Module 7 : Traffic Signal Design

Lecture 36 : Special Requirement in Traffic Signal

1

2

3

4

5

6

7

	Numerical example The traffic flow for a four-legged intersection is as shown in figure 1. Figure 1: Traffic flow for a typical four-legged intersection Given that the lost time per phase is 2.4 seconds, saturation headway is 2.2 seconds, amber time is 3 seconds per phase, find the cycle length, green time and performance measure(delay per cycle). Assume critical $v/c$ ratio as 0.9. Solution The phase plan is as shown in figure 2. Figure 2: Phase plan Sum of critical lane volumes is the sum of maximum lane volumes in each phase, $\Sigma V_{Ci}$ = 433+417+233+215 = 1298 vph. Saturation flow rate, $S_i$ from equation= $\frac{3600}{2.2}$ = 1637 vph. $\frac{V_c}{S_i}$ = $\frac{433}{1637} + \frac{417}{1637} + \frac{233}{1637} + \frac{1298}{1637}$ = 0.793. Cycle length can be found out from the equation C= $\frac{4\times 2.4 \times 0.9}{0.9-\frac{1298}{1637}}$ = 80.68 seconds $\approx$ 80 seconds. The effective green time can be found out as $G_i = \frac{V_{Ci}}{V_C}\times(C-L)$ = 80-(4$\times$2.4)= 70.4 seconds, where $L$ is the lost time for that phase = 4$\times$ 2.4. Green splitting for the phase 1 can be found out   as $g_1$ = 70.4 $\times$ [ $\frac{483}{1298}$] = 22.88 seconds. Similarly green splitting for the phase 2, $g_2 = 70.4 \times [\frac{417}{1298}]$ = 22.02 seconds. Similarly green splitting for the phase 3, $g_3 = 70.4 \times [\frac{233}{1298}]$ = 12.04 seconds. Similarly green splitting for the phase 4, $g_4 = 70.4 \times [\frac{215}{1298}]$ = 11.66 seconds. The actual green time for phase 1 from equation$G_1$= 22.88-3+2.4 $\approx$ 23 seconds. Similarly actual green time for phase 2, $G_2$ = 22.02-3+2.4 $\approx$ 23 seconds. Similarly actual green time for phase 3, $G_3$ = 12.04-3+2.4 $\approx$ 13 seconds. Similarly actual green time for phase 4, $G_4$ = 11.66-3+2.4 $\approx$ 12 seconds. Pedestrian time can be found out from as $G_p =4+\frac{6\times3.5}{1.2}$ = 21.5 seconds. The phase diagram is shown in figure 3. Figure 3: Timing diagram The actual cycle time will be the sum of actual green time plus amber time plus actual red time for any phase. Therefore, for phase 1, actual cycle time = 23+3+78.5 = 104.5 seconds. Delay at the intersection in the east-west direction can be found out from equationas $$d_{EW} =\frac{\frac{104.5}{2}[1-\frac{23-2.4+3}{104.5}]^2}{1-\frac{433}{1637}} =42.57sec/cycle. \nonumber$$ Delay at the intersection in the west-east direction can be found out from equation,as $$d_{WE} =\frac{\frac{104.5}{2}[1-\frac{23-2.4+3}{104.5}]^2}{1-\frac{400}{1637}}=41.44 sec/cycle.$$ (1) Delay at the intersection in the north-south direction can be found out from equation, $$d_{NS} = \frac{\frac{104.5}{2}[1-\frac{23-2.4+3}{104.5}]^2}{1-\frac{367}{1637}}=40.36 sec/cycle.$$ (2) Delay at the intersection in the south-north direction can be found out from equation, $$d_{SN} = \frac{\frac{104.5}{2}[1-\frac{23-2.4+3}{104.5}]^2}{1-\frac{417}{1637}}=42.018 sec/cycle.$$ (3) Delay at the intersection in the south-east direction can be found out from equation, $$d_{SE} = \frac{\frac{104.5}{2}[1-\frac{13-2.4+3}{104.5}]^2}{1-\frac{233}{1637}}=46.096 sec/cycle.$$ (4) Delay at the intersection in the north-west direction can be found out from equation, $$d_{NW} = \frac{\frac{104.5}{2}[1-\frac{13-2.4+3}{104.5}]^2}{1-\frac{196}{1637}}=44.912 sec/cycle.$$ (5) Delay at the intersection in the west-south direction can be found out from equation, $$d_{WS} = \frac{\frac{104.5}{2}[1-\frac{12-2.4+3}{104.5}]^2}{1-\frac{215}{1637}}=46.52 sec/cycle.$$ (6) Delay at the intersection in the east-north direction can be found out from equation, $$d_{EN} = \frac{\frac{104.5}{2}[1-\frac{12-2.4+3}{104.5}]^2}{1-\frac{187}{1637}}=45.62 sec/cycle.$$ (7)