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Calculating Linear Force from Motor Torque?
We've been wondering how one goes about calculating linear force from motor torque.
For example, if you have a motor of X torque, with a spindle of radius R attached to it, how can you determine the linear force (ie tension in the string being rolled onto the spindle)? That's just an example. In reality, our setup is such that the motor has an 40 tooth gear to a 80 tooth gear, which is then on the same axle as a pinion gear which moves across a rack for our linear slide. We're trying to find out how much this rack and pinion can lift; the motor's torque is 283 oz-in. |
Re: Calculating Linear Force from Motor Torque?
I'll give a half answer, as I wager you can figure it out from there (and you gain more insight that way!).
Torque = Force x Distance This is the perpendicular distance from force to the point of rotation. The next step of course is to determine whether or not it's appropriate to use the motor's stall Torque for such calculations. And if not, what Torque would you use? |
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So assuming the motor has torque of 283 oz-in, and a shaft diameter of .25 in (thus radius of .125), if you were to simply wrap a string around the shaft and attach a mass to the end of it, would the rotation of the motor shaft produce a linear force of (283 / .125)? It just sounds unreasonable because the result is 141 lbs of force. |
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So your reasoning is valid thus far, but what is the consequence of using stall Torque? What is the current draw? How fast does it move? |
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By what means can I calculate a given torque at a given speed, then? The motor is 152 rpm. If I wanted, say, 80 rpm, could I find the percentage of total RPM (80/152) and multiply that by the stall torque to find corresponding torque at the given speed? I'm assuming speed and torque change by the same factor, proportionally to the drop in voltage. |
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Assuming constant voltage, T = -Ts/wf * w + Ts. T = torque w = speed (in the same units as wf) These are both variables. Ts = Stall Torque wf = Free Speed These are both constants (and linearly scale with voltage). So, to give the last tidbit. An accepted practice is to determine the max load the lift must lift and/or hold. You'd then want to choose your gear ratio and drum diameter such that this corresponds to some percentage of stall torque. The lower the percentage, the less heat you'll make and the longer the motor will last (but the slower you'll go). Depending on the time it needs to run, and how much it has to hold still (Versus just lift up, then down quickly) anywhere from 10%-25% are valid numbers for FRC. Be careful trusting the word of others (including me) too much on that %age though. It varies motor to motor and is a function of how much heat is made to how much heat the motor is able to dissipate from the armature. |
Re: Calculating Linear Force from Motor Torque?
Gotcha, thanks so much!
Just a quick question. Right now we have it set up such that the motor has an 80 tooth gear, turning a 40 tooth gear, which is on the same axle as a rack and pinion gear (32 tooth I believe). I feel like the calculation is a bit different in this case. My assumption is that I can say the motor torque "doubles" from the gear ratio, lending to a 283 * 2 oz-in torque on the axle w/ the 40 tooth and pinion gear. At this point, do I use the radius of the pinion gear to determine linear force on the rack and pinion? |
Re: Calculating Linear Force from Motor Torque?
You can't just use a stall torque calculation for your designs. With electric motors, the torque is much less when the motor is spinning. You need to find a motor curve and estimate how fast/how much torque the motor will generate. If you're using any sort of gear reduction, you need to take into account that some energy will be lost there. Also, you need a significant safety factor.
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For a Given gear ratio N, To = N*Ti and wo = wi/N. To = output Torque Ti = input Torque wo = output speed wi = input speed I'd wager you already knew that, but I figured others reading might not. Also, for multiple stages of gear reduction, lets say N1... N4. The total reduction Ntot = N1*N1*N2*N4. EDIT: Magnets is correct that each stage of gearing will cause some energy loss. However I do disagree that a significant safety factor is always necessary. |
Re: Calculating Linear Force from Motor Torque?
OK, so from the gear ratio, we calculate a new torque of 35.375 lbs-in.
The radius of the pinion gear is ~ 0.4 in, lending to a linear force of 88.4375 lbs at stall Torque. Are there any calculations that need be taken into account regarding the pinion gear teeth number or the length of the rack or whatnot? |
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Re: Calculating Linear Force from Motor Torque?
To find torque at a specific RPM/Amp draw (at least, for a CIM motor), Direct your attention to the bottom left of the AndyMark CIM specs Try searching your motor number and see what you can find
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Re: Calculating Linear Force from Motor Torque?
Completley off topic.. OMG so this is how you use math in FRC???? Sorry our team measures and dose basic math but nothing like this.
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JVN's Mechanical Design Calculator and Ether's Simple Motor Calculator are great tools available on these forums that can help out. |
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That said, understanding physics principles (like the ones above) can be helpful and are often a part of the engineering process, which is one of the key parts of FRC. Keep in mind that the goal of FIRST is to inspire future engineers. You aren't expected to know anything coming in. Learning engineering principles is also a part of the process. |
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Designing your systems to operate at peak effectiveness* is not absolutely needed, but it sure makes the most of what you've got.
For example: for our gathering system in 2012, the conveyer to lift the balls only needed to hoist 3 balls at a time max, and we used a AndyMark motor in its gearbox, massively overpowered for what we needed. We had it geared (post gearbox) 1:1, and it worked just snazzy. If we had done the math, we would have realized it could have lifted the balls 25x faster with ease *not necessarily efficiency - remember: your robot only has to last 2.25 minutes, pull every last amp/hr you can as long as they are doing something. Why not? |
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However, challenging our students to learn about motor curves and apply that knowledge to designing a component on the robot can inspire them, give them a real world experience and result in a more competitive design. They can learn more and be inspired more. Of course there are plenty of competitive robots each year that are built without applying motor curves or math, but the ones that do use these calculations do some pretty inspiring things. If success is defined more narrowly in terms of specific design objectives for a robot and its subsystems, it is very possible to fail if motor curves or calculations are not used. We are also aiming to change culture to celebrate STEM. We should encourage doing the math so everyone can strive for greater achievements, rather than look for reason to justify not doing it because we can be successful that way. |
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This is not even to mention higher-level math. Offensive Power Rankings have been, despite their controversy, an important part of scouting for many years. Today, most teams pull the numbers off of ChiefDelphi rather than do the linear algebra themselves, but I bet 1114 enjoyed a healthy advantage in doing their own calculations before the method became well-known to the public. Another example is the drivetrain testing my own team did earlier this year. Using a graphing calculator and some basic differential calculus, I produced a reasonably accurate equations of velocity and acceleration based on distance-versus-time data. We've been able to use that model to choose gear ratios for our next drivetrain. |
Re: Calculating Linear Force from Motor Torque?
You could build a regional winning FRC bot this year with no one in your team having more than a 8th grade math education, given a few people on the team fairly familiar with FRC rules of thumb. Frankly, I think you might even be able to get away with a 4th grade math education.
However, this fact completely misses the point of FRC. FRC is one of the few places where students are actually asked to challenge themselves and apply what they've learned. It's been awesome for me to see how different things I've learned in math class (trig, differential equations, etc) apply to robot design, and it's motivated me to continue to do well in the classroom. If you want to build a well engineered robot (which isn't necessary to win a regional), these types of calculations are extremely important. In fact, you can argue that this discussion of DC motors only scratches the surface in terms of design for DC motors (current draw, changes in performance at different voltages, factoring in acceleration of the motor under load, etc are all great things to look into). Without this type of basic knowledge, you're flying blind. If you don't want to rebuild gearboxes a few times each season because you're cooking motors, do your homework, and actually look at torque requirements and safety factors. It looks like Adam's presenting the math a bit formally (which is really the right way to do it), but don't get scared by a few equations. It's honestly not that hard. Adam, although I've never built an elevator, a FOS of 4 sounds a bit conservative. I'd assume a FOS of 2 would work well from just general FRC experience. Of course, I'm sure you've had a lot more experience with elevators than me. |
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For example, why design for 25% motor load with a FOS of 2, when you can call that 12.5% motor load. It's a style preference, I just don't use FOS here (Although I use it a lot elsewhere). AS for what % stall is okay, that's tough to say and depends on how often it's used and what motor you're using. |
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As far as safety factor, what a "good" one is varies by what you're dealing with, and other similar things. If I'm designing something to carry people, I'll use a higher safety factor than if people aren't involved, for example. For an FRC robot, the more critical the system, the higher the FoS should be--to a point. (You probably want somewhere between 1 and overkill, but where overkill is depends a lot on your design, where in the design you're working, what material shape you're using, and how well you made your bumpers this year.) |
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You have to be careful with safety factors. Otherwise you start with a mouse built to government specification---A elephant.
Airplanes are another example of competing safety factors. You don't want wings falling off, but you want your lift to weight ratio much greater than 1. |
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A perfect example of where math like this was extremely helpful was the Mini-bot race in 2011. By the halfway point in the season, sub 2 seconds mini-bots began to become the standard for teams playing in the last few matches of any events, with 'traditional' mini-bots becoming almost useless by the championship. The secret* behind them was so simple, and once explained was easy to understand, but it's not something that most people would have come to without understanding the concepts in this thread. (IIRC, 973 - Adam's Team - Had the fastest mini-bot at the championship (or at least their division) that year.) With that being said, this is an awesome thread. I think I'm going to borrow some of these equations for later. :D *The 'secret' was to remove the transmissions, and use a really small diameter roller/wheel - 3/8" OD or so, IIRC. |
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