《汽车理论》课程教学课件（英文讲稿）第5章 汽车操纵稳定性 Handling Performance 5.3 Linear Dual-freedom Model of Turning Vehicle（线性二自由度汽车模型对前轮角输入响应）

(3) Turning model with tire sliptan(S-α).R+ tanα, ·R= LLR=tan(S-α,)+ tan α2Assu ming small ,ar,α2LLR~R~'s-a1P8-(α-α,)(8-α)+α,RoaWhere arslip angle of front wheelslip angle of rear wheel3Steer angle at front wheelsRLRadius of turnWheelbase1)SoNeutral Steer → α=α2|= R=R2)Oversteer → α<α2= R<RUndersteer → α>α2= R>R3)

So 1） Neutral Steer → 2） Oversteer → 3） Understeer → 1 2 1 2 tan( ) tan tan( ) tan R R L L R              1 2  , , 1 2 1 2 ( ) ( ) L R L R                1 2 0    R R 1＞2  R＞R0 1＜2  R＜R0 （3）Turning model with tire slip Assu ming small Where α1—— slip angle of front wheel α2 —— slip angle of rear wheel δ —— Steer angle at front wheels R —— Radius of turn L —— Wheelbase

U+AU1u+uV+AVA0VS40U+Auxuu+AuVyUxuα2yt0rαSBVd215xu1FY2baFy1L

L b a  2 1  r 2 u u  V x y   2  FY 2 FY1 x x y y u     u  u V V V         u u   u  

2.Movement Differential Eguation of Vehicle（运动微分方程）U(1）Acceleration of vehicleu+Au40U+△U（车辆加速度）40u1) speed increase of vehicle1.1) speed increase along the x- coordinateunder the vehicle axis system(u+△u)cos △0-(v+△v)sin△0-u=ucosoAucosAe-vsinAo-AvsinAo-u=u+△u-vAo-△v.△o-u(: smallA0cos△0~1;sin0~△0)=△u-V·A0

（1） Acceleration of vehicle （车辆加速度） 1）speed increase of vehicle 1.1) speed increase along the x- coordinate under the vehicle axis system                                                        u v small u u v v u u u v v u u u v v u ( cos 1;sin ) cos cos sin sin cos ( )sin       u u   u   2. Movement Differential Equation of Vehicle （运动微分方程）

Uu+Au401.2) speed increase along theU+△Uy- coordinate under the20uvehicle axis systemu+u)sin△0+(v+△v)cos-v=usinAo+AusinAo+vcosAo+AvcosAo-y:smallA0= sinA0=0=u·△0+Au·△0+v+△v-v=△v+u·△0

1.2) speed increase along the y- coordinate under the vehicle axis system ( )sin ( )cos sin sin cos cos sin u u v v v u u v v v small u u v v v v u                                                       u u   u  

2) acceleration ofvehicle2.1） acceleration vector(矢量)component on x- axisdirection△0Au-vAe△ua.X△t△t△ta.x=u-V·@2.2) acceleration vector component on y-axis direction40Av+uAe△va+u△t△t△t=u-V·a=v+u·のv+u·o

2）acceleration of vehicle 2.1) acceleration vector(矢量) component on x- axis direction 2.2) acceleration vector component on y-axis direction x x r u v u a v t t t a u v                    y y r v u v a u t t t a v u                    x r y r a u v a v u               