Another true 3D fail ?

[David Makin]

This one's pretty interesting with or without the scale of the power 5 by  (1.0-abs(real(z))/m)...

Code:

 
    m = sqrt(|z|+sqr(real(w)))
    if m>0.0
    complex t = (1.0-abs(real(z))/m)*(imag(z)+flip(real(w)))^5.0 + flip(2.0*real(z)*real(w))
    z = sqr(z)-sqr(real(w))+flip(real(t))
    w = imag(t)
    endif

[DarkBeam]

Everybody want an image David

[David Makin]

Everybody want an image David

I don't wanna spoil the surprise

[M Benesi]

Code:

 
    m = sqrt(|z|+sqr(real(w)))
    if m>0.0
    complex t = (1.0-abs(real(z))/m)*(imag(z)+flip(real(w)))^5.0 + flip(2.0*real(z)*real(w))
    z = sqr(z)-sqr(real(w))+flip(real(t))
    w = imag(t)
    endif

  Is w complex?  Assigning w=imag(t) tends to make me think it isn't- it should only have a real component then. 

  Looks like:
m=...
  if  m!=0 {
   t= (1-abs(x))/m) * (y+z)^5 + i * 2xz;
  k= k^2 - z^2  +  i * real (t) ;

  w= imag(t)
}

  What is it, a "squarey" brot? 

 

[David Makin]

 Is w complex?  Assigning w=imag(t) tends to make me think it isn't- it should only have a real component then.  

  Looks like:
m=...
  if  m!=0 {
   t= (1-abs(x))/m) * (y+z)^5 + i * 2xz;
  k= k^2 - z^2  +  i * real (t) ;

  w= imag(t)
}

  What is it, a "squarey" brot?  

  

No all vars are complex but outside the calculation the imaginary part of w remains zero. It's basically standard quaternion cut down to 3D (i.e. (x,y,z)) but with (y+flip(z))^5 added to the y and z terms - if this part is scaled by (1.0-abs(x)/m) then the x+y and x+z plane slices are both the standard complex Mandy and the y+z plane is the complex degree 5 Mandy i.e. with 4 bulbs around the main.


Here it is anyway...a fail, but a damn interesting one




www.fractalforums.com/index.php?action=gallery;sa=view;id=12983

[M Benesi]

Definitely cool.

  I still find that taking the "xmas tree" formula (semi Julia mode- full xyz seed, followed by addition of ONLY x-axis component *.5), rotating the x-axis to the magnitude axis (-1,-1,-1) to (1,1,1),

 applying the following transform:

x= 2.*abs(x) -1.
y=2.*abs(y) -1.
z=2.*abs(z)-1.

followed by a rotation back to the x axis makes the fractallyest z^2s...

[jehovajah]

The sophistication of these images is amazing!
Davids so called fail compared with what was achieved early in he holy grail hunt really shows how far we have come in tinkering. Matt's solution shows how far we can go towards different design goals.
Great works, guys!