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Author Topic: 3D Koch cube (escape time) + formula  (Read 9321 times)
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DarkBeam
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The spaghetti formula coder


« on: March 29, 2011, 08:38:29 PM »

Anyone knows if this formula is convertible to escape-time ifs for MB3d or something... undecided I'm not good at this, plus I'm trying to do a formula by scratch without success. Please help me sad cry

http://bernhardhaeussner.de/blog/65_Processing%3A_3D-Koch-Kurve_mit_Sierpinski-Dreieck

Source code;

fraqTriang trs[]=new fraqTriang[4];

void setup(){
  size(640, 480, P3D);
  noStroke();
  fill(150,150,150,20);
  int bigrad=200;
  int req=6;
  
  PVector A=new PVector(-bigrad,0,0);
  PVector B=new PVector(bigrad/2,0,-173.21);
  PVector C=new PVector(bigrad/2,0,173.21);
  PVector D=new PVector(0,-bigrad*(sqrt(6)/3),0);
  
  trs[0]=new fraqTriang(A,B,C,req);
  trs[1]=new fraqTriang(A,D,B,req);
  trs[2]=new fraqTriang(A,C,D,req);
  trs[3]=new fraqTriang(B,D,C,req);
}


void draw(){
  background(255);
  lights();
  
  float winkel=(mouseX/float(width))*TWO_PI;
  float xpos=cos(winkel);
  float ypos=sin(winkel);
  float radius=300.000;
  camera(xpos*radius, mouseY, ypos*radius, // eyeX, eyeY, eyeZ
         0.0, -50.0, 0.0, // centerX, centerY, centerZ
         0.0, -1.0, 0.0); // upX, upY, upZ
      
  for (int i=0;i<trs.length;i++) {
    trs[i).display();
  }
  //saveFrame("frames/koch3d-####.png"); //uncomment to record
}

class fraqTriang{
  PVector PointA;
  PVector PointB;
  PVector PointC;
  fraqTriang recTris[]=new fraqTriang[6];
  int rec;
  float scaling;
  
  fraqTriang(PVector A,PVector B,PVector C, int recursion){
    scaling=0.7;
    PointA=A;
    PointB=B;
    PointC=C;
    rec=recursion;
    applyRecursion ();
  }
    
 void applyRecursion () {
    if (rec!=0) {
      PVector PointAB2=PVector.add(PointA,PointB);
      PointAB2.div(2);
      PVector PointAC2=PVector.add(PointA,PointC);
      PointAC2.div(2);
      PVector PointBC2=PVector.add(PointB,PointC);
      PointBC2.div(2);
      
      PVector PointZ=PVector.add(PointA,PointB);
      PointZ.add(PointC);
      PointZ.div(3);
      PVector PointAB=PVector.sub(PointA,PointB);
      PVector PointAC=PVector.sub(PointA,PointC);
      PVector PointH=PointAB.cross(PointAC);
      PointH.normalize();
      PVector PointAAB2=PVector.sub(PointA,PointAB2);
      float a=PointAAB2.mag();
      float pheight=a*(sqrt(6)/3)*scaling;
      PointH.mult(-pheight);
      PVector PointZH=PVector.add(PointZ,PointH);
      
      recTris[0]=new fraqTriang(PointA,PointAB2,PointAC2,rec-1);
      recTris[1]=new fraqTriang(PointB,PointBC2,PointAB2,rec-1);
      recTris[2]=new fraqTriang(PointC,PointAC2,PointBC2,rec-1);
      
      recTris[3]=new fraqTriang(PointZH,PointAC2,PointAB2,rec-1);
      recTris[4]=new fraqTriang(PointZH,PointAB2,PointBC2,rec-1);
      recTris[5]=new fraqTriang(PointZH,PointBC2,PointAC2,rec-1);
    }
  }  

  
  void display () {
    if (rec==0) {
      beginShape();
      vertex(PointA.x, PointA.y ,PointA.z);
      vertex(PointB.x, PointB.y ,PointB.z);
      vertex(PointC.x, PointC.y ,PointC.z);
      endShape(CLOSE);
    } else {
      for (int i=0; i<recTris.length;i++) {
        recTris.display();
      }
    }
  }
  
}

The interesting part is highlighted... wink
« Last Edit: May 09, 2011, 01:40:38 PM by DarkBeam » Logged

Formulas are never too much (?)

DarkBeam
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Posts: 1666


The spaghetti formula coder


« Reply #1 on: March 29, 2011, 08:43:22 PM »

Obviously, I don't ask for the MB formula, only a C-style clean source code will be enough... wink

IN = x,y,z (3d current point) + size
OUT = x,y,z

Like this

http://www.fractalforums.com/3d-fractal-generation/kaleidoscopic-(escape-time-ifs)/
« Last Edit: March 29, 2011, 08:47:35 PM by DarkBeam » Logged

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msltoe
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Posts: 161


« Reply #2 on: April 15, 2011, 02:31:48 PM »

DarkBeam: I've been thinking recently about how to make arbitrary IFS fractals using escape time. It's a little tricky to do, but well worth it when we figure it out.
The problem is that escape time works by taking functions of functions of functions, etc. Each time a function is applied it skews the entire 3-D space. If the function isn't conformal then after a few iterations space gets drawn out into spikes.

Meanwhile, recursive IFS's don't transform the entire space over each iteration, just a few registration points (i.e., A, B, C). The child object has essentially a fresh new Euclidean (orthogonal) space to be generated in.

So, I've concluded that for escape time, if the function has non-uniform scaling, it must be re-scaled on the next few iterations to restore uniform space. E.g. , if a matrix specifies the function, then some variant of the matrix inverse must be used on the next step to restore the space to conformal.

-mike
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DarkBeam
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Posts: 1666


The spaghetti formula coder


« Reply #3 on: April 15, 2011, 06:04:10 PM »

DarkBeam: I've been thinking recently about how to make arbitrary IFS fractals using escape time. It's a little tricky to do, but well worth it when we figure it out.
The problem is that escape time works by taking functions of functions of functions, etc. Each time a function is applied it skews the entire 3-D space. If the function isn't conformal then after a few iterations space gets drawn out into spikes.

Meanwhile, recursive IFS's don't transform the entire space over each iteration, just a few registration points (i.e., A, B, C). The child object has essentially a fresh new Euclidean (orthogonal) space to be generated in.

So, I've concluded that for escape time, if the function has non-uniform scaling, it must be re-scaled on the next few iterations to restore uniform space. E.g. , if a matrix specifies the function, then some variant of the matrix inverse must be used on the next step to restore the space to conformal.

-mike
smiley

That is true! I use a particular trick for those IFS, I do all calculations in a single formula that must be executed once and finds which is the "index" of the point and then does the transform. The depth must be previously specified, unlike other transforms. smiley

The problem here is that I can't find those "inclined axis" (and the formula has many bifurcations), so can you help me on the algorithm?
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DarkBeam
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Posts: 1666


The spaghetti formula coder


« Reply #4 on: April 15, 2011, 06:24:49 PM »

This is a scheme of the transform, but it's hard to write it, I get lost... sad



At first we need to bring all 3D space to the "red area", with a 3D rotation (it's already hard!)

Then, we must do a translate+scaling for regions 1 and 3 and an additional rotation for 2 and 4.

What I miss is the exact values for all vectors and angles! cry
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msltoe
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Posts: 161


« Reply #5 on: April 16, 2011, 12:24:26 AM »

DarkBeam,

 Start with something simpler like an escape time IFS for the 2D snowflake.
 
 Also, remember that escape time algorithms always carve out density from the previous iteration. So you want something big in early iterations that gets smaller and smaller.

-mike

 
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knighty
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Posts: 570


« Reply #6 on: April 30, 2011, 09:31:17 PM »

I'm afraid this is not possible with KIFS. It's not symmetric enought. I've tried the symetric 3D Koch based on a tetrahedra but the result is a simple cube... Maybe I'm wrong. In fact, I hope so.
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DarkBeam
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Posts: 1666


The spaghetti formula coder


« Reply #7 on: May 01, 2011, 09:52:07 AM »

I'm afraid this is not possible with KIFS. It's not symmetric enought. I've tried the symetric 3D Koch based on a tetrahedra but the result is a simple cube... Maybe I'm wrong. In fact, I hope so.

No, it is actually exact! (too bad! sad ) http://www.math.hmc.edu/funfacts/ffiles/30002.2.shtml

But this formula is different, so it works smiley

Any idea for the squarry koch? http://en.wikipedia.org/wiki/File:Quadratic_Koch_3D_(type1_stage2).png

nicer picture http://www.flickr.com/photos/step26/3404694703/in/photostream grin
« Last Edit: May 01, 2011, 09:55:23 AM by DarkBeam » Logged

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knighty
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Posts: 570


« Reply #8 on: May 01, 2011, 06:33:34 PM »

But this formula is different, so it works smiley
Well you can do it with regular IFS. It is possible to do IFS as escape time fractal but you would need a stack (see this paper.). You could also do it with some sort of combination between IFS ans KIFS by using the symmetries of that object.
With KIFS it is possible to reproduce only a subset of regular IFS: Those whose Transformation set is symmetric, that is, there is a set of folding planes that transform (the order is important because of the non commutativity of linear operator algebra) any of the IFS transformation to any other. In fact this condition is not sufficient apparently: I coudn't obtain the squarry Koch but the squarry koch + something...Humm... I'm talking like a mathematician. afro Hope that it makes sens angel.

Kind of! Here is attached a fragmentarium shader. There are some interresting variations by changing offset, rotation, scale and the order of foldings...

* koch01.zip (0.75 KB - downloaded 98 times.)
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DarkBeam
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The spaghetti formula coder


« Reply #9 on: May 01, 2011, 07:24:05 PM »

If it will work I will give you a millon dollars! afro
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DarkBeam
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The spaghetti formula coder


« Reply #10 on: May 01, 2011, 09:06:32 PM »

Ok, I have translated to Mandelbulb, but it is discontinue in some points,  undecided and discontinue figures have a kinda weird look, but it is nice.

Very similar to my Menger Koch. smiley I will see if it's worth to add to my DB!
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knighty
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« Reply #11 on: May 01, 2011, 09:19:19 PM »

If it will work I will give you a millon dollars! afro
Hehe... deal? evil rolling on floor laughing
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knighty
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Posts: 570


« Reply #12 on: May 01, 2011, 09:20:42 PM »

Ok, I have translated to Mandelbulb, but it is discontinue in some points,  undecided and discontinue figures have a kinda weird look, but it is nice.

Very similar to my Menger Koch. smiley I will see if it's worth to add to my DB!
A picture?
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DarkBeam
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Fractal Senior
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Posts: 1666


The spaghetti formula coder


« Reply #13 on: May 02, 2011, 12:38:01 AM »

Too bad they are all weird, due to a branch cut...
I will work on it, and sorry if I havent said THANK YOU! for your precious help friend!
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msltoe
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Posts: 161


« Reply #14 on: May 02, 2011, 03:59:50 AM »

Something similar to the "squarey" Koch using escape time. It's in Tglad's table, too, I think.
The algorithm is you keep clipping off the corners of 3x3 cubes.


* squary_koch.jpg (78.44 KB, 600x600 - viewed 211 times.)
« Last Edit: May 02, 2011, 04:02:47 AM by msltoe » Logged
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