[Hmmm]

Tom Bishop, multiple relaying moons on earth is possible too!

[douglips]

Here you go. There should be a significant crescent.

Just quick and dirty.


Left is the image from the website showing what the moon should look like, Right is a crop from :32 sec into the video, I adjusted contrast and brightness to make it easier to see.  Looks pretty much spot on.

Yes, we all agreed that the moon appears to be in 95% illumination. The problem is that it should not be in so much illumination.

Your calendar is based on what the moon does do, not what it should do.

Your diagram is actually quite good. Looking at it, I think the moon would appear as it does in the video. Remember, the observer is at the part of the horizon line that is touching the earth. If you draw a line from the observer to the bottom of the terminator on the moon, you'd see the fraction of the angle at the fattest part of the moon that would be black.

What do you think the moon should look like to an observer at that place on the earth in your diagram?

[Hmmm]

Tom Bishop, if the moon is actually artificial, just like the sun, and it is illuminating by itself, then using moon phases won't prove FE or oblate spheroid earth so easily!

[3DGeek]

Tom - the problem with your diagram is that you're still thinking like a flat earther.  In RET, the horizon isn't a horizontal line like you've drawn it.  It depends (critically) on the height of the eyepoint above the ground.  I know you claim that the "horizon rises to the eye" - but that's not what RET says.

So if you choose a more reasonable eye height - then the horizon is depressed and your "horizon line" needs to be two lines that tip downwards.

You can't debunk RET using FE principles!

[douglips]

To be fair, at sea level the horizon line is about as close to flat as possible, it's only like 1/25 of a degree, which would not differ from Tom's diagram in any significant way.

If the observer were at the top of a 10000 foot mountain (e.g. Mauna Kea on a clear day) then RET says you get an extra 1.77 degrees in each direction.

In any case, it is likely that the angles given by the Navy on that website are with respect to the astronomical horizon, which is a flat line.

So, listen to these words because I won't say them very often: Tom Bishop is right.

[douglips]

Here's a diagram showing that the area shaded is as expected.

https://imgur.com/a/6AYee

I don't know if a direct image link works:

[mtnman]

I would like to hear a FE explanation of how a completely round/full moon would ever be possible on FE. If sun & moon are both 3000 miles over flat Earth and we are looking up at both, this couldn't happen.

To see a completely round/full moon, we have to be looking at the moon with the sun's illumination behind us. I think I posted this question on another thread, without any answers. So I thought I would re-post it here.

[model 29]

Tom Bishop is right.

About what?  The moon being 11 degrees above the horizon being impossible with the RE moon's 5 degree inclination?

[Tom Bishop]

Here's a diagram showing that the area shaded is as expected.

https://imgur.com/a/6AYee

I don't know if a direct image link works:

The illustration in the middle is not 95% luminosity.

[Rama Set]

Fascinating. Feel free to tell us more.

[douglips]

a) What do you think 95% luminosity looks like?
b) What percentage luminosity does the image in the middle look like to you?

Remember, the observer is from the left on the image in the middle. I've arranged the moons this way so you can compare the bottom of the largest width of the lit portion of the moon in all three images.

I don't know how to easily take a ball model and rotate it until it matches, but the fact that these all line up, and the middle image came from your (very excellent) diagram says a lot to me.

[markjo]

Tom Bishop is right.

About what?  The moon being 11 degrees above the horizon being impossible with the RE moon's 5 degree inclination?

I hate to tell you all this, but it isn't the 5 degree inclination that you should be looking at.  At the time and date given, the moon was a full 2 days before full.  That means that the moon was around 26 degrees away from being directly opposite the sun.

[douglips]

Tom Bishop is right.

About what?  The moon being 11 degrees above the horizon being impossible with the RE moon's 5 degree inclination?

Sorry I wasn't clear about this. He was right about the horizon being essentially flat:

This was 3DGeek:

Tom - the problem with your diagram is that you're still thinking like a flat earther.  In RET, the horizon isn't a horizontal line like you've drawn it.  It depends (critically) on the height of the eyepoint above the ground.  I know you claim that the "horizon rises to the eye" - but that's not what RET says.

My response was that Tom Bishop was right to draw the horizon flat, because if you are standing at the seashore, the horizon is within 1/20 of a degree of flat, so Tom Bishop's diagram is quite accurate.

I think the only confusion is that he is looking at his edge-on diagram of how much moon is illuminated, and states without evidence that it is not 95% illuminated. It isn't clear to me if he thinks it is more than 95% or less than 95% illuminated.

But, he was dead right about the horizon being pretty much perfectly flat, for an observer near sea level.

[Cocknbull]

Does anyone here understand how spheres can look like discs when seen head on? Is this site for real or is it a clever gimmick to sell advert click throughs? I have never seen so much snatch trying to minge up a simple piece of well renowned and easy to understand logic.

[juner]

Is this site for real or is it a clever gimmick to sell advert click throughs?

You don't know how internet advertising works, do you?

(hint: you would have to actually have ads).

[TomInAustin]

Does anyone here understand how spheres can look like discs when seen head on? Is this site for real or is it a clever gimmick to sell advert click throughs? I have never seen so much snatch trying to minge up a simple piece of well renowned and easy to understand logic.

This is the last place on earth I would expect to find any snatch.  Except maybe at a Sys Admin party.

[StinkyOne]

Does anyone here understand how spheres can look like discs when seen head on? Is this site for real or is it a clever gimmick to sell advert click throughs? I have never seen so much snatch trying to minge up a simple piece of well renowned and easy to understand logic.

This is the last place on earth I would expect to find any snatch.  Except maybe at a Sys Admin party.

Am sys admin, can confirm.   

[Nosyfox]

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...)

[J-Man]

Tom's ability to take on all comers is quite impressive with this data/video.

One can see clearly that both the sun and moon are relatively close, proving once again the FE and trust in your senses.

I've yet to find any information on this site that would lead me to RE belief.

Tom should be awarded slayer status. A whole new crew of RE folks should fire up because the regulars have been waylaid and massacred here.

[Nosyfox]

Tom's ability to take on all comers is quite impressive with this data/video.
One can see clearly that both the sun and moon are relatively close, proving once again the FE and trust in your senses.
I've yet to find any information on this site that would lead me to RE belief.
Tom should be awarded slayer status. A whole new crew of RE folks should fire up because the regulars have been waylaid and massacred here.

Is it an answer to or a comment on my post? If yes, it's a very unexpected and surprising one.
I demonstrate marthematically and without any possiblilty of rebuttal that Tom B. was wrong all along this thread, and as an answer (or comment) you sing his praises...
Well it's hard for me to grasp your logic. I see only one solution : your post has to be taken ironically. But in that case it would be very unfair to Tom! ...

One can see clearly that both the sun and moon are relatively close, proving once again the FE and trust in your senses.

I do not see in this video any clue whatsoever on how far or close the sun and moon are. If you did, please feel free to explain : I always enjoy learning new and cleaver things.

And my open question to FE on the compatibility of this video with FET is still valid!