bugman


« Reply #45 on: January 06, 2010, 12:42:50 AM » 

Did you use an orbit trap to get the weaved effect? I am getting a rather boring looking thing when I use the cosine formula.
It would be interesting to see the inside. Can you render a slice showing half of the fractal so we can see the inside?



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JosLeys


« Reply #46 on: January 06, 2010, 12:49:41 AM » 

No orbit trap. The fourth image I posted is the slice.



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bugman


« Reply #47 on: January 06, 2010, 04:02:27 AM » 

I am curious to see if you get something different from me using the sine version of the formula.



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JosLeys


« Reply #48 on: January 06, 2010, 08:38:21 AM » 

The sine version gives this:



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kram1032


« Reply #49 on: January 06, 2010, 10:39:46 AM » 

Also nice but noisy The knotone surprises with rather simple hyperbolicish knotlike geometry



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Nahee_Enterprises


« Reply #50 on: January 06, 2010, 12:54:09 PM » 

While looking for Glynn fractal patterns using the different variations of the spherical coordinate formulas, I stumbled on an interesting shape while using Garth Thornton's variation. See the images below. The last one is a slice. So this is for z^1.50.2.. Very nice and quite interesting!!! The inside view was a bit surprising in some ways.



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Nahee_Enterprises


« Reply #51 on: January 06, 2010, 01:01:39 PM » 

The sine version gives this: Reminds me of the underside of a "Megalopyge opercularis" (or "Asp" as we always call them in Texas).


« Last Edit: January 30, 2010, 07:00:27 PM by Nahee_Enterprises »

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matsoljare


« Reply #52 on: January 06, 2010, 09:36:44 PM » 

Very interesting, try some other values for that formula!



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JosLeys


« Reply #53 on: January 07, 2010, 12:13:11 AM » 

Well here is the Julia for <.22,0,0> Could somebody try reproducing this to make sure I do not have some stupid error somewhere?



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David Makin


« Reply #54 on: January 07, 2010, 01:22:15 AM » 

Well here is the Julia for <.22,0,0> Could somebody try reproducing this to make sure I do not have some stupid error somewhere?
Looks correct to me Jos. Am not posting my rendered version  I can't be bothered waiting for the render to finish, it's so sloooow



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Nahee_Enterprises


« Reply #55 on: January 08, 2010, 05:28:45 PM » 

Well here is the Julia for <.22,0,0> Could somebody try reproducing this to make sure I do not have some stupid error somewhere? Ahhh!!! The skeletal view of this organism, without the fleshy parts.



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Nahee_Enterprises


« Reply #56 on: January 08, 2010, 05:31:17 PM » 

Well here is the Julia for <.22,0,0> Could somebody try reproducing this to make sure I do not have some stupid error somewhere? Looks correct to me Jos. Am not posting my rendered version  I can't be bothered waiting for the render to finish, it's so sloooow And what if you both made the same (or similar errors) ??



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xenodreambuie


« Reply #57 on: January 30, 2010, 02:26:49 AM » 

Jos and/or David, can you clarify whether this cosine version is the flipped or correct one? Ie any parity additions? It is a beauty, and I haven't managed to reproduce it using inverse iteration. It also seems surprising for one with cosine phi to be symmetric in Z with a fractional power, instead of having a similar kind of asymmetry along X and Z.
I had no difficulty in reproducing the sine version with MIIM. It works with independent roots for theta and phi, with one or two each.



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bugman


« Reply #58 on: January 30, 2010, 07:10:44 AM » 

I also haven't been able to reproduce the cosine Glynn with MIIM. This is the formula I used:
{x,y,z}^(1/1.5) = r^(1/1.5)*{cos(theta)*sin(phi),sin(theta)*sin(phi),cos(phi)} where r = sqrt(x²+y²+z²), theta = (atan2(y,x)+ktheta*pi)/1.5, phi = (acos(z/r)+kphi*pi)/1.5;
I always find one valid root when ktheta = kphi = 0. If z>0 and x>0 then that is the only valid root I find.
If z>0 and x<0 then I find another valid root when ktheta = (y<0?2:1), kphi = 0.
If z<0 then I find two other valid roots when ktheta=((x<0 && y<0)?2:0.5), kphi=((x<0 && y<0)?0:1) and when ktheta=((x<0 && y>0)?1:2.5), kphi=((x<0 && y>0)?0:1).



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JosLeys


« Reply #59 on: January 30, 2010, 10:29:25 AM » 

I think you call this formula the 'flipped' one : z= (R^(@pow/2))*sin(@pow*ph)*exp(i*@pow*th)+c1 zz=(R^(@pow/2))*cos(@pow*ph)+cz R=(z+zz*zz) th=atan2(z) ph=atan2(zz+i*cabs(z)) if ph>=#pi/2,ph=#piph,endif



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