How this instrument works
Wrap a string snugly around any circle — a ring, a barrel lid, or Earth's equator — then splice in extra length and pull the loop up into a new, larger, perfectly concentric circle. The added length, ΔC, and the resulting rise in radius, gap, are locked together by the same constant that ties every circumference to its radius: C = 2πr. Subtract the old circle's equation from the new one and the radius r itself cancels out entirely, leaving ΔC = 2π · gap — which is why this puzzle's famous answer never asks how big the original circle was.
That cancellation is what makes the result so unsettling on first encounter. Splice one metre into a string already wrapped around a basketball and the gap it opens is precisely the same 15.9 centimetres as splicing one metre into a string wrapped around the entire Earth, whose equator runs some 40,075 kilometres around. Intuition expects a length added to something planet-sized to vanish into rounding error; the geometry disagrees, because the gap depends only on ΔC, never on the starting circumference or radius.
The relationship is exactly linear, with no curvature or approximation hiding in it: doubling the added length doubles the gap, and shrinking the added length toward zero shrinks it toward zero in the same straight proportion. The one boundary worth flagging is geometric rather than numerical — the ΔC ⁄ 2π answer only holds if the extra length spreads perfectly evenly into a new concentric circle. Bunch that same slack to one side instead and it forms a localized loop there while the string keeps touching the ground almost everywhere else; uniform lift is a specific, idealized case, not something a real piece of string settles into on its own.
- Enter the length you are splicing in — not the string's full length — into Extra string length added.
- Pick whichever unit fits your scenario, metres, feet, or centimetres; the result mirrors it.
- Read Uniform gap all around for the lift, gap, produced at every point around the loop.
- Set the field to 0 to confirm the gap collapses to nothing, or to roughly 6.283 to see it round to a clean 1.
Worked example — one extra metre around the equator
Take a string that fits snugly around Earth's equator, a loop roughly 40,075 kilometres long, and splice in one extra metre: Extra string length added = 1. Uniform gap all around returns gap = 1 ⁄ (2π) = 0.15915494309189535 m — about 15.9 centimetres, lifted evenly off the ground at every point along the entire circle, from Quito to Singapore, without the equator's own length ever entering the arithmetic.
The proportion holds at both ends of the scale. Set addedLength to 0 and gap collapses to exactly 0, the string still lying as snug as before. Scale addedLength up to 2π ≈ 6.283185 metres instead and gap comes out to a clean 1 metre — ten times the added length would give ten times that result, since the whole relationship is one straight division by the constant 2π.
Questions
Why doesn't the gap depend on the Earth's radius?
Because circumference and radius are locked together by the same constant everywhere on a circle: C = 2πr. Add ΔC to the circumference and the radius must grow by exactly ΔC ⁄ 2π to keep that ratio true — a relationship that never mentions the starting radius r, so a basketball and a planet share an identical 0.159 m lift for the identical 1 m of added string.
How old is this puzzle?
It is one of the oldest chestnuts in recreational mathematics, circulated for generations under labels like 'belt around the Earth' or 'string round the world.' Every telling swaps in a different sphere — the Earth, the Moon, a basketball — because the punchline stays the same regardless: the gap depends only on the added length, never on the sphere's original size.
Does the string need to stay perfectly circular for this to work?
Yes — the ΔC ⁄ 2π answer assumes the loop is raised into a new, larger circle concentric with the old one, so every point lifts by the same amount. Pull the slack to one side instead of spreading it evenly, and most of the extra length bunches into a bulge there while the string still touches the ground almost everywhere else; the uniform 15.9 cm answer is the special case where the extra length distributes perfectly.
How much does the gap grow if I add ten times as much string?
Exactly ten times as much, since gap = ΔC ⁄ 2π is a straight proportion, not a diminishing one: 10 m of added length yields 10 ⁄ 2π ≈ 1.5915 m of lift. Doubling, tripling, or halving Extra string length added scales Uniform gap all around by precisely the same factor, with no upper or lower limit built into the formula itself.
Is this the same formula used to convert circumference to radius?
It uses the same constant, 2π, but applied to a difference rather than a whole measurement: converting a full circumference to a radius is r = C ⁄ 2π, while this sheet converts a change in circumference to a change in radius, gap = ΔC ⁄ 2π. Both come from the identical identity C = 2πr — one solves it directly, the other solves it after subtracting the old loop from the new one.
Which units should I use, and does it matter?
Any of the three offered — metres, feet, or centimetres — works, because ΔC ⁄ 2π is a pure ratio: whatever unit goes into Extra string length added comes back out in Uniform gap all around. Enter 1 ft of added string and it reads about 0.159 ft, roughly 1.91 inches; keep both fields in the same unit rather than mixing a metric input with an imperial reading.