String Girdling Earth Calculator
Solve the rope around Earth puzzle: find extra string length or gap height using ΔL = 2πd with step-by-step circle geometry.
The String Girdling Earth Puzzle
Imagine a string tightly wrapped around Earth's equator. If you raise it uniformly 1 meter above the ground, how much longer must the string be? Surprisingly, only about 6.28 meters ($2\pi$ m), regardless of Earth's size. This counterintuitive result is the string girdling Earth (rope around Earth) problem.
See also the Circle Calculator and Circle Equation Calculator.
The Key Formula
For a gap height $d$ and extra string length $\Delta L$:
$$\Delta L = 2\pi d$$
Earth's radius cancels out, so this works for any sphere: Earth, Mars, a basketball, or a coin.
Worked Example
If you splice 1 meter of string into the loop, the gap is:
$$d = \frac{\Delta L}{2\pi} = \frac{1}{2\pi} \approx 0.159 \text{ m} \approx 16 \text{ cm}$$
That is enough room for a cat to pass underneath.
Frequently Asked Questions
Does Earth's size affect the extra string length?
No. The extra length depends only on the gap height: $\Delta L = 2\pi d$. Earth's radius appears in both the original and raised circumferences and cancels out.
Can this apply to other round objects?
Yes. The principle works for any concentric circles, including athletics track lane offsets.
Is the gap ever equal to the splice length?
No. Since $\Delta L = 2\pi d$ and $2\pi \approx 6.28$, the splice length is always about 6.28 times the gap height.
What if I raise the string by 1 foot?
The extra string needed is $2\pi \times 1 \approx 6.28$ feet, about 6 ft 3 in.