[PROJ] Ellipsoidal orthographic: condition of projectability ?

[Charles Karney]

The visibility condition is simple:

Just take the dot product of the normal to the ellipsoid

   [cos(phi)*cos(lambda), cos(phi)*sin(lambda), sin(phi)]

at the center of the projection and at the test point.

   Positive = visible; zero = horizon; negative = hidden.

This is the same as the spherical formula.  But I hope that this
formulation explains why.

   --Charles

On 9/26/20 6:31 AM, Even Rouault wrote:
> Hi,
> 
> In https://github.com/OSGeo/PROJ/pull/2361, I've implemented the 
> ellipsoidal formulation of the orthographic projection from the formulas 
> from EPSG guidance note 7-2, but one missing piece is the condition of 
> projectability.
> 
> In the spherical formulation, there is a test to check that a given 
> (lam, phi) is visible from the projection plane or not.
> 
> https://github.com/rouault/PROJ/blob/2e104e092578347de11208e9a3a80a3bf711265d/src/projections/ortho.cpp#L53
> 
> It seems it comes from https://pubs.usgs.gov/pp/1395/report.pdf equation 
> (5-3), page 149
> 
> Looking at the formula, one can observe that it is actually the 
> (opposite of the) partial derivative of y(lam, phi, phi0) with respect 
> to phi0.
> 
> But perhaps this is just a coincidence due to the spherical case.
> 
> Now, the question... What about the ellipsoidal case:
> 
> - Should the exact same formula (Q->sinph0 * sinphi + Q->cosph0 * cosphi 
> * coslam < - EPS10) of the spherical case be used as well ?
> 
> - Or should we compute the partial derivative of y(lam, phi, phi0) with 
> respect to phi0 with the ellipsoidal formula of y from
> 
> https://github.com/rouault/PROJ/blob/2e104e092578347de11208e9a3a80a3bf711265d/src/projections/ortho.cpp#L136 
> (will not be pretty...)
> 
> - something else ?
> 
> Even
>