Question bank — INS error propagation — error state equations

False. Bias integrates once to a linear velocity error, then a second time to position, so position grows as $t^2$ — quadratic (see the $t^2$ curve above). Count the integrations, not the source.

True in growth order. Gyro drift needs three integrations (drift → tilt → mis-resolved gravity → velocity → position), giving $t^3$, versus $t^2$ for accel bias — so past the crossover in the growth figure, gyro quality dominates.

False. The errors are small, so a first-order Taylor expansion about the true trajectory gives the linear $\dot{\delta x}=F\delta x+Gw$ printed in §0 — exactly the form a linear Kalman filter needs.

False. The off-diagonal block $-[{\mathbf{f}}^n\times ]$ in $F$ couples them: a tilt mis-resolves the specific force into a spurious acceleration of size $\approx g\psi$, so $\mathbf{\psi }$ directly drives $\dot{\delta \mathbf{v}}$.

False (bounded, oscillatory). Gravity feeds back on horizontal errors, producing a bounded Schuler oscillation (≈84 min) as drawn above; only unmodeled biases add a slow ramp on top. See Schuler Tuning and Oscillation.

False. $\dot{\mathbf{r}}=\mathbf{v}$ is already linear, so perturbing it is exact — the top row of $F$ (a clean identity block $I$) carries no truncation error.

False. With $w=0$ you still have $\dot{\delta x}=F\delta x$: any initial error (alignment tilt, initial position uncertainty) still propagates and, via Schuler coupling, oscillates. Only $\delta x(0)=0$ and $w=0$ gives zero error.

False in general, often true for short flights. For a 2-minute flight ${\omega }_{ie}t\approx 0.009$, so it's negligible; but over an hours-long run these Earth-rate (${\mathbf{\omega }}_{ie}$) and transport-rate (${\mathbf{\omega }}_{en}$) terms shape the Schuler behaviour and cannot be dropped.

Signs wrong. The correct equation is $\dot{\mathbf{\psi }}=-{\mathbf{\omega }}_{in}\times \mathbf{\psi }-C_b^n\delta {\mathbf{\omega }}_{ib}^b$ (bottom row of $F$ plus the $-C_b^n$ block of $G$); both terms are negative because the tilt is transported by ${\mathbf{\omega }}_{in}$ and corrupted by gyro error draining into it.

Wrong operand. It is $-[{\mathbf{f}}^n\times ]\mathbf{\psi }$, using the full specific force ${\mathbf{f}}^n$ (size $\sim g$), not the tiny error $\delta {\mathbf{f}}^b$. Multiplying two small quantities ($\psi \cdot \delta f$) is second order and gets dropped — the tilt acts on the large specific force, exactly as the geometry figure shows.

That deletes the dominant coupling. The whole point is ${\hat{C}}_b^n=(I-[\mathbf{\psi }\times ])C_b^n$; keeping the $-[\mathbf{\psi }\times ]$ piece is what produces the $-[{\mathbf{f}}^n\times ]\mathbf{\psi }$ block of $F$. Setting them equal erases attitude→velocity feedback.

Sign of the gyro block wrong. It is $G=\mathrm{diag}(0,C_b^n,-C_b^n)$ (see §0); the attitude equation carries $-C_b^n\delta {\mathbf{\omega }}_{ib}^b$, so its noise gain is negative.

Wrong row. In $G$, accelerometer error enters through the velocity row (via $C_b^n\delta {\mathbf{f}}^b$); the attitude row is driven only by gyro error ($-C_b^n\delta {\mathbf{\omega }}_{ib}^b$). Mixing them confuses which sensor spec limits which state.

Off by one. The top row $\dot{\delta \mathbf{r}}=\delta \mathbf{v}$ makes velocity the first integral; position is the second integral of the bias. This single miscount turns a $t^2$ law into a false $t^1$ law.

A tilted frame mis-projects the large specific-force vector (size $\sim g$) so a horizontal slice $\approx g\psi$ appears as fake acceleration — the $-[{\mathbf{f}}^n\times ]\mathbf{\psi }$ term shown geometrically in figure s02. Perfect accelerometers don't help if the frame they report in is tipped.

Gyro drift sits three integrations from position ($t^3$ growth) and feeds the gravity-mis-resolution coupling, so past the crossover in figure s03 its cubic ramp overtakes the accelerometer's quadratic one.

The nonlinearities (rotations, gravity variation) depend on where the vehicle actually is; linearizing about the true state keeps the neglected terms genuinely second-order-small, which is what justifies the constant-per-instant $F$.

The gravity feedback loop has a restoring "stiffness" set by $g/R$ (Earth radius $R$), and $2\pi \sqrt{R/g}\approx 84$ min — a natural pendulum-of-planet-size frequency baked into the $F$ coupling. See Schuler Tuning and Oscillation.

The linear model $\dot{\delta x}=F\delta x+Gw$ predicts how errors pattern over time; an external fix (e.g. GPS) observes some combination, and the known $F$ lets the filter back out the unobserved states. See Kalman Filter for INS-GPS Integration.

It is the one state that both accumulates gyro error and injects into velocity error via $-[{\mathbf{f}}^n\times ]$, so it is the bridge (the middle-right block of $F$) that links sensor imperfection to position drift.

No. Even stationary, the accelerometers read the specific force opposing gravity, so ${\mathbf{f}}^n\approx [0,0,-g{]}^{\top }$ is nonzero; a tilt still mis-resolves this into horizontal velocity error — that's how gyro drift leaks in even at standstill.

It vanishes: $-[{\mathbf{f}}^n\times ]\mathbf{\psi }=0$. With no specific force to mis-resolve, a tilt no longer corrupts velocity — a genuine degenerate case where the dominant coupling switches off.

Yes. $\dot{\delta x}=F\cdot 0+G\cdot 0=0$, so it is a fixed point. Any real system fails this only because $\delta x(0)\ne 0$ (imperfect alignment) or $w\ne 0$ (real sensors).

Yes — over ~2 min their contribution ($\omega t\approx 0.009$) is negligible; but you must keep the $-[{\mathbf{f}}^n\times ]$ attitude→velocity block and the sensor-error inputs, which carry ~80% of short-flight error.

One extra power beyond a constant bias: a ramp $\propto t$ → velocity $\propto t^2$ → position $\propto t^3$, so cubic — matching a constant gyro drift's order.

No. Reorienting only changes which components of ${\mathbf{f}}^n$ are large; the cross-product $[{\mathbf{f}}^n\times ]$ still couples the perpendicular tilt components into velocity — orientation cannot zero a nonzero specific force.