Autonomous GNC for reusable rockets — SpaceX approach overview: Exercises

This page is a self-test ladder. Each problem states its level, gives clean numbers, and hides a full solution inside a collapsible callout. Try it, then reveal. Everything builds on the parent overview.

Before we start, one picture fixes every sign convention on this page. Read it with the definition below — each arrow is named there.

Figure — Autonomous GNC for reusable rockets — SpaceX approach overview

Definition Sign convention, and how to read Figure s01

How the figure is drawn. In the picture, the tall red shape is the booster high above the black ground line (which is altitude $h = 0$ ). Three black arrows leave the booster, and one short black arrow at the top-left of the frame is the axis:

the arrow pointing up at the far left, labelled "+ (up positive)", is our reference direction — anything pointing that way counts as positive;
the arrow on the left of the booster pointing down, labelled "velocity = −v₀", is the velocity (it falls, so it points down);
the arrow on the right of the booster pointing down, labelled "g", is gravity;
the arrow below the booster pointing up, labelled "Thrust", is the engine push.

So in one glance: velocity and gravity point down, thrust points up, and "up" is the positive direction.

Now the symbols. Up is positive. The booster falls, so its true velocity vector points down; we write that velocity as $- v_{0}$ where $v_{0} > 0$ is the downward speed (a plain positive number, "how fast it's dropping"). Gravity pulls down: $g = + 9.8 m/s^{2}$ used as a magnitude. Engine thrust pushes up.

The net upward acceleration during the burn is $a_{n e t} = \frac{T}{m} - g$ where $T$ is thrust magnitude and $m$ is mass. When $a_{n e t} > 0$ the rocket is decelerating its fall (slowing down), which is exactly what a landing burn does.

Why we drop the signs in the kinematic formulas below. During the whole braking burn only two directions ever occur: the booster moves down, the deceleration acts up. Nothing changes direction mid-burn. So instead of juggling minus signs we work with plain positive magnitudes: $v$ = current downward speed (m/s, always $\geq 0$ here), $a_{n e t}$ = size of the upward deceleration (m/s², positive), and $s$ = stopping distance, the vertical distance still to fall (metres, positive). With those magnitudes the standard "brake to a stop" relations read $v = v_{0} - a_{n e t} t, v^{2} = v_{0}^{2} - 2 a_{n e t} s .$ Here the minus signs simply say "braking reduces speed over time and over distance." We define $s$ = distance travelled during the burn; when the burn covers the whole way to the ground, $s$ equals the ignition altitude $h_{i g n}$ .

Level 1 — Recognition

Problem L1.1

Name the three letters of GNC and, in one phrase each, the question each one answers.

Recall Solution

Guidance — Where do I want to go? (computes the desired trajectory / target state).
Navigation — Where am I now? (estimates position, velocity, attitude from sensors).
Control — How do I move the actuators to close the gap? (turns commands into gimbal / fin / thruster deflections).

Problem L1.2

A booster's minimum engine thrust gives a thrust-to-weight ratio (TWR) greater than 1 when nearly empty. Which single word describes why it cannot hover?

Recall Solution

TWR > 1 means minimum upward thrust force exceeds the weight force ( $T > m g$ , so $T / m g > 1$ ). Even at lowest throttle the net acceleration is upward — the rocket would accelerate up, not float. Hovering needs thrust exactly equal to weight (TWR = 1), which the engine cannot go low enough to reach. Hence it must do a single timed hoverslam instead.

Level 2 — Application

Problem L2.1

A booster is falling at downward speed $v_{0} = 150 m/s$ . The engine can produce a net upward acceleration $a_{n e t} = 25 m/s^{2}$ . At what altitude $h_{i g n}$ must it light the engine so that velocity reaches zero exactly at the ground?

Recall Solution

Why $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ ? This kinematic relation has no time in it — it links speed directly to the distance travelled $s$ . We want the distance over which speed drops to zero, so this is the right tool (not the time-based $v = v_{0} - a_{n e t} t$ ). Recall $s$ = stopping distance (the vertical distance the burn covers); here the burn runs all the way to the ground, so $s = h_{i g n}$ .

Set final speed $v = 0$ over stopping distance $s = h_{i g n}$ : $0 = v_{0}^{2} - 2 a_{n e t} h_{i g n} \Rightarrow h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} .$ Plug in: $h_{i g n} = \frac{15 0 ^{2}}{2 \cdot 25} = \frac{22500}{50} = 450 m .$ Answer: $h_{i g n} = 450$ m.

The next figure shows this exact case as a speed-versus-altitude curve. The horizontal axis is altitude $h$ in metres (0 at the pad, growing to the right); the vertical axis is speed $v$ in m/s. The accent-red line is the booster's speed as it brakes.

Read it from right to left (falling means altitude decreasing): the black dot on the upper right sits at $h = 450$ m, $v = 150$ m/s — that is ignition. Follow the red curve down-and-left and it lands exactly on the origin, the second black dot at $h = 0$ , $v = 0$ — touchdown. Because the curve reaches zero speed at zero altitude and nowhere sooner, that single meeting point is a perfect hoverslam.

Problem L2.2

Same booster and burn ( $v_{0} = 150$ , $a_{n e t} = 25$ ). How long does the burn last?

Recall Solution

Why the time-form now, not the distance-form? Figure s02 plotted speed against altitude — great for "where does speed hit zero" but it has no clock on it, so you cannot read seconds off it. The question asks how long, which is a clock question. The two kinematic tools split cleanly by what they leave out: $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ has no time (use it for distances), while $v = v_{0} - a_{n e t} t$ has no distance (use it for durations). Since we want seconds, we pick the one that actually contains $t$ .

Use $v = v_{0} - a_{n e t} t$ with final $v = 0$ : $t = \frac{v _{0}}{a _{n e t}} = \frac{150}{25} = 6 s .$ Answer: $t = 6$ s. (Sanity check: average speed $\frac{150 + 0}{2} = 75$ m/s over 6 s = 450 m = $h_{i g n}$ . ✓ — this bridge back to distance is exactly how the two forms agree.)

Problem L2.3

The booster mass at ignition is $m = 30, 000$ kg, and $g = 9.8 m/s^{2}$ . Using $a_{n e t} = 25 m/s^{2}$ , what thrust $T$ is required?

Recall Solution

Why $T = m (g + a_{n e t})$ ? From $a_{n e t} = T / m - g$ , solve for $T$ : the engine must first cancel gravity ( $m g$ ) and then supply the extra deceleration ( $m a_{n e t}$ ). $T = m (g + a_{n e t}) = 30000 (9.8 + 25) = 30000 \times 34.8 = 1, 044, 000 N .$ Answer: $T \approx 1.044 \times 1 0^{6}$ N (1044 kN).

Level 3 — Analysis

Problem L3.1

Using the numbers of L2.1 ( $v_{0} = 150$ , correct $h_{i g n} = 450$ m), suppose the computer lights the engine early, at $h = 600$ m, with the same $a_{n e t} = 25$ . Where does the booster's velocity reach zero, and what physically happens next?

Recall Solution

With the same $a_{n e t}$ and same starting speed, the stopping distance $s$ (how far the burn takes to kill the speed) is fixed: $s_{s t o p} = \frac{v _{0}^{2}}{2 a _{n e t}} = \frac{22500}{50} = 450 m .$ Lighting at 600 m, the booster consumes 450 m of altitude braking and stops at $600 - 450 = 150$ m above the ground. Now it is momentarily at rest at 150 m with the engine still on and TWR > 1 → net upward acceleration → it rises back up. It cannot throttle to hover. So early ignition doesn't "add margin" — it strands the rocket in mid-air or forces a wasteful re-fall. Answer: velocity zero at 150 m altitude; then it accelerates upward (cannot hover).

Problem L3.2

Now it lights late, at $h = 300$ m, same $a_{n e t} = 25$ , same $v_{0} = 150$ . What is the impact speed?

Recall Solution

The burn now has only a stopping distance of $s = 300$ m available, but it needed 450 m. Use $v^{2} = v_{0}^{2} - 2 a_{n e t} s$ with $s = 300$ : $v^{2} = 15 0^{2} - 2 (25) (300) = 22500 - 15000 = 7500,$ $v = 7500 \approx 86.6 m/s (downward).$ Answer: it hits the ground at $\approx 86.6$ m/s → crash. This is the razor-thin corridor: $\pm 150$ m of ignition-altitude error is the difference between rising away and slamming in at highway speed.

Problem L3.3

Explain why the IMU alone cannot navigate the whole descent, and give the growth law of a constant accelerometer bias $b$ in the position estimate. Then say what happens to the 3D position-error size if each axis has the same bias.

Recall Solution

First, notation: a bold letter like $r$ means a vector — a quantity with three components (an $x$ , a $y$ , a $z$ ), here the position in space; likewise $a$ is the acceleration vector. Bold just says "this has three numbers and a direction."

The IMU gives acceleration; position is recovered by integrating twice ( $r = \iint a d t$ — first add up acceleration over time to get velocity, then add up velocity over time to get position).

Work one axis explicitly. Say the $x$ -accelerometer reads a fixed wrong offset $b$ (metres per second², constant). Integrate once for the velocity error: $Δ v_{x} = \int_{0}^{t} b d t = b t .$ Integrate that once more for the position error: $Δ x = \int_{0}^{t} b t^{'} d t^{'} = \frac{1}{2} b t^{2} .$ So on each axis the same algebra gives the same shape $\frac{1}{2} b t^{2}$ — nothing about $x$ was special, so $y$ and $z$ produce $\frac{1}{2} b t^{2}$ too. That is why "the argument is identical along each axis": the integration doesn't know or care which axis it runs on.

The 3D error size. If all three axes carry the same bias $b$ , the error vector is $Δ r = (\frac{1}{2} b t^{2}, \frac{1}{2} b t^{2}, \frac{1}{2} b t^{2})$ . Its length (Pythagoras in 3D) is $∣Δ r ∣ = 3 (\frac{1}{2} b t^{2}) = \frac{3}{2} b t^{2},$ still growing like $t^{2}$ — just $3 \approx 1.73$ times bigger than one axis alone.

Why this matters: the error grows with the square of time — small bias, long flight → huge error. So we fuse a drift-free but slow reference (GPS/radar) via a Kalman filter to bound the drift. Fast IMU + slow GPS = best of both.

Level 4 — Synthesis

Problem L4.1

A booster is at altitude $h = 1000$ m descending at $v_{0} = 200$ m/s. The engine can deliver a net upward accel of $a_{n e t} = 30 m/s^{2}$ . (a) Is it already past the point where a $30 m/s^{2}$ burn can save it? (b) What minimum $a_{n e t}$ would exactly stop it at the ground from here?

Recall Solution

(a) Required ignition altitude for $a_{n e t} = 30$ (stopping distance $s = h_{i g n}$ ): $h_{i g n} = \frac{v _{0}^{2}}{2 a _{n e t}} = \frac{20 0 ^{2}}{2 \cdot 30} = \frac{40000}{60} \approx 666.7 m .$ Since the booster is at 1000 m, which is above 666.7 m, it is not past the point — it should keep falling and light later. It has margin.

(b) The minimum $a_{n e t}$ is the one that just barely stops it using the full 1000 m of stopping distance: $a_{n e t, m i n} = \frac{v _{0}^{2}}{2 h} = \frac{40000}{2000} = 20 m/s^{2} .$ Answers: (a) not past the point (h_ign ≈ 666.7 m < 1000 m); (b) minimum $a_{n e t} = 20 m/s^{2}$ . Interpretation: any engine capable of $\geq 20 m/s^{2}$ can land it; a $30 m/s^{2}$ engine just lights lower and burns shorter.

Problem L4.2

For the case above, using $a_{n e t} = 30 m/s^{2}$ and mass $m = 28, 000$ kg at ignition ( $g = 9.8$ ), find (a) the burn duration and (b) the required thrust.

Recall Solution

(a) Ignition happens at $h_{i g n} \approx 666.7$ m, speed still $v_{0} = 200$ m/s. $t = \frac{v _{0}}{a _{n e t}} = \frac{200}{30} \approx 6.67 s .$ (b) Thrust: $T = m (g + a_{n e t}) = 28000 (9.8 + 30) = 28000 \times 39.8 = 1, 114, 400 N .$ Answers: (a) $t \approx 6.67$ s; (b) $T \approx 1.114 \times 1 0^{6}$ N (1114 kN).

Level 5 — Mastery

Problem L5.1 — Live closed-loop recompute

Real thrust isn't perfectly steady. Suppose the true burn under-performs, delivering only $a_{n e t} = 18 m/s^{2}$ instead of a planned 20. The vehicle recomputes the commanded net accel every cycle from live state $(v, h)$ using $a_{c m d} = v^{2} / (2 h)$ . At the instant it lit (planned), it was at $v = 200$ m/s, $h = 1000$ m. (a) What net accel does the closed-loop law demand at that instant? (b) If the engine can only reach 18, what does the guidance do, and is landing still possible? Reason it out.

Recall Solution

(a) At $v = 200$ , $h = 1000$ : $a_{c m d} = \frac{v ^{2}}{2 h} = \frac{20 0 ^{2}}{2 \cdot 1000} = \frac{40000}{2000} = 20 m/s^{2} .$ (b) The engine can only give 18, i.e. it under-brakes. So over the next cycle the vehicle sheds less speed than planned; at the new altitude the ratio $v^{2} / (2 h)$ climbs (speed too high for the shrinking altitude). The closed-loop law therefore demands more accel next cycle. As long as the engine's ceiling ( $a_{n e t}$ it can actually make, given throttle and mass) stays above the rising demand, it catches up and still nails $v = 0$ at $h = 0$ . If the demand ever exceeds the engine ceiling before touchdown, the case becomes the "late ignition" crash of L3.2. This is exactly why guidance is closed-loop: it re-solves $a_{c m d} = v^{2} /2 h$ each cycle instead of trusting one pre-scripted plan. For the fuel-optimal version of this recompute-under-constraints problem, SpaceX-style guidance solves a real-time convex optimization via lossless convexification, guaranteeing a global optimum in bounded time each cycle. Answer: (a) demands 20 m/s²; (b) it up-commands accel each cycle; landing survives iff engine ceiling stays above the rising demand.

Problem L5.2 — Degenerate / edge cases

Discuss what the ignition formula $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ predicts in three limiting cases, and whether each is physical: (a) $v_{0} = 0$ ; (b) $a_{n e t} \to 0^{+}$ ; (c) $a_{n e t}$ very large.

Recall Solution

(a) $v_{0} = 0$ (already at rest). Substitute $v_{0} = 0$ : $h_{i g n} = 0^{2} / (2 a_{n e t}) = 0$ . Physical meaning: if you already have zero speed you need no braking distance ( $s = 0$ ) — kinematically you could touch down right where you are. (Real gravity would then start pulling you down again, so you'd still need thrust to hold, but the pure "distance to kill the speed" is genuinely zero.) The formula is correct and sensible.

(b) $a_{n e t} \to 0^{+}$ (engine barely stronger than gravity). As $a_{n e t}$ shrinks toward zero from the positive side, the denominator $2 a_{n e t} \to 0^{+}$ while the numerator stays fixed, so $h_{i g n} = v_{0}^{2} / (2 a_{n e t}) \to + \infty$ . Physical meaning: a rocket that can barely decelerate needs an infinite runway to stop — i.e. within any real corridor it cannot stop the fall at all. This is the boundary where the engine is too weak (TWR just above 1); landing becomes impossible, and the formula correctly diverges to warn us.

(c) $a_{n e t}$ very large (extremely powerful burn). As $a_{n e t} \to \infty$ , the denominator grows without bound, so $h_{i g n} = v_{0}^{2} / (2 a_{n e t}) \to 0^{+}$ . Physical meaning: an enormously powerful engine needs almost no stopping distance — you fall nearly all the way to the deck, then a violent, very-short burst stops you. Mathematically fine, but it demands near-perfect timing (a vanishingly small ignition window) and punishing g-loads, so real vehicles deliberately pick a moderate $a_{n e t}$ to keep both the timing corridor and the structural loads sane.

Takeaway: the formula behaves sensibly at every extreme — $v_{0} = 0 \Rightarrow h_{i g n} = 0$ ; weak engine $\Rightarrow h_{i g n} \to \infty$ (impossible); strong engine $\Rightarrow h_{i g n} \to 0$ (twitchy). No case surprises us, which is exactly why it's a trustworthy live decision rule.

The final figure plots the whole family: $h_{i g n}$ against the engine's $a_{n e t}$ for a fixed $v_{0} = 200$ m/s. The horizontal axis is net accel $a_{n e t}$ in m/s², the vertical axis is ignition altitude $h_{i g n}$ in m, and the accent-red curve is the ignition law.

Trace the red curve: at the far left (small $a_{n e t}$ ) it shoots up toward infinity — that is case (b), the too-weak engine. At the far right (large $a_{n e t}$ ) it flattens toward zero — that is case (c), the twitchy super-strong engine. The two annotated ends are exactly the limits you reasoned about above.

Active recall

Recall One-line summary of the ladder

Everything above is one equation seen from five angles: $h_{i g n} = v_{0}^{2} /2 a_{n e t}$ . Recognition names it, Application plugs it, Analysis breaks it (early/late/IMU drift), Synthesis combines it with $T = m (g + a_{n e t})$ and burn time $t = v_{0} / a_{n e t}$ , and Mastery makes it live (closed-loop recompute) and robust (edge cases). Prerequisite tools live in Kinematics — v²=v0²+2as, Rocket Equation (Tsiolkovsky), PID Control, Thrust Vectoring (Gimbal), and Attitude Dynamics & Quaternions.

The ignition-altitude law in one line

h_{i g n} = v_{0}^{2} / (2 a_{n e t})

Required landing-burn thrust

T = m (g + a_{n e t})

Burn duration

t = v_{0} / a_{n e t}

Why closed-loop not open-loop

real mass, wind, and thrust vary each flight; recompute each cycle to bound the error