Flat Earth

[notbatman]

Lmao, oh so this is "confirmed" now?  Hilarious.

Google --> search "fly over Antarctica" --> http://www.antarcticaflights.com.au --> book flight --> fly over South Pole --> lol

So why can't I rent a private jet and fly across Antarctica myself?

There's penguins and ice, why am I being met with military force?

[the joint]

Lmao, oh so this is "confirmed" now?  Hilarious.

Google --> search "fly over Antarctica" --> http://www.antarcticaflights.com.au --> book flight --> fly over South Pole --> lol

So why can't I rent a private jet and fly across Antarctica myself?

There's penguins and ice, why am I being met with military force?

I can't enter the premises of the White House without being met by force, either.  Doesn't mean the Earth is flat. I don't know what would possess someone to think that just because you can't go somewhere except under certain circumstances that it necessarily means the earth is flat.  You should see the obvious logical disconnect, there.  The gap is wider than the plane for the flight you can book to actually go there.

But the fact that there are *authorized* flights that will take you directly over and beyond the South Pole creates some serious problems for the flat earth map, wouldn't you agree?  Regular people like you and me do it all the time.

Edit:  A better question you should be asking might be, "Why can I book a flight and travel over and beyond the South Pole if flat earthers claim this is impossible?"

[Lauda]

Edit:  A better question you should be asking might be, "Why can I book a flight and travel over and beyond the South Pole if flat earthers claim this is impossible?"

They have been drinking some fine kool-aid. There are so many ways to prove this yourself, yet people come up with such idiotic theories. We're living in the 21th century, wake up people.

[notbatman]

Lmao, oh so this is "confirmed" now?  Hilarious.

Google --> search "fly over Antarctica" --> http://www.antarcticaflights.com.au --> book flight --> fly over South Pole --> lol

So why can't I rent a private jet and fly across Antarctica myself?

There's penguins and ice, why am I being met with military force?

I can't enter the premises of the White House without being met by force, either.  Doesn't mean the Earth is flat. I don't know what would possess someone to think that just because you can't go somewhere except under certain circumstances that it necessarily means the earth is flat.  You should see the obvious logical disconnect, there.  The gap is wider than the plane for the flight you can book to actually go there.

But the fact that there are *authorized* flights that will take you directly over and beyond the South Pole creates some serious problems for the flat earth map, wouldn't you agree?  Regular people like you and me do it all the time.

Oh so I'm infringing on the Emperor Penguin's bird house am I?

Extrapolating that the Earth is flat from Antarctica's restricted access status was not the point here. There's better points for that argument. I was explaining to Lauda that one simply can't just go there and wonder around on their own in response to her questions about The New Standard Map of the World and the Antarctic ice ring. There's no "logical disconnect" here.

And again, I'm not interested in an "authority" taking me on an "authorized" predetermined fly by, that proves nothing. Since I'm prevented from flying a private jet over myself at gunpoint I might add that kind of puts a stop to using that part of the world to prove anything.

[notbatman]

Edit:  A better question you should be asking might be, "Why can I book a flight and travel over and beyond the South Pole if flat earthers claim this is impossible?"

They have been drinking some fine kool-aid. There are so many ways to prove this yourself, yet people come up with such idiotic theories. We're living in the 21th century, wake up people.

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

 

[Lauda]

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

"Broadly, the method places a lot of emphasis on reconciling empiricism and rationalism, and making logical deductions based on empirical data." There's no empirical proof just made up theories. Anything that doesn't go with the flat Earth theory is either a lie or conspiracy according to the believers; it doesn't surprise me that people call it a joke. Buy a drone, fly it over Antarctica (I'm pretty sure that yours wouldn't be the only one around there).

[notbatman]

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

"Broadly, the method places a lot of emphasis on reconciling empiricism and rationalism, and making logical deductions based on empirical data." There's no empirical proof just made up theories. Anything that doesn't go with the flat Earth theory is either a lie or conspiracy according to the believers, no wonder that people call it a joke. Buy a drone, fly it over Antarctica (I'm pretty sure that yours wouldn't be the only one around there).

"Empirical evidence, data, or knowledge, also known as sense experience, is a collective term for the knowledge or source of knowledge acquired by means of the senses, particularly by observation and experimentation." -- https://en.wikipedia.org/wiki/Empirical_evidence

Now if my camera and telephoto lens observes an object and captures it on film/CCD at a distance that would put it far below the supposed curvature of the Earth how is that not empirical evidence and proof of a flat Earth? I can come up with all sorts of "empirical evidence" that's just one point.

As for the drone, how do I protect it from the missiles that will be fired at it if I try and fly it across Antarctica?

[the joint]

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

"Broadly, the method places a lot of emphasis on reconciling empiricism and rationalism, and making logical deductions based on empirical data." There's no empirical proof just made up theories. Anything that doesn't go with the flat Earth theory is either a lie or conspiracy according to the believers, no wonder that people call it a joke. Buy a drone, fly it over Antarctica (I'm pretty sure that yours wouldn't be the only one around there).

"Empirical evidence, data, or knowledge, also known as sense experience, is a collective term for the knowledge or source of knowledge acquired by means of the senses, particularly by observation and experimentation." -- https://en.wikipedia.org/wiki/Empirical_evidence

Now if my camera and telephoto lens observes an object and captures it on film/CCD at a distance that would put it far below the supposed curvature of the Earth how is that not empirical evidence and proof of a flat Earth? I can come up with all sorts of "empirical evidence" that's just one point.

As for the drone, how do I protect it from the missiles that will be fired at it if I try and fly it across Antarctica?

Several pages back, I did the actual math as to why, for example, you can clearly see the Chicago skyline at a distance of some ~60 miles.  When you take into account the curvature of light at the horizon, only some ~30-something feet of the skyline will be obfuscated by the curvature of the earth.  At a distance of 60 miles, the effect of light's curvature is so great that you can simply pretend a 500 foot-tall building is something like ~6 times it's actual height.  In other words, it's like a 500 foot-tall building is actually 3,000 feet tall, and you're observing that 3000 foot-tall building at a distance of 60 miles, which you would clearly see.

The math used by flat earthers to say "you shouldn't see the skyline at a distance of 60 miles" is flawed as hell as the other factors not taken into account perfectly explain why we see what we see.  Educate yourself, do the math yourself, and be amazed at the round earth.

(For some explanation, light bends about 0.5 degrees of arc at the horizon whereas a 500 foot-tall building occupies only about 0.08 degrees of arc at a distance of 60 miles, or 1/6th as much.  This means you can account for the bending of light by imagining the building is 6 times taller than it is and then calculating how much you would expect to see due to curvature.  For taller buildings which occupy more perceptual scope, it would be less than 6 times; for shorter buildings, it would be more than 6 times.)

Here:https://bitcointalk.org/index.php?topic=1009045.msg12645979#msg12645979

[TheGr33k]

Did they also prove that 3 gigantic elephants riding an enormous turtle have this flat earth on their backs?


You know, 'cause that would've been nice.

[notbatman]

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

"Broadly, the method places a lot of emphasis on reconciling empiricism and rationalism, and making logical deductions based on empirical data." There's no empirical proof just made up theories. Anything that doesn't go with the flat Earth theory is either a lie or conspiracy according to the believers, no wonder that people call it a joke. Buy a drone, fly it over Antarctica (I'm pretty sure that yours wouldn't be the only one around there).

"Empirical evidence, data, or knowledge, also known as sense experience, is a collective term for the knowledge or source of knowledge acquired by means of the senses, particularly by observation and experimentation." -- https://en.wikipedia.org/wiki/Empirical_evidence

Now if my camera and telephoto lens observes an object and captures it on film/CCD at a distance that would put it far below the supposed curvature of the Earth how is that not empirical evidence and proof of a flat Earth? I can come up with all sorts of "empirical evidence" that's just one point.

As for the drone, how do I protect it from the missiles that will be fired at it if I try and fly it across Antarctica?

Several pages back, I did the actual math as to why, for example, you can clearly see the Chicago skyline at a distance of some ~60 miles.  When you take into account the curvature of light at the horizon, only some ~30-something feet of the skyline will be obfuscated by the curvature of the earth.  At a distance of 60 miles, the effect of light's curvature is so great that you can simply pretend a 500 foot-tall building is something like ~6 times it's actual height.  In other words, it's like a 500 foot-tall building is actually 3,000 feet tall, and you're observing that 3000 foot-tall building at a distance of 60 miles, which you would clearly see.

The math used by flat earthers to say "you shouldn't see the skyline at a distance of 60 miles" is flawed as hell as the other factors not taken into account perfectly explain why we see what we see.  Educate yourself, do the math yourself, and be amazed at the round earth.

(For some explanation, light bends about 0.5 degrees of arc at the horizon whereas a 500 foot-tall building occupies only about 0.08 degrees of arc at a distance of 60 miles, or 1/6th as much.  This means you can account for the bending of light by imagining the building is 6 times taller than it is and then calculating how much you would expect to see due to curvature.  For taller buildings which occupy more perceptual scope, it would be less than 6 times; for shorter buildings, it would be more than 6 times.)

Here:https://bitcointalk.org/index.php?topic=1009045.msg12645979#msg12645979

Nice try but you're attempting to use Astronomical Refraction to make your calculations. When Terrestrial Refraction is used the dispersion angle is so small it can't even be measured due to atmospheric turbulence.

[the joint]

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

"Broadly, the method places a lot of emphasis on reconciling empiricism and rationalism, and making logical deductions based on empirical data." There's no empirical proof just made up theories. Anything that doesn't go with the flat Earth theory is either a lie or conspiracy according to the believers, no wonder that people call it a joke. Buy a drone, fly it over Antarctica (I'm pretty sure that yours wouldn't be the only one around there).

"Empirical evidence, data, or knowledge, also known as sense experience, is a collective term for the knowledge or source of knowledge acquired by means of the senses, particularly by observation and experimentation." -- https://en.wikipedia.org/wiki/Empirical_evidence

Now if my camera and telephoto lens observes an object and captures it on film/CCD at a distance that would put it far below the supposed curvature of the Earth how is that not empirical evidence and proof of a flat Earth? I can come up with all sorts of "empirical evidence" that's just one point.

As for the drone, how do I protect it from the missiles that will be fired at it if I try and fly it across Antarctica?

Several pages back, I did the actual math as to why, for example, you can clearly see the Chicago skyline at a distance of some ~60 miles.  When you take into account the curvature of light at the horizon, only some ~30-something feet of the skyline will be obfuscated by the curvature of the earth.  At a distance of 60 miles, the effect of light's curvature is so great that you can simply pretend a 500 foot-tall building is something like ~6 times it's actual height.  In other words, it's like a 500 foot-tall building is actually 3,000 feet tall, and you're observing that 3000 foot-tall building at a distance of 60 miles, which you would clearly see.

The math used by flat earthers to say "you shouldn't see the skyline at a distance of 60 miles" is flawed as hell as the other factors not taken into account perfectly explain why we see what we see.  Educate yourself, do the math yourself, and be amazed at the round earth.

(For some explanation, light bends about 0.5 degrees of arc at the horizon whereas a 500 foot-tall building occupies only about 0.08 degrees of arc at a distance of 60 miles, or 1/6th as much.  This means you can account for the bending of light by imagining the building is 6 times taller than it is and then calculating how much you would expect to see due to curvature.  For taller buildings which occupy more perceptual scope, it would be less than 6 times; for shorter buildings, it would be more than 6 times.)

Here:https://bitcointalk.org/index.php?topic=1009045.msg12645979#msg12645979

Nice try but you're attempting to use Astronomical Refraction to make your calculations. When Terrestrial Refraction is used the dispersion angle is so small it can't even be measured due to atmospheric turbulence.

https://en.m.wikipedia.org/wiki/Atmospheric_refraction

On the horizon refraction is slightly greater than the apparent diameter of the Sun, so when the bottom of the sun's disc appears to touch the horizon, the sun's true altitude is negative. If the atmosphere suddenly vanished, the sun would too. By convention, sunrise and sunset refer to times at which the Sun’s upper limb appears on or disappears from the horizon and the standard value for the Sun’s true altitude is −50′: −34′ for the refraction and −16′ for the Sun’s semi-diameter.

The light from distant objects on the earth is refracted too; the straight line from your eye to a distant mountain might be blocked by a closer hill, but the actual light path may curve enough to make the distant peak visible. A reasonable first guess: a mountain's apparent altitude at your eye (in degrees) will exceed its true altitude by its distance in kilometers divided by 1500. This assumes a fairly horizontal line of sight and ordinary air density; if the mountain is very high (so much of the sightline is in thinner air) divide by 1600 instead.

[connexus]

The earth is only flat to those who don't actually know how to Google search.

[Foxpup]

How does a map of a spherical Earth confirm that the Earth is flat? It is obviously a spherical map, because it doesn't have a scale, just the words "60 Miles to the Degree" (nautical miles, that is). Degrees of what? Curvature of the Earth, of course! You have to take it into account to read the fucking map, and the map itself says so. Flat-Earthers are getting dumber all the time.

[notbatman]

I got Kool-Aid by the barrel full!

You lash out with "idiodoc theories" but there's no substance to your rebuttal. You're attempting to rebut technical arguments and empirical proof with peer pressure.

"Broadly, the method places a lot of emphasis on reconciling empiricism and rationalism, and making logical deductions based on empirical data." There's no empirical proof just made up theories. Anything that doesn't go with the flat Earth theory is either a lie or conspiracy according to the believers, no wonder that people call it a joke. Buy a drone, fly it over Antarctica (I'm pretty sure that yours wouldn't be the only one around there).

"Empirical evidence, data, or knowledge, also known as sense experience, is a collective term for the knowledge or source of knowledge acquired by means of the senses, particularly by observation and experimentation." -- https://en.wikipedia.org/wiki/Empirical_evidence

Now if my camera and telephoto lens observes an object and captures it on film/CCD at a distance that would put it far below the supposed curvature of the Earth how is that not empirical evidence and proof of a flat Earth? I can come up with all sorts of "empirical evidence" that's just one point.

As for the drone, how do I protect it from the missiles that will be fired at it if I try and fly it across Antarctica?

Several pages back, I did the actual math as to why, for example, you can clearly see the Chicago skyline at a distance of some ~60 miles.  When you take into account the curvature of light at the horizon, only some ~30-something feet of the skyline will be obfuscated by the curvature of the earth.  At a distance of 60 miles, the effect of light's curvature is so great that you can simply pretend a 500 foot-tall building is something like ~6 times it's actual height.  In other words, it's like a 500 foot-tall building is actually 3,000 feet tall, and you're observing that 3000 foot-tall building at a distance of 60 miles, which you would clearly see.

The math used by flat earthers to say "you shouldn't see the skyline at a distance of 60 miles" is flawed as hell as the other factors not taken into account perfectly explain why we see what we see.  Educate yourself, do the math yourself, and be amazed at the round earth.

(For some explanation, light bends about 0.5 degrees of arc at the horizon whereas a 500 foot-tall building occupies only about 0.08 degrees of arc at a distance of 60 miles, or 1/6th as much.  This means you can account for the bending of light by imagining the building is 6 times taller than it is and then calculating how much you would expect to see due to curvature.  For taller buildings which occupy more perceptual scope, it would be less than 6 times; for shorter buildings, it would be more than 6 times.)

Here:https://bitcointalk.org/index.php?topic=1009045.msg12645979#msg12645979

Nice try but you're attempting to use Astronomical Refraction to make your calculations. When Terrestrial Refraction is used the dispersion angle is so small it can't even be measured due to atmospheric turbulence.

https://en.m.wikipedia.org/wiki/Atmospheric_refraction

On the horizon refraction is slightly greater than the apparent diameter of the Sun, so when the bottom of the sun's disc appears to touch the horizon, the sun's true altitude is negative. If the atmosphere suddenly vanished, the sun would too. By convention, sunrise and sunset refer to times at which the Sun’s upper limb appears on or disappears from the horizon and the standard value for the Sun’s true altitude is −50′: −34′ for the refraction and −16′ for the Sun’s semi-diameter.

The light from distant objects on the earth is refracted too; the straight line from your eye to a distant mountain might be blocked by a closer hill, but the actual light path may curve enough to make the distant peak visible. A reasonable first guess: a mountain's apparent altitude at your eye (in degrees) will exceed its true altitude by its distance in kilometers divided by 1500. This assumes a fairly horizontal line of sight and ordinary air density; if the mountain is very high (so much of the sightline is in thinner air) divide by 1600 instead.

96.5 / 1500 = 6.4 centimeters

Guess it should actually read it.

96.5 / 1500 = 0.064°

[Foxpup]

The light from distant objects on the earth is refracted too; the straight line from your eye to a distant mountain might be blocked by a closer hill, but the actual light path may curve enough to make the distant peak visible. A reasonable first guess: a mountain's apparent altitude at your eye (in degrees) will exceed its true altitude by its distance in kilometers divided by 1500. This assumes a fairly horizontal line of sight and ordinary air density; if the mountain is very high (so much of the sightline is in thinner air) divide by 1600 instead.

96.5 / 1500 = 6.4 centimeters

[notbatman]

The light from distant objects on the earth is refracted too; the straight line from your eye to a distant mountain might be blocked by a closer hill, but the actual light path may curve enough to make the distant peak visible. A reasonable first guess: a mountain's apparent altitude at your eye (in degrees) will exceed its true altitude by its distance in kilometers divided by 1500. This assumes a fairly horizontal line of sight and ordinary air density; if the mountain is very high (so much of the sightline is in thinner air) divide by 1600 instead.

96.5 / 1500 = 6.4 centimeters

96.5 / 1500 = 0.064°

OK, I'm getting 107.79 meters added height for a distance of 96.5 km according to the divide by 1500 calculation. This is considerably more than the 6.4 cm faceplam number but, it's still not enough.

60 * 60 * 8 = 28800 inches (731.52 meters)

The drop due to the claimed curvature I'm getting is 731.52 meters for a distance of 96.5 km.


A 10 story building is 33 meters tall.

So if the Earth is flat a 10 story building should measure 140.79 m at a distance of 96.5 km.

If the Earth is round it should not be visible.


Do I get another facepalm or is this correct?
   

[Foxpup]

Do I get another facepalm or is this correct?

You get another facepalm.



There are plenty of buildings in Chicago taller than 33 metres. The city itself is 181 metres above sea level. The top of the Sears Tower is 709 metres above sea level, almost high enough to be visible from 60 miles away without atmospheric refraction.

[notbatman]

Do I get another facepalm or is this correct?

You get another facepalm.



There are plenty of buildings in Chicago taller than 33 metres. The city itself is 181 metres above sea level. The top of the Sears Tower is 709 metres above sea level, almost high enough to be visible from 60 miles away without atmospheric refraction.

50% of the Sears Tower is visible at 96.5 km in the photo in question. ~355 - 709 m that's about 364 m above the horizon.

Now according to the globe earth model taking refraction and the curvature into account only ~85 m should be visible above the horizon.

I'm willing to bet that with a more powerful telephoto lens 100% of the tower will be visible. I'm extremely confidant (and the numbers are backing me up here) that perspective is at play here and not earth curvature.

Can I stop facepalming yet?

[Foxpup]

I'm willing to bet that with a more powerful telephoto lens 100% of the tower will be visible. I'm extremely confidant (and the numbers are backing me up here) that perspective is at play here and not earth curvature.

Can I stop facepalming yet?

No.



Perspective does not work that way. Zooming in doesn't allow you to see over obstructions.

[notbatman]

I'm willing to bet that with a more powerful telephoto lens 100% of the tower will be visible. I'm extremely confidant (and the numbers are backing me up here) that perspective is at play here and not earth curvature.

Can I stop facepalming yet?

No.



Perspective does not work that way. Zooming in doesn't allow you to see over obstructions.

You're assuming there's a physical barrier (earth curvature), but the vanishing points physical limitations are the result of the observers perspective matrix. A camera with no Telephoto Lens has a closer point than one with a TL. One camera will see half a ship on the horizon and one will see 100%. This proves that there's no physical barrier in the case of a ship on the ocean. The same would apply to the sears tower.