Solutions Chapter 5 22687

Traffic Engineering, 4th Edition Roess, R.P., Prassas, E.S., and McShane, W.R. Solutions to Homework No. 2 Problem 5‐1 A volume of 1,200 veh/h is observed at an intersection approach. Find the peak flow rate within the hour for the following peak-hour factors: 1.00, 0.90., 0.80, 0.70. Plot and comment on the results. The peak flow rate of flow is computed as v = V/PHF. The table below summarizes the results for the information given. A plot follows.

Even with the same hourly volume, a small difference in PHF leads to an enormous difference in peak flow rates. Traffic engineers must be able to deal with this peaking characteristic on a regular basis.

Problem 5‐2 A traffic stream displays average vehicle headways of 2.4s at 55 mph. Compute the density and rate of flow for this traffic stream. A headway can be converted to a flow rate as follows:

v = 3600 h

= 3600 = 1,500 veh/hr/ln 2.4

Knowing both flow rate and speed (given), the density may now be computed as: D = v = 1500 = 27.3 veh/hr/ln S 55 ________________________________________________________________________ Problem 5‐3 A freeway detector records occupancy of 0.26 for a 15- minute period. If the detector is 3.5 ft long, and the average vehicle has a length of 18 ft. what is the density implied by this measurement? Density is obtained from occupancy as follows:

Such a high value is indicative of highly congested conditions within a queue.

Problem 5‐4 The following traffic count data were taken from a permanent detector location on a major state highway. From this data, determine (a) the AADT, (b) the ADT for each month. (c) the AAWT, and (d) the AWT for each month. From this information, what can be discerned about the character of the facility and the demand it serves? The table below illustrates the computation of monthly ADT and AWT values.

The AADT is computed as the total annual volume divided by 365 days, or: AADT = 2,365,000 = 6,479 veh/day 365 The AAWT is computed as the total weekday volume divided by 260 days, or: AAWT = 2,067,000 = 7,950 veh/day 260

Because the average weekday volume is higher than the total average volume, it is likely that this is a commuter route. The difference is even clearer if the average weekend traffic is computed. The total weekend volume for the year is 2,365,000 – 2,067,000 = 298,000 vehs. There are 365‐260 = 105 Saturdays and Sundays in the year. Then, the average weekend traffic is computed as: AAWT = 298,000 = 2,838 veh/day 105 This is clearly NOT a recreational route, but one that serves a substantial proportion of regular commuters. ________________________________________________________________________ Problem 5‐5 A lane on the freeway displays the following characteristics: (a) the average headway between vehicles is 2.8 s, and (b) the average spacing between vehicles is 235 ft. What is the rate of flow for the lane? What is the average speed (in mph)? Headway and Spacing can be converted to the macroscopic measures of flow rate and density, as follows: v = 3600 = 3600 = 1,286 veh/h/ln ha 2.8 D = 5280 = 5280 = 22.5 veh/mi/ln da 235 Speed is then computed as: S =

v D

= 1286 = 57.2 mi/h 22.5

Problem 5‐6 The following counts were taken on a major arterial during the evening peak period. From this data determine a) The peak hour b) The peak hour volume c) The peak flow rate within the peak hour d) The peak hour factor

a) The highest hourly volume (within the study period) occurs between 4:30 and 5:30 PM. b) The hourly volume is the volume for this hour, or 1,999 vehs/h.

c) The highest flow rate is the 15‐minute interval within the peak hour with the highest 15‐minute volume. This is the period between 5:15 and 5:30 PM. The flow rate within this period is 506 * 4 = 2,024 veh/h. d) The peak hour factor is 1999/2024 = 0.988.

Problem 5‐7 A peak hour volume of 1 200 veh/h is observed on a freeway lane. What is the peak flow rate within this hour if the PHF is 0.87? The peak flow rate is found as: v =

V = PHF

1200 = 1379 veh/h 0.87

Problem 5‐8 The flow rate on an arterial lane is 1,300 veh/h. If the average speed in the same lane is 35 mph, what is the density? The density is found as: D = V = 1,300 = 37.1 veh/mi/ln S 35

Problem 5‐9 The AADT for a section of suburban arterial is 50,000 veh/day. Assuming that this is an urban radial facility, what range of directional design hour volumes would be expected? From textbook Table 5‐2, Page 109, for an urban radial facility, K factors range from 0.07 to 0.12. D factors range from 0.55 to 0.60. Then: DDHV = AADT * K * D DDHVlow = 50,000 * .07 * .55 = 1925 veh/hr DDHVhigh = 50,000 * .12 * .60 = 3,600 veh/hr This is a very broad range, and highlights the danger in using such generalized factors for estimating demand.

Problem 5‐10 The following travel times were measured for vehicles traversing a 1,000 ft. segment of an arterial.

Determine the time mean speed (TMS) and space mean speed (SMS) for these vehicles.

HINT: Convert ft/s to mi/hr by conversion factor of 1.47

The TMS is now merely the average of the vehicle speeds, or 266.8/8 = 33.35 mi/h. The SMS is based upon the average travel speed, or 163.8/8 = 20.475 s/veh. Then: