Turning circles and pivots

If the wheel speeds are steady, the robot's centre drives along a circle. How big is that circle? Forward speed v and turning rate ω together answer it. A point moving at speed v while its direction turns at ω radians per second moves on a circle of radius

R = v / ω

You can see it makes sense: fast forward motion (large v) with a gentle turn (small ω) draws a big circle; a slow crawl with a quick turn draws a tight one.

R from the wheel speeds. Put the two formulas from the last lesson into R = v / ω and the track width appears:

R = (L / 2) × (v_R + v_L) / (v_R − v_L)

Worked example. L = 0.20 m, v_L = 0.20 m/s, v_R = 0.30 m/s: R = 0.10 × (0.50 / 0.10) = 0.5 m. The robot's centre circles a point 0.5 m to its left. Check with v and ω: 0.25 / 0.5 = 0.5 m. Both routes agree.

Another check. v_R = 0.40, v_L = 0.20, L = 0.20: R = 0.10 × (0.60 / 0.20) = 0.3 m. The bigger speed ratio gave a tighter circle.

Special cases.

• v_R = v_L: the denominator is zero. The circle's radius is infinite, which is just another way of saying the path is a straight line.
• v_L = 0 (one wheel stopped): R = (L / 2) × v_R / v_R = L / 2. The robot's centre circles the stopped wheel at a radius of half the track width, 0.10 m for L = 0.20 m.
• v_R = −v_L (pivot): the sum in the numerator is zero, so R = 0. The robot turns in place around its own centre.

How long does a full turn take? One full turn is 2π radians (about 6.28), and a turn at ω takes time t = 2π / ω. For a pivot with v_R = 0.2 m/s and v_L = −0.2 m/s and L = 0.20 m: ω = 0.4 / 0.2 = 2 rad/s, so a 360° pivot takes 6.28 / 2 ≈ 3.14 s. Pivoting the same robot with half the wheel speeds would take twice as long.