paper: 4 CIM versus 6 CIM theoretical calculations

[InFlight]

Originally Posted by GeeTwo

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.

[GeeTwo]

Originally Posted by InFlight

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.

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.

[AdamHeard]

Originally Posted by GeeTwo

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.

[GeeTwo]

Originally Posted by AdamHeard

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

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.

[InFlight]

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.

[InFlight]

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