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I'm trying to grasp the essence of altitude calculations. One thing I can't figure out in my head is this:

In the International Standard Atmosphere (ISA) the temperature of the air decreases with the altitude, which is called the lapse rate. This is an approximate of 1.98 C (degrees Celsius) or an even more approximate 2 C per 1000 feet. This is usually described through dry adiabatic lapse which means simply that when you go up the pressure rises and the air expands thus through the ideal gas equation it makes sense for the temperature to drop as well.

However when calculating the density altitude we need to "correct" the pressure altitude with this formula:

DA = PA + (118.8 ft/C) x (OAT - ISA temperature)

(from https://en.wikipedia.org/wiki/Density_altitude#Calculation) Here the height change per degree Celsius is only 118.8 feet (approx 120 ft).

My question is this: Why is the relation of (change rate) height and temperature different? With just "going up" it changes 2 C / 1000 feet (ie 500 ft/C) but when adjusting for temperature it changes 120 ft/C).

These both are simple linear equations but have different coefficients.

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  • $\begingroup$ You're misinterpreting the calculations, 118.8 ft/C does not mean that the temperature changes 118.8 C per 1000ft. $\endgroup$
    – GdD
    Commented Aug 8, 2018 at 19:54
  • $\begingroup$ GdD, that's not even what I'm claiming. The change in altitude is either 118.8 ft/C or 500 ft/C $\endgroup$
    – user541905
    Commented Aug 9, 2018 at 7:51
  • $\begingroup$ Sorry @user541905, that's actually what I meant, it isn't saying altitude changes one degree in 118.8 ft/C. Also, keep in mind that that formula gives you an approximation. $\endgroup$
    – GdD
    Commented Aug 9, 2018 at 8:16
  • $\begingroup$ Adiabatic Lapse Rate refers to properties of a packet of rising air which differ from the environmental Lapse Rate of the part of the atmosphere it is rising though. The difference between the two is useful for predicting clouding levels and formation of thunderstorms. $\endgroup$
    – skipper44
    Commented Nov 17, 2020 at 20:09

2 Answers 2

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The equation is saying, that for every degree difference from standard temperature, the difference between pressure and density altitude grows by 118.8 ft. A completely different factor from the lapse rate.

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    $\begingroup$ I get this, but the question is why is the difference 118.8 ft and not 500 ft like it is if we approximate the temperature change at different altitudes? If there would be a totally linear relation between height and temperature it should be the same all the time? Or is this about the fact that talking about feet is essentially hiding the fact that we should be talking about pressure? I get the feeling that talking about feet is trying to make physics "easier" for aviators, whereas I would understand talking about pressure better. (Altimeter is essentially a barometer with a weird scale) $\endgroup$
    – user541905
    Commented Aug 9, 2018 at 7:53
  • $\begingroup$ Or maybe the answer I would be looking for (thinking about your answer and looking at this graph en.wikipedia.org/wiki/Density_altitude#/media/…) is that pressure has a bigger effect on density than temperature (that is to say that in the differential equations defining density, the slope or coefficient of the pressure term is higher than the coefficient of the temperature term to make an approximation) $\endgroup$
    – user541905
    Commented Aug 9, 2018 at 8:02
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    $\begingroup$ @user541905, It is effect of temperature, alone, on density, at given pressure. So it is independent of the effect of pressure. $\endgroup$
    – Jan Hudec
    Commented Aug 9, 2018 at 9:18
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Because it's two different things:

  • the lapse rate is how temperature goes down while you go up in the standard atmosphere
  • the 118 figure is how to adjust pressure altitude to obtain density altitude when temperature differs from ISA temperature.

Check this post to see how that 118 is derived.

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