# Modeling and Control of an Articulated Multibody Aircraft
## 2 Development of the Multibody Equations of Motion
### Multibody Equations of Motion
#### 2.6.1 Translational Dynamics  
For a system of \(k\) clustered rigid bodies, the translational dynamic equations are developed as the sum of their individual contributions, considering their respective centers of mass at \(B_{k}\), with respect to the inertial frame \(I\) as follows [38; 125]:  
\[D^{I}\bigg{(}\sum_{k}m^{B_{k}}D^{I}s_{s_{k}I}\bigg{)}=\sum_{k}F_{k} \tag{16}\]  
where \(\sum_{k}F_{k}\) is the contributing sum of all the forces acting on the aircraft from each rigid body, which will be discussed in Section 2.7. Recall three previously introduced reference points \(b,j\) and \(t\). Separating the individual contributions of the central body and the tail with \(s_{{}_{B_{k}I}}=s_{{}_{B_{k}I}}+s_{{}_{B_{k}I}}\) and \(s_{{}_{B_{k}I}}=s_{{}_{B_{k}I}}+s_{{}_{B_{k}I}}\) gives  
\[D^{I}\bigg{(}m^{B}D^{I}s_{{}_{B_{k}I}}+m^{T}D^{I}s_{{}_{B_{k}I}}+m^{T}D^{I}s_{ {}_{B_{k}I}}+m^{T}D^{I}s_{{}_{B_{k}I}}\bigg{)}=\sum_{k}F_{k} \tag{17}\]  
The first term on the left-hand side of Equation (17) represents the contribution of the central body, while the second, third and fourth terms represent the contributions of the tail. Merging the first and fourth terms of the left-hand side of Equation (17) and introducing \(D^{I}s_{bl}=V_{b}^{I}\) results in  
\[\underbrace{m\ D^{I}V_{b}^{I}}_{\text{Term 1}}\ +\underbrace{m^{T}D^{I}s_{i_{j}}}_{ \text{Term II}}\ +\ \underbrace{m^{T}D^{I}D^{I}s_{j_{b}}}_{\text{Term III}}=\underbrace{\sum_{k}F_{k}}_{ \text{Term IV}} \tag{18}\]  
Since _Term I_ of Equation (18) relates to the body frame, we use the Euler transformation to represent this term in the body-fixed frame as  
\[m\ D^{I}V_{b}^{I}=m\ (D^{B}V_{b}^{I}+\Omega^{BI}V_{b}^{I}) \tag{19}\]  
_Term 11_ of Equation (18) relates to the tail frame, therefore, the rotational time derivative \(D^{I}s_{i_{j}}\) is transformed to the tail frame, using Equation (3) as in  
\[m^{T}D^{I}D^{I}s_{i_{j}}=m^{T}D^{I}(D^{T}s_{i_{j}}+\Omega^{TI}s_{i_{j}}) \tag{20}\]  
The vector \(s_{i_{j}}\) is constant in the tail frame, hence, \(D^{T}s_{i_{j}}=0\) in Equation (20). Additionally, rewriting \(\Omega^{TI}=\Omega^{TB}+\Omega^{BI}\) results in  
\[m^{T}D^{I}D^{I}s_{i_{j}}=m^{T}D^{I}(\Omega^{TB}s_{i_{j}}+\Omega^{BI}s_{i_{j}}) \tag{21}\]  
In addition, the rotational time derivatives of \(\Omega^{TB}\) and \(\Omega^{BI}\) should be transformed into their associated frames, which are the tail and body frames, respectively. Therefore, Equation (21) becomes  
\[ \begin{split} m^{T}D^{I}D^{I}s_{i_{j}}&=m^{T}[D^{ T}(\Omega^{TB}s_{i_{j}})+\Omega^{TI}\Omega^{TB}s_{i_{j}}+D^{B}(\Omega^{ BI}s_{i_{j}})+\Omega^{BI}\Omega^{BI}s_{i_{j}}]\\ &=m^{T}\Bigg{[}\begin{array}{c}D^{T}\Omega^{TB}s_{i_{j}}+ \Omega^{TB}D^{T}s_{i_{j}}+\Omega^{TI}\Omega^{TB}s_{i_{j}}\\ +D^{B}(\Omega^{BI}s_{i_{j}})+\Omega^{BI}D^{BS}s_{i_{j}}+\Omega^{BI}\Omega^{BI}s _{i_{j}}\end{array}\Bigg{]}\end{split}  \tag{22}\]  
Again, since \(D^{T}s_{i_{j}}=0\), the second term on the RHS of Equation (22) vanishes. Additionally, rewriting \(\Omega^{TI}=\Omega^{TB}+\Omega^{BI}\) and using the Euler transformation \(D^{B}s_{i_{j}}=D^{W}s_{i_{j}}+\Omega^{TB}s_{i_{j}}\) results in  
\[m^{T}D^{I}D^{I}s_{i_{j}}=m^{T}\Bigg{[} \begin{array}{c}D^{T}\Omega^{TB}s_{i _{j}}+\Omega^{TB}\Omega^{TB}s_{i_{j}}+\Omega^{BI}D^{T}s_{i_{j}}+D^{B}\Omega^{ BI}s_{i_{j}}\\ +2\Omega^{BI}\Omega^{TB}s_{i_{j}}+\Omega^{BI}\Omega^{BI}s_{i_{j}}\end{array}  \Bigg{]} \tag{23}\]  
However, \(D^{T}s_{i_{j}}=0\); therefore, the final expression for _Term 11_ in Equation (18) is  
\[m^{T}D^{I}D^{I}s_{i_{j}}=m^{T}\Bigg{[} \begin{array}{c}D^{T}\Omega^{TB}s_{i _{j}}+\Omega^{TB}\Omega^{TB}s_{i_{j}}\\ +D^{B}\Omega^{BI}s_{i_{j}}+2\Omega^{BI}\Omega^{TB}s_{i_{j}}+\Omega^{BI}\Omega^ {BI}s_{i_{j}}\end{array}