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Velocities due to angular rates

  1. Aug 14, 2008 #1
    Dear all, I am missing some insight concerning dynamic mechanics:

    1. Problem statement
    A angular rate sensor is attached to a massive body. I would like to calculate the velocity of a certain point on that body due to rotational motion of the body itself.

    2. Relevant equations
    The whole problem, equations and question have been described in detail in a pdf file

    3. The attempt at a solution
    I proposed two approaches. Who can tell me which one is wrong and why. I cannot figure it out myselve.

    Many thanks!
     
  2. jcsd
  3. Aug 14, 2008 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    They are both correct. The cross product of two vectors in represented in some frame A is another vector in that frame. Transforming this into another frame B,

    [tex](\mathbf u \times \mathbf v)_B =
    \boldsymbol T_{A\to B} \, (\mathbf u_A \times \mathbf v_A)[/tex]

    Reference frames are just a representation technique. The position vector from point a to point b is the same vector in all reference frames; it just has different representations in difference frames. he cross product is a pseudo vector rather than a true vector. Pseudo transforms from one frame to another given a proper transformation. (They do not transform under reflection, which is why they are called pseudo vectors). Assuming a transformation from one right-handed system to another, we must have

    [tex]
    \mathbf u_B \times \mathbf v_B
    = (\boldsymbol T_{A\to B}\,\mathbf u_A) \times (\boldsymbol T_{A\to B}\,\mathbf v_A)
    = \boldsymbol T_{A\to B}(\mathbf u_A \times \mathbf v_A)
    [/tex]


    One way to look at the cross product is to view [itex]\mathbf u \times[/itex] as an operator that acts on some vector [itex]\mathbf v[/itex] that produces the cross product of the two vectors:

    [tex](\mathbf u \times) \mathbf v = \mathbf u \times \mathbf v[/tex]

    This in turns suggests determining if a corresponding matrix operator exists that yields the exact same vector for all vectors v as does the cross product operator. The cross product matrix generated from some vector

    [tex]\mathbf u = \bmatrix u_x \\ u_y \\u_z\endbmatrix [/tex]

    is

    [tex]\boldsymbol X(\mathbf u) \equiv
    \bmatrix
    0 & -u_z & \phantom{-}u_y \\
    \phantom{-}u_z & 0 & -u_x \\
    -u_y & \phantom{-}u_x & 0\endbmatrix[/tex]

    does exactly that. This matrix transforms the vector cross product into a matrix-vector product:

    [tex]\mathbf u \times \mathbf v
    = (\mathbf u \times) \mathbf v
    = \boldsymbol X(\mathbf u)\,\mathbf v[/tex]

    A column vector transforms from frame A to frame B via

    [tex]\mathbf u_B = \boldsymbol T_{A\to B} \, \mathbf u_A[/tex]

    The cross product matrix, on the other hand, transforms via the matrix transformation rule

    [tex]\boldsymbol X(\mathbf u_B) =
    \boldsymbol T_{A\to B} \,
    \boldsymbol X(\mathbf u_A) \,
    \boldsymbol T_{A\to B}^{\;\;\;T}
    = \boldsymbol X(\boldsymbol T_{A\to B}\,\mathbf u_A)[/tex]

    The proof of the above is left as an exercise to the reader. With this one can examine whether

    [tex]
    (\boldsymbol T_{A\to B}\,\mathbf u_A) \times (\boldsymbol T_{A\to B}\,\mathbf v_A)
    = \boldsymbol T_{A\to B}(\mathbf u_A \times \mathbf v_A)
    [/tex]

    Step by step,

    [tex]
    \begin{aligned}
    (\boldsymbol T_{A\to B}\,\mathbf u_A) \times
    (\boldsymbol T_{A\to B}\,\mathbf v_A)
    &=
    ((\boldsymbol T_{A\to B}\,\mathbf u_A) \times)
    (\boldsymbol T_{A\to B}\,\mathbf v_A) \\
    &=
    (\boldsymbol X(\boldsymbol T_{A\to B}\,\mathbf u_A))\,
    (\boldsymbol T_{A\to B}\,\mathbf v_A) \\
    &=
    (\boldsymbol T_{A\to B} \,
    \boldsymbol X(\mathbf u_A) \,
    \boldsymbol T_{A\to B}^{\;\;\;T}) \,
    (\boldsymbol T_{A\to B}\,\mathbf v_A) \\
    &=
    \boldsymbol T_{A\to B} \,
    \boldsymbol X(\mathbf u_A) \,
    \boldsymbol T_{A\to B}^{\;\;\;T} \,
    \boldsymbol T_{A\to B}\,\mathbf v_A \\
    &=
    \boldsymbol T_{A\to B} \,
    \boldsymbol X(\mathbf u_A) \, \mathbf v_A \\
    &= \boldsymbol T_{A\to B} (\mathbf u_A \times \mathbf v_A)
    \end{aligned}[/tex]
     
  4. Aug 15, 2008 #3
    Thanks.

    However, in your derivation, you say nothing about the necessary orthogonal property of the transformation matrix.

    Just try in matlab:

    C(AxB)
    (CA)x(CB)

    Once with C a non-orthogonal matrix and once with C a orthogonal matrix. In first case both results will be different, in second case both results will be equal.

    [edit]: from the third last line till the second last line of previous reply, the orthogonal property of the transformation matrix is supposed. Conclusion of the story in a pure algebraic way:

    C(AxB)==(CA)x(CB) only if C is orthogonal
     
    Last edited: Aug 15, 2008
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