<p>Assuming the intensity of the sunlight is proportional to the area of the Sun which is showing, we can roughly calculate the intensity of sunlight, <span class="math-container">$I(d)$</span> as:</p>
<p><span class="math-container">$$I(d) = 1 - \frac{\mathrm{Area}_{\mathrm{overlap}}}{\mathrm{Area}_{\mathrm{Sun}}}$$</span></p>
<p>Where <span class="math-container">$\mathrm{Area}_{\mathrm{overlap}}$</span> is the overlap between the Moon and the Sun (the green area).</p>
<p><a href="https://i.sstatic.net/RIbRZ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RIbRZ.jpg" alt="" /></a><br />
<sub>(source: <a href="http://jwilson.coe.uga.edu/EMAT6680Su12/Carreras/EMAT6690/Essay2/circles%5B1%5D.jpg" rel="nofollow noreferrer">uga.edu</a>)</sub></p>
<p>Some calculation finds the overlap to be:</p>
<p><span class="math-container">$$\mathrm{Area}_{\mathrm{overlap}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2}$$</span></p>
<p>Where <span class="math-container">$d$</span> is the distance between the centres of the Moon and the Sun. Similarly, the area of the Sun is:</p>
<p><span class="math-container">$$\mathrm{Area}_{\mathrm{Sun}} = \pi r^2$$</span></p>
<p>Putting this together, you get the graph:</p>
<p><img src="https://i.sstatic.net/4JhJV.jpg" alt="Intensity changing with time" /></p>
<p>Where the Y axis is brightness (ranging from 100% to 0%) and the X axis is <span class="math-container">$d$</span> (ranging from <span class="math-container">$2r$</span> to <span class="math-container">$0$</span>).</p>
<p>I hope this answers your question: the brightness does not change proportionally to coverage/time.</p>
<h3>Note:</h3>
<p>This assumes that <span class="math-container">$d$</span> increases linearly with time (the moon doesn't speed up or slow down as it moves across the Sun).</p>

CC BY-SA 3.0
2017-08-23T15:33:10.787Z
Yes, but sky brightness is caused by an overall scattering of solar light, not be the light only at your current location - this is why the sky does not go pitch black as soon as the Sun sets ... so while the level of brightness is likely proportional to your formula, that won't be the only factor.
adrianmcmenamin
CC BY-SA 3.0
2017-08-23T17:46:37.267Z
@adrianmcmenamin Agreed, but without experimental data, this is the best I can do
Beta Decay
CC BY-SA 4.0
2017-08-23T11:11:57.737Z
2019-07-13T12:48:46.810Z
2020-06-17T09:47:40.653Z
Beta Decay
<p>During a solar eclipse, some of the visible area of the sun is obscured by the moon.  The exact percentage varies with time and with the observer's location.  My question is, does the sky darken by this same percentage?</p>

<p>In other words, suppose the sun is 80% covered.  Does sky light diminish by 80%?  Are objects illuminated by 20% of their normal amount?</p>

<light><solar-eclipse>
Does the sky darken proportionally during a partial solar eclipse?
Dr Sheldon
# Does the sky darken proportionally during a partial solar eclipse?

(Asked by: username_0 on 2017-08-21)

During a solar eclipse, some of the visible area of the sun is obscured by the moon. The exact percentage varies with time and with the observer's location. My question is, does the sky darken by this same percentage?

In other words, suppose the sun is 80% covered. Does sky light diminish by 80%? Are objects illuminated by 20% of their normal amount?

**Tags:** light, solar-eclipse


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## 1 answers

### Answer by username_1 on 2017-08-23. Score: 2

Assuming the intensity of the sunlight is proportional to the area of the Sun which is showing, we can roughly calculate the intensity of sunlight, $I(d)$ as:

$$I(d) = 1 - \frac{\mathrm{Area}\_{\mathrm{overlap}}}{\mathrm{Area}\_{\mathrm{Sun}}}$$

Where $\mathrm{Area}\_{\mathrm{overlap}}$ is the overlap between the Moon and the Sun (the green area).

[![](https://i.sstatic.net/RIbRZ.jpg)](https://i.sstatic.net/RIbRZ.jpg)  
(source: [uga.edu](http://jwilson.coe.uga.edu/EMAT6680Su12/Carreras/EMAT6690/Essay2/circles%5B1%5D.jpg))

Some calculation finds the overlap to be:

$$\mathrm{Area}\_{\mathrm{overlap}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2}$$

Where $d$ is the distance between the centres of the Moon and the Sun. Similarly, the area of the Sun is:

$$\mathrm{Area}\_{\mathrm{Sun}} = \pi r^2$$

Putting this together, you get the graph:

![Intensity changing with time](https://i.sstatic.net/4JhJV.jpg)

Where the Y axis is brightness (ranging from 100% to 0%) and the X axis is $d$ (ranging from $2r$ to $0$).

I hope this answers your question: the brightness does not change proportionally to coverage/time.

### Note:

This assumes that $d$ increases linearly with time (the moon doesn't speed up or slow down as it moves across the Sun).

> **Comments:**
> **username_2**: Yes, but sky brightness is caused by an overall scattering of solar light, not be the light only at your current location - this is why the sky does not go pitch black as soon as the Sun sets ... so while the level of brightness is likely proportional to your formula, that won't be the only factor.  
> **username_1**: @username_2 Agreed, but without experimental data, this is the best I can do  

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