Is the Earth flat, or round, or both, but not spherical?

[mike tesko]

You can tell a great deal from studying the base line of the moon, and an imaginary base line from the point or tips of the crescent moon! In theory, if you were to create an imaginary base line, from tip or the point, to tip or point of a crescent moon, and draw  another imaginary line at a 90° angle from the middle of the imaginary baseline toward the illuminated portion of the moon itself and back upon itself so that it extends away from the moon this should always point or give the exact location of the sun in the sky - but sometimes it doesn't. This is because when the sun is particularly low down in the sky it's rays become warped and distorted! The higher up in the sky less distortion, etc, etc..

[Reader]

. . . the moon's baseline (or the lunar terminator) is totally and completely out of sync' with the positions of the risen sun and the exalted moon as captured by me in photographs!

That's nonsense. The difference in compass direction of the sun and moon when you took the photograph on 7th February was about 94 degrees, which is consistent with your photographed angle of the lunar terminator.

[mike tesko]

That's nonsense. The difference in compass direction of the sun and moon when you took the photograph on 7th February was about 94 degrees, which is consistent with your photographed angle of the lunar terminator. no, it's not nonsense, the moons baseline is out of sync'. The angle of projected sunlight absorbed by the moon is all wrong! That 94° of difference in compass angle between the sun and the moon on the day I took the photographs is reproduced at the wrong angle, the baseline should run slightly lopsided in keeping with the actual position of the sun in relation to the moon! That baseline should be slightly inclined from top left to bottom right (at a 94° angle), but it's not. Instead the baseline is vertical and out of sync'...

[Reader]

The angle of projected sunlight absorbed by the moon is all wrong!

That's also nonsense. The terminator isn't precisely vertical in your photograph. The 94 degrees gives the approximate tilt of the terminator from horizontal, not from vertical. You can demonstrate this easily by simply viewing a tennis ball held just in front of you and lit by an artificial light that's at arm's length to your left.

[mike tesko]

That's also nonsense. The terminator isn't precisely vertical in your photograph. The 94 degrees gives the approximate tilt of the terminator from horizontal, not from vertical. You can demonstrate this easily by simply viewing a tennis ball held just in front of you and lit by an artificial light that's at arm's length to your left. the angle is all wrong, the sun was too low and the moon so high you can't get that shape in those / these circumstances...

[Reader]

the angle is all wrong, the sun was too low and the moon so high you can't get that shape in those / these circumstances

My 3-D visualization was poor, making my previous explanation faulty. However, the following corrected explanation reaches the same conclusion. As the moon's direction is at about 94 degrees from the sun's direction, the lunar terminator is close to halfway across the moon and is only slightly curved. Thus the shape of the terminator in your photograph is what would be expected (a day later, it would have been slightly curved in the opposite direction). In this situation, it's the sun's elevation that is by far the most important factor that affects the tilt of the terminator, causing it to have a slight "clockwise" lean from vertical of about 1.5 degrees. Thus the moon should have appeared as shown below (except that the moon's right side would be too poorly illuminated to be visible and the sky wouldn't be black), which corresponds closely to what is seen in your photograph.

[Reader]

In a flat-earther's description of the sun as "circulating" above earth, along with a "flat" moon, the moon's phases and eclipses of the moon aren't explained in detail. Hence your need to invent bizarre speculation that the moon is an image projected from earth. Previously, you speculated that it was a projected image of the underside of the earth, so as to account for its cratered appearance. The descriptions you are currently giving are different from those given by any other specific flat-earther or flat-earth website.

[mike tesko]

My 3-D visualization was poor, making my previous explanation faulty. However, the following corrected explanation reaches the same conclusion. As the moon's direction is at about 94 degrees from the sun's direction, the lunar terminator is close to halfway across the moon and is only slightly curved. I'm challenging the angle of 94 degrees, since in fairness I took a series of photographs at around the same time...Thus the shape of the terminator in your photograph is what would be expected (a day later, it would have been slightly curved in the opposite direction). In this situation, it's the sun's elevation that is by far the most important factor that affects the tilt of the terminator, Yes, and the sun was really low down in the sky at the time I took the photographs, there was definitely a greater difference than 94 degrees between the position of the sun and the moon which was elevated high up in the sky!causing it to have a slight "clockwise" lean from vertical of about 1.5 degrees. The lean in the shape of the moon should have been anti-clockwise, not clockwise because of the lower position of the sun..Thus the moon should have appeared as shown below (except that the moon's right side would be too poorly illuminated to be visible and the sky wouldn't be black), which corresponds closely to what is seen in your photograph. I disagree, the moons base (terminator) should have been leaning anti-clockwise, showing the illuminated portion of the moon as illuminated by the sun so low down in the sky!

In your diagram, the sun would need to be elevated higher up in the sky (than it was)  to produce that angle of the moons baseline (terminator) as shown in your diagram, my photograph, and my own version of your diagram...

Moons baseline (terminator) Dividing line between yellow and grey portions...

Projected position of the sun (shown by red line), yet sun much lower down in sky...

[mike tesko]

The actual position of the sun in relation to the moon is shown by the green coloured line in the diagram:-

[mike tesko]

The illuminated part of the moons surface in my photograph should have been with the moons baseline (terminator) from 'A' to 'B', in the following diagram...

The sun would be in the general location of the sky by reference to the 'green arrow'...

[Reader]

You're not allowing for distance. If you stand fairly close to a street lamp that is 40 feet above ground level, you may observe that the lamp has an elevation of, say, 80 degrees. If you look at another lamp 100 yards away that is, say, 45 feet above ground level, you may observe its elevation to be only 7 degrees. From your observation position, the more distant lamp, although higher, has a lower elevation angle. If, however, you raised your eye level to 40 feet, the same height as the nearby lamp, the distant lamp would have an elevation of 1 degree, confirming its greater height above ground level than the nearby lamp's. In that analogy, the moon corresponds to the nearby lamp, and the sun corresponds to the distant lamp. In the case of your photograph, the moon's elevation was about 19 degrees, and the sun's about 1.5 degrees, both as seen from your camera's position. The more distant sun can shine slightly "down" on the moon, so that your green arrow, showing the sun's direction, should point slightly "upwards" (overlapping your red line) rather than slightly "downwards".

[mike tesko]

I have got the angle more or less accurate between the sun which had just come up and the moon elevated high in the sky, as per my photograph. It doesn't matter how close I was to the sun or the moon, since at any distance from any observer the sun or the moon was, is, would create an angle relevant to that conjunction of (a) the location of the observer (b) the position of the sun, and (c) the position of the moon - the angle of illumination of say the moon would be relative and proportionate! In my example, the area of the exalted moon illuminated was totally wrong if the sun which was so low down in the sky supposedly was the source of the moon's illuminated surface. .

[mike tesko]

The 94 degrees gives the approximate tilt of the terminator from horizontal, not from vertical.

It doesn't work just like that or as simple as what your saying, the distance between the sun and the moon has a bearing, as well as the difference in altitude of one or other, or both, and the speed at which the sun and the moon move above or just inside the earth's dome ..

[Reader]

It doesn't matter how close I was to the sun or the moon, . . .

I explained in the post immediately before yours why distance does make a difference. The diagram below is not to scale, but suffices to show that the moon can be illuminated from (slightly) above by the sun, yet have a much greater elevation angle (shown in blue) than the sun from the point of view of an observer.

[Reader]

It doesn't work just like that . . .

I know, thanks. I've already corrected my explanation. The conclusion remained the same.