Today, I just tried to calculate the maximum power that the vex 393 motors can output. I’m not sure if I made an incorrect assumption somewhere or whether I just made an error, but the final power seems a bit low to me.

Power = Torque * Revolutions Per Second

Power = 1/2(1.67 N-m) * 1/2(100 RPM * 1M/60s)
(1/2 because speed and torque are directly negatively related)
Power = 5/6 N-m * 5/6RPS

Power = 0.694Wrot

This is for one individual motor.

Edit: I made an incorrect assumption that work was the same thing as torque and that angular velocity was the same thing as time. My bad…

So you are doing something odd here. You are using the stall torque of the motor and the free speed of the motor to calculate power. These two values are taken at two completely different states. The stall torque is measured when there is not rotation (0RPM) and the free speed is when there is no torque on the shaft (Nm). Therefore, at each of these points the motor is outputting 0 Watts to the shaft.

Secondly, I am not sure where the 1/2 is coming from? Power is going to be TorqueAngular velocity.
In this case you should have P=T2PIRPM/60

Lastly, power curves have been made by jpearman in the second post here for the 393 and 293 motors that you can reference, although the torque is measured in in-lbs.

EDIT: I see why you used the 1/2’s now. By doing so, you have selected the point where power will be maximized on a theoretical torque/speed curve. That makes sense now. However, you need to convert your rpm so that you don’t have rotations left in your units. Your answer of 0.694Wrot is actually correct. If you multiply it by 2PI/rot you will get 4.36 Watts. Hope I didn’t confuse you too much!

The SI unit for rotational velocity is radians per second, not revolutions per second. Multiply your answer by 2*pi and you should get 4.37 Watts, which is the theoretical maximum power based on that stall torque and free speed.

To understand why:

power is torque times rotational velocity.
in a situation involving force instead of torque, power is force times linear velocity.

You can see how the two are basically equivalent by imagining that instead of turning something with X Newton metres of torque you are turning a 1 metre lever that pushes an X Newton force at the end. In that case the power would be equal to X Newtons times the linear velocity of the end of the 1 metre lever. The linear velocity of the end of the lever is the radius times the rotational velocity in radians per second.

**power **= **force **x **linear velocity**

=> power = **force **x **radius **x rotational velocity (rad/sec)
=> power = **torque **x rotational velocity (rad/sec)

If torque were defined as the force times the circumference rather than the force times the radius then you would use revolutions instead of radians. But the convention is to define torque in terms of the radius.

It isn’t that he is ignoring the different states as you imply. He is just assuming 1/2 free speed happens at 1/2 stall torque. In other words he is saying it is perfectly linear which obviously it isn’t but as far as rough estimating goes it is not too bad.

I think the best way to approach this is to look at the graph below, which is the torque-speed curve for the 393.

Look at where the peak of the power curve is located and then look below that for the values of the torque and speed. Roughly speaking, the peak power takes place around, oh say, 7 inch-pounds, which is equal to roughly 0.8 newton-meters. The speed at peak power is roughly 50 RPM. Converting 50 RPM to rads/s gives me about 5.24 rads/sec. Angular velocity should be in rads/s.

Now you can calculate Power = (Torque) * (angular velocity)

And I get something like Power = (0.8 NM) * (5.24 rad/s) which equals 4.2 nm/s, which equals 4.2 Joules/s, which is the same thing as 4.2 watts.

I was just roughing out those numbers, but I think that’s consistent with the max output of the motor, which is supposed to be 3.93 watts, which is where the motor derives its designation.