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Thread: Non-Euclidean geometry and the Pythagorean Theorem

  1. #1

    Default Non-Euclidean geometry and the Pythagorean Theorem

    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?

  2. #2
    Guild Novice
    Join Date
    Dec 2011


    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.

  3. #3


    The Good News; There is a Formula

    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.

  4. #4
    Guild Expert Hai-Etlik's Avatar
    Join Date
    May 2009
    48° 28′ N 123° 8′ W
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    Well, assuming a sphere, the answer is here:

    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.
    Last edited by Hai-Etlik; 12-24-2011 at 06:32 PM.

  5. #5
    Guild Adept Slylok's Avatar
    Join Date
    May 2010
    Georgia, USA


    i found THIS SITE to be quite helpful.
    Our sense of the stability of the earth is an illusion due to the shortness of our lives.
    - Neil deGrasse Tyson

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