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Informal working notes dated 1.11.15, part of Phil's Appendix H on Euler angles in a mechanics frames document. He compares active rotation products such as Rz(φ)Rx(θ)Rz(ψ) and their inverses, using Maple, to derive angular velocity components ωφ, ωθ, ωψ in the body frame. He finds some results match the book (page 134) while others are wrong, and compares with his Appendix G result (G.4.22). Many equations are missing from the extracted text.
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This is the Title PhL 1.11.15
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ωφ = // Goldstein also says this
ωθ = '
ωψ = '
' = = "line of nodes" // ξ (xi = "zeye")
' = ' // ζ (zeta)
' = ' = Rz(φ) Rx(θ) // this seems right to me!!!
' = Rx(θ) Rz(φ+ψ)
Comment: If I put in my usual rotation matrices and say = (0,0,1)T, the resulting vector gives the components of ' in (x,y,z)-space coordinates. I am doing an active rotation. Let's do it right here:
Now write the inverse equation
= Rx(-θ) Rz(-φ)'
In x'-space I can say that ' = (0,0,1)T and this equation then reads
But this is the wrong answer! Why is it the wrong answer. I really need to start with
' = ' = Rz(φ) Rx(θ) Rz(ψ)
Then the inverse would be
= Rz(-ψ)Rx(-θ)Rz(-φ)'
and then I would get
I need to understand IN WORDS why these results are different.
Looking at the pictures, where does the product like Rz(φ) Rx(θ) Rz(ψ) "fit in" ?
Start with this conjecture:
ei = R e'i using my standard notation for R
What is R ? On example would be
= R ' or equivalently ' = R-1
But from the pictures I know that
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
This applies to ALL the basis vectors, so write
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
Then right off the bat I know that
ωφ = = Rz(-ψ) Rx(-θ) Rz(-φ) '
Maple then says in x'-space
and therefore,
ωφ = sinθsinψ ' + sinθcosψ ' + cosθ '
This is the correct answer on page 134. I also know that
ωψ = '
and that is correct answer also.
So I only one the middle term to figure out somehow.
= Rz(φ)
= Rz(φ) = Rz(φ) [Rz(-ψ) Rx(-θ) Rz(-φ)] '
Maple says
This then says
ωθ = ' = = times the above mess.
This is the WRONG answer for this term!
On the other hand, looking at the pictures I could also say
' = Rz'(ψ) '
But I don't know the meaning of Rz'(ψ) . If I blindly use it I get
' = Rz'(-ψ) '
and then I would get
ωθ = ' = = times the above
= cosψ ' -sinψ '
and that is the correct answer!
Invert these to get
= Rx(-θ) Rz(-φ)' // this is just the inverse of the above!
= Rz(-φ-ψ)Rx(-θ)'
Note that
= Rz(φ)
' = Rz(φ) Rx(θ) = '
Therefore
= Rx(-θ) Rz(-φ)'
' = = Rz(φ) = Rz(φ)Rz(-φ-ψ)Rx(-θ)' = Rz(-ψ)Rx(-θ) '
Then we find that
ωφ = = Rx(-θ)Rz(-φ)' // just plugging in!
ωθ = ' = Rz(-ψ)Rx(-θ) '
ωψ = '
Now it is time for Maple.
Therefore,
= Rx(-θ)Rz(-φ)' =
STOP. This is completely wrong because the correct terms have sinθsinψ type products!
*************************************************************
Next,
Therefore
=
and finally
ωψ = ' =
Add to get
ω = ωφ + ωθ + ωψ = + +
= [ cosψ ] ' + [sinθ - sinψ] ' + [ cosθ + ] '
and therefore
ωx' = cosψ
ωy' = sinθ - sinψ
ωz' = cosθ +
My answer in Appendix G however is this:
(ω)'1 = - sinθsinψ - cosψ
(ω)'2 = sinθcosψ - sinψ
(ω)'3 = - cosθ - . // our result (G.4.22)
So everything is wrong, but that was my first attempt! Now go back and fix the errors!
************************
Go back to
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
' = Rz(φ) Rx(θ) Rz(ψ) from doing active rotations.
I am quite confident despite all the problems that these three equations are correct.
Can I take these "part-way" somehow ?
= Rx(θ) Rz(ψ)