Flat Earth debunked by measuring angles to the sun

Rory

Senior Member.
Flat Earth debunker Sly Sparkane recently posted a video in which 23 participants in 9 separate countries measured the sun's elevation angle at solar noon on the September equinox. Naturally, when the results are viewed and compared with the two models of the earth, it should appear obvious which one is accurate.

Here is the result for the flat earth:

flat earth sun test.jpg

Which, as expected, demonstrates a complete lack of cohesion for the model.

Meanwhile, for the globe:

globe earth sun test.jpg

Showing that the angles measured by the participants correspond with both a distant sun, and the earth being a spheroid with a mean radius of around 3959 miles.

Given that this experiment is repeatable by anyone, requiring no great technical knowledge or equipment, it would seem difficult to dispute.

The full video is here (set to start with the results from 7:04 - the rest before that is preamble, explaining the experiment, introducing the participants, etc).

 
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Amber Robot

Active Member
Very nice. This is a great example of how the flat earth model is wrong. There is no model of a flat earth that can fit the observed positions and movement of the Sun (and other celestial objects).

This could be coupled with the same people saying what direction the sun set on that day. If they all say due west that's further proof of the round earth/distant sun. The flat earth model doesn't predict that.
 

Rory

Senior Member.
It shows two suns, quite close.
Approximately true, for the northern hemisphere. And another one in the south. Plus, had there been observers at 40-50°S we can tell there would be another one, as the angles are mirrored. So that's four suns, just for starters, at two different altitudes. :)
 
Approximately true, for the northern hemisphere. And another one in the south. Plus, had there been observers at 40-50°S we can tell there would be another one, as the angles are mirrored. So that's four suns, just for starters, at two different altitudes. :)
I missed that.
 

Rory

Senior Member.
That's four suns, just for starters, at two different altitudes.
Sorry, I was woefully off with that - obviously, wherever two different lines intersect, that must be a sun. So a more accurate picture of where the sun is on the flat earth is something like:

flat earth suns.jpg

Now, maybe we can safely assume that people in the southern hemisphere are always wrong about everything - indeed, because the two guys down there's lines don't intersect, one or both of them evidently did something wrong - and also that only people living between about 48 and 56°N know how to measure things, since 17 of the 23 participants agreed approximately on the location of the sun, and they were all in that vicinity.

The sun, therefore, is actually at an altitude of about 1800 miles - and, perhaps most interestingly, not above the equator on the equinox, but over 1200 miles north of it (looks to be above about 18°N, and there are ~69 miles per degree of latitude).
 
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Sorry, I was woefully off with that - obviously, wherever two different lines intersect, that must be a sun. So a more accurate picture of where the sun is on the flat earth is something like:

View attachment 29517

Now, maybe we can safely assume that people in the southern hemisphere are always wrong about everything - indeed, because the two guys down there's lines don't intersect, one or both of them evidently did something wrong - and also that only people living between about 48 and 56°N know how to measure things, since 17 of the 23 participants agreed approximately on the location of the sun, and they were all in that vicinity.

The sun, therefore, is actually at an altitude of about 1800 miles - and, perhaps most interestingly, not above the equator on the equinox, but over 1200 miles north of it.

I think you are putting too much faith in the accuracy of individual bearings. I still think there are two suns with one bearing to be rejected as an outlier.flat earth sun test2.jpg
 

Rory

Senior Member.
I think you are putting too much faith in the accuracy of individual bearings. I still think there are two suns with one bearing to be rejected as an outlier.
The readings match suncalc, etc pretty well.

What about the reading at about 14°N? And, of course, the one that someone on the equator would have taken?

Plus, the sun you've drawn there to cover all the lines has a diameter of about 340 miles - over ten times the size a 3000-mile high sun would be. ;)
 
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Ranon151

New Member
The two southern hemisphere lines seem to diverge on the globe projection. Is that due to measurement error?
 

Rory

Senior Member.
Did anyone report seeing two suns?
Lol. I don't think they did. I guess that maybe casts a sliver of doubt on the flat earth theory.

There's actually quite a simple way to complete the picture, given that we know:
  • At the time of the equinox, the sum of the degree of latitude and the sun's elevation angle equals 90
  • The distance to the equator is approximately 69 miles multiplied by the degree of latitude
For any given latitude, then, we can calculate for a right-angled triangle and work out where the line of elevation angle will intersect the line perpendicular to 0° (ie, the equator), using 69*latitude*tan(radians(90-latitude)). I've done this in excel, but don't know how to generate a chart in the style of the screenshot in the OP. Will attach it, though, in the hope that someone can finish it off.

I think this also very clearly illustrates why Victorian flat earthers hypothesised a sun at an elevation of around 3000 miles - and from that calculated a size of about 32 miles - as, for latitudes in places such as Europe and North America, that's about the point the elevation angle intersects the line perpendicular to 0°.

Anyway, I do hope someone knows how to plot that chart.
 

Attachments

  • complete the picture.xls
    16 KB · Views: 309
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Mick West

Administrator
Staff member
I you just do 30° to 55°, where most of the experimental results were, then you get an illusion of a cluster. But really there's quite a spread.
20171005-164424-qwgt1.jpg
 

Rory

Senior Member.
The two southern hemisphere lines seem to diverge. Is that due to measurement error?
Yep. Their results are in the video at 4:24. The more southerly one's sum total of latitude and elevation is 90.83, while the other's is 89.5. Small errors, but enough to have their lines diverging rather than intersecting.
 
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Rory

Senior Member.
And the chart showing what the angles would be for a flat earth would be so completely different that it would have long ago been blindingly obvious to pretty much anyone who traveled long distances across the Earth in a north-south direction.
It would be difficult fitting it on the screen. If we assume the hypothetical 3000-mile high sun - correct at 47.6°N - then at 80°N the distance to the equator would be just over 17,000 miles, rather than the 5,500-mile distance we rather inconveniently observe in reality. Worse, by the time we get to 89.9°N, where the angle to the sun is 0.1°, the distance to the equator would be a little over 1.7 million miles. On the flat earth model, above about 50°N the distances between lines of latitude increase dramatically.
 
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Mick West

Administrator
Staff member
And with heights in miles, scale being set by the -90 to +90 being half the circumference:

20171005-172155-pinui.jpg
 

Mick West

Administrator
Staff member
Here's a "bendy light" version.
20171005-182153-fd78w.jpg

The black lines are the observed directions. The yellow lines are bent rays of light that converge 20,000 miles above the earth.

You could conceivably imagine some strange bending light mechanism that just happens to make it look like were are on a globe. However that kind of falls down when you add the moon and the stars, and the actual image of the sun with sunspots.
 

Rory

Senior Member.
The black lines are the observed directions. The yellow lines are bent rays of light that converge 20,000 miles above the earth.
Wouldn't they be converging at 3,000 miles in the flat earth model?

That's definitely something I've long wanted to see (if possible, with just the bendy lines please).
 
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Mick West

Administrator
Staff member
Wouldn't they be converging at 3,000 miles in the flat earth model?

That's definitely something I've long wanted to see (if possible, with just the bendy lines please).

A viable bendy light model needs satisfy the criteria that the rays of light hit the earth at the same angle from vertical as the latitude (assuming equinox condition, the sun over the equator).

So this means that the light from the sun spreads out in a trumpet shape, like this.
20171005-211427-8d3lp.jpg

Currently this is just hacked together using a quadratic bezier, I need to improve it.
 

Rory

Senior Member.
I wondered if it might not be as easy for 3000 miles - is it the case that there's the one straight line at 47.6°N, and then the lines either side of that bend in different directions?

Also, I guess we've no real idea where the bend 'begins', given the ad hoc nature of the model.
 

Mick West

Administrator
Staff member
I wondered if it might not be as easy for 3000 miles - is it the case that there's the one straight line at 47.6°N, and then the lines either side of that bend in different directions?

No, the only straight line is straight up, everything else would have to be bent.
 

Ranon151

New Member
This can be debunked by examining the propagation of radio waves. Even over long distances of hundreds or thousands of miles (or in the cases of LF and VLF propagation, thousands), no vertical "bending" has been observed. Rockets flown straight up into the early atmosphere maintain continuous radio communication with the ground, something impossible if light bent like that.
 

Mick West

Administrator
Staff member
20171005-222253-fliue.jpg

This is a better bendy-light model, bezier quads, the end points are the sun and the ground. The control point is the intersection of a ray from the ground at (latitude) degrees from horizontal with a line from the sun at ((90-latitude)/2) degrees from vertical. Sun is at 4,000 miles. It needs to be further than an eighth circumference with this type of curve.
 

Rory

Senior Member.
This is a better bendy-light model, bezier quads.
That's pretty convincing - it's quite something when the debunkers are designing better flat earth models than the flat earthers themselves.

Also, not only does the light bend in that direction, but also as it travels from the sun to the various parts of the earth, like so:

flat earth sun patterns.jpg
Images taken from video at: www.youtube.com/watch?v=S8D-xckbTwE

I especially like how, in the northern hemisphere winter, it can bend around North and South America, but illuminate the Antarctic beyond it. Although the pattern of the spring/fall sun takes some beating too.
 
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Mick West

Administrator
Staff member
Also, not only does the light bend in that direction, but also as it travels from the sun to the various parts of the earth, like so:
Yeah, the bendy light thing I did kind of assumes a simple flared cone of light, accounting for the actual shape of the sunlight through the seasons is theoretically possible, but the bends would be just ridiculous. Even in summer it's not trivial.
 

Amber Robot

Active Member
If light were to bend like that in such a way that you can look at low angle but see the Sun, which is actually higher in the sky, would we still see the horizon where it is or would the light bend in such a way as to put the horizon somewhere else? Or does the bendy light only apply to direct sunlight?
 

Rory

Senior Member.
If light were to bend like that in such a way that you can look at low angle but see the Sun, which is actually higher in the sky, would we still see the horizon where it is or would the light bend in such a way as to put the horizon somewhere else? Or does the bendy light only apply to direct sunlight?
I'd imagine it depends on what result the flat earther would like to see: in his understanding, pretty much anything's possible.
 

Hevach

Senior Member.
If light were to bend like that in such a way that you can look at low angle but see the Sun, which is actually higher in the sky, would we still see the horizon where it is or would the light bend in such a way as to put the horizon somewhere else? Or does the bendy light only apply to direct sunlight?
The horizon would probably bend downward, creating the appearance that the observer were not on a flat plane, but some conic curve. A sphere perhaps.
 

Amber Robot

Active Member
The horizon would probably bend downward, creating the appearance that the observer were not on a flat plane, but some conic curve. A sphere perhaps.
Can someone draw me a diagram of that? What are the equations that govern the bending of light that show the consistency between seeing the sun half hidden by the horizon during sunset and the horizon being at about eye level as you’d expect for a flat plain? How can bendy light actually explain a sunset?
 

Rory

Senior Member.
Can someone draw me a diagram of that? What are the equations that govern the bending of light that show the consistency between seeing the sun half hidden by the horizon during sunset and the horizon being at about eye level as you’d expect for a flat plain? How can bendy light actually explain a sunset?
You're actually asking for equations and explanations for something impossible that doesn't exist? ;)

It's fun that we draw a few simple diagrams showing the contortions sunlight on a flat earth would have to perform - but I don't think we can really start trying to explain it.
 

Ranon151

New Member
There's one additional kind of bend you have to take into account - the bend required to make sure the sun maintains constant angular size as it moves from horizon to horizon. This bend has to work in both vertical and horizontal directions.

Of course, for the sun's rays to bend in that way, so would the light from other objects close to the horizon, even though they clearly get smaller as they go further away.
 

Gabibbo

New Member
Notice a thing, if the light bends we can't use shadows to measure angular height of the sun anymore cause even them would be biased. We should use some other technique to measure the sun height. Said that, there exist a unique model for the light paths to make us see the sun at an apparent angle height (90°-latitude) different than should really be in the flat earth model. Is this



Here i setted a sun at 5000 km height, but there's no need to set this height to exactly be 5000 to preserve the property of angular height of the sun equal to 90°-latitude, instead if i let these function in their natural domain they all tends to infinity for x=0. But in order to flat earth model to work with a 5000 km height sun, this sun should have a radius of at least 3018 km, pretty high, so it seemed more reasonable to picture a disc sun instead than a spherical one.

So this model is consistent with the angular height of the sun as can be observed in reality, but despite it is the unique model with this property it projects shadows of different length than can be actually observed, and i don't really know exactly how i can do this calculation but probably even angular size of the sun wouldn't be the right one. That is, the only reasonable conclusion is that deosn't exist a flat earth model that can take account of all possible light ralated phenomena making the whole theory wrong.
 
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