Worked examples — Augmented proportional navigation — gravity compensation

3.5.51 · D3 · Physics › Guidance, Navigation & Control (GNC) › Augmented proportional navigation — gravity compensation

Yeh page ek shooting gallery hai. Pehle hum har tarah ki situation ka ek map banate hain jo augmented-PN gravity-compensation law face kar sakta hai, phir us map ke har square mein se ek fully-worked example chalate hain. Agar aapne Augmented proportional navigation — gravity compensation padh liya hai, toh neeche ka har symbol already aapka hai; agar koi symbol abhi bhi fuzzy lagta hai, toh parent note use zero se build karta hai.

Woh ek formula jo hum baar baar use karte rahenge:

Har letter ke plain-word meaning ka quick reminder, taaki kuch bhi bina samjhe use na ho:

• $a_c$ — sideways ("lateral") push jo hum command karte hain, metres per second squared mein.
• $N^{\prime }$ — navigation constant, ek plain unitless number, usually $3$ se $5$ tak.
• $V_c$ — closing speed, missile aur target ke beech ka gap kitni tezi se kam ho raha hai (m/s).
• $\dot{\lambda }$ — line-of-sight (LOS) ki direction kitni tezi se rotate ho rahi hai (rad/s); chota dot matlab "rate of change".
• $a_T$ — target ki apni sideways acceleration, woh part jo LOS ke across point kar raha hai (m/s²).
• $g$ — gravity ki strength, yaani koi object freely girte waqt kitni tezi se speed up hota hai. Earth's surface ke paas $g=9.81{\text{m/s}}^2$ (free fall ka har second lagbhag $9.81$ m/s downward speed add karta hai). Yeh akela number hi woh reason hai jis se missile "droops" karta hai. Yeh environment ka ek fixed constant hai, koi aisi cheez nahi jo hum control kar sakein.
• $g_n$ — gravity ka woh hissa jo LOS ke across point karta hai aur isliye hamari flight ko bend karta hai (m/s²).
• $\lambda$ — LOS elevation angle: $0°$ = zameen ke saath flat dekh rahe hain, $90°$ = seedha upar dekh rahe hain.

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Scenario matrix

Neeche har worked example [Cell N] se tagged hai taaki aap dekh sakein ki hum koi square empty nahi chodh rahe.

Cell	Case class	Kya special hai	Kaunsa example
1	Horizontal LOS, non-maneuvering target	$\lambda =0\Rightarrow g_n=g$ (worst gravity)	Ex 1
2	Steep descending LOS, maneuvering target	$\lambda$ large $\Rightarrow g_n$ small	Ex 2
3	Degenerate: $\dot{\lambda }=0$ (momentary collision course)	PN term vanish ho jaata hai, sirf $-g_n$ kaam karta hai	Ex 3
4	Vertical shot, seedha upar ($\lambda =+90°$)	$g_n=0$, gravity LOS ke along	Ex 4
4b	Vertical shot, seedha neeche ($\lambda =-90°$)	$g_n=0$ bhi — $\pm 90°$ edge complete karta hai	Ex 4b
5	Sign flip: LOS doosri taraf rotate ho raha hai ($\dot{\lambda }<0$)	commanded push sign flip kar leta hai	Ex 5
6	Diving / negative elevation ($\lambda <0$)	$\cos \lambda$ abhi bhi positive → $g_n>0$	Ex 6
7	Real-world word problem: dart-thrower geometry	$g_n$ ko ek picture se build karo	Ex 7
8	Exam twist: acceleration limit command ko saturate kar deta hai	$a_c$ ko autopilot cap par clip karo	Ex 8

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Worked examples

Ex 1 — Horizontal LOS, worst-case gravity [Cell 1]

1. PN term. $N^{\prime }{V}_c\dot{\lambda }=4(800)(0.01)=32{\text{m/s}}^2$. Yeh step kyun? Yeh wahi hai jo pure PN akela LOS spin se demand karta hai — hamaara baseline.
2. Gravity term. $\lambda =0\Rightarrow g_n=g\cos 0=9.81{\text{m/s}}^2$. Yeh step kyun? Horizontal LOS ka matlab hai ki gravity poori LOS ke across point karti hai — jitna bend possible hai utna zyada, maximum $g_n$.
3. Assemble karo. $a_c=32+0-9.81=22.19{\text{m/s}}^2$. Yeh step kyun? Hum $g_n$ subtract karte hain taaki jab gravity baad mein $+9.81$ wapas add kare, toh net across-LOS acceleration exactly intended $32$ ho.

Verify: effective normal accel $=a_c+g_n=22.19+9.81=32.00{\text{m/s}}^2$ — precisely pure-PN demand. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$ (radians unitless hote hain), aur $g_n$ already m/s² mein hai, toh har term cleanly m/s² mein add hota hai. Answer 32 se chota, forecast ke saath match karta hai (hum back off karte hain, kyunki gravity pushing ka kuch part khud karti hai hamaare liye — lekin galat direction mein, toh yeh net-cancel ho jaata hai).

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Ex 2 — Steep descent onto a maneuvering target [Cell 2]

Figure s01 — Ex 2 geometry. Yellow arrow steeply-descending LOS hai ($\lambda =-60°$); pink arrow gravity seedha neeche point kar rahi hai; dotted white arrow gravity ka bada hissa hai jo LOS ke along peda hai (harmless — yeh sirf closing speed change karta hai); short blue arrow $g_n$ hai, woh chota hissa jo LOS ke across peda hai. Picture dikhata hai ki steep geometry $g_n$ ko kyun chota banata hai.

1. PN term. $3(1000)(0.004)=12{\text{m/s}}^2$. Yeh step kyun? Chota $\dot{\lambda }$ → modest baseline demand. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
2. Augmentation term. $\frac{N^{\prime }}{2}{a}_T=\frac{3}{2}(30)=45{\text{m/s}}^2$. Yeh step kyun? $\frac{N^{\prime }}{2}$ feed-forward exactly pre-cancel karta hai woh zero-effort-miss jo ek constant target maneuver otherwise cause karta. Yahan $t_{go}$ time-to-go hai — impact tak baache seconds; ek constant maneuver us window mein $\frac{1}{2}{a}_T{t}_{go}^2$ miss build karta hai, isliye $\frac{N^{\prime }}{2}$ feed-forward is tarah tune kiya gaya hai ki jo bhi current $t_{go}$ ho use erase kar sake — dekho Zero-Effort-Miss (ZEM) guidance. Units: $N^{\prime }/2$ unitless hai, toh $\frac{N^{\prime }}{2}{a}_T$ m/s² mein rehta hai.
3. Gravity term. $g_n=g\cos 60°=9.81(0.5)=4.905{\text{m/s}}^2$. Figure dekho: LOS steeply neeche point kar raha hai, toh pink gravity arrow LOS ke mostly along pada hai; sirf chota blue stub $g_n$ hai. Yeh step kyun? Steep geometry gravity ka zyaadatar hissa LOS direction ke andar chupa deta hai, jahan woh $\dot{\lambda }$ ko bend nahi kar sakti.
4. Assemble karo. $a_c=12+45-4.905=52.10{\text{m/s}}^2$. Yeh step kyun? Gravity ka contribution ab teeno mein sabse chota hai — target maneuver rule karta hai.

Verify: $12+45-4.905=52.095\approx 52.10{\text{m/s}}^2$. Forecast check: augmentation term (45) dominate karta hai, jaisa predict kiya tha. Teeno terms m/s² mein hain, toh sum ek valid acceleration hai.

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Ex 3 — Degenerate collision course, $\dot{\lambda }=0$ [Cell 3]

1. Uncompensated PN. $a_c^{\text{PN}}=N^{\prime }{V}_c(0)+0=0$. Yeh step kyun? Ek pure-PN brain koi LOS spin nahi dekhta aur relax ho jaata hai — kuch command nahi karta.
2. The trap. Zero command ke saath, gravity ka $g_n=9.81$ unopposed kaam karta hai → agla instant flight droop karti hai → $\dot{\lambda }$ nonzero ho jaata hai → PN error aane ke baad correct karne ke liye scramble karta hai. Yeh step kyun? Ghost dikhata hai: ek "fixed" LOS gravity handling ke bina stable nahi hai.
3. Compensated APN. $a_c=0+0-9.81=-9.81{\text{m/s}}^2$. Yeh step kyun? Negative sign ka matlab hai "LOS ke across upar push karo", exactly aane wale droop ko cancel karta hai taaki $\dot{\lambda }$ truly $0$ rahe.

Verify: net across-LOS accel $=a_c+g_n=-9.81+9.81=0$. Units: dono terms m/s², difference m/s² → LOS rate ka zero forcing hai → fixed rehta hai → ek genuine, held collision course. Answer: $-9.81{\text{m/s}}^2$, zero nahi — forecast "free to command zero" galat hai.

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Ex 4 — Straight-up shot, gravity fully harmless [Cell 4]

1. PN term. $5(600)(0.006)=18{\text{m/s}}^2$. Yeh step kyun? LOS spin se standard baseline. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
2. Gravity term. $g_n=g\cos 90°=9.81(0)=0$. Yeh step kyun? Seedha upar, gravity LOS ke along point karti hai, toh woh sirf closing speed slow karti hai — LOS bend nahi kar sakti. Cancel karne ko kuch nahi hai.
3. Assemble karo. $a_c=18+0-0=18{\text{m/s}}^2$.

Verify: $g_n=0$ expected tha vertical LOS ke liye. Command raw PN ke barabar hai, m/s² mein. Forecast answer: zero compensation, sahi.

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Ex 4b — Straight-down shot, $\pm 90°$ edge complete karna [Cell 4b]

1. PN term. $5(600)(0.006)=18{\text{m/s}}^2$. Yeh step kyun? LOS-spin term care nahi karta ki LOS kis taraf point kar raha hai; Ex 4 jaisa same baseline.
2. Gravity term. $g_n=g\cos (-90°)=9.81\cos 90°=0$. Yeh step kyun? $\cos$ even hai, toh $\cos (-90°)=\cos (90°)=0$. Chahe hum seedha upar ya seedha neeche shoot karein, gravity entirely LOS ke along hai — koi across-LOS slice exist nahi karti.
3. Assemble karo. $a_c=18+0-0=18{\text{m/s}}^2$.

Verify: $g_n=0$ identically, straight-up case se match karta hai. Units m/s². Yeh geometry map close karta hai: dono $\lambda =+90°$ aur $\lambda =-90°$ par gravity harmless hai. Forecast answer: koi change nahi — up aur down yahan symmetric hain.

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Ex 5 — LOS doosri taraf spin kar raha hai, $\dot{\lambda }<0$ [Cell 5]

1. PN term. $4(750)(-0.008)=-24{\text{m/s}}^2$. Yeh step kyun? Negative $\dot{\lambda }$ ka matlab hai LOS opposite direction mein swing kar rahi hai, toh PN opposite direction mein push karta hai — sign honestly carry through hota hai. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
2. Gravity term. $g_n=9.81{\text{m/s}}^2$ (abhi bhi positive: gravity hamesha same physical direction mein pull karti hai chahe LOS kisi bhi direction mein spin kare). Yeh step kyun? Gravity ki direction duniya set karti hai, $\dot{\lambda }$ ke sign se nahi — ise accidentally flip mat karo.
3. Assemble karo. $a_c=-24+0-9.81=-33.81{\text{m/s}}^2$.

Verify: effective normal accel $=a_c+g_n=-33.81+9.81=-24$, pure-PN demand of $-24$ se match karta hai. Units m/s² throughout. Command sign PN term se dominated hai lekin gravity term flip nahi hoti. Answer $-33.81{\text{m/s}}^2$.

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Ex 6 — Diving engagement, negative elevation [Cell 6]

Figure s02 — Ex 6. Do LOS arrows draw kiye gaye hain, ek $+30°$ horizontal se upar tilted (blue $g_n$ stub) aur ek $-30°$ neeche tilted (pink $g_n$ stub). Kyunki $\cos$ ek even function hai, dono blue/pink stubs exactly same length ke hain: ek downward-tilted LOS ko same angle par ek upward-tilted LOS jitna hi gravity compensation chahiye.

1. Gravity term. $g_n=g\cos (-30°)=g\cos 30°=9.81(0.8660)=8.496{\text{m/s}}^2$. Yeh step kyun? Cosine ek even function hai: $\cos (-30°)=\cos (30°)$. LOS chahe upar ya neeche same angle par tilt kare, gravity ka jo amount iske across pada hai woh identical hai — figure dono up-tilt aur down-tilt ko same stub length deta hua dikhata hai.
2. PN term. $3.5(900)(0.005)=15.75{\text{m/s}}^2$. Yeh step kyun? LOS spin se baseline. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
3. Assemble karo. $a_c=15.75+0-8.496=7.254{\text{m/s}}^2$.

Verify: $g_n$ positive nikla (8.496), kyunki $\cos$ tilt ke sign ko ignore karta hai — forecast "negative $g_n$" galat hai. $15.75-8.496=7.254{\text{m/s}}^2$. Dono terms m/s², difference m/s².

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Ex 7 — Word problem: the falling-dart geometry [Cell 7]

Figure s03 — Ex 7. Yellow arrow shallow climbing LOS hai $\lambda =20°$ par. Gravity (pink, seedha neeche) perpendiculars drop karke do hisson mein split hoti hai: ek slice LOS ke along ($g\sin 20°$, dotted white) aur ek slice LOS ke across ($g_n=g\cos 20°$, blue). Shallow tilt par blue slice lagbhag poora $g$ hai — almost worst-case bending.

1. Geometry padho. Figure mein LOS yellow arrow hai $20°$ tilt par; gravity (pink) seedha neeche point karti hai. LOS se perpendicular drop karo: gravity ka LOS ke across wala piece $g\cos 20°$ hai, iske along wala piece $g\sin 20°$ hai. Yeh step kyun? Sirf across-LOS slice flight bend karti hai, isliye hum wohi cancel karte hain.
2. Gravity term. $g_n=9.81\cos 20°=9.81(0.9397)=9.219{\text{m/s}}^2$. Yeh step kyun? Shallow tilt → cosine near 1 → gravity ka lagbhag sab kuch abhi bhi LOS bend karta hai (near worst case).
3. PN term. $4(500)(0.012)=24{\text{m/s}}^2$.
4. Augmentation term. $\frac{4}{2}(15)=30{\text{m/s}}^2$. Yeh step kyun? Drone maneuvering kar raha hai, toh hum iske miss ko $N^{\prime }/2$ se feed-forward karte hain (same $t_{go}$-independent logic jaise Ex 2 mein).
5. Assemble karo. $a_c=24+30-9.219=44.781{\text{m/s}}^2$.

Verify: $g_n=9.219{\text{m/s}}^2$, full $9.81$ ke close — forecast "nearly full-strength" confirm. $24+30-9.219=44.781{\text{m/s}}^2$. Sab terms m/s².

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Ex 8 — Exam twist: autopilot acceleration limit [Cell 8]

1. Unclipped demand compute karo. $a_c=44.781{\text{m/s}}^2$ (Ex 7 se). Yeh step kyun? Guidance law pehle run karta hai; limiter iske output par kaam karta hai.
2. Cap apply karo. Kyunki $44.781>40$, delivered command clip hokar $a_c^{\text{cmd}}=40{\text{m/s}}^2$ ho jaata hai. Yeh step kyun? Airframe/structure physically sirf itna hi hard pull kar sakta hai; isse zyada impossible hai.
3. Lost authority. Shortfall $=44.781-40=4.781{\text{m/s}}^2$. Yeh step kyun? Yeh deficit wahi hai jo residual miss cause karta hai — aur yeh exactly woh argument hai ki g-capacity ko gravity chase karne mein waste nahi karna chahiye: agar hum compensation skip karte toh gravity ghost cap ka aur bhi zyada hissa kha jaati.

Verify: clipped value $=\min (44.781,40)=40$. Lost $=4.781{\text{m/s}}^2$, m/s² mein. Compensation term ($-9.219$) us number ke andar hai jo clip ho raha hai, toh saturation par gravity ke against ladai baaki sab ke saath partially sacrifice ho jaati hai — forecast "still gets through fully" galat hai. Isliye cap ke neeche rehna (Ex 1–4 style) matter karta hai.

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Recall Fast self-test

Horizontal LOS $g_n$ ko kiske barabar banata hai? ::: Full $g$ (kyunki $\cos 0=1$) — worst case. Negative elevation $\lambda =-30°$ $g_n$ ka kaun sa sign deta hai? ::: Positive, kyunki $\cos$ even hai ($\cos (-30°)=\cos 30°$). Agar $\dot{\lambda }=0$ aur target inertial hai, toh compensated APN horizontally kya command karta hai? ::: $-g_n=-9.81{\text{m/s}}^2$, droop ko pre-empt karta hua. Straight-up AUR straight-down shots ko kitna compensation chahiye? ::: Dono mein zero, kyunki $g_n=g\cos (\pm 90°)=0$. Autopilot limiter kahan act karta hai — guidance law se pehle ya baad? ::: Baad mein: yeh law ke output $a_c$ ko clip karta hai.

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Connections

• Augmented proportional navigation — gravity compensation — woh parent law jise yeh examples exercise karte hain.
• Proportional Navigation (PN) — woh $N^{\prime }{V}_c\dot{\lambda }$ baseline term.
• Zero-Effort-Miss (ZEM) guidance — $N^{\prime }/2$ augmentation factor aur $t_{go}$ picture ka origin.
• Line-of-Sight rate estimation — har example mein $\dot{\lambda }$ kahan se aata hai.
• Missile Autopilot & Acceleration limits — woh cap jo Ex 8 clip karta hai.
• Coordinate frames & projections — $g_n=g\cos \lambda$ kaise $\vec{g}$ ka LOS-normal par ek projection hai.
• Ballistic trajectory & gravity turn — woh "droop" jise yeh examples fight karte hain.