[Proj] Optimal Albers Standard parallels
j.l.h.hartmann at uva.nl
Mon Feb 22 06:04:44 EST 2010
For me, I certainly would be interested in the algorithm.
On 21-2-2010 20:41, strebe wrote:
> From the publication you cited
> "A simple but useful way of appraising the location of the origin of
> an azimuthal projection is to plot a series of concentric circles of
> radii z representing the isograms of maximum Angular Deformation and
> those of Scale Error at the scale of a convenient atlas map and to
> shift this overlay about on the map until a good fit is obtained
> between some of the extreme points of the area of mapped."
> While phrased in an unnecessarily complicated way, the recommendation
> in "Map Projections for Europe" is the equivalent of my
> recommendation. You have a bunch of points you know are within the
> area of interest. Whether manually (as described in that publication)
> or programmatically, you need to find the smallest small circle that
> circumscribes them all. I don't know of any non-iterative method for
> doing that, though possibly one could be devised. As an algorithmic
> process, it is not difficult, but unlike Albers, it is not just a
> matter of observing the most northerly and most southerly points.
> I note that you talk about comparing Albers to LAEA. Once you know the
> optimal LAEA you can do that, since the distortion values at the
> extremes of the region of interest can be computed by known formulæ
> once the parameters for the projection are known.
> Can I interpret your inquiry to mean that you want to know (presumably
> in complete detail) the algorithm for finding the projection center of
> the optimal LAEA for a region, given a list of points in that region?
> Or are you just wondering if one could be devised?
> — daan Strebe
> On Feb 21, 2010, at 8:01:56 AM, OvV_HN <ovv at hetnet.nl> wrote:
> In the article I was referring to, the angular deformation and the
> 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
> I wondered, could the center of the LAEA projection have been
> Still an academic discussion, of course......
> Oscar van Vlijmen
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