cs 775: advanced computer graphicsparagc/teaching/2012/cs775/... · 2012-01-13 · cs 775: lecture...

CS 775: Advanced Computer Graphics

Lecture 3 : Kinematics

CS 775: Lecture 3 Parag Chaudhuri, 2012

Character Animation

● Traditional– Cell Animation, hand drawn, 2D

– Lead Animator for keyframes

http://animation.about.com/od/flashanimationtutorials/ss/flash31detanim2.htm


Character Animation

● Traditional, hand drawn animation– Lead Animator for keyframes and many

secondary animators for the in-betweens



Character Animation

● Traditional, hand drawn animation– Keyframing @24fps – how many drawings for

a 2 hour movie?



Character Animation

● Computer assisted keyframing

– Keyframes created/posed by hand

– In-betweens interpolated by the semi-automatic techniques

Luxo Jr, PIXAR, 1986

How is this done?


Character Animation

● Interpolating Position and Orientation parameters and animating (rigid) transformations.

● Linear interpolation, spline interpolation, quaternions.

Luxo Jr, PIXAR, 1986

How is this done?


Character Animation

● Creating the keyframe pose for the lamp

– Modify the joint angles at each internal joint

– Modify global position and orientation


Character Animation

● Kinematics

– Study of motion of objects by studying the change in their orientation and position. The cause of the motion, i.e., the forces are not studied.

● Dynamics

– Study of motion of objects in relation to the forces and torques that cause the motion.


Character Animation● Kinematics

– Forward and Inverse Kinematics – used for posing characters and interpolation in keyframed animation.

– Faster to compute, easier to control

● Dynamics

– Physically-based animation – used for animations involving simulations of real world physics, for e.g., collisions.

– Harder to compute and control, more realistic (if modelled correctly),

– Can automatically adjust to changes in the environment.


Character Animation● Terminology

Root Joint

Internal Joints

EndeffectorLinks or Bones

Kinematic Chain


Character Animation● Terminology

0,0,0, t x , t y , t z

12

Joint DoF (parameters)

Root DoF


Character Animation● Forward Kinematics

– Given the joint parameters, find the position of the end effector.

– Position of the root is given as a global transformation.

– Joint parameters are given as relative rotations (local transformations).

– We already know how to solve this! (from CS475/CS675)


Character Animation● Inverse Kinematics

– Given the desired end effector position, compute the joint parameters.

– More interesting and easier to animate actions like – reaching, walking, grabbing.

– Much harder to solve.

– Why?

– Demo



1

2

P x , y

l1

l 2

Given the link lengths and the desired position of the endeffector, find the joint parameter and that will make the endeffector reach the goal.

1 2

Pl1 l 2



1

2

P x , y

l1

l 2

x=l1 cos 1l 2 cos 12

xy

y=l1 sin 1l 2 sin 12



P x , y

IK a hard problem because it is an illposed problem.


Character Animation● Posing the Kinematics problem

– Forward

– Inverse

is the vector of end-effector position and orientation. is the vector of joint parameters. is the mapping between the two.

X= f

= f −1X

X

f


Character Animation● Assumptions

– The mapping is usually non-linear in the general case. However, we linearize the problem by localizing around the current position and use the Jacobian, i.e.,

– Note that the Jacobian is always a function of the current set of joint parameters – so it has to recalculated everytime they change!

d X=J d d=J−1d X

f


Character Animation● Assumptions

– We will attempt to solve the Inverse Kinematics problems under the following assumptions:

● All joints are hinge joints.● All links are rigid and do not change in

length.

– Note that these assumptions are not there because the method we will use to solve the Jacobian require these assumptions but instead to simplify our discussion a bit.


Character Animation● Steps:

– Construct the Jacobian

– Invert the Jacobian

– Iterate

– Detect singularities and avoid ill-conditioning


Character Animation● Forming a kinematic chain (Hierarchical modelling revisit)

● Given a kinematic chain with links (local coordinate frames) numbered from {1} to { j-1}.

● The position and orientation of the root is known in the global coordinate frame, {0}.

● Each has its local x-axis oriented along the length of the link and the local z-axis as the axis of rotation about the joint.

{1}

{ j−1}

{0}{2}


Character Animation● Forming a kinematic chain

● Then the jth link can be added as follows:

– Translating along the local x-axis by the length of link { j-1},

{1}

{2}

l j−1

{ j−1}

{0}



● Then the jth link can be added as follows:

– Translating along the local x-axis by the length of link { j-1},

– Rotating at the new joint origin so that the local x-axis lies along the length of the new link.

{ j−1}

{1}

{2}

l j−1

{ j }

{0}



● So if this the transformation between the { j} and { j-1} frame is given by

– Any point in the frame { j} can be moved to a corresponding point in frame { j-1} as

– Therefore, the coordinates of the point in the global frame is given by

{ j−1}

{1}

{2}

{ j }

M jj−1

p j−1= M jj−1 p j

p j

p j−1

p0=M 0 . M 10 M j−1

j−2 . M jj−1 p j

{0}


Character Animation● Moving to global coordinates

● The transformation from any local frame { j} to the global frame is thus, given by

{ j−1}

{1}

{2}

{ j }

T j= M 10 M j−1

j−2 . M jj−1

T j=[r 11 r 12 r13 t1

r 21 r 22 r 23 t 2

r 31 r 32 r 33 t 3

0 0 0 1]

{0}


Character Animation● Constructing the Jacobian

● The Jacobian relates the end-effector velocities to velocities of the joint parameters.

[V nx

0

V ny0

V nz0

nx0

ny0

nz0

]=[∂ f 1

∂1

⋯ ⋯ ⋯ ⋯ ⋯∂ f 1

∂n−1

⋮ ⋱ ⋮

⋮∂ f i

∂ j

⋮

⋮ ⋱ ⋮∂ f 6

∂1

⋯ ⋯ ⋯ ⋯ ⋯∂ f 6

∂n−1

] [1

2

⋮⋮⋮

˙n−2

˙n−1

]d X=J d

X=J



● The Jacobian relates the end-effector velocities to velocities of the joint parameters.

d X=J d

X=J

UnknownKnown

[V nx

V ny

V nz

nx

ny

nz

]=[∂ f 1

∂1

⋯ ⋯ ⋯ ⋯ ⋯∂ f 1

∂n−1

⋮ ⋱ ⋮

⋮∂ f i

∂ j

⋮

⋮ ⋱ ⋮∂ f 6

∂1

⋯ ⋯ ⋯ ⋯ ⋯∂ f 6

∂n−1

] [1

2

⋮⋮⋮

˙n−2

˙n−1

]Simplifying Notation



– The functions relate the joint parameter angular velocities to the linear velocity of the end-effector.

– The functions relate the joint parameter angular velocities to the angular velocity of the end-effector.

– The velocities of the end-effector are known in the global coordinate system. So the Jacobian has to be formed in the global coordinate system.

f 1, f 2, f 3

f 4 f 5, f 6



– The jth column of the Jacobian relates the jth joint parameter angular velocity to the velocity of the end-effector.

J =[∂ f 1

∂1

⋯ ⋯ ⋯ ⋯ ⋯∂ f 1

∂n−1

⋮ ⋱ ⋮

⋮∂ f i

∂ j

⋮

⋮ ⋱ ⋮∂ f 6

∂1

⋯ ⋯ ⋯ ⋯ ⋯∂ f 6

∂n−1

]



● For the angular velocity,

– The contribution of this angular velocity to the angular velocity of the end-effector (in global coordinates) is given by

– So

{ j−1}

{ j }

{n−1}

j=z j j

j

j0 =T j z j j=u j j

∂ f 4

∂ j

=T j z jx=u jx

∂ f 5

∂ j

=T j z jy=u jy

∂ f 6

∂ j

=T j z j z=u jz

z j

{0}



● For the angular velocity,

– The contribution of this angular velocity to the linear velocity of the end-effector (in global coordinates) is given by

– So

{ j−1}

O j

{n−1}

j=z j j

j

V j0 =T j z j ×Pn−P j j=v j j

∂ f 1

∂ j

=v jx

∂ f 2

∂ j

=v jy

∂ f 3

∂ j

=v jz

On

z j

whereP j=T jO j

P n=T nO n

{0}





– Iterate



Character Animation● Inverting the Jacobian

– To solve IK, we must invert the Jacobian

– Note that the Jacobian is a fat matrix, i.e., it has more columns than rows.

– So the system is underconstrained – there is more than one possible solution.

– We choose the simplest solution, i.e., one of minimum norm:

d=J−1d X

=J−1 X

Find the unique for which we have: min∥∥2 s.t. X=J


Character Animation● Linear Least Squares (Digression)

– For the system

– If A has full row rank then the unique X for which we get

is given by

– To see how this is so, solve the Lagrangian,

AX=B

min∥X∥2 s.t. AX=B

X=AT AAT −1B=A+ B

L=∥X∥2T AX−B


Character Animation● Linear Least Squares (Digression)

– For the system

– If A has full row rank then the unique X for which we get

is given by

– To see how this is so, solve the Lagrangian,

AX=B

min∥X∥2 s.t. AX=B

X=AT AAT −1B=A+ B

L=∥X∥2T AX−B

MoorePenrosePseudoinverse


Character Animation● Singular Value Decomposition (Digression)

– Any rectangular, real, mxn matrix A can be decomposed as:

– The pseudoinverse can be computed from the SVD as

A=U V T

A+=V +U T

where U is a m×m orthogonal matrix is a m×n diagonal matrixV is a n×n orthogonal matrix

where + is formed by transposing and taking a reciprocal of every non-zero singular value


Character Animation● Inverting the Jacobian

– So we solve

– By computing the pseudoinverse of the Jacobian using the SVD

d=J−1d X

J +=V

+U T=∑

i=1

r

i-1v i ui

T

where r=m if the Jacobian has full row rank





– Iterate



Character Animation● Starting the IK solver and iterating

● Initialize the linear velocity components of the end-effector to , compute the Jacobian, invert, update the joints parameters and iterate.

● If dX is too large then due to local nature of the solution, tracking errors will occur, given by

● So sometimes we go from E to G in by taking smaller steps in the direction.

E

G

1

2

j

n−1

dX 1. .3=G−E

∥J d−dX∥

G−E


Character Animation● Starting the IK solver and iterating

● Iterate only a fixed number of times and check whether has fallen below some tolerance.

● Also check after a few iterations whether the arm has become fully extended, is parallel to and the goal is still out of reach.

● What about the angular velocity components of the end-effector?

E

G

1

2

j

n−1

∥G−E∥

G−E





– Iterate



Character Animation● Detect singularities and avoid ill-conditioning

– The Jacobian becomes singular when it loses rank.

– This is detectable using the condition number, of the Jacobian matrix.

C=max

min



– The solution also becomes ill-conditioned near a singulartiy, resulting in very high joint space velocities.

dX

d 1

∥d 1∥≫∥dX∥



– To prevent ill-conditioning we damp the joint space velocities by searching for a solution that minimizes the sum:

– The solution to this is given by

– This is known as the damped least squares solution.

∥J d−dX∥22∥d∥2

d=J T J J T2 I −1dX

=V T T2 I −1U T dX

=∑i=1

r i

i2

2 v iu iTdX

cs 775: advanced computer graphicsparagc/teaching/2012/cs775/... · 2012-01-13 · cs 775: lecture...

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