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

All edges of a spherical convex hull can be viewed/treated as great circles (seminally, all edges of a convex hull in euclidean space can be treated as lines (rather than a line segment)). Each one of these great circles cuts the sphere into two hemispheres. You could thus conceive each great circle as a constraint. A point that is within the convex hull will be on each of the hemispheres defined by each constraint.

Each edge of the original polygon is a candidate edge of the convex hull. To verify if it is indeed an edge of the convex hull, you'd simply need to verify if all nodes of the polygon are on the hemisphere defined by the great circle that goes through the two nodes of the edge in question. However, we'd still need to create new edges that surpass the concave nodes of the polygon.

But lets rather shortcut / brute-forces this: Draw a great circle between every pair of nodes in the polygon. Do this in both directions (i.e. the great circle connecting A to B and the great circle connecting B to A). For a polygon with N nodes, you will thus end up with N^2 great circle. Each one of these great circles is a candidate constraint (i.e. a candidate edge of the convex polygon). Some of these great circles will overlap with the edges of the original polygon, but most won't. Now, remember again: each great circle is a constraint that constrains the sphere to one hemisphere. Now verify if all nodes of the original polygon satisfy the constraint (i.e. if all nodes are on the hemisphere defined by the great circle). If yes, then this great circle is an edge of the convex hull. If, however a single node of the original polygon does not satisfy the constraint, then it isn't and you can discard this great circle.

The beauty of this is that once you converted your latitudes and longitudes into cartesian vectors pointing onto the unit sphere, it really just requires dot products and cross products - You find the great circle that passes through two points on a sphere by its cross product - A point is on the hemisphere defined by a great circle if the dot product of the great circle and the point is greater (or equal) to 0. So even for polygons with a large number of edges, this brute force method should work just fine.