Center of mass error - calculating systematic error in change in PE Suppose we have to calculate systematic error in change in PE. Let's suppose systematic error due to scale is 1%. I'm confused about the center of mass error. 
\begin{align}
\Delta PE = m*g*h_1 - m*g*h_2\\
\end{align}  
However, expression for calculating systematic error is
\begin{align}
\sigma_{\Delta PE} = \sqrt{(m*g*\sigma_{h_1})^2 + (m*g*\sigma_{h_2})^2}\\
\end{align}  
Does any error due to center of mass get canceled out because unlike scale, it's not a percentage, but a fixed amount? Or does it get incorporated somehow? If it gets incorporated, how?
I've come up with something like below, but I don't think it's right: 
\begin{align}
\sigma_{h_1} = \sqrt{(\sigma_{h_{scale}})^2+(\sigma_{h_{CofM}})^2}
\end{align}  
 A: You have to be careful to differentiate between uncorrelated and correlated (systematic) errors. Your first analysis is for uncorrelated errors in $h_1$ and $h_2$, e.g.  measurement noise. The effects of uncorrelated errors add in quadrature as you have written since the resulting errors in $\Delta PE$ can either add or subtract. For the correlated error analysis, you need to introduce a common scale factor $\epsilon$~1
$\Delta PE = m g h_1 \epsilon-m g h_2 \epsilon$
which accounts for small correlated scaling error in both $h_1$ and $h_2$ measurements. Finish off with an error analysis on the single variable $\epsilon$ with a nominal value 1 and an uncertainty of $\sigma_\epsilon$=0.01:
$\sigma_{\Delta PE} = \left|\frac{\partial(\Delta PE)}{\partial \epsilon} \sigma_\epsilon\right| = |m g h_1-m g h_2|\sigma_\epsilon = |\Delta PE|\sigma_\epsilon$
So if you have a 1% error in your length scale, then you have 1% error in $\Delta PE$. If there is uncorrelated measurement noise in $h_1$ and $h_2$ in addition to the scaling systematic, the effect of all three noises get added in quadrature.
=== Added the following in response to a comment below ====
The general formula for the variance $\sigma_V^2$ on a function $V(q_1,q_2,...,q_k)$ for $n$ variables $q_k$ with uncorrelated variances $\sigma_k^2$ is:
$\sigma_V^2 = \sum_{k=1}^{n}\left(\frac{\partial V}{\partial q_k}\right)^2 \sigma_{k}^2$
This is for the small $\sigma_k$ limit.
