Terminal descent — velocity vector alignment, touchdown constraints: Question bank

Quick symbol reminder so nothing here is unearned:

$r$ = position, $v = \dot{r}$ = velocity (the arrow of motion), $a$ = acceleration.
$t_{g o} = t_{f} - t$ = time-to-go, the seconds left until touchdown at time $t_{f}$ .
$a_{c m d}$ = the net acceleration the guidance law commands; the engine physically supplies exactly $T / m = a_{c m d}$ .
$v_{t d}$ = downward touchdown speed, $v_{h} = v_{x}^{2} + v_{y}^{2}$ = horizontal drift speed, $θ$ = tilt of the thrust axis away from vertical.

True or false — justify

Simply aiming the trajectory at the pad position guarantees a soft landing.

False — position hitting the pad says nothing about velocity; you can arrive on the pad at 20 m/s sideways. Soft landing is a two-point boundary value problem pinning both

r_{f}

and

v_{f}

Because the guidance formula already contains a

- g

term, the engine must produce

a_{c m d} + g

to fight gravity.

False — the

- g

inside

a_{c m d}

is what cancels the

+ g

\ddot{r} = a_{c m d} + g

. The engine supplies exactly

T / m = a_{c m d}

; re-adding

g

double-counts gravity.

A perfectly vertical velocity vector at touchdown guarantees the lander won't tip.

False — the body can still be tilted (

θ

large) even with zero horizontal drift; tipping depends on tilt and drift together, via

θ + arctan (v_{h} / v_{t d}) ≲ θ_{t i p}

Setting

v_{f} = (0, 0, - v_{t d})

with zero horizontal components is what forces the velocity vector to rotate toward straight-down.

True — the target has no sideways component, so as

t_{g o} \to 0

the velocity-shaping term drives

v_{x}, v_{y} \to 0

, leaving only the downward part.

The touchdown envelope constraints are alternatives — satisfying any one of them is enough.

False — they are AND-constraints: vertical speed, horizontal speed, tilt, angular rate and position must all be inside limits simultaneously at contact.

As the lander approaches, letting

t_{g o}

shrink all the way to zero is harmless.

False — the position term scales as

1/ t_{g o}^{2}

, so the command blows up to infinite thrust, which the engine cannot supply. You floor

t_{g o}

or hand off to a hover-then-drop phase.

Modelling the vehicle as a point mass throws away the attitude requirement.

False — attitude is not ignored, it is handled by a faster inner loop; the point-mass guidance produces

a_{c m d}

, whose direction the inner loop realises by pointing the thrust axis.

A desired vertical touchdown speed of exactly

0

m/s is the ideal target for

v_{f}

False — a tiny positive

v_{t d} \approx 1 - 2

m/s is deliberate; commanding exactly zero would ask the engine to hover indefinitely and never make contact, and any error leaves you drifting up or stalling.

Spot the error

"We hit the pad, so terminal guidance succeeded — velocity is irrelevant."

Error: velocity is a separate boundary condition. The named constraint is velocity vector alignment; hitting position with wrong velocity scrapes a leg or crashes.

"Thrust command

= a_{c m d} + g

because the rocket must overcome weight."

Error: gravity is already folded into

a_{c m d}

via the

- g

term. Correct command is

T / m = a_{c m d}

; adding

g

again over-thrusts.

"A single constant commanded acceleration is enough to pin both final position and final velocity."

Error: that's over-determined per axis. Matching four scalar conditions needs an acceleration that varies linearly in time, which is where the coefficients

6

and

2

come from.

"In Example 1 the answer is negative because the vehicle is falling."

Error: the sign of

a_{c m d}

reflects the braking direction, not the fall. Falling at

- 25

m/s means you must brake upward, so

a_{c m d} = + 14.46

m/s² is positive.

"The formula's

1/ t_{g o}

velocity term is a fixed gain, so it treats early and late descent the same."

Error: the gain grows as

t_{g o} \to 0

, so the correction on

v_{x}, v_{y}

strengthens near touchdown — that increasing authority is exactly what drives drift to zero in time.

"Since gravity always pulls down, we can drop the

- g

term and still command

T / m = a_{c m d}

Error: pick one convention. If you drop

- g

from the formula you must re-add gravity when sizing thrust; mixing the two conventions gives a gravity error every cycle.

Why questions

Why must position, velocity, and attitude converge at the same instant rather than one after another?

Because touchdown is one event: arriving upright but with sideways velocity, or with zero velocity but tilted, still topples or skids. All conditions have to be satisfied at the single time

t_{f}

Why does the guidance need a velocity term at all, unlike simple pursuit guidance?

Pursuit/PN guidance chases only position (line-of-sight). A soft landing additionally specifies

v_{f}

, so the law needs a term that shapes current velocity toward that target.

Why is there a tilt limit

θ_{m a x}

in the envelope?

A tilted thrust axis has a horizontal component you can't cancel before contact, and combined with drift it pushes the center of mass outside the leg footprint, causing a tip-over.

Why does the position-error coefficient carry

1/ t_{g o}^{2}

while the velocity term carries

1/ t_{g o}

Position is a double integral of acceleration (two time integrations), so correcting a position gap over a shrinking window needs a term scaling as time-squared inverse; velocity is a single integral, giving one power of

1/ t_{g o}

Why can constant acceleration hit a position-and-velocity target but a straight line cannot?

A straight line only has enough freedom to aim position. Constant (then linear) acceleration adds the free parameters needed to fix the final velocity as well — more knobs for more conditions.

Why is

v_{t d}

chosen small but nonzero (1–2 m/s) rather than large?

Small protects the legs from crush and keeps within

v_{t d}^{m a x}

; nonzero ensures the vehicle actually descends and makes firm contact instead of hovering or floating up on command error.

Why does approaching with drift

v_{h}

effectively add to the tilt budget?

The incoming velocity makes an angle

arctan (v_{h} / v_{t d})

from vertical; this angle stacks with body tilt

θ

against the tip threshold, so drift eats into the same margin as tilt.

Edge cases

At the exact instant

t_{g o} = 0

, is the guidance formula valid?

No — it is singular there (

1/ t_{g o}^{2} \to \infty

). You must transition out beforehand to a floored-

t_{g o}

or hover-then-drop phase, never evaluate at literally zero.

What happens if

v_{h} = 0

but

v_{t d}

is also near zero at contact?

The incoming-angle term

arctan (v_{h} / v_{t d})

becomes

arctan (0/0)

, undefined; physically it means near-hover with negligible motion, so tip risk is dominated by body tilt

θ

alone rather than by drift.

If horizontal drift

v_{x}

is already zero at the start of terminal descent, does the velocity-shaping term do anything to it?

No — with

v_{x} = 0

and target

v_{f, x} = 0

the term contributes nothing on that axis; it only acts to remove existing drift, keeping an already-vertical velocity vertical.

What if the vehicle arrives on the pad with velocity and tilt each individually within limits but their combined incoming angle exceeds

θ_{t i p}

It can still tip. The stability condition is on the sum

θ + arctan (v_{h} / v_{t d})

; passing each single-limit check is necessary but not sufficient.

If the commanded acceleration saturates the engine near touchdown, what physically happens to the trajectory?

The vehicle can no longer follow the guidance shape, so the boundary conditions are missed — typically a harder or off-target landing; this is why

t_{g o}

is floored and why methods like convex landing enforce thrust bounds explicitly.

Recall One-line self-summary

Everything on this page traces to three facts: soft landing is a two-point (position and velocity) problem, the $- g$ lives inside $a_{c m d}$ so the engine supplies exactly $a_{c m d}$ , and the $1/ t_{g o}$ terms grow near touchdown — powerful for shaping, dangerous if $t_{g o} \to 0$ unbounded.