# Qualitative understanding of Euler rotation on gravitational vector

If I choose the sequence of my Euler rotations to be $$Z\rightarrow Y \rightarrow Z$$ or in terms of matrix multiplication $$R_x(\phi)R_y(\theta)R_z(\psi)$$, a stationary $$3$$-axis accelerometer can measure a gravitational vector through the following equation

$$\left[ \begin{array}{@{\;}c@{\;}} a^i_x \\ a^i_y \\ a^i_z \end{array}\right] = \left[ \begin{array}{@{\;}r@{\;\;\;} @{\;\;\;}r@{\;\;\;} @{\;\;\;}r@{\;}} % \mathrm{c}_{\theta} \mathrm{c}_{\psi} & \mathrm{c}_{\theta} \mathrm{s}_{\psi} & -\mathrm{s}_{\theta} \\ % \mathrm{s}_{\phi} \mathrm{s}_{\theta} \mathrm{c}_{\psi} - \mathrm{c}_{\phi}\mathrm{s}_{\psi} & \mathrm{s}_{\phi} \mathrm{s}_{\theta} \mathrm{s}_{\psi} + \mathrm{c}_{\phi} \mathrm{c}_{\psi} & \mathrm{s}_{\phi} \mathrm{c}_{\theta} \\ % \mathrm{c}_{\phi} \mathrm{s}_{\theta} \mathrm{c}_{\psi} + \mathrm{s}_{\phi} \mathrm{s}_{\psi} & \mathrm{c}_{\phi} \mathrm{s}_{\theta} \mathrm{s}_{\psi} - \mathrm{s}_{\phi} \mathrm{c}_{\psi} & \mathrm{c}_{\phi} \mathrm{c}_{\theta} % \end{array} \right] \left[ \begin{array}{@{\;}c@{\;}} 0 \\ 0 \\ g \end{array}\right] = \left[ \begin{array}{@{\;}r@{\;}} -g\sin({\theta}) \\ g\sin(\phi) \cos({\theta}) \\ g\cos(\phi) \cos(\theta) % \end{array} \right]$$

I can follow the derivation and see how all $$\psi$$ terms are zero-ed out in the matrix multiplication, but I don't grasp the qualitative meaning of it. Gravitational vector is just a vector. Any Euler rotation applied to a vector in 3D space has to be specified by a chosen sequence of three Euler angles $$\psi,\theta$$ and $$\phi$$.

The equation however suggests that any frame (or vector) can rotate to any orientation with just two Euler angles $$\phi$$ and $$\theta$$ instead of three. Visualizing a rotation of any vector confirms this to be true - any vector can be repositioned to a new vector of any coordinate with just two Euler rotations. If we can rotate a vector or the frame with just two Euler angles then why is Euler angle defined by three angles?

A reason I can think of that can make use of a 3rd rotation is when I visualize a vector to be an airplane instead that rolls around its own axis. Other than that, all geometric vectors like gravitational vector are simply a line with an arrow that makes no use of a Third Euler angle.

The vector $$\vec{g}$$ points in the z-direction, and the last angle $$\psi$$ in the sequence of rotations is about the z-axis also.

So the angle $$\psi$$ has no effect on $$\vec{g}$$. Or in mathematical terms

$$R_z(\psi) \vec{g}= \vec{g}$$

If you had chosen a different set of Euler angles, such as the last operation is not a rotation about the z-axis, then the angle $$\psi$$ would not cancel out.

The two angles that orient a vector, are exactly the angles of the spherical coordinate system, similar to the earth's latitude/longitude angles.