Terminal descent — velocity vector alignment, touchdown constraints

WHAT is terminal descent?

WHY these four? A lander with the right position but sideways velocity will scrape a leg and topple. Right velocity but tilted body means thrust has a horizontal component you can't kill in time. So position and velocity and attitude must converge together.

HOW we build the guidance law from first principles

We want the vehicle to reach a desired state at time $t_f$ . Model the vehicle as a point mass (attitude handled by a faster inner loop):

$\ddot{\mathbf{r}} = \mathbf{a}_{cmd} + \mathbf{g}$

Sign / definition convention (fix this in your head now): $\mathbf{a}_{cmd}$ is the net commanded acceleration the vehicle should experience, and gravity $\mathbf g=(0,0,-g)$ is added separately in the dynamics. This means the thrust acceleration the engine must physically produce is $\mathbf a_{thrust}=\mathbf a_{cmd}$ (gravity is NOT re-added here) — the whole point of the $-\mathbf g$ term inside the formula below is that $\mathbf a_{cmd}$ already contains the gravity-cancelling contribution. So: $\mathbf{a}_{cmd}=\frac{\mathbf T}{m},\qquad \ddot{\mathbf r}=\underbrace{\frac{\mathbf T}{m}}_{\mathbf a_{cmd}}+\mathbf g .$ Keep this straight: the engine supplies exactly $\mathbf a_{cmd}$ , no more.

Step 1 — Ask: what trajectory shape can hit both a position and a velocity target?

Why this step? A single straight line only lets us aim position. To pin position AND velocity at $t_f$ we need enough free parameters. A constant acceleration gives us exactly the freedom we need.

Assume constant commanded acceleration $\mathbf{a}$ over the remaining time-to-go $t_{go}=t_f-t$ :

$\mathbf{r}(t_f)=\mathbf{r}+\mathbf{v}\,t_{go}+\tfrac12(\mathbf{a}+\mathbf{g})\,t_{go}^2$ $\mathbf{v}(t_f)=\mathbf{v}+(\mathbf{a}+\mathbf{g})\,t_{go}$

Step 2 — Impose the two boundary conditions $\mathbf{r}(t_f)=\mathbf{r}_f$ , $\mathbf{v}(t_f)=\mathbf{v}_f$ .

From the velocity equation: $\mathbf{a}+\mathbf{g}=\frac{\mathbf{v}_f-\mathbf{v}}{t_{go}}$

Substitute into the position equation and solve. This is over-determined with only a constant $\mathbf a$ , so we allow $\mathbf a$ to vary linearly in time (add one more free vector). Doing the algebra (this is the classic E-guidance / two-point boundary value result):

Formula Terminal guidance acceleration (soft landing)

$\boxed{\;\mathbf{a}_{cmd}=\frac{6}{t_{go}^{2}}\big(\mathbf{r}_f-\mathbf{r}\big)-\frac{2}{t_{go}}\big(2\mathbf{v}+\mathbf{v}_f\big)-\mathbf{g}\;}$

Where each term comes from:

$\frac{6}{t_{go}^2}(\mathbf r_f-\mathbf r)$ → position error correction (kill the gap).
$-\frac{2}{t_{go}}(2\mathbf v+\mathbf v_f)$ → velocity shaping: bleeds current $\mathbf v$ toward desired $\mathbf v_f$ .
$-\mathbf g$ → gravity cancellation, folded into $\mathbf a_{cmd}$ . Because $\ddot{\mathbf r}=\mathbf a_{cmd}+\mathbf g$ , this $-\mathbf g$ is exactly what cancels the $+\mathbf g$ in the dynamics so the trajectory follows the intended shape. The engine then just produces $\mathbf T/m=\mathbf a_{cmd}$ .

Derivation check of the coefficients: With a linear-in-time acceleration $\mathbf a(t)=\mathbf c_0+\mathbf c_1\tau$ ( $\tau$ measured from now), integrating and matching the four scalar boundary conditions per axis gives the numbers $6$ and $2$ . (Verified in the VERIFY block by integrating and confirming $\mathbf r(t_f),\mathbf v(t_f)$ hit targets.)

Step 3 — Velocity vector alignment. For a vertical soft touchdown choose $\mathbf{v}_f=(0,\,0,\,-v_{td}),\qquad v_{td}\approx 1\text{–}2\ \text{m/s}.$ The horizontal components of $\mathbf v_f$ are zero, forcing the velocity vector to rotate to vertical as $t_{go}\to 0$ .

Touchdown constraints (the inequalities that must ALL hold)

WHY a tilt limit? A lander tips if its velocity + tilt push the center of mass outside the footprint of the legs. Geometrically, stability requires $\theta + \arctan\!\frac{v_h}{v_{td}} \lesssim \theta_{tip}$ i.e. both a slow horizontal drift and a small body tilt.

Figure — Terminal descent — velocity vector alignment, touchdown constraints

Worked Examples

Worked example Example 1 — Compute the guidance command

Given (1-D vertical, $g=3.71$ m/s² Mars): altitude to go $r_f-r=-40$ m (need to descend 40 m to $r_f$ ), current $v=-25$ m/s, desired $v_f=-1.5$ m/s, $t_{go}=4$ s. Find $a_{cmd}$ .

Why set up like this? Vertical axis, targets are the final values; sign convention up = +. $a_{cmd}=\frac{6}{16}(-40)-\frac{2}{4}(2(-25)+(-1.5))-(-3.71)$ $=-15 -\tfrac12(-51.5)+3.71 = -15+25.75+3.71=14.46\ \text{m/s}^2$ Why positive? We're falling fast (−25 m/s) and must brake hard, so the net upward acceleration $>0$ . What must the engine produce? Exactly this: $T/m=a_{cmd}=14.46$ m/s² upward — the $-\mathbf g$ term is already inside $a_{cmd}$ , so do not add $g$ again. (Sanity: the resulting net acceleration in the world is $a_{cmd}+g_{world}=14.46+(-3.71)=10.75$ m/s² up, which is the trajectory-shaping deceleration.) ✅

Common Mistakes (Steel-manned)

Active Recall

Why must position AND velocity be targeted in terminal descent?

A right position with wrong velocity vector causes skid/topple; soft landing pins both endpoints — a two-point boundary value problem.

Write the soft-landing guidance acceleration.

\mathbf a_{cmd}=\frac{6}{t_{go}^2}(\mathbf r_f-\mathbf r)-\frac{2}{t_{go}}(2\mathbf v+\mathbf v_f)-\mathbf g

What does the

-\mathbf g

term do and where does it live?

It cancels the

+\mathbf g

\ddot{\mathbf r}=\mathbf a_{cmd}+\mathbf g

; it is folded inside

\mathbf a_{cmd}

, so the engine produces exactly

\mathbf T/m=\mathbf a_{cmd}

(do not add

g

again).

Does the engine supply

a_{cmd}

a_{cmd}+g

Exactly

a_{cmd}

\mathbf T/m=\mathbf a_{cmd}

, because the

-\mathbf g

inside the formula already handles gravity compensation.

Where do the coefficients 6 and 2 come from?

Matching a linear-in-time acceleration to the four position+velocity boundary conditions per axis.

What happens as

t_{go}\to 0

The

1/t_{go}^2

gain diverges → infinite thrust; must floor

t_{go}

or hand off to a hover-drop.

Name the touchdown envelope constraints.

Vertical speed, horizontal speed, tilt angle, angular rate, and position-within-pad — all AND-constraints.

Why a tilt-angle limit?

To keep the CoM inside the leg footprint so combined tilt + drift don't tip the lander.

What is velocity vector alignment?

Rotating velocity to nearly vertical (horizontal components → 0, vertical → small

-v_{td}

) at touchdown.

Recall Feynman: explain to a 12-year-old

Imagine landing a drone on a tiny box. You don't want to slam into it or slide off. So in the last few seconds you slow down until you're barely moving, make sure you're going straight down (not sideways), and keep the drone standing up straight — and you do all three at once, right when you touch. The math just says: "how hard should I push, given how far I still have to go and how fast I'm still moving?" Push harder if far/fast. And here's the key: the formula already includes the extra push needed to fight gravity — so the motor just does what the formula says, you don't tack on gravity a second time.

Connections

Powered Descent Guidance (PDG) — earlier phase feeding into terminal descent
Proportional Navigation — contrast: aims position only, not velocity
Attitude Control & Inner Loop — enforces the tilt/rate limits
Time-to-go Estimation — the $t_{go}$ that appears in every gain
Convex Optimization Landing (lossless convexification) — modern alternative honoring thrust bounds

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, terminal descent matlab landing ke last few seconds. Yahan sirf pad pe pahunchna kaafi nahi hai — vehicle ka velocity vector bhi seedha neeche point karna chahiye aur speed almost zero honi chahiye, warna lander sideways skid karke gir jaayega. Isliye hum ek "two-point boundary value problem" solve karte hain: position bhi target karo aur velocity bhi target karo, dono ek saath $t_f$ pe milne chahiye.

Guidance law ka formula hai $a_{cmd}=\frac{6}{t_{go}^2}(\mathbf r_f-\mathbf r)-\frac{2}{t_{go}}(2\mathbf v+\mathbf v_f)-\mathbf g$ . Pehla term position gap band karta hai, doosra term current velocity ko desired final velocity ki taraf le jaata hai. Ab important baat: $-\mathbf g$ term already $a_{cmd}$ ke andar hai — matlab engine ko exactly $T/m=a_{cmd}$ produce karna hai, gravity ko dobara add mat karo. Dynamics hai $\ddot{\mathbf r}=a_{cmd}+\mathbf g$ , aur formula ka $-\mathbf g$ us $+\mathbf g$ ko cancel karta hai.

Touchdown constraints ek saath satisfy hone chahiye (AND conditions): vertical speed limit se kam, horizontal drift bahut kam, body ka tilt angle chhota, angular rate low, aur pad ke andar. Agar tilt aur sideways drift dono milke center of mass ko legs ke bahar dhakel de to lander topple ho jaayega — isiliye velocity vector ko vertical rotate karna zaroori hai.

Ek important trap: jaise-jaise $t_{go}\to0$ , gain $1/t_{go}^2$ infinity ki taraf jaata hai — thrust saturate ho jaayega. Isliye practical mein hum $t_{go}$ ko floor karte hain ya last thodi height pe ek simple hover-then-drop pe switch kar dete hain. Mantra: SLOW, STRAIGHT, STEADY, ON-TIME, ON-PAD.

Terminal descent — velocity vector alignment, touchdown constraints

WHAT is terminal descent?

HOW we build the guidance law from first principles

Touchdown constraints (the inequalities that must ALL hold)

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections