Previously, a direct treating ofEqns (3.12) and (3.13), which are the basic equations of motion describing the vehicle motion, has been avoided. The characteristics of vehicle transient response to steering input has only been looked at by focusing on the characteristic equation and its roots.
Looking at the basic equations of motion,(3.12) and (3.13), and transforming both sides of equations gives the following:
fmVsỵ2ðKfỵKrịgbðsị ỵ fmVỵ2
VðlfKflrKrịgrðsị ẳ2Kfdðsị (3.75) 2ðlfKflrKrịbðsị ỵ
8<
:Isþ2 l2fKfþl2rKr
V
9=
;rðsị ẳ2lfKfdðsị (3.76) whereb(s),r(s), andd(s) are the Laplace transformation ofb,r, andd. By solving the algebraic equation forb(s) andr(s), the following equations can be obtained:
bðsị dðsịẳ
2Kf mVþ2
VðlfKflrKrị 2lfKf Isþ2 l2fKfþl2rKr
V
mVsỵ2ðKfỵKrị mVỵ2
VðlfKflrKrị 2ðlfKflrKrị Isỵ2 l2fKf ỵl2rKr
V
(3.77)
rðsị dðsịẳ
mVsỵ2ðKfỵKrị 2Kf 2ðlfKflrKrị 2lfKf
mVsỵ2ðKfỵKrị mVỵ2
VðlfKflrKrị 2ðlfKflrKrị Isỵ2 l2fKfỵl2rKr
V
(3.78)
Then, using theunandzderived earlier and rewritingEqns (3.77) and (3.78)gives the following:
bðsị
dðsịẳGbdð0ị 1ỵTbs 1þ2zsu
n þus22 n
(3.77)0 where
Gbdð0ị ẳ1m2lllf
rKrV2 1þAV2
lr
l (3.79)
Tbẳ IV 2llrKr
1 1m2lllf
rKrV2 (3.80)
and
rðsị
dðsịẳGrdð0ị 1ỵTrs 1þ2zsu
nþus22 n
(3.78)0
Grdð0ị ẳ 1 1þAV2
V
l (3.81)
TrẳmlfV 2lKr
(3.82) Gbdð0ịis the side-slip angle gain constant, which is the value ofbin response todduring steady- state cornering, andGrdð0ịis the yaw rate gain constant, which is the value ofrin response to dduring steady-state cornering. Equations (3.77)0and (3.78)0are the transfer functions of the response of side-slip angle,b, and yaw rate,r, to steering input,d.
If the Laplace transformed response ofbandrtodare given, as inEqns (3.77) and (3.78)or (3.77)0and (3.78)0, the response ofbandrto a givendcould be obtained by inverse Laplace transformation. When a vehicle, traveling on a straight line, is suddenly given a step steering input, the vehicle response is as follows:
bðtị ẳL1ẵbðsị ẳL1 2
4Gbdð0ị 1ỵTbs 1þ2zsu
nþus22 n
d0
s 3
5 (3.83)
rðtị ẳL1ẵrðsị ẳL1 2
4Grdð0ị 1ỵTrs 1þ2zsu
nþus22 n
d0
s 3
5 (3.84)
wherebyL1means inverse Laplace transformation andd0/sis the Laplace transformed steering angle,d(s), a step input with a magnitude ofd0. Applying inverse Laplace transformation formula to the previous equations,b(t) andr(t) become the following. Here, the initial value ofbandris zero. When the vehicle shows response without oscillation at z>1, then the following is obtained:
bðtị ẳGbdð0ịd0
2
41þ1 zþ ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p
unTb
2 zþ ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p e
z ffiffiffiffiffiffiffiffi
z21
p
unt
1 z ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p
unTb
2 z ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p e
zþ ffiffiffiffiffiffiffiffi
z21
p
unt
3
5 (3.83)0
rðtị ẳGrdð0ịd0
2
41þ 1 zþ ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p
unTr 2 zþ ffiffiffiffiffiffiffiffiffiffiffiffiffi
z21
p ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p e
z ffiffiffiffiffiffiffiffi
z21
p
unt
1 z ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p
unTr
2 z ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p ffiffiffiffiffiffiffiffiffiffiffiffiffi z21
p e
zþ ffiffiffiffiffiffiffiffi
z21
p
unt
3
5 (3.84)0
When the response is without oscillation atzẳ1, the following results:
bðtị ẳGbdð0ịd0
1þ
u2nTbun
t1 eunt
(3.83)00
rðtị ẳGrdð0ịd0
1þ
u2nTrun t1
eunt
(3.84)00 And, when the response is with oscillation atz<1, the following is true:
bðtị ẳGbdð0ịd0
"
1þ ffiffiffiffiffiffiffiffiffiffiffiffiffiTb
1z2
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
Tbzun
2 þ
1z2 u2n q
ezuntsin
ffiffiffiffiffiffiffiffiffiffiffiffiffi 1z2 q
untþJb
(3.83)000
whereJbẳtan1 ffiffiffiffiffiffiffiffiffi
1z2
p
un
1=Tbzun
! tan1
ffiffiffiffiffiffiffiffiffi
1z2
p
z
!
rðtị ẳGrdð0ịd0
"
1þ ffiffiffiffiffiffiffiffiffiffiffiffiffiTr
1z2
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1=Trzunị2ỵ
1z2 u2n q
ezuntsin
ffiffiffiffiffiffiffiffiffiffiffiffiffi 1z2 q
untþJr
(3.84)000 whereJrẳtan1
ffiffiffiffiffiffiffiffiffi
1z2
p
un
1=Trzun
! tan1
ffiffiffiffiffiffiffiffiffi
1z2
p
z
!
The responsiveness to steering input, besides the response time that is described earlier, could also be expressed by the time to steady-state value,te, and time to the first peak of the response oscillation,tp, for the yaw rate response to step steering input. This is shown inFigure 3.28.
yaw rate
time
FIGURE 3.28
Responsiveness of yaw rate.
Calculatingteandtpgives the following terms:
teẳ 1
u2nTr (3.85)
tpẳ 1 un ffiffiffiffiffiffiffiffiffiffiffiffiffi
1z2 p
(
ptan1
ffiffiffiffiffiffiffiffiffiffiffiffiffi 1z2
p unTr
1zunTr
!)
(3.86) In particular,tecould be defined as the approximated response time of yaw rate.
To see a time history of the vehicle responses to steering input, it is possible to use Eqn (3.83)0–(3.84)000; however, recently, there has been an easier way to see the time historyda numerical simulation using a PC with some software, for example, Matlab-Simulink.
Equations of vehicle motion(3.12) and (3.13)are rewritten as follows:
db
dt ẳ 2ðKfỵKrị
mV b
1þ 2
mV2ðlfKflrKrị
rþ2Kf
mVd (3.12)0
dr
dtẳ 2ðlfKflrKrị
I b2 l2fKfþl2rKr
IV rþ2lfKf
I d (3.13)0
Side-slip angle and yaw rate are obtained by integrating the right-hand side of the previous equations. It is possible to have the integral-type of block diagram of the vehicle motion to steering input as shown inFigure 3.29. This is the basis of the simulation program using Matlab- Simulink software.
Example 3.6
Execute the simulation of the vehicle response to step steering input,dẳ0.04 rad, by Matlab-Simulink at the vehicle speeds 60 km/h, 100 km/h, and 140 km/h, respectively, with the vehicle parameters as mẳ1500 kg,Iẳ2500 kgm2,lfẳ1.1 m,lrẳ1.6 m,Kfẳ55 kN/rad, andKrẳ60 kN/rad.
Solution
The vehicle parameters for the simulation are set as inFigure E3.6(a). The simulation program is shown inFigure E3.6(b), andFigure E3.6(c)is a simulation condition.Figure E3.6(d)is a result of the simu- lation, and all the results are summarized inFigure E3.6(e).
Kf
2
f
fK
l 2
mV 1
s 1
(Kf+Kr)
2
( )
V K l K mV lf f− r r
+2
I 1
s 1
(lfKf−lrKr)
2
( )
V K l K lf2 f r2 r
2 +
β
r +− +
−
+− +
− δ
FIGURE 3.29
Integral-type block diagram of vehicle motion.
FIGURE E3.6(a)
FIGURE E3.6(b)
FIGURE E3.6(d) FIGURE E3.6(c)
If the vehicle motion is described with coordinates fixed on the ground,Eqns (3.21) and (3.22) will be achieved. By Laplace transforms, the following are obtained:
ms2ỵ2ðKfỵKrị
V s
yðsị ỵ
2ðlfKflrKrị
V s2ðKfỵKrị
qðsị ẳ2Kfdðsị (3.21)0 2ðlfKflrKrị
V syðsị ỵ 8<
:Is2þ2 l2fKfþl2rKr
V s2ðlfKflrKrị 9=
;qðsị ẳ2lfKfdðsị (3.22)0 wherey(s) andq(s) are the Laplace transforms ofyandq. Solving the algebraic equations ofy(s) andq(s) gives the following:
yðsị dðsịẳ
2Kf
2ðlfKflrKrị
V s2ðKfỵKrị 2lfKf Is2þ2 l2fKfþl2rKr
V s2ðlfKflrKrị
ms2ỵ2ðKfỵKrị
V s 2ðlfKflrKrị
V s2ðKfỵKrị 2ðlfKflrKrị
V s Is2þ2 l2fKfþl2rKr
V s2ðlfKflrKrị
(3.87)
0 0.5 1 1.5 2 2.5 3 3.5 4
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
time (s) front tire angle (rad) side slip angle (rad)
0 0.5 1 1.5 2 2.5 3 3.5 4-0.6
-0.4 -0.2 0 0.2 0.4 0.6
time (s)
yaw rate (rad/s)
0 0.5 1 1.5 2 2.5 3 3.5 4
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
time (s) front tire angle (rad) side slip angle (rad)
0 0.5 1 1.5 2 2.5 3 3.5 4-0.6
-0.4 -0.2 0 0.2 0.4 0.6
time (s)
yaw rate (rad/s)
0 0.5 1 1.5 2 2.5 3 3.5 4
-0.06 -0.04 -0.02 0 0.02 0.04 0.06
time (s) front tire angle (rad) side slip angle (rad)
0 0.5 1 1.5 2 2.5 3 3.5 4-0.6
-0.4 -0.2 0 0.2 0.4 0.6
time (s)
yaw rate (rad/s)
V= 60 (km/h) V= 100 (km/h)
V= 140 (km/h) δ
β r
δ
β r
δ
β r
FIGURE E3.6(e)
qðsị dðsịẳ
ms2ỵ2ðKfỵKrị V s 2Kf
2ðlfKflrKrị
V s 2lfKf
ms2ỵ2ðKf ỵKrị
V s 2ðlfKflrKrị
V s2ðKfỵKrị 2ðlfKflrKrị
V s Is2þ2 l2fKfþl2rKr
V s2ðlfKflrKrị
(3.88)
Then, usingunandzderived earlier and by rewriting equations, the equations become the following:
yðsị
dðsịẳGyd€ð0ị1ỵTy1sỵTy2s2 s2
1þ2zsu
n þus22 n
(3.87)0
where
G€ydð0ị ẳ 1 1þAV2
V2
l ẳVGrdð0ị (3.89)
Ty1ẳlr
V (3.90)
Ty2ẳ I
2lKr (3.91)
G€ydð0ị is the lateral acceleration gain constant, which is the lateral acceleration value in response todduring steady-state cornering, and the following is derived:
qðsị
dðsịẳGrdð0ị 1ỵTrs s
1þ2zsu
n þus22 n
(3.88)0
Equations (3.87) and (3.88)or(3.87)0and (3.88)0are the transfer functions of the responses of lateral displacement and yaw angle to vehicle steering input. As previously shown, inverse Laplace transformations of y(s) and q(s) in the previous equations can obtain the lateral displacement and yaw angle responses to a given steering input.
Laplace transforms of the vehicle response to steering input and expression of the vehicle motion in the form of transfer functions are convenient when the actual response to a given steering input is desired. It is also suitable in the case of taking the vehicle as the control target in the control system and studying the control of vehicle motions and the controllability of the vehicle.
Furthermore, un;z;Grdð0ị;G€ydð0ị;Tr;te;andtp, which are the coefficients in the transfer functions, are the parameters that determine the vehicle response characteristics toward steering input and are called the response parameters.