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An Interesting Construct In Nature
Ian Beardsley
April 8, 2026
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Introduction
In order to introduce this interesting construct found in Nature, we must first outline my
discovery of a 1-second invariant connected to the atom’s particles and the Solar System. The
interesting construct will follow.
The discovery finds an extraordinary harmony between the base 60 counting system and division
of time by the ancient Sumerians, 24 hour day, 60 minute hour, 60 second minute where 24 is
2(12), 12 evenly divisible by 1, 2, 3, 4, 6, and 60 evenly divisible by 1, 2, 3, 4, 5, 6, 12, 15, 30
(archaeology), perfect eclipses, earth size, mass, and gravity (planetary science), and the radius
and mass of the particles that make the atom: proton, electron, and neutron (particle physics),
among other factors noted in the paper.
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I have found a one second invariant that predicts the proton, electron, and neutron [1]:
is the fine structure constant
is the coupling constant for the proton
is the coupling constant for the neutron
is the coupling constant for the electron
is the radius of the particle
is the mass of the particle, G universal constant gravity, h Planck constant, c light speed.
I have found the same one second invariant in a quantum analog for the Solar System:
This I use as the ground state. is the mass of the Moon. is a Solar system Planck-type
constant given by:
We say is given by:
That is, we use the kinetic energy of the Earth. The quantum analog for Earth is given by the
ground state equation above, the Earth orbit is given by:
is the radius of the Sun. is the radius of the Moon. is the Earth orbit.
t
1
=
r
i
m
i
⋅
πh
Gc
⋅ κ
i
t
1
= 1 second
α = 1/137
κ
p
=
1
3α
2
κ
n
=
1
3α
2
κ
e
= 1
r
i
m
i
ℏ
2
⊙
GM
3
m
1
c
=
3.0281E8 m
299,792,458 m/s
= 1.010 seconds ≈ 1 second
M
m
ℏ
⊙
ℏ
⊙
= (1 second) K E
earth
ℏ
⊙
1.03351 s =
1
3
⋅
h
α
2
c
2
3
⋅
π r
p
Gm
3
p
ℏ
⊙
= (1.03351 s)(2.7396E 33 J) = 2.8314E 33 J ⋅ s
K E
Earth
=
1
2
(5.972E 24 kg)(30,290 m/s)
2
= 2.7396E 33 J
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
R
⊙
R
m
n = 3
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The Interesting Construct
Not only is the Earth rotation right to produce a 1-second invariant when using base 60 counting,
but its mass and size is right for a gravitational acceleration g at its surface that produces close to
a meter, when used with such a 1-second invariant. Further, this involves the eclipse ratio for the
Earth/Moon/Sun System (400) in that:
(24 hours)(60 min)(60 sec)=86,400 seconds/day
Two smallest primes: 2, 3: 2 x 3 =6 (fundamental to doing arithmetic)
(400)(6)(6)(6)=(86,400 seconds)/day
Six is the regular hexagon used by bees to make their honeycombs because the six-sided regular
hexagon tessellates and has it radii equal to its sides, creating mathematical advantage.
The eclipse is near perfect. The condition is
Close to a meter is produced by the half period of a pendulum (1 swing) that is the 1-second
invariant. This was discovered by Christiaan Huygens, and he proposed defining the meter with
it. He found since , that
Further Considerations
We further find the kinetic energy of the Moon to that of the Earth maps the 24 hour day into 1-
second.
Where is the inclination of the Earth to its orbit. We also have for the atom’s particles
, ,
R
⊙
R
moon
=
r
earth
r
moon
= 400
g ≈ π
2
m /s
2
2 seconds ≈ 2π
1 meter
g
K E
moon
K E
earth
(24 hours)cos(θ ) = 1 second
θ = 23.5
∘
m
p
= κ
p
⋅
π r
2
p
F
n
G
F
n
=
h
ct
2
1
t
1
= 1 second
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It is suggested that when we push on a particle we rotate some of its temporal velocity into
spacial velocity and resistance to this rotation is experienced as the normal force pushing back
creating inertia that we experience as mass.
is this normal force.
The normal force has a relationship to the Planck force, the maximum gravity for the minimum
mass. It links the normal force to a full rotation ( ). We have the normal force
We have the Planck force for gravity
Where, is the Planck mass, and is the Planck length. They are given by:
And, Planck time is:
We form the ratios between the normal force and Planck force:
Divide by Planck time squared and we have:
That number is . We have the final equation:
F
n
2π
F
n
=
h
ct
2
1
= 2.21022E − 42N
F
Planck
= G
m
2
P
l
2
P
= (6.674E − 11)
(2.176434E − 8kg)
2
(1.616255E − 35m)
2
= 1.21020E44N
m
P
l
P
m
Planck
=
ℏc
G
= 2.176434E − 8kg
l
Planck
=
ℏG
c
3
= 1.616255E − 35m
t
Planck
=
ℏG
c
5
= 5.391247E − 44s
F
n
F
Planck
= 1.826326E − 86
F
n
F
Planck
1
t
2
P
= 6.2834743s
−2
2π
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From the Planck units we have:
So, it can be written:
We can write
is a full rotation, so we can define an angular frequency, :
Integrating one more time gives the angle over 1-second:
t
1
= 2π
F
Planck
F
n
⋅ t
P
= 1.00seconds
F
Planck
= G
m
2
P
l
2
P
=
c
4
G
t
1
= 2π
c
4
GF
n
⋅ t
P
F
n
= 2πF
Planck
⋅
t
2
P
t
2
1
2π
ω
F
n
= F
Planck
⋅ t
2
P
⋅
dω
dt
F
n
F
Planck
⋅
1
t
2
P
∫
1second
0
dt = ω
1
ω
1
=
2π
secon d
F
n
F
Planck
⋅
t
1
t
2
P
∫
1 second
0
dt = θ
1
F
n
F
Planck
⋅
t
2
1
t
2
P
= θ
1
θ
1
= 2π
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Thus, the normal force is the force that, when scaled by the Planck force and the Planck time,
gives a full angular displacement in one second. This geometric origin explains why
appears as a natural invariant. We see the second arises naturally from Planck-
scale physics through a factor of .
It might make sense to say: One second is the time it takes for the ratio to accumulate a
full of angular phase, closing a loop in the temporal dimension – out of the temporal and
back in again.
This is reminiscent of the idea in some quantum gravity or pre-geometric models that time
emerges from a cyclic variable. The equation may be hinting at exactly that: the normal force
(which was previously linked to inertia and mass) is the “restoring force” that makes the cycle
close after exactly one second.
Considering the Huygens Pendulum:
We have:
It is really worth noting in this kind of work, there is no such a thing a a precise measurement.
You can’t define the exact radius of the Sun, its atmosphere get’s thinner and thinner, may end
out near the a planet, and where it ends you can’t say exactly because it is plasma. Nature, I
think, uses approximations, and they are good, and probably have a function. You need room to
jiggle, a certain amount of play. I think sometimes Nature uses the ratio of the perimeter of a
regular hexagon to its diameter — which is 3, as opposed to the irrational pi. As Alan Alda said
in a Woody Allen movie when describing the art of humor: “If it bends, its is funny; if it breaks,
it is not funny”. So when we say the eclipse ratio is 400, not 400.5, we are talking of this kind of
functionality.
Demonstrating That Our Equations Are Correct
We know that our equations are correct because if the phenomenon of mass is generated by the
hyperbolic rotation of when we push on a particle, out of the temporal into the spacial, giving
some of its temporal velocity to the spacial, creating a push back we experience as mass, then to
have
F
n
2π
t
1
= 1 second
θ
1
= 2π
F
n
F
Planck
2π
2 seconds ≈ 2π
1 meter
g
2 seconds ≈
F
n
F
Planck
⋅
t
2
1
t
2
P
1 meter
g
F
n
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We must have
Thus having derived the 1-second invariant
Which means
Because the resistance to rotation is measured by the stiffness of space ( ) and the cross-
sectional surface area of the particle exposed to , which is , or in general . In
general we have
, , ,
. We can verify this:
Proton: , :
Neutron: :
F
n
=
h
ct
2
1
t
1
= 1 second
t
1
=
r
i
m
i
⋅
πh
Gc
⋅ κ
i
m
p
= κ
p
⋅
π r
2
p
F
n
G
G
F
n
π r
2
p
A
i
= π r
2
i
m
i
= κ
i
⋅
π r
2
i
F
n
G
F
n
=
h
ct
2
1
F
n
=
6.62607015 × 10
−34
J·s
(299,792,458 m/s)(1 s)
2
= 2.21022 × 10
−42
N
t
1
= 1 second
t
1
=
r
i
m
i
⋅
πh
Gc
⋅ κ
i
κ
p
=
1
3α
2
α = 1/137
t
1
=
0.833 × 10
−15
1.67262 × 10
−27
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 6256.33 = 1.00500 seconds
κ
n
=
1
3α
2
t
1
=
0.834 × 10
−15
1.675 × 10
−27
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 6256.33 = 1.00478 seconds
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Electron: :
We suggest for the electron may be because it is the fundamental quanta (does not consist
of further more elementary particles). G has been rounded to 6.674E-11.
. (Neutron radius)
. (Classical electron radius)
Further Indication That We Are Correct
I didn’t arrive at all of this by just guessing. I found two equations where the proton radius to its
mass produces 1-second:
These two directly yield:
If the proton radius is . And, if
,
We know the two equations are correct because they yield the proton radius accurately if we
equate the left hand sides. They give it as:
We know this is correct because it is given by
κ
e
= 1
t
1
=
2.81794 × 10
−15
9.10938 × 10
−31
⋅
π ⋅ 6.62607 × 10
−34
(6.67430 × 10
−11
)(299,792,458)
⋅ 1 = 0.99773 seconds
κ
e
= 1
r
n
= 0.84E − 15m
r
e
= 2.81794E − 15m
ϕ ⋅
π r
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1 second
1
6α
2
⋅
r
p
m
p
4πh
Gc
= 1second
m
p
= κ
p
π r
2
p
F
n
G
r
p
≈ ϕ ⋅
h
cm
p
F
n
=
h
ct
2
1
t
1
= 1second
r
p
≈ ϕ ⋅
h
cm
p
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, , : Planck energy, Einstein mass-energy equivalence,
and the Compton Wavelength, respectively, if we introduce the factor of .
I explain the factor of by invoking Kristin Tynski [2], her paper titled: One Equation, ~200
Mysteries: A Structural Constraint That May Explain (Almost) Everything.
Tynski shows that for any system requiring consistency across multiple scales of observation has
the recurrence relation:
scale(n+2)=scale(n+1)+scale(n)
Which she says leads to
λ²=λ+1
Whose solution is .
We have:
The CODATA value from the PRad experiment in 2019 gives
With lower bound , which is almost exactly what we got.
For the proton radius in our computations we use [3]
"A measurement of the atomic hydrogen Lamb shift and the proton charge radius"
by Bezginov, N., Valdez, T., Horbatsch, M. et al. (York University/Toronto)
Published in Science, Vol. 365, Issue 6457, pp. 1007-1012 (2019).
It has a value of
E
p
= hν
p
E
p
= m
p
c
2
λ
p
= h /(m
p
c) = r
p
ϕ
ϕ
Φ = 1/ϕ
r
p
= ϕ ⋅
h
cm
p
r
p
= (0.618) ⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.8166E − 15m
r
p
= 0.831f m
±
0.014f m
r
p
= 0.817E − 15m
r
p
= 0.833f m
±
0.012f m
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References
[1] Beardsley, I (2026) The Curious 1-Second Structure In Nature, From The Atom To The Solar
System, https://doi.org/10.5281/zenodo.19456584
[2] Tynski, K. (2024). One Equation, ~200 Mysteries: A Structural Constraint That May Explain
(Almost) Everything.
[3] Bezginov, N., Valdez, T., Horbatsch, M. et al. (York University/Toronto)
Published in Science, Vol. 365, Issue 6457, pp. 1007-1012 (2019) "A measurement of the atomic
hydrogen Lamb shift and the proton charge radius"
