3-1. Lateral Vehicle Dynamics

Lateral Vehicle Dynamics

y- lateral

x- longitudinal Modelling of Automotive Systems

1

Kinematic Model of Lateral Vehicle Motion:

Bicycle Model

Instantaneous rolling centre

Circular radius Steering angle

Centre of gravity

Vehicle orientation (yaw angle)

Wheelbase

Major assumption : velocities at A and B are in the direction of wheels orientation, i.e. the wheel's slip angles are 0. Modelling of Automotive Systems

2

Bicycle Model 2 front wheels are represented by a single wheel at A 2 rear wheels are represented by a single wheel at B Steering angles

δf δr Assumed both front and rear wheels can be steered. For only front steering case, δ r = 0 C - centre of gravity L = l f + lr : Wheelbase of the vehicle Ψ : Orientation of the vehicle - heading angle of vehicle V : velocity of c.g.

β : Slip angle of the vehicle (angle between the motion direction and vehicle orientation) R : Circular radius O : Instantaneous rolling centre

Modelling of Automotive Systems

3

Bicycle Model Course angle of the vehicle

γ =ψ + β

Triangle OCA  π sin  − δ f  sin (δ f − β ) 2  =  lf R sin δ f cos β − sin β cos δ f

or

lf

=

cos δ f R

or

tan δ f cos β − sin β =

lf R

Triangle OCB π  sin  δ r +  sin (β − δ r ) 2 =  lr R sin β cos δ r − sin δ r cos β cos δ r = lr R

or

or sin β − tan δ r cos β =

lr R

Add both

(tan δ

f

− tan δ r )cos β =

l f + lr R Modelling of Automotive Systems

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Bicycle Model Slip angle β

From tan δ f cos β − sin β =

lf

R l sin β − tan δ r cos β = r R

(l

f

lr tan δ f cos β − lr sin β = l f sin β − l f tan δ r cos β =

l f lr R lr l f R

tan δ r + lr tan δ f )cos β − (l f + l f )sin β = 0

tan β =

l f tan δ r + lr tan δ f lf +lf  l f tan δ r + lr tan δ f    + l l f f  

β = tan −1 

i.e. the slip angle is represented by steering angles and wheelbases.


5

Kinematic Model of Lateral Vehicle

(tan δ

f

− tan δ r )cos β =

l f + lr R

Angular velocity ψ& =

V R

Therefore ψ& =

V (tan δ f − tan δ r )cos β l f + lr

X& = V cos(ψ + β ) Y& = V sin (ψ + β )


6

Kinematic Model of Lateral Vehicle: consider vehicle width Limitation of the bicycle model: δo and δi in fact are different.


7

Kinematic Model of Lateral Vehicle: consider vehicle width In the bicycle model ψ& =

V (tan δ f − tan δ r )cos β l f + lr

If the slip angle β is small, δ r = 0 ψ& =

Vδ f l f + lr

=

Vδ f L

on the other hand V ψ& = R Therfore

δf =

L R

Let δ f =

δo =

δo + δi 2

L R+

lw 2

,

=

L R

δi =

L R−

lw 2


8

Kinematic Model of Lateral Vehicle: consider vehicle width Trapezoidal tie rod arrangement to realize

δi > δo


9

Dynamics of Bicycle Model


10

Dynamics of Bicycle Model x − vehiclelongitudinal axis y - vehiclelateralposition,measuredalong the vehicle lateralaxis to the point O which is the centreof rotation

X − Y global coordinates ψ - yaw angle Vx − vehicle longitudinal velocity Using Newton's Law ma y = Fyf + Fyr a y = &y& +ψ& 2 R = &y& + Vxψ& m( &y& + Vxψ& ) = Fyf + Fyr I zψ&& = l f Fyf − lr Fyr Experimental shows the lateral tyre force is proportional to the slip angle (when slip angle is small). Modelling of Automotive Systems

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Dynamics of Bicycle Model Experimental shows the lateral tyre force is proportional to the slip angle (when slip angle is small). Slip angle of tyre : α f = δ − θ vf where δ front tyre steering angle α r = −θ vr Rear tyre Forces Fyf = 2Cα f (δ − θ vf

)

Fyr = 2Cα r (− θ vf

x

)

where Cα f , Cα r are cornering stiffness tan θ vf = tan θ vr =

V y + l f ψ& Vx V y − l rψ& Vx

Steering angle

,

Velocity angle

Slip angle α f

,

lf

ψ

y Modelling of Automotive Systems

12

Dynamics of Bicycle Model When θ vf , θ vr are small y& + l f ψ& θ vf = Vx y& − lrψ& θ vr = Vx

m ( &y& + Vxψ& ) = Fyf + Fyr I zψ&& = l f Fyf − lr Fyr y& + l f ψ&  Fyf = 2Cαf  δ − Vx  y& − lrψ& Fyr = −2Cαr Vx

  


13

&y& + Vxψ& =

2Cαf δ m

2Cαf ( y& + l f ψ& ) 2Cαr ( y& − lrψ& ) − − mV x mV x

2Cαf ( y& + l f ψ& )  lr 2Cαr ( y& − lrψ& ) lf   + ψ&& =  2Cαf δ − Iz  Vx Vx  Iz i.e. 2Cαf δ

2(Cαf + Cαr )

(

) ψ&

(

)ψ&

 2 Cαf l f − Cαr lr  &y& = y& − Vx + −  m mV x mV x  2 Cαf l 2f + Cαr lr2 l f 2Cαf δ 2(Cαf l f − Cαr lr ) ψ&& = y& − − Iz I zVx I zVx


 

14

Introduce state space variable,  y  y&      ψ  ψ&  The lateral equations of motion become 1 0  y  2(Cαf + Cαr ) −  y&  0 d    mV x =   0 dt ψ  0 ψ&  0 − 2(Cαf l f − Cαr lr )  I zVx 

  0 y 2 Cαf l f − Cαr lr     2Cαf 0 − Vx −   y&   mV x    +  m 0 0 1  ψ    2 Cαf l 2f + Cαr lr2     2l f Cαf 0 −  ψ&   I I zVx z  

0

(

(


0

)

)

   δ   

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