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

This page drills the PN law until no case can surprise you. We plug numbers, flip every sign, break the inputs (zero range rate, zero LOS rate), push to the limiting range $R \to 0$ , and finish with a word problem and an exam twist. Before each example: Forecast — cover the answer and guess. After: Verify — plug back and check units.

If any symbol below feels unfamiliar, it was built in the parent note and in Line-of-Sight Geometry and Kinematics, Closing Velocity and Range Rate, and Polar Coordinate Kinematics.

The scenario matrix

Every worked example targets one row. The cell column is the label you'll see in each [!example].

Cell	Case class	What it stress-tests
A	Baseline: $V_{c} > 0$ , $\dot{λ} > 0$ , $N$ standard	Plain substitution, units, $g$ -conversion
B	Sign flip: $\dot{λ} < 0$ (LOS clockwise)	Direction of $a_{c}$
C	Sign flip: $V_{c} < 0$ (opening range)	PN on a receding target — what breaks
D	Degenerate: $\dot{λ} = 0$ exactly	The collision-course fixed point
E	Degenerate: $\dot{R} = 0$ (co-orbiting, $V_{c} = 0$ )	Zero closing speed — no command, no hit
F	Limiting: $R \to 0$ with $\dot{λ} \propto R^{N - 2}$	Terminal behaviour, boundary $N = 2$
G	Word problem: side-window / crossing target	Reading $\dot{λ}$ from raw kinematics
H	Exam twist: solve for $N$ (or $\dot{λ}$ ) backwards	Inverting the law under an actuator cap

Recall Which cell is the stability boundary?

Cell F ::: at $N = 2$ the exponent $N - 2 = 0$ , so $\dot{λ} \propto R^{0} =$ constant — the LOS rate never decays. Convergence needs $N > 2$ .

Cell A — Baseline substitution

Worked example A ·

$V_{c} > 0, \dot{λ} > 0$ , standard $N$ A missile closes at $V_{c} = 800 m/s$ , measures LOS rate $\dot{λ} = 0.015 rad/s$ , and flies $N = 4$ . Command $a_{c}$ ? Express in $g$ ( $g = 9.81 m/s^{2}$ ).

Forecast: guess whether this is a gentle or violent maneuver (under or over 10 g).

Write the law: $a_{c} = N V_{c} \dot{λ}$ . Why this step? PN is a pure product — no calculus needed once you have the three inputs live.
Multiply: $a_{c} = 4 \times 800 \times 0.015 = 48 m/s^{2}$ . Why this step? Order doesn't matter; the number is the raw lateral demand.
Convert to $g$ : $48/9.81 = 4.89 g$ . Why this step? Missile structural and actuator limits are quoted in $g$ , so this is the number an engineer sanity-checks against.

Verify: units are $(dimensionless) \cdot (m/s) \cdot (1/s) = m/s^{2}$ ✓. About $4.9 g$ is a modest, realistic command — well inside a typical $\sim 30 g$ airframe limit.

Cell B — LOS rotating the other way

Worked example B ·

$\dot{λ} < 0$ (sightline swinging clockwise) Same missile: $V_{c} = 800 m/s$ , $N = 4$ , but now $\dot{λ} = - 0.015 rad/s$ . Find $a_{c}$ and say which way the missile turns.

Forecast: guess whether the magnitude changes versus Cell A, and predict the sign.

Substitute: $a_{c} = 4 \times 800 \times (- 0.015) = - 48 m/s^{2}$ . Why this step? The law is linear in $\dot{λ}$ , so flipping its sign flips $a_{c}$ but leaves the magnitude identical to Cell A.
Interpret the sign: $a_{c} < 0$ means acceleration in the direction of decreasing $λ$ . Why this step? A negative $\dot{λ}$ means the sightline is drifting toward smaller $λ$ ; to null that drift the missile must push the same way the sightline is going — i.e. also toward smaller $λ$ . See Feedback Control — Nulling an Error Signal: the sign always opposes the drift's growth, driving $\dot{λ} \to 0$ .

Verify: $∣ a_{c} ∣ = 48 m/s^{2} = 4.89 g$ , identical magnitude to Cell A ✓. The sign is opposite ✓ — PN is symmetric under LOS-direction reversal, exactly what a negative-feedback nuller should be.

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

Worked example C · Range

growing — is PN still valid? The target is pulling away: $\dot{R} = + 300 m/s$ so $V_{c} = - \dot{R} = - 300 m/s$ . LOS rate $\dot{λ} = 0.02 rad/s$ , $N = 4$ . Compute $a_{c}$ and interpret.

Forecast: will the command still null $\dot{λ}$ , or make it worse?

Substitute: $a_{c} = 4 \times (- 300) \times 0.02 = - 24 m/s^{2}$ . Why this step? The formula happily accepts a negative $V_{c}$ ; the sign of the command flips.
Check the stability exponent. From the derivation $\dot{λ} \propto R^{N - 2}$ , but that assumed $R$ shrinking. With $\dot{R} > 0$ , $R$ grows, so $R^{N - 2}$ grows for $N > 2$ . Why this step? Convergence relied on $R \to 0$ . If range opens, the same algebra predicts $\dot{λ}$ increasing — PN diverges on a receding target.

Verify: $∣ a_{c} ∣ = 24 m/s^{2} = 2.45 g$ (finite, so the law computes) but the physics warns you: PN is a terminal, closing-geometry law. If $V_{c} < 0$ you cannot intercept — see Closing Velocity and Range Rate. Units: $(- m/s) (1/s) = m/s^{2}$ ✓.

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

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

Worked example E · Co-orbiting target, range frozen

A missile and target circle a common point so that $\dot{R} = 0$ , hence $V_{c} = 0$ , yet $\dot{λ} = 0.05 rad/s$ (sightline sweeping fast). $N = 4$ . Command?

Forecast: big command (fast LOS!) or zero?

Substitute: $a_{c} = 4 \times 0 \times 0.05 = 0 m/s^{2}$ . Why this step? $V_{c}$ is a factor; zero closing speed zeroes the command regardless of how fast the sightline spins.
Interpret: PN commands nothing because there is no closing geometry to exploit. With $V_{c} = 0$ the range never collapses, so no intercept exists to steer toward. Why this step? It shows $\dot{λ} \neq = 0$ alone is not a call to maneuver — PN weights LOS rate by how fast the gap is actually closing. A large $\dot{λ}$ with $V_{c} = 0$ is a miss you cannot fix with PN.

Verify: $a_{c} = 0$ ✓. The concrete justification comes straight from the PN law: $a_{c} = N V_{c} \dot{λ} = - N \dot{R} \dot{λ}$ , so $\dot{R}$ (equivalently $V_{c}$ ) is a multiplicative factor of the command. Set $\dot{R} = 0$ and the whole product is zero — the loop gain $- N \dot{R}$ vanishes exactly. It is not "some vague physical need"; it is that literal factor in the boxed formula going to zero. This is why practical seekers combine PN with a range-rate gate.

Cell F — Terminal limit and the $N = 2$ boundary

Worked example F · Residual LOS rate as

$R \to 0$ Two missiles begin with the same $\dot{λ}_{0} = 0.04 rad/s$ at $R_{0} = 5000 m$ . Missile P uses $N = 3$ , missile Q uses $N = 6$ . Find each $\dot{λ}$ at $R = 500 m$ (one-tenth range). Then state what $N = 2$ would give.

Forecast: which missile ends nearer to $\dot{λ} = 0$ , and by how many orders of magnitude?

Use the derived scaling $\dot{λ} = \dot{λ}_{0} (R / R_{0})^{N - 2}$ . Why this step? Step 4 of the derivation integrated $\ddot{λ} / \dot{λ} = (N - 2) \dot{R} / R$ to $\dot{λ} \propto R^{N - 2}$ .
Ratio $R / R_{0} = 500/5000 = 0.1$ . Why this step? Only the ratio enters, so absolute units cancel.
Missile P ( $N = 3$ ): $\dot{λ} = 0.04 \times 0. 1^{1} = 0.004 rad/s$ .
Missile Q ( $N = 6$ ): $\dot{λ} = 0.04 \times 0. 1^{4} = 4 \times 1 0^{- 6} rad/s$ . Why this step? Bigger $N$ ⇒ bigger positive exponent ⇒ far steeper collapse of $\dot{λ}$ as $R$ shrinks.
Boundary $N = 2$ : exponent $= 0$ , so $\dot{λ} = 0.04 \times 0. 1^{0} = 0.04 rad/s$ — unchanged. Why this step? This is the stability edge: at $N = 2$ the LOS rate never decays, guaranteeing a miss.

Read the figure below as you check these numbers: the horizontal axis is range $R$ running right-to-left (intercept approaches as $R \to 0$ ), the vertical axis is $\dot{λ}$ . Trace the magenta ( $N = 3$ ) and violet ( $N = 6$ ) curves down to the dotted line at $R = 500 m$ — the violet curve has already plunged three decades lower than magenta, exactly the $1 0^{3}$ gap you computed. Watch the orange dashed line ( $N = 2$ ): it stays flat at $0.04$ , never decaying — that is the failure the boundary warns about.

Verify: P gives $0.004$ , Q gives $4 \times 1 0^{- 6}$ — Q is $1 0^{3}$ smaller ✓ (exponent difference $6 - 3 = 3$ ). $N = 2$ stays at $0.04$ ✓. Terminal $\dot{λ} \to 0$ demands $N > 2$ ; see Navigation Constant Selection and Actuator Limits for why we don't just pick huge $N$ .

Cell G — Word problem: crossing target from the side window

Worked example G · Reading

$\dot{λ}$ from raw kinematics A missile flies straight. A target crosses its path with transverse (across-LOS) relative speed $V_{⊥} = 120 m/s$ at the instant range is $R = 3000 m$ , closing at $V_{c} = 850 m/s$ . With $N = 4$ , what lateral acceleration does PN command right now?

Forecast: the LOS rate isn't given directly — guess whether $a_{c}$ lands near a few $g$ or tens of $g$ .

Recover $\dot{λ}$ from the transverse speed: $V_{⊥} = R \dot{λ} \Rightarrow \dot{λ} = V_{⊥} / R = 120/3000 = 0.04 rad/s$ . Why this step? From Step 1 of the derivation the across-LOS relative speed is exactly $R \dot{λ}$ ; a seeker often measures $V_{⊥}$ (or $\dot{λ}$ ) rather than handing you $\dot{λ}$ on a plate. See Polar Coordinate Kinematics.
Now the law: $a_{c} = N V_{c} \dot{λ} = 4 \times 850 \times 0.04 = 136 m/s^{2}$ . Why this step? Convert the geometry into a command via the same product.
In $g$ : $136/9.81 = 13.9 g$ . Why this step? This is the number checked against the airframe.

Read the figure below to see where each number lives geometrically: the navy line is the LOS of length $R = 3000 m$ at angle $λ$ ; the magenta arrow at the target is the across-LOS speed $V_{⊥} = 120 m/s$ — its length over $R$ is $\dot{λ}$ (Step 1); the orange arrow along the sightline is the closing speed $V_{c} = 850 m/s$ (Step 2's second factor). The two arrows are perpendicular because one is "along LOS" and the other "across LOS" — that decomposition is the whole point of working in the LOS frame.

Verify: $\dot{λ} = 0.04 rad/s$ ✓, $a_{c} = 136 m/s^{2} = 13.9 g$ ✓. Cross-check the "constant bearing" intuition: a fast side-drift ( $120 m/s$ across a $3 km$ line) is a real miss threat, so PN correctly demands a stiff $\sim 14 g$ lead. A pure-pursuit missile would ask for even more later, too late.

Cell H — Exam twist: invert the law under an actuator cap

Worked example H · Solve for the largest usable

$N$ A missile's airframe caps lateral acceleration at $a_{c, m a x} = 200 m/s^{2}$ . Early in flight $V_{c} = 1000 m/s$ and $\dot{λ} = 0.05 rad/s$ . What is the largest integer $N$ that stays within the cap at this instant? What residual $\dot{λ}$ -collapse exponent does that $N$ give?

Forecast: guess whether the standard $N = 4$ survives, or gets clipped.

Set the command to the cap and solve: $N_{m a x} = \frac{a _{c, m a x}}{V _{c} λ ˙} = \frac{200}{1000 \times 0.05} = 4.0$ . Why this step? The law is linear in $N$ , so inverting it isolates the gain the actuator can afford at this operating point.
Largest integer $\leq 4.0$ is $N = 4$ . Why this step? You cannot exceed the cap, so round down. $N = 5$ would demand $250 m/s^{2} > 200$ — clipped, which corrupts the guarantee $\dot{λ} \propto R^{N - 2}$ (a saturated actuator breaks the linear ODE).
Convergence exponent: $N - 2 = 4 - 2 = 2$ , so $\dot{λ} \propto R^{2}$ terminally. Why this step? Confirms $N = 4$ is safely above the $N = 2$ boundary from Cell F — it converges.

Verify: $N = 4$ gives $a_{c} = 4 \times 1000 \times 0.05 = 200 m/s^{2}$ , exactly the cap ✓. $N = 5$ gives $250 m/s^{2} > 200$ ✓ clipped. Exponent $N - 2 = 2 > 0$ ✓ converges. This is the real design tension in Navigation Constant Selection and Actuator Limits: bigger $N$ nulls $\dot{λ}$ faster but saturates the fins; the augmented forms in Zero-Effort-Miss and Augmented Proportional Navigation address the same limit against maneuvering targets.

Recall Cell E result and its lesson

With $V_{c} = 0$ , $a_{c} = ?$ and why ::: $a_{c} = 0$ ; because $a_{c} = N V_{c} \dot{λ}$ has $V_{c}$ as a literal multiplicative factor, zero closing speed forces zero command no matter how fast the sightline spins.

Recall Cell F:

$\dot{λ}$ ratio for $N = 6$ at one-tenth range $\dot{λ} / \dot{λ}_{0} = ?$ ::: $(0.1)^{6 - 2} = 1 0^{- 4}$ .