Exercises — Inertial navigation — accelerometer measures non-gravitational specific force

WHAT: z-direction mein $\mathbf{f}=\mathbf{a}-\mathbf{g}$ apply karo. WHY: sensor specific force read karta hai, $\mathbf{a}$ directly kabhi nahi. $f_z=a_z-g_z=0-(-9.81)=+9.81{\text{m/s}}^2$ $+9.81{\text{m/s}}^2$ (upar) read karega. Spring ko mass ko gravity ke against hold karne ke liye upar push karna padta hai, aur wahi push reading hoti hai.

Free fall mein sirf gravity force hai, toh true acceleration gravity ke barabar hai: $\mathbf{a}=\mathbf{g}=(0,0,-9.81)$. $f_z=a_z-g_z=-9.81-(-9.81)=0$ $0$ read karega — weightlessness. Spring ke paas push karne ke liye kuch nahi, toh kuch read nahi hota. (Yeh sensor form mein Equivalence Principle hai.)

L1.1 se, rest mein $\mathbf{a}=0$ toh $\mathbf{f}=-\mathbf{g}$. Kyunki $\mathbf{g}$ neeche point karta hai, $\mathbf{f}$ upar point karti hai. Reading hamesha us contact force ko mirror karti hai jo spring supply karta hai, aur rest mein woh force gravity se ladhta hai → upward.

Step 1 (WHAT/WHY): neeche ki taraf acceleration matlab $a_z=-3$ (negative kyunki z upar hai). Step 2: gravity $g_z=-9.81$. Step 3: $f_z=a_z-g_z=-3-(-9.81)=+6.81{\text{m/s}}^2$ $+6.81{\text{m/s}}^2$ read karega, phir bhi upar lekin $9.81$ se kam. Tum halka feel karte ho — spring kam support karta hai kyunki mass partly uससे "door gir raha" hai.

$f_z=a_z-g_z=39.24-(-9.81)=49.05{\text{m/s}}^2=5g$ Spring ko weight support ($1g$) bhi karna hai aur mass ko accelerate ($4g$) bhi karna hai, toh reading $5g$ hai. Isliye "$4g$ burn" chest par $5g$ force jaisa lagta hai.

Motion horizontal hai, toh gravity ka koi x-component nahi: $g_x=0$. Braking matlab motion ke opposite acceleration, toh $a_x=-6{\text{m/s}}^2$. $f_x=a_x-g_x=-6-0=-6{\text{m/s}}^2$ $-6{\text{m/s}}^2$ read karega (peeche ki taraf). Tum seatbelt ke against aage jhuke; belt ki contact force wahi hai jo sensor equivalently register karta hai.

WHAT/WHY: level turn mein do acceleration ingredients hain — vertical ($a_z=0$, altitude constant) aur horizontal centripetal $a_x=v^2/R$ turn centre ki taraf point karta hai. Vertical: $f_z=a_z-g_z=0-(-9.81)=+9.81$. Horizontal: $g_x=0$, toh $f_x=a_x=\frac{v^2}{R}=\frac{200^2}{3000}=\frac{40000}{3000}=13.33{\text{m/s}}^2$. Magnitude: $|\mathbf{f}|=\sqrt{f_z^2+f_x^2}=\sqrt{9.81^2+13.33^2}=\sqrt{96.24+177.7}=\sqrt{273.98}\approx 16.55{\text{m/s}}^2$ Load factor: $n=|\mathbf{f}|/g=16.55/9.81\approx 1.69g$. Seat ko pilot ko upar hold bhi karna hai aur unka path curve bhi karna hai — vector sum $g$ se exceed karta hai, toh woh heavier feel karte hain (ek "$1.69g$ turn").

WHY tan? Hamare paas ek right triangle ki do legs hain: horizontal leg $f_x$ (vertical se angle ke opposite) aur vertical leg $f_z$ (adjacent). Opposite/adjacent ratio = $\tan \varphi$ encode karta hai ki $\mathbf{f}$ seedha neeche se kitna steeply tilt hai. Woh angle us ratio se recover karne ke liye hum tan ko invert karte hain → arctan. $\tan \varphi =\frac{f_x}{f_z}=\frac{13.33}{9.81}=1.359$ $\varphi =\arctan (1.359)\approx 53.7^{\circ }$ Coordinated turn mein aircraft bank karta hai taaki uske wings $\mathbf{f}$ ke perpendicular hon; yahan woh level se lagbhag $53.7^{\circ }$ hai.

WHAT galat hua: true acceleration hai $a_z=f_z+g_z=9.81+(-9.81)=0$, toh real position $0$ par rehti hai. Lekin buggy code $\tilde{a}=f_z=9.81$ constant integrate karta hai. Ek constant $\tilde{a}$ ka double integration rest se: $\Delta z_{\text{error}}=\frac{1}{2}\tilde{a}{t}^2=\frac{1}{2}(9.81)(10{)}^2=490.5\text{m}$ 10 seconds mein lagbhag aadha kilometre drift ek bhule hue $\mathbf{g}$ se. Yeh $\frac{1}{2}g{t}^2$ blow-up hai (dekho Dead Reckoning and Error Drift).

Consistency check (WHY): rest mein $\mathbf{a}=0$, toh $\mathbf{f}=-\mathbf{g}$, matlab $|\mathbf{f}|$ ko $g$ ke barabar hona chahiye. $|\mathbf{f}|=\sqrt{0^2+4.905^2+8.496^2}=\sqrt{24.06+72.18}=\sqrt{96.24}=9.81{\text{m/s}}^2=g✓$ Rest ke saath consistent hai. Tilt: reading ka z-component woh part hai jo true-up ke saath aligned hai; horizontal part tilt hai. $\cos \theta =\frac{f_z}{|\mathbf{f}|}=\frac{8.496}{9.81}=0.866\Rightarrow \theta =\arccos (0.866)=30^{\circ }$ Unit horizontal se $30^{\circ }$ tilt hai. (Gravity vector slope-parallel aur slope-normal parts mein split hota hai — accelerometers akele static tilt precisely sense kar sakte hain kyunki woh rest mein $-\mathbf{g}$ read karte hain.)

Orbit mein body par sirf gravity force hai, toh true acceleration local gravitational acceleration ke barabar hai: $\mathbf{a}=\mathbf{g}(r)$, jahan $|\mathbf{g}(r)|=8.11$. $|\mathbf{f}|=|\mathbf{a}-\mathbf{g}|=|\mathbf{g}-\mathbf{g}|=0$ $0$ read karega — orbit perpetual free fall hai, toh sensor weightless hai jabki centre $8.11{\text{m/s}}^2$ se Earth ki taraf accelerate kar raha hai. Ek tiny non-zero reading non-gravitational forces se aayegi — atmospheric drag, solar radiation pressure, thruster firings — kyunki woh contact/applied forces hain jo spring feel kar sakta hai. Orbit mein accelerometers isliye drag sensors hain, gravity sensors nahi.

WHAT/WHY: loop bottom par centre seedha upar hai, toh centripetal acceleration upward hai: $a_z=+v^2/R$. $a_z=\frac{180^2}{1200}=\frac{32400}{1200}=27.0{\text{m/s}}^2$ Specific force z-component: $f_z=a_z-g_z=27.0-(-9.81)=36.81{\text{m/s}}^2$ Load factor: $n=f_z/g=36.81/9.81\approx 3.75g$. Seat ko path curve karne ke liye $27$ upar supply karna hai aur weight hold karne ke liye $9.81$ bhi — dono loop bottom par add hote hain, pilot ko $\approx 3.75g$ par neeche pin karte hain.

$f_z=a_z-g_z=-27.0-(-9.81)=-17.19{\text{m/s}}^2$ $-17.19{\text{m/s}}^2$ read karega — negative matlab seat pilot ko neeche push kar rahi hai (yaani pilot inverted hoke seat mein upar press karta hai). Load factor magnitude $|f_z|/g=17.19/9.81\approx 1.75g$, lekin yeh "seat ke relative negative-g" situation hai: required inward force gravity se exceed karti hai, toh restraint pilot ko loop centre ki taraf push karta hai. All-cases check: bottom par terms add hue ($+3.75g$); top par woh vertical sum mein partly cancel hue lekin sign flip ho gaya — sensor vertical sum mein loop ke har quadrant mein contact force ko faithfully track karta hai.