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The values in the table below agree with my calculations based on Wallis, down to millimeter precision. It's a fine validation of both methods.<BR>
<BR>
Regards,<BR>
-- daan Strebe<BR>
<BR>
<BR>
In a message dated 6/27/06 10:40:04, ovv@hetnet.nl writes:<BR>
<BR>
<BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">From: Strebe-aol.com:<BR>
<BR>
You might avail yourself of a copy of L.P. Lee's monograph, "Conformal<BR>
Projections Based on Elliptic Functions", Cartographica, Monograph Number<BR>
16, 1976. Quoting verbatim from p. 97:<BR>
"The positive y-axis represents part of the equator, extending from lambda =<BR>
0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly<BR>
from a straight line to a curve... The projection of the entire spheroid is<BR>
shown in Fig. 46, again using the eccentricity of the International<BR>
(Hayford) Spheroid. It can be seen that the entire spheroid is represented<BR>
withing the finite area without singular points..."<BR>
<BR>
Reply:<BR>
<BR>
Thanks for this explanation!<BR>
The numbers show it too:<BR>
<BR>
International ellipsoid:<BR>
90*(1-eccentricity) = 82.62073 decimal deg<BR>
<BR>
Tranverse Mercator:<BR>
lat0=0; lon0=0; x0=5e5; y0=0; k0=0.9996;<BR>
// International ellipsoid<BR>
lat=0; lon=82.50; x,y = 18712722.276, 0 meters<BR>
lat=0; lon=82.60; x,y = 18840409.942, 0<BR>
lat=0; lon=82.61; x,y = 18853673.034, 0<BR>
lat=0; lon=82.62; x,y = 18867090.964, 0<BR>
lat=0; lon=82.621; x,y = 18868446.553, 0.2947<BR>
lat=0; lon=82.63; x,y = 18880722.285, 107.602<BR>
lat=0; lon=82.64; x,y = 18894438.954, 366.186<BR>
lat=0; lon=82.65; x,y = 18908216.295, 738.078<BR>
lat=0; lon=82.70; x,y = 18977788.411, 3947.057<BR>
lat=0; lon=82.80; x,y = 19119409.657, 15745.905<BR>
<BR>
For what it's worth: alculated with my improved version of 'Dozier'.<BR>
<BR>
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