Question

I’m using CML to manage the 3D math in an OpenGL-based interface project I’m making for work. I need to know the width of the viewing frustum at a given distance from the eye point, which is kept as a part of a 4×4 matrix that represents the camera. My goal is to position gui objects along the apparent edge of the viewport, but at some distance into the screen from the near clipping plane.

CML has a function to extract the planes of the frustum, giving them back in Ax + By + Cz + D = 0 form. This frustum is perpendicular to the camera, which isn’t necessarily aligned with the z axis of the perspective projection.

I’d like to extract x and z coordinates so as to pin graphical elements to the sides of the screen at different distances from the camera. What is the best way to go about doing it?

Thanks!

Answer 1

This seems to be a duplicate of Finding side length of a cross-section of a pyramid frustum/truncated pyramid, if you already have a cross-section of known width a known distance from the apex. If you don’t have that and you want to derive the answer yourself you can follow these steps.

1. Take two adjacent planes and find
   their line of intersection L1. You
   can use the steps here. Really
   what you need is the direction
   vector of the line.
2. Take two more planes, one the same
   as in the previous step, and find
   their line of intersection L2.
3. Note that all planes of the form Ax + By + Cz + D = 0 go through the origin, so you know that L1 and L2
   intersect.
4. Draw yourself a picture of the
   direction vectors for L1 and L2,
   tails at the origin. These form an
   angle; call it theta. Find theta
   using the formula for the angle
   between two vectors, e.g. here.
5. Draw a bisector of that angle. Draw
   a perpendicular to the bisector at
   the distance d you want from the
   origin (this creates an isosceles
   triangle, bisected into two
   congruent right triangles). The
   length of the perpendicular is your
   desired frustum width w. Note that w is
   twice the length of one of the bases
   of the right triangles.
6. Let r be the length of the
   hypotenuses of the right triangles.
   Then rcos(theta/2)=d and
   rsin(theta/2)=w/2, so
   tan(theta/2)=(w/2)/d which implies
   w=2d*tan(theta/2). Since you know d
   and theta, you are done.

Note that we have found the length of one side of a cross-section of a frustrum. This will work with any perpendicular cross-section of any frustum. This can be extended to adapt it to a non-perpendicular cross-section.