IMU — integrated accelerometer + gyroscope: Worked examples

3.5.15 · D3 · Physics › Guidance, Navigation & Control (GNC) › IMU — integrated accelerometer + gyroscope

Yahan sab kuch parent note ke teen facts par tikaa hai. Chalo unhe simple words mein dobara state karte hain taaki kuch bhi bina samjhe use na ho:

Pehle, position error par ek plain-words note. Jab hum dead-reckon karte hain, hum estimate karte hain ki box kahan hai — uski motion ko add karte (integrate karte) jaate hain. Yeh estimate sach se door hoti jaati hai; us gap ko hum position error kehte hain aur likhte hain $Δ p$ (yahan $p$ = position, $Δ$ = "usme error"). Yeh metres mein measure hota hai. Yeh meaning yaad rakho — yeh drift bullet mein aur dobara Examples 7 aur 8 mein aata hai.

Yahan $a$ = true acceleration (box kitna speed up hota hai), $g$ = gravity vector (neeche point karta hai, toh $g = (0, 0, - 9.81)$ agar $z$ upar hai), aur $f$ = "specific force" = woh push per kilogram jo sensor ki internal spring feel karti hai. World axes fixed rakho: $x$ = East, $y$ = North, $z$ = Up.

Scenario matrix

Har IMU problem inhiin cells mein se ek (ya mix) hoti hai. Neeche ke examples cell tag ke saath hain.

Cell	Kya khaas hai	Example
A. Rest, level	$a = 0$ , $ω = 0$ , koi tilt nahi	Ex 1
B. Rest, tilted (sign of tilt)	still, lekin rotated — gravity axes mein split hoti hai	Ex 2
C. Degenerate: free fall	spring force zero hai	Ex 3
D. Linear acceleration, level	true motion, $g$ subtract karna zaroori	Ex 4
E. Pure rotation (gyro only, on axis)	orientation badlti hai, translation nahi	Ex 5
F. Coupled: accelerate while tilted	$g$ hatane se pehle rotate karna zaroori	Ex 6
G. Limiting behaviour: drift over time	error growth $t^{2}$ vs $t^{3}$	Ex 7
H. Real-world word problem	drone ek tunnel mein, GPS-denied	Ex 8
I. Exam twist: sign trap	negative reading, "up" kaunsi taraf hai?	Ex 9
J. Off-axis rotation (centripetal)	sensor spin axis par nahi hai	Ex 10

Example 1 — Cell A: rest aur level

Worked example Bilkul still, perfectly flat

Ek IMU table par still rakha hai, axes world ke saath aligned hain. Gyro aur accelerometer kya padhenge?

Forecast: Padhne se pehle andaaza lagao. Kya accelerometer reading zero, $+ 9.81$ , ya $- 9.81$ hogi? Kaunsi axis mein?

Gyro: box nahi ghoom raha, toh $ω_{b} = 0$ . Yeh step kyun? Angular rate literally "aap kitni tezi se spin kar rahe ho" hai; still box zero par spin karta hai.
Accelerometer: true acceleration $a = 0$ , toh $f = a - g = 0 - (0, 0, - 9.81) = (0, 0, + 9.81)$ . Yeh step kyun? $f = a - g$ ; ek downward vector subtract karne par upar ki taraf wala add hota hai. Physically, table ki normal force hi sirf spring push hai, aur woh upar point karti hai.

Answer: $ω_{b} = (0, 0, 0)$ rad/s, $f_{b} = (0, 0, + 9.81)$ m/s².

Verify: magnitude $∣ f_{b} ∣ = 9.81$ = ek $g$ . Units: m/s². Reading upar point karti hai — yeh free "which way is up" calibration hai jo parent note mein mentioned hai. ✓

Example 2 — Cell B: rest lekin tilted (tilt ka sign matter karta hai)

Pehle neeche ka figure padho. Lavender box tilted IMU hai. Dark grey arrow woh true up-reaction $f$ hai jo gravity table ko supply karne par force karti hai — yeh hamesha world mein seedha upar point karta hai. Coral aur mint arrows sensor ke apne tilted body $x$ aur $z$ axes hain. Notice karo kaise ek grey arrow do tilted axes par "shadows" (dashed lines) daalta hai: coral shadow woh $g sin θ$ hai jo body- $x$ mein leak hoti hai, mint shadow woh $g cos θ$ hai jo body- $z$ ke saath rehti hai. Butter arc $3 0^{\circ}$ tilt ko mark karta hai. Yeh picture steps 1–2 mein derivation hi hai.

Figure — IMU — integrated accelerometer + gyroscope

World frame mein sensor abhi bhi sirf gravity ka up-reaction feel karta hai: $f_{w} = (0, 0, + 9.81)$ . Kyun? Yeh still hai; world frame mein physics Ex 1 jaisi hi hai.

Us world vector ko body ke tilted axes mein $R^{⊤}$ use karke express karo (world → body, definition box se). $y$ ke baare mein $3 0^{\circ}$ ke liye: $f_{b} = R^{⊤} 00 9.81 = g sin θ 0 g cos θ = (4.905, 0, 8.496) .$ Yeh step kyun? Figure dekho: jaise box aage jhukta hai, "up" push ka kuch hissa body ke $x$ -axis ke saath aa jaata hai (coral shadow). $sin$ measure karta hai kitna sideways leak hota hai; $cos$ kitna vertical rehta hai.

Sign check: doosri taraf jhukao, $θ = - 3 0^{\circ}$ , aur $sin$ sign flip karta hai → $f_{b} = (- 4.905, 0, 8.496)$ . $x$ -component ka sign batata hai aap kaunsi taraf jhuke. Yeh kyun matter karta hai? IMU tilt angle bilkul isi tarah derive karta hai: $θ = arctan (f_{x} / f_{z})$ .

Verify: $∣ f_{b} ∣ = 4.90 5^{2} + 8.49 6^{2} = 9.81$ ✓ (rotation length preserve karta hai). Aur $arctan (4.905/8.496) = 3 0^{\circ}$ ✓.

Example 3 — Cell C: free fall (degenerate, spring force zero)

Worked example Giraya hua IMU

IMU ko release karke freely girne diya jaata hai (koi air drag nahi). Fall ke dauraan accelerometer kya padhega?

Forecast: Yeh $g$ par neeche accelerate kar raha hai. Toh surely $9.81$ padhega... na?

Free fall mein proof mass par sirf gravity ka force hai, toh woh case ke saath accelerate karta hai: $a = g = (0, 0, - 9.81)$ . Yeh step kyun? Kuch bhi mass ko spring ke against push nahi karta; case aur mass saath girte hain.
Specific force: $f = a - g = (0, 0, - 9.81) - (0, 0, - 9.81) = (0, 0, 0)$ . Yeh step kyun? Spring unstretched hai — mass per zero force.

Answer: $f_{b} = (0, 0, 0)$ . Zero.

Verify: Yeh famous "weightless" reading hai — orbit mein ek accelerometer (permanent free fall) ~0 padhta hai. Ex 1 (rest → $+ g$ ) se contrast karo. Yeh do extremes ( $0$ aur $+ g$ ) har possible resting/falling case ko bracket karte hain. ✓

Example 4 — Cell D: linear acceleration, level

Worked example Car ki full acceleration

Ek level IMU car mein East ki taraf $3$ m/s² se accelerate kar raha hai. Koi tilting nahi hai. Body frame mein accelerometer kya padhega (abhi bhi world ke saath aligned)?

Forecast: Do cheezein ab spring ko push kar rahi hain — seat aapko aage dhakeel rahi hai, aur ground aapko thaam raha hai. Dono components predict karo.

True acceleration $a = (3, 0, 0)$ . Kyun? Eastward $3$ m/s² par speed up ho raha hai.
Specific force: $f = a - g = (3, 0, 0) - (0, 0, - 9.81) = (3, 0, + 9.81)$ . Yeh step kyun? Horizontal $3$ real acceleration se aata hai; vertical $+ 9.81$ abhi bhi gravity ka reaction hai.

Answer: $f_{b} = (3, 0, 9.81)$ .

Verify: True acceleration recover karne ke liye hum wapas chalte hain: $a = f + g = (3, 0, 9.81) + (0, 0, - 9.81) = (3, 0, 0)$ ✓. Yeh woh round trip hai jis par Dead Reckoning depend karta hai. Reading ki magnitude $= 3^{2} + 9.8 1^{2} = 9 + 96.24 = 10.26$ m/s², $g$ se thodi zyada — car "heavier" feel hoti hai kyunki aapko push kiya ja raha hai. ✓

Example 5 — Cell E: pure rotation (sirf gyro, axis par)

Pehle neeche ka figure padho. Lavender dot IMU hai, vertical spin axis (dashed slate line) par exactly baitha hai. Mint circular arrow spin $ω_{z}$ dikhata hai. Kyunki sensor axis par hai, woh kabhi circle mein travel nahi karta — bas jagah par pivots karta hai, toh koi circular (centripetal) acceleration nahi hai. Uski spring sirf gravity ka up-reaction feel karti hai (grey arrow, Ex 1 jaisa unchanged). Ex 10 se contrast karo, jahan dot ko axis se door move kiya gaya hai aur ek naya inward pull appear hota hai.

Gyro padhta hai $ω_{b} = (0, 0, 0.5)$ rad/s, constant. Kyun? $z$ ke baare mein constant spin.

Angle ghuma $= \int_{0}^{4} ω_{z} d t = 0.5 \times 4 = 2$ rad $\approx 114. 6^{\circ}$ . Yeh step kyun? Constant rate ko integrate karna sirf rate × time hai — strapdown attitude update ka sabse simple case.

Accelerometer abhi bhi $(0, 0, 9.81)$ padhta hai — koi linear acceleration nahi. Kyun? Sensor exactly vertical spin axis par hai, toh uski koi circular motion nahi hai aur gravity doosre axes mein tipped nahi hui. (Ise axis se door move karo aur yeh badal jaata hai — woh Ex 10 hai.)

Answer: $2$ rad ghuma; accel unchanged.

Verify: $0.5 \times 4 = 2$ rad, aur $2 \times 180/ π = 114.5 9^{\circ}$ ✓. Yeh sirf Reference Frames — Body vs World use karta hai: orientation badla, position nahi. ✓

Example 6 — Cell F: tilt ke saath accelerate (coupled case)

Pehle neeche ka figure padho. Fixed world axes (slate) mein true acceleration $a_{w}$ (coral, East ki taraf point karta) aur gravity ka up-reaction $- g$ (mint, upar point karta) tip-to-tail add hoke total specific force $f_{w}$ (grey diagonal) dete hain. Lavender dashed axes tilted body frame hain: wahi grey vector, tilted axes par read karne par, mixed-up numbers $(- 3.173, 0, 9.496)$ mein split hota hai — yahi device output karta hai. Picture dikhati hai kyun aapko grey arrow ko world axes par rotate karna zaroori hai pehle taaki mint gravity part saaf cancel ho sake.

$f_{b}$ compute karo. $y$ ke baare mein $+ 3 0^{\circ}$ ke liye $R^{⊤}$ (world → body) se: $f_{b} = (2 cos 30 - 9.81 sin 30, 0, 2 sin 30 + 9.81 cos 30) = (- 3.173, 0, 9.496) .$ Yeh step kyun? Yahi physical device output karta hai — mix ho gaya kyunki body tilted hai.

Correct recovery: pehle $R$ se world mein rotate karo (body → world), phir gravity subtract karo. $f_{w} = R f_{b} = (2, 0, 9.81), a_{w} = f_{w} + g = (2, 0, 0) .$ Yeh step kyun? Sirf rotate karne ke baad hi axes world-aligned hain, isliye gravity purely $z$ mein hai aur saaf cancel hoti hai.

Naive shortcut (galat): $f_{b}$ ko world ki tarah padho aur $(0, 0, 9.81)$ subtract karo: $(- 3.173, 0, 9.496) - (0, 0, 9.81) = (- 3.173, 0, - 0.314) .$ Yeh galat kyun hai? Aapne kabhi rotate nahi kiya, toh gravity abhi bhi $x$ aur $z$ mein smeared thi. Aapko ab ek phantom $- 3.173$ m/s² eastward acceleration "dikhti" hai.

Answer: sahi $a_{w} = (2, 0, 0)$ ; naive method garbage $(- 3.173, 0, - 0.314)$ deta hai.

Verify: $R R^{⊤} (2, 0, 9.81) = (2, 0, 9.81)$ toh step 2 exactly round-trip karta hai ✓. Naive $x$ -error hai $2 cos 30 - 9.81 sin 30 = 1.732 - 4.905 = - 3.173$ ✓. Lesson: attitude → rotate → $g$ subtract → integrate, hamesha.

Example 7 — Cell G: limiting behaviour, drift over time

Pehle neeche ka figure padho. Horizontal axis elapsed time hai; vertical axis yeh hai ki aapka position estimate kitne metres drift kar gaya. Coral curve accelerometer error hai, ek gentle parabola ( $t^{2}$ ). Lavender curve gyro-tilt error hai, ek steeper cubic ( $t^{3}$ ). Mint dashed line woh crossover moment mark karti hai jahan cubic parabola ko overtake karta hai — pehle accel error dominate karta hai, baad mein gyro error runaway ho jaata hai. Dots woh chaar numbers hain jo hum steps 1–2 mein compute karte hain.

Accel error $Δ p_{a} = \frac{1}{2} ε_{a} t^{2}$ ki tarah badhta hai.

$t = 10$ : $\frac{1}{2} (0.02) (100) = 1.0$ m.

$t = 100$ : $\frac{1}{2} (0.02) (10000) = 100$ m. Yeh step kyun? Ek constant acceleration error do baar integrate hone par $t^{2}$ position error deta hai.

Gyro-tilt error $Δ p_{g} = \frac{1}{6} g b_{g} t^{3}$ ki tarah badhta hai.

$t = 10$ : $\frac{1}{6} (9.81) (1.745 \times 1 0^{- 4}) (1000) = 0.2853$ m.

$t = 100$ : $\frac{1}{6} (9.81) (1.745 \times 1 0^{- 4}) (1 0^{6}) = 285.3$ m. Yeh step kyun? Bias frame ko linearly tilt karta hai, acceleration ka ek ramp leak karta hai; ramp ko do baar integrate karne par $t^{3}$ milta hai.

Crossover: $t = 10$ s par accel error (1 m) gyro (0.29 m) se aage hai; $t = 100$ s tak gyro cubic (285 m) ne accel quadratic (100 m) ko crush kar diya. Kyun? $t^{3}$ eventually $t^{2}$ ko overtake kar leta hai chahe $b_{g}$ kitna bhi chhota ho — isliye GPS-Denied Navigation ko doosre sensors ke saath fuse karna hota hai.

Verify: ratios — accel error $t = 10$ se $t = 100$ tak ×100 scale karta hai (yani $1 0^{2}$ ) ✓; gyro error ×1000 (yani $1 0^{3}$ ) scale karta hai ✓. Cure ke liye Sensor Fusion & Kalman Filter dekho.

Example 8 — Cell H: real-world word problem

Worked example Drone ek tunnel mein (GPS-denied)

Ek drone constant $12$ m/s East par ek tunnel mein fly karta hai jahan GPS drop out ho jaata hai. Uske IMU ka effective accel error $0.05$ m/s² hai. Tunnel $600$ m lamba hai. Nikalne par position estimate kitna off hai, aur woh andar kitni der tha?

Forecast: Constant velocity matlab true acceleration zero hai — toh error aata kahan se hai?

Tunnel mein time: $t = 600/12 = 50$ s. Kyun? Distance / speed, constant velocity.
True acceleration $0$ hai, lekin sensor mein ek residual error $ε_{a} = 0.05$ m/s² hai jo dead reckoning jaanta nahi, toh woh us error ko real motion ki tarah integrate karta hai. Kyun? IMU ke paas koi bahar ka reference nahi (GPS gaya), toh woh har number par trust karta hai — error sahit.
Position error $Δ p = \frac{1}{2} ε_{a} t^{2} = \frac{1}{2} (0.05) (50)^{2} = \frac{1}{2} (0.05) (2500) = 62.5$ m. Yeh step kyun? Ex 7 jaisa $t^{2}$ law — ek constant acceleration error do baar integrate hone par $t^{2}$ ki tarah badhta hai, "boring" constant-velocity flight ke dauraan bhi accumulate hota rehta hai.

Answer: drone $t = 50$ s ke liye andar tha, aur uska exit position estimate $Δ p = 62.5$ m off hai.

Verify: $600/12 = 50$ ✓, $0.5 \times 0.05 \times 5 0^{2} = 62.5$ ✓. Units: $(m/s^{2}) (s^{2}) = m$ ✓. $600$ m tunnel mein $62.5$ m error ( $\sim 10%$ ) isliye real drones GPS wapas aate hi Dead Reckoning ko vision ke saath fuse karte hain. ✓

Example 9 — Cell I: exam twist, sign trap

Worked example "Up" kaunsi taraf hai?

Test mein bataya gaya ki ek stationary IMU $f_{b} = (0, 0, - 9.81)$ m/s² padhta hai. Ek student concludes "yeh $g$ par neeche accelerate kar raha hai." Asal mein kya ho raha hai?

Forecast: Kya box free fall mein hai, ulta hai, ya student convention ke baare mein galat hai?

Agar yeh truly stationary hota aur $z$ upar point karta, toh $z$ mein $- 9.81$ ki reading impossible hai — rest par $+ 9.81$ milta hai (Ex 1). Kyun? $f = a - g$ ; rest par $a = 0$ toh $f = - g = (0, 0, + 9.81)$ .
Sirf consistent reading yeh hai ki sensor ka body $z$ -axis neeche point kar raha hai (box ulta hai). Tab up-reaction $- z_{body}$ ke saath lie karta hai, $- 9.81$ deta hai. Kyun? Accelerometer hamesha up-reaction report karta hai; agar aapka $z$ -axis inverted hai, toh sign bhi inverted hai.
Sach recover karo: flipped orientation $R$ ke saath (yahan $R = diag (1, - 1, - 1)$ , ek body → world flip), $f_{b}$ ko world mein rotate karo → $(0, 0, + 9.81)$ , phir $a_{w} = f_{w} + g = 0$ . Truly at rest. Kyun? Pehle rotate karne se sign confusion hat jaata hai — Ex 6 jaisa hi lesson.

Answer: box ulta hai lekin stationary hai; woh accelerate nahi kar raha. Student ne body-frame sign ko world truth ki tarah padhne ki galti ki.

Verify: world mein flip karne par, $R f_{b} = diag (1, - 1, - 1) (0, 0, - 9.81) = (0, 0, + 9.81)$ ; phir $a_{w} = (0, 0, 9.81) + (0, 0, - 9.81) = 0$ ✓. Resting reading ka sign sirf batata hai aapka axis kaunsi taraf point kar raha hai, kabhi motion nahi. ✓

Example 10 — Cell J: off-axis rotation (woh case jo Ex 5 ne warn kiya tha)

Worked example Sensor spin axis se door mount kiya gaya

IMU world $z$ -axis ke baare mein $ω = 0.5$ rad/s par spin karta hai, lekin axis se $r = 0.2$ m bahar mount kiya gaya hai. Axis khud translate nahi karta. Accelerometer kaunsa extra force feel karta hai, aur kaunsi direction mein?

Forecast: Ex 5 mein (sensor axis par) accel untouched tha. Axis se door hone par, kya woh ab kuch feel karta hai — aur kya woh inward hai ya outward?

Radius $r$ par rate $ω$ se circle mein jaane wale point mein centripetal acceleration $a_{c} = ω^{2} r$ hoti hai, centre ki taraf (axis ki taraf) point karti hai. Yeh step kyun? Circle par rehne ke liye mass ko inward pull karna zaroori hai; inward pull hi ek tarika hai uski straight-line motion ko modte rehne ka.
Magnitude: $a_{c} = ω^{2} r = (0.5)^{2} (0.2) = 0.05$ m/s², inward direction mein (axis ki taraf). Kyun? Numbers $ω^{2} r$ mein plug karo.
Accelerometer specific force report karta hai $f = a - g$ . True acceleration mein ab yeh inward centripetal part plus up-reaction hai, toh horizontal channel $0.05$ m/s² padhta hai (spring ko mass ko inward push karna hota hai taaki woh circle karta rahe, toh reported specific force ki horizontal plane mein magnitude $0.05$ hai), jabki $z$ abhi bhi $+ 9.81$ padhta hai. Kyun? Spring ko inward pull supply karna hota hai; specific force woh spring push per mass hai.

Answer: ek extra horizontal $0.05$ m/s² centripetal reading appear hoti hai; vertical $9.81$ rehti hai. Total horizontal-plane magnitude $0.05$ m/s².

Verify: $ω^{2} r = 0.25 \times 0.2 = 0.05$ ✓. Isliye ek naive strapdown jo assume karta hai ki sensor rotation axis par baitha hai, spins ko galat padhta hai — ek lever-arm correction zaroori hai, jo rotating frames mein Coriolis Effect terms se closely related hai. ✓

Recall Quick self-test

Ek resting accelerometer body $z$ mein $- 9.81$ padhta hai — kya yeh accelerate kar raha hai? ::: Nahi. Uska $z$ -axis neeche point kar raha hai; yeh stationary hai lekin ulta hai. $t^{3}$ error eventually $t^{2}$ error ko kyun beat karta hai? ::: Kyunki bade $t$ ke liye, $t^{3}$ coefficients chahe kuch bhi hon, $t^{2}$ se tezi se badhta hai. Ex 6 mein, naive method ne phantom acceleration kyun invent ki? ::: Rotation skip kar diya, toh gravity body axes mein smeared rahi aur $x$ mein leak ho gayi. Off the spin axis, kaunsa extra reading appear hota hai aur $ω = 0.5$ , $r = 0.2$ ke liye kitna bada? ::: Centripetal $ω^{2} r = 0.05$ m/s² axis ki taraf. Sabhi coupled cases ke fix ko summarise karne wala ek word kya hai? ::: Rotate (world frame mein) pehle, gravity subtract karo baad mein.