Concept explainers
The capacity of the lane group.
Answer to Problem 5P
The capacity of the lane group is
Explanation of Solution
Given:
The base rate is
Lane width is
Heavy vehicles are
Approach grade is
No on-street parking.
No bus stops.
Bicycle and pedestrian traffic conflicting with this lane group is negligible.
Intersection is in a central business district.
The effective green time for the movement is
The total cycle length is
Formula used:
The adjusted saturated flow rate for a lane group is given by
Here,
The Heavy-vehicle adjustment factor is given by
Here,
The grade adjustment factor is given by
Here,
Parking adjustment factor is given by
Here,
The bus blockage adjustment factor is given by
Here,
The lane utilization adjustment factor is given by
Here,
The right turn adjustment factor for protecting movement on exclusive lane is given by
Here,
The left turn adjustment factor for protecting movement on exclusive lane is given by
Here,
The value of
The capacity of the lane group is given by
Here,
Calculation:
The Heavy-vehicle adjustment factor is calculated as
Substitute
The grade adjustment factor is calculated as
Substitute
Parking adjustment factor is calculated as
Substitute
The bus blockage adjustment factor is calculated as
Substitute
The right turn adjustment factor for protecting movement on exclusive lane is calculated as
Substitute
The left turn adjustment factor for protecting movement on exclusive lane is calculated as
Substitute
The adjusted saturated flow rate for a lane group is calculated as
Substitute
The capacity of the lane group is calculated as
Substitute
Conclusion:
The capacity of the lane group is
Want to see more full solutions like this?
Chapter 10 Solutions
Traffic and Highway Engineering
- A signalized intersection has four phases. The yci are 0.15, 0.19, 0.21, and 0.17 for the four phases, respectively. If the cycle length is 108 seconds and the total lost time is 16 seconds, what is the effective green time for phase 2? Group of answer choices 23 sec 25 sec 21 sec 19 secarrow_forwardThe minimum cycle length for an intersection is determined to be 95 seconds. The cirtical lane group flow ratios were calculated as 0.235, 0.250, 0.170 and 0.125 for phases 1- 4 respectively. What Xc was used in the determination of this cycle length, assuming a lost time of 5 seconds per phase?arrow_forwardFind: A- the total hourly volume B- PHF C-flow rate Density Q3) A- List the Importance of Spot Speed Studies.. B- Explain pedestrian control devices..arrow_forward
- Analyze the unsignalized intersection below and calculate the optimum cycle length if the intersection will be signalized. Approach E W N S Maximum no. of lanes 5 5 4 4 Direction Through left Through left Through left Through left N Total volume pcu/hr 1,480 370 860 586 1,360 494 2,030 265 Sat. flow rate pcu/hr/lane 2,400 1,700 2,400 1,700 2,200 1,800 2,200 1,800arrow_forwardA pretimed four-phase signal has critical lane group flow rates for the first three phases of 200, 187, and 210 veh/h (saturation flow rates are 1800 veh/h/In for all phases). The lost time is known to be 4 seconds for each phase. Assuming X₁ = 0.9. If the cycle length is 60 seconds, what is the estimated effective green time of the fourth phase? 6.93 sec O 21.89 sec Ⓒ 7.78 sec Q 7.41 secarrow_forwardAssume h=1.8 sec/veh. Pedestrian demand for all approaches is 120 per hour, pedestrian speed is 4 ft/sec, deceleration rate = 9 ft/s², reaction time is 1.5 sec, vehicle length is 20 ft. Use 12 seconds lost time per phase. Estimate the desired Cycle length using Time Budget Method 375 One way PHF -0.8 Target V/C 0.85 Medium Pedestrian 45 mph WB and NB 60 ft Crosswalk width=11 ft Level grades Phase 2 One way 60ft 1,200 300- 2,105 t Phase 1 ורוarrow_forward
- Given the volumes per movement and phasing pattern, allocate the green time to the 2 phases. Use amber = 3 sec All red = 2 sec Starting loss = 2 sec S = 1,900 pc/hr/lane a.. Compute the y values b. Compute the lost time if L= 2 x starting loss + 2 x all red c. compute the optimum cycle time (the cycle length is normally rounded off to multiples of 5 or 10) d. allocate the green time to phase 1 and 2arrow_forwardTraffic Volume and Capacity If saturation headway of given traffic is 2.5 sec and the cycle time on a particular intersection is 65 sec. With green time as 25 sec and corresponding yellow time as 3.5 sec. If the start-up loss-time is 2 sec and clearance time is 1 sec, then the actual capacity of the vehicle per lane in veh/hour isarrow_forwardThe maximum flow that could pass through an intersection from a given approach, if that approach was allocated all of the cycle time as effective green with no lost time. Saturation flow B) cycle flow Peak flow (D Approach capacityarrow_forward
- Calculate green split for each approach (Eastbound, Westbound, Northbound, Southbound) using Webster method. Given: Two through lanes and one left turn lane; Saturation Flow Rate: Through = 1,900 veh/hr/ln and Left turn = 900 veh/hr/ln. Total lost time = 4 seconds.arrow_forwardThe following calculations are taken at the approach of an intersection during the morning rush hour. Determine (a) the hourly volume, (b) the peak flow rate in one hour, and (c) the peak hour factor.arrow_forward3. Determine the LOS for pedestrians at two-phase signalised intersection with a cycle length of 90 s. the phase serving the major street vehicular traffic gets 50s of green whereas the phase serving the minor street vehicular traffic gets 30 s of green. Useful formula: dp Table for assessing the LOS LOS A B C D E F 0.5(C-g)² C Average delay/pedestrian (s) >5 ≥ 5-10 > 10-20 >20-30 ≥ 30-45 > 45arrow_forward
- Traffic and Highway EngineeringCivil EngineeringISBN:9781305156241Author:Garber, Nicholas J.Publisher:Cengage Learning