Golden polyhedron
Dan in NY
On the NY Times web site, there is a story and picture about a hollow
molecule composed of 16 atoms of gold. It looks like the figure is an
irregular polyhedron where some vertices have six triangles meeting and
others have four. The article is "16 Golden Atoms in Search of a Catchy
Name" By KENNETH CHANG; Published: May 23, 2006; at URL
http://www.nytimes.com/2006/05/23/science/23find.html.
A related web site is named, called "Evidence of Hollow Golden Cages", URL
http://www.nytimes.com/2006/05/23/science/23find.html, Published online
before print May 19, 2006. It calls the molecules '("bucky gold")', a name
that I suggest could catch on. However, the name of NY Times article
suggests that "someone" is still looking for a catchy name.
I am thinking of the geometric aspects of the molecule. It has 16 gold
atoms: 16 vertices. Since I have trouble visualizing a 3-D figure from a
2-D drawing, I have questions: How many faces does it have? Are they all
equilateral triangles? How many meet at each vertex? (If four and six, how
many vertices have four and how many have six triangles?) I presume that
Euler's Formula: V + F = 2 + E applies. Does it? How many edges does it
have? Are there other similar questions to be asked?
--
Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com)
Lee Rudolph
>Greetings,
>
>On the NY Times web site, there is a story and picture about a hollow
>molecule composed of 16 atoms of gold. It looks like the figure is an
>irregular polyhedron where some vertices have six triangles meeting and
>others have four. The article is "16 Golden Atoms in Search of a Catchy
>Name" By KENNETH CHANG; Published: May 23, 2006; at URL
>http://www.nytimes.com/2006/05/23/science/23find.html.
>
>A related web site is named, called "Evidence of Hollow Golden Cages", URL
>http://www.nytimes.com/2006/05/23/science/23find.html, Published online
>before print May 19, 2006. It calls the molecules '("bucky gold")', a name
>that I suggest could catch on. However, the name of NY Times article
>suggests that "someone" is still looking for a catchy name.
>
>I am thinking of the geometric aspects of the molecule. It has 16 gold
>atoms: 16 vertices. Since I have trouble visualizing a 3-D figure from a
>2-D drawing, I have questions: How many faces does it have?
Well, of course, "it" doesn't have "faces" at all. What "it"
(an instance of such a molecule) has, allowing minor idealization,
is vertices (atoms), and then, allowing considerably more idealization,
edges (bonds). There is no physical structure that corresponds
to faces (unless some Idiotic Micromanager has been sneaking around
behind my back), is there? Leaving physics aside, unless you know
that all the vertices lie on the boundary of the convex hull of
the set of vertices, I would again say that "it"--a network of
vertices and edges in space--has no natural features that can
unequivocally be identified as its "faces" (even if you do know
that the vertices all lie on the boundary of the convex hull, I'd
be cautious).
>Are they all
>equilateral triangles? How many meet at each vertex? (If four and six, how
>many vertices have four and how many have six triangles?) I presume that
>Euler's Formula: V + F = 2 + E applies. Does it?
And that's one big reason to *be* cautious. Why should it? That is,
why should you assume that the atoms-and-bonds lie, naturally, on
the boundary of a genus-0 surface, at all? Why not a torus? (I
haven't looked at the actual picture, mind you. Maybe a torus looks
really, really unlikely for those vertices. Still, there *are*
perfectly nice polyhedral tori in R^3 with 16 vertices all on the
boundary of their convex hull--heck, all on a round sphere, if you
like--and 16 plane quadrilateral faces.)
>How many edges does it
>have? Are there other similar questions to be asked?
Lee Rudolph
pyt...@gmail.com
> Greetings,
>
> On the NY Times web site, there is a story and picture about a hollow
> molecule composed of 16 atoms of gold. It looks like the figure is an
> irregular polyhedron where some vertices have six triangles meeting and
> others have four. The article is "16 Golden Atoms in Search of a Catchy
> Name" By KENNETH CHANG; Published: May 23, 2006; at URL
> http://www.nytimes.com/2006/05/23/science/23find.html.
>
If you look again, I think you'll see that it is made up of 6-triangle
vertices and 5-triangle vertices. There are no 4-triangle vertices. I
believe that this is true of any geodesic dome type structure.
Walter Whiteley
5, the other four of degree 6. There ARE geodesic domes with 6
vertices of degree 4, and the rest of degree 6, but this is not one of
them!
Walter
Dan in NY
news:e4v2sf$9av$1...@panix2.panix.com...
> "Dan in NY" <Som...@NY.net> writes:
>
>>Greetings,
>>
>>On the NY Times web site, there is a story and picture about a hollow
>>molecule composed of 16 atoms of gold. It looks like the figure is an
>>irregular polyhedron where some vertices have six triangles meeting and
>>others have four. The article is "16 Golden Atoms in Search of a Catchy
>>Name" By KENNETH CHANG; Published: May 23, 2006; at URL
>>http://www.nytimes.com/2006/05/23/science/23find.html.
Greetings Lee and other readers,
Thank you for your reply to my post. Of course you are correct in what you
say. I am sorry you didn't see the figure. I have replied to my first post
with a figure for you, one posted and one attached. I hope you can look
there and see the figure one way or another.
Maybe I played a fast one trying to "announce" a 16 atom "molecule" of gold
and then switch to a 3-D geometric figure. Maybe, because I could see the
figure, I thought readers would be able to see it as a wire model of a
polyhedron. So here goes to define or fill in some of the steps:-
First there are the positions of the 16 gold atoms. The figure shows a
symmetric convex figure which is nearly sphere (of the proper size) --
similar to the way the carbon atoms of a bucky ball are nearly a sphere.
The figure shows the gold bonds to consist of triangles with a gold atom at
each vertex of the triangle. Some vertices have six triangles meeting
together and some have five (as pointed out by pyt...@gmail.com).
Now switch to think of a polyhedron. Consider the atoms to be points. Then
the polyhedron has 16 vertices. Every combination of three vertices
determines a plane. From the figure, any plane containing more than three
vertices also has vertices on both sides. From all those planes consider
only those which do not have vertices on both sides of the plane. Draw a
line in each plane to the other two vertices. Call each line joining two
vertices an edge.
These lines correspond to the gold bonds in the figure, and all are
triangles. Back to the planes, consider only that part of each plane that
has a line on it or is inside the triangle now drawn on it. Call that
triangle a face. If you wish, fill the inside of the resulting 3-D
irregular polyhedron to make it a "solid".
The angles of six triangles that meet at a vertex can't be 60 degrees
because then the figure wouldn't have the proper convexity. I suggested
that the triangles were equilateral but now I suggest that they are
isosceles and congruent to each other. I don't know whether there is a
mathematical answer for the angles of the triangles.
For the polyhedron, I still wonder, "How many vertices have six triangles
meeting and how many have five?". Now I ask, "How many edges and how many
faces does the polyhedron have? I suspect that Euler's formula, V + F = 2 +
E applies but I still ask, "Does it?" Does the polyhedron have a point
(center) that is the same distance from the vertices where six triangles
meet? The same q where five triangles meet. It looks so in the figure.
The physicist who discovered or invented the gold "molecule" gets some
remaining questions. Is there a geometric answer or only a range of
possible answers for the angles the edges and faces meet? How "close" are
the vertices to lying on a sphere? If the polyhedron has a center, what are
the relative lengths of lines drawn from the center perpendicular to the
faces?
Now it is after 2:30 am. I have tried but I am not sure I have proof-read
this properly.
Dan in NY
news:1148416815.7...@j73g2000cwa.googlegroups.com...
Greetings <pyt...@gmail.com>,
Thank you for your comment. After I read your words, I looked again at the
figure. I agree with you that there are six and five triangle vertices but
no four triangle vertex. I will take your word for other geodesics.
John Berglund
Dan in NY <Som...@NY.net> wrote:
wrote in message
news:1148416815.7...@j73g2000cwa.googlegroups.com...
> Dan in NY wrote:
>> Greetings,
>>
>> On the NY Times web site, there is a story and picture about a hollow
>> molecule composed of 16 atoms of gold. It looks like the figure is an
>> irregular polyhedron where some vertices have six triangles meeting and
>> others have four. The article is "16 Golden Atoms in Search of a Catchy
>> Name" By KENNETH CHANG; Published: May 23, 2006; at URL
>> http://www.nytimes.com/2006/05/23/science/23find.html.
>>
>
> If you look again, I think you'll see that it is made up of 6-triangle
> vertices and 5-triangle vertices. There are no 4-triangle vertices. I
> believe that this is true of any geodesic dome type structure.
&&&
Greetings ,
Thank you for your comment. After I read your words, I looked again at the
figure. I agree with you that there are six and five triangle vertices but
no four triangle vertex. I will take your word for other geodesics.
--
Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com)
Talk is cheap. Use Yahoo! Messenger to make PC-to-Phone calls. Great rates starting at 1¢/min.
Walter Whiteley
It is evident from the picture (included in my earlier post) that the atoms are
on the convex hull and form the 'vertices' of a polyhedron. The shape is:
- truncate a tetrahedron (making four of the triangles). Then add a vertex just
out of the plane over the center of each of the four hexagonal faces just
created. Join this vertex to the six surrounding vertices, as a shallow pyramid.
The object does have all the symmetries of the underlying tetrahedron (in terms
of rotations and reflections).
The triangles at the six-valent vertices are likely to be isosceles, but the
triangles from the truncated vertices of the original tetrahedron are likely to
be equilateral (by symmetry arguments of the molecule and the forces at each
vertex in terms of the bonding).
It is traditional, even among chemists, to connect the symmetry and the
description of such molecules with polyhedral shapes. Yes it does satisfy
Euler's formula in this image (the faces of the convex hull are the faces).
v- e + f = 16 - 42 + 28 = 2
You know this will owrk because it is, topologically, a triangulated sphere.
One by-product of being a triangulated convex sphere is the rigidity of the
shape (which follows from an old theorem of Cauchy, and in terms of static
rigidity and forces - from a version of Max Dehn a century later).
Walter Whiteley
Dan in NY
--
Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com)
Dan in NY
news:44748E63...@mathstat.yorku.ca...
Greetings Walter and other readers,
Thank you so much for this post. It answers all my questions and more. I
never thought of this truncation.
I have thought out the truncation of the tetrahedron. I now understand it,
if you mean to remove a tetrahedron from each vertex -- where the (length
of) the edge of each removed one is one third the edge of the starting
tetrahedron. If so, I see where the hexagons appear. Also I see that the
remaining triangles (trigons?) are equilateral.
The next step is also easy except for the distance of the new vertex from
its hexagon. I will leave the physics and chemistry to you and others but
this distance -- as well as whether the new vertex is above or below its
hexagon -- may depend on them. If there are "reasonable" geometric
assumptions to make to help determine it, I would then like to compare the
result with the scientist's description of the real AU 16 molecule.
I consider your scientific assumptions to be reasonable ones. I am also
looking for reasonable geometric ones. Previously I specified "convex" but
that was my own way of viewing the diagram. You specify "symmetry" and I
almost did that; it is also a "reasonable" assumption. Yet neither (or
both) of these helps determine the distance from the new vertex to its
hexagon. Without another response to this thread, I will leave that
question unanswered.
I have adopted your assumption that the vertex above each hexagon is over
the center. The isosceles triangles over the hexagons have a base length
equal to the edge of the removed polyhedrons. The length of the two equal
sides depends on the distance of that vertex above its hexagon. If that
distance were to be zero, even those triangles would be equilateral. If it
is not zero, their two equal sides are larger. I intend to write a formula
to determine their length.
Yes, you can say I knew Euler's formula would apply. I included it to help
others help me answer the questions. I will say I would have been very
surprised if it didn't apply but I have forgotten the criterion needed. It
always applies whenever I use it.
--
Dan in NY
(for email change to with go in
dKlinkenbert at havoc dot err dot com)
Odysseus
>
> <pyt...@gmail.com> wrote in message
> news:1148416815.7...@j73g2000cwa.googlegroups.com...
> > Dan in NY wrote:
<snip>
> >> http://www.nytimes.com/2006/05/23/science/23find.html.
> > If you look again, I think you'll see that it is made up of 6-triangle
> > vertices and 5-triangle vertices. There are no 4-triangle vertices. I
> > believe that this is true of any geodesic dome type structure.
> Thank you for your comment. After I read your words, I looked again at the
> figure. I agree with you that there are six and five triangle vertices but
> no four triangle vertex. I will take your word for other geodesics.
In his _Polyhedra: a visual approach_ Anthony Pugh classifies it
among geodesic figures as a one-frequency truncated tetrahedron: 16
vertices (here 12 fivefold and 4 sixfold), 28 faces, and 42 edges
satisfy Euler's formula. (In terms of the radius of the circumsphere,
the length of the edges of the original truncated tetrahedron is
~.852802, and of those dividing its faces ~.977512.)
There's no theoretical obstacle to constructing Fulleresque geodesic
figures containing fourfold centres: if you start with a cube or
octahedron you'll always have some. In practice, however, the
fivefold-&-sixfold kinds are usually preferred because they tend to
have fewer distinct edge lengths, so they're correspondingly easier
to manufacture and assemble. (Among Platonic solids the tetrahedron,
cube, & octahedron are 'bumpier' than the dodecahedron & icosahedron,
so some of their faces become more distorted when projected onto a
sphere.) Any such complete sphere will have exactly twelve fivefold
centres interspersed among sixfold ones. Of course a hemispherical or
otherwise truncated dome will have fewer.
For some applications cube- or octahedron-based geometries are
preferable, despite their greater complexity. For example, one might
want to take advantage of their close-packing properties in
constructing clusters of domes or 'bubbles', where little or no
adaptive distortion of the pattern may be required at the points of
intersection or overlap because of the availability of conveniently
oriented 'seams' for truncation or joining.
--
Odysseus