Calculate $\pi$ By Hand? All over the internet the only hand equation i found was 
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?

I skimmed the other answers. I believe this one is different in that it gives a way to rapidly calculate the digits of $\pi$ and proves from scratch that that number in fact is $\pi$.
It's obvious that $\pi = 6 \times \sin^{-1}(\frac{1}{2})$. It turns out that the derivative of $\sin^{-1}$ is elementary so we can take its power series centered at 0 and then integrate it to get the power series of $\sin^{-1}$. In general, $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$. So $\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-\frac{1}{2}}$. Once I get the power series of $(1 + x)^{-\frac{1}{2}}$, I can substitute $-x^2$ for $x$ to get the power series of $(1 - x^2)^{-\frac{1}{2}}$. The first derivative of $(1 + x)^{-\frac{1}{2}}$ is $-\frac{1}{2}(1 + x)^{-1\frac{1}{2}}$. The second derivative is  $-\frac{1}{2}(-1\frac{1}{2})(1 + x)^{-2\frac{1}{2}}$ and so on. Now we divide by the factorials to get the coefficients of the power series so $(1 + x)^{-\frac{1}{2}} = 1 + -\frac{1}{2}x + -\frac{1}{2}(-\frac{3}{4})x^2 + -\frac{1}{2}(-\frac{3}{4})(-\frac{5}{6})x^3 ...$ So the power series of $(1 - x^2)^{-\frac{1}{2}}$ is $1 + \frac{1}{2}x^2 + \frac{1}{2}(\frac{3}{4})x^4 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})x^6 ...$ Then the power series of $\sin^{-1}$ is $x + \frac{1}{2}(\frac{1}{3})x^3 + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})x^5 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})x^7 ...$ Now finally, $\pi = 6 \times \sin^{-1}(\frac{1}{2}) = 6(2^{-1} + \frac{1}{2}(\frac{1}{3})2^{-3} + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})2^{-5} + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})2^{-7} ...)$.
