Geometry question about the radius of star trails

AtomPages

New Member
I have a project going related to Flat Earth debunking, but I want to get some geometry right first.

It is commonly observed that star trails near the poles are circles, while a star trail right on the celestial equator is a straight line, since it is a great circle on the celestial sphere. I have some pictures that demonstrate this, including some basic calculations for predicting the pixel radius of the star trails near the poles: https://astronomy.stackexchange.com...s-the-equation-for-the-radius-of-a-star-trail

However, what I'm missing is that near the equator, my method of predicting the star trail radius breaks down (as the trails approach a straight line, my predictions diverge). I imagine this has something to do with projecting the great circle onto the observer's field of view, but I don't know how to adjust my formula to account for this. Are there any geometry whizzes that could help with a general formula for the radius of a star trail in terms of the star's declination, even down to the celestial equator?
 

Mendel

Senior Member.
I have been studying star paths, and have successfully used the following equations to map star paths in a multiple exposure photograph onto circles of a radius determined by the star's declination and my camera focal length and sensor size:

I reason that the radius of a star trail in a photograph is the complement of the star's declination, or Abs(Nearest Pole Declination - Star Declination).The degrees / pixel of a digital photograph can be determined by the equation: Arctan( (Sensor Height * Object Height) / (Focal Length * Image Height) ), where the Object Height in this case is 1 pixel in order to get the value of 1 pixel in degrees. The derivation of this equation can be found here, and I can go into more detail if needed.

Combining these concepts, I calculate the arc length between the nearest pole and the star, and then multiply by the calculated Degrees / Pixel in order to overlay a circle of this pixel radius on the image.
Content from External Source
1. "degrees per pixel" varies across the picture, it's higher near the center than at the edges because of how rectilinear projection works

2. Distance to the circle matters. The equatorial circle looks like a straight line because we're in the middle of it—kinda how a coin looks straight when you view it sideways.

I'm not entirely clear on what you want to achieve: what is your input data, and what should the output be? Would you like to be able to draw star paths on any photograph?

(I'm curious as to how that'd help with FE debunking.)
 

AtomPages

New Member
1. "degrees per pixel" varies across the picture, it's higher near the center than at the edges because of how rectilinear projection works
Would this also be referring to lens distortion? If so, I agree that there is some minor distortion closer to the edges with a 55mm shot, though this appears to be minimal enough that it shouldn't affect the results significantly. So by "degrees per pixel", perhaps what I mean is an approximate degrees per pixel.
2. Distance to the circle matters. The equatorial circle looks like a straight line because we're in the middle of it—kinda how a coin looks straight when you view it sideways.

I'm not entirely clear on what you want to achieve: what is your input data, and what should the output be? Would you like to be able to draw star paths on any photograph?
Yes, you have it. The inputs are the camera focal length, camera sensor height + image height (which together determine the actual pixel size), and the star's declination. The output would be the pixel radius for those camera settings of the expected star trail, so that I can superimpose a star path circle on any photograph given those inputs.

Incidentally, I was playing around with this some more and found that dividing my previous prediction by the cosine of the complement of the star's declination (the declination compliment is apparently called the Codeclination) actually yields near exact results for every star trail picture I've taken. I'll post some examples later. I'm not sure of the theoretical validity of this, but it's an interesting find. Intuitively, this could make sense, because as the declination approaches 0, 1/cos(Codeclination) approaches infinity, hence a circle of infinite radius, or a straight line. As the declination approaches 90, 1/cos(Codeclination) approaches 1, which would explain why my previous method worked better closer to the poles but diverged closer to the celestial equator.

With this adjustment, my formula becomes:

Circle Radius in pixels = Codeclination / ( cos(Codeclination * π / 180) * (180/π) * atan(Sensor Height in mm / (Focal Length in mm * Image Pixel Height) ) )
(I'm curious as to how that'd help with FE debunking.)
This part of the project involves comparing the shape of star trails on a spinning globe vs. the shape of star trails on a flat plane with a spinning dome. In the former, I believe I can say that every star trail should be a circle except for exactly on the celestial equator, which should appear to be a straight line. In the latter, star trails should be an oval or ellipse unless you're exactly at the north pole, because if you're away from the north pole, the stars will be getting further away from you as they go lower in the sky, and perspective should "squash" the circular trail.

As a matter of demonstration, I hope to overlay circles on star trail images of stars at various declinations. I realize math won't sway most flat earthers, but for my own completeness, I wanted to be able to back up the radius of these circles with actual theory. The thought being that it's one thing to overlay a circle on a star trail photo and be able to explain it away as "refraction" or something, and another to overlay a circle whose radius is actually predicted by an equation that only makes sense on a spinning globe.

I also have a series of photographs taken by just spinning my camera around on the tripod at various angles from horizontal in a room with markings on the wall, demonstrating that the "star trails" of objects in the image also exactly match this approach. The axis of spin is vertical, but the camera is able to be angled from horizontal to vertical, allowing its field of view to simulate various latitudes on the globe. Whatever point is dead center would be the "declination" of that marking. I measured the "declination" of these markings using a rudimentary sextant attached to the camera, and can plug in the codeclination, sensor height, image height, and focal length to predict the circular paths with near exact accuracy. This should demonstrate that the paths things take from the perspective of a spinning observer exactly match what we observe in the star trails.

I'll have to think through whether dividing by the cosine of the codeclination actually makes theoretical sense, because my whole purpose is to back this up in theory. I'd ideally like to be able to derive this part of the equation somehow.
 
Last edited:

Ann K

Senior Member.
Star trails from a globe, OK. But there seems to be no likelihood of a lot of flat-earthers being convinced, because they don't all agree on what the stars are doing.
This part of the project involves comparing the shape of star trails on a spinning globe vs. the shape of star trails on a flat plane with a spinning dome.
Fixed stars on a dome that spins is indeed one of the favorites, but what do you do with the people who never looked at the sky from the Southern Hemisphere and who dig in their heels with idiocy such as this?
In truth, however, the Earth is a prolonged flat aircraft and all the celebrities and also other celestial objects rotate East to West around Polaris, the only non-moving celebrity overhead positioned perfectly in line straight above the North Pole. The so-called "South Pole" and South Pole star "Sigma Octantis" are both myths
Content from External Source
https://www.flatearthscience.org/flat-earth-star-trails-explained/

Using facts to combat people who only believe "their own facts" is a modern problem in many fields, but a very frustrating problem. Fortunately flat-earth nonsense is a relatively minor facet, and yours is a valiant attempt to find a teaching tool. Thanks.
 

Mick West

Administrator
Staff member
This should demonstrate that the paths things take from the perspective of a spinning observer exactly match what we observe in the star trails.

Surely all you need to do is observer that the angle between any star and the celestial pole (approximately Polaris) remains the same? The "radius" is really just that angle. 0° at the celestial poles, 90° at the celestial equator.

But calling it a radius does not really make sense, and what you are doing by dividing by the cosine is essentially projecting it onto 2D - and Flat Earthers generally don't understand projection.
 

AtomPages

New Member
Surely all you need to do is observer that the angle between any star and the celestial pole (approximately Polaris) remains the same? The "radius" is really just that angle. 0° at the celestial poles, 90° at the celestial equator.
What an elegant way of communicating what I was trying to get at, thanks, Mick.

But calling it a radius does not really make sense
Right, good point, it's not actually a radius. In my head I was trying to capture the radius of the 2D circle path, but that's obviously not the same as the angle between the star and the pole, which is I think why my original formula failed.
what you are doing by dividing by the cosine is essentially projecting it onto 2D - and Flat Earthers generally don't understand projection.
Is dividing by the cosine of that angle actually a known projection method? I'd still be interested in capturing for my own thinking exactly what represents the 2D path of the stars.
 

Mick West

Administrator
Staff member
Is dividing by the cosine of that angle actually a known projection method? I'd still be interested in capturing for my own thinking exactly what represents the 2D path of the stars.
Let's say the stars are on a celestial sphere, with a very large radius c (where c is, say 1,000,000,000 miles, not important, just that it's big)

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)

Of course, you don't need to go through that, you can just look at the focal point, and the focal plane (the sensor), in a camera, and you'll see x=f•tan(a) arises. But it might be useful to imagine projecting an actual object, and thinking about where the /cos(a) comes in.

Also, using focal length in pixels simplifies things. A "pixel" is the physical spacing of sensor elements (or that, adjusted for the eventual image resolution)
 

AtomPages

New Member
Mick, thanks for taking the time to write up this example. This makes it amazingly clear to me. It turns out I was using the concept of x = f * tan(a) and the pixel focal length elsewhere in my work, but didn't make the connection here. This generates accurate circle overlays, and I now understand the math behind it.
 

Attachments

  • AludraStarTrail_Comparison-20221020_055132 copy.jpg
    AludraStarTrail_Comparison-20221020_055132 copy.jpg
    103.1 KB · Views: 12
  • BetelgeuseStarTrail_Comparison_20221018_054352 (1).JPG
    BetelgeuseStarTrail_Comparison_20221018_054352 (1).JPG
    122.4 KB · Views: 11
  • CassopeiaStarTrail_Comparison_20221021_051228 copy.jpg
    CassopeiaStarTrail_Comparison_20221021_051228 copy.jpg
    77.7 KB · Views: 9
  • PolarisStarTrail_Comparison_20220429_2146 (1).jpg
    PolarisStarTrail_Comparison_20220429_2146 (1).jpg
    200.5 KB · Views: 9
Top