
Effect on cycle length due to 20% higher pedestrian flow rates.

Answer to Problem 18P
Cycle length value increases with 4.76 %, from 126 to 132.
Explanation of Solution
Given data:
Calculation:
Evaluating equivalent hourly flow −
Equivalent hourly flow = Traffic volumePHF= 1330.95≃140
Similarly, evaluating equivalent hourly flow for all traffic movements −
Table 1
Approach (width) | N (56ft) | S (56ft) | E (68ft) | W (68ft) |
Left turn | 133/0.95 = 140 | 73/0.95 = 77 | 168/0.95 = 177 | 134/0.95 = 142 |
Through movement | 443 | 393 | 593 | 544 |
Right turn | 148 | 143 | 178 | 188 |
Conflicting pedestrian volume | 948 | 1264 | 1264 | 948 |
Assuming lane configuration as one dedicated left turn and combined through and right lane −
Table 2
Approach | N | S | E | W |
Left | 140 | 77 | 177 | 142 |
Through + Right | 589 (442+147) | 535 (393+178) | 771 (593+178) | 730 (543+187) |
Assume a phase scheme and find critical ratios( Yi ) and sum of critical ratios ( ∑Yi ). Assuming phase as follows −
Table 3
- | Phase lE-W (Left) | Phase llE-W (Through) | Phase lllN-S (Left) | Phase lVN-S (Through) |
qij | 177 | 771 | 140 | 589 |
Sij | 1000 | 3000 (1600+1400) | 1000 | 3000 (1600+1400) |
Yi= qijSij | 0.177 | 0.257 | 0.140 | 0.196 |
Sum of critical ratios −
∑Yi = Y1+Y2+Y3+........Yn= .177+0.257+0.140+0.196=0.77
Assuming lost time per phase ( li )= 3 Sec
So, Total lost time −
L = ∑li= 3+3+3+3=12 Sec
Now, determining the optimum cycle length −
Co = 1.5L + 51−∑Yi= 1.5×12 + 51−0.77= 100 Sec
(Cycle lengths are generally multiple of 5sec or 10 sec, Hence OK)
Finding Total effective green time −
Gie = Co−L= 100−12=88 Sec
Effective time for phase i can be calculated as −
Gei = YiY1+Y2+Y3+........YnGie
For Phase l
Gphase l = 0.1770.177+0.257+0.140+0.196×88= 20.23 Sec
(Assuming yellow time as 4 sec )
(G+Y)phase l = 20.23+4= 24.23 Sec≃ 25 Sec
For Phase ll
Gphase ll = 0.2570.177+0.257+0.140+0.196×88= 29.37 Sec
(G+Y)phase ll = 29.37+4= 33.37 Sec≃ 34 Sec
For Phase lll
Gphase lll = 0.1400.177+0.257+0.140+0.196×88= 16 Sec
(G+Y)phase lll = 16+4= 20 Sec
For Phase lV
Gphase lV = 0.1960.177+0.257+0.140+0.196×88= 22.4 Sec
(G+Y)phase l = 22.4+4= 26.4 Sec≃ 27 Sec
Table 4
Phase | Allocated green & yellow time (in sec) |
(G+Y)1 | 25 |
(G+Y)2 | 34 |
(G+Y)3 | 20 |
(G+Y)4 | 27 |
Total cycle length
C= ∑(G+Y)phase i+ Total red time interval= (25+34+20+27) + 4(1.5)= 112 Sec
Green time required for pedestrian crossing can be calculated as following formula:
(Assuming the crosswalk width as 9 ft< 10 ft ).
Gp= 3.2+LccSp+0.27Nped
Where,
Lcc = cross walk length
Sp = pedestrian speed (assuming as 4ftsec )
Nped = number pedestrians crossing during an interval = vpedi3600C
vpedi = pedestrian flow rate in the subject crossing for travel direction i( p/h)
C = total cycle length
Calculating Nped for each direction:
Nped for N direction = 9483600×112 = 29.49≃ 30 sec
Nped for S direction = 12643600×112 = 39.32≃ 40 sec
Nped for E direction = 12643600×112 = 39.32≃ 40 sec
Nped for W direction = 9483600×112 = 29.49≃ 30 sec
Calculating minimum time required ( Gp ) for each approach:
Minimum time required for N approach ( GP1 ): 3.2+564+0.27×30= 25.3≃26 sec
Minimum time required for S approach ( GP2 ): 3.2+564+0.27×40= 28 sec
Minimum time required for E approach ( GP3 ): 3.2+684+0.27×40= 31 sec
Minimum time required for W approach ( GP4 ): 3.2+684+0.27×30= 28.3≃29 sec
Table 5
Phase | Minimum green time (in sec) |
GP1 | 26 |
GP2 | 28 |
GP3 | 31 |
GP4 | 29 |
GP1 is greater than (G+Y)1 so the allocated green and yellow time for phase 1 is from the table 5 26 sec.
Sum of green and yellow time is given by,
G1+G2+G3+G4=26sec+28sec+31sec+29sec=114 sec
Total cycle length is given by,
C = (total green and yellow time) + (total red time)C=(26+34+31+29)+(4×1.5)C=126sec
Now increasing the pedestrian volume with 20% and calculating the minimum time required by pedestrian for each approach:
New pedestrian volume = old pedestrian volume + old pedestrian volume= 948 + 948×20100= 948+189.6≃ 1138
Table 6
New conflicting pedestrian volume | 1138 | 1517 | 1517 | 1138 |
According to the new pedestrian volume calculating minimum time required by pedestrian for each approach:
Calculating Nped for each direction:
Nped for N direction = 11383600×126 = 39.83≃ 40 sec
Nped for S direction = 15173600×126 = 53.09≃ 54 sec
Nped for E direction = 15173600×126 = 53.09≃ 54 sec
Nped for W direction = 11383600×126 = 39.83≃ 40 sec
Calculating new minimum time required ( G*p ) for each approach:
Minimum time required for N approach ( G*P1 )= 3.2+564+0.27×40= 28 sec
Minimum time required for S approach ( G*P2 ) = 3.2+564+0.27×54= 31.78≃32 sec
Minimum time required for E approach ( G*P3 ) = 3.2+684+0.27×54= 34.78=35 sec
Minimum time required for W approach ( G*P4 ) = 3.2+684+0.27×40= 31 sec
Comparing the G*p values with table 4, and selecting new values as shown in table below:
Selecting greater values in between both G*p and (G+Y)i, which is taken as new minimum green time:
Table 7
Phase | New minimum green time (in sec) |
G*1 | 28 |
G*2 | 32 |
G*3 | 35 |
G*4 | 31 |
Sum of green and yellow time is given by,
G*1+G*2+G*3+G*4=28sec+32sec+35sec+31sec=126 sec
Total new cycle length is given by,
C = (total green and yellow time) + (total red time)C=(28+32+35+31)+(4×1.5)C=132sec
Conclusion:
With using pedestrian volume flow rate 20% higher than given, cycle length increases by 4.76 %.
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- Traffic and Highway EngineeringCivil EngineeringISBN:9781305156241Author:Garber, Nicholas J.Publisher:Cengage Learning
