Paper Claim - Uranus creates a perpendicularity on the moon orbit and creates an angle (91.1 degrees) with the moon axial tilt - The moon orbit regresses because of Uranus perpendicularity - Venus motion interacted with the moon motion under Uranus Perpendicularity to create the moon orbit regression - The moon diameter is created as a function of its orbital distance as a result of Uranus perpendicularity on the moon orbit Paper Objective - The paper reviews and correct the moon orbital triangle data And - The paper tries to prove its claim Gerges Francis Tawdrous +201022532292

1. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Moon Orbital Triangle Analysis (VI)
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 – 10th
November 2020
Abstract
Paper Claim
- Uranus creates a perpendicularity on the moon orbit and creates an angle (91.1
degrees) with the moon axial tilt
- The moon orbit regresses because of Uranus perpendicularity
- Venus motion interacted with the moon motion under Uranus Perpendicularity to
create the moon orbit regression
- The moon diameter is created as a function of its orbital distance as a result of
Uranus perpendicularity on the moon orbit
Paper Objective
- The paper reviews and correct the moon orbital triangle data
And
- The paper tries to prove its claim

2. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
2
1- More Analysis In The Moon Orbital Triangle
1-1 Basic Data Correction And Revision
z
Figure No. (1) (my figure)
Please Note
(1) The blue dotted arrow creates a point (Z) between F & S
SZ = 7665 km ZF = 2414 km
CZS = 77.8 degrees CZF =102.195 degrees
(2) The Green arrow creates a point (Y) after the point D
DY = 2513.7 km DYA= 118.92 degrees

3. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
3
Let's review and correct the moon orbital triangle data
(1st
Point)
- The figure I brought from internet to use in the Explanation -
- We have supposed that the inner circle is Perigee orbit and the
outer circle is apogee orbit – and we have calculated the tangent
AB = 181843 km
- AB = 363686 km (= perigee radius approximately)
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, the triangle (ODB) is a specific Pythagoras triangle (1, 2 and 51/2
)
- i.e. the triangle (ODB) angles are 26.564 degrees and 63.435 degrees
NOTE
- for these 2 angles (26.564 deg and 63.435 deg) we have searched, because these 2
angles will correct many data in the orbital triangle.
(2nd
Point) The Moon Orbital Triangle Data Correction
- EB = Perigee radius = 363000 km
- ED = Apogee radius = 406000 km
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- ES = total solar eclipse radius = 373000 km (error 1%)
(EC = 373000 km = Earth moon distance at T. Solar eclipse, BUT point C is NOT
the moon position in T. solar eclipse, because the distance BC= 86000 km but the
distance between perigee point and total solar eclipse point = 11000 km)
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km (BY =47513.7 km)
- BA = BC = 86000 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 km
THE ANGLES
- The angle between the black and red lines (under E) = 1.1 degrees
- (E) = 13.33 degrees (C)= 121.67 degrees (A) = 45 degrees
- (ECB) = 76.67 degrees (BCA) = 45 degrees
- (BCS = 7.23 deg) (BCZ = 12.195 deg) (BCF = 13.72 deg) (BCD = 26.564 deg)
(ACD = 18.435 deg)
- (BSC = 82.7 deg) (BZC = 77.8 deg) (BFC = 76.82 deg) (BDC = 63.434 deg)
- (CSA =97.23 deg) (CZA =102.195 deg) (CFA= 103.7 deg) (CDA = 116.564 deg)
- (CYA = 118.92 deg) (BYC = 61.08 deg)
- (Uranus axial tilt = 97.8 degrees = FSC 0.6 degrees) and
- (CYA +0.6 deg = 119.5 deg where 119.5 x 0.99 = 118.3 deg (Neptune A. T)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

4. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
4
1-2 Basic Data Correction And Revision
What's the useful result we may get from the previous data correction?
Let's summarize that in following….
1-2-1 Venus Effect
Equation No. 1
116.75 days (Venus Day) x π = 366.78 days
But
(366.78 days /365.25 days) = (24 hours /23.9 hours) (Zero Error)
Where
365.25 days = π x 116.26 days
- Equation No.(1) tells that, Venus day period (116.75 days or degrees) is used in
opposite to the value (116.26 degrees) to create Earth rotation period (23.9 h) in
opposite to Earth day period (24 h)
Simply
- It tells that, Venus day period is interacted with the value 116.26 degrees to
produce these 2 values (24 hours and 23.9 hours) which shows Venus Motion
effect to produce 2 motions of Earth (Earth rotation & Earth day motions)
- Equation No. (1) shows 2 values (116.75 and 116.26) as basic values for Earth
motions effected by Venus Motions …
-
- So I have searched for these 2 values in the moon orbital triangle but no value I
have found
- That's why I have to return to the moon orbit basic triangle (1, 2 and 51/2
) to find
the most accurate value because the difference is so small (116.75 and 116.26)
- And I have found neither!
- I found only 116.434 degrees as the most accurate angle I can reach to…
So
- How these data are interacted with each other (116.26, 116.564 and 116.75)?
- Let's try to discover in following…

5. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
5
1-2-2 The Sun Declination angle 63.7 degrees
Equation No. 2
(a) 63.7 degrees – 63.435 degrees = 0.265 degrees
But
(b) (3600 mkm /86400 mkm) = (2.65/63.435)
And
(c) (47.4/4.7) = (2.65/0.265)
- The sun declination angle 63.7 degrees is the key for our angles question…
116.26 degrees +63.7 degrees = 180 degrees
- Any other angle (116.75 deg or 116.434 deg) can't solve this question, so we have
to accept the angle 116.26 degrees where (accurately 116.26 x π =365.25) But
- Venus Effect is so strong in the previous discussion and its day period (116.75
days) has a massive effect on the inner planets motions data …!
Let's discuss Equation No. 2
- Our triangle angle 116.564 deg gives us only (180-116.564= 63.436 deg)
- It's not enough to do the process! We have faced this situation once before! Let's
try to remember …..
o The light beam is created from the source C2
o If the time is (1second) this Value C2
we will see as a distance =90000 mkm
o But
o Jupiter energy which is sent to Pluto and reflected to Neptune and Neptune
consumed 14 % of this whole energy and the rest Neptune had sent to the
inner planets into 2 Trajectory of energy (one Trajectory to Mercury and the
other trajectory to Venus and earth) but Mars lost its original source of
energy because of its migration
o Each energy Trajectory has 86400 mkm (because Space = Energy)
o The energy is not enough to create the distance (90000 mkm) which for 1
second will be equivalent to C2
(the light beam source of energy)
o So, the energy 86400 mkm will be useless because the light is produced
based on quantum and can't be created by less energy…
o Inevitably we need 3600 mkm -
- Equation no.(b) tells this story, (3600 mkm/86400 mkm) =(2.65/63.43), the value
86400 mkm produced the angle 63.435 degrees but still need 2.64
- Equation no. (c) (47.4 Mercury velocity /4.7 Pluto velocity ) = (2.65/0.265)
o We need 0.265 degrees, but the energy rate is 2.643…. how to solve that?
o (Mercury /Pluto) Velocities creates this interaction between 2.64 and 0.265
to create the angle 63.7 degrees which is required
o Mercury and Pluto velocities interaction depends on their motions
interaction which we have discussed deeply in the previous paper.

6. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
6
1-2-3 The Angle CDA =116.564 degrees
Equation No. 3
116.75 degrees – 116.564 degrees = 0.186 degrees
And
102.195 degrees x 0.186 degrees = 19 degrees 2
- We know that 1 day = 1 degree, So 116.75 days = Venus Day Period
- Between our angle (CDA = 116.564 deg) and Venus day (angle) =116.75 deg,
there's a difference = 0.186 degrees
- The moon motion daily =13.18 degrees but we know that the moon motion
original angle = 0.98562 degrees, because the moon original motion is identical to
the Earth motion, means the moon move per solar day a distance =2.58 mkm with
- an angle 0.98562 degrees on the horizontal line (revolving around the sun)
So
- 13.18 degrees – 0.98562 degrees = 12.195 degrees (this is the angle BCZ) and also
the angle 102.195 deg (12.195 deg +90 deg) is the angle (CZF)
- 2 angles multiplication (102.195 degrees x 0.186 degrees = 19 degrees 2
)….What
does that mean?
- 19 degrees the moon orbit regresses per year, this value (19 degrees) is produced
an interaction between 2 angles (2 motions)
- The interaction between Venus Motion (116.75 degrees or days) and the moon
motion (116.564 degrees) produces the value 0.186 degrees which is the factor by
which the angle 19 degrees is produced based on 102.195 degrees…!
- We understand that, Venus has a great effect on the moon motion and its orbit
regression but we see the moon motion does 2 steps in the process… Venus
produced the value 0.186 degrees in interaction with the moon motion… but the
next step is done by the moon angle (12.195 deg +90 deg)… how to explain that?
- In the previous paper (Uranus Is Perpendicular On Earth Moon Orbit (II)), we
have concluded that¸ Because of Uranus perpendicularity on the moon orbit, the
moon orbit regresses yearly 19 degrees… the perpendicularity is seen clearly in the
angle 102.195 degrees ..although the moon motion angle =12.195 degrees but the
equation uses specifically the angle 102.195 degrees to tell that it's the effect of
Uranus Perpendicularity on the moon orbit – please review this interesting paper.
- Why does the moon orbit regress 19 degrees per year?
1. There's an interaction of moon & Venus motions to produce 0.186 degrees
2. Uranus perpendicularly creates the angle 102.195 degrees
3. the interaction is done between the moon (perpendicular) motion with
Venus interaction effect (102.195 deg x 0.186 deg =19 deg)
4. the moon orbit regresses 19 degrees per year.

7. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
7
1-2-4 Uranus Perpendicularity Effect
Equation No. 4
181840 km sin (1.1 degrees) = 3475 km
- 181840 km, this value we have discussed before where I have discovered the
specific triangle (181840 km (1) and Perigee 363000 km (2) and apogee 406000
km (5)1/2
)
- And we have found that this value 181840 km is consisted of (96150 +86000)
where (96150 km2
= 86000 km2
+43000 km2
)…
- i.e.
- The value 181840 km is very essential distance in the moon orbit geometrical
structure
-
- Equation No. (4) tells that, it's really there's an angle =(1.1 degrees) in the moon
orbital triangle which is created as a result of Uranus perpendicularity on the moon
orbit
- The Equation tells that, if the moon diameter and circumference is created as a
function of its orbit distances, that's done based on this angle
- That simply explain why the moon diameter 3475 km x 116.75 = 406000 km
(apogee radius =406000 km) and (116.75 days = Venus Day period) – this
proportionality and many others can be created because the moon diameter is
created as a function of its orbital distances based on this specific angle (1.1 deg)
- Equation No. (4) proves clearly that Uranus perpendicularity on the moon orbit is a
real and fact –
- Because the angle (97.8 deg -6.7 deg )= 91.1 degrees, this angle 1.1 degrees is
consumed in a decline of the moon orbital triangle on the horizontal level as it's
seen in the first figure in this paper
Notice (1)
Uranus axial tilt 97.8 degrees is equal the angle under (CFA) –because – the angle
CDA =97.27 degrees and the (green rectangle) creates and additional angle =0.6
degrees, so the total angles under (CFA =97.27+0.6 = 97.8 degrees Uranus axial tilt)
Notice (2)
Similar to that, the angle (CYB) = 118.92 degrees and the angle under is 118.92+0.6 =
119.5 deg but 119.5 deg x 0.99 = 118.3 deg = 90 deg + 28.3 deg (Neptune axial tilt)
Notice (3)
The black thick line is the moon axial tilt (6.7) the angle on it = 91.1 degrees because
the moon angular diameter uses 0.5 degrees

8. IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
8
References
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
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Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
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