**“….We’ll Be Cruising at 30,000 ft. for the next 4 hours.”**

If the Earth were a sphere, airplane pilots would have to constantly correct their altitudes downwards so as to not fly straight off into “outer space!” If the Earth were truly a sphere 25,000 miles circumference curveting 8 inches per mile squared, a pilot wishing to simply maintain their altitude at a typical cruising speed of 500 mph, would have to constantly dip their nose downwards and descend 2,777 feet (over half a mile) every minute! Otherwise, without compensation, in one hour’s time the pilot would find themselves 166,666 feet (31.5 miles) higher than expected!

A plane flying at a typical 35,000 feet wishing to maintain that altitude at the upper-rim of the so-called “Troposphere” in one hour would find themselves over 200,000 feet high into the “Mesosphere” with a steadily raising trajectory the longer they go. I have talked to several pilots, and no such compensation for the Earth’s supposed curvature is ever made. When pilots set an altitude, their artificial horizon gauge remains level and so does their course; nothing like the necessary 2,777 foot per minute declination is ever taken into consideration.

To maintain a 30,000 ft. altitude around a round Earth, the airplane would have to be angled significantly lower than in the rear of the airplane to maintain a 30,000 foot relationship to the Earth’s curvature.

Yet this never, ever happens. When traveling in an airplane it is level form nose to stern.

This means that the Earth is not a globe but is a level piece of land below us while in flight.

If one says that we are in a vacuum and gravity holds us in, then how is plane able to “escape” Earth’s gravity pull upon reaching cruising altitude when NASA tells us that it would require

From the surface of the Earth, escape velocity (ignoring air friction) is about

7 miles per second, or25,000 miles per hour. Given that initial speed, an object needs no additional force applied to completely escape Earth’s gravity.

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**Basic Geometry on a Sphere**

The Global Earth theorists for 500 years have been telling us the Earth is a sphere. IF the earth is a globe, and is 25,000 English statute miles in circumference, the surface of all standing water must have a certain degree of convexity–every part must be an *arc of a circle*.

From the summit of any such arc there will exist a curvature or declination of 8 inches in the first statute mile. In the second mile the fall will be 32 inches; in the third mile, 72 inches, or 6 feet, as shown in the diagram above. Spherical trigonometry dictates that a ball-Earth 25,000 miles in circumference would curvate 8 inches per mile varying inversely with the square of the mile, so after six miles there would be an easily detectable and measurable 16 feet, 8 inches of downward curvature.

To determine how much the Earth falls away on the curve you take **miles squared X eight inches.** This is an inverse relationship so the farther one travels the greater the distance of feet or miles the Earth will fall away.

Let the distance from T to figure 1 represent 1 mile, and the fall from 1 to A, 8 inches; then the fall from 2 to B will be 32 inches, and from 3 to C, 72 inches. In every mile after the first, the curvature downwards from the point T increases as the square of the distance multiplied by 8 inches. The rule, however, requires to be modified after the first thousand miles. 1

Miles squared X 8 inches

one foot = .000189394 miles

Curvature of Earth

1 mile 5.33 ft. or .12626 mile

10 miles 66.666 ft. or 1.2626 miles

100 miles 6,666.66 ft. or 12.626 miles

So the farther one travels the greater the drop (or rise) in distance.

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Non NASA camera records continual Flat Earth on plane

**Published on Dec 24, 2014**

Check out this excellent amateur balloon footage of our flat, motionless Earth! You can clearly see the Sun is NOT 93 million miles away as we’re told. This is evidenced by the hot-spot seen on the clouds directly underneath the Sun as it moves over the Earth. Over 20 miles high and the horizon remains perfectly flat and rises to the eye-level of the observer all the way up. If the Earth were a ball, no matter how big, the horizon could not rise with the observer like this. On a ball-Earth the horizon would stay where it was and you would have to look DOWN to the horizon further and further the higher you rose.

SkyeJanuary 26, 2016 at 5:52 amLook at footage 9:54…. either that city is flooded catastrophically, or what you’re seeing is the upper parts of those buildings peeking over the curve of the earth. And yet the video stupidly asks, “Do they look slanted?” Completely missing the point that that one video clip completely disproves your entire stupid flat earth idea. Com’on!

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Joe HillJune 13, 2016 at 2:37 amUnless I have this wrong:To figure the amount of distance (in miles)the surface of earth drops (in feet)to account for its curvature, the formula 8″ times the distance(in miles) squared, and then divided by 12 can be used.

1 mile squared= 1

1 x 8″ = 8

8/12 = .66666667 ft

2 miles squared = 4

4 x 8″ = 32

32/12 =2.66666667 ft

3 miles squared = 9

9 x 8″ = 72

72/12 = 6 ft

4 miles squared = 16

16 x 8″ = 128

128/12 = 10.66666667 ft

5•5=25

25•8=200

200/12= 16.66666667 ft

6 miles is 24 ft

7 miles is 32.66666667 ft

8 miles is 42.66666667 ft

9 miles is 54 ft

10 miles is 66.66666667 ft

11 miles is 80.66666667 ft

Interesting to me.

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Andrew HaugenAugust 10, 2016 at 4:03 pmPlease explain this photo I took flying from USA to the uk

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Kevin McMillenOctober 19, 2018 at 11:10 amThe problem is that you’re not doing the correct math equation.You’re taking the drop in 500 miles, 166,667ft and dividing by 60 minutes, or total flight time, to get the correct but ridiculous number of 2,777ft per minute drop.

To simplify, use 600 miles per hour, divide by 60 minutes means you’re flying 10 miles per minute, 10x10x8/12=67ft roughly. So, which is it? Is the plane dropping 2,777ft per minute or 67ft per minute? Neither.

If the plane is flying 10 miles per minute that means it’s going 1 mile in 6 seconds. Is an 8 inch drop in 6 seconds the correct answer? That’s really not much of a drop but that’s still not correct.

A 747 is about 250ft in length. 250ft/5280=.0473 of a mile. The plane is .0473 of a mile long.

.0473x.0473×8=.017 of an inch. In the planes length the earth drops .017 of an inch. That’s just a little over 1/64th of an inch.

Theoretically an airplane could maintain a flight altitude just a little over 1/64th of an inch out of level and circumnavigate the entire globe.

Hopefully facts will prevail in this silly flat earth craze!

Kevin McMillen

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