[Proj] Optimal Albers Standard parallels

OvV_HN ovv at hetnet.nl
Sun Feb 21 11:01:56 EST 2010


In the article I was referring to, the angular deformation and the scale 
error of the Albers and the LAEA projections are compared for the whole of 
the European Union. The Albers has two standard parallels at 38 and 61 d N, 
with a CM at 9 E. The LAEA has its center at 9 E and 53 N.
I got the impression that the author just shifted the LAEA center up and 
down by trial and error until a reasonable comparison could be made with the 
Albers.
I wondered, could the center of the LAEA projection have been determined 
algorithmically?
Still an academic discussion, of course......


Oscar van Vlijmen


----- Original Message ----- 
From: "strebe" <strebe at aol.com>
To: "PROJ.4 and general Projections Discussions" <proj at lists.maptools.org>
Sent: Saturday, February 20, 2010 9:26 PM
Subject: Re: [Proj] Optimal Albers Standard parallels


Presumably "LAEA" = Lambert azimuthal equal-area projection, in which case 
it has no standard parallels. The question would be where to center it. 
That's an easier problem to solve in the typical case than Albers; you 
simply find the small circle that circumscribes the area of interest. That 
small circle's center coincides with the center of the optimal Lamber 
azimuthal equal-area for the region.




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