Hypersport

[Simeon]

Hypersport

Perhaps rather more attention is given to polychora than they deserve. On
account of the fascinating networks of them which can be constructed. No
doubt they possess a symmetry which might startle the horses; but, for sheer
simplicity, and unbelievable *roundness* - and mystique -give me the
unbeatable glome.
For unit radius, the surface of a circle is 2 pi or 6.28. Rising to 4 pi or 12.57
for a sphere, 2 pi^2 or 19.74 for a glome, and a maximum of 16 pi^3/15 or
33.07 in the 7th dimension or heptaspace.
After which it actually decreases for successive dimensions. The content of
the ball performs similarly; rising to a maximum of 8 pi^2/15 or 5.26 units in
5 dimensions, then decreasing. All of which is well-known.
What is more obvious but less well-known, is the increasing corner-wastage
of these balls as a result of their progressive roundness.
For a circle in a square, this is 21.46%; for a sphere in a cube, 47.64%; for a
glome in a tetracube, 69.16%; a pentaglome in a pentacube, 83.55%.
And by the 11th dimension, a hendecaglome in a hendecacube, 99.91%.
This means that if an 11-ball of diameter 1 is exactly fitted in the smallest
possible hypercubical box, the hypervolume of which is L^11 or 1 unit for
unit side, touching all its 22 decacube sides at their exact centres, its content
will be 64 pi^5/10395. And although being the largest ball which can be
fitted into the box, it will occupy less than a thousandth of it.
The moral would seem to be: if travelling in the 11th dimension, think
carefully before posting home a basket-ball. Because the corners of your
parcel (all 2048 of them) are going to be awfully empty. Not to mention the
colossal amount of red tape involved.

Simeon

[Aale de Winkel]

curious interesting point you are raising here, I don't know what this has to do with "sport" though;

With a visit to http://mathworld.wolfram.com/Hypersphere.html I learned that the surface formulae for the hypersphere you are quoting reads:
S[sub]n[/sub] = 2π[sup]n/2[/sup] / Γ(n/2)
a curious formula which has a maximum at n = 7.25695...., for practical purposes thus at n = 7.
The volume (being the integral of the surface area) the reference shows to be given by:
V[sub]n[/sub] = S[sub]n[/sub] R[sup]n[/sup] / n =(R=1)= S[sub]n[/sub] / n

curious, to say the least!

[Polyhedron Dude]

And by the 11th dimension, a hendecaglome in a hendecacube, 99.91%.
This means that if an 11-ball of diameter 1 is exactly fitted in the smallest
possible hypercubical box, the hypervolume of which is L^11 or 1 unit for
unit side, touching all its 22 decacube sides at their exact centres, its content
will be 64 pi^5/10395. And although being the largest ball which can be
fitted into the box, it will occupy less than a thousandth of it.
The moral would seem to be: if travelling in the 11th dimension, think
carefully before posting home a basket-ball. Because the corners of your
parcel (all 2048 of them) are going to be awfully empty. Not to mention the
colossal amount of red tape involved.

Simeon

May need to use a box shaped like a different polydekon - how about the 11-D version of the octagon - the small xennonated hendekeract .

Polyhedron Dude