Exercises — Terminal descent — velocity vector alignment, touchdown constraints

• $\frac{6}{t_{go}^2}({\mathbf{r}}_f-\mathbf{r})$ → the position-error term: closes the gap between where you are and the pad.
• $-\frac{2}{t_{go}}(2\mathbf{v}+{\mathbf{v}}_f)$ → the velocity-shaping term: bleeds your current velocity toward the desired touchdown velocity.
• $-\mathbf{g}$ → the gravity-cancellation term, already inside the command so the engine never re-adds gravity.
• As $t_{go}\to 0$ the $1/t_{go}^2$ factor in the position term blows up fastest → infinite commanded acceleration.

1. Vertical speed $v_{td}\le v_{td}^{\max }$ (legs would crush).
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 within pad radius $R$.

Positive (upward). You are moving down fast and must reduce that downward speed, which needs a net upward acceleration. In our up-$=+$ convention that is a positive number.

$a_{cmd}=\frac{6}{3^2}(-30)-\frac{2}{3}(2(-18)+(-1.5))-(-3.71)$ $=\frac{6}{9}(-30)-\frac{2}{3}(-37.5)+3.71=-20+25+3.71=8.71{\text{m/s}}^2.$ Engine produces exactly $T/m=8.71$ m/s² upward. Do not add $g$ again.

Apply the formula per axis. $\frac{6}{25}=0.24$, $\frac{2}{5}=0.4$. 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. ${\mathbf{a}}_{cmd}=(-0.8,0,22.41){\text{m/s}}^2.$ The negative $x$ command bleeds off the $+4$ m/s eastward drift; the large positive $z$ brakes the fall.

The velocity makes a right triangle: the "opposite" side is the horizontal speed $|v_x|=6$, the "adjacent" side (along the down direction) is $|v_z|=30$. ${\theta }_v=\arctan \frac{|v_x|}{|v_z|}=\arctan \frac{6}{30}=\arctan (0.2)=11.31^{\circ }.$ Guidance targets $v_{f,x}=0$, so this angle must shrink to $0^{\circ }$ by touchdown.

Velocity-induced lean from vertical: $\arctan \frac{v_h}{v_{td}}=\arctan \frac{0.5}{1.8}=\arctan (0.2778)=15.52^{\circ }.$ Total effective lean: $4^{\circ }+15.52^{\circ }=19.52^{\circ }\le 25^{\circ }.$ Safe. Margin $=25-19.52=5.48^{\circ }$. Look at the figure: the combined arrow (tilt + drift) stays inside the leg-footprint cone.

Tip criterion allows angular budget $25^{\circ }-4^{\circ }=21^{\circ }$ for the velocity lean: $\arctan \frac{v_h}{1.8}\le 21^{\circ }\Rightarrow v_h\le 1.8\tan (21^{\circ })=1.8(0.3839)=0.691\text{m/s}.$ The tip criterion caps $v_h$ at $0.69$ m/s, stricter than the raw $1.0$ m/s limit. The tip criterion binds first — you must slow drift below $0.69$ m/s, not merely below $1.0$.

The position term scales as $1/t_{go}^2$. Halving $t_{go}$ quadruples the term.

• $t_{go}=4$: $-15$ 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 is inverse-square in $t_{go}$ — this is exactly why we must floor $t_{go}$ or hand off to a hover phase before it explodes.

Work with the shaping acceleration $\mathbf{s}(\tau )={\mathbf{c}}_0+{\mathbf{c}}_1\tau$ (we add $-\mathbf{g}$ at the end). Integrate velocity: $\mathbf{v}(\tau )=\mathbf{v}+{\mathbf{c}}_0\tau +\frac{1}{2}{\mathbf{c}}_1{\tau }^2.$ Integrate position: $\mathbf{r}(\tau )=\mathbf{r}+\mathbf{v}\tau +\frac{1}{2}{\mathbf{c}}_0{\tau }^2+\frac{1}{6}{\mathbf{c}}_1{\tau }^3.$ Let $T=t_{go}$. 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: ${\theta }_T=\arctan \frac{|a_x|}{a_z}=\arctan \frac{2.0}{8.877}=\arctan (0.2253)=12.70^{\circ }.$ The engine leans $12.7^{\circ }$ off vertical to kill the eastward drift, then rotates upright as $a_x\to 0$.

Set $\left|\frac{6(-60)}{t_{go}^2}\right|=15$: $\frac{360}{t_{go}^2}=15\Rightarrow t_{go}^2=24\Rightarrow t_{go}=\sqrt{24}=4.90\text{s}.$ Below $t_{go}\approx 4.9$ s (with this residual error) the position term alone would demand $>15$ m/s², saturating the engine. Action: floor $t_{go}$, or hand off to a constant-velocity / gravity-turn "hover-then-drop" phase at low altitude, as flagged in the parent note's $t_{go}\to 0$ mistake. In practice you'd design the trajectory so the error is nearly zero by then, keeping demand bounded.

(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 is $v_x=+2$ (east), command $a_x=-3.52$ (west) → opposes drift ✔. $a_z=+6.31$ is upward → braking ✔. Guidance is pointed correctly. (c) Velocity lean $=\arctan \frac{0.6}{1.7}=\arctan (0.3529)=19.44^{\circ }$. Total $=5^{\circ }+19.44^{\circ }=24.44^{\circ }$. $24.44^{\circ }>22^{\circ }\Rightarrow \text{FAILS the tip criterion.}$ Even though each raw limit might pass individually, the coupled lean exceeds the tip cone — the lander would topple. GNC must trim $v_h$ further or reduce tilt before contact.

(i) PDG produces the higher-altitude braking trajectory that hands off to this terminal law near the pad. (ii) Proportional Navigation is the position-only cousin (great for intercepts) — terminal descent extends it to a two-point (position and velocity) problem. (iii) Time-to-go Estimation supplies the $t_{go}$ that every term here divides by; a bad $t_{go}$ corrupts the whole command. (iv) Convex Optimization Landing solves the same soft-landing boundary problem but with thrust bounds and glide-slope constraints enforced rigorously — the closed-form law here is the unconstrained special case. (v) Attitude Control & Inner Loop is the fast loop that actually rotates the body so the thrust axis realizes the direction of ${\mathbf{a}}_{cmd}$ that this outer loop commands.

Why does the position term scale as $1/t_{go}^2$? ::: Because position is a double integral of acceleration; matching it over time $t_{go}$ divides by $t_{go}^2$. Which coefficients come from the linear-shaping derivation and where? ::: 6 (from the cubic $\frac{1}{6}{\tau }^3$ position integral) and 2 (from the velocity mismatch). What single quantity, if mis-estimated, corrupts every term of the command? ::: $t_{go}$, the time-to-go. When two limits both pass but the lander still tips, what did you forget? ::: The coupled tip criterion $\theta +\arctan (v_h/v_{td})\le {\theta }_{tip}$.