Proportional navigation guidance — N·V_c·λ̇, derivation: Foundations

This page assumes nothing. Before you can read the PN derivation, you must own every symbol it fires at you. We build them one at a time, each anchored to a picture, each earning the next.

0 — The two players and the line between them

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

Look at the figure: the missile sits at the corner, the target sits out in the plane, and the burnt-orange line joining them is the LOS. As the target moves, that line does two things at once — it can get shorter/longer and it can swing like a clock hand. The whole topic is about separating those two motions and controlling them.

1 — $R$ : how far away the target is (range)

Picture the orange line in the figure: $R$ is simply how long it is. Nothing more.

2 — The dot: rate of change (a first look at derivatives)

Before the next symbols, we must earn the little dot that sits on top of them.

$\dot{R} > 0$ → range growing → target running away.
$\dot{R} < 0$ → range shrinking → we are closing in.
$\dot{R} = 0$ → range frozen for this instant.

Two dots, like $\ddot{λ}$ , mean the rate-of-change of the rate-of-change (an acceleration-like quantity). We will need that later; for now just know a second dot = "how fast the first-dot quantity is itself changing."

(If the word "derivative" is new, this dot IS a derivative with respect to time — see Polar Coordinate Kinematics for the fuller machinery.)

3 — $V_{c}$ : closing velocity (why we flip the sign of $\dot{R}$ )

Situation	$\dot{R}$	$V_{c} = - \dot{R}$
Closing fast	$- 900$	$+ 900$ (closing)
Opening (missing)	$+ 400$	$- 400$ (not closing)
Range frozen	$0$	$0$

See Closing Velocity and Range Rate for why $V_{c}$ (not raw speed) is the quantity that matters for intercept timing.

4 — $λ$ : the line-of-sight angle (and its sign convention)

In the figure, the dashed grey line is the fixed reference (a direction pinned in space, not attached to the missile). The angle from that dashed line up to the orange LOS is $λ$ . Measured in radians (a full turn = $2 π$ radians).

5 — $\dot{λ}$ : LOS rotation rate (the star of the show)

The figure shows the LOS at two nearby instants (solid orange, then faded orange). The small wedge between them, swept counter-clockwise, divided by the time elapsed, is a positive $\dot{λ}$ . Think of it as "drift of the target across your window."

This is the same "target pinned on the glass vs drifting on the glass" picture the parent note opens with — drift on the glass is $\dot{λ} \neq = 0$ .

6 — Splitting motion into "along" and "across" (unit vectors and signs)

Any relative motion between missile and target can be broken into two perpendicular pieces, and this split is the engine of the whole derivation. To track signs (not just sizes) we need to name the directions of those pieces.

In the figure the target's velocity $V$ (plum arrow) is decomposed onto the teal $\overset{r}{^}$ axis (giving $\dot{R} \overset{r}{^}$ ) and the teal $\hat{θ}$ axis (giving $R \dot{λ} \hat{θ}$ ). Now the signs are unambiguous: a component along $+ \hat{θ}$ is a positive $R \dot{λ}$ , hence a positive $\dot{λ}$ .

7 — The Coriolis surprise (why a factor of 2 shows up)

The parent derivation uses one non-obvious fact you should meet now so it isn't a shock later.

The key setup, which the misleading shortcuts always skip: we are differentiating a vector whose basis rotates. Recall from §6 that $(\overset{r}{^}, \hat{θ})$ is a rotating frame spinning at rate $\dot{λ}$ . When a unit vector turns, its own time-derivative is not zero — and those two derivatives are where the cross-terms come from. The standard rules for our CCW frame are:

\dot{\overset{r}{^}} = \dot{λ} \hat{θ}, \dot{\hat{θ}} = - \dot{λ} \overset{r}{^} .

Read them physically: as the LOS rotates CCW by a tiny angle, the tip of $\overset{r}{^}$ moves sideways in the $+ \hat{θ}$ direction (hence $\dot{\overset{r}{^}} = \dot{λ} \hat{θ}$ ); and $\hat{θ}$ , being $9 0^{\circ}$ ahead, tips back toward $- \overset{r}{^}$ (hence $\dot{\hat{θ}} = - \dot{λ} \overset{r}{^}$ ). These are the engine of the factor of 2.

The figure shows exactly these two unit-vector derivatives: the teal $\overset{r}{^}$ arrow nudging into the orange $+ \hat{θ}$ direction (that nudge is Contribution A below), and the teal $\hat{θ}$ arrow tipping into the plum $- \overset{r}{^}$ direction — its $+ \hat{θ}$ -pointing consequence when fed by $\dot{R}$ is Contribution B below.

Intuition The correct polar-coordinate derivation of the 2

Start from the relative velocity written in the rotating frame (from §6): $V = \dot{R} \overset{r}{^} + R \dot{λ} \hat{θ} .$ Differentiate term by term with the product rule, and never forget $\overset{r}{^}$ and $\hat{θ}$ are themselves changing:

a = \frac{d V}{d t} = (\ddot{R} \overset{r}{^} + \dot{R} \dot{\overset{r}{^}}) + (\dot{R} \dot{λ} \hat{θ} + R \ddot{λ} \hat{θ} + R \dot{λ} \dot{\hat{θ}}) .

Now substitute $\dot{\overset{r}{^}} = \dot{λ} \hat{θ}$ and $\dot{\hat{θ}} = - \dot{λ} \overset{r}{^}$ :

a = (\ddot{R} \overset{r}{^} + \dot{R} \dot{λ} \hat{θ}) + (\dot{R} \dot{λ} \hat{θ} + R \ddot{λ} \hat{θ} - R \dot{λ}^{2} \overset{r}{^}) .

Collect the pieces pointing along $\hat{θ}$ (the sideways ones):

a_{⊥} = A: from \dot{\overset{r}{^}} \dot{R} \dot{λ} + B: from the \dot{R} term of V \dot{R} \dot{λ} + angle speeds up R \ddot{λ} = R \ddot{λ} + 2 \dot{R} \dot{λ} .

Contribution A is the derivative of the radial unit vector $\overset{r}{^}$ turning ( $\dot{R} \dot{\overset{r}{^}} = \dot{R} \dot{λ} \hat{θ}$ ). Contribution B is the product-rule derivative of the radial coefficient inside the transverse velocity term ( $\dot{R} \dot{λ} \hat{θ}$ ). Two separate, honestly-different $\dot{R} \dot{λ}$ 's — that is the factor of 2. (The leftover $- R \dot{λ}^{2} \overset{r}{^}$ is the centripetal term; it points along $\overset{r}{^}$ , so it does not enter $a_{⊥}$ .)

8 — $a_{c}$ and $N$ : the command, its direction, and the gain

Because $a_{c}$ shares the $\hat{θ}$ axis with $\dot{λ}$ , the PN law's signs now read cleanly: if $V_{c} > 0$ (closing) and $\dot{λ} > 0$ (LOS rotating CCW), then $a_{c} = N V_{c} \dot{λ} > 0$ , i.e. the missile accelerates CCW to catch up with and null the sightline's rotation. Flip $\dot{λ}$ to negative and $a_{c}$ flips too — always toward the direction that removes drift.

Assembled, the whole vocabulary produces the law you're about to derive: $a_{c} = N V_{c} \dot{λ} (gain) \times (closing speed) \times (LOS drift) .$

Prerequisite map

Read top-to-bottom: the humble overdot and the range/angle feed the two rates; the CCW-positive angle also warns us about wrapping; the rates feed closing velocity and the transverse split; those plus the Coriolis correction and the gain $N$ assemble the PN law, whose purpose is to zero $\dot{λ}$ .

Equipment checklist

Cover the right side and answer aloud. If any stumps you, re-read its section before opening the parent note.

What does the overdot on a symbol mean?

"Rate of change per second" — a time derivative (e.g.

\dot{R}

= how fast range changes each second).

What is the line of sight (LOS)?

The straight imaginary line from the missile to the target.

What does

R

measure, and what does

R = 0

mean?

The length of the LOS (range);

R = 0

means a hit.

Why is

V_{c} = - \dot{R}

(why the minus)?

Because closing means

R

is decreasing (

\dot{R} < 0

); the minus makes closing velocity a positive number.

What is

λ

measured from, and which direction is positive?

From a fixed (inertial) reference direction, measured counter-clockwise = positive.

What does

\dot{λ} > 0

mean physically?

The sightline is rotating counter-clockwise (with our CCW-positive convention).

What goes wrong at an angle wrap, and the fix?

When

λ

crosses the reference it jumps (e.g.

2 π \to 0

), giving a huge false

\dot{λ}

; fix by unwrapping the angle before differentiating.

What exactly is

V

in §6?

The target's velocity relative to the missile, read in the inertial (non-rotating) frame.

Name the two LOS unit vectors and their directions.

\overset{r}{^}

points missile→target;

\hat{θ}

\overset{r}{^}

rotated

9 0^{\circ}

CCW (direction of increasing

λ

Is the

(\overset{r}{^}, \hat{θ})

frame inertial or rotating?

Rotating — it spins with the LOS at rate

\dot{λ}

, which is why unit-vector derivatives appear.

Write the relative velocity split with directions.

V = \dot{R} \overset{r}{^} + R \dot{λ} \hat{θ}

What are the derivatives of the rotating unit vectors?

\dot{\overset{r}{^}} = \dot{λ} \hat{θ}

and

\dot{\hat{θ}} = - \dot{λ} \overset{r}{^}

Why is the transverse speed

R \dot{λ}

and not just

\dot{λ}

\dot{λ}

is an angular rate; multiplying by the radius

R

converts spin into sideways speed.

What is the full transverse acceleration, and where do the two

\dot{R} \dot{λ}

come from?

a_{⊥} = R \ddot{λ} + 2 \dot{R} \dot{λ}

; one from

\dot{R} \dot{\overset{r}{^}}

(the radial unit vector turning), one from the product-rule derivative of the radial coefficient in the transverse velocity term.

Which way does positive

a_{c}

point?

Along

+ \hat{θ}

(CCW, direction of increasing

λ

What is

N

and its typical range?

The dimensionless navigation gain, usually

3 \leq N \leq 5

State the PN law in words.

Commanded acceleration = gain

\times

closing velocity

\times

LOS rotation rate,

a_{c} = N V_{c} \dot{λ}

Recall Self-test: what single quantity is PN trying to drive to zero, and why?

$\dot{λ}$ , the LOS rotation rate — because a frozen sightline plus shrinking range is exactly a collision course.