leverage formula??

Hey guys.

Im using a servo motor moving a 1 metre arm. Im planning to put a speaker on the end weighing 126grams. I Need to know how much the speaker will weigh with respect to the leverage, if you know what i mean! Any engineering guys know if theres a formula to work this out??

thanks Rich

If your servo torque is measured in kilogram centimeters you need 0.126 kg * 100 cm = 12.6 kg.cm

To get ounce-inches you divide by 0.0720077887081 so about 175 oz.in

You should also account for the mass of the arm. Figure average lever arm for the arm is 50 cm.

You may need more torque to accelerate the arm at a reasonable rate.

thanks a lot john!

much simpler than i expected!


You might also want to consider fitting a balance weight on the opposite side of the arm so that the torque requirement is drastically reduced. The balance weight x arm length should also come to 12.6kgcm. If perfectly balanced the only force experienced is friction in the bearings.

yes excellent idea thanks!