# Motor Torque Question

Our team is working on our winch and are at the motor selection phase. We realize a gearbox will be necessary on any of the motors, but we have a question about torque:

A CIM Motor on the VEX site has a stall torque of 2.41 n/m which is roughly 21 in-lb. Am I correct that this means the CIM motor will stall with a 21lb load 1in from the rotation axis? If we add a gearbox of say 12:1, would we then be able to lift 252lbs max?

I don’t have a computer with Excel at hand right now to check those numbers (thanks Chromebooks), but I’ll link JVN’s gearing calculator for you to check your numbers with. I don’t know if this is the most up to date version available right now, but it should be fine for your purposes (it’s not like motors have drastically changed over the past year or two). Remember that when you’re playing around with gearing ratios, you should aim to lift at 25% of your stall torque to maximize efficiency and minimize the potential for failure.

https://www.chiefdelphi.com/media/papers/3188

You correctly understand how the gear reduction and unit conversion works, in an ideal case at least. But if you have a stalled motor then you are going nowhere - so you will need to consider the torque and current at the speed you are climbing at.

Also (and I am not sure about the impact of this) the breaker for the CIM is rated at 40A, so I think that means you can’t reach the full stall torque for any extended period of time. At 40A the torque is only 6.25 lb-in.

TLDR use the calculator.

Yes, 21 lbs at 1 inch is what it takes to stop the motor itself. Yes, the gear ratio of 12:1 will multiply that torque by 12. This means that an exactly 252 pound robot could theoretically be held in the same spot, but it would not climb. A 251 lb robot might climb, but very very slowly and with very very hot motors.

Remember too that friction is always present. The gearbox, bearings, rope, etc. together will exert some force counteracting motor motion. This lowers the weight liftable by the winch.

In cases like this, I think you usually want to get the best conversion from electrical energy into mechanical potential energy (aka climb). There’s lots that go into this, which is where the calculator really helps. Peak power is the motor operating point that I think most folks shoot for.

Also a very valid constraint. You have some flex here: quarter of a second at 5x that or two seconds at 1.5x will probably be ok (depends on the fuse exactly, but they’re rarely instantaneous-blow). But, if your design eats 100A per motor for ten seconds, that’ll probably make some electrical components very sad.

The fuse datasheet usually has time-to-trip vs. current charts that can be enlightning.

You are correct in your concept that the torque of the motor must be de-rated for the breaker providing the power. Given the fundamentals demonstrated in the OP’s post, I will add that I recommend designing most mechanisms in FIRST to use no more than 50% of the breaker’s current limit in cases where the motor is able to trip the breaker.

If designing a mechanism to perform an action with a single CIM you will be successful if you design for the CIM to provide no more than 0.337 Nm (2.983 inlbf). If you design with a way to adjust your gearing then you may be able to reduce your reduction after testing. It is better to get a mechanism that will work up and running then to optimize and miss.

Shoot for a point lower than peak power. If you need more power to cover for an inefficiency in your design, you won’t have it if you’re already at the peak.

+1

Don’t design a robot believing that it will perform precisely in tune with anything you see on JVN or while planning. There’s always that margin of unpredictability to take into account. So if you’re trying to lift 252, aim for larger than 12:1.

I’ve always done my calculations using stall torque and free speed… and then adding in a safety factor of 50%. That helps account for things like friction, and puts the motor in a better operating range. Based on my experience with climbers in the past, I wouldn’t go any less than 20:1 on a CIM motor with a 1" diameter drum.

This year you may want to hold something in position (certain height/angle). This effectively stalls a motor (which runs current through it, heating it up). Holding at a position usually consumes a proportion of the maximum power. For example, say that the ‘hold’ works at 25% output power when your battery is at 12V. Then the motor will effectively have 3 volts as its input.

Vex has motor data which shows how long a motor can remain stalled at various voltages (see the “locked rotor” graphs for each motor). At 2-3 volts, most motors do not degrade over time while remaining stalled. This is useful.

If we design our mechanisms to “hold” their load at somewhere between 2-3V, we can be reasonably confident the motor will not burn out, ever. This means that, in a design spreadsheet like JVN’s, if your actual load is less than 25% of the stall load, you can be reasonably confident the motor will not burn out, ever. From there you’ll have to play with gearing and motor types/quantities in order to get your desired outcomes. Adding more motors can reduce the amount of torque load on a single motor, therefore increasing the stall:actual load ratio.

I agree with Jon that a 50% margin of error in potentially-high friction mechanisms is a good idea.

I’ll offer the opinion that blindly using a calculator without understanding why it works is a bad idea, both in that it is likely to lead to engineering mistakes and it cheats you out of important foundational knowledge that enables the tackling of more-complicated problems.

The most fundamental property of permanent-magnet DC motors is their linear torque-speed curve:

The above plot is called a motor curve. Learn them, love them, internalize them - they are one of the most important design tools you will encounter in FRC. The x-axis is motor speed. The yellow line represents motor torque.

As speed increases, torque decreases linearly, and vice-versa.

As power is proportional to the product of speed and torque, a little bit of math (hint: look at the behavior of the function f(x)=x(1-x)) reveals that the power is a quadratic function of motor speed (and also of motor torque), and peaks at 50% of free speed (equivalently, 50% of stall torque).

In general, it is good practice to design your mechanisms to place your motor loads on the “rising side” of the power curve - that is to say, less than 50% of stall torque (equivalently, greater than 50% of free speed). This ensures that if you have made a mistake or there is a disturbance and your load increases from what you have calculated around, the motor will output more power, not less.

There is an additional concern: current draw is proportional to motor torque. The stall current of a CIM is exceedingly high - not only will a stalled CIM almost certainly trip the 40A breaker in short order, but due to the limitations of the FRC battery it will almost certainly not actually be able to ever draw full stall current or output full stall torque. In general, you want to keep current draws fairly low; for a CIM, a good rule of thumb is to never design around more than 40A continuous draw (~30% of stall current/torque) unless you’re sure you know what you’re doing.

Meh.

When we try to push the limits (e.g. minimize time), we need to understand things. But for a “just make this thing go without blowing up” kind of situations, calculators are pretty good as-is. They worked well enough back in the day before I understood things, and for most mechanisms they still work well enough to get “very” close to a final design using some simple rules of thumb.

I guess I’m old, too.

I’d rather understand something, and use approximate numbers, than use a calculator to get exact (but probably meaningless) numbers.

Our usual design strategy is to figure that the motor is happy running at around 80% of it’s free speed, producing about 20% of it’s stall torque. So we design the mechanism to run the motor like that. Then we go back and check things, to make sure it’s going to work like we think.

Keeping the motor on the right side of the power curve…yeah, that’s a good idea! but you have to know what the power curve is, to understand it.

That’s a great rule of thumb.

So we design the mechanism to run the motor like that.

Hey Jim, for students who are reading your post and want to take your sage advice, can you provide a few more details on how you do that?

Then we go back and check things, to make sure it’s going to work like we think.

I’m lovin’ what you’re saying.

Keeping the motor on the right side of the power curve…

By “right” side I’m sure you mean “correct” side. On a “standard” speed vs torque curve, the “right” (correct) side is the left side

yeah, that’s a good idea! but you have to know what the power curve is, to understand it

Are there any students (or mentors!) reading this who want more details?