How to build perspective projection matrix (no API)

问题

I develop a simple 3D engine (Without any use of API), successfully transformed my scene into world and view space but have trouble projecting my scene (from view space) using the perspective projection matrix (OpenGL style). I'm not sure about the fov, near and far values and the scene I get is distorted. I hope if someone can direct me how to build and use the perspective projection matrix properly with example codes. Thanks in advance for any help.

The matrix build:

double f = 1 / Math.Tan(fovy / 2);
return new double[,] { 

    { f / Aspect, 0, 0, 0 },
    { 0, f, 0, 0 },
    { 0, 0, (Far + Near) / (Near - Far),  (2 * Far * Near) / (Near - Far) }, 
    { 0, 0, -1, 0 } 
};

The matrix use:

foreach (Point P in T.Points)
{     
    .
    .     // Transforming the point to homogen point matrix, to world space, and to view space (works fine)
    .     

    // projecting the point with getProjectionMatrix() specified in the previous code :      

    double[,] matrix = MatrixMultiply( GetProjectionMatrix(Fovy, Width/Height, Near, Far) , viewSpacePointMatrix );

    // translating to Cartesian coordinates (from homogen):

    matrix [0, 0] /= matrix [3, 0];
    matrix [1, 0] /= matrix [3, 0];
    matrix [2, 0] /= matrix [3, 0];
    matrix [3, 0] = 1;
    P = MatrixToPoint(matrix);

    // adjusting to the screen Y axis:

    P.y = this.Height - P.y;

    // Printing...
}

回答1:

Following is a typical implemenation of perspective projection matrix. And here is a good link to explain everything OpenGL Projection Matrix

void ComputeFOVProjection( Matrix& result, float fov, float aspect, float nearDist, float farDist, bool leftHanded /* = true */ )
{
    //
    // General form of the Projection Matrix
    //
    // uh = Cot( fov/2 ) == 1/Tan(fov/2)
    // uw / uh = 1/aspect
    // 
    //   uw         0       0       0
    //    0        uh       0       0
    //    0         0      f/(f-n)  1
    //    0         0    -fn/(f-n)  0
    //
    // Make result to be identity first

    // check for bad parameters to avoid divide by zero:
    // if found, assert and return an identity matrix.
    if ( fov <= 0 || aspect == 0 )
    {
        Assert( fov > 0 && aspect != 0 );
        return;
    }

    float frustumDepth = farDist - nearDist;
    float oneOverDepth = 1 / frustumDepth;

    result[1][1] = 1 / tan(0.5f * fov);
    result[0][0] = (leftHanded ? 1 : -1 ) * result[1][1] / aspect;
    result[2][2] = farDist * oneOverDepth;
    result[3][2] = (-farDist * nearDist) * oneOverDepth;
    result[2][3] = 1;
    result[3][3] = 0;
}

回答2:

Another function that may be useful.

This one is based on left/right/top/bottom/near/far parameters (used in OpenGL):

static void test(){
    float projectionMatrix[16];

    // width and height of viewport to display on (screen dimensions in case of fullscreen rendering)
    float ratio = (float)width/height;
    float left = -ratio;
    float right = ratio;
    float bottom = -1.0f;
    float top = 1.0f;
    float near = -1.0f;
    float far = 100.0f;

    frustum(projectionMatrix, 0, left, right, bottom, top, near, far);

}

static void frustum(float *m, int offset,
                     float left, float right, float bottom, float top,
                     float near, float far) {

    float r_width  = 1.0f / (right - left);
    float r_height = 1.0f / (top - bottom);
    float r_depth  = 1.0f / (far - near);
    float x =  2.0f * (r_width);
    float y =  2.0f * (r_height);
    float z =  2.0f * (r_depth);
    float A = (right + left) * r_width;
    float B = (top + bottom) * r_height;
    float C = (far + near) * r_depth;
    m[offset + 0] = x;
    m[offset + 3] = -A;
    m[offset + 5] = y;
    m[offset + 7] = -B;
    m[offset + 10] = -z;
    m[offset + 11] = -C;
    m[offset +  1] = 0.0f;
    m[offset +  2] = 0.0f;
    m[offset +  4] = 0.0f;
    m[offset +  6] = 0.0f;
    m[offset +  8] = 0.0f;
    m[offset +  9] = 0.0f;
    m[offset + 12] = 0.0f;
    m[offset + 13] = 0.0f;
    m[offset + 14] = 0.0f;
    m[offset + 15] = 1.0f;

}

来源：https://stackoverflow.com/questions/18404890/how-to-build-perspective-projection-matrix-no-api

标签

math

graphics

transformation

projection