I'm having some trouble making my calculations of pressure altitude match those in apps and text books.

I'm using the formula from Wikipedia:

(29.92 - altimeter setting) * 1000 + airport elevation

In the app I'm using, the altimeter setting is currently reported as 29.70 with an airport elevation of 22. Using the above formula I would assume:

(29.92 - 29.70) * 1000 + 22 = 242

Yet the app is telling me it is 229

Even more confusing, when looking at the Airman Knowledge Testing Supplement for Sport Pilot, Recreational Pilot, Remote Pilot, and Private Pilot (FAA-CT-8080-2H), figure 8, page 2-8, the pressure altitude for altimeter 29.7 is listed as 205. I assume this is using an airport elevation of 0, but then I would expect the answer to be:

(29.92 - 29.7) * 1000 + 0 = 220

Airman Knowledge Testing Supplement for Sport Pilot, Recreational Pilot, Remote Pilot, and Private Pilot (FAA-CT-8080-2H), figure 8, page 2-8

So my questions are:

  1. Which answer is correct? Wikipedia, the app or appendix 2?
  2. Why are there 3 different answers?

2 Answers 2


Given an altimeter setting ($\mathrm{QNH}$) and the elevation of the airport $h$, the exact equation to compute the pressure altitude is:

$$\mathrm{PA}=h+\frac{T_0}{L}\cdot\left[1-\left(\frac{\mathrm{QNH}}{P_0}\right)^\frac{R_s\cdot L}{g}\right]\tag{1}$$


  • $h$: Elevation of the airport in meters
  • $L$: Temperature lapse $=0.0065 \mathrm{~K/m}$
  • $T_0$: Standard temperature $=288.15 \mathrm{~K}$
  • $P_0$: Standard pressure $=101325 \mathrm{~Pa}$
  • $QNH$: Altimeter setting $\left[\mathrm{~Pa}\right]$
  • $g$: Gravitational acceleration $\approx 9.81 \mathrm{~m/s}^2$
  • $R_s$: specific gas constant for dry air $\approx 287.058 \mathrm{~J \cdot kg^{−1}K^{−1}}$

With the above equation, a $\mathrm{QNH}$ of 29.70 corresponds to a pressure altitude of 204.95 ft. So I'd say the most precise is the book. Your app seems to be off by 2 feet (it should report 227 ft).

The well-known formula to compute the pressure altitude is just a rough estimation:

$$\mathrm{PA_{feet}} = (29.92 - \mathrm{QNH}_{inHg}) \cdot 1000 + h_{feet}\tag{2}$$

It can be derived from equation 1 via a Taylor approximation computed at $\mathrm{QNH} = P_0$:

$$\mathrm{PA}(\mathrm{QNH}) = \mathrm{PA}(P_0) + \frac{d\mathrm{PA}}{d\mathrm{QNH}}\cdot(\mathrm{QNH}-P_0)=\tag{3}$$ $$= h+\frac{T_0R_s}{gP_0}(P_0-\mathrm{QNH})=h+0.0832\cdot(P_0-\mathrm{QNH})$$

The constant $0.0832$ has dimensions of $[\frac{m}{Pa}]$ but if we convert $h$ in feet and $P_0$ and $\mathrm{QNH}$ in inHg we get:

$$\mathrm{PA_{feet}} = (29.92 - \mathrm{QNH}_{inHg}) \cdot 924.5 + h_{feet}\tag{4}$$

which should show where equation 2 comes from. Using 1000 instead of 924.5, beside making it easier to compute, also makes it more precise for values of $\mathrm{QNH}$ that are not too close to $P_0$


The formula from wikipedia is a rough estimation. Think of it as a rule of thumb. It's useful if you need numbers for rough calculations. However, the exact formula is as shown below:

enter image description here

According to that formula, I calculated pressure altitude difference for 28.92 and it's 938 feet, as opposed to 1000 feet. Additionally, as you can tell from the formula, the relationship is not a constant linear value. So difference of 1 inch of HG is not always 938 feet, it's changing.

So, the FAA guide is accurate. And I would say your app is accurate enough also, however probably omitting some numbers after decimal, hence the small difference. Of course, you need to add field elevation on top of those numbers as well.

  • $\begingroup$ The formula you show is not for computing the pressure altitude, but to compute the pressure at a given altitude in the standard atmosphere $\endgroup$
    – fab
    Feb 4, 2022 at 18:42

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