A standard question in vector calculus is to calculate the volume of the shape carved out by the intersection of $2$ or $3$ perpendicular cylinders of radius $1$ in three dimensional space. Such shapes are known as Steinmetz Solids, and for two and three cylinders we have $$V_2=4\cdot \frac{4}{3},$$ $$ V_3=6(2-\sqrt{2})\cdot \frac{4}{3}.$$

Here, and throughout, when we say that the cylinders intersect, we mean that the centers of the cylinders (which are lines) all intersect at a single point. Moreton Moore wrote an article where he calculates the volume of the intersection of $4$ and $6$ cylinders which arise from identifying faces of the octahedron and dodecahedron respectively. In this case, we have $$V_4=\frac{4}{3}\cdot9(2\sqrt{2}-\sqrt{6}),$$ $$V_6=\frac{4}{3} 4\left(3+2\sqrt{3}-4\sqrt{2}\right).$$

Define $$K_n=\inf\{V: V \text{ is the volume of the intersection of }n\text{ cylinders of radius }1\}.$$ Since we are assuming that the centers of the cylinders share a common point, the sphere will be contained in any configuration of cylinders. The maximum volume is infinite, and the limiting volume will be the sphere, that is $$K_\infty=\pi\frac{4}{3}.$$ I believe that for reasons of symmetry, for $n=2,3,4,6$ we have $K_n=V_n$. Naturally, this leads to the question:

What is the optimal configuration of cylinders for each $n$, and what is the resulting volume?

While this may be a difficult problem, I am interested in the more approachable problem of evaluating the rate of convergence to the sphere. Can we identify how fast $K_n$ approaches $K_\infty$?

Question: Let $$f(n)=\frac{K_n}{4/3}-\pi.$$ Then $f(n)$ is monotonically decreasing to $0$, and $f(n)$ measures how fast $K_n$ approaches the volume of the sphere. At what rate is $f(n)$ decreasing to $0$? Can we find the order of magnitude of $f(n)$?

  • $\begingroup$ You should make it clear that the axes of these cylinders have exactly one point in common. $\endgroup$ – The Masked Avenger Jan 13 '14 at 1:28
  • $\begingroup$ You might also drop r or set r=1 after stipulating that all the cylinders have the same radius. $\endgroup$ – The Masked Avenger Jan 13 '14 at 1:32
  • $\begingroup$ Do you assume the central axes of the cylinders all intersect at a certain point? $\endgroup$ – Will Sawin Jan 13 '14 at 1:37
  • $\begingroup$ Perhaps finding an optimal configuration is as difficult as finding the optimal packing of $n$ disks on a sphere... $\endgroup$ – Joseph O'Rourke Jan 13 '14 at 1:52
  • 1
    $\begingroup$ @JosephO'Rourke: The minimum volume is of interest since the maximum volume is infinite. $\endgroup$ – Eric Naslund Jan 13 '14 at 2:03

This is a heuristic, suggesting $f(n)=O( 1/n^2)$.

Consider the sphere of radius $r$. Each cylinder intersects this sphere in a great circle. The great circles divide the sphere into a number of regions. By induction we can see that this is at most $n(n-1)+2$. So if the sphere is relatively evenly divided into regions, each region must have diameter asymptotic to $1/n$. The intersection of the $n$ cylinders is a ball, plus a little piece jutting out in each of these regions. The little piece is approximately a cone of radius $O(1/n)$ and slope $O(1/n)$, so it has height $O(1/n^2)$. Since the area of the base is $O(1/n^2)$, it has volume $O(1/n^4)$. So the sum of all the pieces has volume $O(1/n^2)$.

The fact that the cone is on a spherical base and not a flat base cuts its thickness by $1/2$,so is responsible for only a constant factor.

So assuming that:

i. It's possible to choose the great circles to be relatively evenly distributed

ii. Weirdly-shaped regions don't cause any problems

iii. There's not some other error in these heuristic arguments:

the asymptotic is $f(n) = O(1/n^2)$.

Edit: We can make the upper bound part of this more rigorous using the probabilistic method. Place all the cylinders randomly. Fix a point on the sphere. Consider a ray out from that point - how far does that ray travel in the intersection of the cylinders? It is at most $d^2/2+$-higher order terms, where $d$ is the distance to the nearest great circle corresponding to a cylinder. (In fact, if we express the distance as an angle $\theta$, it is exactly $\sec \theta - 1$.)

Instead of finding the nearest great circle corresponding to a cylinder, we can just look in a random direction and find the nearest great circle corresponding to a cylinder in that direction. So we look along some great circle passing through our point. Each great circle corresponding to a cylinder intersects this one at two random antipodal points. So we need to find the distribution of the distance to the nearest point among $n$ randomly chosen points on the unit circle. This converges rapidly to an exponential distribution of mean $1/n$. Because this distribution decays so rapidly, only the lowest-order term for the radius or the local contribution to volume matters when computing the expected value of the volume. The lowest-order term is proportional to the radius, which is proportional to $d^2$, whose expected value is $1/n^2$. Averaging over all the points, the volume is $O(1/n^2)$

Edit 2: We can make the lower bound rigorous using a simple estimate. The extra radius at a given point is $\sec \theta-1$, where $\theta$ is the distance to the nearest great circle corresponding to a cylinder. For any distance $\epsilon$, the points of distance no more than $\epsilon$ from a single great circle form a thin tube of area at most $ 4 \pi \epsilon$. So the total area of points that are within $\epsilon$ of any great circle form an area of at most $ 4 \pi n \epsilon$. The area of points that are not within $\epsilon$ of any great circle has area at least $4 \pi (1- n \epsilon)$. So if we set $\epsilon=2n/3$, say, then one-third the points of the sphere will extend outwards by at least $\sec \epsilon -1 \ geq \epsilon^2/2 = 2/9n^2$. This gives a lower bound of the form $O(1/n^2)$.


The rate of convergence of $f(n)$ to zero seems about the same as the rate of the best approximation of the ball's volume by the volume of a circumscribed centrally-symmetric polyhedron with $2n$ facets, which should be known already. (The problem of approximating the ball by circumscribed polyhedra has been studied extensively.) When $n$ is large, the volume of the intersection of $n$ cylinders, each tangent to two opposite faces of the best approximating $2n$-faced circumscribed polyhedron is very close to the volume of the polyhedron. For example, look at the illustration posted by Joe: already for $n=3$, the intersection of the three cylinders looks a bit like the circumscribed cube. The faces are curved, but for large $n$, as all faces become very small, the curved faces become relatively less curved and extremely close to the flat ones.


Not an answer; just an illustration for $n=3$:
     (Image by A.J. Hildebrand, Lingyi Kong, Abby Turner, Ananya Uppal, from this link.)


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