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• 1. Photon Size and Weight

• 1.2 Viscosity of Light and Photons per Cubical Metre

• 1.3 New Photon Modell and its Prove

• 1.4 ' Golden Ratio plus 1 ' – ( Division Delta of Frequency ) per ( Division Delta of Diametre )

• 1.5 Photon Shell Mass and Photon Core Mass

• 1.6 Past Research : An antique Way to Calculate Light Speed ...

• 1.2.1 Light Particles per cubical metre and ' Ratio ' – Approach 1

1.2.1.1 Light Particles per cubical metre and ' Ratio ' – Approach 1 . 1

• 1.2.2 Light Particles per cubical metre and ' Ratio ' – Approach 2

• 1.2.3 Light Particles per cubical metre and ' Ratio ' – Approach 3

• 1.2.4 Light Particles per cubical metre and ' Ratio ' – Approach 4

• 1. Photon Size and Weight

' Photon Shells orbiting an electromagnetic Core . 2020 / 2021. '

Understanding a Photon as an ' electromagnetic Particle Gemini System ' .

Whereas a shell of negative mass spins around a core of positive mass

and the strength of electromagnetic attraction stucks both the components

together while the speed our universe expands with counteracts the pull

so that adding energy grows the radius and frequency of spins .

The core is pulled by the shell ' s spinning movement also and since the ratio

of positive and negative mass is ' close to ' 1 we might speak about a

' gemini satellites system ' . This knowledge might to be for the benefit of

Quantum Computers Technology and the Progress of Health and Medical Care ...

With your donation to ' Peta ' you are going to contribute to fundamental Light Research !

And that is the reason why your donation does really help to go ahead with true progress :

We have done an experiment during the weekend a few weeks ago

and it seems to be a further proof for the electromagnetic particle

consisting of both , positive ( matter ) and negative mass ( antimatter ) :

We hold a magnet very close to our ( right ) eye with the other eye closed ,

and we hold the magnet very close to the sun visible in our pupil ;

Now , we focus on the little specular highlight of our sun on the magnet :

The sun appears dark , since the photon particles are attracted

of the strong magnetical force and redirected from its natural source ( sun ) .

Also , light is split up to all of the spectral colours at some differing points

which can be seen best with a simple glow or halogen lamp and a magnet .

If we let sunlight shine through a very tiny – but massive – aperture ,

spectral colours appear too , because different masses of light of varying colours

are attracted by the mass of the aperture and light is separated to the spectral colours hence ...

Important Note :

Do not focus your view on the light source directly , because that might damage the retina !

Watching the reflected , indirect or redirected light might not do harm to your eye ...

' Shell Super Quantums consisting of antimatter – the matter of a photon ' s core is spinning with ... '

This new Photon Modell understood as Super - or Sub Quantums spinning around a core

has become more likely during these days ...

Each movement of a medium or object is circular :

From waves at the sea shore

( sometimes broken down to half of a circle

because of gravitational force and inertia )

over atomes with electrons orbiting a core

to satellites , planets , stars , galaxies and even more , like clouds shaped by air movement.

' Photon Size 2021 – Golden Ratio Approach ' .

Another reconstructed piece of the Library of Alexandria ...

A valid physical and mathematical prove for this piece of rework

is in progress , and we got to mention that it poses a reconstruction

based on an assumption of some skills during the High Antique ...

A lot of understandings created during the High Antique have been lost

and are difficult to get in today ' s ones – sometimes completely different working – complexion of thoughts .

With some strong assumption for a remembrance to be able to be inherited to our children by pure genetical information ...

' Photon Size 2020 . Frequency Approach '

When extracting the square root of 2 it is not just to our liking ;)-

Calculating with the Frequency Approach , using light speed , pi and frequency – for blue light :

frequency ( f ) of blue light : 749481145000000

f red : 428274940000000

delta : 1,75

f * pi * c = 705880666723841961469264,62058547

root of 2 / { ( f * pi * c ) squared } = 2,8382616123699438343626193703283e-48 | ( root of ) and ( * 2 )

= 3,3694282080910665008361241174928e-24

delta blue light Wavelength Approach :

2,5374868391178390999596468399708

1,75 square per 2 plus 1 = 2,53125

Wavelength Approach blue light :

116960709528514642620271,49440552

= 8,5498797333834849467655443627193e-24

The ' Frequency Approach ' has led our idea to determine

the value ' 0 , 5 ' for ' E = 0 , 5 * mass * velocity squared ' exactly ,

and that is still a thing we are not a hundred per cent sure

about it to be physically correct , but we are going to give this a trial :

Since we are calculating the size of an electromagnetic particle by

2 * [ root of { ( root of x = ~ root of 2 ) per ( frequency * pi * light speed c ) squared } ] = ( 1 ) ( = photon size )

by the ' Frequency Approach ' we can calculate ' x ' since

the exact Photon Size is known by the ' Wavelength Approach '

and we see ' x ' as 1,9878862577579345578272656386017 ...

The reciprocal value is 0,50304689018166665544978282282439 ...

' Photon Size 2020 . Classic Approach – Graphical Solution ' .

Below , we are going to see the entire way of the Classic Approach :

' Classic Approach '

This ' Classic Approach ' results cubical metres for expanded Light Speed .

By using the cubical root of mass delta Sun as the value for temperature delta ,

we might have a look at other Stars like ' R 136 a1 ' or ' Sirius A ' and ' Rigel ' et cetera :

' Right click on the graphic in the web browser firefox to enlarge '

Using Earth weight we are going to calculate

with the temperature delta

' Approach 2 ' ( see below ) delivers .

The temperature delta of the light emitting zone

should result values exactly .

We just have found to a solution for calculating

with the higher temperature of White Dwarfs

by some owning a mass same like our Sun ...

For example , if we have a look at the White Dwarf Sirius B :

Temperature Delta of Sirius B and Sun is about 4 ;

Mass is the same approximately .

We regard temperature as Energy :

Temperature = Energy = E = 4 times more than our Sun ;

E = m c c = 4

with c = light speed , m ( mass ) is ' very small '

and can be set to 1 since it is a linear parametre .

Energy should be directly proportional

to speed squared , frequency and a squared radius :

E = Amplitude squared = ( Photon ) radius squared ;

( diametre increases by adding energy )

{ E per ( root of radius r ) } squared = spin speed s ( of the electromagnetic shell )

( 4 * Energy = 8 * spin speed )

1 spin = diametre * pi

1 spin * frequency = spin speed

The speed of light does not change at all –

spin speed of the shell does indeed ...

s = spin speed of the Super - or Sub - Quantums

around the Photon Core = 8

because E = 4 and root of 4 ( = radius ) ;

Resulting a four times higher frequency of the spins

( a shorter wavelength , too ) and

Amplitude = Radius of the Photon = 1 / 2 Diametre

E = Radius or Amplitude squared

E = 4 * Energy of our Sun ;

root of 4 = 2

resulting double photon radius and a four times higher frequency for E = 4

Using the higher mass – but lower temperature – of Red Supergiants ,

we might have seen that energy is less what might mean

lower spin-speed accompanied by longer wavelength , resulting : Red Light !

We will see the radius squared with increasing energy

what goes along with a higher frequency , spin speed and a shorter wavelength hence

which is seen as blue light – we even might talk about a ' Blue Shift ' – the opposite of Red Shift ...

' The part responsible for speeding photons up to light speed

while breaking gravitational influence

should to be the star corona .

Photon shell spin speed increases by higher temperature or energy

while light speed stays pretty much the same ... '

' Assuming energy to increase with mass ... '

1.2 Viscosity of Light and Photons per Cubical Metre

Knowing Light Weight per Cubical Metre enables

the calculation of Viscosity of Light and Photons per Cubical Metre ...

Light Weight per Cubical Metre – Graphical Solution .

Weight of light for 1 squared kilometre results 10 exponent 10 ( 1 / kilogramm )

according to the equivalent .

1 metre is not equal to 1 kilometre – but we are still using ' 1 ' ( for squared kilometres )

1 exponent 2 ( kilometres ) = 10 exponent 6 ( metres )

1 ^= 10 exponent 6 .

So we are going to divide the result of 10 exponent 10 through 10 exponent 6

resulting 10 exponent 4 .

As we are using kilometres exponent 4

( because we multiply kilometres squared by squared speed )

we will take the fourth root of 10 exponent 4 to get weight per

1 kilometre exponent 1 resulting 1 kilogramm .

And as we are using kilometres instead of metres

( but both referred to as 1 unit )

we have to divide the exponent 1

per 3 again because

1 ( kilometre ) = 1000 ( metres )

1 ^= 10 exponent 3 .

And than we are going to take the cubical value to get

weight of light per cubical kilometre in dependence of one second

which should be approximately 1 kilogramm ...

For the ' y - value ' of y = a x x + b x + c = 0

we are going to see the pure Light Density !

1.3 New Photon Modell and its prove

' Photon Spin 2020 . Assuming a new photon modell . '

' Dependency of Energy , Frequency and the Radius of the electromagnetic Particle System . '

' Frequency stays the same despite a higher amplitude . '

' Spin Movement . '

' Wave Movement . '

' Spin Movement distorted . '

The faster the linear movement of the spinning ' gemini ' system , the more wavelike the shape ...

' Wave Movement distorted. '

' Photon Size 2020 – Wavelength Approach . '

1.4 ' Golden Ratio plus 1 ' –

( Division Delta of Frequency ) per ( Division Delta of Diametre )

We would like to precise our latest work

about the complexion of a single photon

by adding the term ' Division Delta '

to the Equivalent of the ratio

( Division Delta of spins of the photon shell ) per

( Division Delta of photon diametre ) .

The ' Golden Ratio ' is defined to be

a / b = ( a + b ) / a .

For the ratio of a photon ' s frequency ( f ) per its diametre ( d ) ,

we are going to use

( 1 + a ) = ( a squared + b squared ) / a squared

( 1 + root of a ) = ( a + b ) / a

whereas

red light ( 700 nm )

diametre ( d ) = 5,8875504282979383133964130248092e-24

blue ( 400 nm )

diametre ( d ) = 8,5498797333834849467655443627193e-24

calculated by the ' Wavelength Approach '

division delta of ( d ) = 1,4521964333909260217517351196628 ( = a )

division delta of ( frequency f ) = 1,75 ( = b )

division delta of ( f ) / division delta of ( d ) = 1,2050711320876149930641848338016

( root of delta r ) * ( delta f )

/

10 exponent

{

exponent of ten from ( root of [ delta r ] ) plus

exponent of ten from ( delta f ) minus

exponent of ten from ( delta r ) minus

exponent of ten from ( delta r )

}

=

Delta Energy E ( by delta r squared )

Down here we will see a first , rough dependency of the photon ' s radii

and frequency ( = Photon Shell Spins ) :

Below , we are going to calculate Shell and Core Mass of a Photon

using three Approaches :

1.5 Photon Shell Mass and Photon Core Mass

' Photon Shell and Core Mass 2020 . Approach A '

' Photon Shell and Core Mass 2020 / 2021 . Approach A polished '

For a mathematical Prove , we will see the ' 1.6 Antique Light Speed ' below ...

Photon Shell and Core Mass 2021 . ' Approach A ' polished & corrected finally :

For a mathematical Prove , we will see the ' 1.6 Antique Light Speed ' below ...

During our last days of work , we have found

to this conclusion what goes calculating photon shell mass

by using the ratio of 1 mass unit per speed ( representing an expanded mass unit ) ...

We are going to start with Energy ( E ) of the shell :

2 E = 1 kg / { ( spin speed ) squared ( = 1 expanded common kilogramm ) }

2 E = 1 kg / { 1 light metre ( 1 ) * pi * frequency ( f ) } squared | root of

root of 2 * 1 E ( E = 1 ) = 1 light metre ( 1 ) * pi * frequency ( f ) | / root of 2

1 kg ( shell kilogramm ) / { 428274940000000 * pi / root of 2 = 951387711883399,00746937580503269 }

We will have the value ' 951387711883399,00746937580503269 ' to the power of four ,

times 5th root of this value , times

2048th root of 5th root of this value ( and so on ... )

what is ~ 8,1457311378889680506301406011433e+62 ;

Now we have to multiply with the actual size in metres

because we have calculated with ( 1 ) light metre

( * 5,8875504282979383133964130248092e-24 )

what results

4,795840284967804625956909340793e+39 ;

We are going to take the square root and the reciprocal value

what is 1,444001499709211787429441403832e-20 ;

As we have divided per ( root of 2 ) at the start to get 1 Quantum of Energy

we have to multiply with ( root of 2 ) =

= 2,0421265049758561761166972293155e-20 ;

And be the cause we have expanded by ' to the power of 4 times fifth root of ( ) times ... '

we have to multiply with { ( root of 2 ) to the power of 4 , times fifth root of ( root of 2 ) ,

times 2048th root of fifth root of ( root of 2 ) ( ... and so on ... ) }

{ = * 4,287275225259577281211680733266 }

What results 8,7551583716289170530802972929092e-20 ( pre ) finally

which is very close to the Values of Approaches B , C and D and

what might near the actual mass exactly with a bit of a patience ...

Then , we might have a look at ' Photon Shell and Core Mass 2020 , ' Approaches B and C ' :

' Photon Shell and Core Mass 2020 – Approach B . '

Photon Shell and Core Mass 2021 – ' Approach B ' refined finally :

We are happy to introduce that we were able to precise ' Approach B '

for calculating a ( red light ) photon ' s shell mass .

Here we go right now with the refined ' Approach B ' !

Energy ( E ) of red light ( with a wavelength of 700 nanometres ) =

2,837779795927041e-19 ( e = exponent )

E = 2,837779795927041 / 10 e19

{ Using E = m * c squared ,

with c = 299.792.458 metres / second

and

mass calculated by

m = h / ( c * wavelength ) }

spin speed = 7,921495422766658494420478482012e-9

{ based on the diametre delivered by the Wavelength Approach and an exact frequency value ( f = c / wavelength ) }

1 E = ( 1 ) = 1 / 1 = ' 1 Quantum ' of Energy ( Energy of the shell )

E = 2,837779795927041e-19

( in E is contained a ( pi squared ) because the spinned distance is circular

beside the linear motion of the whole photon system ,

so we have to divide per ( pi squared ) in order to catch the linear movement )

Since we expand ( 1 ) Quantum of Energy we have to ' divide per ' or ' multiply with ' ( root of pi ) .

( pi squared ) is contained in E ( of m photon and is a delta of c squared )

and ( pi squared ) turns to ' ... * ( times ) or just / ( per ) ( ( 1 ) / ( root of pi ) ) ' :

( 1 ) per { [ ( 10 e19 / 2,837779795927041 ) * ( 7,921495422766658494420478482012 ) squared * ( 10 e9 ) squared ] / ( root of pi ) }

= ( m ) shell mass of red light squared as one usual kilogramm per an expanded common kilogramm

And because we use one Quantum of Energy we have to divide the ' ( 1 ) / Energy Value '

per 2,837779795927041 in order to receive ( 1 ) ; Simply see that above ...

Will we believe a result of negative mass for photon shell ,

which would explain super luminary speed well ?

We are going to calculate with :

Photon Core = x kilogramms

Photon Shell = y kilogramms

I )

x + y = Photon Mass ( known )

II )

Photon Energy = x * light speed squared + y * ( light speed + spin speed ) squared

III )

4 * Photon Energy = x * light speed squared + y * { light speed squared + 2 * light speed * ( root of 8 ) * Diametre * pi * frequency + 8 * spin speed squared }

Energy = Light Intensity = Amplitude squared = Photon Radius squared

Diametre * pi * frequency = spin speed

Energy = mass * ( spin ) speed squared

I . b )

( 1 ) = photon mass = x + y

II . b )

E = h * frequency ( f ) = { ( 1 ) - y } * ( c + spin speed of the core ) squared + y * ( c + spin speed s ) squared

III . b )

E ( of maybe blue light ) = h * f = { ( 1 ) - y } * ( c + spin speed of the core ) squared + y * ( c + spin speed s ) squared

{ E per ( root of radius r ) } squared = spin speed s ( of the electromagnetic shell )

' Photon Shell and Core Mass 2020 – Approach C . '

It is only a logic that – starting with 1 kilogramm for shell weight of a photon –

the result is the positive value of x squared negative kilogramms .

The reason is the following :

Starting with 1 shell kilogramm to be an expanded common kilogramm

can only result positive mass , because 1 shell kilogramm is yet positive .

But if we use negative mass , to simplify we use - ( 1 / 3 ) = 1 shell kilogramm

we will see that ( -1 ) = 3 kilogramms

and that ( -1 ) ( - 1 ) = 9 kilogramms ( the squared value )

1.6 Past Research : An antique Way to Calculate Light Speed ...

I have had this idea some time ago and I have worked to my best

to reconstruct this assumed ' ancient ' way to calculate light speed ,

which might had been lost for over 2000 years ...

' An assumed antique Way to calculate Light Speed ...

Right click on the graphic in the web browser firefox to enlarge '

It has been a long way to get up to the results written above :

1.2.1 Light Particles per cubical metre and ' Ratio ' - Approach 1

Assuming an average 'Light Particle' as a cube with lenght '1',

( 1 per Force ) to be ( 1 cubical metre per flexible weight ) times ( 1 second per flexible way ),

there should be about (( number times 10 to the power of number ) to the power of 3 )

Light particles per 1 cubical metre .

It is a rough calculation related to the units meter, second and kilogramm ,

using given Light Speed ( 3 times 10 to the power of 8 metres per second ) ,

Absolute Density of Ratio '1' as Ratio '2', Mass of Sun ( 2 times 10 to the power of 30 kilogramms ) ,

Ratio '1' as 'Lightsize per Lightweight', and Ratio '3'

If a Light Particle would have a weight of 0 Kilogramms,

it would move with a Speed of ' 1 ' within a scale from 0 to 1 .

And each mass should have its relative weight .

Note 1:

Using the term 'Mass' as kilogramm with one kilogramm as a unit to measure Gravitation

and 'Density of Mass' as 1 cubical metre per a variable size of kilogramm .

Note 2:

This calculation has led to an assumed number which is rounded roughly.

Shown by the graphics below :)

Note 3:

Be the cause

a Sun Light Particle needs exactly that impulse to leave gravitation of Sun,

similar to air particles to be bound to gravitation of Earth .

Note 4:

' Impulse making Particles slowing down to the power of two or even to the ' multiplication of Π ' in general '

1.2.1.1 Light Particles per cubical metre and ' Ratio ' - Approach 1 . 1

Approach 1.1 resulting 1 per ( 8 times 10 to the power of 90 ) kilogramms

per 1 cubical metre of Light

and an average Size of about 1 per ( 10 to the power of 40 ) ‚a' lenght of a cube

( the cubical root of about 1 per ( 10 to the power of 120 ) metres )

1.2.2 Light Particles per cubical metre and ' Ratio ' - Approach 2

The whole Approach 2 has led our idea to calculate with :

( Sun weight ) times ( light speed – from Sun – to the power of 3 ) is equal to

( Earth weight ) times ( ( fleeing Velocity of Earth times ( temperature of our Sun ) ) to the power of 3 ) .

What results an average temperature of a value about a million degrees celsius on our Sun , heat meaning particle movement ...

A new idea ( and it should to be the right one :) leads us to use ( 1 per Sun weight ) times ( light speed squared ) is equal to

( 1 per Earth Weight ) times ( fleeing velocity , vertically , of Earth squared ) times temperature of our Sun

because E = Temperature = m c c and an object would have to overcome more force to leave a celestial body vertically ,

our result is something about 400 kilometres per second of an object to speed up to for leaving

gravitational pull of earth vertically , using Sun Corona with a temperature of something between one and two million Kelvin ...

Light Density Calculation

Calculating with Gravitational Force, for light you might use a = 1 because close to sun

I found some ' magical ' 1 metre per 1 second what might be the fleeing Speed of our Sun

from the core of our Milky Way

Using 1 light-kilogramm per one light-metre , which is the very first idea from 2016

and cubical value of kilogramms of an object effecting one cubical metre of another object

Calculating with density of Jupiter would show

that Jupiter might leave Sun orbit about 2 cm a year, using a = 1 .

What is slower than the fleeing speed of Earth from Sun

despite of a smaller density

what would end up in an assumption of planets , that are attracted

by core of Milky Way rather by a star , if once have left gravitational influence of this star .

Note 5:

To get 'f' you might use ' 1 per force = density of mass times (*) density of speed ,

( (1/1/x per 1 kg) * 1 / (3 * 10 to the power of 8) ) '

by using x as light weight and density of average speed of sunlight

instead of calculating with using gravitational weight of sun (in kilogramms)

times speed how shown before by the upper graphics , what was I thought that days -

resulting to calculate with ( density of ) force , ( 1 / f ) and 'ratio 1' ( 1 / 1/x ) .

Note 6:

Or you just go ahead with the good old Isaac Newton

for ' force ' so you will use ' f = maa '

within the calculations described above ,

simply set his ' f ' in the formula with the squared root ( yellow highlighted ) .

Meaning light weight for mass ( 1/x ) , using its speed squared

as ' f ' in order to go with density of force ' f ' = density of mass times density of speed .

Note 7:

You may use speed ( of light ) squared , it is allowed ,

because you might think about unfolding an ' a lenght ' squared area

to get the point pressure a force is acting on by its speed .

This result has led my calculation to something

close to a trillionst part of a kilogramm ( 1 per squared speed )

for the weight of one cubical metre of light ,

using the calculation mentioned by note 6 and described above .

And if you will look through a microscope some day ,

seeing the little craters by ' light particle impacts '

you can check for the size if right or no ;) .

Note 8:

The closest approach to get the weight of one cubical metre of light I have found , is using just

1 cubical metre per Gravitational Pull of Sun in kilogramm seconds ( 2 times 10 to the power of 30 ) =

1 / 1/x ( 1 per light weight , or let us just say ' size per size per weight ' , what would result weight ) times

Speed in metres per second ( of Sunlight, 3 times 10 to the power of 8 ) to the power of 3 ,

see the graphic below .

Note 9:

The final result now for calculating the weight of one cubical metre of Light

should be to use

1 per ( Gravitational Pull of Sun ( 2 times 10 to the power of 30 ) in kg (*) times 1 second )

= equal to

( 1 per ( 1 / (1/x) ) ) as light density (*) times

cubical value of speed ( of Sunlight 2,7 times 10 to the power of 25 cubical meters per second ) .

What keeps gravitation which should to be measured in kilogramms in balance and

regarding unfolding an 'a' length cube

to get the point pressure a force is acting on by its speed , and

be the cause

a Sun Light Particle needs exactly that impulse to overcome gravitation of Sun,

similar to air particles to be bound to gravitational force of Earth .

A stone moving off earth with 3 meters per second

should have a ' gravitational weight ' of 1 per ( 3,6 times 10 to the power of 25 ) kilogramms

and its impulse ( ! should there be a need for an impulse ? )

to keep calculating with gravitational balance , no matter its size .

( see the graphic with the dark sun , shown below )

The Gravitational Force or Pull of Sun is more powerful the closer to Sun ,

and Light particles might react ( or ' being reflected ' ) faster

from Earth because the Earth has less of gravitational force ( measurement in kilogramms ) than sun .

It's the 21th of November 2018 today and I will continue from time

to time with thinking past about the offered ways to calculate ,

but I strongly think that it is done now :) ? !

Note 10 ( old ) :

' Number ay caramba , the way I pray to stay ' -

' Die Zahl zur Qual , der Weg der Steg '

The result should be the ( 5,4 times ten to the power of 55 )th part

of a one kilogramm for light weight per cubical metre

*** and *** the size of an average light particle, taken by 'ratio 1' :) ,

might unlikely (?) be the cubical root of 1 per ( 2,916 times 10 to the power of 111 ) metres

' a lenght ' of a cube .

Because of using ' ratio 1 ' to be 1 ( light size ) per ( 1 / light weight ) -

and ratio super emotio ;)-

A size of 1 cubical metre of light ,

it would have the 1 per ( 5,4 times 10 to the power of 55 ) part of a one kilogramm of weight .

Or let's just stay with a one kilogramm of light should measure about

5,4 times 10 to the power of 55 cubical meters ;)-

' Größe zu Gewicht wie ( 5,4 mal 10 hoch 55 ) zu 1 '

' Size per Weight is equal to ( 5,4 times 10 to the power of 55 ) per 1 '

Relations : Ratio '1' , and Ratio 'B' ( 1 per a variable size )

Note 11 ( new ) :

' Number ay caramba , the way I pray to stay ' -

' Die Zahl zur Qual , der Weg der Steg '

The result should be the cubical root of ( 1,6 times ten to the power of 167 )th part

of a one kilogramm for light weight per cubical metre

*** and *** the size of an average light particle, taken by 'ratio 1' :) ,

might unlikely (?) be ' the cubical root of the cubical root ' of

1 per ( 2,479 times 10 to the power of 334 ) ' a lenght ' metres of a cube .

Because of using ' ratio 1 ' to be 1 ( light size ) per ( 1 / light weight ) -

and ratio super emotio ;)-

An interesting thing I found , is that using Moon or Earth weight and their orbiting speed

returns results which are very close to the approach of using the calculation mentioned by Approach 4 .

Calculating with weight of Earth ( ~ partially ) and Sun ( partially, too ) and

density of Moon should result in speed Moon away from Earth.

Use Sun weight in kilogramms , because the Earth and Moon both are attracted by Sun .

Using Earth weight and density of Moon shows , that

our Moon would move about 2,5 m away from Earth during a year .

Calculating with 1 per kgs of Sun results 0,03745 metres a year of Moon leaving Earth orbit ,

which is very close to a current result of measurement .

Using 1 per Earth weight in kgs is an ' ideal sphere experiment ' with Earth and Moon isolated .

The result is regardless of other forces within our solar system or universe

and doesn't take distance and the by going loss of gravitational force into consideration .

To get a clear result you should make use of the variable 'a' ;)-

What I found too , is that ' speed squared ' per mass ( in kg ) of a moving object

might be of a fixed aspect ratio : 1 per ~lightSpeed squared ;

e.g. speed of Moon around Earth 'squared' per its weight should be about 1 per lightSpeed 'squared' ,

works with Sun and Earth also , approximately .

A later idea went me to use orbiting speed to the power of four

per weight what should result the fixed aspect ratio

of something about 1 per ten to the power of 10 .

I have tested on moon , earth and sun , and it seems

to work out without many doubts - or just limited ! ;)

Try using core of milky way , its moving speed within space

to the power of four per its weight .

Using kilogramms and metres per second .

Note :

I have been looking for a good ' all in one formula '

to describe and calculate objects within our universe ,

and some further ideas are in development ...

' Impulse making particles slowing down to the power of two '

In true , it should be the relation of a quarter circle instead of being squared , if so and if ...

And gravitational force should become less by ' a graph of a quarter circle ' , too

A trial to prove Approach 2

Prove 1 ( within the formula )

According to the ratio used by Approach 2 , an object within space

with a weight of 5 kilogramms would require another object

to be of a density of 0,025 kilogramms per cubical metre ,

in order this another object to move away with a speed of

2 metres per second , what results 8 by its cubical value .

Weight is relative , and the 5 kilogramm space object

might to be in relation of earth ( or earth kilogramms )

what would result the calculation

' ( 0,025 kg per 6 times 10 to the power of 24 , earth weight )

times five kilogramms '

to receive the actual weight in relation to earth

of the 0,025 kg per 1 cubical metre object ,

which is the same density

required to

leave earth at a speed of 2 metres per second

using the calculation with earth weight .

This result actually is closer to the density of vacuum

( using an assumed value )

regarding the 5 kg object to be made of compressed density .

And according to Archimedes Law , things just arrange to

same ( average ) density .

The ratio used by Approach 2 returns Weight per Second

( while one Second can be very relative , too ) .

Surely , the pyramides have been sunken ' a bit '

since the first corner stone was set ? ;)-

And Jupiter might become an ' Earth II ' one day

when it finally has compressed by gravitational force :)

Prove 2 ( external prove )

Think a stone with 1 kilogramm of Weight ,

( and to be about 1 cubical metre of size or just infinitely small ) ,

it would need an impulse or force

higher than 6 times 10 to the power of 24 ( weight of Earth )

times (*) cubical speed

to overcome gravitational force of Earth .

1 kg ^= Earth Weight times cubical Speed

For light, it might be the same :

1 / x / 1 times ( 1 second per ( 1 kg times 1 cubical metre ) , ( light weight ) )

^=

1 second per ( Sun Weight times cubical Light Speed )

( in order to leave gravitational force of Sun )

what results 1 cubical metre of Light to be about

1 per ( 5,4 times 10 to the power of 55 ) kg of weight per second

The temperature of our Sun is about 10 to the power of 6 degrees Celsius

( a value rounded very roughly and used by an assumed average )

That means the impulse of Sun light to be that times stronger

than on Earth because you can use an average temperature of about

1 degree Celsius on our Earth .

For , let's just say shooting a little stone

( or light ball ) from Earth ,

we would need an impulse of 1 per one million because

our Earth is that times lighter than our Sun .

As Earth is something of about a million times lighter than Sun ,

and about a million times colder ,

light would need a 1 per one million impulse to leave Earth

which could be provided by electricity ,

and if our Earth would be of the same temperature like Sun ,

light would leave at a million times faster speed , assuming .

Another thing , which would be in harmony with Prove 1 too ,

is to use Earth weight in kilogramms instead of Sun Weight ,

to calculate Light Weight per cubical metre .

Because if we regard Prove 1 , we would need to divide

the Light Weight of 1 per 5,4 times 10 to the power 55 kilogramms

per 1 cubical metre

by earth weight firstly ,

and then to multiply with Sun Weight ,

because we think and measure Sun Weight in relation to

earth weight and Earth Kilogramms

( like shown by the graphic above and by Prove 1 ) .

Regarding a 1 kilogramm stone on Earth might weight

something of about a million kilogramms on Sun be the Cause

Sun is a million times heavier than Earth , too .

That would result a Weight of something about

1 per ( 1,6 times ten to the power of 50 ) kilogramms of Light Weight

in relation to Earth , using the Ratio and

what would be a million times heavier than

calculating with total Sun Weight ,

and our Sun is about a million times hotter .

We think the 1 kg stone to leave Earth again :

1 kg from Earth ^= 1 per ( 5,4 times 10 to the power of 55 ) kg light Weight

per second , leaving Sun

Impulse on Sun for Light is 10 to the power of 6 times stronger than on Earth ,

surely we are talking about heat , meaning Particle Movement

The 1 kg stone might need an impulse of 6 times ten to the power of 24 ( earth weight )

times

a speed of x to the power of 3 to leave Earth , which should be equal to the light impulse ,

( 5,4 times 10 to the power of 55 ) and we might regard the temperature delta
of earth and sun which is something about 10 to the power of 6

in comparison to weight delta which should to be the same approximately or even exactly :)-

What should result a speed of something about

10 kilometres per second this 1 kg stone needs to be speeded up

in order to leave earth ;

which is a result very close to the speed Isaac Newton

has proven .

Claimed as common ( and assuming ) light to act with same speed more or less ,

but to mind the weight of the space object it leaves , like e.g. earth or sun ,

the calculation should provide a solid value to deal and to work with

if using it for e.g. the calculation a rocket needs to be speeded up to leave our earth .

You are using cubical value of Speed in metres to the power of 3

because of

calculating with e.g. Sun weight to be compressed within 1 cubical metre .

We might see the calculation below ...

Using Ratio '1' meaning 1 for light size as a unit , this are 'y' meters .

Approach 2 resulting 1 per (5,4 times 10 to the power of 55 ) kilogramms per

1 cubical metre of Light

and an average Size of about 1 per ( 10 to the power of 37 ) ‚a' lenght of a cube

( the cubical root of about 1 per ( 10 to the power of 111 ) metres )

Graphical Solution of the Calculation

Light Particle Attraction by Gravitational Force of e.g. Moon , Earth Satellite

An average white Light Particle sent from Earth by an impulse should be slower

and of shorter distance than sent from Moon with the same impulse .

Light should obey physics like any other object .

Just think about a Light Particle to be the same size like Earth or Moon in relation to infinity :)

The upper graphic would show a nice way to calculate the weight

of an assumed average white Light Particle

by knowing force and all other characteristics of a ' stone ' or ' iron ' ball

shot from the near area of e.g. Moon or Earth etc. while comparing that data

by measured Light Particle attraction of a ' light ray ' attracted by an object like Moon etc .

like the experiment of the thrown Paper Jet , see below ;)-

A short story ...

I have been living near a train station .

While I was watching the tree behind a locomotive and its smoke stack

the tree seemed to flicker and to blur behind the chimney .

I suddenly had the idea that the little light molecules

have been thrown out of their natural passes

and to be pushed upward a little bit ,

surely because of the heat meaning particle movement .

Then I started to think about twice such a things :

Might this be an answer to the question of

light molecules reacting like domino particles either

or just behaving like little iron balls , once shot off

reaching their targets in a straight way .

An interesting quiz , which seems to remain a riddle to me up to today .

But keeping that little secret , it told me another one :

Throwing a paper jet above a heating source ,

we all might know speed , weight and hence force , too , of this experiment .

So , if measurement of light molecule distraction above a heating source

like that one above the smoke pipe of the locomotive would be possible

by a strong objective ; we would know speed of light as well and

comparing that data with the thrown paper jet and its force and

degree of distraction , it would be an easy thing

to calculate the weight of a single light particle .

Short words , long mathematics ;)-

Paper Jet weight = 1 / 1000 kilogramms

Paper Jet speed = 4 metres per 1 second

Paper Jet Force = mass times way per time = 1 per 250 kilogramm metres

Paper Jet Distraction by the Heat Source = 1 metre upward

1 metre upward = 1 / 250 kgm

Light Particle weight = x kilogramms

Light Particle speed = 3 times ten to power of eight metres per one second

Light Particle Force = 3 times ten to power of eight metres times x kilogramms

Light Particle distraction by the Heat Source = y metres ; known

1 metre upward of the paper Jet = 'z' times upward of the Light Particle , 'z' times y metres

1 per 'z' = Light Speed times 'x' kilogramms

Nono end ;)-

1.2.3 Light Particles per cubical metre and ' Ratio ' - Approach 3

A Light Particle to be same size and weight like Moon and

to be blown up to 1 kilogramm of weight and 1 metre of size

If a Light Particle would ' weight ' 1 kilogramm , and

our Moon would weight 6 times 10 to the power of 22 kilogramms ,

may it to be a Light Particle the

1 per ( 6 times 10 to the power of 22 ) th part of a one kilogramm firstly ,

and then the ( 9 times 10 to the power of 17 ) th part of that .

Considering force of a Light Particle and force of Moon to be the same ,

and cubical value of gravitational force , kilogramms to the power of 3 ,

acting per cubical metre by cubical speed .

Assuming a Constant of Gravitational Force , meaning each object

to have the same energy or to be influenced by , within our universe .

Resulting the calculation of the weight of an average light molecule

to be estimated about 1 per ( 10 to power of 120 ) kilogramms ,

and about 1 per ( 10 to the power of 37 ) metres of size

regarding our Moon measures about 10 times 10 to the power of 9 cubical metres

which is that times bigger than one cubical metre a Light Particle is set to .

What would provide a value some how in harmony with

the rework of the very first approach from the year 2016

and with the second approach of august 2018 , too .

Approach 3 resulting 1 per ( 10 times 10 to the power of 120 ) kilogramms

per 1 cubical metre of Light

and an average Size of about 1 per ( 10 to the power of 37 ) ‚a' lenght of a cube

( the cubical root of 1 per ( 10 to the power of 111 ) metres again )

Multiplication of the weight with Moon Size

is because of the Light Particle being blown up to all three parametres

Weight , Size and Speed

in order to achieve a balance

And using this calculation with Earth and Moon

it gives something about 4 times ten to the power of three

kilogramms per cubical metre of moon density .

Just use the Division Delta of both Earth and Moon fleeing Speed .

Division Delta means to reduce the ratio of Moon speed per Earth speed

which should to be ( 0,038 metres per year ) per ( 0,044 metres per year )

by using 0,38 per 0,44 ( just get rid of additional potences, meaning

erasing 10 to the power of -9 ( speed per second ) on both top and bottom

of the ratio to get a pure ratio , which is what we call ' Division Delta ' ) .

The whole Approach 3 is based on Delta Force of Space Objects

doesn ' t exist in relation to infinity .

And some idea about 2*2*2 = 8 and 3*3*3 = 27 ,

with a delta of either 1 or 19 ,

and too, because of assuming the cubical value of Speed acting

per cubical metre by cubical value of kilogramms like mentioned above .

1.2.4 Light Particles per cubical metre and ' Ratio ' - Approach 4 ( Mathematical error ? )

Note 1

1 / f = 1 / x * 1/average Lightspeed

isolate f to :

f = 1 per ( 1 / x * 1/average Lightspeed )

and use this 'f' (force or impulse) now

within the calculation described below ,

simply set this 'f' in the formula

with the squared root , see about that above .

Calculation using both formulas in combination ,

1 / force is equal to 1 cubical metre per flexible weight ( not density of mass )

times speed ( 1 second per flexible way )

as force of movement

and

1 / force is equal to squared root of ( ( 1 cubical metre per x kilogramm ( as density of mass ) )

times speed ( 1 second per flexible way )

as gravitational force

and setting values of force equal to each other

what ended up in a mathematical error ( ? ) to receive a ratio

This has led my idea to distinguish between two main kinds of energy :

Gravitational Force beside ( and resulting in ) Force of Movement .

You might use speed squared , too .

A candy I can still show .

At the end , it is the relativity of weight which is causing clutter :)-

as well as the first three approaches result at least similar values .

It might be the cause light speed is used by its average , too ,

which could be very relative instead .

Air and Light !

Vacuum Density Ballooning

' Actually , a rocket drive might to be compared with a heated Vacuum ,

may be at an elephant's level '

Dedication

This piece of research work is dedicated especially to
The Royal Swedish Academy Of Sciences

and to my friends and parents

See my site for more