Floating a cruise ship in a bucket of water

[DaveC426913]

But a problem arises at the bottom, because there is shallow water: a simple thought to the Bernoulli theorem tells that the increase in speed is accompanied by a decrease of pressure. So the bottom of the ship can hit the canal’s bottom very hardly , and this, together with pulling, causes a series of indents and severe damages to the bottom plating.
This isn’t a joke, in the past the problem arised a lot of times, and I’ve repaired many ships, damaged that way.

Yes, I've heard of this. Ships navigating shallow canals ride lower the faster they move.

 

[hutchphd]

I think this is also why floats on seaplanes have the characteristic "step" halfway down the length to create turbulence so they will not be pulled down. Down is the wrong direction for an airplane...

 

[italicus]

So in the case of ships a good provision is to change the trim , if possible, increasing the aft draft and decreasing the fore one. This can be done moving ballast water towards aft.
Of course, everything is to be done “ cum grano salis” ( salt in your brain!)

 

[Charles Link]

That is an interesting point. But remember that 1 atm is 15 PSI. So the cover slip has 5 pounds of force pushing it down as long as there is a water seal (independent of viscosity and surface tension). The details of that seal do depend upon these and geometric factors. Air must invade the interface. This doesn't affect lateral forces, and it does not really affect buoyancy. It greatly affects attempts to lift objects out of a buoyant state until the seal is breached.

See https://www.physicsforums.com/threa...e-surface-of-water.944587/page-3#post-5977523 post 58 by @haruspex . There will likely, IMO, be some thickness at which the water starts to behave like a water-tight seal, and loses some of its buoyancy properties.

 

[hutchphd]

Suppose I have a rubber duck in the bathtub. Everyone agrees it is floating. Now I spray a very high tensile strength (no surface tension) coating on the water (assume it is very duck-phillic and seals to ducky). Is ducky, by your definition, still floating? How come?

 

[DaveC426913]

There will likely, IMO, be some thickness at which the water starts to behave like a water-tight seal, and loses some of its buoyancy properties.

Certainly, if you tried to lift the ship out of its container, it will definitely resist. (Heck, this happens whenever I try to pull the trash bag out of my kitchen trash bin. It takes, like five minutes to come out!)

But how does it follow that that means it's not floating? Just because you can't lift it from its neutrally buoyant position doesn't mean it's stuck there and not floating. The forces keeping it in place are zero when its stationary, but rise rapidly as you attempt to move it, because water has a very tough time flowing into (or out of) the voided space.

 

[haruspex]

See https://www.physicsforums.com/threa...e-surface-of-water.944587/page-3#post-5977523 post 58 by @haruspex . There will likely, IMO, be some thickness at which the water starts to behave like a water-tight seal, and loses some of its buoyancy properties.

Isn't it just a rate issue? The thinner the layer the greater the force needed to get the water to flow at a given rate. You just need to be patient.

 

[Rene Dekker]

We take a typical cruise ship size, with a length of 202m, width of 28m, and draft of 6.3m (I took the first one from https://www.cruisemapper.com/wiki/753-cruise-ship-sizes-comparison-dimensions-length-weight-draft). Then the surface area of the ship under water is 2 x 6.3 x (202 + 28) + 202 x 28 = 8554 m3 . The volume of a bucket of water we assume to be 20L = 0.02 m3. If we assume the water thickness on all sides of the ship to be the same, then this will give a water layer thickness of 2.3 micrometer. That is, if the bucket follows the shape of the ship to an accuracy of about 1 micrometer, then the ship will be surrounded by water on all sides.

Now, can we call this "floating"? We consider three aspects of that. First of all, afaik, the thickness of a water layer sticking to a steel plate due to adhesion forces, is thicker than 2.3 micrometer. That is, when you remove the bucket, then the water would still cling to the ship on all sides.

Secondly, 2.3 micrometer is less than the diameter of capillary tube. Therefore, the water on the sides of the ship will creep up between the ship and bucket wall, due to capillary forces. That will remove water from the bottom, reducing the thickness of the layer on the bottom. I have not done a calculation on how high the water will go up, and of course that also depends on the shape of the bucket above the "water line".

Thirdly, the cohesion forces of the water would give the thin layer at the bottom extra buoyancy force. I have not done the calculations, but there is a chance that if you remove the walls of the "bucket", but leave its bottom in place, then ship would still "float" on the thin film of water underneath it.

So do we still call this "floating" in the sense that the question asks about? I would say no, but you'll be your own judge.

 

[DaveC426913]

That's a really insightful analysis.

 

[italicus]

We take a typical cruise ship size, with a length of 202m, width of 28m, and draft of 6.3m (I took the first one from https://www.cruisemapper.com/wiki/753-cruise-ship-sizes-comparison-dimensions-length-weight-draft). Then the surface area of the ship under water is 2 x 6.3 x (202 + 28) + 202 x 28 = 8554 m3 . The volume of a bucket of water we assume to be 20L = 0.02 m3. If we assume the water thickness on all sides of the ship to be the same, then this will give a water layer thickness of 2.3 micrometer. That is, if the bucket follows the shape of the ship to an accuracy of about 1 micrometer, then the ship will be surrounded by water on all sides.

First of all, a surface is expressed in $m^2$ and not $m^3$ .
But I haven’t understood your determination of the wetted surface area (WSA) of the hull . There are some empirical formulas, see f.i. this , page 6 :

1 Evaluation of wetted surface area of commercial ships as ...https://repository.si.edu › bitstream › handle › Mil...

there are others indeed.
Anyway , the point isn’t this. Do you think that the external hull surface is as smooth as a mirror? It isn’t. You cannot speak of a water layer of some micrometers, uniformly distributed on the whole wetted surface. Have you ever examined the hull surface from a distance , say, of 20 cm ? The surface is not uniform at that level you are considering.
Moreover, the Archimede’s principle , which we can define really a “law of nature” , is a matter of global hydrostatics, doesn’t take into consideration molecular forces of adhesion and cohesion forces of water. This law can be summarised in the vector equation :

$$\ vec\ P + \ vec\ S = 0 $$

nothing more.

 

[Rene Dekker]

First of all, a surface is expressed in $m^2$ and not $m^3$ .

Thanks for pointing out that typo.

But I haven’t understood your determination of the wetted surface area (WSA) of the hull .

I took the simplification that the underside of the ship is simply rectangular. We are talking about an imaginary theoretical exercise anyways, so there is no point in discussing all kinds of practical details. We just need an idea of what order of magnitude we are talking about.

Anyway , the point isn’t this. Do you think that the external hull surface is as smooth as a mirror? It isn’t. You cannot speak of a water layer of some micrometers, uniformly distributed on the whole wetted surface. Have you ever examined the hull surface from a distance , say, of 20 cm ? The surface is not uniform at that level you are considering.

We are talking about an imaginary theoretical exercise anyways, so there is no point in discussing all kinds of practical details. And I stated that the shape of the "bucket" will have to follow the surface of the ship (whatever shape it has) to an accuracy of 1 micrometer.

Moreover, the Archimede’s principle , which we can define really a “law of nature” , is a matter of global hydrostatics, doesn’t take into consideration molecular forces of adhesion and cohesion forces of water. This law can be summarised in the vector equation :

$$\ vec\ P + \ vec\ S = 0 $$

nothing more.

Fully agree with that, Archimede's principle does not take into consideration molecular forces. And 2.3 micrometer is definitely a scale where molecular forces must be taken into consideration.

 

[Charles Link]

The vector formula I believe you are referring to above, (the Latex seems to be missing a couple things), with force(s) per unit volume is $-\nabla P-\delta g \hat{z}=0$, for hydrostatic equilibrium. When integrated over a volume, along with a version of Gauss' law for the gradient, the result is Archimedes principle.

 

[gleem]

What keeps a boat afloat? The force of the water pressure on the bottom surface. What is the pressure depend on? The distance from the surface. The height of the surface of the water above the lowest part of the boat. If you come up underneath a floating boat with a container of arbitrary size but large enough to fit the boat and bring the container top edge to the surface of the water the boat remains buoyant and will float in that container. If you continue to raise the container nothing in the container change, the boat is still floating. If the container is made to fit the hull shape to arbitrary close tolerances then the boat can float in a lot less water than it can displace.

A boat float on its waterline no matter how much water is under or around it.

 

[hutchphd]

absolutely

Here's my method:

 

[italicus]

The vector formula I believe you are referring to above, (the Latex seems to be missing a couple things), with force(s) per unit volume is $-\nabla P-\delta g \hat{z}=0$, for hydrostatic equilibrium. When integrated over a volume, along with a version of Gauss' law for the gradient, the result is Archimedes principle.

Thank you Charles, I am not able with Latex ! The letter “S” I wrote is the initial of the Italian word “ Spinta” ( push), which stays for the Archimede’s force that makes equilibrium to weight P .

 

[italicus]

A boat float on its waterline no matter how much water is under or around it.

Absolutely agreed.

 

[Charles Link]

Thank you Charles, I am not able with Latex ! The letter “S” I wrote is the initial of the Italian word “ Spinta” ( push), which stays for the Archimede’s force that makes equilibrium to weight P .

I'm using $P$ for pressure, $\delta$ for the density of water, and $g$ for the gravitational constant.

 

[haruspex]

Archimede’s principle , which we can define really a “law of nature”

the Archimede’s force that makes equilibrium to weight P .

I would caution against such a view of Archimedes' Principle.

The buoyancy comes from the integral of the upward component of the fluid's pressure over the surface in contact with the fluid. Archimedes' insight, I assume, is that if we replace the body by more of the fluid occupying the same displacement it would be in equilibrium, so the buoyancy force on that would be equal to its weight.

But note that there is a hidden assumption in equating that to the buoyancy force on the object: that the fluid is able to reach all parts of the body's surface below the surface level of the fluid. For example, the Principle breaks down for a rubber suction cap stuck to the bottom of a tank.

 

[italicus]

I would caution against such a view of Archimedes' Principle.

The buoyancy comes from the integral of the upward component of the fluid's pressure over the surface in contact with the fluid. Archimedes' insight, I assume, is that if we replace the body by more of the fluid occupying the same displacement it would be in equilibrium, so the buoyancy force on that would be equal to its weight.

Obvious!

But note that there is a hidden assumption in equating that to the buoyancy force on the object: that the fluid is able to reach all parts of the body's surface below the surface level of the fluid. For example, the Principle breaks down for a rubber suction cap stuck to the bottom of a tank.

Could you explain better your example, please? Maybe I haven’t understood it!
As a marine engineer and naval architect, with 40 years of experience in the field, I’ve always found the buoyancy by $\rho g V$ , where V is the immersed volume of a floating body, or even a totally submerged body ( think of a submarine for example) , that displaces water.

One moment...do you mean something like this in the sketch?

but it is obvious , the surface of the cup inside the tank is to be considered “part “ of the tank bottom, no component of water pressure on it is directed towards the surface of tha water!

 

[haruspex]

but it is obvious , the surface of the cup inside the tank is to be considered “part “ of the tank bottom

You can choose to consider it that way, but if you blindly apply Archimedes' Principle to the cap alone it does not work. Hence the need for caution.

 

[italicus]

You can choose to consider it that way, but if you blindly apply Archimedes' Principle to the cap alone it does not work. Hence the need for caution.

Believe me, I have never been so blind...In any case, caution is often useful or even necessary, in all circumstances of science and technology. I agree on that.

 

[Charles Link]

To expound upon what @haruspex is describing, consider a very flat glass plate that is completely submerged, where it will experience a buoyant force by Archimedes equal to the weight of the water of the same volume as the glass plate. Then consider what occurs if you press it against a very flat bottom as it is submerged in a deep tank. It will be stuck to the bottom with a tremendous pressure from the water above, but, as there is no water underneath it to counter, the buoyant force given by Archimedes simply does not hold for this second case.

 

[hutchphd]

as there is no water underneath it to counter

I would say this a little differently: as there is no possibility of water influx (thereby communicating the external pressure)...
Seems a clearer statement to me and some clarity appears to be required.

 

[haruspex]

Believe me, I have never been so blind...In any case, caution is often useful or even necessary, in all circumstances of science and technology. I agree on that.

The difficulty is that Archimedes' Principle is generally expressed as "the buoyant force is equal to the weight of the fluid displaced". I never see it hedged around with any caveat. Your phrasing in post #70 sought to elevate it to a universal law, but in that form it is not.
You might not make the mistake of applying it where it is invalid, but I have seen students do so on this very forum.

 

[Charles Link]

Keeping post 82 in mind, and looking back on post 64, there will be some thickness (very thin layer) of water $d$ that we lack data for, where the air pressure on the boat from above will not be countered or transferred to the water below the boat, and for this case we can ask whether the boat is truly floating. This makes it into a somewhat nebulous problem, and it may be beyond the scope of what @DaveC426913 is looking for.

 

[italicus]

The difficulty is that Archimedes' Principle is generally expressed as "the buoyant force is equal to the weight of the fluid displaced". I never see it hedged around with any caveat. Your phrasing in post #70 sought to elevate it to a universal law, but in that form it is not.
You might not make the mistake of applying it where it is invalid, but I have seen students do so on this very forum.

Dear haruspex
I am not a student! I have sometimes taught this subject to them, and still consider Archimedes’ principle a law of nature, when free floating bodies are concerned. The rubber cup you have proposed isn’t a free floating body, no?
But a more difficult example which I was used to make is the following: consider a heavy rubber ball, (but lighter than an equal volume of water) , kept under water and tied by a rope to the bottom of a tank at rest on Earth What is the tension in the rope?

 

[haruspex]

still consider Archimedes’ principle a law of nature, when free floating bodies are concerned

1. The principle as commonly stated does not specify free-floating
2. In fact, it is not limited to free-floating. It works perfectly well for a body resting on the bottom provided the fluid can reach under it.
3. Ergo, if you wish to teach it well, mention that proviso.

 

[italicus]

1. The principle as commonly stated does not specify free-floating
2. In fact, it is not limited to free-floating. It works perfectly well for a body resting on the bottom provided the fluid can reach under it.
3. Ergo, if you wish to teach it well, mention that proviso.

I have made an example where the ball is tied by a rope to the bottom of a tank, three forces act on the ball, one of which is Archimedes’ force. But what is your position? Mine is that this principle is a law of nature, Archimede simply discovered it, didn’t invent, when playing with the king’s crown in his bathtube, according to the legend…
If a cubic block of steel rests on the bottom of a pool, a thin layer of water underneath is enough to give a hydrostatic force upwards. But you really know nothing about how many contact points are between the two surfaces, which aren’t as smooth as a mirror, and how many small areas are truly supported by water: you cannot really speak of a uniform water layer, so what are we talking about?

You have to “weld” the block to the bottom, or seal accurately the perimeter of contact to prevent water entering underneath, so that Archimedes’ law isn’t applicable.
If a ship is moored to a berth by means of ropes, one should consider also the forces exerted by the ropes, to determine a correct ship’s weight, but this way takes to incorrect evaluation; better, one has to release ropes and read drafts; but the water level isn’t really stable around a draft mark , you can have excursions of 10 cm or more, due to sea and ship’s motion. So you have to use the famous “grain of salt” , which is another way to mean experience. The mean draft of the ship gives you the displacement, its accuracy is as good as possible, everybody in the shipping world knows that.

But I think we have repeated this concept of Archimedes’ law too many times, nothing more needs to be added.

 

[Rene Dekker]

To expound upon what @haruspex is describing, consider a very flat glass plate that is completely submerged, where it will experience a buoyant force by Archimedes equal to the weight of the water of the same volume as the glass plate. Then consider what occurs if you press it against a very flat bottom as it is submerged in a deep tank. It will be stuck to the bottom with a tremendous pressure from the water above, but, as there is no water underneath it to counter, the buoyant force given by Archimedes simply does not hold for this second case.

I would judge that in this case, where the water cannot get beneath the object, there is still the same buoyancy force, equal to the weight of the displaced water. But there is an extra downwards force on the object, due to the pressure difference of the water above the object with the near-vacuum beneath the object.

For example, if we measure the force that the object exerts on the floor, then that is still equal to the weight of the object minus the weight of the displaced water. That is, the object still tries to pull the floor up with the buoyancy force. It is just that the floor is able to counteract that force to keep the object down.

 

[italicus]

I would judge that in this case, where the water cannot get beneath the object, there is still the same buoyancy force, equal to the weight of the displaced water. But there is an extra downwards force on the object, due to the pressure difference of the water above the object with the near-vacuum beneath the object.

Be careful. If the water cannot get beneath the object, because the object is “welded” or accurately sealed around the contact perimeter, there is no buoyancy force on the object! That was the first example by Haruspex, the one of the rubber cup in a hole of the bottom tank. No buoyancy force means that the object’s weight is entirely supported by the bottom tank.

For example, if we measure the force that the object exerts on the floor, then that is still equal to the weight of the object minus the weight of the displaced water. That is, the object still tries to pull the floor up with the buoyancy force. It is just that the floor is able to counteract that force to keep the object down.

No, read above. The floor ( or the tank bottom) doesn’t have to counteract any buoyant force to keep the object down. For clarity, we are speaking of objects heavier than an equal volume of water, which therefore tend to sink when left free.
Instead, consider the example that I have posted in #86 , of a rubber ball tied to the bottom by means of a rope, and evaluate the tension in the rope. There are 3 forces acting on the ball , which are in equilibrium.