Peter Popov wrote:
> 
> I see, so shearing comes down to scaling along (an) axe(i)s which are
> (is) not perpendicular to the local axes. Am I right in guessing that?
> If so, thanks for clearing it up for me! If not, I'd be thankful if
> you cleared it up for me <grin>
> 

Well, yes. That's one way of looking at it. Of course, matrix shearing
approaches the problem from a different perspective - the scaling of a point
along one global axis is (linearly) bound to the distance of this point along
another global axis. E.g. when Point.x increases, the multiplier (scale) of
Point.y increases.

-- 
Margus Ramst

Personal e-mail: mar### [at] peakeduee
TAG (Team Assistance Group) e-mail: mar### [at] tagpovrayorg

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