Circle areas are not a win for pi

An enlightening discussion about pi and tau.

Circle areas are not a win for pi

by rdococ » Tue Sep 10, 2013 12:23 pm

Take a circle and split it into n sectors, where the circumference of each sector is [tex]\tau / n[/tex]
[tex]C_x = \tau / n[/tex]

Now let's create a parallelogram with our sectors
[tex]P = { C_1, C_2 ... }[/tex]

The base width of a parallelogram with unit sectors is not the number of sectors the parallelogram holds, because two sectors oriented at opposite direction build to a single unit
[tex]P_b = n / 2[/tex]

And as the classic formula for circle area is just a re-arranged parallelogram area formula, modified to make it look like pi wins over circle area
[tex]P_h P_b = \pi r^2[/tex]

Therefore the [tex]\pi[/tex] is covering up the division, and the reasoning of circle area. And [tex]\tau[/tex] shows the nature and reasoning of circle areas.

Edit #1: added an extra step at the end.
Edit #2: changed parallelogram area's multiplication symbol.
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Re: Circle areas are not a win for pi

by 1=0 » Sun Mar 19, 2017 12:19 pm

I'm not sure exactly what you mean about parallelograms, but you are right about circle areas not being a win for pi. You can find the circumference and area of a circle by inscribing regular polygons in it with more and more sides. The perimeter of the polygon approaches tau*r. This is how pi was originally approximated. The area of the polygon can be found by splitting it into n isosceles triangles, each with a base equal to the side length and a height equal to the apothem. Therefore the entire area is 1/2 *side length*apothem*n (n is no. of sides) = 1/2*perimeter*apothem. As the number of sides approaches infinity, the polygon becomes a circle, and the area becomes 1/2*C*r. Since C=tau*r, this is equal to 1/2*tau*r^2.
The factor of 1/2 also comes in via the power rule if you prove the formula via integration, and it is consistent with the sector area formula, of which circular area is a special case, 1/2*θr^2. There is also a factor of 1/n in the n-dimensional hypervolume of any hypersphere, when compared to its surface, so what is so special about the factor of 1/2 in two dimensions that we should use pi instead of tau just to get rid of it? We could just as easily use tau/3 to get rid of the factor of 1/3 in area of a sphere and claim that that makes it better. The only reason area of a circle is the main argument for pi is because it's pretty much the only one.
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