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I'm a computer scientist doing a dissertation in robotics and I'm designing a mechanical arm that mirrors a humans, as such I am wanting to pivot the entire arm from a "shoulder" using a stepper motor and some gearing, I was looking at holding torques for different stepper motors and realised that I'm not actually sure how to calculate the force that is going to be put on the arm and the motor when the arm is straight out perpendicular, is this the same as the holding torque?

I think part of my confusion is I am not to sure what terms to actually search, so sorry if this has already been answered before.

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To paraphrase https://www.controleng.com/articles/stepper-motor-torque-basics/, holding torque is the amount of torque required to turn a stationary rotor. As long as this is higher than the torque provided by your "arm" when it is extended, it should remain in the position you set it. To determine the maximum torque needed, multiply the weight of your arm by the distance from the point of rotation to it's center of mass. This will give the N-m or lbs-ft required to raise the arm (your operating torque), this is different than the holding torque.

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  • $\begingroup$ Brilliant thank you! $\endgroup$
    – M.Powell
    Jan 21 '20 at 20:09
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When the arm is straight out it has a torque $T= mg*L \quad $with L the distance between arm's CG and rotation center.

Your stepper motor should provide this torque to keep the arm in balance.

if you need to rotate the arm up and down you need an additional torque such that if you rotate the arm with angular acceleration, $\alpha$ $$T_{rotation}=I\alpha$$

$$ I= 1/3mL^2$$

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  • $\begingroup$ This is very helpful but I'm just wondering what mg is in the first equation and what m is in the second? $\endgroup$
    – M.Powell
    Jan 24 '20 at 10:37
  • $\begingroup$ As in what do they correspond to. $\endgroup$
    – M.Powell
    Jan 24 '20 at 10:37
  • $\begingroup$ m is mass and g is Earth's gravity acceleration. g = 9.8m/s^2. mg is wight or the force downward on an object of mass m when that object is near the surface of the earth. $\endgroup$
    – kamran
    Jan 24 '20 at 18:24
  • $\begingroup$ Of course I never even thought haha, thank you $\endgroup$
    – M.Powell
    Jan 26 '20 at 18:57

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