
Re: The nature of gravity
Posted:
Jan 2, 2013 2:04 PM


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.

