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Given two infinite cylinders:

  • radius $r_0$, centered at the origin and pointing $\hat u_1=(0,0,1)$
  • radius $r_1$, centered at $(d,0,0)$ and pointing at $u_2$

For a given distance $d$, if $u_2$ is selected uniformly at random over all possible orientations, what is the probability $P(d; r_0,r_1)$ that the two infinite cylinders will overlap?

Trivial limits: $P(d \le r_0+r_1)=1$, $P(d \gg r_0+r1) \approx 0$.

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For what it's worth, the cylinders overlap iff $\lvert(d,0,0)\cdot\hat n\rvert \le r_0 + r_1$, where $\hat n=\hat u_1\times\hat u_2/\lVert\hat u_1\times\hat u_2\rVert$. – Rahul Aug 15 '12 at 19:05
up vote 1 down vote accepted

Assume $r_0 + r_1 < d$ or everything is trivial. It is clear

  • The two cylinder overlaps $\iff$ their projections onto the $xy$-plane overlap.
  • The projection of the $1^{st}$ cylinder is a circle centered at $(0,0)$ with radius $r_0$.
    The projection of the $2^{nd}$ cylinder is a rectangular strip of width $2r_1$ whose center axis passes through $(d,0)$.
  • These two projections overlap

    $\iff$ center axis of $2^{nd}$ projection intersect a circle of radius $r_0 + r_1$ centered at $(0,0)$.
    $\iff$ angle between this center axis and $x$-axis $\displaystyle \le \sin^{-1}(\frac{r_0+r_1}{d})$

Combine these, one find the probability of overlap is $\displaystyle\quad\frac{2}{\pi}\sin^{-1}(\frac{r_0+r_1}{d})$

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We can restrict our attention to the case where $u2$ is a unit vector, and write it as $u2 = (\cos\theta\cos\phi, \sin\theta\cos\phi, \sin\phi)$. Then Rahul's condition gives you a region in $(\theta, \phi)$ space in which overlap occurs. You can calculate the area of this region by doing some integration. On the other hand, the total area of $(\theta, \phi)$ space is $4\pi$ (the surface area of a unit sphere). The probability of overlap is the ratio of these two areas.

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