How this instrument works
Air pressure at a given height is simply the weight of every molecule of atmosphere still stacked above that point, spread over one square metre. Near the ground that column is dense and heavy, so pressure runs high. Climb, and two things happen together: there is less air left above you, and the air that remains is itself thinner than what you just left, because it too is being compressed by less weight from above. That compounding — density falling as the reading falls, which then makes the reading fall faster — is why a plot of pressure against height curves downward rather than running straight, and why this instrument raises a bracket to a power instead of simply multiplying.
The particular figures 2.25577×10⁻⁵ and 5.25588 are not fitted to data; they fall straight out of combining the hydrostatic equation dP = −ρg·dh with the ideal gas law and a fixed temperature lapse rate. The International Civil Aviation Organization's Standard Atmosphere assumes air cools at a steady 6.5 K for every kilometre gained, starting from 288.15 K (15 °C) at the surface, and integrating that assumption from sea level upward produces exactly this power law. It holds through the troposphere, up to roughly 11000 m — above that the model switches to an isothermal layer and a different formula takes over.
Treat the output as a reference figure, not a live forecast. A real column of air rarely matches the standard 15 °C-and-6.5-K-per-km recipe exactly: a heat wave thins the air faster than the model predicts, and a cold snap or a passing high-pressure system shifts every value up or down. This is precisely why aviation keeps three related but distinct terms apart — pressure altitude (what this formula computes, run in reverse from a measured reading), density altitude (which folds in the day's actual temperature), and true altitude (which needs a locally corrected altimeter setting). Mistaking the standard-atmosphere answer for today's actual sky is the single most common misreading of this kind of chart.
- Enter Altitude, the height above sea level in metres, using the unit menu for feet if that suits your source.
- Enter Sea-level pressure — leave it at the standard 101325 Pa, or type today's actual barometric reading to shift the whole curve to real conditions.
- Read Pressure at altitude in pascals, or switch that field to kPa, hPa or inHg to match your instrument.
- Remember the formula is built for the troposphere: entries above roughly 11000 m leave the layer the model assumes.
Worked example — pressure at Denver's mile-high altitude
Set Altitude to 1500 m and leave Sea-level pressure at the standard 101325 Pa — the mile-high elevation of Denver, Colorado, under a textbook atmosphere. First the bracket: 1 − 2.25577×10⁻⁵ × 1500 = 0.9661635. Raise that to the 5.25588 power and multiply by 101325, and Pressure at altitude returns 84555.9905236 Pa.
Rounded, that is 84.556 kPa — about 83.5% of what the same barometer would read at sea level under identical conditions. It matches the 'thinner air' figure pilots, mountaineers and even bag-of-chips manufacturers cite for that elevation: drive a sealed sea-level bag up to Denver's airport and it visibly puffs out, because the air trapped inside is still pushing at 101325 Pa while the air now surrounding it pushes at only 84556 Pa.
Questions
Why does the pressure curve instead of falling in a straight line?
Because air compresses under its own weight. Every metre gained removes a slice of atmosphere from above, but that slice is thinner than the one below it, since it too has less weight pressing on it. Pressure and density fall together, each feeding the other, which produces a curve rather than a ramp. A straight-line approximation is close enough for a few hundred metres but drifts noticeably wrong by the time you reach a few kilometres.
Where do 2.25577×10⁻⁵ and 5.25588 actually come from?
They are not fitted constants — they fall out of the International Standard Atmosphere's assumptions. The first, 2.25577×10⁻⁵, is the temperature lapse rate divided by sea-level temperature: 0.0065 K/m ⁄ 288.15 K. The second, 5.25588, is standard gravity times dry air's molar mass, divided by the gas constant times that same lapse rate: g·M ⁄ (R·L). Both trace back to the hydrostatic equation integrated against a fixed cooling rate.
Does this tell me the exact pressure outside right now?
No — it gives the standard-atmosphere value for that height, a reference model built on a fixed 15 °C sea-level temperature and a steady cooling rate. Real days depart from that recipe: a heatwave thins a column faster than the model assumes, and a cold snap or a high-pressure system shifts every reading. Type today's actual sea-level figure into Sea-level pressure to get closer to the true number for that day.
How high can I push the Altitude field before the answer breaks?
Keep it under roughly 11000 m, the top of the troposphere in the ISA model. That is where the assumed 6.5 K-per-kilometre cooling stops; above it, standard-atmosphere temperature is treated as flat at −56.5 °C, and the reading follows a different, purely exponential formula rather than this power law. Feed in something like 15000 m and the figure returned is no longer the standard-atmosphere answer.
Who actually relies on this formula outside a classroom?
Pilots and avionics technicians, first: an aircraft's altimeter is a barometer scaled by exactly this relationship, which is why every altimeter carries a knob for dialling in the day's sea-level setting. Meteorologists use the same mathematics in reverse to reduce a mountain station's raw reading to a sea-level figure for weather maps. High-altitude balloon and sounding-rocket engineers use it to predict ambient pressure for parachute and burst-diameter calculations before launch.
Why type in a Sea-level pressure instead of leaving the standard value?
Because real sea-level pressure drifts with the weather, typically between about 98000 and 105000 Pa, rather than sitting fixed at 101325 Pa. Entering today's actual reading — from a nearby station or airport report — shifts the whole curve to match current conditions, the difference between computing a textbook pressure altitude and estimating the true pressure a barometer would show at that height today.