# Average caliper diameter (mean width) of a polyhedron

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 $\pi/2$).

• How do you know that your formula is true? – David G. Stork Jul 18 '18 at 20:12
• I was told it's true by an academic but if anyone is able to find a counterexample that would suffice as well – JDoe2 Jul 18 '18 at 20:59
• Also posted to m.se, math.stackexchange.com/questions/2855629/… with no notice to either site. – Gerry Myerson Jul 22 '18 at 13:16
• Apologies @GerryMyerson I didn't realise this was an issue. What would have been the correct site to post to? – JDoe2 Jul 23 '18 at 13:54
• You post to ONE site. If, after, say, a week, you haven't gotten a satisfactory answer, then you post to the other site – but you leave a link at each site to the post at the other. – Gerry Myerson Jul 23 '18 at 21:40

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.

• So would I be correct in thinking that if the shape is a polyhedron it's average curvature is zero $$\langle H\rangle = 0$$ From this this would give that: $$\langle C\rangle =\frac{\pi - \delta }{4\pi }\sum_e L_e$$ This makes sense and agrees with the above... but my main problem is I wanted to apply the more general version of non-regular polyhedrons (where delta isn't the same for angle). Do you have any idea how this proof might be extended? – JDoe2 Jul 18 '18 at 22:04
• * for every angle * – JDoe2 Jul 18 '18 at 22:21
• I have also been unable to access this reference. – JDoe2 Jul 18 '18 at 22:47
• Although it is behind paywall, here is a link in case somebody has access – მამუკა ჯიბლაძე Jul 19 '18 at 2:54
• I added the references for the more general case where also $\delta$ may vary. – Carlo Beenakker Jul 19 '18 at 6:47

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.

• I see! Ah thank you! So for a cube or any polyhedron the left hand side, $\langle \mathcal{K}\rangle A$, should be zero right? – JDoe2 Jul 19 '18 at 10:56
• @JDoe2 That's correct. – j.c. Jul 19 '18 at 11:02

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.

• hmm I see - thank you! Do you know how this might generalise to concave bodies/have any notion of what the equation I originally posted might represent for a concave body? – JDoe2 Aug 6 '18 at 15:41
• for a general body the sum of edge lengths weighted by angle is proportional to the integral over all affine hyperplanes of the Euler characteristic of the body intersected with the hyperplane. If the body is convex, this reduces to the mean width as those intersections are either empty of convex – alesia Feb 24 at 19:56