Variance over two periods with known variances? If period 1 has variance v1 and period 2 has variance v2, what is the variance over period 1 and period 2? (period 1 and period 2 are the same length)
I've done some manual calculations with random numbers, and I can't seem to figure out how to calculate the variance over period 1 and period 2 from v1 and v2.

If you only know the variances of your two sets, you can't compute the variance of the union of the two. However, if you know both the variances and the means of two sets, then there is a quick way to calculate the variance of their union.
Concretely, say you have two sets $A$ and $B$ for which you know the means $\mu_A$ and $\mu_B$ and variances $\sigma^2_A$ and $\sigma^2_B$, as well as the sizes of each set $n_A$ and $n_B$. You want to know the mean $\mu_X$ and variance $\sigma^2_X$ of the union $X=A\cup B$ of the two sets (assuming that the union is disjoint, i.e. that $A$ and $B$ don't have any elements in common).
With a little bit of scribbling and algebra, you can reveal that
$$\mu_X = \frac{n_A\mu_A + n_B\mu_B}{n_A+n_B}$$
and
$$\sigma^2_X = \frac{n_A\sigma^2_A + n_B\sigma^2_B}{n_A + n_B} - \frac{n_An_B}{(n_A+n_B)^2} (\mu_A - \mu_B)^2 $$
As pointed out in the answer by tards, the formula for the variance of the combined set depends explicitly on the means of the sets $A$ and $B$, not just on their variances. Moreover, you can see that adding a constant to one of the sets doesn't change the first term (the variances remain the same) but it does change the second term, because one of the means changes.
The fact that the dependence on the means enters through a term of the form $\mu_A-\mu_B$ shows you that if you added the same constant to both sets, then the overall variance would not change (just as you'd expect) because although both the means change, the effect of this change cancels out. Magic!
