# Too shallow for linear wave theory?

Discussion in 'Boat Design' started by floating, Nov 4, 2011.

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### floatingJunior Member

A couple of references have mentioned that Airy or linear wave theory doesn't apply in 'very shallow' water depth, but they don't say how to calculate the minimum depth in which linear wave theory applies. What is the minimum depth?
And if I'm trying to apply linear wave theory to water that is too shallow, is there a problem with:
1) Using a wave spectrum
2) Estimating the vertical water particle velocity
What wave behavior is most important in making linear wave theory inapplicable? Steepened crests/flat troughs, horizontal >> vertical velocity, wave breaking, or other?
Sorry for the buckshot question but I am trying to get my footing...feel free to point me to a good reference.

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### DCockeySenior Member

My recollection is the wave height needs to be sufficiently small compared to both the wave length and the water depth for the linear theory to be applicable. So there is no set minimum depth.

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### Tim BSenior Member

There are oft-qouted values for what constitutes deep water, based primarily on wavelength, but usually when the wave-height is relatively small. There is a maximum wave slope, which is well documented, and limits the waveheight for a given wave-length before breaking.

Essentially, the effect of the bottom is to slow down the orbital velocity at the bottom of the orbit. This makes the top of the wave "ride over" the bottom, leading to steepening and breaking. Thankfully there are not many vessels that have to deal with surf. The idea of direct superposition of waves (i.e. LWT) does not apply, not because you can't produce the correct shape (you can), but because it doesn't model the underlying physics, particularly the inviscid assumption.

If you are really interested in very shallow water, then you'll have to start looking at CFD methods, to convert a specific deep water sea-state into a suitable shallow water state. This will have significant dependance on the geometry you assume, but should not be too hard to do in 2D.

What is your interest in shallow water? do you have a particular question that needs it?

Tim B.

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### floatingJunior Member

Thanks for the help. It's the velocity squared in viscous drag that you can't represent by superposition, right?
The practical motivation behind my post is that we are using a model based on linear wave theory to look at how a floating structure interacts with waves, and I want to know whether the results apply if the structure is moved to shallower water. If wave height H=2m and T=8s, you'd have to go to a water depth of H=5m for linear theory to be invalid, based on the attached diagram that I copied from Jacobs (2010). That is shallower than we would moor.

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### DCockeySenior Member

The wave height to depth ratio for which linear theory is limited in shallow water is not due to viscous effects. Rather it is because of the assumptions made in linearizing the free surface boundary condition and applying it at the mean surface.

Superpostion works as long as the linearizing assumptions are met. Superposition stops working when the wave motion is not linear due to the wave height being too large compared to water depth or wave length, not because of viscous effects.

My understanding is that waves break as the water shoals because of inviscid effects, not because the bottom slows the water through viscous drag.

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### DCockeySenior Member

In the chart which float posted above, the Stokes 2nd order, Stokes 4th order, 5th order stream function, and cnoidal theory are all examples of non-linear invisicd theories.

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If i understand your question correctly, you're referring to when a wave is affected by the depth?

Lamb's linear wave theory of the celerity = sqr.rt(g.L/2.pi) is the gravity wave.

In shallow water this expression reduces to simply = sqr.tr (g.h)

In other words the waves are dependent upon the depth h.

More noted here: http://www.boatdesign.net/forums/boat-design/wave-pattern-question-28667.html#post290938

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### DCockeySenior Member

Airy theory which the original poster refered to is the version of wave theory with the linearized boundary condition imposed on the mean surface and a constant depth with zero normal velocity on the bottom.

(gh)^0.5 is the limiting celerity for shallow water with the wave length much longer than the water depth. The celerity and other formulas for finite depth are give in this Wikipedia article.

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Airy theory, or Linear theory, is what the poster asked. Lamb's linear theory is the most widely used; and leads to the Froude depth used by naval architects.

Both end up with the same result, Celerity = sqr.rt(gh)

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If the structure is "not moving" ie self propelled like a vessel, then it matters not. What does matter is that your waves characteristics have changed from deep water to shallow water.

You need to establish how a wave 'X' in deep water transforms into another wave 'Y' in shallow water. That resulting profile wave will affect the floating structure and it subsequent motions differently between wave X and wave Y. Most notably the wave period. But that depends upon the RAO, if you have done one, of your floating structure.

Since the natural roll/heave/pitch periods of your structure may become too close to the period of the wave, from deep to shallow water.

If this is your motivation, then a simple sensitivity analysis of different wave periods and heights and their responses on your structure shall yield the result you are after.

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### floatingJunior Member

The wave conditions that I am considering fall into the Stokes 2nd order nonlinear regime in the chart posted earlier. What are the pitfalls of using linear theory?

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### jehardimanSenior Member

Stokes wave theory assumes a small, but finite, wave steepness which does not account for orbital changes due to water depth or limiting wave height. So Stokes linear or 2nd order will not work for storm conditions, shoaling, or limiting cases for inertial load analysis, see De's 1955 work "Contributions to the Theory of Stokes Waves". But a better thing to do would be to get and read a copy of Oceanographical Engineering by Robert Wiegel. The applicable part is Chapter 2 which covers all wave theory in all regimes, which will help you decide if your analysis is within acceptable uncertainty.

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### DCockeySenior Member

Stokes 2nd order theory includes finite water depth if the finite depth version is used. Ocean Engineering Wave Mechanics, Michael E. McCormick, pp 46-47. "The second-order theory yields reasonably good results when the depth-to-wavelength ratio is greater than 1/10, which is the practical range for most engineering applications." It also predicts limiting wave height. Particle paths / orbits can be calculated and will have a net convections.

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### HawkboatJunior Member

Research Ursell number, see if that gives you the guidance you seek. Non-linear Dean-stream wave models seem to be the Cadillac of wave models.

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### DCockeySenior Member

If "superposition" is valid then velocities from superpositon should also be valid, not velocity squared. Determine the magnitude of the superposition of the velocities and square it to get the velocity squared.

If superposition of the waves isn't sufficiently valid then using wave spectras, etc as inputs to a dynamic model will be questionable at best, and a different method, perhaps Monte Carlo simulation or similar should be used. That is a major complication and would require a field-type of solver for the water motion. On the other hand given the other approximations you are probably using superposition should be reasonable.

Last edited: Jan 9, 2012
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