3.1 Introduction
The primary purpose of a vehicle is transportation, which requires longitudinal dynamics. The chapter is organised with one group of functions in each section as follows:
• 3.2 Steady State Function • 3.3 Functions Over (Long) • 3.4 Functions in (Short) Events • 3.5 Control Functions
3.1.1 References for This Chapter
• 2.4 Propulsion System and “Chapter 23. Driveline” in Ref (Ploechl, 2013)
• 2.5 (Wheel) Braking System “Chapter 24. Brake System Dynamics” in Ref (Ploechl, 2013) • “Chapter 27 Basics of Longitudinal and Lateral Vehicle Dynamics” in Ref (Ploechl, 2013) • “Chapter 6: Adaptive Cruise Control” in Ref (Rajamani, 2012)
3.2 Steady State Functions
Functions as top speed and grade-ability are relevant without defining a certain time period. For such functions, it is suitable to observe the vehicle in steady state, i.e. independent of time. Those functions are therefore called steady state functions, in this compendium. The main subsystems that influences here are the propulsion system, see 2.4, and the (Friction) Brake system, see 2.5.
3.2.1 Traction Diagram
The force generated in the prime mover is transmitted through a mechanical transmission to the wheel which then generates the propulsive forces in the contact patch between tyre and road. In an electric in-wheel motor, the transmission can be as simple as a single-step gear. In a conventional vehi- cle, it is a stepped transmission with several gear ratios (i.e. a gearbox). Then, the drivetrain can be drawn as in Figure 3-1. The torque and rotational speed of the engine is transformed into force and velocity curves via the mechanical drivetrain and driven wheel. The result is a Traction diagram. The transformation follows the following formula, if losses are neglected:
= 𝑟 𝑡 𝑜 ∙ 𝑇 𝑊 𝑒𝑒𝑙 𝑢𝑠; 𝑛 = 𝑊 𝑒𝑒𝑙 𝑢𝑠 ∙𝑟 𝑡 𝑜; [3.1] 𝝎 𝑻 Traction Diagram 𝝎 𝒓𝒂𝒕𝒊𝒐⁄ 𝒓𝒂𝒕𝒊𝒐 𝑻 𝝎 𝑻 𝝎 𝒓𝒂𝒕𝒊𝒐⁄ 𝒓𝒂𝒕𝒊𝒐 𝑻 𝒗 =𝒓𝒂𝒅𝒊𝒖𝒔 𝒓𝒂𝒕𝒊𝒐 𝝎 𝑭 = 𝒓𝒂𝒕𝒊𝒐 𝒓𝒂𝒅𝒊𝒖𝒔 𝑻 𝒗 𝑭 prime mover transmission driven wheel (vehicle) body 𝒗 𝑭 Multiply torque by 𝒓𝒂𝒕𝒊𝒐 𝒓𝒂𝒅𝒊𝒖𝒔
Multiply rotational speed by 𝒓𝒂𝒅𝒊𝒖𝒔 𝒓𝒂𝒕𝒊𝒐 𝝎
𝑻
A traction diagram for a truck is given in Figure 3-2, which also shows that there will be one curve for each gear.
lowest gear
highest gear
Figure 3-2: Example of engine map and corresponding traction diagram map from a truck. (D13C540 is an I6 diesel engine of 12.8-litre and 540 hp for heavy trucks.)
Losses in transmission can be included by loss models for transmission, such as: = 𝜂𝑇∙ 𝑟 𝑡 𝑜 ∙ 𝑇 𝑟 𝑢𝑠; 𝑒𝑟𝑒 𝜂𝑇 ≤ 1; = 𝜂𝜔∙ 𝑟 𝑢𝑠 ∙𝑟 𝑡 𝑜; 𝑒𝑟𝑒 𝜂𝜔≤ 1; 𝑒𝑟𝑒 𝜂𝑇∙ 𝜂𝜔= 𝜂𝑡 𝑡𝑎𝑙 = 𝑣𝑒ℎ𝑖 𝑙𝑒 𝑒𝑛𝑔𝑖𝑛𝑒 = ∙ 𝑇 ∙ ≤ 1; [3.2]
This will move the curves in the first quadrant downwards due to ηT< 1 and to the left due to 𝜂𝜔 < 1.
Tyre rolling friction is a torque loss mechanism, which on its own yields 𝜂𝜔 = 1 and ηT< 1. Tyre lon-
gitudinal slip is a speed loss mechanism, which on its own yields 𝜂𝜔< 1 and ηT= 1. See 2.2.1.6. The
multiplication with 𝜂 is only demonstrative and should be seen more generic: in many cases the losses are additional instead of multiplicative, e.g. for rolling resistance: = 𝑟 𝑡 𝑜 (𝑇 − 𝛥𝑇) 𝑟 𝑢𝑠⁄ = 𝑟 𝑡 𝑜 (𝑇 − 𝑓 ) 𝑟 𝑢𝑠⁄ ;. Which wheels’ to use is discussed in 3.2.2.
A traction diagram is a kind of “one degree of freedom graphical model”. The traction diagram is on complete vehicle level, so the force axis represents the sum of forces from all wheels. This can include more than one propulsion system and also brakes.
3.2.2 Power and Energy Losses
There are power losses 𝑙 (and energy losses 𝑙 = ∫ 𝑙 𝑡) which causes an energy consump-
tion 𝑛 for a transport operation. If the operation starts and stops at same speed (often zero) and
same altitude, 𝑛 = 𝑙 . And 𝑙 = ∑ 𝑙 𝑖 = ∑(∫ 𝑙 𝑡) = ∫(∑ 𝑙 𝑖) 𝑡 = ∫ 𝑙 𝑡
where denotes different losses. One can count the energy consumption per distance 𝐷 =
𝑛 ⁄ = (∫ 𝑥 𝑙 𝑡) ∫ ⁄ 𝑡 [𝐽 ⁄ = 𝑁 ⁄ = 𝑁]. The 𝐷 can be seen as a time-averaged re-
sistance force, summed over all “parts”, ∀ , where there are losses. If the operation has same character (hilliness, speed, etc) lover a long distance, ∑ 𝐷𝑖 → 𝐷 when 𝑥 → and 𝑡 → . See also 3.3.4.1.
3.2.2.1 Driving Resistance Force
Some losses can be identified as true forces, visible directly in a free body diagram, i.e. we don’t need to go via a “Energy loss per distance, 𝐷”. Such are forces from gravity due to road grade and aerody- namic resistance.
Another way to approach this is to study Figure 3-2 and extrapolate that a very low transmission ratio, i.e., a very high gear, we would enable infinite speed, which of course is not realistic. This is because
“driving resistance force” is missing in Figure 3-2. Such force can be added to traction diagram as a curve that typically increases with speed. The top speed is found as the intersection between propul- sion curve and driving resistance force curve, see Figure 3-3.
One part of the driving resistance force comes from driving uphill: the grade or gravitational load on the vehicle. This is negative when driving down-hill. There is also aerodynamic driving resistance force, see Eq [2.63]. Grade and aero-dynamic resistance are (vehicle) body forces.
Also rolling resistance is, often, counted as a resistance force. However, it is not a body force, but in- stead it acts as a torque on the wheel, not a longitudinal force, on each wheel : 𝑖 = 𝑇𝑖− 𝑓 𝑖 ;. But
for a non-driven wheel, the rotational equilibrium of the wheel leads to a small negative longitudinal force in tyre-to-road contact. So, rolling resistance can be included in traction diagram in either the “supply” or resistance curve. If one want to use the traction diagram to show the friction limit where a wheel start to spin, 𝑖 > 𝑖 as in 3.2.6, it can be suitable to include rolling resistance in supply
curve for that wheel, but in the resistance curve for the other wheels. In the following, rolling re- sistance is represented as a torque on driven wheels (which lowers the supply curve) and a longitudi- nal force on non-driven wheels (which lifts the resistance curve). The traction diagram can then also host a curve for friction limit for spinning driven wheels, as in Figure 3-8.
Available for acceleration: 𝑣− 𝑒 Supplied (from propulsion system):
𝑣= 𝑇 − 𝑓 𝑣 ;
Note: The rolling resistance on driven axle is
included in the transformation from the prime mover to 𝑣. But the rolling resistance on non-driven axle leads to a force, 𝑙𝑙 𝑛 𝑛 𝑣, in
the diagram.
− 𝑦
(downhill positive, after ISO8855)
road grade resistance: 𝑔 𝑎 𝑒=
sin − 𝑦 ; rolling resistance, drv: 𝑙𝑙 𝑛 𝑛 𝑣= 𝑓 𝑛 𝑛 𝑣; Driving resistance: 𝑒 = = 𝑔 𝑎 𝑒+ 𝑙𝑙 𝑛 𝑛 𝑣+ 𝑎𝑖 ; topspeed air resistance: 𝑎𝑖 = 0.5 𝜌 𝐴 2;
Traction diagram
Figure 3-3: Traction diagram. Head wind speed, 𝑖𝑛 , is assumed to be zero. See also Figure 3-8.
𝑒 = 𝑙𝑙+ ∙ ∙ sin(− 𝑦) + 1 2∙ ∙ 𝜌 ∙ 𝐴 ∙ ( − 𝑖𝑛 ) 2 ; 𝑙𝑙= ∑ 𝑓 ℎ𝑒𝑒𝑙 non-dri en wheels ; 𝑓 𝑙𝑙 𝑒𝑒𝑙𝑠 𝑟𝑒 𝑢𝑛 𝑟 𝑒𝑛: 𝑙𝑙 = 𝑓 ∙ ∙ ∙ cos( 𝑦) ; [3.3]
As seen in Figure 3-3, the supply and resistance curves are drawn in same diagram. The resulting in- tersection identifies the top speed of the vehicle. This is a stable point in the diagram, so the vehicle condition at top speed is for steady state (no acceleration).
Figure 3-3 also shows that the acceleration can be identified as a vertical measure in the traction dia- gram, divided by the mass. However, one should be careful when using the traction diagram for more than steady state driving. We will come back to acceleration performance later, after introducing the two effects “Load transfer” and “Rotating inertia effect”.
3.2.2.2 Losses due to Longitudinal Tyre Slip
Consider a vehicle with 𝑁 ≥ 3 wheels. Assume a certain vehicle speed and a certain desired propul- sion force = ∑𝑖= ..𝑁(𝑇𝑖⁄ ). Also assume that wheel torques 𝑇𝑖 can be distributed according to 𝑁 −
1 equations, e.g. 𝑇 = 𝑇2; 𝑇 = 𝑇4= 0; for a conventional front axle driven 4 wheeled car. Also assume
same longitudinal stiffness coefficient 𝐶𝐶 on all wheels and that out-of-road-plane equilibria and sus- pension equations defines the vertical forces . . . The, the power loss due to longitudinal tyre slip
𝑙 = ∑ (𝑇𝑖 ( 𝑖− )) 𝑖= :𝑁 = ∑ (𝑇𝑖 𝑠𝑖 | 𝑖|) 𝑖= :𝑁 = ∑ ( 𝑖 𝑖 𝐶𝑖 | 𝑖|) 𝑖= :𝑁 = = 1 ∑ ( 𝑖 2 𝐶𝐶𝑖 𝑖 | 𝑖|) 𝑖= :𝑁 ≈ 𝐶𝐶𝑖 ∑ 𝑖 2 𝑖 𝑖= :𝑁 ;
An example with a fore-aft-symmetric 2-axle vehicle, propelled on one axle gives:
𝑙 ≈ 𝐶𝐶 𝑖 ( 0 2 2⁄ + 2 2⁄ ) = 2 2 𝐶𝐶𝑖 ;
The same vehicle, but propelled equally much on both axles gives:
𝑙 ≈ 𝐶𝐶 𝑖 (( ⁄ )2 2 2⁄ + ( ⁄ )2 2 2⁄ ) = 2 𝐶𝐶𝑖 ;
So, twice as much energy is lost due to longitudinal tyre slip if propelling on 1 instead of 2 axles. When negative wheel torque, one can brake with friction brakes and then there is no energy loss. How- ever, if braking with electric propulsion, the loss can be negative, meaning that energy is regenerated to electric energy storage. If braking so much with electric propulsion that wheel rotates rearwards, there would again be an energy loss, 𝑙 = 𝑇𝑖 𝑖 = 𝑛𝑒 𝑡 𝑒 𝑛𝑒 𝑡 𝑒 > 0.
3.2.2.3 Losses due to Lateral Tyre Slip
(This section might require some studying of Chapter 4 for full understanding.)
There are more driving-resistance effects than covered in Equation [3.3]. One example is that none- Ackermann steering geometry (toe or parallel steering on an axle, or two non-steered axles).
Another effect, which appears also for Ackermann steering geometry, is that power is lost due to lat- eral axle slip. Now, we use the same simple model as in Figure 4-18, but additionally use 𝑦= ⁄ ; and define power losses 𝑙 as sliding velocity counterdirected to force:
𝑙 = − 𝑦 𝑦− 𝑦 𝑦 = − 𝑦 𝑠 𝑦 − 𝑦 𝑠 𝑦 ;.
We also define a Cornering Resistance Coefficient, 𝐶 𝐶: 𝐶 𝐶 = 𝑙 ⁄ = 𝑦2 ((𝑙𝑓) 2 1 𝐶𝑟 + (𝑙𝑟) 2 1 𝐶𝑓 ) ≈ 1 𝐶𝐶 ( 𝑦 ) 2 ; [3.4]
𝐶 𝐶 is such that the additional propulsion force due to cornering is ≈ 𝐶 𝐶 or the additional power is ≈ 𝐶 𝐶 . During a transport operation, the cornering in each time instant is typi- cally described by two variables, e.g. ( 𝑝), but only the combined scalar measure 𝑦= 2⁄ 𝑝 influ-
ences 𝐶 𝐶. Hence, we can plot the following graph:
𝐶
𝐶
Figure 3-4: Left: Cornering Resistance Coefficient. Right: Required steer angle. Vehicle data: = 1500 𝑘 ; = 3; 𝑙 = 1.25; 𝐶 = 60 [𝑘𝑁 1]; 𝐶 = 80 [𝑘𝑁 1].
Notes:
• The model used above is not advanced enough to differ between which axle is driven. For such purpose, one would need e.g. the model in Figure 4-15.
• Normal driving is often below 2 or 3 𝑠⁄ 2, so the coefficient typically stays below 0.01. So, the
influence on energy consumption, during such “maximum normal” negotiation of corners, is still of the same magnitude as rolling resistance coefficient 𝐶 ≈ 0.005. .0.010.
• For ideally tracking axles, see 2.2.6, 𝐶 → and 𝐶 → , which gives that 𝑓 𝑅 → 0 and conse- quently no power loss and no required propulsion force. Therefore, high cornering stiffness is fuel efficient when cornering.
• When driving extreme cornering, such as driving as fast as possible in a circle on a test-track, one will experience that the top speed is much lower than driving straight ahead. That is NOT explained by [3.4]. It would require inclusion of a combined tyre slip model.
3.2.3 Functions After Start
Figure 3-5 shows how the functions can be found in a traction diagram.