ISA Pressure✓ verified

Understanding the Parameters

• Static pressure $P$ — the ambient pressure of still air: the weight of the entire air column above, spread over unit area. At sea level that column presses down at $101{,}325\ \text{Pa}$ — about $10\ \text{tonnes}$ over every square metre. It is static to distinguish it from the Dynamic Pressure $\tfrac12\rho V^2$ that a moving aircraft adds; their sum is the total (stagnation) pressure of Bernoulli Total Pressure.

• Altitude $h$ — enters only through $T(h)$. At $5\ \text{km}$, $P \approx 54{,}020\ \text{Pa}$ — just over half of sea level. At the $11\ \text{km}$ tropopause, $P \approx 22{,}632\ \text{Pa}$, 22 % of sea level: a cruising airliner flies in air holding roughly one-fifth the sea-level pressure, which is precisely why the cabin must be pressurised.

• The exponent $g/(LR) = 5.2559$ — a pure number assembled from three constants, and the heart of the formula. It is large, so pressure falls much faster than temperature: at the tropopause, $T$ has dropped to 75 % of its datum but $P$ to 22 %, because $0.75^{5.2559} \approx 0.223$.

• Specific gas constant $R = 287.05\ \text{J}/(\text{kg}\cdot\text{K})$ — the universal gas constant $\mathcal{R} = 8314.5\ \text{J}/(\text{kmol}\cdot\text{K})$ divided by the mean molar mass of dry air ($\approx 28.96\ \text{kg/kmol}$). This is where the composition of the atmosphere enters; a different gas gives a different profile.

• Why pressure, not height, is what altimeters measure — an altimeter is a barometer with a height scale painted on its face, calibrated by inverting exactly this equation (see Pressure Altitude). Aircraft separation is enforced on pressure surfaces, not true heights, so two aircraft with the same setting fly parallel even when the equation is locally wrong.

Derivation (Approaching a Proof)

Start with hydrostatic balance. Take a thin horizontal slab of air of thickness $dh$ and unit area. Three forces act vertically: pressure $P$ pushing up from below, pressure $P + dP$ pushing down from above, and the slab's weight $\rho g\,dh$. In equilibrium the sum vanishes:

$$P - (P + dP) - \rho g\,dh = 0 \quad\Longrightarrow\quad \boxed{\;\frac{dP}{dh} = -\rho g\;}$$

This says only that pressure falls with height at a rate set by the local air density — the fundamental statement of an atmosphere at rest.

Now eliminate $\rho$ with the ideal gas law in specific form, $P = \rho R T$, so $\rho = P/(RT)$:

$$\frac{dP}{dh} = -\frac{P g}{R T}.$$

Separate variables:

$$\frac{dP}{P} = -\frac{g}{R}\,\frac{dh}{T}.$$

We cannot integrate the right side until we know $T(h)$ — this is precisely why the ISA must define a temperature profile before pressure can be found. Insert the ISA troposphere assumption $T = T_0 - L h$. Change the variable of integration from $h$ to $T$: since $dT = -L\,dh$, we have $dh = -dT/L$, giving

$$\frac{dP}{P} = -\frac{g}{R}\cdot\frac{1}{T}\cdot\left(-\frac{dT}{L}\right) = \frac{g}{LR}\,\frac{dT}{T}.$$

Both sides are now exact differentials of logarithms. Integrate from the sea-level datum $(P_0, T_0)$ to the state at altitude $(P, T)$:

$$\int_{P_0}^{P} \frac{dP'}{P'} = \frac{g}{LR}\int_{T_0}^{T} \frac{dT'}{T'} \quad\Longrightarrow\quad \ln\frac{P}{P_0} = \frac{g}{LR}\,\ln\frac{T}{T_0}.$$

Exponentiating both sides yields the power law:

$$P = P_0 \left(\frac{T}{T_0}\right)^{g/(LR)}.$$

Substituting $T = T_0 - Lh$ gives the explicit form in altitude, $P = P_0\left(1 - \tfrac{Lh}{T_0}\right)^{5.2559}$.

The isothermal branch. Above the tropopause $L = 0$, the change of variable above divides by zero and the derivation must restart. With $T = T_1 = 216.65\ \text{K}$ constant, $\int dh/T = h/T_1$ integrates directly:

$$\ln\frac{P}{P_1} = -\frac{g}{R T_1}(h - h_1) \quad\Longrightarrow\quad P = P_1\,\exp\!\left[-\frac{g\,(h - h_1)}{R T_1}\right].$$

The scale height $H = RT_1/g = 287.05 \times 216.65 / 9.80665 = 6341.6\ \text{m}$ — the height over which pressure falls by a factor of $e$. Its reciprocal is $1.5769\times10^{-4}\ \text{m}^{-1}$, which is exactly the $0.0001577$ constant in the implementation's stratosphere branch, and $P_1 = 22{,}632\ \text{Pa}$ is the troposphere formula evaluated at 11 km. The two branches meet continuously at the tropopause.

Dimensional check. The exponent must be dimensionless for the power law to make sense: $$\frac{g}{LR} = \frac{\text{m}/\text{s}^2}{(\text{K}/\text{m})\times \text{J}/(\text{kg}\cdot\text{K})} = \frac{\text{m}/\text{s}^2}{\text{J}/(\text{kg}\cdot\text{m})} = \frac{\text{m}/\text{s}^2}{\text{m}/\text{s}^2} = 1. \checkmark$$ (using $\text{J} = \text{kg}\cdot\text{m}^2/\text{s}^2$). The ratio $T/T_0$ is dimensionless, so $P$ carries the units of $P_0$, namely Pa.

History and Development

• The barometer (1643–1648). Evangelista Torricelli inverted a mercury-filled tube in 1643 and found the column stood about 760 mm, concluding that the air's weight held it up. In 1648 Blaise Pascal had his brother-in-law Florin Périer carry a barometer up the Puy de Dôme in Auvergne; the column fell by roughly three inches over 1000 m of climb. That experiment proved the atmosphere has finite weight and that pressure decreases with height — the founding measurement of this entire calculator, and the reason the SI unit of pressure bears Pascal's name.

• The barometric formula (1800s). Pierre-Simon Laplace developed the hypsometric relation for determining height from pressure; combined with the developing gas laws and the hydrostatic equation, the isothermal exponential form became standard 19th-century meteorology.

• Aviation forces standardisation (1920s). Once aircraft used barometric altimeters for separation, an agreed $P(h)$ became a safety necessity: a shared wrong answer keeps aircraft apart, while individually right answers do not. National standard atmospheres appeared through the 1920s.

• ICAO (1952) and after. ICAO fixed the sea-level datum $101{,}325\ \text{Pa}$ (exactly one standard atmosphere) and the layer structure used here, extended to 32 km in 1964 and 80 km in 1976; ISO 2533:1975 and the U.S. Standard Atmosphere 1976 agree below 32 km. The two-branch structure — power law to 11 km, exponential above — is a direct fossil of the derivation: it is not two models, but one integration performed under two different temperature assumptions.

Related Concepts: ISA Temperature, ISA Density, Pressure Altitude, Density Altitude, Geopotential Altitude, Stratospheric Temperature, Bernoulli Total Pressure, Dynamic Pressure, Ideal Gas Kinetic Theory

Notes: Only $h$ is an input; $P_0$, $T_0$, $L$, $g$, $R$ are hard-coded, and $T$ is computed internally from $h$ (not an input). Displayed equation is the troposphere branch only — above 11 km the implementation uses the isothermal exponential $P = 22{,}632\,e^{-0.0001577(h-11000)}$ (scale height $6341.6\ \text{m}$). Exponent rounded $5.256$ vs exact $5.2559$ ($0.44\ \text{Pa}$ at 11 km, negligible). Requires geopotential altitude — see Geopotential Altitude.