It's straightforward to do the calculation. Let's pick an operating point, say
2000 rpm at
240 ozin total torque of CIMs.
Calculate the CIM operating condition at 2000 rpm and
60 ozin (for
4CIM drive), and 2000 rpm and
40 ozin (for
6CIM drive):
Code:
Motor Calculator build MCALC_2014d 2/3/2014 1255pm
Enter rpm and ozin, separated by a space: 2000 60
CIM @ 6.62 volts:
oz-in Nm rpm rpm% amps watts out watts heat eff%
60.0 0.424 2000 68.3 24.3 88.7 71.7 55.3
Enter rpm and ozin, separated by a space: 2000 40
CIM @ 5.92 volts:
oz-in Nm rpm rpm% amps watts out watts heat eff%
40.0 0.282 2000 76.4 16.5 59.2 38.5 60.6
4 CIM mechanical watts out = 4*88.7 = 355 watts
6 CIM mechanical watts out = 6*59.2 = 355 watts
4 CIM amps total = 4*24.3 = 97.2 amps
6 CIM amps total = 6*16.5 = 99.0 amps
4 CIM electrical power in = 6.62*97.2 = 643 watts
6 CIM electrical power in = 5.92*99.0 = 586 watts
So under certain operating conditions (in this case, 2000 rpm @ 355 watts out), 6 CIM draws less total power.