Robotics Fundamentals 1, Chapter 3: Challenge 1 solution (reading a motor's torque-speed line)

Formulas: V = k w + I R          tau = k I          tau(s) = k V / R          w(nl) = V / k
          tau = tau(s)(1 - w / w(nl))          P = tau w          P(max) = tau(s) w(nl) / 4          w = 2 pi n / 60
(Ideal model: friction ignored.)

(a) R = V / I(s) = 12 / 4 = 3.0 ohm.
    k = tau(s) / I(s) = 0.12 / 4 = 0.030 N m/A  (= 0.030 V s/rad, the back-EMF constant).
    No-load speed: w(nl) = V / k = 12 / 0.030 = 400 rad/s = 400 x 60 / (2 pi) = 3820 rpm.

(b) 0.12 N m / 0.0981 = 1.22 kg cm.

(c) w = 400 x (1 - 0.04 / 0.12) = 400 x 0.6667 = 266.7 rad/s = 2546 rpm.
    P(mech) = 0.04 x 266.7 = 10.67 W.
    I = tau / k = 0.04 / 0.030 = 1.333 A.
    P(elec) = V I = 12 x 1.333 = 16.0 W.
    Efficiency = 10.67 / 16.0 = 0.667 = 66.7 %  (in the ideal model this equals w / w(nl)).

(d) P(max) = 0.12 x 400 / 4 = 12.0 W, at half the no-load speed and half the stall torque:
    w = 200 rad/s = 1910 rpm,  tau = 0.06 N m,  I = 0.06 / 0.030 = 2.0 A.
    P(elec) = 12 x 2.0 = 24 W,  so efficiency = 12 / 24 = 50 %.

(e) At stall nothing is delivered, and the winding dissipates I^2 R = 4^2 x 3.0 = 48 W (equally V I = 12 x 4 = 48 W).
    That is four times the 12 W the motor could ever deliver, so a stalled small motor overheats within seconds. The driver must limit current, and the robot must not push against a wall with its motors on.

(f) Both scale with the voltage: tau(s) = k V / R = 0.030 x 6 / 3.0 = 0.060 N m (half), w(nl) = 6 / 0.030 = 200 rad/s = 1910 rpm (half).
    A battery that sags makes the robot both slower and weaker.

Check on method: in this ideal model the efficiency at any point is simply w / w(nl); a real motor's no-load current (friction) pulls the efficiency down at the light-load end, which is why a real motor's peak efficiency is a little below no-load speed and below 100 %.