Simple Test you can do (mostly) at home.

Quote from: markjo on February 19, 2026, 03:11:26 PM

How does one calculate and measure this acceleration?  By chance, do they roughly follow the results of Newton's gravity formulas?

You are cherry-picking one tiny sentence about "calculation" because you are terrified to address the mountain of evidence I just laid out.

Not as terrified as you appear to be when asked for some maths that support your claims.

By skipping 95% of my post, you have conceded every single point you ignored.

I concede nothing.  I just don't feel like dealing with your gish gallop.  What's wrong with focusing on one point?

Quote from: Erland on February 17, 2026, 11:25:28 AM

Science, including maths, doesn't work so that you have to do everything yourself. You can, and must, use other's results.

Using others' results is fine for a librarian, but for a seeker of truth, it’s a trap. If the foundation is a lie, you are just building a taller tower of garbage.

Then you are a librarian yourself using the results of Rowbotham about the Bedford Canal, talking about the Black Swan (you never saw it yourself ) etc. etc. From now on, you can only talk about measurements and observations made by yourself, everything else is "a trap" and invalid to use in the debate.

Quote from: Erland on February 17, 2026, 11:25:28 AM

I proved that the consistency of the constellations is consistent with the globe and its atmosphere.

You didn't "prove" anything; you confirmed a confirmation bias. Being "consistent with" is not the same as being "the cause of." A shadow is consistent with a ghost, but that doesn't make the ghost real.

I did what you asked me to do. Period.

Quote from: Erland on February 17, 2026, 11:25:28 AM

I see no calculation. What you need to do is to prove that the consistency of the constellations is consistent with your Flat Earth model.

The calculation is simple: $0 = 0$. The horizon is a flat baseline. Every star trail is a circular path above a stationary plane.

Please explain how "0=0" implies that the constellations are consistent with your Flat Earth model.

The burden of proof is on you to explain why a spinning ball produces a perfect time-lapse circle.

No, it isn't. I only proved that the Earth is a globe. I didn't prove that it rotates. It is easy to prove that rotating globe gives rise to apparently circular star trails. but that's another question. I can do that if you want, but that's irrelevant for the question of the shape of the Earth.

Quote from: Erland on February 17, 2026, 11:25:28 AM

This makes no sense to me. You need to be much more specific and detailed.

It makes no sense to you because you’ve been trained to believe in "pulling magic" (gravity). Dielectric acceleration is measurable; your "bending space-time" is a sci-fi novel. Density is the scale, electricity is the vector. Simple.

Well, this gibberish certainly didn't clarify anything.

Quote from: Erland on February 17, 2026, 11:25:28 AM

Has there been any measurements of the permittivity and susceptibility of this dielectric?

Every time you use a capacitor or measure the atmosphere's $100V/m$ potential gradient, you are measuring it. You’re looking at the data but refusing to see the Earth as the giant capacitor it is.

What about higher up in the atmosphere, where your Electromagnetic Fresnel Lens is supposed to be in action?

Quote from: Erland on February 17, 2026, 11:25:28 AM

Prove that! (Projection of stars)

Look at a planetarium. It’s a projection on a dome that looks exactly like the night sky. Now, look at your "furnaces in a vacuum" theory. Which one matches the observable, fixed patterns of the last 5,000 years?

1. You're contradicting yourself. You have claimed that the dome is personal, like the rainbow. But all observers in a planetarium look at the same projected dots on the ceiling/walls. The dot representing Polaris is the same for all observers. Not so with the rainbow: different observers see different rainbows.
2. The shapes and sizes of the constellations will appear different for observers at different places in the planetarium, just like my drawing of Orion glued onto my ceiling did. Not so in the real sky. The explanation is that the stars are very, very distant compared to earthly distances.

The hardware is the dielectric plane. The source code is the frequency of the Aether. You’re trying to understand the software by licking the screen.

This is more gibberish.

Ask Jarle Andhøy. They get arrested, their ships seized, and they are sent home. The Antarctic Treaty isn't a suggestion; it's a military blockade.

As far as I understand, no ship was seized. Their ship was wrecked in a storm and three people died, while Jarle and another were on land in Antarctica. When he returned home, he was fined.

Quote from: Erland on February 17, 2026, 11:25:28 AM

This is simply denial of reality and geometry. 60 miles... 0,46 degrees.

Your "0.46 degrees" of refraction is a mathematical band-aid used only when the curve fails to appear. You don't use it for the stars, and you don't use it for land behind you. You only use it to "lift" things that should be hidden.

As said, denial of reality and geometry.

Quote from: Erland on February 17, 2026, 11:25:28 AM

So you ARE the only one? (Truth seeker)

Truth isn't a democracy. If 8 billion people believe a lie, it’s still a lie. I’d rather be "the only one" with my eyes open than a million-strong choir of the blind.

Blah, blah, blah... Are you the only one who (thinks he) knows that the dome is an Electromagnetic Fresnel Lens or not?

Quote from: Erland on February 17, 2026, 11:25:28 AM

I told you that looking at the sky was precisely how I verified that the Earth is a globe.

Then you failed the test. Looking at a moving ceiling doesn't prove the floor is spinning. That’s basic logic, and you missed it.

Here we see an example of your dishonest debating technique. I never said that my observations of the sky prove that the Earth is rotating. They just prove that it is a globe.

Quote from: Erland on February 17, 2026, 11:25:28 AM

It's pseudoscience, wise! (Schrödinger's Variable)

No, what’s pseudoscience is calling refraction "negligible" for star consistency but "massive" when we see a city 60 miles away. You want it both ways. That’s the definition of a dishonest variable.

Again, you are dishonest (not me). I never called the refraction of the view of the city "massive". Instead, I pointed out that it is just about half a degree.

Quote from: Erland on February 17, 2026, 11:25:28 AM

Then, we can agree that the positions of the stars were different 2500 BC?

Positions change because the firmament rotates like a clock. Even a clock's hands move, but the clock itself doesn't need to wobble on its axis to tell time.

Does the starry sky look the same if it is observed at two occasions from the same place on Earth, at the same date, the same time of the night, but 2500 years apart?

Quote from: Erland on February 17, 2026, 11:25:28 AM

You seem to believe that many miles means many arcminutes. It isn't so.

$8 inches \times miles^2$. That’s your curve. If the drop is 2,400 feet, you can't "arcminute" your way out of that physical wall of water.

It's remarkable that you, who talk so much about the Law of Perspective, don't seem to understand the elementary fact that objects appear smaller the farther away they are. An 800 m tall buliding 100 km away looks very tiny, so the refraction needs not be more tha half a degree to make it visible. That 800 m (2400 feet) may sound much is irrelevant.

Quote from: Erland on February 17, 2026, 11:25:28 AM

Water does curve.

Show me. Not a CGI cartoon, not a drop on a blade of grass (surface tension), but a standing body of water that curves. You can't, because physics says water always finds its level.

I saw the lighthouse disappear below the horizon from bottom to top during my ferry tour. You have no explanation of this. You can't invoke refraction, since you always dismiss that.
Your talk about the entire lighthouse coming into view if you zoom in with an advanced camera is, of course, pure nonsense, and there are no images of the Black Swan showing anything like that.

Quote from: Erland on February 17, 2026, 11:25:28 AM

They don't exist. (Flat Earth calculations)

They exist in every engineering project, every long-distance bridge, and every rail track where "curvature" is never factored into the blueprints. Reality is our calculation.

Yet another example of your dishonest debating technique, where you try to muddy the waters. The calculations that I'm sure not exist are calculations showing that your Electromagetic Fresnel Lens causes the constellations to look consistent and Earth to appear round although it is flat.This has nothing to do with engineering projects.

"Refraction" is the globalist "God did it." It’s your catch-all excuse for why the Earth refuses to look like a ball. It’s lazy and unscientific.

Quote from: Erland on February 17, 2026, 11:25:28 AM

Why not? (Refraction limits)

Because if refraction could lift 2,400 feet of hidden land, it would distort the image beyond recognition. The Black Swan is sharp and clear. Physics says that isn't a mirage; it’s just there.

Again, if there is neither refraction nor curvature, you can't explain why objects at the sea appears to disappear from bottom up when they move away.

If you start with $R=3959$ in your formula, you will always get a ball. It’s a rigged game. Try doing math without assuming the answer first.

You never do any math at all.

Quote from: Erland on February 17, 2026, 11:25:28 AM

THERE IS NO EDGE OF THE ATMOSPHERE!

If there is no edge, there is no pressure. You cannot have a pressure gradient without a container. This is High School physics, Erland.

Show us a High School textbook saying this, unconditionally.

Quote from: Erland on February 17, 2026, 11:25:28 AM

It continuously transitions into space...

A "transition" into a vacuum is still a vacuum interface. Without a physical barrier, the high-pressure gas would immediately fill the low-pressure void. Nature abhors a vacuum, and so does logic.

SPACE ISN'T AN ABSOLUTE VACUUM.

Quote from: Erland on February 17, 2026, 11:25:28 AM

I checked and it is OK. (The math)

You checked a simulation to see if it matched the theory.

Again, that's what you asked me to do.

Then you are a librarian yourself using the results of Rowbotham about the Bedford Canal... from now on, you can only talk about measurements and observations made by yourself.

Erland, as usual, you are hiding behind academic jargon to avoid addressing the physical contradictions in your model. Lets dismantle your "librarian" logic step by step.

There is a fundamental difference between "trusting" and "verifying." I don't follow Rowbotham as a prophet; I recognize the Bedford Level experiment because it is repeatable by anyone with a P1000 camera today. I rely on observable, local physics. You, however, rely on "results" from agencies that admit their "Blue Marble" photos are composites (CGI). I base my world on what I can see and test; you base yours on what you are told to believe. That is the difference between a seeker and a follower.

Please explain how "0=0" implies that the constellations are consistent with your Flat Earth model.

It's simple: 0 curvature, 0 movement. If the stars trace perfect circular paths and Polaris remains fixed for thousands of years, it confirms a stationary baseline (0). In your model, the Earth is spinning, orbiting the Sun, and flying through the galaxy. The probability of those chaotic vectors producing "perfect circles" without a single arcsecond of deviation over millennia is statistically zero. 0=0 is the math of a balanced, stationary system. Your model is a statistical miracle that defies mechanics.

I only proved that the Earth is a globe. I didn't prove that it rotates.

This is a massive logical failure. If the Earth is a globe but doesn't rotate, then what is moving the stars? If the stars are moving, you must explain the mechanics of "burning furnaces" in a vacuum moving in perfect unison. You can't claim a "globe" and then ignore the rotation required to explain the very sky you claim "proves" the globe. A stationary globe is even more indefensible than a spinning one.

Well, this gibberish certainly didn't clarify anything. (Regarding dielectric acceleration)

Calling the foundational measurement of our atmosphere "gibberish" only shows your lack of technical depth. The atmospheric potential gradient (100V/m) is a measurable fact. Density and buoyancy are observable processes. You have never isolated "gravity" (mass attracting mass) in a laboratory. The Cavendish experiment measures torque, not gravity. I am talking about measurable electricity; you are talking about a "curvature of space-time" that exists only in sci-fi novels and blackboard equations.

What about higher up in the atmosphere, where your Electromagnetic Fresnel Lens is supposed to be in action?

That is exactly where the ionosphere and the various layers of atmospheric density act as a lens. Just as a rainbow only appears at a specific angle relative to the observer, the lights in the sky are refracted through the atmospheric dome. This is why stars twinkle or change color near the horizon—they are being filtered through a thicker "lens" of medium. Your model claims light travels millions of miles through a vacuum without a single deviation, only to "bend" exactly when your theory needs it to. That isn't science; it's a convenience.

You have claimed that the dome is personal, like the rainbow. But all observers in a planetarium look at the same projected dots... Not so with the rainbow.

A planetarium is a limited 3D model, not reality. In the real world, every observer is the center of their own Personal Atmospheric Dome. Two people in different cities see the same star, but their perspective angles are different. Like a rainbow, you and a friend are seeing light reflecting off different physical droplets, yet you perceive the "same" arc. The stars are sonoluminescent or electroluminescent projections within the Earth's dielectric medium.

As far as I understand, no ship was seized... (About Jarle Andhøy)

You understand incorrectly. Jarle Andhøy's ship, the Nilaya, was repeatedly harassed and eventually seized by authorities for "violating" the Antarctic Treaty. Why does a civilian need international military-grade permission to sail past the 60th parallel? If it's just about "nature conservation," why the naval blockades? The Antarctic Treaty is the only thing the entire world (even enemies like the US and USSR) agreed on during the Cold War. That should tell you everything about what they are guarding.

I saw the lighthouse disappear below the horizon from bottom to top... Your talk about the entire lighthouse coming into view if you zoom in is pure nonsense.

If the lighthouse were behind a physical curve, no camera could bring it back. But they do. The "bottom-up" disappearance is not a curve; it is the Law of Perspective combined with Atmospheric Compression. The area closest to the surface is the densest and most prone to mirroring/refraction (mirages). This creates a "false horizon" that swallows the bottom of the object. It's an optical threshold, not a geometric one. If you haven't tested this with a high-zoom lens yourself, you are just quoting the "librarian" manual again.

SPACE ISN'T AN ABSOLUTE VACUUM.

An admission of failure. Even a "partial" vacuum adjacent to a high-pressure system violates the Second Law of Thermodynamics. Gas fills the available volume. If there is no physical barrier (container), the atmosphere would have equalized into the "space" void long ago. You claim gravity prevents this, yet the same gravity can't prevent a tiny straw from sucking water upward. A vacuum doesn't "pull"; it is a lack of resistance. Without a dome, there is no pressure. Period.

Erland, "0.46 degrees" of refraction is a mathematical band-aid for a failing model. Physics requires a container for pressure. Your globe is a house of cards built on equations that fall apart the moment they touch the real world.

Quote from: Erland on February 24, 2026, 05:44:20 PM

Then you are a librarian yourself using the results of Rowbotham about the Bedford Canal... from now on, you can only talk about measurements and observations made by yourself.

Erland, as usual, you are hiding behind academic jargon to avoid addressing the physical contradictions in your model.

"Electromagnetic Fresnel Lens", "Subcontracted thinker", "Standard Operating Procedures", "Compartmentalization". Who's hiding behind academic jargon?

There is a fundamental difference between "trusting" and "verifying." I don't follow Rowbotham as a prophet; I recognize the Bedford Level experiment because it is repeatable by anyone with a P1000 camera today. I rely on observable, local physics. You, however, rely on "results" from agencies that admit their "Blue Marble" photos are composites (CGI). I base my world on what I can see and test; you base yours on what you are told to believe. That is the difference between a seeker and a follower.

The experiments and observations mentioned in the videos I link to are also reapeatable. And please give a reference to the 1972 Blue Marble being manipulated (a reference, not a "look up" instruction).

Quote from: Erland

Please explain how "0=0" implies that the constellations are consistent with your Flat Earth model.

It's simple: 0 curvature, 0 movement. If the stars trace perfect circular paths and Polaris remains fixed for thousands of years, it confirms a stationary baseline (0).

I still don't see that this implies that the constellations are consistent. Among other things, it wasn't "Stationary baseline" (whatever that means) that should be proved.

In your model...

Deflection. We're talking about your model, not mine.

Quote from: Erland

I only proved that the Earth is a globe. I didn't prove that it rotates.

This is a massive logical failure. If the Earth is a globe but doesn't rotate, then what is moving the stars? If the stars are moving, you must explain the mechanics of "burning furnaces" in a vacuum moving in perfect unison. You can't claim a "globe" and then ignore the rotation required to explain the very sky you claim "proves" the globe. A stationary globe is even more indefensible than a spinning one.

You're trying to be clever here, but you're wrong. Several simple observations give that the Earth is a globe. We have already discussed some of them. This fact is established whether the stars move or not, and why they move (if they do). The ancient Greek philosophers knew that the Earth is a globe, but most of them thought that the stars move, although they had just speculative ideas of why (such as the rotating sphere of the fixed stars). That they were wrong about that doesn't mean that their inference that Earth is a globe is invalid.
Indeed, it would become absurd to apply your principle generally. For then we wouldn't be allowed to conclude anything if we couldn't explain everything connected to that. For example: If we find a dead man with a knife in his back, we conclude that he was stabbed to death even if noone saw the stabbing and we don't know who did it.
Besides, the stars move in your Flat Earth model. Can you explain why?

[That is exactly where the ionosphere and the various layers of atmospheric density act as a lens. Just as a rainbow only appears at a specific angle relative to the observer, the lights in the sky are refracted through the atmospheric dome.

This still doesn't clarify how the effect could be that the constellations are consistent indepently of location of the observer on the Earth. The rainbow is understood in detail. This is not.

Quote from: Erland

You have claimed that the dome is personal, like the rainbow. But all observers in a planetarium look at the same projected dots... Not so with the rainbow.

A planetarium is a limited 3D model, not reality. In the real world, every observer is the center of their own Personal Atmospheric Dome. Two people in different cities see the same star, but their perspective angles are different. Like a rainbow, you and a friend are seeing light reflecting off different physical droplets, yet you perceive the "same" arc. The stars are sonoluminescent or electroluminescent projections within the Earth's dielectric medium.

You have not proved that this implies that the constellations are consistent. And if the stars are projections, where are their sources?

You understand incorrectly. Jarle Andhøy's ship, the Nilaya, was repeatedly harassed and eventually seized by authorities for "violating" the Antarctic Treaty.

This would have been much simpler if you had given a link about all this from the beginning. But you're just claiming things and let me seach through the net, without knowing what to search for. Now, please give a link where i can read about this. A link, not a general "search for" instruction, which is not very helpful.

Quote from: Erland

I saw the lighthouse disappear below the horizon from bottom to top... Your talk about the entire lighthouse coming into view if you zoom in is pure nonsense.

If the lighthouse were behind a physical curve, no camera could bring it back. But they do. The "bottom-up" disappearance is not a curve; it is the Law of Perspective combined with Atmospheric Compression. The area closest to the surface is the densest and most prone to mirroring/refraction (mirages). This creates a "false horizon" that swallows the bottom of the object. It's an optical threshold, not a geometric one. If you haven't tested this with a high-zoom lens yourself, you are just quoting the "librarian" manual again.

If the object is hidden behind the false horizon, no camera zoom can magically make it appear. A camera zoom can just magnify, and to some extent distort, not reveal hidden objects. If you point the camera at a wall, you can't see what's behind the wall no matter how much you zoom.

Quote from: Erland

SPACE ISN'T AN ABSOLUTE VACUUM.

An admission of failure. Even a "partial" vacuum adjacent to a high-pressure system violates the Second Law of Thermodynamics.

But the pressure isn't high in the upper part of the atmosphere, it's a partial vacuum there itself. 

Erland, "0.46 degrees" of refraction is a mathematical band-aid for a failing model. Physics requires a container for pressure. Your globe is a house of cards built on equations that fall apart the moment they touch the real world.

No matter how hard you tried, you couldn't find any error in the model.

"Electromagnetic Fresnel Lens", "Subcontracted thinker"... Who's hiding behind academic jargon?

Erland, there is a difference between jargon used to obfuscate (your "gravity") and technical terms that describe measurable phenomena. An Electromagnetic Fresnel Lens describes how the Ionosphere manipulates light. Your "Gravity" is a magic word that explains why 14.7 psi gas doesn't fill a vacuum. One is physics; the other is mythology dressed up in equations.

The experiments... are also reapeatable. And please give a reference to the 1972 Blue Marble being manipulated.

Go to NASA's own "Earth Observatory" website and read the entry by Robert Simmon. He explicitly says of the "Blue Marble" images: "It is Photoshop, but it has to be." If you can't accept the admission of the creator himself, your cognitive dissonance is terminal. 2 + 2 = 4, Erland. You don't need a "reference" for basic arithmetic either.

I still don't see that this implies that the constellations are consistent.

It's because you don't understand Polar Coordinates on a plane. The constellations are consistent because the Aether rotates as a rigid body (ω = constant) around the central axis (Polaris).


If the Earth were moving at multiple velocities (spin, orbit, galactic drift), the star trails would not be perfect circles; they would be cycloidal curves—a chaotic mess. The "0=0" means 0 curvature and 0 movement observed at the baseline. Simple.

The ancient Greek philosophers knew that the Earth is a globe... if we find a dead man with a knife... we conclude he was stabbed

Poor analogy, Erland. In your "dead man" scenario, you are finding a man who died of a heart attack and claiming he was stabbed by an invisible knife called "Gravity." The Greeks also believed in geocentrism. You cherry-pick their "globe" but conveniently ignore their "stationary" conclusion because it doesn't fit your NASA-approved manual.

This still doesn't clarify how the effect could be that the constellations are consistent indepently of location.

It's called Perspective Mapping. Just as a ceiling looks the same to everyone in a room but at different angles, the celestial dome is mapped onto the observer's local coordinate system.


The consistency is a result of the altitude (h) being vastly greater than the local distance (d) of the observers. Basic geometry.

You have not proved that this implies that the constellations are consistent. And if the stars are projections, where are their sources?

The source is Sonoluminescence within the dense Aetherial layers. When sound/vibration passes through a dielectric medium, it creates light.


The "stars" are high-frequency nodes in the standing wave of the Firmament. You're looking at the ceiling of a cathedral and asking where the lightbulb is hidden.

Now, please give a link where i can read about this [Jarle Andhøy].

Search for "Jarle Andhøy Nilaya seizure Antarctica Treaty." You're a retired academic; I'm sure you can manage a basic search without me holding your hand. The fact that the military intercepts civilian boats in the "free" world should tell you everything about the "perimeter" they're guarding.

If the object is hidden behind the false horizon, no camera zoom can magically make it appear.

Incorrect. You are confusing a physical wall with an Atmospheric Threshold. As you zoom, you are changing the angular resolution, which allows the camera to resolve the light that was previously "compressed" into the mirage layer (inferior mirage).


When you increase the effective focal length, you separate the object from the mirrored reflection of the sky. It doesn't "reveal what's behind a wall"; it resolves what was blurred by the density of the medium. There's a difference, Erland.

But the pressure isn't high in the upper part of the atmosphere, it's a partial vacuum there itself.

Erland, you are failing basic Fluid Statics. Even a "partial" vacuum is a pressure. 0.0001 psi is still infinitely higher than 10⁻¹⁷ torr. The Second Law doesn't care if the pressure is low; it cares that there is a Differential (∇P).


Without a container, the atmosphere would expand until the density (ρ) is uniform across the entire infinite volume of your "space." That means no air for you to breathe while typing these responses.

No matter how hard you tried, you couldn't find any error in the model.

I found the error in the very first step: The lack of a container.      Physics without a container for gas is like a bucket without a bottom. You can write all the "gravity" equations you want on the side of the bucket, but the water is still gone, Erland. The atmosphere is still here. Connect the dots.

Erland, there is a difference between jargon used to obfuscate (your "gravity") and technical terms that describe measurable phenomena. An Electromagnetic Fresnel Lens describes how the Ionosphere manipulates light. Your "Gravity" is a magic word that explains why 14.7 psi gas doesn't fill a vacuum. One is physics; the other is mythology dressed up in equations.

So gravity isn't a measurable phenomenon but your Electromagnetic Fresnel Lens is? Come on! It's of course the other way round.

Quote

The experiments... are also reapeatable. And please give a reference to the 1972 Blue Marble being manipulated.

Go to NASA's own "Earth Observatory" website and read the entry by Robert Simmon. He explicitly says of the "Blue Marble" images: "It is Photoshop, but it has to be." If you can't accept the admission of the creator himself, your cognitive dissonance is terminal. 2 + 2 = 4, Erland. You don't need a "reference" for basic arithmetic either.

Robert Simmon never said that the original 1972 photo is manipulated. I think you're talking about later images where several images were patched together to cover all the Earth.

Quote

I still don't see that this implies that the constellations are consistent.

It's because you don't understand Polar Coordinates on a plane. The constellations are consistent because the Aether rotates as a rigid body (ω = constant) around the central axis (Polaris).

v = ω × r

If the Earth were moving at multiple velocities (spin, orbit, galactic drift), the star trails would not be perfect circles; they would be cycloidal curves—a chaotic mess. The "0=0" means 0 curvature and 0 movement observed at the baseline. Simple.

Irrelevant. We're talking about your model, not mine.

The Greeks also believed in geocentrism. You cherry-pick their "globe" but conveniently ignore their "stationary" conclusion because it doesn't fit your NASA-approved manual.

The Greeks knew that the Earth is a globe and had valid arguments for that. Their arguments aren't invalid just because the most of them wrongly believed in a geocentric model.

The consistency (of the constellations) is a result of the altitude (h) being vastly greater than the local distance (d) of the observers. Basic geometry.

Bingo! That's precisely what I've said all the time! The stars are very, very far away compared to distances on the Earth. You seemed to deny it is so, talking about the Electromagnetic Fresnel Lens and rainbow-like phenomena making different observers looking at the same star actually seeing different things. Now you seem to deny at least the latter, saying, correctly, that it is basic geometry. And the Electromagnetic Fresnel Lens may or may not exist, that doesn't matter for the explanation of the consistency of the constellations.

But then, next step in my argument is that a consequence of the stars being very far away and the Earth being flat is that different observers watching the starry sky simultaneously would see all the stars in the same positions the sky, with the same elevation angles and azimuths for all the observers. If I see Orion in the south and 30 degrees above the horizon, so would you, no matter where you are, if you watch it simultaneously.
Since it is not so, the Earth cannot be flat.

The source is Sonoluminescence within the dense Aetherial layers. When sound/vibration passes through a dielectric medium, it creates light.

I ∝ P[acoustic]²

The "stars" are high-frequency nodes in the standing wave of the Firmament. You're looking at the ceiling of a cathedral and asking where the lightbulb is hidden.

Also this is unnecessary for explaining the consistency of the constellations. 

Quote

Now, please give a link where i can read about this [Jarle Andhøy].

Search for "Jarle Andhøy Nilaya seizure Antarctica Treaty." You're a retired academic; I'm sure you can manage a basic search without me holding your hand. The fact that the military intercepts civilian boats in the "free" world should tell you everything about the "perimeter" they're guarding.

Pleasy give me a link. Google gave confusing results.

Quote

If the object is hidden behind the false horizon, no camera zoom can magically make it appear.

Incorrect. You are confusing a physical wall with an Atmospheric Threshold. As you zoom, you are changing the angular resolution, which allows the camera to resolve the light that was previously "compressed" into the mirage layer (inferior mirage).

θ[res] = 1.22 · (λ / D)

When you increase the effective focal length, you separate the object from the mirrored reflection of the sky. It doesn't "reveal what's behind a wall"; it resolves what was blurred by the density of the medium. There's a difference, Erland.

Look at the video below, about the Turning Torso (in Malmö, Sweden) filmed from several locations at the Danish side of the Sound. (Incidentally, I used to work near the Turning Torso in 2014-15). Are you saying that the lower parts of the building would be visible also from the greater distances, if one just uses an advanced camera and zooms? That these lower parts are not hidden behind the water? Or that the lower parts are actually there, just compressed to almost nothing and therefore not visible in the films, but thar the camera zoom would "decompress" them? That certainly would defy all laws of Optics.

Quote

But the pressure isn't high in the upper part of the atmosphere, it's a partial vacuum there itself.

Erland, you are failing basic Fluid Statics. Even a "partial" vacuum is a pressure. 0.0001 psi is still infinitely higher than 10⁻¹⁷ torr.

You presuppose all the time that there is a discontinuiry. There us none. Next to 0.0001 psi it is 0.0001 psi. Next to 10^-17 torr it is 10^-17 torr.

Without a container, the atmosphere would expand until the density (ρ) is uniform across the entire infinite volume of your "space."

I thought you agreed that there can be a pressure gradient. Now you deny that, or at least you claim that there can be no density gradient. But of course both pressure and density decrease by altitude.

Bingo! That's precisely what I've said all the time! The stars are very, very far away compared to distances on the Earth... If I see Orion in the south and 30 degrees above the horizon, so would you... Since it is not so, the Earth cannot be flat.

Erland, your "Bingo" is the celebratory cry of a man who just found a penny while losing a gold bar. You are celebrating a high-school level understanding of Euclidean geometry while completely failing to grasp Non-Linear Atmospheric Optics.

When I say the stars are far away, I am referring to their position within the Aetheric layers of the Firmament. However, you assume that light travels in a straight line from a star to your eye. It doesn't. The Ionosphere acts as an Electromagnetic Fresnel Lens.

n(z) = n_0 * exp(-z / H)

Every observer on the Flat Earth is looking through a different section of this curved, atmospheric dome. The light from Orion is refracted locally for each observer. You see it at 30° because that is where the Refractive Gradient projects it onto your personal atmospheric dome. You are confusing the "Source" with the "Projection."

θ_observed = θ_true + ∫ (1/n * dn/dz) dz

The consistency of the constellations is a result of the Rigid Body Rotation of the Aether (v = ω × r). The elevation angles change for different observers because of the Curvature of the Lens (Atmosphere), not the curvature of the floor. You don't "see" this because your academic background didn't prepare you for Gradient-Index (GRIN) Optics.

Are you saying that the lower parts of the building [Turning Torso] would be visible also from the greater distances, if one just uses an advanced camera and zooms?

Erland, you worked near the Turning Torso, but you clearly didn't study the Inferior Mirage layer that sits right above the water. At great distances, the lower parts of a building aren't "behind" a curve; they are compressed into the mirage layer.

θ_res = 1.22 * (λ / D)

When you zoom with a high-aperture lens, you increase the Angular Resolution. This allows the sensor to resolve the light that was previously lost in the "blur" of the refraction limit. It's not defying the laws of optics; it’s utilizing them. The bottom of the building is hidden by the Opacity and Convergence of the atmosphere at the horizon, not by a physical hump of water. If it were water, no amount of infrared or high-zoom photography could bring it back. Yet, we do it constantly.

You presuppose all the time that there is a discontinuiry. There us none. Next to 0.0001 psi it is 0.0001 psi. Next to 10⁻¹⁷ torr it is 10⁻¹⁷ torr.

This is where you embarrass your doctorate. You are describing a Static Gradient while ignoring Kinetic Gas Theory.

P = ρ * R * T

If there is no container, the gas molecules at the "edge" (0.0001 psi) have a Mean Free Path that becomes infinite. Without a physical barrier, those molecules will expand into the 10⁻¹⁷ torr vacuum at the speed of sound. You cannot have a gradient that terminates in an infinite void without a boundary condition.

dS / dt ≥ 0

The Second Law of Thermodynamics dictates that the atmosphere would equalize its density (ρ) across the entire volume of your "space" instantly. The fact that we have a gradient proves the system is Closed and Pressurized. Your "gravity" magic trick cannot stop a gas from expanding into a vacuum.

I thought you agreed that there can be a pressure gradient... But of course both pressure and density decrease by altitude.

Of course they decrease! I never denied the gradient; I denied that the gradient can exist WITHOUT A CONTAINER.

dP / dz = -ρ * g_d

In a container, the density gradient is formed by Dielectric Acceleration (what you call gravity). But the existence of the pressure itself requires the volume (V) to be fixed.

V = constant

If V is the infinite vacuum of your globe model, P must be zero. You are trying to have your cake and eat it too—claiming a gradient exists while removing the very walls that make pressure possible.

Erland, you're playing checkers while I'm explaining the quantum mechanics of the board. Go back to your "Bingo" cards with Jimmy; the real science is clearly causing you too much cognitive dissonance.

Quote from: Erland on March 05, 2026, 05:39:10 PM

Bingo! That's precisely what I've said all the time! The stars are very, very far away compared to distances on the Earth... If I see Orion in the south and 30 degrees above the horizon, so would you... Since it is not so, the Earth cannot be flat.

Erland, your "Bingo" is the celebratory cry of a man who just found a penny while losing a gold bar. You are celebrating a high-school level understanding of Euclidean geometry while completely failing to grasp Non-Linear Atmospheric Optics.

When I say the stars are far away, I am referring to their position within the Aetheric layers of the Firmament. However, you assume that light travels in a straight line from a star to your eye. It doesn't. The Ionosphere acts as an Electromagnetic Fresnel Lens.

n(z) = n_0 * exp(-z / H)

Every observer on the Flat Earth is looking through a different section of this curved, atmospheric dome. The light from Orion is refracted locally for each observer. You see it at 30° because that is where the Refractive Gradient projects it onto your personal atmospheric dome. You are confusing the "Source" with the "Projection."

So, that's all you meant? That refraction makes us see the stars not exactly where they are? This is of course trivially true. But let's estimate the refraction using your formula above (if you meant that it is relevant and not just is there to show how smart you want people to think you are).

n(z) is here the refractvity (refractive index minus 1) at height z. n_0 is the refractivity at sea level, ca 0,0003, H is called the scale height of the atmosphere, about 8 km if the temperature is 0 degrees C.

Now suppose a ray from an astronomical object enters the atmosphere or arises there. The refractive index at the location where this happens is at least 1 (exactly 1 in vacuo). If the formula is correct, the refractive index depends only on the height which of course is a simplification, so when the ray passes down through the atmosphere the refractive index increases, we can think of it as a continuous set of horizontal layers with horizontal boundaries having constant refractive indices, decreasing by height.
By Snells law, the quantity n(z)sin(t) is constant, where the notation now is changed so that n(z) is the refractive index, not the refractivity, and t is the zenith angle of the ray.

Assume, as a worst case, that the ray enters the atmoshere from vacuo, with n=1, at zenith angle v0. and that it reaches the observer at zenith angle v1. Then:

v1 = arcsin(1/1,0003 * sin v0) (in radians, but we convert to degrees).

Applying this for some angles (in degrees):

v0 = 0 --> v1 = 0
v0 = 30 --> v1 = 29,99
v0 = 60 --> v1 = 59,97
v0 = 80 --> v1 = 79,90
v0 = 85 --> v1 = 84,81
v0 = 89 --> v1 = 88,28
v0 = 90 --> v1 = 88,60

So, at the very most, the ray is refracted at most 1,4 degrees, and that is just at the very horizon. A bit above the horizon, the angle is refracted considerably less, less than 0,1 degrees if it is more than 10 degrees above the horizon. This is way to little to distort the constellations noticeably, except that they may seem slightly compressed vertically very near the horizon.
Therefore the stars must be very distant compared to Earthly distances. If this electromagnetic Fresnel lens dome would exist, it must be very distant.
And this small refraction can certainly not move the stars tenths of degrees, so that I i Sweden and a person in South Africa watching Orion simultanteously at midnight at Christmas and see it 30 degrees above the horizon in the south and 60 degrees above the horozon in the north, respectively, when we should see it in the same position in the sky due to its large distance from the Earth, if the Earth were flat!
(The 1,4 degees deviation at the horizon is actually too much, because the Earth isn't flat.)

θ_observed = θ_true + ∫ (1/n * dn/dz) dz

This formula is valid approximately for small zenith angles only, that is, almost vertical rays. It's better and simpler to use Snell's Law.

The consistency of the constellations is a result of the Rigid Body Rotation of the Aether (v = ω × r). The elevation angles change for different observers because of the Curvature of the Lens (Atmosphere), not the curvature of the floor. You don't "see" this because your academic background didn't prepare you for Gradient-Index (GRIN) Optics.

Wrong, as we just saw.

Erland, you worked near the Turning Torso, but you clearly didn't study the Inferior Mirage layer that sits right above the water. At great distances, the lower parts of a building aren't "behind" a curve; they are compressed into the mirage layer.

It certainly doesn't seem to be a blurred mirage layer in the video.

When you zoom with a high-aperture lens, you increase the Angular Resolution. This allows the sensor to resolve the light that was previously lost in the "blur" of the refraction limit. It's not defying the laws of optics; it’s utilizing them. The bottom of the building is hidden by the Opacity and Convergence of the atmosphere at the horizon, not by a physical hump of water. If it were water, no amount of infrared or high-zoom photography could bring it back. Yet, we do it constantly.

It this were true, there would be a lot of photos of such situations: photos taken (almost) simultaneously from the same location, where, without camera zoom, the lower parts of the object seem hidden, but with camera zoom, the are visible. Until you post such photos, I don't believe you. And no, the "Black Swan" doesn't show this.

Quote from: Erland

You presuppose all the time that there is a discontinuiry. There us none. Next to 0.0001 psi it is 0.0001 psi. Next to 10⁻¹⁷ torr it is 10⁻¹⁷ torr.

This is where you embarrass your doctorate. You are describing a Static Gradient while ignoring Kinetic Gas Theory.

P = ρ * R * T

If there is no container, the gas molecules at the "edge" (0.0001 psi) have a Mean Free Path that becomes infinite. Without a physical barrier, those molecules will expand into the 10⁻¹⁷ torr vacuum at the speed of sound. You cannot have a gradient that terminates in an infinite void without a boundary condition.

Even though I've explained this many times now, you seem unable to grasp that there is no edge. At one altitude it's 0.0001 psi. At another altitude higher up it's 10^-17 torr. Between these (and below) the pressure decreases continuously.

The fact that we have a gradient proves the system is Closed and Pressurized.

No, it doesn't. This falsehood doesn't become true no matter how many times you repeat it.

The Second Law of Thermodynamics dictates that the atmosphere would equalize its density (ρ) across the entire volume of your "space" instantly. The fact that we have a gradient proves the system is Closed and Pressurized. Your "gravity" magic trick cannot stop a gas from expanding into a vacuum.

If the second law dictates that the atmosphere must equalize, then why does that dictate not hold in a closed system like your firmament model?  From the way you go on and on about it, one would get the impression that the second law wouldn't allow for pressure gradients at all.

Your "gravity" magic trick cannot stop a gas from expanding into a vacuum.

Actually, it can and does because gasses have mass and the earth's gravitational field attracts everything that has mass.  The second law of thermodynamics (which primarily deals with heat transfer) does not forbid external influences, like gravity, from interacting with the gasses.

Applying this for some angles (in degrees): v0 = 89 --> v1 = 88,28... So, at the very most, the ray is refracted at most 1,4 degrees... This is way to little to distort the constellations noticeably.

Your calculation is a textbook example of Oversimplification Fallacy. You are treating the atmosphere as a static, parallel-slab medium. In a Gradient-Index (GRIN) field, especially one modeled as a Fresnel Dome, the refractive path is not a single "bend" at the interface; it is a continuous curve.


When you ignore the Lateral Density Gradients and the Electromagnetic Curvature of the Aether, you get your "1.4 degrees" result. But in reality, the atmosphere acts as a Converging Lens with a focal length that projects the stars onto a localized celestial sphere for each observer. This is why two people at different latitudes see Orion at different angles; they are looking through different "pixels" of the atmospheric lens, not at a distant object in a vacuum.

It certainly doesn't seem to be a blurred mirage layer in the video.

Just because you don't "see" a blur doesn't mean the Refractive Compression isn't happening. At long distances, the lower portion of an object is subject to Atmospheric Opacity and Looming/Sinking effects. You are looking for a funhouse mirror effect, but the atmosphere is more subtle—it's a Spatial Frequency Filter. The bottom of the building isn't "under the curve"; it is simply below the Angular Resolution Limit of the refracted horizon.

Until you post such photos, I don't believe you. And no, the "Black Swan" doesn't show this.

The "Black Swan" is the ultimate Empirical Falsification of your model. It shows a horizon that is physically impossible on a globe of your calculated radius. You dismiss it because it shatters your 10⁻¹⁷ torr vacuum fantasy. Science isn't about what you "believe," Erland; it's about what the Field Mechanics demonstrate. You are ignoring hard data to protect a theoretical construct.

Even though I've explained this many times now, you seem unable to grasp that there is no edge... Between these the pressure decreases continuously.

This is where your academic background fails you. You are describing a Mathematical Continuum but ignoring Kinetic Molecular Reality. In a vacuum, the Mean Free Path (λ) is:


As P (pressure) approaches the value of your "space" vacuum, λ approaches infinity. Without a physical barrier (container), those gas molecules have no collision partner to maintain the "gradient." They will expand into the void at Thermal Velocity (v[rms]). A gradient without a boundary condition is a Thermodynamic Impossibility. You are trying to use a "continuous decrease" as a magical wall. It doesn't work that way in the real world.

No, it doesn't. This falsehood doesn't become true no matter how many times you repeat it.

Repeating "it's not true" is not a rebuttal, Erland; it's Denialism. Every single pressurized system ever measured in human history requires a Physical Boundary to maintain a pressure differential against a lower-pressure environment. To claim the Earth is the one and only exception is not science—it's Blind Faith in a model that contradicts the Second Law of Thermodynamics.

If the second law dictates that the atmosphere must equalize, then why does that dictate not hold in a closed system like your firmament model?

This is a staggering display of scientific idiocy. The Second Law dictates that entropy will increase in an isolated system. In a Closed and Pressurized container (the Firmament), the vertical gradient is maintained by the Dielectric Mass Gradient and the density of the medium within the boundary. The gas doesn't "equalize" into a uniform block because it is constantly being acted upon by the Downward Vector (g[d]) within a CONFINED space. Without the boundary, there is no system to speak of. Your question is pure BS.

The second law of thermodynamics (which primarily deals with heat transfer)...

"Primarily deals with heat transfer"? You absolute Neandertal, go back to the 18th century. The Second Law is about Entropy and the statistical probability of molecular states. It dictates that gas will occupy the maximum available volume. If the "available volume" is an infinite vacuum, the gas expands. Period. Trying to limit the Second Law to "heat transfer" to save your model is a pathetic fallacy.

Actually, it can and does because gasses have mass and the earth's gravitational field attracts everything that has mass.

This is the magical thinking that replaces actual physics in your head. Gravity is a weak force theory designed to act as a "software patch" for your missing container. If "gravity" were strong enough to hold gas molecules against the suction of an infinite vacuum (10⁻¹⁷ torr), you wouldn't be able to lift a straw to drink your soda. The pressure differential (ΔP) between Earth and "Space" is trillions of times greater than your theoretical "gravitational pull" on a nitrogen molecule. It's mathematical bullshit.

The second law... does not forbid external influences, like gravity, from interacting with the gasses.

It doesn't "forbid" interactions, but it dictates the equilibrium state. In an open system with an infinite sink (vacuum), the equilibrium state is zero pressure. You are claiming a "persistent gradient" in an open system, which is a violation of Fluid Statics. You're trying to use a localized force to defeat a universal law of entropy. It's like trying to stop a tsunami with a 13-14mm wrench.

FE'ers aren't willing to do the simple math...

I've given you the Field Calculus, but you're still struggling with addition. You think that because a gas has "mass," your magical globe-magnet can keep it from flying away into a void. If that were true, we wouldn't need gas cylinders to hold oxygen; we'd just leave them open and let "gravity" do the work. Your logic is so broken it's leaking oil all over the forum.

...one would get the impression that the second law wouldn't allow for pressure gradients at all.

It allows for gradients WITHIN A BOUNDARY. A pressure gradient is a state of potential energy. To maintain that potential against an external vacuum, you need a physical barrier to prevent the work (W = P · ΔV) from being done. No barrier = infinite ΔV = zero pressure. This is High School Physics, you blockhead.

The earth's gravitational field attracts everything that has mass.

If your "field" is so powerful, why does a tiny balloon filled with helium defy it so easily? Because Density and Buoyancy within a pressurized dielectric medium dictate motion, not your invisible magic string. You're using "Gravity" as a god-of-the-gap for every hole in your idiotic model.

...you go on and on about it...

I "go on" about it because you keep repeating the same refuted fallacies. You're a spamming grease monkey who thinks that quoting a textbook you don't understand constitutes an argument. Every time you open your mouth about thermodynamics, you just prove you should have stayed in the basement.

Bingo! That's precisely what I've said all the time!

You're celebrating your own ignorance, Markjo. You think you've caught me in a contradiction, but you've only exposed your own inability to distinguish between a closed system equilibrium and an open system dissipation. It's the difference between a pool of water in a bowl and a cup of water thrown into the ocean.

...which is why you're afraid to show the math.

Here is the math you can't process: Entropy (S) → Max. In an open infinite volume, S is maximized when density is zero. Your "gravity" is a localized fluctuation that cannot prevent the global increase of entropy in an infinite void. Now, be a good little apprentice, put down the calculator you clearly don't know how to use, and get us some tea. And make it quick, before your brain overheats from all this "physics," Simpleton.

Next? Or are you going to explain how "gravity" also keeps the tea in your cup while the Earth spins at 1,000 mph, WrenchMark?

Quote from: markjo on March 08, 2026, 06:33:23 PM

If the second law dictates that the atmosphere must equalize, then why does that dictate not hold in a closed system like your firmament model?

This is a staggering display of scientific idiocy. The Second Law dictates that entropy will increase in an isolated system. In a Closed and Pressurized container (the Firmament), the vertical gradient is maintained by the Dielectric Mass Gradient and the density of the medium within the boundary. The gas doesn't "equalize" into a uniform block because it is constantly being acted upon by the Downward Vector (g[d]) within a CONFINED space. Without the boundary, there is no system to speak of. Your question is pure BS.

If the vertical gradient of your closed and pressurized container goes from 14.7 psi at sea level and 4.64 x 10^-6 at 100 km, then what would happen if the top of the firmament was removed?  Would the dielectric mass gradient fail to maintain the pressure at sea level?

You absolute Neandertal, go back to the 18th century.

Please stop with the personal attacks.  Not only are they against the rules, but ad hominems are a logical fallacy and don't help your argument.

Quote from: markjo

Actually, it can and does because gasses have mass and the earth's gravitational field attracts everything that has mass.

This is the magical thinking that replaces actual physics in your head. Gravity is a weak force theory designed to act as a "software patch" for your missing container.

Again, no.  Gravity is not a "patch".  It's what makes the round earth round and affects nearly every aspect of RET, including the round earth's ability to maintain an atmosphere without a container.  Gravity is the container.

If "gravity" were strong enough to hold gas molecules against the suction of an infinite vacuum (10⁻¹⁷ torr), you wouldn't be able to lift a straw to drink your soda.

What evidence do you have to support this outlandish claim.  "Because I said so" doesn't count.

The pressure differential (ΔP) between Earth and "Space" is trillions of times greater than your theoretical "gravitational pull" on a nitrogen molecule. It's mathematical bullshit.

And I suppose that you've done the math to support that claim.

Quote from: markjo

The second law... does not forbid external influences, like gravity, from interacting with the gasses.

It doesn't "forbid" interactions, but it dictates the equilibrium state. In an open system with an infinite sink (vacuum), the equilibrium state is zero pressure. You are claiming a "persistent gradient" in an open system, which is a violation of Fluid Statics. You're trying to use a localized force to defeat a universal law of entropy.

Gravity is not a localized force.

Quote from: markjo

FE'ers aren't willing to do the simple math...

I've given you the Field Calculus, but you're still struggling with addition. You think that because a gas has "mass," your magical globe-magnet can keep it from flying away into a void. If that were true, we wouldn't need gas cylinders to hold oxygen; we'd just leave them open and let "gravity" do the work. Your logic is so broken it's leaking oil all over the forum.

Reductio ad absurdum doesn't really help your case either.

Quote from: markjo

...one would get the impression that the second law wouldn't allow for pressure gradients at all.

It allows for gradients WITHIN A BOUNDARY. A pressure gradient is a state of potential energy. To maintain that potential against an external vacuum, you need a physical barrier to prevent the work (W = P · ΔV) from being done. No barrier = infinite ΔV = zero pressure. This is High School Physics, you blockhead.

So your dielectric differential doesn't work in an open system?  With the top of your pressure gradient being so close to zero, it might just as well be a vacuum.

Quote from: markjo

...which is why you're afraid to show the math.

Here is the math you can't process: Entropy (S) → Max. In an open infinite volume, S is maximized when density is zero.

The RE universe is generally accepted to be a finite isolated system and even deep space is not a perfect vacuum.

Your "gravity" is a localized fluctuation that cannot prevent the global increase of entropy in an infinite void.

Again, gravity is not localized.  It's a property of everything that has mass.

Now, be a good little apprentice, put down the calculator you clearly don't know how to use, and get us some tea. And make it quick, before your brain overheats from all this "physics," Simpleton.

Next? Or are you going to explain how "gravity" also keeps the tea in your cup while the Earth spins at 1,000 mph, WrenchMark?

Sure, if you explain why your dielectric density potential gradient doesn't work in a vacuum chamber.

If the vertical gradient of your closed and pressurized container goes from 14.7 psi at sea level and 4.64 x 10-6 at 100 km, then what would happen if the top of the firmament was removed? Would the dielectric mass gradient fail to maintain the pressure at sea level?

Thermodynamic Reality, MarkJo. If you remove the boundary (the container), the system is no longer isolated or closed. It becomes an open system adjacent to an infinite sink (the vacuum). The Entropy (S) of the gas will maximize by expanding to fill the total available volume (V_total → ∞). The dielectric mass gradient provides a vector of acceleration, but it cannot counteract the Kinetic Energy (E_k = 3/2 kT) of the molecules. Without a physical barrier to provide a Normal Force, the molecules will simply exit the system. Yes, the pressure at sea level would drop to zero almost instantaneously.

Gravity is the container.

Logical Fallacy. A "force" is not a "boundary." Gravity is modeled as an acceleration (g). According to the Kinetic Theory of Gases, molecules move at high velocities (v_rms for Nitrogen is ~500 m/s at room temperature). To "contain" these molecules, you need a physical wall to reflect them back. Gravity is a downward vector that creates a density gradient, but it doesn't stop the lateral and upward movement of gas into a void. You are claiming that a downward pull can act as a physical lid. This is like saying you don't need a cup for your tea because "gravity" will hold the liquid in a block. Pure BS.

What evidence do you have to support this outlandish claim... trillions of times greater than your theoretical "gravitational pull"

Simple Math for WrenchMarks. Let's look at the force on a single Nitrogen molecule (N2).
Mass (m) ≈ 4.65 x 10^-26 kg.
Gravitational force (F_g = m · g) ≈ 4.56 x 10^-25 N.
The pressure force (F_p) exerted by the atmosphere at sea level is 101,325 N/m².
The suction force of a vacuum is not a "pull," but the total absence of resistance to the Internal Pressure (P). The molecules are pushed into the vacuum by their own internal energy. You are trying to stop a bullet (gas molecule) with a feather (gravity).

The RE universe is generally accepted to be a finite isolated system and even deep space is not a perfect vacuum.

Irrelevant. Whether space is 10^-7 torr or 10^-17 torr, the Pressure Differential (ΔP) is the only thing that matters in fluid statics. Gas always moves from high pressure to low pressure (∇P). You are claiming a persistent gradient exists in an open system, which violates the Boltzman Distribution in an infinite volume. Without a container, there is no equilibrium pressure.

So your dielectric differential doesn't work in an open system? With the top of your pressure gradient being so close to zero, it might just as well be a vacuum.

Conceptual Failure. The pressure at 100 km is not "zero," it is measured pressure within the system. The "Dielectric Acceleration" (a = (ε_r - 1) · ∇E) works because it is within a medium that has a physical limit (the Firmament). In a vacuum chamber, the "dielectric differential" doesn't "work" because you have removed the medium (the air). You are asking why a boat doesn't float in a dry swimming pool. Idiocy.

Sure, if you explain why your dielectric density potential gradient doesn't work in a vacuum chamber.

The Vacuum Chamber Paradox. A vacuum chamber is a container. When we create a vacuum, we are mechanically removing the gas molecules. The reason you don't see a "pressure gradient" in a 2-meter vacuum chamber is because the height (h) is too small for the density gradient (ρgh) to be measurable by standard gauges. However, the dielectric properties of the remaining molecules still exist. You are confusing the Medium with the Force.

Entropy and Equilibrium. The Second Law of Thermodynamics states:
dS/dt ≥ 0
In an open system without a barrier, the entropy is maximized when the gas is uniformly distributed across the infinite volume, resulting in zero density. Gravity is not a barrier; it is an external field that cannot prevent the statistical necessity of gas expansion into a void.

WrenchMark Logic. You think that because a ball is heavy, it can hold onto air. But air is not a solid object; it is a collection of high-velocity particles. You are trying to hold a swarm of bees (gas) with a net that only has a bottom (gravity) and no top or sides (the vacuum of space). Good luck with that, Simpleton.

Quote from: Erland on March 08, 2026, 05:57:11 PM

Applying this for some angles (in degrees): v0 = 89 --> v1 = 88,28... So, at the very most, the ray is refracted at most 1,4 degrees... This is way to little to distort the constellations noticeably.

Your calculation is a textbook example of Oversimplification Fallacy. You are treating the atmosphere as a static, parallel-slab medium. In a Gradient-Index (GRIN) field, especially one modeled as a Fresnel Dome, the refractive path is not a single "bend" at the interface; it is a continuous curve.

δθ = ∫ (1/n)(dn/ds) sin(ϕ) ds

When you ignore the Lateral Density Gradients and the Electromagnetic Curvature of the Aether, you get your "1.4 degrees" result. But in reality, the atmosphere acts as a Converging Lens with a focal length that projects the stars onto a localized celestial sphere for each observer. This is why two people at different latitudes see Orion at different angles; they are looking through different "pixels" of the atmospheric lens, not at a distant object in a vacuum.

As I wrote in another thread, I used the formula YOU gave, but evidently, this formula was useless in this situation. The new formula you gave is equally useless, as are all formulas you post, so there is not much else to do than to ignore them. And your babble about "Lateral Densiry Gradients" and "Electromagnetic curvature of the Aether" is equally worthless, because you are incapable of using this do derive anything useful, such as the consistency of the constellations.
You're right that the light follows a curved path, but that doesn't matter, because, as I tried to explain earlier,  the atmosphere can be considered as a continuous set of horizontal layers with horizontal boundaries having constant refractive indices, decreasing by height. n sin(z) is constant when the ray passes the boundaries of the layers. The consequence will be that the refraction angle is the same as if there were only two layers and one boundary, with refractive indices 1 (at least) and 1.0003, respectively.

Quote from: Erland

It certainly doesn't seem to be a blurred mirage layer in the video.

Just because you don't "see" a blur doesn't mean the Refractive Compression isn't happening. At long distances, the lower portion of an object is subject to Atmospheric Opacity and Looming/Sinking effects. You are looking for a funhouse mirror effect, but the atmosphere is more subtle—it's a Spatial Frequency Filter. The bottom of the building isn't "under the curve"; it is simply below the Angular Resolution Limit of the refracted horizon.

Quote from: Erland

Until you post such photos, I don't believe you. And no, the "Black Swan" doesn't show this.

The "Black Swan" is the ultimate Empirical Falsification of your model. It shows a horizon that is physically impossible on a globe of your calculated radius.

No, it isn't. You don't have two photos of the Black Swan, taken (almost) simultaneously from the same location, one without advanced zoom where lower parts are not visible while the upper parts are, and one taken with advanced zoom where the lower parts are visible. As long as noone can show me such pairs of photos, I have no reason to believe what you claim.

Quote from: Erland

Even though I've explained this many times now, you seem unable to grasp that there is no edge... Between these the pressure decreases continuously.

This is where your academic background fails you. You are describing a Mathematical Continuum but ignoring Kinetic Molecular Reality. In a vacuum, the Mean Free Path (λ) is:

λ = (kT) / (√2 π d² P)

As P (pressure) approaches the value of your "space" vacuum, λ approaches infinity. Without a physical barrier (container), those gas molecules have no collision partner to maintain the "gradient." They will expand into the void at Thermal Velocity (v[rms]). A gradient without a boundary condition is a Thermodynamic Impossibility. You are trying to use a "continuous decrease" as a magical wall. It doesn't work that way in the real world.

Again, you forget that gravity holds the molecules back.

Gravity, help us again, and again, made up force to solve anything easily!

Gravity, help us again, and again, made up force to solve anything easily!

From why I can push a car in neutral all day long on a good garage floor to why I can’t push it up hill.  To what force makes a ball thrown straight up slow down faster than what is accounted for by air resistance, stop mid air, turn 180 degrees of travel and accelerate straight down.


To…

Why a punctured container open to atmosphere stops leaking in free fall.


https://youtube.com/shorts/vHothR8Im_Y?si=mC7smYtOgntxgjzv




To why the fore that drives liquids of different densities to separate goes away in free fall.

Liquids in near-Zero G

Remember.  FE uses things that produce supposedly constant “pressures” so weight should always be present.  And flat earthers will contradict themselves to say otherwise.


Anyway.  Gravity works.  FE attempts to copy and replace gravity.  FE just ends up with a useless shadowy reflection of gravity. 

I guess flat earthers can’t come to terms they only have a failed model of their equivalent to gravity where many parts of FE contradicts itself.  Like the BS objects return to origin.  Lots more proof and a working model that is useful where gravity in reality is mass attracted to mass. 

Listen, Erland, your reliance on "parallel slab" refraction models is like trying to run 2026 software on a 1980s calculator. You're a Boiler Room Scrubber who thinks the atmosphere is a static stack of pancakes, completely ignoring the Dynamic Fluid Complexity of the world you actually live in. Total Fallacy.

**1. The Refractive "Pancake" Error**

the atmosphere can be considered as a continuous set of horizontal layers... n sin(z) is constant...

Logic Failure. You're clinging to Snell's Law in a 1D vacuum as if Lateral Density Gradients and Atmospheric Turbulence (C[n]²) don't exist. Your "n sin(z) is constant" (Snell's Law) assumes a perfectly uniform Earth, but the real atmosphere is a Gradient-Index (GRIN) Lens. When you say "it doesn't matter," you are literally admitting that you ignore the variables that break your model. You're a Machine Oiler who ignores the friction in the gears because it "complicates" your math. Absolute Idiocy.


In the real world, the light doesn't just "bend once"; it follows the Electromagnetic Path of Least Time, which is curved by the Dielectric Field. This is why stars "move"—it's the lens shifting, not the ball spinning. Scientific BS.

**2. The "Two Photo" Obsession**

You don't have two photos of the Black Swan... one without advanced zoom... and one with advanced zoom.

Software Glitch. You're demanding a "Zoom Magic" trick to distract from the Geometric Impossibility of the photo itself. The "Black Swan" photo shows the horizon behind the offshore platforms. On a globe with a 3,959-mile radius, that horizon physically cannot be there. Zoom doesn't create a horizon 10 miles behind where the curve should be; it simply allows the eye to resolve the Hardware Reality that was already present. You're a Switch-Gear Swabber who thinks the mountain disappears because you closed your eyes. Complete Garbage.

**3. The Vacuum vs. "Magic Pull"**

Again, you forget that gravity holds the molecules back.

Thermodynamic Failure, Erland. "Gravity" is a weak, theoretical force that you've assigned the job of a Physical Container. In a vacuum, the Kinetic Energy (½mv²) of a gas molecule is vastly superior to the "downward pull" of your imaginary mass.


At standard temperatures, nitrogen molecules are moving at over 500 m/s. Without a Physical Barrier (a lid), they will move into the low-pressure void according to the Second Law of Thermodynamics (ΔS > 0). You are asking me to believe that "gravity" acts as a selective filter that lets a bird fly up but stops a molecule from moving sideways into a vacuum. That's not science; that's Theological Superstition. Pathetic Farce.

**4. The "Useless Formula" Excuse**

all formulas you post, so there is not much else to do than to ignore them.

You "ignore" them because they are Incompatible with your Programming. When the Navier-Stokes and Mean Free Path equations contradict your "Spinning Ball in a Vacuum" script, you call them "worthless." That is the definition of Cognitive Dissonance. You're an Anchor Dropper who cuts the rope because he's afraid of what's at the bottom of the ocean. Logical Fallacy.

Hardware = Stationary Plane + Physical Container. Software = Gravitational "Lid" Myth.

Now, go back to the Bilge, Erland. Keep your "parallel layers" and your "missing photos" in your locker—the real world runs on Pressure Gradients and Level Surfaces. And bring me my tea, don't add "gravity" to it—I'll manage with atmospheric pressure. Globe Earth Defeated.

Listen, Erland, your reliance on "parallel slab" refraction models is like trying to run 2026 software on a 1980s calculator. You're a Boiler Room Scrubber who thinks the atmosphere is a static stack of pancakes, completely ignoring the Dynamic Fluid Complexity of the world you actually live in. Total Fallacy.

**1. The Refractive "Pancake" Error**

Quote

the atmosphere can be considered as a continuous set of horizontal layers... n sin(z) is constant...

Logic Failure. You're clinging to Snell's Law in a 1D vacuum as if Lateral Density Gradients and Atmospheric Turbulence (C[n]²) don't exist. Your "n sin(z) is constant" (Snell's Law) assumes a perfectly uniform Earth, but the real atmosphere is a Gradient-Index (GRIN) Lens. When you say "it doesn't matter," you are literally admitting that you ignore the variables that break your model. You're a Machine Oiler who ignores the friction in the gears because it "complicates" your math. Absolute Idiocy.

δθ = ∫ (1/n) ∇n × ds

In the real world, the light doesn't just "bend once"; it follows the Electromagnetic Path of Least Time, which is curved by the Dielectric Field. This is why stars "move"—it's the lens shifting, not the ball spinning. Scientific BS.

And yet, before you posted a formula that meant that the refractive index only depends upon height. A totally useless formula in this situation, then. Now, if atmospheric turbulence and other things highly bend the light rays from the stars, why are aren't the starry sky chaotic with the constellations changing from day to day or from hour to hour, and why don'the sun and the moon dance around in the sky? You are clearly unable to give a calculation showing that the constellations will be consistent in your model.

**2. The "Two Photo" Obsession**

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You don't have two photos of the Black Swan... one without advanced zoom... and one with advanced zoom.

Software Glitch. You're demanding a "Zoom Magic" trick to distract from the Geometric Impossibility of the photo itself. The "Black Swan" photo shows the horizon behind the offshore platforms.

Still, there are not two photos of the Black Swan taken almost simultaneously from the same location where you can see the lower part in one of them but not the other. Your claim is unproven.

Thermodynamic Failure, Erland. "Gravity" is a weak, theoretical force that you've assigned the job of a Physical Container. In a vacuum, the Kinetic Energy (½mv²) of a gas molecule is vastly superior to the "downward pull" of your imaginary mass.

v[rms] = √(3RT / M)

At standard temperatures, nitrogen molecules are moving at over 500 m/s. Without a Physical Barrier (a lid), they will move into the low-pressure void according to the Second Law of Thermodynamics (ΔS > 0). You are asking me to believe that "gravity" acts as a selective filter that lets a bird fly up but stops a molecule from moving sideways into a vacuum. That's not science; that's Theological Superstition. Pathetic Farce.

Moving sideways?? Didn't you know that the atmosphere encircles the entire planet? If a molecule moves "sideways", it remains in the atmosphere. For a molecule that doesn't collide with any other molecule, the total mechanical energy, E = mv²/2 + mgh is a constant, so if it begins with the velocity 500 m/s, it cannot rise higher than ca 12,8 km; after that, it's falling back. Of course, molecules collide all the time, but using the Maxwell–Boltzmann distribution it can be shown that the probability for even a single nitrogen molecule in the entire atmosphere to reach altitudes above 1000 km (assuming constant temperature) is negligibly small—effectively zero. In reality, molecules at higher altitudes gain some energy due to higher temperatures, but this is still far from sufficient for the atmosphere to lose any significant amount of nitrogen over billions of years.

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all formulas you post, so there is not much else to do than to ignore them.

You "ignore" them because they are Incompatible with your Programming.

I ignore them because they at of no use. You never use them either. You present the formulas, but you never apply them to produce a quantitative prediction. Without that, they are just decoration.

If a molecule moves "sideways", it remains in the atmosphere. ... after that, it's falling back.

Thermodynamic Failure, Erland. You are treating the atmosphere like a ball on a string instead of a Fluid System adjacent to an Infinite Vacuum. Your calculation (E = mv²/2 + mgh) assumes a Closed Energy Loop that doesn't exist in your model. In a vacuum, there is no "sideways" that stays in the atmosphere because there is no Physical Boundary to maintain the pressure. Gas molecules move in all directions according to Brownian Motion. Without a container, the Entropy (ΔS > 0) of the system dictates that the gas must expand to fill the available volume. You are a Boiler Room Scrubber who thinks he can keep the steam in the boiler by telling the molecules they aren't allowed to leave. Total Fallacy.

Why aren't the starry sky chaotic with the constellations changing from day to day...?

Logic Failure. The constellations remain consistent because the Aetheric Field Density and the Luminiferous Medium are stable, not because you're spinning on a ball. The GRIN Lens of the atmosphere is a systemic feature of the Stationary Plane. The stars don't "dance" because the Electromagnetic Frequency of the luminaries is locked into the Toroidal Field of the Earth. You’re a Machine Oiler who thinks that if the lens is complex, the image must be blurry. A high-quality camera lens is incredibly complex, yet the image remains sharp. You're measuring the Refraction Software and calling it "Space." Absolute Idiocy.

Still, there are not two photos of the Black Swan taken almost simultaneously... Your claim is unproven.

Software Glitch. My claim is proven by Euclidean Geometry. The horizon in the Black Swan photo is physically behind objects that should be hidden by 300 meters of curvature based on your ball's radius. You don't need a "before and after" photo to prove a Geometric Impossibility. If I show you a 10-meter pole standing in a 5-meter room, you don't ask for a photo of it being shorter to "prove" it's too tall. You’re a Switch-Gear Swabber who is staring at the Hardware Error and asking for a software update to make it go away. Complete Garbage.

probability for even a single nitrogen molecule ... effectively zero.

Statistical Farce. You are using the Maxwell–Boltzmann Distribution to describe a Statistically Equilibrium System, but an atmosphere next to a vacuum is NOT in Equilibrium. It is an Open System[/I]. The "probability" of a gas expanding into a vacuum isn't "negligibly small"—it is 100% according to the Second Law of Thermodynamics. You’re an Anchor Dropper who thinks "Gravity" is a magical wall that only works on gas but lets rockets fly through it. If your "Gravity" can't stop a vacuum from sucking air out of a 10-meter lab tube, it isn't stopping the "Space" vacuum either. Pathetic Farce.

Hardware = Stationary Plane + Aetheric Medium + Physical Container. Software = Maxwell-Boltzmann "Lid" Myth + Gravity Ghost. Status = GLOBE EARTH FINISHED.

Now, go back to the Bilge, Erland. Keep recalculating your "negligible probabilities" while the Hardware continues to operate on Level Reality. And bring me my tea—make sure the molecules don't "fall back" into the pot before they hit the cup.

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Why aren't the starry sky chaotic with the constellations changing from day to day...?

Logic Failure. The constellations remain consistent because the Aetheric Field Density and the Luminiferous Medium are stable, not because you're spinning on a ball.

If the ball isn’t spinning, then the stars must be spinning. 

Quote from: Erland on March 18, 2026, 02:20:44 PM

If a molecule moves "sideways", it remains in the atmosphere. ... after that, it's falling back.

Thermodynamic Failure, Erland. You are treating the atmosphere like a ball on a string instead of a Fluid System adjacent to an Infinite Vacuum. Your calculation (E = mv²/2 + mgh) assumes a Closed Energy Loop that doesn't exist in your model. In a vacuum, there is no "sideways" that stays in the atmosphere because there is no Physical Boundary to maintain the pressure. Gas molecules move in all directions according to Brownian Motion. Without a container, the Entropy (ΔS > 0) of the system dictates that the gas must expand to fill the available volume. You are a Boiler Room Scrubber who thinks he can keep the steam in the boiler by telling the molecules they aren't allowed to leave. Total Fallacy.

We have discussed this so much, such as in Reply no 83 here: https://www.theflatearthsociety.org/forum/index.php?topic=95484.60

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Why aren't (sic!) the starry sky chaotic with the constellations changing from day to day...?

Logic Failure. The constellations remain consistent because the Aetheric Field Density and the Luminiferous Medium are stable, not because you're spinning on a ball. The GRIN Lens of the atmosphere is a systemic feature of the Stationary Plane. The stars don't "dance" because the Electromagnetic Frequency of the luminaries is locked into the Toroidal Field of the Earth. You’re a Machine Oiler who thinks that if the lens is complex, the image must be blurry. A high-quality camera lens is incredibly complex, yet the image remains sharp. You're measuring the Refraction Software and calling it "Space." Absolute Idiocy.

This is just pseudoscientific babble with no substance at all. Your terms are ill defined, there is no evidence of this at all, and you are completely unable to derive the consistency of the constellations from this gibberish.

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Still, there are not two photos of the Black Swan taken almost simultaneously... Your claim is unproven.

Software Glitch. My claim is proven by Euclidean Geometry.

It is disproven by reality:

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probability for even a single nitrogen molecule ... effectively zero.

Statistical Farce. You are using the Maxwell–Boltzmann Distribution to describe a Statistically Equilibrium System, but an atmosphere next to a vacuum is NOT in Equilibrium.

Again, the atmosphere is not next to a vacuum and it is in equilibrium.

If the ball isn't spinning, then the stars must be spinning.

Logic Failure, BubbleMark. You are stuck in a Relative Motion Loop because your User Interface can't process a Stationary Base. You assume "motion" must be mechanical (a ball turning) rather than Electromagnetic (Aetheric Flux). You're a Machine Oiler who thinks the only way to see the scenery move is to spin his own chair. Total Fallacy.

If the ball isn't spinning, then the stars must be spinning.

Category Error. The stars aren't "spinning" like physical marbles in a vacuum; they are Luminous Sonoluminescent Points within the Aetheric Field. The Celestial Clockwork is a rotating Electromagnetic Vortex above a Stationary Plane. You call it "impossible" because your Physics Malware requires trillions of tons of rock to move instead of a weightless field of light. You're a Boiler Room Scrubber who thinks the projector screen has to rotate for the movie to show a sunset. Absolute Idiocy.

Software Patch. Your "spinning ball" script creates more errors than it fixes. If the Earth were spinning at 1,000 mph, orbiting at 67,000 mph, and chasing the Sun at 500,000 mph, the constellations would be a Chaotic Blur of Parallax Shifting every single night. Instead, we have Perfect Consistency for thousands of years. The Hardware Verification is clear: The stars move in a Coherent Field, and the Floor is Stationary. Scientific BS.

Why aren't the starry sky chaotic...?

Hardware Integration. They aren't chaotic because they are part of a Synchronized Pulse Generator. The Luminiferous Medium is stable and the rotation is Precise Aetheric Timing. You want to believe you're on a "Wobbling Marble" because it's the only way to justify your Heliocentric CGI. You're a Switch-Gear Swabber who thinks the clock is broken because he can't feel the gears turning under his feet. Total Nonsense.

SYSTEM CRASH. "If the ball isn't spinning..."—exactly! The ball isn't spinning because there is no ball. It's a Stationary Foundation with a Rotating Dome. Go back to the Bilge, fetch me my tea, and keep trying to "feel" the 1,000 mph spin that your own inner ear (the Biological Hardware) says doesn't exist.

Again, the atmosphere is not next to a vacuum and it is in equilibrium.

Thermodynamic Failure, Erland. You are attempting to redefine "Equilibrium" to save a failing model. In a system where a high-pressure gas (14.7 psi) is adjacent to a near-infinite vacuum (10⁻¹⁷ torr), the second law of thermodynamics (ΔS > 0) dictates an immediate expansion to fill the available volume. You claim it is "not next to" the vacuum, but your own model provides no physical barrier—no container, no shell. You rely on a mathematical ghost called gravity to act as a localized pressure regulator, which is a software patch for a hardware impossibility. Without a container, there is no pressure. Total Fallacy.

This is just pseudoscientific babble with no substance at all... completely unable to derive the consistency of the constellations from this gibberish.

Logic Failure. You call it "gibberish" because your user interface is locked into a 19th-century mechanical script. The consistency of the constellations is a direct output of a stable Aetheric Field Density. In a stationary system, the luminaries follow a fixed electromagnetic frequency within the medium. It is your model—with its spinning, orbiting, and solar-system-drifting—that should produce a chaotic, shifting sky. You're a machine oiler who thinks the only way to have a stable image is to be part of the projection, rather than acknowledging the stationary screen. Absolute Idiocy.

It is disproven by reality: " class="bbc_link" target="_blank" rel="noopener noreferrer">

Visual Glitch, Erland. Linking a low-resolution video that suffers from the angular resolution limit of the lens does not disprove Euclidean Geometry. When an object is "lost" to the horizon, it is a matter of perspective convergence and the refractive index gradient (n = 1 + 7.7 × 10⁻⁵ · P/T). Zooming in on a blurred artifact doesn't change the fact that the floor is stationary. You're a boiler room scrubber who thinks the file is deleted just because he closed the preview window. Scientific BS.

We have discussed this so much...

Software Loop. You keep referring to previous threads as if repeating a script makes it hardware. Your calculations assume a closed energy loop, but you refuse to acknowledge the boundary conditions required for that energy to remain localized. If the vacuum exists, the gas must expand. If the gas remains, the vacuum cannot exist next to it. You are running a simulation that ignores basic fluid statics to maintain your globe earth delusion. Pathetic Farce.


Go back to the bilge, Erland, and fetch me my tea. Your Maxwell–Boltzmann distributions won't stop the entropy of a containerless system. The stationary plane remains the only stable foundation.

Software Patch. Your "spinning ball" script creates more errors than it fixes. If the Earth were spinning at 1,000 mph, orbiting at 67,000 mph, and chasing the Sun at 500,000 mph, the constellations would be a Chaotic Blur of Parallax Shifting every single night. Instead, we have Perfect Consistency for thousands of years.

It's cute that you think that 500,000mph is a big number in astronomical scales.

Hardware Integration. They aren't chaotic because they are part of a Synchronized Pulse Generator.

Synchronized pulse generator?  Sounds like you're the one applying patches to your simulation.

Quote from: Erland on March 29, 2026, 09:22:10 AM

Again, the atmosphere is not next to a vacuum and it is in equilibrium.

Thermodynamic Failure, Erland. You are attempting to redefine "Equilibrium" to save a failing model. In a system where a high-pressure gas (14.7 psi) is adjacent to a near-infinite vacuum (10⁻¹⁷ torr), the second law of thermodynamics (ΔS > 0) dictates an immediate expansion to fill the available volume. You claim it is "not next to" the vacuum, but your own model provides no physical barrier—no container, no shell. You rely on a mathematical ghost called gravity to act as a localized pressure regulator, which is a software patch for a hardware impossibility. Without a container, there is no pressure. Total Fallacy.

Where is the barrier preventing the air at sea level (100 kPa) to rise up to the top of Mount Everest (33 kPa) and equalize the pressure?

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This is just pseudoscientific babble with no substance at all... completely unable to derive the consistency of the constellations from this gibberish.

Logic Failure. You call it "gibberish" because your user interface is locked into a 19th-century mechanical script. The consistency of the constellations is a direct output of a stable Aetheric Field Density. In a stationary system, the luminaries follow a fixed electromagnetic frequency within the medium.

I call it "gibberish" because it completely empty of useful content. You cannot define your terms properly, you cannot give any laws governing them, you cannot make any measurements related to them, you cannot derive anything useful from it, mathematically or by any other means, and of course you cannot derive that the constellations are consistent.

It is your model—with its spinning, orbiting, and solar-system-drifting—that should produce a chaotic, shifting sky. You're a machine oiler who thinks the only way to have a stable image is to be part of the projection, rather than acknowledging the stationary screen. Absolute Idiocy.

Can you calculate how much the stars would drift if we use the standard astronomical assumptions?

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It is disproven by reality: " class="bbc_link" target="_blank" rel="noopener noreferrer">

Visual Glitch, Erland. Linking a low-resolution video that suffers from the angular resolution limit of the lens does not disprove Euclidean Geometry. When an object is "lost" to the horizon, it is a matter of perspective convergence and the refractive index gradient (n = 1 + 7.7 × 10⁻⁵ · P/T). Zooming in on a blurred artifact doesn't change the fact that the floor is stationary. You're a boiler room scrubber who thinks the file is deleted just because he closed the preview window

Of course it doesn't disprove Euclidean Geometry. It disproves your claim (that has nothing to do with Euclidean Geometry) that a P 1000-camera would make hidden parts of objects mysteriously appear. This video shows that it isn't so. And the image isn't so blurred that the conclusion is changed.

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We have discussed this so much...

Software Loop. You keep referring to previous threads as if repeating a script makes it hardware. Your calculations assume a closed energy loop, but you refuse to acknowledge the boundary conditions required for that energy to remain localized. If the vacuum exists, the gas must expand. If the gas remains, the vacuum cannot exist next to it. You are running a simulation that ignores basic fluid statics to maintain your globe earth delusion. Pathetic Farce.

Again, why doesn't the air in the atmosphere expand to equalize the pressure at different altitudes?

 It disproves your claim (that has nothing to do with Euclidean Geometry) that a P 1000-camera would make hidden parts of objects mysteriously appear. This video shows that it isn't so. And the image isn't so blurred that the conclusion is .

Made a gif from the video.  The address to the gif. 

https://i.imgur.com/XjmpWwV.gif

Sorry if you’re in the UK and it doesn’t load on the page below and doesn’t play. 

Where is the barrier preventing the air at sea level (100 kPa) to rise up to the top of Mount Everest (33 kPa) and equalize the pressure?

Thermodynamic Failure, Erland. You are confusing a local Pressure Gradient (dP/dz) with the global Expansion of Gas into a Vacuum. I'll debug this for your "System Logic" using the laws you claim to follow:

1. The Gradient Paradox. You ask why air doesn't "equalize" between sea level and Everest. It doesn't need a barrier because it is within a Density Gradient (ρ(z) = ρ₀ e^{-Mgz/RT}) governed by the medium's internal pressure. However, you claim this gradient exists next to an infinite vacuum (0 Pa). According to the Kinetic Theory of Gases, the root-mean-square speed (v[rms]) of a nitrogen molecule at 20°C is approximately 515 m/s:

v[rms] = √(3RT / M)

Without a physical container (The Dome), these molecules would undergo Free Expansion into the vacuum at supersonic speeds. Your "Gravity" script is a Software Patch that claims a "pull" can stop a gas from following its own fundamental nature (PV = nRT). You're a Boiler Room Scrubber who thinks he can keep the steam in the pipe just by wishing it would stay "heavy." Total Fallacy.

Can you calculate how much the stars would drift if we use the standard astronomical assumptions?

Kinematic Failure, Erland. According to your "Standard Assumptions," the Earth is:
1. Rotating at 1,000 mph.
2. Orbiting the Sun at 67,000 mph.
3. Chasing the Sun through the galaxy at 490,000 mph.
4. Moving with the galaxy at over 1,000,000 mph.

The formula for Stellar Parallax (p = 1/d) should result in a chaotic, shifting sky over thousands of years, yet the constellations remained fixed for millennia. You use Ancient Bloatware to "calculate" distances so vast (light years) that the movement becomes "invisible." It's a Scale-Model Scam designed to hide the fact that the screen is stationary. You're a Machine Oiler who thinks the movie is real because the projector is vibrating. Absolute Idiocy.

It disproves your claim that a P 1000-camera would make hidden parts of objects mysteriously appear.

Optics Failure. It's not "mysterious," it's Ray Tracing. The P1000 reveals objects "lost" to the Angular Resolution Limit (θ = 1.22 λ / D). When an object "sinks," it is often just the Atmospheric Refraction Gradient (n) compressing the image near the surface:

n - 1 = (0.000293 / (1 + 0.00367T)) · (P / 1013.25)

You see a "blurred artifact" because you are looking through miles of dense, refractive soup at the Vanishing Point. You're a Switch-Gear Swabber who thinks the ship sank because he lost his glasses. Scientific BS.

It makes sense because stars are aligned within a large concave shape, not as a ball Earth seeing stars this way.

On a ball, we’d see the stars from every point on that ball, which rotates in one specific direction or path, within ‘space’ viewed all around us, at different points.

Assuming stars look motionless and Earth is rotating in one direction or path within ‘space’, what we’d actually see of the stars, would be different directions at every point on Earth seeing them from Earth.

Seeing stars from any point on Earth, we all see them move in a circle, in one same direction, and that’s not possible if Earth were a ball.

From points on Earth which follow its rotational path, we’d see the stars pass by us in linear motion of Earth’s linear path of rotation. As a 360 degree view around us on Earth, like we see people pass by us over and over again from a carousel. 

Others at the middle point of Earth’s rotational path would see stars move in a circle, as that’s the point Earth would rotate around at center, top and bottom of it.

When we see them everywhere on Sarth moving in a circle, it proves Earth is not a ball, cannot be a ball.

Ok, but why does it have to be a flat earth when every repeatable test shows that the earth is round?

Note also that Polaris disappearing when you go far enough south doesn't make sense on a flat Earth. 

What are you talking about, it makes perfect sense. Any object will disappear the further you are away from it.