<p>It depends on the star's brightness. Astronomical darkness (full night, essentially) begins when the sun is <a href="http://aa.usno.navy.mil/faq/docs/RST_defs.php">18 degrees below the horizon</a>. Then, you can see all of the ones you're going to be able to see. </p>

<p>You will see stars before that point; which ones and how many will depend on how bright they are relative to the brightness of the twilight sky.</p>

CC BY-SA 3.0
2016-02-27T23:57:24.077Z
@jwenting I said to ignore all light pollution. Once we have the formula I described you can add in light pollution as an error term.
Double AA
CC BY-SA 3.0
2016-02-27T23:59:05.913Z
@Jeremy There are things you can do that are useful even before astronomical darkness, eg. navigate a boat or take students/clients out to see object X. Maybe intense research needs extra high quality conditions, but this was tagged "amateur observing" for a reason.
Double AA
CC BY-SA 3.0
2014-08-27T13:13:12.403Z
Well, yes, I knew that "it depends on the star's brightness"; I basically said so in the question. I asked for how to calculate for a given brightness and location. I would downvote if I had enough rep.
Double AA
CC BY-SA 3.0
2014-08-27T17:15:49.737Z
It's a decent answer. And he did answer the "degrees" part of the question.
HDE 226868
CC BY-SA 3.0
2014-08-27T22:05:35.627Z
Sorry, I answered from the practical point of view, when can you count on doing something useful with the observation, which is after astronomical darkness. You can't do much useful with an observation when the star is just barely visible. And you could quibble about visibility - naked eye or with a scope? Whose naked eye - different people can see different apparent magnitudes in full darkness, so it is likely different people will see stars appear differently. Perhaps it would help if you explained why you want to know the answer, what is the intended purpose...
Jeremy
CC BY-SA 3.0
2014-09-03T08:59:57.083Z
and where. If there's a lot of artificial light around, you may never see any but the brightest stars.
jwenting
CC BY-SA 3.0
2014-08-27T10:49:49.723Z
2014-08-27T10:49:49.723Z
Jeremy
<p>I think we can work this our pretty well with a few approximations.</p>

<p>To start, let's say that the observable limit of the naked eye under completely dark skies is 6th magnitude in $V$. Let's also say that the human eye has an angular resolution of 1 arcminute, which will subtend $\pi \cdot 1^2 \approx 3$ square arcminutes on the sky.</p>

<p>Actual dark skies on Earth still have some measurable sky brightness - which is one reason we can't see arbitrarily faint stars - of about $21.8 \ \mathrm{mag} \over \text{arcsec}^2$ in V-band. <a href="http://www.eso.org/~fpatat/science/skybright/twilight.pdf" rel="nofollow noreferrer">See Figure 1 here.</a></p>

<p>The integrated brightness, $m$, of the sky background, $S$, will be:</p>

<p>$$m = S - 2.5 \cdot \log(A)$$</p>

<p>where $A$ is the angular area. Note that the integrated sky brightness will just scale along with $S$, the surface brightness of the sky, since $A$ remains constant. </p>

<p>So, if for $S = {21 \ \mathrm{mag} \over \text{arcsec}^2}$ the observable limit is 6th mag in $V$, the limit for arbitrary $S$ will be:</p>

<p>$$m_{\text{limit}} = 6 - [21 - S(z)]$$</p>

<p>where $S(z)$ is the sky brightness when the Sun is at some zenith angle $z$.</p>

<p>From that reference I linked to, $S = 21.8$ at about $z = 105^{\circ}$ (Fig 5), and it goes up pretty much linearly to $S = 10$ in $V$ at $z = 94^{\circ}$. So:</p>

<p>$$S(z) = 1.07 \cdot z - 90.5$$</p>

<p>Thus the limiting visible $V$ magnitude for a star, as function of the Sun's zenith angle ought to be roughly</p>

<p>$$m_{\text{limit}}(z) = 1.07 \cdot z - 105.5$$</p>

<p>But you wanted the altitude of the Sun below the horizon, not away from the zenith, so subtract $90^{\circ}$ from $z$ in the above:</p>

<p>$$m_{\text{limit}} = 1.07 \cdot \theta - 10$$</p>

<p>where $\theta$ is how far the Sun is below the horizon in degrees. </p>

<p>Approximately, at any rate.</p>

CC BY-SA 4.0
2022-07-07T18:17:42.580Z
Order of operations may require some parentheses for these formulas to work as intended.
Mike G
CC BY-SA 3.0
2017-05-04T21:42:26.043Z
2017-05-04T23:57:54.117Z
2017-05-04T23:57:54.117Z
Thomas
<p>Given a star's apparent location and apparent magnitude, how how many degrees below the horizon must the sun be for the star to be visible to the naked eye for an observer on Earth at a specified location and elevation? You can ignore all light pollution, though I understand there will be some error due to air quality/weather/atmospheric aberrations.</p>

<p>Is there such a formula? If not, why not and what methods of approximation are available?</p>

<p>(I figure the brightness of the afterglow of the sun at a certain angle above the western horizon can be calculated for a certain time and place, and if it is less than the apparent brightness of the star in question, the star will be able to be seen.)</p>

CC BY-SA 4.0
2021-06-01T01:48:57.103Z
I was given the following two references which allegedly deal with this question, which I don't know how to find. If someone is able to find and summarize their contents that would be helpful.  1) M. J. Koomen et al., "Measurements of the Brightness of the Twighlight Sky," _J. Opt. Soc. of Amer._ vol 42 (1952), pp. 353-356. 2) R. Tousy and M. J. Koomen, "Visibility of Stars and Planets During Twilight," ibid, vol. 43 (1953), pp. 177-183. Additionally, if someone knows how to follow the trail from those to newer follow up studies, please let me know, as those appear to be from the 1950s.
Double AA
CC BY-SA 4.0
2021-06-01T02:04:20.803Z
Seems like these two behind a paywall https://www.osapublishing.org/josa/abstract.cfm?uri=josa-42-5-353 https://www.osapublishing.org/josa/abstract.cfm?uri=josa-43-3-177
Double AA
CC BY-SA 4.0
2021-06-09T00:58:10.507Z
Also: "a possible secondary seasonal variation of sky lumination (J.F.J. Schmidt, _Astron. Nachrichten_ 63:98 (1865)" I hope since 1865 there has been some progress on the question apparently in that paper, but I don't know how to determine if or where it has.
Double AA
CC BY-SA 4.0
2022-07-07T14:57:08.890Z
I do not understand the Formula ... mlimit = 1.07 ⋅ θ − 10 from the previous answer? It was stated that, the Visual Limit was 6th magnitude. and rounding 1.07 to exactly "1.00" then ... -10 mag limit = 0° - 10 ( at sunset ) 0 mag limit = 10° - 10 ( 45 minutes after sunset ) 10 mag limit = 20° - 10 ( 90 minutes after sunset ) So how can it change from Visible Limit = 6th mag, to now Visible = 10th mag ?
mvas
CC BY-SA 4.0
2022-07-08T02:01:38.157Z
@DoubleAA sci-hub.se has the second of those papers.
Greg Miller
CC BY-SA 3.0
2014-08-27T05:53:10.243Z
Not exacty an answer to your question, but a very nice article about daylight astronomy that shows you actually can see quite something during the day: http://sky.velp.info/daystars.php
agtoever
CC BY-SA 3.0
2014-08-27T06:07:54.823Z
MODTRAN models the light from the sky. Seems a bit more complicated than just a formula. See: http://modtran5.com/ and http://en.m.wikipedia.org/wiki/MODTRAN
agtoever
<amateur-observing>
Time after sunset until star can be seen
Double AA
# Time after sunset until star can be seen

(Asked by: username_0 on 2014-08-27)

Given a star's apparent location and apparent magnitude, how how many degrees below the horizon must the sun be for the star to be visible to the naked eye for an observer on Earth at a specified location and elevation? You can ignore all light pollution, though I understand there will be some error due to air quality/weather/atmospheric aberrations.

Is there such a formula? If not, why not and what methods of approximation are available?

(I figure the brightness of the afterglow of the sun at a certain angle above the western horizon can be calculated for a certain time and place, and if it is less than the apparent brightness of the star in question, the star will be able to be seen.)

**Tags:** amateur-observing

> **Comments:**
> **username_3**: Not exacty an answer to your question, but a very nice article about daylight astronomy that shows you actually can see quite something during the day: http://sky.velp.info/daystars.php  
> **username_3**: MODTRAN models the light from the sky. Seems a bit more complicated than just a formula. See: http://modtran5.com/ and http://en.m.wikipedia.org/wiki/MODTRAN  
> **username_0**: I was given the following two references which allegedly deal with this question, which I don't know how to find. If someone is able to find and summarize their contents that would be helpful.  1) M. J. Koomen et al., "Measurements of the Brightness of the Twighlight Sky," _J. Opt. Soc. of Amer._ vol 42 (1952), pp. 353-356. 2) R. Tousy and M. J. Koomen, "Visibility of Stars and Planets During Twilight," ibid, vol. 43 (1953), pp. 177-183. Additionally, if someone knows how to follow the trail from those to newer follow up studies, please let me know, as those appear to be from the 1950s.  
> **username_0**: Seems like these two behind a paywall https://www.osapublishing.org/josa/abstract.cfm?uri=josa-42-5-353 https://www.osapublishing.org/josa/abstract.cfm?uri=josa-43-3-177  
> **username_0**: Also: "a possible secondary seasonal variation of sky lumination (J.F.J. Schmidt, _Astron. Nachrichten_ 63:98 (1865)" I hope since 1865 there has been some progress on the question apparently in that paper, but I don't know how to determine if or where it has.  
> **username_1**: I do not understand the Formula ... mlimit = 1.07 ⋅ θ − 10 from the previous answer? It was stated that, the Visual Limit was 6th magnitude. and rounding 1.07 to exactly "1.00" then ... -10 mag limit = 0° - 10 ( at sunset ) 0 mag limit = 10° - 10 ( 45 minutes after sunset ) 10 mag limit = 20° - 10 ( 90 minutes after sunset ) So how can it change from Visible Limit = 6th mag, to now Visible = 10th mag ?  
> **username_2**: @DoubleAA sci-hub.se has the second of those papers.  

---

## 2 answers

### Answer by username_4 on 2014-08-27. Score: 7

It depends on the star's brightness. Astronomical darkness (full night, essentially) begins when the sun is [18 degrees below the horizon](http://aa.usno.navy.mil/faq/docs/RST_defs.php). Then, you can see all of the ones you're going to be able to see.

You will see stars before that point; which ones and how many will depend on how bright they are relative to the brightness of the twilight sky.

> **Comments:**
> **username_0**: Well, yes, I knew that "it depends on the star's brightness"; I basically said so in the question. I asked for how to calculate for a given brightness and location. I would downvote if I had enough rep.  
> **username_5**: It's a decent answer. And he did answer the "degrees" part of the question.  
> **username_4**: Sorry, I answered from the practical point of view, when can you count on doing something useful with the observation, which is after astronomical darkness. You can't do much useful with an observation when the star is just barely visible. And you could quibble about visibility - naked eye or with a scope? Whose naked eye - different people can see different apparent magnitudes in full darkness, so it is likely different people will see stars appear differently. Perhaps it would help if you explained why you want to know the answer, what is the intended purpose...  
> **username_6**: and where. If there's a lot of artificial light around, you may never see any but the brightest stars.  
> **username_0**: @username_6 I said to ignore all light pollution. Once we have the formula I described you can add in light pollution as an error term.  
> **username_0**: @username_4 There are things you can do that are useful even before astronomical darkness, eg. navigate a boat or take students/clients out to see object X. Maybe intense research needs extra high quality conditions, but this was tagged "amateur observing" for a reason.  

---

### Answer by username_7 on 2017-05-04. Score: 4

I think we can work this our pretty well with a few approximations.

To start, let's say that the observable limit of the naked eye under completely dark skies is 6th magnitude in $V$. Let's also say that the human eye has an angular resolution of 1 arcminute, which will subtend $\pi \cdot 1^2 \approx 3$ square arcminutes on the sky.

Actual dark skies on Earth still have some measurable sky brightness - which is one reason we can't see arbitrarily faint stars - of about $21.8 \ \mathrm{mag} \over \text{arcsec}^2$ in V-band. [See Figure 1 here.](http://www.eso.org/~fpatat/science/skybright/twilight.pdf)

The integrated brightness, $m$, of the sky background, $S$, will be:

$$m = S - 2.5 \cdot \log(A)$$

where $A$ is the angular area. Note that the integrated sky brightness will just scale along with $S$, the surface brightness of the sky, since $A$ remains constant.

So, if for $S = {21 \ \mathrm{mag} \over \text{arcsec}^2}$ the observable limit is 6th mag in $V$, the limit for arbitrary $S$ will be:

$$m\_{\text{limit}} = 6 - [21 - S(z)]$$

where $S(z)$ is the sky brightness when the Sun is at some zenith angle $z$.

From that reference I linked to, $S = 21.8$ at about $z = 105^{\circ}$ (Fig 5), and it goes up pretty much linearly to $S = 10$ in $V$ at $z = 94^{\circ}$. So:

$$S(z) = 1.07 \cdot z - 90.5$$

Thus the limiting visible $V$ magnitude for a star, as function of the Sun's zenith angle ought to be roughly

$$m\_{\text{limit}}(z) = 1.07 \cdot z - 105.5$$

But you wanted the altitude of the Sun below the horizon, not away from the zenith, so subtract $90^{\circ}$ from $z$ in the above:

$$m\_{\text{limit}} = 1.07 \cdot \theta - 10$$

where $\theta$ is how far the Sun is below the horizon in degrees.

Approximately, at any rate.

> **Comments:**
> **username_8**: Order of operations may require some parentheses for these formulas to work as intended.  

---

