[Request] What is the formula for expected number of balls over time?

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[u/JT_1983]

Just a very naive approach but if there are n balls, there are ~ n² pairs of balls so you could expect the probability of collision to be ~n ² as well. Taking a continuous time approximation this would give:

dn/dt = a n²

For some constant a depending on size and speed for example. The solution with n(0)=2 is

n = -1/(at-1/2)

This quickly goes to infinity at t = 1/(2a). Note that dn/dt ~ n already gives something exponential and ~ n² is clearly 'worse'. Again naive, there is probably existing math on this (which I do not know).

[u/Whiteminusblue]

https://www.reddit.com/r/interestingasfuck/comments/1w6820h/comment/p7l5iqn/

It was found that the derivative of the number of balls with respect to time is:

dn/dt = (n² -n)/(R-r)²

Where n is the number of balls, and R and r are constants.

Therefore we can just write it as:

dn/dt = a(n² -n)

Where a is a constant defined by the sizes of the balls and the circle.

With a bit of work we can get that at+C = the integral of 1/(n² -n)dn.

This can be rewritten as:

at+C = Int((1-n+n)/(n(n-1))dn

which becomes

at+C = Int((n-(n-1))/n(n-1))dn

at+C = Int(n/n(n-1))dn - Int((n-1)/(n(n-1)))dn

at+C = Int(1/(n-1))dn - Int(1/n)dn

at+C = ln|n-1| - ln|n| + C_2

Giving us 

at+C = ln((n-1)/n)

And

C =  ln((n-1)/n)-at

(Removing the absolute values because n is always greater than 1)

Since we know that at T=0, n=2, we can determine C

C = ln((2-1)/2)-a(0)

C = ln(1/2)

Thus giving us

at+ln(1/2) = ln((n-1)/n)

From which we solve for n

e^at+ln(1/2) = e^ln((n-1)/n)

1/2*e^at = (n-1)/n

1/2*e^at = 1-1/n

1/n = 1-1/2*e^at

At last giving us

n = 1/(1-1/2*e^at )

Or

n = 2/(2-e^at )