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

Answer: (b). A set is convex if the straight line joining any two of its points stays inside the set.

• (a) $\Vert \mathbf{T}\Vert \le T_{\max }$ is a solid disk (all thrust vectors no longer than $T_{\max }$) — pick any two points inside, the segment stays inside. Convex.
• (c) is a linear equality — a flat plane in state-space. Convex.
• (d) is a cone opening upward (glide-slope). Convex.
• (b) $\Vert \mathbf{T}\Vert \ge T_{\min }$ is everything outside a circle. Take one point on the left rim and one on the right rim: the segment between them passes straight through the forbidden hole in the middle. That escape from the set is the definition of non-convex. Look at the pink dashed chord in the figure above — its two pink endpoints sit on the allowed ring, but the dashed line between them dives through the yellow inner circle into the forbidden hole.

• $\mathbf{a}=\mathbf{T}/m$ — the acceleration command (thrust per unit mass; what the rocket actually feels).
• $\Gamma =\Vert \mathbf{T}\Vert /m$ — the slack magnitude that upper-bounds $\Vert \mathbf{a}\Vert$; it's how hard we're throttling, in acceleration units.
• $z=\ln m$ — the log of mass, chosen so mass dynamics become linear.
• $\alpha =1/v_e=1/(I_{sp}{g}_0)$ — the fuel-cost rate: kg of mass lost per newton-second of thrust, expressed so $\dot{z}=-\alpha \Gamma$.

Answer line — $\alpha$ in one sentence ::: $\alpha$ converts thrust magnitude into mass-loss rate; it is the reciprocal of exhaust speed $v_e$.

WHAT we do: find the largest net deceleration and use constant-acceleration kinematics. Sign handling first: up-thrust gives acceleration $+T_{\max }/m$; gravity gives $-g$. So the net upward acceleration while braking is $a=\frac{T_{\max }}{m}-g=15-3.71=11.29{\text{m/s}}^2(>0,\text{ so it decelerates the fall}).$ WHY we can use the speed $25$ m/s: the stopping-distance formula $d=v^2/(2a)$ uses the magnitude of velocity because $v^2=(-25{)}^2=25^2$ — the sign disappears when squared. The velocity is signed ($v_z=-25$), but its square (kinetic-energy term) is direction-free, which is exactly why the formula only needs the speed $|v_z|=25$. $d=\frac{|v_z{|}^2}{2a}=\frac{25^2}{2\times 11.29}=\frac{625}{22.58}\approx 27.68\text{m}.$ Since $27.68\text{m}<150\text{m}$, yes — there's altitude to spare, so a feasible (indeed fuel-optimal) solution exists.

Edge case — when stopping is impossible: the net decelerating acceleration is $a=T_{\max }/m-g$. If $T_{\max }/m\le g$, then $a\le 0$: even at full throttle the engine cannot beat gravity, the lander keeps accelerating downward, and $d=v^2/(2a)$ is undefined or negative — the feasibility check fails and no landing exists from that state. So the very first thing G-FOLD needs is a thrust ceiling above local gravity, $T_{\max }/m>g$.

WHAT the constraint says geometrically: at each altitude $r_z$, the term $\Vert [r_x,r_y]\Vert$ is the radius of a horizontal circle the lander is allowed inside. Because that allowed radius grows linearly with height, stacking the circles makes an upward-opening cone whose tip sits at the target. The figure below plots exactly this: the vertical axis is altitude $r_z$, the horizontal axis is offset $\Vert [r_x,r_y]\Vert$, and the shaded wedge is the allowed funnel.

• ${\theta }_{\text{gs}}=45^{\circ }$: $\tan 45^{\circ }=1$, so radius $=1\times 320=320$ m. In the figure, at the dotted $r_z=320$ line the yellow cone edge sits at $320$ m.
• ${\theta }_{\text{gs}}=30^{\circ }$: $\tan 30^{\circ }\approx 0.5774$, so radius $=0.5774\times 320\approx 184.75$ m — the blue cone is narrower.

WHY tan and not sin: on the right triangle (altitude $r_z$ = adjacent vertical side, horizontal offset = opposite side, slant descent line = hypotenuse), $\tan \theta =\text{opposite}/\text{adjacent}=\text{horizontal}/\text{altitude}$. That ratio is the cone's slope, which is what we want to cap. A smaller ${\theta }_{\text{gs}}$ makes a narrower, steeper funnel — safer against ridges but less room to maneuver.

Step — differentiate the log (WHAT): by the chain rule, $\dot{z}=\frac{d}{dt}\ln m=\frac{\dot{m}}{m}.$ Step — substitute the mass law (WHAT): $\dot{z}=\frac{-\alpha \Vert \mathbf{T}\Vert }{m}=-\alpha \frac{\Vert \mathbf{T}\Vert }{m}=-\alpha \Gamma .$ WHY the log works: the acceleration law $\dot{\mathbf{v}}=\mathbf{T}/m$ has $m$ in the denominator — that $1/m$ is the nonlinear coupling we must kill. The mass law's ratio $\dot{m}/m$ is exactly the chain-rule derivative of $\ln m$. Feeding in $\dot{m}=-\alpha \Vert \mathbf{T}\Vert$, the $m$'s combine into the clean control $\Gamma =\Vert \mathbf{T}\Vert /m$, and we get $\dot{z}=-\alpha \Gamma$: constant times control, no state in a denominator. WHY $z=1/m$ fails (the concrete alternative): try it. $\dot{z}=\frac{d}{dt}{m}^{-1}=-m^{-2}\dot{m}=-m^{-2}(-\alpha \Vert \mathbf{T}\Vert )=\alpha \Vert \mathbf{T}\Vert /m^2$. This still carries $m^{-2}$ and $\Vert \mathbf{T}\Vert$ separately — you cannot fold them into the single control $\Gamma =\Vert \mathbf{T}\Vert /m$ without a leftover $1/m$. So $z=1/m$ leaves a state-dependent factor and the dynamics stay nonlinear. Only the log produces the ratio $\dot{m}/m$ that the control $\Gamma$ already matches — that is the sense in which the log is uniquely clean.

Setup: at any instant, the constraints touching $\Gamma$ are $\Gamma \ge \Vert \mathbf{a}\Vert$ and $\Gamma \ge {\Gamma }_{\min }(z)$ (throttle floor). The cost pays $+\Gamma$ per unit time. Argument (WHY $\Gamma$ hugs $\Vert \mathbf{a}\Vert$): suppose at some instant $\Gamma >\max (\Vert \mathbf{a}\Vert ,{\Gamma }_{\min })$ strictly. Then we could lower $\Gamma$ by a tiny amount $\epsilon$ without violating either lower bound. This reduces the cost integral by $\epsilon dt$ and — since $\Gamma$ enters the dynamics only through $\dot{z}=-\alpha \Gamma$ — also saves mass (raises final $z$), which is only better. So any strictly-slack $\Gamma$ is suboptimal: the optimizer presses $\Gamma$ down onto the tighter of its two floors. Where the floor is inactive (${\Gamma }_{\min }<\Vert \mathbf{a}\Vert$), the binding floor is $\Vert \mathbf{a}\Vert$, hence $\Gamma =\Vert \mathbf{a}\Vert$ — the exact engine physics is recovered. That's why relaxing "costs nothing" = lossless. How Pontryagin closes the last gap (sketch, not just a citation): the hand-wave above assumes we may freely nudge $\Gamma$ down at a single instant. But $\Gamma$ is coupled to the state through the dynamics, so a purely local nudge could in principle be blocked. The Pontryagin Minimum Principle handles this properly: it forms the Hamiltonian $H=\Gamma +\mathbf{\lambda }\cdot (\text{dynamics})$ and asks which $(\mathbf{a},\Gamma )$ minimizes $H$ pointwise along the optimal path. Working that minimization out shows the coefficient multiplying $\Gamma$ is strictly positive (it is $1$ plus a mass-costate term that stays positive), so $H$ is minimized by taking $\Gamma$ as small as its lower bounds allow — i.e. $\Gamma =\max (\Vert \mathbf{a}\Vert ,{\Gamma }_{\min })$ everywhere, ruling out any interval where $\Gamma$ floats strictly above. That positivity is the "no pathological case" guarantee.

Forward Euler = "new value = old value + step $\times$ rate." Applying to each continuous law: ${\mathbf{r}}_{k+1}={\mathbf{r}}_k+h{\mathbf{v}}_k$ ${\mathbf{v}}_{k+1}={\mathbf{v}}_k+h({\mathbf{a}}_k+\mathbf{g})$ $z_{k+1}=z_k-h\alpha {\Gamma }_k$ Per-node constraints (the full list): $\Vert {\mathbf{a}}_k\Vert \le {\Gamma }_k\text{(second-order cone — this is the relaxed thrust magnitude)}$ ${\Gamma }_k\ge {\Gamma }_{\min }(z_k)\text{(throttle floor — recovers the }{T}_{\min }\text{ inner circle)}$ ${\Gamma }_k\le {\Gamma }_{\max }(z_k)\text{(throttle ceiling — the }{T}_{\max }\text{ outer circle)}$ $\Vert [r_{x,k},r_{y,k}]\Vert \le \tan {\theta }_{\text{gs}}(r_{z,k}-r_{z,\text{land}})\text{(glide-slope cone)}$ Without the floor and ceiling lines, the discrete model would only cap thrust from above via the cone and lose the annulus entirely — so both are mandatory. Convexity check: the three Euler updates are linear equalities (flat sets → convex). ${\Gamma }_{\min },{\Gamma }_{\max }$ are the convexified (Taylor) bounds from the parent note, so those two are convex. The two norm constraints are second-order cones — convex. Hence the whole discretized problem is an SOCP solvable by Interior-Point Methods. This is the bridge from continuous physics to what Convex Optimization / Second-Order Cone Programming solvers eat.

Step — undo the log (WHY): $z=\ln m$, so $m=e^z$. Final mass: $m(t_f)=e^{6.7}\approx 812.41\text{kg}.$ Initial mass $m(0)=e^{\ln 1000}=1000$ kg. Fuel burned: $\Delta m=1000-812.41\approx 187.59\text{kg}.$ Step — recover the objective: integrating $\dot{z}=-\alpha \Gamma$ over $[0,t_f]$: $z(t_f)-z(0)=-\alpha {\int }_0^{t_f}\Gamma dt\Rightarrow {\int }_0^{t_f}\Gamma dt=\frac{z(0)-z(t_f)}{\alpha }.$ ${\int }_0^{t_f}\Gamma dt=\frac{\ln 1000-6.7}{5\times 10^{-4}}=\frac{6.9078-6.7}{5\times 10^{-4}}\approx 415.5\text{(m/s)}.$ Cross-check: $\int \Gamma dt=(1/\alpha )\ln (m_0/m_f)$ is exactly the Tsiolkovsky $\Delta v$ ideal budget — reassuring that "minimize $\int \Gamma dt$" literally minimizes velocity-change cost.

Let up be positive. Net vertical acceleration under commanded up-thrust $\Gamma$ is $a=\Gamma -g$.

Strategy A — constant thrust $\gamma$: we must kill $30$ m/s of descent over $200$ m of fall. The net decelerating acceleration is $\gamma -g$, and $|v_z{|}^2=2(\gamma -g)d$: $\gamma -g=\frac{|v_z{|}^2}{2d}=\frac{30^2}{2\times 200}=2.25\Rightarrow \gamma =2.25+3.71=5.96{\text{m/s}}^2.$ Descent time $t_A=|v_z|/(\gamma -g)=30/2.25=13.33$ s. Fuel proxy $=\int \Gamma dt=\gamma t_A=5.96\times 13.33\approx 79.47\text{(m/s)}$.

Strategy B — bang-bang. During coast $\Gamma =2$: net accel $a_1=2-3.71=-1.71$ (still speeding downward). During brake $\Gamma =14$: net accel $a_2=14-3.71=10.29$ (decelerating). Let coast cover distance $d_1$ (speed grows from 30 to $v_1$), brake covers $d_2=200-d_1$ (speed $v_1\to 0$). Coast: $v_1^2=30^2+2(1.71)d_1$ (speed magnitude increases). Brake: $0=v_1^2-2(10.29)d_2\Rightarrow v_1^2=2(10.29)(200-d_1)$. Set equal: $900+3.42d_1=20.58(200-d_1)=4116-20.58d_1$. $24d_1=3216\Rightarrow d_1=134.0\text{m},d_2=66.0\text{m}.$ Then $v_1^2=900+3.42(134.0)=1358.3\Rightarrow v_1=36.86$ m/s. Times: coast $t_1=(v_1-30)/1.71=6.86/1.71=4.01$ s. Brake $t_2=v_1/10.29=36.86/10.29=3.58$ s. Fuel proxy $={\Gamma }_1{t}_1+{\Gamma }_2{t}_2=2(4.01)+14(3.58)=8.02+50.17\approx 58.19\text{(m/s)}$.

Verdict: $58.19<79.47$ — bang-bang wins, saving $\approx 21$ units of fuel proxy ($\approx 27\%$). This is exactly the "coast-then-suicide-burn" that G-FOLD rediscovers, and why SpaceX Falcon 9 Landing and Mars EDL delay ignition. Burning early wastes propellant on gravity loss (holding altitude against $g$ for longer).

Idea: Model Predictive Control re-plans repeatedly. At each control tick you measure the actual state $(\mathbf{r},\mathbf{v},m)$, then re-solve the G-FOLD SOCP from that state to the target, apply only the first thrust command, discard the rest, and repeat next tick. Why convexity is the enabler: each re-solve is an SOCP with a unique global optimum reached by Interior-Point Methods in guaranteed bounded time — so within the ~1 s budget you always get a certifiably optimal, feasible plan, never a stuck local minimum. A non-convex re-solve could return garbage or time out mid-descent; convexity is what makes the closed loop safe to fly. This is the design lineage from Apollo Lunar Descent Guidance (open-loop polynomial) to modern re-solving guidance.