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Topic: The nature of gravity
Replies: 28   Last Post: Apr 11, 2014 4:14 PM

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haroldj.l.jones@gmail.com

Posts: 61
Registered: 3/17/12
Re: The nature of gravity
Posted: Jan 2, 2013 2:04 PM
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In every numerical structure of mass we find four separate components that are involved in one way or another with that particular structure.
Take the 2.2687314x10^23 template above. In protonic units it can be seen as the square of 4.76310196x10^11. As local protonic G is 6.23343585x10^-11 we can work out that the Gm product, Gx2.2687314x10^23, equals 1.41419916x10^13. I've written about Gn, the product of the formula h(c/2)^4=0.334517782.
1/(1.414199156x10^13)^2=5.00010187x10^-27. Multiply this by Gn, 0.334517782, and you get 1.672623x10^-27, the proton SI mass.
The inbetween system, mentioned above, has a Planck mass of 2.822570503x10^-8.
(2.822570503x10^-8) multiplied by c equals 8.46185349. Multiply this by 3.62994678, the quantum adjustor, and you have 30.71607783.
c divided by 30.71607783=9.760115197x10^6.
4(1.672623x10^-27)/(9.760115197x10^6)=6.85493138x10^-34 which is the local Planck constant, h.

So, if 1.414199156x10^13 can be seen as the Gm product of Gx(4.763120195x10^11)^2.
And if GxPlanck mass squared is equal to 4.966118653x10^-26, see above, then
(1.414199156x10^13)/(4.966118653x10^-26)is equal to 2.84769506x10^38 which agrees with {(4.763120195x10^11)/(2.822570503x10^-8)}^2. Remember that in the inbetweener model the mass unit is smaller than the kilogram by (x)^0.5 and so
1.672623x10^-27 in inbetweener mass units must be (x)^0.5 times smaller than 1.672623x10^-27kg. And because 4.763120196x10^11kg is (x) times larger than the the SI system's proton opposite number then the same number, 4.763120196x10^11,
in inbetweener mass units, must be (x)^0.5 times bigger and therefore must be the opposite number to the inbetweener's version of 1.672623x10^-27. Which meanst that
2.84769506x10^38, see above, must be (x) times bigger than the SI version of
of M/m, (Proton opposite)/Proton.
There is an interesting numerical development here. Starting from the protonic model the Gm product 1.41419915x10^13 is always 4.763120196x10^11x29.6906036, whatever the mass model. Also, 2x4.763120196x10^11 divided by c is 3.177611789x10^3metres, is the square root of 1.009721668x10^7 which is the difference between h/4 and the proton mass.
This means that the master mass, (c/G)x(30.71607783/G), must be equal to
2.214307982x10^30 units of mass and has a radius of 3.177611789x10^3m.

Back to the beginning where I mentioned the four components of numerical structure; they are:

(1)The Multiplier represented by c/G and works out in this model at
4.64886018x10^18.
(2)The Associate Mass, in this case 4.763120196x10^11.
(3) The Master Mass, the product of the above two, 2.214307981x10^30.
(4) The Proton Monitor and opposite to the associate mass, and in this case
nominally the same as the proton mass, 1.672623x10^-27, but actually less by
(x)^0.5.


Date Subject Author
12/29/12
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