[KrazyCarl92]

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.