How to recover view space position given view space depth value and ndc xy

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3 Solutions to recover view space position in perspective projection

The projection matrix describes the mapping from 3D points of a scene, to 2D points of the viewport. It transforms from view (eye) space to the clip space, and the coordinates in the clip space are transformed to the normalized device coordinates (NDC) by dividing with the w component of the clip coordinates. The NDC are in range (-1,-1,-1) to (1,1,1).

At Perspective Projection the projection matrix describes the mapping from 3D points in the world as they are seen from of a pinhole camera, to 2D points of the viewport.
The eye space coordinates in the camera frustum (a truncated pyramid) are mapped to a cube (the normalized device coordinates).

Perspective Projection Matrix:

r = right, l = left, b = bottom, t = top, n = near, f = far

2*n/(r-l)      0              0               0
0              2*n/(t-b)      0               0
(r+l)/(r-l)    (t+b)/(t-b)    -(f+n)/(f-n)    -1    
0              0              -2*f*n/(f-n)    0

it follows:

aspect = w / h
tanFov = tan( fov_y * 0.5 );

prjMat[0][0] = 2*n/(r-l) = 1.0 / (tanFov * aspect)
prjMat[1][1] = 2*n/(t-b) = 1.0 / tanFov

At Perspective Projection, the Z component is calculated by the rational function:

z_ndc = ( -z_eye * (f+n)/(f-n) - 2*f*n/(f-n) ) / -z_eye

The depth (gl_FragCoord.z and gl_FragDepth) is calculated as follows:

z_ndc = clip_space_pos.z / clip_space_pos.w;
depth = (((farZ-nearZ) * z_ndc) + nearZ + farZ) / 2.0;

1. Field of view and aspect ratio

Since the projection matrix is defined by the field of view and the aspect ratio it is possible to recover the viewport position with the field of view and the aspect ratio. Provided that it is a symmetrical perspective projection and the normalized device coordinates, the depth and the near and far plane are known.

Recover the Z distance in view space:

z_ndc = 2.0 * depth - 1.0;
z_eye = 2.0 * n * f / (f + n - z_ndc * (f - n));

Recover the view space position by the XY normalized device coordinates:

ndc_x, ndc_y = xy normalized device coordinates in range from (-1, -1) to (1, 1):

viewPos.x = z_eye * ndc_x * aspect * tanFov;
viewPos.y = z_eye * ndc_y * tanFov;
viewPos.z = -z_eye;

2. Projection matrix

The projection parameters, defined by the field of view and the aspect ratio, are stored in the projection matrix. Therefore the viewport position can be recovered by the values from the projection matrix, from a symmetrical perspective projection.

Note the relation between projection matrix, field of view and aspect ratio:

prjMat[0][0] = 2*n/(r-l) = 1.0 / (tanFov * aspect);
prjMat[1][1] = 2*n/(t-b) = 1.0 / tanFov;

prjMat[2][2] = -(f+n)/(f-n)
prjMat[3][2] = -2*f*n/(f-n)

Recover the Z distance in view space:

A     = prj_mat[2][2];
B     = prj_mat[3][2];
z_ndc = 2.0 * depth - 1.0;
z_eye = B / (A + z_ndc);

Recover the view space position by the XY normalized device coordinates:

viewPos.x = z_eye * ndc_x / prjMat[0][0];
viewPos.y = z_eye * ndc_y / prjMat[1][1];
viewPos.z = -z_eye;

3. Inverse projection matrix

Of course the viewport position can be recovered by the inverse projection matrix.

mat4 inversePrjMat = inverse( prjMat );
vec4 viewPosH      = inversePrjMat * vec3( ndc_x, ndc_y, 2.0 * depth - 1.0, 1.0 )
vec3 viewPos       = viewPos.xyz / viewPos.w;

See also the answers to the following question:

  • How to render depth linearly in modern OpenGL with gl_FragCoord.z in fragment shader?

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