# Calculating Air Mass at Zenith for observing sites above sea level?

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.

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?

• You should be able to do trigonometry atop of AM 0 AM 1.5 solar spectra – Alchimista Nov 17 '17 at 10:26
• What's the problem with this en.wikipedia.org/wiki/… – Rob Jeffries Nov 17 '17 at 14:23
• Still taking some time to understand the original wikipedia article and consider both answers. Thanks. – Cam_Aust Nov 20 '17 at 14:33
• @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 Nov 21 '17 at 2:25
• @RobJeffries Just so you know. This end ymadte. ("u-matee" - 'you made a difference this end'). As did uhoh. – Cam_Aust Nov 21 '17 at 2:29

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.

A very crude approximation for zenith air mass can be had using a simple 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.

• uhoh you have chosen a real world example so perfectly to my immediate intended site and application. – Cam_Aust Nov 20 '17 at 14:35
• @Cam_Aust glad to hear it! :-) – uhoh Nov 20 '17 at 14:40