Downloaded 11 times
Traffic light design
- 1. TRAFFIC LIGHT DESIGN Example: A study of TWO phase traffic light is proposed for a 4‐legs junction indicated “NSEW”. The traffic study is resulted in saturated flow data for each brand of the road. Given;‐ Interval Time, 𝐼 = 4 𝑠𝑒𝑐𝑜𝑛𝑑 Amber/Yellow Time, 𝑎 = 3 𝑠𝑒𝑐𝑜𝑛𝑑 Lost Time, 𝑙 = 2 𝑠𝑒𝑐𝑜𝑛𝑑 (Table 1.0: Saturated Flow Data) Actual Flow Rate, (pcu/hr) , Q N S E W 750 700 1700 2100 Saturated Flow, (pcu/hr), S 1960 1870 3600 3950 Determine the value of ; a. Total lost time per cycle, L b. Optimum cycle length, Co c. Effective time, g for each phase. d. Actual green, G for each phase. e. Sketched an operation phase diagram.
- 2. The solution Step 1: Saturated flow table. NS EW N S E W Actual Flow Rate, (pcu/hr) , Q 750 700 1700 2100 Saturated Flow, (pcu/hr), S 1960 1870 3600 3950 y value = Q/S 0.38 0.37 0.47 0.53 ymax 0.38 0.53 Critical flow ratio, 𝑌 = 𝑌!"# = 0.38 + 0.53 = 0.91 a. The total lost time per cycle, L 𝐿 = 𝐼 − 𝑎 + 𝑙 = 4 − 3 + 4 − 3 + 2 + 2 = 6 𝑠𝑒𝑐𝑜𝑛𝑑 b. The optimum cycle length, Co 𝐶! = 1.5𝐿 + 5 1 − 𝑌 = !.!(!)!! !!(!.!") = 155.56 second > Co (120 s) Therefore, Co should be 120 second
- 3. c. The effective time, g for each phase 𝑔 = !!"# ! (𝐶!− 𝐿) NS phase EW phase 𝑔!" = !!"# ! (𝐶! − 𝐿) 𝑔!" = !!"# ! (𝐶! − 𝐿) = !.!" !.!" (120 − 6) = !.!" !.!" (120 − 6) = 47.60 𝑠𝑒𝑐𝑜𝑛𝑑 = 66.40 𝑠𝑒𝑐𝑜𝑛𝑑 d. The actual green, G for each phase 𝐺 = 𝑔 + 𝑙 − 𝑎 NS phase EW phase 𝐺!" = 𝑔 + 𝑙 − 𝑎 𝐺!" = 𝑔 + 𝑙 − 𝑎 = 47.60 + 2 − 3 = 66.40 + 2 − 3 = 46.60 𝑠𝑒𝑐𝑜𝑛𝑑 = 65.40 𝑠𝑒𝑐𝑜𝑛𝑑 ~ 47 𝑠𝑒𝑐𝑜𝑛𝑑 ~ 65 𝑠𝑒𝑐𝑜𝑛𝑑
- 4.