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Furan

Joined: 05 Oct 2010
Posts: 64
Location: Prague, Czech Republic Posted: Thu May 16, 2013 11:26 am    Post subject: I'm afraid you made an error in there. I don't even recognize where that alpha is comming from. a=edge length GoldenRatio phi=(1+sqrt(5))/2 Let's look at the top pyramid of 5 triangles: Circle circumscribed onto its base: R = sqrt((5+sqrt(5))/10) * a = sqrt((2+phi)/5) * a Circle inscribed in its base r = sqrt((5+2*sqrt(5))/20) * a = sqrt((3+4*phi)/20) * a In your case, alpha is the angle of the face normal from the z-axis, that is cos(alpha) = r/v; v = sqrt(3)/2 * a = triangle height cos(alpha) = sqrt((5+2*sqrt(5)/15) = sqrt((3+4*phi)/15) = 0.79465 sin(alpha) = sqrt((10-2*sqrt(5)/15) = sqrt((12-4*phi)/15) = 0.60706 Now let's look at the 10 triangles in the middle row: cos(beta) = (R-r) / v = 0.18759 sin(beta) = 1-cos^2(beta) = 0.98225 So the number you were looking for is beta = 1.382086 rad    abdelhamid belaid

Joined: 13 Aug 2009
Posts: 170 Posted: Thu May 16, 2013 10:24 pm    Post subject: Hello my best friends, thanks Gerd, thanks Furan, I would like to add (another method): firstly, for the first angle, according to my calculations it is P=0.652358 for the second angle Gerd asked about, this is a method where we won't use the golden ratio: let A , B , C , D are four vertices as shown on the first picture down, let alpha is the angle made by the vectors OA and OB (or by the origin and any two other vertices for the same edge), the equation of the plane ABC is x*sin(P) + z*cos(P) + c = 0 we have: A(0 , 0 , h) B(h*sin(alpha)*cos(pi/5) , h*sin(alpha)*sin(pi/5) , h*cos(alpha)) C(h*sin(alpha)*cos(pi/5) , -h*sin(alpha)*sin(pi/5) , h*cos(alpha)) D(h*sin(alpha) , 0 , -h*cos(alpha)) A belongs to ABC, that means 0*sin(P) + h*cos(P) + c = 0 so c= - h*cos(P) so the ABC equation becomes x*sin(P) + z*cos(P) - h*cos(P) = 0 now let's calculate alpha, since B belongs to ABC we have h*sin(alpha)*cos(pi/5)*sin(P) + h*cos(alpha)*cos(P) - h*cos(P) = 0 so sin(alpha)*cos(pi/5)*sin(P) + cos(alpha)*cos(P) - cos(P) = 0 here is alpha = 1.107148 (I call it the golden angle) finally we can calculate "u" an orthogonal vector for the BCD plane and then calculate the second angle T = arccos(u*k / ||u||) we can also using "E" the middle of BC, in this case T = 2*(OE , k) - P also we can use "G" the centroid of the triangle BCD, here is T = (OG , k) also we can directly write the BCD plane equation, we have but to solve a system of equations: (a*xB+b*yB+c*zB+1=0 , a*xC+b*yC+c*zC+1=0 , a*xD+b*yD+c*zD+1=0) , there are many other methods in fact.  _________________My YouTube channelLast edited by abdelhamid belaid on Fri May 17, 2013 11:16 am; edited 1 time in total     abdelhamid belaid

Joined: 13 Aug 2009
Posts: 170 Posted: Fri May 17, 2013 1:07 am    Post subject: I must add, I have made a tiny mistake when writing my dodecahedron formulas, the angle I should write is alpha = 1.107148 instead of 1.07498, I'm sorry.
 Code: Name: AB_dodecahedron F(): max(max( sin(1.107148)*sqrt(x*x+y*y)*cos(atan(tan(2.5*atan2(y,x)))/2.5)+cos(1.107148)*z , sin(1.107148)*sqrt(x*x+y*y)*cos(atan(tan(2.5*(atan2(y,x)+pi/5)))/2.5)-cos(1.107148)*z   ) , abs(z) )- 3 [x]: -3.8 , 3.8 [y]: -3.6 , 3.6 [z]: -3.1 , 3.1 ;

Last edited by abdelhamid belaid on Sat May 18, 2013 8:54 pm; edited 1 time in total     abdelhamid belaid

Joined: 13 Aug 2009
Posts: 170 Posted: Fri May 17, 2013 1:35 am    Post subject: I have one more question, as you see we didn't talk about the first angle P, how would us calculate it ? I mean I want to restore our issues to their origins, in our methods we have used known values (I used P and Furan used the golden ratio), how could us calculate P or the golden ratio? if we have alpha we can easily get P, so how would us calculate alpha? I mean here we need to return to the basic definition of the icosahedron or the dodecahedron ? what are they ? _________________My YouTube channel     inode Joined: 27 Jan 2007
Posts: 127
Location: Austria Posted: Mon May 20, 2013 12:52 pm    Post subject: touching balls Quote: abdelhamid belaid wrote... I must add, ... alpha = 1.107148 instead of 1.07498.

Hi all
Using alpha = 1.107148 seems to be the right angle. Rearranging the formula result in 12 touching balls. There you can compare the empty space between the spheres visually. All the holes looks equal now!
 Code: /* Isosurface: PG_DodecaBalls1 5/2013 12 Touching Balls at dodecahedron polygon middle points */ max(max(   sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5* atan2(y,x) ))/2.5)+cos(1.107148)*z  ,sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5*(atan2(y,x)+pi/5)))/2.5) -cos(1.107148)*z )  ,abs(z))  -(x*x+y*y+z*z) -0.177  Gerd   abdelhamid belaid

Joined: 13 Aug 2009
Posts: 170 Posted: Mon May 20, 2013 11:15 pm    Post subject: very nice, Gerd some variations:
 Code: Name: AB_dodecahedron7 F(): max(max(   sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5* atan2(y,x) ))/2.5)+cos(1.107148)*z ,sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5*(atan2(y,x)+pi/5)))/2.5) -cos(1.107148)*z ) ,abs(z)) -(x*x+y*y+z*z) [x]: -2 , 2 [y]: -2 , 2 [z]: -2 , 2 ; Code: Name: AB_dodecahedron8 F(): max(max(   sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5* atan2(y,x) ))/2.5)+cos(1.107148)*z ,sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5*(atan2(y,x)+pi/5)))/2.5) -cos(1.107148)*z ) ,abs(z))^3 -(x*x+y*y+z*z) [x]: -2 , 2 [y]: -2 , 2 [z]: -2 , 2 ; Code: Name: AB_dodecahedron9 F(): (max(max(   sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5* atan2(y,x) ))/2.5)+cos(1.107148)*z ,sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5*(atan2(y,x)+pi/5)))/2.5) -cos(1.107148)*z ) ,abs(z) )+0.18)^3 -(x*x+y*y+z*z) [x]: -1.2 , 1.2 [y]: -1.2 , 1.2 [z]: -1.2 , 1.2 ; Code: Name: AB_dodecahedron10 F(): (max(max(  sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5* atan2(y,x) ))/2.5)+cos(1.107148)*z ,sin(1.107148) *sqrt(x*x+y*y) *cos(atan(tan(2.5*(atan2(y,x)+pi/5)))/2.5) -cos(1.107148)*z ) ,abs(z) )+0.2)^3 -(x*x+y*y+z*z) [x]: -1 , 1 [y]: -1 , 1 [z]: -1 , 1 ; _________________     denisc

Joined: 24 Apr 2013
Posts: 92 Posted: Thu May 23, 2013 11:34 pm    Post subject: thanks very much AB_dodecahedron8 look like hyperbolic dodecaedron ! i find that a lot of time. with more points on top perhaps denisc    abdelhamid belaid

Joined: 13 Aug 2009
Posts: 170 Posted: Tue Jun 25, 2013 10:17 pm    Post subject: thanks Denis, anyway it's an old work, let's break the silence  _________________My YouTube channel     Furan

Joined: 05 Oct 2010
Posts: 64
Location: Prague, Czech Republic Posted: Tue Jun 25, 2013 10:56 pm    Post subject: Exquisite ! Made my head spin Anyway, how fine can you make the grid? Could you apply a uniform transformation for that (I posted it here somewhere for the hexgrid toroid) with all the features visible? The features closer to center would get really tiny. Or are there any such options when exporting mesh to POVray? I'm struggling with the same problem with the software that I use. I tried matlab and it was promissing. I managed to visualise much finer grids. Speaking of matlab, this is just a teaser of what I did recently (actually a real science for a change): Once I finish this research, I will create more realistic equations of knitwear.    denisc

Joined: 24 Apr 2013
Posts: 92 Posted: Fri Sep 12, 2014 9:56 am    Post subject: hello thanks for your help.    Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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