Optimal guidance — ZEM - ZEV formulation: Question bank

A quick reminder of the vocabulary these questions lean on, so no symbol appears un-earned:

$r, v$ = the vehicle's current position and velocity vectors (measured in a fixed inertial frame — pick "up" and "downrange" as positive axes and stick to them).
$r_{f}, v_{f}$ = the desired position and velocity at the final time $t_{f}$ .
$g$ = gravitational acceleration = the known external acceleration acting during the coast (e.g. $g = - 1.62 \hat{u}$ m/s² on the Moon, pointing down — its sign follows your chosen frame).
$t_{g o}$ = time-to-go $= t_{f} - t$ , how much flight time remains.
ZEM = Zero-Effort-Miss = the position error at $t_{f}$ if you cut thrust now and coast, measured target minus predicted: a positive ZEM means the coast falls short of the target along that axis. $ZEM = r_{f} - r - v t_{g o} - \frac{1}{2} g t_{g o}^{2} .$
ZEV = Zero-Effort-Velocity = the velocity error at $t_{f}$ under the same coast, again desired minus predicted. $ZEV = v_{f} - v - g t_{g o} .$
$λ_{1}, λ_{2}$ = Lagrange multipliers (costates) — constant vectors introduced to enforce the two constraints while minimizing effort; they have no direct physical reading, they are the "prices" that make the optimum satisfy both endpoint conditions.
$a_{c}$ = the acceleration command applied right now (thrust acceleration, above gravity).
The two headline laws: full $a_{c} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV$ and position-only $a_{c} = \frac{3}{t _{g o}^{2}} ZEM$ .

The figure below fixes the geometry these questions keep referring to — the coast prediction, the two error vectors, and how they combine into the command.

Figure — Optimal guidance — ZEM - ZEV formulation

And this one shows why the late-time story is subtle once a real thruster can only push so hard:

True or false — justify

If ZEM is zero right now, the guidance command is automatically zero.

False for the full law.

a_{c} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV

, so a nonzero ZEV still commands thrust; only in the position-only law does ZEM=0 force

a_{c} = 0

ZEM and ZEV are computed by actually flying the coast and measuring where you land.

False. They come from the closed-form propagation

ZEM = r_{f} - r - v t_{g o} - \frac{1}{2} g t_{g o}^{2}

and

ZEV = v_{f} - v - g t_{g o}

, evaluated instantly each cycle — nothing is physically flown.

The gravity term

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

only matters for landers, not interceptors.

False. Any coast under gravity drifts; dropping

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

biases the predicted miss for interceptors too. It merely looks ignorable when the engagement is short so the drift is small.

The optimal acceleration profile over the whole trajectory is constant.

False. Calculus of variations on the quadratic effort cost gives

a^{*} (τ) = λ_{1} (t_{f} - τ) + λ_{2}

with constant multipliers

λ_{1}, λ_{2}

— an affine (linear-in-time) profile, not a constant.

Making the effort integral

J = \frac{1}{2} \int ∥ a ∥^{2} d τ

smaller always means less fuel.

Roughly but not exactly.

J

penalizes squared acceleration (a smoothness/energy proxy), while fuel tracks

\int ∥ a ∥ d t

; minimizing

J

gives the clean linear law but is not literally minimum-propellant.

The position-only ZEM/ZEV law and Proportional Navigation are genuinely the same law.

True. The position-only form

a_{c} = \frac{3}{t _{g o}^{2}} ZEM

is Proportional Navigation with effective navigation ratio

N = 3

; ZEM/ZEV just derives it as an optimal-control result.

The two ZEM/ZEV coefficients having opposite signs is a sign error someone should fix.

False. With ZEM and ZEV both defined desired minus predicted in one consistent frame, solving the

2 \times 2

Lagrange system yields the minus on the ZEV term; near

t_{f}

the position and velocity fixes pull acceleration in opposite directions.

ZEM/ZEV needs the vehicle model to be a double integrator.

True in the standard derivation. The closed-form coast (

\ddot{r} = a + g

) is what gives the tidy

t_{g o}

polynomials; other dynamics need generalized ZEM/ZEV via a state-transition matrix.

Spot the error

"For a soft landing I'll use

a_{c} = \frac{6}{t _{g o}^{2}} ZEM

since 6 is the position coefficient."

The full landing law also carries

- \frac{2}{t _{g o}} ZEV

; drop it and velocity is never matched, so you arrive at the right spot but at killing speed.

"For pure interception I'll use

\frac{6}{t _{g o}^{2}} ZEM

— same position term as landing."

The interception coefficient is

3/ t_{g o}^{2}

, not

6/ t_{g o}^{2}

. The 6 belongs to the two-constraint problem; removing the ZEV constraint changes the moment equations and halves it.

"Both ZEM and ZEV are errors, so both correction terms should add."

The Lagrange solution yields

+ \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV

. The velocity term subtracts because matching arrival velocity often demands the opposite acceleration direction to closing the position gap.

"I integrated position with weight 1 like velocity:

\int_{t}^{t_{f}} a d τ = ZEM

Position needs the lever-arm weight:

\int_{t}^{t_{f}} (t_{f} - τ) a d τ = ZEM

. Early thrust accumulates through two integrations, so it carries the extra

(t_{f} - τ)

factor.

g

is gravity so I add it to

a_{c}

before commanding the thruster."

Gravity is already inside the coast prediction (the

g t_{g o}

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

terms);

a_{c}

is the thrust acceleration above gravity. Adding

g

again double-counts it.

"Since ZEV is a velocity, dividing it by

t_{g o}

gives nonsense units."

Units check out:

ZEV

is m/s, and

ZEV / t_{g o}

is (m/s)/s = m/s², an acceleration — exactly what a command needs.

"I'll compute ZEM once at the start and reuse it."

ZEM/ZEV are meant as continuous feedback; recomputing every cycle lets small residuals be killed while

t_{g o}

is still large, keeping the command bounded.

"The law diverges as

t_{g o} \to 0

, so a real vehicle's command also diverges."

No — real thrusters saturate at a maximum acceleration. Once the ideal command exceeds that cap it clips, so late-time behavior is set by the actuator limit, not the

1/ t_{g o}^{2}

singularity (which is why you must null the errors early).

Why questions

Why does the position corrector get

1/ t_{g o}^{2}

but the velocity corrector only

1/ t_{g o}

A position error must be fixed over an interval, so its needed acceleration scales as distance

/ t_{g o}^{2}

; a velocity error is a one-integration gap, needing acceleration

\sim

velocity

/ t_{g o}

Why is the optimal control an affine function of time rather than something wilder?

The cost is quadratic in

a

and the two constraints are linear integrals, so pointwise stationarity of the Lagrangian forces

a^{*} = λ_{1} (t_{f} - τ) + λ_{2}

, linear in remaining time.

Why do the coefficients blow up as

t_{g o} \to 0

With almost no time left, only a huge acceleration can erase a residual miss (

1/ t_{g o}^{2} \to \infty

). Good guidance drives ZEM/ZEV to zero early so this singularity is never reached — and if it is, actuator saturation caps the response.

Why express the constraints as ZEM and ZEV instead of raw endpoint conditions?

Subtracting the coast part isolates the correctable error, turning a boundary-value problem into "the control just has to fill these two known gaps" — cheap and real-time.

Why does ZEM/ZEV relate to Powered Descent Guidance (Apollo E-Guidance)?

Apollo's E-Guidance is essentially the same idea — predict the coast endpoint, correct position and velocity with

t_{g o}

-scaled gains — so ZEM/ZEV is a clean modern restatement of that lineage.

Why is Time-to-go estimation a make-or-break subproblem here?

Every gain contains

t_{g o}

; a wrong

t_{g o}

mis-scales both terms and can destabilize the loop, so estimating it accurately is as important as the law itself.

Why doesn't ZEM/ZEV need to know the target's future motion for a maneuvering target?

It uses the current predicted coast miss; recomputing each cycle folds in the target's latest state, so no full future trajectory forecast (unlike solving Lambert's Problem) is required.

Edge cases

What does the law command if the vehicle is already perfectly on the coast trajectory (ZEM=0, ZEV=0)?

a_{c} = 0

: both terms vanish, so it coasts — the coast is optimal when no error exists.

What happens if you plug

t_{g o} = 0

literally into the gains?

You divide by zero; the law is undefined at the instant of arrival. In practice guidance is cut off (or frozen) at a small

t_{g o}

threshold before the singularity.

If ZEM and ZEV are nonzero but the two correction terms exactly cancel, does the vehicle stop steering?

Yes at that instant

a_{c} = 0

, but it is momentary; as

t_{g o}

shrinks the two terms scale differently (

1/ t_{g o}^{2}

1/ t_{g o}

), so the balance breaks and steering resumes.

For a pure interceptor, does matching final velocity matter at all?

No — interception only requires

r (t_{f}) = r_{f}

, so the ZEV constraint is dropped entirely and the law collapses to

\frac{3}{t _{g o}^{2}} ZEM

What if the ideal command exceeds the thruster's maximum acceleration?

It gets clipped to the actuator limit (saturation); the vehicle then under-corrects, so the guidance may not achieve the endpoint — the reason well-designed loops keep the ideal command inside the cap by nulling errors early.

What if gravity is zero (deep-space, coasting between burns)?

The gravity terms drop out of ZEM/ZEV, leaving straight-line ballistic prediction; the guidance structure and coefficients are unchanged.

If your

t_{g o}

estimate is too large, how does the command behave?

The gains

6/ t_{g o}^{2}

and

2/ t_{g o}

come out too small, so the vehicle under-corrects early and must scramble late — the opposite of the smooth profile the law was designed to give.

Recall One-line summary of every trap

Gravity is already inside the prediction; ZEM and ZEV are both desired minus predicted coast in one fixed frame; the ZEV term subtracts; position gains go as $1/ t_{g o}^{2}$ and velocity gains as $1/ t_{g o}$ ; the two distinct gains are $\frac{6}{t _{g o}^{2}}$ and $\frac{2}{t _{g o}}$ for landing versus a single $\frac{3}{t _{g o}^{2}}$ for intercept; recompute every cycle; and a real thruster's saturation cap — not just the $t_{g o} \to 0$ singularity — governs late-time behavior.