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Author Topic: Summary of 3D Mandelbrot Set Formulas  (Read 25546 times)
Description: Summary of many different 3D Mandelbrot set formulas
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FrozenOwl
Guest
« Reply #30 on: October 02, 2010, 10:21:31 PM »

Hi everybody,

This is my first post in this forum and first participation to the quest for the MandelGrail.

Recently, I tried to see how changing the coefficients of a quadratic "multiplication" could modify the shape of a Mandelbrot 3D.

Here is the iteration function z^2 + c :
x = v * mx * v + cx
y = v * my * v + cy
z = v * mz * v + cz
where v = (x, y, z) and mx, my and mz are 3x3 symmetrical matrices, with coefficients mxij . So, this can be written :
x = mx11 * x^2 + mx12 * x*y + mx13 * x*z + mx21 * x*y + mx22 * y^2 + …,

Let us start with the « revolution Mandelbrot », defined by mx11=1, mx22=mx33=-1, my12=my21=1 and mz13=mz31=1 (other coefficients = 0).



From there, we can slightly change the mij coeffcients, to see the effects on the shape. I will pass on all the coefficients that distort the shape in an « unfavourable » way (i.e. that clearly move us away from what a « 3D Mandelbrot » should be)
 
Here are some interesting results :
- changing my22, my23/my32, mz23/mz32, mz33 distorts the shape longitudinaly and creates a lateral « fractal spoke » in directions y and z. Watching from the front of the shape :



- changing my33 kind of « splits » the shape in the yz plane in an interesting way. Actually, when you combine the creation of a fractal spoke (on mz23/mz32) and this splitting (on my33), you get three spokes!
To make them identical and equally distributed around the x axis, the « best » parameters seem to be mz32=mz23=0,5 and my33=1/3 (or their opposites).




From there I tried to create additional spokes but could not do that just by changing the matrices coefficients. Since we want some kind of symmetry around the x axis, I tried to see how the iteration equations looked in the following polar coordinates :
x =  r cos ph
y = r cos th sin ph
z = r sin th sin ph

This gives us, in the iteration equations :
x = r^2 cos 2ph + cx
y = r^2 cos th sin 2ph + 1/6 r^2 sin^2 ph (cos 2th - 1) + cy
z = r^2 sin th cos 2ph + 1/2 r^2 sin^2 ph (sin 2th) + cz

The first coefficients are the ones of the « revolution Mandelbrot ».

The second coefficients in y and z equations are similar to a Mandelbrot equation in the yz plane, multiplied by a coefficient (r*sin ph) (i.e. r projected in the yz plane).
This gives the idea to choose a different power for th, and write :
x = r^p1 cos(p1*ph) + cx
y = r^p1 cos(th)sin(p1*ph) + 1/3 r^p1 sin(p1*ph)*cos(p2*th) + cy
z = r^ p1 sin(th)cos(p1*ph) + 1/2 r^p1 sin(p1*ph)*sin(p2*th) + cz

with p1 the power for ph, i.e. along x (p1=2 if you want a shape similar to the classical Mandelbrot)
and p2 the power for th, for the shape in the yz plane. It is as if you « combined » two Mandelbrot figures with different powers, one along x and the other along y and z.
When you increase p2, the number of spokes increases in a way similar to the way the 2D Mandelbrot evolves when you increase its power.
(Below for p2 = 3, 6, 20)










As you can see the p2=20 shape is quite hairy.

And yet,  still no sign of lateral spheres. Sorry Twinbee!
« Last Edit: October 04, 2010, 10:06:00 PM by FrozenOwl » Logged
Tater
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« Reply #31 on: December 12, 2010, 05:58:34 PM »

Here is a summary of some different 3D Mandelbrot set formulas for your convience. I have tried to include a second order and 8th order rendering for each formula presented here. Additional formulas are welcome.

Hi Bugman. I have been experimenting with a two parameter mandelbulb, where the two parameters p and q multiply the angles theta and phi
 v_{n}=<r \cos\, \theta\, \cos\, \phi,\, r \sin\, \theta\, r \cos\, \phi,\, r \cos\, \phi > \rightarrow v_{n+1}=<r^n \cos\, p \cdot n \theta\, \cos\, q \cdot n\, \phi,\, r^n \sin\, p \cdot n \theta\, \cos\, q \cdot n\, \phi, \cos\, q \cdot n\, \phi>+<c_1,\,c_2,\,c_3>

This variation encompasses some of your special cases and provides more flexibility, because proper choices of p and q (positive and negative) can smooth out or roughen up mandelbulbs.

Some more examples here http://www.aurapiercing.com/gallery/main.php
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jehovajah
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Posts: 2749


May a trochoid in the void bring you peace


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« Reply #32 on: December 17, 2010, 10:47:06 AM »

Hi everybody,

This is my first post in this forum and first participation to the quest for the MandelGrail.

Recently, I tried to see how changing the coefficients of a quadratic "multiplication" could modify the shape of a Mandelbrot 3D.

Here is the iteration function z^2 + c :
x = v * mx * v + cx
y = v * my * v + cy
z = v * mz * v + cz
where v = (x, y, z) and mx, my and mz are 3x3 symmetrical matrices, with coefficients mxij . So, this can be written :
x = mx11 * x^2 + mx12 * x*y + mx13 * x*z + mx21 * x*y + mx22 * y^2 + …,

Let us start with the « revolution Mandelbrot », defined by mx11=1, mx22=mx33=-1, my12=my21=1 and mz13=mz31=1 (other coefficients = 0).

...........
As you can see the p2=20 shape is quite hairy.

And yet,  still no sign of lateral spheres. Sorry Twinbee!


Excellent work FrozenOwl. Love the vectors and the matrices to clarify the kind of products involved.
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koffie
Guest
« Reply #33 on: June 03, 2011, 05:19:49 PM »

Hi everybody,

This is my first post in this forum and first participation to the quest for the MandelGrail.

Recently, I tried to see how changing the coefficients of a quadratic "multiplication" could modify the shape of a Mandelbrot 3D.

Here is the iteration function z^2 + c :
x = v * mx * v + cx
y = v * my * v + cy
z = v * mz * v + cz
where v = (x, y, z) and mx, my and mz are 3x3 symmetrical matrices, with coefficients mxij . So, this can be written :
x = mx11 * x^2 + mx12 * x*y + mx13 * x*z + mx21 * x*y + mx22 * y^2 + …,


Hej Frozen Owl, Somehow I thought of the same concept as you independantly, but to make it a bit more complicated I also wanted the squaring to satisfy |(x,y,z)^2|=|(x,y,z)|^2. Working this out gives you the following set of equations which the mx11,mx12 etc have to satisfy:

since I wrote down the formula's before I was aware that you had the same idea so you have to translate, for example mx11=a00, my13=b13 etc.

a00^2 + b00^2 + c00^2 - 1 = 0
2*a00*a01 + 2*b00*b01 + 2*c00*c01 = 0
2*a00*a11 + a01^2 + 2*b00*b11 + b01^2 + 2*c00*c11 + c01^2 - 2 = 0
2*a01*a11 + 2*b01*b11 + 2*c01*c11 = 0
a11^2 + b11^2 + c11^2 - 1 = 0
2*a00*a02 + 2*b00*b02 + 2*c00*c02 = 0
2*a00*a12 + 2*a01*a02 + 2*b00*b12 + 2*b01*b02 + 2*c00*c12 + 2*c01*c02 = 0
2*a01*a12 + 2*a02*a11 + 2*b01*b12 + 2*b02*b11 + 2*c01*c12 + 2*c02*c11 = 0
2*a11*a12 + 2*b11*b12 + 2*c11*c12 = 0
2*a00*a22 + a02^2 + 2*b00*b22 + b02^2 + 2*c00*c22 + c02^2 - 2 = 0
2*a01*a22 + 2*a02*a12 + 2*b01*b22 + 2*b02*b12 + 2*c01*c22 + 2*c02*c12 = 0
2*a11*a22 + a12^2 + 2*b11*b22 + b12^2 + 2*c11*c22 + c12^2 - 2 = 0
2*a02*a22 + 2*b02*b22 + 2*c02*c22 = 0
2*a12*a22 + 2*b12*b22 + 2*c12*c22 =0
a22^2 + b22^2 + c22^2 - 1 =0

A friend of mine already made a render of one of the solution's I've found
The solution he used is
a12 = 2, b00 =- 4/5, b01 =- 6/5, b11 = 4/5, b22 =- 4/5, c00 =- 3/5, c01 = 8/5, c11 = 3/5, c22 =- 3/5 and all others are zero

It's rendered with mandelbulber (I just replaced one of the formula's in the program with my own)

Of course the set equation has infinitly many more solutions (but it's not very easy to find them). Of course when seeing the "squaring" as a map F from R^3 to R^3 we see that if M1 and M2 are orthogonal 3x3 matrices then if F has the property |F(x,y,z)|=|(x,y,z)|^2 then the composition M1 F M2 also has that property (i.e. |M1(F(M2(x,y,z)))|=|(x,y,z)|^2). This allows you to continuesly deform the thus obtained mandlebulb depending on the rotation matrices M1 and M2. Does someone know if there are already movies of using this idea to get a mandlebulb changing it's shape in the way I just described?
« Last Edit: June 03, 2011, 05:51:10 PM by koffie » Logged
koffie
Guest
« Reply #34 on: June 14, 2011, 11:43:49 AM »

Last weekend my friend made a small video of the fractal formula I wrote down.

It can be found at http://leden.vslcatena.nl/~robert/mandelbulb/Derickx-Equations.mp4
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willclark218
Forums Freshman
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Posts: 11


« Reply #35 on: February 09, 2013, 07:53:03 PM »

thanks very much for this post bugman... the "math" goes over my head  embarrass... but I like pictures... this was very informative...
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