Matching Drill and Fisher Price Motors

[Jon Lawton]

Okay, so I have been racking my brains for quite some time trying to figure out how to match the drill and fisher price motors. I have finally come to a solution, and am seeking peer review.

First, the motor specs:

Units: RPM, Amps, N-m, Amps

Drill Motor: 19670 FreeSpeed, 4.5 FreeCurrent, .87 StallTorque, 127 StallCurrent

Fisher Price: 15000 FreeSpeed, 1.1 FreeCurrent, .38 StallTorque, 57 StallCurrent

Then, I decide where I want to run my motors on the Torque-Speed Curve:

Drill Motor (Maximum Power occurs at > 40A):
Speed at 40A = (40-127)(19270)/(4.5-127)=13969.7 RPM
Torque at 40A = -0.87/19670 * 13969.7 + 0.87 = .252123 N-m

Fisher Price Motor (Maximum Power at 29.05A):
Speed: 15000/2 = 7500 RPM
Torque: 0.38/2 = 0.19 N-m

Then I calculated the total torque I'd need, assuming 130lb robot, 1.0 Coefficient of Friction (yes, I measured thereabouts), and 6in diameter wheels:

130lb *3in = 390In-lbs
578.5N * 0.0762 m = 44.0817N-m

Then I made up a table of successive 2:1 reductions and listed the effective torque and speed at each. I noticed that after 5 stages, the Fisher Price motor would spin at 234.375 RPM when loaded at 6.08 N-m, and after 6 stages the Drill motor would spin at 218.277 RPM when loaded at 16.128. Those speeds almost match exactly, and the torque adds up to right about 44N-m. Bingo!

So I derived:

Speed=7500/RatioFP
Speed=13969.7/RatioDrill
7500/RatioFP=13969.7/RatioDrill

Using simple algebra, I got further:

RatioDrill = 1.86263 * RatioFP
RatioFP = .536876 * RatioDrill

Okay, so since I am using two Drill motors and two Fisher Price motors, I know the total torque of these four motors must add up to 44N-m.

2*(.19*RatioFP)+2*(.232*RatioDrill)=44 N-m

Now the problem becomes simple substitution!

2*(.19*(.536876*RatioDrill))+2*(.252*RatioDrill)=4 4 N-m

RatioDrill = 62.1458:1

And, with the same method:

RatioFP = 33.3646:1

Please note, that this assumes 100% efficiency, which WILL NOT be the case. To yield better ratios, use:

2*(.19*RatioFP*.9^nStages)+2*(.232*RatioDrill*.9^n Stages)=44 N-m

Where nStages is the number of stages of gearing to achieve each overall ratio. 90% per stage is a conservative estimate to account for all of the various losses.

My math also says my wheel will be spinning at 224.789 RPM under this load condition (exerting 130 pounds of force). My wheel has an 18.84 inch circumference, so that comes to 5.88 feet per second.

So once I account for losses, I'd wager I'm somewhere between 4.5 and 5 feet per second pushing with 130 pounds of force.

What a beast! What does everybody think?