Best Proof For A Flat Earth – Modern Airplane Travel
Since the beginning of recorded history, man has known that he
lives on a flat earth. The Sumerians, Babylon’s, Hebrews,
Egyptians, Indians and Chinese all agreed the world did not move
and was flat. They saw the sun, moon and the stars moving on the
sky.
Heliocentric Model of the Earth
About 500 years ago, new theories were presented that the earth is
in fact moving and the sun, moon and stars are all stationary.
This began the advent Copernican heliocentric astronomy which has
been the dominant theory taught. See Appendix 1 – Heliocentric
Model.
500 years ago until this last century, there was no way for anyone
to view the earth at 35,000 feet and see the earth’s physical
structure. Today – it is a common experience – it is called
airplane travel.
Modern Airplane Travel – Last Century
Modern airplane flight has created two (2) new experiences viewed
by a significant part of the earth population that we never had
until the last century:
• Ability to see the earth’s horizon at 229 miles away from a
flight height of 35,000 feet
• Flight speed of 500 – 600 miles per hour.
Physical Sensations
We get into an airplane and feel the thrust of the takeoff and the
acceleration and tilting as we ascend at an angle into the sky. We
are cautioned not to leave our seats until a cruising level
attitude is reached which is normally 35,000 feet. We are usually
told by the pilot that we will flying at this level until we reach
our destination. Once this level is reached – we are told that we
are free to take off our seat belts and move around the cabin. We
feel and experience a level flight until we are told that we will
be descending into our expected destination. As we descend, we can
both see and feel a tilting and descending to ground level.
What We Experience and See
We feel and see the ascent tilt up and the level flight motion
when at the proper cruising attitude and the descent tilt down to
our destination. We see a level horizon is all directions.
If The Earth Were A Globe – What We Would See And
Experience.
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.
We would experience an ascent to flight height and then a
continual descent to correspond with the curvature of the earth.
You would definitely feel this and see this as you are moving at
500 – 600 miles per hour.
At 600 Miles per hour – we cover a significant distance on which
curvature could be examined. The airplane would need to descend
166,666 feet in the one hour time.
Miles squared x 8 inches of curvature divided by 12 inches per
foot. See Appendix 2 – Calculation of Earth Curvature.
((500 x 500) x 8) / 12
500 x 500 x 8 = 2,000,000
2,000,000 /12 = 166,666 feet
View From Within the Plane
What would should see is a very pronounced curvature based on the
exact same calculations you would get from the scientists who
measure the earth’s calculation. See Appendix 2 – Curvature
Calculations.
What you see is flat. But – at 35,000 feet - the horizon is at a
distance of 229 miles.
http://www.ringbell.co.uk/info/hdist.htm
Based the standard model for calculating curvature, you should see
34,960 feet lower views on both the ends of your viewing area.
Each 70 miles equals about 1 degree of the 360 degrees of the
supposed globe we live on.
At 35,000 feet looking out of an airplane – we will calculate the
amount of curvature using the same formula used by professors of
Geodesy – Earth science. ( See Appendix 2 – Calculation of Earths
Curvature)
Miles squared x 8 inches of curvature divided by 12 inches per
foot.
((229 x 229) x 8) / 12
229 x 229 x 8 = 419,528
419,528 /12 = 34,960 feet
The amount of downward curvature would be equal to the height of
one of the highest mountains on our planet. I think we could agree
that we would be able to see this. Instead – we see flat.
Relative Distances and Curvature
Now the distances traveled you necessitate the pilots make some
provision for going around this globe.
If the Earth is 24,900 miles in diameter then if we divided that
number by 360 degrees of a sphere we would come up with a distance
between each of the 360 degrees of about 70 miles.
I have taken these flights and will attest that my experience was
the same on all. Takeoff, ascend to cruising attitude, cruise at
that same height for the duration of the flight and descent to the
location.
Los Angeles to Rome – Distance – 6,382 miles
6,382 / 70 per degree - 91 degrees traveled – West to East
Los Angeles to Bangkok, Thailand – Distance - 8,309
8,309 / 70 per degree - 118 degrees traveled – East to West
Los Angeles to Capetown, South Africa – Distance –
9,958
9,958 / 70 per degree - 143 degrees traveled – West to East and
South.
Never done personally.
Flight Control Manuals
Flight control manuals assume a Flat earth. They know the earth is
flat. They make no provision for the curvature of the earth when
they are flying. They ascend and level out and fly to destination
and descend. There is no auto pilot that constantly adjusts the
level of flight to 35,000.
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.
http://books.google.com/books?id=CqhmXXdTIcIC&pg=PA59&dq=flat+earth&hl=en&sa=X&ei=jVyLU7ShGIaCogTbj4DYBA&ved=0CP4BEOgBMCo#v=onepage&q=flat%20earth&f=false
Conclusions:
We don’t feel or see a constant declination in an airplane ride
that we would need to experience with the curvature of the earth
that we are told we have.
We don’t see a curvature on the horizon at 35,000 feet when if the
earth were curved we would see a curve down 34,900 feet in our
view of the horizon from an airplane – the height of one of the
highest locations on earth.
Flight manuals used to teach air navigation to airline pilots uses
horizontal plane geometry to calculate their flight plans and
makes no provision for the earth’s curvature in their flying.
Ask any pilot – they ascend to their cruising altitude and then
fly at that level until they reach their destination. There is no
autopilot that constantly adjusts their altitude. If this was
happening, you will feel the constant lowering of the flight
altitude to the amount of 2,700 feet per minute while in flight.
You have been presented with proof that you have seen with your
eyes that the earth is flat on every flight you have ever taken.
Based on the curvature calculations presented by Geodesy
professors and Isaac Newton’s own method for calculating curvature
of the earth, you would see a very different sight when flying at
35,000 feet.
What you do see and feel is true – our eyes are telling the truth
– you see flat earth and a horizontal – not curved horizon.
Why we have been told this is another story….
Appendix 1 - Heliocentric Model Of the EarthAppendix 1 -
Heliocentric Model Of the Earth
The Heliocentric Model (Earth spins on axis and rotates once per
day) is built purely on assumptions that deny all observational
and experimental evidence.
Notice these seven assumptions which are indispensable to the
Heliocentric Model in general and are so apparent in the Solar
Eclipse Phenomena.
1) It must be assumed that the Sun is stationary in the "solar"
system relevant to the Earth (and to the Moon) and that it has
never traveled East to West daily across the sky as observed by
everyone on Earth throughout all history.
2) Likewise, it must be assumed that the Earth rotates West to
East ccw (counterclockwise) on an "axis" every 24 hours at an
equatorial speed of c. 1040 MPH in spite of there being nothing
but a mathematical model's "necessary evidence" for this motion
whatsoever.
3) It must be assumed that the Earth is also orbiting the Sun
annually (ccw) at an average speed of c. 67,000 MPH.
4) It must be assumed that the Earth’s axial alleged tilt of 23.5
degrees--in combination with its assumed annual orbit around the
sun--is the only available scientific explanation for the seasons.
5) The Earth’s atmosphere must be assumed to be just an airy,
fixed extension of the alleged rotating Earth. It is assumed and
must be assumed that this atmosphere must have the remarkable
ability to synchronize speeds of objects in it at all
altitudes--birds, clouds, jets, low orbit satellites, alleged
geo-synchronous satellites over 22,000 miles out--and to be
unaffected by alleged Earth movements of speeds ranging from 1000
MPH to 67,000 MPH to 500,000 MPH to 660,000,000 MPH. This
assumption is mandatory once the rotating Earth assumption is made
and cannot be ignored in the helio model of the eclipse phenomena.
6) A particularly fantastic assumption necessary to accommodating
the precise Solar Eclipse Phenomena in the Helio Mathematical
Model involves the bold reversal of the Moon’s observed direction
of travel. Acceptance of this. has no basis in reality, of course.
Rather, it must be coupled with prior acceptance of the other
assumptions of a rotating Earth orbiting a stationary Sun. No moon
reversal means no accurate eclipse forecasts and no accurate
eclipse forecasts means no heliocentricity model.
7) It must be assumed that the Stars do not move around the Earth
diurnally (as they have been observed to do by everyone who has
ever lived).
Each one of these seven assumptions is dependent on the other six.
They are all interdependent and totally without observational or
experimental support. They are solely mathematical models
contrived to account for eclipses and other phenomena and replace
the fact that what we actually see explains the phenomena.
Appendix 2 - Calculations of the Curvature of the Earth
There are two (2) measurements of curvature we will use to
compare.
First - Rule Of Thumb For Calculating Curvature.( Square Of
Distance Multiplied By 8)
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.
The Pythagorean theorem for triangle TPO => (R+h)2 = R2+l2.
So,
h = (R2+l2)0.5 – R
R{[1 +(l/R)2]0.5 -1}
The binomial theorem => (1+x)0.5 = 1 +x/2 –x2/8 +…
Thus, we have h ~ l2/2R, approximately. This shows why your
downward curvature h increases as the square of l, the
distance PT.
The earth surface is close to an ellipsoid and thus the radius
of earth curvature R changes with the latitude of location T.
Near the latitude 45 degrees, R is about 6371 km or 2.0902x107
ft. Using this constant number, you can show that the downward
curvature H is about 8.0025 x L2 inches where L is the
distance PT in (statute) miles.
This formula was given to me by a Professor of Geodesy at Ohio
State University.
Newton’s Principia – Lemma XI - Calculation of
Curvature
(Distance squared) / Diameter = Depression.
This is given as a corollary of Lemma XI. In Newton’s Principia.
It has nothing to do with us: but is what is necessary if the
earth be a globe.
Now, that statement of the law being correct it follows that the
square of the Distance, divided by the Diameter equals the
Depression. Or:
(Distance) Squared divided by Diameter of Earth
_________ = Depression.
Diameter
If therefore we wish to find the Depression for any given distance
we must first square the distance, and divide it by 8,000, (that
being the number o f miles in the earth’s diameter).
This will give us the amount of the Depression. The fact that 8
times this particular distance is 64,000, and that the number of
inches in one mile is very nearly the same, viz.: 63,360, accounts
for the “rule of thumb” which simply multiplies the square of the
distance by 8, and reckons the answer as inches.
It will be seen by working out the Distance experimented on in the
Bedford Level, how nearly this rough and ready rule corresponds
with the exact mathematical calculation. The distance in question
is 6 miles.
Now (6 x 6 ) / 8,000 = Depression.
But we cannot divide 36 miles (6 X 6, or 36) by 8,000 without
reducing the 36 to some smaller dimension. Let us reduce 36 miles
to inches, and then we have
2,280,960 (Six miles) / 8,000
= 285 inches (Depression).
2,280,960 divided by 8,000 = 285 inches
By the rule of thumb method we have
6 x 6 = 36
x 8
______
288
So that by the two methods of calculation, the ‘‘rule of thumb ”
method (288 inches) is only 3 inches in excess of the Newtonian
mathematical method, which is 285 inches.
|
Rule Of Thumb - Distance Squared x 8
|
|
Newton's Curvature of Earth Calculation
|
|
Miles
|
Kilometers
|
Curvature Drop Feet
|
Curvature Drop Meters
|
|
Miles
|
Feet
|
Inches
|
Curvature Drop Feet
|
Curvature Drop Meters
|
|
1
|
0.62
|
0.67
|
0.20
|
|
1
|
5,280
|
63,360
|
0.66
|
0.20
|
|
2
|
1.24
|
2.67
|
0.81
|
|
2
|
21,120
|
253,440
|
2.64
|
0.80
|
|
3
|
2
|
6
|
2
|
|
3
|
47,520
|
570,240
|
6
|
2
|
|
4
|
2
|
11
|
3
|
|
4
|
84,480
|
1,013,760
|
11
|
3
|
|
5
|
3
|
17
|
5
|
|
5
|
132,000
|
1,584,000
|
17
|
5
|
|
6
|
4
|
24
|
7
|
|
6
|
190,080
|
2,280,960
|
24
|
7
|
|
7
|
4
|
33
|
10
|
|
7
|
258,720
|
3,104,640
|
32
|
10
|
|
8
|
5
|
43
|
13
|
|
8
|
337,920
|
4,055,040
|
42
|
13
|
|
9
|
6
|
54
|
16
|
|
9
|
427,680
|
5,132,160
|
53
|
16
|
|
10
|
6
|
67
|
20
|
|
10
|
528,000
|
6,336,000
|
66
|
20
|
|
20
|
12
|
267
|
81
|
|
20
|
2,112,000
|
25,344,000
|
264
|
80
|
|
30
|
19
|
600
|
183
|
|
30
|
4,752,000
|
57,024,000
|
594
|
181
|
|
40
|
25
|
1,067
|
325
|
|
40
|
8,448,000
|
101,376,000
|
1,056
|
322
|
|
50
|
31
|
1,667
|
508
|
|
50
|
13,200,000
|
158,400,000
|
1,650
|
503
|
|
60
|
37
|
2,400
|
732
|
|
60
|
19,008,000
|
228,096,000
|
2,376
|
724
|
|
70
|
43
|
3,267
|
996
|
|
70
|
25,872,000
|
310,464,000
|
3,234
|
986
|
|
80
|
50
|
4,267
|
1,300
|
|
80
|
33,792,000
|
405,504,000
|
4,224
|
1,287
|
|
90
|
56
|
5,400
|
1,646
|
|
90
|
42,768,000
|
513,216,000
|
5,346
|
1,629
|
|
100
|
62
|
6,667
|
2,032
|
|
100
|
52,800,000
|
633,600,000
|
6,600
|
2,012
|
|
110
|
68
|
8,067
|
2,459
|
|
110
|
63,888,000
|
766,656,000
|
7,986
|
2,434
|
|
120
|
75
|
9,600
|
2,926
|
|
120
|
76,032,000
|
912,384,000
|
9,504
|
2,897
|
|
130
|
81
|
11,267
|
3,434
|
|
130
|
89,232,000
|
1,070,784,000
|
11,154
|
3,400
|
|
140
|
87
|
13,067
|
3,983
|
|
140
|
103,488,000
|
1,241,856,000
|
12,936
|
3,943
|
|
150
|
93
|
15,000
|
4,572
|
|
150
|
118,800,000
|
1,425,600,000
|
14,850
|
4,526
|
|
160
|
99
|
17,067
|
5,202
|
|
160
|
135,168,000
|
1,622,016,000
|
16,896
|
5,150
|
|
170
|
106
|
19,267
|
5,872
|
|
170
|
152,592,000
|
1,831,104,000
|
19,074
|
5,814
|
|
180
|
112
|
21,600
|
6,584
|
|
180
|
171,072,000
|
2,052,864,000
|
21,384
|
6,518
|
|
190
|
118
|
24,067
|
7,336
|
|
190
|
190,608,000
|
2,287,296,000
|
23,826
|
7,262
|
|
200
|
124
|
26,667
|
8,128
|
|
200
|
211,200,000
|
2,534,400,000
|
26,400
|
8,047
|
|
210
|
130
|
29,400
|
8,961
|
|
210
|
232,848,000
|
2,794,176,000
|
29,106
|
8,872
|
Applying The Curvature to The Earth ( Globe, Ball).
If the earth is a globe as we told via modern day science then the
earth should have a geographical layout that should follow the
rules of a globe.
Remember - we have the following numbers:
Circumference of earth - 24,901 miles divided by 360 degrees
equals 69.16 miles
Circumference of earth - 40,075 kilometer divided by 360
degrees equals 111.31 kilometers.
We should be able to take any point on this globe and find the
following geographical land mass layout.
Let's do one (1) degree of 360 degrees of 24,901 miles. ( 69.16
miles)
Calculate the curvature we should see via the two (2) models are
calculating curvature.
|
Rule Of Thumb - Distance Squared x 8
|
|
Newton's Curvature of Earth Calculation
|
|
Miles
|
Kilometers
|
Curvature Drop Feet
|
Curvature Drop Meters
|
|
Miles
|
Feet
|
Inches
|
Curvature Drop Feet
|
Curvature Drop Meters
|
|
70
|
43
|
3,267
|
996
|
|
70
|
25,872,000
|
310,464,000
|
3,234
|
986
|
Ok - applying this to the physical geographical structure of the
Earth - show me one place on the entire earth that has this
structure. You cannot.
If the Earth is a true globe – there wouldn’t be a single serious
argument against the fact of the Earth’s globular shape
(roundness) ….Not a single argument!
There would be no flat earth arguments as the earth would display
the characteristics of a globe and flatearthers would not be able
to present any arguments as to flatness or lack of globularity of
the earth.
A search of the circumference of the earth on Google will show the
following:
The earth is 24,901 miles (40,075 kilometers) around. This means
if you start at the center of the equator and travel around the
earth you will return to the same place and travel 24,901 miles.
Now - since the earth is presented as a globe - a ball - you
should be able to break this circle into 360 degrees as all
circles (balls) can be divided into 360 parts.
Circumference of earth - 24,901 miles divided by 360 degrees
equals 69.16 miles
Circumference of earth - 40,075 kilometer divided by 360
degrees equals 111.31 kilometers.
All we have done is create a division of how much space would be
between each of these 360 degrees.
So - every 69.16 miles or 111.31 km we have a division which
equals one degree of the total 360 degrees of a circle.
Now - let’s look at a ball ( globe) and compare the distance
between these two (2) points.
If we take any point on a ball and move an equal distance from
that point in 4 directions we will find the first position will be
the top and the other four (4) distances will be lower in
orientation to the first point.
Now we will apply this logic to the curvature of the earth
(globe).
Again - show me one (1) where this is true let alone the entire
globe.
One (1) degree - 69.16 miles - the other 4 locations should be
physically lower in land mass elevation or 3,267 feet below for
rule of thumb or 3,234 feet for Newton's formula.
If the Earth were a globe - then we would see the curvature in the
physical geographical structure of the land where we live
everywhere.
