The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: HorstFue on February 09, 2018, 08:09:42 PM
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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 (http://www.sacred-texts.com/earth/za/za32.htm) ?
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
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.
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Yup. You're right. The horizon gets 10x farther away if you use 10x magnification... Never thought of it that way; I always just used the idea that the objects got hidden by the curvature of the Earth. Clearly the horizon doesn't get 10x farther away. The problem is that FE can simply respond that "waves" block you from seeing anything past a certain point; "waves" can make an "effective horizon" much closer than the actual horizon, which theoretically is extremely large (because we can get very powerful magnification).
The problem is that they haven't quantified any of these things, so there's no point in trying to debate it this way; it's far too prone to disproof by assertion (and those assertions are hard to disprove if done vaguely enough).
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Waves can't explain the horizon in Kansas.
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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 (http://www.sacred-texts.com/earth/za/za32.htm) ?
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
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.
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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.
Except the water is clearly not turbulent in this video
https://youtu.be/MoK2BKj7QYk
He shows where the different shots were taken from, looking at the times when he shows the GPS readings it looks like the same afternoon and the weather doesn't seem significantly different. While the channel does connect with the ocean, you can see that the water is pretty calm in all the shots as you would expect in a fairly narrow channel like this. It's not the open ocean. If waves were a factor you would expect the amount of the building you can see to vary significantly as the swell comes and goes, but you can't.
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Tom, do you even think about what we're saying before just using these trivial dismissals? Your mind is so closed to the Round Earth that after every thing we say, you try these obviously physically wrong rebuttals without even giving a thought to their feasibility and their quantification.
You can't just say the water is turbulent; clearly we aren't seeing waves tens or hundreds of feet tall, which would be required to hide a building by that much. This is addition to the observation that it doesn't look turbulent.
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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.
Except the water is clearly not turbulent in this video
https://youtu.be/MoK2BKj7QYk
He shows where the different shots were taken from, looking at the times when he shows the GPS readings it looks like the same afternoon and the weather doesn't seem significantly different. While the channel does connect with the ocean, you can see that the water is pretty calm in all the shots as you would expect in a fairly narrow channel like this. It's not the open ocean. If waves were a factor you would expect the amount of the building you can see to vary significantly as the swell comes and goes, but you can't.
Those images were taken across an inland sea, which has significant waves on it. Read the chapter Perspective at Sea in Earth Not a Globe. 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.
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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.
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Those images were taken across an inland sea, which has significant waves on it. Read the chapter Perspective at Sea in Earth Not a Globe. 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.
Why would I read anything from a man who thought that the moon was translucent.
And your "Bishop Experiment" which you claim to be able to reproduce any time you like is also on a bay, more exposed to the open sea than the channel in this video.
This is your "heads I win, tails you lose" reasoning again. You constantly contradict yourself.
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Those images were taken across an inland sea, which has significant waves on it. Read the chapter Perspective at Sea in Earth Not a Globe. 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.
Sorry, that's not how "many built up waves" work. You should read up on wave propagation; constructive interference cannot take place the way you say it can. However, this is a quibble. The main point is that there needs to be something extremely large obscuring the building, with the requisite size getting smaller as it gets closer to you. Your example is bad. As long as your eyes are above the dime, the dime cannot obscure the elephant. I think that anyone with common sense can tell you that.
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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.
Except the water is clearly not turbulent in this video
https://youtu.be/MoK2BKj7QYk
He shows where the different shots were taken from, looking at the times when he shows the GPS readings it looks like the same afternoon and the weather doesn't seem significantly different. While the channel does connect with the ocean, you can see that the water is pretty calm in all the shots as you would expect in a fairly narrow channel like this. It's not the open ocean. If waves were a factor you would expect the amount of the building you can see to vary significantly as the swell comes and goes, but you can't.
Those images were taken across an inland sea, which has significant waves on it. Read the chapter Perspective at Sea in Earth Not a Globe. 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.
There are many pictures of the Chicago skyline taken across lake Michigan that show the same effect. The waves on the Great Lakes in fair weather are quite small. A couple of feet.
Please provide evidence that a dime 3 feet below eye level can obscure an elephant.
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The Wiki says:
With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 23 miles away near the lighthouse. The entire beach is visible down to the water splashing upon the shore. Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore and teenagers merrily throwing Frisbees to one another. I can see runners jogging along the water's edge with their dogs. From my vantage point the entire beach is visible.
https://wiki.tfes.org/Experimental_Evidence
Surely any waves over 20 inches would obscure the entire beach and shoreline. Is it really credible that over a span of 23 miles there is not a single wave over this height?
No photographic evidence of his claim is produced.
But the video I posted above? Oh, that's waves...
Heads I win, tails you lose.
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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".
Do you have a video of this occurring? I have yet to find this.
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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.
Those images were taken across an inland sea, which has significant waves on it. Read the chapter Perspective at Sea in Earth Not a Globe. 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.
Sorry, that's not how "many built up waves" work. You should read up on wave propagation; constructive interference cannot take place the way you say it can. However, this is a quibble. The main point is that there needs to be something extremely large obscuring the building, with the requisite size getting smaller as it gets closer to you. Your example is bad. As long as your eyes are above the dime, the dime cannot obscure the elephant. I think that anyone with common sense can tell you that.
Perspective places the horizon at eye level. See: https://wiki.tfes.org/Horizon_always_at_Eye_Level
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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".
Do you have a video of this occurring? I have yet to find this.
There are some Youtube videos. In another thread we are considering adding a list of such videos to our Library.
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It will be interesting to see one. All I've seen so far are videos of small boats on the horizon that are too small to see at low magnification and become visible at higher magnification, but anything that is partially 'sunken' at any magnification does not rise up (restore) as magnification is increased.
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Yes, all magnification does is make things bigger and therefore clearer. If a hull is truly sunken behind the curve of the earth then magnification will not restore it any more than if I see someone go over a hill till I can only see their head will a telescope then allow me to see all of them.
It goes without saying that the "horizon rises to eye level" thing is also incorrect although admittedly you have to get pretty high before the horizon's dip below eye level is noticeable.
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Mr Bishop,
You are quite clearly the authority on the subject here. Would you please tell us, based on the ideas set forth in Perspective at Sea in Earth Not a Globe, just how tall the waves (or the "dime") would have to be in order to produce the effect seen in the video? Of course, we will have to make some assumptions, primarily as to the location of the waves relative to the viewer and the building. Since the waters in the foreground of all of the video shots are quite calm, we'll have to assume that the waves are more distant, closer to the building that the observer, agreed? Since this is video we can also gather some clues regarding the sea state by referring to the Beaufort Scale http://ggweather.com/101/beaufort.htm (http://ggweather.com/101/beaufort.htm) and by observing general weather/atmospheric conditions. I very much look forward to receiving your reply and gaining valuable insight as to how physics behave in the FE model.
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Perspective places the horizon at eye level. See: https://wiki.tfes.org/Horizon_always_at_Eye_Level
Umm sorry Tom, this is some serious unscientific nonsense (especially when you're not at sea level), and even at sea level, it doesn't help you prove anything. You can keep posting links to your silly wiki all you want, it's not gonna convince anyone. In the meantime, check out my other thread which clearly demonstrates the exact opposite of what you're trying to say, maybe you'll learn something: https://forum.tfes.org/index.php?topic=8832.0
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Perspective places the horizon at eye level. See: https://wiki.tfes.org/Horizon_always_at_Eye_Level
Umm sorry Tom, this is some serious unscientific nonsense (especially when you're not at sea level), and even at sea level, it doesn't help you prove anything. You can keep posting links to your silly wiki all you want, it's not gonna convince anyone. In the meantime, check out my other thread which clearly demonstrates the exact opposite of what you're trying to say, maybe you'll learn something: https://forum.tfes.org/index.php?topic=8832.0
Now, now, let's not be so hasty. The idea of the horizon being at eye level isn't in itself unscientific nonsense, it's just not relative to the current conversation. Mr Bishop has been given opportunity to clarify his statements and provide examples of how his theory applies in this particular instance. It's only fair to Mr Bishop and to those participating in this thread to allow him the opportunity to reply before insisting that he pursue another topic/thread.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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https://www.youtube.com/watch?v=MoK2BKj7QYk
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?
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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 (http://www.sacred-texts.com/earth/za/za32.htm) ?
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
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 (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.
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https://www.youtube.com/watch?v=MoK2BKj7QYk
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:
(https://preview.ibb.co/eDULC7/022deg.png) (https://ibb.co/kcZns7)
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:
(https://preview.ibb.co/bvb3zn/043deg.png) (https://ibb.co/m4A7QS)
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:
(https://preview.ibb.co/h2kwZn/45deg.png) (https://ibb.co/fU9OEn)
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.
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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:
(https://preview.ibb.co/g8CePn/0430deg.png) (https://ibb.co/fVcqVS)
Zoomed in a bit:
(https://preview.ibb.co/kEFYjn/0430degx100.png) (https://ibb.co/hdDax7)
Make more sense now assuming round earth assumptions? Can check with your phone calculator.
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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:
(https://preview.ibb.co/g8CePn/0430deg.png) (https://ibb.co/fVcqVS)
Zoomed in a bit:
(https://preview.ibb.co/kEFYjn/0430degx100.png) (https://ibb.co/hdDax7)
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 (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.
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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?
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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 (http://www.sacred-texts.com/earth/za/za32.htm) ?
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
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:
https://www.youtube.com/watch?v=i0ObTd7DLMw
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.
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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.
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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:
(https://preview.ibb.co/mEFKAS/1.png) (https://ibb.co/jZxo4n)
Zommed in:
(https://preview.ibb.co/gwMFjn/2.png) (https://ibb.co/eG3jc7)
Zoomed in more:
(https://preview.ibb.co/nCCAH7/3.png) (https://ibb.co/ih8T4n)
Zoomed in even more:
(https://preview.ibb.co/iVzqH7/4.png) (https://ibb.co/dEcMPn)
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? ???
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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:
https://www.youtube.com/watch?v=i0ObTd7DLMw
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.
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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:
(https://preview.ibb.co/mEFKAS/1.png) (https://ibb.co/jZxo4n)
Zommed in:
(https://preview.ibb.co/gwMFjn/2.png) (https://ibb.co/eG3jc7)
Zoomed in more:
(https://preview.ibb.co/nCCAH7/3.png) (https://ibb.co/ih8T4n)
Zoomed in even more:
(https://preview.ibb.co/iVzqH7/4.png) (https://ibb.co/dEcMPn)
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.
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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?
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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:
(https://preview.ibb.co/iqtQH7/10.png) (https://ibb.co/cRpzc7)
This leave 47.9-4.8 = 43.1 from that point to the tower.
(https://preview.ibb.co/bVB1qS/11.png) (https://ibb.co/hmXVjn)
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
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is limited by visibility of atmosphere
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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?
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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.
(https://preview.ibb.co/mMff0S/n1.png) (https://ibb.co/dz74S7)
"horizon" to target
(https://preview.ibb.co/g3maZn/n2.png) (https://ibb.co/eVA2En)
observer to "horizon". In picture horizon is left at top of circle
(https://preview.ibb.co/mfmyn7/n3.png) (https://ibb.co/dCE7fS)
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
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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.
(https://preview.ibb.co/mMff0S/n1.png) (https://ibb.co/dz74S7)
"horizon" to target
(https://preview.ibb.co/g3maZn/n2.png) (https://ibb.co/eVA2En)
observer to "horizon". In picture horizon is left at top of circle
(https://preview.ibb.co/mfmyn7/n3.png) (https://ibb.co/dCE7fS)
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 ?
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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 ;D Of course plane icon not to scale, but position is.
Here is what looks like with plane at 39,000ft. Plane now 2km long :D, but position and everything else to scale, round earth assumptions. Blue circle of 6371km Radius. (https://www.desmos.com/calculator/k4k0txx9rl) Left and right to about 100km in image. How far can you actually see? I've never seen 100km from plane.
Flat Earth:
(https://preview.ibb.co/nir85S/fe.png) (https://ibb.co/hqdXX7)
Round Earth:
(https://preview.ibb.co/kfA1QS/re.png) (https://ibb.co/c5vCX7)
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:
(https://preview.ibb.co/b95CX7/comp.png) (https://ibb.co/kdMms7)
Or maybe explain a bit more. Mathematics I like but it is also good to make visual.
This is the observer: :D zoomed in of course
(https://image.ibb.co/hTHeen/o.png) (https://imgbb.com/)
blue = earth
red = distance to target
black dotted = line of sight
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Here is what looks like with plane at 39,000ft. Plane now 2km long :D, but position and everything else to scale, round earth assumptions. Blue circle of 6371km Radius. (https://www.desmos.com/calculator/aaw9mhcegu) 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/ (https://www.metabunk.org/curve/)
Flat Earth:
(https://preview.ibb.co/nir85S/fe.png) (https://ibb.co/hqdXX7)
Round Earth:
(https://preview.ibb.co/kfA1QS/re.png) (https://ibb.co/c5vCX7)
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:
(https://preview.ibb.co/b95CX7/comp.png) (https://ibb.co/kdMms7)
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.
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Macarios is right, as usual
Atmospheric perspective (https://en.wikipedia.org/wiki/Aerial_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.
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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.
(https://preview.ibb.co/mMff0S/n1.png) (https://ibb.co/dz74S7)
"horizon" to target
(https://preview.ibb.co/g3maZn/n2.png) (https://ibb.co/eVA2En)
observer to "horizon". In picture horizon is left at top of circle
(https://preview.ibb.co/mfmyn7/n3.png) (https://ibb.co/dCE7fS)
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 (https://pasteboard.co/Her0ToJ.jpg)
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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 (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.)
[img]https://pasteboard.co/HewbXAj.jpg[/img][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/ (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.)
(http://i68.tinypic.com/8z4if4.jpg)
Hope it helped.
For the circle I used brim of small glass. :)
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Yes that same formula like I used, sinus cosinus (https://www.desmos.com/calculator/k4k0txx9rl).
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. (https://www.desmos.com/calculator/k4k0txx9rl) tells us the following:
Now to about 200km left and right:
Flat Earth:
(https://preview.ibb.co/bP9rUn/12.png) (https://ibb.co/jzPfN7)
Round Earth:
(https://preview.ibb.co/m25n27/11.png) (https://ibb.co/mYxd9n)
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"?
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Yes that same formula like I used, sinus cosinus (https://www.desmos.com/calculator/k4k0txx9rl).
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. (https://www.desmos.com/calculator/k4k0txx9rl) tells us the following:
Now to about 200km left and right:
Flat Earth:
(https://preview.ibb.co/bP9rUn/12.png) (https://ibb.co/jzPfN7)
Round Earth:
(https://preview.ibb.co/m25n27/11.png) (https://ibb.co/mYxd9n)
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.
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"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 (https://owlcation.com/stem/How-to-Determin-the-Radius-of-the-Earth-Al-Birunis-Classic-Experiment)
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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 (https://www.desmos.com/calculator/k4k0txx9rl) 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.
(https://preview.ibb.co/mMff0S/n1.png) (https://ibb.co/dz74S7)
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 (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 (https://www.desmos.com/calculator/uxilj5mrlk) got as well. At 30km distance, 51 metre height is hidden.
(https://preview.ibb.co/mu5wen/14.png) (https://ibb.co/gt9NKn)
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 (https://www.youtube.com/watch?v=MoK2BKj7QYk) shows too much, and means earth is flatter than round earth assumptions.
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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 (https://www.desmos.com/calculator/k4k0txx9rl) 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.
(https://preview.ibb.co/mMff0S/n1.png) (https://ibb.co/dz74S7)
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 (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 (https://www.desmos.com/calculator/uxilj5mrlk) got as well. At 30km distance, 51 metre height is hidden.
(https://preview.ibb.co/mu5wen/14.png) (https://ibb.co/gt9NKn)
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 (https://www.youtube.com/watch?v=MoK2BKj7QYk) 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 (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 (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)} (http://y=6371-\sqrt{\left(\left(6371\right)^2-\left(47.9-\sqrt{\left(.002+6371\right)^2-\left(6371\right)^2}\right)^2\right)})
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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/ (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 (https://www.youtube.com/watch?v=MoK2BKj7QYk) 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:
(http://i67.tinypic.com/29wkg2v.png)
The thing calculated hidden parth without, and then with "standard refraction".
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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.
(http://i67.tinypic.com/k1fcz8.png)
Here is the whole video:
https://www.youtube.com/watch?v=PrXw9914uHs
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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.
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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?
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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.
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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.
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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 (http://www.sacred-texts.com/earth/za/za32.htm) ?
(http://www.sacred-texts.com/earth/za/img/fig75.jpg)
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.
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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.
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At 47.9 km by calculation 136m should be hidden.
I tried your calculation, but shows 148m hidden at 47.9km, and 1.6m observer.
Same as my sine and cosine calculation, as expected. Not sure how you worked out the written 'theoretical' values in your link, but they do not come out of your formula.
(https://preview.ibb.co/irhOc7/f.png) (https://ibb.co/gYYkqS)
Anyway, apparently we should allow for round earth refraction also.
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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).
Ok, thank you, easy enough. So, when I include this in my simple Round Earth assumption circle.
We get this result, quite interesting if I may say so myself.
(https://preview.ibb.co/hbQZH7/vid.jpg) (https://ibb.co/czeuH7)
(https://preview.ibb.co/nHTX4n/bluecirclewith6371radiusandstandardrefraction.png) (https://ibb.co/iJHgx7)
Black dotted line is line of sight, rest below is hidden. Blue circle is ground.
Have to say that my calculation comes quite close to the video. Interesting. It simply is a Blue circle and some mathematics applied (https://www.desmos.com/calculator/zfwvutc2ss) and I get result. Have included link so you can zoom in and out and play with it a little for who is interested.
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At 47.9 km by calculation 136m should be hidden.
I tried your calculation, but shows 148m hidden at 47.9km, and 1.6m observer.
Same as my sine and cosine calculation, as expected. Not sure how you worked out the written 'theoretical' values in your link, but they do not come out of your formula.
(https://preview.ibb.co/irhOc7/f.png) (https://ibb.co/gYYkqS)
Anyway, apparently we should allow for round earth refraction also.
I looked at the video again and changed the observer height to an estimated 3m as he is not standing on the ground. I forgot to change it in the link.
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Ah got it. Guy from video says height filmed at 2.1m. Maybe that why difference. So I used observer height 2.1m in last post.
But our mathematics same result I think, but different formula.