How the Sun Disproves Flat Earth

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Taki
captain of 50
Posts: 81

How the Sun Disproves Flat Earth

Post by Taki »

PART ONE:

I. A Discussion of Latitude and Longitude

Let’s examine the Earth Coordinate System that is currently in use today.

Latitude is measured by calculating how far South someone is from the North Pole. Anciently, naval navigators used the angle of Polaris above the horizon to determine their latitude, because a person moves south, Polaris drops further and further in the sky. Very precise measurements revealed that for every arcminute (1/60th of a degree) that Polaris appears to move toward or away from the Horizon, the person measuring Polaris has moved a fixed distance, later standardized as the nautical mile. The benefit of this is that degrees of latitude correspond exactly to a fixed distance, and all distances from the North Pole in nautical miles can be calculated by multiplying the degrees of latitude from North by 60, using the formula below. Since latitude numbers are written as degrees from the equator, we have to change the latitude around by subtracting it from 90.

LAT >= 0 : Distance = 60 * (90 - LAT)

For locations south of the Equator, degrees of latitude are typically represented by Negative numbers, by convention.

If LAT < 0 : Distance = (90 + (-1 * LAT)) * 60

An example - Honolulu, Hawaii, is at 21.3069 degrees North. Its distance from the North Pole is then given by

60 * (90 - 22.3069) = 4121.586 nautical miles.

We can sanity check this against Anchorage, Alaska (latitude 61.2181 degrees North), which should be considerably closer to the North Pole:

60 * (90 - 61.2181) = 1726.914 nautical miles.

Let’s also check Sydney, Australia (latitude 33.8688 degrees South), which is considerably farther away from the North Pole:

(90 + (-1 * (-33.8688)) * 60 = 7432.128 nautical miles

Latitude checks out.

Longitude is considerably tricker to measure, as it requires accurate clocks that can track UTC time (the time at Greenwich, England), but it basically works by calculating the exact time UTC when the sun is at its highest point in the sky (Solar Noon). As the sun is known to move in its arc across the sky by fifteen degrees per hour - a fact which can be measured by anyone on earth at any time using very rudimentary equipment such as a stick and a measuring tape - longitude is given by changing Hours to Degrees:

1 clock hour = 15 degrees longitude

The decimal part of lat/long can then be rendered by changing the fractional minutes and seconds in this fashion:

Decimal degrees = (arcminutes / 60) + (arcseconds/3600)

Add that to the number of degrees and you’re good.

One of the conventions of longitude is that it’s measured East and West, and that degrees West are typically represented by negative values. For example, Honolulu is situated at 157.8583 degrees West, or -157.8583. If we want to compare degrees from the Prime Meridian, it makes it a little bit easier to understand if everything was measured as Degrees East, which can be converted like this:

If LONG > 0: Degrees East = LONG
Else If LONG < 0: Degrees East = 360 + LONG


This is because any angle plus 360 is essentially pointing in the exact same direction as it was before, but this just makes it positive instead of negative. We do this so that the mathematics become simplified when comparing points in the Eastern Hemisphere and points in the Western Hemisphere.

II. Lat/Long on Flat Earth

Now how does all of this relate to Flat Earth? Well, if you have the Latitude and the Longitude of any two points on Flat Earth, then you essentially have two legs of a triangle and the angle between those legs. This means that with some High School trigonometry, you can calculate the distance between those two points as well as the azimuth (compass bearing) from one to the other.

In the diagram below, I have labeled three points, N for the North Pole, T for the target, in this case Honolulu, HI, and O for observer, in this case San Diego, CA. We can use trigonometry to calculate the distance between O and T because the distances to the North Pole are known by virtue of these locations’ latitude, and the angle between them as a function of the difference of their longitudes.

Honolulu, HI is at 21.3069° N, 157.8583° W, which means that it is 4058.346 nautical miles from the North Pole, and 202.1414 degrees East from the Prime Meridian. San Diego, CA is at 32.7157° N, 117.1611° W, which means that it is 3437.058 nautical miles from the North Pole and 242.8839 degrees East from the Prime Meridian. This means that the longitudinal angle between Honolulu and San Diego is 40.7425 degrees (angle O - angle T). The distance we are going to calculate is (for now) unknown, but we will get to that calculation momentarily, because we can form a triangle using the two lengths and the angle between them.
FE_Cropped.png
FE_Cropped.png (933.52 KiB) Viewed 788 times
III. The Law of Cosines

The key to understanding all this is the Law of Cosines, a formula for deriving the lengths / angles of a triangle if only some of them are known. It’s a generalization of the Pythagorean Theorem, and is stated below:

For the Triangle ABC where sides are denoted by a, b, and c:
triangle.png
triangle.png (39.76 KiB) Viewed 788 times
The Law of Cosines is stated thusly:
cosines.png
cosines.png (28.47 KiB) Viewed 788 times
....CONTINUED IN NEXT REPLY....
Last edited by Taki on March 26th, 2021, 1:01 pm, edited 1 time in total.

Taki
captain of 50
Posts: 81

Re: How the Sun Disproves Flat Earth

Post by Taki »

PART TWO

To solve for the length of one side of the triangle if we know the size of an angle and the two legs of the triangle on either side, we can rearrange the terms of our equation thusly:
fe_distance.png
fe_distance.png (22.17 KiB) Viewed 782 times
IV. The Distance Between Points on Flat Earth

Let the value b = the distance from the North Pole to Honolulu, 4058.346 nautical miles
Let the value a = the distance from the North Pole to San Diego, 3437.058 nautical miles
Let the value gamma = the longitudinal difference between these cities, 40.7425 degrees or 0.7110907705 radians (because most math algorithms insist on using radians instead of degrees)

I will present the formula here in Microsoft Excel format so you can copy it into Excel or Google Docs.

The distance c is then calculated - ensuring a conversion of degrees to radians:
=SQRT((4058.346)^2 + (3437.058)^2 - (2 * 4058.346 * 3437.058 * COS(RADIANS(40.7425))))

Giving us the value 2673.379 nautical miles from San Diego to Honolulu.

V. The Azimuth Between Points on Flat Earth

How, then, do we calculate the azimuth from one point on Flat Earth to another? The Law of Cosines to the rescue! Seeing the diagram above, with the points N, O, and T, calculating an azimuth from O to T is done by figuring out its angle from the line heading north, in this case, the angle at point O. We already know the distance to the North Pole, and we’ve just calculated the distance between O and T.
azimuth_angle.png
azimuth_angle.png (21.8 KiB) Viewed 782 times
Let the value a = the distance from the North Pole to San Diego, 3437.058 nautical miles
Let the value b = the Flat Earth distance from San Diego to Honolulu, 2673.379 nautical miles
Let the value c = the distance from the North Pole to Honolulu, 4058.346 nautical miles
Again, I present the Excel formula for calculating this, converting from radians to degrees at the end (since the Excel ACOS function returns values in radians):

=DEGREES(ACOS( (3437.058^2 + 2673.379^2 - 4058.346^2) / (2 * 3437.058 * 2673.379 ) ))

This returns a value of 82.21231 degrees. Since our target T is to the west of O (angle O is greater than angle T), this requires a correction from true north, by subtracting it from 360 degrees, giving us a true azimuth of 277.78769 degrees, or just South of due West. This passes our sanity check, generally speaking.

VI. A Discussion on the Sun

At any time during the day, somewhere on flat earth, there exists a point directly beneath the sun, or in other words, at local solar noon, the sun is exactly overhead. During solar noon at this subsolar point, any vertical object will cast no shadows at all. This is only possible between the tropics of Cancer and Capricorn, and most of the Earth that falls between these tropics is covered by ocean.

If it is daylight where you are, you can (obviously) see the sun. If you can simultaneously find the subsolar point at that exact time, you should be able to calculate your azimuth and distance to the subsolar point. The distance to the subsolar point, combined with the sun’s elevation in degrees above the horizon, should further enable you to calculate the exact distance the sun is over the flat earth. Conversely, if the sun’s altitude is already known, then you should be able to predict how many degrees it is above the horizon from your observation point. See the diagram below for a description of how to calculate this using the tangent property of triangles.

Any observation of the sun’s elevation is made much easier if you have a measuring device like a theodolite or a marine sextant. However, you can also calculate it’s elevation with a stick and a measuring tape.
sun_angles.png
sun_angles.png (55.74 KiB) Viewed 782 times
CONTINUED IN NEXT REPLY...
Last edited by Taki on March 26th, 2021, 2:08 pm, edited 4 times in total.

Taki
captain of 50
Posts: 81

Re: How the Sun Disproves Flat Earth

Post by Taki »

PART THREE

VI. Experiment and Prediction

Flat Earth hypothesizes that the sun moves above the surface of the earth around the North Pole in a (generally) circular motion, at a fixed altitude from the surface. If Flat Earth is true, then we should be able to predict the azimuth from any observation point to the subsolar point, and using the distance from the observer to the subsolar point, predict the elevation of the sun above the horizon (in degrees) given that we know the distance and the altitude of the sun.

Every year, twice a year, on dates a few weeks before and after the Northern Hemisphere Summer Solstice, Hawaii experiences an event they call Lahaina Noon, when the sun passes exactly overhead, making Hawaii the subsolar point at solar noon on that day. We know where and when this phenomenon occurs, so we should be able to observe the sun during Lahaina Noon from any point on Earth where it is daylight. Here is a table of Lahaina Noon dates and times for 2021 (all times given in Hawaii time):
dates_times.png
dates_times.png (66.67 KiB) Viewed 778 times
We can calculate our latitude and longitude for any point on Earth, so we should be able to gather these points and calculate distance and azimuth to the subsolar point. Because we can also see the sun, we should be able to calculate the elevation of the sun using the tangent property of triangles as shown above.

Here’s a screenshot of an Excel spreadsheet detailing out the calculations for a Lahaina Noon over Honolulu. I will happily provide the spreadsheet to anyone who wants a copy of it, so you can check the formulas for doctoring. Any field in RED is for the user to enter their own values, so when you make observations, it will use the tangent property to calculate the expected altitude of the sun. That value will then be averaged and its standard deviation calculated (Gold colored cells).
Spreadsheet.png
Spreadsheet.png (86.97 KiB) Viewed 778 times
(In this Excel example, I have provided many sample locations along with a sample subsolar point. You can access this as a Google Sheet at this link and download a copy for yourself using File->Download->Microsoft Excel (.xlsx) ⇒ https://docs.google.com/spreadsheets/d/ ... sp=sharing )

To make the observation, collect multiple observers across the globe who will be in daylight during Lahaina Noon and have them record their current latitude and longitude. Have them use any method they choose to measure the azimuth to the sun and its elevation above the horizon at the exact moment the sun is directly over the subsolar point. The stick method, a sextant, a theodolite app (which has the advantage of doing both azimuth and elevation simultaneously), a magnetic compass, a gyrocompass, whatever you need. Put these observations in the Grey column and check the observed values against the expected values. The more observers you have, the more accurate your data will be.

As a note on magnetic compasses, it is also well known that magnetic north is not the same as true north. All observers will need to correct their compass’s magnetic declination for deviations from true north at whatever location they’ll be making measurements from (here’s an article from REI for backpackers: https://www.rei.com/learn/expert-advice ... ation.html )

Ideally, all the observers will be in a group phone call or video call with an observer standing under the subsolar point so that the person - in this case, in Hawaii - who sees the no-shadows event can tell everyone else when to make their readings. If you can’t get someone to the subsolar point, then just make sure that your clocks are all synchronized.

The stick method has the advantage of not needing a clear horizon to be usable, so you don’t need to be on the shoreline of the ocean to get a reading. However, for all the observer locations, I recommend being at sea level. All the example cities I’ve cited above are close to the Pacific Ocean, so getting an elevation reading of the sun over Hawaii is (relatively) easier.

VII. Interpreting the Observations

If the observed values do not match the predicted values - and especially if the observed values are WILDLY different from the predicted values - one of two things is true:

1) Earth is not flat
2) Latitude and Longitude values for points on the earth are totally meaningless

This is especially true if every observation location calculates a different altitude to the sun, because that means that on Flat Earth, the sun is in a different position in the sky for every person observing it, and that for everyone on earth except for the person standing under the subsolar point, the sun is NOT over the subsolar point - a logical and physical contradiction.

Because I am a cynic, I do not expect most Flat Earthers to concede point #1. I’m sure they’ll quibble about how “no one actually knows the shape of the earth, no one has actually mapped it accurately” - but then what of Latitude and Longitude? What about all the ways to do Lat/Long math using handheld instruments and clocks? Are all the mathematicians who ever worked the problems of long-distance navigation part of the conspiracy as well? Are Latitude and Longitude just fake, and if so, how is it possible that you can use them to navigate anywhere?

I believe that this experiment, if performed, will conclusively demonstrate the absurdity of Flat Earth as well as the utter recalcitrance of Flat Earthers, who will make every effort (other than checking the mathematics and repeating the experiment themselves) to attempt to toss this out the window.

Because if Lat / Long are real and accurate, and observations don’t match the Flat Earth predictions, then Flat Earth is done for - otherwise, you’re arguing against trigonometry, and that, my friends, is WAAAYYY harder to do without looking completely ignorant.

So, am I wrong? Get some people on the internet to go collect some data. Let's find out!

braingrunt
captain of 1,000
Posts: 2042

Re: How the Sun Disproves Flat Earth

Post by braingrunt »

Sadly Allison is DEAD SET against collecting sun data, and is bothered that we keep asking her to. Tells us to do it.

I've already done some sun observation and it's doing me little good because I already know the earth is proven not to be flat.

Taki
captain of 50
Posts: 81

Re: How the Sun Disproves Flat Earth

Post by Taki »

braingrunt wrote: March 26th, 2021, 2:11 pm Sadly Allison is DEAD SET against collecting sun data, and is bothered that we keep asking her to. Tells us to do it.

I've already done some sun observation and it's doing me little good because I already know the earth is proven not to be flat.
Cuz 1) she's intellectually dishonest or 2) afraid of discovering that she's wrong. She's also too afraid to make mathematical predictions herself, ones that could be independently verified, as I have just done (and assuming a Flat Earth in my mathematical model, to boot!). Besides, if us "globies" collected the data, I know she would immediately suspect us of having falsified it, in the same way that Flat Earthers automatically denounce any curvature photograph as false because they didn't take it themselves. I'm not arguing with her anymore, but with people who are on the fence about Flat Earth.

To all you out there reading this who aren't sure, here's a method of testing it yourselves - that's what Lahaina Noon is all about.

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TheDuke
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Re: How the Sun Disproves Flat Earth

Post by TheDuke »

Don't need to collect complex data. Just get in a plane, any plane and fly straight for a while (depending on accuracy of your measuring ability) and you will see the world is either a sphere or the north pole magically moves around as if the world is a sphere (oid). When I fly my plane, it even matters that the earth is actually a spheriod, not even purely circular. BTW doesn't work if you only fly North-South or East-West, but for all other headings it becomes obvious.

cmichael
captain of 100
Posts: 168

Re: How the Sun Disproves Flat Earth

Post by cmichael »

Taki wrote: March 26th, 2021, 12:53 pm PART ONE:

I. A Discussion of Latitude and Longitude

Let’s examine the Earth Coordinate System that is currently in use today.

Latitude is measured by calculating how far South someone is from the North Pole. Anciently, naval navigators used the angle of Polaris above the horizon to determine their latitude, because a person moves south, Polaris drops further and further in the sky. Very precise measurements revealed that for every arcminute (1/60th of a degree) that Polaris appears to move toward or away from the Horizon, the person measuring Polaris has moved a fixed distance, later standardized as the nautical mile. The benefit of this is that degrees of latitude correspond exactly to a fixed distance, and all distances from the North Pole in nautical miles can be calculated by multiplying the degrees of latitude from North by 60, using the formula below. Since latitude numbers are written as degrees from the equator, we have to change the latitude around by subtracting it from 90.

LAT >= 0 : Distance = 60 * (90 - LAT)

For locations south of the Equator, degrees of latitude are typically represented by Negative numbers, by convention.

If LAT < 0 : Distance = (90 + (-1 * LAT)) * 60

An example - Honolulu, Hawaii, is at 21.3069 degrees North. Its distance from the North Pole is then given by

60 * (90 - 22.3069) = 4121.586 nautical miles.

We can sanity check this against Anchorage, Alaska (latitude 61.2181 degrees North), which should be considerably closer to the North Pole:

60 * (90 - 61.2181) = 1726.914 nautical miles.

Let’s also check Sydney, Australia (latitude 33.8688 degrees South), which is considerably farther away from the North Pole:

(90 + (-1 * (-33.8688)) * 60 = 7432.128 nautical miles

Latitude checks out.

Longitude is considerably tricker to measure, as it requires accurate clocks that can track UTC time (the time at Greenwich, England), but it basically works by calculating the exact time UTC when the sun is at its highest point in the sky (Solar Noon). As the sun is known to move in its arc across the sky by fifteen degrees per hour - a fact which can be measured by anyone on earth at any time using very rudimentary equipment such as a stick and a measuring tape - longitude is given by changing Hours to Degrees:

1 clock hour = 15 degrees longitude

The decimal part of lat/long can then be rendered by changing the fractional minutes and seconds in this fashion:

Decimal degrees = (arcminutes / 60) + (arcseconds/3600)

Add that to the number of degrees and you’re good.

One of the conventions of longitude is that it’s measured East and West, and that degrees West are typically represented by negative values. For example, Honolulu is situated at 157.8583 degrees West, or -157.8583. If we want to compare degrees from the Prime Meridian, it makes it a little bit easier to understand if everything was measured as Degrees East, which can be converted like this:

If LONG > 0: Degrees East = LONG
Else If LONG < 0: Degrees East = 360 + LONG


This is because any angle plus 360 is essentially pointing in the exact same direction as it was before, but this just makes it positive instead of negative. We do this so that the mathematics become simplified when comparing points in the Eastern Hemisphere and points in the Western Hemisphere.

II. Lat/Long on Flat Earth

Now how does all of this relate to Flat Earth? Well, if you have the Latitude and the Longitude of any two points on Flat Earth, then you essentially have two legs of a triangle and the angle between those legs. This means that with some High School trigonometry, you can calculate the distance between those two points as well as the azimuth (compass bearing) from one to the other.

In the diagram below, I have labeled three points, N for the North Pole, T for the target, in this case Honolulu, HI, and O for observer, in this case San Diego, CA. We can use trigonometry to calculate the distance between O and T because the distances to the North Pole are known by virtue of these locations’ latitude, and the angle between them as a function of the difference of their longitudes.

Honolulu, HI is at 21.3069° N, 157.8583° W, which means that it is 4058.346 nautical miles from the North Pole, and 202.1414 degrees East from the Prime Meridian. San Diego, CA is at 32.7157° N, 117.1611° W, which means that it is 3437.058 nautical miles from the North Pole and 242.8839 degrees East from the Prime Meridian. This means that the longitudinal angle between Honolulu and San Diego is 40.7425 degrees (angle O - angle T). The distance we are going to calculate is (for now) unknown, but we will get to that calculation momentarily, because we can form a triangle using the two lengths and the angle between them.

FE_Cropped.png

III. The Law of Cosines

The key to understanding all this is the Law of Cosines, a formula for deriving the lengths / angles of a triangle if only some of them are known. It’s a generalization of the Pythagorean Theorem, and is stated below:

For the Triangle ABC where sides are denoted by a, b, and c:

triangle.png

The Law of Cosines is stated thusly:

cosines.png

....CONTINUED IN NEXT REPLY....
oh, you mean like THIS SUN?

https://www.youtube.com/watch?v=Hs3Xtb3Rr3Q

Did you know Pythagorus was an occultist, just like Newton was? Did you know the theory of gravity has huge problems that modern science can't resolve? Yeah, I guess you would not know because you are focusing your efforts elsewhere.

Taki
captain of 50
Posts: 81

Re: How the Sun Disproves Flat Earth

Post by Taki »

cmichael wrote: March 29th, 2021, 5:09 pm oh, you mean like THIS SUN?

https://www.youtube.com/watch?v=Hs3Xtb3Rr3Q
Ah, Dubay. Dismisses all of the Antarctic research stations as "part of the conspiracy" using the argument from incredulity logical fallacy and totally ignoring the fact that 1) you can read the entire text of the Antarctic treaty online and 2) even get a job working on the continent if you want to, and especially 3) to patrol the entirety of the Flat Earth Antarctic Ice Wall's circumference of at least 60,000 kilometers would take more ships than there are in the combined navies, coast guards, and merchant marine ships of all the nations on the planet and would require the dedicated cooperation of Earth's most bitter military rivals since James Cook first sailed that far south in the 1800's.

Irrelevant to any discussion on the mathematics of the sun's movement in the sky and the observations you can make here, north of the Antarctic circle, to see for yourself.

I'll also allow myself a little ad hominem against him (not that I need it, there are plenty of Dubay takedowns that address the merits of his "arguments" on a mathematical and scientific basis) and say that this "enlightened" former yoga teacher wouldn't hold an intellectual candle to my two year old nephew.
cmichael wrote: March 29th, 2021, 5:09 pm Did you know Pythagorus was an occultist, just like Newton was?
Irrelevant and clearly a case of argumentum ad hominem - you attack Pythagoras as an occultist, but not his mathematics, which have withstood independent review and proof for literally thousands of years. (Side note - the principles of the theorem have been known in ancient Mesopotamia since at least 1600 BC, 1200 years before Pythagoras, he just gets credit for it.) You can personally prove the Pythagorean Theorem and the Law of Cosines yourself. If the Law of Cosines is wrong, why is it wrong, and can you show me mathematically? Also, what does Newton have to do with the Law of Cosines?

Proof of the Pythagorean Theorem
https://www.mathsisfun.com/geometry/pyt ... proof.html

Proof of the Law of Cosines
https://www.mathopenref.com/lawofcosinesproof.html
cmichael wrote: March 29th, 2021, 5:09 pm Did you know the theory of gravity has huge problems that modern science can't resolve?
Red herring. Diversion from the primary discussion here - the mathematics of the sun's motion in the sky - which again, you fail to address. This is bad science, and even worse debate tactics, as you make no attempt at even glancing at the argument at hand.

A+ for histrionics, but a solid F- for logic.
Last edited by Taki on March 29th, 2021, 9:06 pm, edited 5 times in total.

Taki
captain of 50
Posts: 81

Re: How the Sun Disproves Flat Earth

Post by Taki »

One of the things flat earthers believe is that the Sun doesn’t go below the horizon. Even though it’s clearly visible from basically anywhere that the Sun disappears bottom first, they insist that the Sun actually moves so far away that it stops being visible by the naked eye. Though this is clear to anyone that is not blind, for the purposes of this argument, we’re going to imagine that’s the case.

Also, this requires the apparent size of the Sun to change. This doesn’t happen either, but they insist it does. We’re going to imagine that’s the case too.

So, to start with, we need some initial data:
  • Diameter of the Flat Earth… Let’s say it’s the same as the circumference of the globe earth: 40,000 km
  • Distance to the Sun: 3,862 km (according to number of Flat Earth websites)
  • Angular size of the Sun: 32 arc minutes (0.54º)
  • Angular limit of the human eye: 1 arc minute (0.01666º)
According to the flat earth model, the Sun will be rotating above the Earth in some kind of spiral, going in and out to “explain” the equinoxes, but for this, using the average position of the Sun is enough. All we need is the distance between the Sun position at noon and midnight in the same place. That should be half of the flat earth diameter, so 20 000 km.

So, based on this data, we will want to check if it’s possible to have the Sun floating above the flat earth and it’s angular size so small that’s no longer possible for the human eye to see.

So, at noon, the Sun will be right above us. And we’re assuming midnight will be the time the Sun “sets” (though it clearly happens before — just giving the flat earth model far more space than needed to try to make it work).

For starters, the easiest thing to figure out: How far away would the Sun need to be in order to stop being seen? Knowing that the Sun’s angular size is 0.54º (32 arc min), let’s calculate the size of the Sun:
  • α = 2 * arcsin( d / 2D )
  • D = 3,862 km
  • α = 0.54º
  • 0.54º = 2 * arcsin( d / (2 * 3 862) )
  • 0.27º = arcsin( d / 7 724 )
  • 0.00471237 = d / 7 724
  • d = 36.39834588 km
So, at that altitude, for that apparent size, the Sun has to be about 36.4 km in diameter. Knowing that we can now calculate the distance it needs to be from the observer in order to have it’s angular size be 1 arc minute (or 0.02º):
  • α = 2 * arcsin( d / 2D )
  • d = 36.39834588 km
  • α = 0.02º
  • 0.017º = 2 * arcsin( 36.39834588 / 2D )
  • 0.0085º = arcsin( 36.39834588 / 2D )
  • 0.00014835 = 36.39834588 / 2D
  • 2D = 245,354.539130435
  • D = 122,677.269565217 km
Now, that’s the distance at an angle, so we need to calculate it’s projection on the ground using the pythagorean theorem:
  • H² = O² +A²
  • H = 122,677.269565217 km
  • O = 3 862 km
  • 122,677.26956521⁷² = 3 862² + A²
  • A² = 15,049,712,467.9769 -14 915 044
  • A = sqrt(15,034,797,423.9769)
  • A = 122,616.464734459 km
We just placed the Sun outside the Flat Earth, so that’s not going to work. But we do have a maximum distance we want it to be at: 20 000 km. So, at that distance what is the angular position of the Sun?

Again, using the pythagorean theorem:
  • O = 3,862 km
  • A = 20,000 km
  • H² = 3,862² + 20,000²
  • H = sqrt(414 915 044)
  • H = 20 369.4635177267 km
Now, to find the angle:
  • sin(a) = O / H
  • O = 3 862 km
  • H = 20,369.4635177267 km
  • sin(a) = 3,862 / 20,369.4635177267
  • α = sin^-1(0.1895975315)
  • α = 10.92929754º or 655.75785239775 arc minutes
So, if, somehow the Sun disappeared it would still be about 11º above the horizon. Or, in another terms, you could fit the apparent size of the sun at noon nearly 22 times between the position where the Sun would disappear and the horizon… or 656 times what it should look like at that point. Seems too much, right?

Let’s try something else. When would the Sun be so close to the horizon that a human would no longer be able to see it? For that we need to calculate the position of the Sun when it’s perspective would place it at 0.017º (or 1 arc minute) above the horizon:
  • tan(a) = O / A
  • a = 0.017º
  • O = 3,862 km
  • tan(0.017º) = 3 862 / A
  • A = 3,862 / 0.00029671
  • A = 13,016,076.3034613 km
This is impossible, right? There’s no way the Sun could get that far on a flat earth model.

Still, we’re using the distance that many flat earthers say the Sun is at from the ground. Maybe they’re wrong. Maybe there’s a distance that fits, right?

1000 km up? Nope… that would require 3,370,294 km to get close enough to the horizon.

100 km up? Nope… that would still require 337,029 km.

10 km up? Nope… still requires 33,703 km. And that would require the Sun to be below the altitude of flight of commercial planes.

Let’s calculate the actual value:
  • tan(0.017º) = O / 20,000
  • O = 0.00029671 * 20,000
  • O = 5.9342 km
Ok, so we found out how high the Sun needs to be for the flat earth model to work, right? Not quite…

So, at 5.9 km up, with an apparent size of 32 arc min (or 0.54º), what would the distance for the Sun to cross the limit of the human vision (1 arc min, or 0.017º)? Let’s calculate the new sun size:
  • α = 2 * arcsin( d / 2D )
  • D = 5.9342 km
  • α = 0.54º
  • 0.54º = 2 * arcsin( d / (2 * 5.9342) )
  • 0.27º = arcsin( d / 11.8684 )
  • 0.00471237 = d / 11.8684
  • d = 0.05592829211 km (about 56 meters)
And, it would disappear at a distance of…
  • α = 2 * arcsin( d / 2D )
  • d = 0.05592829211 km
  • α = 0.017º
  • 0.017º = 2 * arcsin( 0.05592829211 / 2D )
  • 0.0085º = arcsin( 0.05592829211 / 2D )
  • 0.00014835 = 0.05592829211 / 2D
  • 2D = 377.0023061004
  • D = 188.5011530502 km
So, the Sun wouldn’t be visible after moving about 188.5 km since noon (377 km total of daylight). Those would be some really short days, right?

Well… maybe the apparent size of the Sun is wrong. Everyone is tricking flat earthers these days. Since we already know a lot of data should be easy to extrapolate the diameter of the Sun:
  • α = 2 * arcsin( d / 2D )
  • α = 0.017º
  • D = 20,000 km
  • 0.017º = 2 * arcsin( d / (2 * 20,000) )
  • 0.0085º = arcsin( d / 40,000 )
  • 0.00014835 = d / 40,000
  • d = 5.934 km (oh, boy… this is going to be funny)
And it’s apparent size at noon would be…
  • α = 2 * arcsin( d / 2D )
  • d = 5.934 km
  • D = 5.9342 km
  • α = 2 * arcsin( 5.934 / (2 * 5.9342) )
  • α = 2 * arcsin( 5.934 / 11.8684 )
  • α = 2 * arcsin( 0.4999831485 )
  • α = 2 * 29.99888512
  • α = 59.99777024º (112.5 times the real angular size of the Sun — or… basically take over 33% of the sky)
There you go, ladies and gentleman! You just need to make the Sun have 5.9 km in diameter so it takes up about 33% of the sky, have it float at just 5.9 km high and somehow have it slow down when passing over somebody’s head and you nearly have something that resembles what actually happens in the real globe earth. I mean, ignoring the fact that the sun actually disappears behind the horizon, that it doesn’t change size, and that you can’t actually see it from above when on a plane or on a somewhat tall mountain… Oh, and you’ll also need to make the moon 112.5 times bigger so we don’t lose the eclipses… and have it float around the same place…

Also, the mountain peaks over 3 km high in the path of the sun probably wouldn’t exist as those would probably be destroyed. And you could actually see the sun from above… and probably turn it into an oven.

TL;DR - The very law of perspective as applied to the Sun itself renders Flat Earth a bubbling mud pot of utter drivel.

User avatar
Silver Pie
seeker after Christ
Posts: 8989
Location: In the state that doesn't exist

Re: How the Sun Disproves Flat Earth

Post by Silver Pie »

From here: https://www.dailymail.co.uk/sciencetech ... ROUND.html More pictures and videos at the link.
A Flat Earther spent a staggering $20,000 on a DIY experiment that accidentally proved the planet is round.

He was in the middle of filming for a Netflix documentary titled Behind the Curve when he realised his pricy mistake.

Seeking to disprove the mass of research led by NASA experts, he explained the particulars of the DIY experiment using a laser gyroscope.

The test involved using a camera to film through two holes, with a person standing on the other side and shining a torch back at the camera, according to a report by Unilad.

The Flat Earther professed that if the light can be seen with a camera, the holes in the fence and the torch all at equal differences above the ground, then he could draw a positive conclusion that the planet is flat.
Yet, in a humorous and expensive turn of events, no light could be seen, causing the Flat Earther to mutter an awkward 'interesting' after the mistake.

. . .

The failed experiment has since gone viral on Reddit, with people noticing that even after the experiment, the Flat Earther kept using 'mental gymnastics' to claim his conspiracy theory was still correct.

One person wrote: 'I've seen that clip many times but have yet to see his explanation as to how that happened.'
At first, I thought the drawing below was from the guy doing the experiment, but I noticed it was from a round earth perspective. Decided to keep it in this post anyway.

Image

User avatar
Ymarsakar
captain of 1,000
Posts: 4470

Re: How the Sun Disproves Flat Earth

Post by Ymarsakar »

More than 1 sun in the heavens.

Also i know the documentary behind the curve. The person using an expensive gyro laser was i think either rob skiba or one of the members of globebusters.

Recalling the incident, he did not survey the area to find out if it was level or not and it was a quick test.

The documentary was edited in such a way as to be biasef against the evidence. Many of the experiments disproving a ball earth not a round earth, was not shown.

Ball is sphere. Round can be circle or ball.

The easier experiment that was done was the island of man somewhete between the uk. With a strong camera zoom they were able to see the entire island from the bottom of lightbouse to the sides. Which is a large number of miles away. Sometimes fog covered the bottom but on a ball earth the entire thing should be behind a wall of water and not visible except maybe tallest buildings.

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