Optimal guidance — ZEM - ZEV formulation: Worked examples

Before anything, a one-line refresher so no symbol is used unearned. Each reveal line below hides its answer — quiz yourself, then check:

Recall The two laws (from the parent note)

What is $t_{g o}$ ? ::: time-to-go, $t_{f} - t$ , the seconds remaining until the deadline. What is $ZEM$ ? ::: Zero-Effort-Miss — the position error you'd have if you stopped steering now and coasted under gravity only. What is $ZEV$ ? ::: Zero-Effort-Velocity — the velocity error under that same coast. Soft-landing law (position + velocity)? ::: $a_{c} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV$ Intercept law (position only)? ::: $a_{c} = \frac{3}{t _{g o}^{2}} ZEM$

And the two prediction formulas we will use over and over:

The scenario matrix

Every problem this topic can pose falls into one of these cells. Each row is a "class of case"; the last column names the worked example that nails it. The figure below the table draws each cell as a little picture so you can see the nine situations before you compute them.

#	Case class	What is special about it	Sign / limit behaviour to watch	Covered by
C1	Baseline soft landing, all quantities positive-ish	The "normal" full-law case	ZEM $> 0$ , ZEV $> 0$ , opposite-sign terms	Ex 1
C2	Both errors already zero (degenerate)	You are perfectly on the coast trajectory	$a_{c} = 0$ , no divide-by-zero	Ex 2
C3	Negative ZEM, negative ZEV (coast overshoots and too fast)	Signs flow straight through, command flips	ZEM $< 0$ , ZEV $< 0$ , $a_{c} < 0$	Ex 3
C4	ZEM and ZEV pull opposite ways	Terms nearly cancel	small command from two big numbers	Ex 4
C5	$t_{g o} \to 0$ limit (almost out of time)	Coefficient blow-up $1/ t_{g o}^{2}$	command explodes → thruster saturation	Ex 5
C6	Intercept, full 2-D vector	Position-only law, navigation ratio $N = 3$	geometry, direction of $a_{c}$	Ex 6 (figure)
C7	Intercept where naive gravity-drop matters	Forgetting $g$ gives wrong ZEM	quantify the error	Ex 7
C8	Real-world word problem (Mars descent)	Extract numbers from prose	full soft-landing law	Ex 8
C9	Exam twist: solve for the $t_{g o}$ that makes a command achievable	Invert the law	quadratic in $t_{g o}$	Ex 9

Figure — Optimal guidance — ZEM - ZEV formulation

In the figure each tile is one matrix cell: the cyan arrow is where a pure coast takes you, the amber arrow is the residual error (ZEM/ZEV) the command must kill. Notice C2 has no amber arrow (nothing to fix) and C5 shows the same small miss but almost no time — the reason the command explodes.

The worked examples

Worked example Ex 1 — C1: Baseline vertical soft landing

A Moon lander: altitude $r = 100$ m, velocity $v = - 20$ m/s (minus = downward), targets $r_{f} = 0$ , $v_{f} = 0$ (touch down gently), gravity $g = - 1.62$ m/s², and $t_{g o} = 10$ s.

Forecast: Before computing — will the commanded acceleration be positive (upward, i.e. braking) or negative? Guess now.

ZEM. Why this step? We must first know how badly a pure coast misses the pad. $ZEM = 0 - 100 - (- 20) (10) - \frac{1}{2} (- 1.62) (100) = - 100 + 200 + 81 = 181$ m.
ZEV. Why? Same for the touchdown speed. $ZEV = 0 - (- 20) - (- 1.62) (10) = 20 + 16.2 = 36.2$ m/s.
Command. Why? Feed both into the full law. $a_{c} = \frac{6}{1 0 ^{2}} (181) - \frac{2}{10} (36.2) = 10.86 - 7.24 = 3.62$ m/s².

Verify: $a_{c} > 0$ → upward → braking. Correct: coasting overshoots up by 181 m and arrives too fast, so we thrust up to slow the fall. Units: $\frac{1}{s ^{2}} \cdot m - \frac{1}{s} \cdot \frac{m}{s} = \frac{m}{s ^{2}}$ ✓.

Worked example Ex 2 — C2: You are already on the perfect coast (degenerate)

Same lander, but now it sits exactly on the coast that lands softly: $r = 100$ m, $v = ?$ chosen so that ZEM $= 0$ and ZEV $= 0$ . Take $t_{g o} = 10$ , $g = - 1.62$ , $r_{f} = v_{f} = 0$ .

Forecast: If both errors are zero, what should the guidance command be? Guess.

Find the $v$ that zeros ZEV. Why? ZEV $= 0 \Rightarrow v_{f} - v - g t_{g o} = 0 \Rightarrow v = - g t_{g o} = 16.2$ m/s… but that must also zero ZEM. Solve both together instead: we need $r, v$ on a landing coast. Set ZEM $= 0$ : $0 - 100 - 10 v - \frac{1}{2} (- 1.62) (100) = 0 \Rightarrow - 100 - 10 v + 81 = 0 \Rightarrow v = - 1.9$ m/s.
Check ZEV with that $v$ . Why? A general point won't zero both — this shows only special states are "effort-free." $ZEV = 0 - (- 1.9) - (- 1.62) (10) = 1.9 + 16.2 = 18.1 \neq = 0$ .
Build the state where BOTH are zero. Why this step? To exhibit a genuine C2 point we must coast backward from a perfect soft touchdown — only such a state has both errors vanish, and we need it to test that the law returns exactly zero. At $t_{g o}$ before landing softly, $v = v_{f} - g t_{g o} = 0 - (- 1.62) (10) = 16.2$ m/s (upward), and $r = r_{f} - v t_{g o} - \frac{1}{2} g t_{g o}^{2} = 0 - (16.2) (10) - \frac{1}{2} (- 1.62) (100) = - 162 + 81 = - 81$ m.
Command. Why this step? This is the whole test — plug the zeroed errors into the law and confirm the optimal effort really is nothing. $a_{c} = \frac{6}{100} (0) - \frac{2}{10} (0) = 0$ .

Verify: With ZEM $= ZEV = 0$ the law gives exactly $a_{c} = 0$ — no thrust wasted. This is the whole point of "zero-effort": if the free coast already satisfies the boundary conditions, the optimal effort is literally nothing. (Note: no division by zero occurred; $t_{g o} = 10$ , not $0$ .)

Worked example Ex 3 — C3: Both errors genuinely negative

Lander with the target above it, drifting up too fast: $r = 100$ m, $v = + 40$ m/s (rising), $r_{f} = 200$ m (aim higher), $v_{f} = 0$ , $g = - 1.62$ , $t_{g o} = 10$ .

Forecast: Rising fast toward a target only 100 m higher — will the coast overshoot upward (so ZEM comes out negative), and will the required command therefore point down ( $a_{c} < 0$ )? Guess.

ZEM. Why? Predict where a pure coast leaves us relative to the raised target — this fixes the sign of the position error. $ZEM = 200 - 100 - (40) (10) - \frac{1}{2} (- 1.62) (100) = 100 - 400 + 81 = - 219$ m. Negative — the fast climb sails past and above the target: a coast overshoots, so we must pull down.
ZEV. Why this step? We also need the speed error, because a soft arrival demands $v_{f} = 0$ ; the coast's residual upward speed sets the sign of the velocity term. $ZEV = 0 - (40) - (- 1.62) (10) = - 40 + 16.2 = - 23.8$ m/s.
Command. Why this step? Both errors are known, so we feed them into the full soft-landing law to get the acceleration to apply now. $a_{c} = \frac{6}{100} (- 219) - \frac{2}{10} (- 23.8) = - 13.14 + 4.76 = - 8.38$ m/s².

Verify: $a_{c} < 0$ → downward → we brake the climb, exactly as forecast. Both ZEM and ZEV came out negative, so this genuinely hits cell C3 (contrast Ex 1, where both were positive). The arithmetic carries the sign automatically — you never insert an absolute value; trust the formula, not your gut about "which way." Units ✓.

Worked example Ex 4 — C4: The two terms nearly cancel

$ZEM = 40$ m, $ZEV = 12$ m/s, $t_{g o} = 4$ s. (Given directly — this is a mid-flight feedback tick.)

Forecast: The position term wants a big push; the velocity term subtracts. Will the net command be large or surprisingly small?

Position term. Why compute separately? To expose the tug-of-war between the two corrections before combining them. $\frac{6}{16} (40) = 15$ m/s².
Velocity term. Why separately too? So we can see how big a bite the ZEV correction takes out of the position push — the whole point of C4. $\frac{2}{4} (12) = 6$ m/s².
Net. Why subtract, not add? The soft-landing law carries a minus in front of the ZEV term (opposite-sign corrections near the end), so the two are differenced. $a_{c} = 15 - 6 = 9$ m/s².

Verify: Two sizeable numbers (15 and 6) produced a moderate 9. This is C4's lesson: near the end the ZEV term routinely eats a large chunk of the ZEM term — that opposite sign is doing real work. Units ✓.

Worked example Ex 5 — C5: The

$t_{g o} \to 0$ cliff Same residual errors as Ex 4 but now almost out of time: $ZEM = 40$ m, $ZEV = 12$ m/s, evaluate at $t_{g o} = 4, 2, 1, 0.5$ s.

Forecast: Halving $t_{g o}$ — does the command double, or worse?

$t_{g o} = 4$ : Why start here? It is our reference value from Ex 4, so every later row is compared against it. $\frac{6}{16} (40) - \frac{2}{4} (12) = 15 - 6 = 9$ m/s².
$t_{g o} = 2$ : Why halve to 2? To test the first doubling — if the law were linear in $1/ t_{g o}$ the command would merely double; watch it do more. $\frac{6}{4} (40) - \frac{2}{2} (12) = 60 - 12 = 48$ m/s².
$t_{g o} = 1$ : Why go to 1? To confirm the $1/ t_{g o}^{2}$ scaling of the dominant position term as time runs out. $\frac{6}{1} (40) - \frac{2}{1} (12) = 240 - 24 = 216$ m/s².
$t_{g o} = 0.5$ : Why this final tiny value? To make the blow-up unmistakable — it is where real thrusters saturate and the command becomes physically unrealisable. $\frac{6}{0.25} (40) - \frac{2}{0.5} (12) = 960 - 48 = 912$ m/s².

Verify: From 9 → 48 → 216 → 912: growth is faster than doubling, because the dominant $6/ t_{g o}^{2}$ term scales like $1/ t_{g o}^{2}$ — quarter the time, four-times the position term (15→60→240→960). This is the parent note's warning made numeric: a residual miss with little time left demands unbounded thrust, so real systems must drive ZEM→0 early.

In the figure, the cyan curve is the commanded acceleration versus $t_{g o}$ ; the amber dots mark our four rows. Notice how the curve rockets toward the vertical axis on the left — that steep wall is the saturation cliff you must stay away from by killing miss early.

Worked example Ex 6 — C6: 2-D intercept, position-only — and WHY it

is Proportional Navigation ( $N = 3$ ) An interceptor with predicted miss vector $ZEM = (300, - 450)$ m and $t_{g o} = 6$ s. Velocity is not constrained (we only need to hit).

Forecast: Which law — the $6/ t_{g o}^{2}$ or the $3/ t_{g o}^{2}$ one? And will $a_{c}$ point along $ZEM$ or against it?

Pick the law. Why? No $v_{f}$ requirement → position-only → intercept form $a_{c} = \frac{3}{t _{g o}^{2}} ZEM$ .
Apply componentwise. Why componentwise? Double-integrator dynamics decouple per axis (see Double Integrator Dynamics). $a_{c} = \frac{3}{36} (300, - 450) = (25, - 37.5)$ m/s².
Direction check. Why this step? To reveal a geometric fact that matters for every intercept: a single positive scalar ( $3/ t_{g o}^{2} > 0$ ) cannot rotate a vector, so the command must be parallel to $ZEM$ — you always thrust straight at the predicted miss, never sideways. $a_{c} = \frac{1}{12} (300, - 450)$ is a positive multiple of $ZEM$ . See the figure.
Prove it equals PN with $N = 3$ . Why this step? The parent note claims the intercept form is Proportional Navigation in disguise; we must actually show it, not assert it. In Proportional Navigation the acceleration is $a_{P N} = N V_{c} \dot{λ}$ , where $\dot{λ}$ is the line-of-sight (LOS) turn rate and $V_{c}$ the closing speed. For a constant-velocity target the ZEM measured perpendicular to the LOS obeys $ZEM_{⊥} = R \dot{λ} t_{g o}$ , where $R$ is range and $R = V_{c} t_{g o}$ (range closes at the closing speed). Substitute: $ZEM_{⊥} = V_{c} t_{g o}^{2} \dot{λ}$ . Now feed this into our law: $a_{c} = \frac{3}{t _{g o}^{2}} ZEM_{⊥} = \frac{3}{t _{g o}^{2}} V_{c} t_{g o}^{2} \dot{λ} = 3 V_{c} \dot{λ}$ . That is exactly $a_{P N}$ with $N = 3$ . The coefficient $3$ is the navigation ratio.

Verify: Magnitude $∣ a_{c} ∣ = 2 5^{2} + 37. 5^{2} = 625 + 1406.25 = 2031.25 \approx 45.07$ m/s². And the PN identity checks dimensionally: $[V_{c} \dot{λ}] = \frac{m}{s} \cdot \frac{1}{s} = \frac{m}{s ^{2}}$ ✓ — an acceleration, as it must be. The command is parallel to $ZEM$ because a single positive coefficient can't rotate a vector.

In the figure, the cyan arrow is $ZEM$ (interceptor to predicted miss point) and the amber arrow is $a_{c}$ (drawn six-times longer just to be visible). Observe that the two arrows lie on the same line — the whole content of the "direction check."

Worked example Ex 7 — C7: Forgetting gravity in the ZEM

Same interceptor geometry but now it flies through a gravity field $g = - 9.81$ m/s² acting on the $y$ -axis, $t_{g o} = 6$ s. The true position miss to the target requires the gravity-drop term. Suppose the raw geometric miss (target minus current straight-line coast, gravity ignored) is $(300, - 450)$ m.

Forecast: Adding the gravity-drop term $- \frac{1}{2} g t_{g o}^{2}$ to the $y$ -component — does the $y$ -miss grow or shrink in magnitude?

Gravity drift over the coast. Why? The vehicle falls $\frac{1}{2} g t_{g o}^{2}$ during $t_{g o}$ ; the coast already includes this, so the corrected $y$ -ZEM is the naive value minus $\frac{1}{2} g t_{g o}^{2}$ . $\frac{1}{2} g t_{g o}^{2} = \frac{1}{2} (- 9.81) (36) = - 176.58$ m.
Corrected ZEM. Why? We fold that drift into the position error so the command is aimed at the true miss, not the gravity-free illusion. $ZEM_{y} = - 450 - (- 176.58) = - 273.42$ m. (The $x$ -axis has no gravity: $ZEM_{x} = 300$ .)
Command. Why? Apply the intercept law to the corrected miss. $a_{c} = \frac{3}{36} (300, - 273.42) = (25, - 22.785)$ m/s².

Verify: The $y$ -command dropped from $- 37.5$ (Ex 6, no gravity) to $- 22.785$ — because gravity was already helping pull the vehicle down toward the target, so less commanded thrust is needed. Forgetting the $\frac{1}{2} g t_{g o}^{2}$ term would have over-commanded by $∣ - 37.5 - (- 22.785) ∣ = 14.715$ m/s² — exactly the "forgetting gravity" mistake from the parent note.

Worked example Ex 8 — C8: Real-world word problem (Mars descent)

"A descent stage is 500 m above the landing site, descending at 45 m/s, and drifting horizontally at 8 m/s. It must reach the pad ( $r_{f} = 0$ ) with zero velocity in $t_{g o} = 12$ s. Martian gravity is 3.71 m/s² downward. What is the commanded acceleration vector?"

Set axes (per our conventions): $x$ horizontal, $y$ up. So $r = (0, 500)$ , $v = (8, - 45)$ , $g = (0, - 3.71)$ , $r_{f} = 0$ , $v_{f} = 0$ .

Forecast: The horizontal axis has no gravity and no target offset except zeroing the drift — will its command be small?

ZEM per axis. Why? Coast prediction, axis by axis. $ZEM_{x} = 0 - 0 - (8) (12) - 0 = - 96$ m. $ZEM_{y} = 0 - 500 - (- 45) (12) - \frac{1}{2} (- 3.71) (144) = - 500 + 540 + 267.12 = 307.12$ m.
ZEV per axis. Why axis by axis here too? Because the double integrator decouples the axes (see Double Integrator Dynamics) — each direction has its own independent velocity error, and only $y$ feels gravity, so we must compute them separately to keep the gravity term on the right axis. $ZEV_{x} = 0 - 8 - 0 = - 8$ m/s. $ZEV_{y} = 0 - (- 45) - (- 3.71) (12) = 45 + 44.52 = 89.52$ m/s.
Full law per axis. Why full law? $v_{f} = 0$ is required (soft touchdown). $a_{c x} = \frac{6}{144} (- 96) - \frac{2}{12} (- 8) = - 4 + 1.3333 = - 2.6667$ m/s². $a_{cy} = \frac{6}{144} (307.12) - \frac{2}{12} (89.52) = 12.79667 - 14.92 = - 2.12333$ m/s².

Verify: $a_{c} \approx (- 2.667, - 2.123)$ m/s². The horizontal command is negative → it opposes the $+ 8$ m/s drift, killing it. Interestingly $a_{cy} < 0$ : the ZEV braking demand slightly exceeds the ZEM push here, so the commanded thrust points down a touch — sensible, since the coast was going to overshoot the pad's altitude by 307 m and needs the descent shaped, not just braked. Compare Powered Descent Guidance (Apollo E-Guidance). Units ✓.

Worked example Ex 9 — C9: Exam twist — invert the law for

$t_{g o}$ "Your thruster caps commanded acceleration at $a_{m a x} = 5$ m/s². Given a pure position miss $ZEM = 45$ m along one axis (intercept law, ZEV ignored), what is the smallest $t_{g o}$ at which the command is still achievable?"

Forecast: Since $a_{c} = 3 ZEM / t_{g o}^{2}$ grows as $t_{g o}$ shrinks, the constraint sets a floor on $t_{g o}$ . Bigger or smaller than 5 s? Guess.

Write the constraint. Why? We want the boundary where $a_{c} = a_{m a x}$ . $\frac{3 ZEM}{t _{g o}^{2}} = a_{m a x}$ .
Solve for $t_{g o}$ . Why square-root? It's quadratic in $t_{g o}$ . $t_{g o} = \frac{3 ZEM}{a _{m a x}} = \frac{3 ( 45 )}{5} = 27 \approx 5.196$ s.
Interpret. Why this step? A raw number is useless until we say what it means operationally, which is the whole point of the exam twist. For any $t_{g o} > 5.196$ s the command is below cap; at exactly $5.196$ s it saturates; below it, no achievable command can fix the miss — you must have driven ZEM down earlier.

Verify: Plug back: $\frac{3 ( 45 )}{27} = \frac{135}{27} = 5$ m/s² $= a_{m a x}$ ✓. This is the design principle from Ex 5 turned into a number: thruster saturation defines a deadline for killing residual miss — cross it and no legal command can save the shot, which is precisely why ZEM/ZEV is flown as continuous feedback that shrinks the miss while $t_{g o}$ is still comfortably large.

Recall Self-test

If ZEM = 0 and ZEV = 0, the optimal command is ::: exactly zero — you are already on a boundary-satisfying coast. Which law has no ZEV term, and what is its coefficient? ::: the intercept (position-only) law, $3/ t_{g o}^{2}$ . As $t_{g o}$ halves, the dominant ZEM term multiplies by ::: four (it scales as $1/ t_{g o}^{2}$ ). Why does forgetting $\frac{1}{2} g t_{g o}^{2}$ break a lander? ::: the predicted miss is wrong by exactly the gravity drift, so the command is miscalibrated. The two terms in the soft-landing law have opposite signs because ::: near $t_{f}$ correcting position and correcting velocity demand oppositely-directed acceleration.

See also: Optimal guidance — ZEM - ZEV formulation · Calculus of Variations & Pontryagin's Minimum Principle · Time-to-go estimation · Optimal Control — LQR · Lambert's Problem