| >| > | I see a cube there, not a tetrahedron.  Btw I am
| >| > | a chemist by training, so please don't jump to the
| >| > | conclusion that I am ignorant about the shape of
| >| > | sp3 orbitals.  The point is that you can't "draw-in"
| >| > | the faces of the tetrahedral units in such a way
| >| > | that the faces of the tetrahedra all touch their
| >| > | neighbors with no space left between them.  Try it
| >| > | and perhaps you will see.  Perhaps.
| >| > | 
| >| > If you WERE a chemist, you'd count the little white balls.
| >| > Five at the top, five at the bottom, that makes 10.
| >| > 8 inside the cube, that makes 18.
| >| > Valency of 4, How many green lines are there, and where
| >| > do they connect?
| >| > I conclude that you are indeed ignorant of shape.
| >| > 
| >| > But of course you are NOT a chemist, even by training,
| >| > you are a phuckwit. The best thing for you to do is bail,
| >| > stooopid. 
| >| > Androcles.
| >|
| >| There are only 4 balls inside the cube. The others are 8 at the corners
| >| and 6 in the centres of the faces. The lines connected to those 4 do
| >| correspond to bonds, the others don't. Each is at the centre of a
| >| tetrahedron formed by the 4 balls/atoms it is bonded to, one corner ball
| >| and three face-centre ones. Much clearer pictures can be found at
| >| http://phycomp.technion.ac.il/~nika/diamond_structure.html
| >| David Hartley
| >
| > Each dark blue corner ball has 4 links.
| > Each interior pale blue ball has 4 links.
| > Each green  face-centred ball has 2 links, leaving 12 links open
| > to connect to 12 cubes. Every atom connects to four other
| > atoms. Four atoms form the vertices of a tetrahedron.
| > Therefore the space is tesselated with tetrahedra.
| >
| 
| I am not sure what you mean by links here. There are indeed four lines 
| linked to each dark blue ball in the diagram referenced above, but only 
| one represents a chemical bond; the others are just to show the 
| underlying cubical structure. The four linked balls do, of course, form 
| a tetrahedron, but it is not regular, the corner ball is not inside it, 
| and it overlaps some of the other tetrahedrons. If you mean it has 1 
| bond within the cube, and 3 to balls in other cubes then you are 
| correct.
| 
| If you throw in all the tetrahedrons formed by these sets of 4 atoms 
| linked to a single atom they overlap, as each is in the centre of one 
| tetrahedron and a vertex of 4 others, (but they still don't fill space). 
| If you take just the ones with a pale blue ball at the centre and dark 
| blue or green balls at the vertices (or vice versa) you get a lovely 
| array of regular tetrahedra, connected at each vertex, but otherwise not 
| touching. There are large spaces between them, they are not a 
| tessellation. These spaces cannot be filled in with more regular 
| tetrahedra. (They can of course be filled with irregular tetrahedra, but 
| that's not the issue.)
| 
| > Take a large tetrahedron. Cut the top off at half the height.
| > That leaves a triangular base.
| >
| >             /\
| >            /A \      A is a tetrahedron.
| >           /====\
| >           |    |
| >_____________________________ cut 1.
| >           |    |
| >          /======\
| >         /   R    \   R = remainder.
| >        /==========\
| >
| >          /\ \====\
| >         /B \ \  R \          cut 2.
| >        /====\ \====\     B is a tetrahedron
| >
| >
| >           \====/ /\
| >            \ R/ /C \         cut 3.
| >             \/ /====\      C is a tetrahedron
| >
| >
| >Cut 4 is behind the view.  D is a tetrahedron.
| >
| >What is R after cut 4? An inverted tetrahedron.
 
| No. Its base is a triangle, not a point. 

Okay, yes, you are correct. No fooling you, is there?

Its upper face is also a triangle. That leaves three side faces
extending down to three side faces. Five faces altogether is 
not an octohedron :-)

| Round the middle you're left 

Can't have any rounding, sorry. You can cut a plane, but no rounding allowed.
Isn't this fun, trying to communicate on Usenet? I'm so used to the other person
being an idiot that it just isn't possible for me to be one as well, so you won't mind
if I make ridiculous statements to try to wriggle out of it, will you?


| with one triangle from each side (the bit left in your diagram of cut 
| 3), and a triangle where each of B, C and D attached, and on top you 
| have the triangle where A attached. 8 sides, all triangles, it's an 
| octahedron.
| -- 
| David Hartley

I'm retired now, but I used to work in flight simulation. Most modern
simulator platforms are mounted on 6 extendable hydraulic legs, coupled
to a triangle at the base and a reversed triangle under the platform, making
an octohedron. 
http://www.cadsoft.de/~kls/fltsim/platform.jpg
Needless to say I was fully aware the whole time of the octohedral interior
of the tetrahedron, but please look at the thread title. I was doing a Baez.
When someone intelligent such as yourself broke in, that kinda spoilt the 
fun.
Baez is a typical know-it-all idiot that cannot think, the thread crossed into
sci.physics.relativity and Einstein's idiocy which Baez and his ilk propagate
is no different to me claiming that an octogon is made up of four tetrahedra.
But I admit you caught me. Neither John Baez nor Tom Roberts who idolizes
Baez have the courage to do that. They'd prefer to go on believing in
nonsense for the rest of eternity and continue to spread malicious lies
to young, unsuspecting students for the benefit of their own glory.

[...]

      My apologies for jerking your chain. You didn't deserve it.
      Androcles B.A., M.Sc., Ph.D.