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Chapter 3 Braking Performance

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Chapter 3 Braking Performance CHAPTER 3 BRAKING PERFORMANCE - I ·7.-:_.. ABS test drive. (Photo courtesy ofRobert Bosch GmbH.) BASIC EQUATIONS The general equation for braking performance may be obtained from Newton's Second Law written for the x-direction. The forces on the vehicle are generally of the type shown in Figure 1.6. Then, NSL is: (3-1) where: W =Vehicle weight g =Gravitational acceleration Dx =- ax =Linear deceleration Fxf =Front axle braking force Fxr = Rear axle braking force J)1\ = A 'rodynarrli drag (.) I Jphill gradl· ( 'II!\I' I'l '.I{ \ BI{A K IN ' I'I ~ RI'O J{M/\N\I ;. H IN I ' A MI ' N'I I ,' ~ I II' VI ' I III 'I 1 I ' N!\MII ',: TIIl',IJOIlI a,lId (l'll l hfuk illf ' 1111 ' '( 'rillS ariSl: Irllfllih 'lll('(J"l: llflh ' brak 'S where: : 1 ~ llIl~ Wllllllllhllg rt.:si~ lall ," 'rr ' 'Is, h ';trill' I'd 'lion, and uriv 'lin' drags, A x =Distance traveled during the deceleration IIlllpr,'" 'IISIV ' :I1l ;dYSIS nllh' d <..:c ' Ieralion r<": l(uir 'S detai led knowledge of all Ill!'s ' l or' 'S a 'ling un Ihe vt.: hi ck, In the case where the deceleration is a full stop, then Vf is zero, and X is the stopping distance, SO. Then: ( 'cHlshml Deceleration y2 y2 SD=_o-= - ~- (3-6) Silllpi 'and fundamental relationships can be derived for the case where it 2 Fxt 2 Dx I/'> 1l' ,I SOII ;,lbl ' to assume that the forces acting on the vehicle will be constant M 1I111111 1'. ~1 ( 1~11 a ~rake application. The simple equations that result provide an ,I PI'"' 'IOtJolllor the basic relationships that govern braking maneuvers. From and the time to stop is: 1','1 , (J- I ): Yo Yo (3-7) _ F xt _ dY t =--=- x - - -- - (3-2) s Fxt Dx M dt M w h l'lI': Thus, all other things being equal, the time to stop is proportional to the I'xt - Th 'total ofall longitudinal deceleration forces on the vehicle (+) velocity, whereas the distance is proportional to the velocity squared (i.e., V • r orward velocity doubling the velocity doubles the time to stop. but quadruples the distance required). Tid, l 'qll : lt~ ~ 11 can be ,integrated (because Fxt is constant) for a deceleration ( -,1111\1 ) 11'11\\1 1I1llial velOCity , V0' to final velocity, V( Deceleration with Wind Resistance The aerodynamic drag on a vehic1e is dependent on vehicle drag factors (3-3) and the square of the speed. To determine stopping distance in such cases, a more complicated expression is necessary but can still be integrated. To analyze this case: F (3-8) V - Yr= xtts (3-4) o M' where: IS - Till1l' for the velocity change Fb =Total brake force of front and rear wheels C = Aerodynamic drag factor .. \I ~ : :IIl S ' v 10 i(~ and distance are relaled by V =dx/dl, we can substitute Therefore: 11:1 .II II I J',q, (3-2), tnt 'grat " and obtai nih' r 'lationshi p bel ween v<..:1 (lei ty and d H, l i llll'l': S]) YdY (3-9) dx = M JO o (. -. I Yo I'IINI )AMI' N'I AI ,' (,I , VI 'III( 'I I I >YNAMJ( ',' ('IIAI ' J'b l{ J II1<AKI N i I'I':J{I'ORMAN ·t This llIay h...: illt . 'r:lt .J to obtaill til' stoppin ' distallC ': The.: pararnef.l!J' "1', ," is the rolling resistance coefficient, which will be 2 discussed in the nexl chapter. Note that the total force is independent of the . M Fb+ Vo n · IO) distribution of loads on the axles (static or dynamic): Rolling resis~ance forces SI) = 2 C In I Fb J are nominally equivalent to about 0.01 g deceleratIOn (0.3 ftlsec ). Em'r'gy/Power Th ' c'lll'rgy and/or power absorbed by a brake system can be substantial Aerodynamic Drag t h l! illg a Iypical maximum-effort stop. The energy absorbed is the kinetic The drag from air resistance depends on the dynamic pressure, and is thus l I! l' II 'Y or motion for the vehicle, and is thus dependent on the mass. proportional to the square of the speed, At low speeds it is negligible. At 2 (3-11 ) normal highway speeds, it may contribute a force equivalent to about 0.03 g (1 I ':n(~rgy =~ (V0 - V f1 ftlsec2). More discussion of this topic is presented in the next chapter. Til ' power absorption will vary with the speed, being equivalent to the hl':lk ill ' force times the speed atany instant oftime. Thus, the powerdissipation is ~ n' at cs t at.lhe beginning ofthe stop when the speed is highest. Overthe entire Driveline Drag ,'.llIP. III ' average power absorption will be the energy divided by the time to '. llI)" Thlls: The engine, transmission, and final drive contribute both drag and inertia V 2 effects to the braking action. As discussed in the previous chapter on Power = M ~ (3-12) Accel erati on Performance, the inertia ofthese components adds to the effecti ve 2 ts mass of the vehicle, and warrants consideration in brake sizing on the drive wheels, The drag arises from bearing and gear friction in the transmission and Cal '1I1atioJl o rthe power is informative from the standpoint ofappreciating I hI' IlI'l J'ol'lllan<.;e required from a brake system. A 3000 Ib car in a maximum­ differential, and engine braking. Engine braking is equivalent to the "motor­ ("111111 Slt)P frolll HO mph requires absorption of nearly 650,000 ft-Ib of energy. ing" torque (observed on a dynamometer) arising from internal ~riction a~d air II st()ppl!d in 8 seconds (10 mph/sec), the average power absorption of the pumping losses. (It is worth noting that the pumping losses disappear If ~he brak 'S <.luring this interval is 145 HP. An 80,000 Ib truck stopped from 60 mph engine is driven to a speed high enough to float the valves. Thus, engme Iypi 'ally involves dissipation at an average rate of several thousands of braking disappears when an engine over-revs excessively. This can be a Iu lI'scpower! serious problem on low-speed truck engines where val ve float may occurabove 4000 rpm, and has been the cause of runaway accidents on long grades,) On a manual transmission with clutch engaged during braking, the engine braking BI{AKING FORCES is multiplied by the gear ratio selected. Torque-converter transmission.s are designed for power transfer from the engine to the driveline, but are relat~vely The forces on a vehicle producing a given braking deceleration may arise ineffective in the reverse direction; hence, engine drag does not contrIbute Ifllll1 a number of sources. Though the brakes are the primmy source, others substantially to braking on vehicles so equipped. will he discussed first. Whether or not driveline drag aids in braking depends on the rate of deceleration. If the vehicle is slowing down faster than the driveline compo­ I{ulling Ucsistance nents would slow down under their own friction, the drive wheel brakes must Rollin' r 'sistan " alw:IYs OPpOSl'S vdli ' 1' 1I1(ltioll ; II n ' -" it aids t.he pick up the extra load of decelerating the drivelin~ during t.he ?raking IlI:lk ' ,~ , The rollilll' Il'Si , I:II H'l' (()I ('l' ~, wi ll Iw : maneuver. On the other hand, during low-level deceleratIOns thedrIvelme drag may h' sufficient to d celcrate the rotating driveline components and contrib­ ( \ I ) IIll' In tIll' hr:lkil1l', '('('orl nn th 'driv' wh eoel s as well. I ~ ,I') ( ~l"Ud(' Hrak(' Factor Road 'fad' wi II l:ontri bute uin;cLly tu the braking <.:Ilort, either in a positi ve Brake factor is a mechanical advantage that can be utilized in drum brakes ~ , t.: I1 ~ · (uphill) OJ' negative (downhill), Grade is defined as the rise over the run to minimize the actuation effort required. The mechanism of a common drum (WII i 'al over horizontal distance), The additional force on the vehicle arising brake is shown in simplified form in Figure 3.2, The brake consists oftwo shoes pivoted at the bottom. The application of an actuation force, P a'. push~s the If IIII! t~r~ld " Rg, is given hy: lining against the drum generating a friction force whose magnItude IS the R, = Wsin8 (3-14) normal load times the coefficient of friction (fl) of the lining material against I'll slJIall angles typical of most grades: the drum. Taking moments about the pivot point for shoe A: (3-15) - (radians) == Grade =Rise/run L Mp =ePa + n fl NA - m N A =0 R.., =W sin e == W 8 where: e =Perpendicular distance from actuation force to pivot '1'1111,-; a grade or 4% (0,04) will be equivalent to a deceleration of ± 0.04 g N =Normal force between lining A and drum ( I ,. fI/s c2). n A =Perpendicular distance from lining friction force to pivot m =Perpendicular distance from the normal force to the pivot UI{AKES The friction force developed by each brake shoe is: I\utofilotiv brakes in common usage today are of two types-drum and and \11 I ' II. , . I as shown in Figure 3.1. Then equation (3-15) can be manipulated to obtain: F A _ ~ e and F B _ ~ e (3-16) 'tII) '- , ,,~&,,~ Pa - (m - ~ n) Pa - (m + ~ n) - ... ~ I ,,'I I. e ~,'/: \' '/ ~' ~ .... ­ IlN /' !1 , I, I L rum. brake and disc hrake, (Photo courtesy ofChry,\'[er Corp.) + e Ilisiori 'ally, drum hrak 's have s n common usage in the U.S .
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