I only skimmed the responses so I’m sure this will retread. Here’s the basics.
Torque is equal to Force times distance. To fully get this, you need to treat force and torque as vectors and do cross products, but you can simplify here. For an arm in FRC, all of the torque the arm applies to the gear reduction is the weight of the arm itself times the X component of the distance between the arm’s center of mass and the pivot. So in the worst case scenario (horizontal arm, assuming all of the load of the arm is placed at the end of the arm to be safe), the torque is the weight of the load times the length of the arm. A 100 inch-pound torque would be induced by a 10 pound load applied at the end of a 10 inch long arm at horizontal. For the motor to move the arm, it has to output a torque greater than the torque caused by the load of the arm.
DC motors output their peak torque at stall (zero speed) and their peak “free” speed at no load (zero torque). You can actually plot speed versus torque as a straight line going through the points (free speed, 0 torque) and (0 speed, stall torque). This will give you an approximation of the adjusted speed when you load your motor to any portion of its stall torque.
When using a motor at full throttle you will always be at some point on this curve. When a motor is at stall, it draws its stall current, and when the motor is at free speed, it draws its free current (usually very low relative to stall). This is important because the thermal breakers in the Vex system limit how long you can draw a sustained current, and this limit is below the stall torque of multiple motors. So keep that in mind too. I dont’ know these limitations off the top of my head.
If you had no current limitations, and you wanted to move the arm at the perfect ratio of speed to torque, you would try to gear the arm to operate at max mechanical power (or slightly below this to allow for inefficiency). This would be the speed and torque provided at half of stall torque / free speed. To be safe, maybe 70% of free speed / 30% of stall torque.
So what is this torque? If you have a 7:1 reduction, and 14.76 in-lb of torque at stall, and 100 rpm free speed (I’m not pulling these numbers from anywhere but this thread, check them yourself), then at 30% of stall torque the motor would run at 70 rpm and with 4.43 in-lb torque. With a 7:1 reduction, that results in 10 rpm for the speed and 31 in-lb of torque. We’ve got a 28 inch long arm, so the working load it can easily lift assuming all of it is applied at the very tip of the arm would be 1.1 lbs. It could lift more than this (1.845 lbs at max power, 3.69 lbs at stall torque), but when picking a gear ratio you should try to set the optimal point around here. Keep in mind I didn’t account for the weight of the arm itself in these calculations - you could assume the weight of the arm is uniform throughout the arm and subtract that weight * half the length of the arm from the torque to figure out your actual working torque.
How can you increase the torque available? Either reduce the gearing (slows the arm down, more torque), or add more motors (adds power to the system, basically doubling the stall torque), or counterbalance the arm with springs, reducing the load placed on the motors by applying force / torque in the direction opposite of gravity.
Hope this helps.