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For an animation I'm making I need a special lens-like shape that have
certain properties. It should be kind of like a flattened sphere, or
actually more like a union between a torus and a cylinder (because it should
have a flat area), but without abrupt changes in curvature.
See the image in povray.binaries.images.
I don't think any CSG will do the trick, but maybe someone can come up with
an isosurface function that'll work? Or something different.
Thanks in advance!
Rune
--
3D images and anims, include files, tutorials and more:
Rune's World: http://rsj.mobilixnet.dk (updated Feb 16)
POV-Ray Users: http://rsj.mobilixnet.dk/povrayusers/
POV-Ray Webring: http://webring.povray.co.uk
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I found a function myself that works. I use this function in an isosurface
which I scale by <1,1,0.2>. However, is it possible to make it faster?
(sqrt(max(0,sqrt(x^2+y^2)*2-1)^4+z^2)-1)
Rune
--
3D images and anims, include files, tutorials and more:
Rune's World: http://rsj.mobilixnet.dk (updated Feb 16)
POV-Ray Users: http://rsj.mobilixnet.dk/povrayusers/
POV-Ray Webring: http://webring.povray.co.uk
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I don't see why the union of a cylinder and a torus doesn't meet your
requirements. If I'm not mistaken, the derivatives of the surfaces of the
cylinder and the torus (if done correctly) should be equal at the
intersection, so there shouldn't be any abrupt changes in the smoothness of
both surfaces.
Or perhaps I'm misunderstanding your wishes...
Fernando.
"Rune" <run### [at] mobilixnetdk> wrote in message
news:3c824c95@news.povray.org...
> I found a function myself that works. I use this function in an isosurface
> which I scale by <1,1,0.2>. However, is it possible to make it faster?
>
> (sqrt(max(0,sqrt(x^2+y^2)*2-1)^4+z^2)-1)
>
> Rune
> --
> 3D images and anims, include files, tutorials and more:
> Rune's World: http://rsj.mobilixnet.dk (updated Feb 16)
> POV-Ray Users: http://rsj.mobilixnet.dk/povrayusers/
> POV-Ray Webring: http://webring.povray.co.uk
>
>
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In article <3c82f3e2@news.povray.org>,
> I don't see why the union of a cylinder and a torus doesn't meet your
> requirements. If I'm not mistaken, the derivatives of the surfaces of the
> cylinder and the torus (if done correctly) should be equal at the
> intersection, so there shouldn't be any abrupt changes in the smoothness of
> both surfaces.
But the derivative of the derivative is discontinuous there...the rate
of change in curvature changes abruptly. It transitions from an almost
constantly curving surface to a completely flat surface. This change is
easy to see in the shading.
--
Christopher James Huff <chr### [at] maccom>
POV-Ray TAG e-mail: chr### [at] tagpovrayorg
TAG web site: http://tag.povray.org/
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OK, now I understand that you're speaking about curvature and not
smoothness!
I didn't know that a change in curvature, in smooth surfaces, could be
easily detected in shading.
Pity I couldn't help...
Fernando.
"Christopher James Huff" <chr### [at] maccom> wrote in message
news:chr### [at] netplexaussieorg...
> In article <3c82f3e2@news.povray.org>,
>
> > I don't see why the union of a cylinder and a torus doesn't meet your
> > requirements. If I'm not mistaken, the derivatives of the surfaces of
the
> > cylinder and the torus (if done correctly) should be equal at the
> > intersection, so there shouldn't be any abrupt changes in the smoothness
of
> > both surfaces.
>
> But the derivative of the derivative is discontinuous there...the rate
> of change in curvature changes abruptly. It transitions from an almost
> constantly curving surface to a completely flat surface. This change is
> easy to see in the shading.
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> But the derivative of the derivative is discontinuous there...
I've never seen a proof of this, but I'm 99% sure that it is physically
impossible to have a function with both infinitely continuous derivatives
*and* a perfectly flat spot, unless the entire function itself is flat.
If you only need the first and second derivatives to be continuous, and you
don't care about the third derivative, that can be done...
- Slime
[ http://www.slimeland.com/ ]
[ http://www.slimeland.com/images/ ]
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