# How to calculate the altitude of a star given the hour angle, declination, and latitude?

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?

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.

• 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? Sep 19, 2021 at 18:43
• Included in the answer. Fundamental Astronomy has the derivation. You can read more about it in my answer here. Sep 19, 2021 at 18:52