Worked examples — Mechanization equations — integrating IMU to get position, velocity, attitude

This page is the drill-ground for the parent mechanization note. We will hit every kind of input the mechanization chain can be handed — all signs, the zero cases, the degenerate (nothing-moving) cases, the limits (poles, long time), a real word problem, and one exam-style twist. Nothing here contradicts the parent; we just go slower and cover the corners.

Before any example, one reminder in plain words so no symbol is used unearned:

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The scenario matrix

Every case a mechanization problem can throw at you falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it special	Covered by
A. Zero input	gyro = 0, accel = pure $-g$ → nothing should move	Ex 1
B. Positive linear accel	one axis pushed forward, clean double-integral	Ex 2
C. Negative / braking accel	sign flip, velocity crosses zero	Ex 3
D. Attitude change (rotation ≠ 0)	gyro non-zero, $C_b^n$ actually turns	Ex 4
E. Degenerate: tilt leaks gravity	small attitude error contaminates horizontal	Ex 5
F. Limiting value: near the pole	$\cos \varphi \to 0$ makes longitude rate blow up	Ex 6
G. Real-world word problem	aircraft cruise, transport + Coriolis matter	Ex 7
H. Exam twist: bias growth	separate $t^2$ vs $t^3$ error laws	Ex 8

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Ex 1 — Cell A: the still, level IMU (zero input)

Forecast: guess before reading — will ${\dot{\mathbf{v}}}^n$ be zero, or will one axis show a leftover $9.81$?

1. Attitude. ${\dot{C}}_b^n=C_b^n[{\mathbf{\omega }}_{ib}^b\times ]=C_b^n\cdot 0=0$. Why this step? The gyro feels no rotation, so the rotation table cannot change — orientation is frozen.
2. Velocity. Flat-Earth form: ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$. Level and aligned means $C_b^n=I$ (identity — "no re-writing needed"). So ${\dot{\mathbf{v}}}^n=(0,0,-9.81)+(0,0,+9.81)=(0,0,0)$. Why this step? The table pushes up (that is the $-9.81$ on the Down axis), and gravity pulls down ($+9.81$ Down). They cancel exactly.
3. Position. ${\dot{\mathbf{r}}}^n={\mathbf{v}}^n=0$. Nothing moves. ✔

Verify: Units are ${\text{m/s}}^2$ on every term, so adding them is legal. The sum is the zero vector — physically the object is in equilibrium, exactly what "resting on a table" must give.

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Ex 2 — Cell B: constant forward push (positive linear accel)

Forecast: will position at $t=4$ be nearer $20$ m or $40$ m?

1. ${\dot{v}}_N=f_N=5$. Why? Short run, so no Coriolis; gravity is handled off this axis.
2. Integrate once: $v_N(t)=5t$ (started at rest). Why? Constant acceleration integrates to a straight line in time.
3. Integrate again: $r_N(t)=\frac{1}{2}\cdot 5\cdot t^2=2.5t^2$. Why? Integral of $5t$ is $\frac{1}{2}\cdot 5\cdot t^2$.
4. Plug $t=4$: $v_N=5(4)=20\text{m/s}$, $r_N=2.5(16)=40\text{m}$.

Verify: Check with the kinematic identity $r=\frac{v^2}{2a}=\frac{20^2}{2\cdot 5}=\frac{400}{10}=40\text{m}$. ✔ Same answer two ways.

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Ex 3 — Cell C: braking (negative accel, velocity crosses zero)

Forecast: stop time bigger or smaller than the $4$ s of Ex 2?

1. ${\dot{v}}_N=-4$, so $v_N(t)=20-4t$. Why? The negative sign flips the linear trend downward.
2. Stops when $v_N=0$: $0=20-4t\Rightarrow t=5\text{s}$. Why? "Stopped" means velocity hits zero.
3. Distance: $r_N(t)=20t-2t^2$. At $t=5$: $r_N=100-50=50\text{m}$. Why? Integrate the velocity line.

Verify: Independent check $r=\frac{v_0^2}{2|a|}=\frac{400}{8}=50\text{m}$. ✔ Note the sign of $f_N$ propagated correctly: a braking force gives positive distance but a shrinking velocity — exactly what a negative accel should do.

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Ex 4 — Cell D: a real rotation turns the DCM

Forecast: guess — after $t=\pi$ s at $0.5$ rad/s, how many degrees has it swept?

1. Total swept angle $\theta ={\omega }_{D}t=0.5\cdot \pi =\frac{\pi }{2}=90^{\circ }$. Why? Constant rate integrates to angle = rate × time.
2. A rotation about Down by $\theta$ sends North→East (right-hand rule about Down, look at the figure's orange arc). The nav vector is $(\cos \theta ,\sin \theta ,0)$.
3. Plug $\theta =\frac{\pi }{2}$: $(\cos 90^{\circ },\sin 90^{\circ },0)=(0,1,0)$ — it now points East. Why this step? $\cos 90^{\circ }=0,\sin 90^{\circ }=1$; the vector rotated a full quarter-turn.

Verify: A rotation matrix must keep length $1$: $\sqrt{0^2+1^2+0^2}=1$. ✔ And $\dot{C}=C[\mathbf{\omega }\times ]$ with only ${\omega }_D\ne 0$ produces exactly this in-plane spin — the third (Down) component never changes, as the figure shows.

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Ex 5 — Cell E: degenerate tilt leaks gravity into the horizontal

Forecast: will the leaked accel be closer to $0.1$ or $1.0{\text{m/s}}^2$?

1. A tilted "Down" mis-rotates gravity: the horizontal leak is $\delta a=g\sin (\delta \theta )$. Why? Look at the figure — gravity is the vertical arrow; tilting the frame by $\delta \theta$ projects a slice $g\sin \delta \theta$ onto the horizontal axis.
2. Numbers: $\delta a=9.81\sin (0.1)=9.81(0.09983)=0.9793{\text{m/s}}^2$. Why? $\sin (0.1)\approx 0.09983$ (radians!).
3. This fake accel double-integrates: $\text{drift}=\frac{1}{2}\delta a{t}^2=\frac{1}{2}(0.9793)(20^2)=195.9\text{m}$. Why? Constant fake accel → $\frac{1}{2}a{t}^2$ position error.

Verify: Units: ${\text{m/s}}^2\times {\text{s}}^2=\text{m}$. ✔ And for small angles $\sin \delta \theta \approx \delta \theta$, so $\delta a\approx 9.81(0.1)=0.981$, matching our $0.9793$ to $1\%$ — the exact and small-angle answers agree. This is the dominant INS error path: a tilt of a few degrees corrupts the horizontal channel almost as if $g$ itself were pushing you.

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Ex 6 — Cell F: the limit as latitude approaches the pole

Forecast: as $\varphi \to 90^{\circ }$, does $\dot{\lambda }$ settle to a finite value or blow up?

The rule (from the parent): $\dot{\lambda }=\frac{v_E}{(R_N+h)\cos \varphi }$. Why divide by $\cos \varphi$? Circles of constant latitude shrink toward the poles; the same eastward speed sweeps more longitude on a smaller circle.

1. At $\varphi =60^{\circ }$: $\cos 60^{\circ }=0.5$. $\dot{\lambda }=\frac{200}{6.389\times 10^6\cdot 0.5}=6.261\times 10^{-5}\text{rad/s}$.
2. At $\varphi =89.9^{\circ }$: $\cos 89.9^{\circ }=1.745\times 10^{-3}$. $\dot{\lambda }=\frac{200}{6.389\times 10^6\cdot 1.745\times 10^{-3}}=1.794\times 10^{-2}\text{rad/s}$.
3. Compare: the rate jumped by a factor $\approx 286$ for the same speed. Why? Because $\cos \varphi$ collapses toward zero — the equation is singular at the exact pole.

Verify: Ratio check: ${\dot{\lambda }}_{89.9}/{\dot{\lambda }}_{60}=\cos 60^{\circ }/\cos 89.9^{\circ }=0.5/1.745\times 10^{-3}=286.5$. ✔ This singularity is exactly why NED mechanization fails near the poles and engineers switch to a wander-azimuth or ECEF frame.

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Ex 7 — Cell G: real-world cruise (transport + Coriolis)

Forecast: is the Coriolis correction bigger or smaller than $0.1{\text{m/s}}^2$?

(a) Coriolis on the East channel. Earth's spin in NED is ${\mathbf{\omega }}_{ie}^n={\Omega }_e(\cos \varphi ,0,-\sin \varphi )$. The correction term in the velocity equation is $-2{\mathbf{\omega }}_{ie}^n\times {\mathbf{v}}^n$ with ${\mathbf{v}}^n=(250,0,0)$.

1. Take the cross product. With $\mathbf{v}$ along North, only the Down component of ${\mathbf{\omega }}_{ie}$ and the North velocity produce an East term: East component $=-2(-{\Omega }_e\sin \varphi )(v_N)\cdot (\text{sign from }\hat{N}\times \hat{D})$. Magnitude $=2{\Omega }_e\sin \varphi v_N$. Why this step? $\hat{N}\times \hat{D}$ points along $-\hat{E}$; the two signs combine to a real eastward push — this is why Northbound flights drift East in a rotating frame.
2. Numbers: $2(7.292\times 10^{-5})\sin 45^{\circ }(250)=2(7.292\times 10^{-5})(0.7071)(250)=2.578\times 10^{-2}{\text{m/s}}^2$.

(b) Latitude rate. $\dot{\varphi }=\frac{v_N}{R+h}=\frac{250}{6.371\times 10^6}=3.924\times 10^{-5}\text{rad/s}$. Why? Arc-length rate over meridian radius; on a sphere $R_M=R$.

Verify: Coriolis magnitude order-of-check: $\sim 10^{-4}\times 10^2=10^{-2}{\text{m/s}}^2$ ✔ — small but over an hour it integrates to hundreds of metres, which is why long-haul INS must keep this term. Latitude rate: $3.924\times 10^{-5}\text{rad/s}\times 3600\text{s}\approx 0.141\text{rad}\approx 8.1^{\circ }$ per hour of North travel — sensible for a fast jet.

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Ex 8 — Cell H: exam twist — which error grows faster, gyro or accel bias?

Forecast: the accel bias directly makes accel — surely it wins? Guess, then check.

1. Accel-bias path (Unit A). A constant fake accel $b$ double-integrates: ${\text{err}}_A=\frac{1}{2}b{t}^2$. Why $t^2$? Two integrations of a constant. ${\text{err}}_A=\frac{1}{2}(0.01)(100^2)=50\text{m}$.
2. Gyro-bias path (Unit B). The tilt grows linearly $\delta \theta =\epsilon t$; it leaks gravity as $\delta a\approx g\epsilon t$; that leaks accel triple-integrates once you account for the growing tilt: ${\text{err}}_B=\frac{1}{6}g\epsilon t^3$. Why $t^3$? $\delta a$ itself grows like $t$, and position is two more integrations: $t\to t^3$. ${\text{err}}_B=\frac{1}{6}(9.81)(0.001)(100^3)=1635\text{m}$.
3. Compare: $1635\gg 50$. The gyro bias wins, badly. Why? Its error law is $t^3$ versus the accel's $t^2$; given enough time the higher power always dominates.

Verify: Dimensional check on Unit B: ${\text{m/s}}^2\cdot (1/\text{s per rad-tilt})$… concretely $g\epsilon$ has units ${\text{m/s}}^2\cdot {\text{s}}^{-1}\cdot \text{rad}={\text{m/s}}^3$, times $t^3={\text{s}}^3$ gives metres ✔. Crossover time where the two are equal: solve $\frac{1}{2}b{t}^2=\frac{1}{6}g\epsilon t^3\Rightarrow t=\frac{3b}{g\epsilon }=\frac{3(0.01)}{9.81(0.001)}=3.058\text{s}$ — after only $\sim 3$ s the gyro error overtakes, which is why gyro quality dominates INS grade. This is the reasoning behind fusing with GNSS in the Kalman Filter for INS-GNSS Integration and the whole Strapdown vs Gimbaled INS design trade.

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Recall Which error term has the highest power of time, and why does that decide INS grade?

The gyro-bias-through-tilt term grows as $t^3$ (tilt grows linearly, gravity-leak grows linearly, then two integrations), beating the accel bias's $t^2$. Highest power wins for long runs, so gyro quality dominates.

Recall Why does the longitude-rate equation blow up near the poles?

Because $\dot{\lambda }=v_E/((R_N+h)\cos \varphi )$ and $\cos \varphi \to 0$ at $\varphi =90^{\circ }$: the latitude circles shrink to a point, so any eastward speed sweeps infinite longitude rate — the NED frame is singular there.