True 3D mandelbrot fractal (search for the holy grail continues)

[Timeroot]

David, am I right in thinking that this isn't the whole formula? Is it done after regular Mandelbulb angle multiplying/radius squaring? Pics please! (Mandel and Julia!) 

[David Makin]

David, am I right in thinking that this isn't the whole formula? Is it done after regular Mandelbulb angle multiplying/radius squaring? Pics please! (Mandel and Julia!) 

Nope - that *is* the whole formula

I suddenly wondered what happens if you apply the 2D Mandelbrot but scale the result of that plus the third dimension such that the new magnitude of the triplex is the square of the magnitude of the original triplex, hence:

Get the triplex magnitude, if not zero square the x and y component as if complex then scale the new triplex by dividing by its magnitude and multiplyiing by the magnitude of the original triplex squared. Finally add the constant.

[David Makin]

Pics as requested:

"Electric Mandelbrot"



If I was in the market for a custom-made electric guitar.....

And a test Julia showing I hope why I rather like the Julias for this one

[Timeroot]

Wow, cool pics, they're pretty close to a real grail... cool formula David! The Julias really do sounds exciting, if they can hold all the variety of the regular 2 Mset, preserved well. ...a Douady rabbit (sp?) with long "fur"! 

 

[jehovajah]

  love the palm leaf julia dave.

[kram1032]

Hmm, I didn't hear, how those pics sound (Timeroot ;P) but they indeed look really nice
The Julia resembles certain kinds of leaves
Are there also other leaf-structures in those sets?

[David Makin]

The Julia resembles certain kinds of leaves
Are there also other leaf-structures in those sets?

I don't know, I've only rendered the Mandy and 3 Julias. The formula was just a sudden idea.

I haven't investigated it further as atmo I'm adding some pre-render code for the Mandelbox that automagically finds the correct scale factor for the DE and the extent of the cuboid- I'm doing that after giving up trying to work it out analytically i.e. for all values for xfold,yfold,zfold,spherical fold and scale. So far the result is very promising - the adjustments to the DE it gives agree with my manually discovered values to 4 significant figures, I'm just trying to get it to around 6 and also to have the algorithm calculate what the scale factor should be for the DE value to convert it to a step value i.e. to work out the maximum error in the DE and calculate a scale to apply to convert the DE to a step that allows for the maximum error in the DE.

[KRAFTWERK]

POWER 8 David?, POWER 8!!!! 

[David Makin]

POWER 8 David?, POWER 8!!!! 

OK, OK !!!

Using:

            magn = sqrt(|zri|+sqr(zj))
            if magn>0.0
              zri = zri^@mpwr
              magn = (magn^@mpwr)/sqrt(|zri|+sqr(zj))
              zri = magn*zri + cri
              zj = magn*zj + cj
            else
              zri = cri
              zj = cj
            endif

With seed (-0.05,0.8,-0.20397)

Here's a Jellyfish:

[kram1032]

yay, nice
Not overly fractal though^^

[KRAFTWERK]

 

OK, not too  many fractal details, but nice in its own way...

Thanks David!

( Not going to ask you to zoom in... )

[jehovajah]

The search continues. A signal 2 mandy followed by a signal 8.

[fracmonk]

Before considering "the" true 3d M-set, ask what kind of dimensions you're extending to.  Like dimensions are Euclidean, I think.  Ttwo in complex are algebraically related but unlike.  They do, under multiplication, mutually contribute to each other. More later, badly outta time.

[jehovajah]

After extensive experimentation this is the leading candidates using Nylander like fomulae and QuaSZ by Terry Gintz.

[fracmonk]

To continue my 4-16 comment, it seems many renderings seen here are manifestations of "designer" dimensional systems.  It so much depends on what you might want in dimensional extension.  I already expressed doubts about any single "true" extension beyond the complex, but one good candidate might be one that contains those dimensions along with square roots of i expressed in 2 others, totalling 4.  sqrt(i) would be represented by j, and the other dimension is ij as a multiplicative consequence.  I've worked out the math for all 4 operations (div is complicated, but not necessary here) but it needs to be programmed as a generator on its own, say, in C++.