Solution Manual for Traffic Engineering, 5th Edition

TRAFFIC ENGINEERING 5TH Edition Roger P. Roess, Elena S. Prassas, William R. McShane SOLUTIONS MANUAL March 2018 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Index Solutions to Problems in: Page Chapter 2 ………………………………………………………………………………. 1 Chapter 3 ………………………………………………………………………………. 3 Chapter 4 ………………………………………………………………………………. 9 Chapter 5 ………………………………………………………………………………. 11 Chapter 6 ………………………………………………………………………………. 17 Chapter 7 ………………………………………………………………………………. 23 Chapter 8 ………………………………………………………………………………. 25 Chapter 9 ………………………………………………………………………………. 31 Chapter 10 ………………………………………………………………………………. 37 Chapter 11 ………………………………………………………………………………. 47 Chapter 12 ………………………………………………………………………………. 55 Chapter 13 ………………………………………………………………………………. 67 Chapter 14 ……………………………………………………………………………….. 73 Chapter 15 ……………………………………………………………………………….. 85 Chapter 16 ……………………………………………………………………………….. 103 Chapter 17 ……………………………………………………………………………….. 111 Chapter 18 ……………………………………………………………………………….. 115 Chapter 19 ……………………………………………………………………………….. 123 Chapter 20 ……………………………………………………………………………….. 147 Chapter 21 ……………………………………………………………………………….. 161 Chapter 22 ……………………………………………………………………………….. 169 Chapter 23 ……………………………………………………………………………….. 179 Chapter 24 ……………………………………………………………………………….. 183 Chapter 25 ……………………………………………………………………………….. 185 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Page Chapter 26 ………………………………………………………………………………. 197 Chapter 27 ………………………………………………………………………………. 203 Chapter 28 ………………………………………………………………………………. 213 Chapter 29 ………………………………………………………………………………. 223 Chapter 30 ………………………………………………………………………………. 239 Chapter 31 ………………………………………………………………………………. 259 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Solutions to Problems in Chapter 2 Transportation Modes and Characteristics Problem 2-1 The capacity of a street or highway is affected by a) the physical design of the roadway – such features as the number of lanes, free-flow speed, and geometric design, b) the traffic composition – particularly the presence of trucks and local buses, and c) the control environment – such features as lane use controls, signalization, curb lane controls, etc. Problem 2-2 The capacity of a rapid transit line is affected by: the number of tracks, the personcapacity of each rail car, the length of trains, and the minimum headways at which trains can operate. The latter is limited by either the control system or station dwell times. Problem 2-3 The key element here is that trains may operate 1.8 minutes apart. In this case, the dwell time controls this limit, not the train control system, which would allow closer operation. Thus, one track can accommodate 60/1.8 = 33.3 (say 33) trains/h. Each train has 10 cars, each of which accommodates a total of 50+80 = 130 passengers. The capacity of a single track is, therefore: 33*10*130 = 42,900 people/h Problem 2-4 From Table 2-5 of the text, a freeway with a free-flow speed of 55 mi/h has a vehiclecapacity of 2,250 passenger cars/h. Traffic contains 10% trucks and 2% express buses, each of which displaces 2.0 passenger cars from the traffic stream. At capacity, there are: 2,250*0.10 = 225 trucks 2,250*0.02 = 45 express buses Each of these displaces 2.0 passenger cars from the traffic stream. Thus, the 225+45 = 270 heavy vehicles displace 2*270 = 540 passenger cars from the traffic stream. Thus, the number of passenger cars at capacity is: 2,250 – 540 = 1,710 passenger cars Using the vehicle occupancies given in the problem statement, the person-capacity of one lane is: (1710*1.5)+(225*1.0)+(45*50) = 5,040 persons/h 1 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. As there are 3 lanes in each direction, the capacity of each direction is 3*5040 = 15,120 people/h. Problem 2-5 A travel demand of 30,000 persons per hour is virtually impossible to serve entirely with highway facilities. Even in the best case of a freeway with a 70-mi/h free-flow speed, and an assumed occupancy of 1.5 persons/car, a lane can carry only 3,600 people/h (Table 2-5). That dictates a need for 30,000/3,600 = 8.33 fully-dedicated freeway lanes to serve this demand. While this might be technically feasible if the area were basically vacant land with a new high-density trip generator being built, it would be intractable in most existing development settings. That leaves various public transit options (Table 2-6). Given the observed capacities, it is doubtful that such a demand could be handled by bus transit (either on the street or on a private right-of-way) or light rail. A rapid transit line with one track in each direction would be able to handle the demand. A lot depends on what type of development is spurring the demand. If it is a stadium or entertainment complex that generates high-intensity demand for short periods of time, the solution may be different from a case of a regional shopping mall, where trips are more distributed over time. It is likely that some mix of modes would be needed. Rail transit is expensive, and any new service would have to be linked into a larger rapid transit network to be useful. Auto access is generally preferred by users (except for the traffic it generates), but involves the need to provide huge numbers of parking places within walking distance of the desired destination. A stadium could rely fairly heavily on transit, with heavy rail, light rail, and bus options viable. Some highway access and parking would also be needed. A regional shopping center would have to cater more to autos, as most people would prefer not to haul their purchases on transit. 2 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Solutions to Problems in Chapter 3 Speed, Travel Time, and Delay Studies Problem 3-1 The reaction distance is given by Equation 3-1: d r = 1.47 S t For a speed of 70 mi/h, the result is: d r = 1.47 * 70 * 3.5 = 360.2 ft Other values for the range of speeds specified are shown in Table 1. Figure 1 plots these values. Table 1: Reaction Distance vs. Speed Speed Distance 30 35 40 45 50 55 60 65 70 154.4 180.1 205.8 231.5 257.3 283.0 308.7 334.4 360.2 Figure 1: Reaction Distance vs. Speed 3 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3-2 This problem involves several considerations. At the point when the driver notices the truck, the vehicle is 350 ft away from a collision. To stop, the driver must go through the reaction distance and then the braking distance. The two will be considered separately. Reaction Distance Reaction distance is given by Equation 3-1, and is dependent upon the reaction time, which, for this problem, will be varied from 0.50 s to 5.00 s. A sample solution for 0.50 s is shown, with all results in Table 3. d r = 1.47 S t = 1.47 * 65 * 0.50 = 47.8 ft Table 3: Reactions Distances for Problem 3-2 Speed (mi/h) Reaction Reaction Time Distance (s) (ft) 65 65 65 65 65 65 65 65 65 65 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 47.8 95.6 143.3 191.1 238.9 286.7 334.4 382.2 430.0 477.8 For any result > 350 ft, the driver will not even get his/her foot on the brake before colliding with the truck. Thus, for all reaction times, t ≥ 4.0 s, the collision speed will be 65 mi/h. Braking Distance For all reaction times AAWT, and the monthly ADTs are generally larger than the monthly AWTs, this is likely a recreational route attracting mostly weekend travelers. 12 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2. Traffic peaks in the summer months, for both AWT and ADT. It suggests that during the winter, many commuters may be away on vacation, and that the recreational region served mostly consists of summer activities. Problem 5-4 Density is computed from occupancy measured at a detector as: D= 5,280 * 0 5,280 * 0.15 792 = = = 30.5 veh / mi / ln Lv + Ld 6 + 20 26 Problem 5-5 A spreadsheet is used to determine the total hourly volume for each available hour of four consecutive 15-minute counts. This will identify the peak hour and the peak hour volume of parts a) and b), and permit the computations required in parts c) and d). The spreadsheet is shown below. Time Period Volume (vehs) 4:00-4:15 4:15-4:30 4:30-4:45 4:45-5:00 5:00-5:15 5:15-5:30 5:30-5:45 5:45-6:00 300 325 340 360 330 310 280 240 Cumulative Volume (vehs) 1325 1355 1340 1280 1160 Then: a) The peak hour occurs between 4:15 and 5:15 PM. b) The peak hour volume is 1,355 vehs/h. c) The peak rate of flow within the peak hour is 4*360 =1,440 vehs/h d) The peak hour factor (PHF) is computed as: PHF = V 1,355 1,355 = = = 0.941 4 * V15 4 * 360 1,440 13 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5-6 The trick is that the volume must be divided into two lanes, or 1,800/2 = 900 veh/h/ln. Then, the density is computed as: D= V 900 = = 22.5 vehs / mi / ln S 40.0 Problem 5-7 An estimate of the directional design hour volume (DDHV) is found as: DDHV = AADT * K * D From Table 6-2 of the textbook, the range of K-factors applying to urban radial facilities is 0.07 – 0.12. The range of D-factors applying to urban radial facilities is 0.55 –0.60. Then: DDHVLOW = 150,000 * 0.07 * 0.55 = 5,775 veh / h. DDHVHIGH = 150,000 * 0.12 * 0.60 = 10,800 veh / h Problem 5-8 The time mean speed (TMS) is the arithmetic average of individual vehicle speeds observed. Each speed is the distance (2,000 ft) divided by the travel time (s). This gives a result in ft/s, which should be converted to mi/h. The space mean speed (SMS) is the distance (2,000 ft) divided by the average of the individual travel times. The spreadsheet below helps illustrate these computations: Veh 1 2 3 4 5 6 7 8 9 10 Length (ft) 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 SUM AVG t (s) 40.50 44.20 41.70 47.30 46.50 41.90 43.00 47.00 42.60 43.30 438.00 43.8 S (ft/s) 49.38 45.25 47.96 42.28 43.01 47.73 46.51 42.55 46.95 46.19 457.82 45.8 S (mi/h) 33.59 30.78 32.63 28.76 29.26 32.47 31.64 28.95 31.94 31.42 311.44 31.1 14 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Then: 457.82 = 45.8 ft / s 10 45.8 TMS = = 31.2 mi / h 1.47 TMS = and: 2,000 = 45.7 ft / s 43.8 45.7 SMS = = 31.1 mi / h 1.47 SMS = Problem 5-9 The peak flow rate on the freeway lane is: v= 1200 V = = 1,379 veh / h / ln PHF 0.87 Problem 5-10 The density on the freeway lane is found as: v = S *D D = v = 1300 = 37.1 veh / mi / ln 35 S Problem 5-11 a) The free-flow speed is 71.2 mi/h, the speed when density is “0.” The jam density is 122 pc/mi/ln, the density when speed is “0.” b) To derive the speed-flow curve, substitute D=v/S in the equation: 71.2 v D    v/S  S = 71.21 −  = 71.2 −  = 71.21 − 122S  122   122  71.2v = 71.2 − S 122S 71.2v = (122 * 71.2) S − 122S 2 v = 122S − 1.713S 2 To derive the flow-density curve, substitute S=v/D in the equation: 15 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. D  v  S = 71.21 − =  122  D 71.2 D v = 71.2 − 122 D v = 71.2 D − 0.584 D 2 c) Capacity occurs when both the speed-flow and flow-density curves are at their peak, or when the first derivative of each is 0.0: v = 122S − 1.713S 2 dv = 0 = 122 − 3.426S dS 122 = 35.6 mi / h S= 3.426 v = 71.2 D − 0.584 D 2 dv = 0 = 71.2 − 1.168 D dD 71.2 D= = 61.0 pc / mi / ln 1.168 The capacity is the product of the speed and density at capacity, or: c = 35.6 * 61.0 = 2,172 pc / h / ln 16 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.