Hydrodynamic Similarity of Different Power Levels and Dynamic Analysis of Ocean Current Converter-Platform Systems with a Novel Pulley-Traction Rope Design under Typhoon Irregular Wave and Current

1. Introduction

Global ocean currents are rich in energy. The potential electricity capacity in the Taiwan Kuroshio current is over 4GW [1]. The core technologies of ocean current power generation are being investigated [1,2,3,4,5]. The relevant technologies include (1) the high-efficiency convertor, (2) the deep mooring technology for over 1000m depth seabed beneath the current, (3) the protection from the typhoon wave impact, (4) double main traction ropes for high power convertor, and (5) the investigation of the mathematical model with the fluid-structure interaction (FSI).

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

The deep mooring theory and technology for the ocean current convertor system are important. A few of literatures investigated the dynamic stability of this mooring system under the coupled current-wave effect. WanChi company further developed a 400 kW ocean current convertor following the principle of the 50 kW ones. Lin et al. [4] proposed the mathematical model of the 400 kW ocean convertor-pontoon-traction rope-foundation mooring system under the coupled regular wave-ocean current effects. The motion of this system is four degrees of freedom. The dynamic system include the heaving motion of pontoon and the coupled surging-heaving-pitching motion of the convertor. The dynamic performance and stability of the system under coupled wave-current effect were investigated. Lin and Chen [5] proposed a towed parachute-platform-traction rope-foundation mooring system which protected the submarined platform from the typhoon wave impact. The concentrated mass model was considered. The motion of this system was three degrees of freedom. The coupled motions included surging and heaving of elements. It was theoretically verified the protection function of the proposed methodology under Typhoon wave impact. Lin et al. [6] presented the submarined ocean convertor-surfaced platform-pontoon-traction rope-foundation mooring system. The concentrated mass model was considered. The motion of this system was five degrees of freedom. The coupled motion included surging and heaving of elements. The dynamic stability of the mooring system under the regular wave and steady ocean current was investigated. Lin et al. [7] presented the submarined ocean convertor-submarined platform-2 pontoon-traction rop-foundation mooring system. The concentrated mass model was considered. The motion of this system was six degrees of freedom. The coupled motion included surging and heaving of elements. The dynamic response of this mooring system under the typhoon irregular wave of at Taiwan’s Green Island during the 50-year regression period was investigated. Lin et al. [8] designed a 400kW convertor composed of two rotating horizontal turbines. The hydrodynamic damping and stiffness coefficients were determined by using the computational fluid method. The submarined 400kW convertor-submarined platform-2 pontoon-traction rop-foundation mooring system was considered. The motion of this system was eighteen degrees of freedom. The coupled 3D motion included the translational and rotational of elements. The frequency spectrum of the system with the coupled fluid-structure interaction and regular wave was studied. Further, Lin et al. [9] investigated the transient performance of the 18 DOF mooring system with inital conditions. Pierson and Moskowitz [10] proposed the Jonswap spectrum and confirmed it with experimental measurement results. To simplify actual irregular ocean waves, a limited number of regular wave are usually used to approximate the irregular wave [7]. This study will take several regular waves to simulate irregular waves according to the experimental significant wave height, frequency and the Jonswap wave spectrum. In the literatures [4,5,6,7,8,9], the mooring traction rope is single only. However, for high power convertor its traction force is very great, two traction ropes is necessary. The relavant investgation is helpful for the ocean energy technology.

Theorem and technology of fluid structure interaction (FSI) are widely applied in many different fields including marine engineering [11,12], aerodynamics [13], acoustics [14] and biomechanics [15]. Due to the complexity of the fluid-structure interaction, a few literatures investigated the FSI problem by using analytical method [15], but mostly the numerical methods: (1) the boundary element method [13], (2) the finite-volume method [16], (3) the Coupled SPH–FEM [17].

So far, the relative literatures proposed one single main mooring rope connecting to one foundation. For the high-power ocean current convertor, this study designs a pulley-traction rope set which the rope connects to the two foundations. The similarity law is proposed to calculate the hydrodynamic damping and stiffness coefficient of turbine and platform. The static and dynamic equations of this ocean current power generation anchoring system are proposed to study the rope tension and stability under irregular waves.

2. Mathematical Model

To avoid the wave impact of typhoons, the energy convertor and the floating platform were submerged to a safe depth. For safely mooring a high power ocean convertor this study presents a pulley- traction rope set composed of a pulley and one main rope connected to two separate foundations, as shown in Figure 1. The coupled translational-rotational motion of the mooring system under coupled wave-current effect are considered. The motion is eighteen degrees of freedom. The governing equations of this mooring system are derived as follows:

The global translational and the rotational displacements of the components are.

$$x_i=x_{is}+x_{id},y_i=y_{is}+y_{id},z_i=z_{is}+z_{id},i=1, 2, 3, 4$$

$${\phi }_{jx}={\phi }_{jxs}+{\phi }_{jxd}, {\phi }_{jx}={\phi }_{jxs}+{\phi }_{jxd}, {\phi }_{jx}={\phi }_{jxs}+{\phi }_{jxd}; {\phi }_{jxs}={\phi }_{jxs}={\phi }_{jxs}=0,j=1,2$$

(1)

The total tensions of the ropes are

$$T_i=T_{is}+T_{id},i=A, B, C, D$$

(2)

These displacements and tensions include (1) the static one under the steady current only, (2) the dynamic one due to the wave impact and current. 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 the Steady Current only

3.1. Static Displacements

Under the steady current only, the static displacements of the components are

Foundation:

$${x^{\prime }}_{01}=0,\hspace{1em}{y^{\prime }}_{01}=0,\hspace{1em}\hspace{1em}\hspace{1em}{z^{\prime }}_{01}=L_F/2$$

(3)

,

$${x^{\prime }}_{02}=0,\hspace{1em}{y^{\prime }}_{02}=0,\hspace{1em}\hspace{1em}\hspace{1em}{z^{\prime }}_{02}=-L_F/2,$$

(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 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 the current direction${\varphi }_{cur}$. These parameters can be determined by using the static equilibrium principle 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, 4:

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

(14)

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

(15)

Rotational angles of platform and convertor:

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

(16)

The global setting angle qBs of rope B is

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

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

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

(21)

where the drag of the platform under steady current$f_{Pys}=0.5C_{DPy}\rho A_{PY}{V}^2$, ${\Delta }_1=\left(\pi /2-{\varphi }_{01}\right)$and${\Delta }_2=\left(\pi /2-{\varphi }_{02}\right)-{\varphi }_{cur}$. Because the 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

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

(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

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

(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 the pontoon 3 in the x-direction is

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

(27)

The static equilibrium of the 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), one obtains

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

(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 basic parameters$\left\{{\varphi }_{cur},H_{bed},L_C,L_A,L_F\right\}$are given, the length of rope A₁, L_A1, can be determined by using the bisection method via Equation (29). Substituting L_A1 back into Equations (8-9,12-17), 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

Parameters of the mooring system are listed in Table 1.

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

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

4. Dynamic Analysis

4.1. Similarity of System

4.1.1. Hydrodynamic Similarity of Convertor

Lin et al. [8] presented the hydrodynamic coefficients of fluid-structure interaction of a 400kW turbine and platform determined by using computational fluid dynamics method. An ocean current convertor of MW level is commercial specification. In this study, the dynamic response of a 1 MW turbine mooring system will be investigated. The corresponding hydrodynamic coefficients and other parameters are determined based on the similarity law and the presented parameters [8]. The similarity law is presented as follows:

Similarity of 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 that TSR=3.5 for the 400kW turbine, the efficiency is maximum. Considering the same current velocity V_cur for the model and prototype, Equation (31) becomes

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

(31)

Similarity of power number N_p:

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

(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)

Similarity of hydrodynamic force and moment for convertor:

The hydrodynamic force of convertor is

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

(34)

Dividing Equation(34) by ${\displaystyle \frac{1}{2}}\rho A_{tur}{V}_{cur}^2$, it becomes one in terms of dimensionless variables

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

(35)

where

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

(36)

If the prototype and the model are similar, their dimensionless force coefficients (35) are same

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

(37)

Based on Equations (35-37), one can obtain the following similar parameters

Similarity of drag force coefficient is

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

(38)

Similarity of hydrodynamic damping force coefficient:

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

(39)

Similarity of hydrodynamic stiffness force coefficient:

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

(40)

Based on the similar formula (36, 39, 40), the hydrodynamic damping force relations between the model and prototype

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

(41)

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

(42)

and the hydrodynamic stiffness force relation

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

(43)

Given the coefficients of the model and substituting Equations (31, 33) into Equations (41-43), the hydrodynamic damping and stiffness force coefficients of the prototype can be obtained.

According to the similarity of drag force coefficient (38), the drag force relation is

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

(44)

Based on Equation (44), the similarity of fracture strength of rope is

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

(45)

The hydrodynamic moment of convertor is

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

(46)

Dividing Equation (46) by ${\displaystyle \frac{1}{2}}\rho D_T{A}_{tur}{V}_{cur}^2$, it becomes one in terms of dimensionless variables

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

(47)

where $C_{mTj}=m_{Tj}/\left({\displaystyle \frac{1}{2}}\rho D_T{A}_{tur}{V}_{cur}^2\right)$.

If the prototype and the model are similar, the two dimensionless moments (47) are same

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

(48)

Based on Equations (47-48), one can obtain the following similar parameters

Similarity of drag moment coefficient is

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

(49)

Similarity of hydrodynamic damping in moment:

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

(50)

Similarity of hydrodynamic stiffness in moment:

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

(51)

Based on the similar formula (36, 50-51), the hydrodynamic damping moment relations between the model and prototype

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

(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

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

(54)

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

4.1.2. Hydrodynamic Similarity of Platform

The hydrodynamic force of platform is

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

(55)

Dividing Equation(55) by ${\displaystyle \frac{1}{2}}\rho A_{pla}{V}_{cur}^2$, it becomes in terms of dimensionless variables

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

(56)

where

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

(57)

If the prototype and the model are similar, the two dimensionless force coefficients (56) are same

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

(58)

Based on Equations (56-58), one can obtain the following similar parameters.

Similarity of hydrodynamic damping force coefficient:

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

(59)

Similarity of hydrodynamic stiffness force coefficient:

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

(60)

Based on Equations (57, 59-60), the hydrodynamic damping force relations between the model and prototype

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

(61)

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

(62)

and the hydrodynamic stiffness force relation

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

(63)

Given the coefficients of the model and substituting Equations (31, 33) into Equations (61-63), the hydrodynamic damping and stiffness force coefficients of the prototype can be obtained.

The hydrodynamic moment of platform is

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

(64)

Dividing Equation(64) by ${\displaystyle \frac{1}{2}}\rho D_P{A}_{pla}{V}_{cur}^2$, it becomes one in terms of dimensionless variables

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

(65)

where $C_{mPj}=m_{Pj}/\left({\displaystyle \frac{1}{2}}\rho D_P{A}_{pla}{V}_{cur}^2\right)$.

If the prototype and the model are similar, their dimensionless moments (65) are same

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

(66)

Based on Equations (63-64), one can obtain the following similar parameters.

Similarity of drag moment coefficient is

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

(67)

Similarity of hydrodynamic damping in moment:

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

(68)

Similarity of hydrodynamic stiffness in moment:

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

(69)

Based on the similar formula (36, 68-69), the hydrodynamic damping moment relations between the model and prototype

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

(70)

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

(71)

and the hydrodynamic stiffness moment relation

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

(72)

Given the coefficients of the model and substituting Equations (31, 33) into Equations (70-72), 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 (33) and considering the geometry similarity, one can obtain the similar relations

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

(73)

Similarity of mass:

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

(74)

Similarity of inertia of mass:

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

(75)

Similarity of buoyance:

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

(76)

4.2. Translational Motion in the x-Axis Direction

4.2.1. Equation of Heaving Motion for Pontoon 3

The heaving equation of the pontoon 3 is

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

(77)

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

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

(78)

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

The dynamic tension of the rope C,

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

(79)

where the dynamic elongation between the floating platform 1 and the pontoon 3, ${\delta }_{Cd}=\left(x_{3d}-x_{1d}\right)$.

4.2.2. Equation of Heaving Motion for Pontoon 4

The heaving equation of the pontoon 4 is

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

(80)

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 ${\varphi }_i=2\pi L_E\cos \alpha /{\lambda }_i$ and $L_E=\sqrt{L_B^2-{\left(L_C-L_D\right)}^2}$. α is the relative wave-current angle, ${\lambda }_i$is the wave length, as shown in Figure 5.

The effective spring constant of the rope D- buffer spring connection

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

(81)

in which $K_{D,spring}$ is the constant of the spring connecting with the 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 the rope D, respectively.

The dynamic tension of the rope C

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

(82)

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

4.2.3. Equation of Heaving Motion of the Platform

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

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

(83)

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 Taylor series as follows:

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

(84)

For the 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 the symmetry configuration of the platform is considered, the hydrodynamic force on the platform in the x-direction under the current only$f_{Px}\left(V,0,0,0,0,0,0,0,0,0\right)$=0. Considering small oscillation, the higher order terms are neglected later. The right-handed side second term of Equation (84) is the hydrodynamic force due to the fluid-structure interaction.

The dynamic tensions of the ropes A and B are

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

(85)

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,

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

(86)

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}}}$${\epsilon }_{AZ}={\displaystyle \frac{\left(Z_A-{\epsilon }_{z01}\right)}{r_{A1}}}+{\displaystyle \frac{\left(Z_A-{\epsilon }_{z02}\right)}{r_{A2}}}$, $X_A={\displaystyle \frac{x_{1s}}{L_A}}, Y_A={\displaystyle \frac{y_{1s}}{L_A}}, Z_A={\displaystyle \frac{z_{1s}}{L_A}}$${\epsilon }_{x01}={\displaystyle \frac{x_{01}}{L_A}},{\epsilon }_{y01}={\displaystyle \frac{y_{01}}{L_A}},{\epsilon }_{z01}={\displaystyle \frac{z_{01}}{L_A}},{\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

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

(87)

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 (84-87) into Equation (83), one obtains

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

(88)

4.2.4. Equation of Heaving Motion for the Convertor

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

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

(89)

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

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

(90)

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

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

(91)

where

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

(91)

Substituting Equations (23, 85-87, 91) into Equation (90), one obtains

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

(92)

where

$$\begin{array}{l}{d}_1=T_{As}\left[{\displaystyle \frac{\sin {\theta }_{As1}\cos \left({\Delta }_1+{\varphi }_{cur}\right)}{L_{A1}}}+{\displaystyle \frac{\sin {\theta }_{As2}\cos {\Delta }_2}{L_{A2}}}\right]-{\displaystyle \frac{T_{Bs}\sin {\theta }_{Bs}}{L_B}},\\ d_2=-\left[\cos {\theta }_{As1}\cos \left({\Delta }_1+{\varphi }_{cur}\right)+\cos {\theta }_{As2}\cos {\Delta }_2\right],d_3=-{\displaystyle \frac{T_{Bs}\sin {\theta }_{Bs}}{L_B}}.\end{array}$$

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

The dynamic equilibrium of the convertor in the surging motion is

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

(93)

The hydrodynamic force on the convertor is expressed as

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

(94)

where $f_{Tys}=f_{Ty}\left(V,0,0,0,0,0,0,0,0,0,TRS\right)=C_{DTy}{\displaystyle \frac{1}{2}}\rho A_{TY}{V}^2$, 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 (85, 87, 94) into Equation (93), one obtains

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

(95)

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

The dynamic equilibrium of the pontoon 3 in the surging motion is

$$M_3{\ddot{y}}_{3\mathrm{d}}{ }_3+{\displaystyle \frac{T_{Cs}}{L_C}}\left(y_{3d}-y_{1d}\right)=0$$

(96)

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

The dynamic equilibrium of the pontoon 4 in the surging motion is

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

(97)

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

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

(98)

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

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

(99)

Considering small displacements and based on Equations (79, 85-87, 99), one obtains

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

(100)

4.4.2. Equation of Swaying Motion of Convertor

The dynamic equilibrium of the convertor in the swaying motion is

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

(101)

4.4.3. Equation of Swaying Motion for Pontoon 3

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

$$M_3{\ddot{z}}_{3d}{ }_3+{\displaystyle \frac{T_{Cs}}{L_C}}\left(z_{3d}-z_{1d}\right)=0$$

(102)

4.4.4. Equation of Swaying Motion of Pontoon 4

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

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

(103)

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]

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

(104)

4.5.2. Equation of Rolling Motion of Convertor

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

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

(105)

4.5.3. Equation of Pitching Motion of Convertor

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

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

(106)

4.5.4. Equation of Yawing Motion of Platform

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

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

(107)

where $\Delta {\varphi }_{x1}={\displaystyle \frac{z_{1d}}{L_{A1}\cos {\theta }_{A1s}}},\Delta {\varphi }_{x2}={\displaystyle \frac{z_{1d}}{L_{A2}\cos {\theta }_{A2s}}},\Delta {\theta }_x={\displaystyle \frac{z_{2d}-z_{1d}}{L_B\cos {\theta }_{Bs}}}$. Substituting Equations (85-87) and hydrodynamic moment m_px into Equation (107), one obtains

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

(108)

4.5.5. Equation of Rolling Motion of Platform

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

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

(109)

where $\Delta {\varphi }_{Cy}={\displaystyle \frac{z_{1d}-z_{3d}}{L_{\mathrm{C}}}},\Delta {\varphi }_{A1y}={\displaystyle \frac{z_{1d}}{L_{A1}\sin {\theta }_{A1s}}},\Delta {\varphi }_{A2y}={\displaystyle \frac{z_{1d}}{L_{A2}\sin {\theta }_{A2s}}}$. Substituting Equations (79, 85-86) and hydrodynamic moment m_py into Equation (109), one obtains

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

(110)

4.5.6. Equation of Pitching Motion of Platform

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

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

(111)

where $\Delta {\theta }_{A1}={\displaystyle \frac{x_{1d}}{L_{A1}}}, \Delta {\theta }_{A2}={\displaystyle \frac{x_{1d}}{L_{A2}}},\Delta {\theta }_B={\displaystyle \frac{x_{2d}-x_{1d}}{L_B}}, \Delta {\theta }_C={\displaystyle \frac{y_{2d}-y_{1d}}{L_C}}$. Substituting Equations (79, 85-86) and hydrodynamic moment m_py into Equation (111), one obtains

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

(112)

4.6. Solution Method of Dynamic Displacements

The governing equations (77, 80, 88-89, 92-97, 100-106, 108, 110, 112) can be expressed in matrix format

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

(113)

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

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

(114)

where

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

(115)

According to the FSI parameters of 400kW convertor and platform presented by Lin et al. [8] and considering the similarity of some prototype of different power, for example, 1 MW. The elements of the mass matrix $M={\left[M_{ij}\right]}_{18\times 18}$, the damping matrix $C={\left[C_{ij}\right]}_{18\times 18}$ and the stiffness matrix $K={\left[K_{ij}\right]}_{18\times 18}$ for the prototype of MW-level are determined by using the similar formula in section 4-1 and listed in Appendix B, C and D, respectively. Equation (113) can be rewritten as

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

(116)

where

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

(117)

The solution can be expressed as

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

(118)

where

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

(119)

Substituting the solution (118) into Equation (116), one obtains

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

(120)

By using the balanced method for Equation (120), one obtains

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

(121)

where

$$a_{im}=\left[{\alpha }_{im}\left(M^{-1}K-{\mathrm{\Omega }}_i^2\mathrm{I}\right)-{\beta }_{im}{\mathrm{\Omega }}_i{M}^{-1}C\right], b_{im}=\left[{\beta }_{im}\left(M^{-1}K-{\mathrm{\Omega }}_i^2\mathrm{I}\right)-{\alpha }_{im}{\mathrm{\Omega }}_i{M}^{-1}C\right]$$

${\chi }_{cm}={\displaystyle \sum _{i=1}^N }\left(F_{s,i}{\beta }_{im}+F_{c,i}{\alpha }_{im}\right)$.

and

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

(122)

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

Equations (121-122) can be expressed as

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

(123)

where

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

(124)

The solution of Equation (123) is

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

(125)

4.8. Dynamic Tensions of Ropes

Under irregular wave, the dynamic tension of rope A is

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

(126)

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

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

(127)

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

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

(128)

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

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

(129)

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 the information from the Central Meteorological Bureau Library of Taiwan about the typhoon invading Taiwan from 1897 to 2019 [18] and selecting 150 typhoons greatly affecting Taiwan’s Green Island, the significant wave height H_s during the 50-year regression period H_s =15.4m, the peak period T_w = 16.5sec. Letting N=6, the irregular wave is simulated by six regular waves listed in Table 2 [19]. Figure 6 demonstrates the relation between the significant height H_s and the amplitudes of six regular simulating waves. Table 2 shows that the second wave frequency is completely consistent with the significant frequency. Figure 5 demonstrates that the amplitude of the second wave is significantly larger than that of other waves. The larger the significant wave height H_s is, greatly the larger the amplitudes of the six regular waves are.

The dynamic response of an ocean current convertor of 1 MW power under Typhon irregular wave and current direction is investigated, as shown in Figure 7. The corresponding hydrodynamic coefficients and other parameters are determined based on the similarity law derived in Section 4.1 and listed in Appendix C, D. The parameters of the mooring system are listed in Table 3

Figure 7 demonstrates the effects of the distance of two foundations L_F and the current direction${\phi }_{cur}$on the maximum rope tensions and the displacements of elements under typhon irregular wave. As shown in Figure 7(a), the effect of the current direction on the maximum dynamic tensions of ropes is great. The maximum dynamic tension of rope D, T_D,max, is significantly larger than the others. The larger the distance of two foundations is, the larger the maximum dynamic tensions are. As shown in Figure 7(b), the maximum heaving displacements of all the elements are greatest. The surging displacement are next. The swaying displacements are negligible. The effect of the current direction ${\phi }_{cur}$on displacements is negligible. As shown in Figure 7(c), the pitching angle of the platform${\phi }_{pz}$is about 2^o. The other angular displacement of platform and turbine are very small.

on the maximum rope tensions and the displacements of elements.

Figure 8 demonstrates the comparison of dynamic rope tensions of the traditional mooring system with a single rope A and one with a pulley-rope set under typhoon irregular wave. All the parameters are listed in Table 3. The distance of two foundations L_F=0.2L_A. Figure 8 shows that the dynamic rope tensions of the traditional mooring system with a single rope A are significantly greater than those of the presented one with a pulley- rope set. The dynamic rope tensions T_Ad,max and T_Bd,max of the traditional system are larger than the fracture strength T_frac=2000tons. However, all the dynamic tensions of the pulley- traction rope system is smaller than the fracture strength T_frac=2000tons. Obviously, the pulley-rope design can effectively reduce the 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 f_cur, while the dynamic rope tensions of the pulley-rope system is significantly related to the current direction f_cur.

Figure 9 demonstrates 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 1 MW under typhoon irregular wave. The distance between two foundations L_F=0.2L_A. Other parameters are the same as those of Table 3. As shown in Figure 9(a), if H_bed =1300m, ${\phi }_{cur}=0^o,3.98\text{<}{r}_{AH}\text{<}4.68$, the maximum dynamic tensions T_Dd,max is smaller than the fracture strength T_frac=2000tons. If H_bed =1300m, ${\phi }_{cur}=60^o,$$4.06\text{<}{r}_{AH}\text{<}4.78$, the maximum dynamic tensions T_Dd,max is smaller than the fracture strength T_frac=2000tons. As shown in Figure 9(b), if H_bed =1000m, ${\phi }_{cur}=0^o,$$3.92\text{<}{r}_{AH}\text{<}4.39$, the maximum dynamic tensions T_Dd,max is smaller than the fracture strength T_frac=2000tons. If H_bed =1000m, ${\phi }_{cur}=60^o, 3.99\text{<}{r}_{AH}\text{<}4.51$, the maximum dynamic tensions T_Dd,max is smaller than the fracture strength T_frac=2000tons. It is concluded that for a convertor of 1 MW power and H_bed =1300m, the rope length ratio, $4.1\le r_{AH}\le 4.6$, all the dynamic tensions can be smaller than the fracture strength. If H_bed =1000m, the rope length ratio, $4.0\le r_{AH}\le 4.3$, all the dynamic tensions can be smaller than the fracture strength.

Figure 10a demonstrates the effects of the buffer spring constant g_KB and the current direction${\phi }_{cur}$on the dynamic rope tensions of the pulley-rope system of 1 MW under typhoon irregular wave. The buffer springs A, C, and D are not installed. The distance between two foundations L_F=0.2L_A. Other parameters are the same as those of Table 3. It is found that if the buffer spring constant g_KB<30, the maximum dynamic tensions are greater than the fracture strength of rope T_frac=2000tons.

When the buffer spring constant g_KB>100, i.e., the buffer spring B is not installed, the maximum dynamic tension T_Dd,max (${\phi }_{cur}{=0}^o$)=1259 tons, T_Dd,max (${\phi }_{cur}{=60}^o$)=839 tons are both less than the fracture strength 2000tons. Figure 10b demonstrates the effects of the buffer spring constant g_KD and the current direction${\phi }_{cur}$on the dynamic rope tensions of the pulley-rope system of 1 MW under typhoon irregular wave. The buffer springs A, B, and C are not installed. The distance between two foundations L_F=0.2L_A. Other parameters are the same as those of Table 3. It is found that if the buffer spring constant g_KD<4.6, the maximum dynamic tensions are greater than the fracture strength of rope T_frac=2000tons. When the buffer spring constant g_KD>100, i.e., the buffer spring D is not installed, the maximum dynamic tension T_Dd,max (${\phi }_{cur}{=0}^o$)=1259 tons, T_Dd,max (${\phi }_{cur}{=60}^o$)=839 tons are both less than the fracture strength 2000tons. It is concluded that the pulley-rope system of 1 MW without the buffer springs A, B, C, and D under the typhoon irregular wave is safety.

The dynamic response of an ocean current convertor of 700 kW power under Typhon irregular wave and current direction is investigated, as shown in Figure 11. The corresponding hydrodynamic coefficients and other parameters are determined based on the similarity law derived in Section 4.1 and listed in Appendix C, D. The parameters of the mooring system are listed in Table 4.

Figure 11a demonstrates the effects of the buffer spring constant g_KB and the current direction${\phi }_{cur}$on the dynamic rope tensions of the pulley-rope system of 700 kW under typhoon irregular wave. The ropes A, B, C, and D have the same specification. The fracture strength T_j,frac=1400tons, j=A,B,C,D. It is found that if the buffer spring constant g_KB≅10, the dynamic tension T_Dd,max is minimum and smaller than the fracture strength of rope T_frac=1400tons. When the buffer spring constant g_KD>100, i.e., the buffer spring D is not installed, the maximum dynamic tension T_Dd,max (${\phi }_{cur}{=0}^o$)=1394 tons, T_Dd,max (${\phi }_{cur}{=60}^o$)=1741 tons larger than the fracture strength 1400tons. To overcome this failure, only the specification of rope D is increased to T_D,frac=2000tons as listed in Table 3. As shown in Figure 11b, when the buffer spring constant g_KD>100, i.e., the buffer spring D is not installed, the maximum dynamic tension T_Dd,max (${\phi }_{cur}{=0}^o$)=1392 tons, T_Dd,max (${\phi }_{cur}{=60}^o$)=1738 tons smaller than the fracture strength 2000tons.

6. Conclusions

In this study, the similarity formulas of MW- and KW- level ocean convertor are constructed. According to these, the hydrodynamic damping and stiffness coefficients and other parameters of MW-level system can be determined. The novel pulley-main rope design for MW- level ocean convertor mooring system is proposed. Further, the mathematical dynamic model of a MW- level ocean convertor mooring system under irregular wave and current direction is derived. The analytical solution of this system is presented. The dynamic performance of the proposed 1 MW mooring system under irregular wave impact is investigated and discovered as follows:

1. The dynamic rope tensions T_Ad,max and T_Bd,max of the traditional single-traction rope system are larger than the fracture strength T_frac=2000tons. 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.

2. The dynamic responses of the above two kinds of mooring systems are different. The traditional single traction rope system is not affected by the flow direction, but the pulley-rope system is.

3. The static tension of rope A of the proposed system under the steady current only is close to half of the traditional single traction rope system.

4. For H_bed =1300m, if the rope length ratio, $4.1\le r_{AH}\le 4.6$ , all the dynamic tensions can be smaller than the fracture strength. For H_bed =1000m, if the rope length ratio, $4.0\le r_{AH}\le 4.3$, all the dynamic tensions can be smaller than the fracture strength.

5. In a MW-level power generation system with the pulley-rope design, if buffer spring constant is very small, meanwhile the dynamic tension may become greater than the fracture strength. However, no buffer spring is set, it is reverse.

6. According to the theoretical analysis, the proposed MW-level ocean convertor mooring system with the pulley-rope design under typhoon wave impact is safety.