[Proj] Complex Transverse Mercator
Oscar van Vlijmen
ovv at hetnet.nl
Tue Jun 27 13:34:44 EDT 2006
From: Strebe-aol.com:
You might avail yourself of a copy of L.P. Lee's monograph, "Conformal
Projections Based on Elliptic Functions", Cartographica, Monograph Number
16, 1976. Quoting verbatim from p. 97:
"The positive y-axis represents part of the equator, extending from lambda =
0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly
from a straight line to a curve... The projection of the entire spheroid is
shown in Fig. 46, again using the eccentricity of the International
(Hayford) Spheroid. It can be seen that the entire spheroid is represented
withing the finite area without singular points..."
Reply:
Thanks for this explanation!
The numbers show it too:
International ellipsoid:
90*(1-eccentricity) = 82.62073 decimal deg
Tranverse Mercator:
lat0=0; lon0=0; x0=5e5; y0=0; k0=0.9996;
// International ellipsoid
lat=0; lon=82.50; x,y = 18712722.276, 0 meters
lat=0; lon=82.60; x,y = 18840409.942, 0
lat=0; lon=82.61; x,y = 18853673.034, 0
lat=0; lon=82.62; x,y = 18867090.964, 0
lat=0; lon=82.621; x,y = 18868446.553, 0.2947
lat=0; lon=82.63; x,y = 18880722.285, 107.602
lat=0; lon=82.64; x,y = 18894438.954, 366.186
lat=0; lon=82.65; x,y = 18908216.295, 738.078
lat=0; lon=82.70; x,y = 18977788.411, 3947.057
lat=0; lon=82.80; x,y = 19119409.657, 15745.905
For what it's worth: alculated with my improved version of 'Dozier'.
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