Module 7 : Traffic Signal Design

Lecture 34 : Design Principles of Traffic Signal

1

2

3

4

5

6

7

	Green splitting Green splitting or apportioning of green time is the proportioning of effective green time in each of the signal phase. The green splitting is given by, $$g_i = \left[\frac{V_{c_i}}{\sum_{i=1}^N{V_{c_i}}}\right]\times t_g$$ (1) where $V_{c_i}$ is the critical lane volume and $t_g$ is the total effective green time available in a cycle. This will be cycle time minus the total lost time for all the phases. Therefore, $$t_g = C - N~t_L$$ (2) where $C$ is the cycle time in seconds, $n$ is the number of phases, and $t_L$ is the lost time per phase. If lost time is different for different phases, then effective green time can be computed as follows: $$t_g=C-{\sum_{i=1}^N{t_{L_i}}}$$ (3) where $t_{L_i}$ is the lost time for phase $i$, $N$ is the number of phases and $C$ is the cycle time in seconds. Actual green time can be now found out as, $$G_i=g_i-y_i+t_{L_i}$$ (4) where $G_i$ is the actual green time, $g_i$ is the effective green time available, $y_i$ is the amber time, and $L_i$ is the lost time for phase $i$. Numerical example The phase diagram with flow values of an intersection with two phases is shown in figure 1. Figure 1: Phase diagram for an intersection The lost time and yellow time for the first phase is 2.5 and 3 seconds respectively. For the second phase the lost time and yellow time are 3.5 and 4 seconds respectively. If the cycle time is 120 seconds, find the green time allocated for the two phases. Solution Critical lane volume for the first phase, $V_{C_1}$ = 1000 vph. Critical lane volume for the second phase, $V_{C_2}$ = 600 vph. Total critical lane volumes, $V_C = V_{C_1}+V_{C_2}$ = 1000+600 = 1600 vph. Effective green time can be found out from equation 2 as $T_g$=120-(2.5-3.5)= 114 seconds. Green time for the first phase, $g_1$ can be found out from equation 1 as $g_1 = \frac{1000}{1600}\times 114$ = 71.25 seconds. Green time for the second phase, $g_2$ can be found out from equation 1 as $g_2 = \frac{600}{1600}\times114$= 42.75 seconds. Actual green time can be found out from equation 4. Thus actual green time for the first phase, $G_1$ = 71.25-3+2.5 = 71 seconds (rounded). Actual green time for the second phase, $G_2$ = 42.75-4+3.5 = 42 seconds (rounded). The phase diagram is as shown in figure 2. Figure 2: Timing diagram