Robotics Fundamentals 1, Chapter 3: Challenge 2 solution (gearbox, belt and worm)

Formulas: N = teeth(driven) / teeth(driving)          w(out) = w(in) / N          tau(out) = tau(in) N eta          P = tau w
          belt/gear ratios multiply, efficiencies multiply          v = w r          worm: eta = tan(lambda) / tan(lambda + phi), phi = atan(mu)
(Efficiencies are typical approximate values.)

(a) Stage 1: 48 / 12 = 4:1.   Stage 2: 45 / 15 = 3:1.   Total ratio N = 4 x 3 = 12:1.
    Total efficiency = 0.92 x 0.92 = 0.8464 (84.6 %).

(b) Output speed = 4000 / 12 = 333.3 rpm.
    Output torque = 0.020 x 12 x 0.8464 = 0.2031 N m.

(c) P(in) = tau w = 0.020 x (2 pi x 4000 / 60) = 0.020 x 418.9 = 8.378 W.
    P(out) = 0.2031 x (2 pi x 333.3 / 60) = 0.2031 x 34.91 = 7.091 W  (equally 8.378 x 0.8464 = 7.091 W).
    Lost as heat = 8.378 - 7.091 = 1.287 W (15.4 %).

(d) Wheel radius 0.05 m.   v = w r = 34.91 x 0.05 = 1.745 m/s.
    Push at the rim: F = tau / r = 0.2031 / 0.05 = 4.06 N.
    (Check: F v = 4.06 x 1.745 = 7.09 W, the output power.)

(e) Belt stage 32 / 16 = 2:1, eta = 0.97.   Total ratio = 12 x 2 = 24:1,  total eta = 0.8464 x 0.97 = 0.821.
    Wheel speed = 4000 / 24 = 166.7 rpm.   Torque = 0.020 x 24 x 0.821 = 0.3941 N m.
    v = (2 pi x 166.7 / 60) x 0.05 = 17.45 x 0.05 = 0.873 m/s.   Push F = 0.3941 / 0.05 = 7.88 N.
    Half the speed and 1.94 times the push: the belt halves the speed and doubles the torque, less its 3 % loss.

(f) Ratio = 30 teeth / 2 starts = 15:1.
    phi = atan(0.08) = 4.574 deg.   eta = tan 8 / tan(8 + 4.574) = 0.14054 / 0.22302 = 0.630 (63 %).
    Output torque = 0.020 x 15 x 0.630 = 0.1890 N m.   Output speed = 4000 / 15 = 266.7 rpm.
    Self-locking needs lambda < phi: here 8 deg > 4.57 deg, so it is NOT self-locking (it can be back-driven, though inefficiently).
    Compare with the spur gearbox: the worm gives a similar torque at a lower efficiency (63 % against 85 %) but needs only one stage and turns the axis through a right angle.

(g) Backlash at the rim = r x angle(rad) = 0.05 x (1.5 x pi / 180) = 0.05 x 0.02618 = 0.00131 m = 1.3 mm.

Check on method: always multiply the stage ratios for the total ratio AND multiply the stage efficiencies. A common slip is to apply the efficiency to the speed as well as the torque; the loss comes off the torque (and the power), the speed is fixed by the ratio.