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"An aircraft is cruising at FL170. The altitude shown with QNH set is 17,500 feet. The OAT is -10 deg C. What is the true altitude?"

The ISA temperature deviation is 9°C.

My calculation is:

17,500'
+ 9 °C x 4' x 17,500/1000
= 18,130'

The answer provided is 18,100'. I don't understand where my mistake is.

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  • $\begingroup$ To solve the true altitude, you will need the local barometric pressure, which is not practical. If you are over water, a radar altimeter might give you the most accurate reading (a reasonable approximation of absolute altitude). A GPS might come close, but will likely be off no matter where you are, for lots of reasons. You will need a clarified question, or more data. $\endgroup$
    – mongo
    May 6, 2017 at 12:47
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    $\begingroup$ Apart from knowing a formula or two, this question does give us everything we need to know to answer. The question is clear, and very answerable. $\endgroup$
    – J W
    May 6, 2017 at 13:09
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    $\begingroup$ Ok thanks, the answer provided is 18,100' and I have now calculated 18,130' (17,500' + (ISA temp deviation of 9 deg C x 4 deg C x 17.5))? $\endgroup$
    – uroboros
    May 6, 2017 at 13:45
  • $\begingroup$ What pressure reference datum is QNH using in millibars or inches Hg? That would be needed to evaluate this problem on a flight computer. $\endgroup$
    – Romeo_4808N
    May 6, 2017 at 16:47
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    $\begingroup$ @CarloFelicione That reference datum is available here since it can be determined from the difference between the pressure altitude and the indicated altitude with QNH set. $\endgroup$
    – J W
    May 6, 2017 at 18:16

3 Answers 3

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This can be calculated by knowing a few formulas, or temperature relationships related to altitude. Note that all these calculations are an estimate based on standard ISA temperature and altitude relationships. See this page for other examples of this type of question.

Note the following from the question:

  1. FL170 represents a pressure altitude of 17,000 ft
  2. We know that the OAT is -10°C
  3. Indicated altitude is 17,500 ft

Now, to solve for true altitude, we need to know a few additional items. We need to know that:

  1. ISA at the surface is 15°C
  2. Standard lapse rate is -2°C per thousand feet
  3. To find true altitude, the difference from indicated altitude is 4 ft per 1°C deviation from ISA for every 1,000 ft

Knowing all this, we can calculate the following:

  1. ISA at 17,000 ft (see 4 and 5 above)
  2. Deviation from ISA (see 2 and 7 above)
  3. True altitude (see 6 and 8 above)

This all might sound complicated, so lets run through an example problem. Let's assume a flight at FL190, with an OAT of -15°C, on a standard pressure day.

Since ISA at the surface is 15°C, using the standard lapse rate of -2°C per thousand feet, we know that ISA at 19,000 ft will be -23°C.

Given an ISA of -23°C, we see that our OAT of -15°C shows a difference of +8°C above ISA. Follow that step carefully, because the signs can be confusing.

Given that the temperature is ISA+8°C, and knowing 6 above, we can calculate our true altitude relative to pressure altitude:

$A_T$ is our estimated true altitude, in feet
$\Delta_{ISA}$ is the difference above ISA, in °C
$A_I$ is our indicated altitude, in feet
$A_{Ik}$ is our indicated altitude, in thousands of feet

$A_T=(4 \times A_{Ik} \times \Delta_{ISA}) + A_I$
$A_T=(4 \times 19 \times 8) + 19000$
$A_T=608 + 19000$
$A_T=19608$

In some cases, a question might ask for true altitude based on a pressure altitude, instead of an indicated altitude. In such cases, substitute pressure altitude for indicated altitude in the above formula. Remember, this is all an exercise in estimation which is helpful in reinforcing the relationships between temperature, pressure, and altitude.

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    $\begingroup$ Nice way of educating without providing the final answer. $\endgroup$
    – mins
    May 8, 2017 at 8:20
  • $\begingroup$ IF YOU USE TEMP LAPSE RATE OF -1.98 C per 1000 feet INSTEAD OF -2 you will get the right answer (temp deviation is 8.66 instead of 9 C ) : true alt = 17500 + 4* 8.66 * 17.5 = 18106 very near to 18100 feet $\endgroup$
    – Ali
    Nov 5, 2018 at 12:22
  • $\begingroup$ Thank you for the explanation. How do we know when to use PA or Indicated Altitude in the formula which you have provided(AT=(4×AIk×ΔISA)+AI)? I have noticed some people using Indicated altitude by default and then others use PA by default. So which one is correct? $\endgroup$
    – Skydemon
    Nov 15, 2020 at 7:12
  • $\begingroup$ @Skydemon, The formula that I have provided requires indicated altitude. If working with pressure altitude one must first convert that to indicated. The distinction will typically depend on whether the aircraft is flying above or below the transition altitude, in the flight levels or not. An aircraft flying in the flight levels will be flying reference pressure altitude; and aircraft below the transition level should be flying reference indicated altitude using a QNH or local altimeter setting. Does that answer your question? $\endgroup$
    – J W
    Nov 16, 2020 at 12:00
  • $\begingroup$ @JWalters so if I understand correctly, would we use the Pressure altitude for questions that mention FL and conversely use IA or altitude corrected for pressure error for questions that give the elevation? There is a lot of confusion around which value to use in the formula. Oxford uses only the PA and BGS uses only IA or elevation. Do you have any advice on which value to use for which type of question? $\endgroup$
    – Skydemon
    Nov 17, 2020 at 14:11
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First, True Altitude is applicable to a column of air above a point on the sphere. A QNH is for a station, and in ATC that location is arbitrary, and in places such as the UK is one of 20 or so regions. Such QNH data is accurately called a Regional Pressure Setting, and is the lowest forecast pressure for a defined region within an hour. So for survey work, or precision remote sensing, there would need to be an observation at the point below an aircraft, or at the very least an interpolation (modeled) using known values. In this instance, we don't have that, rather we have a Regional Pressure Setting. Therefore, the resultant figure may be close, but is not the true altitude for the aircraft.

The formula below is from the 1976 International Standard Atmosphere, and reduced by a friend (physicist) who has developed a set of formulas for navigation. The formula provides an accurate answer as it is directly traceable to the 1976ISA, and is not an approximation.

Relationship of true and calibrated (indicated) altitude:

TA= CA + (CA-FE)*(ISADEV)/(273+OAT)

where

TA= True Altitude above sea-level

FE= Field Elevation of station providing the altimeter setting CA= Calibrated altitude= Altitude indicated by altimeter when set to the altimeter setting, corrected for calibration error.

ISADEV= Average deviation from standard temperature from standard in the air column between the station and the aircraft (in C).

OAT= Outside air temperature (at altitude)

Substituting:

ISADEV = (stdISAtemp - (PA * (lapserate)) - differentialfmstdtemp )

TA = 17500 + (17500-0) * (15-17500/1000(2)-(-10)) / (273 - (-10))

TA = 17500 + 17500 * +10 / 283

TA = 17500 + 665.39

TA = 18165

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This question has already been thoroughly answered by other users. I just wanted to mention that the non-approximated version of the formula to compute the true altitude is:

$$h_{true} = h + \frac{h}{T_0}\cdot\left(T_\mathrm{OAT}-T_\mathrm{ISA}\right)\tag{1}$$

where the value usually represented by $\frac{4}{1000}$ is in fact $\frac{1}{T_0}=\frac{1}{288.15}\approx0.00347$

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