## Note: Minibot Normal force requirement

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.

Lets define:

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

so

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.

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