## 3 The Model of Flexible Arm
### Relationship Between Lever Arms and Deformation Angles  
In the coordinate system of \(SRINS_{1}\), it is observed that \(SRINS_{2}\) rotates \(\theta_{x}^{P_{1}}\) around the \(x\)-axis. In the coordinate system of \(SRINS_{2}\), it is sensed that its own p-frame rotates \(\theta_{x}^{P_{2}}\) around the \(x\)-axis, as shown in Figure 3. In Figure 3, \(o_{1}\) represents the center of \(SRINS_{1}\); \(o_{2}^{\prime}\) represents the original center of \(SRINS_{2}\); \(o_{2}\) represents the new center of \(SRINS_{2}\) after undergoing flexural deformation; and \(A\) is the center of the circular arc \(o_{1}o_{2}\) corresponding to the flexural curve, and the \(y\)-axes of both \(SRINS_{1}\) and \(SRINS_{2}\) are tangent to this arc segment. Therefore, the relationship between \(\theta_{x}^{P_{1}}\) and \(\theta_{x}^{P_{2}}\) exists:  
\[\theta_{x}^{P_{2}}=2\theta_{x}^{P_{1}} \tag{15}\]Let the lever arms between two SRINSs be \(P_{0}^{p_{1}}=[ \begin{array}{cc}r_{x0}^{p_{1}}&r_{y0}^{p_{1}}&r_{z0}^{p_{1}}\\ \end{array} ]\), then:  
\[o_{1}C=Bo_{2}=Ao_{2}\sin\theta_{x}^{p_{2}}=\frac{r_{y0}^{p_{1}}}{\theta_{x}^{p _{2}}}\sin\theta_{x}^{p_{2}} \tag{16}\]  
\[AB=Ao_{2}\cos\theta_{x}^{p_{2}}=\frac{r_{y0}^{p_{1}}}{\theta_{x}^{p_{2}}}\cos \theta_{x}^{p_{2}} \tag{17}\]  
Then, the projection of the lever arm component \(r_{y0}^{p_{1}}\) in the \(y\), \(z\) direction under the influence of the flexural angle \(\theta_{x}^{p_{1}}\) can be estimated as:  
\[\left\{ \begin{array}{l}\Delta r_{y}^{p_{1}}\left(\theta_{x}^{p_{1}},r_{y0}^{ p_{1}}\right)=o_{1}C-o_{1}o^{\prime}_{2}=-\Big{(}1-\frac{\sin\theta_{x}^{p_{2}} }{\theta_{x}^{p_{2}}}\Big{)}r_{y0}^{p_{1}}\\ \Delta r_{z}^{p_{1}}\left(\theta_{x}^{p_{1}},r_{y0}^{p_{1}}\right)=Ao_{1}-AB= \frac{1-\cos\theta_{x}^{p_{2}}}{\theta_{x}^{p_{1}}}r_{y0}^{p_{1}}\end{array} \right. \tag{18}\]  
In the same way, the projection of the lever arm component \(r_{z0}^{p_{1}}\) in the \(y\), \(z\) direction under the influence of the flexural angle \(\theta_{x}^{p_{1}}\) can be estimated as:  
\[\left\{ \begin{array}{l}\Delta r_{y}^{p_{1}}\left(\theta_{x}^{p_{1}},r_{z0}^{ p_{1}}\right)=-\frac{1-\cos\theta_{x}^{p_{2}}}{\theta_{x}^{p_{2}}}r_{z0}^{p_{1}} \\ \Delta r_{z}^{p_{1}}\left(\theta_{x}^{p_{1}},r_{z0}^{p_{1}}\right)=\Big{(} \frac{\sin\theta_{x}^{p_{2}}}{\theta_{x}^{p_{2}}}-1\Big{)}r_{z0}^{p_{1}}\end{array} \right. \tag{19}\]  
Then, under the influence of the flexural angle \(\theta_{x}^{p_{1}}\), the change in the lever arms in the \(y\), \(z\) direction is:  
\[\left\{ \begin{array}{l}\Delta r_{y}^{p_{1}}\left(\theta_{x}^{p_{1}}\right)= \Big{(}\frac{\sin\theta_{x}^{p_{2}}}{\theta_{x}^{p_{2}}}-1\Big{)}r_{y0}^{p_{1} }-\Big{(}\frac{1-\cos\theta_{x}^{p_{2}}}{\theta_{x}^{p_{2}}}\Big{)}r_{z0}^{p_{ 1}}\\ \Delta r_{z}^{p_{1}}\left(\theta_{x}^{p_{1}}\right)=\Big{(}\frac{1-\cos\theta_ {x}^{p_{2}}}{\theta_{x}^{p_{2}}}\Big{)}r_{y0}^{p_{1}}+\Big{(}\frac{\sin\theta_ {x}^{p_{2}}}{\theta_{x}^{p_{2}}}-1\Big{)}r_{z0}^{p_{1}}\end{array} \right. \tag{20}\]  
Under the influence of the flexural angle \(\theta_{y}^{p_{1}}\), the change in the lever arms in the \(x\), \(z\) direction is:  
\[\left\{ \begin{array}{l}\Delta r_{x}^{p_{1}}\left(\theta_{y}^{p_{1}}\right)= \Big{(}\frac{\sin\theta_{x}^{p_{2}}}{\theta_{x}^{p_{2}}}-1\Big{)}r_{x0}^{p_{1} }+\Big{(}\frac{1-\cos\theta_{y}^{p_{2}}}{\theta_{x}^{p_{2}}}\Big{)}r_{z0}^{p_{1 }}\\ \Delta r_{z}^{p_{1}}\left(\theta_{y}^{p_{1}}\right)=-\Big{(}\frac{1-\cos\theta_ {x}^{p_{2}}}{\theta_{x}^{p_{2}}}\Big{)}r_{x0}^{p_{1}}+\Big{(}\frac{\sin\theta_ {y}^{p_{2}}}{\theta_{x}^{p_{2}}}-1\Big{)}r_{z0}^{p_{1}}\end{array} \right. \tag{21}\]  
Under the influence of the flexural angle \(\theta_{x}^{p_{1}}\), the change in the lever arms in the \(x\), \(y\) direction is:  
\[\left\{