Human Kinematics Kinematic representation Iterative methods - - PowerPoint PPT Presentation

Human Kinematics

• Kinematic representation
• Iterative methods
• Optimization methods

Maximal coordinates

Motion representation

(x0, R0) (x1, R1) (x2, R2)

Assuming there are m links and n DOFs in the articulated body, how many constraints do we need to keep links connected correctly in maximal coordinates?

Generalized coordinates

θ1, φ1 θ2

state variables: 18 state variables: 9

x, y, z, θ0, φ0, ψ0

constraints: 9

Maximal coordinates

• Direct extension of well understood rigid body

dynamics; easy to understand and implement

• Operate in Cartesian space; hard to
• evaluate joint angles and velocities
• enforce joint limits
• apply internal joint torques
• Inaccuracy in numeric integration can cause body

parts to drift apart

Generalized coordinates

• Joint space is more intuitive when dealing with

complex multi-body structures

• Fewer DOFs and fewer constraints
• Hard to derive the equation of motion

Kinematics

• Forward kinematics
• given a joint configuration, what is the position
• f an end point on the structure?
• Inverse kinematics
• given the position for an end point on the

structure, what angles do the joints need be to achieve that end point?

Quiz

p θ φ

σ θ, φ, σ = f(p)

p θ φ

σ p = f(θ, φ, σ)

Which function is inverse kinematics?

Why inverse?

• More intuitive control
• Maintain environment constraints
• Calculate desired joint angles for control

Motion representation

Global position and orientation are determined by 3 translational and 3 rotational DOFs Each joint can have up to 3 DOFs

1 DOF: knee 2 DOF: wrist 3 DOF: arm

Configurations

θthigh, φthigh, σthigh θknee θankle, φankle

x, y, z, θ, φ, σ

q = x, y, z, θ, φ, σ, θth, φth, σth, θkn, · · ·

Transformation chain

h(q) p θthigh, φthigh, σthigh θknee θankle, φankle

h(q) = T(x, y, z)R(θ, φ, σ)T(d0)R(θth, φth, σth)T(d1)R(θkn)T(d2)R(θan, φan)x a point on foot in local coordinates a point on foot in world coordinates d1 d2 x, y, z, θ, φ, σ d0

Constraints

Position constraint

C(q) = h(q) − p = 0

h(q) p θthigh, φthigh, σthigh θknee θankle, φankle

h(q) = T(x, y, z)R(θ, φ, σ)T(d0)R(θth, φth, σth)T(d1)R(θkn)T(d2)R(θan, φan)x a point on foot in local coordinates a point on foot in world coordinates d1 d2 x, y, z, θ, φ, σ d0

Solutions

• Closed form solutions can only be found for fairly

simple mechanisms

• Numerical solutions
• No solution
• Single solution
• Multiple solution

Under-specified problem

• Multiple solutions
• Mostly bad
• How do we find the optimal solution?
• Heuristics (move the outermost links first)
• Closest to the current configuration
• Energy minimization
• Natural looking motion (whatever it means)

• Kinematic representation
• Iterative methods
• Optimization methods

Iterative method

• Use inverse of Jacobian to iteratively step all

the joint angles towards the goal

• Girard and Maciejewski, Computational

modeling for the computer animation of legged figures, SIGGRAPH 85

Problems

• Pseudo inverse returns the least square minimum

norm solution

• Singularities cause deficient Jacobian
• Linearization about current state causes error

• Kinematic representation
• Iterative methods
• Optimization methods

Optimization method

• Find a solution that optimizes some numeric

metric and satisfies constraints

• Numeric metric
• A function of q that measures the quantity

to be minimized

• Also called “objective function”

Optimization method

min

q G(q)

C(q) = 0

subject to Solve for joint configuration q

Objective functions

• Joint velocity
• Power consumption
• Similarity to the rest pose
• Similarity to the natural pose

Unconstrained optimization

• Treat each constraint as a separate metric and

minimize weighted sum of all metrics

• Also called penalty methods
• each spring pulls on constraint with force

proportional to violation of the constraint

Unconstrained optimization

F(q) = G(q) +

• i

wiCi(q)2

“Softening” constraints by adding them to the

• bjective function

Now we have a new optimization problem: min

q

Gradient descent

For each iteration, move in the direction of negative gradient And update the state

qnew = q − α∂F ∂q

−∂F ∂q = −∂G ∂q − 2 X

i

wi ∂Ci ∂q

T

Ci One of the simplest way to solve this optimization is gradient descent method

compute update while

F(q) > ✏ q = q − α∂F ∂q

Optimization loop

∂F ∂q = ∂G(q) ∂q + 2 X

i

wi(∂Ci ∂q )T Ci

Search step size

The step size in each search iteration can be determined arbitrarily, through a binary search, or solved by another optimization process Step size: where min

α F(q + α∆q)

α

qnew = q + α∆q

Optimize IK problem

We can simplify the problem by ignoring G(q), but we still need to know how to compute constraint derivatives. compute update while

F(q) > ✏ q = q − α∂F ∂q ∂F ∂q = ∂G(q) ∂q + 2 X

i

wi(∂Ci ∂q )T Ci

Constraint derivatives

• We need a direction to move

joints in such way that the handle moves towards the goal

• The derivative of the constraint,

i.e. Jacobian, gives directions where handles move if joints move

h(q) p

C(q) = h(q) − p = 0 ∂C(q) ∂q = ∂h(q) ∂q

Quiz

• What’s the dimension of the Jacobian for a

system with n dofs and m position constraints?

Constraint derivatives

∂C(q) ∂q = ∂h(q) ∂q

h(q) p

x, y, z, θ0, φ0, σ0 θ1

θ2, φ2 q = [x, y, z, θ0, φ0, σ0, θ1, θ2, φ2]

C(q) = h(q) − p = 0

Need to know how to compute derivatives for each transformation

∂h(q) ∂θ1 = T(x, y, z)R(θ0, φ0, σ0)T∂R(θ1) ∂θ1 TR(θ2, φ2)hk h(q) = T(x, y, z)R(θ0, φ0, σ0)TR(θ1)TR(θ2, φ2)hk hk: local coordinate of h

Quiz

• What is the derivative for the translation matrix?
• What is the derivative for a rotation matrix about

z axis?

    1 3 1 5 1 −2 1         cos θ sin θ − sin θ cos θ 1 1    

Constraint derivatives

h(q) p

x, y, z,

θ0, φ0, σ0

θ1

θ2, φ2

hk: local coordinate of h

What is the most efficient way to compute the ?

∂h(q) ∂q

h = T0(x, y, z)R0(θ0, φ0, σ0)T1R1(θ1)T2R2(θ2, φ2)hk

Compute the gradient for the DOF

• n the outermost link first

Unconstrained optimization

• Pros:
• simple and fast, no linear system to solve
• near-singular configurations is less of a problem
• Cons:
• can’t maintain constraints exactly
• constraints fight against each other and the
• bjective function

Constrained optimization

• Treat constraints as “hard constraints”
• Use Lagrangian formulation to solve a

nonlinear optimization

L(q, λ) = G(q) − λ · C min

q,λ L(q, λ)

Constrained method

• Pros
• Enforce constraints exactly
• Quadratic convergence
• Cons
• Large system of equations
• Near-singular configurations cause instability

Leg example

hip knee ankle R(q0)R(q1)R(q2) R(q3) R(q4)R(q5) L0 L1 L2 L3 T0h Th1 T1k Tk2 T2a Ta3 h0

Leg example

C(q) = T0hR0R1R2Th1T1kR3Tk2T2aR4R4Ta3h0 - p q0, q1, q2 q3 q4, q5 q = {q0, q1, q2, q3, q4, q5} p h0 To evaluate gradients of the obj function at each iteration, we need to evaluate constraint and compute Jacobian. How to compute J(q)?

node0 node1 node2 node3 hip knee ankle

Compute Jacobian

Consider the derivative of C with respect to q5

R(q0)R(q1)R(q2) R(q3) R(q4)R(q5) T0h Th1 T1k Tk2 T2a Ta3 h0 joint node

node->getParentBodyNode()->getTransform() joint->getTransformFromParentBodyNode() joint->getTransform(0) joint->getTransformDerivative(1)

∂C ∂q5 = T0hR(q0)R(q1)R(q2)Th1T1kR(q3)Tk2T2aR(q4) ∂R ∂q5 Ta3h0

joint->getTransformFromChildBodyNode().inverse()