Probability that two infinite cylinders overlap 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: $P(d \le r_0+r_1)=1$, $P(d \gg r_0+r1) \approx 0$.
 A: 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 $\displaystyle\quad\frac{2}{\pi}\sin^{-1}(\frac{r_0+r_1}{d})$ 
A: 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.
