Optimal staging — equal mass ratios (for same Isp): Question bank

3.3.46 · D5 · Physics › Rocket Propulsion › Optimal staging — equal mass ratios (for same Isp)

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Definition Characters ki cast

$R_{i}$ — stage $i$ ka mass ratio: ignition par mass ÷ burnout par mass. Hamesha $R_{i} \geq 1$ (tum sirf kho sakte ho mass, burning se). Ye stage ka "leverage" hai.
$R^{*}$ — common optimal mass ratio: jab theorem apply hota hai, sab stages ek hi best value share karte hain, likha jaata hai $R_{1} = R_{2} = \dots = R^{*}$ .
$ϵ$ — structural fraction: structure ÷ (structure + fuel) ek stage ke liye. Dead weight jo tum ek kilogram tankage per carry karne ke liye majboor ho. Dekho Structural fraction.
$v_{e}$ — exhaust velocity; $Δ v = v_{e} ln R$ per stage (ye hai Tsiolkovsky). Same $v_{e}$ = sab stages ke liye same engine tech.
$m_{p}$ — payload mass: useful cargo jo tum actually deliver karna chahte ho (stack ka top). Dekho Payload fraction.
$M_{0}$ — initial total mass: pad par poora rocket — payload plus har stage ka structure aur fuel.
$m_{s i}, m_{f i}$ — stage $i$ ka structure mass aur fuel mass (isliye $ϵ = \frac{m _{s i}}{m _{s i} + m _{f i}}$ ).
$n$ — stages ki sankhya.
$k$ — per-stage constraint factor $k \equiv (\frac{m _{p}}{M _{0}})^{1/ n}$ ka shorthand; ye payload fraction ka $n$ -th root hai, aur boxed result padhta hai $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}$ .

Recall Lagrange condition kahan se aata hai?

Hum $f = \sum_{i} ln R_{i}$ (total $Δ v / v_{e}$ ) ko maximize karte hain constraint $g = \sum_{i} ln \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} = ln \frac{m _{p}}{M _{0}}$ ke subject mein. Lagrange multipliers kehta hai optimum par $\nabla f = λ \nabla g$ , yaani $\partial f / \partial R_{i} = λ \partial g / \partial R_{i}$ har $i$ ke liye. Differentiate karne par: $\frac{\partial f}{\partial R _{i}} = \frac{1}{R _{i}}$ aur $\frac{\partial g}{\partial R _{i}} = \frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}}$ , jo stated condition deta hai. Parent note ke [!formula] box mein poora walk-through dekho.

True ya False — justify karo

Theorem kehta hai equal mass ratios un stages ke liye optimal hain jo SAME

ϵ

aur SAME

v_{e}

share karte hain.

True — Lagrange condition

\frac{1}{R _{i}} = λ (\frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}})

sirf isliye

R_{i}

ka function hai kyunki

ϵ

common hai; ye har

R_{i}

ko same equation solve karne par majboor karta hai.

Agar do stages ka

ϵ

alag ho, toh optimal staging phir bhi equal mass ratios deta hai.

False — per-stage Lagrange equation ab

ϵ_{i}

par depend karta hai, isliye ratios alag hote hain. Jis stage ka

ϵ

chhota hota hai (kam dead weight) usse bada

R

milta hai. Analogous coupling ke liye dekho Variable specific impulse staging.

ϵ \to 0

ke liye total

Δ v

same rehta hai chahe fixed product

\prod R_{i}

ko kisi bhi tarah split karo.

True — objective

v_{e} ln (\prod R_{i})

hai aur sirf product fixed hai, isliye har split identical

Δ v

deta hai. Equal ratios strictly best tabhi hote hain jab

ϵ > 0

ho.

Zyada stages add karna hamesha achievable

Δ v

strictly increase karta hai.

False — pure theory mein (

ϵ \to 0

)

Δ v = v_{e} ln (M_{0} / m_{p})

n

se bilkul independent hai, isliye zyada stages na help karte hain na hurt; aur practically real staging hardware structure add karta hai, isliye 3–4 stages ke baad added dead weight

Δ v

eventually giraa deta hai.

Ek single stage hamesha

n

optimal stages ko match kar sakta hai agar bas uska

R

itna bada kar do.

False — ek stage ka ratio

R_{ma x} = 1/ ϵ

par cap hota hai (

R = \frac{1}{ϵ + ( 1 - ϵ ) k}

se jab

k \to 0

ϵ = 0.1

ke liye wo ceiling

10

hai, orbit ke liye zaruri

\sim 30

se kaafi neeche.

Payload aur initial mass ko link karne wala constraint simply

\prod R_{i} = M_{0} / m_{p}

hai.

Generally False — ye sirf

ϵ \to 0

par hold karta hai. Sahi constraint hai

\frac{m _{p}}{M _{0}} = \prod \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}}

, jo har

R_{i}

mein nonlinear hai.

Kyunki

ln

concave hai, plain AM–GM prove karta hai ki equal ratios kisi bhi

ϵ

ke liye optimal hain.

False — AM–GM sirf tab kaam karta hai jab product fixed ho (

ϵ \to 0

ϵ > 0

ke liye constraint fixed product nahi hai, isliye tumhe genuinely Lagrange multipliers argument chahiye.

R_{1}

(first stage) ko bahut bada aur

R_{2}

ko bahut chhota banana performance waste karta hai jab

ϵ > 0

ho.

True — oversized lower stage wo structure carry karta hai jo use barely chahiye, jabki upper stage starved hota hai; nonlinear constraint imbalance ko punish karta hai,

Δ v

ko equal-ratio value se neeche le jaata hai.

Payload fraction

m_{p} / M_{0}

zyada

Δ v

demand karne par shrink hota hai.

True — zyada

Δ v

ke liye bada

R^{*}

chahiye, aur

\frac{m _{p}}{M _{0}} = (\frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}})^{n}

jaise

R^{*}

badhta hai, girta hai. Dekho Payload fraction.

Error dhundo

Δ v_{t o t a l} = v_{e} ln (R_{1} R_{2})

, aur

R_{1} R_{2}

fixed hai, isliye staging pointless hai."

Product sirf

ϵ \to 0

par fixed hota hai; real

ϵ

ke saath constraint nonlinear hai aur staging mid-flight structure shed karne deta hai, effective ratio ko single stage ki ceiling se upar le jaata hai.

"Stage 2 ka payload sirf

m_{p}

hai."

Nahi — stage 1 ke liye, stage 2 (uska structure aur remaining fuel) us payload ka part hai jo use lift karna hai. Stage 2 akela

m_{p} + m_{s 2}

ko apna burnout mass maanta hai.

"Structure mass

m_{s i} = ϵ m_{f i}

Galat scaling —

ϵ = \frac{m _{s i}}{m _{s i} + m _{f i}}

, isliye

m_{s i} = \frac{ϵ}{1 - ϵ} m_{f i}

ϵ = 0.1

ke liye ye

\frac{1}{9} m_{f i}

hai,

0.1 m_{f i}

nahi.

"Kyunki

R^{*} = 1/ (ϵ + (1 - ϵ) k)

aur

k = (m_{p} / M_{0})^{1/ n}

hai,

ϵ \to 1

lene se bada ratio milta hai."

Bilkul ulta — jaise

ϵ \to 1

R^{*} \to 1

(poora structure, koi fuel nahi), zero

Δ v

deta hai. Chhota

ϵ

hi bade ratios allow karta hai.

"Agli stage ko di gayi mass

M_{i} = M_{i - 1} / R_{i}

hai."

Ye spent structure drop karna ignore karta hai. Sahi:

M_{i} = M_{i - 1} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}}

, jo

M_{i - 1} / R_{i}

se chhota hai kyunki structure bhi jettison hota hai.

Δ v_{1} = v_{e} ln R_{1}

mein burnout mass

= m_{p} + m_{s 2}

hota hai."

Nahi — stage 1 burnout tak

m_{p} + m_{s 2} + m_{f 2} + m_{s 1}

tak jaata hai (stage 2 abhi bhi full hai, stage 1 structure abhi attached hai). Stage 1 ka structure uske burnout ke baad drop hota hai.

"Equal ratios ka matlab hai har stage mein equal fuel mass."

Equal ratios hain, equal masses nahi — lower stage ek bhaari vehicle move karta hai, isliye

R_{1} = R_{2}

ke liye stage 1 mein stage 2 se kaafi zyada fuel chahiye.

Why questions

Ek hi

ϵ

sab stages ke liye kyun equal

R_{i}

force karta hai?

Har stage ke liye Lagrange stationarity condition sirf us stage ke apne

R_{i}

aur shared

ϵ

par depend karti hai; identical equations ke identical solutions hote hain, isliye sab

R_{i}

R^{*}

par milte hain.

ln R

ek stage ke "bang for the buck" ka sahi measure kyun hai?

Tsiolkovsky deta hai

Δ v = v_{e} ln R

, isliye velocity

ln R

mein linearly add hoti hai;

ln

values compare karna actual speed contributions compare karna hai, aur

ln

ki concavity hi balancing ko reward karti hai.

Mass ratio kabhi 1 se kam kyun nahi ho sakta?

R =

ignition mass ÷ burnout mass, aur fuel jalaane se sirf mass remove hota hai, isliye numerator kam se kam denominator ke barabar hai —

R \geq 1

, equality ka matlab koi fuel nahi jala.

Total-

Δ v

formulas natural log ko kisi bhi doosre function se kyun prefer karte hain?

Kyunki variable-mass body par momentum conservation integrate karke

\int v_{e} d m / m = v_{e} ln (mass ratio)

milta hai — log physics se nikalta hai, ye koi chosen convenience nahi hai.

Staging ek monolithic rocket ko kyun beat karta hai jabki ideal

Δ v

same hai?

Monolithic rocket ka

R

structure se

1/ ϵ

par cap hota hai; staging jettison ke baad ek fresh chhote vehicle par shift karti hai, isliye effective ratios compound hote hain (

R^{n}

) us ceiling se kaafi aage.

Equal ratios ka advantage

ϵ \to 0

par kyun khatam hota hai?

Koi structure nahi hone par constraint ek fixed product

\prod R_{i}

tak reduce ho jaata hai, aur

v_{e} ln (\prod R_{i})

phir split ko bilkul ignore karta hai — har partition tie karta hai.

Edge cases

ϵ \to 0

R^{*}

kya ban jaata hai?

R^{*} = (M_{0} / m_{p})^{1/ n}

aur

Δ v = v_{e} ln (M_{0} / m_{p})

n

se independent — idealized "free lunch" limit jahan staging koi ideal penalty ya bonus nahi deta.

ϵ \to 1

: ek stage ka kya hota hai?

Stage pure structure hai koi fuel nahi;

R^{*} \to 1

, ye zero

Δ v

contribute karta hai, aur staging help nahi kar sakti — ye sirf dead weight hai.

n = 1

(single stage): kya general formula phir bhi hold karta hai?

Haan —

\frac{m _{p}}{M _{0}} = \frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}}

, ceiling

R_{ma x} = 1/ ϵ

deta hai jab

m_{p} / M_{0} \to 0

; multi-stage boxed result

n = 1

par isi mein reduce ho jaata hai.

Agar constraint

(⋆)

se nikla

R^{*}

1 se neeche aaye toh kya?

Ye ek impossible mission signal karta hai — tum mass budget se zyada payload maang rahe ho; koi physical fuel split use tak nahi pahunch sakta, kyunki

R < 1

forbidden hai.

m_{p} \to 0

(throw-away probe, koi payload nahi):

R^{*}

kya hai?

k = (m_{p} / M_{0})^{1/ n} \to 0

, isliye

R^{*} \to 1/ ϵ

per stage — har stage apni structural ceiling hit karta hai, maximum jo ek real stage deliver kar sakta hai.

m_{p} \to M_{0}

(payload lagbhag poora rocket):

k \to 1

, isliye

R^{*} \to 1

aur

Δ v \to 0

— lagbhag koi fuel nahi, lagbhag koi acceleration nahi, degenerate "already at your mass budget" case.

Do stages

R_{1} = 8, R_{2} = 2

R_{1} = R_{2} = 4

ϵ = 0

: kaun jeetega?

Koi nahi — dono

v_{e} ln 16

identically dete hain, kyunki

ϵ = 0

par sirf product

R_{1} R_{2} = 16

matter karta hai, split nahi.

Recall Quick self-test

Wo single common quantity jo sab stages mein match karni chahiye taaki equal ratios optimal hoon ::: structural fraction $ϵ$ (same $v_{e}$ ke saath). Real constraint jo " $\prod R_{i}$ fixed" ki jagah leta hai jab $ϵ > 0$ ho ::: $\frac{m _{p}}{M _{0}} = \prod \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}}$ . Constraint factor $k$ ki definition ::: $k = (m_{p} / M_{0})^{1/ n}$ , payload fraction ka $n$ -th root. Ek single stage ke mass ratio par hard ceiling ::: $R_{ma x} = 1/ ϵ$ .