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Author Topic: Ideal 3D mandelbrot?  (Read 11113 times)
Description: An extension of the Mandelbulb formula, which exhibits infinite detail
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Buddhi
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« Reply #30 on: January 10, 2010, 07:34:27 PM »

@twinbee: Beautiful shapes. Could you describe what formulas did you use in post #28?
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gaston3d
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« Reply #31 on: January 10, 2010, 09:52:52 PM »

Here's the z^2 - z^4/2 + c adaption of my above order 2 bulb.

this image shows real advantages of ambient occlusion shading model.
very impressive and inspiring render.
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BradC
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« Reply #32 on: January 10, 2010, 10:00:46 PM »

Excellent images! I guess my images are obviously the ones that don't match everybody else's, lol. I'll try to figure out why...
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twinbee
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« Reply #33 on: January 11, 2010, 01:17:11 PM »

Quote
@twinbee: Beautiful shapes. Could you describe what formulas did you use in post #28?

They're all simply different views of the "Positive Z-component" variation as first shown by Paul found here.

Here's the exponential function I use (p below =2 of course, as it's a quadratic Mandelbulb)

Code:
triplex tpowPOSZ(triplex &a, double p) {
triplex n;

     double r    = sqrt(a.x*a.x + a.y*a.y + a.z*a.z) ;
double theta = atan2(a.y , a.x) ;
double phi = atan2( a.z , sqrt(a.x*a.x + a.y*a.y) ) ;

r=pow(r,p); phi *=p; theta *=p;

n.x = r * cos(theta) * cos(phi) ;
n.y = r * cos(phi) * sin(theta);
n.z = r * sin(phi);
return n;
}


Quote
this image shows real advantages of ambient occlusion shading model.
very impressive and inspiring render.

Thanks!

Quote
Excellent images! I guess my images are obviously the ones that don't match everybody else's, lol. I'll try to figure out why...

One thing I noticed is that I have to use: phi = atan2( a.z , sqrt(a.x*a.x + a.y*a.y) ) ;   
...instead of Paul's recommended: phi = arcsin(a.z/r);

...which for some reason I can't get to work.
« Last Edit: January 11, 2010, 01:22:27 PM by twinbee » Logged
bugman
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« Reply #34 on: January 11, 2010, 10:16:44 PM »

You can only use phi = arcsin(z/r) if you define r = sqrt(x²+y²+z²):
http://www.fractalforums.com/theory/summary-of-3d-mandelbrot-set-formulas/

In order to make the formulas more compact on that post, I had originally defined r = (x²+y²+z²)^(n/2), not realizing that this altered my definition of phi. But since then I have gone back and changed it back to r = sqrt(x²+y²+z²), so everything should be consistent now.
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twinbee
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« Reply #35 on: January 12, 2010, 12:41:49 AM »

That's how I have r defined in my tpowPOSZ function above. Any other idea what I could be doing wrong?
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LesPaul
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« Reply #36 on: January 12, 2010, 12:54:03 PM »

...

<Quoted Image Removed>
Fortunately, I have developed a solution: to make successive powers of q increasingly negligable. An easy way to do this is to use the factorial function:


<Quoted Image Removed>

<Quoted Image Removed>

If you haven't already noticed, this is equivalent to the Maclaurin expansion for the expression q^8 cos q.

...

Wow!  Very promising!

This raises lots of questions...  Some of the big ones:

1) Why start with q8?  Why not something slightly more general, such as:

   q_{n+1} = q_0 + \frac{q_{n}^2}{2!} + \frac{q_{n}^3}{3!} + \cdots

Or, more compactly:

   q_{n+1} = q_0 + \sum_{k=2}^\infty \frac{q_{n}^k}{k!}

It might be most consistent with the traditional Mandelbrot to leave the quadratic term unchanged, in other words, start the series at k = 1, yielding:

   q_{n+1} = q_0 + \sum_{k=1}^\infty \frac{q_{n}^{k+1}}{k!}

    = q_0 + {q_{n}^2} + \frac{q_{n}^3}{2!} + \cdots

2) Has anyone ever tried a similar approach in 2D?

Can't wait to see this explored further!
« Last Edit: January 12, 2010, 01:06:14 PM by LesPaul » Logged
Calcyman
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« Reply #37 on: January 18, 2010, 08:03:38 PM »

Quote
Why start with q^8?

q^8 is the exponent for the original 3D Mandelbulb, so it made sense to begin there.

Quote
Why not something slightly more general, such as:

Some of the images in this thread are generated using q^2 and higher powers. The series you have posted is the logarithmic series expansion. The cosine series is faster to compute, as only even terms are considered.

Quote
Has anyone ever tried a similar approach in 2D?

Not that I know of.


I wonder whether the formula z = z² + c creates a fractal over the Griess algebra (as opposed to the complex plane). If it does, then you have a 196884-dimensional fractal with the same symmetries as the monster group. That would be spectacularly impressive.

http://en.wikipedia.org/wiki/Griess_algebra
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LesPaul
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« Reply #38 on: January 23, 2010, 01:05:17 AM »

Quote
Has anyone ever tried a similar approach in 2D?

Not that I know of.

I went ahead and tried it.  The results are kind of off-topic here, since this is thread is about 3D.  I posted what I found in a new thread here: http://www.fractalforums.com/index.php?topic=2483.msg12057#msg12057
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jehovajah
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« Reply #39 on: February 11, 2010, 05:37:05 AM »

Here's the z^2 - z^4/2 + c adaption of my above order 2 bulb. This unfortunately doesn't seem to match your version at all. I used 15 iterations here...

<Quoted Image Removed>


Seems to me that you have different bailout conditions, and possibly fine tuning settings. Your range seems different also. Stunning images as usual.

Tell me. does x=-.5 have any significance to the quadratic mandelbulb?
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May a trochoid of ¥h¶h iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
jehovajah
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« Reply #40 on: February 11, 2010, 01:34:39 PM »

Excellent images! I guess my images are obviously the ones that don't match everybody else's, lol. I'll try to figure out why...
 
Some differences in the underlying math, plus your rendering algorithm. Here are Terry Gintz shot at rendering
julia zcos(z)+c and mandy zcos(z)+c.

Then z2cosz+c  but based on "pixel"math. Compare with zplot math.I altered the colouring filterand increased iteration to 15. As you can see pixel math seems to give more details in the image.


* juliazcosz+c.png (92.21 KB, 320x240 - viewed 404 times.)

* ejaculation.png (36.11 KB, 320x240 - viewed 412 times.)

* z2cosz+c mandp.png (31.65 KB, 320x240 - viewed 414 times.)

* z2cosz+c mand.png (44.98 KB, 320x240 - viewed 399 times.)
« Last Edit: February 11, 2010, 02:14:23 PM by jehovajah » Logged

May a trochoid of ¥h¶h iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
jehovajah
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« Reply #41 on: February 17, 2010, 05:58:39 AM »

This is a side view.


* z2cosz+c mand2.png (31.24 KB, 320x240 - viewed 376 times.)
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May a trochoid of ¥h¶h iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
Calcyman
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« Reply #42 on: February 20, 2010, 10:37:10 PM »

Based on my idea of multiplying trigonometric and polynomial components, bib discovered the fractal produced by the formula:

z = z * tan(z)

http://www.fractalforums.com/mandelbrot-and-julia-set/calcyman's-idea-(2d)/

The resulting images are unusual even in two dimensions, and I cannot imagine what would happen if it were translated into 3D. The appropriate series expansion is:

z = z^2 + \frac{z^4}{3} + \frac{2z^6}{15} + \frac{17z^8}{315} + \cdots

Unfortunately, the series doesn't converge very quickly, compared with sine and cosine.
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Schlega
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« Reply #43 on: February 23, 2010, 06:14:45 AM »

Hi, I'm a bit new here, so I apologise if this has been covered elsewhere, but I've been thinking about extending the z*tan(z)+c fraction.

My idea was to define (x,y,z)=(a,b,c)/(A,B,C) by

x = (1/R2)(aA+bB)(C/P-c/p)
y = (1/R2)(bA-aB)(C/P-c/p)
z = (1/R2)(cC+pP)

where R2= A*A + B*B +C*C
and p = sqrt(a*a + b*b)

for points off of the z-axis, and treat points on the z-axis as scalars. I know this is not entirely compatible with the mandelbulb definition of exponentiation, but I'm curious to see what the results look like.

From there, I want to define tan(z) by its continued fraction representation.

I'm trying to write the code for it myself, but it's probably not going to be done any time soon since I'm not a programmer. Has anyone tried this yet?
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Timeroot
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« Reply #44 on: February 23, 2010, 06:36:34 AM »

That sounds interesting, I mean using the continued fraction expansion, but I'm confused by what you're doing with x,y,z,a,b,c,A,B,C, and about the difference between p and P. Could you elaborate?  huh?
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