The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Community => Topic started by: Tom Bishop on August 09, 2017, 10:04:28 PM
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According to Round Earth Theory the sun is illuminating the earth and moon as in this top-down view:
(http://i.imgur.com/z2QpoXZ.gif)
However, it has been observed that during the day, on the "day" side of the earth, the moon can be seen with phases which show significant lit area.
It should be impossible for the moon's phase to display significant fullness during the day. But the moon is seen on many occasions to do just that. Here is an example which should be impossible under the Round Earth model:
https://www.youtube.com/watch?v=-u4MtmIuFBw
In the video the author pans across to the sun, showing that it is significantly above the horizon. This person is clearly somewhere on the day side of the earth. How is this observation possible?
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The scale of your diagram is wrong, earth radius 6371km, moon distance 384,400 km, moon radius 1737km.
Redraw!
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The scale of your diagram is wrong, earth radius 6371km, moon distance 384,400 km, moon radius 1737km.
Redraw!
Here is the Earth and Moon to scale in RET:
(https://lh6.googleusercontent.com/-3gl68xQi2vA/Ufa75Wo2ihI/AAAAAAAAAXg/PBeLvzZG9tE/earth_moon_scale_size_distance-fjs-1920x1080.jpg)
How does this help you? It still does not make the phase which was seen possible.
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According to Round Earth Theory the sun is illuminating the earth and moon as in this top-down view:
(http://i.imgur.com/z2QpoXZ.gif)
However, it has been observed that during the day, on the "day" side of the earth, the moon can be seen with phases which show significant lit area.
It should be impossible for the moon's phase to display significant fullness during the day. But the moon is seen on many occasions to do just that. Here is an example which should be impossible under the Round Earth model:
https://www.youtube.com/watch?v=-u4MtmIuFBw
In the video the author pans across to the sun, showing that it is significantly above the horizon. This person is clearly somewhere on the day side of the earth. How is this observation possible?
This isn't impossible. At all. Should I break out the math and attempt a diagram? Once again, out of time at the moment, but I'll get to it later unless someone else tackles this very simple problem first. I just don't have the time to work the math and make a diagram right now.
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Yes, please post us your diagram where a stick figure is 30,000 miles tall and looking around the curvature of the earth. We can't wait to see it!
Please keep in mind that the horizon looks like this in Round Earth Theory:
(http://i.imgur.com/aGcNNZH.png)
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Here is the Earth and Moon to scale in RET:
(https://lh6.googleusercontent.com/-3gl68xQi2vA/Ufa75Wo2ihI/AAAAAAAAAXg/PBeLvzZG9tE/earth_moon_scale_size_distance-fjs-1920x1080.jpg)
How does this help you? It still does not make the phase which was seen possible.
incline the moon by ~5 deg.
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incline the moon by ~5 deg.
Why? It still doesn't help you. There is no possible way for that phase to appear, no matter where the Moon is located over the day side of the earth.
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incline the moon by ~5 deg.
Why? It still doesn't help you. There is no possible way for that phase to appear, no matter where the Moon is located over the day side of the earth.
(http://www.rain.org/~mkummel/stumpers/14jan00b.gif)
imagine a person standing at the "top" of the earth. the sun is in view. the moon is in view. the moon is full. solved.
btw at what time of day was this video captured?
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incline the moon by ~5 deg.
Why? It still doesn't help you. There is no possible way for that phase to appear, no matter where the Moon is located over the day side of the earth.
(http://www.rain.org/~mkummel/stumpers/14jan00b.gif)
imagine a person standing at the "top" of the earth. the sun is in view. the moon is in view. the moon is full. solved.
btw at what time of day was this video captured?
This diagram does not explain what is happening. The sun is not intersecting the horizon in that video, like it is in the diagram you posted, it is in the sky.
And if you rotate the sun upwards to be higher in the sky in that illustration, the blue line of the ecliptic will rotate downwards, and that will cause the moon will move downwards. That explanation is not possible at all (not to mention that the observer is not at the North Pole).
Care to try another explanation?
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no other explanation is required. you haven't shown an anomaly.
watch the video, and then check the diagram. ret predicts that some latitudes should occasionally observe a nearly-full moon around sunset/sunrise when both are near the horizon and opposite one another in the sky. that's exactly what your video shows.
i'm sure there are other videos out there of a full moon during the day. check them out. you'll find that they always show the sun and moon near the horizon and opposite one another in the sky.
also the earth's axis is tilted. read the faq. it's not a coincidence that this event happened at a northern latitude (new york) and not the equator.
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The author of the video states in the description that the observation was taken on April 8, 2017 6:30PM in Lewiston, NY
I plopped that data into http://aa.usno.navy.mil/data/docs/AltAz.php and got the following:
Altitude and Azimuth of the Sun (http://aa.usno.navy.mil/cgi-bin/aa_altazw.pl?form=1&body=10&year=2017&month=4&day=8&intv_mag=10&state=NY&place=Lewiston)
Astronomical Applications Dept.
U.S. Naval Observatory
Washington, DC 20392-5420
LEWISTON, NEW YORK
o , o ,
W 79 02, N43 11
Altitude and Azimuth of the Sun
Apr 8, 2017
Eastern Daylight Time
Time Altitude Azimuth (E of N)
h m o o
...
18:30 13.9 267.4
...
Altitude and Azimuth of the Moon (http://aa.usno.navy.mil/cgi-bin/aa_altazw.pl?form=1&body=11&year=2017&month=4&day=8&intv_mag=10&state=NY&place=Lewiston)
Astronomical Applications Dept.
U.S. Naval Observatory
Washington, DC 20392-5420
LEWISTON, NEW YORK
o , o ,
W 79 02, N43 11
Altitude and Azimuth of the Moon
Apr 8, 2017
Eastern Daylight Time
Time Altitude Azimuth (E of N) Fraction illuminated
h m o o
...
18:30 11.0 94.9 0.95
...
See the bolded.
So, clearly, under the Round Earth model the moon's 5 degree deviation from the ecliptic is not a valid explanation for this observation. Both the sun and moon are more than 5 degrees in the sky under your model. Your illustration just showed a sun at 0 degrees, half intersecting the horizon, and the moon at 5 degrees. This does not reconcile.
Also, under the Round Earth model the earth is tilted with respect to the ecliptic. The moon deviates from the ecliptic, not the earth's tilt. So that will not help you either.
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I don't know what you're thinking...but a simple diagram, drawn to scale, makes it VERY clear:
(https://renaissanceinnovations.com/TFES_eclipse.png)
The 5 degree tilt of the lunar orbit - plus the small sizes of Earth and moon relative to sun make this very, very obvious.
Orange is the sun (too big to draw - but about right) - which would be about 600 feet to the right of your monitor on this scale.
Tiny black dot is the moon (drawn to scale) red dot is Earth (also to scale).
5 degree tilt of the Moon's orbit puts it well off out of the way of Earth's shadow.
This is kiddie stuff.
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That is the SAME diagram that was just posted. Please follow along.
According to the U.S. Naval Observatory the Sun is 13.9 degrees above the horizon and the Moon is 11.0 degrees above the horizon in the video. This makes the 5 degree moon divergence from the ecliptic an impossible explanation for this observation.
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That is the SAME diagram that was just posted. Please follow along.
According to the U.S. Naval Observatory the Sun is 13.9 degrees above the horizon and the Moon is 11.0 degrees above the horizon in the video. This makes the 5 degree moon divergence from the ecliptic an impossible explanation for this observation.
I don't see anything that shows that my diagram is incorrect.
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That is the SAME diagram that was just posted. Please follow along.
According to the U.S. Naval Observatory the Sun is 13.9 degrees above the horizon and the Moon is 11.0 degrees above the horizon in the video. This makes the 5 degree moon divergence from the ecliptic an impossible explanation for this observation.
I don't see anything that shows that my diagram is incorrect.
According to your diagram a person atop the the earth will see the sun at 0 degrees, half intersecting the horizon, and the moon at 5 degrees.
However, the observation which was seen was that both the sun and moon were more than 5 degrees above the horizon. The Sun was 13.9 degrees above the horizon and the Moon was 11.0 degrees.
In Round Earth Theory the ecliptic is the earth's straight line alignment with the sun, and the moon deviates from it by 5 degrees. The observations make the explanation given impossible.
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That is the SAME diagram that was just posted. Please follow along.
According to the U.S. Naval Observatory the Sun is 13.9 degrees above the horizon and the Moon is 11.0 degrees above the horizon in the video. This makes the 5 degree moon divergence from the ecliptic an impossible explanation for this observation.
I don't see anything that shows that my diagram is incorrect.
According to your diagram a person atop the the earth will see the sun at 0 degrees, half intersecting the horizon, and the moon at 5 degrees.
However, the observation which was seen is that both the sun and moon are more than 5 degrees above the horizon. The Sun is 13.9 degrees above the horizon and the Moon is 11.0 degrees.
In Round Earth Theory the ecliptic is the earth's straight line alignment with the sun, and the moon deviates from it by 5 degrees. The observations make the explanation given impossible.
Because that diagram is showing the moon orbit around the Earth from a different angle than is required to view the happening in the video you linked. The moon orbits around the Earth upon that plane, we are looking at the Earth from a 'side on' direction at what would appear to be roughly one of the equinox's. I'm working on a much more long winded reply with a diagram and math, but have to do it in between some things I'm doing over here so it's gonna be a little while.
EDIT: Plus it's been a little while since I did geometry, so I'm having to find some formula again.
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See the bolded.
lol you're now quibbling over ~10 degrees of arc. that's like the width of your hand extended at arm's length.
check out the diagram and keep looking up videos of daytime full moons. you'll notice that they all show the same sun/moon configuration in the sky: the sun and a nearly-full moon close to the horizon and opposite one another in the sky.
you'll never see one at midday.
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See the bolded.
lol you're now quibbling over ~10 degrees of arc. that's like the width of your hand extended at arm's length.
check out the diagram and keep looking up videos of daytime full moons. you'll notice that they all show the same sun/moon configuration in the sky: the sun and a nearly-full moon close to the horizon and opposite one another in the sky.
you'll never see one at midday.
It's more like we are quibbling over 19.775 degrees (13.9 + 11 - 5.145). This is a significant error.
Your expectation that we should look for the moon at midday is fallacious. The moon is only seen when the sky's brightness is below a certain magnitude which is less than the magnitude of the moon. Midday is when the sky is at its brightest. The moon is generally invisible then.
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Yes, please post us your diagram where a stick figure is 30,000 miles tall and looking around the curvature of the earth. We can't wait to see it!
Please keep in mind that the horizon looks like this in Round Earth Theory:
(http://i.imgur.com/aGcNNZH.png)
Your just daft. Go to :44 sec The shadows cast show the sun is the opposite direction. So for a brief time the moon is illuminated by the sun. The moon isn't that far above the horizon and it isn't full. Buy a dam globe, hang a tennis ball and shine a flashlight at it! You draw little silly 2D pictures in paint (just can't shake that flat obsession) that don't represent reality. The difference in the moons orbit and Earths (~5 degrees) is why there isn't a lunar eclipse every month. There's nothing unusual about any of this, we could tell you the next time you'd be able to see one, but you won't look.
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go look up more daytime full moon videos. the sun and nearly-full moon will always be near the horizon and opposite one another in the sky. just as the diagram shows. see for yourself.
It's more like we are quibbling over 19.775 degrees (13.9 + 11 - 5.145). This is a significant error.
lol not really. that's a combined "error" in angular position smaller than my hand held at arm's length.
Your expectation that we should look for the moon at midday is fallacious. The moon is only seen when the sky's brightness is below a certain magnitude which is less than the magnitude of the moon. Midday is when the sky is at its brightest. The moon is generally invisible then.
you'll never see one not opposite to the sun in the sky. you'll never see a video like that where the nearly-full moon is to the north or south.
just like you'll never see a crescent moon opposite the sun in the sky.
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Trying to chase this distraction down I ended up posting in the thread Tom was running from. Copying it here as well. I'll also add, I noticed while moving the time of the video is spot on for the line of site in the picture. Wait? Aren't we supposed to run away scared of these hard questions? I had to chase you down lol.
(https://image.ibb.co/b0xsUv/Moon_phases5_400x400_los.jpg)
There you go, nice paint picture. What do you know, I put your horizon lines, put them on your moon phases picture and Shazam! In the morning and evening you can see a very illuminated moon, who knew? Oh yeah, everyone.
I also checked your date, the moon phase was Waxing gibbous - Visible: 93%
https://www.calendar-12.com/moon_calendar/2017/april
But, SUPER JOB AVOIDING EASILY VERIFIED DISTANCES PROVING YOU WRONG, no one even noticed.
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According to your diagram a person atop the the earth will see the sun at 0 degrees, half intersecting the horizon, and the moon at 5 degrees.
However, the observation which was seen was that both the sun and moon were more than 5 degrees above the horizon. The Sun was 13.9 degrees above the horizon and the Moon was 11.0 degrees.
In Round Earth Theory the ecliptic is the earth's straight line alignment with the sun, and the moon deviates from it by 5 degrees. The observations make the explanation given impossible.
Here's the thing Tom, the answer is in the data you posted. Notice the latitude of Lewiston NY. Have a think about that.
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Trying to chase this distraction down I ended up posting in the thread Tom was running from. Copying it here as well. I'll also add, I noticed while moving the time of the video is spot on for the line of site in the picture. Wait? Aren't we supposed to run away scared of these hard questions? I had to chase you down lol.
https://image.ibb.co/b0xsUv/Moon_phases5_400x400_los.jpg
There you go, nice paint picture. What do you know, I put your horizon lines, put them on your moon phases picture and Shazam! In the morning and evening you can see a very illuminated moon, who knew? Oh yeah, everyone.
I also checked your date, the moon phase was Waxing gibbous - Visible: 93%
https://www.calendar-12.com/moon_calendar/2017/april
But, SUPER JOB AVOIDING EASILY VERIFIED DISTANCES PROVING YOU WRONG, no one even noticed.
(http://i.imgur.com/Kt852Km.png)
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Trying to chase this distraction down I ended up posting in the thread Tom was running from. Copying it here as well. I'll also add, I noticed while moving the time of the video is spot on for the line of site in the picture. Wait? Aren't we supposed to run away scared of these hard questions? I had to chase you down lol.
https://image.ibb.co/b0xsUv/Moon_phases5_400x400_los.jpg
There you go, nice paint picture. What do you know, I put your horizon lines, put them on your moon phases picture and Shazam! In the morning and evening you can see a very illuminated moon, who knew? Oh yeah, everyone.
I also checked your date, the moon phase was Waxing gibbous - Visible: 93%
https://www.calendar-12.com/moon_calendar/2017/april
But, SUPER JOB AVOIDING EASILY VERIFIED DISTANCES PROVING YOU WRONG, no one even noticed.
(http://i.imgur.com/Kt852Km.png)
D-did you try putting your moon over where the moon actually is, instead of creating an optical illusion to attempt to deceive the viewer? Because from what i'm seeing you're pretty damn close if not spot on to what is being seen if you would just put the moon where it's supposed to be. Because the moon doesn't appear full, it's only mostly full. I've still got that post, but your image is so much more useful than mine, so thank you for that.
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D-did you try putting your moon over where the moon actually is, instead of creating an optical illusion to attempt to deceive the viewer? Because from what i'm seeing you're pretty damn close if not spot on to what is being seen if you would just put the moon where it's supposed to be. Because the moon doesn't appear full, it's only mostly full. I've still got that post, but your image is so much more useful than mine, so thank you for that.
Here you go. There should be a significant crescent.
(http://i.imgur.com/ILqulDT.png)
But according to the U.S. Naval Observatory the Moon was 95% full at that time:
Altitude and Azimuth of the Moon (http://aa.usno.navy.mil/cgi-bin/aa_altazw.pl?form=1&body=11&year=2017&month=4&day=8&intv_mag=10&state=NY&place=Lewiston)
Astronomical Applications Dept.
U.S. Naval Observatory
Washington, DC 20392-5420
LEWISTON, NEW YORK
o , o ,
W 79 02, N43 11
Altitude and Azimuth of the Moon
Apr 8, 2017
Eastern Daylight Time
Time Altitude Azimuth (E of N) Fraction illuminated
h m o o
...
18:30 11.0 94.9 0.95
...
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Based on what we know of the scenario it should also be possible to calculate exactly how many degrees of the moon should be in shade, based on the number of degrees the sun needs to descend to match the moon. I will leave that as an exercise for the reader, however.
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Based on what we know of the scenario it should also be possible to calculate exactly how many degrees of the moon should be in shade, based on the number of degrees the sun needs to descend to match the moon. I will leave that as an exercise for the reader, however.
Very carefully worded again, meaning you do not understand how to do it.
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Here you go. There should be a significant crescent.
(http://i.imgur.com/ILqulDT.png)
no it wouldn't. look at your diagram again. it would appear nearly full except for some shadow at the "bottom" of the moon. just like in your video, i might add.
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Tom - you're thinking like a flat-earther (not surprising I suppose) - you need to think three-dimensionally.
If you're looking downward from (say) a million miles above the North Pole - you'd see the moon in some position around it's orbit and the observer watching from some arbitrary point on the surfaces of the Earth.
The elevation is relative to the local horizon - which (depending on latitude/longitude of the observer) can be anywhere imaginable...the earth being a sphere.
So long as the sun, moon and earth are not VERY close to being in a straight line, even a few degrees of misalignment is plenty to produce a full moon.
In the data you have, the moon was only 95% illuminated (it says that in the rightmost column) - so we don't even need a very close alignment to get the result you see.
But you have to think three-dimensionally here...and I know that's hard for you flat-earthers.
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D-did you try putting your moon over where the moon actually is, instead of creating an optical illusion to attempt to deceive the viewer? Because from what i'm seeing you're pretty damn close if not spot on to what is being seen if you would just put the moon where it's supposed to be. Because the moon doesn't appear full, it's only mostly full. I've still got that post, but your image is so much more useful than mine, so thank you for that.
Here you go. There should be a significant crescent.
(http://i.imgur.com/ILqulDT.png)
But according to the U.S. Naval Observatory the Moon was 95% full at that time:
Altitude and Azimuth of the Moon (http://aa.usno.navy.mil/cgi-bin/aa_altazw.pl?form=1&body=11&year=2017&month=4&day=8&intv_mag=10&state=NY&place=Lewiston)
Astronomical Applications Dept.
U.S. Naval Observatory
Washington, DC 20392-5420
LEWISTON, NEW YORK
o , o ,
W 79 02, N43 11
Altitude and Azimuth of the Moon
Apr 8, 2017
Eastern Daylight Time
Time Altitude Azimuth (E of N) Fraction illuminated
h m o o
...
18:30 11.0 94.9 0.95
...
Your image is still a bit off in just where the line of light should be upon the moon. I think the issue here is stemming from looking at this and thinking about it in 2D space, when the world is NOT 2D. Especially so when it comes to looking at the moon. Unfortunately I don't have the software or the know how in order to create anything approaching a rough analogue for this. The angle and calculations for this have to take into account all three axes in order to approach an appropriate model, and I'll admit I don't have the knowledge needed to do it nor properly understand most of the formula I come across. I can look and see that it exists, and it's explained, and these sites all agree with each other. But I'm sorry, I don't know how to make them work, because I don't understand them.
I do personally think, if you match the light side angle correctly, the amount seen by our viewer is reasonably close to what was seen, especially accounting for the increased lightness of the moon because of the time of day with the sun making the entire surface a bit more lit up because of it's interaction with the atmosphere. The same reason why the spots on the moon appear blue. Unfortunately I lack the required mathematics and working technical knowledge of the moon to do more than offer up a logical answer. To me everything looks kosher, it's a sight seen with some frequency, and I don't understand why you seem to find it impossible just because a 2D model and one set of angles can't make it exactly fit.
EDIT: 3DGeek also makes an excellent point down below in your scale being WAY off, which would also attribute to the sight lines not appearing quite right when you draw them to the moon such as you did. Again, most of these errors appear to come from thinking about and attempting to visualize the moon in 2D rather than in 3D. Lemme see if I can dig up the math on area of a half sphere, maybe that can shed some light on the angle needed to create a 95% illumination.
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(http://i.imgur.com/Kt852Km.png)
The scale of the diagram is off by a LOT - and it has one glaring error.
The moon has to be about ten times further from the Earth for the scale to be correct. As it is, it's a bit misleading. The part where it says "How the moon should appear" isn't too far off - but it's a side-view. The little piece of darkness at the bottom of the moon would be a few percent of the diameter of the moon - I dunno - maybe 10%, But because the image of the moon is a circle, the bottom 10% of the diameter is MUCH less than 10% of the area.
The part where it says "How the moon does appear" is incorrect though. It shows a perfect "full moon" with none of the moon in shadow.
The data that Tom found for that situation actually says that the moon was only "0.95" illuminated - 95%. So 5% of the moon's AREA was not bright...but 5% of the area is a lot more than 5% of it's DIAMETER - which is what is so horribly misleading about this picture.
As seen from Earth, the moon is never actually LITERALLY 100% illuminated - but the difference can be very small...to small to be really obvious to the naked eye observer.
This easily fits within "gut feel" expectations if you draw the diagram to scale and don't lie about "How the moon does appear"...that's a downright lie...fake...fake...fake!
So, please FE'ers don't lie and cheat to make a point. Look at the data you're quoting and please don't assume we're stupid...we're not.
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(http://i.imgur.com/Kt852Km.png)
The scale of the diagram is off by a LOT - and it has one glaring error.
The moon has to be about ten times further from the Earth for the scale to be correct. As it is, it's a bit misleading. The part where it says "How the moon should appear" isn't too far off - but it's a side-view. The little piece of darkness at the bottom of the moon would be a few percent of the diameter of the moon - I dunno - maybe 10%, But because the image of the moon is a circle, the bottom 10% of the diameter is MUCH less than 10% of the area.
The part where it says "How the moon does appear" is incorrect though. It shows a perfect "full moon" with none of the moon in shadow.
The data that Tom found for that situation actually says that the moon was only "0.95" illuminated - 95%. So 5% of the moon's AREA was not bright...but 5% of the area is a lot more than 5% of it's DIAMETER - which is what is so horribly misleading about this picture.
As seen from Earth, the moon is never actually LITERALLY 100% illuminated - but the difference can be very small...to small to be really obvious to the naked eye observer.
This easily fits within "gut feel" expectations if you draw the diagram to scale and don't lie about "How the moon does appear"...that's a downright lie...fake...fake...fake!
So, please FE'ers don't lie and cheat to make a point. Look at the data you're quoting and please don't assume we're stupid...we're not.
I linked the phase before, here it is.
I also checked your date, the moon phase was Waxing gibbous - Visible: 93%
https://www.calendar-12.com/moon_calendar/2017/april
The line between dark and Light is a gradient, just like on Earth. I'll try and draw a better scaled, gradient moon later, but it's going to look perfect to me.
BTW, I saw a moon this morning low that was well lit at about 6:30ish, opposite the sun rising, in the east.
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Just quick and dirty.
(https://image.ibb.co/fHdBPv/moons.png)
Left is the image from the website showing what the moon should look like, Right is a crop from :32 sec into the video, I adjusted contrast and brightness to make it easier to see. Looks pretty much spot on.
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Just quick and dirty.
(https://image.ibb.co/fHdBPv/moons.png)
Left is the image from the website showing what the moon should look like, Right is a crop from :32 sec into the video, I adjusted contrast and brightness to make it easier to see. Looks pretty much spot on.
Yes, we all agreed that the moon appears to be in 95% illumination. The problem is that it should not be in so much illumination.
Your calendar is based on what the moon does do, not what it should do.
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Just quick and dirty.
(https://image.ibb.co/fHdBPv/moons.png)
Left is the image from the website showing what the moon should look like, Right is a crop from :32 sec into the video, I adjusted contrast and brightness to make it easier to see. Looks pretty much spot on.
Yes, we all agreed that the moon appears to be in 95% illumination. The problem is that it should not be in so much illumination.
Your calendar is based on what the moon does do, not what it should do.
Why do you use the word 'Your' when referring to a calendar?
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According to your diagram a person atop the the earth will see the sun at 0 degrees, half intersecting the horizon, and the moon at 5 degrees.
However, the observation which was seen was that both the sun and moon were more than 5 degrees above the horizon. The Sun was 13.9 degrees above the horizon and the Moon was 11.0 degrees.
In Round Earth Theory the ecliptic is the earth's straight line alignment with the sun, and the moon deviates from it by 5 degrees. The observations make the explanation given impossible.
Here's the thing Tom, the answer is in the data you posted. Notice the latitude of Lewiston NY. Have a think about that.
I guess you noticed it, as it has since been removed from your posts. Odd.
For anyone wondering, Lewiston NY is at N 43.11.
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According to your diagram a person atop the the earth will see the sun at 0 degrees, half intersecting the horizon, and the moon at 5 degrees.
However, the observation which was seen was that both the sun and moon were more than 5 degrees above the horizon. The Sun was 13.9 degrees above the horizon and the Moon was 11.0 degrees.
In Round Earth Theory the ecliptic is the earth's straight line alignment with the sun, and the moon deviates from it by 5 degrees. The observations make the explanation given impossible.
Here's the thing Tom, the answer is in the data you posted. Notice the latitude of Lewiston NY. Have a think about that.
I guess you noticed it, as it has since been removed from your posts. Odd.
For anyone wondering, Lewiston NY is at N 43.11.
I don't know quite as much about this as I would like, but what does that mean for his statement here?
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I don't know quite as much about this as I would like, but what does that mean for his statement here?
The moon's orbital plane is tilted 5 degrees, but that doesn't mean the moon is always 5 degrees above or below Earth's orbital plane. Also, the 13.9 and 11 degree angles for the sun and moon at that time of day are orientated off Lewiston's latitude, which is basically leaning 43 degrees from the perpendicular orientation of Earth and moon's orbital planes. I'm sure this varies with the time of day and time of season, etc. TB didn't account for everything he should have.
Example: If the moon's plane is tilted 5 degrees, should you see the moon 90 degrees above the horizon on the equator? The 5 degree tilt of the moon's plane should become more of a 'left/right' difference as it nears the horizon when viewed from the equator.
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Just quick and dirty.
(https://image.ibb.co/fHdBPv/moons.png)
Left is the image from the website showing what the moon should look like, Right is a crop from :32 sec into the video, I adjusted contrast and brightness to make it easier to see. Looks pretty much spot on.
Yes, we all agreed that the moon appears to be in 95% illumination. The problem is that it should not be in so much illumination.
Your calendar is based on what the moon does do, not what it should do.
Wrong as you perpetually are. It's just as it should be. This whole thread is a straw man attempt to run from another thread. It looks, EXACTLY as it should and as ALL reputable sources say it should, as all math and accurate diagrams say it should. Trying to explain this to you is like trying to explain to a 5 year old that insists the world is moving and not the car why they are wrong. Shrink your "should be" picture to scale, move it where it should be and render a picture in a 3D graphic program. That's 3 1,2,3 dimensions... Savvy? NVM, obviously, no. Forget it, head to the beach, chant "it's flat, it's flat" while rocking back and forth.
I enjoy honing my skills on you, but really, how the hell do you continue to defile yourself defending a "theory" (laughable to even call it that) that can't predict even 1 natural, observable phenomenon? Your attempting to argue a global conspiracy that has miraculously kept the whole world in the dark about the true nature of the Earth but has somehow failed to notice the FUCKING MOON doesn't look right every month? So while we can predict and explain everything anyone can see looking up, with formulas that have been around for 100's-1000's of years. (Including the phases of the moon) somehow they failed to "bullshit" the moon as well?
So really, PUT UP OR STFU! You and your band of simpletons can't produce a map, the distance between any 2 locations, the direction the sun is going to rise or set tomorrow, what time it is in a given location, the phase of the moon, the next solar or lunar eclipse, why either of them occur. You can't scrounge together enough money or interest to pursue even the simplest of research or experiments humans have been performing for the same 1000's of years while riding horses and sail boats!
We have freaking pictures of the planet and you say the are Photoshoped? Really, well how about you all Photoshop us a picture of your world and put everything we see in it? While your at it, animate the moon, planets and stars and let's check it out.
No one avoided this charade as you said we would, it's just stupid. Something everyone see's all the time is "impossible" because you lack the skill to understand, illustrate or convey it? Pathetic...
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Top rant.
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Tom Bishop, multiple relaying moons on earth is possible too (http://forum.tfes.org/index.php?topic=6625.0)!
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Here you go. There should be a significant crescent.
(http://i.imgur.com/ILqulDT.png)
Just quick and dirty.
(https://image.ibb.co/fHdBPv/moons.png)
Left is the image from the website showing what the moon should look like, Right is a crop from :32 sec into the video, I adjusted contrast and brightness to make it easier to see. Looks pretty much spot on.
Yes, we all agreed that the moon appears to be in 95% illumination. The problem is that it should not be in so much illumination.
Your calendar is based on what the moon does do, not what it should do.
Your diagram is actually quite good. Looking at it, I think the moon would appear as it does in the video. Remember, the observer is at the part of the horizon line that is touching the earth. If you draw a line from the observer to the bottom of the terminator on the moon, you'd see the fraction of the angle at the fattest part of the moon that would be black.
What do you think the moon should look like to an observer at that place on the earth in your diagram?
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Tom Bishop, if the moon is actually artificial, just like the sun, and it is illuminating by itself, then using moon phases won't prove FE or oblate spheroid earth so easily!
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(http://i.imgur.com/Kt852Km.png)
Tom - the problem with your diagram is that you're still thinking like a flat earther. In RET, the horizon isn't a horizontal line like you've drawn it. It depends (critically) on the height of the eyepoint above the ground. I know you claim that the "horizon rises to the eye" - but that's not what RET says.
So if you choose a more reasonable eye height - then the horizon is depressed and your "horizon line" needs to be two lines that tip downwards.
You can't debunk RET using FE principles!
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To be fair, at sea level the horizon line is about as close to flat as possible, it's only like 1/25 of a degree, which would not differ from Tom's diagram in any significant way.
If the observer were at the top of a 10000 foot mountain (e.g. Mauna Kea on a clear day) then RET says you get an extra 1.77 degrees in each direction.
In any case, it is likely that the angles given by the Navy on that website are with respect to the astronomical horizon, which is a flat line.
So, listen to these words because I won't say them very often: Tom Bishop is right.
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Here's a diagram showing that the area shaded is as expected.
https://imgur.com/a/6AYee
I don't know if a direct image link works:
(https://i.imgur.com/qTASfh2.png)
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I would like to hear a FE explanation of how a completely round/full moon would ever be possible on FE. If sun & moon are both 3000 miles over flat Earth and we are looking up at both, this couldn't happen.
To see a completely round/full moon, we have to be looking at the moon with the sun's illumination behind us. I think I posted this question on another thread, without any answers. So I thought I would re-post it here.
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Tom Bishop is right.
About what? The moon being 11 degrees above the horizon being impossible with the RE moon's 5 degree inclination?
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Here's a diagram showing that the area shaded is as expected.
https://imgur.com/a/6AYee
I don't know if a direct image link works:
(https://i.imgur.com/qTASfh2.png)
The illustration in the middle is not 95% luminosity.
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Fascinating. Feel free to tell us more.
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a) What do you think 95% luminosity looks like?
b) What percentage luminosity does the image in the middle look like to you?
Remember, the observer is from the left on the image in the middle. I've arranged the moons this way so you can compare the bottom of the largest width of the lit portion of the moon in all three images.
I don't know how to easily take a ball model and rotate it until it matches, but the fact that these all line up, and the middle image came from your (very excellent) diagram says a lot to me.
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Tom Bishop is right.
About what? The moon being 11 degrees above the horizon being impossible with the RE moon's 5 degree inclination?
I hate to tell you all this, but it isn't the 5 degree inclination that you should be looking at. At the time and date given, the moon was a full 2 days before full. That means that the moon was around 26 degrees away from being directly opposite the sun.
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Tom Bishop is right.
About what? The moon being 11 degrees above the horizon being impossible with the RE moon's 5 degree inclination?
Sorry I wasn't clear about this. He was right about the horizon being essentially flat:
This was 3DGeek:
Tom - the problem with your diagram is that you're still thinking like a flat earther. In RET, the horizon isn't a horizontal line like you've drawn it. It depends (critically) on the height of the eyepoint above the ground. I know you claim that the "horizon rises to the eye" - but that's not what RET says.
My response was that Tom Bishop was right to draw the horizon flat, because if you are standing at the seashore, the horizon is within 1/20 of a degree of flat, so Tom Bishop's diagram is quite accurate.
I think the only confusion is that he is looking at his edge-on diagram of how much moon is illuminated, and states without evidence that it is not 95% illuminated. It isn't clear to me if he thinks it is more than 95% or less than 95% illuminated.
But, he was dead right about the horizon being pretty much perfectly flat, for an observer near sea level.
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Does anyone here understand how spheres can look like discs when seen head on? Is this site for real or is it a clever gimmick to sell advert click throughs? I have never seen so much snatch trying to minge up a simple piece of well renowned and easy to understand logic.
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Is this site for real or is it a clever gimmick to sell advert click throughs?
You don't know how internet advertising works, do you?
(hint: you would have to actually have ads).
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Does anyone here understand how spheres can look like discs when seen head on? Is this site for real or is it a clever gimmick to sell advert click throughs? I have never seen so much snatch trying to minge up a simple piece of well renowned and easy to understand logic.
This is the last place on earth I would expect to find any snatch. Except maybe at a Sys Admin party.
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Does anyone here understand how spheres can look like discs when seen head on? Is this site for real or is it a clever gimmick to sell advert click throughs? I have never seen so much snatch trying to minge up a simple piece of well renowned and easy to understand logic.
This is the last place on earth I would expect to find any snatch. Except maybe at a Sys Admin party.
Am sys admin, can confirm. :(
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(https://i.imgur.com/qTASfh2.png)
The illustration in the middle is not 95% luminosity.
*** I apologize for my rather approximate English. I'll try my best! ***
Why is it not 95% luminosity? the best way to know is to calculate! And it's so easy I wonder why you did not do it yourself already.
If alpha is the angle between the directions Moon-Observer and Moon-Sun, the illumination ratio of the moon is given by : r = (1+cos(alpha))/2 (this is an approximate formula, assuming the light rays coming from the moon are parallel, but the error is very small, and in any case in the right direction...)
Taking into account the data from the Naval Observatory you quoted on your post of Aug. 10, 01:16:53 AM (reply#10), we find that alpha = 26°. This gives an illumination ratio of... 0.9494. Hence the 0.95 given by the National Obervatory.
So I think everything is ok : contrary to your feeling, the moon phase in the video at the begining of the thread is perfectly compatible with Round Earth Theory.
Now the next question from me would be : is it compatible with Flat Earth Theory?...
I would very much appreciate if you - or any other authoritative flatearther - post a similar demo for the case of FET, of course with the same parameters (sun and moon altitudes and azimuts, same place and date,...). Is it possible? I don't know enough FE theory to be able to do it myself (in fact I know nearly nothing concerning FET...)
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Tom's ability to take on all comers is quite impressive with this data/video.
One can see clearly that both the sun and moon are relatively close, proving once again the FE and trust in your senses.
I've yet to find any information on this site that would lead me to RE belief.
Tom should be awarded slayer status. A whole new crew of RE folks should fire up because the regulars have been waylaid and massacred here.
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Tom's ability to take on all comers is quite impressive with this data/video.
One can see clearly that both the sun and moon are relatively close, proving once again the FE and trust in your senses.
I've yet to find any information on this site that would lead me to RE belief.
Tom should be awarded slayer status. A whole new crew of RE folks should fire up because the regulars have been waylaid and massacred here.
Is it an answer to or a comment on my post? If yes, it's a very unexpected and surprising one.
I demonstrate marthematically and without any possiblilty of rebuttal that Tom B. was wrong all along this thread, and as an answer (or comment) you sing his praises...
Well it's hard for me to grasp your logic. I see only one solution : your post has to be taken ironically. But in that case it would be very unfair to Tom! ...
One can see clearly that both the sun and moon are relatively close, proving once again the FE and trust in your senses.
I do not see in this video any clue whatsoever on how far or close the sun and moon are. If you did, please feel free to explain : I always enjoy learning new and cleaver things.
And my open question to FE on the compatibility of this video with FET is still valid!
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http://astro.unl.edu/naap/lps/animations/lps.html (see 95.3% at 7:18am) from https://physics.stackexchange.com/questions/26758/the-moon-during-the-day