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Airspeed describes an aircraft's motion relative to the surrounding air. When the kinematic magnitude is intended, it is the true airspeed,It differs from groundspeed whenever wind changes the relative motion. A ground-station velocity alone cannot establish the aerodynamic airspeed.^[ref]

An airspeed instrument normally infers motion from pressure. At low Mach number, for an aligned Pitot-static tube in the regime of the Bernoulli equation,Here $p_0$ is stagnation pressure, $p$ is static pressure and $\rho$ is air mass density. For a stipulated difference of $245\mathrm{Pa}$, $\rho =1.225kg/{\mathrm{m}}^3$ gives $V=20m/s$; the same difference at $0.6125kg/{\mathrm{m}}^3$ gives $28.28m/s$. The instrument senses a pressure difference, so interpreting its reading requires a density model and an instrument calibration.^[ref]

Indicated airspeed is the displayed reading before instrument and installation corrections. Calibrated airspeed includes those corrections. True airspeed expresses actual motion through the air, rather than simply relabeling either reading. The FAA distinguishes these quantities in its instrument chapter.^[ref]

The square-root calculation above assumes an appropriate low-speed stopping relation. At appreciable Mach number, compressibility changes that relation; a blocked passage or displaced static pressure source introduces a different measurement problem. Report which airspeed convention and corrections accompany a performance curve. Comparing speeds without that information can confound a change of air mass density with a change in the aircraft's loading. Functioning ports and known corrections are part of the inference, not extra measurements of groundspeed.

Groundspeed is the magnitude of an aircraft's horizontal velocity relative to the ground. It determines travel time along a ground track, whereas aerodynamic lift and drag depend on motion relative to the air. Let ${\mathbf{v}}_{g,h}$ and ${\mathbf{v}}_{a,h}$ be the horizontal ground-relative and air-relative velocity vectors, and let ${\mathbf{w}}_h$ be the horizontal wind velocity:For climbing or descending flight this horizontal convention distinguishes groundspeed from the magnitude of the full three-dimensional ground-relative velocity.^[ref]

In an original horizontal-flight example, an airspeed of $50m/s$ directed due north and a wind blowing east at $15m/s$ give ${\mathbf{v}}_{g,h}=(15,50)m/s$ in east/north coordinates. Thus $V_g=\sqrt{2725}\approx 52.20m/s$, with the track about $16.70{}^{\circ }$ east of north. Adding scalar speeds would wrongly give $65m/s$.

To keep a due-north track in the same wind, the aircraft must instead supply an air-relative westward component of $15m/s$. At the same airspeed, its northward component becomes $\sqrt{50^2-15^2}\approx 47.70m/s$, so the on-track groundspeed is smaller than the airspeed. The required wind correction angle is $\arcsin (15/50)\approx 17.46{}^{\circ }$ west of north. Heading and track are distinct directions.

Timed positions from a ground-based tracker estimate groundspeed, not airspeed. Inferring the latter also needs the wind velocity. Conversely, a Pitot-static tube does not measure travel over the ground. The same distinction explains why a headwind changes the ground range of a steady glider without changing its air-relative glide angle.

The wind correction angle is the difference between the aircraft heading and the desired ground track needed to compensate for a crosswind. A heading points the vehicle; a track describes horizontal travel over the ground. In the horizontal model here, the air-relative velocity is aligned with the heading. Wind-direction statements below specify where the air moves, not the meteorological direction it comes from.^[ref]

Let $V$ be true airspeed, let $w_{\parallel }$ be wind along the desired track, and let $w_{\perp }$ be wind to its right. Define a positive correction $\delta$ as heading right of the track. Cancelling cross-track velocity and projecting along the track givesFor the branch with a nonnegative along-track air-relative component,The condition $|w_{\perp }|\le V$ permits crosswind cancellation, but forward progress also requires $V_g>0$.

In an original example, take $V=40m/s$, wind $8m/s$ rightward and $12m/s$ opposite the intended track. Then $\delta \simeq -11.54{}^{\circ }$ and $V_g\simeq 27.19m/s$. A constant-condition $100\mathrm{km}$ ground leg takes about $61.29\min$. With the same headwind, a rightward component of $39m/s$ can still be cancelled, but $\sqrt{40^2-39^2}-12\simeq -3.11m/s$: the resulting track-line motion is backward.

The calculation assumes a uniform horizontal wind and a steady true airspeed. Ground-position measurements can test the resulting groundspeed; they cannot determine the correction from air-relative speed alone without wind information. Wind variation and a difference between heading and air-relative velocity require a fuller vector model.