Transcript Computer Animation and Special Effects

Chap 5
Kinematic Linkages
Animation (U), Chap 5
Kinematic Linkages
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Model Hierarchy vs. Motion Hierarchy

Relative motion


An object’s motion is relative to another object
Object hierarchy + relative motion form
motion hierarchy

Linkages


Components of the hierarchy represent
objects that are physically connected
Motion is often restricted
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Kinematics


How to animate the linkages by specifying or
determining position parameters over time
The branch of mechanics concerned with the
motions of objects without regard to the forces
that cause the motion
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Kinematics
Forward Kinematics

Animator specifies rotation parameters at joints
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Kinematics
Inverse Kinematics

Animator specifies the desired position of hand,
and system solves for the joint angles that satisfy
that desire.
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Hierarchical Modeling
Articulated Model

Represent an articulated figure as a series of links
connected by joints
root
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Degrees of Freedom (DOF)

The minimum number of coordinates required to
specify completely the motion of an object
yaw
roll
pitch
6 DOF: x, y, z, raw, pitch, yaw
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Degrees of Freedom
One DOF Joints

One DOF joint


Allow motion in one direction
Exam: Revolute joint, Prismatic joint
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Degrees of Freedom
Complex Joints

Complex joints

2 DOF joint


Planar joint, universal joint
3 DOF joint
Gimbal
 Ball and socket


Complex joints of DOF n can be modeled as n
one-DOF joints connected by n-1 links of 0
length.
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Degrees of Freedom
Complex Joints
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Degrees of Freedom in Human Model
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Root: 3 translational DOF + 3 rotational DOF
Rotational joints are commonly used
Each joint can have up to 3 DOF


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Shoulder: 3 DOF
Wrist: 2 DOF
Knee: 1 DOF
3 DOF
1 DOF
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Hierarchical Modeling
Data Structure

Hierarchical linkages can be represented by a
tree-like structure




Root node – corresponds to the root object whose
position is known in the global coordinate system
Other nodes – located relative to the root node
Leaf nodes –
Mapping between hierarchy and tree


Node – object part
Arc – joint or transformation to apply to all nodes
between it in the hierarchy
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Hierarchical Modeling
Data Structure
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Hierarchical Modeling
Data Structure
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Hierarchical Modeling
A Simple Example
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Hierarchical Modeling
Tree Structure
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Hierarchical Modeling
Transformations
V0 : a vertex of Link 0
Its world coordinate : V0  T0V0
'
V1 : a vertex of Link 1
Its world coordinate : V1  T0T1V1
'
V1,1 : a vertex of Link 1,1
Its world coordinate : V1,1  T0T1T1,1V1,1
'
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Hierarchical Modeling
Variable rotations at joints
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Hierarchical Modeling
Tree Structure
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Hierarchical Modeling
Transformations
V0 : a vertex of Link 0
Its world coordinate : V0  T0V0
'
V1 : a vertex of Link 1
Its world coordinate : V1  T0T1 R1 (1 )V1
'
V1,1 : a vertex of Link 1,1
Its world coordinate : V1,1  T0T1 R1 (1 )T1,1 R1,1 (1,1 )V1,1
'
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Hierarchical Modeling
With Two Appendages
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Hierarchical Modeling
With Two Appendages
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Joint Space vs. Cartesian Space
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Joint space
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space formed by joint angles
position all joints—fine level control
Cartesian space


3D space
specify environment interactions
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Forward and Inverse Kinematics

Forward kinematics


mapping from joint space to cartesian space
Inverse kinematics

mapping from cartesian space to joint space
1
P
P
2
Forward Kinematics
Inverse Kinematics
P  f (1,2 )
1,2  f 1 ( P)
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Forward Kinematics

Evaluate the tree


Depth-first traversal from the root to leaf
Repeat the following until all nodes and arcs have
been visited



The traversal then backtracks up the tree until an unexplored
downward arc is encountered
The downward arc is then traversal
In implementation

Requires a stack of transformations
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Forward Kinematics
Tree Traversal
M=I
T0
T1.1
M = T0
T 1.2
M = T0*T1.1
M = T0*T1.1*T2.1
T2.1
T2.2
M = T0*T1.1
M = T0
M = T0*T1.2
M = T0*T1.2*T2.2
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Forward Kinematics
Example
?
End Effector
Base
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Inverse Kinematics

The initial pose vector and a desired pose
vector are given, come out the joint values
required to attain the desired pose

Can have zero, one, or more solutions
Over-constrained system
 Under-constrained system

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Singularity problems
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Inverse Kinematics

Example
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2 equations (constraints)
3 unknowns

Infinitely many solutions exist!


This is not uncommon!

See how you can move your elbow while keeping
your finger touching your nose
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Other problems in IK

Infinite many solutions
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Other problems in IK
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No solutions
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Inverse Kinematics

Once the joint values are calculated, the figure
can be animated by interpolating the initial
pose vector values to the final pose vector
values.

Does not provide a precise or an appropriate
control when there is large differences between
initial and final pose vectors. Alternatively,
Interpolate pose vectors
 Do inverse kinematics for each interpolated pose vector

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Inverse Kinematics
Analytic vs. Numerical

Analytic solutions exist only for relatively
simple linkages

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Inspecting the geometry of linkages
Algebraic manipulation of equations that describe
the relationship of the end effector to the base
frame.
Numerical incremental approach


Inverse Jacobian method
Other methods
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Inverse Kinematics
Analytic Method
2
L1
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1
34
L2
Goal
(X,Y)
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Inverse Kinematics
Analytic Method
L1
L2
180- 2
1
(X,Y)
x2  y2
T
Y
X

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Inverse Kinematics
Cosine Law
C
A
a
B
A  B C
cos(a ) 
2 AB
2
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2
2
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Inverse Kinematics
Analytic Method
L2
L1
1
x2  y2

L1  L2  X  Y
2 L1 L2
2
2
2

2
2
2
Animation (U), Chap 5 Kinematic Linkages
X 2 Y 2


2
2 
 X Y 
T  cos 1 

cos(1  T ) 

L1  L2  X 2  Y 2
 2  180  cos (
)
2 L1 L2
1
X

Y
X

cos(180   2 ) 
(X,Y)
180- 2
T
cos(T ) 
1  cos (
1
37
X
L1  X 2  Y 2  L2
2
2
2 L1 X 2  Y 2
L1  X 2  Y 2  L2
2
2 L1 X  Y
2
2
2
)  T
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Inverse Kinematics
More complex joints
End Effector=P
Base
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Why is IK hard?
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Redundancy
Natural motion control

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joint limits
minimum jerk
style?
Singularities


ill-conditioned
Singular
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Solving Inverse Kinematics
Numerically
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

Inverse-Jacobian method
Optimization-based method
Example-based method
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Inverse-Jacobian method
P  f ( )
P  Rn (n  6 usually)
  R (m  DOFs)
m

Jacobian is the n by m matrix relating differential
changes of  d to differential changes of P (dP)
dP
d
 J ( )
dt
dt

f i
where J ij 
 j
So Jacobian maps velocities in joint space to velocities
in Cartesian space
V  J ( )
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Inverse-Jacobian method

Recall the inverse kinematics problem
1
  f ( P)


f is highly nonlinear
Make the problem linear by localizing about the
current operating position and inverting the
Jacobian
  J 1V

and iterating towards the desired post over a series of
incremental steps.
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Inverse-Jacobian method

Given initial pose and desired pose, iteratively
change the joint angles to approach the goal
position and orientation

Interpolate the desired pose vectors for some
intermediate steps

For each step k, find the angular velocities for joints,
and do

 k 1   k  t
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Inverse-Jacobian method
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Computing Jacobian

Compute analytically


Ok for simple joints
Geometric approach


For complex joints
Let’s say we are just concerned with the end
effector position for now.
This also implies that the Jacobian with be an 3×N
matrix where N is the number of DOFs
 For each joint DOF, we analyze how e would change if
the DOF changed

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Computing Jacobian Analytically
An Example
End Effector=P
Base
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Computing Jacobian Analytically
An Example
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Computing Jacobian Geometrically
Jacobian for a 2D arm

Let’s say we have a simple 2D robot arm with
two 1-DOF rotational joints:
• e=[ex ey]
φ2
φ1
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Computing Jacobian Geometrically
Jacobian for a 2D arm

The Jacobian matrix J(Φ) shows how each
component of e varies with respect to each
joint angle
 ex
 
J Φ   1
 e y
 1
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ex 
2 

e y 
2 
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Computing Jacobian Geometrically
Jacobian for a 2D arm

Consider what would happen if we increased φ1 by a
small amount. What would happen to e ?
e  ex

1  1
e y 

1 
•
φ1
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Computing Jacobian Geometrically
Jacobian for a 2D arm

What if we increased φ2 by a small amount?
e  ex

2  2
Animation (U), Chap 5 Kinematic Linkages
e y 

2 
•
φ2
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Computing Jacobian Geometrically
Jacobian for a 2D arm
 ex
 
V  J Φ    1
 e y
 1
ex 
2  

e y 
2 
•
φ2
φ1
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Computing Jacobian Geometrically
Jacobian for a 2D arm
Angular and linear velocities induced by joint axis rotation.
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Computing Jacobian Geometrically
Jacobian for 2D arm


Let’s consider a 1-DOF rotational joint first
We want to know how the global position e
will change if we rotate around the axis.
e  ex

1  1
•
e y 

1 
 a1  (e  r1 )
φ1
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Computing Jacobian Geometrically
Jacobian for 2D arm e
e
 i
•
e  ri
e
 ai  e  ri
i
ri •
ai
a’i: unit length rotation axis in world space
r’i: position of joint pivot in world space
e: end effector position in world space
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Computing Jacobian Geometrically
3-DOF Rotational Joints


Once we have each axis in world space, each
one will get a column in the Jacobian matrix
At this point, it is essentially handled as three
1-DOF joints, so we can use the same formula
for computing the derivative as we did earlier:
e
 ai  e  ri
i

We repeat this for each of the three axes
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Computing Jacobian Geometrically
Jacobian for another 2D arm
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Computing Jacobian Geometrically
Jacobian for another 2D arm
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Computing Jacobian Geometrically
Jacobian for another 2D arm

The problem is to determine the best linear
combination of velocities induced by the
various joints that would result in the desired
velocities of the end effector.

Jacobian is formed by posing the problem in
matrix form.
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Computing Jacobian Geometrically
Jacobian for another 2D arm
ex ex

(G  E ) x 
  
2
V  (G  E ) y   J Φ    1
 e y e y
 (G  E ) z 
 1 2
((0,0,1)  E ) x ((0,0,1)  ( E  P1 ) x

 ((0,0,1)  E ) y ((0,0,1)  ( E  P1 ) y
 ((0,0,1)  E ) z ((0,0,1)  ( E  P1 ) z
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ex 
3  

e y 
3 
((0,0,1)  ( E  P2 ) x 

((0,0,1)  ( E  P2 ) y 
((0,0,1)  ( E  P2 ) z 
CS, NCTU, J. H. Chuang
Inverse-Jacobian method
  f ( P)
1
V  J ( )
1

  J ( )V


Linearize about k locally
Jacobian depends on current configuration
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Inverse-Jacobian method
Problems


Problems in localizing the P = f (θ)
Problems in solving the system

Non-square matrix, but independent rows


Use pseudo inverse
Singularity
When the inverse of Jacobian does not exist
 occurs when any  cannot achieve a given


V
Near singularity

ill-conditioned
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Inverse-Jacobian method
Iterative approach

Given initial pose and desired pose, iteratively
change the joint angles to approach the goal
position and orientation

Interpolate the desired pose vectors for some
intermediate steps

For each step k, compute V, and find the angular
velocities   J 1V for joints, and do
 k 1   k  t
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Inverse-Jacobian method
Iterative approach
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Inverse-Jacobian method
Iterative approach
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Inverse-Jacobian method
Iterative approach

V
No linear combination of vectors gi could produce the desired goal.
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Inverse-Jacobian method
Iterative approach
Frame:

V
Total: 21 frames
No linear combination of vectors gi could produce the desired goal.
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Inverse-Jacobian method
Problems


Problems in localizing the P = f (θ)
Problems in solving the system

Non-square matrix, but independent rows


Use pseudo inverse
Singularity
When the inverse of Jacobian does not exist
 occurs when any  cannot achieve a given



Near singularity

ill-conditioned
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Inverse-Jacobian method
Remedy to Singularity Problems

Problems with singularities can be reduced if
the manipulator is redundancy – when there
are more DOF than there are constraints to be
satisfied


Jacobian is not square
If rows of J are linearly independent (J has full
row rank), then ( JJ T ) 1 exists and instead,
pseudo inverse of Jacobian can be used!
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Inverse-Jacobian method
Pseudo Inverse of the Jacobian
V  J
J V  J J
T
T
T
1
Where J has full row rank, ( J J ) exists!
T
1 T
T
1 T
( J J ) J V  ( J J ) J J
J V  

where J   ( J T J ) 1 J T  J T ( J J T ) 1
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Inverse-Jacobian method
Pseudo Inverse of the Jacobian
J V  
( J T J ) 1 J T V  
J  V  
J T ( JJ T ) 1V  
Let   ( JJ T ) 1V , which implies ( JJ T )   V
and be solved using LU decomposit ion (solve for  ).
J T   
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Inverse-Jacobian method
Dealing with singularities

Physically the singularities usually occur when
the linkage id fully extended or when the axes
of separate links align themselves
V1 and V2 are
in parallel
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Inverse-Jacobian method
Dealing with singularities

Two approaches to the case of full extension


Simply not allow the linkage to become fully
extended
Adopt a different iterative process with the region
of singularity to avoid the case of singularity
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Inverse-Jacobian method
Dealing with near singularities

Small V implies very large 
Two cases
of near
singularity
Near singularity occurs
When 3 links all aligned
And the end is positioned
over the base.
Small V
implies
very large 
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Inverse-Jacobian method
Dealing with near singularities

Damped least square approach

a user-supplied parameter is used to add in a term
that reduces the sensitivity of the pseudoinverse,
T
T
1

  J ( JJ   I ) V
Let   ( JJ   I ) V , and
T
1
solve ( JJ T   I )   V for 
Finally,   J T 
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Adding more control to IK




Pseudoinverse computes one of many possible
solutions that minimizes joint angle velocities
IK using pseudoinverse Jacobian may not
provide natural poses
A control term can be added to the
pseudoinverse Jacobian solution
The control term should not add anything to
the velocities, that is J  V
control term
  J V  ( J  J  I ) z
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Adding more control to IK
Control Term Adds Zero Linear Velocities
A solution of this form
  ( J  J  I ) z
When put into this formula
V  J

V  J (J J  I )z
Like this
V  ( JJ  J  J ) z
V  (J  J )z
After some manipulation,
you can show that it…
V  0z
…doesn’t affect the
desired configuration
V 0
But it can be used to bias
the solution vector
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Adding more control to IK

To bias the solution toward specific joint
angles, such as the middle joint angle between
joint limits, z is defined as
z  a i ( i   ci ) 2
where  i : current joint angles,
 ci : desired joint angles
a : joint gains
This is not a hard constraint s; but the solution
can be biased toward the middle values.
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Adding more control to IK

Joint gain


indicates the relative importance of the associated
desired angle
The higher the gain, the stiff the joint
High: the solution will be such that the joint angle
quickly approach the desired joint angle
 All gians are zero: reduced to the conventional
pseudoinverse of Jacobian.

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Adding more control to IK
How to solve the system?
  J V  ( J  J  I ) z



  J V  J Jz  Iz
T
T 1

  J ( JJ ) (V  Jz )  z


T
T 1

  J ( JJ ) (V  Jz )  z
Let   ( JJ ) (V  Jz ),
T 1
and solve ( JJ )   (V  Jz ) for  .
T

and comoute   J   z
T
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Adding more control to IK
Gain s
Top: {0.1, 0.5, 0.1}
Bottom: {0.1, 0.1, 0.5}
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Null space of Jacobian

The control term  is in the null space of J
  ( J  J  I ) z

The null space of J is the set of vectors
which have no influence on the constraints
  nullspace ( J )  J  0
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Utility of Null Space

The null space can be used to reach
secondary goals
  J V  ( J  J  I ) z
min G ( )
z

Or to find natural pose / control stiffness of joints
G( )  ai (natural(i )   (i ))
2
i
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Optimization-based Method

Formulate IK as an nonlinear optimization problem
 Example
Objective function
 Constraint
 Iterative algorithm
Nonlinear programming method by Zhao & Badler,
TOG 1994


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Objective Function


“distance” from the end effector to the goal
position/orientation
Function of joint angles : G()
Goal
distance
End Effector
Base
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Objective Function
Position Goal
Position/Orientation Goal
Animation (U), Chap 5 Kinematic Linkages
Orientation Goal
weighted sum
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Nonlinear Optimization

Constrained nonlinear optimization problem
limb coordination
joint limits

Solution




standard numerical techniques
MATLAB or other optimization packages
usually a local minimum
depends on initial condition
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Example-based Method


Utilize motion database to assist IK solving
IK using interpolation

Rose et al., “Artist-directed IK using radial basis function
interpolation,” Eurographics’01
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Example-based Method (cont.)

IK using constructed statistical model


Grochow et al., “Style-based inverse kinematics,”
SIGGRAPH’04
Provide the most likely pose based on given constraints
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Example-based Method (cont.)

Constructed pose space (training, extrapolated)
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Videos

“Style-based inverse kinematics”
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