Wave Scattering By A Vertical Circular Cylinder - Revision history

Javier Arcos: /* Problem */

2010-09-09T13:34:22Z

Problem

← Older revision		Revision as of 13:34, 9 September 2010
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	x=R\cos\theta \\		x=R\cos\theta \\
	y=R\sin\theta		y=R\sin\theta
	\end{Bmatrix} j \quad \psi(R,\theta)		\end{Bmatrix} ; \quad \psi(R,\theta)
	</math></center>		</math></center>

Adi Kurniawan: /* Solution */

2009-12-08T17:07:39Z

Solution

← Older revision		Revision as of 17:07, 8 December 2009
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	In the limit as <math> h \to \infty\,, \quad \frac{\cosh k (z+h)}{k h} \to e^{k z} \,</math> and the series expansion solution survives.		In the limit as <math> h \to \infty\,, \quad \frac{\cosh k (z+h)}{k h} \to e^{k z} \,</math> and the series expansion solution survives.

	The total complex potential, incident and scattered, was derived above. The hydrodynamic pressure follows from Bernoulli:		The total complex potential, incident and scattered, was derived above.

			The hydrodynamic pressure follows from Bernoulli:

	<center><math> P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,</math></center>		<center><math> P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,</math></center>

Adi Kurniawan: /* Solution */

2009-12-08T17:07:04Z

Solution

← Older revision		Revision as of 17:07, 8 December 2009
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	<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh k (z+h)}{\cosh k h } (\psi+x) (r,\theta) </math></center>		<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh k (z+h)}{\cosh k h } (\psi+x) (r,\theta) </math></center>

	In the limit as <math> h \to \infty \quad \frac{\cosh k (z+h)}{k h} \~~longrightarrow~~ e^{k z} \,</math> and the series expansion solution survives.		In the limit as <math> h \to \infty\,, \quad \frac{\cosh k (z+h)}{k h} \to e^{k z} \,</math> and the series expansion solution survives.

	The total complex potential, incident and scattered, was derived above. The hydrodynamic pressure follows from Bernoulli:		The total complex potential, incident and scattered, was derived above. The hydrodynamic pressure follows from Bernoulli:

Adi Kurniawan: /* Problem */

2009-12-08T17:05:02Z

Problem

← Older revision		Revision as of 17:05, 8 December 2009
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	or		or

	<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-ikx} \right) = -\frac{\partial}{\partial R} \left( e^{-~~ikE~~\cos R} \right) </math></center>		<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-ikx} \right) = -\frac{\partial}{\partial R} \left( e^{-ikR\cos\theta} \right) </math></center>

	Here we make use of the familiar identity:		Here we make use of the familiar identity:

Adi Kurniawan at 17:01, 8 December 2009

2009-12-08T17:01:24Z

← Older revision		Revision as of 17:01, 8 December 2009
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	{{incomplete pages}}		{{incomplete pages}}

	This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength <math>\lambda\,</math>. This was shown to be the case by [[McCamy and Fuchs 1954]] using separation of variables		== Introduction ==
			This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength <math>\lambda\,</math>. This was shown to be the case by [[McCamy and Fuchs 1954]] using separation of variables.

			== Problem ==

			The incident potential is given as

	<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>		<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>
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	\end{Bmatrix} </math></center>		\end{Bmatrix} </math></center>

			== Solution ==
	Try:		Try:

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	In the limit as <math> h \to \infty \quad \frac{\cosh k (z+h)}{k h} \longrightarrow e^{k z} \,</math> and the series expansion solution survives.		In the limit as <math> h \to \infty \quad \frac{\cosh k (z+h)}{k h} \longrightarrow e^{k z} \,</math> and the series expansion solution survives.

	The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:		The total complex potential, incident and scattered, was derived above. The hydrodynamic pressure follows from Bernoulli:

	<center><math> P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,</math></center>		<center><math> P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,</math></center>
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	<center><math> \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, </math></center>		<center><math> \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, </math></center>

			== Surge exciting force ==
	The surge exciting force is given by		The surge exciting force is given by

Adi Kurniawan at 16:56, 8 December 2009

2009-12-08T16:56:51Z

Adi Kurniawan at 16:47, 8 December 2009

2009-12-08T16:47:41Z

← Older revision		Revision as of 16:47, 8 December 2009
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	In the limit as <math> h \to \infty \quad \frac{\cosh K (z+h)}{K h} \longrightarrow e^{K z} \,</math> and the series expansion solution survives.		In the limit as <math> h \to \infty \quad \frac{\cosh K (z+h)}{K h} \longrightarrow e^{K z} \,</math> and the series expansion solution survives.

	~~== Surge exciting force ==~~

	The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:		The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:

Adi Kurniawan at 16:47, 8 December 2009

2009-12-08T16:47:06Z

← Older revision		Revision as of 16:47, 8 December 2009
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	{{incomplete pages}}		{{incomplete pages}}

	~~== McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder ==~~

	This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength <math>\lambda\,</math>. This was shown to be the case by [[McCamy and Fuchs 1954]] using separation of variables		This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength <math>\lambda\,</math>. This was shown to be the case by [[McCamy and Fuchs 1954]] using separation of variables

Meylan at 23:57, 16 October 2009

2009-10-16T23:57:31Z

← Older revision		Revision as of 23:57, 16 October 2009
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	\| previous chapter = [[Long Wavelength Approximations]]		\| previous chapter = [[Long Wavelength Approximations]]
	}}		}}

			{{incomplete pages}}

	== McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder ==		== McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder ==

Adi Kurniawan: /* McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder */

2009-05-27T03:56:05Z

McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder

← Older revision		Revision as of 16:56, 8 December 2009
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	<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>		<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>

	<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh K(z+h)}{\cosh K h} e^{-~~iKx~~} </math></center>		<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cosh k h} e^{-ikx} </math></center>

	Let the diffraction potential be		Let the diffraction potential be

	<center><math> \phi_7 = \frac{i g A}{\omega} \frac{\cosh K(z+h)}{\cos K h} \psi(x,y) </math></center>		<center><math> \phi_7 = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cos k h} \psi(x,y) </math></center>

	For <math>\phi_7\,</math> to satisfy the 3D Laplace equation, it is easy to show that <math>\psi\,</math> must satisfy the Helmholtz equation:		For <math>\phi_7\,</math> to satisfy the 3D Laplace equation, it is easy to show that <math>\psi\,</math> must satisfy the Helmholtz equation:

	<center><math> \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, </math></center>		<center><math> \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + k^2 \right) \psi = 0\, </math></center>

	In polar coordinates:		In polar coordinates:
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	The Helmholtz equation takes the form:		The Helmholtz equation takes the form:

	<center><math> \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} + \frac{1}{R^2} \frac{\partial^2}{\partial\theta^2} + K^2 \right) \psi = 0 \, </math></center>		<center><math> \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} + \frac{1}{R^2} \frac{\partial^2}{\partial\theta^2} + k^2 \right) \psi = 0 \, </math></center>

	On the cylinder:		On the cylinder:
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	or		or

	<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-~~iKx~~} \right) = -\frac{\partial}{\partial R} \left( e^{-~~iKE~~\cos R} \right) </math></center>		<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-ikx} \right) = -\frac{\partial}{\partial R} \left( e^{-ikE\cos R} \right) </math></center>

	Here we make use of the familiar identity:		Here we make use of the familiar identity:

	<center><math> e^{-~~iKR~~\cos\theta} = \sum_{m=0}^{\infty} \epsilon_m J_m ( K R ) \cos m \theta </math></center>		<center><math> e^{-ikR\cos\theta} = \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \cos m \theta </math></center>

	<center><math> \epsilon_m = \begin{Bmatrix}		<center><math> \epsilon_m = \begin{Bmatrix}
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	Try:		Try:

	<center><math> \psi(R,\theta) = \sum_{m=0}^{\infty} A_m F_m ( K R ) \cos m \theta \, </math></center>		<center><math> \psi(R,\theta) = \sum_{m=0}^{\infty} A_m F_m ( k R ) \cos m \theta \, </math></center>

	Upon substitution in Helmholtz's equation we obtain:		Upon substitution in Helmholtz's equation we obtain:

	<center><math> \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} - \frac{m^2}{R^2} + K^2 \right) F_m ( K R ) = 0 </math></center>		<center><math> \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} - \frac{m^2}{R^2} + k^2 \right) F_m ( k R ) = 0 </math></center>

	This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions		This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions

	<center><math> \begin{Bmatrix}		<center><math> \begin{Bmatrix}
	J_m ( K R ) \\		J_m ( k R ) \\
	Y_m ( K R )		Y_m ( k R )
	\end{Bmatrix} </math></center>		\end{Bmatrix} </math></center>

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	As <math> R \to \infty\,</math>:		As <math> R \to \infty\,</math>:

	<center><math> \psi(R,\theta) \sim e^{-~~iKR~~ + i\omega t} \,</math></center>		<center><math> \psi(R,\theta) \sim e^{-ikR + i\omega t} \,</math></center>

	Also as <math> R \to \infty\, </math>:		Also as <math> R \to \infty\, </math>:

	<center><math> J_m ( K R ) \sim \left( \frac{2}{\pi K R} \right)^{1/2} \cos \left( K R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) </math></center>		<center><math> J_m ( k R ) \sim \left( \frac{2}{\pi k R} \right)^{1/2} \cos \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) </math></center>

	<center><math> Y_m ( K R ) \sim \left( \frac{2}{\pi K R} \right)^{1/2} \sin \left( K R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) </math></center>		<center><math> Y_m ( k R ) \sim \left( \frac{2}{\pi k R} \right)^{1/2} \sin \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) </math></center>

	Hence the Hankel function:		Hence the Hankel function:

	<center><math> H_m^{(2)} ( K R ) = J_m ( K R ) - i Y_m ( K R ) \,</math></center>		<center><math> H_m^{(2)} ( k R ) = J_m ( k R ) - i Y_m ( k R ) \,</math></center>

	<center><math> \sim \left( \frac{2}{\pi K R} \right)^{1/2} e^{-i \left( K R - \frac{1}{2} m \pi - \frac{\pi}{4} \right)} </math></center>		<center><math> \sim \left( \frac{2}{\pi k R} \right)^{1/2} e^{-i \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right)} </math></center>

	Satisfies the far field condition required by <math> \psi(R,\theta) \,</math>. So we set:		Satisfies the far field condition required by <math> \psi(R,\theta) \,</math>. So we set:

	<center><math> \psi(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m A_m H_m^{(2)} ( K R ) \cos m \theta </math></center>		<center><math> \psi(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m A_m H_m^{(2)} ( k R ) \cos m \theta </math></center>

	with the constants <math> A_m \,</math> to be determined. The cylinder condition requires:		with the constants <math> A_m \,</math> to be determined. The cylinder condition requires:

	<center><math> \left. \frac{\partial\psi}{\partial R} \right\|_{R=a} = - \frac{\partial}{\partial R} \sum_{m=0}^{\infty} \epsilon_m J_m ( K R ) \left.\cos m \theta \right\|_{r=a} </math></center>		<center><math> \left. \frac{\partial\psi}{\partial R} \right\|_{R=a} = - \frac{\partial}{\partial R} \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \left.\cos m \theta \right\|_{r=a} </math></center>

	It follows that:		It follows that:

	<center><math> A_m {H_m^{(2)}}^' (K a) = - J_m^' (K a) \,</math></center>		<center><math> A_m {H_m^{(2)}}^' (k a) = - J_m^' (k a) \,</math></center>

	or:		or:

	<center><math> A_m = - \frac{J_m^' ( K a ) }{{H_m^{(2)}}^' (K a)} \,</math></center>		<center><math> A_m = - \frac{J_m^' ( k a ) }{{H_m^{(2)}}^' (k a)} \,</math></center>

	where <math> (')\,</math> denotes derivatives with respect to the argument. The solution for the total velocity potential follows in the form		where <math> (')\,</math> denotes derivatives with respect to the argument. The solution for the total velocity potential follows in the form

	<center><math> (\psi+x)(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m \left[ J_m (K R) - \frac{J_m^'(K a)}{{H_m^{(2)}}^'(K a)} H_m^{(2)} (K a) \right] \cos m \theta </math></center>		<center><math> (\psi+x)(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m \left[ J_m (k R) - \frac{J_m^'(k a)}{{H_m^{(2)}}^'(k a)} H_m^{(2)} (k a) \right] \cos m \theta </math></center>

	And the total original potential follows:		And the total original potential follows:

	<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh K (z+h)}{\cosh K h } (\psi+x) (r,\theta) </math></center>		<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh k (z+h)}{\cosh k h } (\psi+x) (r,\theta) </math></center>

	In the limit as <math> h \to \infty \quad \frac{\cosh K (z+h)}{K h} \longrightarrow e^{K z} \,</math> and the series expansion solution survives.		In the limit as <math> h \to \infty \quad \frac{\cosh k (z+h)}{k h} \longrightarrow e^{k z} \,</math> and the series expansion solution survives.

	The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:		The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:
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	<center><math> X_1 = \iint_{S_B} P n_1 \mathrm{d}S = \mathrm{Re} \left\{ \mathbf{X}_1 e^{i\omega t} \right\} </math></center>		<center><math> X_1 = \iint_{S_B} P n_1 \mathrm{d}S = \mathrm{Re} \left\{ \mathbf{X}_1 e^{i\omega t} \right\} </math></center>

	<center><math> \mathbf{X}_1 = \rho \int_{-\infty}^0 \mathrm{d}Z \int_0^{2\pi} a \mathrm{d}\theta \left( - i \omega \frac{i g A}{\omega} \right) e^{~~K Z~~} n_1 (\psi + x)_{R=a} </math></center>		<center><math> \mathbf{X}_1 = \rho \int_{-\infty}^0 \mathrm{d}z \int_0^{2\pi} a \mathrm{d}\theta \left( - i \omega \frac{i g A}{\omega} \right) e^{k z} n_1 (\psi + x)_{R=a} </math></center>

	Simple algebra in this case of water of infinite depth leads to the expression.		Simple algebra in this case of water of infinite depth leads to the expression.

← Older revision		Revision as of 03:56, 27 May 2009
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	<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>		<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>

	<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cosh K H} e^{-~~iKX~~} </math></center>		<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh K(z+h)}{\cosh K h} e^{-iKx} </math></center>

	Let the diffraction potential be		Let the diffraction potential be

	<center><math> \phi_7 = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cos K H} \psi(X,Y) </math></center>		<center><math> \phi_7 = \frac{i g A}{\omega} \frac{\cosh K(z+h)}{\cos K h} \psi(x,y) </math></center>

	For <math>\phi_7\,</math> to satisfy the 3D Laplace equation, it is easy to show that <math>\psi\,</math> must satisfy the Helmholtz equation:		For <math>\phi_7\,</math> to satisfy the 3D Laplace equation, it is easy to show that <math>\psi\,</math> must satisfy the Helmholtz equation:

	<center><math> \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, </math></center>		<center><math> \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, </math></center>

	In polar coordinates:		In polar coordinates:
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	or		or

	<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-~~iKX~~} \right) = -\frac{\partial}{\partial R} \left( e^{-iKE\cos R} \right) </math></center>		<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-iKx} \right) = -\frac{\partial}{\partial R} \left( e^{-iKE\cos R} \right) </math></center>

	Here we make use of the familiar identity:		Here we make use of the familiar identity:
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	And the total original potential follows:		And the total original potential follows:

	<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh K (Z+H)}{\cosh K H } (\psi+x) (r,\theta) </math></center>		<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh K (z+h)}{\cosh K h } (\psi+x) (r,\theta) </math></center>

	In the limit as <math> H \to \infty \quad \frac{\cosh K (Z+H)}{K H} \longrightarrow e^{K Z} \,</math> and the series expansion solution survives.		In the limit as <math> h \to \infty \quad \frac{\cosh K (z+h)}{K h} \longrightarrow e^{K z} \,</math> and the series expansion solution survives.

	== Surge exciting force ==		== Surge exciting force ==