Archimedes

[boomerangben]

Pye_end,

As I said earlier, the buoyancy force is the resultant of the all the fluid pressures surrounding an object acting on the wetted area. I demonstrated earlier that for a rectangular box shaped vessel, the buoyancy force can be easiliy determined mathematically. It is more complicated for a curved structure, but it works just the same. What Archidemes noticed was that when you do this calculation, you end up finding that the mass of water displaced by a shape is equivalent to the up thrust (buoyancy force) exerted on the shape, ie is a special case and can be used as a short cut.

The density of an object is a property, in the same way that its colour is. It is the same whether the object is underwater, in air or in outer space. No force is generated by a density.

If Naval architects used relative density, can you imagine how hard it would be to calculate the density of that part of the ship which is below the water line? In practice, Naval Architects only use the density of the water their creation is floating in. Only one density, therefore they cannot use relative densities. They don't need to , because of Archimedes Eureka moment.

 

[Pye_End]

So stick a light keel on your box - neutral buoyancy - big surface area - deep - your argument then says that the object will float at a different level?

 

[mel80]

No, because most of the surface area will be vertical, and as such the force from the water pressure will be lateral (though naturally, the force will be cancelled by the pressure on far side of the keel). If you try to avid this with a T sectioned keel, then the pressure on the upper surface of the horizontal section will, to a large extent cancel out the pressure on the lower surface (though obviously not entirely as it is slightly higher in the water column).

The cube example is easy, because the maths is simple, but you can if you want extend it to any underwater shape. The reason people don't is because there is a convenient shorthand; that doesn't however, change the underlying physical processes.

 

[Pye_End]

yes I understand about forces etc. - it is just that I do not think that it is a useful tool to work out whether something floats or sinks or where the waterline lies.

I am not suggesting that every part of a ship has to be worked out for density - you need the overall weight of the vessel, and this will tell you how much water will be displaced, and I am sure that by looking at the underwater profile it can be worked out how much volume/height this represents. They probably do consider variations in water density though!

How would you work it out from a set of drawings?

 

[Pye_End]

ok - put a box under it which is buoyancy neutral - lower down in the water means more upward 'pressure'?

 

[mel80]

Nice try! but if the box is neutraly buoyant , then you've added significantly to the weight. In other words, the forces on the box will act to support only the weight of box, leaving the waterline on the rest of the floating object unchanged.

 

[Pye_End]

But your argument is that lower down there is more head of water and therefore more upward pressure? That means the deeper you go, with the same density, you get more upward force according to you.

Have I missed something here?

 

[jimbaerselman]

Once under water, objects have pressure acting on top surfaces as well as bottom. What matters (and creates bouyancy) is the difference between top and bottom, and that remains constant, whatever the depth.

 

[mel80]

That is true, but in the case of the neutraly buoyant box, the increase in upward force on the floating object (caused by the increase in depth) is exactly equal to the added weight of the box (i.e. the box is neutrally buoyant).

If you consider the case wher the box is not attatched to the floating object, but is sumberged, then yes, increasing the depth will increase the upward force, but it will also incease the downward force caused by pressure on the top of the box. In other words, because the pressure increases linnearly with depth, the net force will always be the same.

Of course, the absolute forces will increase, which is why neutrally buoyant boxes (such as sumbarines) tend to have depth limits!

 

[Pye_End]

ok - stick a long box down on a pole - therefore water pressure difference between top and bottom faces - average density the same as the fluid it is in. What then?

 

[mel80]

Then the buoyancy (caused by the pressure difference) will be identical to the weight of the box, and the box will remain stationary, even when the pole is removed.

 

[jimbaerselman]

Submerge a round box, a long box, a flat box. If its volume is constant, the same vertical bouyancy force will always be generated. Long thin boxes have only a small area for pressures to act on, but the pressures will have a greater differential. Thin flat boxes vice versa. All shapes in between work the same way. Archimedes would merely say 'measure the volume of the water displaced. That'll tell you the volume submerged'.

If the shape is all submerged, the vertical buoyancy force will be the same as the force generated by gravity acting on the same volume of water.

The net force on the shape will be downwards if it's mass is the greater than that of the same volume of water. Upwards other wise. You could, of course, say 'more dense' and 'less dense' if you rally want to talk dirty. Simple really.

Or are you a troll, and me a dense victim?

 

[Pye_End]

So fill that box with air and you get a very different story - ie change its average density.

Glad you now agree that water pressure is inconsequential compared with object density! /forums/images/graemlins/grin.gif

 

[mel80]

Ah, but you have also changed it's weight without changing the net force caused water pressure.

Like I said earlier, you will always get the same answer using relative densities, but they can never explain the physics because a density cannot exert a force and you can't move an object (or counteract gravity) without a force. I will concede that the distinction is largely academic, but it is a crucial difference none the less.

 

[richardabeattie]

It seemed a simple qestion and I have now received 73 emphatic and differing answers. I particularly like the ones which suggest if you lower the barge into the dock when it contains just the tiny amount of water needed to fill the 1" gap the boat will float because other water elsewhere would like to get there if it could!

However a 12oz porridge bowl floats in 4 oz of water in a very slightly larger porridge bowl. So the Panama Canal must work.

 

[Cruiser2B]

/forums/images/graemlins/laugh.gif

The discussion about pressure is quite hilarious; pressure has nothing to do with buoyancy.
The water at the surface exerts one atmosphere of pressure on the hull and the hull exerts one atm of pressure on the water; 33 feet down the water exerts two atms of pressure on the hull and the hull exerts two atms on the water. Water is incompressible - a cubic metre of water at the surface (1 atm) weighs 1 tonne; the same as a cubic metre at 33ft depth - it still weighs 1 tonne, but is now pressurized to two atms. It has everything to do with relative density - take a cubic metre of water (contained within a theoretical shell with no mass or volume). This water is neutrally buoyant - push it down, it will go down (and the pressure will increase); push it up and it will go up. None of it will be supported above the surface of the water. Freeze that water - it still weighs the same (one tonne) but its volume has increased so now it floats - the submerged volume is exactly one cubic metre - the amount of water displaced is one tonne.

 

[Topcat47]

I really find a lot of this quite strange.

Consider a battleship sized hole....better yet one which contains the hull (up to the waterline) plus a cup of water. Then fit the battleship into the hole and pour a cup of water in too. The ship will float, in a cup of water, because that is the amount of water that was NOT displaced when the battleship went into the hole.

As Archimedes noticed, if you step into a full bath, you displace an amount of water equivalent to your own weight. If you then get out of the bath, the level will drop. Get back in and you'll still be in a bath full of water.

The "displaced water" is the volume of water that would have been there if it wasn't full of you, the battleship or whatever.

 

[Stoaty]

Strewth. Try this one. When the weight of water needed to fill the hole your boat leaves in the sea, is heavier than your boat, you float.

 

[duncanmack]

[ QUOTE ]
Strewth. Try this one. When the weight of water needed to fill the hole your boat leaves in the sea, is heavier than your boat, you float.

[/ QUOTE ]

Nope


When the weight of water needed to fill the hole your boat leaves in the sea, is the same weight as your boat, you float.

/forums/images/graemlins/smile.gif

 

[boguing]

Confused enough?

The seals on your under bunk water tanks are of questionable quality.

Which, if either, is worst? Filling the tanks through a 4" pipe up to to deck level, or a 2" pipe.