Does the Chamberlin trimetric projection map (some) great circles to straight lines?

Question

I know that, in general, the Chamberlin trimetric projection maps most great circles on the sphere to some sort of non-straight curve on the flat map.

But what about the 3 special great circles that go through each pair of the 3 control points of the Chamberlin trimetric projection? Does the Chamberlin trimetric projection map those 3 great circles to straight lines on the flat map?

Answer 1

This is not obvious; it requires some analysis. It matters not what properties the overall projection has, because the question only asks about special (measure-zero) parts of the earth. (For example, in the Plate Carree projection, all distances along meridians are correct, but this is definitely not an equidistant projection; all angles between meridians and circles of latitude are correct, but this is not a conformal projection; all areas along the equator are correct, but this is not an equal-area projection.)

Designate the trimetric base points A, B, C (and use the same letters to denote the measures of the triangle's spherical angles at those points). Let Greek letters designate the lengths (in radians) of their opposite sides, so that alpha is the length of BC, beta is the length of AC, and gamma is the length of AB. Without any loss of generality we may rotate the entire situation so that A is at the origin and B is also on the equator. Thus the three points A, B, C have (lat, lon) coordinates equal to (0,0), (gamma, 0), and (phi, lambda), respectively.

As we move along the equator from A to B (and beyond) we will trace out points X with coordinates (mu, 0). The spherical distance from C to X (call this 'h') is given by the spherical law of cosines,

[1] cos(h) = cos(beta)cos(mu) + sin(beta)sin(mu)cos(A).

The trimetric projection, as far as I can tell, lays out a triangle with side lengths alpha, beta, and gamma in the plane. We can locate the corresponding points at (0,0) for A, (gamma,0) for B, and (say) (Cx,Cy) for C. (It's easy to find formulas for Cx and Cy but we won't need them). Therefore, if the great circle AB were to project to a line it would have to be the x axis. Accordingly, by virtue of how the trimetric projection works, X would have to project to coordinates (mu, 0). However, the (map; Euclidean) distance between the projection of C and the projection of X is given by the Pythagorean Theorem as

h' = Sqrt((Cx - mu)^2 + Cy^2)

whence

[2] cos(h') = cos(sqrt((Cx - mu)^2 + Cy^2).

This must equal cos(h), for if not, the trimetric projection would move the projection of X off the x-axis. Because the two formulas [1] and [2] for h and h' in terms of mu are truly different (there is no algebraic equivalence), it is impossible that the great circle uniformly coincides with a straight line on the map.

Answer 2

The Chamberlin trimetric projection is not equidistant, not conformal, inequivalent, and not azimuthal. Rather, the projection strikes a balanced compromise between each of these distortions

Think you want the gnomonic map projection as it displays all great circles as straight lines.

http://en.wikipedia.org/wiki/Gnomonic_projection