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Can The Motion Be Transported From Mercury To The Earth Moon?

I- The Moon Daily Motion Distance

The moon moves per solar day a distance = 2.574 million km = The Earth motion distance per solar day - but we don't see this motion of the moon-

The idea is proved– the Earth and the moon revolve together around the sun and that necessitates the two planets to move equal distances per solar day.

The idea tells us 2 facts

First Fact – the moon moves per solar day a distance = 2.574 million km

Second Fact – we don't see this motion of the moon

The Contraction Effect With Rate 1.0725

There's a contraction effect with the rate 1.0725 in the solar system, I proved it in point no (7-4) 

Shortly

The moon moves per solar day a distance 2.574 million km = the Earth motion distance per solar day- BUT- the rate (1.0725) has effect on the moon motion distance but has No effect on the Earth motion distance –

As a result – the moon distance 2.574 million km be decreased to 2.4 million km –

By that there's a difference =176000 km

Means

The Earth motion distance per solar day is (2.574 million km) and the moon motion distance per solar day is (2.4 million km) and by that there's a difference =176000 km between the two distances which will cause the moon to separate the Earth in their motions

Means-  the moon has to move this 176000 km (on the same solar day) otherwise the moon will be separated from the Earth in their motions    

How can that be possible? Let's answer in the next point

II- The Transported Motion From  Mercury To The Moon

I discovered that–

The distance 176000 km is passed by the moon under effect of Mercury motion-

Means- the motion is transported from Mercury to Venus and from Venus to the moon and this transported motion pushed the moon to pass this required distance 176000 km.

Here I need to suppose that – the transported motion among the 3 planets is a motion different from the planet original motion- means

The planets move their motions normally– but – the transported motion is an addition one- this motion is Pluto transported motion and Mercury transported this motion to Venus and Venus transported it to the moon  

Also I have discovered a perpendicularity in the inner planets (proved in point no.7-4)

(notice we see the moon displacement daily 88000 km in the Earth sky but we don't see the second part of the motion)

DATA

(1)

(90/2π) x 4.095 million km daily = 58.65 million km

(2)

58.65 days x 4.095 million km daily = 243 million km (error 1%)

(3)

243 days x 3.024 million km daily = 2 x 365.25 million km

(4)

4224 days x 176000 km = 2 x 365.25 million km

(5)

116.75 days x 2.082 million km daily = 243 million km

But (116.75/π) = (9010/243)

58.65 days   = Mercury Rotation Period

243 days      = Venus Rotation Period

116.75 days = Venus Day Period

365.25 days = Earth Orbital Period

9010 mkm  = Saturn Orbital Circumference 

DISCUSSION

Let's summarize the idea of  the (three first data)

Each planet motion defines the next planet cycle period but in a distance form

Shortly

Mercury moves in ((90/2π)= 14.3 days) a distance = 58.65 million km but this distance used as Mercury rotation period (58.65 solar days)  

And

Mercury moves during its rotation period (58.65 days) a distance = 243 million km (1%) which is used as (243 solar days = Venus rotation period)

And

Venus moves during its rotation period (243 days) a distance= 2 x 365.25 million km (0.6%) which is used as (365.25 solar days = the  Earth orbital period)

The Earth orbital period is also the moon orbital period around the sun

That shows we have a transported motion among the three planets where the transportation depends on the planets cycles periods

Data no. (4)

4224 days x 176000 km = 2 x 365.25 million km

This data is the important one – because

The value (2 x 365.25 million km) is produced by Venus motion during its rotation period – means – this is the transported value from Venus to the moon – typically – as the motion transportation from Mercury to Venus

And

The distance 176000 km is the required distance for the moon motion- as we have discussed in the previous point-

Shortly- the transportation of  motion is done for this distance (176000 km) basically

And

4224 hours = 176 solar days = Mercury day period – but

One hour of Mercury motion = 24 hours of any planet motion – for that – the period 4224 hours (for Mercury) is seen as (4224 solar days) for the moon

Data No. (5)

116.75 days x 2.082 million km daily = 243 million km But (116.75/π) = (9010/243)

2.082 million km daily = Mars velocity

116.75 days = Venus day period

The data tells the transported motion reach also to Mars – but – we should remember – the energy is reflected from Venus to Mars by that the transported motion among the three planets (Mercury, Venus and the moon) is done on one side but the transportation to Mars is done by reflection of energy – here we have 2 groups

Mercury, Venus and the moon one group where Mars is another group 

III-  The Moon Motion Direction Definition

The next question asks

Let's suppose Mercury motion causes the moon to move this additional distance 176000 km- there are still two questions (first question -  is the motion in the same direction) (second question – why do we see the moon moves 88000 km only?)

Let's answer in details in following

(a)

We agreed that – the moon original motion (2.574 million km) is Not seen by us – but –it's proved fact – because the Earth moves (2.574 million km) and the moon has to move this same distance to save the complying motion with the Earth

Now - this distance 2.574 mkm is decreased by 176000 km and be (2.4 million km)

(b)

Mercury motion causes the moon to move this 176000 km – but – understandable that the motion is (in different direction) from the moon original motion – for that reason we enable to see the moon moves in the Earth sky – we can't see the original motion – that proves the motion of (176000 km) is done in different direction

Now let's – If the motions are done in different directions- how the distances can be added to one another?! 

(c)

Also we don’t see the motion of this distance (176000 km) but only we see the half of it (88000 km)- why can't we see the whole distance??

Shortly

The moon modifies the motion direction during its motion and by that the moon can add the distances to one another- this fact is proved strongly because the moon orbital apogee radius is changed by this modification – I should prove later.

Means - the transported motion is in different direction but the moon modifies this direction by using an intelligent technique in its motion – this technique is explained in the next point – but – for the second question – (why don't we see the distance 176000 km but only 88000 km?) this question I can't answer – because – there are 2 options for solution –

First option – the second part (88000  km) may be passed with the original motion direction and by that we can't see it because we can't see the original  motion     

Second option- Mercury velocity = 2 Mars velocity – it may have some connection between the planets velocities – means – the two planets velocities are defined based on one another and be used together

And based on that the distance 176000 km be passed in the space by the moon motion for a distance =88000 km – now let's explain the moon intelligent technique

IV- The Moon Orbital Apogee Radius Definition

Why Is The Orbital Apogee Radius = 406000 Km?

The moon daily displacement =88000 km and during 29.53 days (the moon day period) the displacements total be = 2.598 million km = 2π x 413600 km

The data tells us the moon orbital apogee radius should be 413600 km and also it tells, because the moon daily displacement (88000 km) is so long, the moon should revolve around the Earth through this apogee orbit its radius (413600 km) only and can't revolve around the Earth through any more near orbit…

Not Facts

The moon orbital apogee radius =406000 km only and the moon revolves around the Earth through near orbits and can reach to perigee radius (363000 km). 

How Can The Moon Do That?

The intelligent moon creates an angle (θ) between its motion direction and its orbit horizontal level by that the real displacement (L) through the orbit be less than (88000 km) because it be (L = 88000 km cos θ), as a result the total displacements be less than (2.598 million km) and that makes the moon orbital apogee radius to be decreased from 413600 km to 406000 km.

We should pay attention to the angle (θ), because this angle controls the moon motion features – where- with the angle (θ) increasing the real displacement (L) be shorter and the moon can revolve around the Earth through more near orbits – but –with the angle (θ) deceasing the real displacement (L) be longer and that pushes the moon far from the Earth to more far orbits.

The moon orbital motion depends on this angle (θ) it tells θ1 = θ0 +1.7

where (θ1) = today angle and (θ0) =yesterday angle

1.7 degrees be used as the moon daily motion degrees for the equation

Notice No.(1)

As a result for the moon using of the angle (θ), the moon orbital radiuses be defined based on Pythagorean rule – because – the moon uses the angle (θ) in its motion – by that – Each point the moon passes shows this fact – and the radiuses be defined based on one another by Pythagorean rule – let's prove that

(363000 km)2 + (86000 km)2 = (373000 km)2

(373000 km)2 + (86000 km)2 = (384000 km)2

(384000 km)2 + (86000 km)2 = (392000 km)2

(392000 km)2 + (86000 km)2 = (406000 km)2      (error 1%)

Where

363000 km = The Moon Orbital Perigee Radius

373000 km = The Total Solar Eclipse Radius

384000 km = The Moon Orbital Distance

406000 km = The Moon Orbital Perigee Radius

Based on that – we see the moon motion uses Pythagorean rule

Notice No.(2)

The using of the angle (θ) causes to modify the motion direction during the month – where we know the directions aren't changed by that the moon technique can modify the angle gradually till enable the distances to be added one another.

Notice No.(3)

149.6 million km cos (26.4 deg) =413.6 million km  = 1000 x 413600 km

And 413600 km cos (26.4 deg) = 363000 km

149.6 million km  = Earth Orbital Distance   

1000  = The rate between Mercury motion and Venus motion   

The data tells the orbital apogee radius 413600 km is defined based on the Earth orbital distance (149.6 million km)- but – the intelligent moon modifies its position to enable itself to revolve around the Earth through more near orbit   

V- The Transported Motion Effect (Planet Diameter Definition)

The analysis proves there's a transported motion from Mercury to the moon- and this  motion is so important for the moon to compensate the contracted distance from the moon orbital motion distance (176000 km)- by that-  the moon motion depends in fact on Mercury motion-

Mercury motion transportation necessitates to depend on Venus and by that a system is created between the three first planets based on this transported motion which helps the moon to move the additional required distance

But

Can we have a description explain how this motion is transported clearly?

Imagine we have a canal of moving water– and – on some point a perpendicular rod pushes the water – that will create A WAVE in water because we have 2 motions perpendicular on one another – then - (the wave is transported through the water) - this wave creates the planet (matter)- means- The matter is consisted of 2 motions perpendicular on one another– by that–This wave creates the planet DIAMETER

The idea is simple but its effect is massive –

The matter is created by 2 motions perpendicular on one another- by that –the perpendicularity on Mercury (point) created Mercury and defined its Diameter - 

This is the main story in this discussion because the transported motion moves through the planets - means – the transported motion for the distance (176000 km) is just a cover for the transported motion among the planets which defines each planet diameter.

We should notice that- because the planets motions use different rates of time- the transported motion transport also the motion rate of time 

For example-Mercury rate of time (1=176) is transported to the moon motion because we need to add the moon velocity (29.8 km/s) and by that the total velocity is 205.8 km/s and this velocity creates the rate (1=205.8)

NOTICE

We review the 3 features for the moon, Saturn and Pluto

(I)

We know

The moon feature (T1=T2) which is (the moon rotation period = the moon orbital period =27.3 days)- this feature creates a proportionality between the moon diameter and its orbital circumference – means – it creates a proportionality between the moon diameter and the planet diameters total (406000 km)

The three planets diameters are defined based on this distance (406000 km) prove the transported motion (2593836 km) created the same one effect on the three planets – even if this effect is seen in different feature for each planet.

(II)

Saturn features is (V1=V2) where

Saturn (9.7 km/s) moves in its rotation period (10.7 h) a distance = Saturn Circumference (error 1.3%)–means- Saturn Rotational Velocity= Saturn Orbital Velocity = 9.7 km/s

Now

Saturn Circumference = 378675 km = 406000 km /1.0725

The rate (1.0725) has a general effect on the solar system I have proved before. 

By that – Saturn Circumference is created based on the same distance =406000 km = the planets diameters total- that shows it's the same one link between the three planets

We note

J (3- (1/2π) = 406000 km (The Planets Diameter Total)

J (1- (1/2π) = 120536 km (Saturn Diameter) –

where

J = Jupiter diameter 142984 km

The data shows Saturn diameter is created by the same rule defines the planets diameters total.

(III)

Pluto feature is (D1=D2)- means – Two Distances Should Be Equal 

What are these 2 distances? Let's review the other planets features

The moon rotation period = the moon orbital period (T1=T2) and

Saturn rotational velocity = Saturn orbital velocity (V 1=V2)

Let's conclude

Pluto motion distance in its rotation period (D1) should be =  Pluto motion distance in its orbital period (D2)! It's illogical

Pluto moves in its rotation period a distance D1=2593836 km and

Pluto moves in its orbital period a distance D2=37100 million km

37100 million km = 2593836 km x 14178

If D1 =  D2 that necessitates (14178 = 1)! How can that be possible?

By the rate of time –

If one second of the moon motion be =  14178 seconds of Pluto motion –

The moon orbital circumference 2593836km will be equal Pluto orbital circumference 37100 million km – so we need the rate of time (1=14178) let's try to find it in following

(IV)

The planets motions use different rates of time–I have proved that in point no.(10)

And

The definition of the rate of time is known –it means – each motion has its own clock- for example – one hour of Mercury motion = 24 hours of any planet motion – means – while Mercury clock moves one hour any other planet clock will move 24 hours –

Here we have a rate of time (1=14178) that means we need 2 clocks – one clock moves (one second) while the other clock moves (14178 seconds) – and this rate of time is used for periods but because of the reflection of the energy we see both as distances – for that – this rate is used between the moon orbit and Pluto orbit.

There are 2 methods to create the rate (1=14178) which are

(First Method)

The final rate of time between Mercury and Mars is (1=2840) means – each second of Mercury clock be = 2840 seconds of Mars clock

Now

Mars  velocity = 5.1 Pluto velocity and the rate of time is produced as velocities rate – means – each one second of Mars = 5.1 seconds of Pluto – this is the basic rule to define the rate of time – by that (14178 = 2840 x 5.1) -  means

(1 second of Mercury Clock be = 14178 seconds of Pluto Clock)

We have succeeded to create the rate of time (1=14178) but between Mercury and Pluto while we needed this rate between the moon and Pluto!

(We Can Consider Mercury Is A Partner In  The  Moon Motion)

(Second Method)

One second of Venus Motion = 48 seconds of Saturn motion

(because 10747 days Saturn orbital period = 48 x 224.7 days Venus orbital period)

Uranus orbital inclination (0.8) has effect on this relationship as following

(48/0.8) =60  

Means –

One second of Venus Motion = 60 seconds of Saturn motion

And the energy is reflected from Venus to the moon and caused to reflect the players 

Means –

One Second Of The Moon Motion = 60 Seconds Of Venus Motion

As a result

One Second Of The Moon Motion = 3600 Seconds Of Saturn Motion

And

14178 = 3600 x 3.93

Shortly

The rate tells (one second of the moon motion = 14178 seconds of Pluto motion) and in details the rate be

one second of the moon motion = 3600 seconds (one hour) of Saturn motion and

one second of Saturn motion = 3.93 seconds of Pluto motion – let's prove that

(i)

as we have seen the distance 2593836 km is seen as a distance for the moon and as a period 2593836 hours for Saturn- means – (one km of the moon = one hour of Saturn)  but the moon daily displacement is 88000 km and its velocity is almost (1 km/s)

means – one km per hours for Saturn is equivalent to one km per second for the moon which necessitates to conclude the rate of time (1s of the moon =3600 s of Saturn)

(ii)

Pluto orbital distance 5906 km = 3.93 x 1503 million km

(Saturn orbital distance is 1433.5 million km has error 5%)

That shows the rate 3.93 works between Saturn and Pluto also

(V)

The three planets diameters are defined based on the distance (406000 km= the planets diameter total) – let's provide the data in following

366556 km = 2390 km (Pluto diameter) x 153.3

366556 km = 3475 km (The moon diameter) x (655.7/2π)

Where

153.3 hours = Pluto Rotation Period

655.7 hours = The Moon Rotation Period

366556 km = The Outer Planets Diameters Total

The data shows the moon and Pluto diameters depends on the 366556 km

And Saturn diameter depends directly on 406000 km  based on the rate 1.0725 as have discussed before.

Thanks a lot – kind regards

Please read

Planet Diameter Equation Shows The Matter Origin (The Proves Discussions)

or

Gerges Francis Tawdrous +201022532292

Physics Department-  Physics & Mathematics  Faculty 

Peoples' Friendship university of Russia – Moscow   (2010-2013)

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Bag It!

Quick math game in GeoGebra. One of the ways we try to teach elementary operations now is to emphasize the story. As classified by the Cognitively Guided Instruction (CGI) researchers, there are two division actions, partitive/fair share and quotative/measure. When kids read a grouping story, like we have 32 marbles in 8 bags of 4, they act it out differently depending on whether we don't know the 32 (multiplication), the 8 (measure) or the 4 (fair share).

Mostly division stories are fair share in traditional instruction, but measure is super helpful for learners seeing the connections between addition and multiplication, and using addition to solving division.

Very little game structure here, so not much replay value, but adds a little motivation to the story string.

War

So I was brainstorming today’s math comic and it gave me an idea for a game. I am not a fan of War, because there are no choices. The number comparison is good, but as a game... I usually have preservice teachers play More or Less, instead. In that variant students draw three cards then take turns declaring “more” or “less.” When you call more, the bigger card wins, less, the smaller card.

Then Nick Smith hopped into a Hangout to quick-test it this new idea with me. Not bad...

Obstacle Course

Divide the deck between the players. Each player draws a hand of 5 cards. 

On your turn, you make the obstacle. Play a card and declare higher or lower. The other players play whichever cards they want that qualify. (Playing a card is not mandatory.) If I play a 10 and call higher, the other player can play whatever Jacks, Queens and Kings they have. (Ace is low.) After you set the obstacle, draw back up to 5 cards. If you’re clearing the obstacle and play all your cards, draw a new 5 and continue play. Then the next player makes an obstacle.

Special play: if you have an equal card, you can reverse the obstacle. Change the challenge from higher to lower or vice versa, and change the order of play if there are more than 2 players.

Derivative Guessing Game

Play in GeoGebra. 1st derivative was a little bit of thinking, but 2nd really got me thinking. 

I’m sure this exists somewhere, but I’ve never seen it. I love the Correlation Coefficient game, though.

The 2048 game is a global phenomenon, captivating millions of players worldwide with its simple yet addictive gameplay. Since its release, this puzzle game has become a staple for both casual and strategic gamers alike. But when was the 2048 game released, and how did it rise to fame? In this article, we’ll dive into the history of the game 2048 online, its development, and why it continues to be a beloved challenge for puzzle enthusiasts.

The Birth of 2048: When Was It First Released?

The 2048 game was first released in March 2014 by Italian web developer Gabriele Cirulli. Originally created as a personal project, the game gained instant popularity due to its simple mechanics and engaging gameplay. It quickly became a viral sensation, spreading across various platforms and sparking numerous clones and variations. The game’s minimalist design and challenging nature are key factors in its widespread appeal.

How 2048 Game Became a Global Sensation

From its humble beginnings, the https://2048game.games/ transformed into a globally recognized puzzle game. Its addictive nature and easy-to-understand rules contributed to its rapid success. What made 2048 so unique was its ability to challenge players to think strategically while keeping the gameplay approachable for all ages. Unlike many other puzzle games, 2048 didn’t require any special skills — just logical thinking and patience.

Why You Should Play 2048 Game

There’s no doubt that the 2048 game is a fun and mentally stimulating experience. But beyond entertainment, playing 2048 can sharpen your strategic thinking. Here’s why it’s worth your time:

Suitable for All Ages: Whether you’re a student, a professional, or a retiree, 2048 is accessible and fun for everyone.

Mental Workout: The game not only entertains but also exercises your mind, making it a great way to enhance your cognitive abilities.

No Downloads Required: Play 2048 directly on your computer or mobile device without the need to download any app.

Challenge Yourself: After reaching the 2048 tile, continue to push yourself with higher numbers like 4096 and 8192.

Gameplay Mechanics: How to Play 2048 Online

One of the key features of the 2048 game is its simplicity. Here’s how to play the game on your computer or mobile device:

Use Arrow Keys: Move tiles up, down, left, or right using your keyboard’s arrow keys.

Merge Tiles: Match two tiles with the same number to combine them into a higher-value tile.

Reach 2048: The goal of the game is to create a tile with the number 2048. But don’t stop there — keep playing to achieve higher scores!

Tips to Master 2048 Game

Mastering 2048 requires both skill and strategy. Here are a few tips to help you achieve high scores:

Focus on One Corner: Always try to keep your largest tile in one corner.

Avoid Vertical Moves: Limit your vertical moves to avoid disrupting your tile arrangements.

Plan Ahead: Think ahead about your moves to prevent the board from filling up.

The Legacy of 2048

Over the years, the 2048 game has maintained its place as one of the most popular online puzzle games. The game has been adapted to various platforms, and many different versions have been created, but the original version remains iconic. Whether you’re playing it on your phone or computer, 2048 continues to provide an endless challenge for players of all skill levels.