# 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 $$\sin a=\sin \phi \sin \delta + \cos \phi \cos \delta \cos H$$ 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 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 $$R=\frac{16.27''\cdot P}{273 + T}$$ 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  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