Talk:Buoyancy

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Untitled[edit]

This article about buoyancy REALLY ain't cool. Shouldn't there be something about the difference between static and dynamic buoyancy etc? Big B.

I (carultch@gmail.com) just added a section about dynamic buoyancy. Let me know if you like it. Currently I am trying to clean up any misconceptions about any dynamic buoyancy calculated via Archimedes. For dynamic buoyancy, remember: Atwood TRUMPS Archimedes. —Preceding unsigned comment added by 75.45.114.217 (talk) 04:54, 22 May 2010 (UTC)

That this section has finally been removed is a good thing. That entire section violated two key Wikipedia precepts, no original research and verifiability. You will not find this nonsense in any physics text, in any civil or mechanical engineering text, or in any oceanography text. You won't find it in any legitimate peer-reviewed journal, either. Why? Because it's nonsense. Look at the sole reference that supports this deleted section, www.physicsmyths.org.uk. This is a crackpot site. — Preceding unsigned comment added by 216.80.140.25 (talk) 12:51, 9 October 2013 (UTC)

No, not crackpot, just an honest attempt to simplify the complexities of fluid dynamics and added mass. Perhaps we should just have a sentence explaining that dynamic buoyancy is more complex, with a link to a full treatment of accelerations in fluids. Dbfirs 17:20, 9 October 2013 (UTC
Crackpots tend to be honest. The removed section is wrong-headed in its premises and modelling. That is what makes it crackpotterish, and has nothing to do with honesty or dishonestyArildnordby (talk) 18:24, 9 October 2013 (UTC)
The model seems to work, at least as a first approximation. The difficulty is in finding a model that gives the correct answers. Have you any better suggestions? Dbfirs 18:35, 9 October 2013 (UTC)
Yawn. Pick up any standard text concerning accelerations in fluids, such as Newman's "Marine Hydrodynamics"Arildnordby (talk) 18:42, 9 October 2013 (UTC)
I don't have access to that text here. Do we have an article covering this? The treatment at added mass seems inadequate. Dbfirs 18:46, 9 October 2013 (UTC)
I agree that added mass is inadequate, in the sense all-too brief and left "mysterious. It is most certainly an article that ought to be expanded; it is ALSO the most natural starting place for treating what might be called hydrodynamic buoyancy. I do not think that the Buoyancy article should be expanded from the static case until solid articles on related concepts (such as added mass) are present.Arildnordby (talk) 18:58, 9 October 2013 (UTC)
Yes, hydrodynamic buoyancy probably doesn't belong in the basic buoyancy article, but it would be useful to insert a brief paragraph with a link to the correct (non-crackpot) treatment so that readers don't think that a polystyrene ball released from rest in water will accelerate upwards at at hundreds of g. (I've just realised that the "crackpot" theory says g but the virtual mass method says 2g for the acceleration (before other forces come into play with increasing velocity.) Dbfirs 19:27, 9 October 2013 (UTC)
As long as a good reference article can be found, I agree to thatArildnordby (talk) 19:33, 9 October 2013 (UTC)

I'm troubled by the opening couple of lines.

"If the buoyancy exceeds the weight, then the object will rise; if the weight exceeds the buoyancy, the object will sink. If the buoyancy equals the weight, the body has neutral buoyancy and may remain at its level."

All of this is true only if the buoyancy and weight are the only two vertical forces acting. In practice, there can be many other forces. There is also a problem that rises can convey the idea of constant velocity rather than acceleration. Shouldn't we say something like:

The buoyancy provides an upward force on the object. According to Newton's first law of motion, if the upward forces (including the buoyancy) balance the downward forces (including the weight) the object will remain at rest. Otherwise, it will accelerate upwards or downwards.
In the most common case, where there are no other forces acting on an object, it will float at the level where it displaces a weight of fluid equal to its own weight. If the object is more dense than the fluid, it cannot displace enough weight of fluid to balance its own weight and will not float.

Stuart White 09:15, 1 January 2006 (UTC)

I have restructured the article a bit, and whilst doing so, tidied up the physics in the introduction. I also amended Newtons to newtons, to match SI conventions, and made some minor edits to the phraseology elsewhere.

Stuart White 09:21, 2 January 2006 (UTC)

The introduction states that buoyancy can be found in fluids. as much as i know, buoyancy is also found in air, as, for example, smoke rises or drops down depending on the weight of the air around it. maybe someone can correct either me or the article?

Thanks, --Abdull 15:13, 18 Dec 2004 (UTC)

Air is a fluid. Any liquid or gas is a fluid; a fluid is a substance that flows. Michael Hardy 00:09, 19 Dec 2004 (UTC)
Shouldn't that be specified for the layman's benefit? Maybe a simple parenthesis? It can be confusing when the technical term is the same as a colloquial term and still diverges. Der Zeitgeist (talk) 18:03, 12 August 2009 (UTC)

On a little side note below Archimedes and his discovery of buoyancy, I provided the reference to the ancient Chinese child prodigy Cao Chong, and his application of earlier Archimedes' principle of buoyancy. I also edited Cao Chong's article a bit (explaining that no official treatise in ancient China had been written about the buoyancy principle, yet several empirical observations were recorded throughout time). In relation to the story of Hieron's crown, Joseph Needham points to a section of the Zhou Li (the Gao Gong Ji section) which would suggest the Chinese during the Han Dynasty (202 BC - 220 AD) were aware of the same principle. He quotes it, stating: "The workers called Li make measures of capacity (liang). They purify (separately) by successive heatings samples of metal (presumably copper) and tin, until there is no further loss of weight. Then they weigh them."

--PericlesofAthens 19:59, 18 March 2007 (UTC)

I suspect that Archimedes did not speak of force and density the way we do today. What I want to know is the reasoning process by which he arrived at the conclusion that an object displaces its weight of fluid exactly, no more, no less. This is not obvious. From what premises did he deduce this? He did not have access to Newtonian Mechanics. —Preceding unsigned comment added by 68.36.254.63 (talk) 18:22, 15 June 2010 (UTC)

Mechanics[edit]

It would be nice to have something about the mechanics of buoyancy in here... what exactly is causing the force? It's not very clear.

If an object has a relatively low density, meaning the volume is large and the mass is small -- as compared to the fluid that surrounds the object, then the fluid will attempt to displace the object. This page states clearly, http://theory.uwinnipeg.ca/physics/fluids/node10.html, that the mass of displaced fluid is equivalent to the net pressure acting on the object. --24.131.137.3 15:53, 21 December 2005 (UTC)
That's like saying "because Archimedes said so". Why does the fluid attempt to displace the object? What exactly is happening at microscopic level?
Consider the hydrostatic head situation in the surrounding liquid. As we go down into the liquid the gravity caused hydrostatic presure in the liquid increases and pushes against the submerged surface of the floating material and will raise the submergence level until the downward weight of the floating object just balances the weight of the mass of the displaced fluid.WFPM (talk) 16:57, 3 September 2009 (UTC)

I am striking the following: "It is obvious that without taking the displaced fluid element into account, energy would not be conserved during the buoyant motion of an object as it would gain both potential and kinetic energy when rising in the fluid." This statement is incorrect. As a buoyant object ascends in a fluid, it loses potential energy, and has zero potential energy at the surface. Energy is conserved. --24.131.137.3 14:42, 21 December 2005 (UTC)

I think the text was referring to the gain of gravitational potential energy which is in fact gained. It is "compensated" for by considering the loss of gravitational potential energy by the fluid that shifts downward as the object shifts upward. Nonetheless, it think the this "energetic" stuff is probably too fine a point for a general article. So I'm OK with it being removed. I do have to wonder if this is one of the "error" that Nature found. Blimpguy 16:10, 21 December 2005 (UTC)
The text was indeed referring to the gravitational potential energy. I can say this because somebody took this passage (as well as the paragraph about acceleration) literally off my webpage http://www.physicsmyths.org.uk/buoyancy.htm (I changed the latter now to include 'gravitational').
The point is that the total energy of the system is
grav.pot. energy of the body + grav.pot. energy of the fluid + kinetic energy of the body + kinetic energy of the fluid
This means that if the body gains grav.pot. energy as well as kinetic energy by rising, the grav.pot. energy of the fluid must decrease accordingly. Otherwise there can be no energy conservation. So there was actually no error here with the original statement.
Thomas

SI units[edit]

Changed the example's units from pounds to Newtons to maintain the use of SI on Wikipedia.

It is conventional in English to write "3 pounds of water" when one means a quantity of water that weighs 3 pounds. But does one follow that usage with Newtons? Michael Hardy 21:48, 12 Apr 2005 (UTC)

Newtons are a unit of force and therefore are a unit of weight, I believe. I'm sure it's correct, but whether or not it's commonly accepted is a bit dubious.--Ryan Hardy 03:30, 15 Apr 2005 (UTC)

It is not conventional to refer to Newtons of water, but then, one asks, why does one normally measure out peas or potatoes by the pound or kilogram, which can be considered (in violation of current standards) force? The answer is that one is measuring, or ought to be measuring mass; weight is being used as a surrogate. If a spring scale is used, one is achieving a measure that depends on the local acceleration of gravity, but if a balance is used, one measures mass. The bouyancy discussion is really almost independent of the local acceleration of gravity, unless that somehow varies grossly over the experimental region. That never holds in common experience with buoyancy, although the variations of gravity and centrifugal force do affect the geoid. When you speak of such and so many Newtons of water, you refer to a mass that depends on local gravity. It is probably better to couch everything as much as possible in terms of masses, or of densities and volumes.

Pdn 05:39, 15 Apr 2005 (UTC)

But buoyant force is a force, nonetheless, and cannot be measured in kilograms. And, since you have to use subtraction attain that measure of force, the units of measure cannot be changed during the operation(kg-kg ≠ kg∙m/s²). Algebraically, it would make sense to use the term "weight" in the context of force.

If Newtons are unsuitable, then is the kilogram-force a possible option?--Ryan Hardy 21:40, 15 Apr 2005 (UTC)

Newtons are SI units. A Kilogram a unit of mass. A Kilogram-force is a force that a body of 1 kg mass exerts under gravity and is approximately 10*1=10 newtons. But newtons are a better option as they are the Internationally accepted. Kilograms are also used to measure force assuming the gravitational force is the same everywhere om earth.--Jetru 11:59, August 23, 2005 (UTC)

Citation Question[edit]

how do i cite this?

Can you express your question a bit more completely? Cite what exactly? Where?
I think he wants to cite this wikipedia article. --PhinnaeusT+Σ 11:40, 2 November 2005 (UTC)
There's a page that explains how to cite Wikipedia articles; I'll see if I can find it. Michael Hardy 21:25, 22 December 2005 (UTC)
Just be careful if you're doing an academic citation. That can be problematic, as a lot of academics don't consider Wikipedia reputable. -Phoenixrod 21:21, 19 August 2006 (UTC)

Nature claims errors[edit]

Nature disputes the accuracy of this article; see http://www.nature.com/news/2005/051212/multimedia/438900a_m1.html and Wikipedia:External_peer_review#Nature. We're hoping they will provide a list of the alleged errors soon. —Steven G. Johnson 01:47, 15 December 2005 (UTC)

Indeed. Disputing the accuracy is not useful in this case without specificity. If it said "buoyancy is the opposite of girlancy", I'd know just how to correct the matter. But Steven, don't you have a PhD in physics? And isn't this freshman-level (or secondary school-level) stuff? If so, I'd think you could straighten out this short article in two minutes (maybe). Michael Hardy 01:52, 15 December 2005 (UTC)

Errors to fix[edit]

Archimedes Principle Reviewer: Prof. Timothy J. Pedley, G. I. Taylor Professor of Fluid Dynamics, University of Cambridge, UK.

  • In the section on acceleration and energy, which discusses how a body moves when it is not neutrally buoyant, it is rightly stated that the acceleration of a body experiencing a non-zero net force is not the same as in a vacuum, because some of the surrounding fluid has to be accelerated as well. However, it is implied that the mass of fluid that has to be added to that of the body, in using Newton's Law to calculate the acceleration, is equal to the mass of fluid displaced. This is not in general true - for example, the added mass for an immersed sphere is half the mass of fluid displaced.
This is in fact incorrect. You seem to be bringing in the drag of the object here (which obviously would depend on its shape). However, the drag is velocity dependent and has nothing to do with the buoyant gravitational acceleration as such. It merely determines the limiting velocity of the buoyant object. The initial acceleration on the other hand depends only on the mass of the object and the displaced fluid and is given by a=g *(m - md)/(m + md) (see my webpage http://www.physicsmyths.org.uk/buoyancy.htm from which much of the acceleration paragraph was lifted anyway (not by me)). Regarding the 'drag' issue, you may also be interested in the page http://www.physicsmyths.org.uk/drag.htm ).
Thomas
  • The entry is rather imprecise. In line 3, for example, the object is said to "float" if the buoyancy exceeds the weight, so here "float" must mean "rise" and not "stay at the same level", which is probably not what was intended because the word has the other meaning in the second paragraph of the section on "Density".
I've only just had my attention drawn to this, and can hardly believe that the reviewers got it so wrong. By their reasoning, an object of mass m, displacing a mass M of fluid, would have a resultant upward force of (Mg − mg), and an "equivalent mass" of (m + M/2). Using Newton's Law as suggested, yields an acceleration of 2g(M − m)/(2m + M). It's easy to see that in the case of a light object in a dense fluid (where m << M), their value for its acceleration is 2g ! If I could get an object to accelerate at 2g in a gravitational field of g, I'd be well on my way to a free energy machine, and wouldn't have to edit here at the current rate of pay ;) --RexxS (talk) 01:51, 23 May 2010 (UTC)
  • the archemedies principal and the quotient of weights hybrid formula is wrong... using values of 1000 and 2500 with a weight of 600, i get -900 submerged. Charlieb000 (talk) 23:47, 24 October 2014 (UTC)
Could you specify which formula you think is incorrect? Dbfirs 08:41, 25 October 2014 (UTC)

I removed the "scale" example.[edit]

Buoyancy is about hydrostatics. In the more complex subject of hydrodynamics, you can try to separate out the inertial and viscous forces involved (ignoring turbulence, etc.), but by the time you present the "scale example" (ignoring mass of scale arms, etc., etc.) and then try to explain why the acceleration of the assembly of connected objects that form the "scale system" is F=ma where m is the mass of the object plus one half the mass of the virtual (and presumably spherical) "fluid element", you have lost 90% the readership. You might as well just refer them to the Reynolds number, but probably you should not do either. -- Pinktulip 03:20, 2 January 2006 (UTC)

You shouldn't bring drag into the buoyancy issue which has nothing to do with it. Buoyancy is merely due to the mechanical coupling of the object with the fluid and the fact that both are subject to gravity, and in this sense the 'scale' example is a good analogon to illustrate what goes on when an object rises in a fluid. It was actually (like some other passages in the article) lifted by somebody off my webpage http://www.physicsmyths.org.uk/buoyancy.htm . As shown there, the force equation is actually F= (m+md)*a where m is the mass of the object and md that of the displaced fluid.
Thomas
Ahh excellent! I found the mastermind behind the original publishing of the information I presented. How do you like the passage I wrote? Any comments about its accuracy/suggestions for improvement? 75.45.114.217 (talk) 19:41, 22 May 2010 (UTC)

Someone confused about density section[edit]

User:24.6.14.204 apparently was confused about that section. I don't see how to make it clearer. TimBentley (talk) 17:50, 3 April 2006 (UTC)

Archimedes discovered buoyancy in his bathtub??[edit]

I have doubts about the statement that Archimedes discovered what is called Archimedes' principle in his bathtub. I've always heard that it was in his bathtub that he realized how he could ascertain whether a goldsmith's workmanship included embezzlement of some of the gold. It was suspected that the smith put in some silver in order to bring a crown's weigh up to the weight of the gold with which he had been entrusted, some of which it was suspected that he kept. Archimedes' solution: see whether the crown, when submerged in water, displaced the same amount of water as an amount of gold equal in weight to the crown.

That is not the same as discovering anything about the amount of buoyancy. Michael Hardy 23:19, 29 May 2006 (UTC)

Oh yes it is! Density is weight per unit volume. Weight is easily determined, volume is more difficult. Volume was determined by the bouyancy (reduction in weight) of the object in a liquid of known density (water). The gold was being stolen by its replacement, in part, by silver. Different densities. Same reduction in weight when immersed but lower overall weight! Paul Beardsell 07:04, 30 September 2006 (UTC)
Following on: Archimedes did not tackle the awkward problem of measuring the crown's volume by collecting water displaced. The density of the crown was calculated thus. Paul Beardsell 13:26, 6 October 2006 (UTC)
Are there any other sources beyond De architectura? Because according to De architectura, he was comparing volumes. See here: http://penelope.uchicago.edu/Thayer/L/Roman/Texts/Vitruvius/9*.html#Intro.12 (there is a link to an English translation as well) rado 11:50, 31 October 2006 (UTC)
As an afterthought, he did determine that the crown was not pure gold by comparing volumes, but he found, by calculation, the quantity of silver mixed with the gold so probably both opinions are correct (though, there is a giant mental leap from calculating the ratio of silver - probably as a ratio of displaced water volumes - to realizing there is som sort of force acting upon the body, not even speaking about finding out the formula for the force) rado 11:59, 31 October 2006 (UTC)
Archimedes Principle is stately pretty plainly in Archimedes' own words in On Floating Bodies, Proposition 5. I am assuming that this work is indeed a genuine work by Archimedes and not a later text attributed to him. Eroica (talk) 20:47, 15 December 2009 (UTC)
If density is determined by d=v/m then both volume and mass must be measured accurately. If density relative to water is determined then only weight need be measured accurately, twice. Density is easier to determine this way as volume need never be measured. I assumed that Archimedes would have realised this and I have asserted that this is the way he measured density of the crown. I have asserted this is what Archimedes did without reference to the original texts. The correct outcome of this is, according to the text quoted, Archimedes did not use buoyancy to determine the density of the crown. This vindicates Michael Hardy's view, earlier, which I contradicted. Paul Beardsell 13:14, 31 October 2006 (UTC)

What remains unresolved (here) is the question of what Archimedes discovered in his bath. Paul Beardsell 13:18, 31 October 2006 (UTC)

What Archimedes discovered in the bath, and what the article does not make clear, is that the weight of water displaced by a floating object is the same weight as the weight of the object itself. --Michael C. Price talk 23:56, 6 March 2010 (UTC)
Which is exactly what his proposition 5 from "On floating bodies" says. I suggest the article should say this as well.--Michael C. Price talk 00:10, 7 March 2010 (UTC)
The main point must be this: It is not necessary to use Archimedes principle concerning buoyancy to do what Archimedes is said to have done in order to find out whether the crown was adulterated. Michael Hardy 16:10, 31 October 2006 (UTC)
A bit more long-windedly: Archimedes could have weighed the crown, NOT submerged in water, then measured the amount of displaced water to get the volume. That tells him the density, and he can likewise get the density of pure gold. There is no need for him to have known anything about buoyancy to do all this. He did formulate "Archimedes' principle", and he could have done so at the time when he solved the problem of the crown, but we cannot conclude as a matter of logical necessity that he MUST have done that. (I haven't yet dug out my copy of Heath's translations to see the details, though.) Michael Hardy 17:11, 31 October 2006 (UTC)

True-ish but not relevant[edit]

The Acceleration section is just not appropriate. And other bits too. What is said is, when true, true for all objects, not just buoyant or floating ones. And other things are not quite right! The section seems to be written by someone who does not understand that -- the author(s) seems to have an imperfect / incomplete understanding of mechanics. I will attempt a re-write. Paul Beardsell 11:42, 6 October 2006 (UTC)

And with some help from others that is complete. Paul Beardsell 14:56, 30 October 2006 (UTC)

I am concerned about this section: "Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum. Suppose that when the rock is lowered by the string into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs will be 10 newtons minus the 3 newtons of buoyant force: 10 − 3 = 7 newtons. This same principle even reduces the apparent weight of objects that have sunk completely to the sea floor, such as the sunken battleship USS Arizona at Pearl Harbor, Hawaii. It is generally easier to lift an object up through the water than it is to finally pull it out of the water."

How is the sunken battleship USS Arizona at Pearl Harbor, Hawaii; relevant to this discussion? Would not the same be true of ANY sunken ship? There are no plans that I am aware of to lift the USS Arizona from the harbor floor, and so the reduction in apparent weight of the sunken ship, THIS SHIP specifically, is of no consequence to this discussion. Furthermore the Arizona's hulk rises above the surface of the harbor sea level at some points, so if it were to be lifted, the apparent reduction in weight from being underwater would rapidly disappear with every bit the sunken ship were raised, since this is not a completely submersed wreck. I recommend the removal of the battleship USS Arizona reference from this article. Seeking comments before performing that edit. —Preceding unsigned comment added by Bugsi (talkcontribs) 21:41, 26 November 2008 (UTC)

(I've since removed the reference to BB Arizona, irrelevant.) —Preceding unsigned comment added by Bugsi (talkcontribs) 18:45, 27 November 2008 (UTC)

Remove diagram[edit]

I'm keen to remove the diagram if a better one is not forthcoming. It's misleading as there is no spring balance weighing the object. All the diagram is doing is showing the measurement of volume. Paul Beardsell 14:56, 30 October 2006 (UTC)

Simple explanation[edit]

The question how boats stay afloat pops up regularly at the reference desk, but the standard explanations don't quite clarify it very well for people who don't get it. So maybe there should be a 'simple explanation' section, however unencyclopedic that may sound. I will give the answer I gave today as a suggestion:

'Displacement' refers not really to water being moved or something (although in a sense it is) but to the volume of the boat below the waterline. The amount of water that would have been there has to be heavier than the boat itself (plus the air, cargo and crew and what have you). The material of the boat itself may be heavier than water (per volume!), but it forms only the outer layer. Inside it is air, which is negligible in weight. But it does add to the volume. The central term here is density, which is mass ('weight') per volume. The thing to keep in mind is that you have to divide the mass of the boat (plus contents) as a whole by the volume below the waterline. If the boat floats than that is equal to the density of water (1 kg/l). You could say that to the water it is as if there is water there because the average density is the same. If you add weight to the boat, the volume below the waterline will have to increase too, to keep the mass/weight balance equal, so the boat sinks a little to compensate. DirkvdM 09:42, 16 November 2006 (UTC)

I can't see how that is simpler than the article which says everything you say. I reckon if a reader can't understand the article then they won't understand your explanation either. The explanation offered also suffers from not being quite correct in that displacement refers precisely and "really" to the weight of water being moved = displaced. As you will read at displacement, displacement is a weight, not a volume. Paul Beardsell 11:22, 16 November 2006 (UTC)
I was wondering the same thing myself. The basic question is of whether the volume of displaced water is measured from the old water level (as the actual volume where the water no longer is) or the new water level (as the volume of the 'hole' in the water where the boat is).
Maybe I just need to experiment ... but either way, the article could use a bit of clarification. -- Smjg (talk) 15:00, 2 November 2009 (UTC)

Boyancy force disappears?[edit]

You write, "The buoyancy force disappears if the fluid is not allowed to flow under the bottom of the object, for example if the object's bottom is fully in contact with the bottom of the container." This needs clarification. What kinds of cases do you have in mind, here? Strictly speaking, the statement is incorrect.

When I take an empty bottle (empty, that is, save the air inside it) with the cap screwed on tight and hold the bottle down at the bottom of a bathtub filled with water, so that the bottom of the bottle is fully in contact with the bottom of the tub, does the buoyancy force disappear? It does not. Rather, the buoyancy is being counteracted by the downward force that I am exerting by holding the bottle down. The net upward force is zero as long as I am holding the bottle down, but the buoyancy force remains. Otherwise, it wouldn't be necessary for me to hold the bottle down. It would stay at the bottom by itself.

So, what would be a real case in which the buoyancy force disappears as the result of an object's bottom being in full contact with the bottom of the container, or as the result of the fluid in some other way not being allowed to flow under the bottom of the object? 66.167.49.46 07:03, 6 December 2006 (UTC)

A real example? A rubber suction cup. Paul Beardsell 08:50, 6 December 2006 (UTC)
Hi, I was the one who had added this sentence. If you place a bucket upside down in an emtpy pool and then fill the pool with water, press the bucket down to prevent water from entering the bucket (no air bubbles seen going up), then you release the bucket, it will stay there by itself. In fact in this case there is a net downward force from water pressure and this will help the remain bucket water-tight. I don't really care that my sentence was deleted, but there is some truth in it so please consider replacing it with a good sentence yourself. Sorry for bad english. Thank you.88.240.9.129 18:54, 9 December 2006 (UTC)
I too have problems with this recent addition to the article. Doubtless the response to your query will be that the bottle is not "fully in contact with the bottom of the container". This does not satisfy me. "Fully in contact with the bottom of the container" must mean there is no space for molecules of the fluid to find their way between object and container. The object thus becomes part of the container. The point is trivial and serves only to confuse. Deleted. Paul Beardsell 08:46, 6 December 2006 (UTC)
In the example of the rubber suction cup, the buoyancy force is being counteracted by the downward pull of the suction, so that the net upward force is again zero. But does this mean that the buoyancy is zero in such a case? It does not. The suction cup still has its buoyancy, but the buoyancy force is being counteracted by the force of the suction. Otherwise, the suction wouldn't be necessary in order to hold the cup in place. For it would then stay on the bottom on its own and under its own weight. 66.167.49.46 11:24, 6 December 2006 (UTC)
I'm not so sure that the ability or inability of molecules of the fluid to get underneath the object bears on the question of whether there is a buoyancy force involved. In the bottle example, it seems to me that even if the bottle were pressed to the bottom of the tub with sufficient force to prevent even a single molecule of water from fitting underneath, thus placing it "fully in contact with the bottom of the container" in the strongest sense of those words, there would still be a buoyancy force in play. One would still be able to feel that force, and to feel one's self counteracting it while pressing down on the bottle to hold it in place. The water molecules would still be pressing differentially on the bottle, with greater pressure toward the bottom and less pressure toward the top. Thus, they would still be "trying" to get underneath, and one would still have to press downward to prevent that from happening. In the event that the bottle became fused to the bottom of the tub as the result of the downward pressure, so that it would now stay on the bottom without being held down, this fusing (or sticking tendency, or structural integrity of the bottle/bottom of the tub, or whatever the right term would be here) would itself represent a force acting counter to the buoyancy that would otherwise push the bottle up. Here again, the buoyancy force would still be present, just counteracted by another force. Just as in the case of the suction cup. No? 66.167.49.46 11:24, 6 December 2006 (UTC)
Well, yes, I agree, I think. But what is the explanation underlying the buoyancy force? I've deleted the one you and I don't like much. Consider if the concrete pool floor were sculpted/moulded to include a hollow concrete bottle just like your glass one. Is there a buoyancy force? Paul Beardsell 21:32, 6 December 2006 (UTC)
In that case, the structure of the pool will include a submerged, hollow projection whose average mass per unit volume (its mean density) will be lower than that of water. So, we can definitely say that if this hollow projection were not attached to the rest of the pool, it would be buoyed up. Is this fact best expressed by saying that there is already a buoyancy force present, but one which is overcome by the strength of the projection's attachment to the rest of the pool? Or, is it best expressed by saying that there is not a buoyancy force in play but that there would be if the projection became detached? I'd say the former. In order for the hollow projection to stay attached indefinitely while immersed, the pool will have to be constructed of a sufficiently strong material. If the pool is made of the sort of stuff that pools normally are, this will not be a problem. But consider a case in which the pool is made of a very weak, thin material. After enough time has passed, the projection might suddenly detach and rise to the surface. This would be due to a buoyancy force acting before the projection detached, as well as after, wouldn't it? (This would of course require that the seam where the projection joins the bottom of the pool break quickly, all the way around, rather than its being breached selectively at one or more weak spots, in which case water would rush in through the breach in the weak areas and fill up the hollow cavity. But given the right material, one could surely construct the pool in such a way that the hollow projection would detach and rise to the surface.) 66.167.49.46 00:00, 7 December 2006 (UTC) Patrick K.
Such a pool costs extra! Consider also a string. One end of the string is "fully in contact with the bottom of the container" - i.e. one end is tied or glued to the container so that no fluid is between it and the container. The other end is tied to a plastic bottle tightly closed and full of air. As by the reasoning now deleted from the article the bottle would not be buoyant and would lie on the floor of the container. But as we know... Paul Beardsell 08:34, 7 December 2006 (UTC)

Well, something similar is the basis of the flush mechanism for a toilet, for example, so there is some basis to it. On the other hand, it's clear that at very least you can't let any water at all displace under the "bucket" or the project is doomed. Gzuckier 20:33, 9 December 2006 (UTC)

You can let some water "displace under the bucket" without the bucket detaching from the bottom. Try it. Not that that proves much of anything. As to toilets I'm sure we're talking sewage. Paul Beardsell 17:46, 16 December 2006 (UTC)

As you lower an inverted bucket into water, the water is going to enter the bucket in order to replace the contraction volume in the water caused by the increased hydrostatic head pressure. The increase in the compressed air volume in the inverted bucket will then transferred to a upward buoyancy pressure on the inverted bottom of the bucket.WFPM (talk) 22:36, 5 September 2009 (UTC)

Most of you seem to be missing the physical mechanics of buoyancy. There is no "buoyant force" (this article notwithstanding). There are two kinds of forces in the universe: body forces (gravity, E&M, Strong, Weak), which act at a distance, and contact forces, which act at the interface of two objects. In actuality contact forces are really electromagnetic forces (body forces) that come into play when the molecules of one object get close to those of the other. So what is "buoyant force"? It is the result of pressure forces (contact forces) of the fluid against the submerged or partially submerged object. Consider then a rigid steel cylinder submerged in water in an upright orientation (i.e. the gravity vector is orthogonal to the flat circular surfaces of the cylinder). The pressure forces on the sides of the cylinder act radially, imparting no force up or down. The pressure on the top surface pushes down, and the pressure on the bottom surface pushes up. Gravity is also acting on both the water and the cylinder. The weight of the water above pushes on the water below, so the pressure on the bottom surface is higher than on the top surface. The weight of the cylinder itself completes the system of forces acting on this body. If the cylinder, whose surfaces are perfectly flat comes to rest on flat plane under the water which is also perfectly flat, then there is no water in contact with the lower surface of the cylinder. Now assume that the cylinder is hollow and that its total density is less than that of water. The forces acting on the body have not changed. There are pressure forces acting on the top surface pushing down. There are pressure forces on the sides pushing radially inward. There is a contact force between the surface of the cylinder and the flat plane. This contact force is a reaction force. Assuming the plane is completely rigid, this force is equal and opposite to the pressure force acting on the top surface plus the gravitational force on the cylinder. Thus the cylinder is in static equilibrium and will NOT rise up due to some mysterious "buoyant force". Lay the cylinder on its side and it will rise up. I'll leave that body diagram as an exercise. This is an idealized example. The example of a suction cup given above is correct. The response that the buoyant force is countered by the "suction force" is incorrect. Like buoyancy, there is no such thing as a "suction force". It too is the result of pressure forces. The response that a body in contact with the container becomes part of the container is also incorrect. Two bodies may be in such close contact to not allow water molecules to pass between them without being molecularly bonded. The suction cup is not bonded to the surface. The "suction force" is simply the result of the difference between the outside pressure forces and the inside pressure forces acting on the flexible cup. The force gets stronger as the volume of the cup increases and the inside pressure decreases. If there were a "buoyant force", then the deformed cup, who's density just decreased, should want to rise even more, but it doesn't, because there is no water getting under the cup to provide a pressure force in the upward direction. This concept of a loss of buoyancy is clear from the equations in the "Forces and Equilibrium" section. Buoyancy is the integral of the fluid stresses around the surface area of the object. If there are no fluid stresses acting upwards, there is no buoyancy. Sorry for the long-winded response here, but this is a pet peeve of mine. The article is not incorrect as it stands now, but the original edit that was deleted four years ago was also correct. Knag (talk) 18:52, 30 May 2010 (UTC)

Provide Proof of Archimedes' Principle[edit]

I'd like to add this simple proof to this article, please comment on this proof:

Consider an infinitesimally thin column of water at (x,y), containing a thin longitudinal section of the submerged object. The column of water above the upper surface z=u(x,y) of the object causes a pressure downwards:

P_\downarrow = \rho g u(x,y) \,\!

In a similar way, the column of water below the low surface z=v(x,y) of the object causes a pressure upwards:

P_\uparrow = \rho g v(x,y) \,\!

Let the infinitesimal transverse area be \Delta A. The infinitesimal resultant force \Delta F exerted on the object is therefore:

\Delta F = \Delta F_\uparrow - \Delta F_\downarrow = P_\uparrow \Delta A - P_\downarrow \Delta A = \rho g [ v(x,y) - u(x,y) ] \Delta A \,\!

Integrating all the infinitesimal forces gives the total force on the object:

F = \int_\Omega{\rho g [ v(x,y) - u(x,y) ] \  dA} \,\!

Assuming \rho and g are constant throughout the fluid, the equation simplifies to:

F = \rho g \int_\Omega{[ v(x,y) - u(x,y) ] \  dA} = \rho g V \,\!

This proves the simple case of Archimedes' Principle when the object can be represented as the volume between two surfaces. --Freiddy 15:42, 30 January 2007 (UTC)

dumbing down / folk story[edit]

This previously good article has been dumbed down and "augmented" by cod science folk stories. I think I'll just revert wholesale to the version of several months ago. Paul Beardsell 21:16, 4 February 2007 (UTC) == == == ==

Although I, unfortunately, agree, I think there should be a more gentle solution, rather than reverting hundreds of edits and the time spent on them. Perhaps some of the relevant stories may stay etc. -- Syphondu 00:29, 3 March 2007 (UTC)

I think we need some part of the story, at least to explain why it is called "Archimedes' principle" and to say something about the history of the problem. It is not just a "folk" story, as it can be traced specifically to Vitruvius' books on architecture. However it might be appropriate to make the description of the story here even shorter and to point to Eureka (word) for a longer version. —David Eppstein 20:02, 18 March 2007 (UTC)

Why not a separate article entitled Archimedes' Principle no? ArdClose (talk) 20:48, 21 November 2007 (UTC)

Image requests[edit]

A diagram showing the forces and volumes in question would be very educational. I'm sure there are enough interesting-looking floating things to make a good photo to liven up the page. -- Beland 06:34, 7 April 2007 (UTC)

ok, added the diagram Yupi666 21:22, 16 June 2007 (UTC)

  • Thanks! Please remember to remove the {{reqphoto}} tag when you add an image to an article. :-) Tim Pierce 13:19, 16 August 2007 (UTC)

measurement[edit]

How is buoyancy measured? In newtons? Widsith 11:56, 4 September 2007 (UTC) Where does the name come from? —Preceding unsigned comment added by 150.7.64.56 (talk) 09:59, 28 November 2007 (UTC)

In newtons or any other unit of force. Michael Hardy 17:31, 4 September 2007 (UTC)

Request Hey, I think a paragraph on how to dtermine buoyancy should be added —Preceding unsigned comment added by Jaskaran3 (talkcontribs) 03:27, 17 October 2007 (UTC)

Now that you know about buoyancy, rock on!

please reply[edit]

if we put an ice cube in a glass of water at room temperature....now ice starts melting....now my question is whether water level will rise or fall..... —Preceding unsigned comment added by 59.93.55.128 (talk) 17:14, 8 February 2009 (UTC)

We are here to discuss improvements in the article, not to discuss standard puzzles based on the ideas in the article. The Wikipedia:Reference Desk is for that sort of thing, assuming it's not homework. —David Eppstein (talk) 17:24, 8 February 2009 (UTC)
....which raises the question of whether this should be in the article (if it isn't already?). Michael Hardy (talk) 21:20, 20 April 2009 (UTC)

The answer to the original question is "no". --Michael C. Price talk 09:09, 7 March 2010 (UTC)

Misspelling??[edit]

I noticed that the page uses the spelling "Archimedes' law". Last I checked, singular names typically tack on an 's' after the apostrophe, even if that name ends in 's'. Is there any specific reason why this rule is being ignored for the page? 97.104.210.67 (talk) 16:11, 20 April 2009 (UTC)

There is a good discussion here. It includes the guidance: "Classical, biblical, and similar names ending in a sibilant, especially if they are polysyllabic, do not take an added s in the possessive". So the current usage in this article seems to comply with that guidance; YMMV --RexxS (talk) 18:11, 20 April 2009 (UTC)

Stability[edit]

I found this in the fluid statics page. I am trying to clean up the page and find this section really has no place in that topic. Since no one is really talking on the talk pages there, I moved it here and have posted this message. Reading the section, it is an essential area to the theory of stability associated with ships and buoyant forces. If someone knows of a better place for this section to land, please enlighten me. I am just making the point that it doesn't belong in fluid statics as it entirely involves the dynamic analysis of a non fluid system operating within a fluidic environment. This is decidedly not a fluid statics topic. It was a copy paste job so if that was bad formatting, please correct it. The section is not my work so feel free to edit it any way you please. —Preceding unsigned comment added by Iron Engineer (talkcontribs) 08:14, 1 July 2009 (UTC)

Although this subtopic is not directly concerned with the mechanism of buoyancy, it does impact on the design of vessels - an application of buoyancy. This is a good place for it. Thank you. --RexxS (talk) 16:55, 1 July 2009 (UTC)

Vandalism??[edit]

Most of the equations do not display: there are syntax errors. I don't know the Wiki math syntax, so I don't know if there are simple misspellings or some more serious errors. There other typographical errors in the current text. I suppose that they weren't there in the past, so maybe the page has been vandalized? Could somebody fix it?--Gautier lebon (talk) 13:49, 7 January 2010 (UTC)

Yes, there was a vandalism/test edit this morning. I've reverted to the previous version and it looks ok to me now. Can you still see any problems that I failed to fix? --RexxS (talk) 15:28, 7 January 2010 (UTC)
Looks OK now, thank you.--Gautier lebon (talk) 12:01, 8 January 2010 (UTC)

Fluid or liquid?[edit]

Liquid, surely? A fluid may be a gas. --Michael C. Price talk 00:29, 7 March 2010 (UTC)

I'm not sure what point you're making, but the effect of buoyancy occurs in all fluids, including gases - helium balloons rely on it. --RexxS (talk) 01:41, 7 March 2010 (UTC)
But Archimede's principle only applies to objects that float on top of a liquid. His treatise was entitled "On floating objects" and is about displacement of liquid volumes (he was in a bath of water, not air, when he had his eureka moment). Perhaps Archimedes' principle and Buoyancy should be separate articles. --Michael C. Price talk 01:51, 7 March 2010 (UTC)
It seems to me that you're assuming Archimedes' principle is solely the law that the Greek mathematician proposed. Looking at sources, I would suggest that what we call Archimedes' principle is actually a more modern generalisation of the original, which applies to "any body wholly or partially immersed in a fluid". A quick Google search shows Britannica and NASA using that definition. There is a wikipedia article on Archimedes which would seem to be the appropriate place to examine On floating objects. I believe that the concept of buoyancy and what sources call "Archimedes' principle" are inextricably linked and belong together in this article. --RexxS (talk) 03:19, 7 March 2010 (UTC)
You have a point, but surely the most natural place to [cover the development] of Archimedes' principle is in its own article? Any objections to such a split? --Michael C. Price talk 09:16, 7 March 2010 (UTC)
I would certainly agree that an article on Archimedes' principle would be encyclopedic in its own right, and could be profitably used to examine its history and development, so there's no reason not to create it. You may want to check the contents of Archimedes as well as this article, and you might want to notify Wikipedia:WikiProject Physics for further input. WP:SPLIT gives guidance on ensuring that the history of contributions is preserved. --RexxS (talk) 14:27, 7 March 2010 (UTC)
Thanks for your support. I see this suggestion has been raised previously, without opposition, so I assume it is only understandable apathy that prevents the split.
I've already has a quick look at Archimedes and Eureka, and I think they would benefit from a split as well. Thanks for the WP:SPLIT tip. --Michael C. Price talk 14:41, 7 March 2010 (UTC)

Atwood machine[edit]

The detailed analogy with Atwood's machine seems a bit strained to me, but I don't want to revert the addition if it is a common analogy or if others find it useful. Dbfirs 06:59, 21 May 2010 (UTC)

Hi: I am the person who added the Atwood's machine analogy. I still need to publish a picture to go with it. It isn't really a common analogy, it is just a good method I feel to think about it.
Many teachers do not even mention the dynamics of buoyant objects and that the displaced fluid also has inertia. This leads students in to thinking buoyancy can cause upward accelerations greater than g and violate conservation of energy. I felt this needed to be addressed.
I made up the Atwood's machine analogy, but it does work. And I did cite my main source of motivation.
If something isn't clear, email me carultch@gmail.com, and I can perhaps improve my posting. —Preceding unsigned comment added by 75.45.114.217 (talk) 04:51, 22 May 2010 (UTC)
(Excuse me tidying the indenting) The explanation and the formulae look a good addition to me. It's notable that it gives a useful result in a limiting case. If you consider the solid object to be very light (a rigid empty container, for example) and the liquid to have a high density, such that m << \rho_f V, then the buoyant force tends toward B = 2 g m and the acceleration tends toward a = g. That shows the acceleration can't exceed g, and the net force (2mg − mg) confirms that. My only small quibble is that I can't quite agree that Archimedes' Principle doesn't apply – as the first sentence suggests. It's rather that when calculating acceleration, AP has to be applied not only to the mass of the solid object, but also to a coupled equal volume of the fluid which must be moved when the solid object rises or sinks. However, that's probably too fine a distinction to worry about. --RexxS (talk) 15:46, 22 May 2010 (UTC)
Hi, I am the author of the Atwood analogy section. The reason why I said "Archimedes principle does not apply", is that we cannot simply equate the (stationary) weight of the fluid displaced (B ≠ \rho_f V g) to the magnitude of the buoyant force. It all depends on what your definition of "weight" is. If you define weight as being the upward force from a scale when measuring your object when stationary, then Archimedes principle "does not apply". However, if you extend the definition of weight to the net force of constraint required to keep a body at rest in a reference frame of identical kinematics as it, THEN it is OK to say that the Archimedes principle still applies, but we just need to be careful about our definition of weight. 75.45.114.217 (talk) 19:32, 22 May 2010 (UTC)
Ah, yes, I see the purpose of the analogy (though we need to be clear that it is only an analogy). I don't think we should say that "Archimedes principle does not apply", but we should explain why the upthrust is not the same as the buoyancy force in a dynamic situation. Dbfirs 22:20, 22 May 2010 (UTC)
An earlier section introduces the concept of "real" and "apparent" weight, as the difference between {m x g} and {m x (net downward force)}, so I don't think there's a problem in illustrating that the apparent weight changes again when the object is moving. However, the new section brings out a key point that does not directly depend on how we view weight: the inertial mass that has to be accelerated is not just the mass of the object, since if the object rises an equal volume of fluid has to fall – and that requires acceleration as well. Perhaps someone who is more expert in Physics can correct any misunderstanding I may have, and devise an appropriate form of words? --RexxS (talk) 22:39, 22 May 2010 (UTC)
Yes, this is the point made in the reference. I've rephrased the paragraph, but I'm not happy with my version. I suggest that we delete the "Dynamic buoyancy" concept (i.e we should mention only real forces), reinstate Archimedes, and just give the acceleration formula, including your explanation and also retaining the Atwood analogy. Dbfirs 23:16, 22 May 2010 (UTC)
(later) ... but we need this "B" for the Atwood equations to work. Is there any sense at all in which "B" is real? Dbfirs 23:31, 22 May 2010 (UTC)
Well, yes. If we were to measure the acceleration of a bubble of air starting to rise in a column of water, then we'd conclude that a buoyant force (B), greater than the force of gravity, was acting on the bubble. As I understand it, that's exactly what the Atwood equation calculates. --RexxS (talk) 00:56, 23 May 2010 (UTC)
Addendum: I just did a couple of algebraic calculations. I calculated the net upforce on the object from (B − m.g); dividing it by m gives the stated acceleration of the object. Similarly, I calculated the net downforce on the "coupled" volume of fluid from (rho.V.g − B); dividing that by rho.V gives the same value for the acceleration of the "coupled" fluid (downwards as expected). This confirms (to me) that B is a useful concept, in that you can use it directly when resolving forces to find the resulting acceleration. --RexxS (talk) 01:25, 23 May 2010 (UTC)
Yes, like you, I'd checked that the stated "B" gives the correct answer for an inverted Atwood's machine, but my point is that this B is not a real force in any sense is it? Dbfirs 09:26, 23 May 2010 (UTC)
It's certainly more real than centrifugal force, for example. At the microscopic level, the molecules of the fluid are under greater pressure towards the bottom of the object and exert a greater force on that part of the object, so the buoyant force is a consequence of the gravitational field. I suppose that someone might consider gravity not to be a real force (as it can be treated as a result of space-time curvature), but I'm not sufficiently expert in the Philosophy of Science to have a strong opinion either way. In any case, I'm supportive of keeping B, as we ought to be telling readers how to calculate the force that provides the acceleration of an object immersed in a fluid. --RexxS (talk) 12:55, 23 May 2010 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── I've done a re-write of the section Beyond Archimedes Principle to simplify and explain where possible, reduce repetition, and try to strike a more encyclopedic tone. I've also added considerations of the limiting cases and of neutral buoyancy. --RexxS (talk) 18:21, 23 May 2010 (UTC)

This paragraph:
  • "Should other forces come in to play in a different situation (such as spring forces, human forces, thrust, drag, or lift), we revert back to the construction of Newton's second law for both bodies, now involving these other forces."
needs to be re-written to clarify what is intended. What does 'revert' mean in this context? We have a convention that encyclopedia articles are not written in the first person, so some re-phrasing is necessary in any case. --RexxS (talk) 20:32, 23 May 2010 (UTC)
RexxS, thank you for your improvements, this article is forming very nicely. Got it, I will be fixing the "we" part. What I meant by "revert" is that the solver of a buoyancy dynamics problem involving non-buoyant and non-gravity forces cannot just use the two derived formulas stated in this article, but rather go back to the free body diagrams and think about the new implications. I'm trying to think of a better way to say that in few words.Carultch (talk) 02:44, 24 May 2010 (UTC)
Yes, thank you to both "carultch" (75.45.114.217) and RexxS for the improvements. I still think that "dynamic bouyancy" is not a "real" force (in the same way that "centrifugal force" is not "real"), but I agree that it is a useful concept (probably more useful than centrifugal force). Dbfirs 07:33, 31 May 2010 (UTC)

The acceleration formula in this article is not quite correct. See the article on added mass to find the correction to the mass of the body. If m is the mass of the body, and b is the mass of the displaced fluid, then the force on the body is g(m-b) due to buoyancy, and the effective inertia is m+b/2 due to added mass. So the acceleration of the body is g(m-b)/(m+b/2). Bo Jacoby (talk) 13:57, 17 August 2011 (UTC).

So for an object whose density is much less than that of the surrounding fluid (m<<b), the acceleration would be asymptotic to g(-b)/(+b/2) = -2g? Think about it: as the light object is accelerated upward, an equal volume of fluid is displaced downward, so by your formula gravity would be accelerating fluid downward at 2g. Check out how the energy of the system (kinetic+potential) increases as the object rises. I reckon I could make a nice perpetual motion machine out of that. --RexxS (talk) 22:42, 17 August 2011 (UTC)

Archimedes' principle; split tag[edit]

This section has a split tag, but no discussion section, so I’m adding one. Moonraker12 (talk) 13:36, 26 May 2010 (UTC)

(Copied from section above)

“…but surely the most natural place to [cover the development] of Archimedes' principle is in its own article? Any objections to such a split? --Michael C. Price talk 09:16, 7 March 2010 (UTC)

I would certainly agree that an article on Archimedes' principle would be encyclopedic in its own right, and could be profitably used to examine its history and development, so there's no reason not to create it. You may want to check the contents of Archimedes as well as this article, and you might want to notify Wikipedia:WikiProject Physics for further input. WP:SPLIT gives guidance on ensuring that the history of contributions is preserved. --RexxS (talk) 14:27, 7 March 2010 (UTC)
Thanks for your support. I see this suggestion has been raised previously, without opposition, so I assume it is only understandable apathy that prevents the split.
I've already has a quick look at Archimedes and Eureka, and I think they would benefit from a split as well. Thanks for the WP:SPLIT tip. --Michael C. Price talk 14:41, 7 March 2010 (UTC) “

(outdent) As the tag is still here, and Archimedes Principle redirects to here, I would like to add my support to the Split idea. Moonraker12 (talk) 13:40, 26 May 2010 (UTC)

Since support seems to be universal, I have splitted it. It would be great if someone could add more information! InverseHypercube (talk) 19:16, 19 March 2011 (UTC)

Coin "floating" on Mercury[edit]

It seems to me that the coin may be floating on the mercury due to surface tension. The way the mercury surface bends and passes UNDER the coin seems to indicate the coin is not displacing any fluid and is therefore not buoyant. Check the article on surface tension or check out this image: http://en.wikipedia.org/wiki/File:Paper_Clip_Surface_Tension_1_edit.jpg —Preceding unsigned comment added by Integracer (talkcontribs) 14:51, 27 May 2010 (UTC)

You're probably right. Mercury has a high surface tension, about seven times that of water (see Surface tension values). If there were no surface tension, we would expect the coin to displace about 60% of its volume of mercury, since brass has a specific gravity of about 8.5 and mercury 13.6. In the image the meniscus is apparent, which means that the coin is displacing some mercury, but obviously not the expected amount. In this case, the total buoyant force is the sum of the force exerted by surface tension and the upthrust from the mercury displaced. I'd have to agree that the dominant force is probably surface tension and it does raise the question of whether the image is misleading in this article. --RexxS (talk) 16:31, 27 May 2010 (UTC)
I'm not an expert on surface tension, but I would imagine that the volume is still displaced, but over a greater area, with an extra depression around the object, otherwise there is no explanation for what happens to the tension force acting on the liquid. (One doesn't see a constant acceleration away from the object!) Dbfirs 17:20, 20 July 2010 (UTC)
Perhaps reading Surface tension might help. A force doesn't produce an acceleration if there is an opposite force acting on it. In this case, the normal reaction on the mercury from its container constrains it from moving. You can see how strong the surface tension of mercury is by observing small droplets, which are almost spherical. The tension force acts normal to the surface of the mercury, so underneath the coin (an almost flat surface) it will produce a vertical force on the coin, which alters the amount of buoyant upthrust needed to balance the weight of the coin. It should not be surprising that the coin does not float with 60% of its volume submerged. --RexxS (talk) 19:11, 20 July 2010 (UTC)
Sorry, yes, I wasn't doubting that the coin itself floats high. I still think that the coin displaces its weight in mercury, as I tried to explain above, just that the mercury is higher at a greater distance from the coin. The surface tension is just a part of the buoyancy force, as it is with any fluid, but a much greater part in the case of mercury. A surface tension normal reaction is transmitted through the fluid in exactly the same way as any other force. The only variation is in the configuration of the fluid near to the interface. I suppose the whole argument hinges on what we mean by "displaced fluid". I propose adding a note to the article explaining what is meant by this. The only consistent meaning (covering both small containers and the effects of surface tension) is the "overflow" from a container that is full before the object is floated on the surface. Dbfirs 05:51, 21 July 2010 (UTC)

Explain Displaced?[edit]

If a object is held down in a container that is only a little larger than itself and the container filled with fluid. then the volume of fluid displaced by the object can be alomst nothing, yet the buoyancy force is unchanged; as the upwards pressure depends only on the height and density of fluid. Is it worth being explicit in the article to say the "submerged volume" even though the displaced volume can be less? This is a somewhat similar problem to the bottle on the bottom the pool discussion? —Preceding unsigned comment added by 129.11.76.230 (talk) 14:33, 20 July 2010 (UTC)

Yes, I see what you mean. I think most people think of submerged volume when they read "displaced", and, in the Archimedes original, the volume was probably the overflow, so it was clearly displaced, but your example highlights the ambiguity in the traditional phrasing. Would it suffice just to add "submerged volume" in parentheses? Dbfirs 17:14, 20 July 2010 (UTC)
Thanks, I think this is a great solution, as I don't think it should be made too much of, because everyone thinks of the classic overflowing bath experiment where they are the same. --92.14.64.241 (talk) 23:47, 20 July 2010 (UTC)
Unfortunately, as pointed out by Integracer and RexxS above, when surface tension comes into play, the submerged volume is not clearly defined (or if you define it simply then it is not the same as the volume of the fluid displaced). I'll have to expand my attempted clarification into a full paragraph as a footnote.

ship listing[edit]

The practical use of this is at the quayside. The one responsible for balancing weight distrabution on a ship to prevent listing, who is he/she and how do they work out their weight? Is that far from the bath experiment, eureka? MacOfJesus (talk) 00:37, 30 July 2010 (UTC)

The rotational stability of a vessel is briefly mentioned in the Stability section. There is a wikilink to Metacentric height, which discusses the issue in far more detail and provides some good references for further reading. --RexxS (talk) 01:14, 30 July 2010 (UTC)
Thank you RexxS, This helps greatly. I presume the Portmaster or Harbourmaster, is ultimately responsible for shipping leaving port in a fit condition. MacOfJesus (talk) 10:07, 10 August 2010 (UTC)

Submarine buoyancy[edit]

Just semantics really: This article says that volume of a submarine is fixed and weight increases/decreases, but that is the opposite of what happens. The weight of the submarine is fixed (whether the submarine is submerged, on the surface, or in dry dock), and the displaced volume alters by allowing sea water in and out of the ballast tanks. I'm altering the text to reflect this. IdreamofJeanie (talk) 10:17, 28 October 2010 (UTC)

No, the overall volume of the submarine really does not change, but water is taken in from the outside to increase its total mass or pumped out to decrease its total mass. The "displaced volume" of water remains the same. I've rewritten the sentence to state that more clearly. --RexxS (talk) 13:19, 28 October 2010 (UTC)
Q. How does the submarine change it's weight? A. By accepting/pumping sea water from outside into and out of its ballast tanks, ie by altering the amount of water it displaces. If I hold a bucket upside down under water it will displace a bucket full of water and tend to rise. When I turn that bucket right way up it now displaces a fraction of the water, and sinks. The weight of the bucket had not been altered in any way. IdreamofJeanie (talk) 13:38, 28 October 2010 (UTC)
The submarine is a sealed fixed volume container, not a bucket. If you sealed the bucket so that water could not enter or leave, it would behave the same whether it was upside down or not. A sealed empty bucket would float either way up, while a sealed bucket with lead weights inside it would sink. Are you suggesting that the contents of a sealed container don't contribute to its total mass? Alternatively, if the submarine were to have an open top, it would indeed behave as any other open container does. --RexxS (talk) 16:56, 28 October 2010 (UTC)
and what would happen if you "sealed the submarine so that water could not enter or leave"? how would it rise/sink then? IdreamofJeanie (talk) 17:46, 28 October 2010 (UTC)
The submarine is sealed (so that its buoyancy remains constant and the internal air pressure is kept around 1 atmosphere). Neither the volume nor mass of a sealed, rigid container alters. Submarines trim their buoyancy by altering the mass of water in their buoyancy tanks, but when running submerged they change depth by using hydrofoils, in the same way as an aircraft uses elevators. However, if a submarine wishes to float on the surface, it reduces its mass by emptying its buoyancy tanks. When it wishes to submerge, it allows enough water into its tanks to increase its total mass to become negatively buoyant. At the desired depth, it will then trim back to neutral buoyancy by expelling a little water from the tanks to reduce its mass so that it becomes neutrally buoyant again. --RexxS (talk) 20:56, 28 October 2010 (UTC)
Rexx, If you were to build a tank just larger than your submarine, and fill it with water, with the submerged submarine on the bottom, then blow the tanks, what would happen to the water level in the tank? Initially the water level has to rise, as the water from the ballast tanks must go somewhere. This means the submarine now displaces more water than before, and has to surface. IdreamofJeanie (talk) 14:41, 28 October 2010 (UTC)
No, it doesn't mean that it displaces more water. The displacement of a vessel fully immersed in water is the mass of water equal to its volume, and is the same whether it's in a tank or in the ocean. --RexxS (talk) 16:56, 28 October 2010 (UTC)
This all depends on whether you think of the tanks as part of the submarine. Dbfirs 07:24, 29 October 2010 (UTC)
If you have a body of water, then the water occupies a physical space. If a submarine arrives on the surface with air in the ballast tanks, a quantity of water is moved away, or displaced. If the submarine then opens its cocks, then some of that displaced water returns to the space it had previously occupied. It is no longer displaced. IdreamofJeanie (talk) 20:23, 1 November 2010 (UTC)
You are still mistaking what is meant by displacement. What if the ballast tanks were filed from a large water tank on the quayside? Do you think that alters the forces acting on the submarine as a whole? The total volume of the submarine still remains unchanged, only its total mass has increased. Given the size of the ocean, it really doesn't matter where the water comes from to fill the ballast tanks. Water has no memory of being displaced, so taking the ballast water from whatever source makes no difference to how buoyancy affects the submarine. --RexxS (talk) 20:51, 1 November 2010 (UTC)

(Outdent) Yes but the maths does not support your interpretation. Displacement (ship) you mentioned earlier refers to the variation of displacement with loading experienced by surface vessels, with no reference to bodies totally immersed in the water. Please see displacement (fluid) which starts "In fluid mechanics, displacement occurs when an object is immersed in a fluid, pushing it out of the way and taking its place. The volume of the fluid displaced can then be measured, as in the illustration, and from this the volume of the immersed object can be deduced (the volume of the immersed object will be exactly equal to the volume of the displaced fluid)." (my emphasis) As I said earlier you can measure the displacement of a submarine in a virtual tank, and, as you said, what applies in a tank also applies in the Atlantic ocean, or anywhere else.

This can be expressed mathamaticaly quite simply, as there are only three variables involved. With W representing the total amount of water in our system (applies equally to a small tank, the Altantic, or "all the oceans, seas, and rivers" in the world), S as the external volume of the submarine, and T as the volume of the ballast tanks.

At any given time, the total contents of the system will be the volume of the water outside the submarine, + the volume of the submarine.

With no submarine present, of course the volume will be W, the volume of the water.

When a submarine arrives on the surface, with air in its tanks, and vents the air, and sinks below the surface, then the submarine will contain T water. As the ammount of water is fixed, then there is now W-T water outside the submarine, and total volume is W-T+S. Displacement is shown to be S-T.

When the submerged submarine expels the water from its tanks, there is once again W water outside the submarine, and total volume is W+S, Displacement is shown to be S

If instead a submarine arrives below the surface, with tanks full of water, then the total volume is W+S, displacement=S.

If that submarine now blows its tanks, there will be W+T water outside the submarine, so total volume is W+T+S, and displacement is shown to be T+S.

In both instances the displacement of the submarine with air in its tanks can be measured to be greater than that of the submarine with water in its tanks, that difference occuring when the ballast tank is blown. IdreamofJeanie (talk) 19:38, 3 November 2010 (UTC)

Except that the displacement of any vessel is a mass, not a volume. Use your three variables plus M as the mass of the submarine with empty tanks, and d the density of the water:
At any given time, the total mass of the system will be the mass of the water outside the submarine, + the mass of the submarine.
With no submarine present, of course the mass will be Wd, the mass of the water.
When a submarine arrives on the surface, with air in its tanks, it displaces a mass of water equal to its own mass (by Archimedes' Principle). Its displacement is M; and that is less than Sd, if it is floating. The total mass of the system is now Wd+M. If the submarine fully vents the air from its tanks, and sinks below the surface, then the submarine will gain a mass Td of water and its mass is now M+Td. As the amount of water is fixed, then there is now a mass Wd-Td of water outside the submarine, and total mass is Wd-Td + M+Td (total mass does not change). Displacement is now Sd, since it is fully immersed; and that is less than M+Td, if it is sinking.
When the submerged submarine expels the water from its tanks, there is once again a mass of Wd water outside the submarine, while the submarine's mass returns to M, and total mass of the system remains Wd+M. Its displacement is still Sd (it didn't change its volume, S), while its mass has reduced from M+Td to M, which we know is less than Sd and so it experiences a net buoyant force (it has a negative total weight), and rises.
If instead, the submarine arrives below the surface with tanks full of water, then its mass is M+Td, and the total mass of the system is Wd+M+Td. Displacement is Sd (which is less than M+Td) and it will sink to the bottom.
If that submarine now blows its tanks, there will be a mass Wd+Td of water outside the submarine, and the submarine's mass has again reduced to M, so total mass is Wd+Td+M (no change there), and displacement is still Sd (it didn't change its volume). Since Sd>M, it will rise, as shown above, until it breaks the surface and displaces less water until once again its displacement becomes equal to M.
Your mistake is to think that volume is a conserved quantity. It isn't in this case, as you've missed the point that compressed air has to be used to expel the water from the buoyancy tanks. It's easy to see that when this happens, the total volume of the system has increased by T, negating your conclusions. Mass is a conserved quantity, and you'll find that the maths supports my explanation of the events in the system you propose. --RexxS (talk) 21:39, 3 November 2010 (UTC)
, no, see the italicised text above, which is taken from the displacement page: the volume of the fluid displaced can be measured. Displacement is defined as a measure of the volume of water being displaced, and when the sub blows its tanks it displaces a greater, measurable, volume. Guess we won't be agereing on this one. IdreamofJeanie (talk) 22:07, 3 November 2010 (UTC)
Sure. Just submerge your submarine in a measuring cylinder (like the picture in the article), and you can measure the volume of the submarine by the increase in volume of (water+submarine). Now expel the water from its ballast tanks and you can measure the volume of the ballast tanks. So what? The displacement of the submarine is exactly the same. Take the submarine out of the measuring cylinder and you can calculate the volume of the submarine again. It is exactly the same volume as it had when you put it in. Its mass has changed of course, because it no longer has water in its tanks. The total mass of the system stays the same, naturally, but the total volume of the system changes whenever water enters or leaves the submarine. A submerged object displaces a mass of water equal to its own volume; A floating object displaces a mass of water equal to its own mass. I'm absolutely certain we won't be agreeing on this one. --RexxS (talk) 00:12, 4 November 2010 (UTC)
my god I don't believe that you can say that with a straight face. "the total volume of the system changes whenever water enters or leaves the submarine" We define displacement as the change in volume of the system. IdreamofJeanie (talk) 09:50, 4 November 2010 (UTC)
Who is this 'we'? The rest of the world defines displacement as the mass of water that would have been there if the vessel wasn't.
"We" is Archimedes et al: see Displacemnent (fluid):In fluid mechanics, displacement occurs when an object is immersed in a fluid, pushing it out of the way and taking its place. The volume of the fluid displaced can then be measured, as in the illustration, and from this the volume of the immersed object can be deduced (the volume of the immersed object will be exactly equal to the volume of the displaced fluid). Take a submarine on the quayside, and open the vents to the ballast tanks. Lower said submarine into a tank, and the ballast tanks will fill as the submarine sinks. water will be displaced by the volume of the pressure hull, as the tanks fill up. Now blow the tanks and water will be displaced by the pressure hull and the ballast tanks combined.
Your mistake is to consider the submarine as a sealed vessel: It is a sealed vessel when the ballast tanks are filled (with either air or water) and the vents are closed but when transitioning from one to the other the submarine is not sealed, there is a hole in the hull which allows water to pass through. As the water is passing through the volume of fluid displaced can be seen to alter. IdreamofJeanie (talk) 15:56, 4 November 2010 (UTC)
You want to exclude the volume of the tanks when full and include them when empty, whenever you calculate the total volume of the submarine. Let me explain why your method is unhelpful. Consider a submarine at a base in freshwater, on a river if you like. It submerges and trims for neutral buoyancy. It then sails underwater into the sea. According to you, the displacement has not changed, since the total volume of the system has not changed. Neither has the mass of the submarine changed, so you would claim it is still neutrally buoyant. Yet, in reality it has become negatively buoyant, because its displacement has actually increased – it is now displacing a greater mass of water (since saltwater has a higher density than fresh by about 2.5%). Let me make that clear: the static buoyancy of a vessel is the resultant force caused by the difference between a vessel's mass and its displacement. --RexxS (talk) 12:52, 4 November 2010 (UTC)
No the displacement hasn't changed. The buoyancy changes as that is a function of the volume displaced X density of dispaced material. IdreamofJeanie (talk) 16:29, 4 November 2010 (UTC)
Exactly as I said, you want the ballast tanks to be part of the volume of the submarine when empty, but not when they are full. How do you want to consider them when they are half-full? The displacement of a vessel is a mass, not a volume. Look at Oyashio class submarine: "Displacement 2,750 tonnes (surfaced) 4,000 tonnes (submerged)"; Typhoon class submarine: "Displacement 23,200–24,500 t (22,830–24,110 long tons) surfaced 33,800–48,000 t (33,270–47,240 long tons) submerged". Are you intending to change all the submarine articles to give displacements in litres? --RexxS (talk) 00:40, 5 November 2010 (UTC)
Actually I do not want to include or exclude anything. I am merely observing what happens to the system as a whole when water enters/leaves the ballast tanks, and offering an explanation, supported by over two thousand years of scientific agreement. (Incidently, when the ballast tanks are half full the displacement of the submarine is still equal to the amount of water displaced by the submarine.) Perhaps you'd care to take the time to read the article on Archimedes, which, under the heading of discoveries and inventions, includes the paragraph: "The most widely known anecdote about Archimedes tells of how he invented a method for determining the volume of an object with an irregular shape. According to Vitruvius, a votive crown for a temple had been made for King Hiero II, who had supplied the pure gold to be used, and Archimedes was asked to determine whether some silver had been substituted by the dishonest goldsmith. Archimedes had to solve the problem without damaging the crown, so he could not melt it down into a regularly shaped body in order to calculate its density. While taking a bath, he noticed that the level of the water in the tub rose as he got in, and realized that this effect could be used to determine the volume of the crown. For practical purposes water is incompressible, so the submerged crown would displace an amount of water equal to its own volume." from this he developed what we now know as the Archimedes Principle, which says that: "Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. With the clarifications that for a sunken object the volume of displaced fluid is the volume of the object, and for a floating object on a liquid, the weight of the displaced liquid is the weight of the object." (and that little quote is directly from this buoyancy article.) You are still failing to understand the distinction drawn between a floating object and one totally immersed. Look at the page you refered me to earlier, Displacement (ship) and note that the second paragraph begins "The word displacement refers to the mass of the water that the ship displaces while floating".
Displacement (fluid) quoted earlier shows that the displacement of a body in a fluid is the volume of that body, and that the displacement will be the same whether the body is submerged in fresh water, sea water, oil or mercury. The buoyancy in each case will be different as buoyancy is a product of displacement multiplied by density, again as shown on the buoyancy page which you are editing. Displacement (ship) as a nautical term is of course the "special" (and most practical) case of a vessel floating in sea water, as historically merchants would be more interested in how much cargo their commercial vessel could carry than in what happens to a stone thrown overboard and allowed to sink. The displacement of a surface vessel is therefore the mass of seawater the ship displaces, and is given as a mass so that the ship's master can effectively load his ship with whatever combination of cargoes are available. For consistacy's sake the dispacament of the submarine is converted to the same units, and this is totally reliable as a submarine will very rarely be operating in a bath of baby oil.
You have repeatedly asserted that displacemnt is a mass - Where is your reliable source that says a body totally immersed in a fluid displaces a fixed mass?
Well if you look at the edit you made: As a submarine blows water from its buoyancy tanks (by pumping them full of air) it rises because its weight is constant as the volume of displaced water is decreased, you are clearly excluding the ballast tanks from the total "weight" of the submarine (although weight vs mass is another debate), otherwise the total mass of the submarine would obviously decrease by the mass of water expelled. You are debating as if the submarine had its ballast tanks permanently open to the outside, which we know isn't true.
I'm fully aware of Archimedes, thanks, and I'd ask you to consider what would be the result of measuring the volume of a submarine by fully immersing it? Whether the tanks were full or empty, the volume of the submarine would be measured the same (assuming you could keep the submarine immersed with empty ballast tanks). You seem to happy with Displacement (fluid) (even though it's unsourced), so how does "the volume of the immersed object will be exactly equal to the volume of the displaced fluid" strike you? The only way that your edit can be read is that you think that the volume of the ballast tanks is not part of the volume of the submarine when full of water - i.e it doesn't displace water when it's full, but it does when empty. I'm saying that is a flawed view: you can make it work when the contents of the tanks are the same as the outside, but it goes wrong when they are different fluids.
Treating the submarine as a fixed volume sealed container that can vary its mass by taking on ballast or jettisoning it is much more consistent, as it doesn't give wrong results when the ballast is different from the fluid it is immersed in (fresh v seawater for example). What you're trying to say is that if the ballast happens to be the same material as the surrounding fluid, then that amount of the fluid is no longer "displaced", and that forces you to redefine "displacement" as a volume. There's really no need for that, and it's not helpful to an understanding of the article.
For what it's worth, Displacement (fluid) never states that displacement is a volume. While I can point to GlobalSecurity.org or Saunders, Stephen (2004). Jane's Fighting Ships 2004-2005. Jane's Information Group. p. 384. ISBN 0710626231.  that gives the displacement of submarines in tonnes, you seem to relying on your own OR explanation of why you think it's really a volume. --RexxS (talk) 17:00, 6 November 2010 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── This is probably going nowhere while we talk in the abstract, so here's an example for a submarine in the ocean by my understanding:

  • Density of seawater = 1.025 tonnes/cubic metre
  • Total volume of submarine = 4,000 cubic metres
  • Displacement in seawater = 4,100 tonnes
  • Mass without ballast = 3,100 tonnes
  • Mass of ballast = 1,000 tonnes
  • Total mass = 4,100 tonnes
  • Net force = (Total Mass - Displacement).g = (4,100 - 4,100).g = 0

Now it empties its ballast tanks:

  • Total volume of submarine = 4,000 cubic metres
  • Displacement in seawater = 4,100 tonnes
  • Mass without ballast = 3,100 tonnes
  • Mass of ballast = 0 tonnes
  • Total mass = 3,100 tonnes
  • Net force = (Total Mass - Displacement).g = (3,100 - 4,100).g = -1,000.g tonne force

Would you be kind enough to show how the same scenario works out, indicating how .. it rises because its weight is constant as the volume of displaced water is decreased. Could you just show me how your method keeps the weight constant and decreases the volume of displaced water, and what the consequential net upward force would be? --RexxS (talk) 18:05, 6 November 2010 (UTC)

quite simple
  • Density of seawater = 1.025 tonnes/cubic metre
  • Total volume of submarine (including ballast tanks) = 4,000 cubic metres
  • Displacement in seawater = 3024.39 cubic meters (envelope of submarine - non-displaced water in tanks)
  • Mass = 3,100 tonnes
  • Net force = (Total Mass - buoyancy).g = (3,100 - 3,100).g = 0
Now it blows its ballast tanks:
  • Total volume of submarine = 4,000 cubic metres
  • Displacement in seawater = 4000 cubic meters (as it now displaces the water that was previously in its tanks)
  • Mass = 3,100 tonnes
  • Total mass = 3,100 tonnes
  • Net force = (Total Mass - buoyancy).g = (3,100 - 4,100).g = -1,000.g tonne force
all neatly squares with Archimedes, whom you are still ignoring.
your interpretation begs the question:how does the submarine "Empty" it's ballast tanks? actually it can't empty them, instead it fills them with compressed air. it uses compressed air to move sea water from a space it previously occupied, ie it displaces the sea water. When the submarine blows its tanks, it is not a sealed unit, it is open to the sea, and is displacing water from its previous position. Were this to happen in a tank, you would see the level of the water rise, and the fact that you can't measure the rise in an ocean does not mean you can just ignore it.
your full/empty displacement scenario is again directly comparable to the bucket referred to earlier. If I immerse an empty bucket in water it will displace the volume of the metal used in making the bucket. If I fill the bucket with water and immerse it it will displace the volume of the metal used, plus the volume of the water that was in it. Does the displacement of the bucket alter? If you open the submarine's valves when immersing your "empty" submarine into a body of water it will displace the volume of its pressure hull, as the tanks fill with water and the submarine sinks. If the tanks are then blown it will displace the volume of pressure hull + Ballast tanks, and submarine will rise, as per our Greek friend. If you now again open the ballast cocks, the water will flow into the tanks, and the water level will again drop. A measurable effect explained two thousand years ago, and still accepted by the scientific community.
again, the buoyancy page (this page you are editing) Says "for a sunken object the volume of displaced fluid is the volume of the object"
Archimedes page says: For practical purposes water is incompressible, so the submerged crown would displace an amount of water equal to its own volume.
the Displacement (fluid) page says: In fluid mechanics, displacement occurs when an object is immersed in a fluid, pushing it out of the way and taking its place. The volume of the fluid displaced can then be measured, as in the illustration, and from this the volume of the immersed object can be deduced (the volume of the immersed object will be exactly equal to the volume of the displaced fluid). How does your change of mass explain the rise and fall of the water level?
the displacement (ship) says:The word displacement refers to the mass of the water that the ship displaces while floating.
Naturally if you are measuring the displacement of a floating ship in tonnes, you would quote the submerged displacement of a submarine the same way, as it would convey the same sort of information to the same people: to say for example that a submarine displaced 2000 tonnes on the surface and 300000 cubic meters (or whatever) submerged would be less clear than to say it displaced 3000 tonnes submerged. As a given volume of sea water would contain a (roughly) constant mass then for practical purposes quoting the displacement in the same terms submerged and floating is a logigal approach (yes that last sentance is OR and that is why I have no thought of adding it into the main text, I merely offer it as an explanation as to why military/shipping sources would quote the figure in tonnes) IdreamofJeanie (talk) 20:51, 7 November 2010 (UTC)
So you think the "displacement in seawater" and the "Net force" are given by this:
  • Displacement in seawater = 3024.39 cubic meters (envelope of submarine - non-displaced water in tanks)
  • Net force = (Total Mass - buoyancy).g = (3,100 - 3,100).g = 0
You calculate your "buoyancy" of 3,100 by multiplying your "displacement" by the density of seawater.
Now it blows its ballast tanks:
  • Net force = (Total Mass - buoyancy).g = (3,100 - 4,100).g = -1,000.g tonne force
  • Displacement in seawater = 4000 cubic meters (as it now displaces the water that was previously in its tanks)
You calculate your "buoyancy" of 4,100 by multiplying your "displacement" by the density of seawater.
Your method keeps the mass of the submarine (excluding ballast) constant, and increases its volume ("displacement" in your terms) by 975.61 cubic metres.
Now here's the problem. It doesn't work unless the ballast is seawater.
You take the volume of the ballast to be 975.61 cubic metres - i.e. seawater.
But the submarine had charged its tanks with freshwater as I outlined earlier. So the ballast in the tanks was 1,000 tons of freshwater and that is 1,000 cubic metres, and your "Displacement in seawater" is actually 3,000 cubic metres. By your method a submarine weighing 3,100 tonnes with 1,000 tonnes of freshwater ballast has a different buoyancy from the same submarine with 1,000 tonnes of seawater ballast. Now do you see why I don't think your perspective is useful? --RexxS (talk) 22:15, 7 November 2010 (UTC)

Air left in tanks[edit]

Just spotted this text which I have removed: "Normally, precautions are taken to ensure that no air has been left in the tanks. If air were left in the tanks and the submarine were to descend even slightly, the increased pressure of the water would compress the remaining air in the tanks, reducing its volume. Since buoyancy is a function of volume, this would cause a decrease in buoyancy, and the submarine would continue to descend". This is clearly not true as the inhabited part of the craft is also full of air and subject to same compression forces. The walls of ballast tanks are metal and resistant to compression by slight changes of depth. The volume of air in the tanks can then be tuned so that the overall displacement at required depth and pressure matches weight of submarine. IdreamofJeanie (talk) 10:40, 28 October 2010 (UTC)

Yes, that text was nonsensical and needed to be removed. The submarine is a sealed, almost incompressible container, and the internal pressure is unaffected by depth. --RexxS (talk) 13:19, 28 October 2010 (UTC)

Gas vs Fluid[edit]

okay what about non fluid buoyancy Drrake (talk) 03:15, 28 November 2010 (UTC)

Gases and liquids are both fluids, as are plasmas. The mechanism of buoyancy applies to all of them. --RexxS (talk) 11:42, 28 November 2010 (UTC)
You are technically correct, sir. However I think most physicists already know what buoyancy is, and most people who would need to look up buoyancy probably don't know that gasses are fluids. In the interest of providing real information to the type of normal person who may benefit from this page I have changed the first instance of the word "fluid" to "liquid, gas or other fluid". I think the "fluid" only description is more suited to a physics textbook than an encyclopedia, and the change while remaining technically correct, will be helpful and well worth the 21 bytes it cost. — Preceding unsigned comment added by 58.105.156.225 (talk) 03:02, 31 January 2012 (UTC)
I was going to object to the addition on the grounds that "fluid" is already linked, but that article starts out with very technical language, so perhaps the redundancy here is helpful to novices. Dbfirs 09:53, 31 January 2012 (UTC)
I linked "fluid" at same time I added the "liquid, gas or other", perhaps the link may have been enough on its own, but i think the extra 4 words earn their keep. — Preceding unsigned comment added by 220.236.90.30 (talk) 13:03, 5 February 2012 (UTC)

Gold Crown[edit]

The story of the gold crown, charming as it is, has nothing to do with buoyancy, so I've removed the story again. More water overflows when a silver/gold object is dropped in, than when a pure gold object of the same weight is immersed. To see that only requires a knowledge of density. Silver is less dense than gold, so a silver/gold object has a greater volume and hence displaces more water. The buoyant force is not relevant to that explanation, and never was – even Vitruvius knew that 2,000 years ago. --RexxS (talk) 00:35, 17 March 2011 (UTC)

what happens in 0 gravity?[edit]

Suppose there is a ball of water in space and a cork made of wood is inserted carefully into the ball. Would it 'rise' from the center of the ball towards the surface or not??? 117.198.122.200 (talk) 14:12, 13 August 2012 (UTC)

No, it will stay where you put it if done carefully enough. No gravity -> no weight -> no buoyancy forces -> no tendency to move towards a surface due to buoyancy. This is assuming it is fully immersed, as surface tension could cause it to move if partly immersed. Cheers, • • • Peter (Southwood) (talk): 18:16, 13 August 2012 (UTC)

Some comments[edit]

When a sinking object settles on the solid floor, it experiences a normal force of:

N=mg-pfVg

This is clearly wrong, the normal force in such a case would be (mg) of the body and the (mg) of the fluidcolumn above it. As there are no fluid beneath the body -> there will be no buoyancy acting on the body.


To prevent confusions like this it should always be stated that archimedes principle can only be used directly to pressure fields that are completly closed. 84.210.26.155 (talk) 21:27, 25 May 2013 (UTC)M

Try putting a (weight-measuring) scale on the solid floor (if it will withstand the effect of the fluid), and you will see that the normal force is indeed N=mg-pfVg. This can be demonstrated to be clearly correct, not clearly wrong. It is very difficult to completely exclude fluid between the floor and the object. If you succeed in doing this, then other forces come into play. Dbfirs 06:43, 26 May 2013 (UTC)

I agree that it is very difficult to exclude fluid between the floor and the object, its also very difficult to create a vacuum, but still much of the calculations in mechanics are done in vacuum for simplicity. Lets do another example; place a scale on the floor, hang a bucket over the scale with a little clearence between the bucket and scale. In the middle of the bucket there is a hole, place a cylindrical rod through the hole down on to the scale. (The rod beeing placed vertical. The hole in the bucket and the rod creats a seal so no water is leaking out of the bucket) If we neglect the friction between the hole and the rod, what will the scale measure? If you start pouring fluid in to the bucket, will this measure change? 84.210.26.155 (talk) 08:52, 26 May 2013 (UTC)

Frictionless seals are difficult to make, but, assuming that this arrangement is possible the scale will initially show the weight of the rod, then, as water is poured in (not directly onto the rod), the reading will gradually reduce by the weight of water corresponding to the volume of rod submerged. I was wrong -- see below. What makes you think that buoyancy can be circumvented? Fluid pressure acts in all directions. You seem to be taking a simplistic viewpoint of the forces involved, perhaps prompted by a simplistic explanation of buoyancy forces that is often given. Dbfirs 11:38, 26 May 2013 (UTC)

Im not trying to circumvent the buoyancy, im stating that it is not there at all in this case. I agree that fluid pressure acts in all directions. But im not sure if we agree on how a pressure creats a force; Lets measure buoyancy in Newtons, this imply that buoyancy is a force. For a pressure (mechanical or fluid) to be able to exert a force on a solid object, it needs a cross-sectional area perpendicular to the pressures direction. Where is that flat on the rod discussed above? 84.210.26.155 (talk) 12:09, 26 May 2013 (UTC)

I have to admit that your rod example is different from your first example because we cannot integrate the pressure forces over the full surface area. Consider the case where the rod has flanges attached, or where a wide rod narrows to one of negligible cross-section to pass through the hole. In these cases we have surfaces for upwards pressure. At a molecular lever, these exist for the smooth rod of constant cross-section, but it's a pity that your experiment is so difficult to carry out in practice. I'd really like to try it to settle my mental conflict between logic and intuition. Dbfirs 16:15, 26 May 2013 (UTC)

In the original example (object on the floor under water) the conflict can maybe be settled like this: if there is no water under the object (or part of the object) then there is no buoyancy. However, to make sure there is no water you'd need to create a vacuum between the object and the floor. Intuitively, that vacuum 'sucks' at the object, exerting a force that compensates the buoyant force. Logically, a vacuum doesn't exert a force, it is just the absence of buoyancy that sucks. (pun unintended but welcome) — HHHIPPO 16:36, 26 May 2013 (UTC)

I have to admit that your rod example is different from your first example because we cannot integrate the pressure forces over the full surface area. It is actually not different. Case 2,with the rod, is just an extension of case 1. In neither of the cases its possible to integrate the pressure forces over the full surface area (when stated that there is no fluid between the floor and object in case 1). If the rod had flanges, there would be a buoyancy force, but it would not be equal to the weight*g of the fluid displaced by the rod. At a molecular lever, these exist for the smooth rod of constant cross-section,.. If the rod has a diameter of 100 meters, will the fluid, on a molecular level, expand through the whole cross-sectional area at the lower end, and thus create a closed pressure field? It's certainly hard to imagine. You seem to be taking a simplistic viewpoint of the forces involved, perhaps prompted by a simplistic explanation of buoyancy forces that is often given. I am actually using your simplistic view of the explanation of buoyancy forces that is often given, to my advantage. People seem to remember the part of Archimedes's Principle; "Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object." fairly well. But what is often forgotten, or even not mentioned, is that this principle comes with some boundary conditions. One of them beeing, as stated in the first post, "archimedes principle can only be used directly to pressure fields that are completly closed". 84.210.26.155 (talk) 17:19, 26 May 2013 (UTC)

Yes, your logic has prevailed over my faulty intuition for the rod example. (I corrected my error in intuition by sliding the rod up and down, then replacing it with a rod of water.) The original solid floor example maintains the buoyancy for the reasons stated by HHIPPO above. Dbfirs 19:38, 26 May 2013 (UTC)

Can you then derive for me what the normal force exerted from the floor on to the object in case 1 is? If i inerpret HHIPPO correct, there is a buoyancy force B, but this is counter acted by the "vacuum force" which is equal to B. Using a free-body diagram (im using a cube as the "object" for simplicity) the normal force exerted from the floor on to the cube will be; N=mg(cube)+mg(water column resting on top of cube)-B(buoyancyforce)+B(vacuum force) (its not easy to write understandable equations on this "talk page", hope your able to follow the logic). The two B's cancel each other and the normal force becomes N=mg(cube)+mg(water column resting on top of cube). 84.210.26.155 (talk) 20:13, 26 May 2013 (UTC)

Yes, if you really could remove all water molecules from under the cube, then it would be like the rod example with the normal reaction force of weight of cube plus weight of column of water. I agree with you that, in this case, it would be better not to mention the buoyancy force at all since it is fictitious. In the real world, such a situation would occur very very rarely. The cube is more easily considered as being part of the floor in these rare circumstances. Dbfirs 20:41, 26 May 2013 (UTC)