Robotics Fundamentals 1, Chapter 2: Challenge 2 solution (centre of gravity and tipping)

Formulas: x(cg) = sum(m x) / M          z(cg) = sum(m z) / M          tan(theta) = d / h          a(tip) = g d / h          slip when tan(theta) = mu

(a) Table (x forward from the axle, z up, metres):
    Part          m (kg)    x         z         m x        m z
    Chassis       0.25      0.000     0.030     0.0000     0.00750
    Motors        0.14      0.000     0.032     0.0000     0.00448
    Wheels        0.08      0.000     0.033     0.0000     0.00264
    Battery       0.30     -0.020     0.040    -0.0060     0.01200
    Controller    0.06      0.030     0.065     0.0018     0.00390
    Sensor mast   0.05      0.120     0.090     0.0060     0.00450
    Caster        0.03      0.110     0.010     0.0033     0.00030
    Total         0.91                          0.0051     0.03532
        M = 0.91 kg
        x(cg) = 0.0051 / 0.91 = 0.0056 m = +5.6 mm (just ahead of the axle)
        z(cg) = 0.03532 / 0.91 = 0.0388 m = 38.8 mm
    The CoG lies between the axle (x = 0) and the caster (x = 0.110), so it is inside the support polygon, but only just.
    Caster share = x(cg) / 0.110 = 0.0056 / 0.110 = 5.1 %.   W = 0.91 x 9.81 = 8.93 N, so the caster carries about 0.45 N: it will lose contact on a bump.

(b) Sideways: tan(theta) = (0.14 / 2) / 0.0388 = 0.070 / 0.0388 = 1.804     theta = 61.0 deg
    Uphill (tips backwards about the axle): d = 0.0056     tan(theta) = 0.0056 / 0.0388 = 0.144     theta = 8.2 deg
    Downhill (tips forwards about the caster): d = 0.110 - 0.0056 = 0.1044     tan(theta) = 2.69     theta = 69.6 deg
    Slipping: tan(theta) = 0.8     theta = 38.7 deg.
    Facing uphill, tipping (8.2 deg) happens long before slipping (38.7 deg): the robot would rear up on even a gentle ramp.

(c) Lifting the front when accelerating: a = g x(cg) / h = 9.81 x 0.0056 / 0.0388 = 1.42 m/s^2.
    That is a modest acceleration: the robot will pop a wheelie if the motors can accelerate it at more than 1.4 m/s^2.

(d) Required total moment: 0.030 x 0.91 = 0.0273.   Moment of everything but the battery: 0.0051 - (-0.0060) = 0.0111.
        x(battery) = (0.0273 - 0.0111) / 0.30 = 0.0162 / 0.30 = +0.054 m (54 mm forward of the axle; it was 20 mm behind, so it moves 74 mm).
    Same height, so z(cg) is unchanged at 38.8 mm.
        New uphill tipping angle: tan(theta) = 0.030 / 0.0388 = 0.773     theta = 37.7 deg   (now about the same as the slipping angle of 38.7 deg)
        New caster share: 0.030 / 0.110 = 27.3 %  (2.4 N): in the usual fifth to a third.

(e) Any two of: lower the CoG (mount the battery lower, shorten or remove the sensor mast, shift weight to the base); move heavy parts further forward (towards the front, so long as the share on the caster stays reasonable); move the caster or drive axle so that the rear wheels are further back; add a rear anti-tip wheel or skid; use a longer wheelbase. Also, use a slower, smoother acceleration ramp on slopes.

Check on method: for tipping uphill, d is measured to the axle (the rear contact), not to the caster. A common slip is to use the wrong edge: forward of the CoG for tipping backwards, behind it for tipping forwards.