Motor Power Calculations

[DMetalKong]

First, a couple of disclaimers:

1. I know there is the JVN design calculator, but as far as I can tell it doesn't work well trying to mix and match motors in a single drivetrain.
2. I know there was a recent thread similar to this a little while ago. I was hoping this could go in a more general direction than slap x gearbox on y to make it match a CIM.
3. I know this is groupsourcing my math, but I think that it could be helpful for everyone to see the math behind the power calculations, if in fact the math is correct.
4. I am by no means confident in my math, I was hoping some more experienced heads could verify if what I am doing in correct.

With that being said
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Let's say that you want to have a drivetrain with 4 CIM motors and 4 RS-550 motors. The max power of the CIM is 340 W, and the max power of the RS-550 is 254 W. The total max power is

4 * 340 W + 4 * 254 W = 2,376 W

Now, there will be losses transmitting the power from the motors to the floor, let's approximate this at 80%. So the power getting to the floor is

2,376 W * 0.8 = 1900 W = 1406 ft-lb/s

(To get from W to ft-lb/s, multiply by 0.74)

Power = Torque * Angular Velocity, so Angular Velocity = Power / Torque.

Torque = Force * Distance. For 4in diameter (2in radius) wheels, the distance is 0.1667ft (2in). The force is equal to the Frictional force, which I'll say is

Ff = u * Fn = 1.0 * 165lb = 165lb

In the above, u is the coefficient of friction for the wheels on carpet (dependent on the wheels, I chose 1.0 as a typical number), and Fn is the normal force, which in most situations is the weight of the robot.

The torque is then

165lb * 0.1667ft = 27.5ft-lb

This means that the angular velocity of the wheels is

(1406 ft-lb/s) / (27.5 ft-lb) = 51.127 rad/s

Converting to rpm

51.127 rad/sec * rev/(2pi rad) * 60 sec/min = 488 rpm

This means that the CIM, which has a free rpm of 5310, should be geared down 5310/488 = 10.88:1.

The RS-550, which has a free rpm of 19300, should be geared down 19300/488 = 39.55:1.

Also, the RS-550 could be geared down to the free speed of the CIM - 19300/5310 = 3.63:1 - and then the two inputs merged, and the final 10.88:1 reduction applied.

[Ether]

Originally Posted by DMetalKong

I was hoping some more experienced heads could verify if what I am doing in correct.

Whether or not it is correct depends on what your design goal was, but you didn't say... so it's not clear (to me) why you used the free speed in your final calculation.

I know there is the JVN design calculator, but as far as I can tell it doesn't work well trying to mix and match motors in a single drivetrain.

As a first approximation, when you combine two motors, you generally want to gear down the faster one to match the slower one's speed, so that each motor is always operating at the same % of its own free speed. So gear down the RS550 19300/5310 = 3.63:1 (as you said).

You can then take that motor combination, and calculate an approximate effective motor curve for the combination, and then treat it as a single motor for purposes of further calculation using JVN's calculator.

If you want to factor in the efficiency of the 3.63:1 gearing it gets a little more complicated, but still doable.

[DMetalKong]

Originally Posted by Ether

You can then take that motor combination, and calculate an approximate effective motor curve for the combination, and then treat it as a single motor for purposes of further calculation using JVN's calculator.

The biggest thing that I am confused about is how to combine the curves of the two motors considering they have such different performance characteristics (max torque, current at stall, free rpm, etc.). What exactly is the process for matching things up?

[Ether]

Originally Posted by DMetalKong

The biggest thing that I am confused about is how to combine the curves of the two motors considering they have such different performance characteristics (max torque, current at stall, free rpm, etc.). What exactly is the process for matching things up?

Gear the higher speed motor down so its free speed matches the free speed of the slower motor. Then scale the geared motor's stall torque up by the same factor and add it to the ungeared motor's data.

Take a CIM and an RS550 for example. Gear the RS550 down by 3.63 so its free speed is 5310, same as CIM. Its stall torque is now 3.63 times greater, or 251 ozin.

The motor curve for the combination would be:

free speed: 5310 rpm
free current: 2.7amps + 1.4amps = 4.1amps

stall torque: 343 + 69.2*3.63 = 594 ozin
stall current: 85amps + 133amps = 219 amps.


The above is a good approximation. If you want to toss in some loss of efficiency for the gearing that complicates things a little. The gear ratio would be slightly different, the free current somewhat higher, and the stall torque less.