Re: Full Moon Impossible on Flat Earth?
« Reply #260 on: July 31, 2018, 08:33:33 AM »
Quote
m = <Dm*cos(ELm)*cos(AZm), Dm*cos(ELm)*sin(AZm), Dm*sin(ELm)>
s = <Ds*cos(ELs)*cos(AZs), Ds*cos(ELs)*sin(AZs), Ds*sin(ELs)>
where:
Dm is distance from earth to moon = 238,900 miles
Ds is distance from earth to sun = 92,960,000 miles
For my example, ELm=20.89, AZm=136.35, ELs=-0.24, AZs=294.62
m = <-160890, 154062, 85186>
s = <38726600, -84508300, -389400>
s cross m = <-7138932301000, -3236313581600, -7630242937800>
p = m cross (s cross m) = <-899841878721166000, -1835766873255628000, 1620528680300286000>
h = m cross z = <154062, 160890, 0>
p dot h = -433987971757638265212000
|p| = 2608805976556893954
|h| = 222757
(p dot h) / (|p|*|h|) = -0.74680042
taking the abs as per the convention in the paper to get the angle we want...
alpha = arccos(0.74680042) = 41.686 degrees!

The fraction of a degree we get in variance is due to different amounts of precision used in the calculations (round-off error).
If you object to the part of the math where we showed the distances are not necessary in this calculation, just forget we ever mentioned it. There's the math done with the actual distances. Can you accept it now?

Why doesn't the angle of the moon's phase change with that math when we place the sun one mile away from the earth and 92,960,000 miles away from the earth?

Surely the angle of the phase must change between those huge ranges.

At this point, it should be painfully clear to all readers that you are deliberately avoiding the suggestion that we move to the "hold up a ball" method instead of this math. You have repeatedly ignored the suggestion, and it should be very obvious to everyone that you are ignoring it. This strikes me as suspiciously deceitful. I trust that our readers see it this way as well.

Upon your insistence, back to the math. You ask, "Why doesn't the angle of the moon's phase change with that math when we place the sun one mile away from the earth and 92,960,000 miles away from the earth?"

Before I go any further, let's be clear. We are talking about the angle of the moon's tilt. We are not discussing the moon's phase.

Let's start with the 3D visualization.

At 41:40, I showed exactly what happens when we change those distances. In the left-hand view, we are looking at the triangle made by the Earth, Moon, and Sun. Right at 41:44, I change the distance to the sun from 100 to 1000 units. Notice that the angles of that triangle in the left-pane DO change. This is exactly what you would expect. (In your sample above, you are using a right-triangle calculator. This triangle is not likely to be a right-triangle. In my example it clearly is not... perhaps we can come back to that.) However, none of the angles of this triangle are the angle that we are looking for. None of these angles is the angle of the moon's tilt.

If we look in the right-hand pane, we'll see the angle we ARE looking for. The angle we are interested in is the blue line in the right-hand diagram. Remember what that blue line in the right-hand pane is? That's the line connecting the sun and the moon. Yes that IS one of the 3 sides of the triangle we were looking at in the left-hand pane. The difference here is that we're looking at that line from a different angle. In the right-hand view, we're looking up at the moon from our observer's location. Scroll back to 41:40 and watch as the distances are changed. Notice that the blue line in the right-hand pane gets skinnier, but its angle in this view does not change.

At 41:55 I increase the distance by another factor of 10 (for a total of 100 times further than it started). Again we see the triangle in the left-hand pane changes, but the angle of the blue line in the right-hand pane does not change.

So we've seen in the simulation that the angle of the moon's tilt doesn't change with those distances, but that answer may be unintuitive. That's understandable. But think about this. Back at 15:48, we took the intersection of 2 planes, and I talked about how that intersection would give us the apparent angle of the light hitting the moon. That may not be obvious, so let me walk through it slowly...

1) The light coming from the sun to the moon must lie within in the Observer-Moon-Sun plane. That seems fairly obvious. That light should be along the Moon-Sun edge of the triangle we used to make the plane.
2) The Observer is ALSO within the Observer-Moon-Sun plane. That also seems obvious. We used the observer as the 3rd point to create that plane.
3) What does a plane look like to a viewer inside the plane? It looks like an edge - a line.
4) So any part of that plane which lies in front of the observer has all collapsed into a single line. Any part of that plane is all the same line from our point of view.
5) If any part of that plane is all the same line, then the line from the sun to the moon looks the same as any other part of that plane when viewed edge-on like this.
6) If we were to move the sun-moon line to another spot within this same plane, that line would look exactly the same - the same as any part of the plane does.

And now with that clear, what happens to the observer-moon-sun plane when we change the distance to either the moon or the sun? If we keep the ANGLE between the observer and the moon steady, but change the distance to the moon. Also if we keep the ANGLE from the observer to the sun steady while changing that distance. Try it out on a table-top. Place 3 objects on a table-top and call the triangle they make the plane. Well the plane is the table-top of course. How does this plane change when you change the distance between your 3 objects? It's still the table-top of course. Changing the distances between the 3 objects does not change the plane that contains them.

7) So changing the distance from the earth to the sun does not affect the observer-moon-sun plane.
8) The plane viewed edge-on still looks exactly the same.
9) That plane viewed edge-on IS the apparent lighting direction from the moon to the sun when viewed by the observer.
10) And it is unaffected by the distances between the bodies - you cannot change the orientation of that plane without changing the angles between the observer and the other bodies.

Hopefully that helps. But we could approach this with pure math as well. Feel free to just stop reading at this point because math is pretty boring to most people, and I think I'm beating a dead horse by now...

Back to this equation:
cos(alpha) = (p dot h) / (|p| * |h|)
We got this far using the full vectors for m and s. We have included all the proper distances up to this point.
Now let's substitute in the values we have for m and s:
m = <Dm*cos(ELm)*cos(AZm), Dm*cos(ELm)*sin(AZm), Dm*sin(ELm)>
s = <Ds*cos(ELs)*cos(AZs), Ds*cos(ELs)*sin(AZs), Ds*sin(ELs)>
Let's pull out those distances as constant factors:
m = Dm*<cos(ELm)*cos(AZm), cos(ELm)*sin(AZm), sin(ELm)> = Dm*mhat
s = Ds*<cos(ELs)*cos(AZs), cos(ELs)*sin(AZs), sin(ELs)> = Ds*shat
Ok? Now let's do the rest of the derivation like that...
s cross m = Dm*Ds*(shat cross mhat)
p = m cross (s cross m) = Dm*mhat cross (Dm*Ds*(shat cross mhat)) = Dm*Dm*Ds*(mhat cross (shat cross mhat))
h = m cross z = Dm*(mhat cross z)    (z already has a unit length)
p dot h = (Dm*Dm*Ds*(mhat cross (shat cross mhat))) dot (Dm*(mhat cross z)) = Dm*Dm*Dm*Ds*( (mhat cross (shat cross mhat)) dot (mhat cross z) )
|p| = |Dm*Dm*Ds*(mhat cross (shat cross mhat))| = Dm*Dm*Ds*|mhat cross (shat cross mhat)|
|h| = |Dm*(mhat cross z)| = Dm*|mhat cross z|

Now let's divide:
(p dot h) / (|p|*|h|) = Dm*Dm*Dm*Ds*( (mhat cross (shat cross mhat)) dot (mhat cross z) ) / (Dm*Dm*Ds*|mhat cross (shat cross mhat)| * Dm*|mhat cross z|)
Quick rearrange:
= (Dm*Dm*Dm*Ds / Dm*Dm*Dm*Ds) * ( (mhat cross (shat cross mhat)) dot (mhat cross z) ) / ( |mhat cross (shat cross mhat)| * |mhat cross z| )
Let's cancel the constants
= ( (mhat cross (shat cross mhat)) dot (mhat cross z) ) / ( |mhat cross (shat cross mhat)| * |mhat cross z| )
And BOOM! proved mathematically.

So let's get real here... that's some fairly advanced math. I don't expect everyone to follow it. What I do expect is for you to VERIFY it EXPERIMENTALLY. Measure the angles to the sun and the moon (or look them up). Crunch these numbers using the final equation from the paper. Then go outside and measure the tilt angle of the moon. Does it match? How close is it? You can do that much with no math background at all.

And please please remember... The answers do not always match up with your first guess. Sometimes nature can be surprising. Embrace it. Science is much cooler when you discover the world is trickier than you thought it was.

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Offline Tumeni

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Re: Full Moon Impossible on Flat Earth?
« Reply #261 on: July 31, 2018, 08:51:59 AM »
Angle b is the top left angle in that image. It changes when we change the distance to the sun (the adjacent side).

But the distance to the sun does not change in reality.

The angle changes when the distance to the sun changes.

Yes, but the distance to the sun does not change in reality.

Quote
Go outside on a day when you can see Sun and Moon. Hold up a ball in front of the Moon, such that the ball is in sunlight. Note how the illuminated part of the Moon and the ball match, showing that they have the same source of illumination, and that that source is a considerable distance away (since the Moon is approx 240k miles away, it must, by definition, be significantly farther than this).

How would that prove how far away the sun is or provide support for a Round Earth?

It proves the Sun to be much farther away than the Moon. If it was not, then the illumination would not match. Of itself, I'm not sure it forms a single definitive proof of a globe. But you shouldn't consider these things in isolation. The combination of many different forms of proof should be considered together. For instance, we see the Moon pass between us and the Sun at solar eclipses; therefore, the Sun is farther away than the Moon. We see Venus and Mercury transit over the Sun. Therefore, the Sun is farther away than both of these bodies at these points in their orbits. 


The moon and sun are often only seen during the day when they are on on opposite skies,

Incorrect. My sample 'ball held up to the Moon' photos were taken with the Moon just West of South, and with the Sun rising in the East. Roughly 90 degrees between.

Holding up a ball between them

Nobody is suggesting that. I said, once again, "hold up a ball IN FRONT OF the Moon"

In terms of your triangle, with Sun, Moon and Earth at the three points, you're holding the ball up along the side connecting Moon and Earth.


 would create some kind of gibbous moon, to which you can angle, rise above or below you, rotate slightly around, to try and match with the moon. It is totally invalid and does nothing to provide insight on the matter.

You have to sight along the line to the ball and the Moon simultaneously, so that you're looking at them from the same angle. OF COURSE, if you look at the ball from somewhere else, it will look different to the Moon. You can make the Moon phase look different as well, IF you look at it from a different angle.

Let's call the line between the ball and Moon the datum, or zero degrees.  Continue that line toward your eye. You should be looking as closely along that line as you can. If you then leave the ball where it is, and move yourself and your eye 20 degrees to one side (with respect to the ball, i.e. the angle is formed AT the ball), the phase on the ball WILL look different, but if you then moved yourself 20 degrees with respect to the Moon, the phase on the Moon would do exactly the same. But you'd have to go out into space to do this, since you can't move on Earth more than a fraction of a degree off this datum line (with the angle formed AT the Moon).

Would you care to address my posts from page 12 now? Specifically, reply #229
« Last Edit: July 31, 2018, 08:54:24 AM by Tumeni »
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Not Flat. Happy to prove this, if you ask me.
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Offline Tumeni

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Re: Full Moon Impossible on Flat Earth?
« Reply #262 on: July 31, 2018, 11:04:43 AM »
Here's mine from two consecutive days. First one was mid-afternoon, second was mid-morning. I failed to note exact positions for the first, but for the second, the Moon was slightly West of South, Sun rising in the East, with about 90 degrees between them.

https://imgur.com/a/Ci10Oo (EDIT - This one seems to have been 404'd - will repost soon)

https://imgur.com/a/7DMpx3L
« Last Edit: July 31, 2018, 11:29:06 AM by Tumeni »
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Nearly?

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Offline Tom Bishop

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Re: Full Moon Impossible on Flat Earth?
« Reply #263 on: July 31, 2018, 11:35:06 AM »
I don't see what the "hold up the ball" method tells us except that the moon is close to the observer like the ball is, as to be able to point into unnatural perspective angles away from the sun.

There certainly is something odd and artificial about this math where the tilt of the moon's phase does not change at all with scenarios where the sun is located one mile from the earth or 92 million miles from the earth. It seems hard to justify that this scenario meets our reality.

The readers can decide for themselves if an arrow in space would not point at what it is pointing at
« Last Edit: July 31, 2018, 04:24:35 PM by Tom Bishop »
"The biggest problem in astronomy is that when we look at something in the sky, we don’t know how far away it is" — Pauline Barmby, Ph.D., Professor of Astronomy

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Offline Tumeni

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Re: Full Moon Impossible on Flat Earth?
« Reply #264 on: July 31, 2018, 11:42:58 AM »
I don't see what the "hold up the ball" method tells us except that the moon is close to the observer like the ball is, as to be able to point into unnatural perspective angles away from the sun.

Is this why you're persistently refusing to actually do it?

It tells us that the Moon and the Ball are illuminated from the same light source. It tells us that the distance between Moon and Ball are very small compared to distance to Sun. If the Sun were close, or between the two, the illumination would differ.

I don't know what you mean by "unnatural perspective angles away from the sun" - are you referring back to that Moon phase photo with the arrow upon it? I tried to get you to progress with an experimental method on that, but you ignored my post. 


The readers can decide for themselves if arrows in space would not point at what they are pointing at

Be specific - arrows pointing where?



Do you agree that if you hold a ball up in front of you, in sunlight, that the centre of the illuminated hemisphere of the ball is directly aligned with the Sun?  Y/N
« Last Edit: July 31, 2018, 11:45:08 AM by Tumeni »
=============================
Not Flat. Happy to prove this, if you ask me.
=============================

Nearly all flat earthers agree the earth is not a globe.

Nearly?

Re: Full Moon Impossible on Flat Earth?
« Reply #265 on: July 31, 2018, 04:20:41 PM »
I don't see what the "hold up the ball" method tells us except that the moon is close to the observer like the ball is, as to be able to point into unnatural perspective angles away from the sun.

There certainly is something odd and artificial about this math where the tilt of the moon's phase does not change at all with scenarios where the sun is located one mile from the earth or 92 million miles from the earth. It seems hard to justify that this scenario  meets our reality.

The readers can decide for themselves if arrows in space would not point at what they are pointing at
I understand that your intuition is telling you that the distance to the sun is a critical part of this issue. I would hope that you can recognize that intuition is often wrong and doesn't prove anything.

Furthermore, you MUST recognize that we are discussing how things work under RE, and in RE, the sun is 93 million miles away from us - It is not 1 mile away, nor is it unknown. We are testing the model as it stands, and the model say the sun averages around 93 million miles away. Nobody cares what happens when the sun is 1 mile away because we're not testing that. Nobody is asking us to determine the distance to the sun and moon here as those are already established under the RE model. We are testing them.

Under the existing RE model, the sun is 93 million miles away, and the moon is 240 thousand miles away. The sun is understood to be MUCH farther from us than the moon is. Because of this, the triangle between the observer, sun, and moon is very nearly a line. The edge of the triangle between the earth and the moon is a fraction of a percent of the length of the other edges. What this means is the angle of the light coming from the sun to the observer is very nearly the same as the light from the sun to the moon. To your eyes, you won't be able to tell the difference.

Because of this, the "hold up a ball" method (which you steadfastly refuse to even try) shows us almost exactly the same lighting as the moon. Hold up a ball and check it. If this works exactly like I've said it does, the lighting on the ball should match the lighting on the moon. If the lighting on the ball is very different, then you have proven that something in this RE model I have described is incorrect.

I'm telling you exactly what the RE model says. What do you suppose gives you the authority to claim that the RE model does not say what the RE model says? That's all we're talking about here. The RE model says the lighting on the ball should match the lighting of the moon very closely. Stop misrepresenting the RE model.

Offline iamcpc

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Re: Full Moon Impossible on Flat Earth?
« Reply #266 on: July 31, 2018, 05:42:11 PM »
Why doesn't the angle of the moon's phase change with that math when we place the sun one mile away from the earth and 92,960,000 miles away from the earth?

Because the angle is based on the relationship between the distances of the right triangle. As long as that relationship is to scale the distances don't matter.


Here are some examples:

A right triangle with two sides of one inch will ALWAYS have two 45° angles and one 90° angle.
A right triangle with two sides of one mile will ALWAYS have two 45° angles and one 90° angle.
A right triangle with two sides of one BILLION miles will ALWAYS have two 45° angles and one 90° angle.

That is why the angle does not change.

Re: Full Moon Impossible on Flat Earth?
« Reply #267 on: August 01, 2018, 04:03:50 PM »
I don't see what the "hold up the ball" method tells us except that the moon is close to the observer like the ball is, as to be able to point into unnatural perspective angles away from the sun.

There certainly is something odd and artificial about this math where the tilt of the moon's phase does not change at all with scenarios where the sun is located one mile from the earth or 92 million miles from the earth. It seems hard to justify that this scenario  meets our reality.

The readers can decide for themselves if arrows in space would not point at what they are pointing at
I understand that your intuition is telling you that the distance to the sun is a critical part of this issue. I would hope that you can recognize that intuition is often wrong and doesn't prove anything.

Furthermore, you MUST recognize that we are discussing how things work under RE, and in RE, the sun is 93 million miles away from us - It is not 1 mile away, nor is it unknown. We are testing the model as it stands, and the model say the sun averages around 93 million miles away. Nobody cares what happens when the sun is 1 mile away because we're not testing that. Nobody is asking us to determine the distance to the sun and moon here as those are already established under the RE model. We are testing them.

Under the existing RE model, the sun is 93 million miles away, and the moon is 240 thousand miles away. The sun is understood to be MUCH farther from us than the moon is. Because of this, the triangle between the observer, sun, and moon is very nearly a line. The edge of the triangle between the earth and the moon is a fraction of a percent of the length of the other edges. What this means is the angle of the light coming from the sun to the observer is very nearly the same as the light from the sun to the moon. To your eyes, you won't be able to tell the difference.

Because of this, the "hold up a ball" method (which you steadfastly refuse to even try) shows us almost exactly the same lighting as the moon. Hold up a ball and check it. If this works exactly like I've said it does, the lighting on the ball should match the lighting on the moon. If the lighting on the ball is very different, then you have proven that something in this RE model I have described is incorrect.

I'm telling you exactly what the RE model says. What do you suppose gives you the authority to claim that the RE model does not say what the RE model says? That's all we're talking about here. The RE model says the lighting on the ball should match the lighting of the moon very closely. Stop misrepresenting the RE model.

This work for you Tom? At this point, I figure your options are:
a) Accept that the Moon Terminator Illusion is exactly what you've been told it is - an illusion. Just admit your mistake and concede that the RE model works really well to explain this particular phenomenon.
b) Find an error in the math.
c) Hold up a ball and show that the model does NOT work as we say it does.
d) Stick your fingers in your ears and keep on repeating, "this is NOT TRUE" over and over until you convince yourself that you are right.

I don't mean to rush you, but I'd like to wrap this up before I leave. You can let the thread fade into oblivion after I'm gone.

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Offline Rama Set

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Re: Full Moon Impossible on Flat Earth?
« Reply #268 on: August 01, 2018, 05:52:36 PM »
Don’t go.
Th*rk is the worst person on this website.

Offline model 29

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Re: Full Moon Impossible on Flat Earth?
« Reply #269 on: August 02, 2018, 02:54:54 AM »
Tom B understands why the moon's phase looks the way it does,
(taken from the other site)
Quote from: Tom Bishop
Nick, in that explanation the video is using the analogy of standing in the middle of a very long room or hallway. When you look down one end of the hallway the corners are angled upwards and when you look down the other end it is angled upwards.


He just keeps making up imaginary or irrelevant factors in order to confuse himself.

Re: Full Moon Impossible on Flat Earth?
« Reply #270 on: August 02, 2018, 04:16:05 AM »
Tom B understands why the moon's phase looks the way it does,
(taken from the other site)
Quote from: Tom Bishop
Nick, in that explanation the video is using the analogy of standing in the middle of a very long room or hallway. When you look down one end of the hallway the corners are angled upwards and when you look down the other end it is angled upwards.


He just keeps making up imaginary or irrelevant factors in order to confuse himself.
One might start to suspect that he is deliberately distracting the conversation in an effort to win a debate at the cost of the truth.

Offline iamcpc

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Re: Full Moon Impossible on Flat Earth?
« Reply #271 on: August 03, 2018, 04:49:34 PM »
Tom B understands why the moon's phase looks the way it does,
(taken from the other site)
Quote from: Tom Bishop
Nick, in that explanation the video is using the analogy of standing in the middle of a very long room or hallway. When you look down one end of the hallway the corners are angled upwards and when you look down the other end it is angled upwards.


He just keeps making up imaginary or irrelevant factors in order to confuse himself.
One might start to suspect that he is deliberately distracting the conversation in an effort to win a debate at the cost of the truth.

I feel like the flat earth wiki should be updated.Instead of "Impossible" it should say "might be impossible"

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Offline Tom Bishop

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Re: Full Moon Impossible on Flat Earth?
« Reply #272 on: August 03, 2018, 05:44:07 PM »
According to that math from the paper, and as admitted by ICST, it doesn't matter how far away the sun is from the moon; the moon will point in the same direction nonetheless.

Consider the following diagram, which replaces the moon with a green arrow in space that points at the sun:



We can see clear away, that the distance between the observer and the sun does matter.

From: https://www.triangle-calculator.com/?what=sas

Triangle Calculator Side Angle Side

Side a: 93000000
Side b: 240000
Angle Y: 110 degrees

Angle A: 69.861°

Triangle Calculator Side Angle Side

Side a: 9300000
Side b: 240000
Angle Y: 110°

Angle A: 68.623°

Triangle Calculator Side Angle Side

Side a: 930000
Side b: 240000
Angle Y: 110°

Angle A: 57.438°

Triangle Calculator Side Angle Side

Side a: 93000
Side b: 240000
Angle Y: 110°

Angle A: 17.824°

Triangle Calculator Side Angle Side

Side a: 9300
Side b: 240000
Angle Y: 110°

Angle A: 2.058°


Yet, if we input 9300 miles into the math of that Author, the moon points in the same direction as if the distance was 93,000,000 miles. A green arrow will point the same way, no matter how far away the target is! This shows that the concept of the paper is fundamentally incorrect.


I understand that your intuition is telling you that the distance to the sun is a critical part of this issue. I would hope that you can recognize that intuition is often wrong and doesn't prove anything.

See above. You are claiming that a green arrow in space will not point at what it is pointing at. This is unjustified. Wonky math doesn't prove that concept.

Quote
Furthermore, you MUST recognize that we are discussing how things work under RE, and in RE, the sun is 93 million miles away from us - It is not 1 mile away, nor is it unknown. We are testing the model as it stands, and the model say the sun averages around 93 million miles away. Nobody cares what happens when the sun is 1 mile away because we're not testing that.

The math in the paper does not assume ANY values for the sun and the moon. The figures for the distance of the sun and moon appear nowhere in the math from the paper, until you attempted to do so in this thread.

Quote
Because of this, the "hold up a ball" method (which you steadfastly refuse to even try) shows us almost exactly the same lighting as the moon. Hold up a ball and check it. If this works exactly like I've said it does, the lighting on the ball should match the lighting on the moon. If the lighting on the ball is very different, then you have proven that something in this RE model I have described is incorrect.

The "hold-up the ball" method is subject to close-range perspective effects. Are you telling us that the moon is also close to the earth?

If the moon is far away, as in the RE model, then perspective should affect it less and less, since the observer isn't able to change distance ratio in viewing positions as much.

Quote
I'm telling you exactly what the RE model says. What do you suppose gives you the authority to claim that the RE model does not say what the RE model says? That's all we're talking about here.

The RE model says the lighting on the ball should match the lighting of the moon very closely. Stop misrepresenting the RE model.

What are you talking about? Show us where in text books about the Round Earth model where it says that all of the celestial bodies are projected onto a screen close above the observer's heads.

No one was ever taught that about the Round Earth model. That is something the author of the paper made up entirely.
« Last Edit: August 03, 2018, 06:21:29 PM by Tom Bishop »
"The biggest problem in astronomy is that when we look at something in the sky, we don’t know how far away it is" — Pauline Barmby, Ph.D., Professor of Astronomy

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Offline Tom Bishop

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Re: Full Moon Impossible on Flat Earth?
« Reply #273 on: August 03, 2018, 05:54:36 PM »
Tom B understands why the moon's phase looks the way it does,
(taken from the other site)
Quote from: Tom Bishop
Nick, in that explanation the video is using the analogy of standing in the middle of a very long room or hallway. When you look down one end of the hallway the corners are angled upwards and when you look down the other end it is angled upwards.


He just keeps making up imaginary or irrelevant factors in order to confuse himself.

The Sun-Earth-Moon system in RET is not like "looking down a hallway." We can't use hallway perspective effects. The sun and the moon are far away, rotating around the observer, and the same distance from the observer throughout the day in RET. They are not changing distances from the observer like the points of a hallway ceiling is. Perspective matters very little.

The various points of a ceiling of a hall way are all at different distances from the observer. This is a significantly different scenario. In fact, this scenario more closely matches the Flat Earth model than the Round Earth model.

The people who are proposing the hallway perspective explanation are pretty much proposing that the close Sun and Moon model of the Flat Earth is correct.
« Last Edit: August 03, 2018, 06:13:07 PM by Tom Bishop »
"The biggest problem in astronomy is that when we look at something in the sky, we don’t know how far away it is" — Pauline Barmby, Ph.D., Professor of Astronomy

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Offline Tumeni

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Re: Full Moon Impossible on Flat Earth?
« Reply #274 on: August 03, 2018, 08:16:19 PM »
The "hold-up the ball" method is subject to close-range perspective effects.

Like what? What are these 'effects'?

Are you telling us that the moon is also close to the earth?

In comparison to the distance to the Sun, it's very close. And that's the point.

If the moon is far away, as in the RE model, then perspective should affect it less and less, since the observer isn't able to change distance ratio in viewing positions as much.

What ARE you talking about?

Show us where in text books about the Round Earth model where it says that all of the celestial bodies are projected onto a screen close above the observer's heads.

Nobody said anything about them being 'projected', but the celestial sphere is such a basic concept, I'm surprised you can't get to grips with it.

Imagine your Earth. It has a horizontal meridian (equator) and a vertical one (Greenwich Meridian). Any point in space can therefore be mapped in terms on angular difference from these meridians, along a line extended from the centre. If you have a celestial object above 0 latitude, 0 longitude, then its distance from the globe does not affect its co-ordinates. An object at 0N, 0W could be at 50 miles above the surface, or 50,000 - the co-ordinates would be the same. From the point of view of the observer, the objects could well all be at the same height, and the maps of the skies would still look the same. Only the intensity of each object would change.

With me?
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Offline Tom Bishop

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Re: Full Moon Impossible on Flat Earth?
« Reply #275 on: August 03, 2018, 08:30:53 PM »
The "hold-up the ball" method is subject to close-range perspective effects.

Like what? What are these 'effects'?

If you have a ball suspended one foot above your head and walk three feet away from it, you can see much more of the ball than if it were suspended at 10,000 feet in the air and you walked three feet away from it. Can we agree with that?

This was a similar example given in another thread:

Quote
Rubix Cube Example

Imagine that we had a giant solved Rubix Cube suspended one foot above our heads. When we look up we can see its white underside. Now imagine that the Rubix Cube slowly recedes away from us into the distance. We will quickly see one of the colored sides of the cube as it recedes and changes angle. The white bottom of the cube will disappear and you will only see it from the colored side.

Now imagine that we have a giant solved Rubix Cube 10,000 feet above us. It is directly over us. When the Rubix Cube recedes away from us into the distance it will take much longer for us to see the colored side of the Rubix Cube and for the white underside to go away.

Agreed so far?

Since the ball is close to you, it is possible to shift your position slightly and see more of the ball or get it to form positions and angles than it would be if the ball were much further away. At closer distances there are greater perspective effects.

Quote from: Tumeni
In comparison to the distance to the Sun, it's very close. And that's the point.

Think about the RET geometry. The moon isn't "moving across a hallway ceiling" to where it is getting closer or further from you. It stays at the same distance from you at all times. So does the sun. In the hallway example every point on the ceiling is a different distance away from you. It's not a good analogy.

The moon barely changes position in relation to you when it passes overhead.

Quote from: Tumeni
Nobody said anything about them being 'projected'

Yes they did. The Authors of the paper in question say that celestial bodies are projected onto a plane above the observer. ICST made a video about that paper too. The moon and sun are projected onto a plane close above the observers head. It is asserted that it doesn't matter how far away the sun is from the earth. The phase of the moon would point in the same direction, no matter if the earth-sun distance were one mile or 93 million miles.

Would a green arrow suspended in the sky which points at the sun point in the same direction whether the sun were one foot away from the earth or 93 million miles away from the earth? According to the math of the paper, it would.

The Authors and ICST seem to assert that the RET geometry doesn't matter at all in the math. The paper does not even use the distance to the moon or the sun when coming up with the moon angles.

If the RET geometry doesn't matter in this math, and it doesn't matter if the sun is 3000 miles away or 93 million miles away, then it is not an RET model. It is simple as that. Whatever math is done to get the result (assuming that it even gives accurate results -- there is no verification), may apply to any model of the earth.
« Last Edit: August 03, 2018, 08:41:06 PM by Tom Bishop »
"The biggest problem in astronomy is that when we look at something in the sky, we don’t know how far away it is" — Pauline Barmby, Ph.D., Professor of Astronomy

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Offline Tumeni

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Re: Full Moon Impossible on Flat Earth?
« Reply #276 on: August 03, 2018, 08:45:08 PM »
If you have a ball suspended one foot above your head and walk three feet away from it, you can see much more of the ball than if it were 10,000 feet in the air and you walked three feet away from it. Can we agree with that?

Yes, but that has no bearing on what the ball experiment is intended to show. Perspective, and how much of each ball you can see, measured to the Nth degree, has no relevance. 

This was a similar example given in another thread: ...
Since the ball is close to you, it is possible to shift your position slightly and see more of the ball or get it to form positions and angles than it would be if the ball were much further away.

That doesn't matter. The amount you can see of each is irrelevant to the purpose of the experiment


The Authors of the paper in question say that celestial bodies are projected onto a plane above the observer.

And you don't get what is being referred to, do you? If you were to draw, describe or imagine a sphere around the Earth, a spherical plane, each object in the sky has a position, a set of co-ordinates on that imaginary sphere. These co-ordinates remain, regardless of the distance at which you imagine the sphere. This could be termed 'projection' onto that sphere, but c'mon, Tom, even school children can get this concept.

ICST made a video about that paper too. The moon and sun are projected onto a plane close above the observers head.

Or far above their head. The height of the spherical plane doesn't matter. The co-ordinates remain the same.

If you want to dispute the maths, take that up with those who presented the maths. I didn't, so I'm not going there. I'm coming at it from a different angle.
« Last Edit: August 03, 2018, 08:46:51 PM by Tumeni »
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Offline Tom Bishop

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Re: Full Moon Impossible on Flat Earth?
« Reply #277 on: August 03, 2018, 09:05:37 PM »
That doesn't matter. The amount you can see of each is irrelevant to the purpose of the experiment

It is perfectly relevant. You can shift the perspective greater and easier when the ball is close than when it is far away. This shows that you need to use the RET geometry rather than a dumb hallway example. The moon isn't going down a hallway in RET. It is the same distance from the observer at all times. It doesn't radically change distances from you.

This is also why the "holding a ball close to you" experiment is invalid to tell us much. Perspective can change drastically, very easily, when things are close to you.

The "math" in the paper does not use the geometry of the RET model to get the result. ICST said it himself that it doesn't matter at all. Therefore it is not an RET model.

The fact that you guys are searching for perspective effects that would not happen in RET, and math that doesn't even use the RET geometry, shows quite clearly and indisputably that this cannot be explained at all.
"The biggest problem in astronomy is that when we look at something in the sky, we don’t know how far away it is" — Pauline Barmby, Ph.D., Professor of Astronomy

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Offline BillO

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Re: Full Moon Impossible on Flat Earth?
« Reply #278 on: August 03, 2018, 09:55:04 PM »
I don't see what the "hold up the ball" method tells us except that the moon is close to the observer like the ball is, as to be able to point into unnatural perspective angles away from the sun.

There certainly is something odd and artificial about this math where the tilt of the moon's phase does not change at all with scenarios where the sun is located one mile from the earth or 92 million miles from the earth. It seems hard to justify that this scenario meets our reality.

The readers can decide for themselves if an arrow in space would not point at what it is pointing at

So, just on the ball experiment and why the light/shadow on it would be the same as on the moon.

Using that nifty calculator you provided, like you said with the moon 240000 miles away the angle to the sun works out to 69.861°, holding a ball 3 feet away in the direction of the moon the angle to the sun works out to 70°.

The ball experiment is valid and would not produce an altogether unnatural angle, only being off by 0.139°.   But maybe I'm missing something in what you are saying.
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Offline Tumeni

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Re: Full Moon Impossible on Flat Earth?
« Reply #279 on: August 03, 2018, 10:13:07 PM »
That doesn't matter. The amount you can see of each is irrelevant to the purpose of the experiment

It is perfectly relevant. You can shift the perspective greater and easier when the ball is close than when it is far away.

If you move around, you're doing the experiment wrong.

This shows that you need to use the RET geometry rather than a dumb hallway example. The moon isn't going down a hallway in RET. It is the same distance from the observer at all times. It doesn't radically change distances from you.

I never said it did, and I've never used a hallway example. Are you confusing me with someone else?

This is also why the "holding a ball close to you" experiment is invalid to tell us much. Perspective can change drastically, very easily, when things are close to you.

If you're moving around enough to see changes on either the ball you're holding, or the Moon, you're doing the experiment wrong.


The "math" in the paper does not use the geometry of the RET model to get the result. ICST said it himself that it doesn't matter at all. Therefore it is not an RET model.

Take that up with ICST, then


The fact that you guys are searching for perspective effects that would not happen in RET, and math that doesn't even use the RET geometry, shows quite clearly and indisputably that this cannot be explained at all.

I stated in my previous post that perspective doesn't come into this experiment, why are you telling me I'm "searching for perspective effects? Could it be that you're just deliberately trying to muddy the waters?
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Not Flat. Happy to prove this, if you ask me.
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Nearly all flat earthers agree the earth is not a globe.

Nearly?