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

3.5.46 · D3 · Physics › Guidance, Navigation & Control (GNC) › Reaction control system — thruster selection, plume impingem

Ye page parent topic ka hands-on lab hai. Hum ek matrix ke through har possible case class solve karte hain — signs, degenerate geometries, limiting distances, ek real docking word problem, aur ek exam-style twist — aur har ek ko scratch se solve karte hain.

Pehli line se pehle, char plain-word reminders taaki koi bhi symbol achanak na aaye:

Agar $\times$ ya "$1/d^2$" thoda shaky lagta hai, to parent note mein unhe build kiya gaya hai; yahan hum sirf use karte hain.

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The scenario matrix

Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai. Milake yeh poori table cover karte hain.

Cell	Case class	Kya cheez tricky hai	Example
A	Pure torque, single axis	cross product ke signs	Ex 1
B	Off-CoM single thruster	torque aur unwanted drift	Ex 2
C	Quadrant sweep of $\mathbf{r}$	${\tau }_z$ ka sign har quadrant mein	Ex 3
C0	Boundary: $r_y=0$ ($x$-axis par)	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, ${\cos }^k\theta$	jab ek big-$k$ plume safe ho jaata hai	Ex 6
G	Real-world word problem: docking	ek thruster inhibit karo, re-allocate karo	Ex 7
H	Limiting values: $d\to \infty$, $\theta \to 90^{\circ }$	pressure $\to 0$	Ex 8
I	Exam twist: force + torque dono commanded	over-determined mini-allocation	Ex 9

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Ex 1 — Pure roll couple (Cell A)

Forecast: compute karne se pehle guess karo — kya ship drift karegi? Woh kis taraf spin karega ($+z$ ya $-z$ ke baare mein)?

Figure padho: do orange arrows force vectors ${\mathbf{F}}_1$ (pointing $+x$) aur ${\mathbf{F}}_2$ (pointing $-x$) hain; notice karo yeh equal length aur opposite hain, isliye tip-to-tail cancel ho jaate hain. Purple curved arrow resulting spin about $z$-axis hai (into the page). Do teal squares thruster mounts ${\mathbf{r}}_1,{\mathbf{r}}_2$ par hain.

1. Har force vector banao. ${\mathbf{F}}_i=F_i{\hat{\mathbf{u}}}_i$ use karke: ${\mathbf{F}}_1=20(1,0,0)=(20,0,0)$ aur ${\mathbf{F}}_2=20(-1,0,0)=(-20,0,0)$. Yeh step kyun? Torque ko vector force chahiye, sirf magnitude nahi, isliye hum pehle ${\mathbf{F}}_i$ assemble karte hain.

2. Net force. Dono force vectors add karo. $\mathbf{F}=(20,0,0)+(-20,0,0)=0.$ Yeh step kyun? Ek couple ka zero net force hona zaroori hai, warna ship rotate karte waqt translate bhi karegi. Yeh figure mein dono opposite orange arrows ke cancel hone se match karta hai.

3. Thruster 1 ka torque. ${\mathbf{\tau }}_1={\mathbf{r}}_1\times {\mathbf{F}}_1$ use karo. $(0,1.5,0)\times (20,0,0)=(1.5\cdot 0-0\cdot 0,0\cdot 20-0\cdot 0,0\cdot 0-1.5\cdot 20)=(0,0,-30).$ Yeh step kyun? Cross product twisting part pick out karta hai; third component $-30$ roll about $z$ hai. Sign negative hai — thumb page ke andar point kar raha hai (purple arrow).

4. Thruster 2 ka torque. $(0,-1.5,0)\times (-20,0,0)=(0,0,(0)(0)-(-1.5)(-20))=(0,0,-30).$ Yeh step kyun? Hume check karna hai ki doosra thruster usi taraf twist kar raha hai, warna torques bhi forces ki tarah cancel ho jaate.

5. Add karo. $\mathbf{\tau }=(0,0,-60)$N·m. Yeh step kyun? Ek sahi couple ke torques add hote hain; yahi couple ka poora point hai.

Verify: magnitude $=2rF=2(1.5)(20)=60$N·m ✓, aur $\mathbf{F}=0$ ✓. Units: N·m (force × lever). Yeh parent note ka "ideal attitude control" hai: zero drift ke saath spin.

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

Forecast: parent ka "steel-manned mistake" kehta hai ek thruster kabhi bhi pure torque nahi hota. Aage padhne se pehle net force predict karo.

1. Force vector banao. $\mathbf{F}=F\hat{\mathbf{u}}=10(1,0,0)=(10,0,0)$N — nonzero. Yeh step kyun? Kuch bhi ise cancel nahi karta, isliye Newton's second law kehta hai CoM accelerate karega: drift.

2. Torque. $\mathbf{\tau }=(0,2,0)\times (10,0,0)=(0,0,-20)$N·m. Yeh step kyun? Yeh confirm karta hai ki hume phir bhi ek twist milta hai, isliye yeh jet ek saath dono kaam karta hai.

Verify: $|\mathbf{\tau }|=rF=2\cdot 10=20$ ✓. Ex 1 (couple) se compare karo, yahan $\mathbf{F}\ne 0$ hai — yeh exactly "one thruster is not a pure torque" trap hai. Fix: $10$N ko khatam karne ke liye ek opposing jet add karo.

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Ex 3 — Quadrant sweep: ${\tau }_z$ ka sign har jagah (Cell C)

Forecast: parent ne kaha tha ki tan/quadrant fixes matter karte hain kyunki ek formula repeat ho sakta hai. Yahan torque ka sign quadrant ke saath flip hota hai. Guess karo kaunse corners negative ${\tau }_z$ dete hain.

Figure padho: har coloured square ek quadrant mein ek thruster mount hai; har arrow ek hi taraf point karta hai ($+x$), kyunki force direction fixed hai. Colour batata hai resulting roll ka sign — orange squares (top half) ${\tau }_z=-1$ dete hain, teal squares (bottom half) ${\tau }_z=+1$ dete hain. Aankh left/right slide karo aur kuch nahi badlta; upar/neeche slide karo aur sign flip ho jaata hai — yeh visual proof hai ki sirf $r_y$ matter karta hai.

General force vector $\mathbf{F}=F\hat{\mathbf{u}}=(F,0,0)$ se shuru karo, to $F_x=F$ aur $F_y=0$. $\mathbf{r}=(r_x,r_y,0)$ ke liye cross product ka $z$-component hai: ${\tau }_z=r_x{F}_y-r_y{F}_x.$ Hamare scenario mein $F_y=0$ aur $F_x=F$ hai, to pehla term khatam ho jaata hai aur yeh collapse ho jaata hai: ${\tau }_z=r_x\cdot 0-r_y\cdot F=-r_{y}F.$ Yeh step kyun? Hum general do-term formula visible rakhte hain, phir $F_y=0$ substitute karte hain taaki reader exactly dekh sake kyun sirf $-r_{y}F$ term survive karta hai: ${\tau }_z$ sirf $r_y$ par depend karta hai, woh lever arm jo $x$-force ke perpendicular hai.

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! Kyun? ${\tau }_z$, $r_x$ ko ignore karta hai; dono top corners $-1$ dete hain.
3. Q III $\mathbf{r}=(-1,-1,0)$: ${\tau }_z=-(-1)\cdot 1=+1$ (counter-clockwise).
4. Q IV $\mathbf{r}=(1,-1,0)$: ${\tau }_z=+1$ — Q III jaisa hi.

Verify: top half ($r_y>0$) ⇒ negative roll, bottom half ($r_y<0$) ⇒ positive roll. $\pm x$ position kabhi matter nahi kiya kyunki force $x$ ke along hai. Isliye designers roll jets ko maximum $|{\tau }_z|$ ke liye $y$ mein far off-axis rakhte hain.

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Ex 3b — Boundary $r_y=0$: $x$-axis par (Cell C0)

Forecast: Ex 3 ne kaha tha $r_y=0$ zero roll dena chahiye. Ise confirm karo, aur geometrically batao kyun.

1. Wohi formula apply karo. ${\tau }_z=-r_{y}F=-(0)(1)=0$. Yeh step kyun? Yeh boundary case hai jo quadrant sweep ne skip kar diya; yahan ${\tau }_z$ ka sign $-$ (upar) se $+$ (neeche) switch hota hai, isliye ise zero se guzarna hi hoga.

2. Pakka karne ke liye poora cross product. $\mathbf{\tau }=(1,0,0)\times (1,0,0)=(0,0,0)$. Yeh step kyun? $\mathbf{r}$ aur $\mathbf{F}$ yahan parallel hain (dono $+x$ ke along), isliye force line seedha CoM se guzarti hai — koi lever arm nahi, koi roll nahi.

Verify: ${\tau }_z=0$ ✓, aur poora vector $0$ ✓. Physically jet ship ko sirf $+x$ ke along straight push karta hai zero roll ke saath — wohi "force through CoM" idea Ex 4 mein hai, ab exactly quadrant boundary par baitha hua. Yeh quadrant discussion ke har cell ko complete karta hai: top, bottom, aur dividing line.

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

Forecast: lever arm aur force ek hi line par hain. Torque kya hona chahiye?

1. Force vector banao, phir cross karo. $\mathbf{F}=F\hat{\mathbf{u}}=50(-1,0,0)=(-50,0,0)$. $\mathbf{\tau }=(3,0,0)\times (-50,0,0)=(0\cdot 0-0\cdot 0,0\cdot (-50)-3\cdot 0,3\cdot 0-0\cdot (-50))=(0,0,0).$ Yeh step kyun? Do parallel (ya anti-parallel) vectors ka cross product zero hota hai — geometrically parallelogram jisko woh span karte hain zero area ka hota hai.

2. Interpret karo. Pure translation: $\mathbf{F}=(-50,0,0)$N, koi spin nahi. Yeh step kyun? Yeh parent ka rule hai "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: yeh jet ek pure translator hai, attitude ke liye useless.

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

Forecast: distance double karne par — kya pressure aadhi ho jaayegi, ya chaar guna giregi?

1. On-axis shape factor. ${\cos }^{4}0=1$. Yeh step kyun? Dead-centre worst case hai — hum constraint hamesha pehle $\theta =0$ par test karte hain.

2. $d=1$m par. $P=500(0.1/1{)}^2=500\cdot 0.01=5$Pa. Yeh step kyun? $1/d^2$ shell-conservation law plug karo; number density thin hoti hai jab same gas baar-baar badi spheres cross karti hai.

3. $d=2$m par. $P=500(0.1/2{)}^2=500\cdot 0.0025=1.25$Pa. Yeh step kyun? Quadratic drop confirm karta hai.

Verify: ratio $P(1)/P(2)=5/1.25=4=(2/1{)}^2$ ✓ — distance double karne par pressure quarter ho jaati hai, $1/d^2$ ki signature. (Shell derivation ke liye Rarefied Gas Dynamics / Plume Modeling dekho.)

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Ex 6 — Off-axis plume: jab tight plume safe ho jaata hai (Cell F)

Forecast: ek tighter plume ($\text{bada }k$) axis par zyada focused hota hai — to off-axis pe woh kamzor hona chahiye. Predict karo kaun sa $k$ panel ko safe rakhta hai.

1. $d=1$ par base on-axis value. Ex 5 se, angle factor se pehle $500(0.1{)}^2=5$Pa. Yeh step kyun? Distance part ko angle part se alag karo taaki sirf $k$ vary kar sakein.

2. $k=2$: ${\cos }^{2}30^{\circ }=(0.8660{)}^2=0.75$. $P=5\cdot 0.75=3.75$Pa $>1$ → violation. Yeh step kyun? Ek broad plume phir bhi $30^{\circ }$ par bahut sa gas dump karta hai.

3. $k=8$: ${\cos }^{8}30^{\circ }=0.75^4=0.3164$. $P=5\cdot 0.3164=1.582$Pa $>1$ → phir bhi violation (barely). Yeh step kyun? Tight plume bhi sirf $1$m par enough nahi; aapko zyada angle ya distance chahiye.

4. $k=8$ ke safe hone ke liye kitna off-axis? $5{\cos }^8\theta =1\Rightarrow \cos \theta =(0.2{)}^{1/8}=0.8177\Rightarrow \theta =35.1^{\circ }$ solve karo. Yeh step kyun? Yeh tight plume ka keep-out half-angle hai — $35.1^{\circ }$ se aage kuch bhi allowed hai.

Verify: ${\cos }^8(35.1^{\circ })=0.8177^8=0.1998\approx 0.2$, aur $5\cdot 0.2=1.0$Pa ✓ (limit bilkul meet karta hai). To bada $k$ danger cone ko shrink karta hai lekin zero nahi karta.

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

Forecast: Ex 5 se aap pehle se jaante ho $d=2$ par $P$ kya hai. Guess karo forward jet pass karta hai ya nahi.

1. Forward jet check karo. $P=500(0.1/2{)}^2{\cos }^{4}0=1.25$Pa. Yeh step kyun? Exact target distance par on-axis worst case.

2. Limit se compare karo. $1.25>1.0$ → forward jet inhibit karo ($F=0$). Yeh step kyun? Hardware safety fuel se zyada important hai; parent ka keep-out rule yeh firing forbid karta hai.

3. Re-allocation setup karo. Do canted jets use karo, har ek ${\theta }_{\mathrm{jet}}=40^{\circ }$ off $+\hat{\mathbf{x}}$, ek $+\hat{\mathbf{y}}$ ki taraf canted aur ek $-\hat{\mathbf{y}}$ ki taraf (taaki unke plume centrelines target ke past sweep karein). Har thrust $F$ ka along-$x$ part $F\cos 40^{\circ }$ hai aur sideways part $F\sin 40^{\circ }$. Pair sum karne par: $\text{along }x:2F\cos 40^{\circ },\text{sideways }y:F\sin 40^{\circ }-F\sin 40^{\circ }=0.$ Yeh step kyun? Hum deliberately symmetric $\pm y$ pair choose karte hain taaki do lateral components equal aur opposite hon — woh cancel ho jaate hain, sirf commanded $+x$ push reh jaata hai. Yeh Ex 1 mein couple ki tarah kisi bhi unwanted drift ko khatam karta hai.

4. Per-jet thrust ke liye solve karo. Along-$x$ sum ko command ke barabar set karo: $2F\cos 40^{\circ }=8\Rightarrow F=\frac{8}{2\cos 40^{\circ }}=\frac{8}{2(0.7660)}=5.22\text{N each}.$ Yeh step kyun? Sirf $x$-projection $8$N command ki taraf useful kaam karta hai, isliye hum magnitude recover karne ke liye ise divide karte hain jo har jet ko fire karna hai. Yeh remaining feasible thrusters par $B\mathbf{f}={\mathbf{w}}_{\mathrm{cmd}}$ solve hai (dekho Control Allocation & Pseudo-inverse).

5. Target par canted-jet plume check karo. Target har canted jet ke centreline se $\theta =40^{\circ }$ angle par hai, phir bhi $d=2$m par. To $P_{\mathrm{canted}}=500(0.1/2{)}^2{\cos }^{4}40^{\circ }=1.25\cdot (0.7660{)}^4=1.25\cdot 0.3444=0.4305\text{Pa}.$ Yeh step kyun? "Target miss karna" proven hona chahiye, assumed nahi — hum off-axis angle par actual impingement pressure compute karte hain.

6. Limit se compare karo. $0.4305<1.0$ → safe ✓. Yeh step kyun? Ab re-allocation legitimate hai: dono canted plumes $1$Pa target limit ke andar rehte hain.

7. Propellant penalty. Total thrust used $=2\times 5.22=10.44$N vs ideal $8$N. Penalty $=(10.44-8)/8=30.5\%$ zyada. Yeh step kyun? Safety ki cost quantify karta hai — canted jets sideways effort waste karte hain.

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 achieve hui, $\approx 31\%$ fuel penalty accepted.

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

Forecast: dono pressure ko zero ki taraf bhejne chahiye — lekin alag alag geometric reasons se.

1. (a) $\theta \to 90^{\circ }$. ${\cos }^{4}90^{\circ }=0\Rightarrow P\to 0$. Yeh step kyun? $90^{\circ }$ par surface plume ke edge/back par hai — is idealized cone mein koi gas sideways nahi jaati.

2. Near-limit numeric. $\theta =89^{\circ }$ par: $\cos 89^{\circ }=0.01745$, ${\cos }^4=9.27\times 10^{-8}$, to $P=5\cdot 9.27\times 10^{-8}=4.6\times 10^{-7}$Pa — practically kuch nahi. Yeh step kyun? Dikhata hai ki approach smooth aur fast hai (fourth power choti cosines ko crush kar deta hai).

3. (b) $d\to \infty$. $P=500(0.1/d{)}^2\to 0$. Yeh step kyun? Infinite door, same gas infinite sphere par spread ho jaati hai — density $\to 0$. Isliye door targets automatically safe hote hain.

Verify: ${\lim }_{\theta \to 90^{\circ }}{\cos }^4\theta =0$ ✓ aur ${\lim }_{d\to \infty }(0.1/d{)}^2=0$ ✓. Dono degenerate limits confirm karte hain ki constraint sirf paas mein aur on-axis binding hoti hai — exactly Ex 7 ka docking regime.

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Ex 9 — Exam twist: force aur torque dono commanded (Cell I)

Forecast: ek couple zero force ke saath torque deta hai; ek common push zero torque ke saath force deta hai. Hume ek mix chahiye. Guess karo dono non-negative rehte hain ya nahi.

1. Do allocation equations likho. Har force vector ${\mathbf{F}}_i=F_i(1,0,0)$ hai. $x$ ke along net force $F_1+F_2$ hai; har ek ka roll ${\tau }_{z,i}=-r_y^{(i)}{F}_i$ hai (Ex 3 ke ${\tau }_z=-r_{y}F$ rule se, jahan $r_y^{(1)}=+1$ aur $r_y^{(2)}=-1$). To: $F_1+F_2=6,{\tau }_z=-(+1)F_1-(-1)F_2=F_2-F_1=4.$ Yeh step kyun? Yeh parent ka $B\mathbf{f}={\mathbf{w}}_{\mathrm{cmd}}$ hai small $2\times 2$ system ke liye likha hua — do equations, do unknowns, to exactly determined hai. (Main $r_y^{(1)},r_y^{(2)}$ use karta hun — ${\mathbf{r}}_1,{\mathbf{r}}_2$ ke $y$-components — bold-vector names ke saath consistent rehne ke liye.)

2. Solve karo. $F_2-F_1=4$ aur $F_1+F_2=6$ se, equations add karo: $2F_2=10\Rightarrow F_2=5$, phir $F_1=6-5=1$. Yeh step kyun? Elementary elimination; $2\times 2$ influence matrix yahan invertible hai, isliye exactly ek solution hai.

3. Feasibility check. $F_1=1\ge 0$ aur $F_2=5\ge 0$ ✓. Yeh step kyun? One-sided constraint ($F_i\ge 0$) matter karti hai — ek jet suck nahi kar sakta. Agar koi value negative hoti to hume opposite jet fire karna padta; yahan dono physical hain, isliye command directly achievable hai.

4. Interpret karo. Solution ek common push hai $F_1=F_2$ worth ... zyada precisely, ise decompose karo: dono jets par shared part $\min (F_1,F_2)=1$N ek pure $2$N push hai zero torque ke saath; jet 2 par extra $4$N akela $+4$N·m roll supply karta hai (aur remaining $4$N push). Force aur torque dono ek asymmetric firing se deliver ho gaye. Yeh step kyun? Answer ko "symmetric push + asymmetric twist" ke roop mein dekhna ise Ex 1 (couple) aur Ex 2 (single off-CoM jet) se connect karta hai — yeh exam twist sirf woh do building blocks superposed hai.

Verify: net force $=F_1+F_2=1+5=6$N ✓; net torque $=F_2-F_1=5-1=4$N·m ✓. Dono magnitudes non-negative ✓. Yeh directly Attitude Dynamics — Euler's Equations se tie karta hai ($4$N·m torque $I\dot{\mathbf{\omega }}$ feed karta hai) aur Rocket Equation & Specific Impulse se (total $F_1+F_2=6$N propellant mass rate set karta hai).

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Recall Self-test

Q: Ex 3 mein, thruster ki $x$-position ${\tau }_z$ ko kyun affect nahi karti? ::: Kyunki force $\hat{\mathbf{x}}$ ke along hai (to $F_y=0$); sirf perpendicular lever arm $r_y$ roll produce karta hai (${\tau }_z=-r_{y}F$). Q: Ex 7 mein, mathematically fuel penalty kya force karta hai? ::: Canted jets apna thrust unke sideways components mein waste karte hain; har ek ka sirf $\cos 40^{\circ }$ commanded axis ki taraf jaata hai. Q: Plume constraint sirf paas mein aur on-axis binding kyun hoti hai? ::: Ex 8 — pressure $\to 0$ as $d\to \infty$ (huge spheres par spread ho jaati hai) aur as $\theta \to 90^{\circ }$ (${\cos }^k\to 0$).