Augmented proportional navigation — gravity compensation: Visual walkthrough

3.5.51 · D2 · Physics › Guidance, Navigation & Control (GNC) › Augmented proportional navigation — gravity compensation

Step 1 — Line of sight, aur woh ek angle jo matter karta hai

KYA HAI. Page par do dots: missile (iske position point ko $M$ label karte hain) aur target (position point $T$ ). Ek seedhi rod kheencho jo unhe connect kare — wahi rod line of sight (LOS) hai. Uski length range $R$ hai (straight-line distance $M$ se $T$ tak, metres mein). Ab yeh rod jo angle ek fixed horizontal reference line se banati hai, use measure karo. Us angle ko $λ$ (Greek "lambda") kaho, horizontal se measure karo, upar ki taraf positive.

YEH ANGLE KYUN, positions themselves kyun nahin? Kyunki missile ko yeh jaanne ki zaroorat nahin ki target absolute coordinates mein kahan hai — use sirf yeh jaanna chahiye ki connecting rod rotate ho rahi hai ya nahin. Ek rod jo fixed direction rakhte hue dono ends ko close karti hai matlab collision. Toh poori guidance mein sabse important number position nahin, balki is rod ka tilt aur yeh kitni tezi se change ho raha hai — yehi important hai.

PICTURE. $M$ se $T$ tak white rod (uski length $R$ labelled), dashed horizontal se uthta amber angle $λ$ , aur ek chhota curved arrow jo dikhata hai $λ$ kis direction mein badhega.

Figure — Augmented proportional navigation — gravity compensation

Step 2 — $\dot{λ} = 0$ collision kyon hoti hai

KYA HAI. Rod ki direction freeze karo lekin dono ends ko ek-doosre ki taraf slide karne do — yaani $R$ shrink ho jabki $λ$ fixed rahe. Rod ko teen moments par draw karo — har baar same tilt, bas choti hoti jaaye.

KYUN. Agar rod kabhi rotate na kare ( $\dot{λ} = 0$ ) phir bhi $R$ shrink ho, toh dono ends ek hi point par ek hi instant mein pohoch rahe hain. Sailors ise constant bearing, decreasing range kehte hain: ek ship jiska compass bearing kabhi change na ho lekin bada lage, matlab woh ship tumse takraane wali hai. Toh guidance ka poora kaam ek nare mein aata hai: $\dot{λ}$ khatam karo.

PICTURE. Decreasing length $R$ ki teen parallel rods, sab ek hi tilt $λ$ par, ek amber collision point par converge karti hui.

Recall Fixed bearing impact kyun deta hai

Non-rotating LOS + shrinking $R$ ::: dono bodies ek hi meeting point ki taraf jaati hain → guaranteed intercept.

Step 3 — Closing velocity, theek se define ki gayi

KYA HAI. Steer karne se pehle, rod ke shrink hone ki rate ko name do. Closing velocity ko define karo is tarah: $V_{c} = - \dot{R},$ range ke rate of change ka negative. Minus sign deliberate hai: $R$ intercept ke dauran decrease ho raha hai ( $\dot{R} < 0$ ), toh $V_{c} = - \dot{R}$ positive nikalta hai. $V_{c} > 0$ matlab "gap closing"; $V_{c} < 0$ matlab target bhaag raha hai.

$- \dot{R}$ se kyun define karein? Kyunki "closing speed" ek positive number honi chahiye jab hum actually close ho rahe hoon — yeh homing missile ke liye useful convention hai. $V_{c}$ ko hum $R$ se build karte hain, jo Step 1 mein already define ho chuka hai, toh kuch bhi naya nahin ghusa raha.

PICTURE. Do instants par rod; $Δ R < 0$ marked, aur $V_{c} = - \dot{R} > 0$ LOS ke along amber "closing" arrow ke roop mein dikhaya gaya.

Step 4 — Rod par do directions set karo (sign convention ke saath)

KYA HAI. Missile par koi bhi push do arrows mein tod sakte hain: ek rod ke along, direction $\hat{t}$ ("tangent", $M$ se $T$ ki taraf pointing), aur ek rod ke across, direction $\overset{n}{^}$ ("normal"). Orientation ek baar fix karo hamesha ke liye: $\overset{n}{^} ko \hat{t} se 9 0^{\circ} counter-clockwise point karna hai (yaani increasing λ ki direction mein).$ Toh acceleration ka positive $\overset{n}{^}$ -component woh hai jo $λ$ increase karta hai.

Sign kyun nail down karein? Baad mein aane wale har term ( $g_{n}$ , $a_{T, n}$ ) ka $\overset{n}{^}$ par ek shadow hoga, aur shadow ka sign tabhi hoga jab tum batao $\overset{n}{^}$ kis taraf point karta hai. " $+ \overset{n}{^}$ = increasing $λ$ " tie karne se poori spin equation ke signs unambiguous ho jaate hain.

PICTURE. LOS rod jisme $\hat{t}$ along (M→T) aur $\overset{n}{^}$ $9 0^{\circ}$ counter-clockwise drawn; increasing $λ$ ka curved arrow $+ \overset{n}{^}$ se agree karta hai; ek generic acceleration apne $\hat{t}$ aur $\overset{n}{^}$ shadows mein decomposed.

Step 5 — Derive karo ki $\dot{λ}$ actually kya drive karta hai

KYA HAI. Ab hum spin equation assert karne ki jagah earn karte hain. Dono bodies ka rod-ke-across separation do cheezein ki wajah se change hota hai: rod turning ( $\dot{λ}$ ) aur rod shrinking ( $\dot{R}$ ). Relative position $R$ ki planar kinematics transverse (across-LOS) relative velocity $R \dot{λ}$ deti hai. Ek baar aur differentiate karo — transverse relative acceleration hai: $a_{n}^{rel} = R \ddot{λ} + 2 \dot{R} \dot{λ} .$

$R \ddot{λ}$ — rod angularly accelerate kar rahi hai
$2 \dot{R} \dot{λ}$ — Coriolis-like term: ek shrinking rod ( $\dot{R} < 0$ ) kisi bhi existing $\dot{λ}$ ko spin up karti hai, bilkul waise jaise ek skater apne arms pull in karke tez spin karta hai.

Woh transverse relative acceleration (target normal accel) minus (missile normal accel) se supply hoti hai. Missile ki normal accel commanded plus gravity likhte hain, $a_{M, n} = a_{c, n} + g_{n}$ : $R \ddot{λ} + 2 \dot{R} \dot{λ} = a_{T, n} - (a_{c, n} + g_{n}) .$ Angular acceleration ke liye solve karo aur $V_{c} = - \dot{R}$ use karo: $\ddot{λ} = \frac{1}{R} [a_{T, n} - a_{c, n} - g_{n}] + \frac{2 V _{c}}{R} \dot{λ}$

YEH KYUN MATTER KARTA HAI — aur $1/ R$ note karo. Forcing terms range $R$ se divide hokar enter hote hain: jitna paas jaate hain (small $R$ ), utna violently koi bhi leftover $g_{n}$ ya $a_{T, n}$ $\dot{λ}$ ko whip karta hai. Yahi derivation hai jo parent ka one-line " $\dot{λ}$ evolves under $(a_{c, n} + g_{n} - a_{T, n})$ " compress kar raha tha — bracket $[a_{T, n} - a_{c, n} - g_{n}]$ exactly wahi combination hai (signs hamare $\overset{n}{^}$ convention se set), aur $1/ R$ woh normalization hai jo us box ne omit ki thi.

PICTURE. $\overset{n}{^}$ ke along transverse velocity $R \dot{λ}$ drawn rod ke saath; shrinking rod $2 \dot{R} \dot{λ}$ spin-up feed kar raha hai; teen forcing arrows $a_{T, n}$ , $a_{c, n}$ , $g_{n}$ $\overset{n}{^}$ par push karte hue ek $1/ R$ amplifier symbol ke saath.

Step 6 — Dekho gravity loop ko kaise poison karti hai

KYA HAI. $a_{c, n} = 0$ set karo aur ek inertial target imagine karo ( $a_{T, n} = 0$ ). Spin equation mein reh jaata hai $\ddot{λ} = - g_{n} / R + (2 V_{c} / R) \dot{λ}$ : ek persistent $- g_{n} / R$ forcing jo kabhi switch off nahin hoti. Rod droop karne lagti hai.

YEH "average out" KYUN NAHIN KARTA. Poore up-and-over ballistic arc mein droop reverse ho jaata hai, toh cancel ho jaata hai — parent ke [!mistake] box mein yeh tempting excuse tha. Lekin terminal homing ek one-way plunge hai kuch seconds ki: gravity poore waqt same direction mein kheenchti hai, aur $1/ R$ ki wajah se uska damage badhta hai jab $R \to 0$ . Toh $\dot{λ}$ par uska effect accumulate ho jaata hai ek real, growing miss mein. Drooping arc khud dekhne ke liye Ballistic trajectory & gravity turn dekho.

PICTURE. Ek seedha intended path (amber dashed) aur neeche sagging real drooping path (cyan solid), gap "accumulating miss" labelled, aur $\overset{n}{^}$ par steady $- g_{n} / R$ forcing.

Step 7 — $g_{n}$ kitna bada hai? Har LOS angle cover kiya

KYA HAI. Gravity vector $g = - g \overset{z}{^}$ hai (seedha neeche, $g \approx 9.81 m/s^{2}$ ). Step 1 se yaad karo ki $λ$ horizontal se measure hota hai, aur Step 4 se ki $\overset{n}{^}$ LOS se $9 0^{\circ}$ counter-clockwise hai. Seedhe-neeche gravity ko us $\overset{n}{^}$ par project karne se milta hai: $g_{n} = g \cdot \overset{n}{^} = g cos λ .$

COSINE KYUN, aur $λ$ horizontal se kyun. Dot product $g \cdot \overset{n}{^}$ poochhta hai "gravity kitni $\overset{n}{^}$ ke along lie karti hai?" — across-direction par gravity ka shadow. Kyunki $λ$ horizontal se tilt hai, jab rod horizontal ho toh $\overset{n}{^}$ seedha upar point karta hai, gravity ko puri tarah oppose karta hai, toh shadow poora $g$ hai; jaise-jaise $λ$ badhta hai, $\overset{n}{^}$ vertical se door jhukta hai aur shadow $cos λ$ se shrink hota hai.

Har case, koi gap nahin ( $λ < 0$ aur $λ > 9 0^{\circ}$ including):

$λ = 0^{\circ}$ (horizontal shot): $g_{n} = g cos 0 = + g$ — worst case, gravity poori rod ko bend karti hai.
$λ = 6 0^{\circ}$ (steep climb): $g_{n} = g cos 6 0^{\circ} = 0.5 g \approx 4.905$ — half strength.
$λ = 9 0^{\circ}$ (seedha upar): $g_{n} = g cos 9 0^{\circ} = 0$ — gravity rod ke along lie karti hai, harmless.
$λ = - 3 0^{\circ}$ (horizontal se neeche diving): $cos (- 3 0^{\circ}) = cos 3 0^{\circ}$ , toh $g_{n} = g cos 3 0^{\circ}$ — cosine even hai, toh $3 0^{\circ}$ ka dive same $∣ g_{n} ∣$ chahta hai jitna $3 0^{\circ}$ ki climb; $g_{n}$ ka sign abhi bhi track karta hai ki $\overset{n}{^}$ gravity ke saath lean karta hai ya against.
$λ = 12 0^{\circ}$ (LOS peeche aur upar point kar raha hai): $cos 12 0^{\circ} = - 0.5$ , toh $g_{n} = - 0.5 g$ — ab gravity ka shadow $- \overset{n}{^}$ ki taraf flip ho jaata hai, aur compensation term accordingly sign flip karti hai. Formula ise automatically handle karta hai.

PICTURE. Rod ke copies $0^{\circ}$ , $6 0^{\circ}$ , $9 0^{\circ}$ , $- 3 0^{\circ}$ , $12 0^{\circ}$ par; har ek par vertical gravity arrow aur uska $\overset{n}{^}$ -shadow, shrinking, vanishing, phir reversing.

Recall Angles ke across gravity shadow

$g_{n}$ at $λ = 0^{\circ}, 6 0^{\circ}, 9 0^{\circ}, 12 0^{\circ}$ ::: $g, 0.5 g, 0, - 0.5 g$ (cosine $λ$ mein even hai, aur $9 0^{\circ}$ ke baad negative ho jaata hai)

Step 8 — Gravity cancel karo: $- g_{n}$ term

KYA HAI. Hum chahte hain ki rod jo net normal acceleration feel kare woh exactly pure-PN demand ke barabar ho. Gravity unavoidably $+ g_{n}$ contribute karegi $a_{M, n} = a_{c, n} + g_{n}$ mein. Toh hum apne commanded normal component ko extra $- g_{n}$ carry karने के लिए design karte hain: $a_{c, n} = PN demand N^{'} V_{c} \dot{λ} - g_{n} ⟹ a_{M, n} jo rod feel kare a_{c, n} + g_{n} = N^{'} V_{c} \dot{λ} .$

ADD KYUN NAHIN, SUBTRACT KYUN? Hum aane wali disturbance ka opposite pre-load karte hain: command $- g_{n}$ ; nature $+ g_{n}$ add karta hai; dono annihilate hote hain; rod sirf clean PN push feel karta hai. Yeh predictive cancellation hai — hum collision course hold karte hain droop appear hone ke baad chase karne ki jagah.

PICTURE. Do equal-length amber arrows opposite directions mein pointing ( $- g_{n}$ commanded, $+ g_{n}$ gravity se) zero net gravity effect mein sum karte hue, ek clean cyan PN arrow chhod jaate hain.

Step 9 — Target move karta hai: $\frac{N ^{'}}{2} a_{T, n}$ term derive karo

KYA HAI. Target ko constant normal acceleration $a_{T, n}$ le jaane do, aur time-to-go ko define karo: $t_{g o} = \frac{R}{V _{c}},$ range divided by closing speed — roughly kitne seconds mein impact hoga. Agar hum kuch bhi na karein, woh constant $a_{T, n}$ target ko collision point se freshman-kinematics amount se drift kar dega: $ZEM = \frac{1}{2} a_{T, n} t_{g o}^{2},$ yahi zero-effort-miss hai (dekho Zero-Effort-Miss (ZEM) guidance).

$\frac{N ^{'}}{2}$ FACTOR KYUN — algebra. Gain $N^{'}$ ke saath PN acceleration command karta hai $a_{c, n} = N^{'} V_{c} \dot{λ}$ . PN ka ek standard ZEM rewrite hai $a_{c, n} = N^{'} \frac{ZEM}{t _{g o}^{2}}$ (loop zero-effort-miss ko remaining flight par gain $N^{'}$ se null karta hai). Target-induced miss ko bhi cancel karne ke liye, same loop ko target ka ZEM contribution feed karo: $a_{c, n}^{(target)} = N^{'} \frac{ZEM _{target}}{t _{g o}^{2}} = N^{'} \frac{\frac{1}{2} a _{T, n} t _{g o}^{2}}{t _{g o}^{2}} = \frac{N ^{'}}{2} a_{T, n} .$ $t_{g o}^{2}$ exactly cancel ho jaata hai, $\frac{N ^{'}}{2}$ bachta hai: $\frac{1}{2}$ kinematics ka $\frac{1}{2} a t^{2}$ hai, $N^{'}$ loop gain hai. Yehi woh jagah hai jahan yeh factor paida hota hai, assume nahin kiya jaata.

PICTURE. Target seedhe predicted path se peelkar ek parabola mein; amber gap $\frac{1}{2} a_{T, n} t_{g o}^{2}$ marked; $t_{g o}^{2}$ cancellation shown; ek feed-forward arrow $\frac{N ^{'}}{2} a_{T, n}$ labelled jo use close karta hai.

Ek-picture summary

KYA HAI. Teen commanded normal terms assemble karo. Kyunki missile ka lateral command LOS ke across apply hota hai, boxed law mein $a_{c}$ ka matlab required normal component $a_{c, n}$ hai — woh number jo autopilot ko line of sight ke perpendicular deliver karna hai: $a_{c} = LOS spin dekho N^{'} V_{c} \dot{λ} + Target chase karo \frac{N ^{'}}{2} a_{T, n} - Gravity se lado g_{n} (a_{c} \equiv a_{c, n})$

PICTURE. Ek missile, teen stacked contribution arrows (cyan PN spin term, white target-lead term, amber $- g_{n}$ gravity term) tip-to-tail add hote hue total command $a_{c}$ mein, LOS, $\overset{n}{^}$ , aur gravity shadow sab labelled.

Recall Feynman: plain words mein poora walkthrough

Apne missile ( $M$ ) se target ( $T$ ) tak ek stick draw karo; uski length range $R$ hai aur uski tilt $λ$ hai. Agar woh stick apna tilt rakhte hue $R$ shrink kare, toh hit hogi — toh missile ka poora kaam hai stick ko spin mat karne do. Careful bookkeeping dikhata hai ki spin sideways pushes ko $R$ se divide karke driven hai, toh jaise close aate ho har leftover push zyada hurt karti hai. Missile koi bhi spin dikhte hi sideways steer karta hai (PN). Gravity ek sideways shadow $g_{n} = g cos λ$ cast karti hai jo dive ke dauran kabhi nahin rukti — toh missile ek equal-and-opposite $- g_{n}$ push karta hai usse pehle hi erase karne ke liye. Aur agar target swerve kare, toh woh time-to-go $t_{g o} = R / V_{c}$ par ek predictable gap $\frac{1}{2} a_{T, n} t_{g o}^{2}$ kholti hai; same loop se chalao aur $t_{g o}^{2}$ cancel ho jaata hai, ek lead term $\frac{N ^{'}}{2} a_{T, n}$ bachta hai. Teen normal components add karo — spin dekho, target chase karo, gravity se lado — aur stick impact tak quiet rehti hai.

Connections

Augmented proportional navigation — gravity compensation — woh parent result jise yeh pictures build karti hain.
Proportional Navigation (PN) — Step 5 ka base law.
Zero-Effort-Miss (ZEM) guidance — Step 9 ka $\frac{N ^{'}}{2}$ yahan se aata hai.
Line-of-Sight rate estimation — Step 1 ka $\dot{λ}$ actually measure kaise hota hai.
Coordinate frames & projections — Step 4 ka $\hat{t} / \overset{n}{^}$ split.
Ballistic trajectory & gravity turn — Step 6 ka physical droop.
Missile Autopilot & Acceleration limits — gravity chase na karke g-capacity save karna kyun matter karta hai.