The summarized question: "Why do you say the FE observations are based off of bad science?"
The reason I say that is because I've gone through EVERY "fact" presented by the Flat Earth Community so far, and not one of them has panned out to be provable scientifically. Several of them are improvable though, because they are based entirely off of perception.
So, that leads me into the topic of this article... One thing that keeps coming up over and over again as a "proof" are the videos taken from various shorelines showing objects at a distance. Time and time again, these videos are presented as proof of a flat earth, but upon detailed analysis most of them actually prove the opposite.
I apologize for the delay in writing this article. I wanted to try to get as much factual information into it as possible and it took much longer than I had anticipated.
The Effect Of Perception
Here is the basic concept of what should be happening (curve exaggerated on purpose):
It seemed pretty simple, you look out straight and level, and measure the drop from where you are looking, and calculate the drop.
People tend to look "down" upon the viewing target without even realizing it, and measuring from the wrong angle. The green arrowed line represents the actual drop, while the blue arrowed line represents the perceived drop.
But what I "know" is irrelevant without proof, I needed to go beyond it being simply what I said, I needed to provide mathematical proof to show what I am talking about.
Calculating The Drop
So.... First of all I wanted to find out what is the proper formula to calculate "the drop".
The simple concept is that you should be able to measure out a certain distance, and whatever formula you use should be able to calculate the drop. Based on that concept, the criteria I used to evaluate each formula was that if you measure out a distance equivalent to the radius of the Earth, the calculated drop should then equal the radius.
To better visual this, I wrote a short computer program that plotted out the drop for each formula into an image, and wrote the actual drop measurements to a text file. The correct formula should plot out a perfect quarter-circle.
Disclaimer: For these formulas, I am using a scenario of the Earth being a perfectly round sphere. I know it is not perfectly round, but the variations in the diameter at the Equator are approximately, 2/10 of 1% of the total size, so for all intents and purposes we an assume it round, at least for these calculations.
The FE Formula
The first formula I tested was the one widely used by the FE community:
"NASA uses spherical trigonometry to determine the curvature of a sphere the size of Earth: 8 X the distance in miles squared."
Or: Drop = 8 + Distance(squared)
(the number 8 representing inches, and the answer is in inches)
First of all, this isn't spherical trigonometry, this is basic math, not even to the Algebra level.
Second, Spherical Trigonometry is only needed when calculating multiple points on sphere. If you're only measuring two points, you don't need to go beyond very basic Geometry, which is taught in pre-Algebra classes.
Anyways, the results of this formula surprised me when plotted out. At a radius of 3959 miles (the officially recognized radius), this formula only curves down 1979 miles at the maximum, almost exactly half of the drop it should have had.
I also found this formula to be tuned a bit to a specific radius, where shorter a shorter radius produces a different shaped curve than a longer one. So, after little bit of experimenting with different values for the radius, I got the formula to finally show the end right drop somewhere around 8000 miles, but the calculated curve by then was not symmetrical, ending up being almost a straight line about halfway through.
The Pythagorean Theorem
The next formula I tested was the well-known Pythagorean Theorem:
Drop = squareroot of Radius(squared) * Radius(squared) + Distance(squared) * Distance(squared)
I knew even before I started that this formula wouldn't plot out correctly, but since a lot of people were talking about it in relation to this topic, I put it in there to see how it would plot out. Besides, for some reason it "seemed" like this formula was involved somehow.
What this formula actually does is to plot the length of the side of a triangle. Since the radius is used for both sides of the equation, we basically have a right triangle when at maximum range.
Subtracting the distance from each result, it does provide a curve though. Surprisingly, it plotted very similar to the formula used by the FE community, for a maximum of 1639 miles, a little less than half the radius,
The Correct Formula?
Now that I had a way to test the various formulas, I started looking for one that would plot out the correct curve.
I spent the next several weeks (not full-time, just when I got some free time) going through all of the formulas I could find for this calculation. What surprised me the most is the similarity between them all, and that none of them seemed to work properly. A lot of people seemed to try to throw trigonometry into the formula for some reason. But even then, their formulas never seemed to plot out a good curve.
Then it finally dawned on me to start looking at computer models since it is a common function of programs to calculate curves in order to draw things correctly on the screen. After a bit of searching, I finally located a few that seemed to work properly. I tweaked them a bit to make them a bit more simple, and finally ended up with this formula:
Drop = Radius - squareroot of (Radius * Radius - Distance * Distance)
Finally I had a proper formula to work with! Does it look familiar? That is because it is a variation of the Pythagorean Theorem, something I vaguely suspected from the beginning.
One added bonus to this formula is that it will also work with any type of numbers. Unlike most of the formulas I found out there, this one returned the same unit of measurement put in. I.e., if you enter miles, the answer will be in miles, if you enter feet, the answer will be in feet, etc.
Note: This is a serious problem with a lot of the formulas published on the Internet, in that you would have to put in feet in some places, miles in another, and it would return something else. Some of the formulas even mixed kilometers and miles in the same equation!
Here is the output image from my program (hard to see at this small size):
The red line across the top is the 8 inches/mile calculation (which turns out to be less than one mile).
The grey diagonal line is simply a 45 degree angle line I used for reference.
The orange line is the Pythagorean Theorem
The yellow line is the FE formula
The blue line is my revised formula
Crunching The Numbers
Now that I had an accurate formula to work with, and could plot them all out, I started crunching the numbers to see what they gave. I had long suspected that the FE formula was incorrect, I was quite sure that it was correct for short distances but exaggerated the curve quite quickly.
I'll say it right up front, I was wrong... The FE formula IS incorrect, but in the opposite direction. It actually dampens the slope the further it goes out (when the radius is close to 3959).
But, for the first several miles, the results were surprisingly similar:
FE Formula:
Mile: 1 Drop: 8 Inches
Mile: 2 Drop: 32 Inches
Mile: 3 Drop: 72 Inches
Mile: 4 Drop: 128 Inches
Mile: 5 Drop: 200 Inches
Mile: 6 Drop: 288 Inches
Mile: 7 Drop: 392 Inches
Mile: 8 Drop: 512 Inches
Mile: 9 Drop: 648 Inches
Mile: 10 Drop: 800 Inches
Pythagorean Theorem:
Mile: 1 Drop: 8.00202057027491 Inches
Mile: 2 Drop: 32.0080808116472 Inches
Mile: 3 Drop: 72.0181760852574 Inches
Mile: 4 Drop: 128.032298726903 Inches
Mile: 5 Drop: 200.050438047037 Inches
Mile: 6 Drop: 288.072580215521 Inches
Mile: 7 Drop: 392.098708463309 Inches
Mile: 8 Drop: 512.128802794323 Inches
Mile: 9 Drop: 648.162840273581 Inches
Mile: 10 Drop: 800.200794883131 Inches
Revised Formula:
Mile: 1 Drop: 8.00202082959004 Inches
Mile: 2 Drop: 32.0080849030637 Inches
Mile: 3 Drop: 72.0181967440294 Inches
Mile: 4 Drop: 128.032364074315 Inches
Mile: 5 Drop: 200.050597583468 Inches
Mile: 6 Drop: 288.072911044001 Inches
Mile: 7 Drop: 392.099321340211 Inches
Mile: 8 Drop: 512.129848381737 Inches
Mile: 9 Drop: 648.164515103563 Inches
Mile: 10 Drop: 800.203347581264 Inches
Where Do We Go From Here?
Now that I've proven that the formulas provided a reasonably reliable measurement in the first several miles (regardless of which one you use), does this make those videos correct?
Not at all. While the estimated drops were not far off from the real ones, the methods used to observe those drops is still quite erroneous.
For example, let's take this video:
At face value, this video seems pretty convincing. However, once you start looking at it closer the evidence falls apart quickly.
I went through the video closer, and if you stop it at 30 seconds in, you will see this image:
This is a view of the Toronto Skyline from across the lake. Take special note of the white object right below the CN Tower. That is the top half of the roof of the Rogers Center.
To better explain this, I searched Google for a picture of Toronto taken from the same direction, and finally found this image:
Of special note, the Rogers Center is not a small building:
Looking back at that first screenshot from the video, you will see that approximately 2/3 of the Rogers Center is hidden below the horizon. This means that the video attempting to prove a flat plane is instead proof of a drop of over 200 feet!
Not to mention that there are many other erroneous claims in video.
But this post is not about picking apart a particular video, this is about the methods used. If you want a particular video dissected, please submit a separate question.
So What Is Happening?
So, what is really happening in these videos to make them so inaccurate? Well, the tests being conducted are not taking all of the various factors into consideration.
Some of the main concerns ignored are:
- Height of the viewing position.
- Height of the target.
- Atmospheric Refraction.
- Height of the viewing position
As was mentioned earlier, these are the IDEAL measurement conditions. The vantage point is directly level with the surface, and the distance measured is perfectly level horizontally from the vantage point.
However, this is nearly impossible to do. Getting a camera right exactly at the water level is very hard to do, and risky to your equipment. Then, maintaining a perfectly level viewing angle is very difficult to do without some sort of leveling equipment. Simply looking out at the horizon forces a downward angle view, regardless of intentions.
Then, once you move your camera ANY distance above the surface, you have to factor in the "Distance to Horizon" (DTH).
From the the viewing position to the horizon, the drop will be zero. The measurement to calculate the real drop is done from the DTH to the target.
Here is the formula to calculate the DTH:
DTH = SquareRoot(height above surface / 0.5736)
(Height is in feet, and output is in miles.)
For example, if the target being viewed is four miles away and the camera is six feet (eye level for a lot of people) above the water level, the DTH is 3.2 miles. That means you have to calculate a drop of .8 miles, for a drop of 6.4 inches.
- Height of the target
The height of the target also has to be taken into account. This is done identically to the distance to horizon, only from the other side.
For example, if the target is 20 miles away and 300 feet tall (such as a tall building), and the viewing point is size feet above the water:
Distance to horizon from the viewing point is 3.2 miles
Distance to horizon from the target is 22 miles.
Calculated drop at 20 miles is 266 feet.
Combined, this gives a total of 25.2 miles, which is well beyond the distance between the two. Thus, the person would be able to see the top 34 feet of the building, while someone at the top of the building would be able to see the entire person.
This scenario is clearly demonstrated in the video posted above. A scenario I used before to describe this is two people standing on opposite side of a very small hill. Since both people can see the feet of the other person, they then deduce that there is no hill.
- Atmospheric Refraction
Nothing gets the people in the FE community riled up faster than referencing atmospheric refraction.
However, it is a well-known and measurable phenomena, easily provable. This phenomena is what is responsible for all the rippling and distortion in videos and photographs. It varies by temperature, humidity, and altitude. At ground level it is the cause for mirages and the weird rippling scene you get when looking out across a parking lot on a hot summer day.
https://en.wikipedia.org/wiki/Atmospheric_refraction
Of key note with this phenomena is the experiment most widely referenced in the FE Community, the Bedford Level Experiment.
https://en.wikipedia.org/wiki/Bedford_Level_experiment
In this experiment, Samuel Rowbotham would demonstrate his perception of a flat earth regularly to people by bringing them out onto a canal, and viewing a distant object (also on the canal) through a telescope placed inches above the water level. He conducted this experiment regularly for a few years, until 1870, when one of his supporters, John Hampden, decided to publish a wager of £500 in the newspaper for anyone who could prove the earth was round.
Alfred Wallace took him up on that bet, and conducted similar experiments on the Bedford Level, but placed them high enough above the ground to correct for refraction. He easily won the bet, and was awarded the money by an agreed-upon referee.
Sadly, Hampden didn't take that well, and brought Wallace to court. The courts agreed with the findings, but had the money returned due to the issue of gambling. Over the next several years, Hampden would repeatedly attack Wallace physically, verbally, and in the courts, over and over. Hampden would be jailed, get released, resume his attacks, and get jailed again. This went on non-stop until Hampden finally died almost twenty years later.
So What Is A Good Experiment?
As you can see, conducting an experiment like this is difficult at best. You cannot simply bring a camera to the shoreline and claim that as proof.
Based on all of the information listed above, I have thought long and hard about how to conduct a reliable experiment, one that is relatively simple to do and not too expensive.
The following requirements have to be met in order to conduct an experiment that will produce accurate results:
- A calm, open, body of water that extends a minimum of a couple of miles, the further the better. Rivers will not work because they have a slope to them, and oceans are not recommended due to the size of the waves.
- A good camera with a powerful optical zoom, positioned several feet above the water level on a stable tripod.
- A bunch of identical floating buoys, the more the better, each with a relatively large object floating above the water level at exactly the same height as the camera. These buoys should positioned in a straight line away from the camera, spaced out evenly to cover as far of a distance as possible.
Once all of the buoys are in place, the view from the camera will show the middle ones slightly higher than either end.
There are other similar experiments possible, I was merely looking for one that someone could recreate with a minimal expense.
Here is a video that shows what I am talking about.
Here is a video that shows what I am talking about.
In Conclusion
I could ramble on about this for many pages, but this is the basic meat of what I had to say.
If there is a particular aspect of this that you disagree with, or want more information on, feel free to ask a question by sending it to Question4XL@gmail.com.
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