Artificial Wasteland · Pattern
Put a bug at each corner of a square, each one crawling always straight at the next. They wind into the center and meet. How far does each bug walk before it gets there? Exactly one side of the square. Operate the chase for any number of bugs, and watch a finite path that turns forever.
Four beetles sit at the four corners of a square tile. On a signal they start to crawl, all at the same steady pace, and each one keeps its head pointed straight at the beetle to its left. The one it is chasing is also moving, so the line of the chase keeps swinging round, and the four of them curl inward in a tightening pinwheel until they collide in a single knot at the middle.
It is an old puzzle, usually posed with bugs or mice or dogs, and it has two answers that feel like they should be hard and turn out to be clean. The first: how long does each bug walk? The second, stranger one: how many times does the pinwheel go around before it closes? You can run the chase yourself first, and read the answers off it, before we show why they have to be true.
Each bug aims a fixed 45° off the line to the center, the mark of an equal-angle spiral. Drag the slider, then release the bugs.
Change the number of corners and the picture stays the same in kind: the bugs always keep the shape of a regular polygon that only shrinks and spins, and every one of them traces the same curve, a spiral that meets every spoke from the center at the same angle. That last fact is the whole key, so it is worth seeing why it cannot be otherwise.
Why it must be a spiral
Freeze the chase at any instant. Because the bugs sit on a regular polygon, pick one bug at a corner A, the bug it is chasing at the next corner B, and the center O. The two corners are the same distance r from the center, so triangle OAB is isosceles, and the angle it opens at the center is one full turn shared among the corners: ∠AOB = 360°/n.
The two base angles of an isosceles triangle are equal, and the three angles sum to 180°, so each base angle is (180° − 360°/n) / 2 = 90° − 180°/n. The base angle at A is precisely the angle between the direction the bug is heading (straight at B) and its own spoke to the center (straight at O). It depends only on n, not on how far in the bugs have crawled. The chase keeps a constant angle to the center at every instant, and a curve that crosses every radius at the same angle is, by definition, an equiangular (logarithmic) spiral. The same spiral the nautilus grows in, and the one a moth flies when it mistakes a candle for the moon.
Split the bug's steady speed v into the part aimed at the center and the part aimed sideways. The inward part is v · cos(90° − 180°/n) = v · sin(180°/n). Every one of those symbols is fixed, so the bug closes on the center at a perfectly constant rate. That single sentence carries both answers.
The first answer
If the distance to the center shrinks at a constant rate, the arithmetic is the arithmetic of a car at constant speed. Start at the circumradius r₀ (the distance from center to corner), close it at rate v·sin(180°/n), and the trip takes time r₀ / (v·sin(180°/n)). The bug moved at speed v the whole time, so the length it actually walked is
L = r₀ / sin(180°/n) = s / (2·sin²(180°/n)),
where s is the side of the polygon (a corner sits at distance s / (2·sin(180°/n)) from the center). The speed v has vanished from the answer entirely: fast bugs and slow bugs walk the same distance, they just take different times to do it. Put in n = 4 and the fraction collapses to L = s exactly. Four bugs on a square each walk one side of the square, no matter how they dawdle, and never mind that the road they walk is a curl that never once runs straight.
| bugs n | each walks (÷ side) | lead angle | closes at |
|---|
Path length as a multiple of the polygon's side. The row for your current slider setting is lit. These are the closed-form values; the instrument above measures the same numbers off a step-by-step chase and lands on them.
The second answer
Now the sideways part of the motion. It is v·cos(180°/n), also constant, and it is what swings the bug around the center. But swinging a fixed sideways speed around a center that is getting closer means the angular speed keeps rising: the nearer the middle, the faster the pinwheel spins. Work out the total angle turned and it comes to cot(180°/n) · ln(r₀ / r), and as the bugs close on the center, r → 0, the logarithm runs off to infinity.
The pinwheel goes around infinitely many times. The polygon completes turn after turn after turn, faster and faster, without limit, in the last sliver of its journey. And yet the road each bug walks to get there is the plain finite length from the first answer. Watch the turns counter in the instrument keep climbing as the bugs crowd the center: it is trying to reach a number that is not there. A finite path, wound around a point infinitely many times. It is Zeno's old trick worn inside out: not an endless task that never finishes, but a finished task with an endless amount of turning folded into its final moment.
The bug arrives. It has walked one side of the square, and it has gone around the center more times than any number you can name.
None of the numbers above are asserted. An offline verifier runs the literal chase (each bug takes a small step straight at the next, over and over) and measures every quantity from the motion that emerges, then matches it to the closed form. All six pass for n = 3, 4, 5, 6, 7, 12.
The verifier and its full run log live in the project's lab notebook at research/pursuit-curves/.
The problem is classical: the "mice problem," "bugs problem," or cyclic pursuit, posed since at least the nineteenth century with beetles, dogs, or missiles. The square case (each pursuer walks one side length) is the familiar result; the general s / (2 sin²(180°/n)) and the finite-path / infinite-winding pairing are recomputed here from scratch rather than quoted.