As I had showed in a previous answer, among the other ways to prove that the Earth is round, we have the triple-right triangle.

The idea is simple:

  1. Starting from point A you move in a straight line for a certain distance.
  2. At point B, turn right 90° degrees, move along the line for the same distance.
  3. At point C, turn again to the right and do the same.
  4. We'll eventually get back at the starting point: point A and C are the same location, thus we just created a triangle with 90° degrees.

This proves that that Earth has a spherical shape (not a perfect sphere), since these movements would only create a square with three sides if we were to do it on a flat surface.

However, the "problem" of this experiment is that it's not really doable on a small scale. The distance must be so much that the curve of our planet can be taken into consideration. Walking 1 meter, then one meter and then another meter won't create a triangle, since the curve of the planet is not that strong.

So my question is: what's the minimum distance we'd need to travel for this experiment to work?

  • $\begingroup$ Perhaps "distance" is not the best term here, so feel free to fix the terminology. Also not sure about the tags to use here. :) $\endgroup$ – Alenanno Jun 11 '13 at 17:51
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    $\begingroup$ Since this question strictly deals with spherical geometry, you might want to move this to the math stackexchange page. $\endgroup$ – abhishek Jun 11 '13 at 17:57
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    $\begingroup$ As an experimentor who works with small effects I'd like to emphasize that you don't need to go pole to equator (to get three right angles) to resolve the effect. You just have the get the measured sum of interior angles above 180 degrees by more than five times the uncertainty on the same figure to claim a "discovery" of spherical curvature on the planet. That's a much small triangle with a much more precise measurement of the angles. $\endgroup$ – dmckee --- ex-moderator kitten Jun 11 '13 at 18:03

When going around a "small" path on a curved surface, the angle you come up short is given by the local Guass-Bonnet theorem. Usually the G-B theorem comes up in connection with topology, involving integrals over the whole manifold. But that is just one facet of it. The local version I find useful at the moment is:

$\iint_{enclosed} K dA = 2\pi - \sum \gamma_i $

Curvature $K$ is 1/(radius of curvature) per dimension, and we multiply together two dimensions. For a sphere, it's constant so the left hand side is just $KA$, $A$ being the area enclosed by your closed-path travel. $\gamma_i$ is the angle of turning at each point $i$ where you make a turn. (For a continuously curving path, there'd be an integral.) Going straight through a point $i$ is $\gamma_i=0$.

If you go around a square of side length L, with L much smaller than R, the area is $L^2$ (pretty close to it.) You think you'd have four turns of 90 degrees, adding to 360 (which is $2\pi radians), but Gauss-Bonnet says no. You'll make three 90 degree turns, and one that's not quite, if you travel a closed path ending at your starting point. Alternatively, you can make strict 90 degree turns four times, going exactly distance L for each side, and then measure how far away you end up from the starting point, and do some extra math. G-B still applies, just so not directly.

At this point, if you have a notion of how small an angular error you can measure, you can plug in numbers. $R_{Earth}$ is about 4000 miles (6400 km). Area $A$ is $L^2$ and $K=1/R^2$. On Earth, if we do this experiment on a reasonably human scale, we hope a few hundred meters or a few kilometers, we expect close to the same value for $A$. Say you can measure, or steer, to about half a degree accuracy. That's 1/100th radian, roughly. We already intend to make four ideal 90 degree turns (the $2\pi$ in the G-B formula) but expect to measure some small angular deficit

$\delta = 2\pi-\sum \gamma$.

when we finish the trip at the start. To get an idea what size trip we must make, just set $\delta = 0.04$ for four steering measurements good to .01 radians. (A naive estimate, not a proper error analysis.) Then,

$ {{L^2}\over{R^2}} = 0.04$

I'll let you grab a calculator and punch in the numbers, and try more accurate angle.

Note this is a rough "back-of-the-envolope" calculation. Better thought-out math would be needed for a real experiment.


On a sphere (Earth isn't a perfect sphere) the minimum size for an edge of your triangle is a line that subtends an angle of 90º measured from the centre of the Earth i.e. the distance from any point on the equator to one of the poles.

The increase of the interior angle over 180º is known as the angular excess or spherical excess. See this article for more info. You could ask on math exchange as abhishek suggests, but I would do some background reading first.

  • $\begingroup$ Yes I used "spherical" for simplicity, but I'll make sure to specify that in the question. :) Thanks for the answer! $\endgroup$ – Alenanno Jun 11 '13 at 18:12

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