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important buoyancy question


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Hi

Not being the acedemic type I have a question regarding bouyancy in boats. Now I know there was a greek fellow who worked all this out. :grin:

If I want to know where the waterline would be in a given hull how do I do that? - I think I am right in that the volume of the displaced water equals the volume of the submerged part. So therefore does it follow that the weight of the object equals the weight of the displaced water in order for the upward force to equal gravity

So all I have to do is find out the weight of my boat, equate that to litres of water and the level where that amount of water would be if it were inside the boat would be the waterline? :yawn:

Or do I sink?

thanks

Wayne

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Hi Wayne,

Yes, that sounds about right, the weight of the water dispalced would = the weight of the boat.

If you added weight to your boat it would go down, as more water would be needed to be displaced to match the new heavier weight, and vise versa..

Just in case you didn't know 1 litre of water weighs 1kg

:Stinky

Edited to say, the dispalced water = weight of whole boat not just the bit under water.

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Coincidentally I was watching a program about the Falkirk Wheel a couple of days ago.

In order to make each Gondola weigh exactly the same as the other, no matter how many boats are in each, they use Archimedes Principle.

Each Gondola is filled to the brim with water, and as the boats move in, they cause their exact weight in water to spill over the side , "a floating object will displace it's own weight of water". Then even if one Gondola has no boats and the other is full of them, they weigh exactly the same, and use minimal power to lift.

I'm still a pounds, shillings and pence man, so my yardsticks are "a pint of water weighs a pound and a quarter", so a gallon (eight pints), weighs 10 lbs, a nice round figure. Another fact (sadly still memorised from my fishtank days), is that a cubic foot contains 6.25 gallons, so a cu ft of water (or displaced volume), weighs 62.5 pounds, or a cu ft of air provides 62.5 pound of buoyancy. :)

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Not being academic myself, I would put the boat in the water and using a pencil draw a line around the waterline

good thinking col....

but.....

When you go onto the boat,, that line will dissapear under the water, unless of course you took out items to your weight.

:Stinky

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Is this the time someone should shout "EUREKA"?

I might be totally wrong but I'm not sure the weight of water displaced exactly equals the weight of the boat. Some boat specs state a weight and also a displacement, which are normally different. I'm trying to think back to O Level physics, but I think the displacement is more about measuring volume isn't it?

For example, a cube of lead will weigh different from a cube of gold the same size in the air. However, drop them in water, they will displace exactly the same volumn of water if they are the same physical size as they will both sink.

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We worked it out based on weight of the boat, therefore volume of water displaced. I still think this is wrong as I am sure it has more to do with the density of the water. For example a boat will sit higher in salt than fresh water.

Should have paid more ayttention at school.

Maybe I'll be having a submarine then :o

rgds

Wayne

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The principle is exact, for anything floating on water on water, (ie supported by it).

It holds true for salt and fresh, because the volume a boat displaces in salt water is less because the water itself is denser and heavier by volume.

The correct definition of the displacement of any floating object is exactly the same as it's weight (as loaded).

This holds true for a yacht or a super tanker.

The cubes of lead or gold are both heavier than water, so do not displace their own weight, but their total volume of water. (It only works for floating objects).

The exceptions are "tonnage" ratings, which are calculations based on volume, rather than weight or displacment.

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