Universit`
a di Pisa
Dipartimento di Ingegneria Civile e Industriale
corso di laurea magistrale in ingegneria meccanica
Constrained Dynamic Simulation and
Trajectory Planning of Road and Race Cars as
a Nonlinear Optimization Problem
Relatori:
Prof. Marco Gabiccini
Prof. Massimo Guiggiani
Ing. Francesco Bucchi
Candidato:
Francesco Perfetti
Abstract
This work aims to create a model of car that can be used both to perform simulations, and to solve optimization problems, such as to travel a certain section of road in a minimum time.
The vehicle parameter considered for the model are taken from a Formula SAE vehicle. In the creation of the model have been considered the main compo-nents of a vehicle: steering, suspension, engine, differential and braking system. Due to the complexity of the problem, we chose to start with the study of a simplified model whose complexity was gradually increased.
Furthermore focusing on the problem of finding the minimum time needed to travel a section of track consisting of a straight road section followed by a left hand turn and then another straight road section, we performed a compar-ison between two models. The first model used an open differential while the second model a limited slip differential. In this way we were able to identify the advantages and disadvantages of one and the other.
Contents
1 Introduction 4 2 Tire Behavior 6 2.1 Tire Structure . . . 6 2.2 Road Surface . . . 6 2.3 Reference System . . . 72.4 Contact Patch and Forces . . . 7
2.5 Slips . . . 9 2.6 Longitudinal Force . . . 9 2.7 Lateral Force . . . 10 2.8 Friction Ellipse . . . 10 2.9 Tire Models . . . 12 2.9.1 Dugoff’s Model . . . 12 2.9.2 Magic Formula . . . 14
2.9.3 Nicolas Comstock Model . . . 14
2.9.4 Models Comparison . . . 15 2.10 Transient Behavior . . . 15 3 Vehicle Model 17 3.1 Reference Systems . . . 17 3.2 Steering . . . 18 3.3 Aerodynamic Forces . . . 20
3.4 Road Tire Friction Forces . . . 20
3.5 Longitudinal and Lateral Slips . . . 21
3.6 Vertical Loads . . . 22
3.7 Suspensions . . . 22
3.7.1 Roll Centers and Roll Axis . . . 22
3.7.2 Lateral Load Transfer . . . 23
3.8 Powertrain . . . 23
3.8.1 Engine . . . 24
3.8.2 Differential . . . 24
3.9 Braking System . . . 30
3.10 Equilibrium Equations . . . 30
4 Optimal Control Problems 32 4.1 Direct Single Shooting . . . 33
4.2 Direct Multiple Shooting . . . 33
CONTENTS CONTENTS
4.4 Objective Function . . . 34
4.5 Constraints . . . 36
4.5.1 Path Constraint . . . 36
4.5.2 Pedal’s Position Constraint . . . 37
4.6 Problem Initialization . . . 37
5 Main Simulations and Results 38 5.1 Single-track Models . . . 38
5.1.1 Linear Force vs Magic Formula . . . 39
5.1.2 Steady State vs Transient Behavior . . . 42
5.2 Four Wheel Models . . . 42
5.2.1 Open Differential . . . 46
5.2.2 Open Differential vs Limited Slip Differential . . . 61
Chapter 1
Introduction
The simulation of the vehicle’s dynamics allows to predict the operating condi-tions in terms of kinematic characteristics (position, velocity and acceleration) and loads (forces and moments applied to the vehicle).
It therefore assumes a very important role in research focused on safety, efficiency and performance.
Many studies have been made in the field of vehicle dynamic simulation. For example paper [5], considers a lane change maneuver for a simple nonlinear single-mass car model with yaw, lateral, and longitudinal freedoms. In [6], a full-lap optimal control calculations for the Barcelona track using a simple four-wheeled car model is described.
The aim of the thesis is to create a mathematical model that simulates the correct behavior of a car. The model will be used to see what are the effects of certain inputs on the behavior of the vehicle. These simulations will be studied as a fake optimization problem. This means that the control inputs will be fixed and the cost function will have a value of zero.
In addition optimization problems will be solved such as for example trav-eling a prescribed distance in minimal time.
In the final part of this work, focusing on the problem of finding the minimum time needed to travel a section of track consisting of a straight road section fol-lowed by a left hand turn and then another straight road section, we performed a comparison between two models. The first used an open differential while the second a limited slip differential.
In this way we were able to identify the advantages and disadvantages of the two different models.
The vehicle parameter considered for the model are taken from a Formula SAE vehicle. The Formula SAE is a competition between university students in which every team has to develop a small Formula-style race car. The geometrical parameters of the car are shown in Figure3.
The front and rear tracks are t1and t2respectively. The wheelbase is l. The
distance between the front axle and the center of gravity is a1. The distance
between the rear axle and the center of gravity G is a2. Of course l = a1+ a2.
Finally H is the height of G from the road plane.
Of course it will be possible to use the model for road cars keeping in mind the assumptions that will be made.
1. Introduction
Figure 1.1 Vehicle basic scheme
It is correct to remember, however, that optimization problems such as trav-eling a prescribed distance in minimal time have a practical interest exclusively for race cars.
The model will be developed using the python programming language and the CasADi open-source tool. CasADi is a symbolic framework for algorithmic (a.k.a. automatic) differentiation and numeric optimization. The main purpose of the tool is to be a low-level tool for quick, yet highly efficient implementation of algorithms for nonlinear numerical optimization.
Chapter 2
Tire Behavior
Tires are one of the most important components of vehicles, since they are the only component keeping the vehicle in contact with the ground. They support the vehicle weight, and any other vertical forces developed such as aerodynamic. Furthermore the interactions between the tires and the road supply the tractive, braking and cornering forces for maneuvering.
In this chapter we will describe the tire structure and its behavior. First we will study the longitudinal dynamics (traction/braking) and then the lateral dynamics (cornering). Finally the above is gathered together in the concept of the friction ellipse.
The last section of this chapter propose some models that describe the tire behavior.
2.1
Tire Structure
The tire is a complex composite structure, made up of many layers of rubberized fabric (plies) with reinforcements cords. The angle between the direction of the cords with respect to the circumferential direction of the tires is usually referred to as the crown angle. As a general rule, plies with a low value of crown angle enhance the handling characteristics of the vehicle, while those with a high value, up to 90 deg, of crown angle enhance ride comfort [10].
With a crown angle of 90 deg, the tire would not be sufficiently rigid at the contact with the ground. To add further stiffness, a belt of several layers of cords of high modulus of elasticity (usually steel or other high-strength materials) is fitted under the tread. The cords in the belt are laid at a low crown angle of approximately 20 deg.
Particularly relevant to tire behavior is the tread of the tire, essentially made of vulcanized filled rubber; it is the contact surface with the ground and determines friction at the tire-ground interface.
2.2
Road Surface
For the study of the wheel kinematics, the road is assumed to have a hard and flat surface like a geometric plane. The type of road surface, its roughness, and
2. Tire Behavior 2.3. Reference System
Figure 2.1 Tire structure.
Figure 2.2 Tire reference system.
whether it is wet or not, play an important role on the grip, so on forces and moments.
2.3
Reference System
It’s important to define the position of the wheel. We will use a moving reference system S = (x, y, z; O) as shown in Figure 2.2. The y-axis is the intersection between a vertical plane containing the rim axis yc and the road plane. The
x-axis is given by the intersection of the road plane with a plane containing Q and normal to yc. Axes x and y define the origin O as a point on the road. The
z-axis is vertical, that is perpendicular to the road, with the positive direction upward. The angle γ between the wheel axis and the road plane is called camber angle.
2.4
Contact Patch and Forces
The area of the tread of a tire that is in contact with the ground at any given moment is called contact patch or footprint. According to the tire’s profile design the contact patch forms a not necessarily coherent area. It is important
2. Tire Behavior 2.4. Contact Patch and Forces
Figure 2.3 Tire footprint.
to note that the shape and size of the contact patch, and also its position with respect to the reference system, depend on the tire operating conditions.
In any point of contact between the tire and the road surface vertical and friction forces are transmitted [4]. The effect of this contact forces can be fully described by a resulting force vector applied at a specific point of the contact patch and a torque vector
F= Fxi+ Fyj+ Fzk
MO= Mxi+ Myj+ Mzk
(2.1) The components of the contact force vector are named according to the direction of the axes:
Fx longitudinal force;
Fy lateral force;
Fz normal force;
Mx overturning moment;
My rolling resistance moment;
Mz self-aligning torque.
The overturning moment and the rolling resistance moment are due to the fact that the normal force is not applied in O but in a point of coordinates (ex,
ey, 0). So they are equal to
Mx= Fzey My =−Fzex (2.2)
If we define a total tangential force having the line of action with distance dtfrom O, the self-aligning torque is
Mz=|Ft|dt (2.3)
with
Ft= Fxi+ Fyj (2.4)
In the study of the dynamics of the whole vehicle, Mx and Mz can be
neglected. The rolling resistance is very low and must be considered only when the braking or the traction torque are low.
2. Tire Behavior 2.5. Slips
2.5
Slips
When we consider a rigid wheel, there is pure rolling if two kinematic conditions are fulfilled
( Vs= 0
Ωz= 0
(2.5) The first condition requires that there is non sliding at the point of contact S, the second that the angular velocity is parallel to the floor.
For a wheel with tire things change. This is because the tire is not rigid and we do not have a contact point but a contact surface. A reasonable definition of pure rolling for a wheel with tire moving on a flat surface is an operating condition in which [4]
(
Ft= Fxi+ Fyj= 0
Mz= 0
(2.6) Because we neglect Mz, we are interested only at the first condition. The
theoretical slip σ tells us how far we are from this condition. If Vsis the difference
between the speed of travel Va and the rolling velocity Vr, σ is defined as
σ = Vs Vr = Va− Vr Vr = σxi+ σyj (2.7) with σx= Vax− ωcRr(γ) ωcRr(γ) σy= Vay ωcRr(γ) (2.8)
If σx (longitudinal slip) and σy (lateral slip) are equal to zero there is pure
rolling. In other words in order to have road-tire friction forces, longitudinal and lateral slips must be non-zero.
In some cases is common the use of the tire slip angle (instead of the lateral slip). It is defined as the angle between the rolling velocity and the speed of travel. Since the rolling velocity has direction i, it is equal to the angle between iand Va tan α =−Vay Vax =− σy 1 + σx (2.9)
2.6
Longitudinal Force
We consider the case of pure braking/traction. Under this condition σy is equal
to zero and the only force is Fx.
As shown in Figure 2.4, Fxis a function of σx. It vanishes when σx= 0 (free
rolling conditions) to increase almost linearly for values of σxup to 0.025:
Fx= Cσxσx (2.10) where constant Cσx= ∂Fp x ∂σx σx=0 (2.11)
2. Tire Behavior 2.7. Lateral Force 0.0 0.2 0.4 0.6 0.8 1.0 −σx 0 200 400 600 800 1000 1200 1400 1600 longitudinal force Fx (N )
Figure 2.4 Longitudinal force Fx due to pure longitudinal slip σx, for constant
vertical load Fzof 1000 N.
can be defined as longitudinal slip stiffness of the tire. Outside this range, which depends on many factors, its value decreases.
It’s useful to define the global longitudinal friction coefficient µx, that is the
ratio between the peak value Fmax
x and the corresponding vertical load
µx(Fz) = Fmax x Fz (2.12)
2.7
Lateral Force
When we consider the case of pure cornering, the only force is Fy. In Figure 2.5
is shown Fy as a function of α. You see clearly that is very similar to Fx as a
function of σx.
Quite relevant are the values of the lateral slip stiffness also called cornering stiffness Cα= ∂Fp y ∂α α=0 (2.13) and the global lateral friction coefficient µy, that is the ratio between the peak
value Fmax
y and the corresponding vertical load
µy(Fz) = Fmax y Fz (2.14)
2.8
Friction Ellipse
The considerations seen in the preceding sections apply only in the case in which longitudinal and lateral forces are generated separately. If the tire produces
2. Tire Behavior 2.8. Friction Ellipse
0 5 10 15 20
slip angle (deg)
0 200 400 600 800 1000 1200 1400 1600 lateral force Fy (N )
Figure 2.5 Lateral force Fy due to pure lateral slip σy, for constant vertical load
Fz of 1000 N.
forces in the x and y directions simultaneously the situation can change, as the traction used in one direction limits that available in the other.
By applying a driving or braking force to a tire that has a certain lateral slip angle, the cornering force reduces. The same applies to the longitudinal force a tire can exert if called to exert a lateral force as well [3].
If Ftis the total force exerted on the wheel by the road while Fxand Fy are
its components, the resultant force coefficient can be expressed as µ = Ft
Fz
=qµ2
x+ µ2y (2.15)
and the following inequality must be respected q
F2
x+ Fy2≤ µFz (2.16)
It is then possible to obtain a polar diagram of the type shown in Figure 2.6, in which the force in the y direction is plotted against the force in the x direction for any given value of σy. Each point of the curves is characterized by
a different value of the longitudinal slip σx. In a similar way it is possible to
plot a curve Fy(Fx) at constant σxas shown in Figure 2.7.
The curves we get, go back as they approach the horizontal axis Figure 2.6 or the vertical axis Figure 2.7. This indicates that points on these curves have exceeded the peak of the longitudinal (lateral) force vs longitudinal (lateral) slip curve shown in Figure 2.4 and in Figure 2.5.
In general, an ellipse is a close approximation to the boundary of the diagram. This ”friction ellipse” represents the maximum force that the tire can generate under these operating conditions.
2. Tire Behavior 2.9. Tire Models −2000 −1500 −1000 −500 0 500 1000 1500 2000 Fx(N ) 0 500 1000 1500 2000 Fy (N ) sy=−0.004 sy=−0.008 sy=−0.012 sy=−0.016 sy=−0.02 sy=−0.03 sy=−0.04
Figure 2.6 Lateral force Fyvs longitudinal force Fxfor−1 ≤ σx≤ 1 and costant
values of σy.
2.9
Tire Models
An extensive amount of work has been done in the field of tire modeling. Many empirical models that can fit measurement data well, have been developed. All these models describe the behavior of the tire in steady-state conditions. Tire’s behavior depends on many factors including longitudinal and lateral slips, vertical load, camber angle, tire pressure etc. In order to keep the model simple, here will be considered only the dependence on the first three factors.
In addition we will only focus on models that consider the simultaneous presence of the longitudinal and lateral force.
2.9.1
Dugoff ’s Model
Dugoffs model provides for calculation of forces under combined lateral and longitudinal tire force generation [9]. The longitudinal tire force is given by
Fx= Cσ σx
1 + σx
f (λ) (2.17)
and the lateral tire force is given by Fy= Cα tan(α) 1 + σx f (λ) (2.18) where λ is given by λ = µFz(1 + σx) 2[(Cσσx)2+ (Cαtan(α))2]1/2 (2.19) and f (λ) = ( (2− λ)λ if λ < 1 1 if λ≥ 1 (2.20)
2. Tire Behavior 2.9. Tire Models 0 500 1000 1500 2000 Fx(N ) −1500 −1000 −500 0 500 1000 1500 Fy (N ) sx=−0.004 sx=−0.008 sx=−0.012 sx=−0.016 sx=−0.02 sx=−0.03 sx=−0.04
Figure 2.7 Lateral force Fyvs longitudinal force Fxfor−1 ≤ σy ≤ 1 and costant
2. Tire Behavior 2.9. Tire Models
2.9.2
Magic Formula
The Magic Formula quickly became the most predominating model. Although, over the years, several versions of the Magic Formula have been developed, they are all based on the following function
y(x) = D sin{C arctan[Bx − E(Bx − arctan(Bx)]} (2.21) where y is Fx or Fy with x being the corresponding practical or theoretical slip
component. The four coefficients have interpretations as stiffness factor (B), shape factor (C), peak factor (D), and curvature factor (E), and are unique for each of Fx and Fy.
Generally the four coefficients, depend on the vertical load. For example the peak factor D can be written as
D = µ(Fz)Fz (2.22)
with
µ(Fz) = a1Fz+ a2 (2.23)
The Magic Formula for the combined slip has many coefficient and it’s a bit complicated. For this reason, here, we will use the Magic Formula and the concept of the friction ellipse in order to create a quite simple model as suggested in [7].
Defining the total slip as
σ =qσ2
x+ σ2y (2.24)
the longitudinal and lateral forces are Fx=− σx σ Ft(Fz, σ) (2.25) Fy =−ks σy σ Ft(Fz, σ) (2.26)
where Ft(Fz, σ) is the resultant force magnitude in the form of eq. (2.21) and
ksis an attenuation factor due to the fact that the maximum friction coefficient
in the lateral direction is smaller than in the longitudinal direction.
2.9.3
Nicolas Comstock Model
According to the Nicolas Comstock Model [1] we have Fx(α, σx) = Fx(σx)Fy(α) q σ2 xFy2(α) + Fx2(σx) tan2(α) (2.27) Fy(α, σx) = q Fx(σx)Fy(α) σ2 xFy2(α) + Fx2(σx) tan2(α) (2.28) with Fx(σx) and Fy(α) respectively the longitudinal force for pure slip and the
2. Tire Behavior 2.10. Transient Behavior 0.0 0.2 0.4 0.6 0.8 1.0 −σx 0 500 1000 1500 2000 longitudinal force Fx (N )
Magic formula model Dugoff model
Nicolas Comstock model
Figure 2.8 Models comparison, for costant vertical load Fzof 1200N and α = 0.
This model gives incorrect result for small slips and a modified version is proposed by Brach and Brach. According to the modified version we have
Fx(α, σx) = Fx(σx)Fy(α)pσ2xCα2+ (1− σx)2cos2(α)Fx2(σx) Cα q σ2 xFy2(α) + Fx2(σx) tan2(α) (2.29) Fy(α, σx) = Fx(σx)Fy(α) q (1− σx)2cos2(α)Fy2(α) + sin2(α)Cσ2x Cσxcos(α) q σ2 xFy2(α) + Fx2(σx) tan2(α) (2.30)
2.9.4
Models Comparison
In Figures 2.8 and 2.9 a comparison between the three models is shown. The first thing we can see is that the Nicolas Comstock model and the magic formula model give quite similar results. However, the Dugoff model is very different. In fact the curve has no peak point and grows continuously with the increment of wheel slip ratio.Therefore has a very different trend compared to that described in section 2.6. For this reason the Dugoff model will not be used.
The friction force-wheel slip curve of the other two models has a similar trend but the magic formula model has simpler and more compact equations. For this reason we will use the magic formula model.
2.10
Transient Behavior
In general, when the vehicle makes a maneuver, tire forces do not develop in-stantaneously, but require the tire to run a certain distance to build up, due to
2. Tire Behavior 2.10. Transient Behavior 0.0 0.2 0.4 0.6 0.8 1.0 −σx 0 500 1000 1500 2000 longitudinal force Fx (N )
Magic formula model Dugoff model
Nicolas Comstock model
Figure 2.9 Models comparison, for costant vertical load Fzof 1200N and α = 7.
the flexible structure. So if we are interested in describing the transient behavior of the tire we can write as follow
˙s∗xij = (sxij − s ∗ xij) u d ˙s∗yij = (syij− s ∗ yij) u d s∗ij = q (s∗ xij) 2+ (s∗ yij) 2 Fxij = Fxij(s∗ij, s∗xij, Zij) Fyij = Fyij(s∗ij, s∗yij, Zij) (2.31)
Chapter 3
Vehicle Model
In this chapter we introduce the components that will constitute the car model which has been used. But before we begin with the detailed description of every part, it’s necessary to focus on the assumptions that we made.
First of all the road is assumed to be perfectly flat like a geometric plane as we said in chapter 2
We will neglect the inertial effects of roll motion but we will consider roll angles that are necessary for evaluating the lateral load transfers. Roll angles are due to suspension deflections and tire vertical deformations that we will assume very small.
This means placing restrictions on the lateral acceleration ay assuming to
take wide-range curves at not too high speed. These restrictions strongly depend on the global roll stiffness of the vehicle studied.
The inertial effect of pitch motion will also be neglected. This means consid-ering small longitudinal acceleration. But as for the roll motion this restriction depend on the suspension’s stiffness.
The steering system is assumed to be perfectly rigid so that the steering angle of each wheel is determined by the angular position δv of the steering
wheel, as controlled by the driver.
The vehicle body is a single rigid body and its mass (sprung mass) is very large if compared to the mass of the wheels (unsprung mass).
With these hypothesis the vehicle body has a planar motion parallel to the road and the wheel centers have a fixed position respect to it.
3.1
Reference Systems
It is necessary to define a body-fixed reference system S = (x, y, z; G) with unit vectors (i, j, k). It has origin in the center of mass G and axes fixed relative to the vehicle. The x-axis marks the forward direction, while the y-axis indicates the lateral direction. The z-axis is vertical, that is perpendicular to the road, with positive direction upward.
It’s important to define a ground-fixed reference system S0= (x0, y0, z0; O0)
with unit vectors (i0, j0, k0) in order to know the position of the vehicle in the
space. Therefore
3. Vehicle Model 3.2. Steering
Figure 3.1 Ground fixed reference system (left side) and body fixed reference system (right side)
where ψ is the vehicle yaw angle. So we can write
VG= ˙xi0+ ˙yj0= ui + vj (3.2)
with
˙x = u cos ψ− v sin ψ ˙y = u sin ψ + v cos ψ ˙
ψ = r
(3.3)
3.2
Steering
The steering system of a vehicle has the main purpose of transmitting the steer-ing inputs done by the driver to the steerable wheels, in order to control the vehicle’s direction. As stated above we consider a perfectly rigid steering system. The steering angle of the two wheels are given by
δ11=−δ01+ τ1δv+ 1 t1 2l(τ1δv) 2 δ12= δ10+ τ1δv− 1t1 2l(τ1δv) 2 (3.4)
The angle δ10is called static toe setting. We have toe-in if δ10> 0 and toe-out
if δ0
1< 0. Of course, δ01= 0 is also possible.
in and toe-out have an effect on the car’s stability and handling. Toe-out increases a cars cornering ability but decreases straight line stability. Toe-in does exactly the opposite of toe-out.
The second term τ1δv is the parallel steering. If we consider only this term,
3. Vehicle Model 3.2. Steering
Figure 3.2 Parallel steering
Figure 3.3 Steering system with Ackermann geometry
The last term is the dynamic toe setting. It is a toe angle that depends on δv and it is is based on the Ackerman principle.
Supposing to have a vehicle with rigid wheels, in Figure 3.2 is shown a parallel steering during a left corner. Point C11 is the instantaneous center of
zero velocity if we consider the inner wheel while C12is the instantaneous center
of zero velocity if we consider the outer wheel.
If both front wheels are free to follow their own natural paths, they would converge and eventually cross each other. Since the vehicle moves along a single mean path, both wheel tracks conflict continuously with each other causing wheel slip.
If now we consider the Ackermann steering we can see that both steering wheels have the same instantaneous center of zero velocity as shown in Figure 3.3
With 1= 1 we have the Ackermann geometry (dynamic toe-out) and with
1= 1 we have the anti-Ackermann geometry (dynamic toe-in). We call 1 the
3. Vehicle Model 3.3. Aerodynamic Forces
Figure 3.4 Aerodynamic forces
3.3
Aerodynamic Forces
The aerodynamic forces depends on many factors. The three components of the total aerodynamic force can be written as
Xa = 1 2ρaV 2C xSa Ya = 1 2ρaV 2C ySa Za = 1 2ρaV 2C zSa (3.5)
where ρa is the air density, V is the speed of the car relative to the air, Sa is
the cross section area of the car and Cx, Cy, Cz are the shape coefficients.
For a normal passenger car, the main influence of aerodynamics in the vehicle dynamics is the aerodynamic resistance or drag force Xa. In race cars also the
vertical force Za plays an important role. The lateral aerodynamic force Ya is
negligible.
In general the total aerodynamic force is not applied at G so we have an aerodynamic moment Ma= Maxi+ Mayj+ Mazk.
It is common practice to do like in Figure 3.4 [4] thus defining the front and rear aerodynamic vertical forces (positive upward) according to
Z1a= 1 l[Zaa2− May + Xah] = 1 2ρaV 2C z1Sa Z2a= 1 l[Zaa1+ May − Xah] = 1 2ρaV 2C z2Sa (3.6)
where Cz1 and Cz2have been introduced.
3.4
Road Tire Friction Forces
In chapter 2 we have chosen a model of tire that allows us to find the longitudinal and lateral forces as a function of the vertical load, the longitudinal slip and the lateral slip.
If Fxij and Fyij are the longitudinal and lateral forces in the tire reference system and δij is the steering angle of a wheel, the component of the tangential
3. Vehicle Model 3.5. Longitudinal and Lateral Slips
force in the vehicle reference system are given by
FTij = Xiji+ Yijj (3.7)
with
Xij = Fxijcos(δij)− Fyijsin(δij) Yij = Fxijsin(δij) + Fyijcos(δij)
(3.8) In order to have more compact equations we define
X1= X11+ X12, X2= X21+ X22 Y1= Y11+ Y12, Y2= Y21+ Y22 ∆X1= X12− X11 2 , ∆X2= X22− X21 2 (3.9)
3.5
Longitudinal and Lateral Slips
Slips depend on the speed of travel and the rolling velocity. If dijis the distance
between G and the center of a wheel the speed of travel is given by
Vij= VG+ rk× dij (3.10)
If ωij is the angular velocity of a wheel and ri its rolling radius, the rolling
velocity is equal to ωijri. So considering also the steering angles δij, according
to eq (2.8) we obtain for each tire • longitudinal slips:
σx11 =
[(u− rt1/2) cos(δ11) + (v + ra1) sin(δ11)]− ω11r1
ω11r1
σx12 =
[(u + rt1/2) cos(δ12) + (v + ra1) sin(δ12)]− ω12r1
ω12r1 σx21 = u− rt2/2− ω21r2 ω21r2 σx22 = u + rt2/2− ω22r2 ω22r2 (3.11) • lateral slips: σy11 = (v + ra1) cos(δ11)− (u − rt1/2) sin(δ11) ω11r1 σy12 = (v + ra1) cos(δ12)− (u + rt1/2) sin(δ12) ω12r1 σy21 = v− ra2 ω21r2 σy22 = v− ra2 ω22r2 (3.12)
3. Vehicle Model 3.6. Vertical Loads
3.6
Vertical Loads
The vertical loads on each tire are [4]
Z11= 0.5(Z10− Z1a+ ∆Z)− ∆Z1 Z12= 0.5(Z10− Z1a+ ∆Z) + ∆Z1 Z21= 0.5(Z20− Z2a− ∆Z) − ∆Z2 Z22= 0.5(Z20− Z2a− ∆Z) + ∆Z2 (3.13) with Z10= mga2 l (3.14) Z20= mga1 l (3.15) Z1a = 1 2ρaSaCz1u 2 (3.16) Z2a = 1 2ρaSaCz2u 2 (3.17) ∆Z =−maxh− Jzxr 2 l ' − maxh l (3.18)
The first terms, Z0
1 and Z20, are the static loads. When the vehicle is
sta-tionary the sum of the vertical loads is equal to the vehicle’s weight. The vertical loads due to the aerodynamic forces are Za
1 and Z2a while ∆Z
is the longitudinal load transfer due to the longitudinal acceleration ax.
The last terms ∆Z1and ∆Z2, are the lateral load transfers due to the lateral
acceleration ay. To get these two loads is necessary to analyze the suspensions.
3.7
Suspensions
As stated before we assume very small suspension deflections and tire defor-mations and so small roll angles. For this reason the vehicle is supposed to be always in its reference configuration. In addition the suspension linkage is supposed to be rigid and planar.
3.7.1
Roll Centers and Roll Axis
For a given suspension scheme we can define two points. Point A is the center of the tire contact patch. Point B is the instantaneous center of rotation of the wheel hub with respect to the vehicle body. The intersection of the straight lines connecting A and B on both sides gives us point Q that is called roll center. The roll center has an important property: a force applied at the roll center does not produce any suspension roll.
The line that connect the roll centers of front and rear suspensions is called roll axis. A force applied to the vehicle body at any point of the roll axis does not produce suspension roll. If q1 is the distance of the front suspension roll
center from the road and q2 is the distance of the rear suspension roll center
from the road we can define the distance of G from the roll axis as
3. Vehicle Model 3.8. Powertrain
Figure 3.5 Roll axis
with
q =a2q1+ a1q2
l (3.20)
3.7.2
Lateral Load Transfer
The lateral load transfer strongly depends on suspension and tire roll stiffness. If ks
φ1and k
s
φ2are the suspension roll stiffness of the front and rear axle respectively and kpφ1 and k
p
φ2 are the tire roll stiffness of the front and rear tires respectively we can write ∆Z1= 1 t1 kφ1 kφY (H− q) + Y1q1+ kφ1kφ2 kφ Y2q2 kφp2 − Y1q1 kpφ1 ∆Z2= 1 t2 kφ2 kφ Y (H− q) + Y2q2+ kφ1kφ2 kφ Y1q1 kφp1 − Y2q2 kpφ2 (3.21) with kφ1 = k p φ1+ k s φ1 kφ2 = k p φ2+ k s φ2 kφ = kφ1+ kφ2 (3.22)
3.8
Powertrain
The powertrain is the set of the elements that generate power and deliver it to the road surface. For a car that use a combustion engine this includes the engine, clutch, transmission and the differential. Here we will use an electric engine so the clutch and the transmission are not necessary.
3. Vehicle Model 3.8. Powertrain
Figure 3.6 Schematic representation of the powertrain
3.8.1
Engine
The engine is the component that provides the power needed to move the vehi-cle. Until a few years ago, cars used only combustion engines. Nowadays, the tendency of car manufacturers is to equip cars in the future years with electric motors.
For our purpose we use a brushless permanent magnet synchronous engine. This type of engine has a torque characteristics shown in Figure 3.7. The red line is the maximum shaft torque while the blue line is the nominal torque and delimits the continuous operating region. Between the two line there is the intermittent operating region.
In the intermittent operating region, the engine can run for a limited time. For this reason our engine will only work in the continuous operating region. As we can see from Figure 3.8, there are two areas. In the first area the engine works at constant torque Tn. In the second area, the engine works at constant
power Pn.
The torque supplied by the motor will be given by
Te= aTlim (3.23)
where a is the throttle pedal position (whose value is between 0 and 1) and Tlim
is Tlim= ( Tn, if ωe≤ ωe,corner Pn/ωe, if ωe≤ ωe,corner. (3.24) The rotational dynamics of the engine is then
Je˙ωe− Te+ Td= 0 (3.25)
where Td is the torque due to the differential.
3.8.2
Differential
The differential is the device that divides the torque input from the propeller shaft between the two output shafts to the wheels, allowing them to rotate at different speeds when necessary, for example during a corner.
3. Vehicle Model 3.8. Powertrain
0 100 200 300 400 500 600 700 800 900
engine speed (rad/s) 0 50 100 150 200 250 300 engine torque
(N) Intermittentoperating region
Continuous operating region
Figure 3.7 Engine torque characteristics: intermittent and continuous operating regions
0 100 200 300 400 500 600 700 800 900
engine speed (rad/s) 0 50 100 150 200 250 300 engine torque (N)
Constant torque Constant power
Figure 3.8 Engine torque characteristics: constant torque and constant power areas
3. Vehicle Model 3.8. Powertrain
Figure 3.9 Open differential
In the case of an ideal differential, also called open differential, the engine torque is divided equally between the two output shafts because the internal friction is negligible.
In some conditions, however, one of the two wheels should receive a higher torque than the other. For example, if one of the two wheels has low grip it should receive a lower torque than the other. Otherwise, the car would not accelerate much, as the maximum longitudinal force would be limited by the wheel with low grip.
This problem can be reduced by using the limited slip differentials. They have a friction torque inside them that allows a torque difference between the two shafts.
When we speak about limited slip differentials, it’s necessary to distinguish between self-locking differentials and controlled differentials. In the first, the locking torque is determined by the mechanical properties of the system while in the second, it is controlled by electronic devices.
The scheme of an open differential is shown in Figure 3.9. It consists of a cage 6, which is supported by the ball bearings CC and is therefore free to rotate about the axis XX of the road wheels. Fixed to the cage 6 is a crown wheel B driven by the bevel pinion A. There are two shafts E1 and E2, the
outer ends of which are connected to the road wheels. Their inner ends pass through the bosses of the differential cage 6 in which they are quite free to turn. Inside the differential cage, their ends are splined into the bevel wheels 1 and 2 with which the bevel pinions 3 and 4 mesh. The pinions 3 and 4 are free to turn on the pin 5 fixed in the differential cage [8].
If τ is the gear ratio between the crown wheel B and the bevel pinion A, the torque and the angular speed of the cage are given by
Th= Td/τ (3.26)
3. Vehicle Model 3.8. Powertrain
Figure 3.10 ZF differential
In each differential mechanism, the relationship between the angular speed of the cage (ωh) and that of the two wheels (ω21and ω22), is given by the Willis
formula
ω21+ ω22= ωh (3.28)
while the rotational dynamics of the cage is
Jh˙ωh= Th− T21− T22 (3.29)
where Jh is the moment of inertia of the cage and T21 and T22 are the torques
that the two wheels exert on the differential. If we define
T =Th− Jh˙ωh
2 (3.30)
For an open differential we have
T21= T
T22= T
(3.31) while for a self-locking differential we have
T21= T /2± ∆T
T22= T /2± ∆T
3. Vehicle Model 3.8. Powertrain 0 50 100 150 200 250 300 350 Th(N) 0 10 20 30 40 50 60 70 ∆ T (N)
Figure 3.11 Differential allowed torque difference between left and right wheels as function of the input torque
where ∆T is generated by the internal friction. During power-on (positive Th)
the slower wheel receives the highest torque while during power-off (negative Th), the slower wheel receives the lowest torque [4].
To better understand how a self-locking differential works, we will briefly describe the ZF differential whose scheme is shown in Figure 3.10.
If compared to the open differential, the main difference is that the shaft 5 around which the conical wheels 3 and 4 rotate, is not directly attached to the cage 8. In fact it is between two pressure plate that rotate with the cage and can freely translate respect to it.
There are two clutch packs. One is between the cage and the pressure disc 6 while the other is between the cage and the pressure disc 7. Each clutch pack can freely translate and is made up of steel plates 9 (integral with the cage) and clutch discs 10 (integral with the shaft).
Pressure plates show a V shaped groove. The shaft 5, pressing against the V-groove, exerts a force on the two pressure discs that is proportional to Th[8].
As a result, pressure plates compress the clutch packs. Thus when the wheels rotate at different speed, clutch packs exert on the shafts E1 and E2 a torque
that is proportional to Th.
In some cases two springs 11 are added to generate a preload. So the locking torque features a minimum locking value also when Th is equal to zero.
Figure 3.11 shows the plot of ∆T vs T for a ZF differential without preload. All the points inside the green region represent possible working conditions. On the red line ∆T assumes its maximum value since ∆ω6= 0 while below the red line, ∆T is unknown (can assume values between 0 and the maximum) since the differential is locked and we have ∆ω = 0. This is due to the value assumed by the friction coefficient of the clutch packs whose trend is shown in Figure 3.12
3. Vehicle Model 3.8. Powertrain −1.0 −0.5 0.0 0.5 1.0 ω (rad/s) −2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 µ µmax (-) real friction approximate friction
Figure 3.12 Real friction and approximate friction The internal efficiency of a differential is given by
µ = |T/2| − ∆T
|T/2| + ∆T (3.33)
so it is possible to obtain
∆T = 1− µ
2(1 + µ)|T | (3.34)
If now we suppose to have Th> 0 (power-on) and ω21> ω22 (so T22> T21)
from eq. 3.32 and eq. 3.34 we obtain
T21= µT22 (3.35)
while if Th> 0 and ω22> ω21 (so T21> T22) we have
T22= µT21 (3.36)
Since in our study Th≥ 0 we can describe the differential behavior with the
following model
T21= T22µff(∆ω) (3.37)
where
ff(∆ω) = tanh(10∆ω) (3.38)
Obviously this is a simplified model since the friction was approximated by the function ff(∆ω) whose trend is shown in Figure 3.12
Summing up, the equations related to the differential mechanism that will be used are
ω21+ ω22− ωh= 0
Th− T21− T22= Jh˙ωh
T21− T22µff(∆ω)= 0
3. Vehicle Model 3.9. Braking System
Figure 3.13
3.9
Braking System
The braking system is necessary to reduce the speed of the car. The total friction torque Tft is divided between the two axes by the braking ratio coefficient kb. Generally kb is greater than 0.5 because of the vertical loads.
In fact during braking, due to the longitudinal load transfer, the vertical loads acting on the front wheels are greater than those that act on the rear wheels. For this reason front tires can exert greater longitudinal forces.
The friction torque of each axle is divided equally between the two wheels. Therefore the friction torque on each wheel is
Tf11 = b kb 2Tftff(ω11) Tf12 = b kb 2Tftff(ω12) Tf21 = b (1− kb) 2 Tftff(ω21) Tf22 = b (1− kb) 2 Tftff(ω22) (3.40)
where b is the braking pedal position (whose value is between 0 and 1) and
ff(ωij) = tanh(10ωij) (3.41)
As we have seen for the differential, even in this case, the function ff(ωij)
is used to approximate the friction.
3.10
Equilibrium Equations
All the aspects we have analyzed converge in the equations of equilibrium. Sum-ming up, we have three equations for the vehicle’s equilibrium
max= X1+ X2−1 2ρaSaCxu 2 may = Y1+ Y2 Jz˙r = Y1a1− Y2a2+ ∆X1t1+ ∆X2t2 (3.42)
3. Vehicle Model 3.10. Equilibrium Equations
with
ax= ˙u− vr
ay= ˙v + ur
(3.43) There are one equation for the equilibrium of the engine and one for that of the differential
Je˙ωe= Te− Td
Jh˙ωh= Th− T21− T22
(3.44) and finally one equation for the equilibrium of each tire
Jt˙ω11=−Tf11− Tr11− Fx11R Jt˙ω12=−Tf12− Tr12− Fx12R Jt˙ω21=−T21− Tf21− Tr21− Fx21R Jt˙ω22=−T22− Tf22− Tr22− Fx22R
(3.45)
where Jtis the moment of inertia of each tire, R is the radius of each tire and
Tr is the rolling resistance.
In addition it is necessary to take in consideration the differential equations that link the speeds in the vehicle’s reference system to those in the fixed refer-ence system.
˙x = u cos ψ− v sin ψ ˙y = u sin ψ + v cos ψ ˙
ψ = r
Chapter 4
Optimal Control Problems
As we sad in the introduction both simulation and optimization will be studied as optimization problems. So in this chapter will be explain how these problems can be solved.
We have a continuous time optimal control problem. The goal is to minimize an objective function by respecting some constraints to which the problem is subject. This kind of problem can be stated as follows [2]
min
x,u
Z T
0
L(x(t), u(t)) dt + E(x(T )) (objective function)
s.t. x(0)− x0= 0, (fixed initial value)
˙x(t)− f(x(t), u(t)) = 0, t ∈ [0, T ], (ODE model) h(x(t), u(t))≤ 0, t ∈ [0, T ], (path constraints)
r(x(T ))≤ 0, (terminal constraints)
(4.1)
where the integral cost contribution L(x, u) is called the Lagrange term and E(x(T )) is called Mayer term. Their sum is called a Bolza objective.
There are three types of approaches to solve these problems: state-space approaches, indirect methods, direct methods.
State-space approaches use the principle of optimality that states that each subarc of an optimal trajectory must be optimal. This leads to the so-called Hamilton-Jacobi-Bellman (HJB) equation, a partial differential equation in the state space. This approach has the great advantage to solve the problem up to global optimality.
Unfortunately it has exponential complexity in the dimension of the state space. So it is restricted to small state dimensions.
Indirect methods use the necessary conditions of optimality of the infinite problem to derive a boundary value problem (BVP) in ordinary differential equations (ODE) that must numerically be solved. The problem of this method is that the ODEs are often difficult to solve due to strong nonlinearity and instability.
Direct methods discretize the continuous time problem, then apply NLP (nonlinear programming problem) techniques to the resulting finite-dimensional optimization problem.
All direct methods discretize the control function in the same way. The difference lies in how they manage the state trajectory.
4. Optimal Control Problems 4.1. Direct Single Shooting
Direct methods are nowadays the most widespread and successfully used techniques. For this reason here below we will focus on three direct methods: direct single shooting, direct multiple shooting, direct collocation.
4.1
Direct Single Shooting
In the direct single shooting method, the control function u(t) is parametrized using some piecewise smooth approximation, typically piecewise constant. If the vector q denote the finite control parameters, the control function becomes u(t; q).
When we use piecewise constant controls, with a fixed grid 0 = t0 < t1 <
... < tn= N and N parameters qi, we can write
u(t; q) = qi if t∈ [ti, ti+1] (4.2)
The single shooting is a sequential approach. This means that the states x(t) on [0, T ] are dependent variables obtained by a forward integration of the dynamic system , starting at x0and using the controls u(t; q).
4.2
Direct Multiple Shooting
Also in the direct multiple shooting the control function u(t) is discretized on a grid. The difference is that it solves the ODE separately on each interval [ti, ti+1], starting with artificial initial values si:
˙xi(t; si, qi) = f (xi(t; si, qi), qi), t∈ [ti, ti+1]
xi(t; si, qi) = si
(4.3) It is then necessary to add equality constraints (xi(ti+1; si, qi)− si+1 = 0)
to ensure continuity of the trajectory. Furthermore for each trajectory piece xi(t; si, qi) we numerically compute the Lagrange term of the cost function
li(si, qi) = Z ti+1 ti L(xi(t; si, qi), qi) dt (4.4) So problem 4.1 becomes min x,u N−1 X i=0
li(si, qi) + E(sn) (objective function)
s.t. x0− s0= 0, (initial value)
xi(ti+1; si, qi)− si+1= 0 (continuity)
h(si, qi)≤ 0 (discretized path constraints)
r(sn)≤ 0, (terminal constraints)
(4.5)
4.3
Direct Collocation
With the direct collocation the control function u(t) is discretized on a grid [t0, t1, ..., tN]. On each interval [ti, ti+1] a set of m collocation points t(1)i , ..., t
(m) i
4. Optimal Control Problems 4.4. Objective Function
is chosen and the trajectory is approximated by a polynomial pi(t; vi) with
coefficient vector vi.
It’s then necessary to respect two conditions: the polynomial must start at si and its derivative at each collocation point must match the function f at
the corresponding value of the polynomial. This result in a system that can be summarized by the vector equation ci(si, vi, qi).
In addition the continuity conditions across interval boundaries require that pi(ti+1; vi)− si+1= 0.
The Lagrange term of the cost function on the collocation intervals is ap-proximated by a quadrature formula using the same collocation points.
Z ti+1 ti L(x, u) dt≈ li(si, vi, qi), i∈ [0, 1, ..., N − 1] (4.6) So problem 4.1 becomes min x,u N−1 X i=0
li(si, vi, qi) + E(sn) (objective function)
s.t. x0− s0= 0, (initial value)
ci(si, vi, qi) = 0 (collocation conditions)
pi(ti+1; vi)− si+1= 0 (continuity conditions)
h(si, qi)≤ 0 (discretized path constraints)
r(sn)≤ 0, (terminal constraints)
(4.7)
The nonlinear program (NLP) in single shooting is small, but often highly nonlinear, whereas the NLP for multiple-shooting is larger, but less nonlinear and more sparse . Finally direct collocation leads to the largest but at the same time most sparse NLP.
Since a sparse structure increase the efficiency and reduce the computation time of the solver, we will use the direct collocation method.
4.4
Objective Function
The number of Lagrangian terms that compose our objective function varies depending on the optimization problem. Each term is penalized with a weight (K) in order to set its influence on the objective function.
In most optimization problems studied, the vehicle is required to run a given trajectory in the shortest time. In order to minimize the total time we have reasoned as follows.
First of all it’s necessary to define the time step h
h = T /N (4.8)
where T is the total time (unknown) and N is the number of nodes. So h is a variable.
We have introduced a constraint that makes the value of h at step k equal to the one at step k+1. So we have
4. Optimal Control Problems 4.4. Objective Function 5.8 5.9 6.0 6.1 6.2 6.3 Tmin(s) −1000 0 1000 2000 3000 4000 5000 6000 1 Ls
Figure 4.1 Variation of the minimum time Tmin for different values of Ks
Then we can compute the Lagrange term related to the minimum time as follow Lt= Kt N−1 X k=0 h2 k (4.10)
Another important requirement is that the steering angle vary gradually. In other words we want the curve of the steering angle as a function of time to be smooth. It is therefore necessary to introduce a Lagrange term related to the changes in the value of the steering angle.
Ls= Ks N−2 X k=0 δ vk+1− δvk hk 2 (4.11) The question is how do we choose the value of Ks? To answer this question
we have to keep in mind that an increase of Ksmakes the curve δvvs t smoother
but increases the total time.
If we compute the total time necessary to run a given trajectory in the minimum time (Tmin) for different values of Kswe obtain a curve like the one
shown in Figure 4.1
For too high values of Ks we get an excessive total time and unrealistic
trajectory. For too low values we get a low total time but an unrealistic trend of the curve δv vs t. Between these two conditions there is a region where we
can pick the value of Ks.
The last two Lagrange terms that we have used are related to the changes in the value of the throttle pedal position (La) and the braking pedal position
4. Optimal Control Problems 4.5. Constraints (Lb). So we have La = Ka N−2 X k=0 ak+1− ak hk 2 (4.12) Lb= Kb N−2 X k=0 bk+1− bk hk 2 (4.13) The considerations made for the choice of Ka and Kb are the same we did
for the choice of Ks.
4.5
Constraints
All the constraints have to be expressed as inequalities. So for each constraint we have to set a lower and an upper bound. Obviously an equality constraint will have the lower and the upper bound equal to zero.
The discretization scheme we will use is the mid-point rule that can be expressed as follow ˙x = f (x, u)→ xk+1h− xk k = f (¯x, u) (4.14) with ¯ x = xk+1+ xk 2 (4.15)
Thus, for example the first of 3.46 becomes 0≤ Je
xk+1− xk
hk − ¯u cos ¯
ψ + ¯v sin ¯ψ≤ 0 (4.16) The simulation problems have been studied as fake optimization problems in which the objective function is set equal to zero and the constraints are the differential equations that describe the dynamic evolution of our system. For the optimization problems, we have added two other very important constraints: the path constraint and the pedal’s position constraint.
4.5.1
Path Constraint
It is important to ensure that the vehicle remains inside the road during the optimization. The road is described using the center line arc length (s). For each value of s a normal vector and a tangent vector can be defined.
If pgis the position vector of the car expressed in the fixed reference system
and ps = f (s) is the position vector of the center line for each value of s,
expressed in the fixed reference system, the path constraint can be written as follow
pg− ps− wn = 0 (4.17)
where n is the normal vector and w is the road width whose value is between wi and we.
4. Optimal Control Problems 4.6. Problem Initialization
Figure 4.2 Path scheme
4.5.2
Pedal’s Position Constraint
In order to reproduce the behavior of a human pilot, the accelerator and brake pedal must not be pressed simultaneously. For this reason we have introduced the following constraint
ab = 0 (4.18)
4.6
Problem Initialization
Due to the complexity of the problem, to find valid initial guess was quite diffi-cult. The path we followed was to use different models of increasing complexity. The solution of the simplest model was used to estimate the initial guess for the second model and so on.
Chapter 5
Main Simulations and
Results
Starting immediately with the study of the complete model resulted too com-plex especially for optimization problems. So we have studied many models increasing the complexity. In this chapter we describe each model and the as-sumptions that have been made for each of them. Moreover, will be presented and discussed the results obtained for the complete model.
5.1
Single-track Models
All these models are united by the following hypothesis: the two tire slip angle on a single axle are the same, so both axles (the front and the rear) are reducible into a single wheel. If we combine eq. (2.9), eq. (3.11) and eq.(3.12) we obtain
α11=− (v + ra1) cos(δ11)− (u − rt1/2) sin(δ11) (u− rt1/2) cos(δ11) + (v + ra1) sin(δ11) α12=− (v + ra1) cos(δ12)− (u + rt1/2) sin(δ12) (u + rt1/2) cos(δ12) + (v + ra1) sin(δ12) α21=− v− ra2 u− rt2/2 α22=− v− ra2 u + rt2/2 (5.1)
Thus the assumption we have made has some consequences. First of all the steering angle is the same for both wheels of the front axle. Furthermore the longitudinal velocity of the vehicle u must be much higher than the vehicle yaw rate.
5. Main Simulations and Results 5.1. Single-track Models
the same angular velocity. So the longitudinal and lateral slips become σx11 = σx12 = [u cos(δ) + (v + ra1) sin(δ)]− ω1r1 ω1r1 σx21 = σx22 = u− ω2r2 ω2r2 σy11 = σy12 = (v + ra1) cos(δ)− u sin(δ) ω1r1 σy21 = σy22 = v− ra2 ω2r2 (5.2)
To make these models very simple, the engine, the braking system and the dynamics of the wheels are not considered. For the same reason we don’t con-sider the effect of the simultaneous presence of longitudinal and lateral force. In other words the longitudinal force can reach its maximum value even if the lateral force is not zero and vice versa. The inputs are the angular speed of the rear wheel ω2 and the steering angle of the front wheel δv
Since we neglect the braking system and the car is rear wheel drive, it is permissible to assume the longitudinal slip of the front wheel equal to zero. So we can just write
σx2 = u− ω2r2 ω2r2 σy1 = (v + ra1) cos(δ)− u sin(δ) u cos(δ) + (v + ra1) sin(δ) σy2 = v− ra2 ω2r2 (5.3)
Finally the last equation of (3.42) becomes clearly
Jz˙r = Y1a1− Y2a2 (5.4)
The first single track model we studied uses the linear approximation of the tire-road friction forces. The second model we studied describes the tire-road friction forces with the magic formula and finally the third model takes into account the transient behavior. The following briefly describes the differences between these three models.
5.1.1
Linear Force vs Magic Formula
A comparison is now proposed between the linear model and the magic formula model. As we can see from Figure5.1, the linear approximation gives reliable results only if the absolute value of both longitudinal and lateral slips is low (less than 0.025).
Suppose we want to find the minimum time required to traverse half circum-ference. We consider a radius of 40 meters and a constant angular speed of 40 rad/s. Of course the initial and final condition for the two models are the same. Figure 5.2 and Figure 5.3 show a comparison of the results obtained. Since the longitudinal and lateral slips are less than 0.025 the two models behave almost the same way.
Let’s look at what happens when the slips are greater than 0.025. Therefore we consider a radius of 40 meters and a constant angular speed of 82 rad/s.
5. Main Simulations and Results 5.1. Single-track Models 0.0 0.1 0.2 0.3 0.4 0.5 −σx 0 200 400 600 800 1000 1200 1400 longitudinal force Fx (N)
Figure 5.1 Comparison between the linear approximation and the magic for-mula. −60 −40 −20 0 20 40 60 x(m) −10 0 10 20 30 40 50 y (m) T rajectorylin T rajectoryM F
Figure 5.2 Comparison between the linear single-track model and the magic formula single-track model for ωh = 40 and circumference radius of
5. Main Simulations and Results 5.1. Single-track Models 0 20 40 60 80 100 120 140 s(m) −0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12 steering angle (rad) δlin v δM F v 0 20 40 60 80 100 120 140 s(m) −1 0 1 2 3 4 5 6 7 lateral acceleration (m/s 2) alin y aM F y 0 20 40 60 80 100 120 140 s(m) −0.025 −0.020 −0.015 −0.010 −0.005 0.000 0.005 lateral slips (-) σlin y1 σM F y1 σlin y2 σM F y2 0 20 40 60 80 100 120 140 s(m) −200 0 200 400 600 800 1000 1200 axle characteristics (N) Ylin 1 YM F 1 Ylin 2 YM F 2 0 20 40 60 80 100 120 140 s(m) −0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ya w rat w (rad/s) rlin rM F 0 20 40 60 80 100 120 140 s(m) −0.05 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 lateral velo cit y (m/s) vlin vM F
Figure 5.3 Comparison between the linear single-track model and the magic formula single-track model for ωh = 40 and circumference radius of
5. Main Simulations and Results 5.2. Four Wheel Models −60 −40 −20 0 20 40 60 x(m) −10 0 10 20 30 40 50 y (m) T rajectorylin T rajectoryM F
Figure 5.4 Comparison between the linear single-track model and the magic formula single-track model for ωh = 40 and circumference radius of
40 m. The point indicates the starting position of the car
As we can see from Figures 5.4 and 5.5 the results we get are very different. The linear model reaches higher lateral forces and therefore follows a narrower trajectory in a shorter time.
5.1.2
Steady State vs Transient Behavior
It is interesting to see the influence of the transient behavior on the vehicle dynamic. As we said in section 2.10 when the vehicle makes a maneuver, tire forces do not develop instantaneously, but require the tire to run a certain distance to build up. So we simulate a step steering input in order to see what happen during the transitory.
The simulation will be made on two models. The first does not consider the transient behavior while the second considers it.
As we can see from Figure 5.6 the two models behave differently. In fact when we consider the transient behavior, a discontinuity in the slips does not produce a discontinuity in the forces.
Of course when the slips have no discontinuity, the differences between the two models aren’t so great. For example if we consider a comparison between the two models for a slow increase in the steering angle we get the results shown in Figure 5.7
Since that circuit driving is known to be smooth, in the optimization prob-lems the transient behavior can be neglect.
5.2
Four Wheel Models
In these models, both the front and the rear axle have two wheels. So all the slip angles can be different.
5. Main Simulations and Results 5.2. Four Wheel Models 0 20 40 60 80 100 120 140 s(m) 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 steering angle (rad) δlin v δM F v 0 20 40 60 80 100 120 140 s(m) −5 0 5 10 15 20 lateral acceleration (m/s 2) alin y aM F y 0 20 40 60 80 100 120 140 s(m) −0.08 −0.06 −0.04 −0.02 0.00 lateral slips (-) σlin y1 σM F y1 σlin y2 σM F y2 0 20 40 60 80 100 120 140 s(m) −500 0 500 1000 1500 2000 2500 3000 axle characteristics (N) Ylin 1 YM F 1 Ylin 2 YM F 2 0 20 40 60 80 100 120 140 s(m) 0.0 0.2 0.4 0.6 0.8 ya w rat w (rad/s) rlin rM F 0 20 40 60 80 100 120 140 s(m) −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0.0 0.1 lateral velo cit y (m/s) vlin vM F
Figure 5.5 Comparison between the linear single-track model and the magic formula single-track model for ωh = 40 and circumference radius of
5. Main Simulations and Results 5.2. Four Wheel Models 0.0 0.2 0.4 0.6 0.8 1.0 t(s) −0.035 −0.030 −0.025 −0.020 −0.015 −0.010 −0.005 0.000 0.005 lateral slips (-) stransient y1 ssteady−state y1 stransient y2 ssteady−state y2 0.0 0.2 0.4 0.6 0.8 1.0 t(s) 0 200 400 600 800 1000 1200 axle characteristics (N) Ytransient 1 Y1steady−state Ytransient 2 Y2steady−state 0.0 0.2 0.4 0.6 0.8 1.0 t(s) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 ya w rat w (rad/s) rtransient rsteady−state 0.0 0.2 0.4 0.6 0.8 1.0 t(s) 0.00 0.02 0.04 0.06 0.08 0.10 0.12 lateral velo cit y (m/s) vtransient vsteady−state
Figure 5.6 Step steering maneuver simulation for two different single-track mod-els. One does not consider the transient behavior while the other considers it
5. Main Simulations and Results 5.2. Four Wheel Models 0.0 0.2 0.4 0.6 0.8 1.0 t(s) −0.007 −0.006 −0.005 −0.004 −0.003 −0.002 −0.001 0.000 lateral slips (-) stransient y1 ssteady−state y1 stransient y2 ssteady−state y2 0.0 0.2 0.4 0.6 0.8 1.0 t(s) −50 0 50 100 150 200 250 300 350 axle characteristics (N) Ytransient 1 Y1steady−state Ytransient 2 Y2steady−state 0.0 0.2 0.4 0.6 0.8 1.0 t(s) −0.05 0.00 0.05 0.10 0.15 0.20 0.25 ya w rat w (rad/s) rtransient rsteady−state 0.0 0.2 0.4 0.6 0.8 1.0 t(s) −0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12 lateral velo cit y (m/s) vtransient vsteady−state
Figure 5.7 Simulation of a slow increase of the steering angle for two different single-track models. One does not consider the transient behavior while the other considers it.
5. Main Simulations and Results 5.2. Four Wheel Models
We started with a four wheel model that does not consider the dynamics of the wheels, the engine ,the braking system, and the differential’s house dynam-ics. So the equations related to the differential become
ω21+ ω22− ωh= 0
Fx21− Fx22µ
ff(∆ω)= 0 (5.5)
As for the single-track model, since we neglect the braking system and the car is rear wheel drive, it is permissible to assume the longitudinal slip of the front wheel equal to zero. So the front lateral slips become
σy11 = (v + ra1) cos(δ11)− (u − rt1/2) sin(δ11) (u− rt1/2) cos(δ11) + (v + ra1) sin(δ11) σy12 = (v + ra1) cos(δ12)− (u + rt1/2) sin(δ12) (u + rt1/2) cos(δ12) + (v + ra1) sin(δ12) (5.6)
The longitudinal and the lateral slips of the rear wheels are that of eq. (3.11) and eq. (3.12).
Finally we studied the complete model that consider all the elements de-scribed in chapter 3.
Here to follow, will be shown and discussed extensively the results obtained with this model related to the problem of finding the minimum time required to travel the following section of track: a straight of 50 m, a left-hand turn with a center-line radius of 20 m and a final straight of 50 m. The track has a width of 10 m.
The car has an initial speed of about 32 m/s, and it is on the outer edge of the track. The only final condition we introduced is that the car must be parallel to the track (ψ = π).
We have studied two cases: in the first the car is equipped with an open differential while in the second is equipped with a limited slip differential.
5.2.1
Open Differential
The results obtained for the car equipped with an open differential are shown in Figure 5.10, Figure 5.8 and Figure 5.9. Considering the accelerator and brake pedal position shown in Figure 5.8 we can identify four phases.
In the first phase (−50 ≤ s < −42) the accelerator pedal is pressed and the car accelerates. In the second phase (−42 ≤ s < −7), that we will call braking phase, the car starts to brake to reduce the speed. In the third phase (−7 ≤ s < 11) both the pedals are not pressed while in the fourth phase (11 ≤ s ≤ 110), that we will call acceleration phase, the accelerator pedal position goes from zero to the maximum value that keeps up to the end.
The first phase is the most simple since the car goes straight and accelerate to increase the speed. The other phases are the most interesting. They are not so simple because the car travels the curve. For this reason we will focus on these phases.
Braking Phase
During this phase, if we see the plot of the longitudinal forces we notice that, during braking, the rear longitudinal forces are very low. This happens for
5. Main Simulations and Results 5.2. Four Wheel Models −40−20 0 20 40 60 80 100 s(m) 0.0 0.5 1.0 1.5 2.0 pe da ls po si ti on (-) accelerator pedal a brake pedal b −40−20 0 20 40 60 80 100 s(m) −0.02 −0.01 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 st ee ri ng an gl e (r ad ) δv −40−20 0 20 40 60 80 100 s(m) −0.2 −0.1 0.0 0.1 0.2 0.3 lo ng it ud in al sl ip s (-) σx11 σx12 σx21 σx22 −40−20 0 20 40 60 80 100 s(m) −1500 −1000 −500 0 500 1000 1500 lo ng it ud in al fo rc es (N ) Fx11 Fx12 Fx21 Fx22 −40−20 0 20 40 60 80 100 s(m) −0.20 −0.15 −0.10 −0.05 0.00 0.05 la te ra ls lip s (-) σy11 σy12 σy21 σy22 −40−20 0 20 40 60 80 100 s(m) −200 0 200 400 600 800 1000 1200 1400 la te ra lf or ce s (N ) Fy11 Fy12 Fy21 Fy22
5. Main Simulations and Results 5.2. Four Wheel Models −40 −20 0 20 40 60 80 100 s(m) 70 80 90 100 110 120 130 140 wheels angular sp eed (rad/s) ω11 ω12 ω21 ω22 −40 −20 0 20 40 60 80 100 s(m) −300 −250 −200 −150 −100 −50 0 T orques (Nm) T21 T22 −40 −20 0 20 40 60 80 100 s(m) 200 400 600 800 1000 1200 1400 1600 V ertical loads (N) Z11 Z12 Z21 Z22 −40 −20 0 20 40 60 80 100 s(m) 20 22 24 26 28 30 32 34 longitudinal velo cit y (m/s) u −40 −20 0 20 40 60 80 100 s(m) −2.5 −2.0 −1.5 −1.0 −0.5 0.0 0.5 lateral velo cit y (m/s) v −40 −20 0 20 40 60 80 100 s(m) −0.2 −0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 ya w rate (rad/s) r
5. Main Simulations and Results 5.2. Four Wheel Models −80 −60 −40 −20 0 20 x (m) −60 −40 −20 0 20 40 y (m) Trajectory
Figure 5.10 Trajectory obtained the for the 90 degree corner. The point indi-cates the starting position of the car
−10 0 10 20 30 40 50 x (m) −60 −50 −40 −30 −20 −10 0 y (m ) brake
5. Main Simulations and Results 5.2. Four Wheel Models
three reasons: Cf1j > Cf2j because kb= 0.75, secondly the inertia of the engine produces the torques T21< 0 and T22< 0 (it is evident in Figure 5.9) that reduce
the braking torque according to eq. (3.45) and finally, due to the longitudinal load transfer, the vertical forces acting on the front wheels are greater than the vertical forces acting on the rear wheels.
In Figure 5.11 is shown in detail the brake pedal position b along the tra-jectory. We can identify two region (A and B). In the first region (A) the value of b is constant and equal to one while in the second region it is constant but lower than one. Until the value of b is one, the trajectory is almost straight and the lateral forces are very low.
When the car enters the corner the lateral forces must increase. We have to remember that since we use a combined model for the tires, the maximum lateral force that a tire can reach is reduced if it has to exert a longitudinal force.
For this reason, the value of b decreases in order to have the optimum between the value of the longitudinal force and the value of the lateral force. Of course since the longitudinal force of the rear wheels are low this argument is referred to the front wheels.
In Figure 5.12 is shown the normalized longitudinal friction (µx/µxmax) vs the normalized lateral friction (µy/µymax) for the four wheels.
We can notice that the front inner wheel reaches the maximum grip (the edge of the circumference). After that, the normalized friction. Instead the outer wheel reaches the maximum grip at the end of the braking phase.
This happen because, the longitudinal slip σx11 (and so the total slip σ11) grows faster than the longitudinal slip σx12 (and so the total slip σ12). This leads to exceed the maximum value of the tire characteristic, going to fall to the right of the maximum point. Area in which you do not use all the available adhesion.
Exceeding the maximum is also due to a reduction of the vertical load Z11.
In fact, when the vertical load is reduced, the maximum point moves to the left (to lower values of the slip).
The effect of all this can be seen graphically in Figure 5.15 where the points represent the value of Ftij(Zij, σij), in three successive time instants of the braking phase. Considering the front inner wheel, it is clear the reduction of the vertical load Z11 (the points move downward) and the increase of the slip
(the points move to the right). Instead, considering the front outer wheel, there is an increase of the vertical load and a lower increase of the slip
We can make the same remarks for the rear wheels with one difference. At the end of the braking phase, the value of b goes to zero. After that both the peals are not pressed until the car enters in the last phase that we will call acceleration phase. Probably this happens because in this part of the curve the lateral forces must be as high as possible.
Acceleration Phase
When the car enters in the acceleration phase, the value of a begins to grow. As we said for the braking phase, since the car is still traveling the curve, the trend of a is the one that guarantees the optimum between the value of the longitudinal force and the value of the lateral force.
5. Main Simulations and Results 5.2. Four Wheel Models −1.0 −0.5 0.0 0.5 1.0 µy/µymax(-) −1.0 −0.5 0.0 0.5 1.0 µx /µ xmax (-) µ11/µ11max µ12/µ12max −1.0 −0.5 0.0 0.5 1.0 µy/µymax(-) −1.0 −0.5 0.0 0.5 1.0 µx /µ xmax (-) µ21/µ21max µ22/µ22max
Figure 5.12 Friction circle for the front wheels (upper) and for the rear wheels (lower) during the braking phase
5. Main Simulations and Results 5.2. Four Wheel Models −40 −35 −30 −25 −20 −15 −10 200 400 600 800 1000 1200 1400 1600 fron t wheels vertical loads (N) Z11 Z12 −40 −35 −30 −25 −20 −15 −10 −1500 −1000 −500 0 500 1000 1500 fron t wheels forces (N) Fx11 Fx12 Fy11 Fy12 −40 −35 −30 −25 −20 −15 −10 70 80 90 100 110 120 130 140 fron t wheels angular sp eed (rad/s) ω11 ω12 −40 −35 −30 −25 −20 −15 −10 −0.2 −0.1 0.0 0.1 0.2 0.3 0.4 fron t wheels slips (-) σx11 σx12 σy11 σy12 σ11 σ12 −40 −35 −30 −25 −20 −15 −10 s(m) 0.0 0.5 1.0 1.5 2.0 brak e p edal p osition (-) brake pedal b
Figure 5.13 Detail of the obtained results for the front wheels during the braking phase.
