A Lighthouse Beam That Sweeps the Shore at Pi Miles a Second

Differentiating y = L tan(wt) gives a spot speed of w R squared over L, so the footprint accelerates with the square of its distance from the lamp: a tenth of pi directly opposite, and exactly pi miles per second nine miles along. The 9 is the along-shore leg, which makes 90 the squared hypotenuse rather than the square of nine, and that misreading is the usual failure. Nothing physical moves at that speed, and a straight coast running 2310 miles would carry a nominally faster-than-light spot carrying no information at all.

A lighthouse stands three miles off a straight coast and turns once a minute. Its beam paints a bright spot on the shore, and that spot slides along the coastline. How fast does it move?

The question has no single answer, which is the first thing to notice. The spot accelerates as it runs out along the coast, and it does so quickly: nine miles from the nearest point it is travelling at exactly $\pi$ miles per second, about 11,310 miles an hour, while directly opposite the lamp it was doing a tenth of that.

One derivative, and where the tangent comes from

Put the origin at the point of the coast nearest the lamp, measure $y$ along the shore, and let $\theta$ be the angle of the beam from the perpendicular. The beam, the perpendicular and the shore make a right triangle with the offshore leg $L$ , so

y = L\tan\theta, \qquad \theta = \omega t, \qquad \omega = \frac{2\pi}{60} = \frac{\pi}{30}\ \text{rad/s}

(1)

Differentiating and using $\sec^2 = 1 + \tan^2$ ,

V = \frac{dy}{dt} = \omega L \sec^2\theta = \frac{\omega\left(L^2 + y^2\right)}{L} = \frac{\omega R^2}{L}

(2)

where $R$ is the distance from the lamp to the spot. The rate at which the beam turns is constant; the rate at which its footprint moves is proportional to the square of how far away that footprint is.

Related rates

Two quantities tied by a geometric relation have their rates of change tied by the derivative of that relation. Here the relation is $y = L\tan\theta$ and the tie is $\dot y = L\sec^2\theta \cdot \dot\theta$ . Nothing physical was added; the acceleration of the spot is a property of the tangent function.

The same answer without differentiating

Equation (2) can be read off a picture, which is worth doing because it explains the square. In a short time $dt$ the beam sweeps through $\omega\,dt$ , so a point on the beam at distance $R$ moves sideways by $R\,\omega\,dt$ , at right angles to the beam. That is one factor of $R$ .

The spot on the shore is not that point, because the shore is oblique to the beam. The angle between them is the same $\theta$ , so a sideways displacement of the beam corresponds to an along-shore displacement larger by $1/\cos\theta = R/L$ . That is the second factor. Multiply: $V = \omega R \cdot R/L$ , which is equation (2) with no calculus at all.

Nine miles along

The 9 in the question is the along-shore leg, not the beam. So the squared distance is

R^2 = L^2 + y^2 = 3^2 + 9^2 = 90 = 10\,L^2

(3)

which is ten times its value at the nearest point. By equation (2) the speed is ten times as well, and the arithmetic closes exactly:

V = \frac{\pi}{30}\cdot\frac{90}{3} = \pi\ \text{miles per second}

(4)

Fig. 1 — The right triangle is the whole calculation. Reading the 9 as the beam length instead of the along-shore leg is the one place this problem is usually misread.

The $\pi$ is not a coincidence that has to be admired. It is the $\pi$ of the full turn, surviving the cancellation $10\,\omega L = 10 \cdot \tfrac{2\pi}{60}\cdot 3 = \pi$ . The numbers 3, 9 and 60 were chosen so it would land there.

Reading the square law

Equation (2) says the speed profile along the coast is a parabola in $y$ with its minimum at the nearest point. Its value there is $\omega L = \pi/10$ , which is the slowest the spot ever moves. The speed doubles by $y = L = 3$ miles out, quadruples by $y = L\sqrt3 \approx 5.2$ , and grows without any upper bound.

Fig. 2 — Speed against position, from equation (2). Most of the growth happens in the last third of the run, which is why the answer feels too large.

There is a scaling reading of equation (2) worth having, because it says which quantity the answer really depends on. At the position $y = 3L$ , whatever $L$ is, the speed is $10\,\omega L$ : it grows in proportion to how far offshore the lamp sits. A lighthouse six miles out, watched eighteen miles along the same coast, throws its spot at $2\pi$ miles per second. So a distant lighthouse is not the gentler one. Distance buys the spot a longer lever, and moving the lamp further out makes every corresponding position along the coast faster rather than slower.

Pi is a speed at a place, not an average

Inverting equation (1) gives the time to reach nine miles: $t = \omega^{-1}\arctan(y/L) = (30/\pi)\arctan 3 = 11.93$ seconds. Nine miles in 11.93 seconds is an average of 0.755 miles per second, which is under a quarter of $\pi$ . The headline figure describes one instant.

That gap between the instantaneous and the average value is what the parabola in figure 2 is telling you, and it is the reason a viewer who tries to sanity-check $\pi$ by dividing a distance by a time will find a contradiction that is not there.

Where the model stops describing the world

Nothing travels at $\pi$ miles per second here. The spot is a place where light happens to land, and consecutive positions of it are lit by different photons that each travelled outward from the lamp. No object accelerates, and no signal runs along the beach, so there is nothing for a speed limit to apply to.

The idealisation does eventually collide with physics, and it is amusing to work out where. Setting $\omega R^2/L$ equal to the speed of light gives $R = \sqrt{cL/\omega} \approx 2310$ miles, so a perfectly straight coast that ran for two thousand miles would carry a nominally faster-than-light spot. It carries no information at that speed, and long before you get there the coast has curved, the earth has curved, and the beam has passed below the horizon.

Two smaller boundaries are worth naming. Equation (2) blows up as $\theta \to \pi/2$ , and at that instant the beam is parallel to the shore and there is no spot at all, so the infinity records a gap in the domain rather than an infinite speed. And the whole setup assumes the lamp turns at a constant rate; a real optic with several panels produces several spots, each obeying equation (2) independently at its own angle.

Sources and further reading

The technique behind equation (2) — Related rates
Where $\sec^2$ comes from — Differentiation of trigonometric functions
The constant in equation (1) — Angular velocity
Why a moving illuminated spot is exempt from the light-speed limit — Faster-than-light

The value $\pi$ was confirmed symbolically and then measured: the beam was swept numerically and the spot's speed read off with a central finite difference, giving 0.314159265 at the nearest point and 3.141593 at nine miles. Equation (2) matched the measurement at 200 positions between 0.15 and 30 miles to better than one part in a million.