# Design and Dynamic Simulation Verification of an On-Orbit Operation-Based Modular Space Robot
## 3 Dynamic Modeling of Spatial Modular Robot Based on R-W Method
### Attitude Motion of Each Rigid Body in the System
The absolute angular velocity vectors of these \(N\) rigid bodies can be combined into a vector matrix formula as follows:  
\[\mathbf{\omega}=-\underline{T}^{T}\mathbf{\omega}_{r}+\mathbf{\omega}_{0}\mathbbm{1}_{N} \tag{4}\]  
\[\mathbf{\omega}_{r}=\underline{H}_{i}^{\Omega T}\underline{j} \tag{5}\]  
The relationship between the absolute angular velocity vector of each rigid body in the system and the generalized coordinate derivative is as follows:  
\[\mathbf{\omega}=-\underline{\beta}\underline{\dot{q}}\mathbf{\omega}_{r}+\mathbf{\omega} _{0}\mathbbm{1}_{N} \tag{6}\]  
The angular velocity of the above equation is changed to:  
\[\Delta\mathbf{\omega}=-\underline{\beta}\underline{\Delta\dot{q}} \tag{7}\]  
In the equation \(\underline{\beta}=-(\underline{H}^{\Omega}\underline{T})\), \(\underline{H}^{\Omega}\) is an \(N\)-order diagonal block vector matrix \(\underline{H}_{i}^{\Omega T}\), with the vector matrix as an element, and \(\underline{\beta}\) is an \(N\times N\) order vector block matrix in which the vector block array in the i-th row and j-th column is \(-T_{ji}\underline{H}_{j}^{\Omega T}\). The transpose of its channel matrix \(\underline{T}\) and vector block matrix \(\underline{\beta}\) are as follows. \[\underline{T}= \begin{pmatrix}-1&0&0&0&0\\ -1&-1&0&0&0&0\\ -1&0&-1&0&0&0\\ -1&0&-1&-1&0&0\\ -1&0&0&0&-1&0\\ -1&0&0&0&0&-1\end{pmatrix} \mathbf{\beta}=\begin{pmatrix}\underline{H}_{1}^{ \Omega T}&0&0&0&0&0\\ \underline{H}_{1}^{\Omega T}&\underline{H}_{2}^{\Omega T}&0&0&0&0  
It is not difficult to see that the vector block \(\underline{\beta}\) is a lower triangular vector block matrix, which is consistent with the position of the transposed non-zero term of the channel matrix \(\underline{T}\). As long as the non-zero item in the \(j\)-th column is replaced by \(\underline{\underline{H}}_{j}^{\Omega T}\), \(\underline{\beta}\) will be generated. The first derivative of the angular velocity on the global coordinate base is  
\[\dot{\mathbf{\omega}}_{i}=-\sum_{k=1}^{N}T_{ki}\dot{\mathbf{\omega}}_{rk}+\mathbf{\omega}_{0 }\quad(i=1,\ldots,N) \tag{8}\]  
Considering \(\dot{\mathbf{\omega}}_{ri}=\overset{\circ}{\mathbf{\omega}}_{ri}+\mathbf{\omega}_{L(i)} \times\mathbf{\omega}_{ri}\), we can get the vector matrix formula composed of the \(N\) rigid body absolute angular acceleration vector, i.e.,  
\[\dot{\mathbf{\omega}}=-\underline{T}^{T}\overset{\circ}{\underline{\omega}}_{r}+ \dot{\mathbf{\omega}}_{0}\underline{1}_{N}-\underline{T}^{T}\underline{\zeta} \tag{9}\]  
where \(\zeta_{i}=\dot{\mathbf{\omega}}_{L(i)}\times\mathbf{\omega}_{ri}\). Considering \(\dot{\mathbf{\omega}}_{r}=\underline{\underline{H}}^{\Omega T}\bar{\underline{q} }+\underline{\mathbf{\eta}}\), the above equation can be rewritten as the final expression of angular acceleration:  
\[\dot{\underline{\mathbf{\omega}}}=\underline{\beta}\bar{\underline{q}}+\dot\mathbf{{\omega}}_{0}\underline{1}_{N}-\underline{T}^{T}(\underline{\zeta}+ \underline{\mathbf{\eta}}) \tag{10}\]