Difference between revisions of "Long Wavelength Approximations"

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= Introduction =
+
{{Ocean Wave Interaction with Ships and Offshore Structures
 +
| chapter title = Long Wavelength Approximations
 +
| next chapter = [[Wave Scattering By A Vertical Circular Cylinder]]
 +
| previous chapter = [[Linear Wave-Body Interaction]]
 +
}}
 +
 
 +
{{incomplete pages}}
 +
 
 +
== Introduction ==
  
 
Very frequently the length of ambient waves <math> \lambda \,</math> is large compared to the dimension of floating bodies.
 
Very frequently the length of ambient waves <math> \lambda \,</math> is large compared to the dimension of floating bodies.
For example the length of a wave with period <math> T=10 \ \mbox{sec}\,</math> is <math> \lambda \simeq T^2 + \frac{T^2}{2} \simeq 150\mbox{m} \,</math>. The beam of a ship with length <math> L=100\mbox{m}\,</math> can be <math>20\mbox{m}\,</math> as is the case for the diameter of the leg of an offshore platform.
+
For example the length of a wave with period <math> T=10 \mbox{s}\,</math> is <math> \lambda \simeq T^2 + \frac{T^2}{2} \simeq 150\mbox{m} \,</math>. The beam of a ship with length <math> L=100\mbox{m}\,</math> can be <math>20\mbox{m}\,</math> as is the case for the diameter of the leg of an offshore platform.
 
 
  
= GI Taylor's formula =
+
== GI Taylor's formula ==
  
 
Consider a flow field given by
 
Consider a flow field given by
  
<math> U(X,t):\ \mbox{Velocity of ambient unidirectional flow} \,</math>
+
<math> U(x,t):\ \mbox{Velocity of ambient unidirectional flow} \,</math>
  
<math> P(X,t):\ \mbox{Pressure corresponding to} \ U(X,t) \,</math>
+
<math> P(x,t):\ \mbox{Pressure corresponding to} \ U(x,t) \,</math>
  
 
<center><math> \lambda \sim \frac{|U|}{|\nabla U|} \gg B \ = \ \mbox{Body characteristic dimension} \,</math></center>
 
<center><math> \lambda \sim \frac{|U|}{|\nabla U|} \gg B \ = \ \mbox{Body characteristic dimension} \,</math></center>
Line 17: Line 24:
 
In the absence of viscous effects and to leading order for <math>\lambda \gg B \,</math>:
 
In the absence of viscous effects and to leading order for <math>\lambda \gg B \,</math>:
  
<center><math> F_X = - \left( \forall + \frac{A_{11}}{\rho} \right) \left. \frac{\partial P}{\partial x} \right|_{X=0} </math></center>
+
<center><math> F_x = - \left( \forall + \frac{A_{11}}{\rho} \right) \left. \frac{\partial P}{\partial x} \right|_{x=0} </math></center>
 
where
 
where
<center><math>  \ F_X: \ \mbox{Force in X-direction} \,</math></center>
+
<center><math>  \ F_x: \ \mbox{Force in x-direction} \,</math></center>
  
 
<center><math>\ \forall: \ \mbox{Body displacement}\,</math></center>
 
<center><math>\ \forall: \ \mbox{Body displacement}\,</math></center>
Line 25: Line 32:
 
<center><math> \ A_{11}: \ \mbox{Surge added mass} \,</math></center>
 
<center><math> \ A_{11}: \ \mbox{Surge added mass} \,</math></center>
  
== Derivation using Euler's equations ==
+
=== Derivation using Euler's equations ===
  
 
An alternative form of GI Taylor's formula for a fixed body follows from Euler's equations:
 
An alternative form of GI Taylor's formula for a fixed body follows from Euler's equations:
  
<center><math> \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial X} \simeq - \frac{1}{\rho} \frac{\partial P}{\partial X} </math></center>
+
<center><math> \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \simeq - \frac{1}{\rho} \frac{\partial P}{\partial x} </math></center>
  
 
Thus:
 
Thus:
  
<center><math> F_X = \left( \rho \forall + A_{11} \right) + \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial X} \right)_{X=0} </math></center>
+
<center><math> F_x = \left( \rho \forall + A_{11} \right) + \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \right)_{x=0} </math></center>
  
If the body is also translating in the X-direction with displacement <math>X_1(t)\,</math> then the total force becomes
+
If the body is also translating in the x-direction with displacement <math>x_1(t)\,</math> then the total force becomes
  
<center><math> \bullet \ F_X = \left( \rho\forall+A_{11} \right) \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial X} \right) - A_{11} \frac{d^2X_1(t)}{dt^2} </math></center>
+
<center><math> \ F_x = \left( \rho\forall+A_{11} \right) \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \right) - A_{11} \frac{\mathrm{d}^2x_1(t)}{\mathrm{d}t^2} </math></center>
  
Often, when the ambient velocity <math> U\,</math> is arising from plane progressive waves, <math> \left| U \frac{\partial U}{\partial X} \right| = 0(A^2) \,</math> and is omitted. Note that <math> U\,</math> does not include disturbance effects due to the body.
+
Often, when the ambient velocity <math> U\,</math> is arising from plane progressive waves, <math> \left| U \frac{\partial U}{\partial x} \right| = 0(A^2) \,</math> and is omitted. Note that <math> U\,</math> does not include disturbance effects due to the body.
  
= Applications of GI Taylor's formula in wave-body interactions =
+
== Applications of GI Taylor's formula in wave-body interactions ==
  
== Archimedean hydrostatics ==
+
=== Archimedean hydrostatics ===
  
<center><math> P=-\rho g Z, \quad \frac{\partial P}{\partial Z} = - \rho g \,</math></center>
+
<center><math> P=-\rho g z, \quad \frac{\partial P}{\partial z} = - \rho g \,</math></center>
  
<center><math> F_Z = - ( \forall + \phi ) \frac{\partial P}{\partial Z} = \rho g \forall </math></center>
+
<center><math> F_z = - ( \forall + \phi ) \frac{\partial P}{\partial z} = \rho g \forall </math></center>
  
 
<center><math> \phi: \ \mbox{no added mass since there is no flow} </math></center>
 
<center><math> \phi: \ \mbox{no added mass since there is no flow} </math></center>
  
So Archimedes' formula is a special case of GI Taylor when there
+
So Archimedes' formula is a special case of GI Taylor when there is no flow. This offers an intuitive meaning to the term that includes the body displacement.
is no flow. This offers an intuitive meaning to the term that includes the body displacement.
 
  
== Regular waves over a circle fixed under the free surface ==
+
=== Regular waves over a circle fixed under the free surface ===
  
<center><math> \Phi_I = \mathfrak{Re} \left\{ \frac{i g A}{\omega} e^{KZ-iKX+i\omega t} \right\}, \quad K=\frac{\omega^2}{g} \, </math></center>
+
<center><math> \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx+i\omega t} \right\}, \quad k=\frac{\omega^2}{g} \, </math></center>
  
<center><math>u=\frac{\partial \Phi_I}{\partial X} = \mathfrak{Re} \left\{ \frac{i g A}{\omega} (-i K) e^{K Z - i K X + i \omega t } \right \} </math></center>
+
<center><math>u=\frac{\partial \Phi_I}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (-i k) e^{k z - i k x + i \omega t } \right \} </math></center>
  
<center><math> \mathfrak{Re} \left\{ \omega A e^{ - K d +i \omega t} \right\}_{X=0,Z=-d} </math></center>
+
<center><math> \mathrm{Re} \left\{ \omega A e^{ - k h +i \omega t} \right\}_{x=0,z=-h} </math></center>
  
 
So the horizontal force on the circle is:
 
So the horizontal force on the circle is:
  
<center><math>F_X = \left( \forall + \frac{a_{11}}{\rho} \right) \frac{\partial u}{\partial t} + O \left( Z^2 \right) </math></center>
+
<center><math>F_x = \left( \forall + \frac{A_{11}}{\rho} \right) \frac{\partial u}{\partial t} + O \left( z^2 \right) </math></center>
  
<center><math> \forall =\pi a^2, \quad a_{11} = \pi \rho a^2 \,</math></center>
+
<center><math> \forall =\pi a^2, \quad A_{11} = \pi \rho a^2 \,</math></center>
  
<center><math> \frac{\partial u}{\partial t} = \mathfrak{Re} \left\{ i\omega^2 e^{-K d + i \omega t} \right\} </math></center>
+
<center><math> \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ i\omega^2 e^{-kh + i \omega t} \right\} </math></center>
  
 
Thus:
 
Thus:
  
<center><math> F_X = - 2 \pi a^2 \omega^2 A e^{-K d} \sin \omega t \,</math></center>
+
<center><math> F_x = - 2 \pi a^2 \omega^2 A e^{-k h} \sin \omega t \,</math></center>
  
We can derive the vertical force along very similar lines. It is simply <math>90^\circ\,</math> out of phase relative to <math>F_X\,</math> with the same modulus.
+
We can derive the vertical force along very similar lines. It is simply <math>90^\circ\,</math> out of phase relative to <math>F_x\,</math> with the same modulus.
  
== Horizontal force on a fixed circular cylinder of draft <math>T\,</math> ==
+
=== Horizontal force on a fixed circular cylinder of draft <math>T\,</math> ===
  
 
This case arises frequently in wave interactions with floating offshore platforms.
 
This case arises frequently in wave interactions with floating offshore platforms.
Line 82: Line 88:
 
Here we will evaluate <math> \frac{\partial u}{\partial t} \,</math> on the axis of the platform and use a strip wise integration to evaluate the total hydrodynamic force.
 
Here we will evaluate <math> \frac{\partial u}{\partial t} \,</math> on the axis of the platform and use a strip wise integration to evaluate the total hydrodynamic force.
  
<center><math> u = \frac{\partial \Phi}{\partial X} = \mathfrak{Re} \left\{ \frac{i g A}{\omega} (- i K) e^{KZ-i K X + i\omega t} \right\} </math></center>
+
<center><math> u = \frac{\partial \Phi}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (- i k) e^{kz-i k x + i\omega t} \right\} </math></center>
  
<center><math> = \mathfrak{Re} \left\{ \omega A e^{KZ+i\omega t} \right\}_{X=0} \,</math></center>
+
<center><math> = \mathrm{Re} \left\{ \omega A e^{kz+i\omega t} \right\}_{x=0} \,</math></center>
  
<center><math> \frac{\partial u}{\partial t} (Z) = \mathfrak{Re} \left\{ \omega A ( i \omega) e^{KZ+i\omega t} \right\} </math></center>
+
<center><math> \frac{\partial u}{\partial t} (z) = \mathrm{Re} \left\{ \omega A ( i \omega) e^{kz+i\omega t} \right\} </math></center>
  
<center><math> = - \omega^2 A e^{KZ} \sin \omega t \,</math></center>
+
<center><math> = - \omega^2 A e^{kz} \sin \omega t \,</math></center>
  
The differential horizontal force over a strip <math> d Z \,</math> at a depth <math> Z \,</math> becomes:
+
The differential horizontal force over a strip <math> \mathrm{d} z \,</math> at a depth <math> z \,</math> becomes:
  
<center><math> dF_Z = \rho ( \forall + a_{11} ) \frac{\partial u}{\partial t} d Z \,</math></center>
+
<center><math> \mathrm{d}F_z = \rho ( \forall + A_{11} ) \frac{\partial u}{\partial t} \mathrm{d} z \,</math></center>
  
<center><math> \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} d Z \,</math></center>
+
<center><math> \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} \mathrm{d} z \,</math></center>
  
<center><math> 2 \pi \rho a^2 \left( - \omega^2 A e^{KZ} \right) \sin \omega t d Z </math></center>
+
<center><math> 2 \pi \rho a^2 \left( - \omega^2 A e^{kz} \right) \sin \omega t \mathrm{d} z </math></center>
  
 
The total horizontal force over a truncated cylinder of draft <math>T\,</math> becomes:
 
The total horizontal force over a truncated cylinder of draft <math>T\,</math> becomes:
  
<center><math> F_X = \int_{-T}^{0} dZ dF = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{KZ} dZ </math></center>
+
<center><math> F_x = \int_{-T}^{0} \mathrm{d}Z \mathrm{d}F = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{kz} \mathrm{d}z </math></center>
  
<center><math> X_1 \equiv F_X = - 2 \pi \rho a^2 \omega^2 A \sin \omega t \cdot \frac{1-e^{-KT}}{K} </math></center>
+
<center><math> X_1 \equiv F_x = - 2 \pi \rho a^2 \omega^2 A \sin \omega t \cdot \frac{1-e^{-kT}}{k} </math></center>
  
This is a very useful and practical result. It provides an
+
This is a very useful and practical result. It provides an estimate of the surge exciting force on one leg of a possibly multi-leg platform as <math> T \to \infty; \quad \frac{1-e^{-kT}}{k} \to \frac{1}{k}\,</math>
estimate of the surge exciting force on one leg of a possibly multi-leg platform
 
As <math> T \to \infty; \quad \frac{1-e^{-KT}}{K} \to \frac{1}{K}
 
\,</math>
 
  
== Horizontal force on multiple vertical cylinders in any arrangement: ==
+
=== Horizontal force on multiple vertical cylinders in any arrangement ===
  
The proof is essentially based on a phasing argument. Relative to the reference frame:
+
The proof is essentially based on a phasing argument. Relative to the reference frame,
  
<center><math> \Phi_I = \mathfrak{Re} \left\{ \frac{i g A}{\omega} e^{KZ-iKX + i\omega t} \right\} \,</math></center>
+
<center><math> \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx + i\omega t} \right\} \,</math></center>
  
* Express the incident wave relative to the local frames by
+
Expressing the incident wave relative to the local frames by introducing the phase factors,
introducing the phase factors:
 
  
<center><math> \mathbf{P}_i = e^{-iKX_i} </math></center>
+
<center><math> \mathbf{P}_i = e^{-ikx_i} </math></center>
  
Let:
+
and letting
  
<center><math> X+X_i + \xi_i \,</math></center>
+
<center><math> x = x_i + \xi_i \,</math></center>
  
Then relative to the i-th leg:
+
Then relative to the i-th leg,
  
<center><math> \Phi_I^{(i)} = \mathfrak{Re} \left\{ \frac{ i g A}{\omega} e^{KZ - iK\xi_i + i\omega t} \mathbf{P}_i \right\} \quad i=1,\cdots,N </math></center>
+
<center><math> \Phi_I^{(i)} = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{kz - ik\xi_i + i\omega t} \mathbf{P}_i \right\} \quad i=1,\cdots,N </math></center>
  
Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is:
+
Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is
  
 
<center><math> \mathbf{X}_1^N = \sum_{i=1}^N \mathbf{P}_i \mathbf{X}_1 \,</math></center>
 
<center><math> \mathbf{X}_1^N = \sum_{i=1}^N \mathbf{P}_i \mathbf{X}_1 \,</math></center>
Line 134: Line 136:
 
The above expression gives the complex amplitude of the force with <math>\mathbf{X}_1\,</math> given in the single cylinder case.
 
The above expression gives the complex amplitude of the force with <math>\mathbf{X}_1\,</math> given in the single cylinder case.
  
* The above technique may be easily extended to estimate the Sway force and Yaw moment on n-cylinders with little extra effort.
+
The above technique may be easily extended to estimate the Sway force and Yaw moment on n-cylinders with little extra effort.
  
== Surge exciting force on a 2D section ==
+
=== Surge exciting force on a 2D section ===
  
<center><math> \Phi_I = \mathfrak{Re} \left\{ \frac{ i g A}{\omega} e^{KZ-iKX+i\omega t} \right\} \,</math></center>
+
<center><math> \Phi_I = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{kz-ikx+i\omega t} \right\} \,</math></center>
  
<center><math> u=\mathfrak{Re} \left\{ \frac{ i g A}{\omega} (- i K ) e^{KZ-iKX+i\omega t} \right\} \,</math></center>
+
<center><math> u=\mathrm{Re} \left\{ \frac{ i g A}{\omega} (- i k ) e^{kz-ikx+i\omega t} \right\} \,</math></center>
  
<center><math> \frac{\partial u}{\partial t} = \mathfrak{Re} \left\{ \frac{ i g A}{\omega} \left(- i \frac{\omega^2}{g} \right) (i\omega) e^{i\omega t} \right\}_{X=0, Z=0} \,</math></center>
+
<center><math> \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ \frac{ i g A}{\omega} \left(- i \frac{\omega^2}{g} \right) (i\omega) e^{i\omega t} \right\}_{x=0, z=0} \,</math></center>
  
<center><math> = \mathfrak{Re} \left\{ i \omega^2 A e^{i\omega t} \right\} = -\omega^2 A \sin \omega t \,</math></center>
+
<center><math> = \mathrm{Re} \left\{ i \omega^2 A e^{i\omega t} \right\} = -\omega^2 A \sin \omega t \,</math></center>
  
 
<center><math> \mathbf{X}_1 = \left( \rho \forall + A_{11} \right) \frac{\partial u}{\partial t} = - \omega^2 A \sin \omega t ( \rho \forall + A_{11} ) \, </math></center>
 
<center><math> \mathbf{X}_1 = \left( \rho \forall + A_{11} \right) \frac{\partial u}{\partial t} = - \omega^2 A \sin \omega t ( \rho \forall + A_{11} ) \, </math></center>
  
* If the body section is a circle with radius <math> a\,</math>:
+
If the body section is a circle with radius <math> a\,</math>,
  
 
<center><math> \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,</math></center>
 
<center><math> \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,</math></center>
Line 154: Line 156:
 
So in long waves, the surge exciting force is equally divided between the Froude-Krylov and the diffraction components. This is not the case for Heave!
 
So in long waves, the surge exciting force is equally divided between the Froude-Krylov and the diffraction components. This is not the case for Heave!
  
== Heave exciting force on a surface piercing section ==
+
=== Heave exciting force on a surface piercing section ===
  
In long waves, the leading order effect in the exciting force is the hydrostatic contribution:
+
In long waves, the leading order effect in the exciting force is the hydrostatic contribution
  
 
<center><math>\mathbf{X}_i \sim \rho g A_w A \,</math></center>
 
<center><math>\mathbf{X}_i \sim \rho g A_w A \,</math></center>
  
where <math>A_w\,</math> is the body water plane area in 2D or 3D. <math>A\,</math> is the wave amplitude. This can be shown to be the leading order contribution from the Froude-Krylov force
+
where <math>A_w\,</math> is the body water plane area in 2D or 3D. <math>A\,</math> is the wave amplitude. This can be shown to be the leading order contribution from the Froude-Krylov force:
  
<center><math> \mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{KZ-iKX} n_3 dS \,</math></center>
+
<center><math> \mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{kz-ikx} n_3 \mathrm{d}S \,</math></center>
  
Using the Taylor series expansion:
+
Using the Taylor series expansion,
  
<center><math> e^{KZ-iKX} = 1 + ( KZ - iKX ) + O ( KB )^2 \,</math></center>
+
<center><math> e^{kz-ikx} = 1 + ( kz - ikx ) + O ( kB )^2 \,</math></center>
  
It is easy to verify that: <math>\mathbf{X}_3 \to \rho g A A_w \,</math>.
+
It is easy to verify that <math>\mathbf{X}_3 \to \rho g A A_w \,</math>.
  
The scattering contribution is of order <math> KB\,</math>. For submerged bodies: <math> \mathbf{X}_3^{FK}=O(KB)\,</math>.
+
The scattering contribution is of order <math> kB\,</math>. For submerged bodies, <math> \mathbf{X}_3^{FK}=O(kB)\,</math>.
  
 
-----
 
-----
  
 
This article is based on the MIT open course notes and the original article can be found
 
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-2002/0B7683D3-9B31-453E-B98F-9F71A3C36C58/0/lecture2.pdf here]
+
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/D28E25F7-9600-48C1-A444-D1F0FA4F0A41/0/lecture11.pdf here]
 +
 
 +
[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]
  
 
[[Category:Linear Water-Wave Theory]]
 
[[Category:Linear Water-Wave Theory]]
 
[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]
 

Latest revision as of 01:55, 12 February 2010

Wave and Wave Body Interactions
Current Chapter Long Wavelength Approximations
Next Chapter Wave Scattering By A Vertical Circular Cylinder
Previous Chapter Linear Wave-Body Interaction



Introduction

Very frequently the length of ambient waves [math]\displaystyle{ \lambda \, }[/math] is large compared to the dimension of floating bodies. For example the length of a wave with period [math]\displaystyle{ T=10 \mbox{s}\, }[/math] is [math]\displaystyle{ \lambda \simeq T^2 + \frac{T^2}{2} \simeq 150\mbox{m} \, }[/math]. The beam of a ship with length [math]\displaystyle{ L=100\mbox{m}\, }[/math] can be [math]\displaystyle{ 20\mbox{m}\, }[/math] as is the case for the diameter of the leg of an offshore platform.

GI Taylor's formula

Consider a flow field given by

[math]\displaystyle{ U(x,t):\ \mbox{Velocity of ambient unidirectional flow} \, }[/math]

[math]\displaystyle{ P(x,t):\ \mbox{Pressure corresponding to} \ U(x,t) \, }[/math]

[math]\displaystyle{ \lambda \sim \frac{|U|}{|\nabla U|} \gg B \ = \ \mbox{Body characteristic dimension} \, }[/math]

In the absence of viscous effects and to leading order for [math]\displaystyle{ \lambda \gg B \, }[/math]:

[math]\displaystyle{ F_x = - \left( \forall + \frac{A_{11}}{\rho} \right) \left. \frac{\partial P}{\partial x} \right|_{x=0} }[/math]

where

[math]\displaystyle{ \ F_x: \ \mbox{Force in x-direction} \, }[/math]
[math]\displaystyle{ \ \forall: \ \mbox{Body displacement}\, }[/math]
[math]\displaystyle{ \ A_{11}: \ \mbox{Surge added mass} \, }[/math]

Derivation using Euler's equations

An alternative form of GI Taylor's formula for a fixed body follows from Euler's equations:

[math]\displaystyle{ \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \simeq - \frac{1}{\rho} \frac{\partial P}{\partial x} }[/math]

Thus:

[math]\displaystyle{ F_x = \left( \rho \forall + A_{11} \right) + \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \right)_{x=0} }[/math]

If the body is also translating in the x-direction with displacement [math]\displaystyle{ x_1(t)\, }[/math] then the total force becomes

[math]\displaystyle{ \ F_x = \left( \rho\forall+A_{11} \right) \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \right) - A_{11} \frac{\mathrm{d}^2x_1(t)}{\mathrm{d}t^2} }[/math]

Often, when the ambient velocity [math]\displaystyle{ U\, }[/math] is arising from plane progressive waves, [math]\displaystyle{ \left| U \frac{\partial U}{\partial x} \right| = 0(A^2) \, }[/math] and is omitted. Note that [math]\displaystyle{ U\, }[/math] does not include disturbance effects due to the body.

Applications of GI Taylor's formula in wave-body interactions

Archimedean hydrostatics

[math]\displaystyle{ P=-\rho g z, \quad \frac{\partial P}{\partial z} = - \rho g \, }[/math]
[math]\displaystyle{ F_z = - ( \forall + \phi ) \frac{\partial P}{\partial z} = \rho g \forall }[/math]
[math]\displaystyle{ \phi: \ \mbox{no added mass since there is no flow} }[/math]

So Archimedes' formula is a special case of GI Taylor when there is no flow. This offers an intuitive meaning to the term that includes the body displacement.

Regular waves over a circle fixed under the free surface

[math]\displaystyle{ \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx+i\omega t} \right\}, \quad k=\frac{\omega^2}{g} \, }[/math]
[math]\displaystyle{ u=\frac{\partial \Phi_I}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (-i k) e^{k z - i k x + i \omega t } \right \} }[/math]
[math]\displaystyle{ \mathrm{Re} \left\{ \omega A e^{ - k h +i \omega t} \right\}_{x=0,z=-h} }[/math]

So the horizontal force on the circle is:

[math]\displaystyle{ F_x = \left( \forall + \frac{A_{11}}{\rho} \right) \frac{\partial u}{\partial t} + O \left( z^2 \right) }[/math]
[math]\displaystyle{ \forall =\pi a^2, \quad A_{11} = \pi \rho a^2 \, }[/math]
[math]\displaystyle{ \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ i\omega^2 e^{-kh + i \omega t} \right\} }[/math]

Thus:

[math]\displaystyle{ F_x = - 2 \pi a^2 \omega^2 A e^{-k h} \sin \omega t \, }[/math]

We can derive the vertical force along very similar lines. It is simply [math]\displaystyle{ 90^\circ\, }[/math] out of phase relative to [math]\displaystyle{ F_x\, }[/math] with the same modulus.

Horizontal force on a fixed circular cylinder of draft [math]\displaystyle{ T\, }[/math]

This case arises frequently in wave interactions with floating offshore platforms.

Here we will evaluate [math]\displaystyle{ \frac{\partial u}{\partial t} \, }[/math] on the axis of the platform and use a strip wise integration to evaluate the total hydrodynamic force.

[math]\displaystyle{ u = \frac{\partial \Phi}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (- i k) e^{kz-i k x + i\omega t} \right\} }[/math]
[math]\displaystyle{ = \mathrm{Re} \left\{ \omega A e^{kz+i\omega t} \right\}_{x=0} \, }[/math]
[math]\displaystyle{ \frac{\partial u}{\partial t} (z) = \mathrm{Re} \left\{ \omega A ( i \omega) e^{kz+i\omega t} \right\} }[/math]
[math]\displaystyle{ = - \omega^2 A e^{kz} \sin \omega t \, }[/math]

The differential horizontal force over a strip [math]\displaystyle{ \mathrm{d} z \, }[/math] at a depth [math]\displaystyle{ z \, }[/math] becomes:

[math]\displaystyle{ \mathrm{d}F_z = \rho ( \forall + A_{11} ) \frac{\partial u}{\partial t} \mathrm{d} z \, }[/math]
[math]\displaystyle{ \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} \mathrm{d} z \, }[/math]
[math]\displaystyle{ 2 \pi \rho a^2 \left( - \omega^2 A e^{kz} \right) \sin \omega t \mathrm{d} z }[/math]

The total horizontal force over a truncated cylinder of draft [math]\displaystyle{ T\, }[/math] becomes:

[math]\displaystyle{ F_x = \int_{-T}^{0} \mathrm{d}Z \mathrm{d}F = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{kz} \mathrm{d}z }[/math]
[math]\displaystyle{ X_1 \equiv F_x = - 2 \pi \rho a^2 \omega^2 A \sin \omega t \cdot \frac{1-e^{-kT}}{k} }[/math]

This is a very useful and practical result. It provides an estimate of the surge exciting force on one leg of a possibly multi-leg platform as [math]\displaystyle{ T \to \infty; \quad \frac{1-e^{-kT}}{k} \to \frac{1}{k}\, }[/math]

Horizontal force on multiple vertical cylinders in any arrangement

The proof is essentially based on a phasing argument. Relative to the reference frame,

[math]\displaystyle{ \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx + i\omega t} \right\} \, }[/math]

Expressing the incident wave relative to the local frames by introducing the phase factors,

[math]\displaystyle{ \mathbf{P}_i = e^{-ikx_i} }[/math]

and letting

[math]\displaystyle{ x = x_i + \xi_i \, }[/math]

Then relative to the i-th leg,

[math]\displaystyle{ \Phi_I^{(i)} = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{kz - ik\xi_i + i\omega t} \mathbf{P}_i \right\} \quad i=1,\cdots,N }[/math]

Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is

[math]\displaystyle{ \mathbf{X}_1^N = \sum_{i=1}^N \mathbf{P}_i \mathbf{X}_1 \, }[/math]

The above expression gives the complex amplitude of the force with [math]\displaystyle{ \mathbf{X}_1\, }[/math] given in the single cylinder case.

The above technique may be easily extended to estimate the Sway force and Yaw moment on n-cylinders with little extra effort.

Surge exciting force on a 2D section

[math]\displaystyle{ \Phi_I = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{kz-ikx+i\omega t} \right\} \, }[/math]
[math]\displaystyle{ u=\mathrm{Re} \left\{ \frac{ i g A}{\omega} (- i k ) e^{kz-ikx+i\omega t} \right\} \, }[/math]
[math]\displaystyle{ \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ \frac{ i g A}{\omega} \left(- i \frac{\omega^2}{g} \right) (i\omega) e^{i\omega t} \right\}_{x=0, z=0} \, }[/math]
[math]\displaystyle{ = \mathrm{Re} \left\{ i \omega^2 A e^{i\omega t} \right\} = -\omega^2 A \sin \omega t \, }[/math]
[math]\displaystyle{ \mathbf{X}_1 = \left( \rho \forall + A_{11} \right) \frac{\partial u}{\partial t} = - \omega^2 A \sin \omega t ( \rho \forall + A_{11} ) \, }[/math]

If the body section is a circle with radius [math]\displaystyle{ a\, }[/math],

[math]\displaystyle{ \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \, }[/math]

So in long waves, the surge exciting force is equally divided between the Froude-Krylov and the diffraction components. This is not the case for Heave!

Heave exciting force on a surface piercing section

In long waves, the leading order effect in the exciting force is the hydrostatic contribution

[math]\displaystyle{ \mathbf{X}_i \sim \rho g A_w A \, }[/math]

where [math]\displaystyle{ A_w\, }[/math] is the body water plane area in 2D or 3D. [math]\displaystyle{ A\, }[/math] is the wave amplitude. This can be shown to be the leading order contribution from the Froude-Krylov force:

[math]\displaystyle{ \mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{kz-ikx} n_3 \mathrm{d}S \, }[/math]

Using the Taylor series expansion,

[math]\displaystyle{ e^{kz-ikx} = 1 + ( kz - ikx ) + O ( kB )^2 \, }[/math]

It is easy to verify that [math]\displaystyle{ \mathbf{X}_3 \to \rho g A A_w \, }[/math].

The scattering contribution is of order [math]\displaystyle{ kB\, }[/math]. For submerged bodies, [math]\displaystyle{ \mathbf{X}_3^{FK}=O(kB)\, }[/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