Much of this is discussed
HERE.
The F-P at 12V with a 180:1 gear ratio and 5 stages works out to:
Freespeed: 20,000RPM/180 = 130RPM
Stall Torque: .65*180*.9^4 = 77N-m
BUT... ...general concensus is that you dare not run the motor near stall very long or you will have a F-P Flambo! AND the F-P transmission is designed for 6V so you are probably going to strip gear teeth if you repeatedly try to pull anything like 77N-m out of that thing.
My "official" recommendation*: Design the mechanism to load your motor (worst case) to 1/4 stall (roughly 20N-m if you use the F-P gearbox).
Reasoning: In this way, you are essentially limiting your power to about 300W which is a lot but you are doing it with your motor spinning 18,000 RPM and your motor will be at or near its peak electical efficiency -- the high RPM gives the fan what it needs to blow away a lot of heat and by operating near peak efficiency there is less heat to blow away in the first place.
300 Watts is not Hay!
More free advice from THE Dr. Joe... ...accept no substitutes...
Joe J.
*read reply #11 of
this thread to see my 6 point plan for getting THE OPTIMUM out of these motors -- but upon reflection, it is too much for too little. Besides, I say
here that "
FIRST robot design is not an optimization process" so get to "good enough" and move on. 300W is good enough in my book.
P.S. Reading my note, I can see that I am going to get a lot of questions to the effect of "how did you come up with the 300W number?" Well, you can get it from knowing that the F-P will put out a peak power of 400W if you load it at 1/2 the Stall Torque, resulting in 1/2 the Free Speed. By loading at 1/4 the Stall Torque, the motor spins at 3/4 the Free Speed.
(1/2*1/2) = 1/4 so the Peak Power is the Stall Torque * Free Speed / 4.
(1/4*3/4) = 3/16 which is 75% of 1/4 so you will get 75% of the Peak Power out of the motor at this operating point or 300W. JJ