Eclipse proportions refute RET

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Re: Eclipse proportions refute RET
« Reply #60 on: January 25, 2018, 03:49:00 AM »
Here is the entire book by Knight/Butler:

https://contraeducacao.files.wordpress.com/2012/09/who-built-the-moon_-knight-christopher.pdf

Do a search with the word astonishing, three occurrences, none of which match the quote.

The quote is from Dr. Farouk El-Baz. The website (quora) simply misplaced the names of the authors.

You're correct that the quote in your source doesn't appear verbatim in the pdf of the book you link. Parts of it do though:

p 31: "... when the Sun is at its lowest and weakest in midwinter, the [full] Moon is at its highest and brightest, and ..."
p 32: "... both [go down] at the same point on the horizon at the equinoxes ..."

It's possible it was referring to a different edition but it's much more likely that the quora 'quote' was paraphrasing - note that it doesn't use quotation marks on this citation, whereas it does on the others.

Further, the quote about water vapour that the quora link ascribes to Dr. El-Baz is also ascribed to him elsewhere, for instance Cosmological Ice Ages, Henry Kroll, p177:

https://books.google.co.uk/books?id=WbyM56LFkzsC&lpg=PA177&vq=baz&pg=PA177#v=onepage&q&f=false

Do you still maintain your source was wrong? If so, no worries, we can go back to my original question: Do you have a reliable source linking Dr. El-Baz to your quote?

Re: Eclipse proportions refute RET
« Reply #61 on: January 25, 2018, 04:51:00 AM »
It is YOU who has committed a tremendous error, by assuming that the direct distance to the sun, explicitly stated as such in the first paragraphs of the thread, is the same as the diagonal distance.
Nope.
He said it was the side of the triangle, that means it has to be the diagonal distance.
So no, YOU have committed a tremendous error by changing it to a right angle triangle.

Both sets of calculations lead to the same conclusion.
And as both are based upon the same BS, both amount to nothing.

But no, they don't lead to the same conclusion.
There is his set of calcs, based upon nothing and just baselessly asserting a value for no reason.
Then there is the calcs I provided based upon the triangles he made up, which then produce a value equivalent to what you had, which disagrees with his claim.

So, the following, then, applies only to you, doesn't it?
Nope, still applies to you and JRowe.

You have both completely failed to show any issue with RET.

The most extraordinary fact is that if THE SUN AND THE MOON HAVE THE SAME DIAMETER
Who gives a shit about that purely fictitious example?
The sun and moon do not have the same diameter, and going down this delusional path doesn't get you any closer to finding a problem with RET.

The OP for some reason (apparently out of complete ignorance and misunderstanding) assumed the sun was 1.3 times the size of the moon.

As that is not RET, anything following from that nonsensical claim doesn't disprove RET.

Here is the entire book by Knight/Butler:
https://contraeducacao.files.wordpress.com/2012/09/who-built-the-moon_-knight-christopher.pdf
Do a search with the word astonishing, three occurrences, none of which match the quote.
So you are now saying you own source is BS and you have no source.

The quote is from Dr. Farouk El-Baz. The website (quora) simply misplaced the names of the authors.
Not quite, they just got the quote wrong, but it is in the book.
Try this one:
Quote
So, when the Sun is at its lowest
and weakest in midwinter, the full
Moon is at its highest and brightest, and
in midsummer, when the Sun is at its
highest and brightest, the Moon is at its
weakest.
If you want to understand how
extraordinary this doppelgänger effect
is, stand on a hilltop or an open plain
and film the Sun at midwinter sunset
(its most southerly point on the
horizon), at the spring equinox, again at
midsummer and again at the autumn
equinox. Then on those same dates film
the Moon setting and you will see that
they both go down at the same point on
the horizon at the equinoxes (March
21
st and September 21
st) but the Moon
will have the opposite setting point to
the Sun at solstices in December and
June.
2

This just shows the problem with using the internet to find quotes.
Some moron claims it and other morons repeat it and then just link to other morons claiming it, with no original source.
« Last Edit: January 25, 2018, 04:53:03 AM by JackBlack »

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #62 on: January 25, 2018, 05:22:10 AM »
He said it was the side of the triangle, that means it has to be the diagonal distance.


The remaining two sides will be k in the Sun triangle, and k+c in the moon triangle. k is the distance to the Sun. c is unknown, defined to just be the difference between the distance to the Sun and the distance to the moon. It will likely be negative.

You used the diagonal distance, which means you had no idea about the subject of this thread.

Then there is the calcs I provided based upon the triangles he made up, which then produce a value equivalent to what you had, which disagrees with his claim.

If we use the correct value for the distances, the calculations provide the very same answer.


The sun and moon do not have the same diameter, and going down this delusional path doesn't get you any closer to finding a problem with RET.


But they do.

The calculations provided in this thread prove directly that the official figures are totally wrong, and moreover, the ONLY set of values which produce meaningful results are p =1 and k the same for both the sun and the moon.

It is very easy to prove this assertion.

K being the same would mean of course that both the sun and the moon are disks, and not spheres.

Here is the proof that the shape of the sun is discoidal and not spherical:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939765#msg1939765

You provided this quote from the book:

"So, when the Sun is at its lowest
and weakest in midwinter, the full
Moon is at its highest and brightest, and
in midsummer, when the Sun is at its
highest and brightest, the Moon is at its
weakest.
If you want to understand how
extraordinary this doppelgänger effect
is, stand on a hilltop or an open plain
and film the Sun at midwinter sunset
(its most southerly point on the
horizon), at the spring equinox, again at
midsummer and again at the autumn
equinox. Then on those same dates film
the Moon setting and you will see that
they both go down at the same point on
the horizon at the equinoxes (March
21
st and September 21
st) but the Moon
will have the opposite setting point to
the Sun at solstices in December and
June."


Here is the quote published by Dr. El-Baz:

The Moon has astonishing synchronicity with the Sun. When the Sun is at its lowest and weakest in mid-winter, the Moon is at its highest and brightest, and the reverse occurs in mid-summer. Both set at the same point on the horizon at the equinoxes and at the opposite point at the solstices. What are the chances that the Moon would naturally find an orbit so perfect that it would cover the Sun at an eclipse and appear from Earth to be the same size? What are chances that the alignments would be so perfect at the equinoxes and solstices?

    Farouk El Baz,
    NASA

Different wordings, order of sentences, not nearly the same.


Your only excuse would be that you did not have your glasses on.
« Last Edit: January 25, 2018, 06:02:42 AM by sandokhan »

Re: Eclipse proportions refute RET
« Reply #63 on: January 25, 2018, 05:24:48 AM »
Yes, if you assume the Sun is that much further than the moon, your results will give you that, it is not in line with observable reality.

We aren’t assuming anything, the distance to the moon has been measured with radar, the distance to Venus has been measured with radar, and the distance to the sun has been calculated from the geometry of the earth-sun-Venus triangle.

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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #64 on: January 25, 2018, 05:26:05 AM »
He said it was the side of the triangle, that means it has to be the diagonal distance.

Different thread, same lying bastard.

The remaining two sides will be k in the Sun triangle, and k+c in the moon triangle. k is the distance to the Sun. c is unknown, defined to just be the difference between the distance to the Sun and the distance to the moon. It will likely be negative.

You used the diagonal distance, which means you had no idea about the subject of this thread.

Then there is the calcs I provided based upon the triangles he made up, which then produce a value equivalent to what you had, which disagrees with his claim.

If we use the correct value for the distances, the calculations provide the very same answer.


The sun and moon do not have the same diameter, and going down this delusional path doesn't get you any closer to finding a problem with RET.


But they do.

The calculations provided in this thread prove directly that the official figures are totally wrong, and moreover, the ONLY set of values which produce meaningful results are p =1 and k the same for both the sun and the moon.

It is very easy to prove this assertion.

K being the same would mean of course that both the sun and the moon are disks, and not spheres.

Here is the proof that the shape of the sun is discoidal and not spherical:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939765#msg1939765

You provided this quote from the book:

"So, when the Sun is at its lowest
and weakest in midwinter, the full
Moon is at its highest and brightest, and
in midsummer, when the Sun is at its
highest and brightest, the Moon is at its
weakest.
If you want to understand how
extraordinary this doppelgänger effect
is, stand on a hilltop or an open plain
and film the Sun at midwinter sunset
(its most southerly point on the
horizon), at the spring equinox, again at
midsummer and again at the autumn
equinox. Then on those same dates film
the Moon setting and you will see that
they both go down at the same point on
the horizon at the equinoxes (March
21
st and September 21
st) but the Moon
will have the opposite setting point to
the Sun at solstices in December and
June."


Here is the quote published by Dr. El-Baz:

The Moon has astonishing synchronicity with the Sun. When the Sun is at its lowest and weakest in mid-winter, the Moon is at its highest and brightest, and the reverse occurs in mid-summer. Both set at the same point on the horizon at the equinoxes and at the opposite point at the solstices. What are the chances that the Moon would naturally find an orbit so perfect that it would cover the Sun at an eclipse and appear from Earth to be the same size? What are chances that the alignments would be so perfect at the equinoxes and solstices?

    Farouk El Baz,
    NASA

Different wordings, order of sentences, not nearly the same.


Your only excuse would be that you did not have your glasses on.

How do a Sun and Moon of identical size produce both total and annular eclipses?
Nullius in Verba

Re: Eclipse proportions refute RET
« Reply #65 on: January 25, 2018, 05:32:31 AM »
How do a Sun and Moon of identical size produce both total and annular eclipses?

The sun has to be slightly larger than the moon in order for annular eclipses to work.  The FE sun and moon are each “about 32 miles across” (emphasis on the “about” term) so if the sun is closer to 35 and the moon closer to 30, then you might have geometry that supports annular eclipses.

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #66 on: January 25, 2018, 06:12:23 AM »
How do a Sun and Moon of identical size produce both total and annular eclipses?

Very good question.

You will find the answer in the oldest treatise on astronomy, The Book of the Luminaries, chapters 71-77:

http://www.johnpratt.com/items/docs/enoch.html#Enoch_71

The distance between the Sun and Moon is not constant, it varies slightly.






Here are the photographs which prove that p = 1 and that k is nearly the same:






http://www.moonglow.net/eclipse/2003nov23/

That many such bodies exist in the firmament is almost a matter of certainty; and that one such as that which eclipses the moon exists at no great distance above the earth's surface, is a matter admitted by many of the leading astronomers of the day. In the report of the council of the Royal Astronomical Society, for June 1850, it is said:--

"We may well doubt whether that body which we call the moon is the only satellite of the earth."

In the report of the Academy of Sciences for October 12th, 1846, and again for August, 1847, the director of one of the French observatories gives a number of observations and calculations which have led him to conclude that,--

"There is at least one non-luminous body of considerable magnitude which is attached as a satellite to this earth."

Sir John Herschel admits that:--

"Invisible moons exist in the firmament."

Sir John Lubbock is of the same opinion, and gives rules and formulæ for calculating their distances, periods.

Lambert in his cosmological letters admits the existence of "dark cosmical bodies of great size."

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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #67 on: January 25, 2018, 06:19:25 AM »
How do a Sun and Moon of identical size produce both total and annular eclipses?

Very good question.

You will find the answer in the oldest treatise on astronomy, The Book of the Luminaries, chapters 71-77:

http://www.johnpratt.com/items/docs/enoch.html#Enoch_71

The distance between the Sun and Moon is not constant, it varies slightly.






Here are the photographs which prove that p = 1 and that k is nearly the same:






http://www.moonglow.net/eclipse/2003nov23/

That many such bodies exist in the firmament is almost a matter of certainty; and that one such as that which eclipses the moon exists at no great distance above the earth's surface, is a matter admitted by many of the leading astronomers of the day. In the report of the council of the Royal Astronomical Society, for June 1850, it is said:--

"We may well doubt whether that body which we call the moon is the only satellite of the earth."

In the report of the Academy of Sciences for October 12th, 1846, and again for August, 1847, the director of one of the French observatories gives a number of observations and calculations which have led him to conclude that,--

"There is at least one non-luminous body of considerable magnitude which is attached as a satellite to this earth."

Sir John Herschel admits that:--

"Invisible moons exist in the firmament."

Sir John Lubbock is of the same opinion, and gives rules and formulæ for calculating their distances, periods.

Lambert in his cosmological letters admits the existence of "dark cosmical bodies of great size."

So they aren't of identical size then?
Nullius in Verba

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #68 on: January 25, 2018, 06:55:05 AM »
A similar question came up in the South Celestial Pole thread: the index of refraction of the ether/aether.

https://www.theflatearthsociety.org/forum/index.php?topic=67769.msg1813267#msg1813267

https://www.theflatearthsociety.org/forum/index.php?topic=67769.msg1813298#msg1813298

Everything depends on the speed of light through the aether in the comparison for the total eclipse/annular eclipse situations.


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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #69 on: January 25, 2018, 06:57:27 AM »
A similar question came up in the South Celestial Pole thread: the index of refraction of the ether/aether.

https://www.theflatearthsociety.org/forum/index.php?topic=67769.msg1813267#msg1813267

https://www.theflatearthsociety.org/forum/index.php?topic=67769.msg1813298#msg1813298

Everything depends on the speed of light through the aether in the comparison for the total eclipse/annular eclipse situations.

Side question: what does aether taste like?
Nullius in Verba

Re: Eclipse proportions refute RET
« Reply #70 on: January 25, 2018, 07:01:03 AM »
Chicken, of course.
Quote from: mikeman7918
a single photon can pass through two sluts

Quote from: Chicken Fried Clucker
if Donald Trump stuck his penis in me after trying on clothes I would have that date and time burned in my head.

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markjo

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Re: Eclipse proportions refute RET
« Reply #71 on: January 25, 2018, 09:01:36 AM »
Quote
I already have. Other people already have. A much smaller moon much closer can produce the same eclipse as a larger moon further away. Your entire "proof" is fundamentally flawed. Anybody who ever owned a baseball cap can figure that out.
Which, still, is empty denial.
I have proven that the proportions necessary do not exist under RET. Get that through your thick skull already. No matter how much you want to insist that it works, YOU HAVE TO FUCKING SHOW HOW IT COULD GIVEN WHAT I HAVE PROVEN
The fact that the sun and moon both have approximately the same angular size proves that the necessary proportions do exist.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

Re: Eclipse proportions refute RET
« Reply #72 on: January 25, 2018, 09:43:20 AM »
How do a Sun and Moon of identical size produce both total and annular eclipses?

Very good question.

You will find the answer in the oldest treatise on astronomy, The Book of the Luminaries, chapters 71-77:

http://www.johnpratt.com/items/docs/enoch.html#Enoch_71

The distance between the Sun and Moon is not constant, it varies slightly.






Here are the photographs which prove that p = 1 and that k is nearly the same:






http://www.moonglow.net/eclipse/2003nov23/

That many such bodies exist in the firmament is almost a matter of certainty; and that one such as that which eclipses the moon exists at no great distance above the earth's surface, is a matter admitted by many of the leading astronomers of the day. In the report of the council of the Royal Astronomical Society, for June 1850, it is said:--

"We may well doubt whether that body which we call the moon is the only satellite of the earth."

In the report of the Academy of Sciences for October 12th, 1846, and again for August, 1847, the director of one of the French observatories gives a number of observations and calculations which have led him to conclude that,--

"There is at least one non-luminous body of considerable magnitude which is attached as a satellite to this earth."

Sir John Herschel admits that:--

"Invisible moons exist in the firmament."

Sir John Lubbock is of the same opinion, and gives rules and formulæ for calculating their distances, periods.

Lambert in his cosmological letters admits the existence of "dark cosmical bodies of great size."

I agree that scientific knowledge from 170 years ago was poor. That’s not news.
"Science is real."
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Re: Eclipse proportions refute RET
« Reply #73 on: January 25, 2018, 01:51:02 PM »
He said it was the side of the triangle, that means it has to be the diagonal distance.
The remaining two sides will be k in the Sun triangle, and k+c in the moon triangle. k is the distance to the Sun. c is unknown, defined to just be the difference between the distance to the Sun and the distance to the moon. It will likely be negative.
You used the diagonal distance, which means you had no idea about the subject of this thread.
You really do like repeating the same ignorant crap.
Yes, k is the distance to the sun, ALONG THE DIAGONAL!!!
Are you aware that you can measure distances along a diagonal?

Again, you were told to draw an isosceles triangle, with one side p M (used p=1.3 as an example), representing the size of the sun, and the other side being k.
That needs to be a diagonal.
There was no line corresponding to the "direct distance" to the sun.

So no, this just shows that yet again, you have no idea what you are talking about.

Do I need to draw you a picture?

Regardless, this is all just bitching about nothing.
Changing it between diagonal distance and direct distance will have no effect on the argument at hand, because they are directly related, cos(x/2)=direct/diagonal.
But of course, you know you need to grasp at whatever straws you can to pretend you are correct. What a shame it just results in you looking even worse.


Then there is the calcs I provided based upon the triangles he made up, which then produce a value equivalent to what you had, which disagrees with his claim.
If we use the correct value for the distances, the calculations provide the very same answer.
No they don't.
His claim:
You will find that the distance to the moon, k+c, is k/p times the square root of (2-p2)
i.e. (k+c)=(k/p)*sqrt(2-p^2))
Mine:
Thus the distance to the sun (c+k), is simply c times p, which you could have easily gotten via rules regarding similar triangles.
Which when using the same variables as he did (i.e. k for the sun distance, (k+c) for the moon distance, simply gives:
k=(k+c)*p
Or (k+c)=k/p. Notice the factor of sqrt(2-p^2) which is missing?
Your claim:
(k - c) = k/p
Which when adjusted to use the same format as above, you have:
(k+c)=k/p

Again, notice the missing term?

So no, the only time you get the same answer is when you have:
sqrt(2-p^2)=1, i.e. p=1.
This would require the sun and moon to be the same size and the same distance from Earth.
This means during an eclipse they are literally in the same location. You would need the moon to pass through the sun. This is impossible.
This would simply be a collision between the sun and moon.

So no, that isn't possible. Using any other value (which would include the correct value) you get completely different numbers.
For example, using his p=1.3 BS, you get them off by a factor of 0.556776436.
When you use the real values, of p=~400, you get a much bigger problem. You need to find the square root of (2-400^2)=-159998
In fact, you start running into issues as soon as p=sqrt(2).
At p=sqrt(2), you end up with that factor being sqrt(2-2)=0. That would indicate the moon is touching Earth.

The sun and moon do not have the same diameter, and going down this delusional path doesn't get you any closer to finding a problem with RET.

But they do.
No they don't.
The fact that you can get annular eclipses proves quite conclusively that the moon is smaller than the sun.

So going down delusional paths of hypotheticals doesn't get you any closer to a problem with RET. All it does is get you a problem with your delusional strawman.

The calculations provided in this thread prove directly that the official figures are totally wrong, and moreover, the ONLY set of values which produce meaningful results are p =1 and k the same for both the sun and the moon.
No. The calculations provided in this thread proves the figures and calcualtions JRowe gave are totally wrong.

He falsely assumed the sun is only 1.3 times the size of the moon rather than using the figure from reality of roughly 400.
As such, the calculations have nothing to do with RET aka reality.

It is very easy to prove this assertion.
No, it is literally impossible as it is false.
You haven't come close to proving it at all.

K being the same would mean of course that both the sun and the moon are disks, and not spheres.
Nope. Even them being discs don't help. Also, the fact they appear as roughly a circle regardless of viewing position shows quite clearly that they are not discs.
If they were discs then viewing them from any angle except directly along their axis, they would appear as ellipses.

So any alleged proof you have for them being discs is clearly BS.

You provided this quote from the book:
Here is the quote published by Dr. El-Baz:
You are yet to establish that quote is by him.
As such I will take it as people miswording the book, or using some other quote.

It still remains that you are falsely attributing this quote to El-Baz.

All you have done is shown how unreliable it is to use random internet sites to try and find quotes like this.

How do a Sun and Moon of identical size produce both total and annular eclipses?
Very good question.
You will find the answer in the oldest treatise on astronomy, The Book of the Luminaries, chapters 71-77:
i.e. you have no answer and thus need to evade the question.

The distance between the Sun and Moon is not constant, it varies slightly.
Yes, and this allows a moon that is smaller than the sun to produce a total or annular eclipse. It does not allow a moon that is as large as the sun to produce an annular eclipse.

If the moon is larger than the sun then it is impossible for their centres to both lie upon the same line of site with the moon in front and the moon to not completely obstruct the view to the sun.

Here are the photographs which prove that p = 1 and that k is nearly the same:
No, they don't.
They show the apparent angular size is the same. They have no indication on the distance to either of them, nor any indication of the real relative size.

All the apparent angular size being roughly the same shows is that the ratio of the sizes is roughly the same as the ratio of the distances.
This means they could hypothetically be the same size at the same distance (ignoring the physical impossibility of that), the sun could be 1.3 times the size of the moon, and 1.3 times as distant, the sun could be 400 times the size and distance of the moon, or some other factor.

It does not show they are the same size, nor does it show they are the same distance.

The fact that it isn't perfect and varies slightly with sometimes the sun appearing smaller than the moon and sometimes larger just shows they are not the same size and same distance and the distances vary.

As the moon obstructs the sun (to produce an eclipse), this shows the moon is closer to Earth than the sun (and thus smaller than the sun). As you can get annular eclipses, this shows the moon is smaller than the sun.

And I will be ignoring your baseless quotes.
Even if they were said by these people, they just amount to a pathetic appeal to authority to attempt to avoid your failings.

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #74 on: January 25, 2018, 02:08:22 PM »
This would require the sun and moon to be the same size and the same distance from Earth.
This means during an eclipse they are literally in the same location. You would need the moon to pass through the sun. This is impossible.
This would simply be a collision between the sun and moon.


Not at all.

As shown in the Fred Bruenjes photographs there is a certain distance between the solar/black sun disks.

Your assertion is refuted.


If they were discs then viewing them from any angle except directly along their axis, they would appear as ellipses.

You are a student of conventional physics.


Here are two of greatest physicists of the 20th century telling you that you are wrong:

“What? Do you mean to tell me that I can tell you how
much magnetic field there is inside of here by measuring
currents through here and here – through wires which
are entirely outside – through wires in which there is no
magnetic field... In quantum mechanical interference experiments
there can be situations in which classically there
would be no expected influence whatever. But nevertheless
there is an influence. Is it action at distance? No, A is
as real as B-realer, whatever that means.”

R. Feynman

“throughout most of 20th century the Heaviside-Hertz form of Maxwell’s equations were taught to college students all over the world. The reason is quite obvious: the Heaviside-Hertz form is simpler, and exhibits an appealing near symmetry between E and H. With the widespread use of this vector-potential-less version of Maxwell’s equations, there arouse what amounted to a dogma: that the electromagnetic field resides in E and H. Where both of them vanish, there cannot be any electromagnetic effects on a charged particle. This dogma explains why when the Aharonov-Bohm article was published it met with general disbelief. . . E and H together do not completely describe the electromagnetic field, and. . . the vector potential cannot be totally eliminated in quantum mechanics. . . the field strengths underdescribe electromagnetism.”

C.N. Yang, Nobel prize laureate

“...the vector potential appears to give the most direct description of the physics. This becomes more apparent the more deeply we go into quantum theory. In the general theory of quantum electrodynamics, one takes the vector and scalar potentials as the fundamental quantities in a set of equations that replace the Maxwell equations: E and B are slowly disappearing from the modern expression of physical laws; they are being replaced by A and φ”

(Feynman et al, 1989, chapter 15, section 5, The Feynman Lecture on Physics (Vol. 2), 1989)


You are relying ONLY on the vector fields, which are created by the POTENTIALS.

Potential = ether = longitudinal waves


Again, R. Feynman:

E and B are slowly disappearing from the modern expression of physical laws; they are being replaced by A and φ.

The existence of the ether changes everything.


E.T. Whittaker proved mathematically the existence of the hidden structure of the potential: bidirectional longitudinal waves.


https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1994059#msg1994059

The achievements of the 1903 and 1904 papers published by Whittaker:

A scalar potential is comprised of a lattice of bidirectional longitudinal waves (ether/Tesla strings).

Electromagnetic or gravitational fields and waves can be decomposed into two scalar potential functions.

The unification of quantum mechanics, general relativity, ether theory into one single subject: ELECTROGRAVITY.

How to construct a scalar interferometer: a standing scalar wave structure.

An extended version of the Aharonov-Bohm effect.

The discovery of the fact that internal EM is generally completely inside the scalar potential, existing as “infolded” harmonic sets of EM antiparallel wave/antiwave pairs.   This internal EM was in Maxwell’s original quaternion equations.

The superluminal speed of gravitational waves.


“Whittaker, a leading world-class physicist himself, single-handedly rediscovered the "missing" scalar components of Maxwell's original quaternions, extending their (at the time) unseen implications for finally uniting "gravity" with the more obvious electrical and magnetic components known as "light."

"In 1903-1904 E.T. Whittaker published a fundamental, engineerable theory of electrogravitation (EG) in two profound papers. The first (W-1903) demonstrated a hidden bidirectional EM wave structure in the scalar potential of vacuum, and showed how to produce a standing scalar EM potential wave -- the same wave discovered experimentally four years earlier by Nikola Tesla.

W-1904 shows that all force field EM can be replaced by interferometry of two scalar potentials, anticipating the Aharonov-Bohm effect by 55 years and extending it to the engineerable macroscopic world. W-1903 shows how to turn EM into G-potential and directly engineer the virtual particle flux of ether. W-1904 shows how to turn G-potential back into force-field EM, even at a distance."

E.T. Whittaker, "On the Partial Differential Equations of Mathematical Physics," Math. Ann., Vol. 57, 1903, p. 333-355 (W-1903)

http://www.cheniere.org/misc/Whittak/ORIw1903.pdf

E.T. Whittaker, "On an Expression of the Electromagnetic Field Due to Electrons by Means of Two Scalar Potential Functions," Proc. Lond. Math. Soc., Series 2, Vol.1, 1904, p. 367-372 (W-1904)

http://hemingway.softwarelivre.org/ttsoares/books_papers_patents/books%20papers%20patents%20(scientis/whittaker/whittaker%20et%20-%20on%20an%20expre.pdf

"In his 1903 paper Whittaker showed that a standing scalar potential wave can be decomposed into a special set of bidirectional EM waves that convolute into a standing scalar potential wave.

The very next year, Whittaker's second paper (cited above) showed how to turn such G potential wave energy back into EM energy, even at a distance, by scalar potential interferometry, anticipating and greatly expanding the Aharonov-Bohm effect. Indeed, Whittaker's second paper shows that the entire present force-field electromagnetics can be directly replaced with scalar potential interferometry. In other words, scalar EM includes and extends the present restricted vector subset of Maxwell's original theory.
 
Specifically, any EM force field can be replaced by two scalar potential fields and scalar interferometry. The combination of this paper and the 1903 Mathematische Annalen paper not only includes the Aharonov-Bohm effect, but specifies a testable method for producing a macroscopic and controlled Aharanov-Bohm effect, even at large distances."


The existence of this hidden structure was proven experimentally by the Aharonov-Bohm effect.

“A new generation of physicists, also educated in the grand assumption that "Heaviside's Equations" are actually "Maxwell's," were abruptly brought up short in 1959 with a remarkable and elegant experiment -- which finally demonstrated in the laboratory the stark reality of Maxwell's "pesky scalar potentials" ... those same "mystical" potentials that Heaviside so effectively banished for all time from current (university-taught) EM theory.

In that year two physicists, Yakir Aharonov and David Bohm, conducted a seminal "electrodynamics" laboratory experiment ("Significance of Electromagnetic Potentials in Quantum Theory," The Physical Review, Vol. 115, No. 3, pp. 485-491; August, 1959). Aharonov and Bohm, almost 100 years after Maxwell first predicted their existence, succeeded in actually measuring the "hidden potential" of free space, lurking in Maxwell's original scalar quaternion equations. To do so, they had to cool the experiment to a mere 9 degrees above Absolute Zero, thus creating a total shielding around a superconducting magnetic ring [for a slightly different version of this same experiment; the oscillation of electrical resistance in the ring (bottom graph) is due to the changing electron "wave functions" -- triggered by the "hidden Maxwell scalar potential" created by the shielded magnet].



Once having successfully accomplished this non-trivial laboratory set up, they promptly observed an "impossible" phenomenon:

Totally screened, by all measurements, from the magnetic influence of the ring itself, a test beam of electrons fired by Aharonov and Bohm at the superconducting "donut," nonetheless, changed their electronic state ("wave functions") as they passed through the observably "field-free" region of the hole -- indicating they were sensing "something," even though it could NOT be the ring's magnetic field. Confirmed now by decades of other physicists' experiments as a true phenomenon, this "Aharonov-Bohm Effect" provides compelling proof of a deeper "spatial strain" -- a "scalar potential" -- underlying the existence of a so-called magnetic "force-field" itself.”

After the first precise experiment carried out by Tonomura and his team at Hitachi using electron holography followed by more experiments using superconducting shields, the Aharonov-Bohm effect is confirmed and that it is a genuine feature of the standard quantum mechanics.


The Allais effect proves immediately that the solar eclipse is not caused by the moon.


*

sandokhan

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Re: Eclipse proportions refute RET
« Reply #75 on: January 25, 2018, 02:15:15 PM »
They show the apparent angular size is the same. They have no indication on the distance to either of them, nor any indication of the real relative size.

They show that Armstrong and Aldrin might as well have used as a vehicle to the moon a simple carriage driven by reindeers: the photographs show both the Black Sun and the Sun itself at a distance of less than 1000 km from the photographer.



No 384,000 km distance to the Moon at all.

« Last Edit: January 26, 2018, 05:53:28 AM by sandokhan »

Re: Eclipse proportions refute RET
« Reply #76 on: January 25, 2018, 02:21:10 PM »
You guys are all so adorable!

Re: Eclipse proportions refute RET
« Reply #77 on: January 25, 2018, 02:33:39 PM »
Not at all.
How about you deal with the math provided just above that.
Show what is wrong with it.
Baselessl dismissal doesn't magically make you correct.


As shown in the Fred Bruenjes photographs there is a certain distance between the solar/black sun disks.
Yes, and that shows your claims BS, because your claims require them to be the same distance and same size.

Your assertion is refuted.
Nope, my argument further backed up.

If they were discs then viewing them from any angle except directly along their axis, they would appear as ellipses.
You are a student of conventional physics.
You mean a student of reality.
Regardless, this is going off topic again.
Deal with the failings of the OP. Stop trying to change the subject to avoid your pathetic failures.

Seriously, in that entire post you completely failed to address the arguments regarding the OP except with your pathetic dismissal.

They show that Armstrong and Aldrin might as well have used as a vehicle to the moon a simple carriage driven by raindeer: the photographs show both the Black Sun and the Sun itself at a distance of less than 1000 km from the photographer.
No they don't.
The moon (which you seem to delusionally call the black sun) has no indication of distance in that photo.

If you think it does, explain how.
Or better still, address the failings of the OP and how their baseless garbage doesn't disprove RET at all.

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #78 on: January 25, 2018, 02:41:33 PM »
because your claims require them to be the same distance and same size.


Yes, the same size.

The distance between the Sun and Moon is not constant, it varies slightly; this distance is infinitesimal compared to the distance between the photographer and the celestial bodies.

If you think it does, explain how.

Sure, here are the ISS/Atlantis solar/lunar transit photographs:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1786946#msg1786946

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1787025#msg1787025


Re: Eclipse proportions refute RET
« Reply #79 on: January 25, 2018, 02:56:55 PM »
because your claims require them to be the same distance and same size.

Yes, the same size.
Thus p=1, thus they are the same distance.
Hence, your claims require them to be the same distance and size. The simple fact is this is impossible (as previously shown), thus you are wrong.

If you think it does, explain how.
Sure, here are the ISS/Atlantis solar/lunar transit photographs:
That is not that photo. Try again.

And deal with the OP's faillings.

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Sentinel

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Re: Eclipse proportions refute RET
« Reply #80 on: January 25, 2018, 03:32:38 PM »
Granted I didn't log in here for quite a while, but boy the bitchslapping of poor Sandy by Jack really never gets old.
Wonder if his FE sun still is at merely 20 miles height above the plane and how his precious Aether is holding up so far though.  :P
"No snowflake in an avalanche ever feels responsible."

Stanislaw Jerzy Lec

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rabinoz

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Re: Eclipse proportions refute RET
« Reply #81 on: January 25, 2018, 05:44:52 PM »
How do a Sun and Moon of identical size produce both total and annular eclipses?

Very good question.
You will find the answer in the oldest treatise on astronomy, The Book of the Luminaries, chapters 71-77:
http://www.johnpratt.com/items/docs/enoch.html#Enoch_71
Really? Where?
Quote from: sandokhan

The distance between the Sun and Moon is not constant, it varies slightly.
No! The distance between the Sun and Moon not being constant does not explain it!

A Sun and Moon of identical size can produce a total eclipse, but the umbra must always be the same size as the moon (or sun).
BUT a Sun and Moon of identical size cannot produce an annular eclipse - whatever the distances from Earth to the Sun and moon.

I cannot imagine anyone being so ignorant of the geometry of shadows to even think that!

Look at this diagram:
That should make it completely obvious that
  • if the light source size is greater than the size of the object, then the umbra is smaller than the size of your object,
  • if the light source size is equal in size to the object, then the umbra is equal in size to the object and
  • if the light source size is less than the size of the object, then the umbra is greater than the size of the object.

To get an annular eclipse, the umbral cone must be shorter than the distance from the moon to the earth and so tapering to a point above the earth's surface.

Hence we must have sun's diameter > moon's diameter.

If you disagree, you draw out your diagram, showing how we can get solar eclipses all the way from an annular eclipse to a total eclipse with an umbra ranging up to 150 km wide at the equator to over 1000 km wide at the poles.

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roy30103

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Re: Eclipse proportions refute RET
« Reply #82 on: January 25, 2018, 08:02:40 PM »
I disagree with the sun being 1.3 times the angular size of the moon. Where did you get that number?

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markjo

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Re: Eclipse proportions refute RET
« Reply #83 on: January 25, 2018, 08:14:49 PM »
This would require the sun and moon to be the same size and the same distance from Earth.
This means during an eclipse they are literally in the same location. You would need the moon to pass through the sun. This is impossible.
This would simply be a collision between the sun and moon.


Not at all.

As shown in the Fred Bruenjes photographs there is a certain distance between the solar/black sun disks.
Are you referring to the "black sun disk" that was poorly photoshopped in?
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

*

roy30103

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Re: Eclipse proportions refute RET
« Reply #84 on: January 25, 2018, 08:20:27 PM »
I am talking about the post that started this thread.

Re: Eclipse proportions refute RET
« Reply #85 on: January 25, 2018, 08:22:13 PM »
I disagree with the sun being 1.3 times the angular size of the moon. Where did you get that number?
He made it up.
But that is the least of the issues.

Re: Eclipse proportions refute RET
« Reply #86 on: January 25, 2018, 10:05:43 PM »
I disagree with the sun being 1.3 times the angular size of the moon. Where did you get that number?
He made it up.
But that is the least of the issues.






The largest angular size of the sun, divided by the smallest angular size of the moon?  That ratio is only 1.11, not 1.3 as he stated.

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #87 on: January 25, 2018, 11:41:13 PM »
Thus p=1, thus they are the same distance.
Hence, your claims require them to be the same distance and size.

No! The distance between the Sun and Moon not being constant does not explain it!

A Sun and Moon of identical size can produce a total eclipse, but the umbra must always be the same size as the moon (or sun).
BUT a Sun and Moon of identical size cannot produce an annular eclipse - whatever the distances from Earth to the Sun and moon.


Again, you are dealing with conventional physics.

If we only use VECTOR FIELDS, then you might have a point.

However, vector fields are caused by POTENTIALS.

And the potential has a hidden substructure.

That is why calculating the Earth-Moon distance using ham radio measurements are erroneous: they fail to take into account the density of ether/aether which modifies the speed of light accordingly.


R. Feynman:

E and B are slowly disappearing from the modern expression of physical laws; they are being replaced by A and φ.

The Aharonov-Bohm effect changed everything.

“A new generation of physicists, also educated in the grand assumption that "Heaviside's Equations" are actually "Maxwell's," were abruptly brought up short in 1959 with a remarkable and elegant experiment -- which finally demonstrated in the laboratory the stark reality of Maxwell's "pesky scalar potentials" ... those same "mystical" potentials that Heaviside so effectively banished for all time from current (university-taught) EM theory."

"Totally screened, by all measurements, from the magnetic influence of the ring itself, a test beam of electrons fired by Aharonov and Bohm at the superconducting "donut," nonetheless, changed their electronic state ("wave functions") as they passed through the observably "field-free" region of the hole -- indicating they were sensing "something," even though it could NOT be the ring's magnetic field. Confirmed now by decades of other physicists' experiments as a true phenomenon, this "Aharonov-Bohm Effect" provides compelling proof of a deeper "spatial strain" -- a "scalar potential" -- underlying the existence of a so-called magnetic "force-field" itself.”





Even in the absence of force fields (conventional physics), the electron will experience a phase shift caused by the potential.


Here is the Aharonov-Bohm effect applied on a grand, global scale:



Curved paths of the ball lightning spheres, where there should be none at all.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1995026#msg1995026


That is why even though p = 1, there can still exist an infinitesimal distance between the Sun and Black Sun, and at the same time have the occurrence of an annular eclipse.

YOU WILL HAVE CURVED PATHS OF LIGHT WHERE THERE SHOULD BE NONE.


The field of mathematics dealing with this kind of situation (nonlinear spherical geometry) is still in its infancy: that is, the mathematical tools needed to describe the phenomenon have not been invented yet.


Both of you are using conventional physics.

But the solar eclipse, as evidenced by the Allais effect, deals only with NONCONVENTIONAL PHYSICS, the physics of the POTENTIAL, and its hidden substructure.


Are you referring to the "black sun disk" that was poorly photoshopped in?

You tried that line of attack before: it won't work now, as it did not at that time.

http://www.moonglow.net/eclipse/2003nov23/

Fred Bruenjes, a world-renowned photographer, has explained clearly the photograph.

The distances and the diameters WERE NOT CHANGED AT ALL.

This photograph proves, once and for all, that the "Moon" is not 384,000km away from the Earth.



The following photographs, taken by another world-renowned photographer, Thierry Legault, prove the same thing:



THE ISS AND THE "MOON": SAME DISTANCE FROM THE SUN, NO 384,000 KM DISTANCE FROM THE EARTH TO THE MOON.



TRANSIT OF MERCURY ACROSS THE SUN




TRANSIT OF ATLANTIS ACROSS THE SUN


We are told that the Earth-Atlantis distance is some 400 km, while the distance between Earth and Mercury is some 77 million km and the Mercury-Sun distance is some 50 million km.

Yet the photographs show the very same distance.




Again, MERCURY TRANSIT ACROSS THE SUN




ISS/ATLANTIS TRANSIT ACROSS THE SUN: NO 150,000,000 KM DISTANCE AT ALL


These are the real dimensions of our solar system. A much smaller sun, and a much smaller Earth-Sun distance.



The OP of this thread is totally correct.

Either using an angular size of 1.3 (or 1.11), the final formula ( k - c = k/p ) shows that the official data on the solar eclipse is totally false.


And deal with the OP's faillings.

Your friends' alts won't help you.

YOU FAILED TO PROPERLY READ THE OP.

You used the diagonal distance (the hypothenuse), instead of the direct distance (the side of the right triangle).

The calculations are correct.

With an angular size of 1.3 (or 1.11), you will get a final formula of k - c = k/p. If p = 1.3 (or 1.11), with the conventional 150,000,000 distance to the Sun, you will get the WRONG Earth-Moon distance.

The only way out is to see that p = 1, and apply the physics of the Aharonov-Bohm effect to properly describe the entire situation.


« Last Edit: January 25, 2018, 11:43:48 PM by sandokhan »

Re: Eclipse proportions refute RET
« Reply #88 on: January 26, 2018, 04:12:00 AM »
Again, you are dealing with conventional physics.
Again, I am dealing with physics based upon reality.
If you want to start a new thread on delusional physics go ahead, but this is not the place for them.

Fred Bruenjes, a world-renowned photographer, has explained clearly the photograph.
The distances and the diameters WERE NOT CHANGED AT ALL.
This photograph proves, once and for all, that the "Moon" is not 384,000km away from the Earth.
The distances are not in the photo. There is no way to tell how far away the moon is in that photo.

Yet the photographs show the very same distance.
No, they don't.
Get this through your thick skull:
SIMPLE PHOTOS LIKE THIS SHOW DIRECTION NOT DISTANCE!!!


The OP of this thread is totally correct.
No, they aren't.
They have baselessly asserted the sun is 1.3 times the size of the moon, and their math was completely wrong.

Either using an angular size of 1.3 (or 1.11), the final formula ( k - c = k/p ) shows that the official data on the solar eclipse is totally false.
No it doesn't.
Using the correct ratio of angular sizes all you get is a ratio of distances, which matches RET.

YOU FAILED TO PROPERLY READ THE OP.
You used the diagonal distance (the hypothenuse), instead of the direct distance (the side of the right triangle).
There you go lying againg.
The triangle representing the sun had 3 sides. One was the sun, with a length of pM, the other 2 were both k.
This is the diagonal distance.

The calculations are correct.
No, they aren't.
He magically got an extra factor of sqrt(2-p^2).

With an angular size of 1.3 (or 1.11), you will get a final formula of k - c = k/p. If p = 1.3 (or 1.11), with the conventional 150,000,000 distance to the Sun, you will get the WRONG Earth-Moon distance.
And you seem to fail to understand yet again.
p is NOT 1.3.
If you want to try comparing it to reality, set p to ~400.
p is not the ratio of angular sizes, itis the ratio of real sizes.

So all you and the OP have shown is that your strawman situation with a sun only 1.3 times the size of the moon doesn't match reality (and that you don't understand the argument at all).

The only way out is to see that p = 1, and apply the physics of the Aharonov-Bohm effect to properly describe the entire situation.
Nope. It works just fine by letting p=400.

*

sandokhan

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Re: Eclipse proportions refute RET
« Reply #89 on: January 26, 2018, 05:43:03 AM »
Again, I am dealing with physics based upon reality.
If you want to start a new thread on delusional physics go ahead, but this is not the place for them.


Question for the moderators: why is this manner of posting allowed here?

He has just called the Aharonov-Bohm effect, "delusional physics".

Let's see how "delusional" it is.

Here is the original paper:

https://journals.aps.org/pr/pdf/10.1103/PhysRev.115.485

The best scientific journals publishing the experimental verification of the Aharonov-Bohm effect:

https://www.nature.com/articles/17755

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3599092/

https://arxiv.org/pdf/cond-mat/0102096.pdf

Confirmation of the Aharonov-Bohm effect:

https://www.physik.uni-muenchen.de/lehre/vorlesungen/wise_09_10/quantum_matter/lecture/Schwarzschild1986.pdf

Neutron Interferometry: the Aharonov-Bohm effect

https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.71.307

https://arxiv.org/pdf/1109.4887.pdf

https://arxiv.org/pdf/1411.3627.pdf

http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/24/003/24003927.pdf

https://authors.library.caltech.edu/7139/1/LEEprl98.pdf

http://www.sciencedirect.com/science/article/pii/S0168900299010384


You should be banned for at least six months for daring to use the word "delusional" in describing the Aharonov-Bohm effect.

"Totally screened, by all measurements, from the magnetic influence of the ring itself, a test beam of electrons fired by Aharonov and Bohm at the superconducting "donut," nonetheless, changed their electronic state ("wave functions") as they passed through the observably "field-free" region of the hole -- indicating they were sensing "something," even though it could NOT be the ring's magnetic field. Confirmed now by decades of other physicists' experiments as a true phenomenon, this "Aharonov-Bohm Effect" provides compelling proof of a deeper "spatial strain" -- a "scalar potential" -- underlying the existence of a so-called magnetic "force-field" itself.”

 “...the vector potential appears to give the most direct description of the physics. This becomes more apparent the more deeply we go into quantum theory. In the general theory of quantum electrodynamics, one takes the vector and scalar potentials as the fundamental quantities in a set of equations that replace the Maxwell equations: E and B are slowly disappearing from the modern expression of physical laws; they are being replaced by A and φ”

(Feynman et al, 1989, chapter 15, section 5, The Feynman Lecture on Physics (Vol. 2), 1989)


This is the delusional world in which you live, a clear and terrible sign of cognitive dissonance.


The triangle representing the sun had 3 sides. One was the sun, with a length of pM, the other 2 were both k.
This is the diagonal distance.




The sun has a discoidal shape.

The final formulas, for both sets of calculations, lead to the same result.

He magically got an extra factor of sqrt(2-p^2).

For the correct sign of c, you get the correct formula at the end.


p is NOT 1.3.
If you want to try comparing it to reality, set p to ~400.


But it is approximately 1.3, that is the value obtained through direct observation.

Read the OP:

Step one. Calculate the proportional sizes of the Sun and moon during an annular eclipse. You'll find the Sun appears to be approximately 1.3 times the size of the moon, it will vary depending on which eclipse.


The distances are not in the photo. There is no way to tell how far away the moon is in that photo.

The official claim is 384,000 km. However, both Armstrong and Aldrin should have used a sledge driven by reindeers to get to the Moon: the photograph features a distance of approximately 1000 km.




Nobody has a thicker skull than you: you have just called the Aharonov-Bohm effect as "delusional".

The difference in the Thierry Legault photographs is not measured in the hundreds of kilometers, thousands of kilometers, or even hundreds of thousands of kilometers.

THE DIFFERENCE AMOUNTS TO TENS OF MILLIONS OF KILOMETERS.



MERCURY TRANSIT ACROSS THE SUN

Mercury-Sun distance: ~50,000,000 km

Earth-Mercury distance: ~77,000,000 km

TENS OF MILLIONS OF KILOMETERS.




ATLANTIS TRANSIT ACROSS THE SUN

Atlantis-Sun official distance: 149,999,600 km

Earth-Atlantis official distance: 400 km

Yet, the SAME DISTANCE SEPARATES ATLANTIS FROM THE SUN AS DOES THE MERCURY-SUN DISTANCE.

These are the real dimensions of our solar system. A much smaller sun, and a much smaller Earth-Sun distance.


Nope. It works just fine by letting p=400.

Sorry, you can't use the 400 figure.

Not when, by direct observation during the solar eclipse, in the case of heliocentrism, the angular size of the Sun is 1.3 (or 1.1) times the angular size of the "Moon".



« Last Edit: January 26, 2018, 05:54:10 AM by sandokhan »