Axes to FET

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Axes to FET
« on: December 21, 2009, 05:52:14 AM »
Here's a list of reasons why FET fails, enjoy
no razor's or personal attacks, but solid proof of discrepancies
Universal acceleration:
On Round Earth, things fall, because the earth has mass.
If you go high enough, you have less gravity due to Newton’s inverse square law: F=G*m1*m2/r^2 with r being distance to the center of mass.
If you go in a canyon, you also have less net gravitational force, because there is stuff above you pulling you upwards.

If you are over a dense part of the earth, such as over rock instead of water, you will have more gravity, because you are closer to denser parts of the earth.
Essentially, acceleration is never constant. How does this work under FET, which predicts a constant acceleration?

Gravity varies on different parts of the earth.  This coincides with both altitude and geography
Source from Cornell-

and if you don't trust the Americans, here's a Swede's 2 cents:
Better yet
It can be measured accurately to more than 5 decimal places for you lovers of accuracy

On working model theory of the horizon effect is bendy light or electromagnetic acceleration. This accounts for the sinking ship effect and Bedford level stuff. The problem with it is the math. When a force is enacted on something, it undergoes acceleration proportional to that force, i.e. F=M*A.
EA tells of a force that is equal for all parts of flat earth, and constant. If you have constant force, you have constant acceleration, which means you have quadratic paths for position.
The problem with this is, FE requires light to move in a circular path. Why is this?  Take two points that would be opposite each other on a round earth. No matter how high up you get, you will never be able to see the other point. Now on flat earth, imagine that a laser is pointed parallel to the earth, in the direction of the other city. If the acceleration in the up-down direction is constant, then at some height, you will be able to see this light.

Now according to the FET map on the FAQ page, Sydney should be quite a ways from Santiago, Chile, correct?  When looking at that same map Juneau seems to be about halfway between. Therefore a flight from Santiago to Juneau should take considerably less time than a flight from Santiago to Sydney.

However, in reality, they take much longer.

Your average Santiago-Juneau flight would be about 27 hrs.

Yet, Santiago to Sydney flights take  about 18hrs

First of all, in FET the sun has a shining radius of 10,000km or in RET terms, a quarter of the earth’s circumference.
Now, let’s consider the southern hemisphere’s summer solstice. Antarctica receives 24 hours of sunlight during this day as most of you know. In FET, Antarctica surrounds the Earth, and the distance between opposite point on this circle is a gaping 40,000km. So how is a spotlight with radius 10,000km supposed to shine light at the same time on points that are 40,000km away?
Furthermore, how is the FE Sun supposed to shine on the earth in a manner that matches up with time zones? Take the spring equinox for example. At this time, there exists a point where Washington DC, is in sunset, and Hong Kong is in Sunrise. This is true for all areas along their longitude line. If you take the FE map, and outline these sunny areas, you get a semi circle, with the base along the longitude lines. The FE sun however shines light in circles, so if that were the case, there is no way the parts of the earth that do get light would get light.

Magnetic fields:
On the round earth, the inner and outer core of molten spinning metals create a magnetic field, where the north magnetic pole is near the north geographic pole, and the south off the coast of Antarctica, near Australia.
Proof of Concept experiment:
The first problem is, how is this field created? FET demands the north pole to be near the center, and the south pole in a ring around the earth. It also says that the magnetic field lines of each have to be parallel to each other, since science they are perpendicular to a flat surface. The only object that I can think of that produces such a field would be like a horseshoe magnet rotated around one pole, so that the other forms a ring. How is such a field to be made from the interior of the disk shaped earth? If the disk cores were rotating, the South Pole would be under the earth.
Even if such a field was made, in FET the south magnetic field is supposed to circum navigate the earth in a ring shape. However, in reality the pole has been located at specific point. So FET is inconsistent with observation. And if it is argued that in FET the poles do align where the point 63 S 138E is the south pole, how come if you stand off the coast of Chile, the south pole will be detected south, and not the opposite side of the world to the north?



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Re: Axes to FET
« Reply #1 on: December 21, 2009, 10:40:21 AM »
Nice copypasta. Thanks for contributing. BTW, the Springer links do not work, since you have to have a subscription.



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Re: Axes to FET
« Reply #2 on: December 21, 2009, 12:20:42 PM »

Please address only one point at a time to start a topic.  Otherwise the ensuing discussion will quickly become unfocused and useless.
Quote from: Roundy the Truthinessist
Yes, thanks to the tireless efforts of Euclid and a few other mathematically-inclined members, electromagnetic acceleration is fast moving into the forefront of FE research.

Re: Axes to FET
« Reply #3 on: December 21, 2009, 12:28:07 PM »
Please address only one point at a time to start a topic.  Otherwise the ensuing discussion will quickly become unfocused and useless.

The assumption there being that that isn't what happens anyway.