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.