No subject
Wed Feb 3 18:16:50 EST 2010
=20
=20
"A simple but useful way of appraising the location of theorigin of an azi=
muthal projection is to plot a series of concentriccircles of radii z repr=
esenting the isograms of maximum AngularDeformation and those of Scale Err=
or 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."
=20
=20
=20
While phrased in an unnecessarily complicated way, therecommendation in "M=
ap Projections for Europe" is the equivalent of myrecommendation. You have=
a bunch of points you know are within the areaof interest. Whether manual=
ly (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 point=
s.
=20
=20
=20
I note that you talk about comparing Albers to LAEA. Once youknow the opti=
mal LAEA you can do that, since the distortion values atthe extremes of th=
e region of interest can be computed by known formul=C3=A6once the paramet=
ers for the projection are known.
=20
=20
=20
Can I interpret your inquiry to mean that you want to know(presumably in=
complete detail) the algorithm for finding theprojection center of the op=
timal LAEA for a region, given a list ofpoints in that region? Or are you=
just wondering if one could bedevised?
=20
=20
=20
Regards,
=20
=E2=80=94 daan Strebe
=20
=20
=20
On Feb 21, 2010, at 8:01:56 AM, OvV_HN <ovv at hetnet.nl> wrote:
=20
In the article I wasreferring to, the angular deformation and the scale=20
error of the Albers and the LAEA projections are compared for the wholeof=
=20
the European Union. The Albers has two standard parallels at 38 and 61d N,=
=20
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=
=20
down by trial and error until a reasonable comparison could be madewith th=
e=20
Albers.
I wondered, could the center of the LAEA projection have been determined=
=20
algorithmically?
Still an academic discussion, of course......
=20
=20
Oscar van Vlijmen
=20
=20
=20
=20
=20
_______________________________________________
Proj mailing list
Proj at lists.maptools.org
http://lists.maptools.org/mailman/listinfo/proj
=20
----------MB_8CC9FF68E498677_41E0_36026_Webmail-m124.sysops.aol.com
Content-Transfer-Encoding: quoted-printable
Content-Type: text/html; charset="utf-8"
<font color=3D'black' size=3D'2' face=3D'arial'>
<div style=3D"font-family: helvetica,arial; font-size: 10pt; color: black;=
"><font color=3D"black" face=3D"arial" size=3D"2">
<div> <font size=3D"2"><font face=3D"Arial, Helvetica, sans-serif"><br>
Apologies for the long delay in responding. Determining the optimal Lamber=
t azimuthal equal-area is equivalent to solving the "minimum covering circ=
le problem" described here:<br>
<br>
http://en.wikipedia.org/wiki/Smallest_circle_problem<br>
<br>
Apparently it can be solved in linear time by an algorithm due to Meggido.=
This is unexpectedly</font></font><font color=3D"black" face=3D"arial" si=
ze=3D"2"><font size=3D"2"><font face=3D"Arial, Helvetica, sans-serif"> (to=
me) </font></font></font><font size=3D"2"><font face=3D"Arial, Helvetica,=
sans-serif"> efficient. However, I have not examined the algorithm itself=
yet; technically the optimal LAEA problem is not QUITE the equivalent to=
the "</font></font><font size=3D"2"><font face=3D"Arial, Helvetica, sans-=
serif">minimum=20
covering circle problem</font></font><font size=3D"2"><font face=3D"Arial,=
Helvetica, sans-serif">" unless the latter requires in its solution not=
only the radius of the circle, but also its location. The radius is irrel=
evant to the optimal LAEA problem; it is the center point that is required=
.<br>
<br>
Regards,<br>
=E2=80=94 daan Strebe<br>
</font></font></div>
<div style=3D"clear: both;"></div>
<div> <br>
</div>
<div> <br>
</div>
<div style=3D"font-family: helvetica,arial; font-size: 10pt; color: black;=
">-----Original Message-----<br>
From: Jan Hartmann <j.l.h.hartmann at uva.nl><br>
To: PROJ.4 and general Projections Discussions <proj at lists.maptools.org=
><br>
Cc: strebe <strebe at aol.com><br>
Sent: Mon, Feb 22, 2010 3:04 am<br>
Subject: Re: [Proj] Optimal Albers Standard parallels<br>
<br>
<div id=3D"AOLMsgPart_2_3b1f8c21-4c94-442c-97a0-6fa5423c8540">
=20
For me, I certainly would be interested in the algorithm.<br>
<br>
Jan<br>
<br>
On 21-2-2010 20:41, strebe wrote:
<blockquote type=3D"cite">
=20
<div><br>
</div>
=20
<div>Oscar:</div>
=20
<div><br>
</div>
=20
<div>From the publication you cited (<span class=3D"Apple-style-span" styl=
e=3D"font-family: monospace; font-size: 11px;"><a class=3D"moz-txt-link-fr=
eetext" target=3D"_blank" href=3D"http://www.ec-gis.org/sdi/publist/pdfs/a=
nnoni-etal2003eur.pdf">http://www.ec-gis.org/sdi/publist/pdfs/annoni-etal2=
003eur.pdf</a><span class=3D"Apple-style-span" style=3D"font-family: Helve=
tica; font-size: medium;">):</span></span></div>
=20
<div><br>
</div>
=20
<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>
=20
<div><br>
</div>
=20
<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>
=20
<div><br>
</div>
=20
<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 formul=C3=
=A6
once the parameters for the projection are known.</div>
=20
<div><br>
</div>
=20
<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>
=20
<div><br>
</div>
=20
<div>Regards,</div>
=20
<div>=E2=80=94 daan Strebe</div>
=20
<div><br>
</div>
<br>
On Feb 21, 2010, at 8:01:56 AM, OvV_HN <a class=3D"moz-txt-link-rfc2396E"=
href=3D"mailto:ovv at hetnet.nl"><ovv at hetnet.nl></a> wrote:<br>
<blockquote style=3D"border-left: 2px solid blue; padding-left: 5px; mar=
gin-left: 5px; color: blue;"><span class=3D"Apple-style-span" style=3D"bor=
der-collapse: separate; color: rgb(0, 0, 0); font-family: Helvetica; font-=
size: medium; font-style: normal; font-variant: normal; font-weight: norma=
l; letter-spacing: normal; line-height: normal; orphans: 2; text-indent:=
0px; text-transform: none; white-space: normal; widows: 2; word-spacing:=
0px;"><span class=3D"Apple-style-span" style=3D"font-family: monospace;=
font-size: 11px;">In the article I was
referring to, the angular deformation and the scale<span class=3D"Apple-co=
nverted-space"> </span><br>
error of the Albers and the LAEA projections are compared for the whole
of<span class=3D"Apple-converted-space"> </span><br>
the European Union. The Albers has two standard parallels at 38 and 61
d N,<span class=3D"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<s=
pan class=3D"Apple-converted-space"> </span><br>
down by trial and error until a reasonable comparison could be made
with the<span class=3D"Apple-converted-space"> </span><br>
Albers.<br>
I wondered, could the center of the LAEA projection have been determined<s=
pan class=3D"Apple-converted-space"> </span><br>
algorithmically?<br>
Still an academic discussion, of course......<br>
<br>
<br>
Oscar van Vlijmen<br>
</span></span></blockquote>
<br>
=20
<div><br>
</div>
<pre><fieldset class=3D"mimeAttachmentHeader"></fieldset><br>
<br>
_______________________________________________<br>
<br>
Proj mailing list<br>
<br>
<a class=3D"moz-txt-link-abbreviated" href=3D"mailto:Proj at lists.maptools.o=
rg">Proj at lists.maptools.org</a><br>
<br>
<a class=3D"moz-txt-link-freetext" target=3D"_blank" href=3D"http://lists.=
maptools.org/mailman/listinfo/proj">http://lists.maptools.org/mailman/list=
info/proj</a></pre>
</blockquote>
</div>
<!-- end of AOLMsgPart_2_3b1f8c21-4c94-442c-97a0-6fa5423c8540 -->
</div>
</font></div>
</font>
----------MB_8CC9FF68E498677_41E0_36026_Webmail-m124.sysops.aol.com--
More information about the Proj
mailing list