# How to compute linear acceleration in 3D from change in roll, pitch and yaw angles?

We know that if a body is rotating only about $$z$$-axis along a circle of radius $$R$$ with an angular rate of $$\omega$$, then the acceleration of the body in 3D is $$a = [0.0\ \ \omega^2R \ \ 0.0]$$. Now let us assume that the body is rotating in 3D such that the change in its roll, pitch and yaw angles in time interval dt is $$\left[d\phi\ \ d\theta \ \ d\psi \right]$$, respectively. The corresponding radii are $$R_1, R_2 , R_3$$. Let us also define the following: $$\omega_x = \frac{d\phi}{dt}, \omega_y = \frac{d\theta}{dt}, \omega_z = \frac{d\psi}{dt}$$.

There is no other motion in any direction. How do we compute the instantaneous acceleration in 3D? Is it simply $$a = [\omega_y^2R_2 \ \ \omega_z^2R_3 \ \ \omega_x^2R_2]$$?

$$\def \b {\mathbf}$$

Start with the Position vector $$~\b P$$ \begin{align*} &\b P=\b S\,\b R\tag 1 \end{align*} where $$~\b S=\b S_z(\psi)\,\b S_y(\theta)\,\,\b S_x(\phi)~$$ is the rotation matrix and $$~\psi,\theta,\phi~$$ are the yaw, pitch , roll angles

the acceleration

\begin{align*} &\b{\ddot{P}}=\b{\ddot{ S}}\,\b R\quad\text{with}\quad \b{\dot{S}}=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \b S \end{align*} \begin{align*} &\b{\ddot{P}}=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \,\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \b S(\psi,\theta,\phi)\,\b R\tag 2 \end{align*} to solve equation (2) you need additional equation which is: \begin{align*} \begin{bmatrix} \dot{\psi} \\ \dot{\theta} \\ \dot{\phi} \\ \end{bmatrix}= \left[ \begin {array}{ccc} {\frac {\cos \left( \psi \right) \sin \left( \theta \right) }{\cos \left( \theta \right) }}&{\frac {\sin \left( \psi \right) \sin \left( \theta \right) }{\cos \left( \theta \right) }}&1\\ -\sin \left( \psi \right) &\cos \left( \psi \right) &0\\ {\frac {\cos \left( \psi \right) }{\cos \left( \theta \right) }}&{\frac {\sin \left( \psi \right) }{\cos \left( \theta \right) }}&0\end {array} \right] \,\begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{bmatrix} \end{align*}

for small angles

\begin{align*} \b S&\mapsto \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} +\begin{bmatrix} 0 & -\psi & \theta \\ \psi & 0 & -\phi \\ -\theta & \phi &0 \\ \end{bmatrix}\quad, \begin{bmatrix} \dot{\phi} \\ \dot{\theta} \\ \dot{\psi} \\ \end{bmatrix}\mapsto\begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{bmatrix} \quad\Rightarrow\\ & \begin{bmatrix} {\phi} \\ {\theta} \\ {\psi} \\ \end{bmatrix} = \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{bmatrix} \,t+\begin{bmatrix} {\phi_0} \\ {\theta_0} \\ {\psi_0} \\ \end{bmatrix} \end{align*} thus you obtain analytical solution for equation (2)

• Is $\mathbf{R} = \left[R_1 \ \ R_2 \ \ R_3\right]^T$? Commented Jun 29, 2023 at 20:06
• Yes this is correct
– Eli
Commented Jun 29, 2023 at 20:51