2. REQUIREMENT SPECIFICATION
2.2 Concept phase
2.2.1 Vehicle model
39
In this section we focus on the vehicle model which can allow us to test the control algorithm.
Simulation of the vehicle behaviour is an important issue in today automotive engineering because of the high complexity of the environment. In an electric powertrain new risks can arise especially if we have more than one motor which provide the tractive effort. Hence, engineers need to simulate what happens in every driving situation to avoid problems in the future developments, before starting any production process.
The simulation of the powertrain control concerns many aspects touching the domain of the vehicle dynamic engineering. One of these aspects concerns the torque distribution system which is responsible along with other systems of the vehicle’s driveability and the safety. Since we do not deal with the vehicle lateral motion as mentioned, we will consider only the vehicle longitudinal model without slip. Following there is a summary of the mathematical formulas to calculate the velocity of the vehicle starting from the delivered torques to the drive wheels.
The first step is to produce an equation for the required ‘tractive effort’ which is the force propelling the vehicle forward, transmitted to the ground through the drive wheels.
Figure 21 – Pure rolling physical model. In blue the drive shaft (motor).
To move the vehicle, we have to provide the sum of the following quantities [30]:
= + - + , + , + !
where:
1. is rolling resistance force (or rolling friction): is the force resisting the motion when the tire rolls on the ground. This is a result of non-linearities of the tire because of the mechanical hysteresis of a not rigid body. This complex behaviour is usually described very
40
simply considering this force proportional to the vehicle weight by a constant coefficient so: = . ∙ 01ℎ34516788
2. - is the air resistance, acting opposite to the relative motion of any object which moves through a fluid. Unlike the resistance rolling it is closely related to the vehicle speed because it depends on its square as well as by other factors concerning the geometry of the vehicle and the fluid density. Since we do not know the geometry of vehicle we will use a global air drag coefficient:
- = . - ∙ 901ℎ3451:;11<=
3. , is the force needed to accelerate the vehicle, in addition to the other resistance forces; according to the famous Newton’s Law:
, = 01ℎ34516788 ∙ 7
where ‘a’ is the desired acceleration. (This is true only if consider a pure rolling, neglecting slip).
4. , is the force that the motor has to deliver to accelerate only the rotor mass.
Since a gear reductor can be used in the torque’s transmission, gear ratio has to be considered.
So, if ‘GR’ is the gear ratio of the system connecting the motor to the axle defined as GR =@@ABCCD
EFGFH and ‘T’ the motor torque, the force transmitted by wheel to the ground is:
= KJIJ
where ‘WR’ is the effective wheel radius. Thereby motor angular speed is giving by L = KJIJ , where v is the tangential wheel’s speed corresponding in our case to the vehicle speed (because of pure rolling assumption). Similarly, the motor angular acceleration is the derivative of the previous formula, so we can say: LM = KJIJ 7
Therefore, the required torque for this angular acceleration is given by:
= N × LM = N × KJIJ 7
where ‘J’ is moment of inertia of the rotational parts. Now we can say that the minimum effort needed to accelerate the rotor is:
= KJIJ =KJIJ N × KJIJ 7 = N OKJIJP 7
If we want also to consider the global gear efficiency ‘Q ’ we can incorporate it in the equation in this way:
, = N R
S OKJIJP
41
5. ! stands for the force required to drive a vehicle up a slope. Simply, it corresponds to the component of weight along the inclined plane:
! = 01ℎ34516788 ∙ T ∙ sin 9X)
Where ‘g’ is the gravitational acceleration and ‘theta’ the angle of the inclined plane.
Now, we are able to calculate the acceleration of the vehicle for a given motor torque summing every single contribute to the total tractive effort by the following basic formula:
KJIJ 9. T 83YX = ∙ 06 . - Z06 N R
S KJIJ[[ \ M
How we can see, the contribution of rotor’s inertia can be interpreted as a further increase of total vehicle mass and thus it can be included in it. That is a first-order differential equation in the variable ′ ′. As usual, to solve this differential equation, it is brought in the Laplace’s domain. The entry variable is the algebraic sum of the following two contributions: the overall delivered wheel’s torque by the motors and the total torque provided by foundation braking system.
Figure 22 – Block diagram of the physical layer.
Aside from the analysed resistance forces, we also must consider the contribution of the friction braking to complete the model. Theoretically it does no matter where we put it; however, we want to think in terms of braking torque, so we will add this contribute to the delivered motor torque at level of wheels.
Then, the resulting torque can be converted in the tractive force, dividing it by the wheel radius. This force is transmitted by the drive wheels to the road. If there are more than one motor, the model’s input shall be the sum of all delivered motor torques while the value of friction torque represents the total braking torque provided by foundation brakes.
Hence, from this ‘tractive effort’ we subtract all the resistance forces in order to calculate the vehicle acceleration and thus its velocity. Referring to the figure above, we note that, among these resistance
42
forces, another one can be added in the model. This is because sometimes we might use an equivalent model of all frictions by the identification of three coefficients called ^_, ^, ^ resulting from the well-known experiment ‘coast down’.
Furthermore, we should add to the model a block which decides the sign of the resistance force, if we want to consider both the directions of speed, forward and reverse without having to change the model each time.