I’m working on a project and I’m trying to determine the best motors for my application but I’m a bit clueless as to what the torque measurement actually “means”.

The specifications state that the motors have a stall torque @ 6V of 57 oz·in. This means that the motor can move a load of 57 oz in a one inch radius circle about its shaft.

For a practical explanation, look at it like this. You place an arm on the motor. This arm has no weight (for the sake of example) and is one inch long from the attachment point to the motor’s shaft to the end of the arm. The arm will then be able to lift 57 ounces at its end. Obviously the mass of the arm and torque loss through gearing need to be taken into consideration, but this is the basic idea.

You can also transform the torque value. Something like 57 oz/in would also be 28.5 oz/2in or 19oz/3in and so on. Always try to keep a buffer when looking at the torque ratings of motors. I usually look for the motor and gear box to be at least 10% more than my worst case scenario.

Torque is equal to the distance from the axis of rotation times the force tangent to the radius available at that distance.

for an arm, the distance should be considered to be the distance to the center of mass of the arm.

The torque that the arm applies TO the motor equals the tangential component of the weight times the distance from the center of mass to the axis of rotation.

If the torque applied to the arm BY the motor equals the torque that is applied by the arm, then the arm will be stationary. if you exceed this torque, then it will move up (signs + or -] matter here). If the arm applies more torque than the motor, it will go down.

In general, max power for a motor is achieved at half of stall torque. This means that a given motor will move a given arm or an object fastest when it is geared such that the torque on the motor will be half of the stall torque. If the motor with a stall torque of 57 oz in needs to move an arm that puts 114 oz in of torque on the motor output shaft, a 4:1 gear reduction would be best (this is neglecting efficiency of a gear reduction) because the motor would need to provide half of its stall torque, or 28.5 oz in to move the arm (Because 28.5 oz in times a mechanical advantage of 4 equals the 114 oz in we need to move the arm). This will result in the highest speed possible for that motor to move that arm because this will fall on the peak power part of the motor curve, where the motor moves at half its free speed and providing half of its stall torque.