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As I'm studying for the PPL knowledge exam, I had the following note:

Pressure levels are raised on warm days and the indicated altitude is lower than true altitude.

(source: AOPA)

But I don't quite understand why.

From my understanding, on warm days the air density is lower. If the air density is lower then the plane will think it's at a high elevation than it really is.

Secondly, if you have warm air and it can expand (since the atmosphere isn't a closed container), the pressure levels should decrease.

Hence, the entire statement that pressure levels increase on warm days and indicated altitude is lower seems false. But apparently it is true. Can someone reconcile this?

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    $\begingroup$ I think there may be a misinterpretation of words here: It doesn't say the pressure increases of warm days, in says the pressure levels raise. What is meant by this is: the geometric height of a surface of equal pressure level is higher on warm days than on cold days. $\endgroup$
    – DeltaLima ♦
    Aug 16, 2021 at 8:13
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    $\begingroup$ Does this answer your question? aviation.stackexchange.com/a/43475/19 $\endgroup$
    – DeltaLima ♦
    Aug 16, 2021 at 9:18
  • $\begingroup$ Hot weather can affect flights. Example: Why Delta’s Athens-Atlanta Flight Had A Boston Stopover: youtube.com/watch?v=IYfJzsxR7kI $\endgroup$
    – gsamaras
    Aug 16, 2021 at 15:33

5 Answers 5

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Remember, altimeters sense pressure rather than density. Pressure is essentially a measure of the weight of the atmosphere above a specific point.

You can imagine the pressure being measured as the weight of a column of air, 1×1 inch square at the bottom and extending up to the top of the atmosphere (around 100 miles). At the surface, that column of air weighs about 14.7 pounds. At 18,000 ft, that column weighs about half as much at around 7.4 lbs.

Thus, when the atmosphere warms, the column of air gets taller but weighs the same. Therefore, at sea level the altimeter wouldn't change. Let's imagine we're at 18,000 ft on a standard day with the altimeter set to 29.92 inHg. In this case, the altimeter reads 18000.

Next, imagine the temperature goes up. Therefore the column of air gets less dense and taller. Assuming the alt setting and height of the airplane stay the same, the height of the air column above the plane gets taller and thus heavier, leaving the altimeter to indicate lower than the actual height of the plane.

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  • $\begingroup$ What is the difference between "atmosphere warms" and "temperature goes up?" In both cases, the air column gets taller but in one case the "air gets less dense" (i.e. in the case where the temp goes up)? Secondly, if the air column is taller but less dense, why would the weight be heavier? Wouldn't the growth in the column be equivalent to the loss in density, keeping the weight the same? I think this is where my confusion is stemming from. $\endgroup$
    – Jonathan
    Aug 17, 2021 at 5:12
  • $\begingroup$ And in general, I was told that if the temperature is really hot, the "density" altitude will indicate a higher altitude than true altitude. So does that mean this statement is false? $\endgroup$
    – Jonathan
    Aug 17, 2021 at 5:14
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    $\begingroup$ @Jonathan There is no difference between those scenarios. The total weight of the column of air hasn't changed. However, at the same true altitude, since the warmer column is taller, but less dense with the same total mass, a higher percentage of that mass will be above you than on a cooler day, thus higher pressure and lower pressure altitude. Your density altitude will indeed be higher, though. Altimeters measure pressure altitude, not density altitude. They have no method of measuring density. $\endgroup$
    – reirab
    Aug 17, 2021 at 7:35
  • $\begingroup$ @Jonathan the two cases aren't "atmosphere warms" and "temperature goes up", the two cases are "temperature goes up and you're measuring at sea level" and "temperature goes up and you're measuring at altitude". $\endgroup$
    – hobbs
    Aug 17, 2021 at 14:45
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    $\begingroup$ @reirab For some reason, your comment sort of made the click in my mind, thank you I get it now. $\endgroup$
    – Jonathan
    Aug 19, 2021 at 18:34
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The formula: PV = nRT is what many are taught to help understand the relationship between pressure, volume, density, and temperature.

It is very important to know, from an aviation point of view, how to apply this formula to an open, unenclosed system.

When air is free to expand Pressure stays constant, increasing temperature increases Volume. This is what causes our "column of air" to grow taller. Because it is taller, it's pressure gradient from top to bottom is less per unit of true altitude, resulting in a higher pressure at a given True Altitude above a ground reference pressure. The altimeter will give a reading lower than true altitude.

This is not to be confused with calculation of Density altitude on the ground for a given pressure and temperature. Because the higher temperature increases the height of our "column", it decreases the Density (same amount of air in a larger volume).

A hot air balloon is an excellent example of how this works. Heating the air inside the ballon allows some of it to escape. The pressure on the inside is the same as on the outside, but the Density of the heated air is much less, giving the balloon enough bouyancy to rise.

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This question (or similar) has been asked many times and in many ways on this site. Unfortunately, an explanation that may be perfectly clear to one person may be confusing to another. Here is my explanation, structured considering you are just working on your PPL, that makes sense to me in response to your question:

Why is it that:

Pressure levels are raised on warm days and the indicated altitude is lower than true altitude.

Air pressure gets lower the higher in altitude you are. So, for example, if you are flying at an altitude where the pressure (level) is 25.00 Hg and the pressure at mean sea level (msl) is 30.00 Hg (and this is what you have set in your altimeter) your "indicated" altitude will be 5000 ft. (Roughly 1 Hg = 1000 ft).

For simplicity, let's assume your "true" altitude is also 5000 ft. Now, as you are flying along, if you encounter warmer air the pressure level of 25.00 Hg that you are flying at (which keeps your altimeter indicating 5000 ft) gets higher in distance (true altitude) above msl.

Since you are flying along keeping your airplane at a pressure level of 25.00 (still indicating 5000 ft) and the pressure level of 25.00 is now truly/actually higher above msl than it was previously before encountering the warmer air, your "true" (actual) altitude above msl will be higher than your "indicated" altitude of 5000 ft.

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The relationship among pressure, density, and temperature is not simple. It is described by the ideal gas law:

$$ P=\frac{\rho k T}{\bar{m}}$$

where $P$ is the pressure, $\rho$ is the density, $T$ is the absolute temperature (in Kelvin or Rankine, not degrees Celsius or Fahrenheit), $k$ is the Boltzmann constant ($1.38\times 10^{-23}~\rm J/K$) and $\bar{m}$ is the average mass of a molecule of air (approximately $4.8\times10^{-23}~\rm g$ for dry air).

Notably, this contains pressure, density, and temperature. So it's not as simple as "if temperature goes up, pressure goes up," since it is equally possible that "temperature goes up, pressure stays the same, so density must go down."

By considering the forces on a parcel of air along with the ideal gas law, and assuming the air is not moving, we can find a pressure lapse rate $dP/dz$ that turns out to be given by:

$$ \frac{1}{P}\frac{dP}{dz} = -\frac{\bar{m}g}{kT}$$

where $g$ is the acceleration of gravity, approximately $9.8~\rm m/s^2$.

Notice that as temperature increases, $dP/dz$ decreases. In other words, the pressure changes less quickly with altitude. This means even if the sea level pressure is standard, the pressure at 1000 feet will be higher than usual. Or, in other words, the pressure levels 29.92 (corresponding to sea level) and 28.86 (corresponding to a pressure altitude of 1000 feet) will be farther apart. So if you climb to 1000 feet of true altitude, you haven't yet reached that level, and your altimeter will indicate lower than 1000 feet.

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It's best to make this very simple. If I'm five feet tall (and I'm not by the way), and I stand in a room that has a 10 foot ceiling, half of the atmosphere in that room is over my head. If I move to a room with a 20 foot ceiling, 3/4 of the atmosphere is over my head. If I am outside, on a warm day the temperature expands the atmosphere and the "ceiling" of the atmosphere goes up. On cooler days the "ceiling" drops. If you pick a fixed elevation, a larger or smaller percentage of the atmosphere is above that point depending on the overall height of the "ceiling" in relation to the fixed point. The more atmosphere that is over that fixed point, the more weight of atmosphere is pushing down on that point, hence a higher pressure. (Changes in air density due to temperature does not offset this in order to simply balance it out as was stated in a comment here.)

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  • $\begingroup$ Suggested edit-- ". If I move to a room with a 20 foot ceiling, 3/4 of the atmosphere in that room is over my head." May seem totally obvious from context, but one possible problem with the "room" analogy is that in the real world, all the atmosphere above the ceiling of the room still has an affect on the barometric pressure at floor level. Your explanation still works, but best to make as clear as possible, so this tiny mod may help. $\endgroup$
    – quiet flyer
    Apr 3, 2022 at 18:31
  • $\begingroup$ The mechanics of this analogy don't match up with the actual atmosphere in at least 2 relevant ways. First, air is far from uniform. At about 18,000' MSL, half the air molecules are below you, and the other half is spread out in the 100,000'+ above you, but the analogy posits uniformity. Second, if things got really cold & your ceiling lowered to 5'1", so barely above your head, you'd have almost no air above you in the analogy, and your altimeter would think you're really high and you'd dive to maintain altitude - the opposite of what actually happens. DV due to deeply flawed analogy. $\endgroup$
    – Ralph J ♦
    Apr 4, 2022 at 0:20

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