Very simple formula for fractal patterns

[Jesse]

Well, I discovered a new strange property of the original formula. In order to obtain the Mandelbrot map of patterns, we must swap real and imag values of C after starting the iterations (or cx, cy in the real number version).

Interesting, so you see the julia shapes before you choose a location for the julia seed?  Have to try it out.
Cool results so far!

I included now the modified formula with folding:
http://www.fractalforums.com/index.php?topic=6061.msg28944#msg28944

[Kali]

Thanks Jesse!! I attached a quick render of the new KaliDucks... 

[Kali]

I mixed Kaliset with Kaliducks, I'll render another scene in full size later.
I found that high iteration values helps with noise (>1000)

[Softology]

Using the stalks orbit trap also works for these when in 2D.









Jason.

[trafassel]

The 3D variant of http://www.fractalforums.com/index.php?action=gallery;sa=view;id=7198

m=x*x+y*y+z*z
x=x/m+0.05
y=abs(y)/m-0.04
z=abs(z)/m-0.16

Outside and inside view.

[trafassel]

The black crescent in the center of the last one is a sort of entangled trees (third pic in the first message of this thread).

(Here I change the julia parameters to (0.05,-0.15,-0.24))

[Kali]

@Trafassel: Thanks for the renders... How did you implement the bailout condition?
I have some ideas to do with Gestaltlupe later, but I will really appreciate if you can add vb.net as you mentioned before

@Jason: Nice traps, I tried some with good results also. Thanks.

---

Well, in the last days I have tried lot of things with variations from my original formula, and played a little in 3D also (you may see some of the images in the gallery). But today I found something that I wanted to show here because I think the images of this new fractal are pretty good and interesting. Also they are quite different from previous ones.

The formula is based on one of the mentioned in fracmonk's thread about his research on connected sets with coexisting multiple-power shapes.

z=abs(z*c+1)+1/abs(z*c+1)

(I added the abs function to the original formula)

This are some quick monochrome renders:







Just a few samples, there's a great variety of shapes...

---

Later I decided to test the original formula (without abs), using also exponential smoothing coloring instead of escapetime (Why I never did that before with fracmonk's formulas?)... I was very intrigued about the strange bifurcation maps I made yesterday for this and other formulas (fracmonk asked me), so I expected to see something different and interesting with this coloring method, and I was right... but...
 I think I will open a new thread for this later...

[fracmonk]

Kali- Very glad you dragged me over here.  A supremely excellent find!  But I'm afraid that because of time constraints on me elsewhere this time of year, I'm so poor, I can barely pay attention...

Just VERY cool...

Later.

[Softology]

Well, in the last days I have tried lot of things with variations from my original formula, and played a little in 3D also (you may see some of the images in the gallery). But today I found something that I wanted to show here because I think the images of this new fractal are pretty good and interesting. Also they are quite different from previous ones.

The formula is based on one of the mentioned in fracmonk's thread about his research on connected sets with coexisting multiple-power shapes.

z=abs(z*c+1)+1/abs(z*c+1)

(I added the abs function to the original formula)

Some more good variations Kali.

A couple of samples I found when exploring.

z=abs(z*c+1)+1/abs(z*c+1) images





And a few without the abs() included
z=(z*c+1)+1/(z*c+1)





Jason.

[Softology]

Adding log into the most recent formulas works too (brings them more back into Ducks level of space filling and complexity).

z=abs(log(z*c+1))+1/abs(log(z*c+1))





z=log(abs(z*c+1))+1/log(abs(z*c+1))





The strange things with these formulas is the "Mandelbrot" versions have no detail, so you don't get a mandelbrot template to find interesting julia coordinates within.  Unless I made a coding error these would have to be the first julia set variations that have no mandelbrot detail.

Jason.

[Kali]

Thanks for your contributions Jason, nice images you found

I'm about to start a new thread on revisiting some escapetime fractals by using abs and inner coloring, I'm counting on you (and others who were interested) for further colaborations.

The strange things with these formulas is the "Mandelbrot" versions have no detail, so you don't get a mandelbrot template to find interesting julia coordinates within.  Unless I made a coding error these would have to be the first julia set variations that have no mandelbrot detail.

I guess that happens if you start Z initial value = (0,0). Try with Z=C, you should get the Mandelbrot version.

[Softology]

The strange things with these formulas is the "Mandelbrot" versions have no detail, so you don't get a mandelbrot template to find interesting julia coordinates within.  Unless I made a coding error these would have to be the first julia set variations that have no mandelbrot detail.

I guess that happens if you start Z initial value = (0,0). Try with Z=C, you should get the Mandelbrot version.

Thanks Kali.  Setting the initial Z to C did the trick.

Jason.

[Softology]

@lkmitch: Very nice, what julia values? (if you saved them )

The Julia parameter is -1 + i.  I think the big difference is how the image was colored.  If you use Ultra Fractal, it was my Statistics coloring, coloring by the mean of imag(z)/real(z).  I've noticed that this brings out different structure from colorings based on the magnitude of the iterate.

Can you give me another tip on how this coloring works?

Keeping track of the average imag(z)/real(z) is easy enough.  Total imag(z)/real(z) during the iterations, and then divide total by the iteration count after all the iterations for the pixel have been calculated.

How do you then index a color palette entry for the resulting average floating point value?  For the curent color methods I use colval=trunc(avmag*colscale)mod 255 to get the floating point value into the 0-255 color palette range, but when using the imag(z)/real(z) value I am not getting those awesome appolonian-like shapes you are?  The colscale is needed (I found) otherwise if you have a very smooth palette, you don't see all the details.

Thanks for any tips.

Jason.

[Softology]

@lkmitch: Very nice, what julia values? (if you saved them )

The Julia parameter is -1 + i.  I think the big difference is how the image was colored.  If you use Ultra Fractal, it was my Statistics coloring, coloring by the mean of imag(z)/real(z).  I've noticed that this brings out different structure from colorings based on the magnitude of the iterate.

Can you give me another tip on how this coloring works?

Keeping track of the average imag(z)/real(z) is easy enough.  Total imag(z)/real(z) during the iterations, and then divide total by the iteration count after all the iterations for the pixel have been calculated.

How do you then index a color palette entry for the resulting average floating point value?  For the curent color methods I use colval=trunc(avmag*colscale)mod 255 to get the floating point value into the 0-255 color palette range, but when using the imag(z)/real(z) value I am not getting those awesome appolonian-like shapes you are?  The colscale is needed (I found) otherwise if you have a very smooth palette, you don't see all the details.

Thanks for any tips.

Jason.

Got it working.  I had to use the average of abs(imag(z)/real(z)).



Jason.

[trafassel]

Softology, Reply #73 looks like a solution of a special circle packing problem.

The are some sphere packing fractal ideas (i.e. some indra pearls), but most with a complicatd generation algorithms.


Cool finding,

Trafassel