Mechanization equations — integrating IMU to get position, velocity, attitude: Exercises

3.5.16 · D4 · Physics › Guidance, Navigation & Control (GNC) › Mechanization equations — integrating IMU to get position, v

Quick symbol reminder, taaki pehli line koi bhi padh sake:

Definition Characters ki cast

$f^{b}$ = specific force jo accelerometer feel karta hai, body axes mein likha gaya (vehicle pe bane axes).
$ω_{ib}^{b}$ = angular rate jo gyro feel karta hai, body vs. inertial (non-rotating star-fixed) space, body axes mein.
$ω_{nb}^{b}$ = body ka angular rate nav frame ke relative, body axes mein. Yeh actually vehicle ko North-East-Down ke relative turn karta hai; yeh gyro rate minus Earth spin aur curved Earth pe travel karne se aane wale parts ke barabar hai. Ek static, short-run flat-Earth problem mein $ω_{nb}^{b} \approx ω_{ib}^{b}$ .
$C_{b}^{n}$ = rotation matrix ("attitude") jo ek body-axis vector ko nav-frame (North-East-Down) vector mein badalta hai.
$[ω \times]$ = skew-symmetric matrix jo vector $ω = (ω_{x}, ω_{y}, ω_{z})$ se banta hai, aur isse multiply karna bilkul wahi kaam karta hai jo cross product $ω \times ()$ karta hai: $[ω \times] = 0 ω_{z} - ω_{y} - ω_{z} 0 ω_{x} ω_{y} - ω_{x} 0, [ω \times] u = ω \times u .$
$v^{n} = (v_{N}, v_{E}, v_{D})$ = nav frame mein velocity; $g^{n}$ = plumb-bob gravity, Down direction mein pointing.
$ϕ$ = latitude (equator se North ka angle), $λ$ = longitude (axis ke around East ka angle), $h$ = height.
$Ω_{e} = 7.292 \times 1 0^{- 5} rad/s$ = Earth ki spin rate.

L1 — Recognition

Problem 1.1

Ek stationary IMU bench pe flat rakha hai (NED, Down $= + z$ ). Har accelerometer axis kya read karta hai, aur gyros kya read karte hain?

Recall Solution 1.1

Gyros: $ω_{ib}^{b} \approx 0$ (chhoti si $Ω_{e}$ Earth-spin term ko ignore karte hue jo note high-grade gyros ke liye flag karta hai). Accels: $f^{b} = (0, 0, - 9.81)$ . Bench upar push karta hai, aur NED mein up-direction $- z$ hai, isliye felt (specific) force Down pe $- 9.81$ baith ta hai. Yeh zero nahi hai: accelerometer gravity feel nahi kar sakta, sirf contact push feel karta hai.

Problem 1.2

Mechanization ke teen "layers" ko us order mein naam do jis order mein inhe compute karna zaroori hai, aur ek-word ka reason batao ki attitude pehle kyun aata hai.

Recall Solution 1.2

Order: Attitude → Velocity → Position. Reason: rotation. Accelerometer ki force body axes mein hoti hai; $C_{b}^{n}$ (attitude) chahiye taaki use nav axes mein rotate kar sako, pehle gravity add karo aur phir velocity integrate karo.

Problem 1.3

Velocity equation mein kaun sa term "gravity wapas add" karta hai, aur yeh add karna zaroori kyun hai?

Recall Solution 1.3

Term $+ g^{n}$ hai. Zaroori isliye hai kyunki accelerometer physically gravitation sense nahi kar sakta (gravity har atom ko equally pull karti hai, isliye koi internal strain nahi hoti). Software ko ise wapas dalna padta hai.

L2 — Application

Problem 2.1

Flat-Earth, no rotation. Ek sled nav- $x$ pe $f_{x} = 3 m/s^{2}$ read karta hai (gravity off-axis handle ki gayi hai). Rest se origin pe start karte hue, $t = 6 s$ pe $v_{x}$ aur $x$ nikalo.

Recall Solution 2.1

Flat-Earth velocity law with no Coriolis, no on-axis gravity: $\overset{v}{˙}_{x} = f_{x} = 3$ . Ek baar integrate kyun karte hain? Acceleration velocity ka rate of change hota hai, isliye velocity acceleration ka running total (integral) hai: $v_{x} (t) = \int_{0}^{t} f_{x} d τ = 3 t \Rightarrow v_{x} (6) = 18 m/s$ . Dobara integrate kyun karte hain? Velocity position ka rate of change hoti hai, isliye position velocity ka running total hai: $x (t) = \int_{0}^{t} v_{x} d τ = \frac{1}{2} (3) t^{2} = 1.5 t^{2} \Rightarrow x (6) = 1.5 (36) = 54 m$ . Yeh do-step "integral ka integral" mechanization ki poori spirit hai: rates andar, positions bahar.

Problem 2.2

$ω = (0, 0, 0.2) rad/s$ ke liye skew-symmetric matrix $[ω \times]$ (symbol reminder mein define kiya gaya) compute karo aur verify karo ki yeh $u = (1, 0, 0)$ ke liye $ω \times u$ reproduce karta hai.

Recall Solution 2.2

$ω_{x} = 0, ω_{y} = 0, ω_{z} = 0.2$ ko skew template mein plug karte hue: $[ω \times] = 0 0.2 0 - 0.2 00 000 .$ $u = (1, 0, 0)^{⊤}$ se multiply karo: $(0, 0.2, 0)^{⊤}$ milta hai. Cross product $(0, 0, 0.2) \times (1, 0, 0) = (0 \cdot 0 - 0.2 \cdot 0, 0.2 \cdot 1 - 0 \cdot 0, 0) = (0, 0.2, 0)$ . ✔ Dono match karte hain — yehi exact reason hai ki $\dot{C}_{b}^{n} = C_{b}^{n} [ω_{nb}^{b} \times]$ mein skew matrix cross product ko replace karta hai.

Problem 2.3

Latitude $ϕ = 4 5^{\circ}$ , height $h = 0$ pe, vehicle due East $v_{E} = 200 m/s$ pe fly karta hai. Prime-vertical radius of curvature $R_{N} = 6.389 \times 1 0^{6} m$ hai. Longitude rate $\dot{λ}$ rad/s mein nikalo.

Recall Solution 2.3

Yeh formula kyun? Longitude $λ$ ek angle hai jo Earth ki spin axis ke around measure hota hai. Jab tum East fly karte ho tum constant latitude ka circle (ek "parallel") trace karte ho. Iska radius poora Earth radius nahi balki shrunken $(R_{N} + h) cos ϕ$ hai, kyunki latitude $ϕ$ pe parallel ek small circle hai jiska radius big circle ke radius ka $cos ϕ$ guna hota hai. Sweep kiya gaya angle equals arc-length traveled divided by circle ka radius, isliye $\dot{λ} = \frac{East arc-rate}{parallel radius} = \frac{v _{E}}{( R _{N} + h ) c o s ϕ} .$ Plug in karo: $\dot{λ} = \frac{200}{6.389 \times 1 0 ^{6} \cdot c o s 4 5 ^{\circ}} = \frac{200}{6.389 \times 1 0 ^{6} \cdot 0.70711} \approx 4.427 \times 1 0^{- 5} rad/s .$ $cos ϕ$ longitude circles ko poles ki taraf shrink karta hai, isliye same East speed se tum longitude ke around jitna upar latitude, utna tez spin karte ho. Dekho Radii of Curvature of the Earth Ellipsoid.

L3 — Analysis

Problem 3.1

Ek gyro mein East axis ke baare mein constant bias $ε = 0.005 rad/s$ hai. Vehicle static hai aur initially level hai. $t = 20 s$ ke baad attitude (tilt) error degrees mein estimate karo.

Recall Solution 3.1

Ek static-vehicle tilt error linearly accumulate hota hai: $δ θ = εt = 0.005 \times 20 = 0.1 rad$ . Convert karo: $0.1 \times \frac{180}{π} \approx 5.7 3^{\circ}$ . Mechanization believe karta hai ki yeh rotate hua jab hua nahi tha, kyunki yeh fake rate integrate karta hai.

Problem 3.2

Problem 3.1 continue karo. Yeh 0.1 rad tilt gravity ko misrotate karta hai. Spurious horizontal acceleration estimate karo jo yeh inject karta hai, $g = 9.81$ use karte hue.

Recall Solution 3.2

Ek tilt $δ θ$ gravity vector ko tilt kar deta hai taaki ek component $g sin (δ θ)$ horizontal channel pe spill ho jaaye: $δ a = g sin (δ θ) = 9.81 sin (0.1) \approx 9.81 \times 0.09983 = 0.9794 m/s^{2} .$ Yeh dominant INS error path hai: ek chhoti attitude error almost poora $g$ wahan leak kar deti hai jahan woh belong nahi karta.

Figure geometry ko concrete banata hai. True gravity (red) straight Down point karti hai. Mechanization sochta hai Down dashed white axis hai, $δ θ$ se tilted (blue arc). Jab yeh apne tilted axis ke along "gravity" remove karta hai, ek yellow horizontal slice $g sin δ θ$ uncancelled reh jaata hai — woh leftover fake horizontal acceleration hai jo tumne compute ki.

Figure — Mechanization equations — integrating IMU to get position, velocity, attitude

Problem 3.3

Ek accelerometer bias $b$ se velocity error $t$ jaisi kyun grow karti hai, jabki same bias se position error $t^{2}$ jaisi kyun grow karta hai?

Recall Solution 3.3

Velocity acceleration ka ek integral hai: $\int_{0}^{t} b d τ = b t$ (linear). Position ek aur integral hai: $\int_{0}^{t} b τ d τ = \frac{1}{2} b t^{2}$ (quadratic). Har integration "time" se multiply karta hai, isliye chain ki har layer error mein $t$ ki ek power add kati hai — yehi reason hai ki INS drift accelerate karta hai aur GNSS correction ki zaroorat hoti hai.

L4 — Synthesis

Problem 4.1

Level IMU, no rotation, flat-Earth. Accels $f^{b} = (0.02, 0, - 9.81)$ m/s $^{2}$ read karte hain (ek chhota $x$ -bias of $0.02$ ). $g^{n} = (0, 0, + 9.81)$ ke saath, $\dot{v}^{n}$ nikalo, phir $v_{N}$ aur North position $t = 100 s$ ke baad rest se.

Recall Solution 4.1

Kyunki attitude identity hai (no tilt, no rotation) $C_{b}^{n} f^{b} = f^{b}$ . Flat-Earth velocity core $\dot{v}^{n} = C_{b}^{n} f^{b} + g^{n}$ use karte hue: $\dot{v}^{n} = f^{b} + g^{n} = (0.02, 0, - 9.81) + (0, 0, 9.81) = (0.02, 0, 0) .$ Down cancel ho jaata hai (static vertical channel ✔). Sirf North bias survive karta hai. $v_{N} (100) = 0.02 \times 100 = 2 m/s$ . North position $= \frac{1}{2} (0.02) (100)^{2} = 100 m$ . Ek $2 mg$ -scale bias ne hume do minute se kam mein 100 m off kar diya.

Problem 4.2

Gyro bias ko accel channel ke saath combine karo. Static, level, gyro bias $ε = 0.002 rad/s$ East ke baare mein $t = 50 s$ ke liye. Nikalo (a) tilt, (b) leaked horizontal acceleration $g sin δ θ$ $t = 50$ s pe tilt ko instantly uss value ka maante hue, aur (c) roughly bound karo horizontal velocity error yeh maante hue ki acceleration $0$ se uss value tak linearly ramp karta hai.

Recall Solution 4.2

(a) $δ θ = εt = 0.002 \times 50 = 0.1 rad$ . (b) $δ a_{m a x} = 9.81 sin (0.1) \approx 0.9794 m/s^{2}$ . (c) Tilt linearly grow karta hai, isliye leaked accel $\approx g sin (εt) \approx g εt$ small angles ke liye. Integrate karo: $δ v \approx \int_{0}^{50} g εt d t = g ε \frac{t ^{2}}{2} = 9.81 \times 0.002 \times \frac{2500}{2} = 24.525 m/s .$ Velocity error yahan $t^{2}$ jaisi grow karta hai kyunki bias ek tilt feed karta hai jo khud $t$ ke saath grow karta hai — ek compounding, faster-than-linear leak.

L5 — Mastery

Problem 5.1

Ek pilot propose karta hai: "30-second drone hop under 1 km ke liye, Earth-rotation aur transport terms skip karo." Dono terms ko quantify karo jo tum drop kar rahe ho. (i) Coriolis/Earth-rotation part $2 Ω_{e} v$ for $v = 30 m/s$ . (ii) Transport-rate part $ω_{e n}^{n}$ : same $30 m/s$ East aur $R_{N} = 6.389 \times 1 0^{6} m$ ke liye iska dominant component $v_{E} / (R_{N} + h)$ use karo, aur ise acceleration $ω_{e n} v$ mein convert karo. Dono ko typical accel-bias-driven acceleration $0.02 m/s^{2}$ ke against compare karo. Kya inhe drop karna justified hai?

Recall Solution 5.1

(i) Coriolis magnitude bound: $2 Ω_{e} v = 2 \times 7.292 \times 1 0^{- 5} \times 30 = 4.375 \times 1 0^{- 3} m/s^{2}$ . (ii) Transport rate: $ω_{e n} \approx \frac{v}{R _{N} + h} = \frac{30}{6.389 \times 1 0 ^{6}} = 4.696 \times 1 0^{- 6} rad/s$ . Frame-correction acceleration ke roop mein $ω_{e n} v = 4.696 \times 1 0^{- 6} \times 30 = 1.409 \times 1 0^{- 4} m/s^{2}$ . Bias-driven $0.02 m/s^{2}$ se compare karo: Coriolis almost $4.6 \times$ chhota hai, aur transport term $100 \times$ se bhi zyada chhota hai. $30 s$ mein dono mein se bada (Coriolis) se zyada se zyada $\frac{1}{2} (4.375 \times 1 0^{- 3}) (30)^{2} \approx 1.97 m$ ka position error milta hai; transport term ka kuch centimetres ka hai. Verdict: Earth-rotation aur transport dono drop karna is short, slow, low-accuracy hop ke liye justified hai (flat-Earth 80/20 core). Yeh ek long-range, high-grade system ke liye justified nahi hoga jahan $Ω_{e}$ dominate karta hai. Dekho Coriolis and Centrifugal Effects aur Strapdown vs Gimbaled INS.

Problem 5.2

Design check. Tum chahte ho ki pure inertial coasting (no GNSS) ke $60 s$ ke baad horizontal position error $10 m$ se kam rahe, ek single accel bias $b$ se dominated ho (level, no tilt maano). Maximum allowable $b$ kya hai?

Recall Solution 5.2

Constant bias se position error: $δ r = \frac{1}{2} b t^{2}$ . $δ r = 10$ , $t = 60$ set karo: $b = \frac{2 δ r}{t ^{2}} = \frac{2 \times 10}{3600} = 5.556 \times 1 0^{- 3} m/s^{2} .$ "Milli-g" mein ( $1 mg = 9.81 \times 1 0^{- 3} m/s^{2}$ ): $b \leq 0.566 mg$ . Toh tumhe half a milli-g se behtar accelerometer chahiye — ek real navigation-grade requirement, aur exactly yehi reason hai ki INS-GNSS fusion exist karta hai is drift ko bound karne ke liye.

Recall Poori ladder ki ek-line summary

Ek IMU rate aur specific force feel karta hai; software rotate karta hai, gravity add karta hai, aur integrate karta hai — aur har integration ek chhote se bias ko ek fast-growing error mein badal deta hai, yehi reason hai ki real systems GNSS ke saath fuse karte hain.