Question bank — Mechanization equations — integrating IMU to get position, velocity, attitude

False. Yeh $-9.81{\text{m/s}}^2$ read karta hai (NED mein, Down $=+z$, reading support ke opposite hai). Table usse upar push karti hai, aur specific force ${\mathbf{f}}^b$ contact force hai, isliye static reading support ke opposite $+g$ magnitude hai — zero sirf free fall mein hota hai.

False. Yeh $0$ read karta hai. Gravity proof mass ke har part ko equally accelerate karti hai, isliye kuch bhi sensor ko squeeze nahi karta — free fall ke dauran koi specific force detect nahi hoti.

False. Isko step out karo: ek gyro bias $\epsilon$ attitude ko linearly tilt karta hai, $\delta \theta =\epsilon t$. $\delta \theta$ ka tilt gravity ko misrotate karta hai isliye ek horizontal acceleration $a\approx g\delta \theta =g\epsilon t$ level channel mein leak hota hai. Use ek baar integrate karne par velocity $\propto t^2$ aur dobara integrate karne par position $\propto t^3$ milta hai — isliye gyro bias se position error $t^3$ ki tarah badhta hai, accelerometer bias ke $t^2$ se bhi bura.

False. Accelerometer output ${\mathbf{f}}^b$ body axes mein rehta hai. Tumhe $C_b^n$ (attitude) chahiye taaki use nav axes mein rotate kar sako (${\mathbf{f}}^n=C_b^n{\mathbf{f}}^b$) gravity add karne se pehle, isliye attitude pehle aani chahiye.

False. Yeh plumb-bob (apparent) gravity hai = gravitational attraction minus Earth ke spin se centrifugal acceleration. Isliye yeh local vertical (Down) ke along point karta hai, na ki exactly Earth ke centre ki taraf.

True. Short baselines ke liye $2{\mathbf{\omega }}_{ie}^n\times {\mathbf{v}}^n$ aur ${\mathbf{\omega }}_{en}^n\times {\mathbf{v}}^n$ corrections negligible hain; flat-Earth trio (NED mein) ${\dot{\mathbf{r}}}^n={\mathbf{v}}^n,{\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n,{\dot{C}}_b^n=C_b^n[{\mathbf{\omega }}_{ib}^b\times ]$ 80/20 core hai. NED convention us trio mein har sign fix karta hai.

False. Finite 3-D rotations commute nahi karte; orientation curved manifold $SO(3)$ par rehti hai. Yahi wajah hai ki hum matrix/quaternion propagate karte hain (${\dot{C}}_b^n=C_b^n[\mathbf{\omega }\times ]$, upar define kiya gaya skew matrix use karke) angles sum karne ki jagah.

True (yeh ek convention hai). NED mein "Down" axis $v_D$ Earth ki taraf point karta hai, isliye positive downward velocity height decrease karti hai; sign purely axis direction ke choice se hai. (ENU mein yeh $\dot{h}=+v_U$ hota.)

Device specific force ${\mathbf{f}}^b$ measure karta hai, acceleration nahi. Static rehne par raw ${\mathbf{f}}^b$ integrate karne par ek fake $9.81{\text{m/s}}^2$ inject hota hai; tumhe nav axes mein rotate karna hoga aur puri ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$ use karni hogi taaki gravity pehle add ho sake.

Tumhe nav frame ke relative rate chahiye, ${\mathbf{\omega }}_{nb}^b={\mathbf{\omega }}_{ib}^b-C_n^b({\mathbf{\omega }}_{ie}^n+{\mathbf{\omega }}_{en}^n)$, aur phir iska skew matrix $[{\mathbf{\omega }}_{nb}^b\times ]$ banao. ${\mathbf{\omega }}_{ib}^b$ raw use karne par attitude Earth aur moving frame ke saath rotate ho jaata hai.

Yeh ek fictitious (frame) correction hai, real force nahi. Yeh sirf isliye appear hota hai kyunki hum Newton's law ko rotating/curving nav frame mein express karte hain; iske peeche koi physical push nahi hai.

$\cos \varphi$ missing hai: correct hai $\dot{\lambda }=v_E/[(R_N+h)\cos \varphi ]$, jahan $v_E$ East velocity component hai aur $R_N$ transverse radius hai. Longitude circles poles ki taraf shrink hote hain, isliye same eastward speed high latitude par zyada longitude sweep karti hai.

Ye alag radii of curvature use karte hain: latitude rate meridian radius $R_M$ use karta hai, longitude rate transverse radius $R_N$ use karta hai. Ellipsoid north-south vs east-west mein alag curve karta hai (north-south mein zyada sharply), isliye $R_M\ne R_N$ — Radii of Curvature of the Earth Ellipsoid dekho.

Nahi — velocity acceleration error ka ek integration hai, position do integrations ka hai. Position error zyada tezi se accumulate hota hai (accel bias se $\propto t^2$) velocity error se ($\propto t$).

Tum ${\mathbf{g}}^n$ add karte ho: equation hai ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$, isliye $+{\mathbf{g}}^n$ term woh jagah hai jahan gravity re-enter karti hai. Accelerometer pehle se hi true gravitational acceleration miss kar chuka tha (woh use feel nahi kar sakta), isliye gravity software mein daali jaati hai remove karne ki jagah.

Gravity ${\mathbf{g}}^n$ naturally local vertical ke along express hoti hai (ek nav-frame direction). Do vectors add karne ke liye unhe same frame share karna hoga, isliye body-frame force pehle $C_b^n$ se rotate ki jaati hai (${\mathbf{f}}^n=C_b^n{\mathbf{f}}^b$).

Coriolis rotating frame mein observe ki gayi velocity differentiate karne se arise hota hai — yeh inherently ek first-derivative-of-velocity effect hai. Attitude aur position kinematics mein woh cross term same tarah involve nahi hoti (dekho Coriolis and Centrifugal Effects — Coriolis $-2\mathbf{\omega }\times \mathbf{v}$ hai, tab sabse bada jab velocity rotation axis ke perpendicular ho).

Ek chota attitude tilt $\delta \theta$ full $g$ ko horizontal channel mein $g\sin \delta \theta \approx g\delta \theta$ ke roop mein leak karta hai. Kyunki $g\approx 9.81$ hai, even $0.1^{\circ }$ tilt zyattar accelerometer biases se bada horizontal acceleration inject karta hai.

Euler angles gimbal lock (singularities) suffer karte hain aur unki rates $\pm 90^{\circ }$ pitch ke paas blow up karti hain. DCM/quaternion har jagah well-behaved rehta hai — lekin tumhe ise ek sahi initial condition dena hoga (ek initial alignment $C_b^n(0)$ ya unit quaternion) aur, kyunki numerical integration constraint se drift karta hai, har step renormalize karna hoga (DCM ko orthonormalize karo, ya quaternion ko unit norm par scale karo). Dekho Direction Cosine Matrix and Quaternions.

Kyunki local "down" direction physically re-tilt hoti hai jab tum curved Earth par travel karte ho. Nav frame rotate hota hai down point karte rehne ke liye, aur woh rotation gyro measurement se remove karni hogi.

Strapdown mein sensors body se bolted hote hain, isliye saari frame rotations (attitude layer) high rate par numerically karni padti hain; ek gimbaled platform frame ko mechanically hold karta hai, rotation hardware mein karta hai (dekho Strapdown vs Gimbaled INS).

Arc length $s=R\theta$, isliye $\theta =s/R$ aur $\dot{\theta }=\dot{s}/R$. Northward speed $v_N$ meridian arc rate hai, isliye $\dot{\varphi }=v_N/(R_M+h)$.

External aids (jaise GNSS) growing INS error reveal karte hain, aur kyunki bias-driven error ka ek known time signature hota hai ($t$, $t^2$, $t^3$), filter drift ko underlying bias se attribute karke correct kar sakta hai (dekho Kalman Filter for INS-GNSS Integration — filter ek drifting-but-smooth INS ko noisy-but-bounded GNSS fix ke saath blend karta hai).

Longitude rate $\dot{\lambda }=v_E/[(R_N+h)\cos \varphi ]$ diverge ho jaata hai kyunki $\cos \varphi \to 0$ — pole par longitude ill-defined hai. Latitude rate $\dot{\varphi }=v_N/(R_M+h)$ perfectly finite rehta hai (meridian radius $R_M+h$ wahan normal positive length hai), isliye sirf east-west channel singular hai; wander-azimuth ya alag frame ise fix karta hai.

Yeh $0$ hona chahiye. Upward support force (${\mathbf{f}}^b$ mein $-9.81$ Down) nav axes mein rotate hoti hai aur add ki gayi $+9.81$ Down gravity ${\mathbf{g}}^n$ ko exactly cancel kar deti hai, isliye ${\dot{\mathbf{v}}}^n=C_b^n{\mathbf{f}}^b+{\mathbf{g}}^n$ ke dono terms zero sum karte hain.

${\dot{C}}_b^n=C_b^n[0\times ]=0$, isliye orientation constant hold hoti hai — non-rotating body ke liye sahi hai, aur iska matlab yeh bhi hai ki koi bhi zero-rate bias frame ko silently tilt karta rehta hai.

Nahi. Low speed par bhi, Earth rate ($7.29\times 10^{-5}\text{rad/s}$) aur transport effects long time par accumulate hote hain, isliye long-duration navigation ko speed se regardless full Earth terms chahiye.

Yeh $\frac{1}{2}b{t}^2$ ki tarah unbounded badhta hai — unaided INS drift signature. Isliye long missions ke liye bounded external aiding (GNSS) zaroori hai.

Nahi — yeh apparent gravity read karta hai, jo gravitational attraction minus equatorial centrifugal term hai ($\approx 0.034{\text{m/s}}^2$ chota). Mechanization ${\mathbf{g}}^n$ yahi apparent value use karta hai (dekho Gravity Model and Geoid).