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

You can also 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.

There is one more thing that makes pi not as good as it seems when it comes to area. Measured in square units, the area of a circle is πr^2, but we could use another shape to measure area. If we measured it using equilateral triangles, the area of a circle would be τ/√3r^2.
τ>π
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