Thirty Seconds From a Cord That Burns Unevenly, on Any Cord At All

Halving the length and timing the flame is not a biased estimator of half the time, it is unrelated to it: across four thousand random cords the midpoint method scattered from under fifteen seconds to over forty-five. Lighting both ends gives exactly thirty on every cord, by an argument that never evaluates the burn rate.

Lambdia
August 10, 2026 · 9 min read

A cord burns from one end to the other in exactly sixty seconds. It does not burn at a steady rate: some stretches race, some crawl, and you have no idea which is which. You have a match and no watch. Measure thirty seconds.

The reflex is to halve the cord. Find its middle, light one end, and call time when the flame reaches the mark. On the cord in Fig. 1 that method reads forty-five seconds, and it would read something else on the next cord you picked up. Half the length is not half the time, and no amount of care measuring the middle will fix that.

What the varying rate actually means

Burn-time density

Treat the cord as the interval $[0,1]$ carrying a strictly positive function $r$ , the local seconds of burning per unit length. A flame starting at $0$ reaches position $t$ at time $T(t) = \int_0^t r(x)\,dx$ , and the end-to-end burn is $T(1) = \int_0^1 r = 60$ . The rate depends on position only, not on which direction the flame arrives from.

Under that model the reflex answer is asking for $T(1/2)$ , and nothing pins it down. The constraint is on the integral of $r$ over the whole cord, and knowing a total tells you nothing about how it splits. The cord drawn below is twelve stretches that each burn for five seconds, with lengths $0.03, 0.04, 0.06$ and so on up to a lazy $0.20$ near the far end. Its first nine stretches happen to fill exactly half the length, so $T(1/2) = 45$ on the nose.

Fig. 1 — One particular cord. Nine of its twelve stretches fit into the first half of its length, which is why the midpoint arrives late.

Rebuild the cord with the slow stretches at the near end instead and $T(1/2)$ jumps above fifty. Across four thousand randomly generated cords, the midpoint method scattered from under fifteen seconds to over forty-five, and landed within a second of thirty less than a third of the time. It is not a slightly biased estimator. It is unrelated to the quantity being measured.

Light both ends and stop thinking about position

Strike the match at both ends at once and wait for the flames to meet. Three facts settle the answer, and none of them mentions where along the cord anything happens.

The flames start at the same instant and stop at the same instant, so they burn for equal times. Between them they consume the entire cord, since one covers $[0,m]$ and the other covers $[m,1]$ for whatever meeting point $m$ they land on. The cord is worth sixty seconds of burning in total. So

\int_0^m r \;=\; \int_m^1 r \;=\; \frac{60}{2} \;=\; 30

(1)

and the clock reads thirty when they meet. On every cord. The rate can be as ugly as you like, because the argument never evaluates it.

The meeting point does exist and is unique, which is worth one sentence rather than an assumption. The function $T$ is continuous and strictly increasing, since $r > 0$ , and it runs from $T(0) = 0$ to $T(1) = 60$ , so it hits the value $30$ exactly once.

Where the flames actually meet

Equation (1) says $m$ is the half-mass point of $r$ , the place that splits the burning time evenly. That is the same as the geometric midpoint only when $r$ is constant, and a smooth example makes the gap concrete. Take

r(x) = 30(1 + 2x), \qquad \int_0^1 r = 30\left[x + x^2\right]_0^1 = 60

(2)

a cord that needs three times as long per unit length at the far end as at the near one. Then $T(t) = 30(t + t^2)$ , and solving $T(m) = 30$ gives $m^2 + m - 1 = 0$ , so

m = \frac{\sqrt5 - 1}{2} = 0.618\ldots

(3)

which is the golden ratio conjugate, sitting a long way from $1/2$ . The flames meet well past the middle, each having burned thirty seconds, and the clock does not care. Seeing this number fall out of a puzzle about string is the sort of thing that makes me trust the model.

Fig. 2 — The same cord as a clock. The meeting point is where the curve crosses 30, and the midpoint is wherever it happens to be, which here is 45.

The uniform cord is the one case where the midpoint method accidentally reads thirty, and that coincidence is the reason the wrong answer feels safe. Anyone who has only ever pictured an even burn has never seen the two quantities come apart.

Two cords measure forty-five seconds

The both-ends trick has a stronger form worth extracting, because the sixty in (1) was never load-bearing.

Halving lemma

A cord with $T$ seconds of burning left in it, lit at both of its free ends, finishes in $T/2$ seconds. The argument is (1) verbatim with $60$ replaced by $T$ .

So take two identical sixty-second cords. Light both ends of the first and one end of the second. When the first burns out, thirty seconds have passed and the second has thirty seconds of cord left. Light its far end now, and by the lemma it finishes fifteen seconds later. Total elapsed: forty-five seconds, with no marks, no watch and no idea how either cord burns.

The chain continues. Light one end of all $n$ cords at time zero and both-end them in turn as each predecessor burns out: the $k$ -th cord has $60/2^{\,k-1}$ seconds left when its turn comes, so it contributes half of that, and the total is

30 + 15 + 7.5 + \cdots = 60\left(1 - 2^{-n}\right)

(4)

which gives thirty seconds from one cord, forty-five from two and fifty-two and a half from three. Every duration reachable this way is a dyadic fraction of a minute, and that is a real ceiling rather than a shortage of cleverness: nothing in the scheme can produce twenty seconds.

The three idealisations

The rate must depend on position only. A cord that burns faster with the grain than against it breaks equation (1), because then the two flames are governed by different densities and the times no longer have to match. Nothing in the puzzle guarantees this, and nothing in the answer survives without it.

The flames must not interact. Real fire preheats what is in front of it, so two approaching flames would speed each other up and the meeting would come slightly early. That is the physical grant the puzzle asks for, and it is the same grant that lets us treat the cord as consumed continuously with no gap and no reignition.

And $r$ must be strictly positive. A stretch of cord that burns in zero time would make $T$ flat there and the meeting point ambiguous; a stretch that never burns at all would stop the flame outright. Both are excluded by the phrase "burns end to end in sixty seconds", which is doing more work than it looks like it is doing.

Sources and further reading

The integral in (1) read as an accumulated quantity — Integral
The existence argument for the meeting point — Intermediate value theorem
Why the half-mass point is not the midpoint — Weighted arithmetic mean
The constant in (3) — Golden ratio

Everything above was checked before publication in two independent ways: the symbolic instance (2) was integrated exactly, and a numerical flame simulation located the meeting time by bisection rather than by the symmetry argument, on the cord in Fig. 1 and on four thousand random cords. The worst deviation from thirty seconds across all of them was $2.4 \times 10^{-12}$ .

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