Bi-Rotor Equations of Motion

As a bi-rotor (quadcopter simplified to 2D) is essentially a rigid body, its equations of motion can be easily derived as $$\begin{array}{rl}m\ddot{x}& =-\sin (\theta )\left(u_1+u_2\right)\\ m\ddot{y}& =\cos (\theta )\left(u_1+u_2\right)-mg\\ I\ddot{\theta }& =a\left(u_1-u_2\right).\end{array}$$

Linearization in a hovering configuration

From the EoM we may see that for the state description $$\begin{array}{rlrl}x& =\left[\begin{array}{c}x\\ y\\ \theta \\ \dot{x}\\ \dot{y}\\ \dot{\theta }\end{array}\right],& \dot{x}& =\left[\begin{array}{c}\dot{x}\\ \dot{y}\\ \dot{\theta }\\ -\frac{1}{m}\sin (\theta )\left(u_1+u_2\right)\\ \frac{1}{m}\cos (\theta )\left(u_1+u_2\right)-mg\\ \frac{a}{I}\left(u_1-u_2\right)\end{array}\right]\\ u& =\left[\begin{array}{c}{u}_1\\ u_2\end{array}\right]& & \end{array}$$ all points $$\begin{array}{rl}{x}^{\ast }& ={\left[\begin{array}{cccccc}x& y& 0& 0& 0& 0\end{array}\right]}^{\top },\{x,y\}\in {\mathbb{R}}^2\\ u^{\ast }& ={\left[\begin{array}{cc}\frac{1}{2}mg& \frac{1}{2}mg\end{array}\right]}^{\top }\end{array}$$ are stationary points, for which the system's linearization is trivial $$\begin{array}{rl}A& =\left[\begin{array}{cccccc}& & & 1& & \\ & & & & 1& \\ & & & & & 1\\ & & -g& & & \\ & & & & & \\ & & & & & \end{array}\right]\\ B& =\left[\begin{array}{cc}& \\ & \\ & \\ & \\ \frac{1}{m}& \frac{1}{m}\\ \frac{a}{I}& \frac{-a}{I}\end{array}\right]\end{array}$$

Derivation in manipulator equation form

For those who want to validate the equations by deriving the manipulator equations, kinetic and potential energy for generalized coordinates $$q=\left[\begin{array}{c}x\\ y\\ \theta \end{array}\right]$$ are $$\begin{array}{rl}T& =\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)+\frac{1}{2}I{\dot{\theta }}^2\\ V& =mgy\end{array}$$ which should yield $$M=\left[\begin{array}{ccc}m& 0& 0\\ 0& m& 0\\ 0& 0& I\end{array}\right]c=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]{\tau }_g=\left[\begin{array}{c}0\\ -mg\\ 0\end{array}\right].$$ The manipulation matrix $$B=\left[\begin{array}{cc}-\sin (\theta )& -\sin (\theta )\\ \cos (\theta )& \cos (\theta )\\ a& -a\end{array}\right]$$ can then be derived separately by evaluating thrust vectors and differentiating moment arms.