03. Fastest descent

Release a ball from rest on a frictionless track. What shape reaches a point 1 m horizontally away fastest? A straight line is shortest, but it is not fastest. Johann Bernoulli posed this classical problem in 1696.

Item	Definition
States	Horizontal position $x$, depth $y$, speed $v$
Control	Track angle $\theta$, measured from downward vertical
Objective	Minimize arrival time $t_f$
Phases	One phase, 20 nodes, free final time
New concepts	Nonlinear dynamics, inequality constraints, a free terminal depth, analytic comparison

Situation

Rather than searching directly for a track, choose the ball’s direction at each instant. A steep descent builds speed but makes little horizontal progress; a shallow track advances horizontally but accelerates slowly. The fastest path balances these effects by descending steeply first, then turning horizontally.

Only terminal horizontal position is fixed. Depth is free, so the target is a vertical line. Gravity and horizontal distance are both set to 1.

Formulation

Dynamics

Depth is positive downward. Angle $\theta=0$ points down, and $\theta=90^\circ$ is horizontal.

$$\dot{x}=v\sin\theta,\qquad\dot{y}=v\cos\theta,\qquad\dot{v}=g\cos\theta.$$

The first two equations resolve velocity into components. Only the component of gravity along the track accelerates the ball. Since $v\dot v=g\dot y$, energy is conserved:

$$\tfrac12v^2=gy,\qquad v=\sqrt{2gy}.$$

Boundary conditions

$$x(0)=y(0)=v(0)=0,\qquad x(t_f)=l.$$

Terminal depth and speed are left to optimization.

Constraints

$$y\ge0,\qquad\theta\ge0,\qquad t_f\ge0.1.$$

The depth inequality excludes the initial node, already fixed by an equality. The angle condition restricts the intended forward-moving solution, and the time bound prevents a degenerate duration. Inequalities are written as nonnegative residuals.

Objective and parameters

$$J=t_f.$$

Symbol	Parameter	Value
$g$	PARAMETERS["g"]	1 m/s²
$l$	PARAMETERS["l"]	1 m

Initial guess

Use time 0 to 2, horizontal position linear from 0 to 1, depth linear from 0 to $1/\sqrt3$, and constant direction $30^\circ$. Speed starts from zero at all nodes. The guess is not dynamically consistent; optimization corrects it.

Results

The track, with depth increasing downward. Blue interpolation through 20 nodes overlaps the analytic cycloid (black dashed). The gray dotted line joins the same endpoints directly.

Speed and track angle in degrees. Dots are node values, blue curves are interpolants, and black dashed curves are analytic.

A ball on the fastest track (blue) compared with a ball on the straight track (white). The straight-track ball is still traveling when the cycloid reaches the endpoint.

• Arrival time: 1.772454 s, agreeing to about seven digits with $\sqrt\pi=$ 1.772454 s.
• The straight track takes 2.101 s. The longer cycloid is about 16% faster.
• Final depth: 0.6365 m, versus $2/\pi=$ 0.6366 m; final speed: 1.128 m/s.
• Direction changes nearly linearly from downward to horizontal: 90.0° at the end.

Explanation

The cycloid

The fastest descent curve is a cycloid, traced by a point on a circle rolling beneath a horizontal ceiling:

$$x=a(\phi-\sin\phi),\qquad y=a(1-\cos\phi).$$

It points down at $\phi=0$ and becomes horizontal at its deepest point, $\phi=\pi$. An optical analogy explains the shape: Snell’s law for a minimum-time path requires $\sin\theta/v$ to remain constant. Since $v=\sqrt{2gy}$, the required curve has $\sin\theta/\sqrt y$ constant, which gives the cycloid.

Why finish horizontally?

Terminal depth is free, so a downward velocity component at arrival contributes nothing to reaching the target line sooner. The best endpoint directs all speed horizontally. Thus $a\pi=l$ and

$$y(t_f)=\frac{2l}{\pi},\qquad t_f=\sqrt{\frac{\pi l}{g}}.$$

For $g=l=1$, $t_f\approx1.7725$. Along the cycloid, $\theta=\phi/2$ and $\phi$ grows linearly with time, so the angle history is a straight line.

Numerical accuracy

The dynamics are nonlinear because of trigonometric terms and products of states and controls. The smooth solution nevertheless gives about seven-digit agreement with 20 nodes (convergence plot).

Try it

• Set l to 2: duration becomes $\sqrt{2\pi}\approx2.507$ s and final depth $4/\pi\approx1.273$ m.
• Set g to 4: the track is unchanged, but duration halves to about 0.886 s.
• Reduce 20 nodes to 8 or 7 and compare errors (roughly $10^{-7}$ and $10^{-5}$). With 5 or 6 nodes, success can be reported despite errors exceeding 0.2 s. Solver success and accuracy are different checks.

Run

uv run python examples/03_brachistochrone.py

View the Python source on GitHub.