 |
|
|
|
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Help
I need the xyz co-ordinates for the vertices of a dodecahedron
or the method for calculating them.
Hope someone can help
Mick Hazelgrove
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Mick Hazelgrove wrote:
> Help
>
> I need the xyz co-ordinates for the vertices of a dodecahedron
> or the method for calculating them.
>
> Hope someone can help
>
> Mick Hazelgrove
I did a quick yahoo search with the keyword in quotes "icosahedron"
and yahoo retuned 2 web sites and an amazing # of web pages.
The second web page had a nightmarish amount of information.
Check out:
http://www.mathconsult.ch/showroom/unipoly/unipoly.html#VertexConfig
The dodecahedron was also mentioned and detailed at the above site.
If that doesn't work for you try the yahoo search engine.
Ken Tyler
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Mick Hazelgrove wrote:
>
> Help
>
> I need the xyz co-ordinates for the vertices of a dodecahedron
> or the method for calculating them.
#local A=sqrt(5)+3;
#local B=sqrt(5)+1;
#local C=2;
#local R=sqrt(A*A+B*B+C*C);
#local A=A/R;
#local B=B/R;
#local C=C/R;
#local Point00=< B, B, B>;
#local Point01=<-B, B, B>;
#local Point02=< B,-B, B>;
#local Point03=<-B,-B, B>;
#local Point04=< B, B,-B>;
#local Point05=<-B, B,-B>;
#local Point06=< B,-B,-B>;
#local Point07=<-B,-B,-B>;
#local Point08=< A, C, 0>;
#local Point09=< 0, A, C>;
#local Point10=< C, 0, A>;
#local Point11=<-A, C, 0>;
#local Point12=< 0,-A, C>;
#local Point13=< C, 0,-A>;
#local Point14=< A,-C, 0>;
#local Point15=< 0, A,-C>;
#local Point16=<-C, 0, A>;
#local Point17=<-A,-C, 0>;
#local Point18=< 0,-A,-C>;
#local Point19=<-C, 0,-A>;
--
"Have the manners not to be hittin' the man until he's your husband, and
entitled to hit back!"
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Is there any way to explain this in more generalized terms? I was looking
for information like this recently... But I don't understand where you are
getting the numbers you're using. Also, is it possible to explain how to get
stuff like tetrahedrons, and octahedrons, etc, etc?
--
Twyst
============================================================
for pov-ray news, reviews, and tutorials
http://twysted.net
e-mail: twy### [at] twysted net
============================================================
John VanSickle wrote in message <362A5D8E.81FDEDFF@erols.com>...
>Mick Hazelgrove wrote:
>>
>> Help
>>
>> I need the xyz co-ordinates for the vertices of a dodecahedron
>> or the method for calculating them.
>
<snipped John's code here>
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Twyst wrote:
> Is there any way to explain this in more generalized terms? I was looking
> for information like this recently... But I don't understand where you are
> getting the numbers you're using. Also, is it possible to explain how to get
> stuff like tetrahedrons, and octahedrons, etc, etc?
>
> --
> Twyst
Yea John, what Twyst said.
The web page link I posted earlier has a considerable amount
of information available about all the known uniform polyhedra
constructs and gives an overly brief explanation on how to
determine the vertices from data represented by the Wythoff symbol.
He explains it as:
All but one of the uniform polyhedra can be described by use of
Wythoff symbol. The Wythoff symbol contains three rational numbers
p, q, and r, all larger than 1. If two of these numbers are equal
to 2, the third one is arbitrary, otherwise only the numerators
2, 3, 4, and 5 can occur, and 4 and 5 cannot occur together. There
are only finitely many ways to choose such p, q, and r. There are
four kinds of Wythoff symbols (plus one exceptional case)
The dodecahedron has the formula 3|2 5, therefore, the vertex
configuration is {2, 5, 2, 5, 2, 5}. After removing trivial
faces, simply {5, 5, 5} remains, that is, 3 pentagons.
The vertex configuration is the sequence of faces arranged around
a vertex. Since vertices are congruent, this sequence is the same for
all vertices. A regular n-sided polygon (an n-gon) is described by n.
Star polygons are described by n/d, where n is the number of vertices,
connected d apart. For example, 5/2 is the pentagram. Some polyhedra
contain retrograde faces. For example, 4/3 is a square, traversed in
the opposite direction. A regular 2-gon is degenerate and can be left
out if it occurs in the formulae.
This all may make perfect sense to the author, and yourself, but
it lost on me. If you could adapt the Wythoff symbols data to your
example and work out one Pov example it would be appreciated.
Ken Tyler
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Twyst wrote:
>
> Is there any way to explain this in more generalized terms? I was looking
> for information like this recently... But I don't understand where you are
> getting the numbers you're using. Also, is it possible to explain how to get
> stuff like tetrahedrons, and octahedrons, etc, etc?
They way I did it was by taking a twelve-sided die and noting how the
points were oriented around each other. There are, of course, an
infinite number of ways to hold the die when doing this, but the easiest
way was to orient it so that eight of the points were are the points of a
cube, and the other points lay in the x-y, y-z, or x-z coordinate planes.
Then remembering that each edge is of the same length, I calculated the
points where they would be if the edges were each one unit in length.
I actually did this while extrapolating the dodecahedron to four dimensions
(which resultant shape contains 600 points, 720 sides, 1200 edges and
120 dodecahedrons).
You can calculate the icosahedron and tetrahedron in a similar manner.
The points for the icosahedron are posted both in this newsgroup and in
my rock include file (http://www.erols.com/vansickl/rock.htm) One set of
points for the tetrahedron works out to:
#local Point0=< .5,0,sqrt(.5)>;
#local Point1=<-.5,0,sqrt(.5)>;
#local Point2=<0, .5,sqrt(.5)>;
#local Point3=<0,-.5,sqrt(.5)>;
(The interested student is invited to check these results.)
The points for the octahedron are simplicity itself,
#local Point0= x;
#local Point1=-x;
#local Point2= y;
#local Point3=-y;
#local Point4= z;
#local Point5=-z;
as are those for the cube:
#local Point0= x+y+z;
#local Point1=-x+y+z;
#local Point2= x-y+z;
#local Point3=-x-y+z;
#local Point4= x+y-z;
#local Point5=-x+y-z;
#local Point6= x-y-z;
#local Point7=-x-y-z;
Somewhere else in this thread someone has quoted some egghead theorems
from some egghead book, but that's not where I got these figures.
Regards,
John
--
"Have the manners not to be hittin' the man until he's your husband, and
entitled to hit back!"
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Ken wrote:
>
> Twyst wrote:
>
> > Is there any way to explain this in more generalized terms? I was looking
> > for information like this recently... But I don't understand where you are
> > getting the numbers you're using. Also, is it possible to explain how to get
> > stuff like tetrahedrons, and octahedrons, etc, etc?
> >
> > --
> > Twyst
>
> Yea John, what Twyst said.
>
> The web page link I posted earlier has a considerable amount
> of information available about all the known uniform polyhedra
> constructs and gives an overly brief explanation on how to
> determine the vertices from data represented by the Wythoff symbol.
>
> He explains it as:
>
> All but one of the uniform polyhedra can be described by use of
> Wythoff symbol. The Wythoff symbol contains three rational numbers
> p, q, and r, all larger than 1. If two of these numbers are equal
> to 2, the third one is arbitrary, otherwise only the numerators
> 2, 3, 4, and 5 can occur, and 4 and 5 cannot occur together. There
> are only finitely many ways to choose such p, q, and r. There are
> four kinds of Wythoff symbols (plus one exceptional case)
>
> The dodecahedron has the formula 3|2 5, therefore, the vertex
> configuration is {2, 5, 2, 5, 2, 5}. After removing trivial
> faces, simply {5, 5, 5} remains, that is, 3 pentagons.
>
> The vertex configuration is the sequence of faces arranged around
> a vertex. Since vertices are congruent, this sequence is the same for
> all vertices. A regular n-sided polygon (an n-gon) is described by n.
> Star polygons are described by n/d, where n is the number of vertices,
> connected d apart. For example, 5/2 is the pentagram. Some polyhedra
> contain retrograde faces. For example, 4/3 is a square, traversed in
> the opposite direction. A regular 2-gon is degenerate and can be left
> out if it occurs in the formulae.
>
> This all may make perfect sense to the author, and yourself, but
> it lost on me. If you could adapt the Wythoff symbols data to your
> example and work out one Pov example it would be appreciated.
It makes vague sense to me; fortunately it isn't necessary to know it
in order to find the points of the regular polyhedra. For that, all
you need to know is how the points lie in relation to each other, and
apply a little Pythagoras.
Regards,
John
--
"Have the manners not to be hittin' the man until he's your husband, and
entitled to hit back!"
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Cool. Thanx! Now I can see about generating some objects for Moray. =)
(CSG-able polyhedra =) )
I just have to figure out how to build it so that it uses planes, instead of
triangles. =)
--
Twyst
============================================================
for pov-ray news, reviews, and tutorials
http://twysted.net
e-mail: twy### [at] twystednet
============================================================
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
On Tue, 20 Oct 1998 10:40:32 -0600, Twyst <twy### [at] twystednet> wrote:
>Cool. Thanx! Now I can see about generating some objects for Moray. =)
>
>(CSG-able polyhedra =) )
>
>I just have to figure out how to build it so that it uses planes, instead of
>triangles. =)
The info you desire can be found in shapes2.inc, included with the POV
distribution.
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Hey,
Here's a site with tons of info on geodesics... not sure if it's what you're
looking for, but check it out.
http://www.li.net/~george/virtual-polyhedra/vp.html
And here's a link to a program that will create geodesic POV files.
http://www.cris.com/~rjbono/html/domes.html
Eric
--------------------------------
http://www.geocities.com/SiliconValley/Heights/2354/
http://www.datasync.com/~reba/
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
|
|
