Proposition: Vanishing periodic transient

The phase-explicit periodic dependency quotient has at most

vertices. Every quotient edge has physical displacement at most $a_n$ and duration $h_n$. Hence

Its limiting scheduled velocity polytope ${K_n}^{\mathrm{vel}}$ satisfies

and therefore converges in Hausdorff distance to $c{\overline{B}}_2$.

Proof. Modulo translations there are one hub type and $S_n{F}_n$ lane types. Including $F_n$ schedule phases gives the vertex bound. The edge displacement bound is Lemma lem:partition. Since $F_n=O(n)$, $S_n=O(n^2)$, and $a_n=O(n^{-9})$, the offset is $O(n^{-5})$.

The quotient is strongly connected. Every graph edge is exposed at some phase, waiting is always possible, and the substrate graph is connected. One can therefore traverse a finite graph route and adjust the terminal phase by waiting.

For completeness, the cycle argument is as follows. Removing cycles from a quotient walk leaves at most $q_n-1$ edges. The removed cycles decompose into simple directed cycles. Their displacement divided by duration is a time-weighted convex combination of simple-cycle velocities. Conversely, strong connectivity supplies fixed connectors between cycle bases, whose cost becomes negligible under repeated cycles. Waiting supplies zero velocity. Thus the limiting velocities form the convex hull of zero and the simple-cycle velocities, and finite-time displacement has offset at most $(q_n-1)D_{\max ,n}$.

The outer inclusion follows from the exact edgewise speed bound. For the inner inclusion, fix $n$ and a unit vector $v$. Choose increasingly distant hubs in direction $v$. Theorem thm:routing reaches them in time at most ${\lambda }_{n}L/c+2U_n+h_n$. Add waiting to make elapsed times asymptotic to ${\lambda }_{n}L/c$. The limiting velocity is $(c/{\lambda }_n)v$. Convexity and waiting give the entire inner ball.