Fundamental Equations of Vehicle Handling

Một phần của tài liệu The science of vehicle dynamics (Trang 142 - 146)

The vehicle has basically only lateral and yaw dynamics (often simply called lateral dynamics), described by the following differential equations (cf. (3.64))

may=Y=Y1+Y2

Jzr˙=N=Y1a1−Y2a2

(6.4) while

max=mvr=X=X1+X2−1

2ρSCxu2 (6.5)

is now an algebraic equation, the unknown being(X1+X2).

2The left and right wheels of the same axle are normally equipped with the same kind of brake.

Therefore, the braking torque is pretty much the same under ordinary operating conditions, and, again, (6.1) holds true. However, there are important exceptions. The left and right braking forces can be different if: (a) the grip is different and one wheel is locked; (b) the friction coefficients inside the two brakes is different (for instance, because of different temperatures, which is often the case in racing cars); (c) some electronic stability system, like ESP or ABS, has been activated.

6.2 Fundamental Equations of Vehicle Handling 133 With an open differential, it is easy to solve (6.4) with respect to the front and rear lateral forces

Y1=ma2

l ay+Jz

l r˙ma2

l ay Y2=ma1

l ayJz

l r˙ma1

l ay

(6.6)

where we took into account that|Jzr˙| |mayai|, since in a carJz< ma1a2 and

rai| |ay|. In a two-axle vehicle with open differential the lateral forces are linear functions of the lateral accelerationay. This is a very peculiar and important result, which greatly impacts on the whole vehicle model, as will be shown.

According to (3.114) and (3.115), the lateral load transfers are linear functions ofY1andY2. Employing (6.6) we obtain the following simplified equations for load transfers in vehicles with open differential

ΔZ1maykφ1kφ2

t1kφ

hq kφ2 +a2q1

lksφ

1

+a2q1

lkφs

2

+a2q1+a1q2

lkpφ

2

=mayη1

ΔZ2may

kφ1kφ2 t2kφ

hq kφ1 +a1q2

lksφ

1

+a1q2 lkφs

2

+a2q1+a1q2 lkpφ

1

=mayη2

(6.7)

The two constantsη1andη2 depend, in a peculiar way, on the roll stiffnesses, on the heights of the no-roll centers3and on the longitudinal position of the center of gravity.

Similarly, the suspension roll angles (3.110) can be set as functions of the lateral acceleration only4

φ1s=may 1 ksφ

1

kφ1kφ2 kφ

hq kφ2 +a2q1

lkφp

1

a1q2

lkφp

2

=mayρ1s

φ2s=may

1 ksφ

2

kφ1kφ2 kφ

hq kφ1 +a1q2

lkφp

1

a2q1 lkφp

1

=mayρ2s

(6.8)

The same applies to tire roll anglesφip φp1 =may 1

kφp

1

kφ1kφ2

kφ

hq kφ2 +a2q1

lkφs

1

+a2q1

lksφ

2

+a2q1+a1q2

lkφp

2

=mayρ1p

φp2 =may 1 kφp

2

kφ1kφ2 kφ

hq kφ1 +a1q2

lkφs

1

+a1q2 lksφ

2

+a2q1+a1q2 lkφp

1

=mayρ2p (6.9)

If, for simplicity, the tires are supposed to be perfectly rigid, that iskpφ

i → ∞,

we haveρ1p=ρ2p=0,ρ1s=ρ2s=(hq)/ kφand the expressions of the lateral load

3We call no-roll center what is commonly called roll center. This aspect is discussed in Sect.3.8.8.

4In this model the roll inertial effects are totally disregarded.

transfers become simpler ΔZ1may1

t1

kφ1(hq) kφ +a2q1

l

=mayη1

ΔZ2may

1 t2

kφ2(hq) kφ +a1q2

l

=mayη2

(6.10)

as in (3.118).

The total vertical loads (3.79) on each tire can also be simplified by discarding the longitudinal load transfer. Moreover, cars with an open differential are not so sporty to have significant aerodynamic vertical loads. Therefore, combining (3.79) and (6.7), we get

Z11=Fz11=mga2

2lmayη1=Z10

2 −ΔZ1(ay) Z12=Fz12=mga2

2l +mayη1=Z10

2 +ΔZ1(ay) Z21=Fz21=mga1

2lmayη2=Z20

2 −ΔZ2(ay) Z22=Fz22=mga1

2l +mayη2=Z20

2 +ΔZ2(ay)

(6.11)

which shows that all vertical loads are (linear) function of the lateral acceleration.

According to (3.123) and taking into account (6.8), we get the following expres- sion for the steering angles of the wheels

δij=δij0 +δvτij+Υijφis(ay)

=δij0 +δvτij+Υijρisay

=δijv, ay) (6.12)

which are functions ofδvand, again, of the lateral accelerationay. More precisely, the termδvτijis the steer angle due to the steering wheel rotationδv, the termδij0 is the toe-in/out angle , and the termΥijφis(ay)is the roll steer angle.

Under the assumed operating conditions (6.3), the tire lateral slips (3.49) become σy11=(v+ra1)11

urt1/2 v+ra1

uδ11

σy12=(v+ra1)12

u+rt1/2 v+ra1

uδ12 (6.13)

6.2 Fundamental Equations of Vehicle Handling 135 σy21=(vra2)21

urt2/2 vra2

uδ21 σy22=(vra2)22

u+rt2/2 vra2

uδ22

sinceu |rti/2|as discussed in (3.4). The lateral slips can be conveniently rewrit- ten taking (6.12) into account

σy11=v+ra1

uδvτ11−δ011−Υ11ρ1say σy12=v+ra1

uδvτ12−δ012−Υ12ρ1say σy21=vra2

uδvτ21−δ021−Υ21ρ2say σy22=vra2

uδvτ22−δ022−Υ22ρ2say

(6.14)

or, more compactly5

σyij =σyij(v, r, u, δv)=σyij(β, ρ, ay, δv) (6.15) Let, γi10 = −γi02=γi0 be the camber angles under static conditions, and let Δγi1=Δγi2=Δγibe the camber variations. The camber angles of the two wheels of the same axle are thus given by

γi1=γi0+Δγi, γi2= −γi0+Δγi (6.16) where the camber variationΔγi, according to (3.83), (6.8) and (6.9), depends on the lateral accelerationay

Δγimay

qibi

bi

ρisρip

=mayχi (6.17)

since the track variation termΔti/(2bi)is usually negligible.

The lateral forces exerted by the tires on the vehicle depend on many quantities, as shown in the second equation in (2.72). For sure, there is a strong dependence on the vertical loadsZij and on the lateral slipsσyij, while, in this model, we can neglect the longitudinal slipsσxij. The camber angles γij need to be considered, since they are quite influential, even if small. According to (3.125), the spin slipsϕij are directly related toγij. Therefore, the simplified model for each lateral force is

Fyij =Fyij

Z0i/2−ΔZi(ay), γi0+Δγi(ay), σyij(v, r, u, δv) cos

δijv) (6.18)

5Here we are abusing the notation: different functions bear the same name. However, the meaning should be sufficiently clear and unambiguous.

The lateral forceYi for each axle is obtained by adding the lateral forces of the left tire and of the right tire (cf. (3.58))

Yi=Fyi1cos δi1v)

+Fyi2cos δi2v)

(6.19) or, more explicitly, taking also (6.15) into account

Y1=Fy11

Z01/2−ΔZ1(ay), γ10+Δγ1(ay), σy11(v, r, u, δv) cos

δ11v) +Fy12

Z10/2+ΔZ1(ay),γ10+Δγ1(ay), σy12(v, r, u, δv) cos

δ12v)

=Y1(v, r, u, δv)=Y1(β, ρ, ay, δv) Y2=Fy21

Z02/2−ΔZ2(ay), γ20+Δγ2(ay), σy21(v, r, u, δv) cos

δ21v) +Fy22

Z20/2+ΔZ2(ay),γ20+Δγ2(ay), σy22(v, r, u, δv) cos

δ22v)

=Y2(v, r, u, δv)=Y2(β, ρ, ay, δv)

(6.20)

The effects ofay on the steering anglesδij can be neglected in the cosine terms because they are very small. On the other hand, these effects are very influential on the congruence equations (6.14).

It must be clearly understood that the functions in (6.20) are known functions.

Một phần của tài liệu The science of vehicle dynamics (Trang 142 - 146)

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