Around 250 B.C., Archimedes calculated the ratio of a circle's circumference to its diameter. Today we know that ratio to be PI, or Math.PI in JAVA, but HOW?
- First, he considered the shapes he already understood and that he knew how to calculate the perimeter for, like triangles, squares, and other polygons.
- Next, Archimedes noticed that if you divide a circle into inscribed polygons (square, pentagon, hexagon, heptagon, octagon, ..., etc.) with an increasing number of sides, each side of the polygon decreased in length, and therefore gave a better and better approximation for the circumference of a circle.
- With this observation, and a little more geometry, we can create an algorithm to calculate the perimeter of larger and larger inscribed polygons.
- Remember, the more sides, (n represents the number of sides in the polygon), the more accurate our perimeter estimation is, so let's do the math...
- From the image above, you can see the octagon is divided into isosceles triangles, the bottom of which (labeled s) is the length we need to know.
- If we draw a line from the circle center perpendicular to the base of the triangle s, we divide s in half, so it becomes
${1 \over 2}s$ as depicted. - From the picture, we can simplify our lives by assuming
$h = 1$ , and that the circle is a Unit circle.- Why can we make this assumption? Remember, we're trying to figure out PI, which we know is the RATIO of
$Circumference \over Diameter$ , and ratios don't care what values we assign as long as the ratio holds!
- Why can we make this assumption? Remember, we're trying to figure out PI, which we know is the RATIO of
- Also, from the diagram we can see 2 angles B and A and because of symmetry
$A = {1 \over 2}B$ - Great, so now what!?!
- WELL, we can calculate B, so let's start there. Remember, a Circle is 360° and we are equally dividing that by n sides (
$n = 8$ in the example). - So...
$B = {360° \over 8}$ or$45°$ and$A = {1 \over 2} * B$ so$A = 22.5°$ in our example. - Now we can calculate
${1 \over 2}s$ if we recall$sin(A) = {{1 \over 2}s \over h}$ as shown in the below diagram of the triangle.
- WELL, we can calculate B, so let's start there. Remember, a Circle is 360° and we are equally dividing that by n sides (
- ...continued...
- Solving for s we see
$s = 2 * h * sin(A)$ - And since we agreed
$h = 1$ , that simplifies to$s = 2 * sin(A)$ - Now we have s, so to get the polygon perimeter, we just multiply by the number of sides to get it.
- Finally, we can calculate PI as
$PI = {(n * s) \over 2h}$
- Solving for s we see
OK, so if you've followed the math, feel free to skip ahead and implement calculating PI as an algorithm, but if not lets lay it out a bit more concisely!
- Step 1: Read into an integer the number of sides for our polygon:
$n = 8$ - Step 2: Calculate the angle
$B = {360.0 \over n}$ - Step 3: Calculate the angle
$A = {1 \over 2} * B$ - Step 4: Calculate the length of 1 triangle base
$s = 2 * sin(A)$ HINT: YOU NEED TO CONVERT A to RADIANS- Math.sin(), and Math.toRadians() will be your friends.
- Step 5: Get the polygon perimeter
$p = n * s$ - Step 6: Estimate PI:
$PI = {p \over 2}$
Finally, your assignment is to use the book and other resources to put the above algorithm into a program to estimate PI!
- Fork my repo
- Using IntelliJ clone your fork locally
- Use IntelliJ to create a feature branch named Feature1
- Implement the algorithm in ArchimedesPiMethod.java
- Just as you did last week (Reference the Lab video in your Week 1 module), create a Feature1 feature branch of your code if you haven't already
- Commit your working code to your local copy
- Push it to your Remote/origin branch (i.e. GitHub: Feature1 -> origin/Feature1)
- Then issue a Pull request to my instructor branch
- Make sure to save the Pull request URL and submit it for the assignment.