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katratzi
October 28th, 2008, 01:49 PM
I have motor/mechanical variable speed pulley/gearbox drive system that I do not know how to calculate the output torque capability of. The system consists of 1700 rpm 12.5kw motor, attached to a Lenze variable speed pulley unit, which the output of is connected to a gearbox with a 31.5 ratio. The Lenze unit has a stated ratio of 20:1, which I am assuming is the max since it is a mechanical variable speed device. Lenze identifies this device as a "Simplabelt compact unit".

I do not know how to calculate output torque of unit like this. Can anyone show me a formula or a good website that has information on this?

HJTRBO
October 28th, 2008, 06:18 PM
T(Nm) = (kW x rpm) / 9549

That will give motor output torque at the shaft. Then your chosen ratio will determine torque multiplier, less any losses through the drive train.

http://craig.backfire.ca/pages/autos/horsepower#torque

Its for automtive applications, but applicable to our industry.

HTH
Nathan

katratzi
October 29th, 2008, 07:44 AM
thanks for link, Nathan. I think the losses you mentioned are the key, especially at the Lenze variable speed pulley unit. Because the torque output capability of the system seems way too high for a 50mm shaft to withstand. This may be ok though, because the roll that the 50mm shaft is attached to is turning a conveyor belt that is tensioned between this roll and another roll at the other end, so maybe (hopefully) it would slip instead of twisting the shaft.

This old Lenze unit is gone, it definitely needs replacing with new motor/ gear unit at this point.

DickDV
October 29th, 2008, 10:06 AM
As a general rule, a variable pitch pulley system such as the Lenze unit you mention is limited to a maximum torque multiplier of 2 due to the tendency of the belt to slip at torque levels higher than that.

katratzi
October 29th, 2008, 11:27 AM
yeah, that would explain a lot. i never considered belt slippage at pulley system, seems obvious now that you point it out. a mechanical idjit, i am.