I was wondering how you calculate the amount of torque required for lifting an arm? What I am trying to do is figure out how much torque I need to lift something at the end of a four-bar. I recall reading this in a post a while back but can’t find the thread now. I think it was something like take your torque and divided it by the length of your arm but I just can’t remember.

Maximum torque created by a gravity load P applied at the end of an arm with length L equals L*P.

You also need to take into account the weight of the arm itself in your torque calculations. Measure the weight of the arm, estimate the location of its center of gravity, then multiply the arm’s weight by the distance from the axis of rotation to the arm’s center of gravity. This gives you the torque due to the weight of the arm. Add the two torque values to get the total torque.

Make sure you pay attention to the length of the axle being subjected to the torque. If the axle is too long and the torque is too high, it could twist (see picture below). This has happened to us couple of times so far.

The amount of torque required to cause the shaft to fail does not depend on the length of the shaft. Having a longer shaft will mean that over its entire length it will twist more under the same load, but it will also twist more before it fails. Regardless of length it will fail at the same torque and the same angle of twist per length.

A shaft subjected to a torque undergoes shear stress and deformation (twisting). The magnitude of the stress is a function of the torque, cross-sectional shape and certain properties of the material. As you correctly stated, stress is not a function of the shaft’s length. Deformation (angle of twist) , on the hand, is a function of the shaft’s length, the longer the member the more it will twist.

If we limit the behavior of the member to the elastic range, then you conclusion is correct; member failure is solely a function of stress. Theoretically speaking , once the load is removed, the shaft assumes its un-deformed shape. If on the other hand, we allow certain amount of inelastic behavior in the shaft, angle of twist, hence the length of the shaft, becomes a design factor. Inelastic behavior does not necessarily mean member (material) failure.

tbh … i never do the calculations, i normally just think about it, like this year, our arm uses lots more metal, the game objects weight more … and because our intake is designed to pick up between 10-15 sacks, i have designed a compound gear box for a 1:25 gear ratio and if this isn’t good enough, i have a design for a 1:35 … and if that doesn’t work … i will go for a 1:49 … but i seriously doubt needing to go higher than a 1:35 … but at gateway, i only used a 1:5 … honestly … you need to prototype everything so start with a minimum of a 1:5 then go up and up … i started with a 1:15 and this failed … btw … using 4 X 269 motors !

The challenge is to figure out a way to distribute the required torque for lifting the arm to axle(s) and motor(s) without causing material failure.

For heavy arms we usually use a turntable or a gear assembly where the arm is directly bolted to the driving gear without the use of axles.

**1.**Determine how much you want to lift

**2.**determine your lift’s length(Lets say 0.381 meters[15 inches])

weight(in Newtons) x length of arm(in meters)=needed torque

so if you wanted to lift 10 pounds(44.5 newtons), that’s

44.5(weight) x 0.381(length)= **16.9** Newtonian meters(necessary torque)

**3.**Then determine your torque(269motors= 0.971 Newtonian meters 393= 1.67 Newtonian meters)

So from there, you need to determine your gearing. Which then, you multiply your torque by the number of motors and gear ratio(ex. take two 393s, thats 3.34, multiply by a 7:1 gear ratio and you get approx **23.38**).

After you find your necessary ratio. Multiply your results by 0.9 if you used chain, 0.8 if you used gears, and 0.7 if you used beveled gears. This will account for skipping, friction, etc. You **must do this for each stage, if you use compound gearing**

After that, find your programming ratio. For ex, if it’s programmed for 63, multiply your final result by 0.5, because you cut the given power in half

**Keep in mind**, Your final result will approximately be your stall torque. Which means, that’s the weight where you will stop lifting. You * must* also use metric