Even Rouault via PROJ <[email protected]> writes: > Or even simpler, compute geodesic distance on ellipsoid (a + h_mean, b > + h_mean) where h_mean is the mean of h_start and h_end. If h_start > and h_end are small compared to a, I would expect whatever mean > formula used to lead to similar results. At least this method is > guaranteed to give the correct result when h_start = h_end = 0 ...
Or go full Pythagoras (geodesic_distance)^2 + (h_start-h_end)^2)^1/2 so you don't need the horiz >> vertical assumption. Bonus points for someone who can do the integrals and see if that's right or not, and if not, if they can find a closed-form solution. Stepping back, I would ask what semantics you want and why they make sense. In 3D, the shortest distance between 2 points is a line in 3d, which when converted back to llh, has h dropping in between. A geodesic is the shortest distance constrained to the ellipsoid. That's longer than the shortest 3d line. But it makes sense if you are navigation at h=0 (or H=0, more likely!, or H "small", constrained to topography). You're asking for a shortest distance that's sort of constrained to the ellipsoid but not entirely. Intuitively I can see the point of your smooth change goal. _______________________________________________ PROJ mailing list [email protected] https://lists.osgeo.org/mailman/listinfo/proj
