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Author Topic: The tetrahedral Mandelbox, formula and some parameters  (Read 1356 times)
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DarkBeam
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Fragments of the fractal -like the tip of it


« on: July 18, 2011, 06:49:16 PM »

Okay! After a long search, and the help of Aexion with his new ideas and Jesse with some technical support (and his fantastic software wink ) I have found a "new" type of fractal.

It is the same as the Amazing box, as the base, but the folding part is much more complicated. I use four foldings instead of 3, and they are "skewed" like the tetrahedron's faces.

Let's define Rotated Folding this formula

- Rotate by the given angles
- Do; x = abs(x+Fold) - abs(x-Fold) - x (double folding) , or x'=1-abs(x-1) (single folding) - Maybe I am using y or z for folding. Can't be 100% sure 'cause assembly code is not clear! (due to fp stack swaps and other hell fiery )
- RotateBack by the given angles

Given a rotation matrix, RotateBack multiplies the axis by the transposed matrix. wink

Okay, now let's see what angles are needed;

 RotatedFolding(+45,+45,+45,Fold) // base
 // other planes. this can be tweaked
 RotatedFolding(-45,-45,+45,Fold)
 RotatedFolding(-45,+45,-45,Fold)
 RotatedFolding(+45,-45,-45,Fold)

I calculate only the 1st rotation matrix. The remaining ones can be obtained changing signs from the previous, following a fixed "geometric rule"

Code:
1st change (change sign of * elements):
.*.
*.*
.*.
2nd change:
..*
..*
**.
3rd change:
.*.
*.*
.*.

Very nice variations have been done changing those rules. Fantasy is the only limit!
The ABox formula is;

Code:
// folding part. This is replaced by my new rule
x = abs(x+Fold) - abs(x-Fold) - x
y = abs(y+Fold) - abs(y-Fold) - y
z = abs(z+Fold) - abs(z-Fold) - z
// folding part ends. The rest of the formula is untouched
rr = x*x + y*y + z*z
if rr < sqr(Min_R) then m = Scale/sqr(Min_R) else
if rr < 1 then m = Scale/rr else m = Scale
x = x * m + Cx
y = y * m + Cy
z = z * m + Cz
« Last Edit: July 18, 2011, 07:09:01 PM by DarkBeam » Logged

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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #1 on: July 18, 2011, 07:04:18 PM »

Preview render of the Julia set in (0,0,0) of those parameters

- Scale = sqrt(3)
- Fold = sqrt(3)/2
- MinR = 2/3
- Single (tetrahedric) folding

Looks absolutely like a Koch snowflake fractal! It's spherical-tetrahedrical.


* kifskochf.jpg (199.64 KB, 900x900 - viewed 150 times.)
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flexiverse
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« Reply #2 on: April 10, 2013, 02:49:16 AM »

just out of interest.  While I've been trying to wrap my head around this..

the mandelbox formula can be applies to any shape right?

e.g.

2D   -    3D

Square   - Cube

Triangle  - Pyramid



So therefore there must be a generic formula  as such:

Pass Number of Points in shape  ( x )
    Do Mandlebox folds.


So  Pass Number of Points in shape ( 4)

    Creates a square mandel  box.  and cube mandel box in 3d.


So  Pass Number of Points in shape ( 3)

    Creates a triangle mandex box, and pyramid mandbox in 3D.


So in terms of generic polyhedra  what is the mandelbox formula ?

e.g. for  convex polyhedron

{3} is an equilateral triangle
{4} is a square

cube has 3 squares around each vertex and is represented by {4,3}.

Tetrahedron {3, 3}   Cube {4, 3}   Octahedron {3, 4}   Dodecahedron {5, 3}   Icosahedron {3, 5}

So there fore  what is the code  to create a mandel box  given  the function header:


Create Mandelbox by Schläfli  {p, q}


  code to create n sided polyheda mandel box?


where Create Mandelbox by Schläfli  {4, 3}

    would create a standard mandel box ?
« Last Edit: April 10, 2013, 02:51:45 AM by flexiverse » Logged
Syntopia
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« Reply #3 on: April 12, 2013, 09:32:02 AM »

the mandelbox formula can be applies to any shape right?

Not sure about that, but Aexion did some interesting experiments some time ago, where he found Mandelbox systems for the platonic systems:
http://www.fractalforums.com/3d-fractal-generation/platonic-dimensions/
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taurus
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« Reply #4 on: April 12, 2013, 12:58:28 PM »

the mandelbox formula can be applies to any shape right?

Might be also of interrest
http://www.fractalforums.com/new-theories-and-research/generalized-box-fold
Buddhi implemented that in mandelbulber
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Alef
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« Reply #5 on: April 17, 2013, 02:33:14 PM »

Maybe too angular. Maybe an Octahedron fold.

Mandelbox probably are just mix of sphere and cube patterns. Maybe it's so popular becouse cubes and spheres constitute most of man made things such as medieval architecture. Maybe alsou spherefold could be switched to some ellipsoid fold, what could generate new patterns.
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