[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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