Displacement of Scooters at Depth - Spun off from the A&I Discussion about Nothernone

[Dan]

No. No. No, no, no. Absolutely no.

Its buoyancy would be equaled to the weight of water that would replaced the air in the cavity, plus the weight of the water displaced by the material.

Dammit, this Archie guy figured it out some 2200 years ago.

I said neutrally buoyant.

 

[Dan]

Problem is that nobody defined buoyancy.
Some discussed of the antiparallel vector.

Here we should be discussing of two forces: hydrostatic force same direction, opposite verse to gravity acceleration and weight parallel and same verse of gravity acceleration.

You are going to confuse more people with such explanation

 

[Dan]

Buoyancy - Wikipedia

 

[fsardone]

You are going to confuse more people with such explanation

I said neutrally buoyant.

I hope not to be confusing anybody.
The volume of the scooter is equal to the volume of the air filled cavities plus the volume of the materials of which the scooter is made up.
That volume multiplyed by the fluif density is what produces the lift, the buoyancy or the hidrostatic force. When the scooter fills up with water only the air volume is lost. The materials’ volume is still providing lift and therefore what you feel being weight is diminisced by the volume of the materials by the density of the fluid.
The weight actually does not change (is mass by gravity acceleration) but it is partly compensated by the lifting hydrostatic force.

Cheers

 

[Dan]

No. No. No, no, no. Absolutely no.

Its buoyancy would be equaled to the weight of water that would replaced the air in the cavity, plus the weight of the water displaced by the material.

Dammit, this Archie guy figured it out some 2200 years ago.

As you say here:

Try re-doing the numbers using a 10L 300 bar steel tank

Mass: about 15kg
Internal volume: 10L
Steel volume: a little less than 2L (IIRC the density of steel is some 7.8 kg/L).
Buoyancy when empty: about 3.5 kp negative.
Buoyancy when flooded (or imploded, but if you can do that, you're pretty damned good!): about 13.5 kp negative.

The 10L 300 bar steel tank has 3.5 kp negative buoyancy when it’s empty, nozzle capped and in the water. When it is flooded, it’ll displace the 10 L of atm air with water, weighing 10 kg. Then the flooded tank would have bouyancy of -13.5 (-3.5 - 10) kp.

If an empty tank with 5L cavity is neutrally buoyant in the water then its buoyancy is 0 kp. If the cavity is displaced with water, then its buoyancy would - 5 kp (0 - 5) kp.

 

[johndiver999]

I think I see where the confusion is coming from. Assuming the AL80 being neutrally buoyant is wrong. Its buoyancy when empty is 2.1 kp positive. That’s why when its 11L cavity is flooded, the buoyancy would be -8.9 kp (-11 + 2.1).

If a DPV is neutrally buoyant, then its buoyancy would be equaled to the weight of water that would replaced the air in the cavity. Therefore it is good to know the actual volume of its cavity for estimating the buoyancy of flooded DPV.

Good to see you have changed your mind. As I’ve tried to say repeatedly when a submerged vessel that is initial filled with air, is vented and replaced with water, then the vessel gets heavier by the amount of water that enters. You had argued against this for quite some time on many posts.

 

[Dan]

I hope not to be confusing anybody.
The volume of the scooter is equal to the volume of the air filled cavities plus the volume of the materials of which the scooter is made up.
That volume multiplyed by the fluif density is what produces the lift, the buoyancy or the hidrostatic force. When the scooter fills up with water only the air volume is lost. The materials’ volume is still providing lift and therefore what you feel being weight is diminisced by the volume of the materials by the density of the fluid.
The weight actually does not change (is mass by gravity acceleration) but it is partly compensated by the lifting hydrostatic force.

Cheers

Understood.

All I’m saying is if the DPV is neutrally buoyant when you put it in the water, the whole DPV is in the water, but not sinking nor floating, then its buoyancy is 0 kp. When it is flooded then it will have negative buoyancy of the weight of the water that displaced the air. If the air cavity is 6 liters, then the buoyancy of the flooded DPV would be - 6 kp or its submerged weight would be 6 kg.

 

[fsardone]

All I’m saying is if the DPV is neutrally buoyant when you put it in the water, the whole DPV is in the water, but not sinking nor floating, then its buoyancy is 0 kp. When it is flooded then it will have negative buoyancy of the weight of the water that displaced the air. If the air cavity is 6 liters, then the buoyancy of the flooded DPV would be - 6 kp or its submerged weight would be 6 kg.

Agree

 

[johndiver999]

There is nothing special about the submerged vessel being neutral or negative or even positive, when water displaces the internal air space the change in buoyancy is equal to the weight of the water admitted to the air space. I’m glad the bc example was finally useful in getting the concept across.

 

[Dan]

Good to see you have changed your mind. As I’ve tried to say repeatedly when a submerged vessel that is initial filled with air, is vented and replaced with water, then the vessel gets heavier by the amount of water that enters. You had argued against this for quite some time on many posts.

That approach only works if the object is neutrally buoyant. The example given, AL80 is not neutrally buoyant. It has 2.1 kp positive buoyancy. So, you can’t just saying that the buoyancy of the flooded AL80 equals to the weight of water in the AL80 tank, which is 11 liters or - 11kp. It will be -8.9 (-11 + 2.1) kp. That’s all what I’m trying to say.