4
$\begingroup$

A motor gives its strength in N-cm or oz-in, but how do you know how much you require? I know you can just calculate the minimum torque you need to move the load, but is there a typical amount over that minimum that is standard? Like 20% or reaching a certain acceleration?

What I'm using a motor for is my electrical engineering capstone project. I need to rotate a mirror about the z axis at between 30/60 Hz. The mirror is the only moving component, so there aren't any other mechanical considerations besides that. I don't know the mirror size yet, as I figured I'd buy as good a motor as I can afford and pick a mirror that's small enough to work with it, but I want to know how to figure out approximately what a given torque will get me. I don't want to spend $50 on a motor with something like "100 oz-in min holding torque" and then get it and find out thats only strong enough for a 2"x2" mirror or something.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

Without more details about your design, the dynamics of your mechanism, or your calculations, its virtually impossible to address your question with any certainty.

However, i offer some advice that may prove useful to you:

If affordable, I would use double the minimum holding torque needed for the application because i usually find that my calculations don't usually account for all the true dynamics of the mechanism. When starting out with an untested design, it's better to err on the side of caution. Having too much torque is usually better than having too little.

Depending on speed requirements, you may be even able to use gears to increase your torque while using a motor with less torque. I may even recommend experimenting with gear ratios since its usually cheaper to buy a different gear than to buy a different motor. Over time, you may find gears to be the better choice until you find the torque that meets your needs.

But if you're really confident that your calculations truly capture the physics, you might be able to get away with 20%-30% more than the minimum torque. Ultimately, it all boils down to how well you can capture the physics of the mechanism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.