Chamberlin Trimetric Inverse

[jasondavies]

It's me again!

I'm wondering if you have any hints regarding an inverse for the Chamberlin Trimetric. According to Christensen (1992):

The inverse transformation is solved rather easily by a rough estimate of the position on the sphere, followed by an iteration of the forward solution.

It's encouraging to know that it's "rather easy", but sadly he neglected to mention the precise details of what initial estimate to use, and indeed the method for converging.

My first thought was to use the azimuthal equidistant in some way for a rough estimate, perhaps centred on the mean of the Trimetric's three points, but the scale of the Trimetric appears to vary depending on the size of the triangle, so I don't think this will work without taking this into account.

Of course, it's technically possible to laboriously differentiate the forward projection and use Newton—Raphson with Jacobians, but Christensen implies this isn't necessary.

Any hints or pointers would be much appreciated.

Thanks!
Jason

[daan]

In the case of Chamberlin, I just use a completely general numerical inverse that I use for all projections I don't have tailored inverses for. You don't need a formal Newton-Raphson, of course; the numerical derivative suffices. You should use the secant method until your iteration is clearly converging.

--daan

[jasondavies]

Thank you, that's very helpful once again.

When you said "secant method", did you mean "bisection method"? Since the secant method is a finite difference approximation to Newton—Raphson, it doesn't guarantee convergence, whereas the bisection method is more robust and suitable for initial rough guesses.

I think perhaps you meant "bisection method" for the initial estimates, followed by secant method for fine tuning (which is essentially Newton—Raphson but with a numerical derivative).

Thanks.
Jason

[daan]

Indeed I meant bisection. With apologies, I responded from a noisy bus. « chuckle » Good luck!

— daan