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I know the latitude and longitude of a city ($41^\circ 87'$ N, $87^\circ 62'$ W), the declination and right ascension of a star($16.51^\circ$, $68.98^\circ$) and the local time, date (27.12.2017, 1:20 AM). Could you please show me the exact calculations for hour angle, altitude?

I will use the formula $LST=100.46+0.98d+15UT$, where d is the number of days since 1.1.2000 at 00:00 and UT is the universal time.

At 1:20 AM, $d=6570.055$.

UT is $UT=7:20$ or in hours $UT=7.3$.

$LST=6648.0639$ or $LST=168.0639$

$HA=LST-RA=168.0639-16.51=151.5539$

How is my calculation?

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  • $\begingroup$ Someone should do an answer-your-own-question here to provide basic astronomy formulas, but, until then, try googling. There should be plenty of resources online. $\endgroup$
    – user21
    Commented Dec 27, 2017 at 13:25
  • $\begingroup$ Minor note: I think you mean 41.87N and 87.62W; if you're using minutes (the apostrophe), they would be between 0 and 59. Also stjarnhimlen.se/comp/tutorial.html is the closest thing I found to a list of formulas that might help you. $\endgroup$
    – user21
    Commented Dec 28, 2017 at 14:36
  • $\begingroup$ Thanks. I found this: stargazing.net/kepler/altaz.html Are the formulas right? $\endgroup$
    – Alex S
    Commented Dec 30, 2017 at 0:08
  • $\begingroup$ Shouldn't $d$ be computed in UTC days, not local time days? $\endgroup$
    – user21
    Commented Dec 30, 2017 at 4:31
  • $\begingroup$ Your formula for LST is missing one important term: the longitude. $\endgroup$
    – JohnHoltz
    Commented Dec 30, 2017 at 15:25

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enter image description here

This should really be a comment, but... you can usually check your calculations using a program like Stellarium (screenshot above).

From the latitude/longitude and right ascension/declination you give, it appears you are viewing Aldebaran from Chicago, IL at 7:20am UTC on 2017-12-27 (since Chicago is on Central Time and not observing Daylight Saving Time, which is 6h behind UTC).

As you can see from the above, the hour angle is about 3h16m19s (or 49.16 degrees), the azimuth (which is what I think you actually wanted instead of hour angle) is about 254 degrees, and the altitude is about 41 degrees.

The process for converting right ascension and declination to azimuth and elevation isn't difficult, and I considered posting a general answer, but I wasn't sure what format to use. Most programming languages have an astronomy library (which will be more accurate), and there are plenty of sites that already walk through how we get the formula, not to mention my https://astronomy.stackexchange.com/a/14508/21 which gives sourced formulas.

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  • $\begingroup$ Your linked answer is a really nice resource; definitely mention it! $\endgroup$
    – uhoh
    Commented Dec 31, 2017 at 2:47

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