I'm going through this person's torque calculation of a simple crane setup. the servo used has a max torque of 57oz.in:

enter image description here

At the top, if Mservo is 34.12oz.in, how is it 62.38% of stall torque when 34.12/57 = 0.5986?

  • $\begingroup$ I do not see the max torque/stall torque of the motor mentioned anywhere! It seems like there is a max torque assumption about the process of 57 oz.in, but not the motor. the max stall torque would be a characteristic of the motor.. 3.62 in * 16 oz gives you about 57 oz.in so I guess that is where the max torque comes from.... $\endgroup$
    – 50k4
    Jan 31, 2020 at 10:50
  • $\begingroup$ @50k4 This is the webpage and it says "Lastly, we assumed that the max torque by the servo was 57 oz-in". $\endgroup$
    – John M.
    Jan 31, 2020 at 11:44

1 Answer 1


I haven't seen the degree symbol $^{\circ}$ to denote a percentage, but both percentages on your drawing result in the same torque: 54.7, not 57. (10.12/0.185) $\approx$ (34.12/0.6238) $\approx$ 54.7.

This, in turn, is about 96 percent of the "rated torque," so I don't know if they de-rated the max torque or not. On the webpage you linked, they said,

Lastly, we assumed that the max torque by the servo was 57 oz-in. After some prototyping, we discovered that we weren't getting anywhere near that torque despite that was what the manufacturer rated the servo for.

So again, they may have de-rated the max torque in the calculations there but failed to denote it (though I wouldn't say 96 percent of datasheet torque is not "anywhere near" the rated torque).

But really, the major flaw with their design is that they're trying to move something while the load is rated at stall torque. They're not left with any additional torque to accelerate the load. If you want to change positions, you have to have speed. If you're starting from a stopped (stalled) position, then you have to accelerate.

$\tau = I\alpha$, so if you're going to accelerate the load (and move from positions) then you need to have some torque reserve.

There's an email address on the page, though! Try contacting the person that posted the page and ask them directly.

  • $\begingroup$ Thanks for your analysis. But they capped the max torque to 60% of stall torque - shouldn't it be enough to accelerate the load? $\endgroup$
    – John M.
    Feb 1, 2020 at 8:55
  • $\begingroup$ @JohnM. - I do not see anywhere that they capped max torque. If they did then I would assume there would be enough margin to accelerate the load, but how quickly it would accelerate would depend on the effective moment of inertia of the load. $\endgroup$
    – Chuck
    Feb 1, 2020 at 10:19
  • $\begingroup$ I guess 'capped' wasn't the right word, but their resulting torque was ~60% of the datasheet stall torque (Mservo = 34.12oz-in). Shouldn't there be plenty left to accelerate the load? $\endgroup$
    – John M.
    Feb 1, 2020 at 17:41

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