# Average Diameter 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 $\pi/2$).

• What is an average diameter? Jul 18, 2018 at 15:08
• @lisyarus OP probably mean mean width. the angular average of the length of projection of an object along all possible directions. See wiki entry for definition. Jul 18, 2018 at 15:26
• @achillehui This makes perfect sense, thank you! (btw, I've read the "mean mean" part several times until I understood this is not a typo) Jul 18, 2018 at 16:38
• @lisyarus Ah sorry - so if the polyhedron was sandwiched tightly between two parallel planes define the calliper distance as the perpendicular distance between these two planes. The average calliper distance over all rotations of the polyhedron would then be what I meant by the average diameter. Apologies I can see this was not clear. I will update the original post. Jul 18, 2018 at 18:48
• Also posted to MO with no notice to either site, mathoverflow.net/questions/306318/… Jul 22, 2018 at 13:17

This answer try to derive the formula from first principle. For an alternative derivation of the formula using relation between mean-width and integral over mean curvature, see the second half of my answer to a related question.

Please note that the formula can fail for non-convex polyhedron (e.g. $[0,1]^3 \setminus [\epsilon,1-\epsilon]^2\times [\epsilon,1]$) We will assume the polyhedron at hand is a convex one.

Let $\pi_1, \pi_2$ be the projections: $\;\;\mathbb{R}^3 \ni (x,y,z) \;\; \overset{\pi_2}{\longmapsto}\;\; \overset{\in \mathbb{R}^2}{(x,y)} \;\; \overset{\pi_1}{\longmapsto}\;\; x \in \mathbb{R}$
Let $\mu O$ be the length/area/volume for $1/2/3$-dim geometric object $O$ in $\mathbb{R}^3$.
Let $R_3$ be a random rotation uniformly sampled over $SO(3)$.

For any convex polyhedron $P$, let $\;\nu P = \verb/E/_{R_3}[\mu \pi_1\pi_2 R_3 P]\;$ be its average caliper diameter.

Let $E(P)$ be the edges of $P$ and $E'(P) = \{ e \in E(P) : \pi_2 e \subset \partial \pi_2 P \}$. When none of the faces of $P$ is tangent to the $z$-direction, edges in $E'(P)$ will get mapped to edges of $\pi_2 P$. Furthermore, the correspondence will be bijective. In the illustration below, $P$ is a dodecahedron and $\pi_2 P$ is the gray polygon on $xy$-plane. $E'(P)$ are those red edges on $P$ and under $\pi_2$, they get mapped to the blue edges of $\pi_2 P$.

$\hspace1in$

If one further project these edges down to $x$-axis using $\pi_1$ and sum over their length, one will notice the sum overs those edges in $E'(P)$ whose $\pi_2 e$ is facing upwards or exactly right equals to the caliper diameter of $P$ along $x$-direction. The same thing happens to the sum over those edges whose $\pi_2 e$ is facing downwards or exactly left. This leads to

$$\mu \pi_1\pi_2 P = \frac12 \sum_{e \in E'(P)} \mu \pi_1 \pi_2 e = \frac12 \sum_{e \in E(P)} \verb/1/_{E'(P)}(e) \mu \pi_1 \pi_2 e$$ where $\displaystyle\;\verb/1/_{E'(P)}(e) = \begin{cases} 1,& e \in E'(P)\\ 0, & \text{ otherwise }\end{cases}\;$ is the indicator function for $e \in E'(P)$.

Replace $P$ by $R_3 P$ and taking expectation value. Notice the probability for $R_3P$ to have any face tangent to $z$-direction is zero, one obtain

$$\nu P = \frac12 \sum_{e\in E(P)} \verb/E/_{R_3}[ \verb/1/_{E'(R_3P)}(R_3e) \mu \pi_1\pi_2 R_3 e ]$$ For any edge $e$, let $L_e = \mu e$ be its length. Let $t_e$ be a unit vector pointing from one end-point of $e$ to another. Let $n_{e,1}$, $n_{e,2}$ be the outward pointing unit normal vectors for the two faces attached to edge $e$. Let $\psi_e = \pi - \delta_e$ be the angle between $n_{e,1}$ and $n_{e,2}$.

If one pick a unit vector $t$ and look at those $R \in SO(3)$ which make $t_{Re} = t$. One will notice $n_{Re,1}$ and $n_{Re,2}$ lies on a circle which lives on a plane normal to $t$. As long as $t$ is not parallel to the $z$-direction, $Re \in E'(RP)$ when the arc between $n_{Re,1}$ and $n_{Re,2}$ cut the $xy$-plane. This leads to

$$\verb/E/_{R_3}[ \verb/1/_{E'(R_3P)}(R_3e) | t_{R_3e} = t ] = \verb/Pr/_{R_3} [ R_3 e \in E'(R_3P) | t_{R_3 e} = t ] = \frac{\psi_e}{\pi} = 1 - \frac{\delta_e}{\pi} \quad$$ Aside from events of probability zero, this conditional expectation is independent of $t$, one find $$\verb/E/_{R_3}[ \verb/1/_{R_3P}(R_3e) \mu \pi_1\pi_2 R_3 e] = \left(1 - \frac{\delta_e}{\pi}\right) \verb/E/_{R_3}[ \mu \pi_1\pi_2 R_3 e ] = \left(1 - \frac{\delta_e}{\pi}\right) \frac{L_e}{2}$$ As a result, $$\nu P = \sum_{e \in E(P)} \frac{\pi - \delta_e}{4\pi} L_e$$

• hmm okay... thank you! Do you have any notion of what this sum represents for concave shapes, if not the average calliper diameter? Aug 6, 2018 at 15:35