bugman
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« Reply #45 on: January 06, 2010, 12:42:50 AM » |
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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 » |
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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 » |
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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 » |
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The sine version gives this:
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kram1032
Fractal Iambus
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« Reply #49 on: January 06, 2010, 10:39:46 AM » |
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Also nice but noisy  The knot-one surprises with rather simple hyperbolic-ish knot-like geometry |
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Nahee_Enterprises
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« Reply #50 on: January 06, 2010, 12:54:09 PM » |
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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!!! 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 » |
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The sine version gives this: Reminds me of the underside of a "Megalopyge opercularis" (or "Asp" as we always call them in Texas). |
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« Last Edit: January 30, 2010, 07:00:27 PM by Nahee_Enterprises »
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matsoljare
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« Reply #52 on: January 06, 2010, 09:36:44 PM » |
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 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 » |
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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
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« Reply #54 on: January 07, 2010, 01:22:15 AM » |
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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
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« Reply #55 on: January 08, 2010, 05:28:45 PM » |
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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
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« Reply #56 on: January 08, 2010, 05:31:17 PM » |
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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
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« Reply #57 on: January 30, 2010, 02:26:49 AM » |
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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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Posts: 122
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« Reply #58 on: January 30, 2010, 07:10:44 AM » |
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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
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« Reply #59 on: January 30, 2010, 10:29:25 AM » |
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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=#pi-ph,endif |
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May 19, 2010, 03:11:42 PM
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