Natural Science Forum / Physics / Particle Physics / January 2007

On Dec 13, 9:27 pm, hjones_...@fsmail.net wrote:

In addition to the above formula the following comments and
explanations are given. The formula is a condensed version taken from a
much larger group of numbers and tables.  At the root of these tables
is the mass of the black hole whose diameter equals one light second,
1.0095x10^35kg.
This is known throughout as M.  Two sets of time scales are involved;
the time scale of one second and the time scale equal to (d)^3 seconds.
The solving of (d) solves everything else.  These two sets of time
scales were tabulated in a lower table, the second, and a higher table,
(d)^3 seconds.  We begin with the formula c^2/h= 1.35639x10^50.  The
square root of this is 1.16464x10^25 which represents the diameter of a
black hole whose mass is equal to the square root of M.  A second
figure is drawn from {(c^2/h)/(8pi)}^0.33rec.=1.754x10^16.  The product
of these two lengths is 2.04287x10^41.
{2.04287x10^41/(d)}^1.7142857=M^2.  At the moment we do not have (d) so
we push ahead with just (2.04287x10^41)^1.7142857 which comes to
6.5698558x10^70 or M^2(d)^1.7142857.  If we reduce this by the power of
0.6 we get 3.0940939x10^42 which is the analogous sister product of
2.04287x10^41.  The higher product belonging, obviously, to the higher
table. Dividing the higher product by the lower gives 15.145813 which
is (d)^2.5.  Therefore, (15.145813)^0.4=(d)=2.96563.
(3.0940939x10^42)^1.7142857=6.932894x10^72=M^2(d)^6; which is the
higher table analogue to M. Therefore, by deduction,
M=1.09499648x10^35kg.  {(c^3)/4}/{1.009499x10^35}=6.67261319x10^-11 =
G.
H Jones