Geodesic distance on Earth

dextercioby

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This is just a mere suggestion for doing such computations without resorting to complicated series expansions of elliptic integrals.

Up to the irregular continental shape & sea floor, we could picture our planet as a revolution ellipsoid. This means that, except for points on the same parallel line, the geodesic is an ellipse and trying to compute the geodesic distance for arbitrary points would lead you to an elliptic integral. Even the setting of such an integral is a difficult task.

One simple way to circumvent this issue is to assume a deformed Earth, and what could be simpler than to assume it spherical ? We know that on the sphere the geodesic line between 2 points is the great circle passing through those points.

So let's say we've got city A with colattitude 23°15' N and longitude 15°34' E and a city B with colattitude 57°15' N and longitude 38°34' W. We assume that these values for the 2 coordinates are the same when we switch to a deformed planet.

It's a simple use of scalar product to get the angle between the 2 radii connecting the 2 points with the planet's center.

Denoting the angle with [itex] \psi [/itex], we find easily

[tex] \cos \psi =\sin\theta_{1} \sin \theta_{2} \cos\left(\varphi_{1}-\varphi_{2}\right) +\cos\theta_{1}\cos\theta_{2} [/tex]

,where getting the spherical angles from the collatitude & the longitude is a trivial matter.

Then the geodesic distance is simply

[tex] D_{\mbox{sph}}= R_{\mbox{sph}} \psi [/tex]

, where [itex] \psi [/itex] must be in radians.

[itex] R_{\mbox{sph}} [/itex] is found easily from the condition that the deformation preserve the volume of the Earth.

[tex] R_{\mbox{sph}}=\sqrt[3]{R_{\mbox{eq}}^{2}R_{\mbox{pole}}} [/tex].

The equatorial & polar radii can be picked up from any atlas.

It would be interesting however to make corrections to this simple computation. One could take into account the average depth of the planetary ocean and the average height of the continents.

Daniel.
 

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