Exercises — Terminal descent — velocity vector alignment, touchdown constraints

• $\frac{6}{t_{go}^2}({\mathbf{r}}_f-\mathbf{r})$ → position-error term: jahan tum ho aur pad ke beech ka gap close karta hai.
• $-\frac{2}{t_{go}}(2\mathbf{v}+{\mathbf{v}}_f)$ → velocity-shaping term: tumhari current velocity ko desired touchdown velocity ki taraf le jaata hai.
• $-\mathbf{g}$ → gravity-cancellation term, already command ke andar hai isliye engine gravity dobara add nahi karta.
• $t_{go}\to 0$ par position term ka $1/t_{go}^2$ factor sabse tezi se blow up hota hai → infinite commanded acceleration.

1. Vertical speed $v_{td}\le v_{td}^{\max }$ (legs crush ho jaate).
2. Horizontal speed $\sqrt{v_x^2+v_y^2}\le v_h^{\max }$ (skid/tip).
3. Tilt angle $\theta \le {\theta }_{\max }$ (thrust axis vs. vertical).
4. Angular rate $|\omega |\le {\omega }_{\max }$.
5. Position pad radius $R$ ke andar.

Positive (upward). Tum neeche tezi se ja rahe ho aur woh downward speed kam karni hai, jiske liye net upward acceleration chahiye. Hamare up-$=+$ convention mein yeh ek positive number hai.

$a_{cmd}=\frac{6}{3^2}(-30)-\frac{2}{3}(2(-18)+(-1.5))-(-3.71)$ $=\frac{6}{9}(-30)-\frac{2}{3}(-37.5)+3.71=-20+25+3.71=8.71{\text{m/s}}^2.$ Engine exactly $T/m=8.71$ m/s² upward produce karta hai. $g$ dobara mat add karo.

Formula har axis par apply karo. $\frac{6}{25}=0.24$, $\frac{2}{5}=0.4$. x: $0.24(10)-0.4(2(4)+0)-0=2.4-3.2=-0.8$ m/s². z: $0.24(-50)-0.4(2(-30)+(-1.5))-(-9.81)=-12+24.6+9.81=22.41$ m/s². ${\mathbf{a}}_{cmd}=(-0.8,0,22.41){\text{m/s}}^2.$ Negative $x$ command $+4$ m/s eastward drift bleed karta hai; bada positive $z$ fall ko brake karta hai.

Velocity ek right triangle banata hai: "opposite" side horizontal speed $|v_x|=6$ hai, "adjacent" side (down direction ke saath) $|v_z|=30$ hai. ${\theta }_v=\arctan \frac{|v_x|}{|v_z|}=\arctan \frac{6}{30}=\arctan (0.2)=11.31^{\circ }.$ Guidance $v_{f,x}=0$ target karta hai, isliye yeh angle touchdown tak $0^{\circ }$ tak shrink hona chahiye.

Velocity-induced lean vertical se: $\arctan \frac{v_h}{v_{td}}=\arctan \frac{0.5}{1.8}=\arctan (0.2778)=15.52^{\circ }.$ Total effective lean: $4^{\circ }+15.52^{\circ }=19.52^{\circ }\le 25^{\circ }.$ Safe. Margin $=25-19.52=5.48^{\circ }$. Figure dekhte hain: combined arrow (tilt + drift) leg-footprint cone ke andar rehta hai.

Tip criterion velocity lean ke liye angular budget $25^{\circ }-4^{\circ }=21^{\circ }$ allow karta hai: $\arctan \frac{v_h}{1.8}\le 21^{\circ }\Rightarrow v_h\le 1.8\tan (21^{\circ })=1.8(0.3839)=0.691\text{m/s}.$ Tip criterion $v_h$ ko $0.69$ m/s par cap karta hai, raw $1.0$ m/s limit se stricter. Tip criterion pehle bind karta hai — drift ko sirf $1.0$ se nahi, $0.69$ m/s se neeche laana hoga.

Position term $1/t_{go}^2$ ke saath scale karta hai. $t_{go}$ adha karne par term chaar guna ho jaati hai.

• $t_{go}=4$: $-15$ m/s².
• $t_{go}=2$: $\times (4/2{)}^2=\times 4\Rightarrow -60$ m/s².
• $t_{go}=1$: $\times (4/1{)}^2=\times 16\Rightarrow -240$ m/s². Growth $t_{go}$ mein inverse-square hai — exactly isliye $t_{go}$ ko floor karna hoga ya hover phase par hand off karna hoga pehle ki yeh explode ho.

Shaping acceleration $\mathbf{s}(\tau )={\mathbf{c}}_0+{\mathbf{c}}_1\tau$ ke saath kaam karo (hum $-\mathbf{g}$ end mein add karenge). Velocity integrate karo: $\mathbf{v}(\tau )=\mathbf{v}+{\mathbf{c}}_0\tau +\frac{1}{2}{\mathbf{c}}_1{\tau }^2.$ Position integrate karo: $\mathbf{r}(\tau )=\mathbf{r}+\mathbf{v}\tau +\frac{1}{2}{\mathbf{c}}_0{\tau }^2+\frac{1}{6}{\mathbf{c}}_1{\tau }^3.$ $T=t_{go}$ lo. Boundary conditions:

$\frac{6}{36}=0.1667$, $\frac{2}{6}=0.3333$. ${\mathbf{r}}_f-\mathbf{r}=(0,0,-60)$. x: $0.1667(0)-0.3333(2(3)+0)-0=-0.3333(6)=-2.0$ m/s². z: $0.1667(-60)-0.3333(2(-22)+(-1.5))-(-3.71)=-10-0.3333(-45.5)+3.71=-10+15.167+3.71=8.877$ m/s². ${\mathbf{a}}_{cmd}=(-2.0,0,8.877){\text{m/s}}^2.$ Thrust tilt from vertical: ${\theta }_T=\arctan \frac{|a_x|}{a_z}=\arctan \frac{2.0}{8.877}=\arctan (0.2253)=12.70^{\circ }.$ Engine eastward drift khatam karne ke liye vertical se $12.7^{\circ }$ lean karta hai, phir $a_x\to 0$ hone par seedha ho jaata hai.

$\left|\frac{6(-60)}{t_{go}^2}\right|=15$ set karo: $\frac{360}{t_{go}^2}=15\Rightarrow t_{go}^2=24\Rightarrow t_{go}=\sqrt{24}=4.90\text{s}.$ $t_{go}\approx 4.9$ s se neeche (is residual error ke saath) position term akela $>15$ m/s² demand karega, engine saturate ho jayega. Action: $t_{go}$ floor karo, ya low altitude par constant-velocity / gravity-turn "hover-then-drop" phase par hand off karo, jaisa parent note ke $t_{go}\to 0$ mistake mein bataya gaya hai. Practice mein tum trajectory aisa design karoge ki tab tak error lagbhag zero ho, demand bounded rakhe.

(a) $\frac{6}{25}=0.24$, $\frac{2}{5}=0.4$. ${\mathbf{r}}_f-\mathbf{r}=(-8,0,-45)$.

• x: $0.24(-8)-0.4(2(2)+0)-0=-1.92-1.6=-3.52$ m/s².
• z: $0.24(-45)-0.4(2(-16)+(-1.5))-(-3.71)=-10.8-0.4(-33.5)+3.71=-10.8+13.4+3.71=6.31$ m/s². ${\mathbf{a}}_{cmd}=(-3.52,0,6.31){\text{m/s}}^2.$ (b) Drift $v_x=+2$ (east) hai, command $a_x=-3.52$ (west) hai → drift oppose karta hai ✔. $a_z=+6.31$ upward hai → braking ✔. Guidance sahi direction mein point kar raha hai. (c) Velocity lean $=\arctan \frac{0.6}{1.7}=\arctan (0.3529)=19.44^{\circ }$. Total $=5^{\circ }+19.44^{\circ }=24.44^{\circ }$. $24.44^{\circ }>22^{\circ }\Rightarrow \text{FAILS the tip criterion.}$ Bhale hi har raw limit individually pass ho jaye, coupled lean tip cone exceed karta hai — lander topple ho jaayega. GNC ko contact se pehle $v_h$ aur trim karna hoga ya tilt reduce karna hoga.

(i) PDG higher-altitude braking trajectory produce karta hai jo pad ke paas is terminal law par hand off karta hai. (ii) Proportional Navigation position-only cousin hai (intercepts ke liye great) — terminal descent ise do-point (position aur velocity) problem tak extend karta hai. (iii) Time-to-go Estimation woh $t_{go}$ supply karta hai jisse yahan har term divide hoti hai; galat $t_{go}$ poora command corrupt karta hai. (iv) Convex Optimization Landing same soft-landing boundary problem solve karta hai lekin thrust bounds aur glide-slope constraints rigorously enforce karte hue — yahan closed-form law unconstrained special case hai. (v) Attitude Control & Inner Loop fast loop hai jo actually body rotate karta hai taaki thrust axis woh direction realize kare jo is outer loop ka ${\mathbf{a}}_{cmd}$ command karta hai.

Position term $1/t_{go}^2$ ke saath kyun scale karta hai? ::: Kyunki position acceleration ka double integral hai; ise time $t_{go}$ par match karna $t_{go}^2$ se divide karta hai. Kaun se coefficients linear-shaping derivation se aate hain aur kahan se? ::: 6 (cubic $\frac{1}{6}{\tau }^3$ position integral se) aur 2 (velocity mismatch se). Kaun si ek quantity, agar galat estimate ho, command ki har term corrupt kar deti hai? ::: $t_{go}$, time-to-go. Jab do limits pass ho jaayein phir bhi lander tip kare, tum kya bhool gaye? ::: Coupled tip criterion $\theta +\arctan (v_h/v_{td})\le {\theta }_{tip}$.