Ocean liner in bucket full of water

[dacruick]

If the hull is completely under water then it's existence is trivial is it not?

Why won't you address my question about the origin of the buoyant force that you claim to be there.

 

[nasu]

No matter what scenario it is, there is the weight of the ship. If the weight of the ship is mass * gravity, and the ship is floating (not accelerating), where is this force causing buoyancy coming from?

From the hydrostatic pressure of the liquid.
A boat 92 m long and 6 m wide (I found these numbers online) has an area of the submerged part of about 1800 m^2.
A column of water 10 cm high will exert a force
F= 1000 kg/m^3 * 10m/s^2* 0.1 m *1800 m^2, about 1.8 MN (or 180 tons).
Allowing that the force exerted by the pressure is not everywhere upward, it gives you an order of magnitude estimate of hundreds of tons.

Is not the compression of water that matters. Is not like sitting on a spring.

 

[A.T.]

If the hull is completely under water

The second hull is not completely under water. It sticks out just as the ship hull does. So again:

When exactly will the ship stop floating?

 

[cmb]

I can't believe that this idea is being proposed without a source of the force to keep the boat afloat.

It wasn't an idea. I was explaining what actaully happens in wet/dry dock.

You seem to be under an illusion that there is a 'pressure' required something like the magnitude of the ship. But if you want to look at it that way, it is the ship's mass all supported on an very large surface area, thus a 'realtviely low pressure' is all that is required. The relative pressure required to 'lift the boat' will match the pressure of however-many-metres of vertical liquid height there is in that small thickness of water.

I suggest you drop the whole idea of 'pressure' and think about 'density'. Once 'afloat' the ships hull below the water line will have exactly the same density as water. Now picture a lump of water floating in that water. Would that body of water sink? See how it is wrong to think it will sink against the water?

 

[Borek]

Why won't you address my question about the origin of the buoyant force that you claim to be there.

You were already told where the buoyant force comes from - it is pressure the water exerts on the ship hull.

And it was also signaled that yes, bucket walls have to survive the pressure as well, yes, bucket walls support the water. But ocean bottom have to survive this pressure as well. There is no difference between both cases. The only difference is amount of water used. Ship doesn't come in contact neither with the sea bottom nor with the bucket walls - it floats in water.

 

[dacruick]

You were already told where the buoyant force comes from - it is pressure the water exerts on the ship hull.

And it was also signaled that yes, bucket walls have to survive the pressure as well, yes, bucket walls support the water. But ocean bottom have to survive this pressure as well. There is no difference between both cases. The only difference is amount of water used. Ship doesn't come in contact neither with the sea bottom nor with the bucket walls - it floats in water.

Then you have water under immense pressure, don't you? And if the water is under immense pressure then it will always try to flow to lower pressure. And unless you've capped your bucket, it will flow outside, to the low pressure.

 

[nasu]

Then you have water under immense pressure, don't you? And if the water is under immense pressure then it will always try to flow to lower pressure. And unless you've capped your bucket, it will flow outside, to the low pressure.

No, you don't. The pressure required to support a 200 ton boat may be of the order of 1/100 atmosphere. Water compression under this pressure is negligible.

Maybe it will help if you explain where do you think the force comes from when the boat floats in the ocean.

 

[cmb]

Then you have water under immense pressure, don't you?

You tell us. What's the psi to lift a ship? - name a ship and give us its mass/hull-surface-area?

(I can tell you - it is the same pressure as that of a given draught of water it sits in! Obviously!)

 

[256bits]

Then you have water under immense pressure, don't you? And if the water is under immense pressure then it will always try to flow to lower pressure. And unless you've capped your bucket, it will flow outside, to the low pressure.

If the water is under immense pressure, instead of flowing out, it could just as well want to push the ship higher up in the bucket. Both scenarios cannot be correct.would you agree.

 

[dacruick]

I can't possibly be right if this many people oppose me. I must have one really huge assumption in my logic. I understand that the force comes from the walls of the ocean, and I can see how that specifically can be extended to the bucket scenarios.

 

[dacruick]

I just don't see how under the weight of the ship, the water does not get displaced. Borek, in your diagram of the three scenarios, what is stopping the water at the bottom of the bucket from being displaced? Intuition (which is not serving me too well today) tells me that it is the water along the edges. But how can that be possible when the volume of water along the edges is so small?

 

[cmb]

I just don't see how under the weight of the ship, the water does not get displaced. Borek, in your diagram of the three scenarios, what is stopping the water at the bottom of the bucket from being displaced? Intuition (which is not serving me too well today) tells me that it is the water along the edges. But how can that be possible when the volume of water along the edges is so small?

The water in the gap is, at that point, the same density as the ship's hull. Why would the ship's hull sink lower if it is not of a greater density than the water?

 

[Borek]

I just don't see how under the weight of the ship, the water does not get displaced. Borek, in your diagram of the three scenarios, what is stopping the water at the bottom of the bucket from being displaced? Intuition (which is not serving me too well today) tells me that it is the water along the edges. But how can that be possible when the volume of water along the edges is so small?

Pretty simple - pressure is a function of density and height of the column, doesn't matter how thin the column is.

Edit: well, at some point it starts to be too thin, and it doesn't behave as a water any longer, but even if it is just a tenth of a millimeter it is still thick enough.

 

[256bits]

I can't possibly be right if this many people oppose me

In your defense I will say: The plate techtonics guy originally came up against stiff opposition to his theory. The quasi-crystal guy was ridiculed at his discovery and what did he just win - a Nobel prize. :)

 

[jmmccain]

Here is a simple, quick, and cheap (no aircraft carriers needed) experiment:

Obtain two stackable drinking glasses, like those used at fast food restaurants. Fill one with water. Set the other, empty glass, in the water. The second glass, being empty (and light) will displace a small amount of water and float. Place coins (or other suitably heavy objects) into the empty glass. Once the second glass is half of the way into the first, you can tell it is still floating. Continue adding coins until the glass is almost all of the way down. This represents the ship in a bucket and it is still easy to observe that it is floating. Continue to add coins and the glass would sink - were it not for the experiment's simple design. Remove the second glass. Note the tiny amount of water that remains in the first. Put the second glass back in. Note that it still floats even though no water was spilled this time.

 

[cmb]

Further still, once the 'inner' glass has been balanced with coins that it is just floating off the bottom of the other glass, note whereabouts the water level comes up to. Then, with that glass still with its coin weights, float it in a big sink of water and note whether it floats to the same depth or not.

 

[jmmccain]

For a real life example of ships in a bucket, see: http://en.wikipedia.org/wiki/Panamax

 

[A.T.]

For a real life example of ships in a bucket, see: http://en.wikipedia.org/wiki/Panamax

Or here, the entire bucket is even lifted:
http://en.wikipedia.org/wiki/Strépy-Thieu_boat_lift

[URL]http://www.canal-du-centre.be/Education/Ast/Images/stcptran.gif[/URL]
From :http://www.canal-du-centre.be/Education/Ast/En/commentcamarche.html

 

[phinds]

Arrrrgggghhhh !

Well I really HATE being wrong but the one thing I hate more than being wrong is being wrong and thinking I'm right, so I thank all the folks in this thread who helped me see the error of my ways.

What I REALLY had a hard time getting my head around was that absent any weirdnesses due to an odd shaped object bottom, no only will the object float, you CAN, despite my previous vehement objection, start with a small amount of water and end up with the object floating. You could put the object into the container, displacing most of the water, then remove the object then place it in again (somewhat carefully so it doesn't rub against the walls) and it will float again.

Color me embarrased.

 

[gulfcoastfella]

The basic situation is played out every day on the Mississippi river. Casinos aren't allowed to operate on land, so they build a 'bucket' inland that's slightly larger than the barge the casino sits on, and then they run a tiny channel from the river into the bucket. Legally, the casinos have defended what they've done by saying the barge the casino sits on is floating on the river, even though it's technically sitting on a layer which is fed by the river.

 

[DaveC426913]

BTW, it is not the one pound of water holding the boat up.The "force" that supplies the lifting for the boat is the rigidity of the very thick walls containing the water and boat. The walls must be thick enough to withstand the huge sideways pressure that the boat will exert on them (via the 1mm of water).

Think about those nested glasses again. When the inner glass is empty, it will float very high. To get it to sink manually, you have to apply pressure to push it down. This will exert pressure on the walls. Enough pressure and you could break the outer glass.

Anyone ever tried to "drown" a beachball? It is virtually impossible. It requires a lot of downward force. In a contained tank, that force gets transmitted to the walls as the water level rises. The walls must be strong.

The walls of a tank to contain an ocean liner must be strong indeed.

 

[A.T.]

The walls of a tank to contain an ocean liner must be strong indeed.

They must be just as strong as to contain just water (on the same level as with the ship). They must be about as strong as the ship hull.

 

[A.T.]

I thank all the folks in this thread who helped me see the error of my ways.

The forum would be useless, if nobody ever learned anything new here. Here is a related question:

Let's say you have an elevated water way on pillars like the upstream entrance to the ship lift:
http://3.bp.blogspot.com/_N7rE_MLfr...1600/Strépy-Thieu+on+the+Canal+du+Centre.jpg
Does the vertical load on the pillars increase, when a heavy ship passes overhead?

 

[xxChrisxx]

The water is all around - in the form of a thin film. What is important is that water exerts pressure on the hull - and it exerts pressure everywhere on the hull surface, just like it would in the case of the boat in the ocean. Yes, in theory it is possible to float an air carrier in a gallon of water.

Archamedes would disagree, or you are using a very odd definition of floating.

Floating is all down to buoyancy (the point that the buoyancy force => the object weight in air). 1 L of freshwater will give 1kgf of upthrust. The upthrust comes from the displacement and there is no appreciable change in pressure of the water. In the case of a 500 ton ship, you must be displace 500000L of water to provide the necessary upthrust.


If water is in thin film it is being used as hydrostatic bearing (which isn't the same as floating), meaning it's simply transferring the load through to an external container due to it being incompressible and the pressure increasing. It also means that the water must be contained to allow an increase in pressure.

In your drawing in a later post, you show a decreasing sized container with the aim of 'proving' that even though you make the container smaller it's still floating. However what you have drawn makes no sense in the 3rd case unless the water is contained somehow. In your drawing the top is shown as being open to air, so the water will not increase in pressure, but flow (in this case over the sides)


They are two different modes of supporting a ship. I would not agree from a technical standpoint that bearing is the same as floating.

(EDIT: I don't know if I'm agreeing with you or not, this thread is pretty hard to follow).

 

[A.T.]

In the case of a 500 ton ship, you must be displace 500000L of water to provide the necessary upthrust.

Displacement is simply the volume of the ship below the water surface. The displacement of the same ship in the tight pool is the same as in the ocean.

In your drawing in a later post, you show a decreasing sized container with the aim of 'proving' that even though you make the container smaller it's still floating. However what you have drawn makes no sense in the 3rd case unless the water is contained somehow.

So where exctly between drawing 2 & 3 does the body stop floating?

In your drawing the top is shown as being open to air, so the water will not increase in pressure,

The pressures at the floating body are the same in all 3 pictures. They must counter the weight which is the same as well.

 

[cmb]

If water is in thin film it is being used as hydrostatic bearing (which isn't the same as floating), meaning it's simply transferring the load through to an external container due to it being incompressible and the pressure increasing.

You don't appear to be acknowledging the caveats everyone has made. We have excluded 'films' in this, for precisely the reason you are repeating.

But you can have a column of water just a few mm diameter and the pressure in that water at 14m depth is the same as that at 14m depth in the ocean - it is at one atmosphere pressure (gauge).

Think about the cuboid boat, as in diagram above. The vertical sides do nothing for 'lift'. The upper side is exposed to 1 atm pressure, the lower is at 2 atm pressure (assume it is 14m down). Let's say it has a 14 million lb displacement, so at a differential pressure above and below of 1 atm it will need a hull surface area of 14 million lb/14psi = 1 million square inches.

If it has, let's say, only 500,000 sq in, then it will not be buoyant until it settles down to 28m depth. (If it is less than 28m high to the gunnels, then it'll sink!)

This is all self-balancing and all in order; a floating object will become 'buoyant' when its average density below the water line is at a sufficient differential less than the water that it can lift the rest of the object above the water line. If it's density is too high to achieve that balance at any depth, then it will sink. If the density is too low to achieve the balance at any depth, it will float out of the water and keep going up!

 

[Borek]

Floating is all down to buoyancy (the point that the buoyancy force => the object weight in air). 1 L of freshwater will give 1kgf of upthrust. The upthrust comes from the displacement and there is no appreciable change in pressure of the water. In the case of a 500 ton ship, you must be displace 500000L of water to provide the necessary upthrust.

I am afraid you are wrong on every account.

Fact that the water overflowed and is no longer present in the bucket doesn't matter. Hull occupies the place water should occupy, so it displaces the water.

Don't be so eager to put any words into Archimedes mouth. After he jumped into his bath water left the tube - but he was still lighter.

You are also wrong about the upthrust source. Upthrust that keeps object floating is a force. Displacement doesn't create force, pressure acting on the surface (ship bottom) does.

Pressure in the water grows approximately by 1 atm for every 10 meters, whether you put a ship in, or not.

 

[rcgldr]

If water is in thin film it is being used as hydrostatic bearing.

The assumption is that container isn't that tight.

...the water will not increase in pressure, but flow (in this case over the sides)

The pressure of water is a function of gravity, density, and depth, not the total volume or mass involved. The pressure of water at the bottom of a filled 30 meter tall, 1 cm diameter cylinder will be the same as the pressure of water at the bottom of a filled 30 meter tall, 1 km diameter cylinder.

 

[cmb]

Let's say you have an elevated water way on pillars like the upstream entrance to the ship lift:
http://3.bp.blogspot.com/_N7rE_MLfr...1600/Strépy-Thieu+on+the+Canal+du+Centre.jpg
Does the vertical load on the pillars increase, when a heavy ship passes overhead?

No. If we view the volume it takes up in the water as 'water-equivalent' (because the ship displaces the water, and only the water, equivalent to its mass), and also that the water level remains the same, so you can see there is no change.

If the vessel is traveling fast, however, the dynamics are more interesting. The water is being pushed up against the bow of the vessel, and therefore the vessel will rise out of the water (this effect directly shows why you can interpret buoyancy in terms of the liquid pressure on the hull). There will also be a bow-wave. We therefore have both more water and less water displaced by the boat (i.e. not only water above the common line, but a boat there too). If the vessel went fast enough and the sensors in your bridge were sensitive enough, you'd see an increase (above the normal) in load as the bow overpasses. A decrease (below the normal) as the stern follows is self-evident because there will have to be an 'average-normal' over the time interval that the vessel passes overhead.

 

[xxChrisxx]

Displacement is simply the volume of the ship below the water surface. The displacement of the same ship in the tight pool is the same as in the ocean.
So where exctly between drawing 2 & 3 does the body stop floating?

The pictures are not helpful becuase they have no information associated with them.


The condition that something does not float is when:
Upthrust < weight in air.
upthrust being = volume displaced * density of water.

So in case 3 there visually appears to be less volume of water necessary to achieve this condition.
If there is, then the pictue is wrong and the object should be touching the bottom.

If there is only 1 gallon (4.5 l) total. The maximum upthrust of fresh water is 4.5 kg. As you can't displace more water than you have. The maximum weight you can support is 4.5kg.


EDIT: This thread keeps changing too quickly, people sticking caveats in everywhere that changed the nature of the question.

 

[phinds]

Chris, you are going through EXACTLY the same thought process that I did. I vehemently opposed the correctness of the concept of floating a battleship in a bucket of water, but when you have thought it all through, you will realize, as I did, that it does indeed work.

The graphics, by the way, are selfcontained and have all the information you need to figure this out.

 

[xxChrisxx]

Chris, you are going through EXACTLY the same thought process that I did. I vehemently opposed the correctness of the concept of floating a battleship in a bucket of water, but when you have thought it all through, you will realize, as I did, that it does indeed work.

The graphics, by the way, are selfcontained and have all the information you need to figure this out.

I'm acutally going back over the thread now with a pen and paper. ;D

 

[xxChrisxx]

Pretty simple - pressure is a function of density and height of the column, doesn't matter how thin the column is.

Edit: well, at some point it starts to be too thin, and it doesn't behave as a water any longer, but even if it is just a tenth of a millimeter it is still thick enough.

It's this that causes the logical gap for me.

I know about increasing pressure and water columns. I know that buoyancy force is related to the pressure on the hull.

Yet if you have a gallon of water you can't practically get enough pressure to support a hull becuase you'd need the head to be very high. Meaning a thin column, and boats aren't that deep.

 

[A.T.]

upthrust being = volume displaced * density of water.

So in case 3 there visually appears to be less volume of water necessary to achieve this condition.

The volume of water is irrelevant to the above condtion, because it doesn't appear in it. I explained that already to you:

volume displaced = volume of the ship below the water surface

In case 3 the volume displaced is exactly the same as in the other cases.

As you can't displace more water than you have.

Wrong, see above

 

[xxChrisxx]

The volume of water is irrelevant to the above condtion, because it doesn't appear in it. I explained that already to you:

volume displaced = volume of the ship below the water surface

In case 3 the volume displaced is exactly the same as in the other cases.

I know that.

Wrong, see above

But that doesn't make sense.

I come from a background of designing buoys in open water. Which is easy becuase volume sat in the water = the volume of water displaced. So for me volume displaces is the volume of water. So I've always worked in terms of total WIA and net and gross buoyancy.

I'm sure there's things I'm missing, but I can't see what.