The FE sun is impossible [revisited]

Tom Bishop among others have fallen back on the support of FE writers to help explain phenomenons. However, supporting these "Sacred Texts" http://www.sacred-texts.com/earth/za/za02.htm means that I can attack them directly and save time.

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I am curious how the sunsets can be explained as "too far to see" or an "optical illusion"

Explained in Chapter 9 of the book Earth Not a Globe:

http://www.sacred-texts.com/earth/za/za27.htm

It states that the sun which supposedly moves parallel to the surface of the Earth in a circle, appears to set because a distancing object will do such as a law of perspective. This would require the velocity of the sun to dramatically increase in speed the closer to the horizon it is in order to look as if it were in fact setting. Moving straight at a constant speed would make it look slower the further away it is.

Just to indulge this little overlooked implication, people the sun is passing over don't notice the sun speeding up but in fact see the opposite: slowing down on this straight line. The only explanation is that the sun actually sets. Besides, if the sun was that small, going such a distance would make it appear to shrink.

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Basically, earthlings would be able to see the sun from anywhere on a flat earth.. even if the sun wasn't shinning directly on them.

The sun is very small and very close to the earth. This allows the sun to only light one section of the world at a time.

Which is why moving that far towards the horizon would make it appear to shrink massively. Only a very large object would appear to remain constant when moving that far away.

These issues were not covered in your "sacred texts".

I once pasted this discussion before, and today I read through every single post on every page. No valid answer was given. I reposted it today for two reasons.
1. To give FE'ers another chance to answer.
2. To become informed of any potential changes in FE theory.

Before Tom pulls out his curved light theory, let me state that that theory as applied to the sun is flawed as well. All light emitted from a flat sun would curve, not just the beams that are conveinient. You'd get a donut shaped light pattern, not an ellipse.


 

And which way do the vertical rays curve?

Quote from: AmatureAstronomer on September 03, 2008, 07:40:10 AM

And which way do the vertical rays curve?

So there would be an intense vertical beam of light which hasn't been bent, at noon, surrounded by a much dimmer donut of light.
...can't say i've ever seen that

So there would be an intense vertical beam of light which hasn't been bent, at noon, surrounded by a much dimmer donut of light.
...can't say i've ever seen that

No, there wouldn't. The vertical rays would not bend at all, those slightly away from vertical would bend by only a tiny bit, those slightly further away would bend by a tiny bit more, etc. The beam of light would remain contiguous.

But the intensity would be different as soon as it passed from directly overhead. Unless the light coming from the side is more intense.

But the intensity would be different as soon as it passed from directly overhead. Unless the light coming from the side is more intense.

No, it wouldn't.

Why not?

The only light not bent is the vertical. So that means for a 32mile wide footprint the light is unbent. Outside this the light is bent more and more. You could argue that point maybe for dawn and dusk but for many hours either side of noon the intensity doesn't vary.

I thought those kind of short responses you just posted were ban worthy?

No, it wouldn't.

There is also the problem of the "spotlight Sun" that directs all of its light straight down.  That does put a problem into the model.

Why not?

Because there is an equal amount of light being emitted from the Sun at 80 degrees to what is being emitted at 90 degrees to the horizontal. Light emitted at 80 degrees is angled at 79.7611025 degrees to the horizontal when it reaches Earth, if Google calculator is to be believed, so there is an equal amount of light being received at 79.7611025 degrees to what is coming straight down. The same argument can be made for any other angle.

Quote from: Osama  bin Laden on September 04, 2008, 01:13:40 AM

No, it wouldn't.

There is also the problem of the "spotlight Sun" that directs all of its light straight down.  That does put a problem into the model.

That was my original point. It's their assertion that all the light begins it's path straight down, and with a flat sun, you can't have it any other way... With a flat sun you either get a spotlight beam of straight down rays, or a donut shaped path. The assertion that only some of those beams curve to form the expected light path... that's just grasping.

Quote from: dyno on September 04, 2008, 01:18:20 AM

Why not?

Because there is an equal amount of light being emitted from the Sun at 80 degrees to what is being emitted at 90 degrees to the horizontal. Light emitted at 80 degrees is angled at 79.7611025 degrees to the horizontal when it reaches Earth, if Google calculator is to be believed, so there is an equal amount of light being received at 79.7611025 degrees to what is coming straight down. The same argument can be made for any other angle.

Like I said the FE model predicts a flat sun. All light leaving it would be at a straight down angle, and should be influenced accordingly. You'd need a round sun to make your assertion work.

Like I said the FE model predicts a flat sun. All light leaving it would be at a straight down angle, and should be influenced accordingly. You'd need a round sun to make your assertion work.

Not necessarily.

Well are you going to explain, or are you exempt from your own prohibition from making short meaningless posts?

I think the onus is on you to explain why a flat Sun would have all of its light directed vertically downwards.

I think the onus is on you to show me even one post in any of the threads that shaped current FET that explain why a flat Sun would NOT have all of its light directed vertically downwards.

The FE theories stated in the Faqs reference a spotlight sun with straight down emittance. Tom's bendy light theory expands that to show angle of light correspondence.

I've read nothing here other than your own posts that state light leaves the flat sun at anything other than a straight downward angle. If you have read something that has influenced your opinion, do me the service of linking me to it.

Because there is an equal amount of light being emitted from the Sun at 80 degrees to what is being emitted at 90 degrees to the horizontal. Light emitted at 80 degrees is angled at 79.7611025 degrees to the horizontal when it reaches Earth, if Google calculator is to be believed, so there is an equal amount of light being received at 79.7611025 degrees to what is coming straight down. The same argument can be made for any other angle.

Then we have to be able to explain the lit areas of the Earth.  Granted the "spotlight" hasn't done that all that well, but it is canon of the FE theory.  Read the FAQ...

A: The sun and moon, each 32 miles in diameter, circle Earth at a height of 3000 miles at its equator, located midway between the North Pole and the ice wall. Each functions similar to a "spotlight," with the sun radiating "hot light," the moon "cold light." As they are spotlights, they only give light out over a certain are which explains why some parts of the Earth are dark when others are light. Their apparent rising and setting are caused by optical illusions.

I have not seen a spotlight that emitted light horizontally.  When I look at a floodlight, there is a focused beam in front of the light, but if I move to the side, the intensity of the light changes, and eventually all I am getting is reflected light.

I've read nothing here other than your own posts that state light leaves the flat sun at anything other than a straight downward angle. If you have read something that has influenced your opinion, do me the service of linking me to it.

What has influenced my opinion is the fact that the notion of a Sun that only emits light straight down is ridiculous.

Then we have to be able to explain the lit areas of the Earth.  Granted the "spotlight" hasn't done that all that well, but it is canon of the FE theory.  Read the FAQ...

Quote from: FAQ

A: The sun and moon, each 32 miles in diameter, circle Earth at a height of 3000 miles at its equator, located midway between the North Pole and the ice wall. Each functions similar to a "spotlight," with the sun radiating "hot light," the moon "cold light." As they are spotlights, they only give light out over a certain are which explains why some parts of the Earth are dark when others are light. Their apparent rising and setting are caused by optical illusions.

I have not seen a spotlight that emitted light horizontally.  When I look at a floodlight, there is a focused beam in front of the light, but if I move to the side, the intensity of the light changes, and eventually all I am getting is reflected light.

The Sun does not need to emit light horizontally; light emitted at 50 degrees to the horizontal will turn around and go the other way before it reaches even the altitude considered to be the boundary of space. According to my calculations, the Sun should light up a circle of radius 7848 kilometres on the surface of the Earth. Obviously, this theory needs a bit of work, but there's a number for you.

Quote from: AmatureAstronomer on September 04, 2008, 04:41:37 AM

I've read nothing here other than your own posts that state light leaves the flat sun at anything other than a straight downward angle. If you have read something that has influenced your opinion, do me the service of linking me to it.

What has influenced my opinion is the fact that the notion of a Sun that only emits light straight down is ridiculous.

Your opinion differs from current FET though.

What has influenced my opinion on the existence of a RE is that the notion of a flat earth is ridiculous. My RET has a round sun that behaves appropriately regardless of current apparent visible position. In conjunction with the Moon and Earth, it accurately predicts tides, eclipses, and stellar precession.

bravo this man ^^^^^

another astronomer that sees how stupid the FET really is compared to waht we see with our scopes and naked eye observations.

eclipses, and occulations and other such astronomical behaviour really do show up FET.

My RET has a round sun that behaves appropriately regardless of current apparent visible position. In conjunction with the Moon and Earth, it accurately predicts tides, eclipses, and stellar precession.

Oh, really?

Well, when's the next solar eclipse in Chicago going to be?

Oh - and show your work instead of blindly trusting what you find on the Internet. "Because the scientists said it was going to be June 2nd" isn't an answer. Show me some math.

Quote from: AmateurAstronomer on September 04, 2008, 05:18:41 AM

My RET has a round sun that behaves appropriately regardless of current apparent visible position. In conjunction with the Moon and Earth, it accurately predicts tides, eclipses, and stellar precession.

Oh, really?

Well, when's the next solar eclipse in Chicago going to be?

Oh - and show your work instead of blindly trusting what you find on the Internet. "Because the scientists said it was going to be June 2nd" isn't an answer. Show me some math.

There is no simple formula to calculate what you are interested in. The calculation of eclipses is tedious work requiring many observations or calculations of the positions of the Moon and the Sun. The lunar orbit data must then be extrapolated to find the months when eclipses might occur and then the exact times determined to see where the eclipse will be visible. The eclipse path can then be projected into the Earth's surface to find if a total eclipse will be visible from a given location at a particular time. This work requires a lot of patience and an understanding of the geometry involved and cannot be reduced to a formula.

Here is a PDF that explains the calculations needed: http://curious.astro.cornell.edu/pdf-files/eclipse.pdf

Here is a PDF that explains the calculations needed: http://curious.astro.cornell.edu/pdf-files/eclipse.pdf

I read the PDF you linked. It is not a method for predicting the Lunar Eclipse. The author of the article does not pretend that it is. The author is answering a question for a software programmer who has been unsuccessful in modeling the RE system (I wonder why). The author gives a few basic introductions for basic orbital mechanics from standard newtonian literature and finishes with the following:

The challenge in eclipse calculation is to determine whether Earth will lie within the
umbra or penumbra of the Moon (vice versa for lunar eclipse) using the three dimensional
positions of the Earth and the Moon. It is quite non-trivial and is a chief ingredient in
the eclipse code. The radii of the Sun, Earth and Moon are given below:

<snip>

I assume that you are not trying to do an eclipse calculation exactly. If so, the problem
becomes much more complicated, as you have to take into account the perturbations of
all planets, the exact relation between θp, θu and the radii R1, R2, etc. In addition, you
will need the initial condition of the Earth-Moon-Sun system, which you can obtain over
the internet. However, to get a rough idea as to modeling the eclipse calculations, you
can possibly approximate the orbits a circles, as the eccentricities are very small. Good
luck on your software development effort!

Regards,
Jagadheep

Basically the author is saying that it's very difficult to calculate a Lunar Eclipse and tells the programmer to "go search the internet" and mentions a few possibilities to look into.

The PDF is not a methodology for predicting the Lunar Eclipse. It's not even a stepping stone for creating a model for predicting the eclipse. The PDF provides nothing of the sort.

Basically the author is saying that it's very difficult to calculate a Lunar Eclipse and tells the programmer to "go search the internet" and mentions a few possibilities to look into.

The PDF is not a methodology for predicting the Lunar Eclipse. It's not even a stepping stone for creating a model for predicting the eclipse. The PDF provides nothing of the sort.

Interestingly, that wasn't how I interpreted that at all.  I thought that it did a good job of laying the groundwork for someone to do the calculations themselves.  Like anything, the introduction often ends with the fact that the actual doing of something is more complex than the introduction.  This is true whether it is fly fishing or predicting eclipses.    It gave enough information that you could start doing the math, assuming that you have the mathematical background in the first place, and get rough ideas of when, and probably where, the eclipse would be occurring.  The fact that there were references to variables at the end that would make the process more complex doesn't diminish the value of the introduction to the topic.

Interestingly, that wasn't how I interpreted that at all.  I thought that it did a good job of laying the groundwork for someone to do the calculations themselves.  Like anything, the introduction often ends with the fact that the actual doing of something is more complex than the introduction.  This is true whether it is fly fishing or predicting eclipses.    It gave enough information that you could start doing the math, assuming that you have the mathematical background in the first place, and get rough ideas of when, and probably where, the eclipse would be occurring. 

Do the math for us then. Some random paper on the internet which gives some vague equations for orbital mechanics and then says "go search the internet" aren't equations for finding the eclipse.

You guys were supposed to show us real equations by which one can use the geometry of the earth, moon and sun to predict the eclipse. I've seen no such method to predict the eclipse.

The only reason the Greeks and ancients (and modern astronomers) are able to predict the Lunar Eclipses was because the predictions are based on recurring charts and tables of past eclipses. It had nothing to do with the shape of the earth or the actual geometry of the cosmos. The Lunar Eclipse is a phenomenon which comes in patterns. By studying these patterns it is possible to predict when the next transit or eclipse will occur. The astronomer can use historic charts and tables with a few equations to predict the time, magnitude, and duration of a future eclipse. There's a good chapter in Zetetic Cosmogony on the subject.

http://books.google.com/books?id=GzkKAAAAIAAJ&pg=PA#PPA74,M1

Do the math for us then. Some random paper on the internet which gives some vague equations for orbital mechanics and then says "go search the internet" aren't equations for finding the eclipse.

Well, the math is above my level of math.  That doesn't make it any less valid for the people that can.  There are plenty of FE and RE that have the math ability (RoboSteve Laden is one) to do the calculations involved, but for most of us, we trust computer programs to do the math for us.

You guys were supposed to show us real equations by which one can use the geometry of the earth, moon and sun to predict the eclipse. I've seen no such method to predict the eclipse.

If you are truly interested find a copy of Newcomb's Tables of the Sun which go into detail of the math used to predict the position of the sun, including eclipses.  These tables were made around the turn of the century, so they don't take into account relativistic effects, but those are smaller than fractions of an arc second, so for normal eclipse viewers they are accurate enough.  They were computed using celestial mechanics (the fact that everything moves around the Sun), so be warned.

Lately, predictions from computers have replaced Newcomb's tables because of the increased accuracy that is possible due to the ability to handle more decimal places and the addition of compensation for relativistic effects.

http://books.google.com/books?id=GzkKAAAAIAAJ&pg=PA#PPA74,M1

Just looked through that.  I never did see the calculations that were going to enable to you to find out when and where future eclipses will be.

Well, the math is above my level of math.

Surely you can Google it up for us if orbital mechanics equations to predict the occurence of the Lunar Eclipse based on the geometry of the earth, moon, and sun truly exist.

Since over the past few years no REer has ever posted such calculations for us when challenged we should assume that they do not exist.

Just looked through that.  I never did see the calculations that were going to enable to you to find out when and where future eclipses will be.

There aren't any equations to predict the eclipse in Zetetic Cosmogony. It's just a description of how astronomers have historically predicted eclipses based on recurring patterns of previous eclipses. By looking at historic charts and tables of past eclipses it's possible to derive an equation to find the time, magnitude, and duration of the next one.

You can find real workable equations to predict the eclipse based on such patterns at the end of Chapter 11 of Earth Not a Globe.

http://www.sacred-texts.com/earth/za/za29.htm

Due to the chaotic orbit of the moon, any eclipses are very hard to accurately predict, in fact, technically impossible. Solving a n-body problem and thus finding an equation to predict lunar eclipses would be impossible, and even if it were, the result would be too complex to be of any practical value.

Due to the chaotic orbit of the moon, any eclipses are very hard to accurately predict, in fact, technically impossible. Solving a n-body problem and thus finding an equation to predict lunar eclipses would be impossible, and even if it were, the result would be too complex to be of any practical value.

Well I guess that means your model isn't as predictive as widely assumed, then.

Quote from: AmateurAstronomer on September 04, 2008, 05:18:41 AM

My RET has a round sun that behaves appropriately regardless of current apparent visible position. In conjunction with the Moon and Earth, it accurately predicts tides, eclipses, and stellar precession.

Oh, really?

Well, when's the next solar eclipse in Chicago going to be?

Oh - and show your work instead of blindly trusting what you find on the Internet. "Because the scientists said it was going to be June 2nd" isn't an answer. Show me some math.

It will probaly be three Saros after the last one.  A saros cycle is 18 years 11 days and 8 hours.  To know for sure you need to know what the last eclipse there was like. 

For more details

http://en.wikipedia.org/wiki/Saros_cycle