I'm posting this on behalf of a friend who needs some help...

Hello... I have been doing some research on motor selection for my application, and I am running into some questions. To preface, I am trying to select a motor that will do the following:

1) Turn a shaft which needs some 200-300 oz-in of torque to even rotate.
2) Turn a shaft at 2000 rpms.

How do I select a motor from that?

What I was trying to do is calculate the kT (kT = 1355/kV) and then apply a speed reduction gearbox to it.

For example, the Rimfire 35-48 1300 Outrunner Brushless Motor has the following specifications:

Max. Constant Current: 70A
Max. Surge Current: 95A
Max. Constant Watts: 777W
Max. Surge Watts: 1050W
No Load Current: 4.2A
Input Voltage: 11.1V
RPM/V (kV Rating): 1300

Looking at this motor, I can calculate a kT value of 1.04 oz-in torque per amp. I see this has a maximum surge current of 95 amps, which would be about 98.8 oz-in of torque (1.04 * 95 = 98.8) at maximum amp and 14,430 rpms (1300 * 11.1 = 14,430 RPM.. well emp?).

This is where I am getting confused.

Is the kT rating a stall torque value, or max efficiency? I assume it is stall torque, for I found another little motor (http://www.robotmarketplace.com/products/0-MSJ.html) and did the math and it came out to what was posted.

I then read somewhere about the maximum power is at 50% of the total rpm. Again, I am confused. Is the total RPM the maximum RPM based on the kV and maximum Current? How does this apply?

If I applied a 4:1 gear reduction to the example motor above, does this compute correctly (ignoring frictional resistances):

- 3607.5 RPM at stall
- 395.2 oz-in torque

I feel like I am missing something with the whole 50% thing.

Help?

Thanks,
Mike