= a a m ; a m m = 1,2,3 Consider : Is it possible that all three - - PowerPoint PPT Presentation

QUESTION: Instead of having to deal with 3-Omegas (one for each element of the

triad), is it possible to find a single Omega (a combined/effective angular rotation rate)

such that:

Smoothly Varying OrthoNormal Triads:

Recall that for any smoothly varying unit vector:

u′ = Ωu×u with

Ωu = u×u′

⇒ Ωu•u = 0 am ′ = Ωm×am with

Ωm = am×am

′ ⇒ Ωm•am = 0 ;

m = 1,2,3

u = u

1

a1 a2 a3

Ωa

am ′ = Ωa×am ;

m = 1,2,3

For a smoothly varying but Rigid-OrthoNormal Basis:

Consider: Is it possible that all three Omegas are EQUAL, i.e. the same vector? Let us assume that they are, so that: Unfortunately, this supposition leads to the conclusion:

Ωa = Ω1 = Ω2 = Ω3

Ωa = Ωa•a1 a1 + Ωa•a2 a2 + Ωa•a3 a3 Ωa = Ω1•a1 a1 + Ω2•a2 a2 + Ω3•a3 a3 Ωa = zero a1 + zero a2 + zero a3 = ZERO Therefore: No simple answer to our original question!

Hence, it follows that the three Omegas are ALL EQUAL, ONLY if they are ALL ZERO! Not a very interesting case.

Ωa = am•Ωa am + am×Ωa ×am ;

m = 1,2,3

Ωa = Ωa•am am + am× Ωa×am ;

m = 1,2,3

Ωa = Ωa•am am + am×am ′ ;

m = 1,2,3

Ωa = Ωa•am am + Ωm ;

m = 1,2,3

Taking a different tack, lets now assume (for the sake of argument) that such a vector exists and see where that leads. Using the parallel / perpendicular decomposition, a vector with the desired property would necessarily satisfy:

Ωa

Ωa = Ωa•a1 a1 + Ω1 ;

m = 1

Ωa = Ωa•a2 a2 + Ω2 ;

m = 2

+ Ωa = Ωa•a3 a3 + Ω3 ;

m = 3

3⋅Ωa = Ωa•a1 a1 + Ωa•a2 a2 + Ωa•a3 a3 + Ω1 + Ω2 + Ω3 3⋅Ωa = Ωa + Ω1 + Ω2 + Ω3 2⋅Ωa = Ω1 + Ω2 + Ω3 Ωa = 1

2 Ω1 + Ω2 + Ω3

Conclude: This is the ONLY vector which COULD have the desired property!

Since is the ONLY vector which could possibly have the desired property, we now adopt it as a TRIAL solution, and then proceed to see if it works.

Ωa = 1

2 Ω1 + Ω2 + Ω3

But first, …some more useful info about our original 3-Omegas.

am•an = const. = 1 ; whenever m=n

0 ; whenever

m≠n d dξ am•an = am ′ •an + am•an ′ = 0

0 = am•an

′ + an•am ′

0 = am• Ωn×an + an• Ωm×am 0 = Ωn• an×am + Ωm• am×an

;

{TSP identity} 0 = − Ωn• am×an + Ωm• am×an 0 = Ωm − Ωn • 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 : Ω2 − Ω3 • a2×a3 = 0

Ω2 − Ω3 •a1 = 0 ⇒

ε1 = Ω2•a1 = Ω3•a1

m=3 & n=1 : Ω3 − Ω1 • a3×a1 = 0

Ω3 − Ω1 •a2 = 0 ⇒

ε2 = Ω3•a2 = Ω1•a2

m=1 & n=2 : Ω1 − Ω2 • a1×a2 = 0

Ω1 − Ω2 •a3 = 0 ⇒

ε3 = Ω1•a3 = Ω2•a3

Ωm − Ωn • am×an = 0

Thus, the 3-Omegas SHARE certain vector components. We will now combine this new observation with the previously derived perpendicularity conditions:

Ω1•a1 = Ω2•a2 = Ω3•a3 = 0

Ωm = Ωm•a1 a1 + Ωm•a2 a2 + Ωm•a3 a3 ; m = 1,2,3 Ω1 = Ω1•a1 a1 + Ω1•a2 a2 + Ω1•a3 a3 Ω2 = Ω2•a1 a1 + Ω2•a2 a2 + Ω2•a3 a3 Ω3 = Ω3•a1 a1 + Ω3•a2 a2 + Ω3•a3 a3 Ω1 = 0 a1 + ε2 a2 + ε3 a3 = ε2a2 + ε3a3 Ω2 = ε1 a1 + 0 a2 + ε3 a3 = ε1a1 + ε3a3 Ω3 = ε1 a1 + ε2 a2 + 0 a3 = ε1a1 + ε2a2 Ω1 + Ω2 + Ω3 = 2ε1a1 + 2ε2a2 + 2ε3a3 Ωa = 1

2 Ω1 + Ω2 + Ω3 = ε1a1 + ε2a2 + ε3a3 =

Ω1 + ε1a1 Ω2 + ε2a2 Ω3 + ε3a3 Ωa = Ωm + εmam ;

m = 1,2,3

+

From here, it is easily shown that our “candidate” vector does indeed have the desired property, namely that:

Ωa×am = Ωm + εmam ×am = Ωm×am + εm am×am Ωa×am = am ′ + εm 0 = am ′

am ′ = Ωa×am ;

m = 1,2,3

Therefore: Trial Vector has the desired property!

ε1 = Ω2•a1 = Ω3•a1 = a2 ′•a3 = −a3 ′•a2 ε2 = Ω3•a2 = Ω1•a2 = a3 ′•a1 = −a1 ′•a3 ε3 = Ω1•a3 = Ω2•a3 = a1 ′•a2 = −a2 ′•a1 Ωa =

1 2 Ω1+Ω2+Ω3 = 3 2ΩAVG

Ω1+ε1a1 Ω2+ε2a2 Ω3+ε3a3 ε1a1+ε2a2+ε3a3

Ω1 = a1×a1

′

Ω2 = a2×a2

′

Ω3 = a3×a3

′

Ωm = am×am

′ ;

m = 1, 2, 3

At any instant during the smooth evolution of a Rigid OrthoNormal triad, there exists a unique combined-effective Angular Rotation Rate vector such that:

am ′ = Ωa×am ;

m = 1,2,3

This vector can be determined from any of the following relations: expressed in terms of the individual omegas: which have the various scalar components:

Ωa

A(ξ) = A 1(ξ)a1(ξ) + A 2(ξ)a2(ξ) + A 3(ξ)a3(ξ) ; A m(ξ) = A(ξ)•am(ξ) ;

m = 1,2,3

d dξA = d dξ A 1a1 + A 2a2 + A 3a3 = d dξ A 1a1 + d dξ A 2a2 + d dξ A 3a3 = A 1 ′ a1 + A 1a1 ′ + A 2 ′ a2 + A 2a2 ′ + A 3 ′ a3 + A 3a3 ′ = A 1 ′ a1 +A 2 ′ a2 +A 3 ′ a3 + A 1a1 ′ + A 2a2 ′ + A 3a3 ′ = A 1 ′ a1 +A 2 ′ a2 +A 3 ′ a3 + A 1 Ωa×a1 + A 2 Ωa×a2 + A 3 Ωa×a3 = A 1 ′ a1 +A 2 ′ a2 +A 3 ′ a3 + Ωa× A 1a1 + Ωa× A 2a2 + Ωa× A 3a3 = A 1 ′ a1 +A 2 ′ a2 +A 3 ′ a3 + Ωa× A 1a1 + A 2a2 + A 3a3 A′ = A 1 ′ a1 +A 2 ′ a2 +A 3 ′ a3 + Ωa×A

a1 a2 a3 Scalar component derivatives Omega correction term

Why? What’s the Big deal?

am ′ = Ωa×am ;

m = 1,2,3