Spatial Vector Algebra - Australian National...

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Spatial Vector Alge bra

The Easy Way to do Rigid Body Dynamics

Roy FeatherstoneDept. Information Engineering, RSISE

The Australian National University

A Short Course on

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Spatial vector algebra is a concise vectornotation for describing rigid−body velocity,acceleration, inertia, etc., using 6D vectorsand tensors.

fewer quantities

fewer equations

less effort

fewer mistakes

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Mathematical Structure

spatial vectors inhabit two vector spaces:

−− motion vectors

−− force vectors

with a scalar product defined between them

m . f = work

‘‘.’’ : M6 F6 R

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m = Σi=1

f = Σi=1

A coordinate vector m=[m1, ...,m6]T

represents a motion vector m in a basis{d1, ...,d6} on M6 if

Likewise, a coordinate vector f =[f1, ..., f6]T

represents a force vector f in a basis{e1, ...,e6} on F6 if

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If {d1, ..., d6} is an arbitrary basis on M6

then there exists a unique reciprocal basis{e1, ..., e6} on F6 satisfying

With these bases, the scalar product of twocoordinate vectors is

di . ej = { 0 : i = j

1 : i = j

m . f = mT f

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Velocity

The velocity of a rigid bodycan be described by

choosing a point, P, in the body

specifying the linear velocity, vP, of thatpoint, and

specifying the angular velocity, ω, of thebody as a whole

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Velocity

The body is then deemed to be

rotating with an angular velocity ω aboutan axis passing through P

while simultaneously

translating with a linear velocity vP

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Velocity

Now introduce a coordinateframe with an origin at anyfixed point O

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Velocity

Define vO to be the velocity ofthe body−fixed point thatcoincides with O at the currentinstant

vO = vP + OP ω

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Velocity

The body can now be regardedas translating with a velocity ofvO while simultaneously rotatingwith an angular velocity of ωabout an axis passing through O

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Introduce the unit vectors i, jand k pointing in the x, y andz directions. O

ω and vO can now beexpressed in terms of theirCartesian coordinates:

ω = ωx i + ωy j + ωz k

vO = vOx i + vOy j + vOz kω =

coordinate vectors what they represent

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The motion of the body can now be expressedas the sum of six elementary motions:

a linear velocity of vOx in the x direction

an angular velocity of ωx about the line Ox

a linear velocity of vOy in the y direction

a linear velocity of vOz in the z direction

an angular velocity of ωy about the line Oy

an angular velocity of ωz about the line Oz

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dOx unit angular motion about the line Ox

unit angular motion about the line Oy

unit angular motion about the line Oz

unit linear motion in the x directiondy

unit linear motion in the y directionunit linear motion in the z direction

Define the followingPlucker basis on M6:

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Velocity

The spatial velocity of the bodycan now be expressed as

v = ωx dOx + ωy dOy + ωz dOz +

+ vOx dx + vOy dy + vOz dz

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Velocity

This single quantity provides acomplete description of thevelocity of a rigid body, and it isinvariant with respect to thelocation of the coordinate frame

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Velocity

ω^= v

The six scalars ωx, ωy,..., vOz are the Pluckercoordinates of v in the coordinate systemdefined by the frame Oxyz

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Velocity

ω^= v

v = ωx dOx + ωy dOy

+ ωz dOz + vOx dx

+ vOy dy + vOz dz

what it represents

coordinate vector

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Now try question set A

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nPA general force actingon a rigid body can beexpressed as the sum of

a linear force f actingalong a line passing throughany chosen point P, and

a couple, nP

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If we choose a differentpoint, O, then the forcecan be expressed as thesum of

a linear force f actingalong a line passing throughthe new point O, and

a couple nO, where nO = nP + OP f

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Now place a coordinateframe at O and introduceunit vectors i, j and k, asbefore, so that

f = fx i + fy j + fz k

nO = nOx i + nOy j + nOz k

nO = f =

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The total force acting on the body can now beexpressed as the sum of six elementary forces:

a linear force of fx acting along the line Ox

a moment of nOx in the x direction

a linear force of fy acting along the line Oy

a linear force of fz acting along the line Oz

a moment of nOy in the y direction

a moment of nOz in the z direction

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eOx unit linear force along the line Ox

unit linear force along the line Oy

unit linear force along the line Oz

unit couple in the x directioney

unit couple in the y directionunit couple in the z direction

eOz ez

Define the followingPlucker basis on F6:

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The spatial force actingon the body can now beexpressed as

f = nOx ex + nOy ey + nOz ez

+ fx eOx + fy eOy + fz eOz

This single quantity provides acomplete description of the forcesacting on the body, and it is invariant withrespect to the location of the coordinate frame

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The six scalars nOx,nOy,..., fz are the Pluckercoordinates of f in thecoordinate system definedby the frame Oxyz

coordinate vector:

fxfyfz

fO =^ nO

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Plucker Coordinates..

..Plucker coordinates are the standardcoordinate system for spatial vectors

a Plucker coordinate system is defined bythe position and orientation of a singleCartesian frame

a Plucker coordinate system has a total oftwelve basis vectors, and covers bothvector spaces (M6 and F6)

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Plucker Coordinates..

so the scalar product between a motionvector and a force vector can be expressedin Plucker coordinates as

which is invariant with respect to thelocation of the coordinate frame

the Plucker basis ex, ey,..., eOz on F6 isreciprocal to dOx, dOy,..., dz on M6

v . f = vOT fO

^^ ^ ^

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Coordinate Transforms

transform from A to B

for motion vectors:

corresponding transformfor force vectors:

BXA = where

* = (BXA)

rz−ry

−rz0

ry−rx0

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Basic Operations with Spatial Vectors

Rigid ConnectionIf two bodies are rigidly connected thentheir velocities are the same

Relative velocityIf bodies A and B have velocities of vA andvB, then the relative velocity of B withrespect to A is

vrel = vB − vA

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If forces f1 and f2 both act on the samebody, then they are equivalent to a singleforce ftot given by

ftot = f1 + f2

If body A exerts a force f on body B,then body B exerts a force −f on body A(Newton’s 3rd law)

Summation of Forces

Action and Reaction

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Scalar Product

If a force f acts on a body with velocity v,then the power delivered by that force is

Scalar Multiples

power = v . f

A velocity of αv causes the same movementin 1 second as a velocity of v in α seconds.A force of βf delivers β times as muchpower as a force of f

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Now try question set B

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Spatial Cross Products

There are two cross product operations:one for motion vectors and one for forces

ω mO + vO mvO mO =^ ^

=*ωvO

ω nO + vO f

ω fvO fO =^ ^

where vO and mO are motion vectors, and fO

is a force.

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Differentiation

The derivative of a spatial vector is itselfa spatial vector

The derivative of a spatial vector that isfixed in a body moving with velocity v is

in general, s = limd

dt δt 0

s(t+δt)−s(t)

if s M6

if s F6*

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Differentiation in Moving Coordinates

velocity of coordinateframe

componentwisederivative of coordinatevector sO

coordinate vectorrepresenting ds/dt

or if s F6*

sO + vO sO

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Acceleration

. . . is the rate of change of velocity:

but this is not the linear acceleration ofany point in the body!

v =^a =^ ω.

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Acceleration

O is a fixed point in space,

and vO(t) is the velocity of the body−fixedpoint that coincides with O at time t,

so vO is the velocity at which body−fixedpoints are streaming through O.

vO is therefore the rate of change ofstream velocity

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Acceleration Example

If a body rotates with constant angular velocityabout a fixed axis, then its spatial velocity isconstant and its spatial acceleration is zero;but each body−fixed point is following a circularpath, and is therefore accelerating.

fixed axis

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Acceleration Formula

Let r be the 3D vector giving the position ofthe body−fixed point that coincides with O atthe current instant, measured relative to anyfixed point in space

ωr.v =^ =we then have

butω.

.a =^ =ω.

r − ω r.. .

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(Look, no Coriolis term!)

Basic Properties of Acceleration

Acceleration is the time−derivative ofvelocity

Acceleration is a true vector, and has thesame general algebraic properties asvelocity

Acceleration formulae are the derivativesof velocity formulae

If vtot = v1 + v2 then atot = a1 + a2

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Now try question set C

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Rigid Body Inertiamass:

CoM:inertiaat CoM:

spatial inertia tensor:

IO =^ IO m c

m cT m1

IO = IC − m c c~~

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Composition

If two bodies with inertias IA and IB arejoined together then the inertia of thecomposite body is

Itot = IA + IB

Basic Operations with Inertias

Coordinate transformation formula

IB = BXA

AXB = (

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Equation of Motion

f = net force acting on a rigid body

I = inertia of rigid body

v = velocity of rigid body

I v = momentum of rigid body

a = acceleration of rigid body

f = (I v) = I a + v I vd

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Motion Constraints

degree of (motion) freedom:

degree of constraint:

S can vary with time

dim(S)

6 − dim(S)

If a rigid body’s motion is constrained, thenits velocity is an element of a subspace,S M6, called the motion subspace

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Motion Constraints

A constraint force does no work againstany motion allowed by the motionconstraint

Motion constraints are caused by constraintforces, which have the following property:

(D’Alembert’s principle of virtual work, and Jourdain’s principle of virtual power)

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Motion Constraints

Constraint forces are therefore elements ofa constraint−force subspace, T F6,defined as follows:

This subspace has the property

dim(T) = 6 − dim(S)

T = { f | f . v = 0 ∀ v S }

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Matrix Representation

The subspace S can be represented byany 6 dim(S) matrix S satisfyingrange(S) = S

Likewise, the subspace T can berepresented by any 6 dim(T) matrix Tsatisfying range(T) = T

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Properties

STT = 0, which implies . . .

STf = 0 and TTv = 0 for all f T and v S

any vectors v S and f T can beexpressed as v = S α and f = T λ, whereα and λ are dim(S) 1 and dim(T) 1

coordinate vectors

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Constrained Motion Analysis

A force, f, is applied to abody that is constrained tomove in a subspace S = range(S) of M6.The body has an inertia of I, and it is initiallyat rest. What is its acceleration?

An Example:

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v = S α

ST fc = 0

Irelevant equations:

v = 0 implies

a = S α + S α. .

f + fc = I a

a = S α.

solution:

a = S (STI S)

−1 S

f + fc = I S α.

ST f = S

TI S α.α = 0

α = (STI S)

−1 S

f + fc = I a + v I v*

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Now try question set D

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Inverse Dynamics

joint 1

link 1

joint 2 joint n

link n

vi, ai link velocity and acceleration

joint velocity, acceleration & axis

fi force transmitted from link i−1 to i

τi joint force variable

Ii link inertia

qi, qi, si. ..

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acceleration is the derivative of velocity

velocity of link i is the velocity of link i−1plus the velocity across joint i

equation of motion

τi = siT fi

active joint force

vi = vi−1 + si qi.

ai = ai−1 + si qi + si qi. ...

fi − fi +1 = Ii ai + vi Ii vi*

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The Recursive Newton−Euler Algorithm

(v0 = 0)

(a0 = 0)

(fn+1 = fee)

τi = siT fi

(Calculate the joint torques τi that willproduce the desired joint accelerations qi.)

fi = fi +1 + Ii ai + vi Ii vi*

vi = vi−1 + si qi.

ai = ai−1 + si qi + si qi. ...

Spatial Vector Algebra - Australian National...

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