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Euler angles and angular velocity

by Curl
Tags: angles, angular, euler, velocity
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Curl
#1
Apr16-12, 08:49 PM
P: 757
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
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#2
Apr17-12, 11:40 AM
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Hi Curl!

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:
$$\frac{ds}{dt}=\frac{r d\phi}{dt}=r \frac{d\phi}{dt} = r \omega$$
Curl
#3
Apr17-12, 09:16 PM
P: 757
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.

D H
#4
Apr17-12, 11:02 PM
Mentor
P: 15,070
Euler angles and angular velocity

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.
FeX32
#5
Apr18-12, 01:22 AM
P: 67
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 think you mean the 'cross product', not times.
FeX32
#6
Apr18-12, 01:38 AM
P: 67
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,
FeX32
#7
Apr18-12, 02:27 PM
P: 67
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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