Augmented proportional navigation — gravity compensation: Visual walkthrough

Step 1 — The line of sight, and the one angle that matters

WHAT. Two dots on a page: the missile (we label its position point $M$ ) and the target (position point $T$ ). Draw the straight rod that connects them — that rod is the line of sight (LOS). Its length is the range $R$ (the straight-line distance $M$ to $T$ , in metres). Now measure the angle this rod makes with a fixed horizontal reference line. Call that angle $λ$ (Greek "lambda"), measured from horizontal, positive turning upward.

WHY this angle and not the positions themselves? Because a missile does not need to know where the target is in absolute coordinates — it only needs to know whether the connecting rod is rotating. A rod that keeps a fixed direction while the two ends get closer means a collision. So the single most important number in all of guidance is not a position, it is the tilt of this rod and how fast it changes.

PICTURE. The white rod from $M$ to $T$ (its length labelled $R$ ), the amber angle $λ$ swept up from the dashed horizontal, and a small curved arrow showing the direction $λ$ would grow.

Figure — Augmented proportional navigation — gravity compensation

Step 2 — Why $\dot{λ} = 0$ is a collision

WHAT. Freeze the rod's direction but let the two ends slide toward each other — that is, let $R$ shrink while $λ$ stays fixed. Draw the rod at three moments — same tilt every time, just shorter.

WHY. If the rod never rotates ( $\dot{λ} = 0$ ) yet $R$ shrinks, the two ends must be racing to the same point at the same instant. Sailors call this constant bearing, decreasing range: a ship whose compass bearing never changes but grows larger is a ship about to hit you. So the entire job of guidance collapses to a slogan: kill $\dot{λ}$ .

PICTURE. Three parallel rods of decreasing length $R$ , all at the same tilt $λ$ , converging on one amber collision point.

Recall Why fixed bearing means impact

Non-rotating LOS + shrinking $R$ ::: both bodies head for the same meeting point → guaranteed intercept.

Step 3 — Closing velocity, defined properly

WHAT. Before we steer, name the rate at which the rod shrinks. Define the closing velocity as $V_{c} = - \dot{R},$ the negative of the range's rate of change. The minus sign is deliberate: $R$ is decreasing during an intercept ( $\dot{R} < 0$ ), so $V_{c} = - \dot{R}$ comes out positive. $V_{c} > 0$ means "gap closing"; $V_{c} < 0$ would mean the target is running away.

WHY define it by $- \dot{R}$ ? Because "closing speed" should be a positive number when we are actually closing — that is the useful convention for a homing missile. We build $V_{c}$ from $R$ , a quantity we already defined in Step 1, so nothing is smuggled in.

PICTURE. The rod at two instants; the shrink $Δ R < 0$ marked, and $V_{c} = - \dot{R} > 0$ shown as an amber "closing" arrow along the LOS.

Step 4 — Set up the two directions on the rod (with a sign convention)

WHAT. Any push on the missile can be broken into two arrows: one along the rod, direction $\hat{t}$ ("tangent", pointing from $M$ toward $T$ ), and one across it, direction $\overset{n}{^}$ ("normal"). We fix the orientation once and for all: $\overset{n}{^} points 9 0^{\circ} counter-clockwise from \hat{t} (i.e. in the direction of increasing λ) .$ So a positive $\overset{n}{^}$ -component of acceleration is the one that increases $λ$ .

WHY nail down the sign? Every later term ( $g_{n}$ , $a_{T, n}$ ) is a shadow onto $\overset{n}{^}$ , and a shadow has a sign only once you say which way $\overset{n}{^}$ points. Tying " $+ \overset{n}{^}$ = increasing $λ$ " makes the whole spin equation's signs unambiguous.

PICTURE. The LOS rod with $\hat{t}$ along it (M→T) and $\overset{n}{^}$ drawn $9 0^{\circ}$ counter-clockwise; the curved arrow of increasing $λ$ agrees with $+ \overset{n}{^}$ ; a generic acceleration decomposed into its $\hat{t}$ and $\overset{n}{^}$ shadows.

Step 5 — Derive what actually drives $\dot{λ}$

WHAT. Now we earn the spin equation instead of asserting it. The across-the-rod separation of the two bodies changes because of two things: the rod turning ( $\dot{λ}$ ) and the rod shrinking ( $\dot{R}$ ). Planar kinematics of the relative position $R$ gives the transverse (across-LOS) relative velocity as $R \dot{λ}$ . Differentiate once more — the transverse relative acceleration is $a_{n}^{rel} = R \ddot{λ} + 2 \dot{R} \dot{λ} .$

$R \ddot{λ}$ — the rod angularly accelerating
$2 \dot{R} \dot{λ}$ — the Coriolis-like term: a shrinking rod ( $\dot{R} < 0$ ) spins up any existing $\dot{λ}$ , just as a skater pulling in her arms spins faster.

That transverse relative acceleration is supplied by (target normal accel) minus (missile normal accel). Writing the missile's normal accel as commanded plus gravity, $a_{M, n} = a_{c, n} + g_{n}$ : $R \ddot{λ} + 2 \dot{R} \dot{λ} = a_{T, n} - (a_{c, n} + g_{n}) .$ Solve for the angular acceleration and use $V_{c} = - \dot{R}$ : $\ddot{λ} = \frac{1}{R} [a_{T, n} - a_{c, n} - g_{n}] + \frac{2 V _{c}}{R} \dot{λ}$

WHY this matters — and note the $1/ R$ . The forcing terms enter divided by the range $R$ : the closer we get (small $R$ ), the more violently any leftover $g_{n}$ or $a_{T, n}$ whips $\dot{λ}$ around. This is the derivation the parent's one-line " $\dot{λ}$ evolves under $(a_{c, n} + g_{n} - a_{T, n})$ " was compressing — the bracket $[a_{T, n} - a_{c, n} - g_{n}]$ is exactly that combination (signs set by our $\overset{n}{^}$ convention), and the $1/ R$ is the normalization that box omitted.

PICTURE. The rod with transverse velocity $R \dot{λ}$ drawn along $\overset{n}{^}$ ; the shrinking rod feeding the $2 \dot{R} \dot{λ}$ spin-up; the three forcing arrows $a_{T, n}$ , $a_{c, n}$ , $g_{n}$ pushing on $\overset{n}{^}$ with a $1/ R$ amplifier symbol.

Step 6 — Watch gravity poison the loop

WHAT. Set $a_{c, n} = 0$ and imagine an inertial target ( $a_{T, n} = 0$ ). The spin equation leaves $\ddot{λ} = - g_{n} / R + (2 V_{c} / R) \dot{λ}$ : a persistent $- g_{n} / R$ forcing that never switches off. The rod droops.

WHY it doesn't "average out." In a full up-and-over ballistic arc the droop reverses, so it cancels — the tempting excuse in the parent's [!mistake] box. But terminal homing is a one-way plunge of a few seconds: gravity tugs the same direction the whole time, and because of the $1/ R$ its damage grows as $R \to 0$ . So its effect on $\dot{λ}$ accumulates into a real, growing miss. See Ballistic trajectory & gravity turn for the drooping arc itself.

PICTURE. A straight intended path (amber dashed) and the real drooping path (cyan solid) sagging below it, with the gap labelled "accumulating miss," and the steady $- g_{n} / R$ forcing on $\overset{n}{^}$ .

Step 7 — How big is $g_{n}$ ? Every LOS angle covered

WHAT. Gravity is the vector $g = - g \overset{z}{^}$ (straight down, $g \approx 9.81 m/s^{2}$ ). Recall from Step 1 that $λ$ is measured from horizontal, and from Step 4 that $\overset{n}{^}$ is $9 0^{\circ}$ counter-clockwise from the LOS. Projecting straight-down gravity onto that $\overset{n}{^}$ gives $g_{n} = g \cdot \overset{n}{^} = g cos λ .$

WHY the cosine, and why $λ$ from horizontal. The dot product $g \cdot \overset{n}{^}$ asks "how much of gravity lies along $\overset{n}{^}$ ?" — the length of gravity's shadow on the across-direction. Because $λ$ is the tilt from horizontal, when the rod is horizontal $\overset{n}{^}$ points straight up, fully opposing gravity, so the shadow is the whole of $g$ ; as $λ$ grows, $\overset{n}{^}$ leans away from vertical and the shadow shrinks by $cos λ$ .

Every case, no gaps (including $λ < 0$ and $λ > 9 0^{\circ}$ ):

$λ = 0^{\circ}$ (horizontal shot): $g_{n} = g cos 0 = + g$ — worst case, all of gravity bends the rod.
$λ = 6 0^{\circ}$ (steep climb): $g_{n} = g cos 6 0^{\circ} = 0.5 g \approx 4.905$ — half strength.
$λ = 9 0^{\circ}$ (straight up): $g_{n} = g cos 9 0^{\circ} = 0$ — gravity lies along the rod, harmless.
$λ = - 3 0^{\circ}$ (diving below horizontal): $cos (- 3 0^{\circ}) = cos 3 0^{\circ}$ , so $g_{n} = g cos 3 0^{\circ}$ — the cosine is even, so a dive of $3 0^{\circ}$ needs the same $∣ g_{n} ∣$ as a climb of $3 0^{\circ}$ ; the sign of $g_{n}$ still tracks whether $\overset{n}{^}$ leans with or against gravity.
$λ = 12 0^{\circ}$ (LOS pointing back and up): $cos 12 0^{\circ} = - 0.5$ , so $g_{n} = - 0.5 g$ — now gravity's shadow flips to $- \overset{n}{^}$ , and the compensation term flips sign accordingly. The formula handles it automatically.

PICTURE. Copies of the rod at $0^{\circ}$ , $6 0^{\circ}$ , $9 0^{\circ}$ , $- 3 0^{\circ}$ , $12 0^{\circ}$ ; on each, the vertical gravity arrow and its $\overset{n}{^}$ -shadow, shrinking, vanishing, then reversing.

Recall The gravity shadow across angles

$g_{n}$ at $λ = 0^{\circ}, 6 0^{\circ}, 9 0^{\circ}, 12 0^{\circ}$ ::: $g, 0.5 g, 0, - 0.5 g$ (cosine is even in $λ$ , and goes negative past $9 0^{\circ}$ )

Step 8 — Cancel gravity: the $- g_{n}$ term

WHAT. We want the net normal acceleration the rod feels to equal exactly the pure-PN demand. Gravity will unavoidably contribute $+ g_{n}$ to $a_{M, n} = a_{c, n} + g_{n}$ . So we design our commanded normal component to carry an extra $- g_{n}$ : $a_{c, n} = PN demand N^{'} V_{c} \dot{λ} - g_{n} ⟹ a_{M, n} the rod feels a_{c, n} + g_{n} = N^{'} V_{c} \dot{λ} .$

WHY subtract, not add? We pre-load the opposite of the coming disturbance: command $- g_{n}$ ; nature adds $+ g_{n}$ ; they annihilate; the rod feels only the clean PN push. This is predictive cancellation — we hold the collision course instead of chasing the droop after it appears.

PICTURE. Two amber arrows of equal length pointing opposite ways ( $- g_{n}$ commanded, $+ g_{n}$ from gravity) summing to zero net gravity effect, leaving one clean cyan PN arrow.

Step 9 — The target moves: derive the $\frac{N ^{'}}{2} a_{T, n}$ term

WHAT. Let the target carry a constant normal acceleration $a_{T, n}$ , and let time-to-go be $t_{g o} = \frac{R}{V _{c}},$ the range divided by closing speed — roughly how many seconds until impact. If we did nothing, that constant $a_{T, n}$ would drift the target off the collision point by the freshman-kinematics amount $ZEM = \frac{1}{2} a_{T, n} t_{g o}^{2},$ the zero-effort-miss (see Zero-Effort-Miss (ZEM) guidance).

WHY the factor $\frac{N ^{'}}{2}$ — the algebra. PN with gain $N^{'}$ commands acceleration $a_{c, n} = N^{'} V_{c} \dot{λ}$ . A standard ZEM rewrite of PN is $a_{c, n} = N^{'} \frac{ZEM}{t _{g o}^{2}}$ (the loop nulls zero-effort-miss with gain $N^{'}$ over the remaining flight). To also cancel the target-induced miss, feed the same loop the target's ZEM contribution: $a_{c, n}^{(target)} = N^{'} \frac{ZEM _{target}}{t _{g o}^{2}} = N^{'} \frac{\frac{1}{2} a _{T, n} t _{g o}^{2}}{t _{g o}^{2}} = \frac{N ^{'}}{2} a_{T, n} .$ The $t_{g o}^{2}$ cancels exactly, leaving $\frac{N ^{'}}{2}$ : the $\frac{1}{2}$ is the $\frac{1}{2} a t^{2}$ of kinematics, the $N^{'}$ is the loop gain. That is where the factor is born, not assumed.

PICTURE. Target peeling off a straight predicted path into a parabola; the amber gap $\frac{1}{2} a_{T, n} t_{g o}^{2}$ marked; the $t_{g o}^{2}$ cancellation shown; a feed-forward arrow labelled $\frac{N ^{'}}{2} a_{T, n}$ closing it.

The one-picture summary

WHAT. Assemble the three commanded normal terms. Since the missile's lateral command is applied across the LOS, in the boxed law $a_{c}$ means the required normal component $a_{c, n}$ — the number the autopilot must deliver perpendicular to the line of sight: $a_{c} = See the LOS spin N^{'} V_{c} \dot{λ} + Chase the target \frac{N ^{'}}{2} a_{T, n} - Fight gravity g_{n} (a_{c} \equiv a_{c, n})$

PICTURE. One missile, three stacked contribution arrows (cyan PN spin term, white target-lead term, amber $- g_{n}$ gravity term) adding tip-to-tail into the total command $a_{c}$ , with the LOS, $\overset{n}{^}$ , and gravity shadow all labelled.

Recall Feynman: the whole walkthrough in plain words

Draw a stick from your missile ( $M$ ) to the target ( $T$ ); its length is the range $R$ and its tilt is $λ$ . If that stick keeps its tilt while $R$ shrinks, you'll hit — so the missile's whole job is don't let the stick spin. Careful bookkeeping shows the spin is driven by the sideways pushes divided by $R$ , so as you close in every leftover push hurts more. The missile steers sideways in proportion to any spin it sees (PN). Gravity casts a sideways shadow $g_{n} = g cos λ$ that never lets up during a dive — so the missile pushes an equal-and-opposite $- g_{n}$ to erase it before it happens. And if the target swerves, it opens a predictable gap $\frac{1}{2} a_{T, n} t_{g o}^{2}$ over the time-to-go $t_{g o} = R / V_{c}$ ; run that through the same loop and the $t_{g o}^{2}$ cancels, leaving a lead term $\frac{N ^{'}}{2} a_{T, n}$ . Add the three normal components — see the spin, chase the target, fight gravity — and the stick stays quiet all the way to impact.

Connections

Augmented proportional navigation — gravity compensation — the parent result these pictures build.
Proportional Navigation (PN) — Step 5's base law.
Zero-Effort-Miss (ZEM) guidance — where Step 9's $\frac{N ^{'}}{2}$ comes from.
Line-of-Sight rate estimation — how $\dot{λ}$ in Step 1 is actually measured.
Coordinate frames & projections — the $\hat{t} / \overset{n}{^}$ split of Step 4.
Ballistic trajectory & gravity turn — the physical droop of Step 6.
Missile Autopilot & Acceleration limits — why saving g-capacity by not chasing gravity matters.