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ME542 Vehicle Dynamics-Lecture 1- 1

ME542 Vehicle Dynamics

Winter 2014

Tu & Th 1:30‐3:00pm

Huei Peng 3012 PMLDepartment of Mechanical Engineering

[email protected], 1-734-769-6553

Office hours: Mon. 4-5pm (Peng), Wed. 1-2pm (Peng)

GSI office hours to be announced


Vehicle Dynamics and Safety

• Accident occurred in May 2006 on I-96, caught by police dash-cam.

• Induced by an initial light side impact, SUV rolled over multiple times.


Lecture 1‐‐Introduction and Motivation(and administrative stuff)

• About this course

– Introduction and administrative information

– Major course content

– Grading policy

– MATLAB/SIMULINK

• Review of rigid body dynamics


Administrative Information• Textbook (not required)

– J.Y. Wong, Theory of Ground Vehicles, John Wiley &Sons, Inc, 4th

edition, 2008. (http://www.wiley.com/WileyCDA/WileyTitle/productCd‐

0470170387.html)

• Course material will be made available on the Ctools site (PPT files should be downloaded and printed before the lectures, HW, solutions, example MATLAB programs will be distributed using this site)


Course Requirements

• Prerequisites

– Knowledge in Newtonian Dynamics (ME240 level) is essential

– That of Automotive Engineering (ME458) and Intermediate Dynamics (ME440) are helpful but not required.

– Familiarity with Matlab/Simulink, since Matlab/Simulink is used extensively in the lecture examples and homework assignments.


Major course content

Background

Tire

Handling

Ride

Lec. Date Lecture contents 1 1/9 Introduction, motivation 2 1/14 Review of Rigid Body Dynamics 3 1/16 MATLAB‐SIMULINK review 4 1/21 Tire Models: Overview, Terminology, Definitions 5 1/23 Brush Tire Model (Lateral) 6 1/28 Brush Tire Model (Lateral) 7 1/30 Brush Tire Model (Longitudinal) 8 2/4 Combined‐Slip Tire Model 9 2/6 (Lateral) Taut String Tire Model, Magic Formula Tire Model 10 2/11 Off‐road tire model 11 2/13 Steady‐State Handling 12 2/18 Steady‐State Handling: Understeering and Oversteering13 2/20 Transient Handling 14 2/25 Lateral‐Yaw‐Roll model 15 2/27 Lateral‐Yaw‐Roll model16 3/11 Four‐Wheel Steering 3/13 Midterm (in class) 17 3/18 Guest lecture: chassis control systems 18 3/20 On‐Center Handling, Steady‐State And Transient Handling of Articulated

Vehicles 19 3/25 Driver‐vehicle interaction 20 3/30 Cross‐wind stability 21 4/1 Ride dynamics—Principle and Vibration Isolation and human perception 22 4/3 Ride dynamics—road excitations 23 4/8 Ride dynamics—Quarter‐car suspension Model 24 4/10 Ride dynamics—Quarter‐car suspension Model 25 4/15 Ride dynamics—Bounce and Pitch Model 26 4/17 Ride dynamics—Active Suspensions 27 4/22 Summary 4/25 Final exam (4‐6pm)


Related Courses

• ME458 Automotive EngineeringEmphasizes the vehicle as an engineering system and

review design consideration associated with all major systems including the vehicle structure, powertrain, suspension, steering and braking.

• ME568 Vehicle Control Systems

Covers control issues for all major vehicle control systems including engine control, cruise control, ABS/traction control, four‐wheel steering, active suspension and advanced control systems for Intelligent Transportation Systems.


Grading Policy

• Grading: 5‐6 Homework 55%

Midterm exam 20%

Final exam 25%

• Homework: Must be handed in on the due date in class (on‐campus

students) or uploaded to Ctools (distance learning students). Latehomework will be accepted up to 48 hours late with a 20% penalty foreach 24 hours (rounded up, i.e., 0 to 24 hours late = 24 hours late). Allproblem sets (home work assignments) are to be completed on yourown. You may discuss homework assignments with your fellowstudents at the conceptual level, but must complete all calculations andwrite‐up, from scrap to final form, on your own. Verbatim copying ofanother student's work is forbidden. If you have any questions aboutthis policy, please do not hesitate to contact the instructor.

• Exams: Midterm: in class, Dynamics, Tire, HandlingFinal exam: 2 hours, Accumulative but more weight on Handling and Ride


MATLAB/SIMULINK Tutorial

http://www.engin.umich.edu/class/ctms/


Getting a Copy of MATLAB/SIMULINK

http://www.mathworks.com/products/studentversion/

CAEN labs

Student version:

Virtual Site

http://virtualsites.umich.edu/


MATLAB/SIMULINK

• Example MATLAB/SIMULINK programs will be distributed, which you can freely use/modify.

• MATLAB is used for simple (usually linear) vehicle dynamic simulations and analysis for this course.

• SIMULINK: A GUI based simulation program. Strength:

– Realistic dynamic phenomenon such as nonlinearity, quantization, noise, switches, look‐up tables, delays, etc. can be simulated easily.

– GUI makes it easy to recycle vehicle simulation modules


Review of Rigid Body Dynamics

• Vehicle Coordinate Systems

• Newton/Euler Formulation

• Lagrange Formulation


Vehicle Coordinate System

• First step in deriving vehicle dynamic equations.

• The Society of Automotive Engineers (SAE) has introduced standard coordinates and notations for describing vehicle dynamics

x

y

z

longitudinal roll

lateral pitch

vertical yaw

ISO coordinate: x is the same but y and z are reversed.


SAE Vehicle‐Fixed Coordinate System‐‐Symbols and Definitions

Pitch angle: the angle between x‐axis and the horizontal plane. Roll angle: the angle between y‐axis and the horizontal plane. Yaw angle: the angle between x‐axis and the X‐axis of a inertia frame

Axis TranslationalVelocity

AngularDisplacement

AngularVelocity

ForceComponent

MomentComponent

x u (forward) p or (roll) Fx Mx

y v (lateral) q or (pitch) Fy My

z w (vertical) r or (yaw) Fz Mz

/ˈθiːtə/, US /ˈθeɪtə/

/ˈfaɪ/

/ˈsaɪ/, /ˈpsaɪ/


Earth Fixed Coordinate and Vehicle Slip

Course angle = +

OXYZ fixed on Earth (does not turn with the vehicle)

Heading angle

Sideslip angle

O

In the left figure, sideslip angle is negative)

tanv

u


Tire Slip

o: Center of tire contact patchZ: Perpendicular to the ground plane.X and Y axes: On the ground plane. Tire camber angle: Angle between the XZ plane and the

wheel plane.

Source: Wong page 7


Ride/Handling Dynamics

SPRUNG MASS RIDE HANDLING longitudinal

lateral vertical pitch roll yaw

UNSPRUNG MASS vertical

wheel rotation

Involves vehicle motions in several directions

( )


Newton/Euler Formulation

Consider a rigid body of mass m, with c.g. at point o, subject to N external forces.

For unconstrained motion, a rigid body possesses 6 DOF, 3 translational (x, y, z) and 3 rotational ().

oi j

k1F

2F

NF


Let V and to denote the absolute velocity of mass center and angular velocity of the rigid body, ( ) be unit vectors of the body-fixed coordinate oxyz. The Newton/Euler equations of motion are then

Newton/Euler Equations

ˆ i , ˆ j , ˆ k

1

N

ii

dV d LF F m

dt dt

ri Fii1

N

Mo d H

dt

Where L is the linear momentum and H is the angular momentum of the rigid body about the mass center o.

Newton’s2nd law:

oi

j

k1F

2F

NF

1r

2r Nr

Euler


Angular Momentum

Let H Hxˆ i Hy

ˆ j Hzˆ k

x

y

z

H

H I

H

wherexx xy xz

xy yy yz

xz yz zz

I I I

I I I I

I I I

2 2( )xx

V

I y z dV xy

V

I xy dV etc.

x xx xy xz xx xy xz

y xy yy yz xy yy yz

z xz yz zz xz yz zz

H I I I p I p I q I r

H H I I I q I p I q I r

H I I I r I p I q I r


Rigid Body Motion of a Vehicle(a key concept: rate of change of a vector with respect to a rotating frame)

Translational motion:

ˆˆ ˆDF ma m ui vj wk

Dt

( )

( )

( )

x

y

z

qw rv

ru pw

p

u

v

w v q

F

F m

F m u

m

F ma

d

dt t

D

D

Cross Product

Change of the unit vectors

Rate of change wrt a fixed frame

Rate of change wrt a rotating frame

Rate of change due to the rotation of the frame

ˆˆˆ ˆˆˆ ui vj wkui vj km w

ud

vdt

w

p u

q v

r

m

w


Rigid Body Motion of a Vehicle (cont.)

Rotational motion:

xx xy xz

xy yy yz

xx xy xz

xy yy yz

xzxz y yz zzz zz

p I p I q I r

q I p I q I r

r I p I q I

I p I q I rd

I p I q I rdt

I rp I q I r

M

2 2

2 2

2 2

xz yz zz xy yy yz

xx x

xx

y xz xz yz zz

xy yy y

x

y

z xx xy xz

xy xz

xy yy yz

xz yzz zz

I pq I q I rq I pr I qr I r

I p

I p I q I r

I r I qr I r I p I qp I rpp I q I r

I p I qp I r p I pq II p I q Ir r

M

M qq

M

I

xx xy xz

xy yy yz

xz yz zz

I p I q I r

H I p I q I r

I p I q I r

D d

Dt dt

M H


Equations of Vehicle Rigid Body Motion

2 2

2 2

2

( )

( )

( )

x

y

z

x xx xy xz xz yz zz xy yy yz

y xy yy yz xx xy xz xz yz zz

z xz yz zz xy yy yz

F m u qw rv

F m v ru pw

F m w pv qu

M I p I q I r I pq I q I rq I pr I qr I r

M I p I q I r I pr I qr I r I p I qp I rp

M I p I q I r I p I qp I r p

2xx xy xzI pq I q I rq


Simplified Equations of Motion• Assume

– vehicle is symmetric wrt the xz plane (Ixy = Iyz =0)

– p,q,r,v, and w are small, i.e., their higher order terms are negligible.

– u=uo + u’, u’ is small compared with uo.

Linear!

Naturally groupsinto 3 sets

'

( )

( )

x

y o

z o

x xx xz

y yy

z xz zz

F mu

F m v ru

F m w qu

M I p I r

M I q

M I p I r

)( rvqwumFx

)( pwruvmFy

)( qupvwmFz

.

.

.


Review of Rigid Body Kinematics

• Kinematics: study of the geometry of motions.

• Quantities: position, velocity and acceleration.

oi j

k

OI J

Kp

or

pr

relr

Given: Kinematics quantities of onepoint (mass center), and the angular motion of the rigid body.

Find: Those at another point on thesame rigid body.

Application: e.g., use motion of c.g.to infer motion of other points on the vehicle


Rigid Body Kinematics

Position:

p o relr r r oi j

k

OI J

K

p

or

pr

relr

p o relr r r Velocity:

p o relV V r

p o rel relV V r r

Acceleration:

p rel rel( )oa a r r


Example 1

A car is traveling in the xz plane, pitching, bouncing

o pGiven:

Find:

o

rel

ˆˆ

ˆ

ˆ

V ui wk

qj

r li

pV pa

l


Example 1 Solution

o p o

rel

ˆˆ

ˆ

ˆ

V ui wk

qj

r li

l

p o rel

0

0 0 0

0 0

ˆˆ ( )

u l u

V V r q

w w ql

ui w ql k


Example 1 Solution

p r rel

2

el ( )

0 0

0

0

0

0

0

0 0

0

0

0 0

oa r

q

ql

u qw q l

w q

a r

l

q

u

u

q

w

q

u

l

w

o

rel

ˆˆ

ˆ

ˆ

V ui wk

qj

r li


Another Kinematic Example (hw1)

• Vehicle Bicycle Model

• Vehicle is driving in the xy plane at a constant forward speed, and turning. All other motions are ignored

• Find the speed and acceleration at the center of the front axle and the rear left tire


Example 2

• Vehicle Bicycle Model

• Vehicle is driving in the xy plane, and yawing. All other motions are ignored

• Find the dynamic equations

vr

a

b

f

u

1xF

2xF

1yF

2yF


Example 2 Solution( )

( )

( )

x

y

z

F m u qw rv

F m v ru pw

F m w pv qu

2 2

2 2

2 2

x xx xy xz xz yz zz xy yy yz

y xy yy yz xx xy xz xz yz zz

z xz yz zz xy yy yz xx xy xz

M I p I q I r I pq I q I rq I pr I qr I r

M I p I q I r I pr I qr I r I p I qp I rp

M I p I q I r I p I qp I r p I pq I q I rq

?


Lagrange’s Formulation • An alternative way to formulate the equations of motion.

• Starts from “energy” (scalar!) rather than vectors.

• We first define three terms

Degrees Of Freedom (DOF):

Constraints

Generalized coordinates

Number of independent coordinates required to uniquely define the motion of a particle, a body or a system of bodies.

Kinematic relationship that limit possible motions.

A set of independent coordinates whose number is equal to the DOF and which uniquely define the position and orientation of the rigid body.


Examples for DOF, Constraints, and GC

Position of a particle in space: 3DOF

Position of a particle on a cylindrical surface

constraint

2DOF, position uniquely described by 2 coordinates, which are referred to as generalized coordinates (q1, q2).

Rigid wheel rolling on flat surface without slip, DOF=? Rigid wheel rolling on flat surface with slip, DOF=?


Lagrange EquationsConsider a system of particles or rigid bodies with n DOFdescribed by n generalized coordinates {q1(t), …, qn(t)}.The Lagrange's equation is then

where

ii i i

d T T VQ

dt q q q

i 1n

( , )i iT T q q

V V(qi )

: kinetic energy

: potential energy

Qi : the ith generalized non‐conservative force.

Qi F jj1

k

r j

qi


Example 3: Compound Pendulum

Derive the equation of motion for the compound pendulum shown below for large angle q.

mb

Ib

L

Is

msr

g


mb

Ib

L

Is

msr

g

DOF: only 1 (q)

Kinetic energy

Potential energy

2 2 2 2 21 1 1 1( ) ( )

2 2 2 2 2s s b b

LT m L r I m I

disk bar

( )(1 cos ) (1 cos )2s b

LV m g L r m g

disk bar

i

d T T VQ

dt

0

2 2( ) ( ) sin ( ) 02 2s b s b s b

L Lm L r m I I g m L r m


Example 4: Inverted Pendulum

• Later in the term, we will apply the Lagrange’s method to derive the lateral‐yaw‐roll model.

• Here we will start from a simpler system, which only involves lateral and roll (yaw is ignored for now)

mc

x

Lp,mp

F


Example 4 Solution

mc

x

Lp,mp

pm g

xF

yF

mc

xF

yF

F

Newtonian Method: for the pendulum

sin cos2 2

p pp p

L Lx x y

2

2( ) ( cos )

2p

x p p p

Ld dF m x m x

dt dt

2( cos sin )2 2

p pp

L Lm x

2

2( ) ( sin )

2p

y p p p p

Ld dF m g m y m

dt dt

2( sin cos )2

pp

Lm

x-direction

y-direction

roll-direction

sin cos2 2

p py x p

L LF F I 21

12p p pI m L

F


Example 4 Solution (cont.)Newtonian Method: for the cart

x cF F m x

2( ) cos sin2 2

p pc p p p

L LF m m x m m

2sin cos

3p p p pm g m x m L

sin cos2 2

p py x p

L LF F I Plug-in equations from x- and y- direction

We wrote down four equations, two of them used to cancel the internal forces Fx and Fy.

mc

xF

yF

F


Example 4 Solution (cont.)

sin cos2 2

cos sin2 2

p pp p

p pp p

L Lx x x x

L Ly y

Lagrange’s method

The kinetic energy and potential energy are then

2 2 2 2 2 2 2 21 1 1 1 1 1[ ] [ 2 cos ]

2 2 2 2 2 2 3p

c p p p p c p p

LT m x m x y I m x m x x L

cos2

pp

LV m g

For x-direction: x

d T T VQ F

dt x x x

2( ) cos sin2 2

p pc p p p

L LF m m x m m

For -direction: 0d T T V

Qdt

2sin cos

3p p p pm g m x m L We worked with two equations


Example 5: Double Pendulum

gl1

m1

l2

k

2

x

1

The double pendulum system consists of a mass‐less rigid bar of length l1, a point‐mass m1, a spring with free length l2 and spring constant k, and another point‐mass m2. Use Lagrange's equation to derive the equations of motion.

2m


DOF: 3 (1, 2, x)

Kinetic energy

Potential energy

2

1 1 1

2 2

2 1 1 2 1 2 2 1 1 2 1

1( )

21

sin( ) ( ) cos( )2

T m l

m x l l x l

gl1

m1

l2

k

2

x

1

21 2 1 1 2 2 2 2 2

1( ) (1 cos ) (1 cos ) cos

2V m m gl m gl m gx kx

A

B

1 1l

2 2( )l x

x

2 1


1 1 1

0d T T V

dt

2 21 1 1 2 1 2 2 1 2 1 2 1 2 1 1

22 1 2 2 2 1 2 1 2 2 2 1 1 2 1 1

2 cos( ) sin( )

( ) sin( ) ( ) cos( ) ( ) sin( ) 0

m l m l x m l x m l

m l l x m l l x m m gl

2 2 2

0d T T V

dt

2 22 2 2 2 1 2 1 2 1 2 1 2 1 2 1

2 2 2 2 2 2

( ) ( ) cos( ) ( ) sin( )

2 ( ) ( )sin( ) 0

m l x m l l x m l l x

m x l x m g l x

0x

d T T V

d xt x

22 2 1 1 2 1 2 1 1 2 1

22 2 2 2 2

sin( ) cos( )

cos ( ) 0

m x m l m l

m g kx m l x


Appendix


Rotation about a fixed axis

Beer, Johnson And Clausen, Vector Mechanics for Engineers, Dynamics, 8th edition, page 920

di dj dk

i j kdt dt dt

ˆˆ ˆu p u

v q v

w

ui v

w

j k

r

w


Cross Product

• An operation on two vectors in the three‐dimensional space

http://en.wikipedia.org/wiki/Cross_product


x

y

ij

k

dij

dt

dji

dt

ˆˆ ˆ

ˆ ˆ0 0

1 0 0

i j kd

i jdt

ˆˆ ˆ

ˆ ˆ0 0

0 1 0

i j kd

j idt

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