Chapter 1
The Profession of Transportation
1-1
To illustrate the importance of transportation in our national life, identify a
transportation-related article that appears in a local or national newspaper. Discuss
the issue involved and explain why the article is newsworthy.
The fire in a rail tunnel under several blocks of Howard Street in
Baltimore, on July 18, 2001, illustrates the impact of transportation on society. A
60-car train derailed inside the tunnel; hazardous materials including hydrochloric
acid ignited, causing the fire that disrupted not only rail service but also power
supply to much of Baltimore. Additionally, the street above the tunnel buckled in
several places due to the intense heat below.
The article from the July 23, 2001 edition of the Washington Post, entitled
“NTSB Wonders If Flooding Caused Baltimore Train Wreck,” indicates that a
break in a water main that runs adjacent to the tunnel may have caused the train
derailment. The National Transportation Safety Board, in its investigation of the
derailment, is examining city records to determine when the break occurred and
whether the water main break caused the derailment, or if heat from the fire after
the derailment caused the break. The impact of a failure in the aging
infrastructure of two modes of transportation (rail and pipeline) demonstrates the
role that transportation can play in daily life.
1-2
Arrange an interview with a transportation professional in your city or state (that is,
someone working for a consulting firm, city, county or state transportation
department, transit or rail agency). Inquire about the job he or she performs, why
he or she entered the profession and what he or she sees as the future challenges in
the field.
Background: Bachelor of Science degree in Civil Engineering. During
undergraduate study, worked part-time for the Traffic Engineering Division of the
Department of Public Works. Earned a Master of Science in Civil Engineering
degree with a major in transportation. Currently employed by a consulting firm,
conducting research in multimodal and intermodal transportation systems.
Envisions the transportation industry becoming cohesive in that modes will
connect with one another at strategic points to allow for the seamless transfer of
people and goods.
Note: This is a brief transcript of an interview with a transportation
professional.
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Chapter 1: Profession of Transportation
1-3
Keep a diary of all trips you make for a period of three to five days. Record the
purpose of the trip, how you traveled, the approximate distance traveled and the
trip time. What conclusions can you draw from the data?
The table shown on the next page provides a list of all trips made by a
student during a five-day period. The following conclusions can be reached about
the student’s travel behavior.
•
24 trips were made (an average a 5.4 trips per day)
• 8 trips by bus
• 9 trips by car
• 7 trips by walking
•
248 minutes were spent traveling (an average of 49.6 minutes per day)
• 77 minutes of travel by bus (31% of total travel time)
• 90 minutes of travel by car (36% of total travel time)
• 81 minutes of travel by walking (33% of total travel time)
•
About an equal amount of time was spent riding buses, walking, and driving a
car.
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Chapter 1: Profession of Transportation
Time (minutes)
Date
Mode
Purpose
Friday,
Jan. 22
Bus
Home to class
10
Walk
Class to class
7
Bus
Class to fraternity house
10
Walk
Fraternity house to class
15
Bus
Class to home
10
Car
Home to store and back
10
Car
Home to fraternity house and
10
Car
Home to store and back
10
Car
To store, library, and home
20
Car
To library and back
10
Sunday,
Jan 24
Car
To library and back
10
Car
To sister's place and back
5
Monday,
Jan 25
Bus
Home to Class
10
Walk
Class to Class
7
Walk
Class halfway home
7
Bus
Rest of way to fraternity house
7
Bus
Fraternity house to class
10
Walk
Class to home
15
Bus
Home to class
10
Walk
Class to fraternity house
15
Walk
Fraternity house to class
15
Bus
Class to home
10
Car
Home to fraternity house and
10
Car
Home to gym and back
5
Saturday,
Jan 23
Tuesday,
Jan 26
TOTAL
248
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Chapter 1: Profession of Transportation
1-4
Identify one significant transportation event that occurred in your city or state.
Discuss the significance of this event.
An important event in transportation in Charlottesville, Virginia occurred
in 1970, when the University of Virginia began its own bus service. The
significance of this event is that the bus service, in a sense, enlarged the
University. That is, students could reside off grounds and still be able to
commute to classes. This opened the door to develop student housing projects off
grounds. In addition, the University now could add new facilities further from
central grounds and still have them accessible to students.
1-5
Describe how transportation influenced the initial settlement and subsequent
development of your home city or state.
Norfolk, Virginia was one of the early settlements in this state due to its
proximity to water and near the Jamestown Settlement. Its transportation
evolution began mainly as water transportation due to its accessibility to the
Elizabeth River. Downtown Norfolk grew around the river, which became a
harbor for the cities of Norfolk and Portsmouth, Virginia. As Norfolk's
population grew, there became a need for more living area and people began to
settle further away from the harbor area. As a result of this migration, horse
drawn vehicles became the choice mode when people traveled to and from the
river. Eventually the mode of transportation that originated Norfolk and
Portsmouth, water, became its main industry for employment. The Norfolk Naval
Shipyard, located in Portsmouth, is a major employer for the area which services
and repairs ships used for the transportation of military goods and service
personnel.
1-6
Describe your state’s transportation infrastructure. Include both passenger and
freight transportation.
Virginia's transportation infrastructure is vast. The highway system, as of
1999, according to data from the Federal Highway Administration, includes
70,325 miles of public roads. Of these 70,325 miles, 57,737 miles are statemaintained, with the remainder maintained by cities, towns, and counties.
Virginia's rail network, excluding yards and sidings, totals 3,295 miles. In
addition, two of the nation's largest railroads, the CSX Corporation and the
Norfolk Southern Corporation, are headquartered within the state. Intercity rail
passenger service is provided by AMTRAK, which operates eight trains with
scheduled stops in Virginia. Rapid rail transit is provided by Metrorail to
commuters in the Virginia suburbs of Washington, D.C. A new component to the
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Chapter 1: Profession of Transportation
commuter rail network is the Virginia Rail Express (VRE), which operates from
Fredericksburg and Manassas to Washington, D.C. Virginia is also served by 13
airports with commercial service to over 600 worldwide destinations. Another 64
airports are licensed for public use, and the majority of these can accommodate
multi-engine aircraft. Virginia also has one of the finest natural ports in the world
as well as three inland ports. The port of Hampton Roads is served by an ice-free
50-ft. deepwater channel capable of handling large volumes of cargo. Virginia
also maintains the Virginia Inland Port, located in Front Royal, which serves the
Ohio Valley and beyond.
1-7
What is the total number of miles in your state’s highway system? What percent of
the highway system is comprised of Interstate highways?
Of Virginia’s 57,737 state-maintained miles of highways in 1999, 1,118
miles are Interstate highways, 1.936% of the state-maintained total. With 70,325
miles of public roads, Interstate highways comprise 1.590% of the total public
road mileage in Virginia.
Note: These data are available from the Internet site of FHWA’s Office of
Highway Policy Information at “http://www.fhwa.dot.gov/ohim/ohimstat.html”
1-8
Estimate the number of personal motor vehicles in your city or state. What is the
total number of miles driven each year? How much revenue is raised per vehicle
per year for each 1 cent/gallon tax? Assume that the average vehicle achieves 25
mpg.
In Virginia, as of 1999, there were 6,083,902 registered motor vehicles.
An estimated 80,197,000,000 vehicle-miles were traveled. Assume the average
vehicle has a gasoline consumption rate of 25 miles per gallon (mpg). Thus the
total amount of gasoline (TG )consumed in one year is:
TG = (miles driven/year)/(mpg)
TG = (80,197,000,000)/25
TG = 3,207,880,000 gallons
The total revenue (TR ) raised by the 1 cent/gallon tax would be:
TR = ($0.01/gallon) * TG
TR = 0.01 * 3,207,880,000
TR = $32,078,800
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Chapter 1: Profession of Transportation
1-9
How many railroad trains pass through your city each week? What percentage of
these are passenger trains?
In Charlottesville, Virginia, an average of 63 freight trains pass through
the city weekly (14 of these stop while 49 pass through). 20 Amtrak trains pass
through weekly.
Percentage passenger trains = (20/(20+63))*100 = 24%
1-10
Review the classified section of the telephone directory and identify ten
different jobs or industries that are related to transportation.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Airlines
Automobile-Dealers
Government - Department of Motor Vehicles
Department of Transportation
Insurance - Automobile
Moving Corporations
Paving Contractors
Railroads
Service Stations
Tire - Dealers
Travel Agencies
1-11
Estimate the proportion of your monthly budget that is spent on
transportation.
AVERAGE SPENDING
(dollars)
PERCENTAGE
Housing (rent and utilities)
$525
47.5%
Food
$350
31.7%
Clothing
$130
11.8%
Transportation (gas, parking,
bus, repairs, etc...)
$100
9.0%
TOTAL
$1105
100%
CATEGORY
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Chapter 1: Profession of Transportation
Approximately 9 percent of the monthly budget is dedicated to
transportation.
1-12
Identify an ITS project or application that is underway in your home state.
Describe the project, its purpose, and the way it is operated.
The Virginia Department of Transportation is implementing a traffic
management system in the Richmond region. This system includes the
installation on Interstates 64, 95, and 295, of permanent changeable message signs
installed prior to major interchanges, video detection equipment at critical
congestion locations, and highway advisory radio throughout the region. This
system is intended to monitor traffic and ultimately ease congestion associated
with major reconstruction projects on the region’s Interstate highways. The
Richmond Smart Traffic Center is located adjacent to an interchange on Interstate
95 and houses the control and communications systems for the traffic
management system.
1-13
Most Departments of Transportation incorporate at least five major transportation
engineering subspecialties within their organization. List and briefly
indicate at least three tasks falling under each specialty.
Most state departments of transportation have at least five major
transportation engineering sub-classes. Five of these subspecialties in the
Virginia Department of Transportation are the planning division, location and
design division, maintenance division, traffic engineering division, and a
construction division. The planning division is primarily concerned developing
long-range transportation plans. This is accomplished by first defining
transportation needs, gathering and analyzing data, and then evaluating
alternatives.
The location and design division is primarily concerned with designing the
transportation system. Usually with the design of highways, this division is
responsible for the selection of dimensions for all geometric features, which
include the longitudinal profiles, vertical curves and elevations, and the right-ofway.
The maintenance division is responsible for maintaining the transportation
system to ensure it is in proper working order. This includes repairing damaged
roadway sections and the scheduling of maintenance operations.
The traffic engineering division is responsible for the integration of the
vehicles, drivers, and pedestrians into the transportation system in a manner that
improves the safety and capacity of streets and highways. This includes
analyzing traffic accidents, design of parking areas, and the design of roadway
traffic signing plans.
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Chapter 1: Profession of Transportation
The construction division is another subspecialty. This division is
responsible for the building of the facilities designed by the location and design
division. The primary tasks of this division include the development of contracts
for highway construction, inspection of highway construction projects as
performed by contractors, and publishing of manuals such as road and bridge
specifications.
1-14
List four major detrimental effects that are directly related to the
construction and use of our highway transportation system.
There are many benefits that have come from the highway system;
however, these benefits have not come without significant costs. The most
obvious cost is that of safety, highway accidents claim approximately 40,000 lives
each year. The transportation system also creates air, water, and even noise
pollution. It also spoils and changes forever the natural beauty of an area, and
consumes vast quantities of precious energy resources.
1-15
Cite four statistics that demonstrate the importance of transportation in the
United States.
The following statistics illustrate the importance of the transportation sector in the
U.S. (data are as of 1997).
• 16.3 % of the United States’ Gross National Product (GNP) is accounted for
expenses related to transportation.
• Approximately 11% of the U.S. workforce is employed by transportation
industries.
• Of all the petroleum used in the U.S., almost 70% is for transportation.
• Expenditures on transportation totaled $1.32 trillion.
1-16
A state has a population of 17 million people and an average ownership of 1.5 cars
per person, each driven an average of 10,000 mi/year, at 20 mi/gal of gasoline (mpg).
Officials estimate that an additional $75 million per year in revenue will be required
to improve the state’s highway system, and they have proposed an increase in the
gasoline tax to meet this need. Determine the required tax in cents per gallon.
First, determine the number of vehicles in the state.
17,000,000 people * 1.5 cars/person = 25,500,000 vehicles in this state.
Next, determine the number of miles driven each year.
25,500,000 cars * 10,000 miles/year/car = 2.55 * 1011 miles driven per year
Now determine the number of gallons consumed each year.
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Chapter 1: Profession of Transportation
(2.55 * 1011 miles/year)/20 mpg = 12,750,000,000 gallons/year
Finally, determine the required tax increase.
(1.275 * 1010 gallons / year) * TAXINCREASE = $75,000,000 / year
TAXINCREASE = $0.00588/gallon
TAXINCREASE = 0.588 cents per gallon
Therefore, impose a 0.6 cents per gallon tax to raise the required revenue.
1-17
Select a single event in Table 1-1 and explain why this is a significant
achievement in the history of transportation.
In my opinion, the completion of the first transcontinental railroad in 1869
was the most significant transportation event in U.S. history. This is because the
completion of the railroad meant that goods and people could now be transported
with relative ease to the western part of the country. This also meant that the
development of the west would become more intense. The completion of the
railroad spawned the development of the remainder of the U.S., which facilitated
trade on both coasts.
1-18
Name and describe the first successful turnpike effort in the newly independent
United States of America.
The Philadelphia and Lancaster Turnpike Road Company was chartered
by Pennsylvania in 1791 to build road between the two cities. This serves as an
early example of a profitable toll road and a roadway with specified design
standards.
1-19
What mode of transportation was the primary contributor to the demise of road
construction in the U.S. in the early 19th century, and what advantages did the new
mode offer?
Canals became popular in the early 19th century through large projects
such as the Erie Canal and several smaller efforts. Recent improvements in
waterway transportation, such as the successful demonstration of the steamboat in
1807, generated interest in use waterways. Waterways provided advantages over
the roadways in their level profiles and relative ease of effort in moving freight
when compared with roadways.
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Chapter 1: Profession of Transportation
1-20
What mode of transportation succeeded the mode noted in Problem 1-19, and what
advantages did it offer?
Canals were succeeded by railways as the primary mode of long-distance
travel. Railroads first appeared in the U.S. around 1830. By 1840, the mileage of
railways was approximately equal that of canals. Railroads continued flourish
while investment in canals declined. Locations for railroads were not confined to
watercourses and therefore could be built almost anywhere.
1-21
The expectations the public has for the transportation system continue to increase.
What is the principal challenge faced by the transportation engineer in meeting
these expectations? What fields of knowledge beyond traditional transportation
engineering are needed?
The public increasingly expects an efficient, effective, long-lasting, and
safe transportation system. This challenge requires a knowledge base beyond
traditional transportation engineering, including an understanding of human
factors, system performance, and technological advances.
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Chapter 2
Transportation Systems and
Organizations
2-1
How would your typical day be changed without availability of your principal mode
of transportation? Consider both personal transportation as well as goods and
services that you rely on.
A typical day in my life would be significantly different without the
airplane. Although I do not use this mode daily, goods and services that I do
purchase are transported via this mode. Other modes, such as trucking, trains, and
the automobile, could serve as replacements to the airplane; however, the airplane
significantly lowers the transit time for shipping goods. For example, I mail a
letter to California. Typical transit time for this letter using the airplane is three
days. By using another mode other than the airplane, the transit time for the same
letter would probably exceed seven days. As for my personal transportation,
long distance travel is accomplished by using the airplane. For example, I take a
vacation to Europe. If I travel using a cruise ship, it would take me in excess of
seven days to reach Europe. However, if I fly, I can arrive in Europe within nine
hours. Having the ability to transport people and goods quickly allows the
international trade market to prosper, which in turn provides me with goods in a
timely and efficient manner.
2-2
What are the most central problems in your state concerning one of the following:
(a) air transportation, (b) railroads, (c) water transportation, (d) highways, or (e)
public transportation. (To answer this question, obtain a copy of the governor’s
plan for transportation in your state or contact a key official in the transportation
department.)
(a)
A problem in Virginia concerning air transportation is the high cost
associated with short haul flights from airports such as Richmond and Norfolk to
connection hubs for major airlines. Another problem is that our air transportation
system is aging while the demand continues to increase; our air transportation
system is approaching capacity and requiring substantial capital investment to
provide modern terminals, increase the number of gates and available parking.
(b)
Virginia is experiencing a new dilemma with its railroads. For the first
time in nearly 30 years, freight railroads are expanding their operations and
growing to serve their market segment. To continue to compete with other
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Chapter 2: Transportation Systems and Organizations
railroads in neighboring states, Virginia must investigate the possibility of
providing rail clearances to facilitate double-stacking of containers into the Port
of Hampton Roads. Another problem associated with the increase in freight rail
transportation is the conflict encountered with passenger trains running on freight
company-owned tracks. As the demand for passenger rail service increases and
the freight market share increases, more conflicts will likely occur and the
passenger services may require parallel or additional track mileage to meet
demand.
(c)
The most central problem concerning water transportation in Virginia is
the increased build-up of silt in our channels. In order for Virginia to remain
competitive, it will have to continue to dredge our navigable waterways. Another
problem is the increased volume of pleasure crafts and cargo vessels. The
increased interaction between these types of vessels will likely result in more
serious accidents. To mitigate this, more boater safety classes should be provided
to ensure all boat operators are responsible on the water.
(d)
Virginia's highways are experiencing increased volumes and delays while
the overall infrastructure is continuing to age. The volume of trucks on Virginia's
highways are significantly increasing annually. As a result, Virginia is
experiencing an accelerated deterioration of our highways as well as more serious
accidents.
(e)
The major problem concerning public transportation is that modern
systems such as the ones in Atlanta and San Francisco are not present in Virginia.
Only Northern Virginia and the suburbs of Washington, D.C. have rapid rail
transit in form of the Metro system that is now facing major renovations. Virginia
does not have a sophisticated rural public transportation system that provides all
individuals with a means of transportation.
2-3
A bridge has been constructed between the mainland and an island. The total cost
(excluding tolls) to travel across the bridge is expressed as C = 50 + 0.5V, where V is
the number of veh/hr and C is the cost/vehicle in cents. The demand for travel
across the bridge is V = 2500 −10C .
(a)
(b)
(c)
(d)
Determine the volume of traffic across the bridge.
If a toll of 25 cents is added, what is the volume across the bridge?
A tollbooth is to be added, thus reducing the travel time to cross
the bridge. The new cost function is C = 50 + 0.2V. Determine the
volume of traffic that would cross the bridge.
Determine the toll to yield the highest revenue for demand and
supply function in part (a), and the associated demand and
revenue.
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Chapter 2: Transportation Systems and Organizations
(a)
Determine the volume of traffic across the bridge.
Substitute the total cost function into the demand function and solve for V.
V = 2500 -10(50 + 0.5V)
V = 2500 - 500 - 5V
6V = 2000
V = 333.33 vehicles/ hour
Therefore, the number of vehicles wanting to cross this bridge is 334
vehicles/hour.
(b)
If a 25 cent toll was added, what is the new volume crossing the
bridge?
Add 25 cents to the original cost function.
C = 50 + 0.5V + 25
C = 75 + 0.5V
Substitute the above cost function into the demand function and solve for V.
V = 2500 - 10(75 + 0.5V)
V = 2500 - 750 - 5V
6V = 1750
V = 291.667
Therefore, the new volume crossing the bridge will now be 292 vehicles / hour.
(c)
An additional toll booth changed the cost function to C = 50 + 0.2V.
Determine the new volume of vehicles wanting to cross this bridge.
Substitute the new cost function into the demand function and solve for V.
V = 2500 -10 (50 + 0.2V)
V = 2500 - 500 - 2V
3V = 2000
V = 666.67 vehicles/ hour
Therefore, the new number of vehicles wanting to cross this bridge is 667
vehicle/hour.
(d)
Determine the toll to yield the highest revenue for part a.
Assume toll rate at T. The new cost function will be C = 50 + 0.5V + T. Since
the revenue generated is the toll rate, T, time the volume, V, first solve for V with
the new cost function.
V = 2500 - 10(50 + 0.5V + T)
V = 2500 - 500 - 5V - 10T
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Chapter 2: Transportation Systems and Organizations
V = (2000 - 10T) / 6
Since the revenue generated is R = T * V , substitute the above expression into the
revenue formula and differentiate with respect to T.
R = T * ((2000 - 10T) / 6)
R = (2000T - 10T2) / 6
dR/dT (2000T - 10T2) / 6 = 0
(2000 - 20T) / 6 = 0
Therefore, the toll which would yield the maximum revenue is T = 100, or T =
$1.00.
R=T*V
R = (2000T - 10T2) / 6
R = (2000(100) - 10(100)2) / 6
R = 16,666.67
Therefore, a toll of $1.00 will yield a revenue of $166.67 per hour.
2-4
A toll bridge carries 10,000 veh/day. The current toll is $3.00/vehicle. Studies have
shown that for each increase in toll of 50 cents, the traffic volume will decrease by
1000 veh/day. It is desired to increase the toll to a point where revenue will be
maximized.
(a)
Write the expression for travel demand on the bridge, related to
toll increase and current volume.
(b) Determine toll charge to maximize revenues.
(c)
Determine traffic in veh/day after toll increase.
(d)
Determine total revenue increase with new toll.
(a)
Write the expression for travel demand on the bridge. Let V = travel
demand.
V = 10000 - 1000(x / 50)
(b)
Determine the toll charge to maximize revenues. Let T = toll charge.
Since the original toll was 300 cents per vehicle, the new toll charge will be
T = 300 + x
The revenue (R) is generated by the equation R = V * T. Substitute the above
expressions into the revenue function and differentiate with respect to x, setting
the derivative equal to zero.
R = (10000 - 1000(x / 50)) * (300 + x)
R = (10000 - 20x) * (300 + x)
R = 3000000 + 10000x - 6000x - 20x2
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Chapter 2: Transportation Systems and Organizations
dR/dx (3000000 + 4000x - 20x2) = 0
4000 - 40x = 0
x = 100
Therefore, an increase in toll of 100 cents will maximize revenues.
(c)
Determine the traffic volume per day after the toll increase.
Now, substitute the new toll, x, into the demand function developed in part a.
V = 10000 - 1000(x / 50)
V = 10000 - 1000(100/50)
V = 10000 - 2000
V = 8000 vehicles per day
The new demand for the bridge will be 8,000 vehicles per day.
(d)
Determine the total revenue with the new toll.
R=V*T
R = 8000 * (300 + 100)
R = 8000 * 400
R = $3,200,000
The total revenue to be generated with the new toll will be $3,200,000 per day.
2-5
Consideration is being given to increasing the toll on a bridge now carrying 4500
veh/day. The current toll is $1.25/veh. It has been found from past experience that
the daily traffic volume will decrease by 400 veh/day for each 25¢ increase in toll.
Therefore, if x is the increase in toll in cents/veh, the volume equation for veh/day is
V = 4500 − 400 ( x / 25) , and the new toll/veh would be T = 125 + x. In order to
maximize revenues, what would the new toll charge be per vehicle and what would
the traffic in veh/day be after the toll increase?
First, solve for the revenue to be generated by the new toll.
R=V*T
R = (4500 - 400(x / 25)) * (125 + x)
R = (562,500 + 4500x - 2000x - 16x2
dR/dT (562,500 + 2500x - 16x2) = 0
32x = 2500
x = 78.125
Therefore, a toll increase of 78.125 cents per vehicle will maximize revenues for
the bridge. For practical purposes and traveler convenience, round the toll
increase to 75 cents.
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Chapter 2: Transportation Systems and Organizations
Next, determine the resulting volume after the new toll increase. Simply
substitute the new toll into the demand function above.
V = 4500 - 400(75/25)
V = 4500 - 1200
V = 3,200 vehicles per day
An increase in toll of 75 cents per vehicle will result in a new demand for the
bridge of 3,200 vehicles per day.
2-6
A large manufacturer uses two factors to decide whether to use truck or rail for
movement of its products to market: cost and total travel time. The
manufacturer uses a utility formula that rates each mode. The formula is
U = 5C + 10T, where C is cost ($/ton) and T is time (hrs). For a given
shipment of goods, a trucking firm can deliver in 16 hrs and charges
$25/ton, whereas a railroad charges $17/ton and can deliver in 25 hrs.
(a)
Which mode should the shipper select?
(b) What other factors should the shipper take into account in making
a decision? (Discuss at least two.)
(a)
Which mode should the shipper select?
Let Utruck be the (dis)utility function for the trucks and Urail the (dis)utility
function for the railroad.
Utruck = 5(25) + 10(16)
Utruck = 285
Next solve the utility formula for shipping via the railroad.
Urail = 5(17) + 10(25)
Urail = 335
Based on the results of the above utility formula, the shipper should ship his
goods by truck since Utruck < Urail.
(b)
List at least two other factors that shippers should take into
consideration when choosing modes to ship products by.
1.
2.
3.
Reliability: Does the mode consistently operate on schedule?
Convenience: Which mode can deliver the freight to a serviceable
location?
Security: Which mode reduces the risk of pilfering.
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Chapter 2: Transportation Systems and Organizations
4.
Rideability: Which mode provides the best ride for the product? In other
words, which mode is less likely to cause damage to the product while in
transit?
2-7
An individual is planning to take an 800-mile trip between two large cities. Three
possibilities exist: air, rail, or auto. The person is willing to pay $25 for every hour
saved in making the trip. The trip by air costs $600 and travel time is 8 hrs, by rail
the cost is $450 and travel time is 16 hrs, and by auto the cost is $200 and travel time
is 20 hrs.
(a) Which mode is the best choice?
(b) What factors other than cost might influence the decision
regarding which mode to use?
Determine the total cost (initial cost plus time cost) for each mode.
Total Cost(air) = 600 + (8 * 25)
Total Cost(air) = $800
Total Cost(rail) = 450 + (16 * 25)
Total Cost(rail) = $850
Total Cost(auto) = 200 + (20 * 25)
Total Cost(auto) = $700
From the above analysis, it appears that the best mode to choose to make this trip
is automobile. Other factors to consider, other than costs, when selecting a mode
to travel might include the following: personal comfort the modes have to offer,
whether additional connections need to be made to reach the final destination, the
level of stress that can be anticipated by traveling by that mode, or whether the
reason for travel is for business or pleasure.
2-8
Name the two key influences on transit system carrying capacity.
Carrying capacity is influenced by headway (the “spacing” in seconds
between each vehicle, and (2) level-of-service (the “comfort factor” experienced
by passengers.
17
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 2: Transportation Systems and Organizations
2-9
What factors affect the long-term viability of fuel taxes as a stable source of revenue
to fund highway system improvements?
The long-term viability of the fuel tax is affected by crude oil shortages,
improved automobile efficiency, use of alternative energy sources, and shifts to
public transportation. Additionally, the availability of revenue is reduced by
diversion of fuel taxes to support transit, ethanol, and government general funds.
2-10
What emerging concepts for financing highway improvements are currently being
explored?
Several financing concepts, including road use metering, electronic toll
collection, and value pricing, may help to offset ineffectiveness in the fuel tax as
vehicles become more fuel-efficient. Road use metering involves charging
highway users according to consumption (vehicle-miles traveled), similar to many
utilities. Electronic tolling is a related concept that can be applied on toll roads.
Value pricing involves payment for values of service rendered as a function fo
demand.
2-11
Describe the organization and function of your state highway/transportation
department.
Under the Governor, responsibility for the general administration of
Virginia's government is distributed among eight cabinet secretaries, one of whom
is the Secretary of Transportation. The Secretary of Transportation is empowered
to oversee Virginia's transportation program. The secretary is also the
chairperson of the Commonwealth Transportation Board, a 16 member policy
board that functions as a board of directors to (i) oversee the construction of
highways and make regulations governing the use of state highways, (ii) ensure
compliance with transportation-related federal laws, (iii) collect transportation
statistics, (iv) regulate the location of outdoor advertising, (v) oversee the
administration of the Transportation Trust Fund, and (vi) generally oversee the
operation of the Virginia Department of Transportation (VDOT). VDOT is
headed by a Commissioner, who is also appointed by the Governor. VDOT's core
function is to construct and maintain the roadways of Virginia. This includes the
daily maintenance and repair, design and engineering of future road projects, and
the long range planning based on future demand projections. It is VDOT's
responsibility to keep the roadways in good working condition throughout the
year. This organization has its central headquarters located in Richmond,
Virginia, and nine other District offices strategically located in other areas of the
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Chapter 2: Transportation Systems and Organizations
state. These District offices employ individuals to maintain the roads and the
right-of-way year round, which includes the mowing of grass in the summer and
plowing snow in the winter. In addition to the roadways, VDOT is also
responsible for all of the roadway signs, signals, and street lighting.
2-12
What are the major activities performed by the highway department in your state
as described by the organization chart and other information furnished on their
website?
The activities focus around the phase of transportation project
development: planning, design, construction, operations, and maintenance.
Organizational structure supports these responsibilities.
2-13
Consult with the U.S. Department of Transportation website and identify the name
and location of highways in your state that are included as part of the National
Highway System.
Virginia has approximately 3,480 miles of the NHS. This includes 1,118
miles of Interstates, including 64, 77, 81, 85, 95, and associated spurs. Also
included are other principal arterials and strategic connectors, such as U.S. 29 and
460.
2-14
List three transportation organizations located in your state. What services do they
provide?
Charlottesville Transit Service: Provides local bus service to the residents of
Charlottesville.
United Airlines: Provides passenger airline service to several cities outside the
state.
Norfolk Southern Corporation: Provides freight rail service throughout the
eastern U.S.
19
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 2: Transportation Systems and Organizations
2-15
Obtain a copy of a Transportation Research Record (published by the
Transportation Research Board) or a CD-ROM of papers from an annual meeting
of TRB.
(a) Select one article and write a short summary of its contents.
(b) Describe the technical area of transportation covered by this article.
Title: Quality Assurance of Hot-Mix Asphalt Comparison of Contractor Quality
Control and Georgia Department of Transportation Data
Authors: Rod E. Turochy, J. Richard Willis, and Frazier Parker
Source: Transportation Research Record: Journal of the Transportation Research
Board No. 1946, pp. 47-54 (2006)
Abstract: Quality assurance is the process by which highway construction
elements are sampled and tested to ensure compliance with specifications and
other project requirements. The results of contractor-performed tests, originally
performed for quality control purposes, are increasingly used in the acceptance
decision in many states. The Georgia Department of Transportation (GDOT) uses
contractor-performed tests in the acceptance decision on acceptable corroboration
of GDOT-performed tests. Statistical analyses were performed to assess
differences between tests conducted on hot-mix asphalt concrete by GDOT and its
contractors during the 2003 construction season. Measurements of gradation and
asphalt content taken by both parties were compared both across all projects and
on a project-by-project basis for projects large enough to meet sample size
requirements for this type of analysis. Both tabular and graphic representations of
data are used to interpret the results. Statistically significant differences occur in
some cases; these differences are much more common when comparing
variability of these measurements than with the means. At the project level, on
most projects in which statistically significant differences occur, the GDOT value
typically is larger.
2-16
What do the following acronyms mean?
AAA, AAR, AASHTO, AI, APTA, ARTBA, FHWA, PCA, TRB
Acronym
AAA
AAR
AASHTO
AI
APTA
ARTBA
FHWA
PCA
TRB
Definition
American Automobile Association
Association of American Railroads
American Association of State Highway and Transportation
Officials
Asphalt Institute
American Public Transit Association
American Road and Transportation Builders Association
Federal Highway Administration
Portland Cement Association
Transportation Research Board
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
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Chapter 2: Transportation Systems and Organizations
2-17
List the seven categories of transportation organizations and cite one example of
each.
There are several organizations and associations active in transportation.
There are seven categories of these organizations. The categories are listed below
along with an example of an organization in each category:
PRIVATE TRANSPORTATION COMPANIES
•
United Airlines
REGULATORY AGENCIES
•
Federal Motor Carrier Safety Administration
FEDERAL AGENCIES
•
U.S. Army Corps of Engineers
STATE AND LOCAL AGENCIES AND AUTHORITIES
•
Virginia Department of Rail and Public Transportation
TRADE ASSOCIATIONS
•
Association of American Railroads
PROFESSIONAL SOCIETIES
•
Institute of Transportation Engineers
CONSUMER ASSOCIATIONS
•
American Automobile Association
2-18
What are the four principal modes for moving freight? Which of these modes
carries the largest share of ton-miles? Which carries the lowest?
The four principal modes for carrying freight are highways, railroads,
water, and pipeline. Of these four modes, railroads carry the highest share of tonmiles, while water transportation carries the lowest share of ton-miles.
21
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 2: Transportation Systems and Organizations
2-19
What are the four principal modes for moving people? Which of these modes
accounts for the largest share of passenger-miles? Which mode accounts for the
lowest?
Air, automobile, bus, and rail are the four principal modes for moving
people. Of these four modes, the automobile accounts for the highest share of
passenger-miles. Conversely, the lowest share of passenger-miles is associated
with the rail mode.
2-20
(a) List four major factors that will determine the future of public transportation in
the United States.
(b) Indicate if the factor is positive, neutral or negative to the success of transit.
According to the TRB report, A Look Ahead: Year 2020, four key trends
that will affect the future of public transportation in the United States are (1)
increasing suburb-to-suburb commuting, (2) increasing legislation to encourage
“livable cities” and “smart growth”, (3) increasing emphasis on improving air
quality, and (4) increasing use of teleworking. Suburb-to-suburb commuting is
difficult for public transit systems to accommodate and therefore will have a
negative impact on mass transit. Popularity of “livable cities” and “smart growth”
will have a positive impact on mass transit, as will an increasing focus on
improving air quality. The reduction in travel associated with teleworking will
have a neutral impact on mass transit.
2-21
What are the advantages and disadvantages of using intercity bus transportation?
There are several advantages and disadvantages of using intercity bus
transportation. The advantages to using this mode are that it is highly energy
efficient. To demonstrate, this mode achieves nearly 300 seat-miles per gallon of
fuel consumed. In addition to its energy efficiency, this mode is very safe. It has
a relatively low crash rate of 12 fatalities per 100 billion passenger miles.
This mode also has disadvantages to using it. For the most part, it is slow
in comparison to other modes. Intercity bus transportation is less convenient, it
lacks through ticketing, less comfortable seats, and its terminating points are
usually located in downtown locations in less active parts of the city.
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
22
Chapter 3
Characteristics of the Driver, the
Pedestrian, the Vehicle and the Road
3-1
Briefly describe the two types of visual acuity.
The two types of visual acuity are static and dynamic. The ability of a
driver to identify an object when both the object and the driver are stationary
depends on one's static acuity. Some factors that affect static acuity include the
background brightness, contrast, and time. The ability of a driver to clearly detect
a moving object depends on the driver's dynamic visual acuity.
3-2
(a) What color combinations are used for regulatory signs (e.g. speed limit signs)
and for general warning signs (e.g. advance railroad crossing signs)
(b) Why are these combinations used?
Regulatory signs use a color combination of black lettering on white
background, and advance warning signs use the color combination of black
lettering on yellow background. These color combinations are used because they
have been shown to be those to which the eye is most sensitive.
3-3
Determine your average walking speed.
Compare your results with that of the suggested walking speed in the MUTCD.
Which value is more conservative and why?
Pass #
Intersection Width (ft)
Walk Time (sec)
Walking Speed (ft/sec)
1
36
7
5.1
2
36
8.2
4.4
3
36
9.5
3.8
4
36
7.6
4.7
5
36
8
4.5
Average
4.5
23
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
In this case, the MUTCD value, 4.0 ft/sec, is more conservative than the
observed speeds. This value is more conservative because it is a slower speed and
it will allow most slower people, such as elderly, individuals with small children,
and handicapped individuals to traverse the intersection safely.
3-4
Describe the three types of vehicle characteristics.
The three types of vehicle characteristics are static, kinematic, and
dynamic. Static vehicle characteristics include the vehicle's weight and size.
Kinematic characteristics involve the motion of the vehicle, and dynamic
characteristics involve the force that causes the motion of the vehicle.
3-5
Determine the maximum allowable overall gross weight of WB-20 Design Vehicle.
(The WB-20 is same as WB-65 and WB-67.)
From Table 3.2, the extreme distance between the axle groups is 43.4 – 45.4 ft
(use 45.4 in this case). The number of axles in the group is 4.
Use Eq. 3.2,
⎡ LN
⎤
⎡ 45.4 × 4
⎤
W = 500 ⎢
+ 12 N + 36 ⎥ = 500 × ⎢
+ 12 × 4 + 36 ⎥ = 72267lb .
⎣ N −1
⎦
⎣ 4 −1
⎦
The maximum allowable overall gross weight is 72267 lb.
3-6
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping
sight distance that should be provided on the road if (a) the road is level and (b) the
road has a maximum grade of 4%? Assume the perception-reaction time = 2.5 sec.
The minimum sight distance required in these cases is the stopping sight distance
(SSD), given by Equation 3.27:
u2
;
a
30( ± G )
g
where u = design speed (mi/h)
t = perception-reaction time (sec)
a = rate of deceleration (taken as 11.2 ft/sec2)
S = 1.47ut +
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
g = gravitational acceleration (taken as 32.2 ft/sec2)
G = grade
a
is typically rounded to 0.35 in calculations.
Note: the term
g
(a)
Determine the minimum sight distance that should be provided for a
level roadway.
Since the roadway is level, G = 0.
S = (1.47)(60)(2.5) + (60)2 / (30)(0.35+0)
S = 220.50 + 345.00
S = 565.50 feet
Therefore, the minimum sight distance for this horizontal roadway is 565 feet.
(b)
Determine the minimum sight distance that should be provided for a
roadway with a maximum grade of 4 percent.
Since this roadway has a maximum grade of 4 percent, G = -0.04. The
downgrade (negative) case provides the most conservative (higher) value for
design.
S = 1.47(60)(2.5) + (60)2 / 30(0.35 – 0.04)
S = 220.50 + 387.10
S = 607.60 feet
Therefore, the minimum sight distance for this roadway should be 608 feet.
3-7
The acceleration of a vehicle can be expressed as:
du
= 3.6 − 0.06u
dt
If the vehicle speed, u, is 45 ft/sec at time T0, Determine:
(a) Distance traveled when the vehicle has accelerated to 55 ft/sec.
(b) Time for vehicle to attain the speed of 55 ft/sec.
(c) Acceleration after 3 seconds.
(a)
Determine the distance traveled by the vehicle when accelerated to 55
ft/sec.
First, determine the time it took for the vehicle to accelerate to 55 ft/sec.
Using a rearrangement of Equation 3.10:
-βt = ln [(α - βut) / (α - βu0)]
t = (-1/β) ln [(α - βut) / (α - βu0)]
25
therefore;
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
t = (-1/0.06) ln [(3.6 - 0.06(55)) / (3.6 - 0.06(45))]
t = 18.33 seconds
Next, determine the distance traveled during this time.
From Equation 3.12:
x = (α / β)t - (α / β2) (1 - e-βt) + (uo / β) (1 - e-βt)
x = (3.6 / 0.06)18.33 - [(3.6 /(0.06)2)(1 - e-0.06(18.33))] + [45 / 0.06 (1 - e0.06(18.33)
)]
x = 933.03 feet.
Therefore, the vehicle traveled 933 feet when accelerating from 45 ft/sec to 55
ft/sec.
(b)
Determine the time it takes for the vehicle to attain the speed of 55
ft/sec.
This time was determined in Part (a) of this problem.
Therefore, it took 18.33 seconds for the vehicle to attain the speed of 55 ft/sec.
(c)
Determine the acceleration of the vehicle after 3 seconds.
First, determine the velocity of the vehicle after 3 seconds.
From Equation 3.11:
ut = (α / β) (1 - e-βt) + uoe-βt
ut = (3.6 /0.06)(1 - e-(0.06)(3)) + 45e-(0.06)(3)
ut = 47.47 ft/sec.
Since acceleration is: a = du/dt = 3.6 - 0.06u
a = 3.6 - 0.06(47.47)
a = 0.75 ft/sec2
Therefore, the acceleration of the vehicle after 3 seconds is 0.75 ft/sec2
3-8
The gap between two consecutive automobiles (distance between the back of a
vehicle and the front of the following vehicle) is 65 ft. At a certain time the front
vehicle is traveling at 45 mph and the following vehicle at 35 mph. If both vehicles
start accelerating at the same time, determine the gap between the two vehicles after
15 sec if the acceleration of the vehicles can be assumed to take the following forms:
du
= 3.4 − 0.07ut (leading vehicle)
dt
du
= 3.3 − 0.065ut (following vehicle)
dt
where ut is the vehicle speed in ft/sec.
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
First, determine the distance each vehicle travels during the elapsed time (15
seconds) using Equation 3.12. For the leading vehicle:
x = (α / β)t - (α / β2) (1 - e-βt) + (uo / β) (1 - e-βt)
x = (3.4 / 0.07)15 - [(3.4 /(0.07)2)(1 - e-0.07(15))] + [((45)(1.47) / 0.07) (1 - e-0.07(15))]
x = 890 feet.
Similarly, for the following vehicle:
x = (3.3 / 0.065)15 - [(3.3 /(0.065)2)(1 - e-0.065(15))] + [((45)(1.47) / 0.065) (1 - e0.065(15)
)]
x = 767 feet.
Since the leading vehicle traveled further, the gap between vehicles increased by
the difference in the distances, 890 – 767 = 123 ft
The initial gap was given as 65 ft, so after 15 sec, the gap is 65 + 123 = 188 ft.
3-9
The driver of a vehicle on a level road determined that she could increase her speed
from rest to 50 mi/hr in 34.8 sec and from rest to 65 mi/hr in 94.8 sec. If it can be
assumed that the acceleration of the vehicle takes the form:
du
= α − βut
dt
determine the maximum acceleration of the vehicle
First, convert miles/hour to feet/second.
50 mi/h = 73.33 ft/sec
65 mi/h = 95.33 ft/sec
Next, use Equation 3.11 to develop the equation for each case as follows:
Case 1 (50 mi/h)
ut = (α / β) (1 - e-βt) + uoe-βt
73.33 = (α / β) (1 - e-β(34.8)) + 0e-β(34.8)
73.33 = (α / β) (1 - e-β(34.8))
Equation 1
Case 2 (65 mi/h)
ut = (α / β) (1 - e-βt) + uoe-βt
95.33 = (α / β) (1 - e-β(94.8)) + 0e-β(94.8)
95.33 = (α / β) (1 - e-β(94.8))
Equation 2
Solve for α in equation 1 and substitute into equation 2.
Equation 3
α = 73.33β / (1 - e-β(34.8))
Substitute this into Equation 2.
95.33 = [(73.33β / (1 - e-β(34.8))) / β) * (1 - e-β(94.8))
27
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
95.33 = (73.33 / (1 - e-β(34.8))) * (1 - e-β(94.8))
(95.33 / 73.33) * (1 - e-β(34.8)) = (1 - e-β(94.8))
1.30 * (1 - e-β(34.8)) = (1 - e-β(94.8))
1.30 - 1.3e-β(34.8) = (1 - e-β(94.8))
1.3e-β(34.8) - e-β(94.8) = 0.30
The above equation can be solved by assuming a value for β, evaluating the left
hand side of the equation, and comparing the result to 0.3 as follows.
β
0.02
0.03
0.04
1.3 e - β (34.8) - e - β (94.8)
0.5
0.4
0.3
As can be seen, the solution is:
β = 0.04
Substituting into Equation 3 gives:
α = 73.33 * 0.04 / (1 - e (-0.04 * 34.8) )
= 3.90 ft/sec2
Since the maximum acceleration is achieved when the velocity of the vehicle is 0,
the value for α determined above is the maximum acceleration.
3-10
If the vehicle in Problem 3-9 is traveling at a speed of 40 mph, how long will it take
after the driver starts accelerating for the vehicle to achieve a speed of 45 mph?
The acceleration model found in Problem 3-9 was
du/dt = 3.90 – 0.04 ut
The values α=3.90 and β=0.04 can then be substituted in Equation 3.10:
t = (-1/β) ln [(α - βut) / (α - βu0)]
t = (-1/0.04) ln [(3.9 – ((0.04)(40)(1.47)) / (3.9 – ((0.04)(45)(1.47))]
t = 5.2 seconds
3-11
Determine the horsepower developed by a passenger car traveling at a speed of 60
mph on an upgrade of 4% with a smooth pavement. The weight of the car is 4500 lb
and the cross-sectional area of the car is 45 ft2.
First, determine all of the resistive forces, air, rolling, and grade, acting on the
vehicle, and then determine its horsepower requirement.
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
The air resistance is (using Equation 3.13):
Ra = 0.5 * [(2.15pCDAu2) / g]
Ra = 0.5 * [(2.15(0.0766)(0.4)(45)(60)2 / 32.2]
Ra = 165.7 lb
The rolling resistance is (using Equation 3.14):
Rr = (Crs + 2.15Crvu2) * W
Rr = (0.012 + 2.15(0.65*10-6)(60)2) * 4500
Rr = 76.6 lb
The grade resistance is:
RG = WG
RG = (4500)(0.04)
RG = 180.0 lb
Since these are the only forces acting on the vehicle, one can now determine the
horsepower requirement. Use Equation 3.17.
P = 1.47Ru / 550
P = [(1.47(165.7 + 76.6 + 180)(60)) / 550]
P = 67.5 hp
3-12
Repeat Problem 3-11 for a 24,000-lb truck with a cross-sectional area of 100 ft2 and
coefficient of drag of 0.5 traveling at 50 mi/hr.
First, determine all of the resistive forces, air, rolling, and grade, acting on the
vehicle, and then determine its horsepower requirement.
The air resistance is (using Equation 3.13)::
Ra = 0.5 * [(2.15pCDAu2) / g]
Ra = 0.5 * [(2.15(0.0766)(0.5)(100)(50)2 / 32.2]
Ra = 319.7 lb
The rolling resistance is (using Equation 3.15):
Rr = (Ca + 1.47Cbu) * W
Rr = (0.02445 + 1.47(0.00044)(50)) * 24,000
Rr = 1361 lb
The grade resistance is:
RG = WG
RG = (24,000)(0.04)
RG = 960.0 lb
29
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
Since these are the only forces acting on the vehicle, one can now determine the
horsepower requirement. Use Equation 3.17.
P = 1.47Ru / 550
P = [(1.47(319.7 + 1361 + 960)(50)) / 550]
P = 352 hp
3-13
A 2500-lb passenger vehicle originally traveling on a straight and level road gets
onto a section of the road with a horizontal curve of radius=850 ft. If the vehicle was
originally traveling at 55 mi/h, determine (a) the additional horsepower on the curve
the vehicle must produce to maintain the original speed, (b) the total resistance force
on the vehicle as it traverses the horizontal curve, and the total horsepower. Assume
that the vehicle is traveling at sea level and has a front cross-sectional area of 30 ft2.
First, determine all of the resistive forces, air and rolling, acting on the vehicle
while it is traveling straight and then determine its horsepower requirement.
The air resistance is (using Equation 3.13):
Ra = 0.5 * [(2.15pCDAu2) / g]
Ra = 0.5 * [(2.15(0.0766)(0.4)(30)(55)2 / 32.2]
Ra = 92.74 lb
The rolling resistance is (using Equation 3.14):
Rr = (Crs + 2.15Crvu2) * W
Rr = (0.012 + 2.15(0.65*10-6)(55)2) * 2500
Rr = 40.57 lb
Since these are the only forces acting on the vehicle, one can now determine the
horsepower requirement on the straight segment. Use Equation 3.17.
P = 1.47Ru / 550
P = [(1.47(92.74 + 40.57)(55)) / 550]
P = 19.60 hp
Now determine the additional resistive force acting on the vehicle due to the
curve.
The curve resistance is (using Equation 3.16):
Rc = 0.5 * [(2.15u2W) / gR]
Rc = 0.5 * [(2.15(55)2(2500) / (32.2)(850)]
Rc = 297.03 lb
Now determine the total additional horsepower for the curve section of roadway.
P = 1.47Ru / 550
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
P = [(1.47(297.03)(55)) / 550]
P = 43.66 hp
Determine the additional required for the vehicle to maintain its original speed
hprequired = hpcurve + hpstraight
hprequired = 43.66 + 19.60
hprequired = 63.26 hp
Therefore, the vehicle will need to produce 63.26 more horsepower to traverse the
curve at its original velocity.
The total resistive force acting on the vehicle while in the curve is:
Rtotal = Ra + Rr + Rc
Rtotal = 92.74 + 40.57 + 297.03
Rtotal = 430.34 lb
Therefore, the total resistive force acting on the vehicle in the curve is 430
pounds.
3-14
A horizontal curve is to be designed for a section of a highway having a design speed
of 60 mi/hr.
(a) If the physical conditions restrict the radius of the curve to 500 ft, what
value is required for the superelevation at this curve?
(b) Is this a good design?
(a) Determine required superelevation
First, determine the coefficient of side friction, fs, from Table 3.3.
fs = 0.12
Next, use equation 3.34 and solve for the superelevation value.
R = u2 / 15(e + fs)
e = [u2 / 15R] - fs
e = [(60)2 / 15(500)] - 0.12
e = 0.36
(b) Is this a good design?
The superelevation for this curve would be 0.36. Since e = 0.36 > 0.10 (allowable
maximum superelevation, this would NOT be a good design.
31
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
3-15
Determine the minimum radius of a horizontal curve required for a highway if the
design speed is 70 mi/hr and the superelevation rate is 0.08.
Determine the minimum radius required for this section of roadway.
First, determine the coefficient of side friction, fs, from Table 3.3. fs = 0.10
Next, use Equation 3.34 to solve for R.
R = u2 / 15(e + fs)
R = [(70)2 / 15(0.08 + 0.10)
R = 1,814.81 feet.
The minimum radius for this curved section of roadway was found to be
approximately 1,815 feet.
3-16
The existing posted speed limit on a section of highway is 55 mph and studies have
shown that that the current 85th percentile speed is 65 mph. If the posted speed limit
is to be increased to the current 85th percentile speed, what should be the increase in
the radius of a curve that is just adequate for the existing posted speed limit?
Assume a superelevation rate of 0.08 for the existing curve and for the redesigned
curve.
For the existing curve, use Equation 3.34 to determine the radius.
R = u2/15(e+fs)
R = (55)2/15(0.08+0.13)
R = 960 ft
Similarly, determine the radius for the curve to be redesigned.
R = (65)2/15(0.08+0.11)
R = 1482 ft
The increase should then be 1482 – 960 = 522 ft
3-17
The radius of a horizontal curve on an existing highway is 750 ft. The superelevation
rate at the curve is 0.08 and the posted speed limit on the road is 65 mi/h. Is this a
hazardous location? If so, why? What action will you recommend to correct the
situation?
Assume that the design speed of this section of roadway is 70 mi/h (5 mi/h above
the posted speed limit.
Next, determine the coefficient of side friction, fs, from Table 3.3. fs = 0.10
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
Next, determine the maximum permissible speed on this existing curve by using
Equation 3.34.
R = u2 / 15(e + fs)
u2 = 15(R)(e + fs)
u = [15(750)(0.08 + 0.10)]1/2
u = 45 mi/h
This curve is hazardous since the speed limit is posted at 65 mi/h yet the
maximum safe speed in the curve is 45 mi/h. One low cost measure to increase
the safety of this curve would be to reduce the speed limit to 45 mi/h, or to post a
curve warning sign with an advisory (maximum safe) speed of 45 mi/h. A longterm solution to improve safety would be to increase the radius of curvature to
permit safe operation at the speed limit. This can be accomplished by using the
above equation (equation 3.34):
R = (65)2 / 15(0.08 +0.10)
R = 1,564.81 ft.
Therefore, to permit safe travel at the maximum speed limit, the radius of the
curve should be increased to 1,565 feet.
3-18
A section of highway has a superelevation of 0.05 and a curve with a radius of only
300 ft. What speed limit will you recommend at this section of the highway?
A trial value for u must be assumed and the corresponding fs found and then
checked for safety. Using Equation 3.34, solve for the value of fs associated with
u = 35 mi/h
R = u2 / 15(e + fs)
300 = 352 / 15(0.05 + fs)
fs = 0.22
Interpolating in Table 3.3, for u =35 mi/h, fs = 0.18 should be assumed.
Therefore, try a lower speed of u = 30 mi/h. Using Equation 3.34, fs = 0.15,
which is less than the assumed to be provided value of 0.20. Therefore,
the speed limit that should be posted on this roadway is 30 mi/h.
33
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
3-19
A curve of radius 250 ft and e = 0.08 is located at a section of an existing rural
highway, which restricts the safe speed at this section of the highway to 50% of the
design speed. This drastic reduction of safe speed resulted in a high accident rate at
this section. To reduce the accident rate, a new alignment is to be designed with a
horizontal curve. Determine the minimum radius of this curve if the safe speed
should be increased to the design speed of the highway. Assume fs = 0.17 for the
existing curve, and the new curve is to be designed with e = 0.08.
First, determine the safe speed on the existing curve, using Equation 3.34.
R = u2 / 15(e + fs)
u2 = 15(R)(e + fs)
u = [15(250)(0.08 + 0.17)]1/2
u = 30.62 mi/h
Next, determine the design speed of the existing roadway. (Remember, the
existing speed is the design speed reduced by 50%.)
Design speed = Existing speed / 0.50
Design speed = 30.62 / 0.50
Design speed = 61.24 mi/h
Next, determine the coefficient of friction, fs, from Table 3.3 using the new design
speed.
fs = 0.12
Now, determine the new radius for this curve.
R = u2 / 15(e + fs)
R = (61.24)2 / 15(0.08 + 0.12)
R = 1,250.11 feet.
Therefore, to permit the safe speed of the curve to be raised to the overall design
speed, the curve's radius must be increased to 1,250 feet.
3-20
What is the distance required to stop an average passenger car when brakes are
applied on a 2% downgrade if that vehicle was originally traveling at 40 mi/hr?
Use equation 3.25 to determine the braking distance.
u2
Db =
a
30( ± G )
g
Note that a is taken as 11.2 ft/sec2; therefore, a/g is equal to 0.35.
Db = 402 / 30(0.35 - 0.02)
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
Db = 161.62 feet
The braking distance required to stop the vehicle is 162 feet.
3-21
A driver on a level two-lane highway observes a truck completely blocking the
highway. The driver was able to stop her vehicle only 20 ft from the truck. If the
driver was driving at 60 mph, how far was she from the truck when she first
observed it?
Use equation 3.35 to determine the stopping sight distance used. Use AASHTO
recommended values, t = 2.5 seconds, a/g = 0.35
u2
(60) 2
SSD = 1.47ut +
= 1.47(60)(2.5) +
= 563 ft
a
30
(
0
.
35
)
30( ± G )
g
Therefore, the distance from the point at which the driver observed the stopped
truck to the truck is 20 + 563 = 583 ft.
3-22
A temporary diversion has been constructed on a highway of +4% gradient due to
major repairs that are being undertaken on a bridge. The maximum speed allowed
on the diversion is 10 mi/hr. Determine the minimum distance from the diversion
that a road sign should be located informing drivers of the temporary change on the
highway.
Maximum allowable speed on highway = 70 mi/hr
Letter height of road sign = 4”
Perception-reaction time = 2.5 sec
Use equation 3.35 to determine the stopping sight distance.
u2
SSD = 1.47ut +
a
30( ± G )
g
While the first term in this equation is simply the distance traveled during the
perception-reaction time, second term is the distance traveled during braking. a/g
is taken as 0.35. Since the vehicle is not stopping (the final speed is not equal to
zero), the equation needs to be modified to take this into consideration. The
second term of equation 3.35 is replaced by equation 3.26 as follows (in which u1
is the initial velocity and u2 is the final velocity):
35
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
2
2
u − u2
SSD = 1.47ut + 1
a
30( ± G )
g
SSD = 1.47(70)(2.5) + (702 - 102) / (30 (0.35 + 0.04))
SSD = 256.67 + 410.26
SSD = 666.93 feet.
Next, determine the readability of the roadway sign.
Readability = (Letter height in inches) * 40 feet / inch of letter height
Readability = 4 inches * 40 feet / inch
Readability = 160 feet
The sign can be read at a distance of 160 feet.
Next, determine the distance from the diversion the sign should be placed.
x = SSD – readability distance
x = 666.93 – 160.00
x = 506.93 feet
The sign should be placed approximately 510 feet prior to the diversion to alert
drivers of the change on the highway.
3-23
Repeat Problem 3-22 for a highway with a down grade of -3.5% and the speed
allowed on the diversion is 15 mph. Assume that a driver can read a road sign
within his or her area of vision at a distance of 40 ft for each inch of letter height.
Use equation 3.35 to determine the stopping sight distance.
SSD = 1.47ut +
u2
a
30( ± G )
g
While the first term in this equation is simply the distance traveled during the
perception-reaction time, second term is the distance traveled during braking. a/g
is taken as 0.35. Since the vehicle is not stopping (the final speed is not equal to
zero), the equation needs to be modified to take this into consideration. The
second term of equation 3.35 is replaced by equation 3.26 as follows (in which u1
is the initial velocity and u2 is the final velocity):
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
36
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
2
2
u − u2
SSD = 1.47ut + 1
a
30( ± G )
g
SSD = 1.47(70)(2.5) + (702 – 152) / (30 (0.35 – 0.035))
SSD = 256.67 + 494.71
SSD = 751.37 ft
Next, determine the readability of the roadway sign.
Readability = (Letter height in inches) * 40 feet / inch of letter height
Readability = 4 inches * 40 feet / inch
Readability = 160 feet
The sign can be read at a distance of 160 feet.
Next, determine the distance from the diversion the sign should be placed.
x = SSD – readability distance
x = 751.37 – 160
x = 591 ft
The sign should be placed approximately 591 feet prior to the diversion to alert
drivers of the change on the highway.
3-24
An elevated expressway goes through an urban area, and crosses a local street as
shown in Figure 3.10. The partial cloverleaf exit ramp is on a 2% downgrade and all
vehicles leaving the expressway must stop at the intersection with the local street.
Determine (a) minimum ramp radius and (b) length of the ramp for the following
conditions:
Maximum speed on expressway = 60 mi/hr
Distance between exit sign and exit ramp = 260 ft
Letter height of road sign = 3”
Perception-reaction time = 2.5 sec
Maximum superelevation = 0.08
Expressway grade = 0%
Assume that a driver can read a road sign within his or her area of vision at a
distance of 50 ft for each inch of letter height, and the driver sees the stop sign
immediately on entering the ramp.
First, determine the readability of the roadway sign.
Readability = (Letter height in inches) * 50 feet / inch of height
Readability = 3 inches * 50 feet / inch
Readability = 150 feet
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(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
Next, determine the speed of the vehicle just prior to it entering the exit ramp
(uexit). Modify equation 3.27 by replacing the second term (braking distance) with
equation 3.26. a/g is taken as 0.35.
2
2
u1 − u 2
SSD = 1.47ut +
– readability distance
a
30( ± G )
g
260 = 1.47(60)(2.5) + [(602 - u2exit) / 30 (0.35 + 0)] - 150
410 = 220 + (602 - u2exit) / 10.5
602 - u2exit = 1995
uexit = (1605)1/2
uexit = 40.06 mi/h
Next, determine the minimum radius required for this exit ramp, using Equation
3.34. For exit speed of 45 mi/h, the new coefficient of side friction should be fs =
0.14.
R = u2exit / 15(e + fs)
R = (40.06)2 / 15(0.08 + 0.14)
R = 486.35 feet.
Next, determine the length required for this exit ramp.
S = 1.47uexitt + u2exit / 30 ((a/g) ± G)
S = 1.47(40.06)(2.5) + (40.06)2 / 30(0.35 - 0.02)
S = 147.23 + 162.10
S = 309.33 feet.
Therefore, the minimum radius for this exit ramp is approximately 490 feet and
the minimum length of the exit ramp was found to be approximately 310 feet.
3-25
Calculate the minimum passing sight distance required for a two-lane rural
roadway that has a posted speed limit of 45 mi/hr. The local traffic engineer
conducted a speed study of the subject road and found the following: average speed
of the passing vehicle was 47 mi/hr with an average acceleration of 1.43 mi/hr/sec,
and the average speed of impeder vehicles was 40 mi/hr.
Time to initiate maneuver, t1 = 4.0 sec
Determine the minimum passing sight distance for this roadway.
First, determine the distance traversed during the perception-reaction time.
d1 = 1.47t1(u - m + (at1 / 2))
d1 = 1.47(4.0)[47 - 7 + ((1.43(4.0)) / 2)]
d1 = 251.44 feet
Determine the distance traveled while passing the vehicle. Table 3.6, t2 = 10.0 sec.
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Chapter 3: Characteristics of the Driver, the Pedestrian, the Vehicle and the Road
d2 = 1.47ut2
d2 = 1.47(47)(10)
d2 = 689.33 feet
Now determine the distance between the passing vehicle and the opposing vehicle
from Table 3.6.
d3 = 180 feet.
Now determine the distance moved by the opposing vehicle.
d4 = (2 / 3) d2
d4 = (2 / 3) 689.33
d4 = 459.55 feet
The minimum passing distance can be found by adding all above distances
together.
d1 + d2 + d3 + d4 = Minimum sight distance
Sight distance = 251.44 + 689.33 + 180 + 459.55
Sight distance = 1,580.32 feet
Therefore, the minimum passing sight distance required for this roadway is
approximately 1,600 feet.
39
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 4
Traffic Engineering Studies
4-1
What are the advantages and disadvantages of machine vision (video image
detection) when compared with other forms of detection?
The greatest advantage of video image detection is that it is non-intrusive;
at detectors need not be installed in the roadway (as with inductive loops or
magnetometers). Systems such as the Autoscope can detect traffic in many
locations within the camera’s field of view. Because of this, a single camera can
replace many inductive loop detectors, thereby providing a wide-area detection
system. Unlike inductive loops, with video image processing traffic need not be
disrupted to install the device and the detection configuration can be changed
either manually or by using a software routine. Disadvantages of this method
include higher initial costs and that cameras must be dedicated to this purpose and
must be detection zones must be placed precisely for accurate detection.
4-2
Select and describe the method and equipment you will recommend for each of the
road sections given below. Give reasons for your recommendations.
(a) A private road leading to an industrial development
(b) A residential street
(c) A rural collector road
(d) A section of an interstate highway
(a)
For a private road leading to an industrial development, which would
likely be low volume and low speed, pneumatic tubes would be an appropriate
choice. Concerns about impact of data collection method on driver behavior
would likely be minimal on such a facility.
(b)
For a residential street, which would likely be low volume and low speed,
radar would be an appropriate choice as concerns may exist about the conspicuity
of pneumatic tubes affecting driver behavior.
(c)
For a rural collector road, where speeds and volumes may be somewhat
high, radar would be appropriate. If expense were not substantial, machine vision
would be worth considering.
(d)
For a section of Interstate highway, given the high speed and high volume
nature of the facility, it would be preferable to not have observers close the
roadway. Therefore, methods such as machine vision or inductive loops (if an
opportunity to temporarily close a lane is available) would be preferable. If
observers can safely be placed, radar would also be acceptable.
41
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 4: Traffic Engineering Studies
4-3
Speed data collected on an urban roadway yielded a standard deviation in speeds of
±4.8 mi/hr.
(a) If an engineer wishes to estimate the average speed on the roadway at
a 95% confidence level so that the estimate is within ±2 mi/hr of the true
average, how many spot speeds should be collected?
(b) If the estimate of the average must be within ±1 mi/hr, what should
the sample size be?
(a) Using Equation 4.5,
N = [(z σ)/d]2
N = [1.96(4.8) / 2]2
N = 22.1 => 23 spot speed observations
Note: z=1.96 for 95% confidence interval
(b)
For speeds within ±1 mi/h:
N = [1.96(4.8) / 1]2
N = 88.5 => 89 spot speed observations
4-4
An engineer wishing to obtain the speed characteristics on a bypass around her city
at a confidence level of 95%, and an acceptable limit of ± 1.0 mi/hr collected a total
of 130 spot speed samples and determined that the variance is 25 (mi/hr)2. Has the
engineer met with all of the requirements of the study?
This can be answered using Equation 4.5 to determine whether minimum sample
size requirements are met.
For the 95% confidence level, z = 1.96
The acceptable margin of error, d =1
The variance, σ2 = 25; σ = 5
2
⎛ (1.96)(5) ⎞
⎛ Zσ ⎞
N =⎜
⎟ = 96.04
⎟ =⎜
1
⎠
⎝
⎝ d ⎠
The minimum sample size is 97 observations. Since 130 were collected, the
data collection requirements are met.
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Chapter 4: Traffic Engineering Studies
4-5
An engineer wishing to determine whether there is a statistically significant
difference between the average speed of passenger cars and that of large trucks on a
section of highway, collected the data shown below. Determine whether the engineer
can conclude that the average speed of large trucks is the same as for passenger
cars.
Trucks
Passenger Cars
Average Speed (mi/hr)
62
59
Standard deviation of speed ±
5.5
6.3
mi/hr
Sample size
275
175
To determine whether the difference in mean speeds was statistically significant,
first, the pooled standard deviation must be determined, using Equation 4.6.
S12 S22
(5.5) 2 (6.3) 2
+
=
+
= 0.58
275
175
n1 n2
Then, compare the absolute value of the difference of the sample means with the
product of the appropriate z-statistic and the pooled standard deviation.
Absolute difference in means = 62 – 59 = 3.
ZSd = (1.96)(0.58)
3 > 1.13
Therefore, a statistically significant difference exists between the two data sets,
and it cannot be concluded that the average speed of large trucks is the same as
that of passenger cars.
Sd =
4-6
Assuming that the data shown in Table 4.2 were collected on a rural road in your
state and consideration is being made to set the speed limit on the road. Speed limits
of 50, 55, 60, and 65 mi/hr are being considered. Plot the expected non-compliance
percentages vs. the associated speed limit on a graph and recommend the speed for
the road. Give reasons for your selection.
Examination of the data in Table 4.2 show that the number of observations
exceeding 50, 55, 60, and 65 mi/hr are 38, 17, 5, and zero, respectively. In
percentages, these are 44.2%, 19.7%, 5.8%, and 0%, respectively. The plot
appears below.
43
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 4: Traffic Engineering Studies
Percent non-compliance
Problem 4-6
50
44.2
40
30
20
19.7
10
5.8
0
0
40
45
50
55
60
65
Possible speed limit (mi/hr)
4-7
The accompanying data show spot speeds collected at a section of highway located
in a residential area. Using the student’s t test, determine whether there was a
statistically significant difference in the average speeds at the 95% confidence level.
Before
After
Before
After
40
23
38
25
35
33
35
21
38
25
30
35
37
36
30
30
33
37
38
33
30
34
39
21
28
23
35
28
35
28
36
23
35
24
34
24
40
31
33
27
33
24
31
20
35
20
36
20
36
21
35
30
36
28
33
32
40
35
39
33
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44
70
Chapter 4: Traffic Engineering Studies
First calculate average and standard deviation for both sets of data.
ub = 35.1 mi/h
sb2 = 3.2092
ua = 27.5 mi/h
sa2 = 5.4442
Next, calculate the pooled standard deviation for both sets of data.
sd = [(sb2/ nb)+(sa2/ na)]½
= [(3.2092 / 30)+(5.4442 / 30)]½
= 1.154
Then compare the difference in average speed with the product of the z-statistic
and standard deviation.
|ub-ua | = 7.6
Zsd = 1.96(1.154)
= 2.262
Since 7.6 > 2.262 therefore the speeds are significantly different.
4-8
Using the data furnished in Problem 4-7, draw the histogram frequency distribution
and cumulative percentage distribution for each set of data and determine (a)
average speed, (b) 85th-percentile speed, (c) 15th-percentile speed, (d) mode, (e)
median, and (f) pace.
(a) average speed = Σui / Σfi
“before” average = 1053/30 = 35.1 mi/h
“after” average = 824/30 = 27.5 mi/h
(b) 85%-ile speeds from cumulative distribution plots
before = 38.4 mi/h
after = 33.6 mi/h
(c) 15%-ile speeds from cumulative distribution plots
before = 30.8 mi/h
after = 20.5 mi/h
(d) mode from histograms
before = 35 mi/h
after = 21 mi/h
(e) median from cumulative distribution plots
before = 34.5 mi/h
after = 31 mi/h
(f) pace from histograms
before = 30 – 40 mi/h
after = 21 – 31 mi/h
45
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 4: Traffic Engineering Studies
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
46
Chapter 4: Traffic Engineering Studies
4-9
Define the following terms and cite examples of how they are used.
Average annual daily traffic (AADT)
Average daily traffic (ADT)
Vehicle-miles of travel (VMT)
Peak hour volume (PHV)
Average annual daily traffic (AADT) is the average of 24-hour traffic counts
collected every day in the year. These counts are used to estimate highway user
revenues, compute accident rates, and establish traffic volume trends.
Average daily traffic (ADT) is the average of 24-hour traffic counts collected
over a number of days greater than one but less than a year. These counts are
used for planning of highway activities, measuring current traffic demand, and
evaluating existing traffic flow.
Vehicle miles of travel (VMT) is a measure of travel usage along a section of
road. It is the product of the volume (ADT) and the length of roadway in miles to
which the volume is applicable. This measure is used mainly as a base for
allocating resources for maintenance and improvement of highways and to
establish highway system usage trends.
Peak hour volume (PHV) is the maximum number of vehicles that pass a point
on a highway during a period of sixty consecutive minutes. This volume is used
for functional classification of highways, geometric design standards selection,
capacity analysis, development of operational programs, and development of
parking regulations.
4-10
Describe the different traffic count programs carried out in your state. What data
are collected in each program?
The Traffic Engineering Division of the Virginia Department of
Transportation conducts an "Interstate, Arterial and Primary Traffic Count
Program." Vehicle classifications consist of passenger cars, 2-axle 4-tire trucks,
2-axle 6-tire trucks, 3-axle 6-10 tire trucks, tractor-trailers, twin trailers, and
buses. There are a total of 1,345 counting stations for this program. Counts are
taken either 2, 4, or 9 times per year at each location. The data collected is then
published in a manual entitled “Average Daily Traffic Volumes on Interstate,
Arterial and Primary Routes” which includes 24-hour average daily traffic (ADT)
by section of roadway, 24-hour vehicle miles traveled (VMT) by route and
county, and 24-hour VMT statewide. In addition, counts are taken on secondary,
and unpaved roads on a less frequent basis. Intersection turning movement counts
are taken at specific locations when needed for detailed traffic operational studies.
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(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 4: Traffic Engineering Studies
4-11
A traffic engineer, wishing to determine a representative value of the ADT on 250
highway links having similar volume characteristics, conducted a preliminary study
from which the following estimates were made: Mean volume = 45,750 veh/day,
Standard deviation = 3750 veh/day. Determine the minimum number of stations for
which the engineer should obtain 24-hr volume counts for a 95-5 precision level.
Use Equation 4.7: n = [t2α/2,N-1(s2/d2)]/ [1+ (1/N)(t2α/2,N-1)(s2/d2)]
α = 100 - 95 = 5
S = 3750
m = 45,750 therefore d = (0.1)(45,750) = 4575
V = N-1 = 250 - 1 = 249
(tα/2,N-1) = 1.96
n = [(1.96)2(37502/45752)]/ [1+(1/250)(1.96)2(37502/45752)]
n = 2.57
Use 3 count stations.
4-12
Describe the following types of traffic counts and when they are used.
(a) screen line counts
(b) cordon counts
(c) intersection counts
(d) control counts
Screen line counts involve dividing the study area into large sections by
drawing imaginary lines (screen lines) across the study area. Counts are then
taken at each place a road crosses this line. This data is then used to detect
variations in traffic volumes and flow direction attributable to changes in land-use
patterns in the area.
Cordon counts are similar to screen line counts with the imaginary line
completely surrounding an area, for example, a CBD. Counts are taken at each
place a road crosses the line, giving information about vehicle accumulations
within the area. The information from cordon counts can be used to plan for
parking facilities, evaluate traffic operational techniques, and long range
infrastructure planning.
Intersection counts are counts of vehicular turning movements and
classifications at an intersection. The data gathered is used to develop signal
timing and phasing plans for signalized intersections and for geometric design
improvements.
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
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Chapter 4: Traffic Engineering Studies
Control counts are taken at strategically located locations chosen to
obtain representative samples of traffic volumes for specific types of highways or
streets. There are two types of control counts; major and minor. Major control
counts are taken monthly with 24-hour directional counts being taken at least
three days during the week
4-13
How are travel time and delay studies used? Describe one method for collecting
travel time and delay data at a section of a highway. Explain how to obtain the
following information from the data collected: (a) travel time, (b) operational delay,
(c) stopped time delay, (d) fixed delay, and (e) travel time delay.
Travel time and delay studies are used to aid the traffic engineer in
identifying problem locations, which may require special attention in order to
improve the overall flow of traffic on the route. Data from these studies may be
used to determine the efficiency of a route with respect to its ability to carry
traffic, identify bottleneck locations with relatively high delays and the causes for
those delays, perform before and after studies to evaluate the effectiveness of
traffic operations improvements, determine the relative efficiency of a route by
developing congestion indices, determine travel times on specific links for use in
assignment models, and perform economic studies in the evaluation of traffic
operation alternatives that reduce travel time. There are several methods for
collecting travel time and delay data including the floating car method, the
average car method, and the moving vehicle technique. In the average car
method, the test car is driven along the length of the test section at a speed that, in
the opinion of the driver, is the average speed of the traffic stream. The time to
traverse the section is noted and the test run is repeated for a minimum number of
times, with the average time then being recorded as the travel time. An additional
stop watch would be used to measure the amount of time the test vehicle is
delayed by impedance of other traffic such as vehicles parking, or by a reduction
in the capacity of the roadway such as a work zone. This value is recorded as the
operational delay. During the test runs, the amount of time the vehicle is stopped
would also be recorded and the average from all runs would be recorded as the
stopped-time delay. The fixed delay would be measured as the time spent waiting
for a traffic signal along the route to turn green. This delay is independent of
other traffic. The travel time delay would be determined by subtracting the travel
time for a vehicle to traverse the study section under uncongested conditions from
the actual travel time.
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(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 4: Traffic Engineering Studies
4-14
Table 4.10 shows data obtained in a travel time study on a section of highway using
the moving-vehicle technique. Estimate (a) the travel time and (b) the volume in
each direction at this section of the highway.
To determine volume using the moving vehicle technique, use Equation 4.9:
VN =
( N S + O N + PN )60
TS + TN
NN = 104.2
ON = 1.6
PN = 1.1
TN = 5.25
NS = 93.9
OS = 1.1
PS = 1.0
TS = 4.90
VN= (93.9 + 1.6 - 1.1)60 / (5.25 + 4.90)
VN= 558 veh
VS= (104.2 + 1.1 - 1.0)60 / (5.25 + 4.90)
VS= 617 veh
To determine travel time using the moving vehicle technique, use Equation 4.10:
TN= TN - [60(ON - PN )/VN ]
TS= TS - [60(OS - PS )/VS ]
TN= 5.25 - [60(1.6-1.1)/558]
TN= 5.20 min
TS= 4.90 - [60(1.1-1.0)/617]
TS= 4.89 min
4-15
An engineer, wishing to determine the travel time and average speed along a section
of an urban highway as part of an annual trend analysis on traffic operations,
conducted a travel time study using the floating-car technique. He carried out 10
runs and obtained a standard deviation of ±3 mi/h in the speeds obtained. If a 5%
significance level is assumed, is the number of test runs adequate?
Use Equation 4.8: N = (
tα (σ ) 2
)
d
For trend analysis, assume ±3 mi/h acceptable error.
N = [1.833(3)/3]2
N = 3.36 runs
Therefore, 10 runs is adequate
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50
Chapter 4: Traffic Engineering Studies
4-16
Briefly describe the tasks you would include in a comprehensive parking study for
your college campus, indicating how you would perform each task and the way you
would present the data collected.
Parking studies, in general, are used to determine the demand and supply
of parking facilities in an area, the projection of the demand into the future, and
the collection of the views of various interest groups on how best to solve any
problems. A comprehensive parking study usually includes an inventory of
existing parking facilities, collection of data on parking accumulation, parking
turnover, and parking duration data, identification of parking generators, and
obtaining information on parking demand. On a college campus, this would
include faculty, staff, and student lots, as well as visitor parking facilities. The
college administration would also have to be interviewed to determine their policy
on parking as it relates to the provision of on-campus parking.
Parking accumulation data would be obtained by checking the amount of
parking at 2-hour intervals from 7:00 am to 5:00 pm on weekdays, when demand
is highest. Parking accumulation would then be plotted against time of day for
each lot. While collecting accumulation data, license plates of vehicles in
selected spaces should be recorded to determine duration and turnover. When the
data is analyzed, the average length of time an individual vehicle occupies a space
can be estimated. The college itself is the parking generator and therefore this
step can be omitted as parking facilities of varying capacity exist throughout
campus; however, the location of the larger parking facilities could be identified
on a map. Parking demand would be collected by interviewing drivers as they
enter the parking facility. The interviewer should attempt to identify the driver’s
trip origin, purpose of their trip and the driver’s destination after parking. This
information could indicate a need for a parking facility in a new location or the
enlargement of an existing facility.
4-17
Select a parking lot on your campus. For several hours, conduct a study of the lot
using the methods described in this chapter. From the data collected, determine the
turnover and duration. Draw a parking accumulation curve for the lot.
To conduct a parking accumulation study of a parking lot, detailed data on
space usage and vehicle turnover must be collected. The number of spaces in use,
as well as license plate numbers of those vehicles and their entry and exit times,
must be collected to monitor turnover and duration.
Turnover is calculated as the ratio of the total number of different vehicles
parked to the number of available spaces. Turnover for the entire study period
can be calculated by dividing the number of vehicles parked during the study
period by the number of spaces available.
51
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Chapter 4: Traffic Engineering Studies
Duration is the average amount of time a vehicle spends in a parking
space, which is a surrogate measure for the availability of parking spaces. By
recording the entry and exit times for each vehicle during the study period,
durations for each of these vehicles can be calculated and then averaged to
determine average duration.
The parking accumulation curve can be drawn to display percent of spaces
used (on the y-axis) as a function of time-of-day (on the x-axis). By noting the
number of spaces that are occupied at discrete intervals (e.g. every hour), the data
for drawing the accumulation curve are obtained.
4-18
Data collected at a parking lot indicate that a total of 300 cars park between 8 a.m.
and 6 p.m. 10% of these cars are parked for an average of 2 hr, 30% for an average
of 4 hr, and the remaining cars are parked for an average of 10 hr. Determine the
space-hours of demand at the lot.
Use Equation 4.12: D = (0.10)(300)(2) + (0.30)(300)(4) + (0.60)(300)(10)
D = 2,220 space-hours
4-19
If 10% of the parking bays are vacant on average (between 8 a.m. and 6 p.m.) at the
parking lot of problem 4-18, determine the number of parking bays in the parking
lot. Assume an efficiency factor of 0.85.
Use Equation 4.12:
2220 + (2220)(0.10) = 2442 space-hours (assuming 10% vacancy)
(0.85)(10)(N) = 2442
N = 288 spaces
4-20
The owner of the parking lot of Problems 4-18 and 4-19 is planning an expansion of
her lot to provide adequate demand for the following 5 years. If she has estimated
that parking demand for all categories will increase by 5% a year, determine the
number of additional parking bays that will be required.
Find space-hours of demand in 5 years:
(1+0.05)5 (2442) = 3,117 space hours
Additional space hours = 3117 – 2442 = 675
Find number of spaces:
(0.85)(10)(N) = 675
N = 79.4
N = 80 spaces
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52
Chapter 5
Highway Safety
5-1
Describe the type of information on a collision diagram.
Collision diagrams include symbols used to represent different types of
maneuvers, types of accidents, and severity of accidents. The date and time the
accident occurs is also included on the diagram. Because the diagrams provide
the information in a pictorial format, the location of accidents is immediately
known.
5-2
A local jurisdiction has determined that for a given set of geometric conditions, a
maximum rate of 8 crashes/million entering vehicles can be tolerated. At an
intersection of 2 roadways with ADTs of 10,000 and 7500, how many crashes can
occur before corrective action must be sought?
Use Equation 5.1
RMEV = A(1,000,000) / (ADT)(365 days)
8 = A(1,000,000) / (17,500)(365)
A = 51.1 crashes
Therefore, 51 crashes per year may occur before corrective action must be sought.
5-3
Studies were conducted at two sites on rural roads with similar characteristics. The
first site was 5.1 miles in length with an ADT of 6500. Over the year-long study
period, 28 crashes occurred on this portion of roadway, five of them resulting in
fatalities. The second site was a 10-mile section with an ADT of 5000. There were 32
crashes in this section with four fatalities. Determine the appropriate crash rates for
both locations, and discuss the implications.
Site 1 – Use Equation 5.2
RMVMT = A(100,000,000) / (ADT)(365 days)(length of road section)
RMVMT = (28)(100,000,000) / (6500)(365)(5.1)
RMVMT = 231.4
RMVMF = RMVMT (# of fatal / # of total)
RMVMF = 231.41(5/28)
RMVMF = 41.3
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Chapter 5: Highway Safety
Site 2 – Use Equation 5.2
RMVMT = A(100,000,000) / (ADT)(365 days)(length of road section)
RMVMT = (32)(100,000,000) / (5000)(365)(10.0)
RMVMT = 175.3
RMVMF = RMVMT(# of fatal / # of total)
RMVMF = (175.3)(4/32)
RMVMF = 21.9
Although site 2 experiences more crashes, with a lower ADT, over the one
year period, the crash rate for that site is lower than for site 1. This is due to the
fact that the roadway section is almost twice as long for site 2, increasing the total
vehicle miles traveled and thereby reducing the rate per 100 million vehicle miles
traveled. Site 1 can be considered less safe than site 2 based on both crash rate
and fatality rate.
5-4
A 5.4-mile segment of highway has the following traffic and safety data:
2005
2006
2007
Annual average daily traffic (AADT)
15,200
16,300
17,400
Number of fatal crashes
3
2
3
Number of crashes
29
31
27
Determine the fatal crash and total crash rates (per 100 million VMT) for
each year and for the three-year period. Comment on and qualify the results
as necessary.
First, determine annual vehicle-miles of travel:
2005 VMT = (15,200 veh/day)(365 day/yr)(5.4 mi) = 29,959,200
Similarly, 2006 VMT = 32,127,300 and 2007 VMT = 34,295,400
Then, determine fatal crash rates:
2005 fatality rate = 3 fatalities / 29,959,200 VMT = 10.1 fatalities / 100 MVMT
Similarly, 2006 fatality rate = 6.2 fatalities per 100 MVMT and
2007 fatality rate = 8.7 fatalities per 100 MVMT
Across all: RMVMF = (3+2+3 fatalities / 96,381,900 VMT) = 8.3
Similarly, determine total crash rates:
2005 crash rate = 29 crashes / 29,959,200 VMT = 96.8 crashes / 100 MVMT
Similarly, 2006 crash rate = 96.5 crashes per 100 MVMT and
2007 crash rate = 78.7 crashes per 100 MVMT
Across all: RMVMT = (29+31+27 crashes / 96,381,900 VMT) = 90.3
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Chapter 5: Highway Safety
Crash rates are slightly below the national average. Fatality rates are well
above the national average; however, the fatality rates are based on a small
sample size of only eight fatalities.
5-5
Describe the categories used to summarize crash data.
Give an example of how each category would be used to evaluate a given location.
The categories used to summarize crash data at a site are (1) type of crash,
(2) severity of crash, (3) contributing circumstances, (4) environmental
conditions, (5) time period of crashes. The type of crash would be used to
determine patterns at a given site. A high number of left turn crashes, for
example, might indicate a sight distance problem. The severity of crashes is
commonly used to compare crashes at different locations by assigning a weighted
scale to each crash based on the severity of the crash. Contributing circumstances
include human factors, environmental factors, or vehicle-related factors. The
assignment of environmental conditions (e.g. daylight, dark, snow, ice, rain)
facilitates the identification of possible causes of crashes and safety deficiencies
that may exist at a particular location. Summarizing crashes by time periods
allows for the identification of periods with rates significantly higher than other
periods.
5-6
Determine the severity number for a section of highway that had 2 fatal crashes, 24
personal injury crashes, and 74 property damage only crashes during a particular
time period. Use a weighting scheme of 20 for fatal crashes, 5 for personal injury
crashes, and 1 for crashes with only property damage. Determine the severity index
by dividing the severity number by the total number of crashes, thereby
normalizing the severity number by the number of crashes.
The severity number is calculated as follows:
(20)(2) + (5)(24) + (1)(74) = 234
The severity index is calculated as follows:
234 / (2 + 24 + 74) = 2.34
5-7
A review of crash records shows that a signalized intersection is a hazardous
location because of an abnormally high number of rear-end collisions. What are the
possible causes of these crashes, and what data should be collected to determine the
actual causes?
Rear-end collisions at a signalized intersection could be caused by a
slippery surface, a large number of turning vehicles, poor visibility of the signals,
55
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Chapter 5: Highway Safety
inadequate signal timings, unwarranted signals, or inadequate roadway lighting.
To determine the cause of the crashes the following studies would be needed; skid
resistance study, weather-related study, traffic conflict study, volume study,
roadway inventory study, delay study, and traffic control devices study.
5-8
The crash rate on a heavily-traveled two-lane rural highway is abnormally high.
The corridor is 14 miles long with an ADT of 34,000. An investigation has
determined that head-on collisions are most common, with an RMVMT of 4.5, and
are caused by vehicles attempting to pass. Determine an appropriate
countermeasure and calculate the estimated yearly reduction in total crashes.
Selected countermeasure - Add passing lanes
Crash reduction factor = 0.25 (from Table 5.13)
RMVMT = A(100,000,000) / (ADT)(365 days)(length of road section)
4.5 = A(100,000,000) / (34,000)(14)(365)
A = 7.82 => 8 crashes per year
(8 crashes/year)(0.25) = 2 accidents per year reduced due to countermeasure
Therefore, the estimated reduction in crashes is 2 crashes per year.
5-9
It is required to test whether large trucks are significantly more involved in serious
crashes on two-lane primary highways than on multilane primary highways. Using
the t test and the data on crash rates for serious crashes given in the table below,
determine whether you can conclude that large trucks are more involved in serious
crashes on two-lane primary roads than on multi-lane primary roads. Use a
significance level of 5 percent.
Truck Crash Rates per 100 MVMT
Two-lane
Multi-lane
Primary Highways
Primary Highways
0.256
0.188
0.342
0.312
0.842
0.421
1.021
0.285
0.361
0.225
0.262
0.183
0.861
0.341
Solution:
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Chapter 5: Highway Safety
Denoting the average crash rates per 100 MVMT on two lane-highways as
−
−
X 1 , and that on multilane highways as X 2 , the hypothesis test will be described
as,
Null Hypothesis:
Alternate Hypothesis:
•
−
−
Ho : X 1 = X 2
−
−
H1 : X 1 > X 2
Determine mean number of crashes:
−
mean number of crashes on two-lane rural roads ( X 1 ) = 0.564
−
mean number of crashes on multi-lane rural roads ( X 2 ) = 0.279
•
Determine crash variances
Crash variance on two-lane rural road (S12) = 0.108
Crash variance on multi-lane rural road (S22) = 0.008
The difference in the variances must be examined for statistical significance to
determine the appropriate case of t-test.
2
2
2
2
Null Hypothesis:
Ho : s1 = s2
Alternate Hypothesis:
H1 : s1 > s2
The F-statistic is 14.29, whereas Fcrit = 4.28 (at α=0.05), therefore, the
difference in variances in statistically significant. The t-test for unequal variances
must be used.
•
Determine t, use Equation 5.4:
−
−
X1 − X 2
T=
1 1
Sp
+
n1 n2
0.564 − 0.279
T=
= 2.221
1 1
0.24
+
7 7
57
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Chapter 5: Highway Safety
•
Determine theoretical t0.05 for degrees of freedom of (7 + 7 – 2) =12:
tcrit = 1.89 for a one-tailed test, therefore reject the null hypothesis, and based on
data given, it cannot be concluded that crash rates tend to be higher on two-lane
highways than on multi-lane highways.
5-10
The table on the next page shows serious (Fatal and Injury, F&I) and non-serious
(Property Damage Only, PDO) crashes involving trucks on interstate highways and
other principal arterial highways at six locations. Using the proportionality test, and
the data given in the table, determine whether it can be concluded that the
probability of large trucks being involved in serious crashes is higher on other
arterial highways than on interstate highways at a 5 percent significance level.
Interstate Highways
F& I Crashes
PDO Crashes
4
22
7
30
6
28
8
35
3
18
6
33
Other Arterials
F& I Crashes
PDO Crashes
10
28
2
27
11
45
6
19
9
27
5
25
Solution:
Denoting the proportion of serious crashes on other arterials as p1, and that
interstate highways as p2, the hypothesis test will be described as follows:
The null hypothesis is that the proportion of fatal and injury crashes on other
arterials is the same as that on interstate highways.
Ho: p1 = p2
The alternative hypothesis is that the proportion of fatal and injury crashes on
other arterials is higher than that on interstate highways..
H1 : p1 > p2 (one-tail test)
•
Determine p1 and p2
p1 = 43/(43 + 171) = 0.20
p2 = 34/(34 + 166) = 0.17
•
Determine Z from Equation 5.6:
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Chapter 5: Highway Safety
p1 − p 2
Z=
⎛1
1 ⎞
p(1 − p )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
(x + x2 )
p= i
(n1 + n2 )
(43 + 34) = 0.186
p=
(214 + 200)
0.20 − 0.17
= 0.785
Z=
1 ⎞
⎛ 1
0.186(1 − 0.186)⎜
+
⎟
⎝ 214 200 ⎠
Zα = 0.05 = 1.645: Z < Zα, so we cannot reject the null hypothesis. It can be
concluded that the proportion of fatal and injury crashes other arterials is not
significantly higher than that interstate highways.
5-11
The numbers of crashes occurring for over a period of three years for different
levels (high, medium, low) of AADT at unsignalized rural intersections are given in
the table below. Using the Kruskal-Wallis H test determine whether it can be
concluded that the distribution of crashes at unsignalized rural intersections is the
same for all AADT levels. Use a significance level of 5 percent.
Low AADT
6
35
3
17
11
30
15
16
25
5
Number of Crashes
Medium AADT
34
28
42
13
40
31
9
32
39
27
High AADT
13
35
19
4
29
0
7
33
18
24
Solution:
•
Rank the number of crashes as shown in Table below.
Low AADT
26
Ranks
Medium AADT
6
59
High AADT
21.5
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Chapter 5: Highway Safety
4.5
29
18
23
10
20
19
14
27
190.5
•
12
1
21.5
2
9
24
8
3
13
99.5
4.5
16
28
11
30
25
7
17
15
175
Determine the statistic H using Eq.5.7
2
Rj
12
H=
− 3(n + 1)
∑
n(n + 1)
nj
H=
12
190.52 99.52 1752
(
∑ 10 + 10 + 10 ) − 3(30 + 1) = 6.12
30(30 + 1)
Determine whether we should accept or reject the null hypothesis
•
Degrees of freedom = (3-1) = 2
α = 0.05
From the χ2 table in Appendix A, we obtain χ20.05,2 =5.991 < H = 6.12
Hence, it cannot be concluded that the distribution of crashes at unsignalized rural
intersections is the same for all AADT levels.
5-12
A rural primary road segment of 2 miles long has an average annual daily traffic
(AADT) of 11,350. The number of crashes that have occurred over the past 5 years
are 5 fatal, 55 injury crashes, and 100 property damage only crashes. Statewide
average crashes for this type of roads are 2 fatal, 130 injury, and 300 property
damage only crashes per 100 MVMT. The weight factors for fatal, injury, and
property damage only crashes are 8, 3, and 1 respectively. Using the critical crash
ratio methodology, determine whether this site can be labeled a hazardous site,
using a 95 percent confidence limit.
Solution:
Step 1. Calculate the traffic base, TB
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60
Chapter 5: Highway Safety
Years * AADT * segmentlength * 365
100 * 106
5 * 11,350 * 2.0 * 365
TB =
= 0.414*100 MVM
100 * 106
TB =
Step 2. Calculate the 3-year average equivalent PDO crash rate for this type of
facility.
AVR = 8*2 + 3*130 + 300 = 706 equivalent PDO crashes per 100 MVMT
Step 3. Select a test factor based on confidence level. Since a confidence level of
95% is specified, the test factor is 1.96.
Step 4. Compute the critical rate:
CR = AVR +
AVR
0.5
+ TF
TB
TB
0.5
706
+ 1.96
0.414
0.414
CR = 788 equivalent PDO crashes per 100 MVMT
CR = 706 +
Step 5. Determine the ratio of actual crash occurrence for the segment with
respect to the critical rate.
8 * 8 + 3 * 55 + 100
= 795
Segment crash history =
0.414
= 795 equivalent crashes per 100 MVM
crash ratio
=
SegmentCrashHistory
StatewideCrashHistory
795
= 1.008
788
Since the ratio is greater than 1, a safety problem is likely to exist.
=
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Chapter 5: Highway Safety
5-13
The Safety Performance functions for injury, fatal, and property damage only
crashes are:
SPinj = 1.602 × 10−6 (length in feet )( AADT )0.491
SPPDO = 3.811 × 10 −6 (length in feet )( AADT ) 0.491
For the data given below, determine:
i. The long term average number of injury crashes per year at the site
ii. The long term average number of fatal crashes per year at the site
iii. The long term average number of property damage only crashes per year at
the site
Data:
Length of road segment = 7585 ft
AADT = 5250 veh/day
Total number of injury crashes for 10 years = 12
Total number of fatal crashes for 10 years = 3
Total number of property only crashes for 10 years = 48
Assume k = 2.51
Solution:
Determine the predicted number of crashes
N ic = SPinj = 1.602 × 10 −6 × 7585 × 52500.491 = 0.440
N fc = SPfatal = 0.03 × 10−6 × 7585 × 52500.491 = 0.008
N Pc = SPPDO = 3.811 × 10 −6 × 7585 × 5250 0.491 = 1.047
•
Estimate w using Eq. 5.22
k
2.51
winj =
=
= 0.363
k + n( N ic ) 2.51 + 10 * 0.440
k
2.51
= 0.969
=
w fatal =
k + n ( N fc ) 2.51 + 10 * 0.008
wPDO =
•
k
2.51
=
= 0.193
k + n( N Pc ) 2.51 + 10 *1.047
Determine long-term average injury crashes (m) using Eq. 5.21
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Chapter 5: Highway Safety
⎛x ⎞
⎛ 12 ⎞
minj = winj (N ic ) + (1 − winj )⎜⎜ inj ⎟⎟ = 0.363(0.440) + (1 − 0.363)⎜ ⎟ = 0.924
⎝ 10 ⎠
⎝ n ⎠
⎛x ⎞
⎛ 3⎞
m fatal = w fatal (N fc ) + (1 − w fatal )⎜⎜ fatal ⎟⎟ = 0.969(0.008) + (1 − 0.969 )⎜ ⎟ = 0.017
⎝ 10 ⎠
⎝ n ⎠
⎛x
⎞
⎛ 48 ⎞
mPOD = wPOD (N Pc ) + (1 − wPOD )⎜ POD ⎟ = 0.193(1.047 ) + (1 − 0.193)⎜ ⎟ = 4.076
⎝ n ⎠
⎝ 10 ⎠
5-14
Compute the PSI index for the site in Problem 5.11 if the relative costs are:
rcinjury = 27.39
rc fatal = 130.35
rcPDO = 1
Solution:
Compute the PSI index for each type of crashes:
PSI injury = minjury − SPinjury = 0.363 – 0.440 = -0.077
PSI fatal = m fatal − SPfatal = 0.017 – 0.008 = 0.009
PSI PDO = mPDO − SPPDO = 4.076 – 1.047 = 3.029
Compute the PSI index for this particular site:
PSIIndex = rcinjury(PSI)Injur + rcfatal(PSI)fatal + rcPDO(PSIPDO)
= -0.077*27.39 + 0.009*130.35 + 3.029
= 2.093
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Chapter 5: Highway Safety
5-15
A state transportation agency has established crash reduction factors shown in the
table. An intersection has been identified as having an abnormally high left-turn
crash rate (17, 20, and 18 crashes in the last three years), attributed to excessive
speeds and the absence of an exclusive left-turn phase. The ADT at the intersection
for the last three years is 8400, and the ADT for after implementation is 9500.
Determine (a) the appropriate countermeasures, and (b) the expected reduction in
crashes if the countermeasures are implemented.
Countermeasure
Reduction Factor
Retime signals
Provide left-turn signal phase
Reduce speed limit on approaches
Provide turning guide lines
Prohibit left turns
0.10
0.30
0.35
0.05
0.75
Since the abnormally high crash rate has been attributed to excessive
speed and the absence of an exclusive left-turn phase, appropriate
countermeasures are:
• Provide exclusive left-turn phase
• Reduce speed limit
Use Equation 5.6
CR = 0.35 + (1-0.35)(0.3)
CR = 0.545
Average crashes/year = (17+20+18)/3 = 18.33
Crashes prevented (per year) = [18.33(0.545)(9500)]/8400 = 11.30 => 11 crashes
5-16
The geometrics along a portion of roadway have been improved to provide wider
travel lanes and paved shoulders. While these improvements have reduced the crash
rate, the local traffic engineer is still concerned that too many crashes are occurring,
particularly those involving collisions with fixed objects along the edge of pavement.
Thirty such crashes have occurred in the past 12 months. What countermeasure
should the engineer undertake to reduce the number of crashes by at least five per
year?
Since the problem has been identified as crashes with fixed objects, an
appropriate countermeasure is to increase roadside recovery distance. For 5 fewer
crashes in a one-year period, a reduction of at least ((30-5)/30) = 17% is required.
From Table 5.12, the roadside recovery distance must be increased to 8
feet to achieve the desired reduction.
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Chapter 5: Highway Safety
5-17
A 10-mile section of a rural two-lane road with 9 ft lanes and 2 ft paved shoulders
has a crash rate of 156/100 million vehicle-miles traveled. The current ADT is
20,000, which, due to a nearby commercial development, is anticipated to increase to
37,700. Assuming the RMVM does not change before the improvement, estimate the
additional lane and shoulder widths required to keep the average number of crashes
per year at the current level.
Current crashes:
A = [RMVM(ADT)(365 days)(length of road)]/100,000,000
A = (156)(20,000)(365)(10)/100,000,000
A = 113.88 => 114 crashes
Future crashes (without improvements):
A = (156)(37,700)(365)(10) / 100,000,000
A = 214.66 => 215 crashes
Calculate percent reduction needed:
215 – 114 = (215)(x)
x = 0.47 = 47%
From Table 5.15,
Increase lane 3 ft and shoulder 4 ft or
Increase lane 2 ft and shoulder 6 ft
5-18
Residents of a local neighborhood have been complaining to city officials that
vehicles are using their side streets as shortcuts to avoid rush hour traffic. Discuss
the options available to the city transportation officials to address the residents’
concerns.
Solutions to cut-through traffic include creating cul-de-sacs, reducing
roadway width at intersections or providing parking at mid-block since narrower
roadways tend to reduce speeds, create one-way streets, install diagonal diverters
at intersections, install speed humps, and increase speed enforcement.
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Chapter 5: Highway Safety
5-19
Estimate the yearly reduction in total and fatal crashes resulting from the upgrade
of an 18.4-mile corridor with partially-controlled access to one with full control. The
road is in an urban area and has an ADT of 62,000.
From Table 5.16
Rate
Full
Partial
Total = 1.86
Total = 4.96
Fatal = 0.02
Fatal = 0.05
Before:[(62000)(4.96)(365)(18.4)] / (1,000,000) = 2065 total crashes
[(62000)(0.05)(365)(18.4)] / (1,000,000) = 21 fatal crashes
After: [(62000)(1.86)(365)(18.4)] / (1,000,000) = 775 total crashes
[(62000)(0.02)(365)(18.4)] / (1,000,000) = 8 fatal crashes
2065 - 775 = 1290 total crashes reduced
21 - 8 = 13 fatal crashes reduced
5-20
An engineer has proposed four countermeasures to be implemented to reduce the
high crash rate at an intersection. CRFs for these countermeasures are 0.25, 0.30,
0.17, and 0.28. The number of crashes occurring at the intersection during the past
three years were 28, 30, and 31, and the AADTs during those years were 8450, 9150,
and 9850, respectively. Determine the expected reductions in number of crashes
during the first three years after the implementation of the countermeasures if the
AADT during the first year of implementation is 10,850 and the estimated traffic
growth rate is 4% per annum after implementation.
CR = CR1+(1-CR1)CR2+(1-CR1)(1-CR2)CR3+(1-CR1)(1-CR2)(1-CR3)CR4
CR = 0.30+(1-0. 30)(0.28)+(1-0.30)(1-0.28)(0.25)
+(1-0. 30)(1-0.28)(1-0.25)(0.17) = 0.69
Year 1:((30+31+28)/3)(0.69)(10850) / ((8450+9150+9850)/3)
24.3 crashes reduced
Year 2:((30+31+28)/3)(0.69)(1+0.04)(10850) / ((8450+9150+9850)/3)
25.3 crashes reduced
Year 3:((30+31+28)/3)(0.69)(1+0.04)2(10850) / ((8450+9150+9850)/3)
26.3 crashes reduced
5-21
Survey your local college campus. What pedestrian facilities are provided? How
might pedestrian safety be improved?
This college campus provides sidewalks for pedestrians as well as
crosswalks at intersections and in the vicinity of bus stops. The campus would
benefit from a dedicated pedestrian-only roadway through the center of campus.
During peak pedestrian times, the sidewalks do not provide sufficient capacity.
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66
Chapter 6
Fundamental Principles
of Traffic Flow
6-1
Observers stationed at two sections XX and YY, 500 ft apart on a highway, record
the arrival times of four vehicles as shown in the accompanying table. If the total
time of observation at XX was 15 sec, determine (a) the time mean speed, (b) the
space mean speed, and (c) the flow at section XX.
Vehicle
Section XX
Section YY
To + 7.58 sec
A
To
B
To + 3 sec
To + 9.18 sec
C
To + 6 sec
To + 12.36 sec
D
To + 12 sec
To + 21.74 sec
First, calculate speeds of individual vehicles:
u1 = L/t = 500 ft / (7.58 sec – 0 sec) = 65.96 ft/sec
u2 = L/t = 500 ft / (9.18 sec – 3 sec) = 80.91 ft/sec
u3 = L/t = 500 ft / (12.36 sec – 6 sec) = 78.62 ft/sec
u4 = L/t = 500 ft / (21.74 sec – 12 sec) = 51.33 ft/sec
a) Time mean speed (TMS), using Equation 6.2, u t =
1 n
∑ ui
n i =1
ut = (65.96+80.91+78.62+51.33) / 4
ut = 69.2 ft/sec
ut = 47.1 mi/h
b) Space mean speed (SMS), using Equation 6.3, u s =
nL
n
∑t
i =1
i
us = (500 ft)(4) / (7.58+6.18+6.36+9.74) sec
us = 67.0 ft/sec
us = 45.6 mi/h
c) Flow at XX, using Equation 6.1, = q = n (3600) / T
q = 4 (3600 sec/h) / 15 sec
q = 960 veh/h
67
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Chapter 6: Fundamental Principles of Traffic Flow
6-2
Data obtained from aerial photography showed six vehicles on a 600 ft-long section
of road. Traffic data collected at the same time indicated an average time headway
of 4 sec. Determine (a) the density on the highway, (b) the flow on the road, and (c)
the space mean speed.
n
L
k = 6 veh / 600 ft
k = 0.01 veh/ft
k = 52.8 veh/mi
a) Density, k =
1
t
q = 1 / (4 sec/veh)
q = 0.25 veh/sec
q = 900 veh/h
b) Flow, q =
q
k
us = (900 veh/h) / (52.8 veh/mi)
us = 17.0 mi/h
c) Space mean speed (SMS), u s =
6-3
Two sets of students are collecting traffic data at two sections, xx and yy, of a
highway 1500 ft apart. Observations at xx show that five vehicles passed that section
at intervals of 3, 4, 3, and 5 sec, respectively. If the speeds of the vehicles were 50, 45,
40, 35, and 30 mi/hr respectively, draw a schematic showing the locations of the
vehicles 20 sec after the first vehicle passed section xx. Also determine (a) the time
mean speed, (b) the space mean speed, and (c) the density on the highway.
The distance traversed by each vehicle 20 seconds after crossing section x-x is
calculated as follows:
Vehicle A:
x A = 50mi / h ×
1.47 ft / sec
× 20 sec = 1470 ft
1mi / h
Similarly:
xB = (45)(1.47)(20-3) = 1125 ft
xC = (40)(1.47)(20-3-4) = 764 ft
xD = (35)(1.47)(20-3-4-3) = 515 ft
xE = (30)(1.47)(20-3-4-3-5) = 221 ft
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68
Chapter 6: Fundamental Principles of Traffic Flow
|---------:------------:---------:---------------:-------------:---|
x-x
E
D
C
B
A y-y
221'
515' 764'
1125'
1470'
a) Time mean speed (TMS), using Equation 6.2, u t =
1 n
∑ ui
n i =1
ut = (50+45+40+35+30) / 5
ut = 40 mi/h
b) Space mean speed (SMS), using Equation 6.3, u s =
n
n
1
∑u
i =1
i
us = 5 / (1/50+1/45+1/40+1/35+1/30)
us = 5 / 0.1291
us = 38.7 mi/h
q
us
k = (5 veh / 15 sec)(3600 sec / hr) / (38.7 mi/hr)
k = 31.0 veh/mi
c) Density, k =
6-4
Determine the space mean speed for the data given in Problem 6-3 using the Garber
and Sankar expression given in Equation 6.5. Compare your answer with that
obtained in Problem 6-3 for the space mean speed and discuss the results.
Equation 6.5 presents the following relationship:
ut = 0.966us + 3.541
First, the individual vehicle speeds must be converted to units of km/h and
then averaged to determine time mean speed, which is 64.4 km/h. Using Eq. 6.5,
64.4 = 0.966u s + 3.541
u s = 63.0 km/h
The value for space mean speed found in Problem 6-3 is 38.7 mi/h, which is equal
to 62.7 km/h. Equation 6.5 predicted a value of 63.0 km/h, 0.3 km/h higher than
the space mean speed calculated directly from the data. This difference is
insignificant for most purposes.
69
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Chapter 6: Fundamental Principles of Traffic Flow
6-5
The data shown below were obtained by time-lapse photography on a highway. Use
regression analysis to fit these data to the Greenshields model and determine (a) the
mean free speed, (b) the jam density, (c) the capacity, and (d) the speed at maximum
flow.
Speed (mi/h)
Density (veh/mi)
14.2
24.1
30.3
40.1
50.6
55.0
85
70
55
41
20
15
A linear regression analysis can be applied to the given data to estimate
parameters in Greenshields’ model of traffic flow. Greenshields’ model
(Equation 6.13) is:
uf
us = u f −
k
kj
A linear regression model takes the form y = a + bx; therefore, in this case, the
given data us and k correspond to y and x respectively. The linear regression
analysis can be performed using equations 6.21, 6.22, and 6.23, or using a
computer software spreadsheet package. Linear regression analysis yields values
of a = 62.8124
and b = -0.56845. Therefore,
a) mean free flow speed, uf = a = 62.8 mi/h
b) jam density, kj
In the regression model, b = uf / kj
b = 0.56845
kj = 62.8 / 0.56845 = 110.49
kj = 110 veh/mi
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70
Chapter 6: Fundamental Principles of Traffic Flow
c) capacity, qmax
Capacity occurs at maximum flow. State flow in terms of density.
uf
k)
q = k × u s = k (u f −
kj
q = 62.8k - (62.8/110.49)k2
q = 62.8k - 0.5684k2
Take the derivative and set equal to zero to maximize flow; solve for
density.
0 = 62.8 – 1.1368k
k = 55.25 when q = qmax
Solve for q
qmax = 62.8(55.25) - 0.5684(55.25)2
qmax = 1735 veh/h
d) speed at maximum flow
Solve for mean speed using k when q = qmax
uf
us = u f −
k
kj
us = 62.8 – 0.5684(55.25)
us = 31.4 mi/h
6-6
Under what traffic conditions will you be able to use the Greenshields model but not
the Greenberg model? Give the reason for your answer.
In light flow conditions, in which the mean speed of traffic is near the mean free
flow speed, the Greenberg model is not appropriate. This is due to mean free flow speed
approaching infinity as density approaches zero in the Greenberg model.
71
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Chapter 6: Fundamental Principles of Traffic Flow
6-7
The table below shows data on speeds and corresponding densities on a section of a
rural collector road. If it can be assumed that the traffic flow characteristics can be
described by the Greenberg model, develop an appropriate relationship between the
flow and density. Also determine the capacity of this section of the road.
Speed (mi/hr)
Density (veh/mi)
60.0
20
46.0
32
40.8
38
39.3
40
35.7
45
32.6
50
30.8
53
28.4
57
24.7
65
18.5
80
A linear regression analysis can be applied to a logarithmic transformation the
given data to estimate parameters in Greenberg’s model of traffic flow.
Greenberg’s model (Equation 6.19) is:
us = c ln k j − c ln k
A linear regression model takes the form y = a + bx; therefore, in this case, the
given data us and ln k correspond to y and x respectively, c ln kj is represented by
a, and c is represented by –b. The linear regression analysis can be performed
using Equations 6.21, 6.22, and 6.23, or using a computer software spreadsheet
package. Linear regression analysis yields values of a = 149.93
and b = - 30.00. Therefore, the jam density, kj can be found as follows:
In the regression model, a = c ln kj and b = -c
c ln kj = 149.9
c = 30.00
ln kj = 4.998
kj = 148.1 veh/mi
The fitted model is therefore:
us = c ln k j − c ln k = 30 ln 148.1 − 30 ln k = 150 − 30 ln k
An relationship between flow and density can be developed by
multiplying both sides by k:
q = 150k − 30k ln k
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72
Chapter 6: Fundamental Principles of Traffic Flow
Using the properties of Greenberg’s model, capacity can be found:
uo = c and ko = kj/e
uo = 30 mi/h
ko = 148.1/2.718 = 54.5 veh/mi
qmax = kouo = (30)(54.5) = 1635 veh/h
6-8
Researchers have used analogies between the flow of fluids and the movement of
vehicular traffic to develop mathematical algorithms describing the relationship
among traffic flow elements. Discuss in one or two paragraphs the main deficiencies
in this approach.
The main deficiency in using fluid flow theory to describe traffic flow is
that, unlike fluids which are continuous, traffic streams are made up of discrete
elements which have the ability to act independently of one another. In using
fluid flow theory, the traffic stream is analyzed at a macroscopic level, without
consideration given to the microscopic interaction between vehicles and the
effects of one vehicle on others in the stream. Average values are used to
describe the traffic stream over a given section (e.g. mean speed); characteristics
and attributes of individual vehicles are not considered.
6-9
Assuming that the expression:
us = u f e
−k / k j
can be used to describe the speed-density relationship of a highway, determine the
capacity of the highway from the data below using regression analysis.
k (veh/mi)
ū (mi/h)
s
43
50
8
31
38.4
33.8
53.2
42.3
Under what flow conditions is the above model valid?
First, to use linear regression, the equation given in the problem must be
converted to the form y = a + bx; this can be done by taking the natural logarithm
of each side of the equation as follows:
us = uf e-k/kj becomes ln(us) = ln(uf) - k/kj
73
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Chapter 6: Fundamental Principles of Traffic Flow
uf and kj are constants; therefore, in the regression,
a = ln(uf); b = -1/kj; x = k; and y = ln(us)
The only transformation needed for the input values is:
ln(us)
us
38.4
3.648057
33.8
3.520461
53.2
3.974058
42.3
3.744787
The linear regression analysis can be performed using equations 6.21, 6.22, and
6.23, or using a computer software spreadsheet package. Linear regression
analysis yields values of a = 4.0626 and b = -0.0103.
a) mean free flow speed, uf
uf = ea = e4.0626
uf = 58.1 mi/h
b) jam density, kj
kj = -1/b = -1/-0.0103
kj = 97.0 veh/mi
c) capacity, qmax
Capacity occurs at maximum flow. State flow in terms of density.
−k / k
q = k × u s = k (u f e j )
q = (58.1)(k)(e-k/97.0)
Take the derivative and set equal to zero to maximize flow; solve for density.
Take the natural logarithm of each side of the equation as follows:
ln(q) = (ln 58.1) - k/97.0 + ln(k)
Then take the derivative and set equal to zero.
1/q = -0.0103 + 1/k
0 = 1/k – 0.0103
k = 97.0 when q = qmax
Solve for q
−k / k
q = k × u s = k (u f e j )
q = 97.0(58.1e-1)
qmax = 2073 veh/h
Because k, at qmax, is approaching kj, this model is valid for high density
conditions only.
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74
Chapter 6: Fundamental Principles of Traffic Flow
6-10
Results of traffic flow studies on a highway indicate that the flow-density
relationship can be described by the expression:
q = uf k −
uf
kj
k2
If speed and density observations give the data shown below, develop an
appropriate expression for speed versus density for this highway, and determine the
density at which the maximum volume will occur as well as the value of the
maximum volume. Also plot speed versus density and volume versus speed for both
the expression developed and the data shown. Comment on the differences between
the two sets of curves.
Speed (mi/h)
Density (veh/mi)
50
45
40
34
22
13
12
18
25
41
58
71
88
99
Note that the given traffic flow relationship,
uf
q = uf k − k2
kj
is equivalent to
uf
q = k (u f − k )
kj
and
us = u f −
uf
kj
k
which is Greenshields’ model.
A linear regression model takes the form y = a + bx; therefore, in this case, the
given data us and k correspond to y and x respectively, while a = uf and b = uf/kj.
The linear regression analysis can be performed using Equations 6.21, 6.22, and
6.23, or using a computer software spreadsheet package. Linear regression
analysis yields values of a = 58.9336 and b = -0.49134. Therefore,
a) mean free flow speed, uf
uf = a = 58.9 mi/h
jam density, kj
75
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Chapter 6: Fundamental Principles of Traffic Flow
uf / kj = b = -0.49134
kj = 58.9 / 0.49134 = 119.88
jam density, kj = 120 veh/mi
b) density at maximum volume
State flow in terms of density.
uf
q = k × u s = k (u f −
k)
kj
q = 58.9k - (58.9/119.88)k2
q = 58.9k - 0.4913k2
Take the derivative and set equal to zero to maximize flow; solve for
density.
0 = 58.9 – 0.9826k
k = 59.94 when q = qmax
Solve for q
q = 58.9(59.94) - 0.4913(59.94)2
q = 1765
(a) maximum volume, qmax = 1765 veh/h
(b) density at maximum volume, k = 60 veh/mi
(c) Plot speed vs. density for data points and equation
(d) Comment on results.
The equation used to describe the speed-density relationship is linear, thereby
yield a straight-line relationship as shown above. The data follow an
approximately linear relationship, the extent of which could be quantified using
the R2 value that would be obtained through regression analysis or by calculating
the sum of squared errors (SSE) between the equation and the data points.
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76
Chapter 6: Fundamental Principles of Traffic Flow
6-11
Traffic on the eastbound approach of a signalized intersection is traveling at 40
mi/hr, with a density of 44 veh/mi/ln. The duration of the red signal indication for
this approach is 30 sec. If the saturation flow is 19500 veh/h/ln with a density of 51
veh/mi/ln, and the jam density is 120 veh/mi/ln, determine the following:
(i) The length of the queue at the end of the red phase
(ii) The maximum queue length
(iii) The time it takes for the queue to dissipate after the end of the red
indication.
Solution:
(i) The length of the queue at the end of the red phase. Determine speed of
backward forming shock wave ω13 when signals turn to red. Use Equation 6.33.
uw =
q2 − q1
k2 − k1
u13 =
q1 − q3
k1 − k3
q1 = (40 mi/hr)(44 veh/mi/ln) = 1760 veh/h/ln
q3 = 0 veh/h/ln
k1 = 44 veh/mi/ln
u13 =
1760 − 0
mi/h = 23.2 mi/hr
44 − 120
= 23.2 mi/h x 1.47 ft/sec/mi/h = 34.0 ft/sec
Length of queue at end of red phase = 30 x 34.0 = 1020 ft
Determine speed of backward recovery wave velocity. Use Equation 6.38:
u34 =
q3 − q4 0 − 1950
=
= 28.3 mi/h = 41.4 ft/sec
k3 − k4 120 − 51
(ii) The maximum queue length. Use Equation 6.40:
Maximum queue length =
rω13ω34
(30)(34.0)(41.4)
=
= 5707 ft
41.4 − 34.0
ω34 − ω13
(iii) The time it takes for the queue to dissipate after the end of the red
indication. Use Equation 6.41:
rω13
(30)(34)
=
= 137 seconds
ω13 − ω34 41.4 − 34.0
77
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Chapter 6: Fundamental Principles of Traffic Flow
6-12
A developer wants to provide access to a new building from a driveway placed 1000
ft upstream of a busy intersection. He is concerned that queues developing during
the red phase of the signal at the intersection will block access. If the speed on the
approach averages 35 mi/hr, the density is 50 veh/mi, and the red phase is 20 sec,
determine if the driveway will be affected. Assume that the traffic flow has a jam
density of 110 veh/mi and can be described by the Greenshields model.
The red phase of the traffic signal creates a stopping shock wave. The speed of a
stopping shock wave is given by Equation 6.47, as follows:
u w = −u f η1
where η1 = k/kj
uw = -(35)(50/110) = -15.9 mi/h
uw = -23.4 ft/s
In 20 seconds, the wave will have traveled backward (toward the driveway in
question)
(23.4)(20) = 468 ft/s, therefore not reaching the driveway.
6-13
Studies have shown that the traffic flow on a two-lane road adjacent to a school can
be described by the Greenshields model. A length of 0.5 mi adjacent to a school is
described as a school zone (see Figure 6.19) and operates for a period of 30 min just
before the start of school and just after the close of school. The posted speed limit
for the school zone during its operation is 20 mph. Data collected at the site when
the school zone is not in operation show that the jam density and mean free speed
for each lane are 118 veh/mi and 63 mph. If the demand flow on the highway at the
times of operation of the school zone is 95% of the capacity of the highway,
determine:
(i)
The speeds of the shock waves created by the operation of the school
zone
(ii)
The number of vehicles affected by the school zone during this 30minute operation
(i)
kj = 118, uf = 63
The traffic flow on this road section can be described by the Greenshields model
So, the capacity of this road section is given by Equation 6.18:
u k 118 × 63
qmax = f j =
= 1858.5vph
4
4
The demand flow is 95% of the capacity, so qA = 0.95 qmax = 1765.575
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78
Chapter 6: Fundamental Principles of Traffic Flow
q = uf k −
uf
kj
k 2 ⇒ q A = 1765.575 = 63k A −
63 2
kA
118
0.534k A2 − 63k A + 1765.575 = 0
kA = 45.822
Given the speed at the school zone uB = 20 mph
u
63
u s = u f − f k ⇒ uB = 20 = 63 −
k B ⇒ k B = 80.540
118
kj
qB = k B uB = 80.540 × 20 = 1610.794
Therefore, the backward shock wave speed is
q − q A 1610.794 − 1765.575
wAB = B
=
= −4.458mph
80.540 − 45.822
kB − k A
(ii)
In the school zone, the mean speed of traffic flow is 20 mph (forward), while the
backward shockwave speed is 4.458 (backward). Therefore, the relative speed of
the traffic flow to the end of shockwave is 20 + 4.458 = 24.458 mph.
During the 30-minute school zone operation, the affected traffic flow travels
24.458 * 30/60 = 12.229 miles.
The density of school zone is kB = 80.540 veh/mile
Therefore, the number of vehicles affected by the school zone operation is
12.229 * 80.540 = 985 vehicles
6-14
Briefly describe the different shock waves that can be formed and the traffic
conditions that will result in each of these shock waves.
Shock waves include frontal stationary, backward forming, backward recovery,
rear stationary and forward recovery shock waves. Frontal stationary shock
waves occur when capacity is reduced to zero and upstream demand continues.
Backward forming shock waves occur when capacity is reduced below the
demand flow rate but not to zero. Backward recovery shock waves form when
capacity is restored or increased to a value greater than the upstream demand. A
rear stationary shockwave occurs when a restricted downstream capacity is
increased to a value above the queued demand, thereby dissipating the queue from
through a forward recovery shock wave.
79
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Chapter 6: Fundamental Principles of Traffic Flow
6-15
Traffic flow on a three-lane (one direction) freeway can be described by the
Greenshields model. One lane of the three lanes on a section of this freeway will
have to be closed to undertake an emergency bridge repair that is expected to take 2
hours. It is estimated that the capacity at the work zone will be reduced by 30% of
that of the section just upstream stream of the work zone. The mean free speed of
the highway is 60 mph and the jam density is 140 veh/mi/ln. If it is estimated that
the demand flow on the highway during the emergency repairs is 80% of the
capacity, using the deterministic approach, determine:
(i)
The maximum queue length that will be formed
(ii)
The total delay
(iii)
The number of vehicles that will be affected by the incident
(iv)
The average individual delay
(i)
uf = 60 mph, and kj = 140 veh/mi/ln
Traffic flow can be described by the Greenshields model, therefore, the capacity
of this freeway is
u k
60 ×140
c = 3× f j = 3×
= 6300 veh/hr
4
4
The demand flow is 80% of the capacity, so it is
v = 80% ⋅ c = 80% × 6300 = 5040 veh/hr
Due to the work zone, the capacity is reduced by 30%, so the reduced capacity
is
cR = 70% ⋅ c = 70% × 6300 = 4410veh/hr
The duration of work zone is 2 hours. The maximum queue length is determined
by Equation 6.58.
qmax = (v − cR )t = ( 5400 − 4410 ) × 2=1980 veh
(ii)
The total delay is determined by Equation 6.61.
t 2 ( v − cR )( c − cR ) 22 ( 5040 − 4410 )( 6300 − 4410 )
=
= 1890 hours
dT =
2 (c − v)
2 ( 6300 − 5040 )
(iii)
The number of vehicles affected by the work zone
= v × t = 5040 × 2 = 10080 vehicles
(iv)
The average individual delay = dT / n = 1890 / 10080 = 0.1875 hours.
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80
Chapter 6: Fundamental Principles of Traffic Flow
6-16
Repeat Problem 6-15 for the expected repair periods of 1 hr, 1.5 hr, 2.5 hr, 2.75 hr,
and 3 hr. Plot a graph of average individual delay vs the repair period and use this
graph to discuss the effect of the expected repair time on the average delay.
Average Delay (hrs)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
2
2.5
3
3.5
Duration Time (hrs)
6-17
Repeat Problem 6-15 for the expected demand flows of 70%, 75%, 85%, and 90%
of the capacity of the highway. Plot a graph of average individual delay vs the
expected demand flow and use this graph to discuss the effect of the expected
demand flow on the average delay.
1.8
Average delay (hrs)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
70%
75%
80%
85%
81
90%
95%
100%
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Chapter 6: Fundamental Principles of Traffic Flow
6-18
Traffic flow on a section of a two-lane highway can be described by the Greenshields
model, with a mean free speed of 55 mph and a jam density of 145 veh/mi/ln. At the
time when the flow was 90% of the capacity of the highway, a large dump truck
loaded with heavy industrial machinery from an adjacent construction site joins the
traffic stream and travels at a speed of 15 mi/hr for a length of 3.5 mi along the
upgrade before turning off onto a dump site. Due to the relatively high flow in the
opposite direction, it is impossible for any car to pass the truck. Determine how
many vehicles will be in the platoon behind the truck by the time the truck leaves
the highway.
Knowing the traffic conditions can be described using Greenshields’ model
allows use of Equation 6.46 which relates speed of the shockwave to the densities
associated with flow before and during the blockage.
uw = u f [1 − (η1 + η 2 )] in which η1 = k1/kj and η2 = k2/kj
To find k1, use the form of Greenshields’ model expressed in equation 6.15:
u
q = uf k − f k2
kj
Given that q1 = 90% of capacity,
ufkj
(55)(145)
q1 = 0.90q max = 0.90
= 0.90
= 1794 veh/h/ln
4
4
q1 = u f k1 −
uf
2
k1 = 1794 = 55k1 −
kj
k1 = 49.6 veh/mi/ln
55 2
k1
145
To find k2, apply the general form of Greenshields’ model, Equation 6.13:
uf
us = u f − k
kj
u2 = u f −
uf
kj
k 2 = 15 = 55 −
55
k2
145
k2 = 105.5 veh/mi/ln
⎡ ⎛ 49.6 105.5 ⎞⎤
u w = u f [1 − (η1 + η 2 )]= 55⎢1 − ⎜
+
⎟⎥ = - 3.8 mi/h
⎣ ⎝ 145 145 ⎠⎦
The growth rate of the platoon is 15 mi/h forward and 3.8 mi/h backward, which
is a total of 18.8 mi/h.
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Chapter 6: Fundamental Principles of Traffic Flow
The truck is on the highway for 3.5 mi / 15 mph = 0.233 hr
The platoon length at the end of the duration is (18.8 mi/h)(0.233 hr) = 4.38 mi
The number of vehicles in the platoon is (4.38 mi)(105.5 veh/mi) = 462 vehicles.
6-19
Briefly discuss the phenomenon of gap acceptance with respect to merging and
weaving maneuvers in traffic streams.
Merging is the process of entering a roadway from a minor road or ramp
and weaving is the process of moving across lanes of a major facility. A vehicle
would merge onto a freeway and weave left to reach a left side exit. In the
process of merging and weaving, drivers must assess available gaps in the traffic
stream and determine if they are large enough to move into safely. A gap can be
defined in terms of time or space, and the minimum time or space acceptable to
drivers varies. Based on observations of accepted and rejected gap sizes on a
roadway, a critical gap can be estimated and from this, the expected number of
acceptable gaps can be determined. This value will allow one to calculate the
storage space required on a ramp, or the point at which a signal should be
installed at an unsignalized intersection.
6-20
The table below gives data on accepted and rejected gaps of vehicles on the minor
road of an unsignalized intersection. If the arrival of major road vehicles can be
described by the Poisson distribution, and the peak hour volume is 1100 veh/hr,
determine the expected number of accepted gaps that will be available for minor
road vehicles during the peak hour.
Number of
Number of
Gap (t) (s)
Rejected Gaps > t
Accepted Gaps < t
1.5
2.5
3.5
4.5
5.5
92
52
30
10
2
3
18
35
62
100
83
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Chapter 6: Fundamental Principles of Traffic Flow
First, determine critical gap, tc , using the algebraic method. Determine the
change in number of accepted and rejected gaps for the gap lengths given, shown
in the following table.
Range of Gap
Lengths (s)
1.5 – 2.5
2.5 – 3.5
3.5 – 4.5
4.5 – 5.5
Change in Number of
Accepted Gaps
92 – 52 = 40
52 – 30 = 22
30 – 10 = 20
10 – 2 = 8
Change in Number of
Rejected Gaps
18 – 3 =15
35 – 18 = 17
62 – 35 = 27
100 – 62 = 38
Difference
40 – 15 = 25
22 – 17 = 5
27 – 20 = 7
38 – 8 = 30
The critical gap occurs in the range exhibiting the smallest difference
between change in number of gap accepted and change in number of gaps
rejected; in this case, this is between 2.5 and 3.5 seconds.
Using Equation 6.49, determine the value of the critical gap.
tc = t1+ [Δt(r-m)] / [(n-p)+(r-m)]
tc= 2.5 + 1(52-18) / [(35-30)+(52-18)]
tc= 3.37 s
Then, using Equation 6.54, determine the expected number of available gaps
during the peak hour.
V= 1100 veh.
T = 3600 sec.
λ = 1100/3600 = 0.306
Freq (h≥t) = (V-1)(e-λt) = (1,100-1)(e-(0.306(3.37))) = 392 gaps
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84
Chapter 6: Fundamental Principles of Traffic Flow
6-21
Using appropriate diagrams, describe the resultant effect of a sudden reduction of
the capacity (bottleneck) on a highway both upstream and downstream of the
bottleneck.
The diagrams below illustrate the impact of a bottleneck on traffic flow.
In the first diagram, the reduction in capacity is shown (C1 reduced to C2) and the
corresponding density at capacity changes from ko1 to ko2. The second diagram
illustrates the effect of the bottleneck in terms of the shock wave that is formed.
When the flow is reduced due to the bottleneck, a queue is formed and continues
to grow as long as the demand flow is greater than the service flow. The rate at
which the queue grows is dependent on the speed of the shock wave, uw.
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Chapter 6: Fundamental Principles of Traffic Flow
6-22
The capacity of a highway is suddenly reduced to 60% of its normal capacity due to
closure of certain lanes in a work zone. If the Greenshields model describes the
relationship between speed and density on the highway, the jam density of the
highway is 112 veh/mi, and the mean free speed is 64.5 mi/hr, determine by what
percentage the space mean speed at the vicinity of the work zone will be reduced if
the flow upstream is 80% of the capacity of the highway.
In Greenshields’ traffic flow model,
qmax = kjuf/4
qmax = (112)(64.5)/4
qmax = 1806 veh/h
Therefore, capacity at the site is normally 1806 veh/h.
Upstream flow is 80% of capacity,
qupstream = (0.80)(1806) = 1444 veh/h
Upstream density can be found the following form of Greenshields’ model:
q = uf k −
uf
kj
k2
q = (64.5)k – (64.5/112)k2
0.57589k2 – 64.5k + 1444 = 0
Using the quadratic formula, k = 30.9 veh/mi
Upstream speed can then be found using q=kus
uupstream = q/k = 1444/30.9
uupstream = 46.7 mi/h
At the work zone, flow is at 60% of capacity,
qworkzone = (0.60)(1806) = 1084 veh/h
Upstream density can be found as:
q = (64.5)k – (64.5/112)k2
0.57589k2 – 64.5k + 1084 = 0
Using the quadratic formula, k = 91.4 veh/mi
Upstream speed can then be found using q=kus
uupstream = q/k = 1084/91.4
uupstream = 11.9 mi/h
The percentage reduction in speed due to the work zone is:
(46.7-11.9)/46.7 = 74.6%
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Chapter 6: Fundamental Principles of Traffic Flow
6-23
The arrival times of vehicles at the ticket gate of a sports stadium may be assumed
to be Poisson with a mean of 30 veh/hr. It takes an average of 1.5 min for the
necessary tickets to be bought for occupants of each car.
(a) What is the expected length of queue at the ticket gate, not including the vehicle
being served?
(b) What is the probability that there are no more than 5 cars at the gate, including
the vehicle being served?
(c) What will be the average waiting time of a vehicle?
q = 30 veh/h (arrival rate)
Q = 40 veh/h (service rate)
a) Expected queue length
Using Equation 6.64,
E(m) = q2/[Q(Q-q)] = (30)2/[40(40-30)] = 2.25 vehicles
b) Probability of no more than 5 cars
Using Equation 6.69,
P(n>N) = (q/Q)N+1 (Probability of more than N)
P(n>5) = (30/40)6 = 0.178
for P(n<=5) = 1-0.178 = 0.822
c) Average waiting time per vehicle
Using Equations 6.66 and 6.65,
E(v) = 1/(Q-q) = 0.1 hr = 6 minutes wait time including queue time and
service time
E(w) = q/(Q(Q-q)) = 0.075 hr = 4.5 minutes wait time in the queue
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Chapter 6: Fundamental Principles of Traffic Flow
6-24
An expressway off-ramp consisting of a single lane leads directly to a tollbooth. The
rate of arrival of vehicles at the expressway can be considered to be Poisson with a
mean of 50 veh/hr, and the rate of service to vehicles can be assumed to be
exponentially distributed with a mean of 1 min.
(a) What is the average number of vehicles waiting to be served at the booth (that is,
the number of vehicles in queue, not including the vehicle being served)?
(b) What is the length of the ramp required to provide storage for all exiting
vehicles 85% of the time? Assume the average length of a vehicle is 20 ft and that
there is an average space of 5 ft between consecutive vehicles waiting to be served.
(c) What is the average waiting time a driver waits before being served at the
tollbooth (that is, the average waiting time in the queue)?
a) Expected queue length
Using Equation 6.64,
E(m) = q2/[Q(Q-q)] = (50)2/[60(60-50)] = 4.17 vehicles
b) Ramp length
0.15 = (0.83)N+1
ln(0.15) = ln(0.83)(N+1)
10.4 = N+1
N = 9.4 vehicles; use 10 vehicles
Ramp length = 10 veh (20 ft/veh) + 9 spaces (5 ft/space) = 245 feet
c) Average waiting time per vehicle
Using Equations 6.66 and 6.65,
E(v) = 1/(Q-q) = 0.1 hr = 6 minutes wait time including queue time and
service time
E(w) = q/(Q(Q-q)) = 0.083 hr = 5 minutes wait time in the queue
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88
Chapter 7
Intersection Design
7-1
Briefly describe the different principles involved in the design of at-grade
intersections.
The fundamental objective in the design of at-grade intersections is to
minimize the severity of potential conflicts both among different streams of traffic
and between pedestrians and turning vehicles while facilitating smooth traffic
flow. The design should therefore incorporate the operating characteristics of
both the vehicles and pedestrians using the intersection. The design of an at-grade
intersection involves the design of the alignment (horizontal and vertical) of the
intersecting roadways, the design of a suitable channeling system, the
determination of the minimum required widths of turning roadways when traffic
is expected to make turns at speeds higher than 15 mi/h, and the assurance that
sight distances are adequate for the type of control at the intersection.
7-2
Describe the different types of at-grade intersections. Also give an example of an
appropriate location for the use of each type.
The basic types of at-grade intersections are T (three-leg) intersections,
four-leg intersections, multi-leg intersections consisting of 5 or more approaches,
and traffic circles. A simple T intersection is suitable for intersections of minor
roads. Four-leg intersections are used mainly at locations where minor or local
roads crossed, although it can be used where a minor road crosses a major
highway. Multi-leg intersections should be avoided whenever possible. When a
multi-leg intersection exists, one leg should be realigned, if possible, into a T
intersection with the minor road, at a distance far enough away from the 4-leg
intersection to allow for independent operation of the intersections. Traffic circles
force traffic to use the intersection in a circular pattern, thereby transforming
crossing conflicts into merging and diverging conflicts. The neighborhood traffic
circle is placed for traffic calming purposes on local streets in residential areas to
reduce travel speeds and cut-through traffic.
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Chapter 7: Intersection Design
7-3
Describe the different types of traffic circles indicating under what conditions you
will recommend the use of each.
According to a Federal Highway Administration report, traffic circles can
be grouped into three categories: rotaries, neighborhood traffic circles, and
roundabouts. A rotary could be where maintaining high speed is important and a
large amount of right-of-way is available to accommodate the large radius
required for a rotary. A neighborhood traffic circle may be suitable at the
intersection of local streets in a low-speed situation where there is interest in
traffic calming measures; typically stop control or no control is used on the
approaches. A roundabout may be suitable for situations intermediate to the two
described above.
7-4
What are the key defining characteristics of roundabouts that distinguish them
from other traffic circles?
Roundabouts, as opposed to other types of traffic circles, have yield
control is used on the approaches, and the roundabout design speed typically does
not exceed 30 miles per hour. Conflicting traffic movements are separated by
pavement markings or channelizing islands such that vehicles are given detailed
guidance as to proper place within the roundabout. Parking is prohibited within
the roundabout to maximize capacity.
7-5
What are the main functions of channelization at an at-grade intersection?
Channelization is intended to separate conflicting traffic movements into
defined paths to facilitate the safe and orderly movements of both vehicles and
pedestrians. This can accomplish many things, including:
• Direct the paths of vehicles so that not more than two paths cross at any
one point
• Control the merging, diverging, or crossing angle of vehicles
• Decrease vehicle wander and the area of conflict among vehicles by
reducing the amount of paved area
• Provide a clear indication of the proper path for different movements
• Give priority to the predominant movements
• Provide pedestrian refuge
• Provide separate storage lanes for turning vehicles, thereby creating space
away from the path of through vehicles for turning vehicles to wait
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Chapter 7: Intersection Design
•
•
•
•
Provide space for traffic control devices so that they can be readily seen
Control prohibited turns
Separate different traffic movements at signalized intersections with
multiple-phase signals
Restrict the speeds of vehicles
7-6
Discuss the fundamental general principles that should be used in designing a
channelized at-grade intersection.
There are several fundamental principles considered in channelized
intersection design. Regarding guidance to the motorist, motorists should not be
required to make more than one decision at a time. Merging and weaving areas
should be as long as possible, but other areas of conflict between vehicles should
be reduced to a minimum. Crossing traffic streams that do not weave or merge
should intersect at 90°, although a range of 60-120° is acceptable; adequate sight
distance is provided regardless of angle. Sharp reverse curves and turning paths
greater than 90° should be avoided. Separate space should be provided for
turning vehicles so that they do not interfere with the movement of through
vehicles. Prohibited turning movements should be blocked with channelized
islands wherever possible. Finally, the location of essential traffic control devices
should be considered in the design process.
7-7
Describe the different types of islands used in channelizing at-grade intersections
indicating the principal function of each type.
Traffic islands can be divided into three categories: channelizing,
divisional, and refuge. Channelizing islands control and direct traffic in merging
and diverging situations to guide motorists into the correct lanes for their
movements at the intersection. Divisional islands to divide traffic inoppposing
directions at intersections to provide clear and separate paths for each direction of
traffic. Refuge islands provide a stopping place out of the path of motorists for
pedestrians crossing wide intersections.
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Chapter 7: Intersection Design
7-8
Figure 7.25a illustrates a three-leg intersection of State Route 150 and State Route
30. Both roads carry relatively low traffic with most of the traffic oriented along
State Route 150. The layout of the intersection, coupled with the high-speed traffic
on State Route 150, have made this intersection a hazardous location. Drivers on
State Route 30 tend to violate the stop sign at the intersection because of the mild
turn onto westbound State Route 150, and they also experience difficulty in seeing
the high-speed vehicles approaching from the left on State Route 150. Design a new
layout for the intersection to eliminate these difficulties for the volumes shown in
Figure 7.25b. Design vehicle is a passenger car.
The intersection shown in the figure is a T-intersection with the two
roadways intersecting at an angle less than 90 degrees. This creates a condition
where the right turning vehicles from the minor road (SR 30) tend to violate the
stop sign because the intersection configuration causes that movement to resemble
a through movement. The intersection should be realigned so that the minor road
(SR 30) intersects SR 150 at a right angle. This will cause the minor road rightturning vehicles to make a much more defined turn, forcing them to slow down
and stop. The realignment will clearly indicate that SR 30 is the minor road, and
by creating a perpendicular intersection, drivers turning either way from the minor
road will have improved sight lines along the major road. The following figures
show the existing intersection and proposed realignment, respectively.
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Chapter 7: Intersection Design
7-9
Figure 7.26a shows the staggered unsignalized intersection of Patton Avenue and
Goree Street. The distance between the T intersections is about 160 ft. The general
layout and striping of the lanes at this intersection result in confusion to drivers and
create multiple conflicts. Design an improved layout for the intersection for the
traffic volumes shown in Figure 7.26b. Design vehicle is a passenger car.
The staggered intersection of Patton Avenue and Goree Street will tend to
cause confusion among drivers who expect to find cross streets intersecting
perpendicularly, without the stagger as shown. The east leg of Patton Avenue
should be realigned to produce a 4-leg intersection. Doing so should reduce the
number of turning accidents, which appear to be the most common type of
accident at this location. The high number of rear end accidents should also be
reduced since motorist confusion will be reduced.
7-10
A ramp from an expressway with a design speed of 35 mi/hr connects with a local
road, forming a T intersection. An additional lane is provided on the local road to
allow vehicles from the ramp to turn right onto the local road without stopping. The
turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle. Determine
the width of the turning roadway if the design vehicle is a single-unit truck. Use 0.08
for superelevation.
The first step is to determine the minimum radius of curvature for the ramp, using
Equation 3.34:
R = u2 / 15(e + fs)
R = (35)2/15(0.08+0.18)
R = 314.10 ft
From Table 7.5, determine appropriate pavement width, W, knowing that:
Operational requirements: Case I
Traffic conditions: Type B
Edge of pavement treatment: Stabilized shoulder
Using the table, interpolating for R = 314 ft, W = 15 ft
Therefore, the width of the turning roadway should be 15 ft.
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Chapter 7: Intersection Design
7-11
Determine the required width of the turning roadway in Problem 7-10 for a twolane operation with barrier curbs on both sides.
The first step is to determine the minimum radius of curvature for the ramp, using
Equation 3.34:
R = u2 / 15(e + fs)
R = (35)/15(0.08+0.18)
R = 314.10 ft
From Table 7.5, determine appropriate pavement width, W, knowing that:
Operational requirements: Case III
Traffic conditions: Type B
Edge of pavement treatment: Barrier curb on both sides
Using the table, interpolating for R = 314 ft, W = 28 ft: for the edge of pavement
treatment specified, add 2 ft.
Therefore, the width of the turning roadway should be 30 ft.
7-12
Repeat Problem 7-10 for a one-lane, one-way operation with provision for passing a
stalled vehicle.
The first step is to determine the minimum radius of curvature for the ramp, using
Equation 3.34:
R = u2 / 15(e + fs)
R = (35)/15(0.08+0.18)
R = 314.10 ft
From Table 7.5, determine appropriate pavement width, W, knowing that:
Operational requirements: Case II
Traffic conditions: Type B
Edge of pavement treatment: Stabilized shoulder
Using the table, interpolating for R = 314 ft, W = 20 ft
Therefore, the width of the turning roadway should be 20 ft
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94
Chapter 7: Intersection Design
7-13
A four-leg intersection with no traffic control is formed by two 2-lane roads with the
speed limits on the minor and major roads being 25 and 35 mi/hr,
respectively. If the roads cross at 90° and a building is to be located at a
distance of 45 ft from the centerline of the nearest lane on the minor road,
determine the minimum distance at which the building should be located
from the centerline of the outside lane of the major road so that adequate
sight distances are provided.
To ensure sufficient sight distances for each intersection approach, a line
of sight must be provided that allows for sufficient time for a driver to perceive
and react to a vehicle on the crossing approach. The required distance (offset of
building from roadway) can be found using Table 7.7 and Equation 7.4.
db
a
=
d a ( d a − b)
Given:
b = 45 ft
From Table 7.7:
da = 165 ft (major approach)
db = 115 ft (minor approach)
115 / 165 = a / (165 – 45)
a = 84 ft
The distance between the centerline of the outside lane on the major road
and the building should be 84 ft.
7-14
What are the main deficiencies of multi-leg intersections? Using a suitable diagram
show how you will correct for these deficiencies.
The main deficiencies of multi-leg intersections are associated with sight
distance and clear assignment of right-of-way. Intersections with more than four
approaches can be corrected by relocating one or more legs to intersect with
another approach away from the main intersection.
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Chapter 7: Intersection Design
7-15
A two-lane minor road intersects a two-lane major road at 90° forming a four-leg
intersection with traffic on the minor road controlled by a yield sign. A
building is located 125 ft from the centerline of the outside lane of the
major road and 35 ft from the centerline of the nearest lane of the minor
road. Determine the maximum speed that can be allowed on the minor
road if the speed limit on the major road is 45 mi/hr.
Given:
a = 125 ft
b = 35 ft
Table 7.9 discusses the different cases when different speed limits are applied on
the minor road.
When v min or = 15 mph, d b = 75 ft < a = 125 ft. Hence, there is no restriction on
the speed limit on the major road.
When v min or = 20 mph, d b = 100 ft < a = 125 ft. Hence, there is no restriction on
the speed limit on the major road.
When v min or = 25 mph, d b = 130 ft > a = 125 ft, and t g = 6.5 sec. From Equation
7.7, the sight distance along the major road should be
d a = 1.47v major t g = 1.47 × 45 × 6.5 = 430 ft
At site, the maximum sight distance along the major road can be computed using
Equation 7.4 :
d a ,mzx
b
=
db
(d b − a)
da, max = (130)(35)/(130-125) = 910 ft > da = 430 ft.
Hence, there is still allowance for increasing speed limit on the minor road.
When v min or = 30 mph, d b = 160 ft > a = 125 ft, and t g = 6.5 sec. From Eq. 7.7,
the sight distance along the major road should be
d a = 1.47v major t g = 1.47 × 45 × 6.5 = 430 ft
At the site, the maximum sight distance along the major road can be computed
using Equation 7.4 :
d a ,mzx
b
=
db
(d b − a )
da, max = (160)(35)/(160-125) = 160 ft < da = 430 ft.
Hence, the maximum speed that can be allowed on the minor road is 25 mph.
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Chapter 7: Intersection Design
7-16
If the speed limit were 40 mi/hr on the major road, and 20 mi/hr on the minor road
in Problem 7-15, determine the minimum distance that the building can be
located from the centerline of the outside lane of the major street.
Given:
b = 35 ft
In Table 7.9 and Table 7.10, the suggested length of leg on the minor road
is d b = 100 ft, and that on the major road is d a = 385 ft, when the speed limits are
40 mph on the major and 20 mph on the minor.
The minimum distance of the building from the centerline of the outside
lane of the major street along the major road can be computed using Eq. 7.4 :
db
a
=
d a ( d a − b)
a = 100*(385 – 35)/385 = 91 ft
The distance from the building to the major road should be at least 91 ft.
7-17
A developer has requested permission to build a large retail store at a location
adjacent to the intersection of an undivided four-lane major road and a
two-lane minor road. Traffic on the minor road is controlled by a stop sign.
The speed limits are 35 and 50 mi/hr on the minor and major roads,
respectively. The building is to be located at a distance of 65 ft from the
near lane of one of the approaches of the minor road. Determine where the
building should be located relative to the centerline of the outside lane of
the major road in order to provide adequate sight distance for a driver on
the minor road to turn right onto the major road after stopping at the stop
line. Design vehicle is a single-unit truck. Lanes on the major road are 12 ft
wide.
Use Equation 7.5:
dISD = 1.47vmajortg
From Table 7.8, tg = (9.5 – 1.0) = 8.5 sec for a single-unit truck. No adjustment
can be made for grade on the minor road since that information is not provided.
For an undivided four-lane road, add 0.7 seconds, such that tg = 9.2 sec
dISD = 1.47vmajortg = (1.47)(50)(9.2) = 676 ft.
For practical purposes, this may be rounded to 676 ft.
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Chapter 7: Intersection Design
7-18
Repeat Problem 7-17 for an intersection of a divided four-lane major road and a
two-lane minor road, if the median width on the major road is 8 ft and the
approach grade on the minor road is 4%.
Use Equation 7.5:
dISD = 1.47vmajortg
From Table 7.8, tg = (9.5-1.0) = 8.5 sec for a single-unit truck. An
adjustment of 0.2 seconds can be made for the 4% grade on the minor road. For a
divided four-lane road, add (0.7)(1.5) = 1.05 seconds, such that tg = 8.5 + 0.2 +
1.05 = 9.75 sec
dISD = 1.47vmajortg = (1.47)(50)(9.75) = 717 ft.
For practical purposes, this may be rounded to 720 ft.
7-19
A minor road intersects a four-lane divided highway at 90° forming a T intersection.
The median width on the major road is 8 ft. The speed limits on the major and
minor roads are 55 and 35 mi/hr, respectively. Determine the minimum sight
distance required for a single unit truck on the minor road to depart from a stopped
position and turn left onto the major road.
Use Equation 7.5:
dISD = 1.47vmajortg
From Table 7.8, tg = 9.5 sec for a single-unit truck. No adjustment can be made
for grade on the minor road since that information is not provided. For a divided
four-lane road, add (0.7)(1.5) = 1.05 seconds, such that tg = 10.55 sec
dISD = 1.47vmajortg = (1.47)(55)(10.55) = 852.97 ft.
For practical purposes, this may be rounded to 860 ft.
7-20
What additional consideration should be given to the sight distances computed in
Problems 7-17 and 7-19 for the design vehicle crossing the intersection?
Minimum requirements determined for right and left turns as presented for Cases
B1 and B2 will usually satisfy the requirement for the crossing maneuver.
AASHTO, however, recommends that the available sight distance for crossing be
checked when the following conditions exist:
• when only crossing maneuver is allowed at the intersection;
• when the crossing maneuver will involve an equivalent width of more than
six lanes, or
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Chapter 7: Intersection Design
• where the vehicle mix of the crossing traffic includes a substantial number
of heavy vehicles and the existence of steep grades that may slow down these
heavy vehicles while their back portion is still in the intersection.
7-21
For the information given in Problem 7-19, determine the sight distance required by
the minor-road vehicle to safely complete a right turn onto the major road.
Use Equation 7.5
dISD = 1.47vmajortg
From Table 7.8, tg = (9.5 – 1.0) = 8.5 sec for a single-unit truck. No adjustment
can be made for grade on the minor road since that information is not provided.
For a divided four-lane road, add (0.7)(1.5) = 1.05 seconds, such that tg = 9.55 sec
dISD = 1.47vmajortg = (1.47)(55)(9.55) = 772 ft.
For practical purposes, this may be rounded to 780 ft.
7-22
Repeat Problem 7-19 for an oblique intersection with an acute angle of 35°.
The acute angle will increase the distance traveled by a vehicle making a left turn.
The original distance W1 = 12*2 + 8 = 32 ft, and the distance actually traveled is
W2 =
W1
32
=
= 55.8 = 56 ft
Sinθ Sin(35 o )
Hence the extra distance traveled is D = 56 – 32 = 24 ft, which is equivalent to
two lanes.
The adjustment in tg for the extra distance is 0.7*2 = 1.4 sec, such that tg = 10.55
+ 1.4 = 11.95 sec.
dISD = 1.47vmajortg = (1.47)(55)(11.95) = 966 ft.
For practical purposes, this may be rounded to 970 ft.
7-23
Briefly discuss what factors are considered in the design of an at-grade railway
crossing in addition to those for an at-grade crossing formed by two highways.
As with intersections of roadways, general design principles include the
crossing occurring at a right angle if possible and far enough from intersections
and entrances so that turning maneuvers at these locations do not affect operations
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at the highway-rail grade crossing. Additional design criteria for the highway-rail
grade crossing include the surface of the crossing roadway at the same level as the
top of the tracks for a distance of 2 ft from the outside of the rails. Also, at a point
on the roadway 30 ft from the nearest track, the elevation of the road should not
be higher or lower than 0.25 ft than the elevation of the tracks unless this cannot
be achieved because of the superelevation of the track.
7-24
A two-lane road crosses an at-grade railroad track at 90°. If the design speed of the
two-lane road is 45 mi/hr and the velocity of the train when crossing the
highway is 80 mi/hr, determine the sight distance leg along the railroad
tracks to permit a vehicle traveling at the design speed to safely cross the
tracks when a train is observed at a distance equal to the sight distance leg.
Use Equation 7.9 to calculate the required sight distance leg along the railroad
tracks.
2
v
v
d T = T (1.47v v t + v + 2 D + L + W )
a
vV
30
g
dT = 80 / 45 [1.47(45)(2.5) + ((45)2 / (30)(0.35)) + 2(15) + 65 + 5]
dT = 814.63 ft
Therefore, the sight distance leg to be provided along the railroad tracks is 815 ft.
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Chapter 7: Intersection Design
7-25
A stop sign controls all vehicles on the highway at a railroad crossing. Determine the
minimum distance a building should be placed from the centerline of the tracks to
allow a stopped vehicle to safely clear the intersection. Assume that the building is
located 40 ft from the centerline of the near lane. The velocity of trains approaching
the crossing is 85 mi/hr.
First, use Equation 7.10 to calculate the required sight distance leg along the
railroad tracks.
d T = 1.47vT [
vg
a1
+ (L + 2D + W − d a )
1
+ J]
vg
dT = 1.47(85) [ 8.8/1.47 + (65 + 2(15) + 5 - 26.3) 1/8.8 + 2]
dT = 2,044 ft
Equation 7.4 can then be used to solve for the distance between the building and
the centerline of the railroad tracks.
db
a
=
d a ( d a − b)
where b = 40 ft, da = 2044 ft, and db is assumed to be 15 ft.
a = (2044-40)(15)/2044
a = 14.7 ft
The building should be located at least 14.7 ft from the railroad tracks.
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Chapter 8: Intersection Control
Chapter 8
Intersection Control
8-1
Using an appropriate diagram, identify all the possible conflict points at an
unsignalized T intersection.
Diverging Conflict
Merging Conflict
Crossing Conflict
8-2
A two-phase signal system is installed at the intersection described in Problem 8-1,
with channelized left-turn lanes and shared through and right-turn lanes. Using a
suitable diagram, determine the possible conflict points. Indicate the phasing
system used.
The recommended phasing for the intersection configuration shown below
consists of two phases: one for westbound traffic (left and through, one for
westbound traffic (through) and eastbound traffic (through-right), and one for
northbound traffic (left and through-right).
Phase 1:
Diverging Conflict
Merging Conflict
Crossing Conflict
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Chapter 8: Intersection Control
Phase 2:
8-3
Using appropriate diagrams, determine the possible conflict points on a four-leg
signalized intersection for a two-phase system. Assume no turn on red.
Phase 1:
Diverging Conflict
Merging Conflict
Crossing Conflict
Phase 2:
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Chapter 8: Intersection Control
8-4
Repeat Problem 8-3 for the following phasing systems:
(a) Four-phase with separate phases for left turns
(b) Four-phase with separate phase for each approach
(a) Four-phase with separate phases for left turns
Phase 1:
Diverging Conflict
Merging Conflict
Crossing Conflict
Phase 2:
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Chapter 8: Intersection Control
Phase 3:
Phase 4:
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Chapter 8: Intersection Control
(b) Four-phase with separate phase for each approach
Phase 1:
Diverging Conflict
Merging Conflict
Crossing Conflict
Phase 2:
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Chapter 8: Intersection Control
Phase 3:
Phase 4:
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Chapter 8: Intersection Control
8-5
Under what conditions would you recommend the use of each of the following
intersection control devices at urban intersections:
(a)
yield sign
(b)
stop sign
(c)
multiway stop sign
(a) A yield sign is used at the junction where a minor road intersects a major
road. The sign would be placed on the minor road if its approach speed is greater
than 10 mi/h. It is also warranted where a separate or channelized right turn lane
exists without an adequate acceleration lane.
(b) A stop sign is used where an approaching vehicle is required to stop before
entering an intersection. The warrants for stop signs suggest that a stop sign may
be used on a minor road when it intersects a major road, at an unsignalized
intersection, and where a combination of high speed, restricted view, and serious
crashes indicate the necessity for such a control.
(c) A multiway stop sign is normally used when the traffic volumes on all
intersection approaches are approximately equal and minimum volume warrants
are met; it is also used as a safety measure at some intersections.
8-6
Both crash rates and traffic volumes at an unsignalized urban intersection have
steadily increased during the past few years. Briefly describe the types of data you
will collect and how you will use those data to justify the installation of a traffic
signal at the intersection.
To justify the installation of a traffic signal, traffic volume data, pedestrian
volumes, and accident data would be collected. These data would be compared
with the signalization warrants in the MUTCD to determine the need for the
signal. The warrants pertain to:
•
•
•
•
•
•
•
•
•
•
•
Minimum vehicular volume
Interruption of continuous traffic
Minimum pedestrian volume
School crossing
Progressive movement
Crash experience
Systems
Combination of warrants
Four-hour volume
Peak-hour delay
Peak-hour volume
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Chapter 8: Intersection Control
8-7
A traffic signal control is being designed for a four-leg intersection on a divided
highway with the characteristics shown in the table below. Determine an
appropriate length of the yellow interval for each approach and how you will
provide it.
N-S Approaches
E-W Approaches
Median width, (ft)
18
10
Number of 12 ft lanes
on each approach
3
2
Design speed, mph
45
35
Grade
0
3.5
Assume the average vehicle length L is 20ft, and the perception-reaction time δ is
1.0 sec. Use the AASHTO recommended deceleration rate a of 11.2 ft/sec2. Use
Equation 8.5.
For N-S approach,
W = 2 × 2 ×12 + 10 = 58 ft,
uo = 45 mph,
G = 0.
uo
W +L
58 + 20
45 ×1.47
+
= 1.0 +
+
= 5.13sec
τ min = δ +
2 ( a + Gg )
45 × 1.47 2 (11.2 + 0 )
uo
A yellow interval of 5.5 seconds will be provided for the N-W approach.
For E-W approach,
W = 3 × 12 × 2 + 18 = 90 ft,
uo = 35 mph,
G = 3.5.
uo
W +L
90 + 20
35 × 1.47
+
= 1.0 +
+
= 5.22sec
τ min = δ +
2 ( a + Gg )
35 × 1.47 2 (11.2 + 0.035 × 32.2 )
uo
A yellow interval of 5.5 seconds will be provided for the E-W approach.
8-8
Determine the minimum green times for each approach in Problem 8-7 if the
effective crosswalk width in each direction is 8 ft and the number of pedestrians
crossing during an interval is 30 in the E-W direction and 25 in the N-S direction.
Since the effective crosswalk width WE for each approach is 8 ft < 10ft, the
minimum green time will be determined by Equation 8.12.
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Chapter 8: Intersection Control
For N-S approach,
L = 2 × 2 ×12 + 10 = 58 ft,
Sp = 4 ft/sec,
Nped = 25.
The minimum green time is
L
58
G p = 3.2 +
+ 0.27 N ped = 3.2 + + 0.27 × 25 = 24.45sec
Sp
4
For E-W approach,
L = 3 × 12 × 2 + 18 = 90 ft,
Sp = 4 ft/sec,
Nped = 30.
The minimum green time is
L
90
G p = 3.2 +
+ 0.27 N ped = 3.2 + + 0.27 × 30 = 33.8sec
Sp
4
8-9
For the geometric and traffic characteristics shown below determine a suitable
signal phasing system and phase lengths for the intersection using the Webster
method. Show a detailed layout of the phasing system and the intersection geometry
used.
Approach (Width)
Peak hour
approach volumes
Left turn
Through movement
Right turn
Conflicting
pedestrian volumes
PHF
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
133
420
140
900
73
373
135
1200
168
563
169
1200
134
516
178
900
0.95
0.95
0.95
0.95
Assume the following saturation flows:
Through lanes:
1600 veh/h/ln
Through-right lanes:
1400 veh/h/ln
Left lanes:
1000 veh/h/ln
Left-through lanes:
1200 veh/h/ln
Left-through-right lanes: 1100 veh/h/ln
Step 1: Calculate equivalent hourly flows
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Chapter 8: Intersection Control
Approach (Width)
Peak hour
approach volumes
Left turn
Through movement
Right turn
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
140
442
147
77
393
142
177
593
178
141
543
187
Step 2: Assume an intersection configuration, assign lane groups, and determine
critical volumes.
In this case, each approach was assumed to have one dedicated left-turn
lane, one through lane, and one through-right lane.
Approach (Width)
Peak hour
approach volumes
Left
Through-right
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
140
589
77
535
177
771
141
730
Step 3: Assume a phasing scheme and determine Yi , sum of critical ratios
Assume four phases as follows:
Phase 1: E-W Phase 2: E-W Phase 3: N-S
Phase 4: N-S
left
thru
left
thru
qij
177
771
140
589
sij
1000
3000
1000
3000
Yi = qij/sij
0.177
0.257
0.140
0.196
Sum Yi = 0.770
Step 4: Calculate lost time per cycle, using Equation 8.8
Assume lost time per phase due to acceleration and deceleration at phase changes
is 3.5 seconds and that an all-red interval of 1.5 seconds is provided at each phase.
Total lost time, L = 20 sec.
Step 5: Calculate cycle length, using Equation 8.6
1.5 L + 5
= ((1.5)(20)+5)/(1-0.770) = 152.2 seconds
C=
1 − ∑ Yi
Use C = 155 seconds
Step 6: Allocate green times
Allocated times are for green and yellow indications; appropriate length of
yellow interval can be subtracted from the total to give green times.
Total effective green time, Gte = C – L =135 seconds
(G+Y)1 = (0.177/0.770)(135) + 3.5 = 34.5 seconds
(G+Y)2 = (0.257/0.770)(135) + 3.5 = 48.5 seconds
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Chapter 8: Intersection Control
(G+Y)3 = (0.140/0.770)(135) + 3.5 = 28.0 seconds
(G+Y)4 = (0.196/0.770)(135) + 3.5 = 37.9 seconds
Step 7: Ensure that green time required for pedestrian movement is provided,
using Equation 8.12.
Gp1 = 3.2 + (56/4) +(0.27)(1200/3600)(155) = 31.1 seconds
Gp2 = 3.2 + (56/4) +(0.27)(1200/3600)(155) = 31.1 seconds
Gp3 = 3.2 + (68/4) +(0.27)(1200/3600)(155) = 34.1 seconds
Gp4 = 3.2 + (68/4) +(0.27)(1200/3600)(155) = 34.1 seconds
Since Gp3 > (G+Y)3 , the allocated sum of green and yellow time for phase
3 should be 34.1 seconds. Times are typically rounded up to the next whole
seconds; therefore, sum of green and yellow times are: G1 = 35 s; G2 = 49 s; G3 =
35 s; G4 = 38 s,
resulting in a total cycle length of C = (35+49+35+38) + (4)(1.5) = 163 seconds.
8-10
Repeat Problem 8-9 using saturation flow rates that are 10% higher. What effect
does this have on cycle length?
Approach (Width)
Peak hour
approach volumes
Left turn
Through movement
Right turn
Conflicting
pedestrian volumes
PHF
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
133
420
140
900
73
373
135
1200
168
563
169
1200
134
516
178
900
0.95
0.95
0.95
0.95
Assume the following saturation flows:
Through lanes:
1760 veh/h/ln
Through-right lanes:
1540 veh/h/ln
Left lanes:
1100 veh/h/ln
Left-through lanes:
1320 veh/h/ln
Left-through-right lanes: 1210 veh/h/ln
Step 1: Calculate equivalent hourly flows
Approach (Width)
North (56 ft) South (56 ft)
Peak hour
approach volumes
Left turn
140
77
Through movement 442
393
Right turn
147
142
113
East (68 ft)
West (68 ft)
177
593
178
141
543
187
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 8: Intersection Control
Step 2: Assume an intersection configuration, assign lane groups, and determine
critical volumes.
In this case, each approach was assumed to have one dedicated left-turn
lane, one through lane, and one through-right lane.
Approach (Width)
Peak hour
approach volumes
Left
Through-right
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
140
295
77
268
177
386
141
365
Step 3: Assume a phasing scheme and determine Yi , sum of critical ratios
Assume four phases as follows:
Phase 1: E-W Phase 2: E-W Phase 3: N-S
Phase 4: N-S
left
thru
left
thru
qij
177
771
140
589
sij
1100
3300
1100
3300
Yi = qij/sij
0.161
0.234
0.127
0.178
Sum Yi = 0.700
Step 4: Calculate lost time per cycle, using Equation 8.8
Assume lost time per phase due to acceleration and deceleration at phase changes
is 3.5 seconds and that an all-red interval of 1.5 seconds is provided at each phase.
Total lost time, L = 20 sec.
Step 5: Calculate cycle length, using Equation 8.6
1.5 L + 5
C=
= ((1.5)(20)+5)/(1-0.700) = 116.7 seconds
1 − ∑ Yi
Use C = 120 seconds
Step 6: Allocate green times
Allocated times are for green and yellow indications; appropriate length of
yellow interval can be subtracted from the total to give green times.
Total effective green time, Gte = C – L =100 seconds
(G+Y)1 = (0.161/0.700)(100) + 3.5 = 26.5 seconds
(G+Y)2 = (0.234/0.700)(100) + 3.5 = 36.4 seconds
(G+Y)3 = (0.127/0.700)(100) + 3.5 = 21.6 seconds
(G+Y)4 = (0.178/0.700)(100) + 3.5 = 28.9 seconds
Step 7: Ensure that green time required for pedestrian movement is provided,
using Equation 8.12.
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Chapter 8: Intersection Control
Gp1 = 3.2 + (56/4) +(0.27)(1200/3600)(120) = 28.0 seconds
Gp2 = 3.2 + (56/4) +(0.27)(1200/3600)(120) = 28.0 seconds
Gp3 = 3.2 + (68/4) +(0.27)(1200/3600)(120) = 31.0 seconds
Gp4 = 3.2 + (68/4) +(0.27)(1200/3600)(120) = 31.0 seconds
Since Gp1 > (G+Y)1 , Gp3 > (G+Y)3 , and Gp4 > (G+Y)4 , the allocated sum
of green and yellow time should be 28.0 seconds for phase 1, 31.0 seconds for
phase 3, and 31.0 seconds for phase 4. Times are typically rounded up to the next
whole seconds; therefore, sum of green and yellow times are: G1 = 28 s; G2 = 37
s; G3 = 31 s; G4 = 31 s, resulting in a total cycle length of C = (28+37+31+31) +
(4)(1.5) = 133 seconds.
By increasing saturation flow rates by 10%, the recommended cycle
length decreased by 18.4% (from 163 seconds to 133 seconds).
8-11
Repeat Problem 8-9 using pedestrian flow rates that are 20% higher. What effect
does this have on cycle length?
Approach (Width)
Peak hour
approach volumes
Left turn
Through movement
Right turn
Conflicting
pedestrian volumes
PHF
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
133
420
140
1080
73
373
135
1440
168
563
169
1440
134
516
178
1080
0.95
0.95
0.95
0.95
Assume the following saturation flows:
Through lanes:
1600 veh/h/ln
Through-right lanes:
1400 veh/h/ln
Left lanes:
1000 veh/h/ln
Left-through lanes:
1200 veh/h/ln
Left-through-right lanes: 1100 veh/h/ln
Step 1: Calculate equivalent hourly flows
Approach (Width)
North (56 ft) South (56 ft)
Peak hour
approach volumes
Left turn
140
77
Through movement 442
393
Right turn
147
142
115
East (68 ft)
West (68 ft)
177
593
178
141
543
187
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 8: Intersection Control
Step 2: Assume an intersection configuration, assign lane groups, and determine
critical volumes.
In this case, each approach was assumed to have one dedicated left-turn
lane, one through lane, and one through-right lane.
Approach (Width)
Peak hour
approach volumes
Left
Through-right
North (56 ft)
South (56 ft)
East (68 ft)
West (68 ft)
140
589
77
535
177
771
141
730
Step 3: Assume a phasing scheme and determine Yi , sum of critical ratios
Assume four phases as follows:
Phase 1: E-W Phase 2: E-W Phase 3: N-S
Phase 4: N-S
left
thru
left
thru
qij
177
771
140
589
sij
1000
3000
1000
3000
Yi = qij/sij
0.177
0.257
0.140
0.196
Sum Yi = 0.770
Step 4: Calculate lost time per cycle, using Equation 8.8
Assume lost time per phase due to acceleration and deceleration at phase changes
is 3.5 seconds and that an all-red interval of 1.5 seconds is provided at each phase.
Total lost time, L = 20 sec.
Step 5: Calculate cycle length, using Equation 8.6
1.5 L + 5
C=
= ((1.5)(20)+5)/(1-0.770) = 152.2 seconds
1 − ∑ Yi
Use C = 155 seconds
Step 6: Allocate green times
Allocated times are for green and yellow indications; appropriate length of
yellow interval can be subtracted from the total to give green times.
Total effective green time, Gte = C – L =135 seconds
(G+Y)1 = (0.177/0.770)(135) + 3.5 = 34.5 seconds
(G+Y)2 = (0.257/0.770)(135) + 3.5 = 48.5 seconds
(G+Y)3 = (0.140/0.770)(135) + 3.5 = 28.0 seconds
(G+Y)4 = (0.196/0.770)(135) + 3.5 = 37.9 seconds
Step 7: Ensure that green time required for pedestrian movement is provided,
using Equation 8.12.
Gp1 = 3.2 + (56/4) +(0.27)(1440/3600)(155) = 33.9 seconds
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Chapter 8: Intersection Control
Gp2 = 3.2 + (56/4) +(0.27)(1440/3600)(155) = 33.9 seconds
Gp3 = 3.2 + (68/4) +(0.27)(1440/3600)(155) = 37.9 seconds
Gp4 = 3.2 + (68/4) +(0.27)(1440/3600)(155) = 37.9 seconds
Since Gp3 > (G+Y)3 , the allocated sum of green and yellow time for phase
3 should be 37.9 seconds. Times are typically rounded up to the next whole
seconds; therefore, sum of green and yellow times are: G1 = 35 s; G2 = 49 s; G3 =
37 s; G4 = 38 s,
resulting in a total cycle length of C = (35+49+37+38) + (4)(1.5) = 165 seconds.
By increasing conflicting pedestrian volumes by 20%, the recommended
cycle length increased by 1.2% (from 163 seconds to 165 seconds).
8-12
Repeat Problem 8-9 using the HCM method and a critical v/c of 0.9.
Step 1: Determine critical ratios
From Problem 8-5:
Phase 1: Y1 = 0.177
Phase 2: Y2 = 0.257
Phase 3: Y3 = 0.140
Phase 4: Y4 = 0.196
Σ ((q/s)i = Σ Yi = 0.770
Step 2: Determine cycle length (using Equation 8.15)
Xc = 0.9 (critical v/c ratio)
L = 20 sec (lost time, from Problem 8-5)
Xc = Σ ((q/s)i (C/(C-L))
0.9 = 0.770 (C/(C-20))
1.169 C – 23.38 = C
C = 138.32 sec; use 140 sec
Step 3: Determine phase lengths
Allocated times are for green and yellow indications; appropriate length of yellow
interval can be subtracted from the total to give green times.
Total effective green time, Gte = C – L =120 seconds
(G+Y)1 = (0.177/0.770)(120) + 3.5 = 31.1 seconds
(G+Y)2 = (0.257/0.770)(120) + 3.5 = 43.5 seconds
(G+Y)3 = (0.140/0.770)(120) + 3.5 = 25.3 seconds
(G+Y)4 = (0.196/0.770)(120) + 3.5 = 34.1 seconds
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Chapter 8: Intersection Control
Step 4: Ensure that green time required for pedestrian movement is provided,
using Equation 8.12.
Gp1 = 3.2 + (56/4) +(0.27)(1200/3600)(140) = 28.1 seconds
Gp2 = 3.2 + (56/4) +(0.27)(1200/3600)(140) = 28.1 seconds
Gp3 = 3.2 + (68/4) +(0.27)(1200/3600)(140) = 30.8 seconds
Gp4 = 3.2 + (68/4) +(0.27)(1200/3600)(140) = 30.8 seconds
Since Gp3 > (G+Y)3 , the allocated sum of green and yellow time for phase
3 should be 30.8 seconds. Times are typically rounded up to the next whole
seconds; therefore, sum of green and yellow times are: G1 = 32 s; G2 = 44 s; G3 =
31 s; G4 = 35 s,
resulting in a total cycle length of C = (32+44+31+35) + (4)(1.5) = 148 seconds.
8-13
Using the results for Problems 8-9 and 8-12, compare the two different approaches
used for computing cycle length.
The HCM method (used in Problem 8-12) is less computationally
intensive than Webster method (used in Problem 8-9). The HCM method allows
for a desired v/c ratio to be a determining factor in cycle length. Webster’s
method yielded a cycle length of 8 seconds longer than the HCM method.
8-14
Briefly describe the different ways the traffic signals at the intersection of an
arterial route could be coordinated, stating under what conditions you would use
each of them.
Traffic signals can be coordinated by several methods: simultaneous
system, alternate system, and progressive system. In a simultaneous system, all
signals have the same cycle length and are in the green phase for the arterial at the
same time. This system works best when intersections are approximately the
same distance apart. In an alternate system, intersections are formed into groups
where successive groups alternate green phases. The alternate system works best
when the intersections within a group are at equal distance from each other. In a
progressive system an offset is introduced between the start of green for the
arterial at one intersection and the start of green for the arterial at the succeeding
intersection. This offset is based on the distance between intersections and the
speed of traffic. This method best accommodates variable spacings between
intersections and heavy directional flows.
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Chapter 8: Intersection Control
8-15
You have been asked to design a simultaneous traffic signal system for six
intersections on a suburban arterial. The distances between consecutive
intersections are:
Intersection A to Intersection B - 3,800 ft
Intersection B to Intersection C - 4,000 ft
Intersection C to Intersection D - 3,900 ft
Intersection D to Intersection E - 3,850 ft
Intersection E to Intersection F - 3,950 ft
Suitable cycle lengths for the intersections are:
Intersection A - 60 sec
Intersection B - 55 sec
Intersection C - 65 sec
Intersection D - 60 sec
Intersection E - 55 sec
Intersection F - 60 sec
If an appropriate progression speed for the arterial is 45 mph, what cycle length
would you use? Give a reason for your choice.
For this arterial, the average distance between intersections is 3,900 ft. By
rearranging Equation 8.19, an appropriate cycle length can be determined:
C=
X
3900
=
= 59.1 seconds
1.47u (1.47)(45)
Since the cycle lengths for the individual signals would be 55 or 60 seconds, and
the computed value for progression is 59.1 seconds, an appropriate cycle length
would be 60 seconds.
8-16
In Problem 8-15, if conditions at intersection C require that the cycle length of 65
sec be maintained, what will be a suitable progression speed?
Equation 8.19 can be used to determine an appropriate progression speed
if the cycle lengths for the system must be 65 seconds.
u=
X
3900
=
= 40.9 mi/h
1.47C (1.47)(65)
The computed value of 40.9 mi/h is approximately equal to 40 mi/h, therefore, a
progression speed of 40 mph is recommended.
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Chapter 8: Intersection Control
8-17
Briefly discuss the different methods by which freeway entrance ramps can be
controlled. Clearly indicate the advantages and disadvantages of each method, and
give the conditions under which each of them can be used.
The methods for controlling freeway entrance ramps are closure, simple
metering, traffic responsive metering, and integrated system control.
Closure entails the physical closure of the ramp by using “Do Not Enter” signs or
by placing barriers at the entrance to the ramp. This form of control is the
simplest, but also the most restrictive and should be used only when absolutely
necessary.
Simple metering consists of setting up a pretimed signal with extremely short
cycles at the ramp entrance. Simple metering can be used to reduce normal ramp
capacity (about 1200 veh/h) to about 250 veh/h, by changing the signal timings of
the ramp meter. This type of metering can be used to improve flow on the
mainline freeway by determining the difference between the downstream volume
and the upstream capacity and setting the metering ramp to match this, or to
improve safety at the merge area by allowing only one vehicle at a time to merge.
Traffic-responsive metering systems are based on the same principles as the
simple metering systems but add the ability to base the metering rate on current
traffic conditions rather than predetermined timing plans based on historic data.
This type of system therefore has the ability to respond to short-term changes in
conditions.
Integrated system control brings several ramps together and controls them as a
group rather than individually, without concern for how they are impacting one
another. This allows the metering rates to be set to maximize the available
mainline capacity and improve overall system flow.
8-18
Compare and contrast the different metering systems that are used in traffic signal
ramp control indicating under what conditions you will use each.
Ramp metering control systems can be divided into two general
categories: pre-timed and traffic response. The traffic response category can be
further divided into local traffic responsive and systemwide traffic responsive.
Pre-timed systems involve the use of traffic data only from a historical
perspective; therefore their operations cannot be altered in an automated real-time
environment and therefore do not require communication with a traffic
management center (TMC). In contrast, traffic response systems utilize current
traffic data as an input to their control algorithms. This category can therefore
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Chapter 8: Intersection Control
respond to changing traffic conditions if traffic data being collected are made
available to the control algorithm. Local traffic responsive control is based only
on conditions immediately upstream and downstream of the ramp junction, while
systemwide traffic responsive systems allow ramp meter control to be based on a
corridor or systemwide optimization of traffic flow.
Pre-timed control would typically be used for an isolated location or in an
area without traffic monitoring or real-time data collection capabilities. Traffic
response control would require real-time data collection; local traffic responsive
control would typically be used for isolated locations or locations without
communications capabilities, while systemwide traffic responsive control would
typically be used in a system with many ramp meters along a corridor and the
ability to communicate with a TMC. In the event of a communications loss or
other failure of a system wide control algorithm, these systems can be
programmed to revert to a local traffic responsive control or to a pre-timed
algorithm.
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Chapter 9
Capacity and Level of Service for
Highway Segments
9-1
Define the elements of a Class I and a Class II highway.
A Class I two-lane highway functions as an arterial or a link to an arterial,
often carrying primarily commuter traffic. Motorists expect relatively high-speed
travel. A Class II two-lane highway may serve as access to a class I two-lane
highway, as a scenic or recreational route, or be located in rugged terrain, thereby
constraining travel speeds. Motorists expect lower speeds on Class II than on
Class I roads. Average trip lengths are shorter on Class II roads than Class I.
9-2
What are the two measures used to describe service quality for a two-lane highway?
Which of the measures are used to describe level-of-service for Class I and Class II
highways?
Service quality on a two-lane highway is described using percent of time
spent following another vehicle (PTSF) and average travel speed (ATS). A
vehicle considered to be following when the time headway between it and the
preceding vehicle is less than 3 seconds. The space mean speed is used as ATS
and can be found by dividing segment length by average travel time of vehicles
traversing the segment in both direction during a designated time interval. PTSF
and ATS are used to determine level-of-service on Class I roads while only PTSF
is used on Class II roads.
9-3
Describe the traffic characteristics associated with each of the six levels of service
for two-lane highways.
In LOS A, motorists are able to travel at their desired speeds, with if any
platoons and minimal passing maneuvers. With LOS B, passing demand
increases significantly and approaches passing capacity at the lower end of LOS
B. In LOS C, formation of platoons and platoon size increase significantly. In
LOS D, flow begins to become unstable, and passing maneuvers are extremely
difficult if not possible to complete. In LOS E, passing is nearly impossible, and
most vehicles are in long platoons; operating conditions are unstable and difficult
to predict. LOS F occurs when demand exceeds capacity and volume fall back
from capacity with highly variable speeds.
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Chapter 9: Capacity and Level of Service for Highway Segments
9-4
The following values of PTSF and ATS have been determined for three separate
two-lane segments. Determine LOS if the segments are: (a) Class I, and (b) Class II.
Segment
10
11
12
PTSF (%)
25
46
67
ATS (mi/h)
52
39
39
(a) For a Class I highway, using Table 9.1, segment 10 operates at LOS A (when
PTSF and ATS correspond to differing levels of service, the lower value of LOS
is used), segment 11 operates at LOS B, and segment 12 operates at LOS D.
(b) For a Class II highway, using Table 9.2, segment 10 operates at LOS A, segment
11 operates at LOS B, and segment 12 operates at LOS C.
9-5
Determine the PTSF for a 4.5 mile two-lane highway segment in level terrain.
Traffic volumes (two-way) are 1100 veh/h. Trucks: 10%; RVs: 7%; PHF: 0.97;
directional split: 60/40; no passing zones: 40%.
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, vp
Determine fg (grade adjustment factor), using Table 9.4; fg = 1.00
Determine ET (PCE for trucks), using Table 9.5; ET = 1.1
Determine ER (PCE for RVs), using Table 9.5; ER = 1.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
f HV =
1
1
= 0.99
=
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.10(1.1 − 1) + 0.07(1.0 − 1)
Determine vp (flow rate), using Equation 9.3
V
vp =
= (1100)/(0.97)(0.99)(1.0) = 1146 pc/h
( PHF )( f HV )( f G )
Step 2: Compute base percent time spent following, BPTSF, using Equation 9.2
BPTSF = 100(1 − e
−0.000879V p
) = 100(1 − e −0.000879 (1146 ) ) = 63.5%
Step 3: Compute percent time spent following, PTSF, using Equation 9.1
Determine fd/np using Table 9.3; by interpolation, fd/np = 7.6%
PTSF = BPTSF + fd/np = 63.5% + 7.6% = 71.1%
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Chapter 9: Capacity and Level of Service for Highway Segments
9-6
Use the data provided in Problem 9-5 to estimate the ATS. Base free flow speed: 55
mi/h; lane width: 11 ft; shoulder width: 3 ft; access points per mile: 15.
Step 1: Compute free flow speed, FFS, using Equation 9.7
Determine fLS (lane and shoulder width adjustment), using Table 9.9; fLS = 3.0
Determine fA (access point density adjustment), using Table 9.10; fA =3.8
FFS = BFFS – fLS – fA = 55 – 3.0 – 3.8 = 48.2 mi/h
Step 2: Compute passenger car equivalent flow rate for peak 15-minute period, vp
Determine fg (grade adjustment factor), using Table 9.7; fg = 1.00
Determine ET (PCE for trucks), using Table 9.8; ET = 1.2
Determine ER (PCE for RVs), using Table 9.8; ER = 1.0
Determine fHV (heavy vehicle adjustment factor); using Equation 9.4
1
1
f HV =
=
= 0.98
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.10(1.2 − 1) + 0.07(1.0 − 1)
Determine vp (flow rate), using Equation 9.3
V
vp =
= (1100)/(0.97)(0.98)(1.0) = 1157 pc/h
( PHF )( f HV )( f G )
Step 3: Compute average travel speed, ATS, using Equation 9.5
Determine fnp (adjustment for effect of passing zones), using Table 9.6; fnp = 1.5
ATS = FFS – 0.00776vp – fnp = 48.2 – 0.00776(1157) – 1.5 = 37.7 mi/h
9-7
Use the results of Problems 9-5 and 9-6 to compute: LOS, v/c, and veh-mi in the
peak 15 minutes and peak hour, and total travel time in the peak 15 minutes.
Results from Problems 9-5 and 9-6:
PTSF = 71.1%
ATS = 37.7 mi/h
Level of service (if a Class I highway), using Table 9.1: LOS D
Level of service (if a Class II highway), using Table 9.2: LOS D
Using Equation 9.8, v/c = 1157/3200 = 0.36
Using Equation 9.9,
V
VMT15 = 0.25(
) Lt = (0.25)(1100)(4.5)/(0.97) = 1275 veh-mi
PHF
Using Equation 9.10,
VMT60 = VLt = (1100)(4.5) = 4950 veh-mi
Using Equation 9.11,
TT15 = VMT15/ATS = 1275/37.7 = 33.8 veh-h
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Chapter 9: Capacity and Level of Service for Highway Segments
9-8
Use the data provided in Examples 9-5 and 9-6 to compute to determine PTSF and
ATS in the peak direction if northbound volume is 1000 veh/h and southbound
volume is 600 veh/h.
Class I, 5 mile, two-lane highway segment in rolling terrain.
Trucks: 14%; RVs: 4%; PHF: 0.95; no passing zones: 50%; BFFS = 60 mi/h.
Lane width = 11 ft; shoulder width = 4 ft; 20 access points per mile.
ATS analysis:
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period in
the peak direction, vd
Determine fg (grade adjustment factor), using Table 9.4; fg = 0.99
Determine ET (PCE for trucks), using Table 9.5; ET = 1.5
Determine ER (PCE for RVs), using Table 9.5; ER = 1.1
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 0.931
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.14(1.5 − 1) + 0.04(1.1 − 1)
Determine vd (flow rate), using Equation 9.3
V
vd =
= (1000)/(0.95)(0.931)(0.99) = 1142 pc/h
( PHF )( f HV )( f G )
Step 2: Compute passenger car equivalent flow rate for peak 15-minute period in
the opposing direction, vo
Determine fg (grade adjustment factor), using Table 9.4; fg = 0.99
Determine ET (PCE for trucks), using Table 9.5; ET = 1.5
Determine ER (PCE for RVs), using Table 9.5; ER = 1.1
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 0.931
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.14(1.5 − 1) + 0.04(1.1 − 1)
Determine vo (flow rate), using Equation 9.3
V
vo =
= (600)/(0.95)(0.931)(0.99) = 685 pc/h
( PHF )( f HV )( f G )
Step 3: Compute free flow speed, FFS, using Equation 9.7
Determine fLS from Table 9.9; by interpolation, fLS = 1.7
Determine fA from Table 9.10; by interpolation, fLS = 5.0
FFS = BFFS – fLS – fA = 60 – 1.7 – 5.0 = 53.3 mi/h
Step 4: Compute average travel speed, ATS, using Equation 9.14
Determine fnp from Table 9.13; by interpolation, fnp = 1.0
ATSd = FFSd – 0.00776(vd + vo) – fnp = 53.3 – 0.00776 (1142+685) – 1.1 = 38.0
mi/h
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Chapter 9: Capacity and Level of Service for Highway Segments
PTSF analysis:
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period in
the peak direction, vd
Determine fg (grade adjustment factor), using Table 9.4; fg = 1.00
Determine ET (PCE for trucks), using Table 9.5; ET = 1.0
Determine ER (PCE for RVs), using Table 9.5; ER = 1.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 1.00
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.14(1.0 − 1) + 0.04(1.0 − 1)
Determine vd (flow rate), using Equation 9.3
V
vd =
= (1000)/(0.95)(1.0)(1.0) = 1053 pc/h
( PHF )( f HV )( f G )
Step 2: Compute passenger car equivalent flow rate for peak 15-minute period in
the opposing direction, vo
Determine fg (grade adjustment factor), using Table 9.4; fg = 1.00
Determine ET (PCE for trucks), using Table 9.5; ET = 1.0
Determine ER (PCE for RVs), using Table 9.5; ER = 1.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 1.0
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.14(1.0 − 1) + 0.04(1.0 − 1)
Determine vo (flow rate), using Equation 9.3
V
vo =
= (600)/(0.95)(1.0)(1.0) = 632 pc/h
( PHF )( f HV )( f G )
Step 3: Compute base percent time spent following, BPTSF, using Equation 9.12
Find values of a and b from Table 9.12; by interpolation,
a = -0.112
b = 0.403
b
0.403
BPTSFd = 100(1 − e avd ) = 100(1 − e (−0.112 )(1053) ) = 84.2%
Step 4: Compute percent time spent following, PTSF, using Equation 9.11
Determine fnp using Table 9.11; by interpolation, fnp = 8.1%
PTSFd = BPTSFd + fd/np = 84.2% + 8.1% = 92.3%
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Chapter 9: Capacity and Level of Service for Highway Segments
9-9
Use the data and results obtained in Problem 9-8 to determine the level of service of
a two-lane section if a passing lane 1.5 mi long is added. The passing lane begins
0.75 mi from the starting point of the analysis segment.
To determine the level of service, lengths of the regions in the segment, PTSFpl
and ATSpl, must first be determined.
From Problem 9-8, PTSFd = 87.6% and ATSd = 38.5 mi/h
Step 1: Determine region lengths.
Region I: Lu =0.75 mi
Region II: Lpl = 1.5 mi
Region III: For PTSF, from Table 9.22, Lde = 3.6 mi; from Table 9.20, fpl = 0.62
For ATS, from Table 9.22, Lde = 1.7 mi; from Table 9.20, fpl = 1.11
Region IV: For PTSF, Ld = 5 – 0.75 –1.5 – 3.6 = -0.85; use Ld = 0
and L’de = 5 – 0.75 – 1.5 = 2.75
For ATS, Ld = 5 – 0.75 – 1.5 – 1.7 = 1.05
Step 2: Compute PTSFpl using Equation 9.16
1 − f pl Lde
′2
PTSFd [ Lu + ( f pl )( L pl ) + (
)(
)]
2
Lde
PTSF pl =
Lt
PTSF pl =
87.6[0.75 + (0.62)(1.5) + (
5
1 − 0.62 2.75 2
)(
)]
2
3.6 = (87.6)(2.079)/5 = 36.4%
Step 3: Compute ATSpl using Equation 9.18
( ATS d )( Lt )
ATS pl =
L pl
2 Lde
+
+ Ld
Lu +
f pl
f pl + 1
ATS pl =
(38.5)(5)
= (38.5)(5)/4.7627 = 40.4 mi/h
1.5 (2)(1.7)
0.75 +
+
+ 1.05
1.11 1.11 + 1
Step 4: Determine level of service from Table 9.1 (for a Class I highway)
Due to PTSF = 36.4%, the level of service is LOS B.
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Chapter 9: Capacity and Level of Service for Highway Segments
9-10
An existing Class I two-lane highway is to be analyzed to determine the two-way
level of service, given the following information:
Traffic data:
PHV = 600 veh/h
60% in the peak direction
8% trucks
2% recreational vehicles
PHF = 0.86
No passing zones: 40%
Geometric data:
Rolling terrain
BFFS = 55 mi/h
Lane width = 11 ft
Shoulder width = 2 ft
8 access points per mile
PTSF analysis:
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period, vp
Determine fg (grade adjustment factor), using Table 9.4; fg = 0.94
Determine ET (PCE for trucks), using Table 9.5; ET = 1.5
Determine ER (PCE for RVs), using Table 9.5; ER = 1.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 0.962
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.08(1.5 − 1) + 0.02(1.0 − 1)
vp =
V
= (600)/(0.86)(0.962)(0.94) = 772 pc/h
( PHF )( f HV )( f G )
Step 2: Compute base percent time spent following, BPTSF, using Equation 9.2
BPTSF = 100(1 − e
−0.000879V p
) = 100(1 − e −0.000879 ( 772 ) ) = 49.3%
Step 3: Compute percent time spent following, PTSF, using Equation 9.1
Determine fd/np using Table 9.3; by interpolation, fd/np = 11.0%
PTSF = BPTSF + fd/np = 49.3% + 11.0% = 60.3%
This corresponds to LOS C (Table 9.1) based solely on PTSF.
ATS analysis:
Step 1: Compute free flow speed, FFS, using Equation 9.7
Determine fLS (lane and shoulder width adjustment), using Table 9.9; fLS = 3.0
Determine fA (access point density adjustment), using Table 9.10; fA = 2.0
FFS = BFFS – fLS – fA = 55 – 3.0 – 2.0 = 50.0 mi/h
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Chapter 9: Capacity and Level of Service for Highway Segments
Step 2: Compute passenger car equivalent flow rate for peak 15-minute period, vp
Determine fg (grade adjustment factor), using Table 9.7; fg = 0.930
Determine ET (PCE for trucks), using Table 9.8; ET = 1.9
Determine ER (PCE for RVs), using Table 9.8; ER = 1.1
Determine fHV (heavy vehicle adjustment factor); using Equation 9.4
1
1
f HV =
=
= 0.931
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.10(1.2 − 1) + 0.07(1.0 − 1)
V
Vp =
= (1100)/(0.86)(0.931)(0.93) = 806 pc/h
( PHF )( f HV )( f G )
Step 3: Compute average travel speed, ATS, using Equation 9.5
Determine fnp (adjustment for effect of passing zones), using Table 9.6; fnp = 2.0
ATS = FFS – 0.00776vp – fnp = 50.0 – 0.00776(806) – 2.0 = 41.9 mi/h
This corresponds to LOS D (Table 9.1) based solely on ATS.
The overall LOS is taken as the poorer of the two (for PTSF and ATS), LOS D.
9-11
An existing Class II two-lane highway is to be analyzed to determine LOS in the peak
direction given the following information:
Peak hourly volume in the analysis direction: 900 veh/h
Peak hourly volume in the opposing direction: 400 veh/h
Trucks: 12% of total volume
Recreational vehicles: 2% of total volume
PHF: 0.95
Lane width: 12 ft
Shoulder width: 10 ft
Access points per mile: 20
Terrain: rolling
Base free flow speed: 60 mi/h
No passing zones: 40% of analysis segment length
PTSF analysis:
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period in
the peak direction, vd
Determine fg (grade adjustment factor), using Table 9.4; fg = 0.94
Determine ET (PCE for trucks), using Table 9.5; ET = 1.0
Determine ER (PCE for RVs), using Table 9.5; ER = 1.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
f HV =
1
1
=
= 1.00
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.12(1.0 − 1) + 0.02(1.0 − 1)
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Chapter 9: Capacity and Level of Service for Highway Segments
vd =
V
= (900)/(0.95)(1.0)(1.0) = 947 pc/h
( PHF )( f HV )( f G )
Step 2: Compute passenger car equivalent flow rate for peak 15-minute period in
the opposing direction, vo
Determine fg (grade adjustment factor), using Table 9.4; fg = 0.94
Determine ET (PCE for trucks), using Table 9.5; ET = 1.5
Determine ER (PCE for RVs), using Table 9.5; ER = 1.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 0.943
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.12(1.5 − 1) + 0.02(1.0 − 1)
vo =
V
= (400)/(0.95)(0.943)(0.94) = 475 pc/h
( PHF )( f HV )( f G )
Step 3: Compute base percent time spent following, BPTSF, using Equation 9.13
Find values of a and b from Table 9.12; by interpolation,
a = -0.073
b = 0.454
b
0.454
BPTSFd = 100(1 − e avd ) = 100(1 − e (−0.073)(947 ) ) = 80.7%
Step 4: Compute free flow speed, FFS, using Equation 9.6
Determine fLS (lane and shoulder width adjustment), using Table 9.9; fLS = 0
Determine fA (access point density adjustment), using Table 9.10; fA = 5.0
FFS = BFFS – fLS – fA = 60 – 0 – 5.0 = 55.0 mi/h
Step 5: Compute percent time spent following, PTSF, using Equation 9.12
Determine fnp using Table 9.11; by interpolation, fnp = 10.1%
PTSFd = BPTSFd + fd/np = 80.7% + 10.1% = 90.8%
This corresponds to LOS E (Table 9.2)
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Chapter 9: Capacity and Level of Service for Highway Segments
9-12
A new section of Richmond Highway is being designed as a six-lane facility (three in
each direction) with a two-way left-turn lane. Determine the peak hour LOS.
Traffic data:
• Directional design hourly volume = 3600 veh/h
• PHF = 0.94
• Assumed base free flow speed = 55 mi/h
Geometric data:
• Urban setting
• Rolling terrain
• Lane width = 11 ft
• Shoulder widths = 4 ft (right side) and 1 ft (left side)
• Average access point spacing = 12 points per mile on each side
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.21.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.25; ET = 2.5
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
f HV =
=
= 0.930
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.05(2.5 − 1) + 0.00
V
= (3600)/(0.94)(3)(0.930)(1.0) = 1373 pc/h/ln
vp =
( PHF )( N )( f HV )( f p )
Step 2: Compute free flow speed using Equation 9.25
FFS = BFFS – fLW – fLC – fM – fA
Determine fLW using Table 9.29, fLW = 1.9
Determine fLC using Table 9.34, fLC = 1.5
Determine fM using Table 9.35, fM = 0.0
Determine fA using Table 9.36, fA = 3.0
FFS = 55 – 1.9 – 1.5 – 0.0 – 3.0 = 48.6 mi/h
Step 3: Compute average passenger car speed and density to determine LOS.
Since FFS = 48.6 mi/h and vp = 1373 pc/h/ln, S = FFS = 48.6 mi/h.
Density = (1373 pc/h/ln) / (48.6 mi/h) = 28.3 pc/mi/ln
This corresponds to LOS D (Table 9.33).
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Chapter 9: Capacity and Level of Service for Highway Segments
9-13
Briefly describe the traffic characteristics associated with each of the levels of
service for basic freeway sections.
At LOS A free flow conditions prevail. Vehicles are not impeded by other
vehicles. Effects of minor incidents or breakdowns are easily absorbed.
At LOS B, reasonably free flow conditions still exist and vehicles continue to
travel at free flow speeds. Ability to maneuver within the traffic stream is slightly
restricted. Effects of minor incidents or breakdowns are easily absorbed.
At LOS C, speeds are at or near the free flow speed, but freedom to maneuver is
noticeably restricted. Lane changes are more difficult. Minor incidents result in
significant deterioration in local level of service.
At LOS D, speeds begin to decline slightly with increasing flows. Freedom to
maneuver is noticeably limited and drivers experience reduced physical and
psychological comfort. Minor incidents will result in queuing as little space is
left in the traffic stream to absorb disruptions.
At LOS E, operations are volatile because there are virtually no gaps; volume is at
or near capacity. Maneuvers such as lane changing or merging traffic at entrance
ramps will result in a disturbance of the traffic flow. Any incident can be
expected to cause extensive queues as the traffic stream has no ability to dissipate
its effects.
At LOS F, breakdown conditions exist and uniform moving flow cannot be
maintained. The flow conditions are such that the number of vehicles that can
pass a point is less than the number of vehicles arriving at that point.
9-14
Describe the factors that affect the level of service of a freeway section and the
impact each has on flow.
Lane width - Traffic is restricted when lane widths are narrower than 12 ft.
Motorists tend to travel more cautiously because of the reduced lateral distance
between vehicles by reducing their speeds.
Lateral clearance - Lateral obstructions tend to have an effect similar to reduced
lane width. Drivers in the lane adjacent to the obstruction will tend to shy away
from the obstruction, moving them closer to vehicles in the adjacent lane and
resulting in a reduction in speeds. Lateral obstructions more than 6 ft from the
edge of the traveled lane have no significant effect of traffic flow.
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Chapter 9: Capacity and Level of Service for Highway Segments
Traffic composition - The effect of large or heavy vehicles in the traffic stream
reduces the maximum flow on the highway because of their operating
characteristics; a heavy vehicle occupies more space in the traffic stream than
does a passenger car.
Grade - The effect of a grade depends on the length and slope of the grade.
Traffic operations are significantly affected when grades of 3% or greater are 0.25
mile or greater in length and when grades of 3% or less are greater than 0.5 mile
in length. The effect of grades on heavy vehicles is much greater than on
passenger cars.
Driver population - A driver population consisting primarily of weekday
commuters will have significantly different behavior than a driver population
consisting of drivers unfamiliar with the roadway. For example, recreational
traffic capacities can be as much as 20% lower than commuter traffic capacities.
Interchange spacing – As interchanges are more closely spaced, the lengths of
basic freeway segments unaffected by interchanges decreases as the weaving
movements at interchanges have an increasing impact on traffic flow resulting in
a reduction of speeds.
9-15
Given: Freeway; design volume of 5000 veh/h; PHF: 0.9; trucks: 10%; design LOS:
C; free flow speed: 70 mi/h; no lateral obstructions; rolling terrain; interchange
spacing: 3 mi.
Determine: Number of 12 ft lanes required in each direction.
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.21.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.25; ET = 2.5
Determine ER (PCE for RVs), using Table 9.25; ER = 2.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
f HV =
1
1
=
= 0.87
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.10(2.5 − 1) + 0.00(2.0 − 1)
Step 2: Assume six lanes (three in each direction)
V
= (5000)/(0.90)(3)(0.87)(1.0) = 2128 pc/h/ln
vp =
( PHF )( N )( f HV )( f p )
Step 3: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29, fLW = 0.0
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Chapter 9: Capacity and Level of Service for Highway Segments
Determine fLC using Table 9.30, fLC = 0.0
Determine fN using Table 9.31, fN = 3.0
Determine fID using Table 9.32, fID = 0.0
FFS = 70 – 0.0 – 0.0 – 3.0 – 0.0 = 67.0 mi/h
Step 4: Compute average passenger car speed and density to determine LOS.
For FFS = 67 mi/h and vp = 2128 pc/h/ln, using Figure 9.9, the average passenger
car speed, S = 60.0 mi/h
Density = (2128 pc/h/ln) / 60.0 mi/h = 35.4 pc/mi/ln
Using Table 9.24, this density corresponds to LOS E, therefore, a three lane
section is inadequate. Repeat steps 2 and 3.
Step 2: Assume eight lanes (four in each direction)
V
= (5000)/(0.90)(4)(0.87)(1.0) = 1597 pc/h/ln
vp =
( PHF )( N )( f HV )( f p )
Step 3: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29 fLW = 0.0
Determine fLC using Table 9.30, fLC = 0.0
Determine fN using Table 9.31, fN = 1.5
Determine fID using Table 9.32, fID = 0.0
FFS = 70 – 0.0 – 0.0 – 1.5 – 0.0 = 68.5 mi/h
Step 4: Compute average passenger car speed and density to determine LOS.
For FFS = 68.5 mi/h and vp = 1597 pc/h/ln, using Figure 9.9, the average
passenger car speed, S = 68.1 mi/h
Density = (1597 pc/h/ln) / 68.1 mi/h = 23.4 pc/mi/ln
Using Table 9.24, this density corresponds to LOS C, therefore, a four lane
section is adequate.
135
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Chapter 9: Capacity and Level of Service for Highway Segments
9-16
An existing rural freeway in rolling terrain is to be analyzed to determine LOS
using the following information:
Number of lanes in each direction: 2
Peak hour volume: 2640 veh/h (in the peak direction)
18% trucks
2% recreational vehicles
PHF = 0.91
Lane width: 12 ft
Lateral clearance: 10 ft
Average interchange spacing: 3 mi
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.21.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.25; ET = 2.5
Determine ER (PCE for RVs), using Table 9.25; ER = 2.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
=
= 0.775
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.18(2.5 − 1) + 0.02(2.0 − 1)
V
= (2640)/(0.91)(2)(0.775)(1.0) = 1871 pc/h/ln
vp =
( PHF )( N )( f HV )( f p )
f HV =
Step 2: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29, fLW = 0.0
Determine fLC using Table 9.30, fLC = 0.0
Determine fN using Table 9.31, fN = 0.0
Determine fID using Table 9.32, fID = 0.0
FFS = 75 – 0.0 – 0.0 – 0.0 – 0.0 = 75.0 mi/h
Step 3: Compute average passenger car speed and density to determine LOS.
For FFS = 75.0 mi/h and vp = 1871 pc/h/ln,
2.6
⎡⎛
160 ⎞⎛ 1871 + (30)(75) − 3400 ⎞ ⎤
⎟⎟ ⎥
S = 75 − ⎢⎜ 75 −
⎟⎜
3 ⎠⎜⎝
(30)(75) − 1000
⎢⎣⎝
⎠ ⎥⎦
The average passenger car speed, S = 69.8 mi/h
Density = (1871 pc/h/ln) / (69.8 mi/h) = 26.8 pc/mi/ln
This corresponds to LOS D.
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Chapter 9: Capacity and Level of Service for Highway Segments
9-17
An existing urban freeway with 4 lanes in each direction has the following
characteristics:
Traffic data:
Peak hour volume (in the peak direction): 7070 veh/h
Trucks: 10% of peak hour volume
PHF = 0.94
Geometric data:
Lane width: 11 ft
Shoulder width: 6 ft
Interchange spacing (average): 1.4 mile
Terrain: rolling
Determine the LOS in the peak hour. Clearly state assumptions used for any values
not given. Show the demand flow rate, mean speed, and density for the given
conditions.
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.21.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.25; ET = 2.5
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
1
1
=
= 0.870
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.10(2.5 − 1) + 0.00
V
= (7070)/(0.94)(4)(0.870)(1.0) = 2163 pc/h/ln
vp =
( PHF )( N )( f HV )( f p )
f HV =
Step 2: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29, fLW = 1.9
Determine fLC using Table 9.30, fLC = 0.0
Determine fN using Table 9.31, fN = 1.5
Determine fID using Table 9.32, fID = 1.1
FFS = 70 – 1.9 – 0.0 – 1.5 – 1.1 = 65.5 mi/h
Step 3: Compute average passenger car speed and density to determine LOS.
For FFS = 65.5 mi/h and vp = 2163 pc/h/ln,
2.6
⎡1
⎛ 2163 + (30)(65.5) − 3400 ⎞ ⎤
⎟⎟ ⎥
S = 65.5 − ⎢ ((7)(65.5) − 340 )⎜⎜
(40)(65.5) − 1700
⎢⎣ 9
⎠ ⎥⎦
⎝
The average passenger car speed, S = 58.3 mi/h
Density = (2163 pc/h/ln) / (58.3 mi/h) = 37.1 pc/mi/ln
This corresponds to LOS E.
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Chapter 9: Capacity and Level of Service for Highway Segments
9-18
An urban freeway is to be designed using the following information:
AADT = 44,000 veh/day
K (proportion of AADT occurring during the peak hour): 0.10
D (proportion of peak hour traffic traveling in the peak direction): 0.60
Trucks: 10% of peak hour volume
PHF = 0.94
Lane width: 12 ft
Shoulder width: 10 ft
Interchange density: 0.5 interchange/mile
Terrain: rolling
Determine the number of lanes required to provide LOS C. Clearly state
assumptions used for any values not given, and show all calculations required.
Step 1: Compute directional design hourly volume (DDHV).
DDHV = (AADT)(K)(D) = (44000)(0.10)(0.60) = 2640 veh/h
Step 2: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.21.
Assume 2 lanes in each direction will be sufficient.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.25; ET = 2.5
Determine ER (PCE for RVs), using Table 9.25; ER = 2.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
f HV =
vp =
1
1
=
= 0.870
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.10(2.5 − 1) + 0.00
V
= (2640)/(0.94)(2)(0.870)(1.0) = 1614 pc/h/ln
( PHF )( N )( f HV )( f p )
Step 3: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29, fLW = 0.0
Determine fLC using Table 9.30, fLC = 0.0
Determine fN using Table 9.31, fN = 4.5
Determine fID using Table 9.32, fID = 0.0
FFS = 70 – 0.0 – 0.0 – 4.5 – 0.0 = 65.5 mi/h
Step 4: Compute average passenger car speed and density to determine LOS.
For FFS = 65.5 mi/h and vp = 1614 pc/h/ln,
2.6
⎡1
⎛ 1614 + (30)(65.5) − 3400 ⎞ ⎤
⎟⎟ ⎥
S = 65.5 − ⎢ ((7)(65.5) − 340 )⎜⎜
(40)(65.5) − 1700
⎢⎣ 9
⎠ ⎥⎦
⎝
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138
Chapter 9: Capacity and Level of Service for Highway Segments
The average passenger car speed, S = 65.3 mi/h
Density = (1614 pc/h/ln) / (65.3 mi/h) = 24.7 pc/mi/ln
This corresponds to LOS C; N = 2 (two lanes in each direction) are required.
9-19
Given: 4-lane freeway (2 lanes in each direction); segment length: 2 mi; sustained
grade: 4%; Design volume of 3000 veh/h; trucks: 10%; buses: 2%; RVs: 3%; PHF:
0.95; free flow speed: 70 mi/h; right side lateral obstruction: 5 ft; design LOS: B.
Determine: number of additional lanes required in each direction.
Step 1: Compute passenger car equivalent flow rate for peak 15-minute period
using Equation 9.21.
Note: this segment is considered to be in mountainous terrain since the maximum
sustained grade is 4%.
Determine fp (driver population factor); assume mainly commuter traffic, fp = 1.0
Determine ET (PCE for trucks), using Table 9.25; ET = 4.5
Determine ER (PCE for RVs), using Table 9.25; ER = 4.0
Determine fHV (heavy vehicle adjustment factor), using Equation 9.4
f HV =
1
1
=
= 0.66
1 + PT ( ET − 1) + PR ( E R − 1) 1 + 0.12(4.5 − 1) + 0.03(4.0 − 1)
Step 2: Assume one additional lane in each direction
V
= (3000)/(0.95)(3)(0.66)(1.0) = 1589 pc/h/ln
vp =
( PHF )( N )( f HV )( f p )
Step 3: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29, fLW = 0.0
Determine fLC using Table 9.30, fLC = 0.4
Determine fN using Table 9.31, fN = 3.0
Determine fID using Table 9.32, fID = 0.0
FFS = 70 – 0.0 – 0.4 – 3.0 – 0.0 = 66.6 mi/h
Step 4: Compute average passenger car speed and density to determine LOS.
For FFS = 66.6 mi/h and vp = 1589 pc/h/ln, using Figure 9.9, the average
passenger car speed, S = 66.4 mi/h
Density = (1589 pc/h/ln) / 66.4 mi/h = 23.9 pc/mi/ln
Using Table 9.24, this density corresponds to LOS C, therefore, a three lane
section is inadequate. Repeat steps 2 and 3.
Step 2: Assume two additional lanes in each direction
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Chapter 9: Capacity and Level of Service for Highway Segments
vp =
V
= (3000)/(0.95)(4)(0.66)(1.0) = 1192 pc/h/ln
( PHF )( N )( f HV )( f p )
Step 3: Compute free flow speed using Equation 9.23
FFS = BFFS – fLW – fLC – fN - fID
Determine fLW using Table 9.29, fLW = 0.0
Determine fLC using Table 9.30, fLC = 0.2
Determine fN using Table 9.31, fN = 1.5
Determine fID using Table 9.32, fID = 0.0
FFS = 70 – 0.0 – 0.2 – 1.5 – 0.0 = 68.3 mi/h
Step 4: Compute average passenger car speed and density to determine LOS.
For FFS = 68.3 mi/h and vp = 1192 pc/h/ln, using Figure 9.9, the average
passenger car speed, S = 68.3 mi/h
Density = (1192 pc/h/ln) / 68.3 mi/h = 17.5 pc/mi/ln
Using Table 9.24, this density corresponds to LOS B, therefore, a four lane
section is adequate, requiring two additional lanes in each direction.
9-20
Given: Roadway segment with 6000 ft of 3% upgrade, followed by 5000 ft of 5%
upgrade; trucks: 8%; RVs: 4%.
Determine: number of PCEs.
Use performance curves provided in Figure 9.13.
Assuming an entry speed of 55 mi/h, at the end of the first grade speed is 38 mi/h.
At end of the second grade, speed is 27 mi/h, which is also the crawl speed for
that grade. Therefore, the effective grade is 5%.
From Table 9.26, for 8% trucks, 5% grade, and greater than 2 miles, ET = 3.5
From Table 9.27, for 4% RVs, 5% grade, and greater than 2 miles, ER = 4.5
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140
Chapter 10
Capacity and Level of Service
at Signalized Intersections
10-1
A two-phase signal system is to be designed for an isolated intersection with a peak
hour factor of 0.95. The critical lane volumes are:
Phase A = 550 veh/hr
Phase B = 600 veh/hr
The total lost time L is 14 sec.
Using the Highway Capacity Manual procedure, determine an appropriate cycle
length for the intersection that will satisfy the following conditions:
minimum cycle length = 45 sec
maximum cycle length = 120 sec
maximum critical v/c ratio (Xc) = 0.85
The saturation flow per lane for each phase is 1900 veh/hr.
Step 1: Determine critical ratios, (v/s)ci
Phase 1: (v/s)c1 = (550/1900)/0.95 = 0.305
Phase 2: (v/s)c2 = (600/1900)/0.95 = 0.332
Σ ((v/s)ci = 0.637
Step 2: Determine cycle length (using Equation 10.3)
Xc = 0.85 (critical v/c ratio)
L = 14 sec
Xc = Σ ((v/s)ci (C/(C-L))
0.85 = 0.637 (C/(C-14))
1.334 C – 18.68 = C
C = 55.9 sec; use 60 sec
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Chapter 10: Capacity and Level of Service at Signalized Intersections
10-2
Figure 10.28 shows peak hour volumes and other traffic and geometric
characteristics of an intersection located in the CBD of a city. Determine the overall
level of service at the intersection. Use operation analysis.
Step 1: Compute minimum green times for pedestrians.
Phase 1 (N/S)
(100 ped/h)(1 h/3600 s)(100 s) = 2.778 ped (per cycle)
Gp1 = 3.2 + (44/4.0) + (0.27)(2.778) = 15.0 s
Phase 2 (E/W)
(100 ped/h)(1 h/3600 s)(100 s) = 2.778 ped (per cycle)
Gp2 = 3.2 + (30/4.0) + (0.27)(2.778) = 11.5 s
Required pedestrian times are less than those provided (G1 = 39 s, G2 = 55 s)
See Exhibits 10.2A through 10.2H for results.
10-3
Figure 10.29 shows an isolated intersection at Third Street and Ellis Avenue outside
the CBD of a city. Third Street is one way in the NB direction with parking allowed
on either side. Ellis Avenue is a major arterial with a separate left-turn lane on the
EB approach. Traffic volume counts at the intersection have shown that the growth
rate over the past three years is 4% per annum, and it is predicted that this rate will
continue for the next five years. The existing peak hour volumes and traffic
characteristics are shown in Figure 10.29. Determine a suitable timing of the existing
three-phase signal that will be suitable for traffic volumes in five years. Assume that,
except for traffic volumes, characteristics will remain the same. Also determine the
LOS at which the intersection will operate. A critical (v/c) ratio of 0.85 or lower is
required at the intersection.
See Exhibits 10.3A through 10.3D for results.
10-4
Figure 10.30 shows projected peak hour volumes in five years and other traffic and
geometric characteristics for the intersection of 3rd and K Streets in the CBD of an
urban area. Determine whether the existing signal timing will be suitable for the
projected demand if the level of service at each approach must be D or better. If the
existing system will not satisfy this requirement, make suitable changes to the phasing
and/or signal timing that will achieve the level of service requirement of D. Determine
the critical volume and the intersection overall level of service, using your phasing
and/or signal timing. A critical (v/c) ratio of 0.85 or lower is required at the
intersection.
See Exhibits 10.4A through 10.4F for results.
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142
Chapter 10: Capacity and Level of Service at Signalized Intersections
10-5
Repeat Problem 10-4, assuming that the intersection is located outside the CBD.
There is adequate right of way available for significant geometric improvement, and
the level of service on each approach should be C or better.
See Exhibits 10.5A through 10.5E for results.
10-6
Figure 10.31 shows the estimated maximum flows at an isolated signalized
intersection planned for an urban area outside the CBD. Using the planning level of
analysis, determine a suitable intersection layout and phasing sequence that will
provide a condition of under capacity.
See Exhibits 10.6A through 10.6E for results.
10-7
It is estimated that traffic will grow at a rate of 4% per annum at the intersection
described in Problem 10-6. Using the planning level of analysis, determine under
what conditions the intersection will be operating after 10 years of operation if your
recommended geometric layout is adopted.
See Exhibits 10.7A through 10.7E for results.
10-8
Using your solution for Problem 10-6 and the operation analysis procedure,
determine the overall level of service at the intersection if it is located within the
CBD. All approaches have 0% grades and 3% heavy vehicles in the traffic stream.
Parking is not allowed, and there is no bus stop on any approach. The PHF at each
approach is 0.90. There are 60 conflicting pedestrians at each approach, and there
are no pedestrian buttons. A critical (v/c) ratio of 0.85 or lower is desired. Use a fourphase signal (one phase for each approach).
See Exhibits 10.8A through 10.8E for results.
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Chapter 10: Capacity and Level of Service at Signalized Intersections
10-9
Figure 10.31 shows existing peak hour volumes at an isolated four-leg intersection,
with each approach consisting of two 10-ft-wide lanes. A signal system is to be
installed at the intersection in three years. Traffic growth is projected at 3% per
annum. Using the planning level of analysis, determine whether it will be necessary to
improve the intersection geometry before the installation of the signals.
See Exhibits 10.9A through 10.9E for results.
10-10
Using the peak hour volumes given for Problem 10-9, determine a suitable
intersection layout if the growth rate is 4% per annum, the signal system is to be
installed in five years, and a condition of under capacity is required.
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144
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2A
Volume Adjustment Module Worksheet
Appr.
Mvt.
Mvt.
Vol.
(veh/h)
LT
75
0.85
88
TH
500
0.85
588
RT
85
0.85
100
LT
75
0.85
88
TH
550
0.85
647
RT
50
0.85
59
LT
20
0.85
24
TH
200
0.85
235
RT
15
0.85
18
LT
25
0.85
29
TH
400
0.85
471
RT
40
0.85
47
PHF
Flow
Rate
(veh/h)
Lane
Group
Flow
rate in
lane
group
(veh/h)
Numbe
r of
lanes
N
Prop. Of LT or
RT
PLT or PRT
LTR
776
2
0.114 LT
0.129 RT
LTR
794
2
0.111 LT
0.074 RT
LTR
277
1
0.085 LT
0.064 RT
LTR
547
1
0.054 LT
0.086 RT
EB
WB
NB
SB
145
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2B
Saturation Flow Rate Module Worksheet
Appr
.
so
N
fw
fHV
fg
fp
fbb
fa
fLU
fRT
fLT
fRpb
fLpb
s
1900
2
0.97
0.95
1.0
1.0
1.0
0.9
0.95
0.98
0.70
0.997
0.988
2023
1900
2
0.97
0.95
1.0
1.0
1.0
0.9
0.95
0.99
0.75
0.997
0.993
2200
1900
1
1.10
0.93
1.0
1.0
1.0
0.9
1.00
0.99
0.82
0.998
0.995
1410
1900
1
1.10
0.93
1.0
1.0
1.0
0.9
1.00
0.99
0.97
0.998
0.993
1664
EB
WB
NB
SB
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146
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2C
Supplemental Worksheet for Permitted Left Turns
For use where the subject approach is opposed by a multilane approach
EB
WB
Enter Cycle Length, C
100
100
Enter Actual Green Time, G
55
55
Enter Effective Green Time, g
55
55
Enter opposing eff. Green time, go
55
55
Enter # of lanes in lane group, N
2
2
Enter # of opposing lane, No
2
2
Enter adjusted left turn flow rate, vLT
88
88
Enter Proportion of lt. turns in lane group, PLT
0.11
0.11
Enter adjusted opposing flow rate, vo
834
815
3
3
Compute left turns per cycle, LTC = vLTC/3600
2.44
2.44
Compute opposing flow per lane, per cycle, volc
12.2
11.9
Determine opposing platoon ratio, Rpo
0.67
1.33
Compute gf
7.3
7.3
Compute opposing queue ratio, qro
0.63
0.27
Compute gq
18.4
9.4
Compute gu
36.7
45.6
Compute PL
0.39
0.35
Determine EL1
3.3
3.3
Compute fmin
0.05
0.05
Compute fm
0.48
0.59
Compute fLT
0.70
0.75
Enter lost time per phase, tL
147
NB
SB
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Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2D
Supplemental Worksheet for Permitted Left Turns
For use where the subject approach is opposed by a single lane approach
EB
NB
SB
Enter Cycle Length, C
100
100
Enter Actual Green Time, G
39
39
Enter Effective Green Time, g
39
39
Enter opposing eff. Green time, go
39
39
Enter # of lanes in lane group, N
1
1
Enter adjusted left turn flow rate, vLT
24
29
Enter Proportion of lt. turns in lane group, PLT
0.09
0.05
Enter Prop. of lt. turns in opposing flow, PLTo
0.05
0.09
Enter adjusted opposing flow rate, vo
547
277
3
3
Compute left turns per cycle, LTC = vLTC/3600
.67
.81
Compute opposing flow per lane, per cycle, volc
15.2
7.7
1
1
Compute gf
17.0
15.4
Compute opposing queue ratio, qro
0.61
0.61
Compute gq
20.3
10.9
Compute gu
18.7
23.6
Compute n
1.62
0
Compute PTHo
0.95
0.91
Determine EL1
2.4
1.9
Compute EL2
1.59
0
Compute fmin
0.06
0.05
Compute gdiff
3.23
0
Compute fLT
0.82
0.97
Enter lost time per phase, tL
Determine opposing platoon ratio, Rpo
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
148
WB
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2E
Supplemental Worksheet for Pedestrian-Bicycle Effects
On Permitted Left Turns
EB
WB
NB
SB
Effective pedestrian green time, gp
55
55
39
39
Conflicting pedestrian volume, vped
100
100
100
100
vpedg
182
182
256
256
OCCpedg
0.091
0.091
0.128
0.128
Opposing queue clearing green, gq
18.4
9.4
20.3
10.9
Consumption of effective pedestrian green, gq/gp
0.335
0.171
0.521
0.279
OCCpedu
0.076
0.083
0.095
0.110
834
815
547
277
0.024
0.027
0.044
0.075
Nrec
1
1
2
2
Nturn
1
1
1
1
ApbT = 1 – OCCr if Nrec = Nturn or 1-0.6(OCCr)
0.976
0.973
0.973
0.955
Proportion of left turns, PLT
0.114
0.111
0.085
0.054
0
0
0
0
0.997
0.997
0.998
0.998
Opposing flow rate, vo
OCCr
PLTA
fLpb = 1.0 - PLT(1-ApbT)(1-PLTA)
149
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2F
Supplemental Worksheet for Pedestrian-Bicycle Effects
On Permitted Right Turns
EB
WB
NB
SB
Effective pedestrian green time, gp
55
55
39
39
Conflicting pedestrian volume, vped
100
100
100
100
0
0
0
0
182
182
256
256
0.091
0.091
0.128
0.128
Effective green, g
55
55
39
39
vbicg
0
0
0
0
OCCbicg
0
0
0
0
0.091
0.091
0.128
0.128
Nrec
1
1
2
2
Nturn
1
1
1
1
ApbT = 1 – OCCr if Nrec = Nturn or 1-0.6(OCCr)
0.909
0.909
0.923
0.923
Proportion of right turns, PRT
0.129
0.074
0.064
0.086
0
0
0
0
0.988
0.993
0.995
0.993
Conflicting bicycle volume, vbic
vpedg
OCCpedg
OCCr
PRTA
FRpb = 1.0 - PLT(1-ApbT)(1-PLTA)
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
150
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2G
Capacity Analysis Module Worksheet
Lane
Group
Mov’ts
Phase
Type
(P,S,T)
Adj.
Flow
Rate
(v)
Adj. Sat.
Flow
Rate
(s)
Flow
Ratio
(v/s)
Green
Ratio
g/C
Lane
Group
Capacity
(c)
Lane
Group
v/c Ratio
(X)
Critical
Lane
Group
815
2023
0.403
0.550
1112
0.733
*
LTR
834
2200
0.379
0.550
1210
0.689
NB
277
1410
0.196
0.390
550
0.504
547
1664
0.329
0.390
649
0.843
EB
LTR
WB
LTR
SB
LTR
Cycle Length, C 100 sec.
Lost time per Cycle, L 6 sec
Y = Sum (v/s)d = 0.732
*
Xc = Y x C / (C-L) = 0.778 sec
151
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.2H
LOS Module Worksheet
Lane Group
Movements
v/c
Ratio
X
Green
Ratio
g/C
Unifor
m
Delay
d1
Delay
Adj.
Facto
r
DF
Lane
Group
Cap.
c
Incr.
Delay
d2
Delay
LOS
EB
LTR
0.733
0.55
17.0
0.68
1112
4.3
15.9
B
LTR
0.689
0.55
16.3
1.31
1210
3.5
24.8
C
LTR
0.504
0.39
23.2
1.0
550
1.5
24.7
C
LTR
0.843
0.39
27.7
1.0
649
12.6
40.3
D
WB
NB
SB
Intersection Delay 25.2 sec/veh
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Intersection LOS C
152
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.3A
Volume Adjustment Module Worksheet
Appr.
Mvt.
Mvt.
Vol.
(veh/h)
LT
122
TH
PHF
Flow
Rate
(veh/h)
Lane
Group
Flow
rate in
lane
group
(veh/h)
Numbe
r of
lanes
N
Prop. Of LT or
RT
PLT or PRT
0.95
128
L
128
1
1 LT
973
0.95
1024
T
1024
2
TH
608
0.95
640
TR
768
2
0.167 RT
RT
122
0.95
128
LT
30
0.95
32
TH
736
0.95
775
LTR
832
2
0.038 LT
0.030 RT
RT
24
0.95
25
EB
RT
LT
WB
NB
LT
SB
TH
RT
153
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.3B
Saturation Flow Rate Module Worksheet
Appr
.
so
N
fw
fHV
fg
fp
fbb
fa
1900
1
1.00
0.93
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1767
1900
2
1.00
0.93
1.0
1.0
0.98
1.0
0.95
0.98
0.70
0.997
0.988
3290
1900
2
1.00
0.93
1.0
1.0
0.98
1.0
0.95
0.98
1.0
0.997
0.993
3207
1900
2
1.00
0.99
1.0
0.9
3
1.0
0.9
0.95
1.0
1.0
0.998
0.995
3306
fLU
fRT
fLT
fRpb
fLpb
s
EB
WB
NB
SB
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
154
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.3C
Capacity Analysis Module Worksheet
Lane
Group
Mov’ts
Phase
Type
(P,S,T)
Adj.
Flow
Rate
(v)
Adj. Sat.
Flow
Rate
(s)
Flow
Ratio
(v/s)
Green
Ratio
g/C
Lane
Group
Capacity
(c)
Lane
Group
v/c Ratio
(X)
Critical
Lane
Group
L
128
1767
0.072
0.099
176
0.727
*
T
1024
3290
0.311
0.425
1398
0.732
*
768
3207
0.239
0.425
1363
0.563
832
3306
0.252
0.343
1134
0.734
EB
WB
TR
NB
LTR
155
*
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.3D
LOS Module Worksheet
Lane Group
Movements
v/c
Ratio
X
Green
Ratio
g/C
Unifor
m
Delay
d1
Delay
Adj.
Facto
r
DF
Lane
Group
Cap.
c
Incr.
Delay
d2
Delay
LOS
L
0.727
0.099
TR
0.732
0.425
17.0
0.68
1112
4.3
15.9
B
TR
0.563
0.425
16.3
1.31
1210
3.5
24.8
C
LTR
0.734
0.343
23.2
1.0
550
1.5
24.7
C
EB
WB
NB
Intersection Delay 25.2 sec/veh
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Intersection LOS C
156
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.4A
Volume Adjustment Module Worksheet
1
Appr.
2
Mvt.
3
Mvt.
Vol.
(veh/h)
4
PHF
5
Flow
Rate
(veh/h)
[3]/[4]
LT
60
.95
63
TH
675
.95
711
RT
30
.95
32
LT
35
.95
37
TH
725
.95
763
RT
25
.95
26
LT
30
.95
32
TH
380
.95
400
RT
40
.95
42
LT
45
.95
47
TH
575
.95
605
RT
30
.95
32
6
Lane
Group
7
Flow
rate in
lane
group
(veh/h)
8
Numbe
r of
lanes
N
9
Lane
Util.
Factor
U
10
Adj.
Flow,
v
(veh/h)
[7]*[9]
11
Prop.
Of LT
or RT
PLT or
PRT
LTR
806
2
1.05
846
.08 LT
.04 RT
LTR
826
2
1.05
867
.04 LT
.03 RT
LTR
474
2
1.05
498
.07 LT
.09 RT
LTR
684
2
1.05
718
.07 LT
.05 RT
EB
WB
NB
SB
157
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.4B
Saturation Flow Rate Module Worksheet
1
Appr.
6
Heav
y
veh.
fHV
7
Grad
e
1.0
.96
1.0
2
1.0
.96
1900
2
1.0
1900
2
1.0
2
Lane
Grou
p
Mvt.
3
Ideal
Sat.
Flow
LTR
4
No.
of
lanes
N
5
Lane
width
fw
1900
2
LTR
1900
LTR
LTR
8
Pkg.
9
Bus
block
.
fbb
10
Area
Type
fa
11
Right
turn
fRT
12
Left
Turn
fLT
13
Adj.
Sat.
Flow
1.0
1.0
1.0
.99
.69
2501
1.0
1.0
1.0
1.0
1.0
.79
2882
.94
1.0
1.0
1.0
1.0
.99
.82
2886
.94
1.0
1.0
1.0
1.0
.99
.86
3049
fp
fg
EB
WB
NB
SB
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
158
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.4C
Supplemental Worksheet for Permitted Left Turns
For use where the subject approach is opposed by a multilane approach
EB
WB
Enter Cycle Length, C
70
70
Enter Actual Green Time, G
30
30
Enter Effective Green Time, g
30
30
Enter opposing eff. Green time, go
30
30
Enter # of lanes in lane group, N
2
2
Enter # of opposing lane, No
2
2
Enter adjusted left turn flow rate, vLT
63
37
Enter Proportion of lt. turns in lane group, PLT
.08
.04
Enter adjusted opposing flow rate, vo
867
846
3
3
Compute left turns per cycle, LTC = vLTC/3600
1.23
.72
Compute opposing flow per lane, per cycle, volc
8.43
8.23
1
1
Compute gf
7.82
11.95
Compute opposing queue ratio, qro
.57
.57
Compute gq
9.69
9.29
Compute gu
20.31
18.05
Compute fs
.33
.35
Compute PL
.29
.17
Determine EL1
9.35
8.3
Compute fmin
.09
.08
Compute fm
.46
.67
Compute fLT
.69
.79
Enter lost time per phase, tL
Determine opposing platoon ratio, Rpo
159
NB
SB
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.4D
Supplemental Worksheet for Permitted Left Turns
For use where the subject approach is opposed by a multilane approach
EB
NB
SB
Enter Cycle Length, C
70
70
Enter Actual Green Time, G
34
34
Enter Effective Green Time, g
34
34
Enter opposing eff. Green time, go
34
34
Enter # of lanes in lane group, N
2
2
Enter # of opposing lane, No
2
2
Enter adjusted left turn flow rate, vLT
32
47
Enter Proportion of lt. turns in lane group, PLT
.07
.07
Enter adjusted opposing flow rate, vo
718
498
3
3
Compute left turns per cycle, LTC = vLTC/3600
.62
.91
Compute opposing flow per lane, per cycle, volc
6.98
4.84
1
1
15.15
11.87
Compute opposing queue ratio, qro
.51
.51
Compute gq
5.97
2.78
Compute gu
180.85
22.13
Compute fs
.43
.56
Compute PL
.25
.21
Determine EL1
5.02
3.09
Compute fmin
.07
.07
Compute fm
.72
.80
Compute fLT
.82
.86
Enter lost time per phase, tL
Determine opposing platoon ratio, Rpo
Compute gf
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
160
WB
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.4E
Capacity Analysis Module Worksheet
1
Lane
Group
Mov’ts
2
Phase
Type
(P,S,T)
3
Adj.
Flow
Rate
(v)
4
Adj. Sat.
Flow
Rate
(s)
5
Flow
Ratio
(v/s)
[3]/[4]
6
Green
Ratio
g/C
7
Lane
Group
Capacity
(c)
[4]*[6]
8
Lane
Group
v/c Ratio
(X)
[3]/[7]
9
Critical
Lane
Group
[*]
EB LTR
846
2501
.338
.429
1072
.789
*
WB LTR
867
2882
.301
.429
1235
.702
NB LTR
498
2886
.173
.486
1402
.355
SB LTR
718
3049
.235
.486
1481
.485
Cycle Length, C 70 sec.
Lost time per Cycle, L 6 sec
Y = Sum (v/s)d = .574
*
Xc = Y x C / (C-L) = .628 sec
161
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.4F
LOS Module Worksheet
2
v/c
Ratio
X
3
Gree
n
Ratio
g/C
4
Unifor
m
Delay
d1
5
Dela
y
Adj.
Facto
r
DF
6
Lane
Group
Cap.
c
7
d2
Cal.
Term
m
8
Incr.
Dela
y
d2
9
Dela
y
[4]x[
5]
+[8]
LTR
.786
.429
13.1
1.0
1072
16
2.8
LTR
.702
.429
12.4
1.0
1235
16
LTR
.355
.486
8.5
1.0
1402
LTR
.485
.486
9.2
1.0
1481
1
Lane Goup
Movements
10
11
12
LOS
Dela
y
LO
S
16.0
C
16.0
C
1.3
13.7
B
13.7
B
16
0.1
8.6
B
8.6
B
16
0.2
9.4
B
9.4
B
EB
WB
NB
SB
Intersection Delay 12.4 sec/veh
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Intersection LOS B
162
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.5A
Planning Method Lane Volume Worksheet
Location: Problem 10.5
Direction: Eastbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
95
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
75
6. RT Volume
510
0
7. RT lanes
8. RT adj. factor
1
RT lane vol:
Cross Product [2]x[1] = 38250
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
450
1
2
112
Protected
Not
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
562
3.98
.68
410
410
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
163
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.5B
Planning Method Lane Volume Worksheet
Location: Problem 10.5
Direction: Westbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
85
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
100
6. RT Volume
620
0
7. RT lanes
8. RT adj. factor
RT lane vol:
Cross Product [2]x[1] = 62000
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
325
1
2
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
1
164
425
5.33
.58
363
363
100
Protected
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.5C
Planning Method Lane Volume Worksheet
Location: Problem 10.5
Direction: Northbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
2. Opposing mainline vol.
3. No of exclusive LT lanes
4. LT adj. factor
130
1100
1
.95
6. RT Volume
7. RT lanes
8. RT adj. factor
RT lane vol:
200
1
0.85
235
Cross Product [2]x[1] = 143000
opposed
Permitted
Protected
5. LT lane volume: [1] / ([3]x[4])
0
137
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
1
Not
875
1
2
875
438
438
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
165
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.5D
Planning Method Lane Volume Worksheet
Location: Problem 10.5
Direction: Southbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
2. Opposing mainline vol.
3. No of exclusive LT lanes
4. LT adj. factor
175
1075
1
.95
6. RT Volume
7. RT lanes
8. RT adj. factor
RT lane vol:
150
1
0.85
176
Cross Product [2]x[1] = 188125
opposed
Permitted
Protected
5. LT lane volume: [1] / ([3]x[4])
0
184
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
166
1
950
1
2
950
475
475
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.5E
Planning Method Signal Operations Worksheet
Phase plan selection from lane vol. w.s.
EB
WB
NB
SB
Critical through-right lane volume
410
363
438
475
137
184
Prot
Prot
LT lane volume
Per
m
Left turn protection
Per
m
Dominant left turn
*
Phase plan selected
1
Min. Cycle length 60
3b
Max cycle length 120
East-West
Phase 1
*
North-South
Phase 1
Phase 2
Phase 3
EWG
NSL
STL
NST
Crit. Phase Vol.
410
137
47
438
Lost time/phase
3
3
0
3
Mvmt. Code
Phase 2
PHF .95
Phase 3
Critical Sum 1032
Lost time/Cycle 9
CBD Adjustment 1
Critical v/c ratio .62
Intersection status Under Capacity
Optional Timing Plan Computation
Reference Sum
Cycle length
East-West
Phase 1
Phase 2
North-South
Phase 3
Phase 1
Phase 2
Phase 3
Green time
167
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.6A
Planning Method Lane Volume Worksheet
Location: Problem 10.6
Direction: Eastbound
Left Turn Movement
Right Turn MovementExclusive
Shared
RT Lane
RT Lane
1. LT Volume
141
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
111
6. RT Volume
755
0
7. RT lanes
8. RT adj. factor
RT lane vol:
Cross Product [2]x[1] = 83805
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
666
1
2
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
9.58
.51
809
809
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
60
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
1
168
832
166
Protected
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.6B
Planning Method Lane Volume Worksheet
Location: Problem 10.6
Direction: Westbound
Left Turn Movement
Right Turn MovementExclusive
Shared
RT Lane
RT Lane
1. LT Volume
126
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
148
6. RT Volume
918
0
7. RT lanes
8. RT adj. factor
1
RT lane vol:
Cross Product [2]x[1] = 135864
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
481
1
2
148
Protected
Not
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
11
.5
625
625
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
60
169
629
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.6C
Planning Method Lane Volume Worksheet
Location: Problem 10.6
Direction: Northbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
2. Opposing mainline vol.
3. No of exclusive LT lanes
4. LT adj. factor
192
1628
1
.95
6. RT Volume
7. RT lanes
8. RT adj. factor
RT lane vol:
296
1
0.85
348
Cross Product [2]x[1] = 312576
opposed
Permitted
Protected
5. LT lane volume: [1] / ([3]x[4])
0
202
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
170
1
1295
1
2
1295
648
648
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.6D
Planning Method Lane Volume Worksheet
Location: Problem 10.6
Direction: Southbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
2. Opposing mainline vol.
3. No of exclusive LT lanes
4. LT adj. factor
259
1591
1
.95
6. RT Volume
7. RT lanes
8. RT adj. factor
RT lane vol:
222
1
0.85
261
Cross Product [2]x[1] = 412069
opposed
Permitted
Protected
5. LT lane volume: [1] / ([3]x[4])
0
273
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
1
Not
1406
1
2
1406
703
703
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
171
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.6E
Planning Method Signal Operations Worksheet
Phase plan selection from lane vol. w.s.
EB
WB
NB
SB
Critical through-right lane volume
809
624
648
703
202
273
Prot
Prot
LT lane volume
Per
m
Left turn protection
Per
m
Dominant left turn
*
Phase plan selected
1
Min. Cycle length 60
3b
Max cycle length 120
PHF .95
East-West
Phase 1
North-South
Phase 1
Phase 2
Phase 3
EWG
NSL
STL
NST
Crit. Phase Vol.
809
202
71
648
Lost time/phase
3
3
0
3
Mvmt. Code
Phase 2
*
Phase 3
Critical Sum 1730
Lost time/Cycle 9
CBD Adjustment 1
Critical v/c ratio 1.04
Intersection status Over Capacity
Optional Timing Plan Computation
Reference Sum
Cycle length
East-West
Phase 1
Phase 2
North-South
Phase 3
Green time
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
172
Phase 1
Phase 2
Phase 3
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.7A
Volume Adjustment Module Worksheet
1
Appr.
2
Mvt.
3
Mvt.
Vol.
(veh/h)
4
PHF
5
Flow
Rate
(veh/h)
[3]/[4]
6
Lane
Group
7
Flow
rate in
lane
group
(veh/h)
8
Numbe
r of
lanes
N
9
Lane
Util.
Factor
U
10
Adj.
Flow,
v
(veh/h)
[7]*[9]
11
Prop.
Of LT
or RT
PLT or
PRT
LTR
689
2
1.05
723
.12 LT
.15 RT
LTR
566
2
1.05
594
.20 LT
.17 RT
1.0 LT
LT
75
.90
83
TH
450
.90
500
RT
95
.90
106
LT
100
.90
111
TH
325
.90
361
RT
85
.90
94
LT
130
.90
144
L
144
1
1.0
144
TH
875
.90
972
T
972
2
1.05
1021
RT
200
.90
222
R
222
1
1.0
222
1.0 RT
LT
175
.90
194
L
194
1
1.0
194
1.0 LT
TH
950
.90
1056
T
1056
2
1.05
1109
RT
150
.90
167
R
167
1
1.0
167
EB
WB
NB
SB
173
1.0 RT
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.7B
Saturation Flow Rate Module Worksheet
1
Appr.
6
Heav
y
veh.
fHV
7
Grad
e
9
Bus
block
.
fbb
10
Area
Type
fa
11
Right
turn
fRT
12
Left
Turn
fLT
13
Adj.
Sat.
Flow
1.0
.97
1.0
1.0
1.0
.9
.98
.72
2330
2
1.0
.97
1.0
1.0
1.0
.9
.98
.57
1833
1900
1
1.0
.97
1.0
1.0
1.0
.9
1.0
.95
1577
T
1900
2
1.0
.97
1.0
1.0
1.0
.9
1.0
1.0
3320
R
1900
1
1.0
.97
1.0
1.0
1.0
.9
0.85
1.0
1411
L
1900
1
1.0
.97
1.0
1.0
1.0
.9
1.0
.95
1577
T
1900
2
1.0
.97
1.0
1.0
1.0
.9
1.0
1.0
3320
R
1900
1
1.0
.97
1.0
1.0
1.0
.9
0.85
1.0
1411
2
Lane
Grou
p
Mvt.
3
Ideal
Sat.
Flow
LTR
4
No.
of
lanes
N
5
Lane
width
fw
1900
2
LTR
1900
L
8
Pkg.
fp
fg
EB
WB
NB
SB
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
174
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.7C
Supplemental Worksheet for Permitted Left Turns
For use where the subject approach is opposed by a multilane approach
EB
WB
Enter Cycle Length, C
60
60
Enter Actual Green Time, G
20
20
Enter Effective Green Time, g
20
20
Enter opposing eff. Green time, go
20
20
Enter # of lanes in lane group, N
2
2
Enter # of opposing lane, No
2
2
Enter adjusted left turn flow rate, vLT
83
111
Enter Proportion of lt. turns in lane group, PLT
.12
.20
Enter adjusted opposing flow rate, vo
594
723
3
3
Compute left turns per cycle, LTC = vLTC/3600
1.38
10.85
Compute opposing flow per lane, per cycle, volc
4.95
6.03
1
1
Compute gf
3.57
2.08
Compute opposing queue ratio, qro
.67
.67
Compute gq
4.90
7.05
Compute gu
15.10
12.95
Compute fs
.50
.42
Compute PL
.32
.59
Determine EL1
.465
8.57
Compute fmin
.13
.16
Compute fm
.53
.22
Compute fLT
.72
.57
Enter lost time per phase, tL
Determine opposing platoon ratio, Rpo
175
NB
SB
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.7D
Capacity Analysis Module Worksheet
1
Lane
Group
Mov’ts
2
Phase
Type
(P,S,T)
3
Adj.
Flow
Rate
(v)
4
Adj. Sat.
Flow
Rate
(s)
5
Flow
Ratio
(v/s)
[3]/[4]
6
Green
Ratio
g/C
7
Lane
Group
Capacity
(c)
[4]*[6]
8
Lane
Group
v/c Ratio
(X)
[3]/[7]
EB LTR
723
2330
.310
.333
777
.931
WB LTR
594
1833
.324
.333
611
.972
NB L
144
1577
.091
.133
210
.685
NB T
1021
3320
.308
.383
1273
.802
NB R
222
1411
.157
.383
541
.410
SB L
194
1577
.123
.133
210
.923
*
SB T
1109
3320
.334
.383
1273
.871
*
SB R
167
1411
.118
.383
541
.309
Cycle Length, C 60 sec.
Lost time per Cycle, L 9 sec
Y = Sum (v/s)d = .781
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Xc = Y x C / (C-L) = .919 sec
176
9
Critical
Lane
Group
[*]
*
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.7E
LOS Module Worksheet
2
v/c
Ratio
X
3
Gree
n
Ratio
g/C
4
Unifor
m
Delay
d1
5
Dela
y
Adj.
Facto
r
DF
6
Lane
Group
Cap.
c
7
d2
Cal.
Term
m
8
Incr.
Dela
y
d2
9
Dela
y
[4]x[
5]
+[8]
LTR
.931
.333
14.7
1.0
777
16
12.8
LTR
.972
.333
15.0
1.0
611
16
L
.685
.133
18.8
1.0
210
T
.802
.383
12.5
1.0
R
.410
.383
10.3
L
.923
1.33
T
.871
R
.309
1
Lane Goup
Movements
10
11
12
LOS
Dela
y
LO
S
27.5
D
27.5
D
21.9
36.9
D
36.9
D
16
6.0
24.8
C
1273
16
2.7
15.2
C
15.5
C
1.0
541
16
0.3
10.6
B
19.5
1.0
210
16
29.3
48.8
E
.383
13.0
1.0
1273
16
4.9
17.9
C
21.1
C
.383
9.8
1.0
541
16
0.1
9.9
B
EB
WB
NB
SB
Intersection Delay 22.6 sec/veh
Intersection LOS C
177
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.8A
Planning Method Lane Volume Worksheet
Direction: Eastbound
Location: Problem 10.8
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
104
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
82
6. RT Volume
557
0
7. RT lanes
8. RT adj. factor
RT lane vol:
Cross Product [2]x[1] = 45674
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
492
1
2
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
1
178
614
3.38
.7
440
440
122
Protected
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.8B
Planning Method Lane Volume Worksheet
Location: Problem 10.8
Direction: Westbound
Left Turn Movement
Right Turn MovementExclusive
Shared
RT Lane
RT Lane
1. LT Volume
93
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
109
6. RT Volume
678
0
7. RT lanes
8. RT adj. factor
1
RT lane vol:
Cross Product [2]x[1] = 73902
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
355
1
2
109
Protected
Not
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
464
4.54
.6
389
389
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
179
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.8C
Planning Method Lane Volume Worksheet
Location: Problem 10.8
Direction: Northbound
Left Turn Movement
Right Turn MovementExclusive
Shared
RT Lane
RT Lane
1. LT Volume
219
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
142
6. RT Volume
1393
0
7. RT lanes
8. RT adj. factor
RT lane vol:
Cross Product [2]x[1] = 197806
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
258
Protected
142
956
1
2
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
180
1
1356
1.0
678
678
Not opposed
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.8D
Planning Method Lane Volume Worksheet
Location: Problem 10.8
Direction: Southbound
Left Turn Movement
Right Turn MovementExclusive
Shared
RT Lane
RT Lane
1. LT Volume
164
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
191
6. RT Volume
1317
0
7. RT lanes
8. RT adj. factor
RT lane vol:
Cross Product [2]x[1] = 251547
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
1
193
Protected
Not opposed
191
1038
1
2
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
1422
1.0
711
711
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
181
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.8E
Planning Method Signal Operations Worksheet
Phase plan selection from lane vol. w.s.
EB
WB
NB
SB
Critical through-right lane volume
440
389
678
711
142
191
Prot
Prot
LT lane volume
Per
m
Left turn protection
Per
m
Dominant left turn
Phase plan selected
Min. Cycle length
1
60
Max cycle length
4
120
East-West
Phase 1
.95
North-South
Phase 1
Phase 2
EWG
NG
SG
Crit. Phase Vol.
440
678
711
Lost time/phase
3
3
3
Mvmt. Code
Phase 2
PHF
Phase 3
Phase 3
Critical Sum 1829
Lost time/Cycle
9
CBD Adjustment
1
Critical v/c ratio
1.10
Intersection status
Over Capacity
Optional Timing Plan Computation
Reference Sum
Cycle length
East-West
Phase 1
Phase 2
North-South
Phase 3
Green time
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
182
Phase 1
Phase 2
Phase 3
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.9A
Planning Method Lane Volume Worksheet
Direction: Eastbound
Location: Problem 10.9
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
116
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
91
6. RT Volume
620
0
7. RT lanes
8. RT adj. factor
1
RT lane vol:
Cross Product [2]x[1] = 56420
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
547
1
2
136
Protected
Not
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
683
5.33
.6
570
570
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
183
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.9B
Planning Method Lane Volume Worksheet
Location: Problem 10.9
Direction: Westbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
103
2. Opposing mainline vol.
3. No of exclusive LT lanes
0.85
4. LT adj. factor
122
6. RT Volume
754
0
7. RT lanes
8. RT adj. factor
RT lane vol:
Cross Product [2]x[1] = 91988
opposed
Permitted
5. LT lane volume: [1] / ([3]x[4])
0
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
395
1
2
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
9.55
.51
506
506
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
60
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
1
184
516
121
Protected
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.9C
Planning Method Lane Volume Worksheet
Location: Problem 10.9
Direction: Northbound
Right Turn MovementExclusive
Shared
RT Lane
Left Turn Movement
RT Lane
1. LT Volume
2. Opposing mainline vol.
3. No of exclusive LT lanes
4. LT adj. factor
158
1339
1
.95
6. RT Volume
7. RT lanes
8. RT adj. factor
RT lane vol:
243
1
0.85
286
Cross Product [2]x[1] = 211562
opposed
Permitted
Protected
5. LT lane volume: [1] / ([3]x[4])
0
166
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
1
Not
1065
1
2
1065
532
532
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
185
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.9D
Planning Method Lane Volume Worksheet
Direction: Southbound
Location: Problem 10.9
Left Turn Movement
RT Lane
1. LT Volume
2. Opposing mainline vol.
3. No of exclusive LT lanes
4. LT adj. factor
Right Turn MovementExclusive
Shared
RT Lane
213
1308
1
.95
6. RT Volume
7. RT lanes
8. RT adj. factor
RT lane vol:
183
1
0.85
215
Cross Product [2]x[1] = 278604
opposed
Permitted
Protected
5. LT lane volume: [1] / ([3]x[4])
0
224
Through Movement
11. Through Volume
12. Parking adj. Factor
13. No. of through lanes including shared lanes
Exclusive LT lane computations
14. Total approach volume ([10]+[11]) / [12]
16. Left turn equivalence
18. Through Lane volume
19. Critical Lane volume
Shared LT lane computations
14. Total approach volume
15. Proportion of left turns in the lane group
16. Left turn equivalence
17. Left turn adj. factor
18. Through lane volume
19. Critical lane volume
Left turn check (if [16]>8)
20. Permitted left turn sneaker capacity
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
186
1
1156
1
2
1156
578
578
Not
Chapter 10: Capacity and Level of Service at Signalized Intersections
Exhibit 10.9E
Planning Method Signal Operations Worksheet
Phase plan selection from lane vol. w.s.
EB
WB
NB
SB
Critical through-right lane volume
570
506
532
578
166
224
Prot
Prot
LT lane volume
Per
m
Left turn protection
Per
m
Dominant left turn
*
Phase plan selected
Min. Cycle length
1
60
Max cycle length
3b
120
East-West
Phase 1
PHF
.95
North-South
Phase 1
Phase 2
Phase 3
EWG
NSL
STL
NST
Crit. Phase Vol.
570
166
58
532
Lost time/phase
3
3
0
3
Mvmt. Code
Critical Sum 1326
Lost time/Cycle
CBD Adjustment
Critical v/c ratio
Intersection status
Phase 2
*
Phase 3
9
1
.79
Under Capacity
Optional Timing Plan Computation
Reference Sum
Cycle length
East-West
Phase 1
Phase 2
North-South
Phase 3
Phase 1
Phase 2
Phase 3
Green time
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Chapter 11
The Transportation Planning Process
11-1
Explain why the transportation planning process is not intended to furnish a
decision or give a single result.
The transportation planning process is a rational one that intends to furnish
unbiased information about the effects that proposed transportation projects will
have on the community, its environmental impacts, and its expected users. This
process is not intended to furnish or to give a single result that must be followed
but to provide information to the general public and to the political bodies
responsible for deciding whether proposed transportation projects should be
developed.
11-2
Describe the steps that an engineer must follow if asked to determine the need for a
grade-separated railroad grade crossing that would replace an at-grade crossing of
a two-lane highway with a rail line.
The first step the engineer will need to follow is to define the situation.
The engineer must fully understand the reason why the grade-separated structure
is said to be needed.
The next step is to define the problem in terms of the objectives to be
accomplished by the project. The objectives for this project might be to improve
safety by reducing or eliminating vehicle-train interaction. Another quantifiable
objective might include the reduction in travel time to the motorists by
eliminating delay due to crossing trains.
The third step for the engineer is to identify solutions to the perceived
problem. Possible solutions to the problem might include the installation of
active warning devices at the grade crossing, the diversion of motorists to other
close-by grade crossings, or to construct a grade separated structure. A “nobuild” or “do-nothing” alternative is typically considered as a possible solution.
The fourth step in the process is to analyze the expected performance of
each of the alternatives under present and future conditions. This analysis should
employ quantified measures of effectiveness such as travel time delay reductions
and crash rate reductions. The analysis should also take into consideration costs
such as the construction, maintenance, and operation costs of each alternative.
The fifth step in the process is to compare the alternatives against the
objectives developed in step 2 of this process. The advantages and disadvantages
of each must be studied.
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The sixth step is to recommend one of the alternatives after considering all
of the factors involved.
The final step in the process is to design the facility after one of the
alternatives has been selected. This step also moves the project out of the
planning process and into project development.
11-3
Describe the basic steps in the transportation planning process.
The basic steps in the transportation planning process are as follows:
Step 1 - Situation definition is where the present situation is analyzed and
described.
Step 2 - Problem definition is where the problem is described in terms of
objectives to be accomplished and establish criteria by which effectiveness of the
project can be measured.
Step 3 - Search for solutions is the brainstorming stage where many options may
be proposed for testing and evaluating. Preliminary feasibility studies are
included to narrow the field to the most promising.
Step 4 - Analysis of performance is used to estimate how the proposed
alternatives, under present and future conditions, operate using the criteria
established earlier.
Step 5 - Evaluation of alternatives is where each alternative is judged by the
criteria as to how the objectives are accomplished. This involves the calculation
of benefits and costs, economic evaluation, and comparison of cost effectiveness.
Step 6 - Choice of project is where the selection is made considering all factors.
It may be based on cost alone in simple cases or it may involve hearings, political
considerations, or other issues.
Step 7 - Specification and construction is the final stage in the process. Once the
project has been selected it moves into the design phase of project development
and ultimately into construction.
11-4
Select a current transportation problem in your community or state. Briefly
describe the situation and the problem. Indicate options available and the major
impacts of each option on the community.
Problem Situation: Traffic congestion and safety on Interstate 81 is widely
identified as a problem needing to be addressed. This aging highway, most of
which was built in the 1960s, carries a relatively high percentage of trucks
through mountainous terrain, resulting in less than desirable service provided by
the facility.
Options available: There are three general categories of options available to
mitigate the problem
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Chapter 11: The Transportation Planning Process
(a)
(b)
(c)
Add lanes to existing highway
Upgrade parallel rail line to improve its capacity, minimizing or
eliminating the need to add lanes
Do nothing alternative
Impacts:
(a)
Increasing the number of travel lanes will decrease the level of
congestion and improve safety. However, this may require additional
right-of-way, result in increased demand and therefore increased air and
noise pollution. It would cause some inconvenience to motorists during
the construction phase.
(b)
Upgrading the parallel rail line will minimize and possibly
eliminate in some rural sections, the need to modify the existing highway.
This would result in increased rail traffic and impacts to communities
along the rail line.
(c)
If no investment is made, congestion would increase and safety
deteriorate; the roadway will operate at a reduced level of service. This
would result in increased vehicle delay and driver dissatisfaction
11-5
Evaluate a proposal to increase tolls on existing roads and bridges. Describe the
general planning analysis used.
In the seven-step process for transportation planning analysis, the following issues
would be addressed:
Situation definition: What is the need for increased tolls? Where are the toll
roads and bridges and what are the traffic volumes on these facilities? Who are
the primary users of these facilities and what are the alternative routes, if any?
Problem definition: Why should tolls be raised and what would be the benefits
of increased revenues? How will the increased tolls help the situation?
Search for solutions: Can tolls be increased only on the roads with existing tolls?
Should more toll roads be added? Can tolls remain constant and other solutions
be sought? There might be viable combinations of increasing certain tolls and
leaving other tolls constant. Determine feasibility of the alternatives developed.
Analysis of performance: Analyze alternatives in terms of cost, travel demand,
time delays, and other measures of effectiveness.
Evaluate alternatives: Compare and the alternatives. Choose the best alternative
after consulting with groups of policy makers.
Implementation: Implement the recommendations.
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Chapter 11: The Transportation Planning Process
11-6
Prepare a study to consider improvements to transportation between an airport and
the city it serves.
Collect data on existing
transportation systems. Determine
destinations of travelers arriving by air.
Inventory of existing facilities
Determine the purpose and need
for potential improvements, including
available resources and time period.
Establish goals and objectives
Determine existing and projected
travel patterns.
Estimate travel demand
Determine existing and future land use
and socioeconomic conditions.
Trip generation
Determine trips made.
Trip distribution
Identify available modes of travel
between airport and city and distribute
trips among them.
Mode choice
Network assignment
Distribute trips across modal networks.
Determine and evaluate alternatives.
Evaluation of alternatives
Select a project or set of projects
for development.
Project selection
11-7
What caused transportation planning to become institutionalized in urban areas,
and what does the process need to be based on?
The 1962 Federal Aid Highway Act institutionalized transportation
planning in urban areas. This act required that all highway projects in urbanized
areas with populations of 50,000 or more be based on a transportation planning
process that is continuous, comprehensive, and cooperative (“3C”). The core of
the process is the travel demand forecasting process, consisting of trip generation,
trip distribution, modal split, and traffic assignment.
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11-8
Explain the three “C”s concept in the transportation planning process, as mandated
in the Federal-Aid Highway Act of 1962.
The three “C”s are continuing, comprehensive, and cooperative term
Continuing implies that the process be revisited frequently and viewed as an
ongoing concern. Comprehensive in this context ensures that all transportation
modes are addressed. A cooperative process indicates that the state (or states) and
all municipalities in an urbanized area work together.
11-9
Urban transportation is concerned with two separate time horizons. Briefly
describe each and provide examples of the types of projects that can be categorized
in each horizon.
The two time horizons that urban transportation planners are concerned
with are short-term projects and long-term projects. The short-term projects are
those that can be implemented within a one- to three-year period. These projects
are designed to provide better management of existing facilities by making them
as efficient as possible. Examples of short-term projects might include re-timing
of traffic signals to improve traffic flow, transit improvements, and car and van
pooling initiatives to reduce traffic congestion.
The second time horizon deals with long-range transportation needs of an
area and identifies the projects to be constructed over a 20- or 25-year period.
Typical long-term projects include the addition of new highway segments,
additional lanes on existing highways, rapid transit systems and extensions, and
access roads to airports.
11-10
An existing highway-rail at-grade crossing is to be upgraded. Plans were developed
in 2001; the cost estimate for that improvement was $230,000 at that time. Due to
funding constraints, construction of the improvement was delayed until 2005. Using
the data given in Table 11.3, estimate the construction cost in 2005 dollars.
Because this is a highway-rail crossing improvement, select the highway cost
index from Table 11.3. Using Equation 11.1:
2005 index
Estimate in 2005 dollars = (Estimate in 2001 dollars)
2001 index
Estimate in 2005 dollars = ($230,000 )
183.6
= $291,630
144.8
Thus, the improvements will cost $291,630 in 2005 dollars.
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Chapter 11: The Transportation Planning Process
11-11
Given the information in Problem 11-10, assume that construction was delayed until
2008. Using the compound growth rate (see Chapter 13) that can be derived from
the data given in Table 11.3, estimate the construction cost in 2008 dollars.
Use the following compound interest equation (see Chapter 13):
F = P(1 + i ) n
Rearrange to solve for i (compound growth rate)
1
⎛ F ⎞n
i = ⎜ ⎟ −1
⎝P⎠
Treat the 2001 index value as P, and the 2005 index value as F.
1
1
⎛ F ⎞n
⎛ 183.6 ⎞ 4
i = ⎜ ⎟ −1 = ⎜
⎟ − 1 = 0.06115
⎝P⎠
⎝ 144.8 ⎠
Now use this compound growth rate to carry forward the 2001 cost
estimate to 2008.
F = P (1 + i ) n = 230,000(1.06115) 7 = $348,470
11-12
Describe the three forms of environmental impact analysis documentation.
There are three forms of environmental impact analysis documentation: a full
environmental impact statement (EIS), a simpler environmental assessment (EA),
or a cursory checklist of requirements known as a categorical exclusion (CE). A
categorical exclusion is typically applicable for small-scale projects that do not
involve substantial new construction or environmental permits. In such cases no
additional analysis or documentation is required. At the other extreme, an EIS is
developed for large-scale projects for which it is apparent that environmental
impacts will occur and therefore need to be documented. On some projects it may
be readily apparent whether significant environmental impacts will occur, in
which case an EA, simpler than an EIS is developed. The findings of the EA will
either indicate that a full EIS is warranted, or that there is no significant impact, in
which case, a FONSI is the final documentation.
11-13
What is the purpose of performing inventories and surveys for each defined
geographic unit or traffic zone within a study area?
Inventories and surveys are made to determine traffic volumes, land-uses,
the origins and destinations of travelers, population, employment, and economic
activity. These inventories are made of existing transportation facilities, both
highway and transit, their capacity, average speeds, travel time, and traffic
volumes.
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11-14
What are the four basic elements that make up the urban transportation forecasting
process?
The four basic elements that make up the urban transportation planning
process are: (1) data collection, (2) analysis of existing conditions and calibration
of forecasting techniques, (3) forecast of future travel demand, and (4) analysis of
results.
11-15
In the data collection phase of the urban transportation forecasting process, what
type of information should the data reveal for a traffic analysis zone?
The data collection phase should provide the transportation planner with
information about the city and its people. The data should reveal, specifically, the
area's economic activity (such as employment, sales volume, and household
income), types of land-uses and densities, travel characteristics (trip and traveler
profiles), and physical transportation facility characteristics (such as capacity,
travel speeds, and travel times).
11-16
List four ways of obtaining origin-destination information. Which method would
produce the most accurate results?
Four methods to obtain origin-destination information are:
(a)
Home interview
(b)
Roadside interview
(c)
Telephone interview
(d)
Return postcard survey
The most accurate method to collect origin-destination is the home interview
method; however, this method is also the most expensive.
11-17
Define the following terms: (a) link, (b) node, (c) centroid, and (d) network.
(a) Link - A link is a portion of the highway system that can be described by its
capacity, lane width, and speed.
(b) Node - A node is the end point of a link and represents an intersection or
location where a link changes direction, capacity, width, or speed.
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(c) Centroid - A centroid is the location within a zone where trips are considered
to begin and end.
(d) Network - A network is a combination of the above elements to represent a
highway system.
11-18
Draw a link-node diagram of the streets and highway within your neighborhood or
campus. For each link show travel times and distances (to the nearest 0.1 mile).
The following link-node diagram is of the roadways surrounding the University of
Virginia. The travel times provided are for automobiles operating at the posted
speed.
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Chapter 11: The Transportation Planning Process
11-19
The initial zone structure for regional travel demand model has been proposed as
summarized below. Name three potential problems with this structure.
Zones:
Employment:
Labor Force:
Population:
Trips produced by the study area:
Trips attracted by the study area:
50
100,000
60,000
200,000
100,000
150,000
The three problems are:
•
•
•
The ratio of labor force to employment is 60,000/100,000 = 0.60, meaning
that a substantial number of workers are imported into the study area but
not considered within the regional model.
With 200,000 people and 50 zones, we have an average of about 4,000
people per zone, whereas an average of 1,000 people per zone is desired.
The number of trips produced and attracted by the area are not equal. (If
students have already discussed productions and attractions, then it may be
noted that the ratio of productions to attractions is 100,000/150,000 or
about 0.67, which is outside of the desired heuristic rule of 0.9 to 1.1.)
11-20
Explain the flaw in each of these traffic analysis zones (TAZs)
(a) TAZ 1 contains dormitories, a research park, and 300 single family
detached dwelling units.
(b) TAZ 2 straddles an interstate highway, with half of the zone east of
the highway and half to the west of the highway.
(c) TAZ 3 contains 5,000 people
(a) The first zone contains multiple land uses—ideally a zone should contain
a single land use.
(b) The second zone includes a major topographical barrier—the interstate
highway. Ideally zones should not include such barriers.
(c) The third zone is probably too large for analysis (containing more than
1,000 people). That in itself is not a firm rule, but in practice such a zone
could easily contain two major parallel routes which may adversely affect
the traffic assignment process.
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Chapter 11: The Transportation Planning Process
11-21
Define these four acronyms and explain how they affect the transportation planning
process: MPO, CLRP, TIP, and STIP. Describe the composition of one of them.
A Metropolitan Planning Organization (MPO) is a transportation policymaking organization made up of representatives from local government and
transportation authorities. For example, the Thomas Jefferson Planning
District Commission generally staffs the Charlottesville-Albemarle MPO.
Serving on the MPO Policy Board are five voting members: two from the
Charlottesville city council, two from the Albemarle County Board of
Supervisors, and one from the Virginia Department of Transportation.
Nonvoting members on the Policy Board include local transit providers
(Jefferson Area United Transportation and Charlottesville Transit Service),
local and regional entities (the Thomas Jefferson Planning District
Commission and the University of Virginia), state agencies (the Virginia
Department of Rail and Public Transportation), and federal agencies (the
Federal Highway Administration, the Federal Aviation Administration, and
the Federal Transit Administration).
MPOs affect the planning process because they are given specific
responsibilities under federal reauthorizations. One of these is to develop and
update the Constrained Long Range Plan (CLRP). This plan has a 20 year
planning horizon and must meet several requirements such as (1) it must
identify the projected demand of persons and goods in the MPO over the
period of the plan and (2) it must identify walkway and bicycle transportation
facilities.
A Transportation Improvement Program (TIP) is a series of projects (four
years out at least) that the MPO approves. The TIP must be financially
constrained, that is, funds must reflect those available. Ideally, the TIP is a
way of allocating limited transportation resources among the various capital
and operating needs of the area. (Under TEA-21 TIPs had a minimum three
year horizon, now under SAFETEA-LU it is four years). Projects must be in
the TIP to receive federal funds. Without the MPOs approval, the project
cannot receive federal funds. Thus if an MPO did not approve a particular
bypass, for example and did not include this project in its TIP, then the project
could not receive federal funds.
The State Transportation Improvement Program (STIP) is a programming
document that includes all projects planned for implementation with the funds
expected from FHWA and FTA for the upcoming three years. The STIP also
includes each MPO’s TIP and all of the projects included in the first (now
four) years of that TIP. Without FHWA approval of the STIP, the project
cannot be funded.
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Chapter 11: The Transportation Planning Process
11-22
Name the steps in the four step travel demand forecasting process where feedback
can occur. Which of these feedback loops directly affects land use? Hint: See
Figure 11.7.
•
•
•
•
•
•
There are three loops where this feedback commonly occurs and three loops
where this feedback may occur:
Congested highway times from traffic assignment affect the trip distribution
and mode choice steps
Transit times and costs from traffic assignment affect the mode choice step
Transit times and costs from traffic assignment affect trip distribution where
transit and travel time form a composite impedance for travel cost
Highway and transit costs may influence employment and residential
locations (possibly in the form of a land use model)
Auto occupancy may be a function of the time of day
Highway and transit times and costs may influence the auto ownership model.
The one feedback loop that directly affects land use is the fourth one named
above: travel costs (whether in the form of monetary costs or travel time) may
ultimately influence employment and residential locations.
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Chapter 12
Forecasting Travel Demand
12-1
Identify and briefly describe the two basic demand forecasting situations in
transportation planning.
There are two basic demand forecasting situations in transportation
planning. The first involves travel demand studies for urban areas, and the second
deals with intercity travel demand. The urban travel demand forecasts, when first
developed in the 1950s and 1960s, required that extensive data sets be prepared
using home interview and/or roadside interview surveys. The information
gathered provided useful insight concerning the characteristics of the trip maker,
such as age, sex, income, auto ownership, the land-use at each end of the trip, and
the mode of travel. Today much of this information can be obtained through U.S.
Census Bureau transportation planning products and from privately assembles
data sets, in addition to the approaches used since the advent of urban travel
forecasting.
In the intercity case, travel patterns are forecast between two cities or
metropolitan areas. Such data generally are aggregated to a greater extent than for
urban travel forecasting, such as city population, average city income, and travel
time or travel cost between city pairs.
12-2
Identify the three factors that affect demand for urban travel.
The three factors that affect the demand for urban travel are: 1) the
location and intensity of land-use, 2) the socio-economic characteristics of the
people living in the area, and 3) the extent, cost, and quality of available
transportation services.
Land-use characteristics are a primary determinant in travel demand. The
amount of traffic generated by a parcel of land depends on how the land is used.
Socio-economic characteristics of the people living within the city also
influence the demand for transportation. Lifestyles and values affect how people
decide to use their resources for transportation.
The availability of transportation facilities and services also affects the
demand for travel. Travelers are sensitive to the levels of service provided by
alternative transportation modes.
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Chapter 12: Forecasting Travel Demand
12-3
Define the following terms:
(a) home-based work (HBW) trips
(b) home-based other (HBO) trips
(c) nonhome-based (NHB) trips
(d) production
(e) attraction
(f) origin
(g) destination
(a) Home-based Work (HBW) trip – a trip for which the purpose is to go from
home to work or from work to home.
(b) Home-based Other (HBO) trip - a trip for which the purpose is to go from
home to another location other than work (e.g., shopping, school, theater) or from
non-work locations to home.
(c) Non-Home Based (NHB) trip - a trip for which neither trip end is at home. A
trip falling into this category might be one that goes from work to shopping, or
school to work.
(d) Production - the ability of a zone to generate trip ends. For all non-home
based trips, productions are synonymous with origins.
(e) Attraction - the ability of a zone to generate trip ends. For non-home based
trips, attractions in a zone can be considered synonymous with trip destinations in
that zone.
(f) Origin - point at which a trip begins.
(g) Destination - point at which a trip ends.
12-4
Given: Cross-classification data for the Jeffersonville Transportation Study Area
Develop the family of cross-classification curves
Determine the number of trips produced (by purpose) for a traffic zone containing
500 houses with an average household income of $35,000. (Use high = 55,000;
medium = 25,000; low = 15,000.)
The following graphs depict the family of cross-classification curves used to
determine the number of trips produced (by purpose).
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Chapter 12: Forecasting Travel Demand
From the graphs above, the following tables can be produced.
Number of HH per Level of Income
Income
Percent
HH/Zone
Total HH
Low
11
500
55
Medium
75
500
375
High
14
500
70
Total
100
500
Percentage of HH Owning x Vehicles
Auto
Ownership
Income
Low
Medium
High
0
19
3
0
1
68
64
14
2
13
32
58
3+
0
1
28
100
100
100
Total
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Chapter 12: Forecasting Travel Demand
Trips per HH per Income Level and Auto Ownership
Auto
Ownership
Income
Low
Medium
High
0
2
3
7
1
7
8
13
2
12
13
18
3+
17
18
23
Next, develop a table to show the number of households owning x vehicles. This
can be accomplished by multiplying the values found in the table of the number of
HH per level of income and the percentage of HH owning x vehicles together.
For example:
For a low income level and 0 auto ownership = (55) * 19% = 11
For medium income level and 0 auto ownership = (375) * 3% = 11
For high income level and 0 auto ownership = (70) * 0% = 0
This procedure is used to generate the following table.
Number of HH Owning x Vehicles
Auto
Ownership
Income
Low
Medium
High
0
11
11
0
1
37
240
10
2
7
120
41
3
0
4
19
Total
55
375
70
The total number of trips made by each income level are then determined as
follows:
Low Income = (2 * 11) + (7 * 37) + (12 * 7) + (17 * 0) = 365
Medium Income = (3 * 11) + (8 * 240) + (13 * 120) + (18 * 4) = 3,585
High Income = (7 * 0) + (13 * 10) + (18 * 41) + (23 * 19) = 1,305
Total Number of Trips = 365 + 3,585 + 1,305 = 5,255
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Chapter 12: Forecasting Travel Demand
Trips by Purpose (%)
Trip
Purpose
Income
Low
Medium
High
HBW
38
37
18
HBO
34
34
38
NHB
28
29
44
TOTAL
100
100
100
The number of trips by purpose can then be calculated as follows:
HBW = (0.38 * 365) + (0.37 * 3585) + (0.18 * 1305) = 1,700
HBO = (0.34 * 365) + (0.34 * 3585) + (0.38 * 1305) = 1,839
NHB = (0.28 * 365) + (0.29 * 3585) + (0.44 * 1305) = 1,716
12-5
Given: A person travels to work in the morning and returns home in the evening.
Determine: Productions and attractions generated in the work and residence zones.
Home - 2 Productions, 0 Attractions.
Work - 0 Productions, 2 Attractions
12-6
Describe and illustrate cross-classification procedures for (a) trip production and
(b) trip attraction.
The FHWA method for estimating trip productions and attractions is based
on the use of cross classification. Cross classification is a method used to
determine the number of trips that begin or end at the home. Relationships are
developed based on socioeconomic data (typically income data) and origindestination surveys. Determining the number of productions and attractions
involves a process of 5 steps which determine the percentage of trips by purpose,
i.e. the number of trips that are home-based work (HBW), home-based other
(HBO), and non-home based (NHB). From these percentages the number of
productions and attractions can be determined.
Step 1: Determine the percentage of households in each economic category (low,
medium, high). Develop a plot of average zonal income versus income
distribution.
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Step 2: Determine the distribution of auto ownership per household for each
category. Draw a curve showing percent of households, at each income level, that
owns x autos.
Step 3: Determine the number of trips per household for each income-auto
ownership category. Draw a curve showing the relationship between trips per
household, household income and ownership.
Step 4: Calculate the total number of trips generated in the zone. This is done by
computing the number of households in each income-auto ownership category
and multiplying this result by the number of trips per household, as determined in
step 3, and summing the results. Thus,
Ti = (HH) * (Ii) * [Σ(Aij) * (T / HH)ij]
Where:
Ti = trips generated by income group i
HH = number of households in the zone
Ii = percentage of households in the zone with income level i
Aij = percentage of households in income level i with j autos per
household
(T / HH)ij = number of trips produced in a household at income level i and
auto ownership j.
Total trips generated: TT = ΣTi
Step 5: Determine the percentage of trips by purpose.
12-7
Given: Socioeconomic data for the Jeffersonville Transportation Study Area.
Determine: The number of home-to-work, home-to-nonwork, and nonhome-based
trips attracted to a zone with the following characteristics: population = 1920;
dwelling units = 800; retail employment = 50; nonretail employment = 820; school
attendance = 0.
Home-to-work trips attracted to zone
(Home-to-work trips in study area) * (Total employ. in zone/Total employ. in
study area)
Home-to-work trips attracted to zone = 36,680 * (870 / (27,324+5,502))
Home-to-work trips attracted to zone = 972 trips
Home-to-nonwork trips attracted to zone
Visiting Friends:
(0.10)*[(Total home-to-nonwork trips)*(Total dwelling units in zone/Total
dwelling units in study area)]
Visiting Friends = (0.10)*[(174,933 * (800 / (15,675+7,567)))]
Visiting Friends = 602 trips
Shopping:
(0.60)*[(Total home-to-nonwork trips)*(Total retail employ. in zone/Total retail
employ. in study area)]
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Chapter 12: Forecasting Travel Demand
Shopping = (0.60)*[(174,933 * (50 / 5,502))]
Shopping = 954 trips
School:
(0.10)*[(Total home-to-nonwork trips)*(School attendance in zone/School
attendance in study area)]
School = (0.10)*[(174,933 * (0 / 28,551))]
School = 0 trips
Non-retail employment:
(0.20)*[(Total home-to-nonwork trips)*(Non-retail employ. in zone/Non-retail
employ. in study area)]
Non-retail employment = (0.20)*[(174,933 * (820 / 27,324))]
Non-retail employment = 1,050 trips
Total home-to-nonwork trips in zone = 602 + 954 + 0 + 1,050 = 2,606 trips
Non-home based trips attracted to zone
Non-retail employment:
(0.60)*[(Total non-home based trips)*(Non-retail employ. zone/Non-retail
employ. in study area)]
Non-retail employment = (0.60)*[(70,537 * (820 / 27,324))]
Non-retail employment = 1,270 trips
Shopping:
(0.40)*[(Total home-to-nonwork trips)*(Total retail employ. in zone/Total retail
employ. in study area)]
Shopping = (0.40)*[(70,537 * (50 / 5,502))]
Shopping = 256 trips
Total non-home based trips in zone = 1,270 + 256 = 1,526
12-8
Given: Small town with three traffic zones, origin-destination survey results.
Provide a trip distribution calculation using the gravity model for two iterations;
assume Kij = 1.
The mathematical formulation for the gravity model as provided as Equation 12.3:
Tij = Pi
( A j Fij K ij )
∑(A F K
j
ij
ij
)
Since Kij = 1, this factor does not affect calculations. The iterative application of
Equation 12.3 is as follows:
207
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Chapter 12: Forecasting Travel Demand
Iteration 1
T11 = 250 * ((395 * 26) / ((395 * 26) + (180 * 41) + (425 * 52)))
T11 = 250 * (10,270 / 39,750)
T11 = 65
T12 = 250 * ((180 * 41) / ((395 * 26) + (180 * 41) + (425 * 52)))
T12 = 250 * (7,380 / 39,750)
T12 = 46
T13 = 250 * ((425 * 52) / ((395 * 26) + (180 * 41) + (425 * 52)))
T13 = 250 * (22,100 / 39,750)
T13 = 139
T21 = 450 * ((395 * 52) / ((395 * 52) + (180 * 13) + (425 * 50)))
T21 = 450 * (20,540 / 44,130)
T21 = 209
T22 = 450 * ((180 * 13) / ((395 * 52) + (180 * 13) + (425 * 50)))
T22 = 450 * (2,340 / 44,130)
T22 = 24
T23 = 450 * ((425 * 50) / ((395 * 52) + (180 * 13) + (425 * 50)))
T23 = 450 * (21,250 / 44,130)
T23 = 217
T31 = 300 * ((395 * 82) / ((395 * 82) + (180 * 50) + (425 * 39)))
T31 = 300 * (32,390 / 57,965)
T31 = 168
T32 = 300 * ((180 * 50) / ((395 * 82) + (180 * 50) + (425 * 39)))
T32 = 300 * (9,000 / 57,965)
T32 = 46
T33 = 300 * ((425 * 39) / ((395 * 82) + (180 * 50) + (425 * 39)))
T33 = 300 * (16,575 / 57,965)
T33 = 86
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Chapter 12: Forecasting Travel Demand
Trip Matrix for Iteration 1
Zone
1
2
3
Productions
1
65
46
139
250
2
209
24
217
450
3
168
46
86
300
Computed
Attractions
442
116
442
Given
Attractions
395
180
425
Next, calculate the adjusted attraction factors using Equation 12.4.
A jk =
Aj
C j ( k −1)
A j ( k −1)
Zone 1
Ajk = (395 / 442) * 395
Ajk = 353
Zone 2
Ajk = (180 / 116) * 180
Ajk = 279
Zone 3
Ajk = (425 / 442) * 425
Ajk = 409
Now apply the gravity model formula for Iteration 2 using the above adjusted
attraction factors.
Iteration 2
T11 = 250 * ((353 * 26) / ((353 * 26) + (279 * 41) + (409 * 52)))
T11 = 250 * (9,178 / 41,885)
T11 = 55
T12 = 250 * ((279 * 41) / ((353 * 26) + (279 * 41) + (409 * 52)))
T12 = 250 * (11,439 / 41,885)
T12 = 68
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Chapter 12: Forecasting Travel Demand
T13 = 250 * ((409 * 52) / ((353 * 26) + (279 * 41) + (409 * 52)))
T13 = 250 * (21,268 / 41,885)
T13 = 127
T21 = 450 * ((353 * 52) / ((353 * 52) + (279 * 13) + (409 * 50)))
T21 = 450 * (18,356 / 42,433)
T21 = 195
T22 = 450 * ((279 * 13) / ((353 * 52) + (279 * 13) + (409 * 50)))
T22 = 450 * (3,627 / 42,433)
T22 = 38
T23 = 450 * ((409 * 50) / ((353 * 52) + (279 * 13) + (409 * 50)))
T23 = 450 * (20,450 / 42,433)
T23 = 217
T31 = 300 * ((353 * 82) / ((353 * 82) + (279 * 50) + (409 * 39)))
T31 = 300 * (28,946 / 58,847)
T31 = 148
T32 = 300 * ((279 * 50) / ((353 * 82) + (279 * 50) + (409 * 39)))
T32 = 300 * (13,950 / 58,847)
T32 = 71
T33 = 300 * ((409 * 39) / ((353 * 82) + (279 * 50) + (409 * 39)))
T33 = 300 * (15,951 / 58,847)
T33 = 81
Trip Matrix for Iteration 2
Zone
1
2
3
Productions
1
55
68
127
250
2
195
38
217
450
3
148
71
81
300
Computed
Attractions
398
177
425
Given
Attractions
395
180
425
Observe that the computed attractions approximately equal the given attractions.
A total convergence would be expected in another iteration.
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
210
Chapter 12: Forecasting Travel Demand
12-9
Given: Study area with four traffic zones, origin-destination survey results.
Provide a trip distribution calculation using the gravity model for two iterations;
assume Kij = 1.
The mathematical formulation for the gravity model as provided as Equation 12.3:
Tij = Pi
( A j Fij K ij )
∑(A F K
j
ij
ij
)
Since Kij = 1, this factor does not affect calculations. The iterative application of
Equation 12.3 is as follows:
Iteration 1
T11 = 3,400 * ((2,800*1.6) /
((2,800*1.6)+(6,500*0.82)+(2,550*0.68)+(4,400*0.86)))
T11 = 3,400 * (4,480 / 15,328)
T11 = 994
T12 = 3,400 * ((6,500*0.82) /
((2,800*1.6)+(6,500*0.82)+(2,550*0.68)+(4,400*0.86)))
T12 = 3,400 * (5,330 / 15,328)
T12 = 1,182
T13 = 3,400 * ((2,550*0.68) /
((2,800*1.6)+(6,500*0.82)+(2,550*0.68)+(4,400*0.86)))
T13 = 3,400 * (1,734 / 15,328)
T13 = 385
T14 = 3,400 * ((4,400*0.86) /
((2,800*1.6)+(6,500*0.82)+(2,550*0.68)+(4,400*0.86)))
T14 = 3,400 * (3,784 / 15,328)
T14 = 839
T21 = 6,150 * ((2,800*0.82) /
((2,800*0.82)+(6,500*1)+(2,550*1)+(4,400*0.9)))
T21 = 6,150 * (2,296 / 15,306)
T21 = 922
T22 = 6,150 * ((6,500*1) /
((2,800*0.82)+(6,500*1)+(2,550*1)+(4,400*0.9)))
T22 = 6,150 * (6,500 / 15,306)
T22 = 2,612
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Chapter 12: Forecasting Travel Demand
T23 = 6,150 * ((2,550*1) /
((2,800*0.82)+(6,500*1)+(2,550*1)+(4,400*0.9)))
T23 = 6,150 * (2,550 / 15,306)
T23 = 1,025
T24 = 6,150 * ((4,400*0.9) /
((2,800*0.82)+(6,500*1)+(2,550*1)+(4,400*0.9)))
T24 = 6,150 * (3,960 / 15,306)
T24 = 1,591
T31 = 3,900 * ((2,800*0.68) /
((2,800*0.68)+(6,500*1)+(2,550*1)+(4,400*0.82)))
T31 = 3,900 * (1,904 / 14,562)
T31 = 510
T32 = 3,900 * ((6,500*1) /
((2,800*0.68)+(6,500*1)+(2,550*1)+(4,400*0.82)))
T32 = 3,900 * (6,500 / 14,562)
T32 = 1,741
T33 = 3,900*((2,550 * 1) /
((2,800 * 0.68)+(6,500 * 1)+(2,550 * 1)+(4,400* 0.82)))
T33 = 3,900 * (2,550 / 14,562)
T33 = 683
T34 = 3,900 * ((4,400 * 0.9) /
((2,800*0.68)+(6,500*1)+(2,550*1)+(4,400*0.82)))
T34 = 3,900 * (3,608 / 14,562)
T34 = 966
T41 = 2,800 * ((2,800*0.86) /
((2,800*0.86)+(6,500*0.9)+(2,550*0.82)+(4,400*1.6)))
T41 = 2,800 * (2,408 / 17,389)
T41 = 388
T42 = 2,800 * ((6,500*0.9) /
((2,800*0.86)+(6,500*0.9)+(2,550*0.82)+(4,400*1.6)))
T42 = 2,800 * (5,850 / 17,389)
T42 = 942
T43 = 2,800 * ((2,550*0.82) /
((2,800*0.86)+(6,500*0.9)+(2,550*0.82)+(4,400*1.6)))
T43 = 2,800 * (2,091 / 17,389)
T43 = 337
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Chapter 12: Forecasting Travel Demand
T44 = 2,800 * ((4,400*1.6) /
((2,800*0.86)+(6,500*0.9)+(2,550*0.82)+(4,400*1.6)))
T44 = 2,800 * (7,040 / 17,389)
T44 = 1,133
Trip Matrix for Iteration 1
Zone
1
2
3
4
Productions
1
994
1,182
385
839
3,400
2
922
2,612
1,025
1,591
6,150
3
510
1,741
683
966
3,900
4
388
942
337
1,133
2,800
Computed
Attractions
2,814
6,477
2,430
4,529
Given
Attractions
2,800
6,500
2,550
4,400
Next, calculate the adjusted attraction factors using Equation 12.4.
A jk =
Aj
C j ( k −1)
A j ( k −1)
Zone 1
Ajk = (2,800 / 2,814) * 2,800
Ajk = 2,786
Zone 2
Ajk = (6,500 / 6,477) * 6,500
Ajk = 6,523
Zone 3
Ajk = (2,550 / 2,430) * 2,550
Ajk = 2,676
Zone 4
Ajk = (4,400 / 4,529) * 4,400
Ajk = 4,274
Now apply the gravity model formula for Iteration 2 using the above adjusted
attraction factors.
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Chapter 12: Forecasting Travel Demand
Iteration 2
T11 = 3,400 * ((2,786*1.6) /
((2,786*1.6)+(6,523*0.82)+(2,676*0.68)+(4,274*0.86)))
T11 = 3,400 * (4,458 / 15,302)
T11 = 990
T12 = 3,400 * ((6,523*0.82) /
((2,786*1.6)+(6,523*0.82)+(2,676*0.68)+(4,274*0.86)))
T12 = 3,400 * (5,349 / 15,302)
T12 = 1,189
T13 = 3,400 * ((2,676*0.68) /
((2,786*1.6)+(6,523*0.82)+(2,676*0.68)+(4,274*0.86)))
T13 = 3,400 * (1,819 / 15,302)
T13 = 404
T14 = 3,400 * ((4,274*0.86) /
((2,786*1.6)+(6,523*0.82)+(2,676*0.68)+(4,274*0.86)))
T14 = 3,400 * (3,676 / 15,302)
T14 = 817
T21 = 6,150 * ((2,786*0.82) /
((2,786*0.82)+(6,523*1)+(2,676*1)+(4,274*0.9)))
T21 = 6,150 * (2,285 / 15,330)
T21 = 916
T22 = 6,150 * ((6,523*1) /
((2,786*0.82)+(6,523*1)+(2,676*1)+(4,274*0.9)))
T22 = 6,150 * (6,523 / 15,330)
T22 = 2,617
T23 = 6,150 * ((2,676*1) /
((2,786*0.82)+(6,523*1)+(2,676*1)+(4,274*0.9)))
T23 = 6,150 * (2,676 / 15,330)
T23 = 1,074
T24 = 6,150 * ((4,274*0.9) /
((2,786*0.82)+(6,523*1)+(2,676*1)+(4,274*0.9)))
T24 = 6,150 * (3,847 / 15,330)
T24 = 1,543
T31 = 3,900 * ((2,786*0.68) /
((2,786*0.68)+(6,523*1)+(2,676*1)+(4,274*0.82)))
T31 = 3,900 * (1,894/14,598)
T31 = 506
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Chapter 12: Forecasting Travel Demand
T32 = 3,900 * ((6,523*1) /
((2,786*0.68)+(6,523*1)+(2,676*1)+(4,274*0.82)))
T32 = 3,900 * (6,523 / 14,598)
T32 = 1,743
T33 = 3,900 * ((2,676*1) /
((2,786*0.68)+(6,523*1)+(2,676*1)+(4,274*0.82)))
T33 = 3,900 * (2,676 / 14,598)
T33 = 715
T34 = 3,900 * ((4,274*0.9) /
((2,786*0.68)+(6,523*1)+(2,676*1)+(4,274*0.82)))
T34 = 3,900 * (3,505 / 14,598)
T34 = 936
T41 = 2,800 * ((2,786*0.86) /
((2,786*0.86)+(6,523*0.9)+(2,676*0.82)+(4,274*1.6)))
T41 = 2,800 * (2,396 / 17,299)
T41 = 388
T42 = 2,800 * ((6,523*0.9) /
((2,786*0.86)+(6,523*0.9)+(2,676*0.82)+(4,274*1.6)))
T42 = 2,800 * (5,871 / 17,299)
T42 = 950
T43 = 2,800 * ((2,676*0.82) /
((2,786*0.86)+(6,523*0.9)+(2,676*0.82)+(4,274*1.6)))
T43 = 2,800 * (2,194 / 17,299)
T43 = 355
T44 = 2,800 * ((4,274*1.6) /
((2,786*0.86)+(6,523*0.9)+(2,676*0.82)+(4,274*1.6)))
T44 = 2,800 * (6,838 / 17,299)
T44 = 1,107
215
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Chapter 12: Forecasting Travel Demand
Trip Matrix for Iteration 2
Zone
1
2
3
4
Productions
1
990
1,189
404
817
3,400
2
916
2,617
1,074
1,543
6,150
3
506
1,743
715
936
3,900
4
388
950
355
1,107
2,800
Computed
Attractions
2,800
6,499
2,548
4,403
Given
Attractions
2,800
6,500
2,550
4,400
Observe that the computed attractions approximately equal the given attractions.
A total convergence would be expected in another iteration.
12-10
Given: Table with production and attraction data.
Determine: The number of productions and attractions that should be used for
each zone in the second iteration.
Calculate the adjusted attraction factors using Equation 12.4.
A jk =
Aj
C j ( k −1)
A j ( k −1)
Zone 1
Ajk = (300 / 250) * 300
Ajk = 360
Zone 2
Ajk = (100 / 150) * 100
Ajk = 67
Zone 3
Ajk = (200 / 300) * 200
Ajk = 133
Zone 4
Ajk = (700 / 600) * 700
Ajk = 817
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Chapter 12: Forecasting Travel Demand
The attractions change with each iteration; there is no change in productions. The
following table summarizes the results.
1
2
3
4
P2
100
200
400
600
A2
360
67
133
817
12-11
Study area with four traffic zones, pertinent data, and first iteration.
Complete the second iteration.
Calculate the adjusted attraction factors using Equation 12.4.
A jk =
Aj
C j ( k −1)
A j ( k −1)
Zone 1
Ajk = (1,000 / 869) * 1,000
Ajk = 1,151
Zone 2
Ajk = (700 / 707) * 700
Ajk = 693
Zone 3
Ajk = (6,000 / 6,110) * 6,000
Ajk = 5,892
Zone 4
Ajk = (500 / 514) * 500
Ajk = 486
Now apply the gravity model formula for Iteration 2 using the above adjusted
attraction factors.
Iteration 2
T11 = 1,000 * ((1,151*1.3) /
((1,151*1.3)+(693 * 0.95)+(5,892*0.8)+(496*0.65)))
T11 = 1,000 * (1,496 / 7,190)
T11 = 208
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Chapter 12: Forecasting Travel Demand
T12 = 1,000 * ((693*0.95) /
((1,151*1.3)+(693*0.95)+(5,892*0.8)+(496*0.65)))
T12 = 1,000 * (658 / 7,190)
T12 = 92
T13 = 1,000 * ((1,151*1.3) /
((1,151*1.3)+(693*0.95)+(5,892*0.8)+(496*0.65)))
T13 = 1,000 * (4,714 / 7,190)
T13 = 655
T14 = 1,000 * ((1,151*1.3) /
((1,151*1.3)+(693*0.95)+(5,892*0.8)+(496*0.65)))
T14 = 1,000 * (322 / 7,190)
T14 = 45
T21 = 2,000 * ((1,151*0.95) /
((1,151*0.95)+(693*1.3)+(5,892*0.85)+(496*0.95)))
T21 = 2,000 * (1,093 / 7,473)
T21 = 293
T22 = 2,000 * ((1,151*0.95) /
((1,151*0.95)+(693*1.3)+(5,892*0.85)+(496*0.95)))
T22 = 2,000 * (901 / 7,473)
T22 = 241
T23 = 2,000 * ((1,151*0.95) /
((1,151*0.95)+(693*1.3)+(5,892*0.85)+(496*0.95)))
T23 = 2,000 * (5,008 / 7,473)
T23 = 1,340
T24 = 2,000 * ((1,151*0.95) /
((1,151*0.95)+(693*1.3)+(5,892*0.85)+(496*0.95)))
T24 = 2,000 * (471 / 7,473)
T24 = 126
T31 = 3,000 * ((1,151*0.80) /
((1,151*0.80)+(693*0.85)+(5,892*1.3)+(496*1)))
T31 = 3,000 * (921 / 9,666)
T31 = 286
T32 = 3,000 * ((693*0.85) /
((1,151*0.80)+(693*0.85)+(5,892*1.3)+(496*1)))
T32 = 3,000 * (589 / 9,666)
T32 = 183
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218
Chapter 12: Forecasting Travel Demand
T33 = 3,000 * ((5,892*1.3) /
((1,151*0.80)+(693*0.85)+(5,892*1.3)+(496*1)))
T33 = 3,000 * (7,660 / 9,666)
T33 = 2,377
T34 = 3,000 * ((496*1) / ((1,151*0.80)+(693*0.85)+(5,892*1.3)+(496*1)))
T34 = 3,000 * (496 / 9,666)
T34 = 154
T41 = 2,200 * ((1,151*0.65) /
((1,151*0.65)+(693*0.95)+(5,892*1)+(496*1.3)))
T41 = 2,200 * (748 / 7,943)
T41 = 207
T42 = 2,200 * ((693*0.95) /
((1,151*0.65)+(693*0.95)+(5,892*1)+(496*1.3)))
T42 = 2,200 * (658 / 7,943)
T42 = 182
T43 = 2,200 * ((5,892*1) /
((1,151*0.65)+(693*0.95)+(5,892*1)+(496*1.3)))
T43 = 2,200 * (5,892 / 7,943)
T43 = 1,632
T44 = 2,200 * ((496*1.3) /
((1,151*0.65)+(693*0.95)+(5,892*1)+(496*1.3)))
T44 = 2,200 * (645 / 7,943)
T44 = 179
Trip Matrix for Iteration 2
Zone
1
2
3
4
Productions
1
208
92
655
45
1,000
2
293
241
1,340
126
2,000
3
286
183
2,377
154
3,000
4
207
182
1,632
179
2,200
994
698
6,004
504
1,000
700
6,000
500
Computed Attractions
Given Attractions
Observe that the computed attractions approximately equal the given attractions.
A total convergence would be expected in another iteration.
219
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Chapter 12: Forecasting Travel Demand
12-12
For the travel pattern illustrated in Figure 12.18, develop the Fratar method of trip
distribution for two iterations.
First Iteration
Zones
Present
Totals
Growth
Factor
Estimated
Future Totals
A
100
3
300
B
250
4
1,000
C
400
2
800
D
300
1
300
Next, use Equation 12.5 to solve the problem.
TAB = 300 * ((25 * 4) / [(25 * 4) + (50 * 2) + (25 * 1)])
TAB = 300 * (100 / 225)
TAB = 133.33
TAC = 300 * ((50 * 2) / [(25 * 4) + (50 * 2) + (25 * 1)])
TAC = 300 * (100 / 225)
TAC = 133.33
TAD = 300 * ((25 * 1) / [(25 * 4) + (50 * 2) + (25 * 1)])
TAD = 300 * (25 / 225)
TAD = 33.33
TBA = 1,000 * ((25 * 3) / [(25 * 3) + (150 * 2) + (75 * 1)])
TBA = 1,000 * (75 / 450)
TBA = 166.67
TBC = 1,000 * ((150 * 2) / [(25 * 3) + (150 * 2) + (75 * 1)])
TBC = 1,000 * (300 / 450)
TBC = 666.67
TBD = 1,000 * ((75 * 1) / [(25 * 3) + (150 * 2) + (75 * 1)])
TBD = 1,000 * (75 / 450)
TBD = 166.67
TCA = 800 * ((50 * 3) / [(50 * 3) + (150 * 4) + (200 * 1)])
TCA = 800 * (150 / 950)
TCA = 126.32
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220
Chapter 12: Forecasting Travel Demand
TCB = 800 * ((150 * 4) / [(50 * 3) + (150 * 4) + (200 * 1)])
TCB = 800 * (600 / 950)
TCB = 505.26
TCD = 800 * ((200 * 1) / [(50 * 3) + (150 * 4) + (200 * 1)])
TCD = 800 * (200 / 950)
TCD = 168.42
TDA = 300 * ((25 * 3) / [(25 * 3) + (75 * 4) + (200 * 2)])
TDA = 300 * (75 / 775)
TDA = 29.03
TDB = 300 * ((75 * 4) / [(25 * 3) + (75 * 4) + (200 * 2)])
TDB = 300 * (300 / 775)
TDB = 116.13
TDC = 300 * ((200 * 2) / [(25 * 3) + (75 * 4) + (200 * 2)])
TDC = 300 * (400 / 775)
TDC = 154.84
Next, calculate the movement between zones as well as the new growth factors.
A - B = (133.33 + 166.67) / 2 = 150
A - C = (133.33 + 126.32) / 2 = 129.83
A - D = (33.33 + 29.03) / 2 = 31.18
TOTAL =
311.01
New Growth factor for A = 300 / 311.01 = 0.965
B - A = (133.33 + 166.67) / 2 = 150
B - C = (666.67 + 505.26) / 2 = 585.97
B - D = (166.67 + 116.13) / 2 = 141.40
TOTAL =
877.37
New Growth factor for B = 1,000 / 877.37 = 1.140
C - A = (133.33 + 126.32) / 2 = 129.83
C - B = (666.67 + 505.26) / 2 = 585.97
C - D = (154.84 + 168.42) / 2 = 161.63
TOTAL =
877.43
New Growth factor for C = 800 / 877.43 = 0.912
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Chapter 12: Forecasting Travel Demand
D - A = (33.33 + 29.03) / 2 = 31.18
D - B = (166.67 + 116.13) / 2 = 141.40
D - C = (154.84 + 168.42) / 2 = 161.63
TOTAL =
334.21
New Growth factor for D = 300 / 334.21 = 0.898
Second Iteration
Zones
Present
Totals
Growth
Factor
Estimated
Future Totals
A
311.01
0.965
300
B
877.37
1.140
1,000
C
877.43
0.912
800
D
334.21
0.898
300
Next, use Equation 12.5 to solve the problem.
TAB = 300 * ((150 *1.14) / [(150 *1.1 4) + (129.83 * 0.912) + (31.18 *
0.898)])
TAB = 300 * (171 / 317.40)
TAB = 161.63
TAC = 300 * ((129.83 * 0.912) / [(150*1.1 4)+(129.83 * 0.912) + (31.18 *
0.898)])
TAC = 300 * (118.40 / 317.40)
TAC = 111.91
TAD = 300 * ((31.18 *0.898) / [(150 *1.1 4) + (129.83 * 0.912) + (31.18 *
0.898)])
TAD = 300 * (28 / 317.40)
TAD = 26.46
TBA = 1,000 * ((150 * 0.965) / [(150 * 0.965) + (585.97 * 0.912)+
(141.4*0.898)])
TBA = 1,000 * (144.75 / 806.13)
TBA = 179.56
TBC = 1,000 * ((585.97*0.912) / [(150 *0.965)+(585.97 * 0.912)+
(141.4*0.898)])
TBC = 1,000 * (534.40 / 806.13)
TBC = 662.92
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Chapter 12: Forecasting Travel Demand
TBD = 1,000 * ((141.40*0.898) / [(150*0.965) + (585.97*0.912)+
(141.4*0.898)])
TBD = 1,000 * (126.98 / 806.13)
TBD = 157.51
TCA = 800 *((129.83*0.965) / [(129.83*0.965)+(585.97*1.14)+(161.63*
0.898)])
TCA = 800 * (125.29 / 938.44)
TCA = 106.80
TCB = 800 * ((585.97 *1.14)/
[(129.83*0.965)+(585.97*1.14)+(161.63*0.898)])
TCB = 800 * (668.01 / 938.44)
TCB = 569.46
TCD = 800 * ((161.63*0.898)/
[(129.83*0.965)+(585.97*1.14)+(161.63*0.898)])
TCD = 800 * (145.14 / 938.44)
TCD = 123.73
TDA = 300 * ((31.18 * 0.965) / [(31.18 * 0.965)+(141.4 * 1.14)+(161.63
*0.912)])
TDA = 300 * (30.09 / 338.69)
TDA = 26.65
TDB = 300 * ((141.4 * 1.14) / [(31.18 * 0.965)+(141.4 * 1.14)+(161.63
*0.912)])
TDB = 300 * (161.2 / 338.69)
TDB = 142.78
TDC = 300 * ((161.63 *0.912) / [(31.18 * 0.965)+(141.4 * 1.14)+(161.63
*0.912)])
TDC = 300 * (147.41 / 338.69)
TDC = 130.57
Next, calculate the movement between zones.
A - B = (161.63 + 179.56) / 2 = 170.60
A - C = (111.91 + 106.80) / 2 = 109.36
A - D = (26.46 + 26.65) / 2 = 26.56
TOTAL =
306.52
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Chapter 12: Forecasting Travel Demand
B - A = (161.63 + 179.56) / 2 = 170.60
B - C = (662.92 + 569.46) / 2 = 616.19
B - D = (157.51 + 142.78) / 2 = 150.15
TOTAL =
936.94
C - A = (111.91 + 106.80) / 2 = 109.36
C - B = (662.92 + 569.46) / 2 = 616.19
C - D = (123.73 + 130.57) / 2 = 127.15
TOTAL =
852.70
D - A = (26.46 + 26.66) / 2 = 26.56
D - B = (157.51 + 142.78) / 2 = 150.15
D - C = (123.73 + 130.57) / 2 = 127.15
TOTAL =
303.86
Results
Zone
New Total
Desired Total
A
306
300
B
937
1,000
C
853
800
D
304
300
These results could serve as the starting point for a third iteration.
12-13
Redo Problem 12-12 using the average growth factor method.
First Iteration
Zones
Present
Totals
Growth
Factor
Estimated
Future Totals
A
100
3
300
B
250
4
1,000
C
400
2
800
D
300
1
300
Next, use Equation 12.5a to solve the problem. In this step, the movement
between zones is calculated.
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224
Chapter 12: Forecasting Travel Demand
TAB = 25 * ((3 + 4) / 2)
TAB = 87.5
TAC = 50 * ((3 + 2) / 2)
TAC = 125
TAD = 25 * ((3 + 1) / 2)
TAD = 50
TBC = 150 * ((4 + 2) / 2)
TBC = 450
TBD = 75 * ((4 + 1) / 2)
TBD = 187.5
TCD = 200 * ((2 + 1) / 2)
TCD = 300
Next, sum the trip ends in each zone and develop the new growth factors.
TA = TAB + TAC + TAD = 87.5 + 125 + 50 = 262.5
TB = TBA + TBC + TBD = 87.5 + 450 + 187.5 = 725
TC = TCA + TCB + TCD = 125 + 450 + 300 = 875
TD = TDA + TDB + TDC = 50 + 187.5 + 300 = 537.5
New growth factor for A =
New growth factor for B =
New growth factor for C =
New growth factor for D =
300 / 262.5 = 1.143
1,000 / 725 = 1.379
800 / 875 = 0.914
300 / 537.5 = 0.558
Second Iteration
Zones
Present
Totals
Growth
Factor
Estimated
Future Totals
A
262.5
1.143
300
B
725
1.379
1,000
C
875
0.914
800
D
537.5
0.558
300
Next, use Equation 12.5a to solve the problem. In this step, the movement
between zones is calculated.
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Chapter 12: Forecasting Travel Demand
TAB = 87.5 * ((1.143 + 1.379) / 2)
TAB = 110.3
TAC = 125 * ((1.143 + 0.914) / 2)
TAC = 128.6
TAD = 50 * ((1.143 + 0.558) / 2)
TAD = 42.5
TBC = 450 * ((1.379 + 0.914) / 2)
TBC = 515.9
TBD = 187.5 * ((1.379 + 0.558) / 2)
TBD = 181.6
TCD = 300 * ((0.914 + 0.558) / 2)
TCD = 220.9
Next, sum the trip ends in each zone and develop the new growth factors.
TA = TAB + TAC + TAD = 110.3 + 128.6 + 42.5 = 281
TB = TBA + TBC + TBD = 110.3 + 515.9 + 181.6 = 808
TC = TCA + TCB + TCD = 128.6 + 515.9 + 220.9 = 865
TD = TDA + TDB + TDC = 42.5 + 181.6 + 220.9 = 445
New growth factor for A =
New growth factor for B =
New growth factor for C =
New growth factor for D =
(1.143)(262.5) / 281 = 1.066
(1.379)(725) / 808 = 1.238
(0.914)(875) / 865 = 0.924
(0.558)(537.5) / 445 = 0.674
Results
Zone
New Total
Desired Total
A
281
300
B
808
1,000
C
866
800
D
445
300
Additionally, these results could serve as the starting point for a third iteration.
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226
Chapter 12: Forecasting Travel Demand
12-14
What data are required in order to use (a) the gravity model and (b) the Fratar
model?
The data needed for the gravity model include:
•
Total number of trips produced in each zone
•
Number of trips attracted to each zone
•
Travel time between each zone (to determine friction factor F)
•
Socioeconomic adjustment factor for interchanges.
The data required to use the Fratar model include:
•
Present trip generation in each zone
•
Growth factors of each zone
•
Present number of trips between zones.
12-15
The amount of lumber produced and consumed by three states is shown in the
following table. Intrastate shipment distances are 200 miles and interstate distances
are 800 miles (between states 1 and 2), 1,000 miles (between states 1 and 3), and 400
miles (between states 2 and 3). Assuming an impedance function of the form 1/d,
estimate the tonnage of lumber that will travel between the three states:
Tons of Lumber Produced and Consumed Per Year (Tons)
State
1
2
3
Lumber Produced
5,880
3,300
9,800
Lumber Consumed
980
10,000
8,000
Impedance factors are established as follows
State
1
2
3
1
1/200 = 0.005
1/800 = 0.00125
1/1000 = 0.001
2
1/800 = 0.00125
1/200 = 0.005
1/400 = 0.0025
3
1/1000 = 0.001
1/400 = 0.0025
1/200 = 0.005
The general form for the gravity model is
Pi A j Fij
Tij = n
∑ (A j Fij )
j=1
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Chapter 12: Forecasting Travel Demand
For example, the number of trips from zone 1 to zone 2, T12, is
T12 =
(5,880)(10,000)(0.00125)
= 2894
(980)(0.005) + (10,000)(0.00125) + (8,000)(0.001)
Thus the initial trip table is
1
2
3
Total
1
1134
57
146
1337
2
2894
2317
3713
8924
3
1852
927
5941
8720
Total
5880
3300
9800
Because computed attractions do not match given attractions, attractions are
adjusted using Equation 12.4:
Aj
A jk =
A j ( k −1)
C j ( k −1)
For example, attractions for the second iteration for zone 2 can be found as =
(10,000)(10,000)/8,924 = 11,206. Similarly, the new attractions for zones 1 and 3
are found to be 719 and 7,340, respectively.
After five iterations, the final flows are found as shown below.
1
2
3
Total
1
835
39
106
980
2
3320
2462
4218
10000
3
1724
799
5476
8000
Total
5880
3300
9800
12-16
Suppose that traffic congestion has rendered the distance-based impedance function
unsuitable for Problem 12-15. A detailed survey yields present-day commodity flows
shown below. In future, state #1 lumber production will increase to 18,000 and state
#2 lumber consumption will increase to 21,800. Estimate the lumber flows between
the three states.
State
1
2
3
Total Consumed
1
200
100
800
1,100
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2
1,800
200
7,000
9,000
228
3
4,000
3,000
1,200
8,200
Total Produced
6,000
3,300
9,000
Chapter 12: Forecasting Travel Demand
Because impedances are not available and an older OD matrix is available, the
Fratar method is appropriate. In the application of Eq. 12.5, note that the growth
factor Gi refers to lumber produced and the growth factor Gj refers to lumber
consumed. Note also that tij is a true from/to matrix (e.g., lumber tons from zone i
to zone j will not equal tons from zone j to zone i).
The growth factors are as follows:
Zone
Production Growth
Factor (Gi)
3.0 (18,000/6,000)
1.0 (No change)
1.0 (No change)
1
2
3
Consumption Growth Factor (Gj)
1.0 (No change)
2.422 (21,800/9,000)
1.0 (No change)
Equation 12.5 may be applied using the tonnage of lumber from zone 1 to zone 2
as an example:
Tij = (t i G i )
t ij G j
∑t
ij
Gj
j
T12 = (t 1G 1 )
t 12 G 2
t 11G 1 + t 12 G 2 + t 13 G 3
T12 = (6,000 * 3)
(1,800)(2.42)
= 9,168
(200)(1) + (1,800)(2.42) + (4,000)(1)
Ten iterations of the Fratar method gives the following trip interchange matrix:
State
1
2
3
Total Consumed
1
532
153
414
1,100
229
2
11,969
767
9,064
21,800
3
5,499
2,379
321
8,200
Total Produced
18,000
3,300
9,800
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Chapter 12: Forecasting Travel Demand
12-17
Survey data suggest the trip interchange matrix shown below. Calibrate the friction
factors for one iteration, assuming travel times are 2.0 minutes for interzonal trips
and 1.0 minute for all intrazonal trips. Assume the friction factor takes the form t-b.
Zone
1
2
3
Given Trips
1
2
10 40
20 50
30 60
3
70
80
90
Based on the trip table, the productions and attractions are:
Zone
1
2
3
Productions
120
150
180
Attractions
60
150
240
The initial guess is to assume b = 1. Thus the intrazonal friction factors
are 1/1 = 1 and the interzonal friction factors are 1/2 = 0.5
The gravity model is then applied with these initial friction factors. For
example, the trips from zone 1 to zone 1 are found to be approximately 28.
T11 =
(120)(60)(1)
P1A1F11
=
= 28
A1F11 + A 2 F12 + A 3F13 (60)(1) + (150)(0.5) + (240)(0.5)
The remaining trips are found in a similar manner.
Zone
1
2
3
1
2
3
28
15
16
35
75
39
56
60
125
We now compare predicted and actual results for each trip impedance. In
this particular example, we only have two trip impedances: impedances for
intrazonal trips and impedances for interzonal trips.
•
For intrazonal trips (T11, T22, T23) a perfect friction factor would have
predicted 10+50+90 = 150 trips. The guess with b = 1 meant that the model
showed 28 + 75 + 125 = 228 trips.
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230
Chapter 12: Forecasting Travel Demand
For interzonal trips (T12,T13, T21, T23, T31, and T33) a perfect friction factor
would have predicted 40+70+20+80+30+60 = 300 trips. The guess with b = 1
predicted, however, 35+56+15+60+16+39 = 222 trips
•
The adjusted friction factors are:
•
•
Intrazonal travel = (1)(150)/228 = 0.658
Interzonal travel = (0.5)(300)/222 = 0.676
The regression equation will be:
F = t-b
Ln(F) = -b ln (t)
Ln (0.658) = -b ln(1)
Ln (0.676) = -b ln(2)
b = 0.565
Thus the friction factor based on one iteration is t-b. For intrazonal travel (with
travel time = 1.0) the friction factor is thus 1.0. For interzonal travel time with
travel time = 2.0, the friction factor is thus 2.0-0.565 = 0.676.
12-18
Determine the share (proportion) of person-trips by each of two modes (private auto
and mass transit) using the multinomial logit model and given the following
information:
Parameter
Ta = access time (min.)
Tw = waiting time (min.)
Tr = riding time (min.)
C = out-of-pocket cost (cents)
Calibration constant, Ak
Private auto
5
0
25
150
-0.01
Mass transit
10
15
40
100
-0.07
Utility function:
Uk = Ak – 0.05 Ta – 0.04 Tw – 0.03 Tr – 0.014 C
Uauto = – 0.01 – 0.05 (5) – 0.04 (0) – 0.03 (25) – 0.014 (150) = – 3.11
Utransit = – 0.07 – 0.05 (10) – 0.04 (15) – 0.03 (40) – 0.014 (100) = – 3.77
Pauto =
eU auto
e −3.11
=
= 0.659
eU auto + eU transit e − 3.11 + e − 3.77
Ptransit =
eU transit
e −3.77
=
= 0.341
eU auto + eU transit e − 3.11 + e − 3.77
231
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Chapter 12: Forecasting Travel Demand
12-19
A mode choice logit model is to be developed based on the following information. A
survey of travelers in an area with bus service found the following data:
Model parameter
Auto
Bus
0
10
X1, waiting time (min.)
X2, travel time (min.)
20
35
5
0
X3, parking time (min.)
X4, out-of-pocket cost (cents)
225
100
Ak, calibration constant
-0.33
-0.27
The following utility function was calibrated based on an observed mode split of
84.9% private auto use and 15.1% bus use.
Utility function: Uk = Ak – 0.10 X1 – 0.13 X2 – 0.12 X3 – 0.0045 X4
After implementing service improvements to the buses, the mode split changed to
81.6% private auto use and 18.4% bus use. Determine a value for the calibration
constant for the bus mode that reflects this shift in mode split.
Using the given information prior to the bus service improvements, the utility
functions are:
Uauto = – 0.33 – 0.10 (0) – 0.13 (20) – 0.12 (5) – 0.0045 (225) =
Ubus = – 0.27 – 0.10 (10) – 0.13 (35) – 0.12 (0) – 0.0045 (100) =
Using the new mode split data, the calibration constant for the bus utility function
will need to be recalculated. The logit function for bus service that reflects this
shift can be written as:
eU bus
eU bus
= 0.184
=
eU bus + eU auto eU bus + e −4.54
Ubus = 6.03
The bus utility function and new value for Abus can now be solved.
Ubus = Abus – 0.10 (10) – 0.13 (35) – 0.12 (0) – 0.0045 (100) = 6.03
Abus = – 0.03
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232
Chapter 12: Forecasting Travel Demand
12-20
A previously developed elasticity model for improvements to a transit system found
a midpoint arc elasticity value of 0.38 to describe the relationship between demand
increase and route mileage. The system is now planning a 60% increase in route
mileage. Estimate the expected demand increase.
Using Equation 12.15,
Midpoint Arc Elasticity, eM =
ΔD / Davg
ΔX / X avg
(1.00 − D x )
(1.00 + D x ) / 2
0.38 =
(1.00 − 1.60)
(1.00 + 1.60) / 2
12-21
Determine the minimum path for nodes 1, 3, and 9 in Figure 12.19. Sketch the final
trees.
233
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Chapter 12: Forecasting Travel Demand
12-22
Assign the vehicle trips shown in the O-D trip table to the network in Figure 12.20,
using all-or-nothing assignment. Make a list of links in the network and indicate the
volume assigned to each. Calculate the total vehicle-minutes of travel. Show the
minimum path and assign traffic for each of the five nodes.
Minimum Paths
Nodes
From - To
1
2
1
3
1
4
1
5
2
1
2
3
2
4
2
5
3
1
3
2
3
4
3
5
Link
Path
1-2
1 - 2, 2 - 3
1 - 5, 5 - 4
1-5
2-1
2-3
2-4
2 - 4, 4 - 5
3 - 2, 2 - 1
3-2
3-4
3 - 4, 4 - 5
Nodes
From - To
4
1
4
2
4
3
4
5
5
1
5
2
5
3
5
4
Link
Path
4 - 5, 5 - 1
4-2
4-3
4-5
5-1
5 - 4, 4 - 2
5 - 4, 4 - 3
5-4
Link
Volume
Travel Time
Veh-Min of Travel
1-2
200
8
1,600
2-1
600
8
4,800
1-5
350
5
1,750
5-1
450
5
2,250
2-5
0
12
0
5-2
0
12
0
2-3
300
3
900
3-2
300
3
900
2-4
600
5
3,000
4-2
250
5
1,250
3-4
250
7
1,750
4-3
350
7
2,450
4-5
1,300
6
7,800
5-4
700
6
4,200
32,650
TOTAL
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234
Chapter 12: Forecasting Travel Demand
Traffic Assigned to Nodes
Node
Volume Assigned
1
1,050
2
750
3
650
4
1,550
5
1,650
Minimum Path Trees
235
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Chapter 12: Forecasting Travel Demand
12-23
Figure 12.21 represents travel times on links connecting six zonal centroids.
Determine the minimum path from zone to zone using all-or-nothing assignment
based on the given trip table.
Link
Volumes
Total Volume for Link
1-2
500
500
2-1
500
500
1-5
0
0
5-1
0
0
1-6
650 + 550 + 200 + 500
1,900
6-1
650 + 200 + 550 + 500
1,900
2-3
0
0
3-2
0
0
2-6
600 + 525 + 550 + 350
2,025
6-2
600 + 525 + 550 + 350
2,025
3-5
575 + 550 + 350 + 500
1,975
5-3
575 + 550 + 350 + 500
1,975
3-6
200+200+550+525+800+350+500+350+550
4,025
6-3
200+200+550+525+800+350+500+350+550
4,025
3-4
600 + 350 + 200 + 200
1,350
4-3
600 + 350 + 200 + 200
1,350
4-5
400
400
5-4
400
400
5-6
0
0
6-5
0
0
Total Trips
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24,350
236
Chapter 12: Forecasting Travel Demand
Minimum Path Trees
12-24
Given the following information, and using the generalized capacity restraint link
performance function, perform two iterations of multipath traffic assignment. A
flow of 10,500 vehicles in the peak hour is to be distributed between three routes
whose properties are given in the following table.
Link Performance Function Component Route 1 Route 2 Route 3
Free flow travel time (min.)
17.0
15.5
12.5
3
3.8
4.2
6.6
Capacity (10 veh/h)
Α
0.3
0.5
0.6
Β
2.5
3.5
4.5
For each path j, the proportion of traffic p(j) and volume v(j) must be found, and
then using Equation 12.13a, the travel time t(j) can be determined.
First Iteration
1
17
= 0.2893
p(1) =
1
1 ⎞
⎛1
+
⎟
⎜ +
⎝ 17 15.5 12.5 ⎠
v(1) = (0.2893)(10,500) = 3038
2.5
⎡
⎛ 3038 ⎞ ⎤
t (1) = 17 ⎢1 + 0.3⎜
⎟ ⎥ = 19.91 min
⎝ 3800 ⎠ ⎦⎥
⎣⎢
1
15.5
= 0.3173
p(2) =
1
1 ⎞
⎛1
+
⎟
⎜ +
⎝ 17 15.5 12.5 ⎠
v(2) = (0.3173)(10,500) = 3332
3.5
⎡
⎛ 3332 ⎞ ⎤
t (2) = 15.5⎢1 + 0.5⎜
⎟ ⎥ = 18.94 min
⎝ 4200 ⎠ ⎦⎥
⎣⎢
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Chapter 12: Forecasting Travel Demand
1
12.5
p(3) =
= 0.3934
1
1 ⎞
⎛1
+
⎜ +
⎟
⎝ 17 15.5 12.5 ⎠
v(3) = (0.3934)(10,500) = 4130
4.5
⎡
⎛ 4130 ⎞ ⎤
t (3) = 12.5⎢1 + 0.3⎜
⎟ ⎥ = 13.41min
⎝ 6600 ⎠ ⎦⎥
⎣⎢
Second Iteration
1
19.91
p(1) =
= 0.2828
1
1 ⎞
⎛ 1
+
+
⎜
⎟
⎝ 19.91 18.94 13.41 ⎠
v(1) = (0.2828)(10,500) = 2969
2.5
⎡
⎛ 2969 ⎞ ⎤
t (1) = 17 ⎢1 + 0.3⎜
⎟ ⎥ = 19.75 min
⎝ 3800 ⎠ ⎦⎥
⎣⎢
1
19.91
p (2) =
= 0.2973
1
1 ⎞
⎛ 1
+
+
⎜
⎟
⎝ 19.91 18.94 13.41 ⎠
v(2) = (0.2973)(10,500) = 3122
3.5
⎡
⎛ 3122 ⎞ ⎤
t (2) = 15.5⎢1 + 0.5⎜
⎟ ⎥ = 18.24 min
⎝ 4200 ⎠ ⎦⎥
⎣⎢
1
19.91
p (3) =
= 0.4199
1
1 ⎞
⎛ 1
+
+
⎜
⎟
⎝ 19.91 18.94 13.41 ⎠
v(3) = (0.4199)(10,500) = 4409
4.5
⎡
⎛ 4409 ⎞ ⎤
t (3) = 12.5⎢1 + 0.3⎜
⎟ ⎥ = 13.72 min
⎝ 6600 ⎠ ⎦⎥
⎣⎢
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Chapter 12: Forecasting Travel Demand
12-25
Consider the user equilibrium assignment and the lowest system cost assignment
discussed in section 12.5.1.4 (Figure 12.16). Zones 1, 2, and 3 represent a company’s
warehouses all located on a privately-owned parcel of land. The vehicles shown are
delivery trucks rather than passenger cars. Which method of traffic assignment is
more appropriate?
The system optimal assignment, where total travel costs are minimized, would be
appropriate assuming the warehouse owner is able to schedule the movement of
delivery trucks on the privately owned parcel of land.
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Chapter 13
Evaluating Transportation
Alternatives
13-1
What is the main objective of conducting a transportation project evaluation?
The main objective of conducting a transportation project evaluation is to
furnish the appropriate information about the outcome of each alternative so that a
selection can be made. This evaluation process should be viewed as an activity in
which information that is relevant to the selection is made available to the person
or group responsible for making a decision.
13-2
Describe four basic issues that should be considered prior to selection of an
evaluation procedure.
The following four basic issues should be considered prior to the selection
of an evaluation procedure:
(a) Who will use the information and what is their viewpoint? A clear definition
of whose viewpoint is being considered in the evaluation is necessary if proper
consideration is to be given as to how the stakeholders will be impacted, either
positively or negatively, by each proposed alternative.
(b) What are the relevant criteria and how should these be measured? A
transportation project is intended to accomplish one or more goals and objectives.
These are measured as criteria, and the numerical or relative results for each
criterion are called measures of effectiveness. It is important that the criteria be
related as closely as possible to the stated objective in order to properly evaluate
each alternative.
(c) What measures of effectiveness are to be used in the evaluation process itself?
One approach is to convert each measure of effectiveness to a common unit, and
then for each alternative compute the summation for all measures. A common
unit is money, and it might be possible to make a transformation of the relevant
criteria to equivalent dollars and then compare each alternative from an economic
viewpoint. Another common unit approach is to convert each measure of
effectiveness to a numeric score. Otherwise, the measures of effectiveness can be
simply reported for each alternative in a matrix form with no attempt made to
combine them. The measures of effectiveness must not only be relevant to the
problems but should be easy to measure and be sensitive to changes made in each
alternative.
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Chapter 13: Evaluating Transportation Alternatives
(d) How well will the evaluation process assist in making a decision? The
decision maker will want to know what the costs of the project will be. This
alone will usually determine the outcome. Also of importance is if the benefits
justify the expenditure of funds for transportation or would the money be best
spent elsewhere. The decision maker will also want to know if the proposed
project is likely to produce the stated results. The evaluation process requires that
engineers have all the appropriate facts about a proposed project and be able to
convey these in a clear and logical manner so that decision making is facilitated.
13-3
List the basic criteria used for evaluating transportation alternatives. What units
are used for measurement?
The basic criteria used in evaluating transportation projects, along with their
corresponding units of used for measurement, are listed below:
(a)
Capital costs (dollars)
Preliminary engineering
Construction
Right-of-way
Vehicles
(b)
Maintenance costs (dollars)
(c)
Facility operating costs (dollars)
(d)
Travel time costs (dollars)
Total hours or cost (dollars)
Average door-to-door speed (miles per hour)
Distribution of door-to-door speed, variance (miles per hour)
(e)
Vehicle operating costs (dollars)
(f)
Crash costs: lives, injuries, property damage (dollars)
13-4
Average demand on a rural roadway ranges from zero to 500 veh/day when the cost
per trip goes from $1.50 to zero.
(a) Calculate the net user benefits per year if the cost decreases from $1.00 to
$0.75/trip (assume a linear demand function).
(b) Compare the value calculated in (a) with the benefits as calculated in typical
highway studies.
(a) Construct the demand curve as shown in the figure below. From this curve,
one can find V1 and V2 from the prices provided.
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Chapter 13: Evaluating Transportation Alternatives
V1 = 167 vehicles per day
V2 = 250 vehicles per day
Determine the net user benefits using Equation 13.1:
Net user benefits = (P1 – P2) (V1 + V2) / 2
Net user benefits = [(1.00 - 0.75)*(162 + 250)] / 2
Net user benefits = $52.13 per day
Next, determine the net user benefits per year.
Yearly net user benefits = 365 * net user benefits
Yearly net user benefits = 365 * 52.13
Yearly net user benefits = $19,027.45
Therefore, the net user benefits per year will be $19,027.
(b)
The benefits calculated in typical highway studies are not based on the
concept of consumer surplus. They are usually based on travel time reduction as
a result of an improvement. In this manner, the annual user benefit would be:
Annual User Benefit = (P1 -P2)*(V2)*365
Annual User Benefit = ($1.00 - $0.75)*(250)*365
Annual User Benefit = $22,812.50
This method used in highway studies will usually overestimate the user benefits
as shown above.
13-5
A ferry is currently transporting 250 veh/day at a cost of $1.25/vehicle. The ferry
can
attract 500 more veh/day when the cost /veh is $0.75. Calculate the net user
benefits/year if the cost /veh decreases from $1.10 to $0.95.
Construct the demand curve as shown in the figure below. From this curve, one
can find V1 and V2 from the prices provided.
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Chapter 13: Evaluating Transportation Alternatives
V1 = 400 vehicles per day
V2 = 550 vehicles per day
Determine the net user benefits using Equation 13.1:
Net user benefits = (P1 – P2) (V1 + V2) / 2
Net user benefits = [(1.10 - 0.95)*(400 + 550)] / 2
Net user benefits = $71.25 per day
Next, determine the net user benefits per year.
Yearly net user benefits = 365 * net user benefits
Yearly net user benefits = 365 * 71.25
Yearly net user benefits = $26,006.25
Therefore, the net user benefits per year will be $26,006.
13-6
What are the two components of the cost of a transportation facility improvement?
Describe each.
The cost of a transportation facility improvement includes two
components: first cost (also called initial cost) and continuing costs for annual
maintenance, operation, and administration. The first cost for a highway or transit
project may include engineering design, right-of-way, and construction. Design
parameters specific to each project will dictate which items will be required and
their costs. Maintenance and operating costs of the facility must also be
determined. These are recurring costs, or continuing costs, that will be incurred
over the life of the facility and are usually based on historical data for similar
projects.
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Chapter 13: Evaluating Transportation Alternatives
13-7
Estimate the average unit costs for (a) operating a standard vehicle on a level
roadway, (b) travel time for a truck company, (c) single-vehicle property damage,
(d) personal injury, and (e) fatality.
The following are estimated average unit costs for:
(a) Operating a standard vehicle on a level roadway.
10 cents per mile for depreciation
4 cents per mile for fuel
8 cents per mile for insurance and registration
3 cents per mile for maintenance
25 cents per mile on level ground
The estimated average unit cost to operate a vehicle on level ground is 25 cents
per mile.
(b) Travel time for a truck company. Using estimated costs of 35 cents per mile
to operate a truck and 25 cents per mile for its driver, the unit cost is
approximately 60 cents per mile. At an average speed of 50 mi/h, the travel time
cost is: ($0.60/mi)(50 mi/h) = $30 per hour of travel.
(c) Single vehicle property damage. According to a sample of crash records in
Virginia, the average property damage per crash is approximately $1,800, and the
average number of vehicles involved in a crash is 1.7. Therefore the average unit
cost for single vehicle property damage is ($1,800 / 1.7) = $1,059.
(d) Personal injury. If a single X-ray costs $225 and a visit to the emergency
room costs $175, it can be estimated that the average cost for a personal injury
(not including costs associated with treatment) is approximately $400.
(e) Fatality. The average unit cost of a fatality is difficult to determine; however,
values between $100,000 and $4.5 million have been used.
13-8
Derive the equation to compute the equivalent annual cost given the capital cost of a
highway, such that A = (A/P) x P, where A/P is the capital recovery factor. Compute
the equivalent annual cost if the capital cost of a transportation project is $100,000,
annual interest = 10%, and n = 15 years.
Knowing that:
F = A(1+i)n-1 + A(1+i)n-2 + … + A(1+i) + A
Multiplying by (1+i),
(1+i)F = A(1+i)n + A(1+i)n-1 + … + A(1+i)
Subtracting the first equation from the second,
i F = A(1+i)n - A
F = A [ (1+i)n – 1 ] / i
Knowing that:
F = P (1+i)n
A [ (1+i)n – 1 ] / i = P (1+i)n
A/P = i (1+i)n / [ (1+i)n – 1 ]
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Chapter 13: Evaluating Transportation Alternatives
From Table 13.3, or using the equation derived above, determine the appropriate
(A/P) term.
EUAC = P(A/P, 10%, 15)
EUAC = 100,000 * (0.1315)
EUAC = $13,150.00
Therefore, the EUAC for this project will be $13,150.00.
13-9
A highway project is expected to cost $1,500,000 initially. The annual operating and
maintenance cost after the first year is $2000 and will increase by $250 each year for
the next 10 years. At the end of the fifth year, the project must be resurfaced at a
cost of $300,000.
(a) Calculate the present worth of costs for this project if the annual interest
rate is 8%.
(b) Convert the value obtained in (a) to equivalent uniform annual costs.
(a) Calculate the present worth using Equation 13.4:
N
Cn
PW = ∑
n
n = 0 (1 + i )
2000
2250
2500
2750
PW = 1500000 +
+
+
+
+
1
2
3
(1 + 0.08)
(1 + 0.08)
(1 + 0.08)
(1 + 0.08) 4
3000
3250
3500
3750
4000
+
+
+
+
+
5
6
7
8
(1 + 0.08)
(1 + 0.08)
(1 + 0.08)
(1 + 0.08)
(1 + 0.08) 9
4250
4500
300000
+
+
10
11
(1 + 0.08)
(1 + 0.08)
(1 + 0.08) 5
PW = 1500000 + 1852 + 1929 + 1985 + 2021 + 2042 + 2048 + 2042 +
2026 + 2001 + 1969 + 1930 + 204175 = 1726020
Therefore, the present worth (cost) for this project is $1,726,020.
(b) Convert the above present worth cost to EUAC using Equation 13.6.
Since the project is analyzed for 11 years, n = 11.
EUAC = PW(A/P, 8%, 11)
EUAC = (1726020) [0.08 (1 + 0.08)11] / [(1 + 0.08)11 - 1]
EUAC = (1726020) (0.1401)
EUAC = 241775
Therefore, the EUAC for this project will be $241,775.
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Chapter 13: Evaluating Transportation Alternatives
13-10
Three transportation projects have been proposed to increase the safety in and
around a residential neighborhood. Each project consists of upgrading existing
street signing to highly retroreflective sheeting to increase visibility. The following
table shows the initial construction costs, annual operating costs, useful life of the
sheeting, and the salvage values for each alternative. Assume that the discount rate
is 10%. Calculate the present worth for each alternative and determine the
preferred project based on the economic criteria.
Use Equation 13.4.
Calculate the present worth for Alternative I.
PWI = 10000 + 2,000(P/A, 10%, 10) + 2,500(P/F, 10%, 10)
PWI = 10000 + 2000(6.145) + 2500(0.3855)
PWI = 21326
Therefore, the present worth of Alternative I is $21,326.
Calculate the present worth for Alternative II.
PWII = 12000 + 1600(P/A, 10%, 10) + 3,000(P/F, 10%, 10)
PWII = 12000 + 1600(6.145) + 3000(0.3855)
PWII = 20675.5
Therefore, the present worth of Alternative II is $20,676.
Calculate the present worth of Alternative III. In order to compare this alternative
to the others, Alternative III must be analyzed using the prevailing useful life
values in the other alternatives (10 years).
PWIII = 5000 + 2500(P/A, 10%, 10) + (5000 + 500)(P/F, 10%, 5) + 500(P/F, 10%,
10)
PWIII = 5000 + 2500(6.145) + 4500(0.6209) + 500(0.3855)
PWIII = 22963.8
Therefore, the present worth of Alternative III is $22,964.
Since Alternative II has the lowest net present worth cost, it is the best alternative
to choose based on the economic criteria provided.
13-11
Two designs have been proposed for a short span bridge in a rural area, as shown in
the following table. The first proposal is to construct the bridge in two phases (Phase
I now and Phase II in 25 years). The second alternative is to construct it in one
phase. Assuming that the annual interest rate is 4%, determine which alternative is
preferred using present worth analysis.
Calculate the present worth of Alternative I. Note that the Phase II has an
additional annual maintenance cost to that of Phase I. This will require that this
annualized costs be "brought" back to the present condition at year 25, and then
this value needs to be brought back to year 0. Using Equation 13.4:
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Chapter 13: Evaluating Transportation Alternatives
PWI = 14200000 + 75000(P/A, 4%, 25) + 12600000(P/F, 4%, 25) +
((25000(P/A, 4%, 25))(P/F, 4%, 25))
PWI = 14200000 + 75000(21.482) + 12600000(0.3751) +
25000(15.622)(0.3751)
PWI = 14200000 + 1611150 + 4726260 + 146495.31
PWI = 20683905.31
Therefore, the present worth of Alternative I is $20,683,905.
Next, calculate the present worth of Alternative II.
PWII = 22400000 + 100000(P/A, 4%, 50)
PWII = 22400000 + 100000(21.482)
PWII = 24548200
Therefore, the present worth of Alternative II is $24,548,200.
Based on the above economic analysis, Alternative I, consisting of a two-phase
construction process should be chosen. Alternative I has a lower net present
worth of costs than does Alternative II.
13-12
Three designs have been proposed to improve traffic flow at a major intersection in
a heavily traveled suburban area. The first alternative involves improved traffic
signaling. The second alternative includes traffic-signal improvements and
intersection widening for exclusive left turns. The third alternative includes
extensive reconstruction, including a grade separation structure. The construction
costs, as well as annual maintenance and user costs, are listed in the following table
for each alternative. Determine which alternative is preferred based on economic
criteria if the analysis period is 20 years and the annual interest rate is 15%. Show
that the result is the same using the present worth, equivalent annual cost, benefit–
cost ratio, and rate-of-return methods.
Calculate the present worth for each of the alternatives.
PWpresent conditions = 0 + 15000(P/A, 15%, 20) + 500000(P/A, 15%,20)
PWpresent conditions = 0 + 15000(6.259) + 500000(6.259)
PWpresent conditions = $3,223,385
PWtraffic signals = 440000 + 10000(P/A, 15%, 20) + 401000(P/A, 15%, 20) 15,000(P/F, 15%, 20)
PWtraffic signals = 440000 + 10000(6.259) + 401000(6.259) - 15000(0.0611)
PWtraffic signals = $3,011,533
PWint. widening = 790000 + 9000(P/A, 15%, 20) + 350000(P/A, 15%, 20) 11000(P/F, 15%, 20)
PWint. widening = 790000 + 9000(6.259) + 350000(6.259) - 11,000(0.0611)
PWint. widening = $3,036,309
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Chapter 13: Evaluating Transportation Alternatives
PWgrade separation = 1250000 + 8000(P/A, 15%, 20) + 301000(P/A, 15%, 20)
PWgrade separation = 1250000 + 8000(6.259) + 301000(6.259)
PWgrade separation = $3,184,031
Based on the present worth analysis, Alternative II (Traffic Signals) is preferred
due to it having the lowest net present worth of costs among all alternatives.
Calculate the equivalent uniform annual cost (EUAC) for each alternative.
EUACpresent conditions = 0 + 15000 + 500000
EUACpresent conditions = $515,000
EUACtraffic signals = 440000(A/P, 15%, 20) + 10000 + 401000 15000(A/F, 15%, 20)
EUACtraffic signals = 440000(0.1598) + 10000 + 401000 - 15000(0.0098)
EUACtraffic signals = $481,165
EUACint. widening = 790000(A/P, 15%, 20) + 9000 + 350000 11000(A/F, 15%, 20)
EUACint. widening = 790000(0.1598) + 9000 + 350000 - 11,000(0.0098)
EUACint. widening = $485,134
EUACgrade separation = 1250000(P/A, 15%, 20) + 8000 + 301000
EUACgrade separation = 1250000(0.1598) + 8000 + 301000
EUACgrade separation = $508,750
Based on the EUAC analysis, Alternative II (Traffic Signals) is preferred. This is
because it has the lowest equivalent uniform annual cost among the alternatives.
Calculate the benefit-cost ratio for each alternative.
First, compare the traffic signal alternative to the present condition alternative.
BCRII/I = (3223385 – (10000 + 401000)(6.259) /
(440000 – (15000)(0.0611) - 0)
BCRII/I = 1.482
Since BCR is greater than 1.00, the traffic signal alternative is preferred to present
conditions.
Next, compare the intersection widening to the traffic signal alternative.
BCRIII/II = [(10000 + 401000)(6.259) – (9000 + 350000)(6.259)] /
(790000 – (11000)(0.0611) - 440000)
BCRIII/II = 0.929
Since BCR is less than 1, the intersection widening is not preferred to the traffic
signal alternative.
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Chapter 13: Evaluating Transportation Alternatives
Next, compare the grade separation alternative to the traffic signal alternative.
BCRIV/II = [(10000 + 401000)(6.259) – (8000 + 301000)(6.259)] /
(1250000 – 439084)
BCRIV/II = 0.787
Since BCR is less than 1, the grade separation alternative is not preferred to the
traffic signal alternative.
From the benefit cost ratio analysis, the traffic signal alternative is preferred.
Calculate the rate of return for each alternative.
First, compare the traffic signal alternative to the present condition alternative.
NPWII/I = 0 = -((440000 - 916.5) - 0) + ((15000 + 500000) (401000 + 10000))(P/A, i, 20)
NPWII/I = 0 = -439084 + 104000(P/A, i, 20)
(P/A, i, 20) = 439084 / 104000
(P/A, i, 20) = 4.222
i = 23.54% (by interpolation or by trial and error)
Since the rate of return is greater than 15%, the traffic signal alternative is
preferred to the present conditions alternative.
Next, compare the intersection widening to the traffic signal alternative.
NPWIII/II = 0 = -((790000 - 672.10) - (440000 - 916.5)) +
((401000 + 10000) – (9000 + 350000))(P/A, i, 20)
NPWII/I = 0 = 350244.4 + 52000(P/A, i, 20)
(P/A, i, 20) = 350244.4 / 52000
(P/A, i, 20) = 6.735
i = 13.82% (by interpolation or by trial and error)
Since the rate of return is less than 15%, the intersection widening alternative is
not preferred.
Next, compare the grade separation to the traffic signal alternative.
NPWIII/II = 0 = -(1250000 - (440000 - 916.5)) + ((401000 + 10000) (8000 + 301000))(P/A, i, 20)
NPWII/I = 0 = - 810916.5 + 102000(P/A, i, 20)
(P/A, i, 20) = 810916.5 / 102000
(P/A, i, 20) = 7.951
i = 11.08% (by interpolation or by trial and error)
Since the rate of return is less than 15%, the grade separation alternative is not
preferred.
Therefore, based on the four different economic analyses performed above, the
Traffic Signal Alternative is the best alternative.
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Chapter 13: Evaluating Transportation Alternatives
13-13
A road is being proposed to facilitate a housing development on a scenic lake. Two
alternatives have been suggested. One of the roadway alignments is to go around the
lake and slightly impact a wetland. The second alternative will also go around the
lake and will significantly impact two wetlands. The following table shows the
anticipated costs for each alternative. Assuming that the annual interest rate is 7%,
determine which alternative is preferred using equivalent annual cost analysis.
First, calculate the EUAC of Alternative I.
EUACI = 75000(A/P, 7%, 15) + 3000 + 7500 + 1500 – 45000
(A/F, 7%, 15)
EUACI = 75000(0.1098) + 12000 – 45000(0.0398)
EUACI = $18,444
Therefore, the EUAC of Alternative I is $18,444.
Next, calculate the EUAC of Alternative II.
EUACII = 125000(A/P, 7%, 15) + 2000 + 2500 + 2500 – 25000
(A/F, 7%, 15)
EUACII = 125000(0.1098) + 7000 – 25000(0.0398)
EUACII = $19,730
Therefore, the EUAC of Alternative II is $19,730.
Alternative I is preferred since it has a lower EUAC.
13-14
Two alternatives are under consideration for maintenance of a bridge. Select the
most cost-effective alternative using present worth analysis. Assume an interest rate
of 10% per year and a design life of 50 years for each alternative.
Alternative A consists of annual maintenance costs of $5,000 per year for the design
life except for:
Year 20, in which bridge deck repairs will cost $20,000
Year 30, in which a deck overlay and structural repairs will cost $105,000
Alternative B consists of annual maintenance costs of $3,000 per year for the design
life except for:
Year 20, in which bridge deck repairs will cost $35,000
Year 30, in which a deck overlay and structural repairs will cost $85,000
NPWA = - {5000(P/A,10%,50) + 15000(P/F,10%,20) + 100000(P/F,10%,30)}
NPWA = - {5000(1/0.1009) + 15000(0.1486) + 100000(0.0573)}
= - $57,513
NPWB = - {3000(P/A,10%,50) + 32000(P/F,10%,20) + 82000(P/F,10%,30)}
NPWB = - {3000(1/0.1009) + 32000(0. 1486) + 82000(0.0573)}
= - $39,186
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Chapter 13: Evaluating Transportation Alternatives
Alternative B is the most cost-effective (higher net present worth).
13-15
Two highway capacity improvement plans have been proposed for a congested
suburban arterial. Select the most cost-effective alternative using present worth
analysis. Assume an interest rate of 5% per year and a design life of 20 years for
each alternative.
Alternative A entails improvements to traffic signals at an initial cost of
$82,000 and a salvage value of $5,000. Annual maintenance costs will be
$700 per year.
Alternative B entails traffic signal improvements and addition of a left-turn
lane at an initial cost of $72,000 and no salvage value. Annual maintenance
costs will be $1,200 per year, except in year 10 in which a rehabilitation will
cost $14,000.
NPWA = - {82000 - 5000(P/F,5%,20) + 700(P/A,5%,20)}
NPWA = - {82000 - 5000(0.3769) + 700(1/0.0802) }
= - $92,613
NPWB = - {72000 + 1200(P/A,5%,20) + 12800(P/F,5%,10)}
NPWB = - {72000 + 1200(1/0.0802) + 12800(0.6139)}
= - $94,820
Alternative A is the most cost-effective (higher net present worth).
13-16
The light-rail transit line described in this chapter is being evaluated by another
group of stakeholders. Using the revised information, determine the weighted score
for each alternative and comment on your result.
First, determine the relative weight. For a ranking of 1, the highest, assign n
where n is the number of objectives. In this case n = 5. Therefore, for a ranking
of 2, the next highest, assign n - 1 = 5 - 1 = 4. This process is then continued for
the remainder of the objectives.
Next, determine the weighting factors using Equation 13.9.
The weighting factor for objective I is:
K1 = relative / sum of relative weights
K1 = [1 / (5 + 4 + 3 + 2 + 1)] * 100
K1 = (1 / 15) * 100
K1 = 6.667 = 7
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Chapter 13: Evaluating Transportation Alternatives
Objective II
K2 = (3 / 15) * 100
K2 = 20
Objective III
K3 = (5 / 15) * 100
K3 = 33
Objective IV
K4 = (4 / 15) * 100
K4 = 26.667 = 27
Objective V
K5 = (2 / 15) * 100
K5 = 13.33 = 13
Objective
Ranking
Relative Weight
Weighting Factor
1
5
1
7
2
3
3
20
3
1
5
33
4
2
4
27
5
4
2
13
15
100
Total
Next, use the estimated values for measures of effectiveness found in Table 13.5
to produce the point score for the candidate transit lines.
Alternatives
MOE
I
II
III
IV
V
1
6.1
6.5
5.1
6.3
7.0
2
20.0
18.4
16.0
14.4
13.6
3
16.5
23.1
26.4
33.0
33.0
4
27.0
23.6
20.3
16.9
16.9
5
13.0
11.1
7.4
5.6
5.6
82.6
82.7
75.2
76.2
76.1
Total
The ranking of alternatives in order of preference is II, I, IV, V, and III.
Alternatives I and II are clearly superior to the others.
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Chapter 13: Evaluating Transportation Alternatives
13-17
Three alternatives to replace an existing two-lane highway with a four-lane highway
that will meet current design standards are proposed. The selected alternative will
provide a more direct route between two towns that are 12.0 miles apart along the
existing highway. With each alternative operating speeds are expected to be at or
near the design speed of 60 miles per hour. Develop the scores for each alternative
and recommend a preferred alternative for development. The following scoring
method, developed by the transportation oversight board, is to be used:
Evaluation
Criterion
Mobility
Safety
Costeffectiveness
Environmental
impacts
Community
impacts
Performance Measure
Weight (%)
Travel time of shortest travel time
alternative divided by travel time of
alternative i
Annual reduction in number of crashes of
alternative i divided by highest annual
reduction in number of crashes among all
alternatives
Project development cost of least expensive
alternative (in $ per mile) divided by project
development cost of alternative i (in $ per
mile)
Area of wetlands impacted of leastimpacting alternative divided by area of
wetlands impacted by alternative i
Number of business and residences
displaced by least-impacting alternative
divided by number of businesses and
residences displaced by alternative i
25
25
20
15
15
The following information has been estimated for each alternative by the planning
staff:
Property
Cost of
development
Length
Annual crash
reduction
Business
displacements
Residential
displacements
Wetlands
impacted
Alt. 1
$10,900,000
Alt. 2
$18,400,000
Alt. 3
$16,900,000
11.2 miles
10
9.8 miles
17
10.1 miles
19
3
4
5
4
3
3
1.5 acres
3.9 acres
3.9 acres
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Chapter 13: Evaluating Transportation Alternatives
Use the approach given in Equation 13.9, in which the score for an alternative is
the sum of the products of the weight for each evaluation criterion and the relative
score or value for each criterion. First, the input values are to be used to calculate
the relative scores, Vij . For example, to evaluate the performance measure for the
criterion mobility, the travel time of each alternative must be calculated. For
example, for alternative 1, the travel time is:
⎛ hr ⎞
⎜
⎟11.2mi = 0.1867hr
⎝ 60mi ⎠
Performance Measure
Alt. 1
Travel time
Annual crash reduction
Cost per mile
Wetlands impacted
Total displacements
Alt. 2
0.1867
10.0000
973214
1.5000
7.0000
Alt. 3
0.1633
17.0000
1877551
3.9000
7.0000
0.1683
19.0000
1673267
3.9000
8.0000
The relative scores, Vij, can then be calculated. For example, for the evaluation
criterion mobility, the travel time of the shortest travel time alternative divided by
travel time of alternative i must be determined for each alternative. The shortest
travel time is for alternative 2, at 0.1633 hours. For alternative 1, the travel time
is 0.1867 hr, and the relative score is 0.1633/0.1867 = 0.8750.
Evaluation Criterion
Mobility
Safety
Cost-effectiveness
Environmental impacts
Community impacts
Kj
25
25
20
15
15
Alt. 1
Alt. 2
Relative Scores (Vij)
0.8750
0.5263
1.0000
1.0000
1.0000
Alt. 3
1.0000
0.8947
0.5183
0.3846
1.0000
0.9703
1.0000
0.5816
0.3846
0.8750
The scores for each alternative i can then be found by finding the products of the
weights (Kj) and relative scores (Vij) and then summing these products.
Alt. 1
Alt. 2
Alt. 3
21.8750
25.0000
24.2574
13.1579
22.3684
25.0000
20.0000
10.3668
11.6325
15.0000
5.7692
5.7692
15.0000
15.0000
13.1250
Scores
85.033
78.504
79.784
The recommended alternative is Alternative 1, with a score of 85.03, the highest
score among the alternatives.
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Chapter 13: Evaluating Transportation Alternatives
13-18
You have been hired as a consultant to a medium-sized city to develop and
implement a procedure for evaluating whether or not to build a highway bypass
around the CBD. Write a short report describing your proposal and
recommendation as to how the city should proceed with this process.
I.
Collection of Required Data and Information
In order for a consultant to develop and evaluate alternatives for a
highway bypass, information is needed to decide upon an evaluation process.
Establishing goals and objectives, identification of stakeholders (e.g., governing
bodies, citizen groups), definition of purpose and need for a bypass, and its
potential impacts (e.g., traffic flow, environmental effects) are all important steps
toward selecting an evaluation process. The data required, such as socioeconomic
data, and traffic data, should also be considered in selecting a suitable evaluation
process.
II.
Use of Information
Upon deciding on the data and information requirements, an evaluation
process will be chosen that best fits the situation and any stated objectives and
goals of the study. If cost is a critical factor, an economic evaluation that includes
a life cycle cost analysis may be followed. The process will then be divided into
components, the data gathered, and alternatives defined and evaluated.
III.
Results
A recommendation would be included with the evaluation of the highway
bypass. If several alternatives are relatively close, additional forms of evaluation
should be considered to ensure a sound decision.
13-19
The following data have been developed for four alternative transportation plans
for a high-speed transit line that will connect a major airport with the downtown
area of a large city. Prepare an evaluation report for these proposals by considering
the cost effectiveness of each attribute. Show your results in graphical form and
comment on each proposal.
Persons Displaced
The existing service moves no one out of their home while plans B, C, and D
displace 3,200 people each. Plan A, however, displaces only 264 people, 92%
less than the other proposals. The graph below depicts the number of people
displaced versus costs. This graph clearly shows that plan A is the preferred
alternative when considering residential displacements.
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256
Chapter 13: Evaluating Transportation Alternatives
Businesses Displaced
Plan A impacts far fewer businesses than does Plans B, C, or D, which will
displace 275 businesses each. As shown in the graph below, Plan A is the
preferred alternative when considering business impacts.
Average Speed
While the existing plan has an average speed of 10.2 mi/h, the four new proposals
increase this value to almost 4 times the current travel speed (Plan A) and in one
case almost 5 times (Plan D) the current rate. Out of all of the proposals, Plan D
has the fastest rate and Plan A has the slowest. When one relates these statistics
to annual passengers carried on each alternative, Plan D is preferred, but only
slightly above the existing and other proposed alternatives. As the graph below
indicates, all of the alternatives are relatively equal. Plan A provides the least
improvement.
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Chapter 13: Evaluating Transportation Alternatives
Costs
From observing the data gathered, Plan A has the lowest cost. Overall, Plan A
fares best in three of the four criteria considered (excepting average speed).
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258
Chapter 13: Evaluating Transportation Alternatives
13-20
A new carpool lane has replaced one lane of an existing six-lane highway. During
peak hours, the lane is restricted to cars carrying three or more passengers. After
five months of operation, the carpool lane handles 800 autos/h, whereas the existing
lanes are operating at capacity levels of 1500 veh/h/ln at an occupancy rate of 1.2.
How would you determine if the new carpool lane is successful or if the lane should
be open to all traffic?
This would be a good case to utilize a control versus experimental group
analysis. By comparing the new level of traffic on the carpool highway with that
of another highway without a carpool lane, it can be determined if the carpool
lane is a success.
In order to do this, one must first have traffic statistics for the number of
automobiles and number of persons per hour on the roads before and after the
modification to the facility. Next, one must establish a minimum acceptable rate
of increase in persons traveling by carpool. If the capacity (persons carried per
lane per hour) with the carpool modification is greater than that with natural
growth plus the minimum acceptable rate, the carpool lane program is considered
successful.
Assume the High Occupancy Vehicles (HOV) in the carpool lane carry 3
passengers per vehicle, the average autos carry 1.2 persons, the acceptable rate of
increase is 10% and the growth rate is 4% in the five months.
800(3) = 2,400 > 1,500(1.2)(1 + 0.04) = 1,872
Therefore, the carpool lane is a success.
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Chapter 14
Highway Surveys and Location
14-1
Describe the three categories of information gathered, in the office prior to any field
survey activity, about the characteristics of the area of a proposed highway location.
The information gathered in an office study of existing information can be divided
into three categories: engineering, environmental, and economic. Engineering
includes topography, geology, climate, and traffic volumes, and social and
demographic, including land use and zoning patterns. Environmental includes
types of wildlife; location of recreational, historic, and archeological sites; and the
possible effects of air, noise, and water pollution. Economic includes unit costs
for construction and the trend of agricultural, commercial, and industrial activities
in the proposed location area.
14-2
Briefly discuss factors that are of specific importance in the location of scenic routes.
The following factors are of specific importance in the location of scenic routes:
•
Special provisions should be provided to discourage fast driving as design
speeds are usually low (e.g. provide narrow road bed).
•
Conflict between driver's attention on the road and the need to enjoy the
scenic view should be minimized. This is achieved by providing turn-outs
with wide shoulders and adequate turning space at regular intervals, or by
providing only straight alignment when the view is exceptional.
•
Only minimum disruption to the area should be caused as a result of the
scenic route construction.
14-3
Describe the factors that significantly influence the location of highways in urban
areas.
Connection to local streets is primarily a factor with design of freeways and
expressways such that traffic flow is made as efficient as possible. Right-of-way
acquisition, particularly in commercial and industrial areas, is substantial expense;
such costs often dictate the available corridor width and the ultimate design. The
interaction between the many travel modes available in urban areas requires
coordination of the highway system with other transportation systems. Adequate
provisions for pedestrians and bicycles is important as increased use of these
modes can reduce energy use and traffic congestion.
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Chapter 14: Highway Surveys and Location
14-4
What are three elements that highway surveys usually involve?
Highway surveys usually involve measuring and computing horizontal and
vertical angles, vertical heights (elevation), and horizontal distances. The surveys
are then used to prepare base maps with contour lines and longitudinal cross
sections.
14-5
Briefly describe the use of each of the following instruments in conventional ground
surveys:
(a) total station
(b) level
(c) measuring tapes
(d) electronic distance measuring devices
(a) The total station is an electronic theodolite and distance measuring device. It
is used for measuring angles in both the vertical and horizontal plane as well
as distances.
(b) A level is used in conjunction with a graduated leveling rod to measure
changes in elevation .
(c) Measuring tapes are used for the direct measurement of horizontal distances.
(d) An electronic distance measuring device (EDM) is used to measure distances.
When an EDM is used in conjunction with a slope reduction calculator, it can
also measure slope and height distances. These devices allow for distances
and direction to be determined from a single instrument setup.
14-6
Briefly compare the factors that should be considered in locating an urban freeway
with those for a rural highway.
Factors that are similar when considering the location for both urban freeways and
rural freeways include:
•
Social and demographic characteristics of the area in which the freeway is
to be located (including land use and zoning patterns).
•
Environmental impacts (including recreation and historic sites, plant and
animal life, wetlands, and air, noise, and water pollution).
•
Serviceability of the route (to industrial and residential areas).
•
Crossing of other transportation facilities.
•
Terrain and soil conditions.
•
Economic feasibility of the location.
•
Directness of route.
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262
Chapter 14: Highway Surveys and Location
Factors that are considered for urban freeways and not usually considered for
rural freeways include:
•
Connection to local streets.
•
Cost of right-of-way acquisition.
•
Coordination of the urban freeway and other urban transportation systems.
•
Adequate provisions for pedestrians.
14-7
Describe how each of the following could be used in highway survey location:
(a) aerial photogrammetry
(b) computer graphics
(c) conventional survey techniques
(a)
Aerial Photogrammetry: Aerial photogrammetry is used in identifying
suitable highway locations and in preparing contour maps suitable for cross
sections. This involves obtaining the aerial photographs and determining
distances and elevations from these photographs.
(b)
Computer Graphics: Computer graphics are used in identifying suitable
highway locations by combining photogrammetry and computer techniques. The
procedure also involves the use of aerial photographs to determine distances and
elevations. In addition, the horizontal and vertical alignment of a proposed
centerline can be obtained and displayed on a monitor. This enables the designer
to make alignment changes and to immediately see the effect of these changes.
(c)
Conventional Survey Techniques: Conventional surveys are used in
identifying suitable locations by determining distances and elevations for all
natural and man-made land features. Contour maps are then obtained which can
be used to identify alternative locations. Profiles, showing the change in elevation
along a proposed centerline, and cross sections at selected stations along the
proposed centerline can also be developed.
14-8
A photograph is to be obtained at a scale of 1:10,000 by aerial photogrammetry. If
the focal length of the camera to be used is 6.5", determine the height at which the
aircraft should be flown if the average elevation of the terrain is 950 ft.
Use Equation 14.1 and solve for H.
f
S=
H −h
(1 / 10,000) = ((6.5 in)/(ft/12in)) / (H – 950 ft)
H - 950 = 5417
H = 6367 feet
Therefore, the plane should fly at 6,367 feet.
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Chapter 14: Highway Surveys and Location
14-9
The distance in the x direction between two control points on a vertical aerial
photograph is 4.5". If the distance between these same two points is 3.6" on another
photograph having a scale of 1:24,000, determine the scale of the first vertical
aerial photograph. If the focal length of the camera is 6"and the average elevation at
these points is 100 ft, determine the flying height from which each photograph was
taken.
Let the actual distance between the points MN be x feet
M'N' = 3.6 in = 0.3 ft
(M'N') / (MN) = (ON') / (ON) = Scale 1:24,000
1 / 24000 = 0.3 / x
x = 7200 feet
Now solve for the scale of the second photo.
M'N = 4.5 in = 0.375 ft
Scale = 0.375 / 7200
Scale = 1:19,200
Now determine the height from which the first photo was
taken (H1).
1 / 19,200 = (6.0/12) / (H1 - 100)
H1 = 9,700 feet
Now determine the height from which the second photo was taken (H2).
1 / 24,000 = (6.0/12) / (H2 - 100)
H2 = 12,100 feet
Therefore, the scale of the first photograph is 1:19,200, the elevation it was taken
from is 9,700 feet, and the second photograph was taken at an elevation of 12,100
feet.
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Chapter 14: Highway Surveys and Location
14-10
The scale at the image of a well-defined object on an aerial photograph is 1:24,000,
and the elevation of the object is 1500 ft. The focal length of the camera lens is 6.5".
If the air base (B) is 250 ft, determine the elevation of the two points A and C and
the distance between them if the coordinates of A and C are as given below.
First determine the flying height, H, using Equation 14.1.
f
S=
H −h
(1 / 24000) = (6.5/12) / (H - 1500)
H - 1500 = 13000
H = 14,500 feet
Now determine the elevation of point A, ha, using Equation 15.1.
(1 / 13000) = (6.5/12) / (14500 - ha)
14500 - ha = 7042
ha = 7,458 feet
Now determine the elevation of point B, hb, using Equation 15.1.
(1 / 17,400) = (6.5/12) / (14500 - hb)
14500 - hb = 9425
hb = 5,075 feet
Now determine the distance between points A and C.
Use Equations 14.4 and 14.5 to solve for the X and Y coordinates, respectively.
XA = (5.5 / 12) / (1 / 13000)
XA = 5958 ft
XB = (6.5 / 12) / (1 / 17400)
XB = 9425 ft
YA = (3.5 / 12) / (1 / 13000)
YA = 3792 ft
YB = (5.0 / 12) / (1/ 17400)
YB = 7250 ft
Now use Equation 14.6 to solve for the distance between these points.
D = ( X A − X B ) 2 + (Y A − YB ) 2
D = (5958 − 9425) 2 + (3792 − 7250) 2
D = 4,897 ft.
Therefore, the elevation at point A is 7,458 ft; point B is 5,075 ft and the distance
between the two points is 4,897 ft.
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Chapter 14: Highway Surveys and Location
14-11
A vertical photo has an air base of 2400 ft. Stereoscopic measurements of parallax at
a point representing the top of a 200 ft tower is 0.278". The camera focal length is
6.5". Photos were taken at an elevation of 7500 ft. Determine the elevation of the
base of the tower.
Use Equation 14.2 to determine the elevation of the base of the tower.
H −h B
=
f
p
7500 − h
200
=
(6.5 / 12) (0.278 / 12)
7500 – h = 4676
h = 2824 ft
Since the elevation of the top of the tower is 2824 ft, the elevation at its base is
2624 ft (200 ft below the top of the tower).
14-12
The length of a runway at a national airport is 7500 ft long and at elevation 1500 ft
above sea level. The airport was recently expanded to include another runway used
primarily for corporate aircraft. It is desired to determine the length of this runway
whose elevation is 1800 ft. An aerial photograph was taken of the airport.
Measurements on the photograph for the national airport runway are 4.80" and for
the corporate runway, are 3.4". The camera focal length is 6". Determine the length
of the corporate runway.
To calculate the length of the corporate runway, the scale of the
photograph at the elevation of the corporate runway must be determined. To
calculate the scale at any point on the photograph, the flying height from which
the photograph was taken must be determined. The scale at the elevation of the
national runway is:
S(1500 ft) = (4.8 in)(1 ft / 12 in) / 7500 ft = 1/18750
Use Equation 14.1 to solve for the flying height,
f
(6 / 12)
1
=
S=
=
H − h H − 1500 18750
H – 1500 = 9375
H = 10,875 ft
Use Equation 14.1 to solve for the scale at the elevation of the corporate runway,
f
(6 / 12)
1
=
S=
=
H − h 10875 − 1800 18150
The length of the corporate runway can then be found,
L = (3.4/12)(18150)
L = 5,143 ft
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266
Chapter 14: Highway Surveys and Location
14-13
Using an appropriate diagram, discuss the importance of side and forward overlaps
in aerial photography.
Side and forward overlaps are critical for stereoscopic viewing of aerial
photographs. In order for the aerial photographs to be viewed in three
dimensions, an object should be viewed by the observer’s left eye on the left
photograph and the same object viewed by the observer’s right eye on the right
photograph on a set of stereopairs. This requires that each object in the area be on
at least two aerial photographs. This is achieved through the side overlap which
provides for approximately 60 percent overlap in the direction of flight, and the
overlap in the direction perpendicular to flight which provides for about 25
percent overlap as shown in Figure 14.3 in the text.
14-14
Under what conditions would the borrowing of new material from a borrow pit for
a highway embankment be preferred over using material excavated from an
adjacent section of the road?
The conditions under which the borrowing of new material from a borrow pit is
preferable to using excavated material from an adjacent section of highway might
include the following:
•
The engineering properties of the material from the adjacent section are
not satisfactory.
•
Excavation of this material may result in serious negative environmental
impacts.
•
Excavation of this material might result in serious drainage problems for
the highway.
•
Excavation of the material might be cost prohibitive.
14-15
Using the data given in Table 14.1, determine the total overhaul cost if the free haul
is 700 ft and the overhaul cost is $7.50 per cubic yard station. Stations of the free
haul lines are 1 + 80 and 8 + 80 and 10 + 20 and 17 + 20.
The first step is to construct the mass diagram shown in Figure 14.17 from the
data in Table 14.1. The data required to solve this problem using the method of
moments are shown below. Note that the ordinate is zero at station 9+63, and
other ordinates as given in Table 14.1 are shown.
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Chapter 14: Highway Surveys and Location
First, find the moments and overhaul distances about stations 1+80 and 8+80.
About sta. 1+80: [(100/2+80)/100](130) + [(80/2)/100](374-130) = 266.6
yd3-sta
Overhaul distance: (266.6 yd3-sta) / 374 yd3 = 0.713 sta
About sta. 8+80: [(63/2+20)/100](299) + [(20/2)/100](374-299) = 161.5
yd3-sta
Overhaul distance: (161.5 yd3-sta) / 374 yd3 = 0.432 sta
The overhaul cost for the first section (between sta. 0+00 and sta. 9+63) can be
calculated as:
Overhaul cost = (0.713 sta + 0.432 sta)(374 yd3)($7.50/yd3-sta) = $3210
Then, find the moments and overhaul distances about stations 10+20 and 17+20.
About sta. 10+20: [(37/2+20)/100](201) + [(20/2)/100](255-201) = 86.2
yd3-sta
Overhaul distance: (86.2 yd3-sta) / 255 yd3 = 0.338 sta
About sta. 17+20: [(49/2)/100](255) = 62.5 yd3-sta
Overhaul distance: (62.5 yd3-sta) / 255 yd3 = 0.245 sta
The overhaul cost for the second section (between sta. 10+20 and sta. 17+20) can
be calculated as:
Overhaul cost = (0.338 sta + 0.245 sta)(255 yd3)($7.50/yd3-sta) = $1115
Total overhaul cost = $3210 + $1115 = $4325.
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268
Chapter 14: Highway Surveys and Location
14-16
The following table shows the stations and ordinates for a mass diagram. The freehaul distance is 600 ft. Overhaul cost is $15 per station yard.
Station
Ordinate (yd3)
0 + 00
0
1 + 00
45
2 + 00
60
2 + 20
90
4 + 00
120
6 + 00
140
7 + 00
110
8 + 20
90
9 + 00
82
10 + 00
60
10 + 30
0
(a) Use the method of movements to compute the additional cost that must
be paid to the contractor.
(b) Sketch the ground profile if the finished grade of this roadway section is
level (0%).
First, find the moments and overhaul distances about stations 2+20 and 8+20.
About sta. 2+20: [(100/2+100+20)/100](45) + [(100/2+20)/100](60-45) +
[(20/2)/100](90-60) = 90 yd3-sta
Overhaul distance: (90 yd3-sta) / 90 yd3 = 1.00 sta
About sta. 8+20: [(80/2)/100](90-82) + [(100/2+80)/100](82-60) +
[(30/2+100+80)/100](60) = 148.8 yd3-sta
Overhaul distance: (148.8 yd3-sta) / 90 yd3 = 1.653 sta
The overhaul cost can be calculated as:
Overhaul cost = (1.000 sta + 1.653 sta)(90 yd3)($15/yd3-sta) = $3582.
269
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Chapter 15
Geometric Design of Highway Facilities
15-1
A rural collector highway located in a mountainous terrain is designed to carry a
design volume of 800 veh/day. Determine the following: (a) a suitable design speed,
(b) lane and usable shoulder widths, (c) maximum desirable grade.
(a) From Table 15.1 (for mountainous terrain)
minimum design speed = 30 mi/h
(b) Lane width = 12 ft, usable shoulder width = 10 ft
(c) From Table 15.4
maximum desirable grade = 10%
15-2
Repeat Problem 15-1 for an urban freeway in rolling terrain.
(a) From Table 15.2, for urban freeway
minimum design speed = 50 mi/h
(b) Lane width = 12 ft, usable shoulder width = 12 ft
(c) From Table 15.4, for freeway in rolling terrain with design speed of 50 mi/h
maximum grade = 5%
15-3
Given: A rural collector is to be constructed in rolling terrain with an ADT of 650
veh/day.
Determine:
(a) minimum design speed
(b) recommended lane width
(c) preferable shoulder width
(d) maximum grade
(a) From Table 15.1, for rolling terrain,
minimum design speed = 40 mi/h
(b) Lane width = 12 ft,
(c) Shoulder width = 10 ft
(d) From Table 15.4
maximum desirable grade = 8%
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Chapter 15: Geometric Design of Highway Facilities
15-4
A +2% grade intersects with a -1% grade at station (535+24.25) at an elevation of
300 ft. If the design speed is 65 mi/h, determine:
(a) the stations and elevations of the BVC and EVC
(b) the elevation of each 100-ft station
(c) the station and elevation of the highpoint
From Table 15.5, k = 193
L = kA = 193 (2 – (-1)) = 579 ft
Station of BVC = (535+24.25) – (579 ft)/2 = 532+34.75
Station of EVC = (535+24.25) + (579 ft)/2 = 538+13.75
Elevation of BVC = 300 – (0.02)(579/2) = 294.21 ft
Elevation at any station on the leading tangent can be found in a similar manner.
The elevation on the curve can be found by subtracting the elevation on the
leading tangent by the offset, which can be found using Equation 15.15,
Y=
A
x2
200 L
Using this procedure, the following table, which tabulates the elevation at 100 ft
stations on the curve, can be generated.
Station
Dist. from BVC
Tangent Elev.
Offset
Curve Elev.
532+34.75
0
294.21
0
294.21
533+00
65.25
295.52
0.11
295.41
534+00
165.25
297.52
0.71
296.81
535+00
265.25
299.52
1.82
297.70
536+00
365.25
301.52
3.46
298.06
537+00
465.25
303.52
5.61
297.91
538+00
565.25
305.52
8.28
297.24
538+13.75
579.00
305.79
8.69
297.10
The distance from the BVC to the high point can be found as:
xhigh = LG1 / (G1 – G2) = (579)(2)/(2-(-1)) = 386 ft
The station of the high point is (532+34.75) + (386 ft) = 536+20.75
The difference between the elevation of the BVC and the elevation of the high
point can be found as:
yhigh = LG12 / (200(G1 – G2)) = (579)(2)2 / (200)(2-(-1)) = 3.86 ft
Therefore, the elevation of the high point is 294.21 + 3.86 = 298.07 ft
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272
Chapter 15: Geometric Design of Highway Facilities
15-5
Determine the minimum length of a crest vertical curve if the grades are +4% and
- 2%. Design speed is 70 mi/h. State assumptions used.
Assumptions used include: perception-reaction time is 2.5 seconds, deceleration
rate is 11.2 ft/sec2, and the case is sight distance is less than the length of the
curve.
Determine required stopping sight distance using Equation 3.27:
u2
(70) 2
SSD = 1.47ut +
= 1.47(70)(2.5) +
a
11.2
30( ± G )
− 0.04)
30(
g
32.2
SSD = 788 ft
Since S<L, use Equation 15.5 to calculate the minimum length of the curve.
AS 2
Lmin =
= (6)(788)2/2158 = 1727 ft
2158
Therefore, for the given design conditions, the minimum length of the curve is
1,727 ft.
15-6
Determine the minimum length of a sag vertical curve if the grades are -4% and
+2%. Design speed is 70 mi/h. State assumptions used. Consider the following
criteria: stopping sight distance, comfort, and general appearance.
Assumptions used include: perception-reaction time is 2.5 seconds, deceleration
rate is 11.2 ft/sec2, and the case is sight distance is less than the length of the
curve.
For the sight distance criterion:
Determine required stopping sight distance using Equation 3.27:
u2
(70) 2
SSD = 1.47ut +
= 1.47(70)(2.5) +
a
11.2
30( ± G )
− 0.04)
30(
g
32.2
SSD = 788 ft
Since S<L, use Equation 15.9 to calculate the minimum length of the curve.
(6)(788) 2
AS 2
=
L=
= 1179 ft
400 + 3.5S 400 + 3.5(788)
For the comfort criterion, use Equation 15.10:
Au 2
L=
= (6)(70)2/46.5 = 633 ft
46.5
273
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Chapter 15: Geometric Design of Highway Facilities
For the general appearance criterion, use Equation 15.11:
L = 100 A = (100)(6) = 600 ft
Therefore, for the given design conditions, the minimum length of the curve is
1,179 ft.
15-7
Show that the offset y of any point of the curve above the BVC is given as:
(g − g2 )x 2
100 y = g1 x − 1
2L
From the properties of a parabola, Y = ax2 , where a is constant
The rate of change of the slope can be written as:
d2y
= 2a
dx 2
If L is the total length of the curve in feet and the total change in slope is A then,
A
2a =
100 L
The equation for the curve can now be written as:
A 2
Y=
x
200 L
At any point on the curve, the vertical offset, y, can be determined by the
equation:
g x
g x g − g2 2
g x
A
y = 1 −Y = 1 −
x2 = 1 − 1
x
100
100 200 L
100
200 L
100 y = g1 x −
( g1 − g 2 ) x 2
2L
15-8
Given a sag vertical curve connecting a -1.5% grade with a +2.5% grade on a rural
arterial highway, use the minimum stopping sight distance and a design speed of 70
mi/h to compute the elevation of the curve at 100 ft stations if the grades intersect at
station (475+00) at an elevation of 300 ft. Locate the low point.
From Table 15.6, k = 181
L = kA = 181 (1.5 – (-2.5)) = 724 ft
Station of BVC = (475+00) – (724 ft)/2 = 471+38.00
Station of EVC = (475+00) + (724 ft)/2 = 478+62.00
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Chapter 15: Geometric Design of Highway Facilities
Elevation of BVC = 300 + (0.015)(724/2) = 305.43 ft
Elevation at any station on the leading tangent can be found in a similar manner.
The elevation on the curve can be found by adding the elevation on the leading
tangent to the offset, which can be found using Equation 15.15,
A 2
Y=
x
200 L
Using this procedure, the following table, which tabulates the elevation at 100 ft
stations on the curve, can be generated.
Station
Dist. from
BVC
Tangent
Elev.
Offset
Curve Elev.
471+38
0
305.43
0
305.43
472+00
62
304.50
0.11
304.61
473+00
162
303.00
0.72
303.72
474+00
262
301.50
1.90
303.40
475+00
362
300.00
3.62
303.62
476+00
462
298.50
5.90
304.40
477+00
562
297.00
8.72
305.72
478+00
662
295.50
12.11
307.61
478+62
724
294.57
14.48
309.05
The distance from the BVC to the low point can be found as:
xlow = LG1 / (G1 – G2) = (724)(1.5)/(1.5 – (-2.5)) = 271.50 ft
The station of the low point is (471+38) + (2+71.50) = 474+09.50
The difference between the elevation of the BVC and the elevation of the low
point can be found as:
ylow = LG12 / (200(G1 – G2)) = (724)(1.5)2 / (200)(1.5 – (-2.5)) = 2.04 ft
Therefore, the elevation of the low point is 305.43 – 2.04 = 303.39 ft
275
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Chapter 15: Geometric Design of Highway Facilities
15-9
A crest vertical curve connects a +4.44% grade and a -6.87% grade. The PVI is at
station 43+50.00 at an elevation of 1240.00 ft. The design speed is 30 mi/h.
Determine:
(a) The length of the vertical curve using the AASHTO method (“K” factors)
(b) The station of the BVC
(c) The elevation of the BVC
(d) The station of the EVC
(e) The elevation of the EVC
(f) The station of the high point
(g) The elevation of the high point
(h) The elevation of station 44+23.23
(a)
(b)
(c)
(d)
(e)
(f)
L = kA = (19)|-4.44-6.87| = (19)(11.31) = 214.89 ft
Station of BVC = 43+50 – (214.89)/2 = 42+42.55
Elevation of BVC = 1240.00 – (0.0444)(214.89/2) = 1235.23 ft
Station of EVC = 43+50 + (214.89)/2 = 44+57.45
Elevation of EVC = 1240.00 – (0.0687)(214.89/2) = 1232.62 ft
Station of the high point = (42+42.55) + (214.90)(4.44)/(11.31) = 43+26.91
(214.90)(4.44) 2
= 1237.10 ft
(g) Elevation of the high point = 1235.23 +
(200)(11.31)
(h) Elevation of station 44+23.23 =
(11.31)(180.68) 2
1235.23 + (0.0444)(180.68) −
= 1234.66 ft
(200)(214.90)
15-10
A horizontal curve is to be designed for a two-lane road in mountainous terrain.
The following data are known: Intersection angle: 40 degrees, tangent length =
436.76 ft, station of PI: 2700+10.65, fs = 0.12, e = 0.08.
Determine:
(a) design speed
(b) station of the PC
(c) station of the PT
(d) deflection angle and chord length to the first 100 ft station
(a) From the given horizontal curve data, the radius can be calculated, from which
design speed for the curve can be derived.
The radius can be found by rearranging Equation 15.22,
R = T / (tan Δ/2) = 436.76 / tan (40°/2) = 436.76 / 0.3640 = 1200 ft
The design speed can then be found:
R = u2 / [15(e + fs)]
(1200) = u2 / [ 15 (0.08+0.12) ]
u2 = 3600
u = 60 mi/h
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276
Chapter 15: Geometric Design of Highway Facilities
(b) Station of the PC can be found by subtracting the tangent length from the
station of the PI.
PC = (2700+10.65) – (4+36.76) = 2695+73.89
(c) The length of the curve can be found using Equation 15.26,
L = πRΔ/180 = (3.1415926)(1200)(40)/180 = 837.76 ft
Station of the PT can be found by adding the length of the curve to the
station of the PC.
PT = (2695+73.89) + (8+37.76) = 2704+11.65
(d) Deflection angle and chord length to the first full station
To find the deflection angle to the first full station, use Equation 15.27,
δ1/2 = 180 l1 / 2πR = (180)(100-73.89)/(2)(3.1415926)(1200) = 0.6233°
The chord to the first full station can be found using Equation 15.28,
C1 = 2R sin (δ1/2) = (2)(1200) sin (1.24666/2) = 26.11 ft
15-11
A proposed highway has two tangents of bearings N 45º54’36” E and N 1º22’30” W.
The highway design engineer, attempting to obtain the best fit for the simple
circular curve to join these tangents, decides that the external ordinate is to be 43.00
ft. The PI is at station 65+43.21
Determine:
(a) The central angle of the curve
(b) The radius of the curve
(c) The length of the tangent of the curve
(d) The station of the PC
(e) The length of the curve
(f) The station of the PT
(g) The deflection angle and chord from the PC to the first full station on the
curve
(a) Δ = 45.91° + 1.375° = 47.285°
(b) R = 43.00
= 469.30 ft
⎞
⎛
1
⎜⎜
−1⎟⎟
⎝ cos(47.285 / 2) ⎠
(c) T = 469.30 tan (47.285/2) = 205.44 ft
(d) Station of the PC = (65+43.21) – 205.44 ft = 63+37.76
(e) L = π(469.30)(47.285)/180 = 387.31 ft
(f) Station of the PT = (63+37.76) + 387.31 ft = 67+25.07
(g) Deflection angle from the PC to the first full station:
DA1 = δ1Δ/2L = (100 – 37.76)(47.285)/(2)(387.31) = 3.7993°
Chord from the PC to the first full station:
C1 = 2(469.30) sin (3.7993) = 62.19 ft
277
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Chapter 15: Geometric Design of Highway Facilities
15-12
A simple circular curve exists with a degree of curve D =12º and e=0.08
A structure is proposed on land on the inside of curve. Assume the road is on level
grade.
Determine:
(a) The radius of the curve
(b) The current maximum safe speed of the curve
(c) The minimum distance allowable between the proposed structure and the
centerline of the curve such that the current maximum safe speed of the
curve would not need to be reduced
Using Equation 15.20:
R = 5729.58 / 12 = 477.47 ft
Next, determine the maximum safe speed for the curve, which can then be used to
determine the required stopping sight distance.
u = (15)(477.47)(0.08 + 0.16) = 41.4 mi/h
(41.4) 2
= 315 ft
(30)(0.35 − 0)
Then, determine the minimum offset distance using Equation 15.43:
(28.65)(315) ⎞
⎛
m = 477.47⎜1 − cos
⎟ = 25.8 ft
477.47 ⎠
⎝
SSD = (1.47)(2.5)(41.4) +
15-13
Given a circular curve connecting 2 tangents that intersect at an angle of 48°. The
PI is at station (948+67.32) and the design speed of the highway is 60 mi/h.
Determine the point of the tangent and the deflection angles to full stations for
laying out the curve.
First determine the radius of the curve:
R = u2 / [15(e + fs)]
For u = 60 mi/h, from Table 3.4, fs = 0.12, e = 0.08
R = (60)2 / [15(0.08+0.12)]
R = 1200 ft
The length of the tangent, T, can be found using Equation 15.22,
T = R tan(Δ/2) = 1200(tan(48˚/2))
T = 534.27 ft
The length of the curve, L, is given by Equation 15.26:
L = RΔπ / 180 = 1200(48)(3.1415926) / 180
L = 1005.30 ft
Station of the PC can be found by subtracting the tangent length from the station
of the PI.
PC = (948+67.32) – (5+34.27) = 943+33.05
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278
Chapter 15: Geometric Design of Highway Facilities
Station of the PT can be found by adding the length of the curve to the station of
the PC.
PT = (943+33.05) + (10+5.30) = 953+38.35
First full station is located at 944+00
δ1 / Δ = l1 / L
δ1 / 48 = 66.95 / 1005.3
δ1 = 3.197°
The first chord can be found using Equation 15.28,
C1 = 2R sin (δ1/2) = 2(1200) sin (3.197˚ / 2)
C1 = 66.94 ft
The first deflection angle = δ1/2 = 1.5985˚
Last full station is located at 953+00
δ2 / Δ = l2 / L
δ2 / 48 = 38.35 / 1005.30
δ2 = 1.831˚
The last chord can be found using Equation 15.28,
C2 = 2R sin ( δ2/2) = 2(1200) sin (1.831˚/ 2)
C2 = 38.35 ft
For other deflection angles between full stations:
δ/48 = 100/1005.3
δ = 4.775˚
δ/2 = 2.3875˚
The chords between full stations can be found using Equation 15.28,
C = 2R sin (δ/2) = 2(1200) sin (2.3875˚) = 99.97 ft
Station
Deflection Angle
Chord Length
PC 943+33.05
0
0
944+00
1.598
66.94
945+00
3.986
99.97
946+00
6.373
99.97
947+00
8.760
99.97
948+00
11.148
99.97
949+00
13.535
99.97
950+00
15.922
99.97
951+00
18.310
99.97
952+00
20.697
99.97
953+00
23.084
99.97
PT 953+38.35
24.000
38.35
279
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Chapter 15: Geometric Design of Highway Facilities
15-14
Given two chords xy and yz of 135 ft each marked on an existing curve to determine
the radius. The perpendicular distance between y and the chord xz is 15 ft.
Determine the radius of the curve and the angle set out from xz to get a line to the
center of the curve.
y
x
w
z
C
Since it is known that the chord xy = 135 ft and yw = 15 ft, The length of xw can
be found using the Pythagorean theorem as follows:
xy2 = xw2 + yw2
xw = xy 2 − yw 2 = 135 2 − 15 2 = 134.16 ft
Solve for the radius, R, using the Pythagorean theorem, knowing the length of xw
and that xc = R and wc = R – 15.
xc2 = xw2 + wc2
R2 = 134.162 + (R-15)2
R2 = 18000 + R2 –30R + 225
30R = 18225
R = 607.50 ft
Solve for the deflection angle from xz to the center of the curve,
cos δ = (xw/xc) = (134.16/607.50) = 0.220846
δ = 77.2412˚
15-15
Given a compound circular curve with radii of 600 ft. and 450 ft. designed to
connect two tangents deflecting by 75°, determine the central angles and the
corresponding chord lengths for setting out the curve if the central angle of the first
curve is 45° and the PCC is at station (675+35.25).
Δ = Δ 1 + Δ2
75 = 45 + Δ2
Δ2 = 30
L1 = RΔ1 (π/180)
L1 = 600 (45) (π/180)
L1 = 471.24
t1 = R1 tan (Δ1/2)
t1 = 600 tan (45/2)
t1 = 248.53
t2 = R2 tan (Δ2/2)
t2 = 450 tan (30/2)
t2 = 120.58
L2 = RΔ2 (π/180)
L2 = 450 (30) (π/180)
L2 = 235.62
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280
Chapter 15: Geometric Design of Highway Facilities
PC = PCC - L1
PT = PCC + L2
PC = (675 + 35.25) - (4 + 71.24)
PT = (675 + 35.25) + 2 + 35.62)
PC = 670 +64.01
PT = 677 + 70.87
For curve 1 (R = 600 ft), using Equation 15.20:
D = 5729.6/R = 5729.6 / 600
D = 9.5493°
D/2 = 4.7747°
L1 = 671 - (670+64.01) = 35.99
δ1/l1 = Δ/L
δ1 = (35.99)(45)/471.24
δ1 = 3.4368°
l2 = 35.25
δ2/l2 = Δ/L
δ2 = (35.25)(45)/471.24
δ2 = 3.3661°
For curve 1:
Station
Deflection Angle
Chord Length
670+64.01
0
0
671
1.7184
35.99
672
6.4931
99.89
673
11.268
99.89
674
16.042
99.89
675
20.817
99.89
675+35.25
22.50
35.25
For curve 2 (R = 450'), using Equation 15.20
D = 5729.6/R
D = 12.732°
D = 5729.6 / 450
D/2 = 6.3662°
281
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Chapter 15: Geometric Design of Highway Facilities
L1 = 676 - (675+35.25) = 64.75
δ1/l1 = Δ/L
δ1 = (70.87)(30)/235.62
δ1 =8.2442
δ1/2 = 4.1221º
l2 = 70.87
δ2/l2 = Δ/L
δ2 = (70.87)(30)/235.62
δ2 = 9.0234º
For curve 2:
Station
Deflection Angle
Chord Length
675+35.25
0
0
676+00
4.1221
64.69
677+00
10.488
99.79
677+70.87
15.00
70.80
15-16
Given an arterial road connected to a frontage road by a reverse curve with parallel
tangents. The distance between the centerline of the two tangent sections is 60 ft.
The PC of the curve is located at station (38+25.31) and the deflection angle is 25°
Determine: the station of PT.
Solve for the radius of the reverse curves, R,
60
d
= 320.20 ft
R=
=
2(1 − cos Δ) 2(1 − cos 25°)
Solve for the length of each of the reverse curves using Equation 15.26,
L = RΔπ / 180 = 320.20(25)(3.1415926) / 180 = 139.71 ft
The station of the PT can be found by adding 2 times the length of each reverse
curve to the station of the PC,
PT = (38+25.31) + (2+79.42)
PT = 41+04.73
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282
Chapter 15: Geometric Design of Highway Facilities
15-17
For the data provided in Problem 15-1, determine (a) minimum radius of horizontal
curvature, (b) minimum length of crest vertical curves, and (c) minimum length of
sag vertical curves, for the maximum grade.
(a) Find the minimum radius
R = u2 / [ 15(e + fs)]
R = (302/15) (1/(e + fs))
R = 60 / (e + fs)
Assuming emax = 0.08 and fs = 0.16
Rmin = 60 / (0.08+0.16) = 250 ft
(b) Using Equation 15.14, L= kA
From Table 15.5, k = 19
L = 19 (10-(-10))
L = 380 ft
(c) Using Equation 15.14, L = kA
From Table 15.6, k = 37
L = 37(10 - (-10))
L = 740 ft
15-18
Repeat Problem 15-17 for an urban freeway in rolling terrain.
(a) Find the minimum radius
R = u2 / [ 15(e + fs)]
R = (502/15) (1/(e + fs))
R = 166.7 / (e + fs)
Assuming emax = 0.08 and fs = 0.14
Rmin = 166.7 / (0.08+0.14) = 758 ft
(b) Using Equation 15.12, L = kA
From Table 15.5, k = 84
L = 84(5- (-5))
L = 840 ft
(c) Using Equation 15.12, L = kA
From Table 15.6, k = 96
L = 96(5- (-5))
L = 960 ft
283
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Chapter 15: Geometric Design of Highway Facilities
15-19
For the data provided in Problem 15-3, determine (a) minimum length of crest
vertical curves, and (b) minimum length of sag vertical curves, for the maximum
grade, (c) maximum superelevation, and (d) maximum degree of curve (use fs =
0.15) (Degrees-minutes-seconds).
(a) Using Equation 15.12, L= kA
From Table 15.5, k = 44
L = 44 (8-(-8))
L = 704 ft
(b) Using Equation 15.12, L = kA
From Table 15.6, k = 64
L = 64(8 - (-8))
L = 1024 ft
(c) Maximum superelevation
Maximum values of superelevation can be as high as 0.12 but vary from
state to state. In Virginia, emax = 0.08 for rural roads.
(d) Maximum degree of curve
Find the minimum radius
R = u2 / [15(e + fs)]
R = (402/15) (1/(e + fs))
R = 106.7 / (e + fs)
Assuming emax = 0.08 and fs = 0.15
Rmin = 106.7 / (0.08+0.15) = 463 ft
15-20
Describe the four methods by which superelevation can be attained on a curved
section of roadway.
One of four methods can be used to achieve superelevation on undivided
highways:
1. A crowned pavement is rotated about the profile of the centerline - the
outside edge of the pavement is raised relative to the centerline until the
outer half of the cross section is horizontal. The outer edge is then raised
by an additional amount to obtain a straight cross section. The whole
cross section is then rotated as a unit about the centerline profile until the
full superelevation is achieved.
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284
Chapter 15: Geometric Design of Highway Facilities
2. A crowned pavement is rotated about the profile of the inside edge - the
centerline profile is raised with respect to the inside pavement edge to
obtain half the required change, while the remaining half is achieved by
raising the outside pavement edge with respect to the profile of the
centerline.
3. A crowned pavement is rotated about the profile of the outside edge similar to method 2 except the change is effected below the outside edge
profile.
4. A straight cross-slope pavement is rotated about the profile of the
outside edge.
Superelevation is achieved on divided highways by one of 3 methods.
1. The whole cross section, including the median, is rotated about the
centerline of the median as a plane section.
2. Each pavement section is rotated separately around the median edges
while keeping the median in a horizontal plane.
3. The two pavements are treated separately, resulting in variable elevation
differences between the median edges.
15-21
Determine the distance required to transition pavement cross-slope from a normal
crown section with a normal crown cross-slope of 2% to superelevation of 6% on a
two-lane highway with a design speed of 50 mi/h.
First, determine the length of superelevation runoff required to transition from a
cross-slope of zero on the outside lane to full superelevation, using Table 15.12.
Lr = 144 ft
Next, determine the tangent runout required using Equation 15.40,
e
⎛ 0.02 ⎞
Lt = NC Lr = ⎜
⎟(144) = 48 ft
ed
⎝ 0.06 ⎠
Then, the determine superelevation transition length required by summing the
superelevation runoff and tangent runout.
Required transition = 144 + 48 = 192 ft.
285
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Chapter 15: Geometric Design of Highway Facilities
15-22
A building is located 19 ft from the centerline of the inside lane of a curved section
of highway with a 400 ft radius. The road is level; e = 0.10. Determine the
appropriate speed limit (to the nearest 5 mi/h) considering the following conditions:
stopping sight distance and curve radius.
To determine sight distance past the building located on the inside of the curve,
use Figure 15.26 and a rearrangement of Equation 15.43.
S=
400
400 − 19
R
R−m
cos −1 [
]=
cos −1 [
] = 248 ft
28.65
400
28.65
R
Use Equation 3.27 for stopping sight distance to determine the appropriate speed.
Assume a = 11.2 ft/sec2, perception-reaction time = 2.5 sec, and that the road is
level.
u2
SSD = 1.47ut +
= 248 = (1.47)(2.5)u + u2 / (30)(0.35)
a
30( ± G )
g
2
0.095238 u + 3.675 u – 248 = 0
u = 35.3 mi/h
Speed limits are posted in increments of 5 mi/h; the appropriate speed is 35 mi/h.
To determine the appropriate speed based on curve radius, use Equation 15.24,
R = u2 / [15(e + fs)]
Assume fs = 0.15 (appropriate for u = 40 mi/h)
400 = u2 / [15(0.10+0.15)]
u = 38.7 mi/h
Speed limits are posted in increments of 5 mi/h; the appropriate speed is 35 mi/h.
15-23
Describe the factors that must be taken into account in the design of bicycle paths.
The design criteria for bicycle paths are somewhat similar to those for
highways, but some of these criteria are governed by bicycle operating
characteristics, which are significantly different from those of automobiles.
Important design considerations for a safe bicycle path include the path width, the
design speed, the horizontal alignment, and the vertical alignment. The minimum
width specified by AASHTO for a 2-way path is 10 ft. and 5 ft. for a 1-way path.
Uniform graded shoulders of at least 2 ft should be provided on either side.
AASHTO recommends a design speed of 20 mph for paved paths or 15 mph for
unpaved paths. Design speeds should be increased for grades greater than 4%.
Superelevation rates for bicycle paths vary from 2% to 5% and coefficients of
side friction should vary from 0.30 to 0.22 for paved paths and 0.15 to 0.11 for
unpaved paths. Grades should not exceed 5%.
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Chapter 15: Geometric Design of Highway Facilities
15-24
Given an average speed of 20 mi/h, a maximum superelevation of 2%, and a total
change in grade of 10%, determine the minimum radius of horizontal curvature and
the minimum length of the vertical curve.
The radius can be found:
R = u2 / [15(e + fs)]
Assume fs = 0.30
R = (20)2 / [15(0.02 + 0.30)]
R = 83.33 ft
From Figure 15.32, for a design speed of 20 mi/h, L = 180 ft
15-25
Given an available area that is 400 ft by 500 ft, design a suitable parking lot layout
for achieving each of the following objectives:
(a) Provide the maximum number of spaces.
(b) Provide the maximum number of spaces while facilitating traffic
circulation by providing one way flow on each aisle.
(a) Provide the maximum number of spaces (accomplished with spaces at a 90°
angle).
Assume space dimensions:
Stall width = 8.5 ft
Stall length = 18 ft
Aisle width = 24 ft
Aisle + row width = 24+18+18 = 60 ft
Allow 25 ft at each row end for traffic circulation,
Number of aisles = 400 ft / 60 ft = 6.67
There can be 6 aisles with 2 rows per aisle
Number of spaces per row = [500 ft – (2)(25 ft)] / (8.5 ft/space) = 52
Total number of spaces = (6 aisles)(2 rows/aisle)(52 spaces/row) = 624 spaces
(b) For optimum traffic circulation, spaces are at a 45° angle (herringbone
design).
Allow 25 ft at each row end for traffic circulation,
Number of aisles with 2 rows each = 500 ft / 45 ft = 11.11
There can be 11 aisles with 2 rows per aisle
Number of spaces per row = [400 ft – (2)(25 ft)] / (14 ft/space) = 25
Total number of spaces = (11 aisles)(2 rows/aisle)(25 spaces/row) = 550 spaces
287
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Chapter 16
Highway Drainage
16-1
What are the two sources of water a highway engineer is primarily concerned with?
Briefly describe each.
The highway engineer is concerned primarily with two sources of water.
The first source, surface water, is precipitation that occurs as rain or snow. Some
of this is absorbed into the soil, and the remainder remains on the surface of the
ground and should be removed from the highway pavement as surface drainage.
The second source, ground water, is that which flows in underground streams;
this is referred to as subsurface drainage. This may become important in
highway cuts or at locations where a high water table exists near the pavement
structure.
16-2
Briefly describe the main differences between surface drainage and subsurface
drainage.
Surface drainage is the system that drains surface water away from the
surface of the pavement. This surface water consists of the precipitation in the
form of rain, ice and snow, less that which is absorbed into the soil. Surface
drainage system features are incorporated into the overall design of the highway
with the objective of ultimately directing all surface runoff to natural waterways,
including:
•
Transverse slopes to facilitate the removal of water from the pavement
surface in the shortest possible time.
•
Longitudinal slopes to facilitate the provision of adequate slopes in the
longitudinal channels.
•
Longitudinal channels or ditches to collect the surface water that runs off
from pavement surfaces, subsurface drains, and other areas of the highway
right-of-way.
•
Curbs and gutters.
•
Drainage structures such as bridges and culverts.
Subsurface drainage is the system that drains groundwater from the
highway pavement structure. The groundwater may be in one or more of the
following forms: (1) surface water that has permeated through cracks and joints in
the pavement to the underlying strata, (2) water moving upward through the
underlying strata due to capillary action, and (3) water existing in the natural
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Chapter 16: Highway Drainage
ground below the water table, usually referred to as groundwater. A subsurface
drainage system usually consists of:
•
Longitudinal drains usually consisting of pipes laid in trenches within the
pavement structure and parallel to the center line.
•
Transverse drains placed below the pavement and in a direction
perpendicular to the center line.
•
Horizontal drains in cuts and embankments to relieve pore pressure.
•
A drainage blanket, which is a layer of material with a high coefficient of
permeability beneath or within the pavement structure placed beneath or
within the pavement structure to facilitate the flow of subsurface water
away from the pavement.
16-3
What are the two main disadvantages of using turf cover on unpaved shoulders?
The two main disadvantages of using turf cover on unpaved shoulders are that turf
cover cannot resist continued traffic and that it loses firmness under conditions of
heavy rain.
16-4
Briefly describe the three properties of rainfall that primarily concern highway
engineers.
Highway engineers are primarily concerned with three properties of
rainfall: intensity, duration, and frequency. The rate of fall (typically expressed in
inches per hour) is known as intensity. The length of time for a given intensity is
known as duration. The probable number of years that will elapse before a given
combination of intensity and duration will be repeated is known as frequency.
16-5
What is meant by a: (a) 10 year storm, (b) 50 year storm, (c) 100 year storm, and (d)
500 year storm?
(a) A 10-year storm is a storm of given intensity and duration for which the
probability that it will occur in a one year period is 1 in 10.
(b) A 50-year storm is a storm of given intensity and duration for which the
probability that it will occur in a one year period is 1 in 50.
(c) A 100-year storm is a storm of given intensity and duration for which the
probability that it will occur in a one year period is 1 in 100.
(d) A 500-year storm is a storm of given intensity and duration for which the
probability that it will occur in a one year period is 1 in 500.
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Chapter 16: Highway Drainage
16-6
Define the following: (a) drainage area, (b) run-off coefficient (C), (c) travel time (ti),
and (d) time of concentration (tc).
(a) Drainage area is that area of land that contributes to the runoff at the point
where the channel capacity is to be determined. This area is normally determined
from a topographic map.
(b) Runoff coefficient (C) is the ratio of runoff to rainfall for the drainage area.
The runoff coefficient depends on the type of ground cover, the slope of the
drainage area, storm duration, prior wetting, and the slope of the ground. For
small drainage areas, typically only type of ground cover and slope of the
drainage area are considered in determination of runoff coefficients.
(c) Travel time (ti) is the ratio of flow length to average flow velocity for a
specific watershed.
(d) Time of concentration (tc) is the time required for runoff to flow from the most
distant point, along hydraulic channels, to the point of interest in the watershed
and is therefore the sum of travel times for the various elements within the
watershed.
16-7
A 170 acre rural drainage area consists of four different watershed areas as follows:
Steep grass covered area = 40%
Cultivated area = 25%
Forested area = 30%
Turf meadows = 5%
Using the rational formula, determine the runoff rate for a storm of 100-year
frequency. Use Table 16.2 for runoff coefficients. Assume that the rainfall intensity
curves in Figure 16.2 are applicable to this drainage area and the following land
characteristics apply. Use Figure 16.4 to calculate average velocity using "fallow or
minimum tillage cultivation" ground cover. Overland flow length = 0.5 miles.
Average slope of overland area = 3%
Determine the runoff rate for a 100-year storm.
Determine the runoff coefficients for each of the four groundcover-type areas
from Table 16.2.
Type of Cover
% Area
Coefficient
Steep grass covered
40
0.60
Cultivated area
25
0.30
Forested area
30
0.20
Turf meadow
5
0.25
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Chapter 16: Highway Drainage
Calculate the weighted run-off coefficient using Equation 16.1.
n
Cw =
∑C A
i =1
n
i
i
∑A
i =1
i
Cw = 170 [(0.40)(0.60) + (0.25)(0.30) + (0.30)(0.20) + (0.05)(0.25)] / 170
Cw = 0.3875
Estimate the average velocity using Figure 16.4. This area can be considered as
Fallow or minimum tillage cultivation.
v = 0.8 ft/sec
Determine travel time using Equation 16.2,
Ti = L / 3600 V
Ti = (0.5)(5280) / (3600)(0.8)
Ti = 0.917 h
Since only one segment is being investigated, Ti = Tc.
Determine the rainfall intensity, using Figure 16.2.
I = 3.0 in/h
Determine the runoff flow rate using the rational formula (Equation 16.4).
Q = CIA
Q = (0.3875)(3.0)(170)
Q = 197.63 ft3/sec
Therefore, the runoff flow rate for this watershed will be 198 ft3/sec.
16-8
Compute rate of runoff using the rational formula for a 225 acre rural drainage
area consisting of two different watershed areas as follows:
Steep grass area = 45%
Cultivated fields = 55%
If the time of concentration for this area is 2.4 hours, determine the runoff rate for a
storm of 50-year frequency. Use the rainfall intensity curves in Figure
16.2. Use Table 16.2 for runoff coefficients.
Determine the runoff coefficients for each of the two groundcover-type areas
from Table 16.2.
Type of Cover
% Area
Coefficient
Steep grass covered
45
0.60
Cultivated fields
55
0.30
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Chapter 16: Highway Drainage
Calculate the weighted run-off coefficient using Equation 16.1,
n
Cw =
∑C A
i =1
n
i
i
∑A
i =1
i
Cw = 225 [(0.45)(0.60) + (0.55)(0.30)] / 225
Cw = 0.435
Determine the rainfall intensity, using Figure 16.2.
I = 1.45 in/h
Determine the runoff flow rate using the rational formula (Equation 16.4).
Q = CIA
Q = (0.435)(1.45)(225)
Q = 141.9 ft3/sec
Therefore, the runoff flow rate for this watershed will be 142 ft3/sec.
16-9
Using the TR-55 method, determine the depth of runoff for a 24-hour, 100-year
precipitation event of 9 inches if the soil can be classified as group B and the
watershed is contoured pasture with good hydrologic condition and an antecedent
soil condition III.
Determine depth of runoff for a 24-hour, 100-year, precipitation of 9 inches.
First determine CNII value from Table 16.3 for contoured pasture with good
hydrologic condition and soil group B.
CNII = 35
Since the antecedent moisture condition of the soil is condition III, now find the
CNIII value for this soil condition from Table 16.5.
CNIII = 55
Determine the potential maximum retention after runoff begins, using Equation
16.7.
S = (1000 / CN) – 10 = (1000 / 55) – 10
S = 8.18 in
Determine depth of runoff using Equation 16.5.
((9) − (0.2)(8.18)) 2
( P − 0.2 S ) 2
=
h=
9 + (0.8)(8.18)
P + 0.8S
h = 3.49 in
Therefore, the depth of runoff will be 3.5 inches.
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Chapter 16: Highway Drainage
16-10
Determine the depth of runoff by the TR-55 method for a 24-hour, 100-year
precipitation of 9 inches for an antecedent moisture condition III, if the following
land uses and soil conditions exist.
Area Fraction
0.30
0.25
0.20
0.15
0.10
Land/Use Condition
Soil Group
Wooded/fair condition
D
Small grain/straight row/good condition
D
Pasture/contoured/fair condition
D
Meadow/good condition
D
Farmstead
D
Since different land uses are present, a weighted CN value must be calculated.
The CNII values can be found in Table 16.3, and the corresponding CNIII values in
Table 16.5, as shown below.
Land-use Type
% Area
CNII
CNIII
Wooded/fair
0.30
79
93.4
Small grain/straight/good
0.25
87
97.4
Pasture/contoured/fair
0.20
83
95.8
Meadow/good
0.15
78
92.8
Farmstead
0.10
86
97.2
Weighted CNIII =
0.30)(93.4)+(0.25)(97.4)+(0.20)(95.8)+(0.15)(92.8)+(0.10)(97.2)
Weighted CNIII = 95.17
Determine the potential maximum retention after runoff begins, using Equation
16.7.
S = (1000 / CN) – 10
S = (1000 / 95.16) – 10
S = 0.51 in
Determine depth of runoff using Equation 16.5.
((9) − (0.2)(0.51)) 2
( P − 0.2 S ) 2
=
h=
9 + (0.8)(0.51)
P + 0.8S
h = 8.42 in
Therefore, the depth of runoff will be 8.42 inches.
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294
Chapter 16: Highway Drainage
16-11
Determine the peak discharge that will occur for the conditions indicated in
Problem 16-10 if the drainage area is 0.5 mi2, and the time of concentration is 1.6
hours.
From Problem 16-10, the depth was found to be h = 8.42 inches.
From Figure 16.5, for a time concentration of 1.6 hours:
q'p = 240 ft3/sec/mi2/in
Determine the peak discharge using Equation 16.6.
qp = q'p (A)(h)
qp = (240)(0.5)(8.42)
qp = 1010 ft3/sec
Therefore, the peak discharge that will occur will be approximately 1010 ft3/sec.
16-12
What is the difference between supercritical and subcritical flow? Under what
conditions will either of these occur?
Flows in channels can be tranquil or rapid. Flow is considered to be
critical when the depth of flow is the critical depth (i.e. the depth at which the
flow in a channel changes from tranquil to rapid). At this depth the specific
energy is minimized. When the flow depth is less than the critical depth, the flow
is known as supercritical. This type of rapid, turbulent flow is prevalent in steep
flumes and mountain streams. When the flow depth exceeds the critical depth,
the flow is subcritical, characterized by slower velocity and tranquil flow.
Subcritical flow is typically found in shallow broad channel of nearly flat slopes.
16-13
A trapezoidal channel of 2:1 side slope and 5 ft bottom width, discharges a flow of
275 ft3/sec. If the channel slope is 2.5% and the Manning coefficient is 0.03,
determine (a) flow velocity, (b) flow depth, and (c) type of flow.
Using the graphical solution of Manning’s equation for a trapezoidal channel with
a side slope of 2:1 and bottom width of 5 ft, provided in Figure 16.6,
(a) Flow velocity = 11 ft/sec
(b) Flow depth = 2.6 ft
(c) Type of flow: The intersection of the discharge flow and the 2.5% slope lies
above the critical curve, thus, the flow is supercritical.
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Chapter 16: Highway Drainage
16-14
A 6 ft wide rectangular channel lined with rubble masonry is required to carry a
flow of 300 ft3/sec. If the slope of the channel is 2% and n = 0.015, determine (a)
flow depth, (b) flow velocity, and (c) type of flow.
Using the graphical solution of Manning’s equation for a rectangular channel of a
width 6 ft and n = 0.015, provided in Figure 16.7,
(a) Flow depth = 2.8 ft
(b) Flow velocity = 18 ft/sec
(c) Type of flow: The intersection of the discharge flow line and the 2% slope
line lies above the critical curve, thus, the flow is supercritical.
16-15
For the conditions given in Problem 16-14, determine the critical depth and the
maximum channel slope at which subcritical flow can occur.
Using Figure 16.7, the critical flow curve intersects the vertical line for a
flow of 300 ft3/sec at a depth of approximately 4.25 ft. The associated channel
slope is approximately 0.007 (0.7%), which is the maximum slope at which
subcritical flow will occur.
16-16
Determine a suitable rectangular flexible lined channel to resist erosion for a
maximum flow of 20 ft3/sec if the channel slope is 2%. Use channel dimensions
given in Problem 16-14.
Using Figure 16.7, the normal depth of flow = 0.45 ft and the flow velocity =
7 ft/sec.
To allow a free board of 1 ft, the revised channel dimension is:
1.45 ft x 6 ft
Determine if a channel lined with jute mesh is suitable to prevent erosion.
Using Figure 16.9, determine dmax
dmax = 0.62 ft
Determine the hydraulic radius, R
R = a /p
R = (0.62)(6.0) / [(2)(0.62) + (6.0)]
R = 0.51 ft
Determine the flow velocity, V, using Figure 16.13.
V = 61.53 R1.028 S00.431
V = 61.53(0.51)1.028(0.02)0.431
V = 5.75 ft / sec
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296
Chapter 16: Highway Drainage
Determine the flow.
Q = VA
Q = (5.75)((0.62)(6.0))
Q = 21.4 ft3 / sec
Since the allowable flow rate for this material is slightly greater than the flow
required, the channel design using jute mesh is adequate.
16-17
A trapezoidal channel of 2:1 side slope and 5 ft bottom width is to be used to
discharge a flow of 200 ft3/sec. If the channel slope is 2% and the Manning
coefficient is 0.015, determine the minimum depth required for the channel. Is the
flow supercritical or subcritical?
Using the graphical solution of Manning’s equation for a trapezoidal channel with
a side slope of 2:1 and bottom width of 5 ft, provided in Figure 16.6,
Qn = flow * manning coefficient
Qn = (200)(0.015)
Qn = 3.0
From Figure 16.6, depth of flow = 1.6 ft.
The intersection of the discharge flow line and the 2% slope line lies above the
critical curve, thus, the flow is supercritical.
Allowing a free board of 1 ft, the minimum depth required for the channel is 2.6
ft.
16-18
Determine whether a 5 ft x 5 ft reinforced concrete box culvert with 45o flared
wingwalls and beveled edge at top of inlet carrying a 50-year flow rate of 200 ft3/sec
will operate under inlet or outlet control for the following conditions. Assume ke =
0.5.
Design headwater elevation (ELhd) = 105 ft
Elevation of stream bed at face of invert = 99.55 ft
Tailwater depth = 4.75 ft
Approximate length of culvert = 200 ft
Slope of stream = 1.5%
n = 0.012
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Chapter 16: Highway Drainage
Determine the required conditions for inlet control.
Calculate the flow rate per ft of width.
Q / NB = 200/5 = 40.0 ft3/sec/ft
Using Figure 16.16, draw a line connecting 5 ft to 40 ft to obtain the headwater
depth at culvert face (HW / D).
HW / D = 1.19 ft/ft
Calculate the required headwater, HW.
HW = (HW / D) * depth of culvert
HW = 1.19 * 5
HW = 5.95 ft
Neglect the approach velocity head in this problem. Therefore;
HWi = 5.95 ft
Calculate the required depression. Use Equation 16.13 to determine the design
headwater depth, HWd.
HWd = ELhd - ELsf
HWd = 105 - 99.55
HWd = 5.45 ft
Determine the fall using Equation 16.14.
Fall = HWi - HWd
Fall = 5.95 - 5.45
Fall = 0.50 ft
Calculate the culvert invert elevation
invert elevation = 99.55 - 0.50 = 99.05 ft
Determine the required conditions for outlet control.
Determine the critical depth with Q / B = 200 / 5 = 40.0, from Figure 16.23.
dc = 3.7 ft
Calculate depth from outlet invert to hydraulic grade line and determine if TW is
greater.
(dc + D) / 2 = (3.7 + 5) / 2 = 4.35 ft
Since TW is not greater, use ho = 4.75 ft
Determine the total head loss (H) from Figure 16.21
H = 2.2 ft
Calculate the required outlet headwater elevation using Equation 16.23.
ELho = Elo + H + ho
ELho = (99.05 - (0.015)( 200)) + 2.2 + 4.75
ELho = 103.00 ft
The required outlet headwater elevation (103 ft) is less than the design headwater
elevation (105 ft); therefore the 5 ft x 5 ft culvert is acceptable and inlet control
governs.
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Chapter 16: Highway Drainage
16-19
Repeat Problem 16-18 using a 6.5 ft diameter circular pipe culvert with ke = 0.5.
Determine the required conditions for inlet control.
Using Figure 16.16, draw a line connecting 78 in to 200 ft3/sec to obtain the
headwater depth at culvert face (HW / D).
HW / D = 0.88 ft/ft
Calculate the required headwater, HW.
HW = (HW / D) * depth of culvert
HW = (0.88)(6.5 ft)
HW = 5.72 ft
Neglect the approach velocity head in this problem. Therefore;
HWi = 5.72 ft
Calculate the required depression. Use Equation 16.13 to determine the design
headwater depth, HWd.
HWd = ELhd - ELsf
HWd = 105 - 99.55
HWd = 5.45 ft
Determine the fall using Equation 16.14.
Fall = HWi - HWd
Fall = 5.72 - 5.45
Fall = 0.27 ft
Calculate the culvert invert elevation
invert elevation = 99.55 – 0.27 = 99.28 ft
Determine the required conditions for outlet control.
Determine the critical depth with Q / B = 200 / 6.5 = 30.8, from Figure 16.24.
dc = 3.75 ft
Calculate depth from outlet invert to hydraulic grade line and determine if TW is
greater.
(dc + D) / 2 = (3.75 + 6.5) / 2 = 5.13 ft
Since TW is greater, use ho = 5.13 ft
Determine the total head loss (H) from Figure 16.22
H = 1.05 ft
Calculate the required outlet headwater elevation using Equation 16.23.
ELho = Elo + H + ho
ELho = (99.28 - (0.015)( 200)) + 1.05 + 5.13
ELho = 102.46 ft
The required outlet headwater elevation (102.46 ft) is less than the design
headwater elevation (105 ft); therefore the 6.5 ft diameter circular culvert is
acceptable and inlet control governs.
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Chapter 16: Highway Drainage
16-20
Determine the ground water infiltration rate for a new two-lane pavement with the
following characteristics:
Lane width = 12 ft
Shoulder width = 10 ft
Length of contributing transverse cracks (Wc) = 20 ft
Rate of infiltration (Kp) = 0.05 ft3/day/ft2
Spacing of transverse cracks = 30 ft
Determine the width of granular base.
W = (number of lanes) * (lane width) + (shoulder width) * 2
W = (2 * 12) + (10 * 2)
W = 44 ft
Since this is a new pavement,
Nc = N + 1
Nc = 3
Use Equation 16.24 to determine infiltration rate (assume Ic = 2.4):
qi = I c (
Nc
W
+ c ) +K p
W WC s
qi = (2.4) [(3 / 44) + (20 / (44)(30))] + 0.05
qi = 0.25 ft3 / day / ft2
Therefore, the ground water infiltration rate will be 0.25 ft3/day/ft2 for this new
pavement section.
16-21
In addition to the infiltration determined in Problem 16-20, ground water seepage
due to gravity also occurs. Determine the thickness of a suitable drainage layer
required to transmit the net inflow to a suitable outlet.
Thickness of subgrade below drainage pipe = 12 feet
Coefficient of permeability of native soil = 0.35 ft/day
Height of water table above impervious layer = 21 feet
Slope of drainage layer = 2%
Permeability of drainage area = 2,000 ft/day
Length of flow path = 44 ft
Calculate the amount of drawdown.
H - Ho = 21 - 12 = 9 ft of drawdown
Determine the radius of influence.
Li = 3.8 (H - Ho)
Li = (3.8) (9) = 34.2 ft
Compute the average inflow rate due to gravity drainage:
((Li + 0.5W) / Ho ) = (34.2 + ((0.5) * (44))) / 12
((Li + 0.5W) / Ho ) = 56.20 / 12
((Li + 0.5W) / Ho ) = 4.68
W/Ho = 44/12 = 3.67
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Chapter 16: Highway Drainage
From Figure 16.34, determine:
k * (H - Ho ) / 2q2 = 1.7
q2 = k * (H - Ho ) / (2 * (1.7))
q2 = (0.35 * 9) / 3.4
q2 = 0.93
qg = q2 / 0.5W
qg = 0.93 / (0.5 * 44)
qg = 0.042 ft3/day/ft2
Calculate the net inflow.
From Problem 16-20, qi = 0.25 ft3 / day / ft2
qn = qi + qg
qn = 0.25 + 0.042
qn = 0.292
p = qn / kd
p = 0.292 / 2,000
p = 1.46 x 10-4
From Figure 16.38:
L / Hm = 200
Hm = L / 200
Hm = 44 / 200
Hm = 0.22 ft = 2.64 in; use 3.0 inches
Therefore, the suitable drainage layer thickness should be 3.0 inches.
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Chapter 17
Soil Engineering for Highway Design
17-1
Determine the void ratio of a soil if its bulk density is 120 lb/ft3 and it has a moisture
content of 25 percent. The specific gravity of the soil particles is 2.7. Also,
determine its dry density and degree of saturation.
Solve for γd, dry density, using Equation 17.11.
γd = γ / (1 + w)
γd = 120 / (1+ 0.25)
γd = 96 lb/ft3
Solve for void ratio, e, using Equation 17.13. This equation can be modified to
use dry density when the degree of saturation is zero (no moisture is present) as
follows:
γd = (Gs γw ) / 1 + e
1 + e = Gs γw / γd
e = ( Gs γw / γd ) - 1
e = ((2.7)(62.4) / 96) - 1
e = 0.755
Solve for the degree of saturation, S.
S = w Gs / e
S = (0.25)(2.7) / 0.755
S = 0.894
17-2
A soil has a bulk density of 135 lb/ft3 and a dry density of 120 lb/ft3, and the specific
gravity of the soil particles is 2.75. Determine: (a) moisture content, (b) degree of
saturation, (c) void ratio, and (d) porosity.
(a) Solve for moisture content, w, using Equation 17.11.
γd = γ / (1 + w)
1 + w = γ / γd
w = (γ / γd ) - 1
w = (135 / 120) - 1
w = 0.125 = 12.5 %
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Chapter 17: Soil Engineering for Highway Design
(b) Solve for void ratio, e, using Equation 17.13. This equation can be modified
to use dry density when the degree of saturation is zero (no moisture is present) as
follows:
γd = (Gs γw ) / 1 + e
1 + e = Gs γw / γd
e = ( Gs γw / γd ) - 1
e = ((2.75)(62.4) / 120) - 1
e = 0.430
(c) Solve for the degree of saturation, S.
S = w Gs / e
S = (0.125)(2.75) / 0.430
S = 0.799 = 79.9%
(d) Solve for porosity, n, using Equation 17.4.
n = e / (1+e)
n = (0.430)/(1+0.430)
n = 0.301
17-3
The weight of a sample of saturated soil before drying is 3 lb and after drying is 2.2
lb. If the specific gravity of the soil particles is 2.7, determine: (a) moisture content,
(b) void ratio, (c) porosity, (d) bulk density, and (e) dry density.
(a) Solve for moisture content, w, using Equation 17.6.
Calculate the weight of the water in the soil sample.
W = Ww + Ws
Ww = 3- 2.2
Ww = 0.8 lbs
w = (Ww / Ws) (100)
w = (0.8 / 2.2) (100)
w = 36.36 %
(b) Solve for void ratio, e, using Equation 17.3. First, the volume of water and the
volume of solid must be determined.
γw = Ww / Vw
Vw = Ww / γw
Vw = 0.8 / 62.4
Vw = 0.0128 ft3
Gs γw = Ws / Vs
Vs = Ws / (Gs γw )
Vs = 2.2 / (2.7)(62.4)
Vs = 0.0131 ft3
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Chapter 17: Soil Engineering for Highway Design
e = Vv / Vs = (Vw + Va) / Vs
Va = 0 (since the sample is saturated, no air voids are assumed to
be present).
e = 0.0128 / 0.0131
e = 0.977
(c) Solve for porosity, n, using Equation 17.4.
n = e / (1+e)
n = (0.977)/(1+0.977)
n = 0.494
(d) Calculate the bulk density, γ, using Equation 17.9
γ=W/V
γ = 3 / (0.0128 + 0.0131)
γ = 3 / 0.0259
γ = 115.8 lb/ft3
(e) Solve for the dry density, γd , using Equation 17.11
γd = Ws / V
γd = 2.2 / (0.0128 + 0.0131)
γd = 84.9 lb/ft3
17-4
A moist soil has a moisture content of 10.2 percent, weighs 40.66 lb, and occupies a
volume of 0.33 ft3. The specific gravity of the soil particles is 2.7. Find: (a) bulk
density, (b) dry density, (c) void ratio, (d) porosity, (e) degree of saturation, and (f)
volume occupied by water (ft3).
(a) Solve for the bulk density, using Equation 17.9.
γ=W/V
γ = 40.66 lb / 0.33 ft3
γ = 123.2 lb/ft3
(b) Solve for the dry density, using Equation 17.11.
γd = γ / (1 + w)
γd = 123.2 / (1 + 0.102)
γd = 111.8 lb/ft3
(c) Solve for void ratio, e, using Equation 17.13.
e = (((1 + w) (Gs)(γw)) / γ) - 1
e = (((1 + 0.102) (2.7) (62.4)) / 123.2) - 1
e = 0.507
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Chapter 17: Soil Engineering for Highway Design
(d) Solve for porosity, n, using Equation 17.4.
n = e / (1 + e)
n = 0.507 / (1 + 0.507)
n = 0.337
(e) Compute the degree of saturation, S.
S = Vw / Vv = (w)(Gs) / e
S = (0.102)(2.7) / 0.507
S = 0.543
(f) Determine the volume occupied by water, using Equation 17.7.
w = Ww / Ws
Ww = (w/(1+w)) W
Ww = (0.102/1.102)(40.66) = 3.76 lb
Vw = Ww / γw = (3.76 lb)/(62.4 lb/ft3)
Vw = 0.0603 ft3
17-5
The moist weight of 0.15 ft3 of soil is 18.6 lb. If the moisture content is 17 percent
and the specific gravity of soil solids is 2.62, find the following: (a) bulk density, (b)
dry density, (c) void ratio, (d) porosity, (e) degree of saturation, and (f) the volume
occupied by water (ft3).
(a) Solve for the bulk density, using Equation 17.9
γ=W/V
γ = 18.6 lb / 0.15 ft3
γ = 124 lb/ft3
(b) Solve for the dry density, using Equation 17.11.
γd = γ / (1 + w)
γd = 124 / (1 + 0.17)
γd = 106 lb/ft3
(c) Solve for void ratio, e, using Equation 17.13.
e = (((1 + w) (Gs)(γw)) / γ) - 1
e = (((1 + 0.17) (2.62) (62.4)) / 124) - 1
e = 0.543
(d) Solve for porosity, n, using Equation 17.4.
n = e / (1 + e)
n = 0.543 / (1 + 0.543)
n = 0.352
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Chapter 17: Soil Engineering for Highway Design
(e) Compute the degree of saturation, S.
S = Vw / Vv = (w)(Gs) / e
S = (0.17)(2.62) / 0.543
S = 0.821
(f) Determine the volume occupied by water, using Equation 17.7.
w = Ww / Ws
Ww = (w/(1+w)) W
Ww = (0.17/1.17)(18.6) = 2.70 lb
Vw = Ww / γw = (2.70 lb)/(62.4 lb/ft3)
Vw = 0.043 ft3
17-6
A liquid limit test conducted in the laboratory on a sample of soil gave the following
results listed below. Determine the liquid limit of this soil from a plot of the flow
curve.
Number of Blows (N)
20
28
30
35
40
Moisture Content (%)
45.0
43.6
43.2
42.8
42.0
The liquid limit is defined as the moisture content after 25 blows of the
standardized testing equipment. From the following figure it can be seen that the
liquid limit is approximately 44%.
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Chapter 17: Soil Engineering for Highway Design
17-7
A plastic limit test for a soil showed that moisture content was 19.2%. Data from a
liquid limit test were as follows:
Number of Blows (N)
14
19
27
(a)
(b)
Moisture Content (%)
42.0
40.8
39.1
Draw the flow curve and obtain the liquid limit.
What is the plasticity index of the soil?
(a) The liquid limit is defined as the moisture content after 25 blows of the
standardized testing equipment. From the following figure it can be seen that the
liquid limit is approximately 39.5%.
(b) Determine the plasticity index using Equation 17.14.
PI = LL - PL
PI = 39.5 - 19.2
PI = 20.3
17-8
The following results were obtained by a mechanical analysis. Classify the soil using
the AASHTO classification system and give the group index.
Sieve Analysis, % Finer than No. 10: 98%, than No. 40: 81%, than No. 200:
38%; liquid limit = 42; plastic limit = 23.
Use Table 17.1 to classify this soil.
Since more than 35% of the material is passing No. 200 sieve, the liquid limit is
greater than 41, the plasticity index is greater than 11, the soil is Group A-7.
Since the plasticity index (19) is greater than the liquid limit minus 30, the
subgroup is A-7-6.
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Chapter 17: Soil Engineering for Highway Design
Determine the Group Index using Equation 17.18.
GI = (F - 35)[0.2 + 0.005(LL - 40)] + 0.01(F - 15)(PI - 10)
GI = (38 - 35)[0.2 + 0.005(42 - 40)] + 0.01(38 - 15)(19 - 10)
GI = 2.7; use GI = 3
The soil can be classified as A-7-6(3).
17-9
The following results were obtained by a mechanical analysis. Classify the soil using
the AASHTO classification system and give the group index.
Sieve Analysis, % Finer than No. 10: 84%, than No. 40: 58%, than No. 200:
8%; soil is not plastic.
Use Table 17.1 to classify this soil.
Since less than 35% of the material is passing No. 200 sieve, there is no liquid
limit, and the soil is not plastic, the soil is Group A-3, and there is no associated
group index.
17-10
The following results were obtained by a mechanical analysis. Classify the soil using
the AASHTO classification system and give the group index.
Sieve Analysis, % Finer than No. 10: 99%, than No. 40: 85%, than No. 200:
71%; liquid limit = 55; plastic limit = 21.
Use Table 17.1 to classify this soil.
Since more than 35% of the material is passing No. 200 sieve, the liquid limit is
greater than 41, the plasticity index is greater than 11, the soil is Group A-7.
Since the plasticity index (34) is greater than the liquid limit minus 30, the
subgroup is A-7-6.
Determine the Group Index using Equation 17.18.
GI = (F - 35)[0.2 + 0.005(LL - 40)] + 0.01(F - 15)(PI - 10)
GI = (71 - 35)[0.2 + 0.005(55 - 40)] + 0.01(71 - 15)(34 - 10)
GI = 23.34
Use GI = 23
The soil can be classified as A-7-6(23).
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Chapter 17: Soil Engineering for Highway Design
17-11
The following results were obtained by a mechanical sieve analysis. Classify the soil
using the Unified Soil Classification System (USCS).
Sieve Analysis, % Passing by Weight than No. 4: 30%, than No. 40: 40%,
than No. 200: 30%; liquid limit = 33; plastic limit = 12.
Use Table 17.3.
Since more than half of the material is larger than No. 200 sieve, the material is
Coarse-grained sand. Since more than half of the soil is smaller than No. 4 sieve,
the soil is either sands with fines or clean sands. Since this soil is above the "A"
line on the Atterberg plot (with LL=33 and PI=21) and the PI is greater than 7,
the group symbol is SC.
Therefore, the soil can be classified as SC - Clayey Sand.
17-12
The following results were obtained by a mechanical sieve analysis. Classify the soil
using the Unified Soil Classification System (USCS).
Sieve Analysis, % Passing by Weight than No. 4: 4%, than No. 40: 44%, than
No. 200: 52%; liquid limit = 29; plastic limit = 11.
Use Table 17.3.
Since the less than half of the material is larger than No. 200 sieve and the Liquid
Limit is less than 50, the soil is in the silts and clays division. Since this soil is
above the "A" line on the Atterberg plot (with LL=29 and PI=11), and the PI is
greater than 7, the group symbol is CL.
The soil can be classified as CL - Sandy Clays.
17-13
The following results were obtained by a mechanical sieve analysis. Classify the soil
using the Unified Soil Classification System (USCS).
Sieve Analysis, % Passing by Weight than No. 4: 11%, than No. 40: 24%,
than No. 200: 65%; liquid limit = 44; plastic limit = 23.
Use Table 17.3.
Since the less than half of the material is larger than No. 200 sieve and the Liquid
Limit is less than 50, the soil is in the silts and clays division. Since this soil is
above the "A" line on the Atterberg plot (with LL=29 and PI=11), and the PI is
greater than 7, the group symbol is CL.
The soil can be classified as CL - Sandy Clays.
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Chapter 17: Soil Engineering for Highway Design
17-14
Following are the results of a sieve analysis:
Sieve
Percent
No.
Finer
4
100
10
92
20
82
40
67
60
58
80
38
100
22
200
6
Pan
-(a) Plot the grain-size distribution curve for this sample.
(b) Determine D10, D30, and D60.
(c) Calculate the uniformity coefficient, Cu.
(d) Calculate the coefficient of gradation, Cc.
(b) From the preceding figure, D10 = 0.100 mm
D30 = 0.160 mm
D60 = 0.279 mm
(c) The coefficient of uniformity can be calculated using Equation 17.19:
Cu = (D60 / D10)
Cu = 0.279 / 0.100
Cu = 2.79
(d) The coefficient of curvature can be calculated using Equation 17.20:
Cc = D302 / (D60)(D10)
Cc = (0.160)2 / (0.279)(0.100)
Cc = 0.92
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Chapter 17: Soil Engineering for Highway Design
17-15
Following are the results of a sieve analysis:
Sieve
Percent
No.
Finer
4
100
10
90
20
80
40
70
60
40
80
29
100
19
200
10
Pan
-(a) Plot the grain-size distribution curve for this sample.
(b) Determine D10, D30, and D60.
(c) Calculate the uniformity coefficient, Cu.
(d) Calculate the coefficient of gradation, Cc.
(b) From the preceding figure, D10 = 0.075 mm
D30 = 0.180 mm
D60 = 0.347 mm
(c) The coefficient of uniformity can be calculated using Equation 17.19:
Cu = (D60 / D10)
Cu = 0.347 / 0.075
Cu = 4.63
(d) The coefficient of curvature can be calculated using Equation 17.20:
Cc = D302 / (D60)(D10)
Cc = (0.180)2 / (0.347)(0.075)
Cc = 1.24
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312
Chapter 17: Soil Engineering for Highway Design
17-16
The results of a compaction test on samples of soil that are to be used for an
embankment on a highway project are listed below. Determine the maximum dry
unit weight of compaction and the optimum moisture content.
Sample
No.
1
2
3
4
5
Moisture
Content
4.8
7.5
7.8
8.9
9.7
Bulk Density
(lb/ft3)
135.1
145.0
146.8
146.4
145.3
For each sample, calculate the dry density and then plot the results to estimate
maximum dry unit weight and optimum moisture content, as shown below.
Dry density can be calculated using Equation 17.11.
γ d = γ / (1 + w)
w
γ
γd
1
4.8
135.1
128.9
2
7.5
145.0
134.9
3
7.8
146.8
136.2
4
8.9
146.4
134.4
5
9.7
145.3
132.5
3
Dry unit weight (lb/ft )
Sample
137
136
135
134
133
132
131
130
129
128
0
5
10
15
Moisture content (%)
From the above graph, the maximum dry unit weight of the soil is approximately
136.2 lb/ft3, and the optimum moisture content is approximately 8.0%.
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Chapter 17: Soil Engineering for Highway Design
17-17
The results of a compaction test on samples of soil that are to be used for an
embankment on a highway project are listed below. Determine the maximum dry
unit weight of compaction and the optimum moisture content.
Sample No.
1
2
3
4
5
6
Bulk Density (lb/ft3)
114.7
118.6
120.4
120.1
118.3
117.1
Moisture Content
12
14
16
18
20
21
For each sample, calculate the dry density and then plot the results to estimate
maximum dry unit weight and optimum moisture content, as shown below.
Dry density can be calculated using Equation 17.11.
γ d = γ / (1 + w)
w
γ
γd
1
12
114.7
102.4
2
14
118.6
104.0
3
16
120.4
103.8
4
18
120.1
101.8
5
20
118.3
98.6
6
21
117.1
96.8
Dry unit weight (lb/ft3)
Sample
105
104
103
102
101
100
99
98
97
96
10
15
20
25
Moisture content (%)
From the above graph, the maximum dry unit weight of the soil is approximately
104.2 lb/ft3, and the optimum moisture content is approximately 15%.
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314
Chapter 17: Soil Engineering for Highway Design
17-18
The results obtained from a seismic study along a section of the centerline of a
highway are shown below. Estimate the depths of the different strata of soil and
suggest the type of soil in each stratum.
Distance of Impulse to Geophone Time for Wave Arrival
(ft)
(10-3 sec)
25
50
75
100
125
150
175
200
225
250
20
40
60
68
74
82
84
86
88
90
Time to arrival (10 -3 sec)
From the data provided, plot the following graph:
100
80
60
40
20
0
0
100
200
300
Distance (ft)
Determine the soil type in each stratum.
u1 = 75 / (60 * 10-3)
u1 = 1,250 ft / sec
u1 = 381 m / sec
The soil in this stratum is probably sand or loess.
u2 = (150 - 75) / ((82 - 60) * 10-3)
u2 = 3,409 ft / sec
u2 = 1,039 m / sec
The soil in this stratum is probably loam or clay.
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Chapter 17: Soil Engineering for Highway Design
u3 = (250 - 150) / ((90 - 82) * 10-3)
u3 = 12,500 ft / sec
u3 = 3,810 m / sec
The soil in stratum is probably sandstone, limestone, or slate and shale.
Next, determine the depths for each layer of soil.
Sin ∝ = u1 / u2
Sin ∝ = 1,250 / 3,409
Sin ∝ = 0.367
∝ = 21.52º, cos ∝ = 0.93
OP = 35 * 10-3 sec (from graph above)
Use Equation 17.21 to determine depth,
H1 = ((35 * 10-3 sec) * 1,250) / (2 * 0.93)
H1 = 23.5 ft
Therefore, the depth for the first stratum is 23.5 ft.
Next, determine the depth for the second layer of soil.
Sin β = u2 / u3
Sin β = 3,409 / 12,500
Sin β = 0.273
β = 15.83º, Cos β = 0.962
PT = 35 * 10-3 sec (from graph above)
Use Equation 17.22 to determine depth,
H2 = ((35 * 10-3 sec) * 3,409) / (2 * 0.962)
H2 = 62.0 feet
Therefore, the depth for the second stratum is 62.0 feet.
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316
Chapter 18
Bituminous Materials
18-1
Briefly describe the process of distillation by which asphalt cement is produced from
crude petroleum. Also describe in detail how to obtain asphalt binders that can be
used to coat highly siliceous aggregates.
Asphalt cement is obtained from the distillation of crude petroleum by
either fractional or destructive distillation. Fractional distillation processes
involve the separation of the different materials in the crude petroleum without
significant changes in the chemical composition of each material. The different
volatile materials are removed at higher and higher temperatures until the
petroleum asphalt is left as residue. Either steam or vacuum is used to increase
the temperature. Destructive distillation, or cracking, involves the application of
high temperatures and pressure and results in chemical changes. Cracking
processes are used when larger amounts of the light fractions (e.g. motor oil) are
needed. The asphaltic material obtained from cracking is not widely used in
paving because it is more susceptible to weather changes.
Asphalt emulsions are produced by breaking asphalt cement into minute
particles and dispersing them in water with an emulsifier. The particles are of like
electrical charge and therefore do not coalesce. Aggregates containing a high
percentage of siliceous materials are electronegative and therefore cationic
emulsions are more effective.
18-2
Describe both the factors that influence the durability of asphalt materials and the
effect of each factor has on the materials.
The durability of an asphaltic material is dependent on its ability to resist
weathering. Some of the factors that influence weathering are oxidation,
volatilization, temperature, exposed surface area, and age hardening.
Oxidation is a chemical reaction that causes gradual and permanent hardening and
eventually considerable loss of the plastic characteristics of the asphalt material.
Volatilization is the evaporation of the lighter hydrocarbons from the asphaltic
materials which causes loss of the plastic characteristics of the material.
As temperatures increase, so do the rates of oxidation and volatilization.
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Chapter 18: Bituminous Materials
The rate of hardening is directly proportional to the ratio of the surface area to the
volume.
Age hardening occurs when an asphaltic material is heated and allowed to cool.
A gel-like structure is formed that hardens over time, significantly within the first
few hours after cooling and decreasing to an almost negligible amount in a year.
18-3
Discuss the factors to consider in producing a rapid-curing cutback asphaltic
material.
The rate of curing indicates the time that should elapse before a cutback
will attain a consistency thick enough for the binder to perform satisfactorily. The
rate of curing is affected by both inherent and external factors. The important
inherent factors are the volatility of the solvent, the quantity of solvent in the
cutback, and the consistency of the base material. In a rapid-curing material,
gasoline or naphtha is used as the solvent because each evaporates quickly and
therefore the material cures faster than with other solvents. External factors
include temperature, ratio of surface area to volume, and wind velocity across the
exposed surface. The higher any of these factors is, the higher the rate of curing.
18-4
Results obtained from laboratory tests on a sample of RC-250 asphalt cement are
given. Determine whether the properties of this material meet the Asphalt Institute
specifications for this type of material; if not, note the differences.
Property
Specification
Test Results
Meet (Y/N)
250-500 centistokes
260 cs
Y
Min 27˚C (80˚F)
75˚F
N
Distillate % to 680˚F
To 437˚F
To 500˚F
To 600˚F
Min 35
Min 60
Min 80
35
54
75
Y
N
N
Residue to 680˚F
Min 65
64
N
Ductility at 77˚F
Min 100 cm
95 cm
N
600-2400 poises
750 poises
Y
Min 99.0
95
N
Kinematic viscosity
Flash Point
Absolute viscosity at
140ºF
Solubility %
While the sample does meet some of the specifications, overall, it does not meet
specifications.
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318
Chapter 18: Bituminous Materials
18-5
Given the specifications for an asphaltic concrete mixture and the results of a sieve
analysis, determine the proportion of different aggregates to obtain the required
gradation.
Coarse aggregates: 60%
Fine aggregates:
35%
Filler:
5%
The following table shows the sieve analysis data and the computations required
to obtain the required gradation.
Passing
sieve
Retained
on sieve
¾ in.
% by Weight
Coarse
Fine
Filler
Total
½ in.
0.6(4) = 2.4
---
---
2.4
½ in.
⅜ in.
0.6(36) = 21.6
---
---
21.6
⅜ in.
No. 4
0.6(40) = 24.0
---
---
24.0
No. 4
No. 10
0.6(15) = 9.0
0.35(6) = 2.1
---
11.1
No. 10
No. 40
0.6(5) = 3.0
0.35(32) = 11.2
---
14.2
No. 40
No. 80
---
0.35(33) = 11.55
0.05(5) = 0.25
11.8
No. 80
No. 200
---
0.35(29) = 10.15
0.05(40) = 2.0
12.15
No. 200
---
---
---
0.05(55) = 2.75
2.75
60.0
35.0
5.0
100.0
Total
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Chapter 18: Bituminous Materials
18-6
Given the particle size distributions of two aggregates and the required limits of
particle size distribution for the mix, determine a suitable ratio for blending the two
aggregates to obtain an acceptable combined aggregate.
Observation of results of sieve analysis indicate that aggregate B appears to more
closely conform to the required mix for most sizes; therefore, the initial mix
design should contain more than 50% aggregate B. The initial trial will be 40%
aggregate A and 60% aggregate B. The following table shows the sieve analysis
data and the computations required to obtain the required gradation.
Sieve Size
Aggregate A
Aggregate B
Mix (%)
Required (%)
OK?
3/4 in.
(0.4)(100)
(0.6)(98)
98.8
96-100
Y
3/8 in.
(0.4)(80)
(0.6)(76)
77.6
65-80
Y
No. 4
(0.4)(50)
(0.6)(45)
47.0
40-55
Y
No. 10
(0.4)(43)
(0.6)(33)
37.0
35-40
Y
No. 40
(0.4)(20)
(0.6)(30)
26.0
15-35
Y
No. 200
(0.4)(4)
(0.6)(8)
6.4
5-8
Y
The proposed design (40% aggregate A and 60% aggregate B) is acceptable.
Note that other combinations are also possible.
18-7
Given four different types of aggregates to be used to produce a blended aggregate
for use in the manufacture of asphaltic concrete, determine the bulk specific gravity
of the aggregate mix.
Material
A
B
C
D
Percent by Weight
35
40
15
10
Bulk Specific Gravity
2.58
2.65
2.60
2.55
Use Equation 18.5 to determine bulk specific gravity.
Pca + Pfa + Pmf
G sb =
Pfa
Pmf
Pca
+
+
Gbca Gbfa Gbmf
Gsb = (0.35)(2.58) + (0.40)(2.65) + (0.15)(2.60) + (0.10)(2.55)
G = 2.61
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320
Chapter 18: Bituminous Materials
18-8
If the specific gravity of the asphalt cement used in a sample of asphalt concrete mix
is 1.01, the maximum specific gravity of the mix is 2.50, and the mix contains 6.5
percent by weight of asphalt cement, determine the effective specific gravity of the
mixture.
Use Equation 18.7.
100 − Pb
G se =
(100 / Gmm ) − ( Pb / Gb )
Pb = 6.5; Gmm = 2.5; Gb = 1.01
100 − 6.5
G se =
(100 / 2.5) − (6.5 / 1.01)
Gse= 2.79
18-9
The table below lists data used in obtaining a mix design for an asphalt paving
mixture. If the maximum specific gravity of the mixture is 2.41 and the bulk
specific gravity is 2.35, determine:
(a) the bulk specific gravity of aggregates in the mix
(b) the asphalt absorbed
(c) the effective asphalt content of the paving mixture
(d) the percent voids in the mineral aggregate VMA.
Material
Asphalt cement
Coarse aggregate
Fine aggregate
Mineral filler
Specific Gravity
1.02
2.51
2.74
2.69
Mix Proportion
6.40
52.35
33.45
7.80
(a) Use Equation 18.5 to determine bulk specific gravity.
Pca + Pfa + Pmf
G sb =
Pfa
Pmf
Pca
+
+
Gbca Gbfa Gbmf
Gsb = (52.35+33.45+7.80) / [(52.35/2.51)+(33.45/2.74)+(7.80/2.69)]
Gsb = 2.60
(b) Use Equation 18.9 to determine asphalt absorption.
G − G sb
Pba = 100 se
Gb
G se G sb
First, Gea must be calculated using Equation 18.7,
100 − Pb
G se =
(100 / Gmm ) − ( Pb / Gb )
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Chapter 18: Bituminous Materials
Gse = (100-6.4) / [(100/2.41)-(6.4/1.02)]
Gse = 2.658
G − G sb
Pba = 100 se
Gb
G se G sb
Paa = 100[(2.658-2.602) / (2.602)(2.658)]1.02
Paa = 0.83%
(c) Use Equation 18.10 to determine effective asphalt content
P P
Pbe = Pb − ba s
100
Pbe = 6.4 – (0.83/100)(93.6)
Pbe = 5.62
(d) Use Equation 18.11 to determine VMA
G P
VMA = 100 − mb s
G sb
VMA = 100 – (2.35)(93.6)/2.602
VMA = 15.5%
18-10
For hot-mix, hot-laid asphaltic concrete mixtures, if the asphaltic content is
specified as 5 to 7 percent, how is the optimum percentage determined?
To determine the optimum asphalt content, each potential mix would be
evaluated. First, the bulk specific gravity and bulk density of each mix would be
determined and then the average bulk density would be plotted against the
corresponding asphalt content. The stability and flow would also be plotted
against asphalt content. Next, the percent voids in the mineral aggregate VMA
and the percent voids in the compacted mixture are computed for each potential
mix. These values are also plotted against the corresponding asphalt percentages.
The five plots are now used to determine the optimum asphalt percentage. From
the plot of unit weight (bulk density) vs. asphalt percentage, the asphalt
percentage resulting in the highest unit weight is chosen. From the plot of
stability vs. asphalt percentage, the asphalt percentage resulting in maximum
stability is chosen. From the specified limits for percent air voids, the mean is
determined and the corresponding asphalt percentage is chosen from that plot.
The optimum asphalt percentage is then determined by averaging these values.
The properties of the mix corresponding to this asphalt percentage are determined
from the five plots and compared to the given criteria.
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Chapter 18: Bituminous Materials
18-11
The aggregate mix used for the design of an asphalt paving concrete consists of 42%
coarse aggregates, 51% fine aggregates, and 7% mineral fillers. If the respective
bulk specific gravities of these materials are 2.60, 2.71, and 2.69, and the effective
specific gravity of the aggregates is 2.82, determine the optimum asphalt content as
a percentage of the total mix using results obtained using the Marshall method as
shown in the following table. The specific gravity of the asphalt cement is 1.02.
First, compute the bulk specific gravity of each possible mix.
Bulk density, γ = Gmb (62.4)
Asphalt %
Gbcm
γ
5.5
1325.3/(1325.3-785.6) = 2.46
153.5
6.0
1330.1/(1330.1-793.3) = 2.48
154.8
6.5
1336.2/(1336.2-800.8) = 2.50
156.0
7.0
1342.0/(1342.0-804.5) = 2.50
156.0
7.5
1347.5/(1347.5-805.1) = 2.48
154.8
Next, compute VMA for each possible mix using Equation 18.11.
G P
VMA = 100 − mb s
G sb
Gbcm and Pta are known for each possible mix; however, Gbam must be computed,
using Equation 18.5, in order to find VMA.
Pca + Pfa + Pmf
G sb =
Pfa
Pmf
Pca
+
+
Gbca Gbfa Gbmf
For 5.5% asphalt content:
Gbcm = 2.46
Pta = 94.5
Pca = 0.42(94.5) = 39.7
Pfa = 0.51(94.5) = 48.2
Pmf = 0.07(94.5) = 6.6
Gbam = (39.7+48.2+6.6) / [(39.7/2.60)+(48.2/2.71)+(6.6/2.69)]
Gbam = 2.66
VMA = 100 – (2.46)(94.5) / 2.66
VMA = 12.61%
For 6.0% asphalt content:
Gbcm = 2.48
Pta = 94
Pca = 0.42(94) = 39.5
Pfa = 0.51(94) = 47.9
Pmf = 0.07(94) = 6.6
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Gbam = (39.5+47.9+6.6) / [(39.5/2.60)+(47.9/2.71)+(6.6/2.69)]
Gbam = 2.66
VMA = 100 - (2.48)(94) / 2.66
VMA = 12.36%
For 6.5% asphalt content:
Gbcm = 2.50
Pta = 93.5
Pca = 0.42(93.5) = 39.3
Pfa = 0.51(93.5) = 47.7
Pmf = 0.07(93.5) = 6.5
Gbam = (39.3+47.7+6.5) / [(39.3/2.60)+(47.7/2.71)+(6.5/2.69)]
Gbam = 2.66
VMA = 100 - (2.50)(93.5) / 2.66
VMA = 12.12%
For 7.0% asphalt content:
Gbcm = 2.50
Pta = 93.0
Pca = 0.42(93) = 39.1
Pfa = 0.51(93) = 47.4
Pmf = 0.07(93) = 6.5
Gbam = (39.1+47.4+6.5) / [(39.1/2.60)+(47.4/2.71)+(6.5/2.69)]
Gbam = 2.66
VMA = 100 - (2.5)(93) / 2.66
VMA = 12.59%
For 7.5% asphalt content:
Gbcm = 2.48
Pta = 92.5
Pca = .42(92.5) = 38.9
Pfa = .51(92.5) = 47.2
Pmf = .07(92.5) = 6.5
Gbam = (38.9+47.2+6.5) / [(38.9/2.60)+(47.2/2.71)+(6.5/2.69)]
Gbam = 2.66
VMA = 100 - (2.48)(92.5) / 2.66
VMA = 13.76%
Now compute % air voids for each mix using Equation 18.12,
G − G mb
Pa = 100 mm
Gmm
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Chapter 18: Bituminous Materials
Gmm must be computed using Equation 18.8,
100
Gmm =
( Ps / G se ) + ( Pb / Gb )
For 5.5% asphalt content:
Gmp = 100 / [(94.5/2.82) + (5.5/1.02)]
Gmp = 2.57
Pav = 100 (2.57-2.46)/2.57
Pav = 4.28%
For 6.0% asphalt content:
Gmp = 100 / [(94.0/2.82) + (6.0/1.02)]
Gmp = 2.55
Pav = 100 (2.55-2.48)/2.55
Pav = 2.75%
For 6.5% asphalt content:
Gmp = 100 / [(93.5/2.82) + (6.5/1.02)]
Gmp = 2.53
Pav = 100 (2.53-2.5)/2.53
Pav = 1.19%
For 7.0% asphalt content:
Gmp = 100 / [(93.0/2.82) + (7.0/1.02)]
Gmp = 2.51
Pav = 100 (2.51-2.50)/2.51
Pav = 0.40%
For 7.5% asphalt content:
Gmp = 100 / [(92.5/2.82) + (7.5/1.02)]
Gmp = 2.49
Pav = 100 (2.49-2.48)/2.49
Pav = 0.40%
From these calculations, plot γ vs. % asphalt, VMA vs. asphalt, Pav vs. asphalt.
Also, using the given data, plot stability vs. % asphalt, flow vs. % asphalt.
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Chapter 18: Bituminous Materials
The maximum stability occurs at 6.5% AC
The maximum unit weight plot occurs at 6.5% AC
For voids, limits are 3% and 5%, average = 4%; at 4% voids we get 5.6% AC
Averaging these values we get (6.5 + 6.5 + 5.6)/ 3 = 6.2 % AC
Checking this against the required values:
At 6.2% AC
Stability = 1850
Flow = 15
Unit weight = 155
VMA = 12.25%
Voids = 2.2%
Assuming medium traffic, % voids is too low.
5.9% AC is needed to obtain minimum voids of 3%
At 5.9% AC
Stability = 1830
Flow = 13.9
Unit weight = 154.5
VMA = 12.4
Voids = 3%
This mix is acceptable.
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Chapter 18: Bituminous Materials
18-12
Determine the asphalt absorption of the optimum mix found in Problem 18-11.
Asphalt absorption is calculated using Equation 18.9
G − G sb
Pba = 100 se
Gb
G se G sb
Pba = 100 [(2.82-2.66) / (2.82)(2.66)]1.02
Pba = 2.18%
18-13
The latitude at the location where a high-speed rural road is to be constructed is 35˚.
The expected ESAL is 32 x 106. The seven-day average high temperature is 53˚C,
and the low air temperature is -18˚C. If the standard deviations for the high and
low temperatures are ±2˚C and ±1˚C respectively, and the depth of the pavement is
155 mm, determine an appropriate asphalt binder for a reliability of 98%.
Use Equation 18.13 to determine the high pavement temperature at a depth of 20
mm,
T20mm = (Tair – 0.00618 Lat2 + 0.2289Lat + 42.2)(0.9545))– 17.78
Note: for Tair, with a reliability of 98%, use 53+(2)(2) = 57
T20mm = (57 – 0.00618(35)2 + 0.2289(35) + 42.2)(0.9545))– 17.78
T20mm = 77.32ºC
From Table 18.12, preliminarily select PG 58-22 as the binder. With ESAL >
10x106, increase temperature grade of binder by one. Therefore, select PG 64-22
as the appropriate binder.
18-14
An urban expressway is being designed for a congested area in Washington, D.C. It
is expected that most of the time traffic will be moving at a slow rate. If the
anticipated ESAL is 8 x 106, determine an appropriate asphalt binder for this
project.
Assume design reliability is 98%. From Table 18.13, preliminarily select
PG 58-16 as the binder. With slow traffic as a design criterion, increase
temperature grade of binder by one. With ESAL < 10x106, no adjustment is
required. Therefore, select PG 64-16 as the appropriate binder.
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Chapter 18: Bituminous Materials
18-15
The table below shows properties of three trial aggregate blends that are to be
evaluated so as to determine their suitability for use in a Superpave mix. If the
nominal maximum sieve of each aggregate blend is 19 mm, determine the initial
trial asphalt content for each of the blends.
Property
Gsb
Gse
Trial Blend 1
2.698
2.765
Trial Blend 2
2.696
2.766
Trial Blend 3
2.711
2.764
Since Gsb and Gse are given, the trial percentage of asphalt binder can be found
using Equations 18.17, 18.18, and 18.19 and the assumed values as indicated in
the textbook:
Pb = 0.05
Ps = 0.95
Gb = 1.02
Va = 0.04
For trial blend 1:
Use Equation 18.17,
P (1 − Va ) 1
1
[
]=
Vba = s
−
Pb
Ps G sb G se
(
)
+
Gb G se
0.95(1 − 0.04)
1
1
[
−
] = 0.0209
0.05 0.95 2.698 2.765
(
+
)
1.02 2.765
Use Equation 18.18,
Vbe = 0.176-0.0675 log Sn = 0.176-0.067 log (19) = 0.0903
Use Equation 18.20,
P (1 − Va )
0.95(1 − 0.04)
= 2.323
Ws = s
=
0.05 0.95
Pb
Ps
(
+
)
+
1.02 2.765
Gb G se
Use Equation 18.19,
Gb (Vbe + Vba )
1.02(0.090 + 0.021)
Pbi = 100
= 100
= 0.0466
(Gb (Vbe + Vba )) + Ws
1.02(0.090 + 0.021) + 2.323
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Chapter 18: Bituminous Materials
For trial blend 2:
Use Equation 18.17,
P (1 − Va ) 1
1
[
]=
Vba = s
−
Pb
Ps G sb G se
(
)
+
Gb G se
0.95(1 − 0.04)
1
1
−
[
] = 0.0218
0.05 0.95 2.696 2.766
(
+
)
1.02 2.766
Use Equation 18.18,
Vbe = 0.176-0.0675 log Sn = 0.176-0.067 log (19) = 0.0903
Use Equation 18.20,
P (1 − Va )
0.95(1 − 0.04)
= 2.324
Ws = s
=
0.05 0.95
Pb
Ps
(
+
)
+
1.02 2.766
Gb G se
Use Equation 18.19,
Gb (Vbe + Vba )
1.02(0.090 + 0.022)
Pbi = 100
= 100
= 0.0469
(Gb (Vbe + Vba )) + Ws
1.02(0.090 + 0.022) + 2.324
For trial blend 3:
Use Equation 18.17,
P (1 − Va ) 1
1
[
]=
Vba = s
−
Pb
Ps G sb G se
(
)
+
Gb G se
0.95(1 − 0.04)
1
1
−
[
] = 0.0164
0.05 0.95 2.711 2.764
(
+
)
1.02 2.764
Use Equation 18.18,
Vbe = 0.176-0.0675 log Sn = 0.176-0.067 log (19) = 0.0903
Use Equation 18.20,
P (1 − Va )
0.95(1 − 0.04)
= 2.322
Ws = s
=
0.05 0.95
Pb
Ps
(
+
)
+
1.02 2.764
Gb G se
Use Equation 18.19,
Gb (Vbe + Vba )
1.02(0.090 + 0.016)
= 100
= 0.0634
Pbi = 100
(Gb (Vbe + Vba )) + Ws
1.02(0.090 + 0.016) + 2.322
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Chapter 19
Design of Flexible Pavements
19-1
Discuss the structural components of a flexible pavement.
A flexible pavement consists of the subgrade, the subbase, the base and
the wearing surface. The subgrade is usually the natural material located along
the horizontal alignment of the pavement and is prepared as a roadbed to serve as
the foundation of the pavement structure. The subbase is located immediately
above the subgrade and consists of material of a superior quality to that which
generally is used for subgrade construction. When the quality of the subgrade
material meets the requirements of the subbase material, the subbase component
may be omitted. The base course lies immediately above the subbase (or
subgrade) and usually consists of granular materials such as crushed stone,
crushed or uncrushed slag, crushed or uncrushed gravel, and sand. The
requirements for plasticity, gradation, and strength increase through the subgrade,
subbase, and base courses. The surface course is the upper course of the road
pavement and is constructed immediately above the base course. The surface
course in flexible pavements usually consists of a mixture of mineral aggregates
and asphaltic materials.
19-2
Describe the purpose of soil stabilization and discuss at least three methods of
achieving it.
Soil stabilization is performed to improve the engineering properties of
natural soils. There are two categories of soil stabilization, mechanical and
chemical. Mechanical stabilization is the blending of different grades of soils to
obtain the required grade. Chemical stabilization is the blending of the natural
soil with chemical agents. Cement stabilization usually involves the addition of
5% to 14% Portland cement by volume of the compacted mixture to the soil being
stabilized. This type of stabilization is usually used to obtain the required
engineering properties for base course materials. The process of stabilizing soils
with cement involves pulverizing the soil, mixing the required quantity of cement
with the pulverized soil, compacting the soil-cement mixture, and curing the
compacted layer. Bituminous stabilization is carried out to achieve waterproofing
of natural materials and/or binding of natural materials. Waterproofing aids in
maintaining the correct moisture content and helps prevent water from seeping
into the subgrade, protecting the subgrade from failure due to an increase in
moisture content. Binding improves the durability characteristics of the natural
soil by providing an adhesive characteristic, whereby the soil particles adhere to
each other, increasing cohesion. Several types of soils can be stabilized with
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Chapter 19: Design of Flexible Pavements
bituminous materials, although it is generally required that less than 25% of the
material passes the No. 200 sieve. Lime stabilization can be used for stabilizing
both base and subbase materials. The materials most commonly used for lime
stabilization are calcium hydroxide and dolomite. Clayey materials are most
suitable for lime stabilization, but the materials should have plasticity index
values less than 10 for the stabilization to be most effective.
19-3
Discuss the process of lime stabilization, and indicate the soil characteristics for
which it is most effective.
Lime stabilization can be used for stabilizing both base and subbase
materials. The materials most commonly used for lime stabilization are calcium
hydroxide and dolomite. Clayey materials are most suitable for lime stabilization,
but the materials should have plasticity index values less than 10 for the
stabilization to be most effective. The reaction of the material with the lime
reduces the tendency to swell as a result of an increase in moisture content; the PI
value of the soil is also reduced. Desirable changes also include a pozzolanic
reaction may also occur resulting in some cementation of the soil and in an
increase in strength; soils with high silica or alumina content may exhibit
significant increases in strength over long periods of time; and, finally,
flocculation occurs, increasing the effective grain size of the soil.
19-4
An axle weight study on a section of highway gave the following data on axle load
distribution. Determine the truck factor for this section of highway. Assume SN =4
and pt = 2.5.
The following table shows the given data and the load equivalency factors taken,
by interpolation if necessary, from Table 19.3.
Axle Load Group:
Single
# of axles/ 1000
vehicles
Factor
Col. (2) x Col. (3)
<4
678
0.0016
1.0848
4-8
775
0.013
10.075
8-12
500
0.102
51
12-16
150
0.388
58.2
16-18
60
0.8225
49.35
18-20
40
1.235
49.4
20-22
7
1.78
12.46
22-24
4
2.49
9.96
24-26
3
3.4
10.2
Total for single axle load group = 251.730
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Chapter 19: Design of Flexible Pavements
Axle Load Group:
Tandem
# of axles/ 1000
vehicles
Factor
Col. (2) x Col. (3)
<6
18
0.0003
0.0054
6-12
236
0.065
15.34
12-18
170
0.045
7.65
18-24
120
0.174
20.88
24-30
152
0.4675
71.06
30-32
66
0.791
52.206
32-34
30
0.9985
29.955
34-36
12
1.245
14.94
36-38
4
1.53
6.12
38-40
1
1.855
1.855
Total for tandem axle load group = 220.011
Total = 251.730 + 220.011 = 471.741
Truck factor = 471.741/1000
Truck factor = 0.4717
19-5
How does an ESAL differ from a truck factor?
A truck factor is defined as the number of 18,000 lb single load
applications caused by a single passage of a vehicle. The ESAL, equivalent
single axle load, is the number of repetitions of an 18,000 lb single-axle load
applied to the pavement on two sets of dual tires.
19-6
A six-lane divided highway is to be designed to replace an existing highway. The
present AADT (both directions) of 6000 vehicles is expected to grow at 5% per
annum. Assume SN =4 and pt =2.5. The percent of traffic on the design lane is 45%.
Determine the design ESAL if the design life is 20 years and the vehicle mix is:
Passenger cars (1000 lb/axle)
= 60%
2-axle single-unit trucks (5000 lb/axle)
= 30%
3-axle single-unit trucks (7000 lb/axle)
= 10%
Use Equation 19.2 to determine ESALs:
ESALi = (fd)(Gjt)(AADTi)(365)(Ni)(FEi)
The design lane use factor, fd, is given as 0.45
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Chapter 19: Design of Flexible Pavements
From Table 19.4, the growth factor, Gjt is 33.06 for a growth rate of 5% and
design life of 20 yr
From Table 19.3, by interpolation, the load equivalency factors are:
2 axle SU truck = 0.008
3 axle SU truck = 0.025
ESAL2-axle SU truck= (0.45)(33.06)(6000)(0.3)(365)(2)(0.008)
ESAL2-axle SU truck = 1.56x105
ESAL3-axle SU truck = (0.45)(33.06)(6000)(0.1)(365)(3)(0.025)
ESAL3-axle SU truck = 2.44x105
Note that the contribution of passenger car traffic to ESALs is negligible and
therefore not considered in the calculations.
Total ESAL = 4.00x105
19-7
A section of a two-lane rural highway is to be realigned and replaced by a four-lane
highway with a full-depth asphalt pavement. The AADT (both ways) on the existing
section can be represented by 500 ESAL. It is expected that construction will be
completed 5 years from now. If the traffic growth rate is 5% and the effective CBR
of the subgrade on the new alignment is 85, determine a suitable depth of the
asphalt pavement using the AASHTO method. Take the design life of the pavement
as 20 years. The resilient modulus of the asphalt (EAC) is 400,000 lb/in2. Assume mi
for the subgrade is 1 and the percent of traffic on the design lane is 45%. Use a
reliability level of 90%, a standard deviation of 0.45, and a design serviceability loss
of 2.0.
First, determine design ESAL
From Table 19.4, the growth factor, Gjt, for 25 years (5 years of construction plus
20 year design life) = 47.73 and the growth factor, Gjt, for 5 years (to subtract the
ESAL for the construction period) = 5.53
Design ESAL = (500)(47.73-5.53)(365)(0.45)
Design ESAL = 3.47x106
Determine Mr from Figure 19.6,
Mr = 19,000 psi
Using the nomograph in Figure 19.8,
Step 1: Connect Reliability of 90% to standard deviation of 0.45 and
extend to first turning line to create “A”
Step 2: Connect “A” to ESAL of 3.47 x 106 and extend to second turning
line to create “B”
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Chapter 19: Design of Flexible Pavements
Step 3: Connect “B” to Mr of 19 x 103 and extend to design serviceability
loss chart to create “C”
Step 4: Connect “C” to ΔPSI of 2.0
Step 5: Draw vertical line to read SN of 3.8
Determine structural coefficients for each layer
From Figure 19.5, with EAC = 400,000 lb/in2
a1= 0.40
Since this pavement is being designed with full-depth asphalt concrete, the asphalt
concrete must provide all of the structural support. Therefore, equation 19.6 can
be modified for use in this problem as:
SN = a1D1 = 3.8 = (0.40)D1
D1 = 9.5 inches.
19-8
The predicted traffic mix of a proposed four-lane urban non-interstate freeway:
Passenger cars
= 69%
Single-unit trucks
2-axle, 5,000 lb/axle
= 20%
2-axle, 9,000 lb/axle
= 5%
3-axle or more, 23,000 lb/axle
= 4%
Tractor semitrailers and combinations
3-axle, 20,000 lb/axle
= 2%
The projected AADT during the first year of operation is 3500 (both directions). If
the traffic growth rate is estimated at 4% and the CBR of the subgrade is 80,
determine the depth of a full-asphalt pavement using the AASHTO method and n =
20 years. The resilient modulus of the asphalt (EAC) is 300,000 lb/in Assume mi for
the subgrade is 1 and the percent of traffic on the design lane is 42%, pt = 2.5 and
SN = 4. Use a reliability level of 90%, a standard deviation of 0.45, and a design
serviceability loss of 2.0. Assume the design lane factor fd is 0.42.
Use Equation 19.2 to determine design ESALs.
ESALi = (fd)(Gjt)(AADTi)(365)(Ni)(fEi)
From Table 19.4, Gjt = 29.78
From Table 19.3, the following truck factors are obtained:
f1 = 0.008 (2-axle, 5,000 lb/axle)
f2 = 0.0715 (2-axle, 9,000 lb/axle)
f3 = 0.2495 (3-axle or more, 23,000 lb/axle)
f4 = 0.141 (3-axle, 20,000 lb/axle)
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Chapter 19: Design of Flexible Pavements
ESAL1 = (0.42)(29.78)(3500)(0.2)(365)(2)(0.008) = 0.511x104
ESAL2 = (0.42)(29.78)(3500)(0.05)(365)(2)(0.0715) = 1.14x105
ESAL3 = (0.42)(29.78)(3500)(0.04)(365)(3)(0.2495) = 4.78x105
ESAL4 = (0.42)(29.78)(3500)(0.02)(365)(3)(0.141) = 1.35x105
Total ESAL = 7.78x105
Determine Mr from Figure 19.6,
Mr = 18,500 psi
Using the nomograph in Figure 19.8,
Step 1: Connect Reliability of 90% to standard deviation of 0.45 and
extend to first turning line to create “A”
Step 2: Connect “A” to ESAL of 7.78 x 105 and extend to second turning
line to create “B”
Step 3: Connect “B” to Mr of 18.5 x 103 and extend to design
serviceability loss chart to create “C”
Step 4: Connect “C” to ΔPSI of 2.0
Step 5: Draw vertical line to read SN of 2.3
Determine structural coefficients for each layer
From Figure 19.5, with EAC = 400,000 lb/in2
a1= 0.40
Since this pavement is being designed with full-depth asphalt concrete, the asphalt
concrete must provide all of the structural support. Therefore, equation 19.6can
be modified for use in this problem as:
SN = a1D1 = 2.3 = (0.37)D1
D1 = 6.2 inches Æ round up to 6.5 inches.
19-9
A rural principal arterial is expected to carry an ESAL of 0.188x106 during the first
year of operation with an expected annual growth of 6% over the 20-year design
life. If the subgrade has a resilient modulus of 15,000 lb/in2 , design a suitable
pavement consisting of a granular subbase with a layer coefficient of 0.13, a
granular base layer with a layer coefficient of 0.14, and an asphalt concrete surface
with an elastic modulus of 400,000 lb/in2 . Assume all mi values =1, the percent of
traffic on the design lane is 47%, and SN = 4. Use a reliability level of 85%, a
standard deviation of 0.45, and a design serviceability loss of 2.0.
First, determine design ESAL
The design lane use factor, fd, is 0.47
From Table 19.4, Gjt = 36.79
Total ESALi = (0.47)(36.79)(0.188x106)
ESAL = 3.25 x106
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Chapter 19: Design of Flexible Pavements
Using the nomograph in Figure 19.8,
Step 1: Connect Reliability of 85% to standard deviation of 0.45 and
extend to first turning line to create “A”
Step 2: Connect “A” to ESAL of 3.25 x 106 and extend to second turning
line to create “B”
Step 3: Connect “B” to Mr of 15 x 103 and extend to design serviceability
loss chart to create “C”
Step 4: Connect “C” to ΔPSI of 2.0
Step 5: Draw vertical line to read SN of 3.9
Determine structural coefficients for each layer
From Figure 19.5, with EAC = 400,000 lb/in2
a1= 0.40
Since this pavement is being designed with three layers, equation 19.6 can be
modified for use in this problem as:
SN = a1D1 + a2D2m2 + a3D3m3
3.9 = (0.40)D1 + (0.14)D2(1) + (0.13)D3(1)
From Table 19.9, for 3.25x106 ESALs, a recommended minimum
thickness of asphalt concrete is 3.5 inches and 6 inches for aggregate base.
The depth of subbase can then be found:
3.9 = (0.40)(3.5) + (0.14)(6)(1) + (0.13)D3(1)
D3 = 12.77 inches Æ round up to 13 inches.
19-10
Using the information given in Problem 19-6, design a suitable pavement consisting
of an asphalt mixture surface with an elastic modulus of 250,000 lb/in2, a granular
base layer with a structural coefficient of 0.14 on a subgrade having a CBR of 10.
Assume all mi values =1, and the percent of traffic on the design lane is 45%. Use a
reliability level of 85%, a standard deviation of 0.45, and a design serviceability loss
of 2.0.
From Problem 19-6, ESAL = 4.00 x 105
Using Equation 19.3, Mr = 1,500(CBR) = 15,000 lb/in2
Using the nomograph in Figure 19.8,
Step 1: Connect Reliability of 85% to standard deviation of 0.45 and
extend to first turning line to create “A”
Step 2: Connect “A” to ESAL of 4.00 x 105 and extend to second turning
line to create “B”
Step 3: Connect “B” to Mr of 15 x 103 and extend to design serviceability
loss chart to create “C”
Step 4: Connect “C” to ΔPSI of 2.0
Step 5: Draw vertical line to read SN of 3.2
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Chapter 19: Design of Flexible Pavements
Determine structural coefficients for each layer
From Figure 19.5, with EAC = 250,000 lb/in2
a1= 0.33
Since this pavement is being designed with two layers, equation 19.6 can be
modified for use in this problem as:
SN = a1D1 + a2D2m2
3.2 = (0.33)D1 + (0.14)D2(1) + (0.13)D3(1)
From Table 19.9, for 4.00x105 ESALs, a recommended minimum
thickness of asphalt concrete is 2.5 inches and 4 inches for aggregate base.
The structural number provided using these minimums can then be found:
SN = (0.33)(2.5) + (0.14)(4)(1) = 1.385
Since this SN is below that required, layer thicknesses will need to be
increased in order to provide sufficient support. If the asphalt concrete
thickness is held at 2.5 inches, the required thickness of the base course
can then be found:
3.2 = (0.33)(2.5) + (0.14)D2(1)
D2 = 16.9 inches Æ round up to 17 inches.
19-11
Repeat Problem 19-10 for a pavement consisting of an asphalt mixture surface with
an elastic modulus of 250,000 lb/in2, and 6” of granular subbase with a Resilient
Modulus of 20 x 103 lb/in2.
From Problem 19-10, SN = 3.2. Since the only change from Problem 19-10 is
that a subbase of 6 inches is added, the same SN value can be used in this
problem.
Using Figure 19.3, the subbase has a structural coefficient of 0.14.
Since this pavement is being designed with three layers, Equation 19.6 can be
modified for use in this problem as:
SN = a1D1 + a2D2m2 + a3D3m3
3.2 = (0.33)D1 + (0.14)D2(1) + (0.14)(6)(1)
From Table 19.9, for 3.42x105 ESALs, a recommended minimum
thickness of asphalt concrete is 2.5 inches and 4 inches for aggregate base.
The structural number provided using these minimums and the prescribed
subbase thickness can then be found:
(0.33)(2.5) + (0.14)(4)(1) + (0.14)(6)(1) =2.25 < 3.2
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Chapter 19: Design of Flexible Pavements
Since this SN is below that required, layer thicknesses will need to be
increased in order to provide sufficient support. If the asphalt concrete
thickness is held at 2.5 inches, the required thickness of the base course
can then be found:
3.2 = (0.33)(2.5) + (0.14)D2(1) + (0.14)(6)(1)
D2 = 10.96 inches Æ round up to 11 inches.
19-12
Repeat Problem 19-7 using two different depths of untreated aggregate bases of 6”
and 12”. Highway contractors in your area can furnish rates for providing and
properly laying an asphalt concrete surface and untreated granular base. Assume a
structural coefficient of 0.12 for the base course. If these rates are available,
determine the cost for constructing the different pavement designs if the highway
section is 5 miles long and the lane width is 12 ft. Which design will you select for
construction?
From Problem 19-7, SN = 3.8. Since the only change from Problem 19-7 is that a
base course of 6 or 12 inches is added, the same SN value can be used in this
problem.
Since this pavement is being designed with two layers, equation 19.6 can be
modified for use in this problem as:
SN = a1D1 + a2D2m2
3.8 = (0.40)D1 + (0.12)D2(1)
For the 6 inch base course:
3.8 = (0.40)D1 + (0.12)(6)(1)
D1 = 7.7 inches.
For the 12 inch base course:
3.8 = (0.40)D1 + (0.12)(12)(1)
D1 = 5.9 inches.
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Chapter 19: Design of Flexible Pavements
19-13
The traffic on the design lane of a proposed four-lane rural interstate highway
consists of 40% trucks. If classification studies have shown that the truck factor can
be taken as 0.45, design a suitable flexible pavement using the 1993 AASHTO
procedure if the AADT on the design lane during the first year of operation is 1000,
pi = 4.2, and pt =2.5 .
Growth rate = 4%
Design life = 20 years
Reliability level = 95%
Standard deviation = 0.45
The pavement structure will be exposed to moisture levels approaching saturation
20% of the time, and it will take about one week for drainage of water. Effective
CBR of the subgrade material is 7. CBR of the base and subbase are 70 and 22,
respectively, and Mr for the asphalt mixture, 450,000 lb/in2.
Calculate ESALs using Equation 19.2,
ESALi = (fd)(Gjt)(AADTi)(365)(fi)
ESAL = (0.40)(29.78)(1000)(0.45)(365)
ESAL = 1.957 x 106
Calculate Mr of subgrade using Equation 19.4,
Mr = 1500 CBR
Mr = (1500)(7) = 1.05 x 104
Calculate serviceability loss (ΔPSI),
ΔPSI = pi - pt
ΔPSI = 4.2 – 2.5 = 1.7
Using the nomograph in Figure 19.8,
Step 1: Connect Reliability of 95% to standard deviation of 0.45 and
extend to first turning line to create “A”
Step 2: Connect “A” to ESAL of 2.0 x 106 and extend to second turning
line to create “B”
Step 3: Connect “B” to Mr of 10 x 103 and extend to design serviceability
loss chart to create “C”
Step 4: Connect “C” to ΔPSI of 1.7
Step 5: Draw vertical line to read SN of 3.7
Determine structural coefficients for each layer
From Figure 19.5, with Mr = 4.5 x 105 (asphalt surface course)
a1= 0.44
From Figure 19.4, with CBR = 70 (base course)
a2= 0.13
From Figure 19.3, with CBR = 22 (subbase)
a3= 0.10
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Chapter 19: Design of Flexible Pavements
Determine drainage index, mi, using Tables 19.5 and 19.6
For fair drainage and 20% saturation, mi = 0.9
Knowing that each course must meet the requirements of the course above it, we
can determine the required depth of each course, using various forms of Equation
19.6. Beginning with the surface course, SN1 can be found using the Mr for the
base course.
From Figure 19.8, SN = 2.7
D1*= SN1/a1 = 2.7/0.44 = 6.14 in. Use 6.5 in.
SN1*= a1D1* = 0.44(6.5) = 2.86
Repeating this procedure for the base course:
Mr = 1.35 x 104 (Subbase)
From Figure 19.8, SN = 3.3
D2* = (3.3-2.86) / [(0.13)(0.9)] = 3.76 in.
From Table 19.9, use recommended minimum of 6 in.
*
SN2 = (a2)(m2)(D2*) + SN1* = (0.13)(0.9)(6) + 2.86 = 3.56
D3* = (3.7-3.56) / [(0.1)(0.9)] = 1.56 in.
From Table 19.9, use recommended minimum of 4 in.
Check design:
SN = 0.44(6.5) + (0.13)(0.9)(6) + (0.1)(0.9)(4) = 3.92 (which is greater
than SN of 3.7), so the design is satisfactory.
The pavement will consist of 6.5 in. of asphalt concrete surface, 6 in. of granular
base, and 4 in. of subbase.
19-14
Repeat Problem 19-13 with the subgrade Mr values for each month from January
through December being 20,000, 20,000, 6000, 6000, 6000, 9000, 9000, 9000, 9500,
9500, 8000, and 20,000 lb/in2, respectively. The pavement structure will be exposed
to moisture levels approaching saturation for 20% of the time, and it will take about
4 weeks for drainage of water from the pavement. Use untreated sandy gravel with
Mr of 15 × 103 lbs/in2 for subbase and untreated granular material with Mr of 28 ×
103 lbs/in2 for the base course.
First determine the relative damage each month.
Using Figure 19.6 or the following equation and table:
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Chapter 19: Design of Flexible Pavements
Month
Mr
uf
Jan.
20000
.01
Feb.
20000
.01
March
6000
.20
April
6000
.20
May
6000
.20
June
9000
.08
July
9000
.08
Aug.
9000
.08
Sept.
9500
.07
Oct.
9500
.07
Nov.
8000
.10
Dec.
20000
.01
Σuf =
1.11
Average uf = Σuf / n = 1.11/12 = 0.093
Roadbed soil, Mr = 8500 lb/in2
From Problem 19-13, ΔPSI = 1.7
Using Figure 19.8, SN = 3.9
Determine the structural layer coefficient for each layer:
From Figure 19.5, Mr = 4.5 x 105 (asphalt concrete, from Problem 20-14)
a1 = 0.44
From Figure 19.4, Mr = 28 x 103 (untreated gravel base)
a2 = 0.13
From Figure 19.3, Mr = 15 x 103 (sandy gravel subbase)
a3 = 0.11
Find layer thicknesses, starting with surface course:
SN of surface course is dependent on base course Mr, therefore enter Figure 19.8
with Mr = 28 x 103 SN = 2.5
D1*= 2.5/0.44 = 5.68 in. Use 6 in.
SN1*= a1D1* = 0.44(6) = 2.64
Repeating this procedure for the base course:
Mr= 1.5 x 104 (Subbase)
From Figure 19.8, SN = 3.2
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Chapter 19: Design of Flexible Pavements
D2* = (3.2-2.64) / [(0.13)(0.7)] = 6.15 in. Use 6.5 in.
SN2* = (a2)(m2)(D2*) + SN1* = (0.13)(0.7)(6.5) + 2.64 = 3.23
D3* = (3.9-3.23) / [(0.11)(0.7)] = 8.7 in. Use 9 in.
Check design:
SN = 0.44(6) + (0.13)(0.7)(6.5) + (0.11)(0.7)(9) = 4.05 (which is greater
than SN of 3.9), so the design is satisfactory
The pavement will consist of 6 in. of asphalt concrete surface, 6.5 in. of untreated
gravel base, and 9 in. of sandy gravel subbase.
19-15
An existing two-lane rural highway is to be replaced by a four-lane divided highway
on a new alignment. Construction of the new highway will commence two years
from now and is expected to take three years to complete. The design life of the
pavement is 20 years. The present ESAL is 0.133175x106. Design a flexible pavement
consisting of an asphalt concrete surface and lime-treated base using the California
(Hveem) method. The results of a stabilometer test on the subgrade soil are as
follows.
Moisture Content (%)
19.8
22.1
24.9
R Value Exudation Pressure Expansion Pressure
(lb/in2)
Thickness (ft)
55
45
16
575
435
165
1.00
0.15
0.10
Total ESAL = 0.133175 x (47.73-5.53) = 5.62 x 106
Calculate traffic index using Equation 19.8,
TI = 9.0(5.62)0.119 = 11.05
Next, calculate thickness above base using Equation 19.9,:
GE = 0.0032(TI)(100-R)
GE = 0.0032(11.1)(100-80)
GE = 0.71 ft.
From Table 19.10, Gf for asphalt concrete with TI of 11.1 = 1.71
Actual depth of AC = 0.71/1.71 = 0.415 ft = 5 in.
Next, calculate base thickness
To determine GE at 300 lb/in2, Plot GE vs. Exudation pressure
At 300 lb/in2, GE = 2.5
From Table 19.10, for lime-treated base, Gf = 1.2
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Chapter 19: Design of Flexible Pavements
GE of base layer = 2.5 - (5/12)(1.71) = 1.7875
Base thickness = 1.7875 / 1.2 = 1.4896 ft. = 17.875 in. Use 18 in.
GE vs. Exudation Pressure
GE vs. Moisture Content
Moisture Content (%)
GE (ft)
4
3
2
1
0
0
200
400
600
800
Exudation Pressure (lb/in2)
25
24
23
22
21
20
19
18
0
1
2
GE (ft)
3
4
Check design:
Plot GE vs. Moisture content to determine moisture content at 300 lb/in2
From plot above, right, Moisture content = 23.5%
19-16
Briefly describe the main steps in the Mechanistic-Empirical Pavement Design
method as given in the MEPDG.
The method of pavement design applied in the MEPDG consists of five basic
steps. The first step is to select a trial design based on another method (such as
the 1993 AASHTO Pavement Design Guide, agency policies or standards, or a
design based on engineering judgment and experience). The second step is
selection of acceptable thresholds for performance criteria related to pavement
distresses of rutting, transverse cracking, longitudinal cracking, alligator cracking,
and smoothness. The third step involves the collection of all input data pertaining
to project information and to traffic pavement structure, material properties, and
climate. Fourth, the trial design is evaluated, typically through the MEPDG
software. Finally, the trial design is revised and then evaluated until an
acceptable design is achieved.
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344
Chapter 19: Design of Flexible Pavements
19-17
Determine the expected rut depth of the fourth layer of a 4” thick HMA flexible
pavement surface divided into four equal sub-layers for the following conditions:
Resilient or elastic strain calculated by the structural response model at the middepth of each HMA sub-layer, in/in = 45 x 10-6
Number of axle load repetitions = 1.5 x 104
Mixture field calibration constants:
β1r,
= 0.99
β2r,
= 0.98
= 0.95
β3
Use Equation 19.10:
Δ p ( HMA) = ε p ( HMA) hHMA = β1r k zε r ( HMA)10k1r n k2 r β2 r T k3 r β3 r
Compute kz = (C1 + C2 D)0.0328196D
C1
= - 0.1039(HHMA)2 + 2.4868HHMA - 17.342
= - 0.1039(4)2 + 2.4868(4) - 17.342 = -9.0572
C2
= 0.01712(HHMA)2 - 1.7331HHMA + 27.428
= 0.01712(4)2 - 1.7331(4) + 27.428 = 20.7695
kz = (C1 + C2 D)0.0328196D
kz = (-9.0572 + 20.7695 x 3)0.03281963 = 0.00199
Δ p ( HMA) = β1r k zε r ( HMA)10k1r n k2 r β2 r T k3 r β3 r
= 0.99 x 0.00199 x 45 x 10-6 x 10-3.35412 x (1.5)(104) x 0.4791 x 0.98 x 851.5606 x 0.95
= 2.59 x 10-6
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Chapter 19: Design of Flexible Pavements
19-18
Repeat problem 19-17 for axle load repetitions of 2 x 104. Based on your answers for
this problem and problem 19-17, discuss the effect of axle load repetitions on
expected rut of HMA flexible pavements.
From Problem 19-17, kz = 0.00199
Δ p ( HMA) = β1r k zε r ( HMA)10k1r n k2 r β2 r T k3 r β3 r
= 0.99 x 0.00199 x 45 x 10-6 x 10-3.35412 x (2)(104) x 0.4791 x 0.98 x 851.5606 x 0.95
= 2.96 x 10-6
Compared with Problem 19-17, an increase of 33% in axle load repetitions (from
15,000 to 20,000) resulted in an increase of 14% in rut depth.
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346
Chapter 20
Design of Rigid Pavements
20-1
Portland cement concrete consists of what four primary elements?
The four primary elements Portland cement concrete consists of are: Portland
cement, coarse aggregate, fine aggregate, and water.
20-2
List and briefly describe the five main types of Portland cement as specified by
AASHTO.
The AASHTO specifications list five main types of Portland Cement:
Type I is suitable for general concrete construction, where no special properties
are required.
Type II is suitable for use in general concrete construction, where the concrete
will be exposed to moderate action of sulphate or where moderate heat of
hydration is required.
Type III is suitable for concrete construction that requires a high concrete strength
in a relatively short time and is sometimes referred to as high early strength
cement.
Type IV is suitable for projects where low heat of hydration is necessary.
Type V is used in concrete construction projects where the concrete will be
exposed to high sulphate action.
20-3
What is the main requirement for the water used in Portland cement?
Water used in Portland cement concrete should also be suitable for drinking. This
requires that the quantity of organic matter, oil, acids, and alkalis should not be
greater than the allowable amount in drinking water.
20-4
Describe the basic types of highway concrete pavements and give the conditions
under which each type will be constructed.
Rigid highway pavements can be divided into three general types: plain concrete
pavements, simply reinforced concrete pavements, and continuously reinforced
concrete pavements.
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Chapter 20: Design of Rigid Pavements
Plain concrete pavements have no temperature steel or dowels for load transfer.
This type of pavement is used mainly on low-volume highways or when cement
stabilized soils are used as subbase material. Joints are typically spaced at
intervals of 10 ft to 20 ft to control cracking. This type is sometimes referred to
as jointed plain concrete pavement (JPCP).
Simply reinforced concrete pavements have dowels for the transfer of traffic loads
across joints, with these joints spaced at larger distances than with plain
pavements, ranging from 30 ft to 100 ft. Temperature steel is used throughout the
slab.
Continuously reinforced concrete pavements have no transverse joints, except
construction joints or expansion joints when they are necessary at specific
positions, such as at bridges. This type of pavement is typically used on highvolume, high-speed roadways. In design this type is often simply referred to as
CRCP.
20-5
List and briefly describe the four type of joints used in concrete pavements.
There are four different types of joints placed in concrete pavements: expansion
joints, contraction joints, hinge joints, and construction joints.
Expansion joints are usually placed transversely, at regular intervals, to provide
adequate expansion space for the slab to expand when the pavement is subjected
to an increase in temperature.
Contraction joints are placed transversely at regular intervals across the width of
the pavement to release some of the tensile stresses that are induced due to a
decrease in temperature.
Hinge joints are placed longitudinally mainly to reduce cracking along the
centerline of highway pavements.
Construction joints are placed transversely across the pavement width to provide
suitable transition between concrete placed at different times or on different days.
20-6
Define the phenomenon of pumping and its affects on rigid pavements.
Pumping is the discharge of water and subgrade (or subbase) material through
joints, cracks, and along the pavement edge. It is primarily caused by the
repeated deflection of the pavement slab in the presence of accumulated water
beneath it. Pumping action results in a relatively large void space formed
underneath the concrete slab. This results in faulting of the joints and eventually
the formation of transverse cracks or the breaking of the corner of the slab. Joint
faulting and cracking is therefore progressive, since formation of cracks facilitates
the pumping action.
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348
Chapter 20: Design of Rigid Pavements
20-7
List and describe the types of stresses that are developed in rigid pavements.
There are typically three categories of stresses developed in rigid
pavements. The first category of stress is induced by traffic wheel loads. There
are three critical locations of the wheel load on the concrete pavements that
induce stress; at the corner, at the interior, and at the edge of the slab. The second
category of stress is due to temperature changes. One type of this stress is caused
by the tendency of the slab edges to curl downward during the day and upward
during the night as a result of temperature gradients (between the base material
and the air) and is resisted by the weight of the slab itself. Another type of this
stress is caused by expansion and contraction of the slab due to daily temperature
gradients. The third category of stress is induced by bending. These stresses are
caused by the yielding of the sub-base or sub-grade supporting the concrete
pavement.
20-8
Determine the tensile stress imposed at night by a wheel load of 750 lb located at the
edge of a concrete pavement with the following dimensions and properties.
Pavement thickness = 8.5 inches
μ = 0.15
Ec = 4.2 x 106 lb/in2
k = 200 lb/in3
Radius of loaded area = 3 inches
Determine the stress on the edge of the slab using Equation 20.8.
0.572 P
l
σe =
[4 log10 ( ) + log10 b]
2
b
h
First, solve for the radius of equivalent distribution of pressure, b.
Since the radius of contact area, a = 3 in. < 1.724h = 1.724(8.5) = 14.654,
b = (1.6a2 + h2)1/2 - 0.675h
b = (1.6(3)2 + (8.5)2)1/2 – (0.675)(8.5)
b = 3.57 inches
Next, solve for the radius of relative stiffness, l.
l = (Ech3 / (12(1 - μ2)k))1/4
l = (((4.2 x 106)(8.5)3 ) / (12(1 - (0.15)2)200))1/4
l = 32.38
Now use Equation 20.8 to determine the tensile stress imposed.
0.572 P
l
σe =
[4 log10 ( ) + log10 b]
2
b
h
σe = ((0.572)(750) / (8.5)2) [4log10(32.38 / 3.57) + log10(3.57)]
σe = 26.0 lb/in2
Therefore, the tensile stress imposed on this slab will be 26.0 lb/in2.
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Chapter 20: Design of Rigid Pavements
20-9
Repeat Problem 20-8 for the load located at the interior of the slab.
Determine the stress at the interior of the slab using Equation 20.10.
0.316 P
l
σi =
[4 log10 ( ) + 1.069]
2
b
h
From Problem 20-8,
b = 3.57 inches
l = 32.38
Use Equation 20.10 to determine the tensile stress imposed.
0.316 P
l
σi =
[4 log10 ( ) + 1.069]
2
b
h
σi = ((0.316)(750) / (8.5)2) [4log10(32.38 / 3.57) + 1.069)]
σi = 3.280 [3.830 + 1.069]
σi = 16.1 lb / in2
Therefore, the tensile stress imposed on this slab will be 16.1 lb/in2.
20-10
Determine the maximum distance of contraction joints for a plain concrete
pavement if the maximum allowable tensile stress in the concrete is 50 lb/in2 and the
coefficient of friction between the slab and the subgrade is 1.7. Assume uniform
drop in temperature. Weight of concrete is 144 lb/ft3.
Use Equation 20.18 to determine length of pavement between contraction joints.
L = 288pc / fγc
L = (288)(50) / (1.7)(144)
L = 58.82 ft
Therefore, the maximum distance between contraction joints should be 58 ft.
20-11
Repeat Problem 20-10, with the slab containing temperature steel in the form of
welded wire mesh consisting of 0.125 in2 steel/ft width. The modulus of steel, Es, is
30 x 106 lb/in2, Ec = 5 x 106 lb/in2, and h = 6 inches.
Use Equation 20.20 to determine length of pavement between contraction joints.
24 p c
L=
(12h + nAs )
hγ c f
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Chapter 20: Design of Rigid Pavements
First, determine the modular ratio, n.
n = Es / Ec
n = (30 x 106) / (5 x 106) = 6
Use Equation 20.20 to determine length of pavement between contraction joints.
24 pc
L=
(12h + nAs )
hγ c f
L = [(24)(50) / (6)(1.7)(144)] [(12)(6) + (6)(0.125)]
L = (1200 / 1468.8) (72.75)
L = 59.4 ft
Therefore, the maximum distance between contraction joints should be 59 ft.
20-12
A concrete pavement is to be constructed for a 4-lane urban expressway on a
subgrade with an effective modulus of subgrade reaction k of 100 lb/in3. The
accumulated equivalent axle load for the design period is 3.25 x 106. The initial and
terminal serviceability indices are 4.5 and 2.5, respectively. Using the AASHTO
design method determine a suitable thickness of the concrete pavement, if the
working stress of the concrete is 600 lb/in2 and the modulus of elasticity is 5 x 106
lb/in2. Take the overall standard deviation, So, as 0.30, the load transfer coefficient,
J, as 3.2, the drainage coefficient as 0.9, and R=95%.
First, determine the design serviceability loss
design serviceability loss = 4.5 - 2.5 = 2.0
From the nomographs provided in Figures 20.13 and 20.14 and using the factors
above, the required thickness of the concrete slab is 9.4 inches, which for design
may be rounded up to 9.5 inches, the nearest half-inch for design.
20-13
A six-lane concrete roadway is being designed being designed for a metropolitan
area. This roadway will be constructed on a subgrade with an effective modulus of
subgrade reaction k of 170 lb/in3. The ESAL's used for the design period is 2.5 x
106. Using the AASHTO design method, determine a suitable thickness of the
concrete pavement (to the nearest 1/2 inch), provided that the working stress of the
concrete to be used is 650 lb/in2 and the modulus of elasticity is 5 x 106 lb/in2.
Assume the initial serviceability is 4.75 and the terminal serviceability is 2.5.
Assume the overall standard deviation, So, is 0.35, the load transfer coefficient J as
3.2, the drainage coefficient, Cd, is 1.15, and R = 95%.
First, determine the design serviceability loss
design serviceability loss = 4.75 - 2.5 = 2.25
351
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Chapter 20: Design of Rigid Pavements
From the nomographs provided in Figures 20.13 and 20.14 and the factors above,
the required thickness of the concrete slab is 7.8 inches, which for design may be
rounded up to 8 inches.
20-14
An existing rural 4-lane highway is to be replaced by a 6-lane divided expressway (3
lanes in each direction). Traffic volume data on the highway indicate that the
AADT (both directions) during the first year of operation is 24,000 with the
following vehicle mix and axle loads.
Passenger cars = 50 percent
2-axle single-unit trucks (12,000 lb/axle) = 40 percent
3-axle single-unit trucks (16,000 lb/axle) = 10 percent
The vehicle mix is expected to remain the same throughout the design life of
20 years, although traffic is expected to grow at a rate of 3.5 percent
annually. Using the AASHTO design procedure, determine the minimum
depth of concrete pavement required for the design period of 20 years.
Pi = 4.5
J = 3.2
Cd = 1.0
Pt = 2.5
S'c = 650 lb/in2
So = 0.3
6
2
Ec = 5 x 10 lb/in
R = 95%
k = 130 lb/in3
First, determine the growth factor for 20 years at 3.5%.
Growth factor = [(1 + 0.035)20 - 1] / 0.035
Growth factor = 28.28
Next, assume a pavement thickness and estimate the ESALs for the 20 year
period.
Assume depth of 9 inches
From Table 20.7:
ESAL of 12 kips = 0.176
ESAL of 16 kips = 0.604
ESAL for 2-axle single unit trucks:
ESAL = 0.176(2)(0.4)(28.28)(24,000)(365)(0.40)
ESAL = 13.95 * 106
ESAL for 3-axle single unit trucks:
ESAL = 0.604(3)(0.1)(28.28)(24,000)(365)(0.40)
ESAL = 17.96 * 106
Total ESAL = 31.91 * 106
Next, determine the design serviceability loss
design serviceability loss = 4.5 - 2.5
design serviceability loss = 2.0
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352
Chapter 20: Design of Rigid Pavements
From the nomographs provided in Figures 20.13 and 20.14 and the factors above,
the required thickness of the concrete slab is 12.3 inches, which is far from the
assumed value of 9 inches. Hence, 9 inches cannot be used as the design depth.
Assume depth of 12 inches
From Table 20.7:
ESAL of 12 kips = 0.174
ESAL of 16 kips = 0.599
ESAL for 2-axle single unit trucks:
ESAL = 0.174(2)(0.4)(28.28)(24,000)(365)(0.40)
ESAL = 13.79 * 106
ESAL for 3-axle single unit trucks:
ESAL = 0.599(3)(0.1)(28.28)(24,000)(365)(0.40)
ESAL = 17.81 * 106
Total ESAL = 31.60 * 106
From the nomographs provided in Figures 20.13 and 20.14 and the factors above,
the required thickness of the concrete slab is 12.2 inches, which is close to the
assumed value of 12 inches. Hence, the rounded value of 12 inches will be used
as the design depth.
20-15
Repeat Problem 20-14 for a pavement containing doweled joints, 6 inch untreated
subbase, and concrete shoulders, using the PCA design method.
First, determine the estimated ESALs:
2-axle (12,000 lb) single unit trucks:
ESAL = 2(0.4)(28.28)(24,000)(365)(0.40)
ESAL = 79.27 * 106
3-axle (16,000 lb) single unit trucks
ESAL = 3(0.1)(28.28)(24,000)(365)(0.40)
ESAL = 29.73 * 106
Subgrade k = 130 lb/in3
For 6 inch untreated subbase, determine effective k from Table 20.10:
k = 140 + [(230 - 140) / 100] * 30 = 167 lb/in3
353
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Chapter 20: Design of Rigid Pavements
Assume a slab 6 inches thick with doweled joints and concrete shoulders.
From Table 20.12, obtain equivalent stress values and then interpolate for
effective k,
Equivalent stress = 304 - ((304 - 289) * (33 / 50)) = 298.9
Stress Ratio = 298.9 / 650 = 0.460
From Table 20.15, determine the erosion factor of 2.72
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions (ESALs):
Load
14,400
19,200
Fatigue analysis
unlimited
7 x 105
Erosion analysis
unlimited
2.6 x 106
The 6 inch slab is inadequate since the allowable ESALs are less than the
expected ESALs.
Assume a slab 7 inches thick with doweled joints and concrete shoulders.
From Table 20.12, Equivalent stress = 248 - ((248 - 236) * (33 / 50)) =
243.9
Stress Ratio = 243.92 / 650 = 0.38
From Table 20.15, determine the erosion factor of 2.54
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions:
Load
Fatigue analysis
14,400
unlimited
19,200
unlimited
Erosion analysis
unlimited
100 x 106
Damage percent = 29.73 x 106 / 100 x 106 = 29.73%
The 7 inch slab is adequate, check a 6.5 inch slab.
Assume a slab 6.5 inches thick with doweled joints and concrete shoulders.
From Table 20.12, equivalent stress = 274 - ((274 - 260) * (33 / 50)) =
269.24
Stress Ratio = 269.24 / 650 = 0.41
From Table 20.15, determine the erosion factor of 2.62
12,000(1.2) = 14,400
16,000(1.2) = 19,200
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354
Chapter 20: Design of Rigid Pavements
Calculate the allowable repetitions:
Load
Fatigue analysis
14,400
unlimited
19,200
unlimited
Erosion analysis
unlimited
10 x 106
The 6.5 inch slab is inadequate, use a 7 inch slab depth.
20-16
Repeat Problem 20-14 using the PCA design method, for a pavement containing
doweled joints and concrete shoulders. The modulus of rupture of the concrete used
is 600 lb/in2.
First, determine the estimated ESALs:
2-axle (12,000 lb) single unit trucks:
ESAL = 2(0.4)(28.28)(24,000)(365)(0.40)
ESAL = 79.27 * 106
3-axle (16,000 lb) single unit trucks
ESAL = 3(0.1)(28.28)(24,000)(365)(0.40)
ESAL = 29.73 * 106
Subgrade k = 130 lb/in3
For 6 inch untreated subbase, determine effective k from Table 20.10:
k = 140 + [(230 - 140) / 100] * 30 = 167 lb/in3
Assume a slab 7 inches thick with doweled joints and concrete shoulders.
From Table 20.12, obtain equivalent stress values and then interpolate for
effective k,
Equivalent stress = 248 - ((248 - 236) * (33 / 50)) = 243.9
Stress Ratio = 243.9 / 600 = 0.41
From Table 20.15, determine the erosion factor of 2.54
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Now, calculate the allowable repetitions (ESALs):
Load
Fatigue analysis
Erosion analysis
14,400
unlimited
unlimited
5
19,200
7 x 10
100 x 106
Damage percent = 29.73 x 106 / 100 x 106 = 29.73%
The 7 inch slab is adequate, check a 6.5 inch slab.
355
(c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 20: Design of Rigid Pavements
Assume a slab 6.5 inches thick with doweled joints and concrete shoulders.From
Table 20.12, Equivalent stress = 274 - ((274 - 260) * (33 / 50)) = 269.24
Stress Ratio = 269.24 / 600 = 0.45
From Table 20.15, determine the erosion factor of 2.62
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Now, calculate the allowable repetitions:
Load
Fatigue analysis
14,400
unlimited
19,200
unlimited
Erosion analysis
unlimited
8 x 106
The 6.5 inch slab is inadequate since the allowable ESALs are less than
the expected ESALs.
The 6.5 inch slab is inadequate, use a 7 inch slab depth.
20-17
Repeat Problem 20-14, using the PCA design method, if the subgrade k value is 50
and a 6-inch stabilized subbase is used. The modulus of rupture of the concrete is
600 lb/in2 and the pavement has aggregate interlock joints (no dowels) and a
concrete shoulder.
First, determine the estimated ESALs:
2-axle (12,000 lb) single unit trucks:
ESAL = 2(0.4)(28.28)(24,000)(365)(0.40)
ESAL = 79.27 * 106
3-axle (16,000 lb) single unit trucks
ESAL = 3(0.1)(28.28)(24,000)(365)(0.40)
ESAL = 29.73 * 106
Subgrade k = 50 lb/in3
For 6 inch stabilized subbase, determine effective k from Table 21.10:
k = 230 lb/in3
Assume a slab 6.5 inches thick with aggregate interlock joints and concrete
shoulders.
From Table 20.12, obtain equivalent stress values and then interpolate for
effective k,
Equivalent stress = 260 - ((260 - 243) * (30 / 100)) = 254.9
Stress Ratio = 254.9 / 600 = 0.42
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356
Chapter 20: Design of Rigid Pavements
From Table 20.16, determine the erosion factor of 2.80
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions (ESALs):
Load
Fatigue analysis
Erosion analysis
14,400
unlimited
60 x 106
19,200
unlimited
1.2 x 106
The 6.5 inch slab is adequate under fatigue analysis but not for the erosion
criterion.
Assume a slab 8 inches thick with aggregate interlock joints and concrete
shoulders.
From Table 20.16, determine the erosion factor of 2.56
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions:
Load
Erosion analysis
14,400
unlimited
19,200
4 x 106
The 8 inch slab is adequate, check a 7.5 inch slab.
Assume a slab 7.5 inches thick with aggregate interlock joints and concrete
shoulders.
From Table 20.16, determine the erosion factor of 2.56
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions:
Load
Erosion analysis
14,400
unlimited
19,200
6 x 106
The 7.5 inch slab is inadequate, use a 8 inch slab depth.
357
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Chapter 20: Design of Rigid Pavements
20-18
Repeat Problem 20-17, using the PCA design method, assuming the pavement has
doweled joints and no concrete shoulders.
First, determine the estimated ESALs:
2-axle (12,000 lb) single unit trucks:
ESAL = 2(0.4)(28.28)(24,000)(365)(0.40)
ESAL = 79.27 * 106
3-axle (16,000 lb) single unit trucks
ESAL = 3(0.1)(28.28)(24,000)(365)(0.40)
ESAL = 29.73 * 106
Subgrade k = 50 lb/in3
For 6 inch stabilized subbase, determine effective k from Table 20.10:
k = 230 lb/in3
Assume a slab 8 inches thick with doweled joints and no concrete shoulders.
From Table 20.11, obtain equivalent stress values and interpolate for k,
Equivalent stress = 242 - ((242 - 225) * (30 / 100)) = 236.9
Stress Ratio = 236.9 / 600 = 0.39
From Table 20.13, determine the erosion factor of 2.80
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions (ESALs):
Load
Fatigue analysis
Erosion analysis
14,400
unlimited
unlimited
19,200
unlimited
20 x 106
The 8 inch slab is adequate under fatigue analysis but not for the erosion criterion.
Assume a slab 8.5 inches thick with aggregate interlock joints and concrete
shoulders.
From Table 20.13, determine the erosion factor of 2.72
12,000(1.2) = 14,400
16,000(1.2) = 19,200
Calculate the allowable repetitions (ESALs):
Load
Erosion analysis
14,400
unlimited
19,200
40 x 106
Damage percent = 29.73 x 106 / 40 x 106 = 74.3%
Use a 8.5 inch slab depth.
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358
Chapter 20: Design of Rigid Pavements
20-19
Briefly describe the steps involved in the MEPDG method for JPCP.
The method of pavement design applied in the MEPDG consists of five
basic steps. The first step is to select a trial design based on another method
(such as the 1993 AASHTO Pavement Design Guide, agency policies or
standards, or a design based on engineering judgment and experience. The
second step is selection of acceptable thresholds for performance criteria related
to pavement distresses of mean transverse joint faulting, transverse slab cracking
(bottom-up and top-down), and smoothness. The third step involves the
collection of all input data pertaining to project information and to traffic,
pavement structure, material properties, and climate. Fourth, the trial design is
evaluated, typically through the MEPDG software, by examining the performance
indicators. Finally, the structural viability of the trial design is evaluated and the
design revised and evaluated until an acceptable design is achieved.
20-20
What are the main input parameters used in the MEPDG for JPCP?
The input parameters for design of jointed plain concrete pavement are
similar to those used in flexible pavement design. These include traffic,
foundation and subgrade soils and material characteristics. Traffic is
characterized by the distribution of axle weights, or loads, also known as load
spectra. This differs from traditional empirical approaches that are based on
converting traffic loads to equivalent single axle loads (ESALs).
20-21
List and define each of the main criteria that are used for evaluating a trial JPCP
structure in the MEPDG method.
The performance criteria used to evaluate a JPCP design are mean
transverse joint faulting, transverse slab cracking (bottom-up and top-down), and
smoothness. Threshold values deemed acceptable for each criterion are selected
initially. The values that result from analysis of a trial design using the MEPDG
are then compared to the acceptable thresholds and the design evaluated
accordingly. Transverse joint faulting refers to the extent to which transverse
joint have faulted such that continuity between slabs is lost. This is evaluated on
a monthly, incremental basis for the pavement design life. Transverse slab
cracking is evaluated as the percentage of slab expected to have transverse cracks
at the end of the design life of the proposed pavement structure. Smoothness
pertains to the profile of the pavement; the greater the deviation in the profile
from its design, the less smooth the pavement is said to be.
359
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Chapter 20: Design of Rigid Pavements
20-22
Determine the expected predicted transverse cracking for the five traffic levels and
associated applied stresses given on Table 20.19 for a given condition of age, season,
and temperature difference if the Modulus of Rupture of the PCC is 650 lb/in2.
Solution:
Load Level Number of
Applications
(n)
Applied
Stress
(σi,j,k,l,m,n)
Allowable
Number of Load
Applications
n/N
(lb/in2) Axle (N)
(1)
1
(2)
(3)
0.02 x 106
300
0.137 x 106
0.15
2
0.05 x 106
280
0.387 x 106
0.13
3
0.08 x 106
275
0.515 x 106
0.16
4
0.10 x106.
270
0.694 x 106
0.14
5
0.12 x106
260
1.308 x 106
0.09
Sum 0.67
•
•
•
•
•
Determine the allowable number of load applications for each axle level load
using Eq.20.24. See column 4 in the Table.
Determine the ratio of number of load applications to allowable number of load
applications for each axle load level. See column 5 in the Table.
Determine total fatigue damage from Equation 20.23
DIF = 0.13 + 0.16 + 0.14 + 0.09 = 0.67
Determine predicted amount of transverse cracking using Equation 20.22
CRK =
1
1
= 0.312
=
−1.98
−1.98
1 + (DI F )
1 + (0.67 )
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360
Chapter 21: Pavement Management
Chapter 21
Pavement Management
21-1
What is meant by the term pavement management? Describe three strategies used by
public agencies to develop restoration and rehabilitation programs.
Pavement management refers to the various strategies that can be used to
select a program for pavement restoration. Three strategies used by public agencies
to develop programs include: a “squeaky wheel” approach in which projects
attracting the greatest attention are selected; a program in which all roads are
repaired on a regular schedule regardless of relative project costs; and a program in
which minimum standards are set and a program of priorities established within
budget limitations.
21-2
What are the three principal uses for pavement condition data?
The three principal uses for pavement condition data are: to establish
project priorities; establish rehabilitation options; and forecast performance.
21-3
What is the difference between PSI and PSR?
PSI (present serviceability index) is a value for pavement condition
determined as a surrogate for PSR (present serviceability rating) and is based on
physical measurements. PSR is a number grade given to a pavement section based
on its ability to serve traffic. It is established by observation and requires judgment
by trained personnel. Therefore, PSR is subjective, while PSI is objective.
21-4
Draw a sketch showing the relationship between pavement condition (expressed as
PSI) and time for a service life of 20 years, if the PSI values range from 4.5 to 2.5 in a
six year period, and then the pavement is resurfaced such that the PSI is increased to
4.0. After another 6 years, the PSI has reached 2.0. With rehabilitation, PSI is
increased to 4.5. At the end of its service life the PSI value is 3.4
The following sketch depicts the relationship between pavement condition
(PSI) and time for a service life of 20 years using the data supplied.
361 (c) 2009 Cengage Learning ALL RIGHTS RESERVED.
Chapter 21: Pavement Management
5
PSI
4
3
2
1
0
0
5
10
15
20
Y ears
21-5
Describe the four characteristics of pavement condition used to evaluate whether a
pavement should be rehabilitated, and if so, determine the appropriate treatment.
The four characteristics of pavement condition used to evaluate whether a
pavement should be rehabilitated are:
Pavement Roughness is a measurement of the extent to which a road surface
deviates from a plane.
Pavement Distress refers to the condition of a pavement in terms of its general
appearance. Distress is a combination of fractured, distorted, or disintegrated
pavement.
Pavement Deflection refers to the structural adequacy of a pavement section. Tests
may be dynamic, static, or destructive.
Skid Resistance describes the effectiveness of a pavement to prevent or reduce skidrelated crashes.
21-6
A given pavement rating method uses six distress types to establish the DR. These
are: corrugation, alligator cracking, raveling, longitudinal cracking, rutting, and
patching. For a section of highway, the number of points assigned to each category
were 6, 4, 2, 4, 3, and 3. If the weighting factors are 2, 1, 0.75, 1, 1, and 1.5,
respectively, determine the DR for the section.
Distress rating (DR) can be calculated using Equation 21.1,
DR = 100 - Σ (di * wi)
DR = 100 - ((6)(2) + (4)(1) + (2)(0.75) + (4)(1) + (3)(1) + (3)(1.5))
DR = 100 - 29
DR = 71
Therefore, the DR for this pavement section is 71.
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362
Chapter 21: Pavement Management
21-7
What are the problems with distress surveys, and how can such problems be solved?
One of the major problems with distress surveys is the variability in results
due to the subjective procedures used, and the concerns about the safety of
pavement surveyors and the hazards associated with their being on the road. The
best solution lies in the use of automated techniques for evaluating distress. This
would remove the survey crew from the hazardous road conditions, and allow for
more consistent results.
Other sources of error are variations in the condition of the highway
segment observed, changes in evaluation procedure, and changes in observed
location from year to year. Some of these variations can be minimized by using the
same pavement location each year by observing pavement condition at regular
intervals of about one mile.
21-8
How are computers used in pavement management?
Computer vision is an area of artificial intelligence (AI) research that
involves the use of sensors and computers to emulate human vision. As applied to
pavement management, the objective is to develop an automated approach capable
of identifying the type, severity, and extent of pavement distresses. The process
generally involves the following four basic steps: image acquisition, image
digitization, image processing, and pattern recognition.
For pavement management, the image is usually captured using a video
camera. Digitization then creates a numerical representation of the image (digitized
image) suitable for computer processing in the form of an array of numbers where
the integer value of each element in the array represents the color or the gray tone
of the corresponding area in the original image. Values range from 0, representing
black, to 255, representing white. The cracked region will have low values due to
crack shadows being much darker than the surrounding background.
The image processing step seeks to extract the different distresses from the
background. The image is first transformed into what is known as a “binary
image”, where the distresses are indicated by black pixels and the background by
white. Algorithms are also applied to clean up the image and remove the “noise.”
Counting the number of black pixels will thus yield an idea of the distressed area.
Finally, the pattern recognition step classifies the distress in a pavement surface
image as a particular type of distress.
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Chapter 21: Pavement Management
21-9
Describe the methods used to determine static and dynamic deflection of pavements.
To what extent are these tests used in pavement rehabilitation management?
The methods used to determine the static and dynamic deflection of
pavements are measurement of static deflections (Benkelman Beam), and
measurement of dynamic or repeated loads (Dynaflect). These methods are rarely
used to evaluate pavement condition, but deflection data are used for design
purposes and to develop strategies.
21-10
A 5000-lb load is placed on two tires, which are then locked in place. A force of 2400
lb is necessary to cause the trailer to move at a speed of 20 mi/h. Determine the value
of the skid number. If treaded tires were used, characterize the pavement type.
First, calculate the skid number using Equations 21.2 and 21.3,
SK = 100 * (L / N)
SK = 100 * (2400 / 5000) = 48
Therefore, the skid number would be 48. Based on the skid number determined
above and Figure 21.13, the pavement can be characterized as having a surface type
(3), fine-textured and gritty.
21-11
The PMS database has accumulated information regarding the performance of a
pavement section before being overlaid as well as the performance of the overlay. The
following data were recorded. Develop a linear prediction model (PCR=a + b*Age)
for the two cases. If a criterion was established that maintenance or rehabilitation
should occur when the PCR reaches a value of 81, when should such actions take
place in the two cases?
Using regression analysis, the prediction model for the pavement section before
being overlaid is:
PCR = 101.47 - 2.31(AGE)
(with an R2 = 0.997)
Using regression analysis, the prediction model for the overlay is:
PCR = 100.96 - 2.72(AGE)
(with an R2 = 0.975)
Therefore, the original pavement section will reach a PCR value of 81 at age:
81 = 101.47 - 2.31(AGE)
AGE = (101.47 - 81) / 2.31
AGE = 8.9 years
The overlay will reach a PCR value of 81 after:
81 = 100.96 - 2.72(AGE)
AGE = (100.96 - 81) / 2.72
AGE = 7.3 years
The first action should take place after 8.9 years for the original section, and after
7.3 years for the overlay.
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364
Chapter 21: Pavement Management
21-12
Referring to the Markovian transition matrix in Table 21.7, what is the probability
that a section with a PCR value of 77 in the current year will have a PCR value
between 60 and 69 (a) one year later, and (b) two years later.
(a) According to the transition matrix, the section is currently in state 7. Therefore
the probability of dropping to state 6 (PCR between 60 and 69) in the next year is
0.30.
(b) To predict the condition two years ahead, one will have to apply the matrix
twice, and consider all possibilities as follows:
A section currently in state 7 can in one year:
•
remain in state 7 with a probability of 0.60
•
drop to state 6 with the probability of 0.30
•
drop to state 5 with a probability of 0.10
For the second year:
•
A section in state 7 can drop to state 6 with a probability of
0.30
•
A section in state 6 can remain in state 6 with a probability
of 0.50
•
A section in state 5 can move to state 6 with a probability of
0.00
Therefore, the probability of that section currently in state 7 of being in state 6 two
years later can be calculated as:
Probability = (0.60)(0.30) + (0.30)(0.50)
Probability = 0.33
The probability of the given section being in state 6 in two years is 0.33.
21-13
Differentiate between corrective and preventive rehabilitation techniques. Cite three
examples of surface treatments in each category. What is the best preventive
maintenance technique for subsurface maintenance?
Corrective rehabilitation involves the permanent or temporary repair of
deficiencies on an as-needed basis. Techniques include patching, crack filling,
joint sealing, and seal coat (with aggregate). Preventive rehabilitation involves
surface applications of either structural or non-structural improvements intended to
keep the quality of the pavement above a predetermined level. Techniques include
fog seal asphalt rejuvenators, joint sealing, and seal coat (with aggregate). The best
preventative technique for subsurface maintenance is proper and functional
drainage.
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Chapter 21: Pavement Management
21-14
Describe the techniques used to repair flexible and rigid pavements and their
effectiveness for the following treatment types: (a) patching, (b) crack maintenance,
and (c) overlays.
(a) Patching
Techniques: Temporary, permanent, spot seal (spray), cold mix, hot mix, and level.
Effectiveness: When properly done, patching is considered to be effective.
Temporary patching is considered moderately effective to serve for a short period
until permanent repairs can be made.
(b) Crack Maintenance
Techniques: Crack cleaning, sealing and filling
Effectiveness: Relatively effective. Fairly short life. Must be repeated often (1- 2
years)
(c) Overlays
Techniques: Thick or thin. Pavement reinforcing, fabric overlays, inverted
overlays
Effectiveness: Considered to be an effective technique.
21-15
This problem is not to be assigned. The problem will be deleted from future printings
of the 4th edition.
21-16
What is the basic difference between an expert system and a conventional computer
program?
The basic difference lies in the fact that an expert system utilizes knowledge
or experience, and therefore they can address problems that can only be solved
through heuristics or subjective knowledge. Conventional computer programs, on
the other hand, are basically quantitative and computational in nature, and address
problems where a defined answer exists.
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366
Chapter 21: Pavement Management
21-17
Discuss the differences between condition and priority assessment models used in
developing pavement improvement programs.
Condition assessment is a method for developing single-year programs.
The pavement condition is compared against an established criterion or “trigger
point” to decide whether a deficiency or need exists. Consequently, all sections are
separated into two groups: “now needs” and “later needs”. The condition
assessment model is only concerned with the “now needs”. Having decided upon
which sections need an action, the next step is to select appropriate treatments.
This could involve the use of simple economic analysis methods. If the needs
exceed the budget, a ranking of the potential projects is performed to determine
which of the needs could be deferred to the next year.
Priority assessment, on the other hand, is a method for developing
multiyear programs, and therefore, may be considered an extension of the condition
assessment method. In this method, performance prediction models are used to
predict when each road section will reach its trigger point. That is, instead of
separating the network into two groups as with the condition assessment method,
this method separates the pavement sections into a number of groups according to
the year where action will be needed. The process from this point is essentially the
same as the condition assessment method.
21-18
Describe six methods that can be used to select a program of pavement rehabilitation.
The six methods that can be used to select a program of pavement
rehabilitation are: 1) Cost effectiveness, 2) Economic cost analysis (Present
Worth), 3) Allocation process, 4) Sufficiency ratings, 5) Visual inspections, and 6)
Management plans. The method selected will depend on data availability and
viewpoints of transportation agencies concerning the relevant factors for selecting
projects.
21-19
This problem is not to be assigned. The problem will be deleted from future printings
of the 4th edition.
21-20
This problem is not to be assigned. The problem will be deleted from future printings
of the 4th edition.
21-21
This problem is not to be assigned. The problem will be deleted from future printings
of the 4th edition.
367 (c) 2009 Cengage Learning ALL RIGHTS RESERVED.