Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site redwood.UUCP Path: utzoo!watmath!clyde!burl!ulysses!allegra!mit-eddie!godot!harvard!seismo!hao!hplabs!hpda!fortune!rhino!redwood!rpw3 From: rpw3@redwood.UUCP (Rob Warnock) Newsgroups: net.physics Subject: Re: Floating a battleship in a gallon of water Message-ID: <97@redwood.UUCP> Date: Wed, 12-Dec-84 04:13:35 EST Article-I.D.: redwood.97 Posted: Wed Dec 12 04:13:35 1984 Date-Received: Fri, 14-Dec-84 05:55:05 EST References: <27@daisy.UUCP> <1296@hao.UUCP> Distribution: net Organization: [Consultant], Foster City, CA Lines: 86 +--------------- | ******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