Density altitude

Skydiving

The density altitude is an important factor in skydiving, and one that can be difficult to judge properly, even for experienced skydivers. In addition to the general change in wing efficiency that is common to all aviation, skydiving has additional considerations. There is an increased risk due to the high mobility of jumpers (who will often travel to a drop zone with a completely different density altitude than they are used to, without being made consciously aware of it by the routine of calibrating to QNH/QFE). Another factor is the higher susceptibility to hypoxia at high density altitudes, which, combined especially with the unexpected higher free-fall rate, can create dangerous situations and accidents. Parachutes at higher altitudes fly more aggressively, making their effective area smaller, which is more demanding for a pilot's skill and can be especially dangerous for high-performance landings, which require accurate estimates and have a low margin of error before they become dangerous.

Calculation

The density altitude can be calculated from the atmospheric pressure and the outside air temperature (assuming dry air) using the following formula:

${\displaystyle \text{DA}\approx \frac{T_{\text{SL}}}{\mathrm{\Gamma }}\left[1-{\left(\frac{P/P_{\text{SL}}}{T/T_{\text{SL}}}\right)}^{{\left(\frac{gM}{\mathrm{\Gamma }R}-1\right)}^{-1}}\right].}$

In this formula,

${\displaystyle \text{DA}}$, density altitude in meters (m);

${\displaystyle P}$, (static) atmospheric pressure;

${\displaystyle P_{\text{SL}}}$, standard sea-level atmospheric pressure, International Standard Atmosphere (ISA): 1013.25 hectopascals (hPa), or U.S. Standard Atmosphere: 29.92 inches of mercury (inHg);

${\displaystyle T}$, outside air temperature in kelvins (K);

${\displaystyle T_{\text{SL}}}$ = 288.15 K, ISA sea-level air temperature;

${\displaystyle \mathrm{\Gamma }}$ = 0.0065 K/m, ISA temperature lapse rate (below 11 km);

${\displaystyle R}$ ≈ 8.3144598 J/mol·K, ideal gas constant;

${\displaystyle g}$ ≈ 9.80665 m/s², gravitational acceleration;

${\displaystyle M}$ ≈ 0.028964 kg/mol, molar mass of dry air.

The National Weather Service (NWS) formula

The National Weather Service uses the following dry-air approximation to the formula for the density altitude above in its standard:

${\displaystyle {\text{DA}}_{\text{NWS}}=145442.16\text{ft}\left(1-{\left[17.326\frac{{}^{\circ }\text{F}}{\text{inHg}}\frac{P}{459.67{}^{\circ }\text{F}+T}\right]}^{0.235}\right).}$

In this formula,

${\displaystyle {\text{DA}}_{\text{NWS}}}$, National Weather Service density altitude in feet (${\displaystyle \text{ft}}$);

${\displaystyle P}$, station pressure (static atmospheric pressure) in inches of mercury (inHg);

${\displaystyle T}$, station temperature (outside air temperature) in degrees Fahrenheit (°F).

Note that the NWS standard specifies that the density altitude should be rounded to the nearest 100 ft.

Approximation formula for calculating the density altitude from the pressure altitude

This is an easier formula to calculate (with great approximation) the density altitude from the pressure altitude and the ISA temperature deviation:

${\displaystyle \text{DA}\approx \text{PA}+118.8\frac{\text{ft}}{{}^{\circ }\text{C}}\left(T_{\text{OA}}-T_{\text{ISA}}\right).}$

In this formula,

${\displaystyle \text{PA}}$, pressure altitude in feet (ft) $\approx \text{station elevation in feet}+27\frac{\text{ft}}{\text{mb}}(1013\text{mb}-\text{QNH})$;

${\displaystyle \text{QNH}}$, atmospheric pressure in millibars (mb) adjusted to mean sea level;

${\displaystyle T_{\text{OA}}}$, outside air temperature in degrees Celsius (°C);

$T_{\text{ISA}}\approx 15{}^{\circ }\text{C}-1.98{}^{\circ }\text{C}\frac{\text{PA}}{1000\text{ft}}$, assuming that the outside air temperature falls at the rate of 1.98 °C per 1,000 ft of altitude until the tropopause (at 36,000 ft) is reached.

Rounding up 1.98 °C to 2 °C, this approximation simplifies to become

${\displaystyle \begin{array}{cc}\text{DA}& \approx \text{PA}+118.8\frac{\text{ft}}{{}^{\circ }\text{C}}\left[T_{\text{OA}}+\frac{\text{PA}}{500\text{ft}}{}^{\circ }\text{C}-15{}^{\circ }\text{C}\right]\\ & =1.2376\text{PA}+118.8\frac{\text{ft}}{{}^{\circ }\text{C}}{T}_{\text{OA}}-1782\text{ft}.\end{array}}$