CEE 3604: Introduction to Transportation Engineering
Fall 2012
Assignment 2: Basic Transportation Data Analysis
Date Due: September 14, 2011
Instructor: Trani
Problem 1
One of the basic problems in transportation engineering is determining travel time and distance for vehicles traveling between
points in a network. Data loggers using Global Positioning System (GPS) in cars and buses provide a good tool to understand
the efficiency of highway networks. Data collected in Blacksburg using an instrumented vehicle is provided in the file called
carDataLog.m and shown in Figure 1.
Figure 1. Sample Car Data Information Collected in Blacksburg.
a)
Import the data using Matlab’s “load” command as demonstrated in class.
b)
Create a new Matlab script to do the following:
b.1) Create three new vector variables label them: time, speed and acceleration. These three vectors contain the
column data in each file.
b.2) Plot time traveled (in x-axis) vs speed (y-axis). Label axes as needed (include units) and change default font sizes
to size 20 for both x and y labels. Change the color of the line to solid blue. Comment on the shape of the profile
observed.
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Figure 1. Speed vs Time Profile.
b.3) Plot time traveled (in x-axis) vs acceleration (y-axis). Label axes and change default font sizes to size 20 for both x
and y labels. Change the color of the line to red. Comment on the shape of the profile observed.
Figure 2. Acceleration vs Time Profile.
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c)
Estimate the average speed for the complete profile. In Matlab we can easily calculate averages using the mean
command. For example:
mean(x) calculates the mean of a vector variable x. Does the average speed seems reasonable?
The average speed of the profile is 11.3 m/s (25.3 mph).
d)
Estimate the total distance traveled by this car. Use a numerical approximation similar to the procedure demonstrated
in the lectures notes Performance of Ground Vehicles (http://128.173.204.63/courses/cee3604/cee3604_pub/
transportation_technology.pdf) pages 21 and 31-31d.
The equation to evaluate numerically this problem is:
(1)
We can setup an Excel spreadsheet (see Figure 3) to solve the problem using a two-second step size (delta t).
Column 1= contains the time of the event
Column 2 = has the velocity of the vehicle directly from the data logged and extracted from the file
Column 3 = Rate of change of distance traveled as a function of time (speed - same as Column 2)
Column 4 = Step size (times between successive points (2 seconds for the data provided)
Column 5 = Product of step size and dS/dt per equation (1).
Column 6 = Distance traveled (assumed zero distance for the first row).
......
!
Figure 3. Excel Spreadsheet to Estimate Distance Traveled.
At the end of 754 second journey, the vehicle has traveled 8,564 meters.
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Problem 2
This past summer Virginia Tech collected traffic data on Highway 460 near Blacksburg. A file called samples of vehicle speed and
highway density data collected. Traffic density and speed are two key variables of interest to transportation engineers to study
the level of service offered by highway. The file contains information similar to that shown below
Figure 2. Sample Traffic Data Collected on Highway 460.
Create a Matlab script to:
a) Load the highway data as demonstrated in class (see script that follows).
b) Create two variables named speed and density to study perform the next steps in the problem.
c) Use Matlab to plot the values of traffic density (x-axis) vs speed (y-axis). Comment of the trend observed. Label the axes
in the plot and use a green marker “^” to indicate each data point in your plot.
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Figure 3. Density vs. Speed Plot.
d) Using the “Basic Fitting” capabilities in Matlab (look at the “Tools” pull down menu in your plot), fit a first degree
polynomial to the data. Indicate the equation of the polynomial and comment on how well the polynomial fits the data.
e) If highway “volume” is measured as the product of density and speed, improve your script to estimate the volume of traffic
for every data point of speed-density recorded.
f) Plot the traffic volume (y-axis) vs. speed (in x-axis). Do you notice any trend? Comment.
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Figure 4. Speed Flow Relationship.
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Problem 3
An empirical formula to estimate the fuel used by a Boeing 747-400 (see Figure 4a) flying international routes is given by:
Fuel = 3181 + 18.4D flown
where:
Fuel is the fuel used in kilograms.
D flown is the distance flown in nautical miles (one nautical mile is 1.15 statute miles)
Figure 4a. Boeing 747-400 at Punta Cana International Airport (A. Trani).
a) Use Matlab or Excel to estimate the fuel used for a given distance ( D flown )
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!
Figure 5. Script to Generate Values of Fuel vs. Distance Flown.
b) Plot the fuel used (in pounds) various distance segments ranging from 1,000 to 5,500 nautical miles (at steps of 100 nm).
Plot the solutions obtained in part (a) and label accordingly.
Figure 6. Plot of Fuel Used vs. Distance Flown for Boeing 747-400.
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c) Find the fuel use for a trip from Punta Cana (airport code PUJ) to Moscow (airport code UUDD) (route distance is 5,030
nautical miles). You can distances between airports at: http://www.gcmap.com.
Based on the plot shown, the aircraft would consume 95,733 kgs.
d) Repeat (c) for a Honolulu-San Francisco flight (2,745 nm).
Based on the plot shown, the aircraft would consume 53,689 kgs.
e) A hybrid car like a Toyota Prius has a fuel economy of 46 miles per gallon under normal highway driving conditions. Find
the number of miles a Prius could travel using the fuel consumed by the Boeing 747-400 in that single flight. A gallon of
jet fuel is equivalent to 3.045 kilograms.
31,439 gallons in the PUJ-UUDD flight translates to 1.45 million miles traveled in the Toyota Prius. This is 6 times the
distance from the Earth to the Moon! Pretty remarkable. Another interesting fact, each kilogram of jet fuel produces about
3.15 kilograms of CO2. This means that the flight from Punta Cana to Moscow will produce 301,560 kg of carbon
emissions (301.6 metric tons). In the European Union Carbon Market, each ton of CO2 is worth $20 per ton. So next time
you travel to Europe, you can offset your carbon footprint for the trip (pay for each ton of CO2) by paying $15 dollars each
way (around 0.75 tons per passenger one way assuming 400 passengers in the plane).
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Problem 4
A new HIgh-Speed Rail System under evaluation for the California Corridor has the following technical characteristics:
S = 20 sq. meters, froll = 0.02, rho=1.225 kg/cu.m., Cd = 0.40 (dim), Cl = 0.1 (dim) and mass = 450,000 kg. The train has the
tractive effort characteristics shown in Table 1.
Table 1. High-Speed Rail Tractive Effort Characteristics.
Speed (m/s)
Tractive Force (N)
0
360,000
30
285,000
60
220,000
90
180,000
a) Derive a 2nd order polynomial (use either Matlab basic curve fitting function or Excel trend line analysis) that relates the
tractive effort and speed. The equation should be of the form:
where TE is the tractive force (N), a, b and c are regression coefficients obtained by regression analysis, and V is the
speed (m/s) of the high-speed rail system. Using Matlab (see Figure 7) we obtain the following quadratic equation:
TE = 3.608e5 − 2892V + 9.722V 2 !
!
!
!
!
!
!
(1)
where TE is the tractive effort in Newtons and V is the train speed in meters per second.
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Figure 7. Tractive Force vs. Speed Diagram for High-Speed Train.
b) Using the equation obtained in part (a) derive the fundamental equation of motion for the vehicle similar to that shown on
page 16 of the course notes Performance of Ground Vehicles (http://128.173.204.63/courses/cee3604/cee3604_pub/
transportation_technology.pdf).
TE = 3.608e5 − 2892V + 9.722V 2
dV 1
= (TE − D − (mg − L) froll )
dt m
dV 1 ⎛
1
1
⎞
= ⎜ TE − ρ SCdV 2 − (mg − ρ SClV 2 ) froll ⎟
⎝
⎠
dt m
2
2
dV 1 ⎛
1
1
⎞
= ⎜ 3.608e5 − 2892V + 9.722V 2 − (1.225)(20)(0.4)V 2 − (450000 * 9.81− (1.225)(20)(0.1)V 2 )* 0.02 ⎟
⎠
dt m ⎝
2
2
dV 1
=
3.608e5 − 2892V + 9.722V 2 − 4.9V 2 − 88290 + 0.0245V 2
dt m
dV 1
=
272510 − 2892V + 4.846V 2
dt m
(
(
)
)
c) For the High-Speed rail repeat the analysis presented in the lecture notes (see pages 21-31d) to estimate components of
travel time between two stations located 100 km away from each other. Assume a deceleration rate of -1 m/s2. State the
acceleration, cruise and deceleration travel time in a simple table. For the 100 km trip, the operator would like to reach
175 mph (78.2 m/s) in the cruise profile. To determine acceleration time use a step size of 1 second. You can do this
analysis either in Excel or Matlab (your choice).
The acceleration profile calculations are shown in Figure 8.
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.....
Figure 8. Numerical Solution for Acceleration and Speed Profiles of High-Speed Train.
Note that in order to reach 78.2 m/s, the train will require 236 seconds of acceleration time from start. At the cruise speed
of 78.2 m/s, the train can reach a maximum acceleration of 0.168 m/s2. The total distance traveled by the train is
calculated using the numerical integration procedure explained in class. The results of the numerical integration of the
train speed are shown in Figure 9. Here we present a solution where the rate of change of distance over time (dS/dt) is
taken as the mid-point of each interval. The solution shown in Figure 10 takes the value of speed in each interval as a
constant (i.e., left hand side calculations in the finite difference equation).
...
Figure 9. Numerical Solution for Speed and Distance Traveled Profiles of High-Speed Train.
Solution with Midpoint Speed Used for Integration.
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...
Figure 10. Numerical Solution for Speed and Distance Traveled Profiles of High-Speed Train.
Solution with Constant Speed per Interval Used for Integration.
The variation of distance traveled using both methods is relatively small (11,170 vs. 11,130 meters). Using a deceleration
rate of 1 m/s2 , the HSR system takes 3,057 meters (see calculation below) to decelerate at a constant rate of -1 m/s2.
The deceleration distance is labeled S3. The deceleration time is calculated to be 78.2 seconds. Recall the equation for
uniformly accelerated motion.
The train cruises for 85,773 meters (53.21 statute miles). The travel time to cover 85.772 km while cruising at 72.8 m/s is
1,178 seconds.
V 2cruise ( 78.2m/s )
=
= 3058 meters
2a
2(1m/s 2 )
2
S3 =
t3 =
S3
3058 m
=
= 78.2 seconds
Vcruise + V final 78.2 / 2 m/s
2
The deceleration time (t3) also can be obtained from the fundamental kinematic relationship,
t3 =
V f 78.2m/s
=
= 78.2 seconds
a
1 m/s 2
Table 1. Summary of Travel Time and Distances by High-Speed Train.
Segment
Acceleration
Cruise
Travel Time (s)
236
11,170
1,178
85,773
78
3,057
Deceleration
Totals
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Travel Distance (m)
1,492
Trani
100,000
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