dbell@daisy.UUCP (David I. Bell) (12/08/84)
| || | A COUNTER-INTUITIVE FACT | /||\ | | ------------------ | Something that I have found amusing | | | | is the fact that a ship can float in a \ \ / / container of water which only contains \~~\ /~~/ a VERY small amount of water. \~~\ SHIP /~~/ \~~\ /~~/ For example, if a container is shaped so \~~\ /~~/ that it is 1/10 inch away from the ship on \~~\ /~~/ all sides (and the bottom), then the water \~~\ /~~/ filling that small gap will hold it up. =====> \~~\--/~~/ \~~~~~~/ The counter-intuitive fact is that the weight of \~~~~/ the water can be *much* less than the ship's weight. \--/ How is it that the water can hold up a ship which weighs more than the water? -dbell-
hull@hao.UUCP (Howard Hull) (12/09/84)
> Something that I have found amusing is the fact that a ship can float in a ******Change "float in" to "can be supported in" and I'll buy it.****** > container of water which only contains a VERY small amount of water. > > For example, if a container is shaped so that it is 1/10 inch away from the ******I will spare us by not bothering to calculate how many gallons of water ******will be in a 1/10 inch thick layer of water the area of the ship's hull. > ship on all sides (and the bottom), > then the water filling that small gap will hold it up. ******There may be situations in which this is true, but usually it wouldn't be. > The counter-intuitive fact is that > the weight of the water can be *much* less than the ship's weight. ******Arrgh. Here we go. The principle of Archimedes would, for the purposes of this discussion, best be stated: "When an object is freely suspended in a liquid, the object will be buoyed up by a force equal to the weight of the displaced liquid." Therefore, if you have a 100,000 ton ship, you are going to need 100,000 tons of water for it to displace; otherwise, a force of some other description will be found to be responsible for supporting the ship. > How is it that the water can hold up a ship which weighs > more than the water? -dbell- The layer of water has to be thin enough that small scale molecular interaction between the water and the container transfers the force represented by the weight of the ship directly to the walls of the container. Then the water is between a rock and a hard place, and has no choice but to support the ship. Howard Hull {ucbvax!hplabs | allegra!nbires | harpo!seismo } !hao!hull
gjk@talcott.UUCP (Greg J Kuperberg) (12/10/84)
> Something that I have found amusing > is the fact that a ship can float in a > container of water which only contains > a VERY small amount of water. ... > The counter-intuitive fact is that the weight of \~~~~/ > the water can be *much* less than the ship's weight. \--/ > > How is it that the water can hold up a ship which weighs > more than the water? -dbell- Leverage. By the same principle, the following system can be in equilibrium: -------------------- | /\ | | || | Big weight Fulcrum Little weight Note that in the case of the toy ship in the sink, shoving the ship down one centimeter causes the water level to move up several centimeters, so that the whole thing works like a lever. --- Greg Kuperberg harvard!talcott!gjk "Strange women lying in ponds distributing swords is no basis for a system of government." -Monty Python
cjh@petsd.UUCP (Chris Henrich) (12/12/84)
[] > > How is it that the water can hold up a ship which weighs > > more than the water? -dbell- > As I stated earlier, it probably can't. However if you are willing to > warp your definition of which water molecules are "holding up the ship" > and if you will also allow me a little poetic license, lookie here: > > || > /||\ > 50,000 tons of water here ------------------ 50,000 tons of water here > |~~~~~~~~~~~~~~~~~~~~~~~~~| 100,000 tons |~~~~~~~~~~~~~~~~~~~~~~~~~| > |~~~~~~~~~~~~~~~~~~~~~~~~~\ /~~~~~~~~~~~~~~~~~~~~~~~~~| > |~~~~~~~~~~~~~~~~~~~~~~/\~~\ /~~/\~~~~~~~~~~~~~~~~~~~~~~| > ----------------------/ \~~\ SHIP /~~/ \--------------------- > \~~\ /~~/ > \~~\ /~~/ > \~~\ /~~/ > \~~\ /~~/ > \~~\--/~~/ > \~~~~~~/<-- 8 pounds of water in trough. > \~~~~/ > \--/ > Howard Hull Where do we start? Let me first observe that the trough is not hanging in midair; it must be pushing up with a force of 100,000 tons to keep the ship stationary against gravity. First reductio ad absurdum: if it is impossible for 8 pounds of water to hold up 100,000 tons of battleship, then it is infinitely more impossible for *ZERO* pounds of water to hold up 100,000 tons of battleship, yet that is just what would happen if the gallon of water were removed and the battleship settled down in the nice dry trough. Second reductio ad absurdum: Let's take that picture and redraw it slightly. || /||\ 50,000 tons of water here ------------------ 50,000 tons of water here |~~~~~~~~~~~~~~~~~~~~~~~|~| 100,000 tons |~|~~~~~~~~~~~~~~~~~~~~~~~| |~~~~~~~~~~~~~~~~~~~~~~~|~\ /~|~~~~~~~~~~~~~~~~~~~~~~~| |~~~~~~~~~~~~~~~~~~~~~~/\~~\ /~~/\~~~~~~~~~~~~~~~~~~~~~~| ----------------------/ \~~\ SHIP /~~/ \--------------------- \~~\ /~~/ \~~\ /~~/ \~~\ /~~/ \~~\ /~~/ \~~\--/~~/ \~~~~~~/<-- 8 pounds of water in trough. \~~~~/ \--/ All I did was to put a thin barrier between the gallon of water in the trough and the 2 X 50,000 tons on either side. If you think that the water on the side is holding up the battleship, then you have to infer that 100,000 tons of force is somehow being transmitted across that thin barrier. Actually, the relative magnitudes of 8 pounds versus 100,000 tons are irrelevant to floating the battleship, because the support provided by the water is not a matter of balancing the water against the ship; it is a matter of the variation of pressure in the water at different depths. This pressure is proportional to the depth; in fact, the force on an area A at depth D is the weight of a column of water, of that depth, over a *horizontal* area A. The subtle thing about pressure is that the same pressure is exerted on the area A, no moatter how A is oriented. Now let's look at the picture again: || /||\ 50,000 tons of water here ------------------ 50,000 tons of water here |~~~~~~~~~~~~~~~~~~~~~~~~~|XX100,000 tonsXX|~~~~~~~~~~~~~~~~~~~~~~~~~| |~~~~~~~~~~~~~~~~~~~~~~~~~\XXXXXXXXXXXXXXXX/~~~~~~~~~~~~~~~~~~~~~~~~~| |~~~~~~~~~~~~~~~~~~~~~~/\~~\XXXXXXXXXXXXXX/~~/\~~~~~~~~~~~~~~~~~~~~~~| ----------------------/ \~~\XXXXSHIPXXXX/~~/ \--------------------- \~~\XXXXXXXXXX/~~/ \~~\XXXXXXXX/~~/ \~~\XXXXXX/~~/ \~~\XXXX/~~/ \~~\--/~~/ \~~~~~~/<-- 8 pounds of water in trough. \~~~~/ \--/ This time, I shaded in the volume of the ship that is below the water level. It can be shown that, because the magnitude of the pressure in the water increases with depth, the net upward force exerted by that pressure on the ship is equal to the volume shaded in times the density of water. This is Archimedes' law. The usual expression of it, involving "displacement," is meant to be a concise way of referring to that volume. One over-interprets the words if one insists on seeing all the water that was supposedly displaced. Regards, Chris -- Full-Name: Christopher J. Henrich UUCP: ..!(cornell | ariel | ukc | houxz)!vax135!petsd!cjh US Mail: MS 313; Perkin-Elmer; 106 Apple St; Tinton Falls, NJ 07724 Phone: (201) 870-5853
fons@mcvax.UUCP (Fons Kuijk) (12/12/84)
In article <27@daisy.UUCP> dbell@daisy.UUCP (David I. Bell) writes: > > | || | > A COUNTER-INTUITIVE FACT | /||\ | > | ------------------ | >Something that I have found amusing | | | | >is the fact that a ship can float in a \ \ / / >container of water which only contains \~~\ /~~/ >a VERY small amount of water. \~~\ SHIP /~~/ > \~~\ /~~/ >For example, if a container is shaped so \~~\ /~~/ >that it is 1/10 inch away from the ship on \~~\ /~~/ >all sides (and the bottom), then the water \~~\ /~~/ >filling that small gap will hold it up. =====> \~~\--/~~/ > \~~~~~~/ >The counter-intuitive fact is that the weight of \~~~~/ >the water can be *much* less than the ship's weight. \--/ > >How is it that the water can hold up a ship which weighs >more than the water? -dbell- Remember this from long ago? | | | | \ / | | \~~~~~~~~~~~~~~~~~~~~/ |~| \~~~~~~~~~~~~~~~~~~/ |~| \~~~~~~~~~~~~~~~~/ |~| \~~~~~~~~~~~~~~/ |~| \~~~~~~~~~~~~/ |~| \~~~~~~~~~~/ |~| \~~~~~~~~/ |~| \~~~~~~/ |~| \~~~~/ |~| \~~/ |~| |~~|_________________|~| |~~~~~~~~~~~~~~~~~~~~~~| ------------------------ What is the difference compared to | || | | /||\ | | ------------------ | | | | | | | \ \ / / | | \~~\ /~~/ |~| \~~\ SHIP /~~/ |~| \~~\ /~~/ |~| \~~\ /~~/ |~| \~~\ /~~/ |~| \~~\ /~~/ |~| \~~\--/~~/ |~| \~~~~~~/ |~| \~~~~/ |~| \~~/ |~| |~~|_________________|~| |~~~~~~~~~~~~~~~~~~~~~~| ------------------------ It is all a matter of balancing forces. ------------------------------------------------------------ |> The counter-intuitive fact is that |> the weight of the water can be *much* less than the ship's weight. |******Arrgh. Here we go. The principle of Archimedes would, for the purposes |of this discussion, best be stated: "When an object is freely suspended in |a liquid, the object will be buoyed up by a force equal to the weight of the |displaced liquid." Therefore, if you have a 100,000 ton ship, you are going |to need 100,000 tons of water for it to displace; otherwise, a force of some |other description will be found to be responsible for supporting the ship. ------------------------------------------------------------ You are not serious about really having to displace 100,000 ton of water? You might as well start with a container containing a small amount of water in it. By inserting the ship, the water will rise. This will cause an upward force due to pressure on the ships surface. Now in this system during insertion, not only the surface of the ship below water level will increase, but also the pressure wil rise due to rising of the water. It is no surprise that the total upward force (being the total sum of local pressure*surface area) equals the weigth of the water that would occupy the same volume that the part of the ship below water level occupies. The system behaves much like a hydrolic lever, which to some people already is thought to be counter intuitive. But apart from that, due to the v-shape of the ship and the container, the system does not behave linear, which makes it amusing indeed. --- No joke Fons Kuijk No graphics ...mcvax!fons No maxim [for insiders:-)]
rpw3@redwood.UUCP (Rob Warnock) (12/12/84)
+--------------- | ******Arrgh. Here we go. The principle of Archimedes would, for the purposes | of this discussion, best be stated: "When an object is freely suspended in | a liquid, the object will be buoyed up by a force equal to the weight of the | displaced liquid." Therefore, if you have a 100,000 ton ship, you are going | to need 100,000 tons of water for it to displace; +--------------- "Arrgh", yourself. ;-} Sorry, you are quite wrong. By floating freely, no matter how shallow the clearance between the hull and the bottom, it has ALREADY displaced the water! The Archimedean "displacement" is the volume of the ship which is below the surface level of the water and has nothing whatsoever to do with the volume of water which is OUTSIDE the ship. In other words, "displacement" is the volume of the hole that would result if you removed the ship, or, in the "form-fitted bathtub" picture (sorry, I can't draw the picture well) it's the amount of water you would have to add to the "bathtub" after removing the ship to restore the water level to the same height it had when the ship was present. In fact, if one could accurately measure it, if you hauled a super-tanker out of the ocean, one would see the water level of the ocean drop as the displacement was (nearly) filled in. The level of the drop would be approximately equal to the volume of the ship divided by the surface area of the ocean. (The "approximately" lets me not worry about such things as the cross-sectional area of the ship at the waterline and the shape of the shoreline -- things you have to consider in the "bathtub" case.) +--------------- | ...otherwise, a force of some | other description will be found to be responsible for supporting the ship. | > How is it that the water can hold up a ship which weighs | > more than the water? -dbell- +--------------- The ship does NOT weigh more than the water that was displaced (see above for the definition of "displaced"). The ship weighs EXACTLY the SAME as the water that is displaced. (That was what made Archimedes shout "Eureka!". The story goes that he had been asked by the king to tell whether a crown given as a gift was real gold or if it had been adulterated with lead. He knew what the densities of gold and lead were, and what the weight of the crown was, but it was such an odd shape that he couldn't accurately calculate its volume, and therefore was unsure of its density. Archimedes supposedly was plopping into a bathtub that was completely full, when he noticed the water spilling over the edge. He put two and two together, got exited, and went running naked down the street to tell the king "Eureka! Eureka!" [supposedly Greek for "I've found it!"]. In those days, running naked down the street must not have been so strange. On the other hand, he was prompt. These days the Special Commission On Weights And Measures would have taken several years to be sure of the results... ;-} ) +--------------- | The layer of water has to be thin enough that small scale molecular interaction | between the water and the container transfers the force represented by the | weight of the ship directly to the walls of the container. Then the water is | between a rock and a hard place, and has no choice but to support the ship. | Howard Hull | {ucbvax!hplabs | allegra!nbires | harpo!seismo } !hao!hull +--------------- Again, sorry. Doesn't work that way. For another way of looking at it, try considering the pressure experienced by any point on the bottom of the "bathtub" when there is only water there. It is supporting EXACTLY the weight of the vertical column of water directly above it. This varies with the depth of the bottom, since water weighs a good deal (about 64 pounds per cubic foot). Now add the ship, removing just enough water to keep the final surface level the same as it was originally. (I assume that the ship doesn't touch bottom anywhere, but floats freely.) Now as odd as it may seem to you at first, each spot on the bottom of the "bathtub" is experiencing EXACTLY the same pressure as before. Moreover, the amount of water you had to remove (or the amount that spilled, if we were as messy as 'ol Archie) would weigh exactly the same as the ship. So, it's not how much water is LEFT, it's how much we SPILLED, that counts. Rob Warnock Systems Architecture Consultant UUCP: {ihnp4,ucbvax!dual}!fortune!redwood!rpw3 DDD: (415)572-2607 Envoy: rob.warnock/kingfisher USPS: 510 Trinidad Ln, Foster City, CA 94404
hull@hao.UUCP (Howard Hull) (12/12/84)
During the night it finally dawned on me what must be done. Assuming that one
gets the battleship to float using any suitable configuration, it is possible
to place a valve in the connecting channel and close it without any resultant
redistribution of the forces. After that is done, then we may indeed remove
the water and the battleship will continue to float in its gallon of water.
One way to avoid problems in teaching this would be to restate the principle
of Archimedes thusly:
"If an object is freely suspended in a liquid, it is buoyed up
by a force equal to the weight of the displaced liquid, or its
virtual equivalent."
So as you can see, I AM STILL serious about requiring the 100,000 tons of
water to be there even if it is in a virtual form, or as you put it, not
"really" there. Please let me apollogize in advance if I confused anyone
by my incomplete statement of the situation.
So now I have a question. If we now wish to squeeze all of the water out of
the gap between the battleship and trough wall, what force pressing down on
the ship will be required to do so? If we can get that question answered,
then we can find an answer to the original question in its proper context...
The original question was "how can the water hold up something greater than
its own weight". This would seem to be a question of fundamental structure
were it not for the degree of freedom allowed by the width of the space
between the ship and the trough.
Regards, Howard Hull
{ucbvax!hplabs | allegra!nbires | harpo!seismo } !hao!hullltn@lems.UUCP (Les Niles) (12/13/84)
[] In article <hao.1296> hull@hao.UUCP (Howard Hull) writes: >... >> The counter-intuitive fact is that >> the weight of the water can be *much* less than the ship's weight. >******Arrgh. Here we go. The principle of Archimedes would, for the purposes >of this discussion, best be stated: "When an object is freely suspended in >a liquid, the object will be buoyed up by a force equal to the weight of the >displaced liquid." Therefore, if you have a 100,000 ton ship, you are going >to need 100,000 tons of water for it to displace; otherwise, a force of some >other description will be found to be responsible for supporting the ship. >... >The layer of water has to be thin enough that small scale molecular interaction >between the water and the container transfers the force represented by the >weight of the ship directly to the walls of the container. Then the water is >between a rock and a hard place, and has no choice but to support the ship. > Howard Hull > {ucbvax!hplabs | allegra!nbires | harpo!seismo } !hao!hull Wrong, wrong, wrong. You don't need 100,000 tons of water, and you don't have to appeal to any "small scale molecular interactions." Sure, it might take more than a gallon to get a 1/10" layer with a large surface area, but the amount of water is still relatively small. The ship's hull *is* "displacing" a large volume of water, even though that much water really isn't there. (You could also work out the potential energy of the water-ship system and find out that it does have a minimum, and therefore a stable equilibrium, with the ship "floating.") It shouldn't be too hard to find an object around the house (kitchen?) and a bowl of water to float it in such that the floating object weighs significantly more than the total amount of water in the bowl. -les niles
moore@ucbcad.UUCP (12/13/84)
> ******Arrgh. Here we go. The principle of Archimedes would, for the purposes > of this discussion, best be stated: "When an object is freely suspended in > a liquid, the object will be buoyed up by a force equal to the weight of the > displaced liquid." Therefore, if you have a 100,000 ton ship, you are going > to need 100,000 tons of water for it to displace; otherwise, a force of some > other description will be found to be responsible for supporting the ship. > > How is it that the water can hold up a ship which weighs > > more than the water? -dbell- > The layer of water has to be thin enough that small scale molecular interaction > between the water and the container transfers the force represented by the > weight of the ship directly to the walls of the container. Then the water is > between a rock and a hard place, and has no choice but to support the ship. > Howard Hull > {ucbvax!hplabs | allegra!nbires | harpo!seismo } !hao!hull RESOLVED : You can float a 100,000 ton ship in an arbitrarily small amount of water. PROOF : A gedanken experiment. Take your battleship and float it in the ocean (We agree that is possible). Now replace the replace the water more than an inch away from the hull with a wall instead. The remaining water is unaffected, since the wall will keep it in place as well as the replaced water did. The battleship is still floating (how would even `know' that the water not touching it has been removed?). So now the ship is floating on a inch thick layer of water. We can of course change the inch to an arbitrarily small thickness (at least until we start getting to thicknesses on the order of inter-molecular spacing), thus we can float the ship on an arbitrarily small amount of water. EXPLAINATION : I think part of the trouble is a instinctual belief in 'conservation of force'. We feel since there is be 100,000 tons of force holding up the ship, there must me 100,000 tons of something around applying the force. This just ain't so. As another author so nicely pointed out, the important concept is leverage; which allows a small force (the weight of the remaining water) to be translated into a large force (the 100,000 tons of the battleship). As to the Archimedes' principle, you place to much weight (excuse me) on the word displaced. Imagine our battleship floating in the ocean once again, and suppose we wish to figure out the forces applied to it by the surrounding water. One could do the proper vector intergals over the surface of the ship, but there is a much neater method. Imagine replacing the ship with a volume of water shaped like the portion of the ship below the water-line. Since this would result in just a flat expanse of water, this replacement shouldn't cause any re-arraignment of the water. Thus the net force on the replacement volume of water (and thus on the battleship it replaced) should be exactly equal to the volume's weight. Notice this argument does not require the replacement volume of water to actually exist, it only uses it as an imaginary artifice. EXPERIMENT : Get two nesting pots and try it. Peter Moore moore@Berkeley, ...!ucbvax!moore
jss@sftri.UUCP (J.S.Schwarz) (12/14/84)
> > How is it that the water can hold up a ship which weighs > more than the water? -dbell- When you figure it out, you'll probably also be able to explain how, at this very moment, I'm being supported by a chair that weighs less than I do. :-) Jerry Schwarz ihnp4!btlunix!jss
fons@mcvax.UUCP (Fons Kuijk) (12/18/84)
<line eater, this is another torpedo of the still floating battleship> In article <1301@hao.UUCP> hull@hao.UUCP (Howard Hull) writes: > "If an object is freely suspended in a liquid, it is buoyed up > by a force equal to the weight of the displaced liquid, or its > virtual equivalent." As far as I am concerned that's about the same as I wrote in my article. >So as you can see, I AM STILL serious about requiring the 100,000 tons of >water to be there even if it is in a virtual form, or as you put it, not >"really" there. Please let me apollogize in advance if I confused anyone >by my incomplete statement of the situation. Call it virtual and everything turns out to be possible! >So now I have a question. If we now wish to squeeze all of the water out of >the gap between the battleship and trough wall, what force pressing down on >the ship will be required to do so? If we can get that question answered, >then we can find an answer to the original question in its proper context... >The original question was "how can the water hold up something greater than >its own weight". This would seem to be a question of fundamental structure >were it not for the degree of freedom allowed by the width of the space >between the ship and the trough. So you are looking for the two balancing forces. That should not be to diffecult. Assume the ship is 1/10 inch above the trough. Then it takes 1/10 inch of lowering the ship. The force needed for that equals the gravitational force of the extra volume of the ship below the water level in the lowered position. In our case that is exactly the water that was in the trough. You can imagine that this is what you'd expect, because in that case, the total vertical force on the wall of the trough has not changed! Of course you can reduce the force after sqeezing out the water, provided the water cannot return. About the question of fundamental structure, well we've seen enough of that by now. --- No joke Fons Kuijk No graphics ...mcvax!fons No maxim [for insiders:-)]
jlg@lanl.ARPA (12/19/84)
> So as you can see, I AM STILL serious about requiring the 100,000 tons of > water to be there even if it is in a virtual form, or as you put it, not > "really" there. Introducing the VIRTUAL TIME OPERATING SYSTEM (VTOS). That's right, now you can use up all of that idle time on your machine without having to work night shift. With VTOS you can increase the throughput of you computer by as much as 50%!! Here is how it works: you think your program is running NOW, but in reality, it's running in the next idle time slice (probably next week some time). But you still get you answers back NOW ... ------------------------------------------------------------------------------ The greatest derangement of the mind is to believe in something because one wishes it to be so - Louis Pastuer James Giles
hull@hao.UUCP (Howard Hull) (12/20/84)
Yes. We have VTOS up and running here between the two VAXes (It's on the
Ethernet). Just "rlogin vtos" and execute anything you desire. Yesterday's
results are delivered to the line printer. Only trouble is, you have to
believe tomorrow exists before it will accept input (which must be in raw mode
with "stty noecho").
{ucbvax!hplabs | allegra!nbires | harpo!seismo } !hao!hull