[Proj] Optimal Albers Standard parallels

[strebe at aol.com]

Apologies for the long delay in responding. Determining the optimal Lambert azimuthal equal-area is equivalent to solving the "minimum covering circle problem" described here:

http://en.wikipedia.org/wiki/Smallest_circle_problem

Apparently it can be solved in linear time by an algorithm due to Meggido. This is unexpectedly (to me)  efficient. However, I have not examined the algorithm itself yet; technically the optimal LAEA problem is not QUITE the equivalent to the "minimum covering circle problem" unless the latter requires in its solution not only the radius of the circle, but also its location. The radius is irrelevant to the optimal LAEA problem; it is the center point that is required.

Regards,
— daan Strebe


 

 

-----Original Message-----
From: Jan Hartmann <j.l.h.hartmann at uva.nl>
To: PROJ.4 and general Projections Discussions <proj at lists.maptools.org>
Cc: strebe <strebe at aol.com>
Sent: Mon, Feb 22, 2010 3:04 am
Subject: Re: [Proj] Optimal Albers Standard parallels


  For me, I certainly would be interested in the algorithm.

Jan

On 21-2-2010 20:41, strebe wrote:
  

  
  
Oscar:
  

  
  
>From the publication you cited (http://www.ec-gis.org/sdi/publist/pdfs/annoni-etal2003eur.pdf):
  

  
  
"A simple but useful way of appraising the location of theorigin of an azimuthal projection is to plot a series of concentriccircles of radii z representing the isograms of maximum AngularDeformation and those of Scale Error at the scale of a convenient atlasmap and to shift this overlay about on the map until a good fit isobtained between some of the extreme points of the area of mapped."
  

  
  
While phrased in an unnecessarily complicated way, therecommendation in "Map Projections for Europe" is the equivalent of myrecommendation. You have a bunch of points you know are within the areaof interest. Whether manually (as described in that publication) orprogrammatically, you need to find the smallest small circle thatcircumscribes them all. I don't know of any non-iterative method fordoing that, though possibly one could be devised. As an algorithmicprocess, it is not difficult, but unlike Albers, it is not just amatter of observing the most northerly and most southerly points.
  

  
  
I note that you talk about comparing Albers to LAEA. Once youknow the optimal LAEA you can do that, since the distortion values atthe extremes of the region of interest can be computed by known formulæonce 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 theprojection center of the optimal LAEA for a region, given a list ofpoints in that region? Or are you just wondering if one could bedevised?
  

  
  
Regards,
  
— daan Strebe
  

  
  
On Feb 21, 2010, at 8:01:56 AM, OvV_HN <ovv at hetnet.nl> wrote:
  
In the article I wasreferring to, the angular deformation and the scale 
error of the Albers and the LAEA projections are compared for the wholeof 
the European Union. The Albers has two standard parallels at 38 and 61d 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 madewith the 
Albers.
I wondered, could the center of the LAEA projection have been determined 
algorithmically?
Still an academic discussion, of course......
    
    
Oscar van Vlijmen
    
  
  

  
  



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