Without any reference to the book (or anything else) at all....
My guess is that he's thinking of it as projecting things onto z = 0 (with
visible things at z = 0 falling in the rectangle of xl, xr, yt, yb), but his
near plane just happens to be closer to the centre of projection than that.
Jamie
Phil Yard wrote:
> I have just been looking through the definition of the Cyrus-Beck line
> clipping algorithm, as defined in 'Procedural elements for computer
> graphics' by David Rogers.
>
> On talking about deriving the normals to the top, bottom and side planes of
> the viewing frustrum, he talks about using the 'cross products of the
> vectors from the centre of projection to the corners at z=0, the plane of
> projection'.
>
> However, he previously defines the view frustrum as a 'perspective volume
> with (xl, xr, yb, yt, zh, zy ) = (-1, 1, -1, 1, 1, -1), with a centre of
> projection at zcp = 5'.
>
> He lists the vectors as:
> v1 = [ 1, 1, -5 ]
> v2 = [ -1, 1, -5 ]
> v3 = [ -1, -1, -5 ]
> v4 = [ 1, -1, -5 ]
>
> Surely the plane of projection should be z = -1, and therefore the z
> component in the above vectors should be -6?
>
> Or am I (more likely!) missing something obvious here?
>
> _____________________________________
> Phil Yard, Lead Programmer, Climax (Brighton) Ltd
> 01273 764109 mob: 0771 258 1164 web: www.climax.co.uk
>
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