Euler angles and angular velocity

by Curl
Tags: angles, angular, euler, velocity
 HW Helper P: 6,187 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: $$\frac{ds}{dt}=\frac{r d\phi}{dt}=r \frac{d\phi}{dt} = r \omega$$
 P: 67 Well that sentence is flawed actually. Say you have a vector $A=A_x\hat{i}+A_y\hat{j}+A_z\hat{z}$ in a Cartesian frame. Knowledge of its velocity in this frame is obtained through its time derivative. This yields $$\frac{d\hat{A}}{dt}=\frac{dA_x}{dt}\hat{i}+\frac{dA_y}{dt}\hat{j}+\frac {dA_z}{dt}\hat{z}$$ which can then be deduced to $$\frac{d\hat{A}}{dt}=\dot{A} \vec{A}+\vec{\omega} \times \vec{A}$$ In the above, $\vec{\omega}$ is what you call angular velocity. Cheers,
 P: 67 Oh jeeze. I came to check this just now and I realized I forgot one line. :) Here it is: $$\frac{d\hat{A}}{dt}=\frac{dA_x}{dt}\hat{i}+\frac{d A_y}{dt}\hat{j}+\frac{dA_z}{dt}\hat{z}+\left( A_x \frac{d\hat{i}}{dt}+A_y\frac{d\hat{j}}{dt}+A_z \frac{d \hat{z}}{dt}\right)$$ Change this line in my above post. This can then deduce the last equation after realizing that the unit vectors have a fixed length.