Honing an approach for analyzing the angular speed of the moon during an eclipse

[AtomPages]

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

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)

The idea here is:

• 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.)

Here are the two images:

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?

 

[Mendel]

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.

40 minutes elapsed, yet your maths does not take the motion of the observer due to Earth's rotation (15⁰/hour) into account.

 

[AtomPages]

40 minutes elapsed, yet your maths does not take the motion of the observer due to Earth's rotation (15⁰/hour) into account.

Interesting, so the physical distance of the observer to the moon? I rotated the second image to match the sun spot location on the sun, so that at least accounted for the rotation of the observer, but I didn't think about how the transposition in space might affect the calculations.

 

[Mendel]

Interesting, so the physical distance of the observer to the moon? I rotated the second image to match the sun spot location on the sun, so that at least accounted for the rotation of the observer, but I didn't think about how the transposition in space might affect the calculations.

Not just the distance, but also the parallax effect. The Earth's surface in effect "overtakes" the moon, so parallax would appear to slow the apparent speed of the moon down.

(I can sketch it for you if it's still unclear.)

 

[AtomPages]

(I can sketch it for you if it's still unclear.)

Yes, please. I think I have an idea of what you mean, but that would help. I'm also not sure how to factor it into the math yet.

 

[AtomPages]

Not just the distance, but also the parallax effect. The Earth's surface in effect "overtakes" the moon, so parallax would appear to slow the apparent speed of the moon down.

Also, if my math is correct, there should only be an 85.95 mile chord length difference between the observer's location on the surface of the earth, given 40 minutes of rotation. So it does seem the parallax angle should be fairly small at first glance.

 

[Mendel]

Also, if my math is correct, there should only be an 85.95 mile chord length difference between the observer's location on the surface of the earth, given 40 minutes of rotation. So it does seem the parallax angle should be fairly small at first glance.

your number for circumference is actually the radius, you need ×2π



That's with the moon set to 10,000 km, at 370,000 km the angle is ~0.3⁰, which is big enough to be a problem (but I did not use exact values).

Orbital motion of Earth vs. the sun adds another ~0.03⁰, but I think in the other direction.

You may be able to use the attached file with https://www.geogebra.org/calculator .

 

[AtomPages]

your number for circumference is actually the radius, you need ×2π

Thanks, good catch. That makes it 540 mi chord length.

That's with the moon set to 10,000 km, at 370,000 km the angle is ~0.3⁰, which is big enough to be a problem (but I did not use exact values).

With the above adjustments, I am also getting about 0.28° parallax angle. Maybe this plays into the angular speed in a way that reduces the apparent angular speed by as much as 4° / hour, but it's not intuitive to me. (For example, just adding the parallax angle to 9.98° / day is still nearly 5° / day off of the expected angular speed)

 

[Mendel]

With the above adjustments, I am also getting about 0.28° parallax angle. Maybe this plays into the angular speed in a way that reduces the apparent angular speed by as much as 4° / hour, but it's not intuitive to me.

Well, in your OP you have the moon diameter at 0.56⁰. Now consider that the moon actually moved 0.28⁰ degrees further than you saw, i.e. half a diameter more—won't that boost her speed considerably?

 

[AtomPages]

Well, in your OP you have the moon diameter at 0.56⁰. Now consider that the moon actually moved 0.28⁰ degrees further than you saw, i.e. half a diameter more—won't that boost her speed considerably?

Oh, of course. In my head I was adding the parallax angle to the daily rate, but that's nonsensical. It's added to the observation angle.

Also, I think I had accidentally doubled the parallax angle -- I calculate it to be atan(540 mi / 220,752.35 mi) = 0.14°.

Using the above calculations, the moon had appeared to move 0.2796° roughly Westward across the sun during the 40 minutes between the photos. If I assume the parallax applied directly against this motion (it wouldn't be, but for simplicity), and add 0.14° to this, I estimate that the moon actually moved 0.4198°. This translates to a daily motion of 14.98°, which is within reasonable limits of error from the expected ~14.8° / day, especially since the moon's motion was not directly against the rotation of the earth.

 

[JMartJr]

For those who want to skip the math, there is a more direct proof that it is in fact the moon blocking the Sun during an eclipse: photos exist of the eclipse with the moon visible by Earthshine:

Earthshine during the Total Solar Eclipse of August 2017 The Moon lit up due to earthshine, parts of the Earth not experiencing the eclipse reflecting back on the Moon. Taken from Right Lane Ranch, Spray, Oregon.

Content from External Source

Source: https://www.flickr.com/photos/rubixdead/36787233095​

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But enjoying the math part, sorry to interrupt... carry on.​

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[AtomPages]

photos exist of the eclipse with the moon visible by Earthshine

I'd say "Flat earthers wouldn't give this picture a second glance because CGI", but then I suppose doing math as a proof is also fruitless

Also, this math was all because I wasn't in the path of totality. It would have been my first choice to have taken a shot myself of the earthshine during a solar eclipse.

 

[FatPhil]

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 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

Why do you take the atan to find an angle that you are only going to take the tan of? tan(a)=2159.2/220752.35, just use that.

? 2750 * 2159.2 / 220752.35
 = 26.89801

Just from eyeballing (and if you'd have used radians rather than degrees this would have been obvious), atan(2xxx/2xxxxx) is going to be about 1/100 rad, and tan(about 1/100 rad) is going to be about 1/100. So you know your answer should be about 27. You've probably divided rather than multiplied.

Note, for angles as small as these, the approximation angle=atan(d/r) is worse than the do-nothing operation of angle=d/r. The correct value is 2*atan(d/2r), and the errors are both third order, so you're good for 4 significant digits, which is sufficient (atan(d/r) is 1/4x^3 low, d/r is 1/12x^3 high, so with the latter you might keep 5 digits of precision). So there's no need to bring tan/sin/cos into things at this point. And of course you can always drop any tan(atan()) operation, no matter what the magnitude of the angle.
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