Den Pressure - A Definable Hypothesis & Experiments (Scepti, iWitness)

Why can’t the one side just stay compressed and the other less compressed?

Because you've taken away the central divider and allowed the higher pressure to expand and the lower pressure to compress by that expansion to create equilibrium.

I'm not entirely sure what you're getting at.
If you've got something to add then come out with it.

Any idea, what?
You havent made a point.
You said you had a method.
Im curious where your thought process goes with this.
The hypo rock weighs hypo 6jugs.
Then what do we do with info?

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

We need some measured weight  reading on land to measure mass. How are you going to do it?

Quote from: Themightykabool on January 17, 2019, 02:09:05 PM

Any idea, what?
You havent made a point.
You said you had a method.
Im curious where your thought process goes with this.
The hypo rock weighs hypo 6jugs.
Then what do we do with info?

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

We need some measured weight  reading on land to measure mass. How are you going to do it?

What was your point?
Yes
We equated a unit of measure (jugs) as a std unit of weight.
Then what?
What is your point?

Are you confused about arbitrary units of measure?
If you were to sell me a chocolate bar.
Would taht be in usd? Cad? Peso? Euro?
Bench marked at 1990?
2019?
How far is the washrooom from your bedroom?
20ft?
Whose foot?
My foot?
Shaquiloneils foot?

Is that what youre hung up on?

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

You can say, "6 jugs," pointing to the 6 jugs.

Quote from: sceptimatic on January 17, 2019, 10:26:50 PM

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

You can say, "6 jugs," pointing to the 6 jugs.

How much does 6 jugs weigh?

It does push back if its pushed into.
However a mass on a rope statically above the understack will not push down onto it and so, will not compress into it for it to push back.

Then why should the air anywhere else push?
The object isn't moving. It isn't pushing into the air above or to the side.

Even if you did have it magically push into the air above, as soon as it starts to fall, that is no longer the case and instead it is pushing into the air below and thus should be pushed up. In your model, if anything, the air should act like a solid and prevent any motion at all.

It's raised up on a rope and displaces atmosphere above it.

Repeating the same claim which has been questioned doesn't help you.
Again, why is it only displacing the atmosphere above it?

All that atmosphere above is pushing down against the mass compressing into it

Why isn't all the atmosphere below pushing up against the mass compressing into it?

All the below atmosphere is, is a resistance to the above atmosphere in the stack but not to the person/object hanging, because there's no push down by the person or object.

Then why isn't the atmosphere above just a resistance to the below atmosphere in the stack but not to the person/object because there's no push up by the person/object?

Again, you have provided no justification for the directionality.

Again, a nice simple example, you have a ball, like a tennis ball. You position your hand such that your thumb is on the left side and your fingers are on the right side, with nothing above or below.
You hold it at shoulder height, close to your shoulder and then extend out your arm so the ball moves straight out horizontally. You now release the ball.
How should the ball move and why?
The ball is not pushing into or displacing the atmosphere above any more than it is doing to the atmosphere below. As such, that cannot provide a basis for the directionality.

How much does 6 jugs weigh?

Considering you want to go down this stupid rabbit hole how about we deal with a more basic question?
How many is 6?
What is a jug?
What is weight?
What is much?
What does any of those strange collection of dots on the screen actually mean?

This is in no way relevant to the discussion at hand.
In order to have a discussion words must be used which have meaning.
Discarding that meaning doesn't help.
The same applies when discussing physical quantities like weight. You have to have some defined unit to describe it.
It doesn't matter if you use kgf, N, jugs, or anything else (other than with regards to adhering to standards which are agreed for ease of communication). None of this has any bearing on if weight exists or not before you have quantified it accurately.

Quote from: sobchak on January 17, 2019, 01:54:02 PM

Why can’t the one side just stay compressed and the other less compressed?

Because you've taken away the central divider and allowed the higher pressure to expand and the lower pressure to compress by that expansion to create equilibrium.

Can you describe the final equilibrium state in terms of pressure and density?

Quote from: Stash on January 17, 2019, 10:57:32 PM

Quote from: sceptimatic on January 17, 2019, 10:26:50 PM

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

You can say, "6 jugs," pointing to the 6 jugs.

How much does 6 jugs weigh?

6 jugs weighs exactly 6 jugs. Am I wrong?

Quote from: sceptimatic on January 17, 2019, 10:12:58 PM

It does push back if its pushed into.
However a mass on a rope statically above the understack will not push down onto it and so, will not compress into it for it to push back.

Then why should the air anywhere else push?
The object isn't moving. It isn't pushing into the air above or to the side.

The object is pushing into the atmosphere above and to the side, because the atmosphere is pushing into the mass as the mass displaces it.
It is not displacing the stack below. It's a stack.

Even if you did have it magically push into the air above, as soon as it starts to fall, that is no longer the case and instead it is pushing into the air below and thus should be pushed up.

If it falls then your rope has snapped and now the mass is pushed down against the resistance of the stack below which is in no way dense enough to counteract that push enough to do anything other than friction grip it and compress until the object hits a foundation.

 

In your model, if anything, the air should act like a solid and prevent any motion at all.

I have no clue why you even think this. It baffles me.

Quote from: sceptimatic on January 17, 2019, 10:12:58 PM

It's raised up on a rope and displaces atmosphere above it.

Repeating the same claim which has been questioned doesn't help you.
Again, why is it only displacing the atmosphere above it?

Simple answer is a foundation allowing it to, whether its on a solid ground directly beneath it or on a suspended rope or whatever above the atmospheric stack below it.

Quote from: sceptimatic on January 17, 2019, 10:12:58 PM

All that atmosphere above is pushing down against the mass compressing into it

Why isn't all the atmosphere below pushing up against the mass compressing into it?

Because it's stacked from below and cannot push up unless something forces it up, which would be any mass under energy push.

A stack is a stack. It stays as literally a stack. It doesn't matter how high you go you will always be on top of a stack of atmosphere that is doing nothing more than being a ready under resistance to a release of mass withing it under energy which has pushed through it from below.

Quote from: sceptimatic on January 17, 2019, 10:12:58 PM

All the below atmosphere is, is a resistance to the above atmosphere in the stack but not to the person/object hanging, because there's no push down by the person or object.

Then why isn't the atmosphere above just a resistance to the below atmosphere in the stack but not to the person/object because there's no push up by the person/object?

It is but it's a resistance by compression of the mass into it and that compression is placed right back onto the mass along with the atmosphere above and around.

Again, you have provided no justification for the directionality.

I have given you plenty. You not understanding it is your issue.

Again, a nice simple example, you have a ball, like a tennis ball. You position your hand such that your thumb is on the left side and your fingers are on the right side, with nothing above or below.
You hold it at shoulder height, close to your shoulder and then extend out your arm so the ball moves straight out horizontally. You now release the ball.
How should the ball move and why?
The ball is not pushing into or displacing the atmosphere above any more than it is doing to the atmosphere below. As such, that cannot provide a basis for the directionality.

You are simply holding the ball above the stack. If you keep hold of the ball you are compressing the atmosphere your hand and ball are in, above and around.
Below is stacked atmosphere that does nothing to the underside of that ball, in terms of the very point of the underside at the absolute very bottom.
the rest of the underside of that ball will be compressed as it pushes into the stack of resistance due to its shape.

If you were holding a flat plate level above that stack then you would not be compressing that stack, only resting on it whilst being compressed against it's thickness at the sides and pushing up against the atmosphere above that is pushing/compressing/crushing down on it.

Drop your ball and it's mass has ro be resisted by the stack along with being pushed down by the atmosphere above along with the mass of atmosphere the ball already displaces by it structure.

The object is pushing into the atmosphere above and to the side, because the atmosphere is pushing into the mass as the mass displaces it.
It is not displacing the stack below. It's a stack.

Why isn't it displacing it below but it is displacing it above?
It is below the stack above so how it it displacing it?

If it falls then your rope has snapped and now the mass is pushed down against the resistance of the stack below which is in no way dense enough to counteract that push enough to do anything other than friction grip it and compress until the object hits a foundation.

Again why?
You have the push and resistance be enough to move it, being the only thing that causes the movement, so why isn't it enough to stop the movement?

I have no clue why you even think this. It baffles me.

It is quite simple, your repeatedly claims of the air resistance and equal and opposite reaction and all that nonsense.
You push into the air, it pushes back with an equal and opposite force preventing you from displacing it.

Simple answer is a foundation allowing it to, whether its on a solid ground directly beneath it or on a suspended rope or whatever above the atmospheric stack below it.

How?
How does the atmospheric stack below it magically make it only displace atmosphere above, not below?

The actual simple answer is that the atmosphere has very little to do with falling objects.

Because it's stacked from below and cannot push up unless something forces it up, which would be any mass under energy push.

It is also stacked above, so how does that push it down?

A stack is a stack. It stays as literally a stack.

No it doesn't. It can get moved around quite easily.
We can move air from below and put it up high.
You fundamentally require it to not stay literally as a stack.
In order for it to stay literally as a stack it would stay and not allow the object through.

It doesn't matter how high you go you will always be on top of a stack of atmosphere

And likewise you will be below a stack.
So that clearly doesn't explain it either.

Quote from: JackBlack

Then why isn't the atmosphere above just a resistance to the below atmosphere in the stack but not to the person/object because there's no push up by the person/object?

It is but it's a resistance by compression of the mass into it and that compression is placed right back onto the mass along with the atmosphere above and around.

I notice you failed to address the "why" part.
If you agree that it is, then that means the atmosphere above and below act the same and thus no net force.

I have given you plenty. You not understanding it is your issue.

No you haven't. Every single one of your "justifications" has been fundamentally flawed.

You are simply holding the ball above the stack.

No, in your model, you are holding the ball inside the stack. There is air above and below, or there is stack above and below.
Ignoring the stack above when it suits you is not providing an explanation.

If you keep hold of the ball you are compressing the atmosphere your hand and ball are in, above and around.

Again, HOW? How does it magically compress above and around, but not below? Especially given the fact that ignoring the slight density gradient pressure in fluids will equalise, which you have already admitted.
That means unless the top is isolated from the bottom, compressing the top will necessarily compress the bottom, with the sole exception of complex fluid dynamics which is well beyond the basics.

You repeatedly asserting that it magically compresses above but not below does not help your case at all.
You need to provide a justification for how this magical feat is achieved.

Below is stacked atmosphere that does nothing to the underside of that ball, in terms of the very point of the underside at the absolute very bottom.
the rest of the underside of that ball will be compressed as it pushes into the stack of resistance due to its shape.

Again, why just the stack below? Why does the exact same thing not happen for the stack above?

Quote from: sceptimatic on January 17, 2019, 10:18:24 PM

Quote from: sobchak on January 17, 2019, 01:54:02 PM

Why can’t the one side just stay compressed and the other less compressed?

Because you've taken away the central divider and allowed the higher pressure to expand and the lower pressure to compress by that expansion to create equilibrium.

Can you describe the final equilibrium state in terms of pressure and density?

If you mean overall pressure, it would be around 21.4 pounds per square inch.

Quote from: sceptimatic on January 17, 2019, 11:17:40 PM

Quote from: Stash on January 17, 2019, 10:57:32 PM

Quote from: sceptimatic on January 17, 2019, 10:26:50 PM

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

You can say, "6 jugs," pointing to the 6 jugs.

How much does 6 jugs weigh?

6 jugs weighs exactly 6 jugs. Am I wrong?

How do you weigh them?

Quote from: sceptimatic on January 18, 2019, 02:12:58 AM

The object is pushing into the atmosphere above and to the side, because the atmosphere is pushing into the mass as the mass displaces it.
It is not displacing the stack below. It's a stack.

Why isn't it displacing it below but it is displacing it above?
It is below the stack above so how it it displacing it?

It's in the stack above its own mass.It's displacing it.

Quote from: sobchak on January 18, 2019, 12:41:20 AM

Quote from: sceptimatic on January 17, 2019, 10:18:24 PM

Quote from: sobchak on January 17, 2019, 01:54:02 PM

Why can’t the one side just stay compressed and the other less compressed?

Because you've taken away the central divider and allowed the higher pressure to expand and the lower pressure to compress by that expansion to create equilibrium.

Can you describe the final equilibrium state in terms of pressure and density?

If you mean overall pressure, it would be around 21.4 pounds per square inch.

Okay, so it's just a uniform, constant pressure, constant density fluid? 

If so,  what physical rules cause this to occur in your model?  Why in this ‘stacked’ and ‘attached’ atmosphere (where these directional attachments surely must add directional forces?) does pressure and density distribute evenly over time?

In the ‘accepted’ conceptual description, this breakdown of pressure and density gradients falls cleanly and clearly out of the kinetic theory of gases and the laws of thermodynamics.  But in your framework, I’m pretty sure the former is invalid, and I’m not even sure about the latter, so I see no logical reason why this should occur. 

You can ‘borrow’ the results from accepted theories and certainly claim axiomatically that they hold in yours if you want, but again, as with your directional displacement of atmosphere, this is just a pragmatic choice of rules for your model, rather than a logical extension and conclusion from first principle concepts.

And of course, while this is fine and you are welcome to do whatever you would like with your model, the fundamental problem with pragmatic model choices such as the ones you incorporate is they greatly diminish the power of the model to be predictive.  You can use them to be descriptive, such that you can describe any observed and reported phenomena by a selective choice of pragmatic constraints on your model, but in an unknown situation, you may not know a priori which choices to make, and will be unable to make a concrete prediction.

I think you are smart enough to sense this, which is I guess why you don't want to go down the route of any hard quantitative predictions based on first principles from your model, no matter how basic and simple they are. 

Quote from: sceptimatic on January 18, 2019, 03:44:34 AM

Quote from: sobchak on January 18, 2019, 12:41:20 AM

Quote from: sceptimatic on January 17, 2019, 10:18:24 PM

Quote from: sobchak on January 17, 2019, 01:54:02 PM

Why can’t the one side just stay compressed and the other less compressed?

Because you've taken away the central divider and allowed the higher pressure to expand and the lower pressure to compress by that expansion to create equilibrium.

Can you describe the final equilibrium state in terms of pressure and density?

If you mean overall pressure, it would be around 21.4 pounds per square inch.

Okay, so it's just a uniform, constant pressure, constant density fluid? 

If so,  what physical rules cause this to occur in your model?  Why in this ‘stacked’ and ‘attached’ atmosphere (where these directional attachments surely must add directional forces?) does pressure and density distribute evenly over time?

What you need to understand, is my model.
It's ok adding pressure to a container and trying to marry that up with atmospheric stacking...but you would skew your whole thinking.
There's a reason why I mention the basics of the basics.

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.
With the Earth dome there is natural stacking and it isn't uniform.

I think you are smart enough to sense this, which is I guess why you don't want to go down the route of any hard quantitative predictions based on first principles from your model, no matter how basic and simple they are.

I go down my route. The basic route because I feel it's the best way to take it for people to understand.
Those who are willing to understand it will be able to piece it together much quicker than those who try to complicate the issue.
The best way to build something is to strip it down to its basic components,a s far down as you can and see how simple it is in its raw state.

If someone gives you a model and it looks complicated, they can tell you anything about how that model was made and explain as to why it works like it does.
All you  can do is accept their explanations or you can start to strip down their model.

That's what I'm doing and it is not what it said on the tin.

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within that same container on a spring.  There is no stacking here.  Why is there a net downward force on the spring? 

Quote

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within a container on a string.  There is no stacking here.  Why is there a net downward force on the spring?

The atmosphere is still stacked inside the container, so suspending the object inside on a spring, the object still displaces the atmosphere and sits on the stack.

Quote from: Stash on January 18, 2019, 01:39:38 AM

Quote from: sceptimatic on January 17, 2019, 11:17:40 PM

Quote from: Stash on January 17, 2019, 10:57:32 PM

Quote from: sceptimatic on January 17, 2019, 10:26:50 PM

You need something to measure mass on land in order for you to have a weight measurement of that mass.
What do you use.

You've managed 6 jugs of water to balance your rock.
So, if someone comes along and says, " how much does that rock weight" you can say, 6jugs.

They say, " ok but how much does the 6 jugs weigh?" You can say " one rock, this size" pointing to your rock.

You can say, "6 jugs," pointing to the 6 jugs.

How much does 6 jugs weigh?

6 jugs weighs exactly 6 jugs. Am I wrong?

How do you weigh them?

The balance scale.
Whats your point?
You seem to be caught in a loop.
Are you a robot?

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Quote from: scepitmatic

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within a container on a string.  There is no stacking here.  Why is there a net downward force on the spring?

The atmosphere is still stacked inside the container, so suspending the object inside on a spring, the object still displaces the atmosphere and sits on the stack.

Okay. 

The balance scale.
Whats your point?
You seem to be caught in a loop.
Are you a robot?

Does a balance scale give you a weight?

Quote from: sceptimatic on January 18, 2019, 05:24:03 AM

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Quote

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within a container on a string.  There is no stacking here.  Why is there a net downward force on the spring?

The atmosphere is still stacked inside the container, so suspending the object inside on a spring, the object still displaces the atmosphere and sits on the stack.

Okay.

I didn't see you mention pressurising a container to suspend something on a spring. Did you mention that or would you like to amend your question?

Quote from: Themightykabool on January 18, 2019, 05:31:26 AM

The balance scale.
Whats your point?
You seem to be caught in a loop.
Are you a robot?

Does a balance scale give you a weight?

Ugh...
Please get to your point.

The balance scale gives you a comparative weight between to things.
RockA = 6jugs
RockB = 12jugs.
RockB = two RockA

Whats your point.

Quote from: sobchak on January 18, 2019, 05:33:23 AM

Quote from: sceptimatic on January 18, 2019, 05:24:03 AM

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Quote

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within a container on a string.  There is no stacking here.  Why is there a net downward force on the spring?

The atmosphere is still stacked inside the container, so suspending the object inside on a spring, the object still displaces the atmosphere and sits on the stack.

Okay.

I didn't see you mention pressurising a container to suspend something on a spring. Did you mention that or would you like to amend your question?

It was amended before you replied, but in case you didn't see it -

Okay, suspend an object within that same container on a spring.  There is no stacking here.  Why is there a net downward force on the spring? 

Quote from: sceptimatic on January 18, 2019, 05:35:13 AM

Quote from: sobchak on January 18, 2019, 05:33:23 AM

Quote from: sceptimatic on January 18, 2019, 05:24:03 AM

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Quote

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within a container on a string.  There is no stacking here.  Why is there a net downward force on the spring?

The atmosphere is still stacked inside the container, so suspending the object inside on a spring, the object still displaces the atmosphere and sits on the stack.

Okay.

I didn't see you mention pressurising a container to suspend something on a spring. Did you mention that or would you like to amend your question?

It was amended before you replied, but in case you didn't see it -

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Okay, suspend an object within that same container on a spring.  There is no stacking here.  Why is there a net downward force on the spring? 

You said the same container, you didn't mention anything about it being pressurised.

Quote from: sobchak on January 18, 2019, 05:46:38 AM

Quote from: sceptimatic on January 18, 2019, 05:35:13 AM

Quote from: sobchak on January 18, 2019, 05:33:23 AM

Quote from: sceptimatic on January 18, 2019, 05:24:03 AM

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Quote

In the container you are not stacking naturally. You are forcing a compression by trapping it inside the container. Therefore there is no natural stacking. It uniform, like you say.

Okay, suspend an object within a container on a string.  There is no stacking here.  Why is there a net downward force on the spring?

The atmosphere is still stacked inside the container, so suspending the object inside on a spring, the object still displaces the atmosphere and sits on the stack.

Okay.

I didn't see you mention pressurising a container to suspend something on a spring. Did you mention that or would you like to amend your question?

It was amended before you replied, but in case you didn't see it -

Quote from: sobchak on January 18, 2019, 04:51:27 AM

Okay, suspend an object within that same container on a spring.  There is no stacking here.  Why is there a net downward force on the spring? 

You said the same container, you didn't mention anything about it being pressurised.

Got it. Another layer of pragmatism. I’m okay with that. Can you describe what happens in the container and the resulting stacking during pressurization in the presence and absence of a suspended object contained within?

Got it. Another layer of pragmatism. I’m okay with that. Can you describe what happens in the container and the resulting stacking during pressurization in the presence and absence of a suspended object contained within?

I've never hung any spring from a container with a mass on it and pressurised the container.

What do I think would happen?
It depends on the object and the container itself I suppose.

What would happen to a near perfectly round object inside a near perfectly spherical container under forced pressurisation?

Does it suspend in the middle of it?

Quote from: sceptimatic on January 18, 2019, 05:33:42 AM

Quote from: Themightykabool on January 18, 2019, 05:31:26 AM

The balance scale.
Whats your point?
You seem to be caught in a loop.
Are you a robot?

Does a balance scale give you a weight?

Ugh...
Please get to your point.

The balance scale gives you a comparative weight between to things.
RockA = 6jugs
RockB = 12jugs.
RockB = two RockA

Whats your point.

Whats your point?

Quote from: Themightykabool on January 18, 2019, 05:31:26 AM

The balance scale.
Whats your point?
You seem to be caught in a loop.
Are you a robot?

Does a balance scale give you a weight?

Go and buy one to find out.

I've never hung any spring from a container with a mass on it and pressurised the container.

What do I think would happen?
It depends on the object and the container itself I suppose.

What would happen to a near perfectly round object inside a near perfectly spherical container under forced pressurisation?

Does it suspend in the middle of it?

sure!

Quote from: sceptimatic

I've never hung any spring from a container with a mass on it and pressurised the container.

What do I think would happen?
It depends on the object and the container itself I suppose.

What would happen to a near perfectly round object inside a near perfectly spherical container under forced pressurisation?

Does it suspend in the middle of it?

sure!

The prediction based on stacking theory is that the ball will sag down due to increased pressure above the ball, causing air to stack and push down, but magically the pressure below the ball does stack up to counter.