# How long do the phases of sunset take? [duplicate]

How long does it take from the moment the Sun star 'touches' the horizon, to the time it totally disappear? (I'm not talking about the radiation (rays), but just about the appearance of the star by itself. I don't care of the rays). If I were near a beach or plane I'd check it, but I'm in crowded urban region so it's difficult to check it I believe).

• – JohnHoltz Feb 25 at 4:15

@Calc-You-Later's answer is correct in places where the path of the Sun is perpendicular to the horizon (at the equator during the equinoxes, somewhere between the tropics the rest of the time). There, the sunset lasts 2 minutes. However, at other latitudes, the sunset may last a bit longer.

Let $$\alpha$$ be the angle the path of the Sun makes with the horizon. This angle depends on your latitude and the date.

The speed of the Sun along its path is $$\vec{v_s} =$$ 0.25°/min. This speed can be projected along the vertical and horizontal axes. The vertical speed is now $$\vec{v_s} . \sin (\alpha)$$. The sunset length is still the apparent angle of the Sun divided by the vertical speed of the Sun.

So at higher latitudes, if the Sun sets, for instance, at a 45° angle, then the sunset will last 2 min / sin(45°) = 2.85 minutes = 2 minutes 51 seconds. If the Sun sets with a 60° angle, the sunset will last 4 minutes.

To get the value of $$\alpha$$, you can add your latitude and the solar declination:

• Determine how many days have passed since December 21st (i.e since the winter solstice). Let's call this number $$x$$
• Multiply this number by the amount of turning the Earth does in one day $$y = x * 360/365.25$$
• Get the cosine of this number and multiply it by the tilt of the Earth $$z = -23.5 * cos(y)$$. This is the solar declination
• Add the latitude $$\theta$$ of your location $$\alpha = z + \theta$$

This does not account for atmospheric effects. The atmosphere will bend the rays of light, and can make it possible to see the Sun even though it is under the horizon.

The atmospheric effects, however, depend on the atmospheric conditions (temperature gradient,...), and are more difficult to quantify.

• Great explanation! However, how cam we know what's the angle in the reality? Is it difference from place to place? – Ubiquitous Student Feb 24 at 15:15
• α is equal to the parallactic angle NCP-Sun-zenith. – Mike G Feb 24 at 17:47
• If you happen to be near the Arctic circle on the summer solstice, "sunset" can last for hours. – jamesqf Feb 24 at 18:21
• Relevant XKCD: what-if.xkcd.com/42 – Mark Feb 25 at 1:23
• @UbiquitousStudent The angle is different at different latitudes and also different times of the year. – Dronir Feb 25 at 13:12

The Sun's apparent angular diameter is approximately 0.5 degrees - though atmospheric conditions can make this vary.

We can calculate how fast the Earth rotates in degrees per minute like so:

$$24h\: /\: 360^{\circ} = 1h\: /\: 15^{\circ} \rightarrow 15^{\circ}\: /\: h$$

$$15^{\circ}\:/\:h = 15^{\circ}\:/\:60sec=0.25^{\circ}\:/\:min$$

If the Earth is rotating at $$0.25^{\circ}\:/\:min$$ and the Sun's apparent angular diameter is 0.5$$^{\circ}$$, then we can see that the the time elapsed from the Sun touching the horizon to the Sun fully dipping below the horizon is:

$$1 min\: /\: 0.25^{\circ} \: *\: 0.5^{\circ}= 2\:\textrm{minutes}$$

However, this varies based on atmospheric factors, since the Sun can be distorted based on the thickness of the atmosphere (as I mentioned with the apparent angular diameter). This is purely rotational - see here for more details on sunset length, and see here for the exact values for the angular size of the Sun. The value is a bit above 0.5 degrees, so in essence, accounting for atmospheric variation and the bending of light (and in agreement with the first link I have put here), the time elapsed from the Sun touching the horizon to the Sun fully dipping below the horizon can range from 2 to 3 minutes.

(Edit: As Mike G noted in his comment, this does not account for different latitudes that the observer may see the sunset from)

• Also consider that the Sun crosses the horizon at an oblique angle unless the observer is at the equator. – Mike G Feb 24 at 4:51
• @Calc-You-Later, Great explanation! Thank you. I'm just not sure how many degrees the Sun moves from the point it starts to touch the horizon until the moment it disappears. According to what I remember from a simple observation that I did in past, for sure it was more than 5 minutes, which maybe says that the Sun moves more than what you suggested? (also, the formula you showed 1min/0.25∘∗0.5∘=2minutes. Maybe we should multiply it by two? (2-3 minutes by 2 and then it more reasonable). I'd like to see your opinion. – Ubiquitous Student Feb 24 at 4:52
• @usernumber's answer perfectly accounts for that discrepancy - I did not include the fact that different observers see the sunset at different angles. which can cause the Sun to take longer to set. – Calc-You-Later Feb 24 at 16:30
• @MikeG It's more complicated than just "the observer is on the equator". Consider the situation in mid December when the plane of the equator is tilted 23 degrees relative to the sun position. The angle will vary up to 23 degrees either side of vertical, depending on your longitude on the equator. At higher latitudes when the day length is significantly different from 12 hours, the angles at sunrise and sunset at a fixed location on the same day can be very different. – alephzero Feb 24 at 20:38
• @alephzero All parallels of declination intersect all hour circles at right angles. At the equator, the horizon coincides with the ±6h circles. Daily changes in solar declination alter the Sun's apparent trajectory by ±0.06° at the equinoxes. – Mike G Feb 24 at 21:02

For a quick and dirty approximation, let's assume a constant Sun angular diameter of 32' and no atmospheric refraction. Let $$\varphi$$ be the observer's geographic latitude, and $$\delta$$ be the Sun's declination (±23.4° at solstices, 0° at equinoxes), which you can get from the NOAA solar calculator. The time between the Sun's lower and upper limbs crossing the horizon is

$$\Delta t \approx \frac{128~\mathrm{s}}{\sqrt{\cos^2 \delta - \sin^2 \varphi}} \tag{1}$$

Now let's account for atmospheric refraction and the eccentricity of Earth's orbit. Not knowing air pressure and temperature, it's standard to assume that points appearing on the horizon are refracted upward by 34'. Let $$\sigma$$ be the Sun's apparent angular radius (15.7' at aphelion, 16.3' at perihelion). The geometric hour angles where the Sun's lower and upper limbs appear to cross the horizon are

$$H_1, H_2 = \cos^{-1} \frac{\sin(-34' \pm \sigma) - \sin \varphi \sin \delta}{\cos \varphi \cos \delta} \tag{2}$$

Then

$$\Delta t = (240~\mathrm{s}/^{\circ}) (H_2 - H_1)$$

Here is a table of sunset duration (mm:ss) at various latitudes on day 15.0 (UT) of each month in 2020, using equation (2). For latitudes between 60°N and 50°S, approximation (1) is within 3% of these values.

      Jan   Feb   Mar   Apr   May   Jun   Jul   Aug   Sep   Oct   Nov   Dec
70N    --  8:06  6:17  7:19 29:04    --    --  9:03  6:18  6:51 14:40    --
60N  6:08  4:47  4:18  4:33  5:39  7:08  6:20  4:52  4:16  4:27  5:29  6:50
50N  4:03  3:34  3:20  3:26  3:50  4:12  4:01  3:33  3:19  3:25  3:50  4:13
40N  3:12  2:56  2:48  2:51  3:03  3:13  3:08  2:54  2:46  2:50  3:05  3:17
30N  2:45  2:35  2:29  2:30  2:38  2:44  2:41  2:32  2:27  2:30  2:40  2:48
20N  2:30  2:22  2:17  2:18  2:24  2:28  2:26  2:19  2:16  2:18  2:26  2:32
10N  2:22  2:15  2:11  2:11  2:16  2:20  2:18  2:12  2:09  2:12  2:19  2:24
0   2:20  2:13  2:09  2:09  2:14  2:17  2:15  2:10  2:07  2:10  2:16  2:21
10S  2:22  2:15  2:11  2:11  2:16  2:20  2:18  2:12  2:09  2:12  2:19  2:24
20S  2:30  2:22  2:17  2:18  2:23  2:27  2:25  2:19  2:16  2:18  2:26  2:33
30S  2:46  2:35  2:29  2:30  2:37  2:43  2:40  2:32  2:27  2:31  2:41  2:49
40S  3:14  2:57  2:48  2:50  3:02  3:11  3:06  2:53  2:46  2:51  3:06  3:19
50S  4:07  3:36  3:21  3:25  3:46  4:06  3:56  3:31  3:18  3:26  3:53  4:19
60S  6:28  4:53  4:18  4:30  5:26  6:38  6:00  4:45  4:15  4:30  5:42  7:21
70S    --  8:38  6:19  7:03 15:53    --    --  8:23  6:15  7:04 21:57    --


Derivation of (1)

If the angular distance AB is $$\frac{32'}{\sin q}$$ and the Sun moves at a rate of $$\frac{1' \cos \delta}{4~\mathrm{s}}$$, then it gets from A to B in $$\Delta t \approx \frac{128~\mathrm{s}}{\cos \delta \sin q}$$

Angle $$q$$ is the Sun's parallactic angle. In the special case of an object on the horizon, Meeus §14 gives

$$\cos q = \frac{\sin \varphi}{\cos \delta}$$

Using $$\sin q = \sqrt{1 - \cos^2 q}$$,

$$\Delta t \approx \frac{128~\mathrm{s}}{\cos \delta \sqrt{1 - \cos^2 q}} = \frac{128~\mathrm{s}}{\cos \delta \sqrt{1 - \left(\frac{\sin \varphi}{\cos \delta}\right)^2}}$$

which simplifies to approximation (1).

Derivation of (2)

The dotted circles show the Sun's geometric positions when its refracted positions are as in the first figure. Let $$h$$ be the geometric altitude of the center of the Sun. Meeus §13 gives

$$\sin h = \sin \varphi \sin \delta + \cos \varphi \cos \delta \cos H$$

Then

$$\cos H = \frac{\sin h - \sin \varphi \sin \delta}{\cos \varphi \cos \delta}$$

and we can substitute $$h = -34' \pm \sigma$$ to get equation (2).

• Neither formula is valid near the Earth's poles. Where the Sun's declination determines its altitude, the change in $\delta$ during sunset is significant. – Mike G Mar 2 at 17:51