Repeat Problem 8-16 using the HCM method and a critical v/c of 0.9.
What will be an ideal response?
Step 1: Determine critical ratios
From Problem 8-16:
Phase 1: Y1 = 0.177
Phase 2: Y2 = 0.257
Phase 3: Y3 = 0.140
Phase 4: Y4 = 0.196
? ((q/s)i = ? Yi = 0.770
Step 2: Determine cycle length (using Equation 8.15)
Xc = 0.9 (critical v/c ratio)
L = 20 sec (lost time, from Problem 8-5)
Xc = ? ((q/s)i (C/(C-L))
0.9 = 0.770 (C/(C-20))
1.169 C – 23.38 = C
C = 138.32 sec; use 140 sec
Step 3: Determine phase lengths
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 =120 seconds
(G+Y)1 = (0.177/0.770)(120) + 3.5 = 31.1 seconds
(G+Y)2 = (0.257/0.770)(120) + 3.5 = 43.6 seconds
(G+Y)3 = (0.140/0.770)(120) + 3.5 = 25.3 seconds
(G+Y)4 = (0.196/0.770)(120) + 3.5 = 34.0 seconds
Step 4: Ensure that green time required for pedestrian movement is provided,
using Equation 8.12.
Gp1 = 3.2 + (56/3.5) + (0.27)(1200/3600)(140) = 31.8 seconds
Gp2 = 3.2 + (56/3.5) + (0.27)(1200/3600)(140) = 31.8 seconds
Gp3 = 3.2 + (68/3.5) + (0.27)(1200/3600)(140) = 35.2 seconds
Gp4 = 3.2 + (68/3.5) + (0.27)(1200/3600)(140) = 35.2 seconds
Since Gp1 > (G+Y)1 , Gp3 > (G+Y)3 , and Gp4 > (G+Y)4 , the allocated sum
of green and yellow time should be 31.8 seconds for phase 1, 35.2 seconds for
phase 3, and 35.2 seconds for phase 4. Times are typically rounded up to the next
whole seconds; therefore, sum of green and yellow times are: G1 = 32 s; G2 = 44
s; G3 = 36 s; G4 = 36 s,
resulting in a total cycle length of C = (32 + 44 + 36 + 36) + (4)(1.5) = 154
seconds.
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