The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: zorbakim on October 09, 2018, 08:06:36 AM

Because the waves are near and the building is far.
The waves on the horizon block the sight.
https://youtu.be/ZPgq4QILC0

Depends on viewer height. There are 3 scenarios.
If your viewer height is above wave height then the wave will block less of the distant object than its own height because you are looking down and over it:
(https://image.ibb.co/iL8rC7/waves.jpg)
If your eye is at wave height then the wave blocks as much of the distant object as its own height as you are looking straight across the top of it:
(https://image.ibb.co/f804X7/waves_b.jpg)
If your eye is below wave height then yes, the wave will then block more of the distant object than the wave's own height:
(https://image.ibb.co/nhHUzn/waves_c.jpg)
And the closer the wave in this scenario, the more of the building it will block:
(https://image.ibb.co/n6Df5S/waves_d.jpg)
So, if your viewer height is low enough then waves would be an explanation for you not being able to see distant objects.
If you're at any reasonable altitude then unless there are massive waves  which there typically are not  then waves cannot be the explanation as you are above wave height and thus the waves cannot block the distant object. A dime CAN block your view of a distant elephant but only if you hold the dime close to your eye. Put the dime on the ground and you'd have to lie on the ground and close your other eye to get low enough that it can block anything.

A dime CAN block your view of a distant elephant but only if you hold the dime close to your eye.
Perspective attempts to bring the surface in the distance to eye level. Unless you have a solid demonstration on how perspective works, I don't see how your argument holds any weight.

A dime CAN block your view of a distant elephant but only if you hold the dime close to your eye.
Perspective attempts to bring the surface in the distance to eye level. Unless you have a solid demonstration on how perspective works, I don't see how your argument holds any weight.
Do some experiments and show me a photo of a dime hiding an elephant with the dime on the ground and the eye/camera height more than a couple of centimetres off the ground.
As elephants are not native to California I'll accept an elephant size object.

While you're doing that, here are the results of the experiment I did previously to demonstrate the effect I outline above:
The set up:
(https://image.ibb.co/kimGTc/setup.jpg)
Eye above the level of the obstacles, less of the Jenga block height is occluded:
(https://image.ibb.co/mm7C1x/above.jpg)
Eye At the level of the obstacles, exactly the Jenga block height is occluded:
(https://image.ibb.co/cyGuEH/level2.jpg)
Eye below the level of the obstacles, more than the Jenga block height is occluded:
(https://image.ibb.co/gXGi8c/below1.jpg)
I look forward to your experiment results.

A dime CAN block your view of a distant elephant but only if you hold the dime close to your eye.
Perspective attempts to bring the surface in the distance to eye level. Unless you have a solid demonstration on how perspective works, I don't see how your argument holds any weight.
Perspective is nothing else than a decreasing angle of sight lines as a result of a increasing distance. That decreasing angle can be calculated with simple trigonometry. When the viewing angles decrease, then the projection of the observed object on our retina becomes smaller.
(https://docs.google.com/drawings/d/e/2PACX1vRyK1YCrlddcTNkcxX5oXZJqAKqG_0HK9MFM1DGPv_4ZOMMAxuzppdZw9aa1nkm1WiKuLF7S9a1XX7/pub?w=385&h=281)
If the angle of sight line of a nearby dime is the same as the angle of sight lines of a far away elephant, both appear on our retina as equal in size.
(https://docs.google.com/drawings/d/e/2PACX1vRQzNCSUODZfpEHImr9uU9phZtXyy9QHoROf89ADm6POk95Mx6X8jdqcS0T1xuThJJtrMVQYv1nqbeX/pub?w=477&h=335)

A dime CAN block your view of a distant elephant but only if you hold the dime close to your eye.
Perspective attempts to bring the surface in the distance to eye level. Unless you have a solid demonstration on how perspective works, I don't see how your argument holds any weight.
Tom, I'm not sure if I understand what you're getting at. Exactly what part of perspective do you think needs to be demonstrated?

The author in the OP has composed several videos demonstrating that perspective is nonlinear (https://www.youtube.com/channel/UCW04DL2HRPulwGWursWCc6Q). If correct, it shows that perspective is not as it has been taught.
Where can we see something which demonstrates that perspective is as it is taught in schools?

Where can we see something which demonstrates that perspective is as it is taught in schools?
(https://upload.wikimedia.org/wikipedia/commons/4/4f/Staircase_perspective.jpg)

The author in the OP has composed several videos demonstrating that perspective is nonlinear (https://www.youtube.com/channel/UCW04DL2HRPulwGWursWCc6Q). If correct, it shows that perspective is not as it has been taught.
Where can we see something which demonstrates that perspective is as it is taught in schools?
And have shown some diagrams showing how a 1m wave cannot block a 100m building unless the viewer height is less than a meter.
I have also shown some pictures of an experiment I did some time back which shows my diagrams reflect reality.
If you feel there is some error in my conclusions then feel free to conduct your own experiment and publish the results for review.

The author in the OP has composed several videos demonstrating that perspective is nonlinear (https://www.youtube.com/channel/UCW04DL2HRPulwGWursWCc6Q).
ICanScienceThat already demonstrated the same thing but you dismissed it.

Where can we see something which demonstrates that perspective is as it is taught in schools?
"something which"? You mean an object, a thing, a device?
There are millions if not billions of such devices, that automatically produce perspective images. Nearly everyone possesses at least one of these devices:
A camera
A camera produces a 2D projection of 3D real world on the image plane, using perspective.
If you don't like lenses, use a "camera obscure (https://en.wikipedia.org/wiki/Camera_obscura)" or a "pinhole camera"
The camera obscura was used as a means to study eclipses, without the risk of damaging the eyes by looking into the sun directly. As a drawing aid, the camera obscura allowed tracing the projected image to produce a highly accurate representation, especially appreciated as an easy way to achieve a proper graphical perspective.

Because the waves are near and the building is far.
The waves on the horizon block the sight.
A 1m wave near me can block a view of a 100m tall object.
How can a 1m wave on the horizon block a 100m tall object beyond the horizon? I can see how that could be on a curved, convex surface but not on a flat surface.
(http://oi65.tinypic.com/331pp5e.jpg)
Here's a 1m wave near  not on the horizon but about 30m away. It's not blocking the 200m island in the distance that is beyond the globe earth horizon. The further away from me and closer to the horizon and island such a wave is, the smaller it appears to me due to perspective and the less of an obstacle it becomes. It only starts becoming an obstacle and blocking the more distant 100m island when it gets close to me.

If correct, it shows that perspective is not as it has been taught.
Sure, but the problem is, his nonlinear perspective is demonstrably not correct.
Where can we see something which demonstrates that perspective is as it is taught in schools?
In the world around you.
But do not take that for granted; compare what you see with what is taught in schools.

Where can we see something which demonstrates that perspective is as it is taught in schools?
Ah sorry, forgot another 14 billion of "devices", that easily and effortless produce perspective images: Everyone (nearly) has two of them. Oh sorry again, must be many, many more: some insects have 8 of them.

About 8" of resolution creates a horizontal line at seven kilometers.
The waves on the horizon are at eye level.
The waves on the horizon look small, but they are enough to cover a farther building.
The secret lies at a angle.
Our eyes can't sense such a small angle.
That's why it's so hard to believe it.
In addition, the swells is higher than the waves.
Not only the waves, but the swells enough to block the sight.

About 8" of resolution creates a horizontal line at seven kilometers.
The waves on the horizon are at eye level.
The waves on the horizon look small, but they are enough to cover a farther building.
The secret lies at a angle.
Our eyes can't sense such a small angle.
That's why it's so hard to believe it.
In addition, the swells is higher than the waves.
Not only the waves, but the swells enough to block the sight.
I totally appreciate the sentiment of waves (swells) being an obstruction. But they don't account for the fact that disappearing ships, hull first, disappear...completely. Unless your swells are, in every instance observed, growing at the rate, suspiciously, in exact occurrence, symbiotically, with the 'sinking' object, I see no reason to entertain the notion that your premise explains the effect.

About 8" of resolution creates a horizontal line at seven kilometers.
The waves on the horizon are at eye level.
The waves on the horizon look small, but they are enough to cover a farther building.
The secret lies at a angle.
Our eyes can't sense such a small angle.
That's why it's so hard to believe it.
In addition, the swells is higher than the waves.
Not only the waves, but the swells enough to block the sight.
OK, so what you're saying amounts to in this diagram, let's imagine that last wave, the one nearest the building, is the wave at the horizon.
(https://image.ibb.co/iL8rC7/waves.jpg)
What your claim amounts to is that you can see the 1m wave, but you can't see the 100m building just beyond it.

A dime CAN block your view of a distant elephant but only if you hold the dime close to your eye.
Perspective attempts to bring the surface in the distance to eye level. Unless you have a solid demonstration on how perspective works, I don't see how your argument holds any weight.
Waves generally are only a few feet high. If those feet are close to the eye (like if you were floating in the ocean) they can block the view of hundreds of feet of far off buildings like so:
(https://i.imgur.com/xUPov5J.jpg)
The problem is, during these pictures where buildings are obscured, they are standing on the land or shore above sea level like this:
(https://i.imgur.com/mUERfoP.jpg)

Except that the surface is approaching eye level at the horizon where the diagram is marked.

Except that the surface is approaching eye level at the horizon where the diagram is marked.
...and the perceived size of waves gets smaller at the same rate the perceived size of larger things becomes smaller.
It doesn't work. 1m things only eclipse 100m things when they are much closer, not farther.
I was just down at the shore watching and photographing the 45' SSW swell hitting Baja and Socal coastal waters. I watched a Navy cruiser head west on the horizon. It was not being hidden be any waves. The more distant Coronado Islands looked the same as they do when sea surface is calm.
This "waves+perspective" rationale makes no sense, but the zetetic proof is in the observation, right? Should I post pics and video of 1m waves NOT obscuring 100m object on rhe horizon to prove it?
(Where the heck IS the horizon on a flat earth? I've asked that several times before and have had to come up with my own formula which had never been confirmed or denied.)

Except that the surface is approaching eye level at the horizon where the diagram is marked.
I don't understand. This is not looking uphill. How is the surface approaching eye level?
Are you saying that looking out over the ocean actually goes uphill?
(https://i.imgur.com/X8Tlada.png)
Even if the water did go uphill under normal viewing conditions, such as standing on a beach, a 6 foot wave would not obstruct hundreds of yards of a building.
(https://i.imgur.com/89I2UXq.jpg)

I'm more than willing to be wrong, but in this example, it seems the waves had to have been inordinately high, like tsunami high, to obscure 84.26 ft at an observation height of 12.1 ft.
(https://i.imgur.com/2N7NpRJ.jpg?1)

I wonder why nobody got away from this view, to consider only a single wave. There are many more waves, hundreds, thousands, millions.
Waves are all around the area between the observer, the horizon and the observed object.
Now, as you cannot look through the waves, it would not make any difference, if you filled up (virtually) the space between the waves with water. This will define a new surface, which is a distance of half the wave height above the level defined in calm weather with no waves, or the normal ground level.
The tops of the waves define a new surface, that reduces the height of the observer and the observed object (measured against ground level) by half the wave height. With these reduced heights you can now go on and construct your "perspective" views.
(https://i.postimg.cc/DyYWC08c/waves.png)

I'm more than willing to be wrong, but in this example, it seems the waves had to have been inordinately high, like tsunami high, to obscure 84.26 ft at an observation height of 12.1 ft.
(Turning Torso image)
When we had that Turning Torso discussion before, "waves" wasn't the flat earth rationale; it was optical compression.
Here is a view of North Coronado Island about 16 miles away from this spot (https://goo.gl/maps/TJHVpu9tUwT2) today, almost high tide with a 45' primary ground swell coming from the SSW.
https://www.youtube.com/watch?v=3RRoIWu5tpM
The island is about 425' at its peak (130m). That spur on the far left is 100 and 120' (3040m) high.
1 meter waves are not blocking 100 meters of that island.

I'm more than willing to be wrong, but in this example, it seems the waves had to have been inordinately high, like tsunami high, to obscure 84.26 ft at an observation height of 12.1 ft.
(Turning Torso image)
When we had that Turning Torso discussion before, "waves" wasn't the flat earth rationale; it was optical compression.
Here is a view of North Coronado Island about 16 miles away from this spot (https://goo.gl/maps/TJHVpu9tUwT2) today, almost high tide with a 45' primary ground swell coming from the SSW.
Not only would the waves have to be hundreds of meters high but also we have to understand that these buildings are many times several feet above sea level. No matter how you spin it i really struggle to see how waves accounting for hundreds of feet of a building disappearing is even possible unless you are floating in the ocean.
How on earth is this wave in any way obstructing the view when both the observer and the building are significantly above sea level?
(https://i.imgur.com/CdQEu76.jpg)

The physical parallel plane rises to an eye level.
It draws a curve(I called it ID graph)
ID graph does not exactly match the eye level as I have shown.
But We can't distinguish it with eyes because of the resolution limit.
Therefore, waves on the horizon can interfere with our vision.
We must distinguish between math and reality.
The reality consists of complex visual phenomena.

The physical parallel plane rises to an eye level.
No, it doesn't. Bobby has shown that result very clearly in the other thread about the horizon rising to eye level but you are free to repeat his experiments or design your own if you dispute his findings.
Therefore, waves on the horizon can interfere with our vision.
No, they can't. I have shown with diagrams above and some real life photos why they can't.
The horizon is basically you seeing the edge of the earth. You can't see the surface of the earth (or sea) beyond that because it is behind the curve of the earth.
You CAN, however, see tall objects beyond the horizon if you have clear line of sight to them and the curve isn't occluding the whole object and the atmosphere is clear enough.
The pictures of the Turning Torso building show that very clearly.
We must distinguish between math and reality.
If the math didn't accurately represent reality then any computer 3D rendering which uses that math would not look real. But it does look real, because it is an accurate reflection of reality.
EDIT: Maybe it would help if you showed some diagrams explaining how light from a tall building is blocked by a small wave on the horizon. I'm struggling to picture it.

The physical parallel plane rises to an eye level.
It draws a curve(I called it ID graph)
ID graph does not exactly match the eye level as I have shown.
But We can't distinguish it with eyes because of the resolution limit.
Therefore, waves on the horizon can interfere with our vision.
We must distinguish between math and reality.
The reality consists of complex visual phenomena.
Did you see my video above? What you're saying can happen isn't happening. I've never seen it happen.
Here's an old discussion topic from this past summer (https://forum.tfes.org/index.php?topic=10077.msg157902) where we talked about whether or not waves were the reason for "sinking ship effect" that ended with a link to a participationexperiment proposed by ICanScienceThat addressing the very thing you're trying to address.

The physical parallel plane rises to an eye level.
It draws a curve(I called it ID graph)
ID graph does not exactly match the eye level as I have shown.
But We can't distinguish it with eyes because of the resolution limit.
Therefore, waves on the horizon can interfere with our vision.
We must distinguish between math and reality.
The reality consists of complex visual phenomena.
Even if the plane did rise to eye level a 3 foot wave is not blocking hundreds of feet of a building. I don't see it.
In this diagram the viewer is standing at sea level looking at something that is very far away with ground level rising to eye level as you have claimed it does. I'm still not seeing a 3 foot wave blocking out more than a few feet of a building which is currently in the ocean.
Please keep in mind that many of these building observations made by Bobby were made with both the viewer and the building several yards above sea level.
(https://i.imgur.com/89I2UXq.jpg)

And how do you know that our perspective lines recede infinitely and continuously into the distance like the classical mathematical model?
According to the quantized paradigms of Quantum Mechanics, very few things in nature are infinite or continuous. In fact, the continuous universe time and space of the Ancient Greeks, where things are infinitely divisible without discrete units of measurement, has increasingly been shown to be false.

And how do you know that our perspective lines recede infinitely and continuously into the distance like the classical mathematical model?
Where in this thread did someone make the claim that perspective lines recede infinitely into the distance? Please try to keep things on topic.
What does this have to the claim that a viewer, standing on the shore with eyes 5 feet above sea level (not floating in the ocean) can have 100's of feet of a building (which is also 5 or more feet above sea level) obstructed by a 3 foot wave?
According to the quantized paradigms of Quantum Mechanics, very few things in nature are infinite or continuous. In fact, the continuous universe time and space of the Ancient Greeks, where things are infinitely divisible without discrete units of measurement, has increasingly been shown to be false.
Ok. What does this have to do with this topic again? We are trying to understand how a normal 34 foot wave could account for an entire skyscraper being obscured from view (without floating in the water and having the wave be very close to your face). The only way something small has been able to effectively obscure something much larger is if it's very close to your eye (like holding up a dime to obscure an elephant).

You are assuming a model of perspective which stretches into infinity continuously without meeting, rather than the finite perspective model observed where the lines appear to meet.
I am asking why you are waving your hands frantically and asserting that there is an unobservable stretch of infinity where the perspective lines meet. You are sounding like someone who is trying to convince others of the existence of ghosts.

You are assuming a model of perspective which stretches into infinity continuously without meeting, rather than the finite perspective model observed where the lines appear to meet.
No I'm not. I never made such an assertion or assumption. To clarify for you, because you seem so hung up on it.
I am officially assuming a model of perspective which does NOT stretch into infinity.
Furthermore such an assumption even if it was made (which it wasn't). What does that have to do with helping me understand how a 6 foot wave can obscure hundreds of feet of a building when both the viewer and building are 10 or more feet above sea level?
I am asking why you are waving your hands frantically and asserting that there is an unobservable stretch of infinity where the perspective lines meet. You are sounding like someone who is trying to convince others of the existence of ghosts.
I never asserted than there is an unobservable stretch of infinity. I'm asking you why you are waving your hands frantically saying that someone asserted some sort of model of perception.
What does this have to do with helping me understand how a 6 foot wave can obscure hundreds of feet of a building when both the viewer and building are 10 or more feet above sea level?
Please try to stay on topic here.

You are assuming a model of perspective which stretches into infinity continuously without meeting, rather than the finite perspective model observed where the lines appear to meet.
I am asking why you are waving your hands frantically and asserting that there is an unobservable stretch of infinity where the perspective lines meet. You are sounding like someone who is trying to convince others of the existence of ghosts.
Zorbakim is assuming the same model. It's the basis of his "side perspective" argument. Similar (but with a drastically different conclusion) to the ICanScienceThat argument that Tom rejected.
I don't want to stifle discussion, but I'd like to see Tom and Zorbakim hash this out. Seems like something FE theory needs to resolve.

You are assuming a model of perspective which stretches into infinity continuously without meeting, rather than the finite perspective model observed where the lines appear to meet.
Though they appear to meet, from a model of perspective looking out toward the endpoint of rails on the horizon, do the train tracks actually meet at a convergence or do they continue to maintain their gauge spread?

You are assuming a model of perspective which stretches into infinity continuously without meeting, rather than the finite perspective model observed where the lines appear to meet.
I am asking why you are waving your hands frantically and asserting that there is an unobservable stretch of infinity where the perspective lines meet. You are sounding like someone who is trying to convince others of the existence of ghosts.
What we're assuming is that if we can see a 1m wave which is, say, 5 miles away on the horizon then I can also see the SODDING GREAT BUILDING which is a few meters beyond it. I'd like to see you or someone else show a diagram demonstrating how a ray of light from a 100m building can be blocked by a 1m wave unless the wave is close and your eye is below 1m.
How perspective works in the real world has been explained to you multiple times. The size you perceive something depends entirely on the angle the light rays from either end of it meet at your eye. The light from either track at A meets at a bigger angle than the light from either track at B, thus you perceive A as being bigger than B even though in reality they are the same size:
(https://image.ibb.co/bHMkTx/Tracks.jpg)
That's it. At what distance does that angle become 0, given that you can see this is a triangle? At infinity.
BUT, we don't have to wait till infinity, in real life we can't perceive things if that angle is below a certain size.
That is where the rails would actually appear to meet at a finite distance simply because we could no longer distinguish them.
Optical zoom would then restore them so we can perceive them as separate rails again because zoom increases that angle again such that we can perceive things more clearly.
But, even without optical zoom, if there was a big building beyond where we can no longer distinguish the two rails then of course we could see it, it's a lot bigger than the distance between the rails, the light from the top and bottom meets at our eye at a much bigger angle than that of the two rails so we can clearly see it.
It's interesting how objects being occluded by the horizon (which in the real world is the edge of the earth) are explained by merging perspective lines or things on the horizon blocking them  ergo, it's not because the earth is curved. But then long distance photography is used as proof that the earth is flat because if the earth is curved then you shouldn't be able to see them. Well...if your model of perspective is correct then you shouldn't be able to see them either. You can't have it both ways...

That's it. At what distance does that angle become 0, given that you can see this is a triangle? At infinity.
That's the rub. No one has ever proven this continuous perspective model. Its part of the Continuous Universe model of the Ancient Greeks who imagined a perfect universe, and which is still used in Classical Mechanics. This Ancient Greek concept of a perfect universe assumes the following:
 That perfect circles can exist
 That one could zoom into a circle forever and see a curve
 That any length of space can be divided into infinitely smaller parts
 That the space can be infinitely long
 Time can likewise be infinitely divided, or infinitely long
 The Perspective Lines receded infinitely and continuously into the distance
Quantum Mechanics has increasingly discredited this Continuous Universe model of Classical Mechanics. We now know that reality is not continuous in this manner. Perfect circles do not exist  there is evidence that there is a fundamental unit of space. Time is likely not infinitely divisible, either. We do not live in this mathematical fantasy of the Ancient Greeks.

That's it. At what distance does that angle become 0, given that you can see this is a triangle? At infinity.
That's the rub. No one has ever proven this continuous perspective model. Its part of the Continuous Universe model of the Ancient Greeks who imagined a perfect universe, and which is still used in Classical Mechanics. This Ancient Greek concept of a perfect universe assumes the following:
 That perfect circles can exist
 That one could zoom into a circle forever and see a curve
 That any length of space can be divided into infinitely smaller parts
 That the space can be infinitely long
 Time can likewise be infinitely divided, or infinitely long
 The Perspective Lines receded infinitely and continuously into the distance
Quantum Mechanics has increasingly discredited this Continuous Universe model of Classical Mechanics. We now know that reality is not continuous in this manner. Perfect circles do not exist  there is evidence that there is a fundamental unit of space. Time is likely not infinitely divisible, either. We do not live in this mathematical fantasy of the Ancient Greeks.
Why would this matter if, say, I am measuring a bathroom cupboard at the shop? The ruler is of course discontinuous, and if you look at it through a microscope it will appear all jagged and nothing like a straight line. Likewise the cabinet. Yet I can measure it perfectly well. So the geometry of the Greeks (which is our geometry) works perfectly well, despite your qualifications.
Also, Greek geometry does not assume that "the space can be infinitely long". Quite the opposite*.
*See "Euclid and Infinity", S. T. Sanders, Mathematics News Letter, Vol. 4, No. 7 (May, 1930), pp. 1522

Why would this matter if, say, I am measuring a bathroom cupboard at the shop? The ruler is of course discontinuous, and if you look at it through a microscope it will appear all jagged and nothing like a straight line. Likewise the cabinet. Yet I can measure it perfectly well.
We cannot "measure perfectly well" that the little area where perspective lines meet is actually an infinite space where the lines continuously approach each other forever, or whether they simply meet.

Why would this matter if, say, I am measuring a bathroom cupboard at the shop? The ruler is of course discontinuous, and if you look at it through a microscope it will appear all jagged and nothing like a straight line. Likewise the cabinet. Yet I can measure it perfectly well.
We cannot "measure perfectly well" that the little area where perspective lines meet is actually an infinite space where the lines continuously approach each other forever, or whether they simply meet.
We have discussed this many times before. The perspective lines are on paper, and of course they do meet if continued.

I don't understand why continuous or quantized space matters.
I showed an example of 1m waves NOT obscuring 100m of something at or beyond the horizon.
https://www.youtube.com/watch?v=3RRoIWu5tpM
Doesn't the observation trump diagrams and rationalizations of how perspective should apply? It's not happening. 100m of that distant island are not being obscured by the swell (which was actually greater than 1m and on a high tide to boot.)

Doesn't the observation trump diagrams and rationalizations of how perspective should apply? It's not happening. 100m of that distant island are not being obscured by the swell (which was actually greater than 1m and on a high tide to boot.)
We have seen that sometimes bodies are hidden and that sometimes they are not hidden. Rowbotham reports that at various times the bodies he was observing were hidden and revealed. What are we to make of this?
This phenomenon, whatever it might be, is what is being discussed. In previous threads you had started we had attempted to calculate whether the sinking phenomenon matched with the globe earth predictions, and it did not.

We have seen that sometimes bodies are hidden and that sometimes they are not hidden. Rowbotham reports that at various times the bodies he was observing were hidden and revealed. What are we to make of this?
That it's not waves? I've never seen that body (island) hidden by waves. It's been hidden by fog banks and haze. That little spur becomes hidden when inferior mirage conditions are present.
I've seen waves and swells much larger than the 45' waves present on the day of that video. Never seen a single case of the horizon pulsing up even a little due to those waves to obscure even close to 100m of that island's elevation.
What are we to make of that? Doesn't that refute any rationalization of 1m at the horizon being able to obscure 100m bodies beyond the horizon? Doesn't it substantiate that 1m waves become small due to distance just like 100m bodies? Classical Greeks reasoning or not, I've yet to EVER see a 1m wave in the distance hide a 100m object. If the 1m wave is up close, sure. But not the reasoning proposed in the opening post or by Rowbotham. Doesn't happen.
This phenomenon, whatever it might be, is what is being discussed.
What I understand the topic to be isn't "phenomenon, whatever is," but specifically "waves."
In previous threads you had started we had attempted to calculate whether the sinking phenomenon matched with the globe earth predictions, and it did not.
Please don't distract from the topic. That's a separate issue (which you mischaracterize, btw but let's not get distracted.). This is about if waves are responsible for hiding of objects on the horizon that are 10 fold the size of the waves.

I don't understand why continuous or quantized space matters.
I showed an example of 1m waves NOT obscuring 100m of something at or beyond the horizon.
https://www.youtube.com/watch?v=3RRoIWu5tpM
Doesn't the observation trump diagrams and rationalizations of how perspective should apply? It's not happening. 100m of that distant island are not being obscured by the swell (which was actually greater than 1m and on a high tide to boot.)
Where is the horizon in your video?
How far is it from the horizon to the island?
It's not good visibility.
according to my calculation, the horizon is at 6~7 kilometers away at 2m eye level.
But in your video, the distance to the horizon and the island is unclear.
The island in your video doesn't look very far.
Then of course it will not be obscured.
My calculation is done under the assumption that visibility and resolution is good.
So a long distance is needed.
If the distance is close, anything will not be obscured.

Where is the horizon in your video?
I don't know where it is on a flat earth. On a globe, it was about 6 1/4 kms away or about 6 3/4 km factoring in standard atmospheric refraction.
How far is it from the horizon to the island?
Since the island is about 26 1/2 kilometers from that viewing spot, that means the island is about 19 3/4 km further than the visible horizon.
It's not good visibility.
Not bad, actually; but what does that matter as long as you can see the island and see if it's being blocked by waves?
according to my calculation, the horizon is at 6~7 kilometers away at 2m eye level.
I'd be interested how you calculated that for a flat earth.
But in your video, the distance to the horizon and the island is unclear.
I provided sufficient detail the first time I posted the video, but now I hope I've made it clear.
The island in your video doesn't look very far.
Then of course it will not be obscured.
20 kilometers beyond the horizon? How far does it need to be? On a flat plane, a 1m obstacle 6500m away forms an angle of 0.009°.
A 100m object must be 650,000m or 650km away to form that same angle. 400 miles!!!
On a globe (with standard atmosphere), a 100m object will be completely hidden by the curve of the earth from that viewing height when it's 45 kilometers away.
My calculation is done under the assumption that visibility and resolution is good.
So a long distance is needed.
If the distance is close, anything will not be obscured.
What's the distance needed if not 650 kilometers? Is it something realistic?

I'm more than willing to be wrong, but in this example, it seems the waves had to have been inordinately high, like tsunami high, to obscure 84.26 ft at an observation height of 12.1 ft.
(Turning Torso image)
When we had that Turning Torso discussion before, "waves" wasn't the flat earth rationale; it was optical compression.
Here is a view of North Coronado Island about 16 miles away from this spot (https://goo.gl/maps/TJHVpu9tUwT2) today, almost high tide with a 45' primary ground swell coming from the SSW.
https://www.youtube.com/watch?v=3RRoIWu5tpM
The island is about 425' at its peak (130m). That spur on the far left is 100 and 120' (3040m) high.
1 meter waves are not blocking 100 meters of that island.
You said that its peak is 130m high and the spur on the far left is 3040m high.
The pixels of island image in your video is as in the following.
The peak is 273 pixels and the spur is 31 pixels from horizon.
(https://blog.naver.com/zorbakim/221377931333)
Then the obscured height is about 20m.
If eye level is 2m and wave height is 1m and the horizon distance is about 7.4km
then the obscured height is about 78m.
But If eye level is 5m and wave height is 1m and the horizon distance is about 11.7km
then the obscured height is about 50m.
But if the horizon distance is farther away, the obscured height is even less.
In short, The obscured height depends on eye level and wave height and horizon distance.
Horizon distance depends on the resolution and ID curve accuracy.
Reflection of light should also be considered.
So it is complex visual phenomena.
More research is needed in the future.
Anyway your video can be fully explained on the flat earth.

You said that its peak is 130m high and the spur on the far left is 3040m high.
The pixels of island image in your video is as in the following.
The peak is 273 pixels and the spur is 31 pixels from horizon.
There's a little bit of inferior mirage so the horizon is appears slightly lower than it actually is; but only slightly so okay. Good enough.
Then the obscured height is about 20m.
I don't know how you did that. The refracted hidden calculation on a globe earth viewing from 3m at a distance of 26500m is 26.4m, so however it is you measured 20m as the obscured height, it under calculates by about 6m. Not bad, actually.
If eye level is 2m and wave height is 1m and the horizon distance is about 7.4km
then the obscured height is about 78m.
The eye level height was 3m, making the globe earth refracted horizon 6.7km and obscured height (another ~20km beyond the horizon) 26m.
How are you calculating horizon distance and obscured height on your flat earth?
But If eye level is 5m and wave height is 1m and the horizon distance is about 11.7km
then the obscured height is about 50m.
But if the horizon distance is farther away, the obscured height is even less.
Globe earth horizon distance and obscured heights are different from what you calculated for your flat earth, but the trend is the same. Increasing eye level height increases horizon distance and reduces obscured height. Wave height does not factor in the equation. I'd still like to know how you are arriving at your horizon distance and obscured height figures.
In short, The obscured height depends on eye level and wave height and horizon distance.
But how? The obscured height depends on eye level, which determines horizon distance. But where does wave height factor in?
In short, The obscured height depends on eye level and wave height and horizon distance.
Horizon distance depends on the resolution and ID curve accuracy.
But how? Are you saying the horizon distance changes with improved resolution? Can you zoom the horizon out to 26+ km from a viewing height of 3m? I dare anyone to show me that.
Wave height and resolution are essentially nonfactors in what is or is not hidden. I can take photos of that view from that vantage point at different zoom lengths and under varying wave conditions and it won't change. Visibility and atmospheric refractive conditions (like mirage or looming/sinking) will change what we see, but not waves on the horizon nor resolution. I can't zoom the hidden part of that island back into view, and in all my life I've never seen waves of any size hiding that island or even the spur on the south end. The proof of your claim is not in the pudding.
Reflection of light should also be considered.
So it is complex visual phenomena.
More research is needed in the future.
Anyway your video can be fully explained on the flat earth.
I'm not seeing it. No 1m wave on the horizon is obscuring anywhere close to 100m of that island. The wave heights are less today. If I go out and photograph/video the island from the same location, it won't appear any differently. There won't be any less obscured. Wave heights have so little to do with how much of that island is obscured it is undiscernable

We have seen that sometimes bodies are hidden and that sometimes they are not hidden. Rowbotham reports that at various times the bodies he was observing were hidden and revealed. What are we to make of this?
That it's not waves? I've never seen that body (island) hidden by waves. It's been hidden by fog banks and haze. That little spur becomes hidden when inferior mirage conditions are present.
I've seen waves and swells much larger than the 45' waves present on the day of that video. Never seen a single case of the horizon pulsing up even a little due to those waves to obscure even close to 100m of that island's elevation.
I agree with Bobby's confusion. I also REALLY struggle with understanding how waves or swells can explain some of the things that we have seen with objects being obscured behind "sea level" when both the observer and the object.
I understand refraction.
I understand chaotic atmospheric conditions.
I understand optics.
I don't understand this:
In this situation how could a wave or swell or whatever possibly be used to account for what is obscuring the building?
(https://i.imgur.com/CdQEu76.jpg)

I don't understand this:
In this situation how could a wave or swell or whatever possibly be used to account for what is obscuring the building?
(https://i.imgur.com/CdQEu76.jpg)
Quite. Not possible. I've provided similar diagrams.
This is also why explaining sunset by perspective doesn't work.
The idea is that perspective lines merge at a finite distance and you can't see beyond that, but then long distance photography is used to "prove" a flat earth. So why can you see those distant objects? Why can you see any of the island in Bobby's photos and video when it is 20km beyond the horizon?

I don't understand this:
In this situation how could a wave or swell or whatever possibly be used to account for what is obscuring the building?
(https://i.imgur.com/CdQEu76.jpg)
Quite. Not possible. I've provided similar diagrams.
This is also why explaining sunset by perspective doesn't work.
The idea is that perspective lines merge at a finite distance and you can't see beyond that, but then long distance photography is used to "prove" a flat earth. So why can you see those distant objects? Why can you see any of the island in Bobby's photos and video when it is 20km beyond the horizon?
The OP diagram does quite well to explain how a wave, at eye level (such as if you were floating in the ocean), which is closer to your eye could obstruct hundreds of feet from a far off building above sea level.
The problem that I see, quite clearly, is that with dozens and dozens of observations provided by Bobby these criteria simply don't hold true. Even if the horizon always rises to eye level as suggested previously the observations simply don't match.

Same island, but on a day with smaller waves, from a viewpoint of about 3' above low tide:
https://www.youtube.com/watch?v=9xCqCLF6i6c
I have no idea how to calculate distance to horizon or predict how much of a distant object is hidden based on a flat earth+perspective+wave height+resolution+et cetera.
With a globe curve calculator, wave height and resolution are not factors. But figuring on a spherical earth with a standard atmosphere, comparing images from viewing heights of 3' and 415', I think narrowing the height estimate of that island's southern spur to 110120' (3436m) is reasonable given that from 3' the spur is completely hidden and the earth curve refracted hidden value is 115'.
(http://oi66.tinypic.com/2ztdeu8.jpg)
(http://oi64.tinypic.com/2bqpg0.jpg)

Same island, but on a day with smaller waves, from a viewpoint of about 3' above low tide:
https://www.youtube.com/watch?v=9xCqCLF6i6c
This video shows well.
The horizon is not a clear mathematical line.
So it is not a clear line.
It is a complex visual line.
The horizon and waves block the sight.
It is difficult to explain the calculation process that I did here one by one.
Instead, I explained it briefly in the video.
It comes from experience and experimentation.
It is not just mathematical reasoning like Earth curvature.

The horizon is not a clear mathematical line.
So it is not a clear line.
It is a complex visual line.
The horizon and waves block the sight.
It is difficult to explain the calculation process that I did here one by one.
Instead, I explained it briefly in the video.
It comes from experience and experimentation.
It is not just mathematical reasoning like Earth curvature.
The horizon definitely blocks line of sight. We're just at odds over what (and where) that horizon is.
I also agree that a vertical "bump" (wave, hill) on the horizon will contribute to the blocking. And it will be more pronounced on a globe by a globe's horizon caused by curvature since objects beyond the curvature's horizon will be tilted and sloping away from the observer.
So, on a globe without that 1m "bump" from a viewing height of 1 meter, the horizon will be around 3900 meters away and a 100 meter tall object will be completely obscured around 43000 meters away. Add a 1 meter "bump" to the horizon at the 3900 meter mark and a 100m tall object only needs to be around 40000 meters away to be completely obscured. That 1m obstacle on the horizon adds another 0.15° of "shadow" to the already hiddenbycurvature horizon. It's just geometry (with the influence of atmospheric refraction).
What I don't see, however, is a 100m tall object that is beyond the horizon (whatever that is) and is fully or even half visible, then becoming fully obscured as waves pass along a horizon between me and the object. "My" island in the video is 26000 meters away. Whatever's hidden can be explained by curvature calculations and not waves, and the height of waves doesn't discernibly alter what is being hidden, even with 215mm of "zoom." A 1 meter wave at 3900m translates geometrically to 67m at the distance of the island. If there's some arcane reason why that vertical obscuration height should be amplified due to "ID curve" perspective, it's not manifesting itself in what I can observe and detect, even with resolution better than my naked eye.

My camera doesn't have a timelapse function, but testing out a hack on that target island feature that reveals distortions that you don't notice in real time. Not as extreme as the Skunk Bay video, but it's still fascinating (at least to me):
https://www.youtube.com/watch?v=b7l5F5R3Zvc
23 frames, 1 minute apart.
This is probably tangent to the topic since it doesn't really address the wavesonhorizonobscuringobjects claim, but since I've been using that island feature as an observation target, I thought it might be interesting and related since I figure I should do this from a vantage point at an elevation near sea level to see if ocean swells and/or waves are detectable. I'm guessing the distorting effects of the atmosphere will be more pronounced.

I guess that's from altitude, Bobby? In that footage it looks like the island is in front of the horizon? I guess because you're higher so the horizon is further away?

Yes. Sorry. Should have mentioned that. View from 360' (https://goo.gl/maps/NsNAUvn71JC2).

For zorbakim: Try this interactive Advanced Earth Curvature Calculator (http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator) by Walter Bislins which has a flat earth model.
Set it up at follows:
Basics Tab
Observer Height: 1m
Target Distance: 50000m
Target Size: 100m
Refraction: 0 (doesn't matter)
Zoom f: 2000mm
View∠ 1.23754° (linked to zoom)
Model: FE
Camera Aim: EyeLevel
Units: metric
View Tab
Height: 1m
Zoom f: 2000mm
View∠: 1.23754°
Pan: 0°
Tilt: 0°
Target 2
NObjects: 200
Distance 1000m
DeltaDist 0
SidePos 0
SideVar 40m
ObjSize 1m
SizeVar 0
ObjType MRod
SideVar Lin
SizeCar Lin
Target 1
NObjects: 1
Distance: 50000m
DeltaDist: 300
SidePos: 0m
SideVar: 0m
ObjSize: 100m
SizeVar: 0%
ObjType: MRod
SideVar: Lin
SizeVar: Lin
You should end up with an image like this: a "wall" of 1m objects in front of a more distant 100m object
(http://oi67.tinypic.com/1zlx4ro.jpg)
Now play with zoom in the view tab and distances in the respective target tabs and see if you can get the 1m wall to obscure the 100m rod.
It won't. Ever. Why? Is there something wrong with the parameter setup? Or is the model not applying perspective correctly as you think it should work? (Or something else?)
Because, if you switch the model option to GE (Globe Earth) it behaves as I see it in the real world. In the FE mode, perspective doesn't create a curveaway appearance and the 1m "wall" never obscures the 100m object.

It's so complicated that I don't know what it is.
(Advanced Earth Curvature Calculator by Walter Bislins)
On the other hand, my math is simple.
Truth is simple, not complicated.
Obviously, the effects of the waves exist on both model.
It will be much more obscured on the round earth.
That's because waves are added to the curvature.
But things are different on the flat earth.
Because it is a visual phenomenon.
Look at the horizon.
It is not a simple line drawn by pencil.
Is there a distinct line between water and air?
If so, that's what our eyes tell us.
Anyway, I have shown that the Flat earth can explain why the building is covered.
It is difficult to get the exact figures.
Visual phenomena are influenced by many factors.
That's the reality.

It's so complicated that I don't know what it is.
(Advanced Earth Curvature Calculator by Walter Bislins)
All you have to do is put in the numbers into the form fields. And then move the sliders and watch what happens. The user need not perform any mathematical calculations. The model does the work.
Try it.
There will be no combination of distances or zoom (resolution) at which point the 1m "wave" wall will obscure the 100m tower on a flat surface.

Truth is simple, not complicated.
There speaks a man who hasn't read up on Relativity or Quantum Theory :)
But in this case, it is pretty simple. On a flat plane a smaller object can only obscure a more distant, bigger object if your eye level is below the height of the smaller object.
Otherwise you are looking over the bigger object and will be able to see the distant one.
I have provided diagrams and photos which demonstrate this.

Anyway, I have shown that the Flat earth can explain why the building is covered.
It is difficult to get the exact figures.
Visual phenomena are influenced by many factors.
That's the reality.
I disagree strongly. Did you not look at my diagrams? The observer is standing 10 feet above sea level. The building that is being obscured is 10 feet above sea level. In both the horizon at eye level model and the flat horizon model there is not a situation where a 3 foot wave can obscure these things like you have claimed when both the observer and the obscured object are above sea level.
Can you help me understand by drawing a diagram of an observer 10 feet above sea level and a wave blocking the vision of something far away 10 feet above sea level?

Anyway, I have shown that the Flat earth can explain why the building is covered.
It is difficult to get the exact figures.
Visual phenomena are influenced by many factors.
That's the reality.
I disagree strongly. Did you not look at my diagrams? The observer is standing 10 feet above sea level. The building that is being obscured is 10 feet above sea level. In both the horizon at eye level model and the flat horizon model there is not a situation where a 3 foot wave can obscure these things like you have claimed when both the observer and the obscured object are above sea level.
Can you help me understand by drawing a diagram of an observer 10 feet above sea level and a wave blocking the vision of something far away 10 feet above sea level?
I can help you.
Don't forget that you are an observer.
Now you are seeing it in the eyes of a third party, not as an observer.
The sea rises at an eye level to the observer's eye.
But the sea never rise at an eye level to the third party's eye.

I can help you.
Don't forget that you are an observer.
Now you are seeing it in the eyes of a third party, not as an observer.
The sea rises at an eye level to the observer's eye.
But the sea never rise at an eye level to the third party's eye.
Yes, I am the observer.
Now I'm seeing through the eyes of a third party, standing off to my right? With a side view?
The sea rises up to my eye, the observer, not the other me, the one off to my right, but the sea doesn't rise to his eye, I mean my eye, the other me, the third party's eye.
All that aside, quite simply, show how your model explains this:
Example A: 3.7m wave required
Example F2: 17.6m wave required
(https://i.imgur.com/iy1lvOS.jpg)
For reference, here's what waves look like:
(https://i.imgur.com/dQKWOPp.jpg)

I can help you.
Don't forget that you are an observer.
Now you are seeing it in the eyes of a third party, not as an observer.
The sea rises at an eye level to the observer's eye.
But the sea never rise at an eye level to the third party's eye.
I don't understand. a person is standing 10 feet above sea level and sees far building which is also 10 feet above sea level.
The bottom part of that building is obscured to the person. Even if the horizon rises to the level of person it seems that a 3 foot wave would not block the view of the building.
Could you correct my diagram?
(https://i.imgur.com/CdQEu76.jpg)

I can help you.
Don't forget that you are an observer.
Now you are seeing it in the eyes of a third party, not as an observer.
The sea rises at an eye level to the observer's eye.
But the sea never rise at an eye level to the third party's eye.
I don't understand. a person is standing 10 feet above sea level and sees far building which is also 10 feet above sea level.
The bottom part of that building is obscured to the person. Even if the horizon rises to the level of person it seems that a 3 foot wave would not block the view of the building.
Could you correct my diagram?
(https://i.imgur.com/CdQEu76.jpg)
If the distance is short, you are right.
But if the distance is long, horizon is up to the eye level in your diagram.
and then, Add the waves to the horizon.

If the distance is short, you are right.
But if the distance is long, horizon is up to the eye level in your diagram.
and then, Add the waves to the horizon.
Can you please show a diagram? I'm still not understanding how the 1m wave, even if that wave was at your eye level, would block the 100m building just behind it.

This is zorbakim's image:
(https://i.imgur.com/xUPov5J.jpg)
But that's not reality. That triangle representing a 1m wave on the (flat earth) horizon is at eye level to a 1m high eye. It never rises above eye level to obscure anything more than 1m behind it...unless the earth curves and "dips" away. But if the earth is flat, that diagram is false.
Here, a model of flat earth shows that a 1m object can never obscure a more distant 100m object from a viewpoint of 1m, no matter where you place them.
https://www.youtube.com/watch?v=TwyqNs_u5Gs
And here is a real life example of 1m ocean swells not obscuring 100m of an object well beyond the 1m view horizon:
https://www.youtube.com/watch?v=3RRoIWu5tpM
Those are two zetetic refutations of a rationalized  a flawed one at that  argument for how something can happen that doesn't. It's fanciful thinking, like this nonsense (https://www.youtube.com/watch?v=jMhia62aJ_k).

If the distance is short, you are right.
But if the distance is long, horizon is up to the eye level in your diagram.
and then, Add the waves to the horizon.
Can you please show a diagram? I'm still not understanding how the 1m wave, even if that wave was at your eye level, would block the 100m building just behind it.
In the diagram below someone is floating in the ocean and a wave comes crashing toward them. This wave could obscure an entire building if it was close enough to the eye. (much like cell phone when close enough to my eyes can obscure an elephant)
The problem is that this does not apply to literally ANY of Bobby's observations because none of them were made while he was floating in the ocean.
(https://i.imgur.com/cHNz8iW.jpg)

If the distance is short, you are right.
But if the distance is long, horizon is up to the eye level in your diagram.
and then, Add the waves to the horizon.
Can you please show a diagram? I'm still not understanding how the 1m wave, even if that wave was at your eye level, would block the 100m building just behind it.
In the diagram below someone is floating in the ocean and a wave comes crashing toward them. This wave could obscure an entire building if it was close enough to the eye. (much like cell phone when close enough to my eyes can obscure an elephant)
The problem is that this does not apply to literally ANY of Bobby's observations because none of them were made while he was floating in the ocean.
(https://i.imgur.com/cHNz8iW.jpg)
Exactly... that diagram is possible and shows how a large object can be blocked by a smaller wave IF your face is below the wave. It is common sense. Why a wave below your line of sight could ever block more then the height of the wave on an object behind it.... ever... makes no sense at all.

Exactly... that diagram is possible and shows how a large object can be blocked by a smaller wave IF your face is below the wave. It is common sense. Why a wave below your line of sight could ever block more then the height of the wave on an object behind it.... ever... makes no sense at all.
Well there is the claim that horizon is always at eye level or always rises to eye level. Which is where the original claim was made.
So if the horizon rises to eye level and the observer and the building are both at sea level then this is what i imagine. Even putting both the observer and the building at sea level, and raising the horizon to eye level it still does not add up to me.
(https://i.imgur.com/89I2UXq.jpg)

Imagine a complete flat plane.
The flat plane almost rises up to eye level in the distant.
But there are no obstacles on the flat plane, so it doesn't cover the building.
However, there are many kinds of waves in the sea.
wind wave, swell, tide etc.
These serve as obstacles on the flat sea.
So they can cover the building.

Imagine a complete flat plane.
The flat plane almost rises up to eye level in the distant.
But there are no obstacles on the flat plane, so it doesn't cover the building.
However, there are many kinds of waves in the sea.
wind wave, swell, tide etc.
These serve as obstacles on the flat sea.
So they can cover the building.
Can you please provide a diagram showing how the light rays from the top of a 100m building are blocked by 1m waves when the observer's eye level is above 1m.
I honestly can't picture what you're trying to describe.

Imagine a complete flat plane.
The flat plane almost rises up to eye level in the distant.
But there are no obstacles on the flat plane, so it doesn't cover the building.
However, there are many kinds of waves in the sea.
wind wave, swell, tide etc.
These serve as obstacles on the flat sea.
So they can cover the building.
If your eye level is at the same level of something else, that something else will never block line of sight to yet something else that's above eye level as long all those somethings reside on a flat plane. Never. Doesn't matter where you place them. You need light to bend upward or the surface to curve down (convex) in order for a 1m "wave" to obscure anything above 1m when viewed from a height of 1m.
https://www.youtube.com/watch?v=TwyqNs_u5Gs

Can you please show a diagram? I'm still not understanding how the 1m wave, even if that wave was at your eye level, would block the 100m building just behind it.
I'm pretty sure I got this figured out. The first premise is that the horizon is always at eye level. I'm going to operate under the assumption that he is adhering to the claim that the horizon always rises to eye level. The horizon rises to eye level the, at the point where the horizon has reached eye level there is a lone wave which is capable of obscuring more of the building than the height of the wave..
Here's an example (with observer and building both at sea level) with a wave on the horizon with the horizon rising to eye level with the building far beyond the horizon.
This helps me understand the original post but still baffles me as to what is happening when Bobby has documented this when both the camera/eye and the building are many feet above sea level.
(https://i.imgur.com/xvYYyeA.jpg)
One thing I noticed is that, even with no wave, part of the building is obscured. Leading me to believe that if something other than optics/environmental things were to blame for part of the building being obscured it would be the horizon and not waves.
(https://i.imgur.com/JYoVVtr.jpg)
Bobby,
This is something that you might be interested in. Inside of the claim that the horizon always raises to eye level who's to say that "raising to eye level" is linear.
(https://i.imgur.com/xIZPr0g.jpg)

Imagine a complete flat plane.
The flat plane almost rises up to eye level in the distant.
But there are no obstacles on the flat plane, so it doesn't cover the building.
However, there are many kinds of waves in the sea.
wind wave, swell, tide etc.
These serve as obstacles on the flat sea.
So they can cover the building.
If your eye level is at the same level of something else, that something else will never block line of sight to yet something else that's above eye level as long all those somethings reside on a flat plane. Never. Doesn't matter where you place them. You need light to bend upward or the surface to curve down (convex) in order for a 1m "wave" to obscure anything above 1m when viewed from a height of 1m.
https://www.youtube.com/watch?v=TwyqNs_u5Gs
That video is different from the ocean situation.
The sea is not an ideal plane.
Imagine an ideal flat plane.
The plane rises to an eye level at a distance.
Imagine adding waves to that.
It's not the same, but the sea is a similar situation.

Imagine an ideal flat plane.
The plane rises to an eye level at a distance.
Imagine adding waves to that.
It's not the same, but the sea is a similar situation.
OK. So first, a plane does not rise to eye level at a distance
You will always be looking down at an angle because no matter how far you can see, the angle theta in this diagram is not 0
(https://image.ibb.co/fxUSB7/horizon.jpg)
Only at infinity would the angle be zero. Secondly, is the situation you're talking about the one shown in iamcpc's diagram?
(https://i.imgur.com/xvYYyeA.jpg)
If so then you have a problem. Even if the 1m wave did rise above your eye line then it will do so at a very shallow angle.
If the building was twice as far as the horizon then because of similar triangles the wave would only obscure 2m of it.
If that diagram is not what you mean then if you could supply one then it would help us all understand your ideas.

Only at infinity would the angle be zero.
Who proved this infinity? Claiming that an infinity exists in something without evidence sounds like a pretty stupid claim to me. Much like an unsubstantiated claim promoting the existence of ghosts. Totally undetected and unobserved.

That video is different from the ocean situation.
The sea is not an ideal plane.
Imagine an ideal flat plane.
The plane rises to an eye level at a distance.
Imagine adding waves to that.
It's not the same, but the sea is a similar situation.
No. The smaller 1m obstacle is the wave. If your eye is at 1m and the wave top is at 1m, it can never be above eye level unless you lower your eye. The wave/obstacle top doesn't rise above eye level as it is moved into the distance. You can't "add the wave" to the eye level horizon. It doesn't work. Even another acclaimed FEer said:
"Find an obstacle of a known height, set it a distance away, and then set the height of your [eye] to the height of the top of the obstacle. [Your eye], object, and horizon should make a straight line. The further the objects are located from each other, the better. The horizon should line up with to top of the object..."
The 1m wave is the obstacle of known height. If the height of your eye is 1m, then the horizon is along the straight line formed by your eye and the top of the wave. If the wave is on the horizon, then it's merged into the vanishing point of the horizon. It's not "added to the horizon." For that to happen, the wave would have to defy the socalled Natural Laws of Perspective.

Claiming that an infinity exists in something without evidence sounds like a pretty stupid claim to me.
Not as stupid as claiming that parallel lines meet at a finite distance as you are effectively doing
Hint: look up the definition of parallel.
Note: I mean actually meet, not "appear to".

Bobby,
This is something that you might be interested in. Inside of the claim that the horizon always raises to eye level who's to say that "raising to eye level" is linear.
Yeah, I picked up on that. It doesn't change my opposition to what zorbakim is promoting, but I recognize that he's applying inverse proportionality of apparent size to distance, and that that is what he says creates our perception of depth. And he does say that as a result, "sideview" perspective reveals the rising plane of the earth presents as a curve. I've redrawn it in your graphic:
(http://oi65.tinypic.com/2ex0xfs.jpg)
And he conveys the concept like this, although he says the horizon only "nearly" rises to eye level (which, I think, is his acknowledgement of infinity in the analysis, in contradiction to Tom Bishop's version):
(http://oi64.tinypic.com/2lk8hoh.jpg)
but Zorbakim's point being that the inverse ratio of apparent size vs distance creates what he says is a sloped rise of the ground plane from a forwardlooking perspective and not the linear slope we, including Rowbotham in "Earth Not a Globe" depict it:
(http://oi68.tinypic.com/mb3lv8.jpg)
I take issue with that analysis, but it is irrelevant to the 1m wave on the horizon argument that is the prime focus of this topic. Whether it's sloped or linear, whatever is at eye level remains at eye level, UNLESS either light is "bendy" or the earth surface slopes away. Perspective doesn't cause something at eye level closer to the horizon to ascend above eye level at or beyond the horizon.
Even if I add the supposed curve and concede, for the sake of argument, that the horizon does rise to eye level, the summit of a 1m obstacle will never "add to" the level of the horizon. It will simply diminish in apparent size with it's peak never elevating to obscure anything. It's lost to the vanishing line of the horizon.
(http://oi67.tinypic.com/sop8pt.jpg)
Similarly, the more distant 100m object (in this graphic: a building)? The most of it that could ever be obscured by a 1m object on a flat surface from a vantage point of 1m is 1m of it's 100m height.
You certainly can  and do  lose resolution with distance so that you can't distinguish 1m from from 10m or 50m. Increasing resolution through telescopic magnification is how many claim they are able to bring objects lost to the horizon back into view. But if 1m is obscured by a 1m obstacle on the horizon, it is never brought back into view by a telescope, at least not if the earth is flat.
On a globe, with an atmosphere, there is "bendy light" of a sort that is less explicable if on a flat earth with an atmoplane. And it's "bendy" in the opposite direction (usually) from the "bendy light" of EAT. So being able to see something that should be obscured were it not for atmoSPHERE is entirely possible. Perhaps not to the point of making the earth appear flat but a least somewhat less convex than an earth with a radius of 3959 miles.
But I digress. Zorbakim's pitch that his sideview perspective, which is markedly different from (and even, in parts, contradictory to) Rowbotham's explanation of perspective, explains how a 1m wave on the horizon can obscure anything beyond the horizon that is much higher than 1m is fatally flawed. Somehow, his 1m wave has elevated above the 1m eye level and defies the inverseproportion concept that zorbakim, himself, explains as if it's a revelation. Perspective can't do that. It doesn't do that.
On a flat plane (without EAT), "what happens at eye level stays at eye level" (to borrow a slogan).
By comparison, on a convex surface, the story is different and what would be at eye level at one distance drops away below eye level at greater distances. On a convex surface, even with the horizon not at eye level, 1m distant objects CAN obscure more distant objects of greater height. But that's academic because the greater factor to hiding distant objects is the curvature and not 1m obstacles on the horizon.

And while I'm pontificating, I'd like to add that I still have difficulty with understanding how distance to a horizon is calculated on a flat earth, whether it be zorbakim's version or Rowbotham's. If resolution is a factor, then the horizon with a 200m telephoto lens should be, what, 4x more distance than with my naked eye? I would love to (again) invite zorbakim to explain the distance to the horizon calculation methodology. To date, the best he's offered is that it's complex and takes experience.

The sea is not an ideal plane.
I hope this video would help.
https://www.youtube.com/watch?v=CyFmkJVcOe8

The sea is not an ideal plane.
I hope this video would help.
It doesn't.
I have several questions. Here's one.
Why don't the waves get smaller with distance?
(http://oi66.tinypic.com/wjev6b.jpg)

The sea is not an ideal plane.
I hope this video would help.
In your video, why doesn't perspective apply to the waves?
(https://i.imgur.com/3H3gNWR.png)
Bottom right is not a 1 meter wave, it's a tsunami. Of the 2012 caliber.

The sea is not an ideal plane.
I hope this video would help.
https://www.youtube.com/watch?v=CyFmkJVcOe8
No, why would a wave grow in size to be as big as the building? The wave never gets taller then 1 meter. That 1 meter high wave loses visual resolution at a distance just like the building does. The building is still 99 meters taller then that 1 meter wave. This would be even more apparent on a flat earth.
Conflating a proven incorrect theory about the horizon "rising" to eye level with perspective and resolution loss of distant objects does not work to explain the appearance of curvature and hidden objects. Not in theory, or by observation.

The picture of wave is just for convenience and for visual effect.
The calculation was done according to perspective.

The picture of wave is just for convenience and for visual effect.
The calculation was done according to perspective.
Are you saying that perspective has an opposite effect on buildings and waves, that buildings shrink while waves enlarge?
Such an effect has never been observed. How would perspective differentiate between the two?
Can you provide the calculation?

The sea is not an ideal plane.
I hope this video would help.
It doesn't.
I have several questions. Here's one.
Why don't the waves get smaller with distance?
(http://oi66.tinypic.com/wjev6b.jpg)
Yes. I think you have found the tiny flaw in the argument.
There is a school of thought which says that in the visual diagram if the building gets smaller because of distance then the waves should too.
This is the stupidest thing I ever saw.

Yeah, I picked up on that. It doesn't change my opposition to what zorbakim is promoting, but I recognize that he's applying inverse proportionality of apparent size to distance, and that that is what he says creates our perception of depth. And he does say that as a result, "sideview" perspective reveals the rising plane of the earth presents as a curve. I've redrawn it in your graphic:
(http://oi65.tinypic.com/2ex0xfs.jpg)
The first big problem with that concept is that, based on the distance from the observer, that wave is WAY more than 3 feet tall. A 3 foot wave that far in the distance would still be negligible in obscuring the building.
The second big problem is that, in this situation, both the observer and the building are at sea level. I still really struggle with how I could make this diagram (which is still very difficult for me to comprehend) when both the observer and the building are 30 feet above sea level.