motion.ppt

7/29/2019 motion.ppt

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Marcelo H. Ang Jr. 1

CHAPTER 2

Rigid Body Motion ,Robot Kinematics of Velocity,

and Robot Statics


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After this chapter, the students are expected

to learn the following:

1. Relate time derivatives of position and

orientation representations with translational

and angular velocities.

2. Transform velocities in different spaces

3. Relate joint velocities with end-effector

velocities


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4. Understand the concept of Jacobians

5. Solve the forward and inverse kinematics ofvelocity

6. Understand robot singularities


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7. Static force/torque transformations between

frames

8. Static force/torque transformations between

task space and joint space

9. Understand redundancy and how to deal with

them


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Translational Velocities

AuB3x1 = translational velocity of frame B(i.e., origin of frame B) relative of

frame A

t(t)-t)(t

0tlim

dtd

B

A

B

A

B

A

B

A pppu

frame of differentiation is A

Velocity, like any vector may be expressed in another

frame, say W

B

A

A

W

BA

Wuru


6/70



In general, velocity vector depends on 2 frames: frames of differentiationAleading subscript

this is the frame where the velocity of B is

instantaneously computed from (can be thought of

as velocity reference point) frame resulting vector is expressed inWleading

superscript

When leading subscript is omitted, it is implied that thevelocity is relative to some understood universal frame of

reference.

A missing leading superscript implies a generic frame of

expression.


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Rotational VelocitiesAwB

{B}

{A}

At a certain instant, frame B has an

orientation ARB and its rotational motion

may be represented by the rotational

(angular) velocity vectorA

w03x1

unit vector alongA

wB= instantaneous = AkB

axis of rotation

magnitude ofAwB

= speed of rotation

AwBis related to t

(t)-t(t

0t

lim

dt

d

B

A

B

A

B

A

B

A R)RRR


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Rotational Velocities

Rot ( k, ) =

cosverskksinkverskksink-verskk

sink-verskkcosverskksinkverskk

sinkverskksink-verskkcosverskk

zzxzyyzx

xyzyyzyx

yxzzxyxx

where vers = (1 - cos)

Let Rot ( k, d ) = incremental change in rotation.AwB = AkB = AkB ARB + ARB = Rot ( k , d )ARB

dt

d

pre-multiplication since d isexpressed in the frame A (base)

B

A

k


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For a differential change (small) d

cos( + d) = cos + d = cos + (-sin)d

for = 0, d small

cos(d) = 1 (approx)

sin( + d) = sin + d = sin + cos(d)

cos

sin


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for = 0, d = small

sin(d) = 0 + 1d

= d (approx)

for = 0, d = small

vers(d) = 1cos(d) = 0 (approx)


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1dkdk-

dk-1dkdkdk-1

xy

xz

yz

)dk,(Rot

Back to:

BA

B

A

B

A

B

A

RI-)d,(Rot

R-R)d,(RotR

k

k

R

dkdk-

dk-dk

dkdk-

R BA

xy

xz

yz

B

A


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Rotational Velocitiesdividing by dt:

B

AB

A

RR

dt

dk

dt

dk-

dt

dk-

dt

dk

dtdk

dtdk-

dt

d

xy

xz

yz

zB

A

yB

A

xB

A

z

y

x

w

w

w

k

k

k

But BA

B

A kw


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B

A

B

A RR

0ww-

w-0w

ww-0

xB

A

yB

A

xB

A

zB

A

yB

A

zB

A

Let this be AwBx =

angular velocity tensor ofA

wB

ARB = AwBxARB


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As with any vector, the rotational velocity vectorAwBmay be expressed in another frame C:

B

A

A

C

BA

C wRw

leading subscript A: frame the body is rotating relative to

(frame & differentiation)

leading superscript C: frame of Expression


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The equivalent matrix product representation is:

xBA

Cw = CRAAwBx ARC

{B} & {C} are attached to the samerigid body which is rotating

{C}

{B}

{A}

ARC

= ARB

BRC

Diff with time ARC =

ARBBRC +

ARBBRC


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AwCxARC = AwBxARBBRC

AwCxARC = AwBxARCAwC =

AwB

rotational velocity of rigid body is equal to rot. velocity

of any frame attached to the rigid body.


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The Orthonormal (Rotation) Matrix

& Skew Symmetric Matrices

RRT = I

RRT + RRT = 0

(RRT)T

+ RRT = 0

Define S = RRT = wx

Then S + ST = 0

S = a skew symmetric matrix


18/70


The Orthonormal (Rotation) Matrix

& Skew Symmetric Matrices

Skew symmetric matrix as vector cross product:

z

y

x

xy

xz

yz

w

w

w

and

0ww-

w-0w

ww-0

Let wS

ThenSp = w x p

where p3x1 vector


19/70


Rotation MatrixR= wx R or

3x3 (time derivative of Rot. Matrix)

or

wax-ox-

nx-

ao

n

xr

= Er(x

r)w

9x1 representation

of orientation 9x3 (a kind of Jacobian associated with

representation)

3x1 angular velocity


20/70


Rotation Matrix

Trajectories are typically specified in terms ofxr, and it

is important to determine the angular velocities

To solve forw given xr, need to solve 9 Equations

with 3 unknowns overdetermined system Solution that minimizes Er w - xr is theleft pseudo inverse, Er

+

w = Er+ xr

Er+ = (ErT Er)-1 ErT

always exists


21/70


Rotation Matrix

But from definition ofErEr

T Er = ( -nxT -oxT -axT )

= +nxTnx + oxTox + axTax= 2I3

Er

+ = (ErT Er)-1 ErT

= (2I3)-1Er

T = ErT

ax-

ox-

nx-

very simple


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Rotation Matrix

w = ErT = ( nxT n + oxT o + axT a

Actually,

R= wx Rwx = R RT

Note: Free of Math. Singularities

xr w

w xr

a

o

n

always possible

Exactly the same


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Euler Angle Rates &

Angular VelocitiesARB = Rot ( z, ) Rot ( y, ) Rot ( z, )

1

0

0

0

1

0

1

0

0

)y,(Rot)z,(Rot)z,(Rotw

cos01

sinsincos0

sincossin-0

w

w = Er-1xr

Jacobian transformation


24/70


Euler Angle Rates &

Angular Velocities

or xr = Er w note that Er does not always exist

wxr

0sin

sin

sin

cos

0cossin-

1sin

cossin-

sin

coscos-

If sin = 0, matrix does not existMath. Singularity

w xr not always possible

Not all possible angular matrices can be represented

Problem with 3 parameter representations for orientation


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Roll Pitch Yaw Rates &

Angular Velocities

ARB = Rot ( z, ) Rot ( y, ) Rot ( x, )

00

1

)y,(Rot)z,(Rot01

0

)z,(Rot10

0

w

sin-01

cossincos0

coscossin-0

w

Er-1RPY

w = Er-1xr

OR

xr = Er w


26/70


Roll Pitch Yaw Ratio &

Angular Velocities

w

0cossin

coscos

0cossin-

1cossin

cossincos

x

2

r

If cos = 0, matrix does not exist

Math. Singularityw xr not always possible

Not all possible angular matrix can be represented

This is a problem with 3 parameter representations for

orientation


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Quaternion Rates &

Angular Velocities

R RT = wx

f ( 0, 1, 2, 3 )f ( 0, 1, 2, 3 )

OR

0 = cos /21 = kx sin /22 = ky sin /2

3 = kz sin /2

itdt

d

z

y

x

012

103

230

321

w

w

w

--

-

---

2

1

01

23

=

xr = Er(x)w

Quaternion

Q i &


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Quaternion Rates &

Angular Velocities

w = Er(x)+ xr

Er(x)+ = [ Er(x)

T Er(x) ]-1 Er

T(x) (left pseudo inverse)

--

----

2

0123

1032

2301

=

always exist

Note: Free of Math. Singularities

w xrxr w

always possible

Si l R i l &


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Simultaneous Rotational &


Given: Frames A, B & C

{A}

{B}

{C}

{B} & {C} in motion

with respect to {A}

Find: Relationships between velocitiesApC =

ApB +ARB

BpC

Si l R i l &


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DifferentiatingAUC =

AUB +ARB

BUC +ARB

BpC

Contribution of rotational velocity of frame B to the

translational velocity of C =ARB

BpC = AwBxARBBpC = AwB x (ARBBpC)

= -A

RBB

pC xA

wB= AwB x (ApC

ApB)

AUC = AUB + ARBBUC + AwB x (ARBBpC)= AUB +

ARBBUC +

AwB x (ApC

ApB)

Si lt R t ti l &


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ARC =ARB

BRC

ARC =ARB

BRC +ARB

BRC

AwCxARC = AwBxARBBRC + ARBBwCxBRC

= AwBxARC + ARBBwCxBRAARC

AwCxARC = AwBxARC + ARBBwCxBRAARC

AwCx = AwBx + ARBBwCxBRA

Si lt R t ti l &


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xCBAw

But

= ARBBwCxBRA

= ARBBw

Cexpressing vector in

diff frame

AwCx = AwBx + OR in vector form:

A

wC =A

wB +A

RBB

wCNote also (in homogeneous transformation)

xCBAw

00

x BA

B

A

B

A

B

A URwT

CB

A

w

C t ti Of


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Computation Of

End-Effector Velocity

(6x1) )qq,(fw

u

N

N

Nv

joint positionjoint velocities

C t ti Of


34/70


Computation Of


Let us examine the contribution of the ith joint motion toend-effector velocity. We set all other joint velocities :qC 0 q1 = q2= = qi-1 = qi+1= qN = so motion is occurring with respect to zi-1 axis

For joint i rotational

wi = zi-1 qiui = wi x Ri-1

i-1pN = zi-1 qi x ( pNpi-1 )

= zi-1

x ( pNp

i-1) q

iNote that oi-1 has no translational velocity

origin of frame i-1 w/c contains zi-1

since joint is rotational

C t ti Of


35/70


Computation Of


N

1i

i

N

1i

i

w

u

N

N

Nw

uv

For a translational joint i,

wi = 0

ui = zi-1 qi

The total velocity of the end-effector during coordinatedmotion is the superposition of all the elementary velocities

that represent single joint motion:

C t ti Of


36/70


Computation Of


N

2

1

N321

q

q

q

)J...JJJ(Nv

6x1

6xN J(q)

Column Ji represents motion contribution of joint i

J(q) = Jacobian matrix

Cartesian joint space

C t ti Of


37/70


Computation Of


For a translational joint i

0

z

1-i

i

J

For a rotational joint i

1-i

1-iN1-i

z

)p-p(ziJ


38/70


Jacobian Transformations

N

B

N

B

B

A

B

A

N

A

N

A

N

A

w

u

R

R

w

uv 0

0

JBvN

Velocities expressed in different frames

AvN BvN

ForARB andApB constants

N = End Effector

B = may be a link coord

frame that is held

instantaneously constant


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40/70



AuN =AuB + AwBx ( ApNApB)

AwN =AwB

B

A

B

A

B

A

N

A

N

A

N

A

N

A

w

u

I

ppI

w

uv

0

x)-(-

anotherJ


41/70



Most simplified form of Manipulator Jacobian is:

when it is computed at the mid-coord. frame

i.e.,2

1NL

All vectors are expressed in frame L

The origin of Frame L is taken as the velocity pt

for the end-effector


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{O}

{L}

{N}

end-effector

Note that although L is moving, it is taken as

instantaneously fixed when computing the Jacobian


43/70



In computing the Jacobian LJL, each column is

jointrot.for

)-(

i-1

L

i-1Li-1

L

i

z

ppz

J

jointationalfor transl0

i-1

L

izJ


44/70



OR LJL can be computed from 0JN:

N

0

N

0

N

0

L

0

0

L

0

L

L

L

L

L

w

u

I

ppI

R

R

w

u

0

)x-(-

0

0

LJL q =0JN q

LJL =0JN


45/70


Robot Kinematics of Velocity

0

R(J)

N(J)

R(J) = Range Space

ofJ= or column

space ofJ

qnJoint Space

xmEnd-Effector Space

N(J) = Null space

of Jq produces nomotion x


46/70



J = Jacobian

(q)GJ iijjq

x = G(q)

x = J qx = Jq

x = representation for E-E configuration

w

u

)(xE0

0)(xE

rr

ppx vx

v = E - Angular velocityv = J0 q = Basic Jacobian


47/70



x = J q

q(q)Jx 0)(xE0

0)(xE

rr

pp

J = Jacobian = E J0(q)


48/70


Inverse Kinematics of Velocity

Solution to x = Jq [ i.e., given x, Find q ]mx1 nx1

Exists if & only ifRankJ = Rank ( J | x )

m x n m x ( n + 1 ) matrix obtained by

augmenting J with column xMeaning x must be in the subspace spanned by thecolumns ofJ


49/70



Solution exists if & only if RankJ0 = min( m0, n )

i.e., columns ofJ0 span the space Rmin( mo,n )

First: Convert x to x0Rmo( velocity, basic kinematicmodel)x0 = J0 c ( 0vN = J0 q )

Rmo Rn m0 6


50/70



General Solution:

q = J0#(q) x0 + [ In - J0#(q)J0(q) ] q0generalized Inverse

nxn Identityany arbitrary disp

Operates on q0 to produce vectorqnN(J)qn = [ In - J0#(q) J0(q) ] q0The mapping by J0 ofqn results in zero vector in Rmo

J0 qn = [J0 - J0 J0#(q) J0(q) ] q0 = 0


51/70



Case 1: m0 = n 6J#(q) = J-1 (possible problem with singularity,

J-1 may not exist)

Case 2: m0 > n, m0 6 (not interesting/useful case,task shall be n)

overdetermined system: more eqns than unknowns.

J#

= (JT

J-1

)JT

= left pseudo inverse= exists only if RankJ = n

Soln minimizes || Jq - x0 ||2


52/70



Case 3: m0 < n, m0 6 (Redundant Robots)underdetermined system = less eqns than unknowns

J# = JT(J JT)-1 = right pseudo inverse

= exists only if RankJ = m0Soln minimizes || q ||2


53/70


Static Forces in Manipulators

Link i+1

Link i

Link i-1

Let fi = force exerted on link i

by link i-1 at coordframe (xi-1, yi-1, zi-1)

ni = moment exerted on link i


54/70



zi-1

ni

fi xi-1

yi-1

PiPi-1

-ni+1

-fi+1

zi zi

ni+1fi+1

oioi


55/70



F = 0 fifi+1 = 0Torques about origin of frame i-1 = 0nini+1 + (pipi-1) x (-fi+1) = 0

If we start with a description of the force and momentapplied by the hand, we can calculate the force and

moment applied by each link working from the last

link down to the base, link.

fn+1nn+1

Force exerted by the manipulator hand on

its environment.


56/70



Recursive Equations:

fi = fi+1ni = ni+1 + (pipi-1) x fi+1

all vectors

expressed in

same frame

(e.g. base frame )What forces are Needed at the Joints in order to

Balance the Reaction Forces & Moments acting in the link

Ti =

niT zi-1 for a rotational link i

fiT zi-1 for a translational link i


57/70


Jacobians In Force Domain

When forces act on a mechanism, work (in the

technical sense) is done if the mechanism moves

through a displacement

Principle if VIRTUAL WORK allows us to make

certain statements about the static case by defining a

VIRTUAL DISPLACEMENT x that is experienced

without passage of time dt = 0(Not only infinitesimal, dx x)


58/70



Since work has units of energy, it must be the same

measured in any set of generalized coordinates

F x = q6x1

Cartesian

Force-Moment

Vector

6x1

Virtual

displ. In

Cartesian space

Torque/Forces at joints

6x1

6x1 virtual joint disp.


59/70



But x = Jq Therefore FT [Jq] = Tq

FT J = Tx

= JT Fexpressed in the same (consistent) Frame


60/70



When the Jacobian loses full rank, there are certaindirections in which the end-effector cannot exert

static forces (through joint actuation) as desired

That is, ifJ is singular, the equation is not valid

F could be increased or decreased in certain

directions with no effect on the value calculated

for These directions are in the null-space of theJacobian


61/70



This also means that near singular configuration,mechanical advantage tends towards infinity, such

that with small joint torques, large forces could be

generated at the end-effector


62/70



Note that a Cartesian space quantity can be convertedinto a joint space quantity without calculating any

inverse kinematic functions.

Cartesian Transformation Of


63/70



Static Force

{B}

AnRAfR

(Reaction forces

for equilibrium)

{C}

AnC AfC

External forces/

moments applied

on frames {C}{A}

Given: AfCAnC

Find: AfBAnB

(the force/moment

experienced at B if

force/moment is

exerted on C)



64/70



Static Force

Why is this important?

CfC

& CnC

can be force sensor readings

But our primary interest is BfB &BnB

(force/moments at tool tip)

Force sensor

Tool{C}

{B}



65/70



Static Force

Equilibrium:

F = 0 AfC + AfR= 0AfR= -

AfC

N = 0A

nC + (

A

pB

A

pC) x

A

fR+ nR= 0AnR= -AnC(

ApBApC) x

AfR

But AfB = -AfR

AfB =AfC



66/70



Static ForceA

nB = -A

nR=A

nC + (A

pBA

pC) xA

fRAnB =AnC + (

ApBApC) x (-

AfC)

AnB =AnC + (

ApCApB) x

AfC

ORAnB =

AnC + ARBBpCx AfC

in Matrix Form

C

A

C

A

C

B

B

A

B

A

B

A

n

fIxpR

Inf 0

Force torque Jacobian transformation



67/70



Static Force

But in typical applications, we would like to relate

B

B

B

B

C

C

C

C

n

f

n

fwith

[e.g. sensor readings will be expressed in local frame

of sensor]

We can transform vectors f& n like any other vector

via Rotation Matrices

C

C

C

C

C

A

C

A

C

A

C

A

n

f

R

R

n

f

0

0



68/70



Static Force

C

C

C

C

C

A

C

A

C

B

B

A

B

A

B

A

n

f

R

R

IxpR

I

n

f

0

00

B

A

B

A

A

B

A

B

B

B

B

B

n

f

R

R

n

f

0

0

C

C

CC

C

A

C

A

C

B

B

A

CA

B

A

BA

n

f

RRxpR

R

n

f

0

Also



69/70



Static Force

C

C

C

C

C

A

C

A

C

B

B

A

C

A

A

B

A

B

B

B

B

B

n

f

RRxpR

R

R

R

n

f

0

0

0

C

C

C

C

C

A

A

B

C

A

C

B

B

A

A

B

C

A

A

B

n

f

RRRxpRR

RR 0

Therefore

(BRAARBBpC xARC) CfC = BRA [(ARBBpC) x (ARCCfC)]= (BRA

ARBBpC) x (

BRAARC

CfC)

= BpC x (BRC

CfC)

= BpC

xBRC

CfC



70/70


Static Force

This is the form given in

Craigs Book

C

C

C

C

C

B

C

B

C

B

C

B

B

B

B

B

n

f

RRxp

R

n

f

0

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