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

3.5.46 · D4 · Physics › Guidance, Navigation & Control (GNC) › Reaction control system — thruster selection, plume impingem

Shuru karne se pehle, ek shared symbols ki list (parent mein earn ki, yahan restate ki gayi taaki pehli line se hi readable ho):

Definition Symbols jo aap baar baar use karenge

$r_{i}$ — thruster $i$ ki position center of mass (CoM) se measure ki gayi, ek arrow jo ship ke balance-point se nozzle tak jaata hai.
$\hat{u}_{i}$ — unit thrust direction: ship par push kis direction mein point karta hai (gas jahan se nikle uske ulte). "Unit" matlab length exactly $1$ , toh yeh sirf direction carry karta hai.
$F_{i} \geq 0$ — thrust ki magnitude. Ek jet sirf push kar sakta hai, kabhi suck nahi, isliye yeh kabhi negative nahi hota.
$τ_{i} = r_{i} \times F_{i}$ — thruster $i$ ka torque (twist). Symbol $\times$ cross product hai: yeh lever-arm aur force leta hai aur return karta hai woh axis jis par body spin karti hai.
$w = [F; τ]$ — wrench, chhe numbers jo net force (3) aur net torque (3) stack karte hain.
$B$ — allocation matrix; column $i$ thruster $i$ ka unit wrench hai, toh $w = B f$ .
$P_{imp} (d, θ) \approx P_{0} (d_{0} / d)^{2} cos^{k} θ$ — woh plume pressure jo ek surface distance $d$ aur off-axis angle $θ$ par feel karti hai.
$θ$ — off-axis angle, plume centerline aur nozzle se surface tak ki line ke beech measure kiya jaata hai. Yeh $0$ (dead-center) se lekar $9 0^{\circ}$ tak forward hemisphere ke edge par jaata hai; $9 0^{\circ}$ se aage ek surface nozzle exit ke peeche baith'ti hai aur practically koi plume receive nahi karti, isliye hum model ko wahan clamp kar dete hain (Ex 5.2 ka domain note dekho).
$α$ — surface-tilt angle, incoming flow direction aur surface normal-flip ke beech measure kiya jaata hai — yaani patch flow ke kitna square-on face karta hai. $α = 0$ matlab patch plume ko head-on face kar raha hai (maximum force); $α = 9 0^{\circ}$ matlab flow edge-on graze karta hai (koi push nahi). Yeh impingement force mein $cos α$ ke roop mein enter karta hai.

Neeche ke teen figures decoration nahi hain — har ek ek derivation step carry karta hai, aur har exercise aapko us arrow ya curve ki taraf point karta hai jo aapko chahiye. Inhe order mein padho.

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

Figure s01 woh picture hai jo Level 2 ke peeche hai. Yellow dot CoM hai; blue arrow lever arm $r$ hai; red arrow force $F$ hai. Dekho kaise red force yellow dot se offset baith'ta hai — woh offset exactly woh hai jo cross product green curved spin arrow $τ$ mein convert karta hai. Jab aap Ex 2.1 karte hain, toh aap woh green arrow compute kar rahe hote hain.

Level 1 — Recognition

Exercise 1.1

Ek single thruster fire karta hai jiska push direction $\hat{u}_{i}$ exactly CoM ke through point karta hai (toh $r_{i}$ aur $\hat{u}_{i}$ parallel hain). Kya yeh koi torque produce karta hai? Cross product use karke ek line mein explain karo.

Recall Solution

Koi torque nahi. Torque hai $τ_{i} = F_{i} (r_{i} \times \hat{u}_{i})$ . Cross product lever arm ka perpendicular part measure karta hai — $r_{i}$ aur $\hat{u}_{i}$ ke beech angle ka sine. Jab woh parallel hain toh angle $0$ hai aur $sin 0 = 0$ , isliye $r_{i} \times \hat{u}_{i} = 0$ . Saari effort pure translation ban jaati hai (ship drift karta hai, spin nahi karta). Figure s01 imagine karo jahan red arrow seedha yellow dot ke through point karne ke liye rotate ho — green spin arrow kuch nahi rah jaata.

Exercise 1.2

Ek spacecraft ke liye allocation matrix $B$ ka size $6 \times N$ hota hai. $6$ ka kya matlab hai, $N$ ka kya matlab hai, aur $B$ ka ek column physically kya represent karta hai?

Recall Solution

$6$ = rigid body ke chhe degrees of freedom: 3 translation directions + 3 rotation axes. Isliye wrench $w$ mein chhe entries hain.
$N$ = ship par thrusters ki sankhya.
Har column ek thruster ka unit wrench hai $[\hat{u}_{i}; r_{i} \times \hat{u}_{i}]$ : woh force aur torque jo thruster $i$ magnitude $1$ par fire karne par banata hai. $B f$ multiply karna sirf har column ko uske on-time $F_{i}$ se scale karta hai aur unhe add kar deta hai.

Level 2 — Application

Exercise 2.1

Ek thruster $r = (0, 0.5, 0)$ m par mount hai aur direction $\hat{u} = (1, 0, 0)$ mein magnitude $F = 20$ N se push karta hai. Force vector aur torque vector compute karo. (Yeh exactly Figure s01 mein draw ki gayi geometry hai.)

Recall Solution

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

Torque: hume cross product chahiye kyunki torque lever arm perpendicular to force hota hai. $r \times F = (0, 0.5, 0) \times (20, 0, 0) .$ Determinant rule $a \times b = (a_{y} b_{z} - a_{z} b_{y}, a_{z} b_{x} - a_{x} b_{z}, a_{x} b_{y} - a_{y} b_{x})$ use karke:

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

$τ = (0, 0, - 10) N\cdotm .$ Push $+ x$ along hai, arm $+ y$ along hai, toh twist $- z$ ke baare mein hai (right-hand rule: arm se force ki taraf fingers curl karke $- z$ ki taraf jaate hain) — yeh Figure s01 mein green curved arrow hai.

Exercise 2.2

Plume model $P_{imp} = P_{0} (d_{0} / d)^{2} cos^{k} θ$ use karke, $P_{0} = 500$ Pa at $d_{0} = 0.1$ m aur $k = 4$ : on axis ( $θ = 0$ ) distance $d = 1$ m par ek surface par kitna pressure lagta hai?

Recall Solution

On axis, $cos^{4} 0 = 1$ , toh poora angular factor $1$ hai (worst case, dead-center). $P_{imp} = 500 (\frac{0.1}{1})^{2} \cdot 1 = 500 \times 0.01 = 5 Pa .$ $1/ d^{2}$ kyun? Utnhi gas fly out hote waqt baar baar bade spherical caps par spread ho jaati hai; area $d^{2}$ ki tarah badhta hai, toh pressure per area $1/ d^{2}$ ki tarah girta hai. Yeh Figure s03 mein blue curve hai jo $d = 1$ m par read ki gayi.

Level 3 — Analysis

Exercise 3.1 — pure roll couple, dono same taraf twist karte hain

Do thrusters ek wheel par $r = 0.5$ m radius par baithe hain: $r_{1} = (0, + r, 0)$ firing $\hat{u}_{1} = + \hat{x}$ , aur $r_{2} = (0, - r, 0)$ firing $\hat{u}_{2} = - \hat{x}$ , har ek $F = 20$ N par. Dikhao ki net force zero hai aur net torque dhundho. (Figure s02 dono thrusters aur shared spin draw karta hai.)

Recall Solution

Net force — kyunki opposite directions cancel ho jaate hain: $F = 20 (1, 0, 0) + 20 (- 1, 0, 0) = (0, 0, 0) N .$ Zero drift: yeh ek couple hai. Figure s02 mein, notice karo ki do red force arrows opposite ways point karte hain — isliye woh sum ho ke kuch nahi bante.

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

Thruster 2 ka torque: $τ_{2} = (0, - 0.5, 0) \times (- 20, 0, 0)$ .

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

$τ_{2} = (0, 0, - 10)$ N·m. Dono $- z$ ke baare mein same taraf twist karte hain — flipped arm aur flipped force same sign multiply karte hain. Yeh Figure s02 mein single green spin arrow hai: dono thrusters ise feed karte hain.

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

Figure s02 ek couple ki picture hai: do blue arms, do opposing red forces, aur ek green spin arrow jise dono drive karte hain. Jab bhi aapko "torque with zero drift" chahiye, yeh woh shape hai jise reach karna hai.

Exercise 3.2 — panel safe rehne ke liye kitni door honi chahiye?

Panel limit $P_{m a x} = 1$ Pa hai. On axis ( $θ = 0$ ), $P_{0} = 500$ Pa, $d_{0} = 0.1$ m ke saath, minimum safe distance $d_{m i n}$ dhundho jahan $P_{imp}$ exactly $1$ Pa tak drop ho.

Recall Solution

$P_{imp} = P_{m a x}$ set karo aur $d$ ke liye solve karo. Kyunki pressure $1/ d^{2}$ ki tarah girta hai, distance double karne se pressure quarter ho jaata hai, toh hum invert karte hain: $1 = 500 (\frac{0.1}{d})^{2} \Rightarrow (\frac{0.1}{d})^{2} = \frac{1}{500} .$ $\frac{0.1}{d} = \frac{1}{500} \Rightarrow d = 0.1 500 = 0.1 \times 22.36 \approx 2.236 m .$ On axis $\approx 2.24$ m se closer koi bhi surface limit se upar hai. Yeh distance exactly on axis keep-out cone ka radius hai — yeh woh yellow dot hai jahan Figure s03 mein blue curve red dashed limit line ko cross karti hai.

Figure s03 on-axis plume pressure (blue) ko log scale par distance ke against plot karta hai, $1$ Pa panel limit red dashed line ke roop mein. Jahan woh cross karte hain (yellow dot, green dotted line) woh Ex 3.2 se $d_{m i n} \approx 2.24$ m hai — us line ke left sab kuch forbidden hai.

Level 4 — Synthesis

Exercise 4.1 — inhibit karein ya allow?

Docking ke dауран ek target $d = 1.5$ m aage $+ x$ axis par baithe hai. Ek forward thruster ka plume centerline seedha uski taraf point karta hai, lekin sensitive patch (ek camera window) us centerline se $θ = 3 0^{\circ}$ off hai. $P_{0} = 500$ Pa, $d_{0} = 0.1$ m, $k = 4$ , limit $P_{m a x} = 1$ Pa use karo. Kya yeh thruster allowed hai?

Recall Solution

Hume dono distance factor aur angular factor chahiye, kyunki patch off-axis hai. Note karo $θ = 3 0^{\circ}$ model ke valid domain $0 \leq θ \leq 9 0^{\circ}$ ke andar safely hai, toh $cos θ > 0$ aur formula directly apply hota hai.

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

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

Pressure: $P_{imp} = 500 \times 0.004444 \times 0.5625 \approx 1.25 Pa .$ Kyunki $1.25 > 1.0$ , thruster inhibited hai ( $F_{i}$ locked to $0$ ). Optimizer ko $B f = w_{cmd}$ remaining thrusters par re-solve karna hoga, likely ek off-axis pair jinka plume window clear karta hai — thodi si propellant cost par, kyunki hardware safety fuel se upar hai.

Exercise 4.2 — hidden disturbance torque

4.1 ka plume (inhibition se pehle) camera window par ek force deliver karta hai (area $A = 0.02$ m², flow ko head-on face karte hue, toh surface-tilt angle $α = 0$ hai aur $cos α = 1$ ). Agar woh window CoM se $r_{s} = (1.5, 0.3, 0)$ m par baith'ti hai, toh estimate karo ki yeh impingement $z$ ke baare mein kitna disturbance torque magnitude add karta hai. $P_{imp} = 1.25$ Pa use karo, aur assume karo ki impingement force $+ x$ mein point karti hai.

Recall Solution

$α$ recall karo (symbol list se): yeh patch ka flow ke relative tilt hai. Yahan window plume ko squarely face karti hai, $α = 0$ , toh $cos α = 1$ aur patch puri pressure catch karta hai. Agar yeh edge-on hoti ( $α = 9 0^{\circ}$ ) toh flow graze past hota aur $cos α = 0$ — koi force nahi.

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

Disturbance torque — CoM se off ek force twist banati hai, toh phir 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. $∣ τ_{dist, z} ∣ = 0.0075 N\cdotm .$ Akele mein chota — lekin yeh ek torque hai jo aapne kabhi command nahi kiya. Yeh aapke attitude loop se ladhta hai aur, ek lambe burn ke baad, counter-fire karna padta hai. Yeh exactly "hidden disturbance torque" warning hai: plume sirf ek heat threat nahi hai, yeh ek control error hai.

Level 5 — Mastery

Exercise 5.1 — plume veto ke saath full selection

Ek ship ko pure roll torque $τ_{z} = - 20$ N·m aur zero net force produce karna hai. Chaar candidate thrusters (har ek max $20$ N, arm $r = 0.5$ m) yeh torque bana sakte hain:

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

Pair A ka plume sab kuch clear karta hai. Pair B ka plume on-axis pressure $P = 500 (0.1/0.4)^{2} = 31.25$ Pa se panel hit karta hai (panel limit $1$ Pa). Dono pairs, $F = 20$ N par fire karne par, same $- 20$ N·m couple dete hain. Minimum-propellant selector kaun sa pair choose karta hai, aur Pair A only feasible choice kyun hai?

Recall Solution

Dono pairs command achieve karte hain: har ek ek couple hai jo $τ = (0, 0, - 2 r F) = (0, 0, - 2 \cdot 0.5 \cdot 20) = (0, 0, - 20)$ N·m deta hai zero net force ke saath (yeh Figure s02 shape hai). Propellant par akele ( $\sum F_{i} = 40$ N each) woh tie karte hain.

Plume constraint tie todhta hai. Pair B ka on-axis pressure hai $P_{B} = 500 (0.1/0.4)^{2} = 500 \times 0.0625 = 31.25 Pa ≫ 1 Pa (limit) .$ Yeh ek massive violation hai — panel limit se $31 \times$ se zyada — toh Pair B infeasible hai: iske thrusters optimizer ke dekhne se pehle hi inhibit ho jaate hain. Pair A ( $P$ saari limits clear karta hai) feasible set mein akela pair bacha hai.

Selector ka jawab: Pair A. Minimum-propellant linear program sirf feasible (non-inhibited) thrusters par search karta hai, toh woh roll couple return karta hai jo har keep-out cone respect karta hai — tab bhi jab koi violating option identical fuel use karta. Safety pehle search space filter karti hai, phir propellant uske andar optimize karta hai.

Exercise 5.2 — keep-out half-angle design karna (aur uska domain)

Pair B ki geometry ke liye hum dhundna chahte hain ki $d = 0.4$ m par ek surface kitni off-axis honi chahiye taaki safe rahe, given $P_{0} = 500$ Pa, $d_{0} = 0.1$ m, $k = 4$ , limit $1$ Pa. Keep-out half-angle $θ_{KO}$ dhundho: woh angle jis par pressure exactly $1$ Pa tak drop ho jaata hai. Phir batao ki model $θ > 9 0^{\circ}$ ke liye kya karta hai.

Recall Solution

Hum $P_{imp} = 1$ Pa ko angle ke liye solve karte hain, $d = 0.4$ m fixed rakh ke. Distance factor hai $(0.1/0.4)^{2} = 0.0625$ , toh: $1 = 500 \times 0.0625 \times cos^{4} θ_{KO} = 31.25 cos^{4} θ_{KO} .$ $cos^{4} θ_{KO} = \frac{1}{31.25} = 0.032 \Rightarrow cos θ_{KO} = 0.03 2^{1/4} .$ $0.03 2^{1/4} = (0.032)^{0.25} \approx 0.4229 \Rightarrow θ_{KO} = arccos (0.4229) \approx 65. 0^{\circ} .$ Interpretation: us range par centerline ke $6 5^{\circ}$ ke andar koi bhi sensitive surface limit se upar hai — ek bahut wide keep-out cone. $3 0^{\circ}$ par panel is mein deep andar hai, jo 5.1 ke veto ko confirm karta hai. Cone ko shrink karne ke liye ya toh thruster door move karo (bada $d$ ) ya iska plume panel se dur angle karo.

$θ$ ka domain — edge case. Formula $cos^{k} θ$ sirf $0 \leq θ \leq 9 0^{\circ}$ ke liye physical hai, woh forward hemisphere jo plume actually fill karta hai. $θ = 9 0^{\circ}$ par, $cos^{4} 9 0^{\circ} = 0$ : bilkul side mein ek surface koi plume nahi dekhti. $θ > 9 0^{\circ}$ ke liye naive formula ek negative base deta ( $cos θ < 0$ ) — aur even exponent $k = 4$ ke saath yeh galat tarike se phir positive ho jaata, nozzle ke peeche plume pressure predict karta jahan koi hai hi nahi. Physical fix: model clamp karo, $P_{imp} (d, θ) \equiv 0$ for $θ > 9 0^{\circ}$ . Exit plane ke peeche ek surface simply plume ke bahar hai — toh woh is axis par hamesha safe hai, aur keep-out cone kabhi $9 0^{\circ}$ se wrap nahi karta.