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Sec G_5 chunk rewrite INSTALLED
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Draft rewrite of Section G.5 in Phil's new-frames document, marked as already installed so it should not be edited here. It interprets Goldstein's three concatenated rotations by ψ, θ, φ using the document's Section 1 formalism of back-rotations and passive transformations. Two exercises give explicit formulas for transforming components of a vector between frames S and S' with Euler angles, plus applications to unit vectors.
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Interpretation of the Goldstein's Triple Concatenation
In our Section 1 formalism, we discuss the idea of three concatenated transformations near (1.1.41). We can compare the equations there to those of (G.5.6),
e"'n = U-1e''n e"n = S-1e'n e'n = R-1en
(',',') = Rζ'(ψ) (',',') (',',') = Rξ(θ) (,,) (,,) = Rz(φ) (,,) .
It follows that the three "back-rotations" are
U-1 = Rζ'(ψ) S-1 = Rξ(θ) R-1 = Rz(φ)
In Fig (1.3.4) we show a back-rotation example where R-1 = Rz(-α) = "back-rotation" and we drew the figure for some small α > 0. In Goldstein's back rotations, the role of α is played by -ψ ,-θ and -φ. Figure (G.5.5) shows that the basis vector "back rotations" are really a forward rotations by ψ, θ and φ.
Doing the concatenation one gets,
e"'n = U-1S-1R-1en = Rζ'(ψ)Rξ(θ) Rz(φ)en
or
(',',') = Rζ'(ψ) Rξ(θ) Rz(φ) (,,)
in agreement with (G.5.7).
Transformation of Kinematic Vectors
The corresponding Passive View transformations of Kinematic Vectors for the three concatenations are,
(V)''' = UV" (V)" = SV' (V)' = RV .
Doing the concatenation and then changing the Section 1 triple-prime to Goldstein's single-prime, we get
(V)' = USR V
= Rζ'(-ψ)Rξ(-θ)Rz(-φ) V .
If we now redefine R to be our notation for Goldstein's overall transformation, we then have
(',',') = R-1(,,) R-1 = Rζ'(ψ) Rξ(θ) Rz(φ)
= Rζ'(ψ) Rξ(θ) Rz(φ) (,,)
(V)' = R V R = [Rζ'(ψ) Rξ(θ) Rz(φ)]-1
= [Rζ'(ψ) Rξ(θ) Rz(φ)]-1 V
We can rewrite the above lines making use of Theorem 3 item (3) to get
(',',') = R-1(,,) R-1 = Rz(φ)Rx(θ)Rz(ψ)
= Rz(φ)Rx(θ)Rz(ψ) (,,)
(V)' = R V R = [ Rz(φ)Rx(θ)Rz(ψ)]-1
= [ Rz(φ)Rx(θ)Rz(ψ)]-1 V
= Rz(-ψ)Rx(-θ)Rz(-φ) V
This last item is the rule for finding the Frame S' components (V)'i of a vector V in terms of the Frame S components Vi .
Figure (G.5.5) shows the Kinematic Vector r which is the position of some point in Frame S .
Exercise 1:
Compute the Frame S' components of the vector r which in Frame S has components (x,y,z) .
(r)' = R r = Rz(-ψ)Rx(-θ)Rz(-φ) r
Therefore
(r)' = = R
and so
x' = (cosψcosφ - sinψcosθsinφ) x + (cosψsinφ + sinψcosθcosφ) y + sinψsinθ z
y' = (- sinψcosφ - cosψcosθsinφ) x + (-sinψsinφ + cosψcosθcosφ) y + cosψsinθ z
z' = sinθsinφ x - sinθcosφ y + cosθ z .
Application: What are the Frame S' components of ?
Apply the previous equation to (x,y,z) = (1,0,0):
x' = cosψcosφ - sinψcosθsinφ = ()'1
y' = - sinψcosφ - cosψcosθsinφ = ()'2
z' = sinθsinφ = ()'3
We now reverse the above Exercise.
Exercise 2:
Compute the Frame S components of the vector r which in Frame S' has components (x',y',z') .
r = R-1(r)' = Rz(φ)Rx(θ)Rz(ψ)(r)'
The matrix R-1 = RT is just the transpose of the matrix shown above; Maple computes it anyway :
Therefore
r = = R-1
and so
x = (cosφcosψ-cosθsinφsinψ) x' + (-cosφsinψ-cosθsinφcosψ) y' +sinθsinφ z'
y = (sinφcosψ + cosθcosφsinψ) x' + (-sinφsinψ+cosθcosφcosψ) y' - sinθcosφ z'
z = sinθsinψ x' + sinθcosψ y' + cosθ z'
Application: What are the Frame S components of ' ?
Apply the previous equation to (x',y',z') = (1,0,0):
'x = cosψcosφ - sinψcosθsinφ = (')1
'y = sinφcosψ + cosθcosφsinψ = (')2
'z = sinθsinψ = (')3