How many CIMs in your drivetrain?

[MrRiedemanJACC]

Originally Posted by Ether

Take a careful look at what I posted:



332+388 = 720 total watts consumed (one CIM)

447+273 = 720 total watts consumed (two CIMs)

They both consume the same power @ 12 volts @ 60 amps, but you get more output mechanical power and less waste heat with two CIMs.

So you don't need as many amps with the 2 CIMs to get the same output power as 1 CIM.

So if I may add to that and continue the idea with a 3 CIM gearbox and 60 amps total load on the the three motors, you use MCALC at 20 amps per CIM and get this:
CIM FR801 001, AM802 001A @ 12.00 volts
oz-in=45.6, Nm=0.322, rpm=86.7, amps=20, watts out = 155.3, watts heat is 84.7 and eff% = 64.7


So I'm seeing the three CIMs each drawing 20 amps at 12 volts:
465.9 watts output mechanical power
254.1 watts waste heat

vs these two choices that Ether's calculations from before.

one CIM drawing 60 amps at 12 volts:
332 watts output mechanical power
388 watts waste heat

two CIMs each drawing 30 amps at 12 volts:
447 watts output mechanical power
273 watts waste heat

So not as much of a gain in efficiency improvement as I had hoped, but the package is still capable of much more.