FUNDAMENTALS OF TRANSPORTATION ENGINEERING By Jon D. Fricker and Robert K. Whitford

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1 FUNDAMENTALS OF TRANSPORTATION ENGINEERING By Jon D. Fricker and Robert K. Whitford This table includes typos Dr. Saito found besides the ones listed in the authors official errata sheet. Please note that Dr. Fricker already added some of these typos to his errata sheet. Page Correction Example 1.2. Question C should read C. At what value of time (VoT) would the two flight 17 Figure nd level. Airplane should be Air because we are talking about the way. Also, non-motorized modes do not belong in the 2 nd level. Non-motorized transportation modes, walking and biking, are modes that use highways (or streets) which are the way. 30 Table 1.6. Column 10 Lane Miles. For 1990, it is 49.2 miles instead of 49 miles (12.3 miles x 4 lanes = 49.2 miles) 30 Trend discussion 2. The freeway expansion by about 29 percent. Actually if you compute it, it is 31 percent [( )/49.2x100 = 31.3%] miles = 12.3 miles x 4 lanes lanes x 2 lanes (widening). 31 In the first paragraph right above SUMMARY. derived from the Table 1.5 and should be derived from the Table 1.6 The solution section to Example 2.9. The first D formula should look like this. D = 13 veh/( mi)/3 lanes = (13/.27 vpm)/3 lanes 79 Variable definition section right below equation (2.24). (also q or V ) is not needed because these variables are not used in this book with the meaning of λ th line from the bottom of the second paragraph. [B]reakdown occurs should read Breakdown occurs 110 Exercise nd sentence. a cumulative apparent presence time value t(p) = seconds. should read a cumulative apparent presence time value Σ t(p) = seconds. 114 Exercise 2.50, 2 nd line. The adjacent major street rate of 1200 vehicles in the lane should read The adjacent major street rate of 1200 vehicles per hour in the lane st paragraph, 2 nd sentence. Tables 3.13 to 3.20 show should read Tables 3.15 to 3.18 show st line. FFS = FFS i - f LW - should be FFS = BFFS - f LW Solution step 3 of B. Right below the f HV calculation. v p = V/(PHF)(N)(f HV )(f p ) = should be v p = V/[(PHF)(N)(f HV )(f p )] = to make the math work. 146 Inside the Think About It box. and by computing density for use with Table 3.7. should be and by computing density for use with Table 3.12.

2 151 In Table FFS mph should read Speed mph. 179 Exercise 3.13 is exactly the same as Exercise Exercise nd sentence from the last. What is the directional flow rate of traffic on this highway? should read What is the directional hourly volume of traffic on this highway? No typo found in section 4.1 covered by CE361 besides the ones given 4 in Dr. Fricker s errata sheet. No typo found in section 4.1 covered by CE361 besides the ones given in Dr. Fricker s errata sheet. 3 lines below equation (6.3). The Z value is If you test only the hazardous segment or not, the z value should be that reflects a one-way test. If you use z = 1.96, your test is whether the segment is under or over represented in the hazardousness nd line from the bottom. was about 1.2 seconds. Figure 6.13 shows 1.12 seconds. Example 6.6 uses 1.12 seconds. On urban surface streets a reaction time can be 1.0 second because drivers are typically alert th line above the black solid line in the middle of page 334. =$14.81 million word per life. should read =$14.81 million worth per life. 341 Equation (6.15). A simpler method is t vo to vf = (v o - v f )/(f*g) because a constant deceleration rate is assumed in either case. 341 Equation (6.16). Replace + between f and F with ±. 345 Table In the Variable f column. Replace 7.89 with 8.12 and 367 with Just above (Note: ). Example 6.16 should be Example Figure 6.32; Caption. The errata sheet asks you to change Dodge St. with Evergreen St. But, Dodge St. is the correct one because Evergreen St. is one way whereas the figure shows a two-way street. Furthermore, there is no stop signs on EB/WB directions on Dodge St. as shown in the diagram in Figure Figure The direction of the arrow above Evergreen Street should point to Garfield Street instead of Allen Street according to the descriptions because Table 6.15 shows Evergreen as EB only street with (S-S) control. Also, because Evergreen St is a one-way street there shouldn t be any stop sign in the WB direction. 365 Exercise 6.4 (b). Z = 1.96 should be Z = if one-way analysis is intended. The problem statement sounds like a one-way analysis. However, if it is meant for a double-sided analysis, then, that is fine. 368 Exercise The line AD should be a double solid centerline G2 G1 L x AL x Ax Equation 7.6. y = = = 2 L 2 L 2L It is not incorrect, but what could be the meaning of (x/l) 2? If we combine equations 7.3 and 7.4, it is naturally should look like this. 2 G2 G1 2 Ax y = x = 2L 2L

3 392 1 st line, 2 nd sentence. The maximum rate of curvature should read The minimum rate of curvature th line from the bottom. 7.6 gives us maximum values of should read 7.6 gives us minimum values of st sentence in the Solution to Example the K values for the maximum length of curve for SSD should read the K values for the minimum length of curve for SSD nd line in the Solution to Example = should read. = rd line from the bottom. the central angle Δ SSD for an arc should read the central angle Δ S for an arc if we consider notations in the Figure 7.17 are the correct ones st line from the bottom. In the equation. Δ SSD should be Δ S. 396 Equation Δ should be Δ S. 396 Equation M SSD should be M s. 396 In the Solution to Example R vehicle and M SSD should read R v and M s. 407 Section st sentence. The change from a superelevated roadway normal tangent roadway is called superelevation runoff. should read The change from a superelevated roadway normal tangent roadway is called transition section consisting of superelevation runoff and tangent runout. 407 Section Beginning of the 2 nd paragraph. Like the superelevation runoff, the spiral has about one third of the length should read The superelevation runoff has about one third of the length 407 Section Last sentence of the 2 nd paragraph. Remove this sentence. 408 Figure The extension line going through point B should be extended to go through the dimension line. Add arrows there. Superelevation runoff is the distance between B and E. 2/3 of the superelevation is the distance between B and D. See below.

4 409 The bottom diagram in Figure See the corrections in read ink nd paragraph, 2 nd line. contains 11 possible conditions should read contains 8 possible conditions. There used to be 11 warrants, but three volume-related warrants were merged first paragraph. Starting from the last part of the 6 th line. Replace the word phase with interval. 432 Figure Add arrows to show the range of X c and X o. 433 The solution to Example 8.4. It is lot easier to understand if the solution is presented like the following because the way X c and X o are presented in the solution is different from the way X c and X o are presented in

5 Figure Solution to Example 8.4 The minimum distance needed to stop X c is, from Equation 8.3a, computed as t s = time to stop = v o /d brake =(30*1.47)/10=4.41 sec X c = v o t r + v o t s (1/2)d brake t s 2 = (30*1.47*1.5) + (30*1.47*4.41) (1/2)(10)(4.41) 2 = ft If the driver was any closer to the intersection than the X c distance just calculated, he must use a higher deceleration rate or less perceptionreaction time (i.e., uncomfortable deceleration, less safe). Now we calculate the distance he must travel, X Y, to completely clear the intersection within the yellow interval 4.0 seconds, using Equation 8.3b with a acc = 0.0. X Y = v o t r + v o (t Y t r ) + (1/2)a acc (t Y t r ) 2 = (30*1.47*1.5) + (30*1.47)( ) + 0 = ft Hence, the maximum distance from the stop line within which he can safely clear the intersection, X o = X Y (W + L) = ( ) = ft Hence, X c > X o. Because X c > X o, a dilemma zone exits. Its length is X c > X o = = The driver s claim cannot be dismissed, but now the issue becomes whether the vehicle was in the dilemma zone when the light changed to yellow. 434 The solution to Example 8.5. It is lot easier to understand if the solution is presented like the following because the way X c and X o are presented in the solution is different from the way X c and X o are defined in Figure The last sentence of the solution, however, is incorrect as you see below. Solution to Example 8.5 A. In Example 8.4, the stopping sight distance, X c, with v o = 30 mph was ft (Eq. 8.3a). Let us name the total interval of yellow and all red that eliminate dilemma zone as t ND. ND for no dilemma zone. t allred = time for eliminating the dilemma zone t ND length of yellow interval t Y

6 From Eq. 8.3b, X o = v o t Y (W + L) = 30*1.47*t ND ( ). Dilemma zone is avoided when X c = X o = 44.1* t ND ( ) Solve for t ND results in, t ND = ( )/44.1 = 5.43 seconds Hence, t allred = = 1.43 seconds. B. Use the definitions of X c and X o. The unknown variable is v o. X c = v o t r + v o 2 /(2d brake ) = v o *1.5 + v o 2 /(2*10) X o = v o t ND (W + L) = v o *(4 + 2) ( ) Just like A, we assume we are talking about the case where we eliminate the dilemma zone. Hence, X c = X o. v o * *v o 2 = v o * *v o 2 4.5* v o + 76 = 0 Solve this for v o. v o = 67.5 fps (45.9 mph) or 22.5 fps (15.3 mph). At these speeds dilemma zone is avoided. (The case in the middle in Figure 8.10) Let v o = 70 fps, then X c = 70* /20 = 350 ft and X o = 70*6-76 = 344 ft. Hence, X c > X o ; therefore, dilemma zone exists when v o > 67.5 fps. (The case in the bottom in Figure 8.10) Let v o = 20 fps, then X c = 20* /20 = 50 ft and X o = 20*6-76 = 44 ft. Hence, X c > X o ; therefore, dilemma zone exists when v o < 22.5 fps. (The case in the bottom in Figure 8.10) Let v o = 40 fps (about the mid-point speed of the boundaries), then X c = 40* /20 = 140 ft and X o = 40*6-76 = 164 ft. Hence, X c < X o ; therefore, dilemma zone does not exist between v o = 67.5 fps and 22.5 fps. (The case in the top in Figure 8.10) 435 In section The term phase was incorrectly used at times, though not all of them. They are supposed to be interval. 436 Definition of Phase. It is slightly incorrect. Phase is typically defined as A signal phase consists of a green interval (G), plus the change (Y) and clearance intervals (AR) that follow it. It is a set of intervals that allows a designated movement or set of movements to follow and to be

7 safely halted before release of a conflicting set of movements. Table 9.5. in the ESAL per Truck column should be 2.44 (see page 466 right above Figure 9.7) should be 1.43 (this is a given value shown in Example 9.2). We cover only sections 10.1 through Figure The last time in the bottom graph seconds should be seconds. Most likely this is a result of miscalculation (44.59 ft/sec)/(2.85) = 15.6 sec = sec mph/s must be converted first into 4.19 ft/s /4.19 = 10.6 sec = sec. 533 Solution to Example nd sentence. The service form of Equation should read The service form of Equation Solution to Example B. Cost/TVM should be Cost/mi. Also, % Labor = ($4.285 M)/($5.808 M) x 100 = 74% instead of There is no extra typo in the sections covered by this course (11.1 and 11.5). Solution to Example 3.1 should be Solution to Example Table Dr. Fricker gave me the following corrections. In the Year 1960 row of Table 13.4, "6.03" should be "1.62" and "17.31" should be "9.82". 723 Section th line. carries 9.32 persons per mile. should be carries 9.32 persons per transit bus. This was computed by dividing 21,205 million passenger miles by 2,275 million vehicle miles of travel. 21,205,000,000/2,275,000,000 = The unit is passenger per vehicle (transit bus) because miles cancel out each other. By the way, passenger miles/1000 BTU was calculated b y dividing 21,205 million passenger miles by 0.1 quads (i.e., 0.1 x BTU). 723 Solution to Example 13.1 (actually Example 13.2) has a couple of typos. Here is my version. Do not try to compare quads values in Section with the values in Table Solution to Example 13.2 The enlarged bus fleet would deliver 30 passengers/bus x 150,000 buses (because the fleet was doubled) x 30,662 miles = 138,000,000,000 passenger miles for 0.2 quads of oil (because the 4 th sentence in the first paragraph of Section says annual oil consumption by transit buses is about 0.1 quads, although Table 13.6 shows quads for buses). The number of passengers moved would be 4,500,000 (= 30 x 150,000) compare to 699,000 (= 9.32 passengers per transit bus x 75,000 buses), representing an increase of persons moved of 3,800,000 (= 4,500, ,000).

8 The single driver automotive equivalent is 30,662 x 3,800,000 = 116,526,000,000 = x passenger miles. At 0.2 passenger miles per 1000 BTU (note a value of is given in the last sentence of the first paragraph of Section ), the single-occupancy autos would consume quads (1.165 x passenger miles x 1000 BTU / 0.2 passenger mile = x ). Thus, the saving would be quads. Applying that saving to automobile use would reduce it from quads (note that Table 13.6 shows quads for automobiles) to (= ), or 7.0 percent. 725 Table Dr. Fricker gave me the following extra corrections. In the Year 1990 column, "2,211,503" should be "2,511,503". In the Year 2000 column, "4,326,667'" should be "4,328,667". Relabel the "Fleet" column as "Vehicles in use, 2000" and change "85,879,000" to "85,579,000". Add to the Source ORNL 2001, Tables 7.6, 7.5, and 6.3". 731 Solution to Example passengers should be 1 van/7 passengers. 3 passengers should be 1 car/3 passengers. 731 Example to a level equivalent to 70 percent of our domestic production in 2000 (Table 13.4) should read to a level equivalent to 50 percent of our domestic production in 2001 (Table 13.4) 736 Table Units? Gram/km most likely according to the EPA website. 747 Solution to Example It is not a typo, but vpm here means vehicle per minute and not a typical vpm meaning vehicle per mile (density).

h CIVIL ENGINEERING FLUID MECHANICS section. ± G = percent grade divided by 100 (uphill grade "+")

h CIVIL ENGINEERING FLUID MECHANICS section. ± G = percent grade divided by 100 (uphill grade +) FLUID MECHANICS section. TRANSPORTATION U.S. Customary Units a = deceleration rate (ft/sec ) A = absolute value of algebraic difference in grades (%) e = superelevation (%) f = side friction factor ± G