blob: b81456b0a65dc068a7d1e9077c36e5ba37e7fa56 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
|
/* 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, const double ts0, const 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)
*******************************************************************************/
const double eccnth = .5 * e;
double ts = ts0;
#ifdef no_longer_used_original_convergence_on_exact_dphi
double Phi = M_HALFPI - 2. * atan(ts);
#endif
int i = N_ITER;
for(;;) {
/*
* sin(Phi) = sin(PI/2 - 2* atan(ts))
* = cos(2*atan(ts))
* = 2*cos^2(atan(ts)) - 1
* = 2 / (1 + ts^2) - 1
* = (1 - ts^2) / (1 + ts^2)
*/
const double sinPhi = (1 - ts * ts) / (1 + ts * ts);
const double con = e * sinPhi;
double old_ts = ts;
ts = ts0 * pow((1. - con) / (1. + con), eccnth);
#ifdef no_longer_used_original_convergence_on_exact_dphi
/* The convergence criterion is nominally on exact dphi */
const double newPhi = M_HALFPI - 2. * atan(ts);
const double dphi = newPhi - Phi;
Phi = newPhi;
#else
/* If we don't immediately update Phi from this, we can
* change the conversion criterion to save us computing atan() at each step.
* Particularly we can observe that:
* |atan(ts) - atan(old_ts)| <= |ts - old_ts|
* So if |ts - old_ts| matches our convergence criterion, we're good.
*/
const double dphi = 2 * (ts - old_ts);
#endif
if (fabs(dphi) > TOL && --i) {
continue;
}
break;
}
if (i <= 0)
pj_ctx_set_errno(ctx, PJD_ERR_NON_CON_INV_PHI2);
return M_HALFPI - 2. * atan(ts);
}
|