Difference between revisions of "Linear Wave-Body Interaction"

From WikiWaves
Jump to navigationJump to search
m (Reverted edits by Meylan (Talk) to last revision by Adi Kurniawan)
Line 1: Line 1:
 
{{Ocean Wave Interaction with Ships and Offshore Structures
 
{{Ocean Wave Interaction with Ships and Offshore Structures
 
  | chapter title = Linear Wave-Body Interaction
 
  | chapter title = Linear Wave-Body Interaction
  | next chapter = [[Long Wavelength Approximations]]
+
  | next chapter = [[Added-Mass, Damping Coefficients And Exciting Forces]]
 
  | previous chapter =  [[Ship Kelvin Wake]]
 
  | previous chapter =  [[Ship Kelvin Wake]]
 
}}
 
}}
 
{{incomplete pages}}
 
 
[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]
 
 
We consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] in the
 
[[Frequency Domain Problem|Frequency Domain]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem).
 
  
 
== Introduction ==
 
== Introduction ==
  
We derive here the equations of motion for a body in [[Linear Plane Progressive Regular Waves]] in the frequency domain in
+
We derive here the equations of motion for a floating body in [[Linear Plane Progressive Regular Waves]] in two dimensions.
two dimensions. We begin with the equations in the time domian. The simplest problems is [[Waves reflecting off a vertical wall]]
 
  
== Equations for a Floating Body in the Time Domain ==
+
== Linear wave-body interactions ==
  
We begin with the equations for a floating two-dimensional body in the time domain.
+
[[Image:Rigid_body.jpg|thumb|right|600px|Rigid body motions]]
  
{{equations of motion time domain without body condition}}
+
Consider a [[Linear Plane Progressive Regular Waves|Linear Plane Progressive Regular Wave]] interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem). We begin by defining the following,
  
{{two dimensional floating body time domain}}
+
<center><math> \zeta(t): \quad \mbox{ambient wave elevation. Regular or random with definitions to be given below}. \,</math></center>
  
More details can be found in [[:Category:Time-Dependent Linear Water Waves|Time-Dependent Linear Water Waves]]
+
<center><math> \xi_1(t): \quad \mbox{Body surge displacement} \,</math></center>
  
== Equations for a Floating Body in the Frequency Domain ==
+
<center><math> \xi_3(t): \quad \mbox{Body heave displacement} \,</math></center>
  
The dynamic condition is the equation of motion for the structure in the [[Frequency Domain Problem|frequency domain]]
+
<center><math> \xi_4(t): \quad \mbox{Body roll displacement} \,</math></center>
can be found from the time domain equations and we introduce the following notation
 
<center><math>
 
\xi_{\nu} = \zeta_{\nu}e^{\mathrm{i}\omega t}\,
 
</math></center>
 
This give us
 
{{standard linear wave scattering equations without body condition}}
 
<center><math>
 
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega}\phi n_{\mu}\, dS
 
- \sum_{\nu} C_{\mu\nu}\zeta_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
 
</math></center>
 
The equations of motion for <math> \zeta_\nu\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions. We use the standard numbering of the modes of motion.
 
  
== Equations for a Fixed Body in Frequency Domain ==
+
where the surge, heave and roll are the three [http://en.wikipedia.org/wiki/Rigid_body_dynamics rigid body] motions which are possible in two dimensions.
  
The equations for a fixed body are
+
== Linear theory ==
  
{{standard linear wave scattering equations without body condition}}
+
We assume that we can apply linear theory to the motions, which are an extension of the [[Linear and Second-Order Wave Theory| linear equations]] for a free-surface. We assume that
{{frequency domain equations for a rigid body}}
+
<center><math> \left| \frac{\partial\zeta}{\partial x} \right| = O(\varepsilon) \ll 1 \, </math></center>
plus the radiation conditions.
+
This is an assumption of small wave steepness which is a reasonable assumption for gravity waves in most cases, except when waves are near breaking conditions. Furthermore we assume
 +
<center><math> \left| \frac{\xi_1}{A} \right| = O(\varepsilon) \ll 1 \, </math></center>
 +
<center><math> \left| \frac{\xi_3}{A} \right| = O(\varepsilon) \ll 1 \, </math></center>
 +
<center><math> \left| \xi_4 \right| = O(\varepsilon) \ll 1 \, </math></center>
 +
These assumptions are valid in most cases and most bodies of practical interest, unless the vessel response at resonance is highly tuned or lightly damped. This is often the case for roll when a small amplitude wave interacts with a vessel weakly damped in roll.
  
We decompose the potential as
+
== Linear systems theory ==
<center>
 
<math>
 
\phi = \phi^{\mathrm{I}} + \phi^{\mathrm{D}}
 
</math>
 
</center>
 
where <math>\phi^{\mathrm{I}}</math> is the incident potential and <math>\phi^{\mathrm{D}}</math>
 
is the diffracted potential.  The boundary condition for the diffracted potential is
 
<center><math>
 
\Delta\phi^{\mathrm{D}}=0, \, -h<z<0,\,\,\,\mathbf{x} \in \Omega
 
</math></center>
 
<center><math>
 
\partial_n\phi^{\mathrm{D}} = 0, \, z=-h,
 
</math></center>
 
<center><math>
 
\partial_n \phi^{\mathrm{D}}  = \alpha \phi,\,z=0,\,\,\mathbf{x} \in \partial\Omega_{F},
 
</math></center>
 
plus
 
<center><math>
 
\partial_n \phi^{\mathrm{D}}  = - \partial_n \phi^{\mathrm{I}},\,\, \mathbf{x} \in \partial\Omega,
 
</math></center>
 
  
Code to calculate the solution (using a slighly modified method) can be found in
+
[[Image:Linear_systems_theory.jpg|thumb|right|600px|Linear systems theory]]
[[Boundary Element Method for a Fixed Body in Finite Depth]]
 
  
== Equations for the Radiation Potential in Frequency Domain ==
+
The vessel dynamic responses in waves may be modelled according to linear system theory. By virtue of linearity, a random seastate may be represented as the linear superposition of [[Linear Plane Progressive Regular Waves]] (see [[Waves and the Concept of a Wave Spectrum]])
 +
<center><math> \zeta(x,t) = \sum_j A_j \cos ( K_j x - \omega_j t + \epsilon_j ) \,</math></center>
 +
where in deep water: <math> K_j = \frac{\omega_j^2}{g} \,</math>. Note that the sum here can be replace by an integral in many formulations. According to the theory of St. Denis and Pierson, the phases <math> \epsilon\, </math>, are random and uniformly distributed between <math> ( - \pi, \pi ] \, </math>. For now we assume them known constants:
  
We decompose the body motion into the rigid body modes of motion. Associated with
+
At <math> X=0\,</math>:
each of these modes is a potential which must be solved for.
 
The equations for the radiation potential in the frequency domain are
 
  
{{standard linear wave scattering equations without body condition}}
+
<center><math> \zeta(t) = \sum_j A_j \cos ( \omega_j t - \epsilon_j ) \,</math></center>
{{frequency domain equations for the radiation modes}}
 
  
{{sommerfeld radiation condition two dimensions for radiation}}
+
<center><math> = \mathrm{Re} \sum_j A_j e^{i\omega_j t - i \epsilon_j} </math></center>
  
Code to calculate the radiation potential can be found in  
+
The corresponding vessel responses follow from linearity in the form:
[[Boundary Element Method for the Radiation Potential in Finite Depth]]
 
  
We denote the solution for each of the radiation potentials by
+
<center><math> \xi_K (t) = \mathrm{Re} \sum_j \Pi_K (\omega_j) e^{i\omega_j t - i\epsilon_j},
<math>\phi_\nu^{\mathrm{R}}</math> and the total potential is written as
 
<center><math>
 
\phi = \phi^{\mathrm{I}} +  \phi^{\mathrm{D}} +
 
\sum_\nu \zeta_\nu \phi_\nu^{\mathrm{R}}
 
</math>
 
</center>
 
  
== Final System of Equations ==
+
\qquad K = 1, 3, 4 </math></center>
  
We substitute the expansion for the potential into the equations in the frequency domain and we obtain
+
Where <math> \Pi_K (\omega) \, </math> is the complex RAO ([[Response Amplitude Operator]]) for mode <math> K\,</math>. It is the object of linear seakeeping theory to derive equations for <math>\Pi\omega)\,</math> the frequency domain. The treatment in the stochastic case is then a simple exercise in linear systems.
<center><math>
 
-\omega^2 \sum_{\nu} M_{\mu\nu}\zeta_{\nu}=-\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
 
\left(\phi^{\mathrm{I}} +  \phi^{\mathrm{D}} +
 
\sum_{\nu} \zeta_\nu \phi_{\nu}^{\mathrm{R}}\right) n_{\mu}\, dS
 
- \sum_{\nu}  C_{\mu\nu}\xi_{\nu},\quad \textrm{for} \qquad \mu=1,3,5,
 
</math></center>
 
  
We then define the matrices
+
== Calculation of the RAO ==
<center><math>
 
A_{\mu\nu} = \mathrm{Re} \left\{ -\frac{\mathrm{i}}{\omega}\rho\iint_{\partial\Omega_{B}}
 
\phi_{\nu}^{\mathrm{R}} n_{\mu}\, dS \right\}
 
</math></center>
 
which is called the added mass matrix and
 
We then define the matrices
 
<center><math>
 
B_{\mu\nu} = \mathrm{Im} \left\{ \rho\iint_{\partial\Omega_{B}}
 
\phi_{\nu}^{\mathrm{R}} n_{\mu}\, dS \right\}
 
</math></center>
 
which is called the damping matrix and the forcing vector is
 
<center><math>
 
f_{\mu} = -\mathrm{i}\omega\rho\iint_{\partial\Omega_{B}}
 
\left(\phi^{\mathrm{I}} +  \phi^{\mathrm{D}} \right) n_{\mu}\, dS
 
</math></center>
 
then the equations can be expressed as follows.
 
<center><math>  \left[-\omega^2 \left(\mathbf{M} + \mathbf{A} \right) +
 
\mathrm{i}\omega \mathbf{B} + \mathbf{C} \right] \vec{\zeta} = \mathbf{f} </math></center>
 
where <math>\mathbf{M}</math> is the mass matrix,  <math>\mathbf{A}</math> is the added mass matrix,
 
<math>\mathbf{B}</math> is the damping matrix, <math>\mathbf{C}</math> is the hydrostatic matrix,
 
<math>\vec{\zeta}</math> is the vector of body displacements and <math>\mathbf{f}</math> is the force.
 
  
 +
The equations of motion for <math> \xi_K(t)\,</math> follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions
  
The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the two dimensional problem for the sake of clarity.
+
We begin by considering the equation in surge.
 +
<center><math> \mathbf{M} \frac{\mathrm{d}^2\xi_1}{\mathrm{d}t^2} = F_{1\omega} ( \xi_1, \dot\xi_1, \ddot\xi_1, t) </math></center>
 +
where <math> \frac{\mathrm{d}\xi_1}{\mathrm{d}t} = \dot\xi_1 \, </math> and <math> F_{1\omega} \, </math> is the force on the body due to the fluid pressures, by virtue of linearity, <math> F_{1\omega} \,</math> will be assumed to be a linear functional of <math> \xi_1, \dot\xi_1, \ddot\xi_1 \, </math>. [[Memory effects]] exist when surface waves are generated on the free surface, so <math> F_{1\omega} \,</math> depends in principle on the entire history of the vessel displacement. We adopt here the [[Frequency Domain Problem|frequency domain]] formulation where the vessel motion has been going on over an infinite time interval, <math> (-\infty, t)\,</math> with <math> e^{i\omega t}\,</math> dependence.
  
== Symmetric body ==
+
We will therefore set:
 +
<center><math> \xi_K(t) = \mathrm{Re} \left\{ \Pi_K e^{i\omega t} \right\}, \qquad K=1,3,4 </math></center>
 +
In this case we can linearize the water induced force on the body as follows:
 +
<center><math> F_{1\omega}(t) = X_1(t) - A_{11} \ddot \xi_1 - A_{13} \ddot \xi_3 - A_{14} \ddot \xi_4 </math></center>
  
For a body which is [[:Category:Symmetry in Two Dimensions|Symmetric in Two Dimensions]]
+
<center><math> - B_{11} \dot \xi_1 - B_{13} \dot \xi_3 - B_{14} \dot \xi_4 </math></center>
the Heave is decoupled from Surge and Roll.
 
In other words the Surge and Roll motions do not influence Heave and vice versa.
 
  
== Estimating the Heave and Pitch resonant frequencies ==
+
<center><math> - C_{11} \xi_1 - C_{13} \xi_3 - C_{14} \xi_4 \,</math></center>
  
Often it is useful to estimate the Heave and Pitch resonant frequencies in terms of the principal dimensions of a body.
+
<center><math> = X_1(t) - \sum_j \left[ A_{1j} \ddot \xi_j + B_{1j} \dot \xi_j + C_{1j} \xi_j \right] </math></center>
Consider a box-like section with beam <math>2L\,</math> and draft <math>d\,</math>
 
The Heave added mass <math> A_{33}, \, </math> is often not too far from its infinite frequency limit, or:
 
<center><math> A_{33} \simeq A_{33}(\infty) = \frac{1}{2} \rho (2L)d = \frac{1}{2} M</math></center>
 
  
<center><math> \omega^2 = \frac{\rho g B}{\rho BT + \frac{1}{2} \rho B T }= \frac{3}{2} \frac{g}{T} \, </math></center>
+
The same expansion applies for other modes, namely Heave (<math> K = 3 \, </math>) and Roll (<math> K=4 \, </math>). In sum:
  
or:
+
<center><math> F_{K\omega} (t) = X_K - \sum_j \left[ A_{Kj} \ddot \xi_j + B_{Kj} \dot \xi_j + C_{Kj} \xi_j \right], \qquad K = 1,3,4 </math></center>
  
<center><math> \omega^* \simeq \left( \frac{3}{2} \frac{g}{T} \right)^{\frac{1}{2}}</math></center>
+
* The added-mass matrix <math> A_{Kj} \,</math> represents the added inertia due to the acceleration of the body in water with acceleration <math>\ddot\xi_j\,</math>.
  
We can derive a corresponding result for Roll.
+
* The damping matrix <math> B_{Kj}\,</math> governs the energy dissipation into the fluid domain in the form of surface waves.
  
== Matlab Code ==
+
* The hydrostatic restoring matrix <math> C_{Kj} \, </math> represents the system stifness due to the hydrostatic restoring forces and moments.
  
 +
For harmonic motions, the matrices <math> A_{Kj} \, </math> and <math> B_{Kj} \, </math> are functions of <math> \omega\,</math>, so we write <math> A_{Kj} (\omega), \ B_{Kj} (\omega)\, </math>. This functional form will be discussed below. The hydrostatic matrix <math> C_{Kj} \, </math> is independent of <math>\omega\,</math> and many of its elements are identically equal to zero. Collecting terms in the left-hand side and denoting by <math> M_{Kj}\,</math> the body inertia matrix:
  
*A program to solve for pitch and heave and only for two geometries can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/rigid_body_motions.m rigid_body_motions.m]
+
=== Surge ===
  
* a program to calculate the solution for a specific geometry (with plot as output as shown) can be found here [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/wave_bem_example_floating.m wave_bem_example_floating.m]
+
<center><math> \sum_j \left[ -\omega^2 \left( M_{1j} + A_{1j} \right) + i\omega B_{1j} + C_{1j} \right] \Pi_j = \mathbf{X}_1 (\omega), \quad j=1,3,4 </math></center>
  
[[Image:Wave_bem_example_floating_RT.jpg|300px|right|thumb|The reflection (solid line) and transmission (dashed line)
+
=== Heave ===
for a dock for heave and pitch (red), heave only (blue) and pitch only (black)]]
 
  
=== Additional code ===
+
<center><math> \sum_j \left[ -\omega^2 \left( M_{3j} + A_{3j} \right) + i\omega B_{3j} + C_{3j} \right] \Pi_j = \mathbf{X}_3 (\omega), \quad j=1,3,4 </math></center>
  
This program requires
+
=== Roll ===
* A program to calculate the geometery [http://www.math.auckland.ac.nz/~meylan/code/boundary_element/circlebody_twod.m  circlebody_twod.m]
 
* {{fixed body bem code}}
 
* {{floating body radiation code}}
 
* {{free surface dispersion equation code}}
 
* {{boundary element code}}
 
  
== Symmetry-reciprocity relations ==
+
<center><math> \sum_j \left[ -\omega^2 \left( I_G + A_{4j} \right) + i\omega B_{4j} + C_{4j} \right] \Pi_j = \mathbf{X}_4 (\omega)</math></center>
  
It will be shown that
+
The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the 2D Problem for the sake of clarity.
<center><math> A_{ij}(\omega) = A_{ji}(\omega) \,</math></center>
 
<center><math> B_{ij}(\omega) = B_{ji}(\omega) \,</math></center>
 
Along the same lines it will be shown that the exciting force <math>\mathbf{X}_j\,</math> can be expressed in terms of <math> \psi_j\,</math> circumventing the solution for the diffraction potential.
 
The core result needed for the proof of the above properties is [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
 
<center><math> \iint_S \left( \psi_1 \frac{\partial\psi_2}{\partial n} - \psi_2 \frac{\partial\psi_1}{\partial n} \right) \mathrm{d}S = 0 \,</math></center>
 
where <math>\nabla^2 \psi_i=0</math>.
 
  
[[Image:Symmetry_boundary.jpg|thumb|right|600px|Boundary]]
+
== Symmetric body ==
  
In the surface wave-body problem define the closed surfaces as shown in figure on the right.
+
Consider a body symmetric about the <math> X = 0\,</math> axis.
Let <math> \phi_j\,</math> be rediation or diffraction potentials. Over the boundaries <math>S^\pm\,</math> we have:
 
  
<center><math> S^+: \quad \phi_j \ \sim \ \frac{igA_j^+}{\omega} e^{Kz-iKx} \,</math></center>
+
[[Image:Symmetric.jpg|thumb|right|600px|Symmetric body]]
  
<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial \phi_j}{\partial x} \ \sim \ -iK\phi_j \,</math></center>
+
For a body symmetric port/starboard:
  
<center><math> S^-: \quad \phi_j \ \sim \ \frac{igA_j^-}{\omega} e^{Kz+iKx} \,</math></center>
+
* Verify that Heave is decoupled from Surge and Roll. In other words the Surge and Roll motions do not influence Heave and vice versa:
  
<center><math> \frac{\partial\phi_j}{\partial n} = - \frac{\partial \phi_j}{\partial x} \ \sim \ - iK\phi_j \,</math></center>
+
<center><math> \left[ -\omega^2 \left( M + A_{33} \right) + i\omega B_{33} + C_{33} \right] \Pi_3 = A </math></center>
  
On <math> S_F: \qquad \frac{\partial\phi_j}{\partial z} = K\phi_j, \qquad \frac{\partial \Phi_j}{\partial n} = \frac{\partial \phi_j}{\partial z} </math>
+
* The only nonzero hydrostatic coefficients are <math> C_{33} \, </math> and <math> C_{44} \, </math>. Verify that this is the case even for non-symmetric sections.
  
On <math> S_\infty: \qquad \left| \phi_j \right|, \quad \left| \nabla \phi_j \right| \to 0 </math>.
+
* Surge and Roll are coupled for symmetric and non-symmetric bodies. The coupled equation of motion becomes:
  
Applying Green's identity to any pair of the radiation potentials <math> \psi_i, \psi_j \,</math>:
+
<u>Surge-Roll</u>
  
<center><math> \iint_{S_B} \left[ \psi_i \frac{\partial\psi_j}{\partial n} - \psi_j \frac{\partial\psi_i}{\partial n} \right] \mathrm{d}S = - \iint_{S_F} \left[ \psi_i \frac{\partial\psi_j}{\partial z} - \psi_j \frac{\partial\psi_i}{\partial z} \right] \mathrm{d}S </math></center>
+
<center><math> \sum_{j=1,4} \left[ -\omega^2 \left( M_{ij} + A_{ij} \right) + i\omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_i, \quad i,j = 1,4 </math></center>
  
<center><math> - \iint_{S_+} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S
+
* When Newton's law is expressed about the center of gravity:
+ \iint_{S_-} \left[ \psi_i \frac{\partial\psi_j}{\partial x} - \psi_j \frac{\partial\psi_i}{\partial x} \right] \mathrm{d}S = 0 </math></center>
 
  
It follows that:
+
<center><math> M_{14} = M_{41} = 0, \ M_{11} = M, \ M_{44} = I_G </math></center>
  
<center><math> \iint_{S_B} \psi_i \frac{\partial\psi_j}{\partial n} \mathrm{d}S = \iint_{S_B} \psi_j \frac{\partial\psi_i}{\partial n} \mathrm{d}S </math></center>
+
where <math> I_G\,</math> is the body moment of inertia about the center of gravity. If the equations are to be expressed about the origin of the coordinate system, then the formulation must start with respect to <math> G\,</math> and expressions derived w.r.t. <math> O \,</math>, via a coordinate transformation.
or
 
<center><math> A_{ij}(\omega) = A_{ji}(\omega), \qquad B_{ij}(\omega) = B_{ji}(\omega). \,</math></center>
 
  
== Haskind relations of exciting forces ==
+
* The exciting forces <math> \mathbf{X}_1, \mathbf{X}_3 \,</math> are defined in an obvious manner along the X- and Z-axis. The Roll moment <math> \mathbf{X}_4 \, </math> is defined initially about <math> G\,</math>.
  
<center><math> \mathbf{X}_i(\omega) = - i\omega\rho\iint_{S_B} (\phi_I + \phi_7) n_i dS </math></center>
+
Need to derive definitions for the coefficients that enter the Heave & Surge-Roll equations of motion:
  
<center><math> - \rho \iint_{S_B} (\phi_I + \phi_7) \frac{\partial \phi_i}{\partial n} dS </math></center>
+
<center><math> M = \rho \forall, \qquad \forall = \ </math> volume of water displaced by body (archimedian principle of buoyancy) </center>
  
where the radiation velocity potential <math> \phi_i \,</math> is known to satisfy:
+
<center><math> C_{33} = \rho g A_\omega = \rho g B, </math></center>
  
<center><math> \frac{\partial\phi_i}{\partial n} = i\omega n_i, \quad \mbox{on} \ S_B </math></center>
+
<center><math> A_\omega = </math> body waterplane area  = <math>B</math> (Beam in two dimensions)</center>
  
and
+
<center><math> \left( C_{44} \right)_G = \rho g \frac{B^3}{12} = \ </math> Roll restoring moment due to a small angular displacement </center>
 +
<center> about the center of gravity. Verify for all wall and non-wall sided sections </center> 
  
<center><math> \frac{\partial\phi_7}{\partial n} = \frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \ S_B </math></center>
+
<center><math> C_{44} \equiv \left( C_{44} \right)+G \ne \left( C_{44} \right)_O; \quad </math> Derive an expression for <math> \ \left( C_{44} \right)_O \ </math> in terms of <math> \ \left( C_{44} \right)_G. </math></center>
  
Both <math> \phi_i\,</math> and <math> \phi_7\,</math> satisfy the condition of outgoing waves at infinity. By virtue of [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
+
-----
 
 
<center><math> \iint_{S_B} \phi_7 \frac{\partial\phi_i}{\partial n} dS = \iint_{S_B} \phi_i \frac{\partial\phi_7}{\partial n} dS = -\iint_{S_B} \phi_i \frac{\partial\phi_I}{\partial n} dS </math></center>
 
 
 
The Haskind expression for the exciting force follows:
 
 
 
<center><math> \mathbf{X}_i(\omega) = \rho \iint_{S_B} \left[ \phi_I \frac{\partial\phi_i}{\partial n} - \phi_i \frac{\partial\phi_I}{\partial n} \right] dS </math></center>
 
 
 
The symmetry of the <math> A_ij(\omega), B_ij(\omega) \,</math> matrices applies in 2D and 3D. The application of Green's Theorem in 3D is very similar using the far-field representation for the potential <math> \phi_j\,</math>
 
 
 
<center><math> \phi_j \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O\left(\frac{1}{R^{3/2}}\right) </math></center>
 
 
 
<center><math> \frac{\partial\phi_j}{\partial n} = \frac{\partial\phi_j}{\partial R} \sim - i K \phi_j + O\left(\frac{1}{R^{3/2}}\right) </math></center>
 
 
 
where <math> R \,</math> is a radius from the body out to infinity and the <math> R^{-\frac{1}{2}} \,</math> decay arises from the energy conservation principle. Details of the 3D proof may be found in [[Mei 1983]] and [[Wehausen and Laitone 1960]]
 
 
 
The use of the Haskind relations for the exciting forces does not require the solution of the diffraction problem. This is convenient and often more accurate.
 
 
 
The Haskind relations take other forms which will not be presented here but are detailed in [[Wehausen and Laitone 1960]]. The ones that are used in practice relate the exciting forces to the damping coefficients.
 
 
 
These take the form:
 
 
 
<u>2D</u>:      <math> B_ii = \frac{\left| \mathbf{X}_i \right|^2}{2\rho g V_g}, \quad V_g = \frac{g}{2\omega}, </math>      Deep water
 
 
 
<u>3D</u>:      <math> B_33 = \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2 \,</math>      --- Heave
 
 
 
(Axisymmetric bodies)      <math> B_22 = \frac{K}{8\rho g V_g} \left| \mathbf{X}_2 \right|^2 \,</math>      --- Sway
 
 
 
So knowledge of <math> \mathbf{X}_i(\omega)\,</math> allows the direct evaluation of the diagonal damping coefficients. These expressions are useful in deriving theoretical results in wave-body interactions to be discussed later.
 
 
 
The two-dimensional theory of wave-body interactions in the frequency domain extends to three dimencions very directly with little difficulty.
 
 
 
The statement of the 6 d.o.f. seakeeping problem is:
 
 
 
<center><math> \sum_{j=1}^6 \left[ - \omega^2 \left( M_{ij} + A_{ij} \right) + i \omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_j, \quad i=1,\cdots,6 </math></center>
 
  
Where:
+
This article is based on the MIT open course notes and the original article can be found
 
+
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here]
<center><math> \mbox{Body inertia matrix including moments of inertia for rotational modes. For details refer to MH} \, </math></center>
 
 
 
<center><math> A_{ij}(\omega): \mbox{Added mass matrix} \,</math></center>
 
 
 
<center><math> B_{ij}(\omega): \mbox{Damping matrix} \, </math></center>
 
 
 
<center><math> C_{ij}: \mbox{Hydrostatic and static inertia restoring matrix. For details refer to MH} \, </math></center>
 
 
 
<center><math> \mathbf{X}_i(\omega): \mbox{Wave exciting forces and moments} </math></center>
 
 
 
At zero speed the definitions of the added-mass, damping matrices and exciting forces are identical to those in two dimensions.
 
 
 
The boundary value problems satisfied by the radiation potentials <math>\phi_j, \ j=1,\cdots,6 \,</math> and the diffraction potential <math> \phi_7 \,</math> are as follows:
 
 
 
Free-surface condition:
 
 
 
<center><math> -\omega^2 \phi_j + g \frac{\partial\phi_j}{\partial Z} = 0, \quad z=0 \quad j=1,\cdots,7 </math></center>
 
 
 
Body-boundary conditions:
 
 
 
<center><math> \frac{\partial\phi_7}{\partial n} = -\frac{\partial\phi_I}{\partial n}, \quad \mbox{on} \quad S_B </math></center>
 
 
 
<center><math> \phi_I = \frac{i g A}{\omega} e^{KZ-iKX\cos\beta-iKY\sin\beta+i\omega t} </math></center>
 
 
 
<center><math> \frac{\partial\phi_j}{\partial n} = i\omega n_j, \quad j=1,\cdots,6 \, </math></center>
 
 
 
<center><math> n_j = \begin{Bmatrix}
 
  & n_j, \qquad & j=1,2,3 \\
 
  & \left( \vec{X} \times \vec{n} \right)_{j+3}, \quad j=4,5,6
 
\end{Bmatrix} </math></center>
 
 
 
<center><math> i=1: \ \mbox{Surge} \qquad i=2: \ \mbox{Sway} \qquad i=3: \ \mbox{Heave} </math></center>
 
<center><math> i=4: \ \mbox{Roll} \qquad i=5: \ \mbox{Pitch} \qquad i=6: \ \mbox{Yaw} </math></center>
 
 
 
At large distances from the body the velocity potentials satisfy the radiation condition:
 
 
 
<center><math> \phi_j (R,\theta) \sim \frac{A_j(\theta)}{\sqrt{R}} e^{KZ-iKR} + O \left( \frac{1}{R^{3/2}} \right) </math></center>
 
 
 
with
 
 
 
<center><math> K = \frac{\omega^2}{g}. \,</math></center>
 
 
 
This radiation condition is essential for the formulation and solution of the boundary value problems for <math>\phi_j\,</math> using panel methods which are the standard solution technique at zero and forward speed.
 
 
 
Qualitative behaviour of the forces, coefficients and motions of floating bodies
 
 
 
<center><math> - \omega^2 \phi + g \phi_Z =0 \quad \begin{cases}
 
  \phi_Z=0,\quad \omega=0  \\
 
  \phi=0, \quad \omega \to \infty
 
\end{cases} </math></center>
 
 
 
<center><math> B_{33}(\omega), \sim \omega, \mbox{at low} \ \omega \,</math></center>
 
 
 
The 2D Heave added mass is singular at low frequencies. It is finite in 3D
 
 
 
The 2D Heave damping coefficient is decaying to zero linearly in 2D and superlinearly in 3D. A two-dimensional section is a better wavemaker than a three-dimensional one
 
 
 
A 2D section oscillating in Sway is less effective a wavemaker at low frequencies than the same section oscillating in Heave
 
 
 
The zero-frequency limit of the Sway added mass is finite and similar to the infinite frequency limit of the Heave added mass.
 
 
 
In long waves the Heave exciting force tends to the Heave restoring coefficient times the ambient wave amplitude the free surface behaves like a flat surface moving up and down.
 
 
 
In long waves the Sway exciting force tends to zero. Proof will follow
 
 
 
In short waves all forces tend to zero.
 
 
 
Pitch exciting moment (same applies to Roll) tends to zero. Long waves have a small slope which is proportional to <math> KA</math>, where <math> K\,</math> is the wave number and <math> A\,</math> is the wave amplitude.
 
 
 
Prove that to leading order for <math>KA\to 0 \,</math>:
 
 
 
<center><math> \left| X_S(\omega) \right| \sim KA C_{55}\,</math></center>
 
 
 
where <math>C_{55}\,</math> is the Pitch (<math> C_{44} \,</math> for Roll) hydrostatic restoring coefficient. [NB: very long waves look like a flat surface inclined at <math> KA\,</math> ].
 
 
 
== Body motions in regular waves ==
 
 
 
Heave:
 
 
 
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega)}{-\omega^2(A_{33} + M) + i\omega B_{33} +C_{33} } </math></center>
 
 
 
Resonance:
 
 
 
<center><math> \omega^2 = \frac{C_{33}}{M+A_{33}} = \frac{\rho g A \omega}{M + A_{33} (\omega)} </math></center>
 
 
 
In principle the above equation is nonlinear for <math>\omega\,</math>. Will be approximated below
 
 
 
At resonance:
 
<center><math> \Pi_3 = \frac{\mathbf{X}_3(\omega^*)}{i\omega^* B_{33}(\omega^*)} \,</math></center>
 
 
 
Invoking the relation between the damping coefficient and the exicting force in 3D:
 
 
 
<center><math> \frac{\left| \Pi_3 \right|}{A} = \frac{\left| \mathbf{X}_3(\omega) \right|}{\omega \frac{K}{4\rho g V_g} \left| \mathbf{X}_3 \right|^2}, \quad V_g=\frac{g}{2\omega} </math></center>
 
 
 
<center><math> =\frac{2\rho g}{\omega^3 \left|\mathbf{X}_3(\omega)\right|}, \quad \mbox{at resonance} </math></center>
 
 
 
This counter-intuitive result shows that for a body undergoing a pure Heave oscillation, the modulus of the Heave response at resonance is inversely proportional to the modulus of the Heave exciting force.
 
 
 
Viscous effects not discussed here may affect Heave response at resonance
 
 
 
The behavior of the Sway response can be found in an analagous manner,
 
 
 
 
 
-----
 
  
This article is based on the MIT open course notes and the original articles can be found
+
[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/666E84F4-5679-47FD-BD7B-9D39877DE5A1/0/lecture9.pdf here] and
 
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C5323823-0180-45EA-B165-15856948A0A2/0/lecture10.pdf here]
 

Revision as of 20:00, 24 April 2010

Wave and Wave Body Interactions
Current Chapter Linear Wave-Body Interaction
Next Chapter Added-Mass, Damping Coefficients And Exciting Forces
Previous Chapter Ship Kelvin Wake


Introduction

We derive here the equations of motion for a floating body in Linear Plane Progressive Regular Waves in two dimensions.

Linear wave-body interactions

Rigid body motions

Consider a Linear Plane Progressive Regular Wave interacting with a floating body in two dimensions (the main concepts survive almost with no change in the more practical three-dimensional problem). We begin by defining the following,

[math]\displaystyle{ \zeta(t): \quad \mbox{ambient wave elevation. Regular or random with definitions to be given below}. \, }[/math]
[math]\displaystyle{ \xi_1(t): \quad \mbox{Body surge displacement} \, }[/math]
[math]\displaystyle{ \xi_3(t): \quad \mbox{Body heave displacement} \, }[/math]
[math]\displaystyle{ \xi_4(t): \quad \mbox{Body roll displacement} \, }[/math]

where the surge, heave and roll are the three rigid body motions which are possible in two dimensions.

Linear theory

We assume that we can apply linear theory to the motions, which are an extension of the linear equations for a free-surface. We assume that

[math]\displaystyle{ \left| \frac{\partial\zeta}{\partial x} \right| = O(\varepsilon) \ll 1 \, }[/math]

This is an assumption of small wave steepness which is a reasonable assumption for gravity waves in most cases, except when waves are near breaking conditions. Furthermore we assume

[math]\displaystyle{ \left| \frac{\xi_1}{A} \right| = O(\varepsilon) \ll 1 \, }[/math]
[math]\displaystyle{ \left| \frac{\xi_3}{A} \right| = O(\varepsilon) \ll 1 \, }[/math]
[math]\displaystyle{ \left| \xi_4 \right| = O(\varepsilon) \ll 1 \, }[/math]

These assumptions are valid in most cases and most bodies of practical interest, unless the vessel response at resonance is highly tuned or lightly damped. This is often the case for roll when a small amplitude wave interacts with a vessel weakly damped in roll.

Linear systems theory

Linear systems theory

The vessel dynamic responses in waves may be modelled according to linear system theory. By virtue of linearity, a random seastate may be represented as the linear superposition of Linear Plane Progressive Regular Waves (see Waves and the Concept of a Wave Spectrum)

[math]\displaystyle{ \zeta(x,t) = \sum_j A_j \cos ( K_j x - \omega_j t + \epsilon_j ) \, }[/math]

where in deep water: [math]\displaystyle{ K_j = \frac{\omega_j^2}{g} \, }[/math]. Note that the sum here can be replace by an integral in many formulations. According to the theory of St. Denis and Pierson, the phases [math]\displaystyle{ \epsilon\, }[/math], are random and uniformly distributed between [math]\displaystyle{ ( - \pi, \pi ] \, }[/math]. For now we assume them known constants:

At [math]\displaystyle{ X=0\, }[/math]:

[math]\displaystyle{ \zeta(t) = \sum_j A_j \cos ( \omega_j t - \epsilon_j ) \, }[/math]
[math]\displaystyle{ = \mathrm{Re} \sum_j A_j e^{i\omega_j t - i \epsilon_j} }[/math]

The corresponding vessel responses follow from linearity in the form:

[math]\displaystyle{ \xi_K (t) = \mathrm{Re} \sum_j \Pi_K (\omega_j) e^{i\omega_j t - i\epsilon_j}, \qquad K = 1, 3, 4 }[/math]

Where [math]\displaystyle{ \Pi_K (\omega) \, }[/math] is the complex RAO (Response Amplitude Operator) for mode [math]\displaystyle{ K\, }[/math]. It is the object of linear seakeeping theory to derive equations for [math]\displaystyle{ \Pi\omega)\, }[/math] the frequency domain. The treatment in the stochastic case is then a simple exercise in linear systems.

Calculation of the RAO

The equations of motion for [math]\displaystyle{ \xi_K(t)\, }[/math] follow from Newton's law applied to each mode in two dimensions. The same principles apply with very minor changes in three dimensions

We begin by considering the equation in surge.

[math]\displaystyle{ \mathbf{M} \frac{\mathrm{d}^2\xi_1}{\mathrm{d}t^2} = F_{1\omega} ( \xi_1, \dot\xi_1, \ddot\xi_1, t) }[/math]

where [math]\displaystyle{ \frac{\mathrm{d}\xi_1}{\mathrm{d}t} = \dot\xi_1 \, }[/math] and [math]\displaystyle{ F_{1\omega} \, }[/math] is the force on the body due to the fluid pressures, by virtue of linearity, [math]\displaystyle{ F_{1\omega} \, }[/math] will be assumed to be a linear functional of [math]\displaystyle{ \xi_1, \dot\xi_1, \ddot\xi_1 \, }[/math]. Memory effects exist when surface waves are generated on the free surface, so [math]\displaystyle{ F_{1\omega} \, }[/math] depends in principle on the entire history of the vessel displacement. We adopt here the frequency domain formulation where the vessel motion has been going on over an infinite time interval, [math]\displaystyle{ (-\infty, t)\, }[/math] with [math]\displaystyle{ e^{i\omega t}\, }[/math] dependence.

We will therefore set:

[math]\displaystyle{ \xi_K(t) = \mathrm{Re} \left\{ \Pi_K e^{i\omega t} \right\}, \qquad K=1,3,4 }[/math]

In this case we can linearize the water induced force on the body as follows:

[math]\displaystyle{ F_{1\omega}(t) = X_1(t) - A_{11} \ddot \xi_1 - A_{13} \ddot \xi_3 - A_{14} \ddot \xi_4 }[/math]
[math]\displaystyle{ - B_{11} \dot \xi_1 - B_{13} \dot \xi_3 - B_{14} \dot \xi_4 }[/math]
[math]\displaystyle{ - C_{11} \xi_1 - C_{13} \xi_3 - C_{14} \xi_4 \, }[/math]
[math]\displaystyle{ = X_1(t) - \sum_j \left[ A_{1j} \ddot \xi_j + B_{1j} \dot \xi_j + C_{1j} \xi_j \right] }[/math]

The same expansion applies for other modes, namely Heave ([math]\displaystyle{ K = 3 \, }[/math]) and Roll ([math]\displaystyle{ K=4 \, }[/math]). In sum:

[math]\displaystyle{ F_{K\omega} (t) = X_K - \sum_j \left[ A_{Kj} \ddot \xi_j + B_{Kj} \dot \xi_j + C_{Kj} \xi_j \right], \qquad K = 1,3,4 }[/math]
  • The added-mass matrix [math]\displaystyle{ A_{Kj} \, }[/math] represents the added inertia due to the acceleration of the body in water with acceleration [math]\displaystyle{ \ddot\xi_j\, }[/math].
  • The damping matrix [math]\displaystyle{ B_{Kj}\, }[/math] governs the energy dissipation into the fluid domain in the form of surface waves.
  • The hydrostatic restoring matrix [math]\displaystyle{ C_{Kj} \, }[/math] represents the system stifness due to the hydrostatic restoring forces and moments.

For harmonic motions, the matrices [math]\displaystyle{ A_{Kj} \, }[/math] and [math]\displaystyle{ B_{Kj} \, }[/math] are functions of [math]\displaystyle{ \omega\, }[/math], so we write [math]\displaystyle{ A_{Kj} (\omega), \ B_{Kj} (\omega)\, }[/math]. This functional form will be discussed below. The hydrostatic matrix [math]\displaystyle{ C_{Kj} \, }[/math] is independent of [math]\displaystyle{ \omega\, }[/math] and many of its elements are identically equal to zero. Collecting terms in the left-hand side and denoting by [math]\displaystyle{ M_{Kj}\, }[/math] the body inertia matrix:

Surge

[math]\displaystyle{ \sum_j \left[ -\omega^2 \left( M_{1j} + A_{1j} \right) + i\omega B_{1j} + C_{1j} \right] \Pi_j = \mathbf{X}_1 (\omega), \quad j=1,3,4 }[/math]

Heave

[math]\displaystyle{ \sum_j \left[ -\omega^2 \left( M_{3j} + A_{3j} \right) + i\omega B_{3j} + C_{3j} \right] \Pi_j = \mathbf{X}_3 (\omega), \quad j=1,3,4 }[/math]

Roll

[math]\displaystyle{ \sum_j \left[ -\omega^2 \left( I_G + A_{4j} \right) + i\omega B_{4j} + C_{4j} \right] \Pi_j = \mathbf{X}_4 (\omega) }[/math]

The extension of these equations to six degrees of freedom is straightforward. However before discussing the general case we will study specific properties of the 2D Problem for the sake of clarity.

Symmetric body

Consider a body symmetric about the [math]\displaystyle{ X = 0\, }[/math] axis.

Symmetric body

For a body symmetric port/starboard:

  • Verify that Heave is decoupled from Surge and Roll. In other words the Surge and Roll motions do not influence Heave and vice versa:
[math]\displaystyle{ \left[ -\omega^2 \left( M + A_{33} \right) + i\omega B_{33} + C_{33} \right] \Pi_3 = A }[/math]
  • The only nonzero hydrostatic coefficients are [math]\displaystyle{ C_{33} \, }[/math] and [math]\displaystyle{ C_{44} \, }[/math]. Verify that this is the case even for non-symmetric sections.
  • Surge and Roll are coupled for symmetric and non-symmetric bodies. The coupled equation of motion becomes:

Surge-Roll

[math]\displaystyle{ \sum_{j=1,4} \left[ -\omega^2 \left( M_{ij} + A_{ij} \right) + i\omega B_{ij} + C_{ij} \right] \Pi_j = \mathbf{X}_i, \quad i,j = 1,4 }[/math]
  • When Newton's law is expressed about the center of gravity:
[math]\displaystyle{ M_{14} = M_{41} = 0, \ M_{11} = M, \ M_{44} = I_G }[/math]

where [math]\displaystyle{ I_G\, }[/math] is the body moment of inertia about the center of gravity. If the equations are to be expressed about the origin of the coordinate system, then the formulation must start with respect to [math]\displaystyle{ G\, }[/math] and expressions derived w.r.t. [math]\displaystyle{ O \, }[/math], via a coordinate transformation.

  • The exciting forces [math]\displaystyle{ \mathbf{X}_1, \mathbf{X}_3 \, }[/math] are defined in an obvious manner along the X- and Z-axis. The Roll moment [math]\displaystyle{ \mathbf{X}_4 \, }[/math] is defined initially about [math]\displaystyle{ G\, }[/math].

Need to derive definitions for the coefficients that enter the Heave & Surge-Roll equations of motion:

[math]\displaystyle{ M = \rho \forall, \qquad \forall = \ }[/math] volume of water displaced by body (archimedian principle of buoyancy)
[math]\displaystyle{ C_{33} = \rho g A_\omega = \rho g B, }[/math]
[math]\displaystyle{ A_\omega = }[/math] body waterplane area = [math]\displaystyle{ B }[/math] (Beam in two dimensions)
[math]\displaystyle{ \left( C_{44} \right)_G = \rho g \frac{B^3}{12} = \ }[/math] Roll restoring moment due to a small angular displacement
about the center of gravity. Verify for all wall and non-wall sided sections
[math]\displaystyle{ C_{44} \equiv \left( C_{44} \right)+G \ne \left( C_{44} \right)_O; \quad }[/math] Derive an expression for [math]\displaystyle{ \ \left( C_{44} \right)_O \ }[/math] in terms of [math]\displaystyle{ \ \left( C_{44} \right)_G. }[/math]

This article is based on the MIT open course notes and the original article can be found here

Ocean Wave Interaction with Ships and Offshore Energy Systems