The mechanical structure of the proposed continuum robot is shown in Fig. 2. The backbone material is a super-elastic rod. The robot is shaped by tendons passing through the disks. One segment of the continuum robot has two degrees of freedom (DOFs), i.e., rotation around *z*-axis (\(\phi\)) and *y*-axis (\(\theta\)), based on the constant curvature kinematics frameworks. The number of tendons is at least three for driving one segment, because tendons must be work in tension. This property brings a disadvantage for the kinematic control of continuum robot with three tendons. Because the shape of one segment is determined by two tendons, the tension of the third tendon is extremely large as the translation error of the third tendon is positive. The third tendon would be snapped. The designed continuum robot with each segment driven by four tendons and controlled by two linear actuators is developed to compensate this disadvantage.

One of the differences between the continuum robot developed in this paper and the continuum robots driven by three tendons is arrangement of actuators. Two common arrangements of actuators are shown in Fig. 3. The second one is selected to arrange the tendons, because the two pairs (\(H_{x}\) and \(H_{y}\)) of tendons are uncoupled with each other. Here, linear motor is selected because it can provide large range of motion. Furthermore, each pair of tendons is connected to a single linear motor. One segment of the continuum robot is driven by six motors. The manufacturing cost is thus lower. The question is whether it is possible to steer four tendons by two linear actuators only. This question will be answered by addressing the kinematic model of continuum robot.

The configuration of each segment robot is defined by three parameters: the curvature (\(k\left( {\varvec{\uprho}} \right)\)), the angle of the plane containing the arc (\(\phi \left( {\varvec{\uprho}} \right)\)), and arc length (\(l\)), as shown in Fig. 2, where \({\varvec{\uprho}}\) is the tendon length and \(k\left( {\varvec{\uprho}} \right) = \theta /l\). By comparing the varied length of tendons in each pair, one can find

$$\Delta l = 2\left( {l - 2n\frac{l}{\theta }\sin \left( {\frac{\theta }{2n}} \right)} \right) \ge 0,$$

(1)

where *n* is the number of disks. Equation (1) indicates that one tendon works in tension and the other one is slack in each pair, which is exactly the requirement of the continuum robot control. Therefore, one just needs to change the moving direction of linear motor based on \(\phi\) for the operation of the continuum robot. On the other side, multiple segments are employed to provide sufficient DOF for accomplishing complex surgical tasks. Twelve tendons and six linear motors are used to operate the robot finally. The tendons are divided into three groups (\(H_{1}\), \(H_{2}\), and \(H_{3}\)), and each group has two pairs (\(H_{ix}\), \(H_{iy}\), and \(i = 1, 2, 3\)). The length of each segment (\(l_{i}\)) is defined by the cable nodes. Thus, the length of each segment can be adjusted based on the surgical requirement.

### Dimensional synthesis

The three-segment continuum robot developed in this paper has six DOFs. The workspace of each segment is determined by three parameters: \(\phi_{i}\), \(\theta_{i}\), and \(l_{i}\), \(i = 1,2,3\). The range of rotation angle \(\phi_{i}\) is 0°–360°. Note that the rotation angle \(\theta_{i}\) is limited because of the constant curvature kinematics. In general, the range of rotation angle \(\theta_{i}\) is ranged 0°–90° or 0°–120°. The entire attitude space is independent of the arc length \(l_{i}\). However, the position of end effector depends on \(l_{i}\). The arc length of each segment should be optimized based on surgical workspace.

In this paper, the range of \(\theta_{i}\) is set as 0°–90°. The reachable workspace of the continuum robot is a circular symmetry in geometry, so it can be defined by the cross section in plane. The approximate boundary of the cross section can be formulated. For the three-segment continuum robot, its left approximate boundary of the cross section can be divided into four sections:

$$s_{1} = {\mathbf{T}}_{1} \left( {\phi_{1} ,\theta_{1} } \right)\left[ {0 0 l_{2} + l_{3} 1} \right]^{\text{T}} ,$$

(2)

$$s_{2} = {\mathbf{T}}_{1} \left( {\phi_{1} ,\theta_{1} } \right){\mathbf{T}}_{2} \left( {\phi_{2} ,\frac{ - \pi }{2}} \right){\mathbf{T}}_{3} \left( {\phi_{3} ,\frac{ - \pi }{2}} \right)\left[ {0 0 0 1} \right]^{\text{T}} ,$$

(3)

$$s_{3} = {\mathbf{T}}_{1} \left( {\phi_{1} , \frac{ - \pi }{2}} \right){\mathbf{T}}_{2} \left( {\phi_{2} ,\theta_{2} } \right)\left[ {0 0 l_{3} 1} \right]^{\text{T}} ,$$

(4)

$$s_{4} = {\mathbf{T}}_{1} \left( {\phi_{1} , \frac{ - \pi }{2}} \right){\mathbf{T}}_{2} \left( {\phi_{2} ,\frac{ - \pi }{2}} \right){\mathbf{T}}_{3} \left( {\phi_{3} ,\theta_{3} } \right)\left[ {0 0 0 1} \right]^{\text{T}} ,$$

(5)

where \(\phi_{1} ,\phi_{2} ,\phi_{3} = 0\), \(\theta_{1} ,\theta_{2} ,\theta_{3} \in \left( {0, \pi /2} \right]\), \({\mathbf{T}}_{i}\) is the transform matrix of the \(i\)th segment, and \(x \le 0\). If \(\phi_{1}\) changes from 0 to \(2\pi\), the cure of each section turns into a surface, which forms the approximate boundary of workspace.

Now the dimensional synthesis of continuum robot can be analyzed based on the requirement of surgical workspace. Suppose that the surgical workspace is a cuboid, i.e., \(x \in \left[ { - x_{r} ,x_{r} } \right]\), \(y \in \left[ { - y_{r} ,y_{r} } \right]\) and \({\text{z}} \in \left[ {z_{rd} ,z_{ru} } \right]\), and \(x_{r} \ge y_{r}\). The case of \(x_{r} \le y_{r}\) is similar, because the workspace is circular symmetry. The cross section of surgical workspace is a rectangle in the \(xz\) plane. Denote the vertexes as \({\mathbf{V}}_{j}\), \(j = 1, \ldots ,4\), and the middle point of \({\mathbf{V}}_{3} \mathbf{V}_{4}\) as \({\mathbf{V}}_{0}\). Based on the geometric property of boundary, the cuboid is in the workspace of continuum robot by providing that \({\mathbf{V}}_{0}\), \({\mathbf{V}}_{1}\), …, \({\mathbf{V}}_{4}\) are all in the workspace. If the robot length is very long, the robot will easily go out of the channel with a small rotation angle *θ*. Therefore, the length should be minimized to improve the dexterity of the continuum robot. The dimensional synthesis can be described as the following optimization problem:

$$\begin{array}{*{20}l} { \hbox{min} } \hfill & {l_{1} + l_{2} + l_{3} } \hfill \\ {{\text{s}} . {\text{t}} .} \hfill & {\left\{ {\begin{array}{*{20}l} {T\left( {\phi_{1j} ,\theta_{1j} ,l_{1} , \ldots ,\phi_{3j} ,\theta_{3j} ,l_{3} } \right){\mathbf{d}}^{e} = {\mathbf{V}}_{j} , \quad j = 0, \ldots ,4} \hfill \\ { - \frac{\pi }{2} \le \theta_{ij} \le \frac{\pi }{2}, \quad l_{i} > 0,\quad i = 1,2,3, \quad j = 0, \ldots ,4} \hfill \\ \end{array} } \right.} \hfill \\ \end{array}$$

(6)

where \({\mathbf{T}} = {\mathbf{T}}_{1} {\mathbf{T}}_{2} {\mathbf{T}}_{3}\) is the transformation from the base coordinate frame \(F_{\text{b}}\) to the frame \(F_{\text{e}}\) of the end effector, and \({\mathbf{d}}^{e}\) represents the end point of the end effector in the frame \(F_{\text{e}}\). Therefore, the length of each segment can be determined through optimization.