Augmented proportional navigation — gravity compensation: Question bank

Setting the stage — every symbol earned first

Before any trap, here is the picture the whole page lives in. Two things move in a vertical plane: the missile and the target. The straight arrow from missile to target is the line of sight (LOS) — imagine a taut string joining them.

Figure — Augmented proportional navigation — gravity compensation

Now build the second-order dynamics that make "cancel $a_{n}$ " mean something.

Formula Why acceleration

$⊥$ LOS controls $\dot{λ}$ — the differential equation In polar (LOS) coordinates the acceleration perpendicular to the sightline is the classic transverse term $a_{n} = r \ddot{λ} + 2 \overset{r}{˙} \dot{λ},$ where $r$ is range and $a_{n}$ is the net perpendicular acceleration (missile minus target). Rearranging for the sightline's angular acceleration: $\ddot{λ} = \frac{a _{n} - 2 r ˙ λ ˙}{r} = \frac{a _{n}}{r} + 2 \frac{V _{c}}{r} \dot{λ},$ using $\overset{r}{˙} = - V_{c}$ . Read it as a driven equation for $\dot{λ}$ : the term $2 \frac{V _{c}}{r} \dot{λ}$ (positive while closing) is destabilising, so a nonzero $\dot{λ}$ grows unless $a_{n}$ pushes back. Substitute the net perpendicular accel $a_{n} = a_{c, n} + g_{n} - a_{T, n}$ and choose the PN command $a_{c, n} = N^{'} V_{c} \dot{λ}$ : the loop then forces $\dot{λ} \to 0$ (this is the standard result quoted in Proportional Navigation (PN)). So driving $a_{n}$ to the pure-PN value is exactly what makes $\dot{λ}$ collapse — and any uncancelled $g_{n}$ or $a_{T, n}$ inside $a_{n}$ keeps $\dot{λ}$ from ever settling.

Formula Why the half-factor

$\frac{N ^{'}}{2}$ — shown, not quoted Zero-effort-miss (ZEM) is where the target would be missed by if both sides stopped steering now. For a constant target accel $a_{T}$ acting over the remaining time $t_{g o}$ , kinematics (distance $= \frac{1}{2} a t^{2}$ ) gives the lateral miss it builds: $ZEM_{target} = \frac{1}{2} a_{T} t_{g o}^{2} .$ PN's own feedback removes a fraction of any ZEM at rate set by $N^{'}$ ; to fully null a maneuver-driven ZEM the feed-forward command must be $a_{c, aug} = \frac{N ^{'}}{r _{g o}} \cdot ZEM = \frac{N ^{'}}{V _{c}^{2} t _{g o}^{2}} \cdot \frac{1}{2} a_{T} t_{g o}^{2} \cdot V_{c}^{2} = \frac{N ^{'}}{2} a_{T},$ where the $t_{g o}^{2}$ from the miss cancels the $t_{g o}^{2}$ hidden in the ZEM-to-command gain, leaving the clean $\frac{N ^{'}}{2}$ . So the $\frac{1}{2}$ is the $\frac{1}{2} a t^{2}$ of freshman kinematics surviving the cancellation (full derivation in Zero-Effort-Miss (ZEM) guidance).

Quick refresher of every symbol used below:

$\dot{λ}$ = LOS angular rate; $\ddot{λ}$ = its rate of change; $λ$ = LOS elevation angle (up from horizontal).
$\overset{n}{^}$ = LOS-normal direction; $g_{n} = ∣ g \cdot \overset{n}{^} ∣ = g cos λ$ = magnitude of perpendicular gravity.
$a_{c, n} = a_{c} \cdot \overset{n}{^}$ = commanded accel projected onto $\overset{n}{^}$ ; $a_{T, n} = a_{T} \cdot \overset{n}{^}$ = target accel projected onto $\overset{n}{^}$ .
$N^{'}$ = navigation constant; $V_{c} = - \overset{r}{˙}$ = closing speed; $a_{T}$ = target accel $⊥$ LOS; $t_{g o}$ = time-to-go.
Net perpendicular dynamics: $a_{n} = r \ddot{λ} + 2 \overset{r}{˙} \dot{λ}$ , with $a_{n} = a_{c, n} + g_{n} - a_{T, n}$ .
Full law: $a_{c} = N^{'} V_{c} \dot{λ} + \frac{N ^{'}}{2} a_{T} - g_{n}$ .

True or false — justify

The commanded acceleration

a_{c}

already includes the effect of gravity, so we do not need the

- g_{n}

term.

False —

a_{c}

is only what the autopilot commands; gravity acts on top of it. The physical normal accel is

a_{c, n} + g_{n}

, so we must pre-subtract

g_{n}

to make that sum equal the pure-PN demand

N^{'} V_{c} \dot{λ}

Gravity compensation makes the missile fly a straight line.

False — it makes the effective guidance loop behave as if gravity-free; the missile still physically droops, but the guidance no longer misreads that droop as target motion.

If the target never maneuvers, the augmentation term

\frac{N ^{'}}{2} a_{T}

vanishes but the

- g_{n}

term generally does not.

True —

a_{T} = 0

kills augmentation, yet gravity keeps bending the LOS unless the shot is vertical, so

g_{n}

compensation is still required.

Over a full ballistic arc gravity averages out, so ignoring it costs nothing during homing.

False — during terminal homing the missile is actively closing; the continuous downward

g_{n}

produces a persistent

\dot{λ}

that accumulates into a steady miss, it does not average away.

A larger navigation constant

N^{'}

increases the target-maneuver term but leaves the gravity term untouched.

True — the gravity term is

- g_{n}

, independent of

N^{'}

; only the PN and augmentation terms scale with

N^{'}

Compensation is reactive: it fixes the miss after the droop appears.

False — it is predictive; by supplying

- g_{n}

in advance it holds

\dot{λ} \equiv 0

so the droop-induced error is never allowed to appear.

For a purely vertical (straight-up) engagement the gravity-compensation term is zero.

True — gravity then lies entirely along the LOS (

∥ \overset{r}{^}

), so

g_{n} = g cos 9 0^{\circ} = 0

and nothing perpendicular needs cancelling.

The component of gravity along the LOS is harmless to guidance.

True (for

\dot{λ}

) —

g \cdot \overset{r}{^}

only changes closing speed

V_{c} = - \overset{r}{˙}

; it does not rotate the sightline, so it does not corrupt the LOS-rate that PN steers on.

In the transverse equation

a_{n} = r \ddot{λ} + 2 \overset{r}{˙} \dot{λ}

, leaving

g_{n}

inside

a_{n}

still lets

\dot{λ}

settle to zero.

False — an uncancelled

g_{n}

acts as a constant forcing term, so

\ddot{λ}

(hence

\dot{λ}

) never reaches a steady zero; that is precisely the persistent rate PN wastefully chases.

Spot the error

"Set

a_{c} = N^{'} V_{c} \dot{λ} + \frac{N ^{'}}{2} a_{T} + g_{n}

Wrong sign — gravity must be subtracted. Adding

g_{n}

would double the droop instead of cancelling it, since real gravity already contributes

+ g_{n}

to the physical normal accel.

"Compensate with the full

g

: write

a_{c} = N^{'} V_{c} \dot{λ} - g

Over-correction — only the LOS-normal part

g_{n} = g cos λ

bends the sightline. Subtracting full

g

would fight a phantom perpendicular pull that isn't there and waste control authority.

"Since

g \cdot \overset{n}{^} = - g cos λ

is negative, the compensation term should be

+ g cos λ

from a plain dot product."

Half-right, half-trap — the dot product is negative because

g

points opposite

\overset{n}{^}

's tilt, but

g_{n}

is defined as the projection's magnitude

g cos λ

, and the command carries gravity's downward direction through the explicit minus sign

- g_{n}

; mixing the two conventions double-counts the sign.

"At LOS elevation

λ

, use

g_{n} = g sin λ

Wrong trig function — in this convention (

λ

measured up from horizontal)

g_{n} = g cos λ

: horizontal LOS (

λ = 0

) gives the maximum

g_{n} = g

since

cos 0 = 1

, whereas

sin 0 = 0

would wrongly say no compensation is needed.

"Because

\dot{λ} = 0

right now, no command is needed at all, gravity term included."

Error — with an inertial target and momentarily fixed LOS, pure PN gives

0

, but gravity would immediately create a droop; the compensation still commands

- g_{n}

to keep

\dot{λ}

at zero.

"APN's

\frac{N ^{'}}{2} a_{T}

handles gravity too, since gravity is just another acceleration."

Error — that term feed-forwards the target's known maneuver via zero-effort-miss; gravity is a self acceleration acting on the missile and needs its own separate

- g_{n}

term.

"The

\frac{1}{2}

\frac{N ^{'}}{2} a_{T}

comes from averaging up and down gravity."

Error — it comes from the

\frac{1}{2} a_{T} t_{g o}^{2}

kinematic ZEM of the target maneuver; the

t_{g o}^{2}

cancels against the command gain to leave

\frac{1}{2}

, nothing to do with gravity.

t_{g o}

can be treated as a fixed constant throughout the engagement."

Error —

t_{g o} \approx r / V_{c}

shrinks to zero at impact; treating it as fixed would mis-scale any ZEM-based feed-forward as the range collapses.

Why questions

Why does only the perpendicular component of gravity affect the LOS rate?

Because in

a_{n} = r \ddot{λ} + 2 \overset{r}{˙} \dot{λ}

only the transverse accel

a_{n}

drives

\ddot{λ}

; a force along

\overset{r}{^}

enters the radial equation and only lengthens/shortens the range, never rotating the sightline.

Why is horizontal LOS the worst case for gravity compensation?

With

λ = 0

the entire gravity vector is perpendicular to the LOS, so

g_{n} = g cos 0 = g

— the maximum possible bending of the sightline.

Why does uncompensated gravity waste the missile's g-capacity?

PN continuously commands lateral accel to null the gravity-induced

\dot{λ}

, so precious acceleration budget is spent fighting a ghost instead of the real target.

Why must

g_{n}

be recomputed as the engagement evolves rather than fixed once?

The LOS elevation

λ

changes throughout the flight, so

g_{n} = g cos λ

changes; a stale value would over- or under-compensate as the geometry rotates.

Why do we express gravity via a projection onto

\overset{n}{^}

instead of just using its vertical value?

Because guidance acts in the LOS frame

(\overset{r}{^}, \overset{n}{^})

; only the projection onto the LOS-normal direction

\overset{n}{^}

enters the

\dot{λ}

dynamics — see Coordinate frames & projections.

Why does the half-factor survive as a clean number, free of

t_{g o}

Because the target miss grows as

\frac{1}{2} a_{T} t_{g o}^{2}

while the ZEM-to-command gain carries a compensating

1/ t_{g o}^{2}

; the two

t_{g o}^{2}

cancel, leaving only

\frac{N ^{'}}{2} a_{T}

Why does making the loop "effectively gravity-free" guarantee intercept on a non-maneuvering target?

Because pure PN in a gravity-free frame drives

\dot{λ} \to 0

, which is the constant-bearing, decreasing-range collision condition — the sightline stops rotating so the paths meet.

Edge cases

LOS elevation exactly

9 0^{\circ}

(target directly overhead, straight-up shot).

g_{n} = g cos 9 0^{\circ} = 0

; no compensation term acts, but the along-LOS gravity still slows closing speed

V_{c}

— harmless to

\dot{λ}

but real for energy.

Closing velocity

V_{c} \to 0

(missile no longer closing,

\overset{r}{˙} \to 0

The PN term

N^{'} V_{c} \dot{λ}

collapses and

t_{g o} \approx r / V_{c} \to \infty

, so the command is dominated by

\frac{N ^{'}}{2} a_{T} - g_{n}

; geometrically this is a degenerate near-parallel engagement where PN loses its intercept guarantee. If

V_{c} < 0

the range is growing — an overshoot/opening pass.

Target maneuver unknown / unestimated (

a_{T}

set to

0

though the target is turning).

The law reduces to PN plus gravity compensation; it will still lag the true maneuver and leave a residual miss, since the feed-forward can only cancel what estimation provides.

\dot{λ} = 0

with an inertial target and horizontal LOS.

Pure PN and augmentation both give

0

, yet

g_{n} = g

; the command is exactly

- g

, the maximal predictive push that keeps the true collision course from drooping.

Descending LOS where the target is below the horizontal (

λ < 0

Since

cos

is even,

g_{n} = g cos λ = g cos ∣ λ ∣

— the sign of

λ

does not change the magnitude of compensation, only the geometry; a shallow depression (

∣ λ ∣

small) still needs nearly full

g

, a steep dive (

∣ λ ∣ \to 9 0^{\circ}

) needs almost none.

LOS elevation beyond

9 0^{\circ}

(

λ \in (9 0^{\circ}, 18 0^{\circ})

, target above-and-behind).

cos λ

turns negative, so the projection

g cos λ

flips direction along

\overset{n}{^}

; carried through the

- g_{n}

convention the compensation push reverses sign automatically — the geometry is handled by the cosine, not by hand.

Very short time-to-go (

t_{g o} \to 0

at impact).

The ZEM-based augmentation

\frac{N ^{'}}{2} a_{T}

stays finite while

\dot{λ}

can spike; gravity's contribution over the vanishing interval becomes negligible, so the

- g_{n}

term matters least right at the end.