# 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

How do I do it? Floating on water (please include examples)

btw - I'm making a boat out of a motorbike

Play with Math: Make Animated GIF and HTML5

by
laxap

How to Construct a Simple Boat

by
notjustsomeone

Cyclicity, Numbers Maths tricks CAT GMAT GRE Prezi

by
nconsul

Maths Tricks, HOW TO find squares in seconds. GRE CAT APTITUDE, GMAT SPEED MATHEMATICS

by
nconsul

Penny Barge Activity

by
kinetike

Incredible Soda Bottle Pontoon Boat

by
deceiver

Control Your Robot Using a Wii Nunchuck (and an Arduino)

by
oomlout

© 2015 Autodesk, Inc.

forgot your password or username?

it happens.

it happens.

Enter the email associated with your account and we will send you your username and a temporary password.

Not a member? Sign Up »

We have sent you an email with a password reset code. Please enter it below.

Not a member? Sign Up »

active| newest | oldestThe wood is less dense than water, so it will float. It weighs 10x20x20x0.6=2400 grams. That is the

massof water it will displace when it floats.The volume of the wood is 4000 cm^3. That volume of

waterwould 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 and booties all the way

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.

_{water}* g * V_{d}This force acts upwards.The weight of the object, W = m*g, pulls downwards.

The sum of these forces is ΣF = B -W = ρ

_{water}* g * V_{d}- m*gAnd in the case where object is floating on the water, ΣF = 0. Then solving for V

_{d}gives you:V

_{d}= m/ρ_{water}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 v

_{total}. So V_{d}kind of has to be somewhere in the range: 0 < V_{d}< V_{total}Also it makes sense to think about what fraction of the object is submerged, i.e the ratio: V

_{d}/ V_{total}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.V

_{d}/ V_{total}= (m/V_{total})/ρ_{water}< 1This suggests another trick. Supposing you know m, i.e. how much the whole thing is going to weigh. You just make V

_{total}big enough that your average density is less than that of water m/V_{total}< ρ_{water}Steve