<p>I'm guessing what you see is the moonlight being scattered by the hexagonal ice crystals in cirrus or cirrostratus clouds, which lie at very high altitudes, 5-6 km and above. The light is scattered by roughly 22º, and because of the slight wavelength dependence, the halo actually has rainbow-like colors, although often they are so faint that you just perceive it as white.</p>

<p>Read more <a href="http://en.wikipedia.org/wiki/22%C2%B0_halo" rel="nofollow noreferrer">here</a>.</p>

<p><a href="https://i.sstatic.net/F2XXY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/F2XXY.png" alt="Moonhalo"></a></p>

CC BY-SA 3.0
2015-04-20T21:09:15.093Z
I think you labelled the wrong angles as 22 degrees. This is the minimum deviation of light passing through the crystals. Larger angles are possible.
ProfRob
CC BY-SA 3.0
2015-04-20T21:45:34.900Z
@RobJeffries: Wow…
pela
CC BY-SA 3.0
2015-04-20T21:55:28.587Z
@RobJeffries: Corrected. Thanks a lot for pointing this out.
pela
CC BY-SA 3.0
2015-04-20T22:07:52.763Z
I don't understand why don't you simply mark the outbound angle (from the observer towards the halo) as 22° one on your drawing? That's the angle we're measuring here. The two angles you marked now could be anything, it's just a coincidence that the Sun and the Moon cover more or less the same solid angle (their angular diameter is 31.6 to 32.7 and 29.3 to 34.1 minutes, respectively) when observed from the surface of the Earth. Plus, the angles you mark aren't 22°. Considering Sun's and Moon's distance, they would be nearly identical to half their angular diameter.
TildalWave
CC BY-SA 3.0
2015-04-20T22:13:02.217Z
@TildalWave: The 22º is the amount that the light is deflected. My first drawing was wrong, but is now corrected. The angle we would measure (for the diameter of the halo) is 2x22º, i.e. 44º. The angle doesn't have anything to do with the angle suspended by neither the Moon, nor the Sun. It's just the angle of deflection. Hopefully my new drawing makes this clearer. Apologies for the confusion.
pela
CC BY-SA 3.0
2015-04-20T22:31:18.803Z
@TildalWave: Ah okay, I understand. You're right; in reality, the part of the solid lines between the Moon and the clouds should be drawn as parallel. I think I can live with my drawing not being to scale, though. If you think it's very confusing, I can make another one tomorrow. But now it's 0:30 in Denmark, so I gotta sleep :)
pela
CC BY-SA 3.0
2015-04-21T06:29:01.117Z
@TildalWave: I'm absolutely okay with that, but if the light is scattered by 22º, then the diameter of the halo measured by the observer would be twice that.
pela
CC BY-SA 3.0
2015-04-21T06:29:23.157Z
We're being asked to move this discussion to chat :)
pela
CC BY-SA 3.0
2015-04-20T18:56:05.243Z
2016-11-15T10:52:04.540Z
2016-11-15T10:52:04.540Z
pela
<p>That sounds very much like a <a href="http://en.wikipedia.org/wiki/22%C2%B0_halo" rel="noreferrer">22° halo</a>.</p>

<p>It's not an astrophysical phenomenon; it results from the refraction of light by ice crystals in Earth's atmosphere.</p>

<p>If this is what it is, it should be a fuzzy but regular circle centered on the Moon, with a radius of about 22 degrees. If you hold your fist out at arm's length with the thumb extended, the angular distance between the base of your fist and the tip of your thumb is typically close to 15 degrees; you can use that for visual comparison. Remember that 22° is the radius of the halo; the diameter is 44°, close to three time the width of your fist+thumb.</p>

CC BY-SA 3.0
2015-04-20T18:59:06.667Z
Aw man, you beat me to it. But +1 for the guide to measuring its size.
pela
CC BY-SA 3.0
2015-04-21T01:47:01.200Z
thanks for the clear description.  I will try to measure the way you said when I see the halo again.
Ganapathy C
CC BY-SA 3.0
2015-04-20T18:48:01.680Z
2015-04-21T23:47:29.050Z
2015-04-21T23:47:29.050Z
Keith Thompson
<p>I have seen a large circular formation of cloud around the full moon sometimes.</p>

<p>Have anyone seen the same? What is the reason behind this formation?
Due to lunar attraction?</p>

<p>Most of the sky is empty and clear, and no clouds at all, only around the moon, far from the moon, not just very close, that cloud circle formed. 
I saw this many times, like the below drawing:</p>

<p><img src="https://i.sstatic.net/G1Euy.jpg" alt="Enter image description here"></p>

CC BY-SA 3.0
2015-04-20T18:52:39.500Z
Maybe a [moonbow](https://en.wikipedia.org/wiki/Moonbow)? Even I have created faint versions of them at night in the moonlight, with a garden hose.
LocalFluff
CC BY-SA 3.0
2015-04-21T14:03:05.223Z
I see these all the time, including around streetlights (even close up). I always figured it was caused by caused by my perscriptive lenses, or maybe the weather (it's constantly very humid where I live).
QuestionC
<the-moon><astrophysics>
Circular formation around the moon
Ganapathy C
# Circular formation around the moon

(Asked by: username_0 on 2015-04-20)

I have seen a large circular formation of cloud around the full moon sometimes.

Have anyone seen the same? What is the reason behind this formation?
Due to lunar attraction?

Most of the sky is empty and clear, and no clouds at all, only around the moon, far from the moon, not just very close, that cloud circle formed.
I saw this many times, like the below drawing:

![Enter image description here](https://i.sstatic.net/G1Euy.jpg)

**Tags:** the-moon, astrophysics
> **Comments:**
> **username_1**: Maybe a [moonbow](https://en.wikipedia.org/wiki/Moonbow)? Even I have created faint versions of them at night in the moonlight, with a garden hose.  
> **username_2**: I see these all the time, including around streetlights (even close up). I always figured it was caused by caused by my perscriptive lenses, or maybe the weather (it's constantly very humid where I live).  

---

## 2 answers

### Answer by username_6 on 2015-04-20. Score: 17 | Accepted answer

That sounds very much like a [22° halo](http://en.wikipedia.org/wiki/22%C2%B0_halo).

It's not an astrophysical phenomenon; it results from the refraction of light by ice crystals in Earth's atmosphere.

If this is what it is, it should be a fuzzy but regular circle centered on the Moon, with a radius of about 22 degrees. If you hold your fist out at arm's length with the thumb extended, the angular distance between the base of your fist and the tip of your thumb is typically close to 15 degrees; you can use that for visual comparison. Remember that 22° is the radius of the halo; the diameter is 44°, close to three time the width of your fist+thumb.

> **Comments:**
> **username_3**: Aw man, you beat me to it. But +1 for the guide to measuring its size.  
> **username_0**: thanks for the clear description.  I will try to measure the way you said when I see the halo again.  

---

### Answer by username_3 on 2015-04-20. Score: 11

I'm guessing what you see is the moonlight being scattered by the hexagonal ice crystals in cirrus or cirrostratus clouds, which lie at very high altitudes, 5-6 km and above. The light is scattered by roughly 22º, and because of the slight wavelength dependence, the halo actually has rainbow-like colors, although often they are so faint that you just perceive it as white.

Read more [here](http://en.wikipedia.org/wiki/22%C2%B0_halo).

[![Moonhalo](https://i.sstatic.net/F2XXY.png)](https://i.sstatic.net/F2XXY.png)

> **Comments:**
> **username_4**: I think you labelled the wrong angles as 22 degrees. This is the minimum deviation of light passing through the crystals. Larger angles are possible.  
> **username_3**: @RobJeffries: Wow…  
> **username_3**: @RobJeffries: Corrected. Thanks a lot for pointing this out.  
> **username_5**: I don't understand why don't you simply mark the outbound angle (from the observer towards the halo) as 22° one on your drawing? That's the angle we're measuring here. The two angles you marked now could be anything, it's just a coincidence that the Sun and the Moon cover more or less the same solid angle (their angular diameter is 31.6 to 32.7 and 29.3 to 34.1 minutes, respectively) when observed from the surface of the Earth. Plus, the angles you mark aren't 22°. Considering Sun's and Moon's distance, they would be nearly identical to half their angular diameter.  
> **username_3**: @TildalWave: The 22º is the amount that the light is deflected. My first drawing was wrong, but is now corrected. The angle we would measure (for the diameter of the halo) is 2x22º, i.e. 44º. The angle doesn't have anything to do with the angle suspended by neither the Moon, nor the Sun. It's just the angle of deflection. Hopefully my new drawing makes this clearer. Apologies for the confusion.  
> **username_3**: @TildalWave: Ah okay, I understand. You're right; in reality, the part of the solid lines between the Moon and the clouds should be drawn as parallel. I think I can live with my drawing not being to scale, though. If you think it's very confusing, I can make another one tomorrow. But now it's 0:30 in Denmark, so I gotta sleep :)  
> **username_3**: @TildalWave: I'm absolutely okay with that, but if the light is scattered by 22º, then the diameter of the halo measured by the observer would be twice that.  
> **username_3**: We're being asked to move this discussion to chat :)  

---

