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For me, I certainly would be interested in the algorithm.<br>
<br>
Jan<br>
<br>
On 21-2-2010 20:41, strebe wrote:
<blockquote cite="mid:43EBAC27.A67E.46DF.8B8E.5369260BAF49@aol.com"
type="cite">
<div><br>
</div>
<div>Oscar:</div>
<div><br>
</div>
<div>From the publication you cited (<span class="Apple-style-span"
style="font-family: monospace; font-size: 11px;"><a class="moz-txt-link-freetext" href="http://www.ec-gis.org/sdi/publist/pdfs/annoni-etal2003eur.pdf">http://www.ec-gis.org/sdi/publist/pdfs/annoni-etal2003eur.pdf</a><span
class="Apple-style-span"
style="font-family: Helvetica; font-size: medium;">):</span></span></div>
<div><br>
</div>
<div>"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."</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>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?</div>
<div><br>
</div>
<div>Regards,</div>
<div>— daan Strebe</div>
<div><br>
</div>
<br>
On Feb 21, 2010, at 8:01:56 AM, OvV_HN <a class="moz-txt-link-rfc2396E" href="mailto:ovv@hetnet.nl"><ovv@hetnet.nl></a> wrote:<br>
<blockquote
style="border-left: 2px solid blue; padding-left: 5px; margin-left: 5px; color: blue;"><span
class="Apple-style-span"
style="border-collapse: separate; color: rgb(0, 0, 0); font-family: Helvetica; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px;"><span
class="Apple-style-span"
style="font-family: monospace; font-size: 11px;">In the article I was
referring to, the angular deformation and the scale<span
class="Apple-converted-space"> </span><br>
error of the Albers and the LAEA projections are compared for the whole
of<span class="Apple-converted-space"> </span><br>
the European Union. The Albers has two standard parallels at 38 and 61
d N,<span class="Apple-converted-space"> </span><br>
with a CM at 9 E. The LAEA has its center at 9 E and 53 N.<br>
I got the impression that the author just shifted the LAEA center up and<span
class="Apple-converted-space"> </span><br>
down by trial and error until a reasonable comparison could be made
with the<span class="Apple-converted-space"> </span><br>
Albers.<br>
I wondered, could the center of the LAEA projection have been determined<span
class="Apple-converted-space"> </span><br>
algorithmically?<br>
Still an academic discussion, of course......<br>
<br>
<br>
Oscar van Vlijmen<br>
</span></span></blockquote>
<br>
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