Optimal staging — equal mass ratios (for same Isp)

3.3.46 · Physics › Rocket Propulsion

The setup

Humare paas ek rocket hai jisme:

Payload mass $m_{p}$ (woh cheez jo hum space mein pohonchaana chahte hain)
Multiple stages, har ek ke paas structural mass aur fuel hai
Same specific impulse $I_{s p}$ saare stages ke liye (same engine technology)

Goal: Final velocity $Δ v_{t o t a l}$ maximize karna by choose karna ki kitna fuel har stage mein jaata hai.

Deriving the optimization

Two-stage example (then generalize)

Stage 2 (upper stage):

$m_{p} + m_{s 2} + m_{f 2}$ mass ke saath ignite hoti hai (payload + structure + fuel)
$m_{p} + m_{s 2}$ tak burn out hoti hai
Deti hai $Δ v_{2} = v_{e} ln (R_{2}) = v_{e} ln (\frac{m _{p} + m _{s 2} + m _{f 2}}{m _{p} + m _{s 2}})$

Stage 1 (lower stage):

$m_{p} + m_{s 2} + m_{f 2} + m_{s 1} + m_{f 1}$ ke saath ignite hoti hai
$m_{p} + m_{s 2} + m_{f 2}$ tak burn out hoti hai (stage 2, stage 1 ke liye "payload" ban jaati hai)
Deti hai $Δ v_{1} = v_{e} ln (R_{1}) = v_{e} ln (\frac{m _{p} + m _{s 2} + m _{f 2} + m _{s 1} + m _{f 1}}{m _{p} + m _{s 2} + m _{f 2}})$

Total velocity: $Δ v_{t o t a l} = Δ v_{1} + Δ v_{2} = v_{e} (ln R_{1} + ln R_{2}) = v_{e} ln (R_{1} R_{2})$

The key insight: structural fraction

Maano structural fraction $ϵ_{i} = \frac{m _{s i}}{m _{s i} + m _{f i}}$ (us stage ke "dry+fuel" ka fraction jo structure hai).

Ek given total initial mass $M_{0}$ aur payload $m_{p}$ ke liye, constraint yeh hai: $M_{0} = m_{p} + \sum_{i} (m_{s i} + m_{f i})$

Hum maximize karna chahte hain $Δ v_{t o t a l} = v_{e} \sum_{i} ln R_{i}$ is constraint ke subject to.

Formula Why equal mass ratios are optimal

Claim: $n$ stages ke liye jinka same $ϵ$ aur same $v_{e}$ hai, maximum $Δ v$ tab milta hai jab: $R_{1} = R_{2} = \dots = R_{n} = R^{*}$

Careful proof (Lagrange multipliers). Ek stage $i$ ke liye jiska initial mass $M_{i - 1}$ hai (mass jo stage $i$ mein enter kar rahi hai) aur burnout mass hai, hum mass ratio ko structural fraction use karke likhte hain. Agar ek stage fuel $m_{f i}$ burn karti hai aur uski structure $m_{s i} = \frac{ϵ}{1 - ϵ} m_{f i}$ hai, toh: $R_{i} = \frac{M _{i - 1}}{M _{i - 1} - m _{f i}}$ aur mass jo agli stage ko pass hoti hai (is stage ki structure drop karne ke baad) woh hai $M_{i} = M_{i - 1} - m_{f i} - m_{s i} .$ Sab kuch $R_{i}$ ke terms mein express karo. Kyunki $m_{f i} = M_{i - 1} (1 - 1/ R_{i})$ aur $m_{s i} = \frac{ϵ}{1 - ϵ} m_{f i}$ : $M_{i} = M_{i - 1} (\frac{1}{R _{i}} - \frac{ϵ}{1 - ϵ} (1 - \frac{1}{R _{i}})) = M_{i - 1} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} .$ $M_{0}$ se $M_{n} = m_{p}$ tak chain karo: $\frac{m _{p}}{M _{0}} = \prod_{i = 1}^{n} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} . (⋆)$ Yeh asli constraint hai — yeh " $\prod R_{i} = const$ " NAHI hai. Hum maximize karte hain $f = \sum_{i} ln R_{i}$ subject to $g = \sum_{i} ln \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} = ln \frac{m _{p}}{M _{0}}$ . $\nabla f = λ \nabla g$ set karke: $\frac{1}{R _{i}} = λ (\frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}}) for every i .$ Right-hand side sirf $R_{i}$ ka function hai (kyunki $ϵ$ saare stages ke liye same hai), toh har $R_{i}$ ko same equation satisfy karni hogi ⇒ saare $R_{i}$ equal hain, $R_{i} = R^{*}$ .

Limit $ϵ \to 0$ : constraint $(⋆)$ ban jaata hai $\frac{m _{p}}{M _{0}} = \prod (1/ R_{i})$ , yaani $\prod R_{i} = M_{0} / m_{p}$ fixed. Sirf is special case mein product fixed hota hai, aur tab AM–GM (concavity of $ln$ ) directly deta hai equal $R_{i} = (M_{0} / m_{p})^{1/ n}$ as optimal. $ϵ > 0$ ke liye coupling nonlinear hai, lekin upar wala Lagrange argument phir bhi equal ratios force karta hai.

Yeh intuitively kyun samajh aata hai?

$ln R$ ko fuel burn karne ka "bang for your buck" socho. Agar ek stage ka $R$ doosre se bahut zyada hai, toh iska matlab hai woh stage proportionally kam dead weight carry kar raha hai. Lekin stages coupled hain: stage 1 ko stage 2 lift karni padti hai. Agar stage 2 heavy hai (chhota $R_{2}$ ), stage 1 zyada kaam karti hai. Optimal balance yeh hai ki "efficiency gain" ko evenly saare stages mein spread karo.

Worked example Two-stage rocket with

$ϵ = 0.1$ Payload $m_{p} = 1000$ kg, total initial mass $M_{0} = 10000$ kg, structural fraction $ϵ = 0.1$ .

Step 1 — equal ratios set up karo. Theorem ke mutabik, hum chahte hain $R_{1} = R_{2} = R^{*}$ . Constraint $(⋆)$ use karo $n = 2$ ke saath: $\frac{m _{p}}{M _{0}} = (\frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}})^{2} = \frac{1000}{10000} = 0.1.$

Step 2 — $R^{*}$ solve karo. Square root lo: $\frac{1 - 0.1 R ^{*}}{0.9 R ^{*}} = 0.1 = 0.3162$ . Toh $1 - 0.1 R^{*} = 0.2846 R^{*} \Rightarrow 1 = 0.3846 R^{*} \Rightarrow R^{*} = 2.60.$

Step 3 — masses reconstruct karo (check).

Stage 1 ignition: $M_{0} = 10000$ kg. Fuel burned $m_{f 1} = M_{0} (1 - 1/ R^{*}) = 10000 (1 - 0.3846) = 6154$ kg. Burnout mass $= 3846$ kg.
Stage 1 structure $m_{s 1} = \frac{ϵ}{1 - ϵ} m_{f 1} = \frac{0.1}{0.9} \cdot 6154 = 684$ kg. Drop kar do.
Stage 2 ignition: $3846 - 684 = 3162$ kg. Check: $M_{1} = M_{0} \cdot \frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}} = 10000 \cdot 0.3162 = 3162$ kg. ✓
Stage 2 fuel $m_{f 2} = 3162 (1 - 1/2.60) = 1946$ kg → burnout $1216$ kg. Structure $m_{s 2} = \frac{0.1}{0.9} \cdot 1946 = 216$ kg → final $1216 - 216 = 1000$ kg $= m_{p}$ . ✓
$R_{2} = 3162/1216 = 2.60 = R_{1}$ . ✓ Equal ratios, constraint satisfied.

Step 4 — the payoff. $v_{e} = 3000$ m/s ke saath: $Δ v = 2 v_{e} ln R^{*} = 2 \cdot 3000 \cdot ln 2.60 = 5734$ m/s.

Worked example Numerical comparison — equal beats unequal

Lo $v_{e} = 3000$ m/s aur negligible structure ( $ϵ \to 0$ ) taaki constraint sirf $R_{1} R_{2} = M_{0} / m_{p}$ ho. Choose karo $M_{0} / m_{p} = 16$ , toh product fixed hai $16$ par.

Case A: Equal ratios $R_{1} = R_{2} = 16 = 4$ : $Δ v = v_{e} (ln 4 + ln 4) = 3000 \cdot 2 ln 4 = 8318 m/s .$ Case B: Unequal $R_{1} = 8$ , $R_{2} = 2$ (same product $8 \times 2 = 16$ ): $Δ v = v_{e} (ln 8 + ln 2) = 3000 \cdot (2.079 + 0.693) = 8318 m/s .$

Yeh identical hain kyunki $ϵ \to 0$ mein objective hai $v_{e} ln (R_{1} R_{2})$ aur product fixed hai — har split same $Δ v$ deta hai! Equal ratios ka advantage tabhi strict hota hai jab $ϵ > 0$ : tab constraint $(⋆)$ nonlinear hai, unequal ratios structural mass waste karte hain, aur equal ratios genuinely $Δ v$ maximize karte hain. (Isi liye theorem ko Lagrange argument chahiye, sirf AM–GM nahi.)

General result for $n$ stages

Interpretation: Zyada stages → har stage ka $R$ chhota hota hai lekin tum zyada $ln$ terms add karte ho. Ideal limit mein product same rehta hai, lekin practically:

Zyada stages → zyada complexity, zyada structural mass (staging mechanisms, interstages)
3-4 stages se aage diminishing returns
Real rockets: 2-3 stages optimal hote hain

Common mistake "Bas ek giant stage banao with huge

$R$ " Kyun sahi lagta hai: Ek single stage with $R = M_{0} / m_{p}$ deta hai $Δ v = v_{e} ln (M_{0} / m_{p})$ , ideal limit mein $n$ optimal stages ke barabar!

The catch: Us single stage ko poore fuel load ke liye saari structural mass carry karni padti hai. Realistic $ϵ$ ke saath, ek single stage ka mass ratio capped hota hai: $R = \frac{1}{ϵ + ( 1 - ϵ ) k}$ se $k \to 0$ mein milta hai $R_{ma x} \to 1/ ϵ$ . $ϵ = 0.1$ ke liye, $R_{ma x} \approx 10$ . Lekin orbit reach karne ke liye shayad tumhe $M_{0} / m_{p} = 30$ chahiye. Ek stage se impossible. Staging tumhe mid-flight mein structure shed karne deta hai, effective mass ratios achieve karta hai jo ek stage ki ceiling se kaafi zyada hain.

The fix: Staging ko mass ratios ka "compounding" socho. Do stages with $R \approx 5.5$ each effective $R_{t o t a l} = 5. 5^{2} \approx 30$ dete hain.

Recall Kisi 12-saal ke bacche ko explain karo

Socho tum ek pahaad chadh rahe ho ek backpack ke saath jisme paani ki botlein bhari hain. Har ghante tum thoda paani peete ho aur khaali bottle phenk dete ho taaki halka ho jao. Sawaal hai: kitna paani har ghante ke liye rakhna chahiye taaki jitna ho sake upar pohoncho?

Tum soch sakte ho "shuru mein saara paani rakho!" lekin tab tum poora time heavy rahoge. Ya "sab end ke liye bachao!" lekin tab shuruaat mein energy nahi hogi jab tum sabse zyada weight carry kar rahe ho.

Best strategy: Har ghante apne backpack ka same fraction shed karo. Agar tum hamesha, maan lo, jo carry kar rahe ho uska aadha drop karte ho (10 kg → 5 kg → 2.5 kg), tum weight apne current load ke relative same rate par shed kar rahe ho. Rockets ke liye "equal mass ratios" ka yahi matlab hai: har stage apne weight ka same proportion shed karta hai, jo maximize karta hai ki tum kitni tez jaate ho.

Connections

Tsiolkovsky rocket equation — $Δ v = v_{e} ln R$ ki foundation
Staging basics — hum stage kyun karte hain
Structural fraction — $ϵ$ high $Δ v$ ka dushman hai
Payload fraction — ultimately $M_{0} / m_{p}$ determine karta hai
Variable specific impulse staging — jab $I_{s p}$ differ kare, equal $R$ optimal NAHI hai
Lagrange multipliers — proof mein use ki gayi formal optimization technique

#flashcards/physics

Jab saare stages ka same

I_{s p}

aur structural fraction ho, toh har stage ka optimal mass ratio kya hota hai?

Saare stages ke equal mass ratios hone chahiye

R_{1} = R_{2} = \dots = R_{n} = R^{*}

. Negligible structure ke liye

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

; structure ke saath

R^{*} = 1/ (ϵ + (1 - ϵ) k)

jahan

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

Asli optimization constraint sirf "mass ratios ka product = constant" kyun NAHI hai?

Kyunki stages ke beech structure drop hoti hai. Asli constraint hai

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

, jo nonlinear hai. Sirf

ϵ \to 0

ki limit mein yeh reduce hota hai

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

mein.

Lagrange condition se equal mass ratios kyun milte hain?

\nabla (\sum ln R_{i}) = λ \nabla g

set karne par, har stage ki condition same function of $R_{i}$ alone hoti hai (kyunki

ϵ

aur

v_{e}

identical hain). Isliye har

R_{i}

same equation satisfy karta hai ⇒ saare equal hain.

Agar ek two-stage rocket mein

M_{0} = 8000

kg aur

m_{p} = 1000

kg hai aur structure negligible hai, toh har stage ka optimal

R

kya hai?

R^{*} = (8000/1000)^{1/2} = 8 = 2.83

har stage ke liye.

Kya mass ratio

R

1 se kam ho sakta hai?

Nahi.

R = m_{b e f or e} / m_{a f t er} \geq 1

hamesha, kyunki fuel jalaane se sirf mass remove hoti hai.

ln R \geq 0

, toh

Δ v \geq 0

Agar upper stages ka

I_{s p}

lower stages se zyada ho, toh optimal staging strategy par kya asar padta hai?

Equal mass ratios ab optimal nahi rahte. Lagrange conditions har stage ke liye alag hoti hain; higher-

I_{s p}

stages ko bade mass ratios milne chahiye.

Practically ek single stage multiple staged rockets jaisi effective mass ratio kyun achieve nahi kar sakti?

Ek single stage ka mass ratio

R_{ma x} \to 1/ ϵ

par capped hota hai (poori structure throughout carry hoti hai). Staging mid-flight mein structure shed karta hai, ratios compound karta hai: do stages with

R \approx 5.5

each effective

R_{t o t a l} \approx 30

dete hain.