從線性深度重建世界位置

Question

我有問題從glsl中先前存儲的線性深度重建世界位置。我讀了大量的信息在網上，但無法找到我的問題...... 所以這是我得到：

VS (storing depth to 32F): 
float linDepth(float z) { 
    return (-z-zNear)/(zFar-zNear); 
} 
void main() { 
    vec4 position = uViewMatrix * uModelMatrix * vec4(aPosition, 1); 
    depth = linDepth(position.z); //store linear view-depth 
} 

FS (reconstuction): 
void main() { 
    vec3 vUV = vec2(0..1, 0..1); (from screen aligned quad) 
    vec3 ndc = vec3(vUV*2-1, linearViewDepth*2-1); 
    vec4 v0 = inverse(uProjectionMatrix)*vec4(ndc, 1); 
    vec3 reconViewPos = v0.xyz/v0.w; 
    vec3 reconWorldPos = inverse(uViewMatrix) * v0; 
} 

...，結果是完全關閉。 儘管我通過使用不變的線性視圖深度作爲ndc z來感知問題。最後，我想申請光線插值方法：

VS (reconstruction, aligned screenquad): 
out vec3 vViewRay; 
void main() { 
    gl_Position = aPosition; 
    vec4 v = vec4(aPosition.x, aPosition.y, 1, 1); //ndc (at the back of cube) 
    v = inverse(uProjectionMatrix) * v; 
    v /= v.w; //view coordinate 
    v /= v.z; //normalize by z for scaling 
    vViewRay = v.xyz; 
} 

FS(reconstruction): 
in vec3 vViewRay; 
float delinDepth(float z) { 
    return -(z*(zFar-zNear)+zNear); 
} 
void main() { 
    vec3 reconViewPos = vViewRay * delinDepth(linearViewDepth); 
} 

Answer 1

在第二重建代碼沒有工作，我想考研，過別的地方結束...