Worked examples — Augmented proportional navigation — gravity compensation

This page is a shooting gallery. We first draw a map of every kind of situation the augmented-PN gravity-compensation law can face, then we walk one fully-worked example through each square of that map. If you have read Augmented proportional navigation — gravity compensation, every symbol below is already yours; if a symbol still feels fuzzy, the parent note builds it from zero.

The one formula we keep firing:

Quick reminder of the plain-word meaning of each letter, so nothing is used unearned:

• $a_c$ — the sideways ("lateral") push we command, in metres per second squared.
• $N^{\prime }$ — the navigation constant, a plain unitless number, usually $3$ to $5$.
• $V_c$ — closing speed, how fast the gap between missile and target shrinks (m/s).
• $\dot{\lambda }$ — how fast the line-of-sight (LOS) direction is rotating (rad/s); the little dot means "rate of change".
• $a_T$ — the target's own sideways acceleration, the part pointed across the LOS (m/s²).
• $g$ — the strength of gravity, i.e. how fast an object speeds up as it falls freely. Near Earth's surface $g=9.81{\text{m/s}}^2$ (every second of free fall adds about $9.81$ m/s of downward speed). This single number is the whole reason the missile "droops". It is a fixed constant of the environment, not something we control.
• $g_n$ — the slice of gravity that points across the LOS and therefore bends our flight (m/s²).
• $\lambda$ — the LOS elevation angle: $0°$ = looking flat along the ground, $90°$ = looking straight up.

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The scenario matrix

Every worked example below is tagged [Cell N] so you can see we leave no square empty.

Cell	Case class	What is special	Which example
1	Horizontal LOS, non-maneuvering target	$\lambda =0\Rightarrow g_n=g$ (worst gravity)	Ex 1
2	Steep descending LOS, maneuvering target	$\lambda$ large $\Rightarrow g_n$ small	Ex 2
3	Degenerate: $\dot{\lambda }=0$ (momentary collision course)	PN term vanishes, only $-g_n$ acts	Ex 3
4	Vertical shot, straight up ($\lambda =+90°$)	$g_n=0$, gravity along LOS	Ex 4
4b	Vertical shot, straight down ($\lambda =-90°$)	$g_n=0$ too — completes the $\pm 90°$ edge	Ex 4b
5	Sign flip: LOS rotating the other way ($\dot{\lambda }<0$)	commanded push flips sign	Ex 5
6	Diving / negative elevation ($\lambda <0$)	$\cos \lambda$ still positive → $g_n>0$	Ex 6
7	Real-world word problem: dart-thrower geometry	build $g_n$ from a picture	Ex 7
8	Exam twist: acceleration limit saturates the command	clip $a_c$ to autopilot cap	Ex 8

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Worked examples

Ex 1 — Horizontal LOS, worst-case gravity [Cell 1]

1. PN term. $N^{\prime }{V}_c\dot{\lambda }=4(800)(0.01)=32{\text{m/s}}^2$. Why this step? This is what pure PN alone demands from the LOS spin — our baseline.
2. Gravity term. $\lambda =0\Rightarrow g_n=g\cos 0=9.81{\text{m/s}}^2$. Why this step? Horizontal LOS means all of gravity points across the LOS — the most bending possible, the maximum $g_n$.
3. Assemble. $a_c=32+0-9.81=22.19{\text{m/s}}^2$. Why this step? We subtract $g_n$ so that when gravity later adds $+9.81$ back on, the net across-LOS acceleration is exactly the intended $32$.

Verify: effective normal accel $=a_c+g_n=22.19+9.81=32.00{\text{m/s}}^2$ — precisely the pure-PN demand. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$ (radians are unitless), and $g_n$ is already m/s², so every term adds cleanly in m/s². Answer smaller than 32, matching the forecast (we back off, because gravity does part of the pushing for us — but in the wrong direction, so it net-cancels).

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Ex 2 — Steep descent onto a maneuvering target [Cell 2]

Figure s01 — Ex 2 geometry. The yellow arrow is the steeply-descending LOS ($\lambda =-60°$); the pink arrow is gravity pointing straight down; the dotted white arrow is the large slice of gravity lying along the LOS (harmless — it only changes closing speed); the short blue arrow is $g_n$, the small slice lying across the LOS. The picture shows why steep geometry makes $g_n$ small.

1. PN term. $3(1000)(0.004)=12{\text{m/s}}^2$. Why this step? Small $\dot{\lambda }$ → modest baseline demand. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
2. Augmentation term. $\frac{N^{\prime }}{2}{a}_T=\frac{3}{2}(30)=45{\text{m/s}}^2$. Why this step? The $\frac{N^{\prime }}{2}$ feed-forward exactly pre-cancels the zero-effort-miss that a constant target maneuver would otherwise cause. Here $t_{go}$ is the time-to-go — the seconds remaining until impact; a constant maneuver builds a miss $\frac{1}{2}{a}_T{t}_{go}^2$ over that window, so the $\frac{N^{\prime }}{2}$ feed-forward is tuned to erase it whatever the current $t_{go}$ happens to be — see Zero-Effort-Miss (ZEM) guidance. Units: $N^{\prime }/2$ is unitless, so $\frac{N^{\prime }}{2}{a}_T$ stays in m/s².
3. Gravity term. $g_n=g\cos 60°=9.81(0.5)=4.905{\text{m/s}}^2$. Look at the figure: the LOS points steeply down, so the pink gravity arrow lies mostly along the LOS; only the short blue stub is $g_n$. Why this step? Steep geometry hides most of gravity inside the LOS direction, where it can't bend $\dot{\lambda }$.
4. Assemble. $a_c=12+45-4.905=52.10{\text{m/s}}^2$. Why this step? Gravity's contribution is now the smallest of the three — target maneuver rules.

Verify: $12+45-4.905=52.095\approx 52.10{\text{m/s}}^2$. Forecast check: the augmentation term (45) dominates, as predicted. All three terms are in m/s², so the sum is a valid acceleration.

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Ex 3 — Degenerate collision course, $\dot{\lambda }=0$ [Cell 3]

1. Uncompensated PN. $a_c^{\text{PN}}=N^{\prime }{V}_c(0)+0=0$. Why this step? A pure-PN brain sees no LOS spin and relaxes — commands nothing.
2. The trap. With zero command, gravity's $g_n=9.81$ acts unopposed → next instant the flight droops → $\dot{\lambda }$ becomes nonzero → PN scrambles to correct after the error appeared. Why this step? Shows the ghost: a "fixed" LOS is not stable without gravity handling.
3. Compensated APN. $a_c=0+0-9.81=-9.81{\text{m/s}}^2$. Why this step? The negative sign means "push upward across the LOS", exactly cancelling the impending droop so $\dot{\lambda }$ stays truly $0$.

Verify: net across-LOS accel $=a_c+g_n=-9.81+9.81=0$. Units: both terms m/s², difference m/s² → the LOS rate has zero forcing → stays fixed → a genuine, held collision course. Answer: $-9.81{\text{m/s}}^2$, not zero — the forecast "free to command zero" is wrong.

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Ex 4 — Straight-up shot, gravity fully harmless [Cell 4]

1. PN term. $5(600)(0.006)=18{\text{m/s}}^2$. Why this step? Standard baseline from LOS spin. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
2. Gravity term. $g_n=g\cos 90°=9.81(0)=0$. Why this step? Straight up, gravity points along the LOS, so it only slows the closing speed — it cannot bend the LOS. Nothing to cancel.
3. Assemble. $a_c=18+0-0=18{\text{m/s}}^2$.

Verify: $g_n=0$ as expected for a vertical LOS. Command equals raw PN, in m/s². Forecast answer: zero compensation, correct.

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Ex 4b — Straight-down shot, completing the $\pm 90°$ edge [Cell 4b]

1. PN term. $5(600)(0.006)=18{\text{m/s}}^2$. Why this step? The LOS-spin term does not care which way the LOS points; same baseline as Ex 4.
2. Gravity term. $g_n=g\cos (-90°)=9.81\cos 90°=0$. Why this step? $\cos$ is even, so $\cos (-90°)=\cos (90°)=0$. Whether we shoot straight up or straight down, gravity lies entirely along the LOS — no across-LOS slice exists.
3. Assemble. $a_c=18+0-0=18{\text{m/s}}^2$.

Verify: $g_n=0$ identically, matching the straight-up case. Units m/s². This closes the geometry map: at both $\lambda =+90°$ and $\lambda =-90°$ gravity is harmless. Forecast answer: no change — up and down are symmetric here.

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Ex 5 — LOS spinning the other way, $\dot{\lambda }<0$ [Cell 5]

1. PN term. $4(750)(-0.008)=-24{\text{m/s}}^2$. Why this step? A negative $\dot{\lambda }$ means the LOS is swinging the opposite way, so PN pushes the opposite direction — the sign is carried through honestly. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
2. Gravity term. $g_n=9.81{\text{m/s}}^2$ (still positive: gravity always pulls the same physical way regardless of how the LOS spins). Why this step? Gravity's direction is set by the world, not by $\dot{\lambda }$'s sign — do not accidentally flip it.
3. Assemble. $a_c=-24+0-9.81=-33.81{\text{m/s}}^2$.

Verify: effective normal accel $=a_c+g_n=-33.81+9.81=-24$, matching the pure-PN demand of $-24$. Units m/s² throughout. The command sign is dominated by the PN term but the gravity term is not flipped. Answer $-33.81{\text{m/s}}^2$.

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Ex 6 — Diving engagement, negative elevation [Cell 6]

Figure s02 — Ex 6. Two LOS arrows are drawn, one tilted $+30°$ above horizontal (blue $g_n$ stub) and one tilted $-30°$ below (pink $g_n$ stub). Because $\cos$ is an even function, the two blue/pink stubs are exactly the same length: a downward-tilted LOS needs the same gravity compensation as an upward-tilted one at the same angle.

1. Gravity term. $g_n=g\cos (-30°)=g\cos 30°=9.81(0.8660)=8.496{\text{m/s}}^2$. Why this step? Cosine is an even function: $\cos (-30°)=\cos (30°)$. Whether the LOS tilts up or down by the same angle, the amount of gravity lying across it is identical — the figure shows both the up-tilt and down-tilt giving the same stub length.
2. PN term. $3.5(900)(0.005)=15.75{\text{m/s}}^2$. Why this step? Baseline from LOS spin. Units: $(\text{rad/s})\cdot (\text{m/s})={\text{m/s}}^2$.
3. Assemble. $a_c=15.75+0-8.496=7.254{\text{m/s}}^2$.

Verify: $g_n$ came out positive (8.496), because $\cos$ ignores the sign of the tilt — forecast "negative $g_n$" is wrong. $15.75-8.496=7.254{\text{m/s}}^2$. Both terms m/s², difference m/s².

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Ex 7 — Word problem: the falling-dart geometry [Cell 7]

Figure s03 — Ex 7. The yellow arrow is the shallow climbing LOS at $\lambda =20°$. Gravity (pink, straight down) is split by dropping perpendiculars into a slice along the LOS ($g\sin 20°$, dotted white) and a slice across the LOS ($g_n=g\cos 20°$, blue). At shallow tilt the blue slice is nearly the full $g$ — almost worst-case bending.

1. Read the geometry. In the figure the LOS is the yellow arrow tilted $20°$; gravity (pink) points straight down. Drop a perpendicular from the LOS: the piece of gravity across the LOS is $g\cos 20°$, the piece along it is $g\sin 20°$. Why this step? Only the across-LOS slice bends the flight, so that is what we cancel.
2. Gravity term. $g_n=9.81\cos 20°=9.81(0.9397)=9.219{\text{m/s}}^2$. Why this step? Shallow tilt → cosine near 1 → almost all of gravity still bends the LOS (near worst case).
3. PN term. $4(500)(0.012)=24{\text{m/s}}^2$.
4. Augmentation term. $\frac{4}{2}(15)=30{\text{m/s}}^2$. Why this step? The drone is maneuvering, so we feed-forward its miss with $N^{\prime }/2$ (same $t_{go}$-independent logic as Ex 2).
5. Assemble. $a_c=24+30-9.219=44.781{\text{m/s}}^2$.

Verify: $g_n=9.219{\text{m/s}}^2$, close to the full $9.81$ — forecast "nearly full-strength" confirmed. $24+30-9.219=44.781{\text{m/s}}^2$. All terms m/s².

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Ex 8 — Exam twist: autopilot acceleration limit [Cell 8]

1. Compute the unclipped demand. $a_c=44.781{\text{m/s}}^2$ (from Ex 7). Why this step? The guidance law runs first; the limiter acts on its output.
2. Apply the cap. Since $44.781>40$, the delivered command is clipped to $a_c^{\text{cmd}}=40{\text{m/s}}^2$. Why this step? The airframe/structure can physically pull only so hard; anything beyond is impossible.
3. Lost authority. Shortfall $=44.781-40=4.781{\text{m/s}}^2$. Why this step? This deficit is what causes a residual miss — and it is exactly the argument for not wasting g-capacity chasing gravity: had we skipped compensation the gravity ghost would eat even more of the cap.

Verify: clipped value $=\min (44.781,40)=40$. Lost $=4.781{\text{m/s}}^2$, in m/s². The compensation term ($-9.219$) is inside the number being clipped, so at saturation the fight against gravity is partially sacrificed along with everything else — forecast "still gets through fully" is false. This is why staying below the cap (Ex 1–4 style) matters.

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Recall Fast self-test

Horizontal LOS makes $g_n$ equal to what? ::: The full $g$ (because $\cos 0=1$) — worst case. A negative elevation $\lambda =-30°$ gives what sign of $g_n$? ::: Positive, since $\cos$ is even ($\cos (-30°)=\cos 30°$). If $\dot{\lambda }=0$ and target is inertial, what does compensated APN command horizontally? ::: $-g_n=-9.81{\text{m/s}}^2$, pre-empting the droop. Straight-up AND straight-down shots need how much compensation? ::: Zero in both, because $g_n=g\cos (\pm 90°)=0$. Where does the autopilot limiter act — before or after the guidance law? ::: After: it clips the law's output $a_c$.

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Connections

• Augmented proportional navigation — gravity compensation — the parent law these examples exercise.
• Proportional Navigation (PN) — the $N^{\prime }{V}_c\dot{\lambda }$ baseline term.
• Zero-Effort-Miss (ZEM) guidance — origin of the $N^{\prime }/2$ augmentation factor and the $t_{go}$ picture.
• Line-of-Sight rate estimation — where $\dot{\lambda }$ comes from in each example.
• Missile Autopilot & Acceleration limits — the cap that clips Ex 8.
• Coordinate frames & projections — how $g_n=g\cos \lambda$ is a projection of $\vec{g}$ onto the LOS-normal.
• Ballistic trajectory & gravity turn — the "droop" these examples fight.