[stanlee]

is limited by visibility of atmosphere

[jimbob]

is limited by visibility of atmosphere

Just parts of it? The bottoms of the towers not the top. With a clear dividing line which RE call the horizon?

[Treep Ravisarras]

This however doesn’t work because it is the fall from the surface of the circle.
You only need to raise the observation point slightly and this goes way out.

Eg Horizon at zero elevation is zero, raise it by 10cm and it become 1.1km
The height a person looks at is 1.6m, this equates to a distance of 4.4km this makes a massive difference to the calculation.

I propose to do as follows. (I use sin calculation with unit circle, but result is the same.)

We use 90° point top of the earth where the circle proceeds horizontal briefly as our "horizon". The observer is to the right of it, the target to the left. Makes easy.



"horizon" to target


observer to "horizon". In picture horizon is left at top of circle


What you say?

You can check with your circle formula or with your website that has round earth assumptions.

p.s. now using exact distance along circle, rather than "cosine" distance which technically is not completely accurate, but that is beside the point we trying making

[Macarios]

This however doesn’t work because it is the fall from the surface of the circle.
You only need to raise the observation point slightly and this goes way out.

Eg Horizon at zero elevation is zero, raise it by 10cm and it become 1.1km
The height a person looks at is 1.6m, this equates to a distance of 4.4km this makes a massive difference to the calculation.

I propose to do as follows. (I use sin calculation with unit circle, but result is the same.)

We use 90° point top of the earth where the circle proceeds horizontal briefly as our "horizon". The observer is to the right of it, the target to the left. Makes easy.



"horizon" to target


observer to "horizon". In picture horizon is left at top of circle


What you say?

You can check with your circle formula or with your website that has round earth assumptions.

p.s. now using exact distance along circle, rather than "cosine" distance which technically is not completely accurate, but that is beside the point we trying making

On your first image I see airplane at the height of approximately 10 000 m.
If we take standard Boeing 737-700 that is 12 m high, we can calculate vanishing point.
Wings are wider than that, tail is higher than that, but they are thin and they will vanish much earlier.
Anyway, we can be generous and calculate this as if the plane is 16 meters big.

Observer's eyes at 1.6 meters high will have horizon at SQRT(1.6*(1.6 + 2*6371000)) = 4515.22 m (4.5 km).
With standard refraction horizon will be at SQRT(1.6*(1.6 + 2*7432833)) = 4877 m (4.9 km)

If your target is at 47.9 km, then the distance from horizon to target will be 47900 - 4515.22 = 43384.78 m
With standard refraction target will be behind horizon by 47900 - 4877 = 43023 m

Hidden part of the target will be SQRT(43384.78² + 6371000²) - 6371000 = 147.7 m
With standard refraction hidden part will be SQRT(43023² + 7432833²) - 7432833 = 124.5 m

Drop from observer's local horizontal will be SQRT(6371000² + 47900²) - 6371000 = 180.1 m
With standard refraction it won't change.
Refraction can bend our view, but not local horizontal line.

Now, about the airplane:
To reach vanishing point it has to reach angular size of about one arc minute (0.0167 degrees).
For that it has to be from observer's eye at (16/2) / tan(0.0167/2) = 54894 m
Ground distance for that will be SQRT(54894² - 10 000²) = 53975 m

At 53975 m distance behind horizon will be 53975 - 4515.22 = 49459.78 m
With standard refraction it will be 53975 - 4877 = 49098 m

Hidden part of those 10 000 m will be SQRT(49459.78² + 6371000²) - 6371000 = 192 m and the plane will be 9808 m above line of sight.
With standard refraction it will be SQRT(49098² + 7432833²) - 7432833 = 162 m and the plane will be 9838 m above line of sight.

It means our "generously" 16 meters big plane will vanish at ARCTAN(9808 / 53975) = 10.3 degrees above horizon.
With standard refraction plane will vanish at ARCTAN(9838 / 53975) = 10.33 degrees above horizon.

Ofcourse, 12 m standard Boeing 737-700 will vanish earlier, at (12/2) / tan(0.0167/2) = 41 170.5 m from eye, which is ground distance of 39 937.6 m.
Vanishing elevation will be roughhly 13.92 degrees above horizon.
(13.94 degrees with standard refraction.)

Was that your question about ?

[Treep Ravisarras]

On your first image I see airplane at the height of approximately 10 000 m.

Yes, correct, at 39,000 feet actually the maximum airliner flying height thereabouts.

If we take standard Boeing 737-700 that is 12 m high, we can calculate vanishing point.
Wings are wider than that, tail is higher than that, but they are thin and they will vanish much earlier.
Anyway, we can be generous and calculate this as if the plane is 16 meters big.

Plane in image is actually 4km big as drawn in picture Of course plane icon not to scale, but position is.

Here is what looks like with plane at 39,000ft. Plane now 2km long , but position and everything else to scale, round earth assumptions. Blue circle of 6371km Radius. Left and right to about 100km in image. How far can you actually see? I've never seen 100km from plane.

Flat Earth:


Round Earth:

Was that your question about ?

Yes, I'm not sure I follow your mathematics but they sound impressive. What exactly does mean what your final answer is? I think you arrive at same answer? Observer of 1.60m height is 4.5km from horizon and tower is 43.4km from horizon, total distance 47.9km. Like so:


Or maybe explain a bit more. Mathematics I like but it is also good to make visual.

This is the observer: zoomed in of course


blue = earth
red = distance to target
black dotted = line of sight

[Macarios]

Here is what looks like with plane at 39,000ft. Plane now 2km long , but position and everything else to scale, round earth assumptions. Blue circle of 6371km Radius. Left and right to about 100km in image. How far can you actually see? I've never seen 100km from plane.

If you are at 800 meters high, you will see SQRT(800*(800+2*6371000)) = 101 km away.
From airplane at 10 000 meters, your horizon will be at SQRT(10 000*(10 000 + 2*6371000)) = 357.1 km.

Open this link, wait for the diagram to load and see the difference between "drop" and "hidden".
Also the difference between "surface level", "eye level", and "line of sight" (unmarked line from eye to horizon).
https://www.metabunk.org/curve/

Flat Earth:


Round Earth:

What is the reason for Flat Earth to limit visibility here?
We know clear air won't block the view, because you can, for example, see Aconcagua from Valparaiso, 160 km away.

If the Earth was flat, your eye level will still be 1.6 m above the surface, and objects bigger than 14 meters will still be visible for you.
(13.9 meters big objects will reach the resolution of your eye at about one arc minute, which is 0.0167 degrees.)
Object 50 meters in size you will see at distance of 50 / tan(0.0167) = 171 544 m = 171.5 km.
Ofcourse, not through city air, and not objects in sky blue color.

Was that your question about ?

Yes, I'm not sure I follow your mathematics but they sound impressive. What exactly does mean what your final answer is? I think you arrive at same answer? Observer of 1.60m height is 4.5km from horizon and tower is 43.4km from horizon, total distance 47.9km. Like so:


Or maybe explain a bit more. Mathematics I like but it is also good to make visual.

(If the visual from the link above is not enough, then please accept my appologies, and let me know, I might try to draw something.)

I'm not sure what was the original question.
And this time I will skip standard refraction for the sake of simplicity.

In celestial navigation one of the most important parameters is Apparent Horizon Dip.
It changes with altitude of navigator and influences the measured height of the observed celestial body above horizon.
Apparent Horizon Dip is angle for which the line of sight towards horizon drops from the local horizontal line.

For example:
If you are 100 meters high, your horizon dip will be ARCTAN(100 / horizon-distance).
From 100 meters horizon-distance is SQRT(100*(100+2*6371000)) = 35696.1 m, so the horizon dip will be ARCTAN(100 / 35696.1) = 0.16 degrees (9.6 arc minutes).
Your horizon will be for 0.16 degrees below your local horizontal.
If your sextant then shows you Arcturus at 12 degrees 14.4 minutes above horizon, and horizon is 9.6 minutes below horizontal, then Arcturus is at elevation of 12 degrees 4.8 minutes.

With eye at 1.6 m horizon dip will be 0.041 degrees (2.46 minutes).
At the distance of 47.9 km drop below your eye level will be 180.1 m , and drop below your line of sight will correctly be those 148 m (147.7).
You will be able to see top of an object taler than 148 m.


So, please tell me, what was the original question.
I was reading the thread again, but haven't found it.

Thanks.

[xenotolerance]

Macarios is right, as usual

Atmospheric perspective is a totally real thing, and it affects visibility without influence from the shape of the Earth. E.g. yesterday was pretty foggy where I live, and I couldn't see the World Trade Center from 6th & 8th. On a clear day with low humidity and clean air, atmospheric perspective has a negligible presence: With a decent telescope you can see the very top of WTC1 from Philly.

If the Earth were flat, there would be no such thing as a horizon. We see a clear horizon line strictly because we can see clearly all the way until the curvature of the Earth hides distant lands. Boats and skyscrapers do not gradually get hazy and disappear, they remain visible and disappear from the bottom up. Without curvature, the only thing that keeps us from seeing literally everything is atmospheric perspective; which involves getting bluer and hazier with distance, and never a clear horizon.

[jimbob]

This however doesn’t work because it is the fall from the surface of the circle.
You only need to raise the observation point slightly and this goes way out.

Eg Horizon at zero elevation is zero, raise it by 10cm and it become 1.1km
The height a person looks at is 1.6m, this equates to a distance of 4.4km this makes a massive difference to the calculation.

I propose to do as follows. (I use sin calculation with unit circle, but result is the same.)

We use 90° point top of the earth where the circle proceeds horizontal briefly as our "horizon". The observer is to the right of it, the target to the left. Makes easy.



"horizon" to target


observer to "horizon". In picture horizon is left at top of circle


What you say?

You can check with your circle formula or with your website that has round earth assumptions.

p.s. now using exact distance along circle, rather than "cosine" distance which technically is not completely accurate, but that is beside the point we trying making

Here is the equation I worked out,  calculating the "Fall" ie part of object not visible considering height of observer and distance to object ie tower. It is general geometry so it works with any ball shape and observer.
https://pasteboard.co/Her0ToJ.jpg

[Macarios]

Here is the equation I worked out,  calculating the "Fall" ie part of object not visible considering height of observer and distance to object ie tower. It is general geometry so it works with any ball shape and observer.
https://pasteboard.co/Her0ToJ.jpg

You can use BB code like this:
(Instead of "url" tag, you can use "img" tag, with or without "width" parameter.)

Code: [Select]

[img]https://pasteboard.co/HewbXAj.jpg[/img]

Code: [Select]

[img width=400]http://i68.tinypic.com/8z4if4.jpg[/img]
If you add "width" parameter, then click on "Preview" to see if the value has to be adjusted.

I tried with "PasteBoard" and it didn't work.
So I used "TinyPic" host at the http://tinypic.com/

And the result is this:
(C is circumference of Earth, mean value would be 40 030 km.
R is radius of Earth, 6371 km.)


Hope it helped.

For the circle I used brim of small glass.

[Treep Ravisarras]

Yes that same formula like I used, sinus cosinus.

What is the reason for Flat Earth to limit visibility here?
We know clear air won't block the view, because you can, for example, see Aconcagua from Valparaiso, 160 km away.

No reason really, I don't think I have ever seen farther than 100km, but if you say you did, the simple mathematics of a Blue circle of 6371km Radius. tells us the following:

Now to about 200km left and right:

Flat Earth:


Round Earth:


How can we tell the difference between flat earth and round earth from an airplane?

Also I do not understand what you mean "horizon dip"?

[Macarios]

Yes that same formula like I used, sinus cosinus.

What is the reason for Flat Earth to limit visibility here?
We know clear air won't block the view, because you can, for example, see Aconcagua from Valparaiso, 160 km away.

No reason really, I don't think I have ever seen farther than 100km, but if you say you did, the simple mathematics of a Blue circle of 6371km Radius. tells us the following:

Now to about 200km left and right:

Flat Earth:


Round Earth:


How can we tell the difference between flat earth and round earth from an airplane?

Also I do not understand what you mean "horizon dip"?

From 1.6 meters above the ground you can tell the difference by observing object of 50 meters in size from a distance.
Let it be bright orange object, clearly visible in blue-ish air.

If the Earth is flat, the object will disappear at vanishing point (point where object's angular size gets as small as one arc minute.
Distance where it happens is 50 / tan(0.0167) = 171 km.
People with somewhat lower vision index will surely see the object at 100 km.
Going higher or lower the angular size of the object won't change and we won't be able to see it without zoom.

If the Earth is globe, the object will hide behind the bulge, and it will happen at 30 km (with standard refraction at 33 km).
At 22 km you will see upper half of the object and lower half at the same distance will remain hidden.
Going higher or lower, withiut any zoom, will change the amount of the object we will be able to see.

Fom an airplane at 10 000 meters we will see it by the distance to horizon.
If visibility is limited by perspective or atmospheric conditions, the conditions won't change (much) when we go higher.
If visibility is limited by Earth's curve, when we go higher, we will simply see farther over the bulge.

---------------

"Horizon dip" (or "angle of depression") is the angle for which line of sight towards horizon drops from your local horizontal, depending on the altitude you are looking from.
If your eye is at the ground level, you will have horizon at the eye level.
When you go higher, your horizon gets lower than horizontal by some small angle.
People on foot, on horse, or in small boats can't see that angle by naked eye.
To have horizon dip of 0.5 degrees you have to be 250 meters high.

If you look from 1.6 meters above sea level, your horizon will drop from horizontal line by 0.041 degrees.
If you look from 16 meters above sea level, your horizon will drop from horizontal line by 0.128 degrees.
If you look from 160 meters above sea level, your horizon will drop from horizontal line by 0.406 degrees.
From airplane at 10 000 meters, horizon will drop by 3.2 degrees.

Al-Biruni, when measuring the Earth's size 100 years ago, used horizon dip from known height.
First he measured the height of his observing point from sea level (using two angles and distance), then he measured horizon dip from that observing point.
And he got the measure with error of less than 1%.

Here (scroll down), horizon dip is marked yelow and named Alpha.
https://owlcation.com/stem/How-to-Determin-the-Radius-of-the-Earth-Al-Birunis-Classic-Experiment

[Treep Ravisarras]

If you look from 1.6 meters above sea level, your horizon will drop from horizontal line by 0.041 degrees.
If you look from 16 meters above sea level, your horizon will drop from horizontal line by 0.128 degrees.
If you look from 160 meters above sea level, your horizon will drop from horizontal line by 0.406 degrees.
From airplane at 10 000 meters, horizon will drop by 3.2 degrees.

Ah yes, that exactly what my calculation based on Round Earth assumption Blue circle of 6371km Radius comes at also. If you input Observer Height (h) in here: α (alpha) observertohorizon becomes exactly your numbers. Got it.

So "horizon dip" is α (alpha) observertohorizon.

My α (alpha) really is from origin, green line, but ends up the same because my line of sight is always horizontal (clever trick makes mathematics simpler). "top of bulge" in your words is always top of circle, place where round earth gets in the way of the line of sight. This case at 4.5km. (we then call this "horizon of 1.6m observer")

Here is the equation I worked out,  calculating the "Fall" ie part of object not visible considering height of observer and distance to object ie tower. It is general geometry so it works with any ball shape and observer.
https://pasteboard.co/Her0ToJ.jpg

Impressive mathematics. Different way of calculating than me, using circle equation rather than sine cosine. Do you get same result as me though? Should be.

If the Earth is globe, the object will hide behind the bulge, and it will happen at 30 km (with standard refraction at 33 km).
At 22 km you will see upper half of the object and lower half at the same distance will remain hidden.
Going higher or lower, withiut any zoom, will change the amount of the object we will be able to see.

That my calculation got as well. At 30km distance, 51 metre height is hidden.



But what you mean by 'standard refraction'? Is there simple formula for that theory?

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible. Video shows too much, and means earth is flatter than round earth assumptions.

[jimbob]

If you look from 1.6 meters above sea level, your horizon will drop from horizontal line by 0.041 degrees.
If you look from 16 meters above sea level, your horizon will drop from horizontal line by 0.128 degrees.
If you look from 160 meters above sea level, your horizon will drop from horizontal line by 0.406 degrees.
From airplane at 10 000 meters, horizon will drop by 3.2 degrees.

Ah yes, that exactly what my calculation based on Round Earth assumption Blue circle of 6371km Radius comes at also. If you input Observer Height (h) in here: α (alpha) observertohorizon becomes exactly your numbers. Got it.

So "horizon dip" is α (alpha) observertohorizon.

My α (alpha) really is from origin, green line, but ends up the same because my line of sight is always horizontal (clever trick makes mathematics simpler). "top of bulge" in your words is always top of circle, place where round earth gets in the way of the line of sight. This case at 4.5km. (we then call this "horizon of 1.6m observer")

Here is the equation I worked out,  calculating the "Fall" ie part of object not visible considering height of observer and distance to object ie tower. It is general geometry so it works with any ball shape and observer.
https://pasteboard.co/Her0ToJ.jpg

Impressive mathematics. Different way of calculating than me, using circle equation rather than sine cosine. Do you get same result as me though? Should be.

If the Earth is globe, the object will hide behind the bulge, and it will happen at 30 km (with standard refraction at 33 km).
At 22 km you will see upper half of the object and lower half at the same distance will remain hidden.
Going higher or lower, withiut any zoom, will change the amount of the object we will be able to see.

That my calculation got as well. At 30km distance, 51 metre height is hidden.



But what you mean by 'standard refraction'? Is there simple formula for that theory?

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible. Video shows too much, and means earth is flatter than round earth assumptions.

Here is the data worked out from the equation and calculated from the video
https://pasteboard.co/HetXwJD.jpg

As you can see there is little difference between theoretical and actual.

If you paste this formula into the calculator, you can check the calculated data

Calculator
https://www.desmos.com/calculator

Formula
y=6371-\sqrt{\left(\left(6371\right)^2-\left(47.9-\sqrt{\left(.002+6371\right)^2-\left(6371\right)^2}\right)^2\right)}

[Macarios]

But what you mean by 'standard refraction'? Is there simple formula for that theory?

Standard refraction is average atmospheric refraction for common conditions, like average temperature distribution in air layers above average temperature of the ground/sea.

Obvious cases of superior or inferior mirage are caused by extreme temperature distribution in the air, causing unusually high refraction upwards or downwards.
Standard refraction occurs in average non-extreme case, bending light slightly downwards.
It is calculated simply by replacing Earth's radius in formulas by 7/6 * R.
So, instead of mean radius of 6371 km, you use 7432.833 km (or 7433 km).

The Metabunk calculator does the whole work easily.
(It was https://www.metabunk.org/curve/)

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible. Video shows too much, and means earth is flatter than round earth assumptions.

Turning Torso is 190 meters tall.
It is divided into 9 blocks (segments), so each block is 21 meters tall.
Google Earth gave me ground elevation there to be 1 ft, so I decided to ignore it.

What I did next was input the distance and eye level into Metabunk calculator and got this:



The thing calculated hidden parth without, and then with "standard refraction".

--------------------------------------------------------------------------------------------------------------------------------------------

I also found another video in which the guy measured by theodolite where will horizontal line from some tower across the water hit the Turning Torso.

He estimated where it should be in FE model, where in GE model without refraction, and where in GE model with refraction.



Here is the whole video:

[jimbob]

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible.


At 47.9 km 130m is hidden from the video
By calculation  136m should be hidden. The 6m discrepancy can be accounted by my assumption that the observation was at 2m from ground level, it could be less. It is possible that refraction could account for some depending on meteorological conditions.

[Macarios]

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible.

At 47.9 km 130m is hidden from the video
By calculation  136m should be hidden. The 6m discrepancy can be accounted by my assumption that the observation was at 2m from ground level, it could be less. It is possible that refraction could account for some depending on meteorological conditions.

If you read my post again, you will see those red markings in the data snap.
If we ignore refraction horizon would hide 148 meters, then 190-148 = 42 meters would be visible, and it would be 2 blocks of the building.
But in the air there is refraction.
If refraction is within standard specification, horizon will hide 124.5 meters, then 190-124.5 = 65.5 meters will be visible, and that is little over three blocks of the building.
130 meters hidden in the video is pretty close to pre-calculated 124.5 meters.

Please tell us again what is the problem with that?

[jimbob]

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible.

At 47.9 km 130m is hidden from the video
By calculation  136m should be hidden. The 6m discrepancy can be accounted by my assumption that the observation was at 2m from ground level, it could be less. It is possible that refraction could account for some depending on meteorological conditions.

If you read my post again, you will see those red markings in the data snap.
If we ignore refraction horizon would hide 148 meters, then 190-148 = 42 meters would be visible, and it would be 2 blocks of the building.
But in the air there is refraction.
If refraction is within standard specification, horizon will hide 124.5 meters, then 190-124.5 = 65.5 meters will be visible, and that is little over three blocks of the building.
130 meters hidden in the video is pretty close to pre-calculated 124.5 meters.

Please tell us again what is the problem with that?

There is no problem with it! It was a quote from someone else that didn't come up as a Quote.
I was highlighting that from my calculation everything was as predicted and there was no problem.

[Macarios]

We still have problem that with 148m hidden, there should only be little bit of tower "turning torse" be visible.

At 47.9 km 130m is hidden from the video
By calculation  136m should be hidden. The 6m discrepancy can be accounted by my assumption that the observation was at 2m from ground level, it could be less. It is possible that refraction could account for some depending on meteorological conditions.

If you read my post again, you will see those red markings in the data snap.
If we ignore refraction horizon would hide 148 meters, then 190-148 = 42 meters would be visible, and it would be 2 blocks of the building.
But in the air there is refraction.
If refraction is within standard specification, horizon will hide 124.5 meters, then 190-124.5 = 65.5 meters will be visible, and that is little over three blocks of the building.
130 meters hidden in the video is pretty close to pre-calculated 124.5 meters.

Please tell us again what is the problem with that?

There is no problem with it! It was a quote from someone else that didn't come up as a Quote.
I was highlighting that from my calculation everything was as predicted and there was no problem.

My apologies.
I could understand if I gave it a thought, but I didn't.

Well, I'm sometimes smart, but obviously not every time.

EDIT: About those "2m from the sea level", as far as I could understand, Mr. Ravisaras used eye elevation of 1.6 m.

[Tontogary]

It took it's time, to follow these argument, but in the end - for me - it appears quite simple.
You know this diagram from http://www.sacred-texts.com/earth/za/za32.htm ?


The distance of the vanishing point - aka the horizon - is defined by the limited resolution of the naked eye, where lines from the observers eye E to the vanishing point H and the surface C to H build an angle less than 1 minute of degree.
Rising observers position will broaden this angle and move point H (horizon) farther away, until the 1 minute criteria is met again.

But what, if observer has "hypervision" and could resolve angles less than 1 minute.

Yes. There are many accounts in the Flat Earth literature of telescopes restoring half sunken ships across calm bodies of water, showing that they are not really behind a "hill of water".

In cases where the water is turbulent, the shinking ship effect cannot be restored, showing waves to be the cause.

Ok then Tom, if that is the case then please can you explain this;
We have on my ship 2 different sextant so, one has an eyepiece with no magnification, the other a monocular with 4 times magnification. Taking a sextant altitude of a star or a heavenly body requires measuring the angle between the apparent horizon and the body, accuracy is grater than 0.1 minute of arc.
When using the 2 separate instruments the same value is obtained, even when accounting for index (instrument) error. The apparent horizon is constant dependant on the observers hieght of eye.

Waves cannot be interfering with the horizon, as navigation relies upon measuring the difference between the actual Zenith distance (from a point 90 degrees above the observers head with the calculated Zenith distance (calculated using an assumed position on the earth. 3 or more of these observations give a reasonably accurate position, and this method of navigation has been practiced for hundreds of years, and was used to map the world accurately (within a few miles, and mostly within a few metres) for hundreds of years before the introduction of Satellite navigation.

If magnification altered the horizon, then the calculations would be inaccurate, and no positions could be obtained. The same if waves were present, the measured altitude would vary, resulting in inaccurate results.

One last point here. If satellites are a hoax, how do we navigate using sat navs on ships, cars, and trucks? How do we communicate with shore when we are hundreds, sometimes thousands of miles from the nearest land?

Waiting in anticipation for your science based responses, and not just vague references to 18th century accounts from obscure texts.

[AATW]

Tom mixes up two things:
1) A ship which is far away but not as far as the horizon and cannot be seen clearly - so he claims that magnification "restores" the hull of the ship. Clearly it doesn't, magnification simply makes things bigger and clearer. So details of the ship which can't be distinguished with the naked eye - like a dark hull against a dark sea - can be seen with magnification. So the hull is not "restored", it was never occluded in the first place.
2) A ship which is beyond the horizon and some of it is occluded by the curve of the earth - in this case no amount of magnification will restore it. Tom fudges this by claiming it's "waves" in this case but I have shown in the other thread about waves that it isn't.

This has all been explained to him many times so either
1) Tom is suffering from a serious case of cognitive dissonance, his identity is so wrapped up in the flat earth society that he cannot admit he is wrong or
2) He's just a troll who doesn't believe any of this nonsense and is just here for the lolz.

Kinda hope it's the latter, the former would be pretty sad.