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Thread: Submarine Buoyancy

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    Submarine Buoyancy

    How does the stability of a submarine’s buoyancy vary with depth?

    http://en.wikipedia.org/wiki/Buoyancy#Submarines states “Submarines rise and dive by filling large tanks with seawater. To dive, the tanks are opened to allow air to exhaust out the top of the tanks, while the water flows in from the bottom. Once the weight has been balanced so the overall density of the submarine is equal to the water around it, it has neutral buoyancy and will remain at that depth.”

    To illustrate my question, consider a five metre high submarine at the surface in calm shallow water. Assume the water is five metres deep at low tide, and that the sub is at rest on the bottom, and then will float one metre above the floor with a one metre tide, so the top of the sub is always exactly at the water surface. Now, consider if the sub is connected to a plastic bag containing liquid underneath it and equal to it in size. This bag is also fastened to the sea floor, and has inlet and outlet valves and pipes, and is able to expand and contract like a bellows. So when the sub is resting on the floor the bag beneath it is squeezed empty, and when the tide rises, the submarine will rise, and the connected bag below it will therefore expand like a bellows, sucking in additional liquid through the inlet valve and pipe, say from a fresh water bag at the surface. As the tide falls, the submarine will also fall, expelling liquid from the bag beneath it, forming a tidal pump powered by the weight of the submarine on top of it. Assume the outlet pipe is wide enough to fully empty the bag over the 6.5 hour period of a falling tide.

    If we then shift this scenario to an ocean location of depth one hundred metres, and balance the submarine flotation tanks so that it is again sitting on the floor at low tide and floating one metre above the floor at high tide, how will the power of pumping be affected? If the same plastic bag of liquid with inlet and outlet valves is fastened both to the ocean floor and to the underside of the sub, will it have the same pumping power as in the shallow water situation? If less by how much?

    If we extend this to depths of one or two kilometres (using a sealed bag of brackish water in place of the manned submarine) will the forces at that depth reduce this tidal pumping power further, and by how much?

    Many thanks for any information on how to understand this problem.

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    There are a lot of submariners on this forum that can probably speak to the realities of neutral buoyancy and depth, but generally speaking a neutrally buoyant object will stay at whatever depth it is moved to. So a fully submerged, neutrally buoyant object would not exert much force against the pump or do much work.

    By contrast, a float in shallow water stays at the surface and can convert the motion of waves and the tides into work.

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    Quote Originally Posted by VQkr View Post
    There are a lot of submariners on this forum that can probably speak to the realities of neutral buoyancy and depth, but generally speaking a neutrally buoyant object will stay at whatever depth it is moved to. So a fully submerged, neutrally buoyant object would not exert much force against the pump or do much work.

    By contrast, a float in shallow water stays at the surface and can convert the motion of waves and the tides into work.
    Thanks. Your second sentence does not seem to follow. If the object stays at the same depth, then it will push out of the way anything that is under it on a falling tide, and pull up anything hooked to it on a rising tide. To "stay at whatever depth it is moved to" means to exert force against anything that resists its constant depth. I am wondering how fast this force reduces with increased depth.

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    A few factors to consider: salinity and compressability. If you are transiting the coast and cross a river of fresh water,
    the boat will set deeper: fresh water is lighter than salt water.
    Compressability: water is incompressable. No question.
    So...... displacement remains the same, but the medium varies with the varying weight of the medium as it is changed by
    salinity.
    Note: Ballast tanks are totaly flooded when diving. Additional trim is acquired through the trim and drain system, which employs hard tanks (made of really thick HY-80 high tensile steel ) which can sustain sea pressure. Too heavy? Blow the
    hard tank. Too heavy forward? Pump from a tank fwd to a tank aft.Too light? Flood additional water into the hard tank.And there are many options , and cross-connects etc .
    Basic stuff of submarines.

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    Water is compressible; if it weren't, the speed of sound in water would be infinite. Usually, water is incompressible enough so that its compressibility can be ignored.

    On the other hand, submarines are probably more compressible than water.

    This means that submarines would tend to lose buoyancy as they descend. On the other hand, I believe submarines usually operate with slight positive buoyancy, and use usually use forward motion to remain submerged.
    Last edited by swampyankee; 2015-Mar-01 at 05:30 AM.

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    Quote Originally Posted by danscope View Post
    Compressability: water is incompressable. No question.
    I suspect you know this is incorrect. While water has a low compressibility, and for some practical applications can be considered to be incompressible, it definitely can be compressed. This is a short extract from the relevant wikipedia page.

    The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1. At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 MPa.
    The bulk modulus of water is 2.2 GPa. The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.


    Source: http://en.wikipedia.org/wiki/Propert...ompressibility

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    As danscope mentions below, a submarine is a complex system of parts to make it go where commanded. Let's stick to a neutrally buoyant bag for simplicity. If we put this bag against a lever connected to a generator, no substantial work is done and no power is generated because it the only force generated against the lever is the hydrodynamic drag against the tidal currents. Since the bag is neutrally buoyant, it has the same amount of potential energy at any depth. For a truly neutrally buoyant object, it will take an arbitrarily small amount of energy to slowly move the object to a different depth.

    Compare this to a buoyant object like a beach ball - it takes force to push it under the water, and more work to push it deeper per F x D. For a lead ball, it takes more work to lift it to shallower depths, again per F x D.

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    Quote Originally Posted by VQkr View Post
    Let's stick to a neutrally buoyant bag for simplicity. If we put this bag against a lever connected to a generator, no substantial work is done and no power is generated
    Thanks again. Let me amplify your simplified example to show how work could be done.

    Consider such a neutrally buoyant bag on a large scale, say one square kilometre and ten metres deep, in water ten metres deep at low tide. With tide of one metre, a bag underneath as described in the opening post would suck in one gigalitre on a rising tide and pump out one gigalitre on a falling tide.

    The pumping movement of this water is work. No electrical power generation is required in order for energy to be obtained. My question is whether this surface model would also work at depth, or whether the neutral buoyancy of a bag would actually become less neutral at depth.

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    The mechanics and procedures relating to how an actual modern submarine maintains and changes depth is irrelevant to Robert Tulip's question, I think.

    Robert, I think you need to key in on what VQkr posted above. Your submarine is neutrally buoyant. Therefore, there will be little force required to get it to change depth up or down. That is what your plastic pumping bag below the submarine is trying to do - exert a force on the submarine against its buoyancy. However, I do not believe VQkr is entirely correct in saying that the submarine will be neutrally bouyant at any depth. Buoyancy does change with depth due to the increase or decrease in the volume of trapped air in the ballast tanks caused by the variable seawater pressure exerted on the tanks.

    You have a five meter high pumping bag. The amount of variable buoyancy in five meters depth change is not zero, certainly, but probably not great. I don't think you'll get much pumping force.

    But your real question is, I believe, what is the effect on this five meter pumping force if we move that five meter depth change from near surface to much deeper, a hundred meters below surface? The answer to that is, I don't know.

    Or, I missed the point of your question entirely.

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    Let's do a thought experiment by pulling back from the complications of a real submarine and the surrounding sea water with its density variations with temperature and salinity. We will keep the system at constant temperature and have the water of the same composition from top to bottom.

    As has been pointed out, both the water and the pressure hull are compressible. Let us adjust the mass of the hull to make it neutral at a chosen depth. If the hull is less compressible than the water it will have positive buoyancy if forced deeper and negative if forced shallower. When released it would return to the original depth, displaying a stable equilibrium. It would rise and fall with the tide much as would a surface ship.

    If the hull is more compressible than the water we have an unstable equilibrium at the chosen depth. If the hull is moved up the least bit it will expand more than the water and float to the surface. If it is moved down the least bit it will contract more than the water and sink.

    If the hull and the water have exactly the same compressibility, the hull will stay at whatever depth we place it. If the tide causes any vertical motion of the water it will drag the hull a bit, but there will be no buoyancy effect to move it in unison with the water.

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    Where and what sort of tides are assumed. Large differences in tide height are caused by geography and bathymetry. If you put it somewhere the increase is great, you'll get large lateral forces. If you put it in the depths, will the tidal change be great enough?
    Et tu BAUT? Quantum mutatus ab illo.

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    Quote Originally Posted by Robert Tulip View Post
    Thanks again. Let me amplify your simplified example to show how work could be done.

    Consider such a neutrally buoyant bag on a large scale, say one square kilometre and ten metres deep, in water ten metres deep at low tide. With tide of one metre, a bag underneath as described in the opening post would suck in one gigalitre on a rising tide and pump out one gigalitre on a falling tide.

    The pumping movement of this water is work. No electrical power generation is required in order for energy to be obtained. My question is whether this surface model would also work at depth, or whether the neutral buoyancy of a bag would actually become less neutral at depth.
    If the bag is neutrally buoyant, than it will not change shape with the tide due to buoyant effects (neither at the surface nor at depth). The "pumping movement" of the water is only work if the movement is against a force, since work equals force x distance.

    Quote Originally Posted by geonuc View Post
    However, I do not believe VQkr is entirely correct in saying that the submarine will be neutrally buoyant at any depth.
    Agreed; my statement was a first-order approximation, incorporating the common assumption used whenever someone describes an object as "neutrally buoyant." The steel of the submarine hull is ~100x less compressible than the water around it, and as I understand it the change in volume of the pressure vessel with depth is much smaller than the increase in density of the sea water (again with depth) meaning that the submarine in a fixed ballast configuration is only actually neutrally buoyant at one depth. Using a bag of sea water instead of a submarine simplifies the conversation.

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    Quote Originally Posted by VQkr View Post
    The steel of the submarine hull is ~100x less compressible than the water around it, ...
    I think we agree that an actual submarine's characteristics are not what's important here, but for the record: while the steel itself is not very compressible, the hull definitely is. We used to run a taut string from side to side while near the surface just before going deep to see the effect the increased sea pressure had on the hull. The string drooped significantly as we dove, depending on the location of the hull structural ribs. We also did the opposite - run a taut wire at test depth and wait for it to snap when we came up.

    As I said, not important for this question but thought I'd share that.

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    Quote Originally Posted by Eclogite View Post
    I suspect you know this is incorrect. While water has a low compressibility, and for some practical applications can be considered to be incompressible, it definitely can be compressed. This is a short extract from the relevant wikipedia page.

    The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1. At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 MPa.
    The bulk modulus of water is 2.2 GPa. The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.


    Source: http://en.wikipedia.org/wiki/Propert...ompressibility
    I suspect you have not been on a submarine for any time. You would enjoy running a heavy diesel engine , while snorkeling and accidently swallow a sea-slug ( a quantity of water to you) which after entering the cylinder, will enjoy being compressed
    as it splits open the cylinder and makes a mess of things. I get my information from the USNavy which has a history of
    submarines at sea.
    Last edited by danscope; 2015-Mar-02 at 04:35 AM.

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    Quote Originally Posted by danscope View Post
    I suspect you have not been on a submarine for any time. You would enjoy running a heavy diesel engine , while snorkeling and accidently swallow a sea-slug ( a quantity of water to you) which after entering the cylinder, will enjoy being compressed
    as it splits open the cylinder and makes a mess of things. I get my information from the USNavy which has a history of
    submarines at sea.
    This observation is not incompatible with the physical fact that water (salt or otherwise) is a compressible fluid with a bulk modulus of ~2.2 GPa. Because this is roughly 20,000 times less compressible than air at STP, water is frequently described as effectively incompressible. Your experience with the damage water can do to piping designed for gas is not unique to submarines; for example, steam mains must prevent condensation from blocking the cross-section of the pipe or the water hammer can cause catastrophic failure.

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    Quote Originally Posted by Hornblower View Post
    If the hull and the water have exactly the same compressibility, the hull will stay at whatever depth we place it. If the tide causes any vertical motion of the water it will drag the hull a bit, but there will be no buoyancy effect to move it in unison with the water.
    These two sentences read to me as contradictory. Staying at the same depth means moving in unison with the tide. I am just wondering if a tethered object at sea with neutral buoyancy will move up and down at the same rate as the tide whatever its depth.

    A diagram of this model, using a bag of water in place of a submarine, is attached. My basic question is whether this pumping system, with a closed bag on top of a valved bag, will work, firstly at the surface, and then at depth. I will need to consult a hydrologist and test this in a laboratory.

    Tidal Pump.png.

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    Quote Originally Posted by Robert Tulip View Post
    I am just wondering if a tethered object at sea with neutral buoyancy will move up and down at the same rate as the tide whatever its depth.
    No.

    Quote Originally Posted by Robert Tulip View Post
    A diagram of this model, using a bag of water in place of a submarine, is attached. My basic question is whether this pumping system, with a closed bag on top of a valved bag, will work, firstly at the surface, and then at depth. I will need to consult a hydrologist and test this in a laboratory.
    On the surface, sure. You want a bag of something that floats in seawater, for example fresh water, rather than something that is neutrally buoyant.

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    Quote Originally Posted by VQkr View Post
    No.
    In that case, how do you explain the statement quoted in my opening post "Once the weight has been balanced so the overall density of the submarine is equal to the water around it, it has neutral buoyancy and will remain at that depth.”

    And why is a bag of fresh water (relative density 40/41 of sea water) not neutrally buoyant?
    Last edited by Robert Tulip; 2015-Mar-02 at 11:34 AM.

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    It is necessary to point out that there are no significant vertical tides in deep ocean. Tides are horizontal movements of water which get pushed around by land masses obstructing their flow, So tides are a feature of coastlines. Oceans have waves and major deep sea flows and salinity gradients and thermal inversions but no vertical tides.

    So the neutral balanced submarine will not experience ups and downs in the way a coastal tide system would.
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    Quote Originally Posted by profloater View Post
    there are no significant vertical tides in deep ocean.
    Not true. See http://en.wikipedia.org/wiki/Tide#me...onstituent.jpg which shows global tides, including deep ocean areas with big tides. In any case, I am asking about tides at depths of 100m or maybe 600 m, noting the average depth of the ocean is 4000 m.

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    Quote Originally Posted by danscope View Post
    I suspect you have not been on a submarine for any time. You would enjoy running a heavy diesel engine , while snorkeling and accidently swallow a sea-slug ( a quantity of water to you) which after entering the cylinder, will enjoy being compressed
    as it splits open the cylinder and makes a mess of things. I get my information from the USNavy which has a history of
    submarines at sea.
    danscope,

    This posts seems excessively snarky and rude. If you wish to share information about your submarine experiences, that's fine, though this is primarily a scientific discussion. But you've been warned repeatedly about rude posts. Please tone it down.
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    well that diagram shows tides less than 60 cm over most of the oceans. The high tides are on the continental shelves. Do you want to work with 60 cm movement? Ok I thought you were aiming for deep ocean. Carry on the good work, a neutral buoyant object will move somewhat with the tides, but not as much I guess as a surfaced buoyant object. It is a question of inertia and applied forces. A large object will not move much at all as its mass rises with the cube and its surface are rises with the square.
    Last edited by profloater; 2015-Mar-02 at 02:15 PM.
    sicut vis videre esto
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    Quote Originally Posted by Robert Tulip View Post
    In that case, how do you explain the statement quoted in my opening post "Once the weight has been balanced so the overall density of the submarine is equal to the water around it, it has neutral buoyancy and will remain at that depth.”
    "Neutral buoyancy" is a term that inherently implies first-order approximation. A trimmed submarine's hull flexes with changes in pressure and will "settle" at a specific depth under the surf, but the force necessary to change the depth of the submarine is still very small when compared to the weight of the submarine itself. Similarly, NASA training pools are an imperfect but adequate simulation of microgravity - a very small force can move a massive object. For large objects, the hydrodynamic forces on the system (drag) will dominate.

    Quote Originally Posted by Robert Tulip View Post
    And why is a bag of fresh water (relative density 40/41 of sea water) not neutrally buoyant?
    You answered your own question. A bag of fresh water is ~2.5% less dense than sea water, so it floats. On the scale of a 1000 m x 1000 m x 10 m bag, the force necessary to push the bag completely under the surface is 2.6 x 109 N, and with a 1 m tide the bag tethered to a stationary object is capable of doing 10.6 GJ of work over each 24-hour period.

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    Quote Originally Posted by profloater View Post
    a neutral buoyant object will move somewhat with the tides, but not as much I guess as a surfaced buoyant object. It is a question of inertia and applied forces. A large object will not move much at all as its mass rises with the cube and its surface area rises with the square.
    This is getting to the point of my question. The ocean itself is a large object, about a billion teralitres in size, and overall it moves up and down quite a bit all the time with the tides. My understanding is that any object with neutral buoyancy floating at a specific depth will have a sinusoidal acceleration with respect to the ocean floor, moving up and down as part of the ocean water column at the same speed as the up and down tidal movement of the surface water, unless there is some other force acting upon it. It will stop moving and reverse direction at the high and low tide turning points of the sine curve, and go fastest at the mid tidal inflection points of the curve. I don’t see that inertia would mean its motion would be delayed to after the tidal motion.
    Quote Originally Posted by VQkr View Post
    "Neutral buoyancy" is a term that inherently implies first-order approximation. A trimmed submarine's hull flexes with changes in pressure and will "settle" at a specific depth under the surf, but the force necessary to change the depth of the submarine is still very small when compared to the weight of the submarine itself. Similarly, NASA training pools are an imperfect but adequate simulation of microgravity - a very small force can move a massive object. For large objects, the hydrodynamic forces on the system (drag) will dominate.
    So, with my model of a submarine resting on a connected bag of water that has inlet and outlet valves, and which 'concertinas' up and down with the tides as it is pulled up and compressed by the tidal movement of the sub, this issue of the force necessary to change the depth of the sub is what I am asking about. Assuming the valves are big enough that the bag can fill and empty in the period of the tide, I am wondering how much a sub (or a bag of water) will change in depth with this applied force, especially considering any hydrodynamic drag, which I don’t understand at all how to calculate. Will the work from the sub in the form of pumping energy be too small to be economically useful? (That is entirely assuming the sub has no better things to do, so we are really better off thinking of the enclosed bag of water).
    Quote Originally Posted by VQkr View Post
    A bag of fresh water is ~2.5% less dense than sea water, so it floats. On the scale of a 1000 m x 1000 m x 10 m bag, the force necessary to push the bag completely under the surface is 2.6 x 109 N, and with a 1 m tide the bag tethered to a stationary object is capable of doing 10.6 GJ of work over each 24-hour period.
    This 97.5% submerged bag of fresh water maintains that constant depth, which I thought was the definition of neutral buoyancy. The ten gigalitre bag you described is a useful model. I am not talking at all about a force needed to push the bag under the surface, but about how the inertia of this bag will push out of the way anything underneath it and pull up something connected to it. In water of ten metre depth at low tide, with tidal range of one metre, this bag would cause a connected bag below to fill and empty at a rate of one gigalitre per tide, at no operating cost, assuming the hydrostatic forces on the inlet and outlet valves of the one metre high bag beneath it are zero. It is this assumption I am wondering about, and how its quantum changes with depth and other factors such as pipe length.

    In agriculture, a 2003 estimate of pumping cost is $24,000 per gigalitre, so a method that can achieve that volume twice a day in coastal waters powered only by the orbital gravitational forces of the sun and moon and with minimal operating cost is worth exploring.

    Moving the fixed floating bag below the surface is not achieved by applying force but rather by adding salt, to achieve neutral buoyancy (if that makes sense). If the bag just contained sea water, I assume it would not float but would settle to the floor. However, since it is not nailed down, I would expect it would not just sit on the floor, but would move up and down with the tide like a submarine (tethered in position so it does not drift away). What I don’t understand is the difference between a floating bag of fresh water, which you would expect would pull up anything beneath it in order to stay floating at the same relative depth, and a submerged bag of salt water, which I would have thought would also retain the same relative depth due to tide. Thanks.
    Last edited by Robert Tulip; 2015-Mar-02 at 06:48 PM. Reason: delete "effort", add "tethered"

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    Question: Does the ocean have a scale height? That is to say, ignoring waves, and just focusing on the tide, would something halfway up the column experience the full 60 cm movement, or just half that, or some other %?

    As for subs, my understanding is that they primarily change depth by using the planes and forward motion. The tidal power schemes I've seen also use planes, in the form of vertical axis ducted propellers.
    Et tu BAUT? Quantum mutatus ab illo.

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    Quote Originally Posted by Robert Tulip View Post
    My understanding is that any object with neutral buoyancy floating at a specific depth will have a sinusoidal acceleration with respect to the ocean floor....This 97.5% submerged bag of fresh water maintains that constant depth, which I thought was the definition of neutral buoyancy.
    Both incorrect. Possibly confusing the issue is the word "float", which has multiple definitions. A cork in sea water floats, as in "bobs around at the surface." A neutrally buoyant jellyfish in sea water floats, as in "drifts, with no net buoyant force while submerged." From here forward I suggest we only use the word "float" to refer to the behavior of an object less dense than the surrounding fluid, bobbing around on the surface like a cork or iceberg. The bag of fresh water floats partially submerged at the surface and is definitely not neutrally buoyant. A bag of sea water drifts under the surface, is neutrally buoyant, and will require only an arbitrarily small amount of force to slowly move to a different depth.

    Quote Originally Posted by Ara Pacis View Post
    Question: Does the ocean have a scale height? That is to say, ignoring waves, and just focusing on the tide, would something halfway up the column experience the full 60 cm movement, or just half that, or some other %?

    As for subs, my understanding is that they primarily change depth by using the planes and forward motion. The tidal power schemes I've seen also use planes, in the form of vertical axis ducted propellers.
    Unlike the atmosphere, which varies in density and substantially expands and contracts, the oceanic tides are due to a moving bulge of water not expansion or contraction. Since the column of water under the bulge is not "expanding", the movement cannot be described by looking at a one-dimensional core of the water. There is lateral movement of the water driving the vertical change. Ignoring all the other currents a free-drifting, neutrally buoyant submerged object in the open ocean would move in a roughly elliptical path with the tide, with magnitude reducing with depth. In practice I doubt it is practical to isolate this motion from all the other water movement happening in the open ocean.

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    So, to generate pumping energy from tide at depth, a floating vessel (a ship or bag of fresh water) would need to be anchored to a large heavy platform, and the inlet/outlet chamber would be on the ocean floor beneath this platform. The energy would be derived on a rising tide from the surface vessel lifting the platform, and on a falling tide by the weight of the platform compressing the i/o bag underneath it.

    I was not considering properly that with a bag of sea water there would be no upward power, because on a rising tide the ocean water from elsewhere rushes in, and there is no force to cause a bag underneath a bag of sea water to fill.

    So, if the surface vessel was a one cubic kilometer floating island (say a bag of fresh water with an airport on top), it could be tethered to a platform and bag on the ocean floor beneath to generate pumping energy of one gigalitre per tidal motion per metre of tide.

    Thanks for helping me to understand this.

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    The bag under your submarine would need to be secured to both the submarine and the ocean floor across its entire horizontal surface to prevent water from simply filling and leaving the unenclosed space.

    I'm not certain, but I believe that at a significant depth, the proportion of weight of all the water beneath the submarine to all of the weight above it will reduce it effectiveness.
    Depending on whom you ask, everything is relative.

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    Lifting the freshwater bag 1 cm further out of the water than where it floated would take 98 MN of force. No heavy weight would be needed to compress the pump bag; the freshwater bag is plenty heavy enough.

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    Quote Originally Posted by VQkr View Post
    Lifting the freshwater bag 1 cm further out of the water than where it floated would take 98 MN of force. No heavy weight would be needed to compress the pump bag; the freshwater bag is plenty heavy enough.
    What I am thinking about is a way to situate a pump on the edge of the continental shelf, at depth of 600 metres. This could be used to pump nutrient-rich water from below the thermocline to the surface. So there would be a gap between the floating fresh water bag and the pump bag. In order to transfer the tidal energy from the floating bag to the pump bag, at gigalitre scale a million tonne weight will be needed on top of the pump bag, connected by strong cables to the floating bag, which would need air chambers in it to lift the weight. The weight is lifted on a rising tide and falls on a falling tide. The weight is needed unless the pump system is in shallow enough water for the floating bag to be in physical contact with the pump bag. I have not suggested any scenario in which the floating bag would change its depth. Thanks again.

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