Terminal descent — velocity vector alignment, touchdown constraints: Worked examples

This page is the exhaustive drill for the parent topic. We take the one guidance formula and the touchdown envelope and push them through every case a problem can present: every sign, the degenerate inputs (zero altitude gap, zero drift), the dangerous limit ( $t_{g o} \to 0$ ), a real word problem, and an exam twist.

Before anything, we re-anchor the two objects we will reuse everywhere, in plain words.

Definition The symbols, spelled out

$r$ = current position, $r_{f}$ = target position (the pad). $r_{f} - r$ = the "gap" arrow pointing from us to the pad.
$v$ = current velocity (an arrow: how fast and which way we move), $v_{f}$ = desired touchdown velocity.
$t_{g o} = t_{f} - t$ = time-to-go (seconds left until touchdown). See Time-to-go Estimation.
$g = (0, 0, - g)$ = gravity, an arrow pointing straight down. Its three components are written $(g_{x}, g_{y}, g_{z})$ ; since gravity has no sideways part, $g_{x} = g_{y} = 0$ and the vertical component is $g_{z} = - g$ (negative because down $= - z$ ). On Mars $g = 3.71 m/s^{2}$ , so $g_{z} = - 3.71$ . Whenever you see $g_{z}$ below, it just means "the $z$ -component of $g$ ", which equals $- g$ .
$T$ = thrust force from the engine (an arrow), $m$ = the vehicle's mass in kg. So $T / m$ is the acceleration the engine produces (force per kilogram). The law commands the engine to make exactly $T / m = a_{c m d}$ .
$v_{t d}$ = vertical touchdown speed (how fast we sink at contact, m/s, e.g. 1.5). $v_{h} = v_{x}^{2} + v_{y}^{2}$ = horizontal touchdown speed (sideways drift at contact). $θ$ = body tilt angle from vertical.
Sign convention everywhere on this page: up $= + z$ , down $= - z$ . A downward speed is a negative number.

Formula Why tilt and drift add — the tip-angle rule

A lander topples when its centre of mass swings past the edge of the leg footprint. Two separate things lean the vehicle that way at contact:

the body tilt $θ$ — the thrust axis is already off-vertical by $θ$ ;
the velocity lean — the velocity arrow makes a right triangle with vertical sink $v_{t d}$ as the adjacent side and horizontal drift $v_{h}$ as the opposite side, so it points $arctan (v_{h} / v_{t d})$ off vertical.

Why arctan, and why $v_{h} / v_{t d}$ ? $tan (angle) = \frac{opposite}{adjacent} = \frac{v _{h}}{v _{t d}}$ measures how far the arrow leans off vertical; $arctan$ is the inverse question "which angle has that ratio?" — it converts a slope back into an angle. Both leans push the same way, so they add: $θ_{eff} = θ + arctan \frac{v _{h}}{v _{t d}} \leq θ_{t i p}$ The landing is tip-safe only if this sum stays under the tip threshold $θ_{t i p}$ . Example 7 works a case where it fails.

The scenario matrix

Every solvable question about this topic falls into one of these cells. The examples below are tagged with the cell(s) they cover, and together they fill the whole table.

Cell	What makes it distinct	Example
A — Vertical, braking	Falling fast, must slow: $a_{c m d}$ comes out positive (up)	Ex 1
B — Vertical, mild	Slow descent, gap small: $a_{c m d}$ near hover	Ex 2
C — Full 3-D, mixed signs	Horizontal + vertical, negative gaps both ways	Ex 3 (figure)
D — Degenerate: zero gap	$r_{f} = r$ : position term vanishes	Ex 4
E — Degenerate: zero drift	$v_{x} = v_{y} = 0$ already: alignment term does nothing horizontally	Ex 4
F — Limiting: $t_{g o} \to 0$	The $1/ t_{g o}^{2}$ blow-up; the floor/handoff fix	Ex 5 (figure)
G — Envelope AND-check	All 5 touchdown inequalities, one fails	Ex 6
H — Tilt + drift geometry	Combined tip angle $θ + arctan (v_{h} / v_{t d})$	Ex 7 (figure)
I — Word problem (real)	Translate English → numbers → command	Ex 8
J — Exam twist	Solve backwards for a required $t_{g o}$	Ex 9

Worked Examples

Worked example Example 1 — Cell A: vertical braking (positive command)

1-D vertical, Mars $g = 3.71 m/s^{2}$ . Gap $r_{f} - r = - 40$ m (descend 40 m), current $v = - 25$ m/s, target $v_{f} = - 1.5$ m/s, $t_{g o} = 4$ s. Find $a_{c m d}$ .

Forecast: we're plunging at 25 m/s — will the engine command be up (positive) or down? Guess before reading.

Position term $\frac{6}{t _{g o}^{2}} (r_{f} - r) = \frac{6}{16} (- 40) = - 15 m/s^{2}$ . Why this step? It measures how badly we're mispositioned; here the pad is below us so it pulls the command down.
Velocity term $- \frac{2}{t _{g o}} (2 v + v_{f}) = - \frac{2}{4} (2 (- 25) + (- 1.5)) = - \frac{1}{2} (- 51.5) = + 25.75 m/s^{2}$ . Why this step? We move far faster than desired; this term pushes back up to bleed the speed off.
Gravity cancel $- g_{z} = - (- 3.71) = + 3.71 m/s^{2}$ (recall $g_{z} = - g$ is the vertical component of $g$ ). Why this step? Removes the $+ g$ the world adds, so the trajectory follows the intended braking shape.
Sum: $a_{c m d} = - 15 + 25.75 + 3.71 = 14.46 m/s^{2}$ (up). Why add them? The three terms are independent jobs (position, velocity, gravity) acting on the same axis, so the total acceleration the engine must produce is simply their sum — superposition on one line of motion.

Verify: the braking term (+25.75) dominates the position term (−15) because we're overspeeding hard — physically sensible. Net world acceleration $= a_{c m d} + g_{w or l d} = 14.46 - 3.71 = 10.75 m/s^{2}$ up, i.e. real deceleration. Units: all terms m/s². ✅

Worked example Example 2 — Cell B: mild descent near hover

Vertical, Moon $g = 1.62 m/s^{2}$ . Gap $r_{f} - r = - 2$ m, $v = - 1.0$ m/s, $v_{f} = - 1.0$ m/s, $t_{g o} = 2$ s.

Forecast: we're already at the target speed and almost there — should the command be close to just cancelling gravity? Guess.

Position term $\frac{6}{4} (- 2) = - 3.0 m/s^{2}$ . Why? Small gap, small correction — but note it's still non-zero, tugging us down onto the pad.
Velocity term $- \frac{2}{2} (2 (- 1.0) + (- 1.0)) = - (1) (- 3.0) = + 3.0 m/s^{2}$ . Why? $2 v + v_{f} = - 3$ isn't zero even though $v = v_{f}$ : the " $2 v$ " weights current speed twice, so it still commands braking.
Gravity cancel $- g_{z} = - (- 1.62) = + 1.62$ .
Sum: $a_{c m d} = - 3.0 + 3.0 + 1.62 = 1.62 m/s^{2}$ (up). Why add them? Same superposition: position pull (−3) and velocity push (+3) cancel exactly, so what's left is pure gravity cancellation — hover.

Verify: it came out exactly $g$ — the position pull and the velocity push cancelled, leaving pure hover. That's the "mild" signature. ✅

Worked example Example 3 — Cell C: full 3-D, mixed signs (figure)

Earth-like $g = 9.81$ , $g = (0, 0, - 9.81)$ . Current $r = (30, - 10, 100)$ m, target $r_{f} = (0, 0, 0)$ (pad at origin, ground level). $v = (- 4, 2, - 20)$ m/s, $v_{f} = (0, 0, - 1.5)$ m/s, $t_{g o} = 5$ s. Find $a_{c m d}$ .

Forecast: in $x$ we're at $+ 30$ moving at $- 4$ (already heading back). Will the $x$ -command be positive or negative? Guess.

The figure below projects this onto the $x$ – $z$ plane: the yellow dot is the pad, the blue dot is the vehicle, the green arrow is the position gap $r_{f} - r$ , the blue arrow the current velocity, and the red arrow the resulting command $a_{c m d}$ — watch how it leans back toward the vertical pad axis while pointing up.

Figure s01 — Ex 3: the command arrow (red) leans back toward the pad's vertical axis and upward, killing both the sideways offset and the fall.

$x$ -axis: gap $= 0 - 30 = - 30$ . $\frac{6}{25} (- 30) = - 7.2$ . Velocity: $- \frac{2}{5} (2 (- 4) + 0) = - \frac{2}{5} (- 8) = + 3.2$ . No gravity in $x$ . $a_{x} = - 7.2 + 3.2 = - 4.0$ m/s². Why? We're right of the pad, so position pulls left (−); we already drift left, so the velocity term nudges right (+). Position wins → net left.
$y$ -axis: gap $= 0 - (- 10) = + 10$ . $\frac{6}{25} (10) = + 2.4$ . Velocity: $- \frac{2}{5} (2 (2) + 0) = - 1.6$ . $a_{y} = + 2.4 - 1.6 = + 0.8$ m/s². Why? Symmetric logic, opposite sign — this axis pushes $+ y$ .
$z$ -axis: gap $= - 100$ . $\frac{6}{25} (- 100) = - 24$ . Velocity: $- \frac{2}{5} (2 (- 20) + (- 1.5)) = - \frac{2}{5} (- 41.5) = + 16.6$ . Gravity: $- g_{z} = - (- 9.81) = + 9.81$ . $a_{z} = - 24 + 16.6 + 9.81 = + 2.41$ m/s².
Result: $a_{c m d} = (- 4.0, + 0.8, + 2.41) m/s^{2}$ . Why keep the axes separate? The formula acts componentwise — each axis has its own gap, velocity and (only for $z$ ) gravity — so we solve three independent 1-D problems and stack the answers into one vector.

Verify: each axis is independent (the formula acts componentwise), so we can check them separately as in Ex 1. Look at the figure: the red command arrow leans back toward the pad axis and up — exactly what a lander needs. ✅

Worked example Example 4 — Cells D & E: zero gap and zero drift

Vertical, $g = 3.71$ . We are exactly over the pad at zero altitude gap: $r_{f} - r = 0$ . Also zero horizontal drift ( $v_{x} = v_{y} = 0$ ). Vertical $v = - 2$ m/s, $v_{f} = - 1.5$ m/s, $t_{g o} = 2$ s.

Forecast: with no position gap, does the position term disappear entirely, leaving only braking + gravity? Guess.

Position term $\frac{6}{4} (0) = 0$ . Why? This is the degenerate cell D: multiply anything by a zero gap and it vanishes. The formula gracefully drops that job.
Horizontal axes (cell E): $v_{x} = v_{y} = 0$ and $v_{f} = 0$ , so velocity term $= - \frac{2}{2} (0) = 0$ , and no gravity horizontally → $a_{x} = a_{y} = 0$ . Why? Nothing to align; the alignment term is silent when there is no drift. Good — it won't invent a sideways kick.
Vertical velocity term $- \frac{2}{2} (2 (- 2) + (- 1.5)) = - (- 5.5) = + 5.5$ .
Gravity $- g_{z} = + 3.71$ . $a_{z} = 0 + 5.5 + 3.71 = 9.21$ m/s² up. Why add them? Same superposition: with the position term zeroed out, only the braking push (+5.5) and gravity cancellation (+3.71) remain, and they sum on the vertical axis.

Verify: command is purely vertical, no phantom horizontal thrust — confirms both degenerate cells behave. Units m/s². ✅

Worked example Example 5 — Cell F: the

$t_{g o} \to 0$ blow-up (figure) Same as Ex 1 numbers ( $r_{f} - r = - 40$ ) but now imagine we let $t_{g o}$ shrink toward 0. Show why the command explodes, and apply a floor at $t_{g o}^{m i n} = 0.5$ s with a thrust ceiling of $150 m/s^{2}$ .

Forecast: the position term is $\frac{6}{t _{g o}^{2}} (- 40)$ . As $t_{g o} \to 0$ , does it grow like $1/ t_{g o}$ or $1/ t_{g o}^{2}$ ? Guess — the answer tells you how violent the blow-up is.

The figure below plots the magnitude of that position term against $t_{g o}$ : the blue curve rockets upward as $t_{g o} \to 0$ , the yellow dashed line marks the thrust saturation ceiling of $150 m/s^{2}$ the engine physically cannot exceed, and the red dashed line marks the $t_{g o} = 0.5$ s floor where we stop trusting the law and hand off.

Figure s02 — Ex 5: the position term grows like $1/ t_{g o}^{2}$ ; halving $t_{g o}$ from 1 s to 0.5 s quadruples it (240 → 960), punching through the 150 m/s² saturation ceiling.

At $t_{g o} = 1$ s: position term $= 6 (- 40) = - 240$ m/s². At $t_{g o} = 0.5$ : $= \frac{6}{0.25} (- 40) = - 960$ m/s². Why this step? Halving $t_{g o}$ quadrupled the term — that's the $1/ t_{g o}^{2}$ signature. This is thousands of $g$ : physically impossible, the engine (ceiling $150 m/s^{2}$ ) saturates.
The fix — floor $t_{g o}$ and hand off by altitude. Clamp $t_{g o} \leftarrow max (t_{g o}, 0.5 s)$ so the denominator never collapses, and switch guidance mode when altitude drops below $h_{h o} = 5$ m (equivalently when the E-guidance command would exceed the $150 m/s^{2}$ ceiling). Below that, run a constant-velocity gravity-turn drop: hold vertical speed at $v_{t d} = 1.5$ m/s and command just enough thrust to cancel gravity plus a small settling term. Why these thresholds? The $5$ m altitude / $150 m/s^{2}$ ceiling / $0.5$ s floor are the three equivalent triggers; whichever fires first ends the $t_{g o}$ -driven law before it demands the impossible.
Command at the floor using Ex 1's other terms scaled to $t_{g o} = 0.5$ : velocity term $- \frac{2}{0.5} (2 (- 25) + (- 1.5)) = - 4 (- 51.5) = + 206$ . Position $- 960$ . Gravity $+ 3.71$ . Sum $= - 750.3$ m/s² — still saturating (far past the $150 m/s^{2}$ ceiling), which is why we hand off rather than trust the formula here. Why report the sum anyway? To show that even at the floor the formula demands the impossible, proving the floor alone isn't enough — you must switch guidance laws, not just clamp the number.

Verify: the figure's curve rises without bound as $t_{g o} \to 0$ ; the flat dashed line is the $150 m/s^{2}$ thrust saturation ceiling; the vertical marker at $0.5$ s is where we stop trusting the law. The $4 \times$ jump ( $240 \to 960$ ) numerically confirms the quadratic growth. ✅

Worked example Example 6 — Cell G: touchdown envelope, one constraint fails

At contact: $v_{t d} = 2.7$ m/s, $v_{h} = 0.4$ m/s, tilt $θ = 3^{\circ}$ , angular rate $∣ ω ∣ = 1^{\circ} / s$ , position $0.5$ m from pad centre. Limits (from the touchdown-envelope definition above): $v_{t d}^{m a x} = 2.5$ , $v_{h}^{m a x} = 1.0$ , $θ_{m a x} = 8^{\circ}$ , $ω_{m a x} = 5^{\circ} / s$ , pad radius $R = 2$ m.

Forecast: four numbers look fine; one is over. Which? Guess before checking.

Vertical: $2.7 \leq 2.5$ ? NO — leg-crush limit violated. Why this step? This is the AND envelope defined above: every inequality must hold; one failure = unsafe.
Horizontal: $0.4 \leq 1.0$ ✅. Tilt: $3 \leq 8$ ✅. Rate: $1 \leq 5$ ✅. Position: $0.5 \leq 2$ ✅. Why check the other four anyway? To locate exactly which clause fails — engineers need the specific violated constraint to fix the guidance, not just a pass/fail flag.

Verify: exactly one failure ( $v_{t d}$ ), so the verdict is unsafe touchdown — a single broken AND-clause condemns the whole landing. This is why guidance targets $v_{t d} \approx 1.5$ with margin below $2.5$ . ✅

Worked example Example 7 — Cell H: combined tip geometry (figure)

Body tilt $θ = 4^{\circ}$ from vertical, horizontal speed $v_{h} = 0.9$ m/s, vertical $v_{t d} = 1.8$ m/s. Tip threshold $θ_{t i p} = 2 0^{\circ}$ . Is the combined effect inside the margin?

Forecast: the velocity contributes an effective extra lean. Will $θ$ + that lean stay under 20°? Guess.

The figure below draws the velocity as a right triangle: the vertical sink $v_{t d}$ is the adjacent side, the drift $v_{h}$ the opposite side, and the blue velocity arrow the hypotenuse. The red arc is the lean angle $arctan (v_{h} / v_{t d})$ ; the green dashed line is the $2 0^{\circ}$ tip limit — see the velocity arrow already leaning past it.

Figure s03 — Ex 7: with slow descent (1.8 m/s) even a 0.9 m/s drift leans the velocity 26.6°, and adding the 4° body tilt pushes the combined angle well past the 20° tip line.

Velocity lean $= arctan \frac{v _{h}}{v _{t d}} = arctan \frac{0.9}{1.8} = arctan (0.5) = 26.56 5^{\circ}$ . Why arctan, and why $v_{h} / v_{t d}$ ? The velocity arrow makes a right triangle: horizontal $v_{h}$ is the opposite side, vertical $v_{t d}$ the adjacent side. $tan$ = opposite/adjacent encodes how far the arrow leans off vertical; $arctan$ answers "which angle has that ratio?" — see the figure's triangle.
Combined $θ_{eff} = θ + arctan (v_{h} / v_{t d}) = 4^{\circ} + 26.56 5^{\circ} = 30.56 5^{\circ}$ . Why add? Body tilt and velocity lean both push the centre of mass the same way, toward the leg footprint edge — exactly the tip-angle rule derived in the envelope section.
Compare $30.56 5^{\circ}$ vs $θ_{t i p} = 2 0^{\circ}$ : exceeds → tips over. Why compare, not just compute? The tip test is itself an inequality (like the envelope): the number only matters relative to the threshold, and here it's over.

Verify: even with a modest $4^{\circ}$ body tilt, a $0.9$ m/s drift on a slow $1.8$ m/s descent leans the velocity a huge $26. 6^{\circ}$ (because the ratio $0.5$ is large). The figure shows the red combined angle punching past the green tip line. The lesson: at slow descent, small horizontal drift becomes a large angle. ✅

Worked example Example 8 — Cell I: real-world word problem

"A Mars lander is $60$ m above the pad, descending at $18$ m/s straight down with $2$ m/s of eastward drift. Guidance wants a $1.5$ m/s vertical touchdown, no drift, in $t_{g o} = 6$ s. Along the east axis, what does it command?" ( $g = 3.71$ , east $= + x$ .)

Forecast: eastward drift $+ 2$ m/s and target $0$ — the east command should be negative (westward). By how much?

Translate to numbers. East axis: $r_{f} - r = 0 - 0 = 0$ (we assume already lined up east–west; only drift matters). $v_{x} = + 2$ , $v_{f, x} = 0$ . Why? Word problems: extract gap, current velocity, target velocity, $t_{g o}$ per axis. No east position error stated → gap 0.
Position term $\frac{6}{36} (0) = 0$ .
Velocity term $- \frac{2}{6} (2 (2) + 0) = - \frac{1}{3} (4) = - 1.333$ m/s². Why? Only job on this axis is killing the $+ 2$ drift → command points west (−), as forecast.
No gravity in $x$ . $a_{x} = - 1.333$ m/s² (westward). Why add (here, just one non-zero term)? Position and gravity are both zero on this axis, so the sum reduces to the velocity term alone — the drift-killer.

Verify: sign is negative (west), matching the physical need to erase eastward drift. Magnitude modest because $t_{g o} = 6$ s is generous. Units m/s². ✅

Worked example Example 9 — Cell J: exam twist, solve backwards

"Given a vertical overspeed problem where the velocity term alone must supply $+ 206$ m/s² of braking, with $v = - 25$ , $v_{f} = - 1.5$ , find the required $t_{g o}$ ." (This is the inverse of the usual computation.)

Forecast: smaller $t_{g o}$ → bigger braking. So a large $206$ demands a small $t_{g o}$ . Roughly how small?

Write the velocity term as an equation: $- \frac{2}{t _{g o}} (2 v + v_{f}) = 206$ . Why? Exam twists invert the formula — instead of plugging $t_{g o}$ in, we solve for it.
Plug knowns: $2 v + v_{f} = 2 (- 25) + (- 1.5) = - 51.5$ . So $- \frac{2}{t _{g o}} (- 51.5) = \frac{103}{t _{g o}} = 206$ .
Solve: $t_{g o} = \frac{103}{206} = 0.5$ s. Why does dividing work here? Rearranging $\frac{103}{t _{g o}} = 206$ means $t_{g o}$ is whatever makes 103 fit into 206 once — a single-unknown linear solve.

Verify: plug back: $- \frac{2}{0.5} (- 51.5) = - 4 (- 51.5) = + 206$ ✅. Note this matches the floor value from Ex 5 — a nice consistency check across cells. ✅

Active Recall

Recall Which matrix cell is "the command comes out equal to

$g$ , pure hover"? Cell B (mild descent) — when the position pull and velocity push cancel. ::: Cell B, Example 2.

Recall Why does the

$t_{g o} \to 0$ term blow up as $1/ t_{g o}^{2}$ and not $1/ t_{g o}$ ? Because the position term carries the factor $6/ t_{g o}^{2}$ , so $t_{g o}$ appears squared in the denominator: halving $t_{g o}$ divides by $(1/2)^{2} = 1/4$ , i.e. multiplies the term by 4. The velocity term only has $1/ t_{g o}$ , so it grows slower and the position term wins the blow-up. ::: The $6/ t_{g o}^{2}$ position term dominates; halving $t_{g o}$ quadruples it (240 → 960 in Ex 5).

Recall Which five inequalities make up the touchdown envelope?

Vertical speed $v_{t d} \leq v_{t d}^{m a x}$ , horizontal speed $v_{h} \leq v_{h}^{m a x}$ , tilt $θ \leq θ_{m a x}$ , angular rate $∣ ω ∣ \leq ω_{m a x}$ , and position within pad radius $R$ . ::: All five must hold at once (AND) — one failure condemns the landing.

Recall In the tip geometry, why does slow descent make small drift dangerous?

The lean is $arctan (v_{h} / v_{t d})$ ; a small $v_{t d}$ in the denominator makes the ratio — and the angle — large. ::: Slow $v_{t d}$ ⇒ big lean angle for the same drift.

Recall At what conditions does guidance hand off from the

$t_{g o}$ -driven law? When altitude drops below about $5$ m, or equivalently when $t_{g o}$ hits its $0.5$ s floor or the command would exceed the $150 m/s^{2}$ thrust ceiling — then switch to a constant-velocity gravity-turn drop at $v_{t d} = 1.5$ m/s. ::: Below ~5 m / $t_{g o} = 0.5$ s floor / $150$ m/s² ceiling → constant-velocity drop.

Cross-links: Powered Descent Guidance (PDG), Proportional Navigation, Attitude Control & Inner Loop, Time-to-go Estimation, Convex Optimization Landing (lossless convexification).