Autonomous GNC for reusable rockets — SpaceX approach overview: Worked examples

The tools, re-earned from zero

Before any example, let us make sure every symbol has a picture.

How to read Fig 1 (below): the lavender arrow is altitude $h$ (up-positive); the coral arrow is the downward velocity of a falling booster; the slate arrow is gravity $- g$ ; the mint arrow is the net braking acceleration $a_{n e t}$ that appears only when the engine fires. Notice coral (fall) and mint (brake) point opposite ways — that opposition is the whole hoverslam.

Figure — Autonomous GNC for reusable rockets — SpaceX approach overview

Definition The sign convention (memorise this before any example)

Up is positive, down is negative — always. From this one rule everything follows:

A quantity's sign tells you its direction: $v > 0$ ⇒ climbing, $v < 0$ ⇒ falling.
In $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ , the letter $s$ is the distance travelled along the motion (always a positive number of metres covered), and the sign in front of $2 a_{n e t}$ encodes whether the acceleration helps or fights that motion.
- Braking descent: motion is downward, $a_{n e t}$ points up (opposing) ⇒ use the minus sign, $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ (speed drops).
- Accelerating upward: motion is upward, $a_{n e t}$ also up (helping) ⇒ use the plus sign, $v^{2} = v_{0}^{2} + 2 a_{n e t} s$ (speed grows). Every time we switch which case we are in (as in Ex 4), we will restate which sign we picked and why.

The scenario matrix

Every hoverslam question is one (or a blend) of these cells. Each row below is covered by at least one worked example.

#	Case class	What makes it different	Example
A	Nominal fall	falling fast, solve for $h_{i g n}$	Ex 1
B	Solve the other unknown	given $h$ , find required $a_{n e t}$ / thrust $T$	Ex 2
C	Zero / degenerate input	$v_{0} = 0$ (already at rest) — what does the formula say?	Ex 3
D	Ignite too HIGH (sign of leftover velocity $> 0$ going up)	stops above pad, then rises	Ex 4
E	Ignite too LOW (leftover velocity still downward at $h = 0$ )	crash speed	Ex 5
F	Limiting value	throttle floor: minimum $a_{n e t}$ ⇒ maximum survivable $v_{0}$	Ex 6
G	Degenerate: engine too weak	$a_{n e t} \leq 0$ (TWR $\leq 1$ ) — formula breaks down	Ex 7
H	Real-world word problem	crosswind → sideways PID Control correction	Ex 8
I	Exam twist	mass changes during burn (link to Rocket Equation (Tsiolkovsky))	Ex 9

Ex 1 — Cell A: the nominal hoverslam

Worked example Statement

A booster falls at $v_{0} = 250 m/s$ . Its engine can produce $a_{n e t} = 25 m/s^{2}$ . At what altitude must it light the engine?

Forecast: guess — will $h_{i g n}$ be bigger or smaller than the 1000 m from the parent note? (Faster fall but stronger brake... guess before reading.)

Pick the timeless equation. Use $0 = v_{0}^{2} - 2 a_{n e t} h$ (minus sign: descent braked by upward $a_{n e t}$ ). Why this step? We want the altitude where speed reaches zero — time is irrelevant, so the no- $t$ relation is the right tool.
Solve for $h$ . $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} = \frac{25 0 ^{2}}{2 \cdot 25} = \frac{62500}{50} = 1250 m$ . Why this step? Rearranging isolates the one unknown we were asked for.
Burn duration (bonus): from the time relation $v = v_{0} - a_{n e t} t$ , set $v = 0$ : $t = v_{0} / a_{n e t} = 250/25 = 10 s$ . Why this step? It sanity-checks feasibility — 10 s of fuel must exist. We switch to $v = v_{0} - a_{n e t} t$ here because the question is now when, not where.

How to read Fig 2: the coral curve is speed vs altitude during the burn. It starts at 250 m/s up at $h = 1250$ m and touches zero exactly at the pad — the signature of a perfect hoverslam. The curve is a parabola in $v$ because $v^{2}$ (not $v$ ) is linear in $h$ .

Verify: plug $h = 1250$ back: $v^{2} = 25 0^{2} - 2 \cdot 25 \cdot 1250 = 62500 - 62500 = 0$ ✓. Speed truly hits zero at the pad. Units: $(m/s)^{2} / (m/s^{2}) = m$ ✓.

Ex 2 — Cell B: given the height, find the thrust

Worked example Statement

Sensors say the engine must light at exactly $h = 800 m$ while falling at $v_{0} = 180 m/s$ . The booster mass is $m = 30, 000 kg$ , $g = 9.8 m/s^{2}$ . What thrust $T$ is required?

Forecast: will the needed thrust be more or less than the booster's weight $m g$ ?

Find the needed $a_{n e t}$ . $a_{n e t} = \frac{v _{0}^{2}}{2 h} = \frac{18 0 ^{2}}{2 \cdot 800} = \frac{32400}{1600} = 20.25 m/s^{2}$ . Why this step? We flipped the same equation to solve for the acceleration this time, because $h$ is now the known.
Convert to thrust. $T = m (g + a_{n e t}) = 30000 (9.8 + 20.25) = 30000 \cdot 30.05 = 901, 500 N$ . Why this step? $a_{n e t} = T / m - g$ , so $T = m (g + a_{n e t})$ — the engine must beat gravity and supply the braking.
Compare to weight. $m g = 30000 \cdot 9.8 = 294, 000 N$ ; so $T / m g \approx 3.07$ — thrust-to-weight $\approx 3$ . Why this step? Confirms the parent-note claim that TWR $> 1$ (indeed $≫ 1$ ) — you truly cannot hover.

How to read Fig 3: the bar chart stacks the two jobs the thrust must do — beat weight ( $m g$ , slate) and supply braking ( $m a_{n e t}$ , mint). Their sum (lavender total) is the required thrust, towering ~3× above the weight bar. That height ratio is the thrust-to-weight ratio.

Verify: $a_{n e t} = T / m - g = 901500/30000 - 9.8 = 30.05 - 9.8 = 20.25$ ✓. Units of $T$ : $kg \cdot m/s^{2} = N$ ✓.

Ex 3 — Cell C: the degenerate input $v_{0} = 0$

Worked example Statement

Suppose the booster is momentarily at rest ( $v_{0} = 0$ ) at some altitude. What does $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ say, and is it physical?

Forecast: will $h_{i g n}$ be zero, infinite, or undefined?

Substitute. $h_{i g n} = \frac{0 ^{2}}{2 a _{n e t}} = \frac{0}{2 a _{n e t}} = 0$ . Why this step? Plugging the degenerate value tells us what the maths thinks should happen.
Interpret. $h_{i g n} = 0$ means "you need zero braking distance," i.e. you are already stopped, so no burn is needed to kill velocity. Why this step? A formula must be read back into words to check it is sane.
But watch gravity. At rest with engine off, gravity immediately makes $v < 0$ again. So "at rest in the air" is not a stable landing — it is the instant between fall and re-fall. The hoverslam idea only applies while you still carry downward speed to brake. Why this step? Covers the trap that " $v_{0} = 0$ ⇒ done" ignores that you cannot hover (TWR $> 1$ ).

Verify: $v^{2} = 0^{2} - 2 a_{n e t} \cdot 0 = 0$ ✓ — consistent but trivially so. The degenerate case is well-behaved (gives 0, not a divide-by-zero).

Ex 4 — Cell D: igniting TOO HIGH (leftover upward velocity)

Worked example Statement

Nominal ignition for $v_{0} = 200, a_{n e t} = 20$ is $h_{i g n} = 1000 m$ (parent note). Suppose the computer lights early, at $h = 1200 m$ . What is the velocity when it reaches $h = 0$ , and what sign does it have?

Forecast: at the pad, will it still be falling, exactly stopped, or moving upward?

Total braking distance is now 1200 m. Speed reaches zero after only $Δ h = v_{0}^{2} /2 a_{n e t} = 1000$ m of braking — that is at altitude $1200 - 1000 = 200 m$ . Here we use the minus sign (descent braked): $0 = v_{0}^{2} - 2 a_{n e t} Δ h$ . Why this step? The brake still needs exactly 1000 m to null the speed; starting 200 m too high means zero-speed happens 200 m up.
Sign flip — now it climbs. Above that point the engine is still firing (TWR $> 1$ ), so from $h = 200 m$ the rocket accelerates upward. Motion and $a_{n e t}$ now point the same way, so we switch to the plus sign, $v^{2} = v_{0}^{2} + 2 a_{n e t} s$ (speed grows). Starting from rest and falling the last $s = 200 m$ back toward the pad: $v^{2} = 0^{2} + 2 a_{n e t} (200) = 2 \cdot 20 \cdot 200 = 8000 \Rightarrow v = 8000 \approx 89.4 m/s (upward) .$ Why this step? We explicitly restate the sign rule: same direction ⇒ plus; $v > 0$ ⇒ climbing.
Consequence. You cannot reach the pad going up — you would sail back into the sky. Real GNC throttles down here, but if it can't throttle below $m g$ , early ignition fails, exactly as the parent note warns. Why this step? Connects the sign of $v$ to the physical failure mode.

How to read Fig 4: follow the coral curve down from 1200 m — it hits zero speed at 200 m (lavender dot, "too high!"), then the mint dashed branch shows the booster speeding up upward, arriving at the pad at ~89 m/s in the wrong direction. The two branches meet at the stop point where speed is momentarily zero.

Verify: stop altitude $= 1200 - 1000 = 200$ ✓; leftover upward speed $2 \cdot 20 \cdot 200 = 8000 \approx 89.44$ m/s, sign $+$ (up) ✓.

Ex 5 — Cell E: igniting TOO LOW (leftover downward velocity = crash)

Worked example Statement

Same booster ( $v_{0} = 200, a_{n e t} = 20$ , need 1000 m). It lights late, at $h = 900 m$ . What is the impact speed at $h = 0$ ?

Forecast: will the rocket stop before the pad, or hit it still moving down?

Available braking distance = 900 m only. Compute the speed after braking through 900 m (descent, minus sign): $v^{2} = v_{0}^{2} - 2 a_{n e t} h = 20 0^{2} - 2 \cdot 20 \cdot 900 = 40000 - 36000 = 4000.$ Why this step? We ask: after using up all 900 m, how much speed is left? Positive $v^{2}$ means it never reached zero.
Take the root. $v = 4000 \approx 63.2 m/s$ , still downward (it never crossed zero, so the sign stays negative). Why this step? $v^{2} > 0$ at the pad ⇒ nonzero impact speed ⇒ crash. The sign is "still falling."
How much altitude was actually needed to stop? $v_{0}^{2} /2 a_{n e t} = 1000 m > 900$ ⇒ shortfall of 100 m — that is why it hits hard. Why this step? Comparing required vs available distance is the clean diagnostic for "too low."

How to read Fig 5: the coral curve starts at 200 m/s and $h = 900$ m but never touches the axis before the pad — it crosses $h = 0$ still at ~63 m/s. The red gap between the curve's foot and the ground is the impact speed: a crash.

Verify: $v = 40000 - 36000 = 4000 \approx 63.25$ m/s, downward ✓. Sanity: less than $v_{0} = 200$ , so it did brake, just not enough ✓.

Ex 6 — Cell F: the limiting value (throttle floor sets max fall speed)

Worked example Statement

An engine cannot throttle infinitely. Suppose its minimum deliverable net acceleration during the landing burn is $a_{n e t}^{m i n} = 18 m/s^{2}$ , and the pad allows a maximum braking distance of $h_{m a x} = 1500 m$ before ground. What is the highest fall speed $v_{0}$ the booster can still stop in time?

Forecast: as the allowed distance grows, does the survivable speed grow linearly, or slower/faster than linear?

Set the boundary case. The worst survivable case uses all the height at the weakest brake: $0 = v_{0}^{2} - 2 a_{n e t}^{m i n} h_{m a x}$ . Why this step? The limit is where equality holds — any faster and you crash, any slower and you have margin.
Solve for $v_{0}$ . $v_{0} = 2 a_{n e t}^{m i n} h_{m a x} = 2 \cdot 18 \cdot 1500 = 54000 \approx 232.4 m/s$ . Why this step? Rearranging the timeless equation gives the max entry speed as a function of the two limits.
Scaling insight. Because $v_{0} \propto h$ , doubling the runway only multiplies survivable speed by $2 \approx 1.41$ — sub-linear. Speed is expensive to kill. Why this step? Reveals why entry burns exist: you can't buy much braking with altitude, so you must arrive slower.

How to read Fig 6: the lavender curve plots survivable $v_{0} = 2 a_{n e t}^{m i n} h$ against available height $h$ . Its flattening shape is the sub-linear $h$ law: to survive twice the speed you would need four times the height (mint guide lines).

Verify: $v_{0} = 2 \cdot 18 \cdot 1500 = 54000 \approx 232.38$ m/s ✓. Check back: $232.3 8^{2} - 2 \cdot 18 \cdot 1500 \approx 0$ ✓.

Ex 7 — Cell G: the engine is too weak ( $a_{n e t} \leq 0$ , TWR ≤ 1)

Worked example Statement

A hypothetical booster (or a badly-throttled one) has thrust $T = 280, 000 N$ , mass $m = 30, 000 kg$ , $g = 9.8$ . It is falling at $v_{0} = 200 m/s$ . What ignition altitude does $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ give, and what does it mean?

Forecast: will the formula give a sensible positive height, or something impossible?

Compute $a_{n e t}$ . $a_{n e t} = \frac{T}{m} - g = \frac{280000}{30000} - 9.8 = 9.333 - 9.8 = - 0.467 m/s^{2}$ . Why this step? Check the sign of the net acceleration before trusting the hoverslam formula — everything depends on it.
$a_{n e t} < 0$ means no braking at all. Thrust ( $9.33 m/s^{2}$ ) is less than gravity ( $9.8$ ), so the engine cannot even hold the booster up — it keeps accelerating downward. Plugging into $h_{i g n} = v_{0}^{2} /2 a_{n e t} = 20 0^{2} / (2 \cdot - 0.467) \approx - 42, 900 m$ gives a negative altitude — physically meaningless. Why this step? A negative $h_{i g n}$ is the maths screaming "impossible": there is no altitude at which this engine stops the fall.
The boundary $a_{n e t} = 0$ (TWR exactly 1). Here $h_{i g n} = v_{0}^{2} /0 = + \infty$ : even infinite height cannot null the speed, because zero net deceleration never slows you. So the hoverslam formula is only valid for $a_{n e t} > 0$ (TWR $> 1$ ) — precisely why real boosters must have thrust exceeding weight, and why they cannot hover. Why this step? Covers the degenerate limit cleanly: $a_{n e t} > 0$ ⇒ finite positive $h_{i g n}$ ; $a_{n e t} = 0$ ⇒ $\infty$ ; $a_{n e t} < 0$ ⇒ negative/nonsense.

How to read Fig 7: the curve $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ is plotted against $a_{n e t}$ . To the right ( $a_{n e t} > 0$ ) it gives sensible positive heights that blow up to infinity as $a_{n e t} \to 0$ (mint asymptote). To the left ( $a_{n e t} < 0$ , coral shaded "no-landing zone") it dives negative — impossible. Only the mint region is a real hoverslam.

Verify: $a_{n e t} = 280000/30000 - 9.8 \approx - 0.4667$ ✓ (negative); $h_{i g n} = 20 0^{2} / (2 \cdot - 0.4667) \approx - 42857$ m ✓ (negative ⇒ non-physical). At $a_{n e t} = 0$ , $h_{i g n} \to \infty$ ✓.

Ex 8 — Cell H: real-world crosswind (sideways PID correction)

Worked example Statement

During the landing burn a steady crosswind pushes the booster sideways. The lateral position error is $e$ (metres to the right of target). Using a PID Control gimbal law $δ = K_{p} e + K_{d} \overset{e}{˙} + K_{i} \int e d t$ , with $K_{p} = 0.5, K_{d} = 1.2, K_{i} = 0.05$ , at one instant $e = 4 m$ , $\overset{e}{˙} = - 1.5 m/s$ (already moving back toward target), and the running integral $\int e d t = 8 m\cdots$ . What gimbal deflection $δ$ is commanded?

Forecast: with error to the right but closing, will the three terms add up or fight each other?

Present term. $K_{p} e = 0.5 \cdot 4 = 2.0$ (deg). Reacts to where the error is now. Why this step? Proportional term pushes against the current 4 m offset.
Damping term. $K_{d} \overset{e}{˙} = 1.2 \cdot (- 1.5) = - 1.8$ . Since $\overset{e}{˙} < 0$ we are already correcting; this term pulls back to avoid overshoot. Why this step? The derivative term is why the rocket doesn't oscillate — it anticipates motion.
Bias term. $K_{i} \int e d t = 0.05 \cdot 8 = 0.4$ . Removes steady offset from the constant wind. Why this step? A persistent crosswind leaves a steady error that only the integral term can erase.
Sum. $δ = 2.0 - 1.8 + 0.4 = 0.6$ deg. Why this step? PID output is the sum of the three; here they partly cancel, giving a gentle net command — exactly the point of $K_{d}$ .

How to read Fig 8: three stacked bars — lavender (present, $+ 2.0$ ), coral (damping, $- 1.8$ , pointing down because it opposes), butter (bias, $+ 0.4$ ) — and their net (mint, $+ 0.6$ ). The near-cancellation shows the controller easing off because the booster is already closing on target.

Verify: $0.5 \cdot 4 + 1.2 \cdot (- 1.5) + 0.05 \cdot 8 = 2.0 - 1.8 + 0.4 = 0.6$ ✓. Sign $> 0$ ⇒ correct toward target; magnitude small because the vehicle is already closing ✓.

Ex 9 — Cell I: exam twist — mass drops during the burn

Worked example Statement

During a 10 s burn the booster's mass falls from $m_{0} = 30, 000$ kg to $m_{1} = 27, 000$ kg as fuel burns. If thrust $T$ is held constant at $T = 900, 000 N$ and $g = 9.8$ , compare $a_{n e t}$ at the start vs the end of the burn. Does the deceleration get stronger or weaker?

Forecast: as mass drops with fixed thrust, does $a_{n e t} = T / m - g$ rise or fall?

Start. $a_{n e t, 0} = \frac{900000}{30000} - 9.8 = 30 - 9.8 = 20.2 m/s^{2}$ . Why this step? Baseline braking at ignition, with full mass.
End. $a_{n e t, 1} = \frac{900000}{27000} - 9.8 = 33.33 - 9.8 = 23.53 m/s^{2}$ . Why this step? Same thrust divided by smaller mass ⇒ larger acceleration; we quantify the change.
Interpret. Braking strengthens by $\approx 3.3 m/s^{2}$ as fuel burns off. So the constant- $a_{n e t}$ assumption in Ex 1–6 is a mild under-estimate of braking late in the burn; the real GNC recomputes $T_{c m d} = m (g + v^{2} /2 h)$ every cycle to absorb this — the exact reason the parent note recomputes live. Mass depletion itself is governed by the Rocket Equation (Tsiolkovsky). Why this step? Ties the numeric drift back to why a live closed loop beats a fixed formula.

How to read Fig 9: two mint bars for $a_{n e t}$ — the taller one (end of burn) sits above the shorter (start), with the coral arrow marking the $\approx 3.3 m/s^{2}$ gain. Same thrust, lighter rocket ⇒ stronger push, which is why a fixed formula slowly drifts and the loop must recompute.

Verify: $a_{n e t, 0} = 900000/30000 - 9.8 = 20.2$ ✓; $a_{n e t, 1} = 900000/27000 - 9.8 \approx 23.533$ ✓; difference $\approx 3.33 m/s^{2}$ , increasing ✓.

Active recall

Recall Which equation, and why no time in it?

Which kinematic relation drives the hoverslam, and why is it preferred over $v = v_{0} - a t$ ? ::: $v^{2} = v_{0}^{2} - 2 a_{n e t} h$ — it links speed directly to altitude with no time term, so we solve for where speed hits zero.

Recall Sign of leftover velocity at the pad

If a booster ignites too high, what is the sign of its velocity when (if) it reaches the pad? ::: Positive (upward) — it stops above ground and, with TWR $> 1$ , climbs again.

Recall When does the hoverslam formula break?

For what values of $a_{n e t}$ does $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ give a physical answer? ::: Only $a_{n e t} > 0$ (TWR $> 1$ ). At $a_{n e t} = 0$ it is $+ \infty$ ; at $a_{n e t} < 0$ it goes negative — no landing possible.

Recall The scaling law

If the allowed braking distance doubles, by what factor does the maximum survivable entry speed grow? ::: By $2 \approx 1.41$ , because $v_{0} \propto h$ — sub-linear.