Re: Gravity pwned
- From: josephus <dogbird@xxxxxxxxxxxxx>
- Date: Sun, 10 Sep 2006 11:30:11 GMT
CoreyWhite@xxxxxxxxx wrote:
I have been working on a theory of gravity, and would like helpmath babble. you can delude your self that you are mkaing sense. This idea is not real. It ignores a lot of physics like the photoelectric effect , relativity and Celestial Maechanics. and it touches on them.
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. If the earth is represented as 10
pennies, then each meter of distance is represented as 0.225876976
pennies. And the equation k(n-k) calculates the average time in actual
seconds for the object to land.
For example 10(10.0451752951-10)=0.451752951
Which is the time it takes for an object 1 meter high to fall and land
on the earth.
But to make things more difficult we are going to use intervals of 1
millionth of a second. So instead of just using the number for one
meter, and multiplying it by 2 in the gravity equation. We devide it
by 1 million first, and that leaves us with an average number of
millionths of a second. That will not be as experientially variable.
josephus
.
- References:
- Gravity pwned
- From: CoreyWhite
- Gravity pwned
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