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Author Topic: Planar triplex algebra  (Read 2105 times)
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M Benesi
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« on: February 10, 2010, 02:42:31 AM »

  Don't think this is going to work as a straight out number system.. rather it's a system of rotations being calculated algebraically, rather than trigonometrically, which might not be too useful.... 

 A new number system based on the 3d fractal formula I posted in the "search for the holy grail" thread that generates images for z^2, such as this one:



  In this system components are divided into planar and linear components as follows:
  x, y, and z are the linear components.

  w, v, and u are the planar components.

  w is the planar component that relates to the linear component x.
  w is the planar magnitude of the y-z plane: w= sqrt(y^2 + z^2)

  v is the planar component that relates to the linear component y.
  v is the planar magnitude of the x-z plane: w= sqrt(x^2 + z^2)

  u is the planar component that relates to the linear component z.
  u is the planar magnitude of the x-y plane: w= sqrt(x^2 + y^2)

  The planar triplex z is noted in this way:

z1= [ (x1,w1) ; (y1,v1) ; (z1,u1) ]
z2= [ (x2,w2) ; (y2,v2) ; (z2,u2) ]

Addition and subtraction follow the same basic format,  here is an addition:

z1 + z2 = [(x1+x2,w1+w2) ; (y1+y2,v1+v2) ;  (z1+z2,u1+u2)

  
Multiplication and division follow slightly different basic formats, here is a multiplication:

  First, we take the magnitude of the system:

|q| = |z1 & z2 *| = sqrt(x1*x2 + w1*w2 + y1*y2 + v1*v2 +  z1*z2 + u1*u2)


  Updated magnitude information will be here later today.

  Then we calculate new components:

x1*2 = |q| * (x1*x2 - w1*w2) / (x1*x2 + w1*w2)
w1*2 = |q| * (x2*w1 + x1*w2) / (x1*x2 + w1*w2)

y1*2 = |q| * (y1*y2 - v1*v2) / (y1*y2 + v1*v2)
v1*2 = |q| * (y2*v1 + y1*v2) / (y1*y2 + v1*v2)

z1*2 = |q| * (z1*z2 - u1*u2) / (z1*z2 + u1*u2)
u1*2 = |q| * (z2*u1 + z1*u2) / (z1*z2 + u1*u2)


  Division will come later.  Movie night...

  Here is an image of a z^13 quadrant of the new formula (all positives for planes, I like the positive quadrant, can set planes according to corresponding xyz signs in order to have 8 different quadrants, but I am still exploring the positive):


  I guess I should include the corresponding z^2 positive quadrant:
« Last Edit: February 21, 2010, 01:57:27 AM by M Benesi » Logged

jehovajah
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« Reply #1 on: February 10, 2010, 07:17:58 AM »

Love it! just love it!  the wave the wave the wave

I've just finished an analysis of twinbee's spherical coordinate formula vis a vis the squaring of the z triplex so i will post that later!

Can you do the planar rotations? I have also gotten round to thinking about what you have been doing and realised that you have been looking at the third rotation v as i described it, but i was not sure because it was buried in the code which was unfamiliar. Listen man way to go this is your thunder and i want you to enjoy it. Hope you loved the movie with your family.

 Repeating Zooming Self-Silimilar Thumb Up, by Craig
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M Benesi
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« Reply #2 on: February 10, 2010, 09:23:56 AM »

  Thanks!  smiley

  I don't really completely understand the math in your posts, as you do things differently than I do (I haven't learned tensor analysis, but I did a search and found a post of yours with a tensor with the third term as v).  I just found your polynomial rotations thread as well.  Need to read it tomorrow (it is almost sleep time now). 

  "Can you do the planar rotations?"   

  If you simply multiply the numbers as I stated in the above post, the linear components rotate around the planar components (I think I wrote it out correctly, otherwise use the trigonometric formulas from the "search for the holy grail" thread).  Since they rotate about the planar components composed of the other 2 linear components, all of them are rotating around one another (in a sense). 

  I've got to post some comments in that thread, as I tried some new things with the formula (I suppose I can say them here as well):

  I tried setting the magnitude (q in this thread, r in the other)

|q| = r = (x^2+w^2)^(n/2) + (y^2+v^2)^(n/2) + (z^2+u^2)^(n/2)     

  In my formula and (of course) ended up with slightly different fractals.  In other words, I am not entirely sure what the magnitude for this number system should be or what magnitude we should use for the fractal (although I am still enjoying the original value, and the tight, yet varied images it creates).  I have this feeling that we should be dividing it by 3^n or 3^(n/2).  So I added an option to my formula that sets a value of <value>^(-n/2), I like to use 9, which basically means I like to use the value 1/3^n to multiply the magnitude by which increases the fractal size to a more manageable level. 

  For now, here is a 13th order 13 iteration fractal that I just got done:


  I'll do the other thread tomorrow.   sleeping
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« Reply #3 on: February 11, 2010, 01:18:57 AM »

Nicely done!!!   And some interesting images as well.    smiley
 
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jehovajah
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« Reply #4 on: February 11, 2010, 06:54:37 AM »

 

  I don't really completely understand the math in your posts, as you do things differently than I do (I haven't learned tensor analysis, but I did a search and found a post of yours with a tensor with the third term as v).  I just found your polynomial rotations thread as well.  Need to read it tomorrow (it is almost sleep time now). 

  "

 cheesy Neither do i! It grows organically from me. Any way I am going to revise some of the notation in polynomial rotations as a later post i did clarified some confusion in notation between orientation and rotation. Any comments are more than welcome.

My thinking about the rotations reflects the degrees of freedom in geometrical space. Tensor math is the only math beside quaternion and clifford algebras that reputedly addresses these transformations full on. Consequently tensor math is hard to get your nogging around! However we now have the luxury of graphical computers and talented programmers such as yourself can write formulae to visualise these transformations.

I note you use chaos pro. I would like to advise you of Terry Gintz programmes as well, which i find intuitive. wink
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May a trochoid of ¥h¶h iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
M Benesi
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« Reply #5 on: February 11, 2010, 08:04:00 AM »

  Thank you Nahee.   

  jehovajah-

  I noticed that at least some (maybe all) of the Terry Glintz' stuff appears to be limited, in the sense that you have to pay a fee to render images larger than 320x200.

  The ChaosPro code is there in the other thread if anyone feels like creating 3d fractals larger than 320x200 without paying a fee.  Not to mention that I learned to program ChaosPro very quickly: I just glanced at the code for one of the pre-compiled fractals and went from there.  Not to mention that Martin told me that the new version will have a feature that really adds something to my fractals.  Why play around when you already have the best? 
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jehovajah
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« Reply #6 on: February 11, 2010, 03:21:22 PM »

You said it! Yes !!
However you may still want to check it out at a later date. I mean did you know Terry's programme has been able to draw mandel bulbs since 2001 for instance?

Anyway,back to the task in hand. Have you had chance to sort out division?

Whatever magnitude|q| you are using now seems fine.
« Last Edit: February 17, 2010, 05:49:28 AM by jehovajah » Logged

May a trochoid of ¥h¶h iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
M Benesi
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« Reply #7 on: February 12, 2010, 05:50:09 AM »

  Wow.  I didn't know Glintz was that far ahead of the curve.  Good for him.  

  As to the division issue, this might be correct (I will have to check it later, but for now I am hungry and want to get this done quickly):

|q| = sqrt(x1*x2 + w1*w2 + y1*y2 + v1*v2 +  z1*z2 + u1*u2)


Found out the magnitudes might need to be separated like so:
linear magnitude= sqrt (x^2+y^2+z^2)
planar magnitude = sqrt (w^2+v^2+u^2)

  Based on using the following magnitude equation in the fractal and getting a clearer, more fractally at lower iteration image:

r= sqrt( (x^2+y^2+z^2)^n +(w^2+v^2+u^2)^mag2 )  

  However, the other formula is still more pleasing in some sense, so.... 


 I use separate magnitudes for the linear and planar variables: lets me make interesting images

  Not sure what the formula should be for 2 variables, yet.  Will figure it out later as I have a bunch of stuff to get done this afternoon.

|q| = sqrt (linear magnitude^2+planar magnitude^2);

x1/2 = |q| * (x1*x2 + w1*w2) / [(x2^2 + w2^2) * (x1*x2 + w1*w2)]
w1/2 = |q| * (x2*w1 - x1*w2) / [(x2^2+ w2^2) * (x1*x2 + w1*w2)]

y1/2 = |q| * (y1*y2 + v1*v2) / [(y2^2+ v2^2) * (y1*y2 + v1*v2)]
v1/2 = |q| * (y2*v1 - y1*v2) / [(y2^2 + v2^2) *(y1*y2 + v1*v2)]

z1/2 = |q| * (z1*z2 + u1*u2) / [(z2^2 + u2^2) * (z1*z2 + u1*u2)]
u1/2 = |q| * (z2*u1 - z1*u2) / [(z2^2 + u2^2) * (z1*z2 + u1*u2)]
« Last Edit: February 12, 2010, 08:22:38 PM by M Benesi » Logged

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