cpp conversion: remove useless pj_, PJ_ and proj_ filename prefixes

1 files changed, 46 insertions, 0 deletions

diff --git a/src/phi2.cpp b/src/phi2.cpp
new file mode 100644
index 00000000..a83302e6
--- /dev/null
+++ b/src/phi2.cpp
@@ -0,0 +1,46 @@
+/* Determine latitude angle phi-2. */
+
+#include <math.h>
+
+#include "projects.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;
+}