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Euler angles and angular velocity

  1. Apr 16, 2012 #1
    How do you prove that angular velocity is just the time derivative of each Euler angle times the basis vector of its respective frame?
    I remember it used to be perfectly clear to me a while back, but now I don't remember how the result was derived, and I couldn't find it in any of my books I looked so far.
    Does anyone remember how the result was derived?

    Thanks
     
  2. jcsd
  3. Apr 17, 2012 #2

    I like Serena

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    Hi Curl! :smile:

    Each euler angle is measured in a plane through the origin.
    In other words, one specific Euler angle behaves exactly as if it was a 2 dimensional polar coordinate.
    If an angle changes by an amount ##d\phi## in an time interval ##dt##, the position changes by ##r d\phi##.
    In other words:
    $$\frac{ds}{dt}=\frac{r d\phi}{dt}=r \frac{d\phi}{dt} = r \omega$$
     
  4. Apr 17, 2012 #3
    the problem I have is that in order to add the 3 angular velocity vectors (one in each frame) implies that the rotation can be represented as the sum of the three individual rotations. But rotation order matters, which makes things confusing.
     
  5. Apr 17, 2012 #4

    D H

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    Exactly. The time derivatives of a set of Euler angles (better said: Tait-Bryan angles, Bryan angles, or Cardan angles; Euler angles are a z-x-z rotation) are not angular velocity.
     
  6. Apr 18, 2012 #5
    I think you mean the 'cross product', not times.
     
  7. Apr 18, 2012 #6
    Well that sentence is flawed actually.
    Say you have a vector [itex] A=A_x\hat{i}+A_y\hat{j}+A_z\hat{z} [/itex] in a Cartesian frame. Knowledge of its velocity in this frame is obtained through its time derivative. This yields
    [tex]
    \frac{d\hat{A}}{dt}=\frac{dA_x}{dt}\hat{i}+\frac{dA_y}{dt}\hat{j}+\frac{dA_z}{dt}\hat{z}
    [/tex]
    which can then be deduced to
    [tex]
    \frac{d\hat{A}}{dt}=\dot{A} \vec{A}+\vec{\omega} \times \vec{A}
    [/tex]
    In the above, [itex] \vec{\omega} [/itex] is what you call angular velocity.

    Cheers,
     
  8. Apr 18, 2012 #7
    Oh jeeze. I came to check this just now and I realized I forgot one line. :)

    Here it is:
    [tex]
    \frac{d\hat{A}}{dt}=\frac{dA_x}{dt}\hat{i}+\frac{d A_y}{dt}\hat{j}+\frac{dA_z}{dt}\hat{z}+\left( A_x \frac{d\hat{i}}{dt}+A_y\frac{d\hat{j}}{dt}+A_z \frac{d \hat{z}}{dt}\right)
    [/tex]

    Change this line in my above post. This can then deduce the last equation after realizing that the unit vectors have a fixed length.
     
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