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

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 (General discussion) (I)
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 –28th
November 2020
Abstract
Paper Argument
- The Moon motion has 2 Orbits – we see one of them only
- The moon second orbit is declined by 0.8 degrees on the first one.
- The 2 orbits are neighbors and define together the Lunar Eclipse Umbra Length
(1.392 mkm)
- Venus axial tilt is changed from 1.774 to 177.4 degrees because of Uranus axial tilt
angle (91.1 degrees) with the moon axial tilt (6.7 deg) and because of Uranus axial
tilt perpendicularity on the moon orbital triangle base.
- Venus original axial tilt (1.774 deg) is created as a rate between its diameter and
Mars diameter – that shows Mars diameter rate effect on the planets motions
- Mars diameter effect explains the reason why Kepler equation (P2
/d3
= Constant)
uses (25) as a constant while (25.2 dg = Mars Axial Tilt)
- The moon uses Pythagoras rule in its motion to define its daily displacement

I do this research
2nd
2
1- The Moon Orbital Triangle Data
1-1 Preface
The Moon Orbital Motion (Revision)
- Why does the moon use Pythagoras Rule in its motion? and why do we need the
moon orbital triangle?
o The moon daily displacement =88000 km but the moon doesn’t use it as its
real displacement but instead the moon uses Pythagoras triangle to define
another real displacement
o Based on that
o The moon uses the right triangle dimension (L= 88000 km Cos θ) where
this (L) is the moon real displacement through its orbit daily
o The angle (θ) is the smallest angle in the right triangle, and it effects on the
moon real displacement and its motion height above perigee radius!
Why?
o the daily displacement = 88000 km and during 29.53 days, the distance the
moon moves = 2.59 mkm which can be done only through apogee orbit
whose radius (r=0.406 mkm) –
That means –
o If the moon uses only 88000 km as its real displacement daily, the moon
would move only through the apogee orbit (r= 0.406 mkm)
o Because the moon tries to move in more near orbits to the Earth, the moon
uses Pythagoras triangle technique to produce smaller displacements though
its orbit while the moon does the same displacement (88000 km) as a real
motion -
i.e.
o This real displacement technique (L= 88000 km Cos θ) gives the moon the
ability to move through lower orbits to the Earth, and the change of the
angle θ changes also its real displacement

I do this research
2nd
3
o The smaller real displacement can be used for more near orbits to Earth (as
perigee radius r=0.363mkm) but the longer real displacement needs more
wide orbit to be passed, which forces the moon to move in higher orbits
above apogee
o That means, the angle θ changes the real displacements and the height of the
moon motion above perigee radius (r= 0.363 mkm).
o
o Then based on that I have suggested the moon motion equation which is
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o θ is the smallest angle in Pythagoras triangle used for the moon real
displacement (L= 88000 km Cos θ),
o 0.985 degrees = Earth daily motion degrees = one of the moon motion
angles –
o With more analysis, a 2nd
force effects on the moon orbital motion is
discovered, this 2nd
force effects on the point (A) in the moon orbital triangle
– where this point is an essential part of the triangle but it's far with 43000
km from apogee radius (r= 0.406 mkm).
o This 2nd
force effects on the moon orbital motion is produced by gravity
forces interaction between the sun and Jupiter on Earth and its moon – thus
2nd
force pulls the moon to apogee radius but Earth gravity tries to keep the
moon at perigee radius. (This 2nd
force effects on the point (A) in the moon
orbital triangle)

I do this research
2nd
4
1-2 The Moon Orbital Triangle Details
Let's review The Earth Moon Orbital Triangle because we use it
Figure No. (1) (my figure)
Let's Review The Moon Orbital Triangle Data
(1st
Point)
- The figure I brought from internet to use in the Explanation -
- We have supposed that the inner circle is Perigee orbit and
the outer circle is apogee orbit – and we have calculated the
tangent DB = 181843 km
- AB = 363686 km (= perigee radius approximately)
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, the triangle (ODB) is a specific Pythagoras
triangle (1, 2 and 51/2
)
- The triangle (ODB) angles are 26.564 deg. and 63.435 deg.

I do this research
2nd
5
(2nd
Point) The Moon Orbital Triangle Data Correction
- EB = Perigee radius = 363000 km
- ED = Apogee radius = 406000 km
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- ES = total solar eclipse radius = 373900 km (error 1%)
(EC = 373000 km = Earth moon distance at T. Solar eclipse, BUT point C is NOT
the moon position in T. solar eclipse, because the distance BC= 86000 km but the
distance between perigee point and total solar eclipse point = 11000 km)
- CX= =87521 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 km CY= 97766 km
- BA = BC = 86000 km
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km
- BY = = 46475 km
- SZ = 7665 km ZF= 2414 km
- DY = 3475 km BX= 16203 km
THE ANGLES
- The angle between the black and red lines (under E) = 1.1 degrees
- (E) = 13.33 degrees (C)= 121.67 degrees (A) = 45 degrees
- (ECB) = 76.67 degrees (BCA) = 45 degrees
- (BCS = 7.23 deg) (BCZ = 12.195 deg) (BCF = 13.72 deg) (BCD = 26.564 deg)
(ACD = 18.435 deg)
- (BSC = 82.7 deg) (BZC = 77.8 deg) (BFC = 76.82 deg) (BDC = 63.434 deg)
- (CSA =97.23 deg) (CZA =102.195 deg) (CFA= 103.7 deg) (CDA = 116.564 deg)
- (CYA = 118.3 deg)
- BCY = 28.39 degrees ECZ= 88.9 degrees
- XCE = 66 degrees
- CZS = 77.8 degrees
- CZF =102.195 degrees
- XCB = 10.67deg
- (Uranus Axial Tilt = 97.8 degrees = FSC 97.2 degrees + 0.6 degrees) (i.e. the
angle under FSC)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

I do this research
2nd
6
1-3 The Moon Diameter Division
Figure No. 2 (my figure) (The Moon Diameter Division)
- Uranus Axial Tilt (97.8 degrees) has an angle (91.1 degrees) with The Moon Axial
Tilt (6.7 degrees), so The blue vertical line is Uranus axial tilt and the angle in the
circles center = 91.1 degrees
- This value 91.1 degrees is decreased with 0.5 degrees which is consumed by the
moon diameter (the circles express the moon diameter with some division), also
0.6 degrees is consumed by the green box (which expresses Saturn and Jupiter
interaction above the moon)
- So the angle between the blue vertical line and the horizontal red line 90 degrees
- The red horizontal line is the moon orbital triangle base (AE = 449197 km)
- The blue vertical line angle above the moon directly =90.6 degrees
- But
- In this figure I cut layers from the moon diameter to create smaller moon diameters
to consume smaller angles than (0.5 degrees), and based on that we measure the
angle of The blue vertical line, let's explain this moon diameter division:
- The Blue Circle its diameter R1= 1390 km and r1 =695 km
- The Red Circle its diameter R2= 2085 km and r2 =1042.5 km
- The Black Circle its diameter R3= 2185 km =π x 695 km and r3 =1092.5 km
- The Brown Circle its diameter R4= 3208 km and r4 =1604 km =5040/π
- The Orange Circle is the moon itself its diameter R5= 3475 km = 5 x 695 km

I do this research
2nd
7
- The moon consumes 0.5 degrees for its diameter (3475 km) …So
- R1 = 1390 km, the consumed angle will be 0.2 degrees i.e. the angle above The
Blue Circle = 90.9 degrees.
- R2= 2085 km (60% of the moon diameter) and because of that, the consumed angle
will be only 0.3 degrees, means, the angle above The Red Circle =90.8 degrees
- Note Please, 90.8 degrees = 90 degrees + 0.8 degrees (Uranus orbital inclination),
by that the rest degree of Uranus axial tilt = its orbital inclination vertically.
- R3= 2185 km (63% of the moon diameter) and because of that, the consumed angle
will be only 0.314 degrees, means, the angle above The Red Circle =90.8 degrees
Notice No. 1
- Pluto Axial Tilt 122.5 degrees = 7.1 x 17.2 degrees (Pluto orbital inclination),
- 7.1 a rate created by Lorentz Length Contraction Phenomenon…
- That tells us, the values (122.5 deg and 17.2 deg) are equivalent values, how? It's a
contraction, the energy of (122.5 deg) in contracted by 7.1 and the rest 17.2 deg is
created from this same energy (122.5 deg). they are 2 equivalent values but one of
them passed through a different frame and faced a contraction phenomenon and
because of that its value (122.5 deg) became (17.2 deg)
- This notice is provided here to show that in the solar system there's some way to
create A Planet Axial Tilt = Its Orbital Inclination…. What's the important
result for this equality? 2 times we have seen this equality in Uranus & Pluto data,
where we know that (122.5 deg x 0.8 deg = 97.8 deg) – there's a geometrical
machine found behind this equality…! We should discuss that later …

I do this research
2nd
8
Notice No. 2
- Are these divided diameters real diameters or invented numbers? Let's test them :
(1)
- 149.6 mkm (Earth Orbital Distance) = 3475 km (the moon diameter) x 43000 km
(the distance from perigee to apogee)
- 149.6 mkm (Earth Orbital Distance) = 2085 km (R2) x 71492 km (Jupiter Radius)
- 149.6 mkm (Earth Orbital Distance) = 1047 x 142984 km (Jupiter diameter) (1047
= The Sun Mass / Jupiter Mass)
(2)
- R4= 3208 km x 466884 km = 149.6 mkm x π2
- 466884 km = Jupiter motion distance during its day period.
(3)
- R3= 2185 km x π x 21346.6 km = 149.6 mkm (2%)
(4)
- R2= 2085 km x 71492 km (Jupiter Radius) =149.6 mkm
(5)
- 4900 mkm = 4222.6 x 1.16 mkm = 10921 km x 449197 km
- 302 mkm = 4222.6 x 71492 km (Jupiter Radius)
- 305 mkm = 88000 km x 3475 km = 25.2 x (3475 km)2
- 177.4 degrees -90 degrees = 87.4 degrees (+0.6 degree = 88 degrees)
- 3475 s x 0.3 mkm/sec =1042.5 mkm but
- 2085 mkm = 3475s x 0.3 mkm/s x2
Discussion
- The points (1-4) show that, Earth orbital distance 149.6 mkm, is produced by the
moon diameter multiplication with other data…. This same using is found for the
smaller diameters, which proves that these diameters are real values
- Point (no.5) shows some light motion behind planet data creation- this data tries to
show how this data is created based on light motion.

I do this research
2nd
9
2-Venus Effect On The Moon Orbital Triangle
2-1 Venus Axial Tilt Creation
Equation No. (1)
90.9 x 1.1 =100 = 177.4 /1.774
(Venus Axial Tilt 177.4 degrees)
Mars Diameter Rate Effect: (Mars diameter=6792 km)
o 6792 / 4879 =1.392 (Mercury Diameter)
o 12104 /6792 =1.774 (Venus Diameter)
o 12756 /6792 =1.878 (Earth Diameter)
o 6792 /3475 =1.95 (Moon Diameter)
o 142984 /21346.6 = 6.7 (Jupiter Diameter)
o 120536 /21346.6 =5.64 (Saturn Diameter)
o 51118 /7070 =7.23 (Uranus Diameter)
o 49528 /7070 =7 (Neptune Diameter)
o 7509 /6792 =1.1 (Pluto Diameter)
o 10921 /6792 =1.6 (Moon Circumference)
The rates result
1.392 mkm = the sun diameter
1.774 1.878 = Mars orbital inclination (1%)
1.95 =(Mars Mass / Mercury Mass) 6.7 deg = The Moon Axial Tilt
5.6 = 0.5 deg (the moon diameter) +5.1 deg (moon orbital inclination)
7.23 = The angle (BCS) in the moon orbital triangle
7 deg = Mercury Orbital Inclination.
1.1 deg = The Angle between the base AE (449197 km) and the horizontal level.
1.6 deg = 6.7 deg (the moon axial tilt) – 5.1 deg (the moon orbital inclination)
Note Please
7070 km = Mars diameter before Mars migration, how to know it?
84 mkm =1092
x 7070 km - Mars ancient orbital distance was 84 mkm

I do this research
2nd
10
Let's return to equation no.(1)
Equation No. (1) (continued)
90.9 x 1.1 =100 = 177.4 /1.774 (Venus Axial Tilt 177.4 degrees)
- Equation no. (1) tells us a clear meaning let's summarize it in following:
o Venus Axial Tilt was 1.744 degrees (Why?)
o Because Mars & Venus Diameters Rate =1.774 caused Venus axial tilt to be
(1.774 deg)! why? and why Mars specifically? Later we should answer…
but suppose … what then…. ?
o 100 is produced by the angles (90.9 and 1.1 degrees), so this 100 is
produced under the moon orbital triangle, almost in the moon diameter, and
specifically above the smallest diameter R1=1390 km (The Blue Circle in
figure no.2)
o That may explain many data for example:
o 38025 km (Venus Circumference) = 1390 km x 27.3, This equation tells
that, if the small moon (1390 km) rotates around its axis once per day, so
during its orbital period (27.3 d) will move a distance = 38025 km (Venus
Circumference) x π (Note (Venus Circumference) x π) = Saturn diameter)
o 12104 km (Venus Diameter)=17.4 x 695 km, where 17.4 deg = the inner
planets orbital inclinations total, (17.4 deg =3.4 deg V. Orb. Incl. x 5.1 deg)
- Simply the equation tells us, some great change is occurred in the solar system
geometry based on the geometrical effect of the moon orbital motion…
- Still the idea is so puzzled … we will accept this conclusion… but … how to
explain the data… Because NOT all planets axial tilts are produced by the
diameters rate with Mars? (only the moon axial tilt is produced)! But all other
results are planets orbital inclinations or angles from the triangle! So why only
Venus and the moon axial tilt are produced? And why Mars diameter specifically?
Still also Uranus and Neptune deal with the ancient Mars diameter!!
- Let's move step by step

I do this research
2nd
11
(1st
Question) Why Mars diameter specifically?
- Why Mars Axial tilt is the constant in Kepler equation?
- Kepler told us, Planets (P2
/d3
)= constant …by actual data the constant =25
Table No.1
Planet P2
* 25 = d3
Error
Mercury (88)2
*25 (57.9)3
0.2%
Venus (224.7)2
*25 (108.2)3
0.3%
Earth (366)2
*25 (149.6)3
0
Mars (687)2
*25 (227.9)3
0.3%
Jupiter (4331)2
*25 (778.6)3
1.4%
Saturn (10474)2
*25 (1433.5)3
1%
Uranus (30589)2
*25 (2872.5)3
1.3%
Neptune (59800)2
*25 (4495.1)3
1.5%
Pluto (90588)2
*25 (5870)3
1.4%
25.2 degrees = Mars Axial Tilt (But the rate 25 = 25.2 degrees/1degrees)
- Why the constant =25? Because
- Planet matter is created as a function in its motion and because of that, Mars
diameter rate defines the planets axial tilts, orbital inclinations and other degrees,
and as a result the planets motions became in consistency with Mars Motion (Data)
and that caused to use Mars axial tilt as a constant
- (1 degree = 1.1 degrees -0.1 degree), Uranus created the decline under the triangle
to produce this angle 1.1 degrees and by that to repair the destruction occurred for
mercury axial tilt (which was 1 degrees) and became (0.01 degree).
(2nd
Question) Why not all planets axial tilts are produced?
- Because there's an interaction is occurred between planet axial tilt and orbital
inclination – as we have seen in Pluto data where (122.5 deg Pluto axial tilt =7.1 x
17.2 deg Pluto orbital inclination) and 7.1 is the contraction rate. Also there's an
interaction between the planets motions data as (97.8 deg Uranus axial tilt = 122.5
deg Pluto Axial Tilt x 0.8 deg Uranus orbital inclination) also as we have seen (5.1
deg (the moon orbital inclination) x 3.4 deg (Venus orbital inclination) = 17.4 deg
(the inner planets orbital inclinations total but 17.4 deg x 0.99 =17.2 deg Pluto
orbital inclination)

I do this research
2nd
12
- So let's conclude what Equation no. 1 tells us…
Equation No. (1) (continued - again)
90.9 x 1.1 =100 = 177.4 /1.774 (Venus Axial Tilt 177.4 degrees)
- When (Uranus Axial Tilt) The Blue Vertical Line Angle =90.9 degrees (directly
above the moon smallest diameter R1 =1390 km) – at this point – with effect of the
declination angle 1.1 degrees (90.9 x1.1) the rate 100 is produced and by this rate
Venus axial tilt 1.774 became 177.4 degrees…
- This interaction helped to produce the moon orbital inclination because (180 deg -
177.4 degrees =2.6 degrees) so through the whole circle (360 degrees) the
difference will be 5.2 degrees (5.1 deg (the moon orbital inclination +0.1 degree),
but
- The value 0.1 degree is used here (1.799 deg +0.1 deg =1.9 deg Mars Orbital
Inclination), where 1.8 deg = Neptune orbital inclination and the value 1.8 degrees
can be produced easily because it = (179.9 degrees /100).

I do this research
2nd
13
2-2 The Triangle BCF Analysis
- We analyze the triangle BCF
- BC =86000 km
- BF = 21000 km
- CF =88527 km
- BCF = 13.7 degrees
- BFC = 76.3 degrees
- The line CB is perpendicular on FB (90 degrees) but
- The line CB has an angle 91.1 degrees on the black line
base (under the blue circle "the moon")
Equation No. (2)
1259.3 degrees = 91.9 degrees x 13.7 degrees
- The angle 13.7 degrees (=BCF)
- We have no 91.9 degrees! We have only 91.1 degrees and that means there's
another angle is added here (0.8 degrees) what's this angle? It's our next question
- We have 2 points here to investigate… which are
- (1st
Point) To analyze the triangle BCF because of the angle 13.7 degrees
And
- (2nd
Point) To analyze the angle 1259.3 degrees other using…
(1st
Point) ….The Triangle BCF
- Its perimeter = 195527 km but (2x 195527 = 391054 km) …..And
- BF = 21000 km and EB = perigee radius = 363000 km means EF = 384000 km
- (384000 km)2
+ (86000 km)2
= 393512 km
- The difference between the 2 values = 0.6% or around 2500 km while the
distance from perigee to apogee =43000 km, means the rate around 5.7%.
- The calculation is clear and puzzled, because the double value of the triangle
perimeter can be produced by using Pythagoras rule for (The Triangle Base and
Perigee Radius)

I do this research
2nd
14
- What geometrical reason behind this calculation? because it tells us that, the
triangle perimeter (knows!!) the perigee radius r= 363000 km, which is not part of
this triangle at all! But…. Let's look at the 2 triangle… (BCF and ECB)
- The dimension (86000 km) is used in both triangles
- And 13.328 degrees = The angle (E) in the triangle EBC
- And 13.7 degrees = The angle (BCF) in the triangle BCF
- If these 2 angles have a relationship, there's some geometrical way to perform this
effect. Let's try to discover this relationship ….
Equation No. (3)
A - 12.195 deg +1.1 deg =13.29 (13. 328)
B- 12.195 deg +1.5 deg = 13.7
C- 2.5 deg x 0.6 deg =1.5 deg
- Equation No (3) tells us why the angles (13.7 and 13.3 degrees are related to each
other), because both are created from the same source (12.195 degrees), this is the
moon motion angle (13.18 deg – 0.98562 deg= 12.195 deg)
- Equation No. (A) added this angle (12.195) to 1.1 degree, which is the angle
between the moon orbital triangle base (EA =449197km) and the horizontal level.
- Equation (A) produces the angle 13.3 degrees and Equation (B) added the angle
(1.5 degrees) to (12.195 deg) to produce the (13.7 deg), but what's the relationship
between 1.1 degrees and 1.5 degrees?
- 180 deg -177.4 deg (Venus Axial Tilt)= 2.6 degrees =1.1 deg + 1.5 deg means
the angle 2.6 deg is produced by Venus axial tilt and that makes one source for
both value (1.1 and 1.5)
- Equation no. (c) shows the interaction between 2.5 deg (Saturn orbital inclination)
and the value (0.6 degrees) produced by the green box (in the figures), where the
result is 1.5 degrees….what does that tell to us??
- 6.7 deg-5.1 deg =1.6 deg=0.1 +1.5 deg i.e. the value0.1 which is used 1.8+0.1=1.9
is found by interaction with Saturn orbital inclination, that causes Mars and Saturn
motions interaction.

I do this research
2nd
15
Equation No. (4)
I - (365.25 days /354.39 days) = 13.736 /13.328
II- (2.6 degrees /2.5 degrees) = 13.7 /13.18
- More equations about theses 2 angles
- Equation (2) (I), shows that, the required rate is used between the lunar synodic
year and the sidereal year
- Equation (2) (II), shows that, 2.5 degrees = Saturn Orbital Inclination) deals with
(2.6 degrees = 180 deg – 177.4 deg (Venus Axial Tilt). We know that, the moon
daily motion =13.18 degrees.
- What's the useful result for this equation no. (2)? Why we searched for it?
- Equation no (20 (I) tells that the value (13.7 deg) depends on the value (13.3)! if
365.25 days depends on 354.39 days or vice versa?
- We accept that the moon motion depends on earth motion and that may cause the
rate (365.25/354.39) is created by some dependency…
- That means, there's some hidden machine connects the angle 13.3 and 13.7 degrees
and that causes perigee radius to be included in the triangle BCF perimeter, may
later we can discover this geometrical machine.
Equation No. (5)
12.195 x 1.09 = 12.195 +1.1 (error 1%)
- In Equation no.(5) please note …. The value 1.09 is very near to 1.1 (1%) and the
equation itself has an error (1%),
- I interest for the error (1%) to explain its reason because this error is used
frequently in the planets data
- Equation No.(5) tells a geometrical rule, by which the values multiplication can be
equal their additions…. We remember here (2x2= 2+2), so the question if this rule
is used in the solar system geometry, I mean, if there's a series of numbers depends
on this rule (2x2 =2+2) and this series is used in the solar system geometry – we
still need more analysis…

I do this research
2nd
16
Equation No. (6)
97.8 deg (Uranus Axial Tilt) = 1.1 deg x 88.9 deg
- Why do we need this equation? Because sin (97.8) = 0.99
- I suppose there's some geometrical machine by which Uranus axial tilt (sin) effect
on the planets data to produce this rate 0.99
o I wish it's not boring to search about this rate (0.99), because hundreds of
important data is created in 2 rates (%100 and 99%) for examples
o 17.4 (The inner planet orbital inclinations total) x 0.99 =17.2 deg (Pluto
orbital inclination)
o 23.6 (The outer planet orbital inclinations total) x 0.99 =23.4 deg (Earth
Axial Tilt)
o (180deg/π)= 28.66 degrees x 0.99 = 28.3 degrees (Neptune Axial Tilt)
- Please remember the angle ECZ =88.9 degrees.
Equation No. (7)
346.6 = 12.195 x 28.4 (where 28.3 degrees = Neptune Axial Tilt) (346.6 days = the
nodal year)
- We still analyze the source (12.195 degrees) trying to see its depth…
- Also I can add the following data ….
o 12.195 deg x 1.9 deg (Mars orb. Inclination) =23.17 =23.4 x0.99
o 12.195 deg =7 (Mercury orb. Inclination) + 5.19
o 12.195 deg = 25.2 – 13 = 28.3- 16.1
o 17.2 - 12.195 = 5
o 17.4 - 12.195 = 5.2 (5.2+5)/2) =5.1
o 63.41 =12.195 deg x 5.2
o 1259.3 =12.195 deg x 103.263 (103.263 -103.177 =0.085)
o 0.085/0.1= 0.85 (= 0.8 deg x1.0725)
- We will use this data in the analyze of the angle 1259.3 degrees

I do this research
2nd
17
(2nd
Point) (The angle 1259.3 degrees analysis)
- 177.4 deg (Venus axial tilt) +2.5 deg (Saturn orbital inclination) = 179.9 degrees
- 179.9 degrees x 7 deg (Mercury orbital inclination) =1259.3 degrees2
- We know this value, let's remember it here
o Because of the planets (Mars And Pluto) migration, the solar system basic
geometrical structure has changed, and because of that, the value 180
degrees is destroyed in the solar system
o Venus and Saturn together unified their powers to create another (180 deg)
but
o The available value was only 179.9 degrees, so the solar system has to use
this value 179.9 degrees in the geometrical process in place of 180 degrees
o As a proof for that
o 180 degrees x 0.8 =225 but 179.9 degrees x 0.8 = 224.7 degrees (224.7 days
= Venus Orbital Period) (0.8 deg = Uranus orbital inclination)
i.e.
o The solar geometrical structure depends on 179.9 degrees in place of
180 degrees.
Equation No. (10)
2094 degrees x 0.6 degree = 1259.3 degrees2
- 0.6 degrees is the angle created by the green box in figures (no. 1 and 2), it's found
above the moon and under the moon orbital triangle base (AE) = 449197 km -
- 2094 mkm = Jupiter Uranus Distance
- We know that 1 mkm = 1 degree, where Mercury orbital circumference =360 mkm
= 360 degrees, and the solar system is one machine, so any planet use this rate.
- i.e. 2094 mkm = 2094 degrees ….But
- The distance 2094 mkm (Jupiter Uranus Distance) is so important in our
discussion about the moon orbital triangle, because the data tells that this distance

I do this research
2nd
18
effects on the triangle data- but how – because it's just a distance?! Let's review its
data analysis as possible to see what effect this distance (2094 mkm) can practice
on the moon orbital motion…
o Light known velocity (0.3 mkm/sec) travels this distance in 6939.75 seconds
o The value 6939.75 seconds = 71 x 97.8 seconds
o We know that 71 is Lorentz Length Contraction Effect Rate
o 97.8 degrees = Uranus Axial Tilt, the blue line in figure no. 2, and we
know that, light motion can use the time periods as distances (because x =ct
and if c=1 so x=t) (where 1 mkm=1 degree, so 97.8 s= 97.8 mkm =97.8 deg)
where (97.8 s x 0.3 mkm/s =29.33 mkm and 29.33 mkm x 97.8 = 2872.5
mkm = Uranus Orbital Distance – that means Uranus orbital distance also is
created base on (97.8s) and not only Uranus axial tilt.
Also
o 2094 seconds x 0.3 mkm /sec (light known velocity) =627 mkm = Earth
Jupiter distance – that means, by using this distance (2094 mkm) as a period
of time (2094 seconds) the light travels from Jupiter to Earth
Notice No. 1
o Based on that, the light almost travels from Earth to Jupiter and then to
Uranus, it's one light beam starts from Earth to Jupiter and then to Uranus,
how we know this is the motion direction and not the reversed one? because
1.2 mkm = 40080 seconds x 29.8 km/sec (Earth velocity), where Earth saves
a value 40080 and rest (1.16 mkm/sec) Earth sent to Jupiter, this is Jupiter
light supposed velocity (1.16 mkm/sec), and also its Jupiter motion disatce
per solar day (error 2.5%) , that means Earth motion is Jupiter source of
energy…

I do this research
2nd
19
Notice No. 2
o Please remember in the moon diameter division we have found (R2) =2085
km and we know that the moon orbital diameter =1 mkm because of that
this value 2085 km can be 2085 million km and be the distance which we
deal with here… Please remember, above this diameter (2085 km) (Uranus
axial tilt) The Blue Line has an angle =90.8 degrees accurately which =
Uranus orbital inclination vertically (0.8 deg + 90 deg) – can be there a
geometrical connection between these 2 values 2085 mkm and 2085 km?
Notice No. 4
o The distance (2094 mkm) is very puzzled one, because it has different
values as (2094 mkm –2088 mkm- 2082 mkm – 2107 mkm), the differences
are so small (less than 1%) but these are different values because they are
used by these values in very accurate equations….
o For example
o Light known velocity (0.3 mkm/sec) during 6939.75 s moves 2082 mkm
But
o Light supposed velocity (1.16 mkm/sec) during 3600 s moves 2088 mkm
o Because of light motion, I prefer to use the value 2082 or 2088 mkm.
But
Equation No. (11)
- 2073 degrees = 2073 mkm = 0.99 x 2094 mkm
o 2073 mkm = 1153 mkm (the planets motion distances total during their
days periods) x 1.8 deg (Neptune orbital inclination)
o 2073 mkm = 175.94 mkm x 11.8 deg (11.8 deg=6.7 deg the moon axial tilt
+ 5.1 deg = the moon orbital inclination) but (175.94 days = Mercury day
period, where Venus moves during 365.25 days a distance=2π x 175.94mkm
o 2073 degrees = 1440 x 1.44 deg (the moon orbit regression monthly)

I do this research
2nd
20
o 2073 degrees=12.195 degrees x 170 where (170 deg = 6.7 deg x 25.2 deg
Mars Axial Tilt = 11.8 deg (the moon axial tilt + orbital inclination) x 14.4
degrees (in the moon orbital triangle this angle = E angle (13.3 deg +1.1
deg) the angle under E
- Please remember, Uranus axial tilt (97.8 degrees) has an angle (91.1 degrees)
with the moon axial tilt (6.7 deg).. but Uranus is perpendicular on the moon orbital
triangle based (EA =449197 km), because Uranus creates an angle 1.1 degrees
between the triangle base and the horizontal level (where this horizontal level is
the moon axial tilt 6.7). and this angle (1.1 degrees) is found under (E) as shown in
the moon orbital triangle. So (E angle =13.3 degrees +1.1 = 14.4 degrees)
Equation No. (12)
2073 degrees/1259.3 degrees = 327.6 /540
- 327.6 days = the lunar sidereal year
- 540 seconds x 1.16 mkm/sec (light supposed velocity) = 627 mkm = Earth Jupiter
distance
- Equation no. (4) makes the days are equivalent to seconds, it's an application my
research hypotheses let's review them here…
o 1st
hypothesis (Planet motion depends on light motion)
o 2nd
hypothesis (Light Motion For 1 Second Cause Planet Motion For 1
Solar Day)
o The motion rule (Light Moves Along A Circle Circumference But Its
Follower Planet Moves Along Its Radius)
- I try to show that, it's a complex machine by which the planets move, i.e. planet
motion can't be considered as result of rigid body mechanics, because the light
motion is part can't be departed from planet motion and planet motion depends on
light motion…
Please Note. the Equation (2073 degrees = 2073 mkm = 0.99 x 2094 mkm)

I do this research
2nd
21
2-3 The Moon 2nd
Orbit of Motion
We still need more analysis for equation No. 2
Equation No. (2) (continued)
- What's the key of this equation? It's the value 91.9 degrees
- We have no this value in the moon orbital triangle, Uranus Axial Tilt (97.8 deg)
creates and angle =91.1 degrees with the Moon Axial Tilt (6.7 deg) and this angle
91.1 deg is decreased by the moon diameter with 0.5 degree and by the green box
(in the figure) by 0.6 degrees and then the angle between the triangle base (EA)
and Uranus axial tilt will be 90 degrees
- So how to find this angle 91.9 degrees we need 0.8 deg +91.1 deg to produce this
angle 91.9 degrees? So from where we can find this angle (0.8 degree )
- From The 2nd
Orbit Of The Moon Motion
The 2nd
Orbit For The Moon Motion:
- The moon moves a motion typical to Earth motion, to avoid the separation from
Earth during their motions…So
- The moon moves per solar day a distance 2.58 mkm with an angle =0.985 degrees
on the horizontal level, along Earth orbital circumference revolving around the
Sun.
- If there's no Lorentz Length Contraction phenomenon in the solar system, the
moon motion would to be a parallel line to Earth motion trajectory
- But because of Lorentz Length Contraction phenomenon with rate (1.0725), the
moon motion distance (2.58 mkm) be contracted to be (2.41 mkm)
- The moon needs to cover the different distance (0.17 mkm), because of that, the
moon moves its daily displacement (88000 km) depending on Earth gravity force
but this displacement isn't enough to cover the different distance

I do this research
2nd
22
- Because the four inner planets masses total = Earth mass, I supposed that, these 4
inner planets practice a gravity equal Earth Mass Gravity on the moon orbital
motion – and because of the distances between these planets and the moon, the 4
inner planets do this effect through Jupiter gravity
- Jupiter effect on the moon motion is done by an interaction between Jupiter gravity
and the sun gravity effect on the moon – and by that- Jupiter creates a 2nd
force
effects on the moon orbital motion- this second force effects on The Point (A) in
the moon orbital triangle – by this 2nd
force Jupiter pulls the moon to apogee radius
(r= 0.406 mkm) while earth gravity tries to save the moon on the perigee point
(r=0.363 mkm).
- Jupiter effect on the Point (A) causes the moon to move another displacement
(88000km) which is required for the moon motion to recover the different distance
(0.17 mkm) – that means – the moon per solar day moves 2 displacements, one of
them (88000 km) is done through the moon orbit around Earth but the second
(88000km) isn't seen by us! why??
- The moon original motion (2.58 mkm) is done on the Earth orbital circumference
trajectory revolving around the sun, and we don't see this motion- but it's a real
regardless our vision because if the moon doesn't move this distance daily it would
be separated from the Earth motion…
- The 2nd
displacement (88000km) the moon moves along this same trajectory of
Earth orbital circumference revolving around the sun – and because of that we
can't see this displacement also.
- Based on that 2 displacements (88000 km x 2) the moon moves per solar day, one
of them we see in the moon orbital triangle and where's the space through which
the moon moves its 2nd
displacement (88000 km) - means – there's one more
orbit for the moon motion through which the moon moves its second displacement
and where's this second orbit of motion? how can we discover this orbit position?

I do this research
2nd
23
Equation No. (2) (continued) (again)
We have 2 features to define this 2nd
orbit
(1st
Feature)
- The angle 91.9 degrees =91.1 degrees +0.8 degrees
- We have the angle 91.1 degrees and then will add also 0.8 degree and this will
be the moon orbit direction …
- Where is this 0.8 degrees? Let's remember figure No.2
o The angle above the diameter R2 =2085 km is equal 90.8 degrees (the
decline angle of Uranus Axial Tilt on the moon diameter)
o But
o The angle between The vertical blue line (Uranus Axial Tilt) and the red
horizontal line (the triangle base 449197 km) =90 degrees
o The difference =0.8 degrees
o So we have 2 moon orbital triangles – from the first one we use 91.1 degrees
– and from the second orbital triangle we use 0.8 degrees – means we stand
on the R2= 2085 km
o This cooperation will produce the angle 91.9 degrees ….means
- The 2nd
orbit of the moon motion declines with 0.8 degrees on the first one and by
that we defines the 2nd
orbit direction…
(2nd
Feature)
- The 2nd
orbit should be a neighbor for the 1st
one – because the 2 moon motion
orbits together define the lunar eclipse umbra length (1.392 mkm)
- In the moon orbital triangle, after the point(A) in the opposite direction to of Earth,
the distance from this point (A) to the end of the lunar eclipse umbra length (1.392
mkm) defines the second orbit…

I do this research
2nd
24
- Shortly
- The distance from the point (A) to the End of the lunar eclipse umbra = 1.392
mkm – 449197 km = 942803 km
- And this value 942803 km = the moon orbital triangle (ABC) Perimeter,
- That means the point (A) separates between 2 orbits of the moon motion,
- That also explains why this triangle (ABC) perimeter x 2 =1.88 mkm which is
equivalent to 1.9 degrees (Mars orbital inclination).
- But why the value 1259.3 degrees is defined by the moon motion?
Equation No. (13)
- 180.8 degrees – 91.9 degrees = 88.9 degrees
- 180.8 degrees = 177.4 deg (Venus axial tilt) + 3.4 deg (Venus orbital inclination)
- 91.9 degrees our angle (Equation n. 2)
- 88.9 degrees is the angle ECZ
- Why this angle 88.9 deg is distinguished? Because the hypotenuse CZ =88000 km
the moon daily displacement –
- That tells, Venus data (177.4+3.4) is created in this interaction under the moon
orbital triangle – basically Venus diameter =1.774 x Mars diameter and Venus
axial tilt be 1.774 x 100 = 177.4 deg because of the interaction occurred under the
moon orbital triangle and by that Venus axial tilt became 177.4 degrees and
because of the Green box in the figure (shows Jupiter and Saturn Interaction),
because of that, a deep interaction is happened between Saturn and Venus based on
it the value 179.9 deg is created (177.4 deg +2.5 deg) – and the moon orbital
inclination be created in consistency with this data 180-177.4 =2.6 x 2=5.2 (where
5.1 the moon orbital inclination + 0.1 degree is consumed for mars orbital
inclination which was 1.8 degrees and became 1.9 degrees)

I do this research
2nd
25
Equation No. (14)
- 95.6 degrees =90+5.6 degrees and 5.6 degrees = 0.5 degree +5.1 degrees (the
moon orbital inclination)
- 5.6 degrees, means, the moon orbital inclination is measured above the moon
diameter (where the moon angular diameter consumes 0.5 degrees)
- 13.18 degrees = The Moon Motion Daily
- Equation no. (14) tells that, the value 1259.3 degrees is defined based on the moon
motion data!
- But
- Vertically…. (90+5.6 =95.6)
- It's the basic feature of this angle (1259.3 degrees), we work here under the moon
orbital triangle, under the angle 90 degrees, and because that, the perpendicularity
is seen in every used data…
Conclusion
- The Moon Has 2 Orbits For Its Motion
- The second orbit is declined by 0.8 degrees on the 1st
orbit (the moon orbital
triangle)
- The 2 orbits are neighbor and define together the lunar eclipse umbra length (1.392
mkm)
- Venus axial tilt is changed from 1.774 to 177.4 degrees because of Uranus axial tilt
angle (91.1 degrees) with the moon axial tilt (6.7 deg) and because of Uranus axial
tilt perpendicularity on the moon orbital triangle base.

I do this research
2nd
26
2-4 Uranus Effect on Venus motion
- The paper conclusion tells us that, Venus axial tilt was 1.774 deg and because
Uranus Axial Tilt effect on the moon axial tilt, Venus axial tilt became 177.4
degrees – so this point of discussion may help us to find some confidence in this
conclusion – because we try see what effect Uranus practices on Venus.
Equation No. (15)
35 km /s x 12104 seconds = 423640 km (421056 km) (0.6%)
- Venus moves during (12104 seconds) a distance = Uranus motion distance during
its day period
- We know that, 12104 km = Venus Diameter
- Equation no. (15) tells us that, there's some consistency between Uranus & Venus
motions – this feature is added to more other features shows a consistency in
motions between Uranus and Venus…
Equation No. (16)
35 km /s x 120536 seconds = 4218760 km (421056 km x 10)
- Venus moves during (120536 seconds) a distance = Uranus motion distance during
its day period x 10 days
- We know that, 120536 km = Venus Diameter
- Equation no. (16) shows more consistency between Venus, Saturn and Uranus
motions
Equation No. (17)
35 km /s x 12756 seconds = 446460 km (449197 km)
- Venus moves during (12756 seconds) a distance = Jupiter Circumference = the
moon orbital triangle base AE
- Equation no. (17) tells us that, there's some consistency between Jupiter & Venus
motions – also – there's a deep consistency between Venus and the moon orbital
triangle data

I do this research
2nd
27
Equation No. (18)
35 km /s x 51118 seconds = 1789130 km (=1.8 mkm) (0.6%)
- Venus moves during (51118 seconds) a distance = 1.8 mkm
- 51118 km = Uranus Diameter
- We know that 1 mkm =1 degree
- 1.8 degrees = Neptune Orbital Inclination
Equation No. (19)
35 km /s x 10921 seconds = 382235 km
- Venus moves during (10921 seconds) a distance = the moon orbital distance
- That shows more consistency between the moon and Venus motions –
Equation No. (20)
177.4 deg (Venus Axial Tilt) / (12.195 deg the moon motion angle) =14.54
We know this rate
Uranus Diameter 51118 km / The Moon Diameter 3475 km = 14.7
Uranus Axial Tilt 97.8 deg / The Moon Axial Tilt 6.7 deg = 14.7
Uranus Mass / Earth Mass = 14.7
Venus Mass /Mercury Mass =14.7
2 Saturn diameters / Uranus diameter =14.7
Jupiter Radius / Mercury diameter =14.7
- Equation No. (2) shows the deep consistency between Venus & Uranus Motions.

I do this research
2nd
28
References
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJ&hl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to The Moon Orbital Triangle (General discussion)(I) (Revised)

Similar to The Moon Orbital Triangle (General discussion)(I) (Revised) (14)

More from Gerges francis

More from Gerges francis (20)

Recently uploaded

Recently uploaded (20)

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