Measuring curvature in Flatland Gauss' Theorema egregium says that

the Gaussian curvature of a surface can be
  determined entirely by measuring angles, distances and their rates on
  the surface itself.

A surface looks like $\mathbb{R}^2$ locally  so the angle sum of arbitrarily small triangles tends to $\pi$, doesn't it? Only when one considers bigger triangles - as Gauss did - one will find angle sums deviating from $\pi$. So I wonder how - concretely - an inhabitant of an arbitrarily (but smoothly) curved surface would measure the Gaussian curvature at a given point. 
 A: Fix an opening angle and draw an isosceles geodesic triangle with the two legs of length $r$ from that opening angle. Measure the angle defect. 
Do this many times for a sequence of $r\to 0$, plot the angle defect versus $r^2$. The limiting slope would be the curvature at the point. As long as the curvature does not change too wildly from point to point, you can get a good approximation for small but finite $r$. 
A: Let's use normal coordinates. The circumference of a circle of radius r around 0 in these coordinates is then given by $$C(r)=\int\limits_{0}^{2\pi}\sqrt{g_{ij}\frac{dx^{i}}{d\phi}\frac{dx^{j}}{d\phi}}d\phi$$ where $x^{1}(\phi)=r \cos(\phi)$, $x^{2}(\phi)=r \sin(\phi)$. In these coordinates we have the expansion $g_{ij}=\delta_{ij}-\frac{1}{3}R_{ikjl}x^{k}x^{l}+O(\vert x \vert^{3})$, so by doing a little Taylor expansion $$C(r) = \int\limits_{0}^{2\pi}\sqrt{r^2-\frac{r^4}{3}R_{1221}+O(r^{5})}d\phi=2\pi\left(r-R_{1221}\frac{r^3}{6}\right)+O(r^{4})$$ with derivative $$\frac{dC(r)}{dr}=2\pi-\pi R_{1221}r^2+O(r^{3})$$
Therefore one could try to measure the sectional curvature (agrees with the Gaussian curvature in an local orthonormal frame like here, $K=R_{1221}$) by measuring the circumferences of circles around some point and investigate how it depends on the radius.
