In article <477b4045@news.povray.org>, pgf### [at] optusnetcomau says...
> Gail Shaw wrote:
> > "Darren New" <dne### [at] sanrrcom> wrote in message
> > news:477ad72c$1@news.povray.org...
> >> You are sitting in a canoe, in a swimming pool, holding a cannon ball 
in
> >> your lap. You throw the cannonball overboard, and it sinks to the
> >> bottom. Does the level of water in the pool go up, go down, or stay th
e
> >> same?
> > 
> > I believe it should stay the same.
> > 
> > 
> 
> Incorrect.  Hint - The cannon ball sinks to the bottom.  Think about 
> what was preventing it doing so before.
> 
Umm. I don't see how that would be incorrect. While in the boat the 
*boat's* hull has to displace an amount of water sufficient to account 
for the added weight of the ball... Ok, you have a point. The 
displacement from the ball is going to be different "in" the boat than 
in the water, since one depends on the *weight* being shifted, while the 
other depends entirely on the *size* of the object. The question then 
is, will the displacement of the boat, from the cannon balls weight, be 
larger or smaller than the displacement of the cannon ball itself? 
Without knowing the size of the boat, the actual weight of the ball, and 
thus what the displacement will be from that, its not possible to 
project if the water will be "more" or "less" displaced by the inclusion 
of an entire cannon ball in the pool, versus a few millimeters of boat 
sinking slightly more into the water from the weight.

I.e., insufficient information to make a ruling.

-- 
void main () {

    if version = "Vista" {
      call slow_by_half();
      call DRM_everything();
    }
    call functional_code();
  }
  else
    call crash_windows();
}

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