TRAFFIC ENGINEERING 5 TH Edition Roger P. Roess, Elena S. Prassas, William R. McShane SOLUTIONS MANUAL March 2018 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 Page Chapter 26 ………………………………………………………………………………. 197 Chapter 27 ………………………………………………………………………………. 203 Chapter 28 ………………………………………………………………………………. 213 Chapter 29 ………………………………………………………………………………. 223 Chapter 30 ………………………………………………………………………………. 239 Chapter 31 ………………………………………………………………………………. 259 1 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 person- capacity 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 vehicle- capacity 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 2 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. Solutions to Problems in Chapter 3 Speed, Travel Time, and Delay Studies Problem 3-1 The reaction distance is given by Equation 3-1: t S d r 47 . 1 = For a speed of 70 mi/h, the result is: ft d r 2 . 360 5 . 3 * 70 * 47 . 1 = = 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 30 154.4 35 180.1 40 205.8 45 231.5 50 257.3 55 283.0 60 308.7 65 334.4 70 360.2 Figure 1: Reaction Distance vs. Speed 3 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. ft t S d r 8 . 47 50 . 0 * 65 * 47 . 1 47 . 1 = = = Table 3: Reactions Distances for Problem 3-2 Speed Reaction Reaction (mi/h) Time Distance (s) (ft) 65 0.50 47.8 65 1.00 95.6 65 1.50 143.3 65 2.00 191.1 65 2.50 238.9 65 3.00 286.7 65 3.50 334.4 65 4.00 382.2 65 4.50 430.0 65 5.00 477.8 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 < 4.0 s, the driver will engage the brake before hitting the truck, and therefore, will at least decelerate somewhat before a collision. How much deceleration will take place depends upon how much braking distance is left when the brake is engaged. In each case, this would be 350 ft – the reaction distance, d r . Once the braking distance available is determined, the braking distance formula of Equation 3-5 is used: ) ( 30 2 2 G F S S d f i b ± − = The braking distance will be determined as indicated. For example, for a reaction time of 2.0 s, the reaction distance (from Table 3) is 191.1 ft. The available braking distance 4 is then 350.0 – 191.1 = 158.9 ft. The initial speed ( S i ) is 65.0 mi/h in a ll cases. The grade is given as level ( G = 0.00), and the friction factor is found from the deceleration rate as: 311 . 0 2 . 32 10 = = = g a F The final speed ( S f ) is the unknown. Then for the example with a 2.0 s reaction time: h mi S S S d f f f r / 4 . 52 5 . 739 , 2 ) 311 . 0 * 30 * 9 . 158 ( 65 ) 000 . 0 311 . 0 ( 30 65 9 . 158 2 2 2 2 = = − = + − = = Table 4 summarizes the results for all reaction times. Table 4: Collision Speed vs. Reaction Time for Problem 3-2 Reaction Braking Collision Time Distance Speed (s) (ft) (mi/h) 0.5 302.2 37.5 1.0 254.5 43.0 1.5 206.7 47.9 2.0 158.9 52.4 2.5 111.1 56.5 3.0 63.4 60.3 3.5 15.6 63.9 4.0 NA 65.0 4.5 NA 65.0 5.0 NA 65.0 Figure 2 shows a plot of these results. Note that in no case is the driver able to stop the vehicle before colliding with the truck. 5 Figure 2: Reaction Time vs. Collision Speed, Problem 3-2 Problem 3-3 In this case, we are dealing with measured skid marks at an accident location. Because skid marks only occur when the brakes are engaged, the reaction time and reaction distance play no role in this solution. The sketch below helps in the understanding of the solution: The only know speed is at the collision point (at the end of the grass skid). The collision speed is 25 mi/h. Using the grass skid, we can work backwards to find the initial speed at the beginning of the grass skid ( S 1 ). This is also the final speed at the end of the pavement skid. Working backwards again, we can find the initial speed ( S i ) at the beginning of the pavement skid. Both solutions use the braking formula of Equation 3-5: 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 0.0 1.0 2.0 3.0 4.0 5.0 Collision Speed (mi/h) Reaction Time (s) Collision with Tree at 25 mi/h Grass Skid = 200 ft F = 0.250 Pavement Skid = 150 ft F = 0.348 S i S 1 6 h mi S S S d h mi S S S d G F S S d i i i PAVE b GRASS b f i b / 3 . 63 006 , 4 305 , 2 ) 378 . 0 * 30 * 150 ( ) 03 . 0 348 . 0 ( 30 0 . 48 150 / 0 . 48 305 , 2 25 ) 280 . 0 * 30 * 200 ( ) 03 . 0 250 . 0 ( 30 25 200 ) ( 30 2 2 2 , 1 2 2 1 2 2 1 , 2 2 = = + = + − = = = = + = + − = = ± − = In an accident investigation, this result would be compared to the speed limit to determine whether excessive speed contributed to the accident. Problem 3-4 This problem involves a reaction distance and a braking distance, as drivers must see a sign and reduce their speed to navigate a hazard. Level terrain is assumed, and standard values for t (2.5 s) F (0.348) and a (10.0 ft/s 2 ) are used. The full distance to respond is the sum of Equation 3-1 for reaction distance and Equation 3-5 or 3- 6 for braking: ( ) ft G S S t S d f i i 1 . 412 6 . 191 5 . 220 348 . 0 * 30 40 60 5 . 2 * 60 * 47 . 1 ) 348 . 0 ( 30 47 . 1 2 2 2 2 = + = − + = ± − + = The sign must be seen a total of 412.1 ft from the hazard. Since the sign can be read from 200 ft, it could be placed as close as 412.1 - 200.0 = 212.1 ft from the hazard. Other considerations, however, would also enter a final decision on the placement of the sign. Problem 3-5 The yellow interval of a traffic signal is designed to let any vehicle that cannot safely stop before entering the intersection safely enter the intersection at the ambient speed, which is generally taken to be an 85 th percentile speed. First, the safe stopping distance must be found for a vehicle traveling at 40 mi/h on a 0.02 downgrade, using Equation 3 -10: ( ) ft G S t S d S 4 . 221 6 . 162 8 . 58 ) 02 . 0 348 . 0 ( 30 40 0 . 1 * 40 * 47 . 1 ) 348 . 0 ( 30 47 . 1 2 2 = + = − + = ± + = As the vehicle is traveling at a speed of 40 mi/h, the yellow must be long enough to allow the vehicle to traverse 221.4 ft at 40 mi/h, or: 7 s y 77 . 3 40 * 47 . 1 4 . 221 = = Problem 3-6 The safe stopping distance is computed using Equation 3-10: ( ) ft G S t S d S 4 . 944 4 . 650 0 . 294 ) 02 . 0 348 . 0 ( 30 80 5 . 2 * 80 * 47 . 1 ) 348 . 0 ( 30 47 . 1 2 2 = + = − + = ± + = Problem 3-7 The minimum radius of curvature is given by Equation 3-3: ft f e S R 7 . 041 , 2 ) 10 . 0 06 . 0 ( 15 70 ) ( 15 2 2 = + = + = 8 9 Solutions to Problems in Chapter 4 Communicating With Drivers: Traffic Control Devices Problem 4-1 A standard in the MUTCD is a mandatory condition, and is accompanied by the words “shall” or “shall not.” Standards must be followed, and failure to do so leaves the agency in charge with potential legal liability for accidents. A guideline in the MUTCD is strongly suggested advice based upon national consensus in the profession. While not legally binding, any deviation should be documented by an engineering study that is kept on file. Legal liability may still exist, especially where no documentation of the deviation is available. A guideline is accompanied by the words “should” or “should not.” An option in the MUTCD is just that – an option. It presents information that may be implemented or not based upon the local judgment of the relevant traffic agency. No legal liability is implied. Support in the MUTCD is simply additional information for the manual user. Problem 4-2 Human eyesight can identify color, then shape or pattern, and finally, specific text. Because color and shape are discernible to most road users at great distances, they are used to code types of information, and to draw the attention of road users requiring this type of information. Thus, the STOP sign has a unique color, shape, and legend. The word “STOP” could easily be omitted, and drivers would still know the meaning of the red octagon. Guide signs are also color- and shape-coded. Directional information is on rectangular signs (long dimension horizontal) with a green background. Services information is on similar-shaped signs, but with a blue background. Cultural or historic directional information is on rectangular signs too, but with a brown background. All warning signs are diamond-shaped with a yellow background. Problem 4-3 Overuse of warning signs is particularly dangerous. If drivers begin to suspect that warning signs are not warning about things that are truly an imminent threat, they may tend to ignore them – which would be a major problem in the case of a real imminent threat. They should be used to bring drivers’ attention to an upcoming hazard that they would not normally be able to discern in time to safely maneuver through it. Regulatory signing should also be used only when needed to inform drivers about a regulation that they would otherwise be unaware of. Overuse again tends to cause drivers to not pay attention to them. Guide signs are unique in that only a small percentage of drivers actually use them: familiar drivers by-and-large know where they are going and how to get there. For others, 10 however, frequent guidance is a comfort, and avoids having confused drivers – who are, by definition – dangerous drivers. Problem 4-4 Table 4-1 is consulted to determine appropriate posting locations. a) For a STOP ahead sign, the advisory speed is an implied “0” mi/h. Using Condition B with a speed limit of 50 mi/h and an advisory speed of 0 mi/h, the sign would be placed 250 ft in advance of the location of the STOP sign. b) For a curve ahead sign, Condition B is used with a speed limit of 45 mi/h and an advisory speed of 30 mi/h. The sign would be placed 100 ft from the curve, which is the minimum advance placement distance permitted. Signs are assumed to be visible for 250 ft away. c) A merge ahead sign may be viewed as a Condition A maneuver. For a speed limit of 35 mi/h, the sign would be placed 565 ft from the merge point. Problems 4-5 and 4-6 Both of these are local projects for students. They require field observation of the students. Instructors should check to see whether students are insured (by the University) for such activities. 11 Solutions to Problems in Chapter 5 Traffic Stream Characteristics Problem 5-1 a) The flow rate is computed as: h veh h v av / 385 , 1 6 . 2 600 , 3 600 , 3 = = = b) The density is computed as: mi veh d D av / 47 . 22 235 280 , 5 280 , 5 = = = c) The average speed is computed as: h mi D v S / 6 . 61 47 . 22 385 , 1 = = = Problem 5-2 The peak hour factor is defined as: v V PHF = where: V = peak hour volume, vehs/h, and v = peak rate of flow within the hour, vehs/h. Therefore: PHF v V * = and: a) V = 5,600*0.85 = 4,760 veh/h b) V = 5,600*0.90 = 5,040 veh/h c) V = 5,600*0.95 = 5,320 veh/h Problem 5-3 The solution is best carried out using a spreadsheet. The spreadsheet that follows finds the desired results as: 12 a) The AADT is the total volume for the year (sum, Col. 4) divided by the total number of days in the year (sum, Col. 3.) day vehs AADT / 071 , 4 365 000 , 486 , 1 = = b) The ADT for each month is the total volume for the month (Col. 4) divided by the total number of days in each month (Col. 3). c) The AAWT is the total weekday volume for the year (sum, Col. 5) divided by the total number of weekdays in the year (sum, Col. 2). day vehs AAWT / 308 , 3 260 000 , 860 = = d) The AWT for each month is the total weekday volume for the month (Col. 5) divided by the number of weekdays in the month (Col. 2). Month No. of Total Total Total ADT AWT Weekdays Days Monthly Weekday Vol Vol Jan 22 31 120,000 70,000 3,871 3,182 Feb 20 28 115,000 60,000 4,107 3,000 Mar 22 31 125,000 75,000 4,032 3,409 Apr 22 30 130,000 78,000 4,333 3,545 May 21 31 135,000 85,000 4,355 4,048 Jun 22 30 140,000 85,000 4,667 3,864 Jul 23 31 150,000 88,000 4,839 3,826 Aug 21 31 135,000 80,000 4,355 3,810 Sep 22 30 120,000 72,000 4,000 3,273 Oct 22 31 112,000 62,000 3,613 2,818 Nov 21 30 105,000 55,000 3,500 2,619 Dec 22 31 99,000 50,000 3,194 2,273 260 365 1,486,000 860,000 The information reveals two things about this facility: 1. Since the AADT > AAWT, and the monthly ADTs are generally larger than the monthly AWTs, this is likely a recreational route attracting mostly weekend travelers.