Weightlessness During Freefall

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NotSoSkeptical

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Weightlessness During Freefall
« on: January 08, 2023, 06:03:42 PM »
Really cool experiment.

Rabinoz RIP

That would put you in the same category as pedophile perverts like John Davis, NSS, robots like Stash, Shifter, and victimized kids like Alexey.

Re: Weightlessness During Freefall
« Reply #1 on: January 08, 2023, 06:39:16 PM »
Cool.  Just another thing flat earther’s will ignore and not taken in account for their arguments/modeling….

Or if they do comment, it will be interesting to see if it adds to their contradictions? 

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Stash

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Re: Weightlessness During Freefall
« Reply #2 on: January 09, 2023, 12:59:06 AM »
I've never seen that one before. Such a simple experiment. The only FE explanation I can think of would be Universal Acceleration (UA): UA asserts that the Earth and the observable universe are accelerating 'upward' at a constant rate of 9.8m/s^2.

But it appears that only tfes believes in that. I've never seen anywhere else out there in the FE community that puts earth in a perpetual upward motion. And I'm not even sure that would cause the effect.

Re: Weightlessness During Freefall
« Reply #3 on: January 09, 2023, 01:03:32 AM »
great example of the P = pgh


would be super good for a super slow mo where they distance that where the v catches up to g and you see the water stops coming out.


bulbma doesn't like formulas though...
too bad

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Stash

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Re: Weightlessness During Freefall
« Reply #4 on: January 09, 2023, 04:38:09 AM »

Re: Weightlessness During Freefall
« Reply #5 on: January 09, 2023, 04:43:57 AM »
Really cool experiment.



Uhhhhhhhh, that's not weightlessness.

If you were actually weightless, wouldn't you be able to stop your fall?

Damn straight you would. Just weightless yourself to a stall in momentum, halt freefall, and stay there or swim gently toward the ground.

But no, you are still falling.

What this is, is terminal velocity. You see, science, real science, doesn't rush to label things incorrectly.

A terminal velocity is when an object reaches the maximum speed it will get.
Quote
Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid. It occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity acting on the object.

Straight from wikipedia. Now, I would ignore the "downward force of gravity" part and the insistence on fluids. There is no downward force of gravity needed, as objects fall when they are heavier than a fluid (or gas). This is just an aspect of buoyancy: it causes floating when object is lighter than a medium, and the inverse is true. Objects travel through a fluid (or a gas), the object attains its max speed, and buoyancy and drag force are equal.

I haven't really changed anything about the definition of terminal velocity, other than applying it to air (which any parachuter can tell you does happen). Whereas you have called it weightlessness.

Wanna see how weightless people are when they hit the ground with no parachute? Then you've changed a term. They aren't weightless, they have achieved terminal velocity and are in a state of equilibrium. This is a blessing from God, as it means the laws of physics make them only kinda sorta smashed instead of totally ground to powder and mush. What? Aren't you happy that when die from a fall, you're less dead than you could be without a max speed?
« Last Edit: January 09, 2023, 04:46:39 AM by bulmabriefs144 »




Re: Weightlessness During Freefall
« Reply #6 on: January 09, 2023, 05:15:26 AM »
Really cool experiment.



Uhhhhhhhh, that's not weightlessness.

If you were actually weightless, wouldn't you be able to stop your fall?

Damn straight you would. Just weightless yourself to a stall in momentum, halt freefall, and stay there or swim gently toward the ground.

But no, you are still falling.

What this is, is terminal velocity. You see, science, real science, doesn't rush to label things incorrectly.

A terminal velocity is when an object reaches the maximum speed it will get.
Quote
Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid. It occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity acting on the object.

Straight from wikipedia. Now, I would ignore the "downward force of gravity" part and the insistence on fluids. There is no downward force of gravity needed, as objects fall when they are heavier than a fluid (or gas). This is just an aspect of buoyancy: it causes floating when object is lighter than a medium, and the inverse is true. Objects travel through a fluid (or a gas), the object attains its max speed, and buoyancy and drag force are equal.

I haven't really changed anything about the definition of terminal velocity, other than applying it to air (which any parachuter can tell you does happen). Whereas you have called it weightlessness.

Wanna see how weightless people are when they hit the ground with no parachute? Then you've changed a term. They aren't weightless, they have achieved terminal velocity and are in a state of equilibrium. This is a blessing from God, as it means the laws of physics make them only kinda sorta smashed instead of totally ground to powder and mush. What? Aren't you happy that when die from a fall, you're less dead than you could be without a max speed?



Your a train wreck.

The topic is why the water lost the motivation to flow out of the punctured container while the container was in free fall.


Dumbed down version..
Quote

Water Doesn’t Leak Out Science Experiment

https://coolscienceexperimentshq.com/water-doesnt-leak/


Gravity is also the reason water doesn’t leak out of the holes when you drop the cup in Step 4. In this step, gravity is still pulling the water back to Earth, but this time the cup and water are falling back to Earth at the same rate when you drop the cup. This doesn’t allow water to leak out of the cup because the water is already taking the path of least resistance back to Earth – straight down with the cup.


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Stash

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Re: Weightlessness During Freefall
« Reply #7 on: January 09, 2023, 05:49:15 AM »
I haven't really changed anything about the definition of terminal velocity, other than applying it to air (which any parachuter can tell you does happen). Whereas you have called it weightlessness.

First off, why in the world would terminal velocity stop the water from flowing out the holes? That, unto itself, makes zero sense.

Secondly, yeah, you totally changed the definition of terminal velocity. Terminal velocity for, let's say, a 1lb object (size of container dropped) is about 130+ MPH. And yeah, for a parachuter jumping out of plane it takes about 12 or so seconds to reach terminal velocity.
When falling in the standard belly-to-Earth position, an average estimate of terminal velocity for skydivers is 120 mph (200 km/h), and a falling person will reach terminal velocity after about 12 seconds, falling some 450 m (1,500 ft) in that time.

You're saying the water container was going 130 mph in under 1 second from a height of maybe a dozen feet? That's just plain lunacy.



Really try and think things through before posting irrelevant, ignorant, and incorrect stuff.

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Slemon

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Re: Weightlessness During Freefall
« Reply #8 on: January 09, 2023, 05:56:31 AM »
I don't get this objection. Gravity isn't magic. Any downwards acceleration is gonna cause pretty much the same effects, whether it's justified with respect to an alt-buoyancy model, UA, modified gravity, subquark strings... You can, for the purposes of this, model it all pretty much the same. All you need is for water to, as Data's link says, take the path of least resistance down to the Earth. That, like, seems completely in line with everything up to and including Turbo's model?
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #9 on: January 09, 2023, 06:23:17 AM »
I don't get this objection. Gravity isn't magic. Any downwards acceleration is gonna cause pretty much the same effects, whether it's justified with respect to an alt-buoyancy model, UA, modified gravity, subquark strings... You can, for the purposes of this, model it all pretty much the same. All you need is for water to, as Data's link says, take the path of least resistance down to the Earth. That, like, seems completely in line with everything up to and including Turbo's model?

It shows the motivation of the water to flow out is gravity. 

When the container is stationary, this is when this would apply.  “ whether it's justified with respect to an alt-buoyancy model, UA, modified gravity, subquark strings..”

So you have the gravity inspired air pressure/pressure gradient motivating the water to spill out.

Then in freefall the water, “back to Earth at the same rate when you drop the cup. This doesn’t allow water to leak out of the cup because the water is already taking the path of least resistance”

Now, in the models absence of gravity. Why would the water stop being motivated to flow out going from standing still to moving downward.

The models that claim absence of gravity while the water is pushed out.  Why would dropping the punctured container in these other models stop the water from flowing.  They don’t depend on the downward force of gravity to motivate the water to flow through the punctures.  The downward force of gravity is “equalized” by being in free fall.

Why would any of the other models that regard gravity as nonexistent care the punctured container is in free fall to motivate the water to flow through the punctures, or not in free fall.


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free fall
noun
1
: the condition of unrestrained motion in a gravitational

https://www.merriam-webster.com/dictionary/free%20fall


It seems free fall would be meaningless in other models where gravity is believed to be totally nonexistent. 

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Stash

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Re: Weightlessness During Freefall
« Reply #10 on: January 09, 2023, 06:26:36 AM »
I don't get this objection. Gravity isn't magic. Any downwards acceleration is gonna cause pretty much the same effects, whether it's justified with respect to an alt-buoyancy model, UA, modified gravity, subquark strings... You can, for the purposes of this, model it all pretty much the same. All you need is for water to, as Data's link says, take the path of least resistance down to the Earth. That, like, seems completely in line with everything up to and including Turbo's model?

Hmmm, not sure if I agree. Maybe the question becomes, "Why is there resistance and what is causing the resistance?"

In Turbo's notion, (I think), everything falls because it's returning to its origin, earth.

In the others, I'm just not sure why the water spouts would stop spouting.

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Slemon

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Re: Weightlessness During Freefall
« Reply #11 on: January 09, 2023, 06:35:20 AM »
I don't get this objection. Gravity isn't magic. Any downwards acceleration is gonna cause pretty much the same effects, whether it's justified with respect to an alt-buoyancy model, UA, modified gravity, subquark strings... You can, for the purposes of this, model it all pretty much the same. All you need is for water to, as Data's link says, take the path of least resistance down to the Earth. That, like, seems completely in line with everything up to and including Turbo's model?

Hmmm, not sure if I agree. Maybe the question becomes, "Why is there resistance and what is causing the resistance?"

In Turbo's notion, (I think), everything falls because it's returning to its origin, earth.

In the others, I'm just not sure why the water spouts would stop spouting.
Honestly I brought up Turbo because, if you grant the premise, it gives the most explicit illustration of why it works - the water is heading right down to the Earth, it just no longer needs to go out the holes to do so. If my understanding is correct, it isn't necessarily that there's resistance to the movement, it's that the downwards force no longer needs to send anything out the holes because the water can go in another direction - so, maybe better to say there is resistance when the container is stationary exerted by the base of the can that as a result pushes the water out the holes, and no resistance when it is falling hence no force out the holes.

In this sort of situation, to me, the best question is "What is the explanation under gravity, and why can it not hold otherwise?"
Beyond Data's circular 'Freefall is defined with reference to gravity,' under FE models I'm not sure I see why this principle would fail.
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Stash

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Re: Weightlessness During Freefall
« Reply #12 on: January 09, 2023, 07:34:03 AM »
I don't get this objection. Gravity isn't magic. Any downwards acceleration is gonna cause pretty much the same effects, whether it's justified with respect to an alt-buoyancy model, UA, modified gravity, subquark strings... You can, for the purposes of this, model it all pretty much the same. All you need is for water to, as Data's link says, take the path of least resistance down to the Earth. That, like, seems completely in line with everything up to and including Turbo's model?

Hmmm, not sure if I agree. Maybe the question becomes, "Why is there resistance and what is causing the resistance?"

In Turbo's notion, (I think), everything falls because it's returning to its origin, earth.

In the others, I'm just not sure why the water spouts would stop spouting.
Honestly I brought up Turbo because, if you grant the premise, it gives the most explicit illustration of why it works - the water is heading right down to the Earth, it just no longer needs to go out the holes to do so. If my understanding is correct, it isn't necessarily that there's resistance to the movement, it's that the downwards force no longer needs to send anything out the holes because the water can go in another direction - so, maybe better to say there is resistance when the container is stationary exerted by the base of the can that as a result pushes the water out the holes, and no resistance when it is falling hence no force out the holes.

In this sort of situation, to me, the best question is "What is the explanation under gravity, and why can it not hold otherwise?"
Beyond Data's circular 'Freefall is defined with reference to gravity,' under FE models I'm not sure I see why this principle would fail.

I'm not clear on the resistance of the base changes when transitioning from stationary to moving when applying Turbo's thing and especially UA.

With Turbo's, all things are desperate to get back home, to the origin, earth. I don't see how that would impart a resistance change in the base of the container. It seems to me that the water would just keep on keepin' on and flow out the holes. I can't think of a reason as to why it would stop flowing. In other words, I don't know why the water's needs would actually change, flowing out the holes seems just as good as riding along with the container.

UA, the earth is rising up to meet the container - But I would think there would be absolutely no change in resistance because nothing is imparting any effect on the container until it meets with the upward accelerating earth. In other words, the container doesn't even know it's falling until it lands.

The terminal velocity bit, just irrelevant and ridiculous.

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Slemon

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Re: Weightlessness During Freefall
« Reply #13 on: January 09, 2023, 07:58:42 AM »
I'm not clear on the resistance of the base changes when transitioning from stationary to moving when applying Turbo's thing and especially UA.
If the base of the container is moving downwards, then necessarily it's exerting less of an upwards force than when it's stationary. The only time I can see that being a problem is with the few 'no reaction force on stationary objects' models that seem to have popped up, but that's a) more an issue with why water goes out the holes in the first place than why it would stop, b) likely resolvable with reference to the properties of liquid.
Like, if we were to keep the can stationary in the air and instead neatly slice off the bottom layer, the water would stop coming out the holes in order to immediately plummet out the bottom. It's taking the only path it can to go down, that's all that's happening in any model. When the container itself is descending, there's an easier path down.

Or, short version: if we accept, for the purposes of discussion, that water wants to go down when it is inside the can, then that is sufficient to explain why water will come out the holes, and why it will stop when the container is dropped. Whether the reason the water stays at the bottom of the can is gravity or buoyancy is irrelevant to the physics of the question.

I can't think of a reason as to why it would stop flowing. In other words, I don't know why the water's needs would actually change, flowing out the holes seems just as good as riding along with the container.
Why do you think it stops flowing under gravity?
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #14 on: January 09, 2023, 08:11:40 AM »
I'm not clear on the resistance of the base changes when transitioning from stationary to moving when applying Turbo's thing and especially UA.
If the base of the container is moving downwards, then necessarily it's exerting less of an upwards force than when it's stationary. The only time I can see that being a problem is with the few 'no reaction force on stationary objects' models that seem to have popped up, but that's a) more an issue with why water goes out the holes in the first place than why it would stop, b) likely resolvable with reference to the properties of liquid.
Like, if we were to keep the can stationary in the air and instead neatly slice off the bottom layer, the water would stop coming out the holes in order to immediately plummet out the bottom. It's taking the only path it can to go down, that's all that's happening in any model. When the container itself is descending, there's an easier path down.

Or, short version: if we accept, for the purposes of discussion, that water wants to go down when it is inside the can, then that is sufficient to explain why water will come out the holes, and why it will stop when the container is dropped. Whether the reason the water stays at the bottom of the can is gravity or buoyancy is irrelevant to the physics of the question.

I can't think of a reason as to why it would stop flowing. In other words, I don't know why the water's needs would actually change, flowing out the holes seems just as good as riding along with the container.
Why do you think it stops flowing under gravity?

Let’s put it this way.  All the other models there is no gravity.  Zero.  Zilch. So by logic, they are already in a state of “weightlessness”.

Notes.  Added.  Maybe the better phrase would be, “by definition things would already be in a state of “weightlessness”….

So, why in models of zero gravity, why would dropping the punctured container have any effect on motivation the water to not flow out the punctures while the container is dropped. 

« Last Edit: January 09, 2023, 08:38:19 AM by DataOverFlow2022 »

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Slemon

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Re: Weightlessness During Freefall
« Reply #15 on: January 09, 2023, 09:06:22 AM »
Let’s put it this way.  All the other models there is no gravity.  Zero.  Zilch. So by logic, they are already in a state of “weightlessness”.
"FE alternatives for gravity don't work."
"Why?"
"Because they're not gravity."

Sheer genius.
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #16 on: January 09, 2023, 09:21:18 AM »
Let’s put it this way.  All the other models there is no gravity.  Zero.  Zilch. So by logic, they are already in a state of “weightlessness”.
"FE alternatives for gravity don't work."
"Why?"
"Because they're not gravity."

Sheer genius.

Gravity is based on a model where mass has a force of attraction.

Quote
Gravity

https://www.merriam-webster.com/dictionary/gravity

3
a
(1)
: the gravitational attraction of the mass of the earth, the moon, or a planet for bodies at or near its surface
(2)
: a fundamental physical force that is responsible for interactions which occur because of mass between particles, between aggregations of matter (such as stars and planets), and between particles (such as photons) and aggregations of matter, that is 10-39 times the strength of the strong force, and that extends over infinite distances but is dominant over macroscopic distances especially between aggregations of matter
 called also gravitation, gravitational force
 compare ELECTROMAGNETISM sense 2a, STRONG FORCE, WEAK FORCE



Quote
weight

https://www.merriam-webster.com/dictionary/weight


6
a
: relative heaviness : MASS
b
: the force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration


So, say in sceptimatic‘a model, what would the definition of gravity, weight, free fall, and weightlessness be?  Would they have any real meaning? 

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Slemon

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Re: Weightlessness During Freefall
« Reply #17 on: January 09, 2023, 10:22:33 AM »
So, say in sceptimatic‘a model, what would the definition of gravity, weight, free fall, and weightlessness be?  Would they have any real meaning?
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume
Free fall - the condition of unrestrained motion in a denpressure system
Weightlessness - lacking an apparent denpressure push

Stop acting like circular arguments are what we need to resort to. They have a real meaning, you just swap out gravity. Congratulations, you have noticed that FET and RET are not the same thing, do you have an actual point to make?
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #18 on: January 09, 2023, 10:32:50 AM »
So, say in sceptimatic‘a model, what would the definition of gravity, weight, free fall, and weightlessness be?  Would they have any real meaning?
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume
Free fall - the condition of unrestrained motion in a denpressure system
Weightlessness - lacking an apparent denpressure push

Stop acting like circular arguments are what we need to resort to. They have a real meaning, you just swap out gravity. Congratulations, you have noticed that FET and RET are not the same thing, do you have an actual point to make?


Sorry.  No gravity, no weight, and free fall becomes meaningless in terms of weight from gravity.

« Last Edit: January 09, 2023, 10:35:09 AM by DataOverFlow2022 »

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Slemon

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Re: Weightlessness During Freefall
« Reply #19 on: January 09, 2023, 10:57:37 AM »
So, say in sceptimatic‘a model, what would the definition of gravity, weight, free fall, and weightlessness be?  Would they have any real meaning?
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume
Free fall - the condition of unrestrained motion in a denpressure system
Weightlessness - lacking an apparent denpressure push

Stop acting like circular arguments are what we need to resort to. They have a real meaning, you just swap out gravity. Congratulations, you have noticed that FET and RET are not the same thing, do you have an actual point to make?


Sorry.  No gravity, no weight, and free fall becomes meaningless in terms of weight from gravity.
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #20 on: January 09, 2023, 11:14:45 AM »
things fall down

makes sense



in case of magnets, i was told to consider the potential energy as if it started at the magnet and was pulled away from it.
"so that metal falls back towards its source"

turb's theory holds (within it's own reality)

Re: Weightlessness During Freefall
« Reply #21 on: January 09, 2023, 11:23:43 AM »
So, say in sceptimatic‘a model, what would the definition of gravity, weight, free fall, and weightlessness be?  Would they have any real meaning?
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume
Free fall - the condition of unrestrained motion in a denpressure system
Weightlessness - lacking an apparent denpressure push

Stop acting like circular arguments are what we need to resort to. They have a real meaning, you just swap out gravity. Congratulations, you have noticed that FET and RET are not the same thing, do you have an actual point to make?


Sorry.  No gravity, no weight, and free fall becomes meaningless in terms of weight from gravity.




With gravity.  The water doesn’t flow out of the punctures while falling because the water and container are falling at the same rate.  While both are accelerating to earth due to gravity, the water follows the path of least resistance.
There is no motivation for the water to overcome and flow out of the restrictions of the punctures.  Essentially, the water and container are converting gravitational potential energy to kinetic energy at the same rate.  When the container was stationary, the potential energy of the water was able to cause the water to flow through the punctures/holes.

Now your definitions.

Quote
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume

Now.  With the dropped container.  Why would the water stop flowing out under denpressure.

With gravity. It makes sense.  The water and bucket fall to earth at the same acceleration and direction.  Taking away the motivation for the water to flow out.

How does denpressure. which has nothing to do with relative acceleration, magically disappear to make the water stop flowing just because the container was dropped.
« Last Edit: January 09, 2023, 11:26:00 AM by DataOverFlow2022 »

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Slemon

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Re: Weightlessness During Freefall
« Reply #22 on: January 09, 2023, 11:28:06 AM »
Now.  With the dropped container.  Why would the water stop flowing out under denpressure.

With gravity. It makes sense.  The water and bucket fall to earth at the same acceleration and direction.  Taking away the motivation for the water to flow out.

How does denpressure. which has nothing to do with relative acceleration, magically disappear to make the water stop flowing just because the container was dropped.
The water doesn’t flow out of the punctures while falling because the water and container are falling at the same rate.  While both are accelerating to earth due to denpressure, the water follows the path of least resistance.
There is no motivation for the water to overcome and flow out of the restrictions of the punctures.  Essentially, the water and container are converting potential energy to kinetic energy at the same rate.  When the container was stationary, the potential energy of the water was able to cause the water to flow through the punctures/holes.
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #23 on: January 09, 2023, 11:38:59 AM »

While both are accelerating to earth due to denpressure,

What “force” in denpressure would make objects accelerate towards earth.  And by you definition “ the heaviness of an object, resulting from denpressure, relative to density and volume” there is no mechanism for turning off “ denpressure” while falling to earth.


While gravity is shown to be a real force by Cavendish experiments.

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Slemon

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Re: Weightlessness During Freefall
« Reply #24 on: January 09, 2023, 11:51:11 AM »
there is no mechanism for turning off “ denpressure” while falling to earth.
So. What you're saying is that the water and container will both be falling at the same rate, thus there's nothing forcing the water out the holes. Great! We're done!

Dude, when you need to just whole-hog replace the argument with 'But FEers don't have a downwards force at all!' and 'we have evidence for gravity!' you're just straight-up gish galloping at that point. It is not a good look. Slow down, take a step back, stop acting like RET is a cult that needs you to blindly defend every bad argument. You are doing way more harm than good here.
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #25 on: January 09, 2023, 11:52:57 AM »

Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume


Dropping the container didn’t change the volume or density of anything.

So.  Why would the water stop flowing under “denpressure”


The real definition of weight.  “the force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration”
https://www.merriam-webster.com/dictionary/weight


It makes sense if the water and container fall at the same rate, the motivation for the water to flow out of the container through the holes is removed.

Gravity and denpressure are not interchangeable.

Re: Weightlessness During Freefall
« Reply #26 on: January 09, 2023, 11:55:50 AM »

Dude, when you need to just whole-hog replace the argument with 'But FEers don't have a downwards force at all!'


What does that have to do with conducting Cavendish experiments and what the Cavendish experiments prove? 

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Slemon

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Re: Weightlessness During Freefall
« Reply #27 on: January 09, 2023, 11:58:23 AM »
Gravity and denpressure are not interchangeable.
No. They're not. You also don't understand denpressure.

Okay. Let's slow down then. Let's look at what is actually required. My position is this: the necessary trait for this experiment to function is for comparable accelerations to be imparted to the water and the can.
When stationary, the reaction force from the bottom of the can induced by a downwards acceleration forces the water out the holes. When in freefall, there is no reaction force between the water and the bottom of the can, hence nothing to force it out. Equally, if we were to yank the can down with our hand faster than gravity would pull it, we'd get the opposite - a reaction force exerted by the top of the can on the water that pushes it out the holes.
Hence, as said, all that is required is a force that sees the water drawn down towards the Earth's surface while it is inside the can, that is the same force that accelerates the can down when it is dropped.

Do you agree with this basic description of the physics at play, independent of any model?
We all know deep in our hearts that Jane is the last face we'll see before we're choked to death!

Re: Weightlessness During Freefall
« Reply #28 on: January 09, 2023, 12:00:21 PM »
Gravity and denpressure are not interchangeable.
No. They're not. You also don't understand denpressure.

Okay. Let's slow down then. Let's look at what is actually required. My position is this: the necessary trait for this experiment to function is for comparable accelerations to be imparted to the water and the can.
When stationary, the reaction force from the bottom of the can induced by a downwards acceleration forces the water out the holes. When in freefall, there is no reaction force between the water and the bottom of the can, hence nothing to force it out. Equally, if we were to yank the can down with our hand faster than gravity would pull it, we'd get the opposite - a reaction force exerted by the top of the can on the water that pushes it out the holes.
Hence, as said, all that is required is a force that sees the water drawn down towards the Earth's surface while it is inside the can, that is the same force that accelerates the can down when it is dropped.

Do you agree with this basic description of the physics at play, independent of any model?


What does any of the have to do with the Cavendish experiments?

And this?

You
Quote
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume

Dropping the container didn’t change the volume or density of anything.



So.  Why would the water stop flowing under “denpressure”

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Slemon

  • Flat Earth Researcher
  • 12330
Re: Weightlessness During Freefall
« Reply #29 on: January 09, 2023, 12:04:41 PM »
Gravity and denpressure are not interchangeable.
No. They're not. You also don't understand denpressure.

Okay. Let's slow down then. Let's look at what is actually required. My position is this: the necessary trait for this experiment to function is for comparable accelerations to be imparted to the water and the can.
When stationary, the reaction force from the bottom of the can induced by a downwards acceleration forces the water out the holes. When in freefall, there is no reaction force between the water and the bottom of the can, hence nothing to force it out. Equally, if we were to yank the can down with our hand faster than gravity would pull it, we'd get the opposite - a reaction force exerted by the top of the can on the water that pushes it out the holes.
Hence, as said, all that is required is a force that sees the water drawn down towards the Earth's surface while it is inside the can, that is the same force that accelerates the can down when it is dropped.

Do you agree with this basic description of the physics at play, independent of any model?


What does any of the have to do with the Cavendish experiments?

And this?

You
Quote
Gravity - a non-existent thing
Weight - the heaviness of an object, resulting from denpressure, relative to density and volume

Dropping the container didn’t change the volume or density of anything.



So.  Why would the water stop flowing under “denpressure”

Cool. Evading.

It has nothing to do with Cavendish, because this argument has nothing to do with Cavendish. Gish galloping is the tactic of throwing a wealth of unrelated arguments at the wall rather than focusing on any one, it is a go-to trick for weak positions and dishonest debaters. Do you think RET is a weak position? It does not need you to use this troll shtick to defend it, it is more than solid enough to stand by its own. You acting like this is the reason we get those posts where FEers assume RET is some cult filled with morons and sheeple, because when you act like this it is genuinely believable.
You are changing the topic, evading questions, and actively misrepresenting FE models.

I told you why the water would stop flowing. You disagreed. I asked a question to see if we can reach a mutual understanding on the physics at play. You evaded, and tried to change the subject to Cavendish.
Please stop resorting to this stupidity unless you genuinely think RET is false and needs you to lie for it.
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