University of Novi Sad FACULTY OF TECHNICAL SCIENCES Basics of Automotive Engineering Part 3: Basics of Vehicle Dynamics Dr Boris Stojić, Assistant Professor Department for Mechanization and Design Engineering Chair for Engines and Vehicles
Basics of Vehicle Dynamics Introduction
Introduction Basic overview Tasks and contents of vehicle dynamics Some basic topics: to study......interaction between vehicle and its surroundings, w/ or w/o driver...how forces influence vehicle motion and vice versa...what the vehicle response will be in certain driving situation...how design changes will affect vehicle behavior etc.
Introduction Basic overview Approaches and assumptions carthrottle.com Full vehicle model, general driving situation: Many degrees of freedom Many inputs and outputs, complex relationships System of coupled non-linear differential equations Not appropriate for analytical study Experimental approach, CAE modeling and simulations newslincolncounty.com carsim.com popularmechanics.com
Introduction Basic overview Approaches and assumptions Simplified models, restricted driving maneuvers: Less DOF Restricted number of I/O s Possibility of ODE linearization Manageable math, appropriate for analytical study Gaining insight into main physical relationships Some aspects of basic engineering analysis carried out easily
Introduction Basic overview Approaches and assumptions Longitudinal vehicle dynamics Forces and motions in longitudinal direction, smooth road surface Predicting top speed, acceleration and braking performances, gradeability, fuel consumption... ni.com
Introduction Basic overview Approaches and assumptions Lateral vehicle dynamics Forces and motions mainly in lateral direction Predicting cornering performances, handling, stability... tut.fi
Introduction Basic overview Approaches and assumptions Vertical vehicle dynamics Forces and motions mainly in vertical direction Ride, vibration behavior, tyre/road contact... scielo.br
Introduction Basic overview Examples of usage in engineering and everyday life Let s name a few... What is the maximum velocity of the vehicle? How many horsepower does the vehicle need? What will be the fuel consumption of the vehicle? How long does it take for the vehicle to come to stop? What happens if shock-absorbers don t work? How to re-gain lost adhesion of the tyre? What happens when the brake is suddenly applied during cornering? Understanding active vehicle safety! And so on and on and on...
Introduction Overview of Newtonian laws of motion 3rd law Reaction forces, Free-body diagram Important application: no action without reaction! 1st law: body equilibrium Net force 2nd law: force, mass, acceleration Rotational motion: torque, moment of inertia, angular acc.
Introduction Overview of Newtonian laws of motion Some important reaction forces: Ground force Friction (adhesion) force F F T T F T
Introduction Engine-to-wheel torque transmission n input ig Definition of gear ratio noutput Transmission power loss: P output = P input g Transmitted torque: T output = T input i g g Vehicles: Input element: ENGINE Transmission i tr, tr Output element: WHEEL n e, T e n w n i e tr T w = T e i tr tr n w, T w
Introduction Engine-to-wheel torque transmission Transmission system of gear pairs connected in series Passenger car: transmission = gearbox + final drive k-m-p.nl otomoto.com.au k-m-p.nl i tr = i g i f tr = g f n w n i e tr car-mri.com T w = T e i tr tr
Introduction Engine-to-wheel torque transmission i g = i I, i II, i III, i IV,... FOR EVERY GEAR APPROPRIATE GEAR RATIO LOWER GEAR LARGER GEAR RATIO E.g. i I = 4.05 i II = 2.82 i III = 1.75 i IV = 1.04 i IV = 0.80
Basics of Vehicle Dynamics Forces acting on the vehicle
Forces acting on the vehicle Overview of forces Gravity effects Aerodynamic forces Tyre-road interaction Tyre peculiarities Side slip very special property of pneumatic tyre Load dependence importance of CG position
Forces acting on the vehicle Gravity force vehicle weight Causes axle loads Motion resistance on the graded road (vector decomposition) h CG W f a W M A = 0 W f l = W cos b W sin h CG Z i = 0 W f + W r = W cos l b A W W f r b h W cosα l l a h W cosα l l CG CG W sinα W sinα = 0: W f W r b W l a W l W r
Forces acting on the vehicle Aerodynamic forces Motion resistance Lift force F W c W A v 2 2 Rill Lateral force racingcardynamics.com
Forces acting on the vehicle Tyre behaviour: rolling resistance Contact pressure distribution of non-moving tyre
Forces acting on the vehicle Tyre behaviour: rolling resistance Rolling tyre: hysteresis effect F X W T r D R Z e R X Internal elastic force Internal friction force
Forces acting on the vehicle Tyre behaviour: rolling resistance Rolling tyre: hysteresis effect F X W T R X = f r W T rolling resistance of a single freerolling tyre f r rolling resistance coefficient r D R Z e R X From: Genta/Morello Dependency of f r on velocity (example)
Forces acting on the vehicle Tyre behaviour: longitudinal slip Definition of slip Theoretical wheel speed: v t = r D Real speed: v v = v t : NO SLIP v < v t : DRIVE WHEEL v v > v t : BRAKE WHEEL r D BRAKE WHEEL s v v v t r 1 D ω v T s=1: car moving, wheel locked DRIVE WHEEL s v t v v t 1 r D v ω T s=1: car standing, wheel sliding s=0: wheel rotating freely
Forces acting on the vehicle Tyre behaviour: longitudinal slip Definition of slip Rigid wheel: just a geometric interpretation! FREE WHEEL BRAKING DRIVING v s =0 v s v s Real tyre: pronounced elasticity far more complex slip mechanism!
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Looking at the single particle of tyre contact patch Undeformed at the beginning Longitudinal deformation increases as the particle travels through the contact patch Particle tip glued to the ground due to adhesion (i) Deformation propagates with the velocity v S (ii) Local longitudinal force increases with the growth of the deformation
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation (i) Deformation u(x) propagates with the velocity v S u(x) v S x r D - v S r D v S No sliding of the particle at the ground but slip exists (vv t )!
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation (ii) Local longitudinal force increases with the growth of the deformation u(x) F tan (x) x F tan (x) Local force distribution x Net longitudinal force sum of elementary (local) tangential forces Proportional to the area under the line
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Low torque Low deformation Low slip Low longitudinal force High torque High deformation High slip High longitudinal force
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Net longitudinal force R X Where does this non- linearity come from?? We said: particle tip remains glued to the road! s Wheel slip This can not be true all the time there is not enough adhesion at the end of the patch!
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Local vertical force distribution Maximum AVAILABLE local longitudinal force = = Local vertical force Friction coeff.
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Required long. force (zone of stick) Available long. force 0 A B 1 2 x Zone of local particle sliding
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Area increases force increases Not linearly with slip anymore! R X Torque increases Deformation increases Slip increases Further slip increase what will happen now? s
Forces acting on the vehicle Tyre behaviour: longitudinal slip Slip mechanism and longitudinal force generation Net longitudinal force R X PEAK POINT Further slip increase whole contact patch slides Tyre friction decreases with sliding velocity increase Net force decreases Wheel slip s Chassis Handbook
Forces acting on the vehicle Tyre behaviour: longitudinal slip Introduction of adhesion coefficient R X W T longitudinal force tyre/axle load (vertical force) Sometimes we use approximation: traction force F T instead of real longitudinal force R X : What s the difference? F T W T T W R X T r W D F T e r D W T f r frwt = Frol R X W T F X r D W T e R X = F T F rol When F T >> F rol R X F T
Forces acting on the vehicle Tyre behaviour: longitudinal slip Introduction of adhesion coefficient MAX s < MAX MAX will decrease with load increase! R XMAX W T s10-15% s=100% s
Forces acting on the vehicle Tyre behaviour: longitudinal slip Some example values of adhesion coefficient s (%) From: Wallentowitz
Forces acting on the vehicle Tyre behaviour: side slip Introduction of tyre side slip angle When subjected to side force, tyre rolls at an angle with respect to longitudinal axis LATERAL FORCE TYRE TRAVELLING DIRECTION LONGITUDINAL DIRECTION
Forces acting on the vehicle Tyre behaviour: side slip Lateral tyre deformation due to side force Vehicle pushes tyre with F Y Tyre particles deform sideways v Net ground lateral force arises R Y R Y acts behind wheel centre t P pneumatic trail F Y - wheel side slip angle z x t P y R Y
Forces acting on the vehicle Tyre behaviour: side slip Lateral tyre deformation due to side force v Tyre structure elasticity additionally affects deformation path F Y R Y
Forces acting on the vehicle Tyre behaviour: side slip Lateral tyre deformation due to side force Side slip angle Lateral deformation distribution v F Y t P R Y Force from vehicle Pneumatic trail Ground reaction R Y t P = ALIGNING MOMENT
Forces acting on the vehicle Tyre behaviour: side slip Pneumatic trail / aligning moment behaviour stanford.edu
Forces acting on the vehicle Tyre behaviour: side slip Lateral force and aligning moment vs. slip angle fromthe Automotive Chassis Vol. 1 M S δ from Chassis Handbook Notice vertical load (F Z ) impact! Pneumatic trail decreases when increases!
Forces acting on the vehicle Tyre behaviour: side slip Non-linear tyre behavior Zone of pronounced non-linearity Linear approximation: F Y = c Applies for small c - tyre lateral stiffness (depends on vertical load!)
Forces acting on the vehicle Tyre behaviour: side slip Factors affecting lateral force / side slip dependency: Vertical load Pressure Camber angle Longitudinal force etc. wikipedia
Forces acting on the vehicle Tyre behaviour: side slip Impact of tyre load W T Source: Wallentowitz Side force F y Vertical load increase Contact length increases larger F y for the same For same F y - decreases when W T increases c increases with W T Side slip angle Relation between W T and c is degressive
Forces acting on the vehicle Tyre behaviour: combined slip Simultaneous presence of longitudinal and side force F R Realized longitudinal force F F X Y F R2 = F X2 + F 2 Y F RMAX = W T MAX F X2 + F Y2 = (W T MAX ) 2 = const F X F Y F R Available side force Wheel lock or full slide: no side force available! Free rolling tyre: maximum side force available!
Forces acting on the vehicle Overall tyre behaviour and modelling issues...
Basics of Vehicle Dynamics Performances: acceleration, top speed, gradeability
Vehicle performances Introduction Summary of motion resistances, impact of velocity Engine torque curve Powertrain parameters
Forces acting on the vehicle Summary of motion resistances GRADE RESISTANCE F = Wsin ROLLING RESISTANCE F rol = f r W WIND RESISTANCE F W c W v A 2 2
Forces acting on the vehicle Summary of motion resistances Velocity impact: comparison rolling vs. aerodynamic resistance (force, power) Impact of grade resistance Definition of power: P F ds dt Infinitesimal work = Force F infinitesimal displacement ds Velocity v P = Fv Rate at which energy is utilized i.e. Velocity at which we can overcome resistance force P = T - for rotary motion = 2n/60 [radians/sec] n [RPM]
Vehicle performances Summary of motion resistances Numerical example: generic car E.g. m = 1400kg W = 14000N c W = 0.3 A = 2.8 m 2 technical-illustration.com
Vehicle performances Summary of motion resistances Numerical example: generic car Net resistance 12% grade Net resistance 5% grade motion resistance for rces (N) 6000 6000 6000 5000 4000 3000 2000 1000 0 Aerodynamic resistance Rolling resistance 0 100 200 300 5000 4000 3000 2000 1000 0 12% Grade resistance 5% Grade resistance 0 100 200 300 5000 4000 3000 2000 1000 0 Net resistance 0% grade 0 100 200 300 velocity (km/h)
Vehicle performances Summary of motion resistances Numerical example: generic car 500 500 kw car motion resistance power (kw W) 450 400 350 300 250 200 150 100 50 0 Net resistance 12% grade Net resistance 5% grade Net resistance 0% grade 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 velocity (km/h) 100 kw car 20 kw car
Vehicle performances Summary of motion resistances Numerical example: generic car 90 80 Net resistance 12% grade 70 motion resistance power (kw W) 60 50 40 30 20 10 0 0 10 20 30 40 50 60 70 80 90 100 110 120 Net resistance 5% grade Net resistance 0% grade velocity (km/h)
Vehicle performances Longitudinal dynamics: equation of motion Applying Newton s Second Law: MASS ACCELERATION = NET FORCE researchgate Actual mass of the vehicle m eff = m + m eq Caution: rotational elements also have to be accelerated! Effective mass that has to be accelerated on account of engine torque Equivalent mass of the rotational inertia lumped at vehicle s CG We adopt approach of effective mass and equivalent mass Equivalent mass i.e. effective mass can be calculated empirically or analytically.
Vehicle performances Longitudinal dynamics: equation of motion Applying Newton s Second Law: m eff a = F i Forces acting in longitudinal direction: F T traction force F W aerodynamic resistance Wsin - grade resistance F rol,f/r front and rear rolling resistance
Vehicle performances Longitudinal dynamics: equation of motion Applying Newton s Second Law: m eff a = F T F W F rol Wsin 1 F rol = f r Wcos F T T r W D T W - net drive torque at all drive wheels F rol = F rol,f + F rol,r net rolling resistance
Vehicle performances Longitudinal dynamics: equation of motion Applying Newton s Second Law: Finally we obtain: F T = F W + F rol + F in + F TRACTION FORCE F T T r W D GRADE RESISTANCE INERTIA RESISTANCE ROLLING RESISTANCE F = Wsin F in = m eff a F rol = f r W WIND RESISTANCE F W c W v A 2 2
Vehicle performances Traction force diagram Starting from engine torque curve Engine torque transforms to traction force Engine RPM transforms to velocity i tr tr Engine Gearbox i g, g Final drive i f, f T e,n e T w,n w Wheel tr = i g i f F n T w T r w D i n g Te igifη r tr = g f v = r D w = r D n w /30 e i f Sum for all drive wheels D = 2n/60 tr
Vehicle performances Traction force diagram 180 160 140 120 100 80 60 40 20 0 Engine torque F T 8000 7000 6000 Engine RPM n e 5000 5000 0 2000 4000 6000 8000 F T Te igifη r g D rd n v 30 i i e f tr 4000 3000 2000 1000 Traction force F T Final drive: i f =3.85 1 st gear: i g =i I =3.3 2 nd gear: i g =i II =1.99 3 rd gear: i g =i III =1.36 4 th gear: i g =i IV =1.00 5 th gear: i g =i V =0.79 0 Vehicle velocity km/h 0 20 40 60 80 100 120 140 160 180 200 220
Vehicle performances Performance determination: graphical approach 8000 7000 6000 Traction force 1 st F T = F W + F rol + F in + F F T 5000 4000 3000 2000 F in 2 nd in (12% grade) 3 rd F 4 th Net resistance 12% grade Net resistance 0% grade 1000 0 F W + F rol v km/h 0 20 40 60 80 100 120 140 160 180 200 220 5 th Maximum velocity at 0% grade
Vehicle performances Acceleration performance m eff a = F T F W F rol Wsin F a T F W F rol m W sin eff 3,00 2,50 a(m/s^2) 2,00 1,50 1,00 0,50 0,00 v (km/h) 0 20 40 60 80 100 120 140 160 180
Vehicle performances Acceleration performance Calculating acceleration time 1/a 2 (s /m) a dv dt IV V dt dv a 1/a 2 (s /m) t v 1 dv a 0 V III III IV II II I v(km/h) I 4 A5 A6 A7 A A A 8 9 1 2 A A A 3 v v v v v v 2 v 4 6 v v 8 v 1 3 5 7 9 10 10 11 v v 12 A A A 13 v 14 A A 12 14 11 v13 v v(km/h) 15
Vehicle performances Dynamic axle loads W f, dyn = W f,stat W in - F Lf W r, dyn = W r,stat W in - F Lr ΔW in h l CG W g a W l f +l r =l l h Wcosα l l r CG f, stat W sinα Neglecting contribution of rotating mass and rolling resistance W l h W cosα l l f CG r, stat W sinα
Basics of Vehicle Dynamics Braking performances
Braking performances Basics of braking process 3 main phases of braking Deceleration (m/ /s 2 ) Time (s) 1 Delay 2 System activation 3 Full deceleration 1 2 3
Braking performances Basics of braking process 3 main phases of braking t 1 t 2 t 3 t S t S Stopping time t L Lost time t B Braking time APPROXIMATION: t L t B
Braking performances Basics of braking process Approximated by 2 phases t L Lost time: v=v 0 =const t B Braking time: a=a F =const - full deceleration t S Stopping time: t S = t L + t B
Braking performances Stopping and braking distance Lost distance: Braking distance: s L =v 0 t L s B 2 0 v 2a F Lost time: 0.51 sec Stopping distance: s S v 0 t L v 2 0 2a F
Braking performances Stopping and braking distance s B v 2 0 2a F s S v 0 t L v 2 0 2a F s L =v 0 t L bbc.co.uk
Braking performances Theoretical maximum deceleration Adhesion coefficient for braking case: = MAX BA W A F BA net brake force on axle (front or rear) W A axle load F s Maximum possible braking force (at axle): F BA,MAX = MAX W A Vehicle equation of motion 2 nd Newton s law for braking: (On level road, neglecting rotational inertia and wind resistance; with maximum braking forces both front and rear; rolling resistance comprised by braking forces) ma MAX = MAX W f + MAX W r a MAX theoretically possible maximum deceleration W f, W r front and rear axle weights Maximum braking force rear juniordesigner.com Maximum braking force front W r W f
Braking performances Theoretical maximum deceleration ma MAX = MAX W f + MAX W r ma MAX = MAX (W f +W r ) = MAX W= MAX mg a MAX = MAX g Braking distance would then be: s B v 2 2 0 MAX g
Braking performances Adhesion utilisation What does full deceleration a F depend on? Maximum braking force maximum deceleration minimum braking distance Two axles braked Optimal goal maximum adhesion utilization on both axles s s This is rarely possible!
Braking performances Adhesion utilisation Brake force distribution should match that of ground forces s s Ground forces distribution varies depending on deceleration Sophisticated control system required to account for this not always present and not always fully efficient
Braking performances Adhesion utilisation Adhesion not fully exploited on both axles! s s Net braking force less than physically possible braking distance increases We introduce braking efficiency B : B a a F MAX a F MAX g 1
Braking performances Braking efficiency impact on braking distance Braking distance is now: s B v MAX 2 0 2 g B Defined by law regulation: Braking efficiency has to provide certain minimum requirements concerning braking performances, in certain driving conditions (e.g. 75%).
Braking performances Optimal distribution of braking forces...