HW8_prob3_GPS15_G19

HW8_prob3_GPS15_G19 - xyz ω θ xyz = D T ω θ = D T ...

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Problem Goldstein 4-19 (3 rd ed. 4.15) The most direct way to do this problem is to take the body frame components ω 0 derived in class or in the text and convert them into space frame components ω with the the help of the transposed Euler matrix ω = A T ω 0 . (1) This is also the most algebraically complicated way. An alternative is to start with ω written in terms of its “natural” component vectors, transform each of these components into the space frame, and add the results. We write ω = ω φ + ω θ + ω ψ , (2) where ω φ = · φ e z , (3) ω θ = · θ e 0 ξ = · θ e ξ , (4) and ω ψ = · ψ e 0 z , (5) where e z is the space frame z basis vector, e 0 z is a body frame z basis vector, and e 0 ξ and e ξ are (the same) intermediate frame x -like basis vectors. Since ω φ is already expressed in the space frame, we need only deal with ω θ and ω ψ . The orthogonal matrix D T will transform the components of ω θ from the intermediate frame to the space f xed frame. Let’s designate those as ( ω θ
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Unformatted text preview: ) xyz ( ω θ ) xyz = D T ω θ = D T    · θ    =    · θ cos φ · θ sin φ    . (6) To obtain the space frame components of ω ψ we can use either A T or D T C T , since their action is the same. We get ( ω ψ ) xyz = A T ω ψ = D T C T    · ψ    =     · ψ sin θ sin φ − · ψ sin θ cos φ · ψ cos θ     . (7) With ( ω φ ) xyz =    · φ    , (8) 1 all of the intermediate results are complete, and we simply add the corresponding components of Eqs.(6), (7), and (8) to obtain   ω x ω y ω z   =     · θ cos φ + · ψ sin θ sin φ · θ sin φ − · ψ sin θ cos φ · φ + · ψ cos θ     , (9) which is the desired result. 2...
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This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.

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HW8_prob3_GPS15_G19 - xyz ω θ xyz = D T ω θ = D T ...

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