I wanted to do a quick calculation for a student Vex robot that has 5 in wheels and is having one rail stalling during a turn. The stalling could have been predicted with some basic torque calculations.
Using static moment equations you can easily derive the minimum torque to just turn for a symmetric robot with four direct drive motors on wheels with radius r . Legacy three-wire 6.5 inlb motors are assumed.
Define the half spacing between axles as “l” and the half width between wheels as “w” , the robot weight as W_lbs and the coefficient of friction for a tire side force as u. Also assume that the weight is evenly distributed on the four wheels and the center of turn is at the cg. Also w > l so turning can occur.
Input turning moment per wheel = w* torque/r and this must be greater than the moment resisting the turn caused by the side friction forces on the wheel.
Resistance moment = l*W_lb*u/4 . The factor of 4 converts the W_lbs to normal force on the wheel.
So torque > r*(l/w)*W_lb*u/4
Lets plug in some typical numbers. r = 2.5 in, l=3.5, w = 7
This gives a required motor torque of 2.5*3.5/7/4*W_lbs*u
or torque >( 2.5/8) * W_lbs* u
or W_lbs < torque*8/2.5/u
Without omni wheels, u is typically around 1 or higher and if the robot W_lb is 12 lbs then the minimum torque
torque_min > 3.75 in lbs
This represents about 60% of the old three wire motors 6.5 in lb spec torque . Clearly there is not much margin for any torque losses in the drive train or extra side friction on the wheels or extra weight from a heaver robot or from game objects. Also any contact forces resisting the turn will easily stall the turn.
Using a 4 in omni wheel in place of the 5 in traction wheels increases the input torque by 5/4 and reduces the u by about 1/3 so we get a combined improvement of 15/4 or 3.75 factor on the required torque.
min_torque_ 4 in omni = 3.75 / 3.75 = 1 in lb or about 15% of the available torque of an old motor. This is where you want to be in your design.