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 Author Topic: Bristorbrot 3D  (Read 11476 times) Description: Mandelbrot 3D, using using Doug Bristor's complex rotational maths 0 Members and 1 Guest are viewing this topic.
dougfractal
Guest
 « on: November 28, 2009, 06:08:15 PM »

Information regarding the Bristorbrot and Doug Bristor's complex rotational maths

http://www.fractaldimension.org.uk/voldsitemirror/

Different Complex Numbers
With the understanding that [imaginary number i] can be represented by rotation of 90° about the origin. I defined [imaginary number j] as a rotation of 90° about the origin and a rotation of 90° with respect to .
So with this in mind I picked up a cube, to help me visualise the rotations, and then recorded the results.

I rotated it upwards 90° , then following it by turning it a different 90° backwards [j], I defined the result:

i × j = -j

Then starting again

I rotated 90° backwards [j], followed by 90° upwards . This gave me the result:

j × i = i

It is the use of these results when applied into the [Mandelbrot algorithm], that produce the [Bristor set]. Imaginary numbers in additional dimensions can be derived by substituting in the new number in the above equations.

The java code includes 4D and 5D functions

but here is my original 3D algorithm

Quote
i.j = -j
j.i =  i

Code:
public int iterate3D(double rec,double imc, double jmc,int max)
{ double re,im,re2,im2,jm,jm2,ij,tmp;
int itr=0;
double mag=.0;
im=re=jm=re2=im2=jm2=.0;
do {
ij=im*jm;
tmp=re2-im2-jm2+rec;
im=2.0f*im*re-ij+imc;
jm=2.0f*jm*re+ij+jmc;
re=tmp;
re2=re*re;
im2=im*im;
jm2=jm*jm;
itr++;
if (itr>max) break;
mag=re2+im2+jm2;
} while (4.0f>mag );
return itr;
}

Bristorbrot rendered by Jos Leys

Bristorbrot rotated. Rendered by Jos Leys

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dougfractal
Guest
 « Reply #1 on: December 04, 2009, 07:07:18 PM »

Thanks twinbee for this render

Re: Has anyone tried this formula?
Quote from: twinbee
Another rendering of the quadratic version (download 10meg 7000x7000 pixels). Here's a preview:

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