Post on 18-Dec-2015
CSCE 641: Forward kinematics and inverse
kinematicsJinxiang Chai
Outline
Kinematics– Forward kinematics– Inverse kinematics
Kinematics
The study of movement without the consideration of the masses or forces that bring about the motion
Degrees of freedom (Dofs)
The set of independent displacements that specify an object’s pose
Degrees of freedom (Dofs)
The set of independent displacements that specify an object’s pose
• How many degrees of freedom when flying?
Degrees of freedom (Dofs)
The set of independent displacements that specify an object’s pose
• How many degrees of freedom when flying?
• So the kinematics of this airplane permit movement anywhere in three dimensions
• Six• x, y, and z positions• roll, pitch, and yaw
Degrees of Freedom
• How about this robot arm?
• Six again
• 2-base, 1-shoulder, 1-elbow, 2-wrist
Work Space vs. Configuration Space
• Work space– The space in which the object exists– Dimensionality
• R3 for most things, R2 for planar arms
• Configuration space– The space that defines the possible object
configurations– Degrees of Freedom
• The number of parameters that are necessary and sufficient to define position in configuration
Forward vs. Inverse Kinematics
• Forward Kinematics– Compute configuration (pose) given individual
DOF values
• Inverse Kinematics– Compute individual DOF values that result in
specified end effector position
Example: Two-link Structure
• Two links connected by rotational joints
BaseEnd Effector
θ2
θ1 X=(x,y)
l2l1
Example: Two-link Structure
• Animator specifies the joint angles: θ1θ2
• Computer finds the position of end-effector: x
BaseEnd Effector
θ2
θ1
x=f(θ1, θ2)
l2l1
(0,0)
X=(x,y)
Example: Two-link Structure
• Animator specifies the joint angles: θ1θ2
• Computer finds the position of end-effector: x
BaseEnd Effector
θ2
θ1
x = (l1cosθ1+l2cos(θ2+ θ2)
y = l1sinθ1+l2sin(θ2+ θ2))
(0,0)
θ2 l2l1
X=(x,y)
Forward kinematics
BaseEnd Effector
θ2
θ1
(0,0)
θ2 l2l1
• Create an animation by specifying the joint angle trajectories
θ1
θ2
X=(x,y)
Inverse kinematics
• What if an animator specifies position of end-effector?
Base
End Effector
θ2
θ1
(0,0)
θ2 l2l1
X=(x,y)
Inverse kinematics
• Animator specifies the position of end-effector: x
• Computer finds joint angles: θ1θ2
Base
End Effector
θ2
θ1
(0,0)
θ2 l2l1
(θ1, θ2)=f-1(x)
X=(x,y)
Inverse kinematics
• Animator specifies the position of end-effector: x
• Computer finds joint angles: θ1θ2
Base
End Effector
θ2
θ1
(0,0)
θ2 l2l1
X=(x,y)
Why Inverse Kinematics?
• Basic tools in character animation - key frame generation
- animation control - interactive manipulation
• Computer vision (vision based mocap)• Robotics• Bioinfomatics (Protein Inverse Kinematics)
Inverse kinematics
• Given end effector position, compute required joint angles
• In simple case, analytic solution exists– Use trig, geometry, and algebra to solve
Inverse kinematics
• Analytical solution only works for a fairly simple structure
• Iterative approach needed for a complex structure
Iterative approach
• Inverse kinematics can be formulated as an optimization problem
Iterative approaches
Find the joint angles θ that minimizes the distance between the character position and user specified position
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(Cx,Cy)
i
ii Cf 2)(minarg
Iterative approaches
Find the joint angles θ that minimizes the distance between the character position and user specified position
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
Iterative approach
Mathematically, we can formulate this as an optimization problem:
The above problem can be solved by many nonlinear optimization algorithms:
- Steepest descent
- Gauss-newton
- Levenberg-marquardt, etc
i
ii cf 2)(minarg
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
0
kkk 1
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
0
kkk 1
How can we decide the amount of update?
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
ii
kki cf
2)(minarg
0
kkk 1
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
ii
kki cf
2)(minarg
0
kkk 1
Known!
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
ii
kki cf
2)(minarg
0
kkk 1
Taylor series expansion
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
ii
kki cf
2)(minarg
2
)(minarg
ii
kkik
i cf
f
0
kkk 1
Taylor series expansion
rearrange
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
ii
kki cf
2)(minarg
2
)(minarg
ii
kkik
i cf
f
0
kkk 1
Taylor series expansion
rearrange
Can you solve this optimization problem?
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
ii
kki cf
2)(minarg
2
)(minarg
ii
kkik
i cf
f
2
))((minarg
i
kii
kki fcf
0
kkk 1
Taylor series expansion
rearrange
This is a quadratic function of
Gauss-newton approach
• Optimizing an quadratic function is easy
• It has an optimal value when the gradient is zero
),...,(
),...,(
),...,(
...
...
1
122
11
2
1
2
2
2
2
1
2
1
2
1
1
1
NiN
N
Ni
N
N
MM
N
M
N
N
fc
fc
fc
fff
fff
fff
2
))((minarg
i
kii
kki fcf
Gauss-newton approach
• Optimizing an quadratic function is easy
• It has an optimal value when the gradient is zero
),...,(
),...,(
),...,(
...
...
1
122
11
2
1
2
2
2
2
1
2
1
2
1
1
1
NiN
N
Ni
N
N
MM
N
M
N
N
fc
fc
fc
fff
fff
fff
2
))((minarg
i
kii
kki fcf
bJ Δθ
Linear equation!
Gauss-newton approach
• Optimizing an quadratic function is easy
• It has an optimal value when the gradient is zero
),...,(
),...,(
),...,(
...
...
1
122
11
2
1
2
2
2
2
1
2
1
2
1
1
1
NiN
N
Ni
N
N
MM
N
M
N
N
fc
fc
fc
fff
fff
fff
2
))((minarg
i
kii
kki fcf
bJ Δθ
Gauss-newton approach
• Optimizing an quadratic function is easy
• It has an optimal value when the gradient is zero
),...,(
),...,(
),...,(
...
...
1
122
11
2
1
2
2
2
2
1
2
1
2
1
1
1
NiN
N
Ni
N
N
MM
N
M
N
N
fc
fc
fc
fff
fff
fff
2
))((minarg
i
kii
kki fcf
bJ Δθ
bJJJ TT 1)(
Jacobian matrix
• Jacobian is a M by N matrix that relates differential changes of θ to changes of C
• Jacobian maps the velocity in joint space to velocities in Cartesian space
• Jacobian depends on current state
N
MMM
N
N
fff
fff
fff
21
2
2
2
1
2
1
2
1
1
1
...
...
Dofs (N)
The
num
ber
of c
onst
rain
ts (
M)
Jacobian matrix
• Jacobian maps the velocity in joint angle space to velocities in Cartesian space
(x,y)
θ1
θ2
Jacobian matrix
• Jacobian maps the velocity in joint angle space to velocities in Cartesian space
(x,y)
θ1
θ2
A small change of θ1 and θ2 results in how much change of end-effector position (x,y)
Jacobian matrix
• Jacobian maps the velocity in joint angle space to velocities in Cartesian space
2
1
2
1
1
2
2
1
1
1
ff
ff
y
x
(x,y)
θ1
θ2
A small change of θ1 and θ2 results in how much change of end-effector position (x,y)
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
f1f2
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
)sin(sin
)cos(cos
212112
212111
llf
llf
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
N
MMM
N
N
fff
fff
fff
21
2
2
2
1
2
1
2
1
1
1
...
...
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
)cos()cos(cos
)sin()sin(sin
21221211
21221211
lll
lllJ
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
)cos()cos(cos
)sin()sin(sin
21221211
21221211
lll
lllJ
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
)cos()cos(cos
)sin()sin(sin
21221211
21221211
lll
lllJ
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
1
21211
1
1 ))cos(cos(
llf
Jacobian matrix: 2D-link structure
2221211
2121211
,))sin(sin())cos(cos(minarg
21
cllcll
)cos()cos(cos
)sin()sin(sin
21221211
21221211
lll
lllJ
Base
θ2
θ1
(0,0)
θ2 l2l1
C=(c1,c2)
2
21211
2
2 ))sin(sin(
llf
Gauss-newton approach
Step 1: initialize the joint angles with
Step 2: update the joint angles:
0
kkk 1
Step size: specified by the user
Another Example
• A 2D lamp character
1
2
3
),,( 0yx
3c
2c
1c
0c
A
48
Another Example
• A 2D lamp character3l
A2l
1l
0l
49
Another Example
• A 2D lamp character
1
2
3
3c
2c
1c
A
Given , , how to compute the global position of the point A?
),,,,,( 3210 yx
?
),,( 0yx0c
50
Composite 2D Transformation
• What’s local coordinate ?
1
2
3
3c
2c
1c
A
?
A
),,( 0yx0c
51
Composite 2D Transformation
• What’s local coordinate ?
1
2
3
3c
2c
1c
A
?
A
1
03
0
l
p
),,( 0yx0c
03221100 )()0,()()0,()()0,()(),( pRlTRlTRlTRyxTp 52
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
),,( 0yx0c
1
03
0
l
p
03221100 )()0,()()0,()()0,()(),( pRlTRlTRlTRyxTp
100
0cossin
0sincos
)( 33
33
3
R
53
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
),,( 0yx0c
1
00
l
p
03221100 )()0,()()0,()()0,()(),( pRlTRlTRlTRyxTp
100
010
01
)0,(2
2
l
lT
54
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
),,( 0yx0c
1
00
l
p
03221100 )()0,()()0,()()0,()(),( pRlTRlTRlTRyxTp
100
0cossin
0sincos
)( 22
22
2
R
55
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
),,( 0yx0c
1
00
l
p
03221100 )()0,()()0,()()0,()(),( pRlTRlTRlTRyxTp
100
010
01
)0,(1
1
l
lT
56
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
),,( 0yx0c
1
00
l
p
03221100 )()0,()()0,()()0,()(),( pRlTRlTRlTRyxTp
100
0cossin
0sincos
)( 11
11
1
R
57
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
),,( 0yx0c
1
00
l
p
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
100
10
001
),0( 00 llT
58
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
?
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
),,( 0yx0c
1
00
l
p
100
cossin
sincos
)(),( 00
00
0 y
x
RyxT
59
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
),,( 0yx0c
60
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
),,( 0yx0c
61
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
),,( 0yx0c
62
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
),,( 0yx0c
63
Composite 2D Transformation
• What’s the current coordinate A ?
1
2
3
3c
2c
1c
A
03221100 )()0,()()0,()(),0()(),( pRlTRlTRlTRyxTp
),,( 0yx0c
64
More Complex Characters?
• A 2D lamp with 6 degrees of freedom
1
2
3
3c
2c
1c
0p
032211003210 )()0,()()0,()(),0()(),(),,,,,( pRlTRlTRlTRyxTyxf ),,( 0yx0c
base
Upper arm
lower arm
middle arm
If you do inverse kinematics with position of this point
What’s the size of Jacobian matrix?
More Complex Characters?
• A 2D lamp with 6 degrees of freedom
1
2
3
3c
2c
1c
0p
032211003210 )()0,()()0,()(),0()(),(),,,,,( pRlTRlTRlTRyxTyxf ),,( 0yx0c
base
Upper arm
lower arm
middle arm
If you do inverse kinematics with position of this point
What’s the size of Jacobian matrix?
2-by-6 matrix
Human Characters
xRRTRTRRRTRRRzyxTf )()()()()()()()()(),,( 332
321
21110
1000000 67
Inverse kinematics
• Analytical solution only works for a fairly simple structure
• Iterative approach needed for a complex structure
Inverse kinematics
• Is the solution unique?
• Is there always a good solution?
Ambiguity of IK
• Multiple solutions
Ambiguity of IK
• Infinite solutions
Failures of IK
• Solution may not exist
Inverse kinematics
• Generally ill-posed problem when the number of Dofs is higher than the number of constraints
Inverse kinematics
• Generally ill-posed problem when the number of Dofs is higher than the number of constraints
• Additional objective– Minimal Change from a reference pose θ0
Inverse kinematics
• Generally ill-posed problem when the number of Dofs is higher than the number of constraints
• Additional objective– Minimal Change from a reference pose θ0
2
0
2)(minarg
iii cf
Inverse kinematics
• Generally ill-posed problem when the number of Dofs is higher than the number of constraints
• Additional objective– Minimal Change from a reference pose θ0
2
0
2)(minarg
iii cf
Satisfy the constraints Minimize the difference between the solution and a reference pose
Inverse kinematics
• Generally ill-posed problem when the number of Dofs is higher than the number of constraints
• Additional objective– Minimal Change from a reference pose θ0
– Naturalness g(θ) (particularly for human characters)
Inverse kinematics
• Generally ill-posed problem when the number of Dofs is higher than the number of constraints
• Additional objective– Minimal Change from a reference pose θ0
– Naturalness g(θ) (particularly for human characters)
)()(minarg2
gcf
iii
Inverse kinematics
• Generally ill-posed problem when the Dofs is higher than the number of constraints
• Additional objective– Minimal Change from a reference pose θ0
– Naturalness g(θ) (particularly for human characters)
)()(minarg2
gcf
iii
Satisfy the constraints How natural is the solution pose?
Interactive Human Character Posing
• video