BasicVehDynamics

ME 379M/397 – Prof. R.G. LongoriaVehicle System Dynamics and Control

Department of Mechanical EngineeringThe University of Texas at Austin

Basic Vehicle Dynamics

Prof. R.G. Longoria

Updated for Spring 2010



Lecture goals

• Review concepts from dynamics

• 2D (planar) dynamics problems

• Discuss cases studies and illustrate need for

introducing 3D dynamic effects



Part 1: Review concepts in dynamics

• Basic coordinate systems

• Free body (or force) diagrams

• How to express position vectors in defined coordinate systems, and how to differentiate them to get velocity and acceleration if needed.

• Relative velocity/acceleration, transferring between coordinate systems.

• Mass properties (e.g., moments of inertia, inertia matrix, etc.)

• Particle and rigid body kinematics

• Newton’s laws, Euler’s equations

• Coordinate transformations are essential for some problems (e.g., turning) – reviewed later

• Bond graphs – optional (*-ed slides)



Vehicle-fixed coordinate system

SAE vehicle axis system

x = forward, on the longitudinal

plane of symmetry

y = lateral out the right side of the

vehicle

z = downward with respect to the

vehicle

p = roll velocity about the x axis

q = pitch velocity about the y axis

r = yaw velocity about the z axis

Consider the standard SAE

coordinate system and

terminology.

Ground vehicle coordinate systems commonly employ a

coordinate system standardized by SAE.



Earth fixed coordinate system

X = forward travel

Y = travel to the right

Z = vertical travel (+down)

ψ = heading angle (between x

and X in ground plane)

ν = course angle (between

vehicle velocity vector and X

axis)

β = sideslip angle (between x

and vehicle velocity vector)



Example: 2 axle vehicle

FBD:

x

z



Relative velocity of particles

It is recommended that you review these basic

kinematic concepts in a reference of your choice.



Relative acceleration of particlesIt is recommended that you review these basic

kinematic concepts in a reference of your choice.



Rigid body velocities

( )0

p o

p

V V V

A A R R

= + Ω×

= + Ω× + Ω× Ω×

ɺ

(cf. Karnopp&Margolis, eqs. 1.18)

(a) Transfer the c.g. velocity to body-fixed directions at the four wheels

(b) If each wheel is constrained to have no velocity perpendicular to the

plane of the wheel, state the kinematic constraints for each wheel.



Translating and rotating ref. framesIt is helpful to have an understanding of the coordinate systems used

for rigid body analysis, and the terminology employed for these

applications. One of the key results is given below.

can be any vector quantity.

XYZ xyz

dV dVV

dt dt

V

= + Ω×

This relationship between

vector quantities in xyz and

XYZ will prove very useful.

Translating reference axes, with body

rotating with angular velocity, ωωωω.

Rotating reference axes rotate with

velocity ΩΩΩΩ, and body rotates with

angular velocity, ωωωω.



Rigid body in 3-D motion

By using body fixed coordinates, the

rotational inertial properties remain fixed.

The products of inertia* are all zero, and this

makes it convenient for our purposes.

*See dynamics text to review inertial properties.

†

†

x y z

x y z

v v v =

= Ω Ω Ω

v

Ω

Newton:

xyz

m

d

dt

= + ×

p v

pF Ω p

≜

With v relative to rotating frame.

(trans. momentum)



Basic equations for a rigid body

x x y z z y

y y z x x z

z z x y y x

F p p p

F p p p

F p p p

= + Ω − Ω

= + Ω − Ω

= + Ω − Ω

ɺ

ɺ

ɺ

x x y z z y

y y z x x z

z z x y y x

T h h h

T h h h

T h h h

= + Ω − Ω

= + Ω − Ω

= + Ω − Ω

ɺ

ɺ

ɺ

xyz

d

dt= + ×

pF Ω p

xyz

d

dt= + ×

hT Ω h

The complete equations for a rigid body are nonlinear,

coupled differential equations.

These are sometimes referred to as the Euler equations, often only when you let the

ref. axes coincide with the principal axes of inertia at the mass center or at a point

fixed to the body so the products of inertia go to zero – this leads to a simpler form.



System dynamics formulation

• State space formulation for the vehicle dynamic

states

• For modeling, use momentum states: p, h

• Can readily switch to velocity states as needed

x x y z z y

y y z x x z

z z x y y x

p F p p

p F p p

p F p p

= − Ω + Ω

= − Ω + Ω

= − Ω + Ω

ɺ

ɺ

ɺ

x x y z z y

y y z x x z

z z x y y x

h T h h

h T h h

h T h h

= − Ω + Ω

= − Ω + Ω

= − Ω + Ω

ɺ

ɺ

ɺ



Value of understanding 6 DOF eqs.

• By having the full dynamic equations at your

disposal, you can:

– Examine effects that might be hard to ‘see’

intuitively or reliably

– You can ‘throw out’ terms that do not apply and

keep those that will impact the problem at hand.

x x y z z y

y y z x x z

z z x y y x

p F p p

p F p p

p F p p

= − Ω + Ω

= − Ω + Ω

= − Ω + Ω

ɺ

ɺ

ɺ

x x y z z y

y y z x x z

z z x y y x

h T h h

h T h h

h T h h

= − Ω + Ω

= − Ω + Ω

= − Ω + Ω

ɺ

ɺ

ɺ



Rigid body - using bond graphs*

Rigid body motion in a body-centered coordinate system:

–vector angular velocity

–coordinate system moves

with body

–3 components of

translational momentum

–3 components of angular

momentum

–6 nonlinearly-coupled

DOF

Karnopp and Rosenberg (1968)



Planar dynamics of a vehicle

Consider a car with total mass,

m, centered at CG reaches

maximum acceleration, and

assume the mass of the wheels

are small compared with the total

mass of the car. The coefficient

of static friction between the

road and the rear driving wheels

is assumed known as µ.

Find relations for the forces at

the front and rear pairs of wheels

the under condition of maximum

acceleration.

Must assume relation for friction force.

FBD:

x

z



planar dynamics (cont.)

0

0

x x y z z y

y y z x x z

z z x y y x

p F p p

p F p p

p F p p

= − Ω + Ω

= = − Ω + Ω

= = − Ω + Ω

ɺ

ɺ

ɺ

0

0

0

x x y z z y

y y z x x z

z z x y y x

h T h h

h T h h

h T h h

= = − Ω + Ω

= = − Ω + Ω

= = − Ω + Ω

ɺ

ɺ

ɺ

( )1 2 10y r r

T W l l W h mglµ+ = = + − −∑0z f r

F W W mg+ = = − −∑

Solve for the forces, apply to x-direction equation.



Example: Deriving Bicycle ModelSymmetric vehicle, No Roll

•Represent the two wheels on the front and rear

axles of a two-axle vehicle by a single equivalent

wheel.

•The bicycle model has at least three states:

–forward CG translational momentum or velocity

–lateral CG translational momentum or velocity

–yaw angular momentum or velocity about CG

( )

( )rear drivefront drive lateral force effect

1 2 1

cos( ) sin( )

cos( ) sin( )

cos( ) sin( )

x y z xf f xr yf f

y x z yr yf f xf f

z z yf f yr xf f

m V V F F F

m V V F F F

I l F l F l F

δ δ

δ δ

δ δ

− Ω = + −

+ Ω = + +

Ω = − +

ɺ

ɺ

ɺ

Wong, Eqs. 5.25 – 5.27:

You should be able to see how the equations

shown here (from Wong) can be derived from the

basic Euler equations.

Note that the right-hand sides are basically just

the ‘external’ forces and torques (here applied by

the tire-surface interaction).



Contrast D’Alembert Formulation

It is common in conducting some

basic rigid body analysis, to

employ kinetic diagrams to

visualize the effect of

translational and rotational

forces.

md

dt

− =

=

∑F a 0

ppɺ

The analysis breaks down, effectively, to a d’Alembert

formulation, where an ‘inertial force’ becomes part of a ‘dynamic

equilibrium’ analysis.

Rate of change of

momentum – ‘inertial

force’

1 I:mpɺ

V

1F

2F

nF

*



Example of D’Alembert Approach

A bicyclist applies the brakes as he descends a

10° incline.

What deceleration a would cause the dangerous

condition of tipping about the front wheel A?

The combined center of mass of the rider and

bicycle is at G. Ans. a = 0.510g

Meriam & Kraige (6/3)



Case study: 2D Vehicle Rollover

• Rollover can occur on flat and level surfaces (on-road). On-

road rollovers typically arise from loss of directional control,

which may result from driver steering actions.

• Off-road rollover may result from the cross-slope effect adding

to lateral forcing from curb impacts, soft ground/soil, or other

obstructions that “trip” the vehicle.

Marine, et al (1999)

Off-road path

Steering input to re-

enter roadway

Loss of directional control

due to excessive corrective

steering



Rollover Classification

• Friction rollover - occurs due to high lateral

friction forces in tire-surface interaction without

any tripping

• Spin out rollover - rear outside tire saturates

before front leading to yaw instability

• Plow out rollover - front outside tire saturates

before rear leading to understeer and possibly

inability to steer out of accident



Quasi-static Rollover

of a Rigid Vehicle

ϕ

cos sin sin cos 02 2

o y zi

t tM ma h mgh mg F tϕ ϕ ϕ ϕ

= − − − + =

∑

Rigid vehicle moving in a steady turn,

and assume there is no roll acceleration.

Take moments about outside wheel,

V

x

Assume that ϕ is small, then you can solve for the

ratio of lateral to gravitational acceleration,

1

2 2

y zia F tt

hg h t mg

ϕϕ

= + − −

Note, you assure tire contact forces are equal, or,1

2

ziF

mg=

by making,1

2

y

y

a g

at

h g

ϕ =

+

(cross-slope design)

‘vehicle-fixed’ axes



Quasi-static Rollover

of a Rigid Vehicle

ϕ

The lateral acceleration at which rollover begins is

the “rollover threshold”.

The point where the inside contact force goes to

zero specifies,

Cross-slope angle can

counter lateral

acceleration

The Static Stability Factor (SSF) is defined for ϕ = 0, or

0, 02

zi

y

F

a tSSF

g hϕ= =

= ≜

This can also be referred to as “rollover threshold”.

0

1

2 2zi

y

F

a th

g h tϕ

ϕ=

= + −



Rollover Threshold

Rollover Threshold (Gillespie, 1992)

Note that these values can exceed the

cornering capabilities that arise from friction

limits (about 0.8).

So vehicle could spin out in such a case,

implying rollover would not occur. We know

this is not true.

y

z

F

Fµ =

We have examined ‘rigid body’ rollover.

The effect of roll angle shows that, at least

for a simple steady-state case, there is more

to rollover prediction than this simple

analysis.

1 2tane

h

tφ −=

Roll Angle, φ

unstableLateral

Accel ay

Rollover threshold

0, 0

tan2

zi

y

F

a t

g hϕ

φ= =

= −

Roll lowers SSH



Summary of dynamics review

• We need models for insight, basic analysis/simulation, and control design.

• Not possible to make comprehensive review. Instead, adopt dynamics concepts for vehicle system modeling on an ‘as needed’ basis, focusing on answering the questions asked.

• Rely on fundamental concepts such as relative velocity/acceleration.

• It can be helpful to understand the basic 3D rigid body equations as a basis for studying simple (e.g., 2D) problems.

• Example given of how vehicle static stability (rollover) can be evaluated with basic planar dynamics.



References

1. J.L. Meriam and L.G. Kraige, Engineering Mechanics: Dynamics (4th

ed.), Wiley and Sons, Inc., NY, 1997.

2. D.T. Greenwood, Principles of Dynamics, Prentice-Hall, 1965.

3. T.D. Gillespie, Fundamentals of Vehicle Dynamics, SAE, Warrendale,

PA, 1992.

4. J.Y. Wong, Theory of Ground Vehicles, John Wiley and Sons, Inc., New

York, 1993 (2nd) or 2001 (3rd) edition.

5. Hibbeler, Engineering Mechanics: Dynamics, 9th ed., Prentice-Hall.

6. J.P. Den Hartog, Mechanics, Dover edition.



Appendix A: Example Problems

1. Anti-rollover control (gyro stabilizer)

2. Bus flywheel

3. Truck with trailer



1. Anti-rollover controlwith gyro stabilizer

An experimental car is equipped with a gyro stabilizer to counteract

completely the tendency of the car to tip when rounding a curve (no change

in normal force between tires and road).

The rotor of the gyro has a mass mo and a radius of gyration k, and is

mounted in fixed bearings on a shaft that is parallel to the rear axle of the

car. The center of mass of the car is a distance h above the road, and the car

is rounding an unbanked level turn at a speed v. At what speed p should the

rotor turn and in what direction to counteract completely the tendency of the

car to overturn for either a right or a left turn? The combined mass of car and

rotor is m.

We introduced this example to motivate the need to review 3D rigid body dynamics: useful for

‘back of the envelope’ analysis but also for building an understanding helpful for more complex

problems.



Dynamics of Spinning Flywheel

With the symmetric flywheel spinning about

the z-axis, if the forces are applied about the

‘torque’ axis, the right-hand rule helps indicate

how the flywheel would precess.

However, we also know that if we spin the

flywheel and precess about the y-axis, a torque

will be applied about the x-axis. The applied

moment would be,

precess velocity

spin velocity

M I p

p

= Ω

Ω =

=Right-hand rule

This concept can be used to solve the gyro-

stabilizer problem.



Solution from Meriam and KraigeThe sense of the spin actually can be inferred

by always making sure you form the right

hand system with spin-precession-torque.

Right-hand

turn

Left-hand

turnRotor should spin in a direction

opposite to rotation of wheels.

Here is the solution from the

Instructor’s manual.



Finding Rollover Torque

Identify the relevant body velocity as you make a left turn, as shown

here, or a right turn. For this case the body has an angular rotational

velocity, ˆzkΩ = −Ω

When you apply Newton’s law for the y direction, you account for

the effect of this rotation (through the Euler equations), showing

how this so-called ‘centrifugal force’ arises,

2

00

y y x y z x

vR

mvF p m v m v

R== =

= − Ω + Ω =ɺ

So in a left-hand turn, this induced force generates a moment about the roll axis (x) that

tends to induce rollover. This rollover torque we are trying to control is,2

x y

mvT F h h

R= =

Note the sign change for a right-hand turn.

Note py here is

momentum, not spin

velocity (as in SAE).



Applying Euler’s Equations

( )rollover

This term must canceltorque

the rollover torque

x x y z z y

x y z z z y y

x y z y z

h T h h

T I I

T I I

= − Ω + Ω

= − Ω Ω + Ω Ω

= + − Ω Ω

ɺ

Assume that Iy−Iz>0.

In a left-hand turn, Ωz<0, the torque to control

is positive, and we require Ωy>0.

In a right-hand turn, Ωz>0, the torque to

control is negative, and we require Ωy>0.

For both, the rotor should spin opposite to the

direction of rolling wheels.

( )2

0x

y z y z x

y y

h

I I T

v hI mv

R R

≈

− − Ω Ω =

Ω =

ɺ

y

y

mvh

IΩ =

Neglecting Iz:

Required spin

velocity:

cf. M&K solution



Visualizing with a Bond Graph*

1

1 1

G

G

G

Tx

TzTy

ωy ωz

hy

hx

hz

I:Ix

I:IyI:Iz

ωxhx

hy

ωy

hzωy

ωz

hyωz

roll

spin

precession

The sum of torques at this 1-

junction reflects the relevant

dynamics. It is just the relation

we wrote before but now we

might write:

rollover spin precessiontorque torque torque

x x y z z yh T h h= − Ω + Ωɺ

Now visualize with a bond

graph.

Causally, you can see that a

torque from the vehicle

body induces precession of

the rotor, but it is the

angular velocity ωz that

leads to a torque about x.

The same can be said for the

torque induced by the spin

velocity ωy.



2. Bus FlywheelAn experimental antipollution bus is

powered by the kinetic energy stored in

a large flywheel that spins at a high

speed p in the direction indicated. As the

bus encounters a short upward ramp, the

front wheels rise, thus causing the

flywheel to precess. What changes occur

to the forces between the tires and the

road during this sudden change?

(Meriam and Kraige, 7/100)



A Bond Graph Perspectiveon the Bus Flywheel

• The bus flywheel is a good example of how rigid body bond

graphs can be used to represent or apply the rigid body

equations.

• One advantage is that graphical modeling can be used for

`intuitive' gain, and for some people this is helpful.

• As previously discussed, a spinning flywheel is mounted in a

bus or cart. The body fixed axes are mounted in the vehicle,

with the convention that z is positive into the ground.

• The bus or cart approaches a ramp, and the questions which

arise include whether any significant loads will be applied, what

their sense will be, and on which parameters or variables they

are dependent.



Flywheel on BusBond Graph Representation



3. Truck with Trailer

A loaded pickup truck which weighs 3600 lb with mass center at G1, is hauling an 1800-lb

trailer with mass center at G2. While going down a 10-percent grade, the driver applies his

brakes and slows down from 60 mi/hr to 30 mi/hr in a distance of 360 ft. For this interval,

compute the x- and y-components of the force exerted on the trailer hitch at D by the

truck. Also find the corresponding normal force under each pair of wheels at B and C.

Neglect the rotational effect of the wheels.


To find the unknown forces, need to determine

acceleration and then the inertial forces.



3. Truck with Trailer (cont)


Compare to

Gillespie

example.

BasicVehDynamics

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