Can a battleship float in a tiny amount of water?











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Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.



Does the battleship float in the tub?



I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.



Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.






buoyancy







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  • Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it.
    – Cosmas Zachos
    48 mins ago















up vote
3
down vote

favorite












Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.



Does the battleship float in the tub?



I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.



Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.






buoyancy







share|cite|improve this question
























  • Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it.
    – Cosmas Zachos
    48 mins ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.



Does the battleship float in the tub?



I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.



Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.






buoyancy







share|cite|improve this question















Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.



Does the battleship float in the tub?



I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.



Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.






buoyancy




buoyancy






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share|cite|improve this question













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edited 21 mins ago









knzhou

40.1k10113194




40.1k10113194










asked 2 hours ago









SlowMagic

433




433












  • Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it.
    – Cosmas Zachos
    48 mins ago


















  • Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it.
    – Cosmas Zachos
    48 mins ago
















Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it.
– Cosmas Zachos
48 mins ago




Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it.
– Cosmas Zachos
48 mins ago










2 Answers
2






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up vote
9
down vote













Yes it floats. And it has displaced its "own weight of water" in the sense that if you had filled the container with water and only then lowered the ship into the container, nearly all that water would have been dispaced and is now sloshing around on the floor.






share|cite|improve this answer





















  • Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
    – SlowMagic
    1 hour ago










  • @SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
    – knzhou
    47 mins ago










  • @SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
    – Neinstein
    46 mins ago












  • @SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
    – knzhou
    44 mins ago


















up vote
2
down vote













The USS Missouri $5.8 times 10^7,rm kg, , 270,rm m$ long with a fully laden draft of $11.5,rm m$ has an underwater surface area in excess of $270times 11.5times 2 approx 6200,rm m^2$ and needs to "displace" $5.8 times 10^7,rm kg$ of salt water (density $approx 1020 ,rm kg , m^{-3}$) to float.



Assume a custom made tank so that an even thickness of water (total volume $1 ,rm litre = 0.001 ,m^{-3}$) surrounds the USS Missouri below its waterline.

This thickness of the water layer would be smaller than $frac{0.001}{6200} approx 1.6 times 10^{-7} rm m$.



So in theory possible but in practice very highly unlikely.




The OP has changed the title from "1 litre" to "a small amount of water".

All that needs to be done is to choose a volume of water such that it is practically possible to float the USS Missouri in a suitably shaped dock and the OP's layer of water 3 cm thick might be possible in practice?



The picture in this answer gives a flavour of an "apparent lack of water" being able to float a ship.






share|cite|improve this answer























  • I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
    – SlowMagic
    1 hour ago











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2 Answers
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up vote
9
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Yes it floats. And it has displaced its "own weight of water" in the sense that if you had filled the container with water and only then lowered the ship into the container, nearly all that water would have been dispaced and is now sloshing around on the floor.






share|cite|improve this answer





















  • Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
    – SlowMagic
    1 hour ago










  • @SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
    – knzhou
    47 mins ago










  • @SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
    – Neinstein
    46 mins ago












  • @SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
    – knzhou
    44 mins ago















up vote
9
down vote













Yes it floats. And it has displaced its "own weight of water" in the sense that if you had filled the container with water and only then lowered the ship into the container, nearly all that water would have been dispaced and is now sloshing around on the floor.






share|cite|improve this answer





















  • Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
    – SlowMagic
    1 hour ago










  • @SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
    – knzhou
    47 mins ago










  • @SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
    – Neinstein
    46 mins ago












  • @SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
    – knzhou
    44 mins ago













up vote
9
down vote










up vote
9
down vote









Yes it floats. And it has displaced its "own weight of water" in the sense that if you had filled the container with water and only then lowered the ship into the container, nearly all that water would have been dispaced and is now sloshing around on the floor.






share|cite|improve this answer












Yes it floats. And it has displaced its "own weight of water" in the sense that if you had filled the container with water and only then lowered the ship into the container, nearly all that water would have been dispaced and is now sloshing around on the floor.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









mike stone

5,9271121




5,9271121












  • Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
    – SlowMagic
    1 hour ago










  • @SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
    – knzhou
    47 mins ago










  • @SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
    – Neinstein
    46 mins ago












  • @SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
    – knzhou
    44 mins ago


















  • Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
    – SlowMagic
    1 hour ago










  • @SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
    – knzhou
    47 mins ago










  • @SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
    – Neinstein
    46 mins ago












  • @SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
    – knzhou
    44 mins ago
















Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
– SlowMagic
1 hour ago




Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"...
– SlowMagic
1 hour ago












@SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
– knzhou
47 mins ago




@SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself.
– knzhou
47 mins ago












@SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
– Neinstein
46 mins ago






@SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = rho_{w} g h$ will push it upwards ($rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all.
– Neinstein
46 mins ago














@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
44 mins ago




@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
44 mins ago










up vote
2
down vote













The USS Missouri $5.8 times 10^7,rm kg, , 270,rm m$ long with a fully laden draft of $11.5,rm m$ has an underwater surface area in excess of $270times 11.5times 2 approx 6200,rm m^2$ and needs to "displace" $5.8 times 10^7,rm kg$ of salt water (density $approx 1020 ,rm kg , m^{-3}$) to float.



Assume a custom made tank so that an even thickness of water (total volume $1 ,rm litre = 0.001 ,m^{-3}$) surrounds the USS Missouri below its waterline.

This thickness of the water layer would be smaller than $frac{0.001}{6200} approx 1.6 times 10^{-7} rm m$.



So in theory possible but in practice very highly unlikely.




The OP has changed the title from "1 litre" to "a small amount of water".

All that needs to be done is to choose a volume of water such that it is practically possible to float the USS Missouri in a suitably shaped dock and the OP's layer of water 3 cm thick might be possible in practice?



The picture in this answer gives a flavour of an "apparent lack of water" being able to float a ship.






share|cite|improve this answer























  • I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
    – SlowMagic
    1 hour ago















up vote
2
down vote













The USS Missouri $5.8 times 10^7,rm kg, , 270,rm m$ long with a fully laden draft of $11.5,rm m$ has an underwater surface area in excess of $270times 11.5times 2 approx 6200,rm m^2$ and needs to "displace" $5.8 times 10^7,rm kg$ of salt water (density $approx 1020 ,rm kg , m^{-3}$) to float.



Assume a custom made tank so that an even thickness of water (total volume $1 ,rm litre = 0.001 ,m^{-3}$) surrounds the USS Missouri below its waterline.

This thickness of the water layer would be smaller than $frac{0.001}{6200} approx 1.6 times 10^{-7} rm m$.



So in theory possible but in practice very highly unlikely.




The OP has changed the title from "1 litre" to "a small amount of water".

All that needs to be done is to choose a volume of water such that it is practically possible to float the USS Missouri in a suitably shaped dock and the OP's layer of water 3 cm thick might be possible in practice?



The picture in this answer gives a flavour of an "apparent lack of water" being able to float a ship.






share|cite|improve this answer























  • I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
    – SlowMagic
    1 hour ago













up vote
2
down vote










up vote
2
down vote









The USS Missouri $5.8 times 10^7,rm kg, , 270,rm m$ long with a fully laden draft of $11.5,rm m$ has an underwater surface area in excess of $270times 11.5times 2 approx 6200,rm m^2$ and needs to "displace" $5.8 times 10^7,rm kg$ of salt water (density $approx 1020 ,rm kg , m^{-3}$) to float.



Assume a custom made tank so that an even thickness of water (total volume $1 ,rm litre = 0.001 ,m^{-3}$) surrounds the USS Missouri below its waterline.

This thickness of the water layer would be smaller than $frac{0.001}{6200} approx 1.6 times 10^{-7} rm m$.



So in theory possible but in practice very highly unlikely.




The OP has changed the title from "1 litre" to "a small amount of water".

All that needs to be done is to choose a volume of water such that it is practically possible to float the USS Missouri in a suitably shaped dock and the OP's layer of water 3 cm thick might be possible in practice?



The picture in this answer gives a flavour of an "apparent lack of water" being able to float a ship.






share|cite|improve this answer














The USS Missouri $5.8 times 10^7,rm kg, , 270,rm m$ long with a fully laden draft of $11.5,rm m$ has an underwater surface area in excess of $270times 11.5times 2 approx 6200,rm m^2$ and needs to "displace" $5.8 times 10^7,rm kg$ of salt water (density $approx 1020 ,rm kg , m^{-3}$) to float.



Assume a custom made tank so that an even thickness of water (total volume $1 ,rm litre = 0.001 ,m^{-3}$) surrounds the USS Missouri below its waterline.

This thickness of the water layer would be smaller than $frac{0.001}{6200} approx 1.6 times 10^{-7} rm m$.



So in theory possible but in practice very highly unlikely.




The OP has changed the title from "1 litre" to "a small amount of water".

All that needs to be done is to choose a volume of water such that it is practically possible to float the USS Missouri in a suitably shaped dock and the OP's layer of water 3 cm thick might be possible in practice?



The picture in this answer gives a flavour of an "apparent lack of water" being able to float a ship.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









Farcher

46.8k33693




46.8k33693












  • I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
    – SlowMagic
    1 hour ago


















  • I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
    – SlowMagic
    1 hour ago
















I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
– SlowMagic
1 hour ago




I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship.
– SlowMagic
1 hour ago


















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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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