[QED]

What data do you imagine might follow from a hypothetical experiment? You do understand what these words mean, yes?

Yes. You want us to explain the results of an experiment that happened in your head.

If you have nothing to actually contribute... we know where this thread is going to end up.

No, Thomas. A theory that describes reality means that it is a theory that can predict what will happen...in reality. I used to live in San Francisco, where I remember seeing a crescent moon one night. What does your theory predict would been seen in the other following locations that I specified? This IS the forum for FET, yes? Where we investigate FET?

This is what scientists do, Thomas. They continuously apply their model to different scenarios to test it.

The more you hide from actually attempting to test your model, the less anyone will ever believe it to be true.

Scientists who believe in their models do not hide like this. It is evident that you do not believe in the FET.

What reality? Test against what? You have not posted any data for what will be seen.

Does FET need data before it can provide an answer to a simple question? I'll ask it again:

I used to live in San Francisco, where I remember seeing a crescent moon one night. What does your theory predict would been seen in the other following locations that I specified?

[Tom Bishop]

Does FET need data before it can provide an answer to a simple question?

The fact is that the orientation that will be seen will be inexplicable in the Round Earth Theory. I have been waiting for you to post some examples to point it out.

[Tumeni]

At 6pm on the 12th, the Moon as viewed from the Greenwich Observatory in London will look like the graphic in this page

https://www.mooncalc.org/#/51.4769,-0.0005,17/2018.08.12/17:58/1/0

It will be in roughly the West-South-Western part of the sky, and will be approx 30 degrees above the horizon.


--

From the Empire State Building in NYC, some four hours behind London, it will be approx 2pm, and the Moon will be in
the south, and look like this

https://www.mooncalc.org/#/40.7485,-73.9858,17/2018.08.12/13:59/1/0

--

From the Griffith Park Observatory in Los Angeles, the Moon will in the East, and will look like this;

https://www.mooncalc.org/#/34.1193,-118.3002,17/2018.08.12/10:52/1/0

--

From Perth, Australia, it will be approximately 2am on the 13th, and the Moon will not be visible at all, as it
will be on the 'wrong' side of the Earth for the Australians to see

https://www.mooncalc.org/#/-31.9505,115.8605,13/2018.08.13/01:53/1/0

---

Convince me or us that this is "inexplicable" ...

[Tom Bishop]

Convince me or us that this is "inexplicable" ...

Real photographs would be better. However, here is a demonstration that those pattern-based moon calculators do not reflect RET.

I took one of your times and went down the Greenwich Longitude (0 degrees E), at 45 degrees North, 0 degrees North, and 45 degrees South. Here are the results:

45 degrees North


Link: https://www.mooncalc.org/#/45,0,3/2018.08.12/17:58/1/2

0 Degrees North


Link: https://www.mooncalc.org/#/0,0,3/2018.08.12/17:58/1/2

45 Degrees South


Link: https://www.mooncalc.org/#/-45,0,3/2018.08.12/17:58/1/2

[Tumeni]

Convince me or us that this is "inexplicable" ...

Real photographs would be better.

Oh, boy. Today is the 6th. I'm predicting what will happen on the 12th. And you want "real photographs" ....

I don't have a time machine.

[Tumeni]

I took one of your times and went down the Greenwich Longitude (0 degrees E), at 45 degrees North, 0 degrees North, and 45 degrees South. Here are the results:

What have you 'demonstrated'?

[Tom Bishop]

I took one of your times and went down the Greenwich Longitude (0 degrees E), at 45 degrees North, 0 degrees North, and 45 degrees South. Here are the results:

What have you 'demonstrated'?

Here is an image which may provide some clarification:



Edit: The image hosting service seems to be working and then not working. Here are direct links:

https://i.imgur.com/FUPmlQf.png
https://preview.ibb.co/gGsd6K/moon_conclusion2.png

[Tom Bishop]

Here is an example, showing that from the North Pole to South Pole (90^o N to 90^o S), that the RET difference is about 2 degrees.

It is easy to see how perspective plays very little part in the Round Earth system.

In RET the distance to the moon is 238,900 miles.

Imagining that distance as a radius of a circle, with observers positioned all around the moon, we can get the circumference of that circle with C=2*pi*R.

The circumference is 1501052.96989 miles

We divide it by 360 to get 4169.59158301 miles per degree

The diameter of the earth is 7,917.5 miles

7,917.5 / 4169.59158301 = a ratio of 1.89, or a little less than 2 degrees.

If we place the earth on the circumference we created above, with two observers standing on direct opposite sides of the earth, about 7917.5 miles apart, looking at the moon, should see a difference in shift of less than 2 degrees.

With 45^o N to 45^o S, instead of 90^o N to 90^o S, we can cut that in half and see that the difference is 0.949 degrees, or about 1 degree.

[Tom Bishop]

Here is a slightly more congruent version of the mooncalc images, with the locations shifted Southward by 10 degrees, where the crescent of the moon is pointing "downwards" on the middle version:

35 degrees North : https://www.mooncalc.org/#/35,0,3/2018.08.12/17:58/1/2
10 degrees South : https://www.mooncalc.org/#/-10,0,3/2018.08.12/17:58/1/2
55 degrees South : https://www.mooncalc.org/#/-55,0,3/2018.08.12/17:58/1/2

Either the moon wasn't exactly over the equator, or the moon wasn't pointing exactly Eastwards. Either way, the orientation of the moon from these different spots more strongly supports the Flat Earth model over the Round Earth model.

[Tumeni]

Here is an image which may provide some clarification:

The differences between the tips of the crescents are 2 x 45 degrees, or 90 degrees in total.

The differences in observer latitude are two sets of 45 degrees, making 90 degrees in total.

What does this indicate to you?

[Tumeni]

Your graphics are faulty.

In the second one, all you've done is draw lines away from the surface. What's the point of view here? Are we behind the observers? To the side of them?

You've taken no account of the position of the Moon in the sky for each observer, and not shown this on your diagram.

 

[Tom Bishop]

Here is an image which may provide some clarification:

The differences between the tips of the crescents are 2 x 45 degrees, or 90 degrees in total.

The differences in observer latitude are two sets of 45 degrees, making 90 degrees in total.

What does this indicate to you?

The various orientations the moon appears at suggests to me that the earth is flat.

The degrees of Latitude were originally defined on the angles celestial bodies at the horizon come out of the horizon. At 0 degrees N the Sun comes straight out of the horizon when it is over the equator. At 45 degrees N and S the sun comes out of the horizon at a 45 degree angle N or S.

https://www.google.com/search?q=how+was+latitude+defined

See the following google definition:

    lat·i·tude
    ˈladəˌt(y)o͞od/
    noun
    noun: latitude; plural noun: latitudes

    the angular distance of a place north or south of the earth's equator, or of a celestial object north or south of the celestial equator, usually expressed in degrees and minutes.

The fact that the moon shifts by that much is not unexpected, since that the longitude lines are based on the changing angles of celestial bodies that come out of the horizon.

That the the shift adds up to 90? Unsurprising, considering how the Latitude was defined.

Your graphics are faulty.

In the second one, all you've done is draw lines away from the surface. What's the point of view here? Are we behind the observers? To the side of them?

You've taken no account of the position of the Moon in the sky for each observer, and not shown this on your diagram.

It's a side view. The 90 degree shift in orientation of the moon is more strongly explained by the Flat Earth model and not the Round Earth model.

[Tumeni]

I've refreshed my screen three times in the last few minutes, and you've edited and changed your post twice, including full delete of your first draft..

Pleas make up your mind about what you want to say before you hit Enter.

[Tumeni]

It's a side view.

Can you do the view along the observer's line of sight, as though you were out in space behind all three?

[Tumeni]

Here is an example, showing that from the North Pole to South Pole (90^o N to 90^o S), that the RET difference is about 2 degrees.

Angles are measured in degrees. Where do you see this angle of 2 degrees? It must be placed somewhere.

It is easy to see how perspective plays very little part in the Round Earth system.

In RET the distance to the moon is 238,900 miles.

Imagining that distance as a radius of a circle, with observers positioned all around the moon, we can get the circumference of that circle with C=2*pi*R.

Observers on the Moon will not be on that circle. You're calculating the circumference of the Moon's orbit with that formula. A number of observers on the Moon at any one time would, at max, be upon a circle of (2*3.14*1079 =) 6780 miles. 

The circumference is 1501052.96989 miles

That's the length of the Moon's orbit, not the circumference of the Moon


We divide it by 360 to get 4169.59158301 miles per degree

So the Moon moves 4170 miles for each degree of its orbit.

The diameter of the earth is 7,917.5 miles

7,917.5 / 4169.59158301 = a ratio of 1.89, or a little less than 2 degrees.

If we place the earth on the circumference we created above, with two observers standing on direct opposite sides of the earth, about 7917.5 miles apart, looking at the moon, should see a difference in shift of less than 2 degrees.

What 'shift' are you talking about? We're talking about how three individuals see the Moon at one moment in time (approximately). What do you think is 'shifting' in this scenario?

[QED]

Does FET need data before it can provide an answer to a simple question?

The fact is that the orientation that will be seen will be inexplicable in the Round Earth Theory. I have been waiting for you to post some examples to point it out.

What I am waiting for is for you to answer the simple question, which I have now explicitly asked twice.

This forum is entitled FET not RET.

[Tumeni]

The degrees of Latitude were originally defined on the angles celestial bodies at the horizon come out of the horizon. At 0 degrees N the Sun comes straight out of the horizon when it is over the equator. At 45 degrees N and S the sun comes out of the horizon at a 45 degree angle N or S.

Degrees of latitude only have meaning as angles. Where would you draw the angle, if not at the centre of a circle or globe?

The fact that the moon shifts by that much is not unexpected, since that the longitude lines are based on the changing angles of celestial bodies that come out of the horizon.

... and these LATITUDE lines correspond exactly with the difference in angle each observer will see in the angle of the Moon. Diagram to follow.

[Tom Bishop]

Here is an example, showing that from the North Pole to South Pole (90^o N to 90^o S), that the RET difference is about 2 degrees.

Angles are measured in degrees. Where do you see this angle of 2 degrees? It must be placed somewhere.

It is easy to see how perspective plays very little part in the Round Earth system.

In RET the distance to the moon is 238,900 miles.

Imagining that distance as a radius of a circle, with observers positioned all around the moon, we can get the circumference of that circle with C=2*pi*R.

Observers on the Moon will not be on that circle. You're calculating the circumference of the Moon's orbit with that formula. A number of observers on the Moon at any one time would, at max, be upon a circle of (2*3.14*1079 =) 6780 miles. 

The circumference is 1501052.96989 miles

That's the length of the Moon's orbit, not the circumference of the Moon


We divide it by 360 to get 4169.59158301 miles per degree

So the Moon moves 4170 miles for each degree of its orbit.

The diameter of the earth is 7,917.5 miles

7,917.5 / 4169.59158301 = a ratio of 1.89, or a little less than 2 degrees.

If we place the earth on the circumference we created above, with two observers standing on direct opposite sides of the earth, about 7917.5 miles apart, looking at the moon, should see a difference in shift of less than 2 degrees.

Moon circumference? I'm talking about the distance around the moon. Here is an illustration:

[Tumeni]

The side view looks like this, surely;



If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.



Tom, do you see how the vertical for each observer matches the dotted red lines you put on the mooncalc screen prints?

Do you see how the yellow lines connecting observer position on the screen prints with the Moon all correspond to the three green lines in the side view?

[Tumeni]

Moon circumference? I'm talking about the distance around the moon. Here is an illustration:
IMG

You're taking radius R as the distance TO the Moon. If you calculate a circle based on that, you're calculating the Moon's orbit length. The circumference of a circle with radius of 240k miles approx.

If you want the "distance around the Moon", or its circumference, you need to take the Moon's radius.