Talk:Air mass (astronomy)

Move proposal
Title says "Airmass", but the text uses "air mass". Make up your mind :). Thue 14:22, 5 Jun 2004 (UTC)

The contraction "Airmass" seems to be frequently (but not universally) used within the astronomy community. This article uses "Airmass" to differentiate itself from the meteorological "Air mass" article. To clarify this further, if others think it's worth doing, would be to change article names to something like the following. This is discussed further at Talk:Air_mass_coefficient. All articles are already cross-linked, so this is probably low priority:
 * Air mass redirects to new name Air mass (meteorology) and remains the primary entry-point for a general readership.
 * Airmass redirects to new name Air mass (optical path)
 * Air mass coefficient redirects to new name Air mass (solar energy)
 * Air mass (astronomy) new article if someone wants to write it - currently exists only as a brief section stub in Airmass

So the trigger to rename the pages would presumably be when someone decides to write an Air mass (astronomy) article. --anmclarke (talk) 23:42, 28 May 2011 (UTC)


 * In my experience, air mass is far more common among astronomers—almost every source cited by this article has is that way. I question whether it’s really necessary to have a separate article Air mass (astronomy)—though this article is primarily concerned with air mass as it relates to optical path, almost all citations are from astronomical publications. I think this article essentially is that article—I’m not sure how much a separate article would cover beyond a slight expansion of the current subsection. JeffConrad (talk) 04:08, 29 May 2011 (UTC)


 * Another possible trigger is Thue’s original issue; if no one objects, I’ll change all instances of airmass to air mass (we already use the latter in the context of “air mass zero”, so we might as well be consistent). I guess we also need to propose page moves on the talk pages of all affected articles. As for the proper title for this one, we use astronomy, optical path, and atmosphere in the first sentence, so any would probably work (perhaps we could include redirects to cover all of them until appropriate separate articles appear). Logically, Air mass would go to disambiguation, but given the number of links and that it has GA class, it may be best to redirect to the article. JeffConrad (talk) 23:56, 5 June 2011 (UTC)


 * I’ve made the change to “air mass” in the text. A question remains whether we need a hyphen in an adjectival context (e.g., “air mass formulas”); my eye says no, but the case could be made for requiring it.
 * We could now propose the page moves suggested above, or we could probably first just make the moves to Air mass (qualifier) and Air mass (solar energy) to keep the title in sync with the text and make for a more obvious coordination between these two articles. I suspect these moves could be done almost immediately with no controversy, but that we’d need a longer, more formal poll for Air mass (meteorology). Of the qualifiers suggested, I prefer atmosphere or astronomy (we use the latter in link from Air mass), though with the latter we might need to tweak the section heading of the same name. JeffConrad (talk) 05:07, 8 June 2011 (UTC)


 * No one has commented in over a week, so I assume there is no objection to the move. Upon further thought, Air mass (astronomy) seems the best choice, being consistent in making the distinctions on the fields of application. Unless someone objects, I will make this change. JeffConrad (talk) 09:47, 19 June 2011 (UTC)

Section Air mass and astronomy
It seems to me that this section is primarily at “seeing” than air mass as such, but I’d rather defer to someone with more direct experience for retitling. JeffConrad (talk) 00:48, 24 June 2011 (UTC)

Airmass Formulae
Luis, do you have a source for the formula
 * $$X =\sec\, z \,(1 - 0.0012 \sec^2 z) $$?

This looks very much like Young and Irvine's 1967 formula


 * $$X = \sec\,z \,(1 - 0.0012 \,(\sec^2 z - 1))$$,

except that the term in the inner parentheses is missing a "&minus;1" (adding the term doesn't really help the accuracy much). I'm going to add mention of formulae by Hardie, Young and Irvine, Kasten and Young, Rozenberg, and perhaps a couple of others, and I don't want to duplicate mention of Young and Irvine if that is the one you had intended. JeffConrad 08:03, 29 October 2006 (UTC)

Edits of 31 October–1 November 2006
I've added several airmass formulae, as well as numerous references. I have assumed that the first correction to the secant formula intended Young and Irvine's 1967 formula, so I have revised it to that form (there is little practical difference).

The formula for airmass at the horizon was added without citing a verifiable source; I have retained it for now, but am inclined to delete it because it does not seem to agree with the accompanying text: for the standard ground temperature of 290 K, the isothermal scale height is approximately 8500 m; using this value and Earth's mean radius of 6371 km, the calculated airmass is 34.3, vs. the accepted value of 38. I fail to see how this can be described as "very good accuracy."

I did not revise the graph to add plots for the additional formulae; I did revise the graph caption to indicate that not all formulae are included. JeffConrad 02:35, 31 October 2006 (UTC)

I've added some material and revised the section on Horizon airmass in attempt to deal with my earlier objection. I've added an External link to Eric Weisstein's article on airmass, which is consistent with my understanding of the topic. I'm not sure it qualifies as a verifiable source, however, so I've added Fact in a couple of places. I don't have easy access to Young's 1976 chapter that Weisstein cites, so I'm reluctant to vouch for the content. JeffConrad 03:51, 1 November 2006 (UTC)

Allen's airmass table
Allen's 1976 airmass table isn't really a 'reprint' of Bemporad's 1904 Table XXIII; the latter was far more expansive, giving values for every 0.1° for zenith angles between 60 and 84°, and for every minute of arc for angles between 84° and 89° (the latter the limit of Bemporad's table). Allen credits Bemporad, Schoenberg (1929), and Snell and Heiser (1968) for the data. I am unable to determine where Allen got the values for 89.51° and 90°; however, my ability to pick details from the text of Bemporad and Schoenberg is severely limited because I don't read German. JeffConrad 07:39, 14 May 2007 (UTC)

Kasten-Young formula
The formula was originally given in terms of altitude $$\gamma$$ as
 * $$X = \frac{1} { \sin\, \gamma + 0.50572 \,(\gamma + 6.07995^\circ )^{-1.6364}}\;;$$

in this article, it is given in terms of zenith angle $$z = 90^\circ - \gamma$$. JeffConrad 06:40, 10 July 2007 (UTC)

Kasten and Young also takes into account the height of the location. Can you post that one. MySchizoBuddy (talk) 23:58, 16 October 2010 (UTC)


 * They actually don't, either in their interpolative formula or in their discussion of the airmass integral at the beginning of the article. Adjustment of the integral for a height other than sea level is easy enough—rather than use zero as the lower limit of integration, use an arbitrary height robs for an observer at an arbitrary height (r is the observer's height plus the radius of the Earth), as is done in the formula given here at the end of Refracting radially symmetrical atmosphere. But it's a numerical integration rather than evaluation of a closed-form solution. JeffConrad (talk) 01:53, 17 October 2010 (UTC)


 * The Kasten and young formula in the book Solar Engineering for Thermal processes (3rd Ed) by duffie and beckman, Page 10 has the numerator as exp(-0.0001184h). I have seen this in other books as well. MySchizoBuddy (talk) 00:22, 22 October 2010 (UTC)


 * I don't have that book, so I can't directly comment. This appears to be a density correction for an isothermal atmosphere with a scale height of about 8435 m. I've seen a few sources include such a correction factor with the “Kasten and Young formula”, but Kasten and Young don't mention it in their 1989 article, which I consider the authoritative source on their formula. The density correction was apparently added by someone else; though it's done fairly commonly for both airmass and refraction, we need to attribute it to someone other than Kasten and Young if we mention it. JeffConrad (talk) 02:15, 22 October 2010 (UTC)


 * Probably a better way of putting it: this density correction is one that could be just as easily applied to any of the interpolative formulas (and to the homogeneous spherical atmosphere) as to Kasten-Young. JeffConrad (talk) 03:17, 22 October 2010 (UTC)

Font size of Notes
Normally, format changes to notes and references are discussed on the talk page before changes are actually made. Is there a reason other than personal preference to reduce the font size for notes? Although I agree that it's normal practice in paginated material, the benefit is less obvious when substantive footnotes essentially are endnotes. One negative is that the notes may be more difficult for readers, especially those over 40 or so, to read, especially when viewed on high-resolution LCD monitors. JeffConrad 08:23, 22 September 2007 (UTC)

Air Mass and Standard Reference Spectra
What is the purpose of this section?
 * Most of the material is already better covered elsewhere in the article.
 * The formula AM = 1/cos u is an approximation that's only valid at small zenith angles; again, this is already better covered elsewhere in the article.
 * Most of the added references don't meet Wikipedia's criteria for verifiable sources.

This section obviously needs some cleanup, but I think it also needs some justification for even existing. Absent this, I think we should remove it. JeffConrad (talk) 02:02, 4 December 2008 (UTC)

Without more material, I'm not sure this section merits a separate article. But if the topic remains in this article, it should be better integrated with the rest of the article.

The parlance of solar power is mixed with that of astronomy; e.g., the designation “AM” is not commonly used in astronomy, which prefers quantity symbols such as X or M. The nomenclature specific to solar power should be limited to that section and indicated as such.

Much of the material in this section is already covered elsewhere. The first sentence is essentially a repeat of what is in the lead section, so it really isn't necessary. The formula for air mass is the same as the simple one given in the next section, except that it uses a different variable for the zenith angle, and gives the air mass as the reciprocal cosine rather than the secant as is commonly done in astronomy. Moreover, other terminology is different, e.g., “angle of the sun to the vertical” and “sun´s position vector” vs. “zenith angle” used in the remainder of the article. This isn't to say that one nomenclature is right and the other is wrong, but simply that using both is apt to confuse.

My familiarity with solar power is minimal, but it probably makes sense to mention ASTM E-490 and that an air mass of 1.5 is the standard reference for PVs. The point of mentioning an air mass of two is less obvious, since we already have it in several graphs.

I could make a stab at cleaning things up, but without a better idea of the overall intent I'd be making some guesses, which I'd rather avoid. JeffConrad (talk) 21:31, 4 December 2008 (UTC)

Edit of 11 December 2008
I've moved and renamed this section, and removed some redundant material, as discussed above. I've omitted the value for the solar constant because the article is about airmass, not solar irradiance. I'm not sure the reference to ASTM E490 is relevant either, but I've included it nonetheless. I moved the section because it really isn't the main topic of this article.

I haven't reviewed either ASTM standard beyond the online abstract, so including them here is marginal. In particular, unlike the earlier ASTM E891 and ASTM E892, ASTM G173 makes no reference to airmass 1.5 in the title. Perhaps someone familiar with ASTM G173 could confirm that the standard still refers to AM 1.5 internally.

I've changed the lead section to make clear the difference in terminology between astronomy and fields such as solar energy.

As proposed for splitting, the section was titled Air Mass and Standard Reference Spectra, and it read:


 * AM (Air Mass) = 1/cosu, where u is the angle of the sun to the vertical . The air mass is an indication of the path length that solar irradiance travels through the atmosphere . The higher the value of airmass, the greater the attenuating effect of the atmosphere.


 * Air Mass Zero (AM 0) or extraterrestrial (there are no effects due to air attenuation immediately outside the earth's atmosphere) reference spectrum (ASTM E-490) has been made to conform to the value of the solar constant accepted by the space community; which is 1366.1 W/m2.
 * In Air Mass 1 solar spectral irradiance, the sun´s position vector is perpendicular to the Earth´s surface . Corresponds to the sun being directly overhead.
 * In Air Mass 1.5 solar spectral irradiance. It is commonly used as reference condition in rating fotovoltaic modules.
 * In Air Mass 2 solar spectral irradiance, the path of the sunlight through the atmosphere is twice AM 1 and the sun´s position vector has rotated 60.1º away from it perpendicular position.

, available from the Sky & Telescope web site at REFR1.BAS. I have no proven means of comparison (in particular, I don't have a copy of Garfinkel's FORTRAN program), so I can't comment on the accuracy. The program appears to slightly underestimate refraction at the horizon, but the results at least are reasonable. And unlike direct application of the integral at the beginning of Refracting radially symmetrical atmosphere, it appears to give reasonable results for zenith angles greater than 90°. Two caveats:


 * 1) The program does not check to see if a given negative altitude is less than or equal to the dip of the horizon for the specified observer's height above sea level.
 * 2) The program gives the airmass relative to the observer's zenith rather than to the zenith airmass at sea level, and values can be surprisingly large. Two things are at work: the increased optical path of the ray of interest, and the decreased optical path at the observer's zenith, so the resulting airmass can be quite large.  Depending upon the application, this may not be the desired result. The program can be modified to give the airmass relative to sea-level zenith by removing the conditional from statement 990 of the program:




 * This actually doesn't work; the optical path at the zenith for a sea-level observer must be calculated separately by running the refraction subroutine again. The change is straightforward, but outside the scope of this Talk page. JeffConrad (talk) 01:41, 23 October 2010 (UTC)


 * If others think it's sufficiently useful, we could briefly mention this program in the article.


 * A comparison between Schaefer's program giving the airmass relative to sea-level zenith and the lowly homogeneous spherical model suggests that, though the latter indeed underestimates the airmass, its results are much better than those from the other simple formulas. I'd mention this in the article were it not for the concern that we'd run afoul of WP:OR. JeffConrad (talk) 10:03, 12 October 2010 (UTC)

If you want to try the homogeneous spherical model the appropriate variant for use at altitude is given at equation A.4 in Air mass coefficient. This is continuous and algebraically correct at and beyond 90°, certainly it doesn't blow-up to infinity or go negative. It doesn't check whether the path intersects the Earth - it assumes uniform "atmosphere" along the entire path, you'd have to check this with some other suitable annular geometry algebra. However whether it predicts anything particularly accurate or useful about the actual atmosphere is unknown - this is where we again stray into WP:OR territory like JeffConrad was concerned about - for example: (a) most of the research underpinning the models is upward looking and away from the horizon; (b) data and models at Air mass coefficient (table and formulae) indicate very great transmittance unpredictability near the horizon due to the wide variability of low altitude atmospheric effects. --anmclarke 02:10, 28 May 2011 (UTC) — Preceding unsigned comment added by Anmclarke (talk • contribs)


 * The spherical model is correct within its fairly simplistic assumptions: finite atmosphere of uniform density, unrefracted optical path, and single attenuation mechanism. How good it is prompts the question, “Compared to what?”. Given that most of the formulas in this article don’t work at all for zenith angles greater than 90°, the spherical homogeneous model is arguably among the more reasonable. As far as I know, the greatest terrestrial elevation above the horizon is the summit of Aconcagua (roughly 7000 m) facing west; with the minimum altitude set to the approximate dip of the horizon (i.e., the ray is nearly tangent to Earth’s surface), a comparison with numerical integration using a two-layer polytropic model gives the following results:


 * {| class="wikitable" style="text-align: right;"

! colspan="3" | Air mass, observer at 7000 m ! App. alt. !! Polytropic !! Homogeneous
 * −2.5°&ensp; || 69.23&emsp; || 69.73&emsp;
 * −2.0°&ensp; || 48.25&emsp; || 57.31&emsp;
 * −1.5°&ensp; || 35.02&emsp; || 45.31&emsp;
 * −1.0°&ensp; || 26.25&emsp; || 34.00&emsp;
 * −0.5°&ensp; || 20.32&emsp; || 23.97&emsp;
 * 0.0°&ensp; || 16.17&emsp; || 16.07&emsp;
 * 0.5°&ensp; || 13.20&emsp; || 10.77&emsp;
 * 1.0°&ensp; || 10.99&emsp; || 7.59&emsp;
 * 2.0°&ensp; || 8.04&emsp; ||  4.51&emsp;
 * 3.0°&ensp; || 6.23&emsp; ||  3.14&emsp;
 * 4.0°&ensp; || 5.04&emsp; ||  2.39&emsp;
 * 5.0°&ensp; || 4.21&emsp; ||  1.93&emsp;
 * 6.0°&ensp; || 3.61&emsp; ||  1.62&emsp;
 * 7.0°&ensp; || 3.15&emsp; ||  1.39&emsp;
 * 8.0°&ensp; || 2.79&emsp; ||  1.22&emsp;
 * 10.0°&ensp; || 2.27&emsp; ||  0.98&emsp;
 * }
 * 2.0°&ensp; || 8.04&emsp; ||  4.51&emsp;
 * 3.0°&ensp; || 6.23&emsp; ||  3.14&emsp;
 * 4.0°&ensp; || 5.04&emsp; ||  2.39&emsp;
 * 5.0°&ensp; || 4.21&emsp; ||  1.93&emsp;
 * 6.0°&ensp; || 3.61&emsp; ||  1.62&emsp;
 * 7.0°&ensp; || 3.15&emsp; ||  1.39&emsp;
 * 8.0°&ensp; || 2.79&emsp; ||  1.22&emsp;
 * 10.0°&ensp; || 2.27&emsp; ||  0.98&emsp;
 * }
 * 7.0°&ensp; || 3.15&emsp; ||  1.39&emsp;
 * 8.0°&ensp; || 2.79&emsp; ||  1.22&emsp;
 * 10.0°&ensp; || 2.27&emsp; ||  0.98&emsp;
 * }
 * 10.0°&ensp; || 2.27&emsp; ||  0.98&emsp;
 * }
 * }


 * Homogeneous atmospheric height is 8440 m; air mass is relative to sea-level zenith.


 * Although the agreement isn’t great, it’s still much better than the results from most other formulas that give closed-form solutions. And this is an extreme case, with the observer close to the limit of the homogeneous atmosphere; agreement with the more rigorous model is much better at lower elevations. Of course, even the “rigorous” model considers only Rayleigh scattering, so even it’s an approximation.


 * It’s a simple matter to ensure that the geometrical path is in the atmosphere by imposing the constraint
 * $$z < 180^\circ - \sin^{-1} \frac {r_\mathrm{E}} {r_\mathrm{E} + h} \,;$$
 * This is the geometrical dip of the horizon; the difference from 180° places the result in the proper quadrant.
 * Whatever the accuracy, I don’t recall ever seeing the spherical homogeneous model used for zenith angles greater than 90°, and though the derivation is straightforward, I don’t think I’ve ever seen the more general formula published, so including it seems pushing it with regard to WP:OR. If we were to include it, we would probably need a source or a derivation. JeffConrad (talk) 08:33, 29 May 2011 (UTC)

Derivation of homogeneous spherical atmosphere with elevated observer
If I were to add a derivation for air mass (yeah, two words) in a homogeneous spherical atmosphere with the observer at an elevation greater than sea level, I’d probably do something like

An observer at O is at an elevation $$y_\mathrm{obs}$$ above sea level in a uniform radially symmetrical atmosphere of height $$y_\mathrm{atm}$$; applying the law of cosines to triangle OAC,


 * $$\begin{align}

{{\left( {{R}_{\text{E}}}+{{y}_{\text{atm}}} \right)}^{2}} & ={{s}^{2}}+{{\left( {{R}_{\text{E}}}+{{y}_{\text{obs}}} \right)}^{2}}-2\left( {{R}_{\text{E}}}+{{y}_{\text{obs}}} \right)s\cos \left( 180{}^\circ -z \right) \\ & ={{s}^{2}}+{{\left( {{R}_{\text{E}}}+{{y}_{\text{obs}}} \right)}^{2}}+2\left( {{R}_{\text{E}}}+{{y}_{\text{obs}}} \right)s\cos z\text{ ;} \end{align}$$

expanding the left- and right-hand sides, eliminating the common terms, and rearranging gives


 * $${{s}^{2}}+2\left( {{R}_{\text{E}}}+{{y}_{\text{obs}}} \right)s\cos z-2{{R}_{\text{E}}}{{y}_{\text{atm}}}-y_{\text{atm}}^{2}+2{{R}_{\text{E}}}{{y}_{\text{obs}}}+y_{\text{obs}}^{2}=0 \,.$$

Solving the quadratic for the path length s, factoring, and rearranging,


 * $$s=\pm \sqrt{{{\left( {{R}_{\text{E}}}+{{y}_{\text{obs}}} \right)}^{2}}{{\cos }^{2}}z+2{{R}_{\text{E}}}\left( {{y}_{\text{atm }}}-{{y}_{\text{obs}}} \right)+y_{\text{atm}}^{2}-y_{\text{obs}}^{2}}-({{R}_{\text{E}}}+{{y}_{\text{obs}}})\cos z \,.$$

The negative sign of the radical gives a negative result, which is not physically meaningful. Using the positive sign, dividing by $$y_\mathrm{atm}$$, and cancelling common terms and rearranging gives the relative air mass:


 * $$X=\sqrt{{{\left( \frac{{{R}_{\text{E}}}+{{y}_{\text{obs}}}} \right)}^{2}}{{\cos }^{2}}z+\frac{2{{R}_{\text{E}}}}{y_{\text{atm}}^{2}}\left( {{y}_{\text{atm}}}-{{y}_{\text{obs}}} \right)-{{\left( \frac{{{y}_{\text{obs}}}} \right)}^{2}}+1}-\frac{{{R}_{\text{E}}}+{{y}_{\text{obs}}}}\cos z \,.$$

When the observer’s elevation is zero, this simplifies to


 * $$X=\sqrt{{{\left( \frac \right)}^{2}}{{\cos }^{2}}z+\frac{2{{R}_{\text{E}}}}+1}-\frac\cos z \,.$$

With the substitutions $$\hat{r} = R_\mathrm{E} / y_\mathrm{atm}$$ and $$\hat{y} = y_\mathrm{obs} / y_\mathrm{atm}$$, this can be given as


 * $$X=\sqrt{{{(\hat{r}+\hat{y})}^{2}}{{\cos }^{2}}z + 2 \hat{r} (1-\hat{y}) - \hat{y}^2 +1} \; - \; (\hat{r}+\hat{y})\cos z \,.$$

At the zenith, z = 0, and cos z = 1, so that


 * $$X=\sqrt{\frac{r_{\text{E}}^{\text{2}}}{y_{\text{atm}}^{\text{2}}}+2\frac+1}-\frac=\frac+1-\frac=1 \,.$$

as expected. At the horizon, z = 90° and cos z = 0, so that


 * $$X=\sqrt{2\frac+1} \,.$$

Maximum zenith angle The maximum possible zenith angle occurs when the ray is tangent to Earth’s surface; from triangle OCG in the figure at right,


 * $$\cos \gamma =\frac{{{r}_{\text{E}}}+{{y}_{\text{obs}}}-h}{{{r}_{\text{E}}}+{{y}_{\text{obs}}}} \,,$$

where $$h$$ is the observer’s height above the horizon. The geometrical dip of the horizon $$\gamma$$ is related to $$z_\mathrm{max}$$ by


 * $$\gamma ={{z}_{\text{max}}}-90{}^\circ \,,$$

so that


 * $$\cos \gamma =\cos \left( {{z}_{\text{max}}}-90{}^\circ \right)=\sin {{z}_{\text{max}}} \,.$$

Then


 * $$\sin {{z}_{\text{max}}}=\frac{{{r}_{\text{E}}}+{{y}_{\text{obs}}}-h}{{{r}_{\text{E}}}+{{y}_{\text{obs}}}} \,.$$

For a non-negative height $$h$$, the angle $$z_\mathrm{max}$$ is always ≥ 90°; however, the inverse sine functions provided by most calculators and programming languages return values in the range ±90°. The value can be placed in the proper quadrant by


 * $${{z}_{\text{max}}}=180{}^\circ -{{\sin }^{-1}}\frac{{{r}_{\text{E}}}+{{y}_{\text{obs}}}-h}{{{r}_{\text{E}}}+{{y}_{\text{obs}}}} \,.$$

If the horizon is at sea level, $$y_\mathrm{obs} = h$$, and this simplifies to
 * $${{z}_{\text{max}}}=180{}^\circ -{{\sin }^{-1}}\frac{{{r}_{\text{E}}}+h}\,.$$

Obviously, the diagram would be needed (I just didn’t feel like uploading it at this point), and a derivation for the maximum possible zenith angle would probably also be a good idea. If this material were added, I’d put it at the end so that only those interested would need to look at it.

Thoughts? Can we do this and still avoid WP:OR (we already give this formula in Air mass coefficient without a source)? Would the benefit of including it justify the considerable space? If necessary, we probably could shorten this by stopping after showing the reduction to the formula in Schoenberg (1929). JeffConrad (talk) 03:11, 1 June 2011 (UTC)


 * My 2 cents worth: I suggest this does not conflict with WP:OR because it is a pretty straightforward application of high school geometry, and if we can't find it as an exercise in a school textbook, then we're certainly not going to find it in a research paper. Also, it is not novel or unique IP: there are many variants of the derivation, e.g. starting with Pythagoras vs starting with law of cosines or starting with triangles looking upwards or downwards; and ending up in expanded or factorized or normalized forms.  As a further example, the law of cosines article cites no source references and instead relies on various standard proofs included within the body of the article.  Thus it seems reasonable to me to include this derivation in the article as you propose.  I also agree with its placement - at the end and in this article - this is further consistent with existing structure where theory is in this Airmass article and practical application is in the companion Air mass coefficient article.


 * As regards the derivation itself, I have two suggestions:


 * You could start by saying "An observer at C is at..." for consistency with the diagram in the article (File:Airmass_geometry.png) which uses "O" for the centre of the Earth and "C" for the observer/Collector;
 * You could take the derivation one step further to the simplified normalized (dimensionless) form as follows:


 * ...and finally normalizing by substituting $$r = R_\mathrm E / y_\mathrm{atm}$$ and $$c = y_\mathrm{obs} / y_\mathrm{atm}$$ gives


 * $$X=\sqrt{{{(r+c)}^{2}}{{\cos }^{2}}z + 2 r (1-c) - c^2 +1} \; - \; (r+c)\cos z \,.$$


 * --anmclarke (talk) 23:45, 3 June 2011 (UTC)


 * I agree that the symbol for radius should be consistent with the rest of the text and with the associated diagram; I had simply used lower case in a derivation I did some time ago. Because Schoenberg (1929) and Young (1974) use upper case (probably the reason I used it here), it probably makes sense to stick with it. I’ve changed the examples above accordingly.
 * I used “O” to indicate the observer’s position because the majority of similar diagrams I’ve seen do the same. We clearly need another diagram for the elevated observer (and probably for the maximum zenith angle), and as you might guess, both of my diagrams put the observer at “O”. I wonder if we need both the more- and less-general diagrams, but even if we decide to include them, the derivation would be at the end of the article, so the two diagrams would not be adjacent.
 * Both forms of the equation are dimensionless, as they must be. The one in Air mass coefficient has fewer symbols under the radical, but its meaning is also more obscure, and to me, the symbol c is confusing because it’s non-mnemonic. I suppose some mention that the equation can be further transformed would support the inclusion of that form in Air mass coefficient; it would be silly to have a separate derivation there.
 * So I would prefer to stay much with what I have here if this doesn’t present a problem.
 * Though deriving the formula for the maximum possible zenith angle (i.e., grazing ray) would make the section even longer, I think it’s important to indicate that using just any arbitrary zenith angle may not give meaningful results. Some years ago, when testing a program that calculated air mass for a refracted path in a variable-density atmosphere, the results kept blowing up, because I had failed to check that the ray path was on or above Earth’s surface. With the homogeneous spherical model, the program would just silently spit out garbage. JeffConrad (talk) 03:07, 4 June 2011 (UTC)


 * All fine by me. --anmclarke (talk) 03:24, 4 June 2011 (UTC)
 * I’ve updated the material above to reflect what I would add. JeffConrad (talk) 10:53, 4 June 2011 (UTC)
 * I fixed the remaining lower-case rs and added the missing zmax to the second figure. I left out the material in strikeout, including the equations, because the section length is already pushing it. Hopefully, this covers it. JeffConrad (talk) 03:21, 5 June 2011 (UTC)


 * I am currently looking around to see if there is a simple formula for air mass as a function of the solar zenith angle as well as the altitude above the sea level. I thought about the formula given here. An important assumption not stated explicitly here is that the density of the air is a constant throughout the thickness $$y_{atm}$$, which is why at $$y_{obs}=\frac{1}{2}y_{atm}$$ and $$z=0^\circ$$, the air mass is exactly 0.5; at $$y_{obs}=y_{atm}$$ and $$z=0^\circ$$, the air mass is 0. This would leave Mount Everest in vacuum. It is fine to assume an isothermal atmosphere, but the density of an isothermal atmosphere decreases exponentially with altitude. --Roland (talk) 08:46, 6 March 2022 (UTC)

Kasten and Young error
The article is claiming (6.07995 + altitude) is equal to (6.07995 - (90-altitude)) where (90-altitude) = zenith. This is true only if it were converting from a sine to cosine, but it's not. It's raised to a power. I'm making the correction and referring to this if anyone disagrees. I found this out because I created a program a couple of years ago based on this article, only now discovering that this article has been wrong due to bad results from my program. Ywaz (talk) 13:38, 11 March 2012 (UTC)

Isothermal Atmosphere
This expression is hard for me to evaluate, as for a typical atmosphere the middle term would entail taking the exp of a fairly large number when we are near the zenith. Is it correcltly quoted from the reference? — Preceding unsigned comment added by 67.6.156.137 (talk) 18:40, 1 August 2015 (UTC)

Citation for "The Earth is not flat"?
We should remove this "citation needed". Sebagr (talk) 18:00, 20 May 2017 (UTC)


 * Agreed. I removed the tag, but since a cite was requested, I added a wikilink to the shape of the Earth. Doublesuited (talk) 22:00, 13 June 2017 (UTC)

Young (1994) formula seems to be incorrect
I'm trying to vectorize the PNG picture with different formulas. And I have a problem with the Young formula: If I define

z(x) = x/180*pi

young(x) = (1.002432*cos(z(x))*cos(z(x)) + 0.148386*cos(z(x)) + 0.0096467)/(cos(z(x))*cos(z(x))*cos(z(x)) + 0.0149864*cos(z(x))*cos(z(x)) + 0.0102963*cos(z(x)) + 0.000303978) and then plot it (along with other models) I get the picture where this estimation is way differs, which seems incorrect. Can you please help me to understand what is wrong? Amishaa (talk) 14:09, 17 January 2023 (UTC)


 * Found a mistake: I used 0.0149864 instead of 0.149864. After the fix it fits reasonably well (especially taking into account 34' difference between zeniths and true zeniths angles). Amishaa (talk) 00:33, 18 January 2023 (UTC)