Mathematics of rotation in 3d space, a lecture that I've prepared.
This presentation is at a Undergraduate in Science (Math, Physics, Engineering) level.
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2. 2
ROTATIONS
TABLE OF CONTENT
SOLO
Rotation of a Rigid Body
Mathematical Computation of a Rotation
Rotation Matrix
Computation of the Rotation Matrix
Consecutive Rotations
Decomposition of a Vector in Two Different Frames of Coordinates
Differential Equation of the Rotation Matrices
Computation of the Angular Velocity Vector from .AB←ω
( ) ( )nRtC x
B
A
ˆ,33 θ−=
Computation of and as functions of .AB←ω
td
dθ
θ =
td
nd
n
ˆ
ˆ =
•
Quaternions
Computation of the Rotation Matrix
Definition of the Quaternions
Product of Quaternions
Rotation Description Using the Quaternions
3. 3
ROTATIONS
TABLE OF CONTENT (continue – 1)
SOLO
Rotation as a Multiplication of Two Matrices
Relations Between Quaternions and Euler Angles
Description of Successive Rotations Using Quaternions
Differential Equation of the Quaternions
Computation of as a Function of the Quaternion and its Derivatives
( )
( )t
B
AB←ω
Computation of as a Function of , and their Derivatives( )
( )t
B
AB←ω
θ nˆ
Differential Equation of the Quaternion Between Two Frames A and B
Using the Angular Velocities of a Third Frame I
Euler Angles
The Piogram
Successive Euler Rotations
321 →→ 231 →→ 312 →→ 132 →→ 213 →→ 123 →→
121 →→ 131 →→ 212 →→ 232 →→ 313 →→ 323 →→
4. 4
ROTATIONS
TABLE OF CONTENT (continue – 2)
SOLO
Cayley-Klein (or Euler) Parameters and Related Quantities
Rotation Matrix in Three Dimensional Space
Euler Parameters
Elementary Features of the 2x2 Rotation Matrix
Gibbs Vector
Differential Equation of Gibbs Vector
References
5. 5
ROTATIONS
Rotation of a Rigid Body
SOLO
23r
31r
12r1
3
2
P
P
1
2
331r
23r
12r
A rigid body in mechanics is defined as a system of mass points subject to the
constraint that the distance between all pair of points remains constant through
the motion.
To define a point P in a rigid body it is enough to specify the distance of this point
to three non-collinear points. This means that a rigid body is completely defined
by three of its non-collinear points. Since each point, in a three dimensional space
is defined by three coordinates, those three points are defined by 9 coordinates.
But the three points are constrained by the three distances between them:
313123231212 && constrconstrconstr ===
Therefore a rigid body is completely defined by 9 – 3 = 6 degrees of freedom.
6. 6
ROTATIONS
Rotation of a Rigid Body (continue – 1)
SOLO
We have the following theorems about a rigid body:
Euler’s Theorem (1775)
The most general displacement of a rigid body with one point fixed is equivalent to
a single rotation about some axis through that point.
Chasles’ Theorem (1839)
The most general displacement of a rigid body is a translation plus a rotation.
Leonhard Euler 1707-1783
Michel Chasles 1793-1880
7. 7
ROTATIONS
Rotation of a Rigid Body (continue – 2)
SOLO
Proof of Euler’s Theorem
O – Fixed point in the rigid body
A,B – Two point in the rigid body at equal
distance r from O.
== rOBOA
__________
A’,B’ – The new position of A,B respectively.
Since the body is rigid rOBOA ==
__________
''
Therefore A,B, A’,B’ are one a sphere
with center O.
α – plane passing through O such that A and A’ are at the same distance from it.
β – plane passing through O such that B and B’ are at the same distance from it.
PP’ – Intersection of the planes andα β
The two spherical triangles APB and A’PB’ are equal.
The arcs AA’ and BB’ are equal. That means that rotation around PP’ that
moves A to A’ will move B to B’.
q.e.d.
8. 8
ROTATIONS
Mathematical Computation of a Rotation
SOLO
A
B
C
O
θ
φφ
nˆ
v
1v
We saw that every rotation is defined by three parameters:
• Direction of the rotation axis , defined by two parameters.nˆ
• The angle of rotation , defines the third parameter.θ
Let rotate the vector around by a large angle , to
obtain the new vector
→
= OAv
nˆ θ→
=OBv1
From the drawing we have:
→→→→
++== CBACOAOBv1
vOA
=
→
( ) ( )θcos1ˆˆ −××=
→
vnnAC
Since direction of is: ( ) ( ) φν sinˆˆ&ˆˆ =×××× vnnvnn
and it’s length is:
AC
→
( )θφ cos1sin −v
( ) θsinˆ vnCB
×=
→
Since has the direction and the
absolute value
CB
→
vn
׈
θφsinsinv
( ) ( ) ( ) θθ sinˆcos1ˆˆ1 vnvnnvv
×+−××+=
9. 9
ROTATIONS
Computation of the Rotation Matrix
SOLO
We have two frames of coordinates A and B defined
by the orthogonal unit vectors and{ }AAA zyx ˆ,ˆ,ˆ { }BBB zyx ˆ,ˆ,ˆ
The frame B can be reached by rotating the A frame
around some direction by an angle .nˆ θ
We want to find the Rotation Matrix
that describes this rotation from A to B.
( )θ,ˆ33 nRC x
B
A =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) θθ
θθ
θθ
sinˆˆcos1ˆˆˆˆˆ
sinˆˆcos1ˆˆˆˆˆ
sinˆˆcos1ˆˆˆˆˆ
AAAB
AAAB
AAAB
znznnxz
ynynnxy
xnxnnxx
×+−××+=
×+−××+=
×+−××+=
Let write those equations in matrix form.
( )
[ ]( )
[ ]( )
( ) [ ]( )
×+
−××+
=
0
0
1
sinˆ
0
0
1
cos1ˆˆ
0
0
1
ˆ θθ
AAAA
B nnnx
[ ]( )
−
−
−
=×
0
0
0
ˆ
xy
xz
yz
A
nn
nn
nn
n [ ] 0ˆ =×ntrace
Rotation Matrix
11. 11
ROTATIONS
Computation of the Rotation Matrix (continue – 2)
SOLO
Ax
Az
Ay
Bz
By
Bx
O
nˆ
θ
θ
θ
θ
[ ] [ ]( )
[ ]( )
( ) [ ]( )
{ } ( )θθθ ,ˆsinˆcos1ˆˆ 3333 nRnnnICC x
AAA
x
A
B
A
B
=×+−××+==↑
The matrix has the following properties:[ ]( )A
n׈
[ ]( )
{ } [ ]( )ATA
nn ×−=× ˆˆ
[ ]( )
[ ]( )
=
−−
−−
−−
=
−
−
−
−
−
−
=××
22
22
22
0
0
0
0
0
0
ˆˆ
yxzyzx
zyzxyx
zxyxyz
xy
xz
yz
xy
xz
yz
AA
nnnnnn
nnnnnn
nnnnnn
nn
nn
nn
nn
nn
nn
nn
T
x
zzyzx
zyyyx
zxyxx
nnI
nnnnn
nnnnn
nnnnn
ˆˆ
000
010
001
33
2
2
2
+−=
+
−= [ ]( )
[ ]( )
( ) 213ˆˆ −=+−=××
AA
nntrace
[ ]( )
[ ] [ ] nn
nn
nn
nn
nnnnn
xy
xz
yz
zyx
AT
ˆˆ000
0
0
0
ˆˆ ×==
−
−
−
=×
[ ]( )
[ ]( )
[ ]( )
( )[ ]( )
[ ]( )
[ ]( )
[ ]( )AATAAT
x
AAA
nnnnnnnnInnn ×−=×+×−=×+−=××× ˆˆˆˆˆˆˆˆˆˆˆ 22
[ ]( )
[ ]( )
[ ]( )
[ ]( )
[ ]( )
[ ]( )
( )T
x
AAAAAA
nnInnnnnn ˆˆˆˆˆˆˆˆ 33 +−−=××−=××××
skew-symmetric
Rotation Matrix (continue – 2)
12. 12
ROTATIONS
Computation of the Rotation Matrix (continue – 3)
SOLO
Ax
Az
Ay
Bz
By
Bx
O
nˆ
θ
θ
θ
θ
[ ] [ ] [ ]( )
( ) [ ]( )
( ) ( ) [ ]( )
( ){ }
[ ] [ ]( )
[ ]( )
( ) [ ]( )
{ }
( ) ( ) B
Axx
AAA
x
TATATA
x
TA
B
CnRnR
nnnI
nnnIC
=−=−=
×−−××+=
×+−××+=
θθ
θθ
θθ
,ˆ,ˆ
sinˆcos1ˆˆ
sinˆcos1ˆˆ
3333
33
33
Note
The last term can be writen in matrix form as
Therefore
In the same way
End Note
In fact is the matrix representation of the vector product:[ ][ ]vnn
×× ˆˆ
( ) ( )vInnvvnn x
T
33ˆˆˆˆ −→−⋅
( ) ( ) ( ) ( ) vvnnnnvvnnvnn
−⋅=⋅−⋅=×× ˆˆˆˆˆˆˆˆ
[ ][ ] T
x nnInn ˆˆˆˆ 33 +−=××
( )[ ] ( )[ ] [ ][ ][ ] [ ]×−=×××→×−=−⋅×=××× nnnnvnvvnnnvnnn ˆˆˆˆˆˆˆˆˆˆˆ
( )[ ]{ } ( ) [ ][ ][ ][ ] [ ][ ]××−=××××→××−=×××× nnnnnnvnnvnnnn ˆˆˆˆˆˆˆˆˆˆˆˆ
Rotation Matrix (continue – 3)
13. 13
ROTATIONS
Computation of the Rotation Matrix (continue – 4)
SOLO
Ax
Az
Ay
Bz
By
Bx
O
nˆ
θ
θ
θ
θ
[ ] [ ]( )
( ) [ ]( )
( ) ( ) [ ]( )
( ){ }
[ ] [ ]( )
[ ]( )
( )( ) [ ]( )
( ) θθ
θθ
θθ
sin0cos123
sinˆcos1ˆˆ
sinˆcos1ˆˆ
33
33
−−−=
=×−−××+=
=×+−××+=
AAA
x
TATATA
x
B
A
ntracenntraceItrace
nnnItracetraceC
Therefore θcos21+=
B
ACtrace
Let compute the trace (sum of the diagonal components
of a matrix) of
B
AC
Also we have
[ ] [ ]( )
( ) [ ]( )
( ) ( ) [ ]( )
( ){ }
[ ] [ ]( )( ) [ ]( )
{ }
[ ] ( ) [ ]( )
{ }=×−−+=
=×−−+−+=
=×+−××+=
θθθ
θθ
θθ
sinˆcos1ˆˆcos
sinˆcos1ˆˆ
sinˆcos1ˆˆ
33
3333
33
AT
x
AT
xx
TATATA
x
B
A
nnnI
nnnII
nnnIC
( ) θθθ sin
0
0
0
cos1cos
000
010
001
2
2
2
−
−
−
−−
+
=
xy
xz
yz
zzyzx
zyyyx
zxyxx
nn
nn
nn
nnnnn
nnnnn
nnnnn
Rotation Matrix (continue – 4)
14. 14
ROTATIONS
Computation of the Rotation Matrix (continue – 5)
SOLO
Ax
Az
Ay
Bz
By
Bx
O
nˆ
θ
θ
θ
θ
Therefore we have
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
−+−−+−
+−−+−−
−−+−−+
=
θθθθθθ
θθθθθθ
θθθθθθ
cos1cossincos1sincos1
sincos1cos1cossincos1
sincos1sincos1cos1cos
2
2
2
zxzyyzx
xzyyzyx
yzxzyxx
B
A
nnnnnnn
nnnnnnn
nnnnnnn
C
We get
( )1
2
1
cos −=
B
AtraceCθ two solutions for θ
If ; i.e. we obtain0sin ≠θ πθ ,0≠
( ) ( )[ ] ( )θsin2/2,33,2
B
A
B
Ax CCn −=
( ) ( )[ ] ( )θsin2/3,11,3
B
A
B
Ay CCn −=
( ) ( )[ ] ( )θsin2/1,22,1
B
A
B
Az CCn −=
Rotation Matrix (continue – 5)
15. 15
ROTATIONS
Consecutive Rotations
SOLO
- Perform first a rotation of the vector , according to the Rotation Matrix
to the vector .
v
( )1133 ,ˆ θnR x
1v
- Perform a second a rotation of the vector , according to the Rotation Matrix
to the vector .
1v
( )2233 ,ˆ θnR x
2v
( )vnRv x
11331 ,ˆ θ=
( ) ( ) ( ) ( )vnRvnRnRvnRv xxxx
θθθθ ,ˆ,ˆ,ˆ,ˆ 3311332233122332 ===
The result of those two consecutive rotation is a rotation defined as:
( ) ( ) ( )1133223333 ,ˆ,ˆ,ˆ θθθ nRnRnR xxx =
Let interchange the order of rotations, first according to the Rotation Matrix
and after that according to the Rotation Matrix .
( )2233 ,ˆ θnR x
( )1133 ,ˆ θnR x
The result of those two consecutive rotation is a rotation defined as:
( ) ( )22331133 ,ˆ,ˆ θθ nRnR xx
Since in general, the matrix product is not commutative
( ) ( ) ( ) ( )2233113311332233 ,ˆ,ˆ,ˆ,ˆ θθθθ nRnRnRnR xxxx ≠
Therefore, in general, the consecutive rotations are not commutative.
Rotation Matrix (continue – 6)
17. 17
ROTATIONS
Decomposition of a Vector in Two Different Frames of Coordinates
SOLO
We have two frames of coordinate systems A and B, with the same origin O.
We can reach B from A by performing a rotation.
Let describe the vector in both frames.v
BzBByBBxBAzAAyAAxA zvyvxvzvyvxvv
111111 ++=++=
( )
=
zA
yA
xA
A
v
v
v
v
( )
=
zB
yB
xB
B
v
v
v
v
&
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) BBABBABBAA
BBABBABBAA
BBABBABBAA
zzzyyzxxzz
zzyyyyxxyy
zzxyyxxxxx
ˆˆˆˆˆˆˆˆˆˆ
ˆˆˆˆˆˆˆˆˆˆ
ˆˆˆˆˆˆˆ1ˆˆ
⋅+⋅+⋅=
⋅+⋅+⋅=
⋅+⋅+⋅=
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]
( ) ( ) ( )[ ] zABBABBABBA
yABBABBABBA
xABBABBABBA
vzzzyyzxxz
vzzyyyyxxy
vzzxyyxxxxv
ˆˆˆˆˆˆˆˆˆ
ˆˆˆˆˆˆˆˆˆ
ˆˆˆˆˆˆˆ1ˆ
⋅+⋅+⋅+
⋅+⋅+⋅+
⋅+⋅+⋅=
from which
Rotation Matrix (continue – 8)
18. 18
ROTATIONS
Decomposition of a Vector in Two Different Frames of Coordinates (continue – 1)
SOLO
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
=
zA
yA
xA
BABABA
BABABA
BABABA
zB
yB
xB
v
v
v
zzzyzx
yzyyyx
xzxyxx
v
v
v
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
( ) ( )AB
A
B
vCv
=
where is the Transformation Matrix
(or Direction Cosine Matrix – DCM) from
frame A to frame B.
B
AC
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
==↑
BABABA
BABABA
BABABA
B
A
B
A
zzzyzx
yzyyyx
xzxyxx
CC
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
:
In the same way ( )
( ) ( ) ( )BA
B
BB
A
A
vCvCv
==
−1
therefore
( ) 1−
=
B
A
A
B CC
Rotation Matrix (continue – 9)
19. 19
ROTATIONS
Decomposition of a Vector in Two Different Frames of Coordinates (continue – 2)
SOLO
( ) ( ) ( )
[ ] ( ) ( )
[ ] ( ) ( ) ( )ATAAB
A
TB
A
TAAB
A
TAB
A
BTB
vvvCCvvCvCvvv
====2
Since the scalar product is independent of the frame of
coordinates, we have
[ ] [ ] [ ] 1−
=→=
B
A
TB
A
B
A
TB
A CCICC
[ ]
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
=
=
100
010
001
3,33,23,1
2,32,22,1
1,31,21,1
3,32,31,3
3,22,21,2
3,12,11,1
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
TB
A
CCC
CCC
CCC
CCC
CCC
CCC
CC
or ( ) ( )
=≠
==
==∑
= 3,2,10
3,2,11
,,
3
1 jji
iji
kjCkiC ij
k
B
A
B
A δ
Those are 9 equations in , but by interchanging i with j we get the same
conditions, therefore we have only 6 independent equations.
( ) 3,2,1,, =jijiC
B
A
We see that the Rotation Matrix is ortho-normal (having real coefficients and the
rows/columns are orthogonal to each other and of unit absolute value.
Rotation Matrix (continue – 10)
This means that the relation between the two coordinate systems is defined by
9 – 6 = 3 independent parameters.
20. 20
ROTATIONS
Differential Equations of the Rotation Matrices
SOLO
We want to develop the differential equation of the Rotation Matrix as a function of
the Angular Velocity of the Rotation.
Let define by:
-the Rotation Matrix that defines a frame of coordinates B at the time t relative
to some frame A.
( )tC
B
A
-the Rotation Matrix that defines the frame of coordinates B at the time t+Δt
relative to some frame A.
( )ttC
B
A ∆+
( )φω ∆−,ˆ33xR -the Rotation Matrix from the frame of coordinates B at the time t to B at time
t+Δt relative to some frame A.
( ) ( ) ( )tCRttC
B
Ax
B
A φω ∆−=∆+ ,ˆ33
and
( ) [ ] [ ] [ ] ( ) [ ]{ }
[ ] [ ] [ ] [ ]
∆
∆
×−
∆
××+=
∆×−∆−××+=∆−
2
cos
2
sinˆ2
2
sinˆˆ2
sinˆcos1ˆˆ,ˆ
2
33
3333
φφ
ω
φ
ωω
φωφωωφω
x
xx
I
IR
Rotation Matrix (continue – 11)
21. 21
ROTATIONS
Differential Equations of the Rotation Matrices (continue – 1)
SOLO
Let differentiate the Rotation Matrix
( ) ( )
( ) ( ) ( ) ( )
( )
( )
( )
( )
( )tC
dt
dIR
tC
t
IR
tC
t
IR
t
tCtCR
t
tCttC
t
C
dt
dC
B
A
xxB
A
xx
t
B
A
xx
t
B
A
B
Ax
t
B
A
B
A
t
B
A
t
B
A
φ
φ
φωφ
φ
φω
φωφω
θ
∆
−∆−
=
∆
∆
∆
−∆−
=
∆
−∆−
=
∆
−∆−
=
∆
−∆+
=
∆
∆
=
→∆→∆
→∆→∆
→∆→∆
3333
0
3333
0
3333
0
33
0
00
,ˆ
lim
,ˆ
lim
,ˆ
lim
,ˆ
lim
limlim
( ) [ ][ ] [ ] [ ]×−=
∆
∆
∆
×−
∆
∆
∆
××=
∆
−∆−
→∆→∆
ω
φ
φ
φ
ω
φ
φ
φ
ωω
φ
φω
θθ
ˆ
2
cos
2
2
sin
ˆ
2
2
2
sin
ˆˆlim
,ˆ
lim 2
2
0
3333
0
xx IR
and
Therefore
( ) [ ] ( )tC
dt
d
dt
tdC B
A
B
A φ
ω ×−= ˆ
Rotation Matrix (continue – 12)
22. 22
ROTATIONS
Differential Equations of the Rotation Matrices (continue – 2)
SOLO
The final result of the Rotation Matrix differentiation is:
Since defines the unit vector of rotation and the rotation rate from B at time t
to B at time t+Δt, relative to A, then is the angular velocity vector of the frame
B relative to A, at the time t
ωˆ
dt
dφ
ω
φ
ˆ
dt
d
( )
ω
φ
ω ˆ
dt
dB
AB =←
( )
( )[ ]( )
( )tCt
dt
tdC B
A
B
AB
B
A
×−= ←ω
By changing indixes A and B we obtain
( ) ( )[ ] ( )
( )tCt
dt
tdC A
B
A
BA
A
B
×−= ←ω
Rotation Matrix (continue – 13)
23. 23
ROTATIONS
Differential Equations of the Rotation Matrices (continue – 3)
SOLO
Let find the relation between and[ ]( )B
AB ×←ω
[ ]( )A
AB ×←ω
For any vector let perform the following computationsv
[ ]( ) ( )
[ ]( )
[ ]( )A
AB
B
A
B
AB
BB
AB vCvv
×=×=× ←←← ωωω
[ ]( ) ( )
[ ]( ) ( )
[ ]( ) ( )BA
B
A
AB
B
A
AB
A
A
B
A
AB
B
A
AA
AB
B
A vCCvCCCvC
×=×=×= ←←← ωωω
Since this is true for any vector we havev
[ ]( )
[ ]( ) A
B
A
AB
B
A
B
AB CC ×=× ←← ωω
Pre-multiplying by and post-multiplying by we get:
A
BC
B
AC
[ ]( )
[ ]( ) B
A
B
AB
A
B
A
AB CC ×=× ←← ωω
Rotation Matrix (continue – 14)
24. 24
ROTATIONS
Differential Equations of the Rotation Matrices (continue – 4)
SOLO
Let differentiate the equation 33x
A
B
B
A ICC =
to obtain
[ ] ( )
[ ] ( )
0=+×−=+×−=+ ←←
dt
dC
C
dt
dC
CCC
dt
dC
CC
dt
dC
A
BB
A
B
AB
A
BB
A
A
B
B
A
B
AB
A
BB
A
A
B
B
A
ωω
Post-multiplying by we get
A
BC
[ ] ( )
[ ] ( )
[ ] ( ) A
B
A
AB
A
B
B
A
B
AB
A
B
B
AB
A
B
A
B
CCCCC
dt
dC
×=×=×= ←←← ωωω
We obtained for the differentiation of the Rotation Matrix
( ) ( )[ ] ( )
( ) ( )[ ] ( )
( ) ( ) ( )[ ] ( )B
AB
A
B
A
B
A
AB
A
B
A
BA
A
B
ttCtCttCt
dt
tdC
×=×=×−= ←←← ωωω
Note
We can see that ( )[ ] ( )
( )[ ] ( )
( ) ( )tttt ABBA
A
AB
A
BA ←←←← =−⇒×=×− ωωωω
End Note
Rotation Matrix (continue – 15)
25. 25
ROTATIONS
Differential Equations of the Rotation Matrices (continue – 5)
SOLO
Suppose that we have a third frame of coordinates I (for example inertial) and we
have the angular velocity vectors of frames A and B relative to I.
We have
[ ]( ) B
I
B
IB
B
I
C
dt
dC
×−= ←ω
[ ]( ) A
I
A
IA
A
I
C
dt
dC
×−= ←ω
A
I
B
A
B
I CCC =
dt
dC
CC
dt
dC
dt
dC
A
IB
A
A
I
B
A
B
I
+=
[ ]( )
[ ]( ) I
A
A
I
A
IA
B
A
I
A
B
I
B
IB
I
A
A
IB
A
I
A
B
I
B
A
CCCCCC
dt
dC
CC
dt
dC
dt
dC
×+×−=−= ←← ωω
or
From which we get:
[ ]( )
[ ]( )A
IA
B
A
B
A
B
IB
B
A
CC
dt
dC
×+×−= ←← ωω
Rotation Matrix (continue – 16)
26. 26
ROTATIONSSOLO
From the equation
Computation of the Angular Velocity Vector from .AB←ω
( ) ( )nRtC x
B
A
ˆ,33 θ−=
( ) ( )[ ]( )
( )tCt
dt
tdC B
A
B
AB
B
A
×−= ←ω
we obtain
( )[ ]( ) ( ) ( )[ ]TB
A
B
AB
AB tC
dt
tdC
t −=×←ω
Since the Rotation Matrix is defined also by and
( ) [ ] ( ) [ ]{ }θθθθ sinˆcos1ˆˆcosˆ, 3333 ×−−+=−= × nnnInRC T
x
B
A
( )tC
B
A nˆθ
we can compute as function of and their derivativesnˆθAB←ω
td
dθ
θ =
td
nd
n
ˆ
ˆ =
•
(this is a long procedure described in the complementary work “Notes on
Rotations”, and a simpler derivation will be given later, we give here the
final result)
[ ] ( ) θθθω sinˆcos1ˆˆˆ
••
← +−×−= nnnnAB
Rotation Matrix (continue – 17)
27. 27
ROTATIONSSOLO
Computation of and as functions of .AB←ω
td
dθ
θ =
td
nd
n
ˆ
ˆ =
•
Let pre-multiply the equation by and use
T
nˆ[ ] ( ) θθθω sinˆcos1ˆˆˆ
••
← +−×−= nnnnAB
[ ] 0ˆˆ,0ˆˆ,1ˆˆ ==×=
•
nnnnnn TTT to obtain
[ ] ( ) AB
TTTT
AB
T
nnnnnnnnn ←
••
← =→+−×−= ωθθθθω
ˆsinˆˆcos1ˆˆˆˆˆˆ
Let pre-multiply the equation by and use[ ]×nˆ[ ] ( ) θθθω sinˆcos1ˆˆˆ
••
← +−×−= nnnnAB
[ ] [ ][ ] ( )
•••
−=−=××=× nnInnnnnnn x
T
ˆˆˆˆˆˆˆ,0ˆˆ 33
to obtain
[ ] [ ] [ ][ ] ( ) [ ] ( ) [ ] θθθθθω sinˆˆcos1ˆsinˆˆcos1ˆˆˆˆˆˆ
••••
← ×+−=×+−××−×=× nnnnnnnnnnn AB
Let pre-multiply the equation by[ ] ( ) [ ] θθω sinˆˆcos1ˆˆ
••
← ×+−=× nnnn AB
[ ]×nˆ
[ ][ ] [ ] ( ) [ ][ ] [ ] ( )θθθθω cos1ˆˆsinˆsinˆˆˆcos1ˆˆˆˆ −×+−=××+−×=××
••••
← nnnnnnnnnn AB
Rotation Matrix (continue – 18)
28. 28
ROTATIONS
Computation of and as functions of (continue – 1)
SOLO
AB←ω
td
dθ
θ =
td
nd
n
ˆ
ˆ =
•
We have two equations:
( ) [ ] [ ] ABnnnn ←
••
×=×+− ωθθ
ˆsinˆˆcos1ˆ
[ ] ( ) [ ][ ] ABnnnnn ←
••
××=−×+− ωθθ
ˆˆcos1ˆˆsinˆ
with two unknowns and
•
nˆ [ ]
•
× nn ˆˆ
From those equations we get:
( )[ ] [ ] ( ) [ ][ ] θωθωθθ sinˆˆcos1ˆsincos1ˆ 22
ABAB nnnn ←←
•
××−−×=+−
or
( ) [ ] ( ) [ ][ ] θωθωθ sinˆˆcos1ˆcos1ˆ2 ABAB nnnn ←←
•
××−−×=−
Finally we obtain:
AB
T
n ←= ωθ
ˆ
[ ] [ ][ ] ABnnnn ←
•
××−×= ω
θ
2
cotˆˆˆ
2
1
ˆ
Rotation Matrix (continue – 19)
29. 29
ROTATIONS
Quaternions
SOLO
The quaternions method was introduced by Hamilton in
1843. It is based on Euler Theorem (1775) that states:
The most general displacement of a rigid body with one point fixed is equivalent to
a single rotation about some axis through that point.
Therefore every rotation is defined by three parameters:
• Direction of the rotation axis , defined by two parameters
• The angle of rotation , defines the third parameter
nˆ
θ
William Rowan Hamilton
1805 - 1865
( ) ( ) ( ) θθ sinˆcos1ˆˆ1 vnvnnvv
×+−××+=
The rotation of around by angle is given by:nˆ θv
A
B
C
O
θ
φφ
nˆ
v
1v
that can be writen
( )[ ] ( ) ( ) θθ sinˆcos1ˆˆ1 vnvvnnvv
×+−−⋅+=
or
( ) ( ) ( ) θθθ sinˆcos1ˆˆcos1 vnvnnvv
×+−⋅+=
30. 30
ROTATIONS
Quaternions (continue – 1)
SOLO
Computation of the Rotation Matrix
We found the Rotation Matrix
that describes this rotation from A to B.
( )θ,ˆ33 nRC x
B
A =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) θθ
θθ
θθ
sinˆˆcos1ˆˆˆˆˆ
sinˆˆcos1ˆˆˆˆˆ
sinˆˆcos1ˆˆˆˆˆ
AAAB
AAAB
AAAB
znznnxz
ynynnxy
xnxnnxx
×+−××+=
×+−××+=
×+−××+=
( )
[ ] [ ]( )
[ ]( )
( ) [ ]( )
{ } ( )A
A
A
B
AAA
x
A
B xCnnnIx ˆ
0
0
1
sinˆcos1ˆˆˆ 33 =
×+−××+= θθ
( )
[ ] [ ]( )
[ ]( )
( ) [ ]( )
{ } ( )A
A
A
B
AAA
x
A
B yCnnnIy ˆ
0
1
0
sinˆcos1ˆˆˆ 33 =
×+−××+= θθ
( )
[ ] [ ]( )
[ ]( )
( ) [ ]( )
{ } ( )A
A
A
B
AAA
x
A
B zCnnnIz ˆ
1
0
0
sinˆcos1ˆˆˆ 33 =
×+−××+= θθ
or
from which
[ ] [ ]( )
[ ]( )
( ) [ ]( )
{ } ( )θθθ ,ˆsinˆcos1ˆˆ 3333 nRnnnICC x
AAA
x
A
B
A
B
=×+−××+==↑
31. 31
ROTATIONS
Quaternions (continue – 2)
SOLO
Definition of the Quaternions
The quaternions (4 parameters) were defined by
Hamilton as a generalization of the complex numbers
( ) 32100 , qkqjqiqqq
+++== ρ
( )2/cos0 θ=q
( ) nˆ2/sin θρ =
( ) ( ) ( ) zyx nqnqnq 2/sin&2/sin&2/sin 111 θθθ ===
where satisfy the relations:kji
,,
1−=⋅=⋅=⋅ kkjjii
kijji
=⋅−=⋅ ijkkj
=⋅−=⋅ jkiik
=⋅−=⋅
1−=⋅⋅ kji
the complex conjugate of is defined asq
( ) 32100
*
, qkqjqiqqq
−−−=−= ρ
32. 32
ROTATIONS
Quaternions (continue – 3)
SOLO
Product of Quaternions
Product of two quaternions andAq Bq
( )( ) ( )( )3210321000 ,, BBBBAAAABBAABA qkqjqiqqkqjqiqqqqq
++++++== ρρ
( ) ( ) ( )3210321033221100 AAABBBBABABABABA qkqjqiqqkqjqiqqqqqqqqq
++++++−−−=
( ) ( ) ( )122131132332 BABABABABABA qqqqkqqqqjqqqqi −+−+−+
therefore
( )( ) ( ) ( )[ ]BAABBABABABBAABA qqqqqqqq ρρρρρρρρ
×++⋅−== 000000 ,,,
Let use this expression to find
( )( ) ( )( ) 2
3
2
2
2
1
2
0
222
000
*
00
*
1ˆˆ
2
sin
2
cos,,,, qqqqnnqqqqqqqqq +++==⋅
+
=⋅+=−==−=
θθ
ρρρρρρ
The quaternion product can be writen in matrix form as:
[ ] [ ]
×−
−
=
×+
−
==
=
A
A
BxBB
T
BB
B
B
AxAA
T
AA
BA
q
Iq
qq
Iq
q
qq
q
q
ρρρ
ρ
ρρρ
ρ
ρ
0
330
00
330
00
1−=⋅⋅=⋅=⋅=⋅ kjikkjjii
kijji
=⋅−=⋅ ijkkj
=⋅−=⋅ jkiik
=⋅−=⋅
38. 38
ROTATIONS
Quaternions (continue – 9)
SOLO
Description of Successive Rotations Using Quaternions
Let describe two consecutive rotations:
- First rotation defined by the quaternion
( )
== 1
11
1101
ˆ
2
sin,
2
cos, nqq
θθ
ρ
- Folowed by the second rotation defined by the quaternion
( )
== 2
22
2202
ˆ
2
sin,
2
cos, nqq
θθ
ρ
After the first rotation the quaternion of the vector is transferred to 1
*
1 qvq
After the second rotation we obtain ( ) ( ) ( )21
*
2121
*
1
*
221
*
1
*
2 qqvqqqqvqqqqvqq ==
Therefore the quaternion representing those two rotation is:
( ) ( )( ) ( )
×
+
+
⋅
−
=
=×++⋅−====
21
21
1
2
2
1
21
2121
21120210212010220110210
ˆˆ
2
sin
2
sinˆ
2
cosˆ
2
cos,ˆˆ
2
sin
2
sin
2
cos
2
cos
,,,,
nnnnnn
qqqqqqqqqq
θθθθθθθθ
ρρρρρρρρρ
( ) ( )210 , qqqq == ρ
21
2121
0
ˆˆ
2
sin
2
sin
2
cos
2
cos
2
cos nnq ⋅
−
=
=
θθθθθ
21
21
1
2
2
1
ˆˆ
2
sin
2
sinˆ
2
cosˆ
2
cosˆ
2
sin nnnnn ×
+
+
=
=
θθθθθ
ρ
39. 39
ROTATIONS
Quaternions (continue – 10)
SOLO
Description of Successive Rotations Using Quaternions (continue – 1)
( ) ( )210 , qqqq == ρ
21
2121
0
ˆˆ
2
sin
2
sin
2
cos
2
cos
2
cos nnq ⋅
−
=
=
θθθθθ
21
21
1
2
2
1
ˆˆ
2
sin
2
sinˆ
2
cosˆ
2
cosˆ
2
sin nnnnn ×
+
+
=
=
θθθθθ
ρ
Two consecutive rotations, followed by , are given by:1q 2q
From those equations we can see that:
0ˆˆˆˆˆˆˆˆˆˆ 21212112211221
=×→×−=×→×=×= nnnnnnnnnnifonlyandifqqqq
The rotations are commutative if and only if are collinear.21
ˆ&ˆ nn
In matrix form those two rotations are given by:
First Rotation: ( ) [ ] ( ) [ ]{ }111111331133 sinˆcos1ˆˆcosˆ, θθθθ ×−−+=− nnnInR
T
xx
Second Rotation: ( ) [ ] ( ) [ ]{ }222222332233 sinˆcos1ˆˆcosˆ, θθθθ ×−−+=− nnnInR
T
xx
Total Rotation:
( ) ( ) ( ) [ ] ( ) [ ]{ }θθθθθθ sinˆcos1ˆˆcosˆ,ˆ,ˆ, 331133223333 ×−−+=−−=− nnnInRnRnR T
xxxx
40. 40
ROTATIONS
Quaternions (continue – 11)
SOLO
Description of Successive Rotations Using Quaternions (continue – 2)
Let find the quaternion that describes the Euler Rotations through the
angles respectively. Let write the rotations according to their order
123 →→
ϕθψ ,,
+
+
+
==
2
sin
2
cos
2
sin
2
cos
2
sin
2
cos
ϕϕθθψψ
ijkqqqq xyz
B
A
−
+
+
+
=
2
sin
2
sin
2
cos
2
sin
2
cos
2
sin
2
cos
2
cos
2
sin
2
cos
ϕθϕθθϕϕθψψ
kjik
+
=
2
sin
2
sin
2
sin
2
cos
2
cos
2
cos
ϕθψϕθψ
−
+
2
cos
2
sin
2
sin
2
sin
2
cos
2
cos
ϕθψϕθψ
i
+
+
2
cos
2
sin
2
cos
2
sin
2
cos
2
sin
ϕθψϕθψ
j
−
+
2
sin
2
sin
2
cos
2
cos
2
cos
2
sin
ϕθψϕθψ
k
41. 41
ROTATIONS
Quaternions (continue – 12)
SOLO
Differential Equation of the Quaternions
Let define
( ) ( )ρ
,0qtq B
A =
- the quaternion that defines the position of B frame
relative to frame A at time t.
( ) ( )tqqttq B
A ∆+∆+=∆+ ρ
,00
- the quaternion that defines the position of B frame
relative to frame A at time t+Δt.
( )
∆
∆
=∆ ∆t
B
A ntq ˆ
2
sin,
2
cos
θθ - the quaternion that defines the position of B frame
at time t+Δt relative to frame B at time t.
We have the relation: ( ) ( ) ( )tqtqttq B
A
B
A
B
A ∆=∆+
or
( ) ( ) ( ) ( )( ) ( ) ( ) ( )[ ] ( ) ( )( )ρρρρρρρρ
θθ
∆∆−+=∆∆+−=∆+∆+−=∆+=
∆
∆
=∆ ∆ ,,0,1,,,,,ˆ
2
sin,
2
cos 00000000
*
qqqqqqqqttqtqntq B
A
B
At
B
A
therefore ( )( )
∆
−
∆
=∆∆− ∆tnqq ˆ
2
sin,1
2
cos,, 00
θθ
ρρ
( ) ( )
∆
−
∆
=∆∆ ∆tnqq ˆ
2
sin,1
2
cos,, 00
θθ
ρρ
or
42. 42
ROTATIONS
Quaternions (continue – 13)
SOLO
Differential Equation of the Quaternions (continue – 1)
( ) ( )
∆
−
∆
=∆∆ ∆tnqq ˆ
2
sin,1
2
cos,, 00
θθ
ρρ
Let divide both sides by and take the limit .0→∆ tt∆
( ) ( ) ( )
=
=
∆
∆
∆
∆
∆
∆
∆
−
∆
==
∆
∆
∆
∆
∆∆∆
→∆
tBttB
t
ntqnqn
tt
tq
td
d
tt
q
ˆ
2
1
,0ˆ
2
1
,0,ˆ
2
2
sin
2
1
,
2
1
2
cos
2
1
,lim 0
0
0
θθρ
θ
θ
θ
θ
θ
θ
ρ
But is the instant angular velocity vector of frame B relative to frame A.tn∆
ˆθ
( )
( ) t
B
AB nt ∆← = ˆθω
( ) ( )
( )( ) ( )
( )ttn
B
AB
B
ABt ←←∆ == ωωθ
,0ˆ,0
So we can write
( ) ( ) ( )
( )ttqtq
td
d B
AB
B
A
B
A ←= ω
2
1
This is the Differential equation of the quaternion that defines the position of B
relative to A, at the time t as a function of the angular velocity vector of frame B relative
to frame A, .
( )tq B
A
( )
( )t
B
AB←ω
43. 43
ROTATIONS
Quaternions (continue – 14)
SOLO
Differential Equation of the Quaternions (continue – 2)
Developing this equation, we get
( ) ( ) ( )
( )ttqtq
td
d B
AB
B
A
B
A ←= ω
2
1
( ) ( )
( )[ ] ( ) ( ) ( )
( )B
AB
B
AB
B
AB
B
AB qtq
dt
d
dt
dq
←←←← ×+⋅−==
ωρωωρωρ
ρ
00
0
,
2
1
,0,
2
1
,
from which
( )B
AB
dt
dq
←⋅−= ωρ
2
10
( ) ( )
( )B
AB
B
ABq
dt
d
←← ×+= ωρω
ρ
0
2
1
or in matrix form
[ ] [ ]
( )
( )t
Iq
q
dt
d B
AB
x
T
←
×+
−
=
ω
ρ
ρ
ρ
330
0
2
1
48. 48
ROTATIONS
Quaternions (continue – 18)
SOLO
Differential Equation of the Quaternion Between Two Frames A and B Using the Angular
Velocities of a Third Frame I
The relations between the components of a vector in the frames A, B and I arev
( ) AB
A
IA
I
B
A
IB
I
B
vCvCCvCv
===
Using quaternions the same relations are given by
( ) B
A
A
I
IA
I
B
A
B
I
IB
I
B
qqvqqqvqv
***
==
Therefore
B
A
A
I
B
I qqq = B
I
A
I
B
A qqq
*
=
Let perform the following calculations
B
A
A
I
B
A
A
I
B
I q
dt
d
qqq
dt
d
q
dt
d
+=
&( )B
IB
B
I
B
I qq
dt
d
←= ω
2
1 ( )A
IA
A
I
A
I qq
dt
d
←= ω
2
1
and use
( ) ( ) B
A
A
I
B
A
A
IA
A
I
B
IB
B
I q
dt
d
qqqq += ←← ωω
2
1
2
1 ( ) ( ) B
A
A
IA
A
I
A
I
B
IB
B
I
A
I
B
A qqqqqq
dt
d
←← −= ωω
1
**
2
1
2
1
to obtain ( ) ( ) B
A
A
IA
B
IB
B
A
B
A qqq
dt
d
←← −= ωω
2
1
2
1
49. 49
ROTATIONS
Quaternions (continue – 19)
SOLO
Differential Equatio of the Quaternion Between Two Frames A and B Using the Angular
Velocities of a Third Frame I (continue – 1)
Using the relations
ABIAIB ←←← += ωωω
and
( ) ( ) B
A
A
IA
B
A
B
IA qq ←← = ωω
* ( ) ( ) B
A
A
IA
B
IA
B
A qq ←← = ωω
we have
( ) ( )
( ) ( ) ( ) ( ) ( )
0
2
1
2
1
2
1
2
1
2
1 B
A
A
IA
B
IA
B
A
B
AB
B
A
B
A
A
IA
B
IA
B
AB
B
A
B
A qqqqqq
dt
d
←←←←←← −+=−+= ωωωωωω
from which ( ) ( ) B
A
A
AB
B
AB
B
A
B
A qqq
dt
d
←← == ωω
2
1
2
1
Since BAAB ←← −= ωω
we get
( ) ( ) ( ) ( ) B
A
A
BA
B
BA
B
A
B
A
A
AB
B
AB
B
A
B
A qqqqq
dt
d
←←←← −=−=== ωωωω
2
1
2
1
2
1
2
1
From we get1
*
== B
A
A
B
B
A
B
A qqqq A
B
B
A qq =
*
Therefore
+
= B
A
A
B
B
A
A
B q
dt
d
qqq
dt
d
0
*B
A
B
A
A
B
A
B qq
dt
d
qq
dt
d
−=
50. 50
ROTATIONS
Euler Angles
SOLO
The orientation of the Body Frame relative to the
Inertial Frame has three degrees of freedom.
We will use 3 Euler Angles that define the orientation
by three consecutive rotations around the consecutive
frame axes.
[ ]
−
=
11
1111
0
0
001
:
θθ
θθθ
cs
sc
[ ]
−
=
22
22
22
0
010
0
:
θθ
θθ
θ
cs
sc
[ ]
−=
100
0
0
: 33
33
33 θθ
θθ
θ cs
sc
The three basic Euler rotations around
axes are described by the rotation matrices:
,3ˆ,2ˆ,1ˆ
51. 51
ROTATIONS
Euler Angles (continue – 1)
SOLO
Introduce the Piogram that represents the following notation:
(from Pio R.L. “Symbolic Representations of Coordinate Transformations”, IEEE
on Aerospace and Navigation Electronics, Vol. ANE-11,June 1964, pp.128-134)
The Piogram
52. 52
ROTATIONSEuler Angles (continue – 2)
SOLO
Rotation Around x Axis by an Angle .ϕ
[ ]
−
==
ϕϕ
ϕϕϕ
cossin0
sincos0
001
x
B
AC
BAAB xx
11 ϕϕω ==←
[ ]
−=−=
ϕϕ
ϕϕϕ
cossin0
sincos0
001
x
A
BC
The Piogram (continue – 1)
53. 53
ROTATIONSEuler Angles (continue – 3)
SOLO
Rotation Around y Axis by an Angle .θ
[ ]
−
==
θθ
θθ
θ
cos0sin
010
sin0cos
y
B
AC
BAAB yy
11 θθω ==←
[ ]
−
=−=
θθ
θθ
θ
cos0sin
010
sin0cos
y
A
BC
The Piogram (continue – 2)
54. 54
ROTATIONSEuler Angles (continue – 4)SOLO
Rotation Around z Axis by an Angle .ψ
[ ]
−==
100
0cossin
0sincos
ψψ
ψψ
ψ x
B
AC
BAAB zz
11 ψψω ==←
[ ]
−
=−=
100
0cossin
0sincos
ψψ
ψψ
ψ x
A
BC
The Piogram (continue – 3)
55. 55
ROTATIONS
Euler Angles (continue – 5)
SOLO
Using the basic Euler Angles we can define the
following 12 different rotations:
(a) six rotations around three different axes:
321 →→ 231 →→ 312 →→ 132 →→ 213 →→ 123 →→
(b) six rotations such that the first and third are around the sam axes, but the second
is different:
121 →→ 131 →→ 212 →→ 232 →→ 313 →→ 323 →→
Suppose that the Transfer Matrix from A to B is defined by three
consecutive Euler Angles: around (unit vector in A Frame),
around (unit vector in intermediate frame), around (unit vector
in B Frame).
B
AC
iθ Iiˆ
jθ
Interjˆ
kθ Bkˆ
[ ] [ ] [ ] [ ] [ ]TB
A
B
A
A
B
A
B
B
Akkjjii
B
A CCCICCC ==→==
−1
&θθθ
56. 56
ROTATIONSEuler Angles (continue – 6)
SOLO
123 →→Euler Angles rotations: using the Piogram
1.Rotation from A to A’ around the third axis by the angle ψ
2. Rotation from A’ to B’ around the third axis by the angle θ
.
3. Rotation from B’ to B around the third axis by the angle ϕ
.
xAv
yAv
zAv
yBv
zBv
xBv
ϕ
θ−
ψ
'xAv
'yAv
'zAv
'xBv
'yBv
'zBv
The Piogram (continue – 4)
57. 57
ROTATIONSEuler Angles (continue – 7)
SOLO
123 →→Euler Angles rotations:
xAv
yAv
zAv
yBv
zBv
xBv
ϕ
θ−
ψ
'xAv
'yAv
'zAv
'xBv
'yBv
'zBv
Piogram:
Example 1: Computation of the relation between to using the PiogramyBv zAyAxA vvv ,,
From the Piogram:
( ) ( ) θϕψθϕψϕψθϕψϕ cossinsinsinsincoscoscossinsinsincos zAyAxAyB vvvv ++++−=
using the Piogram (continue – 1)
The Piogram (continue – 5)
58. 58
ROTATIONSEuler Angles (continue – 8)SOLO
123 →→Euler Angles rotations:
xAv
yAv
zAv
yBv
zBv
xBv
ϕ
θ−
ψ
'xAv
'yAv
'zAv
'xBv
'yBv
'zBv
Piogram:
using the Piogram (continue – 2)
Example 2: Computation of the matrix (1st
column) using the Piogram
B
AC
1=xAv
0=yAv
0=zAv
ψcos' =xAv
ψsin' −=yAv
0' =zAv
ψsin' −=yBv
ψθcossin' =zBv
ψθcoscos=xBv
ψθϕ
ψϕ
cossincos
sinsin
+
+=zBv
ψθϕ
ψϕ
cossinsin
sincos
+
+−=yBv
ψcos
ψcos
ψsin
ψsin
θcos
θsin
θcos
θsin
ϕcos
ϕsin
ϕcos
ϕsin
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
+−+
++−
−
=
θϕψθϕψϕψθϕψϕ
θϕψθϕψϕψθϕψϕ
θψθψθ
ccssccscscss
csssscccsssc
ssccc
CCC
CCC
CCC
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
3,32,31,3
3,22,21,2
3,12,11,1
The Piogram (continue – 6)
59. 59
ROTATIONSEuler Angles (continue – 9)
SOLO
123 →→Euler Angles rotations:
xAv
yAv
zAv
yBv
zBv
xBv
ϕ
θ−
ψ
'xAv
'yAv
'zAv
'xBv
'yBv
'zBv
Piogram:
Example 3: Computation of the angular velocity using the Piogram
using the Piogram (continue – 3)
AB←ω
xAAB←ω
yAAB←ω
zAAB←ω
0
0
0
B
ψ
ψ−
A
θ
θ
'A
ϕ−
ϕ
'B
( ) ( ) ( ) ( )
[ ] [ ] [ ]
−
−
=
−−+
−+
=++=←
ψ
θ
ϕ
θ
ψψθ
ψψθ
ϕθψθψψϕθψω
10
0
0
0
0
1
0
1
0
1
0
0
111 233
'
''
'
''
s
csc
scc
xCyCz
B
B
A
B
A
A
A
A
A
A
A
AB
The Piogram (continue – 7)
73. 73
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities
SOLO
Rotation Matrix in Three Dimensional Space
We saw that the rotation of a vector is given byx
[ ] [ ] [ ] ( )θθ cos1ˆˆsinˆ' −××+×+= xnnxnxx
[ ] [ ] [ ] ( ){ } ( ) xnRxnnnI x
ˆ,cos1ˆˆsinˆ33 θθθ =−××+×+=
The Rotation Matrix has the properties( ) [ ] [ ] [ ] ( ){ }θθθ cos1ˆˆsinˆˆ, 3333 −××+×+=
∆
nnnInR xx
( ) ( )[ ]( )333333
ˆ,ˆ, x
T
xx InRnR =θθOrtho-normal
Unitary ( ) ( )[ ]( )conjugatecomplexInRnR x
T
xx == *
33
*
3333
ˆ,ˆ, θθ
A Theorem from Matrix Algebra states:
Every unitary matrix U can be expressed as an exponential matrix
where H is hermitian (iH is skew-symmetric)
( )iHU exp=
Let find the hermitian matrix that corresponds to the Unitary Rotation Matrix.
74. 74
We found that the matrix has the following properties:[ ]×nˆ
[ ] [ ] T
x nnInn ˆˆˆˆ 33 +−=××
[ ] [ ] [ ] [ ]×−=××× nnnn ˆˆˆˆ
[ ] [ ] [ ] [ ] ( )T
x nnInnnn ˆˆˆˆˆˆ 33 +−−=××××
[ ] [ ]×−=× nn
T
ˆˆ skew-symmetric
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities (continue – 1)
SOLO
Rotation Matrix in Three Dimensional Space (continue – 1)
Define: [ ]×=
∆
nM ˆθ
Therefore [ ][ ] ( )T
x nnInnM ˆˆˆˆ 33
222
+−=××= θθ [ ] MnM 333
ˆ θθ −=×−=
( ) [ ]( )
[ ] [ ] [ ]
[ ][ ] ( ) [ ] θθ
θ
θ
θθ
θ
θ
θ
θθ
θθ
θθ
θ
sinˆcos1ˆˆ
ˆ
!3
ˆ)
!4!2
1(ˆ
!3
1
)
!4!2
1(
11
!3
1)
!4!2
1
(
!4
1
!3
1
!2
1
ˆexpexp
33
3
2
42
2
33
3
2
42
2
2
233
2
2
2
33
432
33
×+−××+=
×
+−+×++−−×+=
+−+++−−+=
+−++−+=
+++++=×=
nnnI
nnnI
MMMI
MMI
MMMMInM
x
x
x
x
x
( ) [ ] [ ] [ ] ( ){ } [ ]( )×=−××+×+=
∆
nnnnInR xx
ˆexpcos1ˆˆsinˆˆ, 3333 θθθθ
75. 75
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities (continue – 2)
SOLO
Rotation Matrix in Three Dimensional Space (continue – 2)
One other way to write this is by using the following matrices:
−=
∆
010
100
000
1E
−
=
∆
001
000
100
2E
−
=
∆
000
001
010
3E
[ ]
EnEnEnEn
nnn
nn
nn
nn
n
zyx
zyx
xy
xz
yz
⋅=++=
−
+
−
+
−=
−
−
−
=×
ˆ
000
001
010
001
000
100
010
100
000
0
0
0
ˆ
321
Therefore
( ) [ ] [ ] [ ] ( ){ } [ ]( ) ( )EnnnnnInR xx
⋅=×=−××+×+=
∆
ˆexpˆexpcos1ˆˆsinˆˆ, 3333 θθθθθ
76. 76
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities (continue – 3)
SOLO
Rotation Matrix in Three Dimensional Space (continue – 3)
One other way to obtain the same result is the following: compute first
[ ] ( ) [ ] [ ] [ ] [ ] ( ){ }
[ ] [ ] [ ] [ ][ ] [ ] ( )
[ ] [ ] [ ] [ ] ( )
[ ] [ ] [ ] θθ
θθ
θθ
θθθ
sinˆˆcosˆ
cos1ˆsinˆˆˆ
cos1ˆˆˆsinˆˆˆ
cos1ˆˆsinˆˆˆ,ˆ 3333
××+×=
−×−××+×=
−×××+××+×=
−××+×+×=×
nnn
nnnn
nnnnnn
nnnInnRn xx
( )
[ ] [ ] [ ]{ }θθ
θ
θ
sinˆˆcosˆ
ˆ,33
××+×= nnn
d
ndR x
Therefore
( )
[ ] ( )nRn
d
ndR
x
x
ˆ,ˆ
ˆ,
33
33
θ
θ
θ
×=
Since is independent of , we can integrate this equation to obtain again:[ ]×nˆ θ
( ) [ ]( ) ( )EnnnR x
⋅=×= ˆexpˆexpˆ,33 θθθ
Secondly compute
77. 77
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities (continue – 4)
SOLO
Euler Parameters
The quaternion that describes the rotation was found to be:
( ) 32100 , qkqjqiqqq
+++== ρ
( )2/cos0 θ=q ( ) nˆ2/sin θρ =
( ) ( ) ( ) zyx nqnqnq 2/sin&2/sin&2/sin 111 θθθ ===
where satisfy the relationskji
,,
1−=⋅⋅=⋅=⋅=⋅ kjikkjjii
kijji
=⋅−=⋅ ijkkj
=⋅−=⋅ jkiik
=⋅−=⋅
The quaternions representing the vector in frames A and B arev
( ) ( )
( ) ( ) ( )
( )BBAA
vvvv
,0&,0 ==
The relation between those quaternions is given by:
( ) ( )
( ) ( )
( )( ) ( ) [ ] [ ][ ]{ } ( )
( )BBBA
vqqqvqqvqv
××+×+⋅+=−== ρρρρρρρ 0
2
000
*
2,0,,0,
We want to perform the same operations using 2x2 matrices with complex entries.
−
=
−
=
=
10
01
,
0
0
,
01
10
321 σσσ
i
i
Pauli Spinor Matrices
For this let introduce the following definitions:
78. 78
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities (continue – 5)
SOLO
Euler Parameters (continue – 1)
Define the operator as the transpose complex conjugate of a matrix A; i.e.:H
T *
( ) ( )TTH
AAA **
==
We can see that
222211 && σσσσσσ === HHH
Matrices having the property are called hermitian.AAH
=
Pauli Spinor Matrices are hermitian with zero trace.
UnitaryII x
H
x
H
122112211
11
10
01
01
10
01
10
σσσσσ
σσ
⇒=⇒=
=
=
=
They have the following properties:
UnitaryII
i
i
i
i
x
H
x
H
222222222
22
10
01
0
0
0
0
σσσσσ
σσ
⇒=⇒=
=
−
−
=
=
UnitaryII x
H
x
H
322332233
33
10
01
10
01
10
01
σσσσσ
σσ
⇒=⇒=
=
−
−
=
=
−
=
−
=
=
10
01
,
0
0
,
01
10
321 σσσ
i
i
W. PAULI
79. 79
ROTATIONS
Cayley-Klein (or Euler) Parameters and Related Quantities (continue – 6)
SOLO
Euler Parameters (continue – 2)
Pauli Spinor Matrices properties (continue – 1):
321
0
0
0
0
01
10
σσσ i
i
i
i
i
=
−
=
−
= 312
0
0
01
10
0
0
σσσ i
i
i
i
i
−=
−
=
−
=
132
0
0
10
01
0
0
σσσ i
i
i
i
i
=
=
−
−
= 123
0
0
0
0
10
01
σσσ i
i
i
i
i
−=
−
−
=
−
−
=
213
0
0
01
10
01
10
10
01
σσσ i
i
i
i =
−
=
−
=
−
=
231
0
0
01
10
10
01
01
10
σσσ i
i
i
i −=
−
−=
−
=
−
=
2233321 xiIi == σσσσσ
From those expressions we found that the relations between Pauli Matrices and
the quaternions are:
or
kijiii
=== 321 && σσσ
321 && σσσ ikijii −=−=−=
−
=
−
=
=
10
01
,
0
0
,
01
10
321 σσσ
i
i