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flailing on the w proof
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Informal, exploratory derivation labeled as a retry on Question 7, from an appendix on Euler angles and angular velocity. It expands d e_n = R(t+dt)e'_n(t+dt) - R(t)e'_n(t), takes components in frames S and S', and defines F = dR/dt. It then tries a z-rotation special case and an A = (dR/dt)R^-1 approach (Plans A-C). The argument is unfinished and questions its own steps.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
Trying again on Question 7.
Review : Write en(t) = R(t)e'n(t) at time t which may be evaluated in any Frame. Then
den(t) ≡ en(t+dt) - en(t)
= R(t+dt)e'n(t+dt) - R(t)e'n(t)
≈ [ R(t) (∂te'n(t)) + (∂tR(t))e'n(t) ] dt . // neglect order (dt)2 etc .
so
(den(t)/dt) = R(t) [∂te'n(t)] + [∂tR(t)] e'n(t) .
[∂ten(t)] = R(t) [∂te'n(t)] + [∂tR(t)] e'n(t) .
[∂ten(t)]X = R(t) [∂te'n(t)]X + [∂tR(t)]X e'n(t) X is some Frame.
One can take components of this equation in any Frame. For example,
( [∂ten(t)]X ) = R(t) ([∂te'n(t)]X) + ([∂tR(t)]X) e'n(t) X is some Frame.
This vector equation is valid in any frame. For example,
( [∂ten(t)]X )i = {R(t) ([∂te'n(t)]X)}i + {([∂tR(t)]X) e'n(t)}i Frame S
( [∂ten(t)]X )'i = {R(t) ([∂te'n(t)]X)}'i + {([∂tR(t)]X) e'n(t)}'i Frame S'
If we choose Frame X = Frame S', we get
( [∂ten(t)]S' )i = {R(t) ([∂te'n(t)]S')}i + {([∂tR(t)]S') e'n(t)}i Frame S
( [∂ten(t)]S' )'i = {R(t) ([∂te'n(t)]S')}'i + {([∂tR(t)]S') e'n(t)}'i Frame S'
Now we know that [∂te'n(t)]S' = 0 I guess because S' is moving but always e'n are constant within S'. So
( [∂ten(t)]S' )i = {([∂tR(t)]S') e'n(t)}i Frame S
( [∂ten(t)]S' )'i = {([∂tR(t)]S') e'n(t)}'i Frame S'
It then seems that the equation continues to be evaluatable in both Frame's components. Again, you just dot with basis vectors! [ agreed! ]
OK, now let's write out the matrix products
( [∂ten(t)]S' )i = ([∂tR(t)]S')ij (e'n(t))j Frame S
( [∂ten(t)]S' )'i = ([∂tR(t)]S')'ij (e'n(t)')j Frame S'
Now define
([∂tR(t)]S' = FS'(t)
Then have
( [∂ten(t)]S' )i = [FS'(t)]ij (e'n(t))j Frame S
( [∂ten(t)]S' )'i = [FS'(t)]'ij (e'n(t)')j Frame S'
How are the two matrices related? I presume like so
[FS'(t)]' = R(t) [FS'(t)]R-1(t)
Now take a closer look at
FS'(t) = [∂tR(t)]S' = [{R(t+dt) - R(t)}/dt ]S'
Take a special case where R(t) = Rz(φ(t)). Then
FS'(t) = [{R(t+dt) - R(t)}/dt ]S' = [ {Rz(φ(t+dt)) - Rz(φ(t)) }/dt ]S'
Are these standard rotation matrices or not???
[FS'(t)]'ij = { [ {Rz(φ(t+dt)) - Rz(φ(t)) }/dt ]S' }'ij
= {dR'z(φ(t+dt)/dt}ij
Plan B. Start over with,
den(t) ≡ en(t+dt) - en(t)
= R(t+dt)e'n(t+dt) - R(t)e'n(t)
≈ [ R(t) (∂te'n(t)) + (∂tR(t))e'n(t) ] dt . // neglect order (dt)2 etc .
Then evaluations:
(den(t))i = [ R(t)ij (∂te'n(t))j + (∂tR(t))ij(e'n(t))j ] dt
(den(t))'i = [ R'(t)ij (∂te'n(t))'j + (∂tR(t))'ij(e'n(t))'j ] dt
So you have to face up to talking about these two objects
(∂tR(t))ijdt = [ R(t+dt) - R(t)]ij = R(t+dt) - R(t)
(∂tR(t))'ijdt= [ R(t+dt) - R(t)]'ij = R'(t+dt) - R'(t) = R(t+dt) - R(t)
Could these really be the same?
Plan C.
A ≡ (dR/dt) R-1(Φ)
A' ≡ (dR'/dt) R'-1(Φ)
Adt ≡ dR R-1(Φ)
A'dt ≡ dR' R'-1(Φ)
Then how about
[A'dt] = R(Φ) [Adt] R-1(Φ) = R(Φ) [dR R-1(Φ)] R-1(Φ)
= R(Φ)dR {R-1(Φ) R(Φ) } R-1(Φ)R-1(Φ)
= [R(Φ)dR R-1(Φ)] [ R(Φ) R-1(Φ)R-1(Φ)]
= [R(Φ)dR R-1(Φ)] R-1(Φ)
= dR' R-1(Φ)