fixed far plane intersection test for fogvolumes#24220
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kfc35
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The math makes sense to me
And I’ve verified the fix in the atmosphere example with free camera enabled
I suggest adding documentation about far_planes for why its of length 6 (greater than length 3) and that it’s padded with vec’s of zeros if there are less than the max of 6
| @@ -144,10 +144,13 @@ fn fragment(@builtin(position) position: vec4<f32>) -> @location(0) vec4<f32> { | |||
| // Calculate the end position of the ray. This requires us to raytrace the | |||
| // three back faces of the AABB to find the one that our ray intersects. | |||
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| // three back faces of the AABB to find the one that our ray intersects. | |
| // back faces of the AABB to find the one that our ray intersects. |
| /// plane and Q is any point in it, in view space. The equation of the plane | ||
| /// for homogeneous point P = (Px, Py, Pz, Pw) is V⋅P = 0. | ||
| far_planes: [Vec4; 3], | ||
| far_planes: [Vec4; 6], |
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Is it possible for this to be 6? The PR description seems to imply the max is 5, and that seems like a correct assumption to make, but 🤷 considering the assumption of 3 far planes was wrong before, I’m not going to advocate for this to be lowered.
However, I do think a doc comment would be useful in describing why this number is greater than 3 now, which some people might expect, or that the extra planes will be appended as Vec4::Zero
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Good point, I initially implemented it with 5 as the max, but I noticed when the camera moves inside the fog volume the plane_coords.w <= 0.0 test doesn't filter any planes since they all face the camera, meaning the (NEG_Z) plane would disappear. I could solve this by filtering based on a dot product between the eye vector and the plane normal instead, however, I thought the current implementation was a bit nicer / cleaner for a few reasons:
- Works correctly for FoV > 90 degrees
- 6 is the total number of planes (no more assumptions based on specific cases)
- Implementation is slightly faster / simpler. I don't have to pass / compute the eye vector and I don't have to perform additional math (on CPU)
ChristopherBiscardi
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Objective
Fixes #22574
Solution
The previous code assumed there are only 3 back faces visible at any time for the FogVolume. This assumption is incorrect and breaks in some instances: e.g. when the camera is looking head-on at one of the faces of a cube. Depending on the distance and size of the cube, 5 back faces will be visible.
The solution is to consider all visible back faces for testing in the shader.
Testing
Tested with this command and on another test scene without atmospherics (see showcase)
cargo run -r --example atmosphere --features free_cameraShowcase
Before:
screenrecording-2026-05-09_19-52-43.mp4
After:
screenrecording-2026-05-09_19-51-07.mp4