[Al Skierkiewicz]

1) What is the CIM's speed?
If free speed is 5310 RPM and stall torque is 343.4 and the calculated speed is the free speed - reduction in speed due to load then speed = free speed - (rated torque * (free speed/max torque)) or
5310 - (100 oz.in. * (5310/343.4)) = 3763.69 RPM

2) What is the CIM's current?
If stall torque is 343.4 oz. in. and stall current is 133, then the current constant is .3873A/oz.in. therefore current at 3763.69 RPM is the constant * RPM + I min
I= (3763.69RPM * .3873A/oz.in.) + 2.7A = 38.73 + 2.7A = 41.43A

3) How much power (watts) is being input to the CIM?
P IN = I * V = 41.43A * 12V = 497.16 W

4) How much power (watts) is the CIM delivering into the load (dynamometer)?
The conversion factor for oz. in and speed is 0.00074 and Pout = Torque * RPM * conversion factor,
Therefore P out = 100 oz. in. * 3763.69 * 0.00074 = 278.51W

5) How do you account for the difference between #4 and #3 above?
As the lost electrical power is the current squared * motor winding resistance, and R motor is 12V/133A (stall current) = .0902 ohms, then electrical power loss is 41.43A^2 * .0902ohms = 154.87W.
If the input power is 497.16W and electrical loss is 154.87 and power out is 278.51W then all other losses amount to 497.16-154.87-278.51=63.78W
These losses are attributable to bearing friction, brush friction, back EMF, friction of armature with the air, etc.