,-1.5) -- (-0.5,1.5);
\draw[thick] (1.5,-1.5) -- (1.5,1.5);
\draw[thick] (-1.5,-1.5) -- (-1.5,1.5);
\draw[thick] (-1.5,1.5) -- (+1.5,1.5);
\draw[thick] (-1.5,0.5) -- (+1.5,0.5);
\draw[thick] (-1.5,-0.5) -- (+1.5,-0.5);
\draw[thick] (-1.5,-1.5) -- (+1.5,-1.5);
\draw [red, very thick] plot [smooth, tension=1] coordinates {(1.5,1.2) (1.2,1.3) (1.1,1.3) (1.0,0.9) (0.8,1.1) (0.5,1.2) (0.1,0.6) (-0.5,1.2) (-1.1,1.1) (-1.5,0.9)};
\draw [red, very thick] plot [smooth, tension=1] coordinates{(1.1,-1.5) (0.9,-1.1) (1.3,-0.9) (1.1, -0.6) (0.7, -0.2) (1.0,0.4) (1.1,1.0) (1.3,1.2)(1.2,1.5)};
\draw [red, very thick] plot [smooth, tension=1] coordinates {(1.5,-1.2) (1.2,-1.3) (1.0,-0.9) (0.8,-1.1) (0.5,-1.1) (0.1,-0.7) (-0.5,-0.7) (-1.1,-1.2) (-1.5,-0.6)};
\draw [red, very thick] plot [smooth, tension=0.5] coordinates{(-0.7,1.5)(-1.2,1.3)(-1.2,1.0)(-1.4,0.6)(-0.7,0.2)(-1,-0.2)(-1.3,-0.6)(-0.6,-1.5)};

\draw[ultra thick, fill = black!20] (-0.5,-0.5) -- (0.5,-0.5) -- (0.5,0.5) -- (-0.5,0.5) -- cycle ;
\draw[decorate, decoration={brace}, yshift=0.5ex]  (-1.5,1.5) -- node[above=0.4ex] {$3mn$}  (1.5,1.5);
\end{tikzpicture}
\caption{Illustration of a good box: The gray box consists of $m^2$ boxes $[0,n]^2$. It is surrounded by left-right and up-down crossings of $n$-good boxes.}
\end{figure}
Now, that the $m$-box is surrounded, we have a joint bound for $\bar R^a_x<2\sqrt{2}mn$ for all $x \in Q_{mn}$. Putting things together, we first pick $n$ sufficiently large and $a$ sufficiently small, such that $\lim_{m\uparrow\infty}\P(\bar R^a(Q_{mn})\ge 2\sqrt{2}mn)=0$. Then, using a union bound and shift invariance, this implies that also $\lim_{m\uparrow\infty}\P(\bar R^a(Q_{mn})\ge mn)=0$ as required.
\end{proof}






\subsection{Regimes for absence of percolation}\label{sec-noperc}
Our first result establishes that indeed, insufficiently many devices and a too slow speed lead to absence of percolation. 
\begin{theorem}\label{thm_lowintense}
For any $\kappa$, $\mathcal S$ and all $r,\rho \ge 0$ the following holds:
\begin{enumerate}
\item[(1)] For all $\lambda\ge 0$ and $\mu_{\mathrm v}$ we have that $T_{\mathrm c}(\lambda,\mu_{\mathrm v},\rho,r)>2\rho$.
\item[(2)] For all $T\ge 0$ and $\mu_{\mathrm v}$ we have that $\lambda_{\mathrm c}(T,\mu_{\mathrm v},\rho,r)>0$.
\item[(3)] For all $T,\lambda\ge 0$ we have that $v_{\mathrm c}(T,\lambda,\rho,r)>0$. 
\end{enumerate}
\end{theorem}
The proof of (2) rests on the idea that we can bound the process by the easier-controllable Cox--Gilbert graph.
For (1) and (3) this comparison fails, and we need to couple our space-continuous model to site-percolation in $\mathbb Z^d$ and employ stabilization arguments. We present the detailed proof in Section~\ref{Sec_Proof_Sub_Rho}.

Let us mention that the comparison to Gilbert graphs also provides rough upper bounds for $\lambda_{\mathrm c}(T,\mu_{\mathrm v},\rho,r)$ and $v_{\mathrm c}(T,\lambda,\rho,r)$ with the help of the numerical analysis provided in \cite{wias9,wias2}. 

\medskip
Let us mention that item (3) in Theorem~\ref{thm_lowintense} does not hold for $T=\infty$. In this case we have $v_{\mathrm c}(\infty,\lambda,\rho,r)=0$ once $\P(g_{\infty,\mu_{\mathrm v},\rho,r}(X^\lambda)\text{ percolates})>0$ for some parameter configuration. Indeed, using Part (2) in Proposition~\ref{prop_scale}, we also have percolation for all $0<a<1$, i.e.,  $0<\P(g_{\infty,\mu_{\mathrm v}^{a},\rho/a,r}(X^\lambda)\text{ percolates})\le \P(g_{\infty,\mu_{\mathrm v}^{a},\rho,r}(X^\lambda)\text{ percolates})$, where we also used monotonicity in $\rho$.


\medskip
Next, note that $r_{\mathrm c}$ can very well be zero, since devices with non-parallel movement on a street share the same position exactly once, leading to a connection even for $r=0$, if $\rho=0$. However, devices that move together in the same direction can not exchange messages if $r=0$. For the consequences of this observation regarding the critical connectivity threshold, see Theorem~\ref{Thm_Rho_1} below. Let us note that, in principle, absence of percolation for $r=0$ could still imply $r_{\mathrm c}(T,\lambda,\mu_{\mathrm v},\rho)=0$. This would be the case if $g_{T,\mu_{\mathrm v},\rho,r}(X^\lambda)$ percolates for all positive $r$, which indicates that the parameters $(T,\lambda,\mu_{\mathrm v},\rho,0)$ are critical for percolation, rendering a mathematical analysis extremely hard.



\medskip
In order to see absence of percolation for large speeds $v_{\rm min}$, first note that any street $s$ such that $2|s|/v_{\rm min}<\rho$ can not be used to exchange messages. However, every street can be used for traveling. More precisely, in order to establish a connection, any device has to remain at least a time of $\rho$ on some street and during that time travels at least a distance of $\rho v_{\rm min}$. As devices reverse their direction when they reach either their destination or their initial position, the minimal length of a street such that communication on it is possible is given by $\rho v_{\rm min}/2$. Although staying on a street of length smaller than $\rho v_{\rm min}/2$ is possible if both the initial position and the destination are on that street, those devices can only connect to the same type of devices that have both their initial position and destination on the same street. In that way, only local connective components are formed, we will ignore this possibility in the following considerations.

We have the following result on absence of percolation for sufficiently high velocities. 
\begin{theorem}\label{thm_largespeeds}
We consider street systems of Poisson--Voronoi or Poisson--Delaunay type.
Then, for any $\kappa$ with bounded support and all $T, \lambda,r,\rho\ge 0$ we have that $v^{\mathrm c}(T,\lambda,\rho,r)<\infty$. 
\end{theorem}
Let us note that we in fact prove that $v^{\mathrm c}(\infty,\lambda,0,r)<\infty$, which implies a uniform bound in $T$ and $\rho$ by the corresponding monotonicites