Difference between revisions of "Long Wavelength Approximations"

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

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