Proportional navigation guidance — N·V_c·λ̇, derivation

WHAT is being guided?

WHY this shape? We will derive that the missile's lateral acceleration must be proportional to $V_c\dot\lambda$ if we want $\dot\lambda \to 0$ . Everything below builds this from geometry — nothing is assumed.

The collision-course idea (build the intuition first)

Figure — Proportional navigation guidance — N·V_c·λ̇, derivation

HOW — derivation from first principles

We work in the plane. Place the missile at the origin and describe the target by polar coordinates $(R, \lambda)$ relative to the missile.

Step 1 — Relative velocity components in the LOS frame

The relative position vector (missile→target) has length $R$ at angle $\lambda$ . Its time derivative gives two components:

\underbrace{\dot R}_{\text{along LOS}}, \qquad \underbrace{R\dot\lambda}_{\text{across LOS}}

Why this step? In polar coordinates any velocity splits into a radial part $\dot R\,\hat r$ and a transverse part $R\dot\lambda\,\hat\theta$ . The transverse relative speed is exactly $V_\perp = R\dot\lambda$ . That transverse component is the "miss-building" motion — it is what makes the sightline rotate.

Step 2 — What "on a collision course" means

If the missile and target are on a perfect collision course, the LOS angle is constant: $\dot\lambda = 0 \;\;\Longrightarrow\;\; V_\perp = R\dot\lambda = 0.$

Why this step? With zero transverse relative velocity, the target closes straight down the sightline — a guaranteed intercept (as $R\to 0$ ). So the goal of guidance is to force $\dot\lambda \to 0$ .

Step 3 — LOS dynamics (the full polar acceleration, with Coriolis)

We need how $\dot\lambda$ itself evolves. In polar coordinates the transverse component of the relative acceleration is not just $R\ddot\lambda$ — it carries the Coriolis term $2\dot R\dot\lambda$ :

a_\perp = R\ddot\lambda + 2\dot R\dot\lambda.

Why this step? Differentiating the transverse velocity $V_\perp = R\dot\lambda$ once gives $\dot V_\perp = \dot R\dot\lambda + R\ddot\lambda$ , but $\dot V_\perp$ is not the physical transverse acceleration — the transverse unit vector $\hat\theta$ is itself rotating, and its rotation contributes a second $\dot R\dot\lambda$ . Adding both gives the standard polar result $a_\perp = R\ddot\lambda + 2\dot R\dot\lambda$ . Missing the factor of 2 is the classic error.

Now set $a_\perp$ equal to the net transverse acceleration (target minus missile). For a non-maneuvering target ( $a_{\perp,\text{target}}=0$ ) with missile lateral command $a_c$ perpendicular to the LOS, $a_\perp = -a_c$ :

R\ddot\lambda + 2\dot R\dot\lambda = -a_c \;\;\Longrightarrow\;\; \ddot\lambda = \frac{-a_c - 2\dot R\,\dot\lambda}{R}.

Why this step? This is the correct LOS dynamics equation — it tells us how our command $a_c$ feeds back onto the very quantity $\dot\lambda$ we want to kill.

Step 4 — Choose $a_c$ so $\dot\lambda$ decays

Substitute the PN law $a_c = N V_c \dot\lambda = -N\dot R\,\dot\lambda$ (since $V_c=-\dot R$ ):

\ddot\lambda = \frac{N\dot R\,\dot\lambda - 2\dot R\,\dot\lambda}{R} = \frac{\dot R\,\dot\lambda\,(N-2)}{R}.

Rearrange into a first-order ODE in $\dot\lambda$ :

\frac{\ddot\lambda}{\dot\lambda} = (N-2)\frac{\dot R}{R}.

Why this step? Integrating both sides:

\ln\dot\lambda = (N-2)\ln R + \text{const} \;\;\Longrightarrow\;\; \dot\lambda \propto R^{\,N-2}.

Step 5 — Read off the stability condition

As the intercept proceeds, $R \to 0$ . Since $\dot\lambda \propto R^{N-2}$ :

N > 2 \;\Longrightarrow\; \dot\lambda \to 0 \text{ as } R\to 0. \quad\checkmark

Why this step? For $N>2$ , the exponent $N-2>0$ , so as range collapses the LOS rate is forced to zero — the sightline stops rotating exactly as we reach the target. That is the collision course from Step 2. In practice $N=3\text{–}5$ (all safely $>2$ ) balances fast $\dot\lambda$ nulling against actuator effort and noise. This closes the loop: the form $N V_c\dot\lambda$ is not a guess — it is the choice that makes $\dot\lambda$ collapse.

Worked examples

Worked example Example 2 — Why a bigger

$N$ ends the engagement "cleaner" Two missiles start with the same $\dot\lambda_0$ at range $R_0$ . One uses $N=3$ , the other $N=5$ . Compare $\dot\lambda$ at $R = R_0/10$ .

Using $\dot\lambda \propto R^{N-2}$ :

$N=3$ : $\dot\lambda/\dot\lambda_0 = (1/10)^{1} = 0.1.$
$N=5$ : $\dot\lambda/\dot\lambda_0 = (1/10)^{3} = 10^{-3}.$

Why this step? Higher $N$ drives the LOS rate to zero far faster as range closes, so the terminal geometry is nearly perfect (tiny residual $\dot\lambda$ = tiny miss). But early in flight higher $N$ demands more acceleration — the design trade-off. (Note: with $N=2$ the exponent is $0$ , so $\dot\lambda$ would not decay at all — right at the stability boundary.)

Common mistakes (steel-manned)

Recall Feynman: explain it to a 12-year-old

Imagine you're on a bike trying to bump into a friend who's also riding. Don't stare at your friend and steer straight at them — you'll always be one step behind. Instead, watch where your friend sits against the background (a tree, a fence). If your friend keeps sliding sideways against the background, you're going to miss — so steer to make them stop sliding. When your friend stays glued to the same spot on the background and just gets bigger and bigger... bonk! — you meet. A guided missile does exactly this: it turns harder when its target slides faster against the sky, and the amount it turns is $N$ times how fast the sky-slide ( $\dot\lambda$ ) is happening times how fast the gap is closing ( $V_c$ ).

Active recall

What does a constant LOS angle with decreasing range imply?

A collision course — the target closes straight down the sightline, guaranteeing intercept.

State the proportional navigation guidance law.

a_c = N\,V_c\,\dot\lambda

(commanded lateral accel = nav constant × closing velocity × LOS rate).

What is the transverse relative velocity in terms of

R

and

\dot\lambda

V_\perp = R\dot\lambda

What is the full transverse component of the relative acceleration in polar form?

a_\perp = R\ddot\lambda + 2\dot R\dot\lambda

(the

2\dot R\dot\lambda

is the Coriolis term).

Starting from the correct LOS dynamics, what proportionality does PN produce for

\dot\lambda

\dot\lambda \propto R^{\,N-2}

What condition on

N

makes

\dot\lambda \to 0

R\to 0

N>2

Typical practical range of the navigation constant

N

3 \le N \le 5

(all safely greater than 2).

Why is the threshold

N>2

and not

N>1

Because the transverse dynamics includes the Coriolis term

2\dot R\dot\lambda

; the factor of 2 shifts the exponent to

N-2

Why not use huge

N

It amplifies measurement noise in

\dot\lambda

and saturates actuators early, despite faster LOS-rate nulling.

Relation between closing velocity

V_c

and range rate

\dot R

V_c = -\dot R

(positive when range decreases).

Why does the command

a_c

fade to zero near intercept (for

N>2

Because

\dot\lambda\to 0

, and

a_c \propto \dot\lambda

How does PN differ from pure pursuit?

PN nulls LOS rotation (aims at future collision point / leads target); pure pursuit points straight at current target position and lags.

Connections

Line-of-Sight Geometry and Kinematics
Coriolis Term in Polar Coordinate Acceleration
Pure Pursuit vs Proportional Navigation
Closing Velocity and Range Rate
Navigation Constant Selection and Actuator Limits
Zero-Effort-Miss and Augmented Proportional Navigation
Feedback Control — Nulling an Error Signal
Polar Coordinate Kinematics

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Proportional Navigation ka core idea bahut simple hai: target ko chase mat karo, uski line of sight (LOS) ki rotation ko rok do. Jab tum kisi moving target ki taraf sightline dekhte ho, agar wo sightline ka angle $\lambda$ constant rehta hai aur distance $R$ kam hoti ja rahi hai, to samajh lo collision pakka hai — ye "constant bearing, decreasing range" wala rule hai. Isliye guidance ka goal hai: $\dot\lambda$ (LOS rotation rate) ko zero karna.

Formula hai $a_c = N\,V_c\,\dot\lambda$ . Yahan $V_c$ closing velocity hai (gap kitni tezi se chhoti ho rahi), $\dot\lambda$ sightline kitni tezi se ghoom rahi, aur $N$ navigation constant (usually 3 se 5). Derivation ka main trick: polar coordinates me transverse relative acceleration $R\ddot\lambda + 2\dot R\dot\lambda$ hota hai — dhyaan do, us Coriolis wale 2 ko mat bhoolna, warna galat answer aayega. LOS dynamics nikalne par milta hai $\ddot\lambda = (-a_c - 2\dot R\dot\lambda)/R$ . Ab PN law daal do to $\dot\lambda \propto R^{N-2}$ aata hai. Matlab jaise-jaise $R\to0$ , agar $N>2$ ho to $\dot\lambda\to0$ — sightline apne aap freeze ho jati hai aur hit ho jata hai.

Kyun important hai? Pure pursuit (seedha target pe aim karna) me missile hamesha peeche reh jati hai aur end me bahut zyada turn chahiye. PN automatically lead leti hai, future collision point pe jaati hai. Aur ek aur mast baat — intercept ke waqt $\dot\lambda\to0$ hone se command $a_c$ bhi apne aap kam ho jata hai, missile smoothly milti hai target se. Yaad rakho: stability ke liye $N$ , 2 se bada hona chahiye (Coriolis 2 ki wajah se), isliye 3-5 use hota hai. Yehi reason hai ye method aaj tak real missiles me use hota hai.

Proportional navigation guidance — N·V_c·λ̇, derivation

WHAT is being guided?

The collision-course idea (build the intuition first)

HOW — derivation from first principles

Step 1 — Relative velocity components in the LOS frame

Step 2 — What "on a collision course" means

Step 3 — LOS dynamics (the full polar acceleration, with Coriolis)

Step 4 — Choose $a_c$ so $\dot\lambda$ decays

Step 5 — Read off the stability condition

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections