## 2 Configuration of the Remote Center Compliance (RCC) Mechanism
### Rotational Precision Analysis  
According to the principle of the RCC mechanism, the motion of the end-effector is expected to rotate around the fixed center without producing any lateral shifts. However, the actual rotation center does not precisely coincide with the theoretical remote rotation center \(O_{\mathrm{RC}}\). In order to reveal the rotational precision of the RCC mechanism precisely, the concept of instantaneous rotational center (which could reflect the essential of rigid-body rotation) is adopted as an evaluation criterion of rotational precision [15]. In Equation (11), six column vectors of the compliance matrix are regarded as six screw motions [23]. For instance, when a unit force \(F_{X}=1\) is applied to the end-effector at Point \(O\), the applied force vector can be written as \(\mathbf{F}=\left[ \begin{array}{cccccc}1&0&0&0&0&0\end{array} \right]^{ \mathrm{T}}\), and the displacement vector of the \(O\)-\(XYZ\) can be calculated as \(\left[ \begin{array}{cccccc}x_{1}&0&0&0&\theta_{1}&0\end{array} \right]^{ \mathrm{T}}=\left[ \begin{array}{cccccc}C_{X,F_{X}}&0&0&0&C_{\theta_{Y},F_{X}}& 0\end{array} \right]^{\mathrm{T}}\), which is the first column vector of the compliance matrix. According to the displacement generated by the applied force \(F_{X}\), the homogeneous representation of the rigid body transformation can be given by  
\[\mathbf{g}_{1}=\left[ \begin{array}{cccc}\mathbf{R}_{1}&\mathbf{p}_{1}\\ 0&1\end{array} \right]=\left[ \begin{array}{cccccc}\cos\theta_{1}&0&\sin \theta_{1}&x_{1}\\ 0&1&0&0\\ -\sin\theta_{1}&0&\cos\theta_{1}&0\\ 0&1&0&1\end{array} \right] \tag{12}\]This relative motion of a rigid body can be given by the exponential map for a twist \(\hat{\xi}_{1}\), which can be represented as [24]  
\[\mathbf{g}_{1}=e^{\hat{\xi}_{1}\theta_{1}}=\left[ \begin{array}{cc}e^{\hat{ \omega}_{1}\theta_{1}}&(\mathbf{I}-e^{\hat{\omega}_{1}\theta_{1}})(\omega_{1} \times\
u_{1})+\omega_{1}\omega_{1}{}^{\mathrm{T}}\
u_{1}\theta_{1}\\ 0&1\end{array} \right] \tag{13}\]  
where the twist \(\hat{\xi}_{1}\) is  
\[\hat{\xi}_{1}=\left[ \begin{array}{cc}\hat{\omega}_{1}&\
u_{1}\\ 0&0\end{array} \right] \tag{14}\]  
The axis \(\omega_{1}\in\mathbf{R}^{3}\) which satisfies \(\exp(\hat{\omega}_{1}\theta_{1})=\mathbf{R}_{1}\) is  
\[\omega_{1}=\left[ \begin{array}{cc}0&1&0\end{array} \right]^{\mathrm{T}} \tag{15}\]  
To find \(\
u_{1}\), we must solve  
\[\left[(\mathbf{I}-e^{\hat{\omega}_{1}\theta_{1}})\hat{\omega}_{1}+\omega_{1} \omega_{1}{}^{\mathrm{T}}\theta_{1}\right]\
u_{1}=\mathbf{p}_{1} \tag{16}\]  
Expanding the left-hand side of the equation yields  
\[\left[ \begin{array}{ccc}\sin\theta_{1}&0&1-\cos\theta_{1}\\ 0&\theta_{1}&0\\ \cos\theta_{1}-1&0&\sin\theta_{1}\end{array} \right]\
u_{1}=\left[ \begin{array} []{c}x_{1}\\ 0\\ 0\end{array} \right] \tag{17}\]  
The solution is given by  
\[\
u_{1}=\left[ \begin{array}{ccc}\frac{\sin\theta_{1}}{2(1-\cos\theta_{1})}& 0&-\frac{1}{2}\\ 0&\frac{1}{\theta_{1}}&0\\ \frac{1}{2}&0&\frac{\sin\theta_{1}}{2(1-\cos\theta_{1})}\end{array} \right] \left[ \begin{array}{c}x_{1}\\ 0\\ 0\end{array} \right]=\left[ \begin{array}{c}\frac{x_{1}\sin\theta_{1}}{2(1- \cos\theta_{1})}\\ 0\\ \frac{x_{1}}{2}\end{array} \right] \tag{18}\]  
Due to the small deformation of the RCC mechanism, the value of \(x_{1}\) and \(\theta_{1}\) can be regarded as nearly zero. The first element in \(\
u_{1}\) can be calculated as  
\[\frac{x_{1}\sin\theta_{1}}{2(1-\cos\theta_{1})}\ =\frac{x_{1}}{2\tan\frac{ \theta_{1}}{2}}\approx\frac{x_{1}}{\theta_{1}}=\frac{C_{X,F_{X}}}{C_{\theta_{Y },F_{X}}} \tag{19}\]  
\(\
u_{1}\) can be rewritten as  
\[\
u_{1}\approx\left[ \begin{array}{ccc}C_{X,F_{X}}\big{/}C_{\theta_{Y},F_{X}} &0&0\end{array} \right]^{\mathrm{T}} \tag{20}\]  
Thus, the twist coordinates for \(\mathbf{g}_{1}\) are  
\[\tilde{\xi}_{1}{=}\Big{[} \begin{array}{cc}\
u_{1}&\omega_{1}\end{array}  \Big{]}^{\mathrm{T}}=\Big{[} \begin{array}{ccc}C_{X,F_{X}}\big{/}C_{\theta_{Y },F_{X}}&0&0&0&1&0\end{array}