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Author Topic: True 3D mandelbrot type fractal  (Read 263320 times)
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David Makin
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« Reply #180 on: September 06, 2009, 06:39:05 PM »

I'm repeatedly disappointed by the quality of the final animation when uploading to DA or YouTube, so here's the original h264 Quicktime 640*480 version of "The Broccoli Virus":

http://www.fractalgallery.co.uk/Broccoli.html

If you'd like to see it more than just once then please save yourself a local copy rather than always going to my page (otherwise I'll start running out of bandwidth).
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cKleinhuis
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« Reply #181 on: September 06, 2009, 07:13:07 PM »

excellent dave, do you render with uf5 ?

i need to try some fancy formulas, have you uploaded the formulas, i would like to play around, and convert standard fractals
for using the 4d math method ...
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« Reply #182 on: September 06, 2009, 07:13:34 PM »

33 seconds per frame?! I would really like to understand your distance estimation method, David.
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David Makin
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« Reply #183 on: September 06, 2009, 08:21:37 PM »

excellent dave, do you render with uf5 ?

i need to try some fancy formulas, have you uploaded the formulas, i would like to play around, and convert standard fractals
for using the 4d math method ...


Yes I did them with UF5 with my WIP version of the rendering method that I posted the pseudo-code for.
I'm still working on my 3D IFS formula so haven't redone the escape-time formula for general release yet.

Note that for the use of bugman's method for general formulas we need to define "a*b" - we have a^n but not the product of two or more different values.

Also this is purely a 3D method - Paul also previously posted a 4D version.

For the 3D method I would suggest that we convert the squared version:

{x,y,z}^2 = r^2{cos(2*theta)cos(2*phi),sin(2*theta)cos(2*phi),-sin(2*phi)}
r=sqrt(x+y+z), theta=atan(y/x), phi=atan(z/sqrt(x+y))

for a*b such that:

{xa,ya,za}*{xb.yb,zb} = r^2{cos(2*theta)cos(2*phi),sin(2*theta)cos(2*phi),-sin(2*phi)}
where:
r = exp(log(sqrt(xa^2+ya^2+za^2))+log(sqrt(xb^2+yb^2+zb^2)))
theta=atan(y/x), phi=atan(z/sqrt(x+y)) where x=(xa+xb), y=(ya+yb) and z=(za+zb)

which I think is a commutative method.

Note that the value for r defined here can be simplified to:  r = exp(0.5*(log(xa^2+ya^2+za^2)+log(xb^2+yb^2+zb^2)))

I'm not sure given that, how to "correctly" define division or even if it is possible ?

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David Makin
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« Reply #184 on: September 06, 2009, 08:27:26 PM »

33 seconds per frame?! I would really like to understand your distance estimation method, David.


I thought the pseudo-code was fairly clear ?

http://www.fractalforums.com/3d-fractal-generation/true-3d-mandlebrot-type-fractal/msg7812/#msg7812

Basically although based on using the Newton to find the distance to the next iteration depth it actually boils down to distance estimation based on iteration density.
Just ask if you need the algorithm clarifying smiley
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David Makin
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« Reply #185 on: September 06, 2009, 08:45:05 PM »

excellent dave, do you render with uf5 ?

i need to try some fancy formulas, have you uploaded the formulas, i would like to play around, and convert standard fractals
for using the 4d math method ...


Yes I did them with UF5 with my WIP version of the rendering method that I posted the pseudo-code for.
I'm still working on my 3D IFS formula so haven't redone the escape-time formula for general release yet.

Note that for the use of bugman's method for general formulas we need to define "a*b" - we have a^n but not the product of two or more different values.

Also this is purely a 3D method - Paul also previously posted a 4D version.

For the 3D method I would suggest that we convert the squared version:

{x,y,z}^2 = r^2{cos(2*theta)cos(2*phi),sin(2*theta)cos(2*phi),-sin(2*phi)}
r=sqrt(x+y+z), theta=atan(y/x), phi=atan(z/sqrt(x+y))

for a*b such that:

{xa,ya,za}*{xb.yb,zb} = r^2{cos(2*theta)cos(2*phi),sin(2*theta)cos(2*phi),-sin(2*phi)}
where:
r = exp(log(sqrt(xa^2+ya^2+za^2))+log(sqrt(xb^2+yb^2+zb^2)))
theta=atan(y/x), phi=atan(z/sqrt(x+y)) where x=(xa+xb), y=(ya+yb) and z=(za+zb)

which I think is a commutative method.

Note that the value for r defined here can be simplified to:  r = exp(0.5*(log(xa^2+ya^2+za^2)+log(xb^2+yb^2+zb^2)))

I'm not sure given that, how to "correctly" define division or even if it is possible ?


Haven't checked but I'm not sure that my suggested method for a*b would give the same result for a*a*a as the original a^3 method ?

Edit: Sorry - I wasn't thinking in terms of magnitudes and angles, for a*b that should probably be:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = exp(log(sqrt(xa^2+ya^2+za^2))+log(sqrt(xb^2+yb^2+zb^2)))
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))

Is that correct ?

« Last Edit: September 06, 2009, 08:55:30 PM by David Makin » Logged

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David Makin
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« Reply #186 on: September 06, 2009, 09:14:08 PM »

excellent dave, do you render with uf5 ?

i need to try some fancy formulas, have you uploaded the formulas, i would like to play around, and convert standard fractals
for using the 4d math method ...


Yes I did them with UF5 with my WIP version of the rendering method that I posted the pseudo-code for.
I'm still working on my 3D IFS formula so haven't redone the escape-time formula for general release yet.

Note that for the use of bugman's method for general formulas we need to define "a*b" - we have a^n but not the product of two or more different values.

Also this is purely a 3D method - Paul also previously posted a 4D version.

For the 3D method I would suggest that we convert the squared version:

{x,y,z}^2 = r^2{cos(2*theta)cos(2*phi),sin(2*theta)cos(2*phi),-sin(2*phi)}
r=sqrt(x+y+z), theta=atan(y/x), phi=atan(z/sqrt(x+y))

for a*b such that:

{xa,ya,za}*{xb.yb,zb} = r^2{cos(2*theta)cos(2*phi),sin(2*theta)cos(2*phi),-sin(2*phi)}
where:
r = exp(log(sqrt(xa^2+ya^2+za^2))+log(sqrt(xb^2+yb^2+zb^2)))
theta=atan(y/x), phi=atan(z/sqrt(x+y)) where x=(xa+xb), y=(ya+yb) and z=(za+zb)

which I think is a commutative method.

Note that the value for r defined here can be simplified to:  r = exp(0.5*(log(xa^2+ya^2+za^2)+log(xb^2+yb^2+zb^2)))

I'm not sure given that, how to "correctly" define division or even if it is possible ?


Haven't checked but I'm not sure that my suggested method for a*b would give the same result for a*a*a as the original a^3 method ?

Edit: Sorry - I wasn't thinking in terms of magnitudes and angles, for a*b that should probably be:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = exp(log(sqrt(xa^2+ya^2+za^2))+log(sqrt(xb^2+yb^2+zb^2)))
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))

Is that correct ?



OK - now I'm embarassed, I missed that Twinbee had actually posted:

**************
// A triplex number is a '3D' number
triplex multiply(triplex a, triplex b) {
    triplex n;
    double pi=3.14159265;
    double r    = pow(a.x*b.x + a.y*b.y + a.z*b.z , 0.5);
    double yang = atan2( pow(a.x*b.x + a.y*b.y, 0.5) ,      a.z  )  ;
    double zang = atan2(a.y , b.x);
    n.x = (r*r) * sin( yang*2 + 0.5*pi ) * cos(zang*2 +pi);
    n.y = (r*r) * sin( yang*2 + 0.5*pi ) * sin(zang*2 +pi);
    n.z = (r*r) * cos( yang*2 + 0.5*pi );
   return n;
}
***********************

Which gives a different a*b - so is bugman's formula more different from Twinbee's than I thought it was ?

Edit:
<blush> OK, my version is actually:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)*(xb^2+yb^2+zb^2))
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))

On balance I think I prefer a combination:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = abs(xa*xb+ya*yb+za*zb)
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))

However we still need division ?

So maybe use this for a*b:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)*(xb^2+yb^2+zb^2))
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))


And this for a/b:

{xa,ya,za}/{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)/(xb^2+yb^2+zb^2))
theta=atan(ya/xa)-atan(yb/xb), phi=atan(za/sqrt(xa+ya))-atan(zb/sqrt(xb+yb))


« Last Edit: September 06, 2009, 09:45:12 PM by David Makin » Logged

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David Makin
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« Reply #187 on: September 06, 2009, 09:54:57 PM »

However we still need division ?

So maybe use this for a*b:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)*(xb^2+yb^2+zb^2))
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))


And this for a/b:

{xa,ya,za}/{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)/(xb^2+yb^2+zb^2))
theta=atan(ya/xa)-atan(yb/xb), phi=atan(za/sqrt(xa+ya))-atan(zb/sqrt(xb+yb))


I don't suppose there's the slightest chance that this is associative as well as the multiply being commutative ?
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David Makin
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« Reply #188 on: September 07, 2009, 12:27:06 AM »

However we still need division ?

So maybe use this for a*b:

{xa,ya,za}*{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)*(xb^2+yb^2+zb^2))
theta=atan(ya/xa)+atan(yb/xb), phi=atan(za/sqrt(xa+ya))+atan(zb/sqrt(xb+yb))


And this for a/b:

{xa,ya,za}/{xb.yb,zb} = r{cos(theta)cos(phi),sin(theta)cos(phi),-sin(phi)}
where:
r = sqrt((xa^2+ya^2+za^2)/(xb^2+yb^2+zb^2))
theta=atan(ya/xa)-atan(yb/xb), phi=atan(za/sqrt(xa+ya))-atan(zb/sqrt(xb+yb))


I don't suppose there's the slightest chance that this is associative as well as the multiply being commutative ?


Just to add I meant is there the slightest chance that the multiplication is associative over addition i.e. a*(b+c) = a*b + a*c.

Just checked and using the above * and / then if c = a/b then a is not necessarily equal to b*c smiley
However the "divide" still makes for an interesting 3D function, I tried redoing my animation for the degree 3 to the degree 10 such that the degree was increased to from 4 to 11 but the result was "divided" by the original value - the result was similar to the original from 3 to 10 but not the same.
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« Reply #189 on: September 07, 2009, 01:44:33 AM »

I would like to see the animation go all the way down to degree 2, or perhaps even to negative powers.
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David Makin
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« Reply #190 on: September 07, 2009, 03:26:20 AM »

I'm repeatedly disappointed by the quality of the final animation when uploading to DA or YouTube, so here's the original h264 Quicktime 640*480 version of "The Broccoli Virus":

http://www.fractalgallery.co.uk/Broccoli.html

If you'd like to see it more than just once then please save yourself a local copy rather than always going to my page (otherwise I'll start running out of bandwidth).

I've added a view of the same animation but using the "division" method added, so the origibal was z^n+c but the small version is (z^(n+1))/z + c where the "divide" is as I outlined in my last posts.
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« Reply #191 on: September 07, 2009, 06:31:24 AM »

Here is my most detailed global illumination rendering so far with 320 samples per pixel:
http://bugman123.com/Hypercomplex/Mandelbrot-White1-large.jpg
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« Reply #192 on: September 07, 2009, 07:53:30 PM »

Here's another variation that Rudy Rucker came up with in 1990:
{x,y,z} = r*{cos(2*phi)*cos(2*theta), cos(2*phi)*sin(2*theta), sin(2*phi)}
where r=sqrt(x+y+z), theta=atan(y/x), phi=atan(z/x)


* Mandelbrot-Rudy, imax=24, GI, AA=80.jpg (127.5 KB, 560x560 - viewed 838 times.)
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twinbee
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« Reply #193 on: September 09, 2009, 03:49:20 PM »

Excellent renders! Rudy's formula actually seems fairly similar to mine. I wonder if it will have the awesome fractal detail of the sea anemone/virus at higher powers when you've zoomed in.

I tried to see if this was the case, but the results seem in between either conclusion:
http://www.skytopia.com/project/fractal/new/3dmand-rudy-rucker.jpg  (1600x800 res)

Here are some other orthographic projection renders. The quality is no where near the images from you guys, but the advantage is speed. Optimization of my main algorithm allowed a speed up of around 100x. All of these together took only around 1-2 days (in total) on a quad core. The images are mapped using a z depth 'ping pong' colour scheme.

Here's a giant 8000x4000 resolution of the old 3d mandy:
http://www.skytopia.com/project/fractal/new/3dmand-standard.jpg

Since higher powers have nice detail, I was seeing if there was any kind of fractal detail in this power 2 version, but it appears not (3000x3000):
http://www.skytopia.com/project/fractal/new/3dmand-bulbs-bulb.jpg

Here's a nice 3000x3000 pixel image from the power 8 (apx) version:
http://www.skytopia.com/project/fractal/new/3dmand-power-apx8.jpg

Finally, here's a nice 3000x3000 pixel image zoom in from the power 8 version:
http://www.skytopia.com/project/fractal/new/3dmand-power8corner.jpg
« Last Edit: September 09, 2009, 03:54:19 PM by twinbee » Logged
David Makin
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« Reply #194 on: September 10, 2009, 10:47:46 AM »

Here's a zoom with a different viewpoint:

http://makinmagic.deviantart.com/art/Amongst-the-Buds-136532213
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