3.2.1 Algorithm description

Each of the HRETR’s parallel modules is abstracted as a geometric line segment and point model in the inverse displacement algorithm. The algorithm can be divided into forward iterations and backward iterations according to the forward approximation and inverse pose adjustment principle. The position vector ${\boldsymbol{r}}_{k}$ of the central link can be obtained by the following algorithm and the inverse displacement of the HRETR can be solved. In order to describe the position change process clearly, the UPS links are omitted. The forward iteration diagram and the flow chart are presented in Figs. 4 and 5.

Figure 4.

Forward iteration. (a) Move the moving platform ${\boldsymbol{P}}_{6}$ to the target point ${\boldsymbol{P}}_{t}$ . (b) Find the joint ${\boldsymbol{P}}'_{\!\!5}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}_{5}{\boldsymbol{P}}'_{\!\!6}}}}$ . (c) Find the joint ${\boldsymbol{P}}'_{\!\!4}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}_{4}{\boldsymbol{P}}'_{\!\!5}}}}$ . (d) Find the joint ${\boldsymbol{P}}'_{\!\!3}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}_{3}{\boldsymbol{P}}'_{\!\!4}}}}$ . (e) Find the joint ${\boldsymbol{P}}'_{\!\!2}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}_{2}{\boldsymbol{P}}'_{\!\!3}}}}$ . (f) Find the joint ${\boldsymbol{P}}'_{\!\!1}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}_{1}{\boldsymbol{P}}'_{\!\!2}}}}$ .

Figure 5.

Forward iteration flow chart.

Assuming that the initial configuration of the HRETR is at downward, as shown in Fig. 4(a), ${\boldsymbol{P}}_{0},{\boldsymbol{P}}_{1}\cdots \cdots {\boldsymbol{P}}_{6}$ denote the position coordinate of each of the robot’s joint points: ${\boldsymbol{P}}_{0}$ is the position of the center of the base platform in the first parallel module, ${\boldsymbol{P}}_{k}$ and ${\boldsymbol{P}}_{k-1}$ are the position of the center of the moving platform and base platform of the $k-\mathrm{th}$ parallel module, respectively, and $d_{k}$ is the central fixed-length link of the $k-\mathrm{th}$ parallel module, as $d_{k}=\left|\left|{\boldsymbol{P}}_{k}-{\boldsymbol{P}}_{k-1}\right|\right|$ . All central fixed-length links mentioned in this paper are of identical length $d$ . ${\boldsymbol{P}}_{t}$ is the target point and ${\boldsymbol{r}}_{7}$ denotes the normal vector of the end platform. Since the central fixed-length link of the first parallel module is connected to its own base platform vertically and fixedly, thus, ${\boldsymbol{P}}_{0}$ is not involved in the iterative process. The forward iteration determines whether the target point is within the maximum range that the robot can initially reach. The judgment method is as follows:

(6) \begin{equation} d_{a}={\sum }_{k=1}^{5}d_{k} \end{equation}

(7) \begin{equation} d_{t}=\left|\left|{\boldsymbol{P}}_{t}-{\boldsymbol{P}}_{1}\right|\right| \end{equation}

where $d_{a}$ denotes the total length of the central fixed-length links of the last five parallel modules of the robot, $d_{t}$ denotes the distance between the center of mass of the moving platform and the target point in the first parallel module. If $d_{a}\lt d_{t}$ , the target point is not within the maximum range that the robot can reach. Otherwise, the target point is within the maximum range that the robot can reach.

The position error $\varepsilon$ is given, and then $d_{f}$ denoting the distance between the end of the robot ${\boldsymbol{P}}''_{\!\!\!6}$ and the target point ${\boldsymbol{P}}_{t}$ is computed by $d_{f}=\left|\left|{\boldsymbol{P}}_{t}-{\boldsymbol{P}}_{6}\right|\right|$ . If $d_{f}\gt \varepsilon$ , then start the forward iteration process is started. Otherwise, the iterative calculation will stop. The forward iteration process is as follows: ${\boldsymbol{P}}'_{\!\!0},{\boldsymbol{P}}'_{\!\!1}\cdots \cdots {\boldsymbol{P}}'_{\!\!6}$ is the position of each point after forward iteration. Moving the end point of the robot to the target point is the first step, as shown in Fig. 4(b), namely ${\boldsymbol{P}}'_{\!\!6}={\boldsymbol{P}}_{t}$ . The second step is finding the line ${\boldsymbol{P}}'_{\!\!6}{\boldsymbol{P}}_{5}$ as shown in Fig. 4(c), and then calculating the length $r_{5}$ of the link ${\boldsymbol{P}}'_{\!\!6}{\boldsymbol{P}}_{5}$ with Eq. (8) and the ratio of $d_{5}$ to $r_{5}$ with Eq. (9). Substituting $r_{5}$ and $\lambda _{5}$ into Eq. (10) and obtaining the coordinate ${\boldsymbol{P}}'_{\!\!5}$ after updating the position is the third step. By repeating the above calculation process, ${\boldsymbol{P}}'_{\!\!4}, {\boldsymbol{P}}'_{\!\!3}, {\boldsymbol{P}}'_{\!\!2}, {\boldsymbol{P}}'_{\!\!1}$ can be obtained. ${\boldsymbol{P}}_{0}$ is not involved in the iterative process, let ${\boldsymbol{P}}'_{\!\!0}={\boldsymbol{P}}_{0}$ . Thus, the forward iteration is complete:

(8) \begin{equation} r_{k-1}=\left|\left|{\boldsymbol{P}}'_{\!\!\!k}-{\boldsymbol{P}}_{k-1}\right|\right| \end{equation}

(9) \begin{equation} \lambda _{k-1}=d_{k-1}/r_{k-1} \end{equation}

(10) \begin{equation} {\boldsymbol{P}}'_{\!\!\!k-1}=(1-\lambda _{k-1}){\boldsymbol{P}}'_{\!\!\!k}+\lambda _{k-1}{\boldsymbol{P}}_{k-1} \end{equation}

The end platform’s mass center ${\boldsymbol{P}}'_{\!\!6}$ coincides with the target point ${\boldsymbol{P}}_{\!t}$ after the forward iteration, but the first parallel module’s base platform’s mass center no longer coincides with ${\boldsymbol{P}}_{0}$ . Therefore, backward iteration is needed to ensure the first parallel module’s moving platform mass center is permanently fixed. Backward iteration is shown in Fig. 6.

Figure 6.

Backward iteration. (a) The configuration of the HRETR after forward iteration and ${\boldsymbol{P}}'_{\!\!\boldsymbol{1}}$ is covered by ${\boldsymbol{P}}''_{\!\!\!1}$ . (b) Find the joint ${\boldsymbol{P}}''_{\!\!\!2}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}''_{\!\!\!1}{\boldsymbol{P}}'_{\!\!2}}}}$ . (c) Find the joint ${\boldsymbol{P}}''_{\!\!\!3}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}''_{\!\!\!2}{\boldsymbol{P}}''_{\!\!\!3}}}}$ . (d) Find the joint ${\boldsymbol{P}}''_{\!\!\!4}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}''_{\!\!\!3}{\boldsymbol{P}}'_{\!\!4}}}}$ . (e) Find the joint ${\boldsymbol{P}}''_{\!\!\!5}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}''_{\!\!\!4}{\boldsymbol{P}}'_{\!\!5}}}}$ . (f) Find the moving platform ${\boldsymbol{P}}''_{\!\!\!6}$ which lies on the line $\overline{\overline{{{\boldsymbol{P}}''_{\!\!\!5}{\boldsymbol{P}}'_{\!\!6}}}}$ .

As shown in Fig. 6(a), the first step of backward iteration is to find the joint ${\boldsymbol{P}}''_{\!\!\!1}$ by moving ${\boldsymbol{P}}'_{\!\!1}$ to the initial point ${\boldsymbol{P}}_{1}$ :

(11) \begin{equation} r'_{\!\!k+1}=\left|\left|{\boldsymbol{P}}'_{\!\!k+1}-{\boldsymbol{P}}''_{\!\!\!k}\right|\right| \end{equation}

(12) \begin{equation} \lambda '_{\!\!k+1}=d_{k+1}/r'_{\!\!k+1} \end{equation}

(13) \begin{equation} {\boldsymbol{P}}''_{\!\!\!k+1}=(1-\lambda '_{\!\!k+1}){\boldsymbol{P}}''_{\!\!\!k}+\lambda '_{\!\!k+1}{\boldsymbol{P}}'_{\!\!k+1} \end{equation}

Then, connect ${\boldsymbol{P}}''_{\!\!\!1}$ and ${\boldsymbol{P}}'_{\!\!2}$ , calculate the distance $r'_{\!\!2}$ between the two points and the ratio $\lambda '_{\!\!2}$ of $d$ to $r'_{\!\!2}$ via Eqs. (11) and (12), and obtain the ${\boldsymbol{P}}''_{\!\!\!2}$ via Eq. (13). ${\boldsymbol{P}}''_{\!\!\!3}, {\boldsymbol{P}}''_{\!\!\!4}, {\boldsymbol{P}}''_{\!\!\!5}, {\boldsymbol{P}}''_{\!\!\!6}$ can be calculated using the same iteration method as shown in Fig. 6(b) to (f). The flow chart of backward iteration is presented in Fig. 7.

Figure 7

Flow chart of backward iteration.

Following a complete iteration, it is necessary to check whether the robot’s end point is closer to the target point to determine whether to proceed with the next iteration. To achieve this, the distance $d_{f}$ between the robot’s end point and the target point is calculated and compared with $\varepsilon$ . If $d_{f}\gt \varepsilon$ , the iterative calculation must be continued. Otherwise, ${\boldsymbol{P}}''_{\!\!\!k+1}$ meets the requirements and the algorithm is complete.

Figure 8.

HRETR inverse displacement algorithm flow chart.

3.2.2 Constraint conditions

Constraint conditions on the mechanical relationships are introduced to the inverse displacement algorithm. According to the mechanical structure of the HRETR, the following physical parameters of each module must be constrained: length of the UPS links, angle of the universal joint of the parallel module’s central link, angle of the universal joint of the UPS links, and angle of the spherical joint of the UPS links.

(1) Length of the UPS links

The lengths of the HRETR’s UPS links can be obtained using Eq. (5). Considering the electric cylinder assembly, the constraint condition on the $i-\mathrm{th}$ UPS link of the $k-\mathrm{th}$ parallel module must be met and can be written as:

(14) \begin{align} \begin{cases} q_{ki,\min }\leq q_{ki}\leq q_{ki,\max }\\[5pt] q_{ki,\min }=l_{bki}\\[5pt] q_{ki,\max }=l_{bki}+l_{rki} \end{cases} \end{align}

where $l_{bki}$ and $l_{rki}$ are the length of the cylinder barrel and the length of the cylinder rod, respectively. If the condition in Eq. (14) is not satisfied, the angle $\theta _{k}$ between the two adjacent central links should be reduced by one step, $\theta _{k}=\theta _{k}-\Delta \theta$ , and the lengths of the parallel module’s links recalculated until the constraint condition is met.

Table I.

Radius of each platform of the HRETR (m)

Table II.

Length parameters for the sleeve and push rod in the UPS link of each module (m)

Table III.

Inverse displacement solutions for different values of position error $\varepsilon$

(2) Angle of the universal joint of the central link

The rotation angle $\theta _{k}$ of the moving platform can be regarded as the angle between the central links of two adjacent parallel modules, as shown in Eq. (15):

(15) \begin{equation} \theta _{k}=\arcsin ({\boldsymbol{r}}_{k}\times {\boldsymbol{r}}_{k+1}) \end{equation}

where ${\boldsymbol{r}}_{k}$ and ${\boldsymbol{r}}_{k+1}$ denote the vector of the central link of the $k-\mathrm{th}$ parallel module and the $(k+1)-\mathrm{th}$ parallel module. Only when $\theta _{k}\leq \theta _{\max }$ , the result can be considered as meeting the constraints. Otherwise, $\theta _{k}$ should be updated using the following methods until Eq. (15) is satisfied:

a. In the forward iteration, ${\boldsymbol{c}}_{k}$ denotes a vector of the same length and opposite direction to ${\boldsymbol{r}}_{k+1}$ , as shown in Eq. (16), and ${\boldsymbol{h}}_{k}$ denotes the plane normal vector by the right-hand screw rule, according to vector ${\boldsymbol{r}}_{k+1}$ and vector ${\boldsymbol{r}}_{k}$ . When $\theta _{k}\leq \theta _{\max }$ is not satisfied, let ${\boldsymbol{h}}_{k}$ be the rotation axis and rotate the vector ${\boldsymbol{v}}_{k}$ by $\theta _{\max }$ according to the right-hand screw rule, to obtain the updated vector ${\boldsymbol{r}}_{k}$ :

(16) \begin{equation} {\boldsymbol{c}}_{k}=-{\boldsymbol{r}}_{k+1} \end{equation}

(17) \begin{equation} {\boldsymbol{h}}_{k}={\boldsymbol{r}}_{k+1}\times {\boldsymbol{r}}_{k} \end{equation}

b. In the backward iteration, ${\boldsymbol{c}}_{k}$ is a vector of the same length and direction as ${\boldsymbol{r}}_{k}$ , as shown in Eq. (18), and ${\boldsymbol{h}}_{k}$ is the plane normal vector by the right-hand screw rule according to vector ${\boldsymbol{r}}_{k}$ and vector ${\boldsymbol{r}}_{k+1}$ . When $\theta _{k}\leq \theta _{\max }$ is not satisfied, set $\theta _{k}=\theta _{\max }$ , let ${\boldsymbol{h}}_{k}$ be the rotation axis, and rotate the vector ${\boldsymbol{v}}_{k}$ by $\theta _{k}$ according to the right-hand screw rule. Calculate the updated vector ${\boldsymbol{r}}_{k+1}$ and ensure the angle constraint of the universal joint is met. Otherwise, let $\theta _{k}=\theta _{k}-\Delta \theta$ and recalculate ${\boldsymbol{v}}_{k}$ and ${\boldsymbol{m}}_{k}$ until the updated vector ${\boldsymbol{r}}_{k+1}$ meets the angle constraint of the universal joint:

(18) \begin{equation} {\boldsymbol{c}}_{k}={\boldsymbol{r}}_{k} \end{equation}

(19) \begin{equation} {\boldsymbol{h}}_{k}={\boldsymbol{r}}_{k}\times {\boldsymbol{r}}_{k+1} \end{equation}

Figure 9.

Configuration of the HRETR’s central link without constraint conditions.

Figure 10.

Configuration of the HRETR’s central link with constraint conditions.

Figure 11.

HRETR configuration with constraint conditions.

(3) Angle of the universal joint of the UPS links

At the center of rotation of the universal joint of the $i-\mathrm{th}$ UPS link of the $k-\mathrm{th}$ parallel module, let the $z-\text{axis}$ point to the moving platform. The two vertical axes of the universal joint are the $x-\text{axis}$ and $y-\text{axis}$ , respectively, where the $x-\text{axis}$ points to the center of the base platform. $\;{_{}^{k-1}}{{\boldsymbol{w}}}{_{ki}^{}}$ is the vector expression of the i-th kinematics chain of the k-th parallel module in the k-th reference coordinate system and can be described by:

(20) \begin{equation} {_{}^{k-1}}{{\boldsymbol{w}}}{_{ki}^{}}={_{}^{k-1}}{{\boldsymbol{R}}}{_{ki}^{}}({_{}^{k-1}}{{\boldsymbol{r}}}{_{k}^{}}/d) \end{equation}

where ${_{}^{k-1}}{{\boldsymbol{R}}}{_{ki}^{}}$ donates the transformation relationship between the direction vector of the i-th UPS link of the k-th parallel module and the central link of the same parallel module and it can be described by:

(21) \begin{equation} {_{}^{k-1}}{{\boldsymbol{R}}}{_{ki}^{}}=\mathrm{Rot}(z,\pi /3)\mathrm{Rot}(y',\varphi _{kiy})\mathrm{Rot}(x'',\varphi _{kix}) \end{equation}

Figure 12.

Trajectory 1: The normal vector at the end remains vertical and downward.

Figure 13.

HRETR configuration along trajectory 1.

Figure 14.

Top view of HRETR configurations along trajectory 1.

Figure 15.

Trajectory 2: The normal vector at the end is divergent all around.

Due to the hyper-redundant mechanical structure, each two adjacent parallel modules are arranged at an interval of angle along the direction of the z-axis and the rotation angle with the z-axis is $\pi /3$ . Then the rotation angle of the universal joint of the $k-\mathrm{th}$ parallel module’s $i-\mathrm{th}$ UPS can be obtained by:

(22) \begin{align} \begin{cases} \varphi _{kix}=\arcsin \left(\dfrac{\sqrt{3}{_{}^{k-1}}{w}{_{kix}^{}}-{_{}^{k-1}}{w}{_{kiy}^{}}}{2}\right)\\[12pt] \varphi _{kiy}=\arctan \left(\dfrac{{_{}^{k-1}}{w}{_{kix}^{}}+\sqrt{3}{_{}^{k-1}}{w}{_{kiy}^{}}}{2{_{}^{i-1}}{w}{_{kiz}^{}}}\right) \end{cases} \end{align}

The rotation range of the universal joint of the $i-\mathrm{th}$ UPS link of the $k-\mathrm{th}$ parallel module should meet the following constraints:

(23) \begin{align} \begin{cases} \varphi _{kix}\leq \varphi _{kix\max }\\[5pt] \varphi _{kiy}\leq \varphi _{kiy\max } \end{cases} \end{align}

when the universal joint’s rotation angle does not satisfy the rotation range, let $\theta _{k}=\theta _{k}-\Delta \theta$ and recalculate $\varphi _{kix}$ and $\varphi _{kix}$ until the angle constraint is satisfied.

Figure 16.

HRETR configuration along trajectory 2.

Figure 17.

Top view of HRETR configurations along trajectory 2.

(4) Angle of the spherical joint of the UPS links

Assuming the initial installation angle of the spherical joint of the UPS links is $\gamma _{0}$ , the normal vector of the spherical joint installation plane can be obtained from the normal vector of the moving platform through the $y-x-y$ Euler angle rotation transformation which can be described by:

(24) \begin{equation} {\boldsymbol{R}}_{in}=\mathrm{Rot}\left(y,-\frac{\pi }{2}\right)\mathrm{Rot}\left(x',-\beta _{i}\right)\mathrm{Rot}\left(y'',\frac{\pi }{2}-\gamma _{0}\right) \end{equation}

where $\boldsymbol{\beta }=\left[\begin{array}{l@{\quad}l@{\quad}l} 0 & 2\pi /3 & 4\pi /3 \end{array}\right]$ denotes the angle between the two adjacent spherical joints on the moving platform. The rotation angle of the spherical joint of the $i-\mathrm{th}$ UPS link of the $k-\mathrm{th}$ parallel module is then given by:

(25) \begin{equation} \eta _{ki}=\arccos (({_{}^{Ok-_{1}}}{{\boldsymbol{R}}}{_{Ok}^{}}{\boldsymbol{R}}_{ki}{\boldsymbol{r}}_{k})^{T}{_{}^{k-1}}{{\boldsymbol{w}}}{_{ki}^{}}) \end{equation}

where ${_{}^{k-1}}{{\boldsymbol{w}}}{_{ki}^{}}$ is the direction vector of the central link vector of the $k-\mathrm{th}$ parallel module in its own base platform coordinate system. The angle constraint of the spherical joint should meet $\eta _{ki}\leq \eta _{ki,\max }$ . Otherwise, let $\theta _{k}=\theta _{k}-\Delta \theta$ and recalculate $\eta _{ki}$ until the angle constraint is satisfied.

The complete flow chart for the HRETR inverse displacement solution algorithm is shown in Fig. 8.

Table IV.

Position error, iteration times, and operating time with trajectory 1