paper: 4 CIM versus 6 CIM theoretical calculations

I was just doing calculations for the resistance limiting the current again, and I got different numbers than Jim/InFlight.

I modeled it as a 0.05 ohm resistor (battery -> PDP) and then 4 or 6 parallel CIM/wiring resistors. That got me a system resistance of 0.087 ohm for a 4-CIM drive and 0.062 ohm for a 6-CIM drive, which leads to overall current draws of only 137 and 193 amps for a 4-CIM or 6-CIM drive respectively. That seems startlingly low for an entire drivetrain’s maximum current. Is 0.3ohms too much to count for a motor + wires + motor controller, or have I done something wrong?

I was using 0.39 ohms for the motor branches circuits, as I computed the CIM resistance separately. But your results are close enough, and just as valid as my assumptions. You come to the same conclusion that the actual system can’t deliver the full stall torque to each motor in either the four or six CIM drives.

Once you get moving the six CIM drive will deliver 150% of the torque; and the acceleration, and time to speed will be much better. The brownout margin is much less, so there’s no free lunch.

Out of curiosity, how would a 2CIM + 2 MiniCIM drive compare? I’ve often wondered if this was a practical weight saving option or if the performance drop would make it not worth the trouble. :confused:

Also, what are the effects of leaving the gearing on an xCIM + xMiniCIM drive identical between all motors? We’ve always just geared MiniCIMs the same as CIMs hoping to get a few extra RPM out of the drive. Is this practical or is there some downside I’m not seeing?

Do you mean 1+1 on each side compared to 2+0, or 2+2 on each side compared to 3+0?

Edit: I’m going to presume the first, as you’re discussing a performance drop. I’ll get back to you, but I seem to recall that it was a performance hit, but much closer to 2 CIMs than 1 CIM.

[strike]Edit2: By assuming a budget of 100A on one side of the drive train, the 2 CIM can deliver 627W at 3365 rpm, the 1+1 494W at 3638 RPM, and a 0+2 can deliver 425W at 3192 rpm. The output power loss is about 21% for 1+1 and 32% for 0+2. A definite hit, but if you’re looking to save weight, it’s a viable way to do it without dropping all the way to 1 CIM (247W at 1287 rpm).
[/strike] - math error!

Edit3: By assuming a budget of 100A on one side of the drive train, the 2 CIM can deliver 627W at 3365 rpm, the 1+1 542W at 3093 rpm, and a 0+2 can deliver 425W at 2648 rpm. The output power loss is about 14% for 1+1 and 32% for 0+2. A definite hit, but if you’re looking to save weight, it’s a viable way to do it without dropping all the way to 1 CIM (247W at 1287 rpm).

The free speeds are only about 10% different, so gearing the same shouldn’t be an issue. It’s not like you’re going to be able to backdrive the CIM at a higher speed than its bushings were designed for, especially after gearbox losses. If you did want to match them even better, you could just use a pinion with one fewer tooth on the mini-CIM.

Matching free speed exactly isn’t nearly as important as is made out to be on Chief, and the motors are designed to be 1:1 drop in replacements for each other without any adjustment in gearing to compensate. The main thing to avoid is one motor forcing the other one to drive faster than its free speed, which doesn’t happen (under load, things even out with these motors nicely). Totally fine to do this.

I meant the first option 1 CIM + 1 MiniCIM on each gearbox. So if I’m reading right, a 1+1 config would be about 79% of the power of a standard 2 CIM setup? If so, not a bad option depending on the game. In 2015 we went this route on our H-Drive since we didn’t need much pushing power and it was all omni wheels anyways.

I’m aware that mechanically it’s not critical, I was just curious as to how it affects the efficiency of the overall system. Presumably the MiniCIMs would be working harder since they would be trying to pull the other motors to a higher RPM, but I’m not sure if this would make enough of a difference in motor current to really matter.

I think a mix of MiniCIM + CIM at a 1:1 ratio could work fine. The MiniCIM having a free speed about 10% higher and having a lower stall current (and therefore a higher resistance) than the CIM means that when they’re running at the same RPM (say, 1,500 rpm) the CIM draws 95 amps and the miniCIM draws only 66 amps. That approximately matches up with their weight proportions, which is analogous to heat capacity, so both of the motors will heat up at the same rate. That means no dead motors.

The closer you are to stall, the more the CIM:MiniCIM heat ratio goes down (which is bad), but not by a large amount. The ratio gets closer to 1:1 as you go nearer to free speed as well, but by the time you are running that fast the heat generated isn’t too much anyway.
It’s possible to gear the MiniCIM such that it’s always running at a higher RPM than the CIM, perhaps by using 11t and 12t pinions, but it probably isn’t necessary. Maybe a team that has done CIM + MiniCIM combos can chime in here?

I wonder if using higher-resistance cables to increase your resistance would be a valid strategy to help prevent brownout of 6+ CIM drivetrains.

EDIT: The numbers for a 2 CIM + 2 MiniCIM drivetrain are as follows:
Approximated as 3.33 CIMs
2,931 theoretical rpm
333 W/CIM
1,108w total, compared to 1,280w for a 4-CIM drive, or about 86% as efficient. Not too great of a drop if you’re running a significantly lighter robot and don’t need the power.

I came up with a peak efficiency of an all-CIM drivetrain as 65%, an all-min-CIM as 57%, and a 1+1 as 61%, so no, not unless I’m missing something. The only possible issue I can think of is that when the speed is greater than the free speed of the CIM, it will be generating current rather than consuming it - I’m not sure what effect that would have on the motor controller.

Motors at or near free speed provide no useful torque. In a real world drive train you have torque losses such as gearbox losses, bearing drag, and chain or belt drive system losses. Thus it’s really not possible, particurily in a FRC limited space to accelerate anywhere near to free speed.

Typically you design your drive system around 80% of free speed which is more realistic. AndyMark always used CIMS at 4455 rpms when providing gearbox performance data in Feet per Second.

A CIM + Mini CIM per side would take about 125% of the time to speed, and distance traveled compared to a two CIM drive.

I agree that there would not be a problem while driving the robot, but there might be an issue when on blocks. I figure (using Vex’s numbers and a bit of linear extrapolation) that putting a CIM and mini-CIM nose to nose at 12V would reach a free speed of 5507 rpm, at which point the mini-CIM would be drawing 7.9A, 46W of mechanical power would be transferred, and the CIM would be generating 1.6A. The CIM would switch from consuming to generating current at about 5442 rpm, so probably even a gearbox on blocks would provide enough drag. Of course, both the CIM and mini CIM free speeds have a +/- 10% variation, so if you put a fast mini and a slow CIM together, you might have some issues when running the motors with the robot on blocks.

This wouldn’t cause any issues, you’d just be pumping some current back to the DC bus and everything would be fine.

Electrical Power and Mechanical Power are not the same.

The Electrical Power into the a DC Motor is Volts x Amps (Watts)

The Mechanical Power Out is:

Power Out = Torque*(0.112985) * Speed * (2*pi/60) in Watts

Note: 1 lb-in = 0.112985 Nm, and radian/sec 2*pi/60 = rpm

The Motor Efficiency is

η=(Power Out) /( Power in)

At near free speed, the torque and Mechanical Power approach zero.

For any of the DC motors used in First Robotics, we can also use the following equations to determine their performance:

Current (Amps) = Torque Load * ((Stall Current- Free Current)/Stall Torque) + Free Current

Torque Load = (Current-Free Current) * Stall Torque/(Stall Current-Free Current)

Speed = Free Speed - (Free Speed / Stall Torque) * Torque Load

If you didn’t have motor controllers, or (illegally) put both motors on the same controller, this would certainly be the case. Are motor controllers designed to handle the reverse bias? - Oh, right. If you decide to change from 100% speed to 50% speed in a short period of time, you’ll create the same situation, so they have to be designed that way. Never mind.

So I’m a dumb Mechanical Engineer and last year we had a 3 Cim per side shifting drivetrain with tracks. Obviously we could have swapped out a Cim for a Minicim to help out with shooter power and other items. (we didn’t and maybe should have). But let me ask this question, How does a Cim plus 2 Minicim drivetrain compare to a 2 cim (per side) in tank drive? Assuming the rules are similar this year with open availability of other motors, would it make sense to keep the 4 cim motors for whatever gamepiece needs there are? (oh and I know I’m asking for everyone to predict the future!)

For those debating the merits of the 1:1 gearing, you might want to take a look at this thread.

To address the benefits of 4 CIM vs 6 CIM vs 4CIM+2 Mini-CIM, 234 has a paper published from a few years back with experimental data.

While 1C+2m would certainly have better performance than a 2 CIM drive train, that performance gain is offset by a greater weight and volume, both of motors and of the gearbox - 3 motor gearboxes usually have a higher center of gravity or more gears than a 2 motor gearbox (or both). There’s also the matter of another motor controller. There are certainly cases where this would be better, but I expect they’re less common than the configurations already discussed.

I’ll second this, it certainly wouldn’t be the first choice for motor configurations IMO. In general there’s rarely a need to use more than 2 CIMs outside of the drive system anyways, so the number of instances you would actually need to swap out to a 1CIM+2MiniCIM should be minimal (short of maybe saving a little weight if you already have a 2CIM+1MiniCIM drive and desperately needed to loose around 1lb)