Christian Froeschlin <chr### [at] chrfrde> wrote:
> > To be more precise: I have a function yielding
> > only 0 or 1 depending on the membership to a given set.
>
> I think the implicit assumption of the iso_surface is that
> the set f(x) == threshold actually represents a 2d surface
> and the numerical solver can creep up to that solution from
> both sides using the 3d function gradient.
>
> What you do seems more like a description of an "inside"
> volume and you expect to render the surface of the volume.
> I would not be surprised at unintuitive behavior here, I'm
> more surprised you got a reasonable result at all :)

See Image at p.b.i

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