[JohnAdams1145]

How in the world are these pages on the same wiki:
https://wiki.tfes.org/High_Altitude_Photographs
https://wiki.tfes.org/Horizon_always_at_Eye_Level


You can't have both ways! If you see the horizon as curved, then how can it always be at eye level? You're clearly looking down on the area that was cast by the "spotlight." Seems like FE has some thinking to do.

[Rama Set]

The dime obscuring an elephant example is not analogous. If a dime is on the ground and you are standing, it does not matter how far away the dime or the elephant is, it cannot obscure it.

Sure, at close distances. However, the horizon line is at eye level.

Incorrect. And the higher your altitude, the more inaccurate your statement is, but yet you can still see skyscrapers in Niagara Falls that are hidden behind the horizon when you on the observation deck of the CN tower.

[douglips]

If the horizon is curved due to the spotlight sun, then a bright light source ought to be visible beyond that circle, especially at night.

[StinkyOne]

The dime obscuring an elephant example is not analogous. If a dime is on the ground and you are standing, it does not matter how far away the dime or the elephant is, it cannot obscure it.

Sure, at close distances. However, the horizon line is at eye level.

This is so obviously false. If a dime is placed on the ground, it will not be visible at the horizon. It is too small. But, in the spirit of being overly generous, let's say you can see the dime at the horizon. This would mean that the elephant is over the horizon and no longer visible. If it was right behind the dime, it would be visible. The dime would never cover the elephant. The only way for your silly statement to work is if you hold the dime in your like of sight.

[Tom Bishop]

How in the world are these pages on the same wiki:
https://wiki.tfes.org/High_Altitude_Photographs
https://wiki.tfes.org/Horizon_always_at_Eye_Level


You can't have both ways! If you see the horizon as curved, then how can it always be at eye level? You're clearly looking down on the area that was cast by the "spotlight." Seems like FE has some thinking to do.

The horizon is a different concept than the sun's circular spotlight.

[Tom Bishop]

The dime obscuring an elephant example is not analogous. If a dime is on the ground and you are standing, it does not matter how far away the dime or the elephant is, it cannot obscure it.

Sure, at close distances. However, the horizon line is at eye level.

Incorrect. And the higher your altitude, the more inaccurate your statement is, but yet you can still see skyscrapers in Niagara Falls that are hidden behind the horizon when you on the observation deck of the CN tower.

At higher altitudes the horizon becomes hazy, and eventually dips, because the edges of the sun's circular spotlight is not the same as the horizon.

[Rama Set]

The dime obscuring an elephant example is not analogous. If a dime is on the ground and you are standing, it does not matter how far away the dime or the elephant is, it cannot obscure it.

Sure, at close distances. However, the horizon line is at eye level.

Incorrect. And the higher your altitude, the more inaccurate your statement is, but yet you can still see skyscrapers in Niagara Falls that are hidden behind the horizon when you on the observation deck of the CN tower.

At higher altitudes the horizon becomes hazy, and eventually dips, because the edges of the sun's circular spotlight is not the same as the horizon.

This is irrelevant, even if it were true.  From the observation deck of the CN tower, the horizon is below eye level and you can see a skyscraper in Niagara Falls partially hidden behind that horizon.

[Treep Ravisarras]

The many built up waves on the horizon line will provide an area for larger bodies to shrink behind, much like how a dime can obscure an elephant behind it.

Yes, absolutely. Nevertheless a very interesting video. I'm not sure what it shows. Let's apply some conjecture and analyse.

Distances in the video are: 6 distances from 25 to 47 km.

At 25km, the ground can be assumed to be 49m lower following simple sine/cosine that you can calculate on your phone calculator
At 28.4km we are assuming 63m lower.
At 34.7km 94m lower
At 40.3km 127m
At 45.1km 159m lower
At the ultimate distance of 47.9km we are assuming to have lost 180m

So, the video shows the Earth is Flatter than what the Round Earth protagonists talk about. Because with a tower of supposedly 190m tall we should only see top 10m. Anyone comment?

[jimbob]

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. Is then the point H also moved farther away?
Anyone can get "hypervision": use a binocular or a telescope.
If e.g. the binocular has a magnification of 7x (standard marine binocular) according to this the horizon should be 7 times farther away.

I frequently use such a binocular at sea, but the horizon always appears to be at the same distance, no significant difference when looking with naked eye or with the binocular - unless details get clearer.

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.

The distance to the horizon and its associated basic maths can be seen on http://www.ringbell.co.uk/info/hdist.htm

Ok simple experiment go stand in the sea or even better a lake, on a very calm day. Stand up to your belly button (0.5m from eyes to surface) and get you girl friend to take the boat out hold an bright coloured object (eg orange stick) horizontal and close to the water line. When she is 2,5 km out, you wont see the stick. This gives the earth radius of 6378137 metres.

[jimbob]

The many built up waves on the horizon line will provide an area for larger bodies to shrink behind, much like how a dime can obscure an elephant behind it.

Yes, absolutely. Nevertheless a very interesting video. I'm not sure what it shows. Let's apply some conjecture and analyse.

Distances in the video are: 6 distances from 25 to 47 km.

At 25km, the ground can be assumed to be 49m lower following simple sine/cosine that you can calculate on your phone calculator as follows:


At 28.4km we are assuming 63m lower.
At 34.7km 94m lower
At 40.3km 127m
At 45.1km 159m lower
At the ultimate distance of 47.9km we are assuming to have lost 180m, I'll show again:


So, the video shows the Earth is Flatter than what the Round Earth protagonists talk about. Because with a tower of supposedly 190m tall we should only see top 10m. Anyone comment?



Just for matter of completion, because the above images have so little angles that you don't really see what it is showing, here with a large angle rather than the minute angles required along the so-called round earth circumference. Here one with 45 degrees:

I teach Alevel maths and your maths makes no sense. Where is your working?assumed variable values? All your diagram shows is that the opposite side of a triangle is equal to the sine of an angle when the hypotenuse is equal to 1.

[Treep Ravisarras]

I teach Alevel maths and your maths makes no sense. Where is your working?assumed variable values? All your diagram shows is that the opposite side of a triangle is equal to the sine of an angle when the hypotenuse is equal to 1.

Yes I am sorry bit confusing. Tidied up a bit:

For longest distance in video 47.9 kilometres:


Zoomed in a bit:


Make more sense now assuming round earth assumptions? Can check with your phone calculator.

[jimbob]

I teach Alevel maths and your maths makes no sense. Where is your working?assumed variable values? All your diagram shows is that the opposite side of a triangle is equal to the sine of an angle when the hypotenuse is equal to 1.

Yes I am sorry bit confusing. Tidied up a bit:

For longest distance in video 47.9 kilometres:


Zoomed in a bit:


Make more sense now assuming round earth assumptions?

OK Let's apply some conjecture and analyse: I will mark your work as I would mark one of my students.

"Yes, absolutely. Nevertheless a very interesting video. I'm not sure what it shows" It shows that at the given distances parts of the building are not visible (kind of obvious). The parts that are not obvious are the bottom bits (also kind of obvious). They have dissapeared over the horizon. Use the horizon calculator http://www.ringbell.co.uk/info/hdist.htm

"At 25km, the ground can be assumed to be 49m lower" That is called curvature, a characteristic of circles not flat surfaces.

why have you divided 6371km by 1000?
What is a and why does it equal 0.34572
Where did you get 45 degrees from?
Sin 45 = Cos 45 =0.7071
Why do you think an isosceles right angled triangle with a hypotenuse (h) of 6371 an opposite and adjacent (a and b) side of length 4504  has any relation to the question. ........I can give you some math lessons but it will cost.... oh and delete that phone calculator, it obviously doesnt work.
 

[Treep Ravisarras]

Have a relook at my tidied up picture.

Using your phone calculator calculate exactly as formula. Easy.

For horizontal distance (b) = 47.9 km you will require angle of 0.43 degrees. It’s the only way. Or do you have another answer?

The 90 degrees are added because a unit circle has 0 degrees at the right just aa my earlier not so tidy graphs. I wanted it to show upright, so I rotate 90 degrees.

For distance of 47.9 metres the height difference assuming round earth principles will be 181m. That is not so easy with the video.

Can you please mark again? 

P.s. your answer of .7071 assumes earth with 1km radius. Have never heard anyone say that. Perhaps new group called Tiny Earthers?

[Macarios]

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. Is then the point H also moved farther away?
Anyone can get "hypervision": use a binocular or a telescope.
If e.g. the binocular has a magnification of 7x (standard marine binocular) according to this the horizon should be 7 times farther away.

I frequently use such a binocular at sea, but the horizon always appears to be at the same distance, no significant difference when looking with naked eye or with the binocular - unless details get clearer.

Yes, the average human eye resolution is roughly one arc minute (0.0167 degrees).
If we look at the Sun during Lahaina Noon in Hawaii, the angular diameter of the Sun and the distance from the eye gives us the Sun diameter:
2 * 5005 * TAN(0.53/2) = 46.3 km
For 46.3 km to reach vanishing point (if Sun wasn't bright) we would need distance of (46.3 / 2) / tan(0.0167 / 2) = 158 850 km
(But Sun is bright, and we would see its light long beyond vanishing point.)

And yet, Sun "vanishes" at the horizon distance of some 10 000 km.
Even if we make it 20 000, it is still much closer than 158 850 km.

Not only that.
Upper half of Sun is 23.15 km high. Vanishing point at (23.15 / 2) / tan(0.0167 / 2) = 79 425 km.
Lower half of Sun is 23.15 km high. Vanishing point at (23.15 / 2) / tan(0.0167 / 2) = 79 425 km.

Still, lower half, at the same distance as upper half vanishes at those 10 000 km, while upper half takes more time

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

In case where the water is turbulent, the whole surface of the water is turbulent.
Except for rare tsunamis, the reason for water to be turbulent is wind.

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".

Such cases are not disputable there (in Flat Earth literature), but here (in reality).
Interesting how this good zoom expanded the ship horizontally all the way to wider than the whole view, and still couldn't bring back the part behind the horizon:



Waves?
What could conveniently stop them from spreading towards us to see them?
Mechanical characteristics of the water surface remains the same all the way.

[jimbob]

Easy fix. You are mistaking my example for the real calculation.

The 45 degree is just the example.

Have a relook at my tidied up picture.

Using your phone calculator calculate exactly as formula. Easy.

For horizontal distance (b) = 47.9 km you will require angle of 0.43 degrees. It’s the only way. Or do you have another answer?

The 90 degrees are added because a unit circle has 0 degrees at the right just aa my earlier not so tidy graphs. I wanted it to show upright, so I rotate 90 degrees.

For distance of 47.9 metres the height difference assuming round earth principles will be 181m. That is not so easy with the video.

Can you please mark again? 

P.s. your answer of .7071 assumes earth with 1km radius. Have never heard anyone say that. Perhaps new group called Tiny Earthers?

Sin45=.7071 as does cos45=.7071 (try this on your calculator. However I still have no idea why you are using 45 degrees as it doesnt relate to anything (why do you need an example of 45 degrees anyway).
What you have found is the angle for a triangle whos opposite side is length 47.9 km and whose adjacent side is 6371. This is a triangle with an opposite side whos length corresponds to the distance to the tower and whos adjacent side corresponds to the radius of the earth. No relation to the number of blocks of the tower that is visible. In fact your calculation doesn't relate to anything relevant to the question at all. Why choose the radius of the earth? why choose the distance to the tower? why calculate an angle that relates to putting a tower 47.9 km high at a distance of half the earths diameter.....you are very confused.

[Treep Ravisarras]

Why choose the radius of the earth? why choose the distance to the tower? why calculate an angle that relates to putting a tower 47.9 km high at a distance of half the earths diameter.....you are very confused.

No, you Alevel maths teacher?
Round earth theory assumes earth has earths radius right 
Furthest distance to tower in video is 47.9 km. Not height of the tower! Height is we assume the internet is 190m tall tower.
But round earth theory "curvature of earth", has a drop of 181metres. So only top 9 meter of tower should be visible. Video shows earth too flat for round earth theory.

Will show you again: Blue is Earth, Red is distance to Tower, Blue dot is height drop:


Zommed in:


Zoomed in more:


Zoomed in even more:


Question for you maths teacher:
What is the circumference of a circle with radius 6371 km?
What angle do you need to travel 47.9 km along the circumference if 360 takes you all around?
Answer: same as my picture 0.43 degrees

Use your phone calculator to fill in angle into formulas in picture. Answer: distance to tower (as we know), but interestingly: height of drop away (h2) due to the supposed "curvature" of the earth: -0.181km or 181m.

That's how easy it is. Maybe study the topic of unit circle.

Please can you mark again this? 

[AATW]

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".

Such cases are not disputable there (in Flat Earth literature), but here (in reality).
Interesting how this good zoom expanded the ship horizontally all the way to wider than the whole view, and still couldn't bring back the part behind the horizon:



Waves?
What could conveniently stop them from spreading towards us to see them?
Mechanical characteristics of the water surface remains the same all the way.

It's interesting that they pretend to care only about empirical measurements and yet just read "accounts" from old books and regard that as good enough.
Meanwhile, the countless photos and videos showing ships sinking beneath the waves or buildings occluded by the curve of the earth are dismissed by "waves".
Although interestingly, waves are never an issue in the "Bishop Experiment" where Tom claims to be able to see a distant beach across a bay at any time and in different atmospheric conditions.

[jimbob]

Why choose the radius of the earth? why choose the distance to the tower? why calculate an angle that relates to putting a tower 47.9 km high at a distance of half the earths diameter.....you are very confused.

No, you Alevel maths teacher?
Round earth theory assumes earth has earths radius right 
Distance to tower is 47.9 km. Not height of the tower! Height is we assume the internet is 190m tall tower.
But round earth theory "curvature of earth", has a drop of 181metres. So only top 9 meter of tower should be visible. Video shows earth too flat for round earth theory.

Will show you again: Blue is Earth, Red is distance to Tower, Blue dot is height drop:


Zommed in:


Zoomed in more:


Zoomed in even more:


Question for you maths teacher:
What is the circumference of a circle with radius 6371 km?
What angle do you need to travel 47.9 km along the circumference?
Answer: same as my picture.

Use your phone calculator to fill in angle into formulas in picture. Answer: distance to tower (as we know), but interestingly: height of drop away (h2) due to the supposed "curvature" of the earth: -0.181km or 181m.

That's how easy it is. Maybe study the topic of unit circle.

Please can you mark again this? 

Check this http://www.ringbell.co.uk/info/hdist.htm

If the observer is an ground level ie 0 meters height, the horizon is at 0 meters
If the observer is 10cm of the ground (.1m) the horizon is now at 1.1km
If the observer is 50cm of the ground (.5m) the horizon is now at 2.5km
If the observer is 180cm of the ground (1.8m) the horizon is now at 4.8km

Your calculation is trying to show (however it is wrong but I can see what your trying to do: you need the equation for a circle x^2+y^2=r) how far the "horizon drops" if you travel 47.9km. This is with a view point at ground level and a horizon at 0 m.

[Treep Ravisarras]

Yes you are correct, my calculation has the viewer at ground level.

However it is not incorrect! Proof is: try to fill in a height of 180.069m (my calculation found) in your website calculator that follows round earth principles.

You will see my calculations are correct.

But yes at ground level.

With distance of 4.8km, my calculation shows height of 1.81m. Reverse this means 0 height horizon at 4.8km distance from observer.
This leave 47.9-4.8 = 43.1 from that point to the tower.

Height difference now 146m following round earth assumptions. If we assume internet is correct 190m tower. This means 44m is visible according to round earth principles. What you think?

[jimbob]

Yes you are correct, my calculation has the viewer at ground level.

However it is not incorrect! Proof is: try to fill in a height of 180.069m (my calculation found) in your website calculator that follows round earth principles.

You will see my calculations are correct.

But yes at ground level.

Now that we are at the same page, maybe we can work together.

With distance of 4.8km, my calculation shows height of 1.81m. Reverse this means 0 height horizon at 4.8km distance from observer. Will show:


This leave 47.9-4.8 = 43.1 from that point to the tower.


Height difference now 146m following round earth assumptions. If we assume internet is correct 190m tower. This means 44m is visible according to round earth principles. What you think?

The equation for the fall (our height in the calculation) is  given by 6371-((6371) ^2-(47.9) ^2) ^0.5=180m

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.

The equation becomes much more complicated when the line is no longer horizontal (ie slopping downward) and “cancels” the “fall of 180m out Quickly with increasing elevation.
One then has to find the gradient of this line as a tangent to the circle and subtract the “fall” of the circle in relation to the “fall” of the line.

If I could upload a photo, I could explain this with more clarity