Geodetic model of MSFS

I’m working on an inertial navigation system, and in order to eliminate rogue
errors between my model and that of the sim, any chance we could get the
following values? Equatorial radius Polar radius Flattening factor

Hello @JB3DG Here are the values we use: Equatorial radius = 6378137.0
Flattening factor = 1./298.257223563 giving a polar radius ~ 6356752.3142
Regards, Sylvain

It’s basically WGS84 :wink: