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Six-Fold Symmetry, or Logos, In Orbiting Bodies!
By !
Ian Beardsley!
Copyright © 2022 by Ian Beardsley"
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I show the earth is the eigenvalue or characteristic root for the matrix that describes the solar
system and to do this I formulate the solar system as six-fold symmetry and suggest this could
be taken as a logos, a formulation of the language of Nature."
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Contents
1.0 The Constant k………………………………………….4
2.0 Orbital Velocity………………………………………….8
3.0 The Earth is the Basis or Eigenvalue…………….20
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1.0 The Constant k: Warren Giordano writes in his paper The Fine Structure Constant And
The Gravitational Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js
And it works, G is:
G=6.67408E-11 N(m2/kg2)
Let us reformulate this as by introducing Avogadro’s number because multiplying Planck’s
constant ‘ ’ by ‘ ’ is G if multiplied by and Avogadro’s number is
:
Equation 1.1
We just need the right units. We can get these:
Since grams and atom cancel we can work in grams even though our equations are in kilograms.
Let us not write H, since formally it is grams per mole of hydrogen but write
We have:
Or,…
Equation 1.2
Where
Equation 1.3
Let us say we were to consider Any Element say carbon . Then in general
h
1 + α
α
10
23
h
1 + α
10
23
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
N
A
H = 6.02E 23
atom s
gr a m
⋅
1gr a m
atom
= 6.02E 23
ℍ = 1
gr a m
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
x = 1.00kg
2
s
m
𝔼
ℂ
of 5 22
Equation 1.4
We have
and
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And,
Therefore we always have:
Equation 1.5
is not molar mass, and that is a variable determined by ; it is a mole of atoms multiplied
by the number of protons in . The reason we point this out, though it may already be clear, is
we wish to find the physical theory behind it. That is we need to find the physical explanation for
equation 1.4
It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in
kilograms, seconds, and meters, it may mean these units of measurement have some kind of a
meaning. We can in fact write it:
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gr a m s
6proton s
N
A
=
6(6E 23pr oton s)
6gr a m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gra m s
𝔼 =
Z ⋅ gra m s
Z ⋅ pr oton s
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
of 6 22
We know that
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared. To begin our search for the meaning of equation 1.4 we convert x, the factor of
1.00 to astronomical units, years, and solar masses, as these are connected to the orbit
of earth as it relates to the sun. We have:
=
We can now write
Eq 1.6.
This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital
velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum
mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We
multiply both sides by and we have earth velocity on the left and the units stay the same on
the right. But what we will do is return to the form in kg-m-s and leave it as an equation but put
in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Eq. 1.7
This brings up an interesting question: while we have masses characteristic of the microcosmos
like protons, and masses characteristic of the macrocosmos, like the minimum mass for a star to
become a neutron star as opposed to a white dwarf after she novas (The Chandrasekhar limit)
which is 1.44 solar masses, we do not have a characteristic mass of the intermediary world where
we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find
that mass let us take the geometric mean between the mass of a proton and the mass of 1.44
solar masses. We could take the average, or the harmonic mean, but the geometric mean is the
squaring of the proportions, it is the side of a square with the area equal to the area of the
rectangle with these proportions as its sides. We have:
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
kg
2
1
⋅
s
m
⋅
(1.98847E 30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
4π
2
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E 30kg
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We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Eq. 1.8
All we really need to do now is divide 1.7 by 1.8 and we get an even number that is the six of our
six-fold symmetry.
Eq. 1.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Equation 1.9
Where k is a constant, given
Equation 1.10
We can explicitly write the constant k:
Equation 1.11
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the degeneracy pressure and collapse. The non-relativistic
equation is:
Equation 1.12
Let us approximate 0.77 with 3/4. Since we have our constant
And
Then
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
800
s
m
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
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Equation 1.13
Since our constant k in terms the Chandrasekhar limit is
Equation 1.14
2.0 Orbital Velocity Earth rotates through 15 degrees in one hour:
The distance then that the earth equatorial surface rotates through is where r is the earth
radius and theta is in radians.
Now we look at how many degrees through which the earth rotates in 1 minute:
For seconds…
Now let’s look at the same for the distance the earth moves around the Sun in an hour, a minute,
and a second as well…
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
360
24
= 15
s = r θ
s = (6,378.14)(0.2618ra d ) = 1669.8k m /hr
24(60) = 1440min
360
1440
= 0.25
∘
= 0.0043633ra d
s = (6,378.14)(0.00436) = 27.83k m /min
24(60)(60) = 86,400s
360
∘
86,400
= 0.004167
∘
= 0.00007272ra d
s = (6,378,14)(0.00007272) = 0.464k m /s
(365.25)(24) = 8766hr /yr
360
8766
= 0.041
∘
= 0.00071678ra d
(149,598,000)(0.000716768) = 107,227k m /hr
(365.25)(24)(60) = 525960
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And finally…
We have the following table:
As we can see I am in good
agreement with Martin
Zombek, Handbook of Space
Astronomy and Astrophysics
which provides data. Notice
27.83 km/min in earth
rotation is approximately the
29.786 km/sec in earth orbit.
That is a valuable clue. Now let
us consider
Earth: 1 AU=149,598,000km (average earth-sun separation).
The earth rotates through about one degree a day.
Earth Orbit:
Earth Rotation: r=6,378.14km
360
525960
= = 0.000684463
∘
= 0.000011946ra d
s = (149,598,000)(0.000011946) = 1,787.1k m /min
(365.25)(24)(60)(60) = 31557600
360
31557600
= 0.000011408
∘
= 0.000000199ra d
(149,598,000)(0.000000199) = 29.786k m /sec
360
365.25
= 0.9856
deg
d ay
≈ 1
deg
d ay
360
∘
= 2π r a dia n s = 6.283
ra d
year
0.9856
∘
= 0.017202424
ra d
d ay
(149,598,000)(0.017202424) = 2,573,448.201
k m
d ay
360
∘
1d ay
= 6.283
ra d
d ay
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Moon: sidereal month =27.83 days, r=405,400 km
The points to be made in this exploration
1. So the earth goes through 1 degree a day and the moon 1 kilometer per second.
2. Earth rotates through 27.83 km/min at its equatorial surface and orbits through 29.786km/s
around the sun. These two are close to the same.
3. The synodic month is 29.53 days approximately equals the 29.786 km/s the Earth moves
around the sun. The sidereal month is 27.83 days is the 27.83 km/min through which the
Earth rotates at its surface.
With that information we have this mystery of sexagesimal in the earth-moon-sun orbital
parameters, solved. We see we can make the following equation in table form:
Consider minute times day:
Thus,..
The circle is divided into 360 units, each unit (each degree) is the distance the earth moves
around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are
360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).
We have:
1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always
close to that because its orbit is approximately circular. We have
s = r θ = (6,378.14)(6.283) = 40,075.0
k m
d ay
(27.83)(24) = 667.92
h ours
m onth
360
667.92
= 0.529
∘
= 0.0094
ra d
hr
(405,400k m)(0.0094r a d /hr) = 3,810.76k m /hr
(3,810.75k m /hr )(hr /60min) = 63.5k m /min = 1.0585k m /s
orbit rotat ion orbit m oon
29.786k m minute 1d ay k ilom eter
secon d 27.83k m degr ee secon d
=
mi n ⋅ d ay
deg
⋅
k m
s
2
(mi n)(d a y) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec
2
mi n ⋅ d ay
deg
⋅
k m
s
2
=
86,400s
2
deg
⋅
k m
s
2
= 86,400 ⋅
k m
deg
86,400 ⋅
k m
deg
⋅ 360
∘
= 311,040,000k m
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This is approximately the diameter of the Earth orbit. We define our variables:
Earth orbits:
Earth rotates:
Earth orbits:
Moon orbits:
Earth completes a 360 degree orbit yields:
Where on the right it is in radians and is the radius of the Earth’s orbit. We have
This is (0.00618)360=2.225
0.00618 is
Where is the inverse of the golden ratio. .
We have:
Equation 2.1
Our base ten counting is defined
is defined
311,040,000k m
149,598,000k m /AU
= 2.079AU
v
e
= 29.786
k m
s
ω
e
=
27.83k m
mi n
=
27.83k m
mi n
⋅
mi n
60sec
= 0.4638
k m
sec
θ
e
=
1d ay
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
v
m
=
1k m
sec
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786)(14,400)(1k m /s)
(0.4638)(149,598,000)
360 = 2.225AU
ϕ
100
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
10
0
= 1,10
1
= 1,10
2
= 100,...
ϕ
of 12 22
, such that
which is given by
Thus since the diameter of the Earth orbit is
Then its radius is
Since we measure time with the Earth orbital period and that period is given by Kepler as
Then
Equation 2.2
Is approximately one year. In this section we set out two show the historical development of the
second by dividing up the motions of the Earth, Moon, and the apparent motion of the Sun into
units of 24, and 60 result in the solar system’s size as based around the Earth orbit giving us
Since we have established a connection between the microcosmos and the macrocosmos we
would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want
to know the universal gravitational constant in these units:
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
ϕ
100
360
∘
= 2AU
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T =
3
2
10 ⋅
ϕ
3
2
10
3
⋅ 360
∘
ϕ
100
360
∘
= 2AU
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
of 13 22
For a year we have
(365.25)(24)(60)(60)=31557600 seconds
And for an AU
1 AU=1.496E11m
We can immediately put this to work. In order for the earth to stay in orbit its centripetal force
must equal the gravitational force. Its orbital velocity must be given by:
Since in our theory we present R as:
Then
Equation 2.3
Which evaluates:
Converting this to meters per second:
Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s
is an average over a varying velocity due to the Earth’s slightly elliptical orbit.
Since we have
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
G
Mm
R
2
=
mv
2
r
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
5
3
GM
ϕ
v =
5
3
(40)(1)
0.618
= 6
AU
year
6
AU
year
⋅
year
3.156E 7
⋅
1.496E11m
AU
= 28,441
m
s
of 14 22
For the earth, we show now it is true for the moon as well, and indeed it would hold for any
system circular, or approximately circular. For the Sun and its planets
If you use astronomical units and Earth years. In general we use
Which has the constant of proportionality
If m is small compared with M, you can write
Since we want to work with Earth and the Moon m is not so small compared to M. So,…
=
=
And these do round neatly to one-exponent like this. The sidereal lunar month is
(27.83)(24)(60)(60)=2,404,513 seconds. Thus,…
The tables give
Which is an average for the lunar orbit in its slightly elliptical orbit.
We need to find G in terms of Lunar Units (LU) as opposed to Astronomical Units (AU), and in
terms of the lunar month as one instead of the earth year. We see G has the same value we had
for solar-earth units:
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
1
k
= G
M + m
4π
2
1
k
= G
M
4π
2
1
k
= (6.674E − 11)
(
5.9722E 24kg + 7.347731E 22kg
4π
2
)
(6.674E − 11)
(
6E 24kg
4π
2
)
4E14
4π
2
=
1E14
π
2
= 1E13
a
3
= (1E13)(2,404,512)
2
= 5.78E 25
a = 3.8668E8m
a = 3.833E8m
of 15 22
=
We only need to convert to m, kg, s by using their relationships with LU, LM, and . That is,
for the orbital velocity is always six:
Where is the orbital velocity of the moon, and is the orbital velocity of the earth. The
difference comes when converting the system from lunar units in the case of the first to kg/m/s
and from astronomical units in the case of the second to kg/m/s. For example the first is:
Which is correct, the tables give that the lunar orbital velocity is 1000 m/s.
To find logos, is to find Nature not just as number, but as dimensionless whole
number, because fractions have inherent in them irrational numbers, which are
unending decimal expressions that at some point must be rounded to so many
figures to put them to use. Logos then is the language of Nature that transcends
conundrum because it would be utilized as relationship between point, plane, and
line as opposed to units. We have found that for the orbital velocity:
Equation 9.1
Where is the earth orbital radius. This results in the orbital velocity for a orbiting body is
Equation 9.2.
Which is logos because it is number, which is 6. The orbital velocity of a body is always 6
because:
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
M
⊕
v
m
=
5
3
(40)(1)
ϕ
= 6
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
v
m
= 6
3.8668E8m
2,404,512s
= 964.886m /s
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
e
v =
5
3
GM
ϕ
v
m
=
5
3
(40)(1)
ϕ
= 6
of 16 22
Where is the orbital velocity of the moon, and is the orbital velocity of the earth, That is G
is always 40, and M is always 1 providing the orbit is circular. Let us show this for Venus. Its
orbital distance VU (Venus units) is 1.082E11meters. It orbital period is 1.94E7s is the Venus
year (VY).
=
Thus we have
We can then express all orbital velocities as 6, but to find there values in a formal system of
units, we need to convert from these natural units to something like kg/m/s. Thus logos
translated into a language we understand can best be done in a square array, or as a matrix
transformation. We have
You will find this gives , ,
All of which are correct within the variations of these velocities in their deviations from a
perfectly circular orbit. Logos is:
Equations 9.3
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
v
=
5
3
(40)(1)
ϕ
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
v
m
= 964.886m /s
v
e
= 28,443m /s
v
v
= 33,463.9m /s
(
5
3
GM
ϕ
5
3
GM
ϕ
5
3
GM
ϕ
)
R
m
T
m
R
e
T
e
R
v
T
v
=
v
m
v
e
v
v
=
6
6
6
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
of 17 22
, such that
which is given by
To say that logos is:
Is to say that for the solar system Kepler’s Law of Planetary Motion holds:
That in general
Because as such for Earth we have
For the Moon we have
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1VU
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
of 18 22
For Venus we have
Making
, ,
This is derived from
Which is to say
Is logos because
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
V U
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
=
5
3
(40)(1)
ϕ
= 6
v
v
=
5
3
(40)(1)
ϕ
= 6
G
Mm
R
2
=
mv
2
R
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1VU
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
of 19 22
such that
which is given by
Which is the relationship between point, plane, and line.
From everything we have said G, the gravitational constant is about 40
We can do this for any planet and get G is approximately 40. We found the orbital velocity of any
planets is 6. This is true because as we have shown, the orbital distance of any planet is
This gives since
That orbital velocities are for the moon, earth, and venus
But what does this mean? It means since
And, v=6, that where the orbital velocities of the planets are 6, their distances from the sun are
all one.
This gives:
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
AU
3
(Ear th Dista nceMeters)
3
⋅
Sol arMa ssKilogra m s
M
⊙
⋅
(Ear thOrbit a l Per iod Secon d s)
2
year
2
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
GM
R
=
5
3
GM
ϕ
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
R =
GM
v
2
R =
40
36
≈ 1
of 20 22
But where six is
3.0 The Earth is the Basis or Eigenvalue
Since equation 1.9 is
Equation 1.9
And since we have said
Which we can write
Equation 3.1
Where r and T are orbital radii and orbital periods subscripted with their designations, then we
see Earth is the eigenvalue because
Equation 3.2.
Which means
Or,…
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
k v
e
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
(
6 6 6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
(
6 k v
e
6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
k v
e
r
e
T
e
= v
e
of 21 22
Equation 3.6.
That is since Eigenvalues are characteristic roots, that is the value such that for the matrix A:
k
r
2
e
T
2
e
= v
e
λ
A
x = λ
x
of 22 22
The Author