Robotics Fundamentals 1, Chapter 5: Challenge 1 solution (time of flight and its errors)

Formulas: d = v t / 2          v = 331.3 + 0.606 T          dv = 0.606 dT          dd / d = sqrt( (dv/v)^2 + (dt/t)^2 )
          laser: d = c t / 2 ;  phase shift: d = c phi / (4 pi f) ;  unambiguous range = c / (2 f) ;  c = 3 x 10^8 m/s (approx)
          beam width at range d for a half-angle a:  w = 2 d tan(a)

(a) T = 25 deg C:  v = 331.3 + 0.606 x 25 = 346.45 m/s.
    d = 346.45 x 0.00840 / 2 = 1.4551 m (1.455 m).

(b) Assuming 20 deg C:  v = 331.3 + 0.606 x 20 = 343.42 m/s.
    d = 343.42 x 0.00840 / 2 = 1.4424 m.
    Error = 1.4424 - 1.4551 = -0.0127 m = -12.7 mm, which is -0.87 % (the ratio 343.42 / 346.45 = 0.9913).
    Too warm a real temperature makes the sound faster, the echo arrives sooner, and a program assuming a cooler day reports too short a distance.

(c) T = 35 deg C:  v = 331.3 + 0.606 x 35 = 352.51 m/s.
    Echo time for 2.00 m:  t = 2 x 2.00 / 352.51 = 0.011347 s = 11.35 ms.
    The program assumes 343.42 m/s:  d = 343.42 x 0.011347 / 2 = 1.9484 m.
    Error = 1.9484 - 2.0000 = -0.0516 m = -51.6 mm = -2.6 %.   (Check: 343.42 / 352.51 = 0.9742, and 0.9742 x 2.00 = 1.948 m.)

(d) At 1.5 m and 20 deg C:  t = 2 x 1.5 / 343.42 = 8.736 ms.
    Timing jitter 30 microseconds:   dt / t = 30e-6 / 8.736e-3 = 0.343 %, which is  v dt / 2 = 343.42 x 30e-6 / 2 = 5.15 mm  (this part of the error is the same at every range).
    Temperature +-2 deg C:           dv = 0.606 x 2 = 1.212 m/s, dv / v = 1.212 / 343.42 = 0.353 %, which at 1.5 m is 5.29 mm.
    Combined (independent errors):   dd / d = sqrt(0.343^2 + 0.353^2) = 0.492 %,   dd = 1.5 x 0.00492 = 7.39 mm   (also sqrt(5.15^2 + 5.29^2) = 7.39 mm).

(e) Laser, 5 mm resolution:  the round trip is 2 x 0.005 = 0.010 m, so t = 0.010 / 3e8 = 33.3 picoseconds.
    Phase shift, f = 10 MHz:
        Unambiguous range = c / (2 f) = 3e8 / (2 x 10e6) = 15 m.
        phi = 60 degrees = pi / 3 rad:   d = 3e8 x (pi / 3) / (4 pi x 10e6) = 3e8 / (12 x 10e6) = 2.50 m.
        One degree of phase = 3e8 x 0.017453 / (4 pi x 10e6) = 0.0417 m = 41.7 mm   (the 20 MHz sensor of the chapter gave 20.8 mm per degree: a higher frequency gives finer resolution
        but a shorter unambiguous range, 7.5 m instead of 15 m).

(f) Beam width at 3 m, half-angle 15 degrees:  w = 2 x 3 x tan(15 deg) = 6 x 0.2679 = 1.61 m.
    The sensor only measures the time of the FIRST strong echo from anywhere inside that 1.6 m wide cone. A thin pole at 3 m reflects the same echo whether it stands on the axis or 0.8 m to either side,
    so the distance is known but the direction is not. Narrow-beam sensors (infrared, laser) or several ultrasonic sensors pointing in different directions are needed to find the side.