# Gear Ratio Math for Cim Motor FP801-005

Please post if this math is correct and if it isn’t please post what is incorrect. Any and all help is appreciated.

First Robotics Cim Motor (FP801-005) Gear Ratio
Formulas:
C = 2 x 3.1415 x R
In/min = RPM x C
MPH = In/min x (1 ft / 12 in) x (1 mi / 5280 ft) x (60 min / hr)
The Gear Ratio = MPH / New MPH
The Gear Ratio = Driven Gear / Driving Gear
Given:
Total Amps = 120 amps (Can’t go over)
Cim Motor FP801-005= 40 amps (Amount Mr. Clark gave us to use)
RPM of Cim Motor FP801-005 = 1600 RPM (Looking under 40 amps)
Diameter of Wheel = 8 in
Radius of Wheel = 4 in
Goal MPH = 7 mph
Circumference:
C = 2 x 3.1415 x R
C = 2 x 3.1415 x 4 in
C = 25.1327 in
Curent In/min:
In/min = RPM x C
In/min = 1600 RPM x 25.1327 in
In/min = 40212.32 in/min
Current MPH:
MPH = In/min x (1 ft / 12 in) x (1 mi / 5280 ft) x (60 min / hr)
MPH = 40212.32 in/min x (1 ft / 12 in) x (1 mi / 5280 ft) x (60 min / hr)
MPH = 38.07984848 mph ( The goal was 7 mph so we need a gear ratio)
Gear Ratio:
Gear Ratio = MPH / New MPH
Gear Ratio = 38.07984848 mph / 7 mph
Gear Ratio = 5.439978354 (This can round down to 5 if that is easier because its mph will be 7.6 (in our goal range))

Yep. Basically correct but kind of hard to follow.

I believe that your math is correct, but your test speed is not. While 40 Amps is the peak current limit the circuit breaker can handle before reseting, this is not the resting speed of the motor.

If you have this motor running with little applied load (torque), it will be running very near free speed, about 2500 RPM. As you apply load to the motor, it will slow down, proportional to the amount of torque required from the motor.

For instance, the stall torque on this motor is 570 oz-in. If the load requires the motor to output 285 (50% of stall torque), then it will run about 1250 RPM (50% of free speed), and about 50% of it’s max current.

That’s theory of course.

If you’re running this for a ball launcher, what will likely happen is you’ll be running this motor at very load load as it spins the wheel, but when the ball comes in contact with it, you will apply a load, slowing down the motor. Depending on how much clearance you have between the wheel and the opposing surface, the amount of torque required will change, and this amount can’t be easily calculated.

As a rule of thumb, I would use about 80% of the free speed as your ‘launching speed’. This may end up being a little high, so you can lower the voltage input as need be to control the exit velocity.

Good luck.

Matt