Convex hull of (longitude, latitude)-points on the surface of a sphere

Question

Standard convex hull algorithms will not work with (longitude, latitude)-points, because standard algorithms assume you want the hull of a set of Cartesian points. Latitude-long

Answer 1

FutureNerd:

You are absolutely correct. I had to solve the exact same problem as Maxy-B for my application. As a first iteration, I just treated (lng,lat) as (x,y) and ran a standard 2D algorithm. This worked fine as long as nobody looked too close, because all my data were in the contiguous U.S. As a second iteration, though, I used your approach and proved the concept.

The points MUST be in the same hemisphere. As it turns out, choosing this hemisphere is non-trivial (it's not just the points' center, as I had initially guessed.) To illustrate, consider the following four points: (0,0), (-60,0), (+60,0) along the equator, and (0,90) the north pole. However you choose to define "center", their center lies on the north pole by symmetry and all four points are in the Northern Hemisphere. However, consider replacing the fourth point with, say (-19, 64) iceland. Now their center is NOT on the north pole, but asymmetrically drawn toward iceland. However, all four points are still in the Northern Hemisphere. Further, the Northern Hemisphere, as uniquely defined by the North Pole, is the ONLY hemisphere they share. So calculating this "pole" becomes algorithmic, not algebraic.

See my repository for the Python code: https://github.com/VictorDavis/GeoConvexHull