Re: Full Moon Impossible on Flat Earth?
« Reply #240 on: July 20, 2018, 01:43:03 AM »
I'm working hard to try and gain a little bit of trust here, so it pains me to have to be blunt. But trust comes from honesty, and I have been nothing but honest with you. Tom I mean no offense, but your statements are nonsensical from a mathematical and physical point of view.

The construct by this author of the paper in question just shows that she was unable to explain the event in any other way.

(There are 2 authors, but whatever.) This is the one, the only, and the true explanation of the event. This is not some pie-in-the-sky what-if type of solution. This is the math that follows from the standard model of the Earth-Moon-Sun system. Nothing outside of the standard Earth-Moon-Sun system are introduced at any point during this paper.

There is some speculation about how observers react when there are insufficient visual cues, but none of that has any bearing on the interesting angle which is "alpha". The angle the moon will be lit from falls out purely from the relative positions and angles of the sun, the moon, and the observer.

Your quote seems to suggest that there was some other way you could interpret this. There is not. If the Earth-Moon-Sun system is physically arranged in the way predicted by science, this is exactly what you get.

She is basically using the inexplicable "celestial sphere" idea where bodies are projected on a celestial sphere like a planetarium and straight lines become curved, which is also described in her work as appearing in astronomical literature.
No they are not. There is no celestial sphere at work. That's not what is happening in the slightest. I want you to start to trust me, so I'll be very patient about this. There is a projection onto the observer's view plane. That is not a "celestial sphere". I tried to explain this in my video by using the marker on a window analogy. Another analogy would be the focal plane at the back of a camera or the retina of your eye. A camera records the image that hits the CCD (or film) at the back. That CCD is a flat plate - a plane. The mathematics allows us to place a plane anywhere in front of the camera and make a perfect scaled image of what the camera will record. The only thing we need in order to prove this is the fact that light travels in straight lines. I know your model says this is not so, but this is the RE model we are testing. In the RE model, light travels in straight lines (since we are not discussing any refraction effects here).

Current thoughts: Her work seems to be mathematical fantasy to try and explain something that is not able to be explained.
It is explained perfectly. The math is really quite boring, but you asked me to do the math. There you are, there is the math.

The physical proof is where I suggest you turn your attention if you still find this hard to believe. Holding a physical ball in your hand and looking at it is much more compelling than a bunch of vector math.

However, you asked me to do this math, and I have done it for you. If there are any specific steps along the way you want more information about, just ask. I would be happy to explain any of it. If you find yourself not completely understanding some of it, please step back and realize that you cannot judge what you don't understand.

*

Offline Tumeni

  • *
  • Posts: 1063
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #241 on: July 20, 2018, 07:34:31 AM »
I may be mistaken, and someone please correct me if I am to be wrong, but this seems to be a bit out of proportion.

Distance from earth to sun: 92,960,000 mi
Distance from earth to moon: 238,900 mi

It has to be to have a meaningful graphic fit onto your screen.

Let's approximate this to 92 'units' distance to the sun, and 0.24 units, distance to the moon, keeping correct proportions

If you represent 92 units across the full width of your 50cm wide monitor, each unit is (50/92) = 0.54cm, or 5mm.

0.24 units, at that scale, is (0.24*5) = 1.3mm. So you'd have to draw Earth and Moon smaller than that to show them on the graphic, with 1.3mm separating them. Seems you'd need a magnifying glass.

No?

 
==============================
==============================
Pete Svarrior "We are not here to directly persuade anyone ... You mistake our lack of interest in you for our absence."

Tom Bishop "We are extremely popular and the entire world wants to talk to us. We have better things to do with our lives than have in depth discussions with every single curious person. You are lucky to get one sentence dismissals from us"

Re: Full Moon Impossible on Flat Earth?
« Reply #242 on: July 20, 2018, 04:18:44 PM »
I may be mistaken, and someone please correct me if I am to be wrong, but this seems to be a bit out of proportion.

Distance from earth to sun: 92,960,000 mi
Distance from earth to moon: 238,900 mi

It has to be to have a meaningful graphic fit onto your screen.

Let's approximate this to 92 'units' distance to the sun, and 0.24 units, distance to the moon, keeping correct proportions

If you represent 92 units across the full width of your 50cm wide monitor, each unit is (50/92) = 0.54cm, or 5mm.

0.24 units, at that scale, is (0.24*5) = 1.3mm. So you'd have to draw Earth and Moon smaller than that to show them on the graphic, with 1.3mm separating them. Seems you'd need a magnifying glass.

No?
Having anticipated this criticism from the previous video, I made you a version to scale for this video. As Tumeni points out, the scale makes all the different parts tiny if they need to share a screen, but with a 3D realtime sim, you can zoom around and look at the different parts from different angles to see it all - it doesn't have to all fit together into the same screen.

In the description of my video: @2:43  New simulation - this time to scale


@41:09 Demonstrating that the answer is independent of the distances. It is based only on the angles

Re: Full Moon Impossible on Flat Earth?
« Reply #243 on: July 25, 2018, 05:09:05 AM »
Tom, you've inspired me. I'll do the math for you. I'll make you a video (100% troll free this time). But in return for all this effort, I'd like something from you. I'd like you to agree in advance that IF I can do the math you ask for, and IF the math shows that the RE model explains this illusion, you will publicly concede the point. You don't have to agree that the world is round, just admit that you made a mistake and the moon terminator illusion is perfectly explained by the standard heliocentric model.

What do you say? Sound fair enough?
If the distances/attributes of the Round Earth Model can explain the moon tilt illusion for gibbous and crescent moons, I have absolutely no problem admitting that. As I have seen, it cannot. It cannot explain it, and this is why the literature is so vague about the matter.

Tom, I've done the math. I've explained it all using a physical model as well as a mathematical one and a 3D simulation. You can verify these against mooncalc or stellarium if you like. Better yet, you can step outside and look in the sky to check it. This is absolutely air tight.

You agreed that if I could provide this, you would concede the point. All you need to do now is to publicly acknowledge that the tilt of the moon is completely consistent with the RE model.

Re: Full Moon Impossible on Flat Earth?
« Reply #244 on: July 25, 2018, 05:52:05 PM »
You agreed that if I could provide this, you would concede the point. All you need to do now is to publicly acknowledge that the tilt of the moon is completely consistent with the RE model.
Such a concession would put serious strain on any flat earth model that accepts that the moon's illumination and phase are due to reflection of the sun's light.  The geometry explaining that illusion doesn't work in any flat earth topology I've seen. The only explanation that might work is if the moon is self-luminescent, which is why (I think) Tom introduced the conundrum in the first place. But as you've explained, you can explain it with moon reflecting sunlight in a round earth model. That doesn't solve the puzzle for flat earth though; thus the theory that the moon generates its own light.

*

Offline Tom Bishop

  • Zetetic Council Member
  • **
  • Posts: 5491
  • Flat Earth Believer
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #245 on: July 25, 2018, 06:24:00 PM »
That's not the Round Earth Model. Since when in RET are images of bodies in space projected onto a plane close above the observer's heads?

Re: Full Moon Impossible on Flat Earth?
« Reply #246 on: July 25, 2018, 06:47:36 PM »
That's not the Round Earth Model. Since when in RET are images of bodies in space projected onto a plane close above the observer's heads?
When you take a photo of something, an image is projected onto the film inside the camera. Use a digital camera, and the film is replaced by a CCD plate. Use your own eyes, and the film is replaced by the retina at the back of your eye. That's right, every time you take a photo of something, images of that something are projected onto a plane inside the camera. Don't try to tell us that cameras don't work under RE next. Last I checked, cameras worked on flat earth too.

Stop trying to make excuses. It's time to own up to it. You were wrong. Just be big enough to admit it.

*

Offline Tom Bishop

  • Zetetic Council Member
  • **
  • Posts: 5491
  • Flat Earth Believer
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #247 on: July 26, 2018, 04:42:46 AM »
Okay, I finished watching the video. I kept waiting with anticipation for math to be done under the dimensions of the Round Earth System, as advertised. However, that did not happen. The video mostly consisted of you reading the equations from the paper. The dimensions of the Round Earth System appear nowhere in it. In fact, the author says in the work that they avoided using the RET numbers.

From p. 9:

Quote
The value of the angle α is the same for the vectors m, s and z or their corresponding unit vectors, which are used in Eq. (11) to avoid having to know the actual distances of the moon and the sun from the observer.

The dimensions for the Round Earth System are nowhere in the math by the authors of that paper, nor is it in your video when describing the matter.

Secondly, it is apparent to all that the author needs to project images onto a plane close above the observer's head in order to attempt to describe this. It is entirely apparent that the authors cannot "really" explain it.

We may as well just say that the moon and sun are a close distance above the observer's head, as to entertain that.

Thirdly, I wanted to point out that at the 42 minute mark you claim that the distance from the earth to the sun doesn't matter, and the moon will point in the same direction regardless.

Will a green arrow that points at the sun, located at the height of the moon, as seen from earth, point in the same direction regardless of whether the sun was one foot away from the earth or if it were 100,000,000 miles away? Clearly not.

All-in-all the marks for the "Round Earth explanation" are poor, and I intend to point these things out when I get around to making the Wiki article on the subject.
« Last Edit: July 26, 2018, 04:51:12 AM by Tom Bishop »

Re: Full Moon Impossible on Flat Earth?
« Reply #248 on: July 26, 2018, 06:24:24 AM »
Okay, I finished watching the video. I kept waiting with anticipation for math to be done under the dimensions of the Round Earth System, as advertised. However, that did not happen. The video mostly consisted of you reading the equations from the paper.
The honorable thing to do is admit you do not understand the math and concede the point.

The dimensions of the Round Earth System appear nowhere in it. In fact, the author says in the work that they avoided using the RET numbers.

From p. 9:

Quote
The value of the angle α is the same for the vectors m, s and z or their corresponding unit vectors, which are used in Eq. (11) to avoid having to know the actual distances of the moon and the sun from the observer.

The dimensions for the Round Earth System are nowhere in the math by the authors of that paper, nor is it in your video when describing the matter.
You want the dimensions of the moon and sun to be used? Here they are. You cannot deny it anymore:
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).

So for the last time, stop going on and on about distances. We can easily prove that the distances cancel out of the equation anyway. If you can't follow the math, there are the numbers using the distances to the moon and sun. Stop whining about it already.

Secondly, it is apparent to all that the author needs to project images onto a plane close above the observer's head in order to attempt to describe this. It is entirely apparent that the authors cannot "really" explain it.

We may as well just say that the moon and sun are a close distance above the observer's head, as to entertain that.
What is apparent to all is that you simply do not understand the math. You went on and on shouting "DO THE MATH." Well I did the math. It isn't my fault you don't understand it. I'm happy to explain any individual step you can't understand.

This isn't that hard to understand really. A camera takes a snapshot of the 2D image it sees in front of it. There is a 3D world in front of the camera, but the image is 2D. We call this capture of a 3D scene onto a 2D image a "projection". That's what we have done here. That's how you explain this phenomenon. What is happening is a 3D direction (the light that hits the moon coming from the direction of the sun) is resolved by your eyes into a 2D image. Your brain judges the direction of that light based on the 2D image.

Thirdly, I wanted to point out that at the 42 minute mark you claim that the distance from the earth to the sun doesn't matter, and the moon will point in the same direction regardless.

Will a green arrow that points at the sun, located at the height of the moon, as seen from earth, point in the same direction regardless of whether the sun was one foot away from the earth or if it were 100,000,000 miles away? Clearly not.
This is your distances complaint AGAIN! I've addressed it several times now. Just above here, I worked the math with all the distances included.

I have shown mathematically that those distances cancel.

I showed in the 3D simulation that those distances do not matter.

Now I've worked the math with the distances included, and guess what? I got the right answer AGAIN!

As long as you can create the proper angles, the distances are not part of this illusion. That's why you can hold a ping-pong ball at arm's length and get the same lighting as the moon has on it.

All-in-all the marks for the "Round Earth explanation" are poor, and I intend to point these things out when I get around to making the Wiki article on the subject.
The only thing poor here is your ability to admit it when you are wrong.

Re: Full Moon Impossible on Flat Earth?
« Reply #249 on: July 26, 2018, 06:34:25 AM »
Show us a photo of yourself holding a ball up to the moon. Show us how the sun's light hitting the ball is significantly different from the sun's light hitting the moon.

I have done this. Others have done this. We are not liars.

This entire thread about "DO THE MATH" was nothing more than a distraction to avoid that simple exercise. You don't have to work the math, just take a photo of a ball. But you won't show us the ball will you? Don't worry. We all know why you won't do it. It's because you know that the results will show you are wrong.

You are hoping that your quote mining and mathematical hand-waving will distract your audience from the fact that the light hitting the moon is perfectly consistent with the direction of the sun. The only reason you refuse to hold a ball up to the moon is because you know what that would show.

Break that down. You know what holding up a ball would show. Instead you continue insisting the opposite of what you know the ball would show. What does that say about your position?

*

Offline Jura-Glenlivet

  • *
  • Posts: 1420
  • Life is meaningless & everything dies.
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #250 on: July 26, 2018, 11:02:22 AM »
We did this a while ago see (https://forum.tfes.org/index.php?topic=6056.msg113954#msg113954) Tom refused to do the zetetic thing then too, Junker did promise to go out and give it a go but never got back to us, Junker did you, will you, try the string test?
« Last Edit: July 26, 2018, 01:10:13 PM by Jura-Glenlivet »
Just to be clear, you are all terrific, but everything you say is exactly what a moron would say.

Offline iamcpc

  • *
  • Posts: 292
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #251 on: July 26, 2018, 04:49:09 PM »
You agreed that if I could provide this, you would concede the point. All you need to do now is to publicly acknowledge that the tilt of the moon is completely consistent with the RE model.
Such a concession would put serious strain on any flat earth model that accepts that the moon's illumination and phase are due to reflection of the sun's light.

The work put in and the math done is very impressive. I seems that someone would have a VERY hard time not admitting that the moon tilt is mathematically possible in the round earth model. I'm consistently amazed that it is.  It's not like people are saying the earth is round. people are saying that the earth COULD BE round and we could still have the observations of the moon that we currently have.

That doesn't solve the puzzle for flat earth though

We have already discussed that if the moon was hundreds of thousands of miles above the sun the earth could be flat and we could be within the angle of a full moon. I know this contradicts the wiki but so does the self lit moon.
« Last Edit: July 26, 2018, 04:50:42 PM by iamcpc »

*

Offline Tom Bishop

  • Zetetic Council Member
  • **
  • Posts: 5491
  • Flat Earth Believer
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #252 on: July 26, 2018, 06:23:21 PM »
I will come back to your other arguments in a bit. This is the bit that makes it clear:

Thirdly, I wanted to point out that at the 42 minute mark you claim that the distance from the earth to the sun doesn't matter, and the moon will point in the same direction regardless.

Will a green arrow that points at the sun, located at the height of the moon, as seen from earth, point in the same direction regardless of whether the sun was one foot away from the earth or if it were 100,000,000 miles away? Clearly not.
This is your distances complaint AGAIN! I've addressed it several times now. Just above here, I worked the math with all the distances included.

I have shown mathematically that those distances cancel.

I showed in the 3D simulation that those distances do not matter.

It is pretty obvious that distances do matter.

Your position: In RET a green arrow suspended in the air at the height of the moon which points at the sun will point in the same direction, regardless of whether the sun is 1 foot from the earth, 1 million miles from the earth, or 100 billion miles from the earth. It will point in the same direction nonetheless!

This is what you are saying, this is what the shady math is saying, and everyone knows that this is NOT TRUE.

Re: Full Moon Impossible on Flat Earth?
« Reply #253 on: July 26, 2018, 06:36:56 PM »
I will come back to your other arguments in a bit. This is the bit that makes it clear:

Thirdly, I wanted to point out that at the 42 minute mark you claim that the distance from the earth to the sun doesn't matter, and the moon will point in the same direction regardless.

Will a green arrow that points at the sun, located at the height of the moon, as seen from earth, point in the same direction regardless of whether the sun was one foot away from the earth or if it were 100,000,000 miles away? Clearly not.
This is your distances complaint AGAIN! I've addressed it several times now. Just above here, I worked the math with all the distances included.

I have shown mathematically that those distances cancel.

I showed in the 3D simulation that those distances do not matter.

It is pretty obvious that distances do matter.

Your position: In RET a green arrow suspended in the air at the height of the moon which points at the sun will point in the same direction, regardless of whether the sun is 1 foot from the earth, 1 million miles from the earth, or 100 billion miles from the earth. It will point in the same direction nonetheless!

This is what you are saying, this is what the shady math is saying, and everyone knows that this is NOT TRUE.
If you can find any error anywhere in the math, please share. If you cannot, then you must accept that the math is correct. It's either correct or it's not. If it's not, please point out the mistake.

Edit: Yes I noticed that you have conveniently ignored where I just worked all the math using the actual distances and still got the correct answer. It's all right there. Stop ignoring it.

Re: Full Moon Impossible on Flat Earth?
« Reply #254 on: July 29, 2018, 04:59:50 AM »
Come on Tom. It's time to make it right. The RE model explains the tilt of the moon just fine. At this point it's entirely clear. I'm not expecting you to give up on FE over it, but I do expect you to admit that the RE model has a perfectly accurate explanation for the angle of the moon's tilt.

Offline iamcpc

  • *
  • Posts: 292
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #255 on: July 30, 2018, 11:35:07 PM »
Come on Tom. It's time to make it right. The RE model explains the tilt of the moon just fine. At this point it's entirely clear. I'm not expecting you to give up on FE over it, but I do expect you to admit that the RE model has a perfectly accurate explanation for the angle of the moon's tilt.

It seems to be that you have shown very strong evidence which suggest that a full moon is possible in the round earth model. Will the corresponding flat earth wiki be updated to suggest that a full moon MIGHT be possible in the existing round earth model?

*

Offline Tom Bishop

  • Zetetic Council Member
  • **
  • Posts: 5491
  • Flat Earth Believer
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #256 on: July 31, 2018, 05:59:19 AM »
ICST, you and the author of that paper say that the 238,900 mile high moon points the same way, whether the sun is one foot away from the earth or 92.96 million miles away from the earth. The angle of the phase will never change.

Right Angled Triangle Calculator: https://www.easycalculation.com/trigonometry/triangle-angles.php

Select "I want to calculate Angle and Hypotenuse side."



Opposite Side: 238900
Adjacent Side: 92960000
angle b = 89.8528 degrees

Opposite Side: 238900
Adjacent Side: 9296000
angle b = 88.5279 degrees

Opposite Side: 238900
Adjacent Side: 929600
angle b = 75.5873

Opposite Side: 238900
Adjacent Side: 92960
angle b = 21.261799999999994

Opposite Side: 238900
Adjacent Side: 9296
angle b = 2.2283999999999935

What am I doing incorrectly here?
« Last Edit: July 31, 2018, 06:58:52 AM by Tom Bishop »

Re: Full Moon Impossible on Flat Earth?
« Reply #257 on: July 31, 2018, 06:45:47 AM »
ICST, you and the author of that paper say that the 238,900 mile high moon points the same way, whether the sun is one foot away from the earth or 92.96 million miles away from the earth. The angle of the phase will never change.

Right Angled Triangle Calculator: https://www.easycalculation.com/trigonometry/triangle-angles.php

Select "I want to calculate Angle and Hypotenuse side."



Opposite Side: 238900
Adjacent Side: 92960000
angle b = 89.8528 degrees

Opposite Side: 238900
Adjacent Side: 9296000
angle b = 88.5279 degrees

Opposite Side: 238900
Adjacent Side: 929600
angle b = 75.5873

Opposite Side: 238900
Adjacent Side: 92960
angle b = 21.261799999999994

Opposite Side: 238900
Adjacent Side: 9296
angle b = 2.2283999999999935

What am I doing incorrectly here?
Tom, I'm glad you aren't abandoning the discussion.
As you already saw, I did the calculations with the full numbers. I'll repeat it here for you:
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?

I'd like to end the discussion of the math now. There is the tilt angle (not the phase - the tilt) calculated using the RE distances and predicted angles for the sun and moon on July 24th at 8pm Los Angeles. Having trouble with the math? Not an issue. Just go outside when the moon and sun are both visible and hold up a ball. Isn't that better than the math anyway?

As I've said, I'd prefer this be the end of the "DO THE MATH" dialog. However, if you insist, I can show you your mistake. Let's just drop it though huh? We don't need the math, and even if we want the math to provide accurate tilt angle predictions, we can skip the part where we drop the distances.

I want to make it perfectly clear to you as well as to all readers here, that I will not lie to you. I will not deceive. I can help people who genuinely want to understand the physics and the math behind this stuff. So if you insist, yes I can point out where you are making some mistakes with the math here. But I sincerely think it's just a waste of time. Let's just accept the ball method instead. Yeah?

*

Offline Tumeni

  • *
  • Posts: 1063
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #258 on: July 31, 2018, 06:56:14 AM »


What am I doing incorrectly here?

1. Using a diagram which does not have Angle b marked upon it? I can see a Greek symbol, but no b.

2. Using a right-angle triangle, when the relationship between Earth, Moon and Sun (assuming you mean each is at a point of the triangle) is mostly never this type of triangle.

3. Doing calculations for the Sun being a short distance away, which it is not. You can prove this for yourself, and prove your calculation of this to be moot by ...

Going 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).
« Last Edit: July 31, 2018, 06:58:14 AM by Tumeni »
==============================
==============================
Pete Svarrior "We are not here to directly persuade anyone ... You mistake our lack of interest in you for our absence."

Tom Bishop "We are extremely popular and the entire world wants to talk to us. We have better things to do with our lives than have in depth discussions with every single curious person. You are lucky to get one sentence dismissals from us"

*

Offline Tom Bishop

  • Zetetic Council Member
  • **
  • Posts: 5491
  • Flat Earth Believer
    • View Profile
Re: Full Moon Impossible on Flat Earth?
« Reply #259 on: July 31, 2018, 07:04:02 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.



What am I doing incorrectly here?

Using a diagram which does not have Angle b marked upon it? I can see a Greek symbol, but no b.

Angle b is the top left angle in that image. It changes when we change the distance to the sun (the adjacent side).

Quote
2. Using a right-angle triangle, when the relationship between Earth, Moon and Sun (assuming you mean each is at a point of the triangle) is mostly never this type of triangle.

The angle changes when the distance to the sun changes. It doesn't matter if the moon is directly overhead, or at 45 degrees into the opposite horizon from the sun.

Quote
3. Doing calculations for the Sun being a short distance away, which it is not. You can prove this for yourself, and prove your calculation of this to be moot by ...

Going 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?

The moon and sun are often only seen during the day when they are on on opposite skies, since the moon disappears when it gets too close to the sun. Holding up a ball between them 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. Close up perspective effects can simulate many shapes.
« Last Edit: July 31, 2018, 07:29:59 AM by Tom Bishop »