<p>The mean angular radius of the Sun in radians is</p>

<p><span class="math-container">$$ \frac{R_\odot}{\mathrm{au}}
= \frac{6.96 \times 10^5~\mathrm{km}}{1.496 \times 10^8~\mathrm{km}}
= 4.65 \times 10^{-3} $$</span></p>

<p>and its mean angular diameter is twice that, 0.00930 radian or 0.533&deg;.
As the Earth's distance from the Sun annually varies by &plusmn;1.67%,
the Sun's angular diameter varies between 0.524&deg; in July and 0.542&deg; in January.</p>

<p>During a sunset, the Sun's <a href="https://en.wikipedia.org/wiki/Horizontal_coordinate_system" rel="nofollow noreferrer">altitude</a> relative to the horizon decreases by that same angle.
If the observer is on the equator, the Sun goes straight down in just over 2 minutes as you would expect.
Otherwise the Sun descends at an oblique angle <span class="math-container">$q$</span>, extending sunset by a factor of <span class="math-container">$1 / \sin q$</span>.
For example, if <span class="math-container">$q$</span> is 30&deg;, sunset takes twice as long as if <span class="math-container">$q$</span> were 90&deg;.
If the observer's latitude is <span class="math-container">$\varphi$</span> and the Sun's declination is <span class="math-container">$\delta$</span>, then</p>

<p><span class="math-container">$$ q = \cos^{-1} \frac{\sin \varphi}{\cos \delta} $$</span></p>

<p>If <span class="math-container">$\delta \approx 0^\circ$</span>, then <span class="math-container">$q \approx 90^\circ - \varphi$</span>.</p>

<p>What changes at a nearly constant rate of 15&deg;/hour is the Sun's <a href="https://en.wikipedia.org/wiki/Hour_angle" rel="nofollow noreferrer">hour angle</a> relative to the meridian, measured around the celestial equator.
One degree of hour angle at declination <span class="math-container">$\delta$</span> spans only <span class="math-container">$\cos \delta$</span> degrees of sky, so the Sun appears to set about 8% slower in June and December than it does in March and September.</p>

CC BY-SA 4.0
2020-03-01T20:08:47.977Z
2020-03-07T02:30:37.697Z
2020-03-07T02:30:37.697Z
Mike G
<p>The easiest way to measure the sun would be to use a pin-hole camera.</p>

<p>Use a piece of card with a pin hole. Hold it up so the sun shines through the pin hole and onto a piece of paper.  (Don't look at the sun through the hole - eye damage)</p>

<p>You will see a circle of light on the paper, this is the image of the sun. If the paper is 100cm from the card and the size of the image of the sun is <span class="math-container">$x$</span> cm, then you can work out the angular size of the sun.  It is about <span class="math-container">$\tan^{-1}(x/100)$</span> which is approximately <span class="math-container">$\frac{180x}{100\pi}$</span> You should get an angle of about 0.5 degrees.</p>

CC BY-SA 4.0
2020-03-02T08:57:00.257Z
Answering the question by suggesting an experiment that can be done at home: I'd upvote this twice if I could.
usernumber
CC BY-SA 4.0
2020-03-01T10:12:10.300Z
2020-03-01T10:12:10.300Z
James K
<p>According to what I what told <a href="https://astronomy.stackexchange.com/questions/35230/how-long-do-the-phases-of-sunset-take">here</a> there is 0.5 degrees from the beginning the end of the sunset. </p>

<p>Unfortunately I didn't understand two things: </p>

<blockquote>
  <p>1) How can I prove in the simplest way that there's 0.5 degrees only
  from the beginning to the end of the sunset. </p>
  
  <p>2) Why it is not adjust to my past observation on these phases (which
  was much longer (range from 5-7 minutes). which makes it between 1-2
  degrees.</p>
</blockquote>

CC BY-SA 4.0
2020-03-01T10:03:21.717Z
I don't understand what you mean.  The sun is 0.5 degrees.  But the sun doesn't go straight down so sunsets tend to last longer.  That doesn't make the sun any bigger. But the **sunset** isn't one or two degrees, nor is it 0.5 degrees.  The sunset doesn't have a size. It does have a time, but that depends on your latitude (and date)
James K
CC BY-SA 4.0
2020-03-01T12:08:53.123Z
Maybe I'm confused ad that's why I'm not understood. Is the sun by itself has a measure of 0.5 degrees (and it's constant)?
Reckless Glacier
CC BY-SA 4.0
2020-03-01T12:25:11.437Z
The angular size of the sun in the sky is 0.5 degrees.  My answer describes how you can test this.  The size of the sun in the sky is constant and doesn't change. It is the same at sunset as at noon (though the "moon illusion" may cause it to appear bigger at sunset)
James K
<sunset>
How to find how many degrees from the beginning to the end of the sunset?
Reckless Glacier
# How to find how many degrees from the beginning to the end of the sunset?

(Asked by: username_0 on 2020-03-01)

According to what I what told [here](https://astronomy.stackexchange.com/questions/35230/how-long-do-the-phases-of-sunset-take) there is 0.5 degrees from the beginning the end of the sunset.

Unfortunately I didn't understand two things:

> 1) How can I prove in the simplest way that there's 0.5 degrees only
> from the beginning to the end of the sunset.
>
> 2) Why it is not adjust to my past observation on these phases (which
> was much longer (range from 5-7 minutes). which makes it between 1-2
> degrees.

**Tags:** sunset

> **Comments:**
> **username_1**: I don't understand what you mean.  The sun is 0.5 degrees.  But the sun doesn't go straight down so sunsets tend to last longer.  That doesn't make the sun any bigger. But the **sunset** isn't one or two degrees, nor is it 0.5 degrees.  The sunset doesn't have a size. It does have a time, but that depends on your latitude (and date)  
> **username_0**: Maybe I'm confused ad that's why I'm not understood. Is the sun by itself has a measure of 0.5 degrees (and it's constant)?  
> **username_1**: The angular size of the sun in the sky is 0.5 degrees.  My answer describes how you can test this.  The size of the sun in the sky is constant and doesn't change. It is the same at sunset as at noon (though the "moon illusion" may cause it to appear bigger at sunset)  

---

## 2 answers

### Answer by username_2 on 2020-03-01. Score: 3

The mean angular radius of the Sun in radians is

$$ \frac{R\_\odot}{\mathrm{au}}
= \frac{6.96 \times 10^5~\mathrm{km}}{1.496 \times 10^8~\mathrm{km}}
= 4.65 \times 10^{-3} $$

and its mean angular diameter is twice that, 0.00930 radian or 0.533°.
As the Earth's distance from the Sun annually varies by ±1.67%,
the Sun's angular diameter varies between 0.524° in July and 0.542° in January.

During a sunset, the Sun's [altitude](https://en.wikipedia.org/wiki/Horizontal_coordinate_system) relative to the horizon decreases by that same angle.
If the observer is on the equator, the Sun goes straight down in just over 2 minutes as you would expect.
Otherwise the Sun descends at an oblique angle $q$, extending sunset by a factor of $1 / \sin q$.
For example, if $q$ is 30°, sunset takes twice as long as if $q$ were 90°.
If the observer's latitude is $\varphi$ and the Sun's declination is $\delta$, then

$$ q = \cos^{-1} \frac{\sin \varphi}{\cos \delta} $$

If $\delta \approx 0^\circ$, then $q \approx 90^\circ - \varphi$.

What changes at a nearly constant rate of 15°/hour is the Sun's [hour angle](https://en.wikipedia.org/wiki/Hour_angle) relative to the meridian, measured around the celestial equator.
One degree of hour angle at declination $\delta$ spans only $\cos \delta$ degrees of sky, so the Sun appears to set about 8% slower in June and December than it does in March and September.


---

### Answer by username_1 on 2020-03-01. Score: 3

The easiest way to measure the sun would be to use a pin-hole camera.

Use a piece of card with a pin hole. Hold it up so the sun shines through the pin hole and onto a piece of paper. (Don't look at the sun through the hole - eye damage)

You will see a circle of light on the paper, this is the image of the sun. If the paper is 100cm from the card and the size of the image of the sun is $x$ cm, then you can work out the angular size of the sun. It is about $\tan^{-1}(x/100)$ which is approximately $\frac{180x}{100\pi}$ You should get an angle of about 0.5 degrees.

> **Comments:**
> **username_3**: Answering the question by suggesting an experiment that can be done at home: I'd upvote this twice if I could.  

---

