Calculating surface of ellipsoid visible from any point outside of it

Question

I have recently become interested in calculating the surface of Earth visible from any point in space. In previous questions such as Visible Portion of the Earth's Surface here and What is the name of the area on Earth which can be observed from a satellite? on Space Exploration, it seems to be assumed that the Earth is spherical, and therefore the visible surface from any point in space would be a spherical cap. In the discussion for this question at Geographic Information Systems, it is shown how to obtain an expression defining the limits of the spherical cap that would be visible from any point in space.

However, in order to be completely accurate for calculating such visible surface, one would need to consider that the Earth's shape is not a perfect sphere, but rather an ellipsoid.

My question is how can we calculate the visible portion of an ellipsoid surface (I suppose this might be called an ellipsoidal cap?) from any point outside of it? The answer to the previously mentioned question on Geographical Information Systems states

The solution for an ellipsoid is pretty messy--it is an irregular shape, not a circle--and is best computed numerically rather than with a formula.

Is this really the case? Are there no closed-form solutions to this problem? If so, how would it be computed numerically?

If it makes the problem easier, we can assume the ellipsoid to be biaxial.

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Other related questions limited to a spherical Earth:

• Describing points on Earth's surface visible from space
• Scope of View of Earth Station by Latitude
• What is the percentage of Earth's hemisphere seen from different orbital heights or distances?

Answer 1

Getting the boundary of the visible region is much easier than calculating its area. The boundary curve is an ellipse. It is the intersection of the ellipsoid and the so-called polar plane of the view-point.

If the ellipsoid has equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ Then the polar plane of the point $(u,v,w)$ has equation $$ \frac{ux}{a^2} + \frac{vy}{b^2} + \frac{wz}{c^2} = 1 $$ And, actually, I have now found that everything you need is probably in this paper. The details of the ellipse “horizon” are worked out, and the enclosed area is computed by numerical methods.

Answer 2

Assuming you are using an actual spheroid (which WGS84 is as far as I understand) and not some other kind of geoid, your spheroidal geoid can be mapped to a sphere via an affine transformation that "stretches" the Earth along the polar axis.

An affine transformation maps lines that are tangent to the spheroid to lines that are tangent to the sphere, and maps concurrent lines to concurrent lines, so whatever the lines of sight are to the "horizon" from the position of your satellite, they map to the lines tangent to the sphere from whatever point in space the affine transformation takes your satellite to.

Since the horizon on the sphere is a circle in a plane, the horizon on the spheroid (after reversing the affine transformation) is also in a plane, though it is not generally a circle since that shape is not preserved.

The horizon will be an ellipse, which seems less ugly than the GIS.se answer makes it out to be. They may have been thinking of the geodetic curves on the spheroid, which do not map to geodetics on the sphere under the affine transformation. In addition, the lines of latitude are not preserved by the affine transformation either, so the determination of latitudes and longitudes along the horizon may require an extra calculation.

The real problem is that relative surface areas are not preserved by the affine transformation between a sphere and a spheroid, so to get the total visible area you would have to integrate an ellipsoidal surface within an arbitrary elliptical boundary. There is not even a closed formula for the length of an arc of an ellipse using anything more commonplace than elliptical functions, so I'd be surprised if there is a simple closed formula for the area you want.

Perhaps someone reading this answer will recognize the appropriate integral to set up in some kind of "closed form" (though perhaps using functions that most people know little about).