Tglad
Fractal Molossus
Posts: 684


« on: February 15, 2015, 05:56:15 AM » 

[INFO]moved by moderator ckleinhuis to here, splitted original topic here: info from tglad afterwards: Did someone move this into its own thread? Its an interesting formula but it is an extension of the transform in these threads: http://www.fractalforums.com/the3dmandelbulb/anew3dmandelbrotlikefractalhttp://www.fractalforums.com/the3dmandelbulb/anewtetrahedralmandelbulbThe formula does the same fold but is smooth, and conformal on the sphere surface. [/INFO]I have a function which I think is right. It isn't quite the same as from the table, and it isn't minimally conformal in 3d, however it is fully conformal in 2d (on the surface of the sphere or on a plane) and is smooth. To perform the function on a sphere, we do it using complex maths on the complex plane: http://davidbau.com/conformal/#0.25*(2%5E0.5)*(z*(z2%5E0.5)*(z(0.5%2B0.866*i)*2%5E0.5)*(z(0.50.866*i)*2%5E0.5))%2F((z(0.5%5E0.5)*(0.5%2B0.866*i))*(z(0.5%5E0.5)*(0.50.866*i))*(z(0.5%5E0.5)*(1)))then convert the point onto the Riemann sphere, just as is often done with mandelbulb functions. I have a feeling it will make interesting 2d fractals too, but the 3 poles will make it a bit harder to visualise.


« Last Edit: February 16, 2015, 04:20:10 PM by DarkBeam »

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DarkBeam
Global Moderator
Fractal Senior
Posts: 2496
Fragments of the fractal like the tip of it


« Reply #1 on: February 15, 2015, 06:20:09 PM » 

The idea is just cool but do you have any finished render (and script)  so I can make some confront I do tons of coding mistakes



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Tglad
Fractal Molossus
Posts: 684


« Reply #2 on: February 16, 2015, 12:24:44 AM » 

Did someone move this into its own thread? Its an interesting formula but it is an extension of the transform in these threads: http://www.fractalforums.com/the3dmandelbulb/anew3dmandelbrotlikefractalhttp://www.fractalforums.com/the3dmandelbulb/anewtetrahedralmandelbulbThe formula does the same fold but is smooth, and conformal on the sphere surface. I updated the formula above to have what should be the correct scale. Looks right now, it turns the unit circle (the equator on the Riemann sphere) into an octahedron on the sphere, as seen by the grey circles. The 0.866 in the formula is more precisely asin(120). I haven't done anything with the formula yet, apart from try it in 2d, the mandelbrot set of it isn't much use, just lots of scattered blobs... possibly there's a better way to visualise it but not sure... it should produce a 'void sponge'like fractal I think, like the menger carpet.



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DarkBeam
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Fractal Senior
Posts: 2496
Fragments of the fractal like the tip of it


« Reply #4 on: February 16, 2015, 09:34:08 AM » 

Thank you a lot I report the formula here for an immediate reference: 0.25*(2^0.5)*(z*(z2^0.5)*(z(0.5+0.866*i)*2^0.5)*(z(0.50.866*i)*2^0.5))  ((z(0.5^0.5)*(0.5+0.866*i))*(z(0.5^0.5)*(0.50.866*i))*(z(0.5^0.5)*(1))) It can be placed as a Difs transform for sure will see when Ofc I need to expand the product and find re() and im() parts before Ok this is the expansion finally got right!!! sqrt(2)/4*z*(sqrt(8)z^3)/(z^3+1/sqrt(8))


« Last Edit: February 16, 2015, 05:23:24 PM by DarkBeam »

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Tglad
Fractal Molossus
Posts: 684


« Reply #5 on: February 16, 2015, 10:40:00 AM » 

I have made the code for you, in fragmentarium (attached). With a bit of simplifying, the function is: Thanks Syntopia for the rendering without distance estimator, which made this easier to prototype, and thanks DarkBeam for coming back to this one, it is nice to get a continuous function.



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DarkBeam
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Posts: 2496
Fragments of the fractal like the tip of it


« Reply #6 on: February 16, 2015, 10:52:31 AM » 

Looking nice thanks mr Mandelbox



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cKleinhuis


« Reply #7 on: February 16, 2015, 01:31:52 PM » 

thank you for the formula, i am typing it into ultrafractal right now, but what is the "i" variable in your formula? the j belongs to the multiplication operatzor, but what is the i from?



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cKleinhuis


« Reply #8 on: February 16, 2015, 01:56:51 PM » 

i typed it in to ultrafractal this is the core of the ultrafractal loop function, use high bailout loop: int itercount=0; counter for the mul sequence operator complex tempz=(1,0) ;init with unit number while(itercount<3) ; play with itercountmax value, 1,2,3 z=z^2+ #pixel complex t1=z(sqrt(2)*#e^((2*(itercount1)*#pi)/3)) complex t2=sqrt(2)*z (#e^((2*itercount+#pi)/3)) tempz=t1/t2 itercount=itercount+1 endwhile z=z*tempz +@seed ; include julia seed



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cKleinhuis


« Reply #9 on: February 16, 2015, 01:59:18 PM » 

itercount=3 impressions. zoomed into spine



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cKleinhuis


« Reply #10 on: February 16, 2015, 01:59:53 PM » 

itercount3 mandelbrots found, into elephant valley



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cKleinhuis


« Reply #11 on: February 16, 2015, 02:00:40 PM » 

itercount3 minibrot



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knighty
Fractal Iambus
Posts: 807


« Reply #12 on: February 16, 2015, 02:04:48 PM » 

That's awesome! Just wondering, why not use stereographic projection instead of spherical coordinates? IIRC, the Riemann sphere is obtained from the complex plane (+ point at infinity) by using the inverse stereographic projection.



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cKleinhuis


« Reply #13 on: February 16, 2015, 03:22:40 PM » 

errh, forget the previous postings, they did not multiply in the multiply loop, fixing it now



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cKleinhuis


« Reply #14 on: February 16, 2015, 03:40:49 PM » 

corrected formula loop: int itercount=0; counter for the mul sequence operator complex tempz=(1,0) ;init with unit number while(itercount<3) ; play with itercountmax value, 1,2,3 z=z^2+ #pixel complex t1=z(sqrt(2)*#e^((2*(itercount1)*#pi)/3)) complex t2=sqrt(2)*z (#e^((2*itercount+#pi)/3)) tempz=tempz*(t1/t2) ; fix for missing mul itercount=itercount+1 endwhile z=z*tempz +@seed ; include julia seed
following are images with "itercount=2,3,4"



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