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# near and far clipping plane

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I'm trying to calculate the size of the far clipping plane but my brain is at a standstill. Someone who could give me a hint? /trysil "A witty and slightly sarcastic quote from an unkown source" -- The generic SIG /trysil Edited by - trysil on February 3, 2002 2:09:42 PM Edited by - trysil on February 3, 2002 3:15:38 PM

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What do you mean by the size of the plane? Planes are infinte in 2 dimensions. Do you mean the distance to the plane, or the size of the far end of the frustrum?

If you mean the size of the far end of the frustrum you can figure this out with a little trig. Draw the right triangle formed by the eye point (1), normalized look at vector times the distance to the far plane (2), and the intersection of the edge of the frustrum (3) and the left,right,top, or bottom plane (it doesn't matter which because they are all the same distance).

  Intersection |-----------2-----3---FAR PLANE----------- | / | / | / <--Hyp | / 1/ look at

you know the length of line 1-2 -- its the distance from the eye to the fustrum.

you know angle formed by 2-1-3 -- its half the FOV

Length of hyp = Length Of line 12 / cos ( FOV / 2)

Length of line 23 = Length of hyp * sin (FOV / 2)

Area of far end of frustrum = width far frustrum * hieght far frustrum

width far frustrum = hieght far frustrum = 2 * length line 23

area far frustrum = 4 * length line 23

Edited by - invective on February 3, 2002 10:47:37 PM

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Thanks!
Sorry about the confusion, my vocabulary became a bit mumbled up

"A witty and slightly sarcastic quote from an unkown source"
-- The generic SIG

/trysil