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Author Topic: Polyhedrons, many many polyhedrons...  (Read 11537 times)
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DarkBeam
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The spaghetti formula coder


« Reply #30 on: January 30, 2012, 05:46:18 PM »

Uh the Hypertile3D code;

Code:
int PluginVarPrepare(Variation* vp)
{
    double pa = 2*M_PI / VAR(hypertile3D_p),
           qa = 2*M_PI / VAR(hypertile3D_q);

    double r = -(cos(pa) - 1) / (cos(pa) + cos(qa));
    if (r > 0)
        r = 1 / sqrt(1 + r);
    else
        r = 1;

    double na = VAR(hypertile3D_n) * pa;

    VAR(cx) = r * cos(na);
    VAR(cy) = r * sin(na);

    VAR(c2) = sqr(VAR(cx)) + sqr(VAR(cy));

    VAR(c2x) = 2 * VAR(cx);
    VAR(c2y) = 2 * VAR(cy);

    VAR(s2x) = 1 + sqr(VAR(cx)) - sqr(VAR(cy));
    VAR(s2y) = 1 + sqr(VAR(cy)) - sqr(VAR(cx));
    VAR(s2z) = 1 - sqr(VAR(cy)) - sqr(VAR(cx));

    // Always return TRUE.
    return TRUE;
}

// You must call the argument "vp".
int PluginVarCalc(Variation* vp)
{
    double r2 = sqr(FTx) + sqr(FTy) + sqr(FTz);

    double x2cx = VAR(c2x)*FTx,
           y2cy = VAR(c2y)*FTy;

    double d = VVAR / (
        VAR(c2) * r2 + x2cx - y2cy + 1
        );

    FPx += d * (
        FTx * VAR(s2x) - VAR(cx) * ( y2cy - r2 - 1)
    );
    FPy += d * (
        FTy * VAR(s2y) + VAR(cy) * (-x2cx - r2 - 1)
    );
    FPz += d * (
        FTz * VAR(s2z)
    );

    return TRUE;
}

That of course is by Zueuk and released under GPL. wink
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knighty
Fractal Bachius
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Posts: 570


« Reply #31 on: January 30, 2012, 08:59:59 PM »

Thank blob.

Thank you DarkBeam for the code. Unfortunately I'm not used with Apophysis 7x and don't know how to use Zueuk's plugin. Do you have an example script?
I'll try to understand his code but at a first sight it seems to me it is a M÷bius transformation or something close.

The method I'm intending to use is almost exactly the same as the 2D one I presented in this thread about using foldings to generate triangular groups tesselations. But the key to do it successfully in 3D is in the painful details, mainly DE and solving for the folding planes (and or spheres). A solution is to go in 4D (using -inverse- stereographic projection) then back in 3D. That's just what I did with the polychora script. For the hyperbolic tesselation that will be a Minkowski spacetime. It's not a joke  wink : The hyper-sphere will be replaced with Minkowski hyper-sphere (hyper-hyperbola) and sine and cosine will be replaced with their hyperbolic cousins. It's really wonderful how it is related to special relativity.  nerd head batting canadian educated cool
 
Here is the updated script. If you want to dive inside the hypersphere just use 'travel' preset. You can chage direction of sight just as usual. in order to go forward change the 'RotAngle' parameter.

* polychora06.zip (3.05 KB - downloaded 54 times.)
« Last Edit: January 30, 2012, 09:04:52 PM by knighty, Reason: That was Minkowski not Lorentz » Logged
DarkBeam
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The spaghetti formula coder


« Reply #32 on: January 30, 2012, 10:23:51 PM »

I can not really understand a single letter of what you said cheesy
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knighty
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Posts: 570


« Reply #33 on: January 31, 2012, 12:17:56 PM »

Lol!  grin
I was a little over excited. maybe the cofee. I was referring to the hyperboloid model.
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eiffie
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Posts: 501


« Reply #34 on: January 31, 2012, 08:06:58 PM »

I always enjoy Knighty's scripts. I was wondering what it would look like "fractalized" with each vertice as another polyhedron. This is the result.
<a href="http://www.youtube.com/v/-eo6ImWMcWg&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/-eo6ImWMcWg&rel=1&fs=1&hd=1</a>
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knighty
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Posts: 570


« Reply #35 on: January 31, 2012, 09:08:32 PM »

Wohooow!

Repeating Zooming Self-Silimilar Thumb Up, by Craig the wave the wave the wave

I especially like how it grows up. Nicely done. With a good music it should be an instant win demo. smiley

Here is attached the 24-cell symmetry group polychoron script. (both "fat" and "thin" versions)

Now I'll go to watch eiffie's video one more time  grin.

* polychora-special.zip (4.91 KB - downloaded 51 times.)
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kram1032
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Posts: 1588


« Reply #36 on: January 31, 2012, 09:50:21 PM »

whoa, really nice fractal polyhedron growth thing!

I wonder... How would that look like with a 16-cell...
http://en.wikipedia.org/wiki/16-cell
That polytope has the remarkable property of being self-dual despite being neither a polygon nor a simplex. It's the sole exception.
Besides that, it's also the base of the lattice of the closest possible packing of 3-Spheres
It's one of the reasons, why 4D space is very special... Or perhaps it exists for the 4D-space being special.

Also interesting on the topic of close sphere packing but slight overkill (yes, this is a slight understatement)
http://en.wikipedia.org/wiki/4_21_polytope

and probably the most famous (and most likely most overkill to be found any time soon) would be
http://en.wikipedia.org/wiki/Leech_lattice

Thanks for bringing me here, knighty smiley
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marius
Iterator
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Posts: 197


« Reply #37 on: February 01, 2012, 12:18:39 AM »

I always enjoy Knighty's scripts. I was wondering what it would look like "fractalized" with each vertice as another polyhedron. This is the result.

fantastic. Love the serene b&w and music sweet music
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DarkBeam
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Fractal Senior
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Posts: 1666


The spaghetti formula coder


« Reply #38 on: February 01, 2012, 11:27:17 AM »

Eiffie, my head hurts but you did a stunning work! Great. A Beer Cup

kram1032 - 16 cell is already possible with Knighty's script (I think) - only he uses a (much!) more appealing projection style than Wiki (some edges go straight to infinity). wink

The only things that makes me a bit sad is that the complex 4D poly are slightly slow. This limits a bit the usability. Anyway a "forest of tubes" can not be used as a base for fractals so not a big problem probably? wink I love this stuff kiss
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kram1032
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Posts: 1588


« Reply #39 on: February 01, 2012, 11:39:48 AM »

I know that the 4D variants already are possible.
Which is exactly why I wondered how the per-vertex-repeat-Sierpinski-style-fractal of it would look like....
Also considering 3D-cross-sections next to shadows or stereo-graphic projections...
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DarkBeam
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Posts: 1666


The spaghetti formula coder


« Reply #40 on: February 01, 2012, 12:38:29 PM »

Once I will implement more 3D foldings on MB3D it will be straightforward. wink
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knighty
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Posts: 570


« Reply #41 on: February 01, 2012, 08:03:28 PM »

I know that the 4D variants already are possible.
Which is exactly why I wondered how the per-vertex-repeat-Sierpinski-style-fractal of it would look like....
Also considering 3D-cross-sections next to shadows or stereo-graphic projections...
3D cross sections would be quite easy to do but I'm afraid that rendering shadow and stereographic projection of such thing is not so obvious.
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kram1032
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Posts: 1588


« Reply #42 on: February 01, 2012, 08:08:49 PM »

Shadow, to my understanding, would be plain projection as already done by the script?
At least it uses SOME projection...
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Syntopia
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syntopiadk
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« Reply #43 on: February 05, 2012, 12:48:57 PM »

I finally had some time to play around with Knighty's new polychora scripts, and I am even beginning to understand a little of it.
These videos were a god help to understand stereographic projection of polychora: http://www.dimensions-math.org/
(I think one of the authors, Jos Leys, is a Fractal Forums member)

Now, for the original 3D polyhedron script, it is quite easy to find the settings for the five convex, regular polyhedrons (the platonic solids):

Type,U,V,W (<- Knighty's variable)
3,0,1,0 Tetrahedron
4,0,0,1 Cube
3,1,0,0 Octahedron
5,0,0,1 Dodecahedron
5,0,1,0 Icosahedron

But for the 4D polychora it is more difficult to find the 6 possible regular and convex polychora.
I tried using the 'polychora06.frag' script, and have found the following candidates:

Type,U,V,W,T
3,0,1,0 0 - 5-cell, 4-Simplex  [Tetra cells]
4,0,1,0 0 - 8-cell, Hypercube [Cube cells]
4,0,0,1 0 - 24-cell * [Octa cells]
4,0,0,0 1 - 16-cell [Tetra cells]
5,0,1,0 0 - 120-cell [Dodeca cells]
5,0,0,0 1 - 600-cell [Tetra cells]

But I would like to have them confirmed, if other tried this - it is not easy to identify these for high vertex counts :-)

In the comments for the script it says the 24-cell symmetry group is not present, and Knighty posted a special script for this.
But in this "polychora-special.frag" fragment, the "24-cell" preset has 8 vertices and 24 edges, so it cannot be a 24-cell? (It could be the 16-cell, though). However, setting (U=1, and V,W,T=0) gives an object with 24 vertices, and 96 edges, which could be the 24-cell?
It does however look extremely similar to the one I got above using the 'polychora06.frag' script - see the attached picture, comparing the two scripts.



* 24-cell.jpg (72.57 KB, 1390x560 - viewed 74 times.)
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knighty
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Posts: 570


« Reply #44 on: February 05, 2012, 04:22:49 PM »

Shadow, to my understanding, would be plain projection as already done by the script?
At least it uses SOME projection...
Yes, That's what is done by the script. But that is a stereographic projection of the projection of the polytope on the unit sphere. It is a projection of a projection. In fact what we see is the stereographic projection of a tiling of the 3-sphere.

@ Syntopia:

Well, I'm also still learning this stuff and still not understanding it thoroughly. Sorry for any inaccuracy or errors. smiley

You are right, the 24-cell can be obtained with both the first and second scripts. I was thinking that the 24-cell was'nt possible with the first script because I remembered reading about it being special.  undecided
It is special in the sens that it is the only regular polytope that is self dual beside the n-simplex.

Also, it seems to me that you gave the correct values for the convex regular polychora.


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