another slice of pi
TRANSCRIPT
A long time ago I wrote an article on how it was possible to determine the value of pi by squeezing a circle between two polygons and increasing the number of sides of those polygons so that they eventually merged with the circle. The original article made reference to sines and tangents, which unfortunately can’t be calculated unless you already know the value of pi ahead of time. Catch 22. I later added to that article to show how it was possible to bypass trigonometry and get the required stepping-stone values by a long line of calculations involving square roots. Yesterday it occurred to me that there might be a slightly tidier way of approaching the problem. It’s the same basic idea, just explained differently. And yes it still requires a long line of calculations involving square roots. Nevertheless I offer this to you as a way of demonstrating how people who lived a long time ago who had endless amounts of paper, ink and time (and probably no social life) could calculate pi as precisely as was needed for the engineering tasks of the day. -DC
The circumference of a circle is said to be 2πr. Let’s pretend we don’t know that and try to work out the circumference the hard way.
We could turn the circumference into a bunch of line segments and measure the length of each line.
We get a better result if we calculate the lengths, instead of measuring them. This can be done with a bit of geometric reasoning.
Let me start with a much easier shape.
If the radius of the circle is 1 then the length of each side of the box is square-root two, and the periemeter of the box is four of those.
22 11
1 1
So I could say that the circumference of the circle is somewhat bigger than four root two, which equals 5.65.
22 11
1 1
I’m going to double the number of sides again....
22 11
1 1
1
... and zoom in on the bit at the top.
My objective is to use this information to work out the sidelength for an octagon.
1
2
221
221
1
Pythagoras gives the sides of the inner triangle
2121
221
12
21
2121
221
12
21
221
2
21 21 2 L 22 ...7653.0
Therefore the circumference of the circle is a bit bigger than 8 x 0.7653
What I have to do is keep doubling the number of sides to the polygon and work out the length of each side in terms of what I had previously.
This gives me a pretty good way to estimate the circumference of the circle.
Suppose Ln represents the length of the side of the polygon with half as many sides. We can use this to determine the length (Ln+1) of the new polygon with twice as many sides. 1
nL
1nL
L21
1
1nL
L21
1
2211 L
1nL
L21
1
2211 L
22111 L
1nL
L21
2
2
212
21
1 1 1
nnn LLL
22111 L
1nL
That ugly thing simplifies to this ugly thing: 2
41
1 1 2 2 nn LL
2
2
212
21
1 1 1
nnn LLL
This means I can go from a four-sided polygon to an eight-sided polygon, to a sixteen-sided polygon, to a thirty-two-sided polygon, to ... as far as I want without ever drawing a picture and using a ruler.
2
41
1 1 2 2 nn LL
I can have so many sides to the polygon that it becomes impossible to see it against the circle, and the perimeter of the polygon becomes the circumference of the circle.
Remember that the values for pi are obtained by halving the values for the circumference.
I set this up on a spreadsheet, starting with a square and doubling the number of sides until I got down to a polygon with so many sides I have no name for it.
By the time I get to two million sides, the estimate for pi is as good as what your calculator gives you.
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