## 72 Tractive Effort

7.2.1 Introduction

The first step in vehicle performance modelling is to produce an equation for the tractive effort. This is the force propelling the vehicle forward, transmitted to the ground through the drive wheels.

Consider a vehicle of mass m, proceeding at a velocity u, upa slope of angle f, as in Figure 7.1. The force propelling the vehicle forward, the tractive effort, has to accomplish the following:

• overcome the rolling resistance;

• overcome the aerodynamic drag;

• provide the force needed to overcome the component of the vehicle's weight acting down the slope;

• accelerate the vehicle, if the velocity is not constant.

We will consider each of these in turn.

### 7.2.2 Rolling resistance force

The rolling resistance is primarily due to the friction of the vehicle tyre on the road. Friction in bearings and the gearing system also play their part. The rolling resistance is approximately constant, and hardly depends on vehicle speed. It is proportional to vehicle weight. The equation is:

where firr is the coefficient of rolling resistance. The main factors controlling firr are the type of tyre and the tyre pressure. Any cyclist will know this very well; the free-wheeling performance of a bicycle becomes much better if the tyres are pumped up to a high pressure, though the ride may be less comfortable.

The value of ¡irr can reasonably readily be found by pulling a vehicle at a steady very low speed, and measuring the force required.

Typical values of firr are 0.015 for a radial ply tyre, down to about 0.005 for tyres developed especially for electric vehicles.

### 7.2.3 Aerodynamic drag

This part of the force is due to the friction of the vehicle body moving through the air. It is a function of the frontal area, shape, protrusions such as side mirrors, ducts and air passages, spoilers, and many other factors. The formula for this component is:

where p is the density of the air, A is the frontal area, and v is the velocity. Cd is a constant called the drag coefficient.

The drag coefficient Cd can be reduced by good vehicle design. A typical value for a saloon car is 0.3, but some electric vehicle designs have achieved values as low as 0.19. There is greater opportunity for reducing Cd in electric vehicle design because there is more flexibility in the location of the major components, and there is less need for cooling air ducting and under-vehicle pipework. However, some vehicles, such as motorcycles and buses will inevitably have much larger values, and Cd figures of around 0.7 are more typical in such cases.

The density of air does of course vary with temperature, altitude and humidity. However a value of 1.25 kg.m-3 is a reasonable value to use in most cases. Provided that SI units are used (m2 for A, m.s-1 for v) then the value of Fad will be given in Newtons.

### 7.2.4 Hill climbing force

The force needed to drive the vehicle up a slope is the most straightforward to find. It is simply the component of the vehicle weight that acts along the slope. By simple resolution of forces we see that:

7.2.5 Acceleration force

If the velocity of the vehicle is changing, then clearly a force will need to be applied in addition to the forces shown in Figure 7.1. This force will provide the linear acceleration of the vehicle, and is given by the well-known equation derived from Newton's second law,

However, for a more accurate picture of the force needed to accelerate the vehicle we should also consider the force needed to make the rotating parts turn faster. In other words, we need to consider rotational acceleration as well as linear acceleration. The main issue here is the electric motor, not necessarily because of its particularly high moment of inertia, but because of its higher angular speeds.

Figure 7.2 A simple arrangement for connecting a motor to a drive wheel motor torque = T

motor torque = T

gear ratio = G

tractive effort = Fte gear ratio = G

tractive effort = Fte

Figure 7.2 A simple arrangement for connecting a motor to a drive wheel

Referring to Figure 7.2, clearly the axle torque = Fter, where r is the radius of the tyre, and Fte is the tractive effort delivered by the powertrain. If G is the gear ratio of the system connecting the motor to the axle, and T is the motor torque, then we can say that:

We will use this equation again when we develop final equations for vehicle performance. We should also note that:

Similarly, motor angular acceleration a 2

The torque required for this angular acceleration is:

T = IG-r where I is the moment of inertia of the rotor of the motor. The force at the wheels needed to provide the angular acceleration (Fma) is found by combining this equation with equation (7.5), giving:

We must note that in these simple equations we have assumed that the gear system is 100% efficient, it causes no losses. Since the system will usually be very simple, the efficiency is often very high. However, it will never be 100%, and so we should refine the equation by incorporating the gear system efficiency ng. The force required will be slightly larger, so equation (7.7) can be refined to:

ngr 2

Typical values for the constants here are 40 for G/r and 0.025 kg.m2 for the moment of inertia. These are for a 30 kW motor, driving a car which reaches 60 kph at a motor speed of 7000rpm. Such a car would probably weigh about 800 kg. The IG2¡r2 term in equation (7.8) will have a value of about 40 kg in this case. In other words the angular acceleration force given by equation (7.8) will typically be much smaller than the linear acceleration force given by equation (7.4). In this specific (but reasonably typical) case, it will be smaller by the ratio:

It will quite often turn out that the moment of inertia of the motor I will not be known. In such cases a reasonable approximation is to simply increase the mass by 5% in equation (7.4), and to ignore the Fma term.

7.2.6 Total tractive effort

The total tractive effort is the sum of all these forces:

where:

• Frr is the rolling resistance force, given by equation (7.1);

• Fad is the aerodynamic drag, given by equation (7.2);

• Fhc is the hill climbing force, given by equation (7.3);

• Fla is the force required to give linear acceleration given by equation (7.4);

• Fma is the force required to give angular acceleration to the rotating motor, given by equation (7.8).

We should note that Fla and Fma will be negative if the vehicle is slowing down, and that Fhc will be negative if it is going downhill.

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