Imagine that in a future era humans decide to build a high speed train circumnavigating the globe at the equator. Ignore the impracticalities of such an undertaking (e.g. building railroad tracks across an ocean) and think about this question: How long does the track need to be? This is easy to answer if you know the circumference of the Earth, which is 24,902 (or roughly 25,000) miles at the equator.

Now, imagine our engineering team determines that, for technical reasons, the track needs to be elevated two feet off the ground. We were already planning to acquire 25,000 miles of track so the question is this: how much additional track do we need in order to build our equatorial railroad two feet above the Earth’s surface?

Leave your answer in a comment below. I’ll post a solution and the names of all puzzle solvers on Monday.

**Solution:** Think of the equator as a giant circle. Let’s call the radius of that circle R, the distance from the center of the Earth to any point on the equator. The elevated track can also be thought of as a giant circle, one with radius R+2 (since the track is two feet off the ground). Now the question of how many additional feet of track we need boils down to subtracting the circumference of the elevated track from the circumference of the equator.

The circumference of any circle is given by 2*Pi*r, where r is the circle’s radius so with a bit of algebra we get:

more track needed | = track circumference – equator circumference | |

= 2*Pi*(R+2) – 2*Pi*R | ||

= 2*Pi*(R+2-R) | ||

= 2*Pi*(2) = 4*Pi ~= 12 |

So, to raise the track by two feet all 25,000 miles around the Earth, you only need to add approximately 12 feet of track! Even more amazing, because the solution is independent of the starting radius, the same answer applies to every possible circle. For example, if you tied a string around a basketball and then decided to raise that string two feet above the surface of the basketball, you’d need precisely the same amount of additional string: 12 feet.

This is a problem I first heard as a kid and to this day it still strikes me as incredibly surprising. Hat’s off to Mudassir Ansari, Dan Stoops, John Baldi, Ricardo Agudo and Jim Goss for coming up with the correct answer!

If i go by the basic simplified calculation

Circumference of earth = 2*pi*r which is 24,902 miles

now elevation of 2 feet means 2/5280 miles ..call it y

so new circumference would be 2*pi*(r+y) = 2*pi*r + 2*pi*y

so additional length required is 2*pi*y = 2*22/7*2/5280 = 1/420 miles or 88/7 feet

Alas, i would undoubtedly fail a Google job interview. Even the marketing interview. But, that being said:

D = C / pie = 7957.75 miles = 42,016,905 feet

New D = D + 4 [2 feet each side] = 42,016,909 feet

New C = D * pie = 25000.00238 miles

D – New D = .00238 = 12.57 feet

Thanks for the chance to flex my indolent math muscles.

Depends a little on what number to use as the actual circumference. If you were using 25000, then the diameter of the earth is 25000/pi. If you’re gonna ride 2 ft over that, then you increase the diameter by 4 feet (2 feet on each side of the circle’s diameter). So the new circumference is ((25000/pi)+4)*pi. So you need another ~12.57 feet.

I really have no idea where this one going, so here’s an off the wall answer.

The 2 feet above the ground is a red herring.

Since a train requires 2 tracks, you need 50,000 miles of track, so you need 25,000 more.

The answer is 2 x 3.1416.

It would be the same if the train would be build around the Sun or if a nanoscopic train would be build around an atom.

I am an engineer and this is the puzzle that teached me to check what seems to be obvious.

12.566 feet

First for ease I converted to meters

24902 * 1609.334 = 40075884.288 = 2 * pi * r

r = 20037942.144/3.14159265

r = 6378275.0905022 metres

1 foot = 0.3048 metres 2 feet = 0.6096

6378275.0905022 + 0.6096 = 6378275.7001022648 metres

2 * pi * r = 2 * 6378275.7001022648 = 40075888.11822975888

40075888.11822975888 – 40075884.288

= 3.83022975888 metres

3.83022975888 * 0.3048 = 12.5663706 = 12 feet 7inches

25400 miles

Seriously, who the hell uses miles and feet these days? It’s 2011!