Newton- Euler Rigid Body

Hello i am busy using Newton-Euler equations to get the dynamic of a quad drone. However, as I am looking online at papers I see mostly 2 different kinds of equations.

\begin{equation} \begin{bmatrix} mI & 0 \\ 0 & J \end{bmatrix} \begin{bmatrix} \dot{v}_G \\ \dot{\omega}_G \end{bmatrix} + \begin{bmatrix} m\omega_G\times v_G \\ \omega_G\times J\omega_G \end{bmatrix} = \begin{bmatrix} f \\ \tau_G \end{bmatrix} \tag1 \end{equation}

\begin{equation} \begin{bmatrix} mI & 0 \\ 0 & J \end{bmatrix} \begin{bmatrix} \dot{v}_G \\ \dot{\omega}_G \end{bmatrix} + \begin{bmatrix} 0 \\ \omega_G\times J\omega_G \end{bmatrix} = \begin{bmatrix} f \\ \tau_G \end{bmatrix} \tag2 \end{equation}

Even equation 2 is found on wikipedia but most papers use equation 1 and my question is, what is the difference and when are both of them used??

• Don't the online papers/wikipedia tell you how to use these equations? – sammy gerbil May 1 '18 at 23:32
• It would be helpful if you could link to some of the papers that you have been reading. – rob May 1 '18 at 23:51
• It depends if $\dot{v}_G$ is material or spatial acceleration. – ja72 May 2 '18 at 1:00
• Related question – ja72 May 7 '18 at 0:00

The proper equation at the center of mass is

\begin{aligned} \hat{f}_G &= \hat{\rm J}_G\, \dot{\hat{v}_G} + \hat{v}_G \times \hat{\rm J}_G\,\hat{v}_G \\ \begin{bmatrix}f\\ \tau_{G} \end{bmatrix} & =\begin{bmatrix}m\\ & J \end{bmatrix}\begin{bmatrix}\dot{v}_{G}\\ \dot{\omega} \end{bmatrix}+\begin{bmatrix}\omega\times & 0\\ v_{G}\times & \omega\times \end{bmatrix}\begin{bmatrix}m\\ & J \end{bmatrix}\begin{bmatrix}v_{G}\\ \omega \end{bmatrix} \\ & = \begin{bmatrix}m\\ & J \end{bmatrix}\begin{bmatrix}\dot{v}_{G}\\ \dot{\omega} \end{bmatrix} + \begin{bmatrix}\omega\times m\,v_{G}\\ \omega\times J\omega \end{bmatrix} \end{aligned}

where $v_G$ is the velocity at the center of mass, and $\dot{v}_G$ is the spatial acceleration at the center of mass.

The material acceleration of the center of mass is

$$a_G = \dot{v}_G + \omega \times v_G$$ as well as the identity $$\alpha = \dot{\omega}$$

Proof

The standard form of the equations is

\begin{aligned} f & = m a_G \\ \tau_G & = J \alpha + \omega \times J \omega \end{aligned}

and $$f = m (\dot{v}_G + \omega \times v_G) = m \dot{v}_G + \omega \times (m v_G)$$

At some other location A, other than the center of mass, where $c$ is vector from that location to the CM the NE equations of motion are

\begin{aligned} \hat{f}_A &= \hat{\rm J}_A\, \dot{\hat{v}_A} + \hat{v}_A \times \hat{\rm J}_A\,\hat{v}_A \\ \begin{bmatrix}f\\ \tau_{A} \end{bmatrix} & =\begin{bmatrix}m & -m [c \times] \\ m [c \times] & J-m [c\times][c\times] \end{bmatrix}\begin{bmatrix}\dot{v}_{A}\\ \dot{\omega} \end{bmatrix}+\begin{bmatrix}\omega\times & 0\\ v_{A}\times & \omega\times \end{bmatrix}\begin{bmatrix}m & -m [c \times] \\ m [c \times] & J-m [c\times][c\times] \end{bmatrix}\begin{bmatrix}v_{A}\\ \omega \end{bmatrix} \end{aligned}

I leave it up to the reader the prove this, based on the standard transformation equations for torque, velocity and spatial acceleration.

Please read this answer for the geometric foundation of the above equations

• I edited the answer with more details. – ja72 May 6 '18 at 23:59