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I was in the middle of writing the same old geographic distance calculation using the Haversine formula when it occurred to me: shouldn't there be simpler way to do this? Haversine is of course derived from the Law of Cosines. But in thinking about this problem, I ran across Napier's Rules for right-angled spherical triangles. It seems like Napier's Rules should apply, after all Latitude small circles and Longitude great circles always intersect at right angles, so you should be able to draw a right-angled spherical triangle where the hypotenuse connects any two points (unless the case devolves into a line or single point).

So if I'm applying Napier's Rules correctly, if our delta latitude and delta longitude in radians are $a$ and $b$ respectively, the angle $c$ should trace the arc between the two points. So Napier says:

$\sin(\pi-c) = \cos(a) \cos(b)$

Which should then simplify to:

$c = \arcsin(\cos(a) \cos(b))$

But when I try to verify this with a couple test points, the result doesn't match the Haversine formula. Is there a mistake in my (admittedly rusty) algebra or is my mistake in assuming I can apply Napier's Rules to this problem?

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2 Answers 2

up vote 4 down vote accepted

Latitude "lines" are not spherical-geometry lines, as they are not great circles, so the "triangles" you are getting are not spherical triangles.

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I think I understand: you're saying Napier's doesn't apply unless all of the arcs involved are sections of great circles? –  JavadocMD Sep 28 '10 at 23:33
    
Yes, the Napier formulae are meant only for spherical triangles, triangles whose sides are arcs of great circles. Remember that only the equator and the prime meridian are the great circles on the globe. –  J. M. Sep 29 '10 at 1:23
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Correction to that last comment of mine: all the meridians/longitudes are great circles, but among the latitudes, only the equator is a great circle. –  J. M. Sep 29 '10 at 2:24
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@JavadocMD: The shortest distance along the surface of a sphere between two points is along the great circle through those two points, so line segments in spherical geometry have to be arcs of great circles. The three sides of a triangle have to be line segments, so they must be arcs of great circles. And, as J.M. said, all lines of longitude are great circles, but the equator is the only line of latitude that is a great circle. –  Isaac Sep 29 '10 at 2:33
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@JavadocMD: You'll sometimes see "The shortest distance along the surface of a sphere between two points is along the great circle through those two points" stated instead as "segments of great circles are geodesics on a sphere." –  J. M. Sep 29 '10 at 7:55

Napier's rule can be applied to solve problems in spherical geometry and I suggest you may apply your angles in such an equation. Sine of the middle part = product of cosines of opposite parts.So $\sin(90-c)=\cos a x \cos b$, considering a as the angle to be found. $\cos a = \cos c$ divided by $\cos b$. where the angle between 100 and 110 is $b$ , and the distance in terms of an angle between the centre of the circle 001 and 110 is $c$. Keep in mind the principles of stereographic projection while converting angles to distance and vice-versa.

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