Solution Manual for Traffic Engineering, 5th Edition

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TRAFFIC ENGINEERING5THEditionRoger P. Roess, Elena S. Prassas, William R. McShaneSOLUTIONS MANUALMarch 2018

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IndexSolutions to Problems in:PageChapter 2……………………………………………………………………………….1Chapter 3……………………………………………………………………………….3Chapter 4……………………………………………………………………………….9Chapter 5……………………………………………………………………………….11Chapter 6……………………………………………………………………………….17Chapter 7……………………………………………………………………………….23Chapter 8……………………………………………………………………………….25Chapter 9……………………………………………………………………………….31Chapter 10……………………………………………………………………………….37Chapter 11……………………………………………………………………………….47Chapter 12……………………………………………………………………………….55Chapter 13……………………………………………………………………………….67Chapter 14…………………………………………………………………………….....73Chapter 15………………………………………………………………………………..85Chapter 16……………………………………………………………………………….. 103Chapter 17……………………………………………………………………………….. 111Chapter 18……………………………………………………………………………….. 115Chapter 19……………………………………………………………………………….. 123Chapter 20……………………………………………………………………………….. 147Chapter 21……………………………………………………………………………….. 161Chapter 22……………………………………………………………………………….. 169Chapter 23……………………………………………………………………………….. 179Chapter 24……………………………………………………………………………….. 183Chapter 25……………………………………………………………………………….. 185

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PageChapter 26……………………………………………………………………………….197Chapter 27……………………………………………………………………………….203Chapter 28……………………………………………………………………………….213Chapter 29……………………………………………………………………………….223Chapter 30……………………………………………………………………………….239Chapter 31……………………………………………………………………………….259

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1Solutions to Problems in Chapter 2Transportation Modes and CharacteristicsProblem 2-1The 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) thetraffic composition – particularly the presence of trucks and local buses, and c) the controlenvironment – such features as lane use controls, signalization, curb lane controls, etc.Problem 2-2The 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 trainscan operate. The latter is limited by either the control system or station dwell times.Problem 2-3The key element here is that trains may operate 1.8 minutes apart. In this case, the dwelltime 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/hProblem 2-4From 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, eachof whichdisplaces 2.0passenger cars from the traffic stream. At capacity, there are:2,250*0.10 = 225 trucks2,250*0.02 =45 express busesEach 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 carsUsing the vehicle occupancies given in the problem statement, the person-capacity ofone lane is:(1710*1.5)+(225*1.0)+(45*50) = 5,040 persons/h

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2As there are 3 lanes in each direction, the capacity of each direction is 3*5040 = 15,120people/h.Problem 2-5A travel demand of 30,000 persons per hour is virtually impossible to serve entirely withhighway facilities. Even in the best case of a freeway with a 70-mi/h free-flow speed, andan assumed occupancy of 1.5 persons/car, a lane can carry only 3,600 people/h (Table2-5). That dictates a need for 30,000/3,600 = 8.33 fully-dedicated freeway lanes to servethis demand.While this might be technically feasible if the area were basically vacantland with a new high-density trip generator being built, it would be intractable in mostexisting development settings.That leaves various public transit options (Table 2-6).Given the observed capacities, itis doubtful that such a demand could be handled by bus transit (either on the street or ona private right-of-way) or light rail.A rapid transit line with one track in each directionwould be able to handle the demand.A lot depends on what type of development is spurring the demand.If it is a stadium orentertainment complex that generates high-intensity demand for short periods of time, thesolution may be different from a case of a regional shopping mall, where trips are moredistributed over time.It is likely that some mix of modes would be needed.Rail transit is expensive, and anynew service would have to be linked into a larger rapid transit network to be useful. Autoaccess is generally preferred by users (except for the traffic it generates), but involves theneed to provide huge numbers of parking places within walking distance of the desireddestination.A stadium could rely fairly heavily on transit, with heavy rail, light rail, andbus options viable. Some highway access and parking would also be needed. A regionalshopping center would have to cater more to autos, as most people would prefer not tohaul their purchases on transit.

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Solutions to Problems in Chapter 3Speed, Travel Time, and Delay StudiesProblem 3-1The reaction distance is given by Equation 3-1:tSdr47.1=For a speed of 70 mi/h, the result is:ftdr2.3605.3*70*47.1==Other values for the range of speeds specified are shown in Table 1. Figure 1 plotsthese values.Table 1: Reaction Distance vs. SpeedSpeedDistance30154.435180.140205.845231.550257.355283.060308.765334.470360.2Figure 1: Reaction Distance vs. Speed3

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Problem 3-2This problem involves several considerations.At the point when the driver notices thetruck, the vehicle is 350 ft away from a collision. To stop, the driver must go through thereaction distance and then the braking distance. The two will be considered separately.Reaction DistanceReaction 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 sis shown, with all results in Table 3.fttSdr8.4750.0*65*47.147.1===Table 3: Reactions Distances for Problem 3-2SpeedReactionReaction(mi/h)TimeDistance(s)(ft)650.5047.8651.0095.6651.50143.3652.00191.1652.50238.9653.00286.7653.50334.4654.00382.2654.50430.0655.00477.8For any result > 350 ft, the driver will not even get his/her foot on the brake before collidingwith the truck.Thus, for all reaction times, t ≥ 4.0 s, the collision speed will be 65 mi/h.Braking DistanceFor all reaction times < 4.0 s, the driver will engage the brake before hitting the truck, andtherefore, will at least decelerate somewhat before a collision.How much decelerationwill take place depends upon how much braking distance is left when the brake isengaged. In each case, this would be 350 ft – the reaction distance,dr.Once the brakingdistance available is determined, the braking distance formula of Equation 3-5 is used:)(3022GFSSdfib±−=The braking distance will be determined as indicated. For example, for a reaction timeof 2.0 s, the reaction distance (from Table 3) is 191.1 ft. The available braking distance4

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is then 350.0 – 191.1 = 158.9ft.The initial speed (Si) is 65.0 mi/h in all cases. Thegrade is given as level (G= 0.00), and the friction factor is found from the decelerationrate as:311.02.3210===gaFThe final speed (Sf) is the unknown. Then for the example with a 2.0 s reaction time:hmiSSSdfffr/4.525.739,2)311.0*30*9.158(65)000.0311.0(30659.1582222==−=+−==Table 4 summarizes the results for all reaction times.Table 4: Collision Speed vs. Reaction Time for Problem 3-2ReactionBrakingCollisionTimeDistanceSpeed(s)(ft)(mi/h)0.5302.237.51.0254.543.01.5206.747.92.0158.952.42.5111.156.53.063.460.33.515.663.94.0NA65.04.5NA65.05.0NA65.0Figure 2 shows a plot of these results. Note that in no case is the driver able to stop thevehicle before colliding with the truck.5

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Figure 2: Reaction Time vs. Collision Speed, Problem 3-2Problem 3-3In this case, we are dealing with measured skid marks at an accident location. Becauseskid marks only occur when the brakes are engaged, the reaction time and reactiondistance 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 collisionspeed is 25 mi/h.Using the grass skid, we can work backwards to find the initial speedat the beginning of the grass skid (S1).This is also the final speed at the end of thepavement skid.Working backwards again, we can find the initial speed (Si) at thebeginning of the pavement skid.Both solutions use the braking formula of Equation 3-5:0.010.020.030.040.050.060.070.00.01.02.03.04.05.0Collision Speed (mi/h)Reaction Time (s)Collision withTree at 25 mi/hGrass Skid = 200 ftF = 0.250Pavement Skid = 150 ftF = 0.348SiS16

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hmiSSSdhmiSSSdGFSSdiiiPAVEbGRASSbfib/3.63006,4305,2)378.0*30*150()03.0348.0(300.48150/0.48305,225)280.0*30*200()03.0250.0(3025200)(30222,1221221,22==+=+−====+=+−==±−=In an accident investigation, this result would be compared to the speed limit to determinewhether excessive speed contributed to the accident.Problem 3-4This problem involves a reaction distance and a braking distance, as drivers must see asign and reduce their speed to navigate a hazard. Level terrain is assumed, and standardvalues fort(2.5 s)F(0.348) anda(10.0 ft/s2) are used.The full distance to respond isthe sum of Equation 3-1 for reaction distance and Equation 3-5 or 3-6 for braking:()ftGSStSdfii1.4126.1915.220348.0*3040605.2*60*47.1)348.0(3047.12222=+=−+=±−+=The sign mustbe seena total of 412.1 ft from the hazard.Since the sign can be readfrom 200 ft, it could be placed as close as 412.1-200.0 = 212.1 ft from the hazard. Otherconsiderations, however, would also enter a final decision on the placement of the sign.Problem 3-5Theyellowinterval of a traffic signal is designed to let any vehicle that cannot safely stopbefore entering the intersection safely enter the intersection at the ambient speed, whichis generally taken to be an 85thpercentile speed.First, the safe stopping distance mustbe found for a vehicle traveling at 40 mi/h on a 0.02 downgrade, using Equation 3-10:()ftGStSdS4.2216.1628.58)02.0348.0(30400.1*40*47.1)348.0(3047.122=+=−+=±+=As the vehicle is traveling at a speed of 40 mi/h, theyellowmust be long enough to allowthe vehicle to traverse 221.4ft at 40 mi/h, or:7

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sy77.340*47.14.221==Problem 3-6The safe stopping distance is computed using Equation 3-10:()ftGStSdS4.9444.6500.294)02.0348.0(30805.2*80*47.1)348.0(3047.122=+=−+=±+=Problem 3-7The minimum radius of curvature is given by Equation 3-3:ftfeSR7.041,2)10.006.0(1570)(1522=+=+=8

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9Solutions to Problems in Chapter 4Communicating With Drivers: Traffic Control DevicesProblem 4-1Astandardin 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 agencyin charge with potential legal liability for accidents.Aguidelinein the MUTCD is strongly suggested advice based upon national consensusin the profession.While not legally binding, any deviation should be documented by anengineering study that is kept on file.Legal liability may still exist, especially where nodocumentation of the deviation is available.A guideline is accompanied by the words“should” or “should not.”Anoptionin the MUTCD is just that – an option.It presents information that may beimplemented or not based upon the local judgment of the relevant traffic agency. No legalliability is implied.Supportin the MUTCD is simply additional information for the manual user.Problem 4-2Human 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 areused to code types of information, and to draw the attention of road users requiring thistype of information.Thus, the STOP sign has a unique color, shape, and legend.Theword “STOP” could easily be omitted, and drivers would still know the meaning of the redoctagon.Guide signs are also color- and shape-coded.Directional information is onrectangular signs (long dimensionhorizontal) withagreenbackground.Servicesinformation is on similar-shaped signs, but with a blue background.Cultural or historicdirectional information is on rectangular signs too, but with a brown background.Allwarning signs are diamond-shaped with a yellow background.Problem 4-3Overuse of warning signs is particularly dangerous.If drivers begin to suspect thatwarning signs are not warning about things that are truly an imminent threat, they maytend to ignore them – which would be a major problem in the case of a real imminentthreat.They should be used to bring drivers’ attention to an upcoming hazard that theywould 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 aregulation that they would otherwise be unaware of.Overuse again tends to causedrivers 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,

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10however, frequent guidance is a comfort, and avoids having confused drivers – who are,by definition – dangerous drivers.Problem 4-4Table 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 ConditionB with a speed limit of 50 mi/h and an advisory speed of 0 mi/h, the sign would beplaced 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 anadvisory speed of 30 mi/h. The sign would be placed 100 ft from the curve, whichis the minimum advance placement distance permitted. Signs are assumed to bevisible for 250 ft away.c)A merge ahead sign may be viewed as a Condition A maneuver. For a speed limitof 35 mi/h, the sign would be placed 565 ft from the merge point.Problems 4-5 and 4-6Both of these are local projects for students.They require field observation of thestudents.Instructorsshouldchecktoseewhetherstudentsareinsured(bytheUniversity) for such activities.

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11Solutions to Problems in Chapter 5Traffic Stream CharacteristicsProblem 5-1a)The flow rate is computed as:hvehhvav/385,16.2600,3600,3===b)The density is computed as:mivehdDav/47.22235280,5280,5===c)The average speed is computed as:hmiDvS/6.6147.22385,1===Problem 5-2The peak hour factor is defined as:vVPHF=where:V=peak hour volume, vehs/h, andv=peak rate of flow within the hour, vehs/h.Therefore:PHFvV*=and:a)V = 5,600*0.85 = 4,760 veh/hb)V = 5,600*0.90 = 5,040 veh/hc)V = 5,600*0.95 = 5,320 veh/hProblem 5-3The solution is best carried out using a spreadsheet. The spreadsheet that follows findsthe desired results as:

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12a)The AADT is the total volume for the year (sum, Col. 4) divided by the total numberof days in the year (sum, Col. 3.)dayvehsAADT/071,4365000,486,1==b)The ADT for each month is the total volume for the month (Col. 4) divided by thetotal number of days in each month (Col. 3).c)The AAWT is the total weekday volume for the year (sum, Col. 5) divided by thetotal number of weekdays in the year (sum, Col. 2).dayvehsAAWT/308,3260000,860==d)The AWT for each month is the total weekday volume for the month (Col. 5) dividedby the number of weekdays in the month (Col. 2).MonthNo. ofTotalTotalTotalADTAWTWeekdaysDaysMonthlyWeekdayVolVolJan2231120,00070,0003,8713,182Feb2028115,00060,0004,1073,000Mar2231125,00075,0004,0323,409Apr2230130,00078,0004,3333,545May2131135,00085,0004,3554,048Jun2230140,00085,0004,6673,864Jul2331150,00088,0004,8393,826Aug2131135,00080,0004,3553,810Sep2230120,00072,0004,0003,273Oct2231112,00062,0003,6132,818Nov2130105,00055,0003,5002,619Dec223199,00050,0003,1942,2732603651,486,000860,000The information reveals two things about this facility:1.Since the AADT > AAWT, and the monthly ADTs are generally larger than themonthly AWTs,thisis likely arecreational route attracting mostly weekendtravelers.

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