Optimal guidance — ZEM - ZEV formulation: Worked examples

3.5.52 · D3 · Physics › Guidance, Navigation & Control (GNC) › Optimal guidance — ZEM - ZEV formulation

Definition Is poore page pe use hone wale coordinate aur sign conventions

Upar positive hai. Position $r$ pad ke upar ki altitude hai; positive $r$ ka matlab "target ke upar."
Velocity ka sign position follow karta hai: $v < 0$ matlab neeche ki taraf ja raha hai, $v > 0$ upar.
Gravity negative hai (yeh neeche point karta hai): Moon pe $g = - 1.62$ m/s², Mars pe $g = - 3.71$ m/s², Earth ke paas $g = - 9.81$ m/s².
2-D mein, $x$ horizontal hai aur $y$ upar hai; gravity sirf $y$ pe act karta hai.
Navigation ratio $N$ ek plain number hai jo Proportional Navigation mein line-of-sight correction ko multiply karta hai. ZEM/ZEV ka intercept form exactly PN nikalta hai $N = 3$ ke saath — hum Ex 6 mein woh equivalence derive karte hain. Jahan bhi " $N = 3$ " dikhta hai iska matlab hai "position-only law with coefficient $3/ t_{g o}^{2}$ ."

Kuch bhi shuru karne se pehle, ek ek-line ka refresher taaki koi bhi symbol bina samjhaye use na ho. Neeche har reveal line apna jawab chhupaati hai — khud se quiz karo, phir check karo:

Recall Dono laws (parent note se)

$t_{g o}$ kya hai? ::: time-to-go, $t_{f} - t$ , deadline tak baaki seconds. $ZEM$ kya hai? ::: Zero-Effort-Miss — woh position error jo tumhare paas hota agar tum abhi steering band kar do aur sirf gravity ke saath coast karo. $ZEV$ kya hai? ::: Zero-Effort-Velocity — usi coast ke under woh velocity error. Soft-landing law (position + velocity)? ::: $a_{c} = \frac{6}{t _{g o}^{2}} ZEM - \frac{2}{t _{g o}} ZEV$ Intercept law (sirf position)? ::: $a_{c} = \frac{3}{t _{g o}^{2}} ZEM$

Aur woh dono prediction formulas jo hum baar baar use karenge:

Scenario matrix

Is topic pe jo bhi problem aa sakti hai woh inhi cells mein se ek mein aati hai. Har row ek "class of case" hai; aakhri column us worked example ka naam deta hai jo use nail karta hai. Table ke neeche figure har cell ko ek chhoti picture ke roop mein dikhata hai taaki tum compute karne se pehle nawon situations ko dekh sako.

#	Case class	Isme kya special hai	Sign / limit behaviour jo dekhni hai	Covered by
C1	Baseline soft landing, saari quantities positive-ish	"Normal" full-law case	ZEM $> 0$ , ZEV $> 0$ , opposite-sign terms	Ex 1
C2	Dono errors already zero (degenerate)	Tum perfectly coast trajectory pe ho	$a_{c} = 0$ , koi divide-by-zero nahi	Ex 2
C3	Negative ZEM, negative ZEV (coast overshoots aur too fast)	Signs seedhe through flow hote hain, command flip hoti hai	ZEM $< 0$ , ZEV $< 0$ , $a_{c} < 0$	Ex 3
C4	ZEM aur ZEV opposite direction mein pull karte hain	Terms almost cancel ho jaate hain	do bade numbers se chhoti command	Ex 4
C5	$t_{g o} \to 0$ limit (almost out of time)	Coefficient blow-up $1/ t_{g o}^{2}$	command explode karta hai → thruster saturation	Ex 5
C6	Intercept, full 2-D vector	Position-only law, navigation ratio $N = 3$	geometry, $a_{c}$ ki direction	Ex 6 (figure)
C7	Intercept jahan naive gravity-drop matter karta hai	$g$ bhoolne se wrong ZEM aata hai	error quantify karo	Ex 7
C8	Real-world word problem (Mars descent)	Prose se numbers extract karo	full soft-landing law	Ex 8
C9	Exam twist: woh $t_{g o}$ solve karo jo command achievable banata hai	Law ko invert karo	$t_{g o}$ mein quadratic	Ex 9

Figure — Optimal guidance — ZEM - ZEV formulation

Figure mein har tile ek matrix cell hai: cyan arrow dikhata hai ki pure coast tumhe kahan le jaata hai, amber arrow woh residual error hai (ZEM/ZEV) jo command ko kill karna hai. Notice karo C2 mein koi amber arrow nahi (fix karne ko kuch nahi) aur C5 same small miss dikhata hai lekin almost koi time nahi — yahi wajah hai command explode karti hai.

Worked examples

Worked example Ex 1 — C1: Baseline vertical soft landing

Ek Moon lander: altitude $r = 100$ m, velocity $v = - 20$ m/s (minus = neeche), targets $r_{f} = 0$ , $v_{f} = 0$ (gently touch down), gravity $g = - 1.62$ m/s², aur $t_{g o} = 10$ s.

Forecast: Compute karne se pehle — kya commanded acceleration positive hogi (upar, yaani braking) ya negative? Abhi guess karo.

ZEM. Yeh step kyun? Pehle hume jaanna chahiye ki pure coast pad ko kitna miss karta hai. $ZEM = 0 - 100 - (- 20) (10) - \frac{1}{2} (- 1.62) (100) = - 100 + 200 + 81 = 181$ m.
ZEV. Kyun? Touchdown speed ke liye bhi same. $ZEV = 0 - (- 20) - (- 1.62) (10) = 20 + 16.2 = 36.2$ m/s.
Command. Kyun? Dono ko full law mein feed karo. $a_{c} = \frac{6}{1 0 ^{2}} (181) - \frac{2}{10} (36.2) = 10.86 - 7.24 = 3.62$ m/s².

Verify: $a_{c} > 0$ → upar → braking. Sahi hai: coasting 181 m upar overshoot karti hai aur bahut fast pohonchti hai, isliye hum fall slow karne ke liye upar thrust karte hain. Units: $\frac{1}{s ^{2}} \cdot m - \frac{1}{s} \cdot \frac{m}{s} = \frac{m}{s ^{2}}$ ✓.

Worked example Ex 2 — C2: Tum already perfect coast pe ho (degenerate)

Same lander, lekin ab woh exactly us coast pe baitha hai jo softly land karti hai: $r = 100$ m, $v = ?$ aisa choose kiya ki ZEM $= 0$ aur ZEV $= 0$ . Lo $t_{g o} = 10$ , $g = - 1.62$ , $r_{f} = v_{f} = 0$ .

Forecast: Agar dono errors zero hain, toh guidance command kya honi chahiye? Guess karo.

Woh $v$ dhundho jo ZEV zero karta hai. Kyun? ZEV $= 0 \Rightarrow v_{f} - v - g t_{g o} = 0 \Rightarrow v = - g t_{g o} = 16.2$ m/s… lekin woh ZEM bhi zero karna chahiye. Dono ko saath solve karo: hume $r, v$ chahiye ek landing coast pe. ZEM $= 0$ set karo: $0 - 100 - 10 v - \frac{1}{2} (- 1.62) (100) = 0 \Rightarrow - 100 - 10 v + 81 = 0 \Rightarrow v = - 1.9$ m/s.
Us $v$ se ZEV check karo. Kyun? Ek general point dono zero nahi karegi — yeh dikhata hai ki sirf special states "effort-free" hain. $ZEV = 0 - (- 1.9) - (- 1.62) (10) = 1.9 + 16.2 = 18.1 \neq = 0$ .
Woh state banao jahan DONO zero hon. Yeh step kyun? Genuine C2 point exhibit karne ke liye hume perfect soft touchdown se backward coast karna hoga — sirf aisi state mein dono errors vanish hoti hain, aur hume ise test karne ki zaroorat hai ki law exactly zero return karta hai. $t_{g o}$ pehle softly land karne ke time pe, $v = v_{f} - g t_{g o} = 0 - (- 1.62) (10) = 16.2$ m/s (upar), aur $r = r_{f} - v t_{g o} - \frac{1}{2} g t_{g o}^{2} = 0 - (16.2) (10) - \frac{1}{2} (- 1.62) (100) = - 162 + 81 = - 81$ m.
Command. Yeh step kyun? Yahi poora test hai — zeroed errors ko law mein plug karo aur confirm karo ki optimal effort sach mein kuch nahi hai. $a_{c} = \frac{6}{100} (0) - \frac{2}{10} (0) = 0$ .

Verify: ZEM $= ZEV = 0$ ke saath law exactly $a_{c} = 0$ deta hai — koi thrust waste nahi. Yahi "zero-effort" ka poora point hai: agar free coast already boundary conditions satisfy karti hai, toh optimal effort literally kuch nahi hai. (Note: koi division by zero nahi hua; $t_{g o} = 10$ , 0 nahi.)

Worked example Ex 3 — C3: Dono errors genuinely negative

Lander jiske saath target upar hai, bahut fast upar drift kar raha hai: $r = 100$ m, $v = + 40$ m/s (rising), $r_{f} = 200$ m (upar aim karo), $v_{f} = 0$ , $g = - 1.62$ , $t_{g o} = 10$ .

Forecast: Target se sirf 100 m upar fast rise ho raha hai — kya coast upar overshoot karegi (toh ZEM negative aayega), aur kya required command isliye neeche point karegi ( $a_{c} < 0$ )? Guess karo.

ZEM. Kyun? Predict karo ki pure coast hume target ke relative kahan chodti hai — yeh position error ka sign fix karta hai. $ZEM = 200 - 100 - (40) (10) - \frac{1}{2} (- 1.62) (100) = 100 - 400 + 81 = - 219$ m. Negative — fast climb target se aage aur upar nikal jaati hai: coast overshoot karti hai, isliye hume neeche pull karna chahiye.
ZEV. Yeh step kyun? Hume speed error bhi chahiye, kyunki soft arrival ke liye $v_{f} = 0$ chahiye; coast ki residual upward speed velocity term ka sign set karti hai. $ZEV = 0 - (40) - (- 1.62) (10) = - 40 + 16.2 = - 23.8$ m/s.
Command. Yeh step kyun? Dono errors known hain, isliye hum unhe full soft-landing law mein feed karte hain ab lagane wali acceleration pane ke liye. $a_{c} = \frac{6}{100} (- 219) - \frac{2}{10} (- 23.8) = - 13.14 + 4.76 = - 8.38$ m/s².

Verify: $a_{c} < 0$ → neeche → hum climb brake karte hain, exactly forecast ke anusaar. ZEM aur ZEV dono negative aaye, isliye yeh genuinely cell C3 hit karta hai (Ex 1 se contrast karo, jahan dono positive the). Arithmetic sign automatically carry karta hai — tum kabhi absolute value insert mat karo; formula pe trust karo, apni gut pe nahi. Units ✓.

Worked example Ex 4 — C4: Dono terms almost cancel karte hain

$ZEM = 40$ m, $ZEV = 12$ m/s, $t_{g o} = 4$ s. (Directly diya gaya — yeh ek mid-flight feedback tick hai.)

Forecast: Position term ek badi push chahta hai; velocity term subtract karta hai. Kya net command badi hogi ya surprisingly chhoti?

Position term. Alag kyun compute karein? Combine karne se pehle dono corrections ke beech tug-of-war expose karne ke liye. $\frac{6}{16} (40) = 15$ m/s².
Velocity term. Alag kyun? Taaki hum dekh sakein ZEV correction position push ka kitna bada hissa le jaati hai — yahi C4 ka poora point hai. $\frac{2}{4} (12) = 6$ m/s².
Net. Subtract kyun, add nahi? Soft-landing law ZEV term ke aage minus carry karta hai (end ke paas position correct karna aur velocity correct karna opposite directions mein acceleration demand karte hain), isliye dono difference hote hain. $a_{c} = 15 - 6 = 9$ m/s².

Verify: Do sizeable numbers (15 aur 6) ne moderate 9 produce kiya. Yahi C4 ka lesson hai: end ke paas ZEV term routinely ZEM term ka bada chunk kha jaata hai — woh opposite sign real kaam kar raha hai. Units ✓.

Worked example Ex 5 — C5:

$t_{g o} \to 0$ cliff Ex 4 jaisi hi residual errors lekin ab almost out of time: $ZEM = 40$ m, $ZEV = 12$ m/s, $t_{g o} = 4, 2, 1, 0.5$ s pe evaluate karo.

Forecast: $t_{g o}$ half karne par — kya command double hogi, ya aur bhi buri?

$t_{g o} = 4$ : Yahan se kyun shuru karein? Yeh Ex 4 se hamara reference value hai, isliye baad ki har row iske saath compare hoti hai. $\frac{6}{16} (40) - \frac{2}{4} (12) = 15 - 6 = 9$ m/s².
$t_{g o} = 2$ : 2 tak half kyun karein? Pehla doubling test karne ke liye — agar law $1/ t_{g o}$ mein linear hota toh command sirf double hoti; dekho yeh aur kya karta hai. $\frac{6}{4} (40) - \frac{2}{2} (12) = 60 - 12 = 48$ m/s².
$t_{g o} = 1$ : 1 tak kyun jaayein? Time khatam hone par dominant position term ki $1/ t_{g o}^{2}$ scaling confirm karne ke liye. $\frac{6}{1} (40) - \frac{2}{1} (12) = 240 - 24 = 216$ m/s².
$t_{g o} = 0.5$ : Yeh final tiny value kyun? Blow-up ko unmistakable banane ke liye — yahi woh jagah hai jahan real thrusters saturate hote hain aur command physically unrealisable ho jaati hai. $\frac{6}{0.25} (40) - \frac{2}{0.5} (12) = 960 - 48 = 912$ m/s².

Verify: 9 → 48 → 216 → 912: growth doubling se tez hai, kyunki dominant $6/ t_{g o}^{2}$ term $1/ t_{g o}^{2}$ jaisi scale karti hai — time quarter karo, position term chaar guna ho jaata hai (15→60→240→960). Yeh parent note ki warning hai numeric form mein: thode time mein residual miss unbounded thrust demand karta hai, isliye real systems ko ZEM→0 jaldi drive karna chahiye.

Figure mein, cyan curve $t_{g o}$ ke against commanded acceleration hai; amber dots hamare chaar rows mark karte hain. Notice karo ki curve left par vertical axis ki taraf kaise rocket karta hai — woh steep wall saturation cliff hai jisse tum miss jaldi maarne ke zariye door rehte ho.

Worked example Ex 6 — C6: 2-D intercept, position-only — aur YEH Proportional Navigation (

$N = 3$ ) kyun hai Ek interceptor jiska predicted miss vector $ZEM = (300, - 450)$ m aur $t_{g o} = 6$ s hai. Velocity constrained nahi hai (sirf hit karna hai).

Forecast: Kaun sa law — $6/ t_{g o}^{2}$ wala ya $3/ t_{g o}^{2}$ wala? Aur kya $a_{c}$ $ZEM$ ke saath ya uske against point karega?

Law choose karo. Kyun? Koi $v_{f}$ requirement nahi → position-only → intercept form $a_{c} = \frac{3}{t _{g o}^{2}} ZEM$ .
Componentwise apply karo. Componentwise kyun? Double-integrator dynamics har axis pe decouple hoti hain (dekho Double Integrator Dynamics). $a_{c} = \frac{3}{36} (300, - 450) = (25, - 37.5)$ m/s².
Direction check. Yeh step kyun? Ek geometric fact reveal karne ke liye jo har intercept ke liye matter karta hai: ek single positive scalar ( $3/ t_{g o}^{2} > 0$ ) vector rotate nahi kar sakta, isliye command parallel honi chahiye $ZEM$ se — tum always predicted miss ki taraf seedhe thrust karte ho, kabhi sideways nahi. $a_{c} = \frac{1}{12} (300, - 450)$ $ZEM$ ka ek positive multiple hai. Figure dekho.
Prove karo ki yeh PN with $N = 3$ ke barabar hai. Yeh step kyun? Parent note claim karta hai ki intercept form Proportional Navigation disguise mein hai; hume ise actually dikhana hai, assert nahi. Proportional Navigation mein acceleration hai $a_{P N} = N V_{c} \dot{λ}$ , jahan $\dot{λ}$ line-of-sight (LOS) turn rate hai aur $V_{c}$ closing speed. Constant-velocity target ke liye LOS ke perpendicular measure kiya gaya ZEM obeys karta hai $ZEM_{⊥} = R \dot{λ} t_{g o}$ , jahan $R$ range hai aur $R = V_{c} t_{g o}$ (range closing speed pe close hoti hai). Substitute karo: $ZEM_{⊥} = V_{c} t_{g o}^{2} \dot{λ}$ . Ab ise apne law mein feed karo: $a_{c} = \frac{3}{t _{g o}^{2}} ZEM_{⊥} = \frac{3}{t _{g o}^{2}} V_{c} t_{g o}^{2} \dot{λ} = 3 V_{c} \dot{λ}$ . Yeh exactly $a_{P N}$ hai $N = 3$ ke saath. Coefficient $3$ hi navigation ratio hai.

Verify: Magnitude $∣ a_{c} ∣ = 2 5^{2} + 37. 5^{2} = 625 + 1406.25 = 2031.25 \approx 45.07$ m/s². Aur PN identity dimensionally check karti hai: $[V_{c} \dot{λ}] = \frac{m}{s} \cdot \frac{1}{s} = \frac{m}{s ^{2}}$ ✓ — ek acceleration, jaise hona chahiye. Command $ZEM$ ke parallel hai kyunki ek single positive coefficient vector rotate nahi kar sakta.

Figure mein, cyan arrow $ZEM$ hai (interceptor se predicted miss point tak) aur amber arrow $a_{c}$ hai (sirf visible hone ke liye chhe-guna lamba draw kiya gaya). Observe karo ki dono arrows same line pe hain — yahi "direction check" ka poora content hai.

Worked example Ex 7 — C7: ZEM mein gravity bhoolna

Same interceptor geometry lekin ab woh ek gravity field mein uda raha hai $g = - 9.81$ m/s² jo $y$ -axis pe act karta hai, $t_{g o} = 6$ s. Target ke liye true position miss gravity-drop term require karta hai. Suppose raw geometric miss (target minus current straight-line coast, gravity ignored) hai $(300, - 450)$ m.

Forecast: $y$ -component mein gravity-drop term $- \frac{1}{2} g t_{g o}^{2}$ add karne par — kya $y$ -miss magnitude mein badhegi ya ghategi?

Coast ke dauran gravity drift. Kyun? Vehicle $t_{g o}$ ke dauran $\frac{1}{2} g t_{g o}^{2}$ girta hai; coast mein yeh already include hai, isliye corrected $y$ -ZEM naive value minus $\frac{1}{2} g t_{g o}^{2}$ hai. $\frac{1}{2} g t_{g o}^{2} = \frac{1}{2} (- 9.81) (36) = - 176.58$ m.
Corrected ZEM. Kyun? Hum us drift ko position error mein fold karte hain taaki command true miss ki taraf aimed ho, gravity-free illusion ki taraf nahi. $ZEM_{y} = - 450 - (- 176.58) = - 273.42$ m. ( $x$ -axis pe gravity nahi hai: $ZEM_{x} = 300$ .)
Command. Kyun? Corrected miss pe intercept law apply karo. $a_{c} = \frac{3}{36} (300, - 273.42) = (25, - 22.785)$ m/s².

Verify: $y$ -command $- 37.5$ (Ex 6, no gravity) se $- 22.785$ tak drop hui — kyunki gravity already vehicle ko target ki taraf neeche pull kar rahi thi, isliye kam commanded thrust chahiye. $\frac{1}{2} g t_{g o}^{2}$ term bhool jaana $∣ - 37.5 - (- 22.785) ∣ = 14.715$ m/s² ka over-command karta — exactly parent note se woh "gravity bhoolna" waali galti.

Worked example Ex 8 — C8: Real-world word problem (Mars descent)

"Ek descent stage landing site se 500 m upar hai, 45 m/s neeche descend kar raha hai, aur horizontally 8 m/s drift kar raha hai. Use pad ( $r_{f} = 0$ ) tak $t_{g o} = 12$ s mein zero velocity ke saath pohonchna hai. Martian gravity 3.71 m/s² downward hai. Commanded acceleration vector kya hai?"

Axes set karo (hamare conventions ke anusaar): $x$ horizontal, $y$ upar. Toh $r = (0, 500)$ , $v = (8, - 45)$ , $g = (0, - 3.71)$ , $r_{f} = 0$ , $v_{f} = 0$ .

Forecast: Horizontal axis mein koi gravity nahi aur koi target offset nahi sirf drift zero karne ke alawa — kya iska command chhoti hogi?

ZEM per axis. Kyun? Coast prediction, axis by axis. $ZEM_{x} = 0 - 0 - (8) (12) - 0 = - 96$ m. $ZEM_{y} = 0 - 500 - (- 45) (12) - \frac{1}{2} (- 3.71) (144) = - 500 + 540 + 267.12 = 307.12$ m.
ZEV per axis. Yahan bhi axis by axis kyun? Kyunki double integrator axes decouple karta hai (dekho Double Integrator Dynamics) — har direction ka apna independent velocity error hota hai, aur sirf $y$ gravity feel karta hai, isliye hume gravity term sahi axis pe rakhne ke liye unhe alag compute karna chahiye. $ZEV_{x} = 0 - 8 - 0 = - 8$ m/s. $ZEV_{y} = 0 - (- 45) - (- 3.71) (12) = 45 + 44.52 = 89.52$ m/s.
Full law per axis. Full law kyun? $v_{f} = 0$ required hai (soft touchdown). $a_{c x} = \frac{6}{144} (- 96) - \frac{2}{12} (- 8) = - 4 + 1.3333 = - 2.6667$ m/s². $a_{cy} = \frac{6}{144} (307.12) - \frac{2}{12} (89.52) = 12.79667 - 14.92 = - 2.12333$ m/s².

Verify: $a_{c} \approx (- 2.667, - 2.123)$ m/s². Horizontal command negative hai → yeh $+ 8$ m/s drift ko oppose karta hai, ise kill karta hai. Interestingly $a_{cy} < 0$ : ZEV braking demand yahan ZEM push se thodi exceed karti hai, isliye commanded thrust thodi neeche point karta hai — sensible hai, kyunki coast pad ki altitude ko 307 m overshoot karne wala tha aur descent shape karne ki zaroorat hai, sirf brake karne ki nahi. Compare Powered Descent Guidance (Apollo E-Guidance). Units ✓.

Worked example Ex 9 — C9: Exam twist — law ko

$t_{g o}$ ke liye invert karo "Tumhara thruster commanded acceleration ko $a_{m a x} = 5$ m/s² pe cap karta hai. Ek axis ke along pure position miss $ZEM = 45$ m diya gaya hai (intercept law, ZEV ignored), toh woh smallest $t_{g o}$ kya hai jis par command still achievable hai?"

Forecast: Kyunki $a_{c} = 3 ZEM / t_{g o}^{2}$ $t_{g o}$ ghatne par badha hai, constraint $t_{g o}$ pe ek floor set karti hai. 5 s se bada ya chhota? Guess karo.

Constraint likho. Kyun? Hum woh boundary chahte hain jahan $a_{c} = a_{m a x}$ ho. $\frac{3 ZEM}{t _{g o}^{2}} = a_{m a x}$ .
$t_{g o}$ ke liye solve karo. Square-root kyun? Yeh $t_{g o}$ mein quadratic hai. $t_{g o} = \frac{3 ZEM}{a _{m a x}} = \frac{3 ( 45 )}{5} = 27 \approx 5.196$ s.
Interpret karo. Yeh step kyun? Ek raw number tab tak useless hai jab tak hum yeh nahi kehte ki iska operationally kya matlab hai, jo exam twist ka poora point hai. Kisi bhi $t_{g o} > 5.196$ s ke liye command cap ke neeche hai; exactly $5.196$ s pe woh saturate hoti hai; uske neeche, koi achievable command miss fix nahi kar sakti — tumhe ZEM pehle hi drive karna chahiye tha.

Verify: Plug back karo: $\frac{3 ( 45 )}{27} = \frac{135}{27} = 5$ m/s² $= a_{m a x}$ ✓. Yeh Ex 5 se design principle hai ek number mein turned: thruster saturation residual miss kill karne ke liye ek deadline define karta hai — ise cross karo aur koi bhi legal command shot nahi bacha sakti, yahi exactly wajah hai ZEM/ZEV continuous feedback ke roop mein fly kiya jaata hai jo miss shrink karta hai jab $t_{g o}$ still comfortably large ho.

Recall Self-test

Agar ZEM = 0 aur ZEV = 0 hain, toh optimal command hai ::: exactly zero — tum already ek boundary-satisfying coast pe ho. Kaun se law mein koi ZEV term nahi hai, aur iska coefficient kya hai? ::: intercept (position-only) law, $3/ t_{g o}^{2}$ . Jab $t_{g o}$ half ho, dominant ZEM term multiply hota hai ::: chaar se (yeh $1/ t_{g o}^{2}$ jaisi scale karti hai). $\frac{1}{2} g t_{g o}^{2}$ bhoolna lander ko kyun break karta hai? ::: predicted miss exactly gravity drift se wrong hota hai, isliye command miscalibrated hoti hai. Soft-landing law mein dono terms ke opposite signs hain kyunki ::: $t_{f}$ ke paas position correct karna aur velocity correct karna opposite-directed acceleration demand karte hain.

See also: Optimal guidance — ZEM - ZEV formulation · Calculus of Variations & Pontryagin's Minimum Principle · Time-to-go estimation · Optimal Control — LQR · Lambert's Problem