Talk:Density of air

Fusion of articles

(this heading was missing and needed) Kurt Artindagi (talk) 02:44, 30 November 2013 (UTC)

Should this page be merged with Atmospheric pressure (or contrawise)? Duk 07:25, 28 Oct 2004 (UTC)

I suspect that they shouldn't be merged, because pressure and density are different concepts. As a quick example, please note that the density calculations require that local pressure be known. Alternately, pressure is an integration of density over altitude. However, the distinction may be more technical than useful, so if you would care to mention why you think there should be a merge, I may find myself enlightened.

Cheers, One-dimensional Tangent 00:36, 30 Oct 2004 (UTC)

Tangent, at first glance I thought these articles should be merged. Because in my mind I see the two concepts as mostly (but not completely) the same. But this is a mild opinion. The impetus was a desire to add information on various atmosphere models (MSIS, NRLMSISE ...) , but I wasn't sure were it should go. Maybe a dedicated page is better. Duk 01:45, 30 Oct 2004 (UTC)

Hmm... If you are planning on doing a page on the various models, or a page for each model, then the Density article would likely be redundant. In that case I definately see your point, and certainly have no objections.

The only thing I see in favor of keeping the Density article is that it might be simpler than MSIS-1986 and NRLMSIS-00, especially for people who only want a rough model and who don't need to know temperature, etc. However, all I know about atmospheric modelling is due to a few hours of light research, so I figure you're probably more qualified to judge this than I am.

So... I'll not take any offense at whatever you choose to do. -grin-

One-dimensional Tangent 02:40, 30 Oct 2004 (UTC)

The density formula had the wrong units, so I multiplied by M (to balance the mol and the Temperature) and divided by 1000 (for the gram to kilogram ratio). It now reads:

Density can then be calculated according to the original formula:

$\rho ={\frac {p\cdot M}{1000\cdot R\cdot T}}$

However, the external link http://wahiduddin.net/calc/density_altitude.htm also looks confused with its units, the gas constant starts off as

R = gas constant , J/(kg*degK) = 287.05

but later uses:

R = 8.31432 gas constant, J/ mol*deg K

This works out as the link uses M / 1000 to correct the different units, but this article appears to have forgotten them. -Wikibob | Talk 22:21, 2005 Jan 17 (UTC)

Since, in this article, you forgot to specify the units which you end up with, that 1000 is irrelevant.

Furthermore, make sure you specify which of the several units given will be used for the pressure (pascals, hectopascals, kilopascals), and check that kelvins per kilometre and see if that denominator fits in too. (Don't remember if units for height are specified in earlier formula, either—should be kilometres, I suppose, if you are using kelvins per kilometre. Gene Nygaard 22:32, 17 Jan 2005 (UTC)

Good point, I've removed the need for the 1000 by adjusting all the units except pressure to SI base units. This means pressure is in pascals, lapse rate in kelvin per metre (looks odd), and height in metres. Although standard units, the lapse rate, height and molecular weight may look odd to some, so here is an alternative with kilometres, and grams (this brings 1000 back, but that could be removed by using kPa for pressure.)

You should point out what mean air molecular mass you are using in the number of moles -- mass conversion. I agree with the statement directly above this. There is no such thing as an 'air molecule", air is 78+% Nitrogen, 20.9% Oxygen, etc... and both N and O exist in air as N2 and O2. Thus the discussion of the effects of humidity in the air is a bit misleading, and should say an "average" value for a hypothetical "air molecule" is calculated from the relative proportions of nitrogen and oxygen in the air, and the fact that they exist as two-atom molecules, thus yielding an average molecular weight for a hypothetical air molecule of 29. — Preceding unsigned comment added by Njaimo (talk • contribs) 22:19, 26 September 2014 (UTC)

I do not think this article needs to be merged with the one on Atmospheric pressure. Pressure and density are indeed, as already said, totally different concepts. This article maybe should be more on the concept of density of "air" as the compound of gasses commonly found in "some reference" atmosphere, i.e. sea level and some common model and in controlled systems. It may be talking about base topics and redirecting to other articles for air in complex systems. It should also treat the sea level density of air in static conditions. It could redirect to Atmospheric pressure for air in the most common earth atmospheric system (and define the system with all the temperature, gravity gradients, composition, ecc.) to explain why density is such in atmosphere. Meanwhile it could redirect to fluid flows for the concepts of density of air (as any fluid) is a fluid dynamics system. And finally redirect to density of air in chemistry and in general the concept of density for compounds of gasses (but I leave this to others as it's not my field anymore). This should turn this article in a nice, correct and tidy one that works as an access point for the many involved concepts of why air and density.

As for an example, a good one could be the classic pressurized piston in static condition with some defined compound of gasses we define as air and showing some calculations of elementary thermodynamics that show the link of density to temperature, (partial) pressure(s) ecc.. So the reader get the idea of why density is such and why of air. Then three links: a links to density, a link to air and a link to thermodynamics models (and maybe chemical, for partial pressure part).

Remove "air molecule" and any reference to it. Else the next step would be students looking for "air" on the periodic table for an "air atom" (and I'm citing personal experience of a student looking for water on it...) (and when you say something they will say you "It's on THE Wikipedia". Oh dear. Let's have the Wikipedia as an help and not as an issue).

Done with August rant :)
Regards Gabriele Dini Ciacci (talk) 11:41, 11 August 2015 (UTC)

---

Effects of altitude

To calculate the density of air as a function of altitude, one requires additional parameters. They are listed below, along with their values according to the International Standard Atmosphere, using the universal gas constant instead of the specific one:

• sea level atmospheric pressure p₀ = 101325 Pa (= 101325 (kg/m·s²) = 1013.25 mbar or hPa = 101.325 kPa)
• sea level standard temperature T₀ = 288.15 K
• Earth-surface gravitational acceleration g = 9.80665 m/s².
• dry adiabatic lapse rate L = −6.5 K/km
• universal gas constant R = 8.31432 J/(mol·K)
• molecular weight of dry air M = 28.9644 g/mol

Temperature at altitude h kilometres above sea level is given by the following formula (only valid below the tropopause):

$T=T_{0}+L\cdot h$

The pressure at altitude h is given by:

$p=p_{0}\cdot (1+{\frac {L\cdot h}{T_{0}}})^{\frac {g\cdot M}{R\cdot -L}}$

Density can then be calculated according to a molar form of the original formula:

$\rho ={\frac {p\cdot M}{1000\cdot R\cdot T}}$

---

Finally for fun I've uploaded a javascript calculator that uses the above parameters here: http://www.geocities.com/robojojoba/air_density.html (source code in http://wikisource.org/wiki/Air_density_calculator) -Wikibob | Talk 00:24, 2005 Jan 22 (UTC)

In school i am designing a fictional planet that is completely water except for a few scattered volcanoes. Would the air be too dense for earth like animals to survive?

---

The text says:

molecular weight of dry air 28.9644 g/mol

the number is correct, and the unit is, but that MOLAR weight each mole of that element containing N (avogadro) molecules weight 28 grams or 0.028 Kg, that not the molecular weight but "molar weight" or "molal weight". Also naturally that mol means mole and not molecule, 28 g would be very heavy one :D

Awaiting international opinions before correcting (maybe international naming is different from my national one)

Gabriele Dini Ciacci 00:03, 13 November 2007 (UTC)

I agree. It should read "molar mass", and be linked to the appropriate wikipedia page. And 0.0289644 kg/mol should be restated as 28.9644 g/mol as you have above.Djd sd 06:13, 14 November 2007 (UTC)

Done. Djd sd 07:54, 16 November 2007 (UTC)

---

The link to the: eXtreme High Altitude Calculator is broken at today. if it keeps to be broken it needs to be removed

Gabriele Dini Ciacci 00:33, 13 November 2007 (UTC)

Let's take that whole line out. It does not really fit into the article. External links should be in the "External Links" section anyway.Djd sd 06:13, 14 November 2007 (UTC)

Done. Djd sd 07:54, 16 November 2007 (UTC)

Relative air density

\rho~_{_{humid~air}} = \frac{p_{d}}{R_{d} \cdot T} + \frac{p_{v}}{R_{v} \cdot T} —Preceding unsigned comment added by 134.221.122.71 (talk) 11:33, 17 February 2009 (UTC)

In my research on corona discharge, I find Peek's law (which I am writing). His book has a variable δ:

δ is the air density factor. It is calculated by the equation:

$\delta ={3.92b \over 273+t}$

where

• b = pressure in centimeters of mercury
• t = temperature in celsius

At SATP (25°C and 760 mmHg):

$\delta ={3.92\cdot 76 \over 273+25}=1$

I also come across this version of "relative air density":

$\delta ={b \over 101.3}{293 \over T}$

which looks at first glance to be the same thing with different units (kPa and K), but I am too tired to do the math. Then I find this formula which has several other variables:

RAD = Relative Air Density

BP = Barometric Pressure (inches??)

RH = Relative Humidity (percent)

T = Air Temperature (°F)

SP = Saturation Pressure (this value comes from a table included in the spreadsheet)

RAD = (1737.97 x (BP - (SP x RH/100))/(460 + T))

What is the relationship between these? Is one a standard formula that would be used in chemistry or whatever? Are b and T standard variable names for these quantities? Do the humidity values affect air density negligibly in most cases, so they are left out of the first ones? - Omegatron 02:54, August 8, 2005 (UTC)

Effects of temperature and pressure

Does anyone think the following line should be in slugs rather then lb_m?

• At standard ambient temperature and pressure (70 °F and 14.696 psia), dry air has a density of ρ_STP = 0.075 lb_m/ft³.

If we went to slugs, then the density would be around .00237 slugs/ft³ according to NASA

Regards Strefli3 21:23, 22 October 2006 (UTC)

There are many "standard" temperatures and pressures used. A user recently changed one of the given standards from 100 kPa to 1 Atm, but without any references or changes to the resulting value of air density. Since you can't change one without changing the other, I reverted the change. According to the linked "standard temperature and pressure" page, there are many standards. As the values listed on the page are just examples anyway, I feel the link to the appropriate Wikipedia page provides enough information to those looking for other standards. Maybe the most common should be listed, but they vary by country and industry. Otherwise we would have to list them all. Djd sd 16:04, 28 April 2007 (UTC)

Effect of water vapor

It seems that the following statement in the article is wrong: "pv = is also known as satuated preaure." Spelling mistakes aside, pv represents the partial pressure of water vapor. Saturation pressure and partial pressure should only be equal if relative humidity is 100%. I believe the correct relationship is pv=saturation pressure*relative humidity fraction. See http://wahiduddin.net/calc/density_altitude.htm as a reference. Djd sd 22:22, 19 March 2007 (UTC)

I made the change. I'll add references when I figure out how. I tried to make the section more clear, but it still needs a cleanup. Djd sd 06:50, 26 March 2007 (UTC)

You are correct about the saturated pressure. I have added the reference to the formula as well. Also in the first section, previously marked as confusing, the units were wrong, I fixed that. Generally it is my opinion that when the math is done using SI units, the result should be presented in those as well. . (Larkuur 09:34, 26 March 2007 (UTC))

Actually, it's not clear to me that units like slugs and pound-feet even belong in this article. Does anyone still learn physics in Imperial units? I realize that Americans still use sort-of-Imperial units in daily life, but I notice that car engines are specified in litres these days, not cubic inches; US scientists certainly use SI units, as does the country's military; and even drug dealers talk in grams and kilos. To me the old units seem useful for a pretty tiny number of readers, and a distraction for the large majority. Oxymetheus —Preceding unsigned comment added by 65.38.32.131 (talk) 00:00, 1 April 2009 (UTC)

The second sentence in section Humid air states This occurs when the molar mass of water vapor (18 g/mol) is less than the molar mass of dry air[note 2] (around 29 g/mol). I think it would be better to change the word when with because because the molar mass of water is always 18 g/mol so, in this context, it is always less than the molar mass of dry air. However, English isn't my mother tongue so I would prefer an English editor changes it if my suggestion seems appropriate enough. --149.34.8.36 (talk) 10:14, 27 October 2021 (UTC)

Need more real-world data points

This article needs more real-world data points. It is fine to be told the density of "dry air", but many readers are likely to want to know the density of the actual air around them -- so, please give some actual densities, at typical humidity, temp, elevation, pressure, etc. -69.87.199.142 23:48, 23 August 2007 (UTC)

Agree with comment above. Something practical might be, how much does the air in my lungs shrink when I go down to the bottom of a 12' deep pool. Or, at what depth of water does a one gallon plastic jug of air become a half gallon? JCW 108.23.6.178 (talk) 05:18, 12 September 2013 (UTC)

Agree with general comment above. I came to Wikipedia (as an engineer) looking for a quick overview and some rough numbers. I'm finding this article is very comprehensive (which is fine); HOWEVER, it too quickly bogs down in detail. Worse however, are too many spelling and grammatical errors, suggesting lack of consensus review (and perhaps technical errors?) -- all reducing my overall confidence in what's here. Perhaps, what I want is in some other "Density of Gasses" or "Properties of Air" article -- leaving the comprehensive discussion "Density of Air" here. I would edit more, but it's not my own expertise. Thanks for what we have. HalFonts (talk) 19:49, 1 March 2015 (UTC)

Standard temperature and pressure

Let's stay consistent with the standard temperature and pressure values. Every so often these data points are changed, sometimes incorrectly. I just looked up the International Standard Atmosphere (as referenced in the "Effects of altitude" section), sea level pressure is 101325 Pa, and sea level temperature is 15° C (288.15 K). There are references to other standards in the "Effects of temperature and pressure" section. We should probably standardize those, too. Or we should at least provide references as to which standards are being used so that the values can be verified. Djd sd 11:40, 30 October 2007 (UTC)

Implementing the formula

I don't know if my understanding is wrong or if the article is not clear enough but to get the "density of dry air calculated using the ideal gas law" formula to work I had to make a some changes. I see that other people in this discussion page have also done similar things ie. multiplying by 1000 etc. I probably didn't read/think about the formula enough, but perhaps a worked through example would help instead of just the formula and the result.

To get this to work:

p = p / (R * T)

for: "At 20 °C and 101.325 kPa, dry air has a density of 1.2041 kg/m3"

ie. 20 °C, 101.325 kPa and 287.05 J/(kg·K) do not just slot straight in to the above formula to get 1.2041. I convert °C to Kelvins and I also multiply by 1000. I think I am multiplying to 1000 to get Pa as opposed to kPa so I have thrown it in with the p component, however it is entirely possible this incorrect.

C/C++/Javaish/Javascriptish:

float DegreesCelciusToKelvin(float fDegreesCelcius) { return fDegreesCelcius + 273.15f; }

float GetDensityOfAirKgPerCubicMeterForPressureKPAAndTemperatureDegreesCelcius(float fPressureKPA, float fTemperatureDegreesCelcius)
{
  const float p = 1000.0f * fPressureKPA; // absolute pressure must be in Pa not KPA
  const float R = 287.05f; // specific gas constant for dry air is 287.05 J/(kg·K) in SI units
  const float T = DegreesCelciusToKelvin(fTemperatureDegreesCelcius); // absolute temperature

  return (p / (R * T));
}

Hullo exclamation mark (talk) 21:11, 9 October 2009 (UTC)

ISA density for aeronautics?

I came to this page looking to confirm the ISA air density (1.225 kg/m3) but couldn't find it. I see this figure for the ISA temperature of 15 deg C but this requires remembering ISA conditions. Would be handy to have some aeronautics-related info if anyone has the time to put it in. Good reference is atmosphere chapter of Shevell's Fundamentals of Flight. Cheers 203.129.23.146 (talk) 22:15, 7 May 2011 (UTC)

Why are properties of sound in dry air mentioned?

Properties of sound are interesting but I see no reason why are they mentioned in the table that should show the density of air. I suppose the extra columns could be moved elsewhere. Simbulu (talk) 23:04, 22 October 2011 (UTC)

Bot amok

The bot reverts the corrections someone else did to the formula - I just lost hours because the formula was wrong. Someone please correct this:

The pressure at altitude h is given by:

$p=p_{0}\cdot \left(1-{\frac {L\cdot h}{T_{0}}}\right)^{\frac {-g\cdot M}{R\cdot L}}$

to

The pressure at altitude h is given by:

$p=p_{0}\cdot 1000\cdot \left(1-{\frac {L\cdot h}{T_{0}}}\right)^{\frac {g\cdot M}{R\cdot L}}$

and

Density can then be calculated according to a molar form of the original formula:

$\rho ={\frac {p\cdot M}{R\cdot T}}\,$

where M is molar mass, R is the ideal gas constant, and T is absolute temperature.

to

Density can then be calculated according to a molar form of the original formula:

$\rho ={\frac {p}{R\cdot T}}\,$

R is the specific gas constant ( J/(kg*degK) = 287.05 for dry air), and T is temperature in deg K. — Preceding unsigned comment added by 94.222.121.251 (talk) 23:29, 7 November 2011 (UTC)

The right formula is

$p=p_{0}\cdot \left(1-{\frac {L\cdot h}{T_{0}}}\right)^{\frac {-g\cdot M}{R\cdot L}}$

The introduction of a constant factor like 1,000 arbitrarily, without including the units that would make it equivalent to the unit, 1, is incorrect. To show why this is so, let us look at a well-known formula for the distance d run at a constant speed v during a time period t:

d = v · t

This formula is right no matter what units you use for speed and for time, provided they are units of speed and of time, respectively. Now suppose you have been talking of metres per second and of seconds, so that the formula gives you distance in metres directly. But now you want to help the reader calculate the distance in feet. The right way to express that in the formula is

d = ( 1 ft / 0.3048 m )· v · t

or

d = 3.28 ft/m · v · t

but not

d = 3.28 · v · t

because the factor ( 1 ft / 0.3048 m ) = 3.28 ft/m is equivalent to 1, the unit, as one foot is equivalent to 0.3048 metres, and multiplying by 1 changes nothing. But the factor 3.28, without units, is clearly not equivalent to 1 and would change all. Kurt Artindagi (talk) 02:36, 30 November 2013 (UTC)

g/cc

I can't think of ANY good reason why air's density in g/cc isn't provided here. A large fraction of the human race uses this as the common units for density. Its absence is embarrassing. I'm adding it.Abitslow (talk) 13:11, 12 November 2013 (UTC)

Helicopter

What is the helicopter doing on this page? Besides the explanation is probably incorrect as well... Kotika98 (talk) 21:17, 25 March 2014 (UTC)

Helicopter rotor blades

are functionally similar to the blades of axial compressors and aircraft wings.

All generate compression in contact surface with a fluid.

Please Kotika98

read Axial compressor#Kinetics and energy equations

and lift (force)#A more detailed physical description

remember (isentropic process):

$\left({\frac {p_{1}}{p_{2}}}\right)^{\frac {1}{\gamma }}$

$=\,\!$

${\frac {\rho _{1}}{\rho _{2}}}$

RookTorre (talk) 16:14, 26 March 2014 (UTC)

Hi, I am the annon that has just removed the helicopter picture from the article. Agreeing with Kotika98, a helicopter (or any wing) flies due to pressue not density differences between upper and lower surfaces (density differences may appear above say Mach 0.6 but are not critical to the process). Regardless, I think the generation of lift is a complex topic best left to the lift (force) article. 217.109.123.82 (talk) 09:10, 23 April 2014 (UTC)

Disagree , idealized incompressible regimes are not realistic. Hand clapping generates compression in the air. Disagree ?

It is not a question of the limits of use of calculations, is conceptual.

Please read: Anderson, John David . (1989) Introduction to Flight, Third Edition , ISBN 0-07-001641-0

(4.2 Incompressible and compressible flow - pg 86) ; (cited : "incompressible flow is a myth" pg 87)

RookTorre (talk) 20:20, 23 April 2014 (UTC)

Hi, fair point the flow will not be perfectly incompressible like in theory. Nevertheless a helicopter is not flying because of any density differences between the upper and lower surfaces, but because of a pressure difference. Hypothetically, if it were possible for a helicopter to have high density air beneath the rotor and low density air above the rotor, but both at the same pressure- it would not fly. Therefore I propose to remove the picture again:

• The picture is misleading- it implies a helicopter flies because of density differences between the upper and lower rotor surfaces.
• The picture is inaccurate- it shows something like a x3 increase in air density below the rotor, which is far too high
• The picture is outside the scope of an otherwise good article- it could be moved to Helicopter or lift (force)

217.109.123.82 (talk) 08:52, 25 April 2014 (UTC)

Disagree,

• The picture is misleading- it implies a helicopter flies because of density differences between the upper and lower rotor surfaces.
  • No, only: "rotor blade pushes the air down increasing your density" cited.(no more)

• The picture is inaccurate it shows something like a x3 increase in air density below the rotor, which is far too high.
  • What would be the appropriate scope an limit layer in blade surface? Submit verifiable references.

• The picture is outside the scope of an otherwise good article it could be moved to Helicopter or lift (force)
  • This case: "The picture is misleading- it implies a helicopter flies because of density differences between the upper and lower rotor surfaces.", your words.

Only air density increasing

Please, illuminates me

what is the real issue?

RookTorre (talk) 16:09, 25 April 2014 (UTC)

"The movement of the helicopter rotor leads to a difference in pressure between the upper and lower blade surfaces, allowing the helicopter to fly. A consequence of the pressure change is local variation in air density, strongest in the (blade) boundary layer (good, its ok. But:) or at transonic speeds."

...or at transonic speeds?

to Bell 47G at 360 rpm and 11.3 m of disc diameter...

the spin velocity(max) is only 213 m.s^-1 (0.63)

better to remove or modify this comment, agree?

RookTorre (talk) 02:18, 29 April 2014 (UTC)

I meant generally compressibility is most noticable at transonic speeds+, as it is I don't think it reads as if I am talking specifically about the particular helicopter (Bell 47G), but if you think otherwise feel free to rephrase. For reference 213m/s + 169km/h Bell-47 top speed = 0.78 Mach on the advancing blade and that is for a slow helicopter. 217.109.123.82 (talk) 12:07, 29 April 2014 (UTC)

ok

the question is conceptual

I leave to others

ty

RookTorre (talk) 20:38, 29 April 2014 (UTC)

Do remove the Helicopter.

1. It has nothing to do with the article topic.
2. The article is about thermodynamic pressure in a static or quasi-static system. The helicopter example talks about "pressure" and being in fluid dynamic it's not clear which one. It is also calling in velocity so implicitly the linking of pressure and momentum change of air molecules. To not be long, it's a messy example and has nothing to do with the original topic since it's analyzing a much more complex system. If it was in a book it would be the stereotypical "wrong and cool" example that helps you to sell books and not make science.
3. If you want to put in a pressure example in fluid dynamics (albeit it not being "air" but any fluid), use a laminar flow around an airfoil at very low Reynolds where there is no dissipation and full pressure recovery. I still think it's a bad example but if you want fluid dynamics, that's it.
4. If you want a good example make the example of a pressurized piston in equilibrium. That does allow you to talk naively about "pressure" without any specification.
5. Compressiblity is noticeable from Mach 0.3. And depends on what "noticeable" means. If by it we mean as commonly implied "noticeable on drag effects" then from 0.3 to up 0.7 the Prandtl–Glauert correction works decently. So you notice it but it's easy to calculate a "reasonably precise" value for most of the involved quantities.
6. Calling in an helicpoter would imply specifying it's advance ratio that is better than blade mach number cause indeed density of air changes the speed of sound and so changes the mach. This is out of the article scope. With this point I mean that a proper example with an helicopter will drift the article out of topic or simplify too much to the extent of getting it wrong.
7. Why the example talks about boundary layer, while in an helicopter the bl is totally a crazy thing cause the blade flaps, twists and moves and you may have any sort of detachment going on and defining pressure in the bounday layer is a topic per se. In the boundary layer, in a static case, at best you have a pressure gradient. So there is not one pressure in the boundary layer. And again "pressure" is vague in fluid dynamics, depends about what we are speaking about.

Sorry if I was polemic (I was... but it's upsetting when you see such a thing there and there is a discussion if to remove it. It's addition was more like a very successful troll edit. It did trolled me. I bear no offense to the author for the nice artwork and I really respect him. But this article is not the right place for that example. The whole point is, please, it's wrong, remove that example.

Also be careful when citing Anderson's books. He is an exceptional teacher and I know Anderson's book are very authoritative in the US, but the rest of the world also have exceptional fluid dynamicists especially on fundamental topics that maybe it's worth reading. Notice that "Introduction to Flight" is a book on that title and not about compressible and incompressible fluid flows.

Ok sorry I am trolling myself (but for the good) I really hope the Helicopter gets removed... I'll not sleep for days now...

Gabriele Dini Ciacci (talk) 11:08, 11 August 2015 (UTC)

Air density value revision 611188337 by 72.69.255.121

Hi 72.69.255.121

Please see U.S. Standard Atmosphere, 1976; Table I - Geopotential Altitude, Metric units; value for: Altitude Z(m) = 0; same for ICAO Standard Atmosphere.

note: "The International Organization for Standardization (ISO) publishes the ISA as an international standard, ISO 2533:1975.[1] Other standards organizations, such as the International Civil Aviation Organization (ICAO) and the United States Government, publish extensions or subsets of the same atmospheric model under their own standards-making authority.

...

The International Civil Aviation Organization (ICAO) published their "ICAO Standard Atmosphere" as Doc 7488-CD in 1993. It has the same model as the ISA, but extends the altitude coverage to 80 kilometres (262,500 feet)." see International Standard Atmosphere

RookTorre (talk) 02:07, 3 June 2014 (UTC)

First section readability

I am by no means a style editor, but I find the second paragraph really hard to read. I made a couple changes already for misuse of plurals, "a/an", and "the" today. Since this page is frequently edited, and I don't have much expertise in Project Physics, posting it here to get some feedback before someone moves an improved version over to the main page.

Currently the paragraph reads:

The air density is a property used in many branches of science as aeronautics;[1][2][3] gravimetric analysis;[4] the air-conditioning[5] industry; atmospheric research and meteorology;[6][7][8] the agricultural engineering in their modeling and tracking of Soil-Vegetation-Atmosphere-Transfer (SVAT) models;[9][10][11] and the engineering community that deals with compressed air[12] from industry utility, heating, drying and cooling processes[12] in industries like cooling towers, vacuum and deep vacuum processes,[5] high pressure processes,[5] gas and light oil combustion processes[5][12] that power turbine-powered airplanes, gas turbine-powered generators and heating furnaces, and air conditioning[5] from deep mines to space capsules.

I am suggesting:

Air density is a property used in many branches of science such as aeronautics,[1][2][3] gravimetric analysis,[4] atmospheric research and meteorology,[6][7][8] agricultural engineering Soil-Vegetation-Atmosphere-Transfer (SVAT) models.[9][10][11] It is used by the engineering community that deals with compressed air[12] from industrial utility, heating, drying and cooling processes,[12] vacuum and deep vacuum processes,[5] high pressure processes,[5] and gas and light oil combustion processes[5][12]. Air density is used in industrial applications like cooling towers, turbine-powered airplanes, gas turbine-powered generators, heating furnaces, and air conditioning.[5]

Wysockat85 (talk) 17:49, 23 September 2016 (UTC)

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Cheers.—InternetArchiveBot (Report bug) 23:49, 8 September 2017 (UTC)

Incorrect value for altitude-dependent formula?

The formula $p=p_{0}\left(1-{\frac {Lh}{T_{0}}}\right)^{\frac {gM}{RL}}$ makes no sense, since the result becomes imaginary as ${h}$ approaches ${L}/{T_{0}}$. Therefore, I changed it to $p=p_{0}\left(1-{\frac {Lh}{T}}\right)^{\frac {gM}{RL}}$ where $T=T_{0}-Lh$. Please note I don't have any prior knowledge and haven't done research on this topic whatsoever, but this seemed to be the only appropriate change, seeing as the formula for $T$ wouldn't be there if it wasn't relevant. Feel free to change it back if you can find proof that this is incorrect.

Units are wrong

If you go to the subheading dry air, it originally uses metric units, then without warning switches to imperial? This is confusing! — Preceding unsigned comment added by 99.43.252.75 (talk) 19:01, 9 December 2020 (UTC)

Dry air — fact or fiction?

This section presents formulae, then values for the titular property, with various units. As appearing, this is not trustworthy content: by Wikipedia's own rules, sources are required for all information that is not common knowledge (and these formulae and values are not common knowledge).

Where are the formulae from? Where are the values from? (The fact that the units of the values bulleted change from line to line—why?—suggests their being drawn from sources, or possibly that plus some added calculations.)

The guidelines and policies also state that original research is prohibited (which, even if sources for the formulae were given, would prohibit writers here from the research activity of creating spreadsheets/writing code to run numbers through formulae to calculate values—for the expertise and periodic interpretive decision-making required in calculating, rather than just reporting, property values). — Preceding unsigned comment added by 2601:246:C700:558:64C9:EAA7:4A93:8547 (talk) 19:53, 3 November 2021 (UTC)