Question

A simple four-leg intersection (a single lane per approach) needs a 2-phase fixed-time signal. The critical flows in the N-S and E-W directions are 600 and 400 veh/hr, respectively, with 0.95 peak-hour factor. Saturation flow is 1800 veh/hr and the total lost time per phase is observed to be 5.2 seconds. Amber time and all-red time are 4 and 1 seconds, respectively. Assume that the intersection should operate at 90% of its capacity. Determine the cycle length and distribution of green (green splits). Calculate also effective green times.

Hint: round the cycle length to the next higher 5-sec increment (e.g. 86 -> 90); then round splits to the next integer values.

Answer 1

Following are the solution to this question:

Step-by-step explanation:

`Y=Y_(NS)+Y_(ew)`

`=(q_(NS))/(S_(NS))+(q_(ew))/(S_(ew))\\\\=(600)/(1800)+(400)/(1800)\\\\=(6)/(18)+(4)/(18)\\\\=(3)/(9)+(2)/(9)\\\\=(1)/(3)+(2)/(9)\\\\=(3+2)/(9)\\\\=(5)/(9)\\\\=0.55555555556\\\\=0.57`

calculating length
`(C_D)=(1.5 \ L +5)/(1-y)\\\\`

`=(1.5 * (1.2) +5)/(1-0.57)\\\\=(1.8 +5)/(0.43)\\\\=(6.8)/(0.43)\\\\= 15.8 \ sec`

critical value:

`\to V_(C_1)= 600 \ (veh)/(hr)\\\\\to V_(C_2)= 400 \ (veh)/(hr)\\\\`

Total critical value:

`V_C= V_(C_1)+ V_(C_2)\\\\`

`= 600 + 400\\\\= 1 000 \ (veh)/(hr)\\\\`

Active green time
`= 15.8-1.2=14.6`

`g_1=(600)/(1000)* 14.6= 8.76 \ sec\\\\g_2=(400)/(1000)* 14.6= 5.84 \ sec`

Native green time:

`G_1= 8.76 -1.2+4 =11.56 \sec\\\\ G_2= 5.84 -1.2+4 =8.64 \sec`