I've posted this on this thread more than once already but it bears repeating hear at what seems to be the end of the thread....
Proof That the Earth Cannot Be Flat
The last few days I've been playing around with some math and thought I'd post some of it here to see if it might move some of the flat earthers maybe an inch or two back toward reality...
Let's put some of the sunset images we've taken to good use and see if what was observed can be made to fit with the FET (Flat Earth Theory).
FET claims the Sun is approximately 3000 miles above the Earth and they do not dispute well established distances between points on the surface of the Earth. I'm going to be using these two presuppositions in my calculations and you'll want to refer to the following diagram to keep track of the variables...
View attachment 26417
Side a is the distance from the ground to the Sun (3000 mi).
Side b is the distance from an observer to a point on the Earth where it is high noon (the point at which the Sun is at it's highest point in the sky).
Side c (a.k.a. the hypotenuse) is the distance from the observer to the Sun itself.
Angle A is the height of the Sun above the horizon in degrees as seen from the observer.
Angle C is always 90°
Angle B is not relevant to this discussion.
Note from the start that if the Earth is flat and the Sun is 3000 miles up (or any number of miles up for that matter) that angle A can never ever get to 0°. The Sun would never set because no matter how long you make side b of that triangle, angle A is always a positive number. The only way for the Sun to set on a flat Earth is if the Sun dipped below the plane of the flat Earth. If that were to happen, then it would be night everywhere on Earth at once, which we know does not happen. It's always high noon somewhere on the Earth.
That, by itself, ought to be enough to convince anyone that the Earth cannot be flat but there's more. Let's take a look at some of these photos we took last week...
So, since we're assuming a flat Earth, I'm going to focus on just a couple of photos that both show the position of the Sun in degrees above the horizon. I should point out that you don't have to trust the numbers generated by the app on the phones used to take these photos. The numbers can be confirmed by anyone by simply fashioning a simple sextant from a cheap plastic protractor.
I'll use these two photos...
View attachment 26418 View attachment 26419
On the left is the Sun's position as seen from my house on May 8th, 2018 at 01:00 UCT (8:00:01pm central time)
On the right is the Sun's position as seen from Knight's house on the same day just 38 seconds later (7:00:39pm mountain time).
The position of the Sun at my house is just .1° above the horizon while at Knight's it was 10.2° (This information is displayed just to the right of the Sun position indicator. It shows the Sun's heading and then it's elevation in degrees. On Knight's photo it's sort of hidden a little by the NW direction indicator but it reads "Sun 284.0 W 10.2°" The 10.2 is the elevation above the horizon in degrees)
So, let's look at Knight's first...
How far West would you have to go from Knight's house (where sides b and c meet) to get to a place on a flat Earth where it was high noon (where sides a and b meet)?
It turns out that when dealing with right triangles if you have the length of any one side and either angle A or B, you can know everything about the whole triangle!
The math is boring and so I'm not going to show all that. Just go
HERE and plug in the numbers for side b (3000) and angle A (10.2).
You get the following results...
Someone 16,700 miles (length of side b) to his west would see the Sun at it's highest point in the sky for that day.
There is no point on Earth 16,700 miles from Denver Colorado.
Still not convinced? Well just wait till you plug in the numbers from my house!
At my house the Sun was only .1 degrees above the horizon. So plugging in the numbers from my house (side b = 3000 and angle A = .1) we get the following results...
You have to go 1,720,000 miles to my West to find high noon beneath a Sun that was 3000 miles above the surface of a flat Earth.
That's One MILLION seven hundred twenty THOUSAND miles!
(That's more that 7 times the real distance to the Moon!)
Now seriously folks! What more proof could you possibly need? How are you going to possibly refute this?
Are you going to deny that the Sun is about 10° further above the horizon in Denver than it is in Houston? Even if you did that, the distance to noon calculations aren't dependent on that!
Are you going to challenge the validity of the
Pythagorean Theorem?
It seems that's your only option! It's either refute the Pythagorean Theorem or you must reject the notion that the Earth is flat based on the mere fact that the Sun gets to within .1° of the horizon at one point on the Earth while at the same time being directly over head at another.
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So let's do some more math!
This time let's assume that the distances on Earth as reported by Google Earth are accurate but that the Earth is flat.
To make the numbers easy, lets assume a location on the equator on an equinox.
And we'll use the same diagram as before...
View attachment 26421
When it is Noon (90° over head (angle C) in one place it is Sunset or Sunrise 6225.25 miles away (side b).
For our Sunset angle (angle A) we'll stick with .1° because any angle below that produces numbers that are even more embarrassing for the FET.
So, plugging in the numbers
HERE, we get the following results....
The Sun would have to be a mere 10.865122 miles above the surface of a flat Earth.
If you use a smaller number for angle a, then the Sun has to be closer and closer to the surface. An angle of .01 would require the Sun to be just over one mile above the surface of the Earth.
Clete :Clete: