No subject
Wed Feb 3 18:16:50 EST 2010
"A simple but useful way of appraising the location of the origin of an az=
imuthal projection is to plot a series of concentric circles of radii z re=
presenting the isograms of maximum Angular Deformation and those of Scale=
Error at the scale of a convenient atlas map and to shift this overlay ab=
out on the map until a good fit is obtained between some of the extreme po=
ints 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 kn=
ow of any non-iterative method for doing that, though possibly one could=
be devised. As an algorithmic process, it is not difficult, but unlike Al=
bers, it is not just a matter of observing the most northerly and most sou=
therly points.
I note that you talk about comparing Albers to LAEA. Once you know the opt=
imal LAEA you can do that, since the distortion values at the extremes of=
the region of interest can be computed by known formul=E6 once the parame=
ters 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?
Regards,
=97 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 scale=
=A0
error of the Albers and the LAEA projections are compared for the whole of=
=A0
the European Union. The Albers has two standard parallels at 38 and 61 d=
N,=A0
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=
=A0
down by trial and error until a reasonable comparison could be made with=
the=A0
Albers.
I wondered, could the center of the LAEA projection have been determined=
=A0
algorithmically?
Still an academic discussion, of course......
Oscar van Vlijmen
--=_1950DD6C.B1FC.434E.84A3.43E4F0F7202A at aol.com
Content-Transfer-Encoding: quoted-printable
Content-Type: text/html; charset="windows-1252"
<html><head></head><body name=3D"Mail Message Editor"><div><br></div><div>=
Oscar:</div><div><br></div><div>From the publication you cited (<span clas=
s=3D"Apple-style-span" style=3D"font-family: monospace; font-size: 11px;=
">http://www.ec-gis.org/sdi/publist/pdfs/annoni-etal2003eur.pdf<span clas=
s=3D"Apple-style-span" style=3D"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 pl=
ot 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 co=
nvenient 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 circums=
cribes 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 no=
t 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 formul=E6 once the parameters=
for the projection are known.</div><div><br></div><div>Can I interpret yo=
ur 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 wonderin=
g if one could be devised?</div><div><br></div><div>Regards,</div><div>=97=
daan Strebe</div><div><br></div><br>On Feb 21, 2010, at 8:01:56 AM, OvV_H=
N <ovv at hetnet.nl> wrote:<br><blockquote style=3D"padding-left: 5px;=
margin-left: 5px; border-left-width: 2px; border-left-style: solid; borde=
r-left-color: blue; color: blue; "><span class=3D"Apple-style-span" style=
=3D"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-al=
ign: auto; text-indent: 0px; text-transform: none; white-space: normal; wi=
dows: 2; word-spacing: 0px; -webkit-border-horizontal-spacing: 0px; -webki=
t-border-vertical-spacing: 0px; -webkit-text-decorations-in-effect: none;=
-webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; "><span=
class=3D"Apple-style-span" style=3D"font-family: monospace; font-size: 11=
px; ">In the article I was referring to, the angular deformation and the=
scale<span class=3D"Apple-converted-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-converte=
d-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=3D"Apple-converted-space"> </span><br>down=
by trial and error until a reasonable comparison could be made with the<s=
pan class=3D"Apple-converted-space"> </span><br>Albers.<br>I wondered=
, could the center of the LAEA projection have been determined<span class=
=3D"Apple-converted-space"> </span><br>algorithmically?<br>Still an=
academic discussion, of course......<br><br><br>Oscar van Vlijmen<br></sp=
an></span></blockquote><br><div><br></div><div class=3D"aol_ad_footer" id=
=3D"u43EBAC27A67E46DF8B8E5369260BAF49"></div></body></html>
--=_1950DD6C.B1FC.434E.84A3.43E4F0F7202A at aol.com--
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