Analysis of Underlapped Symmetrically Ported Valve-Controlled Asymmetric Cylinder Drive

Abstract

The valve-controlled cylinder drive system is the most common type among hydraulic applications. Nonlinear behaviour in such systems is inevitable when the valve spool is around its null position. We utilised the component linking method to investigate the nonlinearities in a Moog valve-controlled asymmetric cylinder drive system by simulation in Fortran, in which a generalised concept is introduced and validated by comparing to the experimental results. An X factor is proposed in the generalised concept to describe the asymmetric cylinder state, which is a constant when the cylinder is extending or retracting, but numerically calculated when the valve spool is in the underlap region. This analytical solution is approximately 200 times more computationally efficient than the numerical solution method. This paper utilises the component linking method to simulate the Moog valve-controlled asymmetric cylinder drive system in Matlab Simulink, and proposes an analytical solution for the X factor when the valve spool is in the underlap region. This analytical solution is approximately 200 times more computationally efficient than the numerical solution method.

1. Introduction

Hydraulic power systems [1] are widely used in industrial manufacture, building construction, port transportation, funfairs, etc. They use fluid to achieve energy transmission without a rigid joint. Hydraulic power is able to produce massive force and torque within a compact and lightweight package, and conventional rigid components like gears and joints can be avoided.

Hydraulic linear actuators are mainly of the symmetric cylinder or asymmetric cylinder type. The symmetric cylinder type has equal piston areas and the asymmetric type (also called single rod) cylinder comes with different piston areas. The symmetric cylinder is usually utilised for high-performance applications, due to its linear performance with a symmetric ported control valve. This type of cylinder is widely applied in aerospace and military contexts for its high performance and dynamic response [2]. However, the asymmetric cylinder [3] drive constitutes 80% of contemporary hydraulic applications [4], such as cranes, injection moulding systems, hydraulic press machines, etc. The asymmetric cylinder drive is usually controlled by a symmetric ported valve, meaning an unequal flow rate will be delivered into the cylinder chambers with the same valve opening in the extending or retracting state, which means system parameters do not remain the same after the valve crosses its null position [5]. Leaney [6] studied these nonlinearities, but how the system performs when the valve is in the underlap region is not explicitly explained. Viersma [7] indicates that pressure jumps will occur when the valve spool is around its null position, causing ‘implosion’ or ‘explosion’ of oil owing to its compressibility, so that smooth operation around the valve’s null position is impossible. Therefore, this article conducted a nonlinear study on valves in an underlap state and proposed an analytical solution for X factor when the valve spool is in the underlap region to improve computational efficiency.

2. Theory

A test rig of a four-way servovalve-controlled asymmetric cylinder hydraulic system was constructed by Leaney [8]. The system is a typical symmetric ported valve-controlled asymmetric cylinder system, which is depicted in Figure 1. The Moog series 76 model 102 four-way valve is used to control the asymmetric cylinder system to achieve the desired output.

Leaney [6] simulated his test rig with the ‘Component Linking’ method, which models each component such as the valve, actuator/cylinder, and load, then combines them to perform the whole-system simulation. This method aims at simplifying the ‘Small Perturbation’ model [9], which utilised a single transfer function to describe a whole hydraulic system. However, as not all parameters of a hydraulic system remain constant during operation and a single transfer function is mainly designed for a linear time-invariant (LTI) system, a single transfer function [10] is inadequate to capture the behaviours of a hydraulic system. Modelling the components separately is easier for the analysis and adjustment of the system parameters.

The components of the asymmetric cylinder system are connected by the concept of ‘power bonding’, in other words, the components are connected by the power transfer; such a concept is still utilised nowadays for the hydraulic system of a 50-wheel loader [11]. In Leaney’s paper [8], three components are the major concerns, which are the valve, the actuator, and the load model. Their power transfers are indicated in Figure 2. The voltage is input to the valve model to control the valve opening, after which the flow passes through the valve and drops to a certain value. Pressure and flow are delivered to the actuator model, which outputs the force/torque and velocity. Finally, the load model outputs the velocity and force.

2.1. Valve Model

The control valve (electrohydraulic valve) is a Moog 76 model 102 flow control four-way servovalve. If current is applied to the armature coil in the torque motor, the T-shaped flapper rotates slightly. Assuming the flapper rotates clockwise, the offset between the left nozzle and flapper decreases, which leads to the decrease of flow passing through the left nozzle and causes the pressure to increase in the left control line. A similar process occurs in the right control line, leading to a decrease of pressure.

The pressure difference in the control line moves the spool, therefore bends the feedback spring to balance the torque generated by the torque motor, so that the spool displacement is dependent on the rotation of the torque motor; furthermore, the spool displacement is dependent on the current.

Appropriate transfer functions, Equations (1) and (2), are given by Moog [12]. The expressions are linear, empirical relationships that approximate the responses of the actual servovalve. For the frequency range below 50 Hz, the valve dynamic can be approximated by a first-order transfer function given below [12]:

$$\frac{Q\left(s\right)}{I\left(s\right)}=K\frac{1}{1+{\tau }_{s}s}$$

(1)

where $K$ is the flow gain $\left[\raisebox{1ex}{${\mathrm{m}}^3$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s} · \mathrm{amp}$}\right.\right]$ and ${\tau }_s$ is the servovalve time constant $\left[\sec \right]$.

A second-order transfer function is necessary to represent a wider frequency range response as given below [12]:

$$\frac{Q\left(s\right)}{I\left(s\right)}=K\frac{1}{1+\left(\frac{2\zeta }{{\omega }_n}\right)s+{\left(\frac{s}{{\omega }_n}\right)}^2}$$

(2)

where ${\omega }_n$ is the natural frequency [$\mathrm{rad}/\mathrm{s}$], and $\zeta$ is the damping ratio. In a limit cycle oscillation study of a low-frequency phenomenon, a higher-order valve model does not show significant difference compared to a lower-order model [13]. Hisa [14] stated that a reduced-order transfer function can be used to approximate a higher-order transfer function, and a first-order model is used as a reference model by Persson et al. [15].

As all of the valve operational frequencies in the industrial sector are much lower than 50 Hz, the first-order transfer function Equation (1) is chosen for the valve model. As the displacement of the spool, $X\left(s\right)$, controls the flow rate of the valve, Equation (1) can be modified as follows:

$$\frac{X_s\left(s\right)}{I\left(s\right)}=K_s\frac{1}{1+{\tau }_{s}s}$$

(3)

where $K_s$ is the gain of the system [$\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{amp}$}\right.$].

The current is generated by an amplifier. Assuming the amplifier is ideal, then current is proportional to the input voltage as given by Equation (4):

$$\frac{I\left(s\right)}{V_v\left(s\right)}=K_A$$

(4)

where $K_A$ is the gain of voltage to current [$\raisebox{1ex}{$\mathrm{amp}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{volt}$}\right.$]. Hence the valve dynamics are represented by:

$$\frac{X_s\left(s\right)}{V_v\left(s\right)}=\frac{K_A{K}_s}{1+{\tau }_{s}s}$$

(5)

A block diagram for this valve model is depicted in Figure 3. Valve model block diagram.

2.2. Flow Gain

The flow gain is the gain that describes the relationship between the valve opening and the oil flow across it. In order to interpret it clearly, a physical layout of the four-way valve-controlled [16] asymmetric cylinder is presented in Figure 4. Physical layout of the four-way valve-controlled asymmetric cylinder [6].

$Q_1$ is the flow rate that is delivered into the piston side chamber, $V_1$ is the volume of fluid that is trapped in the piston chamber, $A_1$ is the piston side area of the cylinder, and the subscript 2 indicates corresponding parameters in the rod side chamber. $P_s$ is the supply pressure, and $P_T$ is the return line pressure of the oil tank, which is zero. $A_V$ is the opening area of the corresponding port of the valve, and subscripts 1, 2, 3, and 4 indicate different ports of the valve which are depicted in Figure 4. Physical layout of the four-way valve-controlled asymmetric cylinder [6].

$x_P$ is the cylinder displacement. $Q_{P1}$ is the flow passing through the $A_{V1}$ port, $Q_{1T}$ is the flow passing through the $A_{V2}$ port, $Q_{2T}$ is the flow passing through the $A_{V3}$ port, and $Q_{P2}$ is the flow passing though the $A_{V4}$ port.

The flow delivered into the chambers can be described as:

$$Q_1=Q_{P1}-Q_{1T}=C_D\sqrt{\frac{2}{\rho }}\left[A_{V1}{\left(P_s-P_1\right)}^{\frac{1}{2}}-A_{V2}{\left(P_1\right)}^{\frac{1}{2}}\right]$$

(6)

$$Q_2=Q_{2T}-Q_{P2}=C_D\sqrt{\frac{2}{\rho }}\left[A_{V3}{\left(P_2\right)}^{\frac{1}{2}}-A_{V4}{\left(P_s-P_2\right)}^{\frac{1}{2}}\right]$$

(7)

For a symmetric ported valve, $A_{V1}=A_{V3}$ and $A_{V2}=A_{V4}$, so that Equations (6) and (7) can be simplified as:

$$Q_1=Q_{P1}-Q_{1T}=C_D\sqrt{\frac{2}{\rho }}\left[A_{V1}{\left(P_s-P_1\right)}^{\frac{1}{2}}-A_{V2}{\left(P_1\right)}^{\frac{1}{2}}\right]$$

(8)

$$Q_2=Q_{2T}-Q_{P2}=C_D\sqrt{\frac{2}{\rho }}\left[A_{V1}{\left(P_2\right)}^{\frac{1}{2}}-A_{V2}{\left(P_s-P_2\right)}^{\frac{1}{2}}\right]$$

(9)

These two equations can be applied when the valve is in the extending, retracting, and underlap region, the underlap region being the valve spool position where all port opening areas are non-zero. Even for a zero-lapped valve, due to the manufacturing tolerances [17], there will be a small underlap when the spool is placed in the null position. Figure 5 depicts the theoretical and actual situation for a nominally zero-lapped valve.

The manufacturing tolerances can be identified as:

i.

Radial clearance between spool and sleeve.

ii.

Tolerance on lap.

Such tolerance [1] will lead to variation of flow gain when the valve is in its null region. The flow gain model block diagram is depicted in Figure 6.

2.3. The Actuator Model

The actuator model connects the valve and load model, which accepts the information of flow and pressure as input, then outputs force to the load model. The types of hydraulic actuators are mainly symmetric cylinder, asymmetric cylinder, and hydraulic motor. Due to the focus of this research, the asymmetric cylinder is the only one under analysis.

Whether the asymmetric cylinder is in the extending or retracting state, the flow in its chamber consists of input flow, displacement flow, and compressibility flow. However, under different operational states, the flow balance equations in the different chambers are not the same.

For the cylinder extending state:

$$Q_1=Q_{D1}+Q_{C1}$$

(10)

$$Q_2=Q_{D2}-Q_{C2}$$

(11)

For the cylinder retracting state:

$$Q_2=Q_{D2}+Q_{C2}$$

(12)

$$Q_1=Q_{D1}-Q_{C1}$$

(13)

where

$$Q_{D1}=v · A_1 Q_{D2}=v · A_2$$

(14)

$$Q_{C1}=\frac{s{P}_1 · V_1}{B} Q_{C2}=\frac{s{P}_2 · V_2}{B}$$

(15)

$\mathrm{A}$ is the area of the cylinder piston side area or rod side area [${\mathrm{m}}^2$], $\mathit{v}$ is the cylinder velocity [$\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$], $\mathit{V}$ is the cylinder chamber volume [${\mathrm{m}}^3$], $\mathit{P}$ is the cylinder chamber pressure [$\raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^2$}\right.$] and B is the fluid Bulking Modulus [$\raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^2$}\right.$]. The subscript 1 and $2$ indicate the piston side chamber and rod side chamber, respectively.

A generalised definition of these equations is proposed by Leaney [8], which uses effective parameters to describe the system behaviours rather than analyse them separately in different cylinder chambers. The generalised equations are depicted as follows:

$$\begin{gathered} P_{LE}=\frac{P_1{A}_1-P_2{A}_2}{A_E} Q_{LE}=\frac{Q_1+Q_2}{2} \\ {A}_E=\frac{A_1+A_2}{2} Q_{DE}=\frac{Q_{D1}+Q_{D2}}{2} \\ {Q}_{CE}=\frac{Q_{C1}+Q_{C2}}{2}=Q_{LE}-Q_{DE} \end{gathered}$$

(16)

Substituting these effective parameters into the flow and pressure equations reveals the following:

For cylinder extending ($A_{V2}$ = 0 and $A_{V1}$ > 0), the service line pressures can be rewritten as follows [8]:

$${\left(P_1\right)}_{ext}=\frac{P_s+{\gamma }^2\frac{1}{2}\left(1+\gamma \right)P_{LE}}{1+{\gamma }^3}$$

(17)

$${\left(P_2\right)}_{ext}=\frac{\gamma P_s-\frac{1}{2}\left(1+\gamma \right)P_{LE}}{1+{\gamma }^3}$$

(18)

For cylinder retracting ($A_{V2}$ > 0 and $A_{V1}$ = 0), the service line pressures can be rewritten as follows [8]:

$${\left(P_1\right)}_{ret}=\frac{{\gamma }^2{P}_s+{\gamma }^2\frac{1}{2}\left(1+\gamma \right)P_{LE}}{1+{\gamma }^3}$$

(19)

$${\left(P_2\right)}_{ret}=\frac{{\gamma }^3{P}_s-\frac{1}{2}\left(1+\gamma \right)P_{LE}}{1+{\gamma }^3}$$

(20)

where $\gamma$ is the area ratio of the cylinder. Leaney [8] introduced the $X$ factor to unify the expression for $P_1$, $P_2$ and utilised generalised concepts in the component linking method. The $X$ factor value depends on the region of the spool operation: $X$ = 1 for cylinder extending state and $X$ = ${\gamma }^2$ for cylinder retracting state. The service line pressure $P_1$ and $P_2$ can be written as follows:

$$P_1=\frac{X{P}_s+{\gamma }^2\frac{1}{2}\left(1+\gamma \right)P_{LE}}{1+{\gamma }^3}$$

(21)

$$P_2=\frac{\mathrm{X}\gamma P_s-\frac{1}{2}\left(1+\gamma \right)P_{LE}}{1+{\gamma }^3}$$

(22)

$$P_{LE}=\left(\frac{4B}{s{V}_{TE}}\right)Q_{CE}$$

(23)

where $\gamma$ is the area ratio, $V_{TE}$ is effective total volume [${\mathrm{m}}^3$], s is the Laplace operator, and P_LE is average pressure. When X = 1 for cylinder extending state, X = γ² for cylinder retracting state.

$$V_{TE}=\frac{2A_E}{A_1}\left[\frac{V_1+{\left(\frac{A_2}{A_1}\right)}^2{V}_2}{1+{\left(\frac{A_2}{A_1}\right)}^3}\right]$$

(24)

The $X$ factor value is known and clear when the valve is outside the underlap region. However, when the valve spool is in the underlap region, the $X$ value is difficult to find explicitly; it can be numerically solved by an implicit Equation (25) [8] as shown below:

$$A_{v1}{\left(1+{\gamma }^3-X\right)}^{\frac{1}{2}}-A_{v2}{\left(X\right)}^{\frac{1}{2}}=\gamma \left[A_{v1}{\left(X\gamma \right)}^{\frac{1}{2}}-A_{v2}{\left(1+{\gamma }^3-X\gamma \right)}^{\frac{1}{2}}\right]$$

(25)

Now the effective load flow can be revealed [8]:

$$Q_{LE}=\frac{C_D}{\sqrt{2\rho }}\sqrt{\frac{{\left(1+\gamma \right)}^2}{1+{\gamma }^3}}\left[A_{v1}{\left(X\gamma P_s-\frac{1}{2}\left(1+\gamma \right)P_{LE}\right)}^{\frac{1}{2}}-A_{v2}{\left[\left(1+{\gamma }^3-X\gamma \right)P_s+\frac{1}{2}\left(1+\gamma \right)P_{LE}\right]}^{\frac{1}{2}}\right]$$

(26)

The actuator model block diagram is depicted in Figure 7.

2.4. The Load Model

The load model varies depending on the test rig setup. In order to simplify the analysis afterwards, the load setup is a simple asymmetric cylinder coupled with a load, which is placed on a surface, so that the load force balance can be depicted by Equation (27) and Figure 8:

$$P_1{A}_1-P_2{A}_2=P_{LE}{A}_E=m · a+friction force+F_{external}$$

(27)

where $m$ is the load mass [$\mathrm{kg}$] and $\mathrm{a}$ is the acceleration [$\mathrm{m}/{\mathrm{s}}^2$].

3. Materials and Methods

3.1. Numerical Solution of Chamber Pressure Behaviour in Underlap Region

Leaney [8] stated that the cylinder chamber pressures $P_1$ and $P_2$ must be numerically solved when the valve is in the underlap region, due to the derived implicit Equation (25). The chamber pressures are the power source to drive the cylinder, so that the system behaviours are decided by chamber pressures. The way the pressures perform when the valve is outside the underlap region is explicit. The chamber pressures’ behaviour in the underlap region are solved numerically as shown in Figure 9.

An input signal is set to drive the valve to move across the underlap region in a small time interval with a constant speed. The supply pressure is 70 bar and all the modelling parameters are the same as those in the Moog valve-controlled asymmetric cylinder system in the last section.

The chamber pressures of the asymmetric cylinder when the valve is in the underlap region are modelled, and they are numerically solved. The key part is the $X$ factor Equation (25) from [8]. The solving command in Matlab is fsolve and it uses the rust-region algorithm [14] to find the root of factor $X$ with given valve opening areas $A_{v1}$, $A_{v2}$ and area ratio $\gamma$. Numerically solving Equation (25) takes a lot of computing time; if the $X$ value can be analytically solved, simulation time cost can be massively reduced. A trial is given in the rest of this section.

3.2. Derivation of Analytical Solution for the Value of X Factor

Squaring both sides of Equation (25) leads to:

$$\begin{gathered} \frac{{A_{V1}}^2}{{\gamma }^2}\left(1+{\gamma }^3-X\right)+\frac{{A_{V2}}^{2}X}{{\gamma }^2}-\frac{2A_{V1}{A}_{V2}}{{\gamma }^2}{\left(X+{\gamma }^{3}X-X^2\right)}^{\frac{1}{2}} \\ ={A_{V1}}^{2}X\gamma +{A_{V2}}^2\left(1+{\gamma }^3-X\gamma \right)-2A_{V1}{A}_{V2}{\left(X\gamma +X{\gamma }^4-X^2{\gamma }^2\right)}^{\frac{1}{2}} \end{gathered}$$

(28)

Rearranged:

$$\begin{gathered} \left[\frac{A_{V1}}{2A_{V2}}\left(1+{\gamma }^3\right)-\frac{A_{V2}}{2A_{V1}}\left({\gamma }^2+{\gamma }^5\right)\right]-\left[\frac{A_{V1}}{2A_{V2}}\left(1+{\gamma }^3\right)-\frac{A_{V2}}{2A_{V1}}\left(1+{\gamma }^3\right)\right]X \\ ={\left[\left(1+{\gamma }^3\right)X-X^2\right]}^{\frac{1}{2}}-{\gamma }^2{\left[\left(\gamma +{\gamma }^4\right)X-{\gamma }^2{X}^2\right]}^{\frac{1}{2}}. \end{gathered}$$

(29)

Let $m$ be given as follows:

$$m=\frac{A_{V1}}{2A_{V2}}\left(1+{\gamma }^3\right)-\frac{A_{V2}}{2A_{V1}}\left({\gamma }^2+{\gamma }^5\right)=\left(1+{\gamma }^3\right)\left(\frac{A_{V1}}{2A_{V2}}-\frac{A_{V2}{\gamma }^2}{2A_{V1}}\right)$$

(30)

Let $n$ be given as follows:

$$n=\frac{A_{V1}}{2A_{V2}}\left(1+{\gamma }^3\right)-\frac{A_{V2}}{2A_{V1}}\left(1+{\gamma }^3\right)=\left(1+{\gamma }^3\right)\left(\frac{A_{V1}}{2A_{V2}}-\frac{A_{V2}}{2A_{V1}}\right)$$

(31)

Substituting $m$ and $n$ into Equation (31) reveals the following:

$$m-nX={\left[\left(1+{\gamma }^3\right)X-X^2\right]}^{\frac{1}{2}}-{\gamma }^2{\left[\left(\gamma +{\gamma }^4\right)X-{\gamma }^2{X}^2\right]}^{\frac{1}{2}}$$

(32)

Squaring both sides of Equation (32) leads to:

$$\begin{array}{ll}{m}^2+n^2{X}^2-2mnX& =\left(1+{\gamma }^3\right)X-X^2+{\gamma }^5\left(1+{\gamma }^3\right)X-{\gamma }^6{X}^2\\ & -2{\gamma }^2{\left[\gamma {\left(1+{\gamma }^3\right)}^2{X}^2-\gamma \left(1+{\gamma }^3\right)X^3-{\gamma }^2\left(1+{\gamma }^3\right)X^3+{\gamma }^2{X}^4\right]}^{\frac{1}{2}}\end{array}$$

(33)

Rearranging Equation (33) reveals the following:

$$\begin{array}{ll}\left(n^2+1+{\gamma }^6\right)X^2& -\left[2mn+\left(1+{\gamma }^5\right)\left(1+{\gamma }^3\right)\right]X+m^2\\ & =2{\gamma }^2{\left[\gamma {\left(1+{\gamma }^3\right)}^2{X}^2-\gamma \left(1+{\gamma }^3\right)X^3-{\gamma }^2\left(1+{\gamma }^3\right)X^3+{\gamma }^2{X}^4\right]}^{\frac{1}{2}}\end{array}$$

(34)

Let

$$\begin{gathered} p=n^2+1+{\gamma }^6 \\ q=2mn+\left(1+{\gamma }^5\right)\left(1+{\gamma }^3\right) \\ h=m^2 \end{gathered}$$

(35)

Equation (34) becomes the following:

$$p{X}^2-qX+h=2{\gamma }^2{\left[\gamma {\left(1+{\gamma }^3\right)}^2{X}^2-\gamma \left(1+{\gamma }^3\right)X^3-{\gamma }^2\left(1+{\gamma }^3\right)X^3+{\gamma }^2{X}^4\right]}^{\frac{1}{2}}$$

(36)

Squaring both sides of Equation (36) reveals the following:

$$\begin{array}{ll}{p}^2{X}^4+q^2{X}^2+h^2& -pq{X}^3+ph{X}^2-pq{X}^3-qhX+ph{X}^2-qhX\\ & =4{\gamma }^6{X}^4-4{\gamma }^5\left(1+\gamma \right)\left(1+{\gamma }^3\right)X^3+4{\gamma }^5{\left(1+{\gamma }^3\right)}^2{X}^2\end{array}$$

(37)

Rearranging the above equation reveals the following:

$$\begin{array}{ll}\left(p^2-4{\gamma }^6\right)X^4& +\left[4{\gamma }^5\left(1+\gamma \right)\left(1+{\gamma }^3\right)-2pq\right]X^3+\left[\left(2ph+q^2\right)-4{\gamma }^5{\left(1+{\gamma }^3\right)}^2\right]X^2\\ & -2qhX+h^2=0\end{array}$$

(38)

Let

$$\begin{gathered} a_4=p^2-4{\gamma }^6 \\ {a}_3=4{\gamma }^5\left(1+\gamma \right)\left(1+{\gamma }^3\right)-2pq \\ {a}_2=\left(2ph+q^2\right)-4{\gamma }^5{\left(1+{\gamma }^3\right)}^2 \\ {a}_1=-2qh \\ {a}_0=h^2 \end{gathered}$$

(39)

Thus Equation (38) becomes the following:

$$a_4{X}^4+a_3{X}^3+a_2{X}^2+a_{1}X+a_0=0$$

(40)

Let

$$\begin{gathered} a=\frac{a_3}{a_4},b=\frac{a_2}{a_4} \\ c=\frac{a_1}{a_4},d=\frac{a_0}{a_4} \end{gathered}$$

(41)

As a result, Equation (40) becomes the following:

$$X^4+a{X}^3+b{X}^2+cX+d=0$$

(42)

The value of the $X$ factor is a problem of finding the root of a quartic Equation (42).

3.3. Root Form and Identification

A general form of roots can be depicted as below in Equation (43):

$$\left\{\begin{array}{c}{X}_1=-\frac{a}{4}+\frac{{\left(U+V\right)}^{\frac{1}{2}}}{2}-\frac{{\left(2U-V-W\right)}^{\frac{1}{2}}}{2}\\ X_2=-\frac{a}{4}-\frac{{\left(U+V\right)}^{\frac{1}{2}}}{2}-\frac{{\left(2U-V+W\right)}^{\frac{1}{2}}}{2}\\ \begin{array}{c}{X}_3=-\frac{a}{4}+\frac{{\left(U+V\right)}^{\frac{1}{2}}}{2}+\frac{{\left(2U-V-W\right)}^{\frac{1}{2}}}{2}\\ X_4=-\frac{a}{4}-\frac{{\left(U+V\right)}^{\frac{1}{2}}}{2}+\frac{{\left(2U-V+W\right)}^{\frac{1}{2}}}{2}\end{array}\end{array}\right.$$

(43)

where

$$\begin{gathered} U=\frac{a^2}{4}-\frac{2b}{3} \\ V=\frac{L}{3}+\frac{Z_0}{3L} \\ W=\frac{\left(a^3-4ab+8c\right)}{4{\left(U+V\right)}^{\frac{1}{2}}} \end{gathered}$$

(44)

where

$$\begin{gathered} Z_0=b^2-3ac+12d \\ {Z}_1=2b^3-9abc+27a^{2}d+27c^2-72bd \\ L={\left[\frac{Z_1+{\left({Z_1}^2-4{Z_0}^3\right)}^{\frac{1}{2}}}{2}\right]}^{\frac{1}{3}} \end{gathered}$$

(45)

Actually, there is only one value of factor X with a given set of parameters. However, there are four roots from a quartic equation and the value of X is one of them.

To identify the correct root, the method is depicted as the following steps:

(1)

Numerically solving the X factor in the underlap region.

(2)

Identify one of the roots in a certain valve displacement in the underlap region corresponding to the same value from the numerical solution.

(3)

Repeat step 1 and 2 with different area ratio $\gamma$

(4)

Repeat step 1, 2, and 3 with different underlap regions.

(5)

Estimate the final analytical solution of the X factor.

Starting with the parameter from Moog’s valve-controlled asymmetric cylinder drive system, for the area ratio $\gamma$ = 11.43/5.8 = 1.9707 and underlap region from 0.0045 mm to −0.0045 mm, root identification is listed in

Table 1. Identifying the root from numerically solved results when the area ratio is 1.9707.

Valve Disp Xs (mm)	$\mathit{X}$ Factor	Root
−0.0045	3.883617722	$X_4$
−0.004404	3.867534491	$X_4$
−0.004296	3.848941124	$X_4$
−0.004101	3.814000308	$X_4$
$⋮$ $⋮$
−0.0009	2.963117621	$X_4$
−0.000753	2.911892307	$X_4$
−0.000735	2.905559937	$X_1$
0	2.637188916	$X_1$
0.0012	2.172727301	$X_1$
0.00255	1.65272018	$X_1$
$⋮$ $⋮$
0.004101	1.119970977	$X_1$
0.0042	1.0894822	$X_1$
0.004302	1.058563937	$X_1$
0.00441	1.026380398	$X_1$
0.0045	1	$X_1$

.

A new phenomenon can be noticed: the $X$ factor values in the underlap region contain two roots of Equation (25), $X_1$ and $X_4$. When valve displacement is from −0.0045 mm to −0.00075 mm, the value of the $X$ factor is the root $X_4$, and when valve displacement is from −0.00074 mm to 0.0045 mm, the value of the $X$ factor is the root $X_1$.

All four root values in the underlap region are described as in Figure 10a, different line styles corresponding to different roots. The values of the roots in the underlap are compared with the numerically solved $X$ values, as in Figure 10b. It can be noticed that a part of $X_4$ and a part of $X_1$ are merged into the numerically solved curve, and they meet at some specific point.

Another interesting phenomenon can be observed: that some parts of the root curve can be combined with parts of other root curves to generate a smoothed curve. The point where $X_4$ and $X_1$ curves meet may change with some parameters. We can change the form of valve opening area $A_{v1}$ and $A_{v2}$ in Equation (25) into the equation below.

$$\begin{gathered} \pi D\left(x_s+0.0000045\right){\left(1+{\gamma }^3-X\right)}^{\frac{1}{2}}-\pi D\left(x_s-0.0000045\right){\left(X\right)}^{\frac{1}{2}} \\ =\gamma \left[\pi D\left(x_s+0.0000045\right){\left(X\gamma \right)}^{\frac{1}{2}}-\pi D\left(x_s-0.0000045\right){\left(1+{\gamma }^3-X\gamma \right)}^{\frac{1}{2}}\right] \end{gathered}$$

(46)

where $A_{v1}=\pi D\left(x_s+0.0000045\right)$, $A_{v2}=\pi D\left(x_s-0.0000045\right)$ [${\mathrm{m}}^2$].

Cancelling the $\pi D$ from the above equation, and dividing both sides by $0.0000045$ reveals

$$\begin{gathered} \left(1+\frac{x_s}{0.0000045}\right){\left(1+{\gamma }^3-X\right)}^{\frac{1}{2}}-\left(1-\frac{x_s}{0.0000045}\right){\left(X\right)}^{\frac{1}{2}} \\ =\gamma \left[\left(1+\frac{x_s}{0.0000045}\right){\left(X\gamma \right)}^{\frac{1}{2}}-\left(1-\frac{x_s}{0.0000045}\right){\left(1+{\gamma }^3-X\gamma \right)}^{\frac{1}{2}}\right] \end{gathered}$$

(47)

The $\frac{{\mathrm{x}}_{\mathrm{s}}}{0.0000045}$ in Equation (47) can be regarded as normalized spool travel from −1 to +1 in underlap. It can be observed that for a certain area ratio γ, the X factor is only affected by the travel percentage. An assumption can be made that the intersection point in Figure 10a is affected by the area ratio; different area ratios $\gamma$ are applied into Equation (47) to verify this. After running the simulation with different area ratios $\gamma$ in the underlap region, the root $X_4$ and $X_1$ distribution is depicted in

Table 2. Root distribution of X₄ and X₁ with different area ratios.

Area Ratio $\mathit{\gamma }$	Percentage of x4	Percentage of x1
1.5	46%	54%
1.9707	41.67%	58.33%
2.5	39%	61%
3	37%	63%
4	33.3%	66.7%

.

Inputting these data into the Matlab CFTOOL (https://www.mathworks.com/products/curvefitting.html) reveals the relationship for $X_1$ percentage with different area ratio $\gamma$ as shown in Equation (48).

$$X_1 Percentage=0.00999{\gamma }^3-0.09352{\gamma }^2+0.3228\gamma +0.2328$$

(48)

The percentage of root $X_4$ is simply obtained by $1-X_1 \mathrm{Percentage}$, so that the $X$ factor root distribution when the valve is in the underlap region is obtained.

3.4. Validation of Analytical Solving Method

The overall component linking model block diagram for this Moog valve-controlled asymmetric cylinder drive is depicted in Figure 11.

The overall block model of the Moog valve-controlled asymmetric cylinder drive system is simulated in Matlab Simulink as shown in Figure 12.

The Simulink model with analytical solution simulation results is compared with the numerical solution simulation results in Fortran and the experimental test results in Figure 13 and Figure 14.

The modelling results from Simulink show consistency with the modelling results from Fortran and experimental results [8], which indicates that the analytical solving method is validated.

4. Results and Discussion

The analytical solution for the X factor in the component linking method is able to save computing time during simulation. The x_s valve stroke keeps changing during operation, which indicates the computer must solve Equation (25) with different values of x_s. If the step size of the simulation is small, for instance 1∗10⁻⁴ s, the computer will need to solve thousands of implicit equations in one second; this process will consume a lot of computing power. The analytical solution in the second section improves this situation.

Simulation tests of the Moog valve-controlled asymmetric cylinder drive system are carried out to compare the time consumed between the implicit solving and analytical solving methods, including a square wave test, a sawtooth test, and a sine wave test.

There is no observable difference between the two results, which indicates both solutions have the same performance. The numerical solution takes 5.49 s, while the analytical solution takes 1.07 s, therefore the numerical solution consumes nearly five times as much time to finish the simulation. However, the valve in the square wave test in Figure 15 is operating outside the underlap region most of the time, therefore the advantage of the analytical solution is not highlighted here.

The sawtooth command test is employed to compare the time cost between the numerical solution and the analytical solution when the valve spool sweeps across its underlap region. A total of 2.31% of the valve spool’s travel in the underlap region is considered in the sawtooth test, and the simulation results of the sawtooth command are depicted in Figure 16.

The analytical solution in the sawtooth test produces the same results as the numerical solution, but the analytical solution costs 1.23 s to finish and the numerical method costs 8.26 s to finish. Nonlinear behaviours can be observed in the results in Figure 16, which include the gradient change of the velocity curve after the system switches its state and some observed oscillations when the valve is in the underlap region around its null position.

Now, assuming the underlap region of the Moog four-way is 20%, a sine wave motion command is sent to the four-way valve so that it only operates in the underlap region. Its pressure behaviour under numerical and analytical simulation results are depicted in Figure 17.

The modelling time costs of calculating chamber pressures when valve travel is in the underlap region are compared in Figure 17. The numerically solved simulation used 1647.17 s to finish, while the analytical solution only uses 8.50 s, which is a significant reduction by a factor of 200.

The time consumptions of the three tests are depicted in

Table 3. Simulation time cost of numerical, analytical, and fundamental methods.

Square Wave Simulation
Numerical solved time cost	Analytical time cost
5.49 s	1.07 s
Underlap Region Simulation
Numerical solved time cost	Analytical time cost
1647.17 s	8.50 s
Sawtooth Command Simulation
Numerical solved time cost	Analytical time cost
8.26 s	1.23 s

.

5. Conclusions

The component linking method is validated in Matlab Simulink (https://www.mathworks.com/products/simulink.html), and its simulation results show consistency with the published results. Therefore, it can be utilised to achieve further research when the valve of the asymmetric cylinder drive system is in the underlap region. How the chamber pressures behave when the valve is in the underlap region is revealed by Simulink modelling, and the inevitable pressure jump process when the valve switches its state is numerically solved with the X factor method. To reduce the computing power required, an analytical solution for the X factor is developed. Equation (47) proved the X factor value is only related to the percentage of valve spool travel in the underlap region, and offered a solution to identify the root distribution percentage in the underlap region. This analytical solution is approximately 200 times more computationally efficient than the numerical solution method.

Underlapped symmetrically ported valves are widely used in systems that require precise flow control, such as chemical processing, liquid distribution systems, etc. By adjusting the opening of the valve, the flow rate of the medium can be precisely controlled. By testing the pressure process formula, it is possible to infer how much displacement the valve core has produced, thereby correcting the valve’s lifespan status.