cpp conversion: minimal steps to fix compilation errors, not warnings

1 files changed, 0 insertions, 46 deletions

diff --git a/src/pj_phi2.c b/src/pj_phi2.c
deleted file mode 100644
index a83302e6..00000000
--- a/src/pj_phi2.c
+++ /dev/null
@@ -1,46 +0,0 @@
-/* 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;
-}