The Moon Orbital Motion Geometry (Revised) (3)

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 Motion Geometry (Revised) (3)
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 –16th
February 2021
Abstract
Paper hypotheses
Hypothesis no. 1
Pluto motion effects on the moon orbital motion and causes to create the moon orbital
inclination (5.1 degrees)
As a result of this effect the moon orbital circumference at apogee orbit be shorter
than the moon displacements total during 29.53 days with 1%
Hypothesis no. 2
Pluto and Uranus Axial Tilts Interaction causes the moon orbit regression (19 degrees
per year) and also causes Venus Axial Tilt Regression.
Paper Conclusions
(1st
)
Earth Cycle 1461 days is created as a result of the moon orbit regression
(2nd
)
Uranus Motion effect on the moon orbital motion and causes to create Metonic Cycle.
(3rd
)
A 2nd
force effect on the moon orbital motion.
Paper Question
How can the far planets effect on the moon orbital motion spite of the huge distance?

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Contents
Subject Page N
1- Introduction 3
2-The Moon Orbital Triangle Description
2-1 Preface
2-2 The Moon Orbital Triangle Description
2-3 The Moon Orbital Triangle Data Analysis
2-4 The Moon Orbital Triangle Major Points
4
3-The Paper Hypotheses Proves Discussion
3-1 The Paper Hypotheses Revision
3-2 Why the moon apogee orbital circumference doesn't = 2.598693 mkm?
3-3 Why does the moon orbit regress? 3-4 The Earth Cycle 8 years
3-5 Why the moon daily displacement =88000 km?
3-6 Why the moon day period =29.53 solar days? 3-7 The Triangle (M1RB) Data Analysis
3-8 The angle 1.1 deg effect on the moon orbit geometry 3-9 4 Planets Motions Interaction
3-10 How can the far planets effect on the moon orbital motion?
36
4- The Moon Orbital Motion Analysis
4-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
4-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
4-3 The Moon Orbital Motion Analysis
4-4 The Moon Orbital Motion Equation
77
5-The Moon Orbit Geometrical Design
5-1 Preface
5-2 The Triangle Geometrical Design
5-3 The moon motion angle (12.195 deg) Analysis
5-4 The Perpendicular Line BC (=86000 km)
5-5 Jupiter Motion effect on the moon orbital motion
93
6- The Moon orbital triangle modification
6-1 Preface 6-2 The Moon orbital triangle modification
105
7- The Moon Orbital Inclination Creation
7-1 The Moon orbital inclination creation geometrical process
7-2 Planets motions effect on the moon orbital inclination creation
7-3 The Moon Orbit Regression
7-4 Planets motions cause The Moon Orbit Regression
7-5 The Moon Orbit Regression Effect on The Earth Motion
125
8- The Moon Orbital Triangle Geometrical Benefits
8-1 Preface
8-2 The Moon orbital triangle shows that (2nd force effect on the moon motion)
8-3 The Moon orbital triangle shows that (There's 2nd Orbit for the moon motion)
8-4 The Moon orbital triangle shows that Uranus effects on the moon motion
138
9- Metonic Cycle Is A Proof of Uranus Effect On The Moon Motion
9-1 Preface
9-2 Uranus Effect On The Moon Orbital Motion
9-3 The Angle 71.9 Degrees Analysis
9-4 The Moon Orbital Triangle Angles Discussions
143
10- Uranus Motion Analysis
10-1 Uranus Motion During 1440 Of Its Days Period
10-2 Uranus Motion During 8 Pluto Days period
10-3 Uranus 144 days Cycle
10-4 The Moon Diameter Creation.
160
11- Appendix No.1 171

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1- Introduction
- The apogee point (r=0.406 mkm) is the most far point the moon can reach from
Earth, the moon apogee orbital circumference =2.550973 mkm. But this distance is
shorter than the moon displacements total during its day period which =2.598 mkm
(Where 88000 km the moon daily displacement x 29.53 days = 2.598693 mkm)
- This interesting data tells us if the moon uses its daily displacement (88000 km) as
a real displacement the moon would revolve around Earth through its apogee orbit
only (even far from its apogee where 2.5986 =2π x 0.4135 mkm)
- The intelligent moon uses an angle (θ) between its displacement motion direction
and its orbit horizontal level, by that the real displacement through the moon orbit
will be (L=88000 km Cos (θ)) which creates a real displacement (L) shorter than
88000 km enables the moon to revolve around Earth Through more near orbits
- This led to conclude, the moon use Pythagorean triangle in its orbital motion
- By the moon using of Pythagorean triangle, The moon orbit be in a triangle form
- The paper uses and analyzes this triangle to prove the paper hypotheses.
- The question is valid (Why the moon orbital circumference at apogee radius
doesn’t =2.598 mkm= the moon displacements total during its day period?)
- The Pythagorean triangle technique using by the moon motion provides 2 new
tools are useful for the moon motion study which are the moon orbital triangle and
the moon orbital motion equation – The paper discusses how to use them –
- In Point No. 2, the paper discusses Major Points in the moon orbital triangle which
we will need to prove the paper hypotheses
- In point No. 3 the paper discuss its hypotheses proves (and answers the question)
- In point No.4 the paper analyzes the moon motion & discuss its suggested equation
- In point No. 5 It discuss the moon orbital triangle Geometrical Design
- In point No. 6 the paper discuss how the moon orbital inclination creation
- In point No. 7 the paper discuss the moon orbital triangle benefits
- In point No. 8 the paper discuss Metonic Cycle origin
- In point No. 9 the paper analyzes Uranus Motion

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2-The Moon Orbital Triangle Description
2-1 Preface

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2-1 Preface
- In this point we discuss how to create this moon orbital triangle and to define its
distances and angles.
- This triangle is created by creation a vertical line BC perpendicular on the triangle
base. This vertical line should be used 2 times in the triangle, one time when the
moon be in perigee and the second time when the moon be in apogee. For that
reason the triangle creates one form for each case and then created also one
combined form for both cases.
- The moon using of Pythagorean triangle is discovered by analyze the moon motion
basic points which are
o Perigee point (r=363000 km), the nearest point the moon can reach to Earth
o Pongee point (r=406000 km), the far point the moon can reach from Earth
o T.S. Eclipse (r=373000 km), the moon creates total solar eclipse at it
o The distance (r=384000 km) which is registered as the moon orbital distance
The following data proves their using Pythagorean rule.
These 4 points are defined based on each other by Pythagorean rule:
o (363000 km)2
+ (86000 km)2
= (373000 km)2
o (373000 km)2
+ (86000 km)2
= (384000 km)2
o (384000 km)2
+ (86000 km)2
= (393000 km)2
o (393000 km)2
+ (86000 km)2
= (406000 km)2
(Error 1%)
- By this data it's discovered the moon using of Pythagorean triangle in its motion
Notice
- The perpendicular Line (BC) which we use to create the moon orbital triangle its
length =86000 km.
Let's know how to create the moon orbital triangle

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- When we use the vertical line BC to be perpendicular on the moon in the perigee
point, the triangle form be as following..
- When we use the vertical line BC to be perpendicular on the moon in the apogee
point, the triangle form be as following..
- The combined form be as following..

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The Triangle Data Summary
- The moon moves on its orbital plane (the Red Line) from Perigee to apogee
- This distance is defined by M1 and M2 distance (=43000 km) and the distance BD
=42800 km be a very similar to it
- The line BC is perpendicular on a point parallel to the perigee point
- So the triangle CBD expresses the moon motion from perigee to apogee
- This triangle data is
o The angle BCD = 26.46 degrees
o The line BC = 86000 km
o The hypotenuse CB = 96062 km
Notice
- This figure I have brought from internet to use in the Explanation -
- We have supposed, the inner circle is the Perigee orbit and the outer circle is the
apogee orbit, And we have calculated the tangent DB = 181843 km
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, The triangle (ODB) angles are 26.564 deg. and 63.435 deg.
- But the triangle (BCD) in our triangle is a similar to this triangle (ODB), their
dimensions are rated and their angles are equal, both are created as a specific
Pythagorean triangle (1, 2 and 51/2
).
Why is this specific Pythagorean triangle (1,2 and 51/2
) is a necessary tool for the
moon orbital motion? The paper answers this question.

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The Moon Orbital Triangle Building
(1st
Point) The Earth Position (Point E)
- The Point (T) refers to The Earth Center
- The Point (M1) refers to The Moon Center (The moon in Perigee Point).
- The Points (T, Q and Y) are on The Earth Ecliptic Line
- The Red Line (TM) is the moon orbit plane with an inclination 5.1 degrees on the
Earth ecliptic line.
- The Green Line (BE) is the moon triangle base, the distance BE = 363000 km, I
choose it and accordingly I have to define the point (E) position.
- The line BC is a perpendicular on the triangle base (BE), its length =86000 km
- The line BC is perpendicular on the triangle base (BE) on the point (B), parallel to
the moon perigee point. (The 1st
Case).
- The angle CBE =90 degrees but the angle CYT = 89.557 degrees.
- The points (Q and P) are the intersection points of CE with the ecliptic and the
moon orbit plane respectively.
- The line TX is a perpendicular from the Earth Center on the base BE
- K is the intersection point between the triangle base (BE) & the moon orbit plane.
- The angle is Zero between the points ( A, B , K , X and E).
- The line EC connects between the points C & E where BC =86000 km and BE =
363000 km (As The Triangle Creation Requirements).

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(2nd
Point) The Moon Motion (From Perigee To Apogee)
- The moon moves on its orbit planet (MT) with an inclination 5.1 degrees on the
ecliptic, from Perigee (M1) (r=363000 km) to Apogee (M2) (r=406000 km).
- The distance M1 M2 = 43000 km (=The Perigee Apogee Distance)
- The line M1B is perpendicular on the triangle Base (EA) on The perigee point.
Notice
- M1B and M2D are perpendicular on the moon orbital triangle base (EA) (the
Green Line) …… BUT
- M1B and M2D are perpendicular on the triangle Base EA on (x-y plain) but the
line BC is perpendicular on the base (EA) on the (z-axis)
- Based on that
- The distance BD is parallel to M1R, and the moon motion from perigee to apogee
(M1M21) can be expressed on the triangle base by the distance (BD) where the
distance (M1M2) =43000 km and the distance BD =42800 km (error 0.4%)
- The blue line is the moon equator line, where the triangle Base (EA) has 1.1
degrees above the moon equator and has 0.443 degrees under the ecliptic.

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- Let's define the Earth Point in following:
(1) In the Triangle ATK
o The angle ATK = 5.1 degrees (the moon orbital inclination)
o The angle TAK =0.443deg (an angle between the base and ecliptic)
o The angle AKT = 174.457 degrees
o The angle BKM1 = 5.543 degrees
(2) In the Triangle M1BK
o The angle M1KB = 5.543 degrees
o The angle KM1B = 84.457 degrees
o The angle RM1M2 = 5.543 degrees
o The distance M1B = 31604 km
o The distance M1K = 327188 km
o The distance BK = 325658 km
o The distance KT = 35812 km
o The distance BX = 361300 km
(3) In the Triangle RM1M2
o The angle M2M1R = 5.543 degrees
o The angle RM2M1 = 84.457 degrees
o The angle M1M2N = 6.643 degrees
o The distance M2R = 4153 km
o The distance M1R = 42800 km
(4) In the Triangle KTX
o The angle XKT = 5.543 degrees
o The distance TX = 3460 km
o The distance KX = 35644 km

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(5) In the Triangle TM1Y
o The angle TM1Y = 84.457 degrees
o The angle TYM1 = 90.443 degrees
o The angle M1TY =5.1 degrees
o The distance TM1 = 363000 km
o The distance YT = 361313 km
o The distance M1Y = 32269.5 km
o The distance YB = 665 km
o The distance M1B = 31604 km
(6) In the Triangle KTE
o The angle E = 63.87 degrees
o The angle ETK = 110.6 degrees
o The angle ETQ = 115.7 degrees
o The distance TX = 3460 km
o The distance TE = 3854 km
o The distance XE = 1700 km (to make the distance BE =363000 km)
o The distance KE = 37344 km (= 35644+1700)
(7) In the Triangle EPK
o The angle EPK = 161.1 degrees
o The angle EKP = 5.543 degrees
o The angle PEK = 13.328 degrees
o The distance PK = 26604 km
o The distance PE = 11147 km
(8) In the Triangle EPT
o The angle TEP = 50.54 degrees
o The angle ETP = 110.57 degrees (84.457+26.12)

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o The angle EPT = 18.89 degrees
o The distance TP = 9190 km
(9) In the Triangle QTP
o The angle TPQ = 161.1 degrees
o The angle T = 115.72 degrees
o The angle PTQ = 5.1 degrees
o The angle TQP = 13.78 degrees
o The distance TQ = 12491 km
o The distance QP = 2529 km
o The distance EQ = 13673 km = 11144 + 2529
Data Analysis
(1)
o The Triangle TXE
o The distance TX = 3460 km The distance XE =1700 km
o The moon diameter =3475 km and the moon radius =1737.5 km, both are
equal the triangle 2 dimensions (error around 2%). That shows geometrical
interaction in this distances definition.
(2)
o The Point (E) is found inside the Earth but far from its center with 3854 km
with an angle 63.8 degrees where its level is far from the Earth center with a
perpendicular distance =1700 km.
(3)
o The line M1B has an angle 90 degrees (M1BK) but the angle M1YT
=90.443 degrees.

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(3rd
Point) The Point (A)
- The Point (A) is a point on the Ecliptic Line I have choose and caused to create it
with an angle =0.443 degrees under the ecliptic line. By that the triangle base (AB)
be found under the Ecliptic with 0.443 degrees and above the moon equator line
(the blue line) with 1.1 degrees.
- That means, the triangle base (AB) depends on the Earth ecliptic line.
- The triangle ABC is a closed triangle where the point (A) is the intersection point
between the ecliptic line, the triangle base AB and the triangle dimension AC
- I choose the distance AB =86000 km.
- The line BC is a perpendicular on the point B, (which is parallel to the perigee
point M1 with a radius r=363000 km). (1st
Case)
- The line BC length =86000 km (I choose it).
Notice
- The moon equator line (the blue line) doesn't intersect neither with the ecliptic nor
the moon orbital triangle AB on the point (A),
- The moon equator line (the blue line) will intersect the ecliptic line beyond the
point (A) with a long distance

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- Let's define this intersection point position in following:
o The moon orbit plane declines on the Ecliptic line with 5.1 degrees, means,
far distance be found between the Earth and moon will cause longer
perpendicular distance between the moon center and the ecliptic line
o For that, we use the moon distance on a apogee because it's the most far
point the moon can reach from Earth
o ON APOGEE …
o Earth moon distance on apogee point = 406000 km
o The perpendicular distance from the moon center to the ecliptic line = 36091
km, because of the moon orbital inclination (5.1 degrees)
o But
o The angle between the ecliptic line and the moon equator line =1.543 deg
o So these 2 lines will be intersected each other at a distance =1340318 km
o i.e.
o The ecliptic line will intersect with the moon equator line after the apogee
point with a distance =1340318 km
o but the distance from perigee to apogee =43000 km
o i.e. The ecliptic line will intersect with the moon equator line after the
perigee point with a distance =1383318 km
o Notice, the lunar eclipse umbra length =1392000 km (error 0.6%)
The Useful Result :
The triangle base (AE) has an angle = 1.1 degrees with the moon equator line.

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(4th
Point) The Line BC
- The line BC is perpendicular on the triangle base on the point (B), so, the angle
ABC =90 degrees. The blue line is the moon equator line and the red line is the
moon orbit plane – the green line is the triangle Base (BA).
- Based on that,
o The angle BYA =89.557 degrees
o The angle CYA =90.443 degrees
o The angle M1NV =91.1 degrees
o The angle M2NM1 =88.9 degrees
o The angle M1NM2 =6.643 degrees
o The angle between the blue line (the moon equator) and the green line
(the triangle Base BA) = 1.1 degrees
o The distance BC = 86000 km (I have choose it)
o The distance AB = 86000 km (I have choose it)
o The distance AY = 86009 km
o The distance YB = 665 km
o The distance MB = 31604 km

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(1st
Question)
- The moon orbital triangle geometrical structure depends on 3 points (E, C and A),
- The Point (E) (found inside Earth)
- The point (C) (found on z-axis)
- But
- What's the point (A)? how this point can be created and effect on the moon orbital
motion and triangle?! Because this point is far from apogee radius with 43000 km
and the moon can't move beyond the apogee radius, means, this point (A) is found
in space and should have no effect on the moon orbital motion! so to find this point
(A) in the moon orbital triangle geometrical structure that creates a question needs
to be solved!
- Geometrically the point (A) is one pillar of the moon orbital triangle pillars,
means, the geometrical structure forces us to accept the massive importance of the
point (A).
- The paper claims that (Another force effects on the moon orbital motion in
addition to Earth gravity force and this point (A) refers to this 2nd
force)
- Our investigation in this study tries to discover if this claim can be proved based
on the moon orbital triangle geometrical design analysis.
(2nd
Question)
- The moon daily displacement 88000 km during 29.53 days creates a total distance
= 2598693 km
- But The moon orbital circumference at apogee orbit =2550973 km
- Where The apogee point is the most far point the moon can reach from Earth, that
means, the moon orbital circumference is shorter than the moon displacements
total during the moon day period (29.53 solar days) with a distance = 47720 km
- Why the moon orbital circumference at apogee doesn't =2598693 km?

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The following major points are selected from the moon orbital geometrical design
discussion because we need them to prove the paper hypotheses – let's refer to these
points in following:
2-4-1 The Necessity of Pythagorean Triangle (1, 2, 51/2
)
2-4-2 The Triangle Data (The Combination Form)
2-4-3 The Value 1290 degrees
2-4-4 The Trapezoid CDM2M1
2-4-5 The Triangle CDM2
2-4-6 The angle 17.4 degrees

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2-4-1 The Necessity of Pythagorean Triangle (1, 2, 51/2
)
(1st
Point) The Moon Motion Limits Definition
- In this moon orbital triangle I have added the line CA2 to create a total angle =137
degrees – based on that
(A)
- The angle ECA2 =137 degrees
- The distance BA2 = 150628 km
- The distance A2A = 64628 km
- The hypotenuse C A2 = 173450 km
- The perimeter of the triangle BCA2 = 173450 +150628 +86000 = 410080 km
- The triangle perimeter (BCA2) =410080 km= the apogee radius (406000 km)
(error 1%)
(B)
- The perimeter of the triangle (A CA2) =121622 + 173450 +64628 = 359700 km
- Perigee radius = 363000 km (error 1%)
A Conclusion
- The triangle BCA2 defines the moon motion limits from perigee to apogee by a
geometrical mechanism depends on The angle 137 degrees……. Why & How?

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(2nd
Point) The Rate 0.08
Why Pythagorean Triangle (1,2, 51/2
) Is Required?
This figure is discussed before.
- The inner circle refers to the perigee orbit
- The outer circle refers to the apogee orbit
- OB = 406000 km = Apogee Radius
- OR = 363000 km = Perigee Radius
- DB = 181843 km
- Perigee Orbital Circumference = 2.28 mkm
- Apogee Orbital Circumference = 2.55 mkm
I - Data
(1)
(DB / Perigee Orbital Circumference) = (181843 km/2.28 mkm) = 0.08
(2)
10.96 = 137 (The basic Angle) x 0.08
(3)
Sin (10.96 degrees) x 406000 km = 77237 km
(4)
Cos (10.96 degrees) 88000 km = 86400 km
II – Discussion
- Why is the Pythagorean triangle (1,2,51/2
) required for the moon orbital motion?
- Because, the rate (0.08) is required to create interaction with the angle (137 deg),
and based on this interaction, the valuable angle (10.96 degrees) will be created,
and based on this angle (10.96 degrees) most of the moon orbital motion data will
be created.
- That answers the question why the rates (1,2,51/2
) were required necessary for the
moon orbital motion? because based on these rates the rate (0.08) will be produced
which will be used to produce the angle (10.96 degrees)…… So

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- Based on the angle (CA2E =137 degrees), the moon orbital motion receives 3
basic data which are
o The apogee point radius (r=0.406 mkm) which is defined by the triangle BC
A2) Perimeter
o The Perigee point radius (r=0.363 mkm) which is defined by the triangle AC
A2) Perimeter
o And the rate (0.08) which is defined between the tangent DB (181843 km)
and the perigee orbital circumference (2.28 mkm)…….. then
o 10.96 = 137 x 0.08
o The valuable angle (10.96 degrees) is created.
Equation No. (3)
Sin (10.96 degrees) x 406000 km = 77237 km
- This equation tells the story in more clear way….
- The value 77237 km is very important…. If the moon moves daily a displacement
= 77237 km, during 29.53 days, the total distance will be = 2.28 mkm = the moon
orbital circumference at perigee orbit (r= 363000 km)
- Means, the perigee orbital circumference = 29.53 displacements each =77237 km,
that tells the value (77237 km) is defined by perigee radius (r=0.363 mkm) and the
moon day period (29.53 solar days), whatsoever the moon apogee radius be ….
Now the angle (10.96 deg) is defined before (10.96 = 137 x 0.08), and by that the
apogee radius is defined….
- I try to show that, we deal here with few players are created depending on each
other , all of them has one origin which is the angle 137 degrees, and has one
result which is the angle (10.96 deg)… what I try to do here is to show how the
data is arranged in a clear direction, by that, I may prove this is Directed Data.
Equation No. (4)
Cos (10.96 degrees) 88000 km = 86400 km
- The analysis is still complex and we need to consider it deeply in following…..

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- Where
o The moon orbital circumference at apogee radius (r=0.406 mkm) equals
only 2.55 mkm and this distance is short!
o Because
o The moon daily displacement =88000 km and during 29.53 solar days the
total displacements will be = 2.598 mkm …..if this distance be the moon
orbital circumference the radius will be = 0.4135 mkm
o Means, The apogee radius will not be 0.406 mkm but 0.4135 mkm !
o Which proves the conclusion, that, the moon uses Pythagorean triangle in its
motion,
o But Why the moon orbital circumference at apogee is not = 2.598 mkm?
o The angle (10.96 degrees) shows that the 2 values are created by
geometrical interaction because
Cos (10.96 degrees) 2.598 mkm =2.55 km
- This is the 2 discussed values (2.598 mkm = the moon displacements total during
29.53 days) and (2.55 mkm = the moon apogee orbital circumference), and the
equation tells that the angle (10.96 degrees) defines them based on each other (for
some geometrical reason). We have to find out what's this geometrical reason for
which the moon apogee orbital circumference is created shorter than its
displacements total.
Notice
137 =95.1 x 1.44
- We still don't know why this angle 137 degrees has so massive effect on the moon
orbital motion…? The previous data is
o 95.1 degrees = 90 degrees + 5.1 degrees (the moon orbital inclination)
o 1.44 degrees = the moon orbit regression degrees per month

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- The angle 137 degrees, is created by the moon orbit motion effect,
- 2 features of the moon orbit motion are unified together to produce this angle (137
degrees) which is the origin of the moon motion distance from perigee to apogee..
which are
o The moon orbital inclination 5.1 degrees
o The moon orbit regression 1.44 degrees per Month.
These 2 features of the moon orbital motion creates together the angle 137 degrees as
their platform to create the moon orbital motion in harmony with these 2 features…

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2-4-2 The Triangle Data (The Combination Form)
- The triangle data is referred before we add the new data only
- The distance CL = 12250.2 km
- The distance CN = 121758.2 km
- The distance CM1 = 117605 km
- The distance CB = 86000 km
- The hypotenuse CM2 = 129064 km
- The hypotenuse Cr = 124660 km (rM2 = 4404 km)
- The hypotenuse CS = 91158.3 km (SM2 = 37905.7 km)
- The distance rM1 = 41339 km (Rr=1461 km)
- The distance SB =30229.7 km (SD= 12570.3 km)
- The hypotenuse BM2 =53204.5 km
- The angle BRM1 =36.44 degrees The angle LM2N =1.1 degrees
- The angle RM1M2 =5.543 degrees The angle M2CN=19.367 degrees

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2-4-3 The Value 1290 degrees
- How to create the value 1290 degrees (or days)?
- Imagine the moon moves in a vertical motion from perigee to apogee and return
back consuming a distance 43000 km x 2 =86000 km in this vertical motion,
where no any distance is done on the orbit horizontal level, means the moon is still
in its original position in its revolution around Earth and the distance 86000 km the
moon consumes in a vertical motion from, perigee to apogee (43000 km) and
return back
- But the moon daily displacement =88000 km
- Means, the moon still have only 2000 km can be passed
- Now
- Imagine that the moon will use this 2000 km only in its horizontal motion
revolving around Earth
- The moon apogee orbital circumference =2550973 km, and if the moon moves
only 2000 km through this orbit, the moon would complete its revolution around
Earth through its apogee orbit in a period =1290 days (error 1%)
- Because of this interesting idea, I searched behind the value 1290 trying to find out
if it's an effective value in the moon orbital motion and found the following:
I-Data
(a)
254 x 5.08 degrees = 1290 degrees
(5.1 deg= the moon orbital inclination) (254 =6939.75 days /27.32 days)
(b)
175.94 x 1.44 =253.3 = 1290 /5.1
(c)
719.76 x 1.79 = 1290 degrees
(d)
7 x 29.2 x 2π =1290 degrees

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(e)
13.177 x 97.8 = 1289 degrees
(f)
17.4 x 11.8 x π = 1290 degrees (11.8 deg =5.1 deg +6.7 deg)
II-Discussion
Equation No. (a)
254 x 5.08 degrees = 1290 degrees
(5.1 deg= the moon orbital inclination) (254 =6939.75 days /27.32 days)
- Equation no. (a) tells us a very interesting new data let's summarize it
- Metonic Cycle (6939.75 solar days) = 254 lunar sidereal month (27.32 days)
- The moon orbit revolve around Earth one time per month and that means the moon
creates its inclination angle (5.1 degrees) by its motion during this month
- That means,
- The value 1290 degrees = the total degrees the moon creates by its motion during
Metonic Cycle –
- This is a simple idea and we know one similar to it
- The moon orbit regresses 1.44 degrees per month and by that the moon orbit total
regression per a year =19 degrees and during 19 years (6939.75 days) the total
degrees will be 361 degrees (full revolution).
- The equation no. (a) tells us that, not only the moon orbit regression is registered
per month (1.44 deg) but also the moon orbital inclination (5.1 deg), and as the
moon regression creates 361 degrees during Metonic Cycle the moon orbital
inclination creates 1290 degrees during Metonic Cycle.

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Equation No. (b)
175.94 x 1.44 =253.3 = 1290 /5.1
- Equation no. (b) tries to know if there's a relationship between the value 1290
degrees and the value 1.44 degrees (the moon orbit regression per month)
- We have found that the value 254 (The Months Number In Metonic Cycle) = the
moon regression value per month (1.44 deg) multiply with 175.94
- And what's this value 175.94
- Mercury Day Period =4222.6 hours =175.94 solar days
- Equation no. (b) tells that, 1.44 deg (the moon orbit regression per month) x 5.1
deg (the moon orbital inclination per month) x 175.94 (Mercury day period) =1290
- Why Mercury day period?
- The other values are acceptable, the data tells that, the moon regression per month
is interacted with the moon orbital inclination per month and both are controlled
by the value 1290 degrees (which express Metonic Cycle and because of that it
controls both values)…
- But why Mercury day period (175.94 solar days) is used as their platform?!

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Equation No. (c)
719.76 x 1.79 = 1290 degrees
- What's 719.76 degrees? It's Mercury Day Period
- Mercury revolve around the sun 2 times to create one day – means the total
degrees should be 360 degrees x 2 =720 degrees
- But
- Mercury day period doesn't = 2 mercury orbital periods perfectly, instead it less
with a value 5040 seconds, for that reason the total degrees doesn't =720 degrees
but equal = 719.76 degrees
- 1.79 degrees = Neptune orbital inclination (1.8 degrees)
- The value 1290 degrees is inherited from Mercury…
- The moon motion is controlled by the value 1290 degrees to create Metonic Cycle
where this value the moon has inherited from Mercury motion – Mercury creates
this value 1290 degrees by its motion interaction with Neptune and then the moon
has to move under its control.
- For that reason, Equation no. (b) shows Mercury day period (175.94 solar days)
because the value 175.94 days is used as a period of time for Mercury but for the
moon it's used as the period in prison, under which the moon has to live.

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Equation No. (d)
7 x 29.2 x 2π =1290 degrees
- As a result the moon live under this value control
- Earth moves during (29.53 solar days) a value = (29.2 degrees)
- The moon moves during (29.53 solar days) a value = (360 deg+ 29.2 degrees)
- 7 degrees = Mercury Orbital Inclination
- Earth and the moon motions are done based on Mercury orbital inclination
interaction with the value 1290 degrees.
Equation No. (e)
13.177 x 97.8 = 1289 degrees
- 13.177 deg = The moon daily motion degrees
- 97.8 deg = Uranus Axial Tilt
Equation No. (f)
17.4 x 11.8 x π = 1290 degrees
- 11.8 deg =5.1 deg (the moon orbital inclination) +6.7 deg (the moon axial tilt)
- 17.4 deg = the inner planets orbital inclinations total (7+3.4+5.1+1.9)
- Notice
- 17.4 deg x 0.99 =17.2 deg (Pluto orbital inclination) =17.2 deg +0.2 deg

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2-4-4 The Trapezoid CDM2M1
I-Data
- The hypotenuse CD = 96061.6 km
- The distance DM2 = 35759 km
- The distance M2M1 = 43000 km
- The distance CM1 = 117605 km
- The angle DM2M1 = 84.457 degrees
- The angle M2M1C = 95.543 degrees
- The angle M1CD = 26.57 degrees
- The angle CDM2 = 153.4 degrees
- The perimeter of the trapezoid CDM2M1= 292426 km
(g)
Tan (17.2 deg) x 943819 km = 292426 km
(h)
Sin (17.1 deg) x 292426 km = 86000 km

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Discussion
Equation no. (g)
Tan (17.2 deg) x 943819 km = 292426 km
- 17.2 degrees (Pluto orbital inclination)
- 943819 km = The Perimeter Of The Triangle AEC (discussed in 1st
Case)
(The triangle AEC dimensions are AE =449197 km, AC =121622
km and CE =373000 km)
- Pluto Orbital Inclination (17.2 degrees) effects on the moon orbital triangle
dimensions and data – we should know why and how?
Equation no. (h)
Sin (17.1 deg) x 292426 km = 86000 km
- The line BC =86000 km
- The angle 17.1 degrees = approximately 17.2 deg (Pluto orbital inclination)
- Why and how Pluto orbital inclination can effect on the moon orbital triangle
dimensions and creation.
Equation no. (i)
Tan (23.4) x 292426 km = 127757 km
- The perimeter of the triangle RM1B =127757 km, that tells us, the value 292426
km is effective value and used by Pluto orbital inclination (17.2 deg) and by Earth
axial tilt (23.4 deg) – that refers to some relationship between Earth and Pluto
which we need to discover it
Notice
- The Perimeter of the triangle CM2N = 293662 km = approximately the perimeter
of the trapezoid CDM2M1= 292426 km

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2-4-5 The Triangle CDM2
I-Data
- The Triangle CDM2 – its dimensions are
- The hypotenuse CM2 = 129064 km
- The distance DM2 = 35759 km
- The hypotenuse CD = 96061.6 km
- The angle DM2C = the angle M2CB =19.367 degrees
- The angle DC M2 = 7.25 degrees
- The angle M2DC = 153.4 degrees
- The angle CDB = 63.4 degrees
(j)
97.8 deg (Uranus axial tilt) =5.1 deg (the moon orbital inclination) x 19.17 deg
(Where 19.637 deg the angle DM2C= M2CM1 x 0.99 = 19.17 deg)
(k)
(153.3 degrees x 8) + 63.6 degrees =1290 degrees

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Equation no. (j)
97.8 deg (Uranus axial tilt) =5.1 deg (the moon orbital inclination) x 19.17 deg
(Where 19.637 deg the angle DM2C= M2CM1 x 0.99 = 19.17 deg)
- Equation no. (j) tells that, the line CM2 express Uranus Motion effect on the
moon orbital motion, and the angle 19.367 degrees shows that clearly
- We should consider that this triangle M2CM1 is the one shows Uranus effect
on the moon orbital motion!
- Let's review some data to prove this point
o 29.2 degrees x 0.8 = 23.4 degrees
o We know that earth moves during 29.53 days a value 29.2 degrees (because
29.53 days x 0.98562 deg per day=29.2 deg) and the moon moves during
this same period a value = (360 deg +29.2 deg) (because 29.53 days x 13.17
degrees per day = 360 deg +29.2 degrees) ………..And
o 0.8 degrees = Uranus orbital inclination
o 23.4 degrees = Earth Axial Tilt
o By Uranus effect Earth axial tilt is created from the value 29.2 degrees
(please note, almost of the Earth and its moon motions data is defined based
on a defined period of time which is one month 29.53 days- based on this
period the data is created)
o 36.44 degrees x 0.8 = 29.2 degrees
o The angle M1RB =36.44 degrees
o That shows the interactions found through the triangle.
Notice
- The Triangle CDM2 Perimeter = 260885 km
- Tan (26.3 deg) x 260885 km = 129064 km (the hypotenuse CM2)
- The angle DCB =26.57 degrees (difference 1%) with 26.3 deg

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Equation no. (k)
(153.3 degrees x 8) + 63.6 degrees =1290 degrees
- The angle M2DC =153.4 degrees
- The angle CDB = 63.4 degrees
- What does this equation tells us?
- We know the value 1290 degrees which we have discussed before, and we know
now that this value express Metonic Cycle period 6939.75 days because each 5.1
degrees express a lunar sidereal month (27.32 days). So this value express Metonic
Cycle (19 years =6939.75 days)
- (153.4 degrees x 8) + 63.4 =1290
- What's this value 8 ? why we need it here?
- It's a cycle
- Earth has a cycle of 8 years (2922 days = 2 x 1461 days)
- Where 1461 days = (365 +365+ 365 +366 days)
But
- 2922 days = 107.4 x 27.2 days
- 27.2 days is the nodal month during which the moon orbit regresses 1.44 degrees
- 107.4 =90 +17.4 degrees (the inner planets orbital inclinations total)
- Also
- 17.4 deg =0.2 deg + 17.2 deg (Pluto orbital inclination)
Let's add some more data for better explanation
- Pluto moves during its day period (153.3 hours) a distance = Earth motions
distance during its day period (24 hours) = the moon displacements total
during 29.53 solar days (error 1%) why?

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- Let's ask a simple question in following
- Why Pluto day period =153.3 hours?
- But
- Uranus day period =17.2 hours
- Neptune day period =16.1 hours
- Saturn day period =10.7 hours
- Jupiter day period =9.9 hours
- Pluto is absolute exceptional between the outer planets, why its day period so long
in comparison with the other planets?
- Our triangle can help us
- The cycle which is consisted of 8 years (for Earth) is used for Pluto as a cycle of
(8 days of Pluto days) but this same cycle is used for Jupiter as 64 days of Jupiter
days, and for Saturn as 80 days of Saturn days and for Neptune as 100 days of
Neptune days
- This cycle is discussed deeply in Uranus Motion Analysis (Point No. 8 of this
paper)
But
- Pluto orbital inclination effect on the moon orbital motion is so massive effect, the
data shows a great effect by Pluto motion on the moon motion
- The next point should be the last point we need to prove the paper hypotheses, let's
see it.

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2-4-6 The angle 17.4 degrees
I-Data
- I have used the angle BCU =17.4 degrees
- The angle C = the angle CAB = 45 degrees because AB = BC =86000 km
- The angle UCA =27.6 degrees, where The Anomalistic month = 27.55 days
- Means if 1 day = 1 degree
- So, this angle 17.6 degrees may express the Anomalistic month
- let's examine the triangle UCA
- The distance BU = 26951 km and so the distance UA = 59050 km
- The hypotenuse CU = 90125 km The hypotenuse AC = 121622 km
- The perimeter of the triangle UCA = 270797 km
But
- 86200 km x π = 270797 km
- The line BC =86000 km = 2 x 43000 km (Perigee apogee distance)
The data shows that, the angle 17.4 deg creates data similar to the moon orbital
motion data (notice 17.4 deg x 0.99 =17.2 deg Pluto orbital inclination)
This data also supports the paper hypotheses that Pluto motion effects on the moon
orbital motion – let's try to prove this fact in the next point.

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3-The Paper Hypotheses Proves Discussion
3-3 Why does the moon orbit regress?
3-4 The Earth Cycle 8 years
3-6 Why the moon day period =29.53 solar days?
3-7 The Triangle (M1RB) Data Analysis
3-8 The angle 1.1 deg effect on the moon orbit geometry
3-9 4 Planets Motions Interaction
3-10 How can the far planets effect on the moon orbital motion?

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Hypothesis no. 1
Pluto motion effects on the moon orbital motion and causes to create the moon orbital
As a result of this effect the moon orbital circumference at apogee orbit be shorter
than the moon displacements total during 29.53 days with 1%
Hypothesis no. 1
Pluto and Uranus Axial Tilts Interaction causes the moon orbit regression and Venus
Axial Tilt Regression
Conclusions
(1st
)
Earth Cycle 1461 days is created as a result to the moon orbit regression
(2nd
)
Uranus Motion effect on the moon orbital motion and causes to create Metonic Cycle.
(3rd
)
A 2nd
force effect on the moon orbital motion.

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The Point Claim
Pluto motion effects on the moon orbital motion and because of that, Pluto orbital
inclination (17.2 degrees) causes to create the moon orbital inclination (5.1 degrees)
As a result of this creation, the moon orbital circumference at apogee orbit be shorter
than the moon displacements total during the moon day period 29.53 days with 1%
The data analysis proves that
- The moon orbital inclination (5.1 degrees) is created depends on Pluto orbital
- The difference in the moon apogee orbital circumference with the moon total
displacements is created by Pluto motion effect.

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I- Data
Group No. 1 (The Difference In Distances)
(1)
2598693 km – 2550973 km =47667 km = 2π x 7595 km (Pluto Circumference) (1%)
(2)
407188 km x tan (10.96 deg) = 88000 km 407188 km-406000 km =1195 km
(3)
Tan (1.44 degrees) x 47667 = 1195 km (1195 km =Pluto Radius)
Group No. 2 (The Orbital Inclination Creation)
(4)
1290 degrees =75 x 17.2 degrees (but 75 deg = 98.6 deg - 23.6 deg)
108 degrees = 75 x 1.44 degrees
(5)
137 deg = 95.1 deg x 1.44 deg = 119.6 +17.4
(6)
17.2 deg (Pluto orbital inclination) = 2 x 5.1 deg (the moon orbital inclination) +7 deg
(7)
122.5 degrees = 95.1 degrees +27.4 degrees

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II- Discussion
Group No. 1 (The Difference In Distances)
(1)
2598693 km – 2550973 km =47667 km = 2π x 7595 km (Pluto Circumference) (1%)
(2)
407188 km x tan (10.96 deg) = 88000 km 407188 km-406000 km =1195 km
(3)
Tan (1.44 degrees) x 47667 = 1195 km (1195 km =Pluto Radius)
Equation no. 1
- The moon orbital circumference at apogee = 2.550973 mkm
- The moon displacements total = 2.598693 mkm
- The difference = 47667
- Where
- 47667 = 2π x 7595 km (Pluto Circumference) (error 1%)
- it's Pluto circumference which causes the difference between the 2 distances
Equation no. 2
- We remember that, cos (10.96 deg) x 406000 km =86400 km but not 88000 km
(the moon daily displacement), that creates the difference between the distances
- By our angle (10.96 deg) the displacement 88000 km can be produced if the
apogee radius be 407188 km
- The apogee radius =406000 km The difference = Pluto Radius
Equation no. 3
Tan (1.44 degrees) x 47667 (the distances difference) = 1195 km
(The moon orbit regresses 1.44 degrees per month)
- Again the result is Pluto Radius
- The data shows that Pluto data is mentioned by this difference of the 2 distances,
it's a result of Pluto motion effect on the moon orbital motion –
- Notice /Pluto moves per solar day 406000 km = Apogee Radius

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Group No. 2 (The Orbital Inclination Creation)
Equation no. 4
1290 degrees =75 x 17.2 degrees (but 75 deg = 98.6 deg - 23.6 deg)
108 degrees = 75 x 1.44 degrees (the moon orbit regression per month)
- 1290 degrees, this angle we know, the rate 75 is created with Pluto orbital
inclination (17.2 deg) but the rate 75 degrees is created as the difference between 2
values (98.6 deg -23.6 deg) what are these 2 values?
o 98.6 deg = 97.8 deg Uranus axial tilt +0.8 deg Uranus orbital inclination
o 23.6 deg= the outer planets orbital inclinations total = 23.4 deg +0.2
o 17.4 deg = the inner planets orbital inclinations total = 17.2 deg +0.2
And
o 17.2 deg = Pluto Orbital Inclination and 23.4 = Earth Axial Tilt
o That tells the value 75 degrees is a difference between Uranus data and
Earth data – this is the basic point behind – because Pluto motion effect on
Earth motion is done by the interaction between Uranus and Pluto motions
That means
o Pluto effect on the Earth moon motion is found by Uranus and Pluto
motions interaction which makes their data to be seen together among the
moon orbital motion data.
Notice
o 108 degrees = 107.2 degrees +0.8 deg (Uranus orbital inclination)
o And 107.2 degrees =90 +17.2 degrees (Pluto orbital inclination)
- Equation no. (4) shows Pluto effect on the Earth and its moon motion data
- Please remember Pluto moves during 365.25 solar days a distance = Earth orbital
distance.

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Equation no. 5
137 deg = 95.1 deg x 1.44 deg = 119.6 +17.4
- 119.6 degrees = 90 degrees +29.6 degrees (= the moon day period 29.53 days)
- 95.1 degrees = 90 degrees + 5.1 degrees (The Moon Orbital Inclination)
- The angle 137 degrees is the basic one in the moon orbit geometrical design, and
this angle is created depends on (95.1 x 1.44) as we have discussed before
- But the other part (119.6+17.4) shows that, this same angle is interacted with Pluto
orbital inclination (17.2 degrees) which shows how Pluto data effects greatly on
the moon orbital motion data.
Notice
o 23.6 deg= the outer planets orbital inclinations total = 23.4 deg +0.2
o 17.4 deg = the inner planets orbital inclinations total = 17.2 deg +0.2
- Please pay attention to this data because it's created based on geometrical
mechanism. And this mechanism itself shows how Earth and Pluto data is
produced by similar ways.

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Equation no. 6
17.2 deg (Pluto orbital inclination) = 2 x 5.1 deg (the moon orbital inclination) +7 deg
7 deg = Mercury Orbital Inclination
- The equation shows an interaction between Pluto, Mercury and the moon data
where we have seen how the value 1290 degrees is created by Mercury motion
interaction with Neptune and the moon has to live under this value 1290 deg
controls –
- So this equation is a logical one – because if Pluto motion effect on the moon
motion that necessitates to create an interaction with Mercury
Equation no. 7
122.5 degrees = 95.1 degrees +27.4 degrees
- 122.5 degrees = Pluto Axial Tilt
- 95.1 degrees = 90 deg + 5.1 deg The Moon Orbital Inclination
- 27.4 degrees = the moon month 27.3 days be near to this value f 1 deg =1 day
- The equation shows more interaction between Pluto and the moon data

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3-3 Why does the moon orbit regress?
The Point Claim
The interaction between Pluto and Uranus Axial Tilts causes the moon orbit
regression and also causes Venus Axial Tilt regression
Means
The Axial Tilts Regression in the solar system is one motion done by an interaction
between Pluto and Uranus Axial Tilts –
As a result for this interaction – The 2 Planets Axial Tilts effect on the moon orbit and
causes its regression 19 degrees per year
Also they effect on Venus axial tilt and causes its regression.
The data analysis proves that
- Pluto and Uranus Axial Tilts Are Interacted
- The moon orbit regression is done by Uranus and Pluto Axial Tilts Effect
- Venus Axial Tilt Also Effected By Uranus And Pluto Interaction.

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I- Data
(8)
122.5 degrees x 0.8 =97.8 degrees
(9)
93.4 +1.1 = 94.5
94.5 +1.1 = 95.6
95.6+1.1 = 96.7
96.7 +1.1 = 97.8
97.8 +1.1 = 98.9
98.9 +1.1 =100
(10)
137 = 14.5 +122.5 but 122.5 =14.5 +108
(11)
97.8 = 19.2 x 5.1 but 2872.5 = 19.2 x 149.6
(12)
176.4 =122.5 x 1.44
But
177.4 = 98.6 x 1.8

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II- Discussion
Equation no. 8
122.5 degrees x 0.8 =97.8 degrees
o 122.5 deg = Pluto Axial Tilt
o 97.8 deg = Uranus Axial Tilt
o 0.8 deg = Uranus Orbital Inclination
- Equation no. (8) shows clearly that both axial tilts are created rated to each other,
the equation shows the interaction between the 2 Axial Tilts
Equation no. 9
93.4 +1.1 = 94.5
94.5 +1.1 = 95.6
95.6+1.1 = 96.7
96.7 +1.1 = 97.8
97.8 +1.1 = 98.9
- This equation tries to show Uranus effect on the moon (and Venus) data,
o 93.4 deg = 90 deg +3.4 deg (Venus orbital inclination)
o 95.6 deg = 90 deg +5.6 deg,
o the moon orbital inclination 5.1deg and the moon angular diameter =0.5 deg
means,
o 5.6 deg = the moon orbital inclination measured above the moon diameter
o 95.6+1.1 = 96.7 (=90 +6.7 deg The Moon Axial Tilt)
o 96.7 +1.1 = 97.8 deg (= Uranus Axial Tilt)
- The equation no.9 shows that, an effect moves through the planets data to create
many planets based on the same rate (1.1). because of that, this data is created
based on geometrical rule and effect, not only that, also created by one effect – one
effect moves through the data to create it.

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Equation no. 10
137 = 14.5 +122.5 but 122.5 =14.5 +108
- We remember the angle 108 deg we discussed in equation no. 4 (108 =107.2+0.8)
- Also we know the angle 137 deg and Pluto axial tilt 122.5 deg
- Equation no. 10 shows that the angle 137 degrees depends on 122.5 deg which
depends on 108 deg where (108 deg =90 +17.2 deg +0.8) – that shows both
important angles (137 and 122.5 degrees) depends on 108 deg which is created by
Pluto & Uranus orbital inclinations total
Equation no. 11
97.8 = 19.2 x 5.1 but 2872.5 = 19.2 x 149.6
- This equation shows that, the rate between Uranus axial tilt and the moon orbital
inclination = the rate between Uranus orbital distance and Earthen orbital distance
- That shows a deep geometrical mechanism is found behind this rate (19.2) which
is used for 2 different data
Equation no. 12
(12)
176.4 =122.5 x 1.44 But 177.4 = 98.6 x 1.8
- 177.4 deg (Venus axial tilt) = 1.8 deg Neptune orbital inclination x 98.6 (where
98.6 degrees = 97.8 deg Uranus axial tilt +.8 deg Uranus orbital inclination)
- Pluto axial tilt 122.5 deg x 1.44 deg the moon orbital regression per month = 176.4
- What's this value 176.4 degrees?
- 177.4 deg (Venus axial tilt) – 1 deg = 176.4 degrees
- How to create that?
- The value 8 deg = 1.3 deg (Jupiter orbital inclination) +6.7 deg the moon axial tilt
- This 8 degrees is used by Mercury and from it Mercury created its orbital
inclination and left 1 degree for geometrical interaction whose result causes Venus
axial tilt to be 176.4 deg

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3-4 The Earth Cycle 8 years
The Point Claim
The Earth Cycle 8 years is created as a result for the moon orbit regression
And because
The moon orbital regression is created by the interaction between Uranus and Pluto
axial tilts effect
The Earth Cycle is effected by this same interaction
And because
Uranus & Pluto axial tilts interaction effect on Venus axial tilt and causes its
regression
Earth and Venus Periods be in harmony as a result
I-Data
(13)
2922 = 107.4 x 27.2
(14)

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II-Discussion
- Please remember the discussion of 8 Earth Cycle in ( 3-4-5 The Triangle CDM2)
- In it we have found this equation (153.3 deg x 8) + 63.6 deg =1290 deg
- The value 153.3 deg is used as 153.3 hours by Pluto (Pluto day period) and the rate
8 is used as 8 Earth years =2922 solar days (2 x 1461 solar days)
- In Uranus motion analysis (point no.8) we discuss the cycle of 8 days, just I bring
a small part of its data to make a reference about it in following:
o Uranus (6.8 km/sec) moves during Pluto day period (153.3 h) a distance =
Jupiter motion distance during 8 of its days period (79.2 h) + 17695 km
o During 8 Pluto days period (153.3 h x 8) Uranus moves a distance = Jupiter
motion distance during 64 Jupiter days (64x 9.9 h) +Jupiter diameter (1%)
o An equal distance is passed by Saturn in 80 of its days
o And
o An equal distance is passed by Neptune in 100 of its days
- The 8 days cycle is a cycle done by all planets, Uranus uses Pluto day period, so
the cycle is done by Uranus motion during 8 Pluto days. But Jupiter uses its day so
this same distance is passed by Jupiter motion during 64 of its days and Saturn
during 80 of Saturn days and Neptune during 100 of Neptune days.
- Based on this cycle Pluto day period became so long in comparison with the other
outer planets, that because this cycle contains the Earth and its moon and seen in
the equation (153.3 deg x 8) + 63.6 deg =1290 deg
- By that Pluto motion data effects on the Earth and its moon motions data
- And because of that, Pluto moves during its day period a distance = Earth
motion distance during its day period = the moon motion distance during its
day period. For better discussion we have to know why the moon daily
displacement =88000 km? let's first discuss the equations in following..

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Equation no. 13
2922 = 107.4 x 27.2
- 107.4 degrees = 90 +17.4 degrees
- 27.2 days = the nodal year
- 2922 days = 2 x 1461 days (365+365+365+366)
- The value 17.4 deg (17.2 deg Pluto orbital inclination) +0.2
- This value effects on the cycle 2922 days
Equation no. 14
- 137 deg = the important angle
- 17.12 deg = very near to Pluto orbital inclination 17.2 deg
- 8 The cycle
Notice (1)
- The moon orbital inclination 5.1 degrees and the moon orbital regression 1.44
degrees per month, these 2 values are created by one process
- So the data shows as following
Old Equation
(π)1/2
x 1.44 degrees x 2 = 5.1 degrees
Notice (2)
- The discussion can be more with answer the question why the moon daily
displacement =88000 km? let's answer in the following point.

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Earth motion distance during its day period = the moon displacements total
during its day period = Pluto motion distance during its day period (error 1%)
I - Data (1st
Group)
(1)
Earth moves during (6939.75 solar days) a distance = 17859.325 mkm
(2)
Pluto moves during (6939.75) x (153.3 hours) a distance = 18000.57 mkm
(3)
The moon displacements total during (6939.75) x (29.53 days) = 18034.278 mkm
(4)
Uranus Orbital Circumference = 18048.449 mkm
Where 153.3 hours = Pluto day period 29.53 days = the moon day period
The moon daily displacement =88000 km
Data Analysis
(4) – (1) = 189.124 mkm
(4) – (2) = 47.879 mkm
(4) – (3) = 14.171 mkm
(3) – (1) = 174.953 mkm
(3) – (2) = 33.708 mkm
(2) – (1) = 141.245 mkm
- Sin (17.2) x 47.879 mkm = 14.171 mkm
- Tan (10.96) x 174.953 mkm = 33.708 mkm
- Tan (13.3) x 141.245 mkm = 33.708 mkm (error 1%)
- 0.8 x 174.953 mkm = 141.245 mkm (error 1%)
- Sin (4.63) x 174.953 mkm = 14.171 mkm
o 0.8 degrees = Uranus Orbital Inclination. Sin (4.6) =0.08
o 17.2 degrees = Pluto Orbital Inclination
o 13.3 degrees = The angle of (E) in the moon orbital inclination
o 10.96 degrees (Cos 10.96 degrees x 88000 km = 86400 km)

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Discussion
- The previous 4 angles are basic data for their planets, let's try to show that
o 0.8 degrees = Uranus Orbital Inclination
o 122.5 deg (Pluto Axial Tilt) x 0.8 = 97.8 deg (Uranus Axial Tilt)
o Pluto orbital inclination 17.2 degrees = 0.99 x 17.4 deg (The inner planets
orbital inclinations total) … also
o Pluto orbital inclination 17.2 deg x 7.1 = 122.5 deg (Pluto Axial Tilt)
o 13.3 degrees is the angle of point (E) (Earth) in the moon orbital triangle
(Earth Orb. Period 365.25 d = The moon Orb. Period 27.3 d x 13.3)
o The angle 10.96 degrees is the valuable angle we have discussed deeply
where (Cos 10.96 degrees x 88000 km = 86400 km).
o Sin (4.6) = 0.08 This rate effects on the moon orbit geometrical design
(where 4.6 deg =5.1 deg "the moon orbital inclination" – 0.5 deg "the moon
diameter angle")
There's an interaction occurred between these 4 planets (Uranus, Pluto, Earth and its
moon), and in this interaction, these 4 basic values are created and based on these 4
values many other data of these planets is created … means, this interaction forms the
geometrical structure of these planets motions …. And if we limited our discussion
for the moon orbit structure, that tells us (The Moon Orbit Geometrical Structure
Is Effected By These 4 Planets Motions Interaction)

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II- Data (Group No.2)
1- The Moon Orbital Circumference at apogee radius = 2550973 km (100%)
2- Earth Daily Motion Distance = 2573483 km (101%)
3- Pluto moves during 153.3 hours =2593836 (102%)
4- The displacements 88000 km total during (29.53 days) = 2598693 km (102%)
5- Uranus motion distance (during 378675 seconds) = 2574990 km (101%)
5-1 = +24017 km
5-2 = +1507 km
5-3 = - 18846 km
5-4 = - 23703 km
4-1 = 47720 km
4-2 = 25210 km
4-3 = 4857 km
3-2 = 20353 km
3-1 = 42863 km
2-1 = 22510 km
Data Analysis
(1st
point)
Sin (71.9) x 23703 km = 22510 km
Sin (72.3) x 25210 km = 24017 km (error 0.6%)
(Cos (71.9) = tan (17.25))
- The angle (71.9 degrees) connects 5 basic values which are:
o 17.2 deg (Pluto orbital inclination)
o – 23703 km =The difference (the moon displacements & Uranus motion)
o 22510 km = The difference (the moon orbit & Earth motion)
o 24017 km = The difference (the moon orbit & Uranus motion)
o 25210 km = The difference (the moon displacements & Earth motion)
- It's a clear interaction between the 4 planets motions, because it's directed data….
This data is not random but directed, for that the same angle (71.9 degrees) is used
Because this angle is Found In The Interaction Point.

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(2nd
point)
(42863/25210) =1.7 (The constant =1.7 deg)
(42863/22510) = 1.9 (Mars Orbital Inclination = 1.9 deg)
- The distance 42863 km should get our attention because many of the moon data
depends on it
o 42863 km= (Pluto motion distance) – (the moon orbit circumference)
o That shows Pluto motion effect on the moon orbital motion
(3rd
point)
Tan (29.53) x 18846 km = 23703 km
o - 18846 km = the difference (Uranus motion & Pluto motion)
o – 23703 km = the difference (the moon displacements & Uranus motion)
- We have found the moon day period (29.53 days), it's created as an angle (29.53
degrees) in this same interaction ….
- The moon orbit regresses 19 degrees per year and causes to change the eclipse
calendar by 19 days by this regression , showing that, 1 degree = 1 day
- The previous interaction effects clearly on the interacting planets data, let's see the
proportionality in Earth and Pluto Data to prove this claim.
Earth and Pluto Data
1- (Pluto day period / Earth day period) = (Earth velocity / Pluto velocity)
2- Pluto orbital distance 5906 mkm= Earth orbital Circumference 940 mkm x 2π
3- Pluto orbital period 90560 solar days= 1461 solar days x 2π3
4- Pluto moves during 365.25 days a distance = 149.6 mkm = Earth orbital dist.
5- Pluto orbital inclination 17.2 deg = 99% the inner planets orbital inclinations
total (17.4 deg)
6- Earth Axial Tilt 23.4 deg= 99% the outer planets orbital inclinations total (23.6
deg)

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(3rd
point)
We will add Uranus motion distance differences together as following:
5-1 = +24017 km
5-2 = +1507 km
5-3 = - 18846 km
5-4 = - 23703 km
The total will be = (5-1) + (5-2) +(5-3) +(5-4) = -17025 km
- What does this value (-17025 km) means?!
- It's Uranus effect on the 3 Planets (Earth, its moon and Pluto), Why?
- Because these 3 planets move during their cycles periods a distances = Uranus
orbital circumference, because their (Metonic) Cycles depends on their motions
distances which equal Uranus Orbital Circumference, that shows Uranus effect on
these 3 planets, where their motions are interacted because of that.
- For that, the distance (-17025 km) should show Uranus effect on these 3 planets!!
- What's This Distance (-17025 km)?!
- 17025 km x 2 = 34309 = π x 10921 km (the moon Circumference). (Error 0.7)
More Data
(a)
5040 sec. x 6.8 km /s (Uranus velocity) = 34309 km (= 10921 km x π)
(b)
34309 sec x 13.1 km/s (Jupiter velocity) = 449197 km = Jupiter Circumference
(c)
34309 sec x 4.7 km/s (Pluto velocity) = 160592 km = Uranus Circumference
(d)
34309 sec x 27.78 km/s (the moon velocity) = 943817 km
(943817 km = the perimeter of the moon orbital triangle ACE) (error 1%)

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(e)
10921 sec x 35 km/s (Venus velocity) = 120536 km (Saturn diameter) (error 1%)
(f)
10921 sec x 29.8 km/s (Earth velocity)=321183 km (2 Uranus Circumferences) (1.3%)
(g)
1153 sec x 29.8 km/s (Earth velocity)= 34309 km = 5040 sec. x 6.8 km /s
Discussion
- The data shows clearly that the value (34309) is used widely in the solar planets
motions data, showing that there's a deep effect practiced from this interaction on
almost all solar planet… but why the used value is a double value?
The moon double cycle
The period (6939.75) x (29.53 solar days) of the moon = 204931 days = 561.07 years
But
30589 days (Uranus Orbital period) = 27.3 days (the moon orbital days) x 1120
561.07 years x 2 = 1123
2 moon Cycles are required to produce the rate (1123) which is very near to the rate
between both cycles (1120)… these (1123) and (1120) aren't 2 random number found
by chance, these are 2 geometrical values are produced by one machine depends on
the planets motions as motions of gears, for that reason, double Cycle is required, and
the moon data will show its cycle (561 years) but many other planets motions will
deal with the double cycle (1123 years), this also we have seen in the angle 12.195
degrees analysis where the triangle hypotenuse was = 2 x 29.53 km, and the value
29.53 solar days is the moon day period, but the value 59 days is used more widely,
for that, the Earth moves during 59 days a distance = its orbital distance.

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3-6 Why the moon day period =29.53 solar days?
I-Data
Equation No. (A)
Tan (12.195 deg) x 708.7 hours (the moon day period) = 153.3 h (Pluto day period)
Tan (13.177 deg) x 655.7 h. (the moon rotation period) = 153.3 h (Pluto day period)
- The angle 12.195 deg. is the moon angle (12.195 deg. = 13.177 deg. - 0.9856 deg),
Based on this angle the moon & Pluto days periods are defined relative to each
other… Why?
- The angle 13.177 degrees is the moon motion daily angle (360 =13.17 deg x 27.3)
and based on this angle the moon & Pluto rotations periods are defined relative to
each other… Why?
- Why the moon day period =29.53 solar days? Because the moon day period is
created in proportionality with Pluto day period and both are created relative to
each other…..But the better question is ….
Why Earth day period =24 hours?
Equation No. (B)
Tan (8.9 deg) x 153.3h (Pluto day period) = 24 hours
- The angle 8.9 degrees =98.9 degrees – 90 degrees
- By this angle Earth and Pluto days periods are created relative to each other!
- Pluto, Earth and the moon motions are interacted because of their motion distances
relative to Uranus orbital circumference…
- So this is more data tries to prove the interaction occurred between these 4 planets.
Shortly
- The moon day period (= 29.53 solar days) because it's created by 2 motions effect
on the moon orbital motion (Earth & Uranus motions) through the 4 planets
motions interaction. (Metonic Cycle is discussed in Point No. 7)
(In that discussion we should discuss, Why "Earth velocity/ Pluto velocity" = Pluto
day period / Earth day period?).

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3-7 The Triangle (M1RB) Data Analysis
The shaded triangle is our aim in this point to be analyzed, let's mention to its data in
following…
The Triangle Data
The hypotenuse RB = 53204.5 km
The distance RM1 =42800 km
The distance BM1 =31605 km
The angle M1BR = 53.56 degrees
The angle BR M1 = 36.44 degrees
The perimeter of the triangle M1RB = 127610 km
Please remember
The perimeter of the trapezoid CDM1M2 = 292425 km

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I-Data
Group No. 1
(A)
36.44 deg x 0.8 = 29.2 deg But 29.2 deg x 0.8 =23.4 deg
(B)
Tan (23.6 deg) x 53.56 deg = 23.4 deg
(C)
Cos (47.2 deg) x 53.56 deg =36.44 deg (where 47.2 = 2 x 23.6 deg)
Group No. 2
(D)
413595 km x tan (17.15 deg) = 127610 km
(E)
406000 km x tan (17.45 deg) = 127610 km
(F)
363000 km x tan (19.367 deg) = 127610 km
(G)
363000 km x sin (53.666 deg) =292425 km
(H)
292425 km x tan (23.6 deg) = 127610 km
(I)
449197 km x sin (34.6) = 127610 km x 2 (where 34.6 =2 x 17.3)
(J)
127610 km x tan (36.76) = 2 x 47667 km
(K)
160592 km x sin (17.26 deg) = 47667 km

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Group No. 3
(L)
127610 km = 53.4 x 2390 km (Pluto Diameter)
= 17 x 7510 km (Pluto Circumference)
(M)
127610 km = 36.7 x 3475 km (The Moon Diameter)
= 73.5 x 1737.5 km (The Moon Radius)
(N)
127610 km = 19 x 6716 km
Group No. 4
(P)
Cos (8.6) x 129065 km = 127610 km (where 8.6 deg x 2 =17.2 deg)
Group No. 5
(Q)
3033 mkm x tan (17.2) = 940 mkm (Earth Orbital Circumference)
(S)
17.2 hours (Uranus Day) x 8.9 = 153 hours (Pluto day period =153.3 hours)

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II-Discussion
Group No. 1
(A)
36.44 deg x 0.8 = 29.2 deg But 29.2 deg x 0.8 =23.4 deg
(B)
Tan (23.6 deg) x 53.56 deg = 23.4 deg
(C)
Cos (47.2 deg) x 53.56 deg =36.44 deg (where 47.2 = 2 x 23.6 deg)
- The data group no. 1 tells a direct information, that Earth axial tilt is created by
some interaction is happened based on this triangle data (The triangle M1RB)
- The rate 0.8 is Uranus orbital inclination 0.8 degrees and shows Uranus effect on
the triangle data
- The data uses the triangle 2 angles (53.56 deg and 36.44 deg) to create Earth
Axial Tilt 23.4 degrees.
- Earth axial tilt creation must be done based on the following proportionality
o 23.4 deg +0.2 deg = 23.6 deg (= the outer planets orbital inclinations total)
o 17.2 deg +0.2 deg = 17.4 deg (= the inner planets orbital inclinations total)
o 17.2 deg = Pluto orbital inclination
- Based on this proportionality Earth Axial Tilt (23.4 deg) must be created. And that
means, Earth axial tilt and Pluto orbital inclination are created together by the
same geometrical process, and based on that, the moon orbital inclination is
created depending on Pluto orbital inclination where Pluto orbital inclination is
effected already by Earth axial tilt, that shows the interaction between the three
planets.

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Group No. 2
- This group of data tries to prove the following claim…
o Because Earth axial tilt and Pluto orbital inclination both are created based
on this triangle angles (or by their help), and the moon orbital inclination is
created as a result for this process – So Based on that –
o The moon apogee orbital circumference be shorter with 47667 km than the
moon displacements total during 29.53 days, by this triangle effect on the
moon orbital motion.
o Means, the process by which the moon orbit be shorter is done here and may
this triangle shows us the full details of this geometrical process – so let's
analyze its data one by one in following…
Equation No. (D)
413595 km x tan (17.15 deg) = 127610 km
- The moon displacements total during 29.53 solar days =(88000 km x 29.53 days) =
= 2598693 km = 2π x 413595 km
- This value based on 17.15 degrees (Pluto orbital inclination 17.2 deg) produces
the perimeter of our triangle (M1RB).
- Let's suppose that, we start from here and no any other radius is found for the
moon orbital circumference – means- we suppose that- the moon orbital radius
only one = 413595 km – and the moon moves on one trajectory where neither
perigee nor apogee is found in its motion – we suppose the moon moves on one
trajectory as a railway – let's see what would happen later..!

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Equation No. (E)
406000 km x tan (17.45 deg) = 127610 km
- From the Group no. 1 discussion, we concluded that, the values 17.2 and 17.4
degrees are created together with the value 23.4 and 23.6 degrees. That means,
these 4 values are created by one geometrical process
- The perimeter of our triangle (M1RB). (127610 km) is with us produced by the
previous equation no. (D)
- Based on that, the value 406000 km is created
- The moon apogee orbit radius 406000 km is created based on the value 17.4 deg
Equation No. (F)
363000 km x tan (19.367 deg) = 127610 km
- The angle BCS =19.367 degrees
- Based on the perimeter of our triangle (M1RB). (127610 km), the perigee radius
363000 km is created.
Notice No. 1
- The angle 19.367 degrees expresses Uranus effect because the dimension CM2
(=129064 km) also expresses Uranus effect as we have discussed before (where
97.8 deg = 19.2 x 5.1 deg and 2872.5 mkm= 19.2 x 149.6 mkm)
- That tells us, by Uranus effect the perigee radius (r=363000 km) is created. So
Pluto data effected to create the apogee radius (r=406000 km) and decreased the
moon orbital circumference from 2598693 km to 2550973 km – where Uranus
effects on the moon motion to create the perigee point radius (r=363000 km).
Notice No. 2
- 2.573482 mkm (Earth motion distance per a solar day) = 7.1 x 0.363 mkm
- 0.363 mkm (Perigee Radius) = 7.1 x 51118 km (Uranus Diameter)

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Equation No. (G)
363000 km x sin (53.666 deg) =292425 km
- The angle (RBM1) = 53.56 degrees (error 0.2%)
- 292425 km = The perimeter of the trapezoid CDM1M2
- The angle RBM1 = 53.56 degrees id very near to this used one (53.666 deg). So
we conclude that, the trapezoid CDM1M2 is created by effect of 2 factors (the
angle RBM1 53.56 deg and Perigee radius 363000 km).
Equation No. (H)
292425 km x tan (23.6 deg) = 127610 km
- The trapezoid CDM1M2 which is created in the previous equation (no. G) and we
have our triangle perimeter 127610 km, By these 2 values the angle 23.6 degrees
is created or defined. That tells the triangle perimeter which is used to produce the
moon orbital different radiuses (0.413 mkm, 0.406 mkm and 0.363 mkm).
- The conclusion should be that (Earth Axial Tilt Is Created In Interaction With
The Moon Orbital Motion).
Equation No. (I)
449197 km x sin (34.6) = 127610 km x 2 (where 34.6 =2 x 17.3)
- The distance 449197 km is the moon orbital triangle base (EA).
- This equation tells that, the triangle BRM1 is interacted directly with the moon
orbital triangle base (EA =449197 km) and that shows why this triangle has so
specific effect.
- The important observation is the angle 34.6 deg =2 x 17.3 deg, I try to show that,
the geometrical design uses the data in different forms which causes a great rich
for the general geometrical description.

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Equation No. (J)
127610 km x tan (36.76) = 2 x 47667 km
- The angle BRM1 = (36.44 degrees) which is very near to this used one (36.67),
and the distance 47667 km is the difference in distance between the moon
displacements total during 29.53 days (2598693 km) and the moon apogee orbital
circumference (2550973 km). that tells us, the triangle BRM1 is used to create the
difference in these 2 distances and based on this interaction the moon apogee
orbital circumference be created shorter than the moon displacements total with
the distance 47667 km.
Equation No. (K)
160592 km x sin (17.26 deg) = 47667 km
- This equation we have to discuss with the group no. 5 because it tells the
difference in these 2 distances is created by an interaction between Uranus and
Pluto motions –
- Please notice Uranus effect on Pluto motion data in following:
o Uranus causes Pluto orbital inclination to be 17.2 degrees where Uranus day
period =17.2 hours
o Uranus causes Pluto axial tilt to be 122.5 degrees where 122.5 deg = 97.8
deg (Uranus Axial Tilt) /0.8 deg (Uranus orbital inclination)
o Uranus causes Pluto Day period to be 153.3 hours where 153.3 hours =8.9 x
17.2 hours
o Uranus causes Pluto orbital period to be 90560 days (where 90560 days =
tan 71.3 deg x 30589 days Uranus orbital period) (71.33+0.5 =71.9) (the
angle 71.9 deg we should discuss later in this paper).
o Uranus causes Pluto Uranus distances to be 3033 mkm in proportionality
with Earth orbital distance (because of the rate between Uranus and Earth
orbital distances)

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Group No. 3
Equation No. (L)
127610 km = 53.4 x 2390 km (Pluto Diameter)
= 17 x 7510 km (Pluto Circumference)
- The angle RBM1 = 53.56 degrees and Pluto orbital inclination =17.2 degrees,
based on these 2 angles Pluto diameter and circumference are created in
proportionality with the triangle (BRM1) Perimeter, where the planet diameter and
Circumference are rated by (π), that tells these 2 angles also are rated to each
other.
Equation No. (M)
127610 km = 36.7 x 3475 km (The Moon Diameter)
= 73.5 x 1737.5 km (The Moon Radius)
- This data is very similar but used for the moon diameter creation.
- The angle BRM1 = 36.44 degrees which is very near to this used one 36.7 deg,
that tells us ….
o Pluto and the moon diameters are created in proportionality to each other
and the previous data is a proof for that.
o This proportionality is found based on the perimeter of this triangle BRM1
Equation No. (N)
127610 km = 19 x 6716 km
- 19 degrees the moon orbit regresses per year
- Mars diameter =6792 km with a difference (1%) with this used value
- This equation refers to Mars because, Mars orbital inclination =1.9 deg and we
know that Mars ancient orbital inclination was 1.8 degrees where the difference
=0.1 degrees can causes this effect (1.9 deg/0.1 deg)= 19.

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Group No. 5
Equation No. (Q)
3033 mkm x tan (17.2) = 940 mkm (Earth Orbital Circumference)
- 3033 mkm = The distance between Uranus and Pluto
- 17.2 degrees = Pluto orbital inclination
- Equation No. (Q) tells that, Earth orbital circumference is defined by an effect of
Pluto and Uranus motions interaction!
- BUT
o 149.6 mkm (Earth orbital distance) = 1047 x142984 km
o Where
o 142984 = Jupiter Diameter
o 1047 = the sun mass / Jupiter mass
- How to understand this data? Logically who's the tree and who's the branch?!
Earth or Pluto?!
- The data tells that, Jupiter data is created based Uranus and Earth data. Jupiter is
produced later – after Earth, Uranus and Pluto data! But that means,
o (1st
) The Sun Is Created After The Solar System Creation!
o (2nd
) The New Data is created based on the old data.
o (3rd
) The interaction between Uranus and Pluto to produce the orbital
inclination 17.2 degrees was older than this all data.
o (4th
) the planets data almost do a configuration with each change for any
planet data – that means- when Mars orbital inclination changed from 1.8
degrees to 1.9 degrees all planets did suitable changes as a result, to create a
general harmony between their planets.

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Equation No. (S)
17.2 hours (Uranus Day) x 8.9 = 153 hours (Pluto day period =153.3 hours)
- Uranus effect is seen in the rate 8.9
- This rate is discussed in point no. (3-3) (equation no. 9) but that discussion isn't
sufficient for this rate and because of that we need to deepen our discussion as
possible for that we should try to prove how this rate (8.9) shows Uranus effect in
the next point (3-8 The angle 1.1 deg effect on the moon orbit geometry).
- Equation no. (S) tells that, Pluto day period (153.3 hours) is created by an effect of
Uranus day period (17.2 hours) by using the powerful rate 8.9
- Let's try to suggest an idea to see what happening behind the data
o Uranus day period 17.2 hours is changed into 17.2 degrees because 1 degree
=1 hour. (1st
question, where can we find data in which 1 hour =1 degree?)
o 17.2 degrees is used by Pluto orbital inclination (2nd
question, Why not by
Neptune?!)
o 153.3 degrees is created in the moon orbital triangle (The Angle CDM2)
o Pluto orbital inclination (17.2 deg) is shared in the moon orbital triangle and
(put 17.2 degrees) and (received 153.3 degrees)!! (Pluto Day Period 153.3
hours) (where 1 hour = 1 degree)
o Where's (17.2 degrees)? This angle is used in the triangle BCU which we
have discussed in the point (2-4-6). Where the other angle (the angle UCA)
=27.6 degrees which we considered its express the Month (27.55 days –
how to understand that??
o The moon orbital period (27.3 days) must be created as a result for Pluto
orbital inclination effect on the moon motion - we have to discuss the value
1.1 degrees to see as deep as possible …
Equation No. (K) (revision) 160592 km x sin (17.26 deg) = 47667 km
- 160592 km = Uranus Circumference, and 47667 km =the difference in the moon
distances, (this equation shows Uranus effect on the moon motion through 17.2)

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3-8 The Angle 1.1 Deg Effect On The Moon Orbit Geometry
I- Data
(NO. 1)
89 +1.1 = 90.1
90.1 +1.1 = 91.2
91.2 + 1.1 = 92.3
92.3 +1.1 = 93.4
93.4 +1.1 = 94.5
94.5 +1.1 = 95.6
95.6+1.1 = 96.7
96.7 +1.1 = 97.8
97.8 +1.1 = 98.9 BUT 98.9 =90 + 8.9
98.9 +1.1 = 100
(NO. 2)
Sin (1.1) x 181843 km = 3491 km (the moon diameter =3475 km error 0.4%)
Sin (1.1) x 124660 = 2394 km (Pluto diameter =2390 km)
Sin (1.1) x 2573483 km (Earth Motion distance per solar day) = 49405 km.
(NO. 3)
75 deg x 1.1 = 82.5 deg
82.5 deg x 1.1 = 90.75 deg and (98.6 deg -23.6 deg) = 75 deg
But
1290 deg = 17.2 x 75 deg and 108 deg = 1.44 x 75 deg
(NO. 4)
17.27 deg x 1.1 = 19 deg
23.4 deg x (1.1)2
= 28.3 deg

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II- Discussion
(NO. 1)
89 +1.1 = 90.1
90.1 +1.1 = 91.2
91.2 + 1.1 = 92.3
92.3 +1.1 = 93.4
93.4 +1.1 = 94.5
94.5 +1.1 = 95.6
95.6+1.1 = 96.7
96.7 +1.1 = 97.8
97.8 +1.1 = 98.9 BUT 98.9 =90 + 8.9
98.9 +1.1 = 100
- What does this data tell us?
- Uranus effect in seen in the number 1.1 degrees
- Uranus effect moves through the planets data to crate many planets data based on
this direct effect (1.1 degrees)
- By this creation of data Uranus made on thread moves through the solar system
- The direct results of Uranus effect can be summarized in following
o Uranus creates the rate (10) between the value 89 & 8.9 degrees by that
Uranus be powerful to change its orbital inclination (0.8 degrees) to be used
as 0.08 degrees or as 8 degrees (please remember the rate 0.08).
o 93.4 deg = 90 deg +3.4 deg (Venus orbital inclination)
o 95.6 deg = 90 deg +5.6 deg (The Moon orbital inclination+ the moon daim.)
o 96.7 deg = 90 deg +6.7 deg (The Moon Axial Tilt)
o 97.8 deg = Uranus Axial Tilt
o 98.9 deg = 90 deg +8.9 deg (where 8.9 is used as 153.3 h/17.2 h)

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(NO. 2)
Sin (1.1) x 181843 km = 3491 km (the moon diameter =3475 km error 0.4%)
Sin (1.1) x 124660 = 2394 km (Pluto diameter =2390 km)
Sin (1.1) x 2573483 km (Earth Motion distance per solar day) = 49405 km.
(Neptune diameter 49528 km)
- The value 181843 km is the tangent (DB) in the circle figure based on which the
rate (0.08) is created (please review, why the triangle 1.2 and 51/2
is required?)
- The value 124660 km is the hypotenuse Cr in the triangle CM1r,
That shows
- The moon and Pluto diameters are created as functions in the triangle different
dimensions base on this very important angle (1.1 degrees). It's a clear effect of
Uranus motion on the moon and Pluto motions…
- Neptune diameter uses this same angle (1.1 degrees) based on Earth motion
distance during its day period to produce its diameter.
- It's one hand effect moves through the solar system data. This (1.1 degrees)
moves by its power to force Venus, Pluto and the moon data to be created
according to its instruction… the other planets found that, if they don't follow this
same value (1.1) their motions will be in harmony with these planets and by that
Uranus leads the solar system motion harmony.
- 1.1 deg = 97.8 deg – 96.7 deg, That means, Uranus to produce this 1.1 degrees
depended on the moon Axial Tilt (6.7 deg), by that produced (91.1 degrees) and
the removes the perpendicularity to reach to this (1.1 degrees). That tells Uranus
depends on the moon data to create this value (1.1 degrees). But this conclusion
must be incorrect because Uranus isn't a small planet and the moon isn't sufficient
to support Uranus motion. The dependency from Uranus on the moon motion must

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be containing Earth motion, even if the Earth data isn't seen in any equation and
only the moon data is seen, that can't be a proof, because the logical design tells
that Uranus motion must depend on Earth motion to do this effect. It's a unification
between Uranus and Earth motions.
(NO. 3)
75 deg x 1.1 = 82.5 deg
82.5 deg x 1.1 = 90.75 deg (90.8 deg = 90 + 0.8 Uranus orbital inclination)
and (98.6 deg -23.6 deg) = 75 deg
But
5.1 deg x 254 = 1290 deg = 17.2 x 75 deg and 108 deg = 1.44 x 75 deg
- This data power is the number 75 degrees
- The number 75 deg is produced by Uranus effect (90.8 degrees) and through
Uranus effect (1.1 degrees) in one side
- On the other side the value 75 is produced by the moon orbital inclination 5.1 deg
during 254 month (Metonic Cycle) which produced 1290 deg, but Pluto orbital
inclination effect will cause this 1290 deg to be 75 deg
- This value is used again by Pluto & Uranus orbital inclinations total (107.2 deg +
0.8 dg =108 deg) to produce the moon orbit regression 1.44 degrees.
Notice
- 75 degrees x 0.8 deg (Uranus orbital inclination) = 60 degrees
- Cosine 60 degrees = 0.5 why this is specific?
- 406000 = 116.75 x 3475 km
- 116.75 (is Venus day period) and we will discuss it later
- But
- The moon apogee orbital radius =406000 km but the moon diameter =3475 km
- This data shows the rate (0.5) effect

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- That shows why the rate (2 or 0.5) is used frequently in the solar system data
- for example
- Uranus orbital distance = 2 Saturn orbital distances
- Saturn orbital distance = 2 Mercury Jupiter distances
- Mercury day period = 2 Mercury Orbital Periods (1.999)
- Venus day period = 2 Mercury rotation periods (1.990)
- Uranus diameter = 22
Earth Diameter
- Many other data can be added to the previous, it's one hand effect moves through
the solar system data.
- This data can be explained only by our previous conclusion which is that
- The solar system data creates configuration with each change occurred in the solar
system, and that means, after the Earth moon creation all solar planets changed
their motions data to create a configuration for the new born planet. This process
can't be an optional process but must be obligatory one, because of that, this
process should be forced by some stronger force than the planets themselves, they
have to configure their data with each new born planet or each change occurred for
any other planet to save the motions general harmony.
- In point (3-10 how can the far planets effect on the moon orbital motion?) we
should discuss this question and tries to see the force behind these planets which
causes the planets data configuration to save the motions general harmony.

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3-9 Four Planets Motions Interaction
- The previous discussion tells that, there's an interaction done between these 4
planets in the moon orbit
- These 4 planets are (Uranus, Pluto, Earth and its moon).
- We should deepen our discussion about this interaction in Metonic Cycle
Discussion (Point no.8), but here we need to discover a few features of this
interaction
o (1st
feature) Planet diameter is created based on its motion, for that, we see
the moon and Pluto diameters are created by this interaction effect
o Geometrically this feature is not understandable yet, because we don't know
how the matter is created, because of that the data be puzzles before us, in
following I add some of this puzzled data…
o Uranus (6.8 km/sec) moves during 7510 seconds a distance = 51118 km =
Uranus diameter but 7510 km = Pluto circumference …..
Also
o Jupiter (13.1 km/sec) moves during 10921 seconds a distance = 142984 km
= Jupiter diameter but 10921 km = the Earth moon circumference …..
But
o Earth (29.8 km/sec) moves during 1737 seconds a distance = 51777 km =
where 51118 km = Uranus diameter and 1737 km = the moon radius
o That tells, during (10921 sec = where the moon circumference =10921km)
Earth moves a distance = 2 x Uranus Circumferences (error 1.3%)
o Pluto (4.7 km/sec) moves during 160592 seconds (where 160592= Uranus
circumferences) a distance = 2 Saturn Circumferences.
And
o (Earth velocity /Pluto velocity) = (Pluto day period / Earth day period)

The Moon Orbital Motion Geometry (Revised) (3)

The Moon Orbital Motion Geometry (Revised) (3)

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