Gravity solved with the quadratic equation



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Topic: Science > Physics
User: ""
Date: 10 Sep 2006 10:23:50 AM
Object: Gravity solved with the quadratic equation
I have been working on a theory of gravity, and would like help
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 use the quadratic formula to
solve 2*n/9.8 = k(n-k) for k, which makes k=(1/14) (7n +- sqrt(49 n^2
- 40 n)).
So now we have the formula ( (1/14) (7n +- sqrt(49 n^2 - 40 n)) ) * (n
- ((1/14) (7n +- sqrt(49 n^2 - 40 n))) ) , which solves the gravity
problem. And solves the problem of averages in my game of pennies!
.

User: "J. Horta"

Title: Re: Gravity solved with the quadratic equation 10 Sep 2006 11:33:55 AM
On Sun, 10 Sep 2006 08:23:50 -0700, CoreyWhite wrote:

I have been working on a theory of gravity, and would like help
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 use the quadratic formula to
solve 2*n/9.8 = k(n-k) for k, which makes k=(1/14) (7n +- sqrt(49 n^2
- 40 n)).

So now we have the formula ( (1/14) (7n +- sqrt(49 n^2 - 40 n)) ) * (n
- ((1/14) (7n +- sqrt(49 n^2 - 40 n))) ) , which solves the gravity
problem. And solves the problem of averages in my game of pennies!

I would get a physics book and start reading. Just my 2 cents...
.


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