Robotics Fundamentals 1, Chapter 9: Challenge 3 solution (payload, motion, deflection and grip)

Formulas: tau = m g r cos( alpha )  (worst at horizontal);  tau(out) = tau(motor) N eta
          trapezoid:  ta = vmax / amax;  long move (D >= vmax^2 / amax):  T = D / vmax + ta;  short move:  ta = sqrt( D / amax ),  T = 2 ta,  v(peak) = amax ta
          grip:  mu N n >= m ( g + a ) SF ;  vacuum  F = deltaP x A
          deflection  delta = F L^3 / ( 3 E I ),  I = b h^3 / 12 ;   tip error from an angle error = r x delta_theta ( radians )
g = 9.81 m/s^2.

(a) Arm and gripper alone:
        arm:      0.06 x 9.81 x ( 0.22 / 2 ) = 0.06 x 9.81 x 0.11 = 0.0647 N m
        gripper:  0.04 x 9.81 x 0.22 = 0.0863 N m                      total 0.1511 N m
    Working torque of the servo:  0.5 x 1.8 = 0.90 N m.   Left for the payload:  0.90 - 0.1511 = 0.7489 N m.
    Largest payload at the tip:  m = 0.7489 / ( 9.81 x 0.22 ) = 0.7489 / 2.158 = 0.347 kg.

(b) Payload 0.30 kg:   payload torque = 0.30 x 9.81 x 0.22 = 0.6475 N m.
        Horizontal:         0.1511 + 0.6475 = 0.7985 N m      (less than the 0.90 N m working torque: margin 0.90 / 0.7985 = 1.13, which is thin)
        40 deg above horizontal:   0.7985 x cos 40 = 0.7985 x 0.7660 = 0.6117 N m
    The HORIZONTAL pose is the worst and sizes the servo.  The arm must be able to hold it there even if it rarely goes there.  (With so little margin, a stronger servo
    or a shorter arm would also leave room for acceleration torque.)

(c) Output torque = 0.12 x 20 x 0.85 = 2.04 N m, continuous.
    The need is 0.7985 N m (part (b), horizontal, holding): the margin is 2.04 / 0.7985 = 2.55, comfortably inside the continuous rating.  The gearbox has also raised the available torque far above
    the 0.90 N m of the derated servo, at the cost of speed (the output turns 20 times slower than the motor).

(d) 120 deg = 2.0944 rad,  vmax = 1.5 rad/s,  amax = 6 rad/s^2.
        Ramp time  ta = 1.5 / 6 = 0.25 s.     Ramp distance = 1.5^2 / ( 2 x 6 ) = 0.1875 rad;  both ramps = 0.375 rad = vmax^2 / amax.
        D = 2.0944 >= 0.375: there is a cruise.   Cruise time = 2.0944 / 1.5 - 0.25 = 1.3963 - 0.25 = 1.1463 s.
        Total T = 2.0944 / 1.5 + 0.25 = 1.3963 + 0.25 = 1.6463 s   ( 0.25 + 1.1463 + 0.25 = 1.6463 s )

    15 deg = 0.2618 rad:  0.2618 < 0.375, so there is NO cruise (a triangle).
        ta = sqrt( 0.2618 / 6 ) = sqrt( 0.04363 ) = 0.2089 s.       T = 2 x 0.2089 = 0.4178 s.       v(peak) = 6 x 0.2089 = 1.253 rad/s  (below the 1.5 allowed).
        Check:  area = 1/2 x T x v(peak) = 0.5 x 0.4178 x 1.253 = 0.2618 rad.

(e) Friction grip:   load to hold = m ( g + a ) = 0.8 x ( 9.81 + 3 ) = 0.8 x 12.81 = 10.248 N.   With SF = 2:  20.496 N of friction.
        N = 20.496 / ( mu x n ) = 20.496 / ( 0.4 x 2 ) = 25.62 N per finger.
    Vacuum:  A = pi x 0.0125^2 = 4.909e-4 m^2 (a 25 mm cup, radius 12.5 mm).   F = 50000 x 4.909e-4 = 24.54 N.
        The same object needs 20.496 N with SF = 2 ( 10.248 x 2 ), and 24.54 N is more than that: the cup is enough, with a little to spare.
        Largest mass at SF = 2:  24.54 / ( 2 x 12.81 ) = 0.958 kg  ( more than the 0.8 kg held ).

(f) F = 0.30 x 9.81 = 2.943 N,  L = 0.25 m,  E = 69 GPa.
    3 mm vertical (flat):  I = 0.025 x 0.003^3 / 12 = 5.625e-11 m^4.        delta = 2.943 x 0.25^3 / ( 3 x 69e9 x 5.625e-11 ) = 2.943 x 0.015625 / 11.644 = 0.04598 / 11.644 = 3.95 mm.
    On edge (25 mm vertical):  I = 0.003 x 0.025^3 / 12 = 3.906e-9 m^4  ( ( 25 / 3 )^2 = 69.4 times larger ),   delta = 0.04598 / ( 3 x 69e9 x 3.906e-9 ) = 0.04598 / 808.6 = 0.057 mm.
    Turning the bar on edge makes it 69 times stiffer for the same material and weight; the tip droops 0.057 mm instead of 3.95 mm.   (The arm is then weak sideways, so it suits a vertical-plane arm.)
    Backlash:  0.8 deg = 0.01396 rad;  tip error = 0.40 x 0.01396 = 5.6 mm  of free movement at the hand, which no deflection calculation will show because the links are not bending: the gears are loose.