<p>The factor k, the fraction of sky you cannot see, is not adjusted because of the ground or your peripheral vision. When on a flat plain (a "sea-level" horizon) and none of the sky is covered by clouds, buildings, or trees, then k=0. (Therefore, F=1, and the Zenithal Hourly Rate is not diminished.)</p>

<p>I do not know how the light pollution map was created and whether it accounts for the humidity. The humidity can have a significant effect on the limiting magnitude. In other words, if it is a dry day and the air is clear, you can see fainter stars than if it is a humid day and the sky is murky. If your location is clear (so you have blue skies during the day and black skies at night), then assuming m=6.5 is realistic. (Note that when you click on a location on the light pollution map, it will give the Bortle classification. Clicking the link for the Bortle class will give the expected limiting magnitude.) Of course, the limiting magnitude also depends on your eyesight. Younger people can see fainter stars than older people. (I just tried the map for my location, and it predicts 5.5 to 6.0. I can usually only see stars in the range of 5.0 to 5.5 because of the light pollution and humidity.)</p>

CC BY-SA 4.0
2020-01-07T18:28:37.363Z
2020-01-07T18:28:37.363Z
JohnHoltz
<p>I'm trying to realistically approximate the value of meteor sightings during a meteor shower because I don't want to have high expectations going in. I have a couple questions about this formula I found from <a href="https://in-the-sky.org/article.php?term=zenithal_hourly_rate" rel="nofollow noreferrer">this site</a></p>
<blockquote>
<p>Number of meteors=ZHR×sin(θ)/(F×r^6.5−m)</p>
</blockquote>
<p><strong>Question 1</strong></p>
<p>I would like to know for the calculation of F, where</p>
<blockquote>
<p>F=1/(1−k)</p>
<p>where k is the fraction of the sky which is which you cannot see, because it is obscured by cloud or other obstructions such as trees.</p>
</blockquote>
<p>Assuming I am on a flat plain with no trees/building whatsoever and I have a clear view from 0-180 degrees (cross section view of the earths crust). Will k be 0.5 based on the fact that I cannot see whats below ground (180-360 degrees).</p>
<p>Also will the limitation of the human eye affect this value? I.e If I am facing north my peripheral vision can only reach maybe north east to north west hence further increasing k value above?</p>
<p><strong>Question 2</strong></p>
<p>What value should I have for <code>m</code> in the above formula if I am at <a href="https://www.lightpollutionmap.info/#zoom=10&amp;lat=510052&amp;lon=11369094&amp;layers=B0FFFFFTFFFFFFFF" rel="nofollow noreferrer">this</a> location? I currently assume a value of 6 according to this <a href="https://en.wikipedia.org/wiki/Limiting_magnitude#cite_note-3" rel="nofollow noreferrer">wiki</a>. Is this right?</p>

<amateur-observing><star-gazing><meteor-shower>
Actual meteor sightings formula
C. Wagner
# Actual meteor sightings formula

(Asked by: username_0 on 2020-01-06)

I'm trying to realistically approximate the value of meteor sightings during a meteor shower because I don't want to have high expectations going in. I have a couple questions about this formula I found from [this site](https://in-the-sky.org/article.php?term=zenithal_hourly_rate)

> Number of meteors=ZHR×sin(θ)/(F×r^6.5−m)

**Question 1**

I would like to know for the calculation of F, where

> F=1/(1−k)
>
> where k is the fraction of the sky which is which you cannot see, because it is obscured by cloud or other obstructions such as trees.

Assuming I am on a flat plain with no trees/building whatsoever and I have a clear view from 0-180 degrees (cross section view of the earths crust). Will k be 0.5 based on the fact that I cannot see whats below ground (180-360 degrees).

Also will the limitation of the human eye affect this value? I.e If I am facing north my peripheral vision can only reach maybe north east to north west hence further increasing k value above?

**Question 2**

What value should I have for `m` in the above formula if I am at [this](https://www.lightpollutionmap.info/#zoom=10&lat=510052&lon=11369094&layers=B0FFFFFTFFFFFFFF) location? I currently assume a value of 6 according to this [wiki](https://en.wikipedia.org/wiki/Limiting_magnitude#cite_note-3). Is this right?

**Tags:** amateur-observing, star-gazing, meteor-shower


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## 1 answers

### Answer by username_1 on 2020-01-07. Score: 1

The factor k, the fraction of sky you cannot see, is not adjusted because of the ground or your peripheral vision. When on a flat plain (a "sea-level" horizon) and none of the sky is covered by clouds, buildings, or trees, then k=0. (Therefore, F=1, and the Zenithal Hourly Rate is not diminished.)

I do not know how the light pollution map was created and whether it accounts for the humidity. The humidity can have a significant effect on the limiting magnitude. In other words, if it is a dry day and the air is clear, you can see fainter stars than if it is a humid day and the sky is murky. If your location is clear (so you have blue skies during the day and black skies at night), then assuming m=6.5 is realistic. (Note that when you click on a location on the light pollution map, it will give the Bortle classification. Clicking the link for the Bortle class will give the expected limiting magnitude.) Of course, the limiting magnitude also depends on your eyesight. Younger people can see fainter stars than older people. (I just tried the map for my location, and it predicts 5.5 to 6.0. I can usually only see stars in the range of 5.0 to 5.5 because of the light pollution and humidity.)


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