Manipulator Kinem

aticsForw

ard and Inverse Kinematics

2D Manipulator Forw

ardKinem

atics

•Forw

ard Kinematics

•G

iven T, find x

The vector of joint angles

The vector of end effector positions

Shorthand notation

2D Manipulator Inverse

Kinematics

•Inverse Kinem

atics = given x, find TThe joint angles

The end effector positions

Problem is m

ore difficult!

2D Manipulator Inverse Kinem

atics Example

•U

se the Law of Cosines:

•W

e also know that:

•D

o some algebra to get:

Again, note the notation short-hand

Inverse Kinematics Exam

ple Continued

•N

ow solve for c2:

•O

ne possible solution:

•Elbow

up vs elbow dow

n

•M

ay be impossible!

•M

ultiple solutions may exist!

•Fundam

ental problem in robotics!

Manipulator Kinem

aticsThe Jacobian

The Jacobian

•M

atrix analogue of the derivative of a scalar function

•A

llows us to relate end effector velocity to joint velocity

•G

iven

•The Jacobian, J, is defined as:

The Jacobian, a 2D 2-Link Manipulator

Example

•The forw

ard kinematics of a 2 link, revolute

joint manipulator are given as:

•W

hat is the relationship between the end

effector velocity, 𝒙, and the joint velocities,

𝜽?•

First step, take the time derivative of the

forward kinem

atics equations:

The Jacobian, a 2D 2-Link Manipulator

Example Continued

•Put into m

atrix form:

•W

e can further manipulate that to understand how

the relationship of the joints com

es into play:

Velocity of “elbow”

Velocity of “end effector”relative to “elbow

”

Singular Jacobians

•The Jacobian is singular w

hen its determinant is equal to 0.

•W

hen the Jacobian is non-singular the following relationship holds:

•Q

uestion---Intuitively, when is this not the case?

•H

int---Think of a configuration where changing the joints does not

change the end effector velocity in any arbitrary direction.

•W

orkspace boundaries

•D

emonstrate m

athematically?

•D

eterminant of the Jacobian = 0

Singular Jacobians Continued

•Find the determ

inant of the Jacobian:

•U

sing the following trig identities:

•Yields:

•Therefore,

•D

oes this make sense?

Singular Jacobians Continued

•The fact that

implies that all possible end effector

velocities are linear combinations of the follow

ing Jacobian matrix:

•M

atrix rank = # of linearly independent columns (or row

s)

•If the Jacobian is full rank, then the end effector can execute any arbitrary velocity

•Is this the case for our 2D

revolute joint planar manipulator?

Singular Jacobians Continued

•Look at the Jacobian again:

•W

e can see that it is made up of:

•W

here:

•The linear independence is a function of the joint angle

•If theta_2 = 0 or pi, the Jacobian loses rank

What is the physical m

anifestation of singularities?

•Configurations from

which certain instantaneous m

otions are unachievable•

A bounded end effector velocity (i.e. a velocity w

ith some finite value) m

ay correspond to an unbounded joint velocity (i.e. one that approaches infinity)

•if w

e want the end effector to attain a certain velocity, w

e need to input an infinite joint velocity

•N

ot a great idea!

•Bounded joint torques m

ay correspond to unbounded end effector torques and forces

•G

ood for static loading•

Not so good for other situations

•O

ften correspond to boundaries of the workspace

Summ

ary of Manipulator Kinem

atics Introduction

•Forw

ard kinematics is relatively sim

ple

•Inverse kinem

atics is relatively complicated and som

etimes

impossible

•A

Jacobian relates end effector velocity to joint velocity

•W

e typically want to com

pute the inverse of the Jacobian•

Typically we have a desired end effector velocity

•Then w

e need to compute the joint velocities to reach that end effector

velocity•

This requires us to compute the inverse of the Jacobian

•Som

etimes w

e can do this, sometim

es we cannot

•A m

anipulator’s singularities correspond to when the determ

inant of the Jacobian is zero.

Manipulator Kinem

aticsKinem

atic Optim

ization for Redundant Manipulators

Redundant Manipulators

Redundant Manipulators

•A

redundant manipulator is one w

here

•Jacobian represents the relationship betw

een end effector and joint velocities:

•For a given end effector velocity, to find joint velocities w

e can invert J

if it is square.

•W

ith a redundant manipulator, J is non-square so not invertible

•There m

ay exist different joint velocities, 𝜽, that yield the desired end

effector velocity, 𝒙•

Multiple solutions = opportunity to optim

ize!•

Minim

al energy•

Minim

al time

Redundant Manipulator –

Minim

al Energy

•M

inimize kinetic energy

•Kinetic energy is proportional to velocity squared

•Thus, w

e can minim

ize:

•W

here Wis a sym

metric

positive definite matrix

•W

eighting function•

Weighs im

portance of each joint•

Makes problem

more general

•Sym

metric = n

x nm

atrix such that

•A

is positive definite if zTAz

is positive for every non-zero colum

n vector zof n

real num

bers

•Exam

ple:

•If

•Then

Redundant Manipulator –

Minim

al Energy Continued

•M

inimize kinetic energy

•Subject to the kinem

atic constraint:

•To solve this problem

we need to em

ploy Lagrange Multipliers

Joseph-Louis Lagrange (born Giuseppe Lodovico

Lagrangia)

Lagrange Multipliers

•In an unconstrained

optimization problem

, we find the extrem

e points of som

e function:

•These are sim

ply where the partial derivatives are 0, or equivalently

where the gradient of f is 0:

where:

•In a constrained

optimization problem

, we add in another function:

Constrained Optim

ization Problem

•Tw

o methods to solve a constrained optim

ization problem•

Substitution•

Lagrange multipliers

Constrained Optim

ization Problem:

Substitution Example

•O

ptimize:

•Subject to:

•Solve for x

1in the constraint:

•Plug x

1into

to yield:

•W

e have transformed the tw

o dimensional constrained problem

into a one dim

ensional unconstrained problem

•D

ifferentiate as a function of x2 :

•Solve and substitute:

Constrained Optim

ization Problem:

Substitution Example Continued

•That w

as easy! Why do I have to learn about Lagrange m

ultipliers?

•The exam

ple was contrived

•G

enerally difficult•

Generally tim

e consuming

Constrained Optim

ization Problem: Lagrange

Multipliers Exam

ple

•The claim

: for an extremum

(minim

um or m

aximum

) of f(x) to exist on g(x), the gradient of f(x) m

ust be parallel to the gradient of g(x).

Find xand y

to maxim

ize f(x, y)subjectto a constraint (show

n in red) g(x, y) = c.

Images and text from

wikipedia

The red line shows the constraint g(x, y) =

c. The blue lines are contours of f(x, y). The point w

here the red line tangentially touches a blue contour is our solution. Since d

1> d

2 , the solution is a m

aximization of f(x, y)

Constrained Optim

ization Problem: Lagrange

Multipliers Exam

ple

•If the gradient of f(x) is parallel to the gradient of g(x), then the gradient of f(x) is a scalar function of the gradient of g(x):

•This m

eans the magnitude of the norm

al vectors do not need to be the sam

e, only the direction (the scalar, l, is term

ed the Lagrange m

ultiplier)•

This implies that:

•W

here:

Constrained Optim

ization Problem: Lagrange

Method Exam

ple

•O

ptimize:

•Subject to:

•W

e know that

Find thegradientConstrained O

ptimization Problem

: Lagrange M

ethod Example Continued

Plug into second equation, solve for

Plug x2 into third equation, solve for

Repeat for

Applying Lagrange Multipliers to the

Redundant Manipulator

•W

e have:

•W

here:

•Rem

ember that x

is m x 1 and q

is n x 1.

The function (energy) we w

ant to minim

ize, I substituted q for theta to represent a vector

Subject to the constraint(the m

anipulator kinematics)

Example: U

sing Lagrange Multipliers to solve

the Redundant Manipulator Problem

•Back to the problem

:

•W

is symm

etric positive definite, thus:

•Find the gradient:

•W

e want solve for the joint velocity in term

s of the end effector velocity, but w

e can't just invert Jbecause it is not square.

Example: U

sing Lagrange Multipliers to solve

the Redundant Manipulator Problem

Example: U

sing Lagrange Multipliers to solve

the Redundant Manipulator Problem

•Set the w

eighting function W to I

•W

here we can define

•A

s the generalized right-pseudo inverse of J.

Redundant Manipulators Sum

mary

•H

as more joint than endpoint degrees of freedom

•A

llows us to optim

ize the path of the robot

•U

se Lagrange multipliers to solve constrained optim

ization problem

Manipulator Kinem

aticsStatic Force Torque Relationships

Problem Statem

ent

•For a given force, F, acting on a m

anipulator’s end effector, what

torques, T, must be supplied at the joints to m

aintain static equilibrium

.

20th

Century Fox

Static Force/Torque Relationship

•Let:

•Let Gx and Gq represent infinitesim

al displacements in the task and

joint spaces (We w

ill call these virtual displacements) such that:

•The virtual w

ork of the system is given as:

•Rem

ember the Jacobian is:

Generalized force acting on the end effector

Static Force/Torque Relationship

•If the robot is in equilibrium

, then

•Com

bining the above equation with the Jacobian yields:

•Transposing both sides yields:

Static Force/Torque Relationship Summ

ary

•The joint torque required to hold a static force at a m

anipulator’s end effector is linearly proportional to the transpose of the Jacobian

Manipulator Kinem

aticsJoint Com

pliance

Problem Definition

•W

hat is the relationship between endpoint displacem

ent and an applied force at the endpoint m

anipulator for a given configuration of the m

anipulator?

•If I apply a force on the end effector, how

much w

ill the robot move?

•W

e know:

•A

ssume each joint ihas a stiffness, k

i :

Joint Compliance Derivation

Relationship between endpoint

displacement and endpoint force = C