Robotics Fundamentals 1, Chapter 3: Challenge 3 solution (size the drive for a four-wheel robot)

Formulas: F = m (a + g sin(theta) + Crr g cos(theta))          tau(wheel) = F r / n          tau(motor) = margin tau(wheel) / (N eta)
          I = tau / k          V(need) = k w(motor) + I R          duty = V(need) / V          traction limit = s mu m g cos(theta)
(Friction in the motor is ignored; all coefficients are typical approximate values.)

(a) sin 15 deg = 0.25882, cos 15 deg = 0.96593.
    accelerate:  3 x 0.4                     = 1.200 N
    climb:       3 x 9.81 x 0.25882          = 7.617 N
    rolling:     0.04 x 3 x 9.81 x 0.96593   = 1.137 N
    F = 1.200 + 7.617 + 1.137 = 9.954 N.
    Wheel radius r = 0.05 m.   tau(wheel) = 9.954 x 0.05 / 4 = 0.1244 N m per wheel.

(b) Stall torque = k V / R = 0.02 x 12 / 4 = 0.060 N m.   Stall current = V / R = 12 / 4 = 3.0 A.
    No-load speed = V / k = 12 / 0.02 = 600 rad/s = 5730 rpm.
    Wheel speed at 0.8 m/s: w = v / r = 0.8 / 0.05 = 16 rad/s = 152.8 rpm.
    Ratio for 75 % of no-load speed: N = 0.75 x 600 / 16 = 28.1, so choose 30:1 (a whole number near it).

(c) tau(motor) = 1.5 x 0.1244 / (30 x 0.75) = 0.1866 / 22.5 = 0.00829 N m.   (Without the margin: 0.00553 N m.)
    Current (with margin) = 0.00829 / 0.02 = 0.415 A.   (Without the margin: 0.00553 / 0.02 = 0.2765 A.)
    Percentage of stall torque = 0.00829 / 0.060 = 13.8 %: comfortably light, the motor will run cool.

(d) w(motor) = N v / r = 30 x 16 = 480 rad/s.
    V(need) = k w + I R = 0.02 x 480 + 0.2765 x 4 = 9.60 + 1.106 = 10.71 V.   Duty = 10.71 / 12 = 89 %.
    The headroom is thin: a fresh 12 V pack will do it, but a pack sagging under load toward 10 to 11 V will not hold 0.8 m/s on the slope. A lower ratio such as 25:1 would
    need less voltage (and more torque, which the motor has plenty of at 13.8 % of stall), so it is the better choice here.
    (Check 25:1: tau(motor) = 1.5 x 0.1244 / (25 x 0.75) = 0.00995 N m = 16.6 % of stall; V(need) = 0.02 x 400 + 0.2765 x 4 x (30/25) = 8.0 + 1.33 = 9.33 V, duty 78 %.)

(e) All four driven, s = 1:   F(max) = 0.6 x 3 x 9.81 x 0.96593 = 17.06 N.   Needed 9.95 N, so the wheels grip, with a margin of 1.7.
    Two driven wheels carrying half the weight, s = 0.5:   F(max) = 0.6 x 0.5 x 3 x 9.81 x 0.96593 = 8.53 N.   Needed 9.95 N: the driven wheels SLIP and the robot cannot climb.
    Steepest slope (all four driven) from tan(theta) = s mu - Crr = 0.6 - 0.04 = 0.56, so 29.2 deg; with only two driven wheels: 0.3 - 0.04 = 0.26, so 14.6 deg, just below the 15 deg required.
    Lesson: more or bigger motors would not help the two-wheel version, since the limit is grip, not torque. Drive all four wheels, add weight over the driven wheels, or use a grippier tyre.

Check on method: the traction check is separate from the torque check. A common slip is to size the motors for the force needed and forget that the wheels can deliver no more than mu x (load on the driven wheels).