breakwater scaling laws
Patri Friedman
Jeff Chan
1. Would it be anchored or tethered? If so then the cost of
anchoring or tethering may be enormous. Check out the costs of
tethering floating platforms. The costs are in the hundreds of
millions of dollars for the tethering system.
2. If not tethered then it would rise and fall with large waves,
i.e., only small waves would be attenuated; stuff in the middle
would still rise and fall on the larger wavelength waves.
Think of a lifering floating in waves. The small waves smaller
than the ring are attenuated in the middle (i.e., the water is
calm), but the whole ring and enclosed water rise and fall
together on larger waves.
3. Energy applied may be larger with a larger breakwater.
Think of a breakwater like an antenna. The larger the antenna,
the longer the wavelengths it can couple with efficiently. And
it still gets all of the higher frequency waves too.
4. If the breakwater must be rigid then the strength requirements
would seem to go up at least as a cube of the major dimension
since it's a three dimensional object responding to energies in
three dimensions. Waves move in all three directions.
5. If the breakwater can be flexible then perhaps the
requirements are different.
Jeff C.
Patri Friedman
Regarding floating breakwaters, some assumptions to check:
1. Would it be anchored or tethered? If so then the cost of
anchoring or tethering may be enormous. Check out the costs of
tethering floating platforms. The costs are in the hundreds of
millions of dollars for the tethering system.
2. If not tethered then it would rise and fall with large waves,
i.e., only small waves would be attenuated; stuff in the middle
would still rise and fall on the larger wavelength waves.
Think of a lifering floating in waves. The small waves smaller
than the ring are attenuated in the middle (i.e., the water is
calm), but the whole ring and enclosed water rise and fall
together on larger waves.
3. Energy applied may be larger with a larger breakwater.
Think of a breakwater like an antenna. The larger the antenna,
the longer the wavelengths it can couple with efficiently. And
it still gets all of the higher frequency waves too.
4. If the breakwater must be rigid then the strength requirements
would seem to go up at least as a cube of the major dimension
since it's a three dimensional object responding to energies in
three dimensions. Waves move in all three directions.
5. If the breakwater can be flexible then perhaps the
requirements are different.
Jeff C.
--
Patri Friedman, Executive Director
The Seasteading Institute
www.seasteading.org
Wayne C. Gramlich
I originally sent the message below to the tsi-engineering-advisors
list on Monday. Alas, it was probably bounced as spam. Hence,
I am retrying. I'm CC'ing Patri this time, so that if he gets it,
but the list doesn't, he will know that there is still a problem
with me posting to the list.
-Wayne
Wayne C. Gramlich wrote:
> All:
>
> It does not appear that I am actually on
> tsi-enginee...@googlegroups.com. So unless,
> people are CC'ing me, I am missing your message traffic.
>
> ------
>
> I believe the most likely structure for a large ocean
> going breakwater will be circular with a radius of R
> and fairly deep depth D (>100 meters) to block out the
> effects of large waves. I suspect that D may have to
> be significantly larger.
>
> The breakwater will have a thickness of T. Thus, the
> total volume of material is the circumference (2*pi*R)
> times the thickness (T) times the depth (D). The breakwater
> cost should be proportional to the total amount of material
> required:
>
> Cost = 2*pi*R*T*D
>
> For me, the question is how T scales as a function of R. I
> do not think D has to change much as the breakwater gets larger,
> just T.
>
> Materials have a maximum amount of compression before they
> fail. Place a beam across two columns of a material (concrete,
> steel, whatever.) There is a maximum weight W at which the
> column material will suffer from a compressive failure. If I
> want to carry double the weight (2W), I have to double the area
> of the two columns.
>
> Now take our circular breakwater and make it square. If we
> align the breakwater such that one surface is facing the
> direction that the waves are coming from, that surface will
> be taking all of the wave force. That force is being propagated
> down the sides of the breakwater. The cross sectional area
> is T*D on each side. We would select T to wide enough to handle
> maximum expected wave forces.
>
> If R is doubled, it picks up twice the wave forces along the
> surface facing the waves. If T*D is not doubled as well, the
> strength of the material will be exceeded and it will fail.
> If we double T, we are back to being able to withstand the
> force. Thus, T scales (at least) linearly with R:
>
> T = k*R,
>
> where k depends upon the material and maximum expected wave
> force.
>
> When we substitute back into the cost equation:
>
> Cost = 2*pi*R*(k*R)*D = 2*pi*k*D*R*R = (2*pi*k*D) * R^2 = O(R^2)
>
> In fact, I think it is actually worse than R^2 because I think
> the beam taking the force needs to grow faster than R^2. Alas,
> I have reached the limit of my understanding of material strength
> scaling, so I will stop at O(R^2).
>
> My ultimate conclusion is that a free standing breakwater will be
> big and expensive; there is no economy of scale. If you double R,
> you quadruple your area and you quadruple your costs (probably more
> than quadruple costs.)
>
> Respectfully submitted,
>
> -Wayne
>
James Hogan
>> My ultimate conclusion is that a free standing breakwater will be
>> big and expensive; there is no economy of scale. If you double R,
>> you quadruple your area and you quadruple your costs (probably more
>> than quadruple costs.)
If the cost of dealing waves in a non-breakwater-sheltered seastead
city is also O(R^2), it seems like there might still be advantages to
building a breakwater. The cost of dealing with waves gets
transferred from the seastead platforms to the breakwater, potentially
freeing the platforms from issues like: destructive platform
collisions during large waves, resident seasickness, general resident
safety, physical stability for various activities, etc.
James
--
James Hogan, Director of Operations
The Seasteading Institute
www.seasteading.org
Patri Friedman
Wayne C. Gramlich
Patri Friedman wrote:
> It made it to the list.
>
> I don't understand why force needs to be transmitted down those arms.
> If instead of a square breakwater, we used a single wall, wouldn't it
> would work just as well for waves from one direction? Thus, no force
> need be transmitted along the arms to the back wall. What resists the
> waves is the force of the ocean holding the breakwater in place (the
> breakwater projects down into the non-moving part of the ocean to resist
> surge).
Waves do not come from just one direction.
> A span requires increased material strength, but to me the breakwater is
> analogous to a wall, not a span. It is supported, and has resistance,
> along its entire length. Your analysis would apply if the breakwater
> was supported / had resistance to movement only along the back wall, in
> which case it would be like a span. But I do not believe that to be the
> case, because, as I mentioned, it could just be a single wall. And of
> course, a wall can be arbitrarily large.
What happens when wave come from two different directions?
Two walls? Three directions? Three walls. The only way
to be sure is to enclose the space. Also, depending upon
wave length, waves will diffract around the edges.
> It seems like this issue of whether it is more like a wall or a span is key.
>
> This is for a free floating breakwater. I can see how if the breakwater
> was anchored, there is a sense in which there is a span between each
> mooring point, so as the breakwater grows we have to add mooring points.
> But even there, equidistant mooring points on a circle only grow
> linearly with R.
I'll have to think about this one. Deep ocean mooring is pretty pricey.
Shallow ocean mooring really limits the locations of the breakwater and
puts us on a path of governments claiming all shallow water locations.
Lastly, your assertion that the wall cost is O(R) is open to question.
There will bending moments from unequal forces applied to different
ends of the wall. I think it will scale more like a beam span.
My knowledge of structures is weak, but big structures tend not
to scale linearly. I'll let somebody with more knowledge on this
topic try and figure out what the rules are.
[snippage]
-Wayne
Patri Friedman
Waves do not come from just one direction.
Patri Friedman wrote:
> It made it to the list.
>
> I don't understand why force needs to be transmitted down those arms.
> If instead of a square breakwater, we used a single wall, wouldn't it
> would work just as well for waves from one direction? Thus, no force
> need be transmitted along the arms to the back wall. What resists the
> waves is the force of the ocean holding the breakwater in place (the
> breakwater projects down into the non-moving part of the ocean to resist
> surge).
What happens when wave come from two different directions?
> A span requires increased material strength, but to me the breakwater is
> analogous to a wall, not a span. It is supported, and has resistance,
> along its entire length. Your analysis would apply if the breakwater
> was supported / had resistance to movement only along the back wall, in
> which case it would be like a span. But I do not believe that to be the
> case, because, as I mentioned, it could just be a single wall. And of
> course, a wall can be arbitrarily large.
Two walls? Three directions? Three walls. The only way
to be sure is to enclose the space. Also, depending upon
wave length, waves will diffract around the edges.
I'll have to think about this one. Deep ocean mooring is pretty pricey.
> It seems like this issue of whether it is more like a wall or a span is key.
>
> This is for a free floating breakwater. I can see how if the breakwater
> was anchored, there is a sense in which there is a span between each
> mooring point, so as the breakwater grows we have to add mooring points.
> But even there, equidistant mooring points on a circle only grow
> linearly with R.
Shallow ocean mooring really limits the locations of the breakwater and
puts us on a path of governments claiming all shallow water locations.
Lastly, your assertion that the wall cost is O(R) is open to question.
There will bending moments from unequal forces applied to different
ends of the wall. I think it will scale more like a beam span.
My knowledge of structures is weak, but big structures tend not
to scale linearly. I'll let somebody with more knowledge on this
topic try and figure out what the rules are.
[snippage]
-Wayne
e.hoog...@student.utwente.nl
My take:
The case Wayne sketches is valid for a more or less static load. If we regard one end of the wavebreak as fixed, and as a matter of sheer inertia, it seems a fair approximation, then relatively persistent high water on the other side indeed dictates O(R^2) material use.
Yet waves are far from static. For wavelengths small relative to the diameter of the wavebreak, no such situation will occur. Most wave-energy in deep water in a bad storm travels between 150-600m wavelength. A wavebreak is supposedly big compared to 150m, but that doesn’t mean there isn’t any action at 600m and beyond.
Persistent/low frequency pressure to one side will deform the wavebreak (or destroy it: make your choice), and will thus cause a wave to partially propagate inwards. In general, you can not hope to keep all waves out, but you should aim to have the wavebreak act as a low-pass filter, smoothing the water surface.
The flexible connections between segments Patri mentioned would accomplish just that: not because they can counter said force better than one monolithic block, but because they would bend rather than break: in the long run / time averaged, the force is zero anyway, so their static load resistance is fairly irrelevant, as long as they keep the shape more or less circular over time.
Getting more quantitative is difficult: I doubt you could say much more without a computer simulation. I could do some calculations, but the approximations involved would make it rather meaningless I fear. The simplest model I can imagine would be a linear 1-d wave meeting a mass-spring interface. It would be a nice demonstration of the principle of a low-pass filter, but little more than that.
The required depth seems also hard to quantify, but I think the 100m estimate might be excessive. It should be proportional to the wavelength targeted, as penetration depth of waves is proportional to wavelength. But the actual moving of water that makes up a wave falls of exponentially with depth*, so most of the ‘waving’ is going on near the surface. Adding more depth should have strongly declining marginal returns.
But as far as computer simulations go, it doesn’t seem terribly complicated. A company like MI&T, who supposedly have the software and the people familiar with it, should be able to do it rather painlessly, a few days or so. But that has a lot of caveats as well: if their code does not support flexible joints, for instance, you might be looking at months of development.
*for a linear wave in ‘deep water’, the velocity field scales with depth as exp(-depth*2*pi/wavelength)
Gary Root
On the general issue of wave breaks. I have been interested reading the various comments on this issue. I know I have been invited into the discussion about this specific issue, but I am interested in understanding more global issues before I discuss this specific problem. There are actually a large number of available solutions to the breakwater issue. During the process of development of ocean based commerce, most of these solutions will be tried and I believe that all of them will, through practical implementation, be adopted to serve in the applications where they have the greatest utility at lowest cost. The solution I prefer for breakwaters is a hybrid solution -- a combination of a small low density breakwater and biomass. I believe biomass and aquaculture are the keys to sustainable ocean development. We need a wide variety of food and fiber crops to plant around our ocean communities. The aquacultural biomass should be the primary breakwater that we utilize. These ocean growing crops will have to be genetically engineered and bred to provide for nutritional, clothing, energy, and construction resource needs. I do not know what other projects you are pursuing but if you have an aquaculture project, it could potentially serve as a key component in creating sustainable breakwaters as well.
Thanks,
Gary Root
Wayne C. Gramlich
e.hoog...@student.utwente.nl wrote:
> My take:
>
> The case Wayne sketches is valid for a more or less static load. If we
> regard one end of the wavebreak as fixed, and as a matter of sheer
> inertia, it seems a fair approximation, then relatively persistent high
> water on the other side indeed dictates O(R^2) material use.
>
> Yet waves are far from static. For wavelengths small relative to the
> diameter of the wavebreak, no such situation will occur. Most
> wave-energy in deep water in a bad storm travels between 150-600m
> wavelength. A wavebreak is supposedly big compared to 150m, but that
> doesn’t mean there isn’t any action at 600m and beyond.
>
> Persistent/low frequency pressure to one side will deform the wavebreak
> (or destroy it: make your choice), and will thus cause a wave to
> partially propagate inwards. In general, you can not hope to keep all
> waves out, but you should aim to have the wavebreak act as a low-pass
> filter, smoothing the water surface.
If there is flexing or movement of the breakwater, the inside surface
of the break water will start to generate waves. Hopefully, the waves
would be smaller though.
> The flexible connections between segments Patri mentioned would
> accomplish just that: not because they can counter said force better
> than one monolithic block, but because they would bend rather than
> break: in the long run / time averaged, the force is zero anyway, so
> their static load resistance is fairly irrelevant, as long as they keep
> the shape more or less circular over time.
Eventually, it will deform into a some shape that should not
get any smaller (an ovoid ellipse?) and we are basically back
in the static load situation again, which I think is O(R^2).
> Getting more quantitative is difficult: I doubt you could say much more
> without a computer simulation. I could do some calculations, but the
> approximations involved would make it rather meaningless I fear. The
> simplest model I can imagine would be a linear 1-d wave meeting a
> mass-spring interface. It would be a nice demonstration of the principle
> of a low-pass filter, but little more than that.
I think being able to do computer simulations is one of the
highest priority tasks on the engineering side of things.
> The required depth seems also hard to quantify, but I think the 100m
> estimate might be excessive. It should be proportional to the wavelength
> targeted, as penetration depth of waves is proportional to wavelength.
> But the actual moving of water that makes up a wave falls of
> exponentially with depth*, so most of the ‘waving’ is going on near the
> surface. Adding more depth should have strongly declining marginal returns.
There are all sorts of conflicting rules on waves that I do
understand. Some are proportional to wave length, some are
proportional to square root, and some are exponential decay.
I never know which ones apply in which situations.
My concern is that waves that approach a simple barrier can
basically "squirt" under the barrier without much attenuation
at all.
> But as far as computer simulations go, it doesn’t seem terribly
> complicated. A company like MI&T, who supposedly have the software and
> the people familiar with it, should be able to do it rather painlessly,
> a few days or so. But that has a lot of caveats as well: if their code
> does not support flexible joints, for instance, you might be looking at
> months of development.
There are other CFD (Computational Fluid Dynamics) codes out there
that do support flexible joints. I am dubious of the claim of
flexible joints, since continued force from one direction will
eventually collapse the structure to zero area.
> *for a linear wave in ‘deep water’, the velocity field scales with depth
> as exp(-depth*2*pi/wavelength)
Watch out, my Physics roommate at MIT (he did his thesis in the
area of accoustics) told me that ocean scientists were shocked
to discover that sometimes 'shallow water' equations worked better
than deep water equations. Waves that travel over thousands of
miles are actually in a shallow ocean basin of about 5 miles deep.
Some of this stuff is quite tricky, and it is beyond my actual
knowledge.
[snippage]
-Wayne
e.hoog...@student.utwente.nl
I don't think this 3d/pressure argument applies to the 2d/incompressible case we are dealing with. It seems to me that while the pressure in a plane in a pressure vessel is proportional to surface area, the pressure on a line in a breakwater is just proportional to the impact cross-section where the waves come in. But I could easily be wrong, I am not a civil/ocean engineer, and I don't claim to really understand the relevant math, all I have is my physical intuition.
Indeed, the argument Wayne makes is what id call the ‘kettle formula’. The argument applies in 2D as well as in 3D. indeed the surface taking pressure is one dimension lower, but so is the cross-section that needs to bear it. (or you could regard the wavebreak as a cylindrical slice)
For static loads, or quasi-static loads, it most certainly applies. Imagine a planar ocean wave , and draw a circle in it, crest to crest, diameter==wavelength. Imagine the water inside the circle to be flat, as is our goal, so there is a net squeezing on the circle, from both opposite ends.
Resisting said load indefinitely requires O(R^2) material use. I don’t think resisting said load is feasible, and instead we should opt to let those kind of waves pass, by letting the wavebreak go with that kind of flow. By doing so, you would obtain O(R) scaling.
Van:
tsi-engineer...@googlegroups.com
[mailto:tsi-engineer...@googlegroups.com] Namens Gary Root
Verzonden: vrijdag 19 juni 2009 0:23
Aan: tsi-engineer...@googlegroups.com
Onderwerp: Re: breakwater scaling laws
Patri,
e.hoog...@student.utwente.nl
If there is flexing or movement of the breakwater, the inside surface
of the break water will start to generate waves. Hopefully, the waves
would be smaller though.
Yep, when the breakwater fully moves with the waves, it will likewise fully transmit them to the other side. For small waves, it would not move at all though, and even big waves would at least experience some attenuation.
http://en.wikipedia.org/wiki/Low-pass_filter#Continuous-time_low-pass_filters
Eventually, it will deform into a some shape that should not
get any smaller (an ovoid ellipse?) and we are basically back
in the static load situation again, which I think is O(R^2).
Waves are a cyclic phenomena: thats what i meant with 'the time averaged force is zero'. Moving along with the big waves will not cause any shape-change over the long run. Currents might though, but that’s a manageable force.
There are all sorts of conflicting rules on waves that I do understand. Some are proportional to wave length, some are proportional to square root, and some are exponential decay.
I never know which ones apply in which situations.
I have ‘Wave motion’ by j. billingham and a.c. king lying on my desk. It is a good book, in my opinion. Understanding the difference between, and relevance of phase/wave/group velocity is important, and it does a good job of covering that.
My concern is that waves that approach a simple barrier can
basically "squirt" under the barrier without much attenuation
at all.
I understand your concern, but in general, you are not going to reach the ocean floor. There is something that is 'sufficiently deep': a 1cm wave is effectively stopped by a 1-cm deep wavebreak (or so my gut tells me). Qualitatively, the physics are in our favor. The velocity/pressure fields attenuate exponentially, or something near enough, so the amount of pressure fluctuation felt at greater depths rapidly decreases.
There are other CFD (Computational Fluid Dynamics) codes out there
that do support flexible joints.
Could be, but I havnt seen them.
I am dubious of the claim of
flexible joints, since continued force from one direction will
eventually collapse the structure to zero area.
What continued force?
Watch out, my Physics roommate at MIT (he did his thesis in the
area of accoustics) told me that ocean scientists were shocked
to discover that sometimes 'shallow water' equations worked better
than deep water equations. Waves that travel over thousands of
miles are actually in a shallow ocean basin of about 5 miles deep.
Some of this stuff is quite tricky, and it is beyond my actual
knowledge.
5 miles qualifies as 'deep' relative to the wavelength under consideration. But yeah, there is all sorts of things to take into account: jumps in salinity/temperature can make the effective depth much lower. Shallow water waves would more easily sneak below a wavebreak*, but at the same time, they tend to be higher frequency, which makes them easier to deal with in general.
*their attenuation with depth is cosh(depth): exp(-depth) is a deep water approximation thereof. The exp(depth) term doesn’t factor into it significantly, unless the water is really shallow relative to the wavelength, but that’s a rather long mathematical argument that I can not really translate into physical terms.
e.hoog...@student.utwente.nl
Visual illustration:
Drawing of a wavebreak with diameter equal to wavelength. White are crests, black are troughs, grey is neutral water.
Trying to stop these forces is a case of ‘do not want’. The waves are moving, but relative to, say, a km diameter, not that fast. Its essentially a static load, so inertial or drag effects will not save the day. Oscillating along with these waves, which implies them propagating inwards, is the only option. But that’s not really a problem, because waves with 1-km wavelength are not much of an issue to begin with: you hardly even notice they are there.
Same diameter wavebreak, facing shorter waves, zoomed in:
Not nearly as bad. The wave-crest to the top-right applies an inward pressure, but only on a fraction of the entire circumference: a fraction that diminishes along with wavelength. The cross-section parallel to the waves does not face a compressive force proportional to the cross-section: infact, as the wavelength goes to zero, it will feel nothing at all. For short waves, all that matters is that they do not locally damage the wavebreak.
Patri Friedman
1) Any drifting breakwater scales with O(R^2), as Wayne said - it is a span, not a wall. But it may not be that bad, as the component which scales w/ O(R^2) is that which resists the flex, not that which stops the waves. I think the flex will be much less force, because the structure will move very slowly, because it will have such enormous drag, and the drag resistance does not cause flex. If we think of it as C1 * R + C2 * R^2, which amortized per unit area is C1/R + C2, then C2 (flex cost) sets the lower cost bound. If C2 << C1, the cost should not be too bad. I am hopeful that it is, but could very easily be wrong.
2) Any breakwater fixed in place scales with O(R), as I said - it is a wall, not a span. This may be a moored breakwater, one with propellers, kits, whatever.
My analysis for (1) is based on thinking of the breakwater as sitting on land, with rusty, high-friction wheels. Wayne's model of a square is a nice simplification. Imagine the incoming waves as pushing evenly down from the top side of the square. The whole thing slowly, reluctantly moves, with most of the energy dissipated as friction (ie drag). The friction resistance is along the top side of the square, and so imposes no flex on the structure. Each little piece of the top wall feels the same push and the same resistance.
But just as Wayne says, the side walls have to transmit the force to the bottom wall. After all, the top wall is getting pushed, and the bottom wall is moving. That makes the bottom wall like a span (from corner to corner) and causes scaling problems.
However, this analysis also leads me to (2). The whole reason we have this effect is that the bottom wall is moving due to force on the top wall. Suppose (on land) that the force is not enough to overcome the static friction. The bottom wall doesn't move, and no force is transmitted. This can be accomplished by mooring our breakwater (equivalent to attaching the top wall to the ground at equally spaced points). Same goes if the top wall has thrusters which keep it in place - again, ocean pushing the top wall, thrusters pushing the top wall, as long as the top wall doesn't move, there is no flexing stress. (Alternately, thrusters on the bottom wall could move the bottom wall at the same rate as the top wall.)
You need a number of mooring lines / thrusters proportional to the incoming force, which is proportional to cross-sectional area, but that's fine, that's just R. Deformation is only O(R^2) if it is resisted via the central structure, not if it is resisted locally. (Local resistance is the equivalent of supporting a bridge with helium balloons along its length, instead of a span between two pillars).
(also note that currents are irrelevant - if the entire medium moves, there is no differential force).
Anyway, I am happy to finally intuitively understand Wayne's point, and glad that there is a possible way around it, if y'all agree that the stress only occurs if the breakwater moves.
Patri Friedman
But I hope I am wrong :).
This is for compressive forces in the horizontal plane. In terms of forces in the vertical plane, I agree that crests & troughs balance out as wavelength goes down, and only exert local forces.
Wayne C. Gramlich
I love your diagrams, but I think what is going on is more
complicated.
The waves are caused by wind blowing in one direction over an
extended distance. The net result is that there is a surface
current in the direction wave travel. (Patri: remember how
the waves pushed my spar models down the Berkeley wave tank?)
In addition, the wind is blowing against the breakwater sticking
above the surface. There is a net positive force on the
breakwater in the direction of wave travel. This force scales
linearly with R and forces the breakwater to scale as O(R^2).
-Wayne
e.hoog...@student.utwente.nl wrote:
> Visual illustration:
>
> Drawing of a wavebreak with diameter equal to wavelength. White are
> crests, black are troughs, grey is neutral water.
>
> Trying to stop these forces is a case of ‘do not want’. The waves are
> moving, but relative to, say, a km diameter, not that fast. Its
> essentially a static load, so inertial or drag effects will not save the
> day. Oscillating along with these waves, which implies them propagating
> inwards, is the only option. But that’s not really a problem, because
> waves with 1-km wavelength are not much of an issue to begin with: you
> hardly even notice they are there.
>
> Same diameter wavebreak, facing shorter waves, zoomed in:
>
> Not nearly as bad. The wave-crest to the top-right applies an inward
> pressure, but only on a fraction of the entire circumference: a fraction
> that diminishes along with wavelength. The cross-section parallel to the
> waves does not face a compressive force proportional to the
> cross-section: infact, as the wavelength goes to zero, it will feel
> nothing at all. For short waves, all that matters is that they do not
> locally damage the wavebreak.
I think even small waves will have a net force in the direction
of wave travel; it just wont' be as big as big waves. I do not
think it cancels out.
-Wayne
Wayne C. Gramlich
e.hoog...@student.utwente.nl wrote:
> /I don't think this 3d/pressure argument applies to the
> 2d/incompressible case we are dealing with. It seems to me that while
> the pressure in a plane in a pressure vessel is proportional to surface
> area, the pressure on a line in a breakwater is just proportional to the
> impact cross-section where the waves come in. But I could easily be
> wrong, I am not a civil/ocean engineer, and I don't claim to really
> understand the relevant math, all I have is my physical intuition./
>
> Indeed, the argument Wayne makes is what id call the ‘kettle formula’.
> The argument applies in 2D as well as in 3D. indeed the surface taking
> pressure is one dimension lower, but so is the cross-section that needs
> to bear it. (or you could regard the wavebreak as a cylindrical slice)
>
> For static loads, or quasi-static loads, it most certainly applies.
> Imagine a planar ocean wave , and draw a circle in it, crest to crest,
> diameter==wavelength. Imagine the water inside the circle to be flat, as
> is our goal, so there is a net squeezing on the circle, from both
> opposite ends.
>
> Resisting said load indefinitely requires O(R^2) material use. I don’t
> think resisting said load is feasible, and instead we should opt to let
> those kind of waves pass, by letting the wavebreak go with that kind of
> flow. By doing so, you would obtain O(R) scaling.
I am having troubles envisioning the value of a breakwater that
only breaks some of the waves.
If you do not have enough strength to resist the continual force
from the direction of wave travel, it will squish your structure
flat.
[snippage]
-Wayne
Wayne C. Gramlich
e.hoog...@student.utwente.nl wrote:
> /If there is flexing or movement of the breakwater, the inside surface/
> /of the break water will start to generate waves. Hopefully, the waves/
> /would be smaller though.///
>
> Yep, when the breakwater fully moves with the waves, it will likewise
> fully transmit them to the other side. For small waves, it would not
> move at all though, and even big waves would at least experience some
> attenuation.
>
> ___http://en.wikipedia.org/wiki/Low-pass_filter#Continuous-time_low-pass_filters_
> <http://en.wikipedia.org/wiki/Low-pass_filter>
>
> ///Eventually, it will deform into a some shape that should not/
> /get any smaller (an ovoid ellipse?) and we are basically back/
> /in the static load situation again, which I think is O(R^2)./
>
> Waves are a cyclic phenomena: thats what i meant with 'the time averaged
> force is zero'. Moving along with the big waves will not cause any
> shape-change over the long run. Currents might though, but that’s a
> manageable force.
The currents are exactly the force that causes the breakwater
into an O(R^2) material cost.
Also, a cylinder has differing diffraction patterns at different
wavelength in relation to the diameter. They do not cancel out.
> ///There are all sorts of conflicting rules on waves that I do
> understand. Some are proportional to wave length, some are proportional
> to square root, and some are exponential decay./
>
> /I never know which ones apply in which situations./
>
> I have ‘Wave motion’ by j. billingham and a.c. king lying on my desk. It
> is a good book, in my opinion. Understanding the difference between, and
> relevance of phase/wave/group velocity is important, and it does a good
> job of covering that.
You have a better reference than I do. I doubt I could follow
the math.
> ///My concern is that waves that approach a simple barrier can/
> /basically "squirt" under the barrier without much attenuation/
> /at all./
>
> I understand your concern, but in general, you are not going to reach
> the ocean floor. There is something that is 'sufficiently deep': a 1cm
> wave is effectively stopped by a 1-cm deep wavebreak (or so my gut tells
> me). Qualitatively, the physics are in our favor. The velocity/pressure
> fields attenuate exponentially, or something near enough, so the amount
> of pressure fluctuation felt at greater depths rapidly decreases.
I know they attenuate, but I also know they are dependent on wave length
and the big waves have a huge wave length. A short period 1 cm wave
should attenuate on a fairly shallow breakwater. The long period 1 cm
wave will problem shoot right through.
> ///There are other CFD (Computational Fluid Dynamics) codes out there/
> /that do support flexible joints. // /
>
> Could be, but I havnt seen them.
>
> ///I am dubious of the claim of/
> /flexible joints, since continued force from one direction will/
> /eventually collapse the structure to zero area./
>
> What continued force?
Wave induced current and direct windo induced forces.
> ///Watch out, my Physics roommate at MIT (he did his thesis in the/
>
> /area of accoustics) told me that ocean scientists were shocked/
>
> /to discover that sometimes 'shallow water' equations worked better/
>
> /than deep water equations. Waves that travel over thousands of/
>
> /miles are actually in a shallow ocean basin of about 5 miles deep./
>
> /Some of this stuff is quite tricky, and it is beyond my actual/
>
> /knowledge./
>
> 5 miles qualifies as 'deep' relative to the wavelength under
> consideration. But yeah, there is all sorts of things to take into
> account: jumps in salinity/temperature can make the effective depth much
> lower. Shallow water waves would more easily sneak below a wavebreak*,
> but at the same time, they tend to be higher frequency, which makes them
> easier to deal with in general.
My recollection is that they the physics of tsunami waves had
to use shallow wave formulas. His ability to do math was a
couple orders of magnitude better than me. I think a tsunami
is closer to a step function than a wave, hence the really
long wave length.
> *their attenuation with depth is cosh(depth): exp(-depth) is a deep
> water approximation thereof. The exp(depth) term doesn’t factor into it
> significantly, unless the water is really shallow relative to the
> wavelength, but that’s a rather long mathematical argument that I can
> not really translate into physical terms.
I trust you. My math would peter out fairly shortly into your
description.
-Wayne
Wayne C. Gramlich
I think you are getting closer.
Patri Friedman wrote:
> I have thought about this a bit more, and I think come to a better
> understanding which incorporates both my & Wayne's viewpoints.
> Specifically, my conclusion is that:
>
> 1) Any drifting breakwater scales with O(R^2), as Wayne said - it is a
> span, not a wall. But it may not be that bad, as the component which
> scales w/ O(R^2) is that which resists the flex, not that which stops
> the waves. I think the flex will be much less force, because the
> structure will move very slowly, because it will have such enormous
> drag, and the drag resistance does not cause flex. If we think of it as
> *C1 * R + C2 * R^2*, which amortized per unit area is *C1/R + C2*, then
> C2 (flex cost) sets the lower cost bound. If C2 << C1, the cost should
> not be too bad. I am hopeful that it is, but could very easily be wrong.
I do not think beams scale linearly with R. I honestly do not know
how they scale. I do know that we can not scale bridges up indefinitely.
I suspect they go exponential. I just do not know what the exponent is.
> 2) Any breakwater fixed in place scales with O(R), as I said - it is a
> wall, not a span. This may be a moored breakwater, one with propellers,
> kits, whatever.
This is sort of true. The amount of material is actually scaling with
R^2, because the disk of land that it sits on is required. But we don't
pay for the underlying land, only the part that we raise from the ocean
float to slightly above the surface. The cost scales with R because
we don't pay extra for the underlying land.
> My analysis for (1) is based on thinking of the breakwater as sitting on
> land, with rusty, high-friction wheels. Wayne's model of a square is a
> nice simplification. Imagine the incoming waves as pushing evenly down
> from the top side of the square. The whole thing slowly, reluctantly
> moves, with most of the energy dissipated as friction (ie drag). The
> friction resistance is along the top side of the square, and so imposes
> no flex on the structure. Each little piece of the top wall feels the
> same push and the same resistance.
>
> But just as Wayne says, the side walls have to transmit the force to the
> bottom wall. After all, the top wall is getting pushed, and the bottom
> wall is moving. That makes the bottom wall like a span (from corner to
> corner) and causes scaling problems.
>
> However, this analysis also leads me to (2). The whole reason we have
> this effect is that the bottom wall is moving due to force on the top
> wall. Suppose (on land) that the force is not enough to overcome the
> static friction. The bottom wall doesn't move, and no force is
> transmitted. This can be accomplished by mooring our breakwater
> (equivalent to attaching the top wall to the ground at equally spaced
> points). Same goes if the top wall has thrusters which keep it in place
> - again, ocean pushing the top wall, thrusters pushing the top wall, as
> long as the top wall doesn't move, there is no flexing stress.
> (Alternately, thrusters on the bottom wall could move the bottom wall at
> the same rate as the top wall.)
The force is being transmitted to the disk of land underneath
which is attached to the planet, which is pretty massive.
> You need a number of mooring lines / thrusters proportional to the
> incoming force, which is proportional to cross-sectional area, but
> that's fine, that's just R. Deformation is only O(R^2) if it is
> resisted via the central structure, not if it is resisted locally.
> (Local resistance is the equivalent of supporting a bridge with helium
> balloons along its length, instead of a span between two pillars).
>
> (also note that currents are irrelevant - if the entire medium moves,
> there is no differential force).
In a simplified world, yes. In the real world, when the wave direction
changes, it has to change the trajectory of the breakwater. This is
a great deal of force.
> Anyway, I am happy to finally intuitively understand Wayne's point, and
> glad that there is a possible way around it, if y'all agree that the
> stress only occurs if the breakwater moves.
Unfortunately, I do not agree because there is a force difference
between the wall facing the waves and the wall facing way from the
waves. Whether the breakwater is moving or not is irrelevant. It
is the force difference between the two sides that will break it.
Also, to break big long wave length waves, the breakwater has to reach
down deep. Stringing propellers along the edge to try and mitigate the
force is going to require moving a huge amount of water. I do not think
it will be at all practical.
[snippage]
-Wayne
e.hoog...@student.utwente.nl
1) Any drifting breakwater scales with O(R^2), as Wayne said - it is a span,
not a wall. But it may not be that bad, as the component
which scales w/ O(R^2) is that which resists the flex, not that which stops
the waves. I think the flex will be much less force, because the
structure will move very slowly, because it will have such enormous drag,
and the drag resistance does not cause flex. If we think of it as C1
* R + C2 * R^2, which amortized per unit area is C1/R + C2, then C2
(flex cost) sets the lower cost bound. If C2 << C1, the cost should
not be too bad. I am hopeful that it is, but could very easily be wrong.
Wrt the first bolded sentence, im not sure i get you. Resisting the flex and stopping the waves are two sides of the same coin.
As for drag: An object that is small with respect to a certain wavelength, will move along with said wave, without significant relative motion, or drag. A small section of the wavebreak will, in the absence of stresses holding it together with the rest of the wavebreak, move with the waves just like a bottle would. A tsunami with miles-long wavelength, would simply rock the whole wavebreak back and forth as one piece, no stress involved.
When the object is not small relative to the wave, the wave will want to move different parts of the object in different directions. Worst case, it will want to move one end of the object in one, and the other end in the opposite direction.
2) Any breakwater fixed in place scales with O(R), as I said - it is a wall,
not a span. This may be a moored breakwater, one with propellers,
kits, whatever.
Unless it is placed squarely on the ocean floor, waves are going to impart relative (relative to itself) motion on it. I don’t see mooring lines effectively change that.
My analysis for (1) is based on thinking of the breakwater as sitting on land,
with rusty, high-friction wheels. Wayne's
model of a square is a nice simplification. Imagine the incoming
waves as pushing evenly down from the top side of the square. The whole
thing slowly, reluctantly moves, with most of the energy dissipated as friction
(ie drag). The friction resistance is along the top side of the square,
and so imposes no flex on the structure. Each little piece of the top
wall feels the same push and the same resistance.
What drag? It wants to move with the waves, zero relative velocity, drag-free. Not having it move with the waves is the goal, so drag is the enemy in that sense.
However, this analysis also leads me to (2). The whole reason we
have this effect is that the bottom wall is moving due to force on the top
wall. Suppose (on land) that the force is not enough to overcome the
static friction. The bottom wall doesn't move, and no force is
transmitted. This can be accomplished by mooring our breakwater
(equivalent to attaching the top wall to the ground at equally spaced
points). Same goes if the top wall has thrusters which keep it in place -
again, ocean pushing the top wall, thrusters pushing the top wall, as long as
the top wall doesn't move, there is no flexing stress. (Alternately,
thrusters on the bottom wall could move the bottom wall at the same rate as the
top wall.)
Not impossible in theory: what you are proposing is essentially active noise cancelling. If the circumference of the wavebreak never moves, then neither do any waves propagate inwards, nor are there any structural stresses. Cancelling a few milliwatts of acoustic energy has been done before, but mega(giga?)watts of oceanic energy?
Anyway, I am happy to finally intuitively understand Wayne's point, and glad
that there is a possible way around it, if y'all agree that the stress only
occurs if the breakwater moves.
Not sure i understand your latter sentence. Do you mean move, or deform? Moving in one piece has little to do with it. Stress indeed only occurs when the breakwater deforms: stress is proportional to deformation, and stress governs material failure. The question is: what are we going to do about said tendency to deform?
Jeff Chan
> The waves are caused by wind blowing in one direction over an
> extended distance. The net result is that there is a surface
> current in the direction wave travel. (Patri: remember how
> the waves pushed my spar models down the Berkeley wave tank?)
> In addition, the wind is blowing against the breakwater sticking
> above the surface. There is a net positive force on the
> breakwater in the direction of wave travel. This force scales
> linearly with R and forces the breakwater to scale as O(R^2).
> -Wayne
Currents and waves are different things in terms of practical
effects and sometimes sources. Waves are cyclical motion about a
point. Currents are a more or less steady flow of water in some
direction.
A somewhat good analogy is AC vs DC electrical current (except
that electrical current is a moving energy field and not
relevantly moving physical particles). (Electrons/holes in an
electrical DC current do move, but relatively very slowly
compared to the movement of the electrical field which carries
the energy.)
Jeff C.
--
Jeff Chan
mailto:nation-...@jeffchan.com
http://www.jeffchan.com/
e.hoog...@student.utwente.nl
Same diameter wavebreak, facing shorter waves, zoomed in:
Not nearly as bad. The wave-crest to the top-right applies an inward pressure, but only on a fraction of the entire circumference: a fraction that diminishes along with wavelength. The cross-section parallel to the waves does not face a compressive force proportional to the cross-section: infact, as the wavelength goes to zero, it will feel nothing at all. For short waves, all that matters is that they do not locally damage the wavebreak.
I want this to be true, but I am not seeing it. The
waves exert force, not just at the tangent point, but everywhere else they hit
the structure. For example, the next wave southwest of the tangent wave
at the top right. I guess they tend to slide off, and the amount of force
in the southwest direction goes down as the angle of incidence changes, but
they still exert force. As the wavelength goes to zero, you have half
troughs and half crests, but the amount of crest hitting the structure is still
proportional to cross-sectional area.
But I hope I am wrong :).
I think you are: I don’t have the precise math handy right now, but ill try to figure it out.
The amount of crest hitting the structure is indeed proportional to cross-section area, but so is the amount of trough. The dominant terms in these two expressions should cancel when added, and what remains should converge to ‘who cares about R?’ rather rapidly for increasingly short wavelengths.
This is for compressive forces in the horizontal plane. In terms
of forces in the vertical plane, I agree that crests & troughs balance out
as wavelength goes down, and only exert local forces.
That one is intuitively more obvious, and perhaps the cancelation here progresses with a higher power of wavelength, but its essentially the same argument.
e.hoog...@student.utwente.nl
> The waves are caused by wind blowing in one direction over an
> extended distance. The net result is that there is a surface
> current in the direction wave travel. (Patri: remember how
> the waves pushed my spar models down the Berkeley wave tank?)
Currents would indeed be able to deform a wavebreak consisting of
flexibly connected segments with a small number of mooring points.
Depending upon the stiffness of the joint of course, but maintaining its
overall shape by making the joint more stiff is probably not the best
solution. Just add a few tension lines over the cross-section of the
wavebreak, problem solved. Either way, the exact shape of the wavebreak
isn't very important.
> In addition, the wind is blowing against the breakwater sticking
> above the surface. There is a net positive force on the
> breakwater in the direction of wave travel. This force scales
> linearly with R and forces the breakwater to scale as O(R^2).
I don't think the wind is a significant factor, in terms of structural
stresses.
e.hoog...@student.utwente.nl
> I am having troubles envisioning the value of a breakwater that
> only breaks some of the waves.
There is a fairly definite relation between wavelength and wave height.
On one end of the spectrum, short waves tend to break when 5*amplitude >
labda. On the other end: long wavelengths just don't accrue a lot of
amplitude. 1-km wavelength waves 'do exist', but even if they have a
10-m amplitude (which they never even get close to), youd hardly even
notice they are there.
But if the overlap between 'waves that scare us' and 'waves we can
reasonably hope to stop' is perfect, I really don't know. It doesn't
look too bad, I think.
Jeff Chan
>> In addition, the wind is blowing against the breakwater sticking
>> above the surface. There is a net positive force on the
>> breakwater in the direction of wave travel. This force scales
>> linearly with R and forces the breakwater to scale as O(R^2).
> I don't think the wind is a significant factor, in terms of structural
> stresses.
They're different. Air and water have vastly different
densities, etc. Both should be accounted for, though the effect
of wind may be much less most of the time.
e.hoog...@student.utwente.nl
> Also, a cylinder has differing diffraction patterns at different
> wavelength in relation to the diameter. They do not cancel out.
Eh?
> they are dependent on wave length
> and the big waves have a huge wave length. A short period 1 cm wave
> should attenuate on a fairly shallow breakwater. The long period 1 cm
> wave will problem shoot right through.
I meant 1-cm wavelength, but otherwise, we are on the same page.
> My recollection is that they the physics of tsunami waves had
> to use shallow wave formulas. His ability to do math was a
> couple orders of magnitude better than me. I think a tsunami
> is closer to a step function than a wave, hence the really
> long wave length.
Indeed, tsunamis in open ocean have a wavelength of miles, which
disqualifies any water as being considered 'deep'. A step function is
notoriously loaded with high-frequency components, so that probably not
that you meant.
e.hoog...@student.utwente.nl
The picture next to this link is a good one, especially when considering the effects of drag.
http://en.wikipedia.org/wiki/Wave#Characteristics
The point trajectory is what an object small relative to the wave would like to do.
Think of any big structure as a lot of points that you have to try and hold together (a rigid body is a major intuition-disruptor). Drag is not your friend, from that perspective, although it obviously is going to be there in gargantuan quantities, for any functional wavebreak. It is your friend in that sense, but when viewed from the perspective of having to hold everything more or less in one piece, it most certainly isn’t.
Jeff Chan
> I know they attenuate, but I also know they are dependent on wave length
> and the big waves have a huge wave length. A short period 1 cm wave
> should attenuate on a fairly shallow breakwater. The long period 1 cm
> wave will problem shoot right through.
Be careful not to mix up period or wavelength with amplitude. The
latter two are both measured as length and the first two are
related.
A tsunami may have a 1 meter amplitude in the open ocean but a
wavelength of a couple hundred kilometers. It therefore has much
more energy than a conventional wave with a 2 meter amplitude and
a wavelength of a hundred meters.
That said, you're right that there should be a relationship
between wavelength and breakwater response. For ships, where
hull length is some small multiple of wavelength, and the hull is
pointed directly into our away form the wave, the waves impart
little motion (but lots of stress) onto the hull. In that sense
a ship is like a one dimensional version of a circular breakwater.
Cheers,
e.hoog...@student.utwente.nl
Cross-sectional force as a function of wavelength/diameter
It is a numerical integration, so I don’t know the exact order, but ill be damned if the shape of the envelope is not wavelength^(1/2). I was hoping for wavelength^(2) as wavelength -> 0, but not quite. It does go to zero, but Sqrt(wavelength) does not imply what we want it to imply, so your intuition was correct Patri. If wavelength/diameter==1 does not work, then wavelength/diameter==0.1, sqrt(0.1)=0.3 doesn’t make a world of difference.
That doesn’t mean connected segments will not work, but it does mean I no longer have an argument as to why it would work. It does not argue against the principle of a low-pass filter, of flexing when faced with a load you can not stop, but it does mean all kinds of waves would be candidates for said flexing. Which makes the question: how exactly would this circle deform, and what kind of waves would that generate inside?
I really couldn’t convince myself to any kind of conclusion without seeing it moving on my screen.
Wayne C. Gramlich
e.hoog...@student.utwente.nl wrote:
>
>> Also, a cylinder has differing diffraction patterns at different
>> wavelength in relation to the diameter. They do not cancel out.
>
> Eh?
A stream of short waves will be blocked by the break water
and there will be relative calm behind the break water
A stream of waves at approximately the wavelength of the
breakwater will diffract around the breakwater. These waves
will still be present behind the break water (with fun
diffraction grating style interference patterns.) See:
<http://www.meted.ucar.edu/marine/ripcurrents/NSF/print.htm#33>
There are no diffraction effects at large wave lengths.
>> they are dependent on wave length
>> and the big waves have a huge wave length. A short period 1 cm wave
>> should attenuate on a fairly shallow breakwater. The long period 1 cm
>> wave will problem shoot right through.
>
> I meant 1-cm wavelength, but otherwise, we are on the same page.
Yes.
>> My recollection is that they the physics of tsunami waves had
>> to use shallow wave formulas. His ability to do math was a
>> couple orders of magnitude better than me. I think a tsunami
>> is closer to a step function than a wave, hence the really
>> long wave length.
>
> Indeed, tsunamis in open ocean have a wavelength of miles, which
> disqualifies any water as being considered 'deep'. A step function is
> notoriously loaded with high-frequency components, so that probably not
> that you meant.
We are on the same page.
-Wayne
Wayne C. Gramlich
Jeff Chan wrote:
> On Friday, June 19, 2009, 8:32:39 PM, Wayne Gramlich wrote:
>
>> The waves are caused by wind blowing in one direction over an
>> extended distance. The net result is that there is a surface
>> current in the direction wave travel. (Patri: remember how
>> the waves pushed my spar models down the Berkeley wave tank?)
>> In addition, the wind is blowing against the breakwater sticking
>> above the surface. There is a net positive force on the
>> breakwater in the direction of wave travel. This force scales
>> linearly with R and forces the breakwater to scale as O(R^2).
>
>> -Wayne
>
> Currents and waves are different things in terms of practical
> effects and sometimes sources. Waves are cyclical motion about a
> point. Currents are a more or less steady flow of water in some
> direction.
If you place an object in waves, it will drift in the direction
of wave travel. Yes, the most of the wave motion is in circles
and ellipses, but superimposed on top of the waves is a net motion.
This force decreases with depth.
> A somewhat good analogy is AC vs DC electrical current (except
> that electrical current is a moving energy field and not
> relevantly moving physical particles). (Electrons/holes in an
> electrical DC current do move, but relatively very slowly
> compared to the movement of the electrical field which carries
> the energy.)
That is a good analogy.
In a hurricane grade storm, these forces are the ones we must
resist and they are substantial.
-Wayne
Wayne C. Gramlich
e.hoog...@student.utwente.nl wrote:
>
>
>> The waves are caused by wind blowing in one direction over an
>> extended distance. The net result is that there is a surface
>> current in the direction wave travel. (Patri: remember how
>> the waves pushed my spar models down the Berkeley wave tank?)
>
> Currents would indeed be able to deform a wavebreak consisting of
> flexibly connected segments with a small number of mooring points.
> Depending upon the stiffness of the joint of course, but maintaining its
> overall shape by making the joint more stiff is probably not the best
> solution. Just add a few tension lines over the cross-section of the
> wavebreak, problem solved. Either way, the exact shape of the wavebreak
> isn't very important.
Agreed wave break shape is unimportant. Agreed tension lines are
a viable solution. Cost of tension lines scales as O(R^2).
>> In addition, the wind is blowing against the breakwater sticking
>> above the surface. There is a net positive force on the
>> breakwater in the direction of wave travel. This force scales
>> linearly with R and forces the breakwater to scale as O(R^2).
>
> I don't think the wind is a significant factor, in terms of structural
> stresses.
I suspect you are right, but for a different reason. Wind tends to
blow over the top of structures. Thus, the wind will tend to
exert the same force on both the front and back walls of the
breakwater.
I should mention, that the wind effects will not be attenuated
by the breakwater and, thus, the structures inside the breakwater
will have to deal with them. Many a harbor has had it fishing
fleet trashed due to a combination of storm surge and wind forces.
-Wayne
e.hoog...@student.utwente.nl
Yup, agreed.
Wayne C. Gramlich
I do not know much about the wave length vs. wave amplitude.
I'm sure the
The USS Ramapo in 1939 may have the record for worst waves:
<http://hyperphysics.phy-astr.gsu.edu/hbase/watwav.html#ramapo>
encountered 34m waves with a wavelength of 342m. If the break
water wants to stop these waves, it will need to be approximately
half the wavelength deep. The amount of force decreases with
depth, so the breakwater at the surface would be thicker and
taper down as depth increased. Floating breakwaters are going
to be *huge*.
-Wayne
e.hoog...@student.utwente.nl
I do not know much about the wave length vs. wave amplitude.
I'm sure the
The USS Ramapo in 1939 may have the record for worst waves:
<http://hyperphysics.phy-astr.gsu.edu/hbase/watwav.html#ramapo>
encountered 34m waves with a wavelength of 342m. If the break
water wants to stop these waves, it will need to be approximately
half the wavelength deep. The amount of force decreases with
depth, so the breakwater at the surface would be thicker and
taper down as depth increased. Floating breakwaters are going
to be *huge*.
-Wayne
Yup, thats bad stuff.
Overall, im not feeling too good about wavebreaks. Trimming down already reasonably behaved waters seems plausible enough, but in the open ocean, it seems a wavebreak has to worry a lot about saving itself before it can hope to do anything else any good.
Jeff Chan
> Jeff Chan wrote:
>> Currents and waves are different things in terms of practical
>> effects and sometimes sources. Waves are cyclical motion about a
>> point. Currents are a more or less steady flow of water in some
>> direction.
> If you place an object in waves, it will drift in the direction
> of wave travel. Yes, the most of the wave motion is in circles
> and ellipses, but superimposed on top of the waves is a net motion.
> This force decreases with depth.
Yes, but it's a small effect compared to the mostly cyclical wave
motion. There's lots of AC with a relatively very small DC
effect/component. Most the energy is cyclical.
Jeff Chan
> Overall, im not feeling too good about wavebreaks. Trimming down already
> reasonably behaved waters seems plausible enough, but in the open ocean,
> it seems a wavebreak has to worry a lot about saving itself before it
> can hope to do anything else any good.
In the end something that has a minimal coupling with waves such
as a spar or to a lesser extent something like SWATH may be a
better, more modular solution than a breakwater. Spars may deal
better with real world ocean conditions than a breakwater, with
or without respect to cost.
A breakwater is attractive if it's possible, but it's a very big
project. An individual platform is relatively a much smaller
project, more self-sufficient and perhaps more durable.
The spar is individualist ("I'll take care of myself") compared
to the collectivist ("we'll take care of you") breakwater.