Average caliper diameter (mean width) of a polyhedron

Question

Define the caliper diameter of a polyhedron as follows:

Let $P_1$ and $P_2$ be two planes both of which are parallel to the x axis such that the perpendicular distance between $P_1$ and $P_2$ is the smallest possible distance allowing the whole of the polyhedron to lie in the region of space between the two planes. Define the perpendicular distance between the two planes as the caliper diameter of the polyhedron.

How do I prove that the average caliper diameter of the polyhedron across all possible rotations is given by this formula:

$$\sum_{e\in E} L_e(\pi - \delta_e)/(4\pi)$$

Where $E$ is the set of all edges of the polyhedron, $L_e$ is the length of edge $e$ and $\delta_e$ is the interior angle where the two faces forming edge $e$ meet (e.g. for a cube the interior angle between two faces is always $\pi/2$).

Answer 1

The mean caliper formula was derived in 1967 by the metallurgist John Cahn [1]. Cahn allowed for a curvature $K$ of the faces and assumed that all dihedral angles $\pi-\delta_e$ are the same, equal to $\epsilon$. Then the average curvature $\langle K\rangle$ is related to the average caliper $\langle C\rangle$ by $$\langle K\rangle=\frac{2\pi}{A}\left(\langle C\rangle-\frac{\epsilon}{4\pi}\sum_e L_e\right)$$ where $A$ is the total area of the faces and $L_e$ is the length of edge number $e$.

A derivation for $\langle K\rangle=0$ that also allows for varying $\delta_e$ is given in [2] (page 231) and in [3] (page 825).

[1] J.W. Cahn, The significance of average mean curvature and its determination by quantitative metallography, Trans. Met. Soc. AIME 239 (1967) 610-616. _{behind a paywall}

[2] R.E. Miles, Poisson flats in Euclidean spaces. Part I: A finite number of random uniform flats, Adv. App. Prob. 1 (1969) 211-237.

[3] R.E. Miles, Direct Derivations of Certain Surface Integral Formulae for the Mean Projections of a Convex Set, Adv. Applied Prob. 7 (1975), 818-829.

Answer 2

The caliper width is also known as the "mean width" of a body, and is one of the valuations studied in integral geometry / geometric probability. "Cahn's formula" and its generalizations as given in Carlo Beenakker's answer amount to the fact that the mean width of a body is expressible as the "total mean curvature" of its bounding surface -- the left-hand side $\langle \mathcal{K}\rangle A$ is the total mean curvature of the smooth part and the term $\frac{\epsilon}{2}\sum_{e}L_e$ on the right-hand side is the total mean curvature contribution from the (singular) edges. For some more keywords and references see this fairly recent MO question.

A derivation of Cahn's formula is given in the Supplementary Material for MacPherson and Srolovitz's beautiful paper "Generalization of the von Neumann relation to coarsening of cellular microstructures in 3D". Actually, the formula there is stated for the special case where $\epsilon = 2\pi/3$, but it's not hard to adapt their proof to general $\epsilon$ and with a little bit of effort, even to the case where the dihedral angles of each of the edges may be different.

Answer 3

As j.c. mentioned in his answer, the average distance between parallel supporting planes is better known as the mean width. More generally, one can take a convex body in $\mathbb{R}^n$ and consider the average $k$-volume of its projections to $k$-dimensional linear subspaces (the case $k=1$ is the mean width). These quantities are called quermassintegrals, and they are proportional to the coefficients in the Steiner formula for the volume of the parallel body. Your question is about the case $n=3$, $k=1$ of this theorem.

To prove it, start with $n=3$ and $k=2$. This is Cauchy's theorem which says that the area of the boundary of the convex body is one quarter of its average projection area: $$ Area(\partial K) = \frac1{\pi} \int\limits_{\xi} Area(pr_\xi K)\, d\xi, $$ where the integral is taken over all unit vectors $\xi$, and $pr_\xi K$ is the projection of $K$ to $\xi^\perp$. The proof is very nice, just integrate the projection area of each face individually.

Now replace $K$ in the Cauchy formula by the $K_\epsilon$, the $\epsilon$-neighborhood of $K$ and apply the Steiner formula to both sides. The LHS is easy to compute by decomposing the boundary of $K_\epsilon$ into flats, pieces of cylinders, and pieces of spheres: $$ Area(\partial K_\epsilon) = Area(\partial K) + \epsilon\sum_e \ell_e(\pi-\delta_e) + \epsilon^2 \cdot 4\pi. $$ For the RHS observe that $pr_\xi(K_\epsilon) = (pr_\xi K)_\epsilon$ and expand the integral in a similar way: $$ Area(pr_\xi K_\epsilon) = Area(pr_\xi K) + \epsilon L(\partial pr_\xi K) + \epsilon^2 \cdot \pi. $$ Now integrating and comparing the coefficients at $\epsilon$ we obtain $$ \sum_e \ell_e(\pi-\delta_e) = \frac1\pi \int_\xi L(\partial pr_\xi K)\, d\xi =\frac1\pi \int_\xi \frac12 \int_{\eta \in \xi^\perp} L(pr_{\xi\oplus\eta}K)\, d\eta d\xi, $$ where we applied the Crofton formula relating the length of a curve to the average length of its projections. It remains to transform the right hand side: $$ \frac1\pi \int_\xi \frac12 \int_{\eta \in \xi^\perp} L(pr_{\xi\oplus\eta}K)\, d\eta d\xi = \int_\zeta L(pr_{\zeta^\perp}K)\, d\zeta, $$ and we get the integral of lengths of projections of $K$ to lines.

This proof comes from 1920's, I guess, and the procedure is called Kubota's recursion. For exact references see Notes for Section 4.5 of

Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001.