# glClipPlane - Plane equation?

I’ve got some problems with using glClipPlanes(GL_CLIP_PLANEi, GLdouble* equation):
All the docs I found say equation is a GLdouble quadruple specifiying the clipping plane’s equation…
I only know two equations for planes:

1. Specify a vektor which points to a vertex on the planes and specify two non-parallel vectors that lie within the plane.
2. Specify a vector which points to a vertex on the plane an which is a the same time a normal of that plane.

glClipPlane cannot use the first way and the second one doesn’t semm to work either:
I interpreted the 4 coordinates as the normal vector starting at the origin *and starting at the eye position… Both ways didn’t produce the expected results.

Help me please, until know I spent WAY to much time on this topic.

you’ve

x = a + rb + sc

so you’ve got

x1 = a1 + rb1 + sc1
x2 = a2 + rb2 + sc2
x3 = a3 + rb3 + sc3

Put that all into a single equation and you’ll have

Ax1 + Bx2 + C*x3 = D

Where the vector ABC is the normal besides…

So A,B,C and D would be 3-dim vectors, right?
But that would mean I had to specify 12 coordinates.

No.

x1 = a1 + rb1 + sc1
x2 = a2 + rb2 + sc2
x3 = a3 + rb3 + sc3

(x1-a1-s*c1)/b1 = r

x2 = a2 + (x1-a1-sc1)/b1b2 + s*c2

and so on.

With vector ABC I meant (A,B,C)

In my red book, 3rd Ed. (OpenGL 1.2) p140 under glClipPlane() it says that " … equation argument points to the four coefficients of the plane equation, Ax + By + Cz + D = 0 … " just like Michael Steinberg says. I like to think of it (like Michael) as a normal vector (A, B, C) and a point on the plane gives you D.

For example, if the normal vector is (3, 4, 5) and a point on the plane is (1, 1, 1), then substituting in gives 31 + 41 + 5*1 + D = 0 => D = -12.

Yes!
I just wanted to state the correlation between his and the opengl representation of a plane… 