Forward Vehicle Dynamics

2 Forward Vehicle Dynamics Straight motion of an ideal rigid vehicle is the subject of this chapter. We ignore air friction and examine the load variation under the tires to determine the vehicle’s limits of acceleration, road grade, and kinematic capabilities. 2.1 Parked Car on a Level Road When a car is parked on level pavement, the normal force, Fz , under each of the front and rear wheels, Fz1 , Fz2 , are 1 a2 mg 2 l 1 a1 = mg 2 l Fz1 = (2.1) Fz2 (2.2) where, a1 is the distance of the car’s mass center, C, from the front axle, a2 is the distance of C from the rear axle, and l is the wheel base. l = a1 + a2 (2.3) z a2 a1 x C 2Fz2 mg 2Fz1 FIGURE 2.1. A parked car on level pavement. Proof. Consider a longitudinally symmetrical car as shown in Figure 2.1. It can be modeled as a two-axel vehicle. A symmetric two-axel vehicle is equivalent to a rigid beam having two supports. The vertical force under the front and rear wheels can be determined using planar static equilibrium equations. X Fz = 0 (2.4) X My = 0 (2.5) Applying the equilibrium equations 2Fz1 + 2Fz2 − mg = 0 −2Fz1 a1 + 2Fz2 a2 = 0 (2.6) (2.7) provide the reaction forces under the front and rear tires. 1 a2 mg 2 a1 + a2 1 a2 = mg 2 l 1 a1 = mg 2 a1 + a2 1 a1 = mg 2 l Fz1 = Fz2 (2.8) (2.9) Example 39 Reaction forces under wheels. A car has 890 kg mass. Its mass center, C, is 78 cm behind the front wheel axis, and it has a 235 cm wheel base. a1 = 0.78 m l = 2.35 m m = 890 kg (2.10) (2.11) (2.12) The force under each front wheel is 1 a2 mg 2 l 1 2.35 − 0.78 = 890 × 9.81 × = 2916.5 N 2 2.35 Fz1 = (2.13) and the force under each rear wheel is 1 a1 mg 2 l 1 0.78 = 890 × 9.81 × = 1449 N. 2 2.35 Fz2 = (2.14) Example 40 Mass center position. Equations (2.1) and (2.2) can be rearranged to calculate the position of mass center. 2l Fz mg 2 2l a2 = Fz mg 1 a1 = (2.15) (2.16) Reaction forces under the front and rear wheels of a horizontally parked car, with a wheel base l = 2.34 m, are: Fz1 = 2000 N Fz2 = 1800 N (2.17) (2.18) Therefore, the longitudinal position of the car’s mass center is at a1 = 2l Fz mg 2 =2 a2 = 2.34 × 1800 = 1.1084 m 2 (2000 + 1800) (2.19) 2l Fz mg 1 =2 2.34 × 2000 = 1.2316 m. 2 (2000 + 1800) (2.20) Example 41 Longitudinal mass center determination. The position of mass center C can be determined experimentally. To determine the longitudinal position of C, we should measure the total weight of the car as well as the force under the front or the rear wheels. Figure 2.2 illustrates a situation in which we measure the force under the front wheels. Assuming the force under the front wheels is 2Fz1 , the position of the mass center is calculated by static equilibrium conditions X X Fz = 0 (2.21) My = 0. (2.22) Applying the equilibrium equations 2Fz1 + 2Fz2 − mg = 0 −2Fz1 a1 + 2Fz2 a2 = 0 (2.23) (2.24) l a1 a2 x C mg 2Fz2 2Fz1 FIGURE 2.2. Measuring the force under the front wheels. provide the longitudinal position of C and the reaction forces under the rear wheels. 2l Fz mg 2 2l = (mg − 2Fz1 ) mg 1 = (mg − 2Fz1 ) 2 a1 = Fz2 (2.25) (2.26) Example 42 Lateral mass center determination. Most cars are approximately symmetrical about the longitudinal center plane passing the middle of the wheels, and therefore, the lateral position of the mass center C is close to the center plane. However, the lateral position of C may be calculated by weighing one side of the car. Example 43 Height mass center determination. To determine the height of mass center C, we should measure the force under the front or rear wheels while the car is on an inclined surface. Experimentally, we use a device such as is shown in Figure 2.3. The car is parked on a level surface such that the front wheels are on a scale jack. The front wheels will be locked and anchored to the jack, while the rear wheels will be left free to turn. The jack lifts the front wheels and the required vertical force applied by the jacks is measured by a load cell. Assume that we have the longitudinal position of C and the jack is lifted such that the car makes an angle φ with the horizontal plane. The slope angle φ is measurable using level meters. Assuming the force under the front wheels is 2Fz1 , the height of the mass center can be calculated by ( h − R ) sin φ a1 z x a2 h ( h − R ) sin φ C 2Fz1 H h cos φ mg φ 2Fz2 FIGURE 2.3. Measuring the force under the wheels to find the height of the mass center. static equilibrium conditions X X FZ = 0 (2.27) My = 0. (2.28) Applying the equilibrium equations 2Fz1 + 2Fz2 − mg = 0 −2Fz1 (a1 cos φ − (h − R) sin φ) +2Fz2 (a2 cos φ + (h − R) sin φ) = 0 (2.29) (2.30) provides the vertical position of C and the reaction forces under the rear wheels. 1 mg − Fz1 2 Fz (R sin φ + a1 cos φ) + Fz2 (R sin φ − a2 cos φ) h= 1 mg sin φ a1 Fz1 − a2 Fz2 =R+ cot φ mg ¶ µ Fz = R + 2 1 l − a2 cot φ mg Fz2 = (2.31) (2.32) A car with the following specifications m 2Fz1 φ a1 l R = = = = = = 2000 kg 18000 N 30 deg ≈ 0.5236 rad 110 cm 230 cm 30 cm (2.33) has a C at height h. h = 34 cm (2.34) There are three assumptions in this calculation: 1− the tires are assumed to be rigid disks with radius R, 2− fluid shift, such as fuel, coolant, and oil, are ignored, and 3− suspension deflections are assumed to be zero. Suspension deflection generates the maximum effect on height determination error. To eliminate the suspension deflection, we should lock the suspension, usually by replacing the shock absorbers with rigid rods to keep the vehicle at its ride height. Example 44 Different front and rear tires. Depending on the application, it is sometimes necessary to use different type of tires and wheels for front and rear axles. When the longitudinal position of C for a symmetric vehicle is determined, we can find the height of C by measuring the load on only one axle. As an example, consider the motorcycle in Figure 2.4. It has different front and rear tires. Assume the load under the rear wheel of the motorcycle Fz is known. The height h of C can be found by taking a moment of the forces about the tireprint of the front tire. µ ¶ Fz (a1 + a2 ) H Rf + Rr h= 2 + − a1 cos sin−1 (2.35) mg a1 + a2 2 Example 45 Statically indeterminate. A vehicle with more than three wheels is statically indeterminate. To determine the vertical force under each tire, we need to know the mechanical properties and conditions of the tires, such as the value of deflection at the center of the tire, and its vertical stiffness. 2.2 Parked Car on an Inclined Road When a car is parked on an inclined pavement as shown in Figure 2.5, the normal force, Fz , under each of the front and rear wheels, Fz1 , Fz2 , is: a2 Rf C h a1 Fz1 mg H φ Rr Fz2 FIGURE 2.4. A motorcycle with different front and rear tires. 1 a2 mg cos φ + 2 l 1 a1 Fz2 = mg cos φ − 2 l l = a1 + a2 Fz1 = 1 h mg sin φ 2 l 1 h mg sin φ 2 l (2.36) (2.37) where, φ is the angle of the road with the horizon. The horizon is perpendicular to the gravitational acceleration g. Proof. Consider the car shown in Figure 2.5. Let us assume the parking brake forces are applied on only the rear tires. It means the front tires are free to spin. Applying the planar static equilibrium equations X Fx = 0 (2.38) X Fz = 0 (2.39) X My = 0 (2.40) shows that 2Fx2 − mg sin φ = 0 2Fz1 + 2Fz2 − mg cos φ = 0 −2Fz1 a1 + 2Fz2 a2 − 2Fx2 h = 0. (2.41) (2.42) (2.43) a1 z x h a2 C a 2F z1 mg φ 2 F x2 2F z2 FIGURE 2.5. A parked car on inclined pavement. These equations provide the brake force and reaction forces under the front and rear tires. 1 a2 mg cos φ − 2 l 1 a1 = mg cos φ + 2 l 1 = mg sin φ 2 Fz1 = Fz2 Fx2 1 h mg sin φ 2 l 1 h mg sin φ 2 l (2.44) (2.45) (2.46) Example 46 Increasing the inclination angle. When φ = 0, Equations (2.36) and (2.37) reduce to (2.1) and (2.2). By increasing the inclination angle, the normal force under the front tires of a parked car decreases and the normal force and braking force under the rear tires increase. The limit for increasing φ is where the weight vector mg goes through the contact point of the rear tire with the ground. Such an angle is called a tilting angle. Example 47 Maximum inclination angle. The required braking force Fx2 increases by the inclination angle. Because Fx2 is equal to the friction force between the tire and pavement, its maximum depends on the tire and pavement conditions. There is a specific angle φM at which the braking force Fx2 will saturate and cannot increase any more. At this maximum angle, the braking force is proportional to the normal force Fz2 Fx2 = μx2 Fz2 (2.47) where, the coefficient μx2 is the x-direction friction coefficient for the rear wheel. At φ = φM , the equilibrium equations will reduce to 2μx2 Fz2 − mg sin φM = 0 2Fz1 + 2Fz2 − mg cos φM = 0 2Fz1 a1 − 2Fz2 a2 + 2μx2 Fz2 h = 0. (2.48) (2.49) (2.50) These equations provide 1 1 a2 h mg cos φM − mg sin φM 2 l 2 l 1 1 a1 h = mg cos φM + mg sin φM 2 l 2 l a1 μx2 = l − μx2 h Fz1 = (2.51) Fz2 (2.52) tan φM (2.53) showing that there is a relation between the friction coefficient μx2 , maximum inclination φM , and the geometrical position of the mass center C. The angle φM increases by decreasing h. For a car having the specifications μx2 a1 l h = = = = 1 110 cm 230 cm 35 cm (2.54) the tilting angle is φM ≈ 0.514 rad ≈ 29.43 deg . (2.55) Example 48 Front wheel braking. When the front wheels are the only braking wheels Fx2 = 0 and Fx1 6= 0. In this case, the equilibrium equations will be 2Fx1 − mg sin φ = 0 2Fz1 + 2Fz2 − mg cos φ = 0 −2Fz1 a1 + 2Fz2 a2 − 2Fx1 h = 0. (2.56) (2.57) (2.58) These equations provide the brake force and reaction forces under the front and rear tires. 1 a2 mg cos φ − 2 l 1 a1 = mg cos φ + 2 l 1 = mg sin φ 2 Fz1 = Fz2 Fx1 1 h mg sin φ 2 l 1 h mg sin φ 2 l (2.59) (2.60) (2.61) At the ultimate angle φ = φM Fx1 = μx1 Fz1 (2.62) 2μx1 Fz1 − mg sin φM = 0 2Fz1 + 2Fz2 − mg cos φM = 0 2Fz1 a1 − 2Fz2 a2 + 2μx1 Fz1 h = 0. (2.63) (2.64) (2.65) and These equations provide 1 1 a2 h mg cos φM − mg sin φM 2 l 2 l 1 1 a1 h = mg cos φM + mg sin φM 2 l 2 l a2 μx1 = . l − μx1 h Fz1 = (2.66) Fz2 (2.67) tan φM (2.68) Let’s name the ultimate angle for the front wheel brake in Equation (2.53) as φMf , and the ultimate angle for the rear wheel brake in Equation (2.68) as φMr . Comparing φMf and φMr shows that φMf φMr ¢ ¡ a1 μx2 l − μx1 h ¢. ¡ = a2 μx1 l − μx2 h (2.69) We may assume the front and rear tires are the same and so, μx1 = μx2 (2.70) therefore, φMf φMr = a1 . a2 (2.71) Hence, if a1 < a2 then φMf < φMr and therefore, a rear brake is more effective than a front brake on uphill parking as long as φMr is less than the tilting angle, φMr < tan−1 ah2 . At the tilting angle, the weight vector passes through the contact point of the rear wheel with the ground. Similarly we may conclude that when parked on a downhill road, the front brake is more effective than the rear brake. Example 49 Four-wheel braking. Consider a four-wheel brake car, parked uphill as shown in Figure 2.6. In these conditions, there will be two brake forces Fx1 on the front wheels and two brake forces Fx1 on the rear wheels. a1 z x h a2 C a 2F x 1 2F z1 mg φ 2 F x2 2F z2 FIGURE 2.6. A four wheel brake car, parked uphill. The equilibrium equations for this car are 2Fx1 + 2Fx2 − mg sin φ = 0 2Fz1 + 2Fz2 − mg cos φ = 0 −2Fz1 a1 + 2Fz2 a2 − (2Fx1 + 2Fx2 ) h = 0. (2.72) (2.73) (2.74) These equations provide the brake force and reaction forces under the front and rear tires. 1 a2 mg cos φ − 2 l 1 a1 = mg cos φ + 2 l 1 = mg sin φ 2 Fz1 = Fz2 Fx1 + Fx2 1 h mg sin φ 2 l 1 h mg sin φ 2 l (2.75) (2.76) (2.77) At the ultimate angle φ = φM , all wheels will begin to slide simultaneously and therefore, Fx1 Fx2 = μx1 Fz1 = μx2 Fz2 . (2.78) (2.79) The equilibrium equations show that 2μx1 Fz1 + 2μx2 Fz2 − mg sin φM = 0 2Fz1 + 2Fz2 − mg cos φM = 0 ¡ ¢ −2Fz1 a1 + 2Fz2 a2 − 2μx1 Fz1 + 2μx2 Fz2 h = 0. (2.80) (2.81) (2.82) z a2 a1 x C 2Fx2 a mg 2Fz2 h 2Fx1 2Fz1 FIGURE 2.7. An accelerating car on a level pavement. Assuming μx1 = μx2 = μx (2.83) will provide 1 1 a2 h mg cos φM − mg sin φM 2 l 2 l 1 1 a1 h = mg cos φM + mg sin φM 2 l 2 l = μx . Fz1 = (2.84) Fz2 (2.85) tan φM (2.86) 2.3 Accelerating Car on a Level Road When a car is speeding with acceleration a on a level road as shown in Figure 2.7, the vertical forces under the front and rear wheels are Fz1 = Fz2 = 1 a2 1 ha mg − mg 2 l 2 lg 1 a1 1 ha mg + mg . 2 l 2 lg (2.87) (2.88) The first terms, 12 mg al2 and 12 mg al1 , are called static parts, and the second terms ± 12 mg hl ag are called dynamic parts of the normal forces. Proof. The vehicle is considered as a rigid body that moves along a horizontal road. The force at the tireprint of each tire may be decomposed to a normal and a longitudinal force. The equations of motion for the accelerating car come from Newton’s equation in x-direction and two static equilibrium equations. X Fx = ma (2.89) Fz = 0 (2.90) My = 0. (2.91) X X Expanding the equations of motion produces three equations for four unknowns Fx1 , Fx2 , Fz1 , Fz2 . 2Fx1 + 2Fx2 = ma 2Fz1 + 2Fz2 − mg = 0 −2Fz1 a1 + 2Fz2 a2 − 2 (Fx1 + Fx2 ) h = 0 (2.92) (2.93) (2.94) However, it is possible to eliminate (Fx1 + Fx2 ) between the first and third equations, and solve for the normal forces Fz1 , Fz2 . Fz1 = (Fz1 )st + (Fz1 )dyn = Fz2 1 a2 1 ha mg − mg 2 l 2 lg (2.95) = (Fz2 )st + (Fz2 )dyn = 1 a1 1 ha mg + mg 2 l 2 lg (2.96) The static parts (Fz1 )st = (Fz2 )st = 1 a2 mg 2 l 1 a1 mg 2 l (2.97) (2.98) are weight distribution for a stationary car and depend on the horizontal position of the mass center. However, the dynamic parts (Fz1 )dyn (Fz2 )dyn 1 ha = − mg 2 lg 1 ha = mg 2 lg (2.99) (2.100) indicate the weight distribution according to horizontal acceleration, and depend on the vertical position of the mass center. When accelerating a > 0, the normal forces under the front tires are less than the static load, and under the rear tires are more than the static load. Example 50 Front-wheel-drive accelerating on a level road. When the car is front-wheel-drive, Fx2 = 0. Equations (2.92) to (2.88) will provide the same vertical tireprint forces as (2.87) and (2.88). However, the required horizontal force to achieve the same acceleration, a, must be provided by solely the front wheels. Example 51 Rear-wheel drive accelerating on a level road. If a car is rear-wheel drive then, Fx1 = 0 and the required force to achieve the acceleration, a, must be provided only by the rear wheels. The vertical force under the wheels will still be the same as (2.87) and (2.88). Example 52 Maximum acceleration on a level road. The maximum acceleration of a car is proportional to the friction under its tires. We assume the friction coefficients at the front and rear tires are equal and all tires reach their maximum tractions at the same time. Fx1 Fx2 = ±μx Fz1 = ±μx Fz2 (2.101) (2.102) Newton’s equation (2.92) can now be written as ma = ±2μx (Fz1 + Fz2 ) . (2.103) Substituting Fz1 and Fz2 from (2.93) and (2.94) results in a = ±μx g. (2.104) Therefore, the maximum acceleration and deceleration depend directly on the friction coefficient. Example 53 Maximum acceleration for a single-axle drive car. The maximum acceleration arwd for a rear-wheel-drive car is achieved when we substitute Fx1 = 0, Fx2 = μx Fz2 in Equation (2.92) and use Equation (2.88) µ ¶ a1 h arwd μx mg (2.105) + = marwd l l g and therefore, arwd g = = a1 μx l − hμx μx a1 . h l 1 − μx l (2.106) The front wheels can leave the ground when Fz1 = 0. Substituting Fz1 = 0 in Equation (2.88) provides the maximum acceleration at which the front wheels are still on the road. arwd a2 ≤ (2.107) g h a/g /g a rwd afw d /g a1/l FIGURE 2.8. Effect of mass center position on the maximum achievable acceleration of a front- and a rear-wheel drive car. Therefore, the maximum attainable acceleration would be the less value of Equation (2.106) or (2.107). Similarly, the maximum acceleration af wd for a front-wheel drive car is achieved when we substitute Fx2 = 0, Fx1 = μx Fz1 in Equation (2.92) and use Equation (2.87). af wd g = = a2 μx l + hμx ³ μx a1 ´ 1− h l 1 + μx l (2.108) To see the effect of changing the position of mass center on the maximum achievable acceleration, we plot Figure 2.8 for a sample car with μx = 1 h = 0.56 m l = 2.6 m. (2.109) Passenger cars are usually in the range 0.4 < (a1 /g) < 0.6, with (a1 /g) → 0.4 for front-wheel-drive cars, and (a1 /g) → 0.6 for rear-wheel-drive cars. In this range, (arwd /g) > (af wd /g) and therefore rear-wheel-drive cars can reach higher forward acceleration than front-wheel-drive cars. It is an important applied fact, especially for race cars. The maximum acceleration may also be limited by the tilting condition aM a2 ≤ . g h (2.110) Example 54 Minimum time for 0 − 100 km/ h on a level road. Consider a car with the following characteristics: length width height wheel base f ront track rear track net weight h μx a1 = = = = = = = = = = 4245 mm 1795 mm 1285 mm 2272 mm 1411 mm 1504 mm 1500 kg 220 mm 1 a2 (2.111) Assume the car is rear-wheel-drive and its engine can provide the maximum traction supported by friction. Equation (2.88) determines the load on the rear wheels and therefore, the forward equation of motion is 2Fx2 = 2μx Fz2 a1 h1 = μx mg + μx mg a l lg = m a. (2.112) Rearrangement provides the following differential equation to calculate velocity and displacement: a1 l a = ẍ = h1 1 − μx g lg a1 = gμx l − hμx μx g Taking an integral Z 27.78 dv = 0 Z (2.113) t a dt (2.114) 0 between v = 0 and v = 100 km/ h ≈ 27.78 m/ s shows that the minimum time for 0 − 100 km/ h on a level road is t= 27.78 ≈ 5.11 s a1 gμx l − hμx (2.115) If the same car was front-wheel-drive, then the traction force would be 2Fx1 = 2μx Fz1 a2 h1 = μx mg − μx mg a l lg = m a. (2.116) and the equation of motion would reduce to a2 l a = ẍ = h1 1 + μx g lg a2 = gμx . l + hμx μx g (2.117) The minimum time for 0 − 100 km/ h on a level road for this front-wheeldrive car is 27.78 ≈ 6. 21 s. (2.118) t= a2 gμx l + hμx Now consider the same car to be four-wheel-drive. Then, the traction force is 2Fx1 + 2Fx2 = 2μx (Fz1 + Fz2 ) g = m (a1 + a2 ) l = m a. (2.119) and the minimum time for 0−100 km/ h on a level road for this four-wheeldrive car can theoretically be reduced to t= 27.78 ≈ 2.83 s. g (2.120) 2.4 Accelerating Car on an Inclined Road When a car is accelerating on an inclined pavement with angle φ as shown in Figure 2.9, the normal force under each of the front and rear wheels, Fz1 , Fz2 , would be: µ 1 a2 Fz1 = mg cos φ − 2 l µ 1 a1 Fz2 = mg cos φ + 2 l l = a1 + a2 ¶ h sin φ − l ¶ h sin φ + l 1 h ma 2 l 1 h ma 2 l (2.121) (2.122) a1 z x h a2 C a 2F x 1 2F z1 mg φ 2 F x2 2F z2 FIGURE 2.9. An accelerating car on inclined pavement. The dynamic parts, ± 12 mg hl ag , depend on acceleration a and height h of mass center C and remain unchanged, while the static parts are influenced by the slope angle φ and height h of mass center. Proof. The Newton’s equation in x-direction and two static equilibrium equations, must be examined to find the equation of motion and ground reaction forces. X Fx = ma (2.123) X Fz = 0 (2.124) X My = 0. (2.125) Expanding these equations produces three equations for four unknowns Fx1 , Fx2 , Fz1 , Fz2 . 2Fx1 + 2Fx2 − mg sin φ = ma 2Fz1 + 2Fz2 − mg cos φ = 0 2Fz1 a1 − 2Fz2 a2 + 2 (Fx1 + Fx2 ) h = 0 (2.126) (2.127) (2.128) It is possible to eliminate (Fx1 + Fx2 ) between the first and third equations, and solve for the normal forces Fz1 , Fz2 . Fz1 = (Fz1 )st + (Fz1 )dyn µ ¶ a2 1 h 1 h = mg cos φ − sin φ − ma 2 l l 2 l (2.129) Fz2 = (Fz2 )st + (Fz2 )dyn µ ¶ 1 h 1 h a1 = mg cos φ + sin φ + ma 2 l l 2 l (2.130) Example 55 Front-wheel-drive car, accelerating on inclined road. For a front-wheel-drive car, we may substitute Fx1 = 0 in Equations (2.126) and (2.128) to have the governing equations. However, it does not affect the ground reaction forces under the tires (2.129 and 2.130) as long as the car is driven under its limit conditions. Example 56 Rear-wheel-drive car, accelerating on inclined road. Substituting Fx2 = 0 in Equations (2.126) and (2.128) and solving for the normal reaction forces under each tire provides the same results as (2.129) and (2.130). Hence, the normal forces applied on the tires do not sense if the car is front-, rear-, or all-wheel drive. As long as we drive in a straight path at low acceleration, the drive wheels can be the front or the rear ones. However, the advantages and disadvantages of front-, rear-, or all-wheel drive cars appear in maneuvering, slippery roads, or when the maximum acceleration is required. Example 57 Maximum acceleration on an inclined road. The maximum acceleration depends on the friction under the tires. Let’s assume the friction coefficients at the front and rear tires are equal. Then, the front and rear traction forces are Fx1 Fx2 ≤ μx Fz1 ≤ μx Fz2 . (2.131) (2.132) If we assume the front and rear wheels reach their traction limits at the same time, then Fx1 Fx2 = ±μx Fz1 = ±μx Fz2 (2.133) (2.134) and we may rewrite Newton’s equation (2.123) as maM = ±2μx (Fz1 + Fz2 ) − mg sin φ (2.135) where, aM is the maximum achievable acceleration. Now substituting Fz1 and Fz2 from (2.129) and (2.130) results in aM (2.136) = ±μx cos φ − sin φ. g Accelerating on an uphill road (a > 0, φ > 0) and braking on a downhill road (a < 0, φ < 0) are the extreme cases in which the car can stall. In these cases, the car can move as long as μx ≥ |tan φ| . (2.137) Example 58 Limits of acceleration and inclination angle. Assuming Fz1 > 0 and Fz2 > 0, we can write Equations (2.121) and (2.122) as a a2 ≤ cos φ − sin φ g h a1 a ≥ − cos φ − sin φ. g h (2.138) (2.139) Hence, the maximum achievable acceleration (a > 0) is limited by a2 , h, φ; while the maximum deceleration (a < 0) is limited by a1 , h, φ. These two equations can be combined to result in − a1 a a2 cos φ ≤ + sin φ ≤ cos φ. h g h (2.140) If a → 0, then the limits of the inclination angle would be − a2 a1 ≤ tan φ ≤ . h h (2.141) This is the maximum and minimum road inclination angles that the car can stay on without tilting and falling. Example 59 Maximum deceleration for a single-axle-brake car. We can find the maximum braking deceleration af wb of a front-wheelbrake car on a horizontal road by substituting φ = 0, Fx2 = 0, Fx1 = −μx Fz1 in Equation (2.126) and using Equation (2.121) µ ¶ a2 h arwb (2.142) −μx mg − = maf wb l l g therefore, ³ a1 ´ 1− . (2.143) h l 1 − μx l Similarly, the maximum braking deceleration arwb of a front-wheel-brake car can be achieved when we substitute Fx2 = 0, Fx1 = μx Fz1 . af wb =− g μx arwb =− g a1 h l 1 + μx l μx (2.144) The effect of changing the position of the mass center on the maximum achievable braking deceleration is shown in Figure 2.10 for a sample car with μx = 1 h = 0.56 m l = 2.6 m. (2.145) a/g arw d /g /g a fwd a1/l FIGURE 2.10. Effect of mass center position on the maximum achievable decceleration of a front-wheel and a rear-wheel-drive car. Passenger cars are usually in the range 0.4 < (a1 /l) < 0.6. In this range, (af wb /g) < (arwb /g) and therefore, front-wheel-brake cars can reach better forward deceleration than rear-wheel-brake cars. Hence, front brakes are much more important than the rear brakes. Example 60 F A car with a trailer. Figure 2.11 depicts a car moving on an inclined road and pulling a trailer. To analyze the car-trailer motion, we need to separate the car and trailer to see the forces at the hinge, as shown in Figure 2.12. We assume the mass center of the trailer Ct is at distance b3 in front of the only axle of the trailer. If Ct is behind the trailer axle, then b3 should be negative in the following equations. For an ideal hinge between a car and a trailer moving in a straight path, there must be a horizontal force Fxt and a vertical force Fzt . Writing the Newton’s equation in x-direction and two static equilibrium equations for both the trailer and the vehicle X Fx = mt a (2.146) X Fz = 0 (2.147) X My = 0 (2.148) we find the following set of equations: Fxt − mt g sin φ = mt a 2Fz3 − Fzt − mt g cos φ = 0 2Fz3 b3 − Fzt b2 − Fxt (h2 − h1 ) = 0 (2.149) (2.150) (2.151) a1 z x a2 h b1 b2 C b3 φ Ct 2F x 2 a 2F x 1 2F z1 mg 2F z2 mt g 2F z3 FIGURE 2.11. A car moving on an inclined road and pulling a trailer. 2Fx1 + 2Fx2 − Fxt − mg sin φ = ma 2Fz1 + 2Fz2 − Fzt − mg cos φ = 0 2Fz1 a1 − 2Fz2 a2 + 2 (Fx1 + Fx2 ) h −Fxt (h − h1 ) + Fzt (b1 + a2 ) = 0 (2.152) (2.153) (2.154) If the value of traction forces Fx1 and Fx2 are given, then these are six equations for six unknowns: a, Fxt , Fzt , Fz1 , Fz2 , Fz3 . Solving these equations provide the following solutions: a = Fxt = Fzt = Fz1 2 (Fx1 + Fx2 ) − g sin φ m + mt 2mt (Fx1 + Fx2 ) m + mt h1 − h2 2mt b3 (Fx1 + Fx2 ) + mt g cos φ b2 − b3 m + mt b2 − b3 ¶ µ b3 2a2 − b1 a2 = mt + m g cos φ 2l b2 − b3 b3 ∙ ¸ 2a2 − b1 Fx1 + Fx2 + (h1 − h2 ) mt − h1 mt − hm b2 − b3 l (m + mt ) (2.155) (2.156) (2.157) (2.158) z a1 x a2 h b1 a C φ F xt h1 F zt 2F x 2 2F z1 1 2F x mg 2F z2 b2 F zt b3 F xt Ct mt g h1 h2 φ 2F z3 FIGURE 2.12. Free-body-diagram of a car and the trailer when moving on an uphill road. Fz2 Fz3 l ¶ µ b3 a1 − a2 + b1 a1 = mt + m g cos φ 2l b2 − b3 b3 ∙ ¸ a1 − a2 + b1 Fx1 + Fx2 + (h1 − h2 ) mt + h1 mt + hm b2 − b3 l (m + mt ) 1 b2 h1 − h2 mt = mt g cos φ + (Fx1 + Fx2 ) 2 b2 − b3 b2 − b3 m + mt = a1 + a2 . (2.159) (2.160) (2.161) However, if the value of acceleration a is known, then unknowns are: Fx1 + Fx2 , Fxt , Fzt , Fz1 , Fz2 , Fz3 . Fx1 + Fx2 Fxt Fzt 1 (m + mt ) (a + g sin φ) 2 = mt (a + g sin φ) h1 − h2 b3 = mt (a + g sin φ) + mt g cos φ b2 − b3 b2 − b3 = (2.162) (2.163) (2.164) Fz1 = Fz2 = Fz3 = l = ¶ a2 2a2 − b1 mt + m g cos φ (2.165) b2 − b3 b3 ¸ ∙ 1 2a2 − b1 + (h1 − h2 ) mt − h1 mt − hm (a + g sin φ) 2l b2 − b3 ¶ µ b3 a1 − a2 + b1 a1 mt + m g cos φ (2.166) 2l b2 − b3 b3 ¸ ∙ 1 a1 − a2 + b1 + (h1 − h2 ) mt + h1 mt + hm (a + g sin φ) 2l b2 − b3 1 mt (b2 g cos φ + (h1 − h2 ) (a + g sin φ)) (2.167) 2 b2 − b3 a1 + a2 . b3 2l µ Example 61 F Maximum inclination angle for a car with a trailer. For a car and trailer as shown in Figure 2.11, the maximum inclination angle φM is the angle at which the car cannot accelerate the vehicle. Substituting a = 0 and φ = φM in Equation (2.155) shows that sin φM = 2 (Fx1 + Fx2 ) . (m + mt ) g (2.168) The value of maximum inclination angle φM increases by decreasing the total weight of the vehicle and trailer (m + mt ) g or increasing the traction force Fx1 + Fx2 . The traction force is limited by the maximum torque on the drive wheel and the friction under the drive tire. Let’s assume the vehicle is four-wheeldrive and friction coefficients at the front and rear tires are equal. Then, the front and rear traction forces are Fx1 Fx2 ≤ μx Fz1 ≤ μx Fz2 . (2.169) (2.170) If we assume the front and rear wheels reach their traction limits at the same time, then Fx1 Fx2 = μx Fz1 = μx Fz2 (2.171) (2.172) and we may rewrite the Equation (2.168) as sin φM = 2μx (Fz1 + Fz2 ) . (m + mt ) g (2.173) Now substituting Fz1 and Fz2 from (2.158) and (2.159) results in (mb3 − mb2 − mt b3 ) μx cos φM + (b2 − b3 ) (m + mt ) sin φM mt (h1 − h2 ) = 2μx (Fx1 + Fx2 ) . (2.174) m + mt If we arrange Equation (2.174) as A cos φM + B sin φM = C then φM r C C2 = atan2( √ ,± 1 − 2 ) − atan2(A, B) A + B2 A2 + B 2 (2.175) (2.176) and p C , ± A2 + B 2 − C 2 ) − atan2(A, B) φM = atan2( √ 2 2 A +B where (2.177) (2.178) A = (mb3 − mb2 − mt b3 ) μx B = (b2 − b3 ) (m + mt ) (2.179) mt (h1 − h2 ) (Fx1 + Fx2 ) . (2.180) C = 2μx m + mt For a rear-wheel-drive car pulling a trailer with the following characteristics: l w h a1 h1 b1 b2 b3 h2 m mt μx φ a = = = = = = = = = = = = = = 2272 mm 1457 mm 230 mm a2 310 mm 680 mm 610 mm 120 mm 560 mm 1500 kg 150 kg 1 10 deg 1 m/ s2 Fz1 Fz2 Fz3 Fzt Fxt Fx2 = = = = = = 3441.78 N 3877.93 N 798.57 N 147.99 N 405.52 N 2230.37 N. (2.181) we find (2.182) To check if the required traction force Fx2 is applicable, we should compare it to the maximum available friction force μFz2 and it must be Fx2 ≤ μFz2 . (2.183) a cos θ + b sin θ = c. (2.184) Example 62 F Solution of equation a cos θ + b sin θ = c. The first type of trigonometric equation is It can be solved by introducing two new variables r and η such that a = r sin η b = r cos η (2.185) (2.186) and therefore, r η p = a2 + b2 = atan2(a, b). Substituting the new variables show that c sin(η + θ) = rr cos(η + θ) = ± 1 − (2.187) (2.188) (2.189) c2 . r2 (2.190) Hence, the solutions of the problem are r c2 c (2.191) θ = atan2( , ± 1 − 2 ) − atan2(a, b) r r and c p (2.192) θ = atan2( , ± r2 − c2 ) − atan2(a, b). r Therefore, the equation a cos θ + b sin θ = c has two solutions if r2 = 2 a + b2 > c2 , one solution if r2 = c2 , and no solution if r2 < c2 . y Example 63 F The function tan−1 2 x = atan2(y, x). There are many situations in kinematics calculation in which we need to find an angle based on the sin and cos functions of an angle. However, tan−1 cannot show the effect of the individual sign for the numerator and denominator. It always represents an angle in the first or fourth quadrant. To overcome this problem and determine the angle in the correct quadrant, the atan2 function is introduced as below. ⎧ y ⎪ tan−1 if y > 0 ⎪ ⎪ ⎪ x ⎪ ⎨ y tan−1 + π sign y if y < 0 atan2(y, x) = (2.193) x ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ π sign x if y = 0 2 In this text, whether it has been mentioned or not, wherever tan−1 used, it must be calculated based on atan2(y, x). y x is Example 64 Zero vertical force at the hinge. We can make the vertical force at the hinge equal to zero by examining Equation (2.157) for the hinge vertical force Fzt . Fzt = h1 − h2 2mt b3 (Fx1 + Fx2 ) + mt g cos φ b2 − b3 m + mt b2 − b3 (2.194) To make Fzt = 0, it is enough to adjust the position of trailer mass center Ct exactly on top of the trailer axle and at the same height as the hinge. In these conditions we have h1 b3 = h2 = 0 (2.195) (2.196) that makes Fzt = 0. (2.197) However, to increase safety, the load should be distributed evenly throughout the trailer. Heavy items should be loaded as low as possible, mainly over the axle. Bulkier and lighter items should be distributed to give a little positive b3 . Such a trailer is called nose weight at the towing coupling. 2.5 Parked Car on a Banked Road Figure 2.13 depicts the effect of a bank angle φ on the load distribution of a vehicle. A bank causes the load on the lower tires to increase, and the load on the upper tires to decrease. The tire reaction forces are: 1 mg (b2 cos φ − h sin φ) 2 w 1 mg Fz2 = (b1 cos φ + h sin φ) 2 w w = b1 + b2 Fz1 = Proof. Starting with equilibrium equations X Fy = 0 X Fz = 0 X Mx = 0. (2.198) (2.199) (2.200) (2.201) (2.202) (2.203) z y h C 2F y 1 mg φ 2 F y2 2F z1 b1 b2 2F z2 FIGURE 2.13. Normal force under the uphill and downhill tires of a vehicle, parked on banked road. we can write 2Fy1 + 2Fy2 − mg sin φ = 0 2Fz1 + 2Fz2 − mg cos φ = 0 2Fz1 b1 − 2Fz2 b2 + 2 (Fy1 + Fy2 ) h = 0. (2.204) (2.205) (2.206) We assumed the force under the lower tires, front and rear, are equal, and also the forces under the upper tires, front and rear are equal. To calculate the reaction forces under each tire, we may assume the overall lateral force Fy1 +Fy2 as an unknown. The solution of these equations provide the lateral and reaction forces under the upper and lower tires. 1 b2 mg cos φ − 2 w 1 b1 = mg cos φ + 2 w 1 = mg sin φ 2 Fz1 = Fz2 Fy1 + Fy2 1 h mg sin φ 2 w 1 h mg sin φ 2 w (2.207) (2.208) (2.209) At the ultimate angle φ = φM , all wheels will begin to slide simultaneously and therefore, Fy1 Fy2 = μy1 Fz1 = μy2 Fz2 . (2.210) (2.211) The equilibrium equations show that 2μy1 Fz1 + 2μy2 Fz2 − mg sin φ = 0 Assuming (2.212) 2Fz1 + 2Fz2 − mg cos φ = 0 ¡ ¢ 2Fz1 b1 − 2Fz2 b2 + 2 μy1 Fz1 + μy2 Fz2 h = 0. (2.213) μy1 = μy2 = μy (2.215) (2.214) will provide 1 1 b2 h mg cos φM − mg sin φM 2 w 2 w 1 1 b1 h = mg cos φM + mg sin φM 2 w 2 w = μy . Fz1 = (2.216) Fz2 (2.217) tan φM (2.218) These calculations are correct as long as tan φM ≤ μy ≤ b2 h b2 . h (2.219) (2.220) If the lateral friction μy is higher than b2 /h then the car will roll downhill. To increase the capability of a car moving on a banked road, the car should be as wide as possible with a mass center as low as possible. Example 65 Tire forces of a parked car in a banked road. A car having m h w b1 = = = = 980 kg 0.6 m 1.52 m b2 (2.221) is parked on a banked road with φ = 4 deg. The forces under the lower and upper tires of the car are: Fy1 Fz1 Fz2 + Fy2 = 2265.2 N = 2529.9 N = 335.3 N (2.222) The ratio of the uphill force Fz1 to downhill force Fz2 depends on only the mass center position. b2 cos φ − h sin φ Fz1 = Fz2 b1 cos φ + h sin φ (2.223) h=0.6 m w=1.52 m b1=b2 Rolling down angle Fz1 Fz2 φ [rad] φ[deg] FIGURE 2.14. Illustration of the force ratio Fz1 /Fz2 as a function of road bank angle φ. Assuming a symmetric car with b1 = b2 = w/2 simplifies the equation to w cos φ − 2h sin φ Fz1 = . Fz2 w cos φ + 2h sin φ (2.224) Figure 2.14 illustrates the behavior of force ratio Fz1 /Fz2 as a function of φ for h = 0.6 m and w = 1.52 m. The rolling down angle φM = tan−1 (b2 /h) = 51.71 deg indicates the bank angle at which the force under the uphill wheels become zero and the car rolls down. The negative part of the curve indicates the required force to keep the car on the road, which is not applicable in real situations. 2.6 F Optimal Drive and Brake Force Distribution A certain acceleration a can be achieved by adjusting and controlling the longitudinal forces Fx1 and Fx2 . The optimal longitudinal forces under the front and rear tires to achieve the maximum acceleration are Fx1 mg µ ¶2 1 a2 a 1h a + = − 2l g 2 l g 1 2 h 1 a2 = − μx + μx 2 l 2 l (2.225) Fx2 mg = = µ ¶2 1 a1 a 1h a + 2l g 2 l g 1 2 h 1 a1 μ + μx . 2 xl 2 l (2.226) Proof. The longitudinal equation of motion for a car on a horizontal road is 2Fx1 + 2Fx2 = ma (2.227) and the maximum traction forces under each tire is a function of normal force and the friction coefficient. Fx1 Fx2 ≤ ±μx Fz1 ≤ ±μx Fz2 (2.228) (2.229) However, the normal forces are a function of the car’s acceleration and geometry. Fz1 = Fz2 = 1 a2 1 ha mg − mg 2 l 2 lg a1 1 ha 1 mg + mg 2 l 2 lg (2.230) (2.231) We may generalize the equations by making them dimensionless. Under the best conditions, we should adjust the traction forces to their maximum µ ¶ a2 Fx1 1 ha = μx − (2.232) mg 2 l lg µ ¶ a1 1 ha Fx2 = μx + (2.233) mg 2 l lg and therefore, the longitudinal equation of motion (2.227) becomes a = μx . g (2.234) Substituting this result back into Equations (2.232) and (2.233) shows that Fx1 mg Fx2 mg µ ¶2 1 a2 a 1h a + = − 2l g 2 l g µ ¶2 1 a1 a 1h a + = . 2l g 2 l g (2.235) (2.236) Depending on the geometry of the car (h, a1 , a2 ), and the acceleration a > 0, these two equations determine how much the front and rear driving forces must be. The same equations are applied for deceleration a < 0, to Fx2 / mg Fx / mg − a1 h Fx1 / mg a/g Fx2 / mg a2 h Fx1 / mg FIGURE 2.15. Optimal driving and braking forces for a sample car. determine the value of optimal front and rear braking forces. Figure 2.15 represents a graphical illustration of the optimal driving and braking forces for a sample car using the following data: μx h l a1 l = 1 0.56 = = 0.21538 2.6 a2 1 = = . l 2 (2.237) When accelerating a > 0, the optimal driving force on the rear tire grows rapidly while the optimal driving force on the front tire drops after a maximum. The value (a/g) = (a2 /h) is the maximum possible acceleration at which the front tires lose their contact with the ground. The acceleration at which front (or rear) tires lose their ground contact is called tilting acceleration. The opposite phenomenon happens when decelerating. For a < 0, the optimal front brake force increases rapidly and the rear brake force goes to zero after a minimum. The deceleration (a/g) = − (a1 /h) is the maximum possible deceleration at which the rear tires lose their ground contact. The graphical representation of the optimal driving and braking forces can be shown better by plotting Fx1 / (mg) versus Fx2 / (mg) using (a/g) as a parameter. Fx1 = Fx1 Fx2 = a a2 − h g a Fx2 a1 + h g a2 − μx h a1 + μx h (2.238) (2.239) Fx1 / mg Driving Fx2 / mg Braking a2 h a1 h FIGURE 2.16. Optimal traction and braking force distribution between the front and rear wheels. Such a plot is shown in Figure 2.16. This is a design curve describing the relationship between forces under the front and rear wheels to achieve the maximum acceleration or deceleration. Adjusting the optimal force distribution is not an automatic procedure and needs a force distributor control system to measure and adjust the forces. Example 66 F Slope at zero. The initial optimal traction force distribution is the slope of the optimal curve (Fx1 / (mg) , Fx2 / (mg)) at zero. µ ¶2 1h a 1 a2 a Fx1 − + d 2l g 2 l g mg = lim µ ¶2 Fx2 a→0 1 h 1 a a1 a d + mg 2l g 2 l g a2 (2.240) = a1 Therefore, the initial traction force distribution depends on only the position of mass center C. Example 67 F Brake balance and ABS. When braking, a car is stable if the rear wheels do not lock. Thus, the rear brake forces must be less than the maximum possible braking force at all time. This means the brake force distribution should always be in the shaded area of Figure 2.17, and below the optimal curve. This restricts the Fx1 / mg Braking Fx2 / mg a1 h FIGURE 2.17. Optimal braking force distribution between the front and rear wheels, along with a thre-line under estimation. achievable deceleration, especially at low friction values, but increases the stability of the car. Whenever it is easier for a force distributor to follow a line, the optimal brake curve is underestimated using two or three lines, and a control system adjusts the force ratio Fx1 /Fx2 . A sample of three-line approximation is shown in Figure 2.17. Distribution of the brake force between the front and rear wheels is called brake balance. Brake balance varies with deceleration. The higher the stop, the more load will transfer to the front wheels and the more braking effort they can support. Meanwhile the rear wheels are unloaded and they must have less braking force. Example 68 F Best race car. Racecars always work at the maximum achievable acceleration to finish their race in minimum time. They are usually designed with rear-wheeldrive and all-wheel-brake. However, if an all-wheel-drive race car is reasonable to build, then a force distributor, to follow the curve shown in Figure 2.18, is what it needs to race better. Example 69 F Effect of C location on braking. Load is transferred from the rear wheels to the front when the brakes are applied. The higher the C, the more load transfer. So, to improve braking, the mass center C should be as low as possible and as back as possible. This is not feasible for every vehicle, especially for forward-wheel drive street cars. However, this fact should be taken into account when a car is Fx1 / mg Driving 0.2 a2 h 0.1 0 0.2 0.4 0.6 0.8 1.0 1.2 Fx2 / mg FIGURE 2.18. Optimal traction force distribution between the front and rear wheels. Fx2 Fx1 Fy1 v Fx2 Fx1 Fy1 FIGURE 2.19. 180 deg sliding rotation of a rear-wheel-locked car. being designed for better braking performance. Example 70 F Front and rear wheel locking. The optimal brake force distribution is according to Equation (2.239) for an ideal Fx1 /Fx2 ratio. However, if the brake force distribution is not ideal, then either the front or the rear wheels will lock up first. Locking the rear wheels makes the vehicle unstable, and it loses directional stability. When the rear wheels lock, they slide on the road and they lose their capacity to support lateral force. The resultant shear force at the tireprint of the rear wheels reduces to a dynamic friction force in the opposite direction of the sliding. A slight lateral motion of the rear wheels, by any disturbance, develops a yaw motion because of unbalanced lateral forces on the front and rear wheels. The yaw moment turns the vehicle about the z-axis until the rear end leads the front end and the vehicle turns 180 deg. Figure 2.19 illustrates a 180 deg sliding rotation of a rear-wheel-locked car. The lock-up of the front tires does not cause a directional instability, although the car would not be steerable and the driver would lose control. 2.7 F Vehicles With More Than Two Axles If a vehicle has more than two axles, such as the three-axle car shown in Figure 2.20, then the vehicle will be statically indeterminate and the normal forces under the tires cannot be determined by static equilibrium equations. We need to consider the suspensions’ deflection to determine their applied forces. The n normal forces Fzi under the tires can be calculated using the following n algebraic equations. 2 n X Fzi − mg cos φ = 0 (2.241) Fzi xi + h (a + mg sin φ) = 0 (2.242) i=1 2 n X i=1 xi − x1 Fzi − ki xn − x1 µ Fz Fzn − 1 kn k1 ¶ − Fz1 = 0 for i = 2, 3, · · · , n − 1 k1 (2.243) where Fxi and Fzi are the longitudinal and normal forces under the tires attached to the axle number i, and xi is the distance of mass center C from the axle number i. The distance xi is positive for axles in front of C, and is negative for the axles in back of C. The parameter ki is the vertical stiffness of the suspension at axle i. Proof. For a multiple-axle vehicle, the following equations X Fx = ma (2.244) X Fz = 0 (2.245) X My = 0 (2.246) provide the same sort of equations as (2.126)-(2.128). However, if the total number of axles are n, then the individual forces can be substituted by a summation. n X 2 Fxi − mg sin φ = ma (2.247) i=1 2 n X i=1 Fzi − mg cos φ = 0 (2.248) z a3 a1 x a2 h C a 2F x1 2F x3 φ 2F x2 2F z3 2F z1 mg 2F z2 FIGURE 2.20. A three-axle car moving on an inclined road. 2 n X Fzi xi + 2h i=1 n X Fxi = 0 (2.249) i=1 Pn The overall forward force Fx = 2 i=1 Fxi can be eliminated between Equations (2.247) and (2.249) to make Equation (2.242). Then, there remain two equations (2.241) and (2.242) for n unknowns Fzi , i = 1, 2, · · · , n. Hence, we need n − 2 extra equations to be able to find the wheel loads. The extra equations come from the compatibility among the suspensions’ deflection. We ignore the tires’ compliance, and use z to indicate the static vertical displacement of the car at C. Then, if zi is the suspension deflection at the center of axle i, and ki is the vertical stiffness of the suspension at axle i, the deflections are Fz zi = i . (2.250) ki For a flat road, and a rigid vehicle, we must have zi − z1 zn − z1 = xi − x1 xn − x1 for i = 2, 3, · · · , n − 1 (2.251) which, after substituting with (2.250), reduces to Equation (2.243). The n − 2 equations (2.251) along with the two equations (2.241) and (2.242) are enough to calculate the normal load under each tire. The resultant set of equations is linear and may be arranged in a matrix form [A] [X] = [B] (2.252) where [X] = ⎡ 2 ⎢ 2x 1 ⎢ ⎢ xn − x2 ⎢ ⎢ k1 l ⎢ ⎢ ··· [A] = ⎢ x − x n i ⎢ ⎢ k1 l ⎢ ⎢ ··· ⎢ ⎣ xn − xn−1 k1 l l = x1 − xn [B] = £ £ Fz1 Fz2 Fz3 ··· Fzn 2 2x2 1 k2 ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· 1 ki ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· 1 kn−1 ¤T (2.253) 2 2xn x2 − x1 kn l ··· xi − x1 kn l ··· xn−1 − x1 kn l ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ (2.254) ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (2.255) mg cos φ −h (a + mg sin φ) 0 · · · 0 ¤T . (2.256) Example 71 F Wheel reactions for a three-axle car. Figure 2.20 illustrates a three-axle car moving on an inclined road. We start counting the axles of a multiple-axle vehicle from the front axle as axle-1, and move sequentially to the back as shown in the figure. The set of equations for the three-axle car, as seen in Figure 2.20, is 2Fx1 + 2Fx2 + 2Fx3 − mg sin φ 2Fz1 + 2Fz2 + 2Fz3 − mg cos φ 2Fz1 x1 + 2Fz2 x2 + 2Fz3 x3 + 2h (Fx1 + Fx2 + Fx3 ) µ ¶ µ ¶ Fz1 Fz1 Fz2 1 Fz3 1 − − − x2 − x1 k2 k1 x3 − x1 k3 k1 = ma = 0 = 0 (2.257) (2.258) (2.259) = 0 (2.260) 2Fz1 + 2Fz2 + 2Fz3 − mg cos φ = 0 2Fz1 x1 + 2Fz2 x2 + 2Fz3 x3 + hm (a + g sin φ) = 0 (x2 k2 k3 − x3 k2 k3 ) Fz1 + (x1 k1 k2 − x2 k1 k2 ) Fz3 − (x1 k1 k3 − x3 k1 k3 ) Fz2 = 0. (2.261) (2.262) which can be simplified to (2.263) The set of equations for wheel loads is linear and may be rearranged in a matrix form [A] [X] = [B] (2.264) where ⎡ ⎤ 2 2 2 ⎦ (2.265) 2x1 2x2 2x3 [A] = ⎣ k2 k3 (x2 − x3 ) k1 k3 (x3 − x1 ) k1 k2 (x1 − x2 ) ⎡ ⎤ Fz1 [X] = ⎣ Fz2 ⎦ (2.266) Fz3 ⎡ ⎤ mg cos φ [B] = ⎣ −hm (a + g sin φ) ⎦ . (2.267) 0 The unknown vector may be found using matrix inversion −1 [X] = [A] [B] . (2.268) The solution of the equations are 1 Fz k1 m 1 1 Fz k2 m 2 1 Fz k2 m 3 = = = Z1 Z0 Z2 Z0 Z3 Z0 (2.269) (2.270) (2.271) where, Z0 = −4k1 k2 (x1 − x2 )2 −4k2 k3 (x2 − x3 )2 −4k1 k3 (x3 − x1 )2 Z1 Z2 Z3 = g (x2 k2 − x1 k3 − x1 k2 + x3 k3 ) h sin φ +a (x2 k2 − x1 k3 − x1 k2 + x3 k3 ) h ¡ ¢ +g k2 x22 − x1 k2 x2 + k3 x23 − x1 k3 x3 cos φ = g (x1 k1 − x2 k1 − x2 k3 + x3 k3 ) h sin φ +a (x1 k1 − x2 k1 − x2 k3 + x3 k3 ) h ¡ ¢ +g k1 x21 − x2 k1 x1 + k3 x23 − x2 k3 x3 cos φ = g (x1 k1 + x2 k2 − x3 k1 − x3 k2 ) h sin φ +a (x1 k1 + x2 k2 − x3 k1 − x3 k2 ) h ¡ ¢ +g k1 x21 − x3 k1 x1 + k2 x22 − x3 k2 x2 cos φ x1 x2 x3 = a1 = −a2 = −a3 . (2.272) (2.273) (2.274) (2.275) (2.276) (2.277) (2.278) a1 z x v a2 θ C φ mg a 2F x1 2F z1 RH z2 2F x2 2F FIGURE 2.21. A cresting vehicle at a point where the hill has a radius of curvature Rh . 2.8 F Vehicles on a Crest and Dip When a road has an outward or inward curvature, we call the road is a crest or a dip. The curvature can decrease or increase the normal forces under the wheels. 2.8.1 F Vehicles on a Crest Moving on the convex curve of a hill is called cresting. The normal force under the wheels of a cresting vehicle is less than the force on a flat inclined road with the same slope, because of the developed centrifugal force mv 2 /RH in the −z-direction. Figure 2.21 illustrates a cresting vehicle at the point on the hill with a radius of curvature RH . The traction and normal forces under its tires are approximately equal to Fx1 + Fx2 Fz1 1 m (a + g sin φ) 2 ∙µ ¶¸ 1 a2 h ≈ mg cos φ + sin φ 2 l l 2 1 h 1 v a2 − ma − m 2 l 2 RH l ≈ (2.279) (2.280) Fz2 l ∙µ ¶¸ 1 a1 h ≈ mg cos φ − sin φ 2 l l 2 1 h 1 v a1 + ma − m 2 l 2 RH l = a1 + a2 . (2.281) (2.282) Proof. For the cresting car shown in Figure 2.21, the normal and tangential directions are equivalent to the −z and x directions respectively. Hence, the governing equation of motion for the car is X Fx = ma (2.283) X v2 (2.284) − Fz = m RH X My = 0. (2.285) Expanding these equations produces the following equations: 2Fx1 cos θ + 2Fx2 cos θ − mg sin φ = ma −2Fz1 cos θ − 2Fz2 cos θ + mg cos φ = m (2.286) 2 v RH 2Fz1 a1 cos θ − 2Fz2 a2 cos θ + 2 (Fx1 + Fx2 ) h cos θ +2Fz1 a1 sin θ − 2Fz2 a2 sin θ − 2 (Fx1 + Fx2 ) h sin θ = 0. (2.287) (2.288) We may eliminate (Fx1 + Fx2 ) between the first and third equations, and solve for the total traction force Fx1 + Fx2 and wheel normal forces Fz1 , Fz2 . Fx1 + Fx2 Fz1 Fz2 ma + mg sin φ 2∙µ cos θ ¶¸ 1 a2 h (1 − sin 2θ) = mg cos φ + sin φ 2 l cos θ l cos θ cos 2θ h (1 − sin 2θ) 1 v 2 a2 1 − m − ma 2 l cos θ cos 2θ 2 RH l cos θ ∙µ ¶¸ 1 a1 h (1 − sin 2θ) = mg cos φ − sin φ 2 l cos θ l cos θ cos 2θ 1 h (1 − sin 2θ) 1 v 2 a1 + ma − m 2 l cos θ cos 2θ 2 RH l cos θ cos θ = (2.289) (2.290) (2.291) If the car’s wheel base is much smaller than the radius of curvature, l ¿ RH , then the slope angle θ is too small, and we may use the following trigonometric approximations. cos θ sin θ ≈ cos 2θ ≈ 1 ≈ sin 2θ ≈ 0 (2.292) (2.293) Substituting these approximations in Equations (2.289)-(2.291) produces the following approximate results: Fx1 + Fx2 Fz1 Fz2 1 m (a + g sin φ) 2 ∙µ ¶¸ 1 a2 h ≈ mg cos φ + sin φ 2 l l 2 h 1 v a2 1 − ma − m 2 l 2 RH l ∙µ ¶¸ a1 1 h ≈ mg cos φ − sin φ 2 l l 2 1 h 1 v a1 + ma − m 2 l 2 RH l ≈ (2.294) (2.295) (2.296) Example 72 F Wheel loads of a cresting car. Consider a car with the following specifications: l w m h a1 v a = = = = = = = 2272 mm 1457 mm 1500 kg 230 mm a2 15 m/ s 1 m/ s2 (2.297) which is cresting a hill at a point where the road has RH = 40 m φ = 30 deg θ = 2.5 deg . (2.298) The force information on the car is: Fx1 + Fx2 Fz1 Fz2 mg Fz1 + Fz2 v2 m RH = = = = = 4432.97 N 666.33 N 1488.75 N 14715 N 2155.08 N = 8437.5 N (2.299) If we simplifying the results by assuming small θ, the approximate values of the forces are Fx1 + Fx2 Fz1 Fz2 mg Fz1 + Fz2 v2 m RH = ≈ ≈ = ≈ 4428.75 N 628.18 N 1524.85 N 14715 N 2153.03 N (2.300) = 8437.5 N. Example 73 F Losing the road contact in a crest. When a car goes too fast, it can lose its road contact. Such a car is called a flying car. The condition to have a flying car is Fz1 = 0 and Fz2 = 0. Assuming a symmetric car a1 = a2 = l/2 with no acceleration, and using the approximate Equations (2.280) and (2.281) ∙µ 1 a2 mg cos φ + 2 l ∙µ 1 a1 mg cos φ − 2 l ¶¸ h sin φ − l ¶¸ h sin φ − l 1 v2 m 2 RH 1 v2 m 2 RH a2 l a1 l = 0 (2.301) = 0 (2.302) we can find the critical minimum speed vc to start flying. There are two critical speeds vc1 and vc2 for losing the contact of the front and rear wheels respectively. vc1 vc2 s µ ¶ h 1 = 2gRH sin φ + cos φ l 2 s µ ¶ h 1 = −2gRH sin φ − cos φ l 2 (2.303) (2.304) For any car, the critical speeds vc1 and vc2 are functions of the hill’s radius of curvature RH and the angular position on the hill, indicated by φ. The angle φ cannot be out of the tilting angles given by Equation (2.141). − a1 a2 ≤ tan φ ≤ h h (2.305) Figure 2.22 illustrates a cresting car over a circular hill, and Figure 2.23 depicts the critical speeds vc1 and vc2 at a different angle φ for −1.371 rad ≤ φ RH φ FIGURE 2.22. A cresting car over a circular hill. φ ≤ 1.371 rad. The specifications of the car and the hill are: l h a1 a RH = = = = = 2272 mm 230 mm a2 0 m/ s2 100 m. At the maximum uphill slope φ = 1.371 rad ≈ 78.5 deg, the front wheels can leave the ground at zero speed while the rear wheels are on the ground. When the car moves over the hill and reaches the maximum downhill slope φ = −1.371 rad ≈ −78.5 deg the rear wheels can leave the ground at zero speed while the front wheels are on the ground. As long as the car is moving uphill, the front wheels can leave the ground at a lower speed while going downhill the rear wheels leave the ground at a lower speed. Hence, at each slope angle φ the lower curve determines the critical speed vc . To have a general image of the critical speed, we may plot the lower values of vc as a function of φ using RH or h/l as a parameter. Figure 2.24 shows the effect of hill radius of curvature RH on the critical speed vc for a car with h/l = 0.10123 mm/ mm and Figure 2.25 shows the effect of a car’s high factor h/l on the critical speed vc for a circular hill with RH = 100 m. 2.8.2 F Vehicles on a Dip Moving on the concave curve of a hill is called dipping. The normal force under the wheels of a dipping vehicle is more than the force on a flat inclined road with the same slope, because of the developed centrifugal vc [m/s] 30 25 20 vc1 vc2 15 l=2.272 m h=0.23 m a1=a2 RH=40 m 10 5 1 60 0.5 40 0 20 0 -0.5 -20 -40 -1 φ [rad] -60 φ [deg] FIGURE 2.23. Critical speeds vc1 and vc2 at different angle φ for a specific car and hill. vc [m/s] h/l=0.10123 a1=a2 RH=1000 m 500 m 200 m 100 m 40 m 1 60 0.5 40 0 20 0 -0.5 -20 -40 -1 φ [rad] -60 φ [deg] FIGURE 2.24. Effect of hill radius of curvature Rh on the critical speed vc for a car. vc [m/s] RH=100 m a1=a2 h/l=0.1 0.2 0.3 0.4 0.5 1 60 0.5 40 0 20 0 -0.5 -20 -40 -1 φ [rad] -60 φ [deg] FIGURE 2.25. Effect of a car’s height factor h/l on the critical speed vc for a circular hill. force mv 2 /RH in the z-direction. Figure 2.26 illustrates a dipping vehicle at a point where the hill has a radius of curvature RH . The traction and normal forces under the tires of the vehicle are approximately equal to Fx1 + Fx2 Fz1 Fz2 l 1 m (a + g sin φ) 2 ∙µ ¶¸ 1 a2 h ≈ mg cos φ + sin φ 2 l l 2 1 h 1 v a2 − ma + m 2 l 2 RH l ∙µ ¶¸ a1 1 h ≈ mg cos φ − sin φ 2 l l 2 h 1 v a1 1 + ma + m 2 l 2 RH l = a1 + a2 . ≈ (2.306) (2.307) (2.308) (2.309) Proof. To develop the equations for the traction and normal forces under the tires of a dipping car, we follow the same procedure as a cresting car. The normal and tangential directions of a dipping car, shown in Figure 2.21, are equivalent to the z and x directions respectively. Hence, the governing z a1 a2 RH θ a C x v 2F x1 2F z1 φ mg 2Fx2 2F z2 FIGURE 2.26. A dipping vehicle at a point where the hill has a radius of curvature Rh . equations of motion for the car are X X X Fx = ma Fz = m (2.310) v2 RH (2.311) My = 0. (2.312) Expanding these equations produces the following equations: 2Fx1 cos θ + 2Fx2 cos θ − mg sin φ = ma −2Fz1 cos θ − 2Fz2 cos θ + mg cos φ = m 2Fz1 a1 cos θ − 2Fz2 a2 cos θ + 2 (Fx1 + Fx2 ) h cos θ +2Fz1 a1 sin θ − 2Fz2 a2 sin θ − 2 (Fx1 + Fx2 ) h sin θ = 0. (2.313) 2 v RH (2.314) (2.315) The total traction force (Fx1 + Fx2 ) may be eliminated between the first and third equations. Then, the resultant equations provide the following forces for the total traction force Fx1 + Fx2 and wheel normal forces Fz1 , Fz2 : Fx1 + Fx2 = ma + mg sin φ 2 cos θ (2.316) Fz1 Fz2 ∙µ ¶¸ 1 a2 h (1 − sin 2θ) = mg cos φ + sin φ 2 l cos θ l cos θ cos 2θ 1 h (1 − sin 2θ) 1 v 2 a2 − ma + m 2 l cos θ cos 2θ 2 RH l cos θ ∙µ ¶¸ a1 1 h (1 − sin 2θ) = mg cos φ − sin φ 2 l cos θ l cos θ cos 2θ h (1 − sin 2θ) 1 v 2 a1 1 + m + ma 2 l cos θ cos 2θ 2 RH l cos θ cos θ (2.317) (2.318) Assuming θ ¿ 1, these forces can be approximated to Fx1 + Fx2 Fz1 Fz2 1 m (a + g sin φ) 2 ∙µ ¶¸ 1 a2 h ≈ mg cos φ + sin φ 2 l l 2 h 1 v a2 1 − ma + m 2 l 2 RH l ∙µ ¶¸ 1 a1 h ≈ mg cos φ − sin φ 2 l l 1 h 1 v 2 a1 + ma + m . 2 l 2 RH l ≈ (2.319) (2.320) (2.321) Example 74 F Wheel loads of a dipping car. Consider a car with the following specifications: l w m h a1 v a = = = = = = = 2272 mm 1457 mm 1500 kg 230 mm a2 15 m/ s 1 m/ s2 (2.322) that is dipping on a hill at a point where the road has RH = 40 m φ = 30 deg θ = 2.5 deg . (2.323) The force information of the car is: Fx1 + Fx2 Fz1 Fz2 mg Fz1 + Fz2 v2 m RH = = = = = 4432.97 N 4889.1 N 5711.52 N 14715 N 10600.62 N (2.324) = 8437.5 N If we ignore the effect of θ by assuming θ ¿ 1, then the approximate value of the forces are Fx1 + Fx2 Fz1 Fz2 mg Fz1 + Fz2 v2 m RH = ≈ ≈ = ≈ 4428.75 N 4846.93 N 1524.85 N 5743.6 N 10590.53 N (2.325) = 8437.5 N. 2.9 Summary For straight motion of a symmetric rigid vehicle, we may assume the forces on the left wheel are equal to the forces on the right wheel, and simplify the tire force calculation. When a car is accelerating on an inclined road with angle φ, the normal forces under the front and rear wheels, Fz1 , Fz2 , are: µ 1 a2 Fz1 = mg cos φ − 2 l µ 1 a1 Fz2 = mg cos φ + 2 l l = a1 + a2 ¶ h sin φ − l ¶ h sin φ + l 1 h ma 2 l 1 h ma 2 l (2.326) (2.327) (2.328) ¡ ¢ where, 12 mg al1 cos φ ± hl sin φ is the static part and ± 12 mg hl ag is the dynamic part, because it depends on the acceleration a. 2.10 Key Symbols a ≡ ẍ af wd arwd a1 a2 ai aM a, b A, B, C b1 b1 b2 b2 b3 C Ct F Fx Fx1 Fx2 Fxt Fz Fz1 Fz2 Fz3 Fzt g, g h H I ki l m mt M R Rf Rr RH t v ≡ ẋ, v vc acceleration front wheel drive acceleration rear wheel drive acceleration distance of first axle from mass center distance of second axle from mass center distance of axle number i from mass center maximum acceleration arguments for atan2 (a, b) constant parameters distance of left wheels from mass center distance of hinge point from rear axle distance of right wheels from mass center distance of hinge point from trailer mass center distance of trailer axle from trailer mass center mass center of vehicle mass center of trailer force traction or brake force under a wheel traction or brake force under front wheels traction or brake force under rear wheels horizontal force at hinge normal force under a wheel normal force under front wheels normal force under rear wheels normal force under trailer wheels normal force at hinge gravitational acceleration height of C height mass moment of inertia vertical stiffness of suspension at axle number i wheel base car mass trailer mass moment tire radius front tire radius rear tire radius radius of curvature time velocity critical velocity w zi x, y, z X, Y, Z track deflection of axil number i vehicle coordinate axes global coordinate axes θ φ φM μ road slope road angle with horizon maximum slope angle friction coefficient Subscriptions dyn f f wd M r rwd st dynamic front front-wheel-drive maximum rear rear-wheel-drive statics