Some pictures of 3D representations by K3DSurf 0.62
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abdelhamid belaid



Joined: 13 Aug 2009
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PostPosted: Sun Jun 05, 2011 12:40 pm    Post subject: Reply with quote

It's not a new idea, any time my friends, I welcome your ideas and suggestions, I will be very glad to do something for others Smile




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Last edited by abdelhamid belaid on Sun May 19, 2013 5:39 pm; edited 1 time in total
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abdelhamid belaid



Joined: 13 Aug 2009
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PostPosted: Mon Jun 06, 2011 11:41 pm    Post subject: Reply with quote

a torus:


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abdelhamid belaid



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PostPosted: Thu Jun 09, 2011 12:31 am    Post subject: Reply with quote

Hi all,
for me this is the first pattern making over z=x*y surface Very Happy



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abdelhamid belaid



Joined: 13 Aug 2009
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PostPosted: Sat Jun 11, 2011 1:47 pm    Post subject: Reply with quote

here are patterned surfaces ( pattern(x,y)^2+f(x,y,z)^2-R^2=0 or generally F(pattern(x,y) , f(x,y,z))=0 )







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abdelhamid belaid



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PostPosted: Fri Jun 17, 2011 11:44 am    Post subject: Reply with quote

Hi all,
It's just over new surfaces Smile



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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Tue Jul 12, 2011 5:52 pm    Post subject: Reply with quote

a torus:

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Last edited by abdelhamid belaid on Sun May 19, 2013 5:56 pm; edited 3 times in total
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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Sun Jul 24, 2011 5:16 pm    Post subject: Reply with quote

Hi all,
A dodecahedron, of course we can get as much as we want faces (F(x,y,z,n)=0) Smile



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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Sat Aug 06, 2011 2:40 am    Post subject: Reply with quote

taking straight lines instead of a single curved line:


over z = sin(x)





z = x^2


z = x*sin(x/3)/3


z = y*x^2/5





z = y*x^2/10


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abdelhamid belaid



Joined: 13 Aug 2009
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PostPosted: Sat Aug 06, 2011 11:05 pm    Post subject: Reply with quote

this is more explanation about my last method to make patterns over a z = f(x,y) surface:
Let's call our step h, the new normal dimension N or N(f) or N(f, h) , we have z = 3/(1.5+x^2) (of course our formula is N(f, h)^2+y^2-R^2=0) I will every time change h (1, 1/2, 1/6, 1/10, 1/20, 1/100), I will get more and more a smooth shape (curved line), that means I will get a normal dimension (section is always orthogonal to the tangent or we can say to our surface) :

h=1


h=0.5


h=0.25


h=0.166666666


h=0.1


h=0.05


h=0.01



more works:
z = y*(x/4)^3



z= 3/(1.5+x^2+y^2)



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abdelhamid belaid



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PostPosted: Mon Aug 08, 2011 12:06 am    Post subject: Reply with quote

normal dimension:

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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Wed Aug 10, 2011 1:59 am    Post subject: Reply with quote

a tetrahedron Smile :



Code:
max( z-2+tan(pi/3)*abs(x)  ,  tan(pi/3)*abs(y)-z-2  )


[x]: -3 , 3
[y]: -3 , 3
[z]: -2 , 2


I mean generally:
max( z-h+tan(pi/3)*abs(x)  ,  tan(pi/3)*abs(y)-z-h  )

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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Wed Aug 10, 2011 6:14 pm    Post subject: Reply with quote

Also other patterned surfaces by the same method:

z = 2*sin(x)^3*cos(y)



z = sin(x)^3*cos(y)+cos(2*sqrt(x^2+y^2))/2



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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Thu Aug 11, 2011 9:15 pm    Post subject: Reply with quote

another work:



For a better view of the recent work:
http://www.youtube.com/watch?v=zRn0Hq4fAgo
http://www.youtube.com/watch?v=pA3h8EOgK-Q


Some fun:


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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Sat Aug 13, 2011 12:01 am    Post subject: Reply with quote

Now the same method and the same pattern over the surface sqrt(x^2+y^2) - 3 -sin(6*atan2(y,x))*cos(2*z)/3 =0 , as you see I've used the same method in polar coordinates Smile






also over sqrt(x^2+y^2) - 3 - cos(6*atan2(y,x)+z)/3 = 0





You can see them on YouTube:
http://www.youtube.com/watch?v=4AGQOOJvFJU
http://www.youtube.com/watch?v=0Fu66K0XsjM
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Last edited by abdelhamid belaid on Sun May 19, 2013 7:09 pm; edited 1 time in total
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abdelhamid belaid



Joined: 13 Aug 2009
Posts: 170

PostPosted: Tue Aug 16, 2011 1:02 pm    Post subject: Reply with quote

The same pattern but it's around the origin this time Smile
It's over: z - cos(sqrt(x^2+y^2))*sin(x) = 0


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