I'm going through this person's torque calculation of a simple crane setup. the servo used has a max torque of 57oz.in:
At the top, if Mservo
is 34.12oz.in, how is it 62.38% of stall torque when 34.12/57 = 0.5986?
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Sign up to join this communityI'm going through this person's torque calculation of a simple crane setup. the servo used has a max torque of 57oz.in:
At the top, if Mservo
is 34.12oz.in, how is it 62.38% of stall torque when 34.12/57 = 0.5986?
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.
Mservo = 34.12oz-in
). Shouldn't there be plenty left to accelerate the load?
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