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IA: Indicated altitude TA: True altitude

I have seen this picture before: enter image description here

  • Concept 1 and I understand this. On cold days, the pressure gradient is shorter which means the indicated altitude at a particular pressure setting will be lower. And on a warm day, the pressure gradient is steeper i.e. there are larger steps between pressure levels. Thus, the plane will indicate an altitude higher than true altitude. We know that altimeters measure pressure rather than density altitude.

  • Concept 2 But, we also know that pressure levels are raised on warm days and the indicated altitude is lower than true altitude. This also makes sense. On a warm day, temperature is higher cause a column of air to expand. Thus, at a particular true altitude there'll be more air above me than below me. This will exert more pressure on the top of the plane causing IA to be lower than TA.

But these two concepts seem to be contradicting. How do I reconcile these two bits of information??

  • Concept 3 If I'm flying from a region of cold air to warm air and if I'm keeping a constant indicated altitude, the plane will actually climb. This concept assumes that concept 2 is correct because once its in the warm air region the indicated altitude is lower than true altitude. And we've all heard the phrase "hot to cold, high to low, look out below." So this is all congruent. But if I chose to accept concept #1, then when I fly to a warmer region, my indicated altitude would be higher than true altitude and the phrase would be cold to hot, look out below.

Essentially, if I'm going to be flying on a particular day and the temperature is going to be significantly hotter than normal then am I going to plan with the notion :

A. that air density is lower which means the plane is going to think it's higher than it actually is?

B. Or pressure levels have raised causing the plane to think it's lower than it actually is?

mnemonic: I was told that when asked the question what will the IA be, I can ask myself what would I think. For example, if I'm at sea level and the temperature suddenly dropped would I think I'm up in the mountains? Likewise, if I'm at sea level and the pressure level suddenly dropped making it harder to breathe would I think I'm up in the mountains.

By this metric, I'd think that on a warm day, I'm NOT up in the mountains and hence I'd assume option B.

EDIT: I was actually the one who also asked this why are pressure levels raised on warm days but this question is a slightly different question and is not a duplicate.

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    $\begingroup$ The question that I assume you are asking is a duplicate of the question linked by @quietflyer above. But, in hotter than standard temps true altitude will be higher than indicated altitude. The opposite is true in colder than standard temps. $\endgroup$
    – user22445
    Feb 2, 2022 at 15:17
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    $\begingroup$ Does this answer your question? Why are pressure levels raised on warm days? $\endgroup$ Feb 2, 2022 at 15:28
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    $\begingroup$ @757toga Oops, I meant just altitude but accidently typed in IAS/TAS. I fixed the post. $\endgroup$
    – Jonathan
    Feb 2, 2022 at 17:01
  • $\begingroup$ @xxavier --the abbreviations are supposed to mean "Indicated Altitude" and "True Altitude"-- maybe the question should be edited to clarify this. $\endgroup$ Feb 3, 2022 at 23:24

3 Answers 3

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Concept 1:

On cold days, the pressure gradient is shorter which means the indicated altitude at a particular pressure setting will be lower.

This is wrong, as you can see from the illustration--

The airplane in the blue (cold) airmass is only at about 7500', but is indicating 10,000'. The indicated altitude is too high, not too low.

Concept 1 would be more correctly stated as "On colder-than-standard days, when the air column is more compressed, the isobaric surfaces1 are closer together, so the pressure gradient (i.e. change in pressure per unit change in altitude) is larger, which means that when the Kollsman window is set correctly for the pressure at the surface, the indicated altitude at any particular actual altitude is higher than it would be on a standard day. Because there's more air below the aircraft, and therefore less air above the aircraft. And the altimeter is really a pressure sensor, and the actual atmospheric pressure at any given point is based on the weight of the air column above that point. So under these conditions, the aircraft is flying lower than the pilot would believe based on the altimeter. The higher the AGL altitude2, the more pronounced this effect-- this effect vanishes at the surface, because there, all the air is above the aircraft, no matter how expanded or compressed the air column is."

Concept 2 is basically correct, and is just another variation of concept 1 as I've re-stated it above.

PS It's clear, from content posted in comments, that the original questioner is simply mis-reading the figure. The vertical axis of the graph shows the true altitude, which is different for each of the three cases, and smallest for the cold case. Indicated altitude is the same for all three cases.

See answers to related ASE question Why are pressure levels raised on warm days?

Footnotes:

  1. An isobaric surface is a surface of constant pressure. A "pressure level" is basically the same thing, except that when we are talking about "pressure levels" we may be taking a 2-dimensional slice through the atmosphere, while an "isobaric surface" can be mapped out in three dimensions.

  2. Or more precisely, the higher the aircraft is above the level of the station for which the Kollsman setting would be valid.

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    $\begingroup$ AHHH! hahah I meant altitude again!!! My fingers are automatically typing airspeed, sorry about the recurring confusion. I fully meant altitude... my fault. Fixed. $\endgroup$
    – Jonathan
    Feb 2, 2022 at 17:02
  • $\begingroup$ In your answer, you state "On colder-than-standard days, ... the indicated altitude at any particular actual altitude is higher than it would be on a standard day." But this seems to conflict with the image I pasted, doesn't it? Or am I misunderstanding? Because, the picture suggests that the indicated altitude will be lower than TA on a colder than standard day. $\endgroup$
    – Jonathan
    Feb 3, 2022 at 17:17
  • $\begingroup$ I finally understood what you're talking about. My fault fully. I looked up a different version of that image: teterborousersgroup.org/wp-content/uploads/2015/09/… and it made much more sense... You were right, I was completely misreading that figure. Thank you $\endgroup$
    – Jonathan
    Feb 5, 2022 at 5:25
  • $\begingroup$ Note that in theory, an unusually dry airmass would also cause the air column to contract so that the isobaric surfaces were closer together than normal, having the same effect on the altimeter as an unusually cold airmass, and similarly, an unusually humid airmass would have the same effect on the altimeter as an unusually warm airmass. But I suspect the possible effect from humidity variations is much less than the possible effect from temperature variations. $\endgroup$ Feb 5, 2022 at 13:31
  • $\begingroup$ Makes sense. In fairness, the figure from the OP is really confusing due to the large red arrows. I knew I understood the general concepts and that picture just flipped my entire understanding. $\endgroup$
    – Jonathan
    Feb 8, 2022 at 3:17
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I think that it's likely this will be marked a duplicate question, but nevertheless here is a simple response to what I assume is your core question.

An altimeter merely shows the difference between what you have set in the kollsman window and the atmospheric pressure where your aircraft is currently at (approximately 1000 feet for every inch of pressure difference).

So, for example, if you have 29.90 set in your kollsman window and your airplane is flying where the pressure is 25.90 your altimeter will indicate about 4000 feet.

On a hot day the "actual" vertical distance (altitude between these two pressure levels expands because of the warm air) will be more than 4000 feet. So, your "true" (actual) altitude will be higher than your indicated altitude. The opposite is true when it's colder than standard atmosphere and the pressure levels are closer together.

Perhaps your "Concept" ideas make this a bit more complicated than necessary.

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if I am at sea level and the temperature suddenly dropped, would I think.
I'm in the mountains?

if I am at sea level and the pressure level suddenly dropped, ... , would I think I'm in the mountains?

You really need to brush up on the concept of density altitude.

Instead of playing "3 card monte" with these concepts, it is recommended to take a methodical approach.

Your plane "thinks" exactly like you do regarding temperature, pressure, and density, or, better put, the engineers who wrote the POH took great care to make performance charts and graphs based on pressure altitude corrected for temperature.

It is density altitude you are after to answer your questions. Get into your POH and just plug in some numbers. For pressure altitude, set your Kollsman to 29.92, then use OAT in Celsius to determine density altitude. Density altitude is the altitude you and your plane "feel".

Density Altitude = Pressure Altitude + [120 × (OAT - ISA Temp Celsius)]

The bit about IA relative to TA pertains to when you are flying nearer to the ground and use the local ground atmospheric pressure to determine AGL. The errors introduced in your altimeter from temperature variation can become significant when it is very cold (look out below). There are also charts available for these effects.

Pressure is proportional to density AND temperature

It might be best to get away from the "stack of air" analogy and to apply the "triangle" relationship to your altimeter, which is calibrated to ISA.

if temp is higher than ISA, than thinner air will give the same pressure.

This is why you are higher than you think you are.

if temp is lower than ISA, then denser air is needed for the same pressure.

This is why you are lower than you think you are.

So, it comes down to the altimeter, which reads pressure. The corrections for ISA are the same, the only difference is that one uses local pressure to determine AGL, and corrects that for temperature, rather than 29.92.

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  • $\begingroup$ More reading about ISA and lapse rates. $\endgroup$ Feb 4, 2022 at 13:17

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