[9 out of 10 doctors agree]

https://wiki.tfes.org/The_Full_Moon_is_Impossible_in_Round_Earth_Theory

The logic in that page is flawed. It asserts that anytime the Moon is in a full Moon position, the light coming to it will be blocked by the Earth. That would be the circumstances for a lunar eclipse, and any astronomer on the Internet will happily explain to you why that doesn't happen every month.

Since I am also an astronomer on the Internet, I will explain it too. Since the article I'm battling says that I can't refute it no matter how much mental gymnastics I do with scale, I'll spite it by doing mental gymnastics with scale. I'll be using this table from NASA as a reference.

The Moon's orbit is angled about 5° from Earth's orbit around the Sun. At its closest approach, the Moon is about 28 Earth-widths from Earth. At tan(5°)*28 I get 2 1/2 Earth-widths from the center maximum. The eclipse zone, for comparison, is one Earth-width wide.

Furthermore, the components of the Moon's path are sinusoidal, so the centerline where the eclipse zone is happens to be the line that the path crosses the fastest. Even though the eclipse zone is 20% of the possible heights, it spends only 13% of its time in the right latitude.

Please ask an astronomer about this before jumping to conclusions.

[Tom Bishop]

That page needs some work. I don't like how it is presented. I've rephrased the argument here:

According to Round Earth Theory the sun is illuminating the earth and moon as in this top-down view:



However, it has been observed that during the day, on the "day" side of the earth, the moon can be seen with phases which show significant lit area.

It should be impossible for the moon's phase to display significant fullness during the day. But the moon is seen on many occasions to do just that. Here is an example which should be impossible under the Round Earth model:



In the video the author pans across to the sun, showing that it is significantly above the horizon. This person is clearly somewhere on the day side of the earth. How is this observation possible?

Further on in that thread we did the math on the above video in question. We looked at the degrees of the sun and moon in the sky and they don't match up to the 95% of moon visible.

[inquisitive]

That page needs some work. I don't like how it is presented. I've rephrased the argument here:

According to Round Earth Theory the sun is illuminating the earth and moon as in this top-down view:



However, it has been observed that during the day, on the "day" side of the earth, the moon can be seen with phases which show significant lit area.

It should be impossible for the moon's phase to display significant fullness during the day. But the moon is seen on many occasions to do just that. Here is an example which should be impossible under the Round Earth model:



In the video the author pans across to the sun, showing that it is significantly above the horizon. This person is clearly somewhere on the day side of the earth. How is this observation possible?

Further on in that thread we did the math on the above video in question. We looked at the degrees of the sun and moon in the sky and they don't match up to the 95% of moon visible.

It's all explained online in many places, please comment on those.

Your use of the word 'we' is tedious and clearly incorrect.

[xenotolerance]

Everyone reading through, please take a few minutes to review the thread that Tom linked. It's stunning to me that he would bring that specific thread back to the forefront, as his argument was pretty finely ground up and destroyed.

The last pertinent post was this:

The illustration in the middle is not 95% luminosity.

            *** I apologize for my rather approximate English. I'll try my best! ***

 Why is it not 95% luminosity? the best way to know is to calculate! And it's so easy I wonder why you did not do it yourself already.

If alpha is the angle between the directions Moon-Observer and Moon-Sun, the illumination ratio of the moon is given by :  r = (1+cos(alpha))/2  (this is an approximate formula, assuming the light rays coming from the moon are parallel, but the error is very small, and in any case in the right direction...)
Taking into account the data from the Naval Observatory you quoted on your post of Aug. 10, 01:16:53 AM (reply#10), we find that alpha = 26°. This gives an illumination ratio of... 0.9494.  Hence the 0.95 given by the National Obervatory.

So I think everything is ok :  contrary to your feeling, the moon phase in the video at the begining of the thread is perfectly compatible with Round Earth Theory.

Now the next question from me would be : is it compatible with Flat Earth Theory?...
I would very much appreciate if you - or any other authoritative flatearther - post a similar demo for the case of FET, of course with the same parameters (sun and moon altitudes and azimuts, same place and date,...). Is it possible? I don't know enough FE theory to be able to do it myself (in fact I know nearly nothing concerning FET...)

[Tom Bishop]

As I said, the middle illustration is not 95% luminosity. We see everything to the left line of the green line in that side view image. If it were totally white it would be a full moon. The blackness displayed to the left of the green line isn't 5% darkness. A rough illustration:



The math earlier on in that thread demolishes the Round Earth Theory. We should not be able to see a 95% full moon during the day at the angles we discussed.

[9 out of 10 doctors agree]

The math earlier on in that thread demolishes the Round Earth Theory. We should not be able to see a 95% full moon during the day at the angles we discussed.

Your calculations determined that we should not be able to see a 100% full moon during the day. If the Moon was 100% full in the video, that would be different. But it's not; it's 95% full, meaning that it's 18.2° away from the full Moon's position.

Edit: I made an error in those calculations in that I didn't convert the phase range of [0,1] into the cosine range of [-1, 1]. The actual phase angle here would be ±25.8°.

[Tom Bishop]

The math earlier on in that thread demolishes the Round Earth Theory. We should not be able to see a 95% full moon during the day at the angles we discussed.

Your calculations determined that we should not be able to see a 100% full moon during the day. If the Moon was 100% full in the video, that would be different. But it's not; it's 95% full, meaning that it's 18.2° away from the full Moon's position.

No. Look at the numbers in that thread. That 95% illuminated moon should not have been seen.

[xenotolerance]

the angles you've labeled 50% dark and 25% dark do not have the relationship to luminosity you think they do. luminosity is the word for what percentage of the moon's area is seen as lit, so in math terms, 'dark' is the Y coordinate of the end of the line between light and dark expressed as a percentage. i.e. 100 * (y + 1)/2, or (y+1) * 50



A 45 degree angle from the center of the moon, aka your claim of 25% dark, would actually appear to be 15% dark. (-rt2/2 + 1) * 50 = 14.6447%

The angle required for 5% dark / 95% luminous is 64 degrees down from the center of the moon.


(after double checking my drawing, this angle is more like 64 degrees than 60.
the 6.7% for 60° numbers are correct, but what's actually seen is closer to 5% and 64° instead)

Tom can't do math QED

Also

saying 'I win' isn't the same as winning, right? so when I write, 'his argument was pretty finely ground up,' I know that doesn't prove he lost. but I find it worth pointing out that Tom felt the need to respond in kind but with STRONGER LANGUAGE

DEMOLISHED

go ahead, look at the numbers in that thread. tom doesn't know what he's talking about

one hopes some of the smarter flat Earth believers here can do trig. CS PhD Dr. Pseudonym can barely help himself from jumping to nitpick and correct people - pete where you at with helping Tommy B do math

[model 29]

I'm not seeing the problem with seeing a mostly full moon during the day.  I added another moon position just a tad further.

[Tom Bishop]

the angles you've labeled 50% dark and 25% dark do not have the relationship to luminosity you think they do. luminosity is the word for what percentage of the moon's area is seen as lit, so in math terms, 'dark' is the Y coordinate of the end of the line between light and dark expressed as a percentage. i.e. 100 * (y + 1)/2, or (y+1) * 50



A 45 degree angle from the center of the moon, aka your claim of 25% dark, would actually appear to be 15% dark. (-rt2/2 + 1) * 50 = 14.6447%

The angle required for 5% dark / 95% luminous is 64 degrees down from the center of the moon.



Tom can't do math QED

Also

saying 'I win' isn't the same as winning, right? so when I write, 'his argument was pretty finely ground up,' I know that doesn't prove he lost. but I find it worth pointing out that Tom felt the need to respond in kind but with STRONGER LANGUAGE

DEMOLISHED

go ahead, look at the numbers in that thread. tom doesn't know what he's talking about

one hopes some of the smarter flat Earth believers here can do trig. CS PhD Dr. Pseudonym can barely help himself from jumping to nitpick and correct people - pete where you at with helping Tommy B do math

Xeno, what are you talking about? I divided up the left side of the moon by 4 equal pieces.

I'm not seeing the problem with seeing a mostly full moon during the day.  I added another moon position just a tad further.

Check how high in the sky the 95% illuminated moon is from that thread. The numbers don't add up.

[9 out of 10 doctors agree]

The math earlier on in that thread demolishes the Round Earth Theory. We should not be able to see a 95% full moon during the day at the angles we discussed.

Your calculations determined that we should not be able to see a 100% full moon during the day. If the Moon was 100% full in the video, that would be different. But it's not; it's 95% full, meaning that it's 18.2° away from the full Moon's position.

No. Look at the numbers in that thread. That 95% illuminated moon should not have been seen.

Yes, it would have. According to your own numbers, it's about 25° or so off from opposite the sun. A phase angle of 25.8° plus 5° from inclination is enough for that, no?

(Also, could someone just post the link in this thread? It's somewhat difficult to find.)

[xenotolerance]

How is that 15% of the left half of the moon?

it isn't, and how much of the left side of the moon's area is lit is not the point. at all. luminosity is how much of the left edge is lit

if you've been confusing area with circumference the whole time, that explains a lot

***

I double-checked my image, and the way I tried to draw a 60­° angle actually resulted in roughly a 64° angle after all. It's a 1:2:rt5 triangle.

so the image I put up compares 15% and 5% darkness, instead of 15 and 6.7

[model 29]

Check how high in the sky the 95% illuminated moon is from that thread. The numbers don't add up.

Are you referring to the moon being 11 degrees above the horizon while its orbit only has an inclination of 5 degrees?

[xenotolerance]

made another set to clarify the side view w.r.t. area vs circumference. the circumference of the left side in the side view maps to the center of the front view





Tom's position has officially been

GENTLY CARESSED

[Tom Bishop]

made another set to clarify the side view w.r.t. area vs circumference. the circumference of the left side in the side view maps to the center of the front view

https://i.imgur.com/hBUcD0N.png

https://i.imgur.com/kOVQId8.png

Tom's position has officially been

GENTLY CARESSED

I see what you are talking about now. Thank you for clarifying.

I don't think the middle illustration is 64 degrees, either, though. I will get back to you.

[xenotolerance]

yeah it looks a bit different



bear in mind of course that copying and pasting in MS Paint and drawing lines is not an accurate process. if you come back and are like 'it's actually a 57.2 degree angle see, DEMOLISHED' that will be boring

you can discuss OP's actual argument any old time by the way