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Study on change of frame for accelerometer measurements

Author
@traversaro

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:

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