Conference rusure::math

    Do you have any requirements on this "shape"?
    
    There are 5 regular polyhedra, i.e., solid shapes that you can pack
    densely in 3 dimensions. 
    
    Or would you be satisfied with something more irregular, but which
    fit your sphere better? How about something curvy?
    
      John

    regular polyhedra would be fine. Could you please list for me ?
    
    Thanks,
    Frank

The 5 regular polyhedra are
	tetrahedron  (triangular pyramid)
	cube
	octahedron  (2 square pyramids glued base-to-base)
	dodecahedron (12 pentagons)
	icosahedron (20 triangles)

The more sides, the closer fit around the sphere (I assume).
Sorry, I can only derive formulas for the first 3.

Here are the formulas for the 5 regular polyhedra metrics.

Notations for a given polyhedron:

p : 	the number of edges adjacent to each vertex

q :	the number of edges of each regular polygonal sides

m :     the number of edges

a :	the length of any edge

d :	the dihedral angle, that is, the angle between two adjacent sides

r :	the radius of the inscribed sphere, that is, the sphere which is 
	tangential to the center of each polygonal side

R :	the radius of the sphere which contains all the vertice (in French,
	we say "la sph�re circonscrite". Is there any english or american	
	translation for this latin-rooted adjective ?)

v :	the volume of the polyhedron

v_i:	the volume of the inscribed sphere

pi :	the well-known ratio of the length of a circle to its diameter
	(to say nothing about many other definitions ...)


Now the formulas are:



                   pi
	      cos ----
     d             p
sin ---  =  ------------
     2             pi
              sin ----
                   q



       a        pi        d
r  =  ---  cot ----  tan ---
       2        q         2



       a        pi        d
R  =  ---  tan ----  tan ---
       2        p         2



       m     /     pi  \ 2      d    3
v  =  ----  ( cot ----  )  tan ---  a
       12    \     q   /        2

   
       4       3
v_i = --- pi  r
       3


v_i    2 pi       pi   /     d \ 2
--- = ------ cot ---- ( tan --- )
 v       m        q    \     2 /


The last formula I think, answers what was requested in .0, that is, how much
the inscribed sphere fills the polyhedron.

Here are the results of the calculation on my Alpha osf/1 system:

					  v_i
Polyhedron		p	q	  ---
				 	   v

Tetrahedron  		3  	3  	0.302300
Cube  			3  	4  	0.523599
Octahedron  		4  	3  	0.604600
Dodecahedron  		3  	5  	0.754697
Icosahedron  		5  	3  	0.828798

Stereometrically yours,

	Herv�

    

    re .1
    
>>>    There are 5 regular polyhedra, i.e., solid shapes that you can pack
>>>    densely in 3 dimensions. 
    
    That's not the usual definition of the regular polyhedra. Usually they
    are described as having a number of identical faces and vertices, each
    face being a regular polyhedron.
    
    If you want a better shape then you can modify an icosohedron. Imagine
    cutting a tiny bit off each of the corners. You modify each triangular
    side into a hexagon, with 3 large sides and 3 small sides. Each corner
    is now a small pentagon. Now take more off the corners, increasing the
    size of the pentagons, shrinking the large sides of the hexagons, and
    increasing the small sides of the hexagons. When the small sides of the
    hexagons are the same size as the large sides stop. You now have a
    shape with 20 hexagons and 12 pentagons. This shape fits a sphere
    better than the original icosahedron. It should be easy enough to
    calculate the volume of this shape - it is the same as the icosahedron
    less the pentagonal pyramids taken from each corner. I expect the same
    sphere will fit inside this truncated icosahedron.
    
    You can improve on this shape further, by gluing another pyramid on to
    each hexagon and pentagon. These pyramids are very flat, but allow a
    slightly larger sphere to be contained in the shape. I'll leave it to
    someone else to determine how high these pyramids ought to be, and the
    size of the sphere that can be contained.
    
    Andrew

in the u.s., the technical name for the construction in .5 is soccer ball.
in the rest of the world it is a football.
close enough to a sphere for recreational work.

    re .4
    
    The English word that you are looking for is "circumscribed".
    
    re .6
    
    Don't count Australia in with the rest of the world on that one. Here a
    real football is approximately ellipsoidal. ;-)

    Is there not a regular polhedra with 60 faces, 90 edges, and 32
    vertices?  the buckey ball?
    
    

    Re .8:
    
    It's not regular.
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

To be precise, some of the faces (12?) are regular pentagons, and the
rest are regular hexagons.

Re .5
The other way to modify the icosahedron is to connect the midpoints of
adjacent edges, so that each triangular face is replaced with four
triangles:

	    /\
	   /  \
	  /----\
	 / \  / \
	/___\/___\

The center triangle is still tangent to your sphere, but the three original
vertices can be brought closer to the sphere's center, until the three outer
triangles are also tangent to the sphere.

You can repeat this recursively, to get a volume that's arbitrarily close to
the sphere.