Worked examples — Powered descent guidance — G-FOLD algorithm (convex optimization)

This page is a drill room. The parent note built the theory; here we exhaust every shape the problem can take: signs, degenerate inputs, limiting cases, a word problem, and an exam twist. Work each one before reading the steps.

Before we start, a one-line recap of the symbols we reuse everywhere (never used before being named):

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The scenario matrix

Every powered-descent question is one (or a blend) of a small family of cells. Rather than memorise a table, picture the map below. It has two big branches — vertical (1-D) feasibility on the left and glide-slope geometry on the right — each splitting into sign cases, degenerate cases, and limits. Sketch of the map, so you can follow it without opening the image file:

Figure (matrix): the same map drawn to scale — root at top, the blue vertical-feasibility branch (leaves C0–C5) on the left, the green glide-slope branch (leaves C6–C7) on the right, and the orange/red fuel-and-exam leaves (C8–C10) hanging off the bottom. Each leaf is one worked example below.

• C0 — degenerate quadrant $v_z>0$ (still climbing): what happens if we start moving up? → Example 0
• C1 — sign of net acceleration (thrust vs gravity): can we even fight gravity? → Example 1
• C2 — stopping distance vs altitude: does the burn fit? → Example 2
• C3 — degenerate $v_z=0$ (hover): what thrust just holds still? → Example 3
• C4 — limiting $T_{\min }\to 0$ (deep throttle): when does coast-then-burn appear? → Example 4
• C5 — bang-bang timing: when to start the suicide burn? → Example 5
• C6 — glide-slope offset (horizontal position): cone constraint → Example 6
• C7 — degenerate cone ${\theta }_{gs}\to 90°$ and $\to 0°$: limits of the funnel → Example 7
• C8 — fuel via $z=\ln m$: log-mass bookkeeping → Example 8
• C9 — word problem (real Mars/F9 numbers): put it together → Example 9
• C10 — exam twist (infeasible input): recognise "no landing exists" → Example 10

Numbers reused across cells (Mars-ish): $g=-3.71{\text{m/s}}^2$, $T_{\max }/m=12{\text{m/s}}^2$, $T_{\min }/m=3{\text{m/s}}^2$ unless stated.

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Example 0 — Initial upward velocity, the $v_z>0$ edge case (Cell C0)

Forecast: Guess whether the extra climb makes the eventual landing easier or harder.

1. Coast up until $v_z=0$. Using the sign convention, gravity decelerates the climb: extra height $\Delta h=\frac{v_z^2}{2|g|}=\frac{15^2}{2(3.71)}=30.3$ m. Why this step? A positive $v_z$ is a brand-new quadrant — the lander first moves away from the pad. Kinematics gives the turning point where upward speed hits zero.
2. Peak height. $r_{z,\text{peak}}=100+30.3=130.3$ m. Why this step? Peak is where all upward speed has been converted to height; from here the problem becomes an ordinary descent.
3. Speed back at 100 m. By symmetry of free fall, it returns to $r_z=100$ m with $v_z=-15$ m/s. Why this step? Energy conservation (no thrust, no drag) makes the down-trip mirror the up-trip — so an initial $+15$ becomes an ordinary $-15$ descent, which we already know how to land (Examples 2, 5).

Verify: $\Delta h=15^2/(2\cdot 3.71)=30.32$ m ✔; peak $=130.3$ m; return speed $=15$ m/s (magnitude) ✔.

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Example 1 — Sign of net acceleration (Cell C1)

Forecast: Guess whether the lander can decelerate (slow its fall) at both throttle settings.

1. Net at full throttle. $a_{\text{net}}=a_{\text{cmd}}+g=12+(-3.71)=8.29{\text{m/s}}^2$ (upward). Why this step? Newton adds accelerations as vectors; up is $+$, gravity is $-$, so we literally add the signed numbers.
2. Net at min throttle. $a_{\text{net}}=3+(-3.71)=-0.71{\text{m/s}}^2$ (downward). Why this step? At the throttle floor the engine can't out-push gravity — the sign flips negative, meaning it still accelerates downward.

Verify: $12-3.71=8.29$ ✔; $3-3.71=-0.71$ ✔. Units: all ${\text{m/s}}^2$. The min-throttle net being negative confirms the engine's floor cannot hover — see C3.

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Example 2 — Does the burn fit? (Cell C2)

Forecast: Guess the stopping distance — more or less than 100 m?

Here $d$ is the vertical distance the lander travels while braking to a stop — measured downward from where the burn starts. Because the lander is directly above the pad, this $d$ is a drop in height, i.e. a change in $r_z$: if the burn started at $r_z=100$ and used up all $100$ m, it would just touch down at $r_z=0$. So comparing $d$ against the starting height $r_z$ is the whole feasibility test.

1. Switch to magnitudes. Following the sign-convention box above, the signed state $v_z=-20$ becomes entry speed $v_0=|v_z|=20$ m/s, and the signed net $a_{\text{net}}=+8.29$ (up, opposing the fall) becomes braking deceleration $a=8.29{\text{m/s}}^2$. Why this step? Braking is a single straight downward segment where velocity and motion share a direction; the direction is already fixed by "up is $+$", so only sizes matter and we may drop the signs.
2. Pick the tool: kinematics $v^2=v_0^2-2ad$. Here $v$ is the speed at the end of braking (we want $v=0$), $v_0=20$ m/s, $a=8.29{\text{m/s}}^2$, and $d$ is the braking distance defined above. Why this tool? We know a start speed, a constant braking deceleration, and want the distance to reach $v=0$. This equation links exactly those three with no time needed — the cleanest question-answerer here.
3. Set final speed to zero and solve for $d$. $0=20^2-2(8.29)d\Rightarrow d=\frac{400}{16.58}=24.1$ m. Why this step? $v=0$ is the "just stopped" condition; solving isolates the shortest possible stopping distance at max brake.
4. Compare $d$ to the starting height $r_z$. $24.1\text{m}<100\text{m}$. ✔ Feasible. Why this step? $d$ is the height eaten up while stopping; if $d\le r_z$ the lander reaches zero speed at or above the pad, so a soft landing is physically possible. If instead $d>r_z$ it would still be moving at ground level — a crash. The inequality $d\le r_z$ is the feasibility criterion.

Verify: $400/(2\cdot 8.29)=24.126\text{m}$ ✔, and $24.1<100$. Units: $(\text{m/s}{)}^2/({\text{m/s}}^2)=\text{m}$ ✔.

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Example 3 — Hover, the degenerate $v_z=0$ case (Cell C3)

Forecast: Guess the number before computing.

1. Hover means zero net acceleration. Require $a_{\text{cmd}}+g=0\Rightarrow a_{\text{cmd}}=-g=3.71{\text{m/s}}^2$. Why this step? "Hold still" is the degenerate case where velocity never changes, so the net must be exactly zero — thrust cancels gravity.
2. Check the box. Is $3\le 3.71\le 12$? Yes. Why this step? A command outside $[T_{\min }/m,T_{\max }/m]$ is physically impossible; here it sits just above the floor, so hover is achievable.

Verify: $-g=3.71$ ✔ and $3\le 3.71\le 12$ ✔.

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Example 4 — Limiting case $T_{\min }\to 0$ (Cell C4)

Forecast: Does zero-floor throttle make more or less altitude available for coasting?

1. Max brake unchanged. Full-throttle net is still $12-3.71=8.29$; stopping distance still $24.1$ m. Why this step? The upper limit sets braking power; lowering the floor doesn't change max deceleration.
2. What the floor controls. With $T_{\min }/m=0$ the engine may switch fully off, giving a pure coast at net $=g=-3.71$. Why this step? Recall the parent note's obstacle: the constraint $\Vert \mathbf{T}\Vert \ge T_{\min }$ forbids all thrust vectors shorter than $T_{\min }$, i.e. it bans the inside of a ball of radius $T_{\min }$ in thrust-space, leaving only the region outside that sphere — a non-convex hole. Sending $T_{\min }/m\to 0$ shrinks that forbidden ball to a point, so the hole vanishes and $\Gamma$ is free to reach $0$ (engine fully off).
3. Coast distance available. From $r_z=100$, the ship coasts freely, then must have $\ge 24.1$ m left to brake. So it may coast up to $100-24.1=75.9$ m before the mandatory burn. Why this step? This is the geometric meaning of bang-bang: coast as long as possible, brake at the last legal instant.

Verify: $100-24.1=75.9$ m of coast ✔; $75.9+24.1=100$ ✔.

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Example 5 — When to fire: bang-bang timing (Cell C5)

Figure (Example 5): height $r_z$ on the vertical axis (m) versus speed $|v_z|$ on the horizontal axis (m/s). The blue curve is the free-fall coast (speed grows as height drops); the red curve is the full-throttle brake (speed shrinks to zero). They meet at the orange dot — the burn-start handoff at $v\approx 28.1$ m/s, height $\approx 47.6$ m. The blue dot top-right is the start (v=20, r=100); the green dot at the origin is the soft touchdown (v=0, r=0). In words: read it right-to-left in time — the ship slides up the blue coast curve gaining speed, hits the orange corner, then rides the red brake curve down to the green origin.

Forecast: Guess whether the coast lasts more or less than 2 seconds.

1. Speed when braking begins. Let $v_b$ be the speed at the moment the burn starts. The brake segment must bleed off all of $v_b$ over the remaining height: $\frac{v_b^2}{2(8.29)}=$ (height remaining). Why this step? Total altitude is shared between coast and brake segments; we match speed and height at the handoff point (the orange dot in the figure).
2. Set up the two segments. Coast drops height $h_c$ gaining speed $v_b^2=20^2+2(3.71)h_c$. Brake needs $\frac{v_b^2}{2(8.29)}=100-h_c$. Why this step? Two equations, two unknowns ($v_b,h_c$) — the geometry of the red brake curve meeting the blue coast curve in the figure.
3. Solve. Substitute: $\frac{400+7.42h_c}{16.58}=100-h_c\Rightarrow 400+7.42h_c=1658-16.58h_c\Rightarrow 24h_c=1258\Rightarrow h_c=52.42$ m. Then $v_b=\sqrt{400+7.42(52.42)}=\sqrt{788.9}=28.09$ m/s. Why this step? Algebra finds the exact handoff altitude — the corner in the figure.
4. Coast time. $v_b=20+3.71t_c\Rightarrow t_c=(28.09-20)/3.71=2.18$ s. Why this step? Under constant coast accel, speed grows linearly, so time is the speed gain over the accel.

Verify: $h_c=52.42$ m, $v_b=28.09$ m/s, and $52.42+\frac{28.09^2}{2\cdot 8.29}=52.42+47.58=100$ ✔. Coast time $2.18$ s ✔.

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Example 6 — Glide-slope cone, horizontal offset (Cell C6)

Figure (Example 6): side view of the glide-slope funnel. Horizontal axis = horizontal radius $\sqrt{r_x^2+r_y^2}$ (m); vertical axis = height $r_z$ (m). The two blue lines are the $45°$ cone walls; the shaded band between them is the allowed funnel. The red dot marks the lander at radius $200$ m, height $200$ m — sitting exactly on the wall. The green dot at the origin is the target. In words: a $45°$ cone is a perfect "V" whose walls rise one metre up for every metre out, so at height 200 m the widest legal position is exactly 200 m out.

Forecast: Guess whether $(120,160)$ at 200 m clears a 45° funnel.

1. Write the cone constraint. $\sqrt{r_x^2+r_y^2}\le \tan {\theta }_{gs}(r_z)$. Why this tool? $\tan \theta =\frac{\text{horizontal}}{\text{vertical}}$ measures the funnel's steepness; the cone says "your horizontal distance may not exceed what that slope allows at your height." (Same "opposite over adjacent" as the parent's arctan story.)
2. Compute the horizontal radius. $\sqrt{120^2+160^2}=\sqrt{14400+25600}=\sqrt{40000}=200$ m. Why this step? Pythagoras turns two horizontal offsets into one radius — the quantity the cone limits.
3. Compute the allowance. $\tan 45°\cdot 200=1\cdot 200=200$ m. Why this step? At 45° the allowed radius equals the height exactly.
4. Compare. $200\le 200$ — on the cone surface, i.e. just barely legal. Why this step? Equality means the lander sits on the funnel wall; any further out is infeasible.

Verify: $\sqrt{120^2+160^2}=200$ ✔; $\tan 45°\cdot 200=200$ ✔; $200\le 200$ ✔.

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Example 7 — Degenerate cones: ${\theta }_{gs}\to 90°$ and $\to 0°$ (Cell C7)

Figure (Example 7): same side-view axes (horizontal radius in m vs height $r_z$ in m). Three cone walls are drawn from the origin: red = $10°$ (a narrow near-vertical chute), blue = $45°$, green = $80°$ (a wide, near-flat funnel). The orange dot is the test point at radius $200$ m, height $200$ m; the dashed gray line marks radius $=200$. The point sits inside the green cone, outside the red one. In words: shrinking ${\theta }_{gs}$ tips the wall toward vertical (a straw you must descend straight down); growing it tips the wall toward flat (almost anything allowed).

Forecast: Which one traps the lander in a near-vertical chute?

1. Case (a) ${\theta }_{gs}=80°$ (wide funnel). Allowed radius $=\tan 80°\cdot 200=5.671\cdot 200=1134.3$ m. Since $200<1134.3$, deeply feasible. Why this step? As ${\theta }_{gs}\to 90°$, $\tan \to \infty$: the cone opens flat, allowing almost any horizontal position — no protection from hills.
2. Case (b) ${\theta }_{gs}=10°$ (narrow chute). Allowed radius $=\tan 10°\cdot 200=0.1763\cdot 200=35.27$ m. Since $200>35.27$, infeasible — the lander is outside a steep vertical funnel. Why this step? As ${\theta }_{gs}\to 0°$, $\tan \to 0$: the cone collapses to a vertical line, forcing an almost purely vertical descent.

Verify: $\tan 80°\cdot 200\approx 1134.3$ ✔; $\tan 10°\cdot 200\approx 35.27$ ✔; $200<1134.3$, $200>35.27$ ✔.

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Example 8 — Fuel via log-mass (Cell C8)

Forecast: More or less than 40 kg of propellant?

1. Recall the log-mass law. From the parent note, $\dot{z}=-\alpha \Gamma$. With $\Gamma$ constant this integrates to $z(t_f)=z_0-\alpha \Gamma t_f$, where $z_0=\ln m_0$. Why this tool? The substitution $z=\ln m$ turns the messy $\dot{m}/m$ into a straight line — integration is trivial when $\Gamma$ is constant.
2. Assemble the numbers. $\alpha =1/v_e=1/2000=5\times 10^{-4}\text{s/m}$, and $z_0=\ln 1000=6.9078$. So $z(t_f)=6.9078-(5\times 10^{-4})(8)(10)=6.9078-0.04=6.8678$. Why this step? Direct substitution of the three given quantities ($\alpha$ from $v_e$, $\Gamma$, $t_f$) into the linear law.
3. Undo the log. $m(t_f)=e^{z(t_f)}=e^{6.8678}=960.79$ kg, because $z=\ln m$ means $m=e^z$. Propellant used $=m_0-m(t_f)=1000-960.79=39.21$ kg. Why this step? The exponential recovers real mass from log-mass; subtracting gives fuel spent.

Verify: $\ln 1000=6.9078$; $6.9078-0.04=6.8678$; $e^{6.8678}=960.79$ kg; burned $39.21$ kg ✔ (< 40, as forecast). Units: $\alpha \Gamma t_f=\frac{\text{s}}{\text{m}}\cdot \frac{\text{m}}{{\text{s}}^2}\cdot \text{s}=$ dimensionless ✔ (as $z$ must be).

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Example 9 — Word problem, Falcon-9 style (Cell C9)

Forecast: Real boosters land with tiny margins — guess whether 500 m is enough.

1. Net brake. $30-9.81=20.19{\text{m/s}}^2$. Why this step? Same signed-sum as C1 but Earth gravity.
2. Stopping distance. Switching to magnitudes ($v_0=|-60|=60$, $a=20.19$), $d=\frac{60^2}{2(20.19)}=\frac{3600}{40.38}=89.15$ m, the vertical braking drop. Why this tool? Kinematics again — known speed, known brake, want distance.
3. Feasibility. $89.15<500$ — fits comfortably. Why this step? $d$ is the height used up while stopping; $d\le r_z$ means it stops above the pad.
4. Hover check. Need $a_{\text{cmd}}=9.81$; box is $[12,30]$. $9.81<12$, so hover is impossible — the engine's floor exceeds gravity. Why this step? Real high-thrust boosters cannot hover; they must land in one continuous fast burn (this is why Falcon 9 does a hoverslam, never a hover).

Recall Why can't Falcon 9 hover?

Its minimum throttle acceleration ($\ge 12$) exceeds Earth gravity ($9.81$), so any thrust makes it climb — it must time a single decelerating burn to arrive at zero speed exactly at the pad. ::: The engine floor is above $|g|$, so hover thrust $9.81$ is below the throttle box.

Verify: $30-9.81=20.19$; $3600/(2\cdot 20.19)=89.15$ m; $89.15<500$; hover $9.81\notin [12,30]$ ✔.

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Example 10 — Exam twist: recognise infeasibility (Cell C10)

Forecast: Trap alert — does the burn fit in only 20 m?

1. Stopping distance at current max brake. Net $=12-3.71=8.29$; with $v_0=|-25|=25$, $d=\frac{25^2}{2(8.29)}=\frac{625}{16.58}=37.7$ m. Why this step? The feasibility test from C2 — compute the braking drop $d$ first.
2. Compare. $37.7>20$ — infeasible. The rocket needs $37.7$ m to stop but has only $20$ m of height, so it is still moving fast when it reaches the pad: it crashes. Why this step? When the braking drop $d$ exceeds the altitude $r_z$, no thrust profile within limits can land softly — the SOCP returns infeasible.
3. Required max thrust. Demand the burn just fits, $d\le 20$: $\frac{625}{2a_{\text{net}}}\le 20\Rightarrow a_{\text{net}}\ge \frac{625}{40}=15.625{\text{m/s}}^2$. Then $T_{\max }/m\ge a_{\text{net}}+|g|=15.625+3.71=19.335{\text{m/s}}^2$. Why this step? Invert the stopping-distance formula to find the minimum net brake that fits, then add gravity back to convert net acceleration into required thrust.

Verify: $625/(2\cdot 8.29)=37.7$ m $>20$ (infeasible) ✔; required $a_{\text{net}}=625/40=15.625$; $T_{\max }/m\ge 19.335$ ✔.

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Recall

Recall Which cell does each example cover?

C0→Ex0, C1→Ex1, C2→Ex2, C3→Ex3, C4→Ex4, C5→Ex5, C6→Ex6, C7→Ex7, C8→Ex8, C9→Ex9, C10→Ex10. ::: One example per scenario cell — the matrix is fully covered.

Related: Tsiolkovsky Rocket Equation (the $z=\ln m$ mass law), Second-Order Cone Programming (the glide-slope cone), SpaceX Falcon 9 Landing and Mars EDL (Examples 9–10), Apollo Lunar Descent Guidance (bang-bang heritage).