How to find how many degrees from the beginning to the end of the sunset?

According to what I what told here 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.

• 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 Mar 1 '20 at 10:03
• 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 Mar 1 '20 at 12:08
• 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 Mar 1 '20 at 12:25
• I believe (could be wrong) that the Sun's actual angular radius smaller (and the Sun gets more egg-shaped) because refraction increases rapidly near the horizon. Of course, @JamesK is correct in saying the Sun looks bigger near the horizon – user21 Mar 2 '20 at 18:11

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.

• Answering the question by suggesting an experiment that can be done at home: I'd upvote this twice if I could. – usernumber Mar 2 '20 at 8:57

$$\frac{R_\odot}{\mathrm{au}} = \frac{6.96 \times 10^5~\mathrm{km}}{1.496 \times 10^8~\mathrm{km}} = 4.65 \times 10^{-3}$$
During a sunset, the Sun's altitude 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 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.