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I'm making a technological project. The goal is to write a program that gets an image of the nightsky and predicts the hour in which the photo was taken.

I implemented a well-known method using the big dipper, but it's not enough. I don't want to limit my program to only work when the big dipper is in the image.

When I read about right ascension, my gut told me seeing a star in the sky and knowing its right ascension (RA) would help me deduce how many hours ago the sunset happened, though I couldn't explain how.

I'm not comfortable enough with this coordinate system, so I need help answering this question: Say I konw the location a photo was taken, and I can identify the celestial objects in the sky and get their RA, can I make some calculations to konw/approximate the hour? If so, how?

Thanks in advance!

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    $\begingroup$ What other information is included in the photograph? Since a camera can be tilted at any angle, you cannot depend on knowing what direction is "up" just based on the "bottom edge" of the photo (unless you want to make that a requirement). In other words, the photo needs some reference, such as the horizon, to determine which direction is "up". Then you could approximate the time by knowing the orientation of the constellation in the photo. You also need the date and latitude since those are the other variables that lead to an answer about the time. $\endgroup$ – JohnHoltz Mar 31 '20 at 22:53
  • $\begingroup$ And of course a program that needs the Big Dipper is going to be useless for those of us who live in the southern hemisphere. ;) $\endgroup$ – PM 2Ring Mar 31 '20 at 23:12
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    $\begingroup$ Does this answer your question? Is it possible to determine an observer's position on Earth from a photo of the night sky and the time of the shot? $\endgroup$ – usernumber Apr 1 '20 at 7:10
  • $\begingroup$ Hi, your information was relatively helpful, especially konwing that it is possible. About tilted cameras - for now we require the photos to be aligned, though we might make an app for this which will save this angle using the gyro device on the phone. John, you said konwing the date and latitude I can get the answer. Do you know how will I do that? Can you write the equations down or refer me to related sources? Thanks! $\endgroup$ – omri slovatik Apr 1 '20 at 8:46
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    $\begingroup$ Hi, thanks again for the reply. For now let's put aside the engeering challenges such as taking into account the distortion of the photograph, there are algorithms for this. For now I just want to know what are these back-calculations I need to make to get the time of the photo. $\endgroup$ – omri slovatik Apr 1 '20 at 22:50
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Will you know the date and location from where the image is taken? If you are sucessfully using the big dipper already you probably using this info. There is a "degeneracy" between date and time. The sky rotates once a siderial day (23hrs 56 min) and once a siderial year (which is 20min longer than tropical year: sun to same position). So the sky tonight at say 20:00 is the same as the sky at 20:04 the previous day. In two weeks that's 1 hour. So you need the date.

At the same time (say Universal Time == London time) at each longitude the observer is facing a different direction. So 15 deg long is equivalent to 1 hour in time. So you need longitude. Others have alluded to latitude. You see stars diffrently at different latitudes. For example at the equator the pole star is on the northern horizon, and invisible in the sothern hemisphere.

The core relationships is between right ascension (like "longitude" for stars), siderial time and hour angle. The hour angle is the angle between the meridial and position of star. Or the siderial time is the right ascension of stars crossing the meridian. You could work this out from an wide field image using a service like https://nova.astrometry.net/ and then convert "siderial time to local time". There's a stack overflow answer plus others.

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