<p>The answer depends on how detailed you want to be. For example,</p>

<ol>
<li>Do you want (a) an approximate answer by assuming the stars are uniformly distributed, or (b) to use the real distribution of stars?</li>
<li>Do you want to know the answer based on (a) the number of stars visible throughout the entire year, or (b) just a given night, or (c) just an instant of time?</li>
<li>The magnitude limit of your sky affects how many stars are visible. More stars are visible from a dark site in the country than in a city.</li>
<li>The other factor that has a small effect on the answer is whether you want to include atmospheric extinction or not. Stars near the horizon appear to be fainter than their published magnitude due to the atmosphere. Some stars close to the southern horizon (in your example) are not visible because they do not rise high enough.</li>
</ol>

<p>Combination 1(a) and 2(c) [uniform distribution, an instant of time] can be calculated based on the area of sky that is circumpolar versus the area for half of the sky. (As @zephyr indicated, you need to use steradians to calculate the area of the sky.)</p>

<p>Combination 1(a) and 2(a) [uniform distribution, entire year] is also easy to calculate from the area of the sky that is circumpolar, and the area of the entire sky minus the area of the sky that never rises (declination less than -55 for your example).</p>

<p>Any other combination requires using a catalog of star positions to "count" how many stars are visible at a given instant of time. In other words, based on the day and time (or the local sidereal time),  </p>

<ul>
<li>calculate the altitude of each star.</li>
<li>if above the horizon, calculate if it is bright enough to be visible given the sky conditions and (optionally) atmospheric extinction</li>
<li>if visible, tally the star as circumpolar or not</li>
<li>after counting all of the stars, perform the final calculation of the ratio.</li>
</ul>

<p>Such a calculation would be easy to perform in a spreadsheet using a catalog such as the Yale Bright Star Catalog (which may not be complete down to the faintest stars visible). I did a similar calculation many, many years ago. See the May 1994 Sky &amp; Telescope magazine, pages 86-88.</p>

CC BY-SA 3.0
2017-07-12T01:55:24.583Z
2017-07-12T01:55:24.583Z
JohnHoltz
<p>I cannot find my original calculations at the moment. I assume that the Sky &amp; Telescope article is copyrighted, but I found a PDF of it on the internet at <a href="https://archive.org/details/Sky_and_Telescope_1994-05-pdf" rel="nofollow noreferrer">https://archive.org/details/Sky_and_Telescope_1994-05-pdf</a>.</p>

<p>Here is a table from the article which shows the number of stars visible at various times of the night for an observer at 40 N latitude:</p>

<p><a href="https://i.sstatic.net/7XtL1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7XtL1.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2017-07-13T02:12:01.047Z
2017-07-13T02:12:01.047Z
JohnHoltz
<p>We know that circumpolar stars are stars that don't rise or set. If we are in a location with $\phi = 35^o N$. What is the fraction of circumpolar stars to all stars visible in this location?</p>

<p>I know the declination of this stars is $\delta &gt; 90 - 35 = 55$ But I don't know what is the degree of all stars visible in this location and is the fraction as simple as dividing this two angles or not?</p>

CC BY-SA 3.0
2017-07-11T20:15:49.037Z
You should be comparing steradian angles of subtended area, not just declinations.
zephyr
<star><observational-astronomy>
Fraction of circumpolar stars to all stars visible in a location
titansarus
# Fraction of circumpolar stars to all stars visible in a location

(Asked by: username_0 on 2017-07-11)

We know that circumpolar stars are stars that don't rise or set. If we are in a location with $\phi = 35^o N$. What is the fraction of circumpolar stars to all stars visible in this location?

I know the declination of this stars is $\delta > 90 - 35 = 55$ But I don't know what is the degree of all stars visible in this location and is the fraction as simple as dividing this two angles or not?

**Tags:** star, observational-astronomy

> **Comments:**
> **username_1**: You should be comparing steradian angles of subtended area, not just declinations.  

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## 2 answers

### Answer by username_2 on 2017-07-12. Score: 1

The answer depends on how detailed you want to be. For example,

1. Do you want (a) an approximate answer by assuming the stars are uniformly distributed, or (b) to use the real distribution of stars?
2. Do you want to know the answer based on (a) the number of stars visible throughout the entire year, or (b) just a given night, or (c) just an instant of time?
3. The magnitude limit of your sky affects how many stars are visible. More stars are visible from a dark site in the country than in a city.
4. The other factor that has a small effect on the answer is whether you want to include atmospheric extinction or not. Stars near the horizon appear to be fainter than their published magnitude due to the atmosphere. Some stars close to the southern horizon (in your example) are not visible because they do not rise high enough.

Combination 1(a) and 2(c) [uniform distribution, an instant of time] can be calculated based on the area of sky that is circumpolar versus the area for half of the sky. (As @username_1 indicated, you need to use steradians to calculate the area of the sky.)

Combination 1(a) and 2(a) [uniform distribution, entire year] is also easy to calculate from the area of the sky that is circumpolar, and the area of the entire sky minus the area of the sky that never rises (declination less than -55 for your example).

Any other combination requires using a catalog of star positions to "count" how many stars are visible at a given instant of time. In other words, based on the day and time (or the local sidereal time),

* calculate the altitude of each star.
* if above the horizon, calculate if it is bright enough to be visible given the sky conditions and (optionally) atmospheric extinction
* if visible, tally the star as circumpolar or not
* after counting all of the stars, perform the final calculation of the ratio.

Such a calculation would be easy to perform in a spreadsheet using a catalog such as the Yale Bright Star Catalog (which may not be complete down to the faintest stars visible). I did a similar calculation many, many years ago. See the May 1994 Sky & Telescope magazine, pages 86-88.


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### Answer by username_2 on 2017-07-13. Score: 0

I cannot find my original calculations at the moment. I assume that the Sky & Telescope article is copyrighted, but I found a PDF of it on the internet at <https://archive.org/details/Sky_and_Telescope_1994-05-pdf>.

Here is a table from the article which shows the number of stars visible at various times of the night for an observer at 40 N latitude:

[![enter image description here](https://i.sstatic.net/7XtL1.png)](https://i.sstatic.net/7XtL1.png)


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