Barrowman equations — centre of pressure calculation for finned rockets: Exercises

3.4.8 · D4 · Physics › Rocket Flight Mechanics › Barrowman equations — centre of pressure calculation for fin

Intuition Poore rocket ke liye ek hi ruler — origin dhyan se padho!

Is page par saare final CP positions ek fixed origin: nose tip se measure ki gayi hain, aur $x$ tail ki taraf badhta hai. Lekin individual component formulas har ek apne local landmark se likhi hain, aur unhe shared ruler par laane ke liye offset jodhna padega:

Nose: $\overset{ˉ}{X}_{n} = k L_{n}$ pehle se hi tip se measure kiya hua hai — koi offset nahi chahiye.
Transition: formula $X_{t}$ se shuru hota hai, jo jahan transition shuru hoti hai woh point hai (jo khud tip se measure hota hai).
Fins: $\overset{ˉ}{X}_{fins}$ $X_{f}$ se shuru hota hai, jo fin root leading edge hai (jo khud tip se measure hota hai). Parameters $X_{t}$ aur $X_{f}$ exactly woh bridges hain jo har local result ko shared tip-based ruler par le jaate hain. Origins ko mix karna CP ki sabse common galti hai — average karne se pehle hamesha confirm karo ki har $\overset{ˉ}{X}_{i}$ finally tip se quote ki gayi hai.

Quick reference — woh paanch tools jo tum baar baar use karoge, har ek ke saath ek-line ka why:

Formula Toolbox (intuition ke saath)

Nose: slope $C_{N α, nose} = 2$ , CP at $\overset{ˉ}{X}_{n} = k L_{n}$ (cone $k = \frac{2}{3}$ , ogive $k = 0.466$ , parabola $k = 0.5$ ). Why 2: slender-body lift slope $= 2 \times$ (base area/reference area) $= 2$ . Why $k L_{n}$ : CP area-growth loading ka centroid hai. Transition: $C_{N α, trans} = 2 [(\frac{d _{2}}{d})^{2} - (\frac{d _{1}}{d})^{2}]$ . Why squares ka difference: lift sirf wahan generate hoti hai jahan cross-section area change hoti hai, isliye yeh aft area minus fore area count karta hai. Yahan $d$ reference diameter hai (body diameter jo non-dimensionalise karne ke liye use hoti hai), $d_{1}$ fore aur $d_{2}$ cone frustum ka aft diameter hai. Fins ( $n$ fins ke liye): $C_{N α, fins} = K_{f b} \frac{4 n ( s / d ) ^{2}}{1 + 1 + ( \frac{2 ℓ _{f}}{C _{r} + C _{t}} ) ^{2}}, K_{f b} = 1 + \frac{r}{s + r}$ Why yeh shape: thin-airfoil lift se shuru karo, finite aspect-ratio ke liye correct karo (yeh woh $\dots$ denominator hai — lambe slender fins zyada lift karte hain per area, chhote stubby wale kam), $(s / d)^{2}$ se scale karo (bada span vs. body quickly dominant ho jaata hai) aur fins ki sankhya $n$ se; finally $K_{f b} > 1$ woh flow add karta hai jo fat body roots par deflect karta hai. Yahan $d$ phir se reference diameter hai. Combine (normal-force-weighted mean): $\overset{ˉ}{X}_{c p} = \frac{\sum _{i} C _{N α, i} X ˉ _{i}}{\sum _{i} C _{N α, i}}$ Why weighted mean: net moment $=$ net force $\times$ lever arm, yaani $\sum N_{i} \overset{ˉ}{X}_{i} = \overset{ˉ}{X}_{c p} \sum N_{i}$ ; jo piece sabse zyada push karta hai (sabse bada $C_{N α, i}$ ) woh balance point ko apni taraf sabse zyada kheenchta hai — bilkul ek see-saw ki tarah jo unevenly loaded ho. Stability: static margin $= \frac{X ˉ _{c p} - X _{c g}}{d}$ calibres — dekho Static stability and static margin. Dhyan raho: yahan $d$ ek alag role play karta hai — yeh body diameter hai jo length yard-stick ke roop mein use hoti hai ("ek calibre"), cm mein distance ko body-diameters ki count mein convert karti hai. Same symbol, lekin slope formulas mein $d$ area normalise karta hai, jabki yahan yeh length normalise karta hai.

Level 1 — Recognition

Exercise 1.1

Kisi bhi slender nose ke liye nose-cone normal-force slope $C_{N α, nose}$ batao, aur ek sentence mein explain karo ki yeh nose shape par kyun depend nahi karta.

Recall Solution

$C_{N α, nose} = 2$ (per radian). Why shape-independent: slender-body theory kehta hai ki slope $2 \times (base area) / (reference area)$ ke barabar hai; nose base ko hi reference lene par woh ratio exactly $1$ ho jaata hai, isliye sirf base area matter karta hai — beech ka ogive/cone/parabola profile kabhi nahi.

Exercise 1.2

Ek cone nose ki length $L_{n} = 21$ cm hai. Tip se measure karte hue uska centre of pressure kahan hai?

Recall Solution

Cone shape constant $k = \frac{2}{3}$ hai, isliye $\overset{ˉ}{X}_{n} = \frac{2}{3} L_{n} = \frac{2}{3} (21) = 14 cm tip se.$ Why $\frac{2}{3}$ : ek cone ke liye cross-section area $S \propto x^{2}$ ki tarah badhti hai, lift-per-length $\propto d S / d x \propto x$ linearly badhti hai, aur triangle-shaped load ka centroid length ke $\frac{2}{3}$ par hota hai. Kyunki yeh pehle se hi tip se measure kiya gaya hai, ise kisi origin offset ki zaroorat nahi hai.

Level 2 — Application

Exercise 2.1

Ek body transition (shoulder) fore diameter $d_{1} = 3$ cm se aft diameter $d_{2} = 5$ cm tak badhti hai. Reference (body) diameter $d = 5$ cm hai. Transition ka normal-force slope nikalo. Kya yeh stabilising (positive) hai ya destabilising (negative)?

Recall Solution

$C_{N α, trans} = 2 [(\frac{5}{5})^{2} - (\frac{3}{5})^{2}] = 2 [1 - 0.36] = 2 (0.64) = 1.28.$ Positive ⇒ yeh normal force add karta hai. Why: ek flare ( $d_{2} > d_{1}$ ) flow ko outward mod deta hai aur ek chhote extra nose ki tarah hawa pakadta hai; sirf jahan area badhti hai wahan lift generate hoti hai.

Exercise 2.2

Ek boat-tail $d_{1} = 5$ cm se $d_{2} = 3$ cm tak shrink hoti hai, reference $d = 5$ cm ke saath. $C_{N α, trans}$ compute karo aur sign interpret karo.

Recall Solution

$C_{N α, trans} = 2 [(\frac{3}{5})^{2} - (\frac{5}{5})^{2}] = 2 [0.36 - 1] = 2 (- 0.64) = - 1.28.$ Negative — ek boat-tail tail par normal force remove karta hai, jo CP ko forward dhakelta hai aur mildly destabilising hai. Why: area ab shrink hoti hai, isliye flow andar close ho jaati hai aur local pressure loading ek flare ke comparison mein ulti direction mein point karti hai.

Exercise 2.3

Teen fins ( $n = 3$ ) ek body par baithe hain jiska radius $r = 2$ cm hai; har fin ka semi-span $s = 6$ cm hai. Body-interference factor $K_{f b}$ compute karo.

Recall Solution

$K_{f b} = 1 + \frac{r}{s + r} = 1 + \frac{2}{6 + 2} = 1 + 0.25 = 1.25.$ Why $> 1$ : flow fat body ke around bend hoti hai aur fin roots par spill hoti hai, isliye har fin akele ke comparison mein zyada deflected air feel karta hai — yahan $25%$ boost.

Level 3 — Analysis

Neeche ki figure is poore level ka map hai. Yeh ek trapezoidal fin ko flatten karke dikhata hai: horizontal axis chord (nose-to-tail direction) ke along chalti hai, vertical axis span (body se door) ke along baahir jaati hai. Violet outline (fin), magenta double-arrows jo bottom par root chord $C_{r}$ aur top par tip chord $C_{t}$ mark karti hain, orange vertical arrow jo semi-span $s$ mark karta hai, aur top par navy arrow jo sweep $X_{r}$ (leading-edge se leading-edge) mark karta hai dhundho. Do orange dots ko join karne wali dashed orange line mid-chord line hai jiski length $ℓ_{f}$ hai — woh ek quantity jo Exercise 3.1 tum se measure karwata hai. Dhyan do ki mid-chord line bilkul seedhi upar nahi jaati: yeh $X_{r}$ se zyada peeche jhukti hai kyunki chhota tip chord apna mid-point forward shift kar leta hai.

Figure — Barrowman equations — centre of pressure calculation for finned rockets

Exercise 3.1

Ek trapezoidal fin (figure dekho) mein root chord $C_{r} = 8$ cm, tip chord $C_{t} = 4$ cm, semi-span $s = 6$ cm, aur sweep distance $X_{r} = 5$ cm (tip leading edge, root leading edge ke kitna peeche hai) hai. Mid-chord sweep length $ℓ_{f}$ compute karo.

Recall Solution

Mid-chord line root chord ke middle se tip chord ke middle tak jaati hai (figure mein do orange dots). Horizontally yeh $X_{r} + \frac{C _{t} - C _{r}}{2}$ move karta hai (sweep plus woh shift jo isliye aata hai ki tip chhota hai), vertically yeh semi-span $s$ climb karta hai: $ℓ_{f} = s^{2} + (X_{r} + \frac{C _{t} - C _{r}}{2})^{2} = 6^{2} + (5 + \frac{4 - 8}{2})^{2} = 36 + (5 - 2)^{2} = 36 + 9 = 45 = 6.708 cm .$ Why $\frac{C _{t} - C _{r}}{2}$ term: chhote tip chord ka mid-chord point apne leading edge se relatively aage hota hai, isliye mid-chord line ka horizontal run sirf $X_{r}$ nahi hai — exactly woh lean jo dashed orange line mein dikhti hai.

Exercise 3.2

Upar wali fin $n = 3$ , body radius $r = 2$ cm, reference diameter $d = 4$ cm ke saath use karo, $C_{N α, fins}$ compute karo.

Recall Solution

Step 1 — interference: $K_{f b} = 1 + \frac{2}{6 + 2} = 1.25$ . Step 2 — aspect-ratio bracket: $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{2 ( 6.708 )}{8 + 4} = \frac{13.416}{12} = 1.118$ , isliye $1 + 1.11 8^{2} = 2.25 = 1.5$ . Step 3 — geometric core: $\frac{4 n ( s / d ) ^{2}}{1 + 1.5} = \frac{4 ( 3 ) ( 6/4 ) ^{2}}{2.5} = \frac{12 ( 2.25 )}{2.5} = \frac{27}{2.5} = 10.8$ . Step 4 — interference apply karo: $C_{N α, fins} = 1.25 \times 10.8 = 13.5.$ Why fins dominate: $(s / d)^{2} = 2.25$ factor aur $\times n$ jaldi multiply ho jaate hain, isliye fins routinely nose ke slope $2$ ko dabaaa dete hain.

Exercise 3.3

Usi fin ke liye, fin CP location $\overset{ˉ}{X}_{fins}$ nikalo agar root leading edge $X_{f} = 55$ cm par hai.

Recall Solution

$\overset{ˉ}{X}_{fins} = X_{f} + \frac{X _{r}}{3} \cdot \frac{C _{r} + 2 C _{t}}{C _{r} + C _{t}} + \frac{1}{6} [(C_{r} + C_{t}) - \frac{C _{r} C _{t}}{C _{r} + C _{t}}] .$ Yahan $X_{f} = 55$ cm (fin root leading edge, tip se measure kiya hua) woh offset hai jo local fin result ko shared tip-based ruler par le jaata hai. Term 2 (taper centroid): $\frac{5}{3} \cdot \frac{8 + 2 ( 4 )}{8 + 4} = \frac{5}{3} \cdot \frac{16}{12} = \frac{5}{3} (1.3333) = 2.2222$ cm. Term 3 (chord-averaging): $\frac{1}{6} [12 - \frac{8 \cdot 4}{12}] = \frac{1}{6} [12 - 2.6667] = \frac{1}{6} (9.3333) = 1.5556$ cm. $\overset{ˉ}{X}_{fins} = 55 + 2.2222 + 1.5556 = 58.78 cm tip se.$ Why $X_{f}$ ke aft mein: sweep aur taper load ko rearward aur outward push karta hai, isliye CP root leading edge ke karib $3.8$ cm peeche land karta hai.

Level 4 — Synthesis

Neeche ki figure Level 4 ki master picture hai: poore rocket ka side view, tip left par, tail right par, ek shared tip-based ruler par draw kiya gaya hai (horizontal axis "nose tip se distance, cm" hai). Magenta dot (nose CP $12$ cm par), violet dot (fin CP $58.8$ cm par), orange triangle (combined CP jo weighted mean produce karta hai, $52.7$ cm par), aur navy diamond (CG $47$ cm par) dhundho. CG aur CP ke beech double-arrow static margin hai. Jo key cheez samajhni hai: combined CP far right mein hai, fins ke karib — kyunki fins total slope ka zyada hissa carry karti hain, woh balance point ko strongly apni taraf kheenchti hain.

Exercise 4.1

Ek full rocket mein hai:

Cone nose: $L_{n} = 18$ cm ( $k = \frac{2}{3}$ ), isliye $\overset{ˉ}{X}_{n}$ tip se.
Fins: Exercises 3.2–3.3 se, $C_{N α, fins} = 13.5$ at $\overset{ˉ}{X}_{fins} = 58.78$ cm.

Total CP location $\overset{ˉ}{X}_{c p}$ aur total slope $C_{N α, total}$ nikalo.

Recall Solution

Nose CP: $\overset{ˉ}{X}_{n} = \frac{2}{3} (18) = 12$ cm tip se; nose slope $= 2$ . Dono CPs pehle se shared tip-based ruler par hain, isliye hum directly average kar sakte hain. Total slope: $C_{N α, total} = 2 + 13.5 = 15.5$ . Weighted mean: $\overset{ˉ}{X}_{c p} = \frac{2 ( 12 ) + 13.5 ( 58.78 )}{2 + 13.5} = \frac{24 + 793.5}{15.5} = \frac{817.5}{15.5} = 52.74 cm tip se.$ Why fins ke itna paas: fins $13.5/15.5 \approx 87%$ total slope carry karte hain, isliye balance point almost poori tarah unki taraf khiich jaata hai.

Exercise 4.2

Rocket ka CG $X_{c g} = 47$ cm par hai aur $d = 4$ cm hai. Static margin compute karo aur batao ki yeh stable hai ya nahi, aur kya yeh "healthy" $1$ – $2$ calibre band mein hai.

Recall Solution

$static margin = \frac{X ˉ _{c p} - X _{c g}}{d} = \frac{52.74 - 47}{4} = \frac{5.74}{4} = 1.435 calibres .$ Yahan $d = 4$ cm length yard-stick hai (ek calibre $=$ ek body diameter), $5.74$ cm gap ko body-diameters ki count mein convert karta hai. Positive ⇒ stable (CP, CG ke peeche hai). Aur $1.435$ comfortably $1$ – $2$ calibre "stable but not over-stiff" range ke andar hai — dekho Angle of attack and restoring moment ki bahut bada margin kyun rocket ko gusts mein hard weathercock karta hai.

Exercise 4.3

Ab rear mein ek boat-tail add karo: $d_{1} = 4$ cm → $d_{2} = 3$ cm, reference $d = 4$ cm, $X_{t} = 62$ cm se shuru hoti hai, length $L_{t} = 4$ cm. Total CP recompute karo.

Recall Solution

Pehle slope. $C_{N α, trans} = 2 [(3/4)^{2} - (4/4)^{2}] = 2 [0.5625 - 1] = - 0.875$ .

Ab transition CP — formula sirf quote nahi karte, build karte hain. Ek conical transition ek truncated cone (frustum) hai. Nose ki tarah, iska lift-per-length area growth $d S / d x$ follow karta hai, jo cone-shaped wall ke liye position ke saath linearly vary karta hai. CP uss linear loading ka centroid hai $X_{t}$ se $X_{t} + L_{t}$ tak ke frustum par. Wahi centroid integral jo humne cone nose ke liye ki — lekin ab ek aise slice par jo non-zero radius se shuru hoti hai (kyunki fore end par pehle se $d_{1}$ diameter hai) — standard result deta hai $\overset{ˉ}{X}_{trans} = X_{t} + \frac{L _{t}}{3} (1 + \frac{1 - d _{1} / d _{2}}{1 - ( d _{1} / d _{2} ) ^{2}}) .$ Ise kaise padhen: agar $d_{1} = 0$ (ek full cone, koi truncation nahi) toh fraction collapse ho jaata hai aur bracket $1$ ban jaata hai, deta hai $\overset{ˉ}{X} = X_{t} + \frac{L _{t}}{3}$ — lekin wide base se peeche measure karo toh yeh wahi $\frac{2}{3}$ -centroid hai jo humne nose ke liye nikala tha. Extra term $\frac{1 - d _{1} / d _{2}}{1 - ( d _{1} / d _{2} ) ^{2}}$ cone ka tip kaatne ka correction hai: ek truncated frustum ki area-growth alag tarah spread hoti hai, centroid shift karta hai. Aur $X_{t}$ simply woh offset hai jo jawab ko shared tip-based ruler par rakhta hai.

Ratio $d_{1} / d_{2} = 4/3 = 1.3333$ (yahan fore diameter bada hai, kyunki yeh boat-tail hai). $\overset{ˉ}{X}_{trans} = 62 + \frac{4}{3} (1 + \frac{1 - 1.3333}{1 - 1.333 3 ^{2}}) = 62 + 1.3333 (1 + \frac{- 0.3333}{1 - 1.7778}) .$ Inner fraction: $\frac{- 0.3333}{- 0.7778} = 0.4286$ , isliye bracket $= 1.4286$ , aur $\overset{ˉ}{X}_{trans} = 62 + 1.3333 (1.4286) = 62 + 1.905 = 63.905$ cm tip se.

New total slope: $2 + 13.5 + (- 0.875) = 14.625$ . New CP: $\overset{ˉ}{X}_{c p} = \frac{2 ( 12 ) + 13.5 ( 58.78 ) + ( - 0.875 ) ( 63.905 )}{14.625} = \frac{24 + 793.5 - 55.917}{14.625} = \frac{761.58}{14.625} = 52.07 cm.$ Why CP forward gaya: boat-tail bahut peeche ek negative force contribute karta hai — ek rear load subtract karna balance point ko thoda forward kheenchta hai ( $52.74$ se $52.07$ cm). Isse static margin karib $0.17$ calibres kama hua.

Level 5 — Mastery

Exercise 5.1 (design)

Tumhare paas Exercise 4.1 ka nose ( $C_{N α} = 2$ at $12$ cm) aur fins ( $C_{N α, fins} = 13.5$ at $58.78$ cm) hain, jo $\overset{ˉ}{X}_{c p} = 52.74$ cm deta hai. Tumhara CG $X_{c g} = 47$ cm par fixed hai, $d = 4$ cm. Tumhara customer exactly $2.0$ calibres ka static margin maang raha hai. Fin set ko kitna move karna hoga (change $X_{f}$ ), yeh maante hue ki fin CP rigidly $X_{f}$ ke saath move karta hai aur fin slope unchanged hai?

Recall Solution

Required CP: $\overset{ˉ}{X}_{c p} = X_{c g} + 2.0 d = 47 + 2 (4) = 55$ cm tip se. Fin CP $Δ$ se shift ho. New weighted mean $55$ ke barabar hona chahiye: $\frac{2 ( 12 ) + 13.5 ( 58.78 + Δ )}{15.5} = 55.$ Numerator $55 \times 15.5 = 852.5$ hona chahiye. Abhi yeh $817.5$ hai, isliye numerator mein extra $852.5 - 817.5 = 35$ chahiye. Woh extra sirf $13.5 Δ$ se aata hai: $13.5 Δ = 35 \Rightarrow Δ = \frac{35}{13.5} = 2.593 cm .$ Fins ko $2.59$ cm rearward move karo. Why it works cleanly: kyunki sirf fin term $Δ$ carry karta hai aur denominator unchanged hai, $\overset{ˉ}{X}_{c p}$ mein shift sirf $\frac{13.5}{15.5} Δ$ hai — fins apni apni movement ka $87%$ CP tak transmit karte hain.

Exercise 5.2 (limiting behaviour)

Exercise 3.2 ki fin geometry lo lekin socho ki body infinitely thin ho jaaye ( $r \to 0$ ) aur alag se infinitely fat ( $r ≫ s$ ). $K_{f b}$ ki limiting values kya hain, aur physically inका kya matlab hai?

Recall Solution

$K_{f b} = 1 + \frac{r}{s + r}$ .

$r \to 0$ : $K_{f b} \to 1 + 0 = 1$ . Matlab: koi body nahi hai toh fins isolation mein kaam karte hain — koi extra deflected flow nahi, koi interference boost nahi.
$r ≫ s$ : $\frac{r}{s + r} \to 1$ , isliye $K_{f b} \to 2$ . Matlab: bahut fat body par fins maximum possible deflected flow dekhte hain; interference unki effectiveness double kar deta hai. Is model mein factor kabhi $2$ se zyada nahi ho sakta — ek built-in ceiling jo estimate ko physical rakhta hai. Compare karo Rocket flight simulation (6-DOF) ke assumptions se jahan aise closed-form limits sanity checks ke roop mein use hote hain.

Exercise 5.3 (synthesis + judgement)

Ek student "rocket ko aur stable banane ke liye" har fin dimension double kar deta hai ( $C_{r}, C_{t}, s, X_{r}$ sab $\times 2$ ), $n = 3$ , $d = 4$ cm, $r = 2$ cm, $X_{f} = 55$ cm rakhte hue. Dikhao ki $C_{N α, fins}$ ka kya hota hai, ise CP aur static margin mein feed karo, aur explain karo ki bade fins zyada static margin guarantee kyun nahi karte.

Recall Solution

Saari fin lengths double karne par $s \to 12$ , $C_{r} \to 16$ , $C_{t} \to 8$ , $X_{r} \to 10$ ho jaate hain, isliye $ℓ_{f} \to 2 (6.708) = 13.416$ cm.

Bracket: $\frac{2 ℓ _{f}}{C _{r} + C _{t}} = \frac{2 ( 13.416 )}{24} = 1.118$ — unchanged (dono parts saath scale hue). Isliye $1 + 1.11 8^{2} = 1.5$ pehle jaisa.
Core: $\frac{4 n ( s / d ) ^{2}}{2.5} = \frac{12 ( 12/4 ) ^{2}}{2.5} = \frac{12 ( 9 )}{2.5} = 43.2$ .
$K_{f b}$ $1.25$ par unchanged (body nahi chhui). $C_{N α, fins} = 1.25 (43.2) = 54.0 - original 13.5 ka chaar guna.$ Ise CP mein feed karo. Fin CP bhi aft move hota hai kyunki chords aur sweep double ho gaye; Exercise 3.3 ko doubled numbers ke saath recompute karo: $\overset{ˉ}{X}_{fins} = 55 + \frac{10}{3} \cdot \frac{16 + 2 ( 8 )}{16 + 8} + \frac{1}{6} [24 - \frac{16 \cdot 8}{24}] = 55 + 4.444 + 3.111 = 62.56 cm .$ New total slope $= 2 + 54.0 = 56.0$ ; new CP: $\overset{ˉ}{X}_{c p} = \frac{2 ( 12 ) + 54.0 ( 62.56 )}{56.0} = \frac{24 + 3378.2}{56.0} = \frac{3402.2}{56.0} = 60.75 cm .$ Static margin agar CG $47$ cm par raha: $\frac{60.75 - 47}{4} = 3.44$ calibres — kaagaz par bahut bada. Lekin parent note ki Centre of gravity determination warning bite karta hai: fin dimensions double karne par fin mass roughly chaar guna ho jaati hai jo tail par rakhti hai, $X_{c g}$ ko peeche kheenchti hai. Maan lo heavy fins CG ko $X_{c g} = 58$ cm tak peeche le jaate hain; tab static margin collapse ho jaata hai $\frac{60.75 - 58}{4} = 0.69$ calibres tak — original $1.4$ se worse, aur dangerously neutral ke karib. Conclusion: fin slope chaar guna hua aur CP aft gaya, phir bhi static margin shrink ho sakta hai, kyunki margin difference $\overset{ˉ}{X}_{c p} - X_{c g}$ hai aur added tail mass CG ko CP ki same direction mein move karta hai. Hamesha CP aur CG dono saath recompute karo.

Recall Self-test cloze — the essentials

Nose slope hamesha ::: $C_{N α, nose} = 2$ (per radian) hota hai Cone CP ::: nose length ke $\frac{2}{3}$ par tip se hota hai Boat-tail ka transition slope ::: negative (destabilising) hota hai Body interference factor ::: $K_{f b} = 1 + \frac{r}{s + r}$ , $1$ aur $2$ ke beech bounded Saare final CP positions ::: nose tip se measure hote hain (ek shared ruler) CP compute hota hai ::: component CPs ka normal-force-slope-weighted mean ke roop mein Healthy static margin band ::: $1$ – $2$ calibres