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## gdalgorithms-list

 [Algorithms] Cyrus-beck line clipping algorithm From: Phil Yard - 2000-07-25 17:29:47 ```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: http://www.climax.co.uk ```
 Re: [Algorithms] Cyrus-beck line clipping algorithm From: Jamie Fowlston - 2000-07-26 09:26:05 ```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: http://www.climax.co.uk > > _______________________________________________ > GDAlgorithms-list mailing list > GDAlgorithms-list@... > http://lists.sourceforge.net/mailman/listinfo/gdalgorithms-list ```