# Queue Spillover Management in a Connected Vehicle Environment
## 3 Intersection Signal Control Modeling
### Phase Maximum Control Distance  
The maximum control distance between the intersection phase, is related to the maximum queue length of the intersection. The maximum queue length is the length of the road between the intersections. Given the different types of vehicles, speed, and acceleration performance of vehicles passing through intersections, the equivalent cars are taken as the research object. The intersections and related parameters, are shown in Figure 3. In Figure 3, \(L\) denotes the distance between the intersection of roads, \(L_{0}\) denotes the intersection of queuing vehicles in front of the length, \(L_{i}\) denotes the distance between the \(i\)-th vehicle on the road and the stop line of the intersection. And the straight, left-turn lanes are separated. To describe the position of the vehicle on the intersection, we used two coordinates: longitudinal X, lateral Y. X and Y are the local coordinates in the viewpoint of the stop line. In Figure 4, the axis X indicates the road traveled by the car. The origin o indicates the position of the intersection stop line, and the positive direction of the axis X is the direction of driving. The calculation of the phase maximum control distance, is determined by the vehicle traveling model, and the specific steps are as follows. Assume that the green light moment is \(t=0\). The vehicle starting delay is \(T\). The vehicle acceleration is \(a\), and the location of the \(n\)-th car at the time of \(t\) is \(S_{n}(t)\). Thus, the starting time of the \(n\)-th car is \(t_{n}\). The coordinates of the stop line of the intersection are (0, 0). Assume that the coordinates of the vehicle are \((x_{n},y_{n})\), so the position of the \(n\)-th car can be calculated according to (1) before the car startup. \[S_{n}(0)=\sqrt{\left(x_{n}-0\right)^{2}-\left(y_{n}-0\right)^{2}} \tag{1}\]  
Figure 4: Intersection coordinates. Figure 3: Schematic diagram of the intersection. The start of the car is moving backwards from the first car. So, the starting time of vehicles, for different positions is calculated as follows:  
\[t_{n}=(n-1)T \tag{2}\]  
Suppose the vehicle drives at a uniform acceleration after startup. The position of the car at moment \(t\ (t>t_{n})\) can be expressed as follows:  
\[S_{n}(t)=S_{n}(0)+\frac{a(t-t_{n})^{2}}{2} \tag{3}\]  
Then the vehicle's driving model shows as follows:  
\[S_{n}(t)=\left\{ \begin{array}{ll}-S_{n}(0)&\quad 0\leq t\leq t_{n}\\ -S_{n}(0)+\frac{a(t-t_{n})^{2}}{2}&\quad t>t_{n}\end{array} \right. \tag{4}\]  
However, the vehicle's driving, needs to be limited by the maximum speed of the roads. Assuming the road's limitation of maximum speed is \(v_{\max}\ \mathrm{km/h}\), the car will be uniformly accelerated to \(v_{\max}\), and keeping the speed \(v_{\max}\) for forward driving. The time for uniform acceleration of the car is as follows:  
\[t_{n}^{\max}-t_{n}=\frac{v_{\max}}{3.6\cdot a} \tag{5}\]  
where: \(t_{n}^{\max}\) is the time when the \(n\)-th car reaches the speed limitation. When the green light is on, the model of the car driving through the intersection is as follows:  
\[S_{n}(t)=\left\{ \begin{array}{ll}-S_{n}(0)&\quad 0\leq t\leq t_{n}\\ -S_{n}(0)+\frac{a(t-t_{n})^{2}}{2}&\quad t_{n}\leq t\leq t_{n}^{\max}\\ -S_{n}(0)+\frac{a(t-t_{n})^{2}}{2}+\frac{v_{\max}(t-t_{n}^{\max})}{3.6}&\quad t \geq t_{n}^{\max}\end{array} \right. \tag{6}\]  
when meeting the condition that the \(n\)-th vehicle is located at the stop line of the end of \(t_{g}\) (green light). That is, (6) satisfies 1\(S_{n}\big{(}t_{g}\big{)}=0\), 2\(S_{n}(0)=-\sqrt{x_{n}^{2}+y_{n}^{2}}\), 3\(t_{n}=(n-1)T\), then the current phase maximum number of vehicles can be obtained, as shown in (7). The distance between the initial position of the \(n\)-th car and the intersection is the phase maximum control distance. \[n=\left\{