Calculate the slip angle β i for each wheel

From the longitudinal velocity, the lateral velocity and the yaw angle velocity of the center of mass for the vehicle, as shown in Figure 2.1, we would figure out:

the velocity vector of the right-front wheel,

1 2

f x

V = −U T × r ， V_y1 = + × V a r (2-4a)

the velocity vector of the left-front wheel,

2 2

f x

V = +U T × r ， V_y2 = + × V a r (2-4b)

the velocity vector of the right-rear wheel,

3 2

r x

V = − × rU T ， V_y3 = − × V b r (2-4c)

and the velocity vector of the right-rear wheel.

4 2

Then comes the equation,

( )

2.3.2 Calculate the angle of inclinationγ for each wheel

The angle of inclination for each wheel can be obtained by the equation (2-6).

/

i u s u f

γ φ φ= + × γ (2-6)

2.3.3 Torque on the drive wheels

The torque transmission starts from engine torque, clutch, gearbox, transmission shaft, differential mechanism, wheel axle, to wheels. Here come some factors that affect the torque among vehicle wheels, which are accelerating or braking accelerationa′ produced by the driver, the drive design (either driven _x by front wheels, rear wheels, or four wheels), and the feature of braking system on the vehicle (the coefficient of braking equation, Q₀ and Q₁ ) If the

accelerationa′ is positive, the car is braking, while negative, accelerating. The _x factor to dispose torque can be defined by these conditions as follows:

(1) If a′ < 0 (accelerating), cars driven by front wheels, Q = 1. _x (2) If a′ < 0 (accelerating), cars driven by rear wheels, Q = 0. _x (3) If a′ < 0 (accelerating), cars driven by four wheels, Q = Wd_x 4. (4) If 0 ≦a′ ≦ 0.3g (middle braking), Q = _x Q₀.

(5) If a′ > 0.3g (heavy braking), Q = _x Q₀ +Q a₁( _x′ −0.3).

After figuring out the factor to dispose torque, through the accelerating or braking accelerationa′ , we can obtain braking force or accelerating force for _x each wheel by equation as follow,

1 2

2.3.4 Calculate t the normal force for each whee^l

Whether a car is accelerating at longitudinal vector or lateral vector, the inertia force will differentiate each wheel’s vertical reaction. The variation of vertical reaction of wheels is called lateral weight transfer, caused by the lateral acceleration when the car is turning. The variation of vertical reaction of wheels is called longitudinal weight transfer, caused by the longitudinal acceleration when the car is braking or accelerating. We have the equations of actual lateral and longitudinal acceleration as follows:

(2-8)

The following factors affect the amount and distribution of lateral weight transfer.

1. the body of the car rolling,

2. the high variation of the rolling center, and

3. the mass which isn’t supported by springs. The mass not supported by springs.

Here come some equations from which we calculate the amount of lateral weight transfer of the front and rear wheels when vehicles are rolling.

(

^{y s ya}

)

^{ra cos s u/ xa s ra sin s u/} caused by the high variation of the car’s rolling center. Related equations are as follows.

The amount of the lateral weight transfer of the front and rear wheels is caused by the mass, not supported by springs. Related equations are as follow.

uf is the grand total of the three variations above.

TF BF RF UF

W =W +W +W (2-13a)

TR BR RR UR

W =W +W +W (2-13b)

Since we get the lateral weight transfer, we can calculate the normal force of each wheel with the following equations.

1

2.3.5 Calculate the coefficient of friction of vehicle wheels’

After each wheel’s vertical reaction, affected by the longitudinal and lateral weight transfer, is figured out, the coefficient of friction can be calculated.

The peak of longitudinal friction coefficientµ_xpi for wheels can be figured out with the following equation (2-15).

(

0 1 2 ²

)

xpi 0.01 SN P P Fzi P Fzi

µ = × × + × + × (2-15)

The coefficientµ_xsi of sliding friction of wheels can be figured out with the following equation (2-16).

(

0 1 2 ²

)

xsi 0.01 SN S S Fzi S Fzi

µ = × × + × + × (2-16)

The lateral slip ratio of wheels can be figured out with the following equation (2-17), when the car braking.

Spi

The peak lateral frictionµ_yi of wheels can be calculated with the following equation (2-18).

(

^{3 1 4 2}

)

yi 0.01 SN B B Fzi B Fzi

µ = × × + × + × (2-18)

2.3.6 Calculate the maximum allowable longitudinal force generated by wheels

With the known coefficient of friction of wheels, we can calculate the maximum of allowable longitudinal and that of lateral forces generated by wheels. Without cars braking/accelerating, F_ymax means the maximum of allowable lateral forces generated by wheels, as what are shown in the following.

If the braking/accelerating force and the centripetal force exist at the same time, we can calculate the maximum of allowable longitudinal force F^’_xmax with the following equations.

(1) If F_{y max} < F_{x max} , the equation to calculate the eccentricity of an ellipse will

be longitudinal force F^’_xmax will be

(

²

)

(2) If F_{y max} > F_{x max} , the equation to calculate the eccentricity of an ellipse will

be longitudinal force F^’_xmax will be

(

²

)

At this time, the main shaft of an ellipse is the longitudinal axis of a wheel.

2.3.7 Calculate t the longitudinal coefficient of frictionµ_xi and slip ratio for each wheelS_i

If the driver needs greater braking force (F ′ ) than the maximum of braking _xi force ( ) generated by wheels, the wheels will be locked. Meanwhile, the value of slip ratio on the wheel will be 1(

xmax

F′

i 1

S = ). If the driver needs greater accelerating force ( F ′_xi) than the maximum of accelerating force ( F′_xmax ) generated by wheels, the wheels will slip. Meanwhile, the value of slip ratio on the wheel will be -1 ( ). Under the conditions when the wheels are lock or slip ( or

i 1 S = −

i 1

S = S_i = − ), the related equation to figure the longitudinal coefficient 1 of friction of each wheel is as follows.

( )

xi xsi Sgn Fxi

µ =µ × ′ (2-21)

We can figure the slip ratio of wheels with the following equation. S_i +1, if a′ > 0. _x

S_i = Sgn (a′ ) = 0, if _x a′ = 0. _x -1, if a′ < 0 _x

If the driver need smaller braking or accelerating force than the maximum of braking or accelerating force

F ′xi xmax

F′ generated by wheels, which means that the wheels are able to generate the braking or accelerating force the driver needs, in normal condition, the equation to figure the longitudinal coefficient of friction of each wheel is as follow.

xi xi

zi

F

µ = F^′ (2-22)

2.3.8 Calculate the longitudinal forceF_ci for each wheel

With the known longitudinal coefficient of friction, we will calculate the longitudinal force under the two condition as follows.

1. When a′ ≦ 0, _x F_ci = µ_xi×F_zi (2-23a)

2.3.9 Calculate the lateral forceFsi for each wheel

With the known longitudinal force, before we figure the lateral force with equations depending on the following conditions, we must define 1₂

s

ε ⁼ ρ ^.

1. If the wheels turn freely, neither being locked nor being slipping (S_i = 0), the

lateral force can be figured out with the following equation.

2 2 2

si yi zi ci

F = µ ×F − × Fε (2-24)

2. If the wheels are locked because of the braking moment overloaded (S_i = 1), the lateral and longitudinal forces can be figured out with the following equations.

ci xsi zi cos i

F = µ ×F × β (2-25a)

si xsi zi sin

F = µ ×F × β_i (2-25b)

3. If the wheels slip because of the accelerating moment overloaded (S_i = -1), the lateral and longitudinal forces can be figured out with the following equations.

ci xsi zi

F = −µ ×F (2-26a)

si 0

F = (2-26b)

2.3.10 Coordinate system transformations

We can transform the lateral and longitudinal forces from the wheel’s coordinate system into the vehicle’s coordinate system:

(

^{cos sin}

)

^{cos sin} inclination, aligning torque could be figured out with the equation (2-27c), as

follows:

If the wheels are locked or slip because of the braking or accelerating moment overloaded (S_i = 1 or S_i = -1), the aligning torque is zero (M_zi = 0).

2.4 Aerodynamic model of a vehicle

Aerodynamic forces and moments acting on a vehicle are the functions that include longitudinal velocity, lateral velocity, yaw angle velocity, aerodynamic coefficient, and sideslip angle. The sideslip angle could be figured out from longitudinal velocity and lateral velocity, with the equation as follows:

tan 1V

α = ⁻ U (2-28)

With the known sideslip angle, the aerodynamic forces and moments acting on a vehicle can be figured out with the following equations:

1. Aerodynamic drag force:

( )

0.5 2

xa front xa

F = − × ×ρ U ×S ×C ×Sgn U (2-29a)

2. Aerodynamic lateral force:

0.5 2

3. Aerodynamic yawing moment:

( )

0.5 2 sin 2

za P front char na ad

M = × ×ρ V ×S ×H ×C × α −C × r^• (2-29c)

2.5 Dynamic equations of operation model

Apply the foregoing aerodynamic forces/moments and forces/moments of wheels into the dynamic differential equations. The acceleration of each degree-of-freedom can be figured out with the following equations.

4 Then, integrate the acceleration of each degree-of-freedom to get the

velocity. The related equation is as follows:

(2-31a)

each degree-of-freedom, with the related equations as follows.

(2-32a)

(

0

cos sin

t

X =∫ U × Ψ − ×V Ψ)^dt

) ^{t (2-32b)}

(

0

sin cos

t

Y =∫ U× Ψ + ×V Ψ d

(2-32c)

0 t

Ψ =∫rdt

At t = 0, V = r = X = Y = Ψ = 0, and U = U₀.

Chapter 3 Force Feedback System

3.1 The Steering System

3.1.1 Introduction of the Steering System

The design of the steering system has an influence on the direction response behavior of a motor vehicle that is often not fully appreciated. The function of the steering system is to steer the front wheels response to driver command inputs in order to provide overall directional control of the vehicle. However, the actual steer angle achieved are modified by the geometry of the suspension system, the geometry and reactions within the steering system, and in the case of front-wheel-drive (FWD), the geometry and reactions from the drive train.

These phenomena will be examined in this section first as a general analysis of a steering system and then by considering the influence of front- wheel-drive.

3.1.2 The steering linkages

The steering system used on motor vehicles vary widely in design [12, 13, 14], but are functionally quite similar. Figure 3.1 illustrate some of these.

The steering wheel connects by shaft, universal joints, and vibration isolators to the steering gearbox whose purpose is to transform the rotary motion of the steering wheel to a translational motion appropriate for steering the wheels. The rack-and-pinion system consists of a linearly moving rack and pinion, mounted on the firewall or a forward cross member, which steers the left and right wheels directly by a tie-rod connection. The tie-rod linkage connects to

steering arms on the wheels, thereby controlling the steer angle. With the tie-rod located ahead of the wheel center, as shown in Figure 3.1, it is a forward-steer configuration.

The steering gear box is an alternative design used on passenger cars and light trucks. It differs from the rack-and-pinion in that a frame-mounted steering gearbox rotates a pitman arm which controls the steer angle of the left and right wheels through a series of relay linkages and tie rods, the specific configuration of which varies from vehicle to vehicle. A rear-steer configuration is show in the figure, identified by the fact that the tie-rod linkage connects to the steering arm behind the wheel center.

Between these two, the rack-and-pinion system has been growing in popularity for passenger cars because of the obvious advantage of reduced complexity, easier accommodation of front-wheel-drive systems, and adaptability to vehicle without frames. The primary functional difference in the steering systems used on heavy truck is the fact that the frame-mounted steering gearbox steers the left road wheel through a longitudinal drag line, and the right wheel is steered from the left wheel via a tie-rod linkage [12].

The gearbox is the primary means for numerical reduction between the rotational input from the steering wheel and the rotational output about the steer axis. The steering wheel to road wheel angle ratios normally vary with angle, but have nominal values on the order of 15 to 1 in passenger cars, and up to as much as 36 to 1 with some heavy trucks. Initially all rack-and-pinion gearboxes had a fixed gear ratio, in which case any variation in ratio with steer angle was achieved through the geometry of the linkages. Today, rack-and-pinion systems are available that vary their gear ratio directly with steer angle.

3.2 Geometry

3.2.1 Front Wheel Geometry

The important elements of a steering system consist not only of the visible linkages just described, but also the geometry associated with the steer rotation axis at the road wheel. The geometry determines the force and moment reaction in the steering system, affecting its overall performance. The important features of the geometry are shown in Figure 3.2.

The steer angle is achieved by rotation of the wheel about a steer rotation axis. Historically, this axis has the name “kingpin” axis, although it may be established by ball joint or the upper mounting bearing on a strut. The axis is normally not vertical, but may be tipped outward at the bottom, producing a lateral inclination angle (kingpin inclination angle) in the range of 10-15degrees on passenger cars.

It is common for the wheel to be offset laterally from the point where the steer rotation axis intersects the ground. The lateral distance from the ground intercept to the wheel centerline is the offset at the ground and is considered positive when the wheel is outboard of the ground intercept. Offset may be necessary to obtain packaging space for brakes, suspension, and steering components. At the same time, it adds “feel of the road” and reduces static steering efforts by allowing the tire to roll around an arc when it is turned [15].

Caster angle result when the steer rotation axis is inclined in the longitudinal plane. Positive caster places the ground intercept of the steer axis ahead of the center of tire contact. A similar effect is created by including a longitudinal offset between the steer axis and the spin axis of the wheel,

although this is only infrequently used. Caster angle normally ranges from 0 to 5 degrees and may vary with suspension deflection.

3.2.2 Formulation of Geometry

Before we analyze the geometry, we define the geometry meaning of each term in fundamental dynamics equation.

Nomenclature

F_feedback: feedback force on the steering wheel M_zr: aligning torque on right-front wheel M_zl: aligning torque on left-front wheel M_T: moment produced by tractive force M_L: moment produced by lateral force M_V: moment produced by vertical force

MV,inclination: moment produced by vertical force acting on lateral inclination angle

M_V,caster: moment produced by vertical force acting on caster angle M_AT: moment produced by aligning torque

M_ALL: the total moment r: tire radius

R_w: steering wheel radius λ: lateral inclination angle υ: caster angle

δ: steer angle τ: gear ratio

3.3 Steering System Forces and Moments

The forces and moments imposed on the steering system emanate from those generated at the tire-road interface. The SAE has selected a convention by which to describe the force on a tire, as shown in Figure 3.3. Assume that the forces are measured at the center of the contact with the ground and provide a convenient basis by which to analyze steering reactions.

The ground reactions on the tire are described by three forces and moments, as follow:

1. Normal force 2. Tractive force 3. Lateral force 4. Aligning torque 5. Rolling resistance 6. Overturning moment

The reaction in the steering system is described by the moment produced on the steer axis, which must be resisted to control the wheel steer angle.

Ultimately, the sum of moments from the left and right wheels acting through the steering linkages with their associated ratios and efficiencies account for the steering-wheel torque feedback to the driver.

Figure 3.4 shows the three forces and moments acting on a right-hand road wheel. Each will be examined separately to illustrate its effect on the steering system.

3.3.1 Tractive Force

The tractive force, F_x, acts on the kingpin offset to produce a moment as shown in Figure 3.5.

Then net moment is:

M_T =(F_xl −F_xr)⋅d (3-1)

The left and right moments are opposite in direction and tend to balance through the relay linkage. Imbalances, such as may occur with a tire blowout, brake malfunction, or split coefficient surfaces, will tend to produce a steering moment which is dependent on the lateral offset dimension.

3.3.2 Lateral Force

The lateral force, Fy, acting at the tire center produces a moment through the longitudinal offset resulting from caster angle, as shown in Figure 3.6. The net moment produced is:

^M_L =

(

^F_yl + ^F_yr

)

⋅^rtanν (3-2)

The lateral force is generally dependent on the steer angle and cornering condition, and with positive caster produces a moment attempting to steer the vehicle out of the turn.

3.3.3 Vertical Force

The vertical force acting on lateral inclination angle, illustrated in Figure 3.7, results in a sine angle force component, which normally acts lateral on the moment arm when the wheel is steered. The moment is zero at zero steer angle.

When steering, both sides of the vehicle lift, an effect which is often described as the source of the centering moment.

M_{V,inclination} =−

(

F_zl +F_zr

)

⋅d ⋅sinλ⋅sinδ (3-3)

The caster angle results in a sin angle force component, which nominally acts forward on the moment arm as shown in Figure 3.8.

M_V,caster =

(

F_zl −F_zr

)

⋅d ⋅sinν ⋅cosδ (3-4)

Then the net moment produced by vertical force is:

M_V =(F_zl − F_zr)⋅d ⋅sinν ⋅cosδ −(F_zl + F_zr)⋅d ⋅sinλ⋅sinδ (3-5)

When steer angle, one side of the axle lift and the other drops, so that the net moment produce depends also on the roll stiffness of the front suspension as it influences the left and right wheel roads.

3.3.4 Aligning Torque

The aligning torque, M, acts vertically and may be resolved into a component acting parallel to the steering axis. Since moments may be translated without a change in magnitude, the equation for the net moment is:

MAT =Mzl +M zr (3-6)

Under normal driving conditions, the aligning torques always act to resist any turning motion, thus their effect is under steer. Only under high braking conditions do they act in a contrary fashion.

3.3.5 Rolling Resistance and Overturning Moments

These moments at most only have a sine angle component acting about the steer axis. They are second-order effects and are usually neglected in analysis of steering system torques.

3.4 Force Feedback On The Steering Wheel

These equations, (3-1), (3-2), (3-5), (3-6), are the moments of steering axis, we can add direct them to get the total moment, M_ALL, on the kingpin axis:

^M_ALL ⁼

(

^M_T ^+M_L^+M_V ^{+ M}_AT

)

^{cos λ2 +ν2 (3-7)}

Then we can through the steering system model, as shown Figure 3.9, and calculate the force on the steering wheel by the way of the total moment.

However, in our simulation, in order to simplify our analysis and retrench the time to calculate, we usually neglect the rigidity of the steering linkage, K, and

other coefficient of viscosity, and the total moment divides gear ratio to get the torque on the steering wheel. Then the torque on the steering wheel divides the radius of the steering wheel to get the force which driver sustains.

Rw MALL

feedback

F 1

⋅

= τ ^(3-8)

Chapter 4 Hardware Integration

4.1 Hardware Structure

The hardware of the steering wheel system is mainly divided into four parts, a principal mechanism, driving device, angular potential meter, and signal process interface. We use the principle of dc motor controllable system, as figure 4-1, to make the force feedback steering wheel output appropriate forces accurately at real-time. In this chapter, we will describe each device in the steering wheel system and analyze the control principle.

4.2 Principal Mechanism

In figure 4-2, it’s a mechanism of the steering wheel which we design with Internet Motion Navigator Corp. during education and industry cooperation. In the mechanism, we can know the torque produced by motor allows a pulley whose ratio of radius is 3:1 to drive the steering wheel on another axis. It indirectly brings forces for users to achieve the force feedback effects produced in virtual reality. At the same time, we add a mechanism on the front of the pulley to limit rotational angle, and then we can ward off the potential meter breakdown because of rotating over-angle.

4.3 Driving Device

Because the driving device in the force feedback steering wheel needs accurate and stable output torques, we use a dc motor device and combine a PWM motor driver to control output current.

4.3.1 DC Motor

We use a geared motor produced by SHAYANG YE Industrial Co., Ltd. in

We use a geared motor produced by SHAYANG YE Industrial Co., Ltd. in

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