CEE 3604: Introduction to Transportation Engineering
Fall 2009
Assignment 3: Vehicle Forces and Kinematics
Date Due: September 17, 2009
Instructor: Trani
Problem 1
Wind tunnel testing of a new, low-profile high-speed train has the following equivalent Davis equation coefficients:
A = 8.57 ;
% units are kN
B = 0.0983;
% units are kN s/m
C = 0.0129; % units are kN s-s/m-m
The vehicle tested is expected to have a mass of 256,000 kg for a 6-car train unit (includes the locomotive unit to pull the
passenger cars) and maximum axle load of 16 metric tons per axle. The front locomotive has nearly constant power out (P) rated
at 15,000 HP with 0.8 efficiency. To avoid slippage of the rail wheels at low speeds, the Tractive Force (T) of the engine is limited
to a constant 400 kN for all speeds below 20 m/s by an electromechanical system.
Solution
% Wind tunnel testing of a new, low-profile high-speed train has the
% following equivalent Davis equation coefficients:
% A = 8.57 ;
% B = 0.0983;
% units are kN
% units are kN s/m
% C = 0.0129; % units are kN s-s/m-m
% The vehicle tested is expected to have a mass of 256,000 kg for a
% 6-car train unit (includes the locomotive unit to pull the passenger cars)
% and maximum axle load of 16 metric tons per axle. The front locomotive
% has nearly constant power out (P) rated at 15,000 HP with 0.8 efficiency.
% To avoid slippage of the rail wheels at low speeds, the Tractive Force (T)
% of the engine is limited to a constant 400 kN for all speeds below 20 m/s
% by an electromechanical system.
% Programmer: A. Trani
clear
clc
% Coefficients of Davis equation applied to Japanese Shinkansen system
% Series 200 (new system)
A = 8.57;
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% units are kN
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B = 0.0983; % units are kN s/m
C = 0.0129; % units are kN s-s/m-m
% Grade resistance equation
M = 256;
% metric tons
g = 9.81;
% gravity force (m/s-s)
X = 1e12;
% grade (1 in X)
Rg = M * g / X;
% or use the standard analysis with sin(phi) on the equations of motion
w=M*1000*g;
% weight in Newtons
phi = 0;
% angle of railway (radians)
RgMod = w * sin(phi);
% in Newtons
% Create a speed vector
V = 0:1:100;
% speed in meters/second
% Create a speed vector
V = 0:1:100;
% speed in meters/second
% Calculate Resistance (in KiloNewtons) according to
% modified Davis equation
% Added the grade resistance
R = A + B * V + C * V.^2 + RgMod/1000;
% Make a plot of total resistance vs speed
plot(V,R,'o--')
xlabel(' Speed (m/s)')
ylabel('Resistance (kN) or T (kN)')
title('Reisistance of High-Speed Rail System')
grid
hold on
% Calculate the Tractive Effort (T) profile
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P = 15000;
% horsepower (hp)
Vkmhr = V*3.6; % velocity in km/hr (needed in the T equation)
nu = 0.8;
% efficiency
% cap the value of T - tractive force - to 400 kN
npoints = length(Vkmhr);
for p = 1:npoints
T(p) = 2650 * nu * P ./ Vkmhr(p) / 1000;
% in kN
if T(p) > 400
T(p) = 400;
% caps the tractive effort to 400 kN
end
end
plot(Vkmhr/3.6,T,'^-r');
% plot the tractive force vs speed (km/hr)
a) Plot the total resistance and tractive force diagram for this vehicle. In your diagram include the limitation of tractive force.
Plot of T and R vs. Speed. Flat Terrain.
b) Find the maximum speed (in meters/second and in miles per hour) for the new high-speed rail in level ground.
By inspection the maximum speed is 83.25 m/s (299.7 km/hr).
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Point of maximum speed for HS rail. 83.25 m/s (299.7 km/hr) is the maximum speed.
c) If the train negotiates a positive (uphill) 1 in 50 slope (one foot vertical rise for every 50 feet horizontally) estimate the
gravity force (in kN) acting against the vehicle motion.
A 1 to 50 slope is equivalent to (1/50) radians (phi). This is also equal to X = 50 in the handout equation. The new plot is shown
below.
Plot of T and R vs. Speed. Slope 1:50.
d) Find the maximum speed for the train for condition (c). Make a plot of the new resistance curve and tractive force to find
the maximum speed.
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Point of maximum speed for HS rail. Slope 1:50 (uphill). Maximum speed is 69.3 m/s (240 km/hr).
e) Repeat (d) when the train returns and now goes downhill
Plot of T and R vs. Speed. Slope -1:50 (downhill). Balancing Sped is 97.5 m/s (351 km/hr).
Point of maximum speed for HS rail. Slope -1:50 (downhill). Maximum speed is 97.5 m/s (351 km/hr).
f) Comment on the values of maximum speed calculated in parts (d) and (e) and contrast with (b). What is the impact of
topography (or terrain) in travel time for a rail system?
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g) Estimate the travel time between two stations located 40 km apart if the driver uses 90% of the maximum speed as the
cruise speed for the train. Consider the acceleration and deceleration phases in your analysis.
Assume flat terrain so the target cruise speed is 74.9 m/s (262 km/hr).
Acceleration Phase
This requires that we set up the acceleration function of the HS train. Recall:
ma = ∑ Fx = T − R
x
1
(T − R)
m
a=
Both T and R are known functions of velocity (V). So we can setup a simple numerical procedure (see pages 18-34 in handout Transportation_technology.pdf) to calculate speed and distance profile in the acceleration phase of the HS profile.
A simple Matlab script is shown below. This process can be done in Excel following the analysis shown on page 30 of the same
handout.
% Calculation of speed and distance traveled in the acceleration phase of
% the HS rail motion between stations
deltaT = 1;
% step size for numerical integration - seconds
Vx(1) = 0;
% initial speed (m/s)
Sx(1) = 0;
% intial distance traveled (meters)
t(1) = 0;
% time clock starts
for i=1:1:400
% Try 400 steps to integrate numerically the ODE
if Vx(i) < 20;
% limits the tractive force to 400 kN
TractiveForce(i) = 400;
end
TractiveForce(i) = interp1(V,T,Vx(i));
Resistance(i) = interp1(V,R,Vx(i));
t(i+1)=t(i)+deltaT;
% calculates T at any speed Vx - use linear interpolation (interp1)
% calsculates R at Vx - use linear interpolation
% updates the time during the acceleration phase
Ax(i) = 1/(M*1000) * ((TractiveForce(i) - Resistance(i)) * 1000);
% acceleration m/s-s
Vx(i+1) = Vx(i) + Ax(i) * deltaT;
% m/s
Sx(i+1) = Sx(i) + Vx(i) * deltaT;
% meters
end
% Make plots of speed, distance and acceleration
figure
plot(t,Vx)
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grid
xlabel('Time (seconds)')
ylabel('Speed (m/s)')
figure
plot(t,Sx)
grid
xlabel('Time (seconds)')
ylabel('Distance Traveled (meters)')
figure
plot(t(1:400),Ax)
grid
xlabel('Time (seconds)')
ylabel('Acceleration (m/s-s)')
Plot of Speed Profile of the HS Train in the Acceleration Phase.
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Looking at the figure (use the Matlab zoom to determine the approximate time to stop the acceleration phase), the acceleration
phase lasts 144.25 seconds as the train reaches the desired 74.9 m/s cruise speed. This process consumes 7,700 meters (using
the numerical integration).
Plot of Distance vs. Time in the Acceleration Phase.
Deceleration Phase
Assume -1 m/s-s as a constant deceleration rate to provide a nice and gentle deceleration for passengers on board the HS train.
1
St = So + V0t + at 2
2
2
2
V − V0
St = f
2a
The distance traveled in the deceleration phase is: 2,805 meters. The time for deceleration is then 64.6 seconds
Cruise Phase
The distance remaining for the cruise phase is 30,215 meters. At 74.9 m/s this phase lasts 403.4 seconds.
Total Travel Time
The total travel time between stations is then 612.25 seconds. if we had not used the numerical integration procedure and
should we had assumed a constant cruise speed of 74.9 seconds, the travel time would have been 534.1 seconds. This
represents a 12.5% difference.
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Problem 2
Blacksburg Transit wants to re-equip their fleet of 24 Flxible Buses with a new more powerful engine that delivers 15% more
power than the baseline engine shown on page 46 of the handout (transportation_technology.pdf). Refer to table on page 46 to
see the current baseline engine parameters. All other characteristics of the vehicle will remain the same.
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a) Using the Matlab script provided construct a Force vs. Speed diagram. Plot the resistance and tractive force in the same
diagram.
Solution
% Script to estimate Force vs. Speed Diagram for a Bus
% Calculates and resistances for the bus (SAE formulas)
% Uses a linear and quadratic approximation for basic and
% Aero resistances
% Programmer: Toni Trani
%
% Version 1 (01/20/09)
% Define model parameters (bus model)
Ca = 0.020;
% Coefficient for trucks
C1 = 7.6;
%
C2 = 0.056;
%
W = 120;
% Bus weight in kN
A = 6.97;
% Frontal area
i = 0;
% Grade in percent (%)
% Define for Tractive Effort estimation
D
= 0.94;
% Wheel diameter (m)
u(1) = 4.5; u(2) = 2.6; u(3) = 1.5; u(4) = 1.1;
J
% gear ratios in explicit vector form
= 4.6;
% Differential reduction ratio
con1 = .06;
% conversion factor (to convert km/hr to m/s)
con2 = 2650;
% conversion to obtain V in km/hr
% conversion factor (to convert hp to N)
nu
= 0.85;
P=
1.15 * [75
130
165
170
190
203
210];
RPM = [800 1200 1600 1800 2000 2200 2400];
% conversion from Hp to Newtons
% power output vector
% RPM vector
% ______________________________________________
% Compute resistances
for a=1:1:130
V(a)
% Starts loop
= a - 1;
% Speed vector (km/hr)
R_basic(a) = (C1 + C2 * V(a)) * W;
% Basic resistance (N)
R_aero(a) = Ca * A * V(a) ^ 2;
% Aerodynamic resistance (N)
R_grade
% Grade resistance (N)
= 10 * W * i ;
R_total(a) = R_basic(a) + R_aero(a) + R_grade;
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% Total resistance (N)
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end
% Ends the comp. loop
% ______________________________________________
figure
plot(V,R_total,'o-')
% Plot total resistance versus speed
xlabel('Speed (km/hr)')
ylabel('Resistance (N)')
title(['Total resistance diagram for ',num2str(i), '% grade'])
grid
% Extraction of sizes of P and u vectors (power and gear ratio). This makes the code independent of the sizes of
% P and u and thus allow you to enter any type of vector length
[nrowp,ncolp] = size(P);
% Extracts number of columns and rows from vector P
[nrowu,ncolu] = size(u);
% Extracts number of columns and rows from vector u
N_low = RPM(1);
% low RPM value
N_high = RPM(ncolp);
% high RPM value
steps = 100;
% Number of times RPM is cycled
increment = (N_high - N_low) / steps;
% Increment for RPM
% ______________________________________________
for r = 1:1:steps
% Start RPM cycle (engine speed loop)
N(r) = N_low + (r -1)* increment;
% Engine speed (RPM)
for h =1:1:ncolu
% Loops various transmission gears
Power(h,r) = interp1(RPM,P,N(r));
% Computer power output for given engine speed
Vt (h,r)
= con1 * N(r) * D * pi / (J * u(h));
% Vehicle speed (km/hr)
TE (h,r)
= con2 * nu * Power(h,r) / Vt(h,r);
% Tractive effort (N)
end
% End transmission loop
end
% Ends engine speed loop
% Plot the tractive effort diagram vs. speed
figure
plot (Vt',TE',V,R_total)
xlabel('Speed (km/hr)')
ylabel('Tractive Effort (N)')
title(['Tractive effort diagram for ',num2str(i), '% grade'])
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grid
Plot of T and R vs. Speed. Bus Model. Flat Terrain.
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b) Find the maximum speed for the Windsor Hills route with the new engine. Show all your work.
Plot of T and R vs. Speed. Bus Model. 7% Grade (uphill). 44 km/hr is the Maximum Speed.
c) If the driver of the bus is very aggressive (lets hope not in real life), what is the maximum instantaneous acceleration of
the bus when the vehicle travels on flat ground at 40 km/hr.
Detail of T and R vs. Speed Diagram. Bus Model. Zero Grade.
R_total( at 40 km/hr) = 1,404 Newtons
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T ( at 40 km/hr) = 10,445 Newtons
a = (T - R) / m = (10,445 - 1,404) [N] / (120,000/g) [kg] = 0.74 m/s-s
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Problem 3
For a long-range, trans-pacific flight, a fully loaded Airbus A380 weighs 1,230,000 lbs. at takeoff from Los Angeles International
Airport. The aircraft acceleration on the runway can be modeled very closely to a linear decaying deceleration model with
parameters:
dV/dt = k1 - k2 * V
where dV/dt = acceleration (m/s-s), V = vehicle speed (m/s), k1 = maximum acceleration (m/s-s)
(1/s). For the vehicle, k1 = 3.0 and k2 = 0.017.
and k2 = slope of dV/dt vs V
a) The aircraft needs to reach 190 mph to lift-off the ground at that weight at sea level conditions. Find the runway length
needed to perform a successful takeoff. The longest runway at Los Angeles allowing A380 operations is 11,000 feet.
Plot of Distance Traveled vs. Speed. Airbus A380.
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Detail of Distance Traveled vs. Speed for A380 Aircraft. Distance to Lift-off Point is ~5,960 feet. In Real Runway length
calculations, an added distance to account for the initial climb of the vehicle is factored in. This assumes the vehicle would cross
an imaginary obstacle of 35 feet at the of the runway required. Moreover, in real runway performance calculations, the runway
would also be adjusted by an engine failure at the critical point during takeoff maneuver. These two factors would easily add
3,500 feet to the distance calculated in this example which assumed all engines (4) working.
b) If you are passenger on board the aircraft measuring time, how long would it take for the aircraft to lift-off after the pilot
releases the brakes on the runway?
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Plot of Time vs. Speed (mph). Note the Vehicle Lift-offs 38.6 seconds after Brake Release.
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