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

3.3.46 · D2 · Physics › Rocket Propulsion › Optimal staging — equal mass ratios (for same Isp)

Step 1 — Rocket kis cheez se bana hai? (mass, ek stack ki tarah draw kiya hua)

KYA. Kisi bhi formula se pehle, chaliye rocket ko blocks ke ek stack ki tarah draw karte hain. Upar se neeche:

payload $m_{p}$ — satellite, woh cheez jo hum actually orbit mein chahte hain;
har stage ke liye, fuel ka ek chunk (jal jaata hai) aur structure ka ek chunk (khaali tank, engine, plumbing).

KYUN. Is chapter ka har symbol ek mass hai — ek kitna-maal number jo kilograms mein measure hota hai. Agar aap har block ko point kar sakte hain, toh baad ki koi bhi equation aapko confuse nahi kar sakti.

PICTURE. Neeche do coloured towers dekho. Orange blocks fuel hain (jab hum fly karte hain yeh gayab ho jaate hain). Violet blocks structure hain (dead weight — hum inhe phenk denge). Upar ek tiny magenta cap hai jo payload $m_{p}$ hai.

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

Step 2 — Ek stage ka "squeeze factor" (mass ratio $R$ )

KYA. Ek stage lo. Fire karne se theek pehle uski kuch starting mass hoti hai — ise $m_{before}$ kahte hain. Yeh apna fuel jalaata hai aur ek chhoti si burnout mass $m_{after}$ ke saath khatam hota hai. In dono ka ratio is show ka star hai:

$R = kitni baar halka hua hum \frac{m _{before}}{m _{after}}$

KYUN yeh ratio aur difference nahi? Kyunki rocket ki physics (Tsiolkovsky rocket equation) kehti hai ki aapko milne wali speed before-to-after mass ke ratio par depend karti hai, na ki aapne kitne kilograms dump kiye. Mass ko double karna hamesha same speed khareedta hai, chahe aap $2 \to 1$ tonne jaayein ya $200 \to 100$ tonne. Ratios, na ki differences, woh cheez hain jo rockets "feel" karte hain.

PICTURE. Neeche red bracket $m_{before}$ (full tank) ko $m_{after}$ (empty tank) mein shrink hote dikhata hai. $R$ literally kitni baar poora tower khaali tower se zyada lamba hai.

Step 3 — Ratio ko speed mein badalna (kyun $ln$ aata hai)

KYA. Ek stage se milne wali speed hai

$Δ v_{i} = exhaust speed v_{e} \times squeeze ka log ln R_{i} .$

Yahan $v_{e}$ exhaust speed hai — gas kitni tez peeche se nikalti hai (bada engine = bada $v_{e}$ ). $Δ v_{i}$ ("delta-vee") us stage se rocket ki speed mein aaya hua change hai.

KYUN ek logarithm? Poochho: "Agar mujhe double speed chahiye, toh mujhe kitna bhaari shuru karna hoga?" Jawab double bhaari nahi hai — yeh squared hai. Speed add hoti jaati hai jabki mass ratio multiply hota jaata hai. Woh ek function jo multiply karna ko add karne mein convert karta hai woh natural logarithm $ln$ hai: $ln (A \times B) = ln A + ln B$ . Isliye yeh exactly yahan aata hai — yeh woh tool hai jo "ratios stack karna" ko "speeds stack karna" mein badalta hai.

PICTURE. Neeche ka curve $ln R$ hai. Notice karo yeh pehle tezi se utha phir flat ho jaata hai: pehla thoda sa fuel bahut saari speed khareedta hai, baad wala fuel kam khareedta hai. Yeh "diminishing returns" shape hi poori wajah hai ki ek akela giant stage wasteful hota hai.

Step 4 — Stages stack karna: speeds add hoti hain, ratios multiply hote hain

KYA. Do stages ek ke baad ek fire karte hain. Stage 1 ke liye, uske upar ki sab cheez — stage 2 aur payload — sirf "woh payload hai jo ise uthana hai." Toh:

$Δ v_{total} = Δ v_{1} + Δ v_{2} = v_{e} ln R_{1} + v_{e} ln R_{2} = v_{e} ln (ratios multiply hote hain R_{1} R_{2}) .$

KYUN. Sequence mein gain ki gayi speeds simply add hoti hain (aap pehle se move kar rahe the, ab aur tezi se jaate hain). Aur kyunki $ln$ products ko sums mein convert karta hai, do $ln$ 's ko add karna waise hi hai jaisa product $R_{1} R_{2}$ ka ek $ln$ lena. Yahi compounding trick hai: squeeze $R$ ke do stages ek super-stage ki tarah kaam karte hain jiska squeeze $R^{2}$ ho.

PICTURE. Neeche, stage 1 poore grey "upar ki sab cheez" block ko uthata hai; burnout ke baad yeh apna violet structure giraa deta hai aur stage 2 halke load ke saath aage le jaata hai.

Step 5 — Catch: structure stages ko ek saath baandh deti hai

KYA. Agar koi structure na hoti, toh hum saari mass fuel mein daal sakte aur $R_{1}, R_{2}$ ki koi bhi split jo same product $R_{1} R_{2}$ de woh same speed deti. Lekin real stages dead weight lete chalte hain. Hum ise structural fraction se measure karte hain:

$ϵ = \frac{m _{s i}}{m _{s i} + m _{f i}} (har stage ke liye same ϵ, assumption ke hisaab se) .$

$ϵ$ yeh hai ki "ek stage ke wet mass ka kaunsa fraction khaali shell hai." $ϵ = 0.1$ matlab 10% dead weight.

KYUN yeh matter karta hai — chaliye masses khud chain karein. Pehle stack climb karte waqt running mass ko ek naam dein. Maano $M_{0}$ poore rocket ki launch mass hai, aur == $M_{i}$ == woh mass hai jo stage $i$ ke burn aur khaali shell girane ke baad bhi fly kar rahi hai. Toh $M_{i - 1}$ exactly "stage $i$ mein enter karne wali mass" hai — woh payload jo stage $i$ ko uthana hai. Bilkul end mein sirf payload bachta hai, isliye $M_{n} = m_{p}$ .

Ab ek stage track karo. Uska ratio hai $R_{i} = M_{i - 1} / m_{after, i}$ , toh burnout mass hai $M_{i - 1} / R_{i}$ aur fuel jo usne jalaya hai $m_{f i} = mass in M_{i - 1} - burn ke baad mass \frac{M _{i - 1}}{R _{i}} = M_{i - 1} (1 - \frac{1}{R _{i}}) .$ Uska structure uske fuel se $ϵ$ ke zariye banna hai: $ϵ = \frac{m _{s i}}{m _{s i} + m _{f i}}$ ko rearrange karne par $m_{s i} = \frac{ϵ}{1 - ϵ} m_{f i}$ milta hai. Burn ke baad hum woh structure giraa dete hain, isliye upar pass ki gayi mass hai $M_{i} = M_{i - 1} - m_{f i} - m_{s i} = M_{i - 1} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} .$ Har stage running mass ko usi shaped factor se multiply karta hai. $M_{0} \to M_{1} \to \dots \to M_{n} = m_{p}$ chain karne par saare factors multiply ho jaate hain:

$\frac{m _{p}}{M _{0}} = \prod_{i = 1}^{n} stage i ka shrink factor \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} . (⋆)$

Yahi asli constraint hai. Notice karo yeh nahi hai " $\prod R_{i} =$ constant" — $ϵ$ har factor ko curved banata hai, aur yahi curvature hai jo ek particular split ko best banati hai.

PICTURE. Neeche ki bar ek stage ki wet mass ko orange fuel aur violet structure mein split karti hai; violet-to-total ka ratio hi $ϵ$ hai, saare stages ke liye fixed.

Step 6 — Equal ratios kyun jeette hain (balancing picture)

KYA — woh cheez naam do jo chahiye. Maano == $f$ == hamara prize hai: $v_{e}$ ke units mein measured total speed. Step 4 se, total speed hai $v_{e} \sum_{i} ln R_{i}$ , toh common $v_{e}$ hataane par, $f = \sum_{i} ln R_{i} (yahi maximise karna hai),$ constraint $(⋆)$ ke subject to. $(⋆)$ ke dono taraf $ln$ lo taaki uska product sum mein badal jaaye aur ise constraint function $g$ kaho: $g = \sum_{i} ln \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} = ln \frac{m _{p}}{M _{0}} (fixed rehna chahiye) .$

KYUN Lagrange condition, term by term. Lagrange multipliers kehta hai: best split par, kisi bhi $R_{i}$ ko nudge karne se $f$ par milne wala "push" ek fixed multiple $λ$ hona chahiye constraint $g$ par milne wale "push" ka — har stage ke liye same $λ$ . Symbols mein $\partial f / \partial R_{i} = λ \partial g / \partial R_{i}$ . Chaliye har side ko $R_{i}$ ke respect mein differentiate karein:

$\frac{\partial f}{\partial R _{i}} = \frac{\partial}{\partial R _{i}} ln R_{i} = \frac{1}{R _{i}}$ — aasaan.
$g$ ke liye, sirf $i$ -vaan term mein $R_{i}$ hai. Uske $ln$ ko split karo: $ln (1 - ϵ R_{i}) - ln R_{i} - ln (1 - ϵ)$ . Aakhri piece ek constant hai. Pehle do ko differentiate karne par: $\frac{\partial}{\partial R _{i}} ln (1 - ϵ R_{i}) = \frac{- ϵ}{1 - ϵ R _{i}}$ (chain rule $- ϵ$ neeche laata hai), aur $\frac{\partial}{\partial R _{i}} (- ln R_{i}) = - \frac{1}{R _{i}}$ .

Pieces ko jodne par condition milti hai, jisme ab har term explain hai: $\partial f / \partial R_{i} \frac{1}{R _{i}} = λ \partial g / \partial R_{i} (\frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}}) .$

KYUN yeh equality force karta hai. Right side sirf $R_{i}$ par depend karta hai (kyunki $ϵ$ aur $λ$ har jagah same hain). Toh har stage ko identical equation solve karni padti hai — aur isliye same jawaab par aana padta hai, jise hum $R_{⋆}$ kehte hain. Unequal ratios sabh ek identical equation satisfy nahi kar sakte.

PICTURE. Do paani ke tanks imagine karo jo neeche se connected hain: efficiency dono mein same level par settle ho jaati hai; ek taraf zyada push karo aur yeh doosri taraf spill ho jaata hai. Plot total speed dikhata hai ek fixed budget ko split karne ke against — ek smooth hill jiski peak bilkul centre mein $R_{1} = R_{2}$ par baithti hai.

KYUN closed form — equal-ratio constraint solve karo. Kyunki saare $R_{i}$ ek value $R_{⋆}$ ke barabar hain, $(⋆)$ mein $n$ identical factors ek single factor ban jaate hain jo $n$ -vein power tak raise hota hai: $\frac{m _{p}}{M _{0}} = (\frac{1 - ϵ R _{⋆}}{( 1 - ϵ ) R _{⋆}})^{n} .$ Dono taraf ka $n$ -vaan root lo aur us number ko $k$ kaho: $\frac{1 - ϵ R _{⋆}}{( 1 - ϵ ) R _{⋆}} = (\frac{m _{p}}{M _{0}})^{1/ n} \equiv k .$ Ab $R_{⋆}$ ke liye ek normal linear equation ki tarah solve karo: multiply out karo, $1 - ϵ R_{⋆} = k (1 - ϵ) R_{⋆}$ ; $R_{⋆}$ terms gather karo, $1 = (ϵ + k (1 - ϵ)) R_{⋆}$ ; divide karo: $R_{⋆} = \frac{1}{ϵ + ( 1 - ϵ ) k} .$

Step 7 — Degenerate case: koi structure nahi ( $ϵ \to 0$ )

KYA. $ϵ = 0$ set karo. Constraint $(⋆)$ simply $\prod R_{i} = M_{0} / m_{p}$ ban jaata hai — ek fixed product. Ab $Δ v = v_{e} ln (\prod R_{i})$ fixed hai chahe aap koi bhi split karein!

KYUN yeh dikhaaein. Yeh woh boundary case hai jo sach reveal karta hai: zero structure ke saath, saari splits tie hain (parent ka numerical comparison dekho — $R_{1} = R_{2} = 4$ aur $R_{1} = 8, R_{2} = 2$ dono $8318$ m/s dete hain). Equal ratios ka fayda strictly tabhi hoga jab $ϵ > 0$ ho, jahan curved constraint lopsided splits ko punish karti hai. Ise "hamesha ties" mat samajhna — yahi trap hai.

PICTURE. Do curves: flat dashed line (ε = 0, har split equal) aur humped solid line (ε > 0, peak strictly middle mein). Same picture, ek lesson.

Ek-picture summary

Neeche, ek canvas par poora argument: block stack (Step 1) → squeeze $R$ (Step 2) → $ln$ jo ratios ko speed mein badalta hai (Step 3) → stacking (Step 4) → structure unhe couple karna (Step 5) → equal $R$ par balanced peak (Step 6), $ϵ \to 0$ flat case shadow line ki tarah (Step 7).

Recall Feynman retelling — simple words mein wapas bolo

Ek rocket blocks ka ek stack hai: upar ek precious chhota payload, aur neeche stages jo fuel se bane hain jo hum jalaate hain aur shells jo hum phenk dete hain. Har stage ki power uska squeeze hai — yeh burn karte waqt kitni baar halka hota hai. Aapko milne wali speed squeeze itself nahi balki uska logarithm hai, kyunki speeds add hoti hain jabki squeezes multiply hote hain, aur log woh machine hai jo multiply karna ko add karne mein badalta hai. Do stages stack karo aur unki speeds add ho jaati hain, toh unke squeezes multiply hote hain — $R$ ke do squeezes ek $R^{2}$ ki tarah kaam karte hain. Twist hai dead structure: har stage ek fixed fraction $ϵ$ ki khaali shell drag karta hai, aur yeh stages ko ek curved constraint ke zariye baandh deta hai. Kyunki saare stages same $ϵ$ aur same engine share karte hain, woh "best" equation jo har stage ko satisfy karni hoti hai identical hai — isliye sab same squeeze $R_{⋆}$ par settle ho jaate hain, jaise paani connected tanks mein ek level dhundh leta hai. Make-believe world mein jahan koi structure nahi hai wahan har split tie karti hai, lekin jaise hi structure real hoti hai, balanced, equal-ratio split hill ke upar baith jaati hai.

Recall

Ek stage ka mass ratio $R$ kya measure karta hai? ::: Ignition aur burnout ke beech stage kitni baar halka hota hai, $m_{before} / m_{after}$ . Mass ratio ko speed mein convert karte waqt $ln$ kyun aata hai? ::: Speeds add hoti hain jabki mass ratios multiply hote hain; $ln$ multiplication ko addition mein badalta hai, isliye total speed $= v_{e} ln (\prod R_{i})$ . Woh ek fact kya hai jo saare optimal ratios ko equal hone par majboor karta hai? ::: Har stage ke liye Lagrange condition sirf us stage ke $R_{i}$ par depend karti hai, aur kyunki $ϵ$ shared hai, har stage same equation solve karta hai ⇒ same $R_{⋆}$ . Kis case mein saari splits equal $Δ v$ deti hain? ::: Sirf degenerate $ϵ \to 0$ case mein, jahan constraint product $\prod R_{i}$ fix kar deti hai.