OpenGL Calculating Normals (Quads)
Question
My issue is regarding OpenGL, and Normals, I understand the math behind them, and I am having some success.
The function I've attached below accepts an interleaved Vertex Array, and calculates the normals for every 4 vertices. These represent QUADS that having the same directions. By my understanding these 4 vertices should share the same Normal. So long as they face the same way.
The problem I am having is that my QUADS are rendering with a diagonal gradient, much like this: Light Effect - Except that the shadow is in the middle, with the light in the corners.
I draw my QUADS in a consistent fashion. TopLeft, TopRight, BottomRight, BottomLeft, and the vertices I use to calculate my normals are TopRight - TopLeft, and BottomRight - TopLeft.
Hopefully someone can see something I've made a blunder on, but I have been at this for hours to no prevail.
For the record I render a Cube, and a Teapot next to my objects to check my lighting is functioning, so I'm fairly sure there is no issue regarding Light position.
void CalculateNormals(point8 toCalc[], int toCalcLength)
{
GLfloat N[3], U[3], V[3];//N will be our final calculated normal, U and V will be the subjects of cross-product
float length;
for (int i = 0; i < toCalcLength; i+=4) //Starting with every first corner QUAD vertice
{
U[0] = toCalc[i+1][5] - toCalc[i][5]; U[1] = toCalc[i+1][6] - toCalc[i][6]; U[2] = toCalc[i+1][7] - toCalc[i][7]; //Calculate Ux Uy Uz
V[0] = toCalc[i+3][5] - toCalc[i][5]; V[1] = toCalc[i+3][6] - toCalc[i][6]; V[2] = toCalc[i+3][7] - toCalc[i][7]; //Calculate Vx Vy Vz
N[0] = (U[1]*V[2]) - (U[2] * V[1]);
N[1] = (U[2]*V[0]) - (U[0] * V[2]);
N[2] = (U[0]*V[1]) - (U[1] * V[0]);
//Calculate length for normalising
length = (float)sqrt((pow(N[0],2)) + (pow(N[1],2)) + (pow(N[2],2)));
for (int a = 0; a < 3; a++)
{
N[a]/=length;
}
for (int j = 0; i < 4; i++)
{
//Apply normals to QUAD vertices (3,4,5 index position of normals in interleaved array)
toCalc[i+j][3] = N[0]; toCalc[i+j][4] = N[1]; toCalc[i+j][5] = N[2];
}
}
}
No correct solution