Exercises — Optimal guidance — ZEM - ZEV formulation

WHAT: $t_{go}$ is just remaining time. WHY: every formula uses it, so compute it first. $t_{go}=t_f-t=19-4=15\text{s}.$ Check the domain rule: $t_{go}=15>0$, so the law is well-defined. Coast position — put in $r=500$, $v=-30$, $g=-9.81$, $t_{go}=15$: $r_{coast}=500+(-30)(15)+\frac{1}{2}(-9.81)(15^2)=500-450-1103.6=-1053.6\text{m}.$ The negative number means: coast unpowered and you crash 1053.6 m below the target level — gravity dragged you way past. That is exactly the kind of error ZEM will measure.

The rule: do you care about the final velocity? If yes → soft law (both terms). If you only care about being there → intercept law. (a) intercept ($3/t_{go}^2$) — just hit. (b) soft — must arrive at zero speed. (c) soft — must match the station's velocity, see Powered Descent Guidance (Apollo E-Guidance) for the same idea in landing. (d) intercept.

Step 1 — ZEM (where would we land coasting?). Why first: the command needs it. $\text{ZEM}=0-200-(-40)(8)-\frac{1}{2}(-3.71)(8^2)=-200+320+118.72=238.72\text{m}.$ Step 2 — ZEV (how wrong would our speed be?). $\text{ZEV}=0-(-40)-(-3.71)(8)=40+29.68=69.68\text{m/s}.$ Step 3 — Command (apply the soft law). $a_c=\frac{6}{8^2}(238.72)-\frac{2}{8}(69.68)=\frac{6}{64}(238.72)-0.25(69.68).$ $a_c=22.38-17.42=4.96{\text{m/s}}^2.$ This is the acceleration above gravity (gravity is already inside ZEM/ZEV). The upward braking thrust must supply $a_c$ plus fight gravity.

Position-only (intercept) law — why: we only want to hit. ${\mathbf{a}}_c=\frac{3}{5^2}(600,-800)=\frac{3}{25}(600,-800)=(72,-96){\text{m/s}}^2.$ Magnitude: $\sqrt{72^2+96^2}=\sqrt{5184+9216}=\sqrt{14400}=120{\text{m/s}}^2.$ This law is Proportional Navigation with navigation ratio $N=3$ — same physics, ZEM-flavoured.

Why it subtracts: ZEM says "you'll overshoot upward-of-target by 238.7 m worth of coast error, so push to cut that," while ZEV says "you'll arrive 69.7 m/s too fast, so you must brake — i.e. accelerate the opposite way to the position push near the end." The $2\times 2$ Lagrange solution (Step 4 of the parent derivation) encodes this as opposite signs. Wrong ($+$) version: $a_c^{\text{wrong}}=22.38+17.42=39.80{\text{m/s}}^2.$ That is about 8× the correct command $4.96$. Physically you'd be adding braking-hunger to position-hunger, over-decelerate, stop short and high, then fall — a failed landing. The minus sign is what makes "arrive gently" different from "arrive hard."

$a_c(t_{go})=\frac{6}{t_{go}^2}(1)-\frac{2}{t_{go}}(0.5)=\frac{6}{t_{go}^2}-\frac{1}{t_{go}}.$

As $t_{go}\to 0$ the $1/t_{go}^2$ term dominates and $a_c$ explodes. Lesson: a residual that is harmless at $t_{go}=5$ demands a monstrous, thruster-saturating command at $t_{go}=0.1$. So you must drive ZEM/ZEV to zero while $t_{go}$ is still large — which is exactly why ZEM/ZEV is run as continuous feedback.

Read the figure below like this: the magenta curve is $a_c(t_{go})$; the four violet dots are the table rows. Sweep your eye right to left (that is the direction time actually runs). Notice the curve is nearly flat and tiny for $t_{go}>2$, then hooks sharply upward as it approaches the left axis — the orange arrow marks that blow-up. The message is visual: all the danger lives in the last fraction of a second, where a saturation ceiling $a_{max}$ (a horizontal cap you can imagine drawn across the top) would be smashed through.

Forecast: on track, closing the gap, ZEM should shrink. Guess: smaller. Verify: with $\mathbf{ZEM}\propto t_{go}^2$, halving $t_{go}$ from 6 to 3 quarters ZEM: $\mathbf{ZEM}(3)={\left(\frac{3}{6}\right)}^2(300,-450)=\frac{1}{4}(300,-450)=(75,-112.5)\text{m}.$ New command: ${\mathbf{a}}_c=\frac{3}{3^2}(75,-112.5)=\frac{1}{3}(75,-112.5)=(25,-37.5){\text{m/s}}^2.$ Interpretation: the command magnitude is the same as it was at $t_{go}=6$ (there $\frac{3}{36}(300,-450)=(25,-37.5)$ too). On an ideal course ZEM/ZEV self-consistently keeps the command constant and bounded — no blow-up. The $t_{go}^2$ shrink in ZEM exactly cancels the $1/t_{go}^2$ growth in the coefficient. This is the healthy counterpart to L3.2.

Work each axis; gravity only acts on $y$. ZEM:

• $x$: $0-0-(10)(10)-\frac{1}{2}(0)(100)=-100$ m.
• $y$: $0-150-(-25)(10)-\frac{1}{2}(-1.62)(100)=-150+250+81=181$ m.
• $\mathbf{ZEM}=(-100,181)$ m.

ZEV:

• $x$: $0-10-0=-10$ m/s.
• $y$: $0-(-25)-(-1.62)(10)=25+16.2=41.2$ m/s.
• $\mathbf{ZEV}=(-10,41.2)$ m/s.

Command with $\frac{6}{100}=0.06$, $\frac{2}{10}=0.2$:

• $x$: $0.06(-100)-0.2(-10)=-6+2=-4$ m/s².
• $y$: $0.06(181)-0.2(41.2)=10.86-8.24=2.62$ m/s².
• $\boxed{{\mathbf{a}}_c=(-4,2.62){\text{m/s}}^2.}$ The $x$-command is negative — it brakes the sideways drift so we don't land off-pad; the $y$-command is a mild upward push, softened (as always) by the subtracted ZEV term.

Read the figure below like this: the navy dot is where we are now, $(0,150)$; the orange star is the target at the origin. The magenta arrow is the ZEM vector — it points from the coasting-landing spot (magenta dot, down-and-left) back up to the target, showing the miss we must erase. The violet arrow at our current position is the command ${\mathbf{a}}_c=(-4,2.62)$ (drawn scaled-up so it is visible). Look at its direction: it tilts up-and-to-the-left — the leftward tilt is the $x$-brake killing the $+10$ m/s sideways drift, exactly the sign we computed.

WHY ${\mathbf{\lambda }}_2=0$: with only one constraint (ZEM), you only need one price — one multiplier. The velocity constraint that produced ${\mathbf{\lambda }}_2$ is dropped, so the optimal control from Calculus of Variations & Pontryagin's Minimum Principle is ${\mathbf{a}}^{\ast }(\tau )={\mathbf{\lambda }}_1(t_f-\tau )={\mathbf{\lambda }}_{1}s$. Use the ZEM moment equation with ${\mathbf{\lambda }}_2=0$: ${\mathbf{\lambda }}_1\frac{t_{go}^3}{3}=\mathbf{ZEM}\Rightarrow {\mathbf{\lambda }}_1=\frac{3\mathbf{ZEM}}{t_{go}^3}.$ Command now ($s=t_f-t=t_{go}$): ${\mathbf{a}}_c={\mathbf{\lambda }}_1{t}_{go}=\frac{3\mathbf{ZEM}}{t_{go}^3}\cdot t_{go}=\frac{3\mathbf{ZEM}}{t_{go}^2}.■$ That $3$ is the navigation ratio $N=3$ of PN — earned, not memorised.

Each axis is identical, so drop the bold and treat $Z_M,Z_V$ as scalars. Let $T=t_{go}$ for brevity. The two moment equations are $\frac{T^3}{3}{\lambda }_1+\frac{T^2}{2}{\lambda }_2=Z_M,\frac{T^2}{2}{\lambda }_1+T{\lambda }_2=Z_V.$

Step A — isolate ${\lambda }_2$ from the simpler (second) equation. Why this one: it is linear in ${\lambda }_2$ with the smallest powers, so solving it for ${\lambda }_2$ costs least algebra. ${\lambda }_2=\frac{Z_V}{T}-\frac{T}{2}{\lambda }_1.$

Step B — substitute into the first equation. Why: this eliminates ${\lambda }_2$ and leaves one equation in the single unknown ${\lambda }_1$. $\frac{T^3}{3}{\lambda }_1+\frac{T^2}{2}\left(\frac{Z_V}{T}-\frac{T}{2}{\lambda }_1\right)=Z_M.$ Expand the bracket term by term: $\frac{T^2}{2}\cdot \frac{Z_V}{T}=\frac{T{Z}_V}{2}$ and $\frac{T^2}{2}\cdot \left(-\frac{T}{2}{\lambda }_1\right)=-\frac{T^3}{4}{\lambda }_1.$ So $\frac{T^3}{3}{\lambda }_1-\frac{T^3}{4}{\lambda }_1+\frac{T{Z}_V}{2}=Z_M.$

Step C — group the ${\lambda }_1$ terms. Why: collecting like terms lets us read off ${\lambda }_1$. Combine the fractions with common denominator $12$: $\frac{T^3}{3}-\frac{T^3}{4}=\frac{4T^3-3T^3}{12}=\frac{T^3}{12}.$ $\frac{T^3}{12}{\lambda }_1=Z_M-\frac{T{Z}_V}{2}.$

Step D — clear the denominator to get ${\lambda }_1$. Why: multiply both sides by $12/T^3$ to leave ${\lambda }_1$ alone. ${\lambda }_1=\frac{12}{T^3}\left(Z_M-\frac{T{Z}_V}{2}\right)=\frac{12Z_M}{T^3}-\frac{6Z_V}{T^2}.$

Step E — back-substitute to get ${\lambda }_2$. Why: Step A already gave ${\lambda }_2$ in terms of ${\lambda }_1$; plug the result in. ${\lambda }_2=\frac{Z_V}{T}-\frac{T}{2}\left(\frac{12Z_M}{T^3}-\frac{6Z_V}{T^2}\right)=\frac{Z_V}{T}-\frac{6Z_M}{T^2}+\frac{3Z_V}{T}=-\frac{6Z_M}{T^2}+\frac{4Z_V}{T}.$

Step F — form the command applied now, $a_c={\lambda }_{1}T+{\lambda }_2$. Why: the control profile is $a^{\ast }(\tau )={\lambda }_{1}s+{\lambda }_2$, and "now" means $s=t_{go}=T$. $a_c=\left(\frac{12Z_M}{T^2}-\frac{6Z_V}{T}\right)+\left(-\frac{6Z_M}{T^2}+\frac{4Z_V}{T}\right).$ Group the $Z_M$ terms: $\frac{12Z_M}{T^2}-\frac{6Z_M}{T^2}=\frac{6Z_M}{T^2}.$ Group the $Z_V$ terms: $-\frac{6Z_V}{T}+\frac{4Z_V}{T}=-\frac{2Z_V}{T}.$ Therefore $\boxed{a_c=\frac{6Z_M}{T^2}-\frac{2Z_V}{T}=\frac{6}{t_{go}^2}\mathbf{ZEM}-\frac{2}{t_{go}}\mathbf{ZEV}}■$ There it is: the $6$, the $2$, and the crucial minus sign — all falling straight out of the two-price Lagrange solution.

The intercept law coefficient is ::: $3/t_{go}^2$ (navigation ratio $N=3$) The soft law subtracts the ZEV term with coefficient ::: $2/t_{go}$ As $t_{go}\to 0$ with fixed residuals, the command ::: blows up like $1/t_{go}^2$, so kill errors early On an ideal intercept course ZEM scales like ::: $t_{go}^2$, keeping the command bounded Gravity enters ZEM as ::: $\frac{1}{2}\mathbf{g}{t}_{go}^2$ and ZEV as $\mathbf{g}{t}_{go}$ The law is only valid for ::: $t_{go}>0$; at $t_{go}\le 0$ freeze the command or re-plan, and always clip to $a_{max}$