It is fairly easy to show that the normal force is about 2x the weight of the minibot. The normal force is independent of the minibot weight… but we can express the ratio to give an idea of how much the normal force is relative to the weight of the minibot.
torq = max effective torque delivered to the wheels
= number of motors*effectiveness*torq_max_motor*gear_ratio
r_wheel = radius of the wheel
W = weight of the robot
N = normal force
u_static = static friction coefficient
u_drag = coefficent of drag is due to bearing friction
drag_force = u_drag* N
Drag force can be measured by letting the minibot drop down the pole with the motor disconnected (mechanically) and measuring the acceleration. I have measured this for our minibot and it was surprisingly high at u_drag =.22 .
The total force applied to the pole< N*u_static in order not to slip.
torq/r_wheel – drag_force = torq/r_wheel – u_drag*N < N*u_static
N = torq/r_wheel/(u_static + u_drag)
So with r_wheel = 2 in, effectiveness = .95 , u_static = 1.2, u_drag = .22, gear_ratio = 1:2 , W = 4lbs, torq_max = 300/16 in_lb, number of motors = 2
N@12 volts = .95*2*300/16*.5/2/(1.2 + .22)= 6.3 lbs
N@14 volts = 7.2 lbs
N/W = 1.57 @ 12 volts
This is for 12 volts… @ 14 volts, N/W = 1.8
We have data also for the dyno testing giving 480 oz in at 14v
This gives N = 10. lbs , N/W = 2.5
If the u_static is lower than 1.2 say .8 then the normal force increases for the dyno motor to
N = 10*(1.42/1.02) = 14 lbs and N/W = 3.5 . Wow… so we are getting up there with a strong motor and weak friction force.