anubis2003

1st January 2003 16:48 UTC

Truncated Icosahedron
Soccer Balls in avs. I love it! It's just an idea, but I'm sure implementing it would be a huge task. If someone could explain some of the math for it(phi, etc.) I wouldn't mind trying it. The only problem I see is that it might look like just a bunch of lines, which could be solved by changing the color to black for the back faces and using a Max Blend.

FYI, A Truncated Icosahedron is a polyhedron with 32 faces (20 hexagons and 12 pentagons), 60 vertices, and 90 edges. It is also the shape of soccerballs and buckyballs(c-60 compounds).

Nic01

1st January 2003 18:30 UTC

Jheriko made an Icosidodecahedron (Or Triacontadihedron) before, and yes, he used phi a lot. If you want to take a personal look at it, go download his 7th pack (Not meant to be an advertisement, but hey, you'll get to see cool presets in the way). I believe he can help you the most in this topic...

For phi, use (1+sqrt(5))*0.5. I don't know much of other mathematical concepts, so I guess that's all I can do....As for coordinates... Well, do some googling =P

anubis2003

1st January 2003 18:46 UTC

Nic01,
Yeah, I've seen Jheriko's Icosidodecahedron, but the Truncated Icosahedron is different. I know what the numerical value is for phi, but I'm having trouble understanding how to apply it to find the coordinates for polyhedrons. I've tried running internet searches but the sites don't use phi to find the coordinates like Jheriko has in his polygons. I'm going to try to find a different way to do it, but if someone can help me with phi then that would be most appreciated. Wow, that last sentence sounded way too nice.

Jaheckelsafar

1st January 2003 21:04 UTC

Jeriko wrote a quick simple thing unsing spherical co-ordinates here.

anubis2003

1st January 2003 21:24 UTC

Thanx Jaheckelsafar, but I am trying to figure out more about how phi relates to "golden" ratios and coordinates.

nixa

1st January 2003 22:23 UTC

Anubis:
The golden ratio was first used in old Greek arhitecture, the relation is whole/biger part = biger part/smaller part or mathematicly writen 1/x=x/(1-x) when you solve this you will get roots x1=(1+sqrt(5))/2 and x2=(1-sqrt(5))/2 the number (1+sqrt(5))/2 is caled Phi and abs((1-sqrt(5))/2) phi. You can also get Phi by deviding infinite fibonaccy number whith a previous fibonaccy number
F(n)/F(n-1).
For he icohasedron coordinates make a google search.

anubis2003

1st January 2003 22:47 UTC

I was finally able to figure out how to use phi at
http://www.rwgrayprojects.com/Lynn/C...s/coord01.html.
I now have a phi-based tetrahedron.

dirkdeftly

2nd January 2003 07:26 UTC

phi is called the golden ratio because it's the ratio most pleasing to the human eye. it shows up everywhere in the human form; the size of the joints of each finger, the relationship between the width of the nose and the mouth, the width of the teeth (going back in the mouth), etc. it can be described thus: given a rectangle with sides phi and 1, the difference between the rectangle and a unit square is a scale replica of the rectangle. therefore we get this equation:
1/(phi-1)=phi/1
now solve for phi:
1/(phi-1)=phi
phi-1=1/phi
phi^2-1=phi
phi^2-phi-1=0
using the quadratic fomula:
phi=(1+-sqrt(1-4*(1*-1)))/2*1
phi=(1+-sqrt(1+4))/2
phi=(1+-sqrt(5))/2
to make phi positive, phi=(1+sqrt(5))/2.

i don't know how phi is used to find the coordinates of regular shapes (of all dimensions, apparantly).

ujay

2nd January 2003 07:48 UTC

A further note on phi. In a pentagon it is the ratio of a diagonal to a side. It is also the ratio of one of the diagonal 'arms' to the side of the inner 5gon.
This was not lost on the ancients who came to revere the pentaGRAM so formed for it's perfection.

UJ

anubis2003

3rd January 2003 00:34 UTC

Thanx for your help everyone, but I think I understand phi now. Still if anyone knows how to go about creating a truncated icosahedron I am still having trouble figuring it out. The most I have figured out is that regular polyhedrons have coordinates in the set {±phi,±phi^2,±phi^3,and for the 120-Polyhedron ±2phi^2}.

anubis2003

4th January 2003 21:52 UTC

I finally made the truncated icosahedron! It was actually a tad more difficult to make than the 120-polyhedron because of its nasty coordinates (the 120 used phi, phi^2, and phi^3 but the soccer ball uses 1/3+2*phi/3 and 2/3 + phi/3, and others). I am going to work on black and white shading to make it look like a real soccer ball. Any other suggestions are welcome.

nixa

5th January 2003 01:32 UTC

This comes a bit late but you could of done this much easyer. You just needed to make a regular icosahedron and turn the triangles into hexagons whith hexagon side length being half of the triangle side lenght. If you will make a solid version for the coloring you might want to do it this way.

dirkdeftly

5th January 2003 02:15 UTC

What the coordinates are shouldn't matter; just as long as they're inputed right.

anubis2003

5th January 2003 03:12 UTC

For once I agree 100% with you Atero - It might have been easier to find the coordinates originally, but now that I have them it would be pointless to find different ones.

dirkdeftly

5th January 2003 09:38 UTC

I found a site you'd be interested in:
http://astronomy.swin.edu.au/~pbourke/polyhedra/
:)

Yathosho

5th January 2003 13:52 UTC

Re: Truncated Icosahedron

Originally posted by anubis2003
Soccer Balls in avs. I love it!

nixa

5th January 2003 14:21 UTC

What I ment is since Jheriko already made a solid icosahedron you could just use his scopes and replace his triangles whith
init: pi=acos(-1);hnb=1.5;n=500;
point: r=(i+1)*pi;a=abs(tan(asin(sin(r*hnb))/hnb));
d=sqrt(1+sqr(min(10/a,a)))*0.3;u=1-u;
x=sin(r)*d*(u*2-1);y=cos(r)*d;

and than chainge the hnb variable from 1.5 to 3.

If you want to use sceleton object in the final version I know this wont be usefull.

anubis2003

5th January 2003 16:16 UTC

Thanks Nixa. I think I might try and see how that works - I misunderstood what you said.
Killahbite, if I called it football, then everyone from the U.S., which there are a lot in the forums (including myself), would say that it doesn't look at all like a football.

dirkdeftly

6th January 2003 00:10 UTC

and there's just about as many brits on the forum, so...

Raz

6th January 2003 01:49 UTC

so everyone knows what a soccer ball is as opposed to a football

Zevensoft

6th January 2003 01:58 UTC

Pfft, American football is for pansies, with their 'padding' and 'helmets'. Go play AFL.

And the weirdest football is gaelic football. It's like soccer mixed with AFL, ie. weird.

anubis2003

6th January 2003 03:09 UTC

A soccer ball is known by almost all people who know english. A football could be misinterpreted by many people in the forums. And Zevensoft, I would have to agree with you about american football being for pansies.

3dino

6th January 2003 04:27 UTC

Good work! :D:up:
Hope you can make it look like this soccerball (press reload if 403)

if you like to, code an American football with its typical flying rotation ;)

Raz

6th January 2003 04:44 UTC

ok enough about the difference between a football and a soccerball already, the god damned thing is made, who gives a crap what its called.

VisualAgnosia

8th January 2003 21:09 UTC

Volleyball has the best looking girls as a sport .. generally.

If only all footballers spent their time sk8ing...

jheriko

17th January 2003 18:18 UTC

why do i always miss threads which are blatantly aimed at me.

lets say that you can get the co-ords for a truncated icosahedron from a regular icosahedron. the reason that i use phi is because it makes it easier to code the superscopes, there is no more to it than that. phi occurs a lot in maths, especially geometry. the occurance of phi in the dodecahedron makes a lot of sense because of the pentagons, it also occurs in the icosahedron simply because it is the dual to the dodecahedron. It occurs in numerous semi-regular polyhedra because these shapes can often be derived from icosahedra and dodecahedra, sometimes in combination with other polyhedra.

For instance, if you wanted to find the vertex co-ordinates of a truncated icosahedron then you would start with the icosahedron and then figure out how 'far down' each vertex you must truncate in order to get triangles and pentagons of equal side length, once you know this then it boils down to doing 'vector type' math to find the x,y and z components of the line from the orignal vertex to the each of the five new ones in order to get the new co-ordinates from the old ones. What you will need to determine to do this is the internal angle of the icosahedron, which i can't remember of the top of my head but has something to do with phi too.

If you are still stuck then I can demonstrate how to do this for one vertex and then you can do the rest.

You can take the vertex co-ords from my icosahedron preset in my 'Purely Platonic' minipack.

anubis2003

17th January 2003 18:30 UTC

Thanks Jheriko, I finally figured it out though, I posted the truncated icosahedron already in this thread, as well as a 120-polyhedron in another thread.

jheriko

17th January 2003 18:52 UTC

hehe... i just assumed that you still hadn't figured it out. sorry :P