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authorKurt Schwehr <schwehr@google.com>2018-05-04 22:44:44 -0700
committerKurt Schwehr <schwehr@google.com>2018-05-04 22:44:44 -0700
commit8e53bdeae08ef8c1a2f0dbf4921b8aa7cc41aa41 (patch)
tree02d0bbae463d6fdb1434d7fe63e46c69965c3e5d /src
parent3f74407838d61df629e96ad7082db4993d696119 (diff)
downloadPROJ-8e53bdeae08ef8c1a2f0dbf4921b8aa7cc41aa41.tar.gz
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Add documentation to pj_phi2 by cffk@
Diffstat (limited to 'src')
-rw-r--r--src/pj_phi2.c15
1 files changed, 15 insertions, 0 deletions
diff --git a/src/pj_phi2.c b/src/pj_phi2.c
index b55d9690..a83302e6 100644
--- a/src/pj_phi2.c
+++ b/src/pj_phi2.c
@@ -7,7 +7,22 @@
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;