Robotics Fundamentals 1, Chapter 10: Challenge 3 solution (adapt the build to a heavier payload)

Formulas (Robotics Fundamentals 3 section 6 and this chapter):  F = m ( a + g sin(theta) + Crr g cos(theta) );  tau(wheel) = F r / 2;  tau(motor) = margin x tau(wheel) / ( N eta );  I = tau(motor) / k + I0;
          V(need) = k omega(motor) + I R,  omega(motor) = N v / r,  duty = V(need) / V(pack);   pack current of a motor = duty x I;   traction limit = 0.78 x mu x m g cos(theta)
          runtime  t = 0.8 x V x C / P ;   fuse = 125 % to 150 % of the peak working current
m = 2.2 kg, v = 0.3 m/s, a = 0.8 m/s^2, 10 degrees, r = 0.0325 m, N = 50, eta = 0.729, k = 0.006, R = 2.0 ohm, I0 = 0.15 A, stall torque 0.018 N m, Crr = 0.03, mu = 0.8.   sin 10 = 0.17365, cos 10 = 0.98481.

(a) Worst working load:
        F = 2.2 x ( 0.8 + 9.81 x 0.17365 + 0.03 x 9.81 x 0.98481 ) = 2.2 x ( 0.8 + 1.7035 + 0.2898 ) = 2.2 x 2.7933 = 6.145 N
        tau(wheel) = 6.145 x 0.0325 / 2 = 0.0999 N m
        tau(motor) = 1.5 x 0.0999 / ( 50 x 0.729 ) = 0.14979 / 36.45 = 0.00411 N m,   which is 0.00411 / 0.018 = 22.8 % of stall torque (inside the 25 to 30 % guide)
        I = 0.00411 / 0.006 + 0.15 = 0.685 + 0.15 = 0.835 A
        omega(motor) = 50 x 0.3 / 0.0325 = 461.5 rad/s;   V(need) = 0.006 x 461.5 + 0.835 x 2.0 = 2.769 + 1.670 = 4.44 V;   duty = 4.44 / 7.4 = 60 %   (under the 90 % limit: enough voltage left)
        Traction:  0.78 x 0.8 x 2.2 x 9.81 x 0.98481 = 13.26 N, more than the 6.15 N needed:  the wheels do not slip.
    The same 50:1 motors cope: the heavier robot loads them to the same 22.8 % as the 2 kg example of Robotics Fundamentals 3.

(b) Flat cruise:  F = 2.2 x 0.03 x 9.81 = 0.647 N;  tau(wheel) = 0.0105 N m;  tau(motor) = 0.0105 / 36.45 = 0.000289 N m;  I = 0.000289 / 0.006 + 0.15 = 0.198 A
        V(need) = 2.769 + 0.198 x 2.0 = 3.165 V;   duty = 3.165 / 7.4 = 42.8 %;   pack current per motor = 0.428 x 0.198 = 0.0847 A
    Pack current for the motors on average:  1.5 x 2 x 0.0847 = 0.254 A;   losses 5 % = 0.0127 A;   logic = 5 x 0.170 / 0.90 / 7.4 = 0.1276 A
    Total average pack current = 0.1276 + 0.2542 + 0.0127 = 0.3946 A;   average power  P = 7.4 x 0.3946 = 2.92 W.

(c) For 3 hours:   E(usable) = 2.92 x 3 = 8.76 Wh;   E(nominal) = 8.76 / 0.8 = 10.95 Wh;   capacity = 10.95 / 7.4 = 1.48 Ah.
    With the cold and aged factors (0.8 x 0.9 = 0.72):  10.95 / 0.72 = 15.2 Wh;   15.2 / 7.4 = 2.06 Ah.
    Choose a standard pack:  1500 mAh gives 0.8 x 7.4 x 1.5 / 2.92 = 3.04 h on a good day;  2200 mAh covers the 3 hours even cold and aged (0.72 x 0.8 x 7.4 x 2.2 / 2.92 = 3.21 h).   The honest choice is the 2200 mAh pack, or a spare 1500 mAh pack.
    Weight: a larger pack weighs more: recheck the total mass and the centre of mass.

(d) Peak working pack current:  motors 2 x 0.5009 = 1.0017 A  (duty 60 % x 0.835 A = 0.501 A each);   losses 5 % = 0.0501 A;   logic peak = 5 x 0.215 / 0.90 / 7.4 = 0.1614 A;   total 1.213 A.
    Fuse:  1.25 x 1.213 = 1.52 A   to   1.50 x 1.213 = 1.82 A.   No common value sits inside that window (1.5 A is just under the lower bound, 2 A above the upper):  choose 2 A (165 %), still far below the 11 A wire rating.
    The stall is unchanged (the same motors): 4.865 A + 0.161 A = 5.03 A, about 2.5 times the new fuse:  it still blows quickly.

(e) What stays:  the wheels, gearbox, motors, 7.4 V pack chemistry, the driver and its stall rule, the star wiring, the wire gauge, the e-stop, the watchdog, the cutoff thresholds and the control structure of the plan.
    What changes:  the fuse (1 A becomes 2 A), the pack capacity (1000 mAh becomes 1500 mAh or 2200 mAh), the tuning of the speed and line loops (more mass means more inertia and a longer stopping distance:  re-measure the braking deceleration and the stopping distance, and keep the obstacle margin), the frame (stiffer, with the payload low and central, Robotics Fundamentals 2), and the power budget numbers in section 3.