Title: Elliptical Mbulb idea
Post by: kram1032 on March 30, 2010, 03:27:01 PM
I'm still thinking about using different multiples for the angles to get a more symmetric power 2-ish Mbulb.
However I noticed a problem with the direct variant, there:
If you just use [rē;2phi;8theta]+c, you don't actually have the "correct" radius for each angle.
So, instead, I thought, an elliptical "radius" could be used.
trying a bit around, I came to a radius formula, entirely dependend on theta, which basically tells you, at which angle of theta the radius is exponentiated how strong, if that makes any sense...
The formula is derived from the parametric coordinates of a rotational ellipsoid, being rotated around the z-axis. It returns values between 2 and 8, with 2 being for theta = 0 (so, the triplex number is a complex number with the second imaginary part = 0) and 8 for theta = +/-pi/2 (which happens at the two poles, where both the real and the first imaginary parts are 0).
The formula is simple:
[Plot] (http://www.wolframalpha.com/input/?i=plot+sqrt%2834-30+cos%282+theta%29%29%2Ctheta%3D-pi%2F2...pi%2F2)
Now, there are two possible ways to use this:
Either, you just take the angles as usual and use the theta-dependend term as power for r or you also take that term as a factor for theta and phi:
1)![[r^{\sqrt{34-30 \cos{2 \theta}}};2 \phi;8 \theta]](/cgi-bin/mimetex.cgi?[r^{\sqrt{34-30 \cos{2 \theta}}};2 \phi;8 \theta])
2)![[r^{\sqrt{34-30 \cos{2 \theta}}}; {\sqrt{34-30 \cos{2 \theta}}} \phi; {\sqrt{34-30 \cos{2 \theta}}} \theta]](/cgi-bin/mimetex.cgi?[r^{\sqrt{34-30 \cos{2 \theta}}}; {\sqrt{34-30 \cos{2 \theta}}} \phi; {\sqrt{34-30 \cos{2 \theta}}} \theta])
plots for the change of r, phi and theta (phi and theta as in example 2)
[Plot r] (http://www.wolframalpha.com/input/?i=3Dplot+r^sqrt%2834-30+cos%282+theta%29%29%2Cr%3D0...2%2Ctheta%3D-pi%2F2...pi%2F2)
[Plot phi] (http://www.wolframalpha.com/input/?i=3Dplot+phi*sqrt%2834-30+cos%282+theta%29%29%2Cphi%3D-pi...pi%2Ctheta%3D-pi%2F2...pi%2F2)
[Plot theta] (http://www.wolframalpha.com/input/?i=plot+theta*sqrt%2834-30+cos%282+theta%29%29%2Ctheta%3D-pi%2F2...pi%2F2)
However I noticed a problem with the direct variant, there:
If you just use [rē;2phi;8theta]+c, you don't actually have the "correct" radius for each angle.
So, instead, I thought, an elliptical "radius" could be used.
trying a bit around, I came to a radius formula, entirely dependend on theta, which basically tells you, at which angle of theta the radius is exponentiated how strong, if that makes any sense...
The formula is derived from the parametric coordinates of a rotational ellipsoid, being rotated around the z-axis. It returns values between 2 and 8, with 2 being for theta = 0 (so, the triplex number is a complex number with the second imaginary part = 0) and 8 for theta = +/-pi/2 (which happens at the two poles, where both the real and the first imaginary parts are 0).
The formula is simple:
[Plot] (http://www.wolframalpha.com/input/?i=plot+sqrt%2834-30+cos%282+theta%29%29%2Ctheta%3D-pi%2F2...pi%2F2)
Now, there are two possible ways to use this:
Either, you just take the angles as usual and use the theta-dependend term as power for r or you also take that term as a factor for theta and phi:
1)
2)
plots for the change of r, phi and theta (phi and theta as in example 2)
[Plot r] (http://www.wolframalpha.com/input/?i=3Dplot+r^sqrt%2834-30+cos%282+theta%29%29%2Cr%3D0...2%2Ctheta%3D-pi%2F2...pi%2F2)
[Plot phi] (http://www.wolframalpha.com/input/?i=3Dplot+phi*sqrt%2834-30+cos%282+theta%29%29%2Cphi%3D-pi...pi%2Ctheta%3D-pi%2F2...pi%2F2)
[Plot theta] (http://www.wolframalpha.com/input/?i=plot+theta*sqrt%2834-30+cos%282+theta%29%29%2Ctheta%3D-pi%2F2...pi%2F2)
Title: Re: Elliptical Mbulb idea
Post by: kram1032 on April 02, 2010, 01:53:58 PM
is noone interested? :(
Title: Re: Elliptical Mbulb idea
Post by: msltoe on April 02, 2010, 03:45:10 PM
Mr. Kram1032,
I've rendered your version 2. Not bad....
Would you like a copy of my code for future experimenting? It's not as fancy as others, but it is fairly easy to play with different formulae.
-mike
I've rendered your version 2. Not bad....
Would you like a copy of my code for future experimenting? It's not as fancy as others, but it is fairly easy to play with different formulae.
-mike
Title: Re: Elliptical Mbulb idea
Post by: kram1032 on April 02, 2010, 03:50:36 PM
:D thanks
Looks nice :)
Which Program do you use? Would be nice to see :)
If you use UF, I somehow didn't have luck with that sofar :(
Best for me would be an implementation for C/C++ or Processing^^
Looks nice :)
Which Program do you use? Would be nice to see :)
If you use UF, I somehow didn't have luck with that sofar :(
Best for me would be an implementation for C/C++ or Processing^^
Title: Re: Elliptical Mbulb idea
Post by: Timeroot on April 02, 2010, 09:16:11 PM
I don't really understand quite what happens here, but what does it look like if this applied to a plain old 2D Mandelbrot?
Title: Re: Elliptical Mbulb idea
Post by: kram1032 on April 02, 2010, 11:24:50 PM
it wouldn't change at all.
All I did was this:
The usual Mset returns nice results but the 2nd order Mbulb has a lot of strage distortions.
To get rid of them, the much more symmetrical 8th order Mbulb is used (usually).
Now what would happen if you double phi (the angle in the complex plane) but multiply theta by eight?
In theory, if I'm not mistaken, you get a higher grade of symmetry. However, this has one problem:
It's quite undefined, what to do with the radius. Should it be squared? should it be raised to the power of eight?
Now, my idea was, to take an elliptically shaped gradient to descide, which power to use.
At very flat angles, around theta=0, you get powers of 2, which is exactly the behaviour of the 2-Mbulb, exactly where you'd find the 2-Mset but this slowly increases and finally reaches eight at +/-90°, so at the poles, it behaves just like the 8-Mbulb, at least the radius does.
The two variants (the image is from the second one) are where the anglaes are multiplied by a constant factor or by the same gradient term.
So, the image basically shows a smooth gradient from the 2-bulb to the 8-bulb, along the magnitude of theta.
I'd love to see a nice exploration of it.
All I did was this:
The usual Mset returns nice results but the 2nd order Mbulb has a lot of strage distortions.
To get rid of them, the much more symmetrical 8th order Mbulb is used (usually).
Now what would happen if you double phi (the angle in the complex plane) but multiply theta by eight?
In theory, if I'm not mistaken, you get a higher grade of symmetry. However, this has one problem:
It's quite undefined, what to do with the radius. Should it be squared? should it be raised to the power of eight?
Now, my idea was, to take an elliptically shaped gradient to descide, which power to use.
At very flat angles, around theta=0, you get powers of 2, which is exactly the behaviour of the 2-Mbulb, exactly where you'd find the 2-Mset but this slowly increases and finally reaches eight at +/-90°, so at the poles, it behaves just like the 8-Mbulb, at least the radius does.
The two variants (the image is from the second one) are where the anglaes are multiplied by a constant factor or by the same gradient term.
So, the image basically shows a smooth gradient from the 2-bulb to the 8-bulb, along the magnitude of theta.
I'd love to see a nice exploration of it.
Title: Re: Elliptical Mbulb idea
Post by: kram1032 on April 03, 2010, 12:09:11 AM
Attached you see the other version. (Variant 1)
While it is far away from a "true Mandelbulb" and rather flat, I really like the shape :)
While it is far away from a "true Mandelbulb" and rather flat, I really like the shape :)
Title: Re: Elliptical Mbulb idea
Post by: Timeroot on April 03, 2010, 02:14:58 AM
Ah, I see now. That's pretty cool. Personally, I really like the Version 1 image. Now that I understand it, I wondered what happens if you let the second power go arbitrarily high - that is, instead of 8, we'd have 16, or 32, etc... and it's pretty self evident that the ellipse get's stretched out until it just reduces to the secant function (more precisely, 2*abs(secant(theta)) ). This means that, near the poles, the power is approximately equal to the reciprocal of distance to the pole. I don't know how the distortion grows as you near the pole, but if it's just first order, this might (probably won't) end up being the closest to the real Mandelbulb... there will have to be a problem, of course, probably an extreme function like secant will give its own distortion to the poles... I suppose some fine-tuned powers will end up being the answer. This is really cool idea Kram, can't wait to see what else to brings!
Title: Re: Elliptical Mbulb idea
Post by: kram1032 on April 03, 2010, 11:22:10 AM
thanks :D
Well, that idea of an aribitarily high power already was tried, I think...
It was one of the alternative coordinate systems...
If I remember correctly, it ended up like a balloon with the shape of the Mset... you know, the one that looks flat but where you'd search for the Mset, it actually is there...
Well, that idea of an aribitarily high power already was tried, I think...
It was one of the alternative coordinate systems...
If I remember correctly, it ended up like a balloon with the shape of the Mset... you know, the one that looks flat but where you'd search for the Mset, it actually is there...
Title: Re: Elliptical Mbulb idea
Post by: KRAFTWERK on April 06, 2010, 11:03:47 AM
Very interesting shape Kram!...
Title: Re: Elliptical Mbulb idea
Post by: kram1032 on April 06, 2010, 11:28:51 AM
thanks :D
Further explorations would be a pain, though^^
If anyone would try this in any of the UF programs or similar, it would be a lot simpler, I guess.
Further explorations would be a pain, though^^
If anyone would try this in any of the UF programs or similar, it would be a lot simpler, I guess.