《汽车理论》课程教学课件（英文讲稿）第4章 汽车制动性 Braking Performance 4.3 Braking Efficiency and its Stability（制动效能及其恒定性）

4-3BrakingEfficiency1. Braking DecelerationCompared with braking force when brakes work, T, Fw,T,couldbeignoredFw(ignored)Fdu+F=Fm-Xb1Xb2Xbdth8GdugOFXbGdtFxb2baXb1FZ2LF

4-3 Braking Efficiency 1. Braking Deceleration  Compared with braking force when brakes work, could be ignored. Tj Fw Tf , , Xb Xb Xb 1 2 Xb du m F F F dt du g F dt G     FZ1 FZ2 FXb1 FX 2b G hg F ignored W ( ) Fj L a b

4-3BrakingEfficiency1. Braking DecelerationBecause ofFxbmax=F,=Foi+F00=Fz1Φ+Fz2=(Fz1+Fz2)p=GpdugGo=gdtGmaxdu=PsgForgeneralbrakesdtmaxForABSdu=PpgdtmaxABSoperation is used to stayatthepeak braking coefficient.(holding thewheelnearpeakslipconditions

4-3 Braking Efficiency 1. Braking Deceleration Because of X max 1 2 1 2 1 2 max ( ) b Z Z Z Z F F F F F F F F G du g G g dt G                            max S du g dt         ABS operation is used to stay at the peak braking coefficient. (holding the wheel near peak slip conditions) For ABS For general brakes max P du g dt        

2.Braking ProcessFpFp,ababFPFp,abab1a0211LT2TOamplify

3 amplify 2. Braking Process

FpFp,ababgdTh0t2tiit2ToDriver'sBrakesBrakes stableReleasingresponse timeacting timeworking timebrakes timeBraking Process

4 Brakes acting time Driver's response time Brakes stable working time Releasing brakes time Braking Process

2.BrakingProcess(1) The process of driver's responseTComprising ofthe time for the driver to be aware of the braking actionT1Ti-- the time for the driver to move the stepUsually t,=0.9 sec(it has nothing to do with the vehicle)(2) The process of brakes acting-t2Comprising ofT- the time for the brakes to pass through the brakes clearanceTt2" the time for the brakes to gain decelerationUsually t2 <l sec( the period of the brakes from beginning to work to its stableworking condition)

2. Braking Process （1）The process of driver's response—— Comprising of —— the time for the driver to be aware of the braking action —— the time for the driver to move the step Usually =0.9 sec （it has nothing to do with the vehicle） ' 1  1  1   1  （2）The process of brakes acting—— Comprising of —— the time for the brakes to pass through the brakes clearance —— the time for the brakes to gain deceleration Usually <1 sec ( the period of the brakes from beginning to work to its stable working condition) 2 '  2   2  2 

2.BrakingProcess(3) The process of brakes stable working condition1> usually that is the process of vehicle with lock condition.（抱死滑拖时间） It depends on the initial speed value(4)Theprocessof releasingbrakesT4T is the time used to release itself by the brakes. Usually T <0.8secpedal forceduldttouch thegairpedalduldtsth wronglockbeginto1t2t2+1:t.,t2t1t3±4

2.Braking Process （3）The process of brakes stable working condition——  usually that is the process of vehicle with lock condition. （抱死滑拖时间）  It depends on the initial speed value. （4）The process of releasing brakes—— is the time used to release itself by the brakes. Usually <0.8sec 3  4  4  4 

3.StoppingDistance-(1) Definition: The distancewhich vehicle passes throughdu/dtifrom the moment of brakesapplication to the moment ofvehicle's stop.( the distance of t2 and t3)2t2t3t2

3.Stopping Distance——S （1）Definition: The distance which vehicle passes through from the moment of brakes’ application to the moment of vehicle’s stop. ( the distance of τ2 and τ3 )

(2) Calculationk =-Jmaxs’(t2') : s2' = Uot2T2dukt={du=[ktdt(T2)S2dtI"2kt2=u=uo +kt-2一u=uo+2ds1kt-一=uo2dtHC[ds=[(u +↓kt')dtmaxS=uot+=ktu.t6g12+-kt.T1s, =uot,12Amax66

" " 2 2 2 "2 0 0 ( ) : 1 1 2 2 e du s k du k d dt u u k u u k                  1  ' 1  " 1  2  a 4  " 2  ' 2  Fp j t Fp b c d e f g 0 3  j 2 2 2 0 2 s s u     ( ) :    2 0 2 0 3 3 max 0 0 " 2 " " "3 " "2 2 0 2 2 0 2 max 2 1 2 1 ( ) 2 1 1 6 6 1 1 6 6 ds u u k dt ds u k d j s u k u s u k u j                           ＝ － ＝ max " 2 j k    （2） Calculation

s2(t2) : S2 = $, +s, = uot + uotz -=max T,S,(t,) : U frs = Ue, U imnal = O, j= jmaxUot2 + Jmaxt?u.uS3822j.2j.maxmaxu.=uomaxs= s, +s,(Stopping Distance)u.Imax T2+)uos=(t, +2 jmax24t2 →0, u.=ugo/3.6,aoS3.625.92 j7max

' " ' " "2 2 2 2 2 2 0 2 0 2 max 2 3 3 max 2 2 " "2 0 0 2 max 2 3 max max " 2 0 max 2 1 ( ) : 6 ( ) : , 0 2 2 2 8 1 ( ) 2 first e final e e s s s s u u j s u u u j j u u u j s j j u u j                       ， 2 3 " 2 "2 ' 0 max 2 " 2 ' 0 2 2 0 ma 2 2 0 ma " 2 0 x x 2 0 ( ) 2 2 24 0 3.6 1 ( ) 3.6 2 25.92 a a a u s s s u j s u j u u s u j                  （Stopping Distance） ， = /

WaoWaoTST23.625.92 jmax=@gJmaxWaoWaoS73.62254.Φempirical formula:S= Axuao +Bxuao A and B depends on the road condition and the type of vehicle,A and B change with the change of these two conditions.(3)influencingfactoruao ; t2' (adjustment) ; t," (brakes design)> Decreasing t2'is the most effective method to decrease thestopping distance

empirical formula: 2 a0 a0 S  Au  Bu '' 2 0 0 ' 2 2 max max '' 2 0 0 ' 2 2 ( ) 3.6 2 25.92 ( ) 3.6 2 254 a a a a u u S j j g u u S               (3) influencing factor ua0 ; t2 ’ (adjustment) ; t2 ’’ (brakes design)  Decreasing t2’’ is the most effective method to decrease the stopping distance.  A and B depends on the road condition and the type of vehicle, A and B change with the change of these two conditions