Solution Chapter 5 - 1 Traffic Engineering, 4th Edition...

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Unformatted text preview: 1 Traffic Engineering, 4th Edition Roess, R.P., Prassas, E.S., and McShane, W.R. Solutions to Problems in Chapter 5 Problem 5‐1 The peak rate of flow is computed as v = V/PHF. The table below summarized the results for the information given. A plot of the results is also shown. Peak Flow Rate vs. PHF Volume Peak Flow PHF Rate (veh/h) (veh/h) 1200 1.00 1200 1200 0.90 1333 1200 0.80 1500 1200 0.70 1714 Flow Rate vs. PHF 1800 1600 1400 Peak Flow Rate (veh/h) 1200 1000 800 600 400 200 0 0.60 0.65 0.70 0.75 0.80 PHF 0.85 0.90 0.95 1.00 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. 1 2 Problem 5‐2 A headway can be converted to a flow rate as follows: 3600 3600 v= = = 1,500 veh / h / ln h 2 .4 Knowing both flow rate and speed (given), the density may now be computed as: v 1500 D= = = 27.3 veh / h / ln S 55 Problem 5‐3 Density is obtained from occupancy as follows: 5280 * O 5280 * 0.26 D= = = 63.9 veh / mi / ln Lv + Ld 18 + 3.5 Such a high value is indicative of highly congested conditions within a queue. Problem 5‐4 The table on the next page illustrates the computation of monthly ADT and AWT values. The AADT is computed as the total annual volume divided by 365 days, or: 2,365,000 AADT = = 6,479 veh / day 365 The AAWT is computed as the total weekday volume divided by 260 days, or: 2,067,000 AAWT = = 7,950 veh / day 260 2 3 Table: ADT and AWT Computed 1 2 3 4 5 6=4/2 7=5/3 Days Weekdays Total Weekday ADT for AWT for Month in in Volume Volume Month Month Month Month (vehs) (vehs) (veh/day) (veh/day) Jan 31 22 200,000 170,000 6,452 7,727 Feb 28 20 210,000 171,000 7,500 8,550 Mar 31 22 215,000 185,000 6,935 8,409 Apr 30 22 205,000 180,000 6,833 8,182 May 31 21 195,000 172,000 6,290 8,190 Jun 30 22 193,000 168,000 6,433 7,636 Jul 31 23 180,000 160,000 5,806 6,957 Aug 31 21 175,000 150,000 5,645 7,143 Sep 30 22 189,000 175,000 6,300 7,955 Oct 31 22 198,000 178,000 6,387 8,091 Nov 30 21 205,000 182,000 6,833 8,667 Dec 31 22 200,000 176,000 6,452 8,000 Total 365 260 2,365,000 2,067,000 77,868 95,507 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: 298,000 AAWT = = 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 Headway and Spacing can be converted to the macroscopic measures of flow rate and density, as follows: 3 4 2600 3600 = = 1,286 veh / h / ln h 2 .8 5280 5280 D= = = 22.5 veh / mi / ln d 235 Speed is then computed as: v 1286 S= = = 57.2 mi / h D 22.5 Problem 5‐6 The determination of the peak hour is illustrated in the table that follows. Note that the determination is made to the nearest 15 minutes by computing all overlapping hourly volumes for each possible combination of four consecutive 15‐minute periods between 4:00 PM and 6:00 PM. Table: Finding the Peak Hour Time Vol Hourly Period (vehs) Vol (vehs) 4:00‐4:15 450 NA 4:15‐4:30 465 NA 4:30‐4:45 490 NA 4:45‐5:00 500 1905 1958 5:00‐5:15 503 1999 5:15‐5:30 506 5:30‐5:45 460 1969 5:45‐6:00 445 1914 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. v= 4 5 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 506/0.25 = 2,024 veh/h. d) The peak hour factor is 1999/2024 = 0.988. Problem 5‐7 The peak flow rate is found as: V 1200 v= = = 1,379 veh / h PHF 0.87 Problem 5‐8 The density is found as: v 1300 D= = = 37.1 veh / mi / ln 35 S Problem 5‐9 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 * 0.07 * 0.55 = 1,925 veh / h DDHVhigh = 50,000 * 0.12 * 0.60 = 3,600 veh / h This is a very broad range, and highlights the danger in using such generalized factors for estimating demand. Problem 5‐10 The TMS is computed as the arithmetic average of individual vehicle speeds. The SMS is a speed computed using the average travel time of the individual vehicles. These computations are shown in the table that follows. 5 6 Table: TMS and SMS Computed Veh Travel Travel Travel Travel No. Time Distance Speed Speed (s) (ft) (ft/s) (mi/h) 1 20.6 1000 48.54 33.02 2 21.7 1000 46.08 31.35 3 19.8 1000 50.51 34.36 4 20.3 1000 49.26 33.51 5 22.5 1000 44.44 30.23 6 18.5 1000 54.05 36.77 7 19.0 1000 52.63 35.80 8 21.4 1000 46.73 31.79 Total 163.8 392.3 266.8 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: 1,000 48.84 SMS = = 48.84 ft / s = = 33.22 mi / h 20.465 1.47 6 ...
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