View Full Version : clipping the viewing volume

Hi,

I'm trying to implement OpenGL GPU Pipeline and got some troubles in clipping in the viewing volume: e.g.

I have line from V1(x1,y1,z1,w1)&V2(x2,y2,z2,w2) - clip coordinates - :

(ignore the ....)

V1 (x1,w1) o \

...................\ o V2 (x2,w2)

...................W=Xmax

where x1>w1 and x2<w2 so I need to clip the line on x axis. I need to find t={x1-Xmax}/{x1-x2}, but what is W {Xmax}? How can I find him?

Ok, in clipping space (post transformation) the coordinates are always in a cube of edge length 2 (from -1 to +1).

Said that, wouldn't be easier to find the clipping volume using some kind of plane-ray intersection?

I mean, take the ray (line) and test it against the 6 planes of the fustrum. Then you will have where the ray intersects with the fustrum(u should assume an infinite ray, so you should extend your segment to a ray).

Once you have that intersection you can clip as needed.

So, summarizing:

- Take your frustum from opengl (look at google for opengl frustum extraction)

- This frustum is in view space so:

-- Take your ray (in world coordinates) and multiply it by the modelview matrix

-- Test your ray against the frustum (you will get an intersection point P)

-- multiply P by the inverse of the modelview matrix (to return it to World coordinates)

Et voila, now you can do whatever you need with that point.

IG, your artwork leaves very, very much to be desired, if you don't mind my saying so. Try using a "code" tag to preserve the formatting.

The unit cube clip-planes are in post-projective clip space, where they should be. I think you're suffering from a conceptual malady. A line in world-space is a line in post-projective clip space; a plane in world-space is a plane in post-projective clip-space. Either can be transformed into or out of either space. Of interest, consequently, is that they are both in the same space for clipping.

Thanks.

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