2 Forward Vehicle Dynamics

Transcription

1 2 Forward Veice Dynamics Straigt motion of an idea rigid veice is te subject of tis capter. We ignore air friction and examine te oad variation under te tires to determine te veice s imits of acceeration, road grade, and kinematic capabiities. 2.1 Parked Car on a Leve Road Wen a car is parked on eve pavement, te norma force, F z, under eac of te front and rear wees, F z1, F z2,are F z1 = 1 2 mg a 2 (2.1) F z2 = 1 2 mg a 1 (2.2) were, a 1 is te distance of te car s mass center, C, fromtefrontaxe, a 2 is te distance of C from te rear axe, and is te wee base. = a 1 + a 2 (2.3) z a 2 a 1 C x 2F z2 mg 2F z1 FIGURE 2.1. A parked car on eve pavement.

2 40 2. Forward Veice Dynamics Proof. Consider a ongitudinay symmetrica car as sown in Figure 2.1. It can be modeed as a two-axe veice. A symmetric two-axe veice is equivaent to a rigid beam aving two supports. Te vertica force under te front and rear wees can be determined using panar static equiibrium equations. Fz =0 (2.4) My =0 (2.5) Appying te equiibrium equations 2F z1 +2F z2 mg =0 (2.6) 2F z1 a 1 +2F z2 a 2 =0 (2.7) provide te reaction forces under te front and rear tires. F z1 = 1 2 mg a 2 a 1 + a 2 = 1 2 mg a 2 F z2 = 1 2 mg a 1 a 1 + a 2 = 1 2 mg a 1 (2.8) (2.9) Exampe 39 Reaction forces under wees. Acaras890 kg mass.itsmasscenter,c, is78 cm beind te front wee axis, and it as a 235 cm wee base. Te force under eac front wee is a 1 =0.78 m (2.10) = 2.35 m (2.11) m = 890 kg (2.12) F z1 = 1 2 mg a 2 = = N (2.13) and te force under eac rear wee is F z2 = 1 2 mg a 1 = = 1449 N. (2.14)

3 2. Forward Veice Dynamics 41 Exampe 40 Mass center position. Equations (2.1) and (2.2) can be rearranged to cacuate te position of mass center. a 1 = 2 mg F z 2 (2.15) a 2 = 2 mg F z 1 (2.16) Reaction forces under te front and rear wees of a orizontay parked car, wit a wee base =2.34 m, are: F z1 = 2000 N (2.17) F z2 = 1800 N (2.18) Terefore, te ongitudina position of te car s mass center is at a 1 = 2 mg F z = = m (2.19) 2 ( ) a 2 = 2 mg F z = = m. (2.20) 2 ( ) Exampe 41 Longitudina mass center determination. Te position of mass center C can be determined experimentay. To determine te ongitudina position of C, we soud measure te tota weigt of te car as we as te force under te front or te rear wees. Figure 2.2 iustrates a situation in wic we measure te force under te front wees. Assuming te force under te front wees is 2F z1, te position of te mass center is cacuated by static equiibrium conditions Fz =0 (2.21) My =0. (2.22) Appying te equiibrium equations 2F z1 +2F z2 mg =0 (2.23) 2F z1 a 1 +2F z2 a 2 =0 (2.24)

4 42 2. Forward Veice Dynamics a 2 a 1 C x 2F z2 mg 2F z1 FIGURE 2.2. Measuring te force under te front wees. provide te ongitudina position of C and te reaction forces under te rear wees. a 1 = 2 mg F z 2 = 2 mg (mg 2F z 1 ) (2.25) F z2 = 1 2 (mg 2F z 1 ) (2.26) Exampe 42 Latera mass center determination. Most cars are approximatey symmetrica about te ongitudina center pane passing te midde of te wees, and terefore, te atera position of te mass center C is cose to te center pane. However, te atera position of C may be cacuated by weiging one side of te car. Exampe 43 Heigt mass center determination. To determine te eigt of mass center C, we soud measure te force under te front or rear wees wie te car is on an incined surface. Experimentay, we use a device suc as is sown in Figure 2.3. Te car is parked on a eve surface suc tat te front wees are on a scae jack. Te front wees wi be ocked and ancored to te jack, wie te rear wees wi be eft free to turn. Te jack ifts te front wees and te required vertica force appied by te jacks is measured by a oad ce. Assume tat we ave te ongitudina position of C and te jack is ifted suc tat te car makes an ange φ wit te orizonta pane. Te sope ange φ is measurabe using eve meters. Assuming te force under te front wees is 2F z1, te eigt of te mass center can be cacuated by

5 2. Forward Veice Dynamics 43 z a 2 a 1 ( R)sin φ x ( R)sin φ C 2F z1 cosφ φ mg H 2F z2 FIGURE 2.3. Measuring te force under te wees to find te eigt of te mass center. static equiibrium conditions FZ =0 (2.27) My =0. (2.28) Appying te equiibrium equations 2F z1 +2F z2 mg = 0 (2.29) 2F z1 (a 1 cos φ ( R)sinφ) +2F z2 (a 2 cos φ +( R)sinφ) = 0 (2.30) provides te vertica position of C and te reaction forces under te rear wees. F z2 = 1 2 mg F z 1 (2.31) = F z 1 (R sin φ + a 1 cos φ)+f z2 (R sin φ a 2 cos φ) mg sin φ = R + a 1F z1 a 2 F z2 cot φ mg µ = R + 2 F z 1 mg a 2 cot φ (2.32)

6 44 2. Forward Veice Dynamics A car wit te foowing specifications m = 2000 kg 2F z1 = N φ = 30 deg rad (2.33) a 1 = 110 cm = 230 cm R = 30cm as a C at eigt. =34cm (2.34) Tere are tree assumptions in tis cacuation: 1 te tires are assumed to be rigid disks wit radius R, 2 fuid sift, suc as fue, cooant, and oi, are ignored, and 3 suspension defections are assumed to be zero. Suspension defection generates te maximum effect on eigt determination error. To eiminate te suspension defection, we soud ock te suspension, usuay by repacing te sock absorbers wit rigid rods to keep te veice at its ride eigt. Exampe 44 Different front and rear tires. Depending on te appication, it is sometimes necessary to use different type of tires and wees for front and rear axes. Wen te ongitudina position of C for a symmetric veice is determined, we can find te eigt of C by measuring te oad on ony one axe. As an exampe, consider te motorcyce in Figure 2.4. It as different front and rear tires. Assume te oad under te rear wee of te motorcyce F z is known. Te eigt of C can be found by taking a moment of te forces about te tireprint of te front tire. = F z 2 (a 1 + a 2 ) mg µ a 1 cos sin 1 H a 1 + a 2 + R f + R r 2 (2.35) Exampe 45 Staticay indeterminate. A veice wit more tan tree wees is staticay indeterminate. To determine te vertica force under eac tire, we need to know te mecanica properties and conditions of te tires, suc as te vaue of defection at te center of te tire, and its vertica stiffness. 2.2 Parked Car on an Incined Road Wen a car is parked on an incined pavement as sown in Figure 2.5, te norma force, F z, under eac of te front and rear wees, F z1, F z2,is:

7 2. Forward Veice Dynamics 45 a 2 C R f a 1 mg F z1 H R r F z2 φ FIGURE 2.4. A motorcyce wit different front and rear tires. F z1 = 1 2 mg a 2 F z2 = 1 2 mg a 1 = a 1 + a 2 cos φ mg cos φ 1 2 mg sin φ (2.36) sin φ (2.37) were, φ is te ange of te road wit te orizon. Te orizon is perpendicuar to te gravitationa acceeration g. Proof. Consider te car sown in Figure 2.5. Let us assume te parking brake forces are appied on ony te rear tires. It means te front tires are free to spin. Appying te panar static equiibrium equations Fx =0 (2.38) Fz =0 (2.39) My =0 (2.40) sows tat 2F x2 mg sin φ =0 (2.41) 2F z1 +2F z2 mg cos φ =0 (2.42) 2F z1 a 1 +2F z2 a 2 2F x2 =0. (2.43)

8 46 2. Forward Veice Dynamics z a 2 a 1 x a C 2F z1 2F x2 2F z2 φ mg FIGURE 2.5. A parked car on incined pavement. Tese equations provide te brake force and reaction forces under te front and rear tires. F z1 = 1 2 mg a 2 cos φ 1 2 mg sin φ (2.44) F z2 = 1 2 mg a 1 cos φ mg sin φ (2.45) F x2 = 1 mg sin φ 2 (2.46) Exampe 46 Increasing te incination ange. Wen φ =0, Equations (2.36) and (2.37) reduce to (2.1) and (2.2). By increasing te incination ange, te norma force under te front tires of a parked car decreases and te norma force and braking force under te rear tires increase. Te imit for increasing φ is were te weigt vector mg goes troug te contact point of te rear tire wit te ground. Suc an ange is caed a titing ange. Exampe 47 Maximum incination ange. Te required braking force F x2 increases by te incination ange. Because F x2 is equa to te friction force between te tire and pavement, its maximum depends on te tire and pavement conditions. Tere is a specific ange φ M at wic te braking force F x2 wi saturate and cannot increase any more. At tis maximum ange, te braking force is proportiona to te norma force F z2 F = μ F x2 x2 z2 (2.47)

9 2. Forward Veice Dynamics 47 were, te coefficient μ x2 is te x-direction friction coefficient for te rear wee. At φ = φ M, te equiibrium equations wi reduce to Tese equations provide 2μ x2 F z2 mg sin φ M =0 (2.48) 2F z1 +2F z2 mg cos φ M =0 (2.49) 2F z1 a 1 2F z2 a 2 +2μ x2 F z2 =0. (2.50) F z1 = 1 2 mg a 2 cos φ M 1 2 mg sin φ M (2.51) F z2 = 1 2 mg a 1 cos φ M mg sin φ M (2.52) tan φ M = a 1μ x2 μ x2 (2.53) sowing tat tere is a reation between te friction coefficient μ x2,maximum incination φ M, and te geometrica position of te mass center C. Te ange φ M increases by decreasing. For a car aving te specifications μ x2 = 1 a 1 = 110 cm (2.54) = 230 cm = 35cm tetitingangeis φ M rad deg. (2.55) Exampe 48 Front wee braking. Wen te front wees are te ony braking wees F x2 =0and F x1 6=0. In tis case, te equiibrium equations wi be 2F x1 mg sin φ =0 (2.56) 2F z1 +2F z2 mg cos φ =0 (2.57) 2F z1 a 1 +2F z2 a 2 2F x1 =0. (2.58) Tese equations provide te brake force and reaction forces under te front and rear tires. F z1 = 1 2 mg a 2 F z2 = 1 2 mg a 1 cos φ 1 2 mg sin φ (2.59) cos φ mg sin φ (2.60) F x1 = 1 mg sin φ (2.61) 2

10 48 2. Forward Veice Dynamics Atteutimateangeφ = φ M and Tese equations provide F x1 = μ x1 F z1 (2.62) 2μ x1 F z1 mg sin φ M =0 (2.63) 2F z1 +2F z2 mg cos φ M =0 (2.64) 2F z1 a 1 2F z2 a 2 +2μ x1 F z1 =0. (2.65) F z1 = 1 2 mg a 2 cos φ M 1 2 mg sin φ M (2.66) F z2 = 1 2 mg a 1 cos φ M mg sin φ M (2.67) tan φ M = a 2μ x1 μ x1. (2.68) Let s name te utimate ange for te front wee brake in Equation (2.53) as φ Mf, and te utimate ange for te rear wee brake in Equation (2.68) as φ Mr.Comparingφ Mf and φ Mr sows tat φ Mf φ Mr = a 1μ x2 μx1. (2.69) a 2 μ x1 μx2 We may assume te front and rear tires are te same and so, μ x1 = μ x2 (2.70) terefore, φ Mf = a 1. (2.71) φ Mr a 2 Hence, if a 1 <a 2 ten φ Mf <φ Mr and terefore, a rear brake is more effective tan a front brake on upi parking as ong as φ Mr is ess tan te titing ange, φ Mr < tan 1 a 2. At te titing ange, te weigt vector passes troug te contact point of te rear wee wit te ground. Simiary we may concude tat wen parked on a downi road, te front brake is more effective tan te rear brake. Exampe 49 Four-wee braking. Consider a four-wee brake car, parked upi as sown in Figure 2.6. In tese conditions, tere wi be two brake forces F x1 on te front wees and two brake forces F x1 on te rear wees.

11 2. Forward Veice Dynamics 49 z a 2 a 1 x a C 2F x1 2F z1 2F x2 2F z2 φ mg FIGURE 2.6. A four wee brake car, parked upi. Te equiibrium equations for tis car are 2F x1 +2F x2 mg sin φ =0 (2.72) 2F z1 +2F z2 mg cos φ =0 (2.73) 2F z1 a 1 +2F z2 a 2 (2F x1 +2F x2 ) =0. (2.74) Tese equations provide te brake force and reaction forces under te front and rear tires. F z1 = 1 2 mg a 2 cos φ 1 2 mg sin φ (2.75) F z2 = 1 2 mg a 1 cos φ mg sin φ (2.76) F x1 + F x2 = 1 mg sin φ (2.77) 2 Atteutimateangeφ = φ M, a wees wi begin to side simutaneousy and terefore, Te equiibrium equations sow tat F x1 = μ x1 F z1 (2.78) F x2 = μ x2 F z2. (2.79) 2μ x1 F z1 +2μ x2 F z2 mg sin φ M =0 (2.80) 2F z1 +2F z2 mg cos φ M =0 (2.81) 2F z1 a 1 +2F z2 a 2 2μ x1 F z1 +2μ x2 F z2 =0. (2.82)

12 50 2. Forward Veice Dynamics z a 2 a 1 C a x 2F x2 2F z2 mg 2F x1 2F z1 FIGURE 2.7. An acceerating car on a eve pavement. Assuming wi provide μ x1 = μ x2 = μ x (2.83) F z1 = 1 2 mg a 2 cos φ M 1 2 mg sin φ M (2.84) F z2 = 1 2 mg a 1 cos φ M mg sin φ M (2.85) tan φ M = μ x. (2.86) 2.3 Acceerating Car on a Leve Road Wen a car is speeding wit acceeration a on a eve road as sown in Figure 2.7, te vertica forces under te front and rear wees are F z1 = 1 2 mg a 2 F z2 = 1 2 mg a mg a g mg (2.87) a g. (2.88) Te first terms, 1 a2 2mg and 1 a1 2mg, are caed static parts, and te second terms ± 1 2 mg a g are caed dynamic parts of te norma forces. Proof. Te veice is considered as a rigid body tat moves aong a orizonta road. Te force at te tireprint of eac tire may be decomposed to a norma and a ongitudina force. Te equations of motion for te acceerating car come from Newton s equation in x-direction and two static

13 2. Forward Veice Dynamics 51 equiibrium equations. Fx = ma (2.89) Fz =0 (2.90) My =0. (2.91) Expanding te equations of motion produces tree equations for four unknowns F x1, F x2, F z1, F z2. 2F x1 +2F x2 = ma (2.92) 2F z1 +2F z2 mg =0 (2.93) 2F z1 a 1 +2F z2 a 2 2(F x1 + F x2 ) =0 (2.94) However, it is possibe to eiminate (F x1 + F x2 ) between te first and tird equations, and sove for te norma forces F z1, F z2. F z1 = (F z1 ) st +(F z1 ) dyn = 1 2 mg a mg a g (2.95) F z2 = (F z2 ) st +(F z2 ) dyn = 1 2 mg a mg a g (2.96) Te static parts (F z1 ) st = 1 2 mg a 2 (2.97) (F z2 ) st = 1 2 mg a 1 (2.98) are weigt distribution for a stationary car and depend on te orizonta position of te mass center. However, te dynamic parts (F z1 ) dyn = 1 2 mg a g (F z2 ) dyn = 1 2 mg a g (2.99) (2.100) indicate te weigt distribution according to orizonta acceeration, and depend on te vertica position of te mass center. Wen acceerating a>0, te norma forces under te front tires are ess tan te static oad, and under te rear tires are more tan te static oad.

14 52 2. Forward Veice Dynamics Exampe 50 Front-wee-drive acceerating on a eve road. Wen te car is front-wee-drive, F x2 =0. Equations (2.92) to (2.88) wi provide te same vertica tireprint forces as (2.87) and (2.88). However, te required orizonta force to acieve te same acceeration, a, must be provided by soey te front wees. Exampe 51 Rear-wee drive acceerating on a eve road. If a car is rear-wee drive ten, F x1 =0and te required force to acieve te acceeration, a, must be provided ony by te rear wees. Te vertica force under te wees wi sti be te same as (2.87) and (2.88). Exampe 52 Maximum acceeration on a eve road. Te maximum acceeration of a car is proportiona to te friction under its tires. We assume te friction coefficients at te front and rear tires are equa and a tires reac teir maximum tractions at te same time. Newton s equation (2.92) can now be written as F x1 = ±μ x F z1 (2.101) F x2 = ±μ x F z2 (2.102) ma = ±2μ x (F z1 + F z2 ). (2.103) Substituting F z1 and F z2 from (2.93) and (2.94) resuts in a = ±μ x g. (2.104) Terefore, te maximum acceeration and deceeration depend directy on te friction coefficient. Exampe 53 Maximum acceeration for a singe-axe drive car. Te maximum acceeration a rwd for a rear-wee-drive car is acieved wen we substitute F x1 = 0, F x2 = μ x F z2 in Equation (2.92) and use Equation (2.88) µ a1 μ x mg + a rwd = ma rwd (2.105) g and terefore, a rwd g = = a 1 μ x μ x μ x a 1. (2.106) 1 μ x Te front wees can eave te ground wen F z1 =0. Substituting F z1 =0 in Equation (2.88) provides te maximum acceeration at wic te front wees are sti on te road. a rwd g a 2 (2.107)

15 2. Forward Veice Dynamics 53 a/g a rwd /g a fwd /g a 1 / FIGURE 2.8. Effect of mass center position on te maximum acievabe acceeration of a front- and a rear-wee drive car. Terefore, te maximum attainabe acceeration woud be te ess vaue of Equation (2.106) or (2.107). Simiary, te maximum acceeration a fwd for a front-wee drive car is acieved wen we substitute F x2 =0, F x1 = μ x F z1 in Equation (2.92) and use Equation (2.87). a fwd g = = a 2 μ x + μ x μ ³ x 1 a 1 1+μ x (2.108) To see te effect of canging te position of mass center on te maximum acievabe acceeration, we pot Figure 2.8 for a sampe car wit μ x = 1 = 0.56 m (2.109) = 2.6m. Passenger cars are usuay in te range 0.4 < (a 1 /g) < 0.6,wit(a 1 /g) 0.4 for front-wee-drive cars, and (a 1 /g) 0.6 for rear-wee-drive cars. In tis range, (a rwd /g) > (a fwd /g) and terefore rear-wee-drive cars can reac iger forward acceeration tan front-wee-drive cars. It is an important appied fact, especiay for race cars. Te maximum acceeration may aso be imited by te titing condition a M g a 2. (2.110)

16 54 2. Forward Veice Dynamics Exampe 54 Minimum time for km/ on a eve road. Consider a car wit te foowing caracteristics: engt = 4245 mm widt = 1795 mm eigt = 1285 mm wee base = 2272 mm front track = 1411 mm (2.111) rear track = 1504 mm net weigt = 1500 kg = 220 mm μ x = 1 a 1 = a 2 Assume te car is rear-wee-drive and its engine can provide te maximum traction supported by friction. Equation (2.88) determines te oad on te rear wees and terefore, te forward equation of motion is 2F x2 = 2μ x F z2 = μ x mg a 1 + μ x mg 1 g a = ma. (2.112) Rearrangement provides te foowing differentia equation to cacuate veocity and dispacement: Taking an integra μ x g a 1 a = ẍ = 1 μ x g 1 g a 1 = gμ x (2.113) μ x Z dv = Z t 0 adt (2.114) between v =0and v = 100 km/ m/ s sows tat te minimum time for km/ on a eve road is t = gμ x a 1 μ x 5.11 s (2.115)

17 2. Forward Veice Dynamics 55 If te same car was front-wee-drive, ten te traction force woud be 2F x1 = 2μ x F z1 = μ x mg a 2 μ x mg 1 g a = ma. (2.116) and te equation of motion woud reduce to μ x g a 2 a = ẍ = 1+μ x g 1 g a 2 = gμ x. (2.117) + μ x Te minimum time for km/ onaeveroadfortisfront-weedrive car is t = a s. (2.118) gμ x + μ x Now consider te same car to be four-wee-drive. Ten, te traction force is 2F x1 +2F x2 = 2μ x (F z1 + F z2 ) = g m (a 1 + a 2 ) = ma. (2.119) and te minimum time for km/ on a eve road for tis four-weedrive car can teoreticay be reduced to t = s. (2.120) g 2.4 Acceerating Car on an Incined Road Wen a car is acceerating on an incined pavement wit ange φ as sown in Figure 2.9, te norma force under eac of te front and rear wees, F z1, F z2,woudbe: F z1 = 1 µ 2 mg a2 F z2 = 1 µ 2 mg a1 cos φ sin φ 1 2 ma cos φ + sin φ ma (2.121) (2.122) = a 1 + a 2

18 56 2. Forward Veice Dynamics z a 2 a 1 x a C 2F x1 2F z1 2F x2 2F z2 φ mg FIGURE 2.9. An acceerating car on incined pavement. a Te dynamic parts, ± 1 2 mg g, depend on acceeration a and eigt of mass center C and remain uncanged, wie te static parts are infuenced by te sope ange φ and eigt of mass center. Proof. Te Newton s equation in x-direction and two static equiibrium equations, must be examined to find te equation of motion and ground reaction forces. Fx = ma (2.123) Fz =0 (2.124) My =0. (2.125) Expanding tese equations produces tree equations for four unknowns F x1, F x2, F z1, F z2. 2F x1 +2F x2 mg sin φ = ma (2.126) 2F z1 +2F z2 mg cos φ =0 (2.127) 2F z1 a 1 2F z2 a 2 +2(F x1 + F x2 ) =0 (2.128) It is possibe to eiminate (F x1 + F x2 ) between te first and tird equations, and sove for te norma forces F z1, F z2. F z1 = (F z1 ) st +(F z1 ) dyn = 1 2 mg µ a2 cos φ sin φ 1 2 ma (2.129)

19 2. Forward Veice Dynamics 57 F z2 = (F z2 ) st +(F z2 ) dyn = 1 2 mg µ a1 cos φ + sin φ ma (2.130) Exampe 55 Front-wee-drive car, acceerating on incined road. For a front-wee-drive car, we may substitute F x1 = 0 in Equations (2.126) and (2.128) to ave te governing equations. However, it does not affect te ground reaction forces under te tires (2.129 and 2.130) as ong as te car is driven under its imit conditions. Exampe 56 Rear-wee-drive car, acceerating on incined road. Substituting F x2 =0in Equations (2.126) and (2.128) and soving for te norma reaction forces under eac tire provides te same resuts as (2.129) and (2.130). Hence, te norma forces appied on te tires do not sense if te car is front-, rear-, or a-wee drive. As ong as we drive in a straigt pat at ow acceeration, te drive wees can be te front or te rear ones. However, te advantages and disadvantages of front-, rear-, or a-wee drive cars appear in maneuvering, sippery roads, or wen te maximum acceeration is required. Exampe 57 Maximum acceeration on an incined road. Te maximum acceeration depends on te friction under te tires. Let s assume te friction coefficients at te front and rear tires are equa. Ten, te front and rear traction forces are F x1 μ x F z1 (2.131) F x2 μ x F z2. (2.132) If we assume te front and rear wees reac teir traction imits at te same time, ten F x1 = ±μ x F z1 (2.133) F x2 = ±μ x F z2 (2.134) and we may rewrite Newton s equation (2.123) as ma M = ±2μ x (F z1 + F z2 ) mg sin φ (2.135) were, a M is te maximum acievabe acceeration. Now substituting F z1 and F z2 from (2.129) and (2.130) resuts in a M g = ±μ x cos φ sin φ. (2.136) Acceerating on an upi road (a >0,φ>0) and braking on a downi road (a <0,φ<0) are te extreme cases in wic te car can sta. In tese cases, te car can move as ong as μ x tan φ. (2.137)

20 58 2. Forward Veice Dynamics Exampe 58 Limits of acceeration and incination ange. Assuming F z1 > 0 and F z2 > 0, we can write Equations (2.121) and (2.122) as a g a 2 cos φ sin φ (2.138) a g a 1 cos φ sin φ. (2.139) Hence, te maximum acievabe acceeration (a >0) is imited by a 2,, φ; wie te maximum deceeration (a <0) is imited by a 1,, φ. Tesetwo equations can be combined to resut in a 1 cos φ a g +sinφ a 2 cos φ. (2.140) If a 0, ten te imits of te incination ange woud be a 1 tan φ a 2. (2.141) Tis is te maximum and minimum road incination anges tat te car can stay on witout titing and faing. Exampe 59 Maximum deceeration for a singe-axe-brake car. We can find te maximum braking deceeration a fwb of a front-weebrake car on a orizonta road by substituting φ = 0, F x2 = 0, F x1 = μ x F z1 in Equation (2.126) and using Equation (2.121) terefore, μ x mg a fwb g µ a2 a rwb g = μ x 1 μ x = ma fwb (2.142) ³ 1 a 1. (2.143) Simiary, te maximum braking deceeration a rwb of a front-wee-brake car can be acieved wen we substitute F x2 =0, F x1 = μ x F z1. a rwb g = μ x a 1 1+μ x (2.144) Te effect of canging te position of te mass center on te maximum acievabe braking deceeration is sown in Figure 2.10 for a sampe car wit μ x = 1 = 0.56 m (2.145) = 2.6m.

21 2. Forward Veice Dynamics 59 a/g a rwd /g a fwd /g a 1 / FIGURE Effect of mass center position on te maximum acievabe decceeration of a front-wee and a rear-wee-drive car. Passenger cars are usuay in te range 0.4 < (a 1 /) < 0.6. In tis range, (a fwb /g) < (a rwb /g) and terefore, front-wee-brake cars can reac better forward deceeration tan rear-wee-brake cars. Hence, front brakes are muc more important tan te rear brakes. Exampe 60 F Acarwitatraier. Figure 2.11 depicts a car moving on an incined road and puing a traier. To anayze te car-traier motion, we need to separate te car and traier to see te forces at te inge, as sown in Figure We assume te mass center of te traier C t is at distance b 3 in front of te ony axe of te traier. If C t is beind te traier axe, ten b 3 soud be negative in te foowing equations. For an idea inge between a car and a traier moving in a straigt pat, tere must be a orizonta force F xt and a vertica force F zt. Writing te Newton s equation in x-direction and two static equiibrium equations for bot te traier and te veice Fx = m t a (2.146) Fz =0 (2.147) My =0 (2.148) we find te foowing set of equations: F xt m t g sin φ = m t a (2.149) 2F z3 F zt m t g cos φ =0 (2.150) 2F z3 b 3 F zt b 2 F xt ( 2 1 )=0 (2.151)

22 60 2. Forward Veice Dynamics z a 2 a 1 x b 3 b 2 b 1 C t 2F z2 C φ mg 2F x2 2F x1 2F z1 a m t g 2F z3 FIGURE A car moving on an incined road and puing a traier. 2F x1 +2F x2 F xt mg sin φ = ma (2.152) 2F z1 +2F z2 F zt mg cos φ =0 (2.153) 2F z1 a 1 2F z2 a 2 +2(F x1 + F x2 ) F xt ( 1 )+F zt (b 1 + a 2 )=0 (2.154) If te vaue of traction forces F x1 and F x2 are given, ten tese are six equations for six unknowns: a, F xt, F zt, F z1, F z2, F z3. Soving tese equations provide te foowing soutions: a = 2 (F x1 + F x2 ) g sin φ m + m t (2.155) F xt = 2m t (F x1 + F x2 ) m + m t (2.156) F zt = 1 2 2m t (F x1 + F x2 )+ b 3 m t g cos φ b 2 b 3 m + m t b 2 b 3 (2.157) F z1 = b µ 3 2a2 b 1 m t + a 2 m g cos φ 2 b 2 b 3 b 3 2a2 b 1 Fx1 + F x2 + ( 1 2 ) m t 1 m t m b 2 b 3 (m + m t ) (2.158)

23 2. Forward Veice Dynamics 61 z a 1 x a 2 b 1 a F xt 1 C φ mg 2F x1 F zt 2F x2 2F z2 b 3 b 2 F zt 1 2 2F z1 C t F xt m t g φ 2F z3 FIGURE Free-body-diagram of a car and te traier wen moving on an upi road. F z2 = b µ 3 a1 a 2 + b 1 m t + a 1 m g cos φ 2 b 2 b 3 b 3 a1 a 2 + b 1 Fx1 + F x2 + ( 1 2 ) m t + 1 m t + m (2.159) b 2 b 3 (m + m t ) F z3 = 1 b 2 m t g cos φ m t (F x1 + F x2 ) (2.160) 2 b 2 b 3 b 2 b 3 m + m t = a 1 + a 2. (2.161) However, if te vaue of acceeration a is known, ten unknowns are: F x1 + F x2, F xt, F zt, F z1, F z2, F z3. F x1 + F x2 = 1 2 (m + m t)(a + g sin φ) (2.162) F xt = m t (a + g sin φ) (2.163) F zt = 1 2 b 2 b 3 m t (a + g sin φ)+ b 3 b 2 b 3 m t g cos φ (2.164)

24 62 2. Forward Veice Dynamics F z1 = b µ 3 2a2 b 1 m t + a 2 m g cos φ (2.165) 2 b 2 b 3 b a2 b 1 ( 1 2 ) m t 1 m t m (a + g sin φ) 2 b 2 b 3 F z2 = b µ 3 a1 a 2 + b 1 m t + a 1 m g cos φ (2.166) 2 b 2 b 3 b a1 a 2 + b 1 ( 1 2 ) m t + 1 m t + m (a + g sin φ) 2 b 2 b 3 F z3 = 1 m t (b 2 g cos φ +( 1 2 )(a + g sin φ)) (2.167) 2 b 2 b 3 = a 1 + a 2. Exampe 61 F Maximum incination ange for a car wit a traier. For a car and traier as sown in Figure 2.11, te maximum incination ange φ M is te ange at wic te car cannot acceerate te veice. Substituting a =0and φ = φ M in Equation (2.155) sows tat 2 sin φ M = (m + m t ) g (F x 1 + F x2 ). (2.168) Te vaue of maximum incination ange φ M increases by decreasing te tota weigt of te veice and traier (m + m t ) g or increasing te traction force F x1 + F x2. Te traction force is imited by te maximum torque on te drive wee and te friction under te drive tire. Let s assume te veice is four-weedrive and friction coefficients at te front and rear tires are equa. Ten, te front and rear traction forces are F x1 μ x F z1 (2.169) F x2 μ x F z2. (2.170) If we assume te front and rear wees reac teir traction imits at te same time, ten and we may rewrite te Equation (2.168) as F x1 = μ x F z1 (2.171) F x2 = μ x F z2 (2.172) 2μ sin φ M = x (m + m t ) g (F z 1 + F z2 ). (2.173) Now substituting F z1 and F z2 from (2.158) and (2.159) resuts in (mb 3 mb 2 m t b 3 ) μ x cos φ M +(b 2 b 3 )(m + m t )sinφ M = 2μ x m t ( 1 2 ) m + m t (F x1 + F x2 ). (2.174)

25 2. Forward Veice Dynamics 63 If we arrange Equation (2.174) as A cos φ M + B sin φ M = C (2.175) ten r C φ M =atan2( A2 + B, ± 1 C2 2 A 2 ) atan2(a, B) (2.176) + B2 and C φ M =atan2( A2 + B, ±p A 2 + B 2 C 2 ) atan2(a, B) (2.177) 2 were A = (mb 3 mb 2 m t b 3 ) μ x (2.178) B = (b 2 b 3 )(m + m t ) (2.179) m t ( 1 2 ) C = 2μ x (F x1 + F x2 ). m + m t (2.180) For a rear-wee-drive car puing a traier wit te foowing caracteristics: = 2272 mm w = 1457 mm = 230 mm we find a 1 = a 2 1 = 310 mm b 1 = 680 mm b 2 = 610 mm b 3 = 120 mm (2.181) 2 = 560 mm m = 1500 kg m t = 150 kg μ x = 1 φ = 10deg a = 1m/ s 2 F z1 = N F z2 = N F z3 = N F zt = N (2.182) F xt = N F x2 = N.

26 64 2. Forward Veice Dynamics To ceck if te required traction force F x2 is appicabe, we soud compare it to te maximum avaiabe friction force μf z2 anditmustbe F x2 μf z2. (2.183) Exampe 62 F Soution of equation a cos θ + b sin θ = c. Te first type of trigonometric equation is a cos θ + b sin θ = c. (2.184) It can be soved by introducing two new variabes r and η suc tat and terefore, Substituting te new variabes sow tat a = r sin η (2.185) b = r cos η (2.186) r = p a 2 + b 2 (2.187) η = atan2(a, b). (2.188) sin(η + θ) = c r r (2.189) cos(η + θ) = ± 1 c2 r 2. (2.190) Hence, te soutions of te probem are θ =atan2( c r r, ± 1 c2 ) atan2(a, b) r2 (2.191) and θ =atan2( c r, ±p r 2 c 2 ) atan2(a, b). (2.192) Terefore, te equation a cos θ + b sin θ = c astwosoutionsifr 2 = a 2 + b 2 >c 2,onesoutionifr 2 = c 2,andnosoutionifr 2 <c 2. Exampe 63 F Te function tan 1 y 2 x =atan2(y, x). Tere are many situations in kinematics cacuation in wic we need to find an ange based on te sin and cos functions of an ange. However, tan 1 cannot sow te effect of te individua sign for te numerator and denominator. It aways represents an ange in te first or fourt quadrant. To overcome tis probem and determine te ange in te correct quadrant, te atan2 function is introduced as beow. tan 1 y if y > 0 x atan2(y,x) = tan 1 y + π sign y if y < 0 (2.193) x π sign x if y =0 2

27 2. Forward Veice Dynamics 65 In tis text, weter it as been mentioned or not, werever tan 1 y x is used, it must be cacuated based on atan2(y, x). Exampe 64 Zero vertica force at te inge. We can make te vertica force at te inge equa to zero by examining Equation (2.157) for te inge vertica force F zt. F zt = 1 2 b 2 b 3 2m t m + m t (F x1 + F x2 )+ b 3 b 2 b 3 m t g cos φ (2.194) To make F zt =0, it is enoug to adjust te position of traier mass center C t exacty on top of te traier axe and at te same eigt as te inge. In tese conditions we ave tat makes 1 = 2 (2.195) b 3 = 0 (2.196) F zt =0. (2.197) However, to increase safety, te oad soud be distributed eveny trougout te traier. Heavy items soud be oaded as ow as possibe, mainy over te axe. Bukier and igter items soud be distributed to give a itte positive b 3. Suc a traier is caed nose weigt at te towing couping. 2.5 Parked Car on a Banked Road Figure 2.13 depicts te effect of a bank ange φ on te oad distribution of a veice. A bank causes te oad on te ower tires to increase, and te oad on te upper tires to decrease. Te tire reaction forces are: F z1 = 1 mg 2 w (b 2 cos φ sin φ) (2.198) F z2 = 1 mg 2 w (b 1 cos φ + sin φ) (2.199) w = b 1 + b 2 (2.200) Proof. Starting wit equiibrium equations Fy =0 (2.201) Fz =0 (2.202) Mx =0. (2.203)

28 66 2. Forward Veice Dynamics z y C 2F y1 2F z1 2F y2 2F z2 φ mg b 2 b 1 FIGURE Norma force under te upi and downi tires of a veice, parked on banked road. we can write 2F y1 +2F y2 mg sin φ =0 (2.204) 2F z1 +2F z2 mg cos φ =0 (2.205) 2F z1 b 1 2F z2 b 2 +2(F y1 + F y2 ) =0. (2.206) We assumed te force under te ower tires, front and rear, are equa, and aso te forces under te upper tires, front and rear are equa. To cacuate te reaction forces under eac tire, we may assume te overa atera force F y1 +F y2 as an unknown. Te soution of tese equations provide te atera and reaction forces under te upper and ower tires. F z1 = 1 2 mg b 2 w cos φ 1 2 mg sin φ (2.207) w F z2 = 1 2 mg b 1 w cos φ mg sin φ (2.208) w F y1 + F y2 = 1 mg sin φ (2.209) 2 At te utimate ange φ = φ M, a wees wi begin to side simutaneousy and terefore, F y1 = μ y1 F z1 (2.210) F y2 = μ y2 F z2. (2.211)

29 2. Forward Veice Dynamics 67 Te equiibrium equations sow tat 2μ y1 F z1 +2μ y2 F z2 mg sin φ =0 (2.212) 2F z1 +2F z2 mg cos φ =0 (2.213) 2F z1 b 1 2F z2 b 2 +2 μ y1 F z1 + μ y2 F z2 =0. (2.214) Assuming wi provide μ y1 = μ y2 = μ y (2.215) F z1 = 1 2 mg b 2 w cos φ M 1 2 mg w sin φ M (2.216) F z2 = 1 2 mg b 1 w cos φ M mg w sin φ M (2.217) tan φ M = μ y. (2.218) Tese cacuations are correct as ong as tan φ M b 2 (2.219) μ y b 2. (2.220) If te atera friction μ y is iger tan b 2 / ten te car wi ro downi. To increase te capabiity of a car moving on a banked road, te car soud be as wide as possibe wit a mass center as ow as possibe. Exampe 65 Tire forces of a parked car in a banked road. A car aving m = 980 kg = 0.6m (2.221) w = 1.52 m b 1 = b 2 is parked on a banked road wit φ =4deg. Te forces under te ower and upper tires of te car are: F z1 = N F z2 = N (2.222) F y1 + F y2 = 335.3N Te ratio of te upi force F z1 to downi force F z2 depends on ony te mass center position. F z1 F z2 = b 2 cos φ sin φ b 1 cos φ + sin φ (2.223)

30 68 2. Forward Veice Dynamics =0.6 m w=1.52 m b 1 =b 2 F F z z 1 2 Roing down ange φ[rad] φ[deg] FIGURE Iustration of te force ratio F z1 /F z2 ange φ. as a function of road bank Assuming a symmetric car wit b 1 = b 2 = w/2 simpifies te equation to F z1 F z2 = w cos φ 2 sin φ w cos φ +2 sin φ. (2.224) Figure 2.14 iustrates te beavior of force ratio F z1 /F z2 as a function of φ for =0.6mand w =1.52 m. Te roing down ange φ M =tan 1 (b 2 /) = deg indicates te bank ange at wic te force under te upi wees become zero and te car ros down. Te negative part of te curve indicates te required force to keep te car on te road, wic is not appicabe in rea situations. 2.6 F Optima Drive and Brake Force Distribution A certain acceeration a can be acieved by adjusting and controing te ongitudina forces F x1 and F x2. Te optima ongitudina forces under te front and rear tires to acieve te maximum acceeration are F x1 mg = 1 2 = 1 2 μ2 x µ 2 a + 1 g μ x a 2 a 2 a g (2.225)

31 2. Forward Veice Dynamics 69 F x2 mg = 1 2 = 1 2 μ2 x µ 2 a + 1 g μ x a 1 a g a 1. (2.226) Proof. Te ongitudina equation of motion for a car on a orizonta road is 2F x1 +2F x2 = ma (2.227) and te maximum traction forces under eac tire is a function of norma forceandtefrictioncoefficient. F x1 ±μ x F z1 (2.228) F x2 ±μ x F z2 (2.229) However, te norma forces are a function of te car s acceeration and geometry. F z1 = 1 2 mg a mg a (2.230) g F z2 = 1 2 mg a mg a (2.231) g We may generaize te equations by making tem dimensioness. Under te best conditions, we soud adjust te traction forces to teir maximum F x1 = 1 µ mg 2 μ a2 x a (2.232) g F x2 = 1 µ mg 2 μ a1 x + a (2.233) g and terefore, te ongitudina equation of motion (2.227) becomes a g = μ x. (2.234) Substituting tis resut back into Equations (2.232) and (2.233) sows tat F x1 mg F x2 mg = 1 2 = 1 2 µ 2 a + 1 a 2 a g 2 g (2.235) a 1 a 2 g. (2.236) µ a g Depending on te geometry of te car (, a 1,a 2 ), and te acceeration a> 0, tese two equations determine ow muc te front and rear driving forces must be. Te same equations are appied for deceeration a<0, to

32 70 2. Forward Veice Dynamics F x / mg F x2 / mg a 1 F x1 / mg F x2 / mg a / g a 2 F x1 / mg FIGURE Optima driving and braking forces for a sampe car. determine te vaue of optima front and rear braking forces. Figure 2.15 represents a grapica iustration of te optima driving and braking forces for a sampe car using te foowing data: μ x = 1 = a 1 = a 2 = (2.237) = 1 2. Wen acceerating a>0, te optima driving force on te rear tire grows rapidy wie te optima driving force on te front tire drops after a maximum. Te vaue (a/g) =(a 2 /) is te maximum possibe acceeration at wic te front tires ose teir contact wit te ground. Te acceeration at wic front (or rear) tires ose teir ground contact is caed titing acceeration. Te opposite penomenon appens wen deceerating. For a<0, te optima front brake force increases rapidy and te rear brake force goes to zero after a minimum. Te deceeration (a/g) = (a 1 /) is te maximum possibe deceeration at wic te rear tires ose teir ground contact. Te grapica representation of te optima driving and braking forces can be sown better by potting F x1 / (mg) versus F x2 / (mg) using (a/g) as a parameter. F x1 = a 2 a g a 1 + a g F x 2 (2.238) F x1 = a 2 μ x F x2 a 1 + μ x (2.239)

33 2. Forward Veice Dynamics 71 F x1 / mg Driving a 2 F x2 / mg Braking a 1 FIGURE Optima traction and braking force distribution between te front and rear wees. Suc a pot is sown in Figure Tis is a design curve describing te reationsip between forces under te front and rear wees to acieve te maximum acceeration or deceeration. Adjusting te optima force distribution is not an automatic procedure and needs a force distributor contro system to measure and adjust te forces. Exampe 66 F Sope at zero. Te initia optima traction force distribution is te sope of te optima curve (F x1 / (mg),f x2 / (mg)) at zero. d F x 1 mg d F x 2 mg = im a 0 µ 2 1 a g 2 µ 1 a 2 g a 2 a g a 1 a 2 g = a 2 (2.240) a 1 Terefore, te initia traction force distribution depends on ony te position of mass center C. Exampe 67 F Brake baance and ABS. Wen braking, a car is stabe if te rear wees do not ock. Tus, te rear brake forces must be ess tan te maximum possibe braking force at a time. Tis means te brake force distribution soud aways be in te saded area of Figure 2.17, and beow te optima curve. Tis restricts te

34 72 2. Forward Veice Dynamics F x1 / mg F x2 / mg Braking a 1 FIGURE Optima braking force distribution between te front and rear wees, aong wit a tre-ine under estimation. acievabe deceeration, especiay at ow friction vaues, but increases te stabiity of te car. Wenever it is easier for a force distributor to foow a ine, te optima brake curve is underestimated using two or tree ines, and a contro system adjusts te force ratio F x1 /F x2. A sampe of tree-ine approximation is sown in Figure Distribution of te brake force between te front and rear wees is caed brake baance. Brake baance varies wit deceeration. Te iger te stop, te more oad wi transfer to te front wees and te more braking effort tey can support. Meanwie te rear wees are unoaded and tey must ave ess braking force. Exampe 68 F Best race car. Racecars aways work at te maximum acievabe acceeration to finis teir race in minimum time. Tey are usuay designed wit rear-weedrive and a-wee-brake. However, if an a-wee-drive race car is reasonabe to buid, ten a force distributor, to foow te curve sown in Figure 2.18, is wat it needs to race better. Exampe 69 F Effect of C ocation on braking. Load is transferred from te rear wees to te front wen te brakes are appied. Te iger te C, te more oad transfer. So, to improve braking, te mass center C soud be as ow as possibe and as back as possibe. Tis is not feasibe for every veice, especiay for forward-wee drive street cars. However, tis fact soud be taken into account wen a car is

35 2. Forward Veice Dynamics 73 F x1 / mg Driving a 2 F x2 / mg FIGURE Optima traction force distribution between te front and rear wees. F x2 F x1 F y1 v F x2 F x1 F y1 FIGURE deg siding rotation of a rear-wee-ocked car. being designed for better braking performance. Exampe 70 F Front and rear wee ocking. Te optima brake force distribution is according to Equation (2.239) for an idea F x1 /F x2 ratio. However, if te brake force distribution is not idea, ten eiter te front or te rear wees wi ock up first. Locking te rear wees makes te veice unstabe, and it oses directiona stabiity. Wen te rear wees ock, tey side on te road and tey ose teir capacity to support atera force. Te resutant sear force at te tireprint of te rear wees reduces to a dynamic friction force in te opposite direction of te siding. A sigt atera motion of te rear wees, by any disturbance, deveops a yaw motion because of unbaanced atera forces on te front and rear wees. Te yaw moment turns te veice about te z-axis unti te rear end eads te front end and te veice turns 180 deg. Figure 2.19 iustrates a 180 deg siding rotation of a rear-wee-ocked car.

36 74 2. Forward Veice Dynamics Te ock-up of te front tires does not cause a directiona instabiity, atoug te car woud not be steerabe and te driver woud ose contro. 2.7 F Veices Wit More Tan Two Axes If a veice as more tan two axes, suc as te tree-axe car sown in Figure 2.20, ten te veice wi be staticay indeterminate and te norma forces under te tires cannot be determined by static equiibrium equations. We need to consider te suspensions defection to determine teir appied forces. Te n norma forces F zi under te tires can be cacuated using te foowing n agebraic equations. 2 2 n F zi mg cos φ =0 (2.241) i=1 n F zi x i + (a + mg sin φ) =0 (2.242) i=1 F zi k i x i x 1 x n x 1 µ Fzn F z 1 F z 1 =0 for i =2, 3,,n 1 k n k 1 k 1 (2.243) were F xi and F zi are te ongitudina and norma forces under te tires attaced to te axe number i, andx i is te distance of mass center C fromteaxenumberi. Tedistancex i is positive for axes in front of C, and is negative for te axes in back of C. Teparameterk i is te vertica stiffness of te suspension at axe i. Proof. For a mutipe-axe veice, te foowing equations Fx = ma (2.244) Fz =0 (2.245) My =0 (2.246) provide te same sort of equations as (2.126)-(2.128). However, if te tota number of axes are n, ten te individua forces can be substituted by a summation. n 2 F xi mg sin φ = ma (2.247) 2 i=1 n F zi mg cos φ =0 (2.248) i=1

37 2. Forward Veice Dynamics 75 z a 3 a 2 a 1 x C a 2F x1 2F x3 2F z3 2F x2 2F z2 φ mg 2F z1 FIGURE A tree-axe car moving on an incined road. n n 2 F zi x i +2 F xi =0 (2.249) i=1 Te overa forward force F x =2 P n i=1 F x i can be eiminated between Equations (2.247) and (2.249) to make Equation (2.242). Ten, tere remain two equations (2.241) and (2.242) for n unknowns F zi, i =1, 2,,n. Hence, we need n 2 extra equations to be abe to find te wee oads. Te extra equations come from te compatibiity among te suspensions defection. We ignore te tires compiance, and use z to indicate te static vertica dispacement of te car at C. Ten, if z i is te suspension defection at te center of axe i, andk i is te vertica stiffness of te suspension at axe i, te defections are z i = F z i. (2.250) k i For a fat road, and a rigid veice, we must ave i=1 z i z 1 x i x 1 = z n z 1 x n x 1 for i =2, 3,,n 1 (2.251) wic, after substituting wit (2.250), reduces to Equation (2.243). Te n 2 equations (2.251) aong wit te two equations (2.241) and (2.242) are enoug to cacuate te norma oad under eac tire. Te resutant set of equations is inear and may be arranged in a matrix form [A][] =[B] (2.252)

38 76 2. Forward Veice Dynamics were [] = F z1 F z2 F z3 F zn T (2.253) [A] = = x 1 x n x 1 2x 2 2x n x n x 2 1 x 2 x 1 k 1 k 2 k n x n x i 1 x i x 1 k 1 k i k n x n x n 1 1 x n 1 x 1 k 1 k n 1 k n (2.254) (2.255) [B] = mg cos φ (a + mg sin φ) 0 0 T. (2.256) Exampe 71 F Wee reactions for a tree-axe car. Figure 2.20 iustrates a tree-axe car moving on an incined road. We start counting te axes of a mutipe-axe veice from te front axe as axe-1, and move sequentiay to te back as sown in te figure. Te set of equations for te tree-axe car, as seen in Figure 2.20, is 2F x1 +2F x2 +2F x3 mg sin φ = ma (2.257) 2F z1 +2F z2 +2F z3 mg cos φ = 0 (2.258) 2F z1 x 1 +2F z2 x 2 +2F z3 x 3 +2 (F x1 + F x2 + F x3 ) = 0 (2.259) µ 1 Fz2 F µ z 1 1 Fz3 F z 1 = 0 (2.260) x 2 x 1 k 2 k 1 x 3 x 1 k 3 k 1 wic can be simpified to 2F z1 +2F z2 +2F z3 mg cos φ = 0 (2.261) 2F z1 x 1 +2F z2 x 2 +2F z3 x 3 + m (a + g sin φ) = 0 (2.262) (x 2 k 2 k 3 x 3 k 2 k 3 ) F z1 +(x 1 k 1 k 2 x 2 k 1 k 2 ) F z3 (x 1 k 1 k 3 x 3 k 1 k 3 ) F z2 = 0. (2.263) Te set of equations for wee oads is inear and may be rearranged in a matrix form [A][] =[B] (2.264)

39 2. Forward Veice Dynamics 77 were [A] = [] = x 1 2x 2 2x 3 k 2 k 3 (x 2 x 3 ) k 1 k 3 (x 3 x 1 ) k 1 k 2 (x 1 x 2 ) F z 1 F z2 (2.265) (2.266) [B] = F z3 mg cos φ m (a + g sin φ) 0. (2.267) Te unknown vector may be found using matrix inversion [] =[A] 1 [B]. (2.268) Te soution of te equations are 1 k 1 m F z 1 = Z 1 (2.269) Z 0 1 k 2 m F z 2 = Z 2 (2.270) Z 0 1 k 2 m F z 3 = Z 3 (2.271) Z 0 were, Z 0 = 4k 1 k 2 (x 1 x 2 ) 2 4k 2 k 3 (x 2 x 3 ) 2 4k 1 k 3 (x 3 x 1 ) 2 (2.272) Z 1 = g (x 2 k 2 x 1 k 3 x 1 k 2 + x 3 k 3 ) sin φ +a (x 2 k 2 x 1 k 3 x 1 k 2 + x 3 k 3 ) +g k 2 x 2 2 x 1 k 2 x 2 + k 3 x 2 3 x 1 k 3 x 3 cos φ (2.273) Z 2 = g (x 1 k 1 x 2 k 1 x 2 k 3 + x 3 k 3 ) sin φ +a (x 1 k 1 x 2 k 1 x 2 k 3 + x 3 k 3 ) +g k 1 x 2 1 x 2 k 1 x 1 + k 3 x 2 3 x 2 k 3 x 3 cos φ (2.274) Z 3 = g (x 1 k 1 + x 2 k 2 x 3 k 1 x 3 k 2 ) sin φ +a (x 1 k 1 + x 2 k 2 x 3 k 1 x 3 k 2 ) +g k 1 x 2 1 x 3 k 1 x 1 + k 2 x 2 2 x 3 k 2 x 2 cos φ (2.275) x 1 = a 1 (2.276) x 2 = a 2 (2.277) x 3 = a 3. (2.278)

40 78 2. Forward Veice Dynamics z a 2 a 1 x v a θ C 2F x1 2F z1 φ mg R H 2F x2 2F z2 FIGURE A cresting veice at a point were te i as a radius of curvature R. 2.8 F Veices on a Crest and Dip Wen a road as an outward or inward curvature, we ca te road is a crest or a dip. Te curvature can decrease or increase te norma forces under te wees F Veices on a Crest Moving on te convex curve of a i is caed cresting. Te norma force under te wees of a cresting veice is ess tan te force on a fat incined road wit te same sope, because of te deveoped centrifuga force mv 2 /R H in te z-direction. Figure 2.21 iustrates a cresting veice at te point on te i wit a radius of curvature R H. Te traction and norma forces under its tires are approximatey equa to F x1 + F x2 1 m (a + g sin φ) (2.279) 2 F z1 1 µ 2 mg a2 cos φ + sin φ 1 2 ma 1 2 m v2 a 2 R H (2.280)

41 F z2 1 2 mg µ a1 2. Forward Veice Dynamics 79 cos φ sin φ ma 1 2 m v2 a 1 R H (2.281) = a 1 + a 2. (2.282) Proof. For te cresting car sown in Figure 2.21, te norma and tangentia directions are equivaent to te z and x directions respectivey. Hence, te governing equation of motion for te car is Fx = ma (2.283) F z = m v2 R H (2.284) My =0. (2.285) Expanding tese equations produces te foowing equations: 2F x1 cos θ +2F x2 cos θ mg sin φ = ma (2.286) 2F z1 cos θ 2F z2 cos θ + mg cos φ = m v2 (2.287) R H 2F z1 a 1 cos θ 2F z2 a 2 cos θ +2(F x1 + F x2 ) cos θ +2F z1 a 1 sin θ 2F z2 a 2 sin θ 2(F x1 + F x2 ) sin θ =0. (2.288) We may eiminate (F x1 + F x2 ) between te first and tird equations, and sove for te tota traction force F x1 + F x2 and wee norma forces F z1, F z2. ma + mg sin φ F x1 + F x2 = 2cosθ F z1 = 1 µ 2 mg a2 (1 sin 2θ) cos φ + cos θ cos θ cos 2θ sin φ 1 (1 sin 2θ) ma 2 cos θ cos 2θ 1 2 m v2 a 2 R H cos θ F z2 = 1 µ 2 mg a1 (1 sin 2θ) cos φ cos θ cos θ cos 2θ sin φ (2.289) (2.290) + 1 (1 sin 2θ) ma 2 cos θ cos 2θ 1 2 m v2 a 1 R H cos θ cos θ (2.291) If te car s wee base is muc smaer tan te radius of curvature, R H, ten te sope ange θ is too sma, and we may use te foowing trigonometric approximations. cos θ cos 2θ 1 (2.292) sin θ sin 2θ 0 (2.293)

42 80 2. Forward Veice Dynamics Substituting tese approximations in Equations (2.289)-(2.291) produces te foowing approximate resuts: F x1 + F x2 1 m (a + g sin φ) (2.294) 2 F z1 1 µ 2 mg a2 cos φ + sin φ 1 2 ma 1 2 m v2 a 2 R H F z2 1 µ 2 mg a1 cos φ sin φ ma 1 2 m v2 a 1 R H (2.295) (2.296) Exampe 72 F Wee oads of a cresting car. Consider a car wit te foowing specifications: = 2272 mm w = 1457 mm m = 1500 kg = 230 mm (2.297) a 1 = a 2 v = 15m/ s a = 1m/ s 2 wic is cresting a i at a point were te road as Te force information on te car is: R H = 40m φ = 30deg (2.298) θ = 2.5deg. F x1 + F x2 = N F z1 = N F z2 = N mg = N (2.299) F z1 + F z2 = N m v2 R H = N

43 2. Forward Veice Dynamics 81 If we simpifying te resuts by assuming sma θ, te approximate vaues of te forces are F x1 + F x2 = N F z N F z N mg = N (2.300) F z1 + F z N m v2 R H = N. Exampe 73 F Losing te road contact in a crest. Wen a car goes too fast, it can ose its road contact. Suc a car is caed a fying car. Te condition to ave a fying car is F z1 =0and F z2 =0. Assuming a symmetric car a 1 = a 2 = /2 wit no acceeration, and using te approximate Equations (2.280) and (2.281) 1 2 mg 1 2 mg µ a2 µ a1 cos φ + sin φ 1 2 m v2 a 2 R H cos φ sin φ 1 2 m v2 a 1 R H = 0 (2.301) = 0 (2.302) we can find te critica minimum speed v c to start fying. Tere are two critica speeds v c1 and v c2 for osing te contact of te front and rear wees respectivey. v c1 = v c2 = s2gr H µ sin φ + 12 cos φ s 2gR H µ sin φ 12 cos φ (2.303) (2.304) For any car, te critica speeds v c1 and v c2 arefunctionsoftei s radius of curvature R H and te anguar position on te i, indicated by φ. Te ange φ cannot be out of te titing anges given by Equation (2.141). a 1 tan φ a 2 (2.305) Figure 2.22 iustrates a cresting car over a circuar i, and Figure 2.23 depicts te critica speeds v c1 and v c2 at a different ange φ for rad

44 82 2. Forward Veice Dynamics φ R H φ FIGURE A cresting car over a circuar i. φ rad. Te specifications of te car and te i are: = 2272 mm = 230 mm a 1 = a 2 a = 0m/ s 2 R H = 100 m. At te maximum upi sope φ =1.371 rad 78.5deg, te front wees can eave te ground at zero speed wie te rear wees are on te ground. Wen te car moves over te i and reaces te maximum downi sope φ = rad 78.5deg te rear wees can eave te ground at zero speed wie te front wees are on te ground. As ong as te car is moving upi, te front wees can eave te ground at a ower speed wie going downi te rear wees eave te ground at a ower speed. Hence, at eac sope ange φ te ower curve determines te critica speed v c. To ave a genera image of te critica speed, we may pot te ower vaues of v c as a function of φ using R H or / as a parameter. Figure 2.24 sows te effect of i radius of curvature R H on te critica speed v c foracarwit/ = mm/ mm and Figure 2.25 sows te effect of a car s ig factor / on te critica speed v c for a circuar i wit R H = 100 m F Veices on a Dip Moving on te concave curve of a i is caed dipping. Te norma force under te wees of a dipping veice is more tan te force on a fat incined road wit te same sope, because of te deveoped centrifuga

45 2. Forward Veice Dynamics 83 v c [m/s] v c v c2 =2.272 m =0.23 m a 1 =a 2 R H =40 m φ [rad] φ [deg] FIGURE Critica speeds v c1 and v c2 at different ange φ for a specific car and i. R H =1000 m 500 m 200 m 100 m 40 m v c [m/s] /= a 1 =a φ φ [rad] [deg] FIGURE Effect of i radius of curvature R on te critica speed v c for a car.

46 84 2. Forward Veice Dynamics /= v c [m/s] R H =100 m a 1 =a φ φ [rad] [deg] FIGURE Effect of a car s eigt factor / on te critica speed v c for a circuar i. force mv 2 /R H in te z-direction. Figure 2.26 iustrates a dipping veice at a point were te i as a radius of curvature R H. Te traction and norma forces under te tires of te veice are approximatey equa to F x1 + F x2 1 m (a + g sin φ) (2.306) 2 F z1 1 µ 2 mg a2 cos φ + sin φ 1 2 ma m v2 a 2 R H F z2 1 µ 2 mg a1 cos φ sin φ (2.307) ma m v2 a 1 R H (2.308) = a 1 + a 2. (2.309) Proof. To deveop te equations for te traction and norma forces under te tires of a dipping car, we foow te same procedure as a cresting car. Te norma and tangentia directions of a dipping car, sown in Figure 2.21, are equivaent to te z and x directions respectivey. Hence, te governing

47 2. Forward Veice Dynamics 85 z a 2 a 1 R H x v C a θ 2F x1 2F z1 φ mg 2F x2 2F z2 FIGURE A dipping veice at a point were te i as a radius of curvature R. equations of motion for te car are Fx = ma (2.310) Fz = m v2 R H (2.311) My =0. (2.312) Expanding tese equations produces te foowing equations: 2F x1 cos θ +2F x2 cos θ mg sin φ = ma (2.313) 2F z1 cos θ 2F z2 cos θ + mg cos φ = m v2 (2.314) R H 2F z1 a 1 cos θ 2F z2 a 2 cos θ +2(F x1 + F x2 ) cos θ +2F z1 a 1 sin θ 2F z2 a 2 sin θ 2(F x1 + F x2 ) sin θ =0. (2.315) Te tota traction force (F x1 + F x2 ) may be eiminated between te first and tird equations. Ten, te resutant equations provide te foowing forces for te tota traction force F x1 + F x2 and wee norma forces F z1, F z2 : F x1 + F x2 = ma + mg sin φ 2cosθ (2.316)