String girdling Earth is a mathematical puzzle with a counterintuitive solution. In a version of this puzzle, string is tightly wrapped around the equator of a perfectly spherical Earth. If the string should be raised off the ground, all the way along the equator, how much longer would the string be?
Alternatively, of string is spliced into the original string, and the extended string rearranged so that it is at a uniform height above the equator. The question that is then posed is whether the gap between string and Earth will allow the passage of a car, a cat or a thin knife blade.
As the string must be raised all along the entire 40000km (20,000miles) circumference, one might expect several kilometres of additional string. Surprisingly, the answer is 2 m or around .
In the second phrasing, considering that is almost negligible compared with the 40000km (20,000miles) circumference, the first response may be that the new position of the string will be no different from the original surface-hugging position. The answer is that a cat will easily pass through the gap, the size of which will be metres or about 16cm (06inches).
Even more surprising is that the size of the sphere or circle around which the string is spanned is irrelevant, and may be anything from the size of an atom to the Milky Way - the result depends only on the amount it is raised. Moreover, as in the coin-rolling problem, the shape the string girdles need not be a circle: 2 times the offset is added when it is any simple polygon or closed curve which does not intersect itself. If the shape is complex, 2 times the offset, times the absolute value of its turning number must be added.[1]
This diagram gives a visual analogue using a square: regardless of the size of the square, the added perimeter is the sum of the four blue arcs, a circle with the same radius as the offset.
More formally, let c  be the Earth's circumference, r  its radius, Δc  the added string length and Δr  the added radius. has a circumference of 2R ,
\begin{align} c+\varDeltac&=2\pi(r+\varDeltar) \ 2\pir+\varDeltac&=2\pir+2\pi\varDeltar \ \varDeltac&=2\pi\varDeltar \ \therefore \varDeltar&=
| \varDeltac | |
| 2\pi |
\end{align}
regardless of the value of c .
This observation also means that an athletics track has the same offset between starting lines on each lane, equal to 2 times the width of the lane, whether the circumference of the inside lane is the standard 400m (1,300feet) or the size of a galaxy.