Floating a Battleship in a Bathtub

daniel_i_l

Gold Member
865
0
I had always thought that things float when they displace an amount of water that weighs more than them. Bur obviously this is wrong because I recently read about "Floating a Battleship in a Bathtub" were the author of a physics book said that as long as there's enough water to fit between the boat and the tub, even a millimeter thick, the boat will float.
So what determines if it can float or not? He mentioned pressure on the boat.
Also, what is hydrostatic pressure and how does it have to do with this?
Thanks.
 

Doc Al

Mentor
44,642
966
You have to understand what is meant by amount of water "displaced". It means the volume taken up by the object compared to the surface of the water, wherever that happens to be. So, assuming the bathtub is big enough to float the battleship, the water level will rise to some point on the tub. The volume of water that you'd have to replace if you removed the ship (to keep the water level at the same raised point in the tub)--that's the amount of water "displaced". That (large) amount of water will weigh as much as the ship.
 

daniel_i_l

Gold Member
865
0
Why does that amount of water keep the ship up?
 

Doc Al

Mentor
44,642
966
It can be shown that the net upward force due to the water pressure on the object--the buoyant force--will equal the weight of the displaced fluid. That's called "Archimedes' Principle". If the buoyant force equals the weight of the object, the object floats. (If the buoyant force is less than the object's weight--it sinks like a stone.)

Read more here: http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html" [Broken]
 
Last edited by a moderator:

Doc Al

Mentor
44,642
966
Here's another way to picture the amount of water displaced that might make more sense. Imagine your giant bathtub filled to the brim with water. You carefully lower your battleship into it. The amount of water that overflows the tub--that's the amount of water displaced by the ship. Note that the amount of water that remains in the tub can be quite small--but the amount displaced must weigh as much as the ship does if the ship floats.
 
2,193
2
I am not convinced of this so-called effect of floating a battleship with, say, only 1 gallon of water, even after "filled bathtub displacement"
Here's why:

A bathtub IS NOT a closed system. As the battleship is lowerd into the water, water moves upwards along the sides of the tub. In order for the battleship to not touch the bottom of the tub, there must be an equal force opposing it.

Where is that force? I don't see it. IF something were to STOP the water from rising, than yes. But there is nothing but atmospheric pressure and the capillary effect to impede the rising water; surely much less than the force of a ship lowering.

So, the bottom of the heavy battleship will touch the bottom of the tub.

But, what about "special" circumstances? I've heard that a "V" shaped hull in water-filled "V" shaped container will produce pressure forces which allow this to happen with a minimal amount of water. Not sure if it's true.
 

rcgldr

Homework Helper
8,522
447
pallidin said:
Where is that force?
It's the pressure differential within the water versus the lower pressure of the air at the surface. Water pressure applies a force on the sides and bottom of the ship, but not on the top, so you have a net upwards force due to the pressure differential. The pressure in the water increases with depth as well, which also increases the pressure diffferential, and the net upwards force.

Gravity is supplying the original force that produces the pressure in the water and the downforce of the ship.

I've heard that a "V" shaped hull in water-filled "V" shaped container will produce pressure forces which allow this to happen with a minimal amount of water.
Pressure of water is related to the depth of the water, not the volume.
 

Doc Al

Mentor
44,642
966
pallidin said:
A bathtub IS NOT a closed system. As the battleship is lowerd into the water, water moves upwards along the sides of the tub. In order for the battleship to not touch the bottom of the tub, there must be an equal force opposing it.
The rising water, and thus increased water pressure difference along the surface of the ship, provides that buoyant force. The tub need only be deep enough so that the ship can float and not touch bottom.

Realize that this is just a thought experiment to dramatize Archimedes' principle. Lots of luck arranging things so that you can float a real battleship in a gallon of water. But the principle is quite real.

Looks like Jeff Reid beat me too it!​
 

DaveC426913

Gold Member
17,752
1,501
pallidin said:
Where is that force? I don't see it. IF something were to STOP the water from rising, than yes. But there is nothing but atmospheric pressure and the capillary effect to impede the rising water; surely much less than the force of a ship lowering.

So, the bottom of the heavy battleship will touch the bottom of the tub.

Just to clear up a possible point of miscommunication/misunderstanding - we are all agreed that:

for a hypothetical battleship that typically has (say) 10 feet of draft in open water, you WILL need a bathtub that is 10 feet deep

right?

As Doc put it:
Doc Al said:
The tub need only be deep enough so that the ship can float and not touch bottom.
 
Last edited:

rcgldr

Homework Helper
8,522
447
Doc Al said:
Lots of luck arranging things so that you can float a real battleship in a gallon of water. But the principle is quite real.
Taking a USA aircraft carrier through the Panama Canal with about 1 foot of clearance on the sides and not much on the bottom (by design), comes about as close as you'll get to a battleship in a bathtub.
 

Doc Al

Mentor
44,642
966
An excellent example!
 
2,193
2
I have seen this debated all over the web. Yet, both DocAl and Jeff Reid seem to provide both the theoretical dynamics and real-world example of this effect. Nice job.

I shall now meditate upon this issue while floating in my bathtub filled with only 16oz of beer.
Ok, I probably WILL sink, 'cause I will drink the beer while floating. ICK!!!
 
anyone out there have the math on this?
 
a formal physics solution. relevant forces in terms of densities, gravity, etc.
 

russ_watters

Mentor
17,945
4,446
buoyant force = weight of fluid displaced

Not sure what exactly you are looking for beyond that....

Did you read the link Doc posted?
 
Did you read the link Doc posted?
The link actually has some physics.........Shouldn't The battleship in the tub solution calculate the approx pressure gradient due to gravity and show how the differential pressure on the top versus the bottom of the ship will float it. Relevant parameters include ship mass, density of fluid, fluid volume, bathtub volume, and idealized ship dimensions (like a box: x by y by z).

What am I missing here?
 

Borek

Mentor
27,852
2,424
Could make an interesting exercise, but is not necessary since Archimedes.

Besides, you will get the same result.
 
ah well........I am not interested in answers. My interest lies in how to get them.

Just saying F=MA or in this case B= W displaced is very boring (to me) and provides zero insight. :zzz:It is just a workbook exercise. As in: "I have a formula. That's all I need."

It leaves out the beauty of physics.:smile:

Kepler told us how to compute lots of stuff about planetary motion....useful but BORING.
Newton provided the insight and the "why" behind the answers (ie. the beauty).
 

russ_watters

Mentor
17,945
4,446
This issue is so simple (a certain volume of water weighs what it weighs) that I don't see how anything is gained by your approach. But if you want, I'm sure you could calculate the weight of a block of water by integrating pressure gradient with depth, if you want. All that tells you though is the same thing that stacking weights on a scale tells you: the force on the scale is the sum of the weights of the weights stacked on it. By stacking 1lb weights one at a time, you can read the scale and verify that 1=1, 1+1=2, 2+1=3, etc. That's an integration force vs height of the stack. It really seems to straightforward of a concept to bother with to me.

But here's how to do it: multiply the cross sectional area (a single number if uniform, a function if not) by the integral of the weight density vs depth.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top