I am trying to do some preliminary calculations for a basic lifting device but some of my numbers seem to be out-of-wack. Could some kind soul take a look if you have a minute spare?(adsbygoogle = window.adsbygoogle || []).push({});

Motor Data (max efficiency)

RPM: 8311 = 870 radians/sec

Torque: 9.213N-cm = 0.09213N-m

Output Power = 870 * 0.09213 = 80.1531N-m/sec

The motor is driving a 12tooth gear meshed with a 72tooth gear for a 6:1 rpm reduction. The 72tooth gear is driving a 12mm trapezoidal thread (vertically mounted).

The RPM on the threaded shaft will be ~8311 / 6 = 1385.16RPM

The Torque will be 0.09213 * 6 = 0.55278N-m

The equation from the thread manufacturer for drive torque is:

T = (F * dm / 2) * (cos b * tan a + u) / (cos b - u * tan a) (kgf mm)

F is the thrust load (kgf) - I believe this is the weight on the nut that the thread is driving (for me, 160kg).

dm is the pitch circle diameter (12mm for my thread)

b is the Flank angle (given as 15 deg in the data sheet)

a is the lead angle (given as 3.31 deg in the data sheet)

u is the friction factor (I am using 0.20 as a generic steel-on-steel coeff)

Therefore:

T = (160 * 12 / 2) * (cos(15deg) * tan(3.31deg) + 0.2) / (cos(15deg) - 0.2 * tan(3.31deg)

T = 960 * (0.2558640807677) / (0.9543588758572)

T = 257.376 kgf-mm

converted to N-m

T = 257.376 * 9.81 * 1000

T = 2524858.56N-m

This number seems huge to me (2.5million Newtons/metre). Can anyone explain how this relates to the motor's output power. I know from practical testing that this motor can perform the task, therefore I think it's more likely that I don't understand the result from the second formula.

Could someone explain what the output from the second formula actually relates too?

Thanks

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# Thread drive torque under load

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