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

3.3.46 · D1 · Physics › Rocket Propulsion › Optimal staging — equal mass ratios (for same Isp)

Parent page pe jo bhi hai woh symbols ki ek choti si pile pe tika hai. Agar inme se koi bhi fuzzy ho, toh Lagrange proof meaningless squiggles mein collapse ho jaata hai. Isliye hum har ek ko zero se define karte hain, us order mein jisme ek doosre pe depend karte hain, aur har ek ko ek picture se jodenge.

1. Mass — woh cheez jo hum push kar rahe hain

Sabse pehle: mass ka matlab sirf "kitna stuff hai" hai. Hum isse kilograms (kg) mein measure karte hain. Ek full fuel tank mein ek empty se zyada mass hoti hai; ek rocket apne payload ke saath akele payload se zyada mass rakhta hai.

Hum ek rocket ki mass ko lagaatar teen tarah ke stuff mein split karte rahenge:

Figure dekho: ek tall bar, teeno pieces mein slice ki hui.

Figure — Optimal staging — equal mass ratios (for same Isp)

Hume saare teen naam kyun chahiye? Kyunki poori optimisation inke beech ek tension ke baare mein hai: fuel speed deta hai, lekin fuel ko hold karne ke liye tanks (structure) chahiye, aur woh tanks dead weight hain. Payload woh prize hai jo tum protect kar rahe ho. Teeno ko alag kiye bina tum problem bhi state nahi kar sakte.

2. Exhaust speed $v_{e}$ aur specific impulse $I_{s p}$ — engine kitna hard push karta hai

Ek rocket gas ko peeche phenk ke move karta hai. Jitna fast woh gas phenke, utni zyada speed har kilogram fuel dilata hai.

Tumhe specific impulse $I_{s p}$ bhi milega. Yeh sirf $v_{e}$ ko seconds mein rewrite kiya hua hai: $v_{e} = I_{s p} g_{0}, g_{0} = 9.81 m/s^{2} .$

Parent ko "same $I_{s p}$ " ek hypothesis ke roop mein chahiye: agar engines alag hote, toh stages ko balance karna ek alag (mushkil) problem hota — dekho Variable specific impulse staging.

3. Logarithm $ln$ — speed slowly kyun badhti hai jab tum fuel add karte ho

Yeh woh symbol hai jisse zyattar readers darte hain, isliye hum isse poora build karte hain.

Ek sawaal se shuru karo: agar pehle kg fuel phenka toh kuch speed boost mila, kya sauwaan kg bhi same boost dega? Nahi — aur reason yeh hai. Har kg fuel ko sirf rocket hi nahi push karna, balki board pe maujood saara fuel bhi push karna hai. Early fuel ek heavy rocket push karta hai (chhota boost); late fuel ek light rocket push karta hai (bada boost). Total gain ek seedhi line nahi hai — yeh curve karta hai aur flat hota jaata hai.

Jo function exactly is "flattening" ko capture karta hai woh hai natural logarithm, likha jaata hai $ln$ .

Curve dekho: pehle tezi se upar jaata hai, phir bend ho kar flat ho jaata hai.

Yeh tool kyun aur, say, square root ya straight line kyun nahi? Kyunki rocket equation (agla section) literally ek logarithm produce karta hai mass ko peeche phenke ki physics se. Hum $ln$ convenience ke liye nahi choose kar rahe — physics khud hume deta hai.

4. Mass ratio $R$ — ek burn ka "leverage"

Ab central character aata hai.

$R = 1$ ka matlab "koi fuel nahi jala, koi change nahi." $R = 4$ ka matlab rocket end mein jitna heavy tha usse chaar guna heavy shuru hua. Do stacked bars dekho — same rocket, ek burn se pehle aur baad mein.

Yahan $Δ v$ (padho "delta-vee") ka matlab change in speed hai. Greek $Δ$ ("delta") "mein change" ka standard shorthand hai. $Δ v$ wohi hai jo hum maximise karna chahte hain.

5. Ek single burn ke do useful re-expressions

Parent burn ko do tarike se slice karta hai. Dono seedhe $R$ ki definition se aate hain.

Extremes check karo: $R = 1$ se $m_{f} = 0$ milta hai (koi burn nahi), aur bade $R$ se $m_{f} \to M$ (almost sab kuch jala diya). Dono sensible hain.

6. Structural fraction $ϵ$ — har tank par "tax"

Iske baare mein aur: Structural fraction. Is definition se tum structure ko fuel se rearrange kar sakte ho: $m_{s} = \frac{ϵ}{1 - ϵ} m_{f},$ jo exactly woh expression hai jo parent's proof "maine jo fuel choose kiya jalane ke liye" ko "woh structure jo mujhe zaroor le jaana hai" mein convert karne ke liye use karta hai.

7. Payload fraction aur master constraint

Pehle, woh number jiske against hum score kar rahe hain.

Ab parent ka central rule. Maano $n$ number of stages hai (ek plain counting number: $n = 2$ two-stage rocket ke liye, $n = 3$ teen ke liye, aur aise hi). Stages ko $i = 1, 2, \dots, n$ bottom se top tak label karo, aur maano $R_{i}$ stage $i$ ka mass ratio hai.

Exactly yeh product kyun? Chaliye ek factor build karte hain, phir unhe chain karte hain — yeh woh sketch hai jo parent skip karta hai.

Mass ko follow karo jaise woh rocket ke through neeche flow karta hai, ek stage at a time. $M_{i - 1}$ likho stage $i$ mein enter karne wali mass ke liye (toh $M_{0}$ woh hai jisse poora rocket shuru hota hai, aur last stage jalne aur drop hone ke baad hum sirf $M_{n} = m_{p}$ ke saath bache hain).

Stage $i$ fuel jalata hai $m_{f i} = M_{i - 1} (1 - \frac{1}{R _{i}})$ — seedha §5 se, $M = M_{i - 1}$ ke saath.
Woh fuel structure force karta hai $m_{s i} = \frac{ϵ}{1 - ϵ} m_{f i}$ — seedha §6 se.
Burn aur khaali structure drop karne ke baad, agle stage ko di jaane wali mass hai $M_{i} = M_{i - 1} - m_{f i} - m_{s i} = M_{i - 1} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} .$

Toh har stage mass ko factor $\frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}}$ se multiply karta hai. Yeh $n$ baar karo, $M_{0}$ se $m_{p}$ tak, aur har factor pichle par multiply hota hai — jo exactly product $(⋆)$ hai. Yehi wajah hai ki yeh product hai aur sum nahi: har stage jo bhi neeche wale stage ne choda use act karta hai.

8. Lagrange multipliers — "ek rule ke saath maximise karo" ka tool

Parent poochta hai: maximise total $Δ v$ jabki fixed payload fraction $(⋆)$ ko obey karo. Yeh ek constrained optimisation hai — tum sirf sabse bada number chase nahi kar sakte, tumhe constraint surface par rehna hai.

Do ingredients is method mein naam paate hain:

Symbol $\nabla$ ("nabla" ya "gradient") shorthand hai "har direction mein ek quantity ki slopes ki list jis direction mein tum $R_{i}$ ko nudge kar sakte ho." Isse ek aise arrow ki tarah sochh jo us direction mein point karta hai jis se quantity fastest badhti hai.

Intuition Yeh specific tool kyun, aur

$λ$ ka matlab kya hai Ordinary "slope ko zero set karo" ek free landscape ke peaks dhundta hai. Lekin yahan tum ek fence (constraint) se lock ho. Best allowed point par, woh direction jisme tum climb karna chahte ( $\nabla f$ ) seedha fence mein ( $\nabla g$ ) point karna chahiye — warna tum fence ke saath slide kar sakte ho aur improve kar sakte ho. Lagrange isse $\nabla f = λ \nabla g$ likhta hai: do arrows parallel hain, aur number $λ$ ("lambda") sirf hai kitna zyada lamba ek arrow doosre se. Full method: Lagrange multipliers.

$λ$ meaning bhi carry karta hai: yeh exchange rate hai — kitna extra $Δ v$ tum gain karte agar constraint thoda loosen hota (e.g. agar tumhe thoda chhota payload allowed hota). Bada $λ$ matlab constraint "expensive" hai; payload sach mein tumhe hold back kar raha hai.

Machinery master karna zaroori nahi parent ke punchline par trust karne ke liye: kyunki har stage same $ϵ$ dekhta hai, har stage ki optimality equation identical hai, isliye saare $R_{i}$ equal aane chahiye. Yahi poora theorem hai, aur ab tum jaante ho iske statement mein har symbol ka matlab kya hai.

Prerequisite map

Ise upar se neeche padho: raw mass ideas $R$ aur $ϵ$ ko feed karte hain; $R$ plus $ln$ plus $v_{e}$ rocket equation dete hain; constraint plus Lagrange theorem dete hain.

Equipment checklist

Right side cover karo aur khud test karo. Agar koi bhi answer surprises kare, toh parent page se pehle us section ko dobara padho.

Natural log

ln x

kya poochta hai?

e \approx 2.718

ko kis power tak raise karna padega taaki

x

mile?" — aur yeh kabhi zyada slowly badhta hai.

Mass ratio

R

hamesha

\geq 1

kyun hona chahiye?

Fuel jalaana sirf mass remove karta hai, toh "after" mass kabhi "before" se badi nahi — ratio 1 se neeche nahi ja sakta.

Ek burn ke liye Tsiolkovsky rocket equation state karo.

Δ v = v_{e} ln R

v_{e}

aur

I_{s p}

mein kya difference hai?

Same physical fact;

v_{e} = I_{s p} g_{0}

v_{e}

ek speed hai (m/s),

I_{s p}

isse seconds mein rewrite kiya hua hai.

Structural fraction

ϵ

ko words mein define karo.

Stage ki structure-plus-fuel mass ka woh share jo dead structure hai:

ϵ = m_{s} / (m_{s} + m_{f})

ϵ > 0

staging ko kyun matter karata hai?

ϵ = 0

ke saath har fuel split same speed deta hai; sirf structural "tax" optimal split ko genuinely doosron se better banata hai.

\prod

ka matlab kya hai,

\sum

se alag?

\prod

= saare terms ko multiply karo;

\sum

= saare terms ko add karo.

Master constraint product kyun hai aur sum kyun nahi?

Har stage incoming mass ko apne surviving-fraction factor se multiply karta hai;

n

stages ko chain karne se

n

factors multiply hote hain.

Start-mass

M

aur ratio

R

wale stage ka fuel burned?

m_{f} = M (1 - 1/ R)

Lagrange's method mein

f

aur

g

kya hain?

f

woh objective hai jo tum maximise karte ho (

\sum ln R_{i}

);

g

woh constraint hai jo tum fixed rakhte ho (

\sum ln

of the per-stage factors

= ln (m_{p} / M_{0})

Multiplier

λ

kya measure karta hai?

Exchange rate — agar constraint thoda loosen hota toh kitna extra

Δ v

milta.

Yahan plain calculus ki jagah Lagrange multipliers kyun use karo?

Hume

Δ v

maximise karna hai jabki ek fixed constraint surface par rehna hai; Lagrange enforce karta hai "gradients

\nabla f

aur

\nabla g

best allowed point par line up karte hain."

Stages mein equal

ϵ

equal

R_{i}

force kyun karta hai?

Har stage ki optimality equation sirf apne

R_{i}

aur shared

ϵ

par depend karti hai, isliye saare stages identical equation solve karte hain — hence equal ratios.