Problem 3.1 The rolling resistance force is reduced on a slope by a cosine factor ( cos ). On the other hand, on a slope the gravitational force is added to the resistive forces. Assume a constant rolling resistance force and write the parametric forms of the total resistive force for both cases of level and sloping roads. At a given speed v, a) Write an expression that ensures an equal resistive force for both cases. b) Solve the expression obtained in (a) for the parametric values of corresponding slopes. c) For the coefficient of rolling resistance equal to., evaluate the values of the slopes obtained in (b) and discuss the result. Result: (a) f R (1- cos ) = sin Problem 3. For the vehicle of Example 3.4., a) Calculate the overall aerodynamic coefficient for the same temperature at altitude of 1 m. b) Repeat (a) for the same altitude at temperature 3 C. c) At the same altitude of (a) at what temperature the drag force increases by %? d) At the same temperature of (a) at what altitude the drag force reduces by %? Problem 3.3 In order to have a rough estimation for the performance of a vehicle, it is proposed to ignore the resistive forces to obtain the No-Resistive-Force (NRF) performance. a) Derive the governing equations of vehicle longitudinal motion for speed v(t) and distance S(t) by neglecting all resistive forces for the CPP (see Section 3.5).
b) For a vehicle of mass 1. ton, determine the required engine power P for achieving acceleration performance of -1 km/h during 1, 8 or 6 seconds. c) Evaluate the power increase factors from 1 seconds to t * seconds defined as [P/P 1 =[P(t * )-P(1)]/P(1)], for t * =8 and 6. Results: (a) Pt v v, m m( v 3 3 v ) S, (b) 46.3, 57.9 and 77. kw, (c).5 and.67 3P Problem 3.4 Use the results of Problem 3.3 and, a) Write the expression for the specific power P s (in W/kg) of a vehicle to reach a certain speed v (km/h) from the rest at a certain acceleration time t. b) Plot the variation of P s versus t from 6 to 1 seconds. Repeat the result for three speeds of 8, 9 and 1 km/h. c) Are the results dependent on the vehicle properties?
Specific power (W/kg) 65 6 8 km/h 9 km/h 1 km/h 55 5 45 4 35 3 5 6 6.5 7 7.5 8 8.5 9 9.5 1 Time of travel (s) Figure S3.4 Specific power requirements at no resistive force Problem 3.5 At very low speeds the aerodynamic force is small and may be neglected. For example, at speeds below 3 km/h, the aerodynamic force is one order of magnitude smaller than the rolling resistance force. For such cases categorised as Low-Speed (LS), ignore the aerodynamic force and for the CPP assume a constant rolling resistance force F, and, a) Integrate the equation of motion (Equation 3.58 with c=) and use the initial condition of v=v at t=t to obtain an expression for the travel time in terms of speed. b) For a vehicle of 1 kg mass and total rolling resistance force of N, when starting to move from standstill, plot the variation of vehicle speed against elapsed time up to 1 seconds and compare it with the results of NRF model (Problem 3.3). The engine power is 5 kw. Results: (a) m( v v) P P Fv t t m ln F F P F v
Velocity (m/s) 3 5 15 LS model NRF model 1 5 4 6 8 1 Figure S3.5 The time history of vehicle speed Problem 3.6 For the vehicle of Problem 3.5 and using the LS method, find the required power for the -1 km/h acceleration to take place in 7 seconds. Result: 58,849 W. Hint: The following statements in MATLAB can be used with a proper initial guess for x. fun=inline( 7-1*x*log(x/(x-(1/3.6/)))+1*(1/3.6/) ); x=fsolve(fun, x, optimset('display','off')); (x = P/F^) Problem 3.7 For the vehicle of Problem 3.5 using the LS method determine the power requirements for a performance starting from rest to reach speed v at time t, for 3 cases of v=8, 9 and 1 km/h for accelerating times varying from 6 to 1 seconds. Plot the results in a single figure.
Power required (kw) 7 65 v=8 v=9 v=1 6 55 5 45 4 35 3 5 6 6.5 7 7.5 8 8.5 9 9.5 1 to reach a certain speed v (km/h) Figure S3.7 Power requirements for different acceleration times Problem 3.8 The power evaluation for the NRF case (Problem 3.3) is a simple closed-form solution but it is not accurate. The LS method (problems 3.5-3.7) produces more accurate results especially in the low speed ranges. By generating plots similar to those of Problem 3.7 show that an approximate equation of P=P NRF +.75F v can generate results very close to those of LS method.
Power required (kw) 7 65 6 v=8 v=9 v=1 Approximate 55 5 45 4 35 3 5 6 6.5 7 7.5 8 8.5 9 9.5 1 to reach a certain speed v (km/h) Figure S3.8 Comparison between the LS and approximate models Problem 3.9 For the LS case use vdv ads acceleration in terms of speed and, that relates the speed to acceleration and distance, substitute for a) Integrate to obtain an expression for travel distance S in terms of velocity v. b) Derive the equation for a motion starting at a distance S from origin with velocity v. c) Simplify the expression for a motion stating from rest at origin. p1 Results: (a) S C mf [ p1 ln( P Fv ).5v 1 p1v 1], (c) S mf ( p1 ln.5v1 p1v 1) p v 1 1 P With p1 and F v v1. F
Power required (kw) Problem 3.1 A vehicle of 1 kg mass starts to accelerate from the rest at origin. If power is constant at 6 kw, for a LS model with F = N, determine the travel time and distance when speed is 1 km/h. Compare your results with those of NRF model. Results: t= 8.3 s, S=153.65 m for LS and t= 7.7 s and S=14.9 m for NRF. Problem 3.11 In Problem 3.8 a close approximation was used for the power estimation of LS method. For the general case including the aerodynamic force, the approximation given by P = P NRF +.5F R v is found to work well. For the vehicle of Example 3.5.3 plot the variations of power versus acceleration times similar to those of Problem 3.8 and compare the exact solutions with those obtained from the proposed method. 75 7 65 v=8 v=9 v=1 Approximate 6 55 5 45 4 35 3 5 6 6.5 7 7.5 8 8.5 9 9.5 1 to reach a certain speed v (km/h)
Figure S3.8 Comparison between the exact and approximate models Problem 3.1 According to the solutions obtained for CTP (see Section 3.6) it turned out that at each gear, the acceleration is constant to a good degree of approximation (see Figure 3.5). Thus a simpler solution can be obtained by considering an effective resistive force for each gear that reduces the problem to a Constant Acceleration Approximation (CAA). In each gear assume the resistive force acting on the vehicle is the average of that force at both ends of the constant torque range. Write the expressions for the average speed at each gear v av, the average resistive force R av and, 1 a) Show that the acceleration, velocity and distance at each gear are ai ( FTi Rav), m v ( t) ai ( t t) v and S i i.5ai ( t t ) v ( t t ) S i Oi i, in which v i vmax ( i 1) and S i S ( i 1) are the initial speed and distance from origin for each gear for i>1 and v and S max for i=1. b) Repeat Example 3.6.1 by applying the CAA method. Table S3.1 Results for Problem 3.1 v av (m/s) F av (N) F T (N) a (m/s^) v max (m/s) t i (s) S i (m) Gear 1 3.14 397.34 14667 7.135 4.398.617 1.356 Gear 4.987 44.83 94 4.418 6.981 1.1 3.37 Gear 3 7.915 43.73 581.699 11.79.719 13.76 Gear 4 1.566 471.36 3667 1.598 17.593 6.796 58.453
Speed (m/s) Acceleration (m/s) 8 7 6 5 4 3 1 1 3 4 5 6 7 Figure S3.1a Variation of acceleration 18 16 14 1 1 8 6 4 1 3 4 5 6 7 Figure S3.1b Variation of speed
Speed (m/s) Distance (m) 8 7 6 5 4 3 1 1 3 4 5 6 7 Figure S3.1c Variation of distance Problem 3.13 A 5 th overdrive gear with overall ratio of 3.15 is considered for the vehicle in Problem 3.1, and the torque is extended to 34 rpm. Obtain the time variations of acceleration, velocity and travel distance for the vehicle by both CAA and numerical methods and plot the results. 35 3 5 15 1 5 5 1 15 5
Distance (m) Figure S3.13a Comparison between speeds of CAA (solid) and numerical (dashed) methods 6 5 4 3 1 5 1 15 5 Figure S3.13b Comparison between distances of CAA (solid) and numerical (dashed) methods Problem 3.14 In Example 3.5. impose a limit for the traction force of F T <.5 W and compare the results.
Velocity (m/s) 5 45 4 35 3 With limit No limit 5 15 1 5 1 3 4 5 6 7 8 Figure S3.14 Comparison of vehicle speed with and without traction limit Problem 3.15 For a vehicle with transmission and engine information given in Example 3.7., include a onesecond torque interruption for each shift and plot similar results. To this end, include a subprogram with listing given below at the end of loop for each gear: % Inner loop for shifting delay: if i<5 % No delay after gear 5! t=max(t); tf=t+tdelay; x=[v(end) s(end)]; p=[ ]; % No traction force [t,x]=ode45(@fixed_thrt, [t tf], x); v=x(:,1); s=x(:,); end % Now plot the results p=[p1 p p3 p4]; % Set back the engine torque
Acceleration (m/s) Distance (m) Velocity (m/s) 5 4 3 1 1 3 4 5 6 5 15 1 5 1 3 4 5 6 Figure S3.15a Speed and distance outputs of Problem 3.15 7 6 5 4 3 1-1 1 3 4 5 6 Figure S3.15b Acceleration output of Problem 3.15
Distance (m) Velocity (m/s) Problem 3.16 Repeat Problem 3.15 with a different shifting delay for each gear of the form 1.5, 1.5, 1. and.75 second for 1-, -3, 3-4 and 4-5 shifts respectively. (For this you will need to change the program). 5 4 3 1 1 3 4 5 6 5 15 1 5 1 3 4 5 6 Figure S3.16a Speed and distance outputs of Problem 3.16
Acceleration (m/s) 7 6 5 4 3 1-1 1 3 4 5 6 Figure S3.16b Acceleration output of Problem 3.16 Problem 3.17 Repeat Example 3.7. for a different shifting rpm. a) Shift all gears at times when the engine speed is 45 rpm. b) Shift the gears at 45, 4, 35 and 3 rpm for shifting 1-, -3, 3-4 and 4-5 respectively. (For this part you will need to change the program).
Acceleration (m/s) Distance (m) Velocity (m/s) 5 4 3 1 1 3 4 5 6 5 15 1 5 1 3 4 5 6 Figure S3.17a Speed and distance outputs of Problem 3.17, part (a) 7 6 5 4 3 1 1 3 4 5 6 Figure S3.17b Acceleration output of Problem 3.17, part (a)
Acceleration (m/s) Distance (m) Velocity (m/s) 5 4 3 1 1 3 4 5 6 15 1 5 1 3 4 5 6 Figure S3.17c Speed and distance outputs of Problem 3.17, part (b) 7 6 5 4 3 1 1 3 4 5 6 Figure S3.17d Acceleration output of Problem 3.17, part (b) Problem 3.18 In Example 3.7.3, investigate the possibility of having a dynamic balance point at gear 4. In case no steady state point is available, find a new gear ratio to achieve a steady-state.
Acceleration (m/s) Distance (m) Velocity (m/s) Problem 3.19 Repeat Example 3.7. with transmission ratios 3.5, 1.77, 1.194,.96 and.711. 5 4 3 1 1 3 4 5 6 5 15 1 5 1 3 4 5 6 Figure S3.19a Speed and distance outputs of Problem 3.19 6 5 4 3 1 1 3 4 5 6 Figure S3.19b Acceleration output of Problem 3.19
Engine speed (rpm) Problem 3. In the program listing given for Example 3.7. no constraint is imposed for the lower limit of engine speed and at low vehicle speeds the engine rpm will attain values less than its working range of 1 rpm. a) For the existing program try to find out at what times and vehicle speeds the engine speed is below 1 rpm. b) Modify the program to ensure a speed of at least 1 rpm for the engine. How are the results affected? 6 5 4 3 1 1 3 4 5 Vehicle speed (m/s) Figure S3.a Variation of engine speed with vehicle speed (No rpm limit)
Distance (m) Velocity (m/s) 15 Without rpm limit With rpm limit 1 5 1 3 4 5 6 4 1 3 4 5 Figure S3.b Comparison of the output results with and without rpm limit Problem 3.1 In a vehicle roll-out test on a level road the variation of forward speed with time is found to be of the form: v a tan( b dt), where a, b and d are three constants. a) Assume an aerodynamic resistive force in the form of for the rolling resistance force F RR. b) Write an expression for the total resistive force acting on the vehicle. Result: (b) F R = md (a+v /a) F A cv and derive an expression
Problem 3. Two specific tests have been carried out on a vehicle with 13 kg weight to determine the resistive forces. In the first test on a level road and still air the vehicle reaches a maximum speed of 195 km/h in gear 5. In the second test on a road with slope of 1%, the vehicle attains maximum speed of 115 km/h in gear 4. In both tests the engine is working at WOT at 5 rpm, where the torque is 1 Nm. a) If the efficiency of the driveline is 9% and 95% at gears 4 and 5 respectively, determine the overall aerodynamic coefficient and the rolling resistance coefficient. b) If the gearbox ratio at gear 5 is.711 and the wheel effective radius is 3 mm, assume a slip of.5% at first test and determine the final drive ratio. c) Calculate the ratio of gear 4 (ignore the wheel slip). Results: (a) c=.314, f R =.14, (b) n f =4.4, (c) n 4 =1.6 Problem 3.3 For a vehicle with specifications given in table below, engine torque at WOT is of the following form: T e =1+a( e -1)-b( e -1), a=.4, b= 8 1-5, e < 6 rpm The driveline efficiency is approximated by.85+i/1 in which i is the gear number. Table P3.3 Vehicle information 1 Vehicle mass 1 kg Rolling Resistance Coefficient. 3 Tyre Rolling Radius.35 m 4 Final drive Ratio 3.5 5 Transmission Gear Ratio 1 4. 6 Gear Ratio.63 7 Gear Ratio 3 1.73 8 Gear Ratio 4 1.14 9 Gear Ratio 5.75 1 Aerodynamic Coefficient C D.4
11 Frontal Area A f. m 1 Air density A 1. kg/m 3 a) Determine the maximum engine power. b) What is the maximum possible speed of the vehicle? c) Calculate the maximum vehicle speed at gears 4 and 5. Results: (a) 69,173 W, (b) 17.6 km/h, (c) 169.9 and 14.6 km/h Problem 3.4 For the vehicle of Problem 3.3, a) For a constant speed of 6 km/h over a slope of 1% which gears can be engaged? b) For case (a) in which gear the input power is minimum? c) On this slope what would be the maximum vehicle speed in each gear? Hint: The following table is useful for solving this problem. Table P3.4 Parameter Gear 1 Gear Gear 3 Gear 3 Gear 5 1 Engine speed rpm Engine torque Nm 3 Vehicle maximum speed km/h
Problem 3.5 The vehicle of Problem 3.3 is moving on a level road at the presence of wind with velocity of 4 km/h. Assume C D =C D +.1 sin, in which is the wind direction relative to the vehicle direction of travel. Determine the maximum vehicle speed in gear 4 for: a) A headwind ( =18) b) A tailwind ( =) c) A wind with =135 degree. Results: (a) 14.5, (b) 19., (c) 14.5 km/h Table S3.4 Parameter Gear 1 Gear Gear 3 Gear 3 Gear 5 1 Engine speed 6366. 4185.8 753.4 1814.4 1193.7 rpm Engine torque 44.74 67.3 11.1 151.7 8. Nm 3 Vehicle maximum speed 56.6 86. 115.1 - - km/h Problem 3.6 Two similar vehicles with exactly equal properties are travelling on a level road but in opposite directions. Their limit speeds are measured as v 1 and v respectively. Engine torque at WOT is approximated by following equation: T e =15-1.14 1-3 ( -314.16) Determine the aerodynamic drag coefficient C D and wind speed in direction of travel v w by: a) Writing a parametric tractive force equation in terms of vehicle speed for both vehicles. b) Then write a parametric resistive force equation in terms of speed for both vehicles. c) Equate the two equations for each vehicle and use the numerical values of Problem 3.3 for m, f R, A f, A, r W and the additional information given in table below,
Table P3.6 1 Transmission Gear Ratio.9 v 1 18 3 v Results:.5 and 19.3 km/h Problem 3.7 While driving uphill in gear 4 on a road with constant slope, the vehicle of Problem 3.3 reaches its limit speed v U at an engine speed of at a still air. The same vehicle is then driven downhill on the same road in gear 5, while keeping the engine speed same as before. Engine powers for uphill and downhill driving are P and U P D respectively. The tyre slip is roughly estimated from equation S x =S +P 1 - (%) where S is a constant and P is power in hp. Assume a small slope angle and use the additional data given in the table to determine: a) Uphill and downhill driving speeds b) Road slope Table P3.7 1 Uphill power P U 9 hp Downhill power P D 1 hp 3 Tyre basic slip S. % 4 Gear progression ratio n 4 /n 5 C 4 1.4 Results: (a) 116.9 and 165.1 km/h, (b) 9.4 % Problem 3.8 For the vehicle of Problem 3.3, a) Derive a general parametric expression for the value of speed v * at the maximum attainable acceleration.
b) Use the numerical values and determine the values of v * at each gear. c) Calculate the maximum accelerations at each gear. Results: (a) v * 3ni d ( a b), (b) 9.6, 13.38, 18.51, 1.45 and 17.88 m/s, (c) 4.7, 3 rw 3 ni ( c db ) 3 r W.59, 1.55,.796 and.98 m/s. Problem 3.9 In Section 3.9 the effect of rotating masses were discussed and equations for including this effect in the acceleration performance of a vehicle were developed. From an energy consumption point of view, when vehicle is accelerated to the speed of v, the rotating inertias will be at rotational speeds related to v (ignore the tyre slip). a) Write the kinetic energies for the vehicle body mass m and rotating masses I e, I g and I w b) From the kinematic relations, relate the rotational speeds to the vehicle speed c) Write the energy terms in terms of vehicle speed v d) Write the total energy of vehicle as: E t.5m v eq e) Determine the equivalent mass m eq and compare it with Equation 3.13 Problem 3.3 For a tyre with the Magic Formula information given in Table 3.3, a) Plot the longitudinal force (F) against slip (s) for both traction and brake regions at normal load values 1.,., 3. and 4. kn (all in a single figure)
Longitudinal force (N) b) Plot coefficients of tyre-road friction for case (a) c) At slip ratios 5, 1, and 5%, plot the variation of F x versus F z (max F z =5kN). d) Differentiate the Magic Formula with respect to slip to find the value of slip at which the force is maximum. Verify your results by comparing them with those of case (a). e) In order to have an impression of the influence of different factors in the Magic Formula tyre model, try the following for the above tyre in (a) at a normal load of 3. kn: I. Multiply coefficient B by.8, 1. and 1. while keeping the other coefficients unchanged. Plot all three results in a single figure. II. Repeat I for coefficient C. III. Repeat I for coefficient D. IV. Repeat I for coefficient E. 5 4 3 1 Fz=1. kn Fz=. kn Fz=3. kn Fz=4. kn -1 - -3-4 -5-1 -5 5 1 Longitudinal slip (%) Figure S3.3a Variation of tyre longitudinal force with slip at different normal loads
Longitudinal force (N) Adhesion coefficient 1..8 1. kn. kn 3. kn 4. kn.4 -.4 -.8-1. -1-5 5 1 Longitudinal slip (%) Figure S3.3b Variation of adhesion coefficient with slip at different normal loads 6 5 4 Sx=5% Sx=1% Sx=% Sx=5% 3 1 1 3 4 5 Normal load (N) Figure S3.3c Variation of tyre longitudinal force with normal load at different slips
Longitudinal force (N) Longitudinal force (N) 4 3 Multiplier=.8 Multiplier=1. Multiplier=1. 1-1 - -3-4 -1-5 5 1 Longitudinal slip (%) Figure S3.3d Effect of the Magic Formula s B factor on the F x 4 3 Multiplier=.8 Multiplier=1. Multiplier=1. 1-1 - -3-4 -1-5 5 1 Longitudinal slip (%) Figure S3.3e Effect of the Magic Formula s C factor on the F x
Longitudinal force (N) Longitudinal force (N) 4 3 Multiplier=.8 Multiplier=1. Multiplier=1. 1-1 - -3-4 -1-5 5 1 Longitudinal slip (%) Figure S3.3f Effect of the Magic Formula s D factor on the F x 4 3 Multiplier=.8 Multiplier=1. Multiplier=1. 1-1 - -3-4 -1-5 5 1 Longitudinal slip (%) Figure S3.3g Effect of the Magic Formula s E factor on the F x
Problem 3.31 The vehicle of Problem 3.3 is moving with a constant speed of 1 km/h. Use the tyre data of Table 3.3 for each of the two driving wheels and for a front/rear weight distribution of 6/4, determine a) Longitudinal slip (in percentage) of the tyres for both cases of FWD and RWD b) Repeat (a) for a 5-degree grade (ignore the load transfer) c) Repeat (a) for a level road with adhesion coefficient of.4