speed on level ground without head-wind. For vehicles with automatic transmissions or CVTs the certain time set
can allow automatic gear-shift (or ratio-change) or not. The reserve can also be measured in propulsion force.
In general terms, the lowest consumption is found in high gears. However, the vehicle will then tend to have a very small reserve in acceleration. It will, in practice, make the vehicle less comfortable and less safe to drive in real traffic, because one will have to change to a lower gear to achieve a certain higher acceleration. The gear shift gives a time delay.
Figure 3-20 shows one way of defining a momentary acceleration reserve. The reserve becomes gener- ally larger the lower gear one selects. A characteristic of electric propulsion systems is that an electric motor can be run at higher torque for a short time than stationary, see Figure 2-62. On the other hand, the stationary acceleration reserve is less gear dependent, since an electric motor can work at certain power levels in large portions of its operating range.
One can calculate the acceleration reserve at each time instant over a driving cycle. However, integra- tion of acceleration reserve, as we did with fuel, emissions and wear, makes less sense. Instead, a mean value of acceleration reserve tells something about the vehicleβs driveability. Minimum or maximum values can also be useful measures.
Acceleration reserve was above described as limited by gear shift strategy. Other factors can be limit- ing, such as energy buffer state of charge for parallel hybrid vehicles or how much overload an electric machine can take short term, see right part of Figure 3-20.
Lon gi tud in al for ce, ππ Speed, π 0 0
gear 1
gear 2
0 0gear 1
Vehicle with conventionalpropulsion system Vehicle with electric propulsion system
MassβAcceleration Reserve on gear 1
gear 2
Short term Acceleration Reserve MassβAcceleration Reserve on gear 2 MassβAcceleration Reserve on gear 1 MassβAcceleration Reserve on gear 2 Lon gi tud in al for ce, ππ Speed, π3.3.5 Load Transfer with Rigid Suspension
Longitudinal load transfer redistributes vertical force from one axle to the other. The off-loaded axle can limit the traction and braking. This is because the propulsion and brake systems are normally de- signed such that axle torques cannot always be ideally distributed.
For functions over longer events it is often reasonable to consider the suspension as rigid. We start with the free-body diagram in Figure 3-21, which includes acceleration, .
x
z
y
π¦
Figure 3-21: Free Body Diagram for accelerating vehicle. Rolling resistance in and .
Note that the free-body diagram and the following derivation is very similar to the derivation of Equa- tion [3.5], but we now include the fictive force β .
Moment equilibrium, around rear contact with ground:
β β + β β (π β ππ ( π¦) + β π π( π¦)) β ππ β ππ β β β = 0; β
β = β ( β
π β ππ ( π¦) + β π π( π¦)
β β ) β ππ β ππ ;
Moment equilibrium, around front contact with ground:
+ β β β β (π β ππ ( π¦) β β π π( π¦)) β ππ β ππ β β β = 0; β β = β ( β π β ππ ( π¦) β β π π( π¦) + β ) + ππ β ππ ; [3.13]
These equations confirm what we know from experience, the front axle is off-loaded under accelera- tion with the load shifting to the rear axle. The opposite occurs under braking.
The load shift has an effect on the tyreβs grip. If one considers the combined slip conditions of the tyre (presented in Chapter 2), a locked braking wheel limits the amount of lateral tyre forces. The same is true for a spinning wheel. This is an important problem for braking as the rear wheels become off- loaded. This can cause locking of the rear wheels if the brake pressures are not adjusted appropriately. See more in 3.4.4.
3.3.5.1 Varying Road Pitch
The model in 3.4.4..3.4.5.2 assumes flat but not level road, i.e. π¦ is constant. An example where π¦ varies is when passing a crest or a sag, see Figure 3-22. If negotiating a curve at the same time as a
crest, a vehicle can lose vertical force under tyres so that lateral grip is affected. Moment equilibrium, around rear and front wheel contact with ground gives:
= β (( + β³ 2) β π β cos( ) + h β sin( ) β β L) β ππ β ππ L ; = β (( + β³ 2) β π β cos( ) β h β sin( ) + β ) + ππ β ππ ; [3.14]
Assuming that we have the road as (π ), then = β arctan (
) β β ; and
β³= 2 2;.
Note that this model is assuming that vertical variations of road are larger than wheel base and track width and same on left and right side of the road/vehicle. Else the variation would be called road une- venness, which will be more treated in Chapter 5.
If models with body vertical and pitch motion and suspension springs, such as in 3.4.5 and 3.4.5.2 it is often suitable to express the vertical fictive force, Μ with π¦ instead of β³ 2. The fictive
force downwards will then be β 2 π
π¦= β π¦ instead. This can be understood from basic
geometry, β³ β βπ
π¦, where π π¦ is the road pitch curvature [1 πππ π‘ β ], see Figure 3-22.
crest
sag
Μ
π¦
β’The variable s is the distance along the road. β’Road gradient versus inertial coordinate
system, π¦β β β², is a function of s
β’where β²=
; is a function of s.
β’Vertical acceleration in inertial coordinate system, Μ β 2;, β’where = 22; is a function of s. s s π¦= β β² π¦= β β² s π¦
Figure 3-22: Free Body Diagram for driving over non-flat vertical road profile.
3.3.6 Acceleration
Acceleration performance like, typically, 0-100 km/h over 5..10 s, will be addressed in this section. These accelerations are relatively steady state (vehicle pitch and heave are relatively constant), so the suspension compliance is not considered.
Accelerations will also be covered in 3.4, as being shorter events. The vehicle pitch and heave vary more and, consequently, the suspension compliance becomes important to model. This modelling is also more suited for braking, which typically involve suspension more than propulsion.