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

3.5.16 · D3 · Physics › Guidance, Navigation & Control (GNC) › Mechanization equations — integrating IMU to get position, v

Yeh page parent mechanization note ke liye drill-ground hai. Hum har tarah ka input cover karenge jo mechanization chain ko diya ja sakta hai — saare signs, zero cases, degenerate (kuch bhi nahi hil raha) cases, limits (poles, lamba time), ek real word problem, aur ek exam-style twist. Yahan kuch bhi parent se contradict nahi karta; hum bas dheere chalte hain aur corners cover karte hain.

Kisi bhi example se pehle, ek reminder plain words mein taaki koi bhi symbol bina wajah use na ho:

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Scenario matrix

Har case jo ek mechanization problem throw kar sakti hai, inheen cells mein se ek mein aati hai. Neeche ke examples mein us cell ka label lagaya gaya hai jo woh cover karte hain.

Cell	Kya special hai	Covered by
A. Zero input	gyro = 0, accel = pure $-g$ → kuch bhi move nahi hona chahiye	Ex 1
B. Positive linear accel	ek axis aage push ki, clean double-integral	Ex 2
C. Negative / braking accel	sign flip, velocity zero cross karti hai	Ex 3
D. Attitude change (rotation ≠ 0)	gyro non-zero, $C_b^n$ actually turn karta hai	Ex 4
E. Degenerate: tilt leaks gravity	chhoti attitude error horizontal ko contaminate karti hai	Ex 5
F. Limiting value: near the pole	$\cos \varphi \to 0$ longitude rate ko blow up karta hai	Ex 6
G. Real-world word problem	aircraft cruise, transport + Coriolis matter karte hain	Ex 7
H. Exam twist: bias growth	$t^2$ vs $t^3$ error laws alag karo	Ex 8

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

Forecast: pehle guess karo — kya ${\dot{\mathbf{v}}}^n$ zero hoga, ya kisi axis par $9.81$ leftover dikhega?

1. Attitude. ${\dot{C}}_b^n=C_b^n[{\mathbf{\omega }}_{ib}^b\times ]=C_b^n\cdot 0=0$. Yeh step kyun? Gyro koi rotation feel nahi karta, toh rotation table change nahi ho sakta — orientation frozen hai.
2. Velocity. Flat-Earth form: ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$. Level aur aligned matlab $C_b^n=I$ (identity — "koi re-writing nahi chahiye"). Toh ${\dot{\mathbf{v}}}^n=(0,0,-9.81)+(0,0,+9.81)=(0,0,0)$. Yeh step kyun? Table upar push karta hai (yeh Down axis par $-9.81$ hai), aur gravity neeche pull karti hai ($+9.81$ Down). Yeh exactly cancel ho jaate hain.
3. Position. ${\dot{\mathbf{r}}}^n={\mathbf{v}}^n=0$. Kuch bhi move nahi karta. ✔

Verify: Units ${\text{m/s}}^2$ har term par hain, toh unhe add karna legal hai. Sum zero vector hai — physically object equilibrium mein hai, exactly jo "table par resting" dena chahiye.

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

Forecast: kya $t=4$ par position $20$ m ke kareeb hogi ya $40$ m ke?

1. ${\dot{v}}_N=f_N=5$. Kyun? Short run hai, toh koi Coriolis nahi; gravity is axis par handle ho chuki hai.
2. Ek baar integrate karo: $v_N(t)=5t$ (rest se shuru kiya). Kyun? Constant acceleration time mein straight line mein integrate hoti hai.
3. Phir se integrate karo: $r_N(t)=\frac{1}{2}\cdot 5\cdot t^2=2.5t^2$. Kyun? $5t$ ka integral $\frac{1}{2}\cdot 5\cdot t^2$ hai.
4. $t=4$ plug karo: $v_N=5(4)=20\text{m/s}$, $r_N=2.5(16)=40\text{m}$.

Verify: Kinematic identity se check karo $r=\frac{v^2}{2a}=\frac{20^2}{2\cdot 5}=\frac{400}{10}=40\text{m}$. ✔ Do tareekon se same answer.

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

Forecast: stop time Ex 2 ke $4$ s se bada hoga ya chhota?

1. ${\dot{v}}_N=-4$, toh $v_N(t)=20-4t$. Kyun? Negative sign linear trend ko neeche flip karta hai.
2. Tab rukta hai jab $v_N=0$: $0=20-4t\Rightarrow t=5\text{s}$. Kyun? "Ruka" matlab velocity zero hit kare.
3. Distance: $r_N(t)=20t-2t^2$. $t=5$ par: $r_N=100-50=50\text{m}$. Kyun? Velocity line ko integrate karo.

Verify: Independent check $r=\frac{v_0^2}{2|a|}=\frac{400}{8}=50\text{m}$. ✔ Note karo ki $f_N$ ka sign correctly propagate hua: braking force positive distance deta hai lekin shrinking velocity — exactly wahi jo negative accel karni chahiye.

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Ex 4 — Cell D: ek real rotation DCM ko turn karti hai

Forecast: guess karo — $t=\pi$ s par $0.5$ rad/s se, kitne degrees sweep hue hain?

1. Total swept angle $\theta ={\omega }_{D}t=0.5\cdot \pi =\frac{\pi }{2}=90^{\circ }$. Kyun? Constant rate angle = rate × time mein integrate hoti hai.
2. Down ke baare mein $\theta$ rotation North→East bhejta hai (Down ke baare mein right-hand rule, figure ka orange arc dekho). Nav vector $(\cos \theta ,\sin \theta ,0)$ hai.
3. $\theta =\frac{\pi }{2}$ plug karo: $(\cos 90^{\circ },\sin 90^{\circ },0)=(0,1,0)$ — ab yeh East point kar raha hai. Yeh step kyun? $\cos 90^{\circ }=0,\sin 90^{\circ }=1$; vector ne pura quarter-turn rotate kiya.

Verify: Rotation matrix length $1$ maintain karni chahiye: $\sqrt{0^2+1^2+0^2}=1$. ✔ Aur $\dot{C}=C[\mathbf{\omega }\times ]$ sirf ${\omega }_D\ne 0$ ke saath exactly yahi in-plane spin produce karta hai — teesra (Down) component kabhi nahi badlata, jaisa figure dikhata hai.

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Ex 5 — Cell E: degenerate tilt gravity ko horizontal mein leak karta hai

Forecast: kya leaked accel $0.1$ ke kareeb hogi ya $1.0{\text{m/s}}^2$ ke?

1. Tilted "Down" gravity ko mis-rotate karta hai: horizontal leak hai $\delta a=g\sin (\delta \theta )$. Kyun? Figure dekho — gravity vertical arrow hai; frame ko $\delta \theta$ se tilting karna horizontal axis par ek slice $g\sin \delta \theta$ project karta hai.
2. Numbers: $\delta a=9.81\sin (0.1)=9.81(0.09983)=0.9793{\text{m/s}}^2$. Kyun? $\sin (0.1)\approx 0.09983$ (radians mein!).
3. Yeh fake accel double-integrate hoti hai: $\text{drift}=\frac{1}{2}\delta a{t}^2=\frac{1}{2}(0.9793)(20^2)=195.9\text{m}$. Kyun? Constant fake accel → $\frac{1}{2}a{t}^2$ position error.

Verify: Units: ${\text{m/s}}^2\times {\text{s}}^2=\text{m}$. ✔ Aur small angles ke liye $\sin \delta \theta \approx \delta \theta$, toh $\delta a\approx 9.81(0.1)=0.981$, hamare $0.9793$ se $1\%$ tak match karta hai — exact aur small-angle answers agree karte hain. Yeh dominant INS error path hai: kuch degrees ki tilt horizontal channel ko almost aise corrupt karti hai jaise $g$ khud tumhe push kar raha ho.

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Ex 6 — Cell F: latitude pole ke paas pahunchne par limit

Forecast: jaise $\varphi \to 90^{\circ }$, kya $\dot{\lambda }$ finite value par settle hoga ya blow up karega?

Rule (parent se): $\dot{\lambda }=\frac{v_E}{(R_N+h)\cos \varphi }$. $\cos \varphi$ se kyun divide karte hain? Constant latitude ke circles poles ki taraf shrink hote hain; same eastward speed chhote circle par zyada longitude sweep karta hai.

1. $\varphi =60^{\circ }$ par: $\cos 60^{\circ }=0.5$. $\dot{\lambda }=\frac{200}{6.389\times 10^6\cdot 0.5}=6.261\times 10^{-5}\text{rad/s}$.
2. $\varphi =89.9^{\circ }$ par: $\cos 89.9^{\circ }=1.745\times 10^{-3}$. $\dot{\lambda }=\frac{200}{6.389\times 10^6\cdot 1.745\times 10^{-3}}=1.794\times 10^{-2}\text{rad/s}$.
3. Compare karo: same speed ke liye rate $\approx 286$ factor se jump kar gayi. Kyun? Kyunki $\cos \varphi$ zero ki taraf collapse hota hai — equation exact pole par singular hai.

Verify: Ratio check: ${\dot{\lambda }}_{89.9}/{\dot{\lambda }}_{60}=\cos 60^{\circ }/\cos 89.9^{\circ }=0.5/1.745\times 10^{-3}=286.5$. ✔ Yahi singularity exactly kyun NED mechanization poles ke paas fail hoti hai aur engineers wander-azimuth ya ECEF frame par switch karte hain.

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

Forecast: kya Coriolis correction $0.1{\text{m/s}}^2$ se badi hai ya chhoti?

(a) East channel par Coriolis. Earth ki spin NED mein ${\mathbf{\omega }}_{ie}^n={\Omega }_e(\cos \varphi ,0,-\sin \varphi )$ hai. Velocity equation mein correction term $-2{\mathbf{\omega }}_{ie}^n\times {\mathbf{v}}^n$ hai jahan ${\mathbf{v}}^n=(250,0,0)$.

1. Cross product lo. $\mathbf{v}$ North along hai, toh sirf ${\mathbf{\omega }}_{ie}$ ka Down component aur North velocity ek East term produce karte hain: East component $=-2(-{\Omega }_e\sin \varphi )(v_N)\cdot (\text{sign from }\hat{N}\times \hat{D})$. Magnitude $=2{\Omega }_e\sin \varphi v_N$. Yeh step kyun? $\hat{N}\times \hat{D}$ $-\hat{E}$ along point karta hai; do signs mila ke real eastward push banta hai — yahi reason hai Northbound flights rotating frame mein East drift karti hain.
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. $\dot{\varphi }=\frac{v_N}{R+h}=\frac{250}{6.371\times 10^6}=3.924\times 10^{-5}\text{rad/s}$. Kyun? Meridian radius par arc-length rate; sphere par $R_M=R$.

Verify: Coriolis magnitude order-of-check: $\sim 10^{-4}\times 10^2=10^{-2}{\text{m/s}}^2$ ✔ — chhota hai lekin ek ghante mein hundreds of metres mein integrate ho jaata hai, yahi kyun long-haul INS must is term ko keep kare. Latitude rate: $3.924\times 10^{-5}\text{rad/s}\times 3600\text{s}\approx 0.141\text{rad}\approx 8.1^{\circ }$ North travel ke ek ghante mein — fast jet ke liye sensible hai.

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Ex 8 — Cell H: exam twist — gyro ya accel bias, kaun zyada fast grow karta hai?

Forecast: accel bias directly accel banata hai — yeh surely jitega? Guess karo, phir check karo.

1. Accel-bias path (Unit A). Ek constant fake accel $b$ double-integrate hoti hai: ${\text{err}}_A=\frac{1}{2}b{t}^2$. $t^2$ kyun? Ek constant ki do integrations. ${\text{err}}_A=\frac{1}{2}(0.01)(100^2)=50\text{m}$.
2. Gyro-bias path (Unit B). Tilt linearly badhti hai $\delta \theta =\epsilon t$; yeh gravity leak karti hai $\delta a\approx g\epsilon t$; woh leaked accel triple-integrate hoti hai jab growing tilt account karo: ${\text{err}}_B=\frac{1}{6}g\epsilon t^3$. $t^3$ kyun? $\delta a$ khud $t$ ki tarah barhta hai, aur position do aur integrations hain: $t\to t^3$. ${\text{err}}_B=\frac{1}{6}(9.81)(0.001)(100^3)=1635\text{m}$.
3. Compare: $1635\gg 50$. Gyro bias jeet jaata hai, buri tarah se. Kyun? Uska error law $t^3$ hai accel ke $t^2$ ke mukable mein; kaafi time dene par higher power hamesha dominate karta hai.

Verify: Unit B ka dimensional check: ${\text{m/s}}^2\cdot (1/\text{s per rad-tilt})$… concretely $g\epsilon$ ke units ${\text{m/s}}^2\cdot {\text{s}}^{-1}\cdot \text{rad}={\text{m/s}}^3$ hain, times $t^3={\text{s}}^3$ metres deta hai ✔. Crossover time jahan dono equal hain: solve karo $\frac{1}{2}b{t}^2=\frac{1}{6}g\epsilon t^3\Rightarrow t=\frac{3b}{g\epsilon }=\frac{3(0.01)}{9.81(0.001)}=3.058\text{s}$ — sirf $\sim 3$ s baad gyro error overtake kar leta hai, yahi kyun gyro quality INS grade dominate karti hai. Yeh reasoning Kalman Filter for INS-GNSS Integration ke saath fusing aur Strapdown vs Gimbaled INS design trade ke peeche hai.

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Recall Kaun sa error term time ka highest power hai, aur yeh INS grade kyun decide karta hai?

Gyro-bias-through-tilt term $t^3$ ki tarah badhta hai (tilt linearly badhti hai, gravity-leak linearly badhti hai, phir do integrations), accel bias ke $t^2$ ko beat karta hai. Long runs ke liye highest power jeet jaata hai, toh gyro quality dominate karti hai.

Recall Longitude-rate equation poles ke paas kyun blow up hoti hai?

Kyunki $\dot{\lambda }=v_E/((R_N+h)\cos \varphi )$ aur $\cos \varphi \to 0$ at $\varphi =90^{\circ }$: latitude circles ek point tak shrink ho jaate hain, toh koi bhi eastward speed infinite longitude rate sweep karti hai — NED frame wahan singular hai.