The Moon Orbital Triangle (General discussion) (V) (Revised)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Moon Orbital Triangle (General discussion) (V) (Revised)
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt – 12th
December 2020
Abstract
Paper hypothesis:
- The moon orbital motion causes the moon orbit regression 19 degrees per sidereal
year (365.25 days)
Paper Argument:
The Moon Orbital Triangle Basic Uses:
(1) The moon uses Pythagoras triangle as one of its orbital motion techniques,
based on that, the triangle supports the moon motion equation definition.
(2) There's 2nd
force effects on the moon orbital motion and this force can be
concluded from the moon orbital triangle
(3) There's 2nd
orbit for the moon motion, and this orbit can be discovered by the
moon orbital triangle data analysis
(4) The moon orbital triangle declination angle (1.1 degrees) on the horizontal
level has a massive geometrical effect on the moon motion.
(5) The triangle shows that, the moon daily displacement (88000 km) is defined
based on geometrical rules.
(6) The Triangle shows why the moon can't move beyond apogee orbit (r= 0.406 mkm).
Paper conclusion:
1. Earth Does A Displacement As A Result Of The Moon Orbit Regression.
2. Earth and its moon motions are interacted with each deeply.

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Contents
Subject Page number
1- Pythagoras Triangle Technique 3
2- Another force (2nd
force) effects on the moon orbital motion 10
3- The moon orbital motion has another orbit (2nd
Orbit) 12
4- The moon orbital triangle declination angle (1.1 degrees) effect 13
5- The moon daily displacement (88000 km) definition. 15
6- The Apogee Orbital Radius Definition (r= 0.406 mkm) 16
7- The moon orbital motion causes the moon orbit regression 19
degrees per a sidereal year (365.25 days)
25
8- The Earth an Moon Motions Interaction 39
9- The moon orbital triangle data. 44

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1- Pythagoras Triangle Technique
1-1 Preface
1-2 The moon uses Pythagoras triangle in its orbital motion
1-3 The moon motion general description.
1-1 Preface
- The moon uses Pythagoras rule in its orbital motion, this data I have discovered
from very long time because
- The moon 4 basic points are created based on each other by Pythagoras rule,
- Let's prove that in following….
- The 4 basic points are:
o Perigee radius (r=363000 km), this is the most near point the moon can
reach to Earth, means, the moon can't be nearer to Earth than 363000 km
o Apogee radius (r=406000 km), this is the most far point the moon can reach
from Earth, means, the moon can't be far from Earth than 406000 km
o The total solar eclipse radius (r=373000 km). The total solar eclipse
phenomenon is done when the moon be at distance 373000 km from Earth
or shorter
o The moon orbital distance (r=384000 km). this is the registered distance in
the moon data sheet as the moon orbital distance
- These are the 4 basic points of the moon orbital motion, and any radius of these 4
radius can be produced by Pythagoras rule based on another radius and the
dimension 86000 km – let's see that in following….
o (86000 km)2
+(363000km)2
= (373000 km)2
o (86000 km)2
+ (373000 km)2
= (384000 km)2
o (86000 km)2
+ (384000 km)2
= (393500 km)2
o (86000 km)2
+ (393500 km)2
= (406000 km)2
- As the data shows, the 4 basic points are created by Pythagoras rule depends on
each other by using the dimension (86000 km) as basic dimension in all equations

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- Because of this data, I have concluded that the moon uses Pythagoras rule (or
triangle) in its orbital motion – but ….
- Why does the moon use Pythagoras triangle as one of the moon motion
techniques? Why Pythagoras rule is useful for the moon motion? let's answer this
question in the following point….

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1-2 The moon uses Pythagoras triangle in its orbital motion
- AC = 88000 km = The Moon Daily Displacement
- BC = L = 88000 km Cos (θ)
- Angle (C ) I call angle (θ) which is the smallest angle in the triangle…
- The previous triangle tries to explain the useful technique of Pythagoras rule which
the moon uses in its orbital motion, let's try to summarize this intelligent technique
in following….
o The moon daily displacement = 88000 km, the moon moves this distance
per solar day…. But
o During 29.53 days the total distance will be 88000 km x 29.53 days =2.58
mkm
o Through what orbit the moon can move through by this distance (2.58
mkm)?! This orbital circumference (2.58 mkm) its radius is (406000 km)
o So, if the moon moves per a solar day (88000 km) and this distance is
considered as a displacement for the moon, so the moon will have to move
by this displacement through apogee radius only (r= 406000 km) –
o Shortly
o The moon can moves daily a displacement = 88000 km only if the moon
moves through apogee radius (r= 406000 km) in this case the moon can't
move in any orbit nearer to Earth except apogee radius – means – the moon
will be prisoner in the apogee radius because it moves daily a displacement
= 88000 km.

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o This is the basic point for which the moon uses Pythagoras triangle
technique, because the intelligent moon can move in more nearer orbits by
using Pythagoras triangle technique…
o The moon moves 88000 km and this value doesn't change, the moon moves
it in all cases
o But
o This distance isn't considered as a displacement through the moon orbit,
means… the moon moves 88000 km but the real displacement through the
orbit is less than this value! How
o As in the triangle ….. the moon moves along the right triangle hypotenuse
(AC), which equal the displacement =88000 km, but the moon creates an
angle (θ) between its motion direction and the horizontal level (BC) by that,
the real displacement through the moon orbit is (L) = 88000 Cos (θ)
o That creates another displacement through the orbit different from the moon
motion target per solar day (88000km) and because the real displacement
(L) is shorter than the moon daily target (88000 km), the moon could move
through orbits more near to the Earth
o But,
o The angle (θ) has a massive effect on the moon motion
o Why….?!
o We know that, the full displacement 88000 km per solar day needs to be
passed through a wide orbit (its radius r= 406000 km), and
o If the angle (θ) =28.635 degrees, the real displacement (L) = 88000 Cos (θ)
will be = 77237 km and this displacement can be passed through a narrow
orbit which is perigee orbit (its radius r= 363000 km)
So
o If the angle (θ) be smaller than (28.63 degrees), so the real displacement will
be greater than 77237 km and that needs more wide orbits to be passed

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through, and the wider orbits need the moon to move through higher orbits
above perigee orbit (r= 0.363 mkm).
o That means
o The change in the angle (θ) causes to change the real displacement and the
height of moon motion above perigee radius… and by that the angle (θ)
controls almost the moon orbital motion basic features –because it defines
the real displacement long (L = 88000 Cos (θ)) and based on that change the
moon motion height above perigee radius… i.e.
o The intelligent moon uses Pythagoras triangle in its orbital motion to create
ways to move in more near orbits to Earth without change its motion daily
target 88000 km but in using this intelligent technique the moon put himself
under the angle (θ) control where the angle defines the real displacement
and the moon motion height above perigee.
o Based on that the moon orbital motion equation can be defined based on this
angle (θ)…. The following equation shows that
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o This equation expresses the moon motion changes, let's discuss it …
o The angle (θ) defines the real displacement (L) and the moon motion height
above perigee (r=0.363 mkm), by that, the angle (θ) is the most important
component in Pythagoras triangle which is used by the moon orbital motion,
so the equation defines the change in this angle day after day and by that the
change in the moon motion can be defined by the definition of the angle (θ)
changes… the equation supposes that, the angle is increased by 0.98562
degrees per solar day based on a hypothesis tells us that the moon motion
daily depends on this angle 0.98562 degrees - But why? let's answer in
following…

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1-3 The Moon Motion General Description.
- The moon moves per a solar day a typical motion to Earth motion, to avoid the
separation from the Earth through their motions…i.e.
- The moon moves during a solar day a distance = 2.58 million km with an angle
=0.98562 degrees on the horizontal level revolving around the sun along the Earth
orbital circumference- typically to Earth motion –
- If there's no Lorentz Length Contraction Phenomenon in the solar system motion,
the moon motion would to be seen as a parallel line to Earth motion trajectory.
- But
- Because of Lorentz Length Contraction Phenomenon effected on the moon motion
(2.58 million km) by the rate 1.0725 and causes this distance to be contracted to
be = (2.41 million km)
- The difficulties are started here, because the different distance (0.17 mkm) will
cause inevitably the separation between the moon and Earth motions, and because
of that the moon has to move an additional distance = this different distance = 0.17
mkm to cover this difference on the same day period to avoid the separation from
Earth through their motions…
- So, The moon has found that, there's a target (0.17 mkm) of motion in its waiting.
- Because of that, the moon moves its daily displacement (88000 km) depending on
Earth Gravity Forces …. But
- This displacement (88000 km) isn't enough, and the moon still needs another
displacement (88000 km) to cover the different distance
- Now this description has 3 important information let's write them in following
o (1st
) The angle 0.98562 degrees defines the daily change in the moon orbital
motion because this angle is the moon motion angle and not because it's the
Earth motion angle, the point here is that, the moon moves a motion
typically to Earth motion and because of that, this angle is the moon motion

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angle and because of that this angle 0.98562 deg is used in the moon orbital
motion equation definition, as we have discussed before
o (2nd
) Because the moon needs to move 2 displacement each equal (88000
km) where one displacement only isn't enough… that tells us, there's one
more orbit for the moon motion because where the moon moves this (2nd
)
displacement – the moon must move it through another orbit, but we see
only one displacement (88000 km), so if 2nd
displacement is found, it must
be passed through another orbit of the moon motion !
o But …Why we don't see this 2nd
displacement (88000 km)? why we see
only one displacement if the moon moves 2 of them?!
o In the moon original motion (2.58 mkm), which is contracted to be (2.41
mkm), the moon moves a typical motion to Earth motion, in distance and in
angle where we don't see this motion because the moon moves it’s a long
the Earth orbital circumference revolving around the sun, and we can't see
the motion on this trajectory of motion – any way- this motion (2.58 mkm)
is a fact regardless our vision for it, because Earth moves daily a distance =
2.58 mkm, and if the moon doesn't move this same distance per solar day
the moon would be separated from Earth motion, by that, our vision for
motion can't be the only proof for it, because this motion is not seen but it's
fact more than any other seen motion….
o (3rd
) There's must be 2nd
force effect on the moon orbital motion, that
because the moon daily displacements (2 x 88000 km) is a distance required
to cover one distance and if this motion is done by one force that
necessitates this displacement to be 170000 km in one motion, but it's not by
that, on the contrary, the moon does 2 displacements per solar day each
=88000 km and because of that, each displacement expresses about one
force and these 2 forces effects have an equal effect to each other and
because of that the displacements are equal…

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2- Another force (2nd
force) effects on the moon orbital motion
- The moon orbital triangle is discussed in full details in point no. (6) of this paper
(6- The Moon Orbital Triangle Data) but I have brought the triangle figure here
to facilitate the explanation..
- The apogee point in this triangle at (POINT D) where ED =406000 km
- As we see
- The great triangle 3 points are (E, C and A), point E is Earth and the distance
between E and C = 373000 km = the distance between Earth and moon at total
solar eclipse radius
- But
- What's this point (A)? because this point is found after apogee point (r=406000
km) with a distance =43000 km
- The triangle will be destroyed without this point (A)! but?
- Earth gravity effect reaches to apogee radius only, and the moon can't move
betony apogee radius (r=406000 km),so what is the force which effect on this
point (A)??
- The moon orbital triangle is an easy method to discover this (2nd
) force effects on
the moon orbital motion….

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- Let's try to discover this (2nd
force)…
- We have 2 reasons at least to suppose that, there's 2nd
force effects on the moon
orbital motion these 3 reasons are
o (1st
reason) The Point (A) in the moon orbital triangle tells us that there's
another force effects on the moon orbital motion
o (2nd
reason)The 2 displacements (2 x 88000 km) according to the previous
analysis tells us that there's one more force effect on the moon orbital
motion
o (3rd
reason) Because Earth Mass = the 4 inner planets masses total (error
1%), this equality suggests that, there's one more force equal Earth gravity
can effect on the moon orbital motion- the distances between these inner
planets and the moon may prevent the gravity effect, but I see this is a
difficulty to be solved but the chance still potential because 2 masses are
equal so they can produce similar to gravity forces and can cause motion for
2 displacements each =88000 km.
2nd
force effect on the moon orbital motion
- 149.6 mkm (Earth orbital distance) = 14984 km (Jupiter diameter) x 1047
- 1047 = The Sun Mass / Jupiter Mass
- The equation tells that, Earth orbital distance is defined based on some interaction
found between the sun mass gravity and Jupiter mass gravity, I want to say that,
Jupiter effects on the sun gravity to define Earth orbital distance..
- By this same rule,
- The sun and Jupiter masses gravity has interactive effect on the Earth moon
motion and by this interactive effect with Earth gravity, another force is created
which effects on the moon orbital motion,
- So this second force is created by an interaction between the sun, Earth and Jupiter
masses gravity
- But, Jupiter gravity is supported by the inner planets gravity in this process

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3- The moon orbital motion has another orbit (2nd
Orbit)
- This conclusion is a simple and clear
- Because
- The moon moves 2 daily displacements each =88000 km under effect of 2 forces,
so one displacement we see through the moon orbit around Earth but the other
displacement we don't
- There must be another orbit for the moon motion an the moon moves its second
displacement (88000 km) through this second orbit
- Where is this second orbit?
- We have 2 features to find this second orbit
o (1st
Feature), the distance between The Point (A) (in the moon orbital
triangle) and the end of the lunar umbra length (1.392 mkm) – let's explain
that….
o The lunar umbra length =1.392 mkm = the sun diameter … but
o From E (Earth) to The point (A) = 449197 km = Jupiter Circumference
o The difference 1.392 mkm – 449197 km = 942803 km
o The triangle ACE perimeter = 942803 km
o That means, The point (A) separates between 2 equal values or 2 equal
triangles, one which we know (the moon orbital triangle) and the second is
seen as the triangle perimeter in the distance between these 2 points (The
point (A) and the lunar umbra length end point) – the data tells us that, the
2nd
orbit is found as a neighbor for the 1st
orbit and found between the first
orbit and the lunar umbra length end point…
o (2nd
Feature), There's some angle 0.8 degrees between these 2 orbit, as we
have discussed that before with Venus Axial Tilt discussion.

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4- The moon orbital triangle declination angle (1.1 degrees) effect
- As we know, The base triangle AE declines on the horizontal level with an angle
1.1 degrees, and this angle has many significant effects on the moon orbital
motion, so let's remember them in following
o Earth angle =13.3 degrees and if we add the angle 1.1 deg which is found
under E, so the total will be 14.4 degrees (where this value is effective
greatly in the moon orbital motion)
o Also
o We know the angle 90.9 degrees which is found inside the moon diameter
(Please review the point no. 8-3 for the full details)… so
o 90.9 degrees x 1.1 degrees = 100 degrees the result (100 deg) has a massive
effect on many geometrical interaction occurred during the moon orbital
motion – first let's know how this angle is created (100 deg)?
o 100 degrees = 90.9 degrees +9.09 degrees
o So what's happened is that, 0.1 degree defines the value (9.09 deg) from the
origin 90.9 deg and then both are added to each other…
o Why 100 deg is useful? because Venus axial tilt was 1.774 degrees (where
Venus diameter =1.774 x Mars diameter) but after the creation of the angle
100 deg, Venus axial tilt 1.774 deg became 177.4 degrees and based on that
the moon orbital inclination (5.1 deg) is created because (360 deg –(177.4
deg x 2) = 5.2 deg (5.1 deg + 0.1 deg)

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o Where Mars orbital inclination was 1.8 degrees and became 1.9 degrees, this
information can easily be known also because (Earth diameter/7070 km)
=1.8 but (Earth diameter /6792 km Mars diameter) =1.9, and that means,
Mars ancient diameter (7070 km) creates an orbital inclination =1.8 deg, but
after Mars Migration, and Mars diameter is changed to be (6792 km)
because of the collisions with planets, so the rate became 1.9 and because of
that Mars orbital inclination =1.9 degrees
o Please remember we know that Mars ancient diameter =7070 km because it
meets the equation (D= Rx 1092
) where Mars ancient orbital distance = 84
mkm, so mars ancient diameter =7070 km.
o Please note
o The angle 1.1 degrees has many other massive effects on the moon orbital
motion and other planets motions geometrical interactions.

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5- The moon daily displacement (88000 km) definition.
- In The Triangle (BCZ)
- Please review this triangle in the moon orbital triangle,
- The triangle hypotenuse (CZ) = 88000 km
- The angle (ZCB) = 12.195 degrees
- This data provides some interesting information, because the triangle hypotenuse
(CZ) defines a distance =88000 km where the moon daily displacement = 88000
km …..and
- The angle =12.195 degrees where (13.18 degrees – 0.98562 deg =12.195 deg)
- We know that, 13.18 degrees = the moon motion degrees per a solar day
- And 0.98562 degrees is the daily motion degrees for Earth and the moon motions
- Why this data is interesting?
- Because the moon daily displacement is defined by Pythagoras rule depends on the
angle (12.195 degrees), that shows the moon daily displacements (2 x 88000 km)
is found based on this angle (12.195 deg) where the basic motion distance (2.58
mkm) which is contracted to be (2.41 mkm) depends on the angle (0.98562 deg)
that proves my claim, that the moon moves a typical motion to Earth motion and
the displacements (88000 km x 2) the moon moves by necessary because of the
contraction.

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6- The Apogee Orbital Radius Definition (r= 0.406 mkm)
6-1 The Apogee Radius Analysis
- In the next point (No. 7), this triangle (the moon orbital triangle) is discussed with
full details, but I have removed many details from this one, because we discuss the
triangle (ECD). This is our triangle and we need to consider it for a while..
- We need in this triangle:
o The angle E (Earth) = 13.33 degrees
o The angle (EDC) = 63.4 degrees
o The angle (ADC) = 116.6 degrees
o The angle (ECD) = 103.3 degrees
o The angle under (E) =1.1 degrees
o The distance ED (Apogee Radius) = 406000 km
o EC = 373000 km
o CD = 96151 km
o CB is perpendicular on the Base (ED)
o The distance EB (Perigee Radius) = 363000 km
o CB =86000 km
- Let's remember this point question in following:
- Why the moon can't move beyond Apogee Orbital Radius (r= 0.406 mkm)?
Let's try to answer in following:

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I- Data
(6-1)
0.114373 x 116.6 degrees = 13.33 degrees
(6-2)
(365.25 days /27.32 days) = 13.3693
(6-3)
(13.3693/13.33) = (361/360)
(6-4)
26.6 degrees = 14.4 degrees +12.195 degrees
II- Discussion
Equation No. (6-1)
0.114373 x 116.6 degrees = 13.33 degrees
0.114373 degrees = 1.1 degrees – 0.98562 degrees
- As we know, the triangle base (EA = 449197 km) declines on the horizontal level
with an angle = 1.1 degrees
- Earth moves per solar day = 0.98562 degrees
- (note please, we accept that, the moon moves a typical motion to Earth motion and
because of that, the angle 0.98562 degrees can be used for the moon motion also)
- The difference between these 2 angles = 0.114373 degrees
- What does Equation no. (6-1) tell us?
- The angle (E) = (13.33 deg.) is defined based on the angle (CDA=116.6 deg.
because of the difference (0.114373 deg)! but this angle is complementary to the
angle CDE = 63.4 deg.! how to understand that?
- Because of the difference (0.114373 deg.) these 3 angles are defined based on the
angle (E =13.33 deg)…. The angle (E=13.33 deg) is their master one, and if this
angle is defined as (13.33 deg) because of the difference (0.114373 deg) the
triangle other angles will be defined accordingly!

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- Because the angles are defined obligatory based on geometrical rules, that
necessitates the triangle dimensions to be defined accordingly, and that means, the
distance (ED = 406000 km = apogee radius) is defined by geometrical necessity.
- It's one thread, and its first point is the angle (E=13.33 degrees), if this angle is
defined by this value, so all other angles and dimensions will be defined by the
geometrical necessities
- So, the question should be…
- Why the angle (E) = 13.33 degrees? Let's try to answer in following
Equation No. (6-2)
(365.25 days /27.32 days) = 13.3693
- Earth Orbital Period =365.25 days and The Moon Orbital Period =27.32 days
So
- Between the 2 planets orbital periods there's a rate = 13.3693
Equation No. (6-3)
(13.3693/13.33) = (361/360)
- The rate (13.3693) which we have found between the 2 planets orbital periods is
rated to the angle (13.33 degrees) by the rate (361/360), and what does that mean?
- 360 degrees = any planets revolution round the sun or rotation around its axis
- 361 degrees = the moon orbit regression during Metonic Cycle, because the
moon orbit regresses 19 degrees per sidereal year (365.25 days), so during 19
sidereal years, the moon regresses 361 degrees, so the previous equation shows 2
motions, one of them is the moon regression…
- Equation no. (6-3) tells that, by the moon regression (361 degrees), the rate 13.693
which we have found between the 2 planets orbital periods will be changes to
13.33 which can be used as (13.33 degrees) which is the angle (E) =13.33 deg. in
the triangle (ECD).

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Equation No. (6-4)
26.6 degrees = 14.4 degrees +12.195 degrees
- 13.18 degrees (the moon daily motion degrees) – 0.98562 degree= 12.195 deg.
- The angle (E =13.33 degrees) + the angle under (E) (= 1.1 deg) = 14.4 degrees
- The angle 26.6 degrees = The angle (BCD)
Note Please
- The angle ECD = 103.3 degrees = 90 degrees +13.33 degrees
- What conclusion this triangle (ECD) can provide us?
- I try to prove that, the triangle angles and dimensions are created by geometrical
necessities depend on the angle 13.33 which is defined as a rate between the 2
planets orbital periods, that means, because the 2 planets orbital periods = (365.25
days and 27.32 days), the moon has to move to apogee radius only (r=0.406 mkm)
and can't move beyond this point because the triangle data is created based on each
other by geometrical necessities…. The major point here is that, this conclusion
taken into consideration the moon orbit regression 19 degrees per sidereal year
(365.25 days), that means, because of the moon regression 19 degrees per sidereal
year, the angles (1.1 degrees) and (0.98562 degrees) caused to define the triangle
angles and dimensions based on each other by geometrical necessities which
causes the moon motion to reach only to apogee radius (r=0.406 mkm) and can't be
beyond it any more …
Shortly
- The moon orbit regression causes the moon apogee radius to be =0.406 mkm and
prevents the moon to move to any far point after this point (r=0.406 mkm) but this
effect is done by a contribution from Earth motion – to see that as deep as possible
let's see at first Venus effect on Earth And Its Moon Motions.

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6-2 Venus Effect On The Earth And Moon Motions.
(6-5)
389.2 degrees = 2.195 degrees x 177.4 deg (Venus Axial Tilt)
(6-6)
177.4 degrees = 13.33 x13.30
(6-7)
29.2 degrees = 2.195 degrees x 13.30
(6-8)
12.195 degrees = 10 degrees +2.195 degrees
(6-9)
389.2 degrees = 26.6 x 14.61 (but 14.61 = "14.4/0.98562")
- Earth moves during 29.53 solar days a value =29.2 degrees
- The moon moves during 29.53 solar days a value =389.2 degrees =360+29.2
- This information is an old one, but we didn't know why both planets move the
same number of degrees because Earth daily moves (0.98562 degrees) but the
moon moves daily (13.18 degrees), and where the final value during 29.53 days for
both =29.2 degrees… we hoped to know why?
Equation No. (6-5)
389.2 degrees = 2.195 degrees x 177.4 deg (Venus Axial Tilt)
- 2.195 degrees = 12.195 degrees – 10 degrees (as equation 6-8 shows), the value 10
degrees = (1/0.1 deg), and this 0.1 deg is used from the angle 1.1 degrees
- Equation no. (6-5) tells that, the value 389.2 degrees (= the moon motion degrees
during 29.53 solar days), this value is related to Venus axial tilt, as the Equation
shows… Shortly, Venus axial tilt connects between the values (389.2 degrees) and
(29.2 degrees) in the moon and Earth motions respectively… this is the effect
Venus does on the moon and Earth motions… through Venus Axial Tilt Venus
connects these 2 values (389.2 and 29.2) together which guarantee the harmony in
Earth and Moon motions…

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Equation No. (6-6)
177.4 degrees = 13.33 x13.30
- This equation tells us why there's confusion in the data? Because there are 2 values
(Equal Each Other Approximately 13.33 and 13.30). but both values are rated with
Venus Axial Tilt…we will see that, how Venus axial tilt is so useful in the
following equation
Equation No. (6-7)
29.2 degrees = 2.195 degrees x 13.30
- 29.2 degrees = Earth motion during 29.53 solar days, and this value can be defined
by the angle 2.195 degrees and the rate 13.30,
- The equations order is useful because it shows that the rates (13.33) and (13.30)
are different but rated to Earth another inside Venus Axial Tilt…. Means
- By Venus Axial Tilt 177.4 degrees, the rate 13.33 which is related to the moon
motion degrees during 29.53 solar days (=389.2 degrees) will be rated to the rate
13.30 which is used between Earth motion degrees during 29.53 days (29.2
degrees) and the angle 2.195 degrees
- Simply,
- Venus is the table on which both planets motions share their data and create a
harmony between both and because of that Earth moves during 29.53 solar days a
number of degrees = perfectly the moon motion degrees during the same period
(389.2 degrees = 360 +29.2 degrees).
- Now let's discuss the last equation (6-9) in following

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Equation No. (6-9)
389.2 degrees = 26.6 x 14.61 (but 14.61 = "14.4/0.98562")
14.4 degrees = the angle (E=13.33) + the angle under E (1.1) = 14.4 deg
26.6 degrees= the angle BCD
0.9856 deg = the Earth motion degree per a solar day
- This is 2 equations, let's separate them in following…
389.2 degrees = 26.6 x 14.61
14.61 = 14.4/0.98562
- What does equation no. (6-9) tell us?
- It tells the angle 389.2 degrees is defined depends on the angle 0.98562 degrees
with some complex mechanism…
- The mechanism uses the angle 14.4 degrees = 10 x 1.44 deg (the moon orbital
regress 1.44 degrees per month),
- How to understand that?
- Equation No. (6-9) tells that, the moon motion degrees (389.2 degrees) is create
because of the moon regression yearly 19 degrees
- The point is that, the harmony of the moon and Earth motions depends on the
moon orbit regression and by using Venus Axial Tilt….!
- That tells us, the moon orbit regression 19 degrees per sidereal year depends on
Venus Axial Tilt!
- The question now will be, why and how Venus axial tilt can effect on the moon
orbit regression.

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6-3 Earth Displacement As A Result For The Moon Orbit Regression
(6-10)
(177.4/1.44) = (29.2/ 0.11437)
(6-11)
12.59 x 0.11437 = 1.44
Equation No. (6-10)
(177.4 degrees /1.44 degrees) = (29.2 degrees / 0.11437 degrees)
- 177.4 degrees = Venus Axial Tilt
- 1.44 degrees = the moon orbit regression per month
- 29.2 degrees = Earth motion during 29.53 solar days
- 0.11437 deg = (1.1 degrees) – 0.98562 degrees
- 0.98562 deg = Earth motion per solar day
- 1.1 degrees = the angle under (E)
- Equation no. (6-10) tells us that, Earth motion degrees during 29.53 solar days is
defined based on 3 values!… Let's ask, to whom the value (0.11437 deg) belong?
- Because we know that, the moon moves a typical motion to earth and the moon
uses the angle (0.98562 degrees) as Earth does, so to whom this difference belong?
- The difference 0.11437 degrees is belong to Earth motion! how to know?
- Because of the value 29.2 degrees because the moon moves during this same
period 389.2 degrees… so equation no. (6-10) expresses Earth motion!
- Equation (6-10) tells that, Earth motion during 29.53 days produces 29.2 degrees
and this value depend on Venus axial tilt, the moon orbit regression and the
difference between (1.1 deg -0.98562 deg)! why?
- It's an old question, let's remember it!
- The question is, if Earth is a fixed point in the space can the distances between the
Earth and the moon be without change if the moon orbit regresses 19 deg yearly?

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- We have discussed that before, and we have discovered that, it's not possible,
- If Earth is a fixed point in space and the moon orbit regression will cause the
distances between the Earth and the moon to be changed… what does that mean?
- Earth does some displacement as a result for the moon orbit regression… and
this displacement of Earth is seen in the difference (0.11437 deg= (1.1deg–0.985d
deg) this difference shows The Earth Displacement which is done but hidden
behind different Earth motions
The Conclusion
- Earth does displacement as a result of the moon orbit regression 19 degrees per
sidereal year, and this displacement is part of the Earth general motion and seen in
Earth motion degrees (29.2 deg) during 29.53 days – this displacement is done by
Earth depends on the angle 1.1 degrees which is created between the moon orbital
triangle base (EA=449197 km) and the horizontal level
- Earth depends on its daily degrees (0.98562 deg) and this angle (1.1 degrees) to
perform this displacement which is required necessary from the Earth to save the
distances between the Earth and moon without changes

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7- The Moon orbital motion causes the moon orbit regression 19 degrees per a
sidereal year (365.25 days)
7-1 More Analysis For The Triangle ECD
- We have more points to analyze in this same triangle, so we need it here, let's
remember its data.
o The angle E (Earth) = 13.33 degrees
o The angle (EDC) = 63.4 degrees
o The angle (ADC) = 116.6 degrees
o The angle (BDC) = 26.6 degrees
o The angle (ECD) = 103.3 degrees
o The angle under (E) =1.1 degrees
o The distance ED (Apogee Radius) = 406000 km
o EC = 373000 km
o CD = 96151 km
o CB is perpendicular on the Base (ED)
o The distance EB (Perigee Radius) = 363000 km
o CB =86000 km
- Still there's one more question left in this triangle ECD let's ask it in following:
- Why The angle (BDC) = 26.6 degrees! Because it should be = 28.63 degrees?!
- But why this angle should be 28.63 degrees? Let's explain that in following:

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- We have discussed this figure before (page no. 5), while we have discussed the
angle (θ) effect on the moon motion… this subject still needs a little analysis
- We know that, the moon motion real displacement (L = 88000 cos (θ)), let's
remember the idea shortly in following:
o The moon uses Pythagoras triangle to decrease its real displacement, and
don't use the whole distance (88000 km) as a real displacement – because –
this distance (88000 km during 29.53 days needs an orbital circumference =
2.55 mkm = 2π x0.406 mkm (Apogee Radius) –means if the moon uses this
distance (88000 km) as a real displacement every day the moon will move
only through the apogee orbit because of the distance huge long, for that, the
moon uses Pythagoras triangle to move a distance =88000 km but with an
angle (θ) which causes the real displacement through the orbit to be shorter
than 88000 km – and by this intelligent technique the moon can move
through more near orbits to the Earth – this is the basic idea – as we have
discussed in point No. (1) of this current paper… but some details can help
our discussion greatly
o The min available displacement = 77237 km and angle (θ) = 28.63 degrees,
why? Because
o Through 29.53 solar days by this real displacement (77237 km) the moon
will move a distance = 2.280 mkm = 2π x 0.363 mkm, and we know that the
distance 0.363 mkm = Perigee Radius, and the moon can't be more near to
Earth than to Perigee point…. That means, Perigee point is the most near
point to Earth the moon can reach, as we know, and because of that, the real
displacement can't be shorter than 77237 km and the angle (θ) can't be

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greater than 28.63 degrees – and that means – the angle (θ) = 28.63 degrees
consists the most near point for the moon motion.
o On the other side
o If the moon moves daily 88000 km as a real displacement, the total distance
during 29.53 days will be = 2.598 mkm = 2π x 0.4135 mkm, this point is
more far from the apogee point (r= 0.406 mkm), that shows the moon never
uses its motion distance (88000 km) as a real displacement along a month….
o If we limited our calculation to the apogee registered radius (r=0.406 mkm),
so the angle (θ) should be = 11 degrees
o At angle (θ) = 11 deg., the real displacement (L = 88000 cos θ) = 86386 km
and during 29.53 days the total distance = 2.55 mkm = 2π x 0.406 mkm
o That means, the moon motion is in a range from (11 degrees) to (28.63
degrees) theoretically, but practically, the angle (θ) can be = Zero even,
because the moon in any day can use the full distance 88000 km as a real
displacement and that causes no serious problem because the moon doesn't
do it daily but for several days at maximum, and because of that the moon
doesn't stay in apogee orbit along the month
o I want to say that,
o The angle (θ) creates limitation for the moon motion in a range between
Zero degree to 28.63 degrees, and by these angles, the angle (θ) defines
each day the moon real displacement and based on this real displacement the
moon motion height above perigee radius will be defined because each
greater displacement needs more wide orbit and that forces the moon to
move through higher orbits above the perigee radius (r=0.363 mkm).
o So This range (Zero deg. to 28.63 deg.) controls the moon motion! but in
the triangle (ECD) we see that, the angle which controls the distance from
perigee to apogee (The angle ECD = 26.6 degrees).
So Why These 2 Angles Aren't Equal?

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- So, the question became clear now…
- We have 2 methods to define the moon motion limitation from perigee to apogee,
and these 2 methods depend on Pythagoras triangle, but the final angle in both
methods are different from each other, they are 26.6 and 28.6 degrees why? they
should be equal because the moon orbital triangle is produced by the moon motion
which uses Pythagoras rule – simply no geometrical reason is found to create these
angles are different from each other – so the question is Why?
- Let's write the answer as a hypothesis here and try to prove it in the following sub-
points ….
- The moon orbit regression (1.44 degrees) per month consumes this difference of
degrees – that means
o Both angles = 28.63 degrees but
o One angle is decreased with 2 degrees because the moon orbit regression
monthly (1.44 degrees)…
o In following sub-points we will discuss and try to know if this hypothesis is
a fact and by what mechanism the moon orbit regression consumes this
difference of degrees….

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7-2 Jupiter 8 days Cycle
I-Data
(7-1)
(6939.75 days /3.3 days) = 2103
(7-2)
6939.75 seconds x 0.3 mkm/sec = 2082 mkm
(2103 x 0.99 = 2082)
(7-3)
2082 mkm = 95.1 x 21.9
(7-4)
8 = 6.7 + 1.3
(7-5)
116.6 =8 x 14.575
(7-6)
6939.75 =5.1 x 1.3 x 1047
II-Discussion
- Let's remember Jupiter 8 Days Cycle before to start our discussion
o Jupiter has a cycle for 8 days. Jupiter day =9.9 hours, so 8 days = 79.2 hours
= 3.3 solar days, but how we know that?!
o Jupiter moves during its day period a distance = its circumference, but not
accurately where there's an additional distance =4%......So
o During 8 days, Jupiter moves a distance = 8 Jupiter circumferences + 1
Jupiter diameter, this data tells us there's some cycle for Jupiter during its 8
days – also
o We have discovered that, Jupiter motion is transported to Saturn, where
Jupiter during 8 of its days moves a distance = 10 days of Saturn motion
distance… we have discussed that deeply before … this data told us that
there's some cycle for Jupiter during 8 of its days…

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Equation No. (7-1)
(6939.75 days /3.3 days) = 2103
- Metonic Cycle period =6939.75 Solar Days ………And
- Jupiter (8 days) Cycle period = 3.3 Solar Days
- I suppose some interaction is happened between Jupiter and the Earth moon
motion and because of that the 2 cycles are rated to each other…
- Equation no. (7-1) tells us that, the rate 2103 is produced between these 2 Cycles
periods…
- We remember the distance 2082 mkm (Jupiter Uranus distance)…
Equation No. (7-2)
6939.75 seconds x 0.3 mkm/sec = 2082 mkm
(2103 x 0.99 = 2082)
- Light known velocity (0.3mkm/sec) travels during 6939.75 seconds a distance =
2082 mkm, and this distance equal 99% of the distance 2103 mkm,
- This distance is a significant distance because, the second period of light motion
became one solar day period of the Earth moon motion and this period (6939.75
seconds) became (6939.75 days = Metonic Cycle)…
- This distance we analyze in point no. (8) with the moon diameter analysis…
- But
- Equations (7-1) and (7-2) tell us that, the distance (2082 mkm) which is seen as a
rate (2082mkm/1mkm) =2082 and with error 1% (2103) is produced between the 2
cycles of the moon and Jupiter motions, that tells us, Jupiter motion interaction
with the moon motion causes a great effect on the moon motion….
- I want to say that….
- Jupiter (8 days Cycle) is interacted with the moon Metonic Cycle…
- Why we need that? because the moon orbit regression is produced by this
interaction between Jupiter and the moon motions – and based on that–the 2
degrees is consumed in the triangle angle (BCD)….

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- i.e. the moon orbit regression depends on the moon orbital motion but not directly,
instead the orbit depends on some specific interaction which we try to analyze in
this point – so it's correct to tell that, the moon orbital motion causes the moon
regression but not as a direct result.
Equation No. (7-3)
2082 mkm = 95.1 mkm x 21.9
Where
2082 mkm = Jupiter Uranus distance
95.1 degrees = 90 degrees +5.1 degrees (The Moon Orbital Inclination)
21.9 = Jupiter Mass / Uranus Mass
- let's remember one old equation in following
- 149.6 mkm (earth orbital distance) = 1047 x 142984 km (Jupiter diameter)
- Where
- 1047 = The Sun Mass / Jupiter Mass
- This equation tells that, Jupiter mass effects on the sun gravity in Earth Orbital
Distance Definition.
- This conclusion we have reach before frequently, many times we have found that,
planets masses rates play an important role in the planets orbital distances
definition, and that tells us there's some complex gravity interaction needs to be
defined in addition to the current methods of gravity definition..
- In our equation no. (7-3), The masses rate between Jupiter and Uranus effects on
the distance 95.1 mkm definition relative to (Jupiter Uranus Distance 2082 mkm),
and this masses rate effects on this distance because these 2 planets are the major 2
effective players on the moon orbital motion…
- The distance 95.1 mkm = 95.1 degrees, So….
- Equation no. (7-3) tells that, the moon orbital inclination (5.1 degrees) is defined
as a result of interaction done between Jupiter and Uranus masses gravity, and that
tells, some strong force is used to create the moon orbital inclination…

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Equation No. (7-4)
8 degrees = 6.7 deg. (The Moon Axial Tilt) + 1.3 deg. (Jupiter orbital inclination)
Equation No. (7-5)
The angle (ADC) 116.6 = 8 degrees x 14.575
- Equations (7-4) and (7-5) try to show the interaction which is done between the
Earth moon, Jupiter and Uranus motions… if such interaction of motions is found
that means, the moon will be in some very specific point because it's an interaction
between 2 great planets (Jupiter and Uranus)… in this interaction we depend on
the number (8) because of Jupiter Cycle Of 8 Days which we have discussed.
- The value 116.6 degrees is so strong because it's consisted of 2 basic values which
are (8 degrees x 14.575) Specifically…. Where
- 8 degrees expresses (Jupiter Cycle of 8 Days)
- 14.575 expresses (Uranus Axial Tilt 97.8 deg. / The Moon Axial Tilt 6.7 deg.)
- To see the depth of this rate 14.575 let's provide the following data
o 14.575 = (Uranus Axial Tilt 97.8 deg. / The Moon Axial Tilt 6.7 deg.)
= (Uranus diameter 51118 km / The Moon diameter 3475 km)
= (Uranus Mass / Earth Mass)
= (Venus Mass / Mercury Mass) (error 1.2%)
= (2 Saturn circumferences / Uranus diameter) (error 1.6%)
= (71492 km Jupiter Radius / 4879 km Mercury diameter)
= (177.4 deg Venus axial tilt / 12.195 deg) (error 1.6%)
= (389.2 deg /26.6 degrees)
= 278.4 degrees /19 degrees)
- The previous data is a part of many similar, then rate 14.575 is a continuous rate
used between huge number of planets data, there's of course a geometrical reason
for such repeated using, But, what we need here to consider is, the moon brilliant
position, because the moon is found inside this relationship and because of that the
moon became specific interaction point between Jupiter and Uranus.

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- To save our discussion goal and don't missed our objective, we need to remember
that, we search for the 2 degrees (from 28.63 to 26.6) which are consumed by the
moon orbit regression (as I supposed) and this consumption is done by a specific
interaction done between Jupiter, Uranus and the moon motions – the previous
equations aim to show that, a deep interaction of motion between Jupiter, Uranus
and the moon is seen in the angle CDE =116.6 degrees in the triangle we analyze
(the triangle ECD).
Equation No. (7-6)
6939.75 days =5.1 degrees x 1.3 degrees x 1047
Where
- 6939.75 days = Metonic Cycle Period
- 5.1 degrees = The moon orbital inclination
- 1.3 degrees = Jupiter orbital inclination
- 1047 = the sun mass/ Jupiter mass
- We know, 1 day = 1 degree, in the moon orbital regression, so
- 6939.75 days = 6939.75 degrees
- What does this equation tell us?
- Equation no (7-6) tells that, Metonic Cycle period is defined based on the moon
orbital inclination (5.1 deg), with a direct effect of Jupiter Gravity on the moon
motion, where Jupiter gravity effect is seen by the masses rate interacting with
Jupiter orbital inclination, that because, by that Jupiter effect on the moon orbital
inclination, and Jupiter practice this effect by using its orbital inclination (1.3 deg)
in addition to the masses rate…
Shortly
- Metonic Cycle is a cycle defined based on the moon orbital inclination (5.1 deg)
by a direct effect of (the sun and Jupiter gravities interaction) through Jupiter
orbital inclination.

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7-3 The Moon Orbit Regression
I-Data
(7-7)
(1.44)2
= 2.07 degrees
(7-8)
5.1 degrees – 3.1 degrees = 2 degrees
(7-9)
26.6 degrees = 5.2 x 5.1 degrees (The Moon Orbital Inclination)
(7-10)
5.1 degrees – 1.44 degrees = 3.66 degrees
Equation No. (7-7)
(1.44)2
= 2.07 degrees
- The moon orbit regresses per month 1.44 degrees, this value square equal the (2.07
degrees) which is the difference between the 2 angles (28.61 deg – 26.6 deg), the
difference (2.8%) is not true because the angle BCD doesn’t = 26.6 deg accurately
but = 26.56 degrees, but I use (26.6 deg for simplicity)
Shortly
- The moon orbit regression monthly causes to consume these 2 degrees between the
2 angles… why this isn't seen in the triangle? Because the moon orbit regression
isn't taken into consideration in the triangle, that needs to create a cycle for the
triangle to rotate with it… the triangle angles and dimensions are defined based on
the final fixed values (as perigee 0.363 mkm and apogee 0.406 mkm)… the
dynamic description will shows the moon orbit regression effect on the moon
motion…

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Equation No. (7-8)
5.1 degrees – 3.1 degrees = 2 degrees
- 5.1 degrees = The Moon Orbital Inclination
- 3.1 degrees = Jupiter Axial Tilt
- Equation no. (7-8) shows that this same (2 degrees) is seen as difference between
the 2 orbital inclinations values, that means, the moon orbital inclination (5.1 deg)
is defined based on Jupiter axial tilt (3.1 deg) with addition the angels (2 degrees),
that tells how the moon orbital inclination is created, and this description is logical
because of the great effect of Jupiter on the moon orbital inclination… that tells us
why the 2 angles have been different (28.63 degrees and 26.6 degrees), because the
difference is used for the moon orbital inclination creation…..
- But
- Equation no. (7-8) tells that, these (2 degrees) is consumed because of the moon
orbit regression (1.44 degrees monthly)! So how to explain that?
- That tells us, the moon orbit regression 1.44 degrees monthly is done depends on
the moon orbital inclination (5.1 degrees)…
Equation No. (7-9)
26.6 degrees = 5.2 x 5.1 degrees (The Moon Orbital Inclination)
- 5.2 = Jupiter orbital distance (778.6 mkm) / Earth moon distance (149.6 mkm)
- The angle BCD in the triangle =26.6 degrees is created based on the moon orbital
inclination by a direct effect of Jupiter on Earth orbital distance
- The idea is that, Jupiter causes a great effect on the moon orbital inclination and
because of that, the data shows frequently Jupiter effect

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Equation No. (7-10)
5.1 degrees – 1.44 degrees = 3.66 degrees
- This equation which we search for, because, the previous discussion told us that,
the moon orbit regression depends on the moon orbital inclination and we don't
know how that's happened….
- Equation no. (7-10) tells us some new data to help
- The difference between both angles =3.66 degrees
- and I interest greatly for the rate 3.66 because of its massive effect, this rate 3.66 =
(3.66 degrees /1 degree),
o (Earth Diameter 12756 km / The Moon Diameter 3475 ) = 3.66
o (Earth orbital period 365.25 days / the moon orbital period 27.3 days) =
(3.66)2
= 13.369
o This rate 13.369 we have discussed deeply in the previous point…
- What Does Equation No (7-10) Tell Us?
- The interaction between the moon orbital inclination and the moon orbit regression
depends on the created matter data, that means, Earth and the moon diameters are
created to be in harmony with the motions they will do and the potential
interactions can be created by such motions, so the motions effect is found deeply
inside the planets matters data and because of that the moon and Earth diameters
are created to be in harmony with their motions interaction…. Behind the planets
matters data the motions interactions are hidden
- That means, the motions interactions are found between Earth and moon motions
and these interactions create the moon orbit regression depends on the moon
orbital inclination but the created matter data hides these interactions and shows
the matter data as reason for such motions interactions…
- For that reason, Earth responds to the moon orbit regression because this
regression is done based on interaction of Earth motion. that means a deep
relationship is found between Earth and the moon orbital triangle data..

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7-4 The Moon Orbit Effect On Earth Data
I-Data
(7-11)
26.6 deg = 23.45 deg +3.1 deg
(7-12)
95930 km cos (23.45 degrees) = 88000 km
(7-13)
113.5 degrees x 0.8 degrees =90.8 degrees
(7-14)
406000 km = 116.6 x 3475 km (The Moon Diameter)
II-Discussion
Equation No. (7-11)
26.6 deg = 23.45 deg +3.1 deg
- 26.6 degrees = The Angle BCD
- 23.45 degrees = Earth Axial Tilt
- 3.1 degrees = Jupiter Axial Tilt
- Equation no. (7-11) shows how the triangle data is created based on Earth and
Jupiter motions interaction
Equation No. (7-12)
95930 km cos (23.45 degrees) = 88000 km
- The value 95930 km is the right triangle hypotenuse for the triangle (BCD) if D be
not on 406000 km but on 405500 km, means before apogee point with (500 km)
only…
- 88000 km = the moon daily displacement
- Equation no. (7-12) tells us that, Earth Axial Tilt is defined by some effect of the
moon orbital motion or the moon orbital motion is done based on Earth Axial Tilt..
The equation shows also that, a great harmony of data is found to enable the 2
planets motions interactions…

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Equation No. (7-13)
113.5 degrees x 0.8 degrees =90.8 degrees
- 113.5 degrees = 90 degrees + 23.5 degrees (Earth Axial Tilt =23.45 degrees)
- 0.8 degrees = Uranus Orbital Inclination
- 90.8 degree = 0.8 degrees +90 degrees
- (Please review the moon diameter division in point no. 8)
- Equation no. (7-13) tells that, Earth axial tilt is defined rated with Uranus orbital
inclination which the moon diameter division deal with it deeply and because of
that the equation shows a deep effect of the moon motions interactions on Earth
data..
Equation No. (7-14)
406000 km = 116.6 x 3475 km (The Moon Diameter)
- 406000 km = apogee Radius
- 116.6 degrees = The angle CDA in our triangle,
(we have discussed this angle deeply in the previous point)
- Equation no. (7-14) tells that, the moon diameter is created as a function in its
apogee distance! We know that, the planet data is create as a function in its motion
and because of that planets data is created in harmony with these planets motions.

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8- The Earth and Moon Motions Interaction
- In this point we try to show as deep as possible the motions interaction between
the Earth and its moon, this investigation is significant because we try to see what
effect is found one by the moon orbital triangle. Specially, the triangle angle (1.1
deg) effect on the Earth motion… let's proceed this investigation in questions form
(1st
Question) Why the moon displacement daily =88000 km?
- We concluded that, 1 day of Earth motion = 29.53 days of the moon motion, this
conclusion is created as an answer for the question why the moon orbital
circumference at apogee radius (r=0.406 mkm) equal (2.58 mkm) = Earth motion
distance per solar day….So this time rates interaction causes the moon orbital
circumference to be 2.58 mkm
- So during 29.53 solar days this distance (2.58 mkm) will produce 88000 km per
solar day, that means
- The apogee radius (r=0.406 mkm) is defined based on the moon daily
displacement (88000 km) or vice versa, I want to sat that, these 2 numbers can't be
created independent from each other, but dependent, So Earth gravity effect isn't
seen here because the moon displacement (88000 km) defines the apogee radius
(r= 0.406 mkm) but Earth gravity is seen in perigee radius because the moon can't
be near than perigee radius (r= 0.363 mkm) and that because of the gravity force.

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(2nd
Question) what effect the angle 1.1 degrees does on Earth Motion?
(I)
- 23.45 deg (Earth axial tilt) – 1.44 degrees (the moon orbit regression monthly)
= 22 degrees = 20 x 1.1 degrees
(II)
- 23.45 deg (Earth axial tilt) – 1.1 degrees = 2 x 2 x 5.68 degrees
(5.68 degrees = 5.18 degree the moon orbital inclination +0.5 degrees)
(III)
- 959300 km cos (23.45 degrees) =88000 km
- The previous data tells that, Earth axial tilt takes into consideration the moon data
(1.1 degrees and 88000 km, …..etc) in its definition.. that tells, Earth motion
depends on the moon orbital triangle, but to make this discussion more clear we
have to ask
(3rd
Question) Does Earth axial tilt depends on the angle (1.1 degrees) (the angle
under E)?
- The moon motion from perigee (0.363 mkm) to apogee (0.406 mkm) depends on
the angle (BCD) which is an angle on the horizontal level, but the angle (1.1 deg)
is an angle on the vertical level…. Earth axial tilt takes into consideration the angle
(1.1 deg) as the previous data tells us and that means there's some interaction
between Earth motion and this angle,
- Data no (III) shows 2 x (5.68) because the moon has 2 orbits and because of that
the moon has 2 inclinations each = 5.1 degrees and extend to 5.18 degrees, that
because the moon moves a displacement (= 2 x 88000 km)… because of that …
the 2 displacements needs 2 orbits and 2 orbital inclinations

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(4th
Question) can the angle 1.1 degrees be seen in any other place in the
triangle?
- Please review the triangle details in point No. (9), I write based on these
details
- The triangle BCZ is a significant one because its hypotenuse = 88000 km = the
moon displacement and this triangle starts from the point (B) Which expresses
perigee radius (r=0.363 mkm).
- We know the moon can't be more near than perigee point to Earth …
- But
- We will copy the triangle BCZ on the line BC to create another similar triangle on
the distance before perigee, this new triangle I call BCN, it's a typical triangle and
its hypotenuse = 88000 km (also)
- Based on this triangle (BCN),
o The angle (ECN = 64.5 degrees) and the angle (CDB =63.4 degrees)
o Simply 64.5 degrees = 63.4 degrees +1.1 degrees
Please note
o That tells the declination (1.1 degrees) is a correct angle but the direction
may be more flexible, because we need to change the angle direct to add this
angle (1.1 degrees) + 63.4 degrees = 64.5 degrees, that can be if the angle
changes its direction…. We should here discuss some details of the moon
orbital triangle …let's do that in following…
o The moon orbital triangle is a very useful method to study the moon motion,
also the triangle depends directly on the moon motion data because of that
the moon orbital triangle gives us a good direction to analyze the moon data
but
o The triangle is not so geometrical form perfectly…. The triangle in many
points move as an imaginary tool, let’s use some examples in following:
o The triangle dimension BC =86000 km, this dimension is an imaginary
dimension, where the distance EC = 373000 = the distance between Earth

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and the moon in the total solar eclipse, and the distance EB = 363000 =
perigee radius… but the difference between these 2 points = 10000 km and
not 86000 km and because of that, the triangle should be used as imaginary
tool… because it's not express a direct geometrical rules only, where I have
tried to create a simple vision to express the moon orbital motion data, and
this vision necessitates to put the data beside each other but not correct
geometrically.
o Also the angle (1.1 degrees), the concept is a correct one, that the moon
orbital triangle has an angle =1.1 degrees with the horizontal level but in
what direction because the moon motion is a dynamic form where the
triangle is a static method so the direction of angle can be changed with the
moon motion direction change
o What I try to do is that, to show the moon orbital data is created based on
each other by geometrical rules because the moon uses Pythagoras triangle
in its motion …. So the geometrical rules are found between the moon
orbital motion data by nature… so this description will prove that, the moon
orbital motion is done by some technique extends than the current
description… for that reason I have to put the data together beside each
other but that doesn't guarantee the geometrical rules to be without violation
by such arrangement for the moon motion data
o The moon orbital triangle analysis gives us the chance to discuss if it's
possible for the moon orbit to regress 19 degrees per year while earth does
no motion as a result for this regression, and as we have concluded that, it's
not possible, because if Earth is a point fixed in space and the moon orbit
regresses 19 degrees per year that will necessitate to change the distances
between the Earth and the moon, but because these distances aren't changed
by this regression, we have concluded that Earth does a motion as a result
for the moon orbit regression but because Earth has different motion, this

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motion which is done as a result for the moon orbit regression is hidden
behind different motions of Earth
o The data (23.45 -1.44) = 22 degree = 1.1 deg x 20), this data is a significant
one because it shows the moon orbit regression monthly (1.44 deg) is
interacted with Earth axial tilt and the horizontal declination angle (1.1)

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The Perpendicular Line BC
- We still need to analyze the Perpendicular line (BC), because this line makes the
triangle as an imaginary triangle and that causes a negative effect on the triangle
although its richness of data
- Why the perpendicular line (BC) is an imaginary line?
- Because the distance between Perigee radius (r=0.363 mkm) and total solar eclipse
(r= 0.373 mkm) = 10000 km but the perpendicular line BC =86000 km… So how
to solve this question?
- Let's the BC be a perpendicular on the base EA, means, the base (EA =449197 km)
is found on the horizontal level but the Line (BC) is found on the vertical level and
between them (90 degrees) – as seen in the figure –
- More clear by using xyz coordinates, the point (C) will be on (Z) axis while the
triangle base (EA) will be in x-y plain
- The moon moves on x-y plain…
- So this perpendicular line BC is a vertical axis (z- axis) in the xyz coordinates
- If we accept that, so the triangle will be a real one and all data will be correct and
fact without any change
- So the distance EC =373000 km but the point (C) isn't the total solar eclipse radius
because there's an angle =13.33 degrees between EC & EB (perigee r=0.363 mkm)

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- When the distance (373000 km) is measured from Earth on the horizontal level
where no angle is found, so this distance will = total solar eclipse radius
- What's the useful result of that?
- The moon moves on (x-y plain) from perigee radius (r= 0.363 mkm), to the total
solar eclipse (r= 0.373 mkm) to the moon orbital distance (r=0.384 mkm) and to
apogee radius (r= 0.406 mkm), the moon moves between these points on x-y plain
but these points are defined by Pythagoras triangle by the vertical triangle
perpendicular on the moon orbit… i.e.
- The perpendicular triangle on the moon orbit is the measurement ruler by which
the moon motion points are defined, we now have 2 important results (1st
) The
moon orbital triangle is a fact and real one but it's found vertically on the moon
orbit (2nd
) The vertical triangle defines the distances which the moon uses in its
orbital motion! why? why the moon orbital motion 4 basic points are defined by
this vertical triangle?! Let's try to answer in following
- Suppose this is the vertical triangle, its dimensions are (363000, 86000) and then
the hypotenuse will be = 373000 km
- So the smallest angle (θ) = 13.328 degrees (1)
- Note Please
o In this triangle I started with perigee radius (r= 363000 km) an we know that
the moon can't be more near to Earth than this point, so it's the 1st
point in
the moon orbital motion
o The distance 373000 km =the Earth moon distance while the moon be in the
total solar eclipse radius, so this is the (2nd
) point of the moon orbital motion
- Now let's move to the next point (r= 38400 km)

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- We will use the same method and find that (86000)2
+(373000)2
= (383000)2
- By that we know, the moon motion 4 basic points are defined by Pythagoras rule
- But ….Let's see what happens with the last point (apogee radius r= 406000 km)
- By using this same triangle, where its dimensions are (86000, and 406000 )
- Let's define the angle..
- So the smallest angle (θ) = 12.229 degrees (2)
- The difference (13.328 degrees) – (12.229 degrees) = 1.1 degrees (A Result)
- Notice No. (1)
o The angle (1.1 degrees) defines the distance which the moon moves from
Perigee (r= 0.363 mkm) to apogee (r=0.406 mkm)
o The angle (1.1 degrees) is the declination angle of the moon orbital triangle
on the horizontal level, means, this is the angle between the triangle red base
line (EA =449197 km) and the horizontal level and this angle is defined
vertically (on z-axis)
o This angle is responsible to define the distance (43000 km) from perigee to
apogee through which the moon can move, and that tells us, the moon
orbital triangle which is a vertical triangle on the moon orbit effects on the
moon orbital motion horizontally (on xy- plain)
- Notice No. (2)
- Let's divide this distance (43000 km) (from perigee to apogee) on 4 parts each
equal =10750 km (very near to the moon circumference =10921 km)
- In Pythagoras triangle, with dimensions are (86000 km, 10750 km and 86670 km),
so the angle (θ) = 7.129 degrees, and for 4 parts (7.129 deg x 4 = 28.5 degrees)
and this angle is the very angle which we have discovered in the moon orbital
motion – we may remember that – because the moon doesn't near than perigee
radius r= 0.363 mkm, because of that, during 29.53 days the max available
displacement should be 77237 km, and this displacement will can be defines by
the equal for the real displacement (L = 88000 Cos (θ)) when (θ) = 28.6 degrees
- That creates a harmony for difference calculations

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- But
- The moon orbital triangle (10921 km) creates an angle = 7.23 degrees, and
- 4 x 7.23 degrees = 29.2 degrees
- This angle is so specific in our investigation because Earth moves during 29.53
solar days (29.2 degrees) and the moon moves during this same period (389.2
degrees but 389.2 deg =360 deg +29.2 degrees) and that create the harmony
between Earth and Moon motions
- PLEASE NOTE
- 29.2 degrees – 26.6 degrees = 2.6 degrees
- The angle BCD = 26.6 degrees (This angle defines the moon motion from perigee
to apogee)
- 2.6 degrees x 2 = 5.2 degrees (the moon orbital inclination =5.1 degrees)
- Also Note
- 26 degrees = 1.1 degrees x 23.6 degrees where
o 26 degrees = (5.1 degrees)2
o 1.1 degrees = The declination angle
o 23.6 degrees = the outer planets orbital inclinations total but
(23.6 deg x 0.99 = 23.4 degrees = Earth Axial Tilt
o We have seen that before
o 17.4 deg (the inner planets orbital inclinations total) x 0.99 = 17.2 deg (Pluto
orbital inclination)

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9- The Moon Orbital Triangle Data
9-1 Preface
The Moon Orbital Motion (Revision)
- Why does the moon use Pythagoras Rule in its motion? and why do we need the
moon orbital triangle?
o The moon daily displacement =88000 km but the moon doesn’t use it as its
real displacement but instead the moon uses Pythagoras triangle to define
another real displacement
o Based on that
o The moon uses the right triangle dimension (L= 88000 km Cos θ) where
this (L) is the moon real displacement through its orbit daily
o The angle (θ) is the smallest angle in the right triangle, and it effects on the
moon real displacement and its motion height above perigee radius (r=0.363
mkm)
Why?
o The daily displacement = 88000 km and during 29.53 days, the total
distance the moon passes = 2.59 mkm which can be done only through
apogee orbit whose radius (r=0.406 mkm) –
That means –
o If the moon uses only 88000 km as its real displacement daily, the moon
would move only through the apogee orbit (r= 0.406 mkm)
o Because the moon tries to move in more near orbits to the Earth, the moon
uses Pythagoras triangle technique to produce smaller displacements though
its orbit while the moon move this same distance (88000 km) as a real
motion -
o So, this real displacement technique (L= 88000 km Cos θ) gives the moon
the ability to move through lower orbits to the Earth, and the change of the
angle θ changes also its real displacement

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o The smaller real displacement can be used for more near orbits to Earth (as
perigee radius r=0.363mkm) but the longer real displacements need more
wide orbit to be passed through, which forces the moon to move in higher
orbits above perigee.
o That means, the angle θ changes the real displacements and the height of the
moon motion above perigee radius (r= 0.363 mkm).
o Then based on that I have suggested the moon motion equation which is
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o θ is the smallest angle in Pythagoras triangle used for the moon real
displacement (L= 88000 km Cos θ),
o 0.985 degrees = Earth daily motion degrees.
o With more analysis, a 2nd
force effects on the moon orbital motion is
discovered, this 2nd
force effects on the point (A) in the moon orbital triangle
– where this point is an essential part of the triangle but it's far with 43000
km beyond apogee radius (r= 0.406 mkm).
o This 2nd
force effects on the moon orbital motion is produced by gravity
forces interaction between the sun and Jupiter on Earth and its moon – So
this 2nd
force pulls the moon to apogee radius but Earth gravity tries to keep
the moon at perigee radius. (This 2nd
force effects on the point (A) in the
moon orbital triangle)
Notice
o 149.6 mkm Earth orbital distance = 142984 km (Jupiter diameter) x 1047
(The sun mass/ Jupiter mass), because of this data, we conclude that, Earth
orbital distance is defined by gravity forces interaction between the sun and
Jupiter masses. Which supports the claim that 2nd
force can be produced by
the sun and Jupiter gravities interaction effect on the Earth moon.

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9-2 The Moon Orbital Triangle Details
Let's review The Earth Moon Orbital Triangle because we use it
Figure No. (1) (my figure)
Let's Review The Moon Orbital Triangle Data
(1st
Point)
- The figure I brought from internet to use in the Explanation -
- We have supposed that the inner circle is Perigee orbit and
the outer circle is apogee orbit – and we have calculated the
tangent DB = 181843 km
- AB = 363686 km (= perigee radius approximately)
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, the triangle (ODB) is a specific Pythagoras
triangle (1, 2 and 51/2
)
- The triangle (ODB) angles are 26.564 deg. and 63.435 deg.

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(2nd
Point) The Moon Orbital Triangle Data Correction
- EB = Perigee radius = 363000 km
- ED = Apogee radius = 406000 km
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- ES = total solar eclipse radius = 373900 km (error 1%)
(EC = 373000 km = Earth moon distance at T. Solar eclipse, BUT point C is NOT
the moon position in T. solar eclipse, because the distance BC= 86000 km but the
distance between perigee point and total solar eclipse point = 11000 km)
- CX= =87521 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 km CY= 97766 km
- BA = BC = 86000 km
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km
- BY = = 46475 km
- SZ = 7665 km ZF= 2414 km
- DY = 3475 km BX= 16203 km
THE ANGLES
- The angle between the black and red lines (under E) = 1.1 degrees
- (E) = 13.33 degrees (C)= 121.67 degrees (A) = 45 degrees
- (ECB) = 76.67 degrees (BCA) = 45 degrees
- (BCS = 7.23 deg) (BCZ = 12.195 deg) (BCF = 13.72 deg) (BCD = 26.564 deg)
(ACD = 18.435 deg)
- (BSC = 82.7 deg) (BZC = 77.8 deg) (BFC = 76.82 deg) (BDC = 63.434 deg)
- (CSA =97.23 deg) (CZA =102.195 deg) (CFA= 103.7 deg) (CDA = 116.564 deg)
- (CYA = 118.3 deg)
- BCY = 28.39 degrees ECZ= 88.9 degrees
- XCE = 66 degrees
- CZS = 77.8 degrees
- CZF =102.195 degrees
- XCB = 10.67deg
- (Uranus Axial Tilt = 97.8 degrees = FSC 97.2 degrees + 0.6 degrees) (i.e. the
angle under FSC)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

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9-3 The Moon Diameter Division
Figure No. 2 (my figure) (The Moon Diameter Division)
- Uranus Axial Tilt (97.8 degrees) has an angle (91.1 degrees) with The Moon Axial
Tilt (6.7 degrees), so The blue vertical line is Uranus axial tilt and the angle in the
circles center = 91.1 degrees
- This value 91.1 degrees is decreased by 0.5 degrees which is consumed by the
moon diameter (the circles express the moon diameter with some division), also
0.6 degrees is consumed by the green box (I suppose Saturn & Jupiter motions
interaction effect is done here to create this 0.6 degrees)
- So the angle between the blue vertical line and the horizontal red line 90 degrees
- The red horizontal line is the moon orbital triangle base (AE = 449197 km)
- The blue vertical line angle above the moon directly =90.6 degrees
- But
- In this figure I cut layers from the moon diameter to create smaller moon diameters
to consume smaller angles than (0.5 degrees), and based on that we measure the
angle of The blue vertical line, let's explain this moon diameter division:
- The Blue Circle its diameter R1= 1390 km and r1 =695 km
- The Red Circle its diameter R2= 2085 km and r2 =1042.5 km
- The Black Circle its diameter R3= 2185 km =π x 695 km and r3 =1092.5 km
- The Brown Circle its diameter R4= 3208 km and r4 =1604 km =5040/π
- The Orange Circle is the moon itself its diameter R5= 3475 km = 5 x 695 km

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- The moon consumes 0.5 degrees for its diameter (3475 km) …So
- R1 = 1390 km, the consumed angle will be 0.2 degrees i.e. the angle above The
Blue Circle = 90.9 degrees.
- R2= 2085 km (60% of the moon diameter) and because of that, the consumed angle
will be only 0.3 degrees, means, the angle above The Red Circle =90.8 degrees
- Note Please, 90.8 degrees = 90 degrees + 0.8 degrees (Uranus orbital inclination),
by that the rest degree of Uranus axial tilt = its orbital inclination vertically.
Notice No. 1
- Pluto Axial Tilt 122.5 degrees = 7.1 x 17.2 degrees (Pluto orbital inclination),
- 7.1 a rate created by Lorentz Length Contraction Phenomenon…
- That tells us, the values (122.5 deg and 17.2 deg) are equivalent values, how? It's a
contraction, the energy of (122.5 deg) in contracted by 7.1 and the rest 17.2 deg is
created from this same energy (122.5 deg). they are 2 equivalent values but one of
them passed through a different frame and faced a contraction phenomenon and
because of that its value (122.5 deg) became (17.2 deg)
- This notice is provided here to show that in the solar system there's some way to
create A Planet Axial Tilt = Its Orbital Inclination…. What's the important
result for this equality? We have seen this equality in Uranus & Pluto data, where
we know that (122.5 deg x 0.8 deg = 97.8 deg)– there's some geometrical machine
found behind this equality…
Notice No. 2
- Are these divided diameters real diameters or invented numbers? Let's test them :
(1)
- 149.6 mkm (Earth Orbital Distance) = 3475 km (the moon diameter) x 43000 km
(the distance from perigee to apogee)
- 149.6 mkm (Earth Orbital Distance) = 2085 km (R2) x 71492 km (Jupiter Radius)

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- 149.6 mkm (Earth Orbital Distance) = 1047 x 142984 km (Jupiter diameter) (1047
= The Sun Mass / Jupiter Mass)
(2)
- R4= 3208 km x 466884 km = 149.6 mkm x π2
- 466884 km = Jupiter motion distance during its day period.
(3)
- R3= 2185 km x π x 21346.6 km (Mars Circumference) = 149.6 mkm (2%)
(4)
- R2= 2085 km x 71492 km (Jupiter Radius) =149.6 mkm
(5)
- 4900 mkm = 4222.6 x 1.16 mkm = 10921 km x 449197 km
- 302 mkm = 4222.6 x 71492 km (Jupiter Radius)
- 305 mkm = 88000 km x 3475 km = 25.2 x (3475 km)2
- 177.4 degrees -90 degrees = 87.4 degrees (+0.6 degree = 88 degrees)
- 3475 s x 0.3 mkm/sec =1042.5 mkm but
- 2085 mkm = 3475s x 0.3 mkm/s x2
Discussion
- The points (1-4) show, Earth orbital distance 149.6 mkm, is defined as a function
in the moon diameter, This same feature is used for the smaller diameters, which
proves these diameters are real values
- Point (No.5) shows light motions behind Planet Data – Let's remember 1st
hypothesis (Planet Motion depends On Light Motion), the show some of these
light motions which caused to create the provided data.

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References
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJ&hl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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