For more details please find the attached the motor data sheet and the gearbox is worm kind with ratio is 10:1.
1. Rated torque at 2000 rpm is =4.78 Nw-m
2. Assume gear efficiency is etag=say .9
3. Torque output = 4.78*10* etag = 43 Nw-m
Cheers
The efficiency is typically a "fudge factor" provided to account for the frictional factors in the gear drive. Frictional effects in a gear drive are nonlinear, so the efficiency figure provided is only an approximation. So, approximately:
Power_out = 0.9 * Power_in.
Power = Torque * Speed
However, at stall, mechanical Power_out = 0.
So the efficiency figure doesn't factor into the calculation of the maximum torque that the motor/gear-drive can resist, which should be:
Max Holding Torque = 14.3 Nm * 10
Note that:
- the electrical power_in at stall will be non-zero
- worm drives are typically not back-driveable, although some are. If the drive is not back-driveable, and if is rated for the maximum stall torque, it should hold against the maximum applied load without electrical power in. Which has its own problems in that you now have to drive the motor in the same direction as the apied load in order to vet the drive to comply with the applied force.
For 1:10 you expect 10 times more torque at the output. In single stage spur gears it can be 98%, but 1:10 would normally be a a lower efficient single stage of two stage which is closer to 96%: 9.6 times
However for worm gears have an efficiency of typically 40-80%
http://orientalmotor.com/products/ac-motors/high-torque-brother-ac-gearmotors.html
So only 4-8 times more torque than motor torque.At low torque or low temperature the no-load loss can be high.
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