Do CIM elevator motors overheat?

Dear CD,

I was wondering if any teams that have progressed further into their build so far have determined that the elevator CIM motors overheat due to stalling?
I am assuming you are using a gearmotor attached to a CIM with a gear reduction ratio in the order of 50 to perhaps 90:1.

When the CIM elevator motor stops moving to hold a tote at a given height it is effectively stalled (motor speed nearly zero) and maximum current is drawn.

Have you found that your CIM elevator mount overheats in a two minute match? Were you holding one, two, three, or four totes?

Have you found it necessary to add an active braking mechanism to aid in the problem?

We were worried about this, so we have a clipping device that holds our stack at the desired height while the lifter retracts and “rests” - at least until it lifts again.

This doesn’t sound quite right. A motor’s rated stall torque is not the same as the torque required to hold a specific load in one place. If you figure out the torque required to do that with your design, then you can figure out how much current the motor will require to maintain that torque.

It might make the motor smoke…it might not.

Why use such a big reduction? Unless you are using a pulley that is about a foot in diameter, that will most likely be too much reduction, and too slow.

If an average pulley/sprocket is about 2.5 inches in diameter, a 70:1 reduction off of a CIM will have a free speed of 0.87 ft/s, and a stall force of 1200 lbs. If you had to stall a CIM to hold up 60 lbs at that reduction, you would be using 5% of your stall torque. A CIM could do that basically forever without problems.

Two years ago we were using a single miniCIM for our lifting mechanism to climb the tower. During extended testing, (lift, stall for 5-10 seconds, pause for a minute as we debug, repeat) it got extremely hot, and as a result we added a second miniCIM. The motor itself never burned out. Also, the design was worked to minimize stall time. During competition it was never stalled for more than a second on each level.

The CIMs can take a while lot of abuse, but it’s better to have a design that does not involve stalling the motor. It may work just fine for a few matches, but if you keep stalling the motor it may eventually fail… Probably in the eliminations when match turn around is so short.

If you’re planning to use a CIM for your lift, I suggest not relying on driving the motor to prevent backdriving and find some other solution to stall your lift.

Safety First.

The current draw is only what is required to produce the torque necessary to hold the weight. This in turn depends on the gear ratio.
Here’s an excellent video explaining motor selection and gearing.

I’ve seen perfectly good elevators run by a Mini CIM at a 12:1 reduction driving an 18T #25 sprocket to lift the elevator that could hold ~50 lbs without breaking a sweat. I’d suggest taking a look at JVN’s mechanical design calculator and playing with the numbers in the linear mechanism section.

You likely won’t pull it off without breakers tripping first (at least if you have some sort of gear reduction).

But if your cim’s are running hot, you can always add one of these: (hint, use some thermal paste, its legal), or just a couple of fans.

Last year we stalled a (mini)cim pretty much the entire match, every match. Never tripped a breaker.

It’s all about WHAT voltage you stall at.

Exactly. If you calculate it so you only need 10% of your stall torque to hold the object up, you only need 10% of the motors stall current to do so. Which you get if you input 10% of the nominal voltage. That becomes 13 amps at 1.2 volts, which is only 16 watts. A CIM can easily dissipate that much wattage, and will have no trouble doing that for the entire match.

Are you sure about the voltage thing? I don’t really understand the whole voltage/current/percent power relationship with speed controllers…but my guess is you’re providing it quite a bit more than 1.2 volts if you’re commanding 10% power, using PWM.

Yes, absolutely.

Model motor as a resistor, you can get it’s resistance from the spec voltage of 12V and spec stall current of 133 Amps. V=IR gets you R = .09 Ohms.

Kt is torque/Amp, and therefore stall torque scales with voltage. So, at 10% of spec voltage, you have 10% current draw, which means 10% of spec stall torque.

Stalling a motor in no way leads to failure on it’s own. Putting too high of a continuous current into a motor (whether that’s at stall, or while rotating) is what fails it (due to heat building up wrecking insulation).

You can extrapolate this a tad further and do a BS calc where you say .2 or so of the motors mass is the armature weight in copper, and then do Q = mCdeltaT, but change Q to Power input and deltaT to deltaT/sec. This lets you get an approximate temp rise per second that’s pretty conservative as it neglects ALL heat transfer paths out of the motor.

The old 884’s were not linear, so 10% duty cycle was not 10% voltage, but the new controllers are much better, and are almost 1:1.

Adams post above me does a much better job of showing how voltage, current and torque are all related.

Everything Adam and Thad said, plus this minor item, which may be key to understanding:

If you’re commanding the motor controller via PWM, you’re not commanding power, you’re commanding output PWM duty cycle.

If the motor controller is linear, then 10% command will create 10% output duty cycle which will effectively cause 10% of the battery’s voltage1 to appear at the motor’s input, which – if the motor is stalled – will create 10% of the motor’s stall current to flow through the motor which will create 10% of the motor’s spec stall torque.

The electrical power being put into the motor is Pin = currentvoltage, which will equal (10% of spec stall current)(10% of battery voltage) which will equal 1/100th of the input power at motor stall at spec voltage.

That’s why the motor doesn’t overheat.

1Let’s assume, for sake of simplicity, that the battery is a constant 12 volt source (same as motor spec voltage).

2The output mechanical power at the motor’s output shaft is Pout=shaftSpeed*loadTorque. When the motor is stalled, shaftSpeed is zero so Pout=0. Therefore all the input power Pin is dissipated inside the motor as heat.

Good catch, I overlooked that his question was more related to the motor controller’s “10%” output than to the motor itself.

Unless you can fill the CIM itself with some sort of thermally conductive (but electrically insulating) fluid, putting a heat sink on its case is less effective than you might think. The power is dissipated in parts that have very little ability to transfer the heat to the outside. To make its way to the outside, most of the heat needs to go through the motor shaft.

Richard Wallace has conducted and reported some excellent tests documenting this. I’ll post a link later unless someone else beats me to it.

We tested the CIM coolers last year and found that while they had a small (but consistent and statistically significant) effect in keeping the motor from heating up too much (on the order of 5%), they aided quite a bit in cool down time. Particularly if a fan was included. I wish our design had been better and we had included them on the final drive train. I have a friend who design motors for a living who explained that while most of the heat is dissipated through the shaft for a motor like this, the case obviously heats up so it is dissipating non-trivial amounts heat through the case. And that if you are operating the motor in a way that causes it to heat up excessively, you should try to redesign. But if that isn’t possible even small amounts of increases heat dissipation can have a positive impact on motor performance and longevity.

One concern I would have with a stalled motor is the effect on the commutator. With the shaft even rotating slowly you are spreading the current over all the segments rather than the 2 the brushes are on with a stalled motor. Other that that–what Adam said. :]