1

Flat Earth Theory / Flight Paths are Curved

« on: May 26, 2019, 02:15:07 AM »

I would like to know the explanation for how it is possible for the fastest route to be curved instead of a straight line on a flat plane. How does this make sense on a flat earth? Thanks!

2

Flat Earth Theory / How does the Sun Create Energy on a Flat Earth?

« on: May 14, 2019, 01:56:34 AM »

As we all know, the sun undergoes a thermonuclear process where it converts hydrogen to helium and into other elements which releases energy. How exactly can the Sun create energy on the Flat Earth?

3

Arts & Entertainment / Clash of Clans

« on: April 29, 2019, 12:44:13 AM »

Any fellow COCk players here in this forum? Townhall 8 maxed here.

4

Flat Earth Theory / Confusion about Tides on a Flat Earth

« on: April 26, 2019, 03:09:11 AM »

The Flat Earth Wiki does not mention anything about how spring tides and neap tides can be achieved. I've heard tides are caused from the gravitation of stars, the moon, and the Sun on the flat earth. So I would like to know how can this phenomena be explained on a flat Earth?

5

Suggestions & Concerns / How to Upload Images?

« on: April 20, 2019, 04:41:58 AM »

Hi,

I'm still new to this forum and don't understand how to upload images to my post. I tried using "" and then put the link in between but it still wouldn't work. I don't want a sloppy link in my post, I want the visual to appear. How can I do this?

6

Flat Earth Theory / Sydney to Johannesburg Flights Disprove A Flat Earth

« on: April 19, 2019, 04:55:55 AM »

Need to Know Stuff
- Qantas uses the Boeing 747-400 aircraft https://www.qantas.com/us/en/about-us/our-company/fleet/qantas-b747-400.html
- The 747-400 maximum fuel capacity is 14,200km for travel
- Distance from Sydney to Johannesburg is 11,060km on the globe
- Flight time is 13h 20m
- Therefore, a plane travels 830km/h

Distance on the Flat Earth
-Longitude of Sydney Airport is 151.1 degrees East
-Longitude of Johannesburg Airport is 28.2 degrees East
Difference of 122.9 degrees
-Sydney Airport is 13764km from North Pole
-Johannesburg Airport is 12887km from North pole

Math
Use Rule of Cosines
d^2 = a^2 + b^2 -2ab cos(C)
d = 189447696 + 166074769 -2(12764)(12887)(-.54)
d= 23,090

Conclusion
The distance from those two points on the Azimuthal Projection is 2.08x greater than the distance on the globe, therefore a plane will have to travel 2.08x faster, or 1726km/h. This is impossible because the maximum speed of the aircraft is only 988km/h.

Furthermore, the maximum fuel capacity of the aircraft can travel a distance of 14,200km. That is only 60% of the fuel needed to cross the distance on the Azimuthal Projection.

7

Suggestions & Concerns / Suggestion: Upvote/Downvote Idea

« on: April 18, 2019, 04:19:28 AM »

I think it would be cool if there was a button that could allow us to like/dislike a post we view. Just a thought.

8

Flat Earth Theory / 'Oumuamua Comet

« on: April 10, 2019, 03:43:46 AM »

'Oumuamua is a comet that had a hyperbolic trajectory with an eccentricity of 1.2. It passed the Sun in late of 2017, with the object's velocity at the perihelion (.25 AU) being 87km/s. Far greater than the Sun's escape velocity at that distance, thus a hyperbolic orbit. I was wondering how flat earthers can explain this? According to the FET, everything is around the dome, completing a full rotation once every 24 hours. How can an object break this rule, and have a hyperbolic orbit?

https://en.wikipedia.org/wiki/%CA%BBOumuamua

9

Flat Earth Theory / Size of the Flat Earth

« on: April 10, 2019, 02:31:21 AM »

It is commonly known that Eratosthenes was able to calculate the circumference of the spherical Earth by observing the Sun's rays during summer solstice. He made a slight error in his calculation of only 15%, which was very impressive for the technology they had 2300 years ago. Knowing this, he was also able to calculate the radius of the spherical earth. Knowing the radius, we can accurately calculate the mass of the Earth since we know that g=9.8m/s, and we know the distance between both objects. Which is r^2. (The equation is m=gr^2/G, which is derived from Newton's famous formula.)

For the flat earth, on the Wiki, it gives an estimate of the flat earth radius. I'm not sure if other flat earthers disagree with the mass too. Why haven't experiments been done to calculate the flat earth's radius, and possibly mass too? Am I not looking hard enough?

10

Flat Earth Theory / Phases of the Moon

« on: April 06, 2019, 12:51:53 AM »

So I read the FE Wiki about the lunar phases and I have some questions.

When the moon is below the sun's altitude the moon is dark and a New Moon occurs.

When the moon is above the altitude of the sun the moon is fully lit and a Full Moon occurs.

So how high/low does the Sun change altitude? We should be seeing the Sun's angular size increase/decrease, but this isn't reality.
Also, since the sun is an omni flourescent light (at least I think it is on the flat earth), we should be seeing the Sun's light faintly illuminate the moon's top hemisphere.
With the full moon now being above the Sun's orbital path, what causes it to FULLY illuminate the moon's surface? Because, the Sun should be at the same angle above/below the moon's orbit for the full/new moon for this to happen. Why are the two so drastically different?

I'd also like to hear an explanation how ALL observers on the flat earth can see the same lunar phase. Because, if you are "behind" the moons orbit, it would almost always be a new moon for you.

11

Flat Earth Theory / Law Of Perspective

« on: April 04, 2019, 02:36:26 AM »

Need to know stuff;
-FE Wiki states the Sun's diameter is 32 miles and the Sun is 3200 miles high.
-Lets assume the Sun is circling above the equator during the Equinoxes.
-Circumference of the orbit is 39000 miles
-Sun is moving at 1625 mph

Using the Law of Perspective, a=2*arctan(g/2r) where:
a=angular size
g=size of the object
r=distance to the object

Noon

a=2*arctan(g/2r)
a=2*arctan(32/6400)
a=.57 degrees

Sunrise
Since the observer is on the equator, the radius of the orbit is 6213 miles. We need to find the hypotenuse of a right isosceles triangle. Using pythagorean theorem:
a^2 + b^2 = c^2
c=8786 miles
Concluding, the distance from the observer to the sun during sunrise is 8786 miles.
Now, going back to the Law of Perspective equation,
a=2*arctan(32/17573)
a=.206

.206/.57=2.7

The size of the Sun is supposed to change 2.7x during the day.

That doesn't happen.

Please correct my math if I made a mistake.