Worked examples — Reaction control system — thruster selection, plume impingement limits

This page is the hands-on lab for the parent topic. We march through a matrix of every case class the topic can throw at you — signs, degenerate geometries, limiting distances, a real docking word problem, and an exam-style twist — and solve each one from scratch.

Before line one, four plain-word reminders so no symbol arrives unearned:

If any of $\times$ or "$1/d^2$" feels shaky, the parent note builds them; here we only use them.

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The scenario matrix

Every worked example below is tagged with the cell it fills. Together they cover the whole table.

Cell	Case class	What makes it tricky	Example
A	Pure torque, single axis	signs of the cross product	Ex 1
B	Off-CoM single thruster	torque and unwanted drift	Ex 2
C	Quadrant sweep of $\mathbf{r}$	which sign of ${\tau }_z$ per quadrant	Ex 3
C0	Boundary: $r_y=0$ (on the $x$-axis)	force line through CoM ⇒ zero roll	Ex 3b
D	Degenerate: $\mathbf{r}\parallel \hat{\mathbf{u}}$	zero torque (force through CoM)	Ex 4
E	Plume falloff, on-axis worst case	$1/d^2$ limiting behaviour	Ex 5
F	Plume off-axis, ${\cos }^k\theta$	when a big-$k$ plume becomes safe	Ex 6
G	Real-world word problem: docking	inhibit a thruster, re-allocate	Ex 7
H	Limiting values: $d\to \infty$, $\theta \to 90^{\circ }$	pressure $\to 0$	Ex 8
I	Exam twist: force + torque both commanded	over-determined mini-allocation	Ex 9

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Ex 1 — Pure roll couple (Cell A)

Forecast: guess before computing — will the ship drift? Which way does it spin (about $+z$ or $-z$)?

Read the figure: the two orange arrows are the force vectors ${\mathbf{F}}_1$ (pointing $+x$) and ${\mathbf{F}}_2$ (pointing $-x$); notice they are equal length and opposite, so tip-to-tail they cancel. The purple curved arrow is the resulting spin about the $z$-axis (into the page). The two teal squares are the thruster mounts at ${\mathbf{r}}_1,{\mathbf{r}}_2$.

1. Build each force vector. Using ${\mathbf{F}}_i=F_i{\hat{\mathbf{u}}}_i$: ${\mathbf{F}}_1=20(1,0,0)=(20,0,0)$ and ${\mathbf{F}}_2=20(-1,0,0)=(-20,0,0)$. Why this step? Torque needs the vector force, not just the magnitude, so we assemble ${\mathbf{F}}_i$ first.

2. Net force. Add the two force vectors. $\mathbf{F}=(20,0,0)+(-20,0,0)=0.$ Why this step? A couple must have zero net force or the ship translates while it rotates. This matches the two opposite orange arrows cancelling in the figure.

3. Torque of thruster 1. Use ${\mathbf{\tau }}_1={\mathbf{r}}_1\times {\mathbf{F}}_1$. $(0,1.5,0)\times (20,0,0)=(1.5\cdot 0-0\cdot 0,0\cdot 20-0\cdot 0,0\cdot 0-1.5\cdot 20)=(0,0,-30).$ Why this step? The cross product picks out the twisting part; the third component $-30$ is the roll about $z$. Sign is negative — thumb points into the page (the purple arrow).

4. Torque of thruster 2. $(0,-1.5,0)\times (-20,0,0)=(0,0,(0)(0)-(-1.5)(-20))=(0,0,-30).$ Why this step? We must check the second thruster twists the same way, else torques cancel like the forces did.

5. Add. $\mathbf{\tau }=(0,0,-60)$N·m. Why this step? Torques of a proper couple add; that's the whole point of a couple.

Verify: magnitude $=2rF=2(1.5)(20)=60$N·m ✓, and $\mathbf{F}=0$ ✓. Units: N·m (force × lever). This is the parent note's "ideal attitude control": spin with zero drift.

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Ex 2 — Single off-CoM thruster: torque and drift (Cell B)

Forecast: the parent's "steel-manned mistake" says one thruster is never pure torque. Predict the net force before you read on.

1. Build the force vector. $\mathbf{F}=F\hat{\mathbf{u}}=10(1,0,0)=(10,0,0)$N — nonzero. Why this step? Nothing cancels it, so Newton's second law says the CoM accelerates: drift.

2. Torque. $\mathbf{\tau }=(0,2,0)\times (10,0,0)=(0,0,-20)$N·m. Why this step? Confirms we still get a twist, so this jet does both jobs at once.

Verify: $|\mathbf{\tau }|=rF=2\cdot 10=20$ ✓. Compared to Ex 1 (couple), here $\mathbf{F}\ne 0$ — this is exactly the "one thruster is not a pure torque" trap. Fix: add an opposing jet to kill the $10$N.

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Ex 3 — Quadrant sweep: sign of ${\tau }_z$ everywhere (Cell C)

Forecast: the parent said tan/quadrant fixes matter because a formula can repeat. Here the sign of the torque flips with the quadrant. Guess which corners give negative ${\tau }_z$.

Read the figure: each coloured square is a thruster mount in one quadrant; every arrow points the same way ($+x$), because the force direction is fixed. The colour tells the sign of the resulting roll — orange squares (top half) give ${\tau }_z=-1$, teal squares (bottom half) give ${\tau }_z=+1$. Slide your eye left/right and nothing changes; slide up/down and the sign flips — that is the visual proof that only $r_y$ matters.

Start from the general force vector $\mathbf{F}=F\hat{\mathbf{u}}=(F,0,0)$, so $F_x=F$ and $F_y=0$. For $\mathbf{r}=(r_x,r_y,0)$ the $z$-component of the cross product is: ${\tau }_z=r_x{F}_y-r_y{F}_x.$ In our scenario $F_y=0$ and $F_x=F$, so the first term dies and this collapses to ${\tau }_z=r_x\cdot 0-r_y\cdot F=-r_{y}F.$ Why this step? We keep the general two-term formula visible, then substitute $F_y=0$ so the reader sees exactly why only the $-r_{y}F$ term survives: ${\tau }_z$ depends only on $r_y$, the lever arm perpendicular to the $x$-force.

1. Q I $\mathbf{r}=(1,1,0)$: ${\tau }_z=-1\cdot 1=-1$ (clockwise, into page).
2. Q II $\mathbf{r}=(-1,1,0)$: ${\tau }_z=-1\cdot 1=-1$ — same sign! Why? ${\tau }_z$ ignores $r_x$; both top corners give $-1$.
3. Q III $\mathbf{r}=(-1,-1,0)$: ${\tau }_z=-(-1)\cdot 1=+1$ (counter-clockwise).
4. Q IV $\mathbf{r}=(1,-1,0)$: ${\tau }_z=+1$ — same as Q III.

Verify: top half ($r_y>0$) ⇒ negative roll, bottom half ($r_y<0$) ⇒ positive roll. The $\pm x$ position never mattered because the force is along $x$. This is why designers place roll jets far off-axis in $y$ for maximum $|{\tau }_z|$.

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Ex 3b — The boundary $r_y=0$: on the $x$-axis (Cell C0)

Forecast: Ex 3 said $r_y=0$ should give zero roll. Confirm it, and say why geometrically.

1. Apply the same formula. ${\tau }_z=-r_{y}F=-(0)(1)=0$. Why this step? This is the boundary case the quadrant sweep skipped over; it's where the sign of ${\tau }_z$ switches from $-$ (above) to $+$ (below), so it must pass through zero.

2. Full cross product to be sure. $\mathbf{\tau }=(1,0,0)\times (1,0,0)=(0,0,0)$. Why this step? $\mathbf{r}$ and $\mathbf{F}$ are parallel here (both along $+x$), so the force line runs straight through the CoM — no lever arm, no roll.

Verify: ${\tau }_z=0$ ✓, and the full vector is $0$ ✓. Physically the jet just pushes the ship straight along $+x$ with zero roll — the same "force through CoM" idea as Ex 4, now sitting exactly on the quadrant boundary. This completes every cell of the quadrant discussion: top, bottom, and the dividing line.

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Ex 4 — Degenerate case: force through the CoM (Cell D)

Forecast: the lever arm and the force are on the same line. What must the torque be?

1. Build the force vector, then cross. $\mathbf{F}=F\hat{\mathbf{u}}=50(-1,0,0)=(-50,0,0)$. $\mathbf{\tau }=(3,0,0)\times (-50,0,0)=(0\cdot 0-0\cdot 0,0\cdot (-50)-3\cdot 0,3\cdot 0-0\cdot (-50))=(0,0,0).$ Why this step? The cross product of two parallel (or anti-parallel) vectors is zero — geometrically the parallelogram they span has zero area.

2. Interpret. Pure translation: $\mathbf{F}=(-50,0,0)$N, no spin. Why this step? This is the parent's rule "force through CoM ⇒ zero torque, pure translation."

Verify: $|\mathbf{r}\times \mathbf{F}|=|\mathbf{r}||\mathbf{F}|\sin \theta$ with $\theta =180^{\circ }\Rightarrow \sin 180^{\circ }=0$ ✓. Degenerate cell confirmed: this jet is a pure translator, useless for attitude.

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Ex 5 — Plume falloff: on-axis worst case (Cell E)

Forecast: doubling the distance — does the pressure halve, or drop by four?

1. On-axis shape factor. ${\cos }^{4}0=1$. Why this step? Dead-centre is the worst case — we always test the constraint at $\theta =0$ first.

2. At $d=1$m. $P=500(0.1/1{)}^2=500\cdot 0.01=5$Pa. Why this step? Plug the $1/d^2$ shell-conservation law; number density thins as the same gas crosses ever-bigger spheres.

3. At $d=2$m. $P=500(0.1/2{)}^2=500\cdot 0.0025=1.25$Pa. Why this step? Confirms the quadratic drop.

Verify: ratio $P(1)/P(2)=5/1.25=4=(2/1{)}^2$ ✓ — doubling distance quarters the pressure, the signature of $1/d^2$. (See Rarefied Gas Dynamics / Plume Modeling for the shell derivation.)

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Ex 6 — Off-axis plume: when a tight plume becomes safe (Cell F)

Forecast: a tighter plume ($\text{bigger }k$) is more focused on axis — so off-axis it should be weaker. Predict which $k$ keeps the panel safe.

1. Base on-axis value at $d=1$. From Ex 5, $500(0.1{)}^2=5$Pa before the angle factor. Why this step? Separate the distance part from the angle part so we vary only $k$.

2. $k=2$: ${\cos }^{2}30^{\circ }=(0.8660{)}^2=0.75$. $P=5\cdot 0.75=3.75$Pa $>1$ → violation. Why this step? A broad plume still dumps lots of gas at $30^{\circ }$.

3. $k=8$: ${\cos }^{8}30^{\circ }=0.75^4=0.3164$. $P=5\cdot 0.3164=1.582$Pa $>1$ → still violation (barely). Why this step? Even a tight plume isn't enough at just $1$m; you'd need more angle or distance.

4. How far off-axis for $k=8$ to be safe? Solve $5{\cos }^8\theta =1\Rightarrow \cos \theta =(0.2{)}^{1/8}=0.8177\Rightarrow \theta =35.1^{\circ }$. Why this step? This is the keep-out half-angle for the tight plume — anything beyond $35.1^{\circ }$ is allowed.

Verify: ${\cos }^8(35.1^{\circ })=0.8177^8=0.1998\approx 0.2$, and $5\cdot 0.2=1.0$Pa ✓ (just meets the limit). So bigger $k$ shrinks the danger cone but doesn't zero it.

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Ex 7 — Docking word problem: inhibit then re-allocate (Cell G)

Forecast: from Ex 5 you already know $P$ at $d=2$. Guess whether the forward jet passes.

1. Check the forward jet. $P=500(0.1/2{)}^2{\cos }^{4}0=1.25$Pa. Why this step? On-axis worst case at the exact target distance.

2. Compare to limit. $1.25>1.0$ → inhibit the forward jet ($F=0$). Why this step? Hardware safety outranks fuel; the parent's keep-out rule forbids this firing.

3. Set up the re-allocation. Use two canted jets, each pointing ${\theta }_{\mathrm{jet}}=40^{\circ }$ off $+\hat{\mathbf{x}}$, one canted toward $+\hat{\mathbf{y}}$ and one toward $-\hat{\mathbf{y}}$ (so their plume centrelines sweep past the target). Each thrust $F$ has an along-$x$ part $F\cos 40^{\circ }$ and a sideways part $F\sin 40^{\circ }$. Summing the pair: $\text{along }x:2F\cos 40^{\circ },\text{sideways }y:F\sin 40^{\circ }-F\sin 40^{\circ }=0.$ Why this step? We deliberately choose a symmetric $\pm y$ pair so the two lateral components are equal and opposite — they cancel, leaving only the commanded $+x$ push. This kills any unwanted drift, exactly like the couple in Ex 1 cancels net force.

4. Solve for the per-jet thrust. Set the along-$x$ sum equal to the command: $2F\cos 40^{\circ }=8\Rightarrow F=\frac{8}{2\cos 40^{\circ }}=\frac{8}{2(0.7660)}=5.22\text{N each}.$ Why this step? Only the $x$-projection does useful work toward the $8$N command, so we divide it out to recover the magnitude each jet must fire. This is the $B\mathbf{f}={\mathbf{w}}_{\mathrm{cmd}}$ solve over the remaining feasible thrusters (see Control Allocation & Pseudo-inverse).

5. Check the canted-jet plume at the target. The target lies at angle $\theta =40^{\circ }$ off each canted jet's centreline, still at $d=2$m. So $P_{\mathrm{canted}}=500(0.1/2{)}^2{\cos }^{4}40^{\circ }=1.25\cdot (0.7660{)}^4=1.25\cdot 0.3444=0.4305\text{Pa}.$ Why this step? "Missing the target" must be proven, not assumed — we compute the actual impingement pressure at the off-axis angle.

6. Compare to limit. $0.4305<1.0$ → safe ✓. Why this step? Now the re-allocation is legitimate: both canted plumes stay under the $1$Pa target limit.

7. Propellant penalty. Total thrust used $=2\times 5.22=10.44$N vs the ideal $8$N. Penalty $=(10.44-8)/8=30.5\%$ more. Why this step? Quantifies the cost of safety — the canted jets waste effort sideways.

Verify: along-$x$ force $=2(5.22)\cos 40^{\circ }=8.00$N ✓; sideways $=5.22\sin 40^{\circ }-5.22\sin 40^{\circ }=0$ ✓; canted plume pressure $0.43$Pa $<1$Pa ✓. Safety achieved, $\approx 31\%$ fuel penalty accepted.

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Ex 8 — Limiting values: pressure vanishes (Cell H)

Forecast: both should send the pressure toward zero — but for different geometric reasons.

1. (a) $\theta \to 90^{\circ }$. ${\cos }^{4}90^{\circ }=0\Rightarrow P\to 0$. Why this step? At $90^{\circ }$ the surface is on the plume's edge/back — no gas travels sideways in this idealized cone.

2. Numeric near-limit. At $\theta =89^{\circ }$: $\cos 89^{\circ }=0.01745$, ${\cos }^4=9.27\times 10^{-8}$, so $P=5\cdot 9.27\times 10^{-8}=4.6\times 10^{-7}$Pa — effectively nothing. Why this step? Shows the approach is smooth and fast (fourth power crushes small cosines).

3. (b) $d\to \infty$. $P=500(0.1/d{)}^2\to 0$. Why this step? Infinitely far, the same gas is spread over an infinite sphere — density $\to 0$. This is why far targets are automatically safe.

Verify: ${\lim }_{\theta \to 90^{\circ }}{\cos }^4\theta =0$ ✓ and ${\lim }_{d\to \infty }(0.1/d{)}^2=0$ ✓. Both degenerate limits confirm the constraint is only ever binding close and on-axis — precisely the docking regime of Ex 7.

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Ex 9 — Exam twist: force and torque both commanded (Cell I)

Forecast: a couple gives torque with zero force; a common push gives force with zero torque. We need a mix. Guess whether both stay non-negative.

1. Write the two allocation equations. Each force vector is ${\mathbf{F}}_i=F_i(1,0,0)$. The net force along $x$ is $F_1+F_2$; the roll of each is ${\tau }_{z,i}=-r_y^{(i)}{F}_i$ (from the ${\tau }_z=-r_{y}F$ rule of Ex 3, where $r_y^{(1)}=+1$ and $r_y^{(2)}=-1$). So: $F_1+F_2=6,{\tau }_z=-(+1)F_1-(-1)F_2=F_2-F_1=4.$ Why this step? This is the parent's $B\mathbf{f}={\mathbf{w}}_{\mathrm{cmd}}$ written out for the small $2\times 2$ system — two equations, two unknowns, so it's exactly determined. (I use $r_y^{(1)},r_y^{(2)}$ — the $y$-components of ${\mathbf{r}}_1,{\mathbf{r}}_2$ — to stay consistent with the bold-vector names.)

2. Solve. From $F_2-F_1=4$ and $F_1+F_2=6$, add the equations: $2F_2=10\Rightarrow F_2=5$, then $F_1=6-5=1$. Why this step? Elementary elimination; the $2\times 2$ influence matrix is invertible here, so there is exactly one solution.

3. Feasibility check. $F_1=1\ge 0$ and $F_2=5\ge 0$ ✓. Why this step? The one-sided constraint ($F_i\ge 0$) matters — a jet can't suck. Had a value gone negative we would have had to fire the opposite jet instead; here both are physical, so the command is directly achievable.

4. Interpret. The solution is a common push of $F_1=F_2$ worth ... more precisely, decompose it: the shared part $\min (F_1,F_2)=1$N on both jets is a pure $2$N push with zero torque; the extra $4$N on jet 2 alone supplies the $+4$N·m roll (and the remaining $4$N of push). Force and torque both delivered by one asymmetric firing. Why this step? Seeing the answer as "symmetric push + asymmetric twist" connects it back to Ex 1 (the couple) and Ex 2 (the single off-CoM jet) — this exam twist is just those two building blocks superposed.

Verify: net force $=F_1+F_2=1+5=6$N ✓; net torque $=F_2-F_1=5-1=4$N·m ✓. Both magnitudes non-negative ✓. This ties directly to Attitude Dynamics — Euler's Equations (the $4$N·m torque feeds $I\dot{\mathbf{\omega }}$) and Rocket Equation & Specific Impulse (the total $F_1+F_2=6$N sets the propellant mass rate).

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Recall Self-test

Q: In Ex 3, why does the $x$-position of the thruster not affect ${\tau }_z$? ::: Because the force is along $\hat{\mathbf{x}}$ (so $F_y=0$); only the perpendicular lever arm $r_y$ produces roll (${\tau }_z=-r_{y}F$). Q: In Ex 7, what mathematically forces the fuel penalty? ::: The canted jets waste thrust in their sideways components; only $\cos 40^{\circ }$ of each goes toward the commanded axis. Q: Why is the plume constraint only ever binding close and on-axis? ::: Ex 8 — pressure $\to 0$ as $d\to \infty$ (spread over huge spheres) and as $\theta \to 90^{\circ }$ (${\cos }^k\to 0$).