Calculating rotation matrix efficiently

Question

I'm trying to efficiently calculate the vertical offset of each corner of a rectangular base. I have an accelerometer mounted in the middle of the base, ADXL345. The steps I have taken, and seems to work are below. Can anyone advise if this is the most efficient way of doing this. I'm using the raspberry pi pico, with MicroPython:

1. convert the accelerometer into a unit vector
2. use the dot product and cross products ( between X =0, y= 0, z= 1 and the accelerometer unit vector) to calculate the quaternion
3. calculate the quaternion rotation matrix
4. calculate the rotation of the four corners using the rotation matrix, but only for the z component, as I'm only interested in vertical offset of the corners

Answer 1

The accelerometer measures the gravity vector in the body frame, call it $a$. If you normalize that, say $\hat a$, it's the third row of the rotation matrix $R$ that represents the rotation of the body. i.e. $$ R^\top \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} = \frac{a}{||a||} = \hat a = z_{\mathcal{W}}^{\mathcal{B}} $$

So $$ R = \begin{pmatrix} ? \\ ? \\ \hat a \end{pmatrix} $$

Now to get the world coordinates of a corner point $p^{\mathcal{B}}$ expressed in the body frame, $$ p^{\mathcal{W}} = R p^{\mathcal{B}} = \begin{pmatrix} ? \\ ? \\ \hat a \end{pmatrix} p^{\mathcal{B}} = \begin{pmatrix} ? \\ ? \\ \hat a ^\top p^{\mathcal{B}} \end{pmatrix} $$

Thus the world $z$ coordinate of the corner point is the dot product between the normalized accelerometer vector and the corner point expressed in the body frame.

If you just need the $z$ coordinate, you do not need to construct the rotation matrix.