# Sea and air pressure

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#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #480 on: September 14, 2020, 05:54:29 AM »
For a tree to grow up it has to grow down. This means it uses the ground as its foundation from where it came from under.
Nothing has changed on Earth due to that tree being a seedling or whether it's 500 feet tall.
The mass was always there....it's just been transferred.
I must admit, Stash is wrong on this tree part. (But I understand the point he is making)
The tree doesn't just magically add more mass.
Instead it takes the mass from elsewhere.
The big problem for you is that it takes in mass from the atmosphere.

That means the amount the atmosphere is pushing down should reduce as the tree grows, meaning the tree (and every other object) should weigh less as the tree grows.
Instead, if you try to weigh the tree as it grows, it gets heavier, which in your model requires the air to push the tree down more.
But according to you that should only happen if you add more mass into the system to make the air push down harder by displacing the air.

I think I need to bone up on my Dendrology, Metallurgy and where it meets Denpressure.

Is it that the displacement caused by the growing tree is exactly equal to the amount of whatever it needs to grow taken from the atmosphere? So within the closed cell of earth nothing is ever added and nothing is ever subtracted?
Yep....essentially.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #481 on: September 14, 2020, 06:33:56 AM »

This also relates to the first problem I raised as soon as I heard about your model, taking a sealed container, weighing it, then pumping all the air out of it.
In reality, it weighs less. But in your model, you have removed the atmosphere from inside it, making it displace more atmosphere, meaning the atmosphere should be pushing it down more making it weigh more.
If the container is sat on a scale then the air inside is already part of the atmospheric volume. It does not count towards the dense mass of the container.
If you allow the atmosphere within the container to be released by pushing away the external atmosphere to allow expansion of the molecules inside the container to start pushing into the lower pressure created by that pump, then you are adding internal atmosphere to external.

All you've done is transferred the pressure.

However...if you were to put scales inside the container, reading zero under normal atmospheric conditions, inside and then allowed evacuation of that contained atmosphere, you would notice a change on the scale to the negative.

Quote from: JackBlack
As for how to weigh the plant, how about just growing it in a pot, with the entire pot on a scale?
The only source of mass then is water that you water it with and the air, as otherwise it is a closed system.

If the plant grows after adding water and nutrients as part of that water then you should know the plant will become more dense and with more volume. The volume means nothing in terms of weighing the plant but the more added density will.
More density means more atmospheric displacement.

Quote from: JackBlack
Why does assembling the tower cause the water to push down more?
It has the same area exposed at the top and thus the same push down.
The force on the side is balanced by the force on the other side.
The more dense mass you add to that tower the more the tower displaces the atmosphere it is being assembled, in.

Quote from: JackBlack
It doesn't make its way through the tower's dense mass. It crushes the tower's dense mass at every point by what that tower pushes into the atmosphere by displacement of its (the tower) own dense mass.
How does it crush it at every point?
Especially if it isn't going through it.
Any part of the tower's dense mass that pushes into and against atmosphere will be acted upon by that as all that's happened is, it's been pushed away.
As for atmosphere going through it. That's volume. It plays no part on displacement as it's already part of that atmosphere.

Quote from: JackBlack
The only location for a downwards force is the top of the tower, which raises the issue of why the force increases as you go down the tower, and why you can place a fragile object at the top without crushing it but you cannot do the same at the bottom.
Can you remember me explaining the stack?
I think I do need to do a crude drawing but I'm not sure you'll get it, as simple as it will be for me, I understand that you are trying to understand it from your side.
I'd be impressed if you actually started to grasp some of it before I stick a drawing up, just so I don't waste more of my time.

Quote from: JackBlack
In addition it also raises the issue of why that downwards force is based upon mass at all. Why don't 2 objects with the same cross sectional area get pushed down with the same force?
If they are exact in dense mass and cross sectional area, they will.

Quote from: JackBlack
No. The water you placed into the pot was already part of the atmosphere. All you've done is poured it in and then discarded it by boiling it out back into that atmosphere. No change. Equal and opposite reaction to action.
No, it was water, a liquid, not part of the atmosphere.
When I boiled it it became part of the atmosphere, adding to it.
Atmosphere plays a part in everything that pushes into it.
In terms of man made ability to weigh dense mass by using a scale plate that must be placed in atmosphere...this is what we need to deal with.
You cannot scale weight a brick buried in the ground and nor at the bottom of a pool.

Quote from: JackBlack
What if I froze it into ice instead? Is it still part of the atmosphere?
Ice like water is displacing atmosphere and that amount is returned/pushed back.

Quote from: JackBlack
Or are you now saying everything is part of the atmosphere?
Atmosphere is part of everything it pushes against. It's all about measuring that push for whatever mass is pushed against it, which basically means a man made scale measurement of it by using a foundation to allow a scale plate to show a reading.

Quote from: JackBlack
That would mean it is no longer just the air, but now everything pushes down, which is a definite improvement as you no longer need to bother with contradicting so many properties of how the air works and can have air pressure act as normal, even providing buoyancy.

But then you are left with why things push down or more importantly, move down to create that downwards push.
Gravity in no way provides any answer. It makes zero sense. It cannot be explained as to what it is and does in any reality.
It is utter nonsense as far as I'm concerned.

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #482 on: September 14, 2020, 02:19:00 PM »
This also relates to the first problem I raised as soon as I heard about your model, taking a sealed container, weighing it, then pumping all the air out of it.
In reality, it weighs less. But in your model, you have removed the atmosphere from inside it, making it displace more atmosphere, meaning the atmosphere should be pushing it down more making it weigh more.
If the container is sat on a scale then the air inside is already part of the atmospheric volume. It does not count towards the dense mass of the container.
If you allow the atmosphere within the container to be released by pushing away the external atmosphere to allow expansion of the molecules inside the container to start pushing into the lower pressure created by that pump, then you are adding internal atmosphere to external.
All you've done is transferred the pressure.
And that in no way addresses the issue.
It does not explain why the weight has lowered when more atmosphere is displaced.
I made it clear that I understood that the air inside doesn't count as part of the container.
But if weight is about displacing atmosphere then displacing it from the inside of the container should add to the weight.
But instead, the weight is lowered.

Your model indicates it should become heavier (i.e. the weight reading increases).
Reality indicates it becomes lighter (i.e. the weight reading decreases).

If the plant grows after adding water and nutrients as part of that water then you should know the plant will become more dense and with more volume. The volume means nothing in terms of weighing the plant but the more added density will.
More density means more atmospheric displacement.
And I made it clear that that water would be accounted for in the weight.
It wont become more dense, it will become more massive. But the added mass of the water doesn't account for all the added mass of the tree/plant.

But you have made it clear that it isn't more atmosphere displaced as the water was already displacing the atmosphere.

Again, by mass, it is actually less atmosphere displaced as the tree absorbs the atmosphere to grow.

The more dense mass you add to that tower the more the tower displaces the atmosphere it is being assembled, in.
Again, you have made it clear that the mass is already displacing the atmosphere.
All we are doing is moving it around.
No extra atmosphere is being displaced.

Any part of the tower's dense mass that pushes into and against atmosphere will be acted upon by that as all that's happened is, it's been pushed away.
As for atmosphere going through it. That's volume. It plays no part on displacement as it's already part of that atmosphere.
How?
Where is the air actually applying the force?
Is it only on the interface between the tower and the air, or is it throughout the tower?

I think I do need to do a crude drawing but I'm not sure you'll get it, as simple as it will be for me, I understand that you are trying to understand it from your side.
I'd be impressed if you actually started to grasp some of it before I stick a drawing up, just so I don't waste more of my time.
The issue with me "grasping" it is that it literally makes no sense.
You claim the air is pushing the tower down. But the only spot for it to do so is at the top.
This is why a diagram would be so helpful, clearly showing just where you claim the force is being applied in your model.

Quote from: JackBlack
In addition it also raises the issue of why that downwards force is based upon mass at all. Why don't 2 objects with the same cross sectional area get pushed down with the same force?
If they are exact in dense mass and cross sectional area, they will.
I meant the more general case where they are not.
Everything we know about air indicates that the force applied is proportional to the area.
For example, if you have a syringe (or other plunger) with the end sealed, the force you need to apply to compress it is based upon the area.
If you connect 2 syringes together, with different areas you can amplify the force at the cost of distance, or vice versa.
Suction caps apply a force based upon the area.

It isn't based upon mass.

So why don't 2 objects in the atmosphere work the same?
2 objects, with the same cross sectional area, with the same atmosphere, yet they are pushed down differently as they have a different mass.

Quote from: JackBlack
No. The water you placed into the pot was already part of the atmosphere. All you've done is poured it in and then discarded it by boiling it out back into that atmosphere. No change. Equal and opposite reaction to action.
No, it was water, a liquid, not part of the atmosphere.
When I boiled it it became part of the atmosphere, adding to it.
Atmosphere plays a part in everything that pushes into it.
In terms of man made ability to weigh dense mass by using a scale plate that must be placed in atmosphere...this is what we need to deal with.
You cannot scale weight a brick buried in the ground and nor at the bottom of a pool.
Again, that in no way addresses what I said.
Once more, you have water in a pot. This weighs some amount.
You now boil the water and thus increase the atmosphere of atmosphere, increasing the pressure in the system.
But now the pot weighs less. Why?

Gravity in no way provides any answer. It makes zero sense. It cannot be explained as to what it is and does in any reality.
It is utter nonsense as far as I'm concerned.
Gravity works wonderfully. It actually makes sense, being internally consistent and consistent with reality (with the exception of galactic rotation curves), and is far more explained than your model, being consistent with the other fundamental forces of nature. You not liking it does not change those facts.

But as you aren't claiming everything is part of the atmosphere, that little tangent is irrelevant.

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#### Themightykabool

• 5358
##### Re: Sea and air pressure
« Reply #483 on: September 14, 2020, 02:34:07 PM »
Grasp the basics of your own model and draw some diagrams.

You need to be able to differentiate between regular air and the sponge air that causes things to go "down".

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #484 on: September 14, 2020, 10:12:44 PM »
This also relates to the first problem I raised as soon as I heard about your model, taking a sealed container, weighing it, then pumping all the air out of it.
In reality, it weighs less. But in your model, you have removed the atmosphere from inside it, making it displace more atmosphere, meaning the atmosphere should be pushing it down more making it weigh more.
If the container is sat on a scale then the air inside is already part of the atmospheric volume. It does not count towards the dense mass of the container.
If you allow the atmosphere within the container to be released by pushing away the external atmosphere to allow expansion of the molecules inside the container to start pushing into the lower pressure created by that pump, then you are adding internal atmosphere to external.
All you've done is transferred the pressure.
And that in no way addresses the issue.
It does not explain why the weight has lowered when more atmosphere is displaced.
I made it clear that I understood that the air inside doesn't count as part of the container.
But if weight is about displacing atmosphere then displacing it from the inside of the container should add to the weight.
But instead, the weight is lowered.

Your model indicates it should become heavier (i.e. the weight reading increases).
Reality indicates it becomes lighter (i.e. the weight reading decreases).

If the plant grows after adding water and nutrients as part of that water then you should know the plant will become more dense and with more volume. The volume means nothing in terms of weighing the plant but the more added density will.
More density means more atmospheric displacement.
And I made it clear that that water would be accounted for in the weight.
It wont become more dense, it will become more massive. But the added mass of the water doesn't account for all the added mass of the tree/plant.

But you have made it clear that it isn't more atmosphere displaced as the water was already displacing the atmosphere.

Again, by mass, it is actually less atmosphere displaced as the tree absorbs the atmosphere to grow.

The more dense mass you add to that tower the more the tower displaces the atmosphere it is being assembled, in.
Again, you have made it clear that the mass is already displacing the atmosphere.
All we are doing is moving it around.
No extra atmosphere is being displaced.

Any part of the tower's dense mass that pushes into and against atmosphere will be acted upon by that as all that's happened is, it's been pushed away.
As for atmosphere going through it. That's volume. It plays no part on displacement as it's already part of that atmosphere.
How?
Where is the air actually applying the force?
Is it only on the interface between the tower and the air, or is it throughout the tower?

I think I do need to do a crude drawing but I'm not sure you'll get it, as simple as it will be for me, I understand that you are trying to understand it from your side.
I'd be impressed if you actually started to grasp some of it before I stick a drawing up, just so I don't waste more of my time.
The issue with me "grasping" it is that it literally makes no sense.
You claim the air is pushing the tower down. But the only spot for it to do so is at the top.
This is why a diagram would be so helpful, clearly showing just where you claim the force is being applied in your model.

Quote from: JackBlack
In addition it also raises the issue of why that downwards force is based upon mass at all. Why don't 2 objects with the same cross sectional area get pushed down with the same force?
If they are exact in dense mass and cross sectional area, they will.
I meant the more general case where they are not.
Everything we know about air indicates that the force applied is proportional to the area.
For example, if you have a syringe (or other plunger) with the end sealed, the force you need to apply to compress it is based upon the area.
If you connect 2 syringes together, with different areas you can amplify the force at the cost of distance, or vice versa.
Suction caps apply a force based upon the area.

It isn't based upon mass.

So why don't 2 objects in the atmosphere work the same?
2 objects, with the same cross sectional area, with the same atmosphere, yet they are pushed down differently as they have a different mass.

Quote from: JackBlack
No. The water you placed into the pot was already part of the atmosphere. All you've done is poured it in and then discarded it by boiling it out back into that atmosphere. No change. Equal and opposite reaction to action.
No, it was water, a liquid, not part of the atmosphere.
When I boiled it it became part of the atmosphere, adding to it.
Atmosphere plays a part in everything that pushes into it.
In terms of man made ability to weigh dense mass by using a scale plate that must be placed in atmosphere...this is what we need to deal with.
You cannot scale weight a brick buried in the ground and nor at the bottom of a pool.
Again, that in no way addresses what I said.
Once more, you have water in a pot. This weighs some amount.
You now boil the water and thus increase the atmosphere of atmosphere, increasing the pressure in the system.
But now the pot weighs less. Why?

Gravity in no way provides any answer. It makes zero sense. It cannot be explained as to what it is and does in any reality.
It is utter nonsense as far as I'm concerned.
Gravity works wonderfully. It actually makes sense, being internally consistent and consistent with reality (with the exception of galactic rotation curves), and is far more explained than your model, being consistent with the other fundamental forces of nature. You not liking it does not change those facts.

But as you aren't claiming everything is part of the atmosphere, that little tangent is irrelevant.
I suggest you go back and read through what I've said because you are confusing yourself.
I explain it all and you do the exact same thing in saying I don't.
Sped more time absorbing what's been said.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #485 on: September 14, 2020, 10:14:04 PM »
Grasp the basics of your own model and draw some diagrams.

You need to be able to differentiate between regular air and the sponge air that causes things to go "down".
No I don't. You need to understand analogies and the reasons I give them, instead of creating your own issues of confusion.

#### Stash

• 7615
##### Re: Sea and air pressure
« Reply #486 on: September 14, 2020, 10:27:22 PM »
For a tree to grow up it has to grow down. This means it uses the ground as its foundation from where it came from under.
Nothing has changed on Earth due to that tree being a seedling or whether it's 500 feet tall.
The mass was always there....it's just been transferred.
I must admit, Stash is wrong on this tree part. (But I understand the point he is making)
The tree doesn't just magically add more mass.
Instead it takes the mass from elsewhere.
The big problem for you is that it takes in mass from the atmosphere.

That means the amount the atmosphere is pushing down should reduce as the tree grows, meaning the tree (and every other object) should weigh less as the tree grows.
Instead, if you try to weigh the tree as it grows, it gets heavier, which in your model requires the air to push the tree down more.
But according to you that should only happen if you add more mass into the system to make the air push down harder by displacing the air.

I think I need to bone up on my Dendrology, Metallurgy and where it meets Denpressure.

Is it that the displacement caused by the growing tree is exactly equal to the amount of whatever it needs to grow taken from the atmosphere? So within the closed cell of earth nothing is ever added and nothing is ever subtracted?
Yep....essentially.

This is one I think you definitely need a diagram for. Basically what I was asking about. From JB:

"Once more, you have water in a pot. This weighs some amount.
You now boil the water and thus increase the atmosphere of atmosphere, increasing the pressure in the system.
But now the pot weighs less. Why?"

If you could wireframe this scenario out, it would help a lot. I'm personally struggling with why in a closed system, if atmosphere is added to the atmosphere why doesn't pressure build within the cell causing more pressure to be pushed/stacked upon an object? Hence the pot should weigh more when in fact it weighs less. And even with some notion of an equal and opposite reaction, one part out for one part in, if that were the case here, the pot should just remain the same weight. But it doesn't.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #487 on: September 14, 2020, 11:11:38 PM »
For a tree to grow up it has to grow down. This means it uses the ground as its foundation from where it came from under.
Nothing has changed on Earth due to that tree being a seedling or whether it's 500 feet tall.
The mass was always there....it's just been transferred.
I must admit, Stash is wrong on this tree part. (But I understand the point he is making)
The tree doesn't just magically add more mass.
Instead it takes the mass from elsewhere.
The big problem for you is that it takes in mass from the atmosphere.

That means the amount the atmosphere is pushing down should reduce as the tree grows, meaning the tree (and every other object) should weigh less as the tree grows.
Instead, if you try to weigh the tree as it grows, it gets heavier, which in your model requires the air to push the tree down more.
But according to you that should only happen if you add more mass into the system to make the air push down harder by displacing the air.

I think I need to bone up on my Dendrology, Metallurgy and where it meets Denpressure.

Is it that the displacement caused by the growing tree is exactly equal to the amount of whatever it needs to grow taken from the atmosphere? So within the closed cell of earth nothing is ever added and nothing is ever subtracted?
Yep....essentially.

This is one I think you definitely need a diagram for. Basically what I was asking about. From JB:

"Once more, you have water in a pot. This weighs some amount.
You now boil the water and thus increase the atmosphere of atmosphere, increasing the pressure in the system.
But now the pot weighs less. Why?"
If you could wireframe this scenario out, it would help a lot. I'm personally struggling with why in a closed system, if atmosphere is added to the atmosphere why doesn't pressure build within the cell causing more pressure to be pushed/stacked upon an object? Hence the pot should weigh more when in fact it weighs less. And even with some notion of an equal and opposite reaction, one part out for one part in, if that were the case here, the pot should just remain the same weight. But it doesn't.
Ok let's carefully go through and stick with the water and the pot so we don't get sidetracked.
We won't move on until you grasp this bit....ok?

The pot's own dense mass is already displacing that amount of atmosphere, disregarding the volume it has within it...in its structure as well as inside the pot, because this would be known as absorption.

This means any part of the structure that can hold or absorb the atmosphere....however minute.

Ok, so the pot alone, in that situation, we can weigh.... and find how much atmosphere it displaces by using a man made scale.
You pick up that pot and place it on a scale plate that is set to zero.
The atmospheric displacement of that pot's dense mass (only) is now transferred back onto the pot's dense mass (not volume).
The pot itself is the resistance to that pressure it's pushing against/into and it requires a foundation to resist that pressure, which is something capable of resisting, fully, which can be either, the ground or even water....etc.......or the scale plate.
In terms of gaining a man made measurement to create a weight, the scales act like a buffer to the pressure, which the pot is using to resist the displacement of its own dense mass of atmosphere.

Basically it's being crushed down onto the scale plate and the amount of resistance to that crush is measured by that scale plate on a spring (as an instance)....just like water will part if the same pot was placed in it.... and if that water can resist the atmospheric crush of the pot by crushing back, the pot will eventually float at whatever depth most of it ends up in.

However, let's go back onto land and fill the pot with water, on that scale.
We know what the pot weighed without the water and in spite of absorbed atmosphere.

Now add water to the pot and what happens?
The water now pushes out the atmosphere (volume) from inside the pot back into the atmosphere outside of that pot but it has compressed it out and the water itself has displaced its own mass of that atmosphere.
Now that atmosphere not only reacts to the dense mass of the pot, it also reacts to the dense mass of the water inside of that pot, which means more push against that atmosphere and the equal push back...plus the amount that's already above the pot. all the way up....but we'll come to that bit.

The pot itself is still holding volume but the volume is in the structure and is a barrier to the water, along with the dense mass of the structure itself...which is why structures differ massively in density and the displacement of atmosphere to the weigh scale.

I'm more than willing to stick to this bit so don't change it and create any confusion.
Keep at this and nit pick as much as you need to.

#### Stash

• 7615
##### Re: Sea and air pressure
« Reply #488 on: September 14, 2020, 11:24:46 PM »
For a tree to grow up it has to grow down. This means it uses the ground as its foundation from where it came from under.
Nothing has changed on Earth due to that tree being a seedling or whether it's 500 feet tall.
The mass was always there....it's just been transferred.
I must admit, Stash is wrong on this tree part. (But I understand the point he is making)
The tree doesn't just magically add more mass.
Instead it takes the mass from elsewhere.
The big problem for you is that it takes in mass from the atmosphere.

That means the amount the atmosphere is pushing down should reduce as the tree grows, meaning the tree (and every other object) should weigh less as the tree grows.
Instead, if you try to weigh the tree as it grows, it gets heavier, which in your model requires the air to push the tree down more.
But according to you that should only happen if you add more mass into the system to make the air push down harder by displacing the air.

I think I need to bone up on my Dendrology, Metallurgy and where it meets Denpressure.

Is it that the displacement caused by the growing tree is exactly equal to the amount of whatever it needs to grow taken from the atmosphere? So within the closed cell of earth nothing is ever added and nothing is ever subtracted?
Yep....essentially.

This is one I think you definitely need a diagram for. Basically what I was asking about. From JB:

"Once more, you have water in a pot. This weighs some amount.
You now boil the water and thus increase the atmosphere of atmosphere, increasing the pressure in the system.
But now the pot weighs less. Why?"
If you could wireframe this scenario out, it would help a lot. I'm personally struggling with why in a closed system, if atmosphere is added to the atmosphere why doesn't pressure build within the cell causing more pressure to be pushed/stacked upon an object? Hence the pot should weigh more when in fact it weighs less. And even with some notion of an equal and opposite reaction, one part out for one part in, if that were the case here, the pot should just remain the same weight. But it doesn't.
Ok let's carefully go through and stick with the water and the pot so we don't get sidetracked.
We won't move on until you grasp this bit....ok?

The pot's own dense mass is already displacing that amount of atmosphere, disregarding the volume it has within it...in its structure as well as inside the pot, because this would be known as absorption.

This means any part of the structure that can hold or absorb the atmosphere....however minute.

Ok, so the pot alone, in that situation, we can weigh.... and find how much atmosphere it displaces by using a man made scale.
You pick up that pot and place it on a scale plate that is set to zero.
The atmospheric displacement of that pot's dense mass (only) is now transferred back onto the pot's dense mass (not volume).
The pot itself is the resistance to that pressure it's pushing against/into and it requires a foundation to resist that pressure, which is something capable of resisting, fully, which can be either, the ground or even water....etc.......or the scale plate.
In terms of gaining a man made measurement to create a weight, the scales act like a buffer to the pressure, which the pot is using to resist the displacement of its own dense mass of atmosphere.

Basically it's being crushed down onto the scale plate and the amount of resistance to that crush is measured by that scale plate on a spring (as an instance)....just like water will part if the same pot was placed in it.... and if that water can resist the atmospheric crush of the pot by crushing back, the pot will eventually float at whatever depth most of it ends up in.

However, let's go back onto land and fill the pot with water, on that scale.
We know what the pot weighed without the water and in spite of absorbed atmosphere.

Now add water to the pot and what happens?
The water now pushes out the atmosphere (volume) from inside the pot back into the atmosphere outside of that pot but it has compressed it out and the water itself has displaced its own mass of that atmosphere.
Now that atmosphere not only reacts to the dense mass of the pot, it also reacts to the dense mass of the water inside of that pot, which means more push against that atmosphere and the equal push back...plus the amount that's already above the pot. all the way up....but we'll come to that bit.

The pot itself is still holding volume but the volume is in the structure and is a barrier to the water, along with the dense mass of the structure itself...which is why structures differ massively in density and the displacement of atmosphere to the weigh scale.

I'm more than willing to stick to this bit so don't change it and create any confusion.
Keep at this and nit pick as much as you need to.

I'm following all the way through and you're describing it as I understand your theory. But you left off the important bit: The boiling off of the water. Keep stepping through this but through that part. Why doesn't the pot weigh the same or more as the atmosphere above it is receiving more atmosphere with which to increase the downward pressure of the stack, as it were.

(If this part is the explanation: "The pot itself is still holding volume but the volume is in the structure and is a barrier to the water, along with the dense mass of the structure itself...which is why structures differ massively in density and the displacement of atmosphere to the weigh scale" that doesn't mean anything in the context.)

?

#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #489 on: September 14, 2020, 11:54:21 PM »
I suggest you go back and read through what I've said because you are confusing yourself.
I explain it all and you do the exact same thing in saying I don't.
Sped more time absorbing what's been said.
I suggest you actually read what I have said and respond to it, preferably by providing a diagram showing where the force is acting, because as much you say you have explained it, you haven't.

I am not confusing myself at all, and I am paying attention to what you have said (paying attention/reading it is quite different to just accepting an "explanation", and I'm not sure what you would count "absorbing" as. I would say I have absorbed it, but again, that doesn't mean accepting it).
You just aren't explaining how your system works, you are not addressing the issues raised.

You haven't explained why when a tree absorbs atmosphere it weighs more even though it displaces less atmosphere due to the amount it has absorbed; nor why in the opposite situation of an airtight vessel weighs less when it is evacuated even though it has displaced more air, nor why a pot of water once boiled weighs less even though it has effectively displaced more atmosphere (and added to the atmosphere) by turning the water into atmosphere.

Likewise you haven't explained why the weight of an object is not based upon the cross sectional area like all other interactions with air pressure are and thus why the weight of a tower increases as the tower is made taller (even though no additional atmosphere is displaced to increase the pressure pushing down), now how the pressure/force increases as you go down the tower when the air can only apply a downwards force at the top of the tower.

The atmospheric displacement of that pot's dense mass (only) is now transferred back onto the pot's dense mass (not volume).
Why?
Why is it only the pot, rather than everything in the atmosphere?
If you take a sealed container and measure the pressure (the force acting on anything in there) and then displace it, such as by compressing the container, or by putting an extra item in without letting any of the air out, then that pressure increases.
A pressure sensor doesn't just measure the pressure of the air it displaces, but all the air in the system, regardless of what displaces it.

Now add water to the pot and what happens?
The water now pushes out the atmosphere (volume) from inside the pot back into the atmosphere outside of that pot but it has compressed it out and the water itself has displaced its own mass of that atmosphere.
Now that atmosphere not only reacts to the dense mass of the pot, it also reacts to the dense mass of the water inside of that pot, which means more push against that atmosphere and the equal push back...plus the amount that's already above the pot. all the way up....but we'll come to that bit.
And why does boiling the water change that?
If anything, boiling the water just displaces the atmosphere more as it adds to the atmosphere. So why does the pot now go back to the weight of before it had the water in it?

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #490 on: September 15, 2020, 12:07:56 AM »
I'm following all the way through and you're describing it as I understand your theory. But you left off the important bit: The boiling off of the water. Keep stepping through this but through that part. Why doesn't the pot weigh the same or more as the atmosphere above it is receiving more atmosphere with which to increase the downward pressure of the stack, as it were.
I apologise for that. I forgot to add the boiling off of water.

Ok....if the water in the pot is how I described before boiling off and the entire pot full is pushing away its own mass, as I said, them when you boil off that water you lose the water inside the pot to the atmosphere, not back to the pot, which means the pot is now filling back up with atmosphere against the water left in at every stage, meaning the atmosphere is now being pushed into much less by the every decreasing water due to this boiling.
Now the boiling part is created by the massive atmospheric agitation of that water which is consistently crushed up due to it being trapped within.
It's crushed up by the every increasing push back of the water by the crash of atmosphere within that pot.
I'm likely over explaining this...but maybe you get the essentials.

Quote from: Stash
(If this part is the explanation: "The pot itself is still holding volume but the volume is in the structure and is a barrier to the water, along with the dense mass of the structure itself...which is why structures differ massively in density and the displacement of atmosphere to the weigh scale" that doesn't mean anything in the context.)
The volume doesn't have to mean anything to you in this context. It's massively relevant but we'll deal with it all as we move through.
Just keep volume in mind and how it works from my side.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #491 on: September 15, 2020, 12:20:03 AM »
The atmospheric displacement of that pot's dense mass (only) is now transferred back onto the pot's dense mass (not volume).
Why?
Why is it only the pot, rather than everything in the atmosphere?
Because that's where the water was placed. It becomes part of that pot's dense makeup against the overall atmosphere. It doesn't become part of any other dense make up other than the exact place it is in.

Quote from: JackBlack
If you take a sealed container and measure the pressure (the force acting on anything in there) and then displace it, such as by compressing the container, or by putting an extra item in without letting any of the air out, then that pressure increases.
Yes, the pressure increases inside the container. I've explained this but I'll do it again.
If you place an object inside the container you displace atmosphere already in that container.
If that atmosphere cannot be pushed outside of it, it means the atmosphere inside of it becomes a bit more compressed and that compression will be placed right back on the object.
That pressure will be readable with a gauge.

Quote from: JackBlack
A pressure sensor doesn't just measure the pressure of the air it displaces, but all the air in the system, regardless of what displaces it.
Yep, so if the pressure inside the container is 14.7 psi and you add in a certain object, you increase that psi by what the object displaces. Basically its own dense mass of it which is naturally compressed back onto the object.

Quote from: JackBlack
If anything, boiling the water just displaces the atmosphere more as it adds to the atmosphere. So why does the pot now go back to the weight of before it had the water in it?
Because the pot becomes empty and back to being part of the atmosphere it was already in before.

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #492 on: September 15, 2020, 12:41:40 AM »
meaning the atmosphere is now being pushed into much less by the every decreasing water due to this boiling
The problem is that by volume the atmosphere is being pushed into more as the water boils and by mass, the same amount.

Because that's where the water was placed. It becomes part of that pot's dense makeup against the overall atmosphere. It doesn't become part of any other dense make up other than the exact place it is in.

Yes, the pressure increases inside the container. I've explained this but I'll do it again.
If you place an object inside the container you displace atmosphere already in that container.
If that atmosphere cannot be pushed outside of it, it means the atmosphere inside of it becomes a bit more compressed and that compression will be placed right back on the object.
The point is that it isn't ONLY that object, it is EVERY object inside it.
That compression is placed on EVERY object inside the sealed container.
The pressure on each object in the container is based upon the air displaced by EVERY object in the container, not just itself.

So why when it comes to weight is it only the air displaced by the object, rather than everything inside your dome?
The water still displaces the atmosphere before it is put into the pot, and it is the same atmosphere.

Because the pot becomes empty and back to being part of the atmosphere it was already in before.
But it isn't. The atmosphere before didn't have all the water vapour in it from the water that you boiled.
Why doesn't that increase the force, like pumping more air into a sealed container increases the pressure?

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #493 on: September 15, 2020, 01:38:31 AM »
The point is that it isn't ONLY that object, it is EVERY object inside it.
That compression is placed on EVERY object inside the sealed container.
The pressure on each object in the container is based upon the air displaced by EVERY object in the container, not just itself.
You mentioned putting an object in the container.
Ok, so add in 2 or 10 or 50. It's the same thing. Each object still displaces it's own dense mass of that container, meaning the container does rise in pressure of air but that pressure return is collectively on all of the objects by their own collective densities.

Quote from: JackBlack
So why when it comes to weight is it only the air displaced by the object, rather than everything inside your dome?
The water still displaces the atmosphere before it is put into the pot, and it is the same atmosphere.
Because the weight is determined by a man made scale and the actual object itself, to be measured, is placed upon that scale and that scale reacts to the atmospheric displacement by the object upon that scale as the object now uses that scale plate as it's resistant foundation.

Quote from: JackBlack
Because the pot becomes empty and back to being part of the atmosphere it was already in before.
But it isn't. The atmosphere before didn't have all the water vapour in it from the water that you boiled.
It didn't have any water in it...just volume/atmosphere.

Quote from: JackBlack
Why doesn't that increase the force, like pumping more air into a sealed container increases the pressure?
Because the sealed container is a sealed barrier to external atmosphere, so by you adding atmosphere to it, you increase the pressure within and lower the pressure externally......but....judging by the size of the atmosphere, you'll hardly notice that for something so minor.

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #494 on: September 15, 2020, 02:04:30 PM »
The point is that it isn't ONLY that object, it is EVERY object inside it.
That compression is placed on EVERY object inside the sealed container.
The pressure on each object in the container is based upon the air displaced by EVERY object in the container, not just itself.
You mentioned putting an object in the container.
Ok, so add in 2 or 10 or 50. It's the same thing. Each object still displaces it's own dense mass of that container, meaning the container does rise in pressure of air but that pressure return is collectively on all of the objects by their own collective densities.
You are still dodging the issue.
When you place an extra object in the container it isn't just it that receives the force, every object does.
The pressure of the air inside the container increases and thus the force on every object increases.

But when it comes to weight you are saying every other object is irrelevant.

Quote from: JackBlack
So why when it comes to weight is it only the air displaced by the object, rather than everything inside your dome?
The water still displaces the atmosphere before it is put into the pot, and it is the same atmosphere.
Because the weight is determined by a man made scale and the actual object itself, to be measured, is placed upon that scale and that scale reacts to the atmospheric displacement by the object upon that scale as the object now uses that scale plate as it's resistant foundation.
Which is completely different to pressure, without any explanation of why.
Why does pressure and weight work so fundamentally differently if they are both coming from the air?
Why is it that pressure cares about all the objects, while weight only cares about itself?
It is as if weight has nothing at all to do with the air pushing it down.

Quote from: JackBlack
Because the pot becomes empty and back to being part of the atmosphere it was already in before.
But it isn't. The atmosphere before didn't have all the water vapour in it from the water that you boiled.
It didn't have any water in it...just volume/atmosphere.
But not the same atmosphere.
By boiling the water you have displaced the atmosphere, which should increase the pressure due to your dome and thus increase the downwards force.

Quote from: JackBlack
Why doesn't that increase the force, like pumping more air into a sealed container increases the pressure?
Because the sealed container is a sealed barrier to external atmosphere, so by you adding atmosphere to it, you increase the pressure within and lower the pressure externally......but....judging by the size of the atmosphere, you'll hardly notice that for something so minor.
Then why do you notice the weight for any object?
I believe this objection was also raised to your model quite some time ago.
The amount of air an object displaces is insignificant compared to the atmosphere, so why does it result in a force on it based upon that displacement?
Why does putting the water in the pot, which is an insignificant change to the amount of atmosphere displaced, change the amount of downwards force on the pot?
If the change in atmospheric displacement is insignificant then it should have an insignificant effect.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #495 on: September 15, 2020, 10:46:20 PM »

Quote from: JackBlack
When you place an extra object in the container it isn't just it that receives the force, every object does.
The pressure of the air inside the container increases and thus the force on every object increases.

But when it comes to weight you are saying every other object is irrelevant.

Every object receives the extra force as long as no air is released upon addition of each object. You are compressing it more and more.

But we are talking about scale weighing one object displacing atmosphere, not a collection of objects.
This is where you're likely getting mixed up.

Let's stick to this particular issue and do not move on until you get it.

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #496 on: September 16, 2020, 02:56:36 AM »
Every object receives the extra force as long as no air is released upon addition of each object. You are compressing it more and more.

But we are talking about scale weighing one object displacing atmosphere, not a collection of objects.
This is where you're likely getting mixed up.
And why should that matter?
Note that a simple pressure sensor also works quite similar to a scale.
Again, why when you weigh an object the atmosphere only cares about the atmosphere displaced by that object, but if you measure the pressure acting on an object the atmosphere cares about the atmosphere displaced by all objects?

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #497 on: September 16, 2020, 10:18:30 PM »
Every object receives the extra force as long as no air is released upon addition of each object. You are compressing it more and more.

But we are talking about scale weighing one object displacing atmosphere, not a collection of objects.
This is where you're likely getting mixed up.
And why should that matter?
Note that a simple pressure sensor also works quite similar to a scale.
Again, why when you weigh an object the atmosphere only cares about the atmosphere displaced by that object, but if you measure the pressure acting on an object the atmosphere cares about the atmosphere displaced by all objects?
The atmosphere is acting on all objects within it....but you are measuring that on one specific object to gain a reading, whether that's by scale or pressure sensor. It doesn't matter. The object is still displacing it's very own dense mass of that atmosphere.

If you want to add in another object inside this container of pressure, then that object also displaces it's own dense mass of it and also have to resist the dense mass of atmosphere the other object displaced...and vice versa.

A pressure sensor would measure that extra compression of one or/and both displacements, with that sensor also being set to zero in normal atmosphere before the addition of the objects.

A direct scale plate measurement of one or both will also read that extra pressure build, whether the objects were placed on it, or not.

Now this is inside a container with localised pressure.
In external atmosphere your pressure is different, because your objects are already within it and are only displacing their own dense mass of that pressure.

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #498 on: September 17, 2020, 02:37:18 AM »
The atmosphere is acting on all objects within it....but you are measuring that on one specific object to gain a reading, whether that's by scale or pressure sensor. It doesn't matter. The object is still displacing it's very own dense mass of that atmosphere.
But it does matter, because only the pressure is based upon the presence of other objects while the weight is effectively unchanged (and can change in the wrong direction to what would be expected with pressure).

A direct scale plate measurement of one or both will also read that extra pressure build, whether the objects were placed on it, or not.
So are you claiming that if you take a sealed container, with a scale inside, with a 1 kg object on it, currently reading 1 kg, and then place another 1 kg weight inside the sealed container but not on the scale, without letting out any air, the weight reading will increase to 2 kg, due to the extra displaced atmosphere?

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #499 on: September 17, 2020, 08:41:20 AM »
So are you claiming that if you take a sealed container, with a scale inside, with a 1 kg object on it, currently reading 1 kg, and then place another 1 kg weight inside the sealed container but not on the scale, without letting out any air, the weight reading will increase to 2 kg, due to the extra displaced atmosphere?
No, I'm not.
The second 1kg is using the base of the container as its foundation to push against the compressed air inside the container. It is not using the scales.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #500 on: September 17, 2020, 08:45:49 AM »
I would be happy for someone to try and understand what I'm saying. Someone who is impartial, who could try and map what I'm saying.
Jane was the closest to ever getting it.
If any person feels they can grasp what I'm saying but maybe feels they might get intimidated, then pm me and try and figure it out without any pressure.
Nitpick and do what you need. All I ask is....be genuine.

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#### Themightykabool

• 5358
##### Re: Sea and air pressure
« Reply #501 on: September 17, 2020, 10:55:48 AM »
Youd probably be happier if you drew some more diagrams

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #502 on: September 17, 2020, 01:51:20 PM »
No, I'm not.
The second 1kg is using the base of the container as its foundation to push against the compressed air inside the container. It is not using the scales.
So are you saying the scales still measure 1 kg?

I would be happy for someone to try and understand what I'm saying.
Stop acting like we aren't trying to understand.
The issue is not the understanding, the issue is your model's inability to match and explain reality.

Nitpick and do what you need. All I ask is....be genuine.
I am.
The main difference between my approach and Jane's, is that Jane didn't care if your model didn't work at all or didn't explain reality. I do care as you are claiming it works and claiming that current models which actually explain reality are wrong.
« Last Edit: September 17, 2020, 01:54:01 PM by JackBlack »

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #503 on: September 17, 2020, 11:01:22 PM »
Grasp the basics of your own model and draw some diagrams.

You need to be able to differentiate between regular air and the sponge air that causes things to go "down".
No I don't.
There is no difference. It's all atmosphere, so understand that and try and grasp it.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #504 on: September 17, 2020, 11:03:42 PM »
Youd probably be happier if you drew some more diagrams
If I honestly thought you grasped anything I would put a few crude drawings in but I tried it before and you went into raptures.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #505 on: September 17, 2020, 11:32:48 PM »
No, I'm not.
The second 1kg is using the base of the container as its foundation to push against the compressed air inside the container. It is not using the scales.
So are you saying the scales still measure 1 kg?
There would possibly be fluctuations due to internal scales requiring recalibration because those scales are set in external atmospheric conditions and are now placed inside a sealed container with one 1kg object directly on that scale...which is fine, because it's that displacement that counts.
If the second 1kg object is added, it will add extra pressure onto itself and transfer that displacement all through the container.....including through the scales as well as on them.

So you can see why the scales would likely read some change but not the change you think of the extra displacement of the second object's 1kg displacement...because that displacement is primarily being resisted by the floor of the container the second object sits on....not the scale plate.

Can you understand this?

Can anyone?

Quote from: JackBlack
Nitpick and do what you need. All I ask is....be genuine.
I am.
The main difference between my approach and Jane's, is that Jane didn't care if your model didn't work at all or didn't explain reality. I do care as you are claiming it works and claiming that current models which actually explain reality are wrong.
I'm not asking you to care whether you think it works. I'm explaining it so people can grasp it.
You can dismiss it until you're blue in the face but my explanations may just peak the interest of some, which means they can take the time to understand it.

You seem to be having a hard time grasping it so I'll simply keep finding ways to explain.

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#### JackBlack

• 15785
##### Re: Sea and air pressure
« Reply #506 on: September 18, 2020, 01:14:53 AM »
If the second 1kg object is added, it will add extra pressure onto itself and transfer that displacement all through the container.....including through the scales as well as on them.

So you can see why the scales would likely read some change but not the change you think of the extra displacement of the second object's 1kg displacement...because that displacement is primarily being resisted by the floor of the container the second object sits on....not the scale plate.
So how much would it change and in what direction?
Would the weight increase or decrease, and by a tiny portion of the 1 kg, or a large portion of it, and why?

Also there is a massive difference between pressure and weight.
With pressure, the second objects displacement does create additional pressure, even though it doesn't use the pressure sensor as "resistance". But with weight, it doesn't.
Why the distinction?

How come with a pressure gauge it does not matter what causes the displacement of the atmosphere (including if it is the pressure gauge itself or some completely different object, not even touching the pressure guage), the pressure reading changes the same, but for weight, it is only if the object displacing the atmosphere is on the scale that the full effect is achieved?

How come for weight it isn't all objects, or how come for pressure it isn't only objects on or part of the pressure gauge?

I'm not asking you to care whether you think it works. I'm explaining it so people can grasp it.
You can dismiss it until you're blue in the face but my explanations may just peak the interest of some, which means they can take the time to understand it.
You seem to be having a hard time grasping it so I'll simply keep finding ways to explain.
I never said you were asking. I am just stating that I care if your model works because you claim it does.
I don't dismiss your model, I repeatedly point out how it contradicts itself and fails to actually explain reality.
Again, it isn't an issue of not grasping it. It is an issue of your model contradicting itself and reality.
Stop acting like the only reason people don't accept your model is because they don't understand.

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #507 on: September 18, 2020, 05:02:39 AM »
If the second 1kg object is added, it will add extra pressure onto itself and transfer that displacement all through the container.....including through the scales as well as on them.

So you can see why the scales would likely read some change but not the change you think of the extra displacement of the second object's 1kg displacement...because that displacement is primarily being resisted by the floor of the container the second object sits on....not the scale plate.
So how much would it change and in what direction?
Would the weight increase or decrease, and by a tiny portion of the 1 kg, or a large portion of it, and why?
To find out how much it would change, try it out.
Make sure you set the experiment up properly or your readings will be skewed.

The weight may possibly increase a little depending on the scale set up and how much play it gives, because, what you have to remember is, the second object is applying it's own dense mass of pressure to the whole container, which includes the whole scale, not just the top of the scale plate.

Try it out and see what you get.

Quote from: JackBlack
Also there is a massive difference between pressure and weight.
With pressure, the second objects displacement does create additional pressure, even though it doesn't use the pressure sensor as "resistance". But with weight, it doesn't.
Why the distinction?
How come with a pressure gauge it does not matter what causes the displacement of the atmosphere (including if it is the pressure gauge itself or some completely different object, not even touching the pressure guage), the pressure reading changes the same, but for weight, it is only if the object displacing the atmosphere is on the scale that the full effect is achieved?
Weight is a man made scale measurement of dense mass placed upon that scale and displacing the atmosphere by that amount, which is measured by the pressure on the spring and the resistance of it to gain a reading.
Atmospheric pressure is also weighed on a scale but it's a pressure scale. It has to only have a one way resistance, which is why they have diaphragms.

#### NotSoSkeptical

• 6950
• Flatness as in the shape of a water droplet.
##### Re: Sea and air pressure
« Reply #508 on: September 18, 2020, 09:53:47 AM »
If the second 1kg object is added, it will add extra pressure onto itself and transfer that displacement all through the container.....including through the scales as well as on them.

So you can see why the scales would likely read some change but not the change you think of the extra displacement of the second object's 1kg displacement...because that displacement is primarily being resisted by the floor of the container the second object sits on....not the scale plate.
So how much would it change and in what direction?
Would the weight increase or decrease, and by a tiny portion of the 1 kg, or a large portion of it, and why?
To find out how much it would change, try it out.
Make sure you set the experiment up properly or your readings will be skewed.

The weight may possibly increase a little depending on the scale set up and how much play it gives, because, what you have to remember is, the second object is applying it's own dense mass of pressure to the whole container, which includes the whole scale, not just the top of the scale plate.

Try it out and see what you get.

Quote from: JackBlack
Also there is a massive difference between pressure and weight.
With pressure, the second objects displacement does create additional pressure, even though it doesn't use the pressure sensor as "resistance". But with weight, it doesn't.
Why the distinction?
How come with a pressure gauge it does not matter what causes the displacement of the atmosphere (including if it is the pressure gauge itself or some completely different object, not even touching the pressure guage), the pressure reading changes the same, but for weight, it is only if the object displacing the atmosphere is on the scale that the full effect is achieved?
Weight is a man made scale measurement of dense mass placed upon that scale and displacing the atmosphere by that amount, which is measured by the pressure on the spring and the resistance of it to gain a reading.
Atmospheric pressure is also weighed on a scale but it's a pressure scale. It has to only have a one way resistance, which is why they have diaphragms.

Hate to break this to you, but all forms of measurement are man made.  Mass, height, width, length, volume, density, etc.  All created by man.
Rabinoz RIP

#### sceptimatic

• Flat Earth Scientist
• 28092
##### Re: Sea and air pressure
« Reply #509 on: September 18, 2020, 10:35:36 AM »
Hate to break this to you, but all forms of measurement are man made.  Mass, height, width, length, volume, density, etc.  All created by man.
You're not breaking anything to me. I'm well aware of it.