Instructables

How do I calculate buoyancy?


How do I do it?  Floating on water (please include examples)
btw - I'm making a boat out of a motorbike

Prfesser2 years ago
Example:  Consider a piece of wood 10x20x20 cm with a density of 0.6 g/cm^3, in water that has a density of 1.0 g/cm^3.

The wood is less dense than water, so it will float.  It weighs 10x20x20x0.6=2400 grams.  That is the mass of water it will displace when it floats.

The volume of the wood is 4000 cm^3.  That volume of water would weigh 4000 grams.  How much additional weight must be added to the wood before the combination weighs 4000 g? (Answer: 1600 g).  So a piece of, say, lead weighing 1599 g could be glued to the wood and it would still float---just barely.

Hope this clarifies.
Butts_booties4 months ago

butts and booties all the way

kelseymh2 years ago
Why don't you search for "buoyancy"? Then you could do your homework yourself.
SirNoodlehe (author)  kelseymh2 years ago
im making a boat out of a motorbike
Add that to the text of your comment. It makes a difference.
rickharris2 years ago

In simplistic terms

An object displaces the same weight of water as it weighs

If you calculate in Kg 1 Kg of water is the same as 1 Ltr volume of water.

So an object of 100 ltr volume will when totally immersed displace 100 Kg of water.

When the amount of water displaced = the weight of the object it will be floating and will not go deeper into the water.

So a 50 KG object will displace 50 ltr of water if the volume of that object is 100 Kg it will only be half way into the water.

For mathematical calculations see other answers.
For an object partially, or totally, submerged in water, there is an upward buoyant force, equal to the volume of displaced water times the density of the water times g. Call it B = ρwater * g * Vd This force acts upwards.

The weight of the object, W = m*g, pulls downwards. 

The sum of these forces is ΣF = B -W = ρwater * g * Vd - m*g

And in the case where object is floating on the water, ΣF = 0. Then solving for Vd gives you:

Vd = m/ρwater

Which basically says that the object displaces a volume of with the same weight as itself.

By the way, this displaced volume is some number less than the total volume of the object. Call that vtotal.  So Vd kind of has to be somewhere in the range:  0 < Vd < Vtotal

Also it makes sense to think about what fraction of the object is submerged, i.e the ratio:  Vd / Vtotal  If this number is much smaller than one, then most of this object's volume is out of the water, and that's probably a good a thing if you're designing a boat.   Also notice that for the case of an object floating on water,  this ratio is equal to the ratio of the (average) density of the object to the density of water.

Vd / Vtotal = (m/Vtotal)/ρwater < 1

This suggests another trick. Supposing you know m, i.e. how much the whole thing is going to weigh.  You just make Vtotal big enough that your average density is less than that of water m/Vtotal < ρwater

Of what, and in what ?

Steve