The Fastest Way Down

What it is

Three frictionless ramps from the same point to the same point. A straight line, a circular arc, and a cycloid, which is the curve traced by a point on the rim of a rolling wheel. Beads released together. Drag to move the finish.

The cycloid wins every time, and on a shallow slope it wins by a lot.

How it works

The bead’s speed at depth y is √(2gy) regardless of the path, so the time down a curve is the integral of ds/√(2gy).

Steeper early means faster sooner, which buys speed to spend on the flatter part later, and the cycloid is the exact shape where that trade is optimal. Bernoulli posed it in 1696 and Newton reportedly solved it overnight.

Finding the right cycloid for a given endpoint has no closed form, so the wheel’s sweep angle comes out of bisection on the ratio x/y and the radius follows.

What surprised me

Not the result, which is famous. It’s that the integral is singular and my first instinct for handling it would have been wrong.

The bead starts at rest, so at the first step the speed is zero and ds/√(2gy) is infinite. Evaluating the integrand at each segment’s start produces Infinity on the very first term and poisons everything downstream. Evaluating at the segment’s midpoint depth instead is finite everywhere and converges properly, with the coarse estimate within 2% of the fine one.

So the calibration test mattered more than usual. A full cycloid arch, cusp to lowest point, takes exactly π√(r/g) whatever its size, so I could check the integrator against a closed form before trusting it to judge a race.

Without that, a subtly biased integrator would have produced a plausible-looking result and I’d have had no way to know. All three times would have been wrong together, and the ordering might still have come out right by luck.

What I would do next

The tautochrone property, where a bead released anywhere on a cycloid reaches the bottom in the same time. Same curve, considerably harder to believe.