CE 345 Home Works
Home Work #1:
Q1) The following relationship between the traffic speed, and density was obtained from
observation on a freeway (U=52-0.35K). Plot the relationship between U & K, Q & K, Q & U
giving the equations for each?
solution:
Figure is similar to figure 7.7 in the slides.
Q2) A number of 500 vehicles passed a point on the roadway during 20 minutes. Find the mean
flow rate and the traffic density if the average travel speed is 50 mph?
Solution:
Q3) A number of 100 vehicles passed through a roadway section over a time period 10 minutes.
Find the mean density for that section if it has a length of 100 ft, and the average time for the
crossing vehicles was 1.12 second?
Solution:
Q4) The following vehicles speeds were measured by radar at certain point on the highway: 30,
45, 35, 40, 50, 38, 42, 39, 51, & 47 mph. Calculate their time mean speed and space mean
speed?
Solution:
∑
√ √
Q5) If the ADT on a highway was 30000 vph in the year 2000, and became 40000 vph in the year
2010, estimate the predicted value in the year 2015?
Solution:
First use the ADT between 2000 and 2010 to get the growth factor value, then sub between
2010 and 2015
( ) ( )
( ) ( )
Q6) A peak hour traffic count was made at a roadway during 6:00-7:00 AM and the following
result were obtained. Find the peak hour factor?
Time Period 6:00-6:15 6:15-6:30 6:30-6:45 6:45-7:00
# of vehicles 300 290 330 280
Flow Rate
1200 1160 1320 1120
( )
∑
Home Work #2:
Q1) The maximum capacities that can be reached at a different highway types are: freeway
(2400 pcphpl), multilane highway (2200 pcphpl), and two-lane two-way (2800 vph total from
both directions). Explain the major reasons for these differences.
Solution:
The Major reasons are the number of lanes and existing of median.
Q2) Given the following conditions for suburban highway: 10000 current ADT, peak hour volume
is 10% of daily traffic volume (K-factor), 50/50 directional split (D-factor), Estimate the future
(after 20 years) directional design hourly volume (DDHV) assuming a growth rate of 3%?
Solution:
( ) ( ) ( )
Q3) Given the following conditions for a rural highway, 60000 future AADT, level terrain, 100
KM/hr basic free flow speed, 10 percent trucks, no recreational vehicles, peak hour volume is
10% of daily traffic, 50/50 directional split, 0.95 peak hour factor. What is the required number
of lanes for level of service C if you design it as a multi-lane highway. Use the ideal conditions for
any missing information ?
Solution:
( ) ( ) ( )
Home Work #3:
Q1) Calculate the minimum radius for a simple circular curve assuming a design speed of 50
mph, maximum superelevation rate of 8%, and side friction factor of 0.14?
Solution:
Q2) Calculate the required spiral length for a simple curve that has a design speed of 50 mph
and a curve radius of 1000 ft. Assume C=1?
Solution:
( )
Q3) Draw a profile showing the super elevation transition from crown 2% until full
superelevation is reached 8% assuming 2-lane 2-way highway. Show all important distance?
Solution:
Q4) Check the need for climbing lane on a 2-lane 2-way highway assuming an entering speed of
70 mph and a grade of 5% with a grade length of 900 ft?
Solution:
Since they intersect above the 15 mph
line then climbing lane is needed
Q5) Calculate the stopping sight distance for a highway that has a design speed of 60 mph,
assuming an upgrade of 3% and friction coefficient of 0.3?
Solution:
( ) ( )
Q6) How many cars cab be placed in a commercial parking lot on land that is 40 m wide (street
frontage) and 100 m deep. Draw a sketch showing the design assuming two driveways and one
way circulation?
Solution:
Home Work #4:
Q1) What is the sign type (regulatory, warning or guide) for each of the following traffic signs?
At grade rail crossing Stop sign Amman 30 Km
Warning Regulatory Guide
Q2) draw a sketch showing the numbers & types of conflict points at a T-intersection?
Solution:
Q3) The intersection of first street (NB/SB) and second street (EB/WB) is located in the central
business district CBD of a town with a population of 50000, and 40 Km/hr approach speeds, on a
level terrain with the following geometric and traffic data. Check all the vehicle volume related
traffic signal warrant?
EB WB NB SB
Item
L T R L T R L T R L T R
PHV (vph) 20 450 30 70 350 80 50 160 40 50 100 40
# of lanes 1 1 1 1 1 1 1 1 1 1 1 1
Solution:
Firstly we can notice from the table that EB/WB is the major street with 3 lanes in each
direction, while NB/SB is the minor street with also 3 lanes in each direction.
( )
For warrant 1:
1000>600 & 250>200 then intersection is warranted according to A 100%
For warrant 2: since above the 2 & 2 line then warranted For warrant 3: since below the 2 & 2 line then not warranted
Q4) Design the following intersection using Pignataro’s method for a three phase system where:
1- Traffic Volumes as shown on the sketch (vph).
2- a=0.27g
3- Lane width = 12 ft 250
4- Pedestrian start up time = 5 sec
5- Average length of vehicle = 20 ft 90
350
6- Average walking speed = 4 ft/sec
7- Average headway =2.5 seconds 70
8- Perception reaction time = 2.5 seconds
9- Peak hour volume = 0.95
10- Left turn factor = 1.6
11- Truck factor = 1.4
12- Speed = 30 mph
13- Trucks = 5% on all approaches
145 270
Solution:
Procedure Corrected volume
Volume/PHF = A A*left turn factor of exist = B B*%Truck = C (vph)
C*Truck Factor = D Corrected volume = B-C+D
Note: don’t round any number till the end, and round to the largest
number (ex:70.1 71, 70.9 71)
250/0.95 = 263.158 263.158*0.05 = 13.158 13.158*1.4 = 18.42 vph 269
263.158-13.158+18.42 = 268.42 = 269 vph
90/0.95 = 94.73 94.73*1.6 = 151.568 151.568*0.05 = 7.5784
7.5784*1.4 = 10.609 vph
151.568-7.5784+10.609 = 154.6 = 155 vph
350/0.95 = 368.42 368.42*0.05 = 18.42 18.42*1.4 = 25.788 vph 376
368.42-18.42+25.788 = 375.788 = 376 vph (This is critical
volume)
70/0.95 = 73.68 73.68*1.6 = 117.888 117.888*0.05 = 5.8944 121
5.8944 *1.4 = 8.25 vph
117.888-5.8944+8.25 = 120.24 = 121 vph
270/0.95 = 284.21 284.21*0.05 = 14.21 14.21*1.4 = 19.894 vph 290
284.21-14.21+19.894 = 289.89 = 290 vph (This is critical
volume)
145/0.95 = 152.63 152.63*1.6 = 244.208 244.208*0.05 = 12.21 249
12.21*1.4 = 17.09 vph
244.208-12.21+17.09 =249.09 = 249 vph
269
155
376
121
249 290
∑( )
∑( ) ( )
( ∑( )) ( )
∑
( ∑( )) ( )
∑
( ∑( )) ( )
∑
