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The aircraft yaw axis is the body-fixed axis about which the nose rotates from side to side. In a right-handed body coordinate system with $x$ forward and $y$ toward the right wing, it is the $z$ axis directed downward through the centre of mass. It completes the orthogonal set with the aircraft roll axis and aircraft pitch axis. The rudder can generate a side force aft of the centre of mass, producing a yawing torque.^[ref]

For a force $\mathbf{F}$ acting at displacement $\mathbf{r}$ from the reference point, the yaw component of $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$ isIn an original example, a rightward $80\mathrm{N}$ tail force at $\mathbf{r}=(-3,0,0)\mathrm{m}$ gives ${\tau }_z=-240\mathrm{N}\mathrm{m}$. With positive yaw defined nose-right, this force turns the nose left. If the axis is a principal axis with moment of inertia $I_z=1200\mathrm{kg}{\mathrm{m}}^2$ and the initial motion is pure yaw, the instantaneous angular acceleration is $\dot{r}={\tau }_z/I_z=-0.20rad/{\mathrm{s}}^2$. A tail's displacement matters as well as the force magnitude.

The aircraft yaw rate $r$ is a component of body angular velocity; it is not always the rate of change of aircraft heading. A banked vehicle projects its rotation onto several body axes, and its ground track also depends on wind. General coupled motion uses Euler's equations, rather than three independent scalar torque laws. Measuring a heading change alone therefore cannot establish the applied yawing moment.

True airspeed is the magnitude of an aircraft's velocity relative to the surrounding air,It determines the relative motion used in aerodynamic models, while groundspeed concerns horizontal travel over the ground. Indicated airspeed and calibrated airspeed describe stages in a pressure-instrument reduction, rather than another reference frame.^[ref]

For an aligned low-Mach number pressure measurement in the regime of the Bernoulli equation,An equivalent airspeed uses the same dynamic pressure with a standard reference mass density, $V_{\mathrm{EAS}}=\sqrt{2q/{\rho }_0}$. HenceOnly where the low-speed pressure relations justify $V_{\mathrm{CAS}}\simeq V_{\mathrm{EAS}}$ may calibrated speed be substituted in this last expression without a separate compressibility correction.

For an original low-speed example, use $V_{\mathrm{CAS}}\simeq 30m/s$, ${\rho }_0=1.225kg/{\mathrm{m}}^3$ and ambient $\rho =0.800kg/{\mathrm{m}}^3$. The corrected difference is approximately $551.25\mathrm{Pa}$, giving $V_{\mathrm{TAS}}\simeq 37.12m/s$. At a fixed pressure difference, halving ambient mass density increases the inferred true airspeed by $\sqrt{2}$, rather than doubling it. This follows from the square-root dependence, not a wind correction.

Obtaining the ambient mass density can require independent static pressure and thermodynamic temperature measurements, with a suitable equation of state. For fixed $q$, a small relative density error changes the inferred speed by approximately $\delta V/V=-\delta \rho /(2\rho )$. Wind information is needed separately to predict the ground track. A single ground-travel speed cannot supply all of these quantities. The pressure inference also depends on clear passages and valid stopping-flow assumptions.^[ref]

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.