Using moon viewing angles to demonstrate a distant moon

AtomPages

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Not that this directly addresses the shape of the earth, but a close moon (~3000 miles) is often associated with the overall theory of Flat Earth.

It is well known that two distant observers can estimate the distance to the moon using lunar parallax:

http://www.etwright.org/astro/moonpar.html
External Quote:
Parallax is the apparent shift caused by viewing an object from two different vantage points. You can see it easily just by alternately blinking your left and right eye. Parallax is also evident in the apparent position of the Moon viewed from two distant points on the Earth, or from the same point six hours apart. Hipparchus, in the second century BC, derived a very good estimate of the distance to the Moon using lunar parallax.
However, I propose a second way to confirm this is with two observers with high powered zoom (telescope or Nikon Coolpix).

Here is a shot taken from Alaska, alternated with one taken from Brisbane, Australia, within an hour of each other.

AlaskaBrisbane.gif


First, notice that the depth of the moon can be detected by observing that the craters on the edges only move about 3 pixels per frame, while the ones near the middle move about 14 pixels. Photoshop actually tells me 14.87 pixels if I use the ruler tool. This would indicate a depth to the moon.

Now, the diameter of the moon in this image is 1131 pixels, making its circumference 3553 pixels. Now, my thinking is that if you find a point near the center of the moon and note how many pixels it moves between the two shots, you can calculate the angle difference between two observers as:

Angle difference = pixel shift * 360° / pixel circumference. I think this should be essentially the parallax angle, and then using the parallax distance method, you should be able to estimate the distance to the moon. Let’s check.

The angle I calculate here is 1.506610506°, between two cities that are 6898mi apart. Using that distance as the arc length, we find that the chord length is 6058mi for the base line of our isosceles triangle. Using the parallax method, we find our estimated distance to the moon is:


Distance to moon = 1/2 base line distance / tan(1/2 parallax angle)
Distance to moon = 1/2 * 6068 mi / tan(0.7533052532° * ∏ / 180°)
Distance to moon = 230,369.98 mi

This is 3.57% away from the current estimated distance of 238,900 mi.

Using a second comparison from Santiago, Chile, to Maryland, I tried the technique again:

MarylandSantiago.gif


Circumference = 8507.43 px
Shift near middle of the moon = 28.07 px
Distance between observers = 5091 mi
Chord distance between observers = 4747 mi
Calculated parallax angle = 1.187808368°

Distance to moon = 228,970.71 mi
Error of 4.16%
 
I like that your axes of comparison are roughly north-south aligned, which means the distances can also be determined via Gleeson's map. In that way, the distance calculation doesn't depend on the shape of the Earth like the 6-hour method does.

If you know the moon's distance, you get to the shape of the Earth by comparing moon elevations from various places.
 
Jos Leys has a series of elegant and beautiful videos based on bearings to the sun as seen from a flat Earth, demonstrating the the local Sun is nonsense. The same sort of thing can be done with the moon, as bearings taken at the same time point in different directions and do not converge, if a flat Earth is assumed. Leys seems to prefer doing the Sun, and just for the beauty of the patterns created I recommend checking out his Sun-based videos. However, here's one where he does the moon during an eclipse, in one minute of video -- not as beautiful as some of the Sun ones, but since this is a moon-based thread, here ya go.

Source: https://youtu.be/EixzcOdp1uo


Teed up to the beginning, as it is short and the whole thing is needed to make the point strongly. The "Too Long, Didn't Watch" instant version is this image:
Capture.JPG

Huh. The lines of sight to the moon to not converge at all on a flat Earth.


Info on time, bearings used, etc for those wanting those details without following the link.
1677684429942.png
 
@AtomPages

Sorry, I am not getting it (the calculation in your OP). Could you please elaborate this "Angle difference = pixel shift * 360° / pixel circumference"?

I might sound dumb, but I don't understand the calculation and want to know.
 
@AtomPages

Sorry, I am not getting it (the calculation in your OP). Could you please elaborate this "Angle difference = pixel shift * 360° / pixel circumference"?

I might sound dumb, but I don't understand the calculation and want to know.
angles can be measured in radians by equating an angle with the length of the subtended arc of a unit circle. that means, if you know the length of an arc with respect to the radius, or with respect to the full circumference of the circle, you can figure out the angle as a proportion.

Since the "pixel shift" is perpendicular to the observer, it's like looking side-on at the circumference, and that pixel shift describes that arc.

The calculation yields the angle at the center of the moon that has its legs pointing at either observer on Earth.
 
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@Mendel

Very clear. I don't know why I was thinking about the circumference of the in-plane circle and not the side-on.

Thanks Mendel
 
Incidentally, for those looking for evidence that the moon is indeed spheroid, these images from different vantage points provide confirmation as well. If you map various points on the surface between images (the red lines), you see that the points near the center move the most, and the points near the edges move the least.

AlaskaBrisbane_sync_annotated2_20230226_0900UTC.gif

AlaskaBrisbaneAnnotations.jpg


Just like on a basketball.
Basketball_annotated_20230301_160528.gif

(I'm not sure why my basketball gif is not animating, but I used the same method to generate those red lines)
BasketballAnnotations.jpg


I was also able to use the above methodology to estimate the distance from the camera to the basketball. I placed the tripod 42" away from the basketball, then moved it 2" to the right. This resulted in an estimate of a 40.45" distance from camera to basketball, 3.69% lower than the actual distance. Pretty similar to the moon estimates.
 
Or just to get folks to see it without the measuring and red lines, I have found this gif useful. Source is somebody posted it somewhere and I saved it.

moon shows it is round.gif
 
I like that one better, it gives you two ways to see the sphere -- the wobble and the lighting following the shape. Saved.
 
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Very nice!

I wonder what chatGPT would give us if you say "please make a sequence of perfect photos of the moon at a fixed place, every night (at XX:XX), for 1 month". Perhaps someone with an account can try :cool:
 
Distance to moon = 1/2 base line distance / tan(1/2 parallax angle)
Distance to moon = 1/2 * 6068 mi / tan(0.7533052532° * ∏ / 180°)
Distance to moon = 230,369.98 mi

This is 3.57% away from the current estimated distance of 238,900 mi.
Actually it is within the range of distances to the moon: that distance is not fixed because its orbit is not a perfect circle. The maximum and minimum distances, according to NASA, are 252,088 miles and 225,623 miles. I'm not sure but I assume that those distances are centre-to-centre, so the distance you are measuring, from surface to surface, will be somewhat less.

Very nice!

I wonder what chatGPT would give us if you say "please make a sequence of perfect photos of the moon at a fixed place, every night (at XX:XX), for 1 month". Perhaps someone with an account can try :cool:
I don't think ChatGPT can do things like that, can it? It could only provide references to things that already exist.
 
Very nice!

I wonder what chatGPT would give us if you say "please make a sequence of perfect photos of the moon at a fixed place, every night (at XX:XX), for 1 month". Perhaps someone with an account can try :cool:
But of course the moon is NOT in a fixed place at the same time every night, so that's doomed to fail! ;) You just want to break chatGPT's little heart, don't you.
 
But of course the moon is NOT in a fixed place at the same time every night, so that's doomed to fail! ;) You just want to break chatGPT's little heart, don't you.
I'd assumed that to mean "from a fixed place," but perhaps I misunderstood.
 
I'd assumed that to mean "from a fixed place," but perhaps I misunderstood.
Well, if you want a picture of the moon at the same time every night for a month, your "fixed place" is going to have to be somewhere out in space, so that doesn't help either. You can get it at the same time every night, but only if you're willing to travel around the world to a different time zone each time.
 
Well, if you want a picture of the moon at the same time every night for a month, your "fixed place" is going to have to be somewhere out in space, so that doesn't help either. You can get it at the same time every night, but only if you're willing to travel around the world to a different time zone each time.
I reluctantly yield the point to you, based on you being right! A most unfair way to argue.
 
Well, if you want a picture of the moon at the same time every night for a month, your "fixed place" is going to have to be somewhere out in space, so that doesn't help either. You can get it at the same time every night, but only if you're willing to travel around the world to a different time zone each time.
Shouldnt there be a time of year when the Moon never sets seen from near one of the poles?
 
Shouldnt there be a time of year when the Moon never sets seen from near one of the poles?
That happens monthly.
A lunar standstill or lunistice is when the moon reaches its furthest north or furthest south point during the course of a month (specifically a draconic month of about 27.2 days). The declination (a celestial coordinate measured as the angle from the celestial equator, analogous to latitude) at lunar standstill varies in a cycle 18.6 years long between 18.134° (north or south) and 28.725° (north or south), due to lunar precession. These extremes are called the minor and major lunar standstills.

Right now we're closer to a major lunar standstill, so any place pole-ward of a polar circle should see a 24-hour moon once a month.
 
Jos Leys has a series of elegant and beautiful videos based on bearings to the sun as seen from a flat Earth, demonstrating the the local Sun is nonsense. The same sort of thing can be done with the moon, as bearings taken at the same time point in different directions and do not converge, if a flat Earth is assumed. Leys seems to prefer doing the Sun, and just for the beauty of the patterns created I recommend checking out his Sun-based videos. However, here's one where he does the moon during an eclipse, in one minute of video -- not as beautiful as some of the Sun ones, but since this is a moon-based thread, here ya go.

Source: https://youtu.be/EixzcOdp1uo


Teed up to the beginning, as it is short and the whole thing is needed to make the point strongly. The "Too Long, Didn't Watch" instant version is this image:
View attachment 57999
Huh. The lines of sight to the moon to not converge at all on a flat Earth.


Info on time, bearings used, etc for those wanting those details without following the link.
View attachment 58000
Those Jos Leys videos are outstanding. Such clear demonstrations.
 
A0BE8693-2832-4EAB-86FF-DA4FF427A5A7.png

So, I made the geometry in the geogebra app. The line segment AB is the distance between the Earth and the Sun. AC is the distance between the Earth and the Moon. BC is therefore the line between the Sun and Moon. I made a line perpendicular to the segment BC, which then is the terminator line on the Moon. There is then a relationship between the angle between the Sun and Moon as seen from Earth (here I put it at 45 degrees) and the position of the lunar terminator as seen from Earth, represented by the angle between AC and the line perpendicular to BC.

I worked out the mathematical relationship to be:

Phi = arctan( 1/sin(theta)*(cos(theta) - AC/AB))

Where phi is the aforementioned angle ( 25.6 degrees in this picture) and theta is the Sun to Moon angle (45 degrees here).

In geogebra one can move the Sun and Moon around and see how this angle changes. If the sun and moon were similar distances to the observer it would look very different than it really does.

For example, when the sun/moon angle is 90
Degrees, the above formula reduces to:

Phi = arctan(-AC/AB)

If AC and AB were equal then phi = -45 degrees and we would see a 3/4 Moon or so. The real distance ratio is close to 400, so phi would be about -8.5 arc minutes, so basically we would see about half moon, which is what is observed.

It takes just a few seconds to set this up in geogebra and convince yourself that the Sun has to be substantially farther than the Moon for us to see what we do. Unless of course the Moon is a self-luminous object whose phases just remarkably happen to coincide with what we would exactly see from a spherical object illuminated by a distant light source as it moved around the Earth.
 
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