I am trying to calculate the torque for the stepper motor I need in my system. I found out this question with a nice answer how to calculate the different parameters (Stepper motor won’t turn once loaded). However, looking in the nominal time for one step I found a contradiction.

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From the provided formula one can conclude that decreasing the motor torque leads to smaller times. Also, having smaller moment of inertia leads to bigger times. Is that correct? In my understanding it is the oposite - smaller load should lead to faster times. I am asking that because in my system the moment of inertia is 0.00012 and I need to have nominal time for a step of about 0.01s. Trying to use the above equation leads to very strange values for torque.

Thank you

  • \$\begingroup\$ Show your calculations and someone will help. \$\endgroup\$ – Transistor Jun 5 '16 at 6:27
  • \$\begingroup\$ Then insert inertia into the equation \$\alpha = \dfrac{T}{J}\$ wher J is inertia, you inserted 1.3, but you say its 0.00012. \$\endgroup\$ – Marko Buršič Jun 5 '16 at 8:45
  • \$\begingroup\$ And the equaiton (I didn't check) seems to be turned over, larger torque->larger time, indeed wrong, larger inertia->shorter time, wrong, larger angle->shorter time,... \$\endgroup\$ – Marko Buršič Jun 5 '16 at 8:50
  • \$\begingroup\$ Thank you! This is what I suspected. The other way around provides much better fit. \$\endgroup\$ – D. K. Jun 5 '16 at 10:58
  • \$\begingroup\$ Ack! That was my answer and I did, in fact, screw up royally. In the original answer the result should be 0.4 seconds rather than 4.9, although this doesn't affect the rest of my answer. Marko got it right. Serves me right for thinking I can do basic math in my head. \$\endgroup\$ – WhatRoughBeast Jun 5 '16 at 16:16

Its turned over: the author probaly meant: \$\Theta=\dfrac{\alpha t^2}{2}\$, then \$t=\sqrt{\dfrac{2 \Theta}{\alpha}}\$, \$T=J\alpha\$, \$\alpha=\dfrac{T}{J}\$, \$t=\sqrt{\dfrac{2 J\Theta}{T}}\$

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