Assuming the beach ball is filled with air and completely submerged at all times (near the surface), the ball will lose buoyancy the deeper it goes. The determining factor on buoyancy is displacement. Displacement is a factor of volume. As a beach ball goes deeper the pressure from the surrounding water will compress the air in the ball, reducing it's volume and displacement. It is the same reason that a submarine hull is compressed when it dives. I spent some time on the USS Baton Rouge in the 80's and they made a point of showing us a string stretched between two bulkheads was taunt on the surface, but there was a noticable sag when we were submerged.
-
Please take a moment and update your account profile. If you have an updated account profile with basic information on why you are on Air Warriors it will help other people respond to your posts. How do you update your profile you ask?
Go here:
Edit Account Details and Profile
beech ball buoyancy
- Thread starter thull
- Start date
- Status
Explanation for everyone BUT Brett:
Its a simple question, but it makes too many assumptions. As an avid scuba diver, and someone that knows about but hates engineering, I can tell you that:
The buoyancy of an object (upward force it creates) is exactly equal to the weight of the water that it displaces. Plain and simple. As you get deeper, the size of the beach ball decreases, and (it displaces less water). Resultantly, it is LESS buoyant.
If you assume that the size of the beach ball doesn't change at any given depth, then it is equally buoyant at all depths.
My explanation for Brett (slide rule required):
Volume of beach ball= Pi*r^3 = V
V/1gal = total gallons
Weight of water it displaces = 8.3lbs * total gallons
Buoyancy of displaced water = -(weight of water it displaces)
Its a simple question, but it makes too many assumptions. As an avid scuba diver, and someone that knows about but hates engineering, I can tell you that:
The buoyancy of an object (upward force it creates) is exactly equal to the weight of the water that it displaces. Plain and simple. As you get deeper, the size of the beach ball decreases, and (it displaces less water). Resultantly, it is LESS buoyant.
If you assume that the size of the beach ball doesn't change at any given depth, then it is equally buoyant at all depths.
My explanation for Brett (slide rule required):
Volume of beach ball= Pi*r^3 = V
V/1gal = total gallons
Weight of water it displaces = 8.3lbs * total gallons
Buoyancy of displaced water = -(weight of water it displaces)
thull
Well-Known Member
MY slide rule says volume of beach ball is (4/3)*Pi*r^3. Brett, what say your slide rule?
bouyancy is a function of density differential. The air in the ball is quite a bit less dense than water, hence it is bouyant. You could look at this a few ways. Assuming you're only going deeper by <10 feet or some small number, the ball isn't going to reduce in volume by much (if you inflated it with a reasonably amount of pressure) and the density of the water over a 10ft depth is fairly constant, so the bouyancy should be equal for both. If, however you're going down 50ft the ball will probably get smaller and the density of the water may change. This situation is FAR too complex for a stupid ASTB question, so my guess is the answer they are looking for is that the bouyancies are equal, under the assumptions mentioned above.
that's my 0.01
that's my 0.01
MasterBates
Well-Known Member
Brett, they just assume anyone with 4K posts is a Mod.
MasterBates
Well-Known Member
Speaking of Mods, anyone see/hear from Fly Navy lately? Last I knew he was going to the boat, but that was around thanksgiving time..
thull
Well-Known Member
MasterBates
Well-Known Member
Actually, 43.
thull
Well-Known Member
Ahem..43..right. YOU STILL CAN DO IT!!
- Status