Proportional navigation guidance — N·V_c·λ̇, derivation: Worked examples

3.5.50 · D3 · Physics › Guidance, Navigation & Control (GNC) › Proportional navigation guidance — N·V_c·λ̇, derivation

Yeh page PN law ko tab tak drill karti hai jab tak koi bhi case aapko surprise na kar sake. Hum numbers plug karte hain, har sign ko ulta karte hain, inputs ko tod-phod ke dekhte hain (zero range rate, zero LOS rate), $R \to 0$ ki limiting range tak jaate hain, aur ek word problem aur ek exam twist ke saath khatam karte hain. Har example se pehle: Forecast — answer cover karo aur guess karo. Baad mein: Verify — wapis plug karo aur units check karo.

Agar koi bhi symbol neeche ajeeb lage, toh woh parent note mein aur Line-of-Sight Geometry and Kinematics, Closing Velocity and Range Rate, aur Polar Coordinate Kinematics mein build kiya gaya tha.

Scenario matrix

Har worked example ek row ko target karta hai. Cell column woh label hai jo aapko har [!example] mein dikhega.

Cell	Case class	Yeh kya stress-test karta hai
A	Baseline: $V_{c} > 0$ , $\dot{λ} > 0$ , $N$ standard	Plain substitution, units, $g$ -conversion
B	Sign flip: $\dot{λ} < 0$ (LOS clockwise)	$a_{c}$ ka direction
C	Sign flip: $V_{c} < 0$ (opening range)	PN ek receding target par — kya toot-ta hai
D	Degenerate: $\dot{λ} = 0$ exactly	Collision-course fixed point
E	Degenerate: $\dot{R} = 0$ (co-orbiting, $V_{c} = 0$ )	Zero closing speed — koi command nahi, koi hit nahi
F	Limiting: $R \to 0$ with $\dot{λ} \propto R^{N - 2}$	Terminal behaviour, boundary $N = 2$
G	Word problem: side-window / crossing target	Raw kinematics se $\dot{λ}$ padhna
H	Exam twist: $N$ (ya $\dot{λ}$ ) ke liye ulta solve karna	Ek actuator cap ke under law ko invert karna

Recall Stability boundary kaunsa cell hai?

Cell F ::: $N = 2$ par exponent $N - 2 = 0$ hai, toh $\dot{λ} \propto R^{0} =$ constant — LOS rate kabhi decay nahi karta. Convergence ke liye $N > 2$ chahiye.

Cell A — Baseline substitution

Worked example A ·

$V_{c} > 0, \dot{λ} > 0$ , standard $N$ Ek missile $V_{c} = 800 m/s$ se close karta hai, LOS rate $\dot{λ} = 0.015 rad/s$ measure karta hai, aur $N = 4$ fly karta hai. Command $a_{c}$ kya hai? $g$ mein express karo ( $g = 9.81 m/s^{2}$ ).

Forecast: guess karo ki yeh gentle hai ya violent maneuver (10 g se kam ya zyada).

Law likhein: $a_{c} = N V_{c} \dot{λ}$ . Yeh step kyun? PN ek pure product hai — ek baar teeno inputs mil jayein toh koi calculus nahi chahiye.
Multiply karo: $a_{c} = 4 \times 800 \times 0.015 = 48 m/s^{2}$ . Yeh step kyun? Order matter nahi karta; yeh number raw lateral demand hai.
$g$ mein convert karo: $48/9.81 = 4.89 g$ . Yeh step kyun? Missile ke structural aur actuator limits $g$ mein quote kiye jaate hain, isliye yeh woh number hai jo ek engineer sanity-check karta hai.

Verify: units hain $(dimensionless) \cdot (m/s) \cdot (1/s) = m/s^{2}$ ✓. Lagbhag $4.9 g$ ek modest, realistic command hai — ek typical $\sim 30 g$ airframe limit ke andar achhe se fit hota hai.

Cell B — LOS doosri taraf rotate kar raha hai

Worked example B ·

$\dot{λ} < 0$ (sightline clockwise swing kar raha hai) Wohi missile: $V_{c} = 800 m/s$ , $N = 4$ , lekin ab $\dot{λ} = - 0.015 rad/s$ . $a_{c}$ find karo aur batao missile kis taraf turn karta hai.

Forecast: guess karo ki magnitude Cell A ke mukable badlegi ya nahi, aur sign predict karo.

Substitute karo: $a_{c} = 4 \times 800 \times (- 0.015) = - 48 m/s^{2}$ . Yeh step kyun? Law $\dot{λ}$ mein linear hai, toh iska sign flip karne se $a_{c}$ flip hota hai lekin magnitude Cell A ke identical rehti hai.
Sign interpret karo: $a_{c} < 0$ matlab decreasing $λ$ ki direction mein acceleration. Yeh step kyun? Ek negative $\dot{λ}$ matlab sightline chote $λ$ ki taraf drift kar rahi hai; us drift ko null karne ke liye missile ko usi taraf push karna padega jis taraf sightline ja rahi hai — yaani chote $λ$ ki taraf bhi. Dekho Feedback Control — Nulling an Error Signal: sign hamesha drift ki growth ko oppose karta hai, $\dot{λ} \to 0$ drive karte hue.

Verify: $∣ a_{c} ∣ = 48 m/s^{2} = 4.89 g$ , magnitude Cell A ke identical ✓. Sign opposite hai ✓ — PN LOS-direction reversal ke under symmetric hai, bilkul wahi jo ek negative-feedback nuller ko hona chahiye.

Cell C — Opening range ( $V_{c} < 0$ )

Worked example C · Range

badh rahi hai — kya PN ab bhi valid hai? Target door ja raha hai: $\dot{R} = + 300 m/s$ toh $V_{c} = - \dot{R} = - 300 m/s$ . LOS rate $\dot{λ} = 0.02 rad/s$ , $N = 4$ . $a_{c}$ compute karo aur interpret karo.

Forecast: kya command $\dot{λ}$ ko null karega, ya aur bura kar dega?

Substitute karo: $a_{c} = 4 \times (- 300) \times 0.02 = - 24 m/s^{2}$ . Yeh step kyun? Formula negative $V_{c}$ khushi se accept karta hai; command ka sign flip ho jaata hai.
Stability exponent check karo. Derivation se $\dot{λ} \propto R^{N - 2}$ , lekin woh assume karta tha ki $R$ simat raha hai. $\dot{R} > 0$ ke saath, $R$ badh raha hai, toh $R^{N - 2}$ $N > 2$ ke liye badh-ta hai. Yeh step kyun? Convergence $R \to 0$ par depend karta tha. Agar range khulti hai, toh wohi algebra predict karta hai ki $\dot{λ}$ badh-ta jaayega — PN ek receding target par diverge karta hai.

Verify: $∣ a_{c} ∣ = 24 m/s^{2} = 2.45 g$ (finite, toh law compute karta hai) lekin physics warn karta hai: PN ek terminal, closing-geometry law hai. Agar $V_{c} < 0$ hai toh aap intercept nahi kar sakte — dekho Closing Velocity and Range Rate. Units: $(- m/s) (1/s) = m/s^{2}$ ✓.

Cell D — Perfect collision course ( $\dot{λ} = 0$ )

Worked example D · Zero LOS rate — fixed point

$V_{c} = 1000 m/s$ , $N = 5$ , $\dot{λ} = 0$ . Command kya hai?

Forecast: zero, ya kuch residual?

Substitute karo: $a_{c} = 5 \times 1000 \times 0 = 0 m/s^{2}$ . Yeh step kyun? $\dot{λ}$ sab kuch multiply karta hai; jab yeh exactly zero hota hai toh poora product vanish ho jaata hai.
Interpret karo: koi command nahi chahiye. Sightline constant bearing hold karta hai jabki $R$ simat-ta hai — derivation ke Step 2 se collision-course condition. Yeh step kyun? $\dot{λ} = 0$ poore loop ka goal state hai; ek baar reach ho jaaye, PN sahi se kuch nahi maangta, jo exactly stable hai.

Verify: $a_{c} = 0$ ✓. Yeh equilibrium hai: ek missile jo pehle se collision course par hai, sidha coast karta hai, koi lateral effort nahi burn karta. Koi bhi perturbation $\dot{λ} \neq = 0$ wapis laata hai aur command phir se appear hota hai — ek self-correcting fixed point.

Cell E — Zero closing speed ( $V_{c} = 0$ )

Worked example E · Co-orbiting target, range frozen

Ek missile aur target ek common point ke around circle karte hain toh $\dot{R} = 0$ , isliye $V_{c} = 0$ , phir bhi $\dot{λ} = 0.05 rad/s$ (sightline tezi se sweep kar raha hai). $N = 4$ . Command kya hai?

Forecast: bada command (fast LOS!) ya zero?

Substitute karo: $a_{c} = 4 \times 0 \times 0.05 = 0 m/s^{2}$ . Yeh step kyun? $V_{c}$ ek factor hai; zero closing speed command ko zero kar deta hai chahe sightline kitni bhi tezi se spin kare.
Interpret karo: PN kuch command nahi karta kyunki exploit karne ke liye koi closing geometry nahi hai. $V_{c} = 0$ ke saath range kabhi collapse nahi hoti, toh koi intercept hi nahi hai jis taraf steer kiya jaaye. Yeh step kyun? Yeh dikhata hai ki $\dot{λ} \neq = 0$ akela maneuver ka call nahi hai — PN, LOS rate ko weight karta hai is hisaab se ki gap actually kitni tezi se close ho raha hai. Ek bada $\dot{λ}$ with $V_{c} = 0$ ek miss hai jise aap PN se fix nahi kar sakte.

Verify: $a_{c} = 0$ ✓. Concrete justification seedha PN law se aata hai: $a_{c} = N V_{c} \dot{λ} = - N \dot{R} \dot{λ}$ , toh $\dot{R}$ (equivalently $V_{c}$ ) command ka ek multiplicative factor hai. $\dot{R} = 0$ set karo aur poora product zero ho jaata hai — loop gain $- N \dot{R}$ exactly vanish ho jaata hai. Yeh "koi vague physical need" nahi hai; yeh literally wahi factor hai jo boxed formula mein zero ho raha hai. Isliye practical seekers PN ko ek range-rate gate ke saath combine karte hain.

Cell F — Terminal limit aur $N = 2$ boundary

Worked example F ·

$R \to 0$ par residual LOS rate Do missiles ek hi $\dot{λ}_{0} = 0.04 rad/s$ se $R_{0} = 5000 m$ par start karte hain. Missile P $N = 3$ use karta hai, missile Q $N = 6$ use karta hai. $R = 500 m$ (ek-daswan range) par har ek ka $\dot{λ}$ find karo. Phir batao $N = 2$ kya deta.

Forecast: kaun sa missile $\dot{λ} = 0$ ke paas khatam hoga, aur kitne orders of magnitude se?

Derived scaling use karo $\dot{λ} = \dot{λ}_{0} (R / R_{0})^{N - 2}$ . Yeh step kyun? Derivation ke Step 4 ne $\ddot{λ} / \dot{λ} = (N - 2) \dot{R} / R$ ko integrate kiya $\dot{λ} \propto R^{N - 2}$ mein.
Ratio $R / R_{0} = 500/5000 = 0.1$ . Yeh step kyun? Sirf ratio enter karta hai, toh absolute units cancel ho jaate hain.
Missile P ( $N = 3$ ): $\dot{λ} = 0.04 \times 0. 1^{1} = 0.004 rad/s$ .
Missile Q ( $N = 6$ ): $\dot{λ} = 0.04 \times 0. 1^{4} = 4 \times 1 0^{- 6} rad/s$ . Yeh step kyun? Bada $N$ ⇒ bada positive exponent ⇒ $\dot{λ}$ ka kaafi zyada steep collapse jab $R$ simat-ta hai.
Boundary $N = 2$ : exponent $= 0$ , toh $\dot{λ} = 0.04 \times 0. 1^{0} = 0.04 rad/s$ — unchanged. Yeh step kyun? Yeh stability edge hai: $N = 2$ par LOS rate kabhi decay nahi karta, miss guaranteed hai.

Neeche ka figure padho jab aap yeh numbers check karo: horizontal axis range $R$ hai jo right-to-left chal rahi hai ( $R \to 0$ par intercept approach karta hai), vertical axis $\dot{λ}$ hai. Magenta ( $N = 3$ ) aur violet ( $N = 6$ ) curves ko $R = 500 m$ par dotted line tak trace karo — violet curve pehle se magenta se teen decades neeche ja chuki hai, bilkul wahi $1 0^{3}$ gap jo aapne compute kiya. Orange dashed line ( $N = 2$ ) dekho: yeh $0.04$ par flat rehti hai, kabhi decay nahi hoti — yeh wahi failure hai jis ke baare mein boundary warn karti hai.

Verify: P deta hai $0.004$ , Q deta hai $4 \times 1 0^{- 6}$ — Q $1 0^{3}$ times chota hai ✓ (exponent difference $6 - 3 = 3$ ). $N = 2$ $0.04$ par rehta hai ✓. Terminal $\dot{λ} \to 0$ ke liye $N > 2$ chahiye; dekho Navigation Constant Selection and Actuator Limits ki kyun hum sirf bada $N$ nahi pick karte.

Cell G — Word problem: side window se crossing target

Worked example G · Raw kinematics se

$\dot{λ}$ padhna Ek missile seedha fly karta hai. Ek target uski path cross karta hai transverse (across-LOS) relative speed $V_{⊥} = 120 m/s$ ke saath us instant par jab range $R = 3000 m$ hai, $V_{c} = 850 m/s$ se close karte hue. $N = 4$ ke saath, PN abhi kaunsi lateral acceleration command karta hai?

Forecast: LOS rate directly diya nahi gaya — guess karo ki $a_{c}$ kuch $g$ ke paas land karega ya tens of $g$ ke.

Transverse speed se $\dot{λ}$ recover karo: $V_{⊥} = R \dot{λ} \Rightarrow \dot{λ} = V_{⊥} / R = 120/3000 = 0.04 rad/s$ . Yeh step kyun? Derivation ke Step 1 se across-LOS relative speed exactly $R \dot{λ}$ hai; ek seeker aksar $V_{⊥}$ (ya $\dot{λ}$ ) measure karta hai bajaaye aapko $\dot{λ}$ seedha dene ke. Dekho Polar Coordinate Kinematics.
Ab law: $a_{c} = N V_{c} \dot{λ} = 4 \times 850 \times 0.04 = 136 m/s^{2}$ . Yeh step kyun? Geometry ko usi product ke zariye command mein convert karo.
$g$ mein: $136/9.81 = 13.9 g$ . Yeh step kyun? Yeh woh number hai jo airframe ke against check hota hai.

Neeche ka figure padho dekho ki har number geometrically kahan hai: navy line $R = 3000 m$ ki length ki LOS hai angle $λ$ par; target par magenta arrow across-LOS speed $V_{⊥} = 120 m/s$ hai — iska length $R$ se divide karna hi $\dot{λ}$ hai (Step 1); sightline ke along orange arrow closing speed $V_{c} = 850 m/s$ hai (Step 2 ka second factor). Dono arrows perpendicular hain kyunki ek "along LOS" hai aur doosra "across LOS" — woh decomposition hi LOS frame mein kaam karne ka poora point hai.

Verify: $\dot{λ} = 0.04 rad/s$ ✓, $a_{c} = 136 m/s^{2} = 13.9 g$ ✓. "Constant bearing" intuition ko cross-check karo: ek fast side-drift ( $120 m/s$ across ek $3 km$ line) ek real miss threat hai, toh PN sahi se ek stiff $\sim 14 g$ lead demand karta hai. Ek pure-pursuit missile baad mein aur zyada maangta, bahut der se.

Cell H — Exam twist: actuator cap ke under law ko invert karo

Worked example H · Sabse bada usable

$N$ find karo Ek missile ka airframe lateral acceleration ko $a_{c, m a x} = 200 m/s^{2}$ par cap karta hai. Flight ke shuruaat mein $V_{c} = 1000 m/s$ aur $\dot{λ} = 0.05 rad/s$ . Is instant par cap ke andar rehne wala sabse bada integer $N$ kya hai? Woh $N$ kaunsa residual $\dot{λ}$ -collapse exponent deta hai?

Forecast: guess karo ki standard $N = 4$ survive karta hai, ya clip ho jaata hai.

Command ko cap par set karo aur solve karo: $N_{m a x} = \frac{a _{c, m a x}}{V _{c} λ ˙} = \frac{200}{1000 \times 0.05} = 4.0$ . Yeh step kyun? Law $N$ mein linear hai, toh ise invert karne se woh gain isolate hota hai jo actuator is operating point par afford kar sakta hai.
$\leq 4.0$ ka sabse bada integer $N = 4$ hai. Yeh step kyun? Cap exceed nahi kar sakte, toh down round karo. $N = 5$ demand karega $250 m/s^{2} > 200$ — clipped, jo guarantee $\dot{λ} \propto R^{N - 2}$ ko corrupt karta hai (ek saturated actuator linear ODE ko tod deta hai).
Convergence exponent: $N - 2 = 4 - 2 = 2$ , toh $\dot{λ} \propto R^{2}$ terminally. Yeh step kyun? Confirm karta hai ki $N = 4$ Cell F ki $N = 2$ boundary se safely upar hai — yeh converge karta hai.

Verify: $N = 4$ deta hai $a_{c} = 4 \times 1000 \times 0.05 = 200 m/s^{2}$ , exactly cap ✓. $N = 5$ deta hai $250 m/s^{2} > 200$ ✓ clipped. Exponent $N - 2 = 2 > 0$ ✓ converges. Yeh Navigation Constant Selection and Actuator Limits mein real design tension hai: bada $N$ $\dot{λ}$ ko tezi se null karta hai lekin fins ko saturate karta hai; Zero-Effort-Miss and Augmented Proportional Navigation mein augmented forms usi limit ko maneuvering targets ke against address karte hain.

Recall Cell E ka result aur uska lesson

$V_{c} = 0$ ke saath, $a_{c} = ?$ aur kyun ::: $a_{c} = 0$ ; kyunki $a_{c} = N V_{c} \dot{λ}$ mein $V_{c}$ ek literal multiplicative factor hai, zero closing speed zero command force karta hai chahe sightline kitni bhi tezi se spin kare.

Recall Cell F: ek-daswan range par

$N = 6$ ke liye $\dot{λ}$ ratio $\dot{λ} / \dot{λ}_{0} = ?$ ::: $(0.1)^{6 - 2} = 1 0^{- 4}$ .