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Unformatted text preview: Introduction to Tansportation Engineering
Fundamentals of Ground Vehicle Performance
Dr. Antonio A. Trani
Professor of Civil and Environmental Engineering
Virginia Polytechnic Institute and State University Blacksburg, Virginia
Fall and Spring 2011 Virginia Tech 1 Introductory Remarks
• Simpliﬁed analysis of various modes of transportation
allow us to integrate the elements of technology needed
to predict cost and performance measures of
effectiveness associated with a mode of transportation
(i.e, speed, travel time, station spacing. level of service,
etc.).
• Need to understand the basic forces that make all
transportation systems move and behave the way they do
• Knowledge of the operational limitations of existing and
future transportation technology can inﬂuence our
decisions with respect to the transportation mode to be
used Virginia Tech 2 Ground Vehicle Performance Basics Virginia Tech 3 Basic Forces Acting on a Ground Vehicle
The following diagram applies to cars, buses, trains, etc.
Lift
Thrust
Drag φ Friction
Force
mg Assumed all forces act through a point mass system (center of gravity)
Thrust is also called Tractive Effort (TE) Virginia Tech 4 Analysis of Individual Forces 12
L =  ρ V SC l
2 (1) 12
D =  ρ V SC D
2 (2) T = f ( V, ρ ) (3) F f = ( mg cos φ – L ) f roll (4) G t = mg sin φ (5) Virginia Tech 5 Nomenclature
T is the net engine thrust force or sometimes called
tractive force or tractive effort (Newtons),
L is the lifting force (Newtons),
V is the vehicle speed (m/s),
D is the drag force (Newtons),
Ff is the friction force (Newtons),
G is the component of the gravity force (Newtons), ρ is the air density (kg/cu.m.), Virginia Tech 6 S is the reference area (m2),
Cl lift coefﬁcient (nondimensional),
CD is the drag coefﬁcient (nondimensional),
froll is the rolling friction coefﬁcient (nondimensional)
m is the vehicle mass
g = gravitational acceleration (9.81 m/s2 or 32.2 ft/s2) Virginia Tech 7 Application of Newton’s Second Law m ax = T – D – Ff – Gf
12
ma x = f ( V, ρ ) –  ρ V SC D – ( mg cos φ – L ) f roll
2 (6)
(7) – m g sin φ
• The gravity term is important and thus has been added
• We could add another term to account for curvature
resistance (good for rail transportation) Virginia Tech 8 Some Remarks
• Acceleration capability of the vehicle decreases as speed
is gained during the acceleration phase
• If expression (7) is integrated twice between an initial
speed, V0 and the desired cruising speed, Vcr the distance
covered during the acceleration phase can be found (this
might require a numerical integration procedure) Virginia Tech 9 Types of Forces
• There are two types of forces acting on the vehicle:
Tractive
Resistance Tractive
Force Resistance
Forces • The following pages explain these forces in more detail Virginia Tech 10 Variations of Tractive Effort and froll
• T decreases with speed for all transportation modes
• froll is a function of the contact material between vehicle
and guideway or road and the vehicle speed
T (N) f roll BiasPly Tire Sea Level
Radial Tire
High Elevation V (m/sec) Virginia Tech Metal Rim
Surface V (m/sec) 11 Rolling Friction Coefﬁcient ( f roll )
• The value of f roll varies with speed, tire construction, tire
pressure, etc.
• f roll is usually determined experimentally
• f roll is associated with a tire rolling free on a road.
• f roll doubles with speed increments from 0 to 200 km/hr. Virginia Tech 12 froll Value (dimensionless) Variation of froll with Speed
Source: W. Hucho (1998)
Data good for radial car tires Speed (km/hr)
Virginia Tech 13 Example Problem # 1
Find the operating speed schedule of a maglev train
operating from Orlando International Airport to the Main
Plaza at Epcot Center in Disneyworld.
The MBB (Krause Maffei) maglev has a linear thrustvelocity diagram as shown in Table 1 (if necessary linear
extrapolation beyond 89.4 m/s is possible).
Using a minimum constant speed segment time of 60
seconds and a ﬁnal deceleration of 0.1 g's to estimate the
velocitytime proﬁle of the vehicle. Assume that grades
are zero throughout the route. Florida is pretty ﬂat like a
tortilla. Virginia Tech 14 Maglev Train Parameters
The following parameters apply to the Maglev train.
S=25 m2, froll=.025 (magnetic rolling friction),ρ=1.225 kg/cum., CD=.55,Cl=.20,g=9.81 m/s2 and m=350,000 kg
MAGLEV Thrust (or tractive effort) Parameters for the Hypothetical
Orlando Line.
Speed (m/s) Thrust Force (N) 0.00 300,000 89.40 150,000 Virginia Tech 15 Analysis of the Maglev Train
Simplify the basic equation of motion by eliminating the
angle between the track and the horizontal, 1
ma x = f ( V, ρ ) –  ρ V 2 SC D – ( mg – L ) f roll
2
Substitute the numerical values for the train in equation
(6) to obtain, dV  ( 214, 163 – 1, 678 V – 8.345 V 2 )
 = a t = 1 m
dt Virginia Tech (8) 16 Integration of Equation of Motion
This can be done both numerically or analytically
Analytical: the equation is a perfect quadratic so there is a
solution to the deﬁnite integral (consult CRC tables for
example)
t V dV
 = ∫ dt
∫ 1
2
 214, 163 – 1, 678 V – 8.345 V 0
Vo m (9) The denominator is of the quadratic form: A + BV + CV 2
Virginia Tech 17 Numerical Integration
A general procedure to solve any complex differential
equation encountered in practice. Start with, 1
dV  ( 214, 163 – 1, 678 V – 8.345 V 2 )
 = a t = m
dt
then take small increments of time into the future and
dV
integrate dt
Breaking down the path into small intervals is the key to
the numerical integration process Virginia Tech 18 How it Works Virginia Tech (A.A. Trani) 19 Graphical Interpretation
• The numerical integration procedure is illustrated
graphically in the diagram
• Note: the Euler integration scheme assumes a constant
value for dV ⁄ dt in the interval ( t – ∆ t , t )
• To achieve good accuracy make the step size ( ∆ t ) small
watching that the roundoff errors in the computation do
not become excessive Virginia Tech 20 Finding the Distance Traveled
The distance traveled is the second integral of the
acceleration function or the integral of the vehicle
velocity function d S
S t = S t – ∆t + ∆ t = S t – ∆t + V t ∆ t dt ( t – ∆t, t )
Note that we are trying to ﬁnd the distance to accelerate
(named S1 hereon) and reach some desired cruise speed
( V cruise ) Virginia Tech 21 More Analysis for Maglev Train
A sketch of the velocity proﬁle is shown in the ﬁgure
illustrating three possible regimes of motion where the
vehicle accelerates, cruises and then decelerates to come
to a full stop at the station
Area under the speed vs time curve is
the distance traveled (S) Speed S2 S1
T1 Virginia Tech S3
T2 T3 Time 22 Cruise and Deceleration Segments
• For second interval (i.e., constant speed segment) we
estimate distances as a function of speed allowing 60
seconds of cruise time. S 2 = t cruise V cruise = 60 V cruise (10) For the deceleration segment assume a uniform
deceleration (assume 0.1g or 1.0 m/s2), V – VS 3 = 2a
2
2 2
0 (11) Virginia Tech 23 Analysis
• To ﬁnd Vcruise and unknown distances S1, S2 and S3 we
use a numerical integration procedure (i.e., Euler,
modiﬁed Euler or RungeKutta of fourth order)
• The pseudocode to ﬁnd distance, velocity and
acceleration of the vehicle is shown in ﬂowchart
corresponding ﬂowchart illustrating the main steps of the
numerical integration procedure
• Note that a maximum speed of 72.25 m/s is reached in the
16 km trajectory followed by the required 60 seconds
cruising time. Note that this problem has a unique
solution for Vcruise and thus the numerical integration
technique advocated here is suitable for any type of
acceleration function
Virginia Tech 24 More Analysis
• The travel time is estimated directly from the integration
procedure or from the solution of basic travel time
kinematic equations once each segment of the trajectory
is known.
• In this particular case the total travel time is known to be
342 seconds from the numerical integration procedure.
The acceleration time is 204 seconds whereas the
deceleration time is 78 seconds
• As a matter of curiosity you can check that the maglev's
maximum speed turns out to be around 90 m/s in this
conﬁguration (i.e., weight and aerodynamic
characteristics) Virginia Tech 25 Numerical Results
1 : V elocit y
1:
2: 2: Dista nce
1 72.25
16090.03 1 1:
2: 2 2 35.75
8045.01
1 2 1
1:
2: 0.75
0.00 2
0.00 100.00 200.00
Time 300.00 400.00 Velocity units are m/s
Distance units are m
Virginia Tech 26 Force Diagram
The force diagram shows all forces acting on the vehicle
Data #2
350000
Te (N) 300000 Treq (N) Force (N) 250000
200000
150000 Maximum Speed 100000
50000 0 20 40 60 80 100 Speed (m/s) Virginia Tech 27 Energy Consumption
The model shown described so far can be modiﬁed to
estimate the energy consumed in the trajectory between
two stations
Knowing that the power required to move the vehicle at
speed V is the product of the tractive force (or thrust)
needed to move the vehicle at V and the speed itself.
Mathematically this becomes,
V
P = TR η (12) where P is the power (Watts), TR is the tractive force
required (in Newtons) to overcome the resistance forces
Virginia Tech 28 opposing the motion of the vehicle,η is an efﬁciency
factor (typically between 0.75 to 0.90 ground vehicles)
and V is the speed (m/s).
Also known is the fact that power is the time rate of doing
work which has units of energy consumption (Joules), t
E= ∫ Pdt (13) 0
where E is the energy consumed in Joules and P is the
power required to overcome the motion of the vehicle.
You can convert energy consumption into KWh by diving
the result for E by 3.6 106. Virginia Tech 29 Numerical Integration Using Excel
Excel or Matlab can be used to numerically integrate the
expressions presented and do the analysis. 1
dV  ( 214, 163 – 1, 678 V – 8.345 V 2 )
 = a t = m
dt
Set several columns to estimate values of time, velocity,
rate of change of velocity, the desired step size interval
(∆t), and the product of the step size and the rate of
change of velocity (in that order) Virginia Tech 30 Excel Representation (Maglev Problem) Sample Formulas
B3 = B2+E2
C3 = 1/350000*(2141631678*B38.345*B3^2)
D3 = A4A3
E3 = D3 * C3 Virginia Tech 31 Observations
• The Maglev train accelerates modestly reaching 5.96 m/s
in about 10 seconds according to the sheet presented
• Note that in order to have a better accuracy in the
predictions we would want to use smaller step sizes
(typically one second or less)
• The acceleration capability of the train diminishes with
speed Virginia Tech 32 General Models for Ground Vehicles
• The characteristics of rail vehicles are very important in
determining travel time, energy consumed and other
measures of effectiveness in intercity travel
• From our previous derivation of the basic equations of
motion for transportation vehicles it is evident that two
types of forces arise while trying to derive the
performance of rail vehicles:
Traction and
Resistance • General models attempt to describe the performance
using these two types of forces in the model Virginia Tech 33 Highway Vehicle Performance Estimation
• Highway vehicles obtain their power from internal
combustion engines. Two types of engines developed
over the last century have dominated: 1) Fourcycle
Internal Combustion (IC) and 2) Diesel engines
• Deﬁne a simple procedure to extract performance from
engine torque and engine speed (in revolutions per
minute or RPM) diagrams. The discussion that follows is
adopted primarily from Vuchic (1982) and Hucho (1998)
• Performance curves for rubbertired vehicles are usually
expressed by manufacturers in terms of torque, power
developed (e.g., shaft horsepower), and fuel consumption
curves versus engine speed (N)
•
Virginia Tech 34 Trends
• The general trend is that at higher N values the torque
decreases and power output of the engine increases.
• The fuel consumption has a minima at intermediate
vehicle speeds (e.g., 45 m.p.h for the average car).
• Since Thrust vs. vehicle speed (not engine speed) is not
given directly it is necessary to derive a diagram that will
approximate the general tractive effort (thrust) vs. speed
function hypothesized in this course Virginia Tech 35 Procedure to Derive Force  Velocity Diagrams
• The procedure requires power information from the
engine or vehicle manufacturer. The speed of the vehicle
(V), just like in rail technology, is related to the
powerplant engine speed (N) by,
V = (N)(D)(π)
Ju i (14) where: J is the differential reduction gearing ratio
(usually varies from 3:1 to 6:1 for typical IC vehicle
applications), u is the transmission gearing ratio for the
ith gear (the value of u is high for low gears and low for
high gears and varies from 5:1 to 1:1), D is the diameter
of the tractive wheels, and N is the engine speed.
i i Virginia Tech 36 Procedure  Force vs. Speed Diagrams
Since N is usually expressed in Revolutions per Minute
(RPM) a conversion will be necessary as D and V will
usually take on units such as meters, and km/hr,
respectively.
Apply a conversion factor to change m/minute to km/hr in
Equation (14) to get:
60 ( N ) ( D ) ( π )
V = 1000 Ju
i (15) where the units of each variable are: N in revolutions per
minute, D in meters, J and u are dimensionless, and V is
expressed in km/hr.
i Virginia Tech 37 Force vs. Speed Diagrams
Using the expression to convert power output to tractive
effort we have, T = ηP
V (16) where: P is the power output, η is the efﬁciency of the
engine ( varies from 0.7 to 0.85 for most IC type
applications), V is the vehicle speed and T is the thrust or
tractive effort produced by the engine.
Since P is usually expressed in horsepower (hp), V in km/
hr and T in Newtons, it is necessary to convert to proper
units before applying this equation to car, truck and bus
vehicles.
Virginia Tech 38 Froce vs. Speed Diagrams
Once the conversion factors are applied we have: T= ηP
V 2650  (17) where:
T is expressed in Newtons
P in horsepower
V in km/hr Virginia Tech 39 Force vs. Speed Diagram
The procedure to estimate force vs. speed performance is:
• Compute V for several values of N and iterate for all
values of u in Equation 15.
i • Substitute the corresponding velocity (V) values into
Equation 17 to estimate T.
Since each vehicles has several feasible transmission
gearing ratios, u , it is necessary to compute several values
of T for each value N in Equation 15.
i The resultant performance of a typical IC engine vehicle
is shown in the following page. Virginia Tech 40 Power (hp) Given Estimate
T (N) First Gear Second Gear Engine Speed  N  (RPM) Third Gear
Fourth Gear Given: Vehicle by Manufacturer
Vehicle Speed (m/s) 60 ( N ) ( D ) ( π )
Use V =  and T = 2650 η P
1000 Ju
V
i
Virginia Tech 41 Estimation of Vehicle Resistance
Many manufacturers ﬁnd useful to distinguish between
the resistance and the propulsive forces required to do
performance analysis
This section presents typical equations found in texts or
vehicle performance tables to familiarize yourself with
some of the basic computations of ground vehicle
performance
The analysis that follows is similar to that explained for
the Maglev vehicle except that resistance forces are
treated independently in the computation. Sometimes this
makes the analysis more clear. Virginia Tech 42 Resistance Formulas for Highway Vehicles
According to the Society of Automotive Engineers (SAE),
the basic resistance of a highway vehicle (in Newtons)
can be represented by,
R = [ c 1 + c 2 V ] W + c a AV 2 b (18) where: c 1, c 2 are coefﬁcients known to have values of 7.6,
and 0.056, respectively for buses and trucks (per SAE
data); W is the total weight of the vehicle (kN), A is the
frontal vehicle area (m2) and V is the vehicle speed in km/
hr., and R is the basic resistance in Newtons.
b Virginia Tech 43 Vehicle Resistance
A typical value for ca is .018 for buses and 0.020 for
streamlined trucks (0.025 for nonstreamlined trucks).
Ground vehicles are also subjected to gradient resistance.
Since equation 18 expresses the weight of the vehicle in
kN the gradient resistance becomes,
R G = 10 Wi (19) where: W is the weight in kN and i is the grade expressed
as a percent. For a 100 kN bus this would imply 1000
Newtons of resistance for each one percent of gradient. Virginia Tech 44 Example # 2  Urban Bus Analysis
To illustrate the use of equations presented let us look at
an urban bus vehicle traveling in hilly terrain (say
Blacksburg). The vehicle considered is a Flxible bus. Flxible Bus (Blacksburg Transit)
Virginia Tech 45 Bus Characteristics
The technical characteristics of the bus are given below;
Table 1: Flxible Bus Power Characteristics.
N
(RPM) 800 1200 1600 1800 2000 2200 2400 P (hp) 75 130 165 170 190 203 210 Transmission gearing ratios are: 4.5:1, 2.6:1, 1.5:1, and
1:1 for the four speed bus gear system. The differential
ratio is constant at 4.6:1. The bus weighs 26,400 lbs (120
kN), the wheel diameter is 37 inches (.94 m.) and the
frontal area of the bus is 6.97 m2 (75 ft2) . The resistance
function for the bus is given by, Virginia Tech 46 R = [ c 1 + c 2 V ] W + c a AV 2 (20) The values for each constant are: ca = 0.020, c1 = 7.6, and
c2 = 0.056;
A surveying team estimates the most demanding grade of
the Windsor Hills Route is 7%. The maximum speeds for
each gear are as follows:
V max for ﬁrst gear = 20 km/hr
V max for second gear = 35 km/hr
V max for third gear = 55 km/hr
V max for fourth gear = 85 km/hr
Virginia Tech 47 With this information we construct a Force vs Speed
diagram using the process explained in previous pages. Virginia Tech 48 Construction of TE vs. V Diagram
a) Start with a low value of N and compute the power
output from the engine (look at the power vs. N table or
data)
b) Estimate V for the value of N selected for every gear
(note that ui changes for every gear) from equation (15)
c) Find the value of T from equation (17) for a given value
of P and V. Do this for every gear.
d) Repeat the process (ac) for other values of N. I
suggest you repeat the process for at least 510 values of
N
A sample T vs. V diagram is shown below.
Virginia Tech 49 Force vs. Speed Diagram for Flxible Bus
x 104 3 First gear T (Newtons)
Tractive Effort (N) 2.5 2 Second gear
1.5 Third gear TextEnd
1 Fourth gear
0.5 Total resistance (0% grade)
0 0 10 20 30 40 50
Speed (km/hr) 60 Virginia Tech 70 80 90 100 50 Notes
• The ﬁrst gear provides the highest value of T as expected
• Higher gears provide modest T values
• The resistance equation for zero grade is shown for
illustration
• Other resistance curves (say for other grade conditions)
can be easily superimposed in the T vs. V diagram to
show the maximum vehicle speed for any gradient
condition
• For example, if the grade is 5% the maximum speed the
bus can achieve is only 43 km/hr as shown in the
following diagram Virginia Tech 51 T vs V Diagram (5% grade resistance forces)
3 x 104 First gear TTractive Effort (N)
(Newtons) 2.5 5
100 2 Second gear
1.5 TextEnd Third gear
Total resistance (5% grade)
Fourth gear 1 0.5
0 10 20 30 40 50
Speed (km/hr) Virginia Tech 60 70 80 90 100 52 Approximation of T vs. V Diagram
x 104 3 First gear Approximation
Profile 2.5 Tractive Effort (N) 2 Second gear
1.5 Third gear TextEnd
1 Fourth gear
0.5 Total resistance (0% grade)
0 0 10 20 30 40 50
Speed (km/hr) 60 70 80 90 100 Use linear segments to approximate the T vs. V curve.
Virginia Tech 53 Finding Acceleration T vs. V Diagram
x 104 3 2.5 Excess Force to
Accelerate Tractive Effort (N) 2 1.5 TextEnd
1 0.5 Total resistance (0% grade)
0 0 10 20 30 40 50
Speed (km/hr) 60 Virginia Tech 70 80 90 100 54 Sample Kinematic Models
Two models are useful in the analysis of ground
transportation systems:
a) Constant acceleration/deceleration)
b) Variable acceleration/deceleration The use of the model depends on the characteristics of the
system studied and thelevel of accuracy needed.
Example 1: in the analysis of deceleration proﬁles for cars
and trains we can use a constant deceleration model
Example 2: in the analysis to determine the acceleration
lane distance of a highway a variable acceleration model
is recommended
Virginia Tech 55 Kinematic Models
dV/dt Constant acceleration/deceleration k V
dV/dt
k1
k2 k1 k1/k2
Virginia Tech V
56 Constant Acceleration Model
Well known equations. the acceleration of the vehicle is
equal to a constant ( k ), V
d  = k
dt
Integrate once to get speed (V) and twice to get distance
traveled (S) V t = V 0 + kt
where: V t is the speed at time t, V 0 is the initial speed of
the vehicle and t is the time from start of motion
Virginia Tech 57 Constant Acceleration Model
The distance traveled (S) is, 1
S t = V 0 t +  kt 2 + S 0
2
where: S 0 is the intial position of the vehicle at time 0
Since k is just the acceleration of the vehicle this equation
is equivalent to the well known relationship, 12
S t = V 0 t +  at + S 0
2
where a is the constant acceleration of the vehicle
Virginia Tech 58 Variable Acceleration Model
Start with, V
d  = k 1 – k 2 V
dt
Integrate once to get speed (V)
–k t
–k t
k
V t = 1 ( 1 – e ) + V 0 e
k2
2 2 where: V t is the speed at time t, V 0 is the initial speed of
the vehicle, k 1 and k 2 are constants and t is the time from
start of motion
Virginia Tech 59 Variable Acceleration Model
The distance traveled (S) is,
–k t
–k t
k
V
 kS t = 1 t – 1 ( 1 – e ) + 0 ( 1 – e )
k2 k2
k2
2
2 Virginia Tech 2 60 Problem 1
The Blacksburg Middle school board hires you as a
transportation engineer to ease complaints from parents
driving vehicles and making a left turn to the school
entrance during the peak hour in the morning (see Figure
below).
The road is divided and has a left turn queueing island
allowing cars to stop before making the turn.
Measurements at the road by the town engineer indicate
that trafﬁc ﬂows in this section at 35.7 km/hr (restricted
by the speed limit). Virginia Tech 61 School
Entrance To Blacksburg R = 12 m. Spacing = S Car Traffic
Counters Car
To Radford FIGURE 1. Blacksburg Middle School Trafﬁc Situation. The typical acceleration model for a car is known to be:
a = 4.0 – 0.1 V where: a is the acceleration of the car (in m/s2) and V is
the vehicle speed in m/s. Virginia Tech 62 a) Find the typical spacing (S) and the average headway
(h) between vehicles traveling from Radford to
Blacksburg during the peak morning period.
b) Find if the average headway (h) allows a typical driver
to make a left turn if the driver has a perception/reaction
time of 0.5 seconds. The radius of the curve to make a left
turn is 12 meters. According to AASHTO standards, the
critical vehicle length is 5.8 meters. Virginia Tech 63 Solution to Part (a)
Find the Spacing (S) between vehicles. Since the density
of the trafﬁc ﬂow is known to be 20 veh/kmla we
compute the spacing as the reciprocal of the density 1S = 1 =  = 0.05 kilometers
k
20
S = 50 meters
To ﬁnd the headway we need to ﬁgure out how fast the
cars are traveling on the road. We use Greenshield’s
model to estimate the speed when k = 20 veh/kmla. Virginia Tech 64 Solution to Part (a)
Traveling at 35.71 km/hr (9.92 m/s) the headway (h) is, 50 h =  = 5.04 seconds
9.92 Virginia Tech 65 Solution to Part (b)
To check if the turning vehicle can make a safe maneuver,
check the time to turn against the headway (h) calculated
in part (a). Account for the reaction time of the turning
vehicle.
The time available to execute a safe turn is (h)  0.5
seconds to account for reaction time, t available = 5.04 – 0.5 = 4.54 seconds
Technically we should use the gap between two
successive vehicles to estimate the time to turn left. In this
case we have to subtract the time traveled by the
oncoming vehicle to cover its car length at 9.92 m/s
Virginia Tech 66 t gap 5.8 = 3.96 seconds
= 5.04 – 0.5 – 9.92 The distance traveled by a vehicle with a nonuniform
acceleration is,
–k t
–k t
k1
V0
k 1 t S =  –  ( 1 – e ) +  ( 1 – e )
2
k2
k2 k2
2 2 Note that t is either 3.96 or 4.54 seconds (depending on
your assumption on when the stopped vehicle starts the
left turn).
Using values of k 1 , k 2 of 4 and 0.1, respectively, the left
turning vehicle travels 35.6 meters in 4.54 seconds and 28
meters in 3.96 seconds.
Virginia Tech 67 A plot of distance traveled vs. time is shown in the
following diagram. The total distance to be traveled in the
left turn maneuver to reach a safe point is, 2 π R + L = 2 π ( 12 ) + 5.8 = 24.65
d = meters
4
4
The vehicle can execute the turn safely. Virginia Tech 68 Distance vs. Time Proﬁle (Turning Car)
The car reaches 24.65 m in 3.73 seconds The car travels 35.65 m in 4.54 seconds
Virginia Tech 69 Problem 2
A new Light Rail Transit (LRT) system to be installed in
San Diego, California has a fully loaded weight (W) of
300 kiloNewtons (kN). The LRT vehicle manufacturer
states that the basic resistance (in Newtons) can be
represented by a quadratic formula,
R B = a + bV + cV 2 where: a , b , and c have been determined to be: 750
(Newtons), 18 (Newtons * hr/km) and 0.14 (Newtons *
hr2/ km2), recpectively. V is the speed of the LRT train in
km/hr. Virginia Tech 70 As usual, the vehicle grade resistance, RG (in Newtons) is
expressed as,
R G = 10 Wi where: W is the weight in kN and i is the grade expressed
as percent (%).
The tractive effort (TE) curve vs. speed for the same
vehicle can be approximated by a line with negative
slope,
TE = 15, 000 – 70 V where: TE is in Newtons and V in km/hr. The tractive
effort and total resistance curves as function of speed are
shown in Figure 2 for ﬂat terrain.
Virginia Tech 71 Tractive Effort (N) Total Resistance (N)
at zero grade FIGURE 2. Tractive Effort and Total Resistance Diagram for LRT Vehicle. Virginia Tech 72 Questions for Problem 2
a) Find the maximum acceleration the vehicle can
generate while traveling at 40 km/hr in a +2% grade in
San Diego.
The basic resistance (RB) at 40 km/hr is,
R B = 750 + 18 ( 40 ) + 0.14 ( 40 ) 2 = 1, 694 Newtons The grade resistance at +2% is estimated to be,
R G = 10 ( 300 ) ( 2 ) = 6, 000 Newtons The tractive force at 40 km/hr is, Virginia Tech 73 TE = 15, 000 – 70 ( 40 ) = 12, 200 Newtons The acceleration is just the sum of forces divided by the
mass of the vehicle, m
12, 200 N – ( 7, 694 N )
a = dV =  = 0.15 2
dt
s
m 9.81  ( 300, 000 N ) ⁄ s2
b) Find the maximum speed the LRT can develop in
another application in Florida where the terrain is ﬂat.
By inspection of Figure 2 the maximum speed is 133 km/
hr. Virginia Tech 74 Plot of TE and Total Resistance vs. Speed Tractive Force Resistance at 2%
grade Virginia Tech 75 ...
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This note was uploaded on 12/31/2011 for the course CEE 3604 taught by Professor Katz during the Fall '08 term at Virginia Tech.
 Fall '08
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