[Proj] Re: Complex Transverse Mercator

[Strebe at aol.com]

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.

Regards,
-- daan Strebe


In a message dated 6/27/06 10:40:04, ovv at hetnet.nl writes:


> 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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