Prev: 9-11 First Responders See Loonylies Loser Tactics
Next: Reaction-Action-Activities relationship?
From: Archimedes Plutonium on 16 Mar 2010 05:39


Archimedes Plutonium wrote:
(snipped)
>
> THEOREM-CONJECTURES:
> (1) Revised Maximum Tiler Conjecture: In 2D, the maximum tiler for
> Euclidean, Elliptic, Hyperbolic
> geometry is the equilateral-triangle.
>
> (2) In 3D, the maximum tiler for Euclidean, Elliptic,
> Hyperbolic geometry is the rectangular-solid-wedge
> which is cut in half at a diagonal, forming long skinny
> wedge shaped sticks. And they are the sort used in
> the Calculus as the picket-fences of integration.
>
> Now what makes these two conjectures work is the
> restrictions involved. There is a number upper limit of
> area and of volume. There is the restriction on the tiler
> candidates in that all have to be a unit area or unit volume. There is
> the restriction of geometrical shapes in that the model of Euclidean
> as square or cube and the model of Elliptic as circle or sphere and
> the model
> of Hyperbolic as hyperbolic triangle or pseudosphere.
>
> So these restrictions pull out a unique maximum
> tiler.
>
> Old math could not do this because they never had a
> precision definition of finite versus infinite. But once you clearly
> define finite from infinite then the Kepler Packing is proved false
> for there is a more dense packing than hexagonal close packing with a
> last row
> shifting in the oblong-hexagonal close packing.
>
> And with a precision definition of infinite, we have packing theorems
> of unique tilers.
>

Now there is an easy proof that the rectangular-solid-wedge is the
maximum tiler
in 3D Euclidean.

And I shall invoke the 100-Model instead of the 10^500 Model simply
because
it is easier for me to write and refer to the boundary of 99.99 rather
than the boundary
of 999..9-9d999..9-9 which has 500 9s to the rightwards and leftwards
of the decimal
point d. It is easier to talk about 99.99.

So in Euclidean 3D, what is the maximum tiler? Well if we use a empty
cube box
container as Euclidean 3D whose sides are 99.99 long and if we had a
unit cube-wedge
which is begot from a cube slicing it at a diagonal. And proceeded to
tile the box with
these unit cube-wedges we end up with a last row-planes of 0.99. Now
since they are wedges
we can fit this 0.99 row-planes with these wedges and the total
density would be somewhere
from 90 to 99%. In the case of the 10^500 Model it would be a density
larger than 99% but
short of 100%. If the Euclidean cube box was exactly 10^500, then the
unit tilers of cube-wedges would be an exact 100% density since no
holes or gaps would exist after the tiling.

But here is where the rectangular-solid-wedge of unit volume excels
over the unit-cube-wedge
in that the rectangular wedge can fill up more of that end row-planes
since the rectangular wedges are long and thin, not short and fat. Now
how long and thin should the rectangular wedges be? That depends on
the Elliptic model and the Hyperbolic Model. Remember, to win
the contest of best tiler, it has to be the maximum in at least two of
the geometries. So if the
rectangular-wedge can beat out the cube-wedge in Euclidean, at what
length of long and thin
does the rectangular-wedge win over the cube-wedge in Elliptic and
Hyperbolic?

Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
 | 
Pages: 1
Prev: 9-11 First Responders See Loonylies Loser Tactics
Next: Reaction-Action-Activities relationship?