I’m requesting a bit of help since I cannot find what exactly is wrong with this bit of code. It seems to render the sphere in exactly one color.

```
/**
* \brief calcTexCoord2 The final step in calculating texture coordinates.
*
* \param texture The texture to use
* \param axis The value of the axis to use.
* \param s_coord The value to plug in for the s equation.
* \param t_coord The value to plug in for the t equation.
* \return Nothing.
*/
void calcTexCoord2(int texture, GLdouble axis, GLdouble s_coord, GLdouble t_coord) {
//cout << "axis: " << axis << " ";
GLdouble s, t; //texture coordinates
s = (s_coord + axis) / (2 * axis);
t = (t_coord + axis) / (2 * axis);
//cout << "axis: " << axis << " s_coord: " << s_coord << " t_coord: " << t_coord << " s: " << s << " t: " << t << endl;
//cout.flush();
glBindTexture(GL_TEXTURE_2D, texture_name[texture]);
glTexCoord2f(s, t);
}
/**
* \brief calcTexCoord Inputs the texture coordinate at that point.
*
* \param vertex The vertex to calculate the texture coordinate for.
* Also, the assumption is that the vertex can be treated as the
* normal, since the sphere is of radius = 1 and centered at origin.
* \param viewer The view vector.
* \return Nothing.
*/
void calcTexCoord(GLdouble vertex[3], GLdouble *viewer) {
//cout << "normal: " << vertex[0] << ", " << vertex[1] << ", " << vertex[2] << " viewer: " << viewer[0] << ", " << viewer[1] << ", " << viewer[2];
//cout << " final: " << dist << endl;
//cout.flush();
//calc reflection vector via 2 * (n dot v) n - v
//start w/ n dot v
GLdouble dot_product = 0;
for (int i = 0; i < 3; i++)
dot_product += vertex[i]*viewer[i];
//2 * (n dot v)
dot_product *= 2;
//2 * (n dot v) n - v
GLdouble reflect[3];
for (int i = 0; i < 3; i++)
reflect[i] = (dot_product * vertex[i]) - viewer[i];
//find the texture to use
int index = 0;
for (int i = 1; i < 3; i++)
if (fabs(reflect[index]) < fabs(reflect[i]))
index = i;
switch (index) {
case 0:
if (reflect[index] > 0) //+x
calcTexCoord2(2, reflect[0], reflect[2], reflect[1]);
else
calcTexCoord2(3, reflect[0], reflect[2], reflect[1]);
break;
case 1:
if (reflect[index] > 0)
calcTexCoord2(0, reflect[1], reflect[0], reflect[2]);
else
calcTexCoord2(1, reflect[1], reflect[0], reflect[2]);
break;
case 2:
if (reflect[index] > 0)
calcTexCoord2(4, reflect[2], reflect[1], reflect[0]);
else
calcTexCoord2(5, reflect[2], reflect[1], reflect[0]);
}
}
/**
* \brief drawReflectivePole Draws the poles of the sphere using triangles.
* Borrowing the code from the book. This one is used
* for the reflective sphere.
* \param pole The pole that we are drawing.
* \param viewer The viewer vector
* \return Nothing.
*/
void drawReflectivePole(GLdouble **pole, GLdouble *viewer) {
glBegin(GL_TRIANGLE_FAN);
for (int i = 0; i < sphere_tri_strip_size; i++) {
calcTexCoord(pole[i], viewer);
glVertex3d(pole[i][0], pole[i][1], pole[i][2]);
}
glEnd();
}
/**
* \brief drawReflectiveSphere Draws a sphere (with reflection).
*
* \param viewer_x The viewer's x coordinate.
* \param viewer_y The viewer's y coordinate.
* \param viewer_z The viewer's z coordinate.
* \return Nothing.
*/
void drawReflectiveSphere(GLdouble viewer_x, GLdouble viewer_y, GLdouble viewer_z) {
if (sphere_quad_strip == NULL)
makeSphere();
//calc viewer vector
GLdouble distance = sqrt(viewer_x*viewer_x + viewer_y*viewer_y + viewer_z*viewer_z);
GLdouble viewer[] = { viewer_x / distance, viewer_y / distance, viewer_z / distance};
glBegin(GL_QUAD_STRIP);
for (int i = 0; i < sphere_quad_strip_size; i++) {
glNormal3dv(sphere_quad_strip[i]);
calcTexCoord(sphere_quad_strip[i], viewer);
glVertex3dv(sphere_quad_strip[i]);
}
glEnd();
drawReflectivePole(sphere_top_tri_strip, viewer);
drawReflectivePole(sphere_bottom_tri_strip, viewer);
}
```

Basically, I’m rendering a sphere, texturing it with a cube map. Since the distance from the origin of all the vertexes are exactly 1, I treat the vertexes the same as the normal.

I’m certain there is a bunch of error in my assignment of the coordinates to calcTexCoord2, but I still cannot see why the sphere is rendered in exactly one color.

Thank you for any help you may provide.