Assuming that such a vector exists, it would have to follow that: 2 $ - - PDF document

Vector Algebra Tools:

For any orthonormal basis am

m= 1 3

and any vectors A,B & C A = A•a1 a1 + A•a2 a2 + A•a3 a3 A× B×C = C•A B ! B•A C A• B×C = C• A×B = B• C×A A = A // + A " ; A // = # #•A # # A " = # #×A ×# # = # #× A×# # A = am•A am + am× A×am ; m = 1,2,3 a1× A×a1 = A ! A•a1 a1 a2× A×a2 = A ! A•a2 a2

+ a3× A×a3 = A ! A•a3 a3

a1× A×a1 + a2× A×a2 + a3× A×a3 = 3$A ! A•a1 a1 + A•a2 a2 + A•a3 a3 a1× A×a1 + a2× A×a2 + a3× A×a3 = 3$A ! A = 2$A

………………………………………………………

2$A = a1× A×a1 + a2× A×a2 + a3× A×a3

Some Calculus Stuff:

unit vector rate: u% = & &u×u with

&

&u = u×u% ' & &u•u = 0 For a smoothly varying but Rigid-OrthoNormal Basis am (

m= 1 3

am

% = &

&m×am with

&

&m = am×am

%

' & &m•am = 0 ; m = 1,2,3 BIG QUESTION: Is it possible that all three Omegas are EQUAL, i.e. the same vector? Let us assume that they are, so that:

& &a = & &1 = & &2 = & &3 .

Unfortunately, this supposition leads to the conclusion

& &a = & &a•a1 a1 + & &a•a2 a2 + & &a•a3 a3 & &a = & &1•a1 a1 + & &2•a2 a2 + & &3•a3 a3 & &a = zero a1 + zero a2 + zero a3 = ZERO

Hence, it follows that the three Omegas are ALL EQUAL ONLY if they are ALL ZERO! Not a very interesting case. NEXT BIG QUESTION: Even though the 3-Omegas are NOT ALL EQUAL, could there still exist a "special" vector (a

combined/effective angular rotation rate) such that:

am % % = & &a ×am ; m = 1, 2, 3

Assuming that such a vector exists, it would have to follow that: 2$& &a = a1× & &a×a1 + a2× & &a×a2 + a3× & &a×a3 = a1× a1 % % + a2× a2 % % + a3× a3 % % = a1× a1 % % + a2× a2 % % + a3× a3 % % 2$& &a = & &1 + & &2 + & &3 & &a ) 1

2 &

&1 + & &2 + & &3 Since one cannot prove the existence of something starting from the supposition that it already exists, we must take care in our interpretation of this result. In fact what has been established here is that, if such a vector exists - it must be this one. In other words, this (above) is the only vector that could possibly have the desired

• property. We shall therefore adopt it as a "trial solution" & check to

see whether it "works." But first, …some more useful info about our 3-Omegas. am•an = const. = 1 ; whenever m=n

0 ; whenever

m*n d d( am•an = am

% •an + am•an % = 0 0 = am•an % + an•am % 0 = am• &

&n×an + an• & &m×am

0 = &

&n• an×am + & &m• am×an ;

{TSP identity} 0 = ! &

&n• am×an + & &m• am×an

0 = &

&m ! & &n • am×an For the choice of "free-indices" with m = n, this identity is trivially

• satisfied. More interesting conclusions result from choices for which

m

* n.

m=2 & n=3 : &

&2 ! & &3 • a2×a3 = 0 & &2 ! & &3 •a1 = 0 '

+1 = &

&2•a1 = & &3•a1

m=3 & n=1 : &

&3 ! & &1 • a3×a1 = 0 & &3 ! & &1 •a2 = 0 '

+2 = &

&3•a2 = & &1•a2

m=1 & n=2 : &

&1 ! & &2 • a1×a2 = 0 & &1 ! & &2 •a3 = 0 '

+3 = &

&1•a3 = & &2•a3 In view of these new "rigidity" restrictions on the 3-Omegas, as well as the previously derived perpendicularity conditions,

& &1•a1 = & &2•a2 = & &3•a3 = 0

the following relationships are easily established, & &m = & &m•a1 a1 + & &m•a2 a2 + & &m•a3 a3 ; m = 1,2,3 & &1 = & &1•a1 a1 + & &1•a2 a2 + & &1•a3 a3 & &2 = & &2•a1 a1 + & &2•a2 a2 + & &2•a3 a3 & &3 = & &3•a1 a1 + & &3•a2 a2 + & &3•a3 a3 & &1 = 0 a1 + +2 a2 + +3 a3 = +2a2 + +3a3 & &2 = +1 a1 + 0 a2 + +3 a3 = +1a1 + +3a3 & &3 = +1 a1 + +2 a2 + 0 a3 = +1a1 + +2a2 & &1 + & &2 + & &3 = 2+1a1 + 2+2a2 + 2+3a3 & &a = 1

2 &

&1 + & &2 + & &3 = +1a1 + +2a2 + +3a3 = & &1 + +1a1 & &2 + +2a2 & &3 + +3a3 & &a = & &m + +mam ; m = 1,2,3

+

From here, it is easily shown that our "candidate" vector does indeed have the desired property, namely that & &a×am = & &m + +mam ×am = & &m×am + +m am×am & &a×am = am % % + +m 0

0 = am

% % am % % = & &a×am ; m = 1,2,3

With the formation of this angular rotation rate vector, it is now a simple matter to differentiate a vector expression written in the component form

A(() = A 1(()a1(() + A 2(()a2(() + A 3(()a3(() ; A k(() = A(()•ak(() ; k=1,2,3

(12) relative to an evolving (not-fixed) orthonormal triad. By following the established rules of calculus, complete differentiation proceeds as follows:

d d(A = d d( A1a1 + A2a2 + A3a3 = d d( A1a1 + d d( A2a2 + d d( A3a3 = d d(A1 a1 + A1 d d(a1 + d d(A2 a2 + A2 d d(a2 + d d(A3 a3 + A3 d d(a3 = d d(A1 a1 + d d(A2 a2 + d d(A3 a3 + A1 & &a×a1 + A2 & &a×a2 + A3 & &a×a3 = d d(A1 a1 + d d(A2 a2 + d d(A3 a3 + & &a× A1a1 + & &a× A2a2 + & &a× A3a3 = d d(A1 a1 + d d(A2 a2 + d d(A3 a3 + & &a× A1a1 + A2a2 + A3a3 d d(A = d d(A1 a1 + d d(A2 a2 + d d(A3 a3 + & &a×A

(13) Thus, the effect of utilizing a set of non-fixed basis vectors can be accounted for through a single "corrective" cross product term. Appropriately, this shall henceforth be referred to as the "omega correction term." This is a fundamental mathematical result which is critical to the development of the general relative motion relations as well as rigid body dynamics.

SUMMARY To find the one and only, combined, effective angular rotation rate vector for a smoothly (-varying rigid orthonormal triad

a1, a2, a3 ,

we may employ any one of the five equivalent expressions

&a = &1 + +1a1 &2 + +2a2 &3 + +3a3 +1a1 + +2a2 + +3a3

1 2 &1 + &2 + &3 = 3 2&AVG

expressed in terms of the individual angular rotation rate vectors

&1 ) a1× d

d(a1

&2 ) a2× d

d(a2

&3 ) a3× d

d(a3

&k ) ak× d

d(ak ;

k=1,2,3.

and scalar factors

+1 = &2•a1 = &3•a1 =

d d(a2 •a3 = ! d d(a3 •a2

+2 = &3•a2 = &1•a2 =

d d(a3 •a1 = ! d d(a1 •a3

+3 = &1•a3 = &2•a3 =

d d(a1 •a2 = ! d d(a2 •a1

As an example, the triads angular rotation rate vector could be computed (one of many possibilities) from the relation

&a = &2 + +2a2 = a2× d

d(a2 + d d(a3 •a1 a2 .

Also note that if the evolution variable is time ((=t), then these angular rotation rate vectors shall be referred to as angular velocities and notationally denoted with lower case omegas (,), and differentiation will be indicated with the standard shorthand “overdot” notation. To illustrate, for time variation the above expression would read

,a = a2×a2

• + a3
• •a1 a2 .