Inversion Ellipse Fold Transformation

[thomas314]

Hi,
the next logical step of a circle fold, i thought, is an elliptic fold of
the z plane. If the 2 axis of the ellipse are equal then the formula behaves
exactly like the circle fold i've done in this thread:

http://www.fractalforums.com/new-theories-and-research/kalis-inversion-circle-fold/msg46047/#msg46047;

Link for ellipse formulas
http://en.wikipedia.org/wiki/Ellipse#Equations

The UF

Code:

class THMInvEllipseFold(common.ulb:UserTransform){
public:
import "common.ulb"

  func THMInvEllipseFold(Generic pparent)
	  UserTransform.UserTransform(pparent)
  endfunc
  
  func Init(complex pz)
	  UserTransform.Init(pz)
	  rot=(0,1)^(@angle/90.0)
  endfunc

  complex func Iterate(complex pz)
    pz=pz*rot                           ;rotate pz
    pz=real(pz)/@a+flip(imag(pz)/@b)    ;Center equation of ellipse
    smz=|pz|                            ;Kalis Inverse Circle Fold {
    if smz<@k
      pz=pz/smz*@k
    endif
    pz=pz*@m+@o                         ; }
    return pz
  endfunc
  private:
    complex rot
default:
  title="InverseEllipseFold"
  float param angle
    caption = "angle"
    default = 0.0
  endparam
  float param a
    caption = "Radius a"
    default = 1.0
  endparam
  float param b
    caption = "Radius b"
    default = 1.0
  endparam
  float param k
    caption = "x/a+y/b"
    default = 1.0
  endparam
  param m
    caption = "faktor z"
    default = (1,0)
  endparam
  param o
    caption = "offset z"
    default = (0,0)
  endparam
}

After this pre-transformation i simply use a Mandelbrot.

Code:

pz=pz^p+c

Some pictures:

Mandelbrot with p= -2.0, 0.0i, Radius a=Radius b, all other parameter have default values



and some Julias with different parameter:







The last 2 ones are rendered with 2 merged elliptic transformations prior the mandel formula like this:

Loop:
z=transformation1(z)
z=transformation2(z)
z=z^p+c
 
The parameter p (power) for this examples are all integer real numbers. If complex floating point numbers are used then
this formula causes symmetry fractures(ugly).
But i found a workaround for that. How?
I'll show that in the next thread (if you want) 

[Alef]

IMHO 4th pics is the best.
Just are waiting for this being uploaded to UF database;)

[tit_toinou]

Good idea and good pictures.
In the 4th picture we can see in the upper left corner a discontinuity..

And yeah I want to know your workaround your discontinuity  !

[s31415]

You can add a mirror on top of the branch cut to get rid of the discontinuity and still get the same type of fractals. The same trick works with logarithms, see this post for instance:
http://algorithmic-worlds.net/blog/blog.php?Post=20110227

Concerning using "elliptic inversion", I find it a bit less desirable, because it stretches the patterns, what ultimately hides the self-similarity. (Although this is not so apparent on your pictures because you probably use an ellipse which is close to be a circle.) The stretching is due to the fact that unlike the circle inversion, it is not a conformal transformation. My personal taste for unstretched patterns makes me stick to conformal transformations with this type of fractals..

Best,

Sam

[DarkBeam]

Stretching can be cool, I love not-exactly-regular figures

[tit_toinou]

But maybe the beauty of this kind of fractal is to compose the non-complex-differentiable-but-continuous folding function (mirrors) with complex-differentiable functions...

[element90]

I been using inverse fold in, inverse fold out and inverse reflect for a while now. All use circles (special case of the ellipse). The following was produced using Mars and Phobos.

z = (inversefold(z))^1.5 + 0.2



As the formula with out "inverse fold in" is the glynn formula there is no surprise that glynns are present here. The image is the same when "circle fold in" is used instead (checked using Saturn & Titan).

[tit_toinou]

Woah. Good finding .

[element90]

Referring back to the original post, I got this in September 2011 which is similar in structure to the second Julia.



For the parameters go to the deviantArt page here http://fav.me/d4a74tn