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Mathematically a conformal map has the property that the Jacobian of the transformation is a scaled version of a rotation matrix. The transformation involved in a spherical Lambert conformal conic map projection doesn't have that property. In what sense is the LCC conformal?

Details

Map projection in pseudocode (x,y are the map coordinates); I'm using the notation as described on pg. 104 of Map Projections - a Working Manual

x=rho*sin(theta)
y=rho0-rho*cos(theta)

Jacobian (phi,lam refer to latitude, longitude respectively):

J =   (drho/dphi)*sin(theta)     rho*cos(theta)*(dtheta/dlam)
     -(drho/dhpi)*cos(theta)     rho*sin(theta)*(dtheta/dlam)          

This would be a scaled rotation matrix if (drho/dphi)==rho*(dtheta/dlam) however, since

 rho=R*F*cot( phi/2+pi/4)**n
 theta=n*(lam-lam0)

(R,F,n, lam0 are constants derived from the parameters defining the projection.) we have

drho/dphi=-n*R*F*cot( phi/2+pi/4)**(n-1)*csc( phi/2+pi/4)*0.5
         =-n*rho*csc(phi/2+pi/4)**2*tan(phi/2+pi/4)*0.5

rho*(dtheta/dlam)=n*rho

Thus, the Jacobian is not a scaled version of a rotation matrix, therefore the LCC is not a conformal map.

  • 2
    You have computed a Jacobian as if the geographic coordinates (phi, theta) were Euclidean--but they are not. Please read the introductory sections of the Manual, especially "Distortion for Projections of the Sphere." The check for conformality is that h = k in formulae (4-4) and (4-6), p. 23. – whuber May 6 '15 at 15:11
  • @whuber - penetrating comment again. Are you hinting that the geographic coordinates(phi,theta) are Riemannian ? – gansub Jan 21 '16 at 9:24
  • @gansub Could you explain what you mean by "Riemannian coordinates," if it's intended to be more specific than any set of coordinate functions for a patch on a Riemannian manifold? – whuber Jan 21 '16 at 14:06
  • @whuber - No I mean you had written that the geographic coordinates(phi, theta) were not Euclidean. So I was wondering whether that was a reference to an alternate space either Riemannian or some arbitrary space. – gansub Jan 21 '16 at 14:54
  • @gansub There is a Riemannian metric (determined by the ellipsoid) but it is not Euclidean, because the ellipsoid is nowhere flat. – whuber Jan 21 '16 at 14:58
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As per whuber's comment: differentiating with respect to the angular coordinates is incorrect, i.e. it is not a conformal map between (lat,lon) and (x,y) map coordinates; instead it is a local mapping from (local) easterly, northerly displacement onto map displacements. Thus, you need to differentiate w.r.t. easterly and northerly displacements in meters (or whatever length unit):

J =  (dx/de)     (dx/dn)   
     (dy/de)     (dy/dn)

  =   rho*cos(theta)*(dtheta/dlam)*(dlam/de)    (drho/dphi)*(dphi/dn)*sin(theta)
      rho*sin(theta)*dtheta/dlam)*(dlam/de)    -(drho/dphi)*(dphi/dn)*(dlam/de)*cos(theta)

with a bit of algebra you end up with:

J = (n*rho)/R/cos(phi)* |  cos(theta)    -sin(theta) |
                        |  sin(theta)     cos(theta) |

It's pretty straightforward to verify that k==h==(n*rho)/R/cos(phi) as a check.

  • 1
    +1 Great self-answer. It's nice to see some rigorous analysis of projections appear on this site. – whuber May 6 '15 at 17:07

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