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Rotation matrix

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    The rotation matrix below describes a beam element which is rotated around three axes x,y and z. Derive the rotation matrix.



    2. Relevant equations

    3. The attempt at a solution
    I can see where the x-values (CXx CYx CZx) come from. They're just the projections of the rotated x-axes (the one with rotation alpha and beta). But I don't understand how the rest is derived can somebody help me?
    Last edited: Mar 30, 2012
  2. jcsd
  3. Mar 30, 2012 #2


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    Rotation about the x-axis through angle [itex]\alpha[/itex] is given by the matrix
    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) \\ 0 & sin(\alpha) & cos(\alpha)\end{bmatrix}[/tex]

    Rotation about the y-axis through angle [itex]\beta[/itex] is given by the matrix
    [tex]\begin{bmatrix}cos(\beta) & 0 & -sin(\beta) \\ 0 & 1 & 0 \\ sin(\beta) & 0 & cos(\beta)\end{bmatrix}[/tex]

    Rotation about the z-axis through angle [itex]\gamma[/itex] is given by the matrix
    [tex]\begin{bmatrix} cos(\gamma) & -sin(\gamma) & 0 \\ sin(\gamma) & cos(\gamma) & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]

    The result of all those rotations is the product of those matrices. Be sure to multiply in the correct order.
  4. Mar 30, 2012 #3
    I suspect that there's a minus sign somewhere wrongly placed in your matrices Halls, am I correct? I moved the minus sign in your second matrix to the lower sine but there's still something wrong for this is my result:

    Code (Text):

    [                        cos(a)cos(b),               -sin(b),                           cos(b)sin(a)                ]
    [ sin(a)sin(c) + cos(a)cos(c)sin(b)         cos(b)cos(c)         cos(c)*sin(a)sin(b) - cos(a)sin(c) ]
    [ cos(a)sin(b)sin(c) - cos(c)sin(a)        cos(b)*sin(c)     cos(a)cos(c) + sin(a)sin(b)sin(c)       ]
    Last edited: Mar 30, 2012
  5. Mar 30, 2012 #4


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    No, all of the minus signs are correctly placed. I am, of course, assuming that a positive angle gives a rotation "counterclockwise" looking at the plane from "above"- from the positive axis of rotation.
  6. Mar 30, 2012 #5
  7. Mar 30, 2012 #6

    D H

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    Staff: Mentor

    Look at your diagram. Are all of those rotations positive by the right hand thumb rule? (Hint: The answer is no.)
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