交通工程课后题答案—Traffic Engineering

Homework of chapter two

2-1. Answer: Basically apply Equation (2-1) on page 33:

dr?1.47St?1.47?55?3.2?258.72ft

2-2. Answer: when t = 0.5s： dr?1.47Sit?1.47?60?0.5?44.10ft

db?si2?s2f2a

So Sf?Si2?2adb??1.47?60?2?2?10??400?44.1??25.71ft/s?17.48mi/h

Also, the same to others：

t = 1.0s: dr?88.20ft Sf?39.28ft/s?26.73mi/h t = 1.5s: dr?132.30ft Sf?49.25ft/s?33.47mi/h t = 2.0s: dr?176.40ft Sf?57.51ft/s?39.12mi/h t = 2.5s: dr?220.50ft Sf?64.72ft/s?44.03mi/h t = 3.0s: dr?264.60ft Sf?71.21ft/s?48.44mi/h t = 3.5s: dr?308.70ft Sf?77.16ft/s?52.49mi/h t = 4.0s: dr?352.80ft Sf?82.68ft/s?56.24mi/h t = 4.5s: dr?396.90ft Sf?87.85ft/s?59.76mi/h t = 5.0s: dr?441.00ft dr?s and Sf?60mi/h So we can see, the faster reaction, the smaller loss. Example for t =2.5s

dr?1.47Sit?1.47?60?2.5?220.5ftdb?Si2?S2f2a?Sf?Si2?2adb??1.47?60?2?2?10??400?220.5??64.72ft/s?44.03mi/h2-3. Answer: Initial speed on the grass-stabilized shoulder:

222 ??????Si2?30dF?0.01G?S?30?250?0.20?0.01?3?35?2500mi/h111fInitial speed just before pavement skid begins:

?mi/h?Si22?30d2?F2?0.01G??Si1?30?100??0.45?0.01?3??2500?3760Si2?S2i22?3760mi/h?61.32mi/h

2-4. Answer: From Table 2-6, we can see that for a Condition C match the problem. Therefore, Equation (2-10) should be used, and the corresponding reaction and maneuver times can be determined.

d?1.47?tr?tm?Si?1.47??11.2?3.5??70?1512.63ft

Thus, we have the distance for the location of the sign:1512.63?100?1412.63ft 2-5. Answer: Apply Equation (2-9), we have:

352?02d?1.47?35?1.0??175.94ft30??0.348?0.01?2?

175.94y??3.42s1.47?35802?022-6. Answer: d?1.47?80?2.5??843.83ft

30??0.348?0.04?2-7. Answer: By applying Equation (2-3), we have:

S2702R???2041.47ft

15?0.01e?f?15??0.06?0.1?Homework of Chapter Three

3-1. Answer: Based on the fundamental relationship of circle, we can obtain the length of the curve:

D =

5729.58= 6.3662° 9002?RL?R?3.1415?900?52??L???816.81ft360?180180 T?Rtan????900?tan26??438.96ft

???2?The station of the P.I. is given as 1,500 + 20.

The distance from the origin toP.C.is : 1520 - T = 1520 - 438.96 = 1081.04 ft Thus, the station ofP.C.is : 1000 + 81.04

The distance from the origin toP.C.is : 1081.04 + L = 1081.04 + 816.81 = 1897.85 ft

Thus, the station ofP.C.is : 1800 + 97.85

???L.C.?2Rsin???2?900?sin26??789.07ft?2??????M?R?1?cos????900?(1?cos26?)?91.09ft

2???????E?Tsin???M?438.96?sin26??91.09?101.34ft?2?5729.585729.58??818.51ft 3-2.Answer: R?D7S3503The length of the spiral transition curve is : Ls?1.6?1.6??244.35ft

R818.512?Sdes???502??Superelevation rate: e?100???6.4% ?15R?fdes??100??15?818.51?0.14?????The lengths of the superelevation and tangent runoffs are :

w?n?ed?bw12?2?6.4?0.75??230.40ft?0.50 eNC1Lt?Lr??230.40?36.00fted6.4Lx?So, the length of the spiral is greater than the length of the superelevation runoff.

??LsD244.35?7??8.6?200200?s???2??45?2?8.6?27.8?????27.8?Lc?100?s??100???397.14ft?7??D??L2???????s?Ts?Rtan????Rtan??R??tan???Ls?Rsin???6R??2???2??244.352??818.51?tan22.5????818.51?cos8.6??818.51?6?818.51???tan22.5??244.35?818.51?sin8.6????462.21ftThe distance between the origin and T.S. is : 1280 - Ts=1280 - 462.21 = 817.79 ft The station of T.S. is 800+17.79

The distance between the origin and S.C. is : 817.79 + Ls= 817.79 + 244.35 = 1062.14 ft

The station of S.C. is 1000 + 62.14

100?s100?27.8??397.14ft D7Thus the station of S.C. is 1400 + 59.28

The distance between S.C. and C.S. is

The distance between C.S. and S.T isLs, thus the station of S.T. is 1700 + 3.63 3-3.

Answer:

602ds?1.47?60?2.5??546.59ft30?(0.348?0.02)M?5729.58?.58??dsD??5629?546.59?5??1?cos??1?cos??????32.44ft???D?5?200???200???

2?Sdes??552????3-4. Answer: e?100??f?100??0.13?15?1000??7.2%?6% ?15Rdes?????Thus the appropriate superelevation rate is 6%.

3-5. Answer: Lr?w?n?ed?bw?12?3?10?0.67?603.00ft

?0.403-6. Answer: (a) 6%, 760ft (b) 5.5%, 875 ft (c) 7%, 630 ft a) From Figure 3.16(a), for a rural freeway in mountainous terrain with a design speed of 60 mi/h, the maximum allowable grade is 6%. Entering Figure 3.19 with 6% on the vertical axis, moving to the “10 mi/h” curve, the critical length of grade is seen to be approximately 760 ft.

b) For a rural arterial in rolling terrain with a design speed of 45 mi/h, the maximum allowable grade is 5.5%, the critical length of grade is seen to be approximately 860ft. c) For an urban arterial in level terrain with a design speed of 40 mi/h, the maximum allowable grade is 7%, the critical length of grade is seen to be approximately 610 ft. 3-7.

Answer: (a) V.P.C. Station : 1555 - 100L/2 = 1055(1000 + 55) V.P.T. Station : 1555 - 100L/2 = 2055(2000 + 55)

10?L?(b)YV.P.C.?YV.P.I?G1???500?4??4802?2??G?G1?2??5?4?22Yx??2?x?G1x?Y0???x?4x?480??0.45x?4x?480?2L??2?10?YV.P.T.??0.45?102?4?10?480?475(c)Y1??0.45?12?4?1?480?483.55x??G1L?4?10??4.44G2?G1?5?4

(d) High point at 444 ft, with the elevation of :

Y4.44??0.45?4.442?4?4.44?480?488.89

5523-8. Answer: (a) ds?1.47?55?2.5??468.88ft

30?(0.348?0.01?3)Assuming that ds?L:

|7?3|?468.882L??407.50?ds

2158The assumption is invalid. Thus,

?2158?2158?L?2ds???2?468.88??398.26ft?|G?G|?|7?3| 1??2602(b) ds?1.47?60?2.5??623.18ft

30?(0.348?0.01?5)Assuming that ds?L:

|2?5|?623.182L??1053.21?ds

400?3.5?623.18702(c) ds?1.47?70?2.5??770.88ft

30?(0.348?0.01?3)Assuming that ds?L:

|?3?2|?770.882L??1376.87?ds

2158

7023-9. Answer: ds?1.47?70?2.5??787.55ft

30?(0.348?0.01?4)Assuming that ds?L:

|1?4|?787.552L??982.50?ds

400?3.5?787.55The minimum length is 983 ft. The station of V.P.C. is 4600 + 31 The station of V.P.T. is 5600 + 14 Elevation of V.P.C. is

Y0?YV.P.I.?G1L9.825?210?4??229.6522?1?4?22Yx???x?4x?4631?0.25x?4x?229.65

?2?9.83?YV.P.T.?0.25?9.832?4?9.83?229.65?214.49High point:

x??G1L4?9.83??7.864G2?G11?4Yhigh?0.25?7.8642?4?7.864?229.65?213.65 Y1?0.25?12?4?1?229.65?225.90Homework of chapter four

4-4. Answer：Warning signs must be placed far enough in advance of the hazard to allow drivers adequate time to perform the required adjustments. The Table 4.2(Guidelines for Advance Placement of Warning Signs) gives the recommended advance placement distances for three conditions. So we can check the Table 4.2. Condition A：High judgment required. Typical applications are warning signs for merging, lane drop, and similar situation.

Condition B：Stop condition. Typical applications are stop ahead, yield ahead, and signal ahead warnings.

Condition C：Deceleration to the listed advisory speed for the condition. (a) Belong to Condition B. 375ft.

(b) Belong to Condition C. 250ft. (c) Belong to Condition A. 400ft.

Homework of chapter five

5-1. Answer：Use the following equation：v?v1?V (V?900veh/h) PHFV900V??900veh/hv2??PHF1.00PHF

V900Vv3???1125veh/hv4??PHF0.80PHF900?1000veh/h0.90

900?128.75veh/h?128v6eh/h0.70So： PHF Peak-hour rate of flow (veh/h) The Comment on the results：

When the Vis fixed, with the decrease of thePHF, thevincreases.

5-2. Answer：Basically apply Equation (5‐9) and (5-11) on Page 145、146：

Rate of flow：v?v3600 Density：D?

Sha1.00 900 0.90 1000 0.80 1125 0.70 1286 Rate of flow：v?Density：D?36003600??1636.36veh/h?1636veh/h ha2.2v1636??32.72veh/mi S505-3. Answer：Basically apply Equation (5‐7) on Page 145： Density：D?5280?O

Lv?LdLv?20ft Ld?3.5ft O?0.255 So the density is：D?5280?O5280?0.255??57.3veh/mi/ln

Lv?Ld20?3.5v1600??40veh/mi/ln S405-4. Answer：Density：D?5-5.Answer：Basically

apply Equation (5‐1) on Page 139：

DDHV?AADT?K?D

Check Table 5-2 to find the “General Ranges for K and D Factors”：

Facility Type is Urban Radial Route, so：K-Factor=0.07 – 0.12 ;D-Factor=0.55 – 0.60

AADT?25000veh/dayDDHVLOW?AADT?Kmin?Dmin?25000?0.07?0.55?962.5veh/h DDHVHIGH?AADT?Kmax?Dmax?25000?0.12?0.60?1800veh/hSo, the range of directional design hour volumes is 962.5 – 1800 veh/h. 5-6. Answer：Basically apply Equation (5‐5) and (5-6) on Page 143： Time mean speed：TSM?d=10560ft n=6

??d/t?iin；Space mean speed：SMS?dnd ?(?ti)/n?tiii105601056010560105601056010560?????156144144168126132?73.5ft/sTSM? 66?10560SMS??72.9ft/s156?144?144?168?126?1325-7. Answer：

Time Period 8:00 – 8:15AM 8:15 – 8:30AM 8:30 – 8:45AM 8:45 – 9:00AM 8:00 – 9:00AM iVolume 150 155 165 160 ??630 (a) The hourly volume：V??V15?150?155?165?160?630veh/h (b) Peak rate of flow：v?V15max?4?660veh/h (c) The peak hour factor：PHF?5-8.Answer： 1. Month 2. No. of Weekdays In Month (days) 22 20 22 22 21 22 3. Total Days in Month (days) 31 28 31 30 31 30 4. Total Monthly Volume (vehs) 200,000 210,000 215,000 205,000 195,000 193,000 5. Total Weekday Volume (vehs) 170,000 171,000 185,000 180,000 172,000 168,000 6. ADT 4/3 (veh/day) 6452 7500 6935 6833 6290 6433 7. AWT 5/2 (veh/day) 7727 8550 8409 8182 8190 7636 V630??0.955 v660Jan(1) Feb(2) Mar(3) Apr(4) May(5) Jun(6) Jul(7) Aug(8) Sep(9) Oct(10) Nov(11) Dec(12) Total 23 21 22 22 21 22 260 31 31 30 31 30 31 365 180,000 175,000 189,000 198,000 205,000 200,000 2365000 160,000 150,000 175,000 178,000 182,000 176,000 2067000 5806 5645 6300 6387 6833 6452 — 6957 7143 7955 8091 8667 8000 — According to the table, we can see:

AADT=Total4/Total3=2365000/365=6479.45veh/day?6479veh/day AAWT=Total5/Total2=2067000/260=7950veh/day 5-9. Answer：

(a) Free-flow speed： When D?0, S?57.5(1-0.008?0)?57.5 mi/h

? SFREE?57.5 mi/h

(b) Jam density： When S?0, 57.5(1-0.008D)?0

? D?57.5/(57.5?0.008)?125 veh/mi

(c)Speed-flow relationship：

S?57.5(1?0.008D)?57.5?0.46D?D??2.17S?125

We can use the following Equation：v?S?D So, v?S?D?S?(?2.17S?125)??2.17S2?125S

(d) Flow-density relationship：S?57.5(1?0.008D)?57.5?0.46D We can use the following Equation：S?So,

v Dv?57.5?0.46D?v??0.46D2?57.5D D(e) Capacity：Because of v??0.46D2?57.5D So,

dv57.5?57.5?0.92D?0?D??62.5veh/mi dD0.92S?57.5?0.46D?57.5?0.46?62.5?28.75mi/hv?S?D?28.75?62.5?1796.875veh/h?1797veh/h

So, C?1797 veh/h

5-10. Answer：(a) Free-flow speed：When D?0, S?61.2e?0.015?0?61.2mi/h ? SFREE?61.2 mi/h

(b) Jam density： When S?0, 61.2e?0.015?D?0 ? D doesn’t exist (c) Speed-flow relationship： S?61.2e?0.015?D?D??We can use the following Equation： v?S?D

11SSln)??lnSo, v?S?D?S?(? 0.01561.20.01561.2(d) Flow-density relationship： S?61.2e?0.015?D We can use the following Equation： S?So,

v?61.2e?0.015D?v?61.2De?0.015D Dv D11ln 0.01561.2(e) Capacity：Because of v?61.2De?0.015D So,

dv?61.2e?0.015D?0.918De?0.015D?0?D?66.67veh/mi dDS?61.2e?0.015?D?61.2?e?0.015?66.67?22.51mi/hv?S?D?22.51?66.67?1501veh/h

So, C?1501 veh/h

Homework of chapter seven

7-1. Answer：From table 7.3, F (1.04) = 0.85 z?1.04?x?57?x?64.904mph?64.9mph 7.6Therefore, 85% of the vehicles are traveling under 64.9 mph.

7-2. Answer：For 95% confidence we can infer that the travel time is bounded between the following interval：

???1.96?,??1.96????152?1.96?17.3,152?1.96?17.3???118.902,185.908???118,186?With the Central Limit Theorem, we can conclude that the average of travel time tends to normal distribution, but it is unable to determine the shape of distribution of travel time. 7-3. Answer：

(a) This table is a set of grouped data： x?Using Equation (7-5), we have：

2?120?3?40?4?30?5?10?2.65

120?40?30?10

3 4 5 317 226 653 1550 1820 1.17 176 136 510 1256 1225 0.98 269 231 498 1700 1750 1.03 589 248 502 2529 2510 0.99 296 170 437 1249 1110 0.89 1647 1011 2600 1500 985 2690 8415 0.91 0.97 1.03 Tj Vj Fj (d) second iteration of O—D matrix Destination Station 1 2 3 4 5 Origin Station Tj 1 151 393 330 242 718 1834 1820 0.99 2 232 201 166 133 513 1245 1225 0.98 3 294 427 261 231 513 1726 1750 1.01 4 418 783 560 243 507 2511 2510 1.00 5 120 211 266 158 420 1175 1110 0.94 1215 2015 1583 1007 2671 Vj Fj 1200 2040 1500 985 2690 8415 0.99 1.01 0.95 0.98 1.01 Tj Vj Fj Homework of chapter ten

10-1. Answer：

(1)、Exposure-based accident and fatality rates：

① Deaths and Accidents per 100,000,000 Vehicle-miles Traveled

?100,000,000?Deaths/100,000,000VMT?15????12512,000,000??

(deathsper100,000,000vehicle?milestraveled)

?100,000,000?Accidents/100,000,000VMT?360????3000

?12,000,000?(accidentsper100,000,000vehicle?milestraveled)

② Deaths and Accidents per 10,000,000 Vehicle-hours Traveled

Vehicle-hours traveled：

AnnualVMT12,000,000mi?400,000h

AnnualSpeed30mi/h?10,000,000?Deaths/10,000,000VHT?15????375

400,000??(deathsper10,000,000vehicle?hourstraveled)

?10,000,000?Accidents/10,000,000VHT?360????9,000

?400,000?(accidentsper10,000,000vehicle?hourstraveled) (2)、Population-based accident and fatality rates：

① Deaths and Accidents per 100,000 Population：

?100,000?Deaths/100,000pop?15????30 deathsper100,000population

50,000???100,000?Accidents/100,000pop?360????720 accidentsper100,000population

?50,000?② Deaths and Accidents per 10,000 Registered Vehicles

?10,000?Deaths/10,000ReVeh?15????4.286

35,000??(deathsper10,000registeredvehicles)

?10,000?Accidents/10,000ReVeh?360????102.857

35,000??(accidentsper10,000registeredvehicles)

Compare：2005 national rates include fatality rates of 1.8 per 100,000,000 Vehicle-miles traveled and 14.3 per 100,000 Population. The rates above are higher

than the nation statistic.

10-2. Answer：Using the normal approximation test, the statistic z is computed as：

z1?fB?fAfA?fB?z1?25?1515?25?1040?1.581

From the standard normal distribution table of Chapter 7： Prob?z?z1?= Prob?z?1.581??0.9429?0.95

The reduction in accidents observed is not statistically significant.

(If Prob?z?z1??0.95, The reduction in accidents observed is statistically significant. Because of the number of accidents is small, sample size may not be sufficient to prove using the normal approximation.) 10-3.

Homework of chapter eleven

11-1. Answer：

From table 11.2 (Typical Parking Generation Rates) for high-rise apartment, the average peak parking occupancy is 0.88 per dwelling unit, or in this case,

0.88?600?528 parking spaces.

A more precise estimate might be obtained using the equation related to facility size：

P?0.34X?105.0?0.34?600?105.0?309.0 spaces

This presents a significant range to the engineer – from 309 to 528 parking spaces needed. 11-2. Answer：

From Table 11.3 (Recommended Parking Ratios (Parking Spaces per 1,000 sq ft of GLA) from a 1998 Study), the center as descried would require a parking ratio of 4.50 spaces per 1,000 sq ft GLA, or 4.50?11-3. Answer：

Zoning requirements are generally set 5% to 10% higher than the 85th percentile demand expectation.

600,000?4.50?600?2700 parking spaces.

1,000① Problem1：

309?(1?0.05)?324.45?325 to 309?(1?0.10)?339.9?340 parking spaces

② Problem2：

2700?(1?0.05)?2835 to 2700?(1?0.10)?2970 parking spaces

11-4. Answer：

Basically apply Equation (11‐1) on Page 316：D?From the issue we know that：

N=2,000 K=35%=0.35 P=100%﹣7%=93%=0.93 O=1.3 R=1.0 pr=1.0

So, the peak parking demand will be estimated as：

D?NKRP*pr2000?0.35?1.0?0.93?1.0??500.769?501 parking spaces O1.3NKRP*pr O11-5. Answer：

??NT???Basically apply Equation (11‐2) on Page 321：P??n?*F

?D???From the issue we know that：

N1?100, N2?150, N3?200, N4?300 T1?14, T2?8, T3?6,T4?10

F?90%?0.90

D?35m?35h 60??????100?14?150?8?200?6?300?10?*0.90?10,491.429?10,492vehs ?*F??35??????60??So, parking supply in this study area is computed as：

??NT?P??n?D?This result means that 10,492 vehicles could be parked in the study area over the 14-hour period of the study. It does not mean that all 10,492 vehicles could be parked at the same time. 11-6. Answer：

(a). Parking Space 1 47 2 25 Number of intervals parked 3 22 4 4 5 3 6 3 7 0 8 1

(b). 7:00 Time Accumulation 5 for internal 11:00 Time Accumulation 14 for internal 7:30 11 11:30 16 8:00 14 12:00 17 8:30 17 12:30 14 9:00 16 1:00 15 9:30 19 1:30 14 10:00 10:30 18 2:00 17 16

(c). Basically apply Equation (11‐3) on Page 324： The average parking duration：D?So：

?(Nxx*X*I)

NT

D???(Nxx*X*I)NT(47*1*0.5)?(25*2*0.5)?(22*3*0.5)?(4*4*0.5)?(3*5*0.5)?(3*6*0.5)?(1*8*0.5)47?25?22?4?3?3?0?123.5?25?33?8?7.5?9?0?4110???1.05h105105

2?0.019?1.9% (d). The overtime rate = 105The parking violation rate =0

(e). Basically apply Equation (11‐4) on Page 325： The parking turnover rate：TR?NT PS*TSTR?

NT105??0.625veh/stall/h

PS*TS24*7Homework of chapter twelve

12-1. Answer：The free-flow speed for a multilane highway is computed using the following Equation：

FFS?BFFS?fLW?fLC?fM?fA As the describe in the 12-1, we can know that：

fM?1.6mi/h, BFFS?55mi/h, fLC?1.8mi/h, fA?5.0mi/h, fLW?1.9mi/h Then：

FFS?55?1.9?1.8?1.6?5?44.7mi/h

12-2. Answer：The free-flow speed of a freeway may be estimated using the following Equation：

FFS?BFFS?fLW?fLC?fN?fID As the describe in the 12-2, we can know that：

fID?1.3mi/h, BFFS?70mi/h, fLC?0.8mi/h, fN?3.0mi/h, fLW?0.0mi/h Then：

FFS?70?0.0?0.8?3.0?1.3?64.9mi/h

12-3. Answer：

(a) Rise on 4% Grade：750*0.04 = 30 ft Rise on 2% Grade：1,500*0.02 = 30 ft Rise on 3% Grade：1,000*0.03 = 30 ft Total Rise on Composite：30 + 30 + 30 = 90 ft

The average grade is then computed as the total rise divided by the total length of the grade, or：

GAV?9090??0.02769?2.77%

750?1,000?1,5003250The appropriate values would now be entered to find passenger – car equivalents using a 2.77% grade, 3250 ft in length. (b)

(c) 3,000 ft of 3% grade is equivalently to 1450 ft of 5% grade. 1450 + 4000 = 5450 ft

The appropriate values would now be entered to find passenger – car equivalents using a 5% grade, 5450 ft in length.

12-4. Answer：Basically apply Equation (12‐9) on Page 21： fHV?1

1?PT(ET?1)?PR(ER?1)Where： PT?0.1, PR?0.05, ET?2.5, ER?2.0 So：

fHV?1?0.833

1?0.1?(2.5?1)?0.05?(2.0?1)3500?4202pc/h 0.833 Vpce?

12-5. Answer： Step 1：

The Free-flow speed of the facility is found using Equation (12-5)：

FFS?BFFS?fLW?fLC?fN?fID

Where：

BFFS?70mi/h (urban freeway)

fID?7.5mi/h fLC?0.8mi/h fN?1.5mi/h fLW?1.9mi/h Then：

FFS?70?1.9?0.8?1.5?7.5?58.3mi/h

Step 2：

Determine the Maximum Service Flow Rates for Each Level of Service. Maximum Service Flow (MSF) rates for each Level of are drawn from Table 12.3. Values are obtained for freeways with Free Flow Speed of 60 mi/h and 55 mi/h Straight-line interpolation may be used to obtain maximum service flow rate for the intermediate free-flow speed of 58.3 mi / h. This is illustrated in Table1

Table 1：Values of MSF for Example 12-5

Level of Service 55 600 990 1430 1910 2250 Free-Flow Speed 58.3 639.6 1080 1515.8 1982.6 2283 60 660 1080 1560 2020 2300 A B C D E Step 3： Determine the Heavy-Vehicle Factor. The heavy-vehicle factor is computed as：

fHV?1

1?PT(ET?1)?PR(ER?1)Where： PT?0.05, PR?0.00, ET1?2.5, ET2?1.5 Then： fHV1?1?0.930

1?0.05?(2.5?1)1?0.976

1?0.05?(1.5?1) fHV2?Step 4：

Determine the Service Flow Rates and Service Volumes for Each Level of Service. Service Flow Rates and Service Volumes are computed using Equations

(12-2) and (12-3)：

SFi?MSFi?N?fHV?fpSVi?SFi?PHF

Where：MSFi=as determined in step 2 N=4(given)

fHV=0.930, fHV=0.976 fp=1.00(regular users) PHF=0.85

These computations are dore in the spreadsheet show in Table Level of MSF N service Cpc/h/ln A B C D E 639.6 1049.4 1515.8 1982.6 2283 4 4 4 4 4 fHV 0.930 0.930 0.930 0.930 0.930 fHV 0.976 0.976 0.976 0.976 0.976 fp 1 1 1 1 1 SF1 SF2 (veh/h) 2497 4097 5918 7740 8913 PHF 0.85 0.85 0.85 0.85 0.85 SV1 (veh/h) 2022 3318 4793 6269 7219 SV2 (veh/h) 2122 3482 5030 6579 7576 (veh/h) 2379 3904 5639 7375 8493 12-6. Answer： Step 1：

Determine the Heavy-Vehicle Factor. The heavy-vehicle factor is computed as： fHV?1

1?PT(ET?1)?PR(ER?1)Where：PT?0.15, PR?0.00, ET?2.5 So：

fHV?1?0.816

1?0.15?(2.5?1)Step 2：

Determine the Demand Flow Rates in Equivalent pce Under Base Conditions The demand volume may be converted to an equivalent flow rate under base conditions using Equation (12-1)： vp?V

PHF?N?fHV?fpWhere：V?4000veh/h, PHF?0.90, N?3, fHV?0.816, fp?1.00 Then：

Step 3：

Find the Level of Service and the Speed and Density of the Traffic Stream.

The demand flow of 1816 veh/h would be used to enter Figure 12.4 to find the Level of Service and speed. It can also be seen that the Level of Service for this multilane highway is E.

The density on the multilane highway maybe estimated as the demand flow rate divided by the speed, or：D?4000?1816pc/h/ln

0.90*3*0.816*1vp?vpS?1816?40.4pc/mi/ln 45 This value can be used to enter Table 12.2 to confirm the Level of Service, which is E, falling within the defined bound areas of 35-45 pc/mi/ln for LOS E. 12-7. Answer： Assure N?2

Step 1：Determine the FFS

FFS?BFFS?fLW?fLC?fN?fID Where：BFFS?75mi/h (suburban freeway) fLW?0.0mi/h (Table 12.5, 12-ft lanes)

fLC?0.0mi/h (Table 12.6, ?6-ft lateral clearance) fN?4.5mi/h (Table 12.7, 2 lanes in one direction) fID?0.0mi/h (Table 12.8, ?0.5 interchanges/mi) FFS?75?4.5?70.5mi/h

Step 2：Determine the MSF for levels of service C, with a minimum acceptable level of service C, it is necessary to determine the maximum service flow rates that would be permitted if these levels of service are to be maintained. These are found from Table 12.3 for a 70.5 mi/h free-flow speed：

MSFC?1770?(1830?1770)?0.1?1776pc/h/ln

Step 3：Determine the Number of lanes required for the Level, Upgrade, and Downgrade Freeway Sections.

The required number of lanes is found using Equation (12-4)： Ni?DDHV

PHF?MSFC?fHV?fpWhere：DDHV?2500veh/h (given) PHF?0.92 (given)

MSFC?1776pc/h/ln (determine in step 2) fp?1.00 (regular users assumed)

There different heavy-vehicle factors must be considered: one for level terrain, one for the upgrade, and one for the downgrade. fHV?1

1?PT(ET?1)?PR(ER?1)Where：PT?0.10(given) PR?0.03(given)

ET(level) = 1.5 (Table 12.13, level terrain) ER(level) = 1.2 (Table 12.13, level terrain)

ET(up) = 2.5 (Table 12.14, 5% grade, 1.5 mile in length) ER(up) = 2.5 (Table 12.15, 5% grade, 1.5 mile in length) ET(down) = 1.5 (Table 12.16, 5% grade, 1.5 mile in length) ER(down) = 1.2 (Table 12.13, level terrain) Then：

fHV(level,down)?1?0.947

1?0.1?(1.5?1)?0.03?(1.2?1) fHV(up)?and：

1?0.837

1?0.1?(2.5?1)?0.03?(2.5?1)2500?1.62 lanes

0.92?1766?0.947?12500?1.83 lanes N(up)?0.92?1766?0.837?1 N(level,down)?This suggests that the level, downgrade and upgrade sections require two lanes in each direction. So the assuming that there are 2 lanes is established. 12-8. Answer：

Step 1：The free-flow speed of the facility is found using Equation (12-5)： FFS?BFFS?fLW?fLC?fN?fID Where：BFFS?70mi/h (urban freeway) fLW?1.9mi/h (Table 12.5, 11-ft lanes) fLC?3.6mi/h (Table 12.6, 0-ft lateral clearance) FFS?70?1.9?3.6?4.5?7.5?52.5mi/h

Step 2：Determine the Maximum Service Flow Rates for Each Level of Service.

MSFrates for each level of service are draw from Table 12.3 Values are obtained for freeways with free-flow speed of 50mi/h, and 55mi/h straight-line interpolation may be used to obtain maximum service flow rates for the intermediate free-flow speed of 64.5mi/h. This is illustrated in Table 12.1：Values of MSF for Example 12-8.

Table 12.1：Values of MSF for Example 12-8

Level of Service A B C D E 50 550 900 1300 1750 2201 Free-Flow Speed 52.5 575 945 1365 1830 2225 55 600 990 1430 1910 2250 Step 3：Determine the Heavy-Vehicle Factor The heavy-vehicle factor is computer as： fHV?1

1?PT(ET?1)?PR(ER?1)Where： PT?0.05(given)

PR?0.00(given) ET?2.5(rolling terrain) Then：

fHV?1?0.930

1?0.05(2.5?1)Step 4：Determine the service volumes for each level of service

Service flow rates and service volumes are computed using Equation (12-2)： And 12-3：

SFi?MSFi?N?fHV?fpSVi?SFi?PHF

Where：MSFi= as determined in the step 2 N?2(given)

fHV= 0.930 (as computed in step3) fp= 1.00 (regular users) PHF?0.90

These computations are done in the spreadsheet shown in table： Level of sevice A B C D E Step 5：

Determine Target-Year peak-Demand volumes. The problem statement indicates that present demand is 3600veh/h and that this volume will increase by 6% per year for the foreseeable future. Future demand volumes may be computed as： Vj?V0(1.03n)

Where：Vj= peak-hour demand volume in target year j V0= peak-hour demand volume in year 0, 2200veh/h

MSFi (pc/h/ln) 575 945 1365 1830 2225 N 2 2 2 2 2 fHV 0.93 0.93 0.93 0.93 0.93 fp 1 1 1 1 1 SF (veh/h) 1070 1758 2539 3404 4139 PHF SV (veh/h) 963 1582 2285 3064 3725 0.9 0.9 0.9 0.9 0.9 N= number of years to target year Then：V0= 2200 veh/h

V5= 2200(1.035) = 2550 veh/h V10= 2200(1.0310) = 2957 veh/h V20= 2200(1.0320) = 3973 veh/h

Step 6：Determine Target Year levels of service The target year demand volumes are stated as full peak-hour volumes. They are, therefore, compared to the service volumes computed in table2 to determine LOS. The results are shown in table 3.

Table 3

Target Year 0 5 10 20 Demand volume (veh/h) 2200 2550 2957 3973 Level of service C D D F As indicated in Table 3, levels of service F prevails in target years 20.In each of these years, demand exceeds capacity. Clearly the point at which capacity is reached occurs in between years 10 and 20.

Capacity, stated in terms of a full peak hour, is 3725 veh/h. The exact year that demand reaches capacity may be found as follows：

3725?2200?(1.03n) n?17.8year s12-9.Answer：(a)

Time 7:00 - 7:15 7:15 - 7:30 7:30 - 8:00 Arrivals (veh/h) 1000 1000 2000 Capacity (veh/h) 450 900 1800 Queue size (veh/h) 1000 – 450 = 550 550 + 1000 – 900 = 650 650 + 2000 – 1800 = 850 8:00 - 9:00 9:00 - 10:00 > 10:00 3900 3500 2800 3600 3600 3600 850 + 3900 – 3600 = 1150 1150 + 3500 – 3600 = 1050 Queue decrease by 800 Queue dissipates 1050/800=1.3125hours after 10:00 AM, or 11:19 AM

① When the road is congest that the queue is generate. ② The maximum lengths of queue is：

LQ?(1150/2)?25?14375ft?14375/5280?7.72mi

So, the maximum queue occurs at 8:00 - 9:00

③ Queue dissipate at 11:19 AM

(b)

Time 7:00 - 7:15 7:15 - 7:30 7:30 - 8:00 8:00 - 9:00 9:00 - 10:00 Arrivals 1000 1000 2000 3900 3500 capacity 500 1000 2000 4000 4000 Queue size 1000 – 500 = 500 500 + 1000 – 1000 = 500 500 + 2000 – 2000 = 500 500 + 3900 – 4000 = 400 400 + 3500 – 4000 = –100 Queue is dissipated at 9:48 AM

① When the road is congest that the queue is generate ② The maximum lengths of queue is：

LQ?(500/2)?25?6250ft?6250/5280?1.18mi

So, the maximum queue occurs at 7:00 - 8:00

③ Queue dissipate at 11:19 AM

12-10. Answer：

haT?(32?4.3)?(8?4.9)?4.42s32?8

(128?3.1)?(32?3.8)??3.24s128?32

haPThen, using Equation (12-16)：EH?haH haP

EH?haH4.42??1.36 haP3.24Homework of chapter thirteen

13-1. 13-2.

Step 1: Determine All Required Input Variables

As all demand are specified as flow rates in pc/h under equivalent base conditions, no further conversion of these is necessary. Key analysis variables are summarized below:

vw=800+1600=2400 pc/h vnm=2000+1000=3000 pc/h V=2400+3000=5400 pc/h V/N=5400/4=1350 pc/h VR=2400/5400=0.444 L=2500 ft

Note that this is a Type B configuration. The weave from left to right can be made with no lane-changes, while the weave from right to left requires one lane-changes. Step 2: Determine the Average Speed of Weaving and Non-weaving Vehicles in the Section equations 13-3 and 13-4 are used to determine the average speeds of weaving and non-weaving vehicles in the section, using the constants of calibration given in Table 13.3.Unconstrained operation is assumed for initial speed estimates.

0.08(1?0.444)2.2(13.5)0.70Ww=25000.5=0.5576 70?10Sw=15+ (1?0.5576) =53.52 mi/h

Wnm0.002(1?0.444)6.0(1350)1.0??55.2725000.5 mi/h

Step 3: Determine the Type of Operation

The assumption of unconstrained operation is checked using the equations and criteria

of Table 13.4 for Type B configurations:

Nw?N*[0.085?0.703VR?(234.8)?0.018(Snw?Sw)]L

234.8?0.018?55.27-53.52?]2500=2.6 lanes

Nw?N*[0.085?0.703*0.444?As this is less than the maximum allowable use of lanes for weaving vehicles in a Type B configuration (3.5), the operation is unconstrained and the speeds determined in step 2 are correct.

Step 4: Determine the Average Seep for All Vehicles in the Weaving Area The space mean seep for all vehicles in the weaving area is given by equation 13-5:

s?2400?3000?54.48mi/h24003000?53.5255.27

Step 5: Determine the Density and Level of Service in the Weaving Area, The density in the weaving area is computed using equation 13-6:

From the criteria of Table13.1,this is level of service C,Again, while not recommended by the HCM2000,separate densities for weaving and non-weaving vehicles can be estimated:

v1350D?N??24.78pc/mi/lnS54.48NWA?2.6lanesNnwA?3?2.6?0.4lanesvw2400NDw?wA?2.6?17.25pc/mi/lnSw53.52Dnwvnw3000N?nwA?0.4?135.70Snw55.27

Actual Value 2400 1350 0.444 2000 ft Step 6: Check other limitations on weaving area operation Parameter Maximum Value Weaving Capacity 4000 Maximum v/N 2400 Maximum VR 0.8 Maximum L 2500 ft None of these limits are exceeded.

Step 7: Determine the capacity of the weaving area VR 0.40 0.444 Capacity 6650 pc/h Cb 0.50 6170 pc/h Using straight-line interpolation: 0.50?0.444Cb?6170?()?(6650?6170)?6170?0.65?480?6439pc/h0.50?0.40 This is a maximum flow rate in pc/h under equivalent base conditions. It could be converted to a flow rate under prevailing conditions and/or a full-hour maximum volume under prevailing conditions. As the demands are given in terms of base flow rates, such conversions would not be useful in this case. The base capacity of 6439 pc/h can be directly compared to the converted total demand flow of 5700pc/h. 13-3. Answer：

Step1: Covert All Demand Volumes to Flow Rates in pc/h Under Equivalent Base Conditions

The freeway and ramp flows approaching the merge area must be converted to flow rates in pc/h under equivalent base conditions using Equation 13-1.In this case, note that the truck percentages and PHF are different for the two. Table 13.8 is used to obtain a value ofEfor rolling terrain (2.5), and Equation12-9 is used to compute the

Theavy-vehicle factor. It is assumed that drivers are familiar users, and that For the ramp demand flow:

f =1.00.

p

fHV?1?0.9301?0.1(2.5?1)

For the freeway demand flow:

vR?800?905pc/h0.95*0.930*1.00

fHV1?1?0.01(2.5?1)?0.870 4000?0.95*0.87*1.00?4840pc/h

vFStep2: Determine the Demand Flow Remaining in Lanes 1 and 2 Immediately

Upstream of the Merge Table 13.6 gives values of

pFM.Equation 13-15 is used to estimate

pFM.This value is

then used in Equation13-11 to determinev12.

ppFM=0.2178－0.000125vR＋0.01115(La/RFFS) =0.2187－0.000125*905＋0.01115*(750/45)=0.2964

FFMFMv=v*p12=4840*0.2964=1434pc/h

Step3: Check Capacity values

To determine whether the section will fail (LOS F), the capacity values of Table 13.8 must be consulted. For a merge section, the critical capacity check is on the downstream freeway section.

vFO=vF?vR=4840+905=5745pc/h

From Table 13.8 the capacity of a four-lane freeway section is 9400pc/h when the FFS is 65mi/h. As 9400>, no failure is expected due to total downstream flow. Total flow entering the merge influence area is:

vR12?vR?v12?905+1434=2339pc/h

As the maximum desirable entering flow for single-lane merge area is4600 ,this element is also acceptable. No failure(LOS F) is, therefore, expected. Step4: Estimate Density and Level of Service in the Ramp Influence Area

As stable Operations are expected, Equation 13-24 may be used to estimate the density in the ramp influence area:

DD

R=5.475+0.00743vR+0.0078v12-0.00627La =18.6pc/mi/ln

RFrom the criteria in Table 13.1, this is level of service B

13-4. Figure 13.20 illustrates two consecutive ramps on an older freeway. It may be assumed that there is a ramp-to-ramp flow of 200 veh/h.

(a) What is the expected level of service for the conditions shown？

(b) Several improvement plans are under consideration： i. Connect the two ramps with a continuous auxiliary lane.

ii. Add a third lane to the freeway and extend the length of acceleration and deceleration lanes to 300 ft.

iii. Provide a lane addition at the on – ramp that continues past the off – ramp on the downstream freeway section. The off – ramp deceleration lane remains 200 ft long.

Which of these three improvements would you recommend？Why？Justify your answer. Answer：

(a) fHV(Freeway) =11+0.1(1.5-1) =11.05=0.95 VF=32000.94*0.95*1.00=3583PC/h

fHV(Ramp1)=11+0.1(1.5-1) =11.05=0.95

vR1=6000.94*0.95*1.00=672PC/h

fHV(Ramp2)=11+0.1(1.5-1) =11.05=0.95

vR2=7000.94*0.95*1.00=784PC/h

Because it has two ways in one direction PFM=1.0 PFD=1.0

v121=VF*PFM=3583*1=3583 pc/h vF2=VF+VR1=3583+672=4255pc/h v122=VR2+VF2-VR2*PFD=4255pc/h

None of demand flows exceed the capacities

DR1=5.475+0.00734*672+0.0078*3583-0.00627*100=37.7pc/mi/ln DR2=4.252+0.0086*4255-0.009*200=6.1pc/mi/ln

So Ramp1 at los E, Ramp2 at los A (b) Choose third plan is reasonable. 13-5.Answer：

vw=400+1200=1600 pc/h vnw=3000+1000=4000 pc/h v=1600+4000=5600pc/h v/N=5600/4=1400pc/h VR=1600/5600=0.286 L=900 ft

a(1?VR)b(Si=15+FFS-101+wi W?vc)N

LdUnder constrained conditions: w=0.351+0.2862.214000.979000.8=2.97

Sw=15+60-101+2.97=27.6mi/h Wnw=0.0021+0.286414001.39000.75=0.41 Snw=15+60-101+0.41=50.5mi/h

S=vw+vnwvwSw+vnwSnw=5600160027.6+400050.5=40.8 mi/h D=v/NS=140040.8=34.3pc/mi/ln Level of service is D

VR 0.20 0.286 0.30 Cb=6680+0.3-0.2860.3-0.2*7310-6680=6768pc/ln

They are not acceptable .because the level of service is too low.

Homework: 1、Update the recent statistics for the highway system development in U.S., China, and other countries. Analyze and compare the statistics and address your own conclusions.

Compare the history of highway system development between U.S. and China.

Compare the legislation and standard system for transportation Engineering in U.S. and China.

2、Compare road user characteristics and related standards in U.S. and China.

3、Compare the characteristics and related standards for highways in U.S. and China. 4、Compare the approaches and related standards for traffic control and management in U.S. and China.

5、Compare the characteristics of traffic volume in China and U.S.; Compare the volume study methodologies in China and U.S.. 6、Compare the study methods and characteristics of speed and travel time in U.S. and China.

7、Compare the differences in traffic accident in China and U.S..

8、Compare the planning and design methods for parking in U.S. and China.

Capacity 7310 pc/h Cb 6680 pc/h

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