## 2 Non-mechanical Remote Center Motion
### Kinematic Modeling  
The kinematic model of a robot is required as shown in Figure 2b. A robot base has a fixed frame \(\{0\}\) and a moving frame \(\{\rm{e}\}\) is attached to a remote center point, and the rotation angle of each joint is \(q_{i}\) and the link parameter is \(r_{i}\). The kinematic model of a robot is derived using a homogeneous transformation matrix between moving frames attached to each joint as shown in Equation (1). The forward kinematics of a robot manipulator can be obtained by a sequential product of transformation matrices as shown in Equation (2). Coordinates of the remote center point in the three-dimensional space, \((x_{RC},y_{RC},z_{RC})\), can be extracted from the fourth column of the matrix \(\mathbf{T}_{0e}\) as shown in Equation (3), and the pan and tilt angles of the end-effector can be calculated from the rotation matrix elements  
Figure 2: Kinematic model of the surgical assistant robot: (**a**) remote center motion; (**b**) configuration of joints and links.  
of the matrix \(\mathbf{T}_{0e}\) as shown in Equation (4). The link parameters of the robot manipulator used in this study are shown in Table 1.  
\[\mathbf{T}_{01}=\left[\begin{array}{cccc}\cos q_{1}&-\sin q_{1}&0&0\\ \sin q_{1}&\cos q_{1}&0&0\\ 0&0&1&r_{1}\\ 0&0&0&1\end{array}\right],\ \ \ \ \ \ \mathbf{T}_{12}=\left[\begin{array}{cccc} \cos q_{2}&0&\sin q_{2}&0\\ 0&1&0&-r_{2}\\ -\sin q_{2}&0&\cos q_{2}&0\\ 0&0&0&1\end{array}\right],\] \[\mathbf{T}_{23}=\left[\begin{array}{cccc}\cos q_{3}&0&\sin q_{3}&0 \\ 0&1&0&0\\ -\sin q_{3}&0&\cos q_{3}&r_{3}\\ 0&0&0&1\end{array}\right],\ \ \ \ \ \ \mathbf{T}_{34}=\left[\begin{array}{cccc} \cos q_{4}&0&\sin q_{4}&0\\ 0&1&0&r_{4}\\ -\sin q_{4}&0&\cos q_{4}&0\\ 0&0&0&1\end{array}\right], \tag{1}\] \[\mathbf{T}_{45}=\left[\begin{array}{cccc}1&0&0&0\\ 0&\cos q_{5}&-\sin q_{5}&-r_{6}\\ 0&\sin q_{5}&\cos q_{5}&0\\ 0&0&0&1\end{array}\right],\ \ \ \ \mathbf{T}_{56}=\left[\begin{array}{cccc} \cos q_{6}&-\sin q_{6}&0&-r_{6}\\ \sin q_{6}&\cos q_{6}&0&0\\ 0&0&1&0\\ 0&0&0&1\end{array}\right],\] \[\mathbf{T}_{6e}=\left[\begin{array}{cccc}-1&0&0&-r_{8}\\ 0&1&0&0\\ 0&0&-1&r_{7}+r_{9}\\ 0&0&0&1\end{array}\right].\]  
\[\mathbf{T}_{0e}=\mathbf{T}_{01}\mathbf{T}_{12}\mathbf{T}_{23}\mathbf{T}_{34} \mathbf{T}_{45}\mathbf{T}_{56}\mathbf{T}_{6e}=\left[\begin{array}{cccc}R_{1 1}&R_{12}&R_{13}&p_{1}\\ R_{21}&R_{22}&R_{23}&p_{2}\\ R_{31}&R_{32}&R_{33}&p_{3}\\ 0&0&0&1\end{array}\right]. \tag{2}\]  
\[x_{RC}=p_{1},\ \ y_{RC}=p_{2},\ \ z_{RC}=p_{3}. \tag{3}\]  
\[Pan=\arctan\left(\frac{R_{22}}{R_{33}}\right),\ \ Tilt=\arctan\left(\frac{-R_{31}}{ \sqrt{R_{32}^{2}+R_{33}^{2}}}\right). \tag{4}\]  
The Jacobian matrix of the differential kinematics is required to find an inverse kinematic solution. The Jacobian matrix can be derived through chain rules as shown in Equation (5), where \(\mathbf{z}_{i}\) and \(\mathbf{o}_{i}\) are vectors that describe the rotation axis and the origin of the \(i\)th moving frame, respectively [15]. Finally, an inverse kinematic solution for the position and orientation of the remote center can be obtained through the pseudoinverse matrix of the Jacobian matrix as shown in Equation (6).  
\[\mathbf{J}=\left[\begin{array}{cccc}\mathbf{z}_{1}\times(\mathbf{o}_{e}- \mathbf{o}_{1})&\mathbf{z}_{2}\times(\mathbf{o}_{e}-\mathbf{o}_{2})&\mathbf{z }_{3}\times(\mathbf{o}_{e}-\mathbf{o}_{3})&\mathbf{z}_{4}\times(\mathbf{o}_{ e}-\mathbf{o}_{4})&\mathbf{z}_{5}\times(\mathbf{o}_{e}-\mathbf{o}_{5})&\mathbf{z}_{6} \times(\mathbf{o}_{e}-\mathbf{o}_{6})\\ \mathbf{z}_{1}&\mathbf{z}_{2}&\mathbf{z}_{3}&\mathbf{z}_{4}&\mathbf{z}_{5}& \mathbf{z}_{6}\end{array}\right] \tag{5}\]  
\[\dot{\mathbf{q}}=\mathbf{J}^{\dagger}\dot{\mathbf{x}} \tag{6}\]