<p>A very crude approximation for <em>zenith</em> air mass can be had using a simple <a href="https://en.wikipedia.org/wiki/Scale_height" rel="nofollow noreferrer">scale height</a> model and assume a uniform temperature and composition (which is not true) so you can treat density as proportional to pressure.</p>

<p>In that case using $P(x) = P_0 exp(-x/H)$ where H is the scale height (very roughly 8,000 meters) and remembering that the integral of $exp(-x)$ is $-exp(-x)$ and that you've normalized sea level to 1.0, the zenith airmass can also be roughly approximated as $exp(-x/H)$. </p>

<p>So just for example, at 5,000 <em>feet</em>, that's $exp(-(5000/3.3)/8000)$ or 0.83.</p>

CC BY-SA 3.0
2017-11-20T14:35:48.680Z
uhoh you have chosen a real world example so perfectly to my immediate intended site and application.
Cam_Aust
CC BY-SA 3.0
2017-11-20T14:40:33.137Z
@Cam_Aust glad to hear it! :-)
uhoh
CC BY-SA 3.0
2017-11-18T12:29:25.467Z
2017-11-18T12:29:25.467Z
uhoh
<p>I would use the wikipedia variable density atmosphere model</p>

<p><a href="https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere</a></p>

<p>Which gives airmass as a function of zenith distance for an exponentially decaying atmosphere of scale-height $H$. This formula is valid for all heights above sea level but the airmass given will be in terms of airmass at zenith from that height. To make your adjustment to an equivalent sea-level airmass then you would multiply by the ratio of the density at your observing altitude to the density at sea level.</p>

CC BY-SA 3.0
2017-11-17T14:31:57.000Z
2017-11-17T14:31:57.000Z
ProfRob
<p>Air mass relates to how much of the Earth's atmosphere the light from an observed object must pass through before being observed.  </p>

<p>At Zenith, and sea level, the value of "air mass" is set to 1. </p>

<p>Light from objects lower to the horizon at an observing site must travel through more atmosphere, and hence the brightness of the object is somewhat diminished. In this situation the value of air mass increases above 1.000, presenting greater dimming for the observed object from increased atmospheric light scattering. </p>

<p>As example, when making precise observations of the magnitude of an object in astronomy, there is a need to calculate the air mass for an object at the time of observation, determined in large part by its altitude angle from zenith, or from the geometric horizon. </p>

<p>There are various formulas for calculating air mass. Precise calculations consider several factors and can be quite complicated.  </p>

<p><a href="https://en.wikipedia.org/wiki/Air_mass_(astronomy)" rel="nofollow noreferrer">For more details see - Air Mass (Astronomy). Wikipedia.</a></p>

<p><strong>My question:</strong>  Is there a simple formula to calculate the air mass at Zenith (only), for observing sites about sea level. </p>

<p>Such a calculation provides a comparison of the merit of an observing site at higher elevation over one at lower elevation in regard to the factor of air mass.  </p>

<p>I imagine for most elevations the factor is quite small. Naturally there are plenty of other factors to consider when it comes the the quality of an observing site - humidity, frequency of clear skies, human light sources and light pollution, etc. </p>

<p>To my understanding, the atmosphere in most calculations is considered to be 9 km in thickness. </p>

<p>A very simple calculation could be the site elevation subtracted from 9 km, divided by 9 km, providing a ratio value somewhat less than 1. However this calculation does not account for the changing density of the atmosphere with increasing elevation.</p>

<p>If there is a simple formula, what is it?</p>

CC BY-SA 3.0
2017-11-17T10:26:45.853Z
You should be able to do trigonometry atop of AM 0  AM 1.5 solar spectra
Alchimista
CC BY-SA 3.0
2017-11-17T14:23:16.677Z
What's the problem with this https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Homogeneous_spherical_atmosphere_with_elevated_observer
ProfRob
CC BY-SA 3.0
2017-11-20T14:33:54.997Z
Still taking some time to understand the original wikipedia article and consider both answers.  Thanks.
Cam_Aust
CC BY-SA 3.0
2017-11-21T02:25:45.243Z
@RobJeffries to answer your comment above. Until one understands the underlying concepts, determining which of the several approaches shown in this article as the simple one in my case was not at all clear. Your direction, as the more experienced, "try here" was most helpful. Once one gains the understandings, it proves more easier to identify that option myself.  In my case wanted to know how to compare sites for quick approximate calculations. A different need to fully understand the physics and full field of methods of calculation. Mind you my interest is now peaked. Many thanks.
Cam_Aust
CC BY-SA 3.0
2017-11-21T02:29:24.480Z
@RobJeffries Just so you know.  This end ymadte. ("u-matee"  - 'you made a difference this end'). As did uhoh.
Cam_Aust
<amateur-observing>
Calculating Air Mass at Zenith for observing sites above sea level?
Cam_Aust
# Calculating Air Mass at Zenith for observing sites above sea level?

(Asked by: username_0 on 2017-11-17)

Air mass relates to how much of the Earth's atmosphere the light from an observed object must pass through before being observed.

At Zenith, and sea level, the value of "air mass" is set to 1.

Light from objects lower to the horizon at an observing site must travel through more atmosphere, and hence the brightness of the object is somewhat diminished. In this situation the value of air mass increases above 1.000, presenting greater dimming for the observed object from increased atmospheric light scattering.

As example, when making precise observations of the magnitude of an object in astronomy, there is a need to calculate the air mass for an object at the time of observation, determined in large part by its altitude angle from zenith, or from the geometric horizon.

There are various formulas for calculating air mass. Precise calculations consider several factors and can be quite complicated.

[For more details see - Air Mass (Astronomy). Wikipedia.](https://en.wikipedia.org/wiki/Air_mass_(astronomy))

**My question:** Is there a simple formula to calculate the air mass at Zenith (only), for observing sites about sea level.

Such a calculation provides a comparison of the merit of an observing site at higher elevation over one at lower elevation in regard to the factor of air mass.

I imagine for most elevations the factor is quite small. Naturally there are plenty of other factors to consider when it comes the the quality of an observing site - humidity, frequency of clear skies, human light sources and light pollution, etc.

To my understanding, the atmosphere in most calculations is considered to be 9 km in thickness.

A very simple calculation could be the site elevation subtracted from 9 km, divided by 9 km, providing a ratio value somewhat less than 1. However this calculation does not account for the changing density of the atmosphere with increasing elevation.

If there is a simple formula, what is it?

**Tags:** amateur-observing

> **Comments:**
> **username_1**: You should be able to do trigonometry atop of AM 0  AM 1.5 solar spectra  
> **username_2**: What's the problem with this https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Homogeneous_spherical_atmosphere_with_elevated_observer  
> **username_0**: Still taking some time to understand the original wikipedia article and consider both answers.  Thanks.  
> **username_0**: @RobJeffries to answer your comment above. Until one understands the underlying concepts, determining which of the several approaches shown in this article as the simple one in my case was not at all clear. Your direction, as the more experienced, "try here" was most helpful. Once one gains the understandings, it proves more easier to identify that option myself.  In my case wanted to know how to compare sites for quick approximate calculations. A different need to fully understand the physics and full field of methods of calculation. Mind you my interest is now peaked. Many thanks.  
> **username_0**: @RobJeffries Just so you know.  This end ymadte. ("u-matee"  - 'you made a difference this end'). As did username_3.  

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## 2 answers

### Answer by username_3 on 2017-11-18. Score: 1

A very crude approximation for *zenith* air mass can be had using a simple [scale height](https://en.wikipedia.org/wiki/Scale_height) model and assume a uniform temperature and composition (which is not true) so you can treat density as proportional to pressure.

In that case using $P(x) = P\_0 exp(-x/H)$ where H is the scale height (very roughly 8,000 meters) and remembering that the integral of $exp(-x)$ is $-exp(-x)$ and that you've normalized sea level to 1.0, the zenith airmass can also be roughly approximated as $exp(-x/H)$.

So just for example, at 5,000 *feet*, that's $exp(-(5000/3.3)/8000)$ or 0.83.

> **Comments:**
> **username_0**: username_3 you have chosen a real world example so perfectly to my immediate intended site and application.  
> **username_3**: @username_0 glad to hear it! :-)  

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### Answer by username_2 on 2017-11-17. Score: 1

I would use the wikipedia variable density atmosphere model

<https://en.wikipedia.org/wiki/Air_mass_(astronomy)#Variable-density_atmosphere>

Which gives airmass as a function of zenith distance for an exponentially decaying atmosphere of scale-height $H$. This formula is valid for all heights above sea level but the airmass given will be in terms of airmass at zenith from that height. To make your adjustment to an equivalent sea-level airmass then you would multiply by the ratio of the density at your observing altitude to the density at sea level.


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