For the geometric and traffic characteristics shown below determine a suitable signal phasing system and phase lengths for the intersection using the Webster method. Show a detailed layout of the phasing system and the intersection geometry used.
Assume the following saturation flows:
Through lanes: 1600 veh/h/ln
Through-right lanes: 1400 veh/h/ln
Left lanes: 1000 veh/h/ln
Left-through lanes: 1200 veh/h/ln
Left-through-right lanes: 1100 veh/h/ln
What will be an ideal response?
Step 1: Calculate equivalent hourly flows
Step 2: Assume an intersection configuration, assign lane groups, and determine
critical volumes.
In this case, each approach was assumed to have one dedicated left-turn
lane, one through lane, and one through-right lane.
Step 3: Assume a phasing scheme and determine Yi , sum of critical ratios
Assume four phases as follows:
Sum Yi = 0.770
Step 4: Calculate lost time per cycle, using Equation 8.8
Assume lost time per phase due to acceleration and deceleration at phase changes
is 3.5 seconds and that an all-red interval of 1.5 seconds is provided at each phase.
Total lost time, L = 20 sec.
Step 6: Allocate green times
Allocated times are for green and yellow indications; appropriate length of
yellow interval can be subtracted from the total to give green times.
Total effective green time, Gte = C – L =135 seconds
(G+Y)1 = (0.177/0.770)(135) + 3.5 = 34.5 seconds
(G+Y)2 = (0.257/0.770)(135) + 3.5 = 48.5 seconds
(G+Y)3 = (0.140/0.770)(135) + 3.5 = 28.0 seconds
(G+Y)4 = (0.196/0.770)(135) + 3.5 = 37.9 seconds
Step 7: Ensure that green time required for pedestrian movement is provided,
using Equation 8.12.
Gp1 = 3.2 + (56/3.5) + (0.27)(1200/3600)(155) = 33.2 seconds
Gp2 = 3.2 + (56/3.5) + (0.27)(1200/3600)(155) = 33.2 seconds
Gp3 = 3.2 + (68/3.5) + (0.27)(1200/3600)(155) = 36.6 seconds
Gp4 = 3.2 + (68/3.5) + (0.27)(1200/3600)(155) = 36.6 seconds
Since Gp3 > (G+Y)3, the allocated sum of green and yellow time for phase
3 should be 36.6 seconds. Times are typically rounded up to the next whole
seconds; therefore, sum of green and yellow times are: G1 = 35 s; G2 = 49 s; G3 =
37 s; G4 = 38 s, resulting in a total cycle length of C = (35 + 49 + 37 + 38) +
(4)(1.5) = 165 seconds.
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