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Thread drive torque under load

  1. Jul 29, 2009 #1
    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?

    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)

    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?

  2. jcsd
  3. Jul 30, 2009 #2


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    Science Advisor
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    Gold Member

    Hi Sentient
    The reason it's huge is because you multiplied by 1000 instead of dividing by 1000. Should be 2.5 N-m. This is still quite a bit higher than the torque available to rotate this shaft (0.55278 N-m). I didn't go through all your math, but I'm assuming the rest was done right.

    You say the motor is reliably turning this shaft, so the question must be how? Lubricant on the threads will reduce the coefficient of friction, and the motor may be putting out more than the rated torque, but it's pretty amazing that so much torque could be made up by those factors alone.
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