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I'm trying to find the altitude of a star for observing, but all I have is the hour angle and declination of the star, along with latitude of the location I'm observing from. How can I find the altitude?

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You can use this fundamental formula in spherical astronomy[1] $$\sin a=\sin \phi \sin \delta + \cos \phi \cos \delta \cos H$$ where

  • $a$ is the wanted altitude,
  • $\phi$ is your latitude,
  • $\delta$ is the declination of the star, and
  • $H$ is the hour angle, measured in the clockwise direction.

Pay attention to the units! (Don't mix degrees, radians and grads. Common cause of error!)

Since I don't know how you are familiar with the trigonometric functions (I believe pretty well), you only get $\sin a$ using that formula. You need to get the $\arcsin$ of that value in order to get wanted altitude $a$.


The solution above is perfectly correct in theory (on competitions, exams, and for personal use), but if you are writing a program on computer, you might find the following useful:

The factor we haven't yet addressed is the atmospheric refraction[2] It causes the star to look higher than in reality. The effect is pretty small, on range of few arc minutes.

First you need to calculate the factor $R$ by the formula[3] $$R=\frac{16.27''\cdot P}{273 + T}$$ where $P$ is the pressure in millibars and $T$ is the temperature in degrees of Celsius. You are perfectly fine using just $R=60''=1'$. Then the apparent altitude of the star is given by $a'=R+a$. Again, pay attention to the units (everything in degrees or everything in minutes ...)


If you are interested in learning about positional or spherical astronomy, then I advise you to visit another Stack Exchange question about this topic. Personaly, I have had great fun with Fundamental Astronomy. But in general, the formula is derived using the spherical law of cosines.

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  • $\begingroup$ Thank you so much! I looked at the link [1] you shared (which is great), but I don't seem to see how that was derived. Would you be able to link me to where I could learn more about htat? $\endgroup$ Sep 19 '21 at 18:43
  • $\begingroup$ Included in the answer. Fundamental Astronomy has the derivation. You can read more about it in my answer here. $\endgroup$
    – User123
    Sep 19 '21 at 18:52

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