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 .

Errata, Fundamentals of Transportation Engineering, 5th Printing Page 2 of 3

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 , 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: . 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
(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 / 21.  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: .
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
(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.
B.  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.
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) =
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 . The in Equation 12.24 was mistaken for a “/”.
699 / (12/30/12) Equation (12.27) should be
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