The Hydrodynamic Similarity between Different Power Levels and a Dynamic Analysis of Ocean Current Energy Converter–Platform Systems with a Novel Pulley–Traction Rope Design for Irregular Typhoon Waves and Currents

Abstract

In the future, the power of a commercial ocean current energy convertor will be able to reach the MW class, and its corresponding mooring rope tension will be very good. However, the power of convertors currently being researched is still at the KW class, which can bear less rope tension. The main mooring rope usually has a single cable and a single foundation. To investigate the dynamic response and rope tension of an MW-class ocean current generator mooring system, here, a similarity rule is proposed for (1) coefficients without any fluid–structure interaction (FSI) using the Buckingham theorem and (2) ones with FSI. The overall hydrodynamic drag and moment including the hydrodynamic coefficients in these two situations are represented in a Taylor series. Assuming similarity between the commercial MW-class and KW-class ocean current convertors, all hydrodynamic parameters of the MW-class system are estimated based on the known KW-class parameters and based on the similarity formula. In order to overcome the extreme tension of the MW-class system and to provide good stability, in this paper, we propose a pulley–rope design to replace the traditional single-traction-rope design. The static and dynamic mathematical models of this mooring system subjected to the impact of typhoon waves and currents are proposed, and analytical solutions are obtained. We find that the pulley–rope design can significantly reduce the dynamic rope tensions of the mooring system. The effect of the length ratio of the main traction rope, rope A, to the seabed depth on the dynamic tension of stabilizing converter rope D is significant. The length ratio is within a safe range, and the maximum rope dynamic tension is less than the fracture strength. In addition, if the rope length ratio is over the critical value, the larger the ratio, the higher the safety factor of the rope. In summary, the pulley–rope design can be safely used in an MW-level ocean current generator system.

1. Introduction

Global ocean currents are rich in energy. The potential electricity capacity in the Taiwan Kuroshio current is over 4 GW [1]. Thus, core technologies for ocean current power generation are being investigated [1,2,3,4,5]. Our investigation includes studying the following: (1) high-efficiency convertors, (2) deep mooring technology for seabeds over 1000 m beneath the current, (3) technologies offering protection from the impact of typhoon waves, (4) double main traction ropes for a high-power convertor, and (5) the mathematical model of fluid–structure interactions (FSIs).

Most of the seabeds in Taiwan Kuroshio are deeper than 1000 m, with the seabed geology being rock, sand, and mud. The design and placement of a mooring foundation therefore need to be safe. The ocean current power generation anchorage system must be able to provide stability to the ocean current turbine so that the ocean current turbine can generate electricity stably and efficiently. For the safety of the anchorage system, a mathematical model of the system needs to be established to investigate its dynamic motion displacement, dynamic rope tension, and stability. However, the convertors currently under study are all small-power generation models, tests in real sea are all conducted in the short term, and the main objective is only to test the performance of an ocean current turbine generator set. Therefore, they all use mooring systems with a single main rope and a single base.

The company WanChi developed a 50 kW ocean current convertor in which the blade is pushed by the current force and moves translationally. Chen et al. [1] successfully tested the 50 kW WanChi convertor moored to the seabed 850 m beneath the Kuroshio current in the Taiwan Pingtung sea area. IHI and NEDO [2] developed a 100 kW ocean current convertor that integrated two sets of rotational turbines. The 100 kW convertor was successfully tested while moored to the seabed 100 m beneath the Japan Kuroshio current. Guo et al. [3] developed a 20 kW ocean current convertor that integrated two sets of rotational turbines. The 20 kW convertor was tested while moored to the seabed 80 m beneath the Taiwan Liuqiu sea area.

The deep mooring theories and technologies for ocean current convertor systems are also important. A few studies investigated the dynamic stability of mooring systems under the coupled current–wave effect. The company WanChi further developed a 400 kW ocean current convertor following the principle of those with 50 kW power. Lin et al. [4] proposed the mathematical model of a 400 kW ocean convertor–pontoon–traction rope–foundation mooring system under the coupled regular wave–ocean current effects. The motion of this system has four degrees of freedom. The dynamic system includes the heaving motion of the pontoon and the coupled surging–heaving–pitching motion of the convertor. The dynamic performance and stability of a system under a coupled wave–current effect were also investigated. Lin et al. [5] presented a submarined ocean convertor–surfaced platform–pontoon–traction rope–foundation mooring system. The concentrated mass model was considered. The motion of this system has five degrees of freedom. The coupled motion included surging and heaving of the elements. Moreover, the dynamic stability of the mooring system under regular waves and steady ocean currents was investigated. Lin et al. [6] presented a submarined ocean convertor–submarined platform–two-pontoon–traction rope–foundation mooring system. The concentrated mass model was considered. The motion of this system has six degrees of freedom. The coupled motion included surging and heaving of the elements. The dynamic response of this mooring system under the influence of an irregular typhoon wave at Taiwan’s Green Island during the 50-year regression period was further investigated. Lin et al. [7] designed a 400 kW converter composed of two rotating horizontal turbines. The hydrodynamic damping and stiffness coefficients were determined using the computational fluid method. A submarined 400 kW convertor–submarined platform–two-pontoon–traction rope–foundation mooring system was considered. The motion of this system has eighteen degrees of freedom. The coupled 3D motion included the translational and rotational motion of the elements. The frequency spectrum of a system with a coupled fluid–structure interaction and a regular wave was then studied. Furthermore, Lin et al. [8] investigated the transient performance of an 18 DOF mooring system with some initial conditions. Pierson and Moskowitz [9] proposed the Jonswap spectrum and confirmed it using experimental measurement results. To simplify actual irregular ocean waves, a limited number of regular waves are usually used to approximate irregular waves [7]. In this study, we will also take several regular waves to simulate irregular waves according to the experimental significant wave height, frequency, and Jonswap wave spectrum. In [4,5,6,7,8], the mooring traction rope is the only rope. However, high-power convertors have large traction forces, so two traction ropes are needed. Such an investigation is therefore helpful for relevant ocean energy technologies.

Theories on fluid–structure interactions (FSIs) and their related technologies are widely applied in many different fields, including marine engineering [10,11,12], aerodynamics [13], acoustics [14], and biomechanics [15,16]. Due to the complexity of fluid–structure interactions, a lot of studies investigating this interaction use numerical methods: (1) the boundary element method [13], (2) the finite-volume method [17], and (3) coupled SPH–FEM [18].

Thus far, studies have proposed a single main mooring rope connecting to one foundation. For a high-power ocean current convertor, in this study, we design a pulley–traction rope set, with the rope connecting to the two foundations. A similarity law is proposed to calculate the hydrodynamic damping and stiffness coefficient of the turbine and platform. Static and dynamic equations of this ocean current power generation anchoring system are proposed to study its rope tension and stability under the influence of irregular waves.

2. Mathematical Model

To avoid the impact of typhoon waves, an energy convertor and its floating platform need to be submerged to safe depths. For safely mooring a high-power ocean convertor, in this study, we present a pulley–traction rope set composed of a pulley and one main rope connected to two separate foundations, as shown in Figure 1. The construction of the pulley–traction rope foundation set took place as follows: Firstly, both ends of rope A were fixed to the two mooring foundations. Secondly, an Abyssal Anchor Base was cast onto the seabed. Finally, a traction pulley fixed to the carrier platform was installed, and then rope A was wrapped around the groove of the pulley to prevent it from loosening. The ocean current energy convertor will tighten rope A under the force of the ocean current. When the direction of the ocean current changes, the convertor will move due to the hydrodynamic force, the pulley will rotate, and lengths L_A1 and L_A2 on both sides of rope A will change accordingly; meanwhile, the tension on both sides will remain consistent. The coupled translational–rotational motion of the mooring system under a coupled wave–current effect was also considered, with the motion having eighteen degrees of freedom. Thus, the governing equations of this mooring system are derived as follows:

The global translational and rotational displacements of the components are

$$\begin{gathered} x_i=x_{is}+x_{id},\hspace{0.5em}{y}_i=y_{is}+y_{id},\hspace{0.5em}{z}_i=z_{is}+z_{id},\hspace{0.5em}i=1, 2, 3, 4 \\ {\phi }_{jx}={\phi }_{jxs}+{\phi }_{jxd}, {\phi }_{jy}={\phi }_{jys}+{\phi }_{jyd}, {\phi }_{jz}={\phi }_{jzs}+{\phi }_{jzd},j=1,2 \end{gathered}$$

(1)

Because the platform and convertor are considered symmetrical, static equilibrium occurs under a steady current only, and their rotational displacements are ${\phi }_{jxs}={\phi }_{jys}={\phi }_{jzs}=0$, j = 1, 2. The total tensions of the ropes are

$$T_i=T_{is}+T_{id},\hspace{0.5em}i=A, B, C, D$$

(2)

These displacements and tensions include (1) the static ones under a steady current only and (2) the dynamic ones due to the impact of waves and currents. Because rope A connects to the pulley fixed to the platform and two foundations, the tension of rope A₁ is equal to that of rope A₂, $T_{A1}=T_{A2}=T_A$, $T_{A1s}=T_{A2s}\equiv T_{As}$, and $T_{A1d}=T_{A2d}\equiv T_{Ad}$.

3. Static Displacements and Equilibrium under a Steady Current Only

3.1. Static Displacements

As shown in Figure 1 and Figure 2, under a steady current only, the static displacements of the components are as follows:

Foundation:

$${x^{\prime }}_{01}=0,\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{y^{\prime }}_{01}=0,\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{z^{\prime }}_{01}=L_F/2,$$

(3)

$${x^{\prime }}_{02}=0,\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{y^{\prime }}_{02}=0,\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{z^{\prime }}_{02}=-L_F/2,$$

(4)

Platform:

$${x^{\prime }}_{1s}=H_{bed}-L_C=L_{A1}\sin {\theta }_{As1}=L_{A2}\sin {\theta }_{As2},$$

(5)

$${y^{\prime }}_{1s}=L_{A1}\cos {\theta }_{As1}\sin {\varphi }_{01}=L_{A2}\cos {\theta }_{As2}\sin {\varphi }_{02}$$

(6)

$${z^{\prime }}_{1s}=L_F/2-L_{A1}\cos {\theta }_{As1}\cos {\varphi }_{01}=-L_F/2+L_{A2}\cos {\theta }_{As2}\cos {\varphi }_{02}$$

(7)

The lengths of rope A₁ and A₂ are

$$L_{A1}=\sqrt{{\left({x^{\prime }}_{1s}-{x^{\prime }}_{01}\right)}^2+{\left({y^{\prime }}_{1s}-{y^{\prime }}_{01}\right)}^2+{\left({z^{\prime }}_{1s}-{z^{\prime }}_{01}\right)}^2}$$

(8)

$$L_{A2}=\sqrt{{\left({x^{\prime }}_{1s}-{x^{\prime }}_{02}\right)}^2+{\left({y^{\prime }}_{1s}-{y^{\prime }}_{02}\right)}^2+{\left({z^{\prime }}_{1s}-{z^{\prime }}_{02}\right)}^2}$$

(9)

$$L_A=L_{A1}+L_{A2}$$

(10)

The static displacements and parameters of the mooring system, $\left\{y_{1s},z_{1s}\right\}$ and $\left\{L_{A1},L_{A2},{\theta }_{As1},{\theta }_{As2},{\varphi }_{01},{\varphi }_{02}\right\}$, change with current direction ${\varphi }_{cur}$. These parameters can be determined using the static equilibrium principle presented later.

Turbine:

$${x^{\prime }}_{2s}=H_{bed}-L_D={x^{\prime }}_{1s}+L_B\sin {\theta }_{Bs},$$

(11)

$${y^{\prime }}_{2s}={y^{\prime }}_{1s}+L_B\cos {\theta }_{Bs}\cos {\varphi }_{cur},$$

(12)

$${z^{\prime }}_{2s}={z^{\prime }}_{1s}+L_B\cos {\theta }_{Bs}\sin {\varphi }_{cur},$$

(13)

Pontoons 3 and 4:

$${x^{\prime }}_{3s}={x^{\prime }}_{1s}+L_C=H_{bed},\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{y^{\prime }}_{3s}={y^{\prime }}_{1s},\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{z^{\prime }}_{3s}={z^{\prime }}_{1s},$$

(14)

$${x^{\prime }}_{4s}={x^{\prime }}_{3s}={x^{\prime }}_{2s}+L_D=H_{bed},\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{y^{\prime }}_{4s}={y^{\prime }}_{2s},\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{z^{\prime }}_{4s}={z^{\prime }}_{2s},$$

(15)

Rotational angles of the platform and convertor:

$${\phi }_{jks}=0, j=1,2; k=x,y,z$$

(16)

The global setting angle θ_Bs of rope B is

$$\sin {\theta }_{Bs}=\left(x_{2s}-x_{1s}\right)/L_{\mathrm{B}}$$

(17)

The relation between the x-y-z and x′-y′-z′ coordinates is

$${\overrightarrow{r}}_p=x_p\overrightarrow{i}+y_p\overrightarrow{j}+z_p\overrightarrow{k}; p=01,02,1s,2s,3s,4s$$

(18)

where

$$\begin{array}{l}{x}_p=\left({x^{\prime }}_p-{x^{\prime }}_{1s}\right),y_p=\left[\left({y^{\prime }}_p-{y^{\prime }}_{1s}\right)\cos {\varphi }_{cur}+\left({z^{\prime }}_p-{z^{\prime }}_{1s}\right)\sin {\varphi }_{cur}\right],\\ z_p=\left[-\left({y^{\prime }}_p-{y^{\prime }}_{1s}\right)\sin {\varphi }_{cur}+\left({z^{\prime }}_p-{z^{\prime }}_{1s}\right)\cos {\varphi }_{cur}\right]\end{array}$$

(19)

3.2. Static Force Equilibrium

Under a steady current only, the static equilibrium of the energy convertor in the current direction is

$$T_{\mathrm{B}s}\cos {\theta }_{Bs}=f_{Tys}$$

(20)

where the drag of the convertor under steady current is $f_{Tys}=0.5C_{DTy}\rho A_{TY}{V}^2$. The static equilibrium of the platform in the current direction is

$$T_{\mathrm{A}s1}\cos {\theta }_{As1}\cos \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)+T_{\mathrm{A}s2}\cos {\theta }_{As2}\cos {\mathsf{\Delta }}_2=f_{Pys}+f_{Tys}$$

(21)

where the drag of the platform under steady current $f_{Pys}=0.5C_{DPy}\rho A_{PY}{V}^2$, ${\mathsf{\Delta }}_1=\left(\pi /2-{\varphi }_{01}\right)$, and ${\mathsf{\Delta }}_2=\left(\pi /2-{\varphi }_{02}\right)-{\varphi }_{cur}$. Because rope A connects to the pulley, as shown in Figure 2, the tensions of ropes A₁ and A₂ are

$$T_{\mathrm{A}s1}=T_{\mathrm{A}s2}=T_{\mathrm{A}s}$$

(22)

Based on Equations (20)–(22), the static tension of rope A is expressed as

$$T_{\mathrm{A}s}={\displaystyle \frac{f_{Pys}+f_{Tys}}{\cos {\theta }_{As1}\cos \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)+\cos {\theta }_{As2}\cos {\mathsf{\Delta }}_2}}={\displaystyle \frac{f_{Pys}+T_{\mathrm{B}s}\cos {\theta }_{Bs}}{\cos {\theta }_{As1}\cos \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)+\cos {\theta }_{As2}\cos {\mathsf{\Delta }}_2}}$$

(23)

The static equilibrium of the platform in the x-direction is

$$T_{Cs}+F_{B1s}=\left[T_{\mathrm{A}s1}\sin {\theta }_{As1}+T_{\mathrm{A}s2}\sin {\theta }_{As2}\right]+T_{\mathrm{B}s}\sin {\theta }_{Bs}+W_1$$

(24)

The static equilibrium of the platform in the z-direction is

$$\cos {\theta }_{As1}\sin \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)-\cos {\theta }_{As2}\sin {\mathsf{\Delta }}_2=0$$

(25)

The static equilibrium of the energy convertor in the x-direction is

$$T_{\mathrm{D}s}=W_2-T_{\mathrm{B}s}\sin {\theta }_{Bs}-F_{B2s}$$

(26)

The static equilibrium of Pontoon 3 in the x-direction is

$$F_{B3s}=W_3+T_{\mathrm{C}s}$$

(27)

The static equilibrium of Pontoon 4 in the x-direction is

$$F_{B4s}=W_4+T_{\mathrm{D}s}$$

(28)

3.3. Solution Method of Static Equilibrium and Displacements

Substituting Equations (7)–(9) into Equation (25), the characteristic equation in equilibrium can be derived:

$$\begin{array}{l}\cos {\theta }_{As1}\left[\sin \left(\pi /2+{\varphi }_{cur}\right)\cos {\varphi }_{01}-\cos \left(\pi /2+{\varphi }_{cur}\right)\sin {\varphi }_{01}\right]\\ -\cos {\theta }_{As2}\left[\sin \left(\pi /2-{\varphi }_{cur}\right)\cos {\varphi }_{02}-\cos \left(\pi /2-{\varphi }_{cur}\right)\sin {\varphi }_{02}\right]=0\end{array}$$

(29)

where ${\theta }_{As1}={\sin }^{-1}{\displaystyle \frac{H_{bed}-L_C}{L_{A1}}}$, ${\theta }_{As2}={\sin }^{-1}{\displaystyle \frac{H_{bed}-L_C}{L_A-L_{A1}}}$, ${\varphi }_{02}={\cos }^{-1}\left({\displaystyle \frac{a_2^2+L_F^2-a_1^2}{2L_F{a}_2}}\right)$, ${\varphi }_{01}={\cos }^{-1}\left({\displaystyle \frac{L_F-a_2\cos {\varphi }_{02}}{a_1}}\right)$, $a_1=L_{A1}\cos {\theta }_{As1}$, $a_2=L_{A2}\cos {\theta }_{As2}$.

If the background parameters $\left\{{\varphi }_{cur},H_{bed},L_C,L_A,L_F\right\}$ are given, only the length of rope A₁, L_A1, in Equation (29) is unknown. Because this equation is implicit, the solution cannot be directly solved. However, it can be determined using the bisection numerical method. Substituting L_A1 back into Equations (8), (9), and (12)–(17), the parameters $\left\{L_{A2},{\theta }_{As1},{\theta }_{As2},\right.$$\left.{\varphi }_{01},{\varphi }_{02},{y^{\prime }}_{1s},{z^{\prime }}_{1s}\right\}$ are obtained.

3.4. Static Numerical Results

The parameters of the mooring system are listed in

Table 1. The parameters of the system.

Parameter	Dimension	Parameter	Dimension
depth of the seabed Hbed	1300 m	current velocity, V	1.5 m/s
length of rope A, LA	5200 m	length of rope B, LB	170 m
length of rope C, LC	60 m	length of rope D, LD	140 m
Static drag of the convertor, FDT	59.35 tons	Static drag of the platform, FDB	0.077 tons

.

Figure 3 shows that the convertor direction is almost the same as the current. The distance L_F between the two foundations increases the deviation of the platform direction from the current slightly.

Figure 4 shows the effects of the current direction and the distance L_F between the two foundations on the static tension of rope A, T_As. If the current velocity is V_cur = 1.5 m/s, the drag of the converter is F_DT = 59.35 tons and the drag of the platform is F_DB = 0.077 tons. If the traditional single traction rope is considered, the corresponding rope length is 2480 m and the tension is T_As = 68.62 tons. However, if L_F = 0.1L_A in this proposed system, the tension T_As is about 34 tons only, which is half of that for the traditional traction rope. The current direction ${\phi }_{cur}$ decreases the static tension T_As. Moreover, the larger the distance L_F, the larger the static tension T_As.

4. Dynamic Analysis

4.1. Similarity of System

To investigate the dynamic response and rope tension of an MW-class ocean current generator mooring system, a similarity rule is proposed. Firstly, for the case without fluid–structure interactions (FSIs), the Buckingham theorem is used to obtain the turbine power number and the tip speed ratio of the stationary ocean current turbine under static conditions, as well as the stable fluid damping coefficient of the turbine and the platform. Secondly, for the case with FSI, the coupling hydraulic resistance and torque of the turbine and the platform are related to the speed and rotation of the motion. The overall hydrodynamic drag and moment, including the hydrodynamic coefficients in the two above situations, is represented in a Taylor series. A similarity formula can then be obtained by making the hydrodynamic coefficients dimensionless. Assuming the similarity of the commercial MW-class ocean current generator and that of the research-type KW class, all hydrodynamic parameters of the MW-class ocean current generator are estimated based on the known KW-class parameters and based on the similarity formula. Lin et al. [7] obtained the hydrodynamic coefficients of a 400 kW turbine and platform using the commercial software Star-CCM. Based on these similarity formulas and the known KW-class parameters, all parameters of the MW-class system are estimated.

4.1.1. Hydrodynamic Similarity for the Convertor

Similarity without FSI

Based on the Buckingham theorem, we can derive the dimensionless parameters.

Similarity for tip speed ratio TSR:

$${\left({\displaystyle \frac{{\omega }_T{D}_T}{V_{cur}}}\right)}_{\mathrm{mod}}={\left({\displaystyle \frac{{\omega }_T{D}_T}{V_{cur}}}\right)}_{pro}$$

(30)

Lin et al. [8] proposed using TSR = 3.5 for the 400 kW turbine, which achieves maximum efficiency. Considering the same current velocity V_cur for the model and prototype, Equation (30) becomes

$${\displaystyle \frac{{\omega }_{T,mod}}{{\omega }_{T,pro}}}={\displaystyle \frac{D_{T,pro}}{D_{T,mod}}}$$

(31)

Similarity for power number N_p:

$${\left({\displaystyle \frac{Power}{\rho D_T^5{\omega }_T^3}}\right)}_{\mathrm{mod}}={\left({\displaystyle \frac{Power}{\rho D_T^5{\omega }_T^3}}\right)}_{pro}$$

(32)

Considering the same current velocity V_cur and sea density r for the model and prototype and substituting Equation (31) into Equation (32),

$${\displaystyle \frac{D_{T,pro}}{D_{T,\mathrm{mod}}}}=\sqrt{{\displaystyle \frac{Powe{r}_{pro}}{Powe{r}_{\mathrm{mod}}}}}.$$

(33)

The similarity of the drag force coefficient without FSI is

$${\left[C_{fTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)\right]}_{\mathrm{mod}}={\left[C_{fTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)\right]}_{pro},j=x,y,z$$

(34)

The similarity of the drag moment coefficient without FSI is

$${\left[C_{mTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)\right]}_{\mathrm{mod}}={\left[C_{mTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)\right]}_{pro},j=x,y,z$$

(35)

Similarity with FSI

Similarity for hydrodynamic force and moment for the convertor:

The total hydrodynamic force of the convertor is

$$\begin{array}{c}{f}_{Tj}\left(V_{cur},{\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T,{\phi }_{Tx},{\phi }_{Ty},{\phi }_{Tz},{\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz},TSR\right)\equiv f_{Tj}\left(V_{cur},s_{T1},s_{T2},s_{T3},s_{T4},s_{T5},s_{T6},s_{T7},s_{T8},s_{T9},TSR\right)\\ =f_{Tj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)+{\displaystyle \sum _{k=1}^9{s}_{Tk}}{\left.{\displaystyle \frac{\partial f_{Tj}}{\partial s_{Tk}}}\right|}_{s_{Ti}=0,i\ne k},j=x,y,z\end{array}$$

(36)

Dividing Equation (36) by ${\displaystyle \frac{1}{2}}\rho A_{tur}{V}_{cur}^2$, the dimensionless variables become one:

$$\begin{array}{c}{C}_{fTj}\left(V_{cur},{\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T,{\phi }_{Tx},{\phi }_{Ty},{\phi }_{Tz},{\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz},TSR\right)\equiv C_{fTj}\left(V_{cur},s_{T1},s_{T2},s_{T3},s_{T4},s_{T5},s_{T6},s_{T7},s_{T8},s_{T9},TSR\right)\\ =C_{fTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)+{\displaystyle \sum _{k=1}^9{\tilde{s}}_{Tk}}{\left.{\displaystyle \frac{\partial C_{fTj}}{\partial {\tilde{s}}_{Tk}}}\right|}_{s_i=0,i\ne k},j=x,y,z\end{array}$$

(37)

where the first term on the right-hand side, $C_{fTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)$, can be derived using the Buckingham theorem without FSI. The other terms are obtained with FSI.

$$C_{fTj}={\displaystyle \frac{f_{Tj}}{\frac{1}{2}\rho A_{tur}{V}_{cur}^2}},\hspace{0.5em}{\tilde{s}}_{Tk}={\displaystyle \frac{s_{Tk}}{{\xi }_T}},\hspace{0.5em}{\xi }_T=\left\{\begin{array}{c}{V}_{cur},\hspace{0.5em}{s}_{Tk}={\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T\hfill \\ 1,\hspace{0.5em}{s}_{Tk}={\phi }_{Tx},{\phi }_{Ty},{\phi }_{Tz}\hfill \\ {\omega }_T,\hspace{0.5em}{s}_{Tk}={\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz}\hfill \end{array}\right.$$

(38)

If the prototype and the model are similar, their dimensionless force coefficients (37) are the same:

$${\left[C_{fTj}\right]}_{pro}={\left[C_{fTj}\right]}_{\mathrm{mod}}$$

(39)

Based on Equations (34) and (37)–(39), we can obtain the following similarity formula.

Similarity for the hydrodynamic damping force coefficient with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{fTj}}{\partial {\tilde{s}}_{Tk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{fTj}}{\partial {\tilde{s}}_{Tk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{pro},j=x,y,z; s_{Tk}={\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T,{\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz}$$

(40)

Similarity for the hydrodynamic stiffness force coefficient with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{fTj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Ti}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{fTj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Ti}=0,i\ne k}\right)}_{pro},j,k=x,y,z$$

(41)

Based on the similarity Formulas (38), (40), and (41), the hydrodynamic damping force relations between the model and prototype are

$${\left({\left.{\displaystyle \frac{\partial f_{Tj}}{\partial s_{Tk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{\mathrm{pro}}={\displaystyle \frac{D_{T, pro}^2}{D_{T,\mathrm{mod}}^2}}{\left({\left.{\displaystyle \frac{\partial f_{Tj}}{\partial s_{Tk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{\mathrm{mod}},j=x,y,z; s_{Tk}={\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T$$

(42)

$${\left({\left.{\displaystyle \frac{\partial f_{Tj}}{\partial {\dot{\phi }}_{Tk}}}\right|}_{{\dot{\phi }}_{Tk}=0,i\ne k}\right)}_{\mathrm{pro}}={\displaystyle \frac{D_{T,pro}^2}{D_{T,mod}^2}}{\displaystyle \frac{{\omega }_{mod}}{{\omega }_{pro}}}{\left({\left.{\displaystyle \frac{\partial f_{Tj}}{\partial {\dot{\phi }}_{Tk}}}\right|}_{{\phi }_{Tk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(43)

and the hydrodynamic stiffness force relation is

$${\left({\left.{\displaystyle \frac{\partial f_{Tj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Tk}=0,i\ne k}\right)}_{\mathrm{pro}}={\displaystyle \frac{D_{T, pro}^2}{D_{T,\mathrm{mod}}^2}}{\left({\left.{\displaystyle \frac{\partial f_{Tj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Tk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(44)

Given the coefficients of the model and substituting Equations (31) and (33) into Equations (42)–(44), the hydrodynamic damping and stiffness force coefficients of the prototype can be obtained.

According to the similarity for drag force coefficient (34), the drag force relation is

$${\displaystyle \frac{{\left(f_{Ty,0}\right)}_{pro}}{{\left(f_{Ty,0}\right)}_{\mathrm{mod}}}}={\displaystyle \frac{D_{T, pro}^2}{D_{T,\mathrm{mod}}^2}}$$

(45)

Based on Equation (45), the similarity for the fracture strength of a rope is

$${\displaystyle \frac{T_{frac,pro}}{T_{frac,\mathrm{mod}}}}={\displaystyle \frac{{\left(f_{Ty,0}\right)}_{pro}}{{\left(f_{Ty,0}\right)}_{\mathrm{mod}}}}={\displaystyle \frac{D_{T, pro}^2}{D_{T,\mathrm{mod}}^2}}$$

(46)

The hydrodynamic moment of the convertor is

$$\begin{array}{c}{m}_{Tj}\left(V_{cur},{\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T,{\phi }_{Tx},{\phi }_{Ty},{\phi }_{Tz},{\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz},TSR\right)\equiv m_{Tj}\left(V_{cur},s_{T1},s_{T2},s_{T3},s_{T4},s_{T5},s_{T6},s_{T7},s_{T8},s_{T9},TSR\right)\\ =m_{Tj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)+{\displaystyle \sum _{k=1}^9{s}_{Tk}}{\left.{\displaystyle \frac{\partial m_{Tj}}{\partial s_{Tk}}}\right|}_{s_{Ti}=0,i\ne k},j=x,y,z\end{array}$$

(47)

Dividing Equation (47) by ${\displaystyle \frac{1}{2}}\rho D_T{A}_{tur}{V}_{cur}^2$, the dimensionless variables become one:

$$\begin{array}{c}{C}_{mTj}\left(V_{cur},{\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T,{\phi }_{Tx},{\phi }_{Ty},{\phi }_{Tz},{\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz},TSR\right)\equiv C_{mTj}\left(V_{cur},s_{T1},s_{T2},s_{T3},s_{T4},s_{T5},s_{T6},s_{T7},s_{T8},s_{T9},TSR\right)\\ =C_{mTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)+{\displaystyle \sum _{k=1}^9{\tilde{s}}_{Tk}}{\left.{\displaystyle \frac{\partial C_{mTj}}{\partial {\tilde{s}}_{Tk}}}\right|}_{s_{Ti}=0,i\ne k},j=x,y,z\end{array}$$

(48)

where the first term on the right-hand side, $C_{mTj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0,TSR\right)$, can be derived using the Buckingham theorem without FSI. The other terms are obtained with FSI. $C_{mTj}=m_{Tj}/\left({\displaystyle \frac{1}{2}}\rho D_T{A}_{tur}{V}_{cur}^2\right)$.

If the prototype and the model are similar, the two dimensionless moments (48) are the same:

$${\left[C_{mTj}\right]}_{pro}={\left[C_{mTj}\right]}_{\mathrm{mod}}$$

(49)

Based on Equations (35), (48) and (49), one can obtain the following similar parameters.

Similarity for hydrodynamic damping in a moment with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{mTj}}{\partial {\tilde{s}}_{Tk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{mTj}}{\partial {\tilde{s}}_{Tk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{pro},j=x,y,z;s_k={\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T,{\dot{\phi }}_{Tx},{\dot{\phi }}_{Ty},{\dot{\phi }}_{Tz}$$

(50)

Similarity for hydrodynamic stiffness in a moment with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{mTj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Ti}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{mTj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Ti}=0,i\ne k}\right)}_{pro},j,k=x,y,z$$

(51)

Based on the similarity Formulas (38), (50), and (51), the hydrodynamic damping moment relations between the model and prototype are

$${\left({\left.{\displaystyle \frac{\partial m_{Tj}}{\partial s_{Tk}}}\right|}_{s_{Tk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{T, pro}^3}{D_{T,\mathrm{mod}}^3}}{\left({\left.{\displaystyle \frac{\partial m_{Tj}}{\partial s_{Tk}}}\right|}_{s_{Tk}=0,i\ne k}\right)}_{\mathrm{mod}},s_{Tk}={\dot{x}}_T,{\dot{y}}_T,{\dot{z}}_T$$

(52)

$${\left({\left.{\displaystyle \frac{\partial m_{Tj}}{\partial {\dot{\phi }}_{Tk}}}\right|}_{{\dot{\phi }}_{Tk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{T, pro}^3}{D_{T,\mathrm{mod}}^3}}{\displaystyle \frac{{\omega }_{T,\mathrm{mod}}}{{\omega }_{T,pro}}}{\left({\left.{\displaystyle \frac{\partial m_{Tj}}{\partial {\dot{\phi }}_{Tk}}}\right|}_{{\dot{\phi }}_{Tk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(53)

and the hydrodynamic stiffness moment relation is

$${\left({\left.{\displaystyle \frac{\partial m_{Tj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Tk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{T, pro}^3}{D_{T,\mathrm{mod}}^3}}{\left({\left.{\displaystyle \frac{\partial m_{Tj}}{\partial {\phi }_{Tk}}}\right|}_{{\phi }_{Tk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(54)

Given the coefficients of the model and substituting Equations (31) and (33) into Equations (52)–(54), the hydrodynamic damping and stiffness moment coefficients of the prototype can be obtained.

4.1.2. Hydrodynamic Similarity for Platform

Similarity without FSI

Based on the Buckingham theorem, we can derive the dimensionless drag coefficient.

The similarity for the drag force coefficient without FSI is

$${\left[C_{fPj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)\right]}_{\mathrm{mod}}={\left[C_{fPj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)\right]}_{pro},j=x,y,z$$

(55)

The similarity for the drag moment coefficient without FSI is

$${\left[C_{mPj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)\right]}_{\mathrm{mod}}={\left[C_{mPj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)\right]}_{pro},j=x,y,z$$

(56)

Similarity with FSI

The hydrodynamic force of the platform is

$$\begin{array}{c} f_{Pj}\left(V_{cur},{\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P,{\phi }_{Px},{\phi }_{Py},{\phi }_{Pz},{\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}\right)\equiv f_{Pj}\left(V_{cur},s_{P1},s_{P2},s_{P3},s_{P4},s_{P5},s_{P6},s_{P7},s_{P8},s_{P9}\right)\\ =f_{Pj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)+{\displaystyle \sum _{k=1}^9{s}_{Pk}}{\left.{\displaystyle \frac{\partial f_{Pj}}{\partial s_{Pk}}}\right|}_{s_i=0,i\ne k},j=x,y,z\end{array}$$

(57)

Dividing Equation (57) by ${\displaystyle \frac{1}{2}}\rho A_{pla}{V}_{cur}^2$, the dimensionless variables become one:

$$\begin{array}{c}{C}_{fPj}\left(V_{cur},{\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P,{\phi }_{Px},{\phi }_{Py},{\phi }_{Pz},{\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}\right)\equiv C_{fPj}\left(V_{cur},s_{P1},s_{P2},s_{P3},s_{P4},s_{P5},s_{P6},s_{P7},s_{P8},s_{P9}\right)\\ =C_{fPj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)+{\displaystyle \sum _{k=1}^9{\tilde{s}}_{Pk}}{\left.{\displaystyle \frac{\partial C_{fPj}}{\partial {\tilde{s}}_{Pk}}}\right|}_{s_{Pi}=0,i\ne k},j=x,y,z\end{array}$$

(58)

where the first term on the right-hand side can be derived using the Buckingham theorem without FSI. The other terms are obtained with FSI.

$$C_{fPj}={\displaystyle \frac{f_{Pj}}{\frac{1}{2}\rho A_{pla}{V}_{cur}^2}},\hspace{0.5em}{\tilde{s}}_{Pk}={\displaystyle \frac{s_{Pk}}{{\xi }_P}},\hspace{0.5em}{\xi }_P=\left\{\begin{array}{c}{V}_{cur},\hspace{0.5em}{s}_{Pk}={\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P\hfill \\ 1,\hspace{0.5em}{s}_{Pk}={\phi }_{Px},{\phi }_{Py},{\phi }_{Pz}\hfill \\ {\omega }_T,\hspace{0.5em}{s}_{Pk}={\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}\hfill \end{array}\right.$$

(59)

If the prototype and the model are similar, the two dimensionless force coefficients (58) are the same:

$${\left[C_{fPj}\right]}_{pro}={\left[C_{fPj}\right]}_{\mathrm{mod}}$$

(60)

Based on Equations (55) and (58)–(60), we can obtain the following similar parameters.

Similarity for the hydrodynamic damping force coefficient with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{fPj}}{\partial {\tilde{s}}_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{fPj}}{\partial {\tilde{s}}_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{pro},j=x,y,z; s_{Pk}={\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P,{\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}$$

(61)

Similarity for the hydrodynamic stiffness force coefficient with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{fPj}}{\partial {\phi }_{Pk}}}\right|}_{{\phi }_{Pi}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{fPj}}{\partial {\phi }_{Pk}}}\right|}_{{\phi }_{Pi}=0,i\ne k}\right)}_{pro},j,k=x,y,z$$

(62)

Based on Equations (59), (61), and (62), the hydrodynamic damping force relations between the model and prototype are

$${\left({\left.{\displaystyle \frac{\partial f_{Pj}}{\partial s_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{\mathrm{pro}}={\displaystyle \frac{D_{P, pro}^2}{D_{P,\mathrm{mod}}^2}}{\left({\left.{\displaystyle \frac{\partial f_{Pj}}{\partial s_{Pk}}}\right|}_{s_{Ti}=0,i\ne k}\right)}_{\mathrm{mod}},j=x,y,z; s_{Pk}={\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P$$

(63)

$${\left({\left.{\displaystyle \frac{\partial f_{Pj}}{\partial {\dot{\phi }}_{Pk}}}\right|}_{{\dot{\phi }}_{Pk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{P,pro}^2}{D_{P,mod}^2}}{\displaystyle \frac{{\omega }_{T,mod}}{{\omega }_{T,pro}}}{\left({\left.{\displaystyle \frac{\partial f_{Pj}}{\partial {\dot{\phi }}_{Pk}}}\right|}_{{\phi }_{Pk}=0,i\ne k}\right)}_{mod},j,k=x,y,z$$

(64)

and the hydrodynamic stiffness force relation is

$${\left({\left.{\displaystyle \frac{\partial f_{Pj}}{\partial {\phi }_{Pk}}}\right|}_{{\phi }_{Tk}=0,i\ne k}\right)}_{\mathrm{pro}}={\displaystyle \frac{D_{P, pro}^2}{D_{P,\mathrm{mod}}^2}}{\left({\left.{\displaystyle \frac{\partial f_{Pj}}{\partial {\phi }_{Pk}}}\right|}_{{\phi }_{Pk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(65)

Given the coefficients of the model and substituting Equations (31) and (33) into Equations (63)–(65), the hydrodynamic damping and stiffness force coefficients of the prototype can be obtained.

The hydrodynamic moment of the platform is

$$\begin{array}{c}{m}_{Pj}\left(V_{cur},{\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P,{\phi }_{Px},{\phi }_{Py},{\phi }_{Pz},{\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}\right)\equiv m_{Pj}\left(V_{cur},s_{P1},s_{P2},s_{P3},s_{P4},s_{P5},s_{P6},s_{P7},s_{P8},s_{P9}\right)\\ =m_{Pj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)+{\displaystyle \sum _{k=1}^9{s}_{Pk}}{\left.{\displaystyle \frac{\partial m_{Pj}}{\partial s_{Pk}}}\right|}_{s_{Pi}=0,i\ne k},j=x,y,z\end{array}$$

(66)

Dividing Equation (66) by ${\displaystyle \frac{1}{2}}\rho D_P{A}_{pla}{V}_{cur}^2$, the dimensionless variables become one:

$$\begin{array}{c}{C}_{mPj}\left(V_{cur},{\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P,{\phi }_{Px},{\phi }_{Py},{\phi }_{Pz},{\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}\right)\equiv C_{mTj}\left(V_{cur},s_{P1},s_{P2},s_{P3},s_{P4},s_{P5},s_{P6},s_{P7},s_{P8},s_{P9}\right)\\ =C_{mPj,0}\left(V_{cur},0,0,0,0,0,0,0,0,0\right)+{\displaystyle \sum _{k=1}^9{\tilde{s}}_{Pk}}{\left.{\displaystyle \frac{\partial C_{mPj}}{\partial {\tilde{s}}_{Pk}}}\right|}_{s_{Pi}=0,i\ne k},j=x,y,z\end{array}$$

(67)

where the first term on the right-hand side can be derived using the Buckingham theorem without FSI. The other terms are obtained with FSI. $C_{mPj}=m_{Pj}/\left({\displaystyle \frac{1}{2}}\rho D_P{A}_{pla}{V}_{cur}^2\right)$.

If the prototype and the model are similar, their dimensionless moments (67) are the same:

$${\left[C_{mPj}\right]}_{pro}={\left[C_{mPj}\right]}_{\mathrm{mod}}$$

(68)

Based on Equations (56), (67), and (68), we can obtain the following similar parameters.

Similarity for hydrodynamic damping in a moment with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{mPj}}{\partial {\tilde{s}}_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{mPj}}{\partial {\tilde{s}}_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{pro},j=x,y,z; s_{Pk}={\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P,{\dot{\phi }}_{Px},{\dot{\phi }}_{Py},{\dot{\phi }}_{Pz}$$

(69)

Similarity for hydrodynamic stiffness in a moment with FSI:

$${\left({\left.{\displaystyle \frac{\partial C_{mPj}}{\partial {\phi }_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{\mathrm{mod}}={\left({\left.{\displaystyle \frac{\partial C_{mPj}}{\partial {\phi }_{Pk}}}\right|}_{s_{Pi}=0,i\ne k}\right)}_{pro},j,k=x,y,z$$

(70)

Based on the similarity formulas in (59), (69), and (70), the hydrodynamic damping moment relations between the model and prototype are

$${\left({\left.{\displaystyle \frac{\partial m_{Pj}}{\partial s_{Pk}}}\right|}_{s_{Pk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{P, pro}^3}{D_{P,\mathrm{mod}}^3}}{\left({\left.{\displaystyle \frac{\partial m_{Pj}}{\partial s_{Pk}}}\right|}_{s_{Pk}=0,i\ne k}\right)}_{\mathrm{mod}},s_{Pk}={\dot{x}}_P,{\dot{y}}_P,{\dot{z}}_P$$

(71)

$${\left({\left.{\displaystyle \frac{\partial m_{Pj}}{\partial {\dot{\phi }}_{Pk}}}\right|}_{{\dot{\phi }}_{Pk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{P, pro}^3}{D_{P,\mathrm{mod}}^3}}{\displaystyle \frac{{\omega }_{T,\mathrm{mod}}}{{\omega }_{T,pro}}}{\left({\left.{\displaystyle \frac{\partial m_{Pj}}{\partial {\dot{\phi }}_{Pk}}}\right|}_{{\dot{\phi }}_{Pk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(72)

and the hydrodynamic stiffness moment relation is

$${\left({\left.{\displaystyle \frac{\partial m_{Pj}}{\partial {\phi }_{Pk}}}\right|}_{{\phi }_{Pk}=0,i\ne k}\right)}_{pro}={\displaystyle \frac{D_{P, pro}^3}{D_{P,\mathrm{mod}}^3}}{\left({\left.{\displaystyle \frac{\partial m_{Pj}}{\partial {\phi }_{Pk}}}\right|}_{{\phi }_{Pk}=0,i\ne k}\right)}_{\mathrm{mod}},j,k=x,y,z$$

(73)

Given the coefficients of the model and substituting Equations (31) and (33) into Equations (71)–(73), the hydrodynamic damping and stiffness moment coefficients of the prototype can be obtained.

4.1.3. Geometrical Inertia and Buoyance Similarities

Based on the similarity formula in (33) and considering the geometric similarity, we can obtain the following similarity relation:

$${\displaystyle \frac{D_{P,pro}}{D_{P,mod}}}={\displaystyle \frac{D_{T,pro}}{D_{T,mod}}}=r_d$$

(74)

where r_d is the scaling factor of the prototype and model dimensions.

Similarity for mass:

$${\displaystyle \frac{m_{pro}}{m_{\mathrm{mod}}}}:{\displaystyle \frac{{\rho }_{pro}Vo{l}_{pro}}{{\rho }_{\mathrm{mod}}Vo{l}_{\mathrm{mod}}}}=r_d^3$$

(75)

Similarity for inertia of mass:

$${\displaystyle \frac{I_{pro}}{I_{\mathrm{mod}}}}:{\displaystyle \frac{R_{pro}^2{m}_{pro}}{R_{\mathrm{mod}}^2{m}_{\mathrm{mod}}}}=r_d^5$$

(76)

Similarity for buoyance:

$${\displaystyle \frac{F_{Bpro}}{F_{B\mathrm{mod}}}}:{\displaystyle \frac{{\rho }_{water}Vo{l}_{pro}}{{\rho }_{water}Vo{l}_{\mathrm{mod}}}}=r_d^3$$

(77)

4.1.4. Some Similarity Results of Convertors

In this study, the similarity accuracy of some similarity formulas for convertors are verified using the commercial software STAR-CCM (CFD) and listed in

Table 2. Comparison of the parameters using similarity formulas and CFD.

Parameters	Similarity Equation	Model [*] [7]	Prototype
			Predicted [**]	CFD	Error (%)
Power (kW)	(33)	394 #	1000	1023.8	2.4
$C_{fTy,0}$	(34)	0.497	0.497	0.468	5.8
$-\partial C_{fTy}/\partial \left({\dot{y}}_{2d}/V_{cur}\right)$	(40)	1.066	1.066	1.017	4.6
$-\partial C_{fTx}/\partial \left({\dot{x}}_{2d}/V_{cur}\right)$	(40)	3.151	3.151	2.898	8.0
$C_{fPy,0}$	(60)	0.032	0.032	0.031	3.1
$-\partial C_{fPy}/\partial \left({\dot{y}}_{1d}/V_{cur}\right)$	(61)	0.034	0.034	0.035	2.9
$-\partial C_{fPx}/\partial \left({\dot{x}}_{1d}/V_{cur}\right)$	(61)	0.449	0.449	0.426	5.1
$-\partial C_{fPy}/\partial \left({\dot{x}}_{1d}/V_{cur}\right)$	(61)	0.0455	0.0455	0.0431	5.3
$-\partial C_{mPz}/\partial \left({\dot{x}}_{1d}/V_{cur}\right)$	(70)	1.417	1.417	1.453	2.5

* Using CFD, **: Using similarity equation, #: by Lin et al. [7].

. Lin et al. [7] presented a 394 kW convertor at V_cur = 1.6 m/s. According to this condition, the difference between the predicted and CFD powers is 2.4%. Moreover, the predicted and CFD hydrodynamic coefficients are very close.

4.2. Translational Motion in the x-Axis Direction

4.2.1. Equation of Heaving Motion for Pontoon 3

The heaving equation of Pontoon 3 is

$$M_3{\ddot{x}}_{3d}-K_{Cd}{x}_{1d}+\left(K_{Cd}+A_{Bx}\rho g\right)x_{3d}=F_{B3d}\left(t\right)={\displaystyle \sum _{i=1}^N }\left[f_{Bs,i}\sin {\Omega }_{i}t+f_{Bc,i}\cos {\Omega }_{i}t\right]$$

(78)

where $F_{B3d}\left(t\right)$ is the irregular wave force applied to Pontoon 3, $f_{Bs,i}=A_{Bx}\rho g{a}_i\cos {\phi }_i$, and $f_{Bc,i}=A_{Bx}\rho g{a}_i\sin {\phi }_i$.

The effective spring constant of the rope C–buffer spring connection is

$$K_{Cd}=K_{\mathrm{rope} \mathrm{C}}/\left(1+K_{\mathrm{rope} \mathrm{C}}/K_{C,spring}\right)$$

(79)

where $K_{C,spring}$ is the constant of the spring connected to rope C. $K_{rope C}=E_C{A}_C/L_C$, where $E_C,A_C,$ and L_C are the Young’s modulus, cross-sectional area, and length of rope C, respectively.

The dynamic tension of rope C is

$$T_{Cd}=K_{Cd}{\delta }_{Cd}$$

(80)

where the dynamic elongation between Floating Platform 1 and Pontoon 3 is ${\delta }_{Cd}=\left(x_{3d}-x_{1d}\right)$.

4.2.2. Equation of Heaving Motion for Pontoon 4

The heaving equation of Pontoon 4 is

$$M_4{\ddot{x}}_{4d}-K_{Dd}{x}_{2d}+\left(K_{Dd}+A_{BT}\rho g\right)x_{4d}=F_{B4d}\left(t\right)={\displaystyle \sum _{i=1}^N }\left[f_{Ts,i}\sin {\Omega }_{i}t+f_{Tc,i}\cos {\Omega }_{i}t\right]$$

(81)

where $F_{B4d}\left(t\right)$ is the irregular wave force applied to Pontoon 4. $f_{Ts,i}=A_{BT}\rho g{a}_i\cos \left({\phi }_i+{\varphi }_i\right)$ and $f_{Tc,i}=A_{BT}\rho g{a}_i\sin \left({\phi }_i+{\varphi }_i\right)$. The phase angle is ${\varphi }_i=2\pi L_E\cos \alpha /{\lambda }_i$ and $L_E=\sqrt{L_B^2-{\left(L_C-L_D\right)}^2}$. α is the relative wave–current angle, and ${\lambda }_i$ is the wavelength, as shown in Figure 5.

The effective spring constant of the rope D–buffer spring connection is

$$K_{Dd}=K_{\mathrm{rope} \mathrm{D}}/\left(1+K_{\mathrm{rope} \mathrm{D}}/K_{D,spring}\right)$$

(82)

in which $K_{D,spring}$ is the constant of the spring connected to rope D. $K_{rope \mathrm{D}}=E_D{A}_D/L_D$, where $E_D,A_D,$ and L_D are the Young’s modulus, cross-sectional area, and length of rope D, respectively.

The dynamic tension of rope C is

$$T_{Dd}=K_{Dd}{\delta }_{Dd},$$

(83)

where the dynamic elongation between Convertor 2 and Pontoon 4 is ${\delta }_{Dd}=\left(x_{4d}-x_{2d}\right)$.

4.2.3. Equation of Heaving Motion of the Platform

The dynamic equilibrium of the floating platform in the heaving motion is

$$-\left(M_1+m_{eff,x}\right){\ddot{x}}_{1d}+f_{\mathrm{Px}}+F_{B1s}-W_1+T_C-\left[T_{A1}\sin {\theta }_{A1}+T_{A2}\sin {\theta }_{A2}\right]-T_B\sin {\theta }_B=0$$

(84)

where the effective mass m_eff,x is derived in Appendix A. The hydrodynamic force on the floating platform due to the fluid–structure interaction (FSI) is expressed in a Taylor series as follows:

$$f_{\mathrm{Px}}\left(V,{\dot{x}}_{1d},{\dot{y}}_{1d},{\dot{z}}_{1d},{\phi }_{1x},{\phi }_{1y},{\phi }_{1z},{\dot{\phi }}_{1x},{\dot{\phi }}_{1y},{\dot{\phi }}_{1z}\right)=f_{Px}\left(V,0,0,0,0,0,0,0,0,0\right)+{\displaystyle \sum _{j=1}^9\frac{\partial f_{Px}}{\partial s_{1j}}{s}_{1j}}+o\left(s_{1m}{s}_{1n}\right)$$

(85)

Briefly, $\left({\dot{x}}_{kd},{\dot{y}}_{kd},{\dot{z}}_{kd},{\phi }_{kx},{\phi }_{ky},{\phi }_{kz},{\dot{\phi }}_{kx},{\dot{\phi }}_{ky},{\dot{\phi }}_{kz}\right)$ $\equiv$ $\left(s_{k1},s_{k2},s_{k3},s_{k4},s_{k5},s_{k6},s_{k7},s_{k8},s_{k9}\right),$ k = 1, 2. When a symmetrical configuration is considered for the platform, the hydrodynamic force on the platform in the x-direction under a current only is $f_{Px}\left(V,0,0,0,0,0,0,0,0,0\right)$ = 0. Considering a small oscillation, the higher-order terms are then neglected. The second term on the right-hand side of Equation (84) is the hydrodynamic force due to the fluid–structure interaction.

The dynamic tensions of ropes A and B are

$$T_{Ad}=K_{Ad}{\delta }_{Ad}, T_{Bd}=K_{Bd}{\delta }_{Bd}$$

(86)

The dynamic elongation is the difference between the dynamic and static lengths, ${\delta }_{\beta d}=L_{\beta d}-L_{\beta s}$, β = A, B. Using the Taylor formula, the dynamic elongation of rope A is derived:

$${\delta }_{Ad}={\epsilon }_{AX}{x}_{1d}+{\epsilon }_{AY}{y}_{1d}+{\epsilon }_{AZ}{z}_{1d}$$

(87)

where

$${\epsilon }_{AX}={\displaystyle \frac{\left(X_A-{\epsilon }_{x01}\right)}{r_{A1}}}+{\displaystyle \frac{\left(X_A-{\epsilon }_{x02}\right)}{r_{A2}}},{\epsilon }_{AY}={\displaystyle \frac{\left(Y_A-{\epsilon }_{y01}\right)}{r_{A1}}}+{\displaystyle \frac{\left(Y_A-{\epsilon }_{y02}\right)}{r_{A2}}},$$

$${\epsilon }_{AZ}={\displaystyle \frac{\left(Z_A-{\epsilon }_{z01}\right)}{r_{A1}}}+{\displaystyle \frac{\left(Z_A-{\epsilon }_{z02}\right)}{r_{A2}}}\hspace{0.5em}{X}_A={\displaystyle \frac{x_{1s}}{L_A}}, Y_A={\displaystyle \frac{y_{1s}}{L_A}}, Z_A={\displaystyle \frac{z_{1s}}{L_A}},$$

$${\epsilon }_{x01}={\displaystyle \frac{x_{01}}{L_A}},{\epsilon }_{y01}={\displaystyle \frac{y_{01}}{L_A}},{\epsilon }_{z01}={\displaystyle \frac{z_{01}}{L_A}},\hspace{0.5em}{\epsilon }_{x02}={\displaystyle \frac{x_{01}}{L_A}},{\epsilon }_{y02}={\displaystyle \frac{y_{01}}{L_A}},{\epsilon }_{z02}={\displaystyle \frac{z_{01}}{L_A}}, r_{A1}={\displaystyle \frac{L_{A1}}{L_A}},r_{A2}={\displaystyle \frac{L_{A2}}{L_A}}$$

The dynamic elongation of rope B

$${\delta }_{Bd}=X_B\left(x_{2d}-x_{1d}\right)+Y_B\left(y_{2d}-y_{1d}\right)+Z_B\left(z_{2d}-z_{1d}\right)$$

(88)

where $X_B={\displaystyle \frac{\left(x_{2s}-x_{1s}\right)}{L_B}}, Y_B={\displaystyle \frac{\left(y_{2s}-y_{1s}\right)}{L_B}}, Z_B={\displaystyle \frac{\left(z_{2s}-z_{1s}\right)}{L_B}}$.

Substituting Equations (83)–(86) into Equation (82), we can obtain the following:

$$\begin{array}{l}-\left(M_1+m_{eff,x}\right){\ddot{x}}_{1d}+\left({\displaystyle \sum _{j=1}^3\frac{\partial f_{Px}}{\partial s_{1j}}{s}_{1j}}+{\displaystyle \sum _{j=7}^9\frac{\partial f_{Px}}{\partial s_{1j}}{s}_{1j}}\right)+{\displaystyle \sum _{j=4}^6\frac{\partial f_{Px}}{\partial s_{1j}}{s}_{1j}}\\ +\left[-K_{Cd}-T_{As}{\displaystyle \sum _{i=1}^2 }\left({\displaystyle \frac{\cos {\theta }_{Ais}}{L_{Ai}}}\right)-K_{Ad}{\epsilon }_{AX}{\displaystyle \sum _{i=1}^2\sin {\theta }_{Ais}}+{\displaystyle \frac{T_{Bs}\cos {\theta }_{Bs}}{L_B}}+K_{Bd}{X}_B\sin {\theta }_{Bs}\right]x_{1d}\\ -\left[{\displaystyle \frac{T_{Bs}\cos {\theta }_{Bs}}{L_B}}+K_{Bd}{X}_B\sin {\theta }_{Bs}\right]x_{2d}+K_{Cd}{x}_{3d}+\left(-K_{Ad}{\epsilon }_{AY}+K_{Bd}{Y}_B\sin {\theta }_{Bs}\right)y_{1d}\\ -K_{Bd}{Y}_B\sin {\theta }_{Bs}{y}_{2d}+\left(-K_{Ad}{\epsilon }_{AZ}{\displaystyle \sum _{i=1}^2\sin {\theta }_{Ais}}+K_{Bd}{Z}_B\sin {\theta }_{Bs}\right)z_{1d}-K_{Bd}{Z}_B\sin {\theta }_{Bs}{z}_{2d}=0\end{array}$$

(89)

4.2.4. Equation of Heaving Motion for the Convertor

The dynamic equilibrium of the convertor in the heaving motion is [8]

$$\begin{array}{c}{M}_2{\ddot{x}}_{2d}-\left({\displaystyle \sum _{j=1}^3\frac{\partial f_{Tx}}{\partial s_{2j}}{s}_{1j}}+{\displaystyle \sum _{j=7}^9\frac{\partial f_{Tx}}{\partial s_{2j}}{s}_{1j}}\right)-{\displaystyle \sum _{j=4}^6\frac{\partial f_{Tx}}{\partial s_{2j}}{s}_{2j}}+\left({\displaystyle \frac{T_{\mathrm{Bs}}\cos {\theta }_{Bs}}{L_B}}+\sin {\theta }_{Bs}{K}_{Bd}{\displaystyle \frac{\left(x_{2s}-x_{1s}\right)}{L_B}}\right)x_{1d}\\ +\left(K_{Dd}-{\displaystyle \frac{T_{\mathrm{Bs}}\cos {\theta }_{Bs}}{L_B}}-\sin {\theta }_{Bs}{K}_{Bd}{\displaystyle \frac{\left(x_{2s}-x_{1s}\right)}{L_B}}\right)x_{2d}\\ -K_{Dd}{x}_{4d}+\sin {\theta }_{Bs}{K}_{Bd}{\displaystyle \frac{\left(y_{2s}-y_{1s}\right)}{L_B}}{y}_{1d}-\sin {\theta }_{Bs}{K}_{Bd}{\displaystyle \frac{\left(y_{2s}-y_{1s}\right)}{L_B}}{y}_{2d}=0\end{array}$$

(90)

4.3. Translational Motion in the y-Direction

4.3.1. Equation of Surging Motion of Platform

The dynamic equilibrium of the floating platform in the surging motion is

$$-\left(M_1+m_{eff,y}\right){\ddot{y}}_{1d}+f_{py}-\left[T_{A1}\cos {\theta }_{A1}\cos \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)+T_{A2}\cos {\theta }_{A2}\cos {\mathsf{\Delta }}_2\right]+T_B\cos {\theta }_B=0$$

(91)

where the effective mass m_eff,y is derived in Appendix A. The hydrodynamic force is

$$f_{\mathrm{Py}}=f_{pys}+f_{pyd}$$

(92)

where $f_{Pys}=f_{Py}\left(V,0,0,0,0,0,0,0,0,0\right)=C_{DPy}{\displaystyle \frac{1}{2}}\rho A_{PY}{V}^2$ and $f_{pyd}={\displaystyle \sum _{j=1}^9\frac{\partial f_{Py}}{\partial s_{1j}}{s}_{1j}}$.

Substituting Equations (23), (84)–(86), and (90) into Equation (89), we can obtain the following:

$$\begin{array}{l}\left(M_1+m_{eff,y}\right){\ddot{y}}_{1d}-\left({\displaystyle \frac{\partial f_{Py}}{\partial {\dot{x}}_{1d}}}{\dot{x}}_{1d}+{\displaystyle \frac{\partial f_{Py}}{\partial {\dot{y}}_{1d}}}{\dot{y}}_{1d}+{\displaystyle \frac{\partial f_{Py}}{\partial {\dot{z}}_{1d}}}{\dot{z}}_{1d}+{\displaystyle \frac{\partial f_{Py}}{\partial {\dot{\phi }}_{px}}}{\dot{\phi }}_{px}+{\displaystyle \frac{\partial f_{Py}}{\partial {\dot{\phi }}_{py}}}{\dot{\phi }}_{py}+{\displaystyle \frac{\partial f_{Py}}{\partial {\dot{\phi }}_{pz}}}{\dot{\phi }}_{pz}\right)\\ -\left({\displaystyle \frac{\partial f_{Py}}{\partial {\phi }_{px}}}{\phi }_{px}+{\displaystyle \frac{\partial f_{Py}}{\partial {\phi }_{py}}}{\phi }_{py}+{\displaystyle \frac{\partial f_{Py}}{\partial {\phi }_{pz}}}{\phi }_{pz}\right)-\left(d_1+d_2{K}_{Ad}{\epsilon }_{AX}-K_{Bd}{X}_B\cos {\theta }_{Bs}\right)x_{1d}\\ -\left(d_3+K_{Bd}{X}_B\cos {\theta }_{Bs}\right)x_{2d}-\left(d_2{K}_{Ad}{\epsilon }_{AY}-K_{Bd}{Y}_B\cos {\theta }_{Bs}\right)y_{1d}-K_{Bd}{Y}_B\cos {\theta }_{Bs}{y}_{2d}\\ -\left(d_2{K}_{Ad}{\epsilon }_{AZ}-K_{Bd}{Z}_B\cos {\theta }_{Bs}\right)z_{1d}-K_{Bd}{Z}_B\cos {\theta }_{Bs}{z}_{2d}=0\end{array}$$

(93)

where $d_1=T_{As}\left[{\displaystyle \frac{\sin {\theta }_{As1}\cos \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)}{L_{A1}}}+{\displaystyle \frac{\sin {\theta }_{As2}\cos {\mathsf{\Delta }}_2}{L_{A2}}}\right]-{\displaystyle \frac{T_{Bs}\sin {\theta }_{Bs}}{L_B}},$

$$d_2=-\left[\cos {\theta }_{As1}\cos \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)+\cos {\theta }_{As2}\cos {\mathsf{\Delta }}_2\right], d_3=-{\displaystyle \frac{T_{Bs}\sin {\theta }_{Bs}}{L_B}}.$$

4.3.2. Equation of Surging Motion of the Convertor in the y-Direction

The dynamic equilibrium of the convertor in the surging motion is

$$-M_2{\ddot{y}}_{2d}+f_{Ty}-T_B\cos {\theta }_B=0$$

(94)

The hydrodynamic force on the convertor is expressed as

$$f_{Ty}=f_{Tys}+f_{Tyd}$$

(95)

where $f_{Tys}=f_{Ty}\left(V,0,0,0,0,0,0,0,0,0,TRS\right)=C_{DTy}{\displaystyle \frac{1}{2}}\rho A_{TY}{V}^2$ and A_Ty is the effective operating area of the convertor. $f_{Tyd}={\displaystyle \sum _{j=1}^9\frac{\partial f_{Ty}}{\partial s_{2j}}{s}_{2j}}$.

Substituting Equations (86), (88), and (95) into Equation (94), we can obtain the following:

$$\begin{array}{l}{M}_2{\ddot{y}}_{2d}-\left({\displaystyle \frac{\partial f_{Tx}}{\partial {\dot{x}}_{2d}}}{\dot{x}}_{2d}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\dot{y}}_{2d}}}{\dot{y}}_{2d}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\dot{z}}_{2d}}}{\dot{z}}_{2d}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\dot{\phi }}_{Tx}}}{\dot{\phi }}_{Tx}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\dot{\phi }}_{Ty}}}{\dot{\phi }}_{Ty}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\dot{\phi }}_{Tz}}}{\dot{\phi }}_{Tz}\right)\\ \hspace{0.5em}\hspace{0.5em}-\left({\displaystyle \frac{\partial f_{Tx}}{\partial {\phi }_{Tx}}}{\phi }_{Tx}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\phi }_{Ty}}}{\phi }_{Ty}+{\displaystyle \frac{\partial f_{Tx}}{\partial {\phi }_{Tz}}}{\phi }_{Tz}\right)\\ \hspace{0.5em}\hspace{0.5em}+K_{Bd}\cos {\theta }_{Bs}\left[-X_B{x}_{1d}+X_B{x}_{2d}-Y_B{y}_{1d}+Y_B{y}_{2d}-Z_B{z}_{1d}+Z_B{z}_{2d}\right]=0\end{array}$$

(96)

4.3.3. Equation of Surging Motion of Pontoon 3 in the y-Direction

The dynamic equilibrium of Pontoon 3 in the surging motion is

$$M_3{\ddot{y}}_{3\mathrm{d}}+{\displaystyle \frac{T_{Cs}}{L_C}}\left(y_{3d}-y_{1d}\right)=0$$

(97)

4.3.4. Equation of Surging Motion of Pontoon 4 in the y-Direction

The dynamic equilibrium of Pontoon 4 in the surging motion is

$$M_4{\ddot{y}}_{4d}+{\displaystyle \frac{T_{Ds}}{L_D}}\left(y_{4d}-y_{2d}\right)=0$$

(98)

4.4. Translational Motion in the z-Direction

4.4.1. Equation of Swaying Motion of Platform

The dynamic equilibrium of the floating platform in the swaying motion is

$$\begin{array}{cc}\left(M_1+m_{eff,z}\right){\ddot{z}}_{1d}-f_{Pz}-[T_{A1}\cos {\theta }_{A1}\sin \hfill & \left({\mathsf{\Delta }}_1+{\varphi }_{cur}\right)-T_{A2}\cos {\theta }_{A2}\sin {\mathsf{\Delta }}_2]\hfill \\ & -T_B\cos {\theta }_B\left(\sin {\varphi }_B\right)-T_C\sin {\varphi }_C=0\hfill \end{array}$$

(99)

where the effective mass m_eff,z is derived in Appendix A. The hydrodynamic force is

$$f_{\mathrm{Pz}}={\displaystyle \sum _{j=1}^9\frac{\partial f_{\mathrm{Pz}}}{\partial s_{1j}}{s}_{1j}}$$

(100)

Considering small displacements and based on Equations (78), (84)–(86), and (98), we can obtain the following:

$$\begin{array}{c}\left(M_1+m_{eff,z}\right){\ddot{z}}_{1d}-\left({\displaystyle \frac{\partial f_{Pz}}{\partial {\dot{x}}_{1d}}}{\dot{x}}_{1d}+{\displaystyle \frac{\partial f_{Pz}}{\partial {\dot{y}}_{1d}}}{\dot{y}}_{1d}+{\displaystyle \frac{\partial f_{Pz}}{\partial {\dot{z}}_{1d}}}{\dot{z}}_{1d}+{\displaystyle \frac{\partial f_{Pz}}{\partial {\dot{\phi }}_{px}}}{\dot{\phi }}_{px}+{\displaystyle \frac{\partial f_{Pz}}{\partial {\dot{\phi }}_{py}}}{\dot{\phi }}_{py}+{\displaystyle \frac{\partial f_{Pz}}{\partial {\dot{\phi }}_{pz}}}{\dot{\phi }}_{pz}\right)\hfill \\ -{\displaystyle \frac{\partial f_{Pz}}{\partial {\phi }_{px}}}{\phi }_{px}-{\displaystyle \frac{\partial f_{Pz}}{\partial {\phi }_{py}}}{\phi }_{py}-{\displaystyle \frac{\partial f_{Pz}}{\partial {\phi }_{pz}}}{\phi }_{pz}+{\displaystyle \frac{T_{As}}{L_A}}\left({\displaystyle \frac{\sin {\theta }_{As1}\sin \left({\mathsf{\Delta }}_1+\varphi \right)}{r_{A1}}}-{\displaystyle \frac{\sin {\theta }_{As2}\sin {\mathsf{\Delta }}_2}{r_{A2}}}\right)x_{1d}\hfill \\ \hfill +\left({\displaystyle \frac{T_{Bs}}{L_B}}+{\displaystyle \frac{T_{Cs}}{L_C}}\right)z_{1d}-{\displaystyle \frac{T_{Bs}}{L_B}}{z}_{2d}-{\displaystyle \frac{T_{Cs}}{L_C}}{z}_{3d}=0\end{array}$$

(101)

4.4.2. Equation of Swaying Motion of Convertor

The dynamic equilibrium of the convertor in the swaying motion is

$$M_2{\ddot{z}}_{2d}-\left({\displaystyle \sum _{j=1}^3\frac{\partial f_{Tz}}{\partial s_{2j}}{s}_{2j}}+{\displaystyle \sum _{j=7}^9\frac{\partial f_{Tz}}{\partial s_{2j}}{s}_{2j}}\right)-{\displaystyle \sum _{j=4}^6\frac{\partial f_{Tz}}{\partial s_{2j}}{s}_{2j}}-{\displaystyle \frac{T_{Bs}}{L_B}}{z}_{1d}+\left({\displaystyle \frac{T_{Bs}}{L_B}}+{\displaystyle \frac{T_{Ds}}{L_D}}\right)z_{2d}-{\displaystyle \frac{T_{Ds}}{L_D}}{z}_{4d}=0$$

(102)

4.4.3. Equation of Swaying Motion for Pontoon 3

The dynamic equilibrium of Pontoon 3 in the swaying motion is [8]

$$M_3{\ddot{z}}_{3d}+{\displaystyle \frac{T_{Cs}}{L_C}}\left(z_{3d}-z_{1d}\right)=0$$

(103)

4.4.4. Equation of Swaying Motion of Pontoon 4

The dynamic equilibrium of Pontoon 4 in the swaying motion is [8]

$$M_4{\ddot{z}}_{4d}+{\displaystyle \frac{T_{Ds}}{L_D}}\left(z_{4d}-z_{2d}\right)=0$$

(104)

4.5. Rotational Motion

4.5.1. Equation of Yawing Motion of Convertor

The dynamic equilibrium of the convertor in the yawing motion is [8]

$$\begin{array}{l}{I}_{Tx}{\ddot{\phi }}_{2x}-\left({\displaystyle \sum _{j=1}^3\frac{\partial m_{Tx}}{\partial s_{2j}}{s}_{2j}}+{\displaystyle \sum _{j=7}^9\frac{\partial m_{Tx}}{\partial s_{2j}}{s}_{2j}}\right)-{\displaystyle \sum _{j=4}^6\frac{\partial m_{Tx}}{\partial s_{2j}}{s}_{2j}}+\left(T_{Bs}\cos {\theta }_{Bs}{R}_{TBx}\right){\phi }_{2x}\\ -\left({\displaystyle \frac{T_{Bs}{R}_{TBx}}{L_B}}\right)z_{1d}+\left({\displaystyle \frac{T_{Bs}{R}_{TBx}}{L_B}}\right)z_{2d}=0\end{array}$$

(105)

4.5.2. Equation of Rolling Motion of Convertor

The dynamic equilibrium of the convertor in the rolling motion is [8]

$$I_y{\ddot{\phi }}_{2y}-\left({\displaystyle \sum _{j=1}^3\frac{\partial m_{Ty}}{\partial s_{2j}}{s}_{2j}}+{\displaystyle \sum _{j=7}^9\frac{\partial m_{Ty}}{\partial s_{2j}}{s}_{2j}}\right)-{\displaystyle \sum _{j=4}^6\frac{\partial m_{Ty}}{\partial s_{2j}}{s}_{2j}}+T_{Ds}{R}_{TDy}{\phi }_{2y}+{\displaystyle \frac{T_{Ds}{R}_{TDy}}{L_D}}{z}_{2d}-{\displaystyle \frac{T_{Ds}{R}_{TDy}}{L_D}}{z}_{4d}=0$$

(106)

4.5.3. Equation of Pitching Motion of Convertor

The dynamic equilibrium of the convertor in the pitching motion is [8]

$$\begin{array}{l}{I}_{Tz}{\ddot{\phi }}_{2z}-\left({\displaystyle \sum _{j=1}^3\frac{\partial m_{Tzd}}{\partial s_{2j}}{s}_{2j}}+{\displaystyle \sum _{j=7}^9\frac{\partial m_{Tzd}}{\partial s_{2j}}{s}_{2j}}\right)-{\displaystyle \sum _{j=4}^6\frac{\partial m_{Tzd}}{\partial s_{2j}}{s}_{2j}}+\left(T_{Bs}{R}_{TBz}\cos {\theta }_B\right){\phi }_{2z}\\ \hspace{0.5em}\hspace{0.5em}-{\displaystyle \frac{T_{Bs}{R}_{TBz}\cos {\theta }_B}{L_B}}{x}_{1d}+{\displaystyle \frac{T_{Bs}{R}_{TBz}\cos {\theta }_B}{L_B}}{x}_{2d}=0\end{array}$$

(107)

4.5.4. Equation of Yawing Motion of Platform

The dynamic equilibrium of the floating platform in the yawing motion is

$$\begin{array}{l}{I}_{Px}{\ddot{\phi }}_{Px}-m_{Px}+T_A{R}_{PAx}\left[\cos \left({\theta }_{A1s}+\mathsf{\Delta }{\theta }_{A1}\right)\sin \left({\phi }_{Px}-\mathsf{\Delta }{\varphi }_{x1}\right)+\cos \left({\theta }_{A2s}+\mathsf{\Delta }{\theta }_{A2}\right)\sin \left({\phi }_{Px}-\mathsf{\Delta }{\varphi }_{x2}\right)\right]\\ +T_B\cos \left({\theta }_{Bs}+\mathsf{\Delta }{\theta }_B\right)R_{PBx}\sin \left({\phi }_{Px}-\mathsf{\Delta }{\theta }_x\right)=0\end{array}$$

(108)

where $\mathsf{\Delta }{\varphi }_{x1}={\displaystyle \frac{z_{1d}}{L_{A1}\cos {\theta }_{A1s}}},\mathsf{\Delta }{\varphi }_{x2}={\displaystyle \frac{z_{1d}}{L_{A2}\cos {\theta }_{A2s}}},\mathsf{\Delta }{\theta }_x={\displaystyle \frac{z_{2d}-z_{1d}}{L_B\cos {\theta }_{Bs}}}$. Substituting Equations (86)–(88) and the hydrodynamic moment m_px into Equation (108), we can obtain the following:

$$\begin{array}{l}{I}_{Px}{\ddot{\phi }}_{Px}-\left({\displaystyle \frac{\partial m_{Px}}{\partial {\dot{x}}_{1d}}}{\dot{x}}_{1d}+{\displaystyle \frac{\partial m_{Px}}{\partial {\dot{y}}_{1d}}}{\dot{y}}_{1d}+{\displaystyle \frac{\partial m_{Px}}{\partial {\dot{z}}_{1d}}}{\dot{z}}_{1d}+{\displaystyle \frac{\partial m_{Px}}{\partial {\dot{\phi }}_{Px}}}{\dot{\phi }}_{Px}+{\displaystyle \frac{\partial m_{Px}}{\partial {\dot{\phi }}_{Py}}}{\dot{\phi }}_{Py}+{\displaystyle \frac{\partial m_{Px}}{\partial {\dot{\phi }}_{Pz}}}{\dot{\phi }}_{Pz}\right)\\ +\left(T_{As}{R}_{PAx}\left(\cos {\theta }_{A1s}+\cos {\theta }_{A2s}\right)+T_{Bs}\cos {\theta }_{Bs}{R}_{PBx}-{\displaystyle \frac{\partial m_{Px}}{\partial {\phi }_{px}}}\right){\phi }_{Px}-{\displaystyle \frac{\partial m_{Px}}{\partial {\phi }_{Py}}}{\phi }_{Py}-{\displaystyle \frac{\partial m_{Px}}{\partial {\phi }_{Pz}}}{\phi }_{Pz}\\ +\left({\displaystyle \frac{T_{Bs}{R}_{PBx}}{L_B}}-T_{As}{R}_{PAx}\left({\displaystyle \frac{1}{L_{A1}}}+{\displaystyle \frac{1}{L_{A2}}}\right)\right)z_{1d}-{\displaystyle \frac{T_{Bs}{R}_{PBx}}{L_B}}{z}_{2d}=0\end{array}$$

(109)

4.5.5. Equation of Rolling Motion of Platform

The dynamic equilibrium of the floating platform in the rolling motion is

$$\begin{array}{l}{I}_{Py}{\ddot{\phi }}_{Py}-m_{Py}+T_A{R}_{PAy}\left[\cos \left({\theta }_{A1s}+\mathsf{\Delta }{\theta }_{A1}\right)\sin \left({\phi }_{Py}+\mathsf{\Delta }{\varphi }_{A1y}\right)+\cos \left({\theta }_{A2s}+\mathsf{\Delta }{\theta }_{A2}\right)\sin \left({\phi }_{Py}+\mathsf{\Delta }{\varphi }_{A2y}\right)\right]\\ +T_C{R}_{PCy}\sin \left({\phi }_{Py}+\mathsf{\Delta }{\varphi }_{Cy}\right)=0\end{array}$$

(110)

where $\mathsf{\Delta }{\varphi }_{Cy}={\displaystyle \frac{z_{1d}-z_{3d}}{L_{\mathrm{C}}}},\mathsf{\Delta }{\varphi }_{A1y}={\displaystyle \frac{z_{1d}}{L_{A1}\sin {\theta }_{A1s}}},\mathsf{\Delta }{\varphi }_{A2y}={\displaystyle \frac{z_{1d}}{L_{A2}\sin {\theta }_{A2s}}}$. Substituting Equations (80), (86), and (87) and the hydrodynamic moment m_py into Equation (110), we can obtain the following:

$$\begin{array}{l}{I}_{Py}{\ddot{\phi }}_{Py}-\left({\displaystyle \frac{\partial m_{Py}}{\partial {\dot{x}}_{1d}}}{\dot{x}}_{1d}+{\displaystyle \frac{\partial m_{Py}}{\partial {\dot{y}}_{1d}}}{\dot{y}}_{1d}+{\displaystyle \frac{\partial m_{Py}}{\partial {\dot{z}}_{1d}}}{\dot{z}}_{1d}+{\displaystyle \frac{\partial m_{Py}}{\partial {\dot{\phi }}_{Px}}}{\dot{\phi }}_{Px}+{\displaystyle \frac{\partial m_{Py}}{\partial {\dot{\phi }}_{Py}}}{\dot{\phi }}_{Py}+{\displaystyle \frac{\partial m_{Py}}{\partial {\dot{\phi }}_{Pz}}}{\dot{\phi }}_{Pz}\right)\\ -{\displaystyle \frac{\partial m_{Py}}{\partial {\phi }_{px}}}{\phi }_{px}+\left[T_{As}{R}_{PAy}\left(\cos {\theta }_{A1s}+\cos {\theta }_{A2s}\right)+T_{Cs}{R}_{PCy}-{\displaystyle \frac{\partial m_{Py}}{\partial {\phi }_{Py}}}\right]{\phi }_{Py}-{\displaystyle \frac{\partial m_{Py}}{\partial {\phi }_{Pz}}}{\phi }_{Pz}\\ +\left(T_{As}{R}_{PAy}\left({\displaystyle \frac{1}{L_{A1}}}+{\displaystyle \frac{1}{L_{A2}}}\right)+{\displaystyle \frac{T_{Cs}{R}_{PCy}}{L_{\mathrm{C}}}}\right)z_{1d}-{\displaystyle \frac{T_{Cs}{R}_{PCy}}{L_{\mathrm{C}}}}{z}_{3d}=0\end{array}$$

(111)

4.5.6. Equation of Pitching Motion of Platform

The dynamic pitching equilibrium of the floating platform about the z-axis is

$$\begin{array}{l}{I}_{Pz}{\ddot{\phi }}_{Pz}-m_{Pz}+T_A{R}_{PAz}\left[\cos \left({\theta }_{As1}+\mathsf{\Delta }{\theta }_{A1}\right)\sin \left({\phi }_{Pz}+\mathsf{\Delta }{\theta }_{A1}\right)+\cos \left({\theta }_{As2}+\mathsf{\Delta }{\theta }_{A2}\right)\sin \left({\phi }_{Pz}+\mathsf{\Delta }{\theta }_{A2}\right)\right]\\ +T_B\cos \left({\theta }_{Bs}+\mathsf{\Delta }{\theta }_B\right)R_{PBz}\sin \left({\phi }_{Pz}+\mathsf{\Delta }{\theta }_B\right)+T_C{R}_{PCz}\sin \left({\phi }_{Pz}+\mathsf{\Delta }{\theta }_C\right)=0\end{array}$$

(112)

where $\mathsf{\Delta }{\theta }_{A1}={\displaystyle \frac{x_{1d}}{L_{A1}}}, \mathsf{\Delta }{\theta }_{A2}={\displaystyle \frac{x_{1d}}{L_{A2}}},\mathsf{\Delta }{\theta }_B={\displaystyle \frac{x_{2d}-x_{1d}}{L_B}}, \mathsf{\Delta }{\theta }_C={\displaystyle \frac{y_{2d}-y_{1d}}{L_C}}$. Substituting Equations (80), (86), and (87) and the hydrodynamic moment m_py into Equation (112), we can obtain the following:

$$\begin{array}{l}{I}_{Pz}{\ddot{\phi }}_{Pz}-\left({\displaystyle \frac{\partial m_{Pz}}{\partial {\dot{x}}_{1d}}}{\dot{x}}_{1d}+{\displaystyle \frac{\partial m_{Pz}}{\partial {\dot{y}}_{1d}}}{\dot{y}}_{1d}+{\displaystyle \frac{\partial m_{Pz}}{\partial {\dot{z}}_{1d}}}{\dot{z}}_{1d}+{\displaystyle \frac{\partial m_{Pz}}{\partial {\dot{\phi }}_{Px}}}{\dot{\phi }}_{Px}+{\displaystyle \frac{\partial m_{Pz}}{\partial {\dot{\phi }}_{Py}}}{\dot{\phi }}_{Py}+{\displaystyle \frac{\partial m_{Pz}}{\partial {\dot{\phi }}_{Pz}}}{\dot{\phi }}_{Pz}\right)\\ -{\displaystyle \frac{\partial m_{Pz}}{\partial {\phi }_{px}}}{\phi }_{px}-{\displaystyle \frac{\partial m_{Pz}}{\partial {\phi }_{Py}}}{\phi }_{Py}+\left(T_{As}{R}_{PAz}\left(\cos {\theta }_{A1s}+\cos {\theta }_{A2s}\right)+T_{Bs}\cos {\theta }_{Bs}{R}_{PBz}+T_{Cs}{R}_{PCz}-{\displaystyle \frac{\partial m_{Pz}}{\partial {\phi }_{Pz}}}\right){\phi }_{Pz}\\ +\left(T_{As}{R}_{PAz}\left({\displaystyle \frac{\cos {\theta }_{A1s}}{L_{A1}}}+{\displaystyle \frac{\cos {\theta }_{A2s}}{L_{A2}}}\right)-{\displaystyle \frac{T_{Bs}\cos {\theta }_{Bs}{R}_{PBz}}{L_B}}\right)x_{1d}\\ +{\displaystyle \frac{T_{Bs}\cos {\theta }_{Bs}{R}_{PBz}}{L_B}}{x}_{2d}+{\displaystyle \frac{T_{Cs}{R}_{PCz}}{L_C}}\left(y_{2d}-y_{1d}\right)=0\end{array}$$

(113)

4.6. Solution Method of Dynamic Displacements

The governing Equations (78), (81), (89), (90), (93)–(98), (101)–(107), (109), (111) and (113) can be expressed in matrix format:

$$M{\ddot{Z}}_d+C{\dot{Z}}_d+K{Z}_d=F_d$$

(114)

where the dynamic displacement vector $Z_d=\left[\begin{array}{ccccccccc}{x}_{1d}& y_{1d}& z_{1d}& x_{2d}& y_{2d}& z_{2d}& x_{3d}& y_{3d}& z_{3d}\end{array}\right.\begin{array}{ccc}{x}_{4d}& y_{4d}& z_{4d}\end{array}$${\left.\begin{array}{cccccc}{\phi }_{Tx}& {\phi }_{Ty}& {\phi }_{Tz}& {\phi }_{Px}& {\phi }_{Py}& {\phi }_{Pz}\end{array}\right]}^T$.

The elements of the force vector are

$$F_d={\left[\begin{array}{cccccccccccccccccc}0& 0& 0& 0& 0& 0& f_7& 0& 0& 0& f_{10}& 0& 0& 0& 0& 0& 0& 0\end{array}\right]}_{18\times 1}^T$$

(115)

where

$$\begin{gathered} f_7={\displaystyle \sum _{i=1}^N }\left[f_{Bs,i}\sin {\Omega }_{i}t+f_{Bc,i}\cos {\Omega }_{i}t\right], f_{10}={\displaystyle \sum _{i=1}^N }\left[f_{Ts,i}\sin {\Omega }_{i}t+f_{Tc,i}\cos {\Omega }_{i}t\right], \\ {f}_{Bs,i}=A_{Bx}\rho g{a}_i\cos {\phi }_i,\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{f}_{Bc,i}=A_{Bx}\rho g{a}_i\sin {\phi }_i, \\ {f}_{Ts,i}=A_{BT}\rho g{a}_i\cos \left({\phi }_i+{\varphi }_i\right),\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}\hspace{0.5em}{f}_{Tc,i}=A_{BT}\rho g{a}_i\sin \left({\phi }_i+{\varphi }_i\right). \end{gathered}$$

(116)

According to the FSI parameters of the 400 kW convertor and platform presented by Lin et al. [8] and considering the similarity between some prototypes with different powers, for example, 1 MW, the elements of the mass matrix $M={\left[M_{ij}\right]}_{18\times 18}$, damping matrix $C={\left[C_{ij}\right]}_{18\times 18}$, and stiffness matrix $K={\left[K_{ij}\right]}_{18\times 18}$ for the an MW-level prototype are determined using the similarity formula in Section 4.1 and listed in Appendix B, Appendix C and Appendix D, respectively. Equation (114) can be rewritten as

$${\ddot{Z}}_d+M^{-1}C{\dot{Z}}_d+M^{-1}K{Z}_d=M^{-1}{F}_d={\displaystyle \sum _{i=1}^N }\left(F_{s,i}\sin {\Omega }_{i}t+F_{c,i}\cos {\Omega }_{i}t\right)$$

(117)

where

$$\begin{gathered} F_{s,i}={\left[\begin{array}{cccccccccccccccccc}0& 0& 0& 0& 0& 0& {\displaystyle \frac{f_{Bs,i}}{M_{77}}}& 0& 0& 0& {\displaystyle \frac{f_{Ts,i}}{M_{1010}}}& 0& 0& 0& 0& 0& 0& 0\end{array}\right]}^T, \\ {F}_{c,i}=\left[\begin{array}{cccccccccccccccccc}0& 0& 0& 0& 0& 0& {\displaystyle \frac{f_{Bc,i}}{M_{77}}}& 0& 0& 0& {\displaystyle \frac{f_{Tc,i}}{M_{1010}}}& 0& 0& 0& 0& 0& 0& 0\end{array}\right] \end{gathered}$$

(118)

The solution can be expressed as

$$\begin{array}{cc}\hfill Z_d& ={\left[\begin{array}{cccccccccccccccccc}{x}_{1d}& y_{1d}& z_{1d}& x_{2d}& y_{2d}& z_{2d}& x_{3d}& y_{3d}& z_{3d}& x_{4d}& y_{4d}& z_{4d}& {\phi }_{Tx}& {\phi }_{Ty}& {\phi }_{Tz}& {\phi }_{Px}& {\phi }_{Py}& {\phi }_{Pz}\end{array}\right]}^T\hfill \\ & ={\displaystyle \sum _{i=1}^N }\left(Z_{dc,i}\cos {\Omega }_{i}t+Z_{ds,i}\sin {\Omega }_{i}t\right)\hfill \end{array}$$

(119)

where

$$\begin{gathered} Z_{dc,i}={\left[\begin{array}{cccccccccccccccccc}{x}_{1dc,i}& y_{1dc,i}& z_{1dc,i}& x_{2dc,i}& y_{2dc,i}& z_{2dc,i}& x_{3dc,i}& y_{3dc,i}& z_{3dc,i}& x_{4dc,i}& y_{4dc,i}& z_{4dc,i}& {\phi }_{Txc,i}& {\phi }_{Tyc,i}& {\phi }_{Tzc,i}& {\phi }_{Pxc,i}& {\phi }_{Pyc,i}& {\phi }_{Pzc,i}\end{array}\right]}^T \\ {Z}_{ds,i}={\left[\begin{array}{cccccccccccccccccc}{x}_{1ds,i}& y_{1ds,i}& z_{1ds,i}& x_{2ds,i}& y_{2ds,i}& z_{2ds,i}& x_{3ds,i}& y_{3ds,i}& z_{3ds,i}& x_{4ds,i}& y_{4ds,i}& z_{4ds,i}& {\phi }_{Txs,i}& {\phi }_{Tys,i}& {\phi }_{Tzs,i}& {\phi }_{Pxs,i}& {\phi }_{Pys,i}& {\phi }_{Pzs,i}\end{array}\right]}^T \end{gathered}$$

(120)

Substituting the solution (119) into Equation (117), we can obtain the following:

$$\begin{array}{l}-{\displaystyle \sum _{i=1}^N{\Omega }_i^2}\left(Z_{dc,i}\cos {\Omega }_{i}t+Z_{ds,i}\sin {\Omega }_{i}t\right)+M^{-1}C{\displaystyle \sum _{i=1}^N }\left(-{\Omega }_i{Z}_{dc,i}\sin {\Omega }_{i}t+{\Omega }_i{Z}_{ds,i}\cos {\Omega }_{i}t\right)\\ +M^{-1}K{\displaystyle \sum _{i=1}^N }\left(Z_{dc,i}\cos {\Omega }_{i}t+Z_{ds,i}\sin {\Omega }_{i}t\right)={\displaystyle \sum _{i=1}^N }\left(F_{s,i}\sin {\Omega }_{i}t+F_{c,i}\cos {\Omega }_{i}t\right)\end{array}$$

(121)

By using the balanced method for Equation (121), we can obtain the following:

$${\displaystyle \sum _{i=1}^N }{a}_{im}{Z}_{dc,i}+{\displaystyle \sum _{i=1}^N }{b}_{im}{Z}_{ds,i}={\chi }_{cm},m=1,2,\dots ,N$$

(122)

where $a_{im}=\left[{\alpha }_{im}\left(M^{-1}K-{\Omega }_i^2\mathrm{I}\right)-{\beta }_{im}{\Omega }_i{M}^{-1}C\right]$, $b_{im}=\left[{\beta }_{im}\left(M^{-1}K-{\Omega }_i^2\mathrm{I}\right)-{\alpha }_{im}{\Omega }_i{M}^{-1}C\right]$, and ${\chi }_{cm}={\displaystyle \sum _{i=1}^N }\left(F_{s,i}{\beta }_{im}+F_{c,i}{\alpha }_{im}\right)$.

And

$${\displaystyle \sum _{i=1}^N }{c}_{im}{Z}_{dc,i}+{\displaystyle \sum _{i=1}^N }{d}_{im}{Z}_{ds,i}={\chi }_{sm},m=1,2,\dots ,N=6$$

(123)

where $c_{im}=\left[{\beta }_{mi}\left(M^{-1}K-{\Omega }_i^2\mathrm{I}\right)-{\gamma }_{im}{\Omega }_i{M}^{-1}C\right]$, $d_{im}=\left[{\gamma }_{im}\left(M^{-1}K-{\Omega }_i^2\mathrm{I}\right)-{\beta }_{mi}{\Omega }_i{M}^{-1}C\right]$, and ${\chi }_{sm}={\displaystyle \sum _{i=1}^N }\left(F_{s,i}{\gamma }_{im}+F_{c,i}{\beta }_{mi}\right)$.

Equations (122) and (123) can be expressed as

$$A{\tilde{Z}}_d=F$$

(124)

where

$$\begin{gathered} A={\left[\begin{array}{cc}{\left[\begin{array}{cccc}{a}_{11}& a_{21}& \cdots & a_{\mathrm{N},1}\\ a_{12}& a_{22}& \cdots & a_{\mathrm{N}2}\\ ⋮& ⋮& \cdots & ⋮\\ a_{1\mathrm{N}}& a_{2N}& \cdots & a_{\mathrm{NN}}\end{array}\right]}_{18\mathrm{N}\times 18\mathrm{N}}& {\left[\begin{array}{cccc}{b}_{11}& b_{21}& \cdots & b_{N1}\\ b_{12}& b_{22}& \cdots & b_{N2}\\ ⋮& ⋮& \cdots & ⋮\\ b_{1,N}& b_{2,N}& \cdots & b_{NN}\end{array}\right]}_{18\mathrm{N}\times 18\mathrm{N}}\\ {\left[\begin{array}{cccc}{c}_{11}& c_{21}& \cdots & c_{N1}\\ c_{12}& c_{22}& \cdots & c_{N2}\\ ⋮& ⋮& \cdots & ⋮\\ c_{1N}& c_{2N}& \cdots & c_{\mathrm{N}N}\end{array}\right]}_{18\mathrm{N}\times 18\mathrm{N}}& {\left[\begin{array}{cccc}{d}_{11}& d_{21}& \cdots & d_{N1}\\ d_{12}& d_{22}& \cdots & d_{N2}\\ ⋮& ⋮& \cdots & ⋮\\ d_{1N}& d_{2N}& \cdots & d_{NN}\end{array}\right]}_{18\mathrm{N}\times 18\mathrm{N}}\end{array}\right]}_{36\mathrm{N}\times 36\mathrm{N}} \\ F={\left[\begin{array}{c}{\left[\begin{array}{c}{\chi }_{c1}\\ {\chi }_{c2}\\ ⋮\\ {\chi }_{c,\mathrm{N}=6}\end{array}\right]}_{18\mathrm{N}\times 1}\\ {\left[\begin{array}{c}{\chi }_{\mathrm{s}1}\\ {\chi }_{\mathrm{s}2}\\ ⋮\\ {\chi }_{\mathrm{s},\mathrm{N}=6}\end{array}\right]}_{18\mathrm{N}\times 1}\end{array}\right]}_{36\mathrm{N}\times 1} \end{gathered}$$

(125)

The solution to Equation (124) is

$${\tilde{Z}}_d=A^{-1}F$$

(126)

4.7. Dynamic Tensions of Ropes

Under the influence of an irregular wave, the dynamic tension of rope A is

$$T_{Ad}={\displaystyle \sum _{i=1}^N }{T}_{Adc,i}\cos {\Omega }_{i}t+T_{Ads,i}\sin {\Omega }_{i}t$$

(127)

where $T_{Adc,i}=K_{Ad}\left({\epsilon }_{AX}{x}_{1dc,i}+{\epsilon }_{AY}{y}_{1dc,i}+{\epsilon }_{AZ}{z}_{1dc,i}\right),$ $T_{Ads,i}=K_{Ad}\left({\epsilon }_{AX}{x}_{1ds,i}+{\epsilon }_{AY}{y}_{1ds,i}+{\epsilon }_{AZ}{z}_{1ds,i}\right)$. The dynamic tension of rope B is

$$T_{Bd}={\displaystyle \sum _{i=1}^N }{T}_{Bdc,i}\cos {\Omega }_{i}t+T_{Bds,i}\sin {\Omega }_{i}t$$

(128)

where $T_{Bdc,i}=K_{Bd}\left[X_B\left(x_{2dc,i}-x_{1dc,i}\right)+Y_B\left(y_{2dc,i}-y_{1dc,i}\right)+Z_B\left(z_{2dc,i}-z_{1dc,i}\right)\right],$

$$T_{Bds,i}=K_{Bd}\left[X_B\left(x_{2ds,i}-x_{1ds,i}\right)+Y_B\left(y_{2ds,i}-y_{1ds,i}\right)+Z_B\left(z_{2ds,i}-z_{1ds,i}\right)\right],$$

$$X_B={\displaystyle \frac{\left(x_{2s}-x_{1s}\right)}{L_B}}, Y_B={\displaystyle \frac{\left(y_{2s}-y_{1s}\right)}{L_B}}, Z_B={\displaystyle \frac{\left(z_{2s}-z_{1s}\right)}{L_B}}.$$

The dynamic tension of rope C is

$$T_{Cd}={\displaystyle \sum _{i=1}^N }{T}_{Cdc,i}\cos {\Omega }_{i}t+T_{Cds,i}\sin {\Omega }_{i}t$$

(129)

where $T_{Cdc,i}=K_{Cd}\left(x_{3dc,i}-x_{1dc,i}\right),T_{Cds,i}=K_{Cd}\left(x_{3ds,i}-x_{1ds,i}\right)$. The dynamic tension of rope D is

$$T_{Dd}={\displaystyle \sum _{i=1}^N }{T}_{Ddc,i}\cos {\Omega }_{i}t+T_{Dds,i}\sin {\Omega }_{i}t$$

(130)

where $T_{Ddc,i}=K_{Dd}\left(x_{4dc,i}-x_{2dc,i}\right),T_{Dds,i}=K_{Dd}\left(x_{4ds,i}-x_{2ds,i}\right)$.

5. Dynamic Response and Discussion

Referring to the information from the Central Meteorological Bureau Library of Taiwan about the typhoon invading Taiwan from 1897 to 2019 [19] and selecting 150 typhoons that greatly affected Taiwan’s Green Island, the significant wave height H_s during the 50-year regression period is set at H_s = 15.4 m and the peak period is set at T_w = 16.5 s. Letting N = 6, the irregular wave is simulated using six regular waves, listed in

Table 3. Irregular waves simulated using six regular waves (H_bed = 1300 m).

Significant Frequency	Parameter	Regular Wave
		1	2	3	4	5	6
Tp = 15.0 s, fp = 0.067 Hz	fi (Hz)	0.043	0.067	0.092	0.115	0.150	0.267
	ki (1/m)	0.0073	0.0179	0.0339	0.0533	0.0906	0.2861
	λi (m)	863.6	350.9	185.6	117.9	69.3	22.0
Tp = 16.5 s, fp = 0.061 Hz	fi (Hz)	0.043	0.061	0.086	0.115	0.150	0.267
	ki (1/m)	0.0073	0.0148	0.0295	0.0533	0.0906	0.2861
	λi (m)	863.6	424.7	212.8	117.9	69.3	22.0
Tp = 17.5 s, fp = 0.057 Hz	fi (Hz)	0.043	0.057	0.082	0.115	0.150	0.267
	ki (1/m)	0.0073	0.0132	0.0272	0.0533	0.0906	0.2861
	λi (m)	863.6	477.6	231.2	117.9	69.3	22.0
	${\phi }_i$ (°)	30	60	90	120	170	300

[20]. Figure 6 presents the relation between the significant height H_s and the amplitudes of the six regular waves. Table 3 shows that the second-wave frequency is completely consistent with the significant frequency. Figure 5 shows that the amplitude of the second wave is significantly larger than that of other waves. The larger the significant wave height H_s, the larger the amplitudes of the six regular waves.

Firstly, the dynamic response of an ocean current convertor with 400 kW power under the influence of irregular typhoon wave and current directions is investigated, as shown in Figure 7. The parameters of the mooring system are listed in

Table 4. The parameters of the system with the 400 kW convertor.

Parameter		Dimension	Parameter	Dimension
length of rope B, LB		152.97 m	length of rope C, LC	100 m
length of rope D, LD		70 m	distance between the two foundations, LF	2 LA
current velocity, V		1.5 m/s	net buoyance of the convertor and platform, FBNT/FBNB	394.9/224.0 tons
static drag of the convertor, FDT		59.35 tons	static drag of the platform, FDB	0.08 tons
mass of the platform, M1		200 tons	mass of the convertor, M2	538 tons
mass of Pontoon 3, M3		100 tons	mass of Pontoon 4, M4	120 tons
cross-sectional area of the surface cylinder of Pontoons 3 and 4, ABX/ABT		1.0/1.0 m2	mass moment of inertia of the convertor about the x, y, and z-axes, $I_{Tx}/I_{Ty}/I_{Tz}$	$8.940\times {10}^{10}$/ $2.712\times {10}^{10}/$ $8.940\times {10}^{10}$ ${\mathrm{kg}-\mathrm{m}}^2$
No buffer springs A, B, C, or D		-	mass moment of inertia of the platform about the x, y, and z-axes, $I_{Px}/I_{Py}/I_{Pz}$	$3.0\times {10}^8$/$5.0\times {10}^6/$$3.0\times {10}^8$ ${\mathrm{kg}-\mathrm{m}}^2$
significant wave height Hs		10 m	significant period Tp	16.0 s
relative angle between the current and wave α		30°	phase angles of the six regular waves used to simulate the irregular waves ${\phi }_i$	30/60/90/120/170/300°
Rope	A, B, C	D	distance from the center of gravity of the platform and convertor	$R_{TBx}=16.5 \mathrm{m}$ $R_{TDy}=12.82 \mathrm{m}$ $R_{TBz}=16.5 \mathrm{m}$ $R_{PAx}=R_{PAy}=R_{PAz}=5 \mathrm{m}$ $R_{PBx}=R_{PBz}=5.8 \mathrm{m}$ $R_{PCy}=R_{PCz}=2.5 \mathrm{m}$
Young’s modulus EPE	100 GPa,	116 GPa,
weight per unit length wPE	16.22 kg/m	26.92 kg/m
diameter DPE	154 mm	187 mm
cross-sectional area APE	0.0186 m2	0.0276 m2
fracture strength Tfrac	759 tons	2200 tons

. All the hydrodynamic damping and stiffness coefficients are determined using the commercial STAR-CCM software [7].

Figure 7 presents the effects of the rope length ratio r_AH = L_A/H_bed and the current direction ${\phi }_{cur}$ on the dynamic rope tensions of the pulley–rope system of the 400 kW convertor under the influence of an irregular typhoon wave. Ropes A, B, and C have the same specifications. Their fracture strength is T_frac = 759 tons. However, the fracture strength of rope D is T_D,frac = 2200 tons. As shown in Figure 7a, if H_bed = 1300 m, ${\phi }_{cur}=0^{°},3.62<r_{AH}<4.20$, the maximum dynamic tension, T_Dd,max, is smaller than the fracture strength of T_frac = 2200 tons. However, if the rope length ratio is r_AH = 4.47 and 4.94, a peak appears in the response spectrum due to resonance. If r_AH > 6.49, T_Dd,max is smaller than the fracture strength of T_frac = 2200 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max. Furthermore, the dynamic tension of the main traction rope A, T_Ad,max, is significantly lower than the fracture strength. If H_bed = 1300 m, ${\phi }_{cur}=60^{°},$$3.69<r_{AH}<4.30$, the maximum dynamic tension, T_Dd,max, is smaller than a fracture strength of T_frac = 2200 tons. If the rope length ratio is r_AH = 4.55, 5.02, and 7.19, a peak appears in the response spectrum due to resonance. If r_AH > 7.27, T_Dd,max is smaller than the fracture strength of T_frac = 2200 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max. As shown in Figure 7b, if H_bed = 1000 m, ${\phi }_{cur}=0^{°},$$3.62<r_{AH}<4.20$, the maximum dynamic tension, T_Dd,max, is smaller than the fracture strength of T_frac = 2200 tons. However, if the rope length ratio is r_AH = 4.47, 4.94, and 7.08, a peak appears in the response spectrum due to resonance. If r_AH > 7.15, T_Dd,max is smaller than a fracture strength of T_frac = 2200 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max. Furthermore, the dynamic tension of the main traction rope A T_Ad,max is significantly lower than the fracture strength. If H_bed = 1000 m, ${\phi }_{cur}=60^{°}, 3.69<r_{AH}<4.28$, the maximum dynamic tension, T_Dd,max, is smaller than a fracture strength of T_frac = 2200 tons. If the rope length ratio is r_AH = 4.55, 5.02, and 7.19, a peak appears in the response spectrum due to resonance. If r_AH > 7.23, T_Dd,max is smaller than the fracture strength of T_frac = 2200 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max.

In summary, if the rope length ratio r_AH is over the critical value, the larger the ratio r_AH, the higher of the safety factor of the rope.

Furthermore, the dynamic response of an ocean current convertor with 1 MW power under the influence of irregular typhoon wave and current directions is investigated, as shown in Figure 8. The corresponding hydrodynamic coefficients and other parameters are determined based on the similarity law derived in Section 4.1 and listed in Appendix C and Appendix D. The parameters of the mooring system are listed in

Table 5. Parameters of the system with the 1 MW convertor.

Parameter		Dimension		Parameter	Dimension
depth of the seabed Hbed		1300 m		length of rope A, LA	5980 m
length of rope B, LB		152.97 m		length of rope C, LC	100 m
length of rope D, LD		70 m		distance between the two foundations, LF	1196 m
current velocity, Vcur		1.5 m/s		net buoyance of the convertor and platform, FBNT/FBNB	1543.6/689.2 tons
static drag of the convertor, FDT		148.3 tons		static drag of the platform, FDB	0.192 tons
mass of the platform, M1		790.6 tons		mass of the convertor, M2	2126.6 tons
mass of Pontoon 3, M3		395.3 tons		mass of Pontoon 4, M4	474.3 tons
cross-sectional area of the surface cylinder of Pontoon 3, ABX		5.75 m2		mass moment of inertia of the convertor about the x, y, and z-axes, $I_{Tx}/I_{Ty}/I_{Tz}$	$8.83\times {10}^{11}$/ $2.68\times {10}^{11}/$ $8.83\times {10}^{11}$ ${\mathrm{kg}-\mathrm{m}}^2$
cross-sectional area of the surface cylinder of Pontoon 4, ABT		5.75 m2		mass moment of inertia of the platform about the x, y, and z-axes, $I_{Px}/I_{Py}/I_{Pz}$	$2.96\times {10}^9$/$4.94\times {10}^7/$$2.96\times {10}^9$ ${\mathrm{kg}-\mathrm{m}}^2$
significant wave height Hs		15.4 m		significant period Tp	16.5 s
relative angle between the current and wave α		30°		phase angles of the six regular waves used to simulate the irregular waves ${\phi }_i$	30/60/90/120/170/300°
HMPE/ Dyneema® SK75	Young’s modulus EPE	116 GPa,		distance from the center of gravity of the platform and convertor	$R_{TBx}=26.1 \mathrm{m}$ $R_{TDy}=20.3 \mathrm{m}$ $R_{TBz}=26.1 \mathrm{m}$ $R_{PAx}=R_{PAy}=R_{PAz}=7.9 \mathrm{m}$ $R_{PBx}=R_{PBz}=9.2 \mathrm{m}$ $R_{PCy}=R_{PCz}=4.0 \mathrm{m}$
	weight per unit length wPE	24.47 kg/m
	diameter DPE	178.9 mm
	cross-sectional area APE	0.0251 m2
	fracture strength Tfrac	2000 tons

.

Figure 8 presents the effects of the distance between the two foundations L_F and current direction ${\phi }_{cur}$ on the maximum rope tensions and the displacements of the elements under the influence of an irregular typhon wave. As shown in Figure 8a, the current direction greatly affects the maximum dynamic tensions of the ropes. The maximum dynamic tension of rope D, T_D,max, is significantly larger than the other tensions. The larger the distance between the two foundations, the larger the maximum dynamic tensions. As shown in Figure 8b, the maximum heaving displacements of all the elements are the greatest, followed by the surging displacements. The swaying displacements are negligible. The effect of current direction ${\phi }_{cur}$ on displacements is negligible. As shown in Figure 8c, the pitching angle of the platform ${\phi }_{pz}$ is about 2°. The other angular displacements of the platform and turbine are very small.

Figure 9 presents a comparison of the dynamic rope tensions of the traditional mooring system with a single rope, rope A, and one with a pulley–rope set under the influence of an irregular typhoon wave. All the parameters are listed in Table 4. The distance between the two foundations is L_F = 0.2L_A. Figure 9 shows that the dynamic rope tensions of the traditional mooring system with a single rope, rope A, are significantly greater than those of the one presented with a pulley–rope set. The dynamic rope tensions, T_Ad,max and T_Bd,max, of the traditional system are larger than a fracture strength of T_frac = 2000 tons. However, all the dynamic tensions of the pulley–traction rope system are smaller than the fracture strength of T_frac = 2000 tons. Obviously, the pulley–rope design can effectively reduce dynamic rope tensions. For the traditional system, the dynamic tension of rope B is the greatest. However, for the pulley–rope system, the dynamic tension of rope D is the greatest. The dynamic tension of the rope in the traditional system has nothing to do with the current direction ${\phi }_{cur}$, while the dynamic rope tension of the pulley–rope system is significantly related to the current direction ${\phi }_{cur}$.

Figure 10 presents the effects of the rope length ratio r_AH = L_A/H_bed and the current direction ${\phi }_{cur}$ on the dynamic rope tensions of the pulley–rope system of the 1 MW convertor under the influence of an irregular typhoon wave. Ropes A, B, and C have the same specifications. Their fracture strength is T_frac = 2000 tons. However, the fracture strength and cross-sectional area of rope D are T_D,frac = 2600 tons and A_PE = 0.0326 m². The distance between the two foundations is L_F = 0.2L_A. The current velocity is V_cur = 1.6 m/s. The other parameters are the same as those in Table 5. As shown in Figure 10a, if H_bed = 1300 m, ${\phi }_{cur}=0^{°}, 3.69<r_{AH}<4.70$, the maximum dynamic tension, T_Dd,max, is smaller than the fracture strength of T_frac = 2600 tons. However, if the rope length ratio is r_AH = 5.02, a peak appears in the response spectrum due to resonance. If r_AH > 6.18, T_Dd,max is smaller than the fracture strength of T_frac = 2600 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max. Furthermore, the dynamic tension of the main traction rope A, T_Ad,max, is significantly lower than the fracture strength. If H_bed = 1300 m, ${\phi }_{cur}=60^{°},$ and $3.77<r_{AH}<4.82$, the maximum dynamic tension, T_Dd,max, is smaller than the fracture strength of T_frac = 2600 tons. If the rope length ratio is r_AH = 5.13, a peak appears in the response spectrum due to resonance. If r_AH > 6.30, T_Dd,max is smaller than the fracture strength T_frac = 2600 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max. As shown in Figure 10b, if H_bed = 1000 m, ${\phi }_{cur}=0^{°},$ $3.73<r_{AH}<4.43$, the maximum dynamic tension, T_Dd,max, is smaller than the fracture strength of T_frac = 2600 tons. However, if the rope length ratio is r_AH = 5.83, a peak appears in the response spectrum due to resonance. If r_AH > 6.19, T_Dd,max is smaller than the fracture strength of T_frac = 2600 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max. Furthermore, the dynamic tension of the main traction rope A, T_Ad,max, is significantly lower than the fracture strength. If H_bed = 1000 m, ${\phi }_{cur}=60^{°}, 3.81<r_{AH}<4.53$, the maximum dynamic tension, T_Dd,max, is smaller than the fracture strength of T_frac = 2600 tons. If the rope length ratio is r_AH = 5.95, a peak appears in the response spectrum due to resonance. If r_AH > 6.34, T_Dd,max is smaller than the fracture strength of T_frac = 2600 tons. Moreover, the larger the ratio r_AH, the lower the maximum dynamic tension T_Dd,max.

In summary, if the rope length ratio r_AH is over the critical value, the larger the ratio r_AH, the higher the safety factor of the rope.

Figure 11a presents the effects of the buffer spring constant γ_KB and current direction ${\phi }_{cur}$ on the dynamic rope tensions of the pulley–rope system of the 1 MW convertor under the influence of an irregular typhoon wave. Buffer springs A, C, and D are not installed. The distance between the two foundations is L_F = 0.2L_A. The other parameters are the same as those in Table 5. If the buffer spring constant is γ_KB < 30, the maximum dynamic tensions are greater than a rope fracture strength of T_frac = 2000 tons. Moreover, if the buffer spring constants are γ_KB = 1.950 and 4.895 for ${\phi }_{cur}=0^{°}$, a peak appears in the response spectrum due to resonance.

When the buffer spring constant is γ_KB > 100, i.e., buffer spring B is not installed, the maximum dynamic tensions, T_Dd,max (${\phi }_{cur}{=0}^{°}$) = 1259 tons and T_Dd,max (${\phi }_{cur}{=60}^{°}$) = 839 tons, are both less than the fracture strength of 2000 tons. Figure 11b presents the effects of the buffer spring constant γ_KD and current direction ${\phi }_{cur}$ on the dynamic rope tensions of the pulley–rope system of the 1 MW convertor under the influence of an irregular typhoon wave. Buffer springs A, B, and C are not installed. The distance between the two foundations is L_F = 0.2L_A. The other parameters are the same as those in Table 4. If the buffer spring constant is γ_KD < 4.6, the maximum dynamic tensions are greater than a rope fracture strength of T_frac = 2000 tons. When the buffer spring constant is γ_KD > 100, i.e., buffer spring D is not installed, the maximum dynamic tensions, T_Dd,max (${\phi }_{cur}{=0}^{°}$) = 1259 tons and T_Dd,max (${\phi }_{cur}{=60}^{°}$) = 839 tons, are both less than the fracture strength of 2000 tons. The pulley–rope system of the 1 MW convertor without buffer springs A, B, C, or D are thus concluded to be safe under the influence of an irregular typhoon wave.

The dynamic response of an ocean current convertor with 700 kW power under the influence of irregular typhoon wave and current directions is investigated, as shown in Figure 12. The corresponding hydrodynamic coefficients and other parameters are determined based on the similarity law derived in Section 4.1 and listed in Appendix C and Appendix D. The parameters of the mooring system are listed in

Table 6. The parameters of the system with the 700 kW convertor.

Parameter		Dimension	Parameter	Dimension
depth of the seabed, Hbed		1300 m	length of rope A, LA	5980 m
length of rope B, LB		152.97 m	length of rope C, LC	100 m
length of rope D, LD		70 m	distance between the two foundations, LF	1196 m
current velocity, V		1.5 m/s	net buoyance of the convertor and platform, FBNT/FBNB	907.4/407.9 tons
static drag of the convertor, FDT		103.9 tons	static drag of the platform, FDB	0.134 tons
mass of the platform, M1		463.0 tons	mass of the convertor, M2	1245.5 tons
mass of Pontoon 3, M3		213.5 tons	mass of Pontoon 4, M4	277.8 tons
cross-sectional area of the surface cylinder of Pontoon 3, ABX		4.03 m2	mass moment of inertia of the convertor about the x, y, and z-axes, $I_{Tx}/I_{Ty}/I_{Tz}$	$3.62\times {10}^{11}$/ $1.10\times {10}^{11}/$ $3.62\times {10}^{11}$ ${\mathrm{kg}-\mathrm{m}}^2$
cross-sectional area of the surface cylinder of Pontoon 4, ABT		4.03 m2	mass moment of inertia of the platform about the x, y, and z-axes, $I_{Px}/I_{Py}/I_{Pz}$	$1.22\times {10}^9$/$2.03\times {10}^7/$$1.22\times {10}^9$ ${\mathrm{kg}-\mathrm{m}}^2$
significant wave height Hs		15.4 m	significant period Tp	16.5 s
relative angle between the current and wave α		30°	phase angles of the six regular waves used to simulate the irregular waves ${\phi }_i$	30/60/90/120/170/300°
HMPE/ Dyneema® SK75	Young’s modulus EPE	116 GPa,	distance from the center of gravity of the platform and convertor	$R_{TBx}=21.8 \mathrm{m}$ $R_{TDy}=17.0 \mathrm{m}$ $R_{TBz}=21.8 \mathrm{m}$ $R_{PAx}=R_{PAy}=R_{PAz}=6.6 \mathrm{m}$ $R_{PBx}=R_{PBz}=7.7 \mathrm{m}$ $R_{PCy}=R_{PCz}=3.3 \mathrm{m}$
	weight per unit length wPE	17.16 kg/m
	diameter DPE	149.7 mm
	cross-sectional area APE	0.0176 m2
	fracture strength Tfrac	1400 tons

.

Figure 12a presents the effects of the buffer spring constant γ_KB and current direction ${\phi }_{cur}$ on the dynamic rope tensions of the pulley–rope system of the 700 kW convertor under an irregular typhoon wave. Ropes A, B, C, and D have the same specifications. The fracture strength is T_j,frac = 1400 tons, j = A,B,C,D. If the buffer spring constant is γ_KB ≅ 10, the dynamic tension T_Dd,max is at a minimum and smaller than a rope fracture strength of T_frac = 1400 tons. When the buffer spring constant is γ_KD > 100, i.e., buffer spring D is not installed, the maximum dynamic tensions, T_Dd,max (${\phi }_{cur}{=0}^{°}$) = 1394 tons and T_Dd,max (${\phi }_{cur}{=60}^{°}$) = 1741 tons, are larger than the fracture strength of 1400 tons. To overcome this failure, only the specification of rope D is increased to T_D,frac = 2000 tons, as listed in Table 4. As shown in Figure 12b, when the buffer spring constant is γ_KD > 100, i.e., buffer spring D is not installed, the maximum dynamic tensions, T_Dd,max (${\phi }_{cur}{=0}^{°}$) = 1392 tons and T_Dd,max (${\phi }_{cur}{=60}^{°}$) = 1738 tons, are smaller than the fracture strength of 2000 tons.

6. Conclusions

In this study, similarity formulas were constructed for MW- and KW-level ocean convertors. Accordingly, the hydrodynamic damping and stiffness coefficients and other parameters of an MW-level system were reasonably estimated. A novel pulley–main rope design was proposed for MW-level ocean convertor mooring systems. Furthermore, a mathematical dynamic model was derived for an MW-level ocean convertor mooring system under irregular wave and current directions. Then, the analytical solution of this system is presented. The dynamic performance of the proposed 1 MW mooring system under the impact of irregular waves was thus investigated, and the following results were discovered:

• The dynamic rope tensions T_Ad,max and T_Bd,max of the traditional single-traction-rope system are larger than the fracture strength of T_frac = 2000 tons. However, all the dynamic tensions of the pulley–rope system are smaller than the fracture strength. Obviously, the pulley–rope design can effectively reduce the dynamic rope tensions.
• The dynamic responses of the above two kinds of mooring systems are different. The traditional single-traction-rope system is not affected by flow direction, but the pulley–rope system is.
• The static tension of rope A of the proposed system under a steady current only is close to half of that for the traditional single-traction-rope system.
• For H_bed = 1300 m, if the traction rope length ratio is $3.77\le r_{AH}\le 4.70$, all the dynamic tensions will be smaller than the fracture strength. For H_bed = 1000 m, if the rope length ratio is $3.81\le r_{AH}\le 4.43$, all the dynamic tensions will be smaller than the fracture strength.
• In this mooring system, the dynamic tension of rope D is at a maximum, but we can increase its fracture strength for improved safety. The length of rope D is short and economic.
• If the traction rope length ratio r_AH is over the critical value, the larger the ratio r_AH, the higher the safety factor of the rope.
• In an MW-level power generation system with the pulley–rope design, if the buffer spring constant is very small, the dynamic tension may become greater than the fracture strength. However, when no buffer spring is set, the opposite occurs.
• According to the theoretical analysis, the proposed MW-level ocean convertor mooring system with the pulley–rope design can be safely used even under the impact of a typhoon wave.