Traffic and Highway Engineering
Traffic and Highway Engineering
5th Edition
ISBN: 9781305156241
Author: Garber, Nicholas J.
Publisher: Cengage Learning
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Chapter 8, Problem 17P
To determine

Effectof cycle length when saturation flow rates at 10% higher from given Problem 8-16.

Expert Solution & Answer
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Answer to Problem 17P

C=133sec

Explanation of Solution

Given information:

Repeat Problem 8-16 using saturation flow rates at 10% higher.

Shows a detailed layout of the phasing system and the intersection geometry used.

  Traffic and Highway Engineering, Chapter 8, Problem 17P

Calculation:

Equivalent hourly flow for north approach at left turn is given by,

Eq=vpPHF.....(1)

Equivalent hourly flow =Eq

Peak hour volume =vp=133 veh/hr

Peak hour factor =PHF=0.95

Substitute the values in equation (1) we get,

Eq=vpPHFEq=1330.95Eq=140 veh/hr

Calculate for other approaches like left turn, right turn and through movement in Table 1.

Table 1 :

ApproachNorth (56ft)South (56ft)East (68ft)East (68ft)
Left turn14077177141
Through movement442393593543
Right turn147142178187

Calculate the critical volume for other approaches like left turn, right turn and through movement in Table 2,

Left to north approach is left turn is 140.

Through movement is 147+442=589.

Table 2 :

ApproachNorth (56ft)South (56ft)East (68ft)East (68ft)
Left turn14077177141
Through movement589535771730

For the saturation flow rates at 10% higher given below.

s11=1600×1.10s11=1760 veh/hr

Similarly calculate the saturation flow rates for other approaches in Table 3,

Approachsaturation flow rates,

veh/h

Through lanes1760
Through right lanes1540
Left lanes1100
Left through lanes1320
Left through right lanes1210

Maximum value of the ratios of approach flow using formula,

Yi=qijsij.......(2)

Maximum value of ratios of approach flow to saturation flow to all lanes =Yi

Flow on lane groups =qij=177 veh/hr

Saturation flow on lane group =sij=1100 veh/hr

Substitute the values in equation (2) we get,

Yi=q ijs ijYi=1771100Yi=0.161

Calculate for other Yi approaches in Table 4.

Table 4 :

ApproachPhase 1:E-W leftPhase 2:E-W throughPhase 4:N-S leftPhase 4:N-S through
qij177771140589
sij1100330011003300
Yi0.1610.2340.1270.178

Sum of critical ratio is given by,

Ysum=0.161+0.234+0.127+0.178Ysum=0.700

Total lost time is given by,

L=i=1ϕli+R.....(3)

Total lost time L.

Number of phases ϕ=4 phases

Lost time for phase li=3.5sec

Total all red time R=1.5sec

Substitute the values in equation (3) we get,

L=i=1ϕli+RL=i=14( 3.5+1.5)L=20sec

Cycle length is given by,

Co=( 1.5×20)+510.700Co=116.67sec

Total effective green time is given by,

Gte=CoLGte=12020Gte=100sec

Allocated green time for phase 1 is given by,

(G+Y)1=( Y 1 Y sum ×G te)+l1(G+Y)1=( 0.161 0.700×100)+3.5(G+Y)1=26.5sec

Allocated green time for other phases is given in the table 5,

Table 5 :

Phase Allocated green time in sec
(G+Y)126.5
(G+Y)236.4
(G+Y)321.6
(G+Y)428.9

Minimum green time for phase 1 is given by,

GP=3.2+LSP+(0.27NPed).....(4)

Crosswalk length =L=56ft

Average speed of pedestrians =SP=4ft/sec

Number of pedestrians crossing during an interval =NPed=(12003600)veh/sec

Substitute the values in equation (4) we get,

GP=3.2+LSP+(0.27N Ped)GP=3.2+564+(0.27( 1200 3600 )120)GP=28sec

Minimum Green time for other phases is given Table 6,

Table 6 :

Phase Minimum green time in sec
GP128
GP228
GP331
GP431

From the Table 5 the GP3 is greater than (G+Y)3, GP1 is greater than (G+Y)1, GP4 is greater than (G+Y)4 is the allocated green and yellow time for phase 3 is 31sec  ,phase 1 is 28sec and phase 4 is 31sec from the table 34.1 sec.

Sum of green and yellow time is given by,

G1=28sec,G2=37sec,G3=31sec,G4=31sec

Total cycle length is given by,

C=(28+37+31+31)+(4×1.5)C=133sec

Conclusion:

Therefore by increasingsaturation flow rates at 10% higher the required cycle length is reduced from 163sec to 133 Sec.

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