Vehicle Dynamics
2.4 Actuating Forces
The basic principle of a vehicle is a steering wheel to control the direction and a motor to produce torque, that is transfered to the vehicle wheels causing the vehicle’s movement. In this section, the work was based on [39], although several relevant modifications were performed in order to achieve a more accurate model.
In this vehicle, it is considered that all the torque produced by the motor is transferred to the rear wheels and related to the force applied by the wheel, this expression is given by the following equation,
T =F Rwheel. (2.3)
meaning that the force(F)induced by the motor on the vehicle wheels is related with the wheels radius (Rwheel)and the torque produced by the motor(T).
The acceleration and angular acceleration produced are given by equation (2.6) and (2.4) respec- tively,
˙ ω= T
Iz, (2.4)
where,
ω= ˙θ= v
Ltan(δ). (2.5)
a=Ftot
M . (2.6)
2.4.1 Forces caused by turning
During turning, while the steering angle is different from zero, there is a lateral force applied on the vehicle provoked by the curve, pushing the car to the center of the curvature circle. The centripetal force, represented asFc in figure2.1, and given by:
Fc= −M v2tan2(δ) L
1
2
cot(δ)
. (2.7)
The centripetal force can be decomposed into two components, front and rear centripetal force, applied in the vehicle wheels,showed in equations (2.8a) and (2.8b) respectively, as in figure2.1.
Fc,f = −M v2tan2(δ) L
1 cot(δ)2
2
. (2.8a)
Fc,r= −M v2tan2(δ) L
0
cot(δ) 2
. (2.8b)
The angular acceleration can be rewritten as:
˙
w= tg(δ) L
V˙ + V Lcos2(δ)
δ˙ (2.9)
When the car is in movement, exists a force that is responsible for the angular movement of the vehi- cle, that can be decomposed in front and rear components, and each one of those can be decomposed into two components, longitudinal and lateral. Those components are given by:
Ftor,f,y =−wI˙ B
L (2.10)
Ftor,f,x=Ftor,f,ytg(δ) (2.11)
Ftor,r,y =IB−Iz IB
Ftor,f,y (2.12)
The force that creates a torque for the rotation of the vehicle in the rear and in the lateral axle is represented in (2.10), meaning that it is directly related with the angular acceleration of the vehicle. So the rotational front force in the longitudinal axle is directly related, only modified by the steering angle as in (2.11). Similar behavior happens on the rear of the vehicle however, in this situation, the lateral force
Figure 2.4:Lateral Vehicle view and traction forces
is represented as in equation (2.12), while the longitudinal component is zero due to the nonexistence of lateral rotation of the rear wheels.
2.4.2 Acceleration forces
The vehicle acceleration exists due to the torque produced by the motor, applied on the rear wheels.
While the torque produced by the motor its only applied on the rear axle, the torque produced by braking is transferred to the rear and front axle. This last relation is naively expressed as in equation (2.13)
Tbrake= 2Te,f+Te,r
1 +cos(δ), (2.13)
Tmov,f =Te,f −Ftrac,f,xRw, (2.14)
Tmov,r=Te,r−Ftrac,r,xRw. (2.15)
When the car is in movement, the total torque responsible for its motion is the one transferred to the wheels by the motor, or the brakes, and also the torque representing the traction force of the tire on the ground as can be seen in figure2.4and2.5. This relation between the torque that is responsible for the motion of the vehicle, the torque transferred to the front and rear wheels and the traction force is given by the expression (2.14) and (2.15), respectively.
Figure 2.5:Traction forces
2.4.3 Dissipative forces
Although the force produced by the vehicle could be considered the total force applied on the vehicle, one needs to consider the dissipative ones.
Fdiss=Fat+Fgrav+Fdrag, (2.16)
Fat=Katr,linV +Katr,quadV2, (2.17)
Fdrag= ρairT hV2
2 , (2.18)
Fgrav=M gsin(φroad). (2.19)
The dissipative forces, given by expression (2.16), which includes the force produced by the tire’s friction on the road, composed by two velocity components, linear and quadratic, and the gainsKatr,lin
andKatr,quad, respectively. The drag induced, related with the velocity and the resistance of the air, and the gravity component that expresses the down force applied in an inclined road, given by expressions
(2.17), (2.18) and (2.19) respectively.
2.4.4 Total force actuating on the vehicle
In the vehicle the rear and front torque responsible for the vehicle’s motion are represented by (2.20) and (2.21) respectively.
Tmov,r=Fprod
M Iw
Rw
(2.20) Tmov,f =Tmov,r
1
cos(δ) (2.21)
Its possible to conclude that the total force that actuates on the vehicle is the produced force, reduced by the dissipative effects, as shown in the following equation,
Ftotal=Fprod−Fdiss (2.22)
The produced forces by the vehicle actuating on the vehicle, in the rear and front of it are given by:
Fprod,r=
Fprod,r,x
Fprod,r,y
=
Ftrac,r,x
0
+Fc,r+Ftor,r+ 0
Ftire,y
(2.23)
Fprod,f =
Fprod,f,x
Fprod,f,y
=Fc,f+Ftor,f+
cos(δ) −sin(δ) sin(δ) cos(δ)
Ftrac,f,x
Ftire,y
(2.24) Concluding, the total produced force to accelerate the vehicle is,
Fprod=Fprod,r,x+Fprod,f,x. (2.25)
As represented in equation (2.22), it is needed to consider the dissipative effects in order to know the total force that is applied to generate movement.
The force that actuates laterally to vehicle is given by,
Flat=Fprod,r,y+Fprod,f,y. (2.26)
Given this, it is now possible to estimate accurately the vehicles’ state. The total force produced given in (2.25) is responsible for the longitudinal acceleration. This relation is expressed by Newton’s first Law expressed in equation2.6.
Flat, given by expression (2.26), is the actuating force perpendicular to the vehicle so, in order to avoid sliding, this one should not overcome the threshold given by:
Fmax= gM
2 µs (2.27)
whereµsis the friction coefficient.
All the results achieved throughout this chapter will be validated in chapter4to guarantee robustness of the model created.