> There is no raytracer that can apply displacement to an analytically
> solved shape.  That's not too bad of course because all non-raytracers
> can't even render those shapes.  At the risk of repeating myself -
> isosurfaces are as close to 'true displacement' as you can get.

Got it.

> This is simply wrong (see above).  What the renderer you have in mind
> probably does is to approximate the shape with a mesh and displace that
> - this is of course much less a 'true displacement' than an isosurface.

That's what it does internally, yes, but the effect has a 1-pixel
resolution, and thus is as close as defining a perfect displaced geometry as
possible. That's what I meant. I may be wrong, though.

> No, you don't have to modify the function as ABX already pointed out:
>
> #declare fn_Shape = function { ... }
>
> #declare fn_Iso = IC_Displace (fn_Shape)
>
> isosurface {
>    function { fn_Iso(x,y,z) }
>    ...
> }

I was not aware of this possibility. My bad.

> Sorry, this will never happen, simply because it is not possible (at
> least not without an approximative root solver like in the isosurface
> shape).

That's sad. I stand corrected then.

Regards,

Roberto

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