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I have gotten conflicting answers from CFI's on this one. In bold are the only things that changed or would potentially change.

Initially:

Aircraft is parked at a very dry (to factor out humidity) airfield

  • Field elevation of 0 ft
  • Local altimeter setting is 29.92
  • Temperature is 15 C
  • 29.92 in the Kollsman window
  • Indicated altitude is 0 ft

Later:

Aircraft is parked at same very dry (to factor out humidity) airfield

  • Field elevation of 0 ft
  • Local altimeter setting is 29.92
  • Temperature is 40 C
  • 29.92 in the Kollsman window
  • Indicated altitude is ? ft

I know the density altitude increased. I know the pressure altitude remained the same. However I would like to know if the indicated altitude is greater than 0 ft, less than 0 ft, or 0 ft.

I have seen diagrams online that try to explain this concept but confusingly enough, change both true altitude and temperature in their explanations. I am keeping true altitude constant here to get to the bottom of it.

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  • $\begingroup$ Robert, I have seen others on stack exchange say this is correct: link and it shows that in warm conditions, indicated altitude reports lower than true. What say you? $\endgroup$
    – user59676
    Commented Aug 4, 2021 at 21:42
  • $\begingroup$ Simply put, your basic altimeter measures the existing ambient air pressure where it is currently located in the atmosphere and mechanically compares that value to what ever is set in the kollsman window using a scale of (roughly) 1 inch hg equals 1000 feet. So, if your altimeter is set to 30.00 hg and the altimeter is physically located in the atmosphere where the ambient pressure is 25.00 hg, the "indicated" altitude will be (roughly) 5000 feet. That's it. $\endgroup$
    – user22445
    Commented Aug 4, 2021 at 23:07
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    $\begingroup$ If "absolutely nothing changes", then pressure doesn't change, then pressure altitude doesn't change by definition. QED. (If we ignore thermal expansion of the internal components of the altimeter, but it is generally unpredictable how this would affect indication). $\endgroup$
    – Zeus
    Commented Aug 5, 2021 at 1:46
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    $\begingroup$ Pressure is a measure of the mass (weight) of air above you. If the only change is that the temperature increases, then the column of air above you is taller (air will expand), but it still masses (and weighs) the same. $\endgroup$
    – Charles Bretana
    Commented Aug 5, 2021 at 3:46
  • $\begingroup$ Charles Bretana, your answer implies that temperature change given constant pressure does not change the indicated altitude. but this link seems to say that it does. It appears to say warmer conditions make indicated altitude report lower than true, given constant pressure. What am I missing? $\endgroup$
    – user59676
    Commented Aug 5, 2021 at 14:21

6 Answers 6

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Using your examples:

If the field elevation is 0 and the altimeter setting is 29.92 then both pressure altitude and indicated altitude will be 0. Remember, pressure altitude is the altitude displayed (indicated) on your altimeter when the altimeter is set to 29.92.

If the temperature increases from 15C to 40C per your examples, and the altimeter setting for your airport (with a field elevation of 0) is still 29.92, the "indicated" altitude will still be 0. (same as pressure altitude)

So, the increase in temperature (15C to 40C) does not change what your altimeter will display (indicated altitude) assuming you have the same setting in the kollsman window of your altimeter. (in your examples where 29.92 is the current "altimeter setting" for the airport. )

What the change in temperature from 15C to 40C does affect is a difference between "Indicated" altitude (what your altimeter is displaying) and "True" altitude (actual altitude above mean sea level -MSL). But in your examples, since you are "parked" at an airport that is at MSL, and the altimeter setting is the same (29.92) then the pressure altitude, indicated altitude, and true altitude are all the same, that is "0."

If you were flying over that airport (using the current altimeter setting) at an "indicated" altitude of 5000 feet and the temperature was 40C (which is much warmer than the standard atmosphere temperature) your "true" altitude (actual altitude above msl) would be higher than your "indicated" altitude of 5000 feet.

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Altimeters are "told" what the pressure is at MSL. They then apply a model of the pressure above that point to map to altitude. The model assumes standard temperature. The model becomes less correct as temperature deviates from standard.

Because your scenario has the plane at a MSL field and the altimeter is set properly for MSL, the indicated altitude will be 0.

For actual altitudes above MSL, but with temperatures above standard, the altimeter will indicate an altitude below the true altitude.

Note that most materials assume you are flying and want to get the true altitude from your indicated. Your question is in the reverse where you want to get the indicated from a known true. On a hot day, indicated elevation is less than true elevation for altitudes above the reference location (which is normally 0MSL).

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  • $\begingroup$ So if you went from parked at a 1000' airport, 29.92 reported and in the window, 13 degrees (ISA for that altitude) and 1000' indicated to: 1000' airport, 29.92 reported and in the window, 40 degrees, there would be a change in the indicated altitude? If so, I regret posing my initial question at 0'. $\endgroup$
    – user59676
    Commented Aug 5, 2021 at 13:56
  • $\begingroup$ @user59676, Right. 0MSL is special because that's the reference. For fields above that, the indicated altitude should be lower when the temperature rises but the altimeter is properly set. $\endgroup$
    – BowlOfRed
    Commented Aug 5, 2021 at 15:51
  • $\begingroup$ @user59676 - the altimeter "only" displays/indicates the difference in pressure (nothing to do with temp) between what is set in the window and the pressure at its current location using a scale of (roughly) 1 hg equals 1000 ft. Ex. set 30.00 and fly where the pressure is 25.00 and it "indicates" 5000 ft. When it is hotter than isa the actual "true" difference will be higher than what is indicated. Because higher temp increases the actual vertical width between 30.00 and 25.00. When it's very hot the actual "true" altitude difference may be 5500 ft not 5000 ft. $\endgroup$
    – user22445
    Commented Aug 5, 2021 at 16:16
  • $\begingroup$ So just so I understand your stance, 757toga. In my scenario posed in these comments, is your answer less than, more than or equal to 1000' on the indicated? $\endgroup$
    – user59676
    Commented Aug 5, 2021 at 16:21
  • $\begingroup$ @user59676 - see the image in this link and associate it with my comment above. This may help in your understanding. 4.bp.blogspot.com/-aLf5lfwqpXk/Uw0Fk-b60yI/AAAAAAAAAEI/… $\endgroup$
    – user22445
    Commented Aug 5, 2021 at 16:26
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Interestingly enough, the FAA needed to put out guidance on this issue you are trying to get to the bottom of.

Here is the scenario. You are flying on a fixed glidepath to the runway. You are exactly in the middle of the localizer and glideslope signal. To put things into perspective, these signals do not change with atmospheric conditions. A glideslope bean will put the aircraft at the same true altitude every single day.

On ILS approaches with quite a few stepdown procedures, the requirements is to remain at or above the MEA for each segment until the final approach fix at which time you will descend on the glideslope to the runway.

Quite a few pilots will capture this glideslope at an altitude higher that what the published altitude is at the final approach fix.

Here is the jist of the InFo from the FAA. In cold than normal weather, those pilots who capture the glideslope high will most likely remain at or above the MEAs for those intermediate segments. IAF --> stepdown --> stepdown --> PFAF.

In warmer than normal weather, the same pilot will most likely descend below the MEA for those segments and could cause traffic issues for other aircraft.

https://www.faa.gov/other_visit/aviation_industry/airline_operators/airline_safety/info/all_infos/media/2011/InFO11009.pdf

The saying "FROM HIGH TO LOW... LOOK OUT BELOW" is true for both pressure and temperature.

The METAR for the airport will automatically compensate for the temperature differences through its use of the altimeter setting. Remember, the altimeter setting is representing the altitude you would be at above sea level.

A really good example of this is in Colorado. Today, the METARS look like this.

Wilkerson Pass (Elev. 11,259 FT) - K4BM 132135Z AUTO 17005KT 10SM SCT032 SCT038 BKN120 12/05 A3077 RMK AO2 LTG DSNT NE

Red Cliff Pass (Elev. 12,047 FT) - KCCU 132135Z AUTO 11013KT 10SM SCT037 SCT055 BKN080 18/04 A3078 RMK AO2

La Veta Pass (Elev. 10,124 FT) - KVTP 132135Z AUTO 10010G14KT 10SM SCT025 SCT034 SCT039 17/08 A3064 RMK AO2

Alamosa Airport (Elev. 7,542 FT) - KALS 132152Z AUTO 11013G25KT 10SM CLR 28/06 A3038 RMK AO2 PK WND 11027/2126 SLP182 T02830061

Colorado Springs Airport (Elev. 6,187 FT) - KCOS 132054Z 14011G18KT 10SM SCT070 SCT250 28/08 A3038 RMK AO2 SLP195 T02830083 56009

Gunnison Airport (Elev. 7,667 FT)- KGUC 132127Z AUTO 08016KT 10SM -RA SCT080 OVC100 22/08 A3042 RMK AO2 PK WND 12029/2105 LTG DSNT E AND SE TSE16RAB24 PRESRR P0000

I have tried to show passes and airport close to each other to preclude the possibility or an airmass or pressure change.

Look at those altimeter settings. They are corrected for the non-standard temperatures to show true altitude at their locations. The altimeter setting must be higher in the passes because it is warmer than standard today in Colorado. If you look at these airports in the winter, you may find the altimeter settings are lower than the corresponding airports.

The elevations listed are for the weather station and not the airport elevation as found in the chart supplements.

https://aviationweather.gov/adds/dataserver_current/httpparam?dataSource=metars&requestType=retrieve&format=xml&stationString=K4BM%20KCCU%20KVTP%20KALS%20KCOS%20KGUC&hoursBeforeNow=1

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  • $\begingroup$ Really good info. But how, on a very hot day, while following the GS inside the FAF for the ILS rwy 25L at LAX do you ensure that you pass LADLE at or above 700 as published? airnav.com/airport/KLAX $\endgroup$
    – user22445
    Commented Aug 13, 2021 at 23:12
  • $\begingroup$ LADLE is only for localizer minimums. Why would you use it on an ILS? TERPS requires there are no waypoints from the PFAF to the MAP on a ILS. This is why FMS's don't show waypoints inside the FAF. $\endgroup$
    – wbeard52
    Commented Aug 14, 2021 at 0:20
  • $\begingroup$ I fully agree. However, if you note other ILS/LOC approaches such as the KPHX ILS/LOC to Rwy 25L (and many others), where there is a LOC only stepdown/crossing altitude inside the FAF there is an asterisk next to the crossing altitude restriction (inside the faf) and a note indicating that the crossing altitude applies only to the LOC approach. There is not a similar note with the KLAX 25L ILS/LOC approach. aeronav.faa.gov/d-tpp/2108/00322IL25L.PDF I agree that on the KLAX approach the 700 applies only to the LOC. But this issue has caused confusion for many years. $\endgroup$
    – user22445
    Commented Aug 14, 2021 at 0:38
  • $\begingroup$ I see your point. I agree with you. I looked for such a note and could not find one. This would be a good thing to report to the FAA. $\endgroup$
    – wbeard52
    Commented Aug 14, 2021 at 0:48
  • $\begingroup$ faa.gov/air_traffic/flight_info/aeronav/aero_data/… $\endgroup$
    – wbeard52
    Commented Aug 14, 2021 at 0:49
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Altimeters work by comparing static air pressure on a sealed metal bellows.

Further research has shown that the bellows is evacuated to eliminate temperature effects and only reads air pressure by measuring the distortion of the metal. If you put the altimeter in an oven (with a venting port) and raised its temperature from 15 C to 40 C, it should read the same altitude! (Question answered right there.)

Above the ground, altitude errors can be introduced by changing barometric pressure caused by highs and lows as well as non standard lapse rates using ground readings as references.

With this particular question, flying locally with temperatures rising from 15 C to 40 C, above the ground the altimeter will read lower than true altitude. But, upon landing, it will read: 0 feet at 29.92.

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  • $\begingroup$ I agree. But taking the question literally may address exactly what the OP is asking. Many answers and comments are saying the same thing in different ways. Making it simple (for me) is the fact that a conventional baro altimeter measures the difference in pressure between what ever is set in the window and what pressure the altimeter is currently exposed to. Ex. 30.00 set in the window and the altimeter (in the aircraft) is currently residing at a position in the atmosphere where the ambient pressure is 25.00 will produce an indicated altitude of (roughly) 5000 ft. $\endgroup$
    – user22445
    Commented Aug 13, 2021 at 16:01
  • $\begingroup$ good comment. The mechanics of a baro altimeter are quite simple. It's when all of the variables are hypothetically insinuated into a question that seems complex that the rather simple operation of an altimeter becomes mired in unnecessary confusion. $\endgroup$
    – user22445
    Commented Aug 13, 2021 at 16:24
  • $\begingroup$ There are ways to compensate for metal expansion or contraction due to changing temperatures. Altimeters are not unlike precision watches. $\endgroup$
    – Robert DiGiovanni
    Commented Aug 14, 2021 at 21:35
  • $\begingroup$ no disagreement. But the functionality of a baro altimeter is nevertheless quite simple. Irrespective of the temperature the altimeter will indicate the difference between what value (e.g. 30.00) is set in the kollsman window and the ambient pressure existing at its current location (e.g. 25.00). In this example the altimeter would indicate 5000 ft. But if colder than isa the true altitude of the altimeter (aircraft) would be lower than the indicated 5000. (Opposite if hotter than isa). $\endgroup$
    – user22445
    Commented Aug 14, 2021 at 22:29
  • $\begingroup$ Yes, it is essentially the same as a mercury baro, it only measures pressure. That really brings the hot and cold "vertical columns" of air in focus as sources of error. $\endgroup$
    – Robert DiGiovanni
    Commented Aug 14, 2021 at 22:35
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An altimeter setting of 29.92 at such airfield (0 ft elevation) means that the air above such airfield, weights the same as a column of mercury of 29.92 inches. The altimeter reads the relative weight of the air above the aircraft compared to the setting (29.92.) Because of the fact that we can assume that the air weights the same in the vicinity of the airfield, the parked aircraft will also read 0 ft. The 29.92 inHg reading is called barometric pressure.

If the temperature rises and the altimeter setting stays constant (barometric pressure constant), the air expands reducing its density so the airfield will report a higher density altitude. A higher density altitude means that the aircraft's performance will be such as if it was at a higher altitude. Human performance is also altered, for example if the density altitude is 2000 ft higher compared to the standard atmosphere and you are travelling in a non pressurized cabin, you will start to experience hypoxia 2000 ft lower compared to the standard atmosphere.

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I share the ops questions, and I don't feel it's been fully answered yet.

If you are flying through air at a constant pressure with the altimeter set properly, at standard temperature, then indicated is showing height above that altimeter setting, which could be 29.92, standard datum plane, or some other setting which is assumed to be mean sea level. Thus indicated shows MSL.

Now as you fly and maintain indicated altitude, you fly into colder air.

The rote rule, which I'm fine with, is that indicated will stay the same, but MSL will be lower. high to low, look out below.

Just how can you possibly be lower, if the altimeter itself didn't change? In this case, it must start to trend down slightly, so you follow it, thinking you are still at the same MSL altitude because it's still indicated, but you're actually lower.

After all nothing else changed, but temperature, and the airplane is flown at a lower MSL than indicated.

Looking at the inverse, say you stayed at the same true altitude (somehow, using a different instrument), then the altimeter would trend down, but you wouldn't follow it, and it would thus show it's errors.

Perhaps the problem we're having is trying to mix high to low, low to high concepts with density altitude concepts. I would be happy if this was true - the altimeter changes a little bit with temperature, it tries to ignore it, but doesn't completely do so. And we as pilots much understand this.

But density altitude is a calculation that is not at all trying to ignore the temperature's effect on altitude. If altimeters did account for temperature, they'd show you density altitude. And you'd have great insight into your aircraft's performance in real time, but you'd have no real clue what your true altitude is.

So calculated density altitudes at one spot are going to be possibly much different than the muted indicated errors on that same spot. But, the errors and density altitude will move in the same direction.

Please feel free to reset my thinking, if it's way off.

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    $\begingroup$ Yours is an interesting viewpoint, but I think it unnecessarily complicates the OPs question and the issue in general. Density altitude is pressure altitude corrected for non-standard temp. The purpose of calculating density altitude is merely (and importantly) to "normalize" the atmosphere for the conditions against which aircraft performance calculations are made (similar to a datum). I'm not disputing what you have said, only I think bringing in density altitude into the question being asked adds a bit more confusing data to a simple concept. $\endgroup$
    – user22445
    Commented Nov 12, 2021 at 20:08
  • $\begingroup$ Fair enough, but since they are related through temperature, I was hoping that thinking of them together would help make sense of the ops question. $\endgroup$
    – agiigi
    Commented Nov 12, 2021 at 22:16

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