Chapter 3

Vehicle Longitudinal Dynamics

3.1 Introduction

The longitudinal motions of a vehicle include accelerating, cruising, uphill and downhill motions. The performance and the driveability of a vehicle including accelerations at various loadings and driving conditions are the main topics of this chapter. Vehicle engineers have long realized the importance of driveability in the commercial success of vehicles. Vehicle longitudinal dynamics involves the study of several issues including engine behaviour, tyre tractive force generation, resistive forces acting on the vehicle and the drivers' gearshifting habits.

In this chapter first, simple vehicle models are developed which are useful for the initial estimation of vehicle longitudinal performance. For more detailed analyses and reliable design, however, more elaborate models including the engine characteristics, tyre slips, shifting delays, rotating masses and driveline losses have to be used. Such models with increasing details are being developed and many examples using typical vehicle data are included throughout and MATLAB® codes for vehicle performance calculations are provided in this chapter.

3.2 Torque Generators

The vehicle is accelerated by a tractive force at the driving wheels which in turn depends on the torque applied at these wheels. In the vehicle longitudinal performance analysis, therefore, the power source which delivers the torque – usually an internal combustion engine (ICE) – plays an important role. Different power sources have different torque generation characteristics. In vehicle performance studies, however, the power source can be regarded as delivering a quasi-steady source of torque (see Section 2.4.3) with predefined characteristics.

Traditionally, IC engines have dominated as power sources in vehicles. However, in spite of all the years of experience using IC engines in vehicles, their torque characteristics and related effects on the vehicle performance still remain an area of discussion. A detailed discussion of the torque generation properties of IC engines was presented in Chapter 2. Electric motors are the main counterparts for engines in hybrid electric vehicles (HEVs) and they have significantly different torque-generating characteristics from engines. In a later section a short review of the performance of electric motors used in vehicle applications will be provided.

3.2.1 Internal Combustion Engines

The torque-speed characteristics of IC engines were discussed in detail in Chapter 2 so only the key features for its application in the current chapter are covered here. The quasi-steady or stationary torque of an ICE is dependent on the speed of rotation and the amount of throttle opening, θ, i.e.:

(3.1)

A typical part throttle engine performance curve representing Equation (3.1) for an SI engine is shown in Figure 3.1. This plot is generated based on engine dynamometer tests.

Using the MT engine formula (see Section 2.6) the performance curves can be represented mathematically by a formula of the form:

(3.2)

in which is the engine full load or wide-open-throttle (WOT) curve and coefficients A, B, C and D are four constants for a specific engine and can be determined from part load tests. For an engine with coefficients given in Table 3.1 and full throttle data of Table 3.2, the part throttle curves are shown in Figure 3.2. In order to produce smooth curves, polynomial curve fitting is applied to the WOT data.

Table 3.1 Coefficients of MT formula for a specific engine.

Table 3.2 Engine full throttle data.

Engines are different in terms of their full load and part load torque-speed characteristics. Diesel engines typically work at larger torques and lower speeds. For diesel engines with common rail fuel injection systems, the WOT torque typically has a flat area of constant torque. Figure 3.3 shows the performance curve of a diesel engine with a common rail system.

3.2.2 Electric Motors

Electric motors play a crucial role in hybrid vehicle powertrain systems (see Chapter 7). Their potential to work as a motor or generator provides opportunities to use them for generation of both traction or braking forces whichever required. There are several types of electric motors suitable for use in hybrid vehicles. Separately excited DC motors are good at speed control and they need relatively simple control electronics. Their brushes need to be changed from time to time and they have high maintenance requirements. Alternating current (AC) motor types include permanent magnet synchronous, induction motors and switched reluctance motors. AC motors are in general less expensive, but require more sophisticated control electronics. Nonetheless, they have higher power density and higher efficiency and in the majority of vehicle applications AC motors are used. The induction AC motors are generally characterized by higher specific powers than permanent magnet motors. One of the inherent properties of electric motors is the production of torque at zero speed. This gives them advantages over IC engines in moving the vehicle from rest without needing a clutch. Power electronics provide the necessary tools for controlling the torque-speed characteristics of the electric motors to suit automotive applications. A constant high torque at low speeds and the retention of maximum power afterwards can be achieved by proper control of the motor voltage, field, flux or frequency depending on the type of the motor. A typical performance curve of an electric motor used in hybrid vehicles is shown in Figure 3.4.

3.3 Tractive Force

The vehicle accelerates through the application of tractive forces. The tractive force of a vehicle is produced at the tyre–road interface and it is therefore a function of both tyre and road properties. Different tyres produce different traction forces on a particular road surface, and a particular tyre produces different traction forces over different road surfaces. The tractive force originates from the torque applied on the axle from the torque source. Thus the tractive force is dependent on both the torque source properties as well as the tyre–road properties.

3.3.1 Tyre Force Generation

The tyre produces a tractive force due to the existence of friction at the road interface. The friction force generation by the tyre, however, is quite complicated owing to its rolling motion. In order to examine the difference between a simple sliding friction and that of a rolling tyre, consider a tyre segment cut out from the tyre contact patch as shown in Figure 3.5. A load W equal to a wheel load is considered to act vertically on the tyre segment. When a force F is applied to the segment in order to move it along the road surface, three cases may exist:

Now consider a tyre with the same axle load W standing on the road surface as shown in Figure 3.6. As long as the tyre does not rotate, the friction force F_f will be similar to that of the segment discussed earlier.

Now if instead of force F, a torque T is applied at the wheel axis as illustrated in Figure 3.7, the direction of friction force at the contact surface will be reversed because the slip direction at the contact surface changes. With the wheel initially at rest, when the torque is applied, the wheel tends to rotate around its axis. This is possible only if the contact patch of the tyre is allowed to slip in reverse direction to the forward motion. The relative motion in the presence of friction at the contact area will result in a friction force opposite to the slip direction. This force will act at the direction of motion of the wheel centre and is called tractive force F_T.

One major difference between the simple slip of the tyre segment or the similar case for a non-rolling tyre with the case of a rolling tyre is the slip phenomenon in the contact patch. In the two former cases, the whole contact area slides relative to the ground surface. All contact points, therefore, will experience similar situations. Each point in contact region will generate a share of the total force proportional to the pressure distribution. In the case of the tyre segment, the pressure distribution will be uniform, so any point will have an equal share. For the non-rotating tyre case, the pressure distribution in the contact area is not uniform, but the slip will be uniform, as illustrated in Figure 3.8.

When the torque is applied at the wheel axis, those elements of the tyre entering the contact region will be forced to stay in contact with the ground and gradually move forward in the direction of the slip. Those close to the leading edge cannot move as fast as those close to the trailing edge. This results in zero slip at leading edge and full slip at the trailing edge. The distribution of slip along the contact length does not have a simple form and an approximation for the slip is illustrated in Figure 3.9. According to this model, a linear slip will exist at the leading region called the ‘adhesion area’ and a uniform slip will exist at the trailing region called the ‘slip area’.

Even this simplified model is difficult to work with, since it needs information regarding the relative width of adhesion and slip lengths, which is dependent on many factors and difficult to determine. In practice, however, a uniform slip is considered in the contact patch of a tyre which means the contact points all have a uniform speed V_s opposite to the direction of wheel travel. If the velocity of wheel centre is denoted by V_w, then the velocity vector, according to Figure 3.10, can be written as:

(3.7)

or in an algebraic form:

(3.8)

in which is the effective radius of the tyre. In a pure rolling case, the wheel centre velocity equals , that means no slip condition with V_S = 0. For the slipping case:

(3.9)

Longitudinal slip S_x of tyre is defined by dividing the slip speed by the rolling speed :

(3.10)

From Equation (3.9):

(3.11)

S_x is sometimes expressed as a percentage. The tractive force of the tyre is dependent on the slip within the contact area. However, there are other factors that control the tractive force, e.g. the normal load F_Z on the tyre and tyre pressure p. Thus, the tractive force of the tyre can be expressed as a function f of the influencing variables:

(3.12)

At a certain tyre pressure, the tractive force will be dependent mainly on the longitudinal slip and normal load.

What has been considered so far is the pure slip of the tyre in the longitudinal direction. In practice, the tyre is subjected to slips with components in both the longitudinal and lateral directions. In these combined slip conditions, the slip in the contact region has a complicated nature, since the contact patch is distorted under the application of loads in different directions. For this reason the tyre will produce a force that is acting somewhere other than on the centre of contact patch and in addition to the forces in three directions, moments are also generated.

The tyre force system can be defined in different ways. According to the SAE definition [1], the origin O of the axis system (see Figure 3.11), is taken on the centre of tyre contact (the intersection of wheel plane W and the projection of spin axis onto the road plane R). The axis is the intersection of the wheel and the road planes. The axis lies in vertical plane V and is perpendicular to the road plane and pointing downward. The axis is therefore in the road plane with positive direction to the right. Only F_x (tractive/brake force) and M_y (rolling resistance torque) will be needed in this chapter and they are discussed in more detail in the next sections.

3.3.2 Mathematical Relations for Tractive Force

A large volume of work is available in the literature dealing with the modelling and prediction of the force generating properties of the tyres. In general, tyre modelling is performed in two basic ways: physical and experimental. In physical modelling a simplified model is assumed for the tyre that accounts for the main physical properties affecting the force generation mechanism. By the application of mechanical laws governing the deformation of tyre elements and friction effects at the contact region, mathematical equations are developed. The accuracy of such models depends on the quality of the assumptions made for the physical model. The advantages of these types of models are their simple nature and that the predicted tyre forces do not necessarily require difficult measurements.

There are also software-based physical models involving considerable complexity. These models usually use Finite Elements methods (FEM). Although these sophisticated models are claimed to generate more accurate results than those of simple physical models, they are unnecessarily complicated for vehicle longitudinal performance predictions. Another approach is based on experimental work to measure the force generation properties of the tyres. Experimental results can also help to explain the physics of tyre behaviour when different loading conditions are tested, and hence influence the further developments of tyre models. The normal approach is to fit curves to the measured tyre data.

Finding a general mathematical relationship applicable to all tyres has been a challenge to the automotive industry. The ‘Magic Formula’ tyre model is the result of work carried out at Delft University. The attraction of this approach is its accuracy in fitting the lateral tyre force and aligning moment data in addition to the longitudinal force data of the tyre. The equations of this tyre model are of basic form [2]:

(3.13)

where:

(3.14)

(3.15)

X stands for either longitudinal slip S_x or sideslip angle α (see Figure 3.11). Y stands for either longitudinal force F_x, side force F_y or self-aligning moment M_z. The coefficients B, C, D, E and the horizontal and vertical shifts S_h and S_v are non-linear functions of the vertical tyre load F_z (and camber angle γ for F_y and M_z). For each tyre, a set of tests must be carried out in order that the dependency of the coefficients on the variables (e.g. F_z) is calculated. Typical relations for longitudinal force of a specific tyre are [2]:

(3.16)

The numerical values for the coefficients a₁ to a₁₀ of the same tyre are given in Table 3.3. F_z is in units of N for these values.

Table 3.3 Magic Formula coefficients for a specific tyre [2].

The advantage of the Magic Formula representation is that once the tyre parameters are available, it will be valid for all loading cases with a large range of parameter variations.

From the results of Example 3.3.1, it can be seen that the overall variation of the tyre longitudinal force with slip can be considered in two linear and non-linear regions as shown in Figure 3.14. The linear region begins from zero slip up to a certain value of 5–10% and the maximum tyre force is developed at slip values around 10–20%. At the full slip (100%) the tyre force is considerably lower than the maximum force. This phenomenon is a typical property of the rolling tyres and if the concept of dry friction is employed, the longitudinal force can be related to the normal load by the coefficient of adhesion:

(3.17)

At a specific normal load, , the coefficient of adhesion μ is a function only of F_x which is already a function of slip S_x i.e.:

(3.18)

In fact, by dividing the values of F_x by the normal load , the trend of variation of μ against the slip will be similar to the variation of F_x with S_x.

From the variation of the coefficient of adhesion, two other issues can be noticed. First, the peak of adhesion coefficient (Figure 3.15) can become even greater than unity at relatively low slips of below 20%. Second, at 100% slip, the value of sliding adhesion coefficient is the lowest value in the non-linear region – in the example tyre in Figure 3.14 it was some 30% lower than . Thus, the behaviour of coefficient of adhesion in rolling tyres is quite different from the coefficient of friction between the two surfaces. Hence, the following conclusions can be drawn:

It should be noted that the adhesion coefficient depends not only on the tyre but on the type of road surface as well. On wet surfaces, the values of will be reduced. Other factors such as tyre treads (shape and depth) will also be influential on the adhesion coefficient. It is worth mentioning that the braking behaviour of the tyre is very similar to its tractive behaviour. Again, to generate braking force at the contact area, slip should take place. This time the rotational speed of tyre is lower than the corresponding straight line speed of the wheel, therefore, the slip will be negative. The braking force can also be determined with the Magic Formula by simply using negative slips. The overall shape of the braking force, therefore, will be similar to that of the tractive force but mirrored relative to the origin.

3.3.3 Traction Diagrams

The tractive force is produced at the tyre–road interface provided that two conditions are fulfilled:

In a quasi-steady condition, with reference to Figure 3.16:

(3.19)

There is, however, a force limitation in the contact area of the tyre depending on the road conditions. Equation (3.19) is valid up to the limit of tractive force:

(3.20)

As long as the wheel torque is lower than , a limited amount of slip will build up, otherwise if a larger torque is applied, the wheel will spin.

Assuming the wheel torque is below the limit, Equation (3.19) means the tractive force is also dependent on the nature of the torque of the wheel. The wheel torque in general is an amplification of the torque of vehicle torque generator (see Section 3.2). Denoting the amplification factor by n (i.e. the overall gear ratio between the torque generator and the wheel):

(3.21)

and from (3.19):

(3.22)

Thus, the traction force will resemble the torque characteristics of the torque generator (T_g) multiplied by an amplification factor.

3.4 Resistive Forces

There are resistive forces opposing the motion of the vehicle; some exist from the start of motion and others build up with speed. Resistive forces consume some part of engine power and reduce the speed and acceleration of the vehicle. In order to analyze the straight line performance of the vehicle, it is necessary to characterize the resistive forces acting on the vehicle during the motion. The resistive forces can be categorized into three types: frictional, air resistance and gravitational. The frictional forces are commonly called ‘rolling resistance’ force. Air resistance forces are known as ‘aerodynamic force’ and the gravitational force is called ‘grade force’.

3.4.1 Rolling Resistance

As the name implies, the resistance to motion of rotating parts is summed up together and builds a total resistive force to slow down the vehicle. The rotating parts work under two conditions: with and without the transmission of torque. The former can be excluded from the current discussion since it deals with the driveline power loss during torque transmission and will be treated separately in Section 3.13. The rolling resistance term, therefore, deals only with those resistive torques during vehicle motion with the driveline freely rolling. This includes all the rotating parts in the driveline components. The driving wheels are constantly connected to the driveline and their connection to the engine is controlled by the clutch. Therefore, even when the gearbox is disengaged and left in the neutral position, the driving wheels will rotate the driveline components except for gears inside the gearbox. The total resistive torques of the free rotating parts can be divided into two main categories: due to friction and tyre deformation resistance.

3.4.1.1 Frictional Torques

The frictional torques in the vehicle driveline consist of three parts:

3.4.1.2 Tyre Deformations

For a non-rolling tyre, the load W will cause a symmetrical pressure distribution around the centre of contact area. The resultant reaction load N will act at the centre of contact region as illustrated in Figure 3.18.

For a rolling tyre, the tyre elements entering the contact region will experience a compression whereas the trailing materials exiting the contact area will tend to stretch. The pressure on the materials in the contact zone, therefore will tend to increase at the leading edge and decrease at the trailing edge. The pressure distribution will be something similar to that shown in Figure 3.19. The resultant reaction force at the ground will act off-centre closer to the leading edge.

With the resultant reaction force moving forward, it is equivalent to an opposing torque acting around the wheel axis. This torque shown in Figure 3.20a is called rolling resistance torque (T_RR). Since the action of the rolling resistance torque is to slow down the wheel (or vehicle), in practice, it is assumed that a resistive force is acting at the ground level with the same effect (see Figure 3.20b). This force is called rolling resistance force (F_RR).

From an energy point of view, the rolling resistance is caused by the deformations of the tyre. The tyre elements require energy to deform when they enter the contact region. Owing to the visco-elastic properties of rubber, the tyre elements are unable to fully restore this energy when they leave the contact region. This phenomenon is called a hysteresis effect. Therefore a net amount of energy is lost to heat during tyre rolling and is equivalent to the work done by the rolling resistance torque or the work done by the rolling resistance force.

3.4.1.3 Other Factors

There are also other factors besides the rolling frictions and tyre deformation that have small influences on the total resistive force:

• Tyre–road friction loss – due to slipping of the tread compound on the ground surface, heat will be generated.
• Aerodynamic resistance due to wheel spin. The aerodynamic resistance for the whole vehicle usually includes the resistive force for a non-spinning wheel. However, to spin a wheel in the presence of air requires a small amount of energy which will add to the other sources of energy loss.
• Propeller effect of wheel rim during spin. Air passing through the spinning wheel rim also adds a small term to the energy losses.

3.4.1.4 Influencing Parameters

Several parameters influence the rolling resistance force. The main parameters are those which affect the tyre or the contact surface including:

• tyre construction:

  type of tyre (radial or bias ply)

  tyre materials

  tread design

  tyre diameter

• tyre operating conditions:

  tyre pressure

  tyre speed

  tyre temperature

  vertical load

  tyre lateral slip

  tyre age

• road surface: it is worth noting that the deformations of the road surface, e.g. on soft deformable ground requires energy and such energy losses are also considered as rolling resistance loss. Some influencing items are:

  road texture

  rigidity

  dryness.

The extent to which the named parameters influence the rolling resistance has been discussed in specialist references (e.g. [3]). An overview of the influence of various parameters qualitatively is shown in Tables 3.4 and 3.5.

Table 3.4 Effect of increasing tyre parameter values on the rolling resistance force.

Table 3.5 Effect of tyre and road types on the rolling resistance force.

3.4.1.5 Mathematical Representation

A mathematical formula for the rolling resistance force will help in the analysis of vehicle motion. To this end, one can write the rolling resistance force as a function of influencing parameters p₁, p₂, etc., that is:

(3.23)

Finding the unknown function f, however, is not a simple task. Of the parameters mentioned earlier, some such as diameter and construction are fixed for an existing tyre, some have little importance such as temperature and some have fixed values, such as pressure. The remaining parameters are speed, load and road type. The type and quality of roads are hard to formulate, thus, for each road type a different formulation is necessary. The dependency of the rolling resistance force on the wheel load is of a linear nature, thus, the ratio of rolling resistance force to the wheel load, f_R is called the rolling resistance coefficient and is independent of load:

(3.24)

The rolling resistance coefficient in general is a function of all the above-mentioned parameters, but according to the justification presented, the only important parameter is the forward speed of the wheel centre, i.e. the vehicle speed:

(3.25)

The variation of rolling resistance coefficient with speed is reported somewhat differently in various references. The general trend nonetheless shows an increase in the rolling resistance coefficient with speed. From an energy point of view this is acceptable since if a particular amount of energy is lost for one revolution of the tyre, then increasing the revolutions in a time interval will increase the lost energy. The variation of rolling resistance coefficient with speed in general can be approximated in the form of a second order polynomial:

(3.26)

where the coefficients f₀, f₁ and f₂ are three constants. The quadratic term, however, can be removed since a similar dependency on speed is available through aerodynamic resistance (see Section 3.4.2). In fact, by eliminating this term we are not disregarding this effect, but leaving it included in the aerodynamic force term. With this justification, the rolling resistance coefficient can be simplified to a first order form:

(3.27)

Often, the dependency of the rolling resistance on speed is also ignored (i.e. f₁ = 0) and only a constant coefficient f₀ is considered for the rolling resistance coefficient. Typical ranges for the coefficient of rolling resistance f₀ can be found in Table 3.6 (additional information may be obtained from [3]).

Table 3.6 Ranges for the rolling resistance coefficient f₀ at low speeds.

3.4.2 Vehicle Aerodynamics

The motion of a vehicle is taking place in the air and the forces exerted by air on the vehicle will influence the motion. Aerodynamics is the study of the vehicle motion in air and the effects include:

• internal flows: air flows through the front grills for the purpose of ventilation and cooling. Air flow through windows and sunroof are other examples.
• ground clearance: in ground vehicles, the ground clearance is very low and causes ground effects.
• streamlining: the design of the exterior in accordance with the streamlines of air flow around the body is of great importance in order to reduce the air resistance. For ground vehicles, however, the exterior shape of body is dependent on several factors and streamlining is only one of them (Figure 3.21).

  Figure 3.21 Streamlines around the body

  

3.4.2.1 Aerodynamic Resistance

Aerodynamic resistance forces result from three basic effects:

The form resistance is responsible for the main part of the aerodynamic force, around 80%, whereas the two others share the remainder at around 10% each.

3.4.2.2 Aerodynamic Forces and Moments

The aerodynamic resistance forces are not in reality discrete forces acting at specific locations, but are actually a summation of infinitesimal forces acting at all points on the vehicle body. The result is a single force R_A acting at a point called the centre of aerodynamic force, or centre of pressure C_P. In general, the aerodynamic forces are described by a three-dimensional force system with three resolved components at the vehicle body axes. The centre of pressure is also different from the vehicle centre of mass C_G and thus the aerodynamic force will have moments around the vehicle axes. Figure 3.22 illustrates the aerodynamic forces and moments.

In aircraft aerodynamics, it is a common practice to define other axes systems such as wind axes, separate from the body axes. Aerodynamic forces – drag, lift and sideforce – are defined as components of the aerodynamic force on the wind axes. In ground vehicle applications, due to small aerodynamic angles (in the absence of side wind), the two axes systems are often taken to be coincident. With this assumption, therefore, the components of force in the negative directions of x, y, and z axes are drag, sideforce and lift respectively. As far as the longitudinal motion of the vehicle is concerned, the most important aerodynamic force is the drag force opposing the motion. The lift component can alter the normal loads of the tyres and will therefore also be indirectly influential.

3.4.2.3 Mathematical Representation

It is common practice to express the aerodynamic forces and moments in terms of dimensionless coefficients. In general, any aerodynamic force is written in the basic form:

(3.28)

where C_F is aerodynamic force coefficient and stands for drag C_D, sideforce C_S and lift C_L coefficients respectively. is the characteristic area for the vehicle and it is taken as the frontal area A_F (also called the projected area). q is called dynamic pressure and is:

(3.29)

in which is the air density that is dependent on the ambient conditions. Under the standard conditions of 1 atmosphere (1.013 bar, 10132.5 Pa) at sea level and a temperature of 15 °C (288.15°K), the air density is 1.225 kg/m³. The air density can be approximately calculated from:

(3.30)

with p in Pascal and T in Kelvin (273.15+°C).

The velocity in Equation (3.29) is called the air speed and is the magnitude of speed of air relative to the vehicle body. In order to calculate the air speed, information on the wind velocity is also necessary. Let us assume a wind speed v_W is acting at angle ψ_W relative to the geographical north (see Figure 3.23). A vehicle is travelling at a speed of v_V in the direction of ψ_V. Vector is the air velocity and can be determined as:

(3.31)

can also be obtained graphically as shown in Figure 3.23.

The aerodynamic moments are expressed as:

(3.32)

where l is the characteristic length and for the vehicle usually the wheelbase is considered. C_M is the moment coefficient and represents C_l, C_m and C_n for roll, pitch and yaw moments respectively.

Aerodynamic forces are dependent on several parameters, such as the body geometry, ambient conditions and airstream properties. The geometry of the vehicle body includes its shape and dimensions. The latter was already considered as the characteristic area, and the vehicle wheelbase was considered as the characteristic length for the aerodynamic moments. The main influential ambient parameter is the air temperature that influences the air density and directly alters the aerodynamic forces.

The airstream properties include air speed and its direction relative to the body coordinates. When the airstream is at an angle to the vehicle forward axis, this angle is called the angle of attack. This angle also has an influence on the magnitude of the aerodynamic force. In order to account for these influential parameters, the dimensionless parameters Reynolds number (Re) and Mach number (M) are used. In general, the force (or moment) coefficients C_F (C_M) are a function of these parameters as well as angle of attack α and sideslip β (see Figure 3.22), i.e.:

(3.33)

In vehicle motion the Mach number is always of a small range of less than 0.2, and the variation of the Reynolds number is also very small. In the case of longitudinal horizontal wind, both angles α and β are small and under these circumstances, the force coefficient is virtually a constant. It is, therefore, practical to consider constant force or moment coefficients for the vehicle.

3.4.3 Slopes

The gravitational force on a slope will act in opposite directions for uphill and downhill motion of the vehicle. With reference to Figure 3.26 for a slope of angle θ relative to horizontal, the gravitational force simply is:

(3.34)

The positive and negative signs are for the downhill and uphill motions respectively. The gravitational force is a constant force as long as the slope is constant.

The slope is often expressed as a percentage rather than an angle. It is the tangent of the angle of the slope multiplied by 100:

(3.35)

It should be noted that in vehicle motion on a slope, the normal reaction force will also change and the rolling resistance force will be altered accordingly:

(3.36)

3.4.4 Resistance Force Diagrams

The total resistive force is the summation of the rolling resistance, aerodynamic and gravity forces:

(3.37)

Of these forces, the gravity force is constant and the two others are speed-dependent.

3.4.5 Coast Down Test

A vehicle coast down test is designed to experimentally measure the resistive forces including the rolling resistance (see Section 3.4.1) and aerodynamic drag. In a coast down experiment the vehicle is accelerated up to a high speed and then with the drivetrain disengaged, it is allowed to slow down under the action of the resistive forces. In general, the resistive forces depend on environmental and road conditions (e.g. temperature, wind, road surface and type). Therefore, under standard conditions, usually in still air and on a level road of sufficient length, useful information of vehicle speed variation with elapsed time can be obtained. This information then can be processed in order to estimate the rolling resistance and air resistance forces. Simple methods for the estimation of the resistive forces from speed-time data are discussed in Section 3.12.

3.5 Vehicle Constant Power Performance (CPP)

Modelling the longitudinal performance of a vehicle in general driving conditions is quite complicated when the engine throttle and gearbox ratios vary during motion. Simple models are used in the first instance to facilitate the analysis and understanding of the vehicle performance. One useful approach is to assume that the vehicle uses constant power to accelerate up to its maximum speed, corresponding in practice to full throttle acceleration. Later on it will be revealed that such acceleration is not a constant power motion, nonetheless, the assumption of constant power simplifies the equations of motion initially and provides a simple workable model for rough calculations.

3.5.1 Maximum Power Delivery

There have been arguments about how a power source can deliver maximum power as a function of its rotational speed. We shall examine this by the following mathematical procedure.

Consider a power source with output speed ω that is a function of time t. As shown in Figure 3.29, the output torque T is a function of speed. Therefore:

(3.38)

In order to obtain maximum power output, we are interested to find a speed at which the power is at maximum. To this end, Equation (3.38) can simply be differentiated with respect to speed ω:

(3.39)

or:

(3.40)

This is a differential equation that relates the output torque to the speed. Integration of Equation (3.40) results in:

(3.41)

where C is the constant of integration. If C = ln P₀ is assumed, then Equation (3.41) can be rewritten as:

(3.42)

or,

(3.43)

Equation (3.43) apparently implies that to deliver a maximum power from a source, it must be a constant power P₀. This statement by itself is false, since the power P₀ does not have to be the maximum power and can have different values. So what is wrong with the mathematical approach for the analysis? One must recall that the differentiation as performed in Equation (3.39) was only a necessary condition to obtain the maximum, and to ensure the existence of the maximum, there is also a sufficient condition, namely:

(3.44)

In the case of our problem, one can simply show by using Equations (3.39) and (3.40) that:

(3.45)

which means this point is not a maximum, nor a minimum, but a critical point. Thus this analytical approach does not lead to any workable conclusion on the way a power source can deliver maximum power. The discussion of this section is aimed at demonstrating that the idea of a constant power delivery does not relate to the maximum power delivery of a power source as is sometimes claimed. The constant power that is considered in the following sections can be any desired value of power.

3.5.2 Continuous Gear-Ratio Assumption

A geared transmission with clutch shifts causes discontinuous torque flow to the wheels and, in turn, delivers discontinuous tractive forces, and hence acceleration. Figure 3.30 shows a typical acceleration time history of a vehicle equipped with a manual transmission. When first gear is engaged, the vehicle accelerates up to a speed at which the engine rotation speed is too high and a gearshift is necessary. During the gearshift from first to second gear, the clutch is first disengaged, gearshifted, and the clutch re-engaged. This process will take some time during which there will be no tractive force available but the resistive forces are still acting on the vehicle. This situation will result in a temporary negative acceleration, as can be seen from Figure 3.30. During other gearshifts the same phenomena will result in discontinuous vehicle accelerations.

We will eventually obtain vehicle performance curves similar to Figure 3.30, but first it is more effective to start with simpler models. The main complexity involved in the general case arises from the gearshift process. One simple and useful assumption is to consider a transmission with continuous gear changing property; in other words, infinite gear ratios and continuous gear ratio changing. This assumption will lead to having a continuous torque flow and corresponding uninterrupted vehicle acceleration.

The constant power assumption may be considered as a single torque-speed point for the engine. In fact, when the power is unchanged, the optimum operating point for the engine can be chosen at this specific power. This means the engine works at a particular torque and speed values denoted by and . At such a point the engine power P_e is:

(3.46)

in which is engine torque in Nm, and P_e and ω* are in Watt and rad/s respectively.

At this engine speed, the vehicle moves ahead with an infinitely variable gear ratio and no shift delays. From a simple kinematic relation one can relate the wheel rotational speed to the engine rotational speed by:

(3.47)

in which n is the overall vehicle gear ratio, that is the gearbox gear ratio n_g times the differential (final drive) gear ratio n_f, i.e. (see Figure 3.31):

(3.48)

Now if it is further assumed that there is no slip for the driving wheels when they roll on the road surface, with reference to Figure 3.32 it is clear that the vehicle speed can be written as:

(3.49)

where r_w is the effective rolling radius of tyres. Equation (3.55) simply shows that the vehicle speed is inversely proportional to the transmission gear ratio.

It should be noted that the assumption of no slip for the tyre is strictly in conflict with its force-generating mechanism which depends on slip (see Section 3.3.1). Nevertheless, including slip in the analysis will cause difficulties we would like to avoid at this stage.

3.5.3 Governing Equations

The mathematical representation of vehicle motion involves the application of Newton's laws of motion to the vehicle. For our case, the motion of interest involves the longitudinal motion properties including the acceleration, speed and distance travelled. Newton's second law of motion is suitable for this problem of a unidirectional motion of the vehicle centre of mass. In Sections 3.3 and 3.4 the tractive and resistive forces acting on the vehicle were explained. The vehicle motion will be the result of interaction between these forces. The longitudinal vehicle motion involves a multitude of situations with different driving and resistive forces. For a moving vehicle the tractive forces are generated at tyre contact patches. The total tractive force F_T is the sum of individual driven tyre forces. The total power consumed due to the forward motion of vehicle with speed v is:

(3.50)

The power available at the driving wheels is:

(3.51)

Due to driveline losses we must have:

(3.52)

in which P_e is the engine power and it was assumed to remain a constant value P*. Equation (3.52) indicates that the power generated by the engine is not fully utilized and some part is lost due to friction in the vehicle driveline system. To include this effect we must consider the driveline efficiencies in our equations. In order to simplify the equations of motion, we will ignore this effect for the time being and it will be included later on in Section 3.13. For an ideal driveline with 100% efficiency, all three power values are equal:

(3.53)

Therefore P* is the constant power delivered to the driving wheels and from Equation (3.50):

(3.54)

This equation shows that the traction force depends only on the forward speed in an inversely proportional fashion. The variation of F_T versus speed in Equation (3.54) at different power outputs is depicted in Figure 3.33.

Note that Equation (3.54) implies that at low vehicle speeds, the tractive force mathematically becomes infinite. In practice, however, the tyre force is dependent on the tyre–road friction coefficient and slip ratio and there will be a maximum tractive force available for the tyre. If this limit is imposed, the total tractive force-speed diagram will be similar to Figure 3.34.

With this constraint in force, the tractive force will no longer remain uniform and it must be divided into two regions of limit adhesion and constant power. For the sake of simplicity this constraint will also be ignored for the time being and the amount of error induced with this assumption will be examined later in this chapter.

Now Newton's Second Law is applied to the longitudinal motion of vehicle with reference to the free body diagram of the vehicle shown in Figure 3.35. The longitudinal equation of motion simply is:

(3.55)

with:

(3.56)

Substituting for the resistive forces F_RR (Section 3.4.1), F_A (Section 3.4.2) and F_G (Section 3.4.3), will lead to:

(3.57)

In most elementary performance analyses, a constant rolling resistance coefficient is considered and with this assumption and using Equation (3.54), the final form of the longitudinal equation of motion of the vehicle will be:

(3.58)

where c and F₀ are two constants defined below:

(3.59)

(3.60)

It should be noted that in Equation (3.58) the air speed v_A is replaced with the vehicle speed v for conditions of still air with no wind.

Equation (3.58) is an ordinary differential equation in terms of the forward speed v of the vehicle. The time history of vehicle velocity will result from the integration of this equation.

It should be noted that Equation (3.58) was obtained under certain assumptions summarized in Table 3.8.

Table 3.8 Assumptions involved in derivation of Equation (3.58).

There is a constant power demand from the engine during entire motion

The vehicle gearbox has infinite gear ratios, changing continuously during vehicle motion

The driving wheels roll and there is no slip at road surface

There is no power loss in the vehicle driveline

Tyres can produce very large tractive forces

Rolling resistance force is a constant in all speeds

There is no wind during vehicle motion

3.5.4 Closed Form Solution

Equation (3.58) can be rearranged as:

(3.61)

Then, integration by separation of variables leads to:

(3.62)

It can be shown that the governing equation of longitudinal vehicle motion will be of the form:

(3.63)

in which v_m is the real root of the third order polynomial:

(3.64)

C_t in Equation (3.63) is the constant of integration and must be calculated by introducing an initial condition (e.g. at t = 0, v = 0), and the three constants a, k₁ and k₂ are given below:

(3.65)

(3.66)

(3.67)

Although Equation (3.63) provides a mathematical solution to the vehicle longitudinal motion, it is in the unusual form of t = f(v) instead of the more useful v = f(t) form. In order to solve for speed vs time, therefore, this non-linear equation must be solved by iteration techniques.

3.5.5 Numerical Solutions

In spite of the availability of a closed form solution for the problem, numerical solutions are often advantageous. Nowadays with the widespread availability of computer software, it is not difficult to set up a program and solve the problem numerically. Equation (3.58) for the vehicle acceleration can be written in the form:

(3.68)

which is an ordinary first order differential equation for the velocity v and can be integrated numerically by using well established numerical methods such as the Runge-Kutta method. Different software packages provide this routine for programmers and MATLAB is a good example which has a family of functions called ‘ode’ (standing for Ordinary Differential Equations). One of these functions called ‘ode45’ is a medium order method suitable for solving non-stiff differential equations. Using this routine is quite straightforward and will suffice for our case. In order to become familiar with using this function, Figure 3.38 shows two ‘m’ files, one as the main program and the other as a function to be called within the main program. As shown in Figure 3.38, the main program consists of five parts, whereas the function consists of three parts. The main program introduces values for the main parameters and sets the initial values (e.g. at time t = 0) for the main variables. MATLAB starts integrating the differential equation given in the separate ‘function’, starting from time ‘t0’ and ending at time ‘tf’. An internal MATLAB function ‘ode’ with input arguments will call the differential equation available in ‘function’ and once the integration process is complete, it will return the result. In other words, the user does not have access to the intermediate steps of the integration process.

A detailed program listing for the solution of the differential Equation (3.68) is presented in Figure 3.39. It is worth noting that the values for the parameters are also needed in the ‘function’, so one way is to provide the vehicle information inside the ‘function’ itself. Another way is to give the information in the main program and then share them with the ‘function’ by using ‘global’ statement. One more point is to note that the value of dv/dt approaches infinity at v = 0, due to division by zero in p/v term. To avoid this, we can set the initial value of velocity to a very small value close to zero. MATLAB uses a variable called ‘eps’ for this purpose (eps = 2.2204e-016).

3.5.6 Power Requirements

The value of the engine power is one of the first requirements in the initial design stages. With a constant power assumption, one can easily evaluate the power necessary for a specific performance requirement. Very rough estimations can be found by using ‘no resistive force’ (NRF) model (see Problem 3.3) or ‘low speed’ (LS) model (see Problem 3.5). For a more accurate power calculation, the general form of the equation of longitudinal motion must be solved.

In the general case, two approaches can be followed; using the closed form equation or numerical integration. In the first approach, Equation (3.63) is used to obtain the power P for given t and v values. Obviously this must be done by applying a trial and error technique. MATLAB's function ‘fsolve’ can be used for this purpose. It finds a solution P^* that satisfies Equation (3.63) for given values of t and v. It should be noted that to initiate the ‘fsolve’ function an initial guess is required. A proper initial value must satisfy (see Equation (3.64)). The following example shows how to use ‘fsolve’ for this problem.

Alternatively, the equation of motion (Equation (3.68)) can be solved by the numerical integration, but as observed in Section 3.6.5 the solution requires the power as an input. In fact, the process examined earlier was to obtain the speed values from a given power value. Here, the process is reversed as the power P is needed for a required speed v* at a specified time t*. Thus in order to calculate the power, an iteration loop must be performed in the numerical integration process. The result in one iteration process will be a value v₁ for speed at time t*; if v₁ happens to be equal to the specified speed v*, then the given power P is the answer. Otherwise P must be modified until the speed obtained from iteration process equals v*.

3.5.7 Time of Travel and Distance

Two important parameters in the vehicle motion are the time duration and the distance the vehicle travels during that time. In this section the equations of motion will be used to find these two parameters. The required time to achieve a specified speed can simply be determined by the direct use of the result of the closed form solution given in Equation (3.63). Alternatively, the numerical integration method can also be used to obtain this time value but the process needs a ‘while’ loop. In fact, the travel time was already obtained from the numerical integration of the equation of motion. The integration methods advance the time by small steps to the final integration time t_f that is the vehicle's total travel time at the end of the integration process. In other words, in such processes the time is pre-selected and velocity is calculated at the end of that specified period. In the numerical method, therefore, the travel time at a desired speed can be obtained from the time history of motion that is in the form of vehicle speed versus time. Alternatively by using a ‘while’ statement, the time can be obtained directly. For the determination of the travelled distance S, the following relation can be used:

(3.69)

and after substituting from Equation (3.58), one obtains:

(3.70)

which is similar to Equation (3.61) and can be integrated similarly. The result reads:

(3.71)

with:

(3.72)

and C_S is the constant of integration. Alternatively, to calculate the distance of travel using numerical integration, instead of using the approach of Equation (3.69), it is recalled that velocity is the derivative of travelled distance, that is:

(3.73)

This adds another ordinary differential equation to the one existing before for the velocity (i.e. Equation (3.68)). For the numerical integration, the number of differential equations does not make any difference, in fact, the procedure of numerical solution is similar when a set of differential equations are present. In the MATLAB program of Section 3.5.5, only a few changes listed below are necessary:

Changes necessary to the program of Section 3.5.5 are listed in Figure 3.43.

3.5.8 Maximum Speed

The maximum speed of vehicle is one of the important factors in the performance analysis. The acceleration of the vehicle shows the increase of the vehicle speed vs time. The maximum speed will occur when the resistive forces equal the tractive force available. Mathematically this means:

(3.76)

This also means a steady-state condition for the vehicle motion. From the equation of motion of the vehicle, it is clear for such a condition that:

(3.77)

Since both tractive force F_T and resistive force F_R are changing by the time, there will be an instant t* in which both forces attain equal values. At this instant the vehicle speed will reach its maximum. If the variation of tractive and resistive forces is drawn in a single figure, the intersection point will indicate the instant t*. The tractive and resistive forces are functions of velocity v, so a diagram can be constructed for the variation of the forces against velocity. The tractive force for CPP was given in Equation (3.54):

(3.78)

and the resistive force F_R(v) depends on the assumptions made for the rolling resistance force. For a constant rolling resistance force f₀W (see Section 3.4.1.5), Figure 3.45 shows the variation of the tractive and resistive forces in a single diagram. The maximum speed of the vehicle is reached where the two forces balance each other. This point is the intersection point of F_T(v) with F_R(v). For the specific condition shown in Figure 3.45, the maximum speed is reached exactly at 50 m/s (180 km/h).

To obtain the maximum speed of the vehicle a mathematical solution can also be obtained from Equation (3.77) after substituting the relevant terms for the tractive and resistive forces:

(3.79)

A closed form solution for Equation (3.79) is in the following form:

(3.80)

in which:

(3.81)

(3.82)

(3.83)

(3.84)

(3.85)

Alternatively, Equation (3.79) can be solved numerically using MATLAB functions ‘fsolve’ or ‘roots’. The following example will use both methods.

3.6 Constant Torque Performance (CTP)

In the previous section we discussed a simplified longitudinal acceleration performance of the vehicle. The major assumption was to consider a constant power demand from the vehicle during the entire motion. This was to simulate a constant load (including the full load) acceleration in which the vehicle is assumed to demand fixed power. In practice, the power demand of the vehicle, however, depends on the torque-speed combination during the acceleration. It is, therefore, necessary to take into the consideration the torque generator's characteristics as well as the vehicle parameters.

There are circumstances in which the torque of the power source is almost constant. Two examples are electric traction motors in their constant torque phase of operation (see Figure 3.4) and modern diesel engines (see Figure 3.3) in their flat torque area. For such cases, the power is not constant and the foregoing discussion cannot be used. Another difference for the constant torque case is the effect of the gearshifts that take place during the acceleration. With a constant torque T from the power source, the tractive force F_Ti of the vehicle in a specified gear ratio n_i is a constant:

(3.86)

in which n_i and r_w are the overall gear ratio and the wheel effective radius respectively. At any speed, the resistive force is known and the difference between the tractive and resistive force will generate the acceleration a(v):

(3.87)

The implication from Equation (3.87) is that any gearshift will change F_Ti and in turn the acceleration.

3.6.1 Closed Form Solution

In order to find the velocity, Equation (3.87) should be integrated with respect to time. In differential equation form:

(3.88)

This is a differential equation that can be integrated by the separation of variables. The closed form solution of Equation (3.88) is of the form:

(3.89)

where:

(3.90)

(3.91)

The time to reach a specified speed v can be obtained from following relation:

(3.92)

The travelled distance S, in terms of vehicle speed is given by:

(3.93)

In order to apply the results to the motion of a vehicle with a number of gears, consider that the torque-speed diagram of the power source is as shown in Figure 3.47. Let us assume that the vehicle accelerates up to the speed of at each gear. In other words, when in first gear the rotational speed reaches , the gear is shifted to gear 2. The same applies for the 2–3 gear and so on.

The maximum speed at each low and middle gear is when the engine rotational speed is :

(3.94)

But this situation for high gears can only happen if the balance of forces is also consistent. The force balance for a high gear n_H is:

(3.95)

which results in:

(3.96)

This result according to Equation (3.90) is simply equal to β and is only valid if the associated engine speed is less than the maximum speed. Therefore the smaller of results obtained from Equation (3.94) and the value of β must be used. The travel time for each gear is calculated from Equation (92), noting that:

(3.97)

The total travel time at the end of gear number N (to reach ) is:

(3.98)

A similar approach must be used for the vehicle speed and travel distance at the end of gear number N.

The maximum power for each gear is:

(3.99)

With the assumption of no power loss in the driveline, the maximum power for all gears at the maximum speed will be equal, as the power at the maximum speed points is:

(3.100)

If the maximum speed at a particular gear is less than the maximum speed , the actual speed must be used in Equation (3.100).

3.6.2 Numerical Solutions

For the constant torque equation of motion 3.87, a closed form solution was presented in the previous section. In order to obtain a numerical solution, the method of Section 3.5.5 can be applied with some modifications. The main difference in this case is that the integration must take place at each gear separately. To this end the process for each gear must be repeated, thus there should be a loop with repetition up to the number of gears. The initial conditions at each gear (except gear 1) will be the final results at the previous gear. Another issue in this case is the need for iteration to find the time at which a gear is engaged. This is because the integration at each gear is taking place between two specified times t₀ and t_f. However, the final integration time t_f is related to the maximum speed of the constant torque phase.

A sample MATLAB program is provided in Figure 3.53 in which a loop is considered for four gears (i = 1: 4), and the iteration loop is done within a ‘while-end’ statement. The function associated with the main program contains the differential equation (3.87). In practice, gearshifts will take some time; nevertheless no delay for gearshifts is included in the program in this case. It would only require a simple modification to include it.

3.7 Fixed Throttle Performance (FTP)

The general longitudinal motion of a vehicle involves the driver's throttle inputs and gearshifts, both of which will have influential effects on the overall performance of the vehicle. Taking into consideration the different inputs from the driver requires a more complicated analysis. One special case in driving already discussed is the full throttle acceleration. In this particular case, the torque-speed characteristics of the engine are usually available and can be used in the analysis. An extension to this case is the fixed throttle acceleration in which instead of 100% throttle input, a fixed part throttle is the input from the driver.

3.7.1 Gearshift and Traction Force

When the throttle is fixed, the quasi-steady engine torque-speed diagram is available and from the MT engine formula (Section 2.6), the part throttle torque-speed relation can be obtained. Full throttle acceleration will be a special case for the analysis and of course, the full throttle curve is usually available for most engines, and analysis can be performed more easily for this case. The engine torque at a constant throttle can be written as:

(3.101)

Assuming no driveline losses (see Section 3.13), the torque at the driving wheels is:

(3.102)

where n_g and n_f are the transmission and final drive gear ratios. According to the free body diagram of the driving wheel shown in Figure 3.55, the tractive force is related to the wheel torque by ignoring the wheel rotational dynamics (Section 3.10):

(3.103)

This tractive force must be produced by the tyre through the generation of slip between the tyre and the road, provided that sufficient friction is present at the surface. For the time being it will be assumed that the tyre is able to generate the necessary tractive force.

In a certain gear, the tractive force will change according to Equation (3.103) with the engine's rotational speed. With the kinematic relation between the engine speed and the wheel rotational speed, the tractive force at each gear can be related to the wheel rotational speed. A no-slip approximation is considered for the tyre rotation in order to directly relate the forward speed of the wheel centre v to its rotational speed by the simple relation:

(3.104)

or,

(3.105)

Therefore, the tractive force at each gear can be directly related to the forward speed of the wheel that is equal to the forward speed of vehicle when no slip is present. At different gear ratios, by using Equations (3.103) and (3.104), different tractive force-speed diagrams can be constructed.

3.7.2 Acceleration, Speed and Distance

The difference between the tractive force of the vehicle F_Ti in a typical gear number i with the resistive force F_R at a specific speed v will generate the acceleration:

(3.106)

This equation can be represented by Figure 3.58 for a specific gear ratio. It is clear that the excess force at each gear varies with vehicle speed and the resulting acceleration will also vary with speed. It is also important to note that at low gears the excess force is greater (see Figure 3.57) and produces greater accelerations.

In order to find the velocity, Equation (3.106) can be integrated with respect to time. In differential form it reads:

(3.107)

Theoretically this is a differential equation that can be integrated by separation of variables. Based on our experience in the previous cases in which the tractive force had simpler forms, and also owing to the fact that the engine torque-speed variation usually takes a complex form, looking for a universal closed form solution is not attractive. Nonetheless, there are cases in which the engine torque-speed variation has a relatively simple form and the integration of Equation (3.107) is possible. In general, however, a numerical solution would be preferred.

In order to solve Equation (3.107) numerically, it can be written as a time-based differential equation of the form:

(3.108)

Since v(t) is obtained through the numerical integration, F_Ti(v) and F_R(v) of Equation (3.106) are available at each instant. The main MATLAB program for this case is similar to the main program of Section 3.6.2, but its accompanying function will differ to some extent. In the main program a similar inner loop for the acceleration up to a certain engine speed (ω_em) will be present. Also, an outer loop for gear selection is present as shown in Figure 3.59. It is assumed that the shifts are performed instantly and no delay is involved.

The function for this case must include the instantaneous engine torque and tractive force calculations. For this reason the engine torque-speed formula is needed in the function statements. This information can be shared with the function through the ‘global’ statement. In the program of Figure 3.60a second order polynomial is assumed for the torque-speed of engine (see Example 3.7.1) and its coefficients are given in a vector ‘p’ that is shared between the main program and the function.

3.7.3 Shift Times

When a gear should be shifted depends on several factors. In automatic transmissions the decision is made by the controller and in manual transmissions by the driver. In the latter case, the gearshift is performed according to the individual driving style and the engine speed at which the gear is shifted differs for different drivers. In theory, when a new gear is engaged, the vehicle can accelerate up to the limit speed of the engine. In practice, however, this is not the case and every driver changes the gear based on their style of driving at speeds below the maximum. Gearshift issues are discussed in more detail in Chapter 4.

In most transmissions, and especially in manual transmissions, there is an engine torque interruption during a shift. This will eliminate the tractive force and with F_T absent, only resistive forces will act on the vehicle during the shifts and negative accelerations will result (see Figure 3.30). This duration is different for different drivers and even for different gears. In the previous analysis, immediate gearshifts with no torque interruption were assumed. In order to consider the torque break involved in the gearshifts, an inner subprogram should be included in the existing MATLAB program. The main idea is to set the tractive force to zero during this ‘tdelay’ time. This is left to the reader as an exercise to apply the necessary changes to the MATLAB program and generate results that are depicted in Figures 3.30 and 3.63 for the vehicle acceleration, speed and distance (see Problem 3.15).

3.7.4 Maximum Speed at Each Gear

The maximum speed at each gear may be reached in two different situations: kinematic limit or dynamic balance point. The kinematic limit is when the tractive force exceeds the resistive force and vehicle can still accelerate but the engine cannot turn faster. A good example is acceleration in low gears at which the tractive force is very high and the resistive forces are very low and the excess force is used to accelerate the vehicle but the engine speed can reach very high values close to its cut-off point. At this point an upshift is needed, as the forward speed is still low but driveline kinematics imposes a constraint on the motion. Upshifts will allow higher kinematic vehicle speeds but at the same time lower tractive forces and reduced accelerations will occur. In high gears, due to high speeds and low traction forces, there are points at which the resistive force can balance the tractive force. These points, called dynamic balance points, represent steady states for the vehicle motion – even though the vehicle may not remain in this steady condition for long.

Figure 3.64 summarizes these issues for an example vehicle on a level road. At low gears (1, 2 and 3) as seen, there are excess tractive forces at the points of engine maximum speeds. These points are kinematic limit points at which the gear must be shifted. At the highest gear (gear 5), the resistive force grows with speed and at a certain point equals the tractive force. At this point, the dynamic balance is met and vehicle attains its final speed. At gear 4, too, it is sometimes possible to have a dynamic balance point, for example on a slight slope. Also in Figure 3.64 an intersection point would be possible for gear 4 if a slightly higher engine speed were attainable.

At the dynamic balance points where the tractive and resistive forces balance, the equation of motion simply is:

(3.109)

This equation will be satisfied at a certain forward speed v* and a corresponding engine speed of ω*. At each gear n_i, Equation (3.109) should be solved for v* where the steady state could be achieved (e.g. high gears), but the speed of engine at v* must be smaller than the engine limit speed , that is:

(3.110)

This process should usually be performed numerically by modifying the existing MATLAB program. However, for some cases where the engine torque-speed relation at a fixed throttle is available, a mathematical solution can also be obtained. Let the engine torque-speed formula be written in the form:

(3.111)

Then the solution is obtained from:

(3.112)

3.7.5 Best Acceleration Performance

A question is often raised of how to perform gearshifts to get the best acceleration performance. In practice, achieving higher velocities in less time means higher average accelerations. From a mathematical view point, velocity is the time integral of acceleration:

(3.113)

In a graphical representation, the area under the acceleration curve versus time gives the velocity. For a fixed throttle performance (FTP) (e.g. WOT), the acceleration behaviour of the vehicle is determined with the gearshifts. To see the effect of the choice of the shift point, consider the acceleration performance of a typical vehicle at two different shift speeds shown in Figure 3.65 in an acceleration-speed diagram. Solid lines represent a shift carried out at a certain engine speed where the traction force is equal at two consecutive gears (corresponding to the point of intersection of traction curves of the two gears), whereas the dashed lines belong to a shift at lower engine speeds.

It is clear that the area under the solid lines (A₁) is larger than the area under the dashed lines (A₂). Also if the shift speeds of the dashed lines are further reduced, the area A₂ will also be decreased. Therefore, as the maximum area is A₁ it can be concluded that the concept of gearshifting at the points of intersection with consecutive traction curves, produces higher accelerations. According to Equation (3.113), the area under the acceleration-time equals the speed of the vehicle at the end of the acceleration phase and thus for the area A₁ the speed is maximum.

In the acceleration-time plot, the acceleration curve of each gear will not remain identical when the shift speed varies. This can be seen in the plots of Figure 3.66 for two different shift speeds of 5500 rpm (solid lines) and 4000 rpm (dashed lines). Nonetheless, the area under the acceleration-time curves is still larger overall as the shift rpm is increased. In order to further clarify the effect of the shift rpm, the time to attain a specific speed is a good measure as it indicates the overall acceleration performance. Figure 3.67 shows the velocity-time curves for four different shift rpm 4000, 4500, 5000 and 5500. It is clear that at any speed (a horizontal line), the time to reach that speed is smaller for larger shift speeds. This is shown for speed of 30 m/s (108 km/h). The time to reach the specified speed reduces in a non-linear fashion for example from 4000 to 4500 rpm the time reduces around 3 seconds, whereas from 5000 to 5500 rpm it is reduced around 1 second.

It should be noted that the overall appearance of the acceleration and speed curves are dependent on the transmission gear ratios as well as the vehicle physical properties. In changing any of the parameters, the details of the figures presented above will change, nevertheless the overall behaviour is similar for different configurations.

3.7.6 Power Consumption

The power requirement during the vehicle motion comprises two parts: power to accelerate; and power to overcome resistive forces. The total power used to move the vehicle at a specified speed v(t) is:

(3.114)

The tractive force F_T can be written as:

(3.115)

Therefore:

(3.116)

The first term P_a is the power needed for the vehicle acceleration and the second term P_R is the power to overcome the resistive forces. Equation (3.116) indicates that the power is an instantaneous quantity that varies during the vehicle motion according to the variations of acceleration or speed. It also signifies that for greater acceleration more power is needed.

During the vehicle motion with a fixed throttle input (e.g. WOT), the variation of power with speed at different gears is shown in Figure 3.68. As the acceleration is higher at lower gears while the speed is low (low resistive forces), the main part of power is used for accelerating and only a small portion is used for resistive forces. Upshifting reduces the acceleration where the speed is high and the resistive forces play an important role; as a result, the share of the power to overcome the resistive forces increases at higher gears. At the final speed (where the acceleration is zero), the total power is devoted to overcoming only the resistive forces.

The time history of power variation at different gears is shown in Figure 3.69 for a full throttle acceleration with gearshifts at the maximum power point. When a gear is upshifted, the acceleration will drop and as the resistive forces remain unchanged at the specific speed, the power demand will be reduced. In less aggressive driving, the acceleration in low gears often ends in lower engine speeds, and as a result, the power use in low gears is usually much lower than the maximum engine power. It should be noted that the selection of transmission gear ratios can change the results by moving the points in the plots. The effect of gear ratios on the power curves, especially at high gears, will be discussed in Chapter 4.

3.8 Throttle Pedal Cycle Performance (PCP)

In real driving situations, the accelerator pedal is depressed by the driver according to driving requirements. In a vehicle equipped with a manual gearbox, the pedal input even in a high acceleration performance is not kept constant, since at least during gearshifts it is necessary to release the pedal and press it once again. In normal driving conditions, the pedal input varies in different driving situations. Thus in the study of vehicle longitudinal performance, one category of problem is when the input variable is the engine throttle and the output is the resulting vehicle motion.

Drivers' pedal inputs can vary largely owing to diverse driving situations and driving habits. The time history of the pedal variations with time may look like that of Figure 3.70 in which the pedal input varies between 0 (fully closed) to 100 (wide open).

In order to simulate a driving condition under pedal inputs, the engine torque-speed- throttle characteristics will be necessary. This can be in the form of look-up tables or the MT engine formula introduced in Chapter 2.

Moreover, during driving, the gearshifts are also carried out according to the vehicle and engine speeds and driver's selection. In order to include the pedal inputs in the MATLAB programs, it is necessary to do it inside the ‘function’ and not inside the main program. The reason is that the pedal inputs are time-dependent and when ‘ode’ is called inside the main, all integration steps are performed by the ‘function’ and the final results are returned. The following example is designed to illustrate this concept.

3.9 Effect of Rotating Masses

When the vehicle accelerates, the wheels, rotating shafts and even the engine accelerates. In other words, in addition to the vehicle body (mass) which needs tractive force to develop its kinetic energy, the rotating inertias also need torques in order that their corresponding kinetic energy is developed. The demanded power produced by vehicle engine is therefore divided and the part that propels the vehicle body is reduced due to the share taken by rotating masses. This will reduce the vehicle acceleration compared with the case with no rotating masses. Consider the torque balance from the engine to the driving wheels through each of the torque-consuming driveline components as shown in Figure 3.74.

The application of Newton's Second Law to the rotational motion of the engine output with regard to Figure 3.74a reads:

(3.117)

This means the clutch torque T_c is what is left from the engine output torque T_e after accelerating the relevant inertias I_e grouped at the engine output shaft (including the engine rotating masses, flywheel, clutch assembly, etc.) with angular acceleration . In Figure 3.74b it is assumed that the inertias of all rotating masses related to the gearbox input and output are summed up as a single inertia I_g at the output shaft of gearbox. Before this inertia, the torque on the output shaft simply is the input torque multiplied by the gearbox ratio n_g:

(3.118)

The output torque after the inertia I_g, therefore, is:

(3.119)

where is the angular acceleration of the output shaft of gearbox. For the final drive, according to Figure 3.74 it is assumed that there is only a single drive axle and all the rotating inertias related to it are collected as a single inertia I_A at the output side of the differential. The governing equation in this case will be:

(3.120)

being the angular acceleration of the final drive output shaft. Finally, the torque output from the final drive reaches the wheel with an overall inertia I_W rotating with the wheel speed and accelerates it. According to Figure 3.74d, the balance of torques around the wheel centre results in:

(3.121)

where F_Ta is the tractive force available at the tyre–road interface. Since all the shafts are connected, their rotational speeds (or accelerations) are related to the engine rotational speed (acceleration), i.e.:

(3.122)

(3.123)

Combining all equations results in:

(3.124)

This is the force propelling the vehicle, whereas in the previous discussions the tractive force was simplified as:

(3.125)

Now, for the longitudinal motion of the vehicle under the action of tractive and resistive forces, the equation of motion reads:

(3.126)

where a is the longitudinal acceleration of the vehicle. F_Ta is produced by the tyre through slip generation at contact patch. With slip at the tyre–road interface, the rotational speed of the tyre will differ slightly from the speed of wheel centre divided by the tyre rolling radius (see Section 3.10). Accepting a small error, the no-slip condition for the tyre will be assumed and that allows the rotational acceleration of the wheel to be directly related to the vehicle acceleration a:

(3.127)

Substituting Equations (3.124), 3.125 and (3.127) into Equation (3.126) yields:

(3.128)

which can be written as:

(3.129)

with:

(3.130)

From Equation (3.130) it is clear that m_eq is always larger than m, therefore the vehicle acceleration in the real case will be smaller than that obtained in previous analyses without accounting for the rotating masses. In other words, when the effect of rotating masses is taken into consideration, it is like considering an extra mass for the vehicle. Alternatively, when no rotating masses are considered, a larger tractive force will be available. Figure 3.75 conveys this concept by assuming a single flywheel equivalent to all rotating masses, attached to the driving wheels.

In general, m_eq can be written as:

(3.131)

where m_r is regarded as the equivalent mass of the rotating inertias:

(3.132)

In Equation (3.132) the inertias are defined as:

It is important to note that the m_eq has different values at different gears, since changes with transmission gear ratio n_g. Larger transmission ratios will make m_eq larger (by power 2 of n_g). This means in the lowest gear (i.e. gear 1), m_eq will have the largest value whereas at highest gear (e.g. gear 5) it will have the smallest value. This is working against the requirements of high accelerations at low gears.

3.9.1 Corrections to Former Analyses

When m_eq is substituted in the equation of vehicle longitudinal motion it will be of the general form:

(3.133)

The gear ratio n for a discrete ratio gearbox is a fixed value at a specific gear. For a CVT, however, its value also depends on the speed. Thus, for the two cases of analyses, the following corrections must be made in order that the effect of rotating masses is included.

3.9.1.1 Discrete Gear Ratios

For all cases of constant torque, fixed throttle and throttle pedal cycle, the condition of gear ratio changing during the vehicle motion was discussed. In order to include the effect of rotating masses, the only required change will be the evaluation of the equivalent mass at each engaged gear, as for a particular gear ratio n the equivalent mass is an unchanging quantity. The method of solution for all cases will be similar once the gear ratio is specified.

3.9.1.2 Continuous Gear Ratios

The continuous variation of gear ratio was considered in the Constant Power Performance analysis. The equation of motion for this case when the rotating masses are included reads:

(3.134)

From the kinematic relation (no slip assumption):

(3.135a)

but since power is kept constant, there are two different cases:

3.10 Tyre Slip

In the foregoing discussions in order to simplify the relation of the vehicle speed to the wheel rotational speed, the driving wheels were assumed to roll without slipping. In reality, however, this is not the case since according to the discussions in Section 3.3, the traction force is produced by the tyre only if longitudinal slip is developed between the tyre and road surface. In fact, in the real driving situation, the process by which the vehicle motion is built up can be described by the flowchart of Figure 3.78. According to this process, the transformation of the wheel torque to the tyre traction force is accomplished through the generation of tyre slip. This process, in addition to the vehicle longitudinal dynamics, involves the wheel rotational dynamics in which the input torque generates the wheel rotational speed. The rotational motion of a driving wheel can be modelled according to the free body diagram of Figure 3.79.

Application of Newton's Second Law to the rotation of a wheel results in:

(3.136)

in which T_w is the wheel torque, ω_w is the rotational speed of the wheel and I_w is the polar moment of inertia of the wheel including the other rotating parts (see Section 3.9). From Equation (3.11), the tyre slip is:

(3.137)

Equations (3.136) and (3.137) show the interdependency between the wheel speed and slip S_x. In fact, Equation (3.136) is a differential equation which, when integrated, leads to values for the wheel rotational speed . To this end, however, S_x must be available beforehand but it is dependent on the wheel rotational speed (Equation (3.137)). With a numerical solution the problem can be resolved by having an initial condition for the vehicle motion, e.g. starting from rest. The wheel rotational speed and vehicle speed are then obtained by simultaneous numerical integration for and v of the vehicle. The solution of the differential Equation (3.136) together with Equation (3.137) involves rapid fluctuations in the wheel slip and its rotational speed values. In MATLAB to solve this type of differential equations one must invoke ‘ode15s’ instead of ‘ode45’. There are also other points that must be taken into consideration:

The programming for this problem is left to the reader and typical speed and acceleration results are given below as a reference. The time history of vehicle speed looks like that shown in Figure 3.80. This figure also shows the equivalent centre of wheel speed that is wheel rotational speed times its radius. It is clear that the equivalent speed must be greater than the vehicle speed due to the wheel slip. The time history of vehicle acceleration is depicted in Figure 3.81 together with the same results for the case with no wheel slip. It is noted that there is a short delay in the response of the vehicle owing to the wheel dynamics. In fact, with the wheel dynamics in effect, the tyre tractive forces build up after the tyre slip is generated.

3.11 Performance on a Slope

So far the vehicle motion was assumed to take place on a level road only for the sake of simplicity. In this section, the effect of motion on a slope will be discussed. It should be noted that the intention is to see what happens to the vehicle performance (e.g. speed or acceleration) when it moves on a sloping road. Another area of interest is to examine the grade capability of a vehicle – which will be discussed in Chapter 4.

The vehicle motion on a slope can be viewed from the following perspectives:

• How will a vehicle accelerate?
• What is the maximum vehicle speed at each gear?
• What is the maximum slope a vehicle can move on at a specified speed?
• What is the vehicle performance on a road of variable slope?

The FBD of a vehicle on a sloping road was shown in Figure 3.36 with all forces acting on it and the governing equation of motion was shown to be of the form:

(3.139)

Depending on the nature of the tractive force F_T the solution method will be different as it was discussed in previous sections.

3.11.1 Constant Power Performance (CPP)

For the constant power case, there is no concern about the gear ratios and with the inclusion of the angle θ in the equation of motion, the performance can be analyzed during vehicle acceleration. The acceleration performance and maximum speed of vehicle on the slope will obviously be reduced due to the increase in resistive forces. The MATLAB program of Figure 3.39 already includes the slope terms and can be used without any change. In order to find the maximum slope on which the vehicle can move with a specified speed, the same MATLAB program for finding maximum speed can be used. In fact, with an iteration procedure the input slope can be varied until the specified maximum speed is reached.

3.11.2 Constant Torque Performance (CTP)

In this case, the gearshifts occur and it is necessary to examine if the vehicle can ascend a slope in a certain gear. In each gear, therefore, there is a certain maximum slope that vehicle can move at a constant speed. In order that the vehicle can maintain the speed in a specific gear n_i right after an upshift, the force balance must be positive at the time of gearshift, i.e.:

(3.140)

where:

(3.141)

and is the maximum speed of previous gear prior to the gearshift. If Equation (3.140) is not satisfied at gear i, the vehicle cannot keep the final speed of previous gear and either a lower maximum speed is achieved or a downshift to a lower gear is required. The maximum achievable vehicle speed in each gear n_i can be written as (Equation (3.96)):

(3.142)

provided that the term in the parenthesis is positive – otherwise gear n_i cannot be maintained and a downshift would be necessary. In addition, the resulting value from Equation (3.142) must be less than or equal to the kinematic speed limit:

(3.143)

In fact, the minimum of the two results obtained from Equations (3.142) and (3.143) is the maximum speed in that gear. The required changes for the inclusion of slopes in the MATLAB programs are straightforward.

3.11.3 Fixed Throttle (FT)

For this case too, the concept is similar to those discussed in previous sections and the force balance condition for any gear to maintain a positive acceleration after an upshift on a slope is:

(3.144)

where is the engine torque at a given throttle input θ and at engine rpm ω_i immediately after shifting to n_i. If this condition is not satisfied for a gear n_i, the resultant force in the direction of motion will be negative after an upshift, and a downshift might be necessary if the engine speed is reduced to a low value ω_L. A steady state maximum vehicle speed is also attainable for those gears not satisfying Equation (3.144) but at a lower value than that of the previous gear. In any case, the constraint for the engine speed (Equation (3.143)) should not be violated.

3.11.4 Variable Slopes

If the vehicle motion involves several piecewise-constant slopes, the solution can be divided into several solutions for each slope. The initial conditions for every slope are the final results of the previous slope. In general, the slope can be expressed as a varying elevation-distance function and it must be defined inside the MATLAB ‘function’.

3.12 Vehicle Coast Down

Vehicle coast down is a condition in which the transmission is disengaged (gear in neutral) and the vehicle is left to roll freely from a particular speed and come to a stop under the action of the resistive forces. A coast down test is very useful when measuring the characteristics of resistive forces acting during vehicle motion. On a level road, the resistive forces comprise the driveline rotational frictions, the rolling resistance and the aerodynamic force. From a coast down test, the time history of vehicle speed can be obtained. In order to exclude the effects of air movements on the results, the tests are usually performed in still, no-wind air conditions. The mathematical modelling of vehicle motion previously developed can be applied to this type of vehicle motion with no motive force from the prime mover and the gradual deceleration of the vehicle under the action of the resistive forces only. The equation of longitudinal motion in a coast down situation will be simpler owing to the absence of the tractive force:

(3.145)

in which F_RR is the rolling resistance force and the term cv² is the aerodynamic force. If the time history of the vehicle speed is available, the two unknowns can be determined. However, the solution depends on the nature of the resistive forces especially the rolling resistance that can be considered a constant or as a function of speed. These will be considered in following sections.

3.12.1 Constant Rolling Resistance

Assuming a constant rolling resistance force is a common practice in many cases, i.e.:

(3.146)

From the time history of vehicle speed it is possible to determine the rolling resistance coefficient f_R and aerodynamic coefficient c.

3.12.1.1 Simple Model

With only a simple test we can estimate the ‘F₀’ and ‘c’ values. This test consists of two high speed and low speed parts. Two sets of data are recorded at high speed and low speed cases. Each set consists of an incremental velocity change defined as v_H1 and v_H2 at high speed and v_L1 and v_L2 at low speed. The incremental velocity can be as small as 5 km/h. The time increments over which the velocity increments occur are called and . The mean velocities at each case are defined as:

(3.147)

(3.148)

The mean decelerations at each velocity are defined as:

(3.149)

(3.150)

The equations of motion (Equation (3.145)) for the low and high speeds can be expressed in terms of the above definitions as:

(3.151)

(3.152)

At low speeds it is quite reasonable to ignore aerodynamic force and Equation (c03-mdis-0173) can be simplified to:

(3.153)

This is an equation suitable for calculating the rolling resistance force. Substituting Equation (3.153) into Equation (3.151) and using Equations (3.148) and (3.150) results in:

(3.154)

If incremental velocities in both low speed and high speed are taken equal, i.e.:

(3.155)

then the results for F₀ and c simplify to:

(3.156)

(3.157)

3.12.1.2 Analytical Model

For the general coast down case, the equation of motion can be integrated analytically. Equation (3.145) can be written as:

(3.158)

in which:

(3.159)

(3.160)

Integration of Equation (3.158) results in:

(3.161)

In general, at time t₀ the speed is v₀, but is always possible to take t₀ = 0, therefore:

(3.162)

where:

(3.163)

If test data for two points, say, points 1 and 2, are available, then:

(3.164)

(3.165)

Equations (3.164) and (3.165) constitute a set of two non-linear equations for the two unknowns e and f. With an iteration technique like MATLAB's ‘fsolve’, the equations can be solved and once e and f are obtained, the aerodynamic coefficient c and rolling resistance coefficient f_R can simply be calculated from:

(3.166)

(3.167)

An interesting result from this method is the evaluation of vehicle mass from a coast down test. Hence, if a further data point is available, then three equations can be solved for three unknowns e, f and m.

3.12.2 Rolling Resistance as a Function of Speed

In general, the resistive force can be written (see Section 3.4.1.5) in the form of:

(3.168)

The two first terms on the right-hand side represent the rolling resistance force which is considered as a linear function of the vehicle speed. It should be noted that the rolling resistance may also involve a speed-squared term, however, that part will be included in the aerodynamic force. In order that the unknowns F₀, f_v and c are evaluated, the speed-time history obtained from the test must produce the variation of total resistive force versus velocity. To this end, the basic form of the equation of motion in a coast down test reads:

(3.169)

Thus by differentiating the time history of vehicle velocity with respect to time, the resistive force versus time will be obtained. If a quadratic curve is fitted to the F_R-v data, it will have the form:

(3.170)

from which the unknowns F₀, f_v and c can be evaluated by a simple comparison of Equations (c03-mdis-0191) and (c03-mdis-0193). It is worth noting that this technique can be applied for powers of speed more than 2. For example, it is possible to include an extra term with a power of 3 for speed v and use the same procedure to obtain the unknown coefficients. The drawback of this method is its sensitivity to the accuracy of test data owing to the differentiation involved in the process. In order to avoid such errors, a more sophisticated non-linear curve fitting techniques must be used.

3.12.3 Inertia of Rotating Masses

At the start of a coast down test, the vehicle has a certain initial velocity and all driveline rotating components have their respective rotational speeds. The total mechanical energy E of the system at the starting instant, therefore, consists of kinetic energies of vehicle body together with those of the rotating inertias:

(3.171)

The second term can be written for the rotating components including wheels, axles, propeller shaft and gearbox output shaft (it is assumed that the gear is in neutral and clutch is disengaged, so the input shaft and gears inside the gearbox are not rotating). Thus:

(3.172)

in which I_wA is the sum of all rotating inertias with the wheel speed and Ip is the sum of those rotating inertias with the speed of differential input shaft (or gearbox output shaft):

(3.173)

Assuming no slip for the tyres, one can write:

(3.174)

Equation (3.171) thus can be simplified to:

(3.175)

This energy will be dissipated by the work done by resistive forces acting on the vehicle, until the vehicle comes to a complete stop. The work of the resistive forces is:

(3.176)

where s is the distance travelled. For the case of a constant rolling resistance force, between any two points 1 and 2, it can be written:

(3.177)

The first integral results:

(3.178)

in which A₁ is the area under the v-t diagram:

(3.179)

and can be determined from summing up the surface between times t₁ and t₂. The second integral in Equation (3.177) can be changed to a time domain integral:

(3.180)

This is the area under the v³-t diagram and can be obtained by processing the original data. Thus the total work of Equation (3.177) is:

(3.181)

The energy change between any two instants is:

(3.182)

This must be equal to the work done by the resistive forces within this period:

(3.183)

In Equation (3.183) all parameters are known but I_eq, therefore:

(3.184)

For the speed-dependent rolling resistance assumption, too, a similar procedure can be followed. Also it should be noted that the equivalent rotating inertia of Equation (3.173) does not include the rotating masses with engine speed (gearbox input shaft). If a special coast down test is carried out with the gear engaged in an arbitrary gear with only the clutch pedal depressed, then from the test results the full value of rotating masses as seen on the driving wheels can be determined by the method explained in this section.

3.13 Driveline Losses

In the preceding discussions the assumption was made that the driveline system is ideal in such a way that all the power transmitted by the driveline will be delivered to the wheels. In practice, however, every component in the driveline will involve losses due to internal friction. The torque necessary to overcome the friction multiplied by speed of rotation of the component results in the power lost to the friction of that component.

3.13.1 Component Efficiencies

Consider a torque-transmitting component shown in Figure 3.87. The input and output powers are:

(3.185)

Due to the rigid connections between the input and the output (flexibilities are ignored), the output rotational speed is related to the input speed by:

(3.186)

The input energy is:

(3.187)

The efficiency of the torque transfer is defined as the ratio of the output to the input energies:

(3.188)

At constant speed and torque inputs, this simplifies to:

(3.189)

in which is the output torque at the ideal (no power loss) condition. Thus, it is concluded that when there is energy loss in a component, the output torque will be reduced with respect to the ideal output torque by a factor smaller than 1, defined as the efficiency of that component, i.e.:

(3.190)

which also leads to:

(3.191)

The torque loss T_L and power loss P_L for the component are:

(3.192)

(3.193)

Figure 3.88 shows the ratio of torque loss T_L to the output torque T_o with the variation of efficiency. For efficiencies less than 0.5 the output torque is less than the torque loss (see Equation (3.192)).

In the above discussions, a constant speed was considered at the time of a constant torque delivery and the dependency of the power loss on the speed of rotation or the value of torque was not considered. In fact, friction in the moving parts is of two different natures: dry and/or viscous. The former is independent of relative speed between the moving parts and once the movement starts, it remains proportional to the load (kinetic friction). Viscous friction, however, is speed-dependent and increases as the relative speed of moving parts increase.

The dry friction is often assumed to be linearly dependent on the load (F = μN). Examples of this type of friction in a driveline are contact friction between gear teeth and friction in bearings. When larger torques are transferred, the contact forces on the gear teeth increase and lead to larger friction forces. The same also goes for bearing loads. Speed-dependent friction is typically due to churning and pumping losses of the lubricant which increase with rotational speed. These losses are not usually affected by the amount of torque that is transmitted. There is also another type of loss that is almost constant, regardless of rotational speed or transmitted torque – sliding friction in seals typically behave this way.

In general, the efficiency of a component, therefore, must be considered to depend on speed and load, i.e.:

(3.194)

In practice, however, it is a common to assume the efficiency is a constant value. In fact, it is an approximation to the real losses in the drivetrain system. This assumption means that increasing the speed of rotation at a specified torque (power increase) will only increase the power loss in the component (Equation (3.193)) and will not influence the torque loss (Equation (3.192)). Conversely at a specified speed, torque increase will increase both the torque and power losses. As Figure 3.89 summarizes, along a line of a certain torque T_i the torque loss is constant but the power loss increases with speed.

For the vehicle driveline with several components, the wheel torque is:

(3.195)

in which is the overall efficiency and subscripts c, g, j, d, a and w stand for clutch, gearbox, joint (universal or constant velocity), differential, axles and wheels respectively. is the ideal wheel torque:

(3.196)

It should be noted that the clutch efficiency treatment for the transient phase is different from what is considered here, nevertheless the concept will be similar (see Section 4.5.5).

3.13.2 Torque Flow Direction

The driveline efficiency depends on the direction of torque flow. The foregoing discussion applies to the case when the engine torque is transmitted by the driveline to turn the wheels. In the cases when the accelerator pedal is suddenly released during the vehicle motion on a level road or when vehicle is moving downhill, the energy flow is reversed. In the former case, the kinetic energy of vehicle and turning parts is used to overcome the resistive forces and in the latter case the gravitational force is also helping vehicle motion. In both cases, with engaged gears and zero throttle, the engine will run in its braking regions. Thus, in these cases the efficiency equation must be written the other way round, that is:

(3.197)

in which,

(3.198)

The notation of using primes in Equation (3.197) emphasizes that in the reverse direction the efficiency is not necessarily the same as the other direction. The torque loss in driveline this time will be:

(3.199)

3.13.3 Effect of Rolling Resistance

In order to move the vehicle from rest, friction in the driveline must be overcome before the vehicle can start to move. Suppose the vehicle is pushed from rest with the gearbox disengaged until it starts to move. The question is whether or not this amount of force is related to the driveline torque or power loss. In fact, this force is just the rolling resistance force that also includes the non-loaded driveline rolling frictions. Therefore, it should not be confused with the driveline losses during the torque delivery.

The objective of the engine torque is to move the vehicle, but throughout the driveline it overcomes friction, and eventually before turning the wheel, it has to overcome the rolling resistance torque. In other words, the total torque loss in the driveline is the summation of component torque losses T_CL and rolling resistance torque T_RR:

(3.200)

The output torque used for moving the vehicle is:

(3.201)

Since the rolling resistance is already included in the resistive forces against the vehicle motion, substituting Equation (3.200) in 3.201 is not correct and the rolling resistance term must be excluded. Thus the output torque at driving wheels is:

(3.202)

For a given overall efficiency the component torque losses can be evaluated as:

(3.202)

This is what was previously considered as the total torque loss. It should also be noted that the unloaded torque losses of the driveline have already been considered in the rolling resistance, and the efficiency of the wheel defined earlier represents only those torque losses in the tyre in addition to rolling resistance torque (e.g. tyre slip).

3.14 Conclusion

In this chapter, the basics of vehicle longitudinal dynamics studies were discussed and various forms of the vehicle motion were studied. The tractive force generation by the tyres and its dependence on tyre slip were discussed. The natures of resistive forces against vehicle motion were also described and characterized. The equations of vehicle motion in different forms were developed, starting with the simplest forms and gradually adding other terms.

Closed form analytical solutions were presented where such solutions are available. These forms are helpful for quick estimations of the approximate performance. For more detailed studies, numerical integrations are preferred and several examples using MATLAB programs are provided. Several example problems were worked out in order to facilitate the understanding of the physics of vehicle motion in different cases. Students are encouraged to become more familiar with the details by reworking the example problems and writing MATLAB programs. Problems in Section 3.16 will further help students build their knowledge on the subjects.

3.15 Review Questions

3.16 Problems

Problem 3.1

The rolling resistance force is reduced on a slope by a cosine factor (). On the other hand, on a slope, the gravitational force is added to the resistive forces. Assume a constant rolling resistance force and write the parametric forms of the total resistive force for both cases of level and sloping roads. At a given speed v₀:

Result:

Problem 3.2

For the vehicle of Example 3.4.2:

Problem 3.3

In order to have a rough estimation of the performance of a vehicle, it is proposed to ignore the resistive forces to obtain the No-Resistive-Force (NRF) performance.

Results:

Problem 3.4

Use the results of Problem 3.3:

Problem 3.5

At very low speeds the aerodynamic force is small and may be ignored. For example, at speeds below 30 km/h, the aerodynamic force is one order of magnitude smaller than the rolling resistance force. For such cases categorized as Low-Speed (LS), ignore the aerodynamic force and for the CPP assume a constant rolling resistance force F₀.

Results:

Problem 3.6

For the vehicle of Problem 3.5 and using the LS method, find the required power for the 0–100 km/h acceleration to take place in 7 seconds.

Result: 58,849 W.

Hint: The following statements in MATLAB can be used with a proper initial guess for x0.

fun=inline(`7−1000*×*log(×/(×−(100/3.6/200)))+1000*(100/3.6/200)');

×=fsolve(fun, ×0, optimset(`Display',`off')); (×= P/F0^2)

Problem 3.7

For the vehicle of Problem 3.5 using the LS method determine the power requirements for a performance starting from rest to reach speed v at time t, for three cases of v = 80, 90 and 100 km/h for accelerating times varying from 6 to 10 seconds. Plot the results in a single figure.

Problem 3.8

The power evaluation for the NRF case (Problem 3.3) is a simple closed-form solution but it is not accurate. The LS method (Problems 3.5–3.7) produces more accurate results especially in the low speed ranges. By generating plots similar to those of Problem 3.7 show that an approximate equation of P = P_NRF + 0.75F₀v can generate results very close to those of LS method.

Problem 3.9

For the LS case, use that relates the speed to acceleration and distance, substitute for acceleration in terms of speed.

Results:

Problem 3.10

A vehicle of 1200 kg mass starts to accelerate from the rest at origin. If power is constant at 60 kW, for a LS model with F₀ = 200 N, determine the travel time and distance when speed is 100 km/h. Compare your results with those of the NRF model.

Results:

Problem 3.11

In Problem 3.8 a close approximation was used for the power estimation of LS method. For the general case including the aerodynamic force, the approximation given by P = P_NRF + 0.5F_Rv is found to work well.

For the vehicle of Example 3.5.3, plot the variations of power versus acceleration times similar to those of Problem 3.8 and compare the exact solutions with those obtained from the proposed method.

Problem 3.12

According to the solutions obtained for CTP (see Section 3.6) it turned out that at each gear, the acceleration is constant to a good degree of approximation (see Figure 3.50). Thus a simpler solution can be obtained by considering an effective resistive force for each gear that reduces the problem to a Constant Acceleration Approximation (CAA). In each gear assume the resistive force acting on the vehicle is the average of that force at both ends of the constant torque range. Write the expressions for the average speed at each gear v_av, the average resistive force R_av.

Problem 3.13

A 5th overdrive gear with overall ratio of 3.15 is considered for the vehicle in Problem 3.12, and the torque is extended to 3400 rpm. Obtain the time variations of acceleration, velocity and travel distance for the vehicle by both CAA and numerical methods and plot the results.

Problem 3.14

In Example 3.5.2 impose a limit for the traction force of F_T < 0.5 W and compare the results.

Problem 3.15

For a vehicle with transmission and engine information given in Example 3.7.2, include a 1-second torque interruption for each shift and plot similar results. To this end, include a subprogram with listing given below at the end of loop for each gear:

% Inner loop for shifting delay:

if i<5 % No delay after gear 5!

t0=max(t);

tf=t0+tdelay;

×0=[v(end) s(end)];

p=[0 0 0]; % No traction force

[t,×]=ode45(@Fixed_thrt, [t0 tf], ×0);

v=×(:,1);

s=×(:,2);

end

%Now plot the results

p=[p1 p2 p3 p4]; % Set back the engine torque

Problem 3.16

Repeat Problem 3.15 with a different shifting delay for each gear of the form 1.5, 1.25, 1.0 and 0.75 seconds for 1–2, 2–3, 3–4 and 4–5 shifts respectively. (Hint: For this you will need to change the program.)

Problem 3.17

Repeat Example 3.7.2 for a different shifting rpm.

Problem 3.18

In Example 3.7.3, investigate the possibility of having a dynamic balance point in gear 4. If no steady state point is available, find a new gear ratio to achieve a steady state.

Problem 3.19

Repeat Example 3.7.2 with transmission ratios 3.25, 1.772, 1.194, 0.926 and 0.711.

Problem 3.20

In the program listing given for Example 3.7.2, no constraint is imposed for the lower limit of engine speed and at low vehicle speeds the engine rpm will attain values less than its working range of 1000 rpm.

Problem 3.21

In a vehicle roll-out test on a level road, the variation of forward speed with time is found to be of the form:

where a, b and d are three constants.

Result:

Problem 3.22

Two specific tests have been carried out on a vehicle with 1300 kg weight to determine the resistive forces. In the first test on a level road and still air, the vehicle reaches a maximum speed of 195 km/h in gear 5. In the second test on a road with slope of 10%, the vehicle attains a maximum speed of 115 km/h in gear 4. In both tests the engine is working at WOT at 5000 rpm, where the torque is 120 Nm.

Results:

Problem 3.23

For a vehicle with specifications given in Table P3.23, engine torque at WOT is of the following form:

Table P3.23 Vehicle information.

Vehicle mass		1200 kg
Rolling Resistance Coefficient		0.02
Tyre Rolling Radius		0.35 m
Final drive Ratio		3.5
Transmission Gear Ratio 1		4.00
Gear Ratio 2		2.63
Gear Ratio 3		1.73
Gear Ratio 4		1.14
Gear Ratio 5		0.75
Aerodynamic Coefficient	CD	0.4
Frontal Area	Af	2.0 m2
Air density	ρA	1.2 kg/m3

The driveline efficiency is approximated by 0.85+i/100 in which i is the gear number.

Results:

Problem 3.24

For the vehicle of Problem 3.23, find the following:

Hint: Table P3.24 is useful for solving this problem.

Table P3.24

Problem 3.25

The vehicle of Problem 3.23 is moving on a level road at the presence of wind with velocity of 40 km/h. Assume C_D = C_D0 + 0.1|sin α|, in which α is the wind direction relative to the vehicle direction of travel. Determine the maximum vehicle speed in gear 4 for:

Results:

Problem 3.26

Two similar vehicles with exactly equal properties are travelling on a level road but in opposite directions. Their limit speeds are measured as v₁ and v₂ respectively. Engine torque at WOT is approximated by following equation:

Determine the aerodynamic drag coefficient C_D and wind speed in direction of travel v_w:

Table P3.26

Transmission Gear Ratio	0.9
v1	180
v2	200

Results:

Problem 3.27

While driving uphill in gear 4 on a road with constant slope , the vehicle of Problem 3.23 reaches its limit speed at an engine speed of at a still air. The same vehicle is then driven downhill on the same road in gear 5, while keeping the engine speed same as before. Engine powers for uphill and downhill driving are and respectively. The tyre slip is roughly estimated from equation S_x = S₀ + P × 10⁻² (%) where S₀ is a constant and P is power in hp.

Assume a small slope angle and use the additional data given in Table P3.27 to determine:

Table P3.27

Results:

Problem 3.28

For the vehicle of Problem 3.23, find the following:

Results:

Problem 3.29

In Section 3.9 the effect of rotating masses were discussed and equations for including this effect in the acceleration performance of a vehicle were developed. From an energy consumption point of view, when the vehicle is accelerated to the speed of v, the rotating inertias will be at rotational speeds related to v (ignore the tyre slip).

Problem 3.30

For a tyre with the Magic Formula information given in Table 3.3:

Problem 3.31

The vehicle of Problem 3.23 is moving with a constant speed of 100 km/h. Use the tyre data of Table 3.3 for each of the two driving wheels and for a front/rear weight distribution of 60/40, determine:

Further Reading

The analysis of the performance of vehicles in the longitudinal direction in terms of accelerations and speeds has been dealt with in many automotive texts. This is not surprising since in practice it has always been one of the most highly debated areas in comparing vehicles – indeed, technical journals and magazines have seemed obsessed with 0 to 60 mile/h times and top speeds even if these aspects of performance cannot actually be used safely on the roads.

A good introduction to the subject which takes a slightly different approach from this book is provided by Gillespie [4]. In Chapter 2, he derives the basic equations and discusses power and traction limited acceleration performance. In Chapter 4, he goes on to discuss in more detail the road loads – both rolling resistance and aerodynamic – which affect longitudinal performance. Gillespie's book, written in the USA in 1992, provides good clear explanations of the fundamentals of vehicle performance – and the only problem for current students is the use of non-SI units which makes the numerical examples somewhat difficult to follow.

Lucas's book, Road Vehicle Performance[5], arose from lecture courses at Loughborough University of Technology. It is interesting because it gives a comprehensive analysis of vehicle performance, and links it to the practical issues of measurements both on the road and on laboratory-based dynamometers. Although it is now rather dated since it was written in 1986, the principles are of course all still relevant and he discusses some useful information on manual vs automatic gearboxes and fuel consumption calculations. It is targeted at undergraduate and graduate students but again has the drawback of using non-SI units.

The text by Guzzella and Sciaretta on vehicle propulsion systems [6] also provides an excellent introduction to vehicle performance in Chapters 2 and 3. It is a modern book, updated in 2007, and based on lectures at the Swiss Federal Institute of Technology (ETH), Zurich. It is useful as a companion to this book because the authors use examples in MATLAB and Simulink, and in particular describe some interesting case studies in the Appendix. They also introduce the QSS toolbox package which was developed at ETH, and which may be downloaded from http://www.idsc.ethz.ch/Downloads/qss.

References

[1] SAE(1978)Vehicle Dynamics Terminology. SAE J670e, Society of Automotive Engineers.

[2] Bakker, E., Nyborg, L. and Pacejka, H.B. (1987) Tyre Modelling for Use in Vehicle Dynamics Studies, SAE Paper 870421.

[3] Wong, J.Y. (2001)Theory of Ground Vehicles,3rd edn.John Wiley & Sons, Inc., ISBN 0-470-17038-7.

[4] Gillespie, T.D. (1992)Fundamentals of Vehicle Dynamics.SAE, ISBN 1-56091-199-9.

[5] Lucas, G.G. (1986)Road Vehicle Performance.Gordon and Breach, ISBN 0-677-21400-6.

[6] Guzzella, L. and Sciarretta, A. (2005)Vehicle Propulsion Systems: Introduction to Modelling and Optimisation.Springer, ISBN 978-3-549-25195.