civil depth notes for mar 31st-transportation

Transportation FE Civil Depth

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Table of Contents 1. Surveying ...................................................................................................................................... 2

1.1 Bearings and Azimuths .......................................................................................................... 21.2 Stationing ............................................................................................................................... 2

2. Driver Performance and Behavior ................................................................................................ 32.1 Information Processing and Perception .................................................................................. 32.2 Driver Expectancy .................................................................................................................. 32.3 Perception-Reaction Time ...................................................................................................... 32.4 Stopping Sight Distance ......................................................................................................... 4

3. Horizontal Curves ......................................................................................................................... 63.1 Circular Curves ...................................................................................................................... 63.2 Superelevation ........................................................................................................................ 83.3 Stopping Sight Distance on Horizontal Curve Section .......................................................... 9

4. Vertical Curves ........................................................................................................................... 104.1 Vertical Curve Elevations .................................................................................................... 104.2 Vertical Curve Design .......................................................................................................... 114.3 Intersection Sight Distance ................................................................................................... 12

5. Speed Characteristics .................................................................................................................. 145.1 Time Mean Speed ................................................................................................................. 145.2 Space Mean Speed ............................................................................................................... 145.3 Spot Speed Data Analysis .................................................................................................... 155.4 Speed, Flow and Density Relationships ............................................................................... 175.5 Speed, Distance, and Time Relationships ............................................................................ 17

6. Signalized Intersections .............................................................................................................. 186.1 Change Interval .................................................................................................................... 186.2 Clearance Interval ................................................................................................................ 19

7. Traffic Safety .............................................................................................................................. 208. Pavement Design ........................................................................................................................ 219. Additional Problems for Self Study ............................................................................................ 22 NOTE

CERM Civil Engineering Reference Manual for PE Exam, 11th Edition by Michael Lindeburg

SRHB Supplied-Reference Handbook, 8th Edition, 2nd Revision by NCEES

ahmed youssef ([email protected])This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.


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1. Surveying

1.1 Bearings and Azimuths See definitions on CERM page 77-11 under section 23. PROBLEM 1 - Convert the following bearings to azimuths from north:

(a) N 740 24 01 W (b) S 850 13 16 W (c) N 840 28 13 E (d) S 080 19 19 E

SOLUTION 1

(a) 2850 35 59 (b) 2650 13 16 (c) 840 28 13 (d) 1710 40 41

1.2 Stationing Stationing concept is used along horizontal alignments for referencing purpose 1 station = 100 feet How do you represent stationing?

o Specific location is represented as Sta 10+00 o Distance is represented as 10.00 sta

PROBLEM 2 What is the station at Point B?

SOLUTION 2 Station at Point A = Sta 22+45 Station at Point B = (Sta 22+45) + 1028 = 2245+1028 =3273

= Sta 32+73



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PROBLEM 3 The survey has identified the beginning and ending points along an urban arterial highway that require new sidewalks. The first station is located at 5+88 and the second station is located at 10+05. What is the difference in length between stations in feet and in stations? SOLUTION 3 Difference in length = (10+05) (5+88) = 1005 588= 417 Difference in stations = 4.17 sta

2. Driver Performance and Behavior

2.1 Information Processing and Perception The time required to respond successfully to any driving situation, such as an emergency, involves four stages:

perception (detection and identification) decision reaction response of the vehicle.

2.2 Driver Expectancy A fundamental component of driver information processing and perception. Drivers operate with a set of expectancies, e.g.:

freeway exits will be on the right side of the road (or the left in Britain, Australia, etc.);

advance warning will be given of hazards in the roadway; other drivers will obey traffic regulations, etc.

2.3 Perception-Reaction Time A significant variable in the successful processing and use of information is the speed with which this is done. Perception-Reaction Time (PRT) is a human factor often cited by traffic engineers concerned with safety. PRT is the interval between the appearance of some object or condition in the drivers field of view and the initiation of a response such as braking or changing course. Note that PRT involves the initiation of a response (e.g. pressing the brake), not the completion of the vehicle maneuver (stopping).



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PRT depends on the situation. Response time is generally quickest when there is one specific response to be made to a single stimulus (brake lights of vehicle ahead). In the case of choice reaction time, in which there is more than one stimulus and/or more than one possible response (e.g. toll plaza), reaction time increases as a function of the number of possibilities. A driver may, for example, have to decide whether to steer or brake, or both, to avoid a pedestrian. The PRT used for design standards by AASHTO includes 1.5 sec for perception and decision, 1.0 sec for making a response, for a total of 2.5 sec, which is generally considered adequate for all but the most complex driving situations. PROBLEM 4 Which of the following factors affect driver performance and behavior?

A. cell phone use B. fatigue C. traffic D. drugs and alcohol E. young / old age F. law enforcement G. All of the above

SOLUTION 4 The correct answer is G. All of the above affect driver performance and behavior.

2.4 Stopping Sight Distance See Chapter 3 in A Policy on Geometric Design of Highways and Streets 2004 by AASHTO. This book is popularly known as Green Book Stopping Sight Distance is the sum of two distances: (1) the distance traversed by the vehicle from the instant the driver sights an object necessitating a stop to the instant the brakes are applied; and (2) the distance needed to stop the vehicle from the instant the brake application begins. These are referred to as brake reaction distance and braking distance, respectively. The AASHTO GB provides the following equations for calculating braking distance and SSD, with and without the effect of grades.



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The following equation includes terms for both the brake reaction distance and braking distance (Use equation 78.43(b) on page 78-9 in CERM. Also see the same equation on page 162 of SRHB):

First part of the equation represents brake reaction distance and the second part represent braking distance PROBLEM 5 A motorist is traveling on a level grade at 50 mph. A tree has fallen across the road and forces the motorist to stop. Assuming a 2.5 sec PRT and 11.2 ft/sec2 deceleration rate, determine the brake reaction distance and braking distance in feet.

A. 147 and 154 B. 165 and 194

C. 184 and 240 D. 165 and 290

SOLUTION 5

The correct answer is C, 184 and 240

2VSSD = 1.47V t +

3032.2

mphmph a G

Brake reaction distance = 1.47Vt = 1.47(50)(2.5) = 184'

where:

SSD = Stopping Sight Distance, ft; V = design speed, mph; t = breaking reaction time, 2.5 sec

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a = deceleration rate, 11.2 ft/sec

G = percent of grade divided by 100, it is in decimal

2 2V 50Braking distance = = = 240' 11.230 30 0

32.2 32.2

mph

a G



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3. Horizontal Curves

3.1 Circular Curves Horizontal circular curve is a circular arc between two straight lines known as

tangents.

See equations 78.1 to 78.12 and Figure 78.1 on pages 78-2 of CERM. Also see page 164 of SRHB.

PROBLEM 6 For the following circular curves having radius R, what is their degree of curve by Arc definition and Chord definition? SOLUTION 6

(a) Roadway curve with 500.00 ft 05729.578' 11 27'33"

500aD

ft

(b) Roadway curve with 1500.00 ft 05729.578' 3 49'10.99"

1500a ftD



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PROBLEM 7 Compute T, L, E, HSO, R, and stations of the PC and PT for the circular curve described below: Highway curve with R = 750.000 ft, I = 180 30, and PI Sta 123+24.80 SOLUTION 7

0 0 0

0

0

0 0

5729.578' 7 38'22"; I = 18 +30' 60' = 18.5 ; 750

I 18.5T = R tan = 750 tan = 122.145 ft2 2

18.5L = 2 R = 2 750 = 242.164 ft360 360

aD ft

I

0 0

0

I I 18.5 18.5E = R tan tan = 750 tan tan 2 4 2 4

= 750 0.1629 0.0809 = 9.881 ftI 18.5HSO = R 1 - cos =750 1-cos = 750(0.013) = 9.72 2

53ftPC Sta. = PI Sta. - T = 12324.800 - 122.145 = 122+02.655PT Sta. = PC Sta. + L = 12202.655 + 242.164 = 124+44.819



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PROBLEM 8 Compute interior (or intersection) angle for the following circular curve: SOLUTION 8

0 0 0 0180 60 60 60I

3.2 Superelevation Used at horizontal curves Use equations 78.37(b) on page 78-7 of CERM. Also see on page 163 of

SRHB. PROBLEM 9 What is the minimum radius, Rmin, that can be used on a horizontal curve with a 70 mph design speed, a maximum superelevation, emax = 0.08, and a side friction factor, f = 0.10? SOLUTION 9

Use equation 78.37(b) emax = 8%; V = 70 mph; fmax = 0.10

2

minmax max

VR =

15(e + )mph

f

2

min

70R = 1814.8015(0.08 + 0.10)

ft



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3.3 Stopping Sight Distance on Horizontal Curve Section Obstructions along the inside of curves can limit the available (chord) sight distance. A curve must be designed that will simultaneously provide the required stopping sight distance while maintaining a clearance from a roadside obstruction. See equation 78.45 and figure 78.9 on page 78-10 in CERM. Also see on page 163 of SRHB. PROBLEM 10 A four-lane undivided highway has a design speed of 40 mph. The lanes are 12 ft wide. The centerline Degree of Curvature, D is 10o 45 Determine the required clearance from the center of the curves inside lane based on Stopping Sight Distance criteria.

A) 22.41 ft B) 21.67 ft C) 305 ft D) 533 ft SOLUTION 10 Using SSD equation (page 162 of SRHB), for V = 40 mph, S = 305 ft. D = 10o 45 is equivalent to R= 532.98 ft (using equation on page 164 of SRHB) The centerline of the inside lane is offset 18 ft (12 ft + 6 ft) from the roadway centerline. Therefore R lane centerline = 532.98 ft 18 ft = 514.98 ft

Using equation 78.45 in CERM (Also see on page 163 of SRHB), 28.65 28.65 3051 cos 514.98 1 cos 22.41

514.98SHSO R ft

R

Answer A



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4. Vertical Curves Vertical curves are used to change the grade of a highway. Most vertical curves take the shape of an equal-tangent parabola. Such curves

are symmetrical about the vertex. Two types of vertical curves Crest and Sag

See figure 78.10 and equations 78.46 to 78.49 on pages 78-11 of CERM. Also see on page 165 of SRHB.

4.1 Vertical Curve Elevations PROBLEM 11 A +3.25% grade intersects a -2.00% grade at Sta. 45+25 and elevation 695.42 ft. A 1000 ft vertical curve connects the two grades. Determine:

a) the station and elevation of turning point b) the elevations along the curve at Sta. 45+00; Sta. 50+25

SOLUTION 11 a) Using equation 78.46 on page 78-11,

2 1 2.00 3.25 0.52510Sta

G GRL

Highest Point Location: x = -G1/R = 6.1905 Sta. (Eqn. 78.48) Sta. = BVC Sta. + x = 4025 ft + 619.05 ft = 4644.05 ft = Sta 46+44.05 Using equation 78.47, Elev. = (R/2)x2+G1(x)+BVC Elev. Elevation at BVC = 695.42-500(0.0325) = 679.17 ft Elev. = (R/2)x2+G1(x)+BVC Elev. = 689.23 ft b)

Using equation 78.47, Elev. = (R/2)x2+G1(x)+BVC Elev., results are tabulated



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At Sta. 45+00 xsta = 4.75 At Sta. 50+25 xsta = 10.00

Sta. (R/2) xsta 2 G1* xsta BVC Elev. Elev.

45+00 -5.92 15.44 679.17 ft 688.69 ft 50+25 -26.25 32.5 679.17 ft 685.42 ft

4.2 Vertical Curve Design Using AASHTO Guidelines Minimum vertical curve length is computed Based on sight distance criteria Use Table 78.4 in CERM (See page 163 of SRHB)

PROBLEM 12 You are designing a vertical curve on a two-lane highway with G1=+3.50%; G2= -2.25%; PVI 85+00; PVI elevation = 457.59 feet; and a Design Speed of 65 mph. What is the most appropriate length of the curve you should design? SOLUTION 12 From Table 78.4 of CERM (Also see page 163 of SRHB), Using SSD equation, for V=65 mph, S=645 ft

Since S = 645 feet < L = 1108.5 feet, our assumption is Good.

Therefore, L =1108.5 ft

2 2 22 1

Assume S < L; ( ) 5.75*645L= = = =1108.5 ft

2,158 2,158 2,158AS ABS G G S



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4.3 Intersection Sight Distance Reference AASHTO Geometric Design of Highways and Streets pages 651-675



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Recommended dimensions of the sight triangles vary with the type of traffic control used at an intersection. But, the equation to calculate side length b which is called Intersection Sight Distance (ISD) is same for all scenarios. The equation is: b = ISD = 1.47Vmajor tg Where Vmajor = Design speed in MPH on major street tg = Time gap in seconds varies by vehicle type and movement type (AASHTO has established guidelines to compute this value) PROBLEM 13 A passenger vehicle is stopped on a minor street at a stop sign of stop controlled intersection. The design speed of the major road is 60 mph. Stopped passenger vehicle is getting ready to make right turn from minor street to major road. If this vehicle requires a time gap of 6.5 to avoid a collision with the vehicle approaching the intersection, at least how far the approaching vehicle has to be from the intersection before the passenger car makes right turn? SOLUTION 13 ISD = 1.47Vmajor tg = 1.47 * 60 * 6.5 = 573.3 ft



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5. Speed Characteristics

5.1 Time Mean Speed From spot speed studies which are measured as vehicles pass a point on the

road It is the average (mean) of all vehicles passing a point on a road over some

specified time period

5.2 Space Mean Speed From travel time runs for a given section of the roadway It is the average (mean) of all vehicles occupying a given section of a highway

over some specified time period PROBLEM 14 Six travel time runs are made on a 1000 feet section and the data is presented in the following table. Compute average speed

Vehicle Measured Time (sec) 1 18 2 20 3 22 4 19 5 20 6 20

SOLUTION 14 6 1000 50.4 fps

119sS

1

Where: speed of the vehicle, and number of vehicles included in the measurement sample

n

ii

t

thi

SS

nS in

1

Where: length of the segment travel time of the vehicle to traverse the section

ns

ii

thi

nLSt

Lt i L



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Example Graphs

5.3 Spot Speed Data Analysis Mean speed or Time mean speed defined in the previous section 85th percentile speed speed at which 85 percent of free-flowing vehicles are

traveling at or below. To establish speed zones nearest 5 mph increment at or below the 85th percentile speed.



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PROBLEM 15 The following speed data was collected using a radar gun on an arterial street:

Estimate the following: a) Mean speed b) 85th percentile speed

SOLUTION 15 Expand the table as shown here:

a) Mean speed: 13820x = = 38.39 mph;

360

b) 85th percentile speed of vehicles: Step 1: Total # observations x 0.85 = 360 x 0.85 = 306

Step 2: The 306th observation is in the 46-50 mph speed group. Therefore the

85th percentile speed is 48 mph.

Speed Group (mph) Frequency

16 to 20 10 21 to 25 17 26 to 30 36 31 to 35 66 36 to 40 84 41 to 45 70 46 to 50 50 51 to 55 21 56 to 60 6

Speed Group Frequency Assumed Speed Total

Speed Cumulative Frequency

16 to 20 10 18 180 1021 to 25 17 23 391 2726 to 30 36 28 1008 6331 to 35 66 33 2178 12936 to 40 84 38 3192 21341 to 45 70 43 3010 28346 to 50 50 48 2400 33351 to 55 21 53 1113 35456 to 60 6 58 348 360

360 13820



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5.4 Speed, Flow and Density Relationships Speed (S) Space mean speed (mph) Flow/Rate of Flow/Volume (v) number of vehicles per hour per lane (vphpl) Density (D) Number of vehicles per mile per lane (vpmpl)

vDS

Spacing Distance between common points (e.g. the front bumper) on

successive vehicles (ft/veh) 5280 ft/mileSpacing in ft

D

Reference - Speed/flow/density relationships graphs from Page 73-7 of CERM (Also see Page 163 of SRHB)

5.5 Speed, Distance, and Time Relationships Distance = Speed * Time PROBLEM 16 Refer to the figure at right. At t = 0, vehicle A, traveling at a speed of 60 km/h, passes a road marker on a straight section of a four-lane divided highway. Vehicle B, traveling with a speed of 90 km/h, passes the marker 2 sec later. Find the time when vehicle B overtakes vehicle A, and the corresponding distance from the road marker. SOLUTION 16 The figure above shows a sketch of the events. S is the distance from the marker where B overtakes A at time t. From the figure, s = sA = sB



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Using s = vt , sA = vAt = 16.7t and sB = vB(t -2) = 25(t-2)

16.7t = 25 (t-2) t = 6.02sec s = sA = 16.7(6.02) = 101 m

6. Signalized Intersections Cycle length the time required for one complete sequence of all signal indications Phase the right-of-way (green), change (yellow), and clearance (all red) intervals in a cycle that are assigned to an independent traffic movement or combination of movements Green interval the right-of-way interval during which the signal indication is green Yellow Change interval the first interval following the green interval or which the signal indication is yellow Clearance interval an interval that follows a yellow change interval and preceds the next conflicting green interval

6.1 Change Interval Also known as Yellow Interval

2Gg2a

v ty Where: y = length of yellow interval (sec) t = driver perception/reaction time (1.0 sec generally used) v = velocity of approaching vehicle (fps) a = deceleration rate (10 fps2 generally used) G = acceleration due to gravity (32.2 fps2 generally used) g = grade of approach (percent/100) Also see page 162 of SRHB for the above equation.



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6.2 Clearance Interval Also known as Red Clearance Interval and All Red

w Lr - If there is no pedestrian trafficv

P Lr - If there is pedestrian traffic or the crosswalk is protected by ped. signalv

Where: r = length of red clearance interval (sec) w = width of intersection; more precisely, the length of the vehicle path from the departure stop line to the far side of farthest conflicting traffic lane (ft) P = width of intersection; more precisely, the length of the vehicle path from the departure stop line to the far side of farthest conflicting ped crosswalk (ft) L = Length of vehicle (20 ft generally used) v = speed of vehicle through intersection (fps)

Also see Page 162 of SRHB for the above equation PROBLEM 17 Estimate change interval for an approach with a grade -2% and an approach speed of 30 mph. SOLUTION 17

For -2% grade, 5280 ft/mile30 miles/hour*

3600 sec/hour1.02*10 2*32.2 (-2 0

)

y /1 0

= 3.4 sec



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7. Traffic Safety Crash rates are normally considered better indicators of risk than crash frequencies.

Crash rates for intersections are normally expressed in terms of crashes per million entering vehicles (MEV), using the following equation:

V*T*365

10*A R6

int

Crash rates for roadway segments are normally expressed in terms of crashes per 100 million vehicle-miles (100 MVM), using the following equation:

L*V*T*365

10*A R8

sec Where: Rint = crash rate for the intersection

Rsec = crash rate for the roadway section A = number of reported crashes T = time period of the analysis (years) V = annual average daily traffic volumes (veh/day) L = length of the segment (miles)

PROBLEM 18 An intersection has a total entering traffic volume of 42,000 vehicles per day. During the past three years, there have been a total of 35 reported intersection-related crashes. What is the crash rate for this intersection? SOLUTION 18

6

int

35 10R 365 3 42,000

= 0.76 crashes per MEV PROBLEM 19 A five-mile long section of two-lane road has an AADT of 8,000. There have been six crashes on this section of road during the past two years. What is the crash rate? SOLUTION 19

8

sec

6 10R 365 2 8,000 5

= 20.5 crashes per 100MVM



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8. Pavement Design Asphalt surface Using AASHTO Guidelines

o Estimate Design Traffic o Determine the Structural Number (SN) for the given traffic, roadway, and

soil characteristics o Compute layer-thickness using the following equation (see on page 166

of SRHB):

SN = a1D1 + a2D2 + +anDn Where SN = structural number of the pavement ai = strength coefficient of the ith layer Di = thickness of the ith layer in inches

PROBLEM 20 Determine the thickness of flexible pavement layer 2 given the following: FACTS: Total Structural Number required = 7.0 Total number of layers = 2 Layer 1 consists of asphalt concrete with a strength coefficient of 0.46 Layer 1 thickness = 6 inches Layer 2 consists of granular base with a strength coefficient of 0.15 SOLUTION 20 Using equation 75.29 in CERM, 1 1 2 2SN D a D a

1 12

2

7.0 6 0.46 28.3"0.15

SN D aDa



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9. Additional Problems for Self Study PROBLEM 21 The local survey crew measures a distance of 1,135 ft of new drainage pipe that is required for the new road construction. The beginning point for the drainage is located at station 8+77. What is the station number of the ending point?

A) 12.20 Sta B) Sta 12+20 C) 20.12 Sta D) Sta 20+12 SOLUTION 21 End point station = (Sta 8+77) + 1,135 = 877 + 1,135 = 2012 = Sta 20+12 Answer D PROBLEM 22 A motorist is traveling down a 6% grade at 65 mph and needs to stop because of a crash scene. Assuming a 2.0 sec PRT and 12.0 ft/sec2 deceleration rate, determine the total SSD in feet.

A) 524 ft B) 642 ft C) 750 ft D) 869 ft

SOLUTION 22

2 2V 65SSD = 1.47Vt + = 1.47(65)2 + 1230 30 0.06

32.2 32.24225 = 191.1+ = 641.52' 9.380

a G

Answer B



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PROBLEM 23 Compute T, L, E, HSO, R, and stations of the PC and PT for the circular curve described below. Highway curve with Da = 30 30, I = 240 45, and PI Sta. 30+00.00. SOLUTION 23

0 0a

0 0

0

0

0 0

5729.578' 5729.578' 5729.578'R = = = = 1637.022 ftD 3 30' 3.5

I = 24 + 45' 60' = 24.75 ;

I 24.75T = R tan = 1637.022 tan = 359.174 ft2 2

24.75L = 2 R = 2 1637.022 = 707.143 ft360 360

E = R t

I

0 0

0

I I 24.75 24.75an tan = 1637.022 tan tan 2 4 2 4

E = 1637.022 0.2194 0.10841 = 38.9368 ft.

I 24.75HSO = R 1 - cos =1637.022 1-cos = 16372 2

.022(0.0232) = 38.0348 ft

PC Sta. = PI Sta. - T = 3000.00 - 359.174 = 2640.826 ~ 26+40.83PT Sta. = PC Sta. + L = 2640.826 + 707.143 = 3347.969 ~ 33+47.97

PROBLEM 24 A -3.00% grade intersects a +1.25% grade at Sta. 85+80 and elevation 210.00 ft. A 350-ft vertical curve connects the two grades. Determine the station and elevation of turning point. SOLUTION 24 R = (1.25+3.00)/3.5= 1.214 Elev. at BVC = 210.00+175(0.03) = 215.25 ft Low Point Station and Elevation Location: x = -G1/R = 2.4706 sta Sta. = BVC Sta. + x = 8405 ft + 247.06 ft= 8652.06 ft = Sta 86+52.06 Elev. = (R/2)x2+G1(x)+BVC Elev. = 211.54 ft



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PROBLEM 25 At a large port facility near New York City tractor-trailer trucks are lined-up in long queues as their drivers wait to have their containers inspected and documents processed. Assume the trucks have an average length of 73.5 ft and the average space between the rear and front bumpers of successive vehicles is 10 feet. What is the best estimate for the jam density (trucks/mile) in one lane of trucks? A) 50 B) 63 C) 77 D) 84

SOLUTION 25 By definition, jam density is the density at zero speed. Based on the information provided the average spacing between the front bumpers of successive vehicles is about 83.5 feet, the jam density can be estimated as follows:

5280 feet vehicleJam density = = 63 vehicles mile1 mile 83.5 feet

Answer B


civil depth notes for mar 31st-transportation

Documents

time mean speed

speed characteristics

space mean speed

vertical curves

intersection sight distance

time relationships

horizontal curves

vertical curve elevations