# Using 2 CIMs to lift 250lbs

With this new gearing I’m not going to be using a work gear. Backdrive is a bit of a concern, but ultimately I’ll just have to find some way around it.

Using 2 CIMs on a 14:40 reduction, at 1750 RPM, with a torque of .9 Nm/s using a 1 inch drum, it should be able to lift without stalling, at it’s peak efficiency.

Using this new layout is it possible, or will it still draw too much juice?

If I added a third CIM would it give it the push it needs? Also is it an electrical problem? It seems like it shouldnt stall since it’s drawing under it’s stall current

Calculate the power required to lift a 250 pound load using a 1 inch diameter drum spinning at 1750 RPM

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Ok lets clear up your units a bit.

0.9 Nm/s is not torque, it’s power. Conveniently 1 Nm/s = 1 watt
0.9 Nm is torque and 1 Nm = 8.85 in-lb = 0.738 ft-lb

Adding more motors will eliminate MOTORS as the tripping point but does not change the overall power requirement. It simply moves the problem to the battery and how much it can handle.

As Ether pointed out you need to change the power requirement.
Power = Work / Time
Work = Force x Distance

For linear items like pulling on a cable, work = force (like weight) * distance
For rotary items like the drum, work = torque = force (like weight) * radius of the drum

So here are the two things you need to consider. How do I get force down to something the motors can handle and how do I get the power down to something the system can handle?

Work will always be the same (before efficiency gets thrown in)
250 lb x 1 ft = 250 ft-lb
125 lb x 2 ft = 250 ft-lb (like when using a pully) Easier on the equipment but twice the distance.

Power then will only change with time (550 ft-lbs = 1hp = 746 watts)
250 ft-lb / 1s = 0.45hp = 339 watts
250 ft-lb / 2s = 0.23hp = 170 watts

How do you double/triple the time?
Twice the length with the same speed (pully system)
Same length with half the speed (gearbox)

JVN Design Calculator to the rescue:

As OP has described their new setup it won’t work:

A significant additional reduction is needed.

The JVN design calculator interface is a great way to describe these systems because you see motor type and quantity, gear ratio, spool diameter, desired travel distance, load, and all the other details that are critical in describing this type of mechanism.

We’re planning to lift one 150 lb robot with one CIM, using 64:1 reduction, and a 1" radius spool.

We expect it to take around 5 seconds to raise the robot a bit more than one foot.

We’re doing two CIMs with a 51:1 reduction and a Ø3" drum. In the magic mathematical universe (assuming 80% efficiency from the 3 stage Toughbox), we pull off a double climb (300 lbs) in one second without blowing our 40A breakers - but we are right at the limit!

Work in rotation is torque times the number of rotations in radians. It is a bit of a trick there, since radians is a dimensionless quantity.

I do not think that you’re at the limit… I think you’re at a good design point.

https://i.imgur.com/kxMWJ1E.png

We’re right at the limit in the real world, I guarantee it

How do you know? I’m curious to hear how you determined that. A calculated 30A is really gentle on breakers, even if you were pulling 60A/breaker for 1s you wouldn’t be close to tripping them.

``````0.2     linear ft/sec
30      ft_lb/sec power required
**40.7	Watts**
``````

1 CIM

can easily provide 40 Watts output power, with lots of margin.

A wide range of gear ratios will work.

Well to be fair, this is my gut instinct/murphy’s law speaking. Mathematically, yes, this is a sound design. But I am anticipating worse losses from friction in our elevator and our gearbox, there’s a solid risk that our motor pinions will jump teeth, our elevator rigging is a likely weak point, the design we have for our robot 2 holder puts us at a mechanical disadvantage, and at the end of the match our battery won’t be giving us 100%. Plus you have to consider that lifting a robot is a dynamic scenario, while the math we use is steady-state. I expect that the compromises we had to make in our actual assembly will quickly eat up any margin we have.

Anyway, I’ve probably just seen too many robots go wrong in my career…

EDIT: The point is that you can’t just say “oh motor gearbox doo doo dah jvn calculator” and spit out a working system. You have to think outside of the (gear)box, as it were.

EXTRA EDIT: I haven’t told my students about any of these fears yet; I’m hoping it just works and then they will believe anything I tell them from here on forward.

Dang, nice catch there.

What I should have said was:
Work = Torque x Linear Travel Dist. / Radius of the wheel

Revs required = LTD / (2pi x Radius)
Work - Torque x Revs x 2pi (2pi cancels out leaving LTD / Radius)

We did the math last year on our rope climber, which was a very rare one that actually climbed the rope (the 1" field rope). We had an incredible amount of friction, so we had to add another motor, even though the math said it would work fine, and hardly load the motor at all.

Yes, you do have to take the real world into account.

I’m curious, were you able to pinpoint a source of the friction, and how are you avoiding it this year?

No one is suggesting that the ‘real world’ be ignored.

However, when I see a design that looks quite reasonable and effective but is being called ‘right on the edge’ it piques my curiosity. Guessing at failure modes like ‘the CIM

pinion will skip’ is even more interesting.

It was a totally fair question and I realized I shouldn’t have said that the design was right on the edge in the same way that we were discussing the original question. It’s also true that I’m making many worst case assumptions that haven’t been proved out yet.

The CIM pinion skipping is because we were forced to put it at the very end of the motor’s output shaft, and I don’t feel that good about the tolerances in our mounting plate. I think it’s a totally valid concern in our case. Oh, also we’re not using a planetary gearbox so it’s not as stable. That was a resource driven constraint.

We are about to test it this afternoon so I’ll come back and let you know how good my gut’s guess was.

Yes, it was due to using the field rope. This year we are using a little tiny string, which has a straight shot. Hopefully we will have enough code in the robot to test it today…

Last year we ran a CIM on a 20:1 reduction on a 1" spool to climb. Our only issue happened when the strap traveled to far to the side and got bound up on itself, jamming the whole thing. oops.

We’re going with the same 20:1 reduction this year, with a much better spool that won’t have that same problem. That gives us a calculated 1 foot travel time of under 1 second. The calculator says it’ll have a current of about 50A, which the breaker datasheet says will take 5 seconds before tripping… that’ll be enough to let us get up and engage our ratchet to lock us in place. If we have any issues, we can easily swap out one of the stages to bring us up to a 25:1 reduction, which brings the current down just under 40A, according to the calculator. I don’t expect to need that, though.

This is going to be the third time we’ve climbed with roughly same reduction and same spool size, so we’ve got a lot of confidence in it!

Alright, the proof is in the pudding: video link

This is our robot. Again, single CIM, 51:1 reduction, 3" drum. We’ll be adding a second motor later so we can do a double lift.

The predicted speed was 12 inches in just under 1 second, you can see that it takes probably twice that. The robot is underweight in this video. I’m running the motor right off of the battery and the battery is definitely on the low side of 12v. There’s some jumpiness because I can’t hold the connector in place. Mmmmaybeeee with a juiced up battery and real motor control we could get 1 second but I really doubt it.

On the other hand, it’s a lot smoother than I expected, so take that, Mr. Murphy!

My conclusion: doing the math right helps a lot. Having margin helps even more.