Brachistochrone: the Fastest Slide

Choose how far across and how far down the finish is, and drag the orange handle to draw a ramp of your own. Four beads are let go together, on the straight ramp, Galileo's circular arc, the cycloid and your curve, with no friction. Watch which arrives first, and read below why it does.

Worked example — Brachistochrone: the Fastest Slide

Straight ramp: time	0.6387 s
Radius of the cycloid's rolling circle	0.5729 m
Cycloid: time	0.5830 s
Galileo's arc (lowest point at the finish): time	0.5921 s
Your curve: time	0.5847 s
Ramp, slower than the cycloid by	9.55%
How far the cycloid goes below the finish	0.000 m
Along the cycloid, sin θ / v is everywhere	0.2109 s/m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Brachistochrone: the Fastest Slide

The brachistochrone (Greek for shortest time) is the curve along which a bead, sliding without friction from rest under gravity, gets from one point to a lower one fastest. It is the cycloid x = R(φ − sin φ), y = R(1 − cos φ) (y measured downwards), the path of a point on the rim of a circle of radius R rolling along the start line, chosen to pass through the finish. Its time is φ_B √(R/g), and along it sin θ / v = 1/(2√(gR)).

What each symbol means

φ: how far the rolling circle has turned, in radians; φ_B at the finish
R: radius of the rolling circle
θ: the slope's angle from the vertical
v: the bead's speed, √(2gy) at depth y on any frictionless track

The longer slide wins 🖖

Press 1 m across, 1 m down and watch the four beads. The straight ramp is the shortest track, 1.414 m, and it comes last at 0.6387 s. The cycloid is 0.060 m longer and arrives first at 0.5830 s, because it drops steeply at the start and the bead spends the rest of the trip moving fast. Every bead arrives at the same speed, 4.429 m/s, the speed a 1 m drop gives on any frictionless track; the shape decides only how soon. Johann Bernoulli put it in his challenge of June 1696: the straight line is the line of shortest distance, "but it is not the one which is traveled in the shortest time". Galileo had already shown in 1638 that a circular arc beats the straight chord, and thought the circle might be the answer. His quarter circle beats the ramp here but not the cycloid: 0.5921 s, 1.56% behind. Notice your orange curve: it only roughly follows the cycloid and is 0.29% behind it. Near the best path the time hardly changes, as it did for the lifeguard.

Far enough away, the fastest path goes below the finish 🖖

Press 4 m across, 1 m down. The cycloid that reaches the finish now dives 0.504 m below it and climbs back up, and still wins: 1.2117 s, against 1.8620 s for the ramp and 1.4847 s for Galileo's arc. Going down further than you need to looks like waste, but the depth buys speed for the long run across, and climbing the last half metre costs less time than crawling across at a shallow slope. The cycloid dips whenever the finish is more than π/2 = 1.5708 times as far across as it is down, because only then has its rolling circle turned more than half a revolution by the time it arrives. At 10 m across, 1 m down it goes 2.312 m below the finish and the ramp takes 115.71% longer. Sussmann and Willems point out that Bernoulli's own 1697 argument has a gap exactly here, at the bottom of the dip, which the later theory of optimal control closes.

The bead obeys Snell's law 🖖

Bernoulli's solution, published in May 1697, did not treat the bead as a bead. Its speed at depth y is √(2gy) whatever track it is on, so he imagined light travelling through a medium whose speed grows with depth in exactly that way, sliced into thin layers. Light crossing the layers keeps sin θ / v the same in every one, Snell's law with speeds, and the least-time path of that light is the least-time path of the bead. The second panel checks it on the bead itself: along the cycloid sin θ / v holds at 0.211 s/m from top to bottom, while on the straight ramp, whose angle never changes, it falls from 0.505 s/m a tenth of the way down to 0.168 s/m nine tenths of the way, as the bead speeds up. The rule that sent the lifeguard into the water at 36.18 m is the rule that draws the cycloid.

Problem solved in full

Let the finish be as low as you like 4 steps
The finish is 4 m across, but you may put it at any depth. How far down should it be for the bead to cover the 4 m sideways in the least time, and how much faster is that than the finish 1 m down? Press 4 m across, 1 m down to check the first step.

Open this problem in the tool
1. With the finish 1 m down, the page gives the cycloid's time:
  
  $H = 1 m : T = φ_{B} R / g = 1.2117 s$
2. If the depth is free, the best finish is where the cycloid itself runs level, at the bottom of its arch, so that the bead crosses the finish line moving straight across it. That is half a turn of the rolling circle:
  
  $\frac{d y}{d x}_{B} = 0 \Rightarrow φ_{B} = π$
3. Half a turn rolls the circle πR along, so R follows from the 4 m, and the drop is the circle's diameter:
  
  $π R = 4 \Rightarrow R = \frac{4}{π} = 1.2732 m, H = 2 R = 2.5465 m$
4. and the time is φ_B √(R/g):
  
  $T = π \frac{1 . 2732}{9 . 80665} = \frac{4 π}{9 . 80665} = 1.1320 s$
Answer

Put the finish 2.5465 m down, 1.5465 m deeper than before, and the bead covers the 4 m in 1.1320 s, 0.0797 s faster than the 1.2117 s to a finish 1 m down. Type 2.5465 into the depth field and the page agrees: φ_B is 180.00° and the cycloid takes 1.1320 s. Fowler's lecture notes reach the same shape in general: the fastest curve to a given distance across is the half-cycloid that arrives level, and its time works out to √(πL/g).

$H = 2.5465 m, T = 1.1320 s, 1.2117 - 1.1320 = 0.0797 s$

Learning path

Least time, least action

References (2)

The history and the mathematics: Bernoulli's challenge of June 1696 and its wording, Galileo's circular arc of 1638, the solutions of 1697, Bernoulli's own solution by Snell's law in thin layers, and the gap in it where the cycloid dips below the finish: H. J. Sussmann and J. C. Willems, "300 years of optimal control: from the brachystochrone to the maximum principle", IEEE Control Systems 17(3), 32–44 (1997).
The brachistochrone derived by the calculus of variations, and the fastest curve to a given horizontal distance when the drop is free, a half-cycloid that arrives level: M. Fowler, "The Calculus of Variations", lecture 2 of the lecture notes for Graduate Classical Mechanics (Physics 7010), University of Virginia.

By the 42Math team · Reviewed against published sources. Corrections welcome.