Brachistochrone Subway Tunnel: Fastest Tunnel Shape Explained

Introduction

A subway tunnel could, in principle, operate without external energy (like electricity) if we assume zero friction everywhere. The train would descend, convert potential energy to kinetic energy, coast through a low point, then climb back up—all by exchanging potential and kinetic energy.

Historical background

What is the tunnel shape that minimizes travel time between two stations separated by a horizontal distance $d$? This is the Brachistochrone problem, first posed by Johann Bernoulli in the late 17th century as “what path gives the shortest time for an object to slide along a smooth curve from point A to point B under the influence of gravity?” Bernoulli, his brother, Leibniz, Newton, and two others submitted solutions—the correct curve is the cycloid. (The last entry was submitted anonymously but was later recognized as Newton’s.) Later, Euler and Lagrange generalized the theory; for more historical info, see:

http://en.wikipedia.org/wiki/Brachistochrone_curve

Problem setup

For a subway we want to minimize the station-to-station travel time. We can apply the Brachistochrone solution to half the journey (from the first station down to the tunnel bottom) and then double the time to get the full transit time. The only fixed parameter is $d$, the above-ground horizontal distance between the stations.

Set up an $x$–$y$ coordinate system with $x$ along the ground and $y$ the depth below ground. The train starts at $x=0,\,y=0$ and ends at $x=d,\,y=0$. Let the maximum depth be $h$; define $h>0$.

Variational approach and equations

The Brachistochrone problem is a special case in the calculus of variations. It extremalizes a definite integral of the form

$$I = \int_{A}^{B} f\!\left(x,y,\frac{dy}{dx}\right)\,dx$$

where A and B are the fixed endpoints along the slide. The Euler differential equation gives the condition for an extremal. The Brachistochrone is solvable because in this case $f = f\bigl(y,\frac{dy}{dx}\bigr)$ only.

The integral to be minimized is

$$I = \int_{A}^{B} \frac{ds}{v}$$

Here $ds$ is the element of arc length and $v$ the speed along the curve. Since $ds^2 = dx^2 + dy^2$ and the speed follows from $\frac{v^2}{2} = g\,y$, we obtain

$$f = \sqrt{\frac{1+\left(\frac{dy}{dx}\right)^2}{2 g y}}.$$

The corresponding Euler equation takes the form

$$y’\frac{\partial f}{\partial y’} – f = c$$

with $y’ = \frac{dy}{dx}$.

In this presentation I omit the intermediate derivation steps, but the integration leads to the cycloid as the time-minimizing path. (I can provide derivation links or references on request.)

Optimal depth and travel time

I calculated the maximum depth required for the minimum station-to-station transit time. The result is

$$h = \frac{d}{\pi},\qquad T = \sqrt{\frac{2\pi d}{g}}.$$

For example, with $d = 1\ \text{mile}$ (given here as 1600 m in the original calculation), this gives $h = 0.32\ \text{miles}$ or 1,681 ft (488 m) and $T = 32.0\ \text{s}$. That is quite deep and steep: the maximum speed at the bottom would be roughly 225 mph and the above-ground average speed about 112 mph.

Of course, air resistance and rail friction would reduce these speeds and would require externally supplied energy in practice, but the idealized values give a useful baseline.

Shallower tunnels and repeated cycloid segments

If a shallower maximum depth is desired—for example $h \approx 800\ \text{ft}$ (244 m)—one option is to preserve the cycloidal shape but shorten the cycloid’s spatial period. In that case the train would descend to 800 ft, ascend back up at $d = 1/2\ \text{mi}$, then repeat the cycloidal arc for the second half of the mile. Transit time scales with the square root of the above-ground distance per cycloid period, so the 1-mile time for that two-cycloid arrangement increases to $T \approx 45.2\ \text{s}$.

Alternative: circular-arc tunnel (no intermediate pop-up)

Another design constraint is to avoid popping back to the surface mid-distance. I considered an arc of a circle spanning $x=0$ to $x=d$ with the arc bottom at 244 m below the x-axis. That circle is centered at $\left(\tfrac{d}{2},\,R-h\right)$, with

$$R = \frac{\left(\tfrac{d}{2}\right)^2}{8h} + \frac{h}{2}.$$

The transit-time expression for a circular arc involves elliptic integrals (“elliptic integral of the first kind with parameter $m=k^2$”) and does not yield a simple closed form in elementary functions, so I evaluated transit times numerically.

Numerical results (Excel simulations)

• d = 1 mile, depth = 244 m: $T = 38.7\ \text{s}$ (this beats the two-cycloid version at 45.2 s).
• d = 1 mile, h = 100 m: $T = 57.2\ \text{s}$.
• d = 1 mile, h = 50 m: $T = 80.0\ \text{s}$.
• d = 1 mile, h = 25 m: $T = 112.7\ \text{s}$.
• d = 1 mile, h = 10 m: $T = 177.3\ \text{s}$.

Conclusion

These results show that a shallow tunnel does not necessarily give acceptable transit times for fast travel, and that the ideal brachistochrone solution (a single deep cycloid) is not likely cost-effective for a real subway system. A compromise—shallower circular arcs or repeated cycloidal segments—can reduce maximum depth while keeping transit time within a useful range.