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