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Corrections to fifth printing of

FUNDAMENTALS OF TRANSPORTATION ENGINEERING

by Jon D. Fricker and Robert K. Whitford

Updated 7 September 2009 to 30 December 2012

|This file contains errors found after changes to the third printing were submitted. If you are using the 4th printing, check to see if those |

|corrections were made. If you find any other contents that are unclear or appear to be in error, please contact Prof. Fricker at |

|fricker@purdue.edu. |

|Page |Correction |

| |CHAPTER 1 TRANSPORTATION BASICS |

| | |

| |CHAPTER 2 TRAFFIC FLOW: THEORY AND ANALYSIS |

|123 |Heading in Table 3.9 should be “Decrease in Average Travel Speed (mi/hr)” |

| | |

| |CHAPTER 3 HIGHWAY DESIGN FOR PERFORMANCE |

| | |

| |CHAPTER 4 MODELING TRANSPORTATION DEMAND AND SUPPLY |

| | |

| |CHAPTER 5 PLANNING AND EVALUATION FOR DECISION-MAKING |

|270 |In Example 5.6, EUAB-EUAC for Project A at d = 5% should be $360,000. At d = 10%, it should be $248,000. In the calculation for |

| |Project A5%, “140,000” should be “240,000”. |

|271 |Project A10% EUAB – EUAC = $148,000 $248,000 |

| | |

| |CHAPTER 6 SAFETY ON THE HIGHWAY |

|312 |Equation (6.3) should be [pic], where m = million entering vehicles or million vehicle-miles. |

|312 |In Example 6.3, insert a second sentence that says “Approximately 9.04 million vehicles enter the intersection in an average year.” |

|313 |In the Solution to Example 6.3, the calculations should be: [pic][pic]. Below the calculations, replace “(9.35)” with “(9.04)” and |

| |“(9.82)” with “(9.84)”. |

|341 |Replace Equation (6.15) and revise the sentence preceding it. “The time needed to go from vo to vf on a level roadway, is |

| |[pic] (6.15)” |

|341 |Just before equation (6.16), “If braking takes place on a hill with a positive (uphill) grade G, the braking distance will be” |

|367 |Stopping on a downhill grade. At one point on SR835, there is a 4.9 percent downhill grade. How long much distance and time will it|

| |take to bring a car traveling at 48 mph to a stop on that downhill segment if the driver’s reaction time is 2.0 seconds and f = 0.29?|

| |CHAPTER 7 HIGHWAY DESIGN FOR SAFETY |

|407 |22 Oct 2012: In Example 7.19, Ls from (7.26) should be 141.75 ft, not 141.5 ft. |

| | |

| |CHAPTER 8 DESIGN OF INTERSECTIONS FOR SAFETY AND EFFICIENCY |

|434 |22 Oct 2012: In Figure 8.11, the entries in cells A11 to A21 should be: |

| |44.00 |

| |v(0) = approach speed (ft/sec) |

| | |

| |66.00 |

| |x(r) = reaction distance (feet) |

| | |

| |4.40 |

| |t(s) = time to stop (sec.) |

| | |

| |96.80 |

| |x(b) = braking distance (feet) |

| | |

| |162.80 |

| |x(s) = stopping distance (feet) |

| | |

| |176.00 |

| |x(Y) = dist. Traveled during Yellow (feet) |

| | |

| |16.00 |

| |L = length of vehicle (ft.), if desired |

| | |

| |238.80 |

| |x(c) = distance to clear intersection (feet) |

| | |

| | |

| |x(c) = x(s) + W + L |

| | |

| |Results: |

| | |

| | |

| |62.80 |

| |x(DZ) = x(c) - x(Y) = [x(s)+W+L] - x(Y) |

| | |

| |CHAPTER 9 HIGHWAY DESIGN FOR RIDEABILITY (PAVEMENT DESIGN) |

|509 |(11/06/2012) Add to REFERENCES, between [2] and [3]: |

| |American Association of State Highway and Transportation Officials (AASHTO), Guide for Design of Pavement Structures, 4th Edition, |

| |1993 |

| | |

| |CHAPTER 10 PUBLIC MASS TRANSPORTATION |

|523 |(4/15/2012) Equation 10.10a includes braking. Some problems – like Exercise 10.8 -- include only acceleration from a stop until Vtop|

| |is reached, then coasting to a complete stop. A new equation (10.10b) is needed for this case: [pic]. |

|524 |(4/16/2012) Add and revise after Equation 10.10 (now 10.10a): If braking is not a factor in the analysis – only acceleration and |

| |coasting occur before Vec is reached -- the corresponding equation is |

| |[pic] (10.10b) |

| |Equations (10.10a) and (10.10b) apply only to cases in which there is no constant speed regime. Equations (10.10a) and (10.10b) are |

| |important calculations to do early in an analysis, if a coasting regime is being considered. |

|528 |(4/14/2012) In Cell C17 of Table 10.3, “15.4” must be “10.6”, because the braking begins after coasting has ended. Use Equation |

| |10.7b, not 10.7a. |

| | |

| |CHAPTER 11 AIR TRANSPORTATION AND AIRPORTS |

|582 |In the solutions to Example 11.4, “1542” should be “1562”, causing the following changes: |

| |Including the time for the first aircraft to take off, the total elapsed time is 1542 1562 seconds or 25.70 26.03 minutes. |

| |To calculate the capacity (ops/hr) of GTN during the period studied, first compute 1542 sec/20 ops = 77.1 sec/op 1562 sec/20 ops = |

| |78.1 sec/op. Then (3600 sec/hr)/(77.1 sec/op) = 46.7 ops/hr (3600 sec/hr)/(78.1 sec/op) = 46.1 ops/hr. |

| | |

| |CHAPTER 12 MOVING FREIGHT |

|687 |Early printings of Edition 1 had Example 12.10 correct. See values in bold font. |

| |Example 12.10 Ruling grade and train speed |

| |Assuming the ruling grade in this series of examples is 0.75 percent, at what speed will a train with four 5000 horsepower |

| |locomotives (efficiency at 85 percent) be able to climb that grade? Use 1200 lbs resistance per locomotive. |

| |Solution to Example 12.10 |

| |From Equation 12.6, the resistance due to the grade G = 0.75 is Rgrade = 20 * 0.75 = 15 pounds per ton. This Rgrade value is added |

| |to the Rtt expression used in Example 12.8, which must be adjusted for a speed V that is now unknown. |

| |[pic]16.6 + (0.01 * V) + (0.002 * V2) |

| |Recalling Equation 12.8, Rtotal = [(Rtt + Rgrade) *(C*w*n)] = TEdrawbar: |

| |TEdrawbar = (0.002 V2 + 0.01 V + 16.6) lbs/tons * 80 cars * 80 tons/car = (12.8 V2 + 64 V + 106,240) lbs |

| |Four 5000 HP locomotives provide power at 85 percent efficiency and speed V (mph), but they also add (1200 * 4) lbs of resistance. |

| |The resistance-propulsion relationship (with Equation 12.11 on the righthand side) becomes |

| |12.8 V2 + 64 V + 106,240+ (1200 * 4) = [pic] |

| |Rearranging terms produces 12.8 V3 + 64 V2 + 111,040 V = 6,375,000. V = 45.4 mph. |

|692 |(11/30/12) In the second line of the Solution to Example 12.13, 201.83 should be 201.825. As a result, Rsf = 292,892 and RWM = |

| |117,108. Raero should be 4896 lb, making Rtotal = 414,896. |

|697 |(12/30/12) In the Solution to Example 12.20, the correct equation for p is [pic]. The [pic]in Equation 12.24 was mistaken for a |

| |“/”. |

|699 |(12/30/12) Equation (12.27) should be [pic] |

|700 |(12/30/12) Example 12.22 should refer to Example 12.20.not Example 12.19. |

| | |

| |CHAPTER 13 Toward A Sustainable Transportation System |

| | |

| |Index |

| | |

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