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Angular orientation.

  1. May 16, 2007 #1
    in kleppner's textbook he expalins quite well why we can't have angular orientation vector (because then we wouldn't have commutavity of vectors), but we can have angular velcoity as a vector, my question is how does this work when we integrate from an angular velocity vector to get an angular orientation which is not a vector? how do we get this?

    thanks in advance.
  2. jcsd
  3. May 16, 2007 #2

    D H

    Staff: Mentor

    You can't integrate angular velocity to get an angular orientation.

    You have to integrate a transformation matrix or a quaternion instead.
  4. May 16, 2007 #3
    could you give an example for this?
    i mean if we have the angular velcoity vector: (w_x,w_y,w_z) how would you get an angular orientation?
  5. May 16, 2007 #4

    D H

    Staff: Mentor

    Here I am assuming that "angular velocity" [itex]\omega_{B\to A}[/itex]means the angular velocity of reference frame A with respect to some other reference frame B, expressed in terms of reference frame A.

    Suppose [itex]T_{A\to B}(t)[/itex] is the transformation matrix from frame A to frame B. Then the time derivative of this matrix is

    [tex]\frac d{dt}T_{A\to B}(t) = T_{A\to B}(t) X(\omega_{B\to A})[/tex]

    where [itex]X(a)[/itex] denotes the skew-symmetric matrix generated from the vector a (i.e., X(a)b = a cross b).
  6. May 16, 2007 #5
    in what course did you learn this?

    i havent yet encouterd that in my studies.

    anyway, shouldn't it be:
    [tex]\frac d{dt}T_{A\to B}(t) = X(\omega_{B\to A})T_{A\to B}(t) [/tex]
  7. May 16, 2007 #6

    D H

    Staff: Mentor

    You probably learned that the time derivatives of some vector quantity [itex]\vect q[/itex] as observed in frames A and B are related via

    \frac {d\vect q}{dt_A} = \frac {d\vect q}{dt_B} + \omega_{A\to B}\times \vect q

    The coordinates of the vector as expressed in frames A and B are

    \vect q_A = \mathbf T_{B\to A} \, \vect q_ B

    Differentiating this with respect to time,

    \frac {d\vect q_A}{dt}
    = \mathbf T_{B\to A} \, \frac {d\vect q_B}{dt} +
    \frac{d}{dt}\mathbf T_{B\to A} \, \vect q_ B

    The first and last equations both relate various expressions of time derivatives of a vector quantity; they just do so differently. The last equation doesn't address the frame of the observer. This is fairly easy: The time derivative of the components of some vector as expressed in some reference frame is the time derivative of that vector as observed in that reference frame.

    The first equation doesn't address the frame in which the quantities are expressed.
    Making these frames explicit,

    \frac {d\vect q_A}{dt_A}
    = \mathbf T_{B\to A}
    \frac {d\vect q_B}{dt_B} + \omega_{A\to B:B}\times \vect q_B
    = \mathbf T_{B\to A}\frac {d\vect q_B}{dt_B}
    + \mathbf T_{B\to A}\mathbf X(\omega_{A\to B:B})\, \vect q_B

    For the above to be true for any vector quantity [itex]\vect q[/itex],

    \frac{d}{dt}\mathbf T_{B\to A} = \mathbf T_{B\to A}\mathbf X(\omega_{A\to B:B})
    Last edited: May 16, 2007
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