Euler angles with "standard" rotations

 

We can specify a general rotation \(R\) of the 3D space by mean of 3 angles \(\alpha\), \(\beta\) and \(\gamma\). Fix three axis \(x\), \(y\) and \(z\), and a rotation will be determined by the image of this three axis, \(X\), \(Y\) and \(Z\). If we call \(N\) to the intersection of the \(xy\) plane with the \(XY\) plane, we can carry the first frame to the second one by means of

  • A rotation along \(z\) of angle \(\alpha\), to carry \(x\) to \(N\). Say \(R_z^{\alpha}\).
  • A rotation along \(N\) (the new \(x\)) of angle \(\beta\), to carry \(z\) to \(Z\). Call it \(R_N^{\beta}\).
  • A rotation along \(Z\) (the new \(z\)) of angle \(\gamma\), to carry \(N\) to \(X\). Say \(R_Z^{\gamma}\).



This way,
\[
R=R_{Z}^{\gamma}R_N^{\beta}R_{z}^{\alpha}
\]

This description has never satisfied me, because the axis of rotation are not defined extrinsically, that is, they are not universal, they do not work for any rotation. They depends on the rotation in particular. But with a little of  algebraic manipulation we can recycle this angle to use them with standard rotations along the axis \(x\), \(y\) and \(z\).

First, observe that
\[
R_N^{\beta}=R_{z}^{\alpha}R_x^{\beta}R_{z}^{-\alpha}
\]
since we can express a rotation in a different axis going and coming back.


Second, by the same reason,
\[
R_{Z}^{\gamma}=(R_N^{\beta}R_{z}^{\alpha}) R_{z}^{\gamma} (R_{z}^{-\alpha}R_N^{-\beta})
\]

But then,
\[
R_{Z}^{\gamma}=((R_{z}^{\alpha}R_x^{\beta}R_{z}^{-\alpha})R_{z}^{\alpha}) R_{z}^{\gamma} (R_{z}^{-\alpha}(R_{z}^{\alpha}R_x^{-\beta}R_{z}^{-\alpha}))=
\]
\[
=R_{z}^{\alpha}R_x^{\beta} R_{z}^{\gamma}R_x^{-\beta}R_{z}^{-\alpha}
\]

And therefore our original rotation can be expressed with standard axis
\[
R=(R_{z}^{\alpha}R_x^{\beta} R_{z}^{\gamma}R_x^{-\beta}R_{z}^{-\alpha})(R_{z}^{\alpha}R_x^{\beta}R_{z}^{-\alpha})R_{z}^{\alpha}=R_{z}^{\alpha}R_x^{\beta} R_{z}^{\gamma}
\]

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