lycium
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« Reply #285 on: October 07, 2009, 11:24:17 PM » |
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for the rendering i've got all sorts of effects going on inside my program (so that it works at very high precision); things like contrast adjustments, blurring and sharpening...
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David Makin
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« Reply #286 on: October 08, 2009, 02:47:53 AM » |
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Like the anim! (still want a zoom in tho). Interesting image too - didn't realise there were big gaping holes in the object as shown. Curious...
I did mention in the DA comments (in reply to Lycium) that I cut the fractal to a plane to create the crater lake. I'll just add something to the main comments just to be sure no-one gets too coinfused
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stigomaster
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« Reply #287 on: October 08, 2009, 04:49:06 PM » |
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Perhaps he was pointing to the gaps beneath the bridges? Golden Gate, go home.
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twinbee
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« Reply #288 on: October 08, 2009, 11:02:40 PM » |
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Stigomaster's right... I meant the gaps below the bridges in David's latest pic.
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David Makin
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« Reply #289 on: October 09, 2009, 05:08:07 AM » |
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Stigomaster's right... I meant the gaps below the bridges in David's latest pic.
Oh - I'm going to do some more examination of the degree 4 to see if that has anything similar
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JosLeys
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« Reply #290 on: October 09, 2009, 06:13:17 PM » |
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I just joined. Hello all. A few weeks ago, Dave Makin helped me get going on 3D fractals. I'm using his Distance Estimate method based on smooth iteration count. I must say it works quite well, although I'm not really sure why! I'm trying to get my hands around the underlying math, as I think it would make a good subject for an article in the appropriate place. Well, thanks to Dave's pseudocode, here is something that I thought turned out quite well, although it has nothing to do with the quest for the elusive 3D Mandelbrot..
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bugman
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« Reply #291 on: October 09, 2009, 06:21:34 PM » |
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Hi Jos! Glad to have you join us! This discussion seems to be attracting quite a few talented individuals. If I may ask, what is this image you have created? Is it a slice of a Julibrot?
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JosLeys
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« Reply #292 on: October 09, 2009, 06:32:40 PM » |
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Well, I guess so. I've seen so many things called a Julibrot that I'm not sure anymore. The z direction (perpendicular to the screen) is c=variable+i.constant), and then the formula is z^2+c (Julia), so in actual fact I'm stacking up all the possible Julia's with seed=variable+i.constant. For the image it is c= var*i.0.1, and var starts at -0.76.
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bugman
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« Reply #293 on: October 09, 2009, 06:57:07 PM » |
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Well, I guess so. I've seen so many things called a Julibrot that I'm not sure anymore. The z direction (perpendicular to the screen) is c=variable+i.constant), and then the formula is z^2+c (Julia), so in actual fact I'm stacking up all the possible Julia's with seed=variable+i.constant. For the image it is c= var*i.0.1, and var starts at -0.76.
That's very nice. I like the embossed detail in the slice. I'm defining the "Julibrot" as the set of all Julia sets (which happens to include the Mandelbrot set). I made a Julibrot animation here: http://bugman123.com/Hypercomplex/Julibrot.m1vIt is a 3D "slice" of the 4D Julibrot starting from y0=0, and then rotating to x0=0, then rotating to xc=0, then rotating to yc=0, and finally rotating back to y0=0 again.
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« Last Edit: October 09, 2009, 07:00:06 PM by bugman »
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David Makin
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« Reply #294 on: October 10, 2009, 01:11:07 AM » |
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Stigomaster's right... I meant the gaps below the bridges in David's latest pic.
I just checked to see that the render was indeed accurate - in the original the default DE divisor parameter to get the step distances was 5 and the initial maximum distances to step was set to 0.1 - to check that it was accurate I first tried changing the maximum step distance allowed to 0.01 and this basically produced the same image (but around 4*slower), I then reset the maximum step distance allowed to 0.1 and changed the DE divisor to 10 (i.e. making all the step distances half the original values) and again this produced the same image (this time about 1.8* slower). So I guess there are gaps under the bridges in places
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David Makin
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« Reply #295 on: October 10, 2009, 02:01:06 AM » |
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I just joined. Hello all. A few weeks ago, Dave Makin helped me get going on 3D fractals. I'm using his Distance Estimate method based on smooth iteration count. I must say it works quite well, although I'm not really sure why! I'm trying to get my hands around the underlying math, as I think it would make a good subject for an article in the appropriate place.
Well, thanks to Dave's pseudocode, here is something that I thought turned out quite well, although it has nothing to do with the quest for the elusive 3D Mandelbrot.. <Quoted Image Removed>
Very nice image - I also like your embossing !
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lycium
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« Reply #296 on: October 10, 2009, 03:07:35 AM » |
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hey cool it's jos leys! welcome to the forums
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xenodreambuie
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« Reply #297 on: October 10, 2009, 03:53:28 AM » |
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There are lots of impressive renders in this thread.
Since I've been experimenting with MIIM (modified inverse iteration method) lately, I tried Paul's formulas for the roots of the quadratic triplex, to get the benefit of the extra detail. It didn't add much worthwhile. On the whole, I was disappointed with the various forms of the quadratic, and the higher powers look far more interesting.
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lycium
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« Reply #298 on: October 10, 2009, 03:58:41 AM » |
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very cool render garth!
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xenodreambuie
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« Reply #299 on: October 10, 2009, 05:41:16 AM » |
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Thanks Lycium! It does have the benefit of soft shadows.
Something I noticed with the quadratic version was that making the third constant nonzero tended to introduce a planar discontinuity in the middle. It could have been due to steps I was taking to avoid invalid roots. Otherwise I guess it's just a result of using spherical coordinates and the z axis being polar. Has anyone found the third constant to give useful variation? (One advantage of inverse iteration is interactive previewing.)
I already deleted my code for the inverse triplex, and I'm not planning to revisit it unless I can do something with the trigonometric version instead. I just looked at the formula Dave provided, and noticed that phi=atan((z/sqrt(x²+y²)). It should be cheaper to use phi=asin(z/r).
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« Last Edit: October 10, 2009, 06:14:17 AM by xenodreambuie, Reason: asin, not acos; I use one or other depending on my purpose »
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