These are the writings of someone that can not come up with a good description for his blog. Everything he thought up at first was far to cliche and sounded like something an elementary school student would write. Being a college student, he found this rather insulting to himself. Therefore, he wrote a description that really doesn't describe anything. But... that was the point. Wasn't it?

  The contents of this blog have been removed from the public eye. I backed up all of its contents for myself, but the stuff that was here before is no longer relevant to the current day (and honestly kind of embarrassing). This was all written at a different time in my life and that time is now gone. Plus, I don't want people who go (or perhaps went) to Chapman University having their name stored in memoria on the Internet for all eternity because somebody they knew five years ago was having an emotional crisis back when they were 19.

For those of you who may have stumbled across this site, or may be looking for me, I will eventually have something set up at http://phnixnet.homelinux.com (though keep in mind I am a conservative now and an entirely different beast than when I wrote this blog) and you can reach me by e-mail at phnix at fastmail dot net or via AIM at FirebirdHL.

I will however, leave one post here and that is the post on the bouncing ball problem. Long after this site died, people still come here looking for the answer that what I wrote provides. It tickles me that other people may be finding it useful and I hope people will use it long after I have forgotten this site ever existed.

So, for those of you who come and find this final note where once stood a place of many angry rantings, farewell and peace be with you.
posted by Kyle at 2/09/2007 02:49:00 AM

  As I was ending the last year of high school, I was presented with a physics problem that not only could I not solve, but could find nobody that could help me solve it. It wasn't until I asked Scott about it a year later that I saw the answer demonstrated to me. Even then, the problem had sort of come to settle in my mind and I never actually solved it myself. However, I'm at work, and we have these smile faced bouncy balls, and so for the first hour of work, I occupied myself by solving this problem. For purpose of record, and to show off, I'll demonstrate it here.

*looks up how to do sub and superscript in html*

The problem starts very simply. Take a ball and some surface, perferably something that will let it bounce when you drop it. Now, bring the ball up to a certain height (we'll call that height h_o) and let it go. The ball will fall, hit the ground and lose energy, and the rise to a height lower than h_o. From there, it will fall again, and the height will continue to drop after each bounce. If you observe this pattern, you'll notice that after each impact, it will lose a percentage of the height it previously had. Now, let's define a number between 0 and 1 and that number will be the percentage of height retained after a bounce. This number is c and is called the coefficent of restitution. For example then, if I drop a ball from 10 meters, and c is .8, the ball should rise to a height of 8 meters after the first bounce. How will I then determine the height of the 10000th bounce? For that, I derive the equation h_n=h_ocⁿ where n is the number of the bounce. So basically, every time I bounce it, I multiply it by c to take away the height it lost and give me the new height.

This is all happy and good, except for one thing. This equation predicts an infinite number of bounces. That equation will give a value for the millionth bounce if I want it to, but I don't think I've ever seen a ball bounce a million times. So, the question I put forward and was unable for the longest time to answer was: Does the ball ever stop bouncing? Will there ever be a point in time at which a ball that is left to bounce will be at rest given that it can concievably bounce an infinite number of times and never quite lose all its height?

In my mind at this point, I came up with two possible answers to my question.

A: The ball will bounce forever. It will just get to a point at which we will be unable to observe the bouncing.
B: The ball will come to rest despite the infinte bouncing and it will be possible to show that this is the case.

As I first began studying this problem, I made some observations that made me perfer B as the answer. The important observation is that if you let a ball go up to some height and come back down and then do the thing at a lower height. The ball bouncing at the lower height will take much less time to complete its trip. So, I could say that, if by the 10th bounce, the height was 1 millimeter, the amount of time it took for it to bounce in that period would barely add any time, and each subsequent bounce would add less time. It made me think that if I took the first 10 bounces, and added them up, and then kept adding on these infinitely small bounces, the amount of time it took to complete bouncing would never approch infinity. This would then mean that it took a finite amount of time to complete an infinite number of bounces. Therefore, if I can show that an infinte number of bounces took a finite amount of time, I would have proved that the ball does in fact come to rest and does not bounce for the entire existance of the universe and so on.

And in high school, I had absolutely no idea how to do this. How to you add up an infinte number of items, and then get a number? So, with no solution in sight, the problem of how long it took for the bouncing ball to come to rest burrowed into the back of my head and for a long time. I lacked the ability to derive some kind of equation for how long it took for the ball to come to rest, despite being fairly confident it did.

So, now I know calculus and all that stuff (even though the problem doesn't require too much calculus) and am able to derive an equation.

So, what we have right now is: h_o=h_ocⁿ, where c is between 0 and 1.

First, I want to find out the amount of time it take for the thing to drop to the ground, since it falls outside of the pattern the rest of the bounces will do. (It doesn't go up, it starts up.) I'll use the kinematic equations to do this.

Btw, in this, due to some issues with square roots, g (the gravitational acceleration constant) will be a positive number (9.8 m/s²) due to some square roots I'll be taking. I'll put a negative sign in front of it when it's suppose to be negative. I'm just doing this 'cause it makes it easier to see... anyway... I get:

0=(1/2)(-g)t² + h_o Solving for t I get

t= √(2h_o/g)

Now, to determine the time each subsequent bounce takes, I'm going to do essentially the same thing. Except this time, h_o will be replaced with h_ocⁿ. I'm going to find the time it takes for the ball to fall from height h_n and then I will double that time since it will take the same time going up as it will going back down.

0 = (1/2)(-g)t² + h_ocⁿ
t =√(2h_ocⁿ/g) (time for a ball to fall from h_n)

t =√(8h_ocⁿ/g) (time for a ball to rise and fall)


Now our total time equation looks something like this:

t = √(2h_o/g) + √(8h_ocⁿ/g)

Technically, there should be some kind of summation on the second term, but I'm lazy and don't want to write it. Bascially, at this point, calculus comes in. At least in terms of infinte series it does.

Say you have the series x=1/a + 1/a^2 + 1/a^3 + ... To find the answer, you would first multiply the sequence by a, leaving you ax = 1 + 1/a + 1/a^2 + ... and then subtract x from ax. Lots of stuff cancels and you get ax - x = 1. Therefore x = 1/(a - 1). This basically shows how we'll deal with that coefficeint of restitution and infinite series still left in the equation. In order to make the geometric series equation work though, we'll have to realise that we'll get the reciprocal of c put into our little a equation since a is the denominator.

What's going to happen is once I deal with all the constants and square roots (which I'm not going to show since I'm short on time. I will replace cⁿ with 1/(1/c - 1). I will do that presently!

t = √(2h_o/g) + √(8h_o(1/(1/c - 1))/g)

And for simplification:

t = √(2h_o/g) + √(8h_o/(g/c - g))

And so, that equation will give you the finite amount of time it will take for a ball to stop bouncing, given a coefficient of restitution and an initial height.

Y'already know hoes down pimps up.

posted by Kyle at 12/12/2004 02:06:00 PM

  Still havn't gotten around to doing Sentience and Der Wille Zur Macht: part 2... might be something to do tonight. Been kinda busy though.

Anyway, in the mean time... I think there are a few people out there that would appriciate the following proofs, since they were not shown by a certain calculus teacher. Hope this makes things easier to really understand. It's really a rather neat proof. Forms a cool relationship. Something you'd NEVER get out of that calculus class. Enjoy!

Proof:
Derivative arcsin(x)=1/sqrt(1-x^2)

Start with y=arcsin(x)
From this, we can say that since arcsin is the inverse of sin...
x=sin(y)
Now, we will take the derivative of this with respect to x of both sides
(d/dx)x=(d/dx)sin(y)
1=sin(y)*dy/dx
1/cos(y)=dy/dy

Now we need to have this derivative in terms of x, we're going to do this by squaring both sides of the derivative and using the pythagorean theorum... like so!

1/(cos(y)^2)=(dy/dx)^2
1/(1-sin(y)^2)=(dy/dx)^2
Since we said above that sin(y)=x... we can then say
1/(1-x^2)=(dy/dx)^2
And finally, we take the square root of both sides, giving us the familiar:
dy/dx=1/sqrt(1-x^2)
QED

Now, let's do the derivative of the arctan(x) using the same methods.
y=arctan(x)
x=tan(y)
Take the derivative with respect to x:
1=(sec(y)^2)dy/dx
dy/dx=1/(sec(y)^2)
Then, since sec(y)^2 is part of that same pythagorean thing....
dy/dx=1/(tan(x)^2 + 1)
Substitute x back into the equation and we have:
dy/dx=1/(x^2 + 1)
QED

Hope this makes things make more sense. It's a pretty cool trick and calculus is full of stuff like that. If you know the tricks and how to derive all these derivatives (and therefore if you can find a derivative you can find an antiderivative) then it will ALWAYS be better than memorising some garbage and spitting it back out. Not only will you do better math, but you'll understand it. That's what makes stuff like this beautiful and enjoyable. I could do a thousand integrals with 1/sqrt(1-x^2) in them, but they are essentially meaningless without this proof above.

Dur wille zur macht will be back soon! Stay tuned! I'll probably do another insane run of Xenosaga tonight, so I'm sure my thoughts will be full.
posted by Kyle at 10/25/2003 10:48:00 AM