I am working on a 6DOF IMU that contains a 3-axis accelerometer and a 3-axis gyroscope. I am building a project to plot the position and orientation of a vehicle/dirt bike in a 3d plane. However, the inertial measurement unit is placed away from the centre of mass of the vehicle i.e. in the boot/taillight of the car/bike. The vehicle will be performing extreme manoeuvres in a terrain. I have the below questions regarding this matter.
As far as I have read, I have to offset accelerometer and gyroscope readings due to its placement away from the centre of mass of the vehicle. In some articles I have read that the placement of the gyroscope does not affect gyroscope readings; however, just to be sure, I am asking once again if gyro measurements needed to be offset.
I have found two articles stating how to offset accelerometer readings with regards to its placement inside the body.
Accelerometer Placement – Where and Why
In this article the below formula is used:
$$ \begin{matrix}{} A_{r’} & = & A + A_r & + \\ & & 2\omega \times V_r & + \\ & & \alpha \times r & + \\ & & \omega \times (\omega \times r) & + &\end{matrix}\begin{aligned} &\leftarrow \text{Inertial Acceleration} \\ &\leftarrow \text{Coriolis Acceleration} \\ &\leftarrow \text{Euler Acceleration} \\ &\leftarrow \text{Centripetal Acceleration} \end{aligned}$$
However, I am not sure how to proceed with this formula. I have $a_x$, $a_y$, and $a_z$ from IMU from a point X (i.e. where the IMU is placed) and using this formula I want to get $a_x$, $a_y$, and $a_z$ at centre of mass of the vehicle; but I am confused as to how to proceed with this formula.
Calculating acceleration offset by Center of Gravity (C.G.)
Starting from the well-known acceleration transformation formula between an arbitrary point A and the center of mass C with $\vec{c} = \vec{r}_C - \vec{r}_A$.
$$ \vec{a}_C = \vec{a}_A + \dot{\vec{\omega}} \times \vec{c} + \vec{\omega} \times \vec{\omega} \times \vec{c} $$
one can use the 3×3 cross product operator to transform the above into
$$ \vec{a}_C = \vec{a}_A + \begin{vmatrix} 0 & -\dot{\omega}_z & \dot{\omega}_y \\ \dot{\omega}_z & 0 & -\dot{\omega}_x \\ -\dot{\omega}_y & \dot{\omega}_x & 0 \end{vmatrix} \vec{c} + \begin{vmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{vmatrix} \begin{vmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{vmatrix} \vec{c} $$
or in the form seen the linked post
$$ \vec{a}_C = \vec{a}_A + \begin{vmatrix} -\omega_y^2-\omega_z^2 & \omega_x \omega_y - \dot{\omega}_z & \omega_x \omega_z + \dot{\omega}_y \\ \omega_x \omega_y + \dot{\omega}_z & -\omega_x^2-\omega_z^2 & \omega_y \omega_z - \dot{\omega}_x \\ \omega_x \omega_z - \dot{\omega}_y & \omega_y \omega_z + \dot{\omega}_x & -\omega_x^2 - \omega_y^2 \end{vmatrix} \vec{c} $$
However I am confused about this formula as well, since none of them give me ax, ay, and az from the centre of mass of the vehicle.
Kindly help me since I am new to this field, I am originally from a computer science background, and I am struggling to understand these concepts.
Also, if you can give me reference documents as to how do I interpret values from accelerometer and gyroscope, rotation and why is it necessary, that would be great.