View Full Version : Clipping triangles in clip space

urquidio

01-27-2007, 06:39 AM

Hi,

I'm working on a software implementation of

opengl and have a question about clipping

triangles.

from what i understand, you can perform the

clipping in clip space by comparing a point

to its w component, e.g.

-w < x < w

-w < y < w

-w < z < w

tells you the point is inside the clip space

if w is positive.

But when i have a triangle that i have

transformed by the projection matrix, the three

vertices will each have different w's. in fact,

their w terms will be equal to the negative of

their original z.

So how do you clip the triangle against the

side of the bounding box if all the w's are

different? which w do you use to set up the

bounding box? i don't understand how to have

different volumes for each point. how do

you clip the edges of the triangle then?

and if you normalize, to get them all

the same, what is the point of this "clip" space

then? after all, every book on the subject

seems to indicate that clipping occurs before

the perspective divide!

thanks for your help in advance!

Flavious

01-27-2007, 07:34 AM

Here's some source code: http://wwwx.cs.unc.edu/~sud/courses/236/a5/softgl_homoclip_smooth.cpp

You might also search for the Sutherland-Hodgman clipping algorithm and Homogeneous polygon clipping in general.

Hope it helps...

Cheers

If you just consider the 1D homogeneous case of (x,w), it's easier to visualize what you're doing.

In this case, you have a 2D plot with x on the horizontal axis and w on the vertical.

The clipping half-spaces are x LEQUAL w, which is everything on or to the left of a diagonal line of slope=1 going through the origin, and x GEQUAL -w, which is everything on or to the right of a diagonal line of slope=-1 also going through the origin.

Now imagine a line segment with endpoints p0 = (x0, w0) and p1 = (x1, w1), where p0 is inside the clip volume and p1 is outside. What you want to do is find the point along the line segment where it intersects the clip plane.

If we assume that the segment is clipped by the x LEQUAL w plane, then there is a point along the line segment that satisfies the plane equation x - w = 0, or {1, -1} when we write just the coefficients (as is typical in OpenGL).

First we compute the signed distance between each point and the plane (just a dot product).

d0 = dot(p0, clip_plane)

d1 = dot(p1, clip_plane)

Then, the lerp factor "a" for the point where distance is zero is

a = d0 / (d0-d1)

regardless of the sign of d0 and d1.

We use "a" to compute the new clipped vertex pc = (xc, yc, zc, wc) as

pc = (1-a) * p0 + a * p1

This new point will lie on the x=w line.

That is, xc == wc.

Does that make sense?

Cass

...ugh, I had to use LEQUAL and GEQUAL because the board rejects the symbols as illegal html tags. :(

[edit: fix lerp equation]

[edit2: fix graph description in paragraph 3]

Flavious

01-27-2007, 09:27 AM

Or if you look at it another way, parametrically, for example with your p0 and p1, and x for left/right:

xc = -/+ wc

or

p0.x + (p1.x - p0.x) * a = -/+ (p0.w + (p1.w - p0.w) * a)

and solve for a.

Same thing... :)

Cheers

Agreed, Flavious.

I actually wrote it down that way first, and changed it to the other because computing the clip distance first makes it clearer how things work for arbitrary clip planes.

Flavious

01-27-2007, 12:06 PM

I actually prefer the geometry you posted, Cass (give me the geometry every time :) ). I only added the other bit to hopefully explain what the linked code I posted is doing.

urquidio

01-30-2007, 09:41 AM

Thanks for your replies. They were very helpful. I was able to implement the clipping in clip-space as you describe.

However, I seem to be running in to a little problem when the triangles have points that are behind the viewer. in these cases, my w has a negative term. this means that my clip test is now wc < xc < -wc, correct? but the "left" plane is still xc = -wc and the right plane is still wc = xc, correct? So i believe that i don't need to change my equations that set xc = wc for the right plane and xc = -wc for the left plane when wc is negative. is this correct?

I think i'm thinking about this correctly, but for some reason i'm getting an incorrect output.

I thought i would simplify things by multiplying by a -1 if w was negative - to make w always positive - but this doesn't work. it seems that the points behind the viewer turn into points in front of the viewer and so i end up getting a portion of the triangle getting mapped to the screen even though it should be completely behind me... So we can't just flip the sign of the transformed point before clipping, correct?

what sort of issues should i be watching out for that could be causing problems with clipping with -w?

thanks

Cases where wc < 0 are guaranteed to fail the inequality because -wc < wc can never be true for negative wc, so no xc can be between them.

You shouldn't be doing any sign flipping of terms.

If you think about it, the "unclipped" part is where a V shape that starts at the origin on the 2D xw graph and goes up, so negative wc will always be clipped away.

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