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I’m interested in finding closed-form inverses for map projection equations in general (of course, this is not always possible). I’m hoping Mathematica can help me with some of the trickier ones.

As a practice run, I thought I would try and use Mathematica to invert Lambert’s Azimuthal Equal-Area Projection. There are of course closed-form inverse equations on MathWorld, but I’m wondering how I would go about finding these using Mathematica.

I’ve tried using Solve and Reduce, for example:

k = Sqrt[2/(1 + s1 Sin[y] + c1 Cos[y] Cos[x])]
Solve[{u, v} == {
  k Cos[y] Sin[x], 
  k (c1 Sin[y] - s1 Cos[y] Cos[x])
}, {x, y}]

But no luck so far. Perhaps I need to specify more constraints?

For example, I know that the longitude and latitude (x and y above) have:

-π ≤ x ≤ π
-π/2 ≤ y ≤ π/2

I wonder if eliminating variables might also help the solving routines.

share|improve this question
    
Have you seen the help pages for Latitude and Longitude? A (minor modification of an) example from that page exhibits an unprojection: Through[{Latitude, Longitude}[GeoGridPosition[{-0.12605573,-0.0464294, 0}, "LambertConformalConic"]]]. –  whuber Feb 16 '13 at 11:56
    
I'm interested in automatically finding closed-form inverse formulæ, given arbitrary forward projection equations, so I'm not sure that really helps me. But thanks anyway! –  Jason Davies Feb 16 '13 at 12:33
    
I've clarified the wording of the question a bit, so it's clear what I'm trying to achieve. –  Jason Davies Feb 16 '13 at 12:35
    
In most cases there's no such thing as closed-form inverse formulae for non-spherical datums, so I presume you mean to apply these equations only to spheres. See pubs.usgs.gov/pp/1395/report.pdf for details. Moreover, only the simpler projections have closed forms, so the problem does not generally have a solution. –  whuber Feb 16 '13 at 12:40
1  
Yes, I’m only dealing with spheres, so that should simplify things. I know there is no general solution, but I'd like to know how to solve the simplest ones using Mathematica. Lambert’s azimuthal equal-area seems relatively simple, yet as you can see I’ve failed so far to find the (already known) result using Solve or Reduce. –  Jason Davies Feb 16 '13 at 12:45
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