Proportional navigation guidance — N·V_c·λ̇, derivation: Question bank

Recall the core law before you start:

Spot the error

"Transverse relative acceleration is

\dot{V}_{⊥} = \dot{R} \dot{λ} + R \ddot{λ}

Error —

\dot{V}_{⊥}

ignores that the transverse unit vector

\hat{θ}

rotates; the true value is

R \ddot{λ} + 2 \dot{R} \dot{λ}

, with a second

\dot{R} \dot{λ}

from the rotating basis. See Coriolis Term in Polar Coordinate Acceleration.

"Since

V_{c} = - \dot{R}

and we are closing,

\dot{R} > 0

Error — closing means the range shrinks, so

\dot{R} < 0

and hence

V_{c} = - \dot{R} > 0

; the sign flip is the whole point of defining

V_{c}

. See Closing Velocity and Range Rate.

"PN sets

a_{c}

along the line of sight to push the missile toward the target."

Error —

a_{c}

is perpendicular (lateral) to the LOS; a purely radial push cannot change

\dot{λ}

, which is the quantity PN exists to control.

"From

\ddot{λ} / \dot{λ} = (N - 2) \dot{R} / R

we integrate to get

\dot{λ} \propto R^{N}

Error — integrating

ln \dot{λ} = (N - 2) ln R + C

gives

\dot{λ} \propto R^{N - 2}

, not

R^{N}

; the exponent is

N - 2

, which is exactly what forces the

N > 2

condition.

"For a non-maneuvering target the transverse relative acceleration equals

+ a_{c}

Error — the missile's lateral command opposes the transverse motion, so

a_{⊥} = - a_{c}

; the sign is what makes the feedback negative and stabilizing.

"PN is open-loop: it fires a preset acceleration profile."

Error — PN is closed-loop feedback that continuously nulls the error signal

\dot{λ}

;

a_{c}

is recomputed every instant from the live LOS rate. See Feedback Control — Nulling an Error Signal.

Why questions

Why is

\dot{λ}

(not

λ

itself) the error signal PN nulls?

Because a constant LOS angle already means collision course; it is the rotation

\dot{λ}

that builds miss distance, so driving the rate — not the angle — to zero is what guarantees intercept.

Why does the transverse velocity

R \dot{λ}

measure "miss-building" motion?

It is the component of relative velocity across the sightline; any nonzero crossing speed swings the LOS and, unchecked, carries the target off the collision line before

R

reaches zero.

Why does PN scale the command by

V_{c}

rather than commanding a fixed turn for a given

\dot{λ}

Because the same

\dot{λ}

at higher closing speed corresponds to a faster-developing miss and less time to correct, so the required acceleration genuinely grows with

V_{c}

— the law encodes that urgency.

Why does the stability condition come out as

N > 2

and not

N > 0

The Coriolis "2" appears in the LOS dynamics, so the effective decay exponent is

N - 2

; only when it exceeds zero does

\dot{λ} \propto R^{N - 2} \to 0

R \to 0

Why does PN "lead" the target automatically without predicting its path?

By nulling

\dot{λ}

it steers toward wherever the sightline points steadily, which is the future collision point; the geometry does the prediction, no explicit forecast needed. See Zero-Effort-Miss and Augmented Proportional Navigation.

Why can the same

\dot{λ}

demand very different accelerations across an engagement?

Because

a_{c} = N V_{c} \dot{λ}

depends on

V_{c}

, which changes with geometry and speed; a modest LOS rate late in flight with high

V_{c}

can command large lateral

g

Edge cases

What does PN command when

\dot{λ} > 0

and the missile is closing (

V_{c} > 0

a_{c} = N V_{c} \dot{λ} > 0

— a lateral acceleration in the direction of increasing

λ

, turning the missile to catch the counterclockwise-drifting sightline and cancel the drift.

What happens to

a_{c}

if the range rate reverses so the target is opening (

V_{c} < 0

The sign of

a_{c}

flips, so the command reverses direction; PN is only designed for closing geometry, and an opening engagement means the intercept is failing.

R \to 0

near intercept, what does the term

2 \dot{R} \dot{λ} / R

do, and why isn't

a_{c}

infinite?

The

1/ R

makes the LOS dynamics stiff, but because

N > 2

forces

\dot{λ} \to 0

faster than

R

shrinks, the product

V_{c} \dot{λ}

stays bounded and

a_{c}

does not blow up in ideal PN.

What is the LOS rate profile if exactly

N = 3

and range drops to one-tenth?

\dot{λ} \propto R^{N - 2} = R^{1}

, so

\dot{λ}

falls to one-tenth of its start — linear decay in range, the slowest of the practical

N \in [3, 5]

band.

What guidance does

\dot{λ} = 0

but

V_{c} = 0

(matched speeds, frozen bearing) imply?

a_{c} = 0

, but with no closing there is no intercept either — a co-moving standoff; PN neither corrects nor collides because the range never shrinks.

If the target maneuvers (nonzero transverse target acceleration), does plain PN still null

\dot{λ}

Not cleanly — plain PN assumes

a_{⊥, target} = 0

, so a maneuver injects a persistent LOS rate that requires augmented PN (adding a target-acceleration term). See Zero-Effort-Miss and Augmented Proportional Navigation.

Recall Self-check before you leave

The error signal PN nulls ::: the LOS rotation rate $\dot{λ}$ , not the LOS angle $λ$ . The exponent in $\dot{λ} \propto R^{?}$ ::: $N - 2$ , so convergence needs $N > 2$ . Direction of the commanded acceleration ::: perpendicular (lateral) to the line of sight. The factor that makes stability $N > 2$ not $N > 1$ ::: the Coriolis $2 \dot{R} \dot{λ}$ term.

Related: Line-of-Sight Geometry and Kinematics, Polar Coordinate Kinematics.

Question bank — Proportional navigation guidance — N·V_c·λ̇, derivation

True or false — justify

Spot the error

Why questions

Edge cases