Equation of motion of a variable mass system3

1

EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEM

LAGRANGIAN APPROACH

SOLO HERMELIN

http://www.solohermelin.com

http://www.solohermelin.com/

2

SOLO EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEM

• Simplified Particle Approach (see Power Point Presentation)

The equations of motion can be developed using

At a given time t the system has

v (t) – system volume.

m (t) – system mass.

S (t) – system boundary surface.

• Reynolds’ Transport Theorem Approach (see Power Point Presentation)

• Lagrangian Approach (this Power Point Presentation)

3

SOLO EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEMLAGRANGIAN APPROACH

TABLE OF CONTENT

• Generalized Forces

Joseph-Louis Lagrange

1736-1813

• Lagrange’s Equations of Motion

• Principal Coordinate Frames

• Inertial Coordinate Frame

• Body Coordinate Frame

• Body Mean System Axes

• Orientation of Body Frame

• Kinetic Energy of the System

• Potential Energy of the System • Elastic Potential Energy

• Gravitational Potential Energy

• Computation of Lagrange’s Equations in Body Coordinates

• Derivation of Equations of Motion

• Summary of the Equations of Motion of a Variable Mass System

• References • Appendix A: Lagrange Equations

4


Lagrange’s Equations of Motion

The Lagrange’s Equations of Motion for a dynamic system are:

iiii

QUTT

td

d =

∂∂+

∂∂−

∂∂

ξξξ

- system kinetic energy. T

- system potential energy. U

- generalized coordinates (i=1,2,…, number of degrees of freedom of the system).iξ

- generalized force along the generalized coordinate given by.iξiQ( )( )i

i

WQ

ξδδ

∂∂=

-virtual work done on the system by all external forces/moments (excluding

those accounted for in the potential energy term) during virtual displacement

along all the generalized coordinates.

Wδ

Table of Contents

5


Principal Coordinate Frames

R

- Position of the mass element dm relative to I.

Itd

RdV

= - Velocity of the mass element dm relative to I.

IItd

Rd

td

Vda

2

2

== - Acceleration of the mass element dm relative to I.

Inertial Coordinate Frame

(vector form) orIzIyIx zRyRxRR ˆˆˆ ++=

IzIyIx

I

zRyRxRtd

RdV ˆˆˆ

++== (vector form) or

(vector form) orIzIyIx

II

zRyRxRtd

Rd

td

Vda ˆˆˆ

2

2

++===

{ }Tzyx RRRR ,,=

(matrix form)

( ) { }T

zyxI RRRV

,,= (matrix form)

( ) { }T

zyxI RRRa

,,= (matrix form)Table of Contents

6


Principal Coordinate Frames (continue - 1)

Cr,

- Position of the mass element dm relative to C.

Body Coordinate Frame

The origin of the Body Frame (B) is located at theinstantaneous Centroid (C) of the system.

0,

=∫m

C mdr

R

- Position of the mass element dm relative to I.

CR

- Position of the centroid C relative to I.

CC rRR ,

+=

0,Cr

- Position of the same mass element dm in the un-deformed system, relative to C.

e

- Change in position of the mass element dm due to elastic deformation of the system.

err CC

+= 0,,

Table of Contents

7



Body Mean System Axes

Mean Body System Axes are defined such that therelative linear and angular momentum, due toelastic deformation, are zero at every instant.

The Body Mean Axes must satisfy the following:

0

=∫ mdtd

ed

Bm

0,

=∫ × md

td

edr

BmC

0, =∫ ⋅ mdtd

ed

td

rd

Bm B

C

Table of Contents

8


Structural Model of the System

Assume that the elastic deformations are small, and can berepresented in terms the normal un-damped modes of vibration.

( ) ( )∑=∞

=1iii tRe ηφ

- are mode shape functions that depend on the position of the mass element of the system.( )Ri

φ

- are generalized coordinates giving the magnitude of the modal displacements and are functions of time.

( )tiη

Structural Dynamic Analysis (e.g. final element method) provides the mode shapefunctions component of each element of the system, as well as the vacuo modalfrequencies ( ) , for a selected number of modes.

( )Ri

φ

iω

iii

td

d ηωη 2

2

2

−=

The mode shape functions are orthogonal.

0ji

m

ji md≠=⋅∫ φφ

i

m

ii Mmd =∫ φφ

MV

Bx

By

BzWz

Wy

Wx

αβ

αβ

Bp

Wp

Bq

WqBrWr

FIRST ELASTIC MODE

SECOND ELASTIC MODE

9



Orientation of the Body Frame

The orientation of the Body Frame relative to theInertial Frame has three degrees of freedom. We will use 3 Euler Angles that define the orientationby three consecutive rotations around the consecutiveframe axes.

[ ]

−=

11

1111

0

0

001

:

θθθθθ

cs

sc

[ ]

−=

22

22

22

0

010

0

:

θθ

θθθ

cs

sc

[ ]

−=

100

0

0

: 33

33

33 θθθθ

θ cs

sc

The three basic Euler rotations aroundaxes are described by the rotation matrices:

,3,2,1

10



Orientation of the Body Frame (continue – 1)

Using the basic Euler Angles we can define the following 12 different rotations:

(a) six rotations around three different axes:

321 →→ 231 →→ 312 →→ 132 →→ 213 →→ 123 →→

(b) six rotations such that the first and third are around the sam axes, but the second is different:

121 →→ 131 →→ 212 →→ 232 →→ 313 →→ 323 →→

Suppose that the Transfer Matrix from Inertia to Body is defined by threeconsecutive Euler Angles: around (unit vector in Inertial Frame),around (unit vector in intermediate frame), around (unit vectorin Body Frame).

BIC

iθ Ii jθInterj

kθ Bk

[ ] [ ] [ ] [ ] [ ] TBI

BI

IB

IB

BIkkjjii

BI CCCICCC ==→==

−1&θθθ

11




The angular velocity vector of rotation of the Body frame relative to Inertia frame is:

kIjIntriBIB kji θθθω ˆˆˆ ++=← In Body frame this is:

( ) ( ) ( ) ( ) ( ) [ ] ( ) [ ] [ ] ( )k

IIjjiij

IntrIntriii

BBk

BIj

BIntri

BB

BIB kjikji θθθθθθθθθω ˆˆˆˆˆˆ ++=++=←

[ ] ( ) [ ] ( ) [ ] [ ] ( )[ ] { } { } { }[ ]

=

=

=←

k

j

i

k

j

i

IIjjii

IntrIntrii

BB

BIB DDDkji

r

q

p

θ

θ

θ

θ

θ

θ

θθθω

321ˆˆˆ:

{ } ( ) ( ){ } ( ) [ ] ( ) ( ){ } ( ) [ ] [ ] ( )

{ } { } { }[ ]321

321

:

ˆˆ:,&ˆˆ:&ˆ:

DDDD

kkDjjDconstiD IIjjii

BIji

IntrIntrii

BIntri

BB

=

======→

θθθθθθ

( ) [ ] [ ] ( ) { } { } { }[ ]

=

==

=

=←

k

j

i

k

j

i

E

k

j

i

BIB DDD

DDD

DDD

DDD

DD

r

q

p

θ

θ

θ

θ

θ

θ

θ

θ

θ

θ

ω

321

333231

232221

131211

:

where:

or

12




The velocity vector of the system centroid C is given by:

( ) [ ] ( )

==

=

=

Cz

Cy

Cx

IC

Cz

Cy

Cx

BI

BC

R

R

R

CCC

CCC

CCC

RC

R

R

R

C

w

v

u

V

333231

232221

131211

:

The following relations will be useful:

and

( )

=

k

j

i

E

θθθ

θ : andwhere [ ] ( )

−−

−=×←

0

0

0

:

pq

pr

qrB

IBω

Table of Contents

( ){ }( ) [ ] ( )B

IBT

E

TBIB

T

Dtd

Dd ×+∂

∂= ←

← ωθ

ω Appendix

[ ] [ ] [ ] ( )BIB

TBI

TBI CC

td

d ×= ←ω Appendix

13


Kinetic Energy of the System

( )∫ ⋅=

tm II

mdtd

Rd

td

RdT

2

1Kinetic Energy of the System

We have:

We can write

( )∫

+⋅

+=

tm I

CC

I

CC md

td

rdV

td

rdVT ,,

2

1

( )( ) ( ) ( )

∫ ⋅+∫+∫⋅=tm I

C

I

C

tm I

CC

tmCC md

td

rd

td

rdmd

td

rdVmdVV ,,,

2

1

2

1

(a) (b) (c)

Let develop each of the three parts of this expression

erRrRR CCCC

++=+= 0,,

I

CC

I

C

I

C

Itd

rdV

td

rd

td

Rd

td

Rd ,,

+=+=

14


Kinetic Energy of the System (continue – 1)

(a) ( )( )

( ) mVVmdVV CCtm

CC

⋅=∫⋅

2

1

2

1

(b)

Use Reynolds’ Transport Theorem when we differentiate( )

0,

=∫tm

C mdr

Therefore( )

∑∫ −=openings

iifluidCiopen

tmB

C mrmdtd

rd

,

, ˆ

and

( ) ( )∫

×+⋅=∫⋅ ←

tmCIB

B

CC

tm I

CC mdr

td

rdVmd

td

rdV ,

,,

ω

( ) ( ) ( )∑∫∑ ∫∫∫∫ +=+=

=openings

ifluidCiopentm

B

C

i SC

tmB

CREYNOLDS

Btm

C mrmdtd

rdmdrmd

td

rdmdr

td

d

iopen

,

,

,

,

,ˆ0

( ) ( )∑⋅−=∫⋅=∫⋅

openingsifluidCiopenC

tm B

CC

tm I

CC mrVmd

td

rdVmd

td

rdV

,

,, ˆ

( ) ( ) ( )∫∫∫ ⋅=

×⋅+⋅= ←tm

B

C

Ctm

CIBCtm

B

C

C mdtd

rdVmdrVmd

td

rdV ,

0

,

,

ω

15



(c)( ) ( )

∫

×+⋅

×+=∫ ⋅ ←←

tmCIB

B

CCIB

B

C

tm I

C

I

C mdrtd

rdr

td

rdmd

td

rd

td

rd,

,,

,,,

2

1

2

1

ωω

( )∫ ⋅=tm B

C

B

C mdtd

rd

td

rd ,,

2

1

(c1)

( )( )∫ ×⋅

+ ←

tmCIB

B

C mdrtd

rd,

,

ω

(c2)

( ) ( )( )∫ ×⋅×+ ←←tm

CIBCIB mdrr ,,2

1 ωω(c3)

(c1)( ) ( )

∫

+⋅

+∫ =⋅

tm BB

C

BB

C

tm B

C

B

C mdtd

ed

td

rd

td

ed

td

rdmd

td

rd

td

rd 0,0,,,

2

1

2

1

( ) ( ) ( )∫

⋅

+∫

⋅

∫ +⋅=

tm BBtm BB

C

tm B

C

B

C mdtd

ed

td

edmd

td

ed

td

rdmd

td

rd

td

rd

2

1

2

1

0

0,0,0,

( )( ) ( )∑ ∫ ×⋅×+∫ ⋅= ←←

rotors mCjrotorBjrotorCjrotorBjrotor

tmFrozenRotors

B

C

FrozenRotorsB

C

jrotor

mdrrmdtd

rd

td

rd,,

,,

2

1

2

1

ωω

( )( )∑ ∫

⋅×+∫

⋅+ ←

rotors m B

CjrotorBjrotortm BFrozenRotors

B

C

jrotor

mdtd

edrmd

td

ed

td

rd

0

,

0

, ω( )∫

⋅

+

tm BB

mdtd

ed

td

ed

2

1

( )∫ ⋅=tm

FrozenRotorsB

C

FrozenRotorsB

C mdtd

rd

td

rd ,,

2

1 ( )∑

∫ ××−⋅+

←⋅

←←rotors

I

mBjrotorCjrotorCjrotorBjrotor

BjrotorCjrotor

jrotor

mdrr

ω

ωω

,

,,2

1

( )∫ ∑∑

∞

=

∞

=+

tm

j

i j

iji md

td

d

td

d ηηφφ1 12

1

16


( )( )∑ ⋅⋅+∫ ∫ ⋅=⋅ ←←

rotorsBjrotorCjrotorBjrotor

tm tmFrozenRotors

B

C

FrozenRotorsB

C

B

C

B

C Imdtd

rd

td

rdmd

td

rd

td

rdωω

,,,,,

2

1

2

1

2

1


(c3)

(c2)

( )∫ ∑

∞

=

+

tm i

ii md

td

d

1

2

2

2

1 ηφ

( )[ ]∫ −⋅=jrotorm

CjrotorCjrotorCjrotorCjrotorCjrotor mdrrrrI ,,,,, 1:

where Second Moment of Inertia Dyadic of the Rotor j Relative to C

( )( )

( )( ) ( )

∫

×⋅=∫

⋅×=∫ ×⋅

←←←

tm B

CCIB

tm B

CCIB

tmCIB

B

C mdtd

rdrmd

td

rdrmdr

td

rd ,,

,,,

,

ωωω

( ) ( )

0

,0,

, ∫

×⋅+∫

×⋅= ←←

tm B

CIBtm B

CCIB md

td

edrmd

td

rdr ωω

( )( )( )∑ ∫ ××⋅+∫

×⋅= ←←←

rotors mCjrotorBjrotorCjrotorIB

tmFrozenRotors

B

CCIB mdrrmd

td

rdr ,,

,,

ωωω

( )[ ] [ ] Bjrotor

rotors mCjrotorCjrotorIB

tmFrozenRotors

B

CCIB mdrrmd

td

rdr ←←← ∑

∫ ××−⋅+∫

×⋅= ωωω

,,,

,

( )[ ]∑⋅+∫

×⋅= ←←←

rotorsBjrotorCrotorjIB

tmFrozenRotors

B

CCIB Imd

td

rdr ωωω

,,

,

17



(c3)

( )[ ]( )∫ −⋅=

tm

OOOOO mdrrrrI ,,,,, 1:

where Second Moment of Inertia Dyadic of the System Relative to O

( ) ( )( )

[ ] [ ]( )

IBtm

CCIBtm

CIBCIB mdrrmdrr ←←←←

∫ ××−⋅=∫ ×⋅× ωωωω ,,,, 2

1

2

1

IBCIB I ←← ⋅⋅= ωω ,2

1

18



To summarize, the Kinetic Energy of the system is given by

( )( ) ( ) ( )

∫ ⋅+∫⋅+∫⋅=tm I

C

I

C

tm I

CC

tmCC md

td

rd

td

rdmd

td

rdVmdVVT ,,,

2

1

2

1

( ) ∑⋅−⋅=openings

iifluidCiopenCCC mrVmVV

,

ˆ2

1IBCIB I ←← ⋅⋅+ ωω

,2

1

( )∑ ⋅⋅+∫ ⋅+ ←←

rotorsBjrotorCrotorjBjrotor

tmFrozenRotors

B

C

FrozenRotorsB

C Imdtd

rd

td

rdωω

,,,

2

1

2

1

∑∞

=

+

1

2

2

1

ii

i Mtd

dη( )

∑ ⋅⋅+∫

×⋅+ ←←←


tmFrozenRotors

B

CCIB Imd

td

rdr ωωω

,,

,

Table of Contents

19


Potential Energy of the System

We consider only

(Electromagnetic, Chemical Potentials are not considered)

ge UUU +=

Elastic Deformation Potential eU

Gravitational Field Potential gU

Table of Contents

20


Potential Energy of the System

Elastic Deformation Potential eU

( ) ( )∑∞

=

=1i

iii tle ηφ

iii

td

d ηωη 2

2

2

−= i

m

ii Mmd =∫ φφ

0ji

m

ji md≠

=∫ φφ

∫ ∑∑∫

−⋅

−=

⋅−=

∞

=

∞

=m iiii

jjj

m B

e mdmdtd

edeU

1

2

12

2

2

1

2

1 ηωφηφ

( ) ( ) ∑∑ ∫∑∑ ∫∞

=

∞

=

∞

=

∞

=

=

⋅=

⋅=

1

22

1

22

1 1

2

2

1

2

1

2

1

iiii

iii

m

iij i

jii

m

ij Mmdmd ηωηωφφηηωφφ

iiii

e MU ηωη

2=∂∂

From

We obtain

From this we obtain

Table of Contents

21


Potential Energy of the System (continue – 1)

Gravitational Field Potential gU

( )2

2

01E

EarthEE

R

RgRRg

−=

220 sec/17.32sec/78.9 ftmg ==

where

mREarth 135.378.6=

( ) ( )

( ) 2/1

2

0

2

02

2

0

2

0

2

2

0

1

1

CECE

Earth

CE

Earth

m

CE

CE

Earth

CE

Earth

m E

EarthECCE

m

CCEg

RR

Rgm

R

RgmmdrR

R

Rg

R

Rgm

mdR

RgRrRmdgrRU

⋅==⋅+=

⋅+−=⋅+=

∫

∫∫

( ) CE

CE

CE

EarthCE

CECE

Earth

C

g

R

R

R

RgmR

RR

Rgm

R

U

2

2

02/3

2

0 22

1 −=⋅

−=∂∂

( )CECE

CE

Earth

C

g RgmRR

Rgm

R

U =−=

∂

∂1

2

2

0

From this we obtain

or

Table of Contents

22


Generalized Forces

- generalized force along the generalized coordinate given by.iξiQ( )( )i

i

WQ

ξδδ

∂∂=

-virtual work done on the system by all external forces/moments (excluding

those accounted for in the potential energy term) during virtual displacement

along all the generalized coordinates.

Wδ

The generalized forces are:

PQ - due to position change, relative to inertial system ( ) [ ] Tzyx

IP RRRR ==

ξ

- due to rotation of the system, around its centroid, relative to inertial systemRQ

( ) [ ] Tkji

ER θθθθξ ==

- due to elastic modal displacementsEQ [ ] TE

4321 ηηηηηξ ==

23


( ) ( ) ( )∑+=openings

iiopenW tStStS

Generalized Forces (continue – 1)

Virtual Work due to Position Change, Relative to Inertial Frame

The virtual work done by change in position is due to pressure distribution and fluid flow through the openings, and to discrete forces applied on the system

( ) RFtd

RdmdstfnpW

openings jj

I

ifluid

ifluidS

P

vehicle

δδ ⋅

+

++−= ∑ ∑∫

ˆ11

where

• Sw(t) the impermeable wall through which the flow can not escape .( )0,

=sV

• Sopen i(t) the openings (i=1,2,…) through which the flow enters or exits .( )0>m ( )0<m

( )2/mN p - pressure on (normal to) the surface .

f - friction force per (parallel to) unit surface .( )2/mN

n1 - outward unit vector normal to the surface element ds

t1 - local unit vector of tangential stress due to flow on the surface element ds

∑j

jF

- discrete forces applied to the system at the position jR

( )N

is the mean position vector of the flow and of the opening on iopenSifluidR

fluidm - fluid rate flowing through the opening iopenS

24



Si

I

CiopenIC

I

iopenfluid

I

Ciopen

I

C

I

ifluid Vtd

rdV

td

rd

td

rd

td

Rd

td

Rd,

,,, ˆˆˆˆˆ

++=++=

- velocity of the centroid C of the system relative to inertiaCV

I

Ciopen

td

rd ,

- mean velocity of the opening i relative to the centroid C

SiV,

- mean velocity of the fluid relative the opening i

Virtual Work due to Position Change, Relative to Inertial Frame (continue – 1)

25



A virtual rotation( )

kIjIntriBE kji θδθδθδθδ ˆˆˆ ++=

will produce a virtual displacement: ( )C

EC rr ,,

×= θδδ

( ) ( )∑∑∑∫ ⋅+⋅+

⋅++−⋅=

k

E

kj

jCjopenings

I

ifluid

ifluidCiopenS

CR MFrtd

RdmrdstfnprW

vehicle

θδδδδδ

,,,

ˆˆ11

( )( ) ( ) ( )( )( )( ) ( )∑∑

∑∫

⋅+⋅×+

⋅×++−⋅×=

k

E

kj

jCj

E

openings

I

ifluid

ifluidCiopen

E

SC

E

MFr

td

Rdmrdstfnpr

vehicle

θδθδ

θδθδ

,

,,

ˆˆ11

( ) ( )

+×+

×++−×⋅= ∑ ∑∑∫

openings kk

jjCj

I

ifluid

ifluidCiopenS

C

E MFrtd

Rdmrdstfnpr

vehicle

,,,

ˆˆ11θδ

The virtual work done along the generalized coordinates (rotation around C relative to inertial frame around Euler axes) is done by the pressure distribution, the flow through the openings and the discrete forces and moments applied on the system:

( )Eθ

Virtual Work due to Rotation, Relative to Inertial Frame

26



- due to elastic modal displacementsEQ [ ] TE


The virtual work done during the elastic deformations along thr generalized coordinates is done by the pressure distribution on the wetted area of the system and the discrete forces and moments applied on the system:

e

( )

( ) ( ) ( ) ( )∑ ∑∑ ∑∫ ∑

∑∑∫

×∇⋅+

⋅+

⋅+−=

×∇⋅+⋅+⋅+−=

∞

=

∞

=

∞

= k iiik

j iiij

S iii

kk

jj

SE

MFdstfnp

eMeFdsetfnpW

W

vehicle

111

11

11

δηφδηφδηφ

δδδδ

( )( ) ( )[ ]

,2,111 =∑ ×∇⋅+∑ ⋅+∫ ⋅+−=

∂∂= ∞ iMFdstfnpp

WQ

kik

lil

Si

i

PEi

W

φφφηδ

δGeneralized Elastic Forces

27



Let add to this equation the following

( )( )

( )01

0

,

5

=∫∫∫ ×∇=∫ ×− ∞∞V

C

GGauss

tSCS dvrpdsnpRR

( ) ( ) ( )0111

0

=⋅∇== ∫∫∫ ∞∞∞tv

Gauss

tStS

dvnpdsnpdsnp

( )[ ] RFtd

RdmdstfnppW

openings jj

I

ifluidifluid

S

P

vehicle

δδ ⋅

+

++−= ∑ ∑∫ ∞

ˆ11

( )[ ] ( )E

openings kk

jjCjSi

I

Ciopen

CifluidCiopenS

CR MFrVtd

rdVmrdstfnpprW

vehicle

θδδ

⋅

+×+

++×++−×= ∑ ∑∑∫ ∞ ,,

,

,,

ˆˆ11

( ) ( ) ( ) ( )∑ ∑∑ ∑∫ ∑

×∇⋅+

⋅+

⋅+−=

∞

=

∞

=

∞

= k iiik

j iiij

S iiiE MFdstfnpW

W111

11 δηφδηφδηφδ

where is the pressure far away from the system, to obtain∞p

28



From those equations we obtain

( ) ( )( )( )

( )I

openings jj

I

iopenI

Cifluidi

TiAI

C

PI

P Ftd

rdVmFF

R

WQ

+

+++=

∂∂= ∑ ∑∑∑

ˆ

δδ

Generalized Position Forces in Inertial Frame

( ) ( )[ ] ( )∫∑ +−= ∞

WsS

IIA dstfnppF

11

Aerodynamic Forces in Inertial Frame

( ) ( )[ ] ( )∑∑ −+= ∞openings

I

iopenSiifluidi

ITi nppSVmF

1, Thrust Forces in Inertial Frame

29



Generalized Moments around Euler Axes (E), relative to C

Aerodynamic Moments relative to C

Thrust Moments relative to C

( ) ( )( )( ) ( )[ ]∫ +−×=

∂∂= ∞

vehicleS

CERE

R dstfnpprW

Q

11,θδδ

∑ ∑∑ +×+

++×+

openings kk

jjCjSi

I

CiopenI

CifluidCiopen MFrVtd

rdVmr

,,

,

,

ˆˆ

( )[ ]∫ +−×= ∞

WS

C dstfnppr11 ( )[ ]∑ +−×+ ∞

openingsSiifluidopeniCiopen VmnppSr ,, 1

∑∑∑ +×+

+×+

kk

jjCj

openings

I

CiopenI

CCiopenifluid MFrtd

rdVrm

,

,

,

ˆˆ

( ) ( )( )( )

( )E

openings kk

jjCj

I

CiopenI

CCiopenifluidi

CTiCAE

RE

R MFrtd

rdVrmMM

WQ

+×+

+×++=

∂∂

= ∑ ∑∑∑∑

,

,

,,,

ˆˆ

θδδ

( )[ ]∫∑ +−×= ∞

WS

CCA dstfnpprM11:,

( )[ ]∑∑ −+×= ∞openings

iopenSiifluidCiopeni

CTi nppSVmrM

1ˆ: ,,,

30



We want to find the Generalized Moments around Body Axes. We must find the transformation from the non-orthogonal Euler Axes (E) to Body Axes (B).

( ) ( ) [ ] ( ) [ ] [ ] ( )[ ] { } { } { }[ ] ( )E

k

j

i

k

j

iI

IjjiiIntr

IntriiB

BB DDDDkji θδ

θδθδθδ

θδθδθδ

θθθθδ

=

=

= 321

ˆˆˆ

( ){ } ( )[ ]( )

( ){ } ( )[ ]( )B

openings kk

jjCj

I

ifluidifluidCiopen

S

CTTE

B

openings kk

jjCj

I

ifluidifluidCiopen

S

C

TER

MFrtd

RdmrdstfnpprD

MFrtd

RdmrdstfnpprDW

vehicle

vehicle

+×+

×++−×=

+×+

×++−×=

∑ ∑∑∫

∑ ∑∑∫

∞

∞

,,,

,,,

ˆˆ11

ˆˆ11

θδ

θδδ

( ) ( )( )( )

( )B

openings kk

jjCj

I

CiopenI

CCiopenifluidi

CTiCA

T

E

RB

R MFrtd

rdVrmMMD

WQ

+×+

+×++=

∂∂= ∑ ∑∑∑∑

,

,

,,,

ˆˆ

θδδ

Table of Contents

31


Computation of Lagrange Equations in Body Frame

The generalized coordinates are [ ]TE

TR

TP ξξξξ

=

where( ) [ ] T

CzCyCxI

CP RRRR ==

ξ

( ) [ ] Tkji

ER θθθθξ ==

[ ] TE


The velocity vector of the system centroid C is given by:

( ) [ ] ( )

==

=

=

Cz

Cy

Cx

IC

Cz

Cy

Cx

BI

BC

R

R

R

CCC

CCC

CCC

RC

R

R

R

C

w

v

u

V

333231

232221

131211

:

The angular velocity of rotation of the Body relative to inertia is:

[ ] [ ] [ ] ( ) { } { } { }[ ]

=

==

=

=←

k

j

i

k

j

i

E

k

j

i

BIB DDD

DDD

DDD

DDD

DD

r

q

p

θ

θ

θ

θ

θ

θ

θ

θ

θ

θ

ω

321

333231

232221

131211

:

32


Computation of Lagrange Equations in Body Frame (continue – 1)

We want to perform a change of coordinates from to( )ξξ , ( )w,ξ

{ } { }ηθηηθθθξ ,,,,,,,,,,: 21321TT

CT

CzCyCx RRRR

==

{ } { }ηωηη ,,,,,,,,,,: 21T

BT

CT Vrqpwvuw ==

- system potential energy. ( )ξU

- generalized force along the generalized coordinate given by.iξiQ

( ) ( )wTT ,, ξξξ = - system kinetic energy.

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( ) ( )( )ηωηθ

ηωθηθηθηθ

,,,,,

,,,,,,,,,, 1

BB

BC

EIC

BB

BC

ETEIC

EIC

EIC

VRT

DVCRTRRT

=

=

−

The coordinates are called quasi-coordinates (see Meirovitch [4], pg. 157),

to differentiate from the Lagrangian’s coordinates that describe the degrees

of freedom.

( )ξξ ,

( )w,ξ

33


Computation of Lagrange Equations in Body Frame (continue – 2)The Lagrange’s Equations of Motion are:

iiii

QUTT

td

d =

∂∂+

∂∂−

∂∂

ξξξ

Let outline (full derivation of the equation on this page is done in Appendix ) the derivation of the Lagrange’s Equations in Body Coordinates

( )

( )

( ) ( )( )( ) ( )

∂∂=

∂∂

∂

∂=

∂

∂B

C

ETB

CI

C

BC

IC

V

TC

V

T

R

V

R

T

θ Appendix

( ) ( )

( ){ }( ) ( )

( ){ }( ) ( )

∂∂

∂∂

+

∂∂

∂∂

+

∂∂=

∂∂

←

←B

CE

TBC

BIB

E

TBIB

EE V

TVTTT

θωθ

ωθθ

Appendix

( ) ( )

∂∂=

∂∂

IC

IC R

T

R

T Appendix

( ) ( )

( ){ }( ) ( )

( ){ }( ) ( )

( )

( ){ }( ) ( ) [ ] ( )

( )

∂∂×−

∂∂

∂∂+

∂∂=

∂∂

∂∂+

∂∂

∂∂+

∂∂=

∂∂

←

←

←

←

BC

B

CT

BIB

E

TBIB

E

BC

E

TBC

BIB

E

TBIB

EE

V

TVD

TT

V

TVTTT

ωθω

θ

θωθω

θθAppendix

34

( ) ( ) ( ) [ ] [ ] ( )IP

CCB

C

TBI

BC

TBII

CI

CI

C

QR

U

R

T

V

T

td

Cd

V

T

td

dC

R

U

R

T

R

T

td

d

=

∂∂+

∂∂−

∂∂+

∂∂=

∂

∂+

∂∂−

∂

∂


Computation of Lagrange Equations in Body Frame (continue – 3)From

( )

( )

( ) ( )( )( ) ( )

∂∂=

∂∂

∂

∂=

∂

∂B

C

ETB

CI

C

BC

IC

V

TC

V

T

R

V

R

T

θ ( ) ( )

∂∂=

∂∂

IC

IC R

T

R

T

( ) ( ) ( )

∂∂+

∂∂−

∂

∂EEE

UTT

td

d

θθθ

( ) ( ) ( ) ( )

∂∂+

∂∂−

∂∂+

∂∂=

←←EEB

IB

T

BIB

T UTT

td

DdT

td

dD

θθωω

( ){ }( ) ( ) [ ] ( )

( )( )E

RBC

BC

TB

IBE

TBIB Q

V

TVD

T

=

∂∂×+

∂∂

∂∂

−←

→

ωθω

From

we obtain

( ) ( )

( ){ }( ) ( )

( ){ }( ) ( )

∂∂

∂∂

+

∂∂

∂∂

+

∂∂=

∂∂

←

←B

CE

TBC

BIB

E

TBIB

EE V

TVTTT

θωθ

ωθθ ( ) ( )

( ){ }( ) ( ) [ ]( )

( )

∂∂×−

∂∂

∂∂+

∂∂=

∂∂

←

←B

C

B

CT

BIB

E

TBIB

EE V

TVD

TTT

ωθω

θθwe obtain

35


Computation of Lagrange Equations in Body Frame (continue – 4)Using

[ ] [ ] [ ] ( )BIB

TBI

TBI Ctd

Cd×= ←ω

and ( ){ }

( ) [ ] ( )BIB

TE

TBIB

T

Dtd

Dd ×+∂

∂= ←

← ωθ

ω

we can compute

( ) ( ) ( )

∂∂+

∂∂−

∂

∂I

CI

CI

CR

U

R

T

R

T

td

d

[ ] ( )[ ]

( ) ( ) ( )

∂∂+

∂∂−

∂∂+

∂∂=

IC

IC

BC

TBI

BC

TBI

R

U

R

T

V

T

td

Cd

V

T

td

dC

[ ] ( ) [ ] [ ] ( )( ) ( ) ( )

( )IPI

CI

CB

C

BIB

TBIB

C

TBI Q

R

U

R

T

V

TC

V

T

td

dC

=

∂∂+

∂∂−

∂∂×+

∂∂= ←ω

36

SOLOEQUATIONS OF MOTION OF A VARIABLE MASS SYSTEM

LAGRANGIAN APPROACH


Finally

( ) ( ) ( )

∂∂+

∂∂−

∂

∂EEE

UTT

td

d

θθθ

( ) ( ) ( ) ( )

∂∂+

∂∂−

∂∂+

∂∂=

←←EEB

IB

T

BIB

T UTT

td

DdT

td

dD

θθωω

[ ] ( )( )

( ){ }( ) ( )

∂∂

∂∂

−

∂∂×+

←

→B

IBE

TBIB

BC

BC

T T

V

TVD

ωθω

( ) [ ] ( )( ){ }

( ) ( ) ( ) ( )

∂∂+

∂∂−

∂∂

∂∂

+×+

∂∂=

←

→←

←EEB

IBE

TBIBB

IBT

BIB

T UTTD

T

td

dD

θθωθωω

ω

[ ]( )( )

( ){ }( ) ( )

( )ER

BIB

E

TBIB

BC

B

CT

Q

T

V

TVD

=

∂∂

∂∂−

∂∂×+

←

→

ωθω

[ ] ( )( ) ( )

( )IP

BII

C

BII

C

BIB

C

BIBB

C

QCR

UC

R

TC

V

T

V

T

td

d

=

∂∂+

∂∂−

∂∂×+

∂∂

←ω

( ) [ ] ( )( ) [ ]( )

( ) ( ) ( )( )E

RT

E

T

E

TB

C

B

CBIB

BIBB

IB

QDU

DT

DV

TV

TT

td

d

−−−

←←

←

=

∂∂+

∂∂−

∂∂×+

∂∂×+

∂∂

θθωω

ω

37



Summarize

( )

( )

[ ] ( )

[ ]( ) [ ] ( )

( )

( ) ( )

∂∂

∂∂

−

∂∂

∂∂

××

×+

∂∂

∂∂

−

←←

←

←E

I

T

BI

BIB

BC

BIB

B

C

BIB

BIB

BC

T

R

T

D

C

T

V

T

VT

V

T

td

d

θωω

ω

ω

0

00

( )

( )

( )

=

∂∂

∂∂

+−− E

R

IP

T

BI

E

I

T

BI

Q

Q

D

C

U

R

U

D

C

0

0

0

0

θ

See Meirovitch and Kwak [6] and Meirovitch [7].

In the same way, for the elastic modes, we have:

,2,1==∂∂+

∂∂−

∂∂=

∂∂+

∂∂−

∂∂

iQUTT

td

dUTT

td

dEi

iiiiii ηηηηηη

Table of Contents

38


Derivation of the Equations of Motion

The Translational Lagrange Equations in Body Coordinates are given by:

( ) [ ] ( )( ) ( ) ( )

( )IP

BII

C

BII

C

BIB

C

BIBB

C

QCR

UC

R

TC

V

T

V

T

td

d

=

∂∂+

∂∂−

∂∂×+

∂∂

←ω

Pre-multiplying by will give the Translational Lagrange Equation in

Inertial Frame.

( ) IB

TBI CC =

( ) ( ) ( )( )I

PIC

IC

IC

QR

U

R

T

V

T

td

d =

∂∂+

∂∂−

∂∂

where we used

( ) ( ) [ ] ( )( )

∂∂×+

∂∂=

∂∂

← BC

BIBB

C

IBI

C V

T

V

T

td

dC

V

T

td

d

ω

39


Derivation of the Equations of Motion (continue – 1)Since the kinetic energy of the system is given by:

( )( ) ( ) ( )

∫ ⋅+∫⋅+∫⋅=tm I

C

I

C

tm I

CC

tmCC md

td

rd

td

rdmd

td

rdVmdVVT ,,,

2

1

2

1



,

ˆ2


,2

1

( )∑ ⋅⋅+∫ ⋅+ ←←


tmFrozenRotors

B

C

FrozenRotorsB

C Imdtd

rd

td

rdωω

,,,

2

1

2

1

∑∞

=

+

1

2

2

1

ii

i Mtd

dη( )

∑ ⋅⋅+∫

×⋅+ ←←←


tmFrozenRotors

B

CCIB Imd

td

rdr ωωω

,,

,

we have:

PmrVmV

T


C

=−=

∂∂ ∑ :ˆ

,

This equation gives the Linear Momentum of the system. The same expression was obtained using Simplified Particles and Reynolds’ Theorem Approaches.

40


Derivation of the Equations of Motion (continue – 2)

we have:

PmrVmV

T


C

=−=

∂∂ ∑ :ˆ

,

This equation gives the derivative of the Linear Momentum of the system. The same expression was obtained using Simplified Particles and Reynolds’ Theorem Approaches if we identify:

∑∑ −−+=

∂∂=

openingsifluidCiopen

openingsifluid

I

CiopenC

I

C

ICI

mrmtd

rdVm

td

Vdm

V

T

td

d

td

Pd

,, ˆ

ˆ

0=

∂∂

CR

T

( )CECE

CE

Earth

C

RgmRR

Rgm

R

U −=−=

∂∂ →

12

2

0

→+=−−+= ∑∑ E

E

EarthP


openingsifluid

I

Ciopen

C

I

C

I

RR

RgmQmrm

td

rdVm

td

Vdm

td

Pd1ˆ

ˆ

2

2

0,

,

Substitute those equation in the Lagrange’s Equation:

( ) ( ) ( )( )I

PIC

IC

IC

QR

U

R

T

V

T

td

d =

∂∂+

∂∂−

∂∂

gmQRR

RgmQF PE

E

EarthPext

+=+=

→

∑ 12

2

0

41


Derivation of the Equations of Motion (continue – 3)

→+=−−+= ∑∑ E

E

EarthP


openingsifluid

I

Ciopen

C

I

C

I

RR

RgmQmrm

td

rdVm

td

Vdm

td

Pd1ˆ

ˆ

2

2

0,

,

Substitute:

( )

( )I

openings jjifluid

I

Ciopen

Ci

TiA

I

P Fmtd

rdVmFFQ

++++= ∑ ∑∑∑

,ˆ

in

to obtain

∑∑∑∑∑ +++++=j

jopenings

ifluid

I

Ciopen


iTiA

I

C Fmtd

rdmrFFgm

td

Vdm

,

,

ˆˆ2

42


Derivation of the Equations of Motion (continue – 4)Rotation Equations

The Rotational Lagrange’s equations in Body Coordinates are given by:

( ) [ ] ( )( ) [ ]( )

( ) ( ) ( )( )E

RT

E

T

E

TB

C

B

CBIB

BIBB

IB

QDU

DT

DV

TV

TT

td

d

−−−

←←

←

=

∂∂+

∂∂−

∂∂×+

∂∂×+

∂∂

θθωω

ω

( )( ) ( ) ( )

∫ ⋅+∫⋅+∫⋅=tm I

C

I

C

tm I

CC

tmCC md

td

rd

td

rdmd

td

rdVmdVVT ,,,

2

1

2

1



,

ˆ2


,2

1

( )∑ ⋅⋅+∫ ⋅+ ←←


tmFrozenRotors

B

C

FrozenRotorsB

C Imdtd

rd

td

rdωω

,,,

2

1

2

1

∑∞

=

+

1

2

2

1

ii

i Mtd

dη( )

∑ ⋅⋅+∫

×⋅+ ←←←


tmFrozenRotors

B

CCIB Imd

td

rdr ωωω

,,

,

( )C

rotorsBjrotorCrotorj

tm RotorsFrozenB

CCIBC

IB

HImdtd

rdrI

T,,

,,, :

=⋅+

×+⋅=

∂

∂ ∑∫ ←←←

ωωω

Since the kinetic energy of the system is given by:

we obtain

43


Derivation of the Equations of Motion (continue – 5)Rotation Equations (continue – 1)

( )C

rotorsBjrotorCrotorj

tm RotorsFrozenB

CCIBC

IB

HImdtd

rdrI

T,,

,,, :

=⋅+

×+⋅=

∂

∂ ∑∫ ←←←

ωωω

we obtain

( ) ( )

⋅×+⋅+

××+

×+

⋅×+⋅+⋅=

×+==

∂

∂

∑∑

∫∫

←←←

←

←←←←

←←

rotorsBjrotorCrotorIB

rotorsBjrotorCrotor

tm RotorsFrozenB

CCIB

Btm RotorsFrozen

B

CC

IBCIBIBCIBC

CIB

I

C

I

C

IIB

II

mdtd

rdrmd

td

rdr

dt

d

III

Htd

Hd

td

HdT

td

d

ωωω

ω

ωωωω

ωω

,,

,,

,,

,,,

,,,

44



we obtain

( ) ( ) { }0 =

∂∂=

∂∂

EE

UT

θθ

( ) [ ] ( )( ) [ ] ( )

( ) ( ) ( )

∂∂+

∂∂−

∂∂×+

∂∂×+

∂∂ −−

←←

←E

T

E

TB

C

B

CBIB

BIBB

IB

UD

TD

V

TV

TT

td

d

θθωω

ω

( ) ( ) [ ] ( ) ( )

( ) ( )

( ) [ ] ( ) ( ) [ ]( )( )B


B

C

B

rotorsBjrotorCrotor

BIB

rotors

BBjrotorCrotor

tmRotorsFrozen

B

CCIB

B

tmRotorsFrozen

B

CC

BIBC

BIB

BIBC

BIBC

mrVmVII

mdtd

rdrmd

td

rdr

dt

d

III

∑−×+∑ ⋅×+∑ ⋅+

∫

××+

∫

×+

⋅×+⋅+⋅=

←←←

←

←←←←

,,,

,,

,,

,,,

ˆωωω

ω

ωωωω

( )E

openings kk

jjCj

I

Ciopen

CCiopenifluidi

CTiCA

T MFrtd

rdVrmMMD

+×+

+×++= ∑ ∑∑∑∑−

,

,

,,,

ˆˆ

( )B

openings kk

jjCj

I

Ciopen

CCiopenifluidi

CTiCA MFrtd

rdVrmMM

+×+

+×++= ∑ ∑∑∑∑

,

,

,,,

ˆˆ

45



Finally

( ) ( ) [ ] ( ) ( )

( ) ( )

( ) [ ] ( ) ( )

( ) ( ) ∑∑∑∑∑∑

∫∫

+×++=

⋅×+⋅+

××+

×+

⋅×+⋅+⋅

←←←

←

←←←←

kk

jjCj

i

BCTi

BCA

B

rotorsBjrotorCrotor

BIB

rotors

BBjrotorCrotor

tm RotorsFrozenB

CCIB

Btm RotorsFrozen

B

CC

BIBC

BIB

BIBC

BIBC

MFrMM

II

mdtd

rdrmd

td

rdr

dt

d

III

,,,

,,

,,

,,

,,,

ωωω

ω

ωωωω

46


Derivation of the Equations of Motion (continue – 8)Elastic Equations

Eiiii

QUTT

td

d =∂∂+

∂∂−

∂∂

ηηη

( )( ) ( ) ( )

∫ ⋅+∫⋅+∫⋅=tm I

C

I

C

tm I

CC

tmCC md

td

rd

td

rdmd

td

rdVmdVVT ,,,

2

1

2

1



,

ˆ2


,2

1

( )∑ ⋅⋅+∫ ⋅+ ←←


tmFrozenRotors

B

C

FrozenRotorsB

C Imdtd

rd

td

rdωω

,,,

2

1

2

1

∑∞

=

+

1

2

2

1

ii

i Mtd

dη( )

∑ ⋅⋅+∫

×⋅+ ←←←


tmFrozenRotors

B

CCIB Imd

td

rdr ωωω

,,

,

Since the kinetic energy of the system is given by:

we have

td

dM

T ii

i

ηη

=∂∂

0=∂∂

i

T

η

2

2

td

dM

TT

td

d ii

ii

ηηη

=∂∂−

∂∂

Therefore

47


Derivation of the Equations of Motion (continue – 9)Elastic Equations (continue – 1)

Eiiii

QUTT

td

d =∂∂+

∂∂−

∂∂

ηηηFrom

∑=∞

=1

22

2

1

iiiie MU ηω

we obtain

iiii

e MU ηωη

2=∂∂

From

( )( ) ( )[ ] ∑∑∫ ×∇⋅+⋅+⋅+−=

∂∂= ∞

kik

jij

Si

i

PEi MFdstfnpp

WQ

W

φφφηδ

δ 11

Therefore

( )[ ] ∑∑∫ ×∇⋅+⋅+⋅+−=

+ ∞

kik

jij

Siii

ii MFdstfnpp

td

dM

W

φφφηωη 112

2

2

Table of Contents

48

SOLOSUMMARY OF EQUATIONS OF MOTION OF

A VARIABLE MASS SYSTEM

( )( )

∑

=∑ ∫∫=∫=

openings

iiopen

openings

i Si

tm td

mdmdmd

td

dtm

iopen

MASS EQUATION

FORCE EQUATION

RIGID-BODY TERMSmVtd

VdCIO

O

C

×+ ←

ω

∑−∑

×+− ←

openings

iiflowiopen

openings

iiflowiopenIO

B

iopen mrmrtd

rd

ˆˆˆ

2 ωFLUID-FLOW TERMS

GRAVITATIONAL, AERODYNAMIC, PROPULSIVE &

∑+∑+=i

TiA FFmg


∑+j

jF

DISCRETE TERMS

49

SOLO

SUMMARY OF EQUATIONS OF MOTION OF A VARIABLE MASS (CONTINUE – 1)

MOMENT EQUATIONS RELATIVE TO A REFERENCE POINT O

RIGID-BODY TERMSIOOIOOIOIOO III ←←←← ⋅+⋅×+⋅ ωωωω

,,,

∑ ⋅×+∑ ⋅+ ←←←

jOjrotorCrotorjIO

jOjrotorCrotorj RjRj

II ωωω

,, ROTORS TERMS

( )

( )

∫

××+

∫

×+

←tm

FrozenRotorO

OOIO

O

tmFrozenRotor

O

OO

dmtd

rdr

dmtd

rdr

td

d

,,

,,

ω

BODY FLUIDS, MOVING PARTS, ELASTICITY,… TERMS

FLUID CROSSING OPENINGS TERMS∑

×+×− ←

openings

iiflowOiopenIO

O

OiopenOiopen mr

td

rdr

,

,,

ˆˆ

ˆ ω

AERODYNAMIC & PROPULSIVE

∑+∑=i

OTiOA MM ,,


( ) ∑+∑ ×−+k

kj

jOj MFRR DISCRETE FORCES

MOMENTS TERMS

−×+

I

OO td

Vdgc

, NON-CENTROIDAL MOMENTS TERMS

50

SOLO

SUMMARY OF EQUATIONS OF MOTION OF A VARIABLE MASS (CONTINUE – 2)

( )[ ]∫∫∑ +−= ∞

WS

A dstfnppF11: AERODYNAMIC FORCES

( )∫∫ −+

−= ∞iopenS

iflowiopeniflowTi dsnppmVVF

1ˆˆ: THRUST FORCES

( ) ( )[ ]∫∫ +−×−=∑ ∞WS

OOA dstfnppRRM11:,

AERODYNAMIC MOMENTS RELATIVE TO O

( ) ( )[ ]∫∫ −×−+

−×

−= ∞iopenS

OiflowiopeniflowOiopenOTi dsnppRRmVVRRM

1ˆˆˆ:,THRUST MOMENTS

RELATIVE TO O


Table of Contents

51


References

2. Meirovitch, L., “Method of Applied Dynamics”, John Wiley & Sons, 1986

3. Goldstein, H., “Classical Mechanics”, 1st, 2nd and 3rd Editions

4. Lanczos, C., “The Variational Principles of Mechanics”, 4th Edition, Dover Publications, 1970

5. Meirovitch, L., “General Motion of a Variable-Mass Flexible Rocket with Internal Flow”, J. Spacecraft, Vol. 7, No. 2, Feb. 1970, pp. 186-195

1. Bilmoria, K.D., Schmidt, D.K., “An Integrated Development of the of Motion for Elastic Hypersonic Flight Vehicles”, AIAA-92-4605-CP, and Journal of Guidance, Control and Dynamics, Vol.18, No.1, Jan.-Feb., 1995, pp. 73-81

6. Meirovitch, L., Kwak, M.K., “Dynamics and Control of Spacecraft with Retargeting Flexible Antennas”, Journal of Guidance, Control and Dynamics, Vol.13, No.2, March-April, 1990, pp. 241-248

52


References (continue – 1)

7. Meirovitch, L., “State Equation of Motion for Flexible Bodies in Terms of Quasi-Coordinates”, Proceedings of the IUTAM/IFAC Symposium on Dynamics of Controlled Mechanical Systems, Switzerland, May-June 1998

8. Weng, S-L., Greenwood, D.T., “General Dynamical Equations of Motion for Elastic Body Systems”, Journal of Guidance, Control and Dynamics, Vol.15, No.6, Nov.-Dec., 1992, pp. 1434-1442

Table of Contents

53


Appendix A: Lagrange Equations

The generalized coordinates are:

iiii

QUTT

td

d =

∂∂+

∂∂−

∂∂

ξξξ

( ) [ ] Tzyx

ICP RRRR ==

ξ Position components relative to Inertial System in Inertial Coordinates

[ ] TkjiR θθθξ =Γ=

Euler Angles around Euler Axes

[ ] TE

4321 ηηηηηξ == Elastic Modes

We want to obtain the Lagrange Equations in Body Coordinates.

( ) ( )wTT ,, ξξξ = Kinetic Energy of the System

( )ξU Potential Energy of the System

kQ Generalized Forces along the Degrees of Freedom Axes

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )ηωηθηωθηθηθηθ

,,,,,,,,,,,,,,, 1 B

B

B

B

EIB

B

B

B

ETEIEIEI VRTDVCRTRRT ==

−

{ } { }ηθηηθθθξ ,,,,,,,,,,: 21321TTT

zyx RRRR

==

{ } { }ηωηη ,,,,,,,,,,: 21T

BT

BT Vrqpwvuw ==

54


The angular velocity vector of rotation of the Body frame relative to Inertia frame is:

kIjIntriBIB kji θθθω ˆˆˆ ++=← In Body frame this is:

( ) ( ) ( ) ( ) ( ) [ ] ( ) [ ] [ ] ( )k

IIjjiij

IntrIntriii

BBk

BIj

BIntri

BB

BIB kjikji θθθθθθθθθω ˆˆˆˆˆˆ ++=++=←

[ ] ( ) [ ] ( ) [ ] [ ] ( )[ ] { } { } { }[ ]

=

=

=←

k

j

i

k

j

i

IIjjii

IntrIntrii

BB

BIB DDDkji

r

q

p

θ

θ

θ

θ

θ

θ

θθθω

321ˆˆˆ:

{ } ( ) ( ){ } ( ) [ ] ( ) ( ){ } ( ) [ ] [ ] ( )

{ } { } { }[ ]321

321

:

ˆˆ:,&ˆˆ:&ˆ:

DDDD

kkDjjDconstiD IIjjii

BIji

IntrIntrii

BIntri

BB

=

======→

θθθθθθ

( ) [ ] [ ] ( ) { } { } { }[ ]

=

==

=

=←

k

j

i

k

j

i

E

k

j

i

BIB DDD

DDD

DDD

DDD

DD

r

q

p

θ

θ

θ

θ

θ

θ

θ

θ

θ

θ

ω

321

333231

232221

131211

:

where:

or

Appendix A: Lagrange Equations (continue – 1)

55



( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

−

−

−

z

BB

BB

EI

y

BB

BB

EI

x

BB

BB

EI

z

BB

EBB

ETEI

y

BB

EBB

ETEI

x

BB

EBB

ETEI

z

y

x

R

VRT

R

VRT

R

VRT

R

DVCRT

R

DVCRT

R

DVCRT

R

T

R

T

R

T

ηωηθ

ηωηθ

ηωηθ

ηωθθηθ

ηωθθηθ

ηωθθηθ

,,,,,

,,,,,

,,,,,

,,,,,

,,,,,

,,,,,

1

1

1

∂∂∂∂∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

=

w

T

v

T

u

T

R

w

R

v

R

u

R

w

R

v

R

u

R

w

R

v

R

u

w

T

R

w

v

T

R

v

u

T

R

u

w

T

R

w

v

T

R

v

u

T

R

u

w

T

R

w

v

T

R

v

u

T

R

u

zzz

yyy

xxx

zzz

yyy

xxx

We have:

In a shorthand notation form:

( )

( )

( ) ( )( )( ) ( )

∂∂=

∂∂

∂

∂=

∂

∂B

B

ETB

BI

BB

I V

TC

V

T

R

V

R

T

θ

( ) [ ] ( )

==

=

=

Cz

Cy

Cx

IC

Cz

Cy

Cx

BI

BC

R

R

R

CCC

CCC

CCC

RC

R

R

R

C

w

v

u

V

333231

232221

131211

:

56



also:


( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

=

∂∂

∂∂∂∂

−

−

−

k

BB

BB

EI

j

BB

BB

EI

i

BB

BB

EI

k

BB

EBB

ETEI

j

BB

EBB

ETEI

i

BB

EBB

ETEI

k

j

i

VRT

VRT

VRT

DVCRT

DVCRT

DVCRT

T

T

T

θηωηθ

θηωηθ

θηωηθ

θηωθηθθ

θηωθηθθ

θηωθηθθ

θ

θ

θ

,,,,,

,,,,,

,,,,,

,,,,,

,,,,,

,,,,,

1

1

1

∂∂∂∂∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

=

r

T

q

T

p

T

rqp

rqp

rqp

r

Tr

q

Tq

p

Tp

r

Tr

q

Tq

p

Tp

r

Tr

q

Tq

p

Tp

kkk

jjj

iii

kkk

jjj

iii

θθθ

θθθ

θθθ

θθθ

θθθ

θθθ

( )

( )

( ) ( )( )( ) ( )

∂∂=

∂∂

∂

∂=

∂

∂

←←

←B

IB

ETB

IBE

BIB

E

TD

TT

ωθ

ωθ

ω

θ

57



In the same way:


( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )

∂∂

∂∂∂∂

=

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

−

−

−

z

y

x

z

BB

EBB

ETEI

y

BB

EBB

ETEI

x

BB

EBB

ETEI

z

y

x

R

T

R

T

R

T

R

DVCRT

R

DVCRT

R

DVCRT

R

T

R

T

R

T

ηωθθηθ

ηωθθηθ

ηωθθηθ

,,,,,

,,,,,

,,,,,

1

1

1

( ) ( )

∂∂=

∂∂

II R

T

R

T

58



and:( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

=

∂∂∂∂∂∂

−

−

−

k

B

B

B

B

EI

j

B

B

B

B

EI

i

B

B

B

BEI

j

B

B

EB

B

ETEI

j

B

B

EB

B

ETEI

i

B

BEB

BETEI

k

j

i

VRT

VRT

VRT

DVCRT

DVCRT

DVCRT

T

T

T

θηωηθ

θηωηθ

θηωηθ

θηωθθηθ

θηωθθηθ

θηωθθηθ

θ

θ

θ

,,,,,

,,,,,

,,,,,

,,,,,

,,,,,

,,,,,

1

1

1

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

+

∂∂

∂∂∂∂

=

r

Tr

q

Tq

p

Tp

r

Tr

q

Tq

p

Tp

r

Tr

q

Tq

p

Tp

w

Tw

v

Tv

u

Tu

w

Tw

v

Tv

u

Tu

w

Tw

v

Tv

u

Tu

T

T

T

kkk

jjj

iii

kkk

jjj

iii

k

j

i

θθθ

θθθ

θθθ

θθθ

θθθ

θθθ

θ

θ

θ

( ){ }( )

( ){ }( )

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+

∂∂

∂∂

∂∂

=

∂∂

∂∂

→

r

T

q

T

p

T

rqp

rqp

rqp

w

T

v

T

u

T

wvu

wvu

wvu

T

T

T

E

TBIB

E

TBBV

θω

θ

θθθ

θθθ

θθθ

θθθ

θθθ

θθθ

θ

θ

θ

333

222

111

333

222

111

3

2

1

59



∂∂

+∂∂+

∂∂

∂∂

+∂∂+

∂∂

∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

∂∂

+∂

∂+

∂∂

∂∂

+∂

∂+

∂∂

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

zk

yk

xk

zk

yk

xk

zj

yj

xj

zj

yj

xj

zi

yi

xi

zi

yi

xi

kkk

jjj

iii

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

RC

wvu

wvu

wvu

θθθθθθ

θθθθθθ

θθθθθθ

θθθ

θθθ

θθθ

232221131211

232221131211

232221131211

∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

∂∂

+∂

∂+

∂∂

zk

yk

xk

zj

yj

xj

zi

yi

xi

RC

RC

RC

RC

RC

RC

RC

RC

RC

θθθ

θθθ

θθθ

333231

333231

333231

( ){ }( )

( ){ }( ){ }( ){ }

{ }{ }{ }

{ } [ ][ ]

{ } [ ][ ]

{ } [ ][ ]

{ } [ ]

{ } [ ]

{ } [ ]

( ){ } [ ]( ){ } [ ]( ){ } [ ]

[ ] [ ] [ ][ ] ( )( ) [ ] ( )( ) TB

B

TB

B

TTB

B

TTB

B

TTB

B

TTT

I

TTT

I

TTT

I

TT

I

TT

I

TT

I

T

k

T

I

T

j

T

I

T

i

T

I

k

TB

B

j

TB

B

i

TB

B

E

TB

B VDVDDD

DV

DV

DV

DCR

DCR

DCR

CDR

CDR

CDR

CR

CR

CR

V

V

V

V

×−=×××−=

×−

×−

×−

=

×−

×−

×−

=

×−

×−

×−

=

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

=∂

∂321

3

2

1

3

2

1

3

2

1

θ

θ

θ

θ

θ

θ

θ


[ ] ( )( ) [ ] ( )( ) [ ] ( )BB

TTBB

TTBB VDVDDV ×−=×=×=

and( ){ }

( )

( ){ }( ){ }( ){ }

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=∂

∂

→

→

→

→

k

TB

IB

j

TB

IB

i

TB

IB

kkk

jjj

iii

E

TB

IB

rqp

rqp

rqp

θω

θω

θω

θθθ

θθθ

θθθ

θω

60


( ){ }( )

( ){ }( )

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+

∂∂

∂∂

∂∂

=

∂∂∂∂∂∂

∂∂

∂∂

→

r

T

q

T

p

T

rqp

rqp

rqp

w

T

v

T

u

T

wvu

wvu

wvu

T

T

T

T

T

T

E

TBIB

E

TBBV

k

j

i

θω

θ

θθθ

θθθ

θθθ

θθθ

θθθ

θθθ

θ

θ

θ

θ

θ

θ

333

222

111

333

222

111

3

2

1


( ){ }( ) [ ] ( )( ) [ ] ( )( ) [ ] ( )B

B

TTB

B

TTB

BE

TB

B VDVDDVV ×−=×=×=∂

∂

θ

We found

( ){ }( )

( ){ }( ){ }( ){ }

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=∂

∂

→

→

→

→

k

TB

IB

j

TB

IB

i

TB

IB

kkk

jjj

iii

E

TB

IB

rqp

rqp

rqp

θω

θω

θω

θθθ

θθθ

θθθ

θω

[ ] ( )

( ){ }( ){ }( ){ }

∂∂∂∂∂∂

∂∂

∂∂

∂∂

+

∂∂∂∂∂∂

×−

∂∂

∂∂∂∂

=

∂∂

∂∂∂∂

→

→

→

r

T

q

T

p

T

w

T

v

T

u

T

VD

T

T

T

T

T

T

k

TBIB

j

TBIB

i

TBIB

BB

T

k

j

i

k

j

i

θω

θω

θω

θ

θ

θ

θ

θ

θ

In a shorthand notation form: ( ) ( ) [ ] ( )( )

( ){ }( ) ( )

∂∂

∂∂

+

∂∂×−

∂∂=

∂∂

←

→B

IBE

TB

IB

B

B

B

B

T

EE

T

V

TVD

TT

ωθω

θθ

61



( ) ( )

∂∂=

∂∂

II R

T

R

T( )

( )

( ) ( )( )( ) ( )

∂∂=

∂∂

∂

∂=

∂

∂B

B

ETB

BI

BB

I V

TC

V

T

R

V

R

T

θUsing and

[ ] [ ] ( )I

P

II

B

B

TB

I

B

B

TB

I

III

QR

U

R

T

V

T

td

Cd

V

T

td

dC

R

U

R

T

R

T

td

d

=

∂∂+

∂∂−

∂∂+

∂∂=

∂∂+

∂∂−

∂

∂we obtain

Using

( ) ( ) [ ] ( )( )

( ){ }( ) ( )

∂∂

∂∂

+

∂∂×−

∂∂=

∂∂

←

→B

IBE

TBIB

BB

BB

T

EE

T

V

TVD

TT

ωθω

θθ

( )

( )

( ) ( )( )( ) ( )

∂∂=

∂∂

∂

∂=

∂

∂

←←

←B

IB

ETB

IBE

BIB

E

TD

TT

ωθ

ωθ

ω

θ

and

we obtain

( ) ( ) ( )

∂∂+

∂∂−

∂

∂EEE

UTT

td

d

θθθ

( ) ( ) ( ) ( ) [ ] ( )( )

( ){ }( ) ( )

( )E

RB

IBE

TB

IB

B

B

B

B

T

EEB

IB

T

B

IB

T QT

V

TVD

UTT

td

DdT

td

dD

=

∂∂

∂∂

−

∂∂×+

∂∂+

∂∂−

∂∂+

∂∂=

←

→

←← ωθω

θθωω

62



Computations of and td

Dd T ( ){ }( )E

TBIB

θω

∂∂ →

Basic Euler Rotations

The three basic Euler rotations, around the axes are described by the Rotation Matrices:

,3,2,1

[ ]

−=

11

1111

0

0

001

:

θθθθθ

cs

sc [ ]

−=

22

22

22

0

010

0

:

θθ

θθθ

cs

sc

[ ]

−=

100

0

0

: 33

33

33 θθθθ

θ cs

sc

Let differentiate with respect to Euler Angles.

[ ] [ ] [ ] 11

11

11

11

11

11

11

11

11 1

0

0

001

010

100

000

0

0

000

0

0

001

θθθθθ

θθθθ

θθθθ

θθθ ×−=

−

−=

−−−=

−=

cs

sc

sc

cs

cs

scd

d

d

d

[ ] [ ] [ ] 22

22

22

22

22

22

22

22

22 2

0

010

0

001

000

100

0

000

0

0

010

0

θθθ

θθ

θθ

θθ

θθ

θθ

θθθ

×−=

−

−=

−

−−=

−=

cs

sc

sc

cs

cs

sc

d

d

d

d

[ ] [ ] [ ] 3333

33

33

33

33

33

33

33 3

100

0

0

000

001

010

000

0

0

100

0

0

θθθθθ

θθθθ

θθθθ

θθθ

×−=

−

−=

−−

−=

−= cs

sc

sc

cs

cs

sc

d

d

d

d

63


Table of Contents



Dd T ( ){ }( )E

TBIB

θω

∂∂ →

Basic Euler Rotations (continue – 1)

is the matrix representation of the cross product of the vector ; i.e. in Cartesian coordinates :

[ ]×A

A

( )zyx 1,1,1

( ) ( )zzyyxxzzyyxx BBBAAABA 111111 ++×++=×

( ) ( ) ( ) zxyyxyxzzxxyzzy

zyx

zyx

zyx

BABABABABABA

BBB

AAA 111

111

−+−+−=

=

[ ] ( ) { } ( )zyxzyx

z

y

x

xy

xz

yz

BA

B

B

B

AA

AA

AA,,,,:

0

0

0

×=

−−

−

=

The matrix is skew-symmetric; i.e.:[ ]×A [ ] [ ]×−=× AA

T

[ ] [ ] [ ] [ ] 11111

1111 1 θθθθ

θθ ×−==d

d

td

d

[ ] [ ] [ ] [ ] 22222

2222 2 θθθθ

θθ ×−==d

d

td

d

[ ] [ ] [ ] [ ] 33333

3333 3 θθθθ

θθ ×−==d

d

td

d

64



Computations oftd

Dd T

=

td

Dd

td

Dd

td

Dd

td

Dd 321

++++++= k

kj

ji

ik

kj

ji

ik

kj

ji

i d

Dd

d

Dd

d

Dd

d

Dd

d

Dd

d

Dd

d

Dd

d

Dd

d

Dd θθ

θθ

θθ

θθ

θθ

θθ

θθ

θθ

θθ

333222111

{ } ( )

( ){ } ( ) [ ] ( )

( ){ } ( ) [ ] [ ] ( )

{ } { } { }[ ]321

3

2

1

:

ˆˆ:,

ˆˆ:

ˆ:

DDDD

kkD

jjD

constiD

IIjjii

BIji

IntrIntrii

BIntri

BB

=

==

==

==→

θθθθ

θθFrom

0111 =∂∂

=∂∂

=∂∂

kji

DDD

θθθ

Using those equations, and

[ ] [ ] ( ) [ ] 0ˆ 22211

2 =∂∂

=∂∂

×−=×−=∂∂

kj

IntrIntrii

i

DDDDjD

D

θθθ

θ

[ ] [ ] [ ] ( ) [ ]

[ ] [ ] ( ) [ ] ( ) [ ] [ ] ( ) [ ] [ ] [ ] ( )

[ ] ( ) [ ] [ ] ( ) [ ]

=∂∂

×−=×−=

−×−=×−=∂∂

×−=×−=∂∂

0

ˆˆ

ˆˆˆˆ

ˆ

3

32

3

3113

k

IIjjii

BIntr

IIjjiiii

Intr

IntriiI

IjjIntr

Intriij

IIjjii

i

D

DDkj

kjkjD

DDkDD

θ

θθ

θθθθθθθ

θθθ

01111 =∂∂

+∂∂

+∂∂

= kk

jj

ii

DDD

td

Dd θθ

θθ

θθ

[ ] i

k

k

j

j

i

i

DD

DDD

td

Dd

θ

θθ

θθ

θθ

21

2222

×−=

∂∂

+∂∂

+∂∂

=

[ ] [ ] ji

k

k

j

j

i

i

DDDD

DDD

td

Dd

θθ

θθ

θθ

θθ

3231

3333

×−×−=

∂∂

+∂∂

+∂∂

=

65


Table of Contents


Computations oftd

Dd T

[ ] [ ] jik

k

j

j

i

i

DDDDDDD

td

Dd θθθθ

θθ

θθ

32313333 ×−×−=

∂∂

+∂∂

+∂∂

=

[ ] ik

k

j

j

i

i

DDDDD

td

Dd θθθ

θθ

θθ

212222 ×−=

∂∂+

∂∂+

∂∂=

01111 =∂∂

+∂∂

+∂∂

= kk

jj

ii

DDD

td

Dd θθ

θθ

θθ

We found:

Therefore:

01 =td

Dd T

[ ] iT

T

DDtd

Dd θ

×= 122

[ ] [ ] jT

iT

T

DDDDtd

Ddθθ

×+×= 23133

66



Computations of( ){ }

( )E

TBIB

θω

∂∂ →

By differentiating [ ] { } { } { }[ ]

=←

k

j

i

BIB DDD

θ

θ

θ

ω

321

we obtain:

( )[ ] [ ][ ]

[ ] [ ] [ ] [ ] kjkj

k

j

i

k

j

i

iiii

BIB

DDDDDDDD

DDDDDDD

θθθθ

θ

θθ

θ

θθ

θθθθω

13123121

3121321 0

×+×=×−×−=

×−×−=

∂∂

∂∂

∂∂=

∂∂ ←

( ) [ ][ ]

[ ] [ ] kk

k

j

i

k

j

i

jjjj

B

IB

DDDD

DDDDD

θθ

θ

θ

θ

θ

θ

θ

θθθθω

2332

32321 00

×=×−=

×−=

∂∂

∂∂

∂∂

=∂

∂ ←

( )[ ] 0000321

=

=

∂∂

∂∂

∂∂

=∂

∂ ←

k

j

i

k

j

i

kkkk

BIB DDD

θ

θ

θ

θ

θ

θ

θθθθω

( ){ } [ ] [ ]{ }[ ] [ ] k

T

j

T

T

kj

i

TB

IB

DDDD

DDDD

θθ

θθθ

ω

×−×−=

×+×=∂

∂ ←

3121

1312

( ){ } [ ]{ } [ ] kTT

kj

TBIB DDDD θθ

θω

×−=×=∂

∂ ←3223

( ){ }0

=∂

∂ ←

k

TBIB

θω

67


Table of Contents


Computations of [ ] ( )B

IB

TD ×→ω

Since we have[ ] [ ] { }( ) 0

≡×=× Tlll

Tl DDDD

[ ] [ ]{ }{ }{ }

[ ] [ ] [ ]( )kji

T

T

T

BIB

T DDD

D

D

D

D θθθω ×+×+×

=×← 321

3

2

1

[ ] [ ]

[ ] [ ]

[ ] [ ]

×+×

×+×

×+×

=

jT

iT

kT

iT

kT

jT

DDDD

DDDD

DDDD

θθ

θθ

θθ

2313

3212

3121


Dd T ( ){ }( )E

TBIB

θω

∂∂ →

68



Computations of( ){ }

( ) [ ] ( )B

IB

T

E

TB

IB D ×+∂

∂→

→ ωθ

ω

Let add and and compare with ( ){ }

( )E

TBIB

θω∂

∂ ←

[ ] ( )BIB

TD ×←ω td

Dd T


Dd T ( ){ }( )E

TBIB

θω

∂∂ →

( ){ } [ ] ( )td

DdD

TB

IBT

i

TBIB 1

1 0 ==×+∂

∂←

←

ωθ

ω

( ){ } [ ] ( ) [ ]td

DdDDD

T

iTB

IBT

j

TBIB 2

122 =×=×+∂

∂←

← θωθ

ω

( ){ } [ ] ( ) [ ] [ ]td

DdDDDDD

T

jT

iTB

IBT

k

TBIB 3

23133 =×+×=×+∂

∂←

← θθωθ

ω

or( ){ }

( ) [ ] ( )BIB

T

E

TBIB

T

Dtd

Dd×+

∂∂

= ←← ω

θω

69


[ ] [ ] [ ] kkjjii

B

IC θθθ=


Computations oftd

Cd B

I

[ ] [ ] [ ] iiiBii i

td

d θθθ ×−= ˆ

[ ] [ ] [ ] jjjr

jj jtd

dθθ

θ ×−= intˆ

[ ] [ ] [ ] kkkIkk k

td

d θθθ ×−= ˆ

we can compute

[ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] ( ) [ ] BI

BI

B

Bkkjjii

B

Bkkjjiiii

BI CDCii

d

dC×−=×−=×−=

=

∂∂

1ˆˆ θθθθθθ

θθ

[ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] kkjj

Intr

Intriikkjjj

iij

BI j

d

dC θθθθθθ

θθ

×−=

=

∂∂ ˆ

[ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] ( ) [ ] B

I

B

I

B

Intrkkjjii

B

Intrkkjjiiii

Intr

Intrii CDCjjj ×−=×−=×−=−×−= 2ˆˆˆ θθθθθθθθ

[ ] [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] kk

I

Ijjiikkk

jjiik

BI k

d

dC θθθθθ

θθθ

×−=

=

∂∂ ˆ

[ ] [ ] [ ] ( ) [ ] [ ] [ ] [ ] [ ] [ ] ( ) [ ] BI

BI

B

Ikkjjiiiijj

I

Ijjii CDCkk ×−=×−=−−×−= 3ˆˆ θθθθθθθ

where are the unit vectors along the three consecutive Euler axes of rotations (in Body, Intermediate and Inertial coordinates).

IIntrB kji ˆ,ˆ,ˆ

we found:

70


Table of Contents


Computations oftd

Cd B

I

we can compute

[ ] [ ] B

I

i

B

I CDC ×−=

∂∂

1θ

[ ] [ ] B

I

j

B

I CDC ×−=

∂∂

2θ

[ ] [ ] B

I

k

B

I CDC ×−=

∂∂

3θ

[ ] [ ] [ ] [ ]k

k

BI

jj

BI

ii

BI

BI CCC

td

Cd θθ

θθ

θθ

∂

∂+

∂∂

+∂

∂=

[ ] [ ] [ ] kB

IjB

IiB

I CDCDCD θθθ ×−×−×−= 321

[ ] [ ] [ ]( ) [ ] ( ) B

I

B

IB

B

Ikji CCDDD ×−=×+×+×−= ←ωθθθ

321

therefore [ ] [ ] ( ) B

I

B

IB

B

I Ctd

Cd×−= ←ω

[ ] ( ) [ ]×−=∂

∂1DC

C TBI

i

BI

θ

[ ] ( ) [ ]×−=∂

∂2DC

C TBI

j

BI

θ

[ ] ( ) [ ]×−=∂

∂k

TBI

k

BI DC

C

θ

January 5, 2015 71

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 –2013

Stanford University1983 – 1986 PhD AA

Equation of motion of a variable mass system3

Science

Transcript of Equation of motion of a variable mass system3