<p>OK, lets start with the simplest mathematical approach to illustrate the path to a fully analytical answer.
The sun presents an angular width of 32 arcminutes to any point on earth.  That is 32/60 or 0.533 degrees of arc or angular span. 
Lets assume the Earth does not have its 23 degrees of tilt, for this first approximation.
Then as a second approximation lets assume the Earth rotates around the sun in 24 hours, you are still on the equator.
Our calculation is as follows;</p>

<p>0.533 degrees/360 degrees) = (hours sunset/24 hours).  </p>

<p>Solve for hours sunset and you get,</p>

<p>24 hrs X (0.533/360) = 0.0355 hrs, which is</p>

<p>0.0355 hrs X 60 min/hrs = 2.13 minutes, which is</p>

<p>2.13 min X 60 secs/min = 128 seconds</p>

<p>OK, now that is the first order approximation only and explains the minima of the nice charts previously provided.  </p>

<p>The first and trivial correction would be to notice that the 24 hour assumption is not accurate, hence leap years!  Beyond that we have actually 23:56 per year.  That will get you 127.56 seconds for sunset.</p>

<p>The real solution for the deep divers out there is to understand that the angular width of the sun in the sky is 32 arcminutes but only for any one instant in time for any one point on the Earth.  So the next calculation would be to integrate over the diameter of the earth to incorporate the angular width of you are traversing during the sunset traverse time.  You the observer are moving, rotating with the surface of the Earth, and hence you are spreading out the apparent angular size of the sun to the extent you are traversing during that sunset period, and this will add time to the sunset period.</p>

<p>Now that is the easier side of all of this.  The next calculation would add the geometric correction for latitude which the observer is located in.  This introduces horizontal relative component of motion of the sun to the observer, greatly increasing the time when one is not at the summer or winter equinoxes. (The prior calculations had the sun directly perpendicular to the rotation of the Earth.)  In the titled Earth Sun systems, this effect is minimized at the equinox positions of the earth sun system and asymptotes toward the prior calculation if one is on the equator and on the equinox twice per year.  Again, this is seen nicely in the charts of the previous answers.</p>

<p>I hope that help folks understand some of basic underpinnings of the math and geometry which the actual calculations must take into effect.</p>

<p>No calculators allowed and you can still get there.</p>

CC BY-SA 4.0
2018-07-06T04:08:23.400Z
Can you clarify what you mean by "the 24 hour assumption is not accurate, hence leap years". The length for 1 year is not related to the length of 1 day regardless of how you measure a day (assuming you want "noon" to be when the Sun or arbitrary star crosses the meridian). Also, I think your statement "Beyond that we have actually 23:56 per year" should read "actually 23:56 per DAY", not year.
JohnHoltz
CC BY-SA 4.0
2018-07-02T03:04:42.210Z
2018-07-02T03:24:43.453Z
2018-07-02T03:24:43.453Z
Matt Dugas
<p>The Sun's diameter being ½ degree out of 360, I figure it is 2 minutes.
Very even precisely two, because the division of time as into minutes, very very long ago, was designed with the movement of the Sun as its base.</p>

CC BY-SA 3.0
2015-12-10T17:43:28.463Z
@barrycarter I agree, I was thinking strictly ecliptical. I tried to downvote my own post, but that is not allowed. I should know better, I've lived in strange places where the Sun never sets, or worse, never rises. Artists have painted themselves to great careers using the strange light that the Sun shows when in limbo at its horizon in between its seasons.
LocalFluff
CC BY-SA 3.0
2015-12-10T21:21:59.750Z
There is also refraction - it is quite often possible to see the Sun, or part of it, when technically it's elevation is below 0 degrees as a result of this - as the atmosphere is thickest at the horizon and the degree of refraction greatest.
adrianmcmenamin
CC BY-SA 3.0
2018-01-06T05:49:26.510Z
@adrianmcmenamin - There's also the *opposite* effect, albedo: https://en.wikipedia.org/wiki/Albedo#Snow .
Rob
CC BY-SA 3.0
2018-01-06T13:40:22.460Z
@barrycarter - That's linked to from: https://en.m.wikipedia.org/wiki/Daytime#Near_the_poles - The idea being that if it was a long time before dawn (expected to still be dark) that the Sun could reflect off the sky (clouds, or to a lesser extent fog/rain or when it's snowing) and then reflect off of the snowy ground resulting in a very bright view - it **doesn't** make the Sun rise earlier (any more than refraction or cloud reflection does) but it's as bright outside as if it were later in the day (throwing off measuring instruments **or** making it bright enough to land a plane).
Rob
CC BY-SA 3.0
2015-12-10T14:35:10.613Z
2015-12-10T14:35:10.613Z
LocalFluff
<p>The time it takes depends on various factors: the angle that the path of the sun makes with the horizon is the main one, though there are also optical effects caused by the atmosphere have an effect too.</p>

<p>Generally the closer to the equator you live, the steeper is the angle, and so the faster is the sunset. </p>

<p>Using Stellarium I did a couple of tests:</p>

<ul>
<li>In the UK (50 degrees North) on 10th December, it took the sun 4min 47s seconds to sink below a simulated horizon.</li>
<li>In Angloa (10 degrees south), on the same day it took 2min 26s for the sun to set.</li>
</ul>

<p>It seems that in most populated regions, a sunset takes between 2 and 5 minutes.</p>

<p>There are locations, close to the Antarctic circle at this time of year, at which the sun merely partially sets, and then rises again. And at the Pole, the sun moves around in horizontal circles in the sky each day. During summer there is a permanent sun, as winter approaches the sun gets closer to the horizon, and then sets over several days. (Randall calculates 38 to 40 hours in the blog that Barry links) </p>

CC BY-SA 3.0
2015-12-12T18:02:05.787Z
Actually, the ecliptic is the Sun's *yearly* path, not daily.
AstroFloyd
CC BY-SA 3.0
2015-12-10T21:47:36.540Z
2017-08-21T06:50:32.707Z
2017-08-21T06:50:32.707Z
James K
<p>From the time that the sun appears on the horizon, or meets it on its setting, to the time that it is fully visible, or no longer visible on its setting, how much time passes?
Secondly, is there a place in the world where a sunrise/sunset occurs over a period of a few days? Meaning, that from the time it begins to appear over the horizon until it is fully visible, a period of a few days pass without night intervening (and the same for the opposite with sunset)?</p>

CC BY-SA 3.0
2016-01-04T01:25:11.713Z
At the equator, the sun seems to rise and set quickly - as Rudyard Kipling says,
"On the road to Mandalay,
Where the flyin'-fishes play,
An' the dawn comes up like thunder outer China 'crost the Bay!"

While the farther North or South you go, sunset lingers more and more. Summer sunsets in Georgia seem to take 15 minutes or so to ease under the covers of the darkening land.
Howard Miller
<the-sun>
How long does a sunrise or sunset take?
user11308
# How long does a sunrise or sunset take?

(Asked by: username_0 on 2015-12-10)

From the time that the sun appears on the horizon, or meets it on its setting, to the time that it is fully visible, or no longer visible on its setting, how much time passes?
Secondly, is there a place in the world where a sunrise/sunset occurs over a period of a few days? Meaning, that from the time it begins to appear over the horizon until it is fully visible, a period of a few days pass without night intervening (and the same for the opposite with sunset)?

**Tags:** the-sun

> **Comments:**
> **username_1**: At the equator, the sun seems to rise and set quickly - as Rudyard Kipling says, "On the road to Mandalay, Where the flyin'-fishes play, An' the dawn comes up like thunder outer China 'crost the Bay!"  While the farther North or South you go, sunset lingers more and more. Summer sunsets in Georgia seem to take 15 minutes or so to ease under the covers of the darkening land.  

---

## 3 answers

### Answer by username_7 on 2015-12-10. Score: 11 | Accepted answer

The time it takes depends on various factors: the angle that the path of the sun makes with the horizon is the main one, though there are also optical effects caused by the atmosphere have an effect too.

Generally the closer to the equator you live, the steeper is the angle, and so the faster is the sunset.

Using Stellarium I did a couple of tests:

* In the UK (50 degrees North) on 10th December, it took the sun 4min 47s seconds to sink below a simulated horizon.
* In Angloa (10 degrees south), on the same day it took 2min 26s for the sun to set.

It seems that in most populated regions, a sunset takes between 2 and 5 minutes.

There are locations, close to the Antarctic circle at this time of year, at which the sun merely partially sets, and then rises again. And at the Pole, the sun moves around in horizontal circles in the sky each day. During summer there is a permanent sun, as winter approaches the sun gets closer to the horizon, and then sets over several days. (Randall calculates 38 to 40 hours in the blog that Barry links)

> **Comments:**
> **username_8**: Actually, the ecliptic is the Sun's *yearly* path, not daily.  

---

### Answer by username_2 on 2018-07-02. Score: 2

OK, lets start with the simplest mathematical approach to illustrate the path to a fully analytical answer.
The sun presents an angular width of 32 arcminutes to any point on earth. That is 32/60 or 0.533 degrees of arc or angular span.
Lets assume the Earth does not have its 23 degrees of tilt, for this first approximation.
Then as a second approximation lets assume the Earth rotates around the sun in 24 hours, you are still on the equator.
Our calculation is as follows;

0.533 degrees/360 degrees) = (hours sunset/24 hours).

Solve for hours sunset and you get,

24 hrs X (0.533/360) = 0.0355 hrs, which is

0.0355 hrs X 60 min/hrs = 2.13 minutes, which is

2.13 min X 60 secs/min = 128 seconds

OK, now that is the first order approximation only and explains the minima of the nice charts previously provided.

The first and trivial correction would be to notice that the 24 hour assumption is not accurate, hence leap years! Beyond that we have actually 23:56 per year. That will get you 127.56 seconds for sunset.

The real solution for the deep divers out there is to understand that the angular width of the sun in the sky is 32 arcminutes but only for any one instant in time for any one point on the Earth. So the next calculation would be to integrate over the diameter of the earth to incorporate the angular width of you are traversing during the sunset traverse time. You the observer are moving, rotating with the surface of the Earth, and hence you are spreading out the apparent angular size of the sun to the extent you are traversing during that sunset period, and this will add time to the sunset period.

Now that is the easier side of all of this. The next calculation would add the geometric correction for latitude which the observer is located in. This introduces horizontal relative component of motion of the sun to the observer, greatly increasing the time when one is not at the summer or winter equinoxes. (The prior calculations had the sun directly perpendicular to the rotation of the Earth.) In the titled Earth Sun systems, this effect is minimized at the equinox positions of the earth sun system and asymptotes toward the prior calculation if one is on the equator and on the equinox twice per year. Again, this is seen nicely in the charts of the previous answers.

I hope that help folks understand some of basic underpinnings of the math and geometry which the actual calculations must take into effect.

No calculators allowed and you can still get there.

> **Comments:**
> **username_3**: Can you clarify what you mean by "the 24 hour assumption is not accurate, hence leap years". The length for 1 year is not related to the length of 1 day regardless of how you measure a day (assuming you want "noon" to be when the Sun or arbitrary star crosses the meridian). Also, I think your statement "Beyond that we have actually 23:56 per year" should read "actually 23:56 per DAY", not year.  

---

### Answer by username_4 on 2015-12-10. Score: -1

The Sun's diameter being ½ degree out of 360, I figure it is 2 minutes.
Very even precisely two, because the division of time as into minutes, very very long ago, was designed with the movement of the Sun as its base.

> **Comments:**
> **username_4**: @barrycarter I agree, I was thinking strictly ecliptical. I tried to downvote my own post, but that is not allowed. I should know better, I've lived in strange places where the Sun never sets, or worse, never rises. Artists have painted themselves to great careers using the strange light that the Sun shows when in limbo at its horizon in between its seasons.  
> **username_5**: There is also refraction - it is quite often possible to see the Sun, or part of it, when technically it's elevation is below 0 degrees as a result of this - as the atmosphere is thickest at the horizon and the degree of refraction greatest.  
> **username_6**: @username_5 - There's also the *opposite* effect, albedo: https://en.wikipedia.org/wiki/Albedo#Snow .  
> **username_6**: @barrycarter - That's linked to from: https://en.m.wikipedia.org/wiki/Daytime#Near_the_poles - The idea being that if it was a long time before dawn (expected to still be dark) that the Sun could reflect off the sky (clouds, or to a lesser extent fog/rain or when it's snowing) and then reflect off of the snowy ground resulting in a very bright view - it **doesn't** make the Sun rise earlier (any more than refraction or cloud reflection does) but it's as bright outside as if it were later in the day (throwing off measuring instruments **or** making it bright enough to land a plane).  

---

