Anyone know the formula for the amount of atmosphere we have to look through at a given angle above the horizon? A relative value is fine (i.e. let's consider the zenith=90°=1; all other angles are a multiple of that). Seems this should be a straightforward trigonometry problem, but I haven't figured it out. Thanks in advance.

# Formula for Amount of Atmosphere Looked Through at Given Angle?

### #1

Posted 21 October 2020 - 12:40 AM

### #2

Posted 21 October 2020 - 12:46 AM

Here's a formula for atmospheric extinction: https://asterism.org...and-refraction/

https://skyandtelesc...ric-extinction/

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### #3

Posted 21 October 2020 - 12:56 AM

Air mass is the term for what you're looking for. I see it's also mentioned in the calculator linked above.

https://en.wikipedia...ass_(astronomy)

### #4

Posted 21 October 2020 - 01:14 AM

* Believe it or not... the air mass goes almost exactly as 1/cos(zenith angle).* One would think that it's far more complex, because the pressure and temperature vary with altitude and the atmosphere is a spherical shell, so the rays follow curved paths, etc. etc. But for angles reasonably above the horizontal, the profoundly simplistic isotropic parallel-plate approximation captures nearly all of the effect, even including the up-refraction differential. [It's easy to show why, but I won't delve into the math.] Tom

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### #5

Posted 21 October 2020 - 01:38 AM

Anyone know the formula for the amount of atmosphere we have to look through at a given angle above the horizon? A relative value is fine (i.e. let's consider the zenith=90°=1; all other angles are a multiple of that). Seems this should be a straightforward trigonometry problem, but I haven't figured it out. Thanks in advance.

I don't know if this is what you're looking for since you may want the actual formula, but if you download Stellarium and choose a target or area of sky, the description on the upper left hand side tells you the amount of atmosphere you're looking thru as "airmasses". Airmasses represents the amount of atmosphere you're looking thru. Hope this helps.

Regards,

Skyhunter1

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### #6

Posted 21 October 2020 - 02:15 AM

Anyone know the formula for the amount of atmosphere we have to look through at a given angle above the horizon?

Don’t have an equation but I was 4000feet on Green Mountain about 50miles east-northeast from where the Moon was setting over the Puget Sound. Not only was I looking UP seeing through the atmosphere, I was also looking down to sea level through 50miles of air. Yes, I was probably looking through 100+ miles of air, much of it near sea level. The image of the crescent Moon took on the strangest aspect of refraction I have ever seen. The Moon’s image didn’t have its normally gittery appearance but like slow moving twisty clear taffy. The lower the Moon sank into the atmosphere, the more twisted did its image appear & ever slower did it appear to distort. With seconds left before parts of the Moon dipped below the horizon the Moon’s appearance lost all visual clues as a crescent. It twisted ever more BUT at all times appeared as one object, not broken into multiple parts.

**Edited by litesong, 21 October 2020 - 02:22 AM.**

### #7

Posted 21 October 2020 - 05:15 AM

Tom's right. I just use these values for air mass:

1 Zenith

2 30˚

3 18˚

4 9˚

5 Horizon

### #8

Posted 21 October 2020 - 05:49 AM

One would think that it's far more complex, because the pressure and temperature vary with altitude and the atmosphere is a spherical shell, so the rays follow curved paths, etc. etc. But for angles reasonably above the horizontal, the profoundly simplistic isotropic parallel-plate approximation captures nearly all of the effect, even including the up-refraction differential. [It's easy to show why, but I won't delve into the math.] TomBelieve it or not... the air mass goes almost exactly as 1/cos(zenith angle).

If you look at the Sky and Telescope article, the have the same basic formula but then go on to include other factors like Aerosol Optical depth.

They provide two examples, a star at 30 degrees altitude on Mouna Kea, 0.3 magnitudes extinction, and a star at 30 degrees altitude in the eastern US on a mediocre night, 0.8 magnitudes extinction.

Jon

### #9

Posted 21 October 2020 - 06:08 AM

Tom's simplification using secant (1/cos) is close enough until you hit about 5 degrees from the horizon. At that point the air mass grows rapidly and is something like 38 air masses near zero degrees.

It gets even trickier when altitude is included. And I am not going to try to touch extinction right now, with all of the smoke at given altitudes varying night-to-night (I have been smoked out for 3 out of 6 DSO observing nights that looked like they *might* work so far this cycle, and completely blanked the prior two Moon cycles. We have had 0.14 inches of rain from May through October.) I have had 4 mags of extinction at zenith in town many nights over the past month.

### #10

Posted 21 October 2020 - 06:25 AM

Tom's simplification using secant (1/cos) is close enough until you hit about 5 degrees from the horizon. At that point the air mass grows rapidly and is something like 38 air masses near zero degrees.

In general, nobody observes stuff within 5 degrees of the horizon if they have any choice about it.

Anyway, formulae don't work well at such low altitudes because refraction varies unpredictably depending on the atmosphere's temperature profile.

### #11

Posted 21 October 2020 - 06:59 AM

Omega Centauri is one of the few things I observe that low, and it is only a shadow of how it appears 10 degrees in the sky, which is already far too low. I really don't like going much below 20 degrees for any serious observing. There is already too much negative stuff happening at 30 degrees (roughly one additional air mass), so by 20 degrees (~two additional air masses) contrast is really suffering. And that is without accounting for the negative aspects to seeing and atmospheric chromatic dispersion.

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