Inertial navigation — accelerometer measures non-gravitational specific force: Question bank

3.5.13 · D5 · Physics › Guidance, Navigation & Control (GNC) › Inertial navigation — accelerometer measures non-gravitation

True or false — justify karo

Ek accelerometer acceleration measure karta hai.

False. Yeh specific force

f = a - g

measure karta hai — contact force per unit mass jo spring supply karti hai. Acceleration tabhi milti hai jab computer

g

wapas add karta hai.

Ek bilkul still accelerometer bench par zero read karta hai.

False. Rest par

a = 0

hai, isliye

f = - g = + g

upar ki taraf hai (

\approx 9.81 m/s^{2}

). Spring actively upar push kar rahi hai proof mass ko gravity ke against hold karne ke liye, aur yahi push reading hai.

Free fall mein accelerometer

g

neeche ki taraf read karta hai.

False. Free fall mein sirf gravity hi force hai, jise spring feel nahi kar sakti, isliye

a = g

aur

f = 0

hota hai. Yeh exactly zero read karta hai — weightlessness.

Do accelerometers, ek Earth par rest par aur ek rocket mein deep space mein

9.81 m/s^{2}

par accelerate karte hue, same reading dete hain.

True. Dono "up" axis par

+ 9.81 m/s^{2}

read karte hain aur inhe alag nahi kiya ja sakta — yeh Equivalence Principle hai physical form mein: gravity aur acceleration identical specific force produce karte hain.

Proof mass ko double karne se reading double ho jaati hai.

False. Reading force per unit mass hai,

f = F_{c} / m

. Zyada bhaari mass ko proportionally zyada spring force chahiye, isliye ratio — aur reading — unchanged rehti hai.

Agar sirf sensor output pata ho, toh usse do baar integrate karke position mil sakti hai.

False. Raw output

f

hai,

a

nahi. Pehle

a = f + g (r)

banana padega, warna missing

g

integrate hokar

\frac{1}{2} g t^{2}

error ban jaata hai. Dekho Dead Reckoning and Error Drift.

Ek accelerometer gravitational field ki presence directly sense kar sakta hai.

False. Gravity proof mass aur case ko equally kheenchti hai, isliye spring kabhi usse fight nahi karti. Gravity sensor ke liye invisible hai; isse bahar se ek gravity model ke zariye supply karna padta hai.

Spot the error

"Gravity proof mass par ek contact force hai, isliye yeh reading mein dikhta hai."

Gravity ek body force hai jo mass aur case ke har atom par equally kaam karti hai — contact force nahi. Sirf contact forces (spring, electrostatic feedback) read hoti hain, isliye gravity exclude hoti hai.

"Rest par spring

- g

read karta hai kyunki gravity neeche point karti hai."

Spring woh force read karta hai jo woh supply karta hai, jo mass ko hold karne ke liye upar hoti hai. Isliye

f = - g = + g

(upar),

- g

nahi. Reading gravity ki direction ke opposite hai.

a = f - g

, isliye true acceleration paane ke liye gravity subtract karo."

Sign error hai.

f = a - g

se hume milta hai

a = f + g

— tumhe modelled gravity add karni hai, subtract nahi.

"Kyunki

g

ek constant

9.81

hai, hum ise ek baar add karke bhool sakte hain."

g = g (r)

position ke saath vary karta hai (altitude, latitude, J2 flattening). Isse loop coupled ban jaata hai:

g

jaanne ke liye position chahiye, aur position update karne ke liye

g

— dekho Gravity Model (WGS-84 / J2).

"Orbit mein ek astronaut ko gravity feel nahi hoti, isliye accelerometer zero read karta hai."

Orbit mein strong gravity hoti hai (wahi path curve karti hai). Accelerometer zero isliye read karta hai kyunki craft free fall mein hai — sab kuch saath accelerate karta hai, isliye koi contact force ki zaroorat nahi. Weight ka absence hai, gravity ka nahi.

"Accelerometer

f

measure karta hai, isliye navigation equation mein gyroscope ki zaroorat nahi."

f

moving body frame mein measure hota hai;

g

add karne ke liye (jo world frame mein jaana jaata hai) device ki orientation jaanni padti hai. Woh gyroscopes aur strapdown equations se aati hai.

Why questions

Accelerometer ka naam logon ko mislead kyun karta hai?

Kyunki yeh

m/s^{2}

ki units output karta hai aur acceleration ke saath share karta hai, lekin quantity specific force

f = a - g

hai — true acceleration ke equal sirf tab hoti hai jab

g = 0

ho (deep space).

Ek single accelerometer "Earth par rest" aur "space mein upar accelerating" mein farq kyun nahi kar sakta?

Dono identical specific force

+ g

dete hain. Koi bhi local measurement uniform gravity ko uniform acceleration se alag nahi kar sakti — yahi Equivalence Principle hai.

Derivation inertial frame mein kyun shuru karni padti hai?

Newton's Second Law (

m a = F_{c} + m g

) sirf inertial frame mein bina fictitious forces ke hold hota hai; rotating/accelerating frame mein clean

f = a - g

paane ke liye extra pseudo-force terms chahiye honge.

Gravity bhoolne se position error itni tezi se kyun badhti hai?

Acceleration mein missing constant

g

ek baar integrate hokar velocity mein

g t

ban jaata hai aur dobara integrate hokar position mein

\frac{1}{2} g t^{2}

— ek quadratic runaway jo kuch seconds mein metres tak pahunch jaata hai.

"Gravity wapas add karna" step sensor ki jagah computer kyun karta hai?

Sensor physically gravity feel nahi kar sakta (mass aur case par equal pull), isliye

g

hardware ke paas kabhi hoti hi nahi. Sirf ek stored model aur jaani hui position hi ise supply kar sakti hai.

Ek coordinated banked turn mein pilot "heavier" kyun feel karta hai?

Seat ki contact force ko weight support karna bhi hai aur centripetal acceleration provide karni bhi hai, isliye

∣ f ∣ = g^{2} + (v^{2} / R)^{2} > g

— accelerometer exactly woh badhaa hua magnitude read karta hai.

Edge cases

Projectile ke arc ke bilkul top par accelerometer kya read karta hai (thrown ball, engine off)?

Poori flight mein zero, top bhi include — ball poore time free fall mein hoti hai, isliye

a = g

aur

f = 0

rehta hai chahe velocity momentarily horizontal ho.

Ek horizontal-axis accelerometer kya read karta hai jab device still aur level rakhi ho?

Us axis par zero — gravity ka koi horizontal component nahi hai, aur still mass ko jagah par rakhne ke liye horizontal contact force zero hoti hai. Sirf vertical axis

+ g

read karti hai.

Agar accelerometer deep space mein hai, kisi bhi mass se door, constant velocity par drift kar raha hai?

Exactly zero read karega —

g \approx 0

aur

a = 0

, isliye

f = 0

. Yahan (aur sirf yahan) "zero reading" sach mein "no acceleration" ka matlab hai.

Deep space mein thruster steady fire karte waqt kya read karta hai?

Thrust magnitude per unit mass:

f = a

(kyunki

g = 0

hai). Yeh ek hi regime hai jahan sensor output true acceleration ke barabar hoti hai.

Supporting cable cut hone ke bilkul us instant mein reading ka kya hota hai?

Yeh discontinuously

+ g

(supported) se

0

(free fall) par jump kar jaati hai jis moment contact force khatam hoti hai — sensor spring force ke sudden loss ko track karta hai, gravity mein kisi bhi change ko nahi.

Ek rotating platform par (vertical axis ke baare mein spin), radius

r

par radially-mounted accelerometer kya read karta hai?

Yeh centripetal specific force

ω^{2} r

inward ki taraf read karta hai, kyunki contact force ko proof mass ka path curve karna padta hai — chahe platform ka centre translate nahi kar raha ho.

Connections

Equivalence Principle — kyun rest-on-Earth aur accelerating-in-space indistinguishable hain.
Newton's Second Law — $f = a - g$ ka inertial-frame starting point.
Strapdown Inertial Navigation System — jahan $a = f + g$ integrate hota hai.
Gyroscope and Attitude Determination — $g$ add karne ke liye zaruri orientation supply karta hai.
Gravity Model (WGS-84 / J2) — $g (r)$ ka source jo tum wapas add karte ho.
Dead Reckoning and Error Drift — gravity mishandle hone par $\frac{1}{2} g t^{2}$ runaway.