| Topic: |
Science > Physics |
| User: |
"Tracker 1972" |
| Date: |
18 Jan 2004 11:49:51 AM |
| Object: |
Simple buoyancy question. |
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
--
mmmmmmmm, tasty.
of course itsnotitsnot :-)
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| User: "tj Frazir" |
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| Title: Re: Simple buoyancy mass displaced |
18 Jan 2004 10:52:43 PM |
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How much water you displace ...
30 feet x 170 feet x 10 feet
1000 tons of water would be in that box.
the weight of the water that is not there .
if you pushed it down to 12 feet it is 1200 tons
you displaced 200 more tons of sea water.
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| User: "Gene Nygaard" |
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| Title: Re: Simple buoyancy mass displaced |
19 Jan 2004 09:38:17 AM |
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(tj Frazir) wrote in message news:<22931-400B629B-228@storefull-3215.bay.webtv.net>...
How much water you displace ...
30 feet x 170 feet x 10 feet
1000 tons of water would be in that box.
the weight of the water that is not there .
if you pushed it down to 12 feet it is 1200 tons
you displaced 200 more tons of sea water.
So what do you call these 1.48 Mg tons, tj?
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| User: "Mark Mallory" |
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| Title: Re: Simple buoyancy mass displaced |
21 Jan 2004 03:37:30 AM |
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tj Frazir wrote:
How much water you displace ...
30 feet x 170 feet x 10 feet
1000 tons of water would be in that box.
the weight of the water that is not there .
Not even close, you illiterate Turd: 30' x 170' x 10' is 1444 cubic meters. 1
cubic meter of water weighs 3184 pounds, or about 1.6 tons.
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| User: "Gene Nygaard" |
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| Title: Re: Simple buoyancy mass displaced |
21 Jan 2004 05:56:43 PM |
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On Wed, 21 Jan 2004 09:37:30 GMT, Mark Mallory <nobody@nowhere.org>
wrote:
tj Frazir wrote:
How much water you displace ...
30 feet x 170 feet x 10 feet
1000 tons of water would be in that box.
the weight of the water that is not there .
Not even close, you illiterate Turd: 30' x 170' x 10' is 1444 cubic meters. 1
cubic meter of water weighs 3184 pounds, or about 1.6 tons.
I already asked tj about what he calls that fraser ton equal to 1.48
Mg; so what do you call your ton to distinguish it from all the other
tons out there?
Maybe a better question is, what do you call these pounds you are
using, equal to about 320 grams if there are 3184 of them in a cubic
meter of sea water?
Gene Nygaard
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| User: "tj Frazir" |
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| Title: Re: Simple buoyancy mass displaced |
21 Jan 2004 11:10:10 PM |
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I pulled mine out of my head without even doing the math in my head and
was close.
but you missed te point anyway.
30 feet by 10 feet by 170 feet is around 51000 cibic feet 1785 tons...
in my head again ,,not even a pencel .
I was thinkng the load with out the ships mass when I said 1000 tons (
ship talk ).
gw 700 hauls 1000 dwt
I know what any ship weighs by lookng at it at a glance.
your 3500 tons is way off.
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| User: "John C. Polasek" |
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| Title: Re: Simple buoyancy question. |
18 Jan 2004 04:14:11 PM |
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On Sun, 18 Jan 2004 17:49:51 GMT, Tracker 1972
<tracker1972@itisnthotmail.com> wrote:
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
We know water has a pressure gradient of .44 psi/foot. Let
R = rho density for water, and kR for the object, k<1.
dP/dD = R(1-k)*g is constant net pressure gradient
where g is 32 fpss.
For force we integrate, using A for Area(D):
Force = INT(R(1-k)*g*AdD = R(1-k)*g*Volume
so you get density x volume x g = Force.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
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| User: "John C. Polasek" |
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| Title: Re: Simple buoyancy question. |
19 Jan 2004 09:27:12 AM |
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On Sun, 18 Jan 2004 22:14:11 GMT, John C. Polasek
<jpolasek@cfl.rr.com> wrote:
On Sun, 18 Jan 2004 17:49:51 GMT, Tracker 1972
<tracker1972@itisnthotmail.com> wrote:
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
JP:
We know water has a pressure gradient of .44 psi/foot. Let
R = rho density for water, and kR for the object, k<1.
dP/dD = R(1-k)*g is constant net pressure gradient
where g is 32 fpss.
For force we integrate, using A for Area(D):
Force = INT(R(1-k)*g*AdD = R(1-k)*g*Volume
so you get density x volume x g = Force.
Maybe I should have added that the force is
Force = RVg - kRVg = Mg - kMg = W - kW or
Net force = weight of water - weight of ball.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
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| User: "Harry Conover" |
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| Title: Re: Simple buoyancy question. |
18 Jan 2004 10:43:31 PM |
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Tracker 1972 <tracker1972@itisnthotmail.com> wrote in message news:<tracker1972-EF2C21.17495018012004@news-text.blueyonder.co.uk>...
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
I believe that the point you may be missing is that buoyancy is a
result of gravitational forces, and with gravitational forces no
ambiguity regarding direction exists. In a gravity free environment
such as space, no buoyancy exists and a cork dropped into water would
not float.
Harry C.
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| User: "Steve Harris" |
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| Title: Re: Simple buoyancy question. |
19 Jan 2004 03:31:01 PM |
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(Harry Conover) wrote in message news:<7ce4e226.0401182043.54426468@posting.google.com>...
Tracker 1972 <tracker1972@itisnthotmail.com> wrote in message news:<tracker1972-EF2C21.17495018012004@news-text.blueyonder.co.uk>...
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
You've got it. Only it's not "slightly greater", it's a lot greater.
More than you think until you do the numbers.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
That's because you didn't do the numbers. A submerged tennis ball
displaces a tennis ball-shaped ball of water, which otherwise would be
motionless. That means the surrounding water is supplying enough
differential pressure to support a ball of water that size, which
means the force is M*g where M is the mass of a ball of water the size
of a tennis ball (134 grams or so). This is opposed by the weight of
the tennis ball m (less than 58.5 g), which significantly less. So the
force is (M-m)g and the mass is m and the acceleration is force/mass =
(M-m)g/m. What is the ratio (M-m)/m = (M/m) - 1? I make it to be
about 2.3 - 1 = 1.3. That's the number of g's initial acceleration of
a submerged tennis ball: 1.3 g's.
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| User: "Richard" |
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| Title: Re: Simple buoyancy question. |
18 Jan 2004 12:30:28 PM |
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Tracker 1972 wrote:
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
--
mmmmmmmm, tasty.
of course itsnotitsnot :-)
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
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| User: "Jim Greenfield" |
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| Title: Re: Simple buoyancy question. |
18 Jan 2004 08:19:14 PM |
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Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Tracker 1972 wrote:
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
--
mmmmmmmm, tasty.
of course itsnotitsnot :-)
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
OK
So if something compressible (water is not), sinks, it will go down
until the specific gravity of the unit sinking equals that of the
water... and remain at that level- correct?
Jim G
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| User: "Richard" |
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| Title: Re: Simple buoyancy question. |
18 Jan 2004 10:03:00 PM |
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Jim Greenfield wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Tracker 1972 wrote:
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
So is it just the pressure difference top to bottom or is it something
that just escapes me?
Thanks in advance,
Tracker1972.
--
mmmmmmmm, tasty.
of course itsnotitsnot :-)
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
OK
So if something compressible (water is not), sinks, it will go down
until the specific gravity of the unit sinking equals that of the
water... and remain at that level- correct?
Jim G
No.
Richard Perry
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| User: "Jim Greenfield" |
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| Title: Re: Simple buoyancy question. |
19 Jan 2004 06:55:38 PM |
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Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
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| User: "John C. Polasek" |
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| Title: Re: Simple buoyancy question. |
19 Jan 2004 07:37:28 PM |
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On 19 Jan 2004 16:55:38 -0800, (Jim
Greenfield) wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
My equation shows that if the ball is brought to a depth where it is
compressed so k is exactly 1, then it will stay at that depth in
unstable equilibrium.
If it is pushed down by minute amount, then k will increase from 1
and the ball will descend and accelerate.
If the ball is lifted by a minute amount, then k will decrease from 1
and the ball will accelerate up.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
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| User: "Richard" |
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| Title: Re: Simple buoyancy question. |
19 Jan 2004 09:39:23 PM |
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"John C. Polasek" wrote:
On 19 Jan 2004 16:55:38 -0800, (Jim
Greenfield) wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
My equation shows that if the ball is brought to a depth where it is
compressed so k is exactly 1, then it will stay at that depth in
unstable equilibrium.
If it is pushed down by minute amount, then k will increase from 1
and the ball will descend and accelerate.
If the ball is lifted by a minute amount, then k will decrease from 1
and the ball will accelerate up.
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
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| User: "tj Frazir" |
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| Title: Re: Simple buoyancy question. |
19 Jan 2004 11:05:50 PM |
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Hay richard ,,,,oil ya fool.
what an idiot
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| User: "Richard" |
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| Title: Re: Simple buoyancy question. |
20 Jan 2004 07:45:40 AM |
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tj Frazir wrote:
Hay richard ,,,,oil ya fool.
what an idiot
Are you saying that oil will sink if you take it down to a certain depth
and release it?
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| User: "" |
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| Title: Re: Simple buoyancy question. |
20 Jan 2004 12:18:28 PM |
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In article <400CA2EB.272CBA3F@yahoo.com>, Richard <no_mail_no_spam@yahoo.com> writes:
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
In order to provide the effect cited, the object must merely
be _less_ compressible than water _at_ its equilibrium depth
There is no need for negative compressibility (c.f. "Flubber") or for
downward curvature in the compressiblity graph.
Take a hypothetical perfectly rigid steel ball and sink it in a
hypothetical infinitely deep tank of water. Since water is
compressible, the density of the water in the tank increases with
depth. So the steel ball will sink until it reaches a depth at
which the density of water is equal to the density of the steel ball.
This will be a stable equilibrium. If the ball is displaced slightly
upwards, it will experience a downward net force. If it is displaced
slightly downwards, it will experience an upward net force.
The equilibrium depth for steel in water would be rather deep.
John Briggs
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| User: "Jim Greenfield" |
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| Title: Re: Simple buoyancy question. |
20 Jan 2004 07:40:25 PM |
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wrote in message news:<gfuslcNSUhBu@eisner.encompasserve.org>...
In article <400CA2EB.272CBA3F@yahoo.com>, Richard <no_mail_no_spam@yahoo.com> writes:
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
In order to provide the effect cited, the object must merely
be _less_ compressible than water _at_ its equilibrium depth
There is no need for negative compressibility (c.f. "Flubber") or for
downward curvature in the compressiblity graph.
Take a hypothetical perfectly rigid steel ball and sink it in a
hypothetical infinitely deep tank of water. Since water is
compressible, the density of the water in the tank increases with
depth. So the steel ball will sink until it reaches a depth at
which the density of water is equal to the density of the steel ball.
This will be a stable equilibrium. If the ball is displaced slightly
upwards, it will experience a downward net force. If it is displaced
slightly downwards, it will experience an upward net force.
The equilibrium depth for steel in water would be rather deep.
John Briggs
My thoughts were based on water being pretty rigid (negligible
compressability).
Apart from that, further complication arises as a weight has to be
attached to the ball to get it to sink in the first place, said weight
having the same compressability as water (to avoid further confusion).
Now to get equilibrium sink/rise, a fluke is needed, as the specific
gravity of the ball would need to be the same as that of the water _at
the level at which the volume of the ball has reduced thus far_ and of
course allowing for the weight (migraine!!!!)
Richard has things arse-up eh?
Jim G
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| User: "tj Frazir" |
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| Title: Re: Simple buoyancy constant D mercury oil |
20 Jan 2004 08:25:54 PM |
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Mercury is the nutral bouyant an ajusted befor the dive, Oil is pumped
into the bag to change the dimentions of the water that is displaced.
the dencity will not increase and the water will not compress.
How many psi push in each direction and how tall the colom pushes down
..
The higher the water tower is the harder the water comes out your hose.
Mass is bouyant in energy in space .
But in water 1000 feet of water pushes one foot less at 999.
In a steel can oil can make it nutral bouy.
Put the oil ina bag in the can with some air to make it nutral.
If there was no air it would sink ...a bit of air and rise. The oil
is the 99 % of the floataton and the air 1 % for the can to be nutral.
Thats good till the air gets smashed then it sinks . To get back up
fry some water .
It will take allot of air at 3500 psi to raise the can from 22,000 feet
, if ou used a can for air the can must hold 10000 psi to fill a void
3 times its sise.
To get the can back up weld the water for a lng time to get a bit of
high presure gasses .
As it comes up the exsess will just bleed out to sea at the lower end
of the can. If it dont the air will be at 3500 psi up top side.
Oil bag can and welder hydroelectrolisis .
You want to use the least amount of air to raise and lower.
You can take a 2000 psi oxygen bottle down and take the cap off
,,nothing will happen if you point it down. some water will go in .
It was at 2000 psi at top side.
But a bit of vaccuume down deap.
You can pour it out the oxygen bottle like it was just air and you were
just 10 feet down.
deap dives not it sinks ,,a dive takes thrust.
you nead to fry enouph water to fill a house with air to fill the 2000
psi bottle.
So you use murcury to ajjust oil to float and hydroelectrolisis to
fine tune .
Pumping oil into the bag is the best way and murcury to find nutral .
Changing
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| User: "Richard" |
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| Title: Re: Simple buoyancy question. |
20 Jan 2004 08:57:07 PM |
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Jim Greenfield wrote:
briggs@encompasserve.org wrote in message news:<gfuslcNSUhBu@eisner.encompasserve.org>...
In article <400CA2EB.272CBA3F@yahoo.com>, Richard <no_mail_no_spam@yahoo.com> writes:
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
In order to provide the effect cited, the object must merely
be _less_ compressible than water _at_ its equilibrium depth
There is no need for negative compressibility (c.f. "Flubber") or for
downward curvature in the compressiblity graph.
Take a hypothetical perfectly rigid steel ball and sink it in a
hypothetical infinitely deep tank of water. Since water is
compressible, the density of the water in the tank increases with
depth. So the steel ball will sink until it reaches a depth at
which the density of water is equal to the density of the steel ball.
This will be a stable equilibrium. If the ball is displaced slightly
upwards, it will experience a downward net force. If it is displaced
slightly downwards, it will experience an upward net force.
The equilibrium depth for steel in water would be rather deep.
John Briggs
My thoughts were based on water being pretty rigid (negligible
compressability).
Apart from that, further complication arises as a weight has to be
attached to the ball to get it to sink in the first place, said weight
having the same compressability as water (to avoid further confusion).
Now to get equilibrium sink/rise, a fluke is needed, as the specific
gravity of the ball would need to be the same as that of the water _at
the level at which the volume of the ball has reduced thus far_ and of
course allowing for the weight (migraine!!!!)
Richard has things arse-up eh?
Jim G
No Jim, you and the others are the ones waving their arses in the air.
What I said was perfectly correct.
Richard Perry
.
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| User: "Richard" |
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| Title: Re: Simple buoyancy question. |
20 Jan 2004 08:53:39 PM |
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wrote:
In article <400CA2EB.272CBA3F@yahoo.com>, Richard <no_mail_no_spam@yahoo.com> writes:
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
In order to provide the effect cited, the object must merely
be _less_ compressible than water _at_ its equilibrium depth
There is no need for negative compressibility (c.f. "Flubber") or for
downward curvature in the compressiblity graph.
Take a hypothetical perfectly rigid steel ball and sink it in a
hypothetical infinitely deep tank of water. Since water is
compressible, the density of the water in the tank increases with
depth. So the steel ball will sink until it reaches a depth at
which the density of water is equal to the density of the steel ball.
This will be a stable equilibrium. If the ball is displaced slightly
upwards, it will experience a downward net force. If it is displaced
slightly downwards, it will experience an upward net force.
The equilibrium depth for steel in water would be rather deep.
John Briggs
That's all well and good, but that wasn't the behavior that he
described. Read it again.
Richard Perry
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| User: "John C. Polasek" |
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| Title: Re: Simple buoyancy question. |
20 Jan 2004 08:31:52 AM |
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On Mon, 19 Jan 2004 21:39:23 -0600, Richard
<no_mail_no_spam@yahoo.com> wrote:
"John C. Polasek" wrote:
On 19 Jan 2004 16:55:38 -0800, (Jim
Greenfield) wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
My equation shows that if the ball is brought to a depth where it is
compressed so k is exactly 1, then it will stay at that depth in
unstable equilibrium.
If it is pushed down by minute amount, then k will increase from 1
and the ball will descend and accelerate.
If the ball is lifted by a minute amount, then k will decrease from 1
and the ball will accelerate up.
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
The ball is neither more nor less compressible: it's compressible. The
compressibiltiy is a constant equal more or less equal to 3/Youngs
modulus.
3Ydx/x = dP
dV = x*dP/3Y
The change in volume with P is proportional to size x and inverse to
the modulus of elasticity.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
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| User: "Richard" |
|
| Title: Re: Simple buoyancy question. |
20 Jan 2004 08:49:48 PM |
|
|
"John C. Polasek" wrote:
On Mon, 19 Jan 2004 21:39:23 -0600, Richard
<no_mail_no_spam@yahoo.com> wrote:
"John C. Polasek" wrote:
On 19 Jan 2004 16:55:38 -0800, (Jim
Greenfield) wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
My equation shows that if the ball is brought to a depth where it is
compressed so k is exactly 1, then it will stay at that depth in
unstable equilibrium.
If it is pushed down by minute amount, then k will increase from 1
and the ball will descend and accelerate.
If the ball is lifted by a minute amount, then k will decrease from 1
and the ball will accelerate up.
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
The ball is neither more nor less compressible: it's compressible. The
compressibiltiy is a constant equal more or less equal to 3/Youngs
modulus.
3Ydx/x = dP
dV = x*dP/3Y
The change in volume with P is proportional to size x and inverse to
the modulus of elasticity.
Provide a single example of an inanimate object exhibiting the behavior
that you described. I've seen rocks hovering above the ocean floor, but
I've never seen anything like you described. If you pull the hovering
rock down it will rise back up, if you pull it up it will descend back
down.
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
|
|
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| User: "John C. Polasek" |
|
| Title: Re: Simple buoyancy question. |
21 Jan 2004 09:02:05 AM |
|
|
On Tue, 20 Jan 2004 20:49:48 -0600, Richard
<no_mail_no_spam@yahoo.com> wrote:
"John C. Polasek" wrote:
On Mon, 19 Jan 2004 21:39:23 -0600, Richard
<no_mail_no_spam@yahoo.com> wrote:
"John C. Polasek" wrote:
On 19 Jan 2004 16:55:38 -0800, (Jim
Greenfield) wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
My equation shows that if the ball is brought to a depth where it is
compressed so k is exactly 1, then it will stay at that depth in
unstable equilibrium.
If it is pushed down by minute amount, then k will increase from 1
and the ball will descend and accelerate.
If the ball is lifted by a minute amount, then k will decrease from 1
and the ball will accelerate up.
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
The ball is neither more nor less compressible: it's compressible. The
compressibiltiy is a constant equal more or less equal to 3/Youngs
modulus.
3Ydx/x = dP
dV = x*dP/3Y
The change in volume with P is proportional to size x and inverse to
the modulus of elasticity.
Provide a single example of an inanimate object exhibiting the behavior
that you described. I've seen rocks hovering above the ocean floor, but
I've never seen anything like you described. If you pull the hovering
rock down it will rise back up, if you pull it up it will descend back
down.
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
I was simply assuming water to be incompressible relative to the
object. Everything is compressible of course. If you have to consider
both compressibilities then you have to deal with the compound
compliance of the two.
The compound bulk modulus would be
Y1*Y2/(Y1 + Y2)
but since this thread is (like most others) only an arm-waving
exercise, it would be fruitless to get fancier.
Any rock I know of will sink like a--well, rock. The rock you feature
is unique indeed because it has the same sp. gr. as the water. Now you
are treating the water as more compressible than the rock. But at your
diving depth, say even 100 feet, I'll bet the density of the water
does not change more than 1 part in 10,000, and the rock not at all.
In any case if the water is more compressible than the rock, then yes
it will behave opositely as you say, rising below the null point and
dropping above the null point.
But you are describing rubber water and really, really light rocks.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
.
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| User: "Jim Greenfield" |
|
| Title: Re: Simple buoyancy question. |
21 Jan 2004 11:58:06 PM |
|
|
John C. Polasek <jpolasek@cfl.rr.com> wrote in message news:<qb4t001k061s1rfpudsm0ftir33rmeboj9@4ax.com>...
On Tue, 20 Jan 2004 20:49:48 -0600, Richard
<no_mail_no_spam@yahoo.com> wrote:
"John C. Polasek" wrote:
On Mon, 19 Jan 2004 21:39:23 -0600, Richard
<no_mail_no_spam@yahoo.com> wrote:
"John C. Polasek" wrote:
On 19 Jan 2004 16:55:38 -0800, (Jim
Greenfield) wrote:
Richard <no_mail_no_spam@yahoo.com> wrote in message news:<400AD0C4.D8C96ECD@yahoo.com>...
Replace the ball with an equal volume of water: Since this water doesn't
flow in any direction wrt the body of water surrounding it, then it must
be that no net force is exerted on this volume of water, i.e. the
gravitational force on this volume is exactly canceled by the upward
water pressure acting on it. Now if you remove the water that surrounds
this volume, then the sphere of water left hanging in space would drop
with a force of gravity acting on it. Thus when the sphere of water is
enclosed by the surrounding body of water, then there is an upward force
on that sphere of water (in the form of water pressure) that is equal
and opposite to the force of gravity on the same volume of water. By
replacing the sphere of water again with the tennis ball, you have now
removed most of the mass in that spherical volume, and thus most of the
gravitational force acting on this volume, but the upward pressure due
to the surrounding water remains the same. Thus there is a net upward
force on the ball equal to the weight of the displaced volume of water
minus the weight of the ball.
This is the very argument that Archimedes used to derive the principle
named for him.
Richard Perry
I thought that you were correct, but still have a problem with the
fact that a compressible ball (under increasing pressure with depth),
displaces less and less water, and therefore should reach an
equilibrium.
Surely fish which live at depth do not have to continually do work to
avoid sinking to the bottom? Or do they alter their own bouyancy by
altering their SG? (produce a bag of gas or oil to suit?)
Jim G
My equation shows that if the ball is brought to a depth where it is
compressed so k is exactly 1, then it will stay at that depth in
unstable equilibrium.
If it is pushed down by minute amount, then k will increase from 1
and the ball will descend and accelerate.
If the ball is lifted by a minute amount, then k will decrease from 1
and the ball will accelerate up.
If the ball remains more compressible than water then its density will
continuously increase with increasing depth. Since it already has a
specific gravity greater than that of the water at the same depth, then
it certainly will not drop to a specific gravity equal to that of the
surrounding water by further compression. Though the water density is
also increasing the ball's density is increasing at a faster rate. It
will continuously sink to any depth available. Only objects that are
less compressible than water can reach an equilibrium level.
In order to provide the effect that you cite above, you must have an
object that is equal in density to the surrounding water at some depth,
but is less compressible than water above that depth, and more
compressible than water below that depth. I doubt that a material with a
negative compressibility curve exists, but I'm sure the Navy would find
it very useful:)
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
The ball is neither more nor less compressible: it's compressible. The
compressibiltiy is a constant equal more or less equal to 3/Youngs
modulus.
3Ydx/x = dP
dV = x*dP/3Y
The change in volume with P is proportional to size x and inverse to
the modulus of elasticity.
Provide a single example of an inanimate object exhibiting the behavior
that you described. I've seen rocks hovering above the ocean floor, but
I've never seen anything like you described. If you pull the hovering
rock down it will rise back up, if you pull it up it will descend back
down.
Richard Perry
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
I was simply assuming water to be incompressible relative to the
object. Everything is compressible of course. If you have to consider
both compressibilities then you have to deal with the compound
compliance of the two.
The compound bulk modulus would be
Y1*Y2/(Y1 + Y2)
but since this thread is (like most others) only an arm-waving
exercise, it would be fruitless to get fancier.
Any rock I know of will sink like a--well, rock. The rock you feature
is unique indeed because it has the same sp. gr. as the water. Now you
are treating the water as more compressible than the rock. But at your
diving depth, say even 100 feet, I'll bet the density of the water
does not change more than 1 part in 10,000, and the rock not at all.
In any case if the water is more compressible than the rock, then yes
it will behave opositely as you say, rising below the null point and
dropping above the null point.
But you are describing rubber water and really, really light rocks.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
Brilliant:
Wife: "Jim, mow the lawns this weekend!"
Jim: " Sorry dear; I have to go fishing"
Wife: (with feeling) "Have to?"
Jim: "Yes- I have to test an equilibrium of floatation hypthesis"
(exits right rapidly)
.
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| User: "tj Frazir" |
|
| Title: Re: Simple buoyancy question. |
20 Jan 2004 09:26:58 PM |
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The water dencity is not increasing.
Oil wount compress either.
.
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| User: "Mark Mallory" |
|
| Title: Re: Simple buoyancy question. |
18 Jan 2004 06:04:06 PM |
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Tracker 1972 wrote:
OK, so I know that the buoyant force generated is equal to the
weight/mass of the displaced liquid.
What I can't get my head around is, well, how does it know what 'up' is.
The pressure applied by the liquid is all over the surface so acts in
all directions at once, I assume.
The only thing I can think of is the slight increase in pressure as you
move down the object, so there is a slightly greater force at the base.
That's exactly what it is.
It just strikes me that the rate at which submerged objects such as a
tennis ball accelerate would require more force compared to the
admittedly small mass.
Work out the numbers for yourself! Calculate the pressure difference due to the
difference in height, multiply by the area of the top and bottom of the
submerged object (easy if it's cube-shaped), and you have the force. Compare
that force to the weight of liquid that the object's volume can hold.
.
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