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?

Jean-Claude Arbaut has reminded us of the identity
$$
\frac\pi4=4\arctan\frac15-\arctan\frac1{239}.
$$
Let us examine that.  You learned in high school that $\tan\dfrac\pi4=1$, and that
\begin{align}
\tan(\alpha+\beta) & = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \tag 1 \\[10pt]
& =\frac{c+d}{1-cd}
\end{align}
Thus
$$
\arctan c+\arctan d=\alpha+\beta=\arctan\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\arctan\frac{c+d}{1-cd}
$$
From $(1)$ we get
$$
\tan(\alpha+\beta+\gamma+\delta)=\frac{c+d+e+f-cde-cdf-cef-def}{1-cd-ce-cf-de-df-ef+cdef}
$$
where $c,d,e,f$ are the respective tangents of $\alpha,\beta,\gamma,\delta$, and hence
$$
\tan(4\alpha) = \frac{4\tan\alpha-4\tan^3\alpha}{1 - 6\tan^2\alpha+\tan^4\alpha}.
$$
Hence
$$
4\arctan c = \arctan\frac{4c-4c^3}{1-6c^2+c^4}.
$$
So
$$
4\arctan\frac15 = \arctan\frac{(4/5)-(4/5^3)}{1-(6/5^2)+ (1/5^4)} = \arctan\frac{480}{476} = \arctan\frac{120}{119}.
$$
Next we look at
\begin{align}
& 4\arctan\frac15 - \arctan\frac{1}{239} = \arctan\frac{120}{119} - \arctan\frac{1}{239} \\[15pt]
= {} & \arctan\frac{(120/119)-(1/239)}{1+(120/119)(1/239)} \\[15pt]
= {} & \arctan\frac{28561}{28561} = \arctan 1 = \frac\pi4.
\end{align}
