
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 %.
Want to see more full solutions like this?
Chapter 8 Solutions
Traffic And Highway Engineering
- 5.27 At a parking lot, vehicles arrive according to a Poisson process and are processed (parking fee collected) at a uniform deterministic rate at a single station. The mean arrival rate is 4.2 veh/min and the processing rate is 5 veh/min. Determine the average length of queue, the average time spent in the system, and the average waiting time in the queue. 5.28 Consider the parking lot and conditions described in Problem 5.27. If the rate at which vehicles are processed became exponentially distributed (instead of deterministic) with a mean processing rate of 5 veh/min, what would be the average length of queue, the average time spent in the system, and the average waiting time in the queue?arrow_forwardDetermine the heel and toe stresses and the factor of safeties for sliding and overturning for the gravity dam section shown in the figure below for the following loading conditions: - Horizontal earthquake (Kh) = 0.1 - Normal uplift pressure with gallery drain working - Silt deposit up to 30 m height - No wave pressure and no ice pressure -Unit weight of concrete = 2.4 Ton/m³ and unit weight of silty water = 1.4 Ton/m³ - Submerged weight of silt = 0.9 Ton/m³ - Coefficient of friction = 0.65 and angle of repose = 25° Solve this question with the presence of gallery and without gallery., discuss the issue in both cases.... 144 m E 4m W 8m 6m 8m7m 120marrow_forwardOn page 1, in the first body paragraph ("In the United States…to be smart."), edit the Kaya source so the Name of Web Page is "Fast Facts 2020: Demographics", the correct Name of the webpage.arrow_forward
- 4.12 A 400 m vertical curve connects a -2.00% grade to a +4.00% grade. The P.I. is located at station 150 + 00 and elevation 60.00 m above sea level. A pipe is to be located at the low point on the vertical curve. The roadway at this point consists of two 3.6 m lanes with a normal crown slope of 2%. If the lowest point on the surface of the road- way must clear the pipe by 0.75 m, what is the station and maximum elevation of the pipe?arrow_forward=7.5 in., d1 = 1.5 in., b2 = 0.75 in., d2= 6.0 in., b3 = 3.0 in., and d3 = 2.0 in. Determine (a) the centroid location in inches from the bottom surface. Round off to two decimal places. M H b₂ y ... d₁ M by dy Xarrow_forwardVehicles arrive at a single park-entrance booth where a brochure is distributed. At 8 A.M., there are 20 vehicles in the queue and vehicles continue to arrive at the deterministic rate of λ(t) = 4.2 − 0.1t, where λ(t) is in vehicles per minute and t is in minutes after 8:00 A.M. From 8 A.M. until 8:10 A.M., vehicles are served at a constant deterministic rate of three per minute. Starting at 8:10 A.M., another brochure-distributing person is added and the brochure-service rate increases to six per minute (stillarrow_forward
- Vehicles arrive at a single park-entrance booth where a brochure is distributed. At 8 A.M., there are 20 vehicles in the queue and vehicles continue to arrive at the deterministic rate of λ(t) = 4.2 − 0.1t, where λ(t) is in vehicles per minute and t is in minutes after 8:00 A.M. From 8 A.M. until 8:10 A.M., vehicles are served at a constant deterministic rate of three per minute. Starting at 8:10 A.M., another brochure-distributing person is added and the brochure-service rate increases to six per minute (still at a single booth). Assuming D/D/1 queuing, determine the longest queue, the total delay from 8 A.M. until the queue dissipates; and the wait time of the 40th vehicle to arrive.arrow_forwardAt 8:00 A.M. there are 10 vehicles in a queue at a toll booth and vehicles are arriving at a rate of λ(t) = 6.9 − 0.2t. Beginning at 8 A.M., vehicles are being serviced at a rate of μ(t) = 2.1 + 0.3t [λ(t) and μ(t) are in vehicles per minute and t is in minutes after 8:00 A.M.]. Assuming D/D/1 queuing, what is the maximum queue length, and what would the total delay be from 8:00 A.M. until the queue clears?arrow_forwardIntroduction: Orifice and Free Flow Jet in Applied Fluid Mechanics' I need to introduction only for answerarrow_forward
- compute the load bearing capacity, displacement, stress distribution, tabulate the answersarrow_forwardcompute the load bearing capacity, displacement, and stress distribution, tabulate the answersarrow_forwardcompute the load bearing capacity, displacement, stress distribution, tabulate the answersarrow_forward
- Traffic and Highway EngineeringCivil EngineeringISBN:9781305156241Author:Garber, Nicholas J.Publisher:Cengage Learning
