INS error propagation — error state equations: Worked examples

Definition The vocabulary we borrow from the parent note

The parent note 3.5.17 defines the objects we reuse here. In plain words:

$δ x$ — the error state: the stack of everything the INS gets wrong. Its three parts are $δ r$ (position error, metres), $δ v$ (velocity error, m/s), and $ψ$ (attitude/tilt error, a small angle in radians saying how far the computed level is tilted from true level).
$F$ — the system matrix: it says how each error feeds the others (e.g. velocity error feeds position error). We do not re-derive $F$ ; we read off its blocks from the parent note.
$G w$ — the noise/error inputs: $w$ is the stack of raw sensor errors (accel bias, gyro drift) and $G$ maps them into the state.
$b_{a}$ — the accelerometer bias: the constant, unwanted acceleration (in $m/s^{2}$ ) that the accelerometer adds to the true specific force. It is often quoted in micro-g ( $μg$ ), where $1 μg = 1 0^{- 6} \times 9.8 m/s^{2}$ . It is the accelerometer part of the error input $w$ .
$ϵ$ — the gyro drift: the constant, unwanted rotation rate (in rad/s) that the gyro adds to the true rate. Often quoted in $^{\circ} / hr$ . It is the gyro part of $w$ .
$f^{n}$ — the specific-force vector expressed in the navigation frame (North-East-Down). "Specific force" is simply the acceleration an accelerometer feels; the superscript $n$ means "written in nav-frame coordinates". In level flight its dominant part is gravity, $∣ f^{n} ∣ \approx g$ .
$ω_{i e}$ — Earth-rate: how fast the Earth spins ( $7.29 \times 1 0^{- 5}$ rad/s). $ω_{e n}$ — transport-rate: how the local nav frame turns as the vehicle moves over the curved Earth. Both are tiny over short flights — Cell G shows exactly when it is safe to drop the $ω_{i e}, ω_{e n}$ terms.

The scenario matrix

Every INS error-growth problem is one of these cells. We will hit all of them.

Cell	What varies	Question it answers
A. Accel bias, sign +	constant $b_{a} > 0$	how fast does a positive accel bias grow position error?
B. Accel bias, sign −	constant $b_{a} < 0$	does the sign just flip, or something subtler?
C. Gyro drift	constant $ϵ$	why is gyro error worse than accel error long-term?
D. Zero input (degenerate)	$b_{a} = 0, ϵ = 0$ , but $δ v_{0} \neq = 0$	what happens with a perfect sensor but a bad initial condition?
E. Limiting / bounded case	gravity coupling ON	why don't horizontal errors blow up — the Schuler bound
F. Real-world word problem	mixed bias + drift	total drift of a real navigation-grade IMU over 1 hour
G. Exam twist	"which term dominates?"	order-of-magnitude reasoning, drop the small terms

Two quick tools we will lean on:

Figure — INS error propagation — error state equations

Figure s01. Horizontal axis = time $t$ . Three curves from the same constant slope: the flat black line is the rate (a constant $c$ ); the dashed black line is its first integral, a straight ramp growing like $t$ ; the red curve is the second integral, a parabola growing like $t^{2}$ . Read it as "each integration bends the line up by one power of $t$ " — exactly what turns a constant accelerometer bias into quadratic position error.

Why we need this: every example is "how many times is the source integrated before it reaches position?" Count the integrations, count the powers of $t$ .

Cell A — Positive accelerometer bias, level flight

Worked example A ·

$b_{a} = + 100 μg$ , level flight, 60 s A navigation-grade accelerometer has a residual bias of $100 μg$ . One $g \approx 9.8 m/s^{2}$ , so $b_{a} = 100 \times 1 0^{- 6} \times 9.8 \approx 9.8 \times 1 0^{- 4} m/s^{2} .$ Ignore Earth rotation and gravity coupling (short, level flight). Initial errors zero. Find the position error at $t = 60$ s.

Forecast: guess now — is the position error more like a few centimetres, or a couple of metres?

Velocity error. With $δ v (0) = 0$ : $\dot{δ v} = b_{a}$ , a constant, so $δ v (t) = b_{a} t$ . Why this step? The velocity equation reduces to "rate = constant"; integrating a constant from zero gives a straight line.
Position error. With $δ r (0) = 0$ : $\dot{δ r} = δ v = b_{a} t$ , a ramp, so $δ r (t) = \frac{1}{2} b_{a} t^{2}$ . Why this step? Position is the second integral. Integrating a ramp gives the parabola from figure s01.
Plug in $t = 60$ . $δ r = \frac{1}{2} (9.8 \times 1 0^{- 4}) (60)^{2} = \frac{1}{2} (9.8 \times 1 0^{- 4}) (3600) \approx 1.76 m .$

Verify: units $= m/s^{2} \cdot s^{2} = m$ ✓. Sanity: velocity error at 60 s is $b_{a} t \approx 0.059 m/s$ — a few cm/s, plausible for a good IMU; over a minute that accumulates to ~1.8 m. Position grew quadratically.

Cell B — Negative bias (does sign matter?)

Cell C — Constant gyro drift (why gyros are worse)

Worked example C · gyro drift

$ϵ = 0.01^{\circ} / hr$ , 60 s A gyro drifts at $0.01^{\circ} / hr$ (navigation grade). Convert: $ϵ = 0.01 \times \frac{π}{180} rad / 3600 s = 4.848 \times 1 0^{- 8} rad/s .$ Gravity coupling ON via $\dot{δ v} = - g ψ$ with $g = 9.8$ (see the definition box just above). Initial errors zero. Find $δ r (60)$ .

Forecast: the drift is tiny. Guess: micrometres, millimetres, or centimetres of position error in a minute?

Drift builds tilt. With $ψ (0) = 0$ : $\dot{ψ} = ϵ \Rightarrow ψ (t) = ϵ t$ . Why this step? Gyro error feeds the attitude equation as a constant rate; the tilt ramps up.
Tilt mis-projects gravity into velocity. $\dot{δ v} = - g ψ = - g ϵ t \Rightarrow δ v = - \frac{1}{2} g ϵ t^{2}$ . Why this step? A tilted platform reads a slice of gravity as horizontal acceleration — the $- [f^{n} \times]$ coupling from $F$ . That's one more integration than the accel case.
Velocity to position. $δ r = - \frac{1}{6} g ϵ t^{3}$ . Why this step? Third integration — drift → tilt → velocity → position. That's the extra power of $t$ .
Plug in. $δ r = - \frac{1}{6} (9.8) (4.848 \times 1 0^{- 8}) (60)^{3} = - \frac{1}{6} (9.8) (4.848 \times 1 0^{- 8}) (216000) \approx - 1.71 \times 1 0^{- 2} m .$

Verify: dimensional check of $\frac{1}{6} g ϵ t^{3}$ , term by term: $[g] [ϵ] [t^{3}] = (\frac{m}{s ^{2}}) (\frac{rad}{s}) (s^{3}) = \frac{m \cdot rad \cdot s ^{3}}{s ^{3}} = m \cdot rad = m$ because a radian is dimensionless ✓. Magnitude ~1.7 cm in one minute. Cubic growth: over an hour this term explodes far faster than the accel bias, which is why gyro spec dominates long flights.

Cell D — Perfect sensors, bad initial velocity (degenerate input)

Worked example D ·

$b_{a} = 0, ϵ = 0$ , but $δ v_{0} = 0.1 m/s$ This is the one cell that breaks our zero-initial-condition convention on purpose. The IMU is ideal — zero bias, zero drift — but the INS was initialized with a $δ v (0) = δ v_{0} = 0.1 m/s$ velocity error (bad GPS handoff); $δ r (0) = 0$ . Ignore coupling. Find $δ r (60)$ .

Forecast: with a perfect sensor, does the error stay put, shrink, or still grow?

Velocity error is frozen. $\dot{δ v} = 0 \Rightarrow δ v (t) = δ v_{0} = 0.1 m/s$ (constant). Why this step? No bias means no forcing; the velocity error just stays at its initial value.
Position still ramps. $\dot{δ r} = δ v_{0} \Rightarrow δ r = δ v_{0} t = 0.1 \times 60 = 6 m .$ Why this step? Even a constant velocity error integrates into a linear position error. Perfect sensors do not stop error growth if the initial state is wrong.

Verify: units $m/s \cdot s = m$ ✓. Note $6 m > 1.76 m$ from Cell A — the initial condition here hurt more than the sensor bias. Lesson: the homogeneous solution ( $F δ x$ acting on a non-zero $δ x_{0}$ ) matters as much as the driving noise.

Cell E — The limiting case: bounded Schuler oscillation

Worked example E · why horizontal error does NOT blow up

In Cells A–D we dropped gravity's feedback. Restore it. A position error $δ r$ (horizontal) changes the direction of local vertical, which changes the tilt $ψ$ , which changes $δ v$ , which changes $δ r$ — a closed loop. Model it as a 1-D oscillator: $\ddot{δ r} = - ω_{s}^{2} δ r$ with $ω_{s}^{2} = g / R$ , $R = 6.371 \times 1 0^{6}$ m the Earth radius.

Forecast: a loop with negative feedback — does the error diverge, settle, or oscillate forever?

Read the equation. $\ddot{δ r} = - ω_{s}^{2} δ r$ is simple harmonic motion. Why this step? Negative feedback proportional to displacement is the signature of an oscillator, not a runaway.
Schuler frequency. $ω_{s} = g / R = 9.8/6.371 \times 1 0^{6} = 1.24 \times 1 0^{- 3} rad/s .$ Why this step? This is the natural frequency of the gravity-error loop.
Schuler period. $T = 2 π / ω_{s} = 2 π /1.24 \times 1 0^{- 3} \approx 5064 s \approx 84.4 min .$ Why this step? $T = 2 π / ω$ converts frequency to period; it matches the famous 84-minute Schuler period.

Verify: $9.8/6.371 \times 1 0^{6} \approx 1.24 \times 1 0^{- 3}$ ✓; $2 π /1.24 \times 1 0^{- 3} \approx 5064$ s $= 84.4$ min ✓. The horizontal error therefore oscillates within a bound instead of running away as $t^{2}$ — this is the limiting behaviour promised in the parent note. See Schuler Tuning and Oscillation.

Figure s02. Horizontal axis = time measured in Schuler periods (1 unit = $T \approx 84.4$ min). Vertical axis = horizontal position error, normalized so the oscillation amplitude is $1$ (in these units the dotted bound lines sit at exactly $+ 1$ and $- 1$ ). The dashed black curve is the uncoupled prediction (gravity feedback off), running away as $t^{2}$ . The red curve is the gravity-coupled truth: it oscillates between the two dotted bound lines ( $\pm 1$ ) instead of diverging. The point of the figure: turning gravity feedback on converts a runaway into a bounded ~84-minute oscillation, capped at its amplitude.

Cell F — Real-world word problem: nav-grade IMU, one hour

Worked example F · combined bias + drift over

$t = 3600$ s A real navigation-grade IMU has accel bias $b_{a} = 50 μg$ and gyro drift $ϵ = 0.01^{\circ} / hr$ . Uncoupled (worst-case bound, ignore Schuler damping). Initial errors zero. Estimate total horizontal position error after 1 hour.

Forecast: for a good IMU over an hour, guess the order of magnitude: metres, hundreds of metres, or kilometres?

Convert. $b_{a} = 50 \times 1 0^{- 6} \times 9.8 = 4.9 \times 1 0^{- 4} m/s^{2}$ ; $ϵ = 4.848 \times 1 0^{- 8} rad/s$ (as Cell C). Why this step? Get both sources into SI before integrating.
Accel term. $δ r_{a} = \frac{1}{2} b_{a} t^{2} = \frac{1}{2} (4.9 \times 1 0^{- 4}) (3600)^{2} \approx 3175 m .$ Why this step? Cell A's quadratic law with the hour-long time.
Gyro term. $δ r_{g} = \frac{1}{6} g ϵ t^{3} = \frac{1}{6} (9.8) (4.848 \times 1 0^{- 8}) (3600)^{3} \approx 3694 m .$ Why this step? Cell C's cubic law applied at $t = 3600$ s: three integrations from drift, so $t^{3}$ . Note this cubic term has now overtaken the quadratic accel term — given enough time, cubic always wins.
Total (linear superposition). $δ r \approx 3175 + 3694 \approx 6869 m \approx 6.9 km .$ Why this step? The error model is linear, so independent contributions simply add.

Verify: each term in metres ✓; gyro > accel confirms cubic dominance ✓. Reality check: an uncoupled pure-inertial run drifts kilometres per hour — which is exactly why real systems fuse GPS via a Kalman filter. The true (Schuler-damped) error is smaller, but this bound shows why aiding is essential.

Cell G — Exam twist: which term can you drop?

Worked example G · 2-minute missile flight — keep only what matters

Recall from the vocabulary box that $ω_{i e}$ is Earth-rate ( $7.29 \times 1 0^{- 5}$ rad/s, the Earth's spin) and $f^{n}$ is the specific-force vector in the nav frame (several $g$ during powered flight). During a 120 s flight, which is safe to neglect: the Earth-rate term, or the $- [f^{n} \times]$ attitude→velocity coupling? Justify by order of magnitude.

Forecast: guess which block of $F$ carries 80% of the short-flight error.

Size the Earth-rate rotation over the flight. $ω_{i e} t = 7.29 \times 1 0^{- 5} \times 120 = 8.75 \times 1 0^{- 3} rad \approx 0. 5^{\circ} .$ Why this step? A term matters only if it rotates the error appreciably during the run; $0.009$ rad is negligible, so the $ω_{i e}, ω_{e n}$ terms in $F$ barely act.
Size the specific-force coupling. In powered flight $∣ f^{n} ∣$ can be several $g$ ; even a $1 0^{- 4}$ rad tilt gives $\dot{δ v} \sim ∣ f^{n} ∣ \cdot ψ \approx 3 \times 9.8 \times 1 0^{- 4} \approx 3 \times 1 0^{- 3} m/s^{2}$ , dominating the tiny Earth-rate cross term. Why this step? Compare magnitudes of the two $F$ blocks acting on the same states; the $[f^{n} \times]$ block is orders of magnitude larger.
Decision. Drop $ω_{i e}, ω_{e n}$ ; keep the $- [f^{n} \times]$ block and the sensor-error inputs. Why this step? 80% of short-flight error lives in that one coupling — the 80/20 rule for $F$ .

Verify: $ω_{i e} t = 8.75 \times 1 0^{- 3}$ rad $≪ 1$ ✓, so neglecting it introduces <1% error. This matches Gyro and Accelerometer Error Models guidance: on short flights, sensor errors + attitude coupling dominate; Earth-rate and transport-rate terms are second order.

Matrix coverage check

Recall Did we hit every cell?

A (+bias) ::: Example A, quadratic, 1.76 m B (−bias) ::: Example B, sign preserved, −1.76 m C (gyro drift) ::: Example C, cubic, −1.71 cm in 60 s D (degenerate: perfect sensor, bad init) ::: Example D, linear from $δ v_{0}$ , 6 m E (limiting/bounded) ::: Example E, Schuler oscillation, 84.4 min F (real-world) ::: Example F, ~6.9 km/hr uncoupled bound G (exam twist / drop terms) ::: Example G, keep $[f^{n} \times]$ , drop Earth rate

Related: Strapdown INS Mechanization Equations · Direction Cosine Matrix and Small-Angle Rotations · Schuler Tuning and Oscillation