Autonomous GNC for reusable rockets — SpaceX approach overview: Visual walkthrough

We build up to one boxed result: $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}}$ Do not worry about any symbol yet — every one is earned below.

Step 1 — A falling thing has a downward speed

WHAT. Picture the booster after its coast, high up, engine off, dropping straight down. At the moment we start watching, it moves downward at some speed we call $v_{0}$ .

WHY. Before we can talk about stopping, we must name the thing we want to stop: the downward speed. Everything else is built to kill this number.

PICTURE. In the figure the rocket is the tall shape; the fat downward arrow is its velocity. Its length is the speed. The dashed line at the bottom is the ground (height $0$ ); the dashed line where the rocket sits is the current height $h$ .

Figure — Autonomous GNC for reusable rockets — SpaceX approach overview

We choose upward as positive. So a downward fall means the velocity points the negative way — but $v_{0}$ itself is written as a positive number (the size of that downward speed). Keeping this convention straight is the whole game.

Step 2 — The engine can only push one way: up

WHAT. The engine, when lit, pushes the rocket up with a force. Gravity always pulls down. The rocket also has a mass. These three — up-push, down-pull, mass — decide how quickly the rocket's speed changes.

WHY. "How quickly speed changes" is exactly what we need, because we want the downward speed to shrink from $v_{0}$ to $0$ . A thing that changes speed is accelerating, so we must build the acceleration.

PICTURE. Two arrows on the rocket: a long burnt-orange arrow up (thrust $T$ ) and a shorter teal arrow down (weight). The up arrow is drawn longer on purpose — the engine out-pushes gravity. The plum arrow is the leftover upward pull, the net effect.

Why this shape and not something fancier? Because Newton's law $force = mass \times acceleration$ rearranged says $acceleration = force / mass$ . We just added up the forces (up minus down) and divided by mass. Nothing else is needed.

Step 3 — Draw the speed dying: the speed-vs-height picture

WHAT. As the rocket rises through its burn (or rather, keeps falling but decelerating), its downward speed drops from $v_{0}$ toward $0$ . We plot speed on the up-axis against height on the side-axis, watching the rocket travel from height $h$ down to height $0$ .

WHY. We care about where the speed hits zero (which height), not when (which second). So we want a picture with height on one axis and speed on the other, with time hidden away. This is the whole reason we will pick a time-free formula in Step 4.

PICTURE. The curve starts at the point $(h, v_{0})$ — high up, moving fast — and must land exactly on $(0, 0)$ — pad height, dead stop. If the curve reaches zero speed before height zero (dotted plum), the rocket stops in the air and falls again. If it still has speed at height zero (dotted teal), it slams the pad. Only the burnt-orange curve is a good landing.

Recall Why "stop early" and "stop late" both fail

Stop early ::: velocity hits 0 while still airborne; TWR>1 means you can't hold there, so you rise or drop again → miss. Stop late ::: still moving at height 0 → crash into the pad.

Step 4 — Choose the tool with no time in it

WHAT. We need an equation linking start speed $v_{0}$ , final speed ( $0$ ), the acceleration $a_{n e t}$ , and the distance travelled $h$ — without time.

WHY. Because the picture in Step 3 has no time axis. Among the standard motion equations, exactly one has no $t$ : $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ This is the constant-acceleration relation. (We wrote a minus because $a_{n e t}$ points up while motion is down — the acceleration is fighting the fall, removing speed.)

PICTURE. A menu of three motion equations; two are crossed out because they contain $t$ . The survivor is circled — it connects speed and distance directly.

Step 5 — Demand a perfect landing and solve

WHAT. Impose the landing condition from Step 3: the speed must be exactly $v = 0$ at the moment the distance travelled equals the full drop, $s = h$ .

WHY. "Zero speed at zero height" is the definition of a soft landing. Plugging it in turns the general relation into the one number the flight computer cares about.

PICTURE. The equation with $v = 0$ and $s = h$ substituted, and arrows peeling it apart to isolate $a_{n e t}$ .

Set $v = 0$ , $s = h$ : $0 = v_{0}^{2} - 2 a_{n e t} h$ Move the acceleration term across (add $2 a_{n e t} h$ to both sides): $2 a_{n e t} h = v_{0}^{2}$ Divide both sides by $2 h$ : $a_{n e t} = \frac{v _{0}^{2}}{2 h}$

Reading it: to stop from speed $v_{0}$ within height $h$ , you need exactly this much deceleration. Fall twice as fast → you need four times the deceleration (because of the square). Give yourself twice the height → you need half.

And since $a_{n e t} = T / m - g$ , the thrust the engine must produce is $T = m (g + \frac{v _{0}^{2}}{2 h}) .$ Here $m (g)$ just holds the rocket up against gravity; the extra $m \cdot \frac{v _{0}^{2}}{2 h}$ is the braking part.

Step 6 — Flip the question: when do I light the engine?

WHAT. In flight the engine's achievable $a_{n e t}$ is roughly fixed. What the computer doesn't know until the moment arrives is the height at which it must ignite. So solve the boxed result for $h$ instead of $a_{n e t}$ .

WHY. The rocket is falling and watching its own speed and height every few milliseconds. It wants a trigger: "the instant my height drops to this value, light up." That trigger is the ignition altitude.

PICTURE. Same rocket falling; a horizontal plum trip-line sits at height $h_{i g n}$ . Above the line, engine off (grey). The moment the rocket crosses it, the engine flashes on (orange).

This is a live, closed-loop decision — recomputed each cycle as $v_{0}$ and mass change. Once lit, the thrust is nudged every cycle by $T_{c m d} = m (g + \frac{v ^{2}}{2 h})$ using the present $v$ and $h$ (PID Control and Convex Optimization handle the fine corrections and fuel-optimality; Thrust Vectoring (Gimbal) aims the push sideways to cancel drift).

Step 7 — The numbers, and the two failure edges

WHAT. Put real values in: falling at $v_{0} = 200$ m/s, engine gives $a_{n e t} = 20$ m/s².

WHY. A formula you can't put numbers into isn't yet yours. And the two "off-by-a-bit" cases show why the timing must be exact.

PICTURE. A number line of altitude. At $1000$ m the correct ignition (orange) lands cleanly at $0$ . The "late by 100 m" case (teal, light at 900 m) still has speed at the pad. The "early by 100 m" case (plum, light at 1100 m) stops airborne.

$h_{i g n} = \frac{20 0 ^{2}}{2 \cdot 20} = \frac{40000}{40} = 1000 m$

Burn time (using the time-ful equation $v = v_{0} - a_{n e t} t$ , set $v = 0$ ): $t = \frac{v _{0}}{a _{n e t}} = \frac{200}{20} = 10 s .$

Light at exactly 1000 m → 10 s later, speed $0$ at height $0$ . Perfect.
Light late (900 m) → not enough braking distance → still moving at touchdown → crash.
Light early (1100 m) → speed hits $0$ at ~100 m up → but TWR>1 means it can't hover → rises again / wastes fuel → miss.

Degenerate check — what if $v_{0} = 0$ ? Then $h_{i g n} = 0$ : if you already have no downward speed you need no braking distance. The formula agrees with common sense. And if $a_{n e t} \to 0$ (engine barely beats gravity), $h_{i g n} \to \infty$ — you'd have to start braking impossibly high; a weak engine can't land, which is exactly why boosters keep TWR comfortably above 1.

The one-picture summary

Everything above, compressed: the falling rocket, the trip-line at $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ , the speed curve gliding into the origin, and the single boxed formula.

Recall Feynman: the whole walkthrough in plain words

You're dropping a ball and you want it to stop exactly on a coin, but your only tool is a spring under it that's too strong to let the ball float. So you can't ease it down. Instead you let it fall, and you fire the spring at one perfect height so the ball's speed dies right as it kisses the coin. To find that height: name the falling speed ( $v_{0}$ ). Figure out how hard you can brake — engine push minus gravity, divided by mass ( $a_{n e t}$ ). Then use the one motion rule that ignores time and only relates speed to distance. Demand "speed zero at height zero," and out pops: brake distance $= v_{0}^{2} /2 a_{n e t}$ . That distance is the height to light the engine. Fall faster and you need four times the room (it's squared). The rocket measures its own speed and height thousands of times a second and pulls the trigger the instant it crosses that line.