How do i mechanically generate pi? Here's a question with very "real-world" implications... I want to produce pi in my basement woodworking shop. (And no, using pi on the calculator or computer is not allowed, nor are books.) The shop is as well equipped as a typical vocational technical high school woodworking shop. I have every conceivable hand tool as well as a table saw, band saw, woodworking lathe, thickness planer, jointer planer, radial arm saw, digital weight scale, digital calipers, a microscope, etc. For materials, I have an extensive collection of woods as well pipes, strings, pulleys, rules, glues, nuts and bolts, etc., etc., etc.
While this may be a thought exercise for many, it is a real-world problem for me -- there are times when I have to work in extreme accuracy for all sorts of math problems (someday, I'd like to start making wooden clocks), and every tip or trick I can acquire with pi will probably help me down the road with other problems.
So far, I'm comfortable mechanically working down to the thousandths, but I'd like to refine that... So as a practice experiment, I'd like to generate pi as accurately as possible in the shop...
Ideas?
Thanks, Alan
P.S. You might think this question would be better served by a machinist, but I used to work with one and have picked his brain clean.
P.P.S. I also have paints/stains/oils that will stabilize wood in the extreme. 
 A: Use your lathe and ruler to build a wooden cylinder of diameter 1. Then by re-enacting this animation you can measure the circumference which is $\pi$. Or you can do  it backwards by wrapping a length of string around the cylinder. Note your figure for $\pi$ will be more accurate the more accurately you can measure the cylinder diameter.
A: Check about this one, and related videos :D
https://www.youtube.com/watch?v=qYAdXm69l8g
A: YIf you want to build $\pi$ physically, you could build a quadratrix compas (a tool that draw a quadratrix of Hippias, a curve that allow you to construct $\pi$) :
https://en.wikipedia.org/wiki/Quadratrix_of_Hippias
Now, I'm not sure it would be more precise than simply taking $\pi = 3.14159265$ and mesuring with high precision tools
A: The mass of a cylinder is given by
$$m = \rho \pi r^2 h$$
where $\rho$ is the density of the material, $r$ is the radius, and $h$ is the height. If the cylinder has a height of $1/\rho r^2$, then its mass will be $\pi$ mass-units (assuming the density is given in mass-units/length-units$^3$).
Now, the uncertainty in $\pi$ from this method will be
$$\frac{\sigma_\pi}{\pi} = \sqrt{\left(\frac{\sigma_\rho}{\rho}\right)^2 + \left(\frac{\sigma_r}{r}\right)^2 + \left(\frac{\sigma_h}{h}\right)^2}$$
where $\sigma_x$ refers to the uncertainty in quantity $x$.
Given your measurement equipment and since different batches of wood have varying densities, the density error is going to dominate. You're going to want to construct this cylinder in two steps. First, form the wood into a rectangular prism with dimensions strictly larger than the final cylinder. Measure the length, width, height, and mass of the prism and calculate its density. Use this density to determine the proper height. Second, form the cylinder.
As an example, say you use a piece of American redwood to create a cylinder 1 foot in radius. The density of American redwood is 28 lbs/ft$^3$, so the height of the cylinder will be $1/28 = 0.03571 ft = 0.4286 in$. The purpose of the rectangular prism construction is to pin down the density of this piece of redwood to a higher accuracy. This piece of wood will have a mass of
$$m = \rho \pi r^2 h = 28\pi(1^2)(0.03571) = \pi\,\,lbs$$
This method has the advantage of using only measurements of straight line dimensions (length, height, diameter, etc.) and other direct measurements (mass) of the object itself rather than proxies (strings wrapped around circumference, e.g.).
As a check on the uniformity of density, you could cut a piece of wood into several identically sized pieces and weigh them to see the variation.
You won't want to use the paints, stains, and oils on the final product since these will add mass, giving incorrect values for $\pi$.
