Determining lifting capacity of a servo

[Jerran]

How can I determine the maximum lifting capacity of a servo motor from its torque rating? Im trying to build a 5 servo arm capable of lifting 1-5 pounds but Im having trouble remembering how to determine how large of servos Ill need from the information specified on their data sheets.

Just a shot in the dark but Im thinking it should be like as follows correct?
Torque = Force x Distance so if the servo has a torque of 200 oz inches and the arm its attached to is 12 inches...
servo T=F*D = t=f*d arm
200oz.in = force* 12in
force = 200oz.in/12 in
force = 16.67 ounces

 

[williB]

i dont think thats how u use distance in the formula..
because you have a pivot point ..with the servo as the fulcrum...
let me check..

 

[williB]

i've looked and looked ...
maybe a ME can help you out..
torque is the cross product of force and distance ,which may or may not derive into simple multiplication .. not sure..

But it seems to me that if you have a weight or force of 200 in oz 12 inches from the fulcrum that you are going to need a lot more force to move it ..

 

[Jerran]

aye thats why I was thinking you could use the length of the arm as the radius of the shaft in the equasion, as you get farther from the fulcrum the force decreases inversly perportionately. I still havent been able to find a cut and dry method on the net to determine it yet though :/

 

[williB]

ok .sorry .. i think you are correct..
16.67 oz. would be the weight that a 200in oz motor could pick up , or hold stationary ,when that 16.67 oz weight was 12 inches from the stepper..
i misread your origional post.