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3D Mandelbrot Set

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Daniel Hulme

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Mar 30, 1999, 3:00:00 AM3/30/99
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I am a second year Computing student from the University of Tasmania,
Australia.

I am trying to find someone who can render a 3D image for me, since I
don't have the knowledge or tools to do so myself.

The image is a 3D mandelbrot set, created within a 3D space which uses
an extension of the algebra for complex numbers. The third dimension
uses a j component, where j^2 = -1 and ij = 1.

The formula f(n) = f(n - 1)^2 + k, f(0) = 0 is still used, but this time
you have f(n) and k as points in 3 dimensions and not 2.

To implement one iteration, lets say f(x-1) = a + bi + cj and k = x + yi
+ zj:

f(x) = (a + bi + cj)^2 + x + yi + zj
= a^2 + 2abi + 2acj - b^2 + 2bc - c^2 + x + yi + zj
= (a^2 - b^2 - c^2 + 2bc + x) + (2ab + y)i + (2ac + z)j

I suspect either Mathematica or POV-Ray would be able to render such an
image, but neither I nor my friends have been able to work out how. I
have been able to produce cross-sections.

Dan.


David Delikat

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Mar 30, 1999, 3:00:00 AM3/30/99
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Daniel Hulme wrote:
>
> I am a second year Computing student from the University of Tasmania,
> Australia.
>
> I am trying to find someone who can render a 3D image for me, since I
> don't have the knowledge or tools to do so myself.
>
> The image is a 3D mandelbrot set, created within a 3D space which uses
> an extension of the algebra for complex numbers. The third dimension
> uses a j component, where j^2 = -1 and ij = 1.
>

doesn't i == j then? I am curious, could you expand on this?

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