Calling all math geniuses

OK so i have been thinking A LOT about torque. what i need to know is how much torque is required to lift an arm like on zippy or green machine. i am a freshmen in HS and am pretty good at math and science so if i am given a formula and what all the symbols are equal to then i should be set.

i am looking for the formula to lift an arm with even weight distribution and also the formula for non even weight distribution. and the axis of rotation is right at the end of the arm.

if there is anything that you need to be clarified i would be happy to try to explain it differently.

thanks guys
~DK :slight_smile:

The torque equation is

torque = r x F x sin(theta)

r = distance along the arm from the pivot point to the center of mass
F = force applied (in this case, weight due to gravity)
theta = angle between the force applied and the vector defining the radius

If you haven’t taken trig/pre-calculus yet, sin(theta) =1 at theta = 90 degrees, sin(theta) =0 at at theta = zero or theta =180 degrees.

What this means is that there is maximum torque at a 90 degree angle. For example, since the force of gravity pulls straight down, if you hold your arm straight to the side, parallel to the ground (at 90 degrees with the pull of gravity), there is maximum torque, and it takes a lot of effort to hold your arm up. If you hold your arm straight up or down (pointing to the sky or ground), there’s minimal torque because your arm (the torque arm) is parallel to the direction of the force of gravity, so it doesn’t take much effort to hold your arm up or down.

If the weight is spread out over the arm, you treat the weight as if it were concentrated at the center of mass. So typically, you use a value of r which is shorter than the actual length of the arm, because the arm’s center of mass is not at the far end. The highest possible value for sin(theta) is 1, so to find the maximum torque possible, substitute 1 for this value. This will make more sense when you learn about cross products in physics.

There’s a better explanation in Wikipedia here:

thank you so much

just to ensure i understand i used a hypothetical situation were…

arm length = 1 meter
pivot to center of mass = 0.6 meters
mass of arm = 1.5 kg

Force on arm = 1.5 kg X 9.80
Force on arm = 14.7 Newtons

i used max theta (1)

torque = 0.6 meters X 14.7 Newtons X 1
Torque = 8.82 newton-meter


also 1 newton-meter is equal to 8.850745792 inch pounds right?

so my situation torque = 78.063577882 inch-pounds

thank you again

Math looks good. A motor module’s stall torque is 6.5 in-lbs.
My advice would be to shoot for a gear ratio that provides a bit more torque than u calculate you need, so you dont have motors giving way when you need precise movements of the arm.

yeah i recently was bored and calculated the max and prefered torque (about 1/2 of the max load) for ever gear ratio, yes even compounds (i only did compounds of 4 gears total. example 12 to 60-12 to 60.

Lol, I actually happened to do the same thing a few days ago, except for compund and simple chain gears. TONS of combos. actually feels like a waste since ive realized how often chains break.

The calculation looks good! As a team we’ve talked about torque in a generic sense (“Don’t make the arm too long, or it’ll increase the torque requirement”) but never actually calculated the value. Together with Dreadnaught’s input on 6.5 in-lb, it makes for a good guideline. Thanks for bringing this up.

also, if you would like to run an arm, you need new-ish motors because my robot has an arm that is extremely taxing on our motors, and our older ones dont have the torque to lift it up.

i would just like to point out with any motor it is most efficient to run the motor at 1/2 of it’s max torque both to ensure the motor is safe (even with clutches) and also because it is most efficient. when you make a power vs speed graph you will see running the motor at 1/2 of its max rpm is most efficient and when you run a vex motor at 50 rpm the torque output is 3.25 in-lbs (gathered from a load speed curve i made). i am on team 1768 for the high school (the big ones not vex) division bots we have our comp. this Friday and Sat. so this is what we do when we design an arm and it is the most efficient use of time (as opposed to guess and check) and the arm will be the most efficient. although vex is really geared towards guess and check in how the pieces always fit together i prefer this way.

To take these calculations to the next level,
consider that you can use some #32 rubber bands or some latex tubing as a spring. The gravitational potential energy of the height of the arm can be stored in the potential energy of a spring, and evenly traded off through the changing radius of a cam. When done correctly, it means that nearly no torque is needed to lift the arm. Torque is still needed to accelerate the arm in either direction, so low mass arms are still a good idea.