Kinematics. CS 448D: Character Animation Prof. Vladlen Koltun Stanford University

Transcription

1 Kinematics CS 448D: Character Animation Prof. Vladlen Koltun Stanford University

2 Kinematics Kinematics: The science of pure motion, considered without reference to the matter of objects moved, or to the force producing or changing the motion. Oxford English Dictionary

3 Skeletal animation We animate skeletons of characters Tree structure Vertices are joints Edges are bones (or links) A mesh is created from a skeleton every frame by skinning A vertex is expressed as a combination of a number of positions relative to respective joints

4 Skeletons Joints and bones define transformations. (Left) Can represent with tree structure that associates joint transformations with edges and bone transformations with vertices. (Right)

5 Human joints Human body contains over 100 types of joints, characterized by the types of induced movement. We represent joint configurations as rigid-body transformations. Types of joints correspond to different sets of allowed transformations. Picture from

6 Kinematics of skeletons Forward Kinematics: Compute configuration of a bone (end effector) in the global coordinate frame given configurations of all ancestor joints. Inverse Kinematics: Given configuration of a bone (end effector) in the global coordinate frame, compute configurations of all ancestor joints.

7 Kinematics of skeletons Forward kinematics forms a basic step for many techniques. Necessary. Always specifying all joint configurations manually can be tedious and unnecessary. Inverse kinematics used for more intuitive, higher-level specification of keys. Walk character by placing feet in the right places on the ground Create boxing animation by positioning fists In practice, animators go back and forth. (Position foot, adjust knee, rotate hip, reposition foot, adjust some more,...)

8 Forward Kinematics Configuration of any bone in the global coordinate frame can be determined by tree traversal.

9 Forward Kinematics What is the configuration of the right hand? C =

10 Forward Kinematics J 0 What is the configuration of the right hand? C = J 0

11 Forward Kinematics J 0 J 1 What is the configuration of the right hand? C = J 0 J 1

12 Forward Kinematics J 0 J 1 What is the configuration of the right hand? C = J 0 J 1 B 1

13 J 1 J 2 Forward Kinematics J 0 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2

14 J 1 J 2 Forward Kinematics J 0 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2 B 2

15 J 1 J 2 Forward Kinematics J 0 J 3 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2 B 2 J 3

16 J 1 J 2 Forward Kinematics J 0 J 3 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2 B 2 J 3 B 3

17 J 1 J 2 Forward Kinematics J 0 J 3 J 4 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2 B 2 J 3 B 3 J 4

18 J 1 J 2 Forward Kinematics J 0 J 3 J 4 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2 B 2 J 3 B 3 J 4 B 4

19 J 1 J 2 Forward Kinematics J 0 J 3 J 4 J 5 What is the configuration of the right hand? C = J 0 J 1 B 1 J 2 B 2 J 3 B 3 J 4 B 4 J 5

20 Inverse Kinematics interactive demonstration

21 Simple example l 1 θ 2 (x, y) l 2 θ 1 Given l 1, l 2, and the position (x, y) of the end effector, compute and θ 1 θ 2

22 Simple example l 2 θ 2 l 1 (x, y) θ 1 l 3 l 3 = x 2 + y 2 θ 2 = arccos l2 1 + l 2 2 (x 2 + y 2 ) 2l 1 l 2 θ 1 = arctan y x + arccosl2 1 +(x 2 + y 2 ) l 2 2 2l 1 x2 + y 2

23 General considerations Closed-form analytical solution generally not available. Solution might not exist. (End effector configuration not reachable.) Number of solutions could be infinite. (Number of variables greater than number of constraints.) We will approximate a solution iteratively. Will try to find small steps, each bringing us closer to the goal.

24 Iterative method t θ 2 θ 3 s θ 1 (θ 1,..., θ n ) (s 1,..., s k ) (t 1,..., t k ) - joint configurations (e.g., angles) - end effector configurations (e.g., positions) - target end effector configurations

25 Iterative method t e θ 2 θ 3 s θ 1 s s(θ) θ Express as a function of We want to take a step θ, so that s(θ + θ) is closer to We will find θ through a differential of a hypothetical motion θ(τ), where τ is time We want to find θ(τ) that yields ṡ = e, where e = t s. θ t

26 t Iterative method θ 2 θ 3 s e ṡ = ( ds1 dτ,..., ds k dτ ) T θ 1 ds i dτ = d dτ s i(θ) = d dτ s i(θ 1 (τ),..., θ n (τ)) = s i dθ 1 θ 1 dτ + + s i dθ n θ n dτ ( si =,..., s )( i dθ1 θ 1 θ n dτ,..., dθ ) T n dτ ds 1 dτ ds 2 dτ. ds k dτ = s 1 s 1 θ 1 s 2 s 2 θ 1. s k θ 1 θ 2... θ s k θ 2... s 1 θ n s 2 θ n. s k θ n dθ 1 dτ dθ 2 dτ. dθ n dτ ṡ = J s (θ) θ

27 t Iterative method J s (θ) θ = e θ 2 θ 3 s e θ 1 We can compute J s (θ) and e directly. We are looking for θ such that If J s (θ) is invertible, the solution is Otherwise the solution is given by the pseudo-inverse:, and then look into the pseudo- Let s see first how to compute inverse. J s (θ) θ = e J s (θ) θ = J 1 s (θ)e θ = J + s (θ)e

28 Computing the Jacobian We consider the case where s are end-effector positions and θ are orientations of hinge joints. s i θ j is the linear velocity of the rotation induced on end-effector i by hinge j. If end-effector i is unaffected by the rotation of j, the velocity is 0. Otherwise, s i θ j = v j (s i p j ) where p j is the position of joint j and v j is the unit vector pointing along the axis of rotation according to the right hand rule. s i θ j s i p j

29 Computing the pseudo-inverse θ Need to find such that J θ = e. If J has rank k (gradients for distinct end-effectors are linearly independent), then is invertible. Consider JJ T J + = J T (JJ T ) 1 and θ = J + e Then J θ = JJ T (JJ T ) 1 e = e

30 Euler integration Choose small constant α. Update θ := θ + αj + e Verify that end-effector moved closer to goal. Otherwise reduce α.

31 Redundant degrees of freedom When the space of joint configurations has higher dimensionality than the space of end-effector configurations, there is generally an infinite number of solutions to IK. We can try to optimize within the space of solutions to get a solution with desirable properties. J(I J + J)z =0 for any vector z. We set where H = H(θ) θ = J + e +(I J + J) H is some function we wish to minimize. Then and every step J θ = e θ := θ + α ( J + e +(I J + J) H ) not only brings us closer to the goal end-effector configuration, but also decreases the secondary error function H.

32 Redundant degrees of freedom If we wish to limit joint angles to a certain range, or keep joint angles close to specific desirable values, we can set H = n β i (θ i θi 0 ) 2 i=1 where θ 0 is the optimal joint configuration and β i are weights that act as stiffness parameters.

33 Singularities and Numerical Instability

34 Other techniques: Optimization Can be formulated as an optimization problem in the joint configuration space. Objective function is distance to target. A wide range of possible constraints. (Joint limits.) At any point in the space, can evaluate objective function and its partial derivatives.