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Author Topic: Novice approach to non-m like colouring formula  (Read 2796 times)
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simon.snake
Fractal Phenom
******
Posts: 494


Experienced Fractal eXtreme plugin creator!


simon.fez SimonSideBurns
« on: February 22, 2012, 06:44:09 PM »

Hi

I've been trying to get my head around using a FractInt formula to colour pixels to make something that doesn't look like a mandelbrot and I've been having some fun experimenting.

One of my more interesting ones is below:


smf-test-10 {
  t = 0
  iter = (0,0)
  flag = 0
  p = pixel
  s = sin(real(p))
  c = cos(imag(p)):
  iter = iter + (1,1)
  si = sin(iter)
  ci = cos(iter)
  t = t + c * p * (1 - s + (si - ci))
  if (t < s*p)
    flag = 1
  else
    s = s + s - sin(iter)
  endif
  flag == 0
  }

And here's the initial picture which isn't particularly exciting.

(image removed)

Zooming in doesn't show any more detail but zooming out starts to get a little better.  Here is the first zoom out:

(image removed)

Another zoom out:

(image removed)

Another zoom out:

(image removed)

Continuing zooming out (I appreciate some of these effects come from aliasing as the distance out gets larger but they are fascinating regardless):

(image removed)

(image removed)

(image removed)

(image removed)

(image removed)

(image removed)

This effect was not at all what I was expecting but interesting (for me).  Has anyone else played with this sort of formula and if so, what have you found out along the way?

Simon
« Last Edit: August 04, 2013, 01:59:31 PM by simon.snake » Logged

MOTEL BRAND = Anagram of MANDELBROT

Can anyone do any better?
simon.snake
Fractal Phenom
******
Posts: 494


Experienced Fractal eXtreme plugin creator!


simon.fez SimonSideBurns
« Reply #1 on: February 25, 2012, 11:00:47 PM »

Having updated my formula a little to the following:

Code:
smf-test-10c {
  t = 0
  iter = (0,0)
  flag = 0
  p = pixel
  s = sin(real(p))
  c = sin(imag(p))
  l = flip(s + c) + s:
  iter = iter + (1,1)
  si = sin(iter)
  ci = cos(iter)
  t = t + c * p * ((1,-1) - s + si + ci)
  if (t > 600)
    flag = 1
    z = 0 - (t * iter)
  else
    s = s + s + c - si
  endif
  flag == 0
  }

I get these new images (progressively zooming out):



















Still haven't really got any idea how this formula makes these images but I think they are a bit more like 'pop art'.
Logged

MOTEL BRAND = Anagram of MANDELBROT

Can anyone do any better?
Ryan D
Alien
***
Posts: 31


WWW
« Reply #2 on: June 22, 2012, 12:03:22 AM »

When I was a young pup, I was interested in mathematically-derived Op Art things like Moiré patterns.  In 1980, the TRS-80 was quite the toy, but the limited pixel resolution (128 x 48!!) stopped you from doing too much.  Still, I tried a few simple things, polar coordinate graphs and so on.  If anything was too complex, it just devolved into an almost random pile of pixels, so for a little bit of time I played with the Moiré patterns you could create.  After a couple of years a dot-matrix printer came on the scene and I soon was spending time waiting for the pins to punch out nifty patterns.

I've edited your formula a bit to allow for some changing parameters, which allows me to create an animation out of it (animation without changing parameters isn't as interesting, I fear).

Code:
smf-test-10var {
;RD added p1 and p2 - originally both were hard coded at (0,0)
  t = 0
  iter = (0,0)
  flag = 0
  p = pixel
  s = sin(real(p)) + p1
  c = cos(imag(p)) + p2:
  iter = iter + (1,1)
  si = sin(iter)
  ci = cos(iter)
  t = t + c * p * (1 - s + (si - ci))
  if (t < s*p)
    flag = 1
  else
    s = s + s - si
  endif
  flag == 0
  }

Implement a nifty Lissajous curve with the parameters (vary them according to sine curves) and you have a set of points that alters almost forever.  The four parameter components follow periods of 3, 5, 7 and 11, and I have also animated the zoom parameter to decrease about 30% from the maximum to the minimum.  Might as well do that according to a prime number too, so that follows a period of 2.  If you were to animate this long enough for the periods to all line up again as they were at the start, it would take just short of 80 hours.  OK, this animation isn't nearly interesting enough for that, but here's a minute of it.

Ryan

<a href="http://vimeo.com/moogaloop.swf?clip_id=49591662&amp;amp;server=vimeo.com&amp;amp;fullscreen=1" target="_blank">http://vimeo.com/moogaloop.swf?clip_id=49591662&amp;amp;server=vimeo.com&amp;amp;fullscreen=1</a>
« Last Edit: September 17, 2012, 04:14:28 PM by Ryan D, Reason: new animation render, updated link » Logged
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