Can $n$ disjoint lines in $\boldsymbol R^3$ be knotted? No... Let $X_n$ be the configuration space of $n$ disjoint lines in $\boldsymbol R^3$. It is not hard to see that $X_n$ is path connected: Let $L^1,...,L^n$ be $n$ disjoint lines. First wiggle them a little so that no two of them are parallel, and choose coordinates so that none of them are parallel to the $y-z$ plane. Parametrize $L^i=L^i_0$ as $s\mapsto (s, b_i + s\cdot v_i)$, where $b_i,v_i\in\boldsymbol R^2$. For each $t\in\boldsymbol R_{\geq 0}$ shift the $L$ to the left by $t$ units and squeeze by $1/(t+1)$ in the $y-z$ directions, so that $L^i_t$ is parametrized by $(s, (b_i+t\cdot v_i)/(t+1) + s\cdot (v_i/(t+1)))$. In the limit $t\mapsto\infty$, $L^i_{\infty}$ is parametrized by $s\mapsto (s,v_i)$, the lines are all parallel, and the problem reduces to path connectivity of the space of $n$ points in the plane. Fair enough.

What happens if the lines are thickened up a little? Let $Y_n$ be the configuration space of cylinders of radius 1 in $\boldsymbol R^3$. The obvious adaptation of the argument above fails because we aren't allowed to shrink the cylinders. Path connectivity of $X_n$ seems to depend on the fact that we can push whatever rats nest $L$ is arranged in out to infinity, but to do it the lines have to become arbitrarily close to one-another.

Questions: Is $Y_n$ connected? How many connected components does it have? I tried for a while to find a naive, combinatorial proof that $X_n$ is connected but everything I try implies connectivity of $Y_n$.