I am confused.

Some sources say it is possible at least theoretically ( http://www.wiskit.com/marilyn/battleship.jpeg ) and some say it is not true ( http://blog.knowinghumans.net/2012/09/a-battleship-would-not-float-in-bathtub.html )

Is it necessary or not that there exists an amount of water around the ship that weights at least the same as the weight of the ship?

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    $\begingroup$ Canal narrow boats in the UK do this "experiment" every day. See luphen.org.uk/images/2004/2004-06/2004-06-05-153833.jpg for example. (image found by google) $\endgroup$ – alephzero Jan 10 '17 at 23:44
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    $\begingroup$ In the second link, the author assumes the mass of water displaced refers only to the small amount that was present before and has moved around. This is not the case. Archimedes principle assumes the water level does not change (in fact, he postulated for a bathtub filled to the brim to make this point explicitly clear). The displaced water we speak of is the mass of water that has the same volume as that of the object below the water line $\endgroup$ – Jim Jan 11 '17 at 14:11

Yes a ship can float in a big bath tub with very little water. No you do not need as much water as the weight of the ship. In theory, you can use less than a cupful!


Suppose the ship is floating in the ocean, far from the sea bed or any shore. The layer of water in contact with the ship (let's say a layer of thickness 1mm all round the submerged surface) is providing enough force ("upthrust") to support the whole weight of the ship. This layer of water is in equilibrium vertically : it is pulled down by gravity (its own weight), pushed down by the weight of the ship above it, and pushed up by the water below it. It is in equilibrium horizontally : it is pushed outwards by the ship, and pushed inwards by the surrounding water.

Suppose the surrounding water beyond the contact layer is replaced by a thick concrete wall like a dam, many meters thick. The concrete wall is pushing both horizontally and vertically on the contact layer of water next to the ship. It makes no difference to the contact layer whether it is being pushed by other water or by the concrete wall. In both cases the contact layer remains in equilibrium and does not move. It does not, for example, get squeezed upwards and outwards between the concrete wall and the ship. That process of adjustment has already taken place, when the ship was launched or loaded.

The pressure in the contact layer of water varies only with its depth, not with its thickness. If there is more or less water between the ship and the wall, the pressure at that depth does not change.

The concrete wall or dam can be replaced by an enormous bath tub, providing that it is strong enough to exert the same force which the wall exerted.

Comment on Explanations by "Marilyn" and Brian Holtz

My explanation is essentially the same as Marilyn's (wiskit.com) - without the benefit of her excellent diagrams. The only difference is that Marilyn starts with water in the dock and replaces most of it with the battleship. I start with the ship floating in the ocean and replace most of the ocean water with dry dock.

Brian Holtz (Knowing Humans blog) is incorrect. His reasoning (accessed 12 January 2017) is not very clear to me, so I apologise to him if I have misunderstood it. His arguments are as follows :

1. The bathtub must initially contain sufficient water to be displaced by the ship when it floats.

eg If the battleship (USS Missouri, for example) weighs 45,000 tonnes then the bathtub must initially contain at least 45,000 tonnes of water.

The volume of the bath tub does need to be at least equal to the volume of 45,000 tonnes of water. However, it is not necessary for water to be actually displaced from the bathtub and to overflow from it. A small amount of water displaced upwards is just as good as an ocean displaced sideways.

If the battleship is lowered gradually into the bathtub and fits snugly into it, the cupful of water at the bottom will be squeezed up into the gap, increasing the depth of water. With only a cupful of water this will happen very quickly when the ship is almost in place. As this water moves upwards the upthrust which it provides increases. Eventually the upthrust is sufficient to support the whole weight of the ship.

As Deep says in his answer, and Jim in his comment, the volume of water displaced in Archimedes' Principle refers to the volume of the ship which is below the final waterline, not below the initial waterline. We cannot, of course, float a battleship in a bathtub which is only 1m deep, however wide it is. The draught of the USS Missouri is 8.8m, so our bathtub must be at least this deep. It must also be at least as wide and long as the ship at this height above the keel.

2. The battleship could not be floated by pouring an arbitrarily small amount of water into the gap between the ship and the bathtub. "You can't do the enormous work of lifting a massive ship merely by balancing it against a small mass of water."

Not correct. Hardly any work needs to be done to float the ship. Floating it is just a question of redistributing the load from direct contact with the bathtub to indirect contact via the layer of water. The ship does not need to be lifted up any further than say 1 micron - just enough to ensure that it is not pressing directly on the bath tub at any point.

To create the narrow gap the ship could rest with its whole weight on the bottom of the bathtub (both the ship and the tub must be enormously strong to do this) while narrow supports, perhaps 1 mm thick, at the sides prevent it from making contact with the sides of the bathtub. Water could easily be poured into this gap under its own gravity - there is nothing to prevent it falling down, until it reaches the level of the water already poured in.

As the water gets deeper it gradually exerts more pressure on the ship, so there is less pressure on keel of the ship (directly from the bathtub) and more on the rest of the underneath surface (from the water, which is in turn pushing on the rest of the bathtub). When the water is deep enough (at least 8.8m for USS Missouri) the pressure from it will be enough to support the whole weight of the ship and it will no longer exert any contact force directly on the bathtub.

To float the battleship any finite distance higher up in the bathtub (say 5mm higher) it is only necessary to add more water to the tub. However, the amount of water required to do this could be very large because raising the ship will increase enormously the volume of the gap which needs to be filled. A gap which is initially "snug" does not remain "snug" as the ship rises vertically.

The work done will of course also be enormous : the weight of the battleship times the distance moved upwards. However, this work is done by gravity, acting on the extra water. If the extra water is already in a reservoir above the current water level, gravity will carry it down into the tub. But if it is necessary to pump this extra water up to the current water level from below the level of the keel, the energy required to do so will be at least equal to the work done in raising the 45,000 tonne battleship by 5mm.

So as Brian Holtz says, There is no free lunch. Lifting the battleship even by 5mm requires enormous energy. But this is not the same as getting it to float, which is just a question of transferring the weight from the keel to the contact layer of water.

3. In canal locks there is sufficient clearance all round (behind and in front as well as at the sides) to hold a volume of water equal to the weight of the ship above the original water level.

Not necessarily true. There is no reason why a rectangular ship (eg a barge) cannot "dock" with clearance of say only 6" on either side and below. The water it displaced has been pushed aside and behind it. A dam can then be erected behind it, again with only 6" clearance. When the dam is strong enough, the water behind it can be pumped away, leaving the barge afloat in an isolated "dock" which contains much less than its own weight of water.

The same thing happens in canal locks when the ship occupies almost the total volume of the lock. After the lower gate is closed, water is allowed to fall in from the lock above, raising the height of the ship. This requires enormous energy, but it is all done by gravity, courtesy of rainwater and reservoirs.

4. In the no-overlow scenario, there is enough room at the top of the bath tub to hold the mass of water which balances the floating object.

It is not clear what this means. If it means that the volume of the empty bath tub must be big enough to hold a volume of water equal to the ship's weight, then this is not in dispute. eg See Marilyn's diagrams. The bath tub needs to be at least as deep as the draught of the battleship. But the final amount of water in it can be very small. It is the weight of the missing ("displaced") water below the waterline which is crucial, not the weight of the water which remains.

If it means that the weight of water in the gap must be at least as much as the weight of the ship, this is false, as argued in #1 and #3.

5. When large machines like telescopes "float" on a thin film of lubricating oil, the oil is kept pressurized in a sealed system, and the pool of oil is not open to the atmosphere.

True, but this is a matter of convenience, not necessity. The telescope could float just as well on a much "deeper" film of oil which is open to the atmosphere. Sealing the vessel enables high pressure to be achieved fairly uniformly and with the minimum amount of oil. When open to the atmosphere, the balancing depth (not mass) of oil provides the required pressure.

  • $\begingroup$ I don't understand how this doesn't violate Archimedes's principle. If any object in a fluid is buoyed up by a force equal to the weight of the displaced fluid, and if the amount of fluid can be arbitrarily low (in theory), then the buoyant force can be arbitrarily low... which means that the ship should sink if there isn't much of the fluid. Where am I going wrong? $\endgroup$ – Mehrdad Jan 11 '17 at 8:03
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    $\begingroup$ @Mehrdad The Archimedes Principle is really a "mnemonic" to calculate the nett upward force that is the sum of all the forces acting across the interface between fluid and floated / steeped object. This force is the same vector value as the weight of the fluid that would fill the volume occupied by the submerged part of the object. It is not a literal weight of displaced fluid, and no fluid has to be actually "displaced" to create it: i.e. there is no process whereby the force arises as the fluid is displaced: the force is simply the sum of forces on the body's boundary from the fluid. $\endgroup$ – WetSavannaAnimal Jan 11 '17 at 9:54
  • $\begingroup$ @Mehrdad See my answer here where I show how to sum these forces in the general case, and show that it leads to Archimedes principle in the case where the pressure varies linearly with depth, which is the case of an incompressible fluid in a uniform gravitational field. Hopefully you can see that Archimedes Principle will actually be incorrect if the pressure varies nonlinearly with depth, e.g. fluid in a nonuniform gravitational field or for a compressible fluid. $\endgroup$ – WetSavannaAnimal Jan 11 '17 at 9:59
  • $\begingroup$ @Mehrdad See my comment about bearings as well. $\endgroup$ – WetSavannaAnimal Jan 11 '17 at 10:28
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    $\begingroup$ This bit about enormous pressure required to pump in the water from below is misleading. The pressure required is the same as the pressure required to pump water up to the top of the waterline, which is enormous only for unrealistically large ships. (We can easily pump water to the top of 10-story buildings, for example; that's how faucets work on the 10th floor.) $\endgroup$ – Peter Shor Jan 11 '17 at 12:42

Yes, a ship can float in a tub. The second link you provided (to this blog post) is wrong. @sammy gerbil wrote a good answer explaining why it's possible, so I will reply to the points made in the incorrect blog post one by one.

They ignore Archimedes' assumption that the body of water has enough water and enough unused volume to combine together to balance the weight of the immersed body.

It's unclear what "unused" water and volume are, or what it means for them to "combine"; this argument is too poorly-specified to be rebutted.

A floating ship is hydraulically balanced against the mass of the top layer of water that the ship has displaced upward in the body of water on which the ship floats. A ship can only float if the body of water can contain that top layer of water and that water has a mass equal to that of the ship.

What is a "top layer" of water? How thick is this "top layer" supposed to be? The claims here are again too vague to be meaningful.

Another way to think about it is to ask whether the battleship in the empty bathtub could be floated simply by pouring in the water to fill the bathtub around it. Battleship floaters claim this would work with an arbitrarily small amount of water. But there is no free lunch -- you can't do the enormous work of lifting a massive ship merely by balancing it against a small mass of water.

So-called "battleship floaters" are correct. A very small mass can absolutely raise a very large mass; for example this happen when you use a lever for mechanical advantage (like playing see-saw with one adult and one child). The battleship, though huge, will only rise a very tiny amount whereas the water will fall a much longer distance when you pour it in.

Let us suppose the battleship has mass $m$ and the volume submerged in water is $v$. Its cross-sectional area at the water line is $a$. This means the average depth of the bottom of the battleship (averaged over unit of cross-sectional area) is $v/a$.

The ship sits in a tank with cross-sectional area $A$ at the water line. Then if the ship were to rise an infinitesimal height $\mathrm{d}h$, its gravitational energy would increase by $m g \mathrm{d}h$. Imagining the water were stationary, the ship would leave an empty volume behind it of $a \mathrm{d} h$. Allowing the water to flow from the surface into the empty volume left behind by the ship, the water's gravitational potential energy would decrease by $\rho a \mathrm{d}h g f$ as it flowed in, with $f$ the mean distance the water falls, but the water simply falls to where the bottom of the ship was, an average fall of $v/a$. So the drop in energy of the water is $\rho g v \mathrm{d}h$. If we are to have equilibrium, we need the energy gained by the ship to equal the energy lost by the water, or $m g \mathrm{d}h = \rho g v \mathrm{d}h$, or

$$ m = \rho v$$

that is the volume of the ship under the water line, times the density of water, must be equal to the mass of the ship. However, nowhere did our calculation involve $A$, the cross-sectional area of the tub or $V$, the volume of water. These things are simply not relevant to the calculation and can be made very small. There is no problem with conservation of energy in floating a battleship. In fact, demanding energy be conserved gives us Archimedes' principle and demonstrates that a small volume of water can float the ship.

Some floaters point to canal locks (e.g. the Miraflores in Panama) that can float a ship with just a foot of clearance on the sides (and allegedly the bottom). However, they ignore the clearance at the front and back. After a ship enters a canal lock, you can bet that there is a new top layer of water (relative to the prior water level) whose mass is equal to that of the ship.

I don't know the details of canal locks in practice, but there is no such requirement in theory and the author does not justify their claim.

Floaters tell naive skeptics that when an object is floated in a full tub, the system doesn't remember that some water overflowed when the object was added, and that it floats just as well when it is taken out and then added back to the tub -- which now will not overflow at all. However, what the floaters don't notice is that the no-overflow system has just enough room at the top to hold the mass of water that balances the object.

This is false. Here is a short video I made that shows a cup with more water floating inside a cup with less water.


The cup with more water can float in the cup with less water with no trouble.

Some floaters point out that large machines like telescopes are often "afloat" on a thin film of oil. However, these lubricants are kept pressurized in a sealed system, and the pool of oil is not open to the atmosphere. It's a safe bet that such a telescope cannot be levitated (i.e. lifted) by a thin film of oil unless there is some balancing mass of oil (or some other way of pressurizing the oil).

Again I wouldn't know the details of actual telescopes, but there's no reason this couldn't work and the author fails to justify his claims.

So yes, the battleship can float, and the page you found claiming it cannot was simply mistaken.

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    $\begingroup$ Beautiful experiment in its simplicity! I think this one deserves the bonus. $\endgroup$ – Pirx Jan 13 '17 at 2:01

No, that's not necessary. You need a bathtub that's big enough to hold the ship, of course, and sufficient water to fill the space between the ship and the tub to a level such that the volume of the submerged portion of the ship times the density of the water equals the mass of the ship. However, if your "bathtub" closely conforms to the contours of the ship, then the amount of water required could be quite small.

Archimedes' Principle

Looking at the answers and discussion above, it appears that Archimedes' Principle has not been described yet in complete clarity, so there still seems to be some confusion. Here is my attempt to remove some of the fog around this:

Step 1

Let us imagine that we replace the submerged part of a floating body (or the volume of a fully submerged body) with the fluid in which it is contained. We will examine the forces on the volume of fluid that now occupies this space. Notice that this volume of fluid is in equilibrium: It will be at rest, with no accelerations; after all, we are just looking at a uniform body of fluid.

Step 2

If we draw a free-body diagram of just this volume of fluid, we will find these forces: (1) A volume force due to gravitation, which when integrated over the volume adds up to the weight of the fluid, and (2) a surface force due to the pressure distribution on the surface of the volume. If we integrate this pressure distribution over the surface, we obtain the buoyancy force. Now, based on our realization that the volume we are looking at is in equilibrium, we conclude from Newton's Second Law that the net force on the volume must vanish. Thus the integral over the pressure forces on our volume's surface is exactly equal but opposite to the weight of the fluid.

Step 3

Now let us go back to the original (floating or submerged) object. We ask: What is the buoyancy force acting on this object? The answer is that, just as before, we obtain that force by integrating the pressure distribution over the surface of the object. However, note that by our design, this surface has the exact same shape and location as the one of our volume of fluid in Step 2. As a consequence, the pressure everywhere on this surface is exactly the same as for our volume of fluid, and consequently the buoyancy force is exactly the same. Therefore we have found that the buoyancy force on the floating or submerged object is the same as the weight of the fluid above. This concludes the proof.

As a corollary, for an object to be "floating", implying the object is in equilibrium with no acceleration, the volume of the submerged portion of the object must therefore be such that the weight of an equal volume of liquid is equal to the weight of the floating object.

  • $\begingroup$ thanks but how can I be sure that this answer is correct? can u please cite something like an official source for this? $\endgroup$ – user40602 Jan 10 '17 at 20:07
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    $\begingroup$ I'm your official source ;-) Seriously, that's a straightforward application of Archimedes' Principle: All it requires is fluid to be present to the level required. $\endgroup$ – Pirx Jan 10 '17 at 21:08
  • $\begingroup$ Although you would also need quite a strong bathtub. The water pressure at the bottom of the battleship is the same as that when it was floating in the sea $\endgroup$ – Martin Beckett Jan 11 '17 at 1:20
  • $\begingroup$ @MartinBeckett, the water pressure at the bottom of the battle ship is the same as the water pressure in any body of water at that same depth... Bathtub, ocean, lake, or any other sized body of water. $\endgroup$ – Solomon Slow Jan 11 '17 at 2:08
  • $\begingroup$ @jameslarge - yes but the draught of your average battleship is generally more than the depth of a typical bathtub. So if you wish to float a battleship in a bathtub the bottom of the tub must be quite strong - even though the boat doesn't actually touch the bottom. I hope this note will prevent any unfortunate household flooding incidents $\endgroup$ – Martin Beckett Jan 11 '17 at 4:15

Some of the comments on the answers here show that the Archimedes Principle is being thought of by many people as a force that somehow arises literally as a direct result of the fluid's being displaced by a boat. In fact, Archimedes Principle is only a mnemonic for the results of the full calculation of what is actually happenning, which is the transmission of normal force across fluid - floated object interface.

The whole story here is the sum of the normal pressure forces on the floated / steeped body over its submerged boundary, and this is quite independent of whether the body of fluid weights more or less than the imaginarily displaced fluid, exactly as described in Sammy Gerbil's Answer. In my answer here I derive the general expression:

$$\mathbf{F} = \int_V\boldsymbol{\nabla}(p(\mathbf{r}))\,\mathrm{d}V\tag{1}$$

for the sum of these normal pressure forces where we are to imagine the pressure field $p(\mathbf{r})$ that would be present in the fluid within the surface if the fluid weren't being displaced by the body taking up the volume $V$. This imaginary displaced fluid device comes from the application of the (Gauss) divergence theorem to the first principles expression for the sum of the pressure normal forces on the body:

$$\mathbf{F} = \int_{\partial V}\,p(\mathbf{r})\, \mathbf{\hat{n}}(\mathbf{r})\,\mathrm{d} S\tag{2}$$

an expression which is patently free of imaginary displaced fluids. (2) and (1) are readily shown to give the Archimedes Principle if we put the equilibrium condition for a fluid in a gravitational field $\nabla p(\mathbf{r}) = \rho\,\mathbf{g}(\mathbf{r})$ into (1).

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    $\begingroup$ No. Archimedes principle is valid for arbitrary pressure distributions. Simply imagine the volume of the submerged portion of the floating body be replaced by the liquid it is floating in. Trivially, that liquid will be in equilibrium, so the integral of the pressure force on its surface must equal the weight of that "ship-shaped" volume of liquid. Now reverse the thought experiment and replace the volume of liquid (or gas) with a solid body: The forces at the interface will not change. If the weight of the body is the same as the weight of the displaced fluid, we have equilibrium. $\endgroup$ – Pirx Jan 11 '17 at 13:04
  • $\begingroup$ +1 For emphasizing that the Archimedes Principle is just a mnemonic rule. Regarding the validity of Archimedes Principle for arbitrary pressure field or non uniform gravitational field, Pirx got it right and a proof in the language of vector calculus can be found in this answer of mine. $\endgroup$ – Diracology Jan 12 '17 at 22:28
  • $\begingroup$ @Diracology Hmm, yes, very nice application of the divergence theorem. +1 on that one ;-) $\endgroup$ – Pirx Jan 13 '17 at 1:43
  • $\begingroup$ @Pirx I must be having a really bad hair day. WHat hope do we have on fake news when I am contributing to it! $\endgroup$ – WetSavannaAnimal Jan 13 '17 at 3:38

Sammy Gerbil and Pirx have already answered the question correctly. I will only include a minor statement here, since the whole confusion seems to revolve around the concept of "weight of displaced water". "Weight of displaced water" is the weight of water that would have to occupy the submerged volume of the body, if the body were to be not present. It has nothing to do with the amount of water that is already there when the floating object is present.


Is it necessary or not that there exists an amount of water around the ship that weights at least the same as the weight of the ship?

Water in a pond weights less than the air above it – yet the air does not sink.

You need water to be more than 10 meters deep to weight at least the same as the weight of the air above.


Yes you can float a battleship in an amount of water weighing less than the battleship. I failed to interpret the conditions required to explain with great simplicity why the answer is yes, you can float a battleship in a properly sized bathtub with less water (wieght )than the wieght of the ship. I retract my incorrect assertion on which my previous answer was based, and offer the following correct answer complete with its simple explanation. Provided the shape of tub closely mirrors that of the hull, the analogy becomes one of a close fitting piston (the battleship) in a cylinder (the bathtub) which has one end sealed and the other open to atmosphere. As the wieght of the piston(a battleship ) press's down on an amount of water weighing less than the piston, the water is displaced up the gap between the piston and the cylinder walls. This elevation increase results in "head pressure ", in feet of water, @27 inches of water elevation is equal to one pound PSI, acting on the piston and counteracting its wieght. With a properly designed close fitting cylinder(the bathtub) with respect to the piston's shape(the battleship ) an arbitrarily small amount of water will do the job. The battleship and bathtub have become a machine of sorts, and hydraulics rules the action. This interpretation of the initial conditions is the correct analysis and the answer pristinely simple. I herein apply for the bounty on this question, and challenge anyone to disprove it. I learned so much following this question and the contributions of all who commented on or answered this question!

  • $\begingroup$ Upon further analysis of my updated answer I came to the realization that it too is flawed with respect to the equal wieght of water aspect. Because of the head pressure increase the (column of water) that surrounds the piston(ie battleship ) weighs more as it rises and in fact the water in the tub(the cylinder ) will equal the wieght of the battleship (the piston ) So, the answer is yes you can float a battleship in a bathtub with a volume less than its displacement but that volume will equal the wieght of the battleship (piston) $\endgroup$ – RaSullivan Jan 14 '17 at 18:42

Well the density of the submerged portion of the ship cannot exceed the density of the fluid. Such is not mass independent so yes, you need sufficient fluid mass to ensure the boat CAN float and not just sit atop a puddle.

I think this question can be understood as: is there a water level when a boat is considered to be floating.

Let me not be an ass. . . The reason one hardly considers the mass of the water to be a factor is that it is readily assumed there is plenty of fluid for the boat or object to either float or sink (be immersed in fluid). Without enough fluid mass the object sits a top the fluid or floats, same with greater fluid mass.

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    $\begingroup$ This answer is not clear. The question is "Do you need the same weight of water as the weight of the ship?" $\endgroup$ – sammy gerbil Jan 11 '17 at 0:27

protected by Qmechanic Jan 11 '17 at 9:17

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