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maybe it's a little impolite to ask but I got stuck with knowing bunch of formulas and not understanding how to use it:

Can you help me on understanding how the correct answer 7300ft came out of the following question:

Given the following information, what is the true altitude? (rounded to the nearest 50 ft) QNH: 983 hPa; Altitude: FL 85; Outside Air Temperature: ISA - 10° (1,00 P.)


My attempt based on the fomula from here:

  • $A_T$ 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

$A_T = (4\times A_{Ik} \times \Delta_{ISA}) + A_I$


  • $A_{Ik} = 8.5$
  • $\Delta_{ISA} = -10° - ( 15° - (8.5 \times 2) ) = -10° - ( 15° - 17°) = -10° - (-2°) = ISA-8$
  • $A_i = 8,500$

Therefore,

$A_t = ( 4 \times 8.5 \times (-8) ) + 8,500 = ( 34 \times (-8) ) + 8,500 = -272 + 8500 = 8288$

So my incorrect answer is 8288'. The correct is 7300' :(

I also realize I did not use QNH in my calculations.

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You need to correct your pressure altitude (FL85) for the barometer setting QNH first before you apply the temperature correction.

I assume a 30ft / hPa offset here:

$\textrm{barometric altitude} = \textrm{pressure altitude} + (\textrm{QNH} - 1013.25)\cdot 30 = 8500 -907.5 = 7592.5 \textrm{ft}$

Based on the pressure difference between sea level and your altimeter's measurement, the estimated altitude is 7592.15, assuming an ISA temperature profile.

Now, due to the fact that the actual temperature is lower, the atmosphere shrinks, and you end up flying lower.

Note that I will use -10° C for $\Delta_{ISA}$, because that was given in the question:

Outside Air Temperature: ISA - 10°

Now applying your correction formula yields:

$A_T = (4\times A_{Ik}\times\Delta_{ISA})+ A_I = (4 \times \frac{7592.5}{1000}\times -10) + 7592.5 = -303.7 + 7592.5 = 7288.8 $

Rounding that to the nearest 50 ft results ins 7300 ft.

This is an approximation which is a bit on the low side. But underestimating true altitude is usually safer than overestimating.

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    $\begingroup$ WOW! Thank you, DeltaLima, for your comprehensive answer and for editing my post with proper formula formatting (it’s much easier to read). That helped me, and that will help people looking for the answer in the future. $\endgroup$ Feb 24 at 9:12

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