Lecture15-hierarchical-animation

# Desired change from this specific pose to reach some

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Desired change from this specific pose to reach some target. Target shown here in red We’ll compute a set of changes to the pose in order to get closer to the target. Then repeat. Solve iteratively – numerically solve for step toward goal E.g., human arm is typically modeled as 3-1-3 or 3-2-2 linkage

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First, we need some math Suppose we have 6 functions of 6 unknowns. ) , , , , , ( ) , , , , , ( ) , , , , , ( ) , , , , , ( ) , , , , , ( ) , , , , , ( 6 5 4 3 2 1 6 6 6 5 4 3 2 1 5 5 6 5 4 3 2 1 4 4 6 5 4 3 2 1 3 3 6 5 4 3 2 1 2 2 6 5 4 3 2 1 1 1 x x x x x x f y x x x x x x f y x x x x x x f y x x x x x x f y x x x x x x f y x x x x x x f y = = = = = = ( 29 X F Y = This is just an arbitrary function, not meant to mean anything We can represent that using vectors and matrices. What is the type of F?
IK – math notation Now let’s take the derivative of y. Which is the sum of the partials of each parameter. 6 6 5 5 4 4 3 3 2 2 1 1 x x f x x f x x f x x f x x f x x f y i i i i i i i + + + + + = X X F Y = And we can represent that using vectors and matrices too. This matrix is called the Jacobian. Each entry in the Jacobian describes how one functions changes wrt 1 variable What are the dimensions of the Jacobian?

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Solving for Change X X F Y = That’s the general equation, rewrite with Jacobian written as J. X X J Y ) ( = And X with a dot on it means “the first derivative of X” So the change in Y is the matrix which describes the rate of change of each function wrt X times the rate of change of X.
Solving for Change X X F Y = That’s the general equation, rewrite with Jacobian written as J. X X J Y ) ( = Now rewrite with a linkage in mind ( 29 θ θ J V =

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Inverse Kinematics - Jacobian ( 29 θ θ J V = [ ] z y x z y x v v v V ϖ ϖ ϖ , , , , , = [ ] 4 3 2 1 , , , θ θ θ θ θ = = 4 1 1 4 2 1 θ ϖ θ ϖ θ θ θ θ z z y x x x v v v v J Change in position Change in orientation Joint position velocities Relates a joint position change to end effector change
Inverse Kinematics - Jacobian ( 29 θ θ J V = [ ] z y x z y x v v v V ϖ ϖ ϖ , , , , , = [ ] 4 3 2 1 , , , θ θ θ θ θ =

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