NASA's page about Earth atmosphere models gives a weird equation about temperature change in different atmospheric layers. One of the formulas looks like this:
Can someone explain to me what this means?
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$\begingroup$ I just thought that these are two different equations $\endgroup$ – mad.redhead Oct 25 '20 at 14:59
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1$\begingroup$ I'm not sure I understand. They are indeed two different equations. One is for temperature vs altitude. The other is for pressure vs temperature. Every equation is described in the above link, with the associated units. $\endgroup$ – Eric Duminil Oct 25 '20 at 15:14
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1$\begingroup$ sorry, I guess my question was really dumb and incorrect. But I just couldnt stand the fact that T = 15.04 + 0.00649*h is that simple formula to calculate temperature with altitude. $\endgroup$ – mad.redhead Oct 26 '20 at 10:15
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2$\begingroup$ Your question wasn't dumb nor incorrect. In order to save time, you could have mentioned what you understand and what not, but that's about it. It's always better to ask when something's not clear. $\endgroup$ – Eric Duminil Oct 26 '20 at 16:54
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There's some kind of explanation about the units under the picture. The formula is based off the average temperature in degrees Celsius (measured over the entire year, and the entire Earth), which is 15.04 at sea level, and the temperature decreases by 0.00649 degrees every meter above sea level (the average lapse rate). The formula is 'valid' until 11 kilometers, where the troposphere ends.
It's just a model of the average temperature; of course, at the poles it will be colder and at the equator it will be warmer. The lapse rate also varies, while the average of 6.5 °C per kilometer is quite well-known, it can vary between 9.8 °C when it's dry and about 5 °C when it's moist; see Wikipedia for the derivation of these rates.
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$\begingroup$ what if I want to give the equation my own temperature? For example, simulate a hotter day or colder, will the change of (15.04) to other temperature fix this? Or should I also add something like dry air coefficient? $\endgroup$ – mad.redhead Oct 24 '20 at 18:25
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1$\begingroup$ But there is no humidity coef in the equation. Is there any alternative options to calculate atmospheric pressure? $\endgroup$ – mad.redhead Oct 24 '20 at 18:43
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1$\begingroup$ I've added some more information about how lapse rates vary with humidity; I doubt I know much more about the subject than you do. If you now have a question about pressure, it's probably better to ask a new question. $\endgroup$ – Glorfindel Oct 25 '20 at 10:49
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1$\begingroup$ @mad.redhead If you want to account for humidity you'll need to distinguish between the dry adiabatic lapse rate when relative humidity is less than 100% and the moist adiabatic lapse rate when relative humidity is 100%. That said, the atmosphere is only approximately adiabatic. How much the troposphere locally deviates from adiabatic (convective available potential energy) is one of the key indicators of whether bad weather is on the way. $\endgroup$ – David Hammen Oct 25 '20 at 20:00
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3$\begingroup$ @mad.redhead If you need anything precise, the approximation by this equation may be insufficient for your needs. In the real world, all kinds of profiles including inversions and unstable distributions are possible, which is why we use radiosondes and remote sensing techniques to measure profiles of T and q worldwide multiple times per day. The average lapse rate is just good enough for a rough approximation, but nothing more. $\endgroup$ – gerrit♦ Oct 26 '20 at 10:06
