mbd week 1 2017 - lth · 2017-01-25 2 7 name copies signature alisic, rijiad bengtsson, frida...
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2017-01-25
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Multibody DynamicsSpring 2017
www.mek.lth.seEducation/Advanced Courses
MULTIBODY DYNAMICS (FLERKROPPSDYNAMIK) (FMEN02, 7.5p)
1
Course responsible and teacher
• Per Lidström, Mechanics, LTH
• 046-2220479
• Aylin Ahadi
2
MBD Class 2017
Alisic, Rijiad Lindqvist, Max
Bengtsson, Frida Lund, Mikael
Björklund, Viktor Nilsson, Eddie
Bring, Björn Niu, Mutong
Dahlström, Linus Svensson, Christoffer
De Groote, Alicia Svensson, Jakob
Gabrielsson, Jonas Thuresson, Pontus
Jiang, Tao Winnykamen, Yoan
Jin, Run Wittig, Cornelius
Johansson, David
Johansson, Jenny
Johari, Daniel
Kalkan, Sven
Khan, Shoajb
Kröger, Dennis
Larsson, Henrik
3 4
Name Mail‐adress Personal ID Signature
Alisic, Rijiad [email protected], Frida [email protected]örklund, Viktor [email protected], Björn [email protected]öm, Linus [email protected] Groote, Alicia [email protected], Jonas [email protected], Tao [email protected], Run [email protected], David [email protected], Jenny [email protected], Daniel [email protected], Sven [email protected], Shoajb [email protected]öger, Dennis [email protected], Henrik [email protected], Maxt [email protected], Mikael [email protected], Eddie [email protected], Mutong [email protected], Christoffer
[email protected], Jakob [email protected], Pontus [email protected], Yoan [email protected], Cornelius [email protected]
COURSE REGISTRATION
Multibody Dynamics (FMEN02) Spring 2017
5
Password: euler
Leonard Euler 1707-1783 6
2017-01-25
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7
Name copies Signature
Alisic, Rijiad
Bengtsson, Frida
Björklund, Viktor
Bring, Björn
Dahlström, Linus
De Groote, Alicia
Gabrielsson, Jonas
Jiang, Tao
Jin, Run
Johansson, David
Johansson, Jenny
Johari, Daniel
Kalkan, Sven
Khan, Shoajb
Kröger, Dennis
Larsson, Henrik
Lindqvist, Maxt
Lund, Mikael
Nilsson, Eddie
Niu, Mutong
Svensson, Christoffer
Svensson, Jakob
Thuresson, Pontus
Winnykamen, Yoan
Wittig, Cornelius
COURSE BOOK RESERVATION (HARD COPY)Fundamentals of Multibody Dynamics, Lecture NotesSEK 300-350
Course program
8
Multibody Dynamics (FMEN02, 7.5p), 2017 Course literature: Lidström P., Nilsson K.: Fundamentals of Multibody Dynamics, Lecture Notes (LN). Div. of
Mechanics, LTH, 2017. The ‘Lecture Notes’ will be on the Course web site. Hand out material: Solutions to selected Exercises, Examination tasks (Assignments), Project specification. Examples of previous written tests.
Teacher: Prof. Per Lidström, Ass. Prof. Aylin Ahadi (Lectures, Exercises, Course coordination) Phone: 046-2220479, email: [email protected] Schedule (w. 1-7): Lectures: Monday 10 - 12 Tuesday 15 - 17 Thursday 8 - 10, room M:IP2 Exercises: Wednesday 8 - 10, room M:X1a-b Schedule (w. 8): Lectures. Monday 8 - 10, 15 - 17, room M:IP2
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Weekly Timeplan 2017
Week 1
Hour
Monday
Tuesday
Wednesday
Thursday
Friday
8
Lecture
M:X1a-b
9
10
Lecture M:IP2
11
12
13
Lecture
M:IP2
14
15
Lecture
M:IP2
16 10
Weekly Timeplan 2017
Week 2
Hour
Monday
Tuesday
Wednesday
Thursday
Friday
8
Exercise
M:X1a-b
9
10
Exercise M:IP2
11
12
13
Lecture
M:IP2
14
15
Lecture
M:IP2
16
11
Weekly Timeplan 2017
Week 3-7
Hour
Monday
Tuesday
Wednesday
Thursday
Friday
8
Exercise
M:X1a-b
9
10
Lecture M:IP2
11
12
13
Lecture
M:IP2
14
15
Lecture
M:IP2
16 12
Weekly Timeplan 2017
week 8
Hour
Monday
Tuesday
Wednesday
Thursday
Friday
8
9
10
Lecture
M:IP2
11
12
13
14
15
Lecture
M:IP2
16
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13 14
Course objectives and contents
15
Course objectives: The objective of this course is to present the basic theoretical knowledge of the Foundations of Multibody Dynamics with applications to machine and structural dynamics. The course gives a mechanical background for applications in, for instance, control theory and vehicle dynamics. Course contents: 1) Topics presented at lectures and in the “Lecture Notes”. 2) Exercises. 3) Examination tasks 4) Application project The scope of the course is defined by the curriculum above and the lecture notes (Fundamentals of Multibody Dynamics, Lecture Notes). The teaching consists of Lectures and Exercises: Lectures: Lectures will present the topics of the course in accordance with the curriculum presented above. Exercises: Recommended exercises worked out by the student will serve as a preparation for the Examination tasks. Solutions to selected problems will be distributed.
Examination
16
Examination: The examination consists of two parts:
1. Hand in solutions to 3 examination tasks (assignments containing 2-4 problems each) and one Application project.
The examination tasks will be handed out starting by w.2. Last date for handing in solution to the examination tasks and the project report:
Examination task 1: Rigid body kinematics February 3rd Examination task 2: Rigid and flexible body dynamics February 17th Examination task 3: Multibody dynamics Mars 3rd Application project (with Adams software) Mars 17th
2. Written examination containing five (5) tasks focusing on the theoretical parts of the course content.
A correctly solved task gives 3p (Total possible score 5x3p = 15p). Date for the written examination: Monday, Mars 13, 14:00-19:00 in room: M:L1-L2. Credit requirements: Credit 3: Solutions to all examination tasks with 65% of the problems correctly solved. Approved project report. Written examination with 50% of the tasks correctly solved (7.5p). Credit 4: Solutions to all examination tasks with 75% of the problems correctly solved. Approved project report. Written examination with 67% of the tasks correctly solved (10p). Credit 5: Solutions to all examination tasks with 90% of the problems correctly solved. Approved project report. Written examination with 83% of the tasks correctly solved (12.5p).
CONTENTS
1. INTRODUCTION 1 2. BASIC NOTATIONS 12 3. PARTICLE DYNAMICS 13
3.1 One particle system 13 3.2 Many particle system 27
4. RIGID BODY KINEMATICS 36 4.1 The rigid body transplacement 36 4.2 The Euler theorem 47 4.3 Mathematical representations of rotations 59 4.4 Velocity and acceleration 72
5. THE EQUATIONS OF MOTION 89 5.1 The Newton and Euler equations of motion 89 5.2 Balance laws of momentum and moment of momentum 94 5.3 The relative moment of momentum 99 5.4 Power and energy 100
6. RIGID BODY DYNAMICS 104 6.1 The equations of motion for the rigid body 104 6.2 The inertia tensor 109 6.3 Fixed axis rotation and bearing reactions 138 6.4 The Euler equations for a rigid body 143 6.5 Power, kinetic energy and stability 158 6.6 Solutions to the equations of motion: An introduction to the case of
7. THE DEFORMABLE BODY 172 7.1 Kinematics 172 7.2 Equations of motion 196 7.3 Power and energy 205 7.4 The elastic body 209
8. THE PRINCIPLE OF VIRTUAL POWER 217 8.1 The principle of virtual power in continuum mechanics 219
17
1. THE MULTIBODY 222 9.1 Multibody systems 222 9.2 Degrees of freedom and coordinates 224 9.3 Multibody kinematics 229 9.4 Lagrange’s equations 232 9.5 Constitutive assumptions and generalized internal forces 252
9.6 External forces 261 9.7 The interaction between parts 267 9.8 The Lagrangian and the Power theorem 295
10. CONSTRAINTS 300 10.1 Constraint conditions 300 10.2 Lagrange’s equations with constraint conditions 305 10.3 The Power Theorem 323
11. COORDINATE REPRESENTATIONS 327 11.1 Coordinate representations 328 11.2 Linear coordinates 331 11.3 Floating frame of reference 359 APPENDICES A1 A.1 Matrices A1 A.2 Vector spaces A6 A.3 Linear mappings A12 A.4 The Euclidean point space A37 A.5 Derivatives of matrices A45 A.6 The Frobenius theorem A54 A.7 Analysis on Euclidean space A56 SOME FORMULAS IN VECTOR AND MATRIX ALGEBRA EXERCISES
18
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Mechanics in perspective
19
Quantum Nano Macro‐ContinuumMicro‐meso ‐
Å nm m mm
lengthscale
Prerequicites
20
Multibody Dynamics
Linear algebra
3D Analysis
Mechanics (Basic course
Continuummechanics
Matlab
Finite element method
Mathematics Mechanics Numerical analysis
Synthesis
Content
21
Multibody Dynamics
Flexible bodies
Rigid bodies
Coordinates
Lagrange’sequations
Constraints
Course objectives
22
an ability to
perform an analysis of a multibody system and to present the result in a written report
read technical and scientific reports and articles
on multibody dynamics some knowledge of
industrial applications of multibody dynamics
computer softwares for multibody problems
23
Industrial applications
24
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Commercial MBS softwares
• Adams
• Dads
• Simpac
• Ansys
• Dymola/Modelinc
• Simulink/Matlab
• RecurDyn
25
Adams Student EditionMultibody Dynamics Simulation
https://www.youtube.com/watch?v=hd-cFu7yMGw
Multibody system
26
Slider crank mechanism
Multibody system
27
Multibody system
Figure 1.3 Topology and coordinates of slider crank mechanism.
B
A C
Cy
Cx
B
A Slider
Crank
Rod
O
Bx
Guider Ground
Ground
O
, , , ,B C Cx x y Configuration coordinates:
4# Parts:
28
Configuration coordinatesRigid part in plane motion, 3DOF
, ,C Cx yConfiguration coordinates:29
Greubler formulaTwo dimensional
( )m
ii 1
f 3 N 1 r
f
N
ir
Number of parts:
Number of degrees of freedom (DOF):
Number of DOF removed by joint i:
Number of joints: m
ir 2Revolute joint: ir 2Translational joint:30
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Joints in three dimentions
( )m
ii 1
f 6 N 1 r
ir 5ir 5
31
Degrees of freedom and constraints
a) b)
Figure 1.5 a) Three-bar mechanism b) The double pendulum.
B
A D
C
Ground
A Ground
B
g
32
Greubler formula
( ) , m
ii 1
f 3 N 1 r f 1
Slider crank mechanism:
Three bar mechanism:
Double pendulum:
,N 4
( )f 3 4 1 4 2 1
( )f 3 4 1 4 2 1
,m 4 ir 2
,N m 4 ir 2
,N 3 ,m 2 ir 2
( )f 3 3 1 2 2 2
B
A D
C
Ground
A
Ground
B
g
33
Biomechanics
,N 11 ,m 10 ir 2 ( )f 3 11 1 10 2 30 20 10
34
Multibody dynamics software Adams
35
The rigid body
ec
ec c c
m
F a
M I ω ω I ω
Figure 1.8 The rigid body.
ω
P
tB
c
cPp
cv
36
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The flexible body
( (tr ) ) ( )0
22e
1U dv X
2 E E
B
37
, , ... , 1 2 nq q q q
The Multibody
( )U U q
( ) , ,...,kk k k
d T T UQ 0 k 1 n
dt q q q
( , )k kQ Q q q
Configuration coordinates:
Kinetic energy: Potential energy:
Equations of motion: (Lagrange’s equations)
38
( , )T T q q
Constraints
cos cos
sin sin
C
C
LR x 0
2L
R y 02
cos
sin
B C
C
Lx x 0
2L
y 02
Constraint at A: Constraint at B:
, , , ,B C Cx x y Configuration coordinates:
39
Equations of motion under constraints
, ,...,n
k0 k
k 1
g g q 0 1 m
( , )k kg g t q
Constraint equations: 1 m n
Equations of motion:
( ) , ,...,
, ,...,
mnonk kk k
1
nk
0 kk 1
d L LQ g 0 k 1 n
dt q q
g g q 0 1 m
L T U
( , , )non nonk kQ Q t q q
( ), ,...,t 1 m
Lagrangian:
Nonconservative forces:
Constraint forces:40
Chapter 3: Particle Dynamics
is left for self study and recapitulation of basic concept i mechanics.
You may try to solve some of the exercises but don’t spend to much
time on this introductory chapter!
m v
1F
41 42
Euclidean space
, , ...x y z E Points in Euclidean space
y x u V Vectors in Translation vector space
y x u
dim( ) 3V 1 2 3e e e e basis
[ ]1
1 1 2 2 3 3 1 2 3 2
3
a
a a a a
a
ea e e e e e e e a
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43
Cartesian coordinate system
oE Fixed point, 1 2 3e e e e Fixed basis
[ ] [ ]ox 1 1 2 2 3 3 oxx o o x x x o o x e er e e e e r e
oE Fixed point, 1 2 3f f f f Fixed basis
[ ]x o x ff [ ]o o o ee
[ ] [ ] [ ] [ ] ([ ] [ ] )x o o x o x x x o e f e f e ee f e f e
, , det3 3B B B 0 f e 44
[ ] ([ ] [ ] ) [ ] [ ] [ ]B x x o B x x o f e e f e ee e
[ ] ([ ] [ ] )1x B x o f e e
Cartesian coordinate system
Transformation of coordinates for points
[ ] ([ ] [ ] )x B x o e f f
45
What about vectors?
[ ] , ea e a [ ] , fa f a Bf e
[ ] [ ] [ ] [ ] [ ]B B f f e e fa f a e a e a a a
[ ] [ ] , [ ] [ ]1B B e f f ea a a a
a V
46
What about tensors?
[ ] , eAe e A [ ] , fAf f A Bf e
[ ] [ ]1B Bf eA A
[ ] [ ] 1B Be fA A
( )EndA V
TB e f
47
1 2 3e e e e
Orthonormal basis
i j ij
1 i j
0 i j
e e
1 1 1 1 2 1 3
T2 1 2 3 2 2 2 2 2 3
3 3 1 3 2 3 3
1 0 0
0 1 0
0 0 1
e e e e e e e
e e e e e e e e e e e e
e e e e e e e
T3 3I e e
48
,1 2 3e e e e
Orthonormal bases
, detT3 3B B I B 1
1 2 3f f f f
Bf e
Orthonormal matrix
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49
( , , )det det
( , , )
V
V
e
Qa Qb QcQ Q
a b c
The determinant
1 2 3e e e eON-basis:
Rigid body kinematics
( ) , ,AB t constant A B p Rigidity condition:
Figure 4.1 Rigid body motion.
0B
B
A
B
( )AB 1tp A
B
( )AB 2tp
50
Rigid body transplacement
O
A
A
Reference placement
Present placement
0B
B
( )A Ap r Ar
Au
( ), A A A p r 51
Rigid body transplacement
( ) : ( ) ( ), A A AR a r a r a V
( ) ( ) , ,B A B A A B r r r r
( ), A A A p r
Rigidity condition:
Transplacement:
( )
( ) ( ) , ,
( ) ,
A
A A
A
R
R R
R
0 0
a b a b a b
a a a
V
V
52Isometry
( ) ( ) ( ), , , ,A A AR R R a b a b a b V
Rigid body transplacement
Linearity:
( )A AR a R a
Isometry implies orthonormality:
,T TA A A A R R R R 1V
1 TA A R R
Notation:
53
Rigid body transplacement
( ) ( ) , A A A r a r R a a V
P A a r r
( ) ( ) ( )P A A P A r r R r r
( )P A A P A p p R r r
AP A APp R r
54
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Rigid body transplacement
AB A ABp R r
A
B
B
A
0B
B
ABr
AB ABp Rr
( )A SOR V
55
( ) ( ) , B B B r a r R a a V
Rigid body transplacement
( ) ( ) , A A A r a r R a a V
?B AR R
Does the rotation tensor depend on the reduction point?
!B A R R R
56
No!
Rigid body transplacement
Theorem 4.1 A rigid body transplacement is uniquely determined by the displacement of (anarbitrary) reduction point A; Au and an orthonormal transformation ( )SOR V which isindependent of the reduction point.
( )P A AP u u R 1 r
57
Example 4.1 (Plane rigid transplacement, cf. section 4.2.2) The same rigid body transplacement is visualized in both Figure 4.6 a) and 4.6 b) below with different choices of reduction point.
Transplacement with reduction point A.
B
B
A A A
Transplacement with reduction point B.
Figure 4.6 Plane rigid body transplacement.
Au
Bu
A A
B
B B
58
Summary
Box 4.1: Rigid body transplacement
:
:
: detA
A reduction point
displacement of the reducion point A
orthogonal transformation with 1u
R R
( ) ( )( )A A u u r u R 1 r r
( ) ( ) ( )A A p r r R r r
59
The translation
, , ,A B 0 A B u u u
Proposition 4.2 A translation is a rigid transplacement determined by a displacement 0u ,
independent of the reduction point, and an orthonormal transformation R equal to the unittensor.
0B
A
B
A
B
ABp
0u
0u
B
ABr
R 1
60
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The rotation
( , )A n
Reference placement Present placement
n
B
0B B
B
A A
a
b
ABr ABp
Rotation axis:
61
The rotation Figure 4.9 Rotation around a fixed axis.
1i
2i ABr
A
B
3 i n A
1e 2e
B
ABp
3 3 e i n 1x
2x
( )AB AB AB r r n n r
AB 1 1 2 2 r i ix x
( )AB AB AB p p n n p
AB 1 1 2 2 p e ex x
( )2 2 2AB AB 1 2 AB r p n rx x
62
The rotation Figure 4.9 Rotation around a fixed axis.
1i
2i ABr
A
B
3 i n A
1e 2e
B
ABp
3 3 e i n 1x
2x
cos sin
( sin ) cos , 1 1 2
2 1 2
3 3
0 2
e i i
e i i
e i n
63
The rotation matrix
i
e i R
cos sin
sin cos
0
0
0 0 1
iR
,T T i i i i
R R R R 1 det 1i
R
cos sin
( ) ( ) sin cos ,1 2 3 1 2 3
0
0
0 0 1
e e e i i i
64
1 1
2 2
3 3 3
R i e
R i e
R i e i n
i
R i e i R
AB ABp Rr
1i
2i ABr
A
B
3 i n A
1e 2e
B
ABp
3 3 e i n 1x
2x
The rotation tensor
65
Summary
Box 4.3: Rotation
( , ) :
:
:
cos sin
: , sin cos
:
( ) :
A
1 2 3
A rotation axis
A reduction point
displacement of the reduction point A
0
orthonormal transformation 0
0 0 1
rotation angle 0 2
referen
i
n
u 0
R R
i i i i , 3 3tial base i n e
66
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The Euler theorem
A 0B
n
axis ( , )A n
C
i
Theorem 4.2 (The Euler theorem, 1775) Every rigid transplacement with a fixed point isequal to a rotation around an axis through the fixed point.
67
The canonical representation
Rn n
cos sin
sin cos
0
0
0 0 1
iR
( ), 1 2 3 3 i i i i i n
68
i
R i i R
cos sin
( ) sin cos1
T1 2 3 2
3
0
0
0 0 1
i
i
R i R i i i i i
i
69
The tensor product
( )End a b V, a b V,
( ) ( ) a b u a b uDefinition:
Notation:
70
The vector product
a b V, a b V,
( ) a 1 u a uDefinition:
Notation:
b
a b
a
( )Skew a 1 V, ( ),End a 1V VNotation:
sin a b a b
71
( cos + sin ) +( ( sin )+ cos ) )1 2 1 1 2 2 3 3 i i i i i i i i
( ) cos ( )sin3 3 1 1 2 2 2 1 1 2 R i i i i i i i i i i
( cos + sin ( sin )+ cos )1
1 2 1 2 3 2
3
i
i i i i i i
i
The canonical representation
cos sin
( ) sin cos1
1 2 3 2
3
0
0
0 0 1
i
i i i i
i
cos ( ) sin n n 1 n n n 1
( ) det( ) det( )p R i1 R 1 R
cos sin
det sin cos ( )(( cos ) sin )2 2
0
0 1
0 0 1
,( ) , i1 2 3p 0 1 e R
The characteristic equation, eigenvalues
72
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Summary
Box 4.5: Rigid transplacement
:
:
:
:
:
A
A
A reduction point
position vector of the reduction point in the reference placement
displacement of the reducion point
position vector of a material point in the reference placement
p
r
u
r
p
:
osition vector of a material point in the present placement
rotation tensorR
( ) ( )A A A
translation rotation
p r r u R r r
73
Screw representation of the transplacement
Screw line
, S A ASS r r r n SL
Su
A
S
nASr
ASr
Theorem 4.3 (Chasles’ Theorem, 1830) Every transplacement of a rigid body can be realized by a rotation about an axis combined with a translation of minimum magnitudeparallel to that axis. The reduction point for this so-called screw transplacement is located onthe screw line given in (4.28).
( ),
( cos )
TA
AS 02 1
R 1 u
r
( )S Ss p r n R r r
74
75
Weekly Timeplan 2017
Week 2-7
Hour
Monday
Tuesday
Wednesday
Thursday
Friday
8
Exercise
M:X1a-b
9
10
Lecture M:IP2
11
12
13
Lecture M:IP2
Lecture
M:IP2
14
15
Lecture
M:IP2
16
The special orthogonal group
The set of all orthonormal tensors on V forms an algebraic group, the orthonormal group,
( ) T TO Q Q Q QQ 1= V
Being a group means that
, ( ) ( )
( )
( )
1 2 2 1
1 T
O O
O
O
Q Q Q Q
1
Q Q
V V
V
V
However, in general we have 1 2 2 1Q Q Q Q and we say that the operation of combining
orthonormal tensors is non-commutative. The set of all rotations forms a subgroup in ( )O V called the special orthonormal group
( ) ( ) detSO O 1 R R= V V
76
Non-commutativity of rotations
77
The main representation theorem
78
This representation is mainly unique. If R 1 then n may be arbitrarily chosen, while( )0 modulo 2 . If R 1 then n is uniquely determined, apart from a factor 1 . For a
given n the angle is uniquely determined ( )modulo 2 . For a definition of the tensor
( )End a 1 V see Appendix A.4.39.
Theorem 4.1 To every rotation tensor ( )SOR V there exists a unit vector n and an angle so that
( , ) cos ( ) sin R R n n n 1 n n n 1
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79
The topology of SO(V )
P
O
x
y
z
P
P
x
y
z
O
( )1 2 3e e e e 1
2
3
n
n
n
en
en e n
The main representation theoremusing components
1 1 1 1 2 1 3
T
2 1 2 3 2 1 2 2 2 3
3 3 1 3 2 3 3
n n n n n n n
n n n n n n n n n n
n n n n n n n
e e en n n n
3 2
3 1
2 1
0 n n
n 0 n
n n 0
e e en 1 n 1
80
The main representation theoremusing components
cos ( ) sin e e e e
R n n 1 n n n 1
cos ( ) sinT T
e e e e e en n 1 n n n 1
cos ( ) sin R n n 1 n n n 1
( )1 2 3e e e e 1
2
3
n
n
n
en
en e n
81
How to calculate the rotation axis and the rotation angle from the rotation tensor
( ) ( ) b b
Rn n R 1 n 0 R 1 n 0
11 12 13
3 321 22 23
31 32 33
R R R
R R R
R R R
bR
, ?R n
1
3 12
3
n
n
n
bn
( )1 2 3b b b b
82
83
The eigenvalue problem
11 12 13 1
21 22 23 2
31 32 33 3
R R R n 0
R R R n 0
R R R n 0
,( ) , i1 2 3p 0 1 e R
cos , sin2 3 2 3
2 2i
( ) det11 12 13
21 22 23
31 32 33
R R R
p R R R
R R R
R
The eigenvalue problem
11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
R 1 R R n 0 n
R R 1 R n 0 n
R R R 1 n 0 n
n
cos11 22 33tr R R R 1 2 R
arccos( ) arccos( )
arccos( )
11 22 33
11 22 33
R R R 1tr 1
2 2R R R 1
22
R
84
1 1
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Coordinate representations of ( )SO V
• Euler angles
• Bryant angles
• Euler parameters
• Quaternions
• ……
85 86
Euler angles: ( , , )
0B
1e
2e 3e
R
A 01e
02e
03e
A
B
Present placementReference placement
o
o o e
e Re e R
87
01e
02e
03 3f e
1f
2f
Factorization:
3g
3f
1 1g f
2f
2g
( , , )
2e
1e
3 3e g
1g
2g
( ) ( ) ( )3 1 3 R R R R
( )3 R ( )1 R ( )3 R
Euler angles: First rotation
01e
02e
03 3f e
1f
2f
cos sin
( ) sin coso3
0
0
0 0 1
eR ( ) ( ) ,o
o o3 3
ef R e e R
( )
( )
0 0 0 01 2 3
1 2 3
e e e e
f f f f
88
Euler angles: Second rotation
3g
3f
1 1g f
2f
2g
( ) cos sin
sin cos1
1 0 0
0
0
fR ( ) ( ) ,1 1
fg R f f R
( )
( )
1 2 3
1 2 3
f f f f
g g g g
89
Euler angles: Third rotation
2e
1e
3 3e g
1g
2g
( ) ( ) ,3 3 g
e R g g R cos sin
( ) sin cos3
0
0
0 0 1
gR
( )
( )
1 2 3
1 2 3
g g g g
e e e e
90
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91
Euler angles: All rotations in combination
( ) ( )
( ) ( )
( ) ( )
o
o o3 3
1 1
3 3
e
f
g
f R e e R
g R f f R
e R g g R
)
( ) ( ) ( ) ( ) ( ) ( )o
o
o o3 1 3 3 1 3
e
g f g e f g
R
e Re g R f R R e R R R
( ) ( ) ( ) ( ) ( ) ( )oo o o
3 1 3 3 1 3 e f g
e Re R R R e e R R R
Euler angles: The rotation
( ) ( ) ( )o o3 1 3 e e f g
R R R R
cos sin cos sin
sin cos cos sin sin cos
sin cos
0 1 0 0 0
0 0 0
0 0 1 0 0 0 1
cos cos sin cos sin cos sin sin cos cos sin sin
sin cos cos cos sin sin sin cos cos cos cos sin
sin sin sin cos cos
o e f g e
92
Euler angles
Theorem 4.2 A rotation R is determined by a set of three real coordinates, called the Euler
angles: ( , , ) , , , 3 0 2 0 0 2 . Conversely these angles are
essentially determined by R . In fact the angle ,0 is uniquely determined by R . If
,0 then also , ,0 2 are uniquely determined by R . When 0 only is
uniquely determined and when only . 93
cos cos sin sin cos sin sin cos
sin cos cos sin sin sin cos coso
0
0
0 0 1
eR
cos( ) sin( )
sin( ) cos( )11 12 13
21 22 23
31 32 33
0 R R R
0 R R R
0 0 1 R R R
Euler angles: Singularity
Only is determined. Singular!
,0
94
Euler angles: Singularity
cos , sin cos
cos , sinsin sin
cos , sinsin sin
233
23 13
32 31
R 1
R R
R R
o
11 12 13
21 22 23
31 32 33
R R R
R R R
R R R
eR
cos cos sin cos sin cos sin sin cos cos sin sin
sin cos cos cos sin sin sin cos cos cos cos sin
sin sin sin cos cos
, singular0
95
The rotation angle
( cos )cos( ) coscos
1 1
2
cos ( cos )cos( ) costr 1 2 1 R
96
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Euler parameters
sin ( ) sin cos22 22 2 2
R 1 n n 1 n 1
cos , sin0e2 2
e n
cos ( ) sin R n n 1 n n n 1
( cos )( ) sin1 1 n n 1 n 1
cossin2 1
2 2
sin sin cos2
2 2
97
Euler parameters
( ), RON-basis,1 2 3b b b b ,1 1 2 2 3 3n n n n b b b
cos , sin , sin , sin
1 1 2 2 3 3
0 1 1 2 2 3 3
e e e
e e n e n e n2 2 2 2
e b b b
3S
0e
3e4
( , , , ) ,
4 2 2 2 20 1 2 3 0 1 2 3
3 4 2 2 2 20 1 2 3
e e e e e e e e e 1
S e e e e e 1
( )( ) 02 2e R 1 e 1 e 1 e 1
98
1n
3 2
0 3 1
2 1
0 e e
2e e 0 e
e e 0
( )3 2 3 2
3 1 3 1
2 1 2 1
1 0 0 0 e e 0 e e
e 0 1 0 2 e 0 e e 0 e
0 0 1 e e 0 e e 0
bR
( ) ( , ) ( ) , ( , )2 30 0 0e e 2 2e e e S R R R e 1 e 1 e 1 e
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
2 20 1 1 2 0 3 1 3 0 2
2 21 2 0 3 0 2 2 3 0 1
2 21 3 0 2 2 3 0 1 0 3
2 e e 1 2 e e e e 2 e e e e
2 e e e e 2 e e 1 2 e e e e
2 e e e e 2 e e e e 2 e e 1
How to calculate matrix elements fromEuler parameters
99
( ), RON-basis,1 2 3b b b b
o
11 12 13
21 22 23
31 32 33
R R R
R R R
R R R
eR
How to calculate Euler parameters from matrix elements
, , ,
2 11 22 330
2 ii ii 11 22 33i
R R R 1tr 1e
4 4R 1 2R R R Rtr 1
e i 1 2 32 4 4
R
R
There is a 2 -1 correspondence between Euler parameters and rotation matrices!
No singularity!
100
( , , , ) ( , , , )2 2 2 20 1 2 3 0 1 2 3e e e e 1 e e e e 0 0 0 0
,32 231
0
R Re
4e
11 22 330
R R R 1e 0
4
,13 312
0
R Re
4e
21 12
30
R Re
4e
How to calculate Euler parameters from matrix elements, condinued
101
,1 2 21 124e e R R
11 22 330
R R R 1e 0
4
,1 3 31 134e e R R 2 3 32 234e e R R
,21 122
1
R Re
4e
, 11 22 330 1
1 R R Re 0 e 0
4
31 133
1
R Re
4e
How to calculate Euler parameters from matrix elements
102
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, , 22 11 330 1 2
1 R R Re 0 e 0 e 0
4
31 133
2
R Re
4e
How to calculate Euler parameters from matrix elements
103
, , , 33 11 220 1 2 3
1 R R Re 0 e 0 e 0 e 0
4
The composition of rotations
Let 1R and 2R be two rotations. The composition of 1R and 2R is given by 2 1R R R and this represents another rotation. We have the following Proposition 4.1 If ,( , )1 1 0 1 1eR R e , ,( , )2 2 0 2 2eR R e and 2 1R R R then ( , )0eR R e
where , ,0 0 1 0 2 1 2e e e e e , , ,2 0 1 1 0 2 1 2e e e e e e e (4.18)
Corollary 4.1 Two rotations commute if and only if their rotation axes are parallel, i.e. 2 1 1 2 1 2 R R R R e e
104
The exponential map
exp( ) : ..., ( )! !
2 31 1End
2 3 A 1 A A A A V
... exp( ) exp( )! !
2 32 3
2 3
R 1 N N N N n 1
( )Skew N n 1 V
( )SOR n V
105
The exponential map:
( ) with corresponding ( )SO Lie group Lie algebra SkewV V
O
A
A
Reference placement
Present placement 0B
tB
( )A A tp pAr
P
P
Pr
( )P P tp p
Time-dependent transplacementMotion
( )
( , ) ( ) ( )( ) ( ) ( )( ),
A
P P A A P A A P A
t
t t t t t t 0
p
p r r u R r r p R r r106
Rigid body velocity
, P A AP P v v ω p
( )Taxω RR
ω
P
tB A
APp
Av
d
AP AP p ω p
TW RRSpin tensor:
Angular velocity:
Rigid body velocity field:
107 108
Rigid body rotation
( )T W RR R WR ω 1 R ω R
( ) (exp( ( ) ))( )
t
0
0 0
t s ds0
R ω RR R ω 1 R
R R
Thus, if the angular velocity is known and the rotation tensoris known at time then the rotation tensor : is known.
( )tω ωt 0 ( )tR R
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α ω
( ), P A AP AP P a a α p ω ω p
α
ω
P
tB A
APp
Aa
APα p
( )AP ω ω p
Rigid body acceleration
Angular acceleration:
109
Tangential acceleration
Centripetal acceleration
α ωSpecial case:
01e
02e
03e
1e
2e3e
ω R
Euler-Poisson velocity formulafor moving bases
, , 0 0 01 1 2 2 3 3 e Re e Re e Re
( )1 2 3e e e e( )0 0 0 01 2 3e e e e
0
e eR R
110
01e
02e
03e
1e
2e3e
ω R
Euler-Poisson velocity formulafor moving bases
Theorem 4.4 (Euler-Poisson velocity formula for moving bases) , , ,i i i 1 2 3 e ω e (4.32)
where
3
i ii 1
1
2
ω e e (4.33)
111
Euler-Poisson velocity formulafor moving vectors and tensors
,
( ) ( ) ( ) ( )3
j iji j 1
t t t A t
iA A e e
( ) rel A ω A A ω 1 A
rel a ω a a
3
reli
i 1
a
iea e a
( ) ( ) ( )3
ii 1
t t a t
ia a e
,
( ) ( ) ( )3
relj ij
i j 1
t t A t
ieA e e A
112
Angular acceleration
3rel
ii 1
iω ω e
rel rel ω ω ω ω ω
3
ii 1
iω e
1e
2e3e
ω
113
Proposition 4.6 The angular velocity ω , expressed in Euler angles, is given by 1 1 2 2 3 3 ω e e e
where
sin sin cos
sin cos sin
cos
1
2
3
(4.39)
and ( )1 2 3e e e e is an RON-basis following the rotation of the rigid body (‘fixed to the
body’).
Angular velocity and Euler angles
114
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Combining rotationsThe addition of angular velocities, version 1
( )T1 1 1axω R R
( )1 1 tR R
2 2 1 ω ω R ω
( )2 2 tR R 2 1R R R
( )T2 2 2axω R R ( )Taxω RR
115
Addition rule 1
Combining rotationsThe addition of angular velocities, version 2
1R
2R
1f
2f 3f
1g
2g
3g
0feω
0geω
gfω
01e
02e
03e
(( ) )T2 2axgf fω R R
1 0f R e 2g R f
( )0
T11 1ax feω ω R R
( )0
Tax geω ω R R
0g Re 2 1R R R
116
Combining rotationsThe addition of angular velocities, version 2
1R
2R
1f
2f 3f
1g
2g
3g
0feω
0geω
gfω
01e
02e
03e
0 0 ge gf feω ω ω
117
Addition rule 2
Summary
Box 4.8: Velocity and acceleration
( )
( )
, ( ) ,
A A
A
T
Rigid body velocity field
Rigid body acceleration field
ax Spin tensor Angular velocity
v v ω p p
a a α p ω ω p
W RR ω W
α ω
, , ,
i i
rel
Angular acceleration
i 1 2 3 Euler Poisson for vectors
Absolute and
e ω e
u ω u u
( )
sin sin cos
sin cos sin
cos
1 1 2 2 3 3
1
2
3
relative velocity
Euler Poisson for tensors
Angular velocity in Euler angles
eA ω A A ω 1 A
ω e e e
e ( ) -
( )
0 0
1 2 3
2 2 1
RON basis fixed to the body
Addition of angular velocities 1
gfge fe
e e e
ω ω R ω
ω ω ω ( ) Addition of angular velocities 2
118
Derive the matrix representation e
a b of the tensor product a b relative to the
ON-basis ( )1 2 3e e e e . The matrix representations of , a b V are given by
1
2
3
a
a
a
ea and
1
2
3
b
b
b
eb (1)
Exercise 0:1
119
Show that if , a b V and ( )EndA V then ( )T a b b a , ( ) T a b A a A b , ( ) ( ) A a b Aa b ( )( ) a b c d a d b c , det( ) 0 a b , ( )tr a b a b (2)
Exercise 0:2
120
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Let (EndA V) and u V with the matrix representations eA and e
u , respect-
tively, in a RON-basis ( )1 2 3e e e e , i.e. e
A e e A and e
u e u . Show that
e e e
Au A u (1)
Exercise 1:1
121
Let ( )SkewA V and let ( )ax a A V be its axial vector, i.e.
, Au a u u V (2)
Assume that T
1 2 3a a ae
a in a RON-basis ( )1 2 3e e e e . Show that
3 2
3 1
2 1
0 a a
a 0 a
a a 0
eA (3)
Exercise 1:2
122
Exercise 1:4
Let , ( )1 2 SOR R V be rotations with the following matrix representations 1 1
eR e e R and 2 2
eR e e R (5)
where ( )1 2 3e e e e is an ON-basis. Show that
2 1 2 1
e e eR R R R (6)
123
Exercise 1:5
Let , ( )1 2 SOR R V be rotations with the following matrix representations
1 1e
R e e R and 2 2f
R f f R (7)
where 1f R e and ( )1 2 3e e e e is a RON-basis. Show that
2 1 1 2e e f
R R R R (8)
124
Consider a rotation tensor ( )SOR V and its characteristic polynomial
( ) det( )p R 1 R . Since this is a third order polynomial there are three roots
, , 1 2 3 to the characteristic equation
( )p 0 R (9) We know that one of the roots is equal to 1, let’s say 3 1 . Show that the rotation
angle is given by
arccos 1 2
2
(10)
Exercise 1:6
125
Exercise 1:8
Consider the transplacement of a rigid body with the rotation tensor R n , . Show that (“Rodrigues formula”) ( ), ,AB AB AB AB A B p r c p r (15) where ( )axc C is the Cayley vector corresponding to the Cayley tensor C defined by ( )( ) ( ) ( )1 1 C R 1 R 1 R 1 R 1 (16) Show that the tensor R 1 is invertible and that C is anti-symmetric.
126
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A rigid translacement of a body is governed by the translation A u V and the rotation ( )SOR V where A is a fixed reduction point. Which of the following statements are true and which are false? Give a simple motivation for your answer!
a) , 0 Ra a a V .
b) , P A AP P u u Rr c) exp( ) R w 1 for some vector w V .
d) T R R 1 . e) T
1 1 2 2 3 3 R f e f e f e f e for some RON-basis ( )1 2 3e e e e and
( )1 2 3f f f f .
Exercise 1:9
127
Given a rotation tensor ( , , ) R R where , , are Euler angles. Calculate the
matrices ( , , ) 015 35 60 e
R and ( , , )0
0115 35 60 e
R e .
Exercise 1:12
128
Exercise 1:12
The rotation matrix expressed in Euler angles:
R ( )
cos ( )
sin ( )
0
sin ( )
cos ( )
0
0
0
1
1
0
0
0
cos ( )
sin ( )
0
sin ( )
cos ( )
cos ( )
sin ( )
0
sin ( )
cos ( )
0
0
0
1
Conversion from degrees to radians:
Rd ( ) R
180
180
180
Calculation of the rotation matrix:
Rd 15 35 60( )
0.299
0.815
0.497
0.943
0.171
0.287
0.148
0.554
0.819
Checking the rotation matrix:
Rd 15 35 60( ) 1Rd 15 35 60( ) Rd 15 35 60( )T
1
0
0
0
1
0
0
0
1
Calculating the rotation of the first basis vector:
Rd 15 35 60( )
1
0
0
0.299
0.815
0.497
129 130
Exercise 1:14
131
Exercise 1:14
01e
02e
03e
1e
2e
3e
n
Example 4:1 The wheel
c vv i p r ω j k 132
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133
1. The relative position vector ABr of material points A and B in the reference configuration of a body is given by
1 2 32 3 4AB r i i i (21)
in the RON-basis system 1 2 3( )i i i i . The body is rotated the angle 45 about an
axis ( , )A n where
1 2 3
1 1 1
3 3 3 n i i i (22)
Determine the matrix i
R of the rotation tensor R and the matrix AB ip of the
transplaced vector AB ABp Rr .
Exercise 1:17
17.