Optimal staging — equal mass ratios (for same Isp): Worked examples

3.3.46 · D3 · Physics › Rocket Propulsion › Optimal staging — equal mass ratios (for same Isp)

Yeh page parent note on optimal staging ki "har case run karo" companion hai. Parent ne prove kiya tha kyun equal mass ratios jeetate hain. Yahan hum us machinery ko har tarah ke input se guzaarte hain — badi aur choti structural fraction, ek stage ya kai stages, degenerate cases jahan maths tootne ki dhamki deta hai, ek real launch problem, aur ek exam twist. Har example batata hai ki woh matrix ke kis cell mein aata hai.

Shuru karne se pehle, do tools ka ek-line reminder — taaki koi bhi symbol bina explanation ke na aaye.

Recall Har symbol ka matlab (parent mein build kiya gaya)

$m_{p}$ ::: payload mass — woh useful cargo jo hum deliver karte hain. $M_{0}$ ::: liftoff par pad par total mass (payload + saari structure + saara fuel). $R_{i}$ ::: stage $i$ ka mass ratio = (jab yeh ignite hota hai tab mass) ÷ (jab yeh burn out hota hai tab mass). Hamesha $R_{i} \geq 1$ . $ϵ$ ::: structural fraction = structure ÷ (structure + fuel) ek stage ke liye. "Dead weight tax." $v_{e}$ ::: exhaust speed. Yeh ek mass ratio ko speed mein convert karta hai $Δ v = v_{e} ln R$ ke zariye. $R^{*}$ ::: woh single value jo optimum par saare stages share karte hain.

Scenario matrix

Is topic se jo bhi problem aa sakti hai woh inhi cells mein se ek hoti hai. Neeche ke examples har ek ko hit karte hain.

Cell	Kya vary karta hai	Kaun si danger test karta hai	Example
A	$ϵ = 0$ (koi dead weight nahi)	Kya product-fixed shortcut kaam karta hai?	Ex 1
B	$ϵ > 0$ , small ( $n = 2$ )	Nonlinear constraint solve karna	Ex 2
C	Equal vs unequal split, $ϵ > 0$	Prove karna ki equal strictly better hai	Ex 3
D	Badhta $n$ (1, 2, 3, 4)	Diminishing returns; $R_{ma x} = 1/ ϵ$ ceiling	Ex 4
E	Degenerate: $R^{*} \to 1/ ϵ$	Kya hota hai jab $m_{p} / M_{0} \to 0$	Ex 5
F	Degenerate: $R_{i} = 1$ (empty stage)	Zero-fuel stage kuch nahi add karta	Ex 6
G	Real-world word problem	Orbit reach karo; $n$ chuno	Ex 7
H	Exam twist: alag $v_{e}$ per stage	Jab "equal $R$ " rule toot jaata hai	Ex 8

Worked example Ex 1 — Cell A: zero structure, product fixed hai

$v_{e} = 3000$ m/s, $M_{0} / m_{p} = 16$ , $ϵ = 0$ . Do stages. Dikhao ki koi bhi split jisme $R_{1} R_{2} = 16$ ho wahi $Δ v$ deta hai, aur equal ratios $R_{1} = R_{2} = 4$ aisa hi ek split hai.

Forecast: Andaza lagao — kya split yahan matter karta hai? (Jawab: nahi, kyunki $ln$ ek product ko sum mein badal deta hai.)

$ϵ = 0$ par constraint likho. Yeh step kyun? Parent se, jab $ϵ \to 0$ tab factor $\frac{1 - ϵ R}{( 1 - ϵ ) R} \to \frac{1}{R}$ ho jaata hai, to $(⋆)$ collapse ho kar $\frac{m _{p}}{M _{0}} = \frac{1}{R _{1} R _{2}}$ ban jaata hai, yaani $R_{1} R_{2} = M_{0} / m_{p} = 16$ .
Objective. Kyun? Total speed add hoti hai: $Δ v = v_{e} (ln R_{1} + ln R_{2}) = v_{e} ln (R_{1} R_{2})$ . Sirf product enter karta hai.
Fixed product plug karo. $Δ v = 3000 ln 16 = 3000 \times 2.7726 = 8318$ m/s. Equal ratios $R_{1} = R_{2} = 4$ dete hain $ln 4 + ln 4 = ln 16$ — same number.

Verify: Lopsided split try karo $R_{1} = 8, R_{2} = 2$ : $ln 8 + ln 2 = 2.0794 + 0.6931 = 2.7726 = ln 16$ . Bilkul same. ✓ Units: $(m/s) \times (dimensionless ln) = m/s$ . ✓

Worked example Ex 2 — Cell B: real structure, nonlinear constraint solve karo (

$n = 2$ ) $m_{p} = 1000$ kg, $M_{0} = 10000$ kg, $ϵ = 0.1$ , $v_{e} = 3000$ m/s. $R^{*}$ aur $Δ v$ nikalo.

Forecast: $ϵ = 0.1$ ke saath, kya $R^{*}$ zero-structure value $10 \approx 3.16$ se bada hoga ya chota? (Dead weight aapko cost karta hai, isliye same payload fraction squeeze karne ke liye $R^{*}$ bada hona chahiye.)

$k$ compute karo. Kyun? Parent ka closed form $k = (m_{p} / M_{0})^{1/ n} = (0.1)^{1/2} = 0.31623$ use karta hai.
$R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}$ mein plug karo. Yeh step kyun? Yeh $(⋆)$ ke $n$ -th-root form ko directly solve karta hai. $R^{*} = \frac{1}{0.1 + 0.9 \times 0.31623} = \frac{1}{0.38461} = 2.600$ .
Speed. $Δ v = n v_{e} ln R^{*} = 2 \times 3000 \times ln 2.600 = 6000 \times 0.9555 = 5733$ m/s.

Verify: Masses rebuild karo. Stage-1 fuel $= M_{0} (1 - 1/ R^{*}) = 10000 (1 - 0.38461) = 6154$ kg; burnout $3846$ ; structure $\frac{0.1}{0.9} \times 6154 = 684$ ; hand-off $3846 - 684 = 3162$ kg. Phir $M_{1} / M_{0} = 3162/10000 = 0.3162 = k$ ✓, aur dobara karne par exactly $1000$ kg payload milta hai ✓ (parent ke worked check se match karta hai).

Worked example Ex 3 — Cell C:

$ϵ > 0$ hone par equal strictly unequal se better hai Wohi rocket jaise Ex 2 mein tha ( $M_{0} = 10000$ , $m_{p} = 1000$ , $ϵ = 0.1$ , $v_{e} = 3000$ ). Optimal equal split ( $R_{1} = R_{2} = 2.600$ ) ko ek forced unequal split se compare karo. Fair comparison ke liye, dono ko sacchi constraint $(⋆)$ satisfy karni chahiye aur exactly $1000$ kg deliver karna chahiye.

Forecast: Ex 1 ne kaha ki $ϵ = 0$ par splits tie karte hain. Ab $ϵ > 0$ hai — andaza lagao ki tie tooti ya nahi.

Constraint par ek unequal pair chuno. Kyun? Hum $R_{1}, R_{2}$ freely nahi chun sakte; inhein $\frac{1 - ϵ R _{1}}{0.9 R _{1}} \cdot \frac{1 - ϵ R _{2}}{0.9 R _{2}} = 0.1$ satisfy karna hoga. $R_{1} = 3.2$ set karo aur $R_{2}$ solve karo: pehla factor hai $\frac{1 - 0.32}{0.9 \times 3.2} = \frac{0.68}{2.88} = 0.23611$ ; doosre ki zaroorat hai $= 0.1/0.23611 = 0.42353$ . To $\frac{1 - 0.1 R _{2}}{0.9 R _{2}} = 0.42353 \Rightarrow 1 - 0.1 R_{2} = 0.38118 R_{2} \Rightarrow R_{2} = 1/0.48118 = 2.0782$ .
Dono $Δ v$ 's compute karo. Kyun? Direct comparison. Equal: $Δ v_{e q} = 6000 ln 2.600 = 5733$ m/s. Unequal: $Δ v_{u n} = 3000 (ln 3.2 + ln 2.0782) = 3000 (1.16315 + 0.73139) = 3000 \times 1.89454 = 5684$ m/s.
Gap padho. $5733 - 5684 = 49$ m/s unequal jaane se lost.

Verify: $5733 > 5684$ , to equal ratios jeetate hain jab $ϵ > 0$ ✓. Advantage $ϵ = 0$ par gaya (Ex 1) aur $ϵ = 0.1$ par dikhta hai — exactly wahi jo Lagrange argument ne predict kiya tha.

Worked example Ex 4 — Cell D:

$n$ ke saath $R^{*}$ aur $Δ v$ kaise change hote hain Fixed $M_{0} / m_{p} = 30$ (yaani $m_{p} / M_{0} = 1/30$ ), $ϵ = 0.1$ , $v_{e} = 3000$ . $n = 1, 2, 3, 4$ tabulate karo.

Forecast: Kya ek giant stage ( $n = 1$ ) kaam bhi karta hai? (Yaad karo $R_{ma x} = 1/ ϵ = 10$ ; humein effective 30 chahiye. Dekho kya toota jaata hai.)

$n = 1$ . Yahan se kyun shuru karein? Yeh ceiling expose karta hai. $k = (1/30)^{1} = 0.03333$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.03333} = \frac{1}{0.13} = 7.6923$ . Ratio $7.6923$ ka ek stage deta hai $Δ v = 3000 ln 7.6923 = 6121$ m/s — orbit ke liye kaafi nahi aur required payload fraction barely reachable hai ( $R < R_{ma x} = 10$ , to yeh thoda sa physically possible hai lekin weak hai).
$n = 2$ . $k = (1/30)^{1/2} = 0.18257$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.18257} = \frac{1}{0.26432} = 3.7833$ , $Δ v = 2 \times 3000 ln 3.7833 = 7982$ m/s.
$n = 3$ . $k = (1/30)^{1/3} = 0.32183$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.32183} = \frac{1}{0.38965} = 2.5664$ , $Δ v = 3 \times 3000 ln 2.5664 = 8481$ m/s.
$n = 4$ . $k = (1/30)^{1/4} = 0.42729$ , $R^{*} = \frac{1}{0.1 + 0.9 \times 0.42729} = \frac{1}{0.48456} = 2.0637$ , $Δ v = 4 \times 3000 ln 2.0637 = 8703$ m/s.

Neeche di gayi figure inhi chaar points ko plot karti hai. Ise left se right padho: blue curve total $Δ v$ hai, har point apne $R^{*}$ ke saath tagged hai; points ke beech orange numbers gains hain $+ 1861, + 499, + 222$ — dikhne mein shrink kar rahe hain. Red dashed line single-stage ceiling $v_{e} ln (1/ ϵ) = 6908$ m/s hai, jiske neeche yahan $n = 1$ bhi baitha hai (kyunki hamaara payload negligible nahi hai).

Verify: $Δ v$ badhta hai $6121 \to 7982 \to 8481 \to 8703$ — increasing lekin shrinking gaps ke saath ( $1861$ , phir $499$ , phir $222$ ). Yahi "3–4 stages se aage diminishing returns" hai jo parent ne claim kiya tha. ✓ Dekho Staging basics.

Worked example Ex 5 — Cell E: ceiling

$R^{*} \to 1/ ϵ$ jab payload shrink karta hai $ϵ = 0.1$ , ek stage ( $n = 1$ ). $m_{p} / M_{0}$ ko $0$ ki taraf push karo aur $R^{*}$ dekho.

Forecast: Jab aap tinier payload fraction demand karte ho, toh kya ek stage ka $R^{*}$ bina bound ke barhta hai, ya kisi wall se takraata hai?

Limit lo. Kyun? $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}$ jahan $k = (m_{p} / M_{0})^{1/1} = m_{p} / M_{0} \to 0$ . Tab $R^{*} \to \frac{1}{ϵ} = 10$ .
Interpret karo. Kyun? Zero payload ke saath bhi, ek single stage ratio $10$ se aage nahi ja sakta: usse apna khud ka $10%$ structure uthana padta hai. Max single-stage speed $= v_{e} ln (1/ ϵ) = 3000 ln 10 = 6908$ m/s.
Consequence. Koi bhi mission jisme $M_{0} / m_{p} > 10$ ho woh ek stage ke saath impossible hai — aapko zaroor stage karna hoga. Yahi core lesson hai.

Verify: $m_{p} / M_{0} = 0.001$ plug karo: $R^{*} = \frac{1}{0.1 + 0.9 \times 0.001} = \frac{1}{0.1009} = 9.911$ — $10$ ki taraf creep kar raha hai lekin kabhi reach nahi karta ✓. Parent ki Structural fraction ceiling se tie karta hai.

Worked example Ex 6 — Cell F: ek degenerate "empty" stage kuch nahi karta

Do stages, lekin maan lo humne accidentally stage 2 mein koi fuel nahi ( $m_{f 2} = 0$ ) daala, to $R_{2} = 1$ . $R_{1} = 4$ , $v_{e} = 3000$ .

Forecast: Kya dead stage $Δ v$ mein help karta hai, hurt karta hai, ya kuch nahi karta?

Empty stage se speed. Kyun? $Δ v_{2} = v_{e} ln R_{2} = 3000 ln 1 = 3000 \times 0 = 0$ . Ratio $1$ kuch contribute nahi karta — $ln 1 = 0$ .
Kuch nahi se bhi bura. Kyun? Uski structure $m_{s 2} > 0$ abhi bhi dead mass hai jise stage 1 ko uthana pada. To fueled stage ka effective $R_{1}$ ab chhota hai agar woh structure wahan hoti hi nahi. Empty stage strictly hurt karta hai.
Lesson. Optimal design mein har stage ka $R_{i} = R^{*} > 1$ hota hai; $R = 1$ wala stage delete karna chahiye, fly nahi karna.

Verify: $ln 1 = 0$ exactly ✓, to $Δ v_{s t a g e 2} = 0$ ; total speed sirf stage 1 se aati hai: $3000 ln 4 = 4159$ m/s, aur dead stage par koi bhi structure ise aur kam kar deti hai. ✓

Worked example Ex 7 — Cell G: real-world — satellite ko LEO par pahunchao

Ek satellite $m_{p} = 1500$ kg ko low Earth orbit tak pahunchana hai jisme $Δ v_{r e q} = 9400$ m/s chahiye (gravity + drag losses including). Engines dete hain $v_{e} = 3200$ m/s, $ϵ = 0.08$ . Kitne equal stages, aur launch mass $M_{0}$ kya hogi?

Forecast: Computing se pehle $n$ guess karo. (Orbit ke liye $M_{0} / m_{p}$ bada chahiye; single stage $R = 1/0.08 = 12.5$ par cap karta hai, yaani $Δ v \leq 3200 ln 12.5 = 8083$ m/s — 9400 se kam. To $n \geq 2$ . Lekin kya $n = 2$ best hai, ya zyada jaana chahiye?)

$n = 1$ rule out karo. Pehle kyun? Ek single stage $Δ v = v_{e} ln (1/ ϵ) = 3200 ln 12.5 = 8083$ m/s $< 9400$ se aage nahi ja sakta. Infeasible. To $n \geq 2$ .
Har $n$ ke liye, woh launch mass nikalo jo woh demand karta hai. Yeh step kyun? "Kitne stages" ek optimization over $n$ hai: har feasible $n$ ko alag $M_{0}$ chahiye, aur hum sabse chhoti $M_{0}$ (sabse sasta rocket) chahte hain. Diye gaye $n$ ke liye, per-stage ratio mission se fix hota hai: $R^{*} = exp (\frac{Δ v _{r e q}}{n v _{e}})$ , aur phir $(⋆)$ deta hai $\frac{M _{0}}{m _{p}} = (\frac{( 1 - ϵ ) R ^{*}}{1 - ϵ R ^{*}})^{n}$ .
Tabulate karo. Kyun? Launch masses directly compare karo.
- $n = 2$ : $R^{*} = exp (9400/6400) = 4.344$ ; feasible since $4.344 < 12.5$ . Factor $\frac{0.92 \times 4.344}{1 - 0.08 \times 4.344} = \frac{3.99648}{0.65248} = 6.1251$ ; $M_{0} / m_{p} = 6.125 1^{2} = 37.52$ ; $M_{0} = 1500 \times 37.52 = 56276$ kg.
- $n = 3$ : $R^{*} = exp (9400/9600) = 2.6635$ ; feasible. Factor $\frac{0.92 \times 2.6635}{1 - 0.08 \times 2.6635} = \frac{2.45042}{0.78692} = 3.1140$ ; $M_{0} / m_{p} = 3.114 0^{3} = 30.20$ ; $M_{0} = 45305$ kg.
- $n = 4$ : $R^{*} = exp (9400/12800) = 2.0844$ ; feasible. Factor $\frac{0.92 \times 2.0844}{1 - 0.08 \times 2.0844} = \frac{1.91765}{0.83325} = 2.30140$ ; $M_{0} / m_{p} = 2.3014 0^{4} = 28.05$ ; $M_{0} = 42069$ kg.
Trade-off padho. Launch mass girti hai jab $n$ badhta hai ( $56.3 \to 45.3 \to 42.1$ t) lekin shrinking savings ke saath, exactly Ex 4 ke diminishing returns. Kyunki har added stage real complexity aur interstage mass laata hai (parent ki caveat), $n = 2$ ya $n = 3$ practical sweet spot hai; $n = 3$ , $n = 2$ ke upar $\approx 11$ t launch mass trim karta hai aur yeh recommended choice hai.

Verify: Check karo ki $n = 2$ design mission satisfy karta hai: $Δ v = 2 \times 3200 \times ln 4.344 = 6400 \times 1.46875 = 9400$ m/s ✓, payload fraction $1500/56276 = 0.02666$ ✓. Aur $n = 3$ : $3 \times 3200 ln 2.6635 = 9600 \times 0.97917 = 9400$ m/s ✓ with $M_{0} = 45305$ kg. Launch masses strictly $n$ ke saath decrease karti hain ✓. Uses Payload fraction.

Worked example Ex 8 — Cell H: exam twist — stages ke

alag $v_{e}$ "Equal $R$ " rule ne same $v_{e}$ assume kiya tha. Ab stage 1 ka $v_{e 1} = 2800$ m/s hai, stage 2 ka $v_{e 2} = 3400$ m/s hai, $ϵ \to 0$ , product fixed $R_{1} R_{2} = 16$ . Kya ratios phir bhi equal hone chahiye?

Forecast: Kaun sa stage bada mass ratio deserve karta hai — efficient wala ya weak wala?

Objective aur constraint likho. Kyun? Ab $Δ v = v_{e 1} ln R_{1} + v_{e 2} ln R_{2}$ (do $ln$ terms par unequal weights), aur $ϵ \to 0$ par constraint hai $ln R_{1} + ln R_{2} = ln 16$ . Variables introduce karo $x = ln R_{1}, y = ln R_{2}$ to objective linear ban jaata hai: $Δ v = v_{e 1} x + v_{e 2} y$ , constraint $x + y = ln 16$ , physical bounds ke saath $x \geq 0, y \geq 0$ (ratios $\geq 1$ ).
Lagrangian build karo. Kyun? Stationary points locate karne ke liye. Equality ke liye multiplier $λ$ ke saath: $L (x, y, λ) = v_{e 1} x + v_{e 2} y - λ (x + y - ln 16) .$ $\partial L / \partial x = v_{e 1} - λ = 0$ aur $\partial L / \partial y = v_{e 2} - λ = 0$ demand karte hain $λ = v_{e 1} = v_{e 2}$ simultaneously — impossible jab $v_{e 1} \neq = v_{e 2}$ . To koi interior stationary point nahi hai; ek line segment par linear objective ka maximum endpoint par hona chahiye (ek KKT corner jahan ek bound $x = 0$ ya $y = 0$ activate hoti hai).
Corners check karo. Kyun? Feasible segment $x + y = ln 16, x, y \geq 0$ ke do endpoints hain. Kyunki $v_{e 2} > v_{e 1}$ , saara "budget" higher- $v_{e}$ stage mein daalo: $x = 0 (R_{1} = 1), y = ln 16 (R_{2} = 16)$ .
Compare karo. Equal ( $R_{1} = R_{2} = 4$ ): $Δ v = 2800 ln 4 + 3400 ln 4 = (2800 + 3400) \times 1.3863 = 8595$ m/s. Corner ( $R_{1} = 1, R_{2} = 16$ ): $Δ v = 2800 \times 0 + 3400 ln 16 = 3400 \times 2.7726 = 9427$ m/s.
Lesson. Alag $I_{s p}$ ke saath, equal ratios ab optimal nahi hain — maximum boundary ki taraf push ho jaata hai, aur high- $v_{e}$ stage ko bada mass ratio carry karna chahiye (yahan yeh poora fuel budget le leta hai). Clean "equal $R^{*}$ " theorem same- $I_{s p}$ assumption ka ek special privilege hai; us assumption ko hatao aur poora answer badal jaata hai. Yeh precisely woh regime hai jo Variable specific impulse staging handle karta hai.

Verify: $9427 > 8595$ ✓ — corner equal split ko beat karta hai, yeh confirm karta hai ki "same $I_{s p}$ " hypothesis parent theorem ke liye essential hai. (Method: Lagrange multipliers with boundary/KKT check.)

Recall Rapid self-test

Agar $ϵ = 0$ aur product $R_{1} R_{2}$ fixed ho, to kya split $Δ v$ change karta hai? ::: Nahi — $ln$ sirf product ko matter karta hai (Ex 1). $ϵ > 0$ hone par equal unequal ko kyun beat karta hai? ::: Constraint $(⋆)$ nonlinear ban jaati hai; unequal ratios structural mass waste karte hain (Ex 3). Structural fraction $ϵ$ wale ek single stage ka maximum mass ratio kya hai? ::: $R_{ma x} = 1/ ϵ$ (Ex 5). $R_{i} = 1$ wala stage kitna $Δ v$ contribute karta hai? ::: Zero, aur uski structure hurt bhi karti hai (Ex 6). "Equal $R$ " rule kab fail hota hai? ::: Jab stages ke alag $v_{e}$ hon — tab high- $v_{e}$ stage ko favour karo (Ex 8).