For the April 8th, 2024 eclipse, I had an idea to track the angular speed of the moon and then compare it with the theoretical average angular speed of the moon given its orbital period. In the FE community, there are some who claim that the object that occludes the sun during a solar eclipse is not the moon, but some other object, and this was intended to be some evidence to the contrary. However, I'm having some trouble getting my observational data to line up with the theoretical data: it's close, but not close enough to be convincing. I infer that there's a flaw in my logic somewhere, and I'm hoping you can help me detect it.

I started with this thread as background: https://www.metabunk.org/threads/ge...-the-radius-of-star-trails.12729/#post-282871

These are the originals, so they are mirror images due to the telescope's mirror. They were taken at 14:20:06 and 15:00:20 EST, a total of 2414 seconds apart.

They were taken freehand with a camera through a telescope eyepiece, so the exact focal length is unknown. However, we can know the expected angular diameter of the moon at the time of the eclipse based on its distance of 220,752.35 mi from earth, and the moon's diameter 2,159.2 mi. Using that, we can reverse engineer the focal length in pixels using the x = f*tan(a) equation.

I used this concept to determine that the angular diameter of the moon at the time of the first image was 0.5603976679° (angular diameter = atan(2,159.2 mi / 220,752.35 mi).

I then resized and overlaid the two images, lining up the sun spots for a guide, resulting in this composite (I also mirrored the image so it would match the non-mirrored sky):

What you see here is the second image overlaid on the first image, and the second image is partially transparent in order to see through the moon.

I then measured the pixel diameter of the moon using Photoshop, and found it to be 2570 px. Plugging this into the formula x = f*tan(a), we get:

f = x / tan(a)

f = 2570 px * tan(0.5603976679°)

f = 262745.4476 px

Then I measured the pixel distance between the centers of the moon in the overlaid image (shown as the line above), and found that to be 1278.73 px.

Therefore, I plugged in:

x = f * tan(a)

a = atan(x/f)

a = atan(1278.73 px / 262745.4476 px)

a = 0.2788449976°

I believe this means the angle that the moon moved in 2414 seconds was 0.2788449976°, which is a daily rate of 9.980202068°.

Now, again, ~10° / day is at least the same order of magnitude as the expected ~14.8° / day of the moon, but it's also outside of the expected range of daily angular motion of the moon, and thus I think is not very convincing. (I do realize that a flat earther would not be swayed by any of this logic, but at this point it's more of an exercise for myself)

I then tried the same methodology using screen captures of Stellarium at the same times, and got a rate of 9.592043251° / day. Stellarium clearly shows the moon traveling around at the correct positions, so I'm assuming my methodology is incorrect somewhere.

Finally, I used a similar methodology on a series of pictures taken directly from my camera without a telescope, using a 55mm focal length (so as not to have to reverse engineer the focal length). With this set, I got an average of 10.741° / day, still seemingly too low.

Can you help me spot the error?

I started with this thread as background: https://www.metabunk.org/threads/ge...-the-radius-of-star-trails.12729/#post-282871

The idea here is:You look at polaris. Say there's a star 20° to the right of it. The angle a = 20°

Imagine from above, the distance directly to the star is c, as it's on the sphere.

The distance of the star from the line between you and Polaris is c•sin(a)

The distance along that line (the perpendicular distance) to that star is c.cos(a)

A standard rectilinear projection of the star onto a camera with a focal length of f pixels, giving a separation of the star from polaris in pixels of x means (similar triangles)

x/f = c•sin(a)/(c•cos(a))

c is irrelevant, so

x = f•sin(a)/cos(a)

x = f•tan*(a)

- I have two eclipse images with timestamps, and if I know the focal length and can measure the pixel distance between the center of the moon in the two images, I should be able to calculate angle "a" between myself and the two locations of the moon in its orbit, using x = f*tan(a).
- I think that this angle divided by the time between the images should result in the same angular speed as that of the moon (the number of degrees the moon travels in its orbit per unit time).
- Given that the moon was at perigee (closest point to earth) just one day before the eclipse, this should be close to the higher value of its expected angular speed (see Kepler's second law)
- The range of the moon's angular speed is said to be 11.6° to 14.8°, so I would expect the result to be close to 14.8° / day. (https://cseligman.com/text/sky/moon... of 13.2 degrees,degrees per day near perigee.)

These are the originals, so they are mirror images due to the telescope's mirror. They were taken at 14:20:06 and 15:00:20 EST, a total of 2414 seconds apart.

They were taken freehand with a camera through a telescope eyepiece, so the exact focal length is unknown. However, we can know the expected angular diameter of the moon at the time of the eclipse based on its distance of 220,752.35 mi from earth, and the moon's diameter 2,159.2 mi. Using that, we can reverse engineer the focal length in pixels using the x = f*tan(a) equation.

I used this concept to determine that the angular diameter of the moon at the time of the first image was 0.5603976679° (angular diameter = atan(2,159.2 mi / 220,752.35 mi).

I then resized and overlaid the two images, lining up the sun spots for a guide, resulting in this composite (I also mirrored the image so it would match the non-mirrored sky):

What you see here is the second image overlaid on the first image, and the second image is partially transparent in order to see through the moon.

I then measured the pixel diameter of the moon using Photoshop, and found it to be 2570 px. Plugging this into the formula x = f*tan(a), we get:

f = x / tan(a)

f = 2570 px * tan(0.5603976679°)

f = 262745.4476 px

Then I measured the pixel distance between the centers of the moon in the overlaid image (shown as the line above), and found that to be 1278.73 px.

Therefore, I plugged in:

x = f * tan(a)

a = atan(x/f)

a = atan(1278.73 px / 262745.4476 px)

a = 0.2788449976°

I believe this means the angle that the moon moved in 2414 seconds was 0.2788449976°, which is a daily rate of 9.980202068°.

Now, again, ~10° / day is at least the same order of magnitude as the expected ~14.8° / day of the moon, but it's also outside of the expected range of daily angular motion of the moon, and thus I think is not very convincing. (I do realize that a flat earther would not be swayed by any of this logic, but at this point it's more of an exercise for myself)

I then tried the same methodology using screen captures of Stellarium at the same times, and got a rate of 9.592043251° / day. Stellarium clearly shows the moon traveling around at the correct positions, so I'm assuming my methodology is incorrect somewhere.

Finally, I used a similar methodology on a series of pictures taken directly from my camera without a telescope, using a 55mm focal length (so as not to have to reverse engineer the focal length). With this set, I got an average of 10.741° / day, still seemingly too low.

Can you help me spot the error?