Opengl 模型视图投影中的第四行是否为查看位置?

Opengl 模型视图投影中的第四行是否为查看位置?,opengl,glsl,shader,coordinate-transformation,perspectivecamera,Opengl,Glsl,Shader,Coordinate Transformation,Perspectivecamera,我想从摄影机的方向照亮顶点级别的平面。当我旋转视图时,我希望照明不会改变,但当我稍微看向别处时,它是最亮的。mvp[3]不是我认为的摄像机坐标吗 #version 450 in vec3 vertex; uniform mat4 mvp; out vec4 color; void main() { gl_Position = mvp * vec4(vertex,1.); vec3 n = vec3(0.,0.,1.); vec3 v = normalize( vec3

我想从摄影机的方向照亮顶点级别的平面。当我旋转视图时,我希望照明不会改变,但当我稍微看向别处时,它是最亮的。mvp[3]不是我认为的摄像机坐标吗

#version 450
in vec3 vertex;
uniform mat4 mvp;
out vec4 color;

void main()
{
    gl_Position =  mvp * vec4(vertex,1.);
    vec3 n = vec3(0.,0.,1.);
    vec3 v = normalize( vec3(mvp[3])-vertex );
    //I tried the other direction in the mat4
    //vec3(mvp[0][3],mvp[1][3],mvp[2][3]);

    color = vec4( dot(v,n) );
}
模型视图投影矩阵中的第4列是否为查看位置

不,不是。视图矩阵的第4列将包含摄影机位置,但模型视图投影矩阵是模型矩阵、视图矩阵和投影矩阵的组合

视图矩阵通常如下所示:

mat4 view;

view[0] : ( X-axis.x, X-axis.y, X-axis.z, 0 )
view[1] : ( Y-axis.x, Y-axis.y, Y-axis.z, 0 )
view[2] : ( Z-axis.x, Z-axis.y, Z-axis.z, 0 )
view[3] : ( trans.x,  trans.y,  trans.z,  1 ) 
r = right, l = left, b = bottom, t = top, n = near, f = far 

mat4 projection;

projection[0] :     2*n/(r-l)      0              0                0
projection[1] :     0              2*n/(t-b)      0                0
projection[2] :     (r+l)/(r-l)    (t+b)/(t-b)    -(f+n)/(f-n)    -1    
projection[3] :     0              0              -2*f*n/(f-n)     0
mat4 matA;
mat4 matB;{

mat4 matC;
for ( int i0 = 0; i0 < 4; ++ i0 )
    for ( int i1 = 0; i1 < 4; ++ i1 )
        matC[i0][i1] = matB[i0][0] * matA[0][i1] + matB[i0][1] * matA[1][i1] + matB[i0][2] * matA[2][i1] + matB[i0][3] * matA[3][i1];
透视投影矩阵可能如下所示:

mat4 view;

view[0] : ( X-axis.x, X-axis.y, X-axis.z, 0 )
view[1] : ( Y-axis.x, Y-axis.y, Y-axis.z, 0 )
view[2] : ( Z-axis.x, Z-axis.y, Z-axis.z, 0 )
view[3] : ( trans.x,  trans.y,  trans.z,  1 ) 
r = right, l = left, b = bottom, t = top, n = near, f = far 

mat4 projection;

projection[0] :     2*n/(r-l)      0              0                0
projection[1] :     0              2*n/(t-b)      0                0
projection[2] :     (r+l)/(r-l)    (t+b)/(t-b)    -(f+n)/(f-n)    -1    
projection[3] :     0              0              -2*f*n/(f-n)     0
mat4 matA;
mat4 matB;{

mat4 matC;
for ( int i0 = 0; i0 < 4; ++ i0 )
    for ( int i1 = 0; i1 < 4; ++ i1 )
        matC[i0][i1] = matB[i0][0] * matA[0][i1] + matB[i0][1] * matA[1][i1] + matB[i0][2] * matA[2][i1] + matB[i0][3] * matA[3][i1];
矩阵乘法的工作原理如下:

mat4 view;

view[0] : ( X-axis.x, X-axis.y, X-axis.z, 0 )
view[1] : ( Y-axis.x, Y-axis.y, Y-axis.z, 0 )
view[2] : ( Z-axis.x, Z-axis.y, Z-axis.z, 0 )
view[3] : ( trans.x,  trans.y,  trans.z,  1 ) 
r = right, l = left, b = bottom, t = top, n = near, f = far 

mat4 projection;

projection[0] :     2*n/(r-l)      0              0                0
projection[1] :     0              2*n/(t-b)      0                0
projection[2] :     (r+l)/(r-l)    (t+b)/(t-b)    -(f+n)/(f-n)    -1    
projection[3] :     0              0              -2*f*n/(f-n)     0
mat4 matA;
mat4 matB;{

mat4 matC;
for ( int i0 = 0; i0 < 4; ++ i0 )
    for ( int i1 = 0; i1 < 4; ++ i1 )
        matC[i0][i1] = matB[i0][0] * matA[0][i1] + matB[i0][1] * matA[1][i1] + matB[i0][2] * matA[2][i1] + matB[i0][3] * matA[3][i1];
mvp[3]是第四列,而不是第四行。请参阅