The Earth is flat? It's a circular flat disk surrounded by a wall of ice?

Let me guess. An OBLATE SPHEROID, right?



Ok, Neil de Grasse.

All those 'supposed' photographs of Earth for the past 4-5 decades that clearly depict a perfectly spherical ball.....



...and NOW, all of a sudden, "NO, it's not perfectly spherical, it's ACTUALLY bulged at the equator and fatter in the southern hemisphere!"



.........Really?



Ok, Tommy, you win.

Thanks for encouraging me to actually do the math, BeyondET. This just gets better and better.

Ok so follow along with your calculators, folks. We'll take out all the guess work, here.

Google tells us that Earth is 24,901 miles in circumference.

Half of that gives us a hemisphere or half circumference of 12,450 miles.

Now, if we want to find out how many inches the Earth curves per mile, all we have to do is take the Radius of the Earth,

Google says it is 3,959 miles,

and one quarter of the circumference,

Google says 6,225.25 miles,

and create a simple algebraic equation for the inches of drop per mile.

For this equation, all we need is the Radius distance 'IN FEET'

and the variable (X) that we are trying to deduce.

So, Earth's radius 'IN FEET' is 20,903,520 FEET.

(X) being our variable for 'inches of drop/curve per mile'

Here's our equation:

6,225.25 miles (one quarter of Earth's circumference) multiplied by (X) should equal the same amount of FEET as Earth's Radius 20,903,520.

6225.25(X) = 20,903,520

Now, if we try 8 for X, which is what 'Science' tells us is the 'inches of drop/curve' per mile of circumference on the Earth, look what we get.

6,225.25 x 8 = 49,802 inches. (Note: 'NOT' 20,903,520 inches)

Now divide that by 12 to get the feet. 49,802/12 = 4,150.16666 feet.

If our Radius is 3,959 miles, then our feet of drop should be 20,903,520 feet.

SO, this tells us that 8 inches per mile is WAY too little for all of "Science's" stated measurements of the Earth to be correct.

In order for Science to be correct about all their stated measurements of the Earth, (X) would need to equal 3,357.8603268945 inches of drop per mile.

Try it for yourself on your very own calculator.

6,225.25(3,357.8603268945) = 20,903,519.999999

If your calculations match these results, and I assure you they do unless you've done something wrong, then that means there are supposedly 3,357.86 inches of drop per mile. That's almost 280 feet or 279.821693907875 feet.

Now, for anyone math-inclined, I welcome any correction to that.

AND, having shown all that, we can see very clearly that there is absolutely no way possible that the amount of curvature this would demand would be very visible in that pic, regardless of all the compressing factors of the image, etc.

And Mt. Rainier would be 54,565.230312035 feet below visible view. That's over TEN MILES below view.

I think the dishes are done, here.

Earth is FLAT!!!

What point do you prove with pictures of the earth shaped like a pear?

Since the Earth is flattened at the poles and bulges at the equator, geodesy represents the shape of the earth with an oblate spheroid. The oblate spheroid, or oblate ellipsoid, is an ellipsoid of revolution obtained by rotating an ellipse about its shorter axis.

My bust on my statement of earth being 3900 miles radius and some change, your correct it's more than 20,000 miles so yea it would be extremely hard to see a 8 inch drop per mile with the naked eye.

Here is Carl Sagan explaining how in 240 BC, Eratosthenes deduced that the earth's surface is curved, that the geometry of the entire planet must be spherical, and calculated very closely the actual diameter of it, all by looking at the shadows of a couple of sticks and noting the passing of the summer solstice.

[video=youtube;0ZLI7WZZRJQ]https://www.youtube.com/watch?v=0ZLI7WZZRJQ[/video]

((from http://twanight.org/?id=3003972))

View attachment 159228

A long exposure photo sequence with an ultra wide-angle lens has captured nearly an entire night over the Teide Observatoryon the high slopes of Mount Teide, Tenerife, Canary islands.
The Earth rotation appears in form of the rotating sky above us. Long exposure night-time images capture this motion in form of star trails around the celestial poles. Here the north celestial pole is on the left, marked by star Polaris, only a degree away from the north direction of the Earth rotation axis. On the right end star trails over the southern horizon are curved around the southern celestial pole (under the horizon of this northern hemisphere location).

Juan Carlos Casado, Starryearth.com

I'm just confused is all.

There are many pictures on the internet and in other media depicting the earth as being round. Why do these not disprove flat earth theory?

In general, we at the Flat Earth Society do not lend much credibility to photographic evidence. It is too easily manipulated and altered. Many of the videos posted here to "prove a round earth" by showing curvature will show no curvature or even convex curvature at parts. The sources are so inaccurate it's difficult to build an argument on them in either case. Furthermore, barrel distortion and other quirks of modern cameras will cause a picture to distort with little or no apparent altercation; especially without references within the picture. Photographs are also prone to distortion when taken through the bent glass of a pressurized cabin as well as atmospheric conditions on the outside. With this litany of problems, it's easy to see why photographic evidence is not to be trusted.
It's also worth noting that if a person were to look down at the earth from high above, they would expect to see a circular shape where the sun's spotlight was shining. This explains why high altitude photographs are generally curved to produce the illusion of a round horizon.

---> Frequently Asked Questions - The Flat Earth Wiki

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The earth is not flat where I live
Or is this the edge of the world?