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Intersection Phase Design Guide

The document designs a four phase signal cycle for an intersection. It calculates the demand and capacity for each phase. Phase 1 has a green time of 23 seconds, Phase 2 has 19 seconds, Phase 3 has 23 seconds, and Phase 4 has 13 seconds. The total cycle length is 96 seconds with 17 seconds for intergreen times.

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محمد المهندس
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40 views4 pages

Intersection Phase Design Guide

The document designs a four phase signal cycle for an intersection. It calculates the demand and capacity for each phase. Phase 1 has a green time of 23 seconds, Phase 2 has 19 seconds, Phase 3 has 23 seconds, and Phase 4 has 13 seconds. The total cycle length is 96 seconds with 17 seconds for intergreen times.

Uploaded by

محمد المهندس
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Example:

Design the four phase for intersection shown below with a (6 sec)
intergreen period on first phase and (5 sec) intergreen for other phases.

Sol.
Phase 1:
q= RT+TH+(1.75LT)
= [35 + 620 + (1.75x65) = 769 veh/hr.
S = SoFwFH.VFgFpedFPLT
1 1
𝐹𝐿𝐹 = = = 0.996
1 + 0.05𝑃𝐿𝑇 65
1 + 0.05 ( )
35 + 620 + 65
S = 1900x2x1x1x0.996x1x1≈ 3785 veh/hr/lane.
Phase 1:
q= RT+TH+(1.75LT) = 20 veh/hr.
S = SoFwFH.VFgFpedFLF = 1900x2x1x1x1x1x1≈ 3800 veh/hr/lane.
Phase 2:
q= RT+TH+(1.75LT)
= 50 + 510 + (1.75x40) = 630 veh/hr.
S = SoFwFH.VFgFpedFLF
1 1
𝐹𝐿𝐹 = = = 0.997
1 + 0.05𝑃𝐿𝑇 40
1 + 0.05 ( )
50 + 510 + 40
S = 1900x2x1x1x0.997x1x1≈ 3789 veh/hr/lane.
Phase 2:
q= RT+TH+(1.75LT) = 20 veh/hr.
S = SoFwFH.VFgFpedFLF = 1900x2x1x1x1x1x1≈ 3800 veh/hr/lane.

Phase 3:
q= RT+TH+(1.75LT) = 700 + (1.75x30) = 753 veh/hr.
S = SoFwFH.VFgFpedFLF
1 1
𝐹𝐿𝐹 = = = 0.998
1 + 0.05𝑃𝐿𝑇 30
1 + 0.05 ( )
700 + 30
S = 1900x2x1x1x0.998x1x1≈ 3793 veh/hr.
Phase 4:
q= RT+TH+(1.75LT) = 370 + (1.75x30) = 423 veh/hr.
S = SoFwFH.VFgFpedFLF
1 1
𝐹𝐿𝐹 = = = 0.997
1 + 0.05𝑃𝐿𝑇 30
1 + 0.05 ( )
370 + 30
S = 1900x2x1x1x0.997x1x1≈ 3789 veh/hr.
Variable Phase 1 Phase 2 Phase 3 Phase 4
q(PCU/hr) 769 20 630 20 753 423
S(PCU/hr) 3785 3800 3789 3800 3793 3789
Y = q/s 0.203 0.005 0.166 0.005 0.199 0.112
Ymax. 0.203 0.166 0.199 0.112
∑ 𝑌𝑚𝑎𝑥. 0.680 sec.

𝐿 = ∑(𝐼𝐺 − 1) = (6 − 1) + (5 − 1) + (5 − 1) + (5 − 1) =
17 𝑠𝑒𝑐.
1.5𝐿+5 1.5𝑥17+5
𝐶𝑜 = = = 96 𝑠𝑒𝑐.
1−∑ 𝑌𝑚𝑎𝑥. 1−0.68

𝐶𝑜 = 96 𝑠𝑒𝑐. 60 𝑠𝑒𝑐. ≤ 𝐶0 ≥ 120 𝑠𝑒𝑐.


𝐺𝑇 = 𝐶𝑜 − 𝐿 = 96 − 17 = 79 𝑠𝑒𝑐.
𝑌𝑚𝑎𝑥. 0.203
𝑔𝑝ℎ.1 = ∑ 𝑌𝑚𝑎𝑥.
𝑥𝐺𝑇 = 𝑥79 = 24 𝑠𝑒𝑐.
0.68

𝐺𝑝ℎ.1 = 𝑔𝑝ℎ.1 − 1 = 24 − 1 = 23 𝑠𝑒𝑐.


𝑌𝑚𝑎𝑥. 0.166
𝑔𝑝ℎ.2 = ∑ 𝑌𝑚𝑎𝑥.
𝑥𝐺𝑇 = 𝑥79 = 20 𝑠𝑒𝑐.
0.68

𝐺𝑝ℎ.2 = 𝑔𝑝ℎ.2 − 1 = 20 − 1 = 19 𝑠𝑒𝑐.


𝑌𝑚𝑎𝑥. 0.199
𝑔𝑝ℎ.3 = ∑ 𝑌𝑚𝑎𝑥.
𝑥𝐺𝑇 = 𝑥79 = 24 𝑠𝑒𝑐.
0.68

𝐺𝑝ℎ.3 = 𝑔𝑝ℎ.3 − 1 = 24 − 1 = 23 𝑠𝑒𝑐.


𝑌𝑚𝑎𝑥. 0.112
𝑔𝑝ℎ.4 = ∑ 𝑌𝑚𝑎𝑥.
𝑥𝐺𝑇 = 𝑥79 = 14 𝑠𝑒𝑐.
0.68

𝐺𝑝ℎ.3 = 𝑔𝑝ℎ.3 − 1 = 14 − 1 = 13 𝑠𝑒𝑐.


Phase 1
G = 23 sec. A = 3 sec. R = 70 sec.

Phase2
R= 69 sec. IG = 5 sec. G=20sec. A=3sec.

Phase3
R= 65 sec. IG = 5 sec. G= 24sec. A=3sec.

Phase4
R= 75 sec. IG = 5 sec. G= 13sec. A=3sec.

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