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 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 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 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)

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o (2nd
feature) There are different rates of time in the solar system.
o That means, the rate of time is not a common feature….Earth day is 24
hours and the moon day is 708.7 hours, then the question is …. Does Earth
hour = the moon hour?! Because
o We have seen that, there's a rate 1 degrees = 1 day and another rate 1 degree
= 1 hour because of that, the period of time is defined based on its
conditions and not through common conditions…
o The different rates of time is a feature can't be understood easily in Rigid
body mechanics, it's clearly a light motion feature –but how to use in the
rigid body mechanics… what clock is used by what planet?!
o The data faces our logic and challenge us to be able to explain them….
Mercury 8 days = 175.941 solar days x 8 = 1407.6 solar days But
o Mercury rotation period = 1407.6 hours
o The machine behind creates the different rates of time but how and why?
Notice
o During Mercury 8 days, Mercury moves a distance = Uranus orbtial
distance (2872.5 mkm x 2 = 5745 mkm).
o Mercury 8 days, that means, the Cycle of 8 reaches also to Mercury
This Cycle is
o 8 Pluto days Uranus uses in its motion
o 8 of its days Jupiter uses in its motion
Jupiter motion is transported to Saturn
Saturn motion is transported to Neptune
o 8 of its years the Earth uses in its motion
Also
o 8 of its Mercury days Mercury uses

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3-10 how can the far planets effect on the moon orbital motion?
- The figure is used for the explanation…
- How can the far planets effect on the moon orbital motion?
- Imagine we have a triangle, whatsoever this triangle dimensions lengths, but its
angles total is =180 degrees, So if one angle =120 degrees and found at a distance
1 million km that doesn't effect, the rest 2 angles total should be =60 degrees.
- The idea depends on the space nature, it supposes that, the solar planets are found
inside Space created geometrically- as seen in the figure – The space is created
based on geometrical design, So when any planet does any motion the whole space
be effected by it. it's not a force or mass gravity which causes the effect – but the
space geometrical structure Reaction – the space is built by geometrical structure,
that means, each motion of any planet creates a reaction in this space.
- That cause the priority for the planets, and explains how a small planet as Pluto
can effect on the moon orbital motion, because this effect doesn't depend on the
mass gravity force instead depend on the position of the effective planet relative to
the effected one –it's the planets positions in the space which creates the effect
according to the space geometrical design – for example – a planet is found in the
light direction of another planet, so this planet motion effects on the other one not
as the other planets effect because of the direction of light motion – the idea
depends on the planets positions relative to one another.

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4- The Moon Orbital Motion Analysis
4-3 The Moon Orbital Motion Analysis

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- Let's summarize this question answer in following:
o The moon uses Pythagorean triangle basically to decrease its displacement
daily through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
would revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Simply, because the moon uses this technique the moon can revolve around
Earth through perigee orbit (r=0.363 mkm)
o Let's explain this intelligent technique with some details to show the useful
result of using Pythagorean triangle by the moon orbital motion….

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- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km, but the real orbital
horizontal displacement be less than (88000 km) and this is the objective for which
the moon uses Pythagorean triangle –
As an example,
- If (θ) =28.63 degrees, the real displacement (L== 88000 km cos (θ)) = 77237 km,
So, if the moon real displacement daily be (77237 km), during 29.53 days the
moon will pass a distance = 2.28 million km and this will be the moon orbital
circumference, where 2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagorean triangle,
- Now let's suppose the moon doesn't use Pythagorean triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagorean triangle is so useful for the moon orbital motion.

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The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees),
but before we need to analyze the moon orbital motion
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee, we will need
this notice later in our discussion

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4-3 The Moon Orbital Motion
- The moon moves per a solar day a motion typical to the Earth motion to avoid the
separation from Earth through their motions, based on this rule, the moon moves
per a solar day 2.573 million km with an angle declines on the horizontal level
0.98562 degrees as typical to Earth motion
- If there's no Lorentz Length Contraction Phenomenon effect on the moon motion,
the moon motion trajectory would to be a parallel line to Earth Motion Trajectory,
But Lorentz Length Contraction effects on the moon motion daily distance (2.573
mkm) with a rate 1.0725 and causes this distance to be contracted (2.399 mkm)
- The moon difficulties are started here, because the difference between both
distances (0.17 mkm) will cause the moon to be separated from Earth motion
inevitably
- We should notice that, these motions are done far from our observation, means, we
see nothing of this motion distance, because the moon moves on the Earth orbital
circumference revolving around the sun, but, even if we can't observe this motion
distance the motion is still fact and proved by its power, because the Earth moves
per a solar day 2.573 mkm and if the moon doesn't move this same distance every
solar day that necessities the moon to be separated from the Earth through their
motions course – based on that- the facts prove this motion regardless our
observation ability for it.
- Now the moon has an additional distance to be passed (0.17 mkm) and the moon
has to pass this distance on the same solar day to avoid the separation from the
Earth during their motions.
- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force (by which we see the moon in the Earth sky), but the different
distance (0.17 mkm) to be covered still needs the moon to move one more
displacement (= 88000 km)

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- The previous explanation tells that, the moon has to move 2 displacements each =
88000 km, while we see one displacement only because it's done through the
moon orbital motion around Earth but the other displacement should be done also
because this total distance (0.17 mkm) is required to cover the different distance
and create the total (2.573 mkm) which saves the moon and Earth motions
accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagorean triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagorean triangle
in its motion, because the moon can't decrease its daily displacement (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.399 mkm) and Earth motion distance (2.573 mkm),
So the moon needs to move this displacement perfectly, but if it's used as a
displacement through the moon orbit, the moon would be always a prisoner in the
apogee orbit (r=0.406 mkm) as we have discussed before, because of that, the
moon creates Pythagorean triangle technique by which the moon moves actually
88000 km daily but the real displacement through the moon orbit became less (L =
88000 Cos θ) and by that the moon can achieve 2 objectives, First to pass the
required distance (88000 km) and Second to move in near orbits to Earth, that
shows the intelligent moon motion technique…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1)

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The Moon Orbital Motion Needs One More Orbit
- The previous explanation tells that, the moon moves 2 displacements each =88000
km, we see one of these 2 displacements but where's the other displacement?!
- We know that, the moon original motion (2.573 mkm) which is contracted to be
(2.399 mkm) isn't seen by us because the moon moves this distance revolving with
Earth around the sun along the Earth Orbital Circumference
- We may accept that, the 2nd
displacement the moon does on this same trajectory
and isn't seen by us.
- So,
- There must be one more orbit for the moon to move through this 2nd
displacement.
means,
- There's 2nd
Orbit For The Moon Motion
- But
- How can we discover this second orbit if we can't observe the 2nd
displacement
motion?
- We can discover this 2nd
orbit by the moon orbit data analysis. So we should
depend on the moon orbital triangle data analysis to define this 2nd
orbit position.
- For that we have to discuss the moon 2nd
orbit in our deep analysis of The Moon
Orbital Triangle Geometrical Structure.

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4-4-1 The Equation Concept 4-4-2 The Equation Test and Accuracy
4-4-1 The Equation Concept
The Moon Orbital Motion Equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept we have
discussed, which is (the moon uses Pythagorean triangle in its orbital motion)
- The moon uses Pythagorean triangle and by this intelligent technique the moon be
under control of the angle (θ) change
- The angle (θ) defines almost all the moon motion features.…
- The moon uses this technique, aiming to create a real displacement shorter than its
actual displacement (88000 km) based on the equation (L =88000 cos (θ)) and by
that while the moon moves a displacement =88000 km but the real displacement
(L) through its orbit be shorter than 88000 km and by that the moon can revolve
around Earth through more near orbits than its apogee orbit (r=0.406 mkm).
- The moon orbital motion equation depends on this concept and, the equation
uses (the constant) 1.7 degrees as the moon daily motion degrees, and the equation
uses the previous day angle (θ0) to produce the today angle (θ1)
- We have 3 questions in this equation study which are:
o How does this equation work?
o Is this equation trustee and correct?
o Why does the equation use the angle 1.7 degrees for the moon daily motion?
Let's try to answer….

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How to use this equation?
- Perigee Radius =0.363 mkm, so Its Orbital Circumference =2.28 mkm
- Suppose the moon will revolve around Earth through perigee orbit only during
29.53 days, so
- (2.28 mkm /29.53 days) = 77237 km
- This is (the real displacement = L = 88000 km Cos θ = 77237 km),
- What's the angle θ value? the angle θ = 28.63 degrees
- Suppose the moon stand on this point yesterday with the angle (θ) =28.63 degrees,
where the moon will move today?
- From Perigee (the most near point to Earth) the moon will move in Ascending
motion because it moves from perigee (0.363 mkm) to apogee (0.406 mkm)
- In Ascending motion we use (-1.7 degrees) because the angle (θ) is decreased
where the real displacement (L) is increased, So let's do that in following
o (θ1= θ0 - 1.7 degrees)
o (θ1= 28.63 degrees - 1.7 degrees) = 26.93 degrees
o L = 88000 Cos (26.93 degrees) = 78454 km
o During 29.53 days so (78454 km x 29.53 days = 2.316 mkm)
o 2.316 mkm = 2π x 368722 km
That means
o The moon was (before motion) on Perigee radius (r=0.363 mkm) and starts
its motion displacement 88000 km. For day motion the equation uses 1.7
degrees, that means, the moon on perigee uses Pythagorean triangle with
angle (28.63 degrees) and during one solar day the moon uses - 1.7 degrees
and by that the angle will be (26.93 degrees)…... The angle 1.7 degrees
expresses The Moon Daily Motion
o By using Pythagorean triangle its angle (θ) = 26.93 deg, the displacement
(88000 km) will create a real displacement through the moon orbit = 78454
km and the moon will finish its motion today at a distance 368722 km

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means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon after 1 day motion will be at the point 368722 km and will
have the Pythagorean triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee (0.406 mkm) to perigee (0.363 mkm),
so the angle (1.7 degrees) will be positive (+1.7 degrees) because the angle
(θ) is increased and the real displacement (L = 88000 Cos (θ)) be shorter.
So
o If the moon in apogee radius (r=0.406 mkm), what's the angle (θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 10.96 degrees (=11 deg approx.)
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) means (θ1= 11 degrees + 1.7 degrees) = 12.7 deg.
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
So
o After one day the moon will be on 403467 km far from apogee (406000 km)
with 2540 km
Now let's see this equation test and efficiency in following

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4-4-2 The Equation Test and Accuracy
- I have tested the Equation with real data for 2 months June 2020 and October 2020
- The results are very good and I provide the results here for better vision
concerning the equation efficiency
1st
Test June 2020
Day Registered Data The Results (1.7) Difference
6-6-2020 369418 km
7-6-2020 373729 km 374772.5 - 1044
8-6-2020 378917 km 378821.5 96
9-6-2020 384534 km 383667.7 867
10-6-2020 390096 km 388890 1206
11-6-2020 395156 km 394000 1156
12-6-2020 399345 km 398604.2 741
13-6-2020 402395 km 402361.3 34
14-6-2020 404153 km 405052.8 -900
15-6-2020 404574 km ---- ---
16-6-2020 403718 km 401848.5 1870
17-6-2020 401733 km 400876.1 857
18-6-2020 398840 km 398640.7 200
19-6-2020 395303 km 395417.4 115
20-6-2020 391409 km 391521.2 -113
21-6-2020 387432 km 387273.4 159
22-6-2020 383607 km 382968.4 639
23-6-2020 380110 km 378852 1258
24-6-2020 377044 km 375107 1937
25-6-2020 374451 km 371836.5 2615
26-6-2020 372338 km 369077 3262
27-6-2020 370703 km 366855.6 3847
[

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The 1st
Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 2 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 3 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 20) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1300
km to 2000 km) = error (4.5%), that means (2 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (3 values of 20) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

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2nd
Test October 2020
Day Registered Data Results (1.7) Difference
5-10-2020 405,690 km --- ---
6-10-2020 404,171 km 403125.3 km 1046 km
7-10-2020 401,649 km 401390 km 259 km
8-10-2020 398,073 km 398545.6 Km - 473 km
9-10-2020 393,464 km 394568.8 km -1105 km
10-10-2020 387,944 km 389510 km -1567 km
11-10-2020 381,763 km 383520 km -1758 km
12-10-2020 375,302 km 376875.3km -1574 km
13-10-2020 369,063 km 369981km -919 km
14-10-2020 363,617 km 363363.4km 254 km
15-10-2020 359,530 km 357612 km 1918 km
16-10-2020 357,269 km 353307 km 3962 km
17-10-2020 357,105 km ---- --
18-10-2020 359,048 km --- --
19-10-2020 362,851 km 364979.7 km - 2129 km
20-10-2020 368,058 km 368579.3 km -522 km
21-10-2020 374,101 km 373492.4 km 609 km
22-10-2020 380,412 km 379168.3 Km 1244 Km
23-10-2020 386,497 km 385059.3Km 1438 km
24-10-2020 391,989 km 390694.3 km 1295 km
25-10-2020 396,659 km 395729.5 km 930 km
26-10-2020 400,395 km 399958.7 km 437 km
27-10-2020 403,181 km 403299 km 112 km
28-10-2020 405,059 km 405738.5 km -680 km
29-10-2020 406,104 km 407359.4 km -1256 km
[

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The Test Results Analysis:
- The Total Results Are 22 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 5 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 2 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 22) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
km to 2000 km) = error (4.5%), that means (5 values of 22) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (2 values of 22) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

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4-4-3 The Value 1.7 degrees
- The 3rd
question was, why the equation uses 1.7 degrees?
Because
1.7 degrees = 0.98562 degrees + 0.712 degrees
Where
- 0.98562 degrees = Earth motion daily degrees, and it equals the moon daily
motion degrees because the moon has to move an equal distance to Earth motion
daily distance to save their motions accompanying
- This question and the angle 0.712 degrees is discussed deeply (Metonic Cycle
Discussion)

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The Moon Motion Difficulties
- There are 2 basic difficulties are observed in the moon orbital motions, let's refer
to them in following:
o (1st
Difficulty) The moon moves per day different distances from perigee to
apogee…..
o We know the moon moves from perigee to apogee (go and back) during
Anomalistic month (27.55 solar days)
o (43000 km x 2) / 27.55 days = 3122 km
o The moon doesn't use this rate (3122 km) in its motion, instead the moon
can move (6000 km) on one day only and on another day may move only
2500 km (or even less)!
o The moon orbital equation tries to solve this difficulty by using the rate 1.7
degrees in the equation (θ1 = θ0 + 1.7 degrees), the value 1.7 degrees is a
great number and enables the moon to move around (5000 km) per solar day
and by that if the moon moves per solar day 4000 km the different distance
will be 1000 km and if the moon moves 6000 km the different will be
– 1000 km, it’s the same difference, and by that, the error be minimized as
possible enabling the equation to be more efficient..
o (2nd
Difficulty) The moon stays in perigee and apogee points long time….
o That means, while the moon be on perigee or apogee, the moon doesn't use
the equation and doesn't change its distance to perigee or apogee for long
days…we may notice that in the equation tests, when the moon reach to
perigee or apogee the equation stops its work and stays 2 or 3 days to return
to its work… because the moon consumes long time to leave the points
(perigee and apogee)…

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5-The Moon Orbit Geometrical Design
5-1 Preface
5-3 The moon motion angle (12.195 deg) Analysis

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5-1 Preface
On What Facts This Study Depend? On The Logical Geometrical Structure
- Example.
- The moon orbital triangle base (The Green Line) (EA) = 449197 km
- In this distance, the point (A) I have concluded and was not found in the moon
motion data sheet, so Can be this point (A) a real point, or it's invented one?
o The distance EA causes the distance BD (43000 km) be = DA (43000 km)
o The distance EA 449197 km = Jupiter Circumference
o The distance BA = 86000 km = BC
o The triangle BCD is a Pythagorean specific triangle (1, 2, 51/2
)
o The perimeter of the triangle (ECA) = the distance from the point (A) to the
end on the lunar eclipse umbra length (1.392 mkm).
If I have invented the point (A), how can I created these relationships with it, where I
depend on the moon orbital motion real data? The main power behind this analytical
study is The Logical Geometrical Structure Of The Moon Orbital Motion Data.

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I-Data
- The angle DM2N = 90 degrees
- The angle NM2L = 1.1 degrees
- The angle DM2C = 19.367 degrees
So
- The angle CM2L =71.7 degrees
- This angle is considered to be =71.9 degrees because of specific effect of the moon
diameter we should discuss in the last point of this paper
(Point No. 8-4 The Moon Diameter Creation)

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The Angle 71.9 Degrees Analysis
(Why we need to discuss this angle 71.9 degrees?)
Because this angle can answer why the moon orbital motion equation uses the
constant 1.7 degrees for the moon daily motion (θ1= θ0 +1.7 degrees).
- The angle CM2L = 71.9 degrees
- The angle M1 L M2 =88.9 degrees
I- Data
(m)
The angle M1 N M2 =88.9 degrees
88.9 degrees – 71.9 degrees = 17 degrees
(n)
(17 degrees /0.8) = 21.25 degrees
(o)
21.25 degrees x 0.08 = 1.7 degrees (the moon motion equation constant)
(p)
17 degrees x 1.7 degrees = 29 degrees
(q)
23.4 deg = 1.8 deg x 13.177 deg x 0.98562 deg
Discussion
Equation No. (o)
- Equations (from m to o) give us a simple geometrical method to change the value
17 degrees into 1.7 degrees, but why this method is useful?
- Because the value 21.25 degrees is one of the moon motion angles which is
- 21.25 degrees = 11.8 degrees x 1.8 degrees
- Where
- 11.8 degrees = 5.1 deg (the moon orbital inclination) +6.7 deg (the moon axial tilt)
- But what's 1.8 degrees?! Let's discover in following…

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o The moon moves from perigee to apogee and return back during its orbital
period.
o The distance from perigee to apogee on the moon orbital triangle (BD)
controlled by the angle (BCD =26.577 degrees)
o The moon go and return during the cycle (26.577 degrees x 2 = 53.15 deg)
o (53.15 degrees /29.53 solar days) =1.8 degrees
o Why I divide this angle 53.15 degrees on 29.53 days?
o Because
o The moon starts a new cycle from perigee to apogee after completes its day
period. Means the angle (53.15 degrees) should be distributed during the
synodic month (29.53 solar days).
- The previous explanation shows that, the angle 21.25 degrees is used in the moon
orbital motion because it depends on 2 angles (11.8 deg) and (1.8 deg) are used in
the moon day motion. based on that, the interaction between the angle 17 degrees
and 21.25 degrees can be created because both angles are used in the same motion
- Then
- The last step is to change the angle 21.25 degrees into 1.7 degrees as following
- 21.25 degrees x 0.08 = 1.7 degrees
- We remember this rate (0.08) based on which the valuable angle (10.96 deg) is
created. (please remember 137 degrees x 0.08= 10.96 degrees
Notice
- The most 3 basic values in the moon motion are (137 deg, 10.96 deg and 0.08)
- As the valuable angle (10.96 deg) is created based on this rate (0.08), the moon
orbital motion equation angle (1.7 deg) is created based on it….BUT
- Why the data shows that, Uranus orbital inclination (0.8 degrees) is used in this
process? The data uses (17 degrees /0.8 degrees) = 21.25 degrees, showing clearly
the using of Uranus orbital inclination (0.8 degrees) Why? because the data tries to
show Uranus effect on the moon orbital motion…. the next points supports it.

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Equation No. (p)
- We know both angles 17 and 1.7 degrees but what's this 29 degrees?!
- The major lunar standstill can be +28.5 = (23.4 deg + 5.1 deg)
- The moon angular diameter = 0.5 degrees, that means, when the moon orbital
inclination is measured above the moon diameter it will be =5.6 degrees
- So the angle 28.55 degrees +0.5 degrees = 29.05 degrees
- That shows Uranus effect on the moon motion during Metonic Cycle, which effect
on the moon daily orbital motion and effect on the moon motion equation by the
constant (1.7 degrees)
Equation No. (q)
23.4 deg = 1.8 deg x 13.177 deg x 0.98562 deg
Where
23.45 deg = Earth Axial Tilt 0.98562 deg = Earth motion daily degrees
13.177 deg = the moon daily motion degrees
1.8 degrees = is the angle we have discussed in the previous equation (no.3), the
angle of the moon motion from perigee to apogee during its day period
(53.15 degrees /29.53 solar days).
Equation no. (q) shows that Earth axial tilt is created depending on the moon motion.
Please Note
The angle 71.9 degrees is rich angle in its discussion, the previous analysis is a part of
its complete discussion, which is written in. (8-3 The Angle 71.9 Degrees Analysis)
Please review the full discussion.

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5-3 The Moon Motion Angle (12.195 degrees) Analysis
I-Data
(I)
Tan (12.195 degrees) x 407188 km = 88000 km
And
13.177 degrees – 0.98562 degrees = 12.195 degrees
(II)
(10.96 degrees) + 1.25 degrees =12.195 degrees
Where
13.177 degrees = the moon daily motion degrees
0.98562 degrees = Earth daily motion degrees
0.8 degrees = Uranus Orbital Inclination
II- Discussion
- The Apogee Orbit (r=0.406 mkm) permits a displacement =86400 km only based
on the valuable angle (10.96 degrees), as maximum displacement during 29.53
days because (86400 km x 29.53 days = 2.55 mkm = 2π x 0.406 mkm)
- But
- What about the actual displacement 88000 km, which angle expresses it?
- The data shows that, the angle 12.195 degrees can define this displacement (88000
km) relative to the radius (40718800 km) which is very near to apogee radius =
(406000 km) (error 0.3%).
- Equation No (II) tells that, Uranus orbital inclination 0.8 degrees is used as (1/0.8),
i.e.
- The angle (10.96 degrees) + (1/0.8 degrees) = 12.195 degrees
- The data shows Uranus effect on the moon orbital motion
NOTICE (1)
Uranus effect on the moon orbital motion is discussed in Metonic Cycle Discussion
(Point no. 7 of this paper)

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NOTICE (2)
The following explanation shows a new geometrical technique is using in the moon
geometrical structure, it's just example using the angle 12.195 deg in this technique
I-Data
- In the triangle ABC
- AB = 12.195 km
- AC = 2 x 29.53 km
- The Angle A = 78.081
- The Angle C = 11.919 degrees
- But
- Cos (12.195 degrees) x 12.195 degrees = 11.919 degrees
1- How This Triangle Is Created?
- The geometrical structure uses the angle 12.195 degrees as a distance= 12.195 km,
and creates the angle (C) depends on the angle 12.195 degrees as the data shows
- So this triangle is created depending on the angle 12.195 degrees
2- This Triangle Purpose
- The triangle aims to create the hypotenuse AC = 59.06 km = 2 x 29.53 km
3- Why This Triangle Is Created?
- To create the value (29.53 km) depends on the value 12.195 degrees geometrically,
both data is the moon motion data, but the triangle tries to connect both data
geometrically, why? because Nothing is independent (the geometrical concept),
because of that, the new data should be created based on the old data, and by that
there's always one line connecting all data
This simple example is for this technique explanation.. and the rate (1km=1degree) is
used here only and not a general rate, although the value (2x 29.53) is used more
widely than (29.53) in all data. (For example, Earth during 59 days moves a distance
= its orbital distance Error 1% ).

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NOTICE (3)
- There's one more geometrical technique is used in the moon orbital motion
geometry let's refer to it in following
- The geometrical design uses The Triangle Perimeter As A Distance
- That means
- The distance in straight line is considered equivalent to a triangle perimeter
- This feature we have faced 2 times in our analysis let's refer to one of them here
Example No. 1
The perimeter of the triangle (ACE) whose dimension (AE =449197 km, AC =121622
km and CE =373000 km) its perimeter = 943819 km
But the lunar total eclipse umbra length = 1.392 mkm from the Earth
1.392 mkm – the distance AE (the triangle base 449197 km) = 943819 km
Why these 2 distances are equal? The geometrical design considers the triangle
perimeter as equivalent to the distance (straight line) and divided it into 2 equal parts.
Example No. 2
Tan (5.1) x 2598693 = 232000 km
Where
5.1 deg = The Moon Orbital Inclination
2598693 km = The Moon Displacements Total During 29.53 Solar Days
In Point No. (6) (The moon orbital inclination creation), we will discuss how the
value 232000 km is created, where the difference between 2598693 km and 2550973
km (the moon orbital circumference at apogee) = 47667 km, the orbit geometrical
design uses this value 47667 km as a right triangle base its angle =29 degrees, by that
the triangle perimeter be 232000 km, causing to create the moon orbital inclination
5.1 deg, in this process the triangle perimeter is considered as equivalent to a straight
line to create the angle 5.1 deg.
(Please review this process details in point no. (6-1The Moon orbital inclination
creation geometrical process)

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- Let's summarize how this triangle idea is created in following:
o Uranus Axial Tilt =97.8 deg and the Earth Moon Axial Tilt =6.7 deg. So
between them (97.8 – 6.7 = 91.1 degrees)
o The number 91.1 degrees gives a reference for some perpendicularity
between the moon axial tilt and Uranus axial tilt, but there's 1.1 deg!
o So, the solution was to decline the triangle base (EA) with 1.1 degrees on
the horizontal level and by that Uranus axial tilt will be perpendicular on the
triangle base (AE) if this triangle based depends on the moon axial tilt…
o This is the original idea of this triangle
o For that reason the line BC is perpendicular on the moon orbital triangle
- Based on this description
- The line BC shows Uranus motion effect on the moon orbital motion.
- Specifically the line BC refers to Uranus Axial Tilt (97.8 deg)
- In Metonic Cycle Discussion we should discuss more effects done by this line BC
on the moon orbital motion trying to prove that Uranus Motion effect on the moon
orbital motion and causes to create the moon Metonic Cycle.

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I-Data
(1)
(Sin (10.96 degrees) x 449197 km = 85403 km
- This equation causes disappointment for the investigation because neither the
value 88000 km nor 86000 km is created based on the triangle base (EA=449197
km= Jupiter Circumference) based on our valuable angle (10.96 deg), so, that tells
something must be un-understandable!
Shortly
How that is happened? As following:
o 137 degrees x 0.08 = 10.96 degrees (our angle)
o (137 degrees +1.543 degrees) x 0.08 =11.084 degrees
o (137 degrees -1.543 degrees) x 0.08 =10.836 degrees
Based on that
o Tan (11.084 degrees) x 449197 km = 88000 km
- Both values (88000 km and 86000 km) are defined based on the triangle base
(EA=449197 km) based on both angles (11.084 and 10.836 degrees) where these 2
angles are created by the original angle 137 degrees (as our angle 10.96 deg).
- But
- The angle 1.543 degrees (found between the ecliptic line an the moon equator line)
effects on our angle (10.96 degrees) to produce these 2 new angles (11.084 and
10.836 degrees) where these 2 angles should be considered as similar forms for our
angle (10.96 degrees).

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- Please note this data importance because the base EA =449197 km = Jupiter
Circumference, because of that, this data may refer to Jupiter effect on the moon
orbital motion.
Notice
- Tan (10.836) x 29.2 = 5.6
- Where
- Earth moves during 29.53 solar days a value 29.2 degrees but the moon moves
during this same period (360 deg + 29.2 deg)
- 5.6 degrees = 0.5 deg +5.1 deg = that means, when the moon orbital inclination be
measure above the moon diameter the value will be 5.6 degrees
- That tells us, the moon orbital inclination is rated to the Earth and moon motions
during 29.53 days by this angle (10.836). That means these 3 values are created
rated to each other.

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6-The Moon Orbital Triangle Analysis
6-1 Preface

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In This Figure We Study
(1) The Shaded Triangle CBK
- CB = 86000 km
- M2B = 128400 km (where BD =AD= AM2 =42800 km)
- M2C = 154554 km
- CM2B = 33.81 degrees
- M2CB = 56.19 degrees
- The perimeter of the triangle (CBM2) = 368954 km

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(2) The Shaded Triangle CNM2…
- The distance CN = 121758.2 km
- The distance M2N = 42800 km
- M2CN = 19.367 degrees
- CM2N = 70.633 degrees
- CM2M1 = 65.09 degrees
- The distance M1M2 =43000 km (the distance from perigee to apogee)
- The perimeter of the triangle CNM2 = 293622.2 km
Let's try to use these 2 shaded triangles in following…

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- We start with the triangle CM2N,
o In the triangle the hypotenuse CM2 (=129064 km) is an interesting piece
o We use it to create the base BK
Notice
o The base BK is consisted of 3 equal parts each part =42800 km (So
BK=128400 km)
But
o The hypotenuse CM2 (129064 km = 3 x 43000 km), where the distance
from perigee (M1) to apogee (M2) = 43000 km
o The difference reason is that, the angle RM1M2 =5.543 degrees and by that
the hypotenuse M1M2 (43000 km) will produce the adjacent RM1
(=428000 km) which is for the base BK (= 3 x 42800 km).
What we try to do here?
- We will move the hypotenuse CM2 (=129064 km) to be used in the triangle base
BK and then depend on it we will move the moon motion from perigee to apogee
(M1M2) to be used on this same base (BK), let's try to do that in following:
1st
Step
o The hypotenuse CM2 (=129064 km) is used on the base BK (=128400 km),
the difference is very small and so we will neglect.
o Based on that the hypotenuse CM2 (= 129064 km) is used to be the base BK
(=128400 km).
o The angle CM2N = 65.1 degrees
o Based on the base BK (128400 km) we create the hypotenuse KC = 154554
km based on the angle BKC = 33.81 degrees (Why?)
o We need an angle =65.1 degrees on this hypotenuse KC
But
o The perpendicular line C3 creates as angle (90 degrees) on the base KB,
where the hypotenuse CK has an angle 33.81 degrees with the base BK
o That means, the angle CKC3 = 56.19 degrees

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o But we need 65.1 degrees, for that we create the line C4 which has an angle
= 8.9 degrees with the vertical line C3, by that, the angle C4KC =65.1
degrees
o Based on that, the moon motion from perigee to apogee (M1 to M2) should
be express on the line C4.
- The previous imaginary geometrical process aimed to deepen our understanding
for the moon orbital triangle – in following we should ask some questions to see if
we get this triangle depth yet ….
(1st
Question)
- The moon motion from perigee to apogee (M1M2 Distance =43000km), and the
triangle in its most simple form uses this distance as seen in the triangle CBD
where DB =42800 km to express the distance from perigee to apogee BUT why
the great triangle (KBC) uses 3 parts of this distance (KB =3 x 42800 km =
128400 km)?
o The distance 449197 km (EA distance) = 88000 km x 5.1
o The distance 492197 km (EK distance) = 88000 km x 5.6
5.1 deg = the moon orbital inclination
0.5 deg = the moon angular diameter
5.6 deg = the moon orbital inclination measured above the moon
diameter.
(2nd
Question)
- What's the importance result of using the angle 8.9 degrees (C4KC3)?
o 95.6 deg +1.1 deg = 96.7 deg
o 96.7 deg +1.1 deg = 97.8 deg
o 97.8 deg +1.1 deg = 98.9 deg (= 90 +8.9 degrees)

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(3rd
Question)
- How the hypotenuse CM2 (=129064 km) can be the base KB = 128400 km)?
- In this triangle
- BC = 121758.2 km AB = 40763.2 km
- AC = 128400 km BAC = 71.49 degrees
Notice
- The distance BC =121758.2 km = the distance CN in the triangle CM2N, based on
this distance the hypotenuse CM2 (=129064 km) is crated but they used an angle
=19.367 degrees…
- In our triangle the angle (C ) =18.51 degrees, so the triangle data is changed based
on that But.
- The base KB (=128400 km) can be produced based on this same distance (BC =
121758.2 km) if we can find an angle =71.49 degrees
- So where this angle can be found?
- It's the angle BC1K = 71.366 degrees
- Where
- The angle isn't used in the same triangle and also the distance 121758.2 km isn't
used in the triangle, that tells is some geometrical interaction is occurred behind
and create this distance (KB = 128400 km) depends on the distance 121758.2 km
and the angle 71.32 degrees before to use this distance 128400 km in the CBK
triangle.

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(4th
Question)
- What specific features concerning this distance 128400 km?
(1)
- If we use this distance (128400 km as adjacent and has an angle 33.81 degrees) so
this triangle hypotenuse will be 154554 km
- Neptune Circumference = 155597 km (error 0.6%)
(2)
- 406000 km (apogee radius) = π x 129234 km
- Where this distance 128400 km is created in parallel to the hypotenuse 129064 km,
- That shows geometrical interactions behind…

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(5th
Question)
How the transportation of the hypotenuse CM2 to be used as KB can be proved?
- The angle KCB = 56.19 degrees
- And
- The angle BCM2 = 19.367 degrees
- So
- The angle M2CK = 36.823 degrees
Notice
- (37 degrees)2
= 10 x 137 degrees (the most important angle in the moon orbital
motion)- w have analyzed the angle 37 degrees deeply in this paper - the point is
that – the main angle 137 degrees depends almost on this angle 37 degrees – but
the produced angle (=36.823 degrees), which may refer to some geometrical
interaction is found to decrease this angle from 37 degrees to 36.832 degrees…

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2-2 The Moon orbital triangle modification
- In this triangle the distance AK is added to the original one, by that the distance
KB = 128400 km (= 3 x 42800 km) – and we can't use the value (43000 km x 3
=129000 km) because of the angle RM1M2 which causes the adjacent RM1 to be
=42800 km where the hypotenuse M1M2 =43000 km (the distance from perigee to
apogee)
- The triangle BCK became so interesting geometrical figure because the distance
BC =86000 km = 2 x 43000 and the distance BK =128400 km = 3 x 42800 km
- Many important changes are occurred in this triangle among which the following:
o The angle CKB =33.81 degrees
o The hypotenuse CM2 =129064 km but the base BK = 128400 km
o The hypotenuse CK = 154554 km
o The angle BCK =71.366 degrees
o The angle C4KC3 =8.9 degrees

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The Modification Objective
- This modification aims to use the hypotenuse CM2 (=129064 km) to be the base
BK (128400 km), based on this base the moon motion from perigee to apogee (the
distance M1M2) will be moves with the hypotenuse CM2
- For that, I have tried to save the angle (65.1 degrees) which = CM2M1 to be used
with the new modification
- But
- The angle 65.1 degrees is used between the line KC4 and the hypotenuse CK in
place of the base BK
- That because, I have found that, the angle 33.81 degrees should be used inevitably
between the 2 dimensions (128400 km and 86000 km) because these 2 dimensions
are rated to each other (3 x 42800 and 2 x 43000) by that, I supposed they have
their own angle (which is 33.81 degrees) and because of that I have used the angle
65.1 degrees above the hypotenuse CK in place of the base BK
- Whether this using is correct or not, it's not so important to know now, because we
have to analyze the geometrical basics on which this triangle is modified to know
if this modification (generally) can be valid geometrically
- Let's write the basic questions we need to answer concerning the triangle
geometrical modification in following…
(1st
Question)
- The triangle base BK =128400 km, where the moon motion from perigee to
apogee (M1 to M2) =43000 km, how to prove that, this base (128400 km) is
belonged to the moon orbital motion?
(2nd
Question)
- Can Jupiter motion effect on the moon orbital motion? prove the answer by this
triangle geometrical design analysis?

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Question No. (1) The Triangle Base BK =128400 km
Let's remember the question …
- The triangle base BK =128400 km, where the moon motion from perigee to
apogee (M1 to M2) =43000 km, how to prove that, this base (128400 km) is
belonged to the moon orbital motion?
Shortly
- How to prove that, this triangle Base BK is belonged to the moon orbital motion
geometrically?
- This proof should be discussed in many points, let's write them in following…
(1st
Point) …Let's remember this triangle basics
- The Point (B) is parallel to the moon perigee point (M1) and this triangle is created
to cause the distance from B to E to be =363000 km = the perigee radius, therefore
- The distance EB =363000 km and the distance ED = 405800 km and the distance
EA= 448600 km and EK =491400 km.

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- These values are changed with the mentioned one because of the difference
between 43000 km and 42800 km, for that, the motioned values were as following
(ED =406000 km apogee radius), (EA =449197 km = Jupiter Circumference) and
(EK=492197 km).
- To make these distances equal accurately we have to make the distance M1M2 to
be 43200 km to cause the distance RM1 =43000 km, and this modification may
cause serious effects on the triangle, for that, it's preferred to use the distances as
them now keeping in our minds that they refer to their original values…
- The question simply is ….
- The points B D are belonged to the moon orbital motion (because they are
perigee and apogee radiuses) but why the points A K are belonged to the
moon orbital motion?
The Point K (EK = 492197 km)
I - Data
(1)
492197 km x 2π = 3092565 km = 29.53 days x 104724 km
But
104724 km x sin (180 deg/π) = 88000 km (the moon daily displacement)
(2)
Tan (42.4) x 492197 km = 449197 km
Tan (42.1) x 449197 km = 406000 km
Tan (42.1) x 142741 km =129064 km
Tan (41.8) x 406000 km = 363000 km

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II – Discussion
Equation No. (1)
492197 km x 2π = 3092565 km = 29.53 days x 104724 km
But
104724 km x sin (180 deg/π) = 88000 km (the moon daily displacement)
- This equation uses the distance (EK =492197 km) as a radius (similar to the
perigee and apogee radiuses), it's circumference divided by lunar synodic month
will be 104724 km
- The equation second part gives the required proof, because the rate (sin (180
deg/π)) is a geometrical rate, means, this value isn't created for any planet or
motion, it's simply a common one, as any geometrical rule (for example, the
triangle angles total =180 degrees, this rule doesn't refer to any specific triangle
but simply it's used by all triangles)… similar to that, the rate (sin (180 deg/π))
shows that it's a constant in any planet motion, means, it refers to the planet motion
limitations and by that it's part of this planet motion …
Equation No. (2)
Tan (42.4) x 492197 km = 449197 km
Tan (42.1) x 449197 km = 406000 km
Tan (42.1) x 142741 km =129064 km
Tan (41.8) x 406000 km = 363000 km
- It's almost the same angle is used between the 4 points (perigee 363000 km,
apogee 406000 km, point A =449197 km, and point K = 492197 km)
- That shows geometrical order between these 4 points shows their interaction
between each other – that tells –these 4 points are created depends on each other.
- The relation (Tan (42.1) x 142741 km =129064 km) We should discuss in the
next point.

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Question No. (2) Jupiter Effect On The Moon Orbital Motion.
Point No. (1)
- The Perimeter Of The Trapezoid BSM1M2 = 142741 km
- (Jupiter diameter =142984 km)
I-Data
(a)
Tan (53.55) x 105430 km = 142741 km
(b)
Sin (17.44) x 142741 km = 42800 km
(c)
Cos (26.6) x 142741 km = 127610 km
(d)
Sin (12.226) x 142741 km = 30230 km
(e)
Cos (52.95) x 142741 km = 86000 km
II-Discussion
Equation (a)
Tan (53.55) x 105430 km = 142741 km
- We notice that, this value 105430 km is very near to our previous discussion value
104724 km (in equation no. 1) (error 0.7%)
- The angle RBM1 =53.55 degrees
- That tells us the value 104724 km is mentioned geometrically in the moon orbital
motion which supports the conclusion that the point 492197 km is belonged to the
moon orbital motion.
Equation (b)
Sin (17.44) x 142741 km = 42800 km

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- Frequently we have found a clear effect of Pluto orbital inclination (17.2 deg) or
its spouse (17.4 degrees) on the moon orbital motion showing that the distance
42800 km (the perigee apogee distance =43000 km) is effected by it.
Equation (C)
Cos (26.6) x 142741 km = 127610 km
- 26.46 degrees is the angle DCB = BC1D
- 142741 km = the perimeter of the trapezoid BSM2M1
- 127610 km = the perimeter of the triangle M1RB (we discuss deeply)
- The equation shows deep geometrical interactions behind
Equation (d)
Sin (12.226) x 142741 km = 30230 km
- The distance SB = 30230 km
- The angle 12.226 degrees is very near to the angle 12.195 degrees where (13.177
deg the moon daily motion – 0.9856262 deg = 12.195 deg).
- That shows the trapezoid perimeter effect on the moon motion.
Equation (e)
Cos (52.95) x 142741 km = 86000 km
- The angle RBM1 = 53.55 degrees, and the error (53.44/52.9 = 1%)
- The value 86000 km = the line BC length on which the triangle is created.

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Notice
- The trapezoid perimeter 142741 km (where Jupiter diameter =142984 km), shows
Jupiter effect on the moon orbital motion
- If this effect isn't seen or proved by the discussion that because we don’t know
how the matter is created, better vision we can get by the following data:
o Jupiter (13.1 km/sec) moves during 10921 seconds a distance =142984 km =
Jupiter diameter (but 10921 km = the moon circumference)
o Uranus (6.8 km/sec) moves during 7510 seconds a distance =51118 km =
Uranus diameter (but 7510 km = Pluto Circumference)
- I have used Uranus data also to show that, it's a general using in the solar system
motions, we don’t know why or how Jupiter can use the moon circumference
(10921 km) as a period of time (10921 seconds) or for Uranus how to use Pluto
data, but the using shows these planets motions effect on the small planets.

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Point No. (2)
Data
o 137 degrees x 0.08 = 10.96 degrees (our angle)
o (137 degrees +1.543 degrees) x 0.08 =11.084 degrees
o (137 degrees -1.543 degrees) x 0.08 =10.836 degrees
Based on that
Discussion
- The distances (88000 km = the moon daily displacement) and (86000 km = the
line BC), these 2 distances are defined based on the triangle base EA =449197 km
= Jupiter Circumference
- Because the angles 11.084 and 10.836 degrees are created by the same method by
which the valuable angle 10.96 degrees is created, that shows Jupiter motion effect
deeply on the moon orbital motion data.

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Extending Discussion
Data
(f)
142984 km x 2 = 51118 km x 5.6
(g)
129064 km x sin (23.4) =51118 km
(h)
6.8 km/sec x 120536 sec. = 2 x 409822.5 km
4.7 km /sec x 86400 sec =406000 km
Discussion
Equation (F)
142984 km x 2 = 51118 km x 5.6
- 142984 km = Jupiter diameter
- 51118 km = Uranus diameter
- 5.6 degrees = 5.1 deg (the moon orbital inclination) + 0.5 deg (the moon
angular diameter), So 5.6 degrees is the moon orbital inclination measured above
the moon diameter.
- Equation no. (F) tells that, the moon and its orbital inclination causes an
interaction between Jupiter and Uranus motions
Equation (g)
129064 km x sin (23.4) =51118 km
- The Earth Axial Tilt (23.4 deg) is created effected by the hypotenuse CM2 and
Uranus diameter…
- The general conclusion is that, Earth axial tilt is created effected by the moon
orbital motion – but the vision can be clearer if the proportionality between Earth
and Uranus axial tilts be analyzed (where 113.45 deg =97.8 x 1.16).

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Equation (h)
6.8 km/sec x 120536 sec. = 2 x 409822.5 km
4.7 km /sec x 86400 sec =406000 km
- Pluto (4.7 km/sec) moves during a solar day (86400 seconds) distance =406000
km = apogee radius
- But
- Uranus moves during 378675 seconds a distance = 2.57499 mkm = Earth motion
distance during a solar day
- Because this distance (2.57499 mkm = Earth motion distance) is greater than the
moon apogee circumference (2.550973 mkm) with 1%
- This distance (2.57499 mkm = 2π x 409822.5 km) where the radius 409822.5 km
is greater than apogee radius (r=406000 km) with 1%
- But
o Uranus uses 378675 seconds (where 378675 km = Saturn Circumference)
o And
o Uranus uses 120536 seconds (where 120536 km = Saturn Diameter)
o Why?
o The distance between the Earth and Moon at total solar eclipse = 378675
km (= Saturn Circumference) Why?

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The line BC
- The basic idea in this modification is to use the hypotenuse CM2 on the base BK
to transport the moon motion from perigee to apogee (43000 km) and to make it
shown on the base BK
- By the transportation of the angle M1M2C (65.1 degrees) to be use as C4KC (65.1
degrees) by that, the moon motion from perigee (M1 to M2) (on the horizontal
level) is used to create the line BC on the vertical level
- The line BC is created basically in form C4K with an angle 8.9 degrees with the
vertical axis (z-axis)
- But the geometrical design uses this value (8.9 degrees) and creates the line BC as
a perpendicular line (90 degrees) on the base BK
- The removing of the angle 8.9 degrees is done by the following interaction
o 95.6 deg +1.1 deg = 96.7 deg
o 96.7 deg +1.1 deg = 97.8 deg
o 97.8 deg +1.1 deg = 98.9 deg (= 90 +8.9 degrees)
In more details
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

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7- The Moon Orbital Inclination Creation
7-1 The Moon orbital inclination creation geometrical process

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7-1 The Moon Orbital Inclination Creation Geometrical Process
In this triangle
- ab = 0.232 mkm
- ac = 2.608975 mkm
- bc = 2.598 mkm (the moon displacements total)
- The angle c = 5.1 degrees (The Moon Orbital Inclination)
- This figure tells us that
o To create the angle 5.1 degrees we need 2 distances (1st
distance) the moon
displacements total during 29.53 days (2.598 mkm) and this distance is
found and defined by the moon daily displacement.
o Also we need the distance ab =232000 km
o This is the factor based on which the moon orbital inclination will be created
o So, we need to produce this distance (232000 km)… so let's try to do
Equation No. (1)
Cos (10.96 degrees) x 2598693 km = 2550973 km
- The angle 10.96 degrees is the most valuable angle in the moon orbital triangle we
should discuss its origin in the moon triangle geometrical design (Point No. 4)
- The distance 2598693 km =the moon displacements total during 29.53 days
- The distance 2550973 km = the moon apogee orbital circumference
- Equation No. (1) tells, the moon apogee orbital circumference 2.55 mkm is created
depending on the moon displacements total by the angle 10.96 degrees
- The difference = 2598693 km – 2550973 km = 47667 km
- Then
- From this difference 47667 km we need to create the distance 232000 km
- How??

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- The moon displacements total during 29.53 days = 2598693 km
- And
- The Earth moves during 29.53 days a value 29.2 degrees
- The moon moves during 29.53 days a value (360 + 29.2 degrees)
- Let's create another triangle, its base = 47667 km and its angle 29
- In this triangle
- The BC = 47667 km
- The angle A = 29 degrees based on that
- The hypotenuse AC = 98321 km
- The distance AB = 86000 km
- The triangle perimeter = 231982 km
- The input data is 47667 km and the angle 29.2 deg (is used as 29 deg)
- The output is the perimeter of triangle (ABC) =231982 km
- Tan (5.1) x 2598693 km (the displacements total) =232000 km
- The previous explanation shows the geometrical mechanism by which the moon
orbital inclination (5.1 degrees) is created depends on the difference between the 2
distances (2598693 km and 2550973 km).

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The Moon Orbital Inclination (5.1 degrees)
- The moon orbital inclination (5.1 degrees) is a geometrical value
- For example
- Sin (5.1) x (180/π) =5.1
o The moon orbital inclination (5.1 degrees) is created based on geometrical
calculations and because of that this value has its geometrical power (as all
the moon other data)
o The planet is a geometrical structure as one building, based on this idea, the
planet data is created based on each other geometrically…. That means, no
data is found without geometrical reason otherwise the building will be
useless – imagine one building is built and has no a door or stair how to use
it- the building is built based on a geometrical concept and similar to that the
plant data is created based on a geometrical concept.
o The moon orbital inclination (5.1 degrees) is created with some geometrical
interaction to cause the moon orbit regression 1.44 degrees per month
o The moon orbit regression is created by the geometrical mechanism based
on which the moon orbital inclination is created, that means the moon
orbital inclination creation process contains both features the inclination
degrees 5.1 degrees and the regression effect 1.44 degrees per month….
o The angle 137 degrees which we will discuss in the moon orbital
geometrical design shows this fact (137 =95.1 degrees x 1.44 degrees),
telling that, from one process the 2 features are created.

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The using of the moon orbital triangle
- let's use the moon orbital triangle…
- The triangle CDb is our triangle, because its perimeter = 232000 km
- The distance CD =86000 km
- The distance Db = 47667 km (please remember DB =42800 km)
- The distance DX = 2598693 km (the moon displacements total)
- The distance bX = 2550973 km (the moon orbital circumference at apogee)
- The angle CXD =1.89 degrees
- The angle DCb =29 degrees So
- The Perimeter of triangle CDb =232000 km
- The Distance DX = 2598693 km (the moon displacements total)
- Tan (5.1) x 2598693 km = 232000 km
- By that the moon orbital inclination is created by the proportionality between the
perimeter of triangle CDb and the distance DX.

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I-Data
Mars Velocity (24.1 km/sec) = Pluto Velocity (4.7 km/sec) x 5.1
II-Discussion
- The data tells that, Mars and Pluto motions effect on the moon orbital motion and
causes to create the moon orbital inclination =5.1 degrees
- That means, the geometrical process which we have studied in the previous point
was the geometrical mechanism by which the moon perform the effect of these 2
planets on the moon orbital motion.
- That explains many data has no explanation before but now we may explain it, for
example
o Pluto moves during a solar day a distance =406000 km = The Earth moon
distance at apogee radius. That tells us Pluto effect on the previous process
is found in the moon orbital circumference creation (at apogee orbit whose
radius =0.406 mkm and circumference =2.55 mkm).
o Mars moves during a solar day a distance =2.082 mkm = 0.8 x 2.609 mkm
(This distance is the length of the hypotenuse CX in the moon orbital
triangle (in the previous page) and = the hypotenuse ac in the discussion
triangle for point (2-1). The data tells that, Mars motion depends on the
moon displacements total –
o These are 2 forces, Pluto works for the moon orbital circumference (2.55
mkm) and Mars works for the moon displacements total and the balance
between these 2 forces create the moon orbital inclination 5.1 degrees.
So
o The moon orbital inclination is created by Pluto and Mars Motions effect
on the moon orbital motion
o If the moon orbit regression is done by the same process by which the moon
orbital inclination is created, one of these 2 planets must be a player.

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(1st
Point)
- Let's return to this triangle again
- The hypotenuse ac =2608975 km
- The moon apogee orbital circumference = 2550973 km
- The difference =58000
- Sin (1.3) x 2550973 km =58000 km
- The data leads us to the angle 1.3 degrees! Why? because
- 8 deg = 1.3 deg (Jupiter orbital inclination) + 6.7 deg (the moon axial tilt)
- The data tells that,
- The moon axial tilt is created with Jupiter orbital inclination in the same process
based on the valuable value (8 degrees)
- 8 degrees expresses Uranus orbital inclination 0.8 degrees, because Uranus uses
this value in different forms as 0.08 or 8
(2nd
Point)
- The previous discussion still has benefits for our analysis….
- The value 47667 km tells us the following
o Tan (1.44 degrees) x 47667 = 1195 km (Pluto Radius)
o Tan (1.44 degrees) x 69118 = 1737.5 km (The Earth Moon Radius)
o But what's the value 69118 km?

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Equation No. (2)
Cos (13.244 degrees) x 2598693 km = 2529522 km
- The difference 2598693 km – 2529522 = 69118 km
- The moon daily motion =13.177 degrees (error 0.5%)
The data tells that,
- Pluto radius is created by the distance 47667 km based on the angle 1.44 deg.
And
- The moon radius is created by the distance 69118 km based on the angle 1.44 deg.
- The moon and Pluto motions interaction should be discussed in Uranus motion
effect on the moon orbital motion – what we need here from this data is the angle
1.44 degrees –
- The moon orbital circumference 2.55 mkm is created based on the moon total
displacements during a month (29.53 solar days) and because of that the angle 1.44
degrees is used because the moon orbit regresses 1.44 degrees Per Month
- The total = 47667 km + 69118 km = 116785 km = 2 x 58000 km (error 0.6%)
where the value 58000 is used with Jupiter orbital inclination creation.
Equation No. (3)
(π)1/2
x 1.44 degrees x 2 = 5.1 degrees
- The moon orbital inclination is a geometrical value, as we have discussed before,
and the value 1.44 degrees is created based on a geometrical interaction as the
equation shows.
- The data tells that, the moon orbit regression (1.44 degrees) is created as a feature
with the moon orbital inclination (5.1 degrees), both are created in the same
process.

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Discussion
- There are many simple and direct questions are needed to be solved – for example
- Why does the moon orbital motion depend on geometrical rules?
- Because many planets effect on the moon orbital motion and the geometrical rules
create balancing points between these planets motions effects, as the paper
supposes that, 2nd
force effect on the moon orbital motion and Uranus motion
effect on the moon orbital motion and causes to Create Metonic Cycle… the idea
is that …. If many planets gravities effect on the moon orbital motion the
geometrical rules will be a necessary tool to create a balancing for the moon orbital
motion….the next question is
- How can many planets motions effect on the moon orbital motion?
- This question we should answer later but the data shows this fact clearly regardless
our explanation, let's see the following data
(1)
- (Jupiter Mass / Earth Mass) x 142984 km = 149.6 mkm (Earth orbital distance)
(where 142984 km = Jupiter diameter)
- The data tells that, Earth orbital distance definition is effected by Jupiter mass rate
to Earth mass, that shows an effect of Jupiter on Earth motion
(2)
- The moon apogee orbital circumference 2550973 km = 21.86 x 116785 km
- Where 21.86 = (Jupiter mass / Uranus mass)
- And 116785 km = 47667 km + 69118 km
- We have used this distance 116785 km in the previous page explanation to show
how the moon motion data depends on the angle (1.44 degree = the moon orbit
regression per month)
- This data shows the force behind causes the interaction from which the data is
created, this force is the masses rate between Jupiter and Uranus which are the
most 2 planets effective on the moon orbital motion data.

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I-Data
(No. 1)
6.8 km /sec (Uranus Velocity) = 4.7 km /sec (Pluto Velocity) x 1.44
II-Discussion
- The moon orbit regresses 1.44 degrees per month
- The data tells that
- Mars and Pluto motions interaction cause the moon orbital inclination to be 5.1
degrees
- And
- Uranus Pluto motions Interaction cause the moon orbit to regress 1.44 degrees
per month.
- That makes Pluto motion as a central point of both planets (Mars and Uranus)
motions effect, and Pluto transports this effect to the moon orbital motion
- Let's discuss more data in following

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The Moon Orbit regression 19 degrees per year
- The idea is a simple one
- Uranus Orbital Distance =19 Earth Orbital Distance
- So, if Uranus motion effects on the Earth moon orbital motion, the number 19
years should be seen in this effect, that lead us to conclude, (Uranus motion effect
on the moon orbital motion causes to create the moon Metonic Cycle)
- The next question should be,
- Is this 19 created by Uranus /Earth orbits rate or it's an independent number?
- Because
- 97 degrees = 5.1 degrees x 19
- 97 deg (= 97.8 deg Uranus Axial Tilt – 0.8 deg Uranus orbital inclination)
- 5.1 deg (= the moon orbital inclination)
- If Metonic Cycle is created by Uranus effect so the number 19 is the orbits rate but
why we see it between the axial tilt and orbital inclination?!
- The data tells that,
- Uranus orbital distance must be created depending on its axial tilt, and Earth
orbital distance is created depending on its axial tilt and Earth axial tilt is rated to
the moon orbital inclination!
- The data my show that in following
More Data
(j)
(149.6/23.45) = (29.8/4.7) = (153.3/24) = (5906/928)
149.6 mkm = Earth Orbital Distance 23.4 deg = Earth Axial Tilt
(29.8 /4.7) = Earth Velocity / Pluto Velocity
(153.3/24) = Pluto day period / Earth day period
5906 mkm = Pluto orbital distance
928 mkm = Earth Jupiter distance when they be on 2 different sides from the sun

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(k)
119.2 deg = 5.1 deg x 23.4 deg
- Earth moves during 29.53 solar day (29.2 degrees) (29.53 x 0.9862 deg)
- The moon moves during 29.53 solar day (360 deg + 29.2 deg) (29.53 x13.17 deg)
- 29.2 + 90 = 119.2 degrees
- So
- The angle which leads both motions is created depending on Earth axial tilt and the
moon orbital inclination.

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- Let's remember an old question,
- If the Earth is a fixed point in Space and the moon orbit regresses 19 degrees per
year, can the distances between the moon and Earth be without change although
the moon orbit regression?
- The answer is Not
- If the Earth is a fixed point in space the distances between the Earth and the moon
must be changed by the moon orbit regression
- That means,
- Earth does some motion as a result for the moon orbit regression to save the
distances between the Earth and moon without changes
- The next question is,
- Does Earth do any motion as a result of the moon orbit regression?
- The answer is yes
- It's the Earth Cycle (365 +365 + 365 +366 days) which is 1461 days
- The Cycle which = 2 x 1461 days is created as a result for the moon orbit
regression (this cycle is discussed in point no. 4 the triangle design)
- The following data supports this conclusion also
Equation No. (4)
1.461 x 0.98562 degrees = 1.44 degrees
- 0.98562 degrees = The Earth motion degrees daily
- 1.44 degrees = The moon orbit regression per month
- 1.461 = 1461 days /1000
- 95.1 degrees = 23.41 deg x 4 +1.44 degrees
o If Earth motion for 1 year depends on its axial tilt 23.45 degrees, so this
equation shows 4 years Cycle. (similar to that, the moon displacements per
year depends on its orbital inclination 5.1 deg)

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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)
8-1 Preface
- The moon orbital triangle geometrical analysis provides a new and effective idea
let's try to summarize it in following
o The moon orbital triangle shows that many forces effect on the moon orbital
motion because of that many geometrical rules are used in this motion to
define each force balancing points
o I refer to Earth gravity force effect on the moon motion as 1st
force
o I refer to all other planets effects on the moon motion as 2nd
force
o The sun gravity force is considered to be including into both forces
- The triangle shows that, many forces (or motions) interaction effects on the moon
motion and by that the moon orbit geometrical design became a specific one,
showing these forces effects.
- The triangle analysis depends on the Logical Geometrical Analysis, for that, the
absent data can be concluded and (more important) the forces created this data can
be discovered
- Based on that, Jupiter and Uranus (in addition to other planets) have effects on the
moon orbital motion. this conclusion can be formed by the moon orbital triangle
data analysis.
- This analysis supports the paper claims are: (1st
) (There's 2nd
force effects on
The Earth Moon Orbital Motion (2nd
) (Uranus Motion effects on the Earth
moon orbital motion and creates Metonic Cycle)

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8-2 The Moon orbital triangle shows (2nd
force effect on the moon motion)
(The Triangle Data Analysis In Discussed In Point No.2 Of This Current Paper)
- What Proves Can Be Provided For The 2nd
Force Hypothesis?
o (1st
Proof) The Point (A) In The Moon Orbital Triangle
o (2nd
Proof) The 2nd
Displacement 88000 km
o (3rd
Proof) Metonic Cycle Creation…. Let's discuss them in following:
(1st
Proof)
- The moon orbital triangle causes to raise the question, because the Point (A) is one
of its 3 basic points and no force we know can create this Point (A) which is found
far from apogee radius (r=0.406 mkm) with a distance =43000 km, because of that
the distance EA =449197 km
- So how this point is found and effect on the moon orbital triangle? We have no
answer except that 2nd
force is found effects on the moon orbital motion, this 2nd
force effects on the Point (A). So Earth gravity force effects on the moon motion
on one side and this 2nd
force effects on the moon motion on other side to create
general balancing of the moon motion.
- Although no clear definition for the force creates the point (A), this force is still
fact because of the geometrical massive significance of the point (A).
- means, the point (A) should be considered as a proof for this force existence

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(2nd
Proof) The 2nd
Displacement 88000 km
- The moon orbital motion story tells us, the moon contracted distance (2.399 mkm)
needs (0.17 mkm) to be = Earth motion distance (2.573 mkm) per solar day, and
the moon has to move this additional distance (0.17 mkm) on the same solar day,
But the moon daily displacement =88000 km, means, the moon has to move one
more displacement (88000 km) which we don't see…
- If this story is real, and the distance 0.17 mkm should be passed, and if 1 force
only effects on the moon, so this force should cause the moon to move 0.17 mkm
completely…but the moon displacement is only (a half) of the required distance…
that tells us there are 2 forces causes 2 equal displacements (regardless our
observation for them).
- The argument here depends on the moon basic motion (2.573 mkm) which creates
the moon daily displacement (88000 km), if the connection between these 2
distances is a real one, so the 2nd
displacement must be a fact and that necessitates
to find 2nd
force effects on the moon orbital motion.
(3rd
Proof) Metonic Cycle Creation.
- Uranus Orbital Circumference =19 Earth Orbital Circumference …… means
- While Uranus revolves around the sun one revolution, Earth (and its moon)
revolve around the sun 19 revolutions (19 years =6939.75 solar days)
- If Uranus motion effect on the Earth moon motion, the period 19 years should be
seen in this effect data because it’s the basic rate between the 2 orbits
- The moon Metonic Cycle (6939.75 solar days=19 years) tells that, there's a
possibility of Uranus motion effect on the moon motion..
- The point is, if Uranus really effects on the moon orbital motion to create Metonic
Cycle, so this will be a solution for the question (What's this 2nd
force effects on
the moon orbital motion), or at least will give us a light to see other players effect
on the moon orbital motion in place of the one planet gravity effect vision.

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8-3 The Moon orbital triangle shows (There's 2nd
Orbit for the moon motion)
I- Data
(1)
The moon orbital triangle (ECA) Perimeter = 943817 km
The Lunar Eclipse Umbra Length = 1.392 mkm
The distance (EA) = 449197 km + (The Perimeter) 943817 km = 1.392 mkm
II- Discussion
- The Point (A) divides (the lunar eclipse umbra length) into 2 equal parts, after the
Point (A) this part is seen in the triangle perimeter (ECA) and
- Before the Point (A) this part is seen in the distance from the Point (A) to the end
of The Lunar Eclipse Umbra Length
- Can This Be A Proof?
- The geometrical division is a proof, because the moon orbit data is created based
on geometrical interactions for that reason the moon orbital triangle shows these
geometrical interactions and rules, and these geometrical rules tell, many players
are interacted here –for that reason, the triangle (ECA) perimeter has a relationship
with The Lunar Eclipse Umbra Length (Where the geometrical necessity of this
relationship still need to be caught, but the mere existence of this relationship is a
proof for different player effect on the moon orbit geometrical creation).
- I want to say, the moon orbit is NOT a trajectory of a rigid body revolves around
Earth, instead, it's a network of forces lines and the moon moves through this
networks taking into consideration these forces lines effects AND shows that in its
motion data.

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- Let's review the triangle concept in following:
o The moon orbital triangle is a vertical triangle effects on the moon orbit,
where the line (BC) is perpendicular on the moon orbital triangle base (EA)
and because of that the point (C) is found on (z-axis), where the moon
orbital motion is done on (x-y plain)
o How That Can Be Possible?
o I supposed Uranus Axial Tilt (97.8 degrees) is the line (BC), the moon axial
tilt =6.7 degrees and the difference =91.1 degrees, for that reason the moon
orbital triangle declines on the moon equator line with 1.1 degrees and the
line (BC) is perpendicular (90 deg) on the moon orbital triangle base (EA).
o I have designed this triangle basically based on this data and the triangle is
used sufficiently for the moon real motion and data.
o Uranus indeed effects on the moon orbital motion in different features, not
only in Metonic Cycle, but also by Uranus axial tilt effect on the moon axial
tilt, not that only…
o Earth moves during its day period a distance = The moon displacements
total during its day period = Pluto motion during its day period, (error 1%),
This feature also is found by Uranus effect on the moon orbital motion (this
feature is discussed is discussed in the point No. 3 of this paper (The Paper
Hypotheses Proves Discussion).
o The moon day period (29.53 solar days) is a piece of gold because this
period of time shows that it's created by 2 motions effect on the moon
orbital motion – shortly – Earth and Uranus motions effect on the moon
orbital motion, forcing the moon day period to be 29.53 solar days.
o This discussion should be completed with Metonic Cycle Discussion (Point
No. 8 of this paper).

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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
9-1 Preface
The claim
- The Earth Moon Metonic Cycle (6939.75 Solar Days) is created by Uranus
Motion Effect On The Moon Orbital Motion.
The Proves
o (1st
Proof) Uranus Orbital Circumference =19 Earth Orbital Circumference,
So while Uranus revolves around the sun 1 complete revolution the Earth
(with its moon) revolve around the sun 19 revolutions..
o If Uranus Motion effects on the moon orbital motion, the number 19 should
be seen in this effect data (This proof is discussed in 8-2)
o (2nd
Proof) Earth Motion Distance During Its Day Period = The Moon Total
Displacements During 29.53 solar days (The Moon Day Period) = Pluto
Motion Distance During 153.3 hours (Pluto Day Period) – this feature of
motion is created by Uranus motion effect on the 3 planets.
(This proof is discussed in 3-5 Why the moon daily displacement =88000 km?)
o (3rd
Proof) Uranus Moves During (1440 Of Its Days Period) A Distance =
The Earth Moon Total Displacements During Metonic Cycle (6939.75 days)
(This proof is discussed in Point no. 9 (Uranus Motion Analysis)
o (4th
Proof) The Moon Orbital Triangle Data Shows Uranus Effect On The
Moon Motion. (This proof is discussed in 8-4)

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9-2 Uranus Effect On The Moon Orbital Motion (1st
Proof)
(1st
Proof Discussion)
In this figure
- The Red Ball Shows Earth
- The Yellow Ball shows The Earth Moon
- The Blue Ball shows Uranus
- (S) is the Sun
- The figure suggests that, a triangle contains these 3 planets together in their
revolutions around the sun
- Let's suppose the three planets, Earth, its moon and Uranus move in parallel to
each other in their revolutions around the sun, and to guarantee this parallelism
between them the figure provides a triangle contains these 3 planets -
- Uranus orbital circumference = Earth orbital circumference x 19
In accurate calculations
- Uranus (18048 mkm) = Earth (940 mkm) x 19 (error 1%)
- This data means, while Earth revolves around the sun 19 times, Uranus revolves
around the sun 1 time only

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- If the 3 planets move in parallel to each other, that means, Uranus will divide its
revolution trajectory around the sun into 19 parts, and each part will be a sufficient
for one Earth orbital circumference (difference 1%)
- Uranus motion trajectory effect is observed on the Earth moon motion trajectory,
let's show how that happens:
- The moon moves through its orbital circumference revolving around the Earth
(while the masses gravity forces imprison the moon inside the range from perigee
(0.363 mkm) to apogee (0.406 mkm) and prevents the moon to move out of this
motion range).
- But
- Uranus motion effects on the Earth moon motion (inside its prison) and forces the
moon to change its motion trajectory through 19 years. Because of that the moon
doesn't move through the same point 2 times during 19 years (6939.75 solar days),
that creates Metonic Cycle, that happens because the moon motion reflects Uranus
Motion Effect revolving around the sun, where Uranus moves on a trajectory
doesn't pass through the same point 2 times during (19 years) (according to the
moon time) similar to that the moon moves through its orbital circumference
doesn't pass through the same point 2 times during 19 sidereal years.
- Shortly
- Metonic Cycle Is Created By Uranus Motion Effect On The Moon Orbital Motion.

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The angle 71.9 degrees is an angle created by the interaction between the 4 planets
Motions (Earth, its moon, Pluto and Uranus).
Notice
The angle 71.9 degrees is discussed (Partially) with this triangle data analysis
(Point 5-2 The Triangle Geometrical Design)
In following the full discussion is inserted with repeat part is discussed in that point.
Please remember
(Why we need to discuss this angle 71.9 degrees?)
Because this angle can answerer why the moon orbital motion equation uses the
constant 1.7 degrees for the moon daily motion (θ1= θ0 +1.7 degrees).

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I- Data (Group No.1)
(1)
The angle M1 L M2 =88.9 degrees
88.9 degrees – 71.9 degrees = 17 degrees
(2)
(17 degrees /0.8) = 21.25 degrees
(3)
21.25 degrees x 0.08 = 1.7 degrees (the moon motion equation constant)
(4)
(5)
23.4 deg = 1.8 deg x 13.177 deg x 0.98562 deg
Discussion
Equation No. (3)
- Equations (from 1 to 3) give us a simple geometrical method to change the value
17 degrees into 1.7 degrees, but why this method is useful?
- Because the value 21.25 degrees is one of the moon motion angles which is
- 21.25 degrees = 11.8 degrees x 1.8 degrees
- Where
- 11.8 degrees = 5.1 deg (the moon orbital inclination) +6.7 deg (the moon axial tilt)
- But what's 1.8 degrees?! Let's discover in following…
o The moon moves from perigee to apogee and return back during its orbital
period.
o The distance from perigee to apogee on the moon orbital triangle (BD)
controlled by the angle (BCD =26.577 degrees)
o The moon go and return during the cycle (26.577 degrees x 2 = 53.15 deg)
o (53.15 degrees /29.53 solar days) =1.8 degrees
o Why I divide this angle 53.15 degrees on 29.53 days?

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o Because
o The moon starts a new cycle from perigee to apogee after completes its day
period. Means the angle (53.15 degrees) should be distributed during the
synodic month (29.53 solar days).
- The previous explanation shows that, the angle 21.25 degrees is used in the moon
orbital motion because it depends on 2 angles (11.8 deg) and (1.8 deg) are used in
the moon day motion. based on that, the interaction between the angle 17 degrees
and 21.25 degrees can be created because both angles are used in the same motion
- Then
- The last step is to change the angle 21.25 degrees into 1.7 degrees as following
- 21.25 degrees x 0.08 = 1.7 degrees
- We remember this rate (0.08) based on which the valuable angle (10.96 deg) is
created. (please remember 137 degrees x 0.08= 10.96 degrees
Notice
- The most 3 basic values in the moon motion are (137 deg, 10.96 deg and 0.08)
- As the valuable angle (10.96 deg) is created based on this rate (0.08), the moon
orbital motion equation angle (1.7 deg) is created based on it….BUT
- Why the data shows that, Uranus orbital inclination (0.8 degrees) is used in this
process? The data uses (17 degrees /0.8 degrees) = 21.25 degrees, showing clearly
the using of Uranus orbital inclination (0.8 degrees) Why? because the data tries to
show Uranus effect on the moon orbital motion…. the next points supports it.
Equation No. (4)
- We know both angles 17 and 1.7 degrees but what's this 29 degrees?!
- The major lunar standstill can be +28.5 = (23.4 deg + 5.1 deg)
- The moon angular diameter = 0.5 degrees, that means, when the moon orbital
inclination is measured above the moon diameter it will be =5.6 degrees

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- So the angle 28.55 degrees +0.5 degrees = 29.05 degrees
- That shows Uranus effect on the moon motion during Metonic Cycle, which effect
on the moon daily orbital motion and effect on the moon motion equation by the
constant (1.7 degrees)
Equation No. (5)
23.4 deg = 1.8 deg x 13.177 deg x 0.98562 deg
Where
23.45 deg = Earth Axial Tilt 0.98562 deg = Earth motion daily degrees
13.177 deg = the moon daily motion degrees
1.8 degrees = is the angle we have discussed in the previous equation (no.3), the
angle of the moon motion from perigee to apogee during its day period
(53.15 degrees /29.53 solar days).
Equation no. (5) shows that Earth axial tilt is created depending on the moon motion.

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II- Data (Group No.2)
(A)
2 x 71.9 degrees = 12.195 degrees x 11.8 degrees
Where
12.195 degrees = The moon motion angle (13.177 deg – 0.98562 deg)
11.8 degrees =6.7 degrees (moon axial tilt) + 5.1 deg (moon orbital inclination)
Why does the data use double values (2 x 71.9 deg)??
(B)
122.5 = 71.9 degrees x 1.7
Where
122.5 degrees = Pluto Axial Tilt
1.7 degrees = The moon motion equation constant ((θ1= θ0 + 1.7 degrees)
- Why does the equation use 1.7 degrees for moon motion daily? (this question is
asked in the moon motion equation discussion), the data tells that the angle 71.9
degrees (the interaction angle) has an effect to do that
- So, the constant (1.7 deg) depends on the interaction angle (71.9 deg) and Pluto
Axial Tilt (122.5 deg)…
BUT
- (122.5 deg -71.9 deg) x 2 = 101.2 degrees
- In the distances data Earth motion distance daily (2573483 km) is considered as
(101%) (in the moon daily displacement dissuasion), If there's a relationship
between this 101% and the value 101 deg, we may conclude, this value also refers
to the using of (2 x 71.9 degrees)! Why?
ALSO
- 71.9 degrees / 101.2 = 0.712 we remember θ1= θ0 + 1.7 degrees where 1.7 deg =
0.98562 deg +0.712 deg, it's another proof that, the constant (1.7 deg) is produced
by the planets interaction (specifically between Pluto and the moon motion).

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(C)
- 14 degrees x 5.1 degrees (the moon orbital inclination) =71.4 degrees
- 71.9 degrees = 71.4 degrees + 0.5 degree (the moon angular diameter)
- And
- 14 degrees = (5.1 degrees (the moon orbital inclination) + 8.9 degrees)
Let's remember 8.9 degrees
o 95.6 deg + 1.1 deg = 96.7 deg
o 96.7 deg + 1.1 deg = 97.8 deg
o 97.8 deg + 1.1 deg = 98.9 deg
Where
o 95.6 deg = 90 deg + 0.5 degrees + 5.1 deg (The Moon orbital inclination)
o 96.7 deg = 90 deg + 6.7 deg (The Moon Axial Tilt)
o 96.7 deg = 90 degrees + 8.9 degrees
o 1.1 deg = the angle of the moon triangle base (EA) moon equator line.
(D)
- 63.7 degrees = (71.9 deg – 8.9 deg) + (71.9 deg – 8 x 8.9 deg)
- Where
- 63.7 deg = The Sun Declination
- Equation no. (D) tells a very important information, which are:
o (1) The interaction angle (71.9 deg) is used in double Value (2 x 71.9),
because of a geometrical necessity.
o (2) The (8 days) Cycle, we have discovered in Jupiter Uranus motions, is
used here to define the interaction angle based on which the most of the
moon data is created – i.e. the cycle (8 days) effects on the moon motion
- The cycle (8 days) is discussed with many details in (Uranus Motion Analysis).

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(4th
Proof Discussion)
We analyze here 2 angles (32.967 degrees 36.912 degrees)
Point No. (A) (The angle 32.967 degrees)
- This is the moon orbital triangle, I have added the triangle CUB
- BCU = 32.967 degrees
- CU = 102500 km
- BU = 55756 km
Let's analyze this data in following
- BU = 55756 km = 43000 km +12756 km (Earth Diameter)
- CU= 102500 km = 2 x 51118 km (Uranus Diameter)
- The angle BCU = 32.967 degrees where 32.967 deg x 0.8 = 26.36 degrees
o 0.8 degrees = Uranus Orbital Inclination
o 26.36 degrees = the angle controls the moon motion from perigee to
apogee as we have seen in the moon orbital triangle original form (BCD),
but the angle (BCD) = 26.56 degrees (Error 1%)

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- The Idea Summary:
o Uranus motion effects on the moon motion revolving around Earth, this is
one of the paper claims– Earth force imprison the moon inside the range
(Perigee and Apogee distance =43000 km) – so – any effect of Uranus
motion on the moon motion will be acceptable if it be in the range (Perigee
And Apogee Distance).
o Uranus Motion creates The Red Perpendicular Line BC in The Moon
Orbital Triangle (BC =86000 km)
o Let's suppose, the line BC receives Uranus Motion effect and provides it to
the moon motion.
Let's imagine how that's doing
o The Line BC moves in angle (32.967 degrees), that means, the line BC
changes the angle (BCU) from Zero degree to (32.967 degrees) and then
return to Zero Again
o It's a cycle, but the line BC moves in its opening for the angle from Zero to
(32.967 degrees) in some way and doesn't return through this same way
when the opened angle (32.967 degrees) be closing to be Zero
o That creates a motion of cycle of this line BC (column).
o This motion is A Waving Motion (going and return back but NOT through
the same way).
o By this motion the line BC effects on the moon motion revolving around
the Earth… now the line BC should be considered as a column built on the
moon body or is connected by it – and that means- if this line BC moves (by
angle opening or closing) the moon will move with it or effected by it.
o The angle is (BCU =32.967 degrees), but the moon doesn't reach to this
angle range for 2 reasons (1st
) Because the moon can't move beyond apogee
radius (0.406 mkm) (2nd
) Because of Uranus Orbital Inclination effect.
o (32.967 degrees) x 0.8 = 26.36 degrees

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o That means, this angle (32.967 deg) is seen in the moon orbital motion as
(26.36 degrees) because of Uranus orbital inclination effect on this angle.
o The angle = (BCD) =26.6 degrees
o The angle 26.36 degrees controls the moon motion during a distance 42500
km (i.e. From perigee to Before apogee point with 500 km) (error 1%).
o Error 1% is found frequently in Uranus effect on Earth moon motions.
Notice
- Uranus effect is seen strongly in the data for example
o CU = 102500 km = 2 Uranus Diameters
o BU = 43000 km (perigee apogee distance) + 12756 km (Earth Diameter)
o AU = 30589 km (error 0.4%) (where 30589 days = Uranus orbital period)
More Data
- (BCU) = (32.967 degrees) x 3 = 98.9 degrees
- Where
o 98.9 degrees = 97.8 degrees (Uranus Axial Tilt) + 1.1 degrees
o 97.8 degrees (Uranus Axial Tilt) = 96.7 degrees + 1.1 degrees
(96.7 degrees =90 degrees +6.7 deg The Moon Axial Tilt).
o 96.7 degrees (Uranus Axial Tilt) = 95.6 degrees + 1.1 degrees
(5.6 deg = 0.5 deg (the moon angular diameter) + 5.1 deg (the moon
orbital inclination)

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Point No. (B) (The angle 36.912 degrees)
- In This moon orbital triangle I have added the line CG
- The angle BCG = 36.912 degrees
- BG = 64600 km
- CG = 107560 km
- The angle ECG = 113.58 degrees
- Note Please
o Cos (36.912 degrees) = 0.8
o Tan (36.912 degrees) = 0.7511
I-Data Analysis
- (97.8 degrees /122.5 degrees) = Cos (36.912 degrees)
o 122.5 deg= Pluto Axial Tilt
o Uranus Orbital Inclination = 0.8 degrees
o Also Cos (36.912 degrees) = 0.8
The data tells that, the angle (36.912 degrees) is used in Uranus Pluto motions
interaction data

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- (13.177 degrees /17.4 degrees) = Tan (36.912 degrees) (error 0.8%)
o 13.177 degrees = The Moon Motion Per Solar Day
o 17.4 degrees = The Inner Planets Orbital Inclinations Total
o 17.2 degrees = Pluto Orbital Inclination
The data tells that, the angle (36.912 degrees) is used for The Moon Daily Motion
- (17.4 degrees/ 23.45 degrees) = Tan (36.6 degrees) (error 1%)
o 17.4 degrees = The Inner Planets Orbital Inclinations Total
o (36.6 degrees) is different with (36.912 degrees) with 1%
The data tells that, the angle (36.912 degrees) is used for Earth Axial Tilt (Please
remember Earth data has always an error =1% concerning Uranus effect).
- (26.3 degrees/ 32.96 degrees) = Cos (36.912 degrees)
o 26.3 degrees = The angle of the moon motion from perigee to apogee
(26.56 degrees error 1%)
o 32.96 degrees = the angle is discussed in the previous triangle
o Where
The data tells that, the angle (36.912 degrees) is used for The moon motion angle
(26.6 deg) From Perigee To Apogee.
- (29.53 days /36.912 degrees) = Cos (36.912 degrees)
o The angle = BCG (-36.912 deg)
o 29.53 days = the moon day period
The data tells, the angle (36.912 degrees) is used for The Moon Day Period (29.53
solar days).

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- (36.912 degrees/ 46.1 degrees) = Cos (36.912 degrees)
o The angle (A) =45 degrees
o The triangle base (EA) is declined with 1.1 degrees on the horizontal level,
so the total angle will be 45 deg +1.1 deg = 46.1 degrees ….. So
o The angle (36.912 deg.) / the total (46.1 deg.) = Cos (36.912 deg.)
The data tells, the angle (36.912 degrees) is used for The angle (A) in the Moon
Orbital Triangle
Notice
37 x π2
=365.25
- This data shows the massive importance of the angle (36.912 degrees)
- Please Note
- Uranus, Pluto and the moon data is controlled by this angle (36.912 deg)
A Conclusion
o It's the same angle (36.912 deg) is used for Uranus, Pluto, the moon and
Earth motions data showing that this angle (36.912 deg) is created inside the
interaction of these planets motions

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The Triangle (ECG) Analysis
(Please review the triangle form of 1st
Case)
- The angle ECG = 113.6 degrees = 90 deg +23.6 degrees
Where
- 23.6 deg (The Outer Planets Orbital Inclinations Total) x 0.99 =23.45 deg (Earth
Axial Tilt)
- 17.4 deg (The Inner Planets Orbital Inclinations Total) x 0.99 =17.2 deg (Pluto
Orbital Inclination).
- The Right Triangle Hypotenuse (CG) = 107560 km = 51118 km +56382 km
- 51118 km = Uranus Diameter
- 56382 km = the distance BU (55756 km) (error 1%)
Note Please
- The Point G divides the distance BA into BT = 3 and TA =1
- means, BG = 43000 km +21500 km
- and , XG =21500 km
- (21500 km = Mars Circumference)
A Comment
- The angle and triangle analysis shows that, Uranus data is used strongly in the
moon orbital triangle, in addition to many other planets, as the distance 449197 km
= Jupiter circumference, or the distance 21500 km = Mars circumference, BUT
- Uranus data is used dominantly along the moon orbital triangle data specially through the
angle (36.912 deg) which should be origin point from which different data is created and
Uranus axial tilt based on which the moon orbital triangle is created
- The angle ECG =113.6 degrees tells us that, Earth axial tilt (23.45 degrees) is
created based on the moon orbital triangle geometrical structure.

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II-Discussion
- I use this data to explain the solar system suggested description in following:
o The solar system is a theater of puppets, all planets are connected with each
other by the same thread, and each planet data is created in harmony with
the general motion of this thread and that forces this data to be
complementary to each other –
o The double production experiment is a good example to explain this idea,
from Gamma ray, electron and positron are created complementary to each
other and so they are equal in mass and opposite in charges
o This is the meaning of (complementary to each other), without observation I
expect that, Gamma rays will produce positron in addition to the electron –
even if I can't catch this positron by observation, simply because of the
charge conservation law –
o It's the concept of the matter creation – the complementary couple – for that
reason – Pluto circumference =7511 km because Uranus velocity =6.8
km/s, It's a geometrical mechanism connects Pluto with Uranus–this
connection is created based on geometrical rules which control planet
motion and creation data.
o Because of that Uranus (6.8 km/sec) moves during 7511 seconds a distance
= 51118 km = Uranus diameter where 7511 km = Pluto circumference,
which shows a deep interaction behind the data creation
o The basic conclusion is (The solar system is a network of motions)
o That answers the question (How the far planets can effect on the moon
orbital motion spite of the huge distances)?
o These planets effect by their motions and not by their masses gravity… we
have discussed that before.

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10- Uranus Motion Analysis
10-2 Uranus Motion During 8 Pluto Days period
(3rd
Proof)
Uranus Moves During (1440 Of Its Days Period) A Distance = The Earth Moon
Total Displacements During Metonic Cycle (6939.75 Solar Days)
I-Data
- Uranus has a cycle with (144 of its days), Where
- Uranus 144 days = 2476.8 hours
- Pluto 16 days = 2452.8 hours
- The difference = 1 Solar Day
- This cycle we should discuss later in details …. Now we try to know if this cycle
effect on the moon motion….
- Uranus moves during 1440 of its days (1440 x 17.2 h = 24768 hours), during this
period Uranus moves a distance = 606.3 mkm
- The Earth moon moves per a solar day a displacement =88000 km,
- During 6939.75 days (Metonic Cycle), the moon moves a distance = 610.7 mkm
And
- Uranus diameter 51118 km x (1092
) = 607.3 mkm
II- Discussion
- The values (606.3 mkm and 610.7 mkm) are different with around (1%)
- The data tells that, the distance Uranus moves during its cycle (1440 Uranus days)
= the moon displacements total during Metonic Cycle, which shows that both
values are related to each other.

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- For more confirmation, Uranus gives us the value (607.3 mkm) as a result of the
equation (51118 km x 1092
), where we remember this equation because
o Mercury orbital distance (57.9 mkm) = Mercury diameter x 1092
o Earth orbital distance (149.6 mkm) = Earth diameter x 1092
o Saturn orbital distance (1433.5 mkm) = Saturn diameter x 1092
- By this same equation Uranus produces the result 607 mkm = the moon total
displacement during Metonic Cycle = Uranus motion distance during 1440 its days
- The data shows, the moon motion is effected by Uranus Motion, Supporting the
hypothesis, (Metonic Cycle is created by Uranus Effect on the moon motion)
Notice
- During (1440 days of Uranus days period) Uranus moves a distance = 606.3 mkm
- 1440 days x 17.2 hours = 24768 hours = 1032 solar days
- The Moon total displacement during (6939.75 solar days) = 610.7 mkm
- i.e.
- Equal distances (error 1%) are passed in 2 different periods of time
o (6939.75 solar days / 1032 solar days) =6.724
o But
o 6.7 degrees = The Moon Axial Tilt (error 0.3%)
o We may remember that, a deep relationship is found between Uranus axial
tilt and the moon axial tilt (97.8 degrees – 6.7 degree = 91.1 degree).
Shortly
Uranus motion effects on the Earth moon motion and forces the moon to move
Metonic Cycle during 19 years

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10-2 Uranus Motion During 8 Pluto Days Period
- Why do we need to remember Uranus 144 days Cycle? We have 2 reasons
o (1st
) Because Uranus moves during 1440 Uranus days (17.2 h) a distance =
The Moon Total Displacements During Metonic Cycle
o (2nd
) Because Uranus Pluto Motions interaction can be seen clearly in
studying Uranus 144 days Cycle….
- How Uranus Motion Cycle (During 8 Pluto Days) Is Discovered?
o Jupiter (13.1 km/s) moves during its day period (9.9 h) a distance = Jupiter
circumference (449197 km) + 17695 km
o During 8 of Jupiter days periods (9.9 x 8 =79.2 h), Jupiter moves a distance
= 8 Jupiter circumferences + Jupiter diameter (error 1%)
o Because of Jupiter diameter I concluded that, Jupiter has a cycle in 8 days
Then
o Jupiter motion distance during 8 of its days (79.2 h) which = 3735072 km,
this same distance = Saturn motion distance during 10 of Saturn days period
o Means, Saturn moves during 10 of its days a distance = 3735072 km,
o I have concluded that, Jupiter motion energy is transported to Saturn motion
energy by the rate 80% Because of that I expected that, Saturn should
transport its motion energy to Uranus by the same 80% (but it's incorrect!)
o Saturn transported the motion energy with this rate 80% to Neptune and that
means the distance Saturn passes during 8 of Saturn days period, Neptune
passes during 10 days of Neptune days period (error 5%)
The question is why Saturn doesn't transport the motion energy to Uranus?!
- A surprise was in our waiting…
- 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
- That means, 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%)

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- It's very similar to Jupiter 8 days Cycle ….
- Easily we have concluded that, Uranus motion creates a cycle of Pluto 8 days
where Uranus moves during this period a distance = Jupiter motion distance during
64 Jupiter days
- Because Jupiter motion energy is transported to Saturn and from Saturn to Neptune
the distance should be equal for all these planets – by that – a chance is available
to compare between these motions distances - let's write them in following
I- Data
1. Uranus (6.8 km/s) moves during 8 days of Pluto days period (1226.4 hours) a
distance = (4415040 seconds x 6.8 km/s) = 30.022272 million km
2. Jupiter (13.1 km/s) moves during 64 days of Jupiter days period (64 x9.9 h =
633.6 hours) a distance = (2280960 seconds x 13.1 km/s) = 29.880576 million km
3. Saturn (9.7 km/s) moves during 80 days of Saturn days period (80 x10.7 h = 856
hours) a distance = (3081600 seconds x 9.7 km/s) = 29.891520 million km
4. Neptune (5.4 km/s) moves during 100 days of Neptune days period (100 x16.1 h =
1610 hours) a distance = (5796000 seconds x 5.4 km/s) = 31.298400 million km
5. Jupiter Circumference (449197 km) x 64 = 28.748639 million km
6. Saturn Circumference (378675 km) x 80 = 30.294001 million km
7. Neptune Circumference x 2 (311193.6 km) x 100 = 31.119360 million km
Data Analysis
- The distances are very near which passed by
o (1) Uranus during 8 Pluto days
o (2) Jupiter during 64 of its days
o (3) Saturn during 80 of its days
o (4) Neptune during 100 of its days
Where
- The planets circumferences effect on their cycles because of that, the distances are
compared including the planets circumferences.

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II-Discussion
(1st
Point)
- Uranus motion distance during 8 of Pluto days period (30.022272 mkm) – Jupiter
motion distance during 64 of Jupiter days period (29.880576 mkm) = 142984 km
(= Jupiter diameter) (error 1%)
- Because of Jupiter diameter value, I have concluded that, Uranus Jupiter Creates
a cycle by their motions interaction.. Where Uranus uses 8 of Pluto days periods
and Jupiter uses 64 of its days period.
(2nd
Point)
- Jupiter motion distance during 64 of its days period (29.880576 mkm) – the total
of 64 Jupiter Circumferences (28.748639 mkm) = 1.1318 m km
- Jupiter moves per a solar day a distance = 1.1318 m km
- That means, these 2 values express 2 motions interacted together to produce this
value (1.1318 m km) which is considered as a value defined based on cycle
period (a solar day).
(3rd
Point)
- Saturn motion distance during 80 of its days (=29.891520 mkm) – Jupiter motion
durance during 64 of its days period (= 29.880576 mkm) = 10921 km
o 10921 km = The Earth Moon Circumference
o The 2 values are considered to be equal, and its almost correct because the
difference between both = 0.036%
o But this very small error = the moon circumference
o Please Note 29891520 km = 10921 x 2737
o But 29880576 km = 10921 x 2736
o Please remember the cycle 2737 years

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(4th
Point)
- Neptune motion distance during 100 of Neptune days periods (=31298400 km) –
Saturn motion distance during 80 of Saturn days periods (=29891520 km) =
1406880 km ………. But
o 1406880 km = The Sun Diameter 139200 km (Error 1%)
o The Sun Diameter 139200 km = 49528 km Neptune diameter x 28.1
o Neptune Axial Tilt =28.3 degrees
(5th
Point)
- Neptune motion distance during 100 of Neptune days periods (=31.298400 mkm)
– Uranus motion distance during 8 of Pluto days periods (=30.022272 mkm) =
1.276128 mkm = π x 406000 km
o Earth Moon Distance at apogee radius = 406000 km, where this is the most
far point the moon can reach from Earth.

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- 144 days of Uranus days periods = 144 x 17.2 hours = 2476.8 hours
- 16 days of Pluto days period = 16 x 153.3 hours = 2452.8 hours
- The Difference = 24 hours = 1 Solar Day
- The data shows these are 3 values (144 Uranus days, 16 Pluto days and the solar
day) they are 3 cycles
- We have seen that before
o 6939.75 solar days = Metonic Cycle
o 6585.36 solar days = Saros Cycle
o 354.39 solar days = The lunar year
And
- The data shows that, there's an interaction of Uranus, Pluto and Earth motions
- This data we have discussed before and reach to the following conclusions
o Uranus motion effect on Pluto motion causes Pluto day period to be =153.3
hours where it's so long day period in comparison with the other outer
planets
o This effect is found because Pluto moves during (6939.75 x 153.3 hours) a
distance = Uranus orbital circumference
o This data also we have discussed before

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Uranus Pluto Motions Interaction
I- Data
(a)
4.7 km / sec x 10921 seconds = 51118 km (Uranus diameter) (error 0.4%)
(b)
4.7 km / sec x 51118 seconds = 2 x 120536 (Saturn Diameter) (error 0.4%)
(c)
4.7 km / sec x 2 x 120536 (Saturn Diameter) =1.1318 mkm (Jupiter daily velocity)
II- Discussion
- The data shows that, the results are created as different values in the cycles of 8
Pluto days cycle and the other planets equal distances
- That tells, these cycles are created by an effect of Pluto motion during different
periods of time
- That shows an effect of Pluto motion more extending than our expectation for its
effect on the solar system motion…
More Data shows Uranus and Pluto Motions Interactions
Equation No. (d)
90560 days = 13.177 x 0.99 x 6939.75 days
- 90560 solar days = Pluto Orbital Period
- Equation (d) shows, Pluto orbital period depends on Metonic Cycle (6939.75 days)
- and on the value 13.177, where the moon moves per solar day 13.177 degrees
Equation No. (e)
Pluto during 6939.75 days moves a distance = 2815 mkm
- The data tells that, Pluto uses also Metonic Cycle (6939.75 solar days) and moves
during this period a distance = Mercury Uranus Distance
Equation No. (f)
21.8 x 0.8 degrees (Uranus orbital inclination) = 17.4 degrees
21.8 = Jupiter Mass / Uranus Mass

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I-Data
- In this figure I cut layers from the moon diameter to create smaller moon diameters
- he Blue Circle its diameter R1= 695 km and r1 =347.5 km
- The Red Circle its diameter R2= 1390 km and r2 =695 km
- The Black Circle its diameter R3= 2085 km and r3 =1042.5 km
- The Brown Circle its diameter R4= 2780 km and r4 =1390 km
- The Orange Circle its diameter R5= 3208 km r5 = 1604 km = (5040/π)
- The moon diameter R6= 3475 km r6 = 1737.5 km
Data Analysis
- Why we need to cut the moon diameter into smaller diameters?
- Let's summarize the idea in following:
o Uranus Axial Tilt = 97.8 degrees
o The moon axial tilt = 6.7 degrees
o The difference = 91.1 degrees
o 1.1 degrees = 0.6 degrees +0.5 degrees
o 0.5 degrees = The moon angular diameter, that means, because the moon
diameter =3475 km it consumes 0.5 degree
o The difference 91.1 degrees will be 90.6 degrees because of the moon
diameter
o But
o 0.8 degrees = Uranus orbital inclination and on the vertical axis it will be =
90 degrees +0.8 degrees =90.8 degrees
o So the difference between 90.6 degrees and 90.8 degrees =0.2 degrees

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o That's why we need to cut the moon diameter into smaller diameter which
consumes 0.3 degrees in place 0.5 degrees
o The moon diameter is divided into 5 equal parts we need to cut 2 parts of it,
so the black circle (R3= 2085) where we moved from the moon surface
inside the moon body a distance = 1390 km, on this point the difference will
be =90.8 degrees
The Angle 0.2 Degrees
- The 0.2 degrees which is created by the moon diameter cutting is a very effective
value in the solar system geometry as seen in following:
o 17.2 deg (Pluto orbital inclination) +0.2 deg = 17.4 deg (the inner planets
orbital inclinations total)
o 23.4 deg (Earth Axial Tilt) +0.2 deg = 23.6 deg (the outer planets orbital
inclinations total)
The Angle 0.6 Degrees
- I claim The angle 0.6 degrees is used to create Mars orbital inclination (1.9 deg) by
interaction done by Jupiter and Saturn… the following data shows that clearly
Jupiter And Saturn Interaction
- 1.3 deg (Jupiter orbital inclination) +0.6 deg = 1.9 deg (Mars orbital inclination)
- 1.9 deg (Mars orbital inclination) + 0.6 deg = 2.5 deg (Saturn orbital inclination)
- The data shows that, Mars orbital inclination is created based on this 0.6 degrees,
this process depends on Uranus and the moon axial tilts interaction which
creates a harmony between the moon and Mars motions as seen in following:
o 1.9 deg (Mars orbital inclination) x 13.177 deg = 25.2 deg (Mars Axial Tilt)
(13.177 degrees = The moon daily motion degrees)
o (13.177 deg / 0.524 deg) =25.2 deg (Mars Axial Tilt)
(0.524 degrees = Mars Motion Daily Degrees)
o 687 days (Mars orbital period) =27.3 days (the moon orbital period) x 25.2
o The moon day period (708.7 h) = Mars day period (24.7 h) x (180/2π)

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Uranus Angle 90.8 Degrees Geometrical Effect
Data
- 29 deg + 90 deg = 28.2 deg + 90.8 deg
Where
- 90.8 degrees is our investigation angle (Uranus Angle)
- 28.3 degrees = Neptune Axial Tilt
- 29 degrees, we know this angle…
o 28.5 deg +0.5 deg (the moon angular diameter)
o 28.5 The moon declination during the major stand still (+28.5 and -28.5)
- The data shows that, Neptune Axial Tilt, is created by an effect on Uranus angle
(90.8 degrees) (this angle 90.8 is called Uranus angle because = 90 +0.8 deg)
- The data tells that, some interaction contains Uranus and Neptune together is
found in the moon orbit ….
- This data is supported by another one which is
- 1.44 deg = 0.8 deg x 1.8 deg
o 1.44 deg = the moon orbit regression angle per month
o 0.8 deg = Uranus Orbital Inclination
o 1.8 deg =Neptune Orbital Inclination
Notice
- (1600 = 88000 km – 86400 km)
- And 5040 km = π x 1604 km
- Where
- 5040 seconds are required for Mercury day period to be 176 solar days
- This notice suggests that, the moon diameter is created based on this value 1604
km by a direct effect of Mercury Motion.

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Appendix No.1
Is There Lorentz Length Contraction Effect In The Solar System?
i.e.
(Are There Relativistic Effects In The Solar System?)
Lorentz Length Contraction Effect is the near possible answer to explain the planets
data, in following I provide one example of such planets data to prove that, this
conclusion is the most near one to explain it.
I- Data (A)
Why These Distances Are Equal?
(1)
Saturn Orbital Distance = Saturn Uranus Distance
= Mars Orbital Circumference
= Pluto Neptune Distance
= Pluto eccentricity Distance
= Neptune Orbital Distance/π
= Uranus Orbital Distance /2
= Mercury Jupiter Distance x 2
(2)
Mercury Neptune Distance = Saturn Pluto Distance
Jupiter Pluto Distance = Uranus Neptune Circumference
Earth Neptune Distance = Mercury Saturn Circumference (0.5%)
(3)
Jupiter Mercury Distance = 2 Mercury Orbital Circumference
Jupiter Venus Distance = Venus Orbital Circumference (1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (1.2%)
(Earth and Jupiter at 2 different sides from the sun)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (0.8%)
Pluto Orbital Distance = Earth Orbital Circumference x 2π
II- Discussion (A)
The previous distances form around 50% of all distances found in the solar system
(All orbital and internal distances)… Why These Distances Are Equal One Other?
We may notice that – the distances equality can be produced more easily by light
motion than the rigid body motion - for example – when we push a ball toward a wall
the ball after collision with the wall will return a distance (NOT) equal the original
one - because the collision causes to decrease the ball motion momentum – but the
light can be reflected at equal distances easily – means – equal distances can be
produced by light motion more easy than the Rigid Body Motion.

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I-Data (B)
Why These Distances Are NOT Equal?
1. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
2. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
3. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
4. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
5. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (0.6%)
6. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (0.8%)
7. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (0.7%)
8. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (0.7%)
9. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4894
nce
Circumfere
Orbital
Jupiter
= (1.5 %)
(10)
I-Discussion (B)
The same rate (1.0725) is used for all equations (around 18 distances = 40% of all
solar system distances) – why?
Suppose the equal distances are produced by light reflection and that cause these
distances to be equal – as I have supposed in the previous point (A).
Now suppose– part of these equal distances – is passed through another frame relative
to us – so this part of distances will suffer from Lorentz Length Contraction Effect
which is seen in the rate 1.0725
(Another frame can be found in the solar system because we deal with light motion) –
This explanation can answer why some distances are equal and others are rated with
the same rate (1.0725) – it's simply a feature of light motion.
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

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References
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
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
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=2944p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics Mathematics Faculty
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=2Y4ZdTUAAAAJhl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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