Reaction control system — thruster selection, plume impingement limits: Exercises

Before we start, one shared cast of symbols (earned in the parent, restated so line one is readable):

Definition Symbols you will reuse

$r_{i}$ — the position of thruster $i$ measured from the center of mass (CoM), an arrow drawn from the balance-point of the ship to the nozzle.
$\hat{u}_{i}$ — the unit thrust direction: which way the push on the ship points (opposite to where gas leaves). "Unit" means length exactly $1$ , so it carries direction only.
$F_{i} \geq 0$ — the thrust magnitude. A jet can only push, never suck, so it is never negative.
$τ_{i} = r_{i} \times F_{i}$ — the torque (twist) from thruster $i$ . The symbol $\times$ is the cross product: it takes lever-arm and force and returns the axis the body spins about.
$w = [F; τ]$ — the wrench, six numbers stacking net force (3) and net torque (3).
$B$ — the allocation matrix; column $i$ is thruster $i$ 's unit wrench, so $w = B f$ .
$P_{imp} (d, θ) \approx P_{0} (d_{0} / d)^{2} cos^{k} θ$ — the plume pressure a surface feels at distance $d$ and off-axis angle $θ$ .
$θ$ — the off-axis angle, measured between the plume centerline and the line from the nozzle to the surface. It runs from $0$ (dead-center) up to $9 0^{\circ}$ at the edge of the forward hemisphere; beyond $9 0^{\circ}$ a surface sits behind the exit and receives essentially no plume, so we clamp the model there (see the domain note in Ex 5.2).
$α$ — the surface-tilt angle, measured between the incoming flow direction and the surface normal-flip — i.e. how square-on the patch faces the flow. $α = 0$ means the patch faces the plume head-on (maximum force); $α = 9 0^{\circ}$ means the flow grazes edge-on (no push). It enters the impingement force as $cos α$ .

The three figures below are not decoration — each one carries a derivation step, and every exercise points you back to the arrow or curve you need. Read them in order.

Figure — Reaction control system — thruster selection, plume impingement limits

Figure s01 is the picture behind all of Level 2. The yellow dot is the CoM; the blue arrow is the lever arm $r$ ; the red arrow is the force $F$ . Look at how the red force sits offset from the yellow dot — that offset is exactly what the cross product converts into the green curved spin arrow $τ$ . When you do Ex 2.1, you are computing that green arrow.

Level 1 — Recognition

Exercise 1.1

A single thruster fires with its push direction $\hat{u}_{i}$ pointing exactly through the CoM (so $r_{i}$ is parallel to $\hat{u}_{i}$ ). Does it produce any torque? Explain in one line using the cross product.

Recall Solution

No torque. The torque is $τ_{i} = F_{i} (r_{i} \times \hat{u}_{i})$ . The cross product measures the perpendicular part of the lever arm — the sine of the angle between $r_{i}$ and $\hat{u}_{i}$ . When they are parallel that angle is $0$ and $sin 0 = 0$ , so $r_{i} \times \hat{u}_{i} = 0$ . All the effort becomes pure translation (the ship drifts, doesn't spin). Picture Figure s01 with the red arrow rotated to point straight through the yellow dot — the green spin arrow collapses to nothing.

Exercise 1.2

The allocation matrix $B$ for a spacecraft has size $6 \times N$ . What does $6$ mean, what does $N$ mean, and what does a single column of $B$ physically represent?

Recall Solution

$6$ = the six degrees of freedom of a rigid body: 3 translation directions + 3 rotation axes. That is why the wrench $w$ has six entries.
$N$ = the number of thrusters on the ship.
Each column is one thruster's unit wrench $[\hat{u}_{i}; r_{i} \times \hat{u}_{i}]$ : the force and torque that thruster $i$ makes when fired at magnitude $1$ . Multiplying $B f$ just scales each column by its on-time $F_{i}$ and adds them.

Level 2 — Application

Exercise 2.1

A thruster is mounted at $r = (0, 0.5, 0)$ m and pushes in direction $\hat{u} = (1, 0, 0)$ with magnitude $F = 20$ N. Compute the force vector and the torque vector. (This is exactly the geometry drawn in Figure s01.)

Recall Solution

Force: $F = F \hat{u} = 20 (1, 0, 0) = (20, 0, 0)$ N — the red arrow in Figure s01.

Torque: we need the cross product because torque is lever arm perpendicular to force. $r \times F = (0, 0.5, 0) \times (20, 0, 0) .$ Using the determinant rule $a \times b = (a_{y} b_{z} - a_{z} b_{y}, a_{z} b_{x} - a_{x} b_{z}, a_{x} b_{y} - a_{y} b_{x})$ :

$x$ : $0.5 \cdot 0 - 0 \cdot 0 = 0$
$y$ : $0 \cdot 20 - 0 \cdot 0 = 0$
$z$ : $0 \cdot 0 - 0.5 \cdot 20 = - 10$

$τ = (0, 0, - 10) N\cdotm .$ The push is along $+ x$ , the arm is along $+ y$ , so the twist is about $- z$ (right-hand rule: fingers from arm to force curl toward $- z$ ) — that is the green curved arrow in Figure s01.

Exercise 2.2

Using the plume model $P_{imp} = P_{0} (d_{0} / d)^{2} cos^{k} θ$ with $P_{0} = 500$ Pa at $d_{0} = 0.1$ m and $k = 4$ : what pressure hits a surface on axis ( $θ = 0$ ) at distance $d = 1$ m?

Recall Solution

On axis, $cos^{4} 0 = 1$ , so the whole angular factor is $1$ (worst case, dead-center). $P_{imp} = 500 (\frac{0.1}{1})^{2} \cdot 1 = 500 \times 0.01 = 5 Pa .$ Why $1/ d^{2}$ ? The same amount of gas spreads over ever-larger spherical caps as it flies out; area grows like $d^{2}$ , so pressure per area falls like $1/ d^{2}$ . This is the blue curve in Figure s03 read at $d = 1$ m.

Level 3 — Analysis

Exercise 3.1 — the pure roll couple, both twist the same way

Two thrusters sit on a wheel of radius $r = 0.5$ m: $r_{1} = (0, + r, 0)$ firing $\hat{u}_{1} = + \hat{x}$ , and $r_{2} = (0, - r, 0)$ firing $\hat{u}_{2} = - \hat{x}$ , each at $F = 20$ N. Show the net force is zero and find the net torque. (Figure s02 draws both thrusters and the shared spin.)

Recall Solution

Net force — because opposite directions cancel: $F = 20 (1, 0, 0) + 20 (- 1, 0, 0) = (0, 0, 0) N .$ Zero drift: this is a couple. In Figure s02, notice the two red force arrows point opposite ways — that is why they sum to nothing.

Torque of thruster 1: $τ_{1} = (0, 0.5, 0) \times (20, 0, 0) = (0, 0, - 10)$ N·m (same computation as 2.1).

Torque of thruster 2: $τ_{2} = (0, - 0.5, 0) \times (- 20, 0, 0)$ .

$z$ : $0 \cdot 0 - (- 0.5) (- 20) = - 10$ .

$τ_{2} = (0, 0, - 10)$ N·m. Both twist the same way about $- z$ — the flipped arm and the flipped force multiply to the same sign. That is the single green spin arrow in Figure s02: both thrusters feed it.

Net: $τ = (0, 0, - 20)$ N·m $= 2 r F$ about $- z$ . Pure torque, zero drift — the ideal attitude actuator.

Figure s02 is the picture of a couple: two blue arms, two opposing red forces, and one green spin arrow they both drive. Whenever you need "torque with zero drift," this is the shape to reach for.

Exercise 3.2 — how far must a panel be to be safe?

The panel limit is $P_{m a x} = 1$ Pa. On axis ( $θ = 0$ ), with $P_{0} = 500$ Pa, $d_{0} = 0.1$ m, find the minimum safe distance $d_{m i n}$ where $P_{imp}$ drops to exactly $1$ Pa.

Recall Solution

Set $P_{imp} = P_{m a x}$ and solve for $d$ . Because pressure falls as $1/ d^{2}$ , doubling distance quarters pressure, so we invert: $1 = 500 (\frac{0.1}{d})^{2} \Rightarrow (\frac{0.1}{d})^{2} = \frac{1}{500} .$ $\frac{0.1}{d} = \frac{1}{500} \Rightarrow d = 0.1 500 = 0.1 \times 22.36 \approx 2.236 m .$ Any surface closer than $\approx 2.24$ m on axis is over the limit. This distance is exactly the radius of the keep-out cone on axis — it is the yellow dot where the blue curve crosses the red dashed limit line in Figure s03.

Figure s03 plots the on-axis plume pressure (blue) against distance on a log scale, with the $1$ Pa panel limit as the red dashed line. Where they cross (yellow dot, green dotted line) is $d_{m i n} \approx 2.24$ m from Ex 3.2 — everything left of that line is forbidden.

Level 4 — Synthesis

Exercise 4.1 — inhibit or allow?

During docking a target sits $d = 1.5$ m ahead on the $+ x$ axis. A forward thruster's plume centerline points straight at it, but the sensitive patch (a camera window) lies $θ = 3 0^{\circ}$ off that centerline. Use $P_{0} = 500$ Pa, $d_{0} = 0.1$ m, $k = 4$ , limit $P_{m a x} = 1$ Pa. Is this thruster allowed?

Recall Solution

We need both the distance factor and the angular factor, because the patch is off-axis. Note $θ = 3 0^{\circ}$ is safely inside the model's valid domain $0 \leq θ \leq 9 0^{\circ}$ , so $cos θ > 0$ and the formula applies directly.

Distance factor: $(d_{0} / d)^{2} = (0.1/1.5)^{2} = (0.0667)^{2} = 0.004444$ .

Angular factor: $cos^{4} 3 0^{\circ}$ . Since $cos 3 0^{\circ} = 3 /2 \approx 0.8660$ , $cos^{4} 3 0^{\circ} = (0.8660)^{4} \approx 0.5625.$

Pressure: $P_{imp} = 500 \times 0.004444 \times 0.5625 \approx 1.25 Pa .$ Since $1.25 > 1.0$ , the thruster is inhibited ( $F_{i}$ locked to $0$ ). The optimizer must re-solve $B f = w_{cmd}$ over the remaining thrusters, likely an off-axis pair whose plumes clear the window — at a small propellant cost, since hardware safety outranks fuel.

Exercise 4.2 — the hidden disturbance torque

The plume in 4.1 (before inhibition) delivers a force on the camera window (area $A = 0.02$ m², facing the flow head-on so the surface-tilt angle is $α = 0$ and $cos α = 1$ ). If that window sits at $r_{s} = (1.5, 0.3, 0)$ m from the CoM, estimate the disturbance torque magnitude about $z$ that this impingement adds. Use $P_{imp} = 1.25$ Pa, and assume the impingement force points in $+ x$ .

Recall Solution

Recall $α$ (from the symbol list): it is the tilt of the patch relative to the flow. Here the window faces the plume squarely, $α = 0$ , so $cos α = 1$ and the patch catches the full pressure. If it were edge-on ( $α = 9 0^{\circ}$ ) the flow would graze past and $cos α = 0$ — no force.

Impingement force: $F_{imp} = P_{imp} A cos α = 1.25 \times 0.02 \times 1 = 0.025$ N, in $+ x$ .

Disturbance torque — a force off the CoM makes a twist, so again the cross product: $τ_{dist} = r_{s} \times F_{imp} = (1.5, 0.3, 0) \times (0.025, 0, 0) .$ $z$ -component: $r_{x} F_{y} - r_{y} F_{x} = 1.5 \cdot 0 - 0.3 \cdot 0.025 = - 0.0075$ N·m. $∣ τ_{dist, z} ∣ = 0.0075 N\cdotm .$ Small in isolation — but this is a torque you never commanded. It fights your attitude loop and, over a long burn, must be counter-fired. That is exactly the "hidden disturbance torque" warning: the plume is not just a heat threat, it is a control error.

Level 5 — Mastery

Exercise 5.1 — full selection with a plume veto

A ship must produce a pure roll torque $τ_{z} = - 20$ N·m and zero net force. Four candidate thrusters (each max $20$ N, arm $r = 0.5$ m) can make this torque:

Pair A: ${$ thruster at $(0, + 0.5, 0)$ pushing $+ \hat{x}$ , thruster at $(0, - 0.5, 0)$ pushing $- \hat{x}}$ .
Pair B: same geometry but $3 0^{\circ}$ toward a nearby deployed panel.

Pair A's plume clears everything. Pair B's plume hits the panel with on-axis pressure $P = 500 (0.1/0.4)^{2} = 31.25$ Pa (panel limit $1$ Pa). Both pairs, fired at $F = 20$ N, give the same $- 20$ N·m couple. Which pair does the minimum-propellant selector choose, and why is Pair A the only feasible choice?

Recall Solution

Both pairs achieve the command: each is a couple giving $τ = (0, 0, - 2 r F) = (0, 0, - 2 \cdot 0.5 \cdot 20) = (0, 0, - 20)$ N·m with zero net force (this is the Figure s02 shape). On propellant alone ( $\sum F_{i} = 40$ N each) they tie.

The plume constraint breaks the tie. Pair B's on-axis pressure is $P_{B} = 500 (0.1/0.4)^{2} = 500 \times 0.0625 = 31.25 Pa ≫ 1 Pa (limit) .$ That is a massive violation — over $31 \times$ the panel limit — so Pair B is infeasible: its thrusters are inhibited before the optimizer even sees them. Pair A ( $P$ clears all limits) is the only pair left in the feasible set.

Selector's answer: Pair A. The minimum-propellant linear program only searches over feasible (non-inhibited) thrusters, so it returns the roll couple that respects every keep-out cone — even when a violating option would have used identical fuel. Safety filters the search space first, then propellant optimizes within it.

Exercise 5.2 — designing the keep-out half-angle (and its domain)

For pair B's geometry we want to find how far off-axis a surface must sit to be safe at $d = 0.4$ m, given $P_{0} = 500$ Pa, $d_{0} = 0.1$ m, $k = 4$ , limit $1$ Pa. Find the keep-out half-angle $θ_{KO}$ : the angle at which pressure just drops to $1$ Pa. Then state what the model does for $θ > 9 0^{\circ}$ .

Recall Solution

We solve $P_{imp} = 1$ Pa for the angle, holding $d = 0.4$ m fixed. The distance factor is $(0.1/0.4)^{2} = 0.0625$ , so: $1 = 500 \times 0.0625 \times cos^{4} θ_{KO} = 31.25 cos^{4} θ_{KO} .$ $cos^{4} θ_{KO} = \frac{1}{31.25} = 0.032 \Rightarrow cos θ_{KO} = 0.03 2^{1/4} .$ $0.03 2^{1/4} = (0.032)^{0.25} \approx 0.4229 \Rightarrow θ_{KO} = arccos (0.4229) \approx 65. 0^{\circ} .$ Interpretation: any sensitive surface within $6 5^{\circ}$ of the centerline at that range is over the limit — a very wide keep-out cone. The panel at $3 0^{\circ}$ is deep inside it, confirming 5.1's veto. To shrink the cone you either move the thruster farther (bigger $d$ ) or angle its plume away from the panel.

Domain of $θ$ — the edge case. The formula $cos^{k} θ$ is only physical for $0 \leq θ \leq 9 0^{\circ}$ , the forward hemisphere the plume actually fills. At $θ = 9 0^{\circ}$ , $cos^{4} 9 0^{\circ} = 0$ : a surface exactly to the side sees no plume. For $θ > 9 0^{\circ}$ the naive formula would give a negative base ( $cos θ < 0$ ) — and with an even exponent $k = 4$ it would wrongly turn positive again, predicting plume pressure behind the nozzle where there is none. The physical fix: clamp the model, $P_{imp} (d, θ) \equiv 0$ for $θ > 9 0^{\circ}$ . A surface behind the exit plane is simply outside the plume — so it is always safe on this axis, and the keep-out cone never wraps past $9 0^{\circ}$ .