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.

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