View Full Version : Non-Euclidean geometry and the Pythagorean Theorem

12-24-2011, 05:12 PM
Hey guys,

So, I'm doing a hand-drawn map- with latitude and longitude- of my main fantasy country for a novel. Here's the problem.

It is possible, in a Euclidean plane, to determine the distance between any two points by determining their up/down and left/right distances. That's called the Pythagorean theorem, and we all know it. But what about on a spherical planet? For example, one of my main cities is at 31.61 degrees south and 0 degrees east; another city is at 9.88 degrees east and 31.88 degrees south. The circumference of the planet is 3491.86 miles. If my calculations are correct, they should be about 700 miles apart. But I'm not sure, and my question is, is there a formula to determine the distance between two points on a planet of a certain radius if their longitude and latitude are given?

12-24-2011, 05:22 PM
I just learned this in my Trig. class this last semester. Unfortunately, since the semester is over, I've forgotten it all already.
In answer- Yes there is a way, using trigonometry, to calculate that. But I don't remember the specifics and I don't have a book. You might try google for trig or spherical measurements.

12-24-2011, 05:59 PM
The Good News; There is a Formula (http://en.wikipedia.org/wiki/Great-circle_distance)

The Bad News... its slightly complex, although it is really just trigonometry, there's a lot of trigonometry. This is especially true if you account for the fact that the Radius of a Planet is not constant, even when you don't account for Mountains and Stuff, the planet tends to be wider around the equator. Of course, for most mapping purposes you can be forgiven for just assuming the Radius is constant when not modified by Mountains/Valleys and estimating.

12-24-2011, 06:11 PM
Well, assuming a sphere, the answer is here: http://en.wikipedia.org/wiki/Great-circle_distance

It gets more complex if you want to use a spheroid, and really complex if you want to use the geoid.

Another option is to use an Equidistant Azimuthal projection centred at one of the endpoints, all distances along lines through that point will be true.

PS: Also, a circumference of 3491.86 miles gives a radius of 894.29 km, That's incredibly tiny. Earth has a mean radius of 6,371.01 km.

Assuming similar density, and that I got my math right, that should mean it has about 1/7 of Earth's surface gravity.

12-24-2011, 09:24 PM
i found THIS SITE (http://www.movable-type.co.uk/scripts/latlong.html) to be quite helpful.