Quote:
Originally Posted by nickcvet89
Hey, I am having a hard time determining output speed and torque from the CIM-U-LATOR gearbox using the 2 RS-550 banebots motors. Can anyone explain what formulas I need to determine this? All help is appreciated.
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Banebots 550-12:
19,300 rpm free speed
70.5 ozin stall torque
85 stall amps
CIM-U-LATOR:
Gear reduction is 2.7:1
... so 2 550's on a CIM-U-LATOR would be
free speed = 19,300/2.7 = 7148 rpm*
stall torque = 2*70.5*2.7 = 381 ozin*
stall amps = 2*85 = 170 amps
Is that what you were asking?
* neglecting gearbox efficiency losses