Question

因此,我转播了一台硬盘(真正的地貌是一根ube,可能出现剪.,而块状的地貌是一门脑pon子),我需要计算深度。

我可以比较容易地计算在世界空间抵消的数额。不幸的是,我花了几个小时时间,却未能赶上它的深度。

我所能取得的唯一正确结果是:

gl_FragDepth = gl_FragCoord.z

基本上,我需要知道如何计算FragCoord.z,以便我能够:

Take the inverse transformation from gl_FragCoord.z to eye space
Add the depth perturbation
Transform this perturbed depth back into the same space as the original gl_FragCoord.z.

如果这似乎是一个重复的问题,我表示歉意;这里还有其他一些处理类似问题的员额。然而,在执行所有这些建议之后,没有任何工作是正确的。此时此刻,我不是试图找人帮助,而是要求制定一部完整的法典。这只是几个方面。

Answer 1

关键守则是:

float far=gl_DepthRange.far; float near=gl_DepthRange.near;

vec4 eye_space_pos = gl_ModelViewMatrix * /*something*/
vec4 clip_space_pos = gl_ProjectionMatrix * eye_space_pos;

float ndc_depth = clip_space_pos.z / clip_space_pos.w;

float depth = (((far-near) * ndc_depth) + near + far) / 2.0;
gl_FragDepth = depth;

Answer 2

就今后的另一个参考而言,这是同文字母一样的公式。

vec4 v_clip_coord = modelview_projection * vec4(v_position, 1.0);
float f_ndc_depth = v_clip_coord.z / v_clip_coord.w;
gl_FragDepth = (1.0 - 0.0) * 0.5 * f_ndc_depth + (1.0 + 0.0) * 0.5;

此处,modelview_projection is 4x4 modelview-projection Format and v_position is Object-space status of the pixel being made (in my case based by a raymarcher).

The equation comes from the window coordinates section of this manual. Note that in my code, near is 0.0 and far is 1.0, which are the default values of gl_DepthRange. Note that gl_DepthRange is not the same thing as the near/far distance in the formula for perspective projection matrix! The only trick is using the 0.0 and 1.0 (or gl_DepthRange in case you actually need to change it), I ve been struggling for an hour with the other depth range - but that is already "baked" in my (perspective) projection matrix.

Note that this way, the equation really contains just a single multiply by a constant ((far - near) / 2) and a single addition of another constant ((far + near) / 2). Compare that to multiply, add and divide (possibly converted to a multiply by an optimizing compiler) that is required in the code of imallett.

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