Corey's Gravity Dynamics Formulas
Corey...@gmail.com
modeling everything else that we know about gravity into the theory.
So I can self publish my work in a journal or on a website like
Wikipedia. Gravity is represented as F= G( (m1*m2) / r^2 ) , and I
would like to change this theory, but make sure that the math still
works exactly the same. The Force Of Gravity is equal to the
Gravitiational Constant multiplide by the masses of two objects, and
divided by their distance appart.
So in my game where you form a circle of 10 pennies that represent the
gravitational pull of one object, and my individual penny which sits
outside of the circle entirely solitary. The odds still remain 10/11,
when you are flipping a fair coin to decide which pile wins each
round. As the piles move there is a .09765625% of flipping 10 wins in a
row for the individual penny, and a 50% chance that the pile of 10
pennies will win on the first round. But my question was, how do I
calculate the average number of coin flips before the larger pile
wins.
And the answer is k(n-k)
That's right, k(n-k). So in my illistration, you can see that the
circle of 10 pennies attracts to lonely penny into its gravitational
field after 10 coin flips on average. But theoretically the number of
rounds in the game could come close to infinity. And in practice you
win after the first round or too. And I think you can see how this
example illistrates a basic understanding of gravity. If we assume
that gravity accelerates everything on earth at 9.8 m/s^2. For example
if we look at the earth as being a mass of 10 pennies, and we look as
the signle penny as being a distance of 4.9 meters, then if we follow
this equation.
t = sqrt( ( 2(4.9 m) ) / ( 9.8 m/s^2 ) ) = 1 s
And if we say the average number of coin flips it takes to produce this
effect is 10, then each coin flip represents 1/10th of a second. So
on average it takes just 1 second. Now obviously with correct
preportions of pennies, and more sophisticated mathematics, and a
better understanding of the physical formulas for gravity. We could do
a lot more. And be far more precise.
So here is my final gravity theory. We using the quadratic formula to
solve 2*n/9.8 = k(n-k) for k:
k=(1/14) (7n +- sqrt(49 n^2 - 40 n)).
So now a final example...
We are dropping a ball from 10 meters above the ground. So we plug 10
meters into n to solve for k.
k=(1/14) (7n +- sqrt(49 n^2 - 40 n))
k=9.791574237
My question to calculate the average number of coin flips in my game
is k(n-k), so we plug in k & n:
k*(10-k) = 2.040816327 = average number of coin flips
Now we take the square root of the average number of flips to get the
actual time it takes to land:
sqrt(avg flips) = 1.428571429 = number of seconds to land.
Now finally to factor in a problem with my equation we say that if k
is 9.791574327, that means our large gravity pile is that many
pennies. And our small gravity pile is exactly 0.208425673 pennies!
Eric Gisse
[snip]
It is crap.
No amount of fiddling Newton will turn it into GR.
AlanS
>So in my game where you form a circle of 10 pennies
Save those and buy yourself a clue.