K3DSurf forum

Furan · 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

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.

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abdelhamid belaid · Joined: 13 Aug 2009 Posts: 170

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.

abdelhamid belaid · Joined: 13 Aug 2009 Posts: 170

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 ? Smile

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inode · Joined: 27 Jan 2007 Posts: 127 Location: Austria

abdelhamid belaid · Joined: 13 Aug 2009 Posts: 170

very nice, Gerd Smile

some variations:

denisc · Joined: 24 Apr 2013 Posts: 92

thanks very much

AB_dodecahedron8

look like hyperbolic dodecaedron !

i find that a lot of time.

Cool

with more points on top perhaps
denisc

abdelhamid belaid · Joined: 13 Aug 2009 Posts: 170

thanks Denis,
anyway it's an old work, let's break the silence Smile

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Furan · Posted: Tue Jun 25, 2013 10:56 pm Post subject:

Exquisite ! Made my head spin Very Happy

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

hello

thanks for your help.