TRAFFICLIGHTDESIGN

Example:

AstudyofTWOphasetrafficlightisproposedfora4‐legsjunctionindicated“NSEW”.Thetrafficstudyisresultedinsaturatedflowdataforeachbrandoftheroad.Given;‐

IntervalTime, 𝐼 = 4 𝑠𝑒𝑐𝑜𝑛𝑑

Amber/YellowTime,𝑎 = 3 𝑠𝑒𝑐𝑜𝑛𝑑

LostTime,𝑙 = 2 𝑠𝑒𝑐𝑜𝑛𝑑

(Table1.0:SaturatedFlowData)

ActualFlowRate,(pcu/hr),Q

N S E W750 700 1700 2100

SaturatedFlow,(pcu/hr),S 1960 1870 3600 3950

Determinethevalueof;

a. Totallosttimepercycle,Lb. Optimumcyclelength,Coc. Effectivetime,gforeachphase.d. Actualgreen,Gforeachphase.e. Sketchedanoperationphasediagram.

Thesolution

Step1:Saturatedflowtable.

NS EWN S E W

ActualFlowRate,(pcu/hr),Q 750 700 1700 2100

SaturatedFlow,(pcu/hr),S 1960 1870 3600 3950

yvalue=Q/S 0.38 0.37 0.47 0.53ymax 0.38 0.53

Criticalflowratio,𝑌 = 𝑌!"#

= 0.38 + 0.53

= 0.91

a. Thetotallosttimepercycle,L

𝐿 =   𝐼 − 𝑎 + 𝑙

= 4 − 3 + 4 − 3 + 2 + 2

= 6 𝑠𝑒𝑐𝑜𝑛𝑑

b. Theoptimumcyclelength,Co

  𝐶! =  1.5𝐿 + 51 − 𝑌

=   !.!(!)!!!!(!.!")

=155.56second>Co(120s)

Therefore,Coshouldbe120second

c. Theeffectivetime,gforeachphase𝑔 = !!"#

!(𝐶! − 𝐿)

NSphase EWphase

𝑔!" =!!"#!

(𝐶! − 𝐿)𝑔!" = !!"#!

(𝐶! − 𝐿)

= !.!"!.!"

(120 − 6)= !.!"!.!"

(120 − 6)

= 47.60 𝑠𝑒𝑐𝑜𝑛𝑑= 66.40 𝑠𝑒𝑐𝑜𝑛𝑑 

d. Theactualgreen,Gforeachphase

𝐺 = 𝑔 + 𝑙 − 𝑎

NSphase EWphase

𝐺!" = 𝑔 + 𝑙 − 𝑎𝐺!" = 𝑔 + 𝑙 − 𝑎

= 47.60 + 2 − 3= 66.40 + 2 − 3

= 46.60 𝑠𝑒𝑐𝑜𝑛𝑑= 65.40 𝑠𝑒𝑐𝑜𝑛𝑑

~ 47 𝑠𝑒𝑐𝑜𝑛𝑑~ 65 𝑠𝑒𝑐𝑜𝑛𝑑

f. Theoperationphasediagram.

G

R

Ra=3s

I=4s

a=3s

I=4s

NSphase

EWphase

Start=0s Co=120s47s 50s

51s 116s 119s

G65s