with the help of the post changing frame, we are now good to use the Euler angle.

recall

$V_R = R_n ( - \theta ) V_S$

for the rotating frame axis is rotating positive with the static frame.

the Euler angle is performed on 3 steps

1. rotate on $Z_S$, the z-axis with $\alpha$, which is $R_{zS} ( - \alpha )$. the x-axis and the y-axis is now different, we notate this frame with a 1 .
2. rotate on $Y_1$, the y-axis in the 1- frame  by angle $\beta$, which is $R_{y1} ( - \beta )$. the new axis is notated by 2.
3. rotate on $Z_2$, the z-axis in the 2-frame by angle $\gamma$, which is $R_{z2} ( - \gamma )$. the new axis is notated by R.

The rotating frame is related with the static frame by:

$V_R = R_{z2} ( - \gamma ) R_{y1} ( - \beta ) R_{zS} ( - \alpha ) V_S$

or

$R_R ( \alpha, \beta, \gamma ) =$$R_{zS} ( - \gamma )$ $R_{y1} ( - \beta )$ $R_{zS} ( - \alpha )$

for each rotation is on a new frame, the computation will be ugly, since, after each rotation, we have to use the rotation matrix in new coordinate.

There is another representation, notice that:

$R_{y1} ( -\beta ) =$ $R_{zS} ( - \alpha )$ $R_{yS} ( - \beta )$ $R_{zS} ( \alpha)$

which mean, the rotating on y1 -axis by $\beta$ is equal to rotate it back to $Y_S$  on zS -axis and rotated it by $\beta$ on yS – axis, then rotate back the $Y_S$ to $Y_1$ on zS – axis.

i use a and b for the axis between the transform.

and we have it for the z2-axis.

$R_{z2} ( -\gamma ) = R_{y1} ( - \beta ) R_{z1} ( - \gamma ) R_{y1} ( \beta )$

by using these 2 equation and notice that the z1-axis is equal to zS-axis.

$R_R ( \alpha , \beta, \gamma ) = R_{zS} ( - \alpha ) R_{yS} (- \beta ) R_{zS} ( - \gamma )$

which act only on the the same frame.