Visual walkthrough — Optimal guidance — ZEM - ZEV formulation

We will use exactly three characters throughout. Let me draw them once, plainly, before any maths:

• $\mathbf{r}$ — where the vehicle is right now (a dot in space, an arrow from the origin to the dot).
• $\mathbf{v}$ — which way and how fast it is moving right now (an arrow attached to the dot).
• $\mathbf{a}$ — the push we control (the thrust arrow — the only thing we get to choose).

The bold-face just means "this is an arrow, it has a direction, not only a size." A smart 12-year-old thinks: position is a place, velocity is a speed-arrow, acceleration is a shove. That is all $\mathbf{r},\mathbf{v},\mathbf{a}$ ever mean.

Step 1 — Draw the world: a dot that coasts under gravity

WHAT. Before we steer anything, let's watch the vehicle do nothing — engine off. Only gravity $\mathbf{g}$ acts. This "do-nothing" future is the whole trick.

WHY. The genius of ZEM/ZEV is to ask "if I quit steering right now, where do I end up?" To answer that we must first know how a coasting dot moves. That motion is double-integrator dynamics: acceleration integrated once gives velocity, integrated twice gives position.

PICTURE. Look at the figure. The white dot starts at $\mathbf{r}$ with speed-arrow $\mathbf{v}$. With the engine off it drifts along the pale-yellow dashed curve, bending because gravity keeps adding downward speed.

The coasting dot lands (at the deadline $t_f$) at a place and a speed we can predict with school kinematics, using the countdown $t_{go}$:

$\underset{\text{where it drifts to}}{\underbrace{{\mathbf{r}}_{coast}(t_f)}}=\underset{\text{start place}}{\underbrace{\mathbf{r}}}+\underset{\text{drift from current speed}}{\underbrace{\mathbf{v}{t}_{go}}}+\underset{\text{gravity sag}}{\underbrace{\frac{1}{2}\mathbf{g}{t}_{go}^2}}$

$\underset{\text{drift speed at end}}{\underbrace{{\mathbf{v}}_{coast}(t_f)}}=\underset{\text{current speed}}{\underbrace{\mathbf{v}}}+\underset{\text{speed gravity adds}}{\underbrace{\mathbf{g}{t}_{go}}}$

Each term is a picture: $\mathbf{v}{t}_{go}$ is "speed times time = distance," the classic straight coast; $\frac{1}{2}\mathbf{g}{t}_{go}^2$ is the extra downward sag gravity produces.

Step 2 — Name the two errors: ZEM and ZEV

WHAT. Compare where the coasting dot ends up with where we want it: the target place ${\mathbf{r}}_f$ and target speed ${\mathbf{v}}_f$. The gaps are our two errors.

WHY. If those gaps are zero, we don't need to do anything — coasting already wins. So the only reason to burn fuel is to close these gaps. Naming them lets us aim the engine at exactly the thing that's wrong.

PICTURE. Two arrows in the figure: the chalk-blue arrow ZEM points from the coast-landing spot to the target spot (a position gap). The chalk-pink arrow ZEV points from the coast-arrival speed to the target speed (a velocity gap, drawn in the speed-diagram inset).

Reading the ZEM formula term by term: start from the target ${\mathbf{r}}_f$, then subtract everything the free coast already gives you ($\mathbf{r}$, the drift $\mathbf{v}{t}_{go}$, the sag $\frac{1}{2}\mathbf{g}{t}_{go}^2$). Whatever is left over is the gap the engine must fill.

Step 3 — Write the true ending, engine included

WHAT. Now turn the engine on. The commanded acceleration can be any wiggle over the remaining seconds. We write it as $\mathbf{a}(\tau )$, where $\tau$ is a dummy time variable that sweeps across the flight — it runs from now, $\tau =t$, to the deadline, $\tau =t_f$, i.e. $\tau \in [t,t_f]$. Add up its effect on the final speed and final place.

WHY. We can't choose a good $\mathbf{a}$ until we know how each instant of pushing changes the endpoint. So we integrate the true dynamics $\dot{\mathbf{r}}=\mathbf{v},\dot{\mathbf{v}}=\mathbf{a}+\mathbf{g}$. The symbol ${\int }_t^{t_f}(\cdots )d\tau$ just means "add up the contribution of every instant $\tau$ from now until the deadline."

PICTURE. The figure slices the remaining time into thin bars. Each bar is a little push $\mathbf{a}(\tau )d\tau$. A push done early (left bars) has a long lever-arm $(t_f-\tau )$ — it keeps moving the position for a long time. A push done late (right bars) barely nudges position. The bar heights times their lever arms is the picture behind the integral.

$\mathbf{v}(t_f)=\underset{\text{the coast part}}{\underbrace{\mathbf{v}+\mathbf{g}{t}_{go}}}+\underset{\text{all the pushes, summed}}{\underbrace{{\int }_t^{t_f}\mathbf{a}(\tau )d\tau }}$

$\mathbf{r}(t_f)=\underset{\text{the coast part}}{\underbrace{\mathbf{r}+\mathbf{v}{t}_{go}+\frac{1}{2}\mathbf{g}{t}_{go}^2}}+\underset{\text{pushes, weighted by remaining time}}{\underbrace{{\int }_t^{t_f}\underset{\text{lever arm}}{\underbrace{(t_f-\tau )}}\mathbf{a}(\tau )d\tau }}$

Step 4 — The constraints become "fill the two gaps"

WHAT. Demand that the true ending equals the target: $\mathbf{r}(t_f)={\mathbf{r}}_f$ and $\mathbf{v}(t_f)={\mathbf{v}}_f$. Move the coast parts to the other side.

WHY. The coast parts on the left are exactly the things we subtracted to define ZEM and ZEV. So the boundary conditions collapse into two clean statements about the pushes.

PICTURE. Two "buckets" in the figure. The velocity bucket must be filled to the level ZEV by the raw sum of pushes. The position bucket must be filled to level ZEM, but each push pours in scaled by its lever-arm $(t_f-\tau )$.

$\underset{\text{lever-weighted pushes}}{\underbrace{{\int }_t^{t_f}(t_f-\tau )\mathbf{a}(\tau )d\tau }}=\mathbf{ZEM},\underset{\text{raw pushes}}{\underbrace{{\int }_t^{t_f}\mathbf{a}(\tau )d\tau }}=\mathbf{ZEV}$

There are infinitely many push-profiles $\mathbf{a}(\tau )$ that fill both buckets. Which one is best?

Step 5 — "Best" means least effort → a straight-line push profile

WHAT. Among all profiles that fill both buckets, pick the one that spends least effort. We measure effort by $J=\frac{1}{2}{\int }_t^{t_f}\Vert \mathbf{a}(\tau ){\Vert }^{2}d\tau .$ Using calculus of variations with two Lagrange multipliers ${\mathbf{\lambda }}_1,{\mathbf{\lambda }}_2$, the answer is forced to be linear in remaining time.

WHY the profile comes out linear (the proof-in-a-picture). This is the key result, so let's earn it rather than assert it. Lagrange's recipe says: to minimise $J$ subject to the two bucket constraints, subtract each constraint times an unknown constant multiplier and minimise the combined quantity freely: $\mathcal{L}={\int }_t^{t_f}[\frac{1}{2}\Vert \mathbf{a}(\tau ){\Vert }^2-{\mathbf{\lambda }}_1\cdot (t_f-\tau )\mathbf{a}(\tau )-{\mathbf{\lambda }}_2\cdot \mathbf{a}(\tau )]d\tau .$ Here ${\mathbf{\lambda }}_1$ pays for the position bucket (so it multiplies the lever weight $(t_f-\tau )$) and ${\mathbf{\lambda }}_2$ pays for the velocity bucket (weight $1$). Now here is the crucial idea: because $\tau$ appears separately inside the integral, we can minimise at each instant $\tau$ on its own — pick the best $\mathbf{a}$ for that one instant. The thing inside the bracket, for a single instant, is a plain upward parabola in $\mathbf{a}$: $f(\mathbf{a})=\frac{1}{2}\Vert \mathbf{a}{\Vert }^2-[{\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2]\cdot \mathbf{a}.$ A parabola's bottom is where its slope is zero. The slope (gradient) is $\mathbf{a}-[{\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2]$, so the minimum sits at $\boxed{{\mathbf{a}}^{\ast }(\tau )={\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2}$ That is a straight line in remaining time — a constant ${\mathbf{\lambda }}_2$ plus a slope ${\mathbf{\lambda }}_1$ times $(t_f-\tau )$ — precisely because the cost was a parabola (quadratic) and the constraint weights ($1$ and $(t_f-\tau )$) were the only $\tau$-shapes available to lean on.

PICTURE. The figure shows candidate profiles (wiggly, chalk-blue) all filling the buckets, and the winner (straight pale-yellow line). The straight line is the flattest way to deliver the required amounts — least squared area.

Here ${\mathbf{\lambda }}_1$ tilts the line and ${\mathbf{\lambda }}_2$ raises it. We don't know their values yet — the buckets will pin them down.

Step 6 — Substitute back and solve the tiny $2\times 2$ system

WHAT. Put the straight-line ${\mathbf{a}}^{\ast }$ into the two bucket-equations. Substitute $s=t_f-\tau$ (remaining time) so the integrals become clean powers of $t_{go}$.

WHY. Now the unknowns are just two constant vectors ${\mathbf{\lambda }}_1,{\mathbf{\lambda }}_2$. Two vector equations, two vector unknowns — plain algebra finishes the job.

PICTURE. The figure plots the two moment integrals ${\int }_0^{t_{go}}{s}^{k}ds$ as coloured areas under $s$ and $s^2$, giving the $\frac{t_{go}^2}{2}$ and $\frac{t_{go}^3}{3}$ factors you literally read off as triangle/curve areas.

${\mathbf{\lambda }}_1\underset{{\int }_0^{t_{go}}{s}^{2}ds}{\underbrace{\frac{t_{go}^3}{3}}}+{\mathbf{\lambda }}_2\underset{{\int }_0^{t_{go}}sds}{\underbrace{\frac{t_{go}^2}{2}}}=\mathbf{ZEM}$

${\mathbf{\lambda }}_1\underset{{\int }_0^{t_{go}}sds}{\underbrace{\frac{t_{go}^2}{2}}}+{\mathbf{\lambda }}_2\underset{{\int }_0^{t_{go}}ds}{\underbrace{t_{go}}}=\mathbf{ZEV}$

Let me name the two equations so the algebra is easy to track:

$\text{(I)}\frac{t_{go}^3}{3}{\mathbf{\lambda }}_1+\frac{t_{go}^2}{2}{\mathbf{\lambda }}_2=\mathbf{ZEM},\text{(II)}\frac{t_{go}^2}{2}{\mathbf{\lambda }}_1+t_{go}{\mathbf{\lambda }}_2=\mathbf{ZEV}$

Eliminate ${\mathbf{\lambda }}_2$. Multiply (I) by $2/t_{go}^2$ and (II) by $3/t_{go}$ so both carry the same ${\mathbf{\lambda }}_2$ coefficient:

$\text{(I)}\times \frac{2}{t_{go}^2}:\frac{2t_{go}}{3}{\mathbf{\lambda }}_1+{\mathbf{\lambda }}_2=\frac{2}{t_{go}^2}\mathbf{ZEM}$ $\text{(II)}\times \frac{3}{t_{go}}:\frac{3t_{go}}{2}{\mathbf{\lambda }}_1+3{\mathbf{\lambda }}_2=\frac{3}{t_{go}}\mathbf{ZEV}$

Subtract $3\times$ the first scaled line from the second to kill ${\mathbf{\lambda }}_2$:

$(\frac{3t_{go}}{2}-2t_{go}){\mathbf{\lambda }}_1=\frac{3}{t_{go}}\mathbf{ZEV}-\frac{6}{t_{go}^2}\mathbf{ZEM}$ $-\frac{t_{go}}{2}{\mathbf{\lambda }}_1=\frac{3}{t_{go}}\mathbf{ZEV}-\frac{6}{t_{go}^2}\mathbf{ZEM}\Rightarrow \boxed{{\mathbf{\lambda }}_1=\frac{12}{t_{go}^3}\mathbf{ZEM}-\frac{6}{t_{go}^2}\mathbf{ZEV}}$

Back-substitute into the scaled first line ${\mathbf{\lambda }}_2=\frac{2}{t_{go}^2}\mathbf{ZEM}-\frac{2t_{go}}{3}{\mathbf{\lambda }}_1$:

${\mathbf{\lambda }}_2=\frac{2}{t_{go}^2}\mathbf{ZEM}-\frac{2t_{go}}{3}(\frac{12}{t_{go}^3}\mathbf{ZEM}-\frac{6}{t_{go}^2}\mathbf{ZEV})=\frac{2}{t_{go}^2}\mathbf{ZEM}-\frac{8}{t_{go}^2}\mathbf{ZEM}+\frac{4}{t_{go}}\mathbf{ZEV}$ $\boxed{{\mathbf{\lambda }}_2=-\frac{6}{t_{go}^2}\mathbf{ZEM}+\frac{4}{t_{go}}\mathbf{ZEV}}$

Step 7 — The command applied right now

WHAT. We only ever apply the push at the current instant, i.e. at $\tau =t$, where $t_f-\tau =t_{go}$. So evaluate ${\mathbf{a}}^{\ast }(t)={\mathbf{\lambda }}_1{t}_{go}+{\mathbf{\lambda }}_2$.

WHY. Guidance is run as feedback: use the command now, then next cycle recompute fresh ZEM/ZEV. We never store the whole profile — only its value at "now."

PICTURE. The figure highlights the left endpoint of the straight-line profile (at $s=t_{go}$) — that single dot is the number the thruster obeys this instant.

Now just add. First scale ${\mathbf{\lambda }}_1$ by $t_{go}$:

${\mathbf{\lambda }}_1{t}_{go}=(\frac{12}{t_{go}^3}\mathbf{ZEM}-\frac{6}{t_{go}^2}\mathbf{ZEV})t_{go}=\frac{12}{t_{go}^2}\mathbf{ZEM}-\frac{6}{t_{go}}\mathbf{ZEV}$

Then add ${\mathbf{\lambda }}_2$ term by term, grouping the $\mathbf{ZEM}$ pieces and the $\mathbf{ZEV}$ pieces:

${\mathbf{a}}_c={\mathbf{\lambda }}_1{t}_{go}+{\mathbf{\lambda }}_2=\underset{\text{ZEM coeff}}{\underbrace{(\frac{12}{t_{go}^2}-\frac{6}{t_{go}^2})}}\mathbf{ZEM}+\underset{\text{ZEV coeff}}{\underbrace{(-\frac{6}{t_{go}}+\frac{4}{t_{go}})}}\mathbf{ZEV}$

Each bracket is now just adding fractions with the same denominator — no ellipsis, no guesswork:

$\frac{12}{t_{go}^2}-\frac{6}{t_{go}^2}=\frac{6}{t_{go}^2},-\frac{6}{t_{go}}+\frac{4}{t_{go}}=-\frac{2}{t_{go}}$

Step 8 — The interception law: why the coefficient becomes $3$, not $6$

WHAT. For a missile we only care about hitting the target's position; we don't care what velocity we arrive with. So we drop the velocity constraint entirely — there is no ZEV bucket to fill.

WHY the coefficient changes. This is the subtle point the reviewer flagged. You do not simply erase the ZEV term from the soft-landing law; that would leave $6/t_{go}^2$, which is wrong. The $6$ was the answer to a two-constraint problem. With only one constraint we redo the algebra: only the position bucket survives, and the optimal profile has no velocity bucket to lean on.

PICTURE. The figure shows the single-bucket problem: only the position bucket exists now, so the optimal straight line is pinned by just one condition.

Redo Lagrange with one multiplier $\mathbf{\mu }$ for the single position constraint. The combined quantity is $\mathcal{L}={\int }_t^{t_f}[\frac{1}{2}\Vert \mathbf{a}(\tau ){\Vert }^2-\mathbf{\mu }\cdot (t_f-\tau )\mathbf{a}(\tau )]d\tau ,$ and minimising the parabola at each instant (slope zero) gives ${\mathbf{a}}^{\ast }(\tau )=\mathbf{\mu }(t_f-\tau ).$ Notice it passes through zero at the deadline ($\tau =t_f$): with no velocity to match, the optimal push naturally fades to nothing at the very end. Now fill the single (position) bucket, using the same $s=t_f-\tau$ substitution as Step 6: ${\int }_0^{t_{go}}s\left(\mathbf{\mu }s\right)ds=\mathbf{\mu }{\int }_0^{t_{go}}{s}^{2}ds=\mathbf{\mu }\frac{t_{go}^3}{3}=\mathbf{ZEM}\Rightarrow \mathbf{\mu }=\frac{3}{t_{go}^3}\mathbf{ZEM}.$ The command applied now is ${\mathbf{a}}_c={\mathbf{a}}^{\ast }(t)=\mathbf{\mu }{t}_{go}$: ${\mathbf{a}}_c=\frac{3}{t_{go}^3}\mathbf{ZEM}\cdot t_{go}=\frac{3}{t_{go}^2}\mathbf{ZEM}.$

Step 9 — The edge case: $t_{go}\to 0$ and negative $t_{go}$

WHAT. Watch the coefficients $6/t_{go}^2$ and $2/t_{go}$ as the clock runs out, and ask what a negative $t_{go}$ would mean.

WHY. Real thrusters saturate; a command that goes to infinity is a design red flag. And a countdown that has passed zero is a mission state we must recognise, not silently feed into the formula.

PICTURE. The figure plots both coefficients against $t_{go}$: two chalk curves rocketing toward the ceiling as $t_{go}\to 0^{+}$. The $1/t_{go}^2$ curve (ZEM) climbs faster than $1/t_{go}$ (ZEV).

${\lim }_{t_{go}\to 0^{+}}\frac{6}{t_{go}^2}=+\infty ,{\lim }_{t_{go}\to 0^{+}}\frac{2}{t_{go}}=+\infty$

The one-picture summary

One image, the whole story: coast the dot (yellow), read the two gaps ZEM (blue) and ZEV (pink), fill the buckets with the straight-line optimal push, evaluate at "now" — and out drops ${\mathbf{a}}_c=\frac{6}{t_{go}^2}\mathbf{ZEM}-\frac{2}{t_{go}}\mathbf{ZEV}$.

Quick self-check

Related tools: Optimal Control — LQR · Lambert's Problem · Time-to-go estimation · Double Integrator Dynamics.