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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Thu Jul 19, 2012 2:01 pm    Post subject: Reply with quote

sure Gerd I do, try this Very Happy
Code:
min(max( abs(y-floor(y/(2*cos(pi/6))+0.5)*(2*cos(pi/6)))-cos(pi/6) , abs(x-3*floor(x/3+0.5))*cos(pi/6)+abs(y-floor(y/(2*cos(pi/6))+0.5)*(2*cos(pi/6)))*sin(pi/6) -cos(pi/6) )^2,max( abs((y-cos(pi/6))-floor((y-cos(pi/6))/(2*cos(pi/6))+0.5)*(2*cos(pi/6)))-cos(pi/6) , abs((x-1.5)-3*floor((x-1.5)/3+0.5))*cos(pi/6)+abs((y-cos(pi/6))-floor((y-cos(pi/6))/(2*cos(pi/6))+0.5)*(2*cos(pi/6)))*sin(pi/6) -cos(pi/6) )^2  )+z^2-0.04
[x]: -4.5 , 4.5
[y]: -4 , 4
[z]: -0.3 , 0.3


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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Mon Jul 23, 2012 5:55 pm    Post subject: Reply with quote

this is a much more better formula Wink
Code:
min(max(abs(y-floor(y/2+0.5)*2),abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6))-1,abs(y-floor(y/2+0.5)*2))^2+z^2-0.0225
[x]: -5 , 5
[y]: -4.5 , 4.5
[z]: -0.5 , 0.5




generally write:
Code:
F( min(max(abs(y-floor(y/2+0.5)*2),abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6))-1,abs(y-floor(y/2+0.5)*2)) , z)=0

for example:
Code:
abs(min(max(abs(y-floor(y/2+0.5)*2),abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6))-1,abs(y-floor(y/2+0.5)*2)) )+abs(z)-0.2
[x]: -5 , 5
[y]: -4.5 , 4.5
[z]: -0.2 , 0.2



also:
Code:
(5*min(max(abs(y-floor(y/2+0.5)*2),abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6))-1,abs(y-floor(y/2+0.5)*2)))^2+(5*z)^20-1
[x]: -5 , 5
[y]: -4.5 , 4.5
[z]: -0.2 , 0.2


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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Fri Aug 24, 2012 11:53 am    Post subject: Reply with quote

Hi, this is the same but without min and max functions Wink
Code:
(((abs(y-floor(y/2+0.5)*2)+ abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6)+abs(abs(y-floor(y/2+0.5)*2) - (abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6))))/2-1 +abs(y-floor(y/2+0.5)*2) -abs(( abs(y-floor(y/2+0.5)*2) +abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6) +abs( abs(y-floor(y/2+0.5)*2) -(abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6)) ))/2-1-abs(y-floor(y/2+0.5)*2)))/2)^2+z^2-0.04

[x]: -5.5 , 5.5
[y]: -6.5 , 6.5
[z]: -0.3 , 0.3




sure, generally you would write:
Code:
F(  (((abs(y-floor(y/2+0.5)*2)+ abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6)+abs(abs(y-floor(y/2+0.5)*2) - (abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6))))/2-1 +abs(y-floor(y/2+0.5)*2) -abs(( abs(y-floor(y/2+0.5)*2) +abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6) +abs( abs(y-floor(y/2+0.5)*2) -(abs(x-3/cos(pi/6)*floor(x/3*cos(pi/6)+0.5))*cos(pi/6)+abs(y-floor(y/2+0.5)*2)*sin(pi/6)) ))/2-1-abs(y-floor(y/2+0.5)*2)))/2)  ,z)=0

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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Mon Oct 29, 2012 9:36 pm    Post subject: Reply with quote

Hello all,
I missed math and k3dsurf, is there anyone else like me? Cool also this is the same but this time without "floor" function Wink
Code:
min(abs(abs(abs(abs(abs(abs(y)-16*cos(pi/6))-8*cos(pi/6))-4*cos(pi/6))-2*cos(pi/6))-cos(pi/6)),max( abs(abs(abs(abs(abs(abs(y)-16*cos(pi/6))-8*cos(pi/6))-4*cos(pi/6))-2*cos(pi/6))-cos(pi/6)), abs(abs(abs(abs(abs(x)-12)-6)-3)-1.5)*cos(pi/6)+abs(abs(abs(abs(abs(abs(y)-16*cos(pi/6))-8*cos(pi/6))-4*cos(pi/6))-2*cos(pi/6))-cos(pi/6))*sin(pi/6) )-cos(pi/6))^2+z^2-0.04
[x]: -6 , 6
[y]: -6 , 6
[z]: -0.7 , 0.7



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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Fri Nov 09, 2012 11:03 pm    Post subject: Reply with quote

also, another formula for the same isosurface, I have no idea what k3dsurf could do recently, I have very silent friends on this forum Twisted Evil
Code:
min(max(abs(2/pi*atan(tan(pi/2*y)))-1,abs(2/pi*atan(tan(pi/2*y)))*cos(pi/3)-(-abs(3/cos(pi/6)/pi*atan(tan(pi*x/3*cos(pi/6))))+1/cos(pi/6))*sin(pi/3)),abs(2/pi*atan(tan(pi/2*y))))^2+z^2-0.04

[x]: -5.5 , 5.5
[y]: -6.3 , 6.3
[z]: -0.4 , 0.4


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Furan



Joined: 05 Oct 2010
Posts: 64
Location: Prague, Czech Republic

PostPosted: Fri Feb 15, 2013 6:38 pm    Post subject: HexGrid Donut Reply with quote

Sorry for being away for so long, I've been hanging around mainly at fractalforums.com Now I'm ready to fuse these two parts of my life together Smile I have an insane idea that I'm working on, it's gonna be awesome Cool
Anyway, this is a toroid (R=2, r=1) made with my hexagonal grid with constant relative grid ratio:
Code:
 Transformation to local orthogonal system with distortion (usable to any other repeating pattern on this toroid)
x1=atan2(y,x)*8
y1=(atan2(z,(sqrt(x^2+y^2)-2)))*8/sqrt(x^2+y^2)
z1=((sqrt(x^2+y^2)-2)^2+z^2-1)*4/sqrt(x^2+y^2)

Good old hexagonal grid:
F=0.25+2*z1^2+(0.25*(cos(4*y1)+cos(2*sqrt(3)*x1-2*y1)+cos(2*sqrt(3)*x1+2*y1))+0.03*(cos(4*sqrt(3)*x1)+cos(6*y1-2*sqrt(3)*x1)+cos(6*y1+2*sqrt(3)*x1))) = 0
Limits:
-3.1..3.1
-3.1..3.1
-1.1..1.1

I'm skeptical about the grid density, it's gonna need something like 500x500x200, probably more.
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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Sat Mar 16, 2013 6:17 pm    Post subject: Reply with quote

Hi all,
I've made some changes to your formula Furan Smile
Code:
0.2/(1+2/(1+3*(atan2(abs(z),sqrt(x*x+y*y)-6)-pi)^2))+0.8*((( ((sqrt(x*x+y*y)-6)^2+z*z)-4))*4/sqrt(x*x+y*y))^2+(0.25*(cos(2*4*((atan2(z,(sqrt(x*x+y*y)-6)))*8/sqrt(x*x+y*y)))+cos(2*sqrt(3)*(atan2(y,x)*8)-4*((atan2(z,(sqrt(x*x+y*y)-6)))*8/sqrt(x*x+y*y)))+cos(2*sqrt(3)*(atan2(y,x)*8)+4*((atan2(z,(sqrt(x*x+y*y)-6)))*8/sqrt(x*x+y*y))))+0.03*(cos(4*sqrt(3)*(atan2(y,x)*8))+cos(6*2*((atan2(z,(sqrt(x*x+y*y)-6)))*8/sqrt(x*x+y*y))-2*sqrt(3)*(atan2(y,x)*8))+cos(6*2*((atan2(z,(sqrt(x*x+y*y)-6)))*8/sqrt(x*x+y*y))+2*sqrt(3)*(atan2(y,x)*8))))   =0
[x]: -8.5 , 8.5
[y]: -8.5 , 8.5
[z]: -4 , 4




We are eager to know your idea Furan, I think it's making 3d fractals by making patterns many times (making patterns over a surface, then making the same patterns over the new surface, then making the same patterns over the new surface, ......) Smile
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inode



Joined: 27 Jan 2007
Posts: 127
Location: Austria

PostPosted: Mon Mar 18, 2013 2:45 pm    Post subject: Hexagonal Grid Torus Reply with quote

Hi all,
I was deriving the formula for a hexagonal grid torus too
and I got a simpler term. Here it is...

Code:
Name: PG_HexaGridTorus1
/* Isosurface: PG_HexaGridTorus1  12/2012  Gerd Platl */
F(): -0.5
     -44*(sqrt( (sqrt(x*x+y*y)-5)^2 +z*z)-2)^2
     -cos(                             48*atan2(y,x))
     -cos(14*atan2(z,sqrt(x*x+y*y)-5) -24*atan2(y,x))
     -cos(14*atan2(z,sqrt(x*x+y*y)-5) +24*atan2(y,x))
-0.12*( cos(               28*atan2(z,sqrt(x*x+y*y)-5))
       +cos(24*atan2(y,x) -14*atan2(z,sqrt(x*x+y*y)-5))
       +cos(24*atan2(y,x) +14*atan2(z,sqrt(x*x+y*y)-5))
      )
[x]: -7.2 , 7.2
[y]: -7.2 , 7.2
[z]: -2.2 , 2.2
;


Last edited by inode on Mon Mar 25, 2013 7:05 pm; edited 1 time in total
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inode



Joined: 27 Jan 2007
Posts: 127
Location: Austria

PostPosted: Mon Mar 18, 2013 2:54 pm    Post subject: Hexagonal Grid Cylinder Reply with quote

Hi all,
I also was creating the formula for the hexagonal grid cylinder...

Code:
Name: PG_HexaGridCylinder1
/* Isosurface: PG_HexaGridCylinder1  12/2012  Gerd Platl
   coat formula taken from PG_HexaGrid1
   4.04 is used to connect coat at 360°
   4.04*2*sqrt(3) =~ 13.995 =~ 14.0
*/
F(): -0.5
     -88*(sqrt(x*x+y*y)-2)^2
             -cos(               8*z)
             -cos(14*atan2(y,x) -4*z)
             -cos(14*atan2(y,x) +4*z)     
-0.12*( cos(      28*atan2(y,x))
       +cos(12*z -14*atan2(y,x))
       +cos(12*z +14*atan2(y,x)))
[x]: -2.1 , 2.1
[y]: -2.1 , 2.1
[z]: 0 , 5
;

and a variation of it...

Code:
Name: PG_HexaGridCylinder2
/* Isosurface: PG_HexaGridCylinder2  12/2012  Gerd Platl */
F(): 1.0
     -88*(sqrt(x*x+y*y)-2)^2
              -cos(               8*z)
              -cos(14*atan2(y,x) -4*z)
              -cos(14*atan2(y,x) +4*z)     
- 0.12*( cos(      28*atan2(y,x))
        +cos(12*z -14*atan2(y,x))
        +cos(12*z +14*atan2(y,x)))
[x]: -2.3 , 2.3
[y]: -2.3 , 2.3
[z]: 0 , 5
;

and a 2nd variation with imaginary circles build from the pillow elements...

Code:
Name: PG_HexaGridCylinder3
/* Isosurface: PG_HexaGridCylinder3  12/2012  Gerd Platl */
F(): -0.55
     -88*(sqrt(x*x+y*y)-2)^2             
            -cos(               8*z)             
            -cos(14*atan2(y,x) -4*z)             
            -cos(14*atan2(y,x) +4*z)     
-0.6*( cos(      28*atan2(y,x))
      +cos(12*z -14*atan2(y,x))       
      +cos(12*z +14*atan2(y,x)))
[x]: -2.2 , 2.2
[y]: -2.2 , 2.2
[z]: 0 , 5
;
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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Thu Apr 25, 2013 10:20 pm    Post subject: Reply with quote

Hi all, it's just a crazy work you would say Smile
Code:
F(): min(min(min( min( min( max(max( (sqrt(x^2+z^2)-1)^2+y^2-0.16,-x),-z),max(abs(x+1)-1,((x+1)^2+y^2+(z-1)^2-0.25)*(y^2+(z-1)^2-0.36)-0.03 )) ,max(sqrt((x-1)^2+y^2)-(0.65+min(-0.05,cos(30*atan2(y,(x-1)))/10) ) ,abs(3*(z+0.2))-1) ),max(max(abs(sqrt((x-1)^2+y^2)-0.35)-0.05,z+0.2),-(z+1.3)) ) ,min(min( max((x+1)^2+(z-2.5)^2-(0.2+0.1*abs(y))^2,abs(y)-0.5),(x+1)^2+y^2+(z-2.5)^2-0.0625 ) ,max(max((x+1)^2+y^2-0.0225,z-2.5),-(z-2.5+1.2)) )  ) ,max(y^2+(z-1)^2-0.25,x+1) )

[x]: -3.5 , 2.1
[y]: -1 , 1
[z]: -1.4 , 3
;


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denisc



Joined: 24 Apr 2013
Posts: 92

PostPosted: Mon Apr 29, 2013 3:15 pm    Post subject: Reply with quote

hello Very Happy
i am new .
thanks to all.
i am very impressive!!!
i don't know , it"s possible to wite min(min(min ect !

Can you explain, a little , how construct dodecaedron and icosaedron with

iso formula ? and also pentagon curve or solid pentagon

like a little of that :
( (abs(y)-1/(1+exp(7*(abs(x)-1))))*cos(atan(-7*exp(7*(abs(x)-1))
/(1+exp(7*(abs(x)-1)))^2)) )^2+z^2-0.04

i am very curious to understand that?

thanks in advance


cheers

denis
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Furan



Joined: 05 Oct 2010
Posts: 64
Location: Prague, Czech Republic

PostPosted: Tue Apr 30, 2013 11:08 am    Post subject: Reply with quote

I'd use something like this:
(Use different approach for tetrahedron.)

F(x,y,z) = max(f1,f2,...,fN)

N=3 for cube
N=6 for dodecahedron
N=10 for icosahedron

f1 = R - abs(X*n1)
f2 = R - abs(X*n2)
f3 = ...

where
X=[x,y,z]
R = half distance between opposite faces (radius of the inscribed sphere)
n_i = normal UNIT vector of one of the face of the i-th pair.

Start by calculating the normal vectors. You can find some help here:
http://en.wikipedia.org/wiki/Dodecahedron
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denisc



Joined: 24 Apr 2013
Posts: 92

PostPosted: Tue Apr 30, 2013 5:25 pm    Post subject: Reply with quote

thank very much mr furan

Smile

it's not very easy to understand in one time.

i think it's near a same

abdelhamid belaid


formula for the Cube

Code:
max(max(abs(x),abs(y)),abs(y))-a =0
or in a mathematical writing: max(|x|,|y|,|z|)-a =0

i am a little trouble by the fact of max is the symbol of booleen intersection,

and i don't know , in final , where is a mathematical thinking and where is a 3D soft thinking specific language .

(it might be interesting to have more information or links for that)

when you write: x^2+y^2-1

the math language and soft is the same , and it"s easy

apart from that, I'll try to understand

cheers

den
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denisc



Joined: 24 Apr 2013
Posts: 92

PostPosted: Wed May 01, 2013 11:32 am    Post subject: Reply with quote

hello furan
what do you mean by

n_i = normal UNIT vector of one of the face of the i-th pair and ? and Start by calculating the normal vectors ?

the normal vector is calculate by/

|u|*|v|*sin of the angle ?

for dodecaedron

2*abs(2 /((1 + √5) / 2))*sin(arcos(-1/√5))

it's right?

in a begining, i supposed you mean length between 2 paire of faces?

another

it's impossible to write in k3d :

max(abs(1-x),abs(1-y),abs(1-z))

i suppose you mean


for the cube

max(max(abs(x),abs(y)),abs(z))

don't work


max(max(abs(x),abs(y)),abs(z))-1

work ?


for
N=3 for cube
X=[x,y,z]

but for
N=6 for dodecahedron

X=[x,y,z,?,?,?]


thanks for your help

denis
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Furan



Joined: 05 Oct 2010
Posts: 64
Location: Prague, Czech Republic

PostPosted: Wed May 01, 2013 9:42 pm    Post subject: Reply with quote

I made a tiny mistake, sorry for that.
F(x,y,z) = min(f1,f2,...,fN)

Normal unit vector is orthogonal to the face surface
n = (nx ny nz)
You should be able to calculate it with a bit of trigonometry. If not, there is an easier way:
http://en.wikipedia.org/wiki/Dodecahedron
Let φ = (1 + √5) / 2 be the golden ratio.

Let us calculate the middle point of the side facing us as the geometric center of its 5 corners:
The pink vertex at [φ,0,-1/φ]
The orange vertices at [1,1,-1] and [1,-1,-1]
The green vertices at [0,1/φ,-φ] and [0,-1/φ,-φ]
middle point M = 1/5 * [2+φ,0,-1/φ-2-2*φ]
n1 = (M-[0, 0, 0])/norm(M-[0, 0, 0])
where norm(M-[0, 0, 0]) = Ri (radius of inscribed sphere)
For points as presented above,
Ri= 0.5*sqrt(5/2 + 11/2*sqrt(5))
n1 = 1/5 * [2+φ, 0, -1/φ-2-2*φ] / Ri
f_i = R - abs(n_i * X)
where R is the inscribed radius you want, X=[x,y,z]

You can start with the cube
R=1
n1 = (1, 0, 0)
n2 = (0, 1, 0)
n3 = (0, 0, 1)
=>
f1 = 1-abs(x)
f2 = 1-abs(y)
f3 = 1-abs(z)
F = min(f1,f2,f3)
It should work now.
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