Optimal guidance — ZEM - ZEV formulation

WHAT are we solving?

We want a thrust/acceleration command $\mathbf{a}(t)$ that flies a vehicle so that at the final time $t_f$:

• position hits a target: $\mathbf{r}(t_f) = \mathbf{r}_f$
• (optionally) velocity matches: $\mathbf{v}(t_f) = \mathbf{v}_f$ (needed for soft landing / rendezvous)

while spending minimum control effort: $J = \tfrac{1}{2}\int_{t}^{t_f} \|\mathbf{a}(\tau)\|^2\, d\tau .$

The dynamics (double integrator + gravity) over the coast: $\mathbf{r}_{coast}(t_f) = \mathbf{r}(t) + \mathbf{v}(t)\,t_{go} + \tfrac12 \mathbf{g}\,t_{go}^2,$ $\mathbf{v}_{coast}(t_f) = \mathbf{v}(t) + \mathbf{g}\,t_{go}.$

So concretely: $\boxed{\mathbf{ZEM} = \mathbf{r}_f - \mathbf{r} - \mathbf{v}\,t_{go} - \tfrac12\mathbf{g}\,t_{go}^2}$ $\boxed{\mathbf{ZEV} = \mathbf{v}_f - \mathbf{v} - \mathbf{g}\,t_{go}}$

HOW: Derivation from first principles

We use the double-integrator model (control is acceleration): $\dot{\mathbf{r}} = \mathbf{v}, \qquad \dot{\mathbf{v}} = \mathbf{a} + \mathbf{g}.$

Step 1 — Write the true final states with control included

Integrate the true dynamics from now ($t$) to $t_f$. Why? We need the actual endpoint as a function of the unknown $\mathbf{a}(\tau)$.

$\mathbf{v}(t_f) = \underbrace{\mathbf{v} + \mathbf{g}\,t_{go}}_{\text{coast part}} + \int_t^{t_f}\mathbf{a}(\tau)\,d\tau.$

Position needs a double integral; integrating by parts, $\mathbf{r}(t_f) = \underbrace{\mathbf{r}+\mathbf{v}t_{go}+\tfrac12\mathbf{g}t_{go}^2}_{\text{coast part}} + \int_t^{t_f}(t_f-\tau)\,\mathbf{a}(\tau)\,d\tau.$

Why the $(t_f-\tau)$ weight? Because acceleration applied early has more time to move the position — it accumulates through two integrations, leaving a lever-arm $(t_f-\tau)$.

Step 2 — Impose the boundary conditions using ZEM/ZEV

Set $\mathbf{r}(t_f)=\mathbf{r}_f$, $\mathbf{v}(t_f)=\mathbf{v}_f$. Subtract the coast parts (= the ZEM/ZEV definitions): $\int_t^{t_f}(t_f-\tau)\,\mathbf{a}(\tau)\,d\tau = \mathbf{ZEM},\qquad \int_t^{t_f}\mathbf{a}(\tau)\,d\tau = \mathbf{ZEV}.$

Why this is beautiful: the two constraints are now expressed purely in terms of the coast errors. The control just has to "fill" ZEM and ZEV.

Step 3 — Minimize effort subject to those two integral constraints

Minimize $J=\tfrac12\int\|\mathbf{a}\|^2 d\tau$ with two vector constraints. Introduce constant Lagrange multipliers $\boldsymbol\lambda_1,\boldsymbol\lambda_2$: $\mathcal L = \int_t^{t_f}\Big[\tfrac12\|\mathbf a\|^2 - \boldsymbol\lambda_1\!\cdot\!(t_f-\tau)\mathbf a - \boldsymbol\lambda_2\!\cdot\!\mathbf a\Big]d\tau.$ Pointwise stationarity $\partial\mathcal L/\partial\mathbf a = 0$ gives the optimal linear-in-time profile: $\mathbf a^*(\tau) = \boldsymbol\lambda_1 (t_f-\tau) + \boldsymbol\lambda_2.$ Why linear? The calculus of variations for a quadratic cost with these constraints forces the control to be an affine function of remaining time — the optimal profile is never "wild."

Step 4 — Solve the two moment equations at the current instant

Plug $\mathbf a^*$ back into the two constraints. Let $s=t_f-\tau$, so $ds=-d\tau$, $s:t_{go}\to 0$: $\int_0^{t_{go}} s(\boldsymbol\lambda_1 s + \boldsymbol\lambda_2)\,ds = \boldsymbol\lambda_1\tfrac{t_{go}^3}{3}+\boldsymbol\lambda_2\tfrac{t_{go}^2}{2}=\mathbf{ZEM},$ $\int_0^{t_{go}} (\boldsymbol\lambda_1 s + \boldsymbol\lambda_2)\,ds = \boldsymbol\lambda_1\tfrac{t_{go}^2}{2}+\boldsymbol\lambda_2\,t_{go}=\mathbf{ZEV}.$

Solve this $2\times2$ system for $\boldsymbol\lambda_1,\boldsymbol\lambda_2$, then the command applied NOW is $\mathbf a^*(t) = \boldsymbol\lambda_1 t_{go} + \boldsymbol\lambda_2$. Doing the algebra:

Why $6$ and $2$? Solving the $2\times2$: $\boldsymbol\lambda_2 = (6\,\mathbf{ZEM} - 4\,t_{go}\mathbf{ZEV})/t_{go}^2$ etc.; combining gives coefficients $6/t_{go}^2$ and $2/t_{go}$. The signs differ because ZEM (position error) and ZEV (velocity error) must be corrected with opposite-tendency accelerations near the end.

WHY these coefficients blow up (the $t_{go}\to 0$ story)

As $t_{go}\to 0$, $1/t_{go}^2\to\infty$. What this means physically: if you still have a miss with almost no time left, you need enormous acceleration to fix it — real thrusters saturate, so a good guidance system drives ZEM→0 early so the command stays bounded. This is why ZEM/ZEV is run as continuous feedback: each cycle recomputes ZEM/ZEV, so tiny residuals are killed while $t_{go}$ is still large.

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

• Proportional Navigation — the $3/t_{go}^2\,\mathbf{ZEM}$ special case.
• Calculus of Variations & Pontryagin's Minimum Principle — source of the linear costate profile.
• Double Integrator Dynamics — the plant model used.
• Time-to-go estimation — computing $t_{go}$ is the practical hard part.
• Powered Descent Guidance (Apollo E-Guidance) — historical ZEM/ZEV lander.
• Lambert's Problem — related boundary-value targeting in orbital mechanics.
• Optimal Control — LQR — same quadratic-effort minimization philosophy.

Concept Map

Hinglish (regional understanding)