Using a AM planetary gearbox, we found it possible to hook up 1 Cim and 1 FisherPrice motor to a super-shifter. While the torque is no where near the same, the rpm is. So theoretically, we could effectively use only 2 Cim motors in our drive system. Right?
Good? Bad? Input? :ahh:
Technically it will work.
I don’t have the numbers to back this up but… I would much rather use a CIM in drive compared to the FP.
The CIMs are tougher and can take the continues duty of drive better then the FP motors.
See this thread: http://www.chiefdelphi.com/forums/showthread.php?threadid=99795
If you, by some chance, stall your drive, you will cook the fisher price motor. E.g. Too high a gear ratio, or get pinned or " bumped". In addition, you lose power as the torque curve characteristics do not match, and present some loss in power in combination. The 775’s would be a better choice.
WHICH FP?
The FP 00801-0673 is a 290W motor.
The FP 00968-9015 is a 180W motor.
The FP 00968-9013 is a 170W motor.
You are pairing this with a CIM which is a 340W motor.
SO… you can match the free speeds but if you do, then at high loads, the FP will be struggling more than the CIM (for struggling read closer to its stall torque… …which means more heat generated which when coupled with the fact that the 550 series can of a FP has much much less surface means HOT HOT HOT – actually, since I think the FPs have thermal protection these days, means that the motor will be cutting out at high loads).
OR… you can match stall torques but if you do, then at low loads (i.e. free speeds) the FP will be slowing down the CIM.
BUT… you can’t do both.
Lower power means that you have to give up on something.
Joe J
Joe, where did you get the specs for the FP-9013 motor??
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I assumed that the 9012 in this sheet was a typo in this motor comparison chart from the FIRST. You are right that is an assumption.
On another point, a PM from Ether helped me realize I was sloppy in my thinking and typing in my post above.
Me: “for struggling read closer to its stall torque”
Not true.
The motors are tied together via a gearing that links speeds (and adds torques). So… as the speed of the two motors changes, they run the same % of their individual free speeds, so each motor is also loaded at the same % of its stall torque (i.e. they are both “equally close” in some sense to their stall torques).
EVEN SO… …I still think it is better to match stall torques.
Why? I guess it is that the lower powered motor in this case is the smaller size motor and I want that smaller motor to be farther away from its stall torque so that under high load conditions, the smaller motor is running at lower current and higher efficiency point than it would be if I matched free speeds.
Joe J.