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Author Topic: Complex Constant 'C' in Mandelbulb  (Read 1963 times)
Description: How do we choose that complex constant
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JVillella
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« on: December 17, 2011, 10:58:44 PM »

Hello everybody. I began learning about fractals a few weeks back and have been extremely fascinated since. I am currently learning about the Mandelbulb, but am stuck on one of the details. With the Mandelbrot set I would choose c, in z = z^2 + c, in the range [-2.5, 1] for the real part and [-1, 1] for the imaginary part. The screen resolution would then be scaled down to this range. With hypercomplex triplex numbers how would I define c? What are the ranges for the x, y, z components. I would truly appreciate a reply. Thank you.
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DarkBeam
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« Reply #1 on: December 18, 2011, 08:43:23 PM »

You are confusing everything imo, smiley you are talking about Julia  set (I think?) and not Mandelbrot set. Anyway no such range exists - but instead you know that the most interesting Julia sets are near to the "contour" of the fractal. In the same moment ... Mandelbrot set has no contours because it's a fractal! So this is the mystery behind all of it wink

How to find this nonexisting boundary you ask? By trial error as always smiley Have fun

Cheers ... Luca
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No sweat, guardian of wisdom!
JVillella
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« Reply #2 on: December 18, 2011, 09:53:50 PM »

You are confusing everything imo, smiley you are talking about Julia  set (I think?) and not Mandelbrot set. Anyway no such range exists - but instead you know that the most interesting Julia sets are near to the "contour" of the fractal. In the same moment ... Mandelbrot set has no contours because it's a fractal! So this is the mystery behind all of it wink

How to find this nonexisting boundary you ask? By trial error as always smiley Have fun

Cheers ... Luca

Hmm, Are your sure? With the complex constant 'C' if we set it to some constant for the entire iteration over every pixel we will get Julia Set. What differentiates this from the Mandelbrot set? Further, that range I spoke of is the range of the complex plane. As we change it do we not get a translated and/or scaled (zoomed) fractal? In addition to this does the Mandelbulb have to raytraced (or some other rendering method)? Or can we simply iterate over every pixel and use the Mandelbulb algorithm + equation (z = z^8 + c)?
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Syntopia
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« Reply #3 on: December 18, 2011, 10:48:07 PM »

Hi JVillella, and welcome to the forums.

If you are asking about where the Mandelbulb is located in triplex space, it is centered at the origin and has approximately unit radius.

But you need to do some kind of 3D -> 2D mapping to draw it (unless you are satisfied with 2D slices), and you'll probably want to use distance estimation.

The following thread might help you get started:
http://www.fractalforums.com/meet-and-greet/new-here-t8773/msg36722/#msg36722
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JVillella
Guest
« Reply #4 on: December 18, 2011, 11:07:47 PM »

Hi JVillella, and welcome to the forums.

If you are asking about where the Mandelbulb is located in triplex space, it is centered at the origin and has approximately unit radius.

But you need to do some kind of 3D -> 2D mapping to draw it (unless you are satisfied with 2D slices), and you'll probably want to use distance estimation.

The following thread might help you get started:
http://www.fractalforums.com/meet-and-greet/new-here-t8773/msg36722/#msg36722


This is perfect! Thank you so much smiley
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