The Moon Orbital Motion Geometry

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
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 –8th
February 2021
Abstract
Paper hypothesis
The Moon Orbit Is Created In A Triangle Form
The hypothesis explanation
- The moon uses Pythagorean triangle as one of its motion techniques because the
moon daily displacement =88000 km, and during 29.5 days the total distance will
=2.598 mkm, which should be = the moon orbital circumference…. But
- (2.598 mkm = 2π x 413000 km) (the moon apogee radius =0.406 mkm) (1%)
- Based on that, if the moon uses its displacement as a real displacement through its
orbit, the moon would revolve around Earth through its apogee orbit only along
month.
- The intelligent moon creates an angle (θ) between its displacement motion
direction and its orbit horizontal level, by that, the real displacement through the
orbit be (L =88000 km cos (θ)), and will be shorter than 88000 km enables the
moon to revolve around Earth through more near orbits.
- The moon using of Pythagorean triangle technique creates the moon orbit in a
triangle form. (The paper hypothesis)
- The using of Pythagorean triangle is a proved fact otherwise the moon apogee
radius should be (r=0.413 mkm) in place of (r=0.406 mkm).
- The moon orbital triangle shows that geometrical mechanism controls the moon
motion

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- For explanation why the moon motion is controlled by a geometrical mechanism,
the answer be, many planets effect on the moon orbital motion and the geometrical
mechanism is the necessary tool to create a balancing for the moon motion.
- For example
- Uranus motion effects on the moon orbital motion and causes to create Metonic
Cycle. also Jupiter effects on the moon orbital motion.
- The question is how can these far planets effect on the moon orbital motion?
- These planets effect on the moon motion by their own motions and not by gravity
force. the paper refers to how this mechanism is done.
- Based on the moon orbital triangle geometrical analysis many questions are
answered for example
- Why the moon apogee orbital circumference=2.55 mkm and doesn't = 2.598
mkm? Because
- The moon uses the difference between these 2 distances (2.598 mkm & 2.55 mkm)
to create its orbital inclination with an angle (5.1 degrees)- the paper explains the
geometrical mechanism to create the moon orbital inclination.
Paper Conclusions
(1st
)
There's a 2nd
force effects on the moon motion in addition to Earth gravity
(2nd
)
Uranus Motion effects on the moon orbital motion and causes Metonic Cycle
(3rd
)
The moon orbit regression (19 degrees per year) is created as a result for the moon
orbital inclination creation.
(4th
)
Earth Cycle (365+365+365+366 days) is created as a result for the moon obit
regression.

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Contents
Subject Page N
1- Introduction 4
2- The Moon Orbital Motion Analysis
2-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
2-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
2-3 The Moon Orbital Motion Analysis
2-4 The Moon Orbital Motion Equation
7
3- The Moon Orbital Triangle Description
3-1 Preface
3-2 The Moon Orbital Triangle Description
3-3 The Moon Orbital Triangle Data Analysis
23
4-The Moon Orbit Geometrical Design
4-1 Preface
4-2 The Triangle Data (The Combination Form)
4-3 The Necessity Of Pythagorean Triangle (1, 2, 51/2)
4-4 The Triangle Geometrical Design
4-5 The moon motion angle (12.195 deg) Analysis
4-6 The Perpendicular Line BC (=86000 km)
4-7 Why the moon day period =29.53 solar days?
4-8 Jupiter Motion effect on the moon orbital motion
39
5- The Moon Orbital Inclination Creation
5-1 Preface
5-2 The Moon orbital inclination creation geometrical process
5-3 Planets motions effect on the moon orbital inclination creation
5-4 The Moon Orbit Regression
5-5 Planets motions cause The Moon Orbit Regression
5-6 The Moon Orbit Regression Effect on The Earth Motion
69
6- The Moon Orbital Triangle Benefits
6-1 Preface
6-2 The Moon orbital triangle shows that (2nd
force effect on the moon motion)
6-3 The Moon orbital triangle shows that (There's 2nd
Orbit for the moon motion)
6-4 The Moon orbital triangle shows that Uranus effects on the moon motion
84
7- Metonic Cycle Is A Proof of Uranus Effect On The Moon Motion
7-1 Preface
7-2 Uranus Effect On The Moon Orbital Motion
7-3 Why The Moon Displacement Daily =88000 Km?
7-4 The Angle 71.9 Degrees Analysis
7-5 The Moon Orbital Triangle Angles Discussions
89
8- Uranus Motion Analysis
8-1 Uranus Motion During 1440 Of Its Days Period
8-2 Uranus Motion During 8 Pluto Days period
8-3 Uranus 144 days Cycle
8-4 The Moon Diameter Creation.
113
8- Appendix No.1 123

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1- Introduction
The suggested moon orbital triangle is inserted in following..
- The black line (AY) is the Earth Ecliptic Line
- The moon moves on its orbital plane (Red line) from perigee(M1) to apogee (M2).
- The triangle base AB has an angle (=0.443 degrees) with the ecliptic line
- The lines M1B and M2D are perpendicular on the triangle base (AB)
- The triangle is formed by the creation of the line BC (the brown Line) which is
created perpendicular on the moon position–means- this brown line (BC) moves
with the moon motion from perigee to apogee and return back. Because of that the
brown line BC is used 2 times in the figure, to be perpendicular on the triangle
base (AB) where the moon be in its perigee point (B) and in its apogee point (D)
- The blue line is the moon equator line, and because there's an angle 1.543 degrees
between the Earth ecliptic line and the moon equator line, the triangle base (AB) is
created between these 2 lines where the base (AB) has 0.443 deg with the ecliptic
line and has 1.1 deg with the moon equator line.

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The moon uses Pythagorean triangle as one of its motion techniques
- Because the moon uses Pythagorean triangle in its motion there are 3 new tools
discovered can be used in the moon motion study, these tools are
o The concept (The Moon Uses Pythagorean Triangle In Its Motion)
which should be discussed in the moon orbital motion analysis (Point no. 2)
o The moon orbital triangle, which we have seen in the previous page, the
paper discusses this triangle geometrical basics deeply, trying to show how
this triangle can be useful in the moon motion study and analysis (Points
No. 3 and 4)
o The moon orbital motion equation (θ1= θ0+1.7 degrees). The paper shows
how to use this equation and test its accuracy in the moon position definition
daily.
- The moon using of Pythagorean triangle is discovered by the moon motion basic
points analysis, where
o Perigee radius (r=0.363 mkm), the most near point the moon can reach to
Earth.
o Apogee radius (r=0.406 mkm), the most far point the moon can reach from
Earth
o Total Solar Eclipse radius (r= 0.373 mkm), the moon creates A total solar
eclipse when the moon be at this distance from Earth or Shorter.
o The Moon Orbital distance (r=0.384 mkm), this value is the registered one
in the moon data sheet as the moon orbital distance.
- 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%)
Based on this data the moon using of Pythagorean triangle is discovered.

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The Paper Contents
- In Point No. 2 The paper analyzes the moon orbital motion, explains the concept
of Pythagorean triangle using and the moon orbital motion equation with its test
- In Point No. 3 The paper provides the moon orbital triangle data (full details)
- In Point No. 4 The paper analyzes the moon orbital triangle geometrical design
- In Point No. 5 The paper discusses Metonic Cycle to prove that, this cycle is a
proof for Uranus motion effect on the moon orbital motion
- In Point No. 6 The paper analyzes Uranus Motion to show that the harmony
which is found between Uranus and the Earth moon motions is created by Uranus
effect on the moon orbital motion.

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2- The Moon Orbital Motion Analysis
2-3 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

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3-3 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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2-4-1 The Equation Concept 2-4-2 The Equation Test and Accuracy
2-4-1 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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2-4-2 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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2-4-3 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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3- The Moon Orbital Triangle Description
3-1 Preface

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3-1 Preface
- In this point we discuss how to create this moon orbital triangle and to define its
distances and angles.
- The previous figure is the combination figure for the triangle 2 cases. These Cases
are created by change of the BC Line Position. We should discuss the triangle
details where the line BC is perpendicular on the point (B) which is parallel to the
moon perigee point (1st
Case) and then where the line BC is perpendicular on the
point (D) which is parallel to the moon apogee point (2nd
Case). (this detailed
discussion be in the point 3-2 The Moon Orbital Triangle Description)
- Then
- There are 4 basic questions we have to refer in the point (3-3 The Moon Orbital
Triangle Data Analysis) – these questions we should discuss and answer in the
point no. (4) The moon orbital triangle geometrical design
- Let's see these 2 cases figures in following..
1st
Case
2nd
Case

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- Let's use the (1st
Case) Triangle to explain the triangle creation process and details
in following

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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 a 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)
- This figure of 2 circles 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
- AB = 363686 km (= Perigee Radius Approximately)
- 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 the moon orbital triangle is a similar to
this triangle (ODB) where their dimensions are rated and their angles are equal,
both are created as a specific Pythagorean triangle (1, 2 and 51/2
)
- In the triangle data analysis we should answer the question (What's The
Geometrical Necessity For Which The Specific Pythagorean Triangle (1, 2
And 51/2
) Is Used For The Moon Orbital Motion?)

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The Point (A)
(2nd
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.

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(3rd
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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(4th
Question)
- What's this line BC (the perpendicular brown line BC =86000 km)? and why
this is used in the moon orbital motion points definition
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%)

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4-The Moon Orbit Geometrical Design
4-1 Preface
4-3 The Necessity Of Pythagorean Triangle (1, 2, 51/2
)
4-5 The moon motion angle (12.195 deg) Analysis
The question
Why The Moon Orbital Circumference doesn't =2598693 km?
Because by the difference between the displacements total 2598693 km and the moon
orbital circumference 2550973 km, the moon motion caused to create its orbital
inclination =5.1 degrees
This answer is discussed deeply in the next point no. 5 (The Moon Orbital
Inclination Creation)

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4-1 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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- 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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4-3 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 CS to create a total angle =137
degrees – based on that
(A)
- The angle ECS =137 degrees
- The distance BS = 150628 km
- The distance SA = 64628 km
- The hypotenuse CS = 173450 km
- The perimeter of the triangle BCS = 173450 +150628 +86000 = 410080 km
- The triangle perimeter (BCS) =410080 km= the apogee radius (406000 km)
(error 1%)
(B)
- The perimeter of the triangle (ACS) = 121622 + 173450 +64628 = 359700 km
- Perigee radius = 363000 km (error 1%)
A Conclusion
- The triangle BCS 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 (CSB =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
BCS) Perimeter
o The Perigee point radius (r=0.363 mkm) which is defined by the triangle
ACS) 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)
- 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….
- This explanation is not so correct because the apogee radius is defined before by
the triangle (BCS) Perimeter and (the rate 0.08) is defined based on it because we
use it in the circles figure.
- I try to show that, we deal here with few players are created depending on each
other , all of them has on 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, and by that, I may prove this is A Directed Data.

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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…..
- 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 = 0.413 mkm
o Means, the apogee radius will not be 0.406 mkm but 0.413 mkm !
o Which proves the paper claim, 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 base 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…?

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- 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
- 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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4-4-1 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
- Because of this interesting idea, I searched behind the value 1290 trying to know if
it's am 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

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(d)
7 x 29.2 x 2π =1290 degrees
(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 but also the moon orbital inclination, and the moon regression creates
361 degrees during Metonic Cycle the moon orbit 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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4-4-2 The Trapezoid CDM2N
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 ca 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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4-4-3 The Triangle CDM2
I-Data
- The Triangle CDM2 – its dimensions are
- The hypotenuse CM2 = 129064 km
- The distance DM2 = 35759 km
- The hypotenuse CD = 96061.6 km
- The angle DM2C = the angle M2CB =19.367 degrees
- The angle DC M2 = 7.25 degrees
- The angle M2DC = 153.4 degrees
(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 that, almost of the Earth and its moon motions data is defined
based on a defined period of time which is one month 29.53 days- based on
this period the data is created)
o 36.44 degrees x 0.8 = 29.2 degrees
o The angle M1RB =36.44 degrees
o That shows the interactions found through the triangle.
Notice
- The Triangle CDM2 Perimeter = 260885 km
- Tan (26.3 deg) x 260885 km = 1290064 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 in point no. (4-4-1 The
Value 1290 degrees) Where 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 creates the interaction between Uranus, Pluto, Earth and its moon, by
this interaction Metonic Cycle is created effective on the 3 planets by Uranus
effect on all other planets – Shortly - Pluto day period be 153.3 hours because of
Uranus motion effect on Pluto motion – this effect in seen in the triangle DCM2
where the angle 71.9 degrees is created inside this interaction
Notice
- This cycle (8 days cycle) is discussed deeply in Uranus Motion Analysis (Point
No. 8 of this paper)
- Let's discuss this angle 71.9 degrees in following

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4-4-4 The angle 71.9 degrees
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)
17 degrees x 1.7 degrees = 29 degrees
- 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. (7-4 The Angle 71.9 Degrees Analysis)
Please review the full discussion.
For better understanding for our discussion we need to discuss the value 17.4 degrees
or 17.2 degrees which we have seen frequently in our discussion

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4-4-5 The angle 17.2 degrees (Pluto orbital inclination)
I-Data
- I have the angle BCU =17.4 degrees
- And because the angle C = the angle CAB = 45 degrees because AB = BC
- The angle UCA =27.6 degrees, and we know the Anomalistic month = 27.55 days
- Means
- this angle 17.6 degrees may express the Anomalistic month is 1 day = 1 deg
- 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 – showing a deep interaction between Pluto and the moon motions

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4-5 The Moon Motion Angle (12.195 degrees) Analysis
I-Data
(I)
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 (407300 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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- 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
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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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
o Tan (10.836 degrees) x 449197 km = 86000 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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5- The Moon Orbital Inclination Creation
5-1 Preface
5-2 The Moon orbital inclination creation geometrical process
5-5 Planets motions cause The Moon Orbit Regression
5-6 The Moon Orbit Regression Effect on The Earth Motion

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5-1 Preface
- The moon daily displacement =88000 km, during the moon day period which is
29.53 solar days, the total displacements be 2598693 km.
- The moon orbital circumference at apogee = 2550973 km
- The apogee orbit is the most far point from the Earth which the moon can reach
- The moon orbit circumference is shorter than the total displacements during 29.53
days with a distance = 47667 km
- That proves (The moon uses Pythagorean triangle rule as one of the moon
orbital motion techniques)
- Then the question be
- Why the moon orbital circumference doesn't = 2598693 km
- The answer is …
- Because the moon uses this difference (47667 km) to create its orbital inclination
angle (5.1 degrees)…
- That means, if the moon orbital circumference be =2598693 km, in this case the
moon orbital inclination will be = Zero
- More conclusions can be raised based on this description, the most important one
is that….
- The moon orbit regression is caused by the same geometrical mechanism by
which the moon orbital inclination is created
- That tells the moon orbital inclination creation process causes also the moon orbit
to be regressed (1.44 degrees) per month.
- These conclusions tell, the process by which the moon orbital inclination is created
is a very effective and significant process found in the moon orbital motion
This point tries to analyze this process details to see how the moon orbital inclination
is created and why does the moon orbit regress?, in addition to we should check the
claim (Earth Cycle 1461 days is created as a result for the moon orbit regression 1.44
degrees per month).

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5-2 The Moon Orbital Inclination Creation Geometrical Process
In this triangle
- ab = 0.232 mkm
- ac = 0.2608975 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.

I do this research
2nd
75
(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?

The Moon Orbital Motion Geometry

The Moon Orbital Motion Geometry

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The Moon Orbital Motion Geometry