This is a comment from Do Carmo's Differential Geometry of Curves and Surfaces:
Equation $\iint n dp d\theta=2l$ (where $n$ is thenumber of intersection points of a curve $C$ with all straight lines in the plane that meet $C$) can also be used to obtain an efficient way of estimating lengths of curves. Indeed, a good approximation for the integral $\iint n dp d\theta$ is given as follows. Consider a family of parallel straight lines such that two consecutive lines are at a distance $r$. Rotate this family by angles of $\pi/4, 2\pi/4, 3\pi/4$ in order to obtain four families of straight lines. Let $n$ be the number of intersection points of a curve $C$ with all these lines. Then $\frac{1}{2}{nr} \frac{\pi}{4}$ is an approximation to the integral $ \iint n dp d\theta=\hbox{length of }C$.
My question: How can I prove it and if it is true that $\lim_{r \to 0}\frac{1}{2}n(r)r\frac{\pi}{4}=\hbox{length of }C$?
I couldn't understand how the author arrived at the formula $\frac{1}{2}{nr} \frac{\pi}{4}$.
In the ultimate reference on integral geometry, Santaló's Integral Geometry and Geometric Probability, he gives this formula, originally due to Steinhaus (one reference is "Length, shape and area," Colloq. Math. 3 (1954), 1-13), on pp. 31-32. If we turning the grid through multiples of $\pi/m$, we obtain $$\ell\approx \frac{N\pi r}{2m}.$$ If we let $r\to 0$ and $m\to\infty$, then we can suspect with some hypotheses that this is calculating the integral $\dfrac12\iint n\,dp\,d\theta$.
Here's a reference for the error estimates: P.A.P. Moran, "Measuring the length of a curve," Biometrika 53 (1966), 359-364.