# [Proj] Ellipsoidal geodesic projections

Karney, Charles ckarney at Sarnoff.com
Mon May 31 07:49:05 EST 2010

```> From: Mikael Rittri [Mikael.Rittri at carmenta.com]
> Subject: RE: [Proj] Ellipsoidal geodesic projections
>
> your documentation of GeographicLib that it was possible
> to construct such a projection with all geodesics exactly
> straight.
>
> > ...geodesic problems ... Bessel ... showed how the problem
> > may be transferred to an "auxiliary sphere" where the latitude
> > phi has been replaced by the reduced latitude beta where
> > tan(beta) = (1 - f)tan(phi). On this sphere, the geodesic
> > is a great circle and the azimuth is the same as on the
> > ellipsoid. However, the ellipsoidal distance is related to
> > the great circle distance by an integral; and the ellipsoidal
> > longitude is similarly related to the longitude on the
> > auxiliary sphere.
> (http://geographiclib.sourceforge.net/html/geodesic.html)
>
> So I must have misunderstood this text. I guess "the geodesic"
> does not stand for an arbitrary geodesic but only one of
> those that intersect an origin point?

It's true that the transformation to the auxiliary sphere converts a
geodesic to a great circle.  However the equation for the longitude on
the auxiliary sphere depends on the azimuth of the geodesic.  Thus the
transformation does not define a map projection.  Symbolically, the
auxiliary sphere transformation is

(beta, omega) = f(phi, lambda; alpha0)

[phi = latitude, lambda = longitude, beta = reduced latitude, omega =
longitude on auxiliary sphere, alpha0 = geodesic azimuth at equator
crossing].  The presence of the alpha0 parameter in the transformation
disqualifies it as the basis for a map projection.

--
Charles Karney <ckarney at sarnoff.com>
Sarnoff Corporation, Princeton, NJ 08543-5300

Tel: +1 609 734 2312
Fax: +1 609 734 2662
```