Introduction to Tansportation Engineering. Fundamentals of Ground Vehicle Performance

Transcription

1 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 201 Virginia Tech 1

2 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

3 Ground Vehicle Performance Basics Virginia Tech 3

4 Basic Forces Acting on a Ground Vehicle The following diagram applies to cars, buses, trains, etc. Lift Thrust Drag Friction Force mg φ Assumed all forces act through a point mass system (center of gravity) Thrust is also called Tractive Effort (TE) Virginia Tech 4

5 Analysis of Individual Forces L D T = = = 1 --ρv 2 SC l ρv 2 SC D 2 f( V, ρ) F f = ( mgcosφ L)f roll G t = mgsinφ (1) (2) (3) (4) (5) Virginia Tech 5

6 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), F f 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

7 S is the reference area (m 2 ), C l lift coefficient (nondimensional), C D is the drag coefficient (nondimensional), f roll is the rolling friction coefficient (nondimensional) m is the vehicle mass g = gravitational acceleration (9.81 m/s 2 or 32.2 ft/s 2 ) Virginia Tech 7

8 Application of Newton s Second Law ma x = T D F f G f ma x = mgsinφ fvρ (, ) 1 --ρv 2 SC D ( mgcosφ L)f roll 2 (6) (7) 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

9 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, V 0 and the desired cruising speed, V cr the distance covered during the acceleration phase can be found (this might require a numerical integration procedure) Virginia Tech 9

10 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

11 Variations of Tractive Effort and f roll T decreases with speed for all transportation modes f roll is a function of the contact material between vehicle and guideway or road and the vehicle speed T (N) Sea Level f roll Bias-Ply Tire High Elevation Radial Tire Metal Rim Surface V (m/sec) V (m/sec) Virginia Tech 11

12 Rolling Friction Coefficient ( ) The value of varies with speed, tire construction, tire pressure, etc. f roll is usually determined experimentally f roll f roll 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

13 Variation of f roll with Speed f roll Value (dimensionless) Source: W. Hucho (1998) Data good for radial car tires Speed (km/hr) Virginia Tech 13

14 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 velocity-time profile of the vehicle. Assume that grades are zero throughout the route. Florida is pretty flat like a tortilla. Virginia Tech 14

15 Maglev Train Parameters The following parameters apply to the Maglev train. S=25 m 2, f roll =.025 (magnetic rolling friction),ρ=1.225 kg/cu-m., C D =.55,C l =.20,g=9.81 m/s 2 and m=350,000 kg MAGLEV Thrust (or tractive effort) Parameters for the Hypothetical Orlando Line. Speed (m/s) Thrust Force (N) , ,000 Virginia Tech 15

16 Analysis of the Maglev Train Simplify the basic equation of motion by eliminating the angle between the track and the horizontal, ma x = fvρ (, ) 1 --ρv 2 SC D ( mg L)f roll 2 Substitute the numerical values for the train in equation (6) to obtain, dv = dt a = 1 t m --- ( 214, , 678V 8.345V ) (8) Virginia Tech 16

17 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) V dv , 163 1, 678V 8.345V 2 Vo m = t 0 dt (9) The denominator is of the quadratic form: A + BV + CV 2 Virginia Tech 17

18 Numerical Integration A general procedure to solve any complex differential equation encountered in practice. Start with, dv = dt a = 1 t m --- ( 214, , 678V 8.345V ) then take small increments of time into the future and integrate dv dt Breaking down the path into small intervals is the key to the numerical integration process Virginia Tech 18

19 How it Works Virginia Tech (A.A. Trani) 19

20 Graphical Interpretation The numerical integration procedure is illustrated graphically in the diagram Note: the Euler integration scheme assumes a constant value for in the interval (, ) dv dt t t t t To achieve good accuracy make the step size ( ) small watching that the round-off errors in the computation do not become excessive Virginia Tech 20

21 Finding the Distance Traveled The distance traveled is the second integral of the acceleration function or the integral of the vehicle velocity function ds S t = S t t t = S t t + V t t Note that we are trying to find the distance to accelerate (named S 1 hereon) and reach some desired cruise speed V cruise ( ) dt ( t t, t) Virginia Tech 21

22 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 Speed Area under the speed vs time curve is the distance traveled (S) S 1 S 2 S 3 T1 T2 T3 Time Virginia Tech 22

23 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 = 60V cruise For the deceleration segment assume a uniform deceleration (assume 0.1g or 1.0 m/s 2 ), 2 2 V S 2 V 0 3 = a (10) (11) Virginia Tech 23

24 Analysis To find V cruise and unknown distances S 1, S 2 and S 3 we use a numerical integration procedure (i.e., Euler, modified Euler or Runge-Kutta 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 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 V cruise and thus the numerical integration technique advocated here is suitable for any type of acceleration function Virginia Tech 24

25 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

26 Numerical Results 1: Velocit y 2: Dista nce 1: 2: : 2: : 2: Time Velocity units are m/s Distance units are m Virginia Tech 26

27 Force Diagram The force diagram shows all forces acting on the vehicle Data #2 Force (N) Speed (m/s) Te (N) Treq (N) Maximum Speed Virginia Tech 27

28 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, P = T R V η (12) where P is the power (Watts), T R is the tractive force required (in Newtons) to overcome the resistance forces Virginia Tech 28

29 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), E = t 0 Pdt 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 (13) Virginia Tech 29

30 Numerical Integration Using Excel Excel or Matlab can be used to numerically integrate the expressions presented and do the analysis. dv = dt a = 1 t m --- ( 214, , 678V 8.345V ) 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

31 Excel Representation (Maglev Problem) Excel solution to the first-order differential equation Virginia Tech (A.A. Trani) 31

32 Solve the Equation Using Numerical Integration Virginia Tech (A.A. Trani) 31a

33 Excel Representation (Maglev Problem) Sample formulas to use C3 =1/350000*( *B *B3^2) D3 = =A4-A3 E3=D3*C3 B4=B3+E3 Virginia Tech (A.A. Trani) 31b

34 Explanations The initial condition of the Maglev train is assumed to be known (IBVP) : at time t=0 the train has zero speed (V) at time t=0 the acceleration is calculated using the known equation of motion m/s 2 Virginia Tech (A.A. Trani) 31c

35 Explanations With the acceleration at time t=0, we calculate the velocity at next time interval ( ) m/s The process is repeated until a desired final speed is reached Virginia Tech (A.A. Trani) 31d

36 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

37 Modeling Rail and Highway Vehicles Simple Performance Models Virginia Tech (A.A. Trani) 32a

38 Why Modeling? We can make informed decisions about what transportation technology is best suited for a given application We understand the tradeoffs between performance and energy consumption We can estimate travel time and travel cost for each mode of transportation Virginia Tech (A.A. Trani) 32b

39 Approach to Modeling Ground Vehicles Equations provided so far give a general perspective of the fundamental forces acting on the vehicle Now we focus on how various organizations and researchers model ground vehicles The fundamental equations differentiate the forces acting on the vehicle as either: a) resistance or b) tractive effort (same as before) Virginia Tech (A.A. Trani) 32c

40 General Representation of Tractive Force Diagrams for Internal Combustion Engine Vehicles Total Resistance Force Virginia Tech (A.A. Trani) 34d

41 General Representation of Tractive Force Diagrams for Rail Technology Vehicles Force (N) Tractive Force Total Resistance Force Speed (m/s) Virginia Tech (A.A. Trani) 34e

42 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

43 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) Four-cycle 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 rubber-tired 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

44 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

45 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 i is the transmission gearing ratio for the ith gear (the value of u i 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. Virginia Tech 36

46 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: V = 60( N) D 1000Ju i ( )( π) (15) where the units of each variable are: N in revolutions per minute, D in meters, J and u i are dimensionless, and V is expressed in km/hr. Virginia Tech 37

47 Force vs. Speed Diagrams Using the expression to convert power output to tractive effort we have, T = ηp V 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. (16) Virginia Tech 38

48 Froce vs. Speed Diagrams Once the conversion factors are applied we have: T = 2650 ηp V (17) where: T is expressed in Newtons P in horsepower V in km/hr Virginia Tech 39

49 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 i in Equation 15. Substitute the corresponding velocity (V) values into Equation 17 to estimate T. Since each vehicles has several feasible transmission gearing ratios, u i, it is necessary to compute several values of T for each value N in Equation 15. The resultant performance of a typical IC engine vehicle is shown in the following page. Virginia Tech 40

50 Power (hp) Given T (N) Estimate First Gear Second Gear Engine Speed - N - (RPM) Third Gear Given: Vehicle by Manufacturer Fourth Gear Vehicle Speed (m/s) 60( N) ( D) ( π) Use V = and T = 2650 ηp V 1000Ju i Virginia Tech 41

51 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

52 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 b = [ c 1 + c 2 V]W + c a AV 2 (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 (m 2 ) and V is the vehicle speed in km/ hr., and is the basic resistance in Newtons. R b Virginia Tech 43

53 Vehicle Resistance A typical value for c a is.018 for buses and for streamlined trucks (0.025 for non-streamlined 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 = 10Wi (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

54 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

55 Bus Characteristics The technical characteristics of the bus are given below; Table 1: Flxible Bus Power Characteristics. N (RPM) P (hp) 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 m 2 (75 ft 2 ). The resistance function for the bus is given by, Virginia Tech 46

56 R = [ c 1 + c 2 V]W + c a AV 2 (20) The values for each constant are: c a = 0.020, c 1 = 7.6, and c 2 = 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 V max for first gear = 20 km/hr 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

57 With this information we construct a Force vs Speed diagram using the process explained in previous pages. Virginia Tech 48

58 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 u i 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 (a-c) for other values of N. I suggest you repeat the process for at least 5-10 values of N A sample T vs. V diagram is shown below. Virginia Tech 49

59 Force vs. Speed Diagram for Flxible Bus 3 x 104 First gear 2.5 T (Newtons) Tractive Effort (N) TextEnd Second gear Third gear Fourth gear 0.5 Total resistance (0% grade) Speed (km/hr) Virginia Tech 50

60 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

61 T vs V Diagram (5% grade resistance forces) 3 x 104 First gear T (Newtons) Tractive Effort (N) TextEnd Second gear Third gear Total resistance (5% grade) Fourth gear Speed (km/hr) Virginia Tech 52

62 2.5 3 x Approximation of T vs. V Diagram First gear Approximation Profile Tractive Effort (N) TextEnd Second gear Third gear Fourth gear 0.5 Total resistance (0% grade) Speed (km/hr) Use linear segments to approximate the T vs. V curve. Virginia Tech 53

63 3 x 104 Finding Acceleration T vs. V Diagram Tractive Effort (N) TextEnd Excess Force to Accelerate 0.5 Total resistance (0% grade) Speed (km/hr) Virginia Tech 54

64 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

65 dv/dt k Kinematic Models Constant acceleration/deceleration dv/dt V k 1 k 2 k 1 k 1 /k 2 V Virginia Tech 56

66 Constant Acceleration Model Well known equations. the acceleration of the vehicle is equal to a constant ( ), d---- V = k dt k Integrate once to get speed (V) and twice to get distance traveled (S) V t = V 0 + kt V t V 0 t where: is the speed at time t, is the initial speed of the vehicle and is the time from start of motion Virginia Tech 57

67 Constant Acceleration Model The distance traveled (S) is, S t = V 0 t kt 2 + S 0 2 S 0 where: 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, S t = V 0 t at 2 + S 0 2 where a is the constant acceleration of the vehicle Virginia Tech 58

68 Variable Acceleration Model Start with, d---- V = k 1 k 2 V dt Integrate once to get speed (V) k V 1 t e k 2t ( ) V 0 e k 2t = + k 2 V t V 0 k 1 k 2 t where: is the speed at time t, is the initial speed of the vehicle, and are constants and is the time from start of motion Virginia Tech 59

69 Variable Acceleration Model The distance traveled (S) is, k S 1 t = ---t k 2 k 1 2 k e k 2t ( ) + V e k 2t ( ) k 2 Virginia Tech 60

70 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

71 School Entrance Spacing = S To Blacksburg R = 12 m. 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 = V where: a is the acceleration of the car (in m/s 2 ) and V is the vehicle speed in m/s. Virginia Tech 62

72 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

73 Solution to Part (a) Find the Spacing (S) between vehicles. Since the density of the traffic flow is known to be 20 veh/km-la we compute the spacing as the reciprocal of the density S = 1 -- = = 0.05 k 20 kilometers 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/km-la. Virginia Tech 64

74 Solution to Part (a) Traveling at km/hr (9.92 m/s) the headway (h) is, h = = seconds Virginia Tech 65

75 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) seconds to account for reaction time, t available = = 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

76 t gap = = seconds The distance traveled by a vehicle with a non-uniform acceleration is, S = k 1 t k 2 k 1 2 k e k 2t ( ) + V e k 2t ( ) k 2 t Note that is either 3.96 or 4.54 seconds (depending on your assumption on when the stopped vehicle starts the left turn). k 1 k 2 Using values of, 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

77 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, d = 2πR L = 2π ( 12) = meters The vehicle can execute the turn safely. Virginia Tech 68

78 Distance vs. Time Profile (Turning Car) The car reaches m in 3.73 seconds The car travels m in 4.54 seconds Virginia Tech 69

79 Problem 2 A new Light Rail Transit (LRT) system to be installed in San Diego, California has a fully loaded weight (W) of 300 kilo-newtons (kn). The LRT vehicle manufacturer states that the basic resistance (in Newtons) can be represented by a quadratic formula, R B a bv cv = where: a, b, and c have been determined to be: 750 (Newtons), 18 (Newtons * hr/km) and 0.14 (Newtons * hr 2 / km 2 ), recpectively. V is the speed of the LRT train in km/hr. Virginia Tech 70

80 As usual, the vehicle grade resistance, R G (in Newtons) is expressed as, R G = 10Wi 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, 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

81 Tractive Effort (N) Total Resistance (N) at zero grade FIGURE 2. Tractive Effort and Total Resistance Diagram for LRT Vehicle. Virginia Tech 72

82 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 (R B ) at 40 km/hr is, R B = ( 40) = 1, 694 ( ) 2 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

83 TE = ( 40) =, 12, 200 Newtons The acceleration is just the sum of forces divided by the mass of the vehicle, a = dv = 12, 200N ( 7, 694N) dt ( 300, 000N) m = m s 2 s 2 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

84 Plot of TE and Total Resistance vs. Speed Virginia Tech 75