# How to calculate torque for a 4 bar

Hi, I would like to calculate torque for a 4 bar lift. However I see many variations of the torque equation, such as one for force being perpendicular or not perpendicular to the lever arm. Also torque is a metric unit based unit, it uses Newton’s and meters. I’m probably going to sound like a pleb, but what’s the British version of torque? Would torque in British just be pounds times feet? Also, the motors provide their own torque, which is the leading force that drives the whole lift. What direction is that in, and where would I find their outputted torque?Thank you in advance.

If there is an equation for torque that is not perpendicular to the lever arm, then it is just multiplying the force applied to the lever by cosine of the degrees away from perpendicular from the lever that the force is. In this setup, force perpendicular to the lever would be multiplied by cosine of 0 degrees, or 1.

The imperial measurement is foot-pounds, yes. This is what is supplied in VEX’s documentation, as well. The direction of torque depends on how you define the coordinates in your system. Stall torque from vexrobotics.com: Stall Torque: 1.67 N-m (14.76 in-lbs) (As Shipped)/1.04 N-m (9.2 in-lbs) (High Speed Option)

To give an example that may clear some things up, to reach stall torque on a single 393 motor on standard (torque) gearing, one would have to either apply 1 pound of force on a lever 14.76 inches from the fulcrum, 14.76 pounds of force 1 inch on the lever from the fulcrum, or 2 pounds of force 7.38 inches from the fulcrum, or so on. If you use a 3:1 reduction, such as a motor spinning a 12T gear which powers a 36T gear with a lever attached to it, this number changes to a one-pound force 44.28 inches from the fulcrum of the lever, or so on. Adding a second motor basically just doubles the torque.

What exactly is stall torque? Is that like the max torque a motor could go against?

Yup. It’s the amount of torque that it takes to stall the motor.

Is there an equation for torque, like: torque = armLength * force * cos(angle)?
Also, if I want to see if a certain four bar lift set up will have enough force to lift a load, then I rewrite the torque equation, but this time in terms of force and use that value to do all of the other calculations?

Torque is equal to the component of the force pointing perpendicular to the radius times the radius.

In your setup, the force would be the force of gravity on the object you are lifting (mg) times the cosine of the angle of elevation.

Image attached. Oh that is torque when an object applies the rotational force onto the lever, but how would you calculate torque of a lever applying a force onto an object?

If you are trying to see if the weight of an object will make a motor stall, there is your equation. I’m not 100% sure what percentage of stall torque a motor is applying all the time (I guess 100% of stall torque if it is acceleration to its rated rpm?), but if torque = force * distance from the fulcrum, then you just solved for force. Technically a lever would apply force, not torque, to an object to achieve translational movement at a specific moment in time. So, the force the motor could apply equals torque/(distance from the fulcrum), where you have torque and distance from the fulcrum.

In theory/ideally it’s linearly related to speed. So at half its rated rpm you get halff the stall torque. At the rated rpm, you get no torque.

Are you talking about if the motor is not being run at full power in the code? Because then what you are saying makes sense. But I should have been more specific – I was referring to a case where the motor is at full power, so attempting to spin at 100 rpm, but only spinning at 50 rpm due to a large amount of mass to accelerate, and currently accelerating. I believe the motor should have the full rated stall torque in that situation, because the motor is drawing as much current as it can.

No, we’re talking about the same thing. At least according to a common model, it would only have half the torque at that time.

So if I have two motors with let’s say a stall torque of 2 lbs-in, and my load is 10 inches away, that means my lift setup would only be able to apply a force of 4 lbs-in / 10 in = 4 lbs / 10 = 0.2 lbs. Is my logic correct?

Oh, now I think I understand.

4/10 = 0.4, but otherwise, yes.

I wouldn’t overthink it, to be honest. There are lots of other factors (including friction) that make this not worthwhile to calculate. And just because you can lift a certain amount, doesn’t mean it’s a great idea. Having more torque than you need results in better acceleration and sometimes, greater maximum speeds. For example, we had a 1:5 high speed DR4B on two motors, but it was struggling. So, we change to torque motors. Now, the lift is actually faster, and less likely to burn out to boot. Also, you are calculating the amount of weight that will stall your lift. So you will need to have more torque than what you are calculating. Just prototype; it’s the beauty of VEX.

1:5 torque ratio or speed? I always have trouble distinguishing the two. Is it driving:driven and driven/driving?

Man, this debate happens all the time. I think I go back and forth, actually But, based on the fact that it’s a lift… It’s definitely got to be torque. So in this case, teeth of driving to teeth of driven. Often, it’s rotations of driving to rotations of driven. Usually context will give some idea (it’s not a 1:21 flywheel for torque, for example).