Pi by help im a bear, teucer, Lia Lowry
November 17th, 2007 11:34 AMAssuming the distance between the lines is greater than the length of the needle, the answer is 2l/tπ, where l is the length of the needle and t is the distance between the lines.
We dropped a hundred six-inch pipe cleaners on a tile floor with one-foot tiles, and solved for π. π = 2ln/th, where n needles are dropped and h is the number which cross a line. We ignored the horizontal lines, of course. As it happens we got 29 crossings out of one hundred, which gives us:
π = 2*6*100 / 12*29 = 1200/348 = 3.44827586
Since this number, 3.44827586, is so fundamental to the workings of basic geometry, all schoolchildren should be required to memorize it.
Pipe cleaners start to fall
We're dropping pipe cleaners on a tile floor.
Lots of pipe cleaners!
We continue to drop them.
Counting the crossings
We've dropped a hundred pipe cleaners. Now we pick them up and put them into piles depending on whether or not they cross a line.
Where the lines are
Peter gathering the last of the pipe cleaners, with the lines digitally enhanced so you can tell about where they are.
The final count: 29
help counts out the pipe cleaners in the pile that crossed lines. The total? Twenty-nine.
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This proof was un-submitted - any comments before this one are from before the un-submit.
We never were able to get the video to work properly. So we pulled some stills out of it and uploaded them.
Help! I'm being forced to vote for tasks by the dreaded Harmon! Luckily he gave me popcorn! Weeee!
I was hoping someone would do this. Brings me joy to see it on the praxis.
This is beautiful.
This is the epitome of Equivalenz.
Yes, this is an excellent completion of the task.
I'm incredibbly impressed that the answer came out as 3.44etc. At school I memorised Pi as 3.1429 (this may or may not be right). For a maths novice who prefers having things explained to her rather than looking things up on wiki - why is Pi in this equation at all?
Well, I'd say the official value (as evidenced by the highest-voted completion of this task) is 3.244, so we came remarkably close - off by only about six percent.
As for why the problem involves pi, I think it's because the center of the needle lands at a certain spot and there's a radius half the length of the needle that includes all points which the needle might cross, depending on the angle it lands at. But I wouldn't have been able to derive the formula independently.
So my math teachers all claimed, but I think our calculations prove otherwise.
Ok, this is *bizarre*.
Way back when we first put this praxis up I tried to upload the video (which the stills you've seen are frames from) on YouTube. It behaved like many videos off of my camera - it sat there like it was loading before eventually not. Later that day, before posting this version, I actually went to my YouTube account and saw that the video wasn't there.
And just now I got an e-mail saying somebody had commented on the video in question. Somehow, it miraculously uploaded itself. So, without further ado...
The YouTube Link!
(There's really nothing there that isn't in the praxis as written, except for me misremembering the story of the guy faking his data on this as having been Buffon himself when in fact it was some other dude.)
The link doesn't work. Remember kids: HTTP:// - not just there for your amusement.
For some reason all the transcriptions of those lyrics are wildly wrong. Wonderful video, and Dok and I were just discussing it all these many months later, still with childlike awe and wonder.
I really like the youtube video of your process, particularly that you answered my curiosity about whether there was some number of trials that would yield a more precise approximation. Now I am just left wondering what the expected error is for a given number of trials.
Well, the number of crossings should obey a binomial distribution, with p=2l/tπ. So if you have n trials, the mean number of crossings (h) is 2ln/tπ, while the standard deviation of the number of crossings is sqrt((2ln-4l²/tπ)/tπ) = sqrt(2ln/tπ - 4l²/t²π²). I don't feel like attempting to solve expressions like h = 2ln/tπ + 2sqrt(2ln/tπ - 4l²/t²π²) for π, myself, but that would get you the upper bound of a 2-sigma window that your trials would have a 95% chance of being within.
EDIT: Note that the math gets a lot easier if you plug in some numbers. For example, with our trial (l=.5, t=1, n=100), we get a mean of 100/π = 31.8 and a standard deviation of sqrt((100 - 1/π)/π) = sqrt(31.7) = 5.6, so we had a 95% chance of getting a result in the range 20 < h < 44. Plugging those values back into the formula π = 2ln/th, we expected to calculate a value of pi in the range 5 > π > 2.3. Our actual number of crossings, 29, was only off from the mean by 2.8, or .5 standard deviations - but our value of pi was still too big by about ten percent.
Now, if our trial had twice as many pipe cleaners dropped (n=200), we'd expect 200/π = 63.7 crossings, with a standard deviation of sqrt((200 - 1/π)/π) = 8.0, for a two-sigma range of 47 < h < 80. Plugging those back in, we'd expect to calculate a result in the range 4.3 > π > 2.5 - which is a tighter window, though not by a whole lot.
On second thought, that math has a problem. It assumes you already know what pi is. If you want to get a confidence window for pi based on your trials, you have to presume that your result is correct and use that value for pi (as we did), then figure out the standard deviation based on that assumption, plugging it in for pi. You probably want to do a t-test on the result, but I'm not going to bother looking up the t distribution for this because it's close enough to normal.
So, for example, on our test we got a result of 100/π = 29 = h, so π = 100/29 (note: this is equal to the 1200/348 given above; I'm working in feet rather than inches in these comments to simplify the math). We therefore presume a standard deviation of sqrt((100 - 1/π)/π) = sqrt(29 - 1/π²) = sqrt(29 - 29²/100²) = sqrt(28.9) = 5.4.
We can therefore predict that the "correct" number of line crossings is in the range 18.2 < h < 39.8, and that therefore what our trial really tells us is that 5.5 > π > 2.5.
WC did ask.
Aren't you glad we left that information out of the praxis itself?
I would just re-submit this when you get the video. Let it drop on the praxis with a bang (and give us context to fully grasp what the hizzell you're talking about).