String Girdling Earth Calculator: Gap and Extra Length
The string girdling Earth puzzle asks: if you wrap a rope tightly around the Earth and splice in a little extra length, how high off the ground does the rope rise? The answer is famously counterintuitive: the gap depends only on the extra length you add, not on the size of the Earth at all. Use this calculator to solve either direction: enter extra string to find the gap, or enter your desired gap to find the extra string needed. Unit switches, show-your-work steps, and examples for other planets are all included.
Formula
Worked example
Adding 1 metre of string: gap = 1 / (2 * pi) = 1 / 6.2832 = 0.1592 m = about 15.9 cm. A house cat (roughly 20 cm tall) would not quite fit, but a large rabbit would. The Earth's radius of 6,371 km plays no role in this calculation.
The string girdling Earth paradox
Imagine a string wrapped tightly all the way around the Earth at the equator. The circumference is about 40,075 km. Now you splice in one extra metre of string and pull it outward until it sits uniformly above the surface everywhere. How high off the ground does it sit?
Most people guess a fraction of a millimetre. The actual answer is about 15.9 cm, roughly the height of a large coffee mug. What makes this surprising is that the answer has nothing to do with the Earth's size. If you did the same experiment with a basketball or the Sun, you would still get exactly the same 15.9 cm gap from that one extra metre of string.
Why the Earth's size does not matter
The mathematics is straightforward. The original string has circumference C = 2 * pi * r. When you raise the string by a uniform gap h, the new circumference is C' = 2 * pi * (r + h). The extra length needed is:
delta_L = C' - C = 2 * pi * (r + h) - 2 * pi * r = 2 * pi * h
The radius r cancels out entirely. The relationship between extra length and gap depends only on the constant 2 * pi, which means the same extra string creates the same gap regardless of whether you wrap it around a marble, the Earth, or a star. Rearranging: h = delta_L / (2 * pi).
This also holds for any closed curve, not just a circle. If the Earth were shaped like a cube, and you raised the string uniformly off every face and edge, the same formula would still apply.
How to use this calculator
Choose one of two modes from the first dropdown:
- Find the gap - enter the length of extra string you are adding and get the resulting gap height.
- Find the extra string - enter the gap height you want and get the extra length to add.
You can also switch between metric (metres, kilometres) and imperial (feet, miles) units. The sphere selector lets you compare Earth, the Moon, Mars, the Sun, a golf ball, or a tennis ball to confirm that the gap really does not change. Enter a custom radius to test any sphere you like. The chart below the inputs shows how gap grows linearly with extra string over a range of values.
Real-world applications
The string girdling Earth result has several surprising applications:
- Athletics tracks: the starting lines in lane 2, 3, and beyond are staggered forward by 2 * pi times the lane width so that each runner covers the same distance. A lane 1.25 m wide means a stagger of about 7.85 m per lane, independent of the track radius.
- Cable sizing: adding one layer of insulation (say 2 mm thick) to a cable increases its circumference by exactly 2 * pi * 0.002 = 12.6 mm of extra material per revolution, regardless of how thick the original cable is.
- Orbit altitude: a spacecraft flying 1 km higher than the International Space Station travels an extra 2 * pi km (about 6.28 km) per orbit compared to one flying directly overhead.
- Engineering tolerances: pipe fitters and boilermakers use the same relationship when calculating the extra plate needed when stepping out one layer on a cylindrical vessel.
Historical context and related puzzles
The string girdling Earth puzzle appears in recreational mathematics books dating back at least to the mid-twentieth century and is frequently used to illustrate how algebraic simplification can overturn geometric intuition. The result is related to:
- The napkin ring problem: if you drill a hole through the centre of any sphere so that the remaining ring has height h, the volume depends only on h, not on the original sphere's radius.
- Cavalieri's principle: volumes can be equal even when shapes look very different.
- Visual calculus: Mamikon's theorem and its generalisations, which sweep tangent lines to find areas without integration.
The puzzle is also used in medical education. The relationship between a tendon's displacement and the radius of the joint it crosses follows the same 2 * pi geometry, making it a teaching tool in orthopaedics and biomechanics.
What fits through the gap?
| Object | Approximate gap | Extra string needed |
|---|---|---|
| Credit card thickness | 0.76 mm | 4.8 mm |
| Pencil diameter | 7 mm | 44 mm |
| House cat | 20 cm | 1.26 m |
| Human adult walking | 2 m | 12.57 m |
| Double-decker bus | 4.2 m | 26.4 m |
| 10-storey building | 30 m | 188.5 m |
Common reference objects and the extra string needed to create a gap they fit through (Earth, metric).
Frequently asked questions
If I add 1 metre of string around Earth, how high does it rise?
Adding 1 metre raises the string by 1 / (2 * pi), which equals approximately 0.1592 metres or 15.92 centimetres, about the height of a coffee mug. This result holds no matter what the size of the Earth is.
Does the size of the Earth matter?
No. The formula is gap = extra string / (2 * pi). The Earth's radius cancels out in the algebra. You get the exact same gap whether you wrap the string around the Earth, the Moon, a golf ball, or a marble. Try switching spheres in the calculator above to see this for yourself.
Can a cat really walk under the raised string?
With 1 metre of extra string, the gap is about 15.9 cm. A typical house cat is about 20-25 cm tall at the shoulder, so it would not quite fit standing up. You would need to add about 2 * pi * 0.2 = 1.26 metres of extra string to create a 20 cm gap that a cat could walk through.
What is the formula for the string girdling Earth problem?
The two forms are: extra length = 2 * pi * gap height, and gap height = extra length / (2 * pi). The key insight is that the sphere's radius does not appear in either equation because it cancels when you subtract the original circumference from the new one.
Does the result change if the Earth is not a perfect sphere?
The result holds for any smooth closed curve or surface, not just circles. Even if the Earth were shaped like a cube or an irregular lump, raising a string uniformly by height h from every point on its surface would still require exactly 2 * pi * h extra length. The proof generalises beyond spheres.
How much string would I need to wrap all the way around the Earth?
The Earth's mean equatorial circumference is approximately 40,075 km, which equals about 40,075,000 metres. This is just the base amount to lie on the surface. Any gap you want on top of that requires an additional 2 * pi * h metres.
What gap does 1 foot of extra string create?
One foot equals 0.3048 metres. Dividing by 2 * pi gives a gap of 0.3048 / 6.2832, which equals about 0.0485 metres or 4.85 centimetres, roughly the height of a golf ball.