汽车理论武汉理工课件

1、4-5Braking Proportioning1.Vertical forces given by the ground when vehicle brakesFZ1 and FZ2FW (ignored)FjhgGaFXb2bFXb1Fwhere z Braking IntensityLZ2FZ1du = zg dtìF= G (b + zh )ï Z1LgíïF= g (a - zh )ïî Z 2Lg Balancing the brake outputs on both the front and rear axles is

2、 achieved by “proportioning” the pressure appropriately for the foundation brakes installed on the vehicle. It is well recognized that the preferred design is to bring both axles up to the lockup point simultaneously.Vertical forces given by the ground when vehicle brakesFZ1 and FZ2FW (ignored)FjhgG

3、aFbFXb1Xb2FZ2LFZ11.Vertical forces given by the ground when vehicle brakesFZ1 and FZ2if front lockup and rear lockup occur at the same time or one after anotherFW (ignored)FjhgGaFbFXb1Xb2FZ2LFZ1ì F= G (b + jh )du = jïZ1LggídtïF= G (a -jh )ïî Z 2Lg2. Ideal proportioning

4、curve”I” curve" I "curveFm 2Fm1Definition: A special curve which present the relationship between front and rear brakes braking force determined by the pressure applied to each brake and the gain of when front and rear lockupsoccur together.How to get the “I” curve?If front and rear brakes

5、 lockups together bothF1 and F2 could meet the demands of these equations.= GjìF+ Fm1m 2= GjFm+ Fmìï12= FZ1jïíFm1ïFí = Fjïîa - jhgFFZ 2m 2Z 2îm 2substitution F= 1 é mgb2 + 4hg L F- æ mgb + 2Föùm 22 ê hGm1çhm1 ÷ú

6、;ëgègøúûSteps:1. From the first equation(1)we can get a group of lines which and the same slope of 45 degree, each line presents the different friction coefficient.2. From the second equation(2)we can get a group of lines starting at the origin and extending upward and to th

7、e right, each line presents the different friction coefficient.3. Finding out the cross points of the first group lines and the second group lines with the same coefficient.4. Connecting the cross points mentioned above all together, we can get the ideal ”I” curve.ØDifferent point of the curve

8、has its own friction coefficient and the corresponding F1and F2 .0.4Fm0.520.60.7j0.8j0.80.6 Fm13. Real proportioning linewhere braking proportioning coefficient（制动分配系数）Fm 2= 1 -1Fm1bb = Fm1=Fm1FmFm1 + Fm 2Ø Generally the traditional vehicle has a fixed, constant, and

9、 straight proportioning line.4. Synchro-adhesion coefficient 0（同步附着系数）How to get 0?（1） Method One Graph Method（2） Method TwoAnalytic ApproachQFm 2a -j0hgfor "b "Fm1= bFm 21- b b + j0hg= bÞ j= Lb - b a -j h1- b0h0ggDefinition : The friction coefficient of the cross point between “I” cu

10、rve and “” line. Synchro frictioncoefficient j0Fm 2bI (fullload)j0 fullI (empty load)j0emptyFm1Note5.Front and rear lockup line clusters f 线组 & r线组FjhgGaFXb2bFXb1FZ2LFZ1How to get “f” ?QGz = G du = F+ Fg dtXb1Xb 2 L F= Gb + F× h+ F× hjXb1Xb1gXb 2gÞ F= L -jhgF- GbXb2jhXb1hggthe rel

11、ationship between FXb1 and FXb22)To Plot the relationship on a graphMaking each “f” line with a different “”FXb2j = 0.40.5FXb1(0, - Gb )hg（2）”r” line cluster2)To Plot the relationship on a graphFXb20.5j = 0.1 0.2j = 0.1(Ga , 0) FXb1hg(0, - Gb )hg“r” clusters（3）To combine “

12、f” graph with “r” graphF2“I” curveFXb20.5j = 0.1 0.2j = 0.1F1(Ga , 0) FXb1hg(0, - Gb )hg“r” clustersØ Connecting the cross point of “r” and “f” with the same “” together makes the ideal proportioning curve ”I” curve.”f” clusters1. There are two means to get “I” curve(1) by the

13、 relationship between F1and F2(2)by the relationship betweenFXb1 and FXb2“f” and “r” cross points ”I” curveUsually the first is the preferred one to get “I” curve.ìFxb1 = 0ìF= Ga(1) ïGaj(2) ï xb1híF=ígï xb2L + jhï F= 0îgîxb2ìFm1 + Fm 2 = Gjï

14、;Fb + jhí m1 =g ïFm 2a - jhgî2. Key Definitions” I ” curve:The idea brake forces proportioning curve inthe sense of front and rear lockup occur together.“” line:The real brake forces proportioning line in thesense of the forces applied on the wheels.“0” :The friction coefficient of th

15、e cross point between “I”line and “” line.“f” line: “f” line describes the relationship between front and rear brake forces given by the ground (FXb2FXb1) when front lockup occurs first.“r” line: “r” line describes the relationship between front and rear brake forces given by the ground (FXb2FXb1) w

16、hen rear lockup occurs first.3. Analyzing of “f” clusters and “r” clusters1） “f” clustersthe intersection of “f ” clusters with x-coordinate2） “r” clusters6. To analyze the braking process under different “”conditions.æ j = 0.3 öjjçj = 0.5÷(1)0è0øFm 2FXb 2b lineB¢r

17、(j = 0.7)BI curveK2j = 0.7Kj = 0.3K1j0= 0.5A¢Af (j = 0.3)FXb1Fm1Fm 2FXb 2b lineB¢r(j = 0.7)BI curveK2j = 0.7Kj = 0.3K1j0= 0.5A¢Af (j = 0.3)FXb1Fm1At the point of “A” :Ø Both F1 ,F2and FXb1, FXb2reach this point at the same time ,front lockup begins to occur.Ø F1= FXb1F2= FXb

18、2 After “A” point:;Ø F1 and F2 change along the” ” line continuouslyØ FXb1and FXb2 change along the “f” line of “” which is equal to 0.3.Ø F1FXb1;F2= FXb2 At the point of “ k1” :Ø rear lockup begins to occur .Ø It is a stable situation in which front and rear lockup exist to

19、gether.Ø It is different from the situation in which front and rear lockup occur together.(front rear)ØVehicle lost the steering ability.ØØdu/dt reaches itsAt the point of “A ” :um value which is equal to ×g=0.3g.when FXb1 and FXb2 reach the point of “K1”, F1 and F2 reach th

20、e point of “A ” at the same time.æ j = 0.7 öjjçèj = 0.5÷(2)0ø0 From (the beginning of braking )”O” to “B” :ØF1and F2 change along the real straight proportioningline ;Ø FXb1 and FXb2 changes along the real straight line line portioningØ F1= FXb1 ;

21、F2 = FXb2“B” is the cross point of “r” line with which is equal to 0.7 and “” line.ØAt the point of “B” :Both F1 ,F2 and FXb1 , FXb2 reach this point at the same time , rearlockup begins to occurØ F1= FXb1F2= FXb2 After “B” point:;Ø F1 and F2 change along the” ” line continuouslyØ

22、; FXb1 and FXb2 change along the “R” line of “” which is equal to 0.7.Ø F1=FXb1 ;F2FXb2 At the point of “ k2” :Ø front lockup begins to occur. It is a unstable situation in which front and rear lockup exist together.Ø It is different from the situation in which front and rear lockup o

23、ccur together.(rear front)Ø It is a dangerous situation because swerving effect would occur easily.Ø du/dt reaches itsum value which is equal to ×g=0.7g.At the point of “B” :Ø when FXb1 and FXb2 reach the point of “K2”, F1 and F2 reach the point of “B” at the same time.æ j =

24、 0.5 öj = jçj = 0.5÷(3)0è0øØFrom (the beginning of braking )”O” to “K” :F1and F2 change along the real straight proportioningline ;Ø FXb1 and FXb2 changes along the real straightproportioningline line too.“K” is the cross point of“I” curve and “” curve.Ø F1= F

25、Xb1 ; F2 = FXb2At the point of “K” :Ø Both Fu1 ,Fu2 and FXb1 , FXb2 reach this point at the same time Front and rear lockup occur together at this point.Ø F1= FXb1 ;F2= FXb2Ø du/dt reaches its ×g=0.5g.um value which is equal toSummarize7. Braking Coefficient In Use（利用附着系数）Analyze

26、jj0æ du ö= z× gç÷AèdtøAæ du ö= æ du ö= z× g = j g = 0.3gç dt ÷ç dt ÷A'èøA'èømaxAnalyzejj0æ du ö= z× gç÷BèdtøBæ du ö= æ du ö× g = j g

27、 = 0.7g= zç dt ÷ç dt ÷B 'èøB 'èømax（2）How to get Braking Coefficient In Use ？（如何求利用附着系数）jj01)Point A ： front lockup barely occursFront Axle Braking Coefficient In Use前轴利用附着系数F= F= b F= b F= b G du = bGzXb11mXbg dtF= G (b + zh )Z1Lgj = FXb1 fFZ1jj02)Poi

28、nt B：rear lockup barely occursRear Axle Braking Coefficient In Use后轴利用附着系数)a - zhgF= (1- b )G du = (1- b )GzXb2g dtF= G (Z 2Lj = FXb2 rFZ 2=ØNote :the two sections of the curves under thebisector have no real meaning, because z.3）zmax (no lockup) calculated by the use of adhesion coefficient由利用

29、附着系数计算车轮不抱死条件下的zmaxj = Lb - b0hg4）Theum deceleration (no wheel lockup)车轮不抱死条件下能达到的最大制动度The shortest braking distance when the wheels without locking in no ABS vehicle没有ABS又不车轮抱死时的最短制动距离=1 æt ¢ + t ¢¢ ö+u2sç2 ÷ u a 03.6 è 22 ø a025.92 jmaxjmax = zmax g例题 已

30、知某轿车：满载质量为1600kg，轴距为2700mm，质心到前轴距离为1250mm，质心高度为500mm，制动器制动力分配系数为0.60，制动系反应时间t ¢ = 0.02s，2制动²度上升时间t 2= 0.02s，行驶车速ua0= 30km/ h。试计算该车在附着系数为0.7 的路面上制动时不出现车轮抱死的制动距离。Braking Coefficient In Use（i）Braking Intensity （Z）8. BrakingEfficient （制动效率）（1）Definition: the ratio of Braking Intensity to Braki

31、ngCoefficient In UseZZ=E =E1)jj0jjfrfr b z1j f =(b + zh )gLzbÞ E=fjLb - j hffg2)jj0 (1- b )z1jr =(a - zh )gLzaÞ E =jL(1- b ) + j hrrrg9. How to choose synchro-friction coefficient0 If braking system works according to the line under the “I” curve, front lockup occurs first definitely. This

32、 is a stable situation with no sideslip caused by side force. On the country if the braking system works according to the line above the “I” curve, rear lockup occurs first definitely. This is an unstable situation with swerving effect. We could make the probability of front lockup occurs first incr

33、ease by the increase of .b1b2j01j02Ø For car：high speedgood road conditionswerving effect is very dangerous So 0 is designed with the big value.Ø For truck：low speedbad road condition （情况复杂，弯道多）steering ability could not be lostSo 0 is designed with the little value for escaping from front

34、 lockup.Ø The principle of determining the 0 depends on which factor should be the important one , which are side slip and the losing ability of steering.10. Adjustment Of Braking Proportioning1. ECE Braking RegulationsTruckCar462.Adjustment Method of Braking Proportioning2.Adjustment Method of Braking Proportioning2.Adjustment Method of Braking Proportioning11. “ABS” Anti-lock Braking SystemThe advantages of ABS:1） No lockup the steering ability is not o