proof of the Kepler Packing Problem is mostly a check on three faces of a cube #689 new book 2nd edition: New True Mathematics
Archimedes Plutonium
you need worry only on three of those faces. These 3 faces or
walls are the the leftmostface, the ceiling face and the backface.
The other three faces are not to worry because they are the ones
controlled by the Hexagonal Close Packing, HCP.
So it comes down in the end to a worry of whether you can fit
some more spheres into the faces, or corners of these three
faces mentioned.
In the proof of KPP, we consider every Real Number as
a length of a cube and then we compute whether that Real
Number is one of these four possible outcomes:
(1) Pole to Pole stacking
(2) HCP-alone stacking
(3) Corner-edge HCP
(4) Oblong-HCP
Those are the four and only four possible stacking patterns
and every Real Number is pegged with one of those four
outcomes. For instance say you give me the Real number
2. So a cube of 2 is volume 8 and the most dense pack
is Pole to Pole with 8 spheres. Say you give me 11, then
the most dense pack is Corner-edge HCP. Say you give
me 11.5 then the most dense pack is HCP-alone. Say
you give me 11.99 then the most dense pack is
Oblong-HCP.
So give me any Real Number and it corresponds to one
of those four patterns and those four patterns cycle through
every N and N+1 integers. So a Mathematical Induction proof
is obtained.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Archimedes Plutonium
Cube
in which there is both Oblong HCP and Corner-Edge HCP?
As the previous post indicated Kepler Packing Problem KPP is nothing
more than a
tally and tabulation of what happens to three walls where the other
three
are the walls to stack with HCP-pure. When you get to the three remote
walls
of far left (I am righthanded and so will start stacking in the lower
righthand corner).
And the ceiling wall and the back wall.
So in a sentence, the KPP is an accounting problem of these three
walls.
Now the Oblong occurrs when you have an Real such as 11.99 or 20.99
or 222.99. When you have almost the next integer but not quite. So the
0.99
is almost 1 but a sliver short. So in those cases we slide the
neighors around
at the wall face and is able to pack another sphere in each gap that
is 0.99
distance to the wall.
But the Corner Edge comes into play when the ceiling wall is high
enough
to change the topmost or second to topmost layers into a pole-to-pole
contact
and opening up a row in the leftmost corner. If a 0.99 exists in the
ceiling wall would force us to use the Oblong since it is more dense.
Sounds like I am a bit ambivalent here.
But I actually need a hands on model to be able to cite with
confidence. Mind
models are never really all that reliable. As best as I can tell there
is the possibility
of a Oblong on the length and width walls with a Corner Edge on the
ceiling wall.
And their is the chance of an Oblong on the ceiling wall concomitant
with
no Oblong on the length and width walls.
So there are cases, as far as I can mind model this stuff, that has
Oblong on two
walls and Corner Edge on ceiling wall.
There is the case of having a Corner Edge on a length wall and width
wall for they
are not confined to ceiling walls only.
So yes, there are mixes of Corner Edge and Oblong. And both are HCP
stacks for
the main body of stacking and only at the end at these three faces are
you forced
to try to improve upon HCP-alone with the techniques of Corner Edge
and Oblong.
I think what will happen is that someone will find a distance
parameter of all the
three walls that matter and if the distance is is 0.99 or more but
less than
the next whole integer it is Oblong stacking and if the distance to
ceiling is
0.2 to 0.83 the ceiling wall and side walls are Corner Edge.
Some may wonder how I am so sure there is another stacking arrangement
that
I may have missed. And the answer is that the lattice density of HCP
is proven
to be a maximum via kissing points and since over 90% of the stacking
body
is HCP, it leaves only the spheres on those three walls to play around
with.
So I am hemmed in to an accounting of three walls and hemmed in by the
accounting
of all the Real Number values between N and N+1. I am as constrained
as the
regular polyhedra are constrained as per existence. There can only be
5 regular
polyhedra for they are constrained by angles and distances. Likewise
KPP is
constrained by three walls and what the HCP-pure brings to those three
walls.
Musatov
Positive real numbers have three cube roots. The roots of eight are 2,
-1 + sqrt(3) i and -1 - sqrt(3)i.
M--m--m
(Musatov)
Archimedes Plutonium
(Core-HCP-pure)
So we have every positive Real Number associated with a Kepler Packing
Density and the four types of packing:
(1) Pole to Pole
(2) HCP-alone
(3) Corner-Edge-HCP
(4) Oblong-HCP
Now the thing about those stackings except for Pole to Pole is that
the majority of stacking of the other three techniques is that of HCP.
So that in Corner Edge and Oblong is just a worry at the end where
the bulk of the stacking -- over 90% has already taken place -- and
now we
are just worried about the last two layers of three walls of a cube.
So let me dwell upon the HCP portion since it is over 90% of all of
Kepler Packing and let me see if some interesting Corollaries of
mathematics accrues.
Let me define Core-HCP-pure as that portion of a cube that is the
bulk of the stacking.
So that in the example of the 11unit Cube there are alternating
11 spheres with 10 spheres and there are 12 rows and 12 layers
and all of those spheres comes to (11 x 6) + (10 x 6) = 126
and 126 x 12 = 1512 spheres.
So in the 11Unit Cube, and according to my marble stacking the
Core-HCP-pure is 1512 (accuracy needs checking)
And the HCP-pure does not start until the 3unit Cube since we cannot
do a HCP in a 2 unit Cube. So the 3unit Cube looks like this for the
basement layer:
OOO
OO
OOO
And it has 8 spheres in basement layer, then 5 in second layer
and then 8 in top layer for a total of 21 spheres.
So we have a function to discover as to what the Core-HCP-pure
is for every Real number beyond 3.
So give me a Real Number of 3 or beyond and there is a Core-HCP-pure.
So we have a Inner Density Kepler Packing. Now here is the intriguing
question
that ties in with the bigger Kepler Packing Problem. Does that
function of
Core-HCP-pure coupled with the leftover space, cycle periodically
through the
Real Numbers. Because I
implied that the overall Kepler Packing Problem function cycles
through
the Reals enabling me to use a Mathematical Induction proof.
So here we have stripped away the nettlesome strange things going on
with the last layers before the three walls of fiddling and playing
around to
garner extra sphere packings. Here we concentrate on the Core or Bulk
of Kepler packing.
So let us just concentrate on that Over 90% of the Kepler Packing
Problem
of its core HCP-pure.
So give me any Real Number and then it should be straightforward to
have
a corresponding number which is the Core-HCP-pure.
For 3 the corresponding f(x) is equal to 21
For 4 the corresponding f(x) is equal to 44
.
.
For 11 the corresponding f(x) is equal to 1512
So the preliminary function looks like something of the type of x^3 +
x^2
as the numbers get larger.
Second Corollary:
what is the function that describes the leftover space as defined by a
rectangle of the wall face with a line drawn parallel to wall face
and furthest reaches of last layer or row of spheres. For example
the Real Number 11.99 has a rectangular solid of 11 by 0.99 by 11.
There usually will be three of these rectangular solids for the three
wall faces whenever the Cube is not an integer cube.
So that I want to prove the Corollary Kepler Packing Problem that
says the Empty Space leftover by the Core HCP-pure periodically
cycles through the Real Numbers.
So in other words proving these Corrolaries is a fantasic aid
into going into the proof of the big overall Kepler Packing Problem.
And it shows us that the leftover space is taken care of completely
by those two techniques of Corner Edge and Oblong.
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
>
> Second Corollary:
> what is the function that describes the leftover space as defined by a
> rectangle of the wall face with a line drawn parallel to wall face
> and furthest reaches of last layer or row of spheres. For example
That is not useful enough; and rather use the furthest reaches, let us
use the minimum tangent of the hollows.
> the Real Number 11.99 has a rectangular solid of 11 by 0.99 by 11.
> There usually will be three of these rectangular solids for the three
> wall faces whenever the Cube is not an integer cube.
>
> So that I want to prove the Corollary Kepler Packing Problem that
> says the Empty Space leftover by the Core HCP-pure periodically
> cycles through the Real Numbers.
Archimedes Plutonium
Archimedes Plutonium
Mistake in that there is not 126 in each layer, and obviously 11^3
has to be larger than above.
>
> For 3 the corresponding f(x) is equal to 21
> For 4 the corresponding f(x) is equal to 44
> .
> .
> For 11 the corresponding f(x) is equal to 1512
>
> So the preliminary function looks like something of the type of x^3 +
> x^2
> as the numbers get larger.
>
Same mistake here in that the f(x) for x = 11 has to be smaller
than 11^3.
And the function has to be x^3 subtract something.
Archimedes Plutonium
I am getting so very close to what that function is.
For the moment I call it x^3 -[[tan(x)]]
Where the [[]] indicates those least or greater integer
and where the "tan" means the tangent trigonometric function.
The Least Integer function [] is where
you have a number such as 11.35 then the [] is 11. Or you have
11.999 then [] is 11. The Greater Integer function is where you
have 3.01 then [[]] is 4.
So now I have to calibrate tangent trig function so that I get
For 3 the corresponding f(x) is equal to 21
For 4 the corresponding f(x) is equal to 44
and I have to work out what the correct answer is for the case of
11Cube.
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
>
> I am getting so very close to what that function is.
>
> For the moment I call it x^3 -[[tan(x)]]
>
> Where the [[]] indicates those least or greater integer
> and where the "tan" means the tangent trigonometric function.
>
> The Least Integer function [] is where
> you have a number such as 11.35 then the [] is 11. Or you have
> 11.999 then [] is 11. The Greater Integer function is where you
> have 3.01 then [[]] is 4.
>
> So now I have to calibrate tangent trig function so that I get
> For 3 the corresponding f(x) is equal to 21
> For 4 the corresponding f(x) is equal to 44
>
> and I have to work out what the correct answer is for the case of
> 11Cube.
>
So obviously it is not tangent function but its inverse of arctan.
Some values are
arctan(3) = 1.24
arctan(4) = 1.32
arctan(5) = 1.37
arctan(6) = 1.40
.
.
arctan(11) = 1.48
So I need for 3Cube to make up for 27-21 = 6
For the 4Cube to make up for 64-44 = 20
So I need a adjustment on arctan so that [?arctan(3)] = 6
and that [?arctan(4)] = 20 etc etc
In my long career in mathematics I was never faced with
having data and then finding out what the function was.
So this is good practice in Function Finding.
Archimedes Plutonium
sooner or later.
First off, I have to correct the data also:
3Cube = 27- 8 = 19
4Cube = 64-20 = 44
5Cube = 125-28 = 97
6Cube = 216-36 = 180
Rather than the arctan the natural-logarithm looks better candidate
Ln(27) = 3.29
Ln(64) = 4.15
Ln(125) = 4.82
Ln(216) = 5.37
So I need something more for the ? to get from
3.29 to 8
from 4.15 to 20
from 4.82 to 28
from 5.37 to 36
remembering that term is [Least Integer]
Perhaps it is a trigonometric function as the ? such as arctan
Archimedes Plutonium
Archimedes Plutonium wrote:
> Well, if I keep changing enough, I will find the correct function
> sooner or later.
>
> First off, I have to correct the data also:
> 3Cube = 27- 8 = 19
> 4Cube = 64-20 = 44
> 5Cube = 125-28 = 97
> 6Cube = 216-36 = 180
>
> Rather than the arctan the natural-logarithm looks better candidate
>
> Ln(27) = 3.29
> Ln(64) = 4.15
> Ln(125) = 4.82
> Ln(216) = 5.37
>
> So I need something more for the ? to get from
>
> 3.29 to 8
> from 4.15 to 20
> from 4.82 to 28
> from 5.37 to 36
>
> remembering that term is [Least Integer]
>
> Perhaps it is a trigonometric function as the ? such as arctan
>
Did the 7 Cube although cannot vouch for my accuracy
3Cube = 27- 8 = 19
4Cube = 64-20 = 44
5Cube = 125-28 = 97
6Cube = 216-36 = 180
7Cube = 343-60 = 283
Curiously the above progression of 8, 20, 28, 36, 60, ...
has two triangular-numbers of 28, 36
Maybe I should get more data, especially after 8 and 9 Cubes since
that is where the layers increase beyond the length of the cube due
to hollow nesting.
When I get this function or equation of Core HCP-Pure, then the
remainder
empty space to be packed cycles through the integer-Cubes and thus
the only three alternatives are:
Corner-edge, Oblong and HCP-pure.
So the final conclusion of the Kepler Packing Problem started in 1611
is that the most dense packing in Euclidean Space involves the
HCP-pure as the bulk of the packing and at the last two layers of
length
width and depth can some alterations be performed so as to raise the
density higher with Corner-edge and Oblong for which HCP-pure would
be of a lower density. The function or equation gives the density of
over 90% of a the Cubes beyond the 3Cube, remember this is all Reals
beyond 3. So the empty space when we reach the three walls of interest
have only the Corner-edge, Oblong or HCP-pure to pack.
In fact as you look at the above examples of the 7Cube = 343-60 = 283
a good chunk of that wasted space of 60 spheres is at the height wall.
So in the future, we can rig a computer to where we feed it a Real
Number
of 3 or greater and out spits a data form that tells us this:
(1) how many spheres compose the Core-HCP-pure
(2) how many unit-cube spaces were lost
(3) how much empty space on the length wall, width wall, and height
wall
defined by "effective hollow to wall distance"
(4) whether there is room for a Corner-edge packing and how much of an
increase in number of spheres
(5) whether there is room for Oblong packing and how many more spheres
added than if not applied.
I have been hunting through the literature and as of yet unable to
spot whether
anyone has the function or equation for Core HCP-pure. Apparently
noone bothered
and perhaps noone ever thought of this parameter as being important.
Archimedes Plutonium
Mathematical
Conjecture of the Kepler Packing and shows it to be wrong. Wrong
because
the most dense packing in both 3D and 2D Euclidean is the Oblong HCP
or Oblong Hexagonal respectively.
Call it the "Education of Archie on KPP" up till now. I had a
mindblock
and bet most everyone else has this same mindblock when it comes to
both KPP and Gauss's 2D version. The mindblock or paradox is that we
think that N^3 number is larger than the number of spheres in KPP
given
any N. Well it is true for small N. Because 3^3 and no more than 27
spheres
can fit in that cube or that 8^3 and no more than 512 spheres can fit
in that
cube. Likewise in 2D that no more than 9 circles can fit in 3^2 or no
more than
64 circles can fit in a square 8 by 8. This is a mindblock that most
everyone
comes into packing problems with this paradox. They think and I
thought
that N^3 or N^2 was always larger than the number of spheres or
circles
in that packing.
Well at 10, which is the magic number, the experiments I conducted
show
quite clearly that the number of spheres in a 10Cube are more than
10^3
and the number of circles in a 10Square is larger than 100. Most
everyone
that enters Packing enters it with the misconception that N^3 or N^2
are upper
limits to how many spheres and circles can be contained in 3D or 2D
respectively.
But at the integer 10 (it maybe 9 since my ruler is not that precise)
that there
is more spheres or circles than the N^3 or N^2 and the reason for this
is due
to the nesting inside of hollows of lower layers. And the reason for
our misconception
is due to our visualization that 10^3 or 1,000 small unit cubes
completely tile the
big cube with no holes or gaps in between those small unit cubes.
Contrarily
we visualize all those holes and gaps in the spheres or circles and we
have a
hard time of realizing that there could be more spheres and circles
than unit
cubes or unit circles.
This maybe the first time a Physics Layout of Data proves a
Mathematical
Conjecture.
Experiment and Results: I have uniform magnetic circles of 25mm
diameter.
Either at 9 or 10 for sure, is there an extra row of circles due to
the hollows
of the hexagonal packing.
I need only show the Gaussian 2D Packing for it proves not only 2D but
also
Keplerian 3D with spheres. The data is transferable from 2D to 3D.
I also have marbles but not of uniform size but close enough to gain
this data:
HEX stacking pattern yield this data:
3Sq = 8 N^2 = 9 circle status = -1
4Sq = 14 N^2 = 16 circle status = -2
5Sq = 23 N^2 = 25 circle status = -2
6Sq = 33 N^2 = 36 circle status = -3
7Sq = 46 N^2 = 49 circle status = -3
8Sq = 60 N^2 = 64 circle status = -4
9Sq = 77 +8 N^2 = 81 circle status = +4
10Sq = 95 +10 N^2 = 100 circle status = +5
11Sq = 116 +10 N^2 = 121 circle status = +5
Notice that the number of circles at 9 changes from a
negative status to a positive status due to the adding up of the
hollow-nesting.
For about a week now, I have been looking for a function or
equation to describe the Core Hex or Core HCP stacking
but in that quest, I was under the misconception that it was
always smaller than N^3 or N^2
I no longer need to find any such function or equation. The physical
data proves the Gaussian 2D and the Keplerian 3D
And all I need to do is show it in Gaussian 2D where it is much
easier.
The end result proof is that there is a more dense packing than
hexagonal
or hexagonal-close-pack. Both Corner-edge Hex and Oblong Hex are more
dense than what Gauss showed and what Kepler conjectured for 3D.
This number data is sufficient as a Proof of both Gaussian 2D and
Keplerian
3D. This maybe the first time in history that Physics data and
experiment
trumps over all the mathematical writeups claiming to be proofs.
In Gaussian 2D with the magnetic circles of 25mm
__ height__ __ length__ Corner-Edge
possibility?
9circlesSQ 201mm 225mm not possible
10circlesSQ 225mm 250mm not possible
11circlesSQ 246mm 275mm enough headroom, yes!
Both Gauss and Kepler were not privy to the possibility of a more
dense stacking than just the Hexagonal or Hexagonal Close Packing
alone. If there is enough headroom in the height then it is possible
to shift all the circles or all the spheres to the rightward most
corner
and thus opening up a entire gap to add one more circle in 2D or
one more half-row in 3D
And if the cube or square in 3D or 2D respectively is a fractional
cube or square of say 11.99 long then the Oblong Hex stacking
occurrs in the length and width end faces so that an entire half-
plane spheres are added in 3D and a entire row in 2D is
added above that of Hex or Hexagonal Close Pack.
Anyone can read alleged proofs of Kepler Packing which says
nothing more than lattice density is pi/sqrt18 is maximum.
And that was proven centuries ago with kissing-points. But that does
not say
anything about Corner Edge or Oblong Hex adding more density
than the lattice density of pi/sqrt18.
The reason no proof surfaced until now for Kepler Packing of year
1611, is the vast confusion of trying to apply a lattice density when
the conjecture requires Container-Density.
Lattice density is inconsistent with a proof involving Euclidean
Space.
When you have Euclidean Space then you need Container Density.
I am proud of myself on this achievement because any person aged
10 and older can run the above experiment. Can see how Corner Edge
and Oblong Hex raises the density.
Here is the illustration:
OOOO
OOOOO when the headroom of the ceiling face is large enough to scoot
that upper layer to the rightward side then we can add a new sphere
shown by the X
XOOOO
OOOOO
And if the upper layer is full then we do the scooting on the second
to top
layer as such:
OOOOO
XOOOO
And here is the illustration of Oblong Hex when the square or cube has
a 0.99 after the integer
O 9
_O 8
O 7
_O 6
O 5
_O 4
The underline indicates a distance of 0.99 to the wall. Those
are either spheres in 3D or circles in 2D and they are numbered
for easy reference.
In the Oblong technique we scoot 5 downwards leaving a gap
large enough to pack a new sphere or circle. We leave 6 undisturbed
but move 7 downwards leaving a gap which we add a new circle or
sphere. And we do that entire face of both the length face and width
face, depending on whether we are in 3D or 2D.