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The aircraft pitch axis is the transverse body axis about which the nose rotates upward or downward. In the convention used by NASA Glenn Research Center, it passes through the centre of mass and points toward the right wing tip; the aircraft roll axis points toward the nose and the aircraft yaw axis toward the underside. These are body-fixed directions, so their orientation relative to the ground changes with the aircraft.^[ref]

Choose right-handed body coordinates $x$ forward, $y$ right and $z$ down. A force $\mathbf{F}$ acting at displacement $\mathbf{r}$ from the centre of mass produces torque $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$. Its pitching moment isFor an original tail-load example, $\mathbf{r}=(-4,0,0)\mathrm{m}$ and $\mathbf{F}=(0,0,100)\mathrm{N}$ give ${\tau }_y=+400\mathrm{N}\mathrm{m}$. The downward tail load lifts the nose in this convention. If the aircraft pitch axis is a principal axis, the motion is initially pure pitch and the corresponding moment of inertia is $I_y=2000\mathrm{kg}{\mathrm{m}}^2$, then the instantaneous angular acceleration is $\dot{q}={\tau }_y/I_y=0.20rad/{\mathrm{s}}^2$. Coupled three-axis rotation requires the full Euler's equations.

The axis is a geometrical reference, not necessarily a physical hinge. Changing a control-surface load changes both forces and torques; a zero net translational force does not establish zero pitching moment. A balance measurement therefore needs the reference point and sign convention alongside its reported moment.^[ref]

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.

In principal body coordinates, Euler's equations express angular momentum balance as $I\dot{\omega }+\omega \times (I\omega )=\tau$, with applied torque $\tau$. Setting the torque to zero gives the following torque-free specialization.

Let $J=I_1=I_2$, ${\omega }_3=c$ andHere $c$ and $\nu$ are constant. The first two of Euler's equations give ${\dot{\omega }}_1=-\nu {\omega }_2+{\tau }_0/J$ and ${\dot{\omega }}_2=\nu {\omega }_1+{\tau }_0/J$. Combining them producesMultiplication by the integrating factor $e^{-i\nu t}$ gives, for $\nu \ne 0$,For completeness, $\nu =0$ instead gives $z(t)=z(0)+{\tau }_0(1+i)t/J$. These formulas describe a symmetric top under constant transverse body torque.

The stated initial conditions require $b\ne 0$ and $I_3\ne J$. They give $c=b/(I_3-J)$, $\nu =b/J$ and $z(0)=i{\tau }_0/b$. Substitution and simplification yieldTaking the real part and imaginary part, the angular velocity in the principal body frame isIts value at $t=0$ is the required initial condition; differentiating the first two components verifies both forced Euler's equations.

For the symmetric top, set $I_1=I_2=J$. The third of Euler's equations reduces to $I_3{\dot{\omega }}_3={\tau }_3=0$. Thus the axial angular velocity component is a conserved quantity:The transverse angular velocity components are generally variable. A torque constant in the principal body frame generally rotates in the inertial frame; constancy in the two frames is different.

For a vector $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$ in the principal body frame, the derivative of a body-fixed basis vector is $(d{\mathbf{e}}_i/dt{)}_{\hat{S}}=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the expansion gives the rotating-frame derivative formula, also called the Coriolis formula:Apply this to angular momentum $\mathbf{L}={\sum }_i{I}_i{\omega }_i{\mathbf{e}}_i$. The principal moments of inertia are constant in the principal body frame, so external torque balance becomesResolving the cross product gives Euler's equations:All components and dotted components in this display refer to the principal body frame.

Choose body-fixed principal axes through a fixed point or the centre of mass, with principal moments $I_1,I_2,I_3$. Angular momentum has components $L_i=I_i{\omega }_i$. The inertial derivative is the body derivative plus $\mathbf{\omega }\times \mathbf{L}$, so Euler's equations readThey avoid differentiating the rotating inertia tensor in laboratory coordinates. Torques must refer to the chosen point, and the displayed fixed moments require a rigid body.

For zero torque, laboratory angular momentum is constant, while its components can change in the body frame. Both $L^2$ and rotational kinetic energy $T=\frac{1}{2}\sum I_i{\omega }_i^2$ are conserved. The intersections of their constant-value surfaces constrain the motion. For distinct $I_1<I_2<I_3$, steady rotation about either extreme principal axis is stable against small perturbations, whereas rotation about the intermediate axis is unstable. For example perturbing rotation $\Omega$ about axis 2 giveswhose coefficient is positive. A symmetric body has additional simplifications and can precess. Conservation of laboratory angular momentum does not imply constant body-frame angular velocity.

Take axes fixed in a rigid body, chosen along its principal axes, with moments of inertia $I_1,I_2,I_3$. The angular momentum components are $H_i=I_i{\omega }_i$. Newton's rotational equation is $\mathbf{\tau }=(d\mathbf{H}/dt{)}_{\mathrm{inertial}}$, while the rotating-frame transport rule givesThus Euler's equations areThe nonlinear cross terms arise because the principal axes rotate. Even with no external torque, the body-frame components of angular momentum can vary while the angular momentum vector stays fixed in inertial space.

With zero torque, the equations conserveUniform rotation about any principal axis is a torque-free solution. If $I_1<I_2<I_3$, linearization about the intermediate-axis solution ${\omega }_2=\Omega$ givesThe positive coefficient gives exponential growth: intermediate-axis rotation is unstable, whereas rotation about the largest or smallest moment is linearly stable. For a symmetric body, Euler's equations also describe the precession of the transverse angular velocity around its symmetry axis. Torques from gravity or an applied couple enter through the right-hand sides rather than being absorbed into the inertial cross terms.

Inertia and principal directions. About the centre of mass, the inertia tensor isIt relates angular momentum and angular velocity by $\mathbf{J}=\mathsf{I}\mathbf{\omega }$ and gives rotational kinetic energy $T=\mathbf{\omega }\cdot \mathsf{I}\mathbf{\omega }/2$. The principal axes are orthogonal eigenvector directions of this real symmetric tensor; in that basis it is diagonal, with entries the principal moments of inertia $I_1,I_2,I_3$.

Derivatives in a rotating frame. Write $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$ in a body-fixed orthonormal basis. Each basis vector rotates according to $(d{\mathbf{e}}_i/dt{)}_0=\mathbf{\omega }\times {\mathbf{e}}_i$. Differentiating the components and basis separately proves the rotating-frame derivative formula,The body derivative differentiates the scalar components in that basis. A fixed pivot can replace the centre of mass when the corresponding angular-momentum balance is valid.

Euler equations. The inertial angular-momentum equation is $d\mathbf{J}/dt=\mathbf{G}$. Since the principal moments are constant in the body frame, substitute $J_i=I_i{\omega }_i$:These are Euler's equations in the principal body frame.

Linear perturbation of free rotation. After the impulse, $\mathbf{G}=0$. Let ${\omega }_3=\Omega$ to leading order and ${\omega }_1,{\omega }_2$ be small. The third Euler equation changes ${\omega }_3$ only at quadratic order, so the linear transverse equations areFor ${\omega }_1=a{e}^{pt}$ and ${\omega }_2=b{e}^{pt}$, eliminate $a,b$ fromThis givesIts units are inverse square seconds. Positive $p^2$ gives a growing real exponential; negative $p^2$ gives bounded oscillatory perturbations.

Inertia is a tensor. For a rigid body, distances between material points are fixed. Write each position relative to the centre of mass as $\mathbf{r}$, so its rotational velocity is $\mathbf{\omega }\times \mathbf{r}$. Integrating $\mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})$ gives angular momentumThe rotational kinetic energy isThe inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity only along a principal axis, or within a subspace of equal principal moments. For translation plus rotation about the centre of mass, $T=M{v}_{\mathrm{CM}}^2/2+T_{\mathrm{rot}}$. Moving the reference to a parallel axis adds $M(d^{2}1-\mathbf{d}{\mathbf{d}}^{\mathsf{T}})$ to the tensor, the tensor form of the parallel-axis theorem.

Euler's equations. The external torque about the centre of mass obeys $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{space}}$. In body axes, the transport rule givesThe principal moments are constant in that frame, henceThese are Euler's equations. With zero torque, spatial angular momentum and kinetic energy are conserved, although the body components generally vary. For distinct moments, small perturbations about rotation around the largest or smallest principal moment oscillate, whereas those about the intermediate moment grow. This is the intermediate-axis instability.

A lamina. Choose the lamina's plane as $z=0$. Then $I_{xz}=I_{yz}=0$, so the normal is a principal axis, andThis perpendicular-axis theorem applies to a planar mass distribution; it is not a general identity for a three-dimensional body. Rotate the in-plane axes to diagonalize the remaining symmetric two-by-two block. For a uniform rectangular lamina with sides $a,b$ and mass $M$, its central principal moments are $I_x=M{b}^2/12$, $I_y=M{a}^2/12$ and $I_z=M(a^2+b^2)/12$. If $a>b$, then $I_x<I_y<I_z$: torque-free rotation about the long in-plane axis or the normal is stable, while that about the short in-plane axis is unstable. Coincident moments in a square require the degenerate, rather than strict intermediate-axis, case. Under a constraint to rotate only about the normal, the motion reduces to the scalar equation ${\tau }_z=I_z{\dot{\omega }}_z$.

A symmetric top without torque. Let $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Euler's equations reduce toThus ${\omega }_1+i{\omega }_2$ rotates at signed body-frame frequency $\nu$, while the axial component stays fixed. The symmetry axis has constant angle to fixed spatial angular momentum. Indeed $\mathbf{L}=I_{\perp }\mathbf{\omega }+(I_{\parallel }-I_{\perp }){\omega }_3{\mathbf{e}}_3$, soIt precesses in space about $\mathbf{L}$ with angular speed $L/I_{\perp }$. This spatial precession and the body-frame frequency $\nu$ describe different motions and should not be identified.

A heavy top. If a symmetric top is pivoted a distance $\ell$ from its centre of mass, gravity gives torque ${\mathbf{r}}_{\mathrm{CM}}\times M\mathbf{g}$. Here the relevant moments are about the fixed pivot. The axial angular momentum and the vertical angular-momentum component are conserved, because gravity supplies neither an axial torque nor a torque about the vertical. Using tilt $\theta$, precession angle $\varphi$ and spin angle $\psi$, its Lagrangian isEliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is nutation. Steady precession at $\dot{\varphi }=\Omega$ and fixed tilt satisfiesFor a rapidly spinning top on the slow branch, the first term is small andGravity then changes the direction of angular momentum rather than simply tipping the top straight down. The approximation fails near conditions where nutation or the fast-precession branch matters; the full equations retain those possibilities.

Let a symmetric top have principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Apply a torque $({\tau }_1,{\tau }_2,0)$ constant in a principal body frame. Then the axial angular velocity $c={\omega }_3$ is a conserved quantity. Define $z={\omega }_1+i{\omega }_2$ and $\nu =(I_{\parallel }-I_{\perp })c/I_{\perp }$. The first two of Euler's equations combine into the equation in the title. This derivation uses constancy in the principal body frame, which generally does not mean constant torque in an inertial frame.

Multiplying by the integrating factor $e^{-i\nu t}$ proves, for $\nu \ne 0$,Equivalently $z$ rotates about the constant point $z_{\ast }=i({\tau }_1+i{\tau }_2)/(\nu I_{\perp })$ in the complex transverse plane. When $\nu =0$, the solution is instead $z(t)=z(0)+({\tau }_1+i{\tau }_2)t/I_{\perp }$. The angular velocity is $(\mathrm{Re}z,\mathrm{Im}z,c)$.