Eclipse proportions refute RET

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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #30 on: January 24, 2018, 06:32:58 PM »
They cannot produce both an annular and total eclipse, because I have already proven the proportions necessary do not work for RET. If you disagree I am waiting for you to stop being such a fucking coward and address the proof.

Your "proof is fatally flawed." I let it slide because you also proved that the Earth can't be flat. The numbers I gave produce an annular eclipse. If you want a total eclipse, use any number between 362600 and 380000 km. It is primarily the difference between the Moon's perigee(362600) and apogee(405400) that determines whether an eclipse is total or annular.

Check the numbers.
If it's flawed then show the flaw. Randomly giving a set of numbers without appealing any real world facts, as I did, is meaningless, all the more so because one set of numbers cannot explain both total and annular eclipses. If you want to know what relationship numbers need in order to explain both then READ MY FUCKING POST ALREADY.
If you are just going to repeat that lie, then you have no understanding of what it says. The shape of the Earth never enters into it. I could have made the entire post and never mentioned it once. RET is brought up only to demonstrate the numbers do not work. They are far more in line with what FET predicts about the nearby nature of the Sun and moon. You have demonstrated yourself to be nothing but an evasive, lying coward.

You have already been shown the flaw in this these. Want to see your fatal flaw with your own eyes? On the next sunny day, block the Sun with you hand.

And stop calling me a coward, you fucking imbecile.

You are a coward. You are refusing to read, refusing to respond. You think repeating bullshit over and over makes it true.
You have not presented any flaw, you have presented denial. You have presented the RE numbers with the insistence that they work, and refused to address the proof that they don't. You are blindly saying it doesn't work, but are unable to say why or how. All you are doing is insisting that what you want to be the case is true, despite your total inability to justify it or deal with anything contrary.
That's the resort of a coward, clinging desperately to fantasy rather than face reality.

I have proven myself. I await any REer to do the same.
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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #31 on: January 24, 2018, 06:41:31 PM »
They cannot produce both an annular and total eclipse, because I have already proven the proportions necessary do not work for RET. If you disagree I am waiting for you to stop being such a fucking coward and address the proof.

Your "proof is fatally flawed." I let it slide because you also proved that the Earth can't be flat. The numbers I gave produce an annular eclipse. If you want a total eclipse, use any number between 362600 and 380000 km. It is primarily the difference between the Moon's perigee(362600) and apogee(405400) that determines whether an eclipse is total or annular.

Check the numbers.
If it's flawed then show the flaw. Randomly giving a set of numbers without appealing any real world facts, as I did, is meaningless, all the more so because one set of numbers cannot explain both total and annular eclipses. If you want to know what relationship numbers need in order to explain both then READ MY FUCKING POST ALREADY.
If you are just going to repeat that lie, then you have no understanding of what it says. The shape of the Earth never enters into it. I could have made the entire post and never mentioned it once. RET is brought up only to demonstrate the numbers do not work. They are far more in line with what FET predicts about the nearby nature of the Sun and moon. You have demonstrated yourself to be nothing but an evasive, lying coward.

You have already been shown the flaw in this these. Want to see your fatal flaw with your own eyes? On the next sunny day, block the Sun with you hand.

And stop calling me a coward, you fucking imbecile.

You are a coward. You are refusing to read, refusing to respond. You think repeating bullshit over and over makes it true.
You have not presented any flaw, you have presented denial. You have presented the RE numbers with the insistence that they work, and refused to address the proof that they don't. You are blindly saying it doesn't work, but are unable to say why or how. All you are doing is insisting that what you want to be the case is true, despite your total inability to justify it or deal with anything contrary.
That's the resort of a coward, clinging desperately to fantasy rather than face reality.

I have proven myself. I await any REer to do the same.

You have proven yourself to be a fucking imbecile.
Nullius in Verba

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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #32 on: January 24, 2018, 06:42:21 PM »
They cannot produce both an annular and total eclipse, because I have already proven the proportions necessary do not work for RET. If you disagree I am waiting for you to stop being such a fucking coward and address the proof.

Your "proof is fatally flawed." I let it slide because you also proved that the Earth can't be flat. The numbers I gave produce an annular eclipse. If you want a total eclipse, use any number between 362600 and 380000 km. It is primarily the difference between the Moon's perigee(362600) and apogee(405400) that determines whether an eclipse is total or annular.

Check the numbers.
If it's flawed then show the flaw. Randomly giving a set of numbers without appealing any real world facts, as I did, is meaningless, all the more so because one set of numbers cannot explain both total and annular eclipses. If you want to know what relationship numbers need in order to explain both then READ MY FUCKING POST ALREADY.
If you are just going to repeat that lie, then you have no understanding of what it says. The shape of the Earth never enters into it. I could have made the entire post and never mentioned it once. RET is brought up only to demonstrate the numbers do not work. They are far more in line with what FET predicts about the nearby nature of the Sun and moon. You have demonstrated yourself to be nothing but an evasive, lying coward.

You have already been shown the flaw in this these. Want to see your fatal flaw with your own eyes? On the next sunny day, block the Sun with you hand.

And stop calling me a coward, you fucking imbecile.

You are a coward. You are refusing to read, refusing to respond. You think repeating bullshit over and over makes it true.
You have not presented any flaw, you have presented denial. You have presented the RE numbers with the insistence that they work, and refused to address the proof that they don't. You are blindly saying it doesn't work, but are unable to say why or how. All you are doing is insisting that what you want to be the case is true, despite your total inability to justify it or deal with anything contrary.
That's the resort of a coward, clinging desperately to fantasy rather than face reality.

I have proven myself. I await any REer to do the same.

You have proven yourself to be a fucking imbecile.

When you can point out any problem in the proof and justify yourself with more than empty denial, I will care about your opinion.
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dualearththeory.proboards.com/
On the sister site if you want to talk.

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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #33 on: January 24, 2018, 06:45:37 PM »
They cannot produce both an annular and total eclipse, because I have already proven the proportions necessary do not work for RET. If you disagree I am waiting for you to stop being such a fucking coward and address the proof.

Your "proof is fatally flawed." I let it slide because you also proved that the Earth can't be flat. The numbers I gave produce an annular eclipse. If you want a total eclipse, use any number between 362600 and 380000 km. It is primarily the difference between the Moon's perigee(362600) and apogee(405400) that determines whether an eclipse is total or annular.

Check the numbers.
If it's flawed then show the flaw. Randomly giving a set of numbers without appealing any real world facts, as I did, is meaningless, all the more so because one set of numbers cannot explain both total and annular eclipses. If you want to know what relationship numbers need in order to explain both then READ MY FUCKING POST ALREADY.
If you are just going to repeat that lie, then you have no understanding of what it says. The shape of the Earth never enters into it. I could have made the entire post and never mentioned it once. RET is brought up only to demonstrate the numbers do not work. They are far more in line with what FET predicts about the nearby nature of the Sun and moon. You have demonstrated yourself to be nothing but an evasive, lying coward.

You have already been shown the flaw in this these. Want to see your fatal flaw with your own eyes? On the next sunny day, block the Sun with you hand.

And stop calling me a coward, you fucking imbecile.

You are a coward. You are refusing to read, refusing to respond. You think repeating bullshit over and over makes it true.
You have not presented any flaw, you have presented denial. You have presented the RE numbers with the insistence that they work, and refused to address the proof that they don't. You are blindly saying it doesn't work, but are unable to say why or how. All you are doing is insisting that what you want to be the case is true, despite your total inability to justify it or deal with anything contrary.
That's the resort of a coward, clinging desperately to fantasy rather than face reality.

I have proven myself. I await any REer to do the same.
j

You have proven yourself to be a fucking imbecile.

When you can point out any problem in the proof and justify yourself with more than empty denial, I will care about your opinion.

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.
Nullius in Verba

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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #34 on: January 24, 2018, 06:49:03 PM »
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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

What is the magical fundamental flaw you keep appealing to? Where is it? All I see if you insisting that the conclusion is wrong, which can't be the case if all the reasoning stands.
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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #35 on: January 24, 2018, 06:56:28 PM »
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

What is the magical fundamental flaw you keep appealing to? Where is it? All I see if you insisting that the conclusion is wrong, which can't be the case if all the reasoning stands.

...how the hell are you able to function?
Nullius in Verba

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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #36 on: January 24, 2018, 06:58:31 PM »
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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

What is the magical fundamental flaw you keep appealing to? Where is it? All I see if you insisting that the conclusion is wrong, which can't be the case if all the reasoning stands.

...how the hell are you able to function?
I will now ignore all your posts in this thread until you are able to point out where the reasoning in my original post failed.
No handwaving, no insisting RET has to work because it just has to! No stupid and irrelevant situations. No open lies about what my argument stated.

I await this evidence you so-clearly have.
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sokarul

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Re: Eclipse proportions refute RET
« Reply #37 on: January 24, 2018, 07:17:15 PM »
Have you teleported to the USS Enterprise yet?
ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

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rabinoz

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Re: Eclipse proportions refute RET
« Reply #38 on: January 24, 2018, 07:19:37 PM »
You say that you "will now ignore all your posts in this thread until you are able to point out where the reasoning in my original post failed."

Well, here is my reasoning again!

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.
Remember that you said, "Calculate the proportional sizes of the Sun and moon during an annular eclipse".

Quote from: JRoweSkeptic
Step two. Create two isoceles triangles, one giving the angular distance of the Sun in the sky, one giving the angular distance of the moon. We can fill in these values.
We want the angles to be equal, to give the situation of a total eclipse.
No, no, no!
You already said that, "during an annular eclipse. You'll find the Sun appears to be approximately 1.3 times the size of the moon".
Now you are saying, "We want the angles to be equal, to give the situation of a total eclipse." You can't have it both ways!.

An annular eclipse is not the same as a total eclipse! Go and read, Types of Solar Eclipse.

Quote from: JRoweSkeptic
The angular distance of both objects will be the same. So in both triangles, the angle will be x. The side opposite this angle will have length M (the apparent size of the moon) in the moon triangle, and 1.3 M in the Sun triangle.
By saying that "the Sun appears to be approximately 1.3 times the size of the moon" you have already said that "angular distance of both objects" can not "be the same".

So I'm not being cowardly or anything like that, I'm simply saying the if the apparent size of the sun is 1.3 times the apparent size of the moon the angles are not the same.

That is why I will not carry on with the rest - as I said in the first place, you are starting from an incorrect premise.
Quote from: JRoweSkeptic
The RE values are dramatically far from what it is RET states.
So you have proved nothing. If you think I am wrong, do what others have asked and draw some diagrams.

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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #39 on: January 24, 2018, 07:21:47 PM »
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An annular eclipse is not the same as a total eclipse!
I KNOW. THAT WAS CENTRAL TO MY WORKINGS.
The sizes of the Sun and moon are not going to magically change between the two though.
Stop acting like you've somehow won when you haven't even understood the basics of how math works.

We calculate the proportion, then determine how far one has to be shifted in order for the angles to be the same.
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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #40 on: January 24, 2018, 07:24:00 PM »
And that's me signing off. Forty posts and everything can be responded to with the OP. Round earthers and all their usual ignorance and total lack of understanding.

Not that any of you are ever going to admit anything. The fact you repeat your bullshit for this long is something you'll take as a victory, I've seen it. Oh, the thread went on over a page, that dumb FEer must have been refuted in that time. It's not like all REers do is repeat the same handful of things and whine like a baby whenever anything goes past that. It's not like you always rely on bullshit.

I win. Again. Goodbye.
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Re: Eclipse proportions refute RET
« Reply #41 on: January 24, 2018, 07:31:52 PM »
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.

1.3 is much larger than what ever actually happens, but, for the sake of discussion, OK.
 
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Step two. Create two isoceles triangles, one giving the angular distance size of the Sun in the sky, one giving the angular distance size of the moon. We can fill in these values.

If we want the angles to be equal, to give the situation of a total eclipse[, the] angular distance size of both objects will be the same. So in both triangles, the angle will be x.

The side opposite this angle will have length M (the apparent size of the moon) in the moon triangle, and [in the case of the annular eclipse] 1.3 M in the Sun triangle if they were at the same 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.

Step three. Use the cosine rule to come up with a quadratic formula relating c and k.

Step four. Solve for c. If you don't know how to solve a quadratic equation, I can't help you.

Some attempts at clarification have been applied to your text. If you think these are incorrect, please discuss.

Solving a quadratic equation: for an equation in the form

A x2 + B x + C = 0

two answers for the value of x will satisfy the equation:

x = (-B + sqrt( B2 - 4 A C )) / (2 A), and

x = (-B - sqrt( B2 - 4 A C )) / (2 A)

sqrt(Q) is the square root of Q. If B2 - 4 A C = 0 (i.e. if B = 2 and A times C = 1), both values for x will be the same, and there is a single solution. Sometimes the solutions will be imaginary.

But none of this matters since it's irrelevant to the problem.

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I cannot write math into the forum, but I expect any capable reader will be able to run through and verify the calculations for themselves. Using p as the proportion (given as 1.3 above, but left open so that you may test with your own).

You will find that the distance to the moon, k+c, is k/p times the square root of (2-p2)

This is not correct.

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Certainly, there is some error in this calculation. According to RET the precise distances to the Sun and moon vary, as does p, but using p=1.3 and k as the distance to the Sun google gives me the distance to the moon as 64 million km. This is substantially different to the value RET gives.

You're right that it's substantially different. It's a clue that your technique is wrong.

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For those interested in the theory behind this calculation, we begin by finding a proportion relating the Sun and moon. We then create, essentially, one large triangle.

There are two different, but similar (in the geometric sense - similar triangles have the same angles, but not necessarily the same side lengths) isosceles triangles. One with its opposite side with a length equal to the diameter of the moon, the other with the opposite side the length of the diameter of the sun.

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At one point is the observer, who looks up during a total eclipse to see the Sun and moon with the same angular size. They are different distances away however, so this triangle (currently a V, with the observer looking up) will have two lines opposite the angle, at varying distances away. One is the moon, the further is the Sun. Thus, there are two triangles in this one, the only differences being a) the size of the object, b) the distance to the object.
We can then use the proportion to relate the two distances, so b is the only unknown left to find. The distance to the Sun gives us the distance to the moon, and vice versa.

No, you can't, law of cosines won't help unless you know the length of at least one side of each of the triangles. Even if you know the distance to the sun but not the moon, you still need to know the size of the moon to determine its distance.

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The RE values are dramatically far from what it is RET states.

No, your calculated values are dramatically wrong because your math is wrong. Your answer is not the "RET" value.

Using numbers provided by Google:

Diameter of sun: 864,575.9 miles.
Diameter of moon: 2,159 miles.

If they have the same apparent size, the sun must be (864,575.9 mi / 2,159 mi) = 400.45 times further away. Call it 400.

Mean distance to sun: call it 93,000,000 miles.

93,000,000 miles / 400 = 232,000 miles.

That sounds about right.

If the moon appears 30% smaller than at mean, it must be about 30% further away. As mentioned earlier, though, the magnitude of an annular eclipse (ratio of apparent diameters) is never as small as 0.77 (30% smaller).

Note: putting these replies together takes time and effort. If you've really already "checked out" of the thread by the time this is posted and won't respond, well, too bad for you!
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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JRoweSkeptic

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Re: Eclipse proportions refute RET
« Reply #42 on: January 24, 2018, 07:43:38 PM »
Quote
This is not correct.
You're right that it's substantially different. It's a clue that your technique is wrong.
If you had an argument, you ought to be able to give it without multiple lines of pure assertion. This was pointless.

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No, you can't, law of cosines won't help unless you know the length of at least one side of each of the triangles. Even if you know the distance to the sun but not the moon, you still need to know the size of the moon to determine its distance.
We have this. The size of the moon, M, and the relative size of the Sun.
The size of the Sun and the size of the moon are constant throughout eclipses.

Quote
That sounds about right.
That was also entirely circular. You took the RE measurements, assumed they were accurate, and worked backwards to determine the numbers you started with were accurate. That proves nothing. 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.

I will give you the 30% point, I chose the best possible value for you but many photos of annular eclispes are not of a perfect annulus. Shrinking p does not help you, you are welcome to do so.


Or, another chance:
We can calculate the relative distance to the moon in terms of the distance to the Sun (or vice versa) by taking two instances of them being in alignment.
That is mathematical fact. It gives us all the data we need.
If I did not successfully do it, what way would work?
No more circular arguments. You have a total eclipse and an annular eclipse. That's all. if what I did failed, how would you calculate the relative distances?
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EvolvedMantisShrimp

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Re: Eclipse proportions refute RET
« Reply #43 on: January 24, 2018, 07:54:31 PM »
Quote
This is not correct.
You're right that it's substantially different. It's a clue that your technique is wrong.
If you had an argument, you ought to be able to give it without multiple lines of pure assertion. This was pointless.

Quote
No, you can't, law of cosines won't help unless you know the length of at least one side of each of the triangles. Even if you know the distance to the sun but not the moon, you still need to know the size of the moon to determine its distance.
We have this. The size of the moon, M, and the relative size of the Sun.
The size of the Sun and the size of the moon are constant throughout eclipses.

Quote
That sounds about right.
j
That was also entirely circular. You took the RE measurements, assumed they were accurate, and worked backwards to determine the numbers you started with were accurate. That proves nothing. 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.

I will give you the 30% point, I chose the best possible value for you but many photos of annular eclispes are not of a perfect annulus. Shrinking p does not help you, you are welcome to do so.


Or, another chance:
We can calculate the relative distance to the moon in terms of the distance to the Sun (or vice versa) by taking two instances of them being in alignment.
That is mathematical fact. It gives us all the data we need.
If I did not successfully do it, what way would work?
No more circular arguments. You have a total eclipse and an annular eclipse. That's all. if what I did failed, how would you calculate the relative distances?

You would either need to know the actual relative sizes to determine the relative distances, or there is a parallax technique that would work without that information if you measured the time and angles of the eclipse from two different locations.
« Last Edit: January 24, 2018, 07:56:15 PM by EvolvedMantisShrimp »
Nullius in Verba

Re: Eclipse proportions refute RET
« Reply #44 on: January 24, 2018, 08:13:10 PM »
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.
Premise denied.
Please provide reference to an observation of an annular eclipse where the angular size of the Sun is ~1.3 times the angular size of the Moon.

I'll wait.

Re: Eclipse proportions refute RET
« Reply #45 on: January 24, 2018, 08:42:35 PM »
Quote
This is not correct.
You're right that it's substantially different. It's a clue that your technique is wrong.
If you had an argument, you ought to be able to give it without multiple lines of pure assertion. This was pointless.

Instead of simply claiming "pure assertion", can we see what you think is baseless? Specifically.

Quote
Quote
No, you can't, law of cosines won't help unless you know the length of at least one side of each of the triangles. Even if you know the distance to the sun but not the moon, you still need to know the size of the moon to determine its distance.
We have this. The size of the moon, M, and the relative size of the Sun.
The size of the Sun and the size of the moon are constant throughout eclipses.

Right. And in a total eclipse, their apparent sizes are the same. Therefore the ratio of their distances must be the same as their ratio of diameters. No need to solve quadratics or invoke the law of cosines.
 
Quote
Quote
That sounds about right.
That was also entirely circular. You took the RE measurements, assumed they were accurate, and worked backwards to determine the numbers you started with were accurate. That proves nothing. 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.

Observable reality tells us the sun is much further than the moon. You accepted the "distance to the Sun google gives me", so that's apparently a given. What distance did you use? Can you show how you calculated "the distance to the moon as 64 million km" from that?

If the moon were 64 million km away, it would have to be about half a million km in diameter to have its apparent size. Measurements of parallax relative to stars make it clear that it's neither that large nor that far away. In fact, they give us a very accurate measure of both its distance and can distinguish surface variations to a few dozen meters.

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I will give you the 30% point, I chose the best possible value for you but many photos of annular eclispes are not of a perfect annulus. Shrinking p does not help you, you are welcome to do so.

Or, another chance:
We can calculate the relative distance to the moon in terms of the distance to the Sun (or vice versa) by taking two instances of them being in alignment.
That is mathematical fact. It gives us all the data we need.

No, it's not. You need the ratio of their sizes and one of the distances to do so. There is no suggestion that you know or have considered the actual sizes in your exercise. If you had, and they were reasonably correct, you should have come up with a reasonably close answer, unless your procedure was wrong.

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If I did not successfully do it, what way would work?
No more circular arguments. You have a total eclipse and an annular eclipse. That's all. if what I did failed, how would you calculate the relative distances?

You need their relative sizes. Your exercise with annular eclipses (with whatever magnitude) only shows a proportional change in the ratio of distances, not the distances themselves, without additional data (like actual sizes). [EMS beat me to this obvious answer.]
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: Eclipse proportions refute RET
« Reply #46 on: January 24, 2018, 09:03:43 PM »
Step two. Create two isoceles triangles, one giving the angular distance of the Sun in the sky, one giving the angular distance of the moon. We can fill in these values.
We want the angles to be equal, to give the situation of a total eclipse. The angular distance of both objects will be the same. So in both triangles, the angle will be x. The side opposite this angle will have length M (the apparent size of the moon) in the moon triangle, and 1.3 M in the Sun triangle.
I read ahead. My bad.

If the apex angle is x, for the Moon triangle the side opposite this angle is length M. This is not the apparent size of the Moon, it is the actual size of the Moon. There is no basis to then say in the Sun triangle that the side opposite angle x is 1.3M. The side opposite angle x is length S.

The ratio of M:S will be equal to the ratio of k+c:k (distance to Moon:distance to Sun) for your limiting case eclipse.

There is no reason to postulate S=1.3M


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rabinoz

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Re: Eclipse proportions refute RET
« Reply #47 on: January 24, 2018, 09:56:31 PM »
Quote
An annular eclipse is not the same as a total eclipse!
I KNOW. THAT WAS CENTRAL TO MY WORKINGS.
The sizes of the Sun and moon are not going to magically change between the two though.
True, "the sizes of the Sun and moon are not going to magically change between the two" but the distances do change.
Even the flat earth, the Sun and possobly the moon change height. Here from "the FAQ"
Quote
What About Seasons?
The radius of the sun's orbit around the Earth's axis symmetry varies throughout the year, being smallest when summer is in the northern annulus and largest when it is summer in the southern annulus. Additionally it also raises and lowers. This causes the effect of the sun appearing to move in a figure eight throughout a year.
On the Globe the distance
        from the earth to the sun ranges from 146 million km to 152 million km and
        from the earth to the sun ranges from 356,500 km at the perigee to 406,700 km.
Quote from: JRoweSkeptic
Stop acting like you've somehow won when you haven't even understood the basics of how math works.
I do believe I have a smattering, but maybe not quite up to the Major General's level:
"I’m very well acquainted, too, with matters mathematical,
I understand equations, both the simple and quadratical,
About binomial theorem I’m teeming with a lot o’ news –
With many cheerful facts about the square of the hypotenuse.
and
I’m very good at integral and differential calculus."
Well, not "very good"!
Quote from: JRoweSkeptic
We calculate the proportion, then determine how far one has to be shifted in order for the angles to be the same.
OK, the distances must change - that's what the Globe model claims, big deal - as I said, you've proven nothing!

Now, run off and try again!

Re: Eclipse proportions refute RET
« Reply #48 on: January 24, 2018, 10:34:31 PM »
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.
Citation needed.
I am yet to hear anyone ever claim such a thing, nor have I ever seen a picture which shows that.

Until you substantiate that the rest of your argument is pure garbage.

But what I will do is see what the maximum ratio could be if HC theory is correct (not RET, as that is separate, and a problem like this would completely destroy FET more so than the existence of annular and total solar eclipse with a shadow larger than the alleged size of the moon).
At apogee, the furthest the moon is from Earth and thus the smallest angular diameter, the moon is 405 400 km away with a mean radius of 1737.1 km, giving an angular size of 0.49 degrees.
Meanwhile, the Earth at perihelion (where it is closest to the sun to give the largest apparent size of the sun) is roughly 147 095 000 km, with the sun having a mean radius of 695 700 km, giving an apparent size of 0.54 degrees, 1.10 times that of the moon.
In the opposite situation, the sun would be 0.96 times that of the moon.

And rather than bother going through all the math your way, I can tell you by using simple trig, with a right angle triangle and tan, to have the sun appear 1.3 times the size of the moon (at perihelion), the moon only needs to be 476 995 km away from Earth, no where near your 64 million.
Likewise, I can tell you that if the moon was 64 million km away it would have an angular size of only 0.003110266 degrees, making the sun around 170 times as large.

So as per usual, you are full of shit.

Re: Eclipse proportions refute RET
« Reply #49 on: January 24, 2018, 11:04:31 PM »
To deal with it in more detail:
Step two. Create two isoceles triangles, one giving the angular distance of the Sun in the sky, one giving the angular distance of the moon. We can fill in these values.
We want the angles to be equal, to give the situation of a total eclipse. The angular distance of both objects will be the same. So in both triangles, the angle will be x. The side opposite this angle will have length M (the apparent size of the moon) in the moon triangle, and 1.3 M in the Sun triangle.
This makes no sense at all.

What you are actually doing here is merely making the sun 1.3 times the distance to the moon, and 1.3 times the size of the moon.
What you actually need to do is have 2 angles, one will be x, the angular size of the moon, and the other will be p x, the angular size of the sun.
This ratio is a ratio of the angles, not a ratio of the sizes of the lines.
The sizes of the lines can be 2M, the diameter of the moon and 2S, the diameter of the sun.

We can also break it into 2 simple right angle triangles rather than 2 isosceles triangles which massively simplifies the math (but means we use x/2 and p x/2 for the angles)

I cannot write math into the forum
So you know it's full of shit and are trying to hide it?

but I expect any capable reader will be able to run through and verify the calculations for themselves.
They sure as hell can verify that what you are saying is pure nonsense.
You don't even need to be good at math to realise what you are saying is crap.
You are trying to determine the distance between 2 objects to give them a ratio of angular size, with no indication of the size of either object (or their relative size).

You will find that the distance to the moon, k+c, is k/p times the square root of (2-p2)
But just to check the math, doing what you actually asked for, where you have the moon and the sun being the same angular size (x), with the sun having a diameter p times that of the moon (which itself is M), where the distance along the diagonal to the moon is c, and to the sun is c+k, using the cosine rule (c^2=a^2+b^2-2*a*b*cos(C), which for an isosceles triangle can be simplified to c^2=a^2+a^2-2*a*acos(C)=2*a^2-2*a^2*cos(C)=2*a^2*(1-cos(C), which can be further simplified to c=2*a*sqrt(1-cos(C))

For the moon we get the equation:
M=2*c*sqrt(1-cos(x))
Which can be rearranged to give:
sqrt(1-cos(x))=M/2c
For the sun we get the equation:
p*M=2*(c+k)*sqrt(1-cos(x))
Subbing in the rearranged equation from the moon we get:
p*M=2*(c+k)*M/2c
Which is simplified to:
p=(c+k)/c
and rearranged to:
c+k=c p

Thus the distance to the sun (c+k), is simply c times p, which you could have easily gotten via rules regarding similar triangles.

But notice how this doesn't match your math?

So even with all the prior false claims, you are still wrong here again.

Doing it properly, where you note the sun is roughly 400 times the size of the moon, this means the sun needs to be roughly 400 times further away. So using the approximate distance of 150 million km to the sun, this would put the moon 375 000 km from Earth.

Certainly, there is some error in this calculation.
Well that is one thing you got right. There is so much error it isn't funny. As some would say "it isn't even wrong."

The RE values are dramatically far from what it is RET states.
No, it is dramatically different from what you state, which is nothing like what RET nor HC theory state.


Edit: Breakdown to demonstrate how many of the responses are either evasion or misunderstanding.
Your entire OP seems to be misunderstanding and evasion.

Their relative sizes stay the same during an annular eclipse. Only the distances change.
Yes, their relative sizes remain the same, with the sun remaining roughly 400 times the size of the moon.
Notice how 400 is nothing like your 1.3?

None of this relies on any assumptions, beyond the basic assumptions of math and logic. It applies to both FET and RET, but if we apply it to the RE numbers, they fail.
You mean except discarding basic math and logic.

And that's me signing off. Forty posts and everything can be responded to with the OP. Round earthers and all their usual ignorance and total lack of understanding.

Not that any of you are ever going to admit anything. The fact you repeat your bullshit for this long is something you'll take as a victory, I've seen it. Oh, the thread went on over a page, that dumb FEer must have been refuted in that time. It's not like all REers do is repeat the same handful of things and whine like a baby whenever anything goes past that. It's not like you always rely on bullshit.

I win. Again. Goodbye.
You mean that's you running away because you have been refuted yet again.
You are the one displaying ignorance and a total lack of understand. You lose yet again.

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NAZA

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Re: Eclipse proportions refute RET
« Reply #50 on: January 25, 2018, 12:50:09 AM »
Not a good week for the Flatters.
First Lackey provides proof that the surface of Lake Pontchartrain is curved, and now another prima donna helps prove the large and distant sun and moon.
Coming soon:  Hoppy discuss satellite TV.

You delusionists should stick to "it looks flat"

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #51 on: January 25, 2018, 01:45:42 AM »
We want the angles to be equal, to give the situation of a total eclipse. The angular distance of both objects will be the same. So in both triangles, the angle will be x. The side opposite this angle will have length M (the apparent size of the moon) in the moon triangle, and 1.3 M in the Sun triangle.
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.




But just to check the math, doing what you actually asked for, where you have the moon and the sun being the same angular size (x), with the sun having a diameter p times that of the moon (which itself is M), where the distance along the diagonal to the moon is c, and to the sun is c+k,


In the first paragraph k denotes the DIRECT DISTANCE TO THE SUN, not the diagonal.

In the second paragraph c becomes the DISTANCE ALONG THE DIAGONAL to the moon, and we add k to it, c + k, as the distance along to the diagonal to the sun.


TWO DIFFERENT SCENARIOS.


However, I see a discrepancy in the defined values, in paragraph 1.

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.

If k is the distance to the Sun, then you cannot have k + c as the distance to the moon: k - c would be the correct value (c is the difference between the distance to the Sun and the distance to the moon).

Yes, it is stated that c might be negative, but it makes more sense to define k (dist. to the sun) and (k - c) (dist. the moon).


Let's do the calculations for both versions.


1.

We have two right triangles.

The subtended angle for both is x/2.

For the triangle involving the moon, we have side 1 = M/2 and side 2 = k - c

For the triangle involving the sun, we have side 1 = pM/2 and side 2 = k

Using the tangent function for angle x, we get:

k/p = (k - c)

If p = 4/3, and k = 150,000,000 km, then (k - c) = 112,500,000 km


2. Now the distance along the diagonal to the sun is k, and the distance to the moon, along the diagonal becomes (k - c), where c is the difference.

Two right triangles.

angle x/2 and side 1 = pM/2, hypothenuse = k for the sun

angle x/2 and side 1 = M/2, hypothenuse = (k - c)

Using the sine function for angle x, we get:

(k - c) = k/p


So comments like the following, simply have no place here at all.

Your entire OP seems to be misunderstanding and evasion.

There is so much error it isn't funny. As some would say "it isn't even wrong."

You mean except discarding basic math and logic.

You mean that's you running away because you have been refuted yet again.
You are the one displaying ignorance and a total lack of understand. You lose yet again.



Doing it properly, where you note the sun is roughly 400 times the size of the moon, this means the sun needs to be roughly 400 times further away. So using the approximate distance of 150 million km to the sun, this would put the moon 375 000 km from Earth.

That is just the hypothesis pur forward by modern science.


However, some of the brightest minds have emitted not only wonder but also doubt towards these numbers.

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


As for the “capture” theory, even scientist Isaac Asimov, well known for his works of fiction, has written, “It’s too big to have been captured by the Earth. The chances of such a capture having been effected and the Moon then having taken up nearly circular orbit around our Earth are too small to make such an eventuality credible.”


It seems impossible that such an oddity could naturally fall into such a precise and circular orbit. It is a fascinating conundrum as articulated by science writer William Roy Shelton, who wrote, “It is important to remember that something had to put the Moon at or near its present circular pattern around the Earth. Just as an Apollo spacecraft circling the Earth every 90 minutes while 100 miles high has to have a velocity of roughly 18,000 miles per hour to stay in orbit, so something had to give the Moon the precisely required velocity for its weight and altitude … The point—and it is one seldom noted in considering the origin of the Moon — is that it is extremely unlikely that any object would just stumble into the right combination of factors required to stay in orbit. ‘Something’ had to put the Moon at its altitude, on its course and at its speed. The question is: what was that ‘something’?”

If the precise and stationary orbit of the Moon is seen as sheer coincidence, is it also coincidence that the Moon is at just the right distance from the Earth to completely cover the Sun during an eclipse? While the diameter of the Moon is a mere 2,160 miles against the Sun’s gigantic 864,000 miles, it is nevertheless in just the proper position to block out all but the Sun’s flaming corona when it moves between the Sun and the Earth. Asimov explained: “There is no astronomical reason why the Moon and the Sun should fit so well. It is the sheerest of coincidences, and only the Earth among all the planets is blessed in this fashion.”

Re: Eclipse proportions refute RET
« Reply #52 on: January 25, 2018, 01:55:59 AM »
However, some of the brightest minds have emitted not only wonder but also doubt towards these numbers.

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

Do you have a reliable source linking Dr. El-Baz to that quote?

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #54 on: January 25, 2018, 02:13:55 AM »
In the first paragraph k denotes the DIRECT DISTANCE TO THE SUN, not the diagonal.
What? Didn't get enough of an ass beating in the last thread so now you come here to get some more.

He is making an isosceles triangle.
There is no line representing the direct distance to the sun there, just the diagonal. i.e. the distance to the sun along the diagonal.

But thanks to this I have noticed another massive problem with his nonsense, he has the moon triangle be bigger than the one for the sun, which makes no sense at all.
Of course, this was just a typo on his part, he just meant it the other way around.

Let's do the calculations for both versions.
How about before doing them you see if they make any sense.

So comments like the following, simply have no place here at all.
No they do, and it applies to you as well.
You seem to not understand at all and only seem capable of just plugging in numbers rather than thinking if the math would apply at all.
Perhaps that explain your extreme ignorance and stupidity regarding the Sagnac effect?

That is just the hypothesis pur forward by modern science.
No, it is what all the evidence points to. It isn't a hypothesis.

However, some of the brightest minds have emitted not only wonder but also doubt towards these numbers.
Nope.
Regardless, this has nothing at all to do with showing that the OP is pure BS.

The Moon has astonishing synchronicity with the Sun
And there you go running off on a tangent. If you want to start a discussion on that topic, do so elsewhere.

Deal with the complete failure of the OP or get lost.

Re: Eclipse proportions refute RET
« Reply #55 on: January 25, 2018, 02:16:35 AM »
https://www.quora.com/Were-there-any-scientists-that-doubt-the-moon-to-be-natural
So that's a no.
You have random guy, but they don't even back you up.
If you notice, the quote is preceeded by who said it.
Your quote wasn't said by NASA, it was said by Christopher Knight and Alan Bulter

Re: Eclipse proportions refute RET
« Reply #56 on: January 25, 2018, 02:17:07 AM »
https://www.quora.com/Were-there-any-scientists-that-doubt-the-moon-to-be-natural

Okay, so from that source, your quote is from the book Who Built the Moon by Christopher Knight and Alan Bulter.

Your link cites Farouk El-Baz with the following quote:
"If water vapour is coming from the Moon’s interior is this serious. It means that there is a drastic distinction between the different phases of the lunar interior – that the interior is quite different from what we have seen on the surface."

I trust you won't associate Dr. El-Baz with your quote again.

Re: Eclipse proportions refute RET
« Reply #57 on: January 25, 2018, 02:20:53 AM »
I disproved the notion that the moon we observe during annular eclipses can be as far away from the Sun during total eclipses as RET predicts. If you disagree, I am still waiting for you to provide a problem with the proof.

I just did. Check the numbers.
No you didn't, you threw out a bunch of random numbers that you want to insist work. I have already shown that, even if they do, they are not the numbers we observe in the real world as the proportion is entirely wrong.

I suggest you at least try to actually read my post.

They aren't random. They are the actual numbers. Check them. See if they work. See if they produce an annular eclipse. If they don't, then we'll go from there.

They cannot produce both an annular and total eclipse, because I have already proven the proportions necessary do not work for RET. If you disagree I am waiting for you to stop being such a fucking coward and address the proof.


Quote
The point to consider is to do a liitle research to find out  why eclipses appear as they do.
It has to do with the sizes of the earth, the moon and the sun and the distances between them.
I know, which is why I used that knowledge in constructing the post, and demonstrating that the distances between them are not in line with RET.
Do you seriously believe you've proven something about the proportions of the sun and moon that nobody in human history has figured out...something every physicist, cosmologist, and astronomer got wrong?  Not only did they get it wrong but they got it wrong exactly the same way.

Logically it just doesn't hold water.  However, if you think you've discovered something unique then by all means, submit it for peer review.

Mike
Since it costs 1.82˘ to produce a penny, putting in your 2˘ if really worth 3.64˘.

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sandokhan

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Re: Eclipse proportions refute RET
« Reply #58 on: January 25, 2018, 02:25:09 AM »
But thanks to this I have noticed another massive problem with his nonsense, he has the moon triangle be bigger than the one for the sun, which makes no sense at all.

But he does not. He stated that c might be negative in value.

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.

Both sets of calculations lead to the same conclusion.


So, the following, then, applies only to you, doesn't it?

Your entire OP seems to be misunderstanding and evasion.

There is so much error it isn't funny. As some would say "it isn't even wrong."

You mean except discarding basic math and logic.

You mean that's you running away because you have been refuted yet again.
You are the one displaying ignorance and a total lack of understand. You lose yet again.



The most extraordinary fact is that if THE SUN AND THE MOON HAVE THE SAME DIAMETER, then p = 1 and k is the same for both the Sun and the Moon.


More quotes relating to the solar eclipse.

To be precise, the Moon is 400 times smaller than the star at the centre of our solar system, yet it is also just 1/400th of the distance between the Earth and the Sun.

Whilst the surprisingly neat number of 400 for relative size and distance is apparently an amusing coincidence of the decimal counting system, the odds against this optical illusion happening at all are huge. Experts are deeply puzzled by the phenomenon. Isaac Asimov, the respected scientist and science-fiction guru, described this perfect visual alignment as being ‘the most unlikely coincidence imaginable’.

C. Knight and A. Butler (Who Built the Moon)




Your quote wasn't said by NASA, it was said by Christopher Knight and Alan Bulter

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.
« Last Edit: January 25, 2018, 02:26:54 AM by sandokhan »

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Ising

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Re: Eclipse proportions refute RET
« Reply #59 on: January 25, 2018, 03:45:19 AM »

However, some of the brightest minds have emitted not only wonder but also doubt towards these numbers.

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


More to the point : so what ? Why should this mean the Moon has been crafted by man ? People used to wonder at the fact the Moon only ever present one face to the Earth, and to conclude that it couldn't possibly be a coincidence, only to later find out that it indeed wasn't, and that it was due to the mutual interaction between Earth and Moon. Likewise, maybe someone will find an explanation to the alignements to which you refer. Until then, no conclusion can be drawn.