# Study on change of frame for accelerometer measurements

#### Author

Unless noted, notation follows what is described in "A Multibody Dynamics Notation – Revision_2" 📚

## Problem description

Assume that we have two accelerometers, that are attached to the same rigid body, but placed in two different frames ($$S$$ and $$B$$). We indicate with:

• $$A$$ the (absolute) inertial frame that we assume to be rigidly attached to the ground (disregarding the rotation of the earth on itself and any other cosmic rotational motion).
• $${}^A o_B \in \mathbb{R}^3$$ the 3D vector that represent the position of the origin of frame $$B$$ w.r.t. to frame $$A$$, such that if $$S$$ is another frame we have:
$\begin{equation} {}^A o_S = {}^A o_B + {}^A R_B {}^B o_S. \tag{1} \end{equation}$
• $${}^A g \in \mathbb{R}^3$$ the gravity vector expressed in the inertial frame.
• The angular velocity $${}^B \omega_{A,B} \in \mathbb{R}^3$$ is defined as:
$\begin{equation} {}^B \omega_{A,B} \times = {}^B R_A {}^A \dot{R}_B \end{equation}$

from which we can derive that:

$\begin{equation} {}^A \dot{R}_B = {}^A R_B {}^B \omega_{A,B} \times \end{equation}$

and

\begin{align} {}^A \ddot{R}_ B &= {}^A \dot{R}_ B {}^B \omega_{A,B} \times + {}^A R_B {}^B \dot{\omega}_{A,B} \times = \\ &= {}^A R_B {}^B \omega_{A,B} \times {}^B \omega_{A,B} \times + {}^A R_B {}^B \dot{\omega}_{A,B} \times. \tag{2} \end{align}

As $$B$$ and $$S$$ are rigidly attached to the same rigid body, we have that $${}^B o_S$$ is constant and so:

$\begin{equation} {}^B \dot{o}_S = 0. \tag{3} \end{equation}$

The measure of each sensor is given by:

$\begin{equation} m_S = {}^S R_A ( {}^A \ddot{o}_S - {}^A g) \tag{4} \end{equation}$
$\begin{equation} m_B = {}^B R_A ( {}^A \ddot{o}_B - {}^A g) \end{equation}$

To express $$m_S$$ in function of $$m_B$$, we can first take $$(1)$$ and take first a derivative:

$\begin{equation} {}^A \dot{o}_S = {}^A \dot{o}_B + {}^A \dot{R}_B {}^B o_S + {}^A R_B {}^B \dot{o}_S \end{equation}$

and then we derivative again:

$\begin{equation} {}^A \ddot{o}_S = {}^A \ddot{o}_B + {}^A \ddot{R}_B {}^B o_S + 2 {}^A \dot{R}_B {}^B \dot{o}_S + {}^A R_B {}^B \ddot{o}_S \end{equation}$

as we know that the $$B$$ and $$C$$ are rigidly attached to the same body (see $$(3)$$), we have:

$\begin{equation} {}^A \ddot{o}_S = {}^A \ddot{o}_B + {}^A \ddot{R}_B {}^B o_S. \tag{5} \end{equation}$

By substituting $$(5)$$ in $$(4)$$ we have:

\begin{align} m_S &= {}^S R_A ( {}^A \ddot{o}_B + {}^A \ddot{R}_B {}^B o_S - {}^A g) = \\ &= {}^S R_A ( {}^A \ddot{o}_B - {}^A g) + {}^S R_A ({}^A \ddot{R}_B {}^B o_S ) = \\ &= {}^S R_B m_B + {}^S R_A ({}^A \ddot{R}_B {}^B o_S ) \end{align}

By using $$(2)$$, we have then:

$\begin{equation} m_S = {}^S R_ B ( m_ B + {}^B \omega_{A,B} \times {}^B \omega_{A,B} \times {}^B o_S + {}^B \dot{\omega}_{A,B} \times {}^B o_S ) \end{equation}.$