Yes, it does have to be lighter than the equivulant volume of water (as I stated in my first post), but you don’t need to have that water available to make the boat float.
Sure. You’re completely wrong.
Let’s start at the beginning. The so-called buoyant force results because an object immersed in a fluid has a greater force on the bottom (directed upward) than the force on the top (directed downward). This is due to the fact that pressure increases with depth in any fluid. This increase in pressure depends only on depth. It does not depend on the diameter of the column of fluid, nor does it depend on how much fluid is below the point at which you are measuring.
Since only depth matters, it matters not a whit how much water surrounds an object that is floating, so long as all surfaces below the waterline are wetted. As others have mentioned, this requires a suitably shaped container that is only slightly larger than the object that is floating (for the extreme case).
I have no idea where you are getting this gibberish about needing “at least twice the below-waterline volume of the carrier”.
No he’s not. You would not have to add “as much water as is equal to its volume.” A floating object displaces a weight of fluid equal to its own weight. In fact, as you pour water into the mold, and the water level creeps up the sides of the ship, the pressure at the bottom will eventually be high enough to float the ship. What matters is the depth of fluid, ONLY. If the mold is close enough to the dimensions of the ship, relatively little water will be required.
(Sigh.) No, Omar, you’re wrong, because you left out one key point. In order for an object to float, it must be lighter than an equal volume of water, WHICH IT HAS DISPLACED. You are confusing this displaced water with the water that remains. The amount of water that remains is irrelevant, so long as the floating object is completely immersed in it.
–robby (former physics instructor)
It doesn’t take much water at all. If you are asking how much water an AA displaces, well fresh water is about 8lb/gallon, slightly more for sea water. Do the math. (for your information, 1 gallon = .1336 ft^3)
Think about it like this. Assume I have a cube with dimension 10’ x 10’ x 10’. Suppose that the cube weighs in at 80lbs. How much water will the cube have to displace in order to float? (assuming water is 8lbs/gallon). Right, 10 gallons. Now, 10 gallons is 1.3 ft^3. So, in order to float the cube I need a vessel with a volume slightly larger than 1.3 ft^3. Knowing that my cube’s side is 100 ft^2, I deduce that I can float my 80lb 10’ cube in a shallow pan that is slightly larger than 10 x 10 x .013 ft.
Ok, not the clearest example I suppose. But I believe it is correct.
Conversion calculator here if you want it
This is a classic physics demonstartion. You can see a version of it here. The wood block weighs a good kilogram or so, but will happily float in about 50 cc’s of water.
The amount is irrelevant, so as long as the object is immersed? Sounds to me that is exactly what makes it relevant.
Let’s start over with Emilio’s Link.
http://www.physics.umd.edu/deptinfo/facilities/lecdem/services/demos/demosf2/f2-08.htm
The block of wood on the right is floating in 300ml of water. Drain it to 1 ml of water. Block of wood isnt floating anymore, is it?
Now, tighten the tolerance, or space, of that container. Now, that 1ml of water may float it again. But now drain that amount to .0001 ml of water. Block of wood doesnt float again, does it?
YES, THE AMOUNT OF WATER THAT REMAINS IS RELEVANT.
Omar, I beleive the point that robby was making is that the amount of water actually displaced doesn’t need to exist. It’s just theoretical and never needs to be placed in the container. So even though the block would displace 300 ML of water if placed in a swiming pool, in this container, you only need 1 ML of water.
Of course the amount of water that remains is relevent, but people are busy talking about the water that is displaced.
Omar, you are either confused or are being disingenuous. In your first post above, you are erroneously comparing the volume of water being displaced to that remaining in the container. In the second, you are apparently only referring to the volume remaining in the container, which is equivalent to the volume of water surrounding the floating object.
THE AMOUNT OF WATER THAT REMAINS (OR SURROUNDS THE FLOATING OBJECT) IS IRRELEVANT.
You just gave three examples, using volumes of 300 mL, 1 mL, and 0.0001 mL, in which your block of wood is floating. Did you notice that the volumes are all different? In other words, what is relevant is not the volume, but whether or not the floating object is completely immersed. With small enough tolerance between the object and the container in which it is immersed, this volume can be very small indeed.
If you can change a variable (volume, in this case) and get the same results, it is not relevant.
I just wanted to take back what I said earlier. Ya’ll have convinced me.
Not to beat a dead horse, but here’s a thought experiment for anyone not convinced by the (quite thorough) explanations already given above. Plus, everyone likes ASCII art…
Take a carrier in the middle of the ocean. Construct a cylindrical, water-tight, steel barrier around it – all the way to the ocean’s floor:
~~~~water level~~~~~~~|~~~~\carrier/~~~~|~~~~~~
| |
water-tight | |
steel barrier -->| |
(cylinder) | |
| |
====ocean floor================================
Now pump all the water out of the ocean, leaving only what’s inside the cylinder. The carrier continues to float. Now construct an artificial floor under the carrier, and pump all the water out from underneath the floor. The carrier continues floating…
|~~~~\carrier/~~~~|
| (water in here) |
artificial floor-->|-----------------|
| |
| |
| |
====ocean floor================================
Continue constructing tighter boundaries (i.e., higher floors and closer walls), pumping away the water from each new separated region. When you reach your goal (1 gallon of water left, say) you can stop, and the carrier was floating the entire time.
-P
You need a better understanding of fuid mechanics. All that counts is the pressure of the water and the pressure of the water is only dependent on the height to the surface. If I show you a spigot coming out of a wall and it has a pressure gauge showing X PSI, the only thing you know from that is the height to the free surface but you have no idea of the amount of water behind the wall. It can be a pint or it can be millions of gallons.
As long as the aircraft carrier is covered with water to the flotation line it had in the open ocean, it will float.
In fact, it is very possible that some dry docks may float ships in amounts of water which are smaller than the ship’s weight.
Not to mention the experiment mentioned by tastycorn. Just do it. To insist on denying something so easily verified is foolish.
Wow! Pretty impressive ASCII art! 
yoyodyne, I think I’ve figured out where your reasoning goes wrong. Where you say “That water has to remain in the container, or the boat touches bottom,” you have gone astray. The displaced water need not remain in the container.
Think about it: in the initial case, the outer layer of water supports a volume of water that one could imagine was in the shape of a ship. Remove this water, and drop in a ship that displaces this same volume of water, and has a weight equivalent to the weight of the displaced water. As far as the surrounding water and the container are concerned, the two situations are identical. The displaced water need not be present.
BTW, if some replies in this thread seem somewhat short, it is no doubt due to the fact that this is basic fluid mechanics. There is no debate here, and to insistently profess statements that contradict physical reality is indeed foolishness.
Ok, let’s start over with the carrier example.
-
You have a carrier.
-
You have a mold of the carrier just a little bigger than the carrier.
-
You drop your carrier in the mold.
-
You fill up the mold until the carrier JUST starts to float. Say this amount is 100 gallons.
-
The carrier is floating. In only 100 gallons of water.
-
You now drain 25 gallons. 75 gallons are left. Is the carrier floating? No.
-
Drain another 25 gallons. 50 gallons are left. Is the carrier floating? No.
-
Drain another 25 gallons. 25 gallons are left. Is the carrier floating? No.
THE AMOUNT OF WATER REMAINING IS RELEVANT.
If you change the container to be smaller, than the amount needed to float a carrier has also gotten smaller. If you change the container to be much larger, than the amount of water needed to float the carrier has also gotten larger.
THE MINIMUM AMOUNT OF WATER NEEDED TO FLOAT A CARRIER, OR A GIVEN OBJECT IS NOT A CONSTANT.
It is forever changing, depending on the object that needs to be floated, and the container.
You cannot pick a set amount of water, say 100 gallons, and float a carrier in ALL AND EVERY scenario, such as a swimming pool ten times the size of the carrier.
YES, THE AMOUNT OF WATER REMAINING IS RELEVANT.
<sigh> “It’s taking longer than we thought…”
DNFTT
Omar, nothing you are saying is wrong. It’s just not what we’ve been talking about since the begining. While there must be enough water to fill the gaps between the battleship and the mold, that is dependent on the shape of the two. The point of contention in this entire thread has been about the displaced water, not the remaining water.
You are correct that if you have too little water remaining the ship won’t float. But all you need to do is adjust the shape of things. No one argues that point because it’s not interesting. The interesting question was whether you need water equivilent to the weight of the ship.
No one is trying to pick a single amount of water necessary to remain in the mold. It’s not important since it’s not related to the weight of the ship. It’s just an artifact of the shapes of the objects and the gaps.
Exactly right. Therefore, the volume of water surrounding the floating object is not a critical variable, or in other words, it is irrelevant.
What is relevant is the pressure of the water beneath the floating object, which is solely dependent on the depth of the water at that point.
Omar, you are taking a very pedantic, narrow view of the term “relevant.” The point is that is does not matter what volume of water surrounds the ship, so long as the ship is completely immersed below the waterline. The ship can be in a matched mold, a narrow canal, or the ocean.
I think I’ve settled on “disingenuous.”
He also appears to be desperately searching for a way to validate his initial erroneous statements.
Incidentally, Omar, you never did explain how your cite on the previous page demonstrates that tastycorn is “pretty much wrong.”
I also find it fascinating that, in the space of a single thread, you have gone from humble tyro as evidenced in the following quotes to the unparalleled mastery of the subject you now enjoy.
Another example is in the design of dams…
The force that the dam must hold back is only a function of how tall the dam is.
a dam 200ft tall and with a reservoir 100 miles long has to be just as strong as a dam 200ft tall but with a reservoir that goes back only a few inches.
the force that water exerts on an object (what keeps a sip afloat) depends only on the pressure of said water, in other words, the height of the column of water, not the volume of it.
Ok, robby. Answer me this.
If you take a ship from a mold, to a narrow canal, to an ocean, does the MINIMUM amount of water needed to keep that ship afloat in each body of water change?
In other words, does 10 gallons work in the mold as well as the narrow canal as well as the ocean. Or in ANY size hole on this planet big enough to fit the said ship.
If your answer is:
“No, the amount of water does not matter. 10 gallons will work in ALL those examples”
Then the amount of water needed to float the said ship is irrelevant.
If your answer is:
“If I go from a mold to a canal, I gotta add more water to keep my ship afloat. And I definitely have to add more water if I go to a bigger size, such as the ocean.”
So if you have to change the amount of water needed to keep your ship afloat, tell me again how the amount of water needed is irrelevant?
Yes, the DISPLACED water is irrelevant.
Being disingenuous is saying that the size of the container does matter. In the fact that said object needs to AT LEAST fit in the container. That is being disingenuous.
robby, the used car seller: "Yeah, my car doesnt care what amount of gas is needed to run. It could be 25 gallons, 30 gallons, 35 gallons, or it could be 40 gallons. The car will run (the ship will float). The amount simply does not matter.
‘Non-proficient in the ways of cars’ buyer: “Sounds great! I will take it”
Buyer than drives the car 100 miles home, to his driveway, until there is .1 gallons left in the car when he shuts it off.(Emptying out water to below the minimum needed to float said ship in said container)
Next morning, car does not start. Buyer gets on the phone.
Buyer: “Hey, I thought you said the amount of gasoline was not relevant.”
robby: “Well, duh. You need enough to reach the fuel pickup, the fuel lines, and the fuel injectors” (Amount needed to at least immerse the ship).
Buyer: “Well, I dont feel that you exactly lied to me, but I feel that you were being very disingenuous. I want my money back. I am taking you to Judge Judy.”
Judge Judy’s courtroom
Judge Judy: (Addressing robby, the defendant) “Why did you tell the plantiff that the amount of gas needed to run the car was irrelevant?”
robby: Your honor, if you read up above, I was talking about DISPLACED gasoline. The tank only holds 15 gallons. Any volume above that is just going to spill onto the street.
Judge Judy: “So you are saying that the amount of displaced gasoline is irrelevant, but the minimum amount of gasoline is ALSO irrelevant?”
robby: “Yes, your honor. The plaintiff could of relocated the fuel tank to the front of the car. He could of used smaller ID fuel lines. Smaller injectors. (changing the size of ship container) Then the car could of ran on .1 gallons again.”
Judge Judy: “But what happens when the plaintiff runs down to .0001 gallons of gas?” (changing minimum amount of water needed to make ship float)
robby: “Ummm…”
Judge Judy: “Guilty of being disingenuous. Give the man his money back”
Yes, I am being comical in that situation.
The fact is, to someone that does not grasp the entire concept at all, you cannot say:
THE AMOUNT OF WATER NEEDED TO FLOAT A GIVEN OBJECT IN A GIVEN CONTAINER IS IRRELEVANT.
Someone is going to read all this. He is gonna go to his neighbor’s house, who has a canoe sitting in an EMPTY olympic size swimming pool. He’s gonna tell his neighbor that he just learned on the internet that is doesnt matter what amount of water it takes to float his canoe in that swimming pool. The amount of water needed is irrelevant. The neighbor is going to hand him a 8oz glass of water and say, “Alright, Einstein, go make my canoe float.”