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Author Topic: Glynn Julia set  (Read 25314 times)
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bugman
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« 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
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« 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
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« 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
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« Reply #48 on: January 06, 2010, 08:38:21 AM »

The sine version gives this:


* 4D_test_139.jpg (888.12 KB, 1024x737 - viewed 404 times.)
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kram1032
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« Reply #49 on: January 06, 2010, 10:39:46 AM »

Also nice but noisy smiley

The knot-one surprises with rather simple hyperbolic-ish knot-like geometry smiley
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Nahee_Enterprises
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« 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^1.5-0.2..

Very nice and quite interesting!!!     smiley

The inside view was a bit surprising in some ways.
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Nahee_Enterprises
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« 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 » Logged

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



Very interesting, try some other values for that formula!
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JosLeys
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« 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?


* 4D_test_141.jpg (317.43 KB, 800x511 - viewed 406 times.)
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David Makin
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« 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 smiley
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Nahee_Enterprises
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« 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.    wink
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Nahee_Enterprises
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« 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   smiley

And what if you both made the same (or similar errors) ??    evil
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xenodreambuie
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« 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
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« 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(x▓+y▓+z▓), 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
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« Reply #59 on: January 30, 2010, 10:29:25 AM »

I think you call this formula the 'flipped' one :
Code:
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=#pi-ph,endif
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