2 plane intersection

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rouncer 104 Nov 18, 2007 at 12:31

You cross the normals to get the intersection line direction
then you find an arbitrary point that the both planes share
and you get an origin for the line.

I read that, but I took to a piece of paper to do the algebra and
im not quite good enough at it to do it, can anyone point me
to a piece of code that does it that I can just insert into my
program?

If its too much trouble dont worry about it, im looking right now
and have been unsuccesful so far.

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roel 101 Nov 18, 2007 at 13:02
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rouncer 104 Nov 18, 2007 at 13:43

thanks alot i used the one there, i had to grab it out of a whole library.
But I found it eventually and it worked.

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_oisyn 101 Nov 18, 2007 at 17:14

Basically:

lineDir = cross(a.normal, b.normal);
p = a.normal * a.distance;  // arbitrary point on plane A
v = cross(lineDir, a.normal); // direction toward plane B, parallel to plane A
lineOrigin = rayPlaneIntersection(b, p, v);
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rouncer 104 Nov 19, 2007 at 06:50

Is that one as quick as the 2 unknowns version?
Ill save it on my computer anyway, pretty clever.

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UnrealSolo 101 Jan 03, 2008 at 14:06

thanks for this, its what I was looking for too,
however im doing polygon-polygon intersection,
as part of polyhedron-polyhedron intersection,

If I get a line that is the intersection, I then have to test for where this line crosses with each wire in each polygon, I think its possibly quicker to leave it as the way I have it at the moment wich is to test each wire in each polygon to see where it intersects the other polygon, as lineIntersectplane is quicker than lineInterscetline.

Unles anyone has any other ideas ?…

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_oisyn 101 Jan 04, 2008 at 23:29

If you’re doing a polyhedron-polyhedron intersection, you should use a separating axis test, it’s a lot quicker than what you’re trying to do.

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UnrealSolo 101 Jan 11, 2008 at 16:05

thanks is there an easy explanation of this ?
or even some code ?

im having quite a hard time doing this csg building,
the libraries seem to stop well short of doing this sort of thing,
or they are part of complete packages wich are hard to follow the bits that I need, and many of the explanations seem to go over my head …

the way Ive done it doesnt seem to take too long, the problem I have atm is that I have a surface wich is intersected by many models and results in 2000 intersection lines, of wich 700 are reflex points and I need to add lines so i can produce a list of convex polygons. (its a realy realy huge maze).

it fixes it but takes about a minute.

its like a massive join the dots puzzle, wich would be easy if I could just do it as I see it …

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_oisyn 101 Jan 11, 2008 at 17:02

Basically

bool SATest(object a, object b, vector3 v)
{
    float minA, maxA, minB, maxB;
    
    a.ProjectOnAxis(v, &minA, &maxA);
    b.ProjectOnAxis(v, &minB, &maxB);

    // test for overlap
    return maxB > minA && maxA > minB;
}

bool IntersectionTest(object A, object B)
{
    // test each plane of A
    for each (Plane p in a.GetPlanes())
    {
        if (!SATest(a, b, p.GetNormal())
            return false;
    }

    // test each plane of B
    for each (Plane p in b.GetPlanes())
    {
        if (!SATest(a, b, p.GetNormal())
            return false;
    }

    // test each combination of edges of A and B
    for each (Edge eA in a.GetEdges())
    {
        for each (Edge eB in b.GetEdges())
        {
            vector3 v = Cross(eA.GetDir(), eB.GetDir());
            if (!SATest(a, b, v))
                return false;
        }
    }

    return false;
}

There’s a lot of info regarding the separating axis test on the internet. Just google for it.

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Nyad 101 Mar 28, 2008 at 12:55

Hey if I’m just doing plane-plane, tri-tri, tri-plane intersection.
Would it be silly to use the separating axis test? since its slower than these types of intersection?

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_oisyn 101 Mar 28, 2008 at 14:23

A plane is not a convex polytope, so you can’t so SATs with planes. Besides, the only thing you need to test with a tri-plane intersection is whether all three verts of a tri are on the same side of the plane.