Powered descent guidance — G-FOLD algorithm (convex optimization): Question bank

Before we start, two words you must have pinned (everything below leans on them):

Recall Convex set / convex function (one-line refresher)

A convex set is one where the straight line between any two points inside it stays inside. A convex function is bowl-shaped: it has one lowest point, no false valleys. See Convex Optimization and Second-Order Cone Programming. The whole point of G-FOLD is to make both the constraints (sets) and the cost (function) convex so a solver hits the global minimum.

Definition Every symbol on this page, defined once

These are the exact quantities from the parent note — pin them before reading any question.

$r$ — position vector of the lander, measured in a fixed target-centred frame where $r_{z}$ is height above the ground and $r_{x}, r_{y}$ are the two horizontal offsets from the target. $r_{z, land}$ is the height of the landing surface (usually $0$ ).
$v = \dot{r}$ — velocity vector (rate of change of position).
$m$ — mass of the lander; $z = ln m$ is the log-mass.
$T$ — the thrust vector we command from the engine; $∥ T ∥$ is its magnitude (push in newtons).
$g$ — the constant gravitational acceleration vector near the planet's surface; its magnitude $g$ points down (e.g. $g \approx 3.71 m/s^{2}$ on Mars).
$T_{m i n}, T_{m a x}$ — the engine's thrust floor and ceiling (newtons): it physically cannot push softer than $T_{m i n}$ or harder than $T_{m a x}$ .
$a = T / m$ — the acceleration command, thrust per unit mass (units $m/s^{2}$ ); this is the new control variable after dividing the dynamics by $m$ .
$Γ$ — the slack variable, a stand-in for $∥ T ∥/ m$ (the acceleration-magnitude), with $∥ a ∥ \leq Γ$ . Fuel cost is $\int Γ d t$ .
$α = 1/ v_{e}$ — the mass-consumption coefficient, the reciprocal of exhaust speed $v_{e} = I_{s p} g_{0}$ ; it converts thrust into mass-burn rate via $\overset{m}{˙} = - α ∥ T ∥$ . Bigger $α$ (slower exhaust) ⇒ more fuel burned per newton.
$θ_{g s}$ — the glide-slope angle defining the safe descent cone; $t_{f}$ — the total flight time.

True or false — justify

The mass equation

\overset{m}{˙} = - α ∥ T ∥

says a lander loses mass even with engines off.

False. With engines off

∥ T ∥ = 0

, so

\overset{m}{˙} = 0

— no exhaust, no mass loss. Mass only drops when thrust is nonzero. (

α = 1/ v_{e} > 0

is a fixed constant, so it never causes loss on its own.)

Minimizing total thrust

\int ∥ T ∥ d t

is the same as minimizing fuel.

True for the true thrust, but the trap is minimizing thrust pointwise (keeping it low everywhere). A long, gentle burn fighting gravity for longer burns more total propellant than a short hard burn — gravity loss. Fuel is the integral, not the peak.

The constraint

∥ T ∥ \leq T_{m a x}

(upper bound) is what makes the problem non-convex.

False. The inside of a sphere is convex, so the upper thrust-ceiling bound is fine. It is the lower bound

∥ T ∥ \geq T_{m i n}

(the thrust floor) — the region outside a sphere — that is non-convex.

After lossless convexification the solver might return a solution with

Γ > ∥ a ∥

(slack unused).

False at the optimum. The cost

\int Γ d t

directly penalizes

Γ

, so the optimizer squeezes

Γ

down until it equals

∥ a ∥

. Any interior point with

Γ > ∥ a ∥

is strictly worse and gets rejected — that is the Pontryagin Minimum Principle guarantee.

The change of variable

z = ln m

makes the cost nonlinear.

False. It makes the mass dynamics linear (

\overset{z}{˙} = - α Γ

). The cost

\int Γ d t

is linear in

Γ

and is untouched by the log substitution.

Fixing flight time

t_{f}

before solving throws away optimality.

False. For each fixed

t_{f}

we get the exact convex optimum; we then sweep

t_{f}

with an outer 1-D search. The fuel-vs-

t_{f}

curve is unimodal, so the sweep finds the true best

t_{f}

without breaking convexity.

A convex problem can still have several equally-good landing solutions.

True. Convex means no worse local minima, but the set of optimal points can be flat (a tie). What is guaranteed is that every returned optimum has the same, globally-best cost — you never get stuck in an inferior valley.

The glide-slope constraint

∥ [r_{x}, r_{y}] ∥ \leq tan θ_{g s} (r_{z} - r_{z, land})

is non-convex because of the

tan

False.

tan θ_{g s}

is just a fixed positive number (a slope), and

r_{x}, r_{y}, r_{z}

are the position components in the target frame. The constraint is a norm

\leq

a linear function — a convex second-order cone. It slots into the SOCP for free.

Spot the error

"We relaxed

∥ a ∥ = Γ

∥ a ∥ \leq Γ

, so the model no longer represents the real engine."

The error: relaxation is only loose off the optimum. At the optimum the inequality is active (

∥ a ∥ = Γ

), recovering the true engine physics exactly. Nothing is lost — hence lossless.

Γ

is defined as

∥ T ∥/ m

, so it is just a fixed function of the state — not a real variable."

The error:

∥ T ∥

is the control we are choosing, so

Γ

(the acceleration-magnitude slack) is a genuine decision variable. After convexification

Γ

is only linked to

a

by the inequality

∥ a ∥ \leq Γ

, giving the optimizer freedom that the proof then pins down.

"Divide

m \dot{v} = T + m g

m

and you still have

m

in the equation, so it's still nonlinear."

The error: dividing gives

\dot{v} = a + g

where

a = T / m

is the new control and

g

is the constant gravity vector. The

m

is absorbed into the definition of

a

; the equation is now linear in the chosen variables.

"The Taylor expansion of

T_{m i n} e^{- z}

is an approximation, so G-FOLD's answer is only approximate."

Partly a trap. The floor

T_{m i n} e^{- z}

is expanded about the reference log-mass

z_{0}

to second order:

T_{m i n} e^{- z_{0}} [1 - (z - z_{0}) + \frac{1}{2} (z - z_{0})^{2}]

. Because

e^{- z}

is convex, its second-order Taylor polynomial is an upward parabola — itself convex — so the constraint

Γ \geq (parabola)

stays convex. It is tight near

z_{0}

, exactly where the true trajectory's mass lives, and can be re-referenced to tighten.

"Minimum-fuel descent should throttle gently the whole way down for a smooth landing."

The error: minimum-fuel is typically bang-bang — coast/min-thrust, then a hard full-thrust brake near the end (the "suicide burn"). Smooth gentle throttling burns extra fighting gravity longer.

"Because interior-point methods are iterative, G-FOLD cannot guarantee the global optimum."

The error: for a convex program, interior-point methods converge to the global optimum in guaranteed polynomial time. Convexity, not the iteration, is what supplies the global guarantee.

Why questions

Why take the logarithm of mass instead of just using

m

Because

\overset{m}{˙} / m

is the log-derivative

\frac{d}{d t} ln m

. Setting

z = ln m

turns the messy

1/ m

mass dynamics into the perfectly linear

\overset{z}{˙} = - α Γ

(with

α = 1/ v_{e}

) — logs convert division into subtraction.

Why is a global optimum (not just a good local one) so important during descent?

You have roughly one second to compute a trajectory before committing. A non-convex solver could return a false valley or fail to converge; a convex solve reliably returns the single best answer every time, which safety demands.

Why does the fuel cost "want"

Γ

small, and why does that fact matter?

The cost is

\int Γ d t

, so smaller

Γ

(the acceleration-magnitude slack) means less fuel. That downward pressure is exactly what drives

Γ

onto its lower bound

∥ a ∥

, making the lossless relaxation collapse back to the true equality.

Why do we search over

t_{f}

from the outside instead of optimizing it inside?

Letting

t_{f}

vary inside makes the discretized dynamics bilinear (products of unknowns), destroying convexity. Kept outside as a 1-D sweep over a unimodal curve, convexity of each inner solve is preserved.

Why is the fuel-versus-flight-time curve unimodal (one minimum)?

Too short a

t_{f}

demands violent, near-max thrust to brake in time — heavy fuel burn (and eventually infeasibility); too long a

t_{f}

wastes propellant hovering against gravity for extra seconds. Fuel falls then rises around one sweet spot, so the curve has a single dip — safe to bracket with a 1-D search. (It is verified numerically per problem rather than proven in closed form.)

Why can't we just cap thrust and forget the lower bound

T_{m i n}

Real engines physically cannot throttle below a floor

T_{m i n}

without flaming out. Ignoring it gives a solution the hardware can't execute; convexification lets us keep the floor while staying convex.

Why is minimizing fuel equivalent to maximizing final mass?

Fuel spent equals initial mass minus final mass, and initial mass is fixed. So less fuel burned ⇔ more mass left at touchdown — the two objectives point the same way.

Why does G-FOLD reproduce Apollo/Falcon-style "burn late" behaviour without being told to?

It falls out of minimizing

\int Γ d t

under the dynamics — delaying thrust cuts the time spent fighting gravity (gravity loss). See Apollo Lunar Descent Guidance and SpaceX Falcon 9 Landing.

Edge cases

What if the available deceleration is less than gravity (

T_{m a x} / m < g

Here

T_{m a x}

is the engine ceiling and

g

the gravity magnitude; the best upward acceleration

T_{m a x} / m

can't overcome the downward

g

, so the lander cannot arrest its fall — the problem is infeasible. G-FOLD reports infeasibility rather than a fake landing, the correct safe outcome.

What happens at the instant thrust is exactly

T_{m i n}

(engine at its floor)?

The throttle-box lower constraint

Γ \geq T_{m i n} e^{- z}

becomes active. This is a valid, feasible edge of the convex box — the solver handles it like any active constraint, no special casing.

What if the target is directly below the lander (purely vertical descent)?

The glide-slope cone degenerates:

∥ [r_{x}, r_{y}] ∥ \leq tan θ_{g s} (r_{z} - r_{z, land})

is trivially satisfied since horizontal offset is ~0. The problem reduces to the 1-D vertical landing — a valid limiting case.

What does

θ_{g s} \to 9 0^{\circ}

mean for the glide-slope cone?

tan 9 0^{\circ} \to \infty

, so the horizontal-offset bound becomes unlimited — the cone opens fully and imposes no restriction. Physically, "any approach angle is allowed."

What does

θ_{g s} \to 0^{\circ}

mean?

tan 0^{\circ} = 0

forces

∥ [r_{x}, r_{y}] ∥ \leq 0

, i.e. purely vertical descent along the target axis. The cone collapses to a line — extremely restrictive but still convex.

What if the initial velocity is already zero at the correct altitude and position?

The trajectory is already a soft pinpoint state; the minimum-fuel solution needs only a hover-then-touch or immediate zero divert, and the cost integral is essentially the propellant to hold against gravity to the ground.

What if two different flight times

t_{f}

give nearly identical fuel?

The unimodal fuel curve is flat near its minimum, so several

t_{f}

tie. Any of them is acceptable; pick the one giving extra margin (e.g. slightly more time for actuator response).

Recall One-line summary of the whole trap set

Non-convexity lives only in the thrust lower bound; $z = ln m$ linearizes mass; the $Γ$ -relaxation is lossless because cost pushes $Γ = ∥ a ∥$ ; $t_{f}$ stays outside; the optimum is bang-bang; and every degenerate cone/thrust case still lands inside a convex, globally-solvable program.