Wave Momentum Flux - Revision history

Xjwiki: /* Simplification */

2019-01-02T08:26:40Z

Simplification

← Older revision		Revision as of 08:26, 2 January 2019
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	We use the following vector theorem		We use the following vector theorem
	<center><math> \nabla ( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>		<center><math> \nabla \cdot( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>
	If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and for solenoidal flow (<math>\nabla \cdot \mathbf{v}=0 </math>), we can use Gauss's vector theorem:		If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and for solenoidal flow (<math>\nabla \cdot \mathbf{v}=0 </math>), we can use Gauss's vector theorem:
	<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}		<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}

Xjwiki: /* Momentum flux in potential flow */ \iiint_{\partial\Omega} \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \mathbf{n} \mathrm{d}S \nequiv \iint_{\partial\Omega} \frac{\partial\Phi}{\partial n} \na

2019-01-02T08:23:40Z

Momentum flux in potential flow: \iiint_{\partial\Omega} \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \mathbf{n} \mathrm{d}S \nequiv \iint_{\partial\Omega} \frac{\partial\Phi}{\partial n} \na

← Older revision		Revision as of 08:23, 2 January 2019
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	<center><math> \iint_{\partial\Omega(t)} ( \mathbf{v} \cdot \mathbf{n} ) \mathbf{v} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S		<center><math> \iint_{\partial\Omega(t)} ( \mathbf{v} \cdot \mathbf{n} ) \mathbf{v} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S
	= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>		= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>
	~~since for <math> \nabla^2 \Phi = 0 \, </math>;~~
	<center><math> \iiint_{\partial\Omega} \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \mathbf{n} \mathrm{d}S \equiv \iint_{\partial\Omega} \frac{\partial\Phi}{\partial n} \nabla\Phi \mathrm{d}S. </math></center>
	In the above <math> \mathbf{n} \, </math> is the unit vector pointing out of the volume <math> \Omega(t) </math> and <math> V_n = \mathbf{n} \cdot \nabla \Phi </math>		In the above <math> \mathbf{n} \, </math> is the unit vector pointing out of the volume <math> \Omega(t) </math> and <math> V_n = \mathbf{n} \cdot \nabla \Phi </math>

Xjwiki: /* Simplification */

2018-12-31T12:26:08Z

Simplification

← Older revision		Revision as of 12:26, 31 December 2018
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	We use the following vector theorem		We use the following vector theorem
	<center><math> \nabla ( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>		<center><math> \nabla ( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>
	If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and for solenoidal flow (<math>\nabla \cdot \mathbf{v} </math>), we can use Gauss's vector theorem:		If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and for solenoidal flow (<math>\nabla \cdot \mathbf{v}=0 </math>), we can use Gauss's vector theorem:
	<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}		<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}
	\left(\mathbf{v} \cdot \mathbf{n} \right) \mathbf{v} \mathrm{d}S </math></center>		\left(\mathbf{v} \cdot \mathbf{n} \right) \mathbf{v} \mathrm{d}S </math></center>

Xjwiki: /* Simplification */

2018-12-31T12:25:42Z

Simplification

← Older revision		Revision as of 12:25, 31 December 2018
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	We use the following vector theorem		We use the following vector theorem
	<center><math> \nabla ( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>		<center><math> \nabla ( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>
	If we have potential ~~and solenoidal~~ flow then <math> \mathbf{v} = \nabla \Phi \,</math> and we can use Gauss's vector theorem:		If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and for solenoidal flow (<math>\nabla \cdot \mathbf{v} </math>), we can use Gauss's vector theorem:
	<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}		<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}
	\left(\mathbf{v} \cdot \mathbf{n} \right) \mathbf{v} \mathrm{d}S </math></center>		\left(\mathbf{v} \cdot \mathbf{n} \right) \mathbf{v} \mathrm{d}S </math></center>

Xjwiki: /* Simplification */ Correcting the formulas

2018-12-30T21:18:09Z

Simplification: Correcting the formulas

← Older revision		Revision as of 21:18, 30 December 2018
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	We use the following vector theorem		We use the following vector theorem
	<center><math> (\mathbf{v} ~~\cdot \nabla )~~ \mathbf{v} = ~~\nabla~~ ( ~~\frac{1}{2}~~ \mathbf{v} \cdot \mathbf{v} ~~) -~~ \mathbf{v} ~~\times~~ (\nabla \~~times~~ \mathbf{v} ) </math></center>		<center><math> \nabla ( \mathbf{v} \mathbf{v} )=(\mathbf{v} \cdot \nabla ) \mathbf{v}+ \mathbf{v} (\nabla \cdot \mathbf{v} ) </math></center>
	If we have potential flow then <math> \mathbf{v} = \nabla \Phi \,</math> and we can use Gauss's vector theorem:		If we have potential and solenoidal flow then <math> \mathbf{v} = \nabla \Phi \,</math> and we can use Gauss's vector theorem:
	<center><math> \iiint_{\Omega(t)} \nabla ( ~~\frac{1}{2}~~ \mathbf{v} ~~\cdot~~ \mathbf{v} ) \mathrm{d}V = ~~\frac{1}{2}~~ \iint_{\partial\Omega(t)}		<center><math> \iiint_{\Omega(t)} \nabla \cdot( \mathbf{v} \mathbf{v} ) \mathrm{d}V = \iint_{\partial\Omega(t)}
	\left(\mathbf{v} \cdot \mathbf{v} \right) \mathbf{n} \mathrm{d}S </math></center>		\left(\mathbf{v} \cdot \mathbf{n} \right) \mathbf{v} \mathrm{d}S </math></center>
	In potential flow it follows that		In potential flow it follows that
	<center><math> \iint_{\partial\Omega(t)~~} \frac{1}{2~~} ( \mathbf{v} \cdot \mathbf{v} ) \mathbf{n} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S		<center><math> \iint_{\partial\Omega(t)} ( \mathbf{v} \cdot \mathbf{n} ) \mathbf{v} \mathrm{d}S = \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi \mathrm{d}S
	= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>		= \iint_{\partial\Omega(t)} V_n \mathbf{v} \mathrm{d}S </math></center>
	since for <math> \nabla^2 \Phi = 0 \, </math>;		since for <math> \nabla^2 \Phi = 0 \, </math>;

Adi Kurniawan: /* Mean horizontal momentum flux due to a Plane Progressive Regular Wave */

2010-11-06T11:35:15Z

Mean horizontal momentum flux due to a Plane Progressive Regular Wave

← Older revision		Revision as of 11:35, 6 November 2010
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	<center><math> P = - \rho \frac{\partial\Phi}{\partial t} - \frac{1}{2} \rho ( u^2 + v^2 ) - \rho g z </math></center>		<center><math> P = - \rho \frac{\partial\Phi}{\partial t} - \frac{1}{2} \rho ( u^2 + v^2 ) - \rho g z </math></center>
	so that the momentum flux is		so that the momentum flux is
	<center><math> \frac{dM_x}{dt} = \rho \int_{-h}^{\zeta} \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} ( v^2 - u^2 ) + g z \right] \mathrm{d}z = \rho \left( \int_{-h}^{0} + \int_{0}^{\zeta} \right) \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} (v^2 - u^2) + gz \right] \mathrm{d}z </math></center>		<center><math>
			\begin{align}
			\frac{dM_x}{dt} &= \rho \int_{-h}^{\zeta} \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} ( v^2 - u^2 ) + g z \right] \mathrm{d}z \\
			&= \rho \left( \int_{-h}^{0} + \int_{0}^{\zeta} \right) \left[ \frac{\partial\Phi}{\partial t} + \frac{1}{2} (v^2 - u^2) + gz \right] \mathrm{d}z
			\end{align}
			</math></center>

	Taking mean values in time and keeping terms of <math> O(A^2) </math> we obtain:		Taking mean values in time and keeping terms of <math> O(A^2) </math> we obtain:

Adi Kurniawan: /* Hydrostatic Term */

2010-11-06T11:33:46Z

Hydrostatic Term

← Older revision		Revision as of 11:33, 6 November 2010
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	Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux\|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:		Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux\|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:
	<center><math> \frac{d \mathcal{M}_{H,S}}{dt} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} \mathrm{d}S = - \rho g V_{\~~mbox~~{Fluid}} \mathbf{k} </math></center>		<center><math> \frac{\mathrm{d} \mathcal{M}_{H,S}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} \mathrm{d}S = - \rho g V_{\text{Fluid}} \mathbf{k} </math></center>
	where <math>V_{\mbox{Fluid}} </math> is the fluid volume.		where <math>V_{\mbox{Fluid}} </math> is the fluid volume.
	This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.		This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.

Adi Kurniawan: /* Hydrostatic Term */

2010-11-06T11:31:55Z

Hydrostatic Term

Meylan: /* Circular object in two dimensions */

2010-04-03T21:55:43Z

Circular object in two dimensions

← Older revision		Revision as of 21:55, 3 April 2010
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	What is the mean horizontal force on the wavemaker? From the momentum conservation theorem the mean horizontal flux of momentum to the left must flow into the wavemaker. This mean flux translates into a mean horizontal force in the same direction, as shown in the figure. Not an easy conclusion without using some basic fluid mechanics!		What is the mean horizontal force on the wavemaker? From the momentum conservation theorem the mean horizontal flux of momentum to the left must flow into the wavemaker. This mean flux translates into a mean horizontal force in the same direction, as shown in the figure. Not an easy conclusion without using some basic fluid mechanics!

	~~=== Circular object in two dimensions ===~~

	~~[[Image:Rotating_circle.jpg\|thumb\|right\|600px\|Rotating circle]]~~

	Consider a circular object in two dimensions rotating at an angular frequency <math>\omega\,</math> under a free surface and with a small amplitude. In other words the circle center rotates as a fluid particle in a regular wave.

	It can be shown that according to linear theory this rotational motion creates regular waves propagating to the right with no waves propagating to the left! What is the mean horizontal force acting on the circle? From the momentum conservation principle the mean force <math> \overline{F}\, </math> acting on the circle must be equal to <math> \overline{L} \, </math>, thus:

	~~<center><math> \overline{F} = - \frac{1}{4} \rho g A^2 </math></center>.~~

	~~[[Image:Dean_circle.jpg\|thumb\|right\|600px\|Dean circle]]~~

	~~[[Dean 1953]] has shown that regular plane progressive waves in deep water propagating over a circle fixed under the free surface are entirely transmitted with a phase shift.~~
	In other words the wave reflection is zero. Note this is the only body in 2 dimensions with this property. Reflection is non-zero in finite depth for all known bodies. We can show by applying the momentum conservation principle that the mean horizontal force <math> \overline{F} = 0 </math>

Meylan: /* Wave maker */

2010-04-03T21:55:13Z

Wave maker

← Older revision		Revision as of 21:55, 3 April 2010
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	So the mean horizontal momentum flux due to a plane progressive wave against its direction of propagation and equal to <math> - \frac{1}{4} \rho g A^2 </math>.		So the mean horizontal momentum flux due to a plane progressive wave against its direction of propagation and equal to <math> - \frac{1}{4} \rho g A^2 </math>.

	=== [[~~Wave maker~~]] ===		=== [[Wavemaker Theory]] ===

	[[Image:Wave_maker_momentum.png\|600px\|right\|thumb\|Wavemaker momentum]]		[[Image:Wave_maker_momentum.png\|600px\|right\|thumb\|Wavemaker momentum]]

← Older revision		Revision as of 11:31, 6 November 2010
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	Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux\|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:		Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial. Note that unlike the [[Wave Energy Density and Flux\|energy conservation principle]], the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component. The integral of the hydrostatic term over the remaining surfaces leads to:
	<center><math> \frac{d \mathcal{M}_{H,S}}{dt} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} dS = - \rho g V_{\mbox{Fluid}} \mathbf{k} </math></center>		<center><math> \frac{d \mathcal{M}_{H,S}}{dt} = - \rho \iint_{\partial\Omega_F+\partial\Omega^+ + \partial\Omega^- +\partial\Omega_{\infty}} gz \mathbf{n} \mathrm{d}S = - \rho g V_{\mbox{Fluid}} \mathbf{k} </math></center>
	where <math>V_{\mbox{Fluid}} </math> is the fluid volume.		where <math>V_{\mbox{Fluid}} </math> is the fluid volume.
	This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{d\mathcal{M}_H}{dt} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.		This is simply the static weight of the volume of fluid bounded by <math> \partial\Omega_F, \partial\Omega^+, \partial\Omega^- \,</math> and <math> \partial\Omega_{\infty}.</math> With no waves present, this is simply the weight of the ocean water "column" bounded by <math> \partial\Omega\, </math> which does not concern us here. This weight does not change in principle when waves are present at least when <math> \partial\Omega^+, \partial\Omega^- \,</math> are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "rational" to apply the momentum conservation theorem over the "linearized" volume <math> V_L(t) \, </math> which is perfectly possible within the framework derived above. In this case <math> \frac{\mathrm{d}\mathcal{M}_H}{\mathrm{d}t} \,</math> is exactly equal to the static weight of the water column and can be ignored in the wave-body interaction problem.

	On <math> S_F; P=P_a=0 \, </math> and hence all terms within the free-surface integral and over <math> S_{\infty} \, </math> (seafloor) can be neglected. It follows that:		On <math> S_F; P=P_a=0 \, </math> and hence all terms within the free-surface integral and over <math> S_{\infty} \, </math> (seafloor) can be neglected. It follows that:
	<center><math> \frac{d \mathcal{M}}{dt} = - \rho \iint_{\partial\Omega^\pm +\partial\Omega_B} \left[ \frac{P}{\rho} \mathbf{n} + \mathbf{v} (V_n -U_n) \right] dS </math></center>		<center><math> \frac{\mathrm{d} \mathcal{M}}{\mathrm{d}t} = - \rho \iint_{\partial\Omega^\pm +\partial\Omega_B} \left[ \frac{P}{\rho} \mathbf{n} + \mathbf{v} (V_n -U_n) \right] \mathrm{d}S </math></center>

	Note that the free surface integrals also vanish for the horizontal component since the hydrostatic force is always vertical. This momentum flux formula is of central importance in wave-body interactions and has many important applications, some of which are discussed bellow. Note that the mathematical derivations involved in its proof apply equally when the volume <math> \Omega </math> and its enclosed surface are selected to be at their linearized positions. In such a case it is essential to set <math> U_n=0\,</math> and <math> V_n \ne 0 </math>. Let the math take over and suggest the proper expression for the force. In the fully nonlinear case, <math>U_n \ne 0 \, </math> on <math> \partial\Omega_F\, </math> and <math> P=0 \,</math> on <math> \partial\Omega_F \, </math>!		Note that the free surface integrals also vanish for the horizontal component since the hydrostatic force is always vertical. This momentum flux formula is of central importance in wave-body interactions and has many important applications, some of which are discussed bellow. Note that the mathematical derivations involved in its proof apply equally when the volume <math> \Omega </math> and its enclosed surface are selected to be at their linearized positions. In such a case it is essential to set <math> U_n=0\,</math> and <math> V_n \ne 0 </math>. Let the math take over and suggest the proper expression for the force. In the fully nonlinear case, <math>U_n \ne 0 \, </math> on <math> \partial\Omega_F\, </math> and <math> P=0 \,</math> on <math> \partial\Omega_F \, </math>!
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	<center><math> \overrightarrow{F}_B = \iint_{S_B} P \vec{n} dS </math></center>		<center><math> \overrightarrow{F}_B = \iint_{S_B} P \vec{n} dS </math></center>
	With <math> \vec{n} \,</math> pointing inside the body. We may therefore recast the momentum conservation theorem in the form:		With <math> \vec{n} \,</math> pointing inside the body. We may therefore recast the momentum conservation theorem in the form:
	<center><math> \overrightarrow{F}_B (t) = - \frac{d \overrightarrow{M}}{dt} - \rho \iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n) \right] dS </math></center>		<center><math> \overrightarrow{F}_B (t) = - \frac{\mathrm{d} \overrightarrow{M}}{\mathrm{d}t} - \rho \iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n) \right] \mathrm{d}S </math></center>
	Where <math> S^\pm \, </math> are fluid boundaries at some distance from the body. If the volume of fluid surrounded by the body, free surface and the furfaces <math> S^\pm \, </math> does not grow in time, then the momentum of the enclosed fluid cannot grow either, so <math> \overrightarrow{M}(t) \, </math> is a stationary physical quantity. It is a well known result that the mean value in time of the time derivative of a stationary quantity is zero. So:		Where <math> S^\pm \, </math> are fluid boundaries at some distance from the body. If the volume of fluid surrounded by the body, free surface and the furfaces <math> S^\pm \, </math> does not grow in time, then the momentum of the enclosed fluid cannot grow either, so <math> \overrightarrow{M}(t) \, </math> is a stationary physical quantity. It is a well known result that the mean value in time of the time derivative of a stationary quantity is zero. So:
	<center><math> {\overline{\frac{d\overrightarrow{M}}{dt}}}^t = 0 </math></center>		<center><math> {\overline{\frac{\mathrm{d}\overrightarrow{M}}{\mathrm{d}t}}}^t = 0 </math></center>

	<u> Proof </u>:		<u> Proof </u>:

	<center><math> {\overline{\frac{dF}{dt}}}^t = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \frac{dF{\tau}}{d\tau} d\tau = \lim_{T\to\infty} \frac{1}{2T} [F(T) - F(-T) ] </math></center>		<center><math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} \frac{\mathrm{d}F{\tau}}{\mathrm{d}\tau} \mathrm{d}\tau = \lim_{T\to\infty} \frac{1}{2T} [F(T) - F(-T) ] </math></center>

	Since <math> F(\pm \tau) \, </math> must be bounded for a stationary signal <math> F(t) \, </math>, it follows that <math> {\overline{\frac{dF}{dt}}}^t = 0 \, </math>.		Since <math> F(\pm \tau) \, </math> must be bounded for a stationary signal <math> F(t) \, </math>, it follows that <math> {\overline{\frac{\mathrm{d}F}{\mathrm{d}t}}}^t = 0 \, </math>.

	Taking mean values, it follows that:		Taking mean values, it follows that:

	<center><math> {\overline{\overrightarrow{F}_B (t)}}^t = - \rho \ {\overline{\iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n ) \right] dS}}^t </math></center>		<center><math> {\overline{\overrightarrow{F}_B (t)}}^t = - \rho \ {\overline{\iint_{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n ) \right] \mathrm{d}S}}^t </math></center>

	This is the fundamental formula underlying the definition of the mean wave drift forces acting on floating bodies. Such forces are very imprtant for stationary floating structures and can be expressed in terms of integrals of wave effects over control surfaces <math> S^\pm \, </math> which may be located at infinity.		This is the fundamental formula underlying the definition of the mean wave drift forces acting on floating bodies. Such forces are very imprtant for stationary floating structures and can be expressed in terms of integrals of wave effects over control surfaces <math> S^\pm \, </math> which may be located at infinity.