2.1. Bionic inspiration and conceptual design

A snail’s body can extend clockwise (as shown by the red arrow in Fig. 1 [Reference Sutcharit, Thach, Chhuoy, Ngor, Jeratthitikul, Siriwut, Srisonchai, Ng, Pholyotha, Jirapatrasilp and Panha24]) or retract counterclockwise along the spiral line of the shell. From the perspective of mechanical kinematics, the movement of the snail’s body is an approximate swing movement guided along the spiral line. Inspired by this, we propose a HyA, whose output movement is similar to that of a snail’s body.

Figure 1.

A snail’s shell [Reference Sutcharit, Thach, Chhuoy, Ngor, Jeratthitikul, Siriwut, Srisonchai, Ng, Pholyotha, Jirapatrasilp and Panha24].

The HyA has a structure similar to that of a snail. First, the HyA has an arc shell, inspired by the snail shell. It has a closed reference end, inspired by the closed center point of the snail’s shell, and has a movable rod, inspired by the snail’s body. It also contains an open end to provide telescopic space, which was inspired by the opening of the snail’s shell. In Fig. 2, we have depicted an outline drawing (Fig. 2(a)) and an internal drawing (Fig. 2(b)) of the conceptual design of the HyA.

Figure 2

Conceptual design of the HyA.

2.2. Motion of the HyA

First, the centerline of the movable rod coincides with that of the arc shell, as shown in Fig. 3. The force generated by the hydraulic oil acting on the big head of the movable rod pushes the R line to rotate around point O. In other words, the movable rod is pushed by the hydraulic oil in the arc shell to produce circular motion, as shown in Fig. 3. The sequence of motion is (a)→(b)→(c). From the other side, the force generated by the hydraulic oil acts on the Rr line by the movable rod, equivalently, and pushes the Rr line to rotate around point O uniformly, as shown in Fig. 3.

Figure 3.

Movement of the HyA.

The motion analysis tells us that the hydraulic oil will directly push the Rr line to rotate around point O, that is, there is no conversion mechanism, just like a vane hydraulic swing cylinder. The Rr line is designed as a torsion arm that connects point A with point O. Finally, we designed a shaft at point O, which outputs outward rotational motion.

2.3. Design parameters and cross-design strategy

The main parameters of the HyA are shown in Table I.

Table I.

Main parameters of the HyA.

Some mechanical dimensions are described in Fig. 4.

Figure 4.

Mechanical dimensions of the HyA. $D_{1}$ , diameter of big head of movable rod; $D_{2}$ , diameter of small head of movable rod; $R$ , radius of HyA; th, thickness of the arc shell.

The output torque, $T_{o}$ , of the HyA can be expressed by Eq. (1):

(1) \begin{equation}T_{o}=P_{s}\left(\int _{R-\frac{D_{1}}{2}}^{R-\frac{D_{2}}{2}}f_{1}\!\left(r\right)dr+\int _{R-\frac{D_{2}}{2}}^{R}f_{2}\!\left(r\right)dr+\int _{R}^{R+\frac{D_{2}}{2}}f_{3}\!\left(r\right)dr+\int _{R+\frac{D_{2}}{2}}^{R+\frac{D_{1}}{2}}f_{4}\!\left(r\right)dr\right)\end{equation}

where $f_{1}\!\left(r\right)=2r\sqrt{\frac{D_{1}^{2}}{4}-\left(R-r\right)^{2}}$ ,

$f_{2}\!\left(r\right)=2r\sqrt{\frac{D_{1}^{2}}{4}-\left(R-r\right)^{2}}-2r\sqrt{\frac{D_{2}^{2}}{4}-\left(R-r\right)^{2}}$ , $f_{3}\!\left(r\right)=2r\sqrt{\frac{D_{1}^{2}}{4}-\left(r-R\right)^{2}}-2r\sqrt{\frac{D_{2}^{2}}{4}-\left(r-R\right)^{2}}$ , $f_{4}\!\left(r\right)=2r\sqrt{\frac{D_{1}^{2}}{4}-\left(r-R\right)^{2}}$ , $P_{s}$ is the rated pressure, and $r$ is the integral variable.

We can see that Eq. (1) is too complex for analytical calculation. The pressure acts uniformly on the surface; we can consider that the hydraulic oil pressure acting on the head of the movable rod is equivalent to a concentrated force. Therefore, we can simplify Eqs. (1) to (2) as:

(2) \begin{equation}T_{o}\approx \frac{\pi }{4}P_{s}R\left(D_{1}^{2}-D_{2}^{2}\right)\end{equation}

Considering the reduction of weight and the convenience of manufacturing, we chose 7075 aluminum alloy to manufacture the arc shell. $th$ can be described by Eq. (3) as:

(3) \begin{equation}th=\frac{\varphi P_{s}D_{1}}{\left(2.3\left[\sigma \right]-P_{s}\right)}\end{equation}

where $[\sigma ]$ is the allowable stress of the material (in this work it is $82\text{ MPa}$ ), and $\varphi$ is the structural safety factor, which is $\varphi =2$ in this study.

Finally, we can get the cross constraints, that is, the torque constraints and dimensional constraints (i.e., height and width) as described by Eqs. (4), (5) and (6), respectively, as:

(4) \begin{equation}500\text{ Nm} < \mathrm{T}_{\mathrm{o}} < 550\text{ Nm}\end{equation}

(5) \begin{equation}h=\frac{D_{2}}{2}+2R+\frac{D_{1}}{2}+th\lt 140\text{ mm}\end{equation}

(6) \begin{equation}w=2\mathrm{*}R+D_{1}+2th\lt 150\text{ mm}\end{equation}

where reasonable parameter ranges of each are

\begin{equation*}D_{1}\epsilon (25\,{\rm mm}, 32\,{\rm mm}),D_{2}\epsilon (12\,{\rm mm}, 18\,{\rm mm}),R\epsilon (45\,{\rm mm}, 52\,{\rm mm}).\end{equation*}

Then, we traverse the space, $D_{1}$ , $D_{2}$ , and $R$ , with a step size of 0.05 mm to obtain the values that meet the constraints. The traversal results are shown in Fig. 5.

Figure 5.

Cross-design strategy.

The cross-design strategy is based on the selection of the data point density in a reasonable numerical region in constraint space, so as to make the design parameters more reasonable.

As can be seen from the traversal results, the density of data points that meet the constraints near $D_{1}=30\,\textrm{mm}$ is large (Fig. 5(a)), so we chose 30 mm as the final size of $D_{1}$ . Then, we used rectangle (1) to cover the area around $D_{1}=30$ mm in Fig. 5(a), and rectangle (3) to cover the area around $D_{1}=30$ mm in Fig. 5(b). We drew a rectangle (2) and moved it along the coordinates in Fig. 5(b). Finally, we can see that in the subgraph of $th$ vs. $D_{1}$ , the overlap between rectangle (2) and rectangle (3) is a maximum near the region $th=7.5$ mm. Meanwhile, we can also get the appropriate values of $R$ and $D_{2}$ , that is, $R=50$ mm and $D_{2}=16\text{ mm}$ . Under this condition, the minimum output torque of the HyA is 531.1 Nm, the size of h is 130.5 mm, and w is 145 mm.

3.1. Manufacture strategy of the arc shell

Figure 6.

Manufacturing diagram of the HyA.

After conceptual and parameter design, we found that there was a major manufacturing problem in the detailed design, whereby the inner surface of the arc shell could not be machined by a milling machine in some places, as shown in the blue circle and red circle areas of Fig. 6. The red, blue, and black arrows in Fig. 6 represent the milling cutter feed path. Thus, we devised a split manufacturing strategy, as shown in Fig. 7(b), where we divided the arc shell into upper and lower parts, and then manufactured them separately. Figure 7(b) shows the manufacturing process of the lower part, where the red, blue, black, and green arrows represent the milling cutter feed path. Finally, bolts were used to assemble the upper and lower parts together. Figure 8 shows a 3D diagram of the HyA.

Figure 7.

Splitting manufacturing strategy diagram.

Figure 8.

3D diagram of the HyA. 1: Hydraulic oil port, 2: back cover, 3: swing bar, 4: sliding bar, 5: movable rod, 6: shaft, 7: front cover, 8: lower housing, 9: upper housing, and 10: sealings.

We have listed a comparison of the HyA and other rotary HyAs in Table II. It can be seen that the torque to weight ratio of our HyA is almost three times that of other products, which means that the weight of a robot joints with the HyA will be greatly reduced. In addition, the HyA can provide a greater joint rotation angle than the other products.

3.2. Mechanical analysis

The movable rod is subjected to the pressure of the hydraulic oil and completes circular motion under the constraint of the arc shell. An external force also acts on the movable rod, suggesting that the force environment of the movable rod is extreme. Here, we carried out a force analysis of the movable rod. As shown in Fig. 9, the hydraulic oil pressure, $P_{s}=21$ MPa, acts on the B plane of the movable rod, where $F_{m}$ is the equivalent force of $P_{s}$ . We decomposed $F_{m}$ into $F_{m1}$ and $F_{m2}$ , where $F_{m1}$ pushes the movable rod to rotate and $F_{m2}$ pushes the movable rod to squeeze the arc shell, which will produce extrusion stress and sliding friction, as shown in the small red circle in Fig. 9. Similarly, extrusion stress and sliding friction will also occur in the big red circled area in Fig. 9.

We used the finite element method to conduct a finite element analysis, which is used to calculate the stress and deformation in detail. The results are shown in Fig. 10. The results show that the maximum stress of the arc shell (7075 aluminum alloy) and movable rod (40Cr) is under the maximum stress that the material can withstand. The total deformation of the arc shell and movable rod is small. Figure 10 also shows the extrusion stress between the movable rod and arc shell. The maximum extrusion stress was 258.7 MPa, which occurs in the contact area between the front cover and the movable rod. Into this area, we inserted a sealing groove, which leads to a reduction in the contact area between the front cover and the movable rod. The blue circle area of Fig. 11 shows the contact between the movable rod and arc shell, where the extrusion stress of this area was almost 100 MPa. The extrusion stress, shown in the blue and red circles of Fig. 11, is moderately large, and there is contact sliding friction. Therefore, we considered anodizing the inner surface of the arc shell to increase its hardness and wear resistance and treated the front cover the same way. Meanwhile, we treated the surface of the movable rod with nitridation technique.

Finally, the assembled HyA is shown in Fig. 12, and the weight of the HyA is 1.93 kg.

4.1. Dynamics model of the HyA

Like other HyAs, the HyA is controlled by a servo valve to realize servo control, as shown in Fig 13. We establish the dynamic model of the servo valve controlling HyA. First, we make the following assumptions: (1) all connecting pipelines are short and thick, and the hydraulic oil weight, friction pressure loss, and pipeline dynamics are ignored; (2) the hydraulic oil flows evenly, and the oil temperature and elastic modulus are constant; and (3) the external leakage is zero, and the internal leakage is a laminar flow.

Table II.

Comparison of HyA and other rotary HyAs.

Figure 9.

Force analysis of the HyA.

We divide the dynamics model into two parts according to the right and left movement of the servo valve core. The core of the servo valve moves to the right ( $X_{v}\gt 0$ ), which leads to the movable rod extending counterclockwise. The hydraulic oil flows through servo valve $q_{r1}$ , and $q_{r2}$ is equal to that flowing into the two cavities of the HyA, that is, $q_{1l}$ and $q_{2l}$ . $q_{r1}$ and $q_{r2}$ can be described by Eq. (7):

(7) \begin{align}\begin{cases} q_{r1}=C_{d}wX_{v}\sqrt{\dfrac{2}{\rho }\left(P_{s}-P_{1}\right)}\\[14pt] q_{r2}=C_{d}wX_{v}\sqrt{\dfrac{2}{\rho }P_{2}} \end{cases}\end{align}

where $C_{d}$ is the flow coefficient of the servo valve, $w$ is the area gradient of the servo valve, $X_{v}$ is the displacement of the servo valve core, $\rho$ is the hydraulic oil density, $P_{1}$ is the pressure acting on the big head of the movable rod, and $P_{2}$ is the pressure acting on the small head of the movable rod.

$q_{1l}$ and $q_{2l}$ can be obtained from the volume conservation of hydraulic oil, as described by Eq. (8):

(8) \begin{align}\begin{cases} q_{1l}=C_{i}\!\left(P_{1}-P_{2}\right)+\dfrac{V_{1}}{\beta _{e}}\dfrac{dP_{1}}{dt}+\dfrac{dV_{1}}{dt}\\[15pt] q_{2l}=C_{i}\!\left(P_{1}-P_{2}\right)-\dfrac{V_{2}}{\beta _{e}}\dfrac{dP_{2}}{dt}+\dfrac{dV_{2}}{dt} \end{cases}\end{align}

where $C_{i}$ is the internal leakage coefficient, $V_{1}$ is the cavity volume of the big head of the movable rod, $V_{2}$ is the cavity volume of the small head of the movable rod, and $\beta _{e}$ is the elastic modulus of the hydraulic oil.

The volume, $\textit{V}$ , of the hydraulic oil of the arc shell can be described by Eq. (9):

(9) \begin{equation}V=AR\theta\end{equation}

where $\theta$ is the rotation angle.

Figure 10.

Finite element analysis results of arc shell and moveable rod. (a) Total stress of the arc shell, (b) total deformation of the arc shell, (c) total stress of the movable rod, and (d) total deformation of the movable rod, all at 210 bar.

Figure 11.

Finite element analysis results of the extrusion stress between the arc shell and moveable rod.

Figure 12.

Assembled HyA.

Figure 13.

Dynamic model of the servo valve controlling the HyA.

By combining the above equations, we can get Eq. (10):

(10) \begin{align}\begin{cases} q_{1l}=C_{i}\left(P_{1}-P_{2}\right)+\dfrac{V_{1}}{\beta _{e}}\dfrac{dP_{1}}{dt}+A_{1}R\dfrac{d\theta }{dt}\\[12pt] q_{2l}=C_{i}\left(P_{1}-P_{2}\right)-\dfrac{V_{2}}{\beta _{e}}\dfrac{dP_{2}}{dt}+A_{2}R\dfrac{d\theta }{dt} \end{cases}\end{align}

where $A_{1}$ is the area of the big head of the movable rod and $A_{2}$ is the area of the small head of the movable rod.

Next, we set $V_{1}=V_{2}=V_{0}$ , where $V_{0}$ is the half total volume of the arc shell. Then, we define the load flow, $Q_{L1}$ (Eq. (11)) and the load pressure, $P_{L}$ , (Eq. (12)) as:

(11) \begin{equation}Q_{L1}=\frac{q_{1}+nq_{2}}{1+n^{2}}\end{equation}

(12) \begin{equation}P_{L}=\frac{T_{ow}}{A_{1}}=R\left(P_{1}-nP_{2}\right)\end{equation}

where $n=\dfrac{A_{2}}{A_{1}}$ and $T_{ow}$ is the output torque of the shaft.

Then, by combining the various terms, we get Eq. (13):

(13) \begin{equation}Q_{L1}=\frac{C_{i}\left(P_{1}-P_{2}\right)\left(1+n\right)}{1+n^{2}}+A_{1}R\frac{d\theta }{dt}+\frac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}\frac{dP_{L}}{dt}\end{equation}

Similarly, we deduce $q_{l1}$ , $q_{l2}$ , and $Q_{L2}$ when the servo valve core moves to the left ( $X_{v}\lt 0$ ), and the movable rod of the HyA retracts clockwise as:

(14) \begin{align}\begin{cases} q_{l1}=C_{d}wX_{v}\sqrt{\dfrac{2}{\rho }P_{1}} \\[14pt] q_{l2}=C_{d}wX_{v}\sqrt{\dfrac{2}{\rho }\left(P_{s}-P_{2}\right)} \end{cases}\end{align}

(15) \begin{equation}Q_{L2}=\frac{C_{i}\left(P_{1}-P_{2}\right)\left(1+n\right)}{1+n^{2}}+A_{1}R\frac{d\theta }{dt}-\frac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}\frac{dP_{L}}{dt}\end{equation}

If we linearize $Q_{L1}$ and $Q_{L2}$ , we get Eq. (16):

(16) \begin{equation}Q_{L}=\begin{cases} Q_{L1}=q_{r1}=K_{q}X_{v}-K_{c}P_{L}\\[5pt] Q_{L2}=q_{l2}=K_{q}^{i}X_{v}-K_{c}^{i}P_{L} \end{cases}\end{equation}

where $K_{rq}=C_{d}w\sqrt{\frac{2}{\left(1+n^{3}\right)\rho }\left(P_{S}-\frac{P_{L}}{R}\right)}$ , $K_{rc}=\frac{C_{d}wX_{v}}{R}\sqrt{\frac{1}{2\left(1+n^{3}\right)\rho \left(P_{S}-\frac{P_{L}}{R}\right)}}$ , $K_{lq}=C_{d}w\sqrt{\frac{2}{\left(1+n^{3}\right)\rho }\left(nP_{S}+\frac{P_{L}}{R}\right)}$ , and $K_{lc}=\frac{C_{d}wX_{v}}{R}\sqrt{\frac{1}{2\left(1+n^{3}\right)\rho \left(nP_{S}+\frac{P_{L}}{R}\right)}}$ .

The load balance equation is described as Eq. (17):

(17) \begin{equation}P_{L}A_{1}R=J\frac{d^{2}\theta }{dt}+C\frac{d\theta }{dt}+K\theta +T_{f}\end{equation}

where $J$ is the equivalent inertia, $C$ is the equivalent damping, $K$ is the equivalent spring stiffness, $T_{f}$ is the equivalent load torque, and $\theta$ is the rotation angle.

Next, we combine Eqs. (8), (10), (13), (15), (16), and (17) and perform a Laplace transform on them. Finally, we get Eq. (18):

(18) \begin{align} \begin{cases} {\unicode[Arial]{x03B8}} _{r}=\dfrac{A_{1}K_{rq}X_{v}-T_{f}\left(\dfrac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}S+\dfrac{1+n}{1+n^{3}}\dfrac{C_{i}}{R}+K_{rc}\right)}{k_{a1}S^{3}+k_{a2}S^{2}+k_{a3}S+k_{a4}},X_{v}\gt 0\\[12pt] {\unicode[Arial]{x03B8}} _{l}=\dfrac{A_{1}K_{lq}X_{v}-T_{f}\left(\dfrac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}S+\dfrac{1+n}{1+n^{3}}\dfrac{C_{i}}{R}+K_{lc}\right)}{k_{a11}S^{3}+k_{a22}S^{2}+k_{a33}S+k_{a44}},X_{v}\lt 0 \end{cases}\end{align}

where $k_{a1}=k_{a11}=\frac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}J$ , $k_{a2}=\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}J+K_{rc}J+\frac{V_{0}C}{R\beta _{e}\left(1+n^{2}\right)}$ , $k_{a22}=\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}J+K_{lc}J+\frac{V_{0}C}{R\beta _{e}\left(1+n^{2}\right)}$ , $k_{a3}=\frac{V_{0}K}{R\beta _{e}\left(1+n^{2}\right)}+\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}C+K_{rc}C+A_{1}^{2}R$ , $k_{a33}=\frac{V_{0}K}{R\beta _{e}\left(1+n^{2}\right)}+\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}C+K_{lc}C+A_{1}^{2}R$ , $k_{a4}=\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}K+K_{rc}K$ , $k_{a44}=\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}K+K_{lc}K$ , and $S$ is the Laplace operator.

When $C, K$ can be neglected, Eq. (18) is simplified to the standard Eq. (19). Note that Eq. (19) only describes the dynamics of the case $V_{1}=V_{2}=V_{0}$ , which makes it sufficient to design the control algorithm:

(19) \begin{align} \begin{cases} {\unicode[Arial]{x03B8}} _{r}=\dfrac{A_{1}K_{rq}X_{v}-T_{f}\left(\dfrac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}S+\dfrac{1+n}{1+n^{3}}\dfrac{C_{i}}{R}+\dfrac{C_{e}}{1+n^{2}}\dfrac{1}{R}+K_{c}\right)}{S\left(\dfrac{S^{2}}{w_{h1}^{2}}+\dfrac{2\varepsilon _{h1}}{w_{h1}}S+1\right)},X_{v}\gt 0\\[30pt] {\unicode[Arial]{x03B8}} _{l}=\dfrac{A_{1}K_{lq}X_{v}-T_{f}\left(\dfrac{V_{0}}{R\beta _{e}\left(1+n^{2}\right)}S+\dfrac{1+n}{1+n^{3}}\dfrac{C_{i}}{R}+\dfrac{C_{e}}{1+n^{2}}\dfrac{1}{R}+K_{c1}\right)}{S\left(\dfrac{S^{2}}{w_{h2}^{2}}+\dfrac{2\varepsilon _{h2}}{w_{h2}}S+1\right)},X_{v}\lt 0 \end{cases}\end{align}

where $w_{h1}=w_{h2}=\sqrt{\frac{k_{a3}}{k_{a1}}}=\sqrt{\frac{k_{a33}}{k_{a11}}}=\sqrt{\frac{R\beta _{e}\left(1+n^{2}\right)\left(A_{1}^{2}R\right)}{V_{0}J}}$ , $\varepsilon _{h1}=\frac{k_{a2}}{2\sqrt{k_{a1}k_{a3}}}=\frac{\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}J+K_{rc}J}{2\sqrt{\frac{V_{0}}{\beta _{e}\left(1+n^{2}\right)}J\left(A_{1}^{2}\right)}}$ , and $\varepsilon _{h2}=\frac{k_{a22}}{2\sqrt{k_{a11}k_{a33}}}=\frac{\frac{1+n}{1+n^{3}}\frac{C_{i}}{R}J+K_{lc}J}{2\sqrt{\frac{V_{0}}{\beta _{e}\left(1+n^{2}\right)}J\left(A_{1}^{2}\right)}}$ .

We can get the minimum natural frequency, $w_{h\min }$ , of the HyA control system from Eq. (19), as described as Eq. (20), where $\theta$ is the maximum rotation angle of the HyA:

(20) \begin{equation}w_{h\min }=\left(1+\sqrt{n}\right)\sqrt{\frac{RA_{1}\beta _{e}}{J\theta }}\end{equation}

Equation (20) tells us that the minimum natural frequency of the HyA control system is affected by the parameters $R, A_{1},A_{2}, \beta _{e},\theta$ , and $\textit{J}$ . So, we can increase the natural frequency by increasing the mechanical parameters of $R, A_{1},A_{2}$ and reduce the parameters of $J$ and $\theta$ , which will increase the response speed of the HyA control system.

The stability of the HyA control system is affected by the parameters $\varepsilon _{h1}$ and $\varepsilon _{h2}$ . We can see that $\varepsilon _{h1}$ and $\varepsilon _{h2}$ are affected by $A_{1},A_{2},C_{i},K_{rc}$ , and $K_{lc}$ so that we can appropriately increase the internal leakage by the mechanical design to increase the stability of the HyA control system. But, doing so will increase the energy consumption of the system. Therefore, we can design a control algorithm to ensure the stability of the HyA control system.

Similarly, we can deduce the output torque (Eq. (21)) of the HyA under the condition of $C, K\approx 0$ . It can be seen that Eq. (21) has similar properties to Eq. (19). In other words, we can design similar control algorithms for the position control and torque control for the HyA as:

(21) \begin{align}\begin{cases} \mathrm{T}_{r}=JS^{2}{\unicode[Arial]{x03B8}} _{r}+\mathrm{T}_{f}\\[5pt] \mathrm{T}_{l}=JS^{2}{\unicode[Arial]{x03B8}} _{l}+\mathrm{T}_{f} \end{cases}\end{align}

4.2. Design of the position control and torque control algorithms

Equations (19) and (21) have similar transfer function forms. We design a simple proportional integral (PI) controller using classical control theory, described as Eqs. (22) and (23):

(22) \begin{equation}u_{\theta }=K_{p}\left(\theta _{o}-\theta _{m}\right)+K_{i}\int \left(\theta _{o}-\theta _{m}\right)\end{equation}

(23) \begin{equation}u_{T}=K_{Tp}\left(T_{ow}-T_{owm}\right)+K_{Ti}\int \left(T_{ow}-T_{owm}\right)\end{equation}

where $u_{\theta }$ is the control input of the position control of the HyA, $u_{T}$ is the control input of the torque control of the HyA, $K_{p}$ is the scale factor of the PI position controller of the HyA, $K_{Tp}$ is the scale factor of the PI torque controller of the HyA, $K_{i}$ is the integral coefficient of the PI position controller of the HyA, $K_{Ti}$ is the integral coefficient of the PI torque controller of the HyA, $\theta _{o}$ is the expectation angle of the shaft, $\theta _{m}$ is the measured angle of the shaft, $T_{ow}$ is the expectation torque of the shaft, and $T_{owm}$ is the measured torque of the shaft.

Figure 14 shows the structure of the control system of the HyA, and Fig. 15 shows the control diagram of the HyA with the PI controller.

Figure 14.

Structure of the control system of the HyA.

Figure 15.

Control diagram of the HyA with the PI controller.

Figure 16.

Basic test system for the HyA.

5.1. Starting pressure experiment

We tested the starting pressure of the HyA in two movement directions; the results are shown in Table III.

We can see from Table III that the starting pressure of extending movement is smaller than that of retracting movement, which is caused by the area of the big head of the moveable rod being larger than that of the small head of the moveable rod. The average starting pressure of extending movement is 0.26 MPa, and that of retracting movement is 0.39 MPa. The rated pressure of the HyA is 21 MPa, and the percentage of the starting pressure in the two movement directions to the rated pressure is 1.24% and 1.86%, which indicates that the starting pressure in the two movement directions is very small.

Table III.

Results of the starting pressure test.

Figure 17.

Pressure sensor.

Figure 18.

Hydraulic oil in the measuring cup.

5.2. Rated pressure

We adjusted the manual pressure regulating valve to make the pressure of the HyA reach 210 bar, as shown in Fig. 17. Under this condition, the HyA could swing normally without external leakage, which shows that the HyA can work normally under the rated pressure.

Table IV.

Results of the internal leakage test of the HyA.

Figure 19.

Movement test of the HyA.

Figure 20.

HyA servo experimental system.

5.3. Leakage

In order to test the internal leakage of the HyA, we exposed the low pressure oil port, as shown in Fig. 16. A timer and measuring cup were used to measure the internal leakage of HyA under the rated pressure of 210 bar, as shown in Fig. 18. The measurement results are shown in Table IV. The leakage recorded in Table IV was timed for 90 s. Figure 18 shows an example measurement. Since the seal on the movable rod is a two-way seal, there is a very little difference in the internal leakage of the movable rod when extending and retracting, with average leakages of 8.8 mL/min and 9 mL/min, respectively. It can be seen that the internal leakage of the HyA was very small, which can be considered as effectively negligible. We observed by a visual inspection that there was no external hydraulic oil leakage of the HyA at 210 bar, which indicated that the external leakage was zero.

Figure 21.

Output torque of the HyA under different pressures. (a) Output torque of the HyA when retracting and (b) output of the HyA when extending.

Figure 22.

Results of 0.5 Hz sinusoidal signal tracking.

Figure 23.

Results of 1 Hz sinusoidal signal tracking.

5.4. Movement

We switched the solenoid directional valve to change the movement direction of the HyA, as shown in Fig. 19. The HyA extended when the indicator light on the solenoid directional valve was on, and retracted when the indicator light on solenoid directional valve was off, as shown by the small red circle in Fig. 19. The movements of the HyA are as shown in the big red circle in Fig. 19. In testing the HyA’s movement, the moveable rod of the HyA was pushed by the hydraulic oil to move smoothly, which made the shaft rotate.

5.5. Servo control

We established a HyA servo experimental system based on the principle shown in Fig. 14, as displayed in Fig. 20. As we could not produce a suitable torque load, we simply tested the static output torque of the HyA under different pressures and measured the subsequent positions of the HyA servo.

Figure 24.

Results of 2 Hz sinusoidal signal tracking.

We can see from Fig. 21 that the actual output torque of the HyA is less than the calculated torque, and there are fluctuations in the actual output torque. At the maximum output torque, the error is about 7.3% in the retraction direction and 4.5% in the extending direction. As shown in the mechanical analysis in Section 3.2, there is sliding friction between the movable rod and the arc shell, which makes the actual output torque of the HyA less than the calculated torque. We also found that the pressure of the oil source was unstable, and we were unable to keep the pressure stable by adjusting the manual pressure regulating valve, which makes the actual output torque of the HyA fluctuate.

Next, we carried out sinusoidal tracking experiments. The results, as shown in Figs. 22, 23, and 24, show that there is a phase difference at different frequencies, and the phase difference increased with an increase of frequency, 6.84 degrees at 0.5 Hz, 11.16 degrees at 1 Hz, and 19.44 degrees at 2 Hz. The phase difference in the retraction direction is larger than that in the extension direction, which is caused by the difference between $K_{rq}$ and $K_{lq}$ in Eq. (19). When a phase difference is not considered, the amplitude tracking error is very small, 0.8% at 0.5 Hz, 1% at 1 Hz, and 1.5% at 2 Hz.