Given the mass properties

of an aircraft about the

b’x, b’y, b’z directions.

Transform them

to the bx, by, bz directions.

bx

by

b’y

b’x

ψ

ψ

bx

by

b’y

b’x

ψ

ψ

5.6.3

Given this:

Find this:

[ ]I =

zyzxz

yzyxy

xzxyx

II I

I II

I I I

[ ]I' =

z'z'y'z'x'

z'y'y'y'x'

z'x'y'x'x'

III

III

III

Solution

bx

by

b’y

b’x

ψ

ψ

bx

by

b’y

b’x

ψ

ψ

bx

by

bz

cψ sψ 0

10 0

-sψ cψ 0

b’x b’y b’z

bx

by

bz

cψ sψ 0

10 0

-sψ cψ 0

b’x b’y b’z

[ I ] = [l B-B’ ] [ I’ ] [l B-B’ ]T

[ ]I' =

z'z'y'z'x'

z'y'y'y'x'

z'x'y'x'x'

III

III

III

where:

[ ]I =

zyzxz

yzyxy

xzxyx

II I

I II

I I I

and

[ ] [ ] [ ]100

0cs

0sc

and

100

0cs

0sc

ψψ

ψψ

BB'

T

B'Bψψ

ψψ

B'B

−

==−= −−− lll

Multiplying the first two matrices: [ l B-B’ ] [ I’ ]:

[ ][ ]

+++

+++

=

=−

z'z'y'z'x'

z'y'ψz'x'ψy'ψy'x'ψy'x'ψx'ψ

z'y'ψz'x'ψy'ψy'x'ψy'x'ψx'ψ

z'z'y'z'x'

z'y'y'y'x'

z'x'y'x'x'

ψψ

ψψ

'

III

IcIs-IcIs-IcIs-

IsIcIsIcIsIc

III

III

III

100

0cs-

0sc

'IBB

l

[ ][ ][ ]

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

{ } { } { }

+−+

+−−+−+−

+−+−++

=

−−

z'z'y'ψz'x'ψz'y'ψz'x'ψ

z'y'ψz'x'ψy'x'ψψy'

2

ψx'

2

ψy'x'

2

ψ

2

ψx'y'ψψ

z'y'ψz'x'ψy'x'

2

ψ

2

ψx'y'ψψy'x'ψψy'

2

ψx'

2

ψ

T

''

IIcIsIsIc

IcIsIcs2IcIsIscIIcs

IsIcIscIIcsIcs2IsIc

'IBBBB

ll