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I try to simulate geodesic on the ellipsoid recently. I have two points on the ellipsoid. After solving the inverse problem, I can get the distance and two azimuths of two points. As the figure https://i.stack.imgur.com/dJnUj.jpg. (I can obtain S12 and alpha 1 2) But how can I draw the geodesic on the ellipsoid? What I can get from the inverse problem doesn't make sense to me to draw this geodesic path. Can somebody help me with this question? Thanks!!

Tuiyo Ty
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  • there is no closed form for this. On a sphere, yes. On an ellipsoid, messy. If it is a spheroid, there is, at least, Clairaut's relation. If all three axes are of different lengths then you cannot expect much formally. – Will Jagy Jan 29 '15 at 05:16
  • the ellipsoid in this problem is like the Earth so it's biaxial ellipsoid(two axes are the same). I review countless references about geodesic on the ellipsoid and all of them only provide the algorithm of distance and azimuths but even if I have those information, I still cannot draw the geodesic line on the surface. In view of the technology of navigation or something else about path guidance, I am sure that there is a way to draw this line. Can somebody give a hint to me about this problem? THX!!! – Tuiyo Ty Jan 29 '15 at 07:44

2 Answers2

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Once you have s12 and α1, you can solve a series of direct problems with a fixed φ1, λ1 and α1 and a varying s13 in the range 0 ≤ s13 ≤ s12 to find a sequence of waypoints 3. Join up these points and you have your geodesic.

The GeodSolve utility from GeographicLib will do the job on the command line (scroll down to the examples section). Alternatively use the GeodesicLine class in GeographicLib (see the example C++ code provided). Equivalent functionality is also provided in C by the proj.4 library, version 4.9.1.

cffk
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If it is a surface of revolution ( two axes of ellipsoid same) by Clairaut's Law it is easy to integrate numerically. The invariant minimum radii are: $r_1 sin \alpha_1 ,r_2 sin \alpha_2 $. If there is no axis-symmetry it is doable, but very cumbersome.

Narasimham
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