Hello..
I need a motor to spin a shaft at the following minimum requirements:
1500-2000 RPM
300 oz-in torque
I found a motor on Robot Market Place:
http://www.robotmarketplace.com/products/0-MSJ.html
This motor produces the following specs:
Speed : 16,000 rpm @ 12V
Angular velocity constant: 1560 rpm/V
Amps @ nominal: 1.2 Amps
Efficiency: 71.4%
Peak Power: 0.36 hp
Stall current: 91.8 A
Stall torque: 78.7 oz-in
If I made a gear ratio reduction of a 8:1, the math would make it produce:
Stall Torque oz-in = 78.7 * 8
Stall Torque oz-in = 629.6
Speed RPM = 16000 / 8
Speed RPM = 2000
Did I caclulate that correctly?
Questions:
1) In my car experience, Torque to relative to RPM. Meaning, if this motor can get up to 2000 RPMs it would then produce 629.6 (assuming calculations are correct). My concern is the shaft I need to spin from this motor has a minimum torque requirement to make it rotate in the first place. The shaft doesn't just freely spin. The minimum is around the 300-400 oz-in. Would this still work?
2) In achieving this gear ratio, I simply would like to put a gear on the shaft that I need to spin, put a gear on the motor, and have some gears inbetween the two (to cover a distance gap and create the gear ratio). How would I do the gearing at this point? If it were just two gears, I could achieve an 8:1 with a 5 tooth and a 40 tooth. If I just used a 5 tooth, 40 tooth, and a 5 tooth; would this cancel out the 8:1?
3) Any other thoughts?
Thanks,
Mike