# Long Wavelength Approximations

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

## GI Taylor's formula

Consider a flow field given by

$U(x,t):\ \mbox{Velocity of ambient unidirectional flow} \,$

$P(x,t):\ \mbox{Pressure corresponding to} \ U(x,t) \,$

$\lambda \sim \frac{|U|}{|\nabla U|} \gg B \ = \ \mbox{Body characteristic dimension} \,$

In the absence of viscous effects and to leading order for $\lambda \gg B \,$:

$F_x = - \left( \forall + \frac{A_{11}}{\rho} \right) \left. \frac{\partial P}{\partial x} \right|_{x=0}$

where

$\ F_x: \ \mbox{Force in x-direction} \,$
$\ \forall: \ \mbox{Body displacement}\,$
$\ A_{11}: \ \mbox{Surge added mass} \,$

### Derivation using Euler's equations

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

$\frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \simeq - \frac{1}{\rho} \frac{\partial P}{\partial x}$

Thus:

$F_x = \left( \rho \forall + A_{11} \right) + \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x} \right)_{x=0}$

If the body is also translating in the x-direction with displacement $x_1(t)\,$ then the total force becomes

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

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

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

### Archimedean hydrostatics

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

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

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

So the horizontal force on the circle is:

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

Thus:

$F_x = - 2 \pi a^2 \omega^2 A e^{-k h} \sin \omega t \,$

We can derive the vertical force along very similar lines. It is simply $90^\circ\,$ out of phase relative to $F_x\,$ with the same modulus.

### Horizontal force on a fixed circular cylinder of draft $T\,$

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

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

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

The differential horizontal force over a strip $\mathrm{d} z \,$ at a depth $z \,$ becomes:

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

The total horizontal force over a truncated cylinder of draft $T\,$ becomes:

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

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 $T \to \infty; \quad \frac{1-e^{-kT}}{k} \to \frac{1}{k}\,$

### Horizontal force on multiple vertical cylinders in any arrangement

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

$\Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx + i\omega t} \right\} \,$

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

$\mathbf{P}_i = e^{-ikx_i}$

and letting

$x = x_i + \xi_i \,$

Then relative to the i-th leg,

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

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

$\mathbf{X}_1^N = \sum_{i=1}^N \mathbf{P}_i \mathbf{X}_1 \,$

The above expression gives the complex amplitude of the force with $\mathbf{X}_1\,$ 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

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

If the body section is a circle with radius $a\,$,

$\rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,$

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

$\mathbf{X}_i \sim \rho g A_w A \,$

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

$\mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{kz-ikx} n_3 \mathrm{d}S \,$

Using the Taylor series expansion,

$e^{kz-ikx} = 1 + ( kz - ikx ) + O ( kB )^2 \,$

It is easy to verify that $\mathbf{X}_3 \to \rho g A A_w \,$.

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