4.1.
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Dynamics model of the HyALike 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.
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.
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)
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\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}
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