Title: LambdaBulb Post by: fractalrebel on November 23, 2009, 11:53:19 PM Here is a bit different type of beast. Its a 4th power lambdabulb with a seed of 1.0625 + i*0.2375 with shadows. The function is
Title: Re: LambdaBulb Post by: cKleinhuis on November 24, 2009, 12:10:59 AM oh yeah, this is what i call a nice object ! O0
Title: Re: LambdaBulb Post by: David Makin on November 24, 2009, 12:45:24 AM Nice one Ron !
I have an idea for doing a pseudo-Newton ;) Title: Re: LambdaBulb Post by: fractalrebel on November 24, 2009, 05:40:42 AM Here is a fractal version of a Segway. This is a lambdabulb that is the standard power of two. The julia seed is 0.964 - i*0.303.
Title: Re: LambdaBulb Post by: JColyer on November 24, 2009, 07:59:51 PM Ohhh pretty... hmmm... have to slice these things open and see what makes 'em tick!
Title: Re: LambdaBulb Post by: JColyer on November 25, 2009, 12:58:50 AM @fractalrebel - mind sharing your expansion (code) for the lambda bulb formula? I'm trying to get it right, but something's really wrong with the pictures I'm generating!!
I tried adapting the 'standard' 3d mandelbulb code, but seems like I'm missing something..... Thanks! JC Title: Re: LambdaBulb Post by: fractalrebel on November 25, 2009, 04:22:23 AM @fractalrebel - mind sharing your expansion (code) for the lambda bulb formula? I'm trying to get it right, but something's really wrong with the pictures I'm generating!! I tried adapting the 'standard' 3d mandelbulb code, but seems like I'm missing something..... Thanks! JC All of my code object oriented Ultrafractal 5 code. I define a function for taking powers in spherical coordinates, and an other for multiplication. The target function is static func Spower(bool altel, float power, Vector a, Vector b) ; Power function for s = x + iy + jz expressed in spherical coordinates float r = sqrt(a.m_x^2 + a.m_y^2 + a.m_z^2)+1e-10 float phi = atan2(a.m_x + flip(a.m_y)) ; azimuth float theta = 0 if altel theta = asin(-a.m_z/r) else theta = asin(a.m_z/r) endif r = r^power phi = power*phi theta = power*theta b.m_x = r*cos(theta)*cos(phi) b.m_y = r*cos(theta)*sin(phi) b.m_z = r*sin(theta) b.m_w = 0 endfunc Here is rhe multiplication function. a and b are the two input vectors and c is the output vector. static func SMult(bool altel, Vector a, Vector b, Vector c) float ra = sqrt(a.m_x^2 + a.m_y^2 + a.m_z^2)+1e-10 float phia = atan2(a.m_x + flip(a.m_y)) ; azimuth float thetaa = 0 if altel thetaa = asin(-a.m_z/ra) else thetaa = asin(a.m_z/ra) endif float rb = sqrt(b.m_x^2 + b.m_y^2 + b.m_z^2)+1e-10 float phib = atan2(b.m_x + flip(b.m_y)) ; azimuth float thetab = 0 if altel thetab = asin(-b.m_z/rb) else thetab = asin(b.m_z/rb) endif float r = ra*rb float phi = phia + phib float theta = thetaa + thetab c.m_x = r*cos(theta)*cos(phi) c.m_y = r*cos(theta)*sin(phi) c.m_z = r*sin(theta) c.m_w = 0 endfunc Title: Re: LambdaBulb Post by: JColyer on November 25, 2009, 01:52:24 PM ron - thanks for the code mine was really close but I had the multiply function implemented as a standard vector cross product and that was screwing me up. should have an image posted later today.
Title: Re: LambdaBulb Post by: fractalrebel on November 27, 2009, 09:08:35 PM The following image is an 8th power lambdabulb with a seed of 0.046411130776+ i*1.2609375008. There appears to be a gryphon on a throne hiding in the fractal.
Title: Re: LambdaBulb Post by: bugman on December 01, 2009, 07:13:21 AM Ron, I really like your rendering of the Lamdabulb at the top of this post. I'd like to see more unusal formulas like this. Here are some of my renderings of it. The lower image is with a negative sign (using your notation altel = true).
Title: Re: LambdaBulb Post by: kram1032 on December 02, 2009, 01:00:44 AM also a great variant :D
I really like them a lot! Title: Re: LambdaBulb Post by: miner49er on December 02, 2009, 09:53:45 PM That one in the lower left looks like some kind of monsters with loads of mouths. I like it! |