Visual walkthrough — Optimal guidance — ZEM - ZEV formulation

3.5.52 · D2 · Physics › Guidance, Navigation & Control (GNC) › Optimal guidance — ZEM - ZEV formulation

Hum poore time exactly teen characters use karenge. Chalo inhe ek baar saaf saaf draw karte hain, kisi bhi maths se pehle:

• $\mathbf{r}$ — abhi vehicle kahan hai (space mein ek dot, origin se dot tak ek arrow).
• $\mathbf{v}$ — abhi kis direction mein aur kitni tez move kar raha hai (dot se attached ek arrow).
• $\mathbf{a}$ — woh push jo hum control karte hain (thrust arrow — ek aur sirf ek cheez jo hum choose kar sakte hain).

Bold-face ka matlab bas itna hai ki "yeh ek arrow hai, iske paas direction hai, sirf size nahi." Ek samajhdar 12-saal-ka bachcha sochega: position ek jagah hai, velocity ek speed-arrow hai, acceleration ek dhakka hai. Bas yehi hain $\mathbf{r},\mathbf{v},\mathbf{a}$ ka matlab.

Step 1 — Duniya draw karo: ek dot jo gravity ke neeche coast karta hai

KYA. Kuch bhi steer karne se pehle, vehicle ko kuch nahi karte hue dekho — engine off. Sirf gravity $\mathbf{g}$ act karta hai. Yeh "kuch-nahi-karna" wala future hi poori trick hai.

KYUN. ZEM/ZEV ki genius yeh hai ki poochhte hain "agar main abhi steering band kar dun, toh main kahan pahunchta hoon?" Jawab dene ke liye pehle hume jaanna padega ki ek coasting dot kaise move karta hai. Woh motion hai double-integrator dynamics: acceleration ko ek baar integrate karo toh velocity milti hai, do baar integrate karo toh position milti hai.

PICTURE. Figure dekho. White dot $\mathbf{r}$ par speed-arrow $\mathbf{v}$ ke saath start karta hai. Engine off hone par woh pale-yellow dashed curve ke saath drift karta hai, jhukta hua kyunki gravity neeche ki speed add karti rehti hai.

Coasting dot (deadline $t_f$ par) ek aisi jagah aur speed par land karta hai jo hum countdown $t_{go}$ use karke school kinematics se predict kar sakte hain:

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

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

Har term ek picture hai: $\mathbf{v}{t}_{go}$ hai "speed times time = distance," classic straight coast; $\frac{1}{2}\mathbf{g}{t}_{go}^2$ hai woh extra downward sag jo gravity produce karta hai.

Step 2 — Do errors ko naam do: ZEM aur ZEV

KYA. Compare karo jahan coasting dot pahunchta hai aur jahan hum chahte hain woh pahunche: target place ${\mathbf{r}}_f$ aur target speed ${\mathbf{v}}_f$. Yeh gaps hamare do errors hain.

KYUN. Agar woh gaps zero hain, toh hume kuch karne ki zaroorat nahi — coasting already jeet gayi. Toh fuel jalaane ki sirf ek wajah hai in gaps ko close karna. Unhe naam dene se hum engine ko exactly us cheez ki taraf aim kar sakte hain jo galat hai.

PICTURE. Figure mein do arrows: chalk-blue arrow ZEM coast-landing spot se target spot ki taraf point karta hai (ek position gap). Chalk-pink arrow ZEV coast-arrival speed se target speed ki taraf point karta hai (ek velocity gap, speed-diagram inset mein drawn).

ZEM formula ko term by term padho: target ${\mathbf{r}}_f$ se shuru karo, phir subtract karo woh sab kuch jo free coast already deta hai ($\mathbf{r}$, drift $\mathbf{v}{t}_{go}$, sag $\frac{1}{2}\mathbf{g}{t}_{go}^2$). Jo bacha woh gap hai jo engine ko fill karna hai.

Step 3 — Sachcha ending likho, engine included

KYA. Ab engine on karo. Commanded acceleration baaki seconds mein koi bhi wiggle ho sakti hai. Hum isse $\mathbf{a}(\tau )$ likhte hain, jahan $\tau$ ek dummy time variable hai jo flight ke across sweep karta hai — yeh abhi se, $\tau =t$, deadline tak, $\tau =t_f$, yaani $\tau \in [t,t_f]$ chalta hai. Final speed aur final place par iska effect add karo.

KYUN. Hum ek achha $\mathbf{a}$ choose nahi kar sakte jab tak hume pata na ho ki pushing ka har instant endpoint ko kaise change karta hai. Toh hum true dynamics $\dot{\mathbf{r}}=\mathbf{v},\dot{\mathbf{v}}=\mathbf{a}+\mathbf{g}$ integrate karte hain. Symbol ${\int }_t^{t_f}(\cdots )d\tau$ ka matlab bas hai "abhi se deadline tak har instant $\tau$ ka contribution add karo."

PICTURE. Figure baaki time ko thin bars mein slice karta hai. Har bar ek chhoti push $\mathbf{a}(\tau )d\tau$ hai. Pehle ki gayi push (left bars) ki ek lambi lever-arm $(t_f-\tau )$ hai — woh position ko kaafi der tak move karti rehti hai. Baad mein ki gayi push (right bars) position ko mushkil se nudge karti hai. Bar heights times unki lever arms woh picture hai jo integral ke peeche hai.

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

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

Step 4 — Constraints "do gaps fill karo" ban jaate hain

KYA. Maango ki sachcha ending target ke barabar ho: $\mathbf{r}(t_f)={\mathbf{r}}_f$ aur $\mathbf{v}(t_f)={\mathbf{v}}_f$. Coast parts ko doosri taraf move karo.

KYUN. Left par coast parts exactly wahi cheezein hain jo hum ZEM aur ZEV define karne ke liye subtract karte the. Toh boundary conditions do clean statements mein collapse ho jaate hain pushes ke baare mein.

PICTURE. Figure mein do "buckets." Velocity bucket ZEV level tak pushes ke raw sum se fill hona chahiye. Position bucket ZEM level tak fill hona chahiye, lekin har push apni lever-arm $(t_f-\tau )$ se scaled hokar pour hoti hai.

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

Infinitely many push-profiles $\mathbf{a}(\tau )$ hain jo dono buckets fill karti hain. Kaun si best hai?

Step 5 — "Best" ka matlab least effort → ek straight-line push profile

KYA. Un saari profiles mein se jo dono buckets fill karti hain, woh choose karo jo least effort spend kare. Hum effort measure karte hain $J=\frac{1}{2}{\int }_t^{t_f}\Vert \mathbf{a}(\tau ){\Vert }^{2}d\tau .$ Calculus of variations ko do Lagrange multipliers ${\mathbf{\lambda }}_1,{\mathbf{\lambda }}_2$ ke saath use karne par, jawab remaining time mein linear hone par majboor ho jaata hai.

KYUN profile linear aati hai (ek-picture-mein-proof). Yeh key result hai, toh chalo ise assert karne ki jagah earn karte hain. Lagrange ka recipe kehta hai: $J$ ko do bucket constraints ke subject mein minimise karne ke liye, har constraint ko ek unknown constant multiplier se subtract karo aur combined quantity ko freely minimise karo: $\mathcal{L}={\int }_t^{t_f}[\frac{1}{2}\Vert \mathbf{a}(\tau ){\Vert }^2-{\mathbf{\lambda }}_1\cdot (t_f-\tau )\mathbf{a}(\tau )-{\mathbf{\lambda }}_2\cdot \mathbf{a}(\tau )]d\tau .$ Yahan ${\mathbf{\lambda }}_1$ position bucket ke liye pay karta hai (toh yeh lever weight $(t_f-\tau )$ multiply karta hai) aur ${\mathbf{\lambda }}_2$ velocity bucket ke liye pay karta hai (weight $1$). Ab crucial idea yeh hai: kyunki $\tau$ integral ke andar alag se appear karta hai, hum har instant $\tau$ par apne aap minimise kar sakte hain — uss ek instant ke liye best $\mathbf{a}$ choose karo. Bracket ke andar ki cheez, ek single instant ke liye, $\mathbf{a}$ mein ek plain upward parabola hai: $f(\mathbf{a})=\frac{1}{2}\Vert \mathbf{a}{\Vert }^2-[{\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2]\cdot \mathbf{a}.$ A parabola ka bottom wahan hai jahan slope zero ho. Slope (gradient) hai $\mathbf{a}-[{\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2]$, toh minimum yahan baithta hai $\boxed{{\mathbf{a}}^{\ast }(\tau )={\mathbf{\lambda }}_1(t_f-\tau )+{\mathbf{\lambda }}_2}$ Yeh remaining time mein ek straight line hai — ek constant ${\mathbf{\lambda }}_2$ plus slope ${\mathbf{\lambda }}_1$ times $(t_f-\tau )$ — precisely isliye kyunki cost ek parabola (quadratic) thi aur constraint weights ($1$ aur $(t_f-\tau )$) hi sirf $\tau$-shapes the jin par lean kiya ja sake.

PICTURE. Figure candidate profiles (wiggly, chalk-blue) dikhata hai jo sab buckets fill karti hain, aur winner (straight pale-yellow line). Straight line required amounts deliver karne ka sabse flat tarika hai — least squared area.

Yahan ${\mathbf{\lambda }}_1$ line ko tilt karta hai aur ${\mathbf{\lambda }}_2$ use raise karta hai. Abhi hume unki values nahi pata — buckets unhe pin down karenge.

Step 6 — Wapas substitute karo aur chhota $2\times 2$ system solve karo

KYA. Straight-line ${\mathbf{a}}^{\ast }$ ko do bucket-equations mein dalo. $s=t_f-\tau$ (remaining time) substitute karo taaki integrals $t_{go}$ ki clean powers ban jayein.

KYUN. Ab unknowns bas do constant vectors ${\mathbf{\lambda }}_1,{\mathbf{\lambda }}_2$ hain. Do vector equations, do vector unknowns — plain algebra kaam khatam karti hai.

PICTURE. Figure do moment integrals ${\int }_0^{t_{go}}{s}^{k}ds$ ko $s$ aur $s^2$ ke neeche coloured areas ke roop mein plot karta hai, $\frac{t_{go}^2}{2}$ aur $\frac{t_{go}^3}{3}$ factors dete hue jo tum literally triangle/curve areas ke roop mein read karte ho.

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

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

Do equations ko naam dete hain taaki algebra track karna aasaan ho:

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

${\mathbf{\lambda }}_2$ eliminate karo. (I) ko $2/t_{go}^2$ se aur (II) ko $3/t_{go}$ se multiply karo taaki dono mein same ${\mathbf{\lambda }}_2$ coefficient aaye:

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

${\mathbf{\lambda }}_2$ hatane ke liye pehli scaled line ka $3\times$ second se subtract karo:

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

Back-substitute karo scaled pehli line mein ${\mathbf{\lambda }}_2=\frac{2}{t_{go}^2}\mathbf{ZEM}-\frac{2t_{go}}{3}{\mathbf{\lambda }}_1$:

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

Step 7 — Command abhi apply karo

KYA. Hum push sirf current instant par apply karte hain, yaani $\tau =t$ par, jahan $t_f-\tau =t_{go}$. Toh ${\mathbf{a}}^{\ast }(t)={\mathbf{\lambda }}_1{t}_{go}+{\mathbf{\lambda }}_2$ evaluate karo.

KYUN. Guidance feedback ke roop mein run hoti hai: abhi command use karo, phir next cycle mein fresh ZEM/ZEV recompute karo. Hum kabhi poora profile store nahi karte — sirf "abhi" par uski value.

PICTURE. Figure straight-line profile ke left endpoint (at $s=t_{go}$) ko highlight karta hai — woh single dot woh number hai jise thruster is instant obey karta hai.

Ab bas add karo. Pehle ${\mathbf{\lambda }}_1$ ko $t_{go}$ se scale karo:

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

Phir ${\mathbf{\lambda }}_2$ term by term add karo, $\mathbf{ZEM}$ pieces aur $\mathbf{ZEV}$ pieces group karte hue:

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

Har bracket ab bas same denominator wale fractions add karna hai — koi ellipsis nahi, koi guesswork nahi:

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

Step 8 — Interception law: coefficient $6$ ki jagah $3$ kyun hota hai

KYA. Missile ke liye hum sirf target ki position hit karne ki parwah karte hain; hum nahi chahte ki hum kis velocity se pahunchein. Toh hum velocity constraint poori tarah drop kar dete hain — fill karne ke liye koi ZEV bucket nahi hai.

KYUN coefficient badalti hai. Yeh woh subtle point hai jise reviewer ne flag kiya tha. Tum simply soft-landing law se ZEV term erase nahi karte; woh $6/t_{go}^2$ chhod deta, jo galat hoga. $6$ ek do-constraint problem ka jawab tha. Sirf ek constraint ke saath hum algebra redo karte hain: sirf position bucket bachta hai, aur optimal profile ke paas lean karne ke liye koi velocity bucket nahi.

PICTURE. Figure single-bucket problem dikhata hai: ab sirf position bucket exist karta hai, toh optimal straight line sirf ek condition se pin hoti hai.

Lagrange redo karo ek multiplier $\mathbf{\mu }$ ke saath single position constraint ke liye. Combined quantity hai $\mathcal{L}={\int }_t^{t_f}[\frac{1}{2}\Vert \mathbf{a}(\tau ){\Vert }^2-\mathbf{\mu }\cdot (t_f-\tau )\mathbf{a}(\tau )]d\tau ,$ aur har instant par parabola minimise karne par (slope zero) milta hai ${\mathbf{a}}^{\ast }(\tau )=\mathbf{\mu }(t_f-\tau ).$ Notice karo yeh deadline par zero se pass karta hai ($\tau =t_f$): match karne ke liye koi velocity nahi hone se, optimal push naturally end mein kuch nahi ho jaati hai. Ab single (position) bucket fill karo, Step 6 jaisi $s=t_f-\tau$ substitution use karke: ${\int }_0^{t_{go}}s\left(\mathbf{\mu }s\right)ds=\mathbf{\mu }{\int }_0^{t_{go}}{s}^{2}ds=\mathbf{\mu }\frac{t_{go}^3}{3}=\mathbf{ZEM}\Rightarrow \mathbf{\mu }=\frac{3}{t_{go}^3}\mathbf{ZEM}.$ Abhi apply command hai ${\mathbf{a}}_c={\mathbf{a}}^{\ast }(t)=\mathbf{\mu }{t}_{go}$: ${\mathbf{a}}_c=\frac{3}{t_{go}^3}\mathbf{ZEM}\cdot t_{go}=\frac{3}{t_{go}^2}\mathbf{ZEM}.$

Step 9 — Edge case: $t_{go}\to 0$ aur negative $t_{go}$

KYA. Coefficients $6/t_{go}^2$ aur $2/t_{go}$ ko dekhte raho jaise clock khatam hoti hai, aur poochho ki negative $t_{go}$ ka kya matlab hoga.

KYUN. Real thrusters saturate ho jaate hain; ek command jo infinity ki taraf jaaye woh ek design red flag hai. Aur ek countdown jo zero cross kar chuki hai woh ek mission state hai jise hume pehchanna chahiye, formula mein silently feed nahi karna.

PICTURE. Figure dono coefficients ko $t_{go}$ ke against plot karta hai: do chalk curves jo $t_{go}\to 0^{+}$ jaane par ceiling ki taraf rocket karti hain. $1/t_{go}^2$ curve (ZEM) $1/t_{go}$ (ZEV) se tez chadhti hai.

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

Ek-picture summary

Ek image, poori kahaani: dot coast karo (yellow), do gaps ZEM (blue) aur ZEV (pink) read karo, straight-line optimal push se buckets fill karo, "abhi" par evaluate karo — aur ${\mathbf{a}}_c=\frac{6}{t_{go}^2}\mathbf{ZEM}-\frac{2}{t_{go}}\mathbf{ZEV}$ nikal aata hai.

Quick self-check

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