A simple Flat Earth proof anyone can verify from a former world-record photo

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A long-distant observation/photograph by world-record holder Gumŕ is shown on his website https://beyondrange.wordpress.com/2017/08/01/the-gulf-of-lion-alps-and-mont-ventoux-on-a-winter-morning-from-the-bastiments-peak-pyriness-436-km/. The photograph in question depicts Doigt de Dieu which has an elevation of 3.9 km from an observation point of Bastiments, Pyrenees at an elevation of around 2.5 km at 436 km.

The calculation below can be used to determine if objects should be visible from the observer at a given distance when factoring in the alleged curvature of the Earth.

How much of an object is hidden behind the curvature of the Earth is called the hidden height, hh. If you know the distance along the surface s, this formula can calculate the hh:



Where the hidden height is hh, the assumed radius of Earth is R (6371 km), the observer’s eye-height is hₒ (in this case 2.5 km) and the distance of the object from the observer is s (436 km).

When applying the hh equation above, the hh comes out at around 5.2 km which should put the entire mountain at 3.9 km behind the curve:



Verify result yourself: (6371)/cos(436/6371-acos(6371/(6371+2.5)))-6371 (change degree to radius)

Copy-n-paste bold into Web 2.0 Caclulator

Naturally, Round-Earthers will argue special atmospheric conditions explain this.
« Last Edit: April 24, 2024, 12:07:35 AM by CHIPPY »

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JackBlack

  • 21893
The calculation below can be used to determine if objects should be visible from the observer at a given distance when factoring in the alleged curvature of the Earth.
And while ignoring refraction. Refraction is well known, and known to cause light to generally bend downwards meaning you can see further than the geometric horizon.
Over such great distances, refraction can be a significant issue.
A simple approximation is to replace the radius by seven 6ths of the radius.
Doing that, and putting in the correct altitude (2.7 km) gives a hidden height of 3.7 km.
That allows the object to be seen.

But the real nail in the coffin for FEers is the other mountains shown.
You have an observation from 2.7 km.
The mountain, with an elevation of 3973 m is observed quite low, in fact it appears lower than closer mountains which are shorter, including those below the altitude of the observer.
How does this happen?

Naturally, Round-Earthers will argue special atmospheric conditions explain this.
You mean you already know the reason why you are wrong, but you are going to ignore it and instead lie to everyone to pretend Earth is flat, even though your evidence, when taken without that part factoring in still shows Earth is round, just larger than the "official" number?

Yet.  When conditions are not supportive of refraction.  A boat going out to sea starts to become blocked from view bottom up after four miles. 


The calculation below can be used to determine if objects should be visible from the observer at a given distance when factoring in the alleged curvature of the Earth.
And while ignoring refraction. Refraction is well known, and known to cause light to generally bend downwards meaning you can see further than the geometric horizon.
Over such great distances, refraction can be a significant issue.
A simple approximation is to replace the radius by seven 6ths of the radius.
Doing that, and putting in the correct altitude (2.7 km) gives a hidden height of 3.7 km.
That allows the object to be seen.

But the real nail in the coffin for FEers is the other mountains shown.
You have an observation from 2.7 km.
The mountain, with an elevation of 3973 m is observed quite low, in fact it appears lower than closer mountains which are shorter, including those below the altitude of the observer.
How does this happen?

Naturally, Round-Earthers will argue special atmospheric conditions explain this.
You mean you already know the reason why you are wrong, but you are going to ignore it and instead lie to everyone to pretend Earth is flat, even though your evidence, when taken without that part factoring in still shows Earth is round, just larger than the "official" number?
The elevation the photo was taken at was 2.5 km, not 2.7 km, they were not at the very top of the mountain. But even so, you need to factor in considerable refraction for it to be visible. Whenever people see further than they should on a globe, Round Earthers invariably resort to the lazy go-to rationalization of "big refraction boyz".
« Last Edit: April 24, 2024, 04:51:00 AM by CHIPPY »


The elevation

It’s called refraction.  Well proven and understood.


Yet.  When conditions are not supportive of refraction.  A boat going out to sea starts to become blocked from view bottom up after four miles.


Another example where effects of refraction can be minimized. 

 
Yes, curvature can be measured and modeled as proven by Blackpool Photo

https://www.theflatearthsociety.org/forum/index.php?topic=92160.0

I was divided on making this its own thread. Or post it in response to the old lie “the curvature can’t be measured”.

I decided this it is a stand alone thread.  Unfortunately, it probably with be ignored by the people who claim a “laser should measure curvature.”  And probably too much background.  Especially on subjects that deserve their own threads. 

So.  The earth is huge, and the amount of curvature is slight to us small humans.

For the claim a laser level should measure curvature?  Not sure how that works on the scale of the earth?  And would be much different than surveying and using line of sight and correcting for the horizontal line? 

The laser does go straight except if bent by atmospheric conditions?  Then you have to somehow measure drop perpendicular to the laser.  But line of sight works the same way in surveying. So the laser level is really a moot point anyway. 

And level surface as defined in surveying follows the sphere of the earth anyway. 



https://gcekbpatna.ac.in/assets/documents/lecturenotes/Surveying_Mod2_Levelling.pdf

So the definition of level surface brings us to the world of surveying.  It was long understood the earth is too big to measure with a flat edge. And the earth can accurately be measured with line of sight (which arguably has the same strengths and problems as “a laser level”), and tweaking with adjusting for refraction.

So surveying ties into the long known dip of the horizon.

Using surveying and the dip of the horizon, it’s long been known that you can calculate the curvature of the earth.



Which can be verified.

Quote

Rainy Lake Experiment: Conclusion

http://walter.bislins.ch/bloge/index.asp?page=Rainy+Lake+Experiment%3A+Conclusion

Summary
All data and observations agree with the predictions of the Globe Model, which includes Terrestrial Refraction. The predictions for the Flat Earth Model, however, contradict the observations.
The Rainy Lake Experiment shows even better than the Bedford Level Experiment

 that the earth is a globe, since we also have GPS measurements that are not influenced by Refraction or Perspective, but are of a pure geometric nature. GPS measurements directly provide the radius of the earth.
Only one conclusion remains:
The earth cannot be flat, but is a globe with a mean radius of 6371 km!

Almost to the point of this thread…. Hold on

So I guess this photo a few years ago caused quite the buzz?  Who knew?

A photo by:
Quote
Kevin Jackson, of Birkdale, Southport, captured the amazing view of the seaside resort from the Sefton coast.

https://www.lancs.live/news/lancashire-news/stunning-picture-blackpool-thats-set-19719171.amp



https://i2-prod.lancs.live/incoming/article19719194.ece/ALTERNATES/s615/0_blackpooljfif.jpg

As presented by Dave McKeegan in this video.



Now the point of all this.  Blackpool tower looks relatively taller than the background hills because the earth is spherical.



We can model the view of Blackpool Tower relative to the distance hills for a flat earth vs spherical earth.

1. The height of the Blackpool Tower is known.
2. The position of the photographer is known.
3. The radius of the earth is measured and known.
4. The distance to the background hills and their height are known.



If the earth was flat.  The background hills would be taller in the photo than the tower.



A rough model of how the photo should look for the tower relative to the background mountains on a flat earth. 



The evidence is pretty clear. And even supposedly converted this person,Ranty-Flat-Earth, back to spherical earth

https://www.reddit.com/r/pics/comments/odpjrm/this_image_converted_me_from_a_very_prominent/

A long-distant observation/photograph

It’s called refraction.

It’s been proven over and over.

So now you have to take in account and explain the Blackpool Photo and why the taller mountain tops appear relatively below the tower height. 

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gnuarm

  • 143
Yet.  When conditions are not supportive of refraction.  A boat going out to sea starts to become blocked from view bottom up after four miles.

Where does the four miles come from?  If you are sitting on the beach by the water, the bottom of the ship will start to disappear very quickly.  At four miles the curvature of the earth would be around ten feet at four miles.  That's not going to be very noticeable.  Is this what you mean? 

Yet.  When conditions are not supportive of refraction.  A boat going out to sea starts to become blocked from view bottom up after four miles.

Where does the four miles come from?  If you are sitting on the beach by the water, the bottom of the ship will start to disappear very quickly.  At four miles the curvature of the earth would be around ten feet at four miles.  That's not going to be very noticeable.  Is this what you mean?

It’s called fudge factor and ensuring a distance when refraction is minimized that the bottom of the boat will most definitely be blocked from view.

If you want more conclusive results that derail the thread…

From another thread.


Other than laser levels, which can measure for level over longer distances, but you say they aren’t accurate for longer distances, of course.




From this video…


Learned about this experiment using a laser tangent to the curved earth with a boat as a target on a lake 3 miles out.


Quote
Where Are We? Ch. 1 The Circumference of the Earth | Genius by Stephen Hawking

https://indiana.pbslearningmedia.org/resource/hawking_genius_ep06_clip01/where-are-we-ch-1-the-circumference-of-the-earth-genius-by-stephen-hawking/



In this clip from Genius by Stephen Hawking, learn how to calculate the circumference of the Earth. Three volunteers learn by measuring the flatness of the lake that they will be able to calculate the size and shape of the Earth. Using a powerful laser that projects a straight beam of light and a boat, the volunteers shoot the beam across the lake. This experiment shows the curvature of the lake. This was first discovered by the ancient Greek philosopher, mathematician and geometer Eratosthenes. He proved the Earth wasn't flat through observing the sun and the direction it cast shadows. If the Earth was flat, the sun would always shine at the same angle no matter what time of day it was. Using all the data collected from the curvature of the lake the volunteers are able to calculate the circumference of the Earth.


The laser at three miles out was about six foot above what should be “level”…..

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JackBlack

  • 21893
The elevation the photo was taken at was 2.5 km, not 2.7 km
Based on what?
I find nothing in your source supporting that.
Instead I find this:
"From the Pic de la Dona, at about 2700 meters"

they were not at the very top of the mountain
And I never said they were.
Those mountains go to 2881 m.

But even so, you need to factor in considerable refraction for it to be visible.
i.e. standard refraction.
Not even extreme refraction.

Whenever people see further than they should on a globe
You mean see further than they should on a globe without atmosphere.

Again, you are showing you know the answer to your lies, yet you are happy to repeat these lies, knowing they are lies.

Do you think refraction does not occur?

I also see you ignored the question about why this high mountain appears lower than lower mountains which are lower than the observation point?

The elevation the photo was taken at was 2.5 km, not 2.7 km
Based on what?
I find nothing in your source supporting that.
Instead I find this:
"From the Pic de la Dona, at about 2700 meters"

they were not at the very top of the mountain
And I never said they were.
Those mountains go to 2881 m.

But even so, you need to factor in considerable refraction for it to be visible.
i.e. standard refraction.
Not even extreme refraction.

Whenever people see further than they should on a globe
You mean see further than they should on a globe without atmosphere.

Again, you are showing you know the answer to your lies, yet you are happy to repeat these lies, knowing they are lies.
My lies? Isn't always the case that ignorant people always accuse others of what they are doing secretly themselves.

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EarthIsRotund

  • 253
  • Earth is round. Yes.
No, the case is always that the earth is round.
I love Mairimashita Iruma Kun

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JackBlack

  • 21893
My lies?
Yes, your lies.
You lie claiming this proves a flat Earth, while knowing the reason it doesn't.

Isn't always the case that ignorant people always accuse others of what they are doing secretly themselves.
No, it is not always the case. But it is some times, like when FEers accuse REers of making up excuses by bringing up facts backed up by mountains of evidence, while the FEers are off making excuses.

Now again, care to tell us how such a tall mountain, which should be above the observe appears lower than a shorter mountain which should appear below the observer?

The elevation the photo was taken at was 2.5 km, not 2.7 km
Based on what?
I find nothing in your source supporting that.
Instead I find this:
"From the Pic de la Dona, at about 2700 meters"

they were not at the very top of the mountain
And I never said they were.
Those mountains go to 2881 m.

But even so, you need to factor in considerable refraction for it to be visible.
i.e. standard refraction.
Not even extreme refraction.

Whenever people see further than they should on a globe
You mean see further than they should on a globe without atmosphere.

Again, you are showing you know the answer to your lies, yet you are happy to repeat these lies, knowing they are lies.
My lies? Isn't always the case that ignorant people always accuse others of what they are doing secretly themselves.

The only reason you are a flat earther, is your calculation skills in this instance in relation to heights of the mountains, distance
between the mountains, earth curve, and refraction index on the day and time the photo was taken, are for want of a better word - SHITTY.

Verify result yourself: (6371)/cos(436/6371-acos(6371/(6371+2.5)))-6371 (change degree to radius)

I find it funny that you have to change from degrees to radians to make the math work how you want it to.
When you put that exact equation in degrees, it comes out to approx 2.29

Its almost like an equation made to be used in degrees is accurate when used in degrees. Who knew?

Before you try spewing some bs about how its not supposed to be in degrees, the below picture (which showed up in an earlier post on this thread) shows that the Radius of the earth was calculated with degrees. As such, you would need to multiply (or divide, I honestly don't remember) with pi to convert that number to be able to be used in radians instead of degrees.

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EarthIsRotund

  • 253
  • Earth is round. Yes.
I think it matters not if degrees or radians were used? sin(90°) is the same as sin(π/2), is it not?
I love Mairimashita Iruma Kun

That is true, but sin(10) is different between degrees and radians.

You would have to change the inside number in order to keep the answer the same.

I was just trying to say that 1 degree is not equal to one radian.

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EarthIsRotund

  • 253
  • Earth is round. Yes.
That is true, but sin(10) is different between degrees and radians.

You would have to change the inside number in order to keep the answer the same.

I was just trying to say that 1 degree is not equal to one radian.


uh, no I understand - but I was saying that 1° = π/180. notice the degree sign convention that is omitted when angles are in radians instead of degrees - because radians are natural way to express angles. what I mean is that in CHIPPY's original expression there's arccos(some_distance/some_other_distance) which then is added (subtracted, really, but does it matter?) by some other number which is also in radians. and then the cosine of result is taken. so if you try to add degrees with radians, your results will be out of whack.


tl;dr learn trigonometry
I love Mairimashita Iruma Kun

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JackBlack

  • 21893
I find it funny that you have to change from degrees to radians to make the math work how you want it to.
If you want to use natural units, you need to use radians, not degrees.

e.g. if you want to use the length of an arc to obtain the angle, as used in that formula, the most straightforward way is to use radians.
i.e. if you have a length along the curve of S, and have a radius of R, then the angle, in radians, is S/R.
If you want to use degrees, then you need to use 360 °*S/(2*pi*R) = 360 °/2pi is the conversion from radians to degrees.

When you put that exact equation in degrees, it comes out to approx 2.29
And how did you do that?
Did you just leave in S/R?
If so, what is this meant to represent?
Do you understand the formula being used?

Here is a simple diagram:

S is the length along the arc/hypothetical level surface, from the point of the observer to the point of the observed object.
R is the radius of Earth.
h0 is the height of the observer above Earth.
hh is the height hidden by Earth.
And for added bits, S0 and Sh break S into 2 pieces, the piece from the observer to the horizon, and the piece from the horizon to the object being observed.
a, a0 and ah are angles for S, S0 and Sh respectively.
i.e.
a = a0+ah
S = S0 + Sh

So what is the math, which holds regardless of if you are using radians or degrees?
Well, we have a right angle triangle such that:
cos(a0)=R/(R+h0)
And another so:
cos(ah)=R/(R+hh)

But we know R, and h0, but not a0, so we need to take the inverse.
This gives us:
a0 = arccos(R/(R+h0))

We don't directly know hh or ah, but we do have a relationship to get ah.
a = a0+ah
Rearranging this we get:
ah = a - a0.
And if we sub in the previous result we get:
ah = a - arccos(R/(R+h0))

Now we can rerrange the equation from the right angle triangle:
cos(ah)=R/(R+hh)
R+hh = R/cos(ah)
hh = R/cos(ah) - R

Now subbing in the result we had for ah we get:
hh = R/cos(a - arccos(R/(R+h0))) - R

But the question is what is a?
Well this depends on if you are using degrees or radians.
a is the angle subtended for an arc length of S.

If you are using radians, this is simple:
a = S/R.
If you are using degrees you instead need:
a = 360 ° * S/(2 * pi * R)

You cannot use a = S/R while using degrees for cos and arccos.

And the obvious giveaway that your reasoning is wrong is:
S/R=0.068
a0 = arccos(R/(R+h0)) = 1.60 °
S/R - arccos(R/(R+h0)) = -1.53 nonsense units.

So the angle to calculate the amount hidden would then be negative.

So congrats on showing you have no idea what you are talking about.

If instead you use the correct degrees method, where you use
a = 360 ° * S/(2 * pi * R)
then you end up with
a - arccos(R/(R+h0)) = 2.32 °,
and then eventually hh = 5.2 km

Before you try spewing some bs about how its not supposed to be in degrees, the below picture (which showed up in an earlier post on this thread) shows that the Radius of the earth was calculated with degrees.
No, it doesn't. And that is not the only formula you can use.

As such, you would need to multiply (or divide, I honestly don't remember) with pi to convert that number to be able to be used in radians instead of degrees.
No, you don't.
The conversion is 180/pi or pi/180, depending on how you go.
« Last Edit: May 01, 2024, 04:17:59 PM by JackBlack »

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gnuarm

  • 143
Yet.  When conditions are not supportive of refraction.  A boat going out to sea starts to become blocked from view bottom up after four miles.

Where does the four miles come from?  If you are sitting on the beach by the water, the bottom of the ship will start to disappear very quickly.  At four miles the curvature of the earth would be around ten feet at four miles.  That's not going to be very noticeable.  Is this what you mean?

It’s called fudge factor

If you are trying to do real science, you don't use "fudge factors". 


Quote
and ensuring a distance when refraction is minimized that the bottom of the boat will most definitely be blocked from view.

If you want more conclusive results that derail the thread…

From another thread.


Other than laser levels, which can measure for level over longer distances, but you say they aren’t accurate for longer distances, of course.




From this video…


Learned about this experiment using a laser tangent to the curved earth with a boat as a target on a lake 3 miles out.


Quote
Where Are We? Ch. 1 The Circumference of the Earth | Genius by Stephen Hawking

https://indiana.pbslearningmedia.org/resource/hawking_genius_ep06_clip01/where-are-we-ch-1-the-circumference-of-the-earth-genius-by-stephen-hawking/



In this clip from Genius by Stephen Hawking, learn how to calculate the circumference of the Earth. Three volunteers learn by measuring the flatness of the lake that they will be able to calculate the size and shape of the Earth. Using a powerful laser that projects a straight beam of light and a boat, the volunteers shoot the beam across the lake. This experiment shows the curvature of the lake. This was first discovered by the ancient Greek philosopher, mathematician and geometer Eratosthenes. He proved the Earth wasn't flat through observing the sun and the direction it cast shadows. If the Earth was flat, the sun would always shine at the same angle no matter what time of day it was. Using all the data collected from the curvature of the lake the volunteers are able to calculate the circumference of the Earth.


The laser at three miles out was about six foot above what should be “level”…..

Yeah.  Interesting that they obtained results that perfectly match the 8 inches per mile squared rule.  Not close, not approximately, but, dead ON

How did they do that?  I'm not even sure how they measured their height above the lake in the helicopter, not to mention the ability to visually tell when the chopper was at just the right position.  That defies belief.