Dynamic Rigid Bodies

8/3/2019 Dynamic Rigid Bodies

1/50

Rigid Body Dynamics

CSE169: Computer Animation

Instructor: Steve Rotenberg

UCSD, Winter 2008


2/50

Cross Product

[ ]xyyxzxxzyzzy

zyx

zyx

babababababa

bbb

aaa

=

=

ba

kji

ba


3/50

Cross Product

[ ]

zyxxzz

zxyxzy

zyyzxx

xyyxzxxzyzzy

bbabac

babbac

bababc

babababababa

++=

+=

+==

=

0

0

0

bac

ba


4/50

Cross Product

=

++=

+=

+=

z

y

x

xy

xz

yz

z

y

x

zyxxzz

zxyxzy

zyyzxx

b

b

b

aa

aa

aa

c

c

c

bbabac

babbacbababc

0

0

0

0

00


5/50

Cross Product

=

=

=

0

0

0

0

00

xy

xz

yz

z

y

x

xy

xz

yz

z

y

x

aa

aa

aa

b

bb

aa

aaaa

c

cc

a

baba


6/50

Derivative of a Rotating Vector

Lets say that vectorr is rotating around theorigin, maintaining a fixed distance

At any instant, it has an angular velocity of

rr

=

dt

d

r

r


7/50

Derivative of Rotating Matrix

If matrix A is a rigid 3x3 matrix rotating withangular velocity

This implies that the a, b, and c axes must be

rotating around The derivatives of each axis are xa, xb, and

xc, and so the derivative of the entire matrix is:

AAA

== dt

d


8/50

Product Rule

( )

( )dt

dcabc

dt

dbabc

dt

da

dt

abcd

dt

dbabdt

da

dt

abd

++=

+=

The product rule defines the derivative ofproducts


9/50

Product Rule

It can be extended to vector and matrix productsas well

( )

( )

( )

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

BAB

ABA

b

ab

aba

bab

aba

+=

+=

+=


10/50

Dynamics of Particles


11/50

Kinematics of a Particle

onaccelerati

ityveloc

position

2

2

dt

d

dt

d

dt

d

xva

xv

x

==

=


12/50

Mass, Momentum, and Force

force

momentum

mass

ap

f

vp

mdt

d

m

m

==

=


13/50

Moment of Momentum

The moment of momentum is a vector

Also known as angular momentum (the twoterms mean basically the same thing, but areused in slightly different situations)

Angular momentum has parallel properties withlinear momentum

In particular, like the linear momentum, angularmomentum is conserved in a mechanical system

prL =


14/50

Moment of Momentum

prL =

p

1r

2r3r

p

p

L is the same for all three of these particles


15/50

Moment of Momentum

prL =

p

1r2r

3r

p

p

L is different for all of these particles


16/50

Moment of Force (Torque)

The moment of force(ortorque) about apoint is the rate of change of the moment

of momentum about that point

dt

dL =


17/50

Moment of Force (Torque)

( )fr

frvv

frpv

p

rp

rL

prL

=+=

+=+==

=

m

dt

d

dt

d

dt

d


18/50

Rotational Inertia

L=rxp is a general expression for themoment of momentum of a particle

In a case where we have a particle

rotating around the origin while keeping afixed distance, we can re-express the

moment of momentum in terms of its

angular velocity


19/50

Rotational Inertia

( )

( ) ( )

rrI

IL

rrL

rrrrL

vrvrLprL

=

=

===

===

m

mmm

mm


20/50

Rotational Inertia

=

=

=

22

22

22

00

0

00

0

yxzyzx

zyzxyx

zxyxzy

xy

xz

yz

xy

xz

yz

rrrrrr

rrrrrr

rrrrrr

m

rrrr

rr

rrrr

rr

m

m

I

I

rrI


21/50

Rotational Inertia

( )( )

( )

IL

I

=

+

+

+

=22

22

22

yxzyzx

zyzxyx

zxyxzy

rrmrmrrmr

rmrrrmrmr

rmrrmrrrm


22/50

Rotational Inertia

The rotational inertia matrix I is a 3x3 matrix thatis essentially the rotational equivalent of mass

It relates the angular momentum of a system toits angular velocity by the equation

This is similar to how mass relates linearmomentum to linear velocity, but rotation addsadditional complexity

IL =

vp m=


23/50

Systems of Particles

momentumtalto

massofcenterofposition

particlesallofmassltota1

==

=

==

iiicm

i

ii

cm

n

i

itotal

m

m

m

mm

vpp

xx


24/50

Velocity of Center of Mass

cmtotalcm

total

cmcm

i

ii

i

ii

cm

i

iicmcm

m

m

m

m

m

dt

dm

m

m

dt

d

dt

d

vp

pv

vx

v

xxv

=

=

==

==


25/50

Force on a Particle

The change in momentum of the center of massis equal to the sum of all of the forces on the

individual particles

This means that the resulting change in the totalmomentum is independent of the location of the

applied force

i

iicm

icm

dt

d

dt

d

dt

d

===

=

fppp

pp


26/50


( )

=

=

icmicm

iicm

pxxL

prL

The total moment of momentum aroundthe center of mass is:


27/50

Torque in a System of Particles

( )

( )

=

=

==

=

iicm

ii

cm

iicmcm

iicm

dt

d

dt

d

dt

d

fr

pr

prL

prL


28/50


We can see that a system of particles behaves a lot like

a particle itself

It has a mass, position (center of mass), momentum,

velocity, acceleration, and it responds to forces

We can also define its angular momentum and relate a

change in system angular momentum to a force appliedto an individual particle

( )

= iicm fr

= icm ff


29/50

Internal Forces

If forces are generated within the particle system(say from gravity, or springs connectingparticles) they must obey Newtons Third Law(every action has an equal and opposite

reaction) This means that internal forces will balance out

and have no net effect on the total momentum of

the system As those opposite forces act along the same line

of action, the torques on the center of masscancel out as well


30/50

Kinematics of Rigid Bodies


31/50

Kinematics of a Rigid Body

For the center of mass of the rigid body:

2

2

dtd

dtd

dtd

cmcmcm

cmcm

cm

xva

xv

x

==

=


32/50

Kinematics of a Rigid Body

For the orientation of the rigid body:

onacceleratiangular

locityangular ve

matrixnorientatio3x3

dt

d

A

=


33/50

Offset Position

Lets say we have a point on a rigid body Ifr is the world space offset of the point relative

to the center of mass of the rigid body, then the

position x of the point in world space is:

rxx += cm


34/50

Offset Velocity

The velocity of the offset point is just thederivative of its position

rvv

rxxv

rxx

+=

+==

+=

cm

cm

cm

dt

d

dt

d

dt

d


35/50

Offset Acceleration

The offset acceleration is the derivative of theoffset velocity

( )rraa

rr

vva

rvv

++=

++==

+=

cm

cm

cm

dt

d

dt

d

dt

d

dt

d


36/50

Kinematics of an Offset Point

The kinematic equations for an fixed pointon a rigid body are:

( )rraa

rvvrxx

++=

+=+=

cm

cm

cm


37/50

Dynamics of Rigid Bodies


38/50

Rigid Bodies

We treat a rigid body as a system of particles, where the

distance between any two particles is fixed

We will assume that internal forces are generated tohold the relative positions fixed. These internal forcesare all balanced out with Newtons third law, so that theyall cancel out and have no effect on the total momentumor angular momentum

The rigid body can actually have an infinite number ofparticles, spread out over a finite volume

Instead of mass being concentrated at discrete points,we will consider the density as being variable over thevolume


39/50

Rigid Body Mass

With a system of particles, we defined the totalmass as:

For a rigid body, we will define it as the integral

of the density over some volumetric domain

= dm

=

=n

i

imm1


40/50

Rigid Body Center of Mass

The center of mass is:

=

d

d

cm

xx


41/50

Rigid Body Rotational Inertia

( )( )

( )

=

+

++

=

zzyzxz

yzyyxy

xzxyxx

yxzyzx

zyzxyx

zxyxzy

III

IIIIII

drrdrrdrr

drrdrrdrr

drrdrrdrr

I

I

22

22

22


42/50

Diagonalization of Rotational Inertial

==

=

z

y

x

T

zzyzxz

yzyyxy

xzxyxx

I

I

I

where

III

IIIIII

00

00

00

00 IAIAI

I


43/50

Derivative of Rotational Inertial

( )

( )

( )

( )

III

IIAIAII

AIAII

AIAAIAI

AIAAI

AAIAI

0

0

00

000

=

+=+=

+=

+=

+=

=

dt

d

dtd

dt

ddt

d

dt

d

dt

d

dt

d

dt

d

TTT

T

TT

T

TT


44/50

Derivative of Angular Momentum

( )

II

III

III

IIL

IL

+=

+=

+=

+===

dt

d

dt

d

dt

d


45/50

Newton-Euler Equations

I

I

af

+=

= m


46/50

Applied Forces & Torques

( )

( )II

fa

fr

ff

==

==

1

1

m

iicg

icg


47/50

Properties of Rigid Bodies

afvp

av

x

mm

m

=

=

IIfr

IL

AI

+==

=


48/50

Rigid Body SimulationRigidBody {

void Update(float time);void ApplyForce(Vector3 &f,Vector3 &pos);

private:

// constants

float Mass;Vector3 RotInertia; // Ix, Iy, & Iz from diagonal inertia

// variables

Matrix34 Mtx; // contains position & orientationVector3 Momentum,AngMomentum;

// accumulators

Vector3 Force,Torque;

};


49/50

Rigid Body Simulation

RigidBody::ApplyForce(Vector3 &f,Vector3 &pos) {

Force += f;

Torque += (pos-Mtx.d) x f

}


50/50

Rigid Body Simulation

RigidBody::Update(float time) {

// Update positionMomentum += Force * time;

Mtx.d += (Momentum/Mass) * time; // Mtx.d = position

// Update orientationAngMomentum += Torque * time;

Matrix33 I = MtxI0MtxT // AI0A

T

Vector3 = I-1L

float angle = || * time; // magnitude of Vector3 axis = ;

axis.Normalize();

Mtx.RotateUnitAxis(axis,angle);

}

Documents

Dynamic Rigid Bodies