Question

The answer needed is only for part C.

A) Willow and Stevens are two-way streets in an urban area. A four-phase operation is used for the intersection of Willow and Stevens with a cycle time of 110 seconds, including 15 sec Green for the left turn (LT) vehicles on each street. If the green time for the through (TH) vehicles on Willow is 30 sec, determine the green time for the through (TH) vehicles on Stevens and construct the signal timing cut-out diagrams for vehicles only without considering pedestrians. Assume a clearance (Yellow 3 sec + All Red 2 sec) time of 5 seconds when the green light is switched between Willow and Stevens (but do not use the clearance time when the signal changes from LT to TH on the same street. B) The width of Stevens Avenue is 52 feet. How long should be the Flash Don’t Walk (FDW) duration if the walking speed of pedestrians is 4 feet/sec?

C) If the minimum headway is 2 sec, delay per phase is 4.6 sec, and there are two through lanes and one left-turn lane on Willow in each direction, how long should be the cycle time if the combined (LT + TH) traffic volume on this street is 3000 vehicles/hour?

Answer 1

its a 4 phase signal

Phase 1 - NB

Phase 2 - SB

Phase 3 - EB

Phase 4 - WB

given minimum headway = 2 seconds and delay per phase

Therefore saturation flow = 3600/2 = 1800 vphpl

so for Willow Street , total saturation flow 1800 *3 = 5400 vphpl

critical flow ratio for Willow Street = 3000/5400 = 5/9 = 0.5555

green time for Willow Street is given to be 15+30 = 45 seconds

yellow time = 3 seconds

Lost time = yellow + allred - 2 = 5-2 = 3 seconds/phase

Effective Green time = Actual green time + yellow time - lost time = 30 + 3 -3 = 30 seconds

Cycle Length is calculated by the formula

where C = cycle length N number of phases, Li = start up lost time = 1 second

where N = number of phases = 3, L = los time per phase = 3 seconds, Xc= V/C ratio = 0.90 in this case and (V/S) is the critical flow ratio

Delay for a phase is given by

So we know d = 4.6 secs

gi = 290 secs

Vi = 3000 veh/r

S = 5400 veh/hr

So we get 4.6 = C/2 * (1-30/C)2 /(1-5/9) So C *(1-30/C)2 = 2.044

Therefore solving for C we get C = 38.8 seconds.

so Cycle time should be 38.8 seconds.