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.
That is something your model cannot address.
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.
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.
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.