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TRACTION-LIMITED ACCELERATION Transverse Weight Shift Due To The sam' 111 'Ihod is liS d to '"kulal!; perforrnan(.:(' in high gear as well: \) 'JIll' 1111 :11 ;1111,11 llll ' II II I, ill IIl1 ilt, III ill lit HI 1.2. is '1IJlV ""I'd illto transla­ d(IJwl ill("ll l I (ll H\ xs ) wlll'fI d lvid 'd II I in Fq, (2- 1) ). W '<;all S" its ma 'nitude Fx = Te Ntf'Tltf"r li S IlIlIows: = 20 I ft-Ib (1 .0 x 2.92) (0.99 x 0.97)/12.59 in x 12 in/ft MMf =1'ff /r- =360.2 in-lb-st.:c2 I 12.5 92 in 2 = 2.27 Ih-scc2lin = 537 Ib I' 'rhaps Iht' ilion:: familiar form is the effective weight: ffiw= we I(Nt Nf) = 4400 rev/min . 2 1t rad/rev . I minl60 sec I( 1.0 x 2.92) W t; ff = M e!"/" g = 2.27 Ib-sec2lin x 386 in/sec2 = 877 Ib = 157.8 rad/sec 'omparing this figure to the weight of a typical passenger car (2500 Ib), Vx = Ww' r= 157.8 radlsec x 12.59 in= 1987 in/sec = 165 ftlsec = 1 13 mph W I: S" that it adds about 35% to the effective weight of the car during :t ' , 'I 'ral ion in first gear. The inertia of the non-driven wheels will add another 27 Ib to 111(' effecti ve weight (I %). TRACTION-LIMITED ACCELERATION 2) akulat~ the maximum tractive effort and corresponding road speed in fi l'sl and fifth gears of the car described above when inertial losses are Presuming there is adequate power from the engine, the acceleration may 11 ' gl 'cted. be Iimited by the coefficient of friction between the tire and road. In that case Fx is limited by: Sulution: (2-13) Maximum tractive effort will coincide with maximum torque, which (I(' 'urs at 4400 rpm. So the problem reduces to finding the tractive effort from where: till' lirst term in Eq. (2-9) for that value of torque. Il = Peak coefficient of friction Fx= Tc Ntf'Tltf/r W = Weight on drive wheels = 20 J ft-Ib (4.28 x 2.92) (0.966 x 0.99)112.59 in x 12 in/ft The weight on a drive wheel then depends on the static plus the dynamic = 2290 Ib load due to acceleration, and on any transverse shift ofload due to drive torque. TIll.' road speed is determined by use of the relationships given in Eq. X). Although the equation is written in terms of acceleration, the same Il'laliollships hold true for speed. That is: Transverse Weight Shift due to Drive Torque I"J\J = Nf Ww and we = Nt wd = Nt Nf Ww (2-8a) Transverse weight shift occurs on all solid drive axles, whether on the front or rear of the vehicle. The basic reactions on a rear axle are shown in Figure 'I'll, wheel rotational speed will be: 2.10. The drives haft into the differential imposes a torque T d on the axle. As will be seen, the chassis may roll compressing and extending springs on (.ow =we/(Nt Nf)=4400rev/min ' 21trad/rev' I minl60sec/(4.28 x 2.92) opposite sides ofthe vehicle such that a torque due to suspension roll stiffness, = 36.87 rad/sec T s' is produced. Any difference between these two must be absorbed as a 'rhe corresponding ground speed will be found by con verti ng the rotational difference in weight on the two wheels. If the axle is of the non-locking type, spc"d to translational speed at the circumference of the tire. then the torque delivered to both wheels will be limited by the traction limit on the most lightly loaded wheel. V x =ffiw' r=36.87 rad/secx 12.59 in =464.2 inlsec=38.7 ftlsec =26.4 mph 34 Tsf =K<1>f <1> (2-16a) Tsr = K<1>r <1> (2-16b) K<1> = K<1>f+ K<1>r (2-16c) where: Tsf = Roll torque on the front suspension T sr = Roll torque on the rear suspension W -L + W K<1>f = Front suspension roll stiffness 2 y K<1>r = Rear suspension roll stiffness Fig . 2. 10. j;'ret>-body diagram oja solid drive axle, K<1> = Total roll stiffness Writ in g NSL for rotation of the axle about its centerpoint allows the 1t' 1I ·tiolls to be related. When the axle is in equilibrium: L To = (W12 + W Y- W12 + W y) tl2 + Ts - T d = 0 (2-14) I II" W Y =(Td - T s)/ t III th ' above equation, Td can be related to the drive forces because: ~. y (2-15) wh'r': Fx =Total drive force from the two rear wheels r = Tire radius Nr =Final drive ratio 11owcver, it is necessary to determine the roll torque produced by the :,IISP 'nsion, which requires an analysis of the whole vehicle because the H·lt ,tion of the drive torque on the chassis attempts to roll the chassis on both Iht' fro nt and rear su spensions. The entire system of interest is illustrated in 1; i,'.1l 1' ' 2.1 I for the case of a rear-wheel-drive car. Th ' drive torque react ion at the engine/transmission is transferred to the IraiIK' and distrihut (I bctwe 'n the front and rem suspensions. It is generally asH lIlIll'd that til' roll torqu ' produ . .t\ hy a SII SP 'I1SiOIl is proportional (() roll all/'.k Ilollkl" S I.aw) III' tht' ,,·hassis. Till-II: Fig. 2. J J Diagram ojdrive torque reactions on the chassis. It. 7 Now 'I' ~ r (' III Ill' 11'1 11 1, '.1 tn III ' IIdl :111/',1' , IIllJ 111'11111 all ',I' 'all h ' I' 'Ial -II 1111111' driw l!lIqllt' ;t,o.; foll()ws, 'I'll 'n,11 all ,I ,is sinlply Ih l: driv 'lor'll! - divided Traction Limits I, /III' IlIl:d roll slilTII 'ss: Solving {'or Fx gives the final expression for the maximum tractive force (I) ='I'd IK</> =1'<1 I(K(prf- K</>r) (2-17) that can be developed by a solid rear axle with a non-locking differential: 'lit '11' 1'11 1'1' , ,~llhslill!lin g in E l), (2- 16b): fl Wb F - L (2-23) 'l'sr = K(pr T d I(K(pf + K</>r) xmax - h 2 fl r K<j>f l--fl+---­ , ' ~ ' h i s ill ,llIl'lI ~<.In be substituted into Eq, (2-14), along with the expression L Nft Kq> 1.1/ 1.1 ohl:1I11 'J from Ell, (2-15): For a solid rear axle with a locking differential, additional tractive force can F r K<j>r (2- 18a) Wy = x I 1- - 1 be obtained from the other wheel up to its traction limits such that the last term Nfl K<j>r+-K<j>f in the denominator of the above equation drops out. This would also be true 'I'll(' I 'nil in the brackets collapses to yield: in the case of an independent rear suspension because the driveline torque reaction is picked up by the chassis-mounted differential. In both ofthese cases W _ Fx r K~)r (2-18b) the expression for the maximum tractive force is: y - Nrl K<j> fl Wb Thist:qllalioll gi ves the magnitude of the lateral load transfer as a function F - L (2-24) III IIw Ir' ~ 1 'I iv " f'or(.; e and a number of vehicle parameters such as the final drive xmax - h I --fl ,.11 II', Ir('ad uflk axl e, lire radius, and suspension roll stiffnesses, The net load L 1'/1 III I.' r ';u',axl during acceleration will be its static plus its dynamic compo­ 111" 111 (s ' , Jo.q, ( 1-7», For a rear axle: FinaJJy, in the case ofa front axle, the fore/aft load transfer is opposite from the rear axle case, Since the load transfer is reflected in the second term of the (2-19) denominator, the opposite direction yields a sign change. Also, the term "W blL" arose in the earlier equations to represent the static load on the rear drive axle, For a front-wheel-drive vehicle the term becomes "W clL." For the solid ,Nl'/;I - ,ling the rolling resistance and aerodynamic drag forces, the accel­ front drive axle with non-locking differential: (" .1 111111 IS simply (he t.ractive force divided by the vehicle mass, (2-20) fl We F xmax .....,;L=- ____ =_ ___ (2-25) h 2 fl r Kq>r I +-fl+----­ 'l'IIl'IIII1' w -i ghl on the rig ht. rear wheel, WIT' will be W/2 - W , or: y L Nft K<j> (2-21 ) F x h And for the solid front drive axle with locking differential, or the indepen­ 2 I dent front drive axle as typical of most front-wheel-drive cars today: IIIIII (2-22) /I WI' II ('; h (2-26) , I. 'xmax = IH 1'1 H INIlI\MINJ' I , ' ( II ' VI 'IIII 'I 1, 1I 't'N Mil ': . EXAMPI ,E PROBLEMS Wilh a lOCking Jifferential the thinj term in the dl;nominator disappears CEq. (2 24» so that we obtain : I) JoiIlU Ih' Ira ·tion-lilllilt;U C1cccll; ratio/l fOf (he fl;ur-drivc passl; nger car wi lllll llU wilhoul a 10 kin J differential on a surface of rnoueralt' friction level. F _ (0,62) 1850 Ib ' ('it' ill flll'llIalioll thai will he needed is as follows: 11471b = 1147 = 1305 lb xmax - 21 1-0.121 0.879 1 - 108 0.62 Wl'il\ hl.~ Front ­ 21()) Ib Rear - 1850 Ib Total ­ 3950 Ib 'G h 'i ·hl 21,0 in Wheelbase - 108 in a -~ xmax = 1305 lb =0 330 g' s = 10 64 ~ x- Mg 39501b' .
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