Inertial navigation — accelerometer measures non-gravitational specific force: Worked examples

3.5.13 · D3 · Physics › Guidance, Navigation & Control (GNC) › Inertial navigation — accelerometer measures non-gravitation

Yeh page ek drill hai har us case ki jo accelerometer equation $f = a - g$ tumhare saamne rakh sakti hai. Agar tumne abhi tak nahi dekha ki yeh equation kahan se aati hai, toh pehle use parent topic mein build karo.

Kisi bhi number se pehle, aao hum picture aur sign rule par agree kar lein — neeche koi bhi symbol aisa nahi hai jo humne pin down nahi kiya ho.

Recall Equation, words mein

Accelerometer specific force $f = a - g$ read karta hai. Yahan $a$ box ki actual acceleration hai (jis tarah uski velocity sach mein change hoti hai), aur $- g$ matlab hai "gravity wapis subtract kar di". Component by component: $f_{x} = a_{x} - g_{x}$ aur $f_{z} = a_{z} - g_{z}$ . Hamare convention se $g_{x} = 0$ aur $g_{z} = - g$ hai, isliye $f_{x} = a_{x}$ aur $f_{z} = a_{z} + g$ .

Figure dekho: true acceleration $a$ (blue) aur minus-gravity $- g$ (green, hamesha upar point karta hai length $g$ ke saath) tip-to-tail add hote hain; unka sum orange reading $f$ hai. Neeche ka har example bas yahi triangle hai alag-alag blue arrows ke saath.

The scenario matrix

Har row ek case class hai. Jo examples follow karte hain unhe us cell se label kiya gaya hai jo woh cover karte hain.

#	Case class	Kya special hai	Example
C1	$a_{z} > 0$ (upar accelerate)	reading $g$ se zyada	Ex 1
C2	$a_{z} < 0$ (neeche accelerate, phir bhi move kar raha)	reading $g$ se kam, positive rehti hai	Ex 2
C3	$a_{z} = - g$ bilkul (free fall)	zero reading — degenerate	Ex 3
C4	$a_{z} < - g$ (gravity se zyada neeche throw kiya)	reading negative ho jaati hai	Ex 4
C5	Pure horizontal $a_{x} > 0$ , $a_{z} = 0$ (forward)	do non-zero components, ek taraf tilt	Ex 5
C6	Combined $a_{x} \neq = 0, a_{z} \neq = 0$ (banked/climbing turn)	full vector, magnitude & tilt	Ex 6
C7	Limiting values ( $R \to \infty$ , aur $a_{x} \to \infty$ )	reading mein kya dominate karta hai	Ex 7
C8	Real-world word problem (rocket liftoff)	words → vector mein translate karo	Ex 8
C9	Exam twist (reading di gayi hai, motion dhundho)	equation ko invert karo	Ex 9
C10	Pure horizontal $a_{x} < 0$ , $a_{z} = 0$ (braking)	doosri taraf tilt	Ex 10

Hum poore mein $g = 9.81 m/s^{2}$ use karte hain, aur do decimals tak round karte hain.

Examples

Worked example Ex 2 — Elevator upar jaate waqt decelerate kar raha hai (Cell C2)

Wohi lift, ab top ke paas aate waqt slow ho rahi hai: $a_{z} = - 2 m/s^{2}$ (velocity abhi bhi upar, lekin kam ho rahi hai). Reading?

Forecast: $g$ se upar ya neeche? Positive ya negative?

True acceleration: $a = (0, - 2)$ . Kyun? "Upar jaate waqt slow hona" matlab velocity change neeche point karta hai, isliye $a_{z} = - 2$ chahe abhi kisi bhi direction mein move kar raha ho.
Apply karo: $f_{z} = a_{z} + g = - 2 + 9.81 = 7.81 m/s^{2}$ . Kyun? Spring thoda relax kar sakti hai kyunki gravity ki pull ka kuch hissa mass ko neeche accelerate karne ke liye "allow" kiya gaya hai.
Result: $f_{z} = + 7.81 m/s^{2}$ , abhi bhi upar ki taraf.

Verify: yeh $g$ se neeche gaya lekin positive raha — tum lighter feel karte ho, weightless nahi, kyunki $∣ a_{z} ∣ = 2 < g = 9.81$ hai. Sanity: reading $= g$ minus $2$ ✓.

Worked example Ex 4 — Gravity se

zyada neeche push kiya (Cell C4) Ek rocket sled apna thrust seedha neeche push karne ke liye aim karta hai $a_{z} = - 15 m/s^{2}$ par (free fall se bhi tez). Reading?

Forecast: kya reading negative ho sakti hai? Physically iska kya matlab hoga?

True acceleration: $a = (0, - 15)$ . Kyun? Diya gaya: vehicle neeche $15$ par accelerate karta hai, $g$ se zyada.
Apply karo: $f_{z} = a_{z} + g = - 15 + 9.81 = - 5.19 m/s^{2}$ . Kyun? Gravity akeli sirf $9.81$ ka downward pull supply kar sakti hai; extra $5.19$ spring se aana chahiye jo mass ko neeche kheeche — ek downward contact force.
Result: $f_{z} = - 5.19 m/s^{2}$ (yaani $5.19$ neeche ki taraf).

Verify: sign flip hua kyunki $∣ a_{z} ∣ = 15 > g$ hai. Physically mass case ke dwara neeche drag kiya ja raha hai, isliye reading neeche point karti hai. Boundary check: $a_{z} = - 9.81$ par yeh exactly $0$ hota (Ex 3), aur aur negative hone par reading negative ho jaati hai ✓.

Worked example Ex 5 — Pure horizontal acceleration, forward (Cell C5)

Ek car flat road par $a_{x} = + 4 m/s^{2}$ se aage accelerate karti hai; koi vertical motion nahi, $a_{z} = 0$ . Ek 2-axis accelerometer kya read karta hai, aur woh kis direction mein tilt karta hai?

Forecast: kya reading purely horizontal hogi, purely vertical, ya slanted?

True acceleration: $a = (+ 4, 0)$ . Kyun? Forward speed up, level ground.
Components: $f_{x} = a_{x} = 4$ aur $f_{z} = a_{z} + g = 0 + 9.81 = 9.81$ . Kyun? Spring abhi bhi weight support karta hai ( $f_{z} = g$ ) aur mass ko aage dhakhelna bhi hai ( $f_{x} = 4$ ).
Magnitude: $∣ f ∣ = 4^{2} + 9.8 1^{2} = 16 + 96.24 = 112.24 = 10.59 m/s^{2}$ . Square root kyun? $f$ ek vector hai; iska length uske perpendicular parts ka Pythagorean combination hai.
Tilt angle vertical se: $θ = arctan (f_{x} / f_{z}) = arctan (4/9.81) = 22.1 9^{\circ}$ $+ x$ ki taraf. $arctan$ kyun? Hum poochh rahe hain "kis angle ka yeh horizontal-to-vertical ratio hai?" — $arctan$ tangent ko undo karke woh angle deta hai.

Verify: $∣ f ∣ > g$ jaise expected tha (extra horizontal shove), aur reading acceleration ke relative backward lean karti hai — isliye jab tum accelerate karte ho toh hanging pendulum peeche swing karta hai. Units sab $m/s^{2}$ ✓.

Worked example Ex 6 — Aircraft ek banked, climbing situation mein (Cell C6)

Ek plane simultaneously aage-horizontally $a_{x} = + 6 m/s^{2}$ accelerate karta hai aur vertical acceleration $a_{z} = + 2 m/s^{2}$ se climb karta hai. Reading vector, uski magnitude, aur tilt dhundho.

Forecast: Ex 5 ki car se bada ya chhota?

True acceleration: $a = (6, 2)$ . Kyun? Dono components diye gaye hain.
Components: $f_{x} = 6$ , $f_{z} = 2 + 9.81 = 11.81$ . Kyun? Horizontal part seedha pass hota hai; vertical part mein hamesha present $g$ add hota hai.
Magnitude: $∣ f ∣ = 6^{2} + 11.8 1^{2} = 36 + 139.48 = 175.48 = 13.25 m/s^{2}$ .
Tilt from vertical: $θ = arctan (6/11.81) = 26.9 4^{\circ}$ .

Verify: $f_{x}$ aur $f_{z}$ dono car ki values se zyada hain (extra climb ne $f_{z}$ ko $g$ se upar uthaya), isliye magnitude badi hai — consistent. Yeh total woh load factor hai jo pilot feel karta hai: $∣ f ∣/ g = 13.25/9.81 = 1.35$ g's. ✓

Worked example Ex 7 — Limiting behaviour (Cell C7)

Do limits. (a) Ek coordinated level turn mein, $f_{x} = v^{2} / R$ , $f_{z} = g$ , jahan $R$ upar define ki gayi turn radius hai. Reading ka kya hoga jab $R \to \infty$ (seedha aur seedha)? (b) Jab forward acceleration $a_{x} \to \infty$ ho, toh $∣ f ∣$ mein kya dominate karta hai?

Forecast: (a) kya reading $g$ ya $0$ approach karta hai? (b) kya vertical $g$ abhi bhi matter karta hai?

(a) Set karo $f_{x} = v^{2} / R$ . Kyun? Circular path ki centripetal acceleration $v^{2} / R$ hoti hai, horizontal.
Lo $R \to \infty$ : $f_{x} = v^{2} / R \to 0$ , isliye $∣ f ∣ = (v^{2} / R)^{2} + g^{2} \to 0 + g^{2} = g$ . Kyun? Infinite radius ek straight line hai — koi centripetal demand nahi — isliye sirf weight support bachta hai.
(b) Bade $a_{x}$ ke liye: $∣ f ∣ = a_{x}^{2} + g^{2}$ . Jab $a_{x} \to \infty$ , constant $g^{2}$ negligible ho jaata hai, isliye $∣ f ∣ \approx ∣ a_{x} ∣$ aur tilt $arctan (a_{x} / g) \to 9 0^{\circ}$ (reading lagbhag flat/horizontal ho jaati hai).

Verify: limit (a) correctly straight-and-level answer $g$ recover karta hai (Ex 5 se $a_{x} = 0$ ke saath match karta hai). Limit (b): agar tum insanely hard accelerate karo, toh gravity ek rounding error hai aur $f$ tumhari acceleration ki taraf point karta hai ✓.

Worked example Ex 8 — Real-world word problem: rocket liftoff (Cell C8)

Ek rocket vertically lift off karta hai. Burnout par uske engines true upward acceleration $a_{z} = + 40 m/s^{2}$ dete hain. Iska onboard accelerometer (z-up) Strapdown Inertial Navigation System ke dwara use hota hai. (i) Woh kya read karta hai? (ii) Navigation computer ko true $a$ recover karna hai — dikhaao ki yeh kaam karta hai.

Forecast: kya computer $40$ wapis paane ke liye $g$ add karega ya subtract karega?

Reading: $f_{z} = a_{z} + g = 40 + 9.81 = 49.81 m/s^{2}$ . Kyun? Spring weight support karta hai ( $9.81$ ) aur $40$ upward push provide karta hai.
g's mein: $49.81/9.81 = 5.08$ g's crew ko feel hote hain. $g$ se divide kyun? Reading ko normal gravity ke multiple ke roop mein express karne ke liye.
Computer recover karta hai $a_{z} = f_{z} + g_{z} = 49.81 + (- 9.81) = 40 m/s^{2}$ . $g_{z} = - 9.81$ kyun add karte hain? Equation $a = f + g$ kehti hai modeled gravity vector wapis add karo; yahan woh model $z$ mein $- 9.81$ contribute karta hai.

Verify: step 3 exactly woh $40$ return karta hai jisse humne start kiya tha — loop self-consistent hai. $+ g$ term bhool jaana $49.81$ chhod dega aur, Dead Reckoning and Error Drift ke anusaar, trajectory bigad jaegi. ✓

Worked example Ex 9 — Exam twist: reading di gayi, motion dhundho (Cell C9)

Ek accelerometer (z-up, single axis) $f_{z} = 7.00 m/s^{2}$ output karta hai. Yeh jaana hua hai ki vehicle purely vertically move karta hai. Iska true acceleration kya hai, aur motion describe karo.

Forecast: kya vehicle upar ya neeche accelerate kar raha hai?

Equation invert karo: $a_{z} = f_{z} - g = 7.00 - 9.81 = - 2.81 m/s^{2}$ . $g$ subtract kyun? $f_{z} = a_{z} + g$ se, $a_{z}$ ke liye solve karne par $a_{z} = f_{z} - g$ milta hai.
Sign interpret karo: $a_{z} < 0$ ⇒ true acceleration neeche $2.81 m/s^{2}$ par point karta hai. Kyun? Negative $z$ -component matlab velocity neeche ki taraf change ho rahi hai (ya toh tez gir raha hai, ya upar jaate waqt slow ho raha hai).
Regime par sanity: $∣ a_{z} ∣ = 2.81 < g$ , isliye yeh Cell C2 territory hai (below $g$ , positive reading) — occupants lighter feel karte hain lekin weightless nahi.

Verify: plug back karo $f_{z} = a_{z} + g = - 2.81 + 9.81 = 7.00$ ✓ — given reading se exactly match karta hai. $0$ aur $g$ ke beech ki reading ka matlab hamesha $g$ se kam magnitude ki downward true acceleration hoti hai. ✓

Worked example Ex 10 — Pure horizontal acceleration, braking (Cell C10)

Ex 5 wali wohi car ab brake karti hai, $a_{x} = - 4 m/s^{2}$ par decelerate karti hai; abhi bhi level hai, $a_{z} = 0$ . Reading, uski magnitude, aur woh kis taraf tilt karti hai dhundho — Ex 5 se compare karo.

Forecast: Ex 5 jitni hi magnitude? Same tilt direction?

True acceleration: $a = (- 4, 0)$ . Kyun? Aage jaate waqt brake karna matlab velocity change peeche point karta hai, isliye $a_{x} = - 4$ .
Components: $f_{x} = a_{x} = - 4$ aur $f_{z} = a_{z} + g = 0 + 9.81 = 9.81$ . Kyun? Horizontal part seedha pass hota hai (ab negative); vertical part abhi bhi weight support karta hai.
Magnitude: $∣ f ∣ = (- 4)^{2} + 9.8 1^{2} = 16 + 96.24 = 112.24 = 10.59 m/s^{2}$ . Kyun? Squaring sign mitaata hai, isliye length Ex 5 ke barabar hai.
Tilt from vertical: $θ = arctan (f_{x} / f_{z}) = arctan (- 4/9.81) = - 22.1 9^{\circ}$ — yaani Ex 5 ki opposite side ki taraf $22.1 9^{\circ}$ ( $- x$ ki taraf).

Verify: Ex 5 se identical magnitude lekin mirror-image tilt — reading ab forward lean karti hai, isliye brake karte waqt tum seatbelt ke against throw hote ho. $a_{x}$ ka sign horizontal component aur isliye tilt direction ka sign flip karta hai, horizontal quadrant complete karta hai. ✓

Recall Quick self-test

Ek sensor exactly $0$ read karta hai. Vehicle kya kar raha hai? ::: Free fall mein — true acceleration $g$ ke barabar hai (Cell C3). Ek sensor $- 5.19$ (neeche) read karta hai. $∣ a_{z} ∣$ ke baare mein kya sach hona chahiye? ::: $∣ a_{z} ∣ > g$ : case mass ko gravity se zyada neeche kheench raha hai (Cell C4). Jab turn radius $R \to \infty$ ho, toh level turn mein $∣ f ∣$ kya approach karta hai? ::: $g$ — straight path mein koi centripetal demand nahi (Cell C7). Same rate par braking aur accelerating dono same reading magnitude dete hain — kya differ karta hai? ::: Tilt direction flip hoti hai (forward vs backward lean); magnitude identical hai (Ex 5 vs Ex 10).

Connections

Newton's Second Law — yahan har inversion $m a = F_{c} + m g$ se start hoti hai.
Equivalence Principle — Ex 3 mein zero reading.
Strapdown Inertial Navigation System — jahan Ex 8 ka $a = f + g$ integrate hota hai.
Gravity Model (WGS-84 / J2) — woh $g$ supply karta hai jo computer wapis add karta hai.
Dead Reckoning and Error Drift — kya galat hota hai agar tum woh step skip karo.
Gyroscope and Attitude Determination — orientation provide karta hai taaki $g$ sahi axes mein resolve ho sake.