Proportional navigation guidance — N·V_c·λ̇, derivation: Visual walkthrough

We will draw a pursuer (the missile) and a target (a plane). The whole story is about one line joining them — the line of sight — and whether that line rotates.

Step 1 — Two dots and the line between them

WHAT: we replaced "a missile and a plane" with just two numbers, $R$ and $λ$ .

WHY: every question about hitting the target is really a question about how these two numbers change over time. If we track $R$ and $λ$ , we track everything.

PICTURE: in the figure, the blue dot is the missile, the orange dot is the target, the gray line is the LOS. The angle $λ$ is measured from the dashed reference line up to the LOS. The little dot over a symbol means "rate of change per second" — so $\dot{λ}$ is how fast the gray line swings, and $\dot{R}$ is how fast the two dots get closer or farther.

See Line-of-Sight Geometry and Kinematics for the full frame setup.

Step 2 — Splitting the target's motion into two arrows

WHAT: we broke one velocity arrow into a red "along" arrow and a green "across" arrow.

WHY: we care about these two directions specifically because each controls one of our two numbers. The along part changes $R$ ; the across part changes $λ$ . Any velocity can be written this way — it is just choosing perpendicular axes that ride along with the LOS instead of fixed East/North axes. This is exactly Polar Coordinate Kinematics.

Term by term, in the across-LOS piece:

$\dot{λ}$ = radians/second the line turns.
$R$ = length of the line. A point at the end of a rod turning at $\dot{λ}$ rad/s moves sideways at (length)×(turn rate) = $R \dot{λ}$ metres/second. That is why the across-speed is $R \dot{λ}$ , not just $\dot{λ}$ .

We name the across-speed $V_{⊥} = R \dot{λ}$ .

Step 3 — What "on a collision course" looks like

WHAT: we asked "when do the dots collide?" and found the answer is $V_{⊥} = 0$ .

WHY: collision means the gap $R$ goes to zero without the line ever sliding off its bearing. Sliding bearing ( $\dot{λ} \neq = 0$ ) means the target is crossing in front of or behind you — a miss.

PICTURE: left panel — the green across-arrow is nonzero, so the LOS sweeps to a new angle each frame (the target "drifts on the glass"): miss. Right panel — the across-arrow is gone, all motion is red/along, the sightlines stay parallel and stacked: collision.

This is the error signal we want to null: our error is $\dot{λ}$ itself.

Step 4 — How fast does $\dot{λ}$ itself change? (the hidden factor of 2)

To drive $\dot{λ}$ to zero we must know how $\dot{λ}$ responds to a push. That means differentiating the across-velocity $V_{⊥} = R \dot{λ}$ — but carefully.

WHAT: the picture shows why one term is missing. The green "across" direction $\hat{θ}$ is itself rotating as the LOS swings. Between one instant and the next, the across-arrow points a slightly different way.

WHY the extra term: the radial motion $\dot{R}$ carries velocity, and because the across-direction has rotated by $\dot{λ} d t$ , part of that radial velocity leaks into the across-direction. That leak adds a second $\dot{R} \dot{λ}$ . This leaked term is the Coriolis term — see Coriolis Term in Polar Coordinate Acceleration.

PICTURE: the two curved arrows in the figure each represent one copy of $\dot{R} \dot{λ}$ ; together they make the $2$ .

Step 5 — Wiring in the missile's command

The missile fires its thrusters sideways to the LOS. That commanded acceleration is $a_{c}$ . For a target that is not itself maneuvering across the LOS, the only across-acceleration is the missile's own, pointed to oppose the drift:

WHAT: we set the physical across-acceleration $a_{⊥}$ equal to $- a_{c}$ (minus because the command pushes against growing $λ$ ).

WHY: this is the moment the control input enters the geometry. Before this line, $a_{⊥}$ was pure kinematics; now we say "the missile supplies it."

PICTURE: the orange arrow is $a_{c}$ pushing the missile off its current heading, bending the LOS back toward a fixed bearing.

Step 6 — Plug in the PN law and watch $\dot{λ}$ collapse

Now substitute the PN law $a_{c} = N V_{c} \dot{λ}$ . Since $V_{c} = - \dot{R}$ (see Closing Velocity and Range Rate), that is $a_{c} = - N \dot{R} \dot{λ}$ .

\ddot{λ} = \frac{- ( - N R ˙ λ ˙ ) - 2 R ˙ λ ˙}{R} = \frac{R ˙ λ ˙ ( N - 2 )}{R}

WHAT: the command term and the Coriolis term merged into a single clean factor $(N - 2)$ .

WHY: dividing through by $\dot{λ}$ turns this into a story about ratios:

\frac{λ ¨}{λ ˙} = (N - 2) \frac{R ˙}{R} .

Each side is "rate of change ÷ current value" — exactly the shape whose integral is a logarithm. That is why a logarithm appears next: it is the natural tool for "growth proportional to current size."

PICTURE: curves of $\dot{λ}$ versus $R$ for several $N$ , all pinned to the same start on the right, read right-to-left as the engagement runs ( $R$ shrinking toward 0).

Step 7 — Read the stability condition off the graph (all cases)

As the missile arrives, $R \to 0$ . What happens to $\dot{λ} \propto R^{N - 2}$ depends entirely on the sign of the exponent $N - 2$ :

Every case shown:

$N > 2$ (exponent positive): $R^{N - 2} \to 0$ as $R \to 0$ . The swing-rate is crushed to zero — collision course achieved. ✓ (green curve dives to zero)
$N = 2$ (exponent zero): $R^{0} = 1$ , so $\dot{λ}$ stays constant — never decays. Right on the boundary, no good. (gray flat line)
$N < 2$ (exponent negative): $R^{N - 2} \to \infty$ as $R \to 0$ . The swing-rate blows up — the miss grows near the target. ✗ (red curve shoots up)

WHY this closes everything: back in Step 3 we said "the job is to force $\dot{λ} \to 0$ ." Steps 4–7 show that the specific form $N V_{c} \dot{λ}$ — with $N > 2$ — does exactly that automatically, driven by the geometry itself. The law is not a guess; it is the choice that makes the graph dive.

The one-picture summary

This single panel stacks the whole chain: two dots → split the motion → collision means the green across-arrow dies → the loop from command $a_{c}$ back to $\dot{λ}$ → the power law → the dive to zero for $N > 2$ .

Recall Feynman retelling — say it in plain words

Imagine a plane out your side window. Draw the invisible line from you to it. Two things can happen to that line: it can get shorter (you're closing in) or it can swing across your window (the plane is drifting forward or back). If the line only gets shorter and never swings, you two are headed for the same spot — a crash. So the whole trick to hitting something is: kill the swing.

To kill the swing you need to know how a sideways nudge changes it. When you work that out you find a sneaky extra term — the Coriolis one, a factor of two — because the "sideways" direction is itself turning while you measure. Miss that and your whole answer is off.

Then you make your missile push sideways in proportion to the swing-rate, scaled by how fast you're closing, times a number $N$ . Do the algebra and the swing-rate ends up tied to the remaining distance by a power, $N - 2$ . As you close the last few metres, if that power is positive (that is, $N$ bigger than 2), the swing gets strangled to nothing right as you arrive — a clean hit. If $N$ were 2 the swing would just coast; less than 2 and it would explode in your face. That's why every real missile uses $N$ between 3 and 5.

Recall Quick self-check

What physically is $R \dot{λ}$ ? ::: The across-LOS relative speed — how fast the sightline sweeps sideways; it is the miss-building motion. Why does the across-acceleration have a factor of 2? ::: One $\dot{R} \dot{λ}$ from the product rule, one more because the transverse basis vector itself rotates (Coriolis). Why must $N > 2$ ? ::: Because $\dot{λ} \propto R^{N - 2}$ ; only a positive exponent $N - 2 > 0$ sends $\dot{λ} \to 0$ as $R \to 0$ . What is the error signal PN nulls? ::: The LOS rotation rate $\dot{λ}$ .