But the point is that pov is doing SOMETHING.  If it finds a definite
isosurface for a mathematically undefined region of space, then povray is in
reality not computing the given function, but converting it to some defined
function. Or there is an error in how it has defined some functions.

perhaps
    x^a = sign(x)^int(a) * abs(x)^a

because
    noise3d(x^3.1, y, z)
gives "something" for negative values of x !

I'm not complaining as much as striving to solve the mystery . . .
Chris Huff wrote:

> In article <387EA67E.7CE91E84@faricy.net>, David Fontaine
> <dav### [at] faricynet> wrote:
>
> > What I think POV *should* do is not include these areas in the graph. Say
> > I
> > want to graph a hemisphere on my TI-89, I do x^2+y^2+z^2=1,
> > z^2=-x^2-y^2+1,
> > z=sqrt(1-x^2-y^2), and graph it and it gives me my hemisphere, and the
> > area
> > *outside* the hemisphere is ignored (not rendered) because it is
> > sqrt(neg).
> > Because if people want it to work the other way they can always just make
> > it sqrt(abs(x)) instead of sqrt(x).
>
> I think this might be difficult to do, because the isosurface is solved
> by sampling the function at different points. I don't see how it could
> be done.
> Then again, it is after midnight now, it is probably just that I am
> tired and not thinking straight.
>
> --
> Chris Huff
> e-mail: chr### [at] yahoocom
> Web page: http://chrishuff.dhs.org/

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