Greg M. Johnson <pte### [at] thecommononethatstartswithycom> wrote:
> That's where I was headed. Does the cannonball displace more
> water "underwater" than in the boat

  No, it displaces less when underwater because its volume is "less" than
its weight, when compared to the water.

-- 
                                                          - Warp

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Warp wrote:
> Greg M. Johnson <pte### [at] thecommononethatstartswithycom> wrote:
>> That's where I was headed. Does the cannonball displace more
>> water "underwater" than in the boat
> 
>   No, it displaces less when underwater because its volume is "less" than
> its weight, when compared to the water.

Right. And you know that *because* it sinks.  The water level wouldn't 
change if the thing you threw over floated.

-- 
   Darren New / San Diego, CA, USA (PST)
     It's not feature creep if you put it
     at the end and adjust the release date.

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Darren New nous apporta ses lumieres en ce 2008/01/01 19:13:
> 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 the 
> same?
> 
> (I've asked this of probably a dozen or more scuba dive instructors, and 
> only one has gotten it right. The reasoning behind the correct answer is 
> obvious once you hear it. I don't remember if I got it right when I 
> heard it.)
> 
The level will go down.
When in your hand, you displace a volume of water that have the mass of the 
cannonball. When you drop it, the displace a volume of water egual to it's own 
volume. The cannonball have a higher density than the water as it sink.

-- 
Alain
-------------------------------------------------
You know you've been raytracing too long when you wonder if ground fog or 
athmosphere will look better for your company's market share pie chart.
Christoph Rieder

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Warp <war### [at] tagpovrayorg> wrote:
> Greg M. Johnson <pte### [at] thecommononethatstartswithycom> wrote:
> > That's where I was headed. Does the cannonball displace more
> > water "underwater" than in the boat
>
>   No, it displaces less when underwater because its volume is "less" than
> its weight, when compared to the water.
>
> --
>

Rethinking.
Using symbols:
Ph20= density of water
Pfe= density of cannonball
Mball= mass of canonball


IN CANOE:
displacement = Mball /(Ph2o)

UNDERWATER:
displacement = Mball / (Pfe)



But I'm not even sure if that's completely right now.

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On Wed,  2 Jan 2008 12:32:33 EST, "gregjohn" <pte### [at] yahoocom> wrote:

>
>But I'm not even sure if that's completely right now.

In the boat the cannonball displaces its own mass of water. In the water the
cannonball displaces its own volume of water.
Do I need to go on? :)


Regards
	Stephen

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gregjohn <pte### [at] yahoocom> wrote:
> IN CANOE:
> displacement = Mball /(Ph2o)

  No. In the canoe the ball displaces an amount of water equal to the
weight of the ball.

> UNDERWATER:
> displacement = Mball / (Pfe)

  In the water the ball displaces an amount of water equal to the
volume of the ball.

  The amount of water which weights the same as the ball is larger than
the amount of water which has the same volume as the ball.

-- 
                                                          - Warp

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Warp <war### [at] tagpovrayorg> wrote:
> gregjohn <pte### [at] yahoocom> wrote:
> > IN CANOE:
> > displacement = Mball /(Ph2o)
>
>   No. In the canoe the ball displaces an amount of water equal to the
> weight of the ball.
>


That's what I said, only more precisely.

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Darren New wrote:
> 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 the 
> same?
> 
> (I've asked this of probably a dozen or more scuba dive instructors, and 
> only one has gotten it right. The reasoning behind the correct answer is 
> obvious once you hear it. I don't remember if I got it right when I 
> heard it.)

It goes down as measured against the edge of the pool.  While in the 
canoe, the cannonball displaces its mass in water, but in the water it 
displaces only its volume.  Since the cannonball is denser than water, 
it displaces more while in the canoe than in the water.  Since there is 
less displaced water, the water level of the pool goes down.

Regards,
John

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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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Patrick Elliott <sel### [at] rraznet> wrote:
> 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.

  The size of the boat nor the weight of the ball make any difference on
the original question: The original question didn't ask how much the water
level changes, it only asks if the water raises or lowers. To answer this
question the only information needed is whether the cannonball has higher
density than water or not. Since cannonballs are made of metal, the answer
is yes. Thus its weight displaces more water than its volume, and thus
the answer is that the water level lowers.

-- 
                                                          - Warp

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