup>. Single/multiple Peltier modules will be used to cool a insulated box with water inside. The box will be kept outside the room. A pump will circulate the cold water of the box in a pipe which will go inside the room. Inside the room, The pipe will be connected to something similar to a radiator, but rather than throwing heat into the air, it will throw cool air.
Also, I want to achieve maximum efficiency, by correctly balancing the current, voltage and temp difference.
Pumping heat out of the Peltiers will most probably not be a problem as they will be water cooled.
Is this practically possible?</p>
| |mechanical-engineering|thermodynamics|hvac|refrigeration|thermocouple| | <p>The simple answer: YES, a compressor will still be slightly more effective (2022) and noisier, but yes it can work well if well designed.</p>
<p>It will draw more power compared to a compressor but will be almost silent and super compact.</p>
<p><em><strong>Proof of concept</strong></em>.</p>
<p>Built one out of an old plastic container, 4 old pc fans (low speed) an old PSU and some cabling, separate the box into 2 compartments: Hot and Cold.</p>
<p>Peltiers are super cheap now days, just add some ventilation ducts and ... voila your own portable AC, it's too bad power prices today make it more costly in the long run than a compressor design, but in summer it's sure a blessing to have.</p>
| 33476 | Can Peltier Modules be practically used to cool a room? |
2020-03-03T09:12:18.067 | <p>As you can see in the below shematic, the magnetic water conditioners can be used for the pipes of incoming process water on sinter plants to reduce surface tension of water and get fuel consumption benefits. But I researched and couldn't find any type of this conditioners/armatures .</p>
<p><a href="https://i.stack.imgur.com/wHutw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wHutw.jpg" alt="enter image description here"></a></p>
<p>Look at first article about this: <a href="https://www.steeltimesint.com/contentimages/features/KumarIronMaking2.pdf" rel="nofollow noreferrer">https://www.steeltimesint.com/contentimages/features/KumarIronMaking2.pdf</a>
This looked unrealistic to some engineers that magnetic water conditioner is not really exist thing.</p>
<p>But.. </p>
<p>Please also read: <a href="https://www.sciencedirect.com/science/article/pii/S187538921400532X/pdf?md5=3846d794d248117cceb8e706f4b1489f&pid=1-s2.0-S187538921400532X-main.pdf" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/pii/S187538921400532X/pdf?md5=3846d794d248117cceb8e706f4b1489f&pid=1-s2.0-S187538921400532X-main.pdf</a></p>
<p>The above article looks more realistic, so is this magnetic charging chute can reproduce for water as well?</p>
<p>Is there any industrial-scale magnetic water conditioner exist?</p>
| |mechanical-engineering|fluid-mechanics|chemical-engineering|mechanical| | <p>The mystical claims for magnetic water conditioning have been given a special technical term to describe them, used only by real hip engineers. </p>
<p>It is <em>jive- _ss baloney</em>. </p>
| 33479 | What kind of magnetic water conditioner can industrial sinter plant use? |
2020-03-03T18:04:50.500 | <p>Does anyone know where I can find Heating tape that has these same leads or know what they are called? It is meant to be plugged into the second picture attatched.<a href="https://i.stack.imgur.com/ahQQf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ahQQf.jpg" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/G5ybU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G5ybU.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/LXsZz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LXsZz.jpg" alt="enter image description here"></a></p>
| |electrical-engineering|heating-systems|heat-treatment| | <p>When searching for 1100 watt heating tape, a number of returns match almost exactly the image you've provided. A few of the manufacturers provide NEMA separable molded plugs and also offer bare leads. The bare leads would be useful in this case, as you can add your own 90° banana plugs.</p>
<p><a href="https://i.stack.imgur.com/Dn8aO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dn8aO.png" alt="right angle enclosed banana plug"></a></p>
<p>To locate the correct plug, I used "right angle enclosed banana plug" as the search term. The image and source information came from <a href="https://www.zoro.com/zoro-select-banana-plug-right-angle-1000vac-red-5txc0/i/G1470156/feature-product?gclid=EAIaIQobChMIrMzDg5b_5wIViMDICh1c_wWLEAQYAiABEgIGnPD_BwE" rel="nofollow noreferrer">Zoro</a> online. I had better returns using shrouded in place of enclosed. Many sources. Be sure to check the current draw for the tape you'll use and the connector you'll select.</p>
| 33480 | Heating tape that has leads like picture below |
2020-03-06T10:13:05.997 | <p>I'm translating a document in which there is a mention of <strong>обрешетки</strong> used for preparing goods (medical equipment of large size) for shipment. This term derives from reshetka (решетка) - lattice, and means "custom-made frame of wooden planks built around a load destined for shipping". It's kind of a rough crate built around the product intended for shipping, not pre-made.</p>
<p>I wonder what the proper term in English might be. Multitran <a href="https://www.multitran.com/m.exe?l1=2&l2=1&s=%D0%BE%D0%B1%D1%80%D0%B5%D1%88%D0%B5%D1%82%D0%BA%D0%B0" rel="nofollow noreferrer">provides a long list of options</a>, but without describing them much, so it's hard to distinguish.</p>
<p>There's a whole <a href="http://docs.cntd.ru/document/1200011144" rel="nofollow noreferrer">National Standard document in Russian</a> that describes how these "frames" must be built, with many pictures. However, it provides the wrong translation there in English: "roof boarding" (because the same Russian term is also used for roof lattice). </p>
<p><a href="https://i.stack.imgur.com/NOewG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NOewG.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/fHhF6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fHhF6.jpg" alt="enter image description here"></a></p>
<p>I'm submitting my translation in an hour, so I'll stick with descriptive "wooden frames" but I'd be glad to learn the formal term if it exists. </p>
| |terminology|transportation| | <p>The English term for this is definitely <strong><em>crate</em></strong>.</p>
| 33522 | What do we call these "transport frames made of wood" (обрешетки in Russian) |
2020-03-05T17:31:26.770 | <p>I keep seeing these large rectangular structures on job sites they put in the ground. </p>
<p><a href="https://i.stack.imgur.com/uJSXm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uJSXm.jpg" alt="image"></a></p>
<p>I'm guessing something for strengthening the surface structures?</p>
| |civil-engineering| | <p>It is a trench shield. It gets placed in a trench after the trench is dug to prevent workers from being hurt or killed in the event of a trench collapse.</p>
<p><a href="https://i.stack.imgur.com/rL7Mc.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/rL7Mc.jpg" alt="enter image description here"></a></p>
<p>This picture from GMC trench shield shows a partially collapsed trench with a shield installed that would protect the workers installing the blue brute pipe.</p>
| 33530 | What is this large steel structure used in excavation? |
2020-03-07T06:59:59.267 | <p><a href="https://i.stack.imgur.com/1BLFk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1BLFk.jpg" alt=""></a></p>
<p>How would i resolve the forces so they are acting on the bottom joints.</p>
| |mechanical-engineering|statics| | <p>We consider each member that has mid-span loads as a simply supported beam and resolve the loads on the beam to vertical reactions and those reactions will be loads at those joints. </p>
<p>Then solve the truss with the midspan loads removed and reactions added as joint loads.</p>
<p>After we find all the member axial forces and supports' reaction we Solve that member as a simply supported beam with appropriate axial compression or tension and find the moments and shears on that member.</p>
<p>For example, the first bottom member on the right, say we call the support B and the joint to its right A:</p>
<p><span class="math-container">$ \Sigma M_B=0 \ 312*5.05/6.7= 235kN =P_A \rightarrow \ and \ 312-235= 77=P _B $</span></p>
<p>So we add these two loads to joints A and B and remove them from mid-span of this member. </p>
<p>Then we solve the truss and find the axial load for AB which should be tension and solve it. </p>
<p>We need to pay extra attention to those members with compression, checking for buckling with both the secant method and Euler's critical buckling load.</p>
| 33543 | What to do when the forces on a truss is not on the joints |
2020-03-07T17:11:21.620 | <p>This kind of a sleeve from cardboard put upon a cardboard box is called <strong><a href="https://www.multitran.com/m.exe?l1=2&l2=1&s=%D0%BE%D0%B1%D0%B5%D1%87%D0%B0%D0%B9%D0%BA%D0%B0" rel="nofollow noreferrer">обечайка</a></strong> (obechaika) in Russian. What do we call them in English? I was told on Proz.com that we call them "cardboard shells", is that so?</p>
<p>I was translating a text in which medical products were packaged into cardboard boxes with such additional shells to provide additional protection.</p>
<p>Such shells provide additional strength but sometimes such shells are made from decorative paper with fancy design and are put around, for instance, a box of chocolates to make it look more refined. These are also <strong>obechaikas</strong> in Russian.</p>
<p>Here, a cardoard box is slid inside an <strong>obechaika</strong>:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/tEOHf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tEOHf.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/m6Ak3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6Ak3.jpg" alt="enter image description here" /></a></p>
</blockquote>
<p>Here's another use of the term <strong>obechaika</strong>, a decoratice paper shell around a food product in a plastic container:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/z9RSc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z9RSc.jpg" alt="enter image description here" /></a></p>
</blockquote>
<p>In this usage <strong>obechaika</strong> is more frequently called <strong>shuber</strong>, for some reason. The word is yet absent from Russian dictionaries, might be a new coinage.</p>
| |terminology| | <p>I would use “cardboard sleeve”, and it would apply to both your examples.</p>
| 33549 | What do we call these corrugated board sleeves put upon cardboard boxes to provide additional strength? |
2020-03-08T16:57:03.927 | <p>I am designing phone cases for 3D Printing. I am fairly new to the field, however I cannot find design specifications for any phone at all, i.e. exact measurements for the positioning of cameras, buttons, ports, etc. Can anybody recommend somewhere where I can find these documents? Also, if this question is in the wrong forum, can a link to the correct forum be linked please.</p>
| |design| | <p>You'll highly unlikely find any exactly drawings, as mentioned in the comments. These are usually trade secrets. However, you can pull out a caliber and measure the phone you have in mind yourself and make a few prototypes, continually reiterating until you get the results you want.</p>
<p>Classic <a href="https://en.wikipedia.org/wiki/Reverse_engineering" rel="nofollow noreferrer">Reverse Engineering</a> :)</p>
| 33560 | Where can I get design specifications for phones? |
2020-03-08T21:18:07.367 | <p>I've found this video on YouTube:</p>
<p><a href="https://youtu.be/hWK8Gv0JHqk?t=220" rel="nofollow noreferrer">https://youtu.be/hWK8Gv0JHqk?t=220</a></p>
<p>Watch from 3:40 to 4:20.</p>
<p>I'd like to learn more about this linkage that makes that link alternate between hitting and missing the sliding link.</p>
<p>I've searched for the animated movements from <a href="http://507movements.com/" rel="nofollow noreferrer">507 Mechanical Movements</a> superficially and found nothing.</p>
<p>Thanks in advance.</p>
| |linkage| | <p>I believe what you are seeing is a pendulum mounted on a sliding bracket. The weight with the thumbscrew indicates that it can be tuned to match the required movement to engage the V-shaped "socket" into which it engages on alternating strokes. </p>
<p>Slowed down to one-quarter speed, the video shows that on one stroke, the pendulum is swinging left, while the sliding mechanism is pushing right, with inertia causing the engaging lever to rise. In the next half-stroke, the sliding mechanism pushes left with the lever in the socket, preventing inertial movement of the lever.</p>
<p>The next half-stroke to the right disengages the lever from the socket, gravity causing the lever to fall simultaneously with the pendulum causing the lever to fall with sufficient momentum to prevent the next left sliding stroke from engaging the socket.</p>
<p>I searched for various terms attempting to locate a specific term with no success.</p>
<p>I believe one can reproduce this mechanism by experimenting with varying angles between the lever and the pendulum arm and by varying distance on the pendulum weight.</p>
<p>I'd love to know what purpose the alternating actuation serves. One thinks it could be a two stroke engine system of sorts.</p>
| 33563 | What is this bistate linkage? |
2020-03-10T12:24:24.293 | <p>Not sure if anyone would know this, but was discussing it with friends, and it seems like quite the puzzle. In short, how is nuclear fuel in reactors handled?</p>
<p>Logically, the nuclear fuel must be transported to the site, then placed into the reactor, then - once fully used - removed. But given the strength of the radioactivity of the elements, how on earth is this done?</p>
<p>Using the Chernobyl disaster as an example, the guys who had to clear up the mess weren't able to be close to the source for more than ca. 60 seconds without it being deadly. Even with lead protection, electrical circuits broke, so robots seem not to be an option. Just seems like a really difficult task to be able to transport, install, remove and dispose of the uranium without there being major hurdles.</p>
<p>Anyone know who it is done?!</p>
<p>Also, on a related topic: if a nuclear power station is decommissioned, how is the core dismantled? How can you safely remove the irradiated graphite, for example?</p>
<p>My friends and I spent ages debating how it could be done and we would love to know the answer!!</p>
<p>Thanks in advance.</p>
| |nuclear-engineering| | <p>I work at a small research reactor---power reactors are significantly different, and I'm not qualified to say much about them. Since we don't operate constantly, our fuel burns up very slowly (the reactor's been there for ~50 years and the only change was when we swapped our original fuel for fuel from a decommissioned research reactor, for reasons unrelated to burnup. When we handle fuel we basically just make sure that it stays under a lot of water. Think "I wouldn't touch it with a ten-foot pole", except we do. </p>
<p>Chernobyl is a glaring exception to normal procedures. Part of the problem at Chernobyl was that all of the design features built into the fuel to make it safer to handle were useless, as they were all melted into a lump of corium. Not the normal state of affairs.</p>
<p>For decommissioning, the usual procedure is to just let everything sit underwater for a while to let stuff cool off a bit. After that waste is often enclosed in some sort of solid which shields against the radiation, and then left to sit some more. Looking at graphite specifically, there's recently been some fascinating research aimed at using the waste graphite as a power source in batteries!</p>
<p>-sol</p>
| 33584 | How is nuclear fuel dealt with? |
2020-03-10T18:12:38.150 | <p>I'm trying to calculate the bending moment distribution on a fixed-fixed Euler-Bernoulli beam. It is known that interelement nodes show different values for the bending moment. What is customary to do with this discrepancy?</p>
<p>Im doing the following:</p>
<p><span class="math-container">$$
M_z =\begin{bmatrix}-6/L^2 + 12x/L^3 \\-4/L + 6x/L^2 \\ 6/L^2 - 12x/L^3 \\-2/L + 6x /L^2 \end{bmatrix}^T\begin{bmatrix} w_1 \\ \theta_1 \\w_2 \\ \theta_2\end{bmatrix}
$$</span></p>
<p>This returns the moment distribution along the element, where the 1st vector is the second derivative of the shape functions and the 2nd one is the computed displacements.</p>
| |mechanical-engineering|structural-analysis|stresses|beam|finite-element-method| | <p>If you have a fixed-fixed beam then you always get zero internal force and displacement because of zero nodal displacement. If displacement vector is zero then bending moment is also zero, because of approximate nature of finite element. I know two ways to fix this discrepancy:
- split element into two and add node in middle
- consider the distributed load effect in internal force using double integration method etc.. (for example code look at method <code>GetLoadInternalForceAt()</code> at this file: <a href="https://github.com/BriefFiniteElementNet/BriefFiniteElement.Net/blob/master/BriefFiniteElementNet/ElementHelpers/EulerBernoulliBeamHelper.cs" rel="nofollow noreferrer">EulerBernoulliBeamHelper.cs</a></p>
| 33586 | Bending moment in Euler-Bernoulli finite Element |
2020-03-10T22:21:57.303 | <p>I’m building a pitching machine that has an 8” wheel (16” OD tire) that is directly mounted to a 1/2” motor shaft.</p>
<p>Most similar questions reference a gearbox or pulpy system, but my situation is directly mounted to a 12v motor shaft.</p>
<p>Due to motor/fabrication design, the closest I can get my magnetic rpm sensor is a radius of 5” from the shaft.</p>
<p>So known variables for the calculation will be RPM at a diameter of 10” and the OD of the tire at 16” all on the same wheel.</p>
<p>How do I calculate the speed (MPH) of the tire?</p>
<p>I would imagine the OD of the tire runs slower than the shaft.</p>
<p>I hope this makes sense.</p>
| |mechanical-engineering| | <p>Whatever RPM your sensor reads is the RPM of the 16" tire as well. You say the tire is directly mounted on the shaft.</p>
<p>E.g., if your sensor is reading 100RPM your tire is turning the same and the ball is ejected theoretically, depending on geometry of the shoot and ignoring spitting added acceleration,</p>
<p><span class="math-container">$ v= 16/12 *\pi*100= 133*\pi (ft/m) $</span></p>
| 33590 | Calculate MPH on the tire based on RPM of the hub? |
2020-03-11T00:47:32.020 | <p>I have some issues regarding a lead screw.</p>
<p>It's used as the main driving system for a scissor lift which requires about 2Nm torque to operate.</p>
<p>This is the exact part if it matters: </p>
<p><a href="https://www.amazon.co.uk/TOOGOO-Optical-Housings-Aluminum-Coupling/dp/B07PFQKQV6/ref=sr_1_2?keywords=200mm+opticalis&qid=1583887143&sr=8-2-spell" rel="nofollow noreferrer">https://www.amazon.co.uk/TOOGOO-Optical-Housings-Aluminum-Coupling/dp/B07PFQKQV6/ref=sr_1_2?keywords=200mm+opticalis&qid=1583887143&sr=8-2-spell</a></p>
<p>due to space constraints in the design I have to use a universal joint and place the motor at a very slight angle to clear a few things:</p>
<p><a href="https://www.amazon.co.uk/sourcing-map-Rotatable-Universal-Steering/dp/B07H9ZKBSB/ref=sr_1_4?keywords=6mm%2Bto%2B8mm%2Buniversal&qid=1583887292&sr=8-4&th=1" rel="nofollow noreferrer">https://www.amazon.co.uk/sourcing-map-Rotatable-Universal-Steering/dp/B07H9ZKBSB/ref=sr_1_4?keywords=6mm%2Bto%2B8mm%2Buniversal&qid=1583887292&sr=8-4&th=1</a></p>
<p>Regardless of how much I tighten the grub screws, they will eventually come loose (only on the lead screw end, not the motor end) and slide over the the lead screw. I've tightened them so much that even one of them became stripped and I had to replace it.</p>
<p>I have a bottle of medium strength loctite, does using that for the grub screws solve the issue or are there better ways of doing it? </p>
<p>Another idea was using a file to make a flat spot on the lead screw but I'm not sure how well that'd work or if it makes things worse. Since I only have 1 lead screw, I thought it'd be best to ask before destroying it.</p>
| |mechanical-engineering|materials|design| | <p>There are a range of options for such a connection. I would put loctite for a plain grub screw at the bottom of the list. Depending on the safety aspects of the joint breaking loose, it might be very important for it not to slip. Your other options (in rough order of reliability):</p>
<ul>
<li>Grub screw to a flat-spotted shaft</li>
<li>Grub screw into a drilled and tapped hole</li>
<li>Through-hole with a tapered pin</li>
<li>keyway and key</li>
<li>Spline fit or other shape-matched fit</li>
<li>welded-on connectors (like an automobile)</li>
</ul>
| 33592 | Driving a lead screw using a universal joint |
2020-03-13T20:46:02.053 | <p>When I cut aluminum bar or t-slot extrusion on my bandsaw I am not happy with how "square" this edge is. I don't have a milling machine at home, and I have been told grinding aluminum is a very bad idea. </p>
| |aluminum| | <p>Small very sharp teeth, hard blade, not too fast or the blade will flex, using wax was a good idea or plenty of lubricant. Make sure the blade is cutting perfectly square or you'll be chasing your tail. I'm the QM at an aluminum extruder (30 years), we cut millions of lbs a year (6xxx & 3xxx alloys). We have always used a lubricant called Boelube, made by Orelube Corp, it has a picture of a plane on the label. Dealing with burrs is tough. On our automated saws, dedicated to cutting parts for a major auto OEM, the blades are hundreds of dollars and we have to have them sharpened maybe every other month, but we can keep burrs below 0.010" on some of our more "regular" blades, they don't cost as many hundreds as the better blades. If you're just trying to get a flat edge, you might try sandpaper on a perfectly flat surface. Most aluminum sands relatively quickly and you're more protected from over-doing it.</p>
| 33637 | How can I square off a piece of aluminum bar after cutting without a milling machine? |
2020-03-13T20:55:34.880 | <p>I have a real mechanical system (ball & beam) which I want to control. The system is approximated by the second order transfer function: </p>
<p><span class="math-container">$$ T(s) = \frac{b_0}{s^2+a_1s+a_2} $$</span></p>
<p>and in state-space form:</p>
<p><span class="math-container">$$ \dot{x_1} = x_2 $$</span>
<span class="math-container">$$ \dot{x_2} = -a_1x_2-a_2x_2+b_0u $$</span></p>
<p>where the parameters <span class="math-container">$\ b_0,a_1,a_2 $</span> have been identified, so they are known. I have designed two controllers. The first one is a full state feedback controller described by the equation:</p>
<p><span class="math-container">$$ u = -Kx+k_rr $$</span></p>
<p>where the term <span class="math-container">$\ k_rr $</span> is because I want input reference <span class="math-container">$\ (r) $</span> tracking. The other controller is a dynamic state feedback controller desctibed by the equations:</p>
<p><span class="math-container">$$ u = -Kx-k_iz $$</span>
<span class="math-container">$$ \dot{z} = x_1-r $$</span></p>
<p>Both controllers are doing well but the only problem I have is when the ball reaches the reference point there is a bit of oscillation of the system (the beam oscillates) in order to keep the ball at the desired point. I have found out that the oscillations occur due to the measurement of the ball's position and not because of the gains of the controllers because this happens even if I make these gains smaller. The sensor used to measure the ball's position on the beam is a linear potentiometer and it appears that the measurements include some unwanted noise which results to rapid changes of the control signal which leads to the oscillations. I have tried to reduce the noise by using a low pass filter with the below transfer function:</p>
<p><span class="math-container">$$ F(s) = \frac{15}{s+15} $$</span></p>
<p>The block diagram of the control loop is:</p>
<p><a href="https://i.stack.imgur.com/SxGHV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SxGHV.png" alt="enter image description here"></a></p>
<p>The oscillations have been reduced but not achieved a satisfactory level. Is there any way to modify these controllers in a way to affect these oscillations and reduce them ? Would a PID-controller perform better due to the existence of the derivative of the error ? Or should I design a better filter for the noise reduction which should also be fast (because the system is functioning in real time) enough ? </p>
<p>I place some plots for two different cases (PD-controller gains) where the actual position (output) is shown compared to the reference and the control signal. Note that the sudden changes to the real system's output are due to external disturbances I apply in order to check if the system goes back to the reference. </p>
<ul>
<li>First Case (<span class="math-container">$\ K_p = 3.9352 $</span> & <span class="math-container">$\ K_d = 2.248 $</span>):</li>
</ul>
<p><a href="https://i.stack.imgur.com/EBYJC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EBYJC.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/vM2iC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vM2iC.jpg" alt="enter image description here"></a></p>
<ul>
<li>Second Case (<span class="math-container">$\ K_p = 2.3536 $</span> & <span class="math-container">$\ K_d = 1.6423 $</span>):</li>
</ul>
<p><a href="https://i.stack.imgur.com/In7jJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/In7jJ.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/x0I5B.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x0I5B.jpg" alt="enter image description here"></a></p>
| |control-engineering|control-theory|measurements|sensors|noise| | <p>You can minimize these peaks by using a Kalman Filter. Kalman Filter is used to estimate the next step (k+1) in each step at (k). Also, it is used for cases like this, to reduce sensor noise. There are plenty of books out there to learn how to implement a Kalman Filter.</p>
| 33638 | How to reduce oscillations (rapid changes of control signal) when controlling a real system, which occur due to noise in measurement from sensors? |
2020-03-16T09:21:52.693 | <p>I'm looking to put gradual pressure on a syringe.</p>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>Cams offer a simple way to make a variable motion profile.</p>
<p><a href="https://i.stack.imgur.com/swtoS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/swtoS.png" alt="enter image description here"></a></p>
<p><em>Figure 1. A simple cam-operated contraption can be built out of wood.</em></p>
<ol>
<li>Syringe.</li>
<li>Syringe holder. This really needs a slot from one edge to allow easy changing of the syringe.</li>
<li>Drive-rod guide lower.</li>
<li>Drive-rod guide upper.</li>
<li>Drive rod. This and the two guides ensure that the syringe only sees vertical motion on its plugner.</li>
<li>The drive cam. The radius on this changes to give the motion profile you require for a constant velocity rotation of the cam.</li>
<li>Cam axle.</li>
<li>Cam handle.</li>
<li>Assembly supports.</li>
</ol>
<p>Instructions for use:</p>
<ul>
<li>Rotate cam handle anti-clockwise to the horizontal position (9 o'clock).</li>
<li>Lift the drive-rod and insert the syringe.</li>
<li>Rotate cam handle clockwise at even speed to 3 o'clock position.</li>
</ul>
<p>Design:</p>
<ul>
<li>Get out your protractor, some paper and a pencil.</li>
<li>Measure the stroke of your syringe. This is the difference between the cam's minimum and maximum radius.</li>
<li>Draw a semicircle on the paper with a radius of, say 50 mm.</li>
<li>Draw radii every 10° from the centre out past the semi-circle.</li>
<li><p>On each 10° line mark a point as many mm away from the semi-circle as you want the syringe depressed at that angle. So, for an even dispense on a 90 mm syringe you would have a point at 50, 55, 60, ... 135, 140 mm giving you a stroke of 90 mm in 180°. Draw that first as your reference and then you should be able to work out how to get the profile you want. Your profile just has to:</p>
<ul>
<li>start on the same 0° start point as we've constructed above.</li>
<li>be always increasing.</li>
<li>finish on the 180° point we've constructed above.</li>
</ul></li>
</ul>
<p>Have fun.</p>
<p>The image was created with <a href="http://onshape.com" rel="nofollow noreferrer">OnShape</a> which is an online, in-your-browser 3D CAD package. Use is free if you are prepared to make your designs public. It's all rather amazing!</p>
| 33666 | Changing (alternating non linear) screw motion |
2020-03-17T14:38:06.813 | <p>I've been trying, as a thought exercise, to run some calculations on a floor jack-like mechanism that is expected to lift 130kg (287 lbs). Dimensions and setup are as per the image below. </p>
<p><a href="https://i.stack.imgur.com/XpAYi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XpAYi.jpg" alt="enter image description here"></a></p>
<p>The 130kg load is to be positioned at the end of the boom, to simplify things, I assume the load would be at 90 degrees relative to the boom. My understanding, is that the force would be greatest then, anyway...</p>
<p>Here's my approach to try and size the motor:
1. Find the torque at point A, due to the load on the end of the boom. </p>
<p><span class="math-container">$$F = ma = 130\cdot10 = 1300\text{ N}$$</span></p>
<p>I specify <span class="math-container">$A$</span> as 10, rather than 9.8. So, 130 kg load exerts a force of 1300 N.</p>
<p>The torque, then at point A would be </p>
<p><span class="math-container">$$\tau = rF\sin\theta = 0.5 \cdot 1300 \sin(90) = 650\text{ Nm}$$</span></p>
<ol start="2">
<li>I don't know what force the linear actuator needs to produce - x. I proceed to calculate what applied force at point B would yield a torque greater than 650 Nm. Let's assume a 10% safety margin, so 715 Nm needed at point B.</li>
</ol>
<p>The angle will go from 90 degrees down, let's say to 30 degrees at the max lift point of the boom. </p>
<p>With reduced angle, the torque is also diminished, so at the highest point, the linear actuator's force applied will need to be (at 30 degree angle relative to the boom)</p>
<p><span class="math-container">$$715 = rF\sin\theta = 0.1F\sin(30°) \therefore F = 23,833\text{ N}$$</span></p>
<p>This sounds like a ridiculously high force rating for an electric linear actuator. </p>
<p>Is my approach correct? I'm quite certain I'm missing something important. Your help will be greatly appreciated. </p>
<p><strong>EDIT</strong>
Clearly, the above setup isn't ideal for an electrical linear actuator. The other lifting setup I can think of, is a mobile crane, like in the pic below. </p>
<p>Taking the same size and parameters as above, the force needed by the actuator is much more palatable.</p>
<p><a href="https://i.stack.imgur.com/eflbJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eflbJ.jpg" alt="enter image description here"></a> </p>
<p>Let's assume force is applied at a distance of <strong>40cm</strong> from the point of rotation and an initial <strong>30 degree</strong> angle. The lifting mechanism needs to counter a torque of <strong>715Nm</strong> (calculated above). </p>
<p><span class="math-container">$$715 = rF\cos\theta = 0.4F\cos(30°) \therefore F = 2,102.94\text{ N}$$</span></p>
<p>Is this correct? (ignoring the counter-weights on the actual crane) </p>
| |mechanical-engineering|torque|linear-motors| | <p>You determined the torque at A correctly. You can get the force at B using the same equations.</p>
<p>650 Nm / .1 m = 6500 N.</p>
<p>Yes, it's a lot, 5 times the load, but not close to where you ended up.</p>
<p>The force will go down with the angle, meaning you should be dividing the force by the cosine, not multiplying by the sine (.5), but then you actually used .3 instead of sin(30).</p>
<p>You can look at the device as a simple machine. If you want to directly apply force at B, you are on the wrong end of the lever, and you will be giving yourself mechanical <em>dis</em>advantage.</p>
<p>A typical jack like this overcomes the load by using hydraulic pressure to convert long throws of the handle into small movements of the lifted load. Other options for jacks are screw scissors, which use the screw as a simple plane to trade large amounts of rotary motion into linear motion, or ratchet jacks which just use levers to move the load into a higher ratcheted position.</p>
<p>If you want to use some type of motor, you would typically determine a method of mechanical advantage that let you lift the load within the desired time period. A earth-moving crane wouldn't notice your load because it's a huge machine designed for moving tons at a time. Similarly, a clockwork-like mechanism would take years to move such a load.</p>
| 33688 | Sizing a motor for a floor jack-like mechanism |
2020-03-18T02:45:43.473 | <p>I'm looking for a very very thin wire that does not bend easily. Which kind of material should I look for? The thinnest the wire the better but copper <a href="https://imall.com/product/50m-0.01mm-Insulating-Copper-Wire-BGA-Motherboard-Fingerprint-Maintenance-Fly-Line-Solder-for-iPhone/banggood.com/1369196/530-225138/en" rel="nofollow noreferrer">can be found</a> in a very thin format (0.01mm) but it bends with very few force applied to it.</p>
<p>What is the least ductile and thinnest wire material I can buy?</p>
<p>Thanks in advance</p>
| |materials|wire| | <p>For microscopic electrical probe pins, gold-plated <em>tungsten</em> is used because its stiffness modulus is large- which provides best resistance against buckling under compressive stress.</p>
| 33697 | What is the less ductile and thinnest wire material? |
2020-03-18T05:54:03.410 | <p>I'm designing a block for creating turbulence inside a traditional furnace. I would like to know about a material which is economical and can be put into operation for more than 2 years.</p>
| |materials|manufacturing-engineering| | <p>Most industrial furnaces are lined with <a href="https://theconstructor.org/building/fire-bricks-properties-types-uses/29377/" rel="nofollow noreferrer">fire bricks</a> to ensure long operating times at high temperatures. Depending on the properties of the bricks, particularly the grade of the <a href="https://en.wikipedia.org/wiki/Fire_clay" rel="nofollow noreferrer">fire clay</a>, temperatures up to 1640 <span class="math-container">$\small\sf{^o}$</span>C can be tolerated.</p>
<blockquote>
<p>Fire clay contains two major constituents- silica and alumina, of which, the silica percentage varies from 60 to 70% and alumina varies from 25 to 35%</p>
</blockquote>
| 33700 | What material can withstand continuous working temperature of 1000 degree Celsius? |
2020-03-18T07:07:29.163 | <p>I am testing out different materials with different specific heat capacities and their correlation with soundproofing (e.g. wood, cork, styrofoam, etc.) but I can't seem to find the specific heat capacity of cardboard anywhere (the average S.H.C. of regular corrugated cardboard).</p>
| |thermodynamics| | <p>Review this link:
<a href="https://help.iesve.com/ve2018/table_6_thermal_conductivity__specific_heat_capacity_and_density.htm" rel="nofollow noreferrer">https://help.iesve.com/ve2018/table_6_thermal_conductivity__specific_heat_capacity_and_density.htm</a></p>
<p>"carboard" is no such thing - I assume that you are referring to Corrugate = Kraft Liner Board with a Kraft Flute medium. There are many more reliable materials for soundproofing than corrugate.... though there is liner with foam fluting/face stock.
If you're stuck on corrugate, remove one face liner to expose the medium - this will give you the sinusoidal configuration to bounce your sound waves.</p>
| 33702 | What is the specific heat capacity of corrugated cardboard? |
2020-03-18T11:04:41.310 | <p>I am using GA and MCMC for estimating the parameters of a transfer function. GA returns results that each parameter has one value when minimizing the error function, and MCMC returns the whole chain results for each parameter after the burnin, which gives a statistical distribution of the possible values for that parameter. </p>
<p>My first question is: What is the benefit of using MCMC over GA? Which one is better than other in what aspects? </p>
<p>My second question is: Do I get the mean value of each parameter from the distribution of the chain in order to get the fit for the transfer function? If so, do I do that with mean taking the mean value of the distribution? what if the distribution is skewed?</p>
<p>Thank you in advance </p>
<p>sorry I am new here, I didn't know how to tag this question.</p>
| |control-engineering|transfer-function|statistics| | <blockquote>
<p>Question 1: What is the benefit of using MCMC over GA? Which one is better than other in what aspects?</p>
</blockquote>
<p>As you state, if you are using the <em>Markov Chain Monte Carlo (MCMC)</em> properly, it will give you the complete probability distribution of your estimated parameters.
This gives you knowledge about the quality of your estimation, but this is a computationally expensive method.</p>
<p>The <em>Genetic Algorithms (GA)</em> will converge to a minimum of your cost function, which can be local minimum.
Furthermore, there is no knowledge about the quality of the estimation, rather than the cost-function value.</p>
<blockquote>
<p>Question 2: Do I get the mean value of each parameter from the distribution of the chain in order to get the fit for the transfer function? If so, do I do that with mean taking the mean value of the distribution? what if the distribution is skewed?</p>
</blockquote>
<p>If taking the mean-value of the distribution is a good approximation of your parameter depends on the type of distribution. I would state the in general the bin with the highest probability is the best approximation of your parameter.
However, if you applied MCMC to estimate your parameters, you can use this technique again to estimate the system response.</p>
| 33706 | Compare Genetic Algorithms and Markov Chain Monte Carlo in their statistical properties in parameter estimation task |
2020-03-18T13:17:58.200 | <p>I have biomedical data for analysis and I am using Genetic Algorithm for parameter estimation for the model of the data. As I read about the system identification, I saw terms like dynamical system. I read further and I am still confused whether the electrical impedance readings of a muscle tissue is considered a dynamical system or not.</p>
<p>Is it a nonlinear system? how can I tell? is there a direct rule on whether a biological tissue is a nonlinear system or not?</p>
<p>Thanks in advance</p>
| |biomedical-engineering| | <p>Generally speaking, any mathematical model that is time dependent will be considered a 'dynamic' system, some engineering descriptions will say dynamic systems are described as systems having the ability to store and release energy. </p>
<p>From <a href="https://en.wikipedia.org/wiki/Dynamical_system" rel="nofollow noreferrer">Wiki</a></p>
<blockquote>
<p>In mathematics, a dynamical system is a system in which a function describes the time dependence of a point in a geometrical space. Examples include the mathematical models that describe the swinging of a clock pendulum, the flow of water in a pipe, and the number of fish each springtime in a lake.</p>
</blockquote>
<p>As far as nonlinear things go. Most real world systems are nonlinear. However, it's difficult to say immediately whether something is, before one goes on to model it.</p>
<p>But to give an example from the modelling of a tentacle:</p>
<p><a href="https://i.stack.imgur.com/2gUVi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2gUVi.png" alt="enter image description here"></a></p>
<p>Biological real world systems get complicated fast, and end up being inherently, nonlinear.</p>
<p>However we often linearize such systems around a operating point (<a href="https://en.wikipedia.org/wiki/Taylor_series" rel="nofollow noreferrer">Taylor Series</a> for example) once we have a clear understanding of what it may be. Will will drastically reduce your problem with a given amount of (hopefully) small error. </p>
| 33707 | Is biological tissue a Dynamical System and a Nonlinear System? |
2020-03-19T17:12:51.113 | <p><a href="https://i.stack.imgur.com/3mdkW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3mdkW.jpg" alt="enter image description here"></a></p>
<p>When a crack has been initiated in a material, further loading of the material causes stress concentrations at the crack tip. This high stress causes the material to break at the tip and the crack grows. As the material is loaded again, another stress concentration occurs and the crack grows again. <strong>But why do we need to unload the material in between?</strong> Why doesn't the crack just propagate immediately from the initial notch? </p>
| |materials|stresses| | <p>When the material "breaks", it's the inter-molecular bonds that are breaking. Once they are able to form a new bond to the next available molecule the crack propagation stops and stress is reduced. </p>
<p>Once the load is removed, the material is deformed slightly from it's previously unloaded state. Adding load again creates a similar stress compared to the first load cycle causing another cycle of fracture.</p>
<p>Look up brittle vs ductile fracture, those sources will explain why some materials fail instantly while others fail after high number of cycles.</p>
| 33721 | Why does fatigue crack growth require unloading? |
2020-03-20T17:25:01.053 | <p>This is a follow-up on my previous question here:
<a href="https://engineering.stackexchange.com/questions/33688/sizing-a-motor-for-a-floor-jack-like-mechanism">Sizing a motor for a floor jack-like mechanism</a></p>
<p>The floor jack setup is clearly inefficient when it comes to electric actuators. Something similar to a mobile crane would work better, as I'd benefit from the mechanical advantage of the lever formed between point of rotation and actuator.</p>
<p>My question is: can the actuator be anchored behind the point of rotation of the lifting arm, e.g. </p>
<p><a href="https://i.stack.imgur.com/qilFf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qilFf.png" alt="enter image description here"></a></p>
<p>NB: 400 mm is the linear actuator's install length. The stroke is 300mm.</p>
<p>If my understanding is correct, the system should have significant mechanical advantage?</p>
<p><em>EDIT</em>
Would this setup be similar to what you see on skip trucks? E.g.
<a href="https://i.stack.imgur.com/Bpvi2.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bpvi2.gif" alt="enter image description here"></a></p>
| |mechanical-engineering|torque|linear-motion| | <p>Let's call the point where the hinge is welded to the boom D.</p>
<p>Scaling CD from your diagram we pick 30mm.</p>
<p>That means the mechanical leverage of this gadget is approximately </p>
<p><span class="math-container">$$F_{boom}= F_{actuator}\frac{30}{400}= 0.075F{actuator}$$</span></p>
<p>This gets even worse because we need to consider the reduction of leverage due to the location of D, a nearly ratio of 1/20 efficiency.</p>
<p>And on top of that, the boom will lock up after a small rotation where the actuator will hit the boom <span class="math-container">$\theta=\arctan(30/144)= 11.7degrees $</span></p>
<p>By increasing the length of CD and changing the location of the pivots we can improve the leverage. </p>
| 33731 | Linear actuator behind lifting boom |
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