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Chapter 1: Kinematics
Chew Chee Meng
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What you would learn Definitions of positions and orientations of rigid
bodies Analysis of different representations for
orientation
Transformation of coordinates Determination of new positions and
orientations after a sequence of rigid bodymotions
Kinematics modeling of robotic manipulators
Forward and inverse kinematics of position
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Spatial descriptions and Transformations
Need to specify spatial attributes of
various objects with which a manipulationsystem deals
Cartesian coordinate frames are used
Where is the
cup?
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Spatial descriptions and Transformations
Position & orientation of a Rigid Body
Position Orientation
Location of rigid body
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Position & orientation of a Rigid Body
Position: Attribute of a point
Represented
by a position
vector
With reference
to a coordinateframe
3
z
y
x
A
p
p
p
P
xA
zA
yA{A}
AP
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Position & orientation of a Rigid Body
Orientation: Attribute of a body
Represented by Matrices
(Rotation Matrices)
RA
B = [AxB
AyBAzB ]
3x3
A rigid body
body frame
Reference frame
xA
zA
yA
{A} xB
zB yB{B}
By attaching a coordinate
frame to the body and
then give a description of
this coordinate framerelative to the reference
coordinate frame
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Position & orientation of a Rigid Body
Orientation representation
RA
B = [AxB
AyBAzB ]
3x3
Orthogonal unit vectors(orthonormal matrix)
1)det( RAB Multiplication of rotation matricesgenerally not commutativei.e. R1R2R2R1
xA
zA
yA{A} xB
zB yB{B}
Note: AxB means xB expressed in {A}
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Position & orientation of a Rigid Body
Orientation representation
RA
B = [AxB
AyBAzB ]
ABABAB
ABABAB
ABABAB
zzzyzx
yzyyyx
xzxyxx
Remark:
Components are direction cosines
(dot product of two unit vectors yields the cosine of the angle
between them)
Choice of frame is arbitrary for the vectors, but must be the same
xA
zA
yA
{A} xB
zByB{B}
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Position & orientation of a Rigid Body
Orientation
RA
B = [AxB
AyBAzB ]
B T
AT
B T B B B B T
A A A A A
B TA
R
x
y x y z
z
Note: Rows of the matrix are the unit
vectors of {A} expressed in {B}
ABABAB
ABABAB
ABABAB
zzzyzx
yzyyyx
xzxyxxxA
zA
yA
{A} xB
zB
yB{B}
transpose of zA expressed in {B}
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Position & orientation of a Rigid Body
Orientation
RB
A
TA
BR
3100
010
001
IRR BA
B
A
B
A
T
B
A
T
B
A
T
B
A
A
B
TA
B
zyx
z
y
x
Hence, description of frame {A} relative to {B},
This suggests:
Verification:
TA
B
A
B RR 1
I3 denotes the 3x3 identity matrix
Linear algrebra: inverse of a matrix with orthonormal columns is
equal to its transpose.
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Position & orientation of a Rigid Body
Orientation representation
xA
zA
yA{A} xB
zB yB{B}Uniqueness
3 independent
parameters
333231
232221
131211
rrr
rrr
rrr
R
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Position & orientation of a Rigid Body
Elementary (Basic, Fundamental) Rotation
matrices
xAo
zAo
yA1
zA1
Rotation about xAo by angle
cossin0
sin-cos0
001
)(RR XA0
1
A
yAo
, xA1
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Position & orientation of a Rigid Body
Elementary (Basic, Fundamental) Rotation
matrices
xAo
zAo
,yA1
zA1
Rotation about yAo by angle
yAo
cos0sin-
010
sin0cos
)(RR YA0
1
A
xA1
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Position & orientation of a Rigid Body
Description of a frame
A set offour vectors givingposition (typically of the origin) and
orientation information of the frame
relative to the reference frame
xA
zA
yA
{A}xB
zB yB{B}
ApBORG
BORGAABR p,
E.g., Frame {B} relative to frame {A} is described by
Once we have attached a frame to a
rigid body, how to represent the
frame location and orientation?
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Coordinate Transformations (Mappings)
Mapping: Changing descriptions from frame to
frame Expressing the position vector of a point in
space in terms of various reference coordinate
systems
xA
zA
yA{A}
xB
zByB
p
{B}? B PGiven
FindA
P
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Coordinate Transformations (Mappings)
Mappings involving translated frames (i.e.,
frames having same orientations)
AP = BP + APBORG
Only applicable when orientations of frames A and B are the same
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Coordinate Transformations (Mappings)
Mappings involving rotated frames
Above equation assumes origins of Frames A & B are coincident.
What if the two origins are not coincident?
PRPBA
B
A
Coordinates of P in B
Coordinates of P in A
P
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Coordinate Transformations (Mappings)
Example 1-1: Figure 1-1 shows a frame {B} which is
rotated relative to frame {A} about z by 30 degrees.Here, z is pointing out of the page. There is a point Pwhose position vector expressed in {B} is Bp = [0 2 0]T.What is the position vector of the point P expressed in
{A}?
P
xA
xB
yAyB
30oFigure 1-1
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Coordinate Transformations (Mappings)
Solution:
Writing the unit vectors of {B} in terms of {A} and stacking
them as columns of the rotation matrix we obtain:
100
0866.05.0
05.0866.0
RA
B
GivenB
p = [0 2 0]T
,
0
732.1
1
ppBA
B
AR
P
xA
xB
yAyB
30o
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Coordinate Transformations (Mappings)
Solution:
100
0866.05.0
05.0866.0
RA
B
P
xA
xB
yAyB
30o
Remark:
RA
B acts as a mapping which isused to describe p relative to frame{A} given Bp. The original vectorp is notchanged in space. We simply
compute the new description of thevector relative to a new frame.
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Coordinate Transformations (Mappings)
Mappings involving general frames
xA
zA
yA
{A}
xB
zByB
p
{B}
?B PGiven
FindA
P
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Coordinate Transformations (Mappings)
Mappings involving general frames
xA
zA
yA
xA
zA
yA
{A}
xB
zByB
p
{B}
A A A B A A B
BORG B BORG BP P R P P R P
PRP BABA PPP ABORGAA and
3' IRA
A
{A} and {A} same
orientation:
RR ABA
B
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Coordinate Transformations (Mappings)
Mappings involving general frames
PRPP BABBORGAA
Translational
transformation
Rotational
transformation
xA
zA
yA
xA
zA
yA
{A}
xB
zByB
p
{B}
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Homogeneous Transformation
xA
zA
yA
{A}
xB
zByB
p
{B}
PRPPBA
BBORG
AA
PTPBA
B
A Homogeneous transformation
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Homogeneous Transformation
1
z
A
yA
x
A
A
p
p
p
P
1
z
B
y
Bx
B
B
p
p
p
P
10BORG
AA
BAB PRT
Homogeneoustransformation matrix
Augmented vector
4x4 matrix
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Homogeneous Transformation Example 1-2
Figure 1-2 shows a frame {B} which is rotated relative to
frame {A} about z by 30 degrees, and translated 10 units inxA , and 5 units in yA. Find
Ap where Bp = [3 7 0]T.
Figure 1-2
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Homogeneous Transformation Solution:
1000
0100
50866.05.0
1005.0866.0
TA
B
0
562.12
098.9
pAThat is,
1
0
562.12098.9
1
0
73
1000
0100
50866.05.01005.0866.0
pp BABA T
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Homogeneous Transformation
Interpretation 1:
Trepresents coordinate transformations in compactform
xA
zA
yA{A}
xB
zByB
p
{B}
10
BORG
AA
BA
B
PRT
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Homogeneous Transformation
Interpretation 2:
Trepresents position & orientation of the coordinateframe {B} relative to {A}
xA
zA
yA{A}
xB
zB yB
{B}
10
BORG
AA
BA
B
PRT
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Homogeneous Transformation
Interpretation 3:
Trepresents rotation and translation of the coordinateframe {A} to {B}
xA
zA
yA{A}
10
BORG
AA
BA
B
PRT
xB
zByB
{B}
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Homogeneous Transformation
For free vectors (e.g. force, velocity, etc),
augment the vectors with 0 rather than 1:
0
z
A
yA
x
A
A
p
p
p
P
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Homogeneous Transformation
Consecutive transformation:
x0
z0
y0{0}
x1
z1 y1
{1}
xn
znyn
{n}
p
n PGiven
Find0
P
xn-1
zn-1
yn-1{n-1}
Ai
i
1 homogeneous transformationmatrix from frame ito frame i-1
(i= 1,,n)
augmented position vector of P
expressed in frame n
augmented position vector of P
expressed in frame 0
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Homogeneous Transformation
Consecutive transformation:
0 0 1 1 0
1 2
n n n
n nP A A A P A P
Ai
i
1homogeneous transformation
matrix from frame ito frame i-1(i= 1,,n)
Pn
Po
augmented position vector of P
expressed in frame n
augmented position vector of P
expressed in frame 0
0 1 1 0
1 2
n
n n A A A A homogeneous transformation
from frame n to frame 0
where
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Homogeneous Transformation
Inverse of a Homogeneous Transformation
Matrix
TAB TB
A
1T
A
BGiven , find or
ITTA
B
A
B 1
3 3
1 0 0 0
00 1 0 0
0 1 0 10 1 0 0 1 0
0 0 0 1
A A B BORG X Y IR P
IRXAB RRRXB
ATA
BA
B 1
0 BORGAA
B PRYBORG
ATA
BBORG
AA
B PRPRY 1
By definition,
=>
=>
i.e.
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Homogeneous Transformation
Inverse of a Homogeneous Transformation
Matrix
1
0 1
A T A T A
B B BORGB AA B R R PT T
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Homogeneous Transformation Example 1-4
Figure 1-5 shows a frame {B} which is rotated relative to frame {A}
about z by 30 degrees, and translated four units in xA, and threeunits in yA. Find .T
B
A
Figure 1-5
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Homogeneous Transformation Solution:
1000
0100
30866.05.0
405.0866.0
TA
B
1000
0100
598.00866.05.0
964.405.0866.0
11
0
pBORG
ATA
B
TA
BAB
BA
RRTT
E l 1 5
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Example 1-5
Assume we know TBT in Figure 1-6, which describes the frame at the
manipulators fingertips {T} relative to the base of the manipulator, {B}. Also,
we know where the tabletop is located in space relative to the manipulators
base because we have a description of the frame {S} which is attached to the
table as shown, TB
S . Finally, we know the location of the frame attached to the
bolt lying on the table relative to the table frame, that is, TSG . Calculate the
position and orientation of the bolt relative to the manipulators hand, TT
G .
Figure 1-6
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Solution:
1T T B S B B S
G B S G T S GT T T T T T T
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Operators: Translations, Rotations, and
Transformation
Operators :
Another interpretation of the mathematicalforms used for coordinate transformations
Only one coordinate frame is involved
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Operators: Translations, Rotations, and
Transformation
Translational operators: Moves a point in space afinite distance along a given vector direction
xA
zA
yA{A}
AP1 AP2
AQ
How a vectorAP1 is translated by a vectorAQ :
Result of the operation is a new vectorAP2 =AP1 +AQ
12
1000
100
010
001
Pq
q
q
PA
z
y
x
A
To write the translation operation as a matrix operator,
where qx, qy and qz are components of the translation vector Q
x
y
z
q
q
q
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Operators: Translations, Rotations, and
Transformation
Rotational operators:
Operates on a vectorAP1
and changes that vector to a new
vectorAP2, by means of a rotation, R (about origin)
Same as the rotation matrix that describes a frame rotated byR relative to the reference frame
Define a rotational operatorRK():
AP2 = RK()AP1
- Kis the axis of rotation- (in degrees) is the amount of
rotation about K
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Operators: Translations, Rotations, and
Transformation
Rotational operators:
10000100
00cossin
00sincos
)(
ZR
E.g. operator that rotates about the Zaxis by can be written as:
100
0cossin
0sincos
)(
Z
R
homogeneous transformation matrix3x3 rotation matrix
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Operators: Translations, Rotations, and
Transformation Example 1-6
Figure 1-3 shows a vectorAp1 = [0 2 0]
T. We wish to compute the vector
obtained by rotating this vector about z by 30 degrees. Call the new vectorAp2.
Figure 1-3
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Operators: Translations, Rotations, and
Transformation
Solution:
Note: Rotation matrix which rotates vectors by 30 degrees about z =Rotation matrix which describes a frame rotated 30 degrees about zrelative to the reference frame.
100
0866.05.0
05.0866.0
)0.30(zR
0
732.1
1
0
2
0
100
0866.05.0
05.0866.0
)0.30( 12 pp zAA
R
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Operators: Translations, Rotations, and
Transformation
Transformation operators
Rotate and translate
Define operatorT to be one which rotates and translates a vectorAP1 to a new vector
AP2:
AP2= TAP1
(1-3)
The transform that rotates by R and translates by Q is thesame as the transform that describes a frame rotated by R
and translated by Q relative to the reference frame.
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Operators: Translations, Rotations, and
Transformation
Example 1-7
Figure 1-4 shows a vectorAp1. We wish to rotate it about z by 30degrees, and translate it 10 units in xA, and 5 units in yA. Find
Ap2 whereAp1 = [3 7 0]
T.
O T l i R i d
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Operators: Translations, Rotations, and
Transformation
Solution:
The operator T, which performs the rotation and translation, is
1000
0100
50866.05.0
1005.0866.0
T
10
562.12
098.9
10
7
3
10000100
50866.05.0
1005.0866.0
12 ppAA
T
Note that this example is numerically similar to Example 1-2, but the
interpretation is different.
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Other Orientation Representation
Orientation of rigid body represented by 3x3
rotation matrix R
9 elements Subject to:
orthogonalityconditions (3 equations) and
unit length conditions (3 equations)=> only 3 of 9 elements are independent (ie. there is
redundancy in R)
Different representations of orientation whichrequires only three or four parameters
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Other Orientation Representation
X-Y-Z fixed angles
Each of the three rotations takes place about an axis in the
fixed reference frame, {A}
X Y Z(roll) (pitch) (yaw)
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Other Orientation Representation
X-Y-Z fixed angles
X
ZYAW
PITCH
ROLL
Y
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Other Orientation Representation
X-Y-Z fixed angles
( , , ) ( ) ( ) ( )
0 0 1 0 0
0 0 1 0 0
0 0 1 0 0
A
B XYZ Z Y X R R R R
c s c s
s c c s
s c s c
c c c s s s c c s c s s
s c s s s c c s s c c s
s c s c c
where cand s denote cosine and sine functions, respectively
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Other Orientation Representation
X-Y-Z fixed angles
Inverse transformation
Given
333231
232221
131211
rrr
rrr
rrr
RAB , find , ,
Nine equations (6 dependencies) and three unknowns
11 12 13
21 22 23
31 32 33
r r r c c c s s s c c s c s s
r r r s c s s s c c s s c c s
r r r s c s c c
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Other Orientation Representation
X-Y-Z fixed angles
Inverse transformation
11 12 13
21 22 23
31 32 33
r r r c c c s s s c c s c s s
r r r s c s s s c c s s c c sr r r s c s c c
Remark:
Atan2(y,x) computes tan-1(y/x) and uses signs of both x and y to
determine the quadrant in which the angle lies
)1,(2tan
),(2tan
2
3131 rrA
csA
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Other Orientation Representation
X-Y-Z fixed angles
Inverse transformation
21 11
tan 2( , )
tan 2( , )
A s cr r
Ac c
11 12 13
21 22 23
31 32 33
r r r c c c s s s c c s c s s
r r r s c s s s c c s s c c sr r r s c s c c
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Other Orientation Representation
X-Y-Z fixed angles
Inverse transformation
11 12 13
21 22 23
31 32 33
r r r c c c s s s c c s c s s
r r r s c s s s c c s s c c sr r r s c s c c
32 33
tan 2( , )
tan 2( , )
A s c
r rA
c c
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Other Orientation Representation
X-Y-Z fixed angles
Inverse transformation
),(2tan2
21
2
1131 rrrA 32 3321 11tan 2( , ) tan 2( , )
r rr rA A
c c c c
Remark:
2 solutions in general If |r31|=1 (implies r11 = r21 = r32 = r33= 0), the solution degenerates(mathematical singularity) such that only the difference of and may be computed:
If r31 = +1: => + = Atan2 (-r23 , r22)
If r31 = -1: => - = Atan2 (r23 , r22)
12 13
22 23
31
0
0
0 0
r r c c c s s s c c s c s s
r r s c s s s c c s s c c s
r s c s c c
0tan 2( 1,0) 90A 0
tan 2(1,0) 90A
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Other Orientation Representation
Z-Y-X Euler angles
Each rotation is performed about an axis of the movingframe {B},
rather than the fixed reference frame, {A}
ZY X
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Other Orientation Representation
Z-Y-X Euler angles
RRRRB
B
B
B
A
B
A
B
( , , ) ( ) ( ) ( )
0 0 1 0 00 0 1 0 0
0 0 1 0 0
A
B Z Y X Z Y X R R R R
c s c ss c c s
s c s c
c c c s s s c c s c s ss c s s s c c s s c c s
s c s c c
Note:
Same as the result obtained for the same three rotations taken in theopposite order about fixed axes.
Recall:Consecutive
transformation
O O
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Other Orientation Representation
Four-parameter representations for
orientationEquivalent angle-axis
Euler parameters or Unit Quaternion Nonminimal representations of
orientation
Oth O i t ti R t ti
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Other Orientation Representation
Equivalent angle-axis
Anyrotation matrix can berepresented by choosing the
properaxis and angle (Eulers
theorem on rotation)
quadruple of ordered real
parameters consisting ofone
scalar (angle of rotation) and one
unitvector (axis of rotation)(according to right-handrule)
K
AK
Oth O i t ti R t ti
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Other Orientation Representation
Equivalent angle-axis
)( KAR
cvkkskvkksk-vkk
sk-vkkcvkkskvkk
skvkksk-vkkcvkk
zzxzyyzx
xyzyyzyx
yxzzxyxx
=
where v = (1 - cos), ,T
A
x y zK k k k , sign of is determined by right-hand rule
Remark:
Forsmallangular rotations, axis of rotation becomes ill-defined.
Non-unique representation: )()( KK RR
(1-4)
1 (unit vector)AK
Oth O i t ti R t ti
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Other Orientation Representation
Example 1-8
A frame {B} is described as follow: initiallycoincident with {A} we rotate {B} about the vectorAk = [0.707 0.707 0]T (passing through the origin)
by an amount = 30 degrees. Give the framedescription of {B} with reference to {A} after therotation.
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Solution:
x x y x z z x y
x y z y y z y x
x z y y z x z z
k k v30 c30 k k v30 - k s30 k k v30 k s3030 k k v30 k s30 k k v30 c30 k k v30 - k s30
k k v30 - k s30 k k v30 k s30 k k v30 c30
o o o o o o
A o o o o o o o
o o o o o o
R
k
where v30o = 1-cos 300;
Since there is no translation of the origin,
0 0.933 0.067 0.354 0
30 0 0.067 0.933 0.354 0
0 0.354 0.354 0.866 0
0 0 0 1 0 0 0 1
A o
A
B
RT
k
k = [kx, ky, kz]T =[0.707 0.707 0]T
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Example 1-9
A frame {B} is described as follows: initially coincident with {A}, we rotate
{B} about the vector
TA
]0.0707.0707.0[k , passing through the point TA P 321 , by o30 . Give the frame description of {B} with reference to{A}.
Solution:
Before performing the rotation, {A} and {B} are coincident. We define
two new frames, {A} and {B} which are obtained by translating {A} and
{B}, respectively, to point P. {B} and {B} can be treated as if they aremounted on the same rigid body, whereas {A} and {A} are fixed in the
space.
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Solution:
1000
3100
2010
1001
'TA
A
1000
3100
2010
1001
TBB
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Solution:
Note that the point P is on the axis of rotation. Now, lets
rotate {B} relative to {A}. This is a rotation about an axiswhich passes through the origin, so we may use
cvkkskvkksk-vkk
sk-vkkcvkkskvkk
skvkksk-vkkcvkk
zzxzyyzx
xyzyyzyx
yxzzxyxx
k
RA
to compute {B} relative to {A} after the rotation.
1000
0866.0354.0354.0
0354.0933.0067.0
0354.0067.0933.0
1000
0
0
0
30oAAB
RT k
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Solution:
1000
05.0866.0354.0354.0
13.1354.0933.0067.0
13.1354.0067.0933.0
TTTT BBA
B
A
A
A
B
Finally,
Note: After rotating about the axis k , {B}becomes separated from {A} as shown in Fig.
1-9.
Figure 1-9
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Other Orientation Representation
Equivalent angle-axis
Inverse Problem
RAB AK
x x x
A
B y y y
z z z
x x y x z z x y
x y z y y z y x
x z y y z x
n o a
Let R n o a
n o a
k k v c k k v - k s k k v k s
k k v k s k k v c k k v - k s
k k v - k s k k v k s
z zk k v c
given
Given Find ,
where v = (1 - cos)
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Other Orientation Representation
Equivalent angle-axis
Inverse Problem
RA
B
RA
B
From the diagonal elements,
Trace ( ) = 1 + 2cos
=> cos = ( Trace ( ) 1 )
(Two solutions for = A)+A
-A
x x y x z z x yx x x
y y y x y z y y z y x
z z z x z y y z x z z
k k v c k k v - k s k k v k sn o a
n o a k k v k s k k v c k k v - k s
n o a k k v - k s k k v k s k k
A
KR
v c
where v = (1 - cos)
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Equivalent angle-axis
Inverse Problem
x x y x z z x yx x x
y y y x y z y y z y x
z z z x z y y z x z z
k k v c k k v - k s k k v k sn o a
n o a k k v k s k k v c k k v - k s
n o a k k v
- k s
k k v
k s
k k
A
KR
v
c
sin2
n-ak zxy
sin2
a-ok
yz
x
sin2
o-n
k
xy
z
If sin 0 :From : nz & ax elements
From : oz & ay elements
From : ny & ox elements
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Other Orientation Representation
Equivalent angle-axis
Inverse Problem
AK
In general:
2 solutions
(1)
AK
(2)
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Other Orientation Representation
Equivalent angle-axis
Inverse Problem
100
010
001
RAB Identity (null rotation)
2
zzyzx
zy
2
yyx
zxyx
2
x
A
B
2k1-k2kk2k
k2k2k1-k2k
k2kk2k2k1-
R
If = 0
If = 180
( symmetric )
K = any axis (infinite solutions)
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Other Orientation Representation
Equivalent angle-axis
Inverse Problem
x
zz
x
y
y
z
yz
x
2k
ak
2k
nk
2o
nnk
2 solutions for
(If one solution is ,
the other is )
2
x x y x zx x x
A 2
B y y y x y y y z
2z z z x z y z z
-1 2k 2k k 2k k n o a
R n o a = 2k k -1 2k 2k k
n o a 2k k 2k k -1 2k
If = 180 (cont)
K
1K
1K
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Ot e O e tat o ep ese tat o
Equivalent angle-axis
Inverse Problem
RA
B
Ifall off-diagonal terms are 0, one of kx, ky or kz = 1 &
the rest are 0s. The component of that is 1 has avalue of 1 in the diagonal of .
E.g. kx = 1, ky = kz = 0
A
B
1 0 0
R 0 -1 0
0 0 -1
If = 180 (cont)
K
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p
Euler parameters (Unit Quaternion)
Able to solve the non-uniqueness problem
encountered by angle/axis representation
2sin1
xk
2sin2
yk
2sin3
zk
2cos4
K
Euler parameters:
where kx, ky and kz are the coordinates of the equivalent axis
and is the equivalent angle
4 parameters not
independent because:
1242
3
2
2
2
1
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p
Euler parameters (Unit Quaternion)
Based on Eq ( 1-4) (equivalent angle-axis) and the definition of unit
quaternion:
1)2(2)(2)(2
)(21)(2)(2
)(2)(21)(2
2
3
2
441324231
4132
2
2
2
44321
42314321
2
1
2
4
R (1-5)
x x y x z z x y
x y z y y z y x
x z y y z x z z
k k v c k k v - k s k k v k s
k k v k s k k v c k k v - k s
k k v - k s k k v k s k k v c
A
B R
1
sin
2
xk
2
sin
2
yk
3
sin
2
zk
2 22 4cos 2cos 1 2 1
2 22sin 1 cos , etc
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p
Euler parameters (Unit Quaternion)
Inverse Problem
333231
232221
131211
rrr
rrr
rrr
RAB
1)sgn(2
133221123321 rrrrr
1)sgn(2
111332231132 rrrrr
1)sgn(2
122113312213 rrrrr
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1rrr
Given , find i
Note: sgn(x) = 1 for x 0 and sgn(x) = -1 for x < 0
2 2
4 1 1 2 3 4 1 3 2 4
2 2
1 2 3 4 4 2 2 3 1 4
2 2
1 3 2 4 2 3 1 4 4 3
2( ) 1 2( ) 2( )
2( ) 2( ) 1 2( )
2( ) 2( ) 2( ) 1
R
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p
Euler parameters (Unit Quaternion)
Inverse Problem
1)sgn(2
133221123321 rrrrr
1)sgn(2
111332231132 rrrrr
1)sgn(2
122113312213 rrrrr
3322114 12
1rrr
Note:
Implicitly assumed (by considering [-,])04
No singularity occurs (compared to angle/axis representation)