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/* Determine latitude angle phi-2. */
#include <math.h>
#include "proj.h"
#include "proj_internal.h"
static const double TOL = 1.0e-10;
static const int N_ITER = 15;
/*****************************************************************************/
double pj_phi2(projCtx ctx, double ts, double e) {
/******************************************************************************
Determine latitude angle phi-2.
Inputs:
ts = exp(-psi) where psi is the isometric latitude (dimensionless)
e = eccentricity of the ellipsoid (dimensionless)
Output:
phi = geographic latitude (radians)
Here isometric latitude is defined by
psi = log( tan(pi/4 + phi/2) *
( (1 - e*sin(phi)) / (1 + e*sin(phi)) )^(e/2) )
= asinh(tan(phi)) - e * atanh(e * sin(phi))
This routine inverts this relation using the iterative scheme given
by Snyder (1987), Eqs. (7-9) - (7-11)
*******************************************************************************/
double eccnth = .5 * e;
double Phi = M_HALFPI - 2. * atan(ts);
double con;
int i = N_ITER;
for(;;) {
double dphi;
con = e * sin(Phi);
dphi = M_HALFPI - 2. * atan(ts * pow((1. - con) /
(1. + con), eccnth)) - Phi;
Phi += dphi;
if (fabs(dphi) > TOL && --i)
continue;
break;
}
if (i <= 0)
pj_ctx_set_errno(ctx, PJD_ERR_NON_CON_INV_PHI2);
return Phi;
}
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