CS 775: Advanced Computer Graphics. Lecture 3 : Kinematics

Transcription

1 CS 775: Advanced Computer Graphics Lecture 3 : Kinematics

2 Traditional Cell Animation, hand drawn, 2D Lead Animator for keyframes

3 Traditional, hand drawn animation Lead Animator for keyframes and many secondary animators for the in-betweens

4 Traditional, hand drawn animation how many drawings for a 2 hour movie?

5 Computer assisted keyframing Keyframes created/posed by hand In-betweens interpolated by the semi-automatic techniques How is this done? Luxo Jr, PIXAR, 1986

6 Interpolating Position and Orientation parameters and animating (rigid) transformations. Linear interpolation, spline interpolation, quaternions. How is this done? Luxo Jr, PIXAR, 1986

7 Creating the keyframe pose for the lamp Modify the joint angles at each internal joint Modify global position and orientation

8 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.

9 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.

10 Terminology Links or Bones End effector Internal Joints Root Joint Kinematic Chain

11 Terminology Joint DoF (parameters) 1 2 0, 0, 0, t x,t y, t z Root DoF

12 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)

13 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

14 Inverse Kinematics l2 2 l1 1 P x, y Given the link lengths l 1 and l 2 the desired position of the end effector, find the joint P parameter 1 and 2 that will make the end effector reach the goal.

15 Inverse Kinematics 2 l2 l1 1 x P x, y y x =l 1 cos 1 l 2 cos 1 2 y=l 1 sin 1 l 2 sin 1 2

16 Inverse Kinematics P x, y IK a hard problem because it is an ill posed problem.

17 Posing the Kinematics problem Forward X = f Inverse = f 1 X X is the vector of end-effector position and orientation. is the vector of joint parameters. f is the mapping between the two.

18 Assumptions The mapping f 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., d X = J d d =J 1 d X Note that the Jacobian is always a function of the current set of joint parameters so it has to recalculated everytime they change!

19 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.

20 Steps: Construct the Jacobian Invert the Jacobian Iterate Detect singularities and avoid ill-conditioning

21 Steps: Construct the Jacobian Invert the Jacobian Iterate Detect singularities and avoid ill-conditioning

22 Forming a kinematic chain (Hierarchical modelling revisit) {0 } {2 } { j 1 } {1} 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.

23 Forming a kinematic chain {0 } {2 } { j 1 } {1} Then the jth link can be added as follows: Translating along the local x-axis by the length of link { j-1}, l j 1

24 Forming a kinematic chain {0 } {2 } {1} { j 1 } { j} Then the jth link can be added as follows: Translating along the local x-axis by the length of link { j-1}, l j 1 Rotating at the new joint origin so that the local x-axis lies along the length of the new link.

25 Forming a kinematic chain {0 } {2 } {1} { j 1 } { j} So if this the transformation between the { j} and { j-1} frame is given by j 1 M j Any point p j in the frame { j} can be moved to a j 1 corresponding point p j 1 in frame { j-1} as p j 1= M j p j Therefore, the coordinates of the point in the global frame is given by p0 = M 0. 0 M 1 j 2 M j 1. j 1 M j p j

26 Moving to global coordinates {0 } {2 } { j 1 } {1} { j} The transformation from any local frame { j} to the global frame is thus, given by T j =0 M 1 j 2 M j 1. j 1 M j [ r 11 r 12 r 13 r r r T j = r 31 r 32 r t1 t2 t3 1 ]

27 Constructing the Jacobian d X = J d X = J The Jacobian relates the end-effector velocities to velocities of the joint parameters. [ ][ f1 1 f1 n 1 ][ ] 1 2 V nx 0 V ny 0 fi V nz = 0 j nx 0 ny f6 f 6 0 nz n 2 1 n 1 n 1 0

28 Constructing the Jacobian Known d X = J d Unknown X = J The Jacobian relates the end-effector velocities to velocities of the joint parameters. Simplifying Notation [ ][ f1 1 f1 n 1 ][ ] 1 2 V nx V ny fi V nz = nx j ny f6 f 6 nz n 2 1 n 1 n 1

29 Constructing the Jacobian The functions f 1, f 2, f 3 relate the joint parameter angular velocities to the linear velocity of the end-effector. The functions f 4 f 5, f 6 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.

30 Constructing the Jacobian The jth column of the Jacobian relates the jth joint parameter angular velocity to the velocity of the endeffector. J = [ f1 1 f6 1 f1 n 1 fi j f6 n 1 ]

31 Constructing the Jacobian {0 } { j 1 } j {n 1 } { j} zj For the angular velocity, j =z j j The contribution of this angular velocity to the angular velocity of the end-effector (in global coordinates) is given by 0 j =T j z j j =u j j So f4 f5 f6 = T j z j x =u jx = T j z j y =u jy = T j z j z =u jz j j j

32 On Constructing the Jacobian {0 } { j 1 } j zj {n 1 } Oj For the angular velocity, j =z j j The contribution of this angular velocity to the linear velocity of the end-effector (in global coordinates) is given by 0V j = T j z j P n P j j =v j j So f1 =v jx j f2 f3 =v jy =v jz j j where P j =T j O j P n=t n O n

33 Steps: Construct the Jacobian Invert the Jacobian Iterate Detect singularities and avoid ill-conditioning

34 Inverting the Jacobian To solve IK, we must invert the Jacobian d =J 1 d X =J 1 X 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: Find the unique for which we have: min 2 s.t. X =J

35 Linear Least Squares (Digression) For the system AX =B If A has full row rank then the unique X for which we get 2 min X is given by s.t. AX =B X = AT AAT 1 B= A + B To see how this is so, solve the Lagrangian, L= X 2 T AX B

36 Linear Least Squares (Digression) For the system AX =B If A has full row rank then the unique X for which we get 2 min X is given by s.t. AX =B X = AT AAT 1 B= A + B Moore Penrose Pseudoinverse To see how this is so, solve the Lagrangian, L= X 2 T AX B

37 Singular Value Decomposition (Digression) Any rectangular, real, mxn matrix A can be decomposed as: T A=U V where U is a m m orthogonal matrix is a m n diagonal matrix V is a n n orthogonal matrix The pseudoinverse can be computed from the SVD as A+ =V + U T where + is formed by transposing and taking a reciprocal of every non-zero singular value

38 Inverting the Jacobian So we solve d =J 1 d X By computing the pseudoinverse of the Jacobian using the SVD r J =V U = i v i ui + + T -1 T i=1 where r=m if the Jacobian has full row rank

39 Steps: Construct the Jacobian Invert the Jacobian Iterate Detect singularities and avoid ill-conditioning

40 Starting the IK solver and iterating 1 2 G j n 1 E Initialize the linear velocity components of the end-effector to dx 1..3=G E, 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 J d dx So sometimes we go from E to G in by taking smaller steps in the G E direction.

41 Starting the IK solver and iterating 1 2 G j n 1 E Iterate only a fixed number of times and check whether G E has fallen below some tolerance. Also check after a few iterations whether the arm has become fully extended, is parallel to G E and the goal is still out of reach. What about the angular velocity components of the endeffector?

42 Steps: Construct the Jacobian Invert the Jacobian Iterate Detect singularities and avoid ill-conditioning

43 Detect singularities and avoid ill-conditioning The Jacobian becomes singular when it loses rank. This is detectable using the condition number, C = max min of the Jacobian matrix.

44 Detect singularities and avoid ill-conditioning The solution also becomes ill-conditioned near a singulartiy, resulting in very high joint space velocities. d 1 dx dx d 1

45 Detect singularities and avoid ill-conditioning To prevent ill-conditioning we damp the joint space velocities by searching for a solution that minimizes the sum: J d dx 2 2 d 2 The solution to this is given by d =J T J J T 2 I 1 dx =V T T 2 I 1 U T dx r = i=1 i 2 i T v u dx 2 i i This is known as the damped least squares solution.