# How can I calculate ambient air pressure from flight data?

I am playing around with data collected during a flight. I would like to find out the (real, not ISA) ambient air pressure for each data point, since it is not reported in the spread sheet containing my flight data.

I have the following data for different time points, but do not know how to (if even possible) derive from this the ambient air pressure:

• Mach

• Density altitude

• Pressure altitude

• Airspeed ktas and kcas

• Static air temperature

• Total air temperature

• Geometric altitude

• Pressure at the ground / surface pressure

I would be thankful for a solution or any tips for literature helpful to solve this.

Under the following link I found a formula I could rearrange but I believe this would return the pressure in the standard atmosphere... https://www.weather.gov/media/epz/wxcalc/pressureAltitude.pdf

• Not sure what "total air temperature" is. Aug 22, 2021 at 17:24
• @RobertDiGiovanni: Some help here. Jan 19, 2022 at 19:44

All you need is the pressure altitude. Pressure altitude is just a convoluted way to express outside pressure. That's how it's computed:

$$\mathrm{Press.Alt.}=\frac{T_0}{L}\left[1-\left(\frac{P_X}{P_0}\right)^\frac{R_sL}{g}\right]\tag{1}$$

where:

• $$P_X$$: Outside temperature in Pa (what you're interested in)
• $$P_0$$: Standard pressure $$=101325 \mathrm{~Pa}$$
• $$L$$: Temperature lapse $$=0.0065 \mathrm{~K/m}$$
• $$T_0$$: Standard temperature $$=288.15 \mathrm{~K}$$
• $$g$$: Gravitational acceleration $$\approx 9.81 \mathrm{~m/s}^2$$
• $$R_s$$: specific gas constant for dry air $$\approx 287.058 \mathrm{~J \cdot kg^{−1}K^{−1}}$$

You just need to invert it and get:

$$P_X=P_0\left(1-\frac{L\cdot\mathrm{Press.Alt.}}{T_0}\right)^\frac{g}{R_sL}\tag{2}$$

The result will be in Pa, so remember to convert it to inHg or mbar

• I see now the pdf in your question. Yes that's the same equation as equation 1. You already had the answer :)
– fab
Oct 17, 2022 at 3:36

It kind of depends on how accurate you want to be because you could start asking about instrument errors and position errors and stuff like that.

But, assuming you don't want to get into that...

The link you included would indeed show a standard atmosphere pressure but the pressure altitude data you have is also referenced to the same standard atmosphere (it actually depends on the setting of the altimeter... if it is set to 29.92 / 1013 mBar then you're in luck).

Ori

• The list of available data includes “pressure altitude”, which is reading of altimeter set to 29.92 inHg (1013 hPa). As opposed to “baro altitude”, which is the reading of altimeter set to whatever appropriate local altimeter setting. Aug 22, 2021 at 16:28

"real" air pressure

Real air pressure is measured the old fashioned way with mercury in a tube. Air pressure will lift mercury a certain height in a tube against gravity. The space above the mercury is a vacuum. The ratio of the weight of the mercury to the area of the tube cross section is lbs per inch$$^2$$.

The altimeter is designed to do the same thing, and by setting it to 29.92 one can compare the altimeter reading to standard sea level pressure. From a graph of pressure vs altitude, "real" pressure in lbs per inch$$^2$$ can be derived. The altimeter reading is from static pressure.

So from your list, use "pressure altitude" to compare with "geometric altitude". Pressure altitude is based on a Kollsman setting of 29.92. If pressure altitude is higher than geometric, then your ambient pressure will be lower than standard pressure for that geometric altitude.