field guide

Body rate cross-coupling and when you can ignore it

How measured body rates couple into Euler-angle rates, and the conditions under which the cross-coupling terms can be safely ignored.

updated 2026-06-28 state-estimation kinematics

See background: rotating-coordinate-frames — where the $\boldsymbol\omega \times (I\boldsymbol\omega)$ term comes from.

This note documents when that approximation breaks.

Euler’s equations for a rotating rigid body, diagonal inertia tensor:

$$I \dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times (I \boldsymbol{\omega}) = \boldsymbol{\tau}$$

Expanded:

$$\begin{aligned} I_{xx}\dot{p} + (I_{zz} - I_{yy}) q r &= \tau_x \\ I_{yy}\dot{q} + (I_{xx} - I_{zz}) p r &= \tau_y \\ I_{zz}\dot{r} + (I_{yy} - I_{xx}) p q &= \tau_z \end{aligned}$$

Rearranged for the rate derivatives:

$$\begin{aligned} \dot{p} &= \frac{\tau_x}{I_{xx}} + \frac{(I_{yy} - I_{zz})}{I_{xx}}\, q r \\ \dot{q} &= \frac{\tau_y}{I_{yy}} + \frac{(I_{zz} - I_{xx})}{I_{yy}}\, p r \\ \dot{r} &= \frac{\tau_z}{I_{zz}} + \frac{(I_{xx} - I_{yy})}{I_{zz}}\, p q \end{aligned}$$

Each axis has a “direct” term (torque / inertia) and a “cross” term from $\boldsymbol{\omega} \times (I\boldsymbol{\omega})$ — gyroscopic coupling between axes.

What the symmetric quad’s code does

p_dot = tau_roll  / Ixx;
q_dot = tau_pitch / Iyy;
r_dot = tau_yaw   / Izz;

The cross terms are dropped entirely.

Which terms really drop for Ixx = Iyy

Plug in $I_{xx} = I_{yy}$:

$$\begin{aligned} \dot{p} &= \frac{\tau_x}{I_{xx}} + \frac{(I_{xx} - I_{zz})}{I_{xx}}\, q r \\ \dot{q} &= \frac{\tau_y}{I_{yy}} + \frac{(I_{zz} - I_{xx})}{I_{yy}}\, p r \\ \dot{r} &= \frac{\tau_z}{I_{zz}} + \frac{(0)}{I_{zz}}\, p q = \frac{\tau_z}{I_{zz}} \end{aligned}$$

What’s truly decoupled: yaw ($r$). The $(I_{xx} - I_{yy})$ factor is zero, so r_dot has no cross term from roll/pitch rates.

What’s NOT decoupled but is treated as if it were: roll and pitch. The $p$ and $q$ equations still have nonzero cross terms with $q r$ and $p r$ — unless $I_{zz} = I_{xx}$ (spherical symmetry, not true for any quadcopter).

Why you can drop them anyway (the justification)

Near hover — second-order small. The cross term is a product of two angular rates ($q \cdot r$, $p \cdot r$). When the quad is trying to stay level, these rates are small, so their product is very small.

Control torques dominate. $\tau_x / I_{xx}$ is the commanded signal from the PID. The cross term is a disturbance. For a 250-class quad at moderate rates, the cross term is 1-2 orders of magnitude below the commanded angular acceleration.

Integral term eats the error. Any unmodelled disturbance from dropping the term gets absorbed by the I-term of the rate PID. It sees a steady offset in the tracking error and integrates it out.

When it breaks

Asymmetric build. Battery strapped off-centre, mismatched motors, an ESC fails mid-flight. Now $I_{xx} \neq I_{yy}$. The r_dot cross term pops back: $(I_{xx} - I_{yy})/I_{zz} \cdot p q$ means rolling the quad also induces yaw. This is real and you’ll see it as unwanted yaw drift during roll manoeuvres.

Aggressive aerobatics. High roll/pitch rates make $q r$ and $p r$ large. A flip or snap roll pushes the cross term from “negligible” to “noticeable torque disturbance.” The copilot (or rate PID) has to overcome it.

Heavy frame with large $I_{zz}$. Longer arms or a big payload increase the $I_{xx} - I_{zz}$ gap, scaling the cross term even at moderate rates.

In short: the symmetric-inertia, near-hover, low-rate assumption is valid for a well-built quad doing normal flight. The cross terms code is needed when the build is asymmetric, the flight is aggressive, or both.