Ocean liner in bucket full of water

[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.

 

[A.T.]

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.

Correct

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.

Interesting effect when you go with a ship through a narrow channel, with reed at the bank. The water in front of the ship seems calm, but the reed bends towards the ship, long before the bug wave reaches it.

 

[A.T.]

volume displaced = volume of the ship below the water surface

I know that.

Then stick to it and don't confuse it with the volume of water outside of the ship.

 

[gulfcoastfella]

View the film of water holding the boat as being surrounded by boundary conditions. It doesn't matter what is holding the water in place, as long as the water experiences pressure on it's outside surface. The boundary condition means it can be water, steel, concrete, anything, as long as it is fortified enough to withstand the pressure.

 

[jmmccain]

boats aren't that deep.

Only as deep as they need to be... A boat doesn't care how close to shore it is - only that it has enough draft (or head, or whatever else floats your boat).

 

[DaveC426913]

View the film of water holding the boat as being surrounded by boundary conditions. It doesn't matter what is holding the water in place, as long as the water experiences pressure on it's outside surface. The boundary condition means it can be water, steel, concrete, anything, as long as it is fortified enough to withstand the pressure.

Yes. This is important. The .1mm layer of water next to the the ship needs to be held in place by something. In the open ocean it's more water; in a tank it's the walls. This is what ultimately holds up the 100,000 tonne vessel.

 

[dacruick]

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

Chris, I as well went through the exact same thing (as I'm sure you are finding out as you read over the thread), I realized I was wrong but I'm still not sure why. If I think about the concept of floating and extend it to this thought experiment, I show up with misconceptions, an intuitive crux. But if I start thinking about the water holding the ship in place in this bucket (pretty much the same thing) it helped me think in a less limited way about the situation. I'm still ironing it out so I'd like to hear what conclusion you come to.

 

[nasu]

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.

Maybe if you do it very fast and the water has no time to escape, you may create a large pressure for a while and break the glass. I think this is what you have in mind, right?
Otherwise, at equilibrium, the pressure on the walls of the outer glass should not be more than the pressure when you have a glass full of water.

 

[mrspeedybob]

Some real calculations might be helpful to this thread.

Suppose I have a rectangular boat that is 100 meters wide and 1,000 meters long. It weighs 1,000,000 metric tons. The density of water is about 1 metric ton per cubic meter, for the sake of this calculation we will assume it is exactly that. Since the boat weighs 1,000,000 metric tons it will displace 1,000,000 cubic meters of water. We know the width and length of the boat multiply to 100,000 square meters so the depth of the hull in the water must be 10 meters. 10x100x1000=1,000,000. Since the sides of our boat are vertical there is no upward thrust on them, the only pressure that is relevant is the pressure at the bottom of the hull. The only surface it acts on is the flat surface at the bottom. As established earlier the density of water is 1 metric ton per cubic meter so the pressure at 1 meter of depth is 1 metric ton per square meter, this pressure arises from nothing more then the weight of the water pushing down from above. At a depth of 10 meters the pressure is 10 tons per square meter. Now suppose the column of water is not 1 square meter but 1 square cm. A column of water 1 cm x 1 cm x 1000 cm weighs 1 kg. The pressure at the bottom is 1 kg per square cm. There are 10,000 square cm in a square m so that is equivalent to a pressure of 10,000 kg per square meter, 10 tons per square meter. So the pressure is the same regardless of how wide or long the column of water is.

Now let's try to float our boat in a container that is 100.02 meters wide, 1000.02 meters long, and 10.01 meters deep. Each side of the boat is up against a column of water that is 1000.01 meters long, 10 meters deep, and .01 meters thick. That is 100.001 cubic meters of water on each side. Each end is a column of water 100x10x.01= 10 cubic meters. Since the boat has 2 sides and 2 ends the total volume of water (excluding the water that is under the hull) is 220.002 cubic meters. The surface area of this column of water is (1000.01x.01x2)+(100x.01x2)=22.0002 square meters. 220.002 metric tons per 22.0002 square meters = 10 tons per square meter.

10 tons per square meter of pressure applied to the 100,000 square meters of the bottom of the boats hull = 1,000,000 tons of upward force. This exactly equals the weight of the boat so it moves neither up, nor down. It floats with 1 cm of water beneath it.

That 1 cm layer of water is a total volume of 1000.01x100.01x.01=1000.110001 cubic meters. So our 1,000,000 ton ship is floating in only 1220.112001 tons of water.

Narrowing the margins from 1 cm to 1 mm or .1 mm does not change the physics.

 

[cmb]

Some real calculations might be helpful to this thread.

D'you mean like in #61?

 

[cmb]

the only pressure that is relevant is the pressure at the bottom of the hull.

No air pressure on the top?

 

[jmmccain]

Take all of the air away and the boat would still float.

 

[cmb]

Take all of the air away and the boat would still float.

I thought you'd have explained where/how the air pressure acts, if you wanted to give a thorough calculation.

The pressure in the water is the water column height and the air pressure above it, and on top of the ship is just the air pressure.

(I suspect that some of those new super-cruise ships, the air pressure is significantly higher in the water (it will carry sea level pressure) than that on top of its 40m height!)

 

[logics]

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..

There is no arguing: the principle is true and anyone can verify it, on a smaller scale.
If you put a heavy steel pan in any slightly-bigger vessel you'll see it floating, as long as water surrounds it up to *its waterline. [*where weight of submerged volume of water (displaced water)= its total weight]
But you get a most striking demonstration when you put it on the bottom of the empty vessel: you won't believe your eyes seeing it rise as you pour just a little water.

btw: probably it takes at least two buckets to float a liner

 

[DaveC426913]

Yeah, I'd love to have someone build a model and verify this for all. I know it works but it is extremely hard to explain to someone who can't see it.

Too small scale doesn't work. A teaspoon of water in a tumbler won't prove anything to a skeptic.

Question: how big a scale do you think a demo would need to be to prove the point?

I wonder what Mythbusters is doing these days?

 

[phinds]

Yeah, I'd love to have someone build a model and verify this for all. I know it works but it is extremely hard to explain to someone who can't see it.

Too small scale doesn't work. A teaspoon of water in a tumbler won't prove anything to a skeptic.

Question: how big a scale do you think a demo would need to be to prove the point?

I wonder what Mythbusters is doing these days?

Dave, I think an example with just a couple of drinking glasses would do the trick. I think this was mentioned somewhere above. Fill one with water and put the other one in, weighing it down with coins or something until it sinks down almost to the bottom. Now THAT is not going to prove it to a skeptic, but you then take the upper glass out, note that there is VERY little water in the lower glass and then put the upper one in again and note that it DOES float. If that won't convince'm probably nothing would. Of course, when this thread started, *I* probably would not have been convinced by that. I just was not willing to "get" it at first.

I love your idea of getting the mythbusters to do it with something REALLY large.

 

[DaveC426913]

Dave, I think an example with just a couple of drinking glasses would do the trick. I think this was mentioned somewhere above. Fill one with water and put the other one in, weighing it down with coins or something until it sinks down almost to the bottom. Now THAT is not going to prove it to a skeptic, but you then take the upper glass out, note that there is VERY little water in the lower glass and then put the upper one in again and note that it DOES float. If that won't convince'm probably nothing would. Of course, when this thread started, *I* probably would not have been convinced by that. I just was not willing to "get" it at first.

I love your idea of getting the mythbusters to do it with something REALLY large.

I can see skeptics believing a 150g glass can float on a teaspoon of water while still not believing that a 100,000 tonne ocean liner can float on a bucket of water. A teaspoon of water compared to the weight of the glass is just just the right amount. It just doesn't scale down in a way that's incontrovertably convincing.

It wouldn't have to be anything near full scale (that could be a bit impractical ), but you'd want it large enough that the weight of the object is orders of magnitude greater than the weight of the water.

 

[phinds]

... It just doesn't scale down in a way that's incontrovertably convincing.

It wouldn't have to be anything near full scale (that could be a bit impractical ), but you'd want it large enough that the weight of the object is orders of magnitude greater than the weight of the water.

Yeah, I can't argue with that.

I logged onto the Myth Buster's site to see about suggesting it, but it seems to require setting up a password with them, and I don't do that. It just gets you spam.

 

[DaveC426913]

Yeah, I can't argue with that.

I logged onto the Myth Buster's site to see about suggesting it, but it seems to require setting up a password with them, and I don't do that. It just gets you spam.

You've got a password here.


I'll write to them.

 

[phinds]

You've got a password here.

Yes but I CARE about being here (and the one other forum I'm on) ... don't care about being there.

I'll write to them.

Excellent

 

[jmmccain]

Perhaps a pair of nestable garbage cans would be large enough? That's a fair amount of water to be displaced and it's still cheap. Except for the coins...