Long Wavelength Approximations

Analytical solutions of the wave-body problem formulated above are rare. The few exceptions which find frequent use in practice are:

• Wavemaker theory (Studied)

• Diffraction by a vertical circular cylinder (Studied below)

• Long-wavelength approximations (Studied next)

Long-wavelength approximations

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{sec}\, }[/math] is [math]\displaystyle{ \lambda \simeq T^2 + \frac{T^2}{2} \simeq 150m \, }[/math]. The beam of a ship with length [math]\displaystyle{ L=100m\, }[/math] can be [math]\displaystyle{ 20m\, }[/math] as is the case for the diameter of the leg of an offshore platform.

GI Taylor's formula

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

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{ \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]

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

A) 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.

B) Regular waves over a circle fixed under the free surface

[math]\displaystyle{ \Phi_I = \mathfrak{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} = \mathfrak{Re} \left\{ \frac{i g A}{\omega} (-i K) e^{K Z - i K X + i \omega t } \right \} }[/math] [math]\displaystyle{ \mathfrak{Re} \left\{ \omega A e^{ - K d +i \omega t} \right\}_{X=0,Z=-d} }[/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} = \mathfrak{Re} \left\{ i\omega^2 e^{-K d + i \omega t} \right\} }[/math]

Thus:

[math]\displaystyle{ F_X = - 2 \pi a^2 \omega^2 A e^{-K d} \sin \omega t \, }[/math]

• 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.

C) 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} = \mathfrak{Re} \left\{ \frac{i g A}{\omega} (- i K) e^{KZ-i K X + i\omega t} \right\} }[/math] [math]\displaystyle{ = \mathfrak{Re} \left\{ \omega A e^{KZ+i\omega t} \right\}_{X=0} \, }[/math] [math]\displaystyle{ \frac{\partial u}{\partial t} (Z) = \mathfrak{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{ d Z \, }[/math] at a depth [math]\displaystyle{ Z \, }[/math] becomes:

[math]\displaystyle{ dF_Z = \rho ( \forall + a_{11} ) \frac{\partial u}{\partial t} d Z \, }[/math] [math]\displaystyle{ \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} d Z \, }[/math] [math]\displaystyle{ 2 \pi \rho a^2 \left( - \omega^2 A e^{KZ} \right) \sin \omega t 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} dZ dF = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{KZ} dZ }[/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]

D) 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 = \mathfrak{Re} \left\{ \frac{i g A}{\omega} e^{KZ-iKX + i\omega t} \right\} \, }[/math]

• Express the incident wave relative to the local frames by

introducing the phase factors:

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

Let:

[math]\displaystyle{ X+X_i + \xi_i \, }[/math]

Then relative to the i-th leg:

[math]\displaystyle{ \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]

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.

E) Surge exciting force on a 2D section

[math]\displaystyle{ \Phi_I = \mathfrac{Re} \left\{ \frac{ i g A}{\omega} e^{KZ-iKX+i\omega t} \right\} \, }[/math] [math]\displaystyle{ u=\mathfrak{Re} \left\{ \frac{ i g A}{\omega} (- i K ) e^{KZ-iKX+i\omega t} \right\} \, }[/math] [math]\displaystyle{ \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] [math]\displaystyle{ = \mathfrak{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!

F) 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 dS \,\lt math\gt \lt /center\gt Using the Taylor series expansion: \lt center\gt \lt math\gt 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].