Barrowman equations — centre of pressure calculation for finned rockets: Question bank

3.4.8 · D5 · Physics › Rocket Flight Mechanics › Barrowman equations — centre of pressure calculation for fin

Definition Vocabulary — in questions se pehle inhe pakka karo

Angle of attack $α$ : woh chota angle jo rocket ki pointing axis aur hawa ke actual direction ke beech hota hai. Jab rocket bilkul seedha fly karta hai, $α = 0$ ; ek jhonka use thoda tilt kar ke koi chota $α$ deta hai. Barrowman mein poora radians mein measure hota hai. Dekho Angle of attack and restoring moment.
Normal force $N$ : sideways (axis ke perpendicular) air-force jo $α \neq = 0$ hone par aati hai.
Normal-force coefficient $C_{N} = \frac{N}{\frac{1}{2} ρ v ^{2} A _{r e f}}$ : $N$ ko dimensionless banaya gaya ( $ρ$ = air density, $v$ = speed, $A_{r e f}$ = ek chosen reference area).
Normal-force slope $C_{N α} \equiv \frac{\partial C _{N}}{\partial α}$ (units: per radian): woh coefficient kitni tezi se har unit tilt par badhta hai. Yahi — force khud nahi — woh cheez hai jo har Barrowman term deliver karta hai.
CP (centre of pressure): woh single point jahan saari sideways air-force imagine ki ja sakti hai ki act kar rahi hai.
CG (centre of gravity): woh single point jahan rocket ka poora weight imagine kiya ja sakta hai ki act kar raha hai. Dekho Centre of gravity determination.
Static margin: $(\overset{ˉ}{X}_{c p} - X_{c g})$ ka gap body-diameters ("calibres") mein measure kiya gaya. Positive = CP behind CG = stable. Dekho Static stability and static margin.

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

True or false — justify

Kya CP hamesha ek well-designed rocket par CG ke peeche hota hai?

True. Tabhi air-force ek lever par pivot (CG) ke peeche act karta hai — yaani

\overset{ˉ}{X}_{c p} > X_{c g}

— ek aisa torque create karta hai jo tail ko wapas line mein push karta hai: ek positive static margin.

Barrowman ka nose-cone normal-force slope

C_{N α} = 2

per radian cone ki sharpness par depend karta hai.

False. Slender-body theory mein slope sirf base area par depend karta hai (base ko reference maan ke 2 per radian deta hai); shape sirf yeh change karta hai ki CP kahan baithta hai, force kitna aata hai nahi.

Ek boat-tail (rear taper jo body ko narrow karta hai) hamesha total CP ko forward move karta hai.

False generally. Ek boat-tail ek negative

C_{N α}

deta hai jo tail ke paas located hota hai; ek bade

\overset{ˉ}{X}

par ek negative weight actually weighted-mean CP ko forward (chote

X

ki taraf) kheench sakta hai, stability kam karta hai — tumhe compute karna hoga, guess nahi.

Zero angle of attack par ek symmetric rocket ki net side force zero hai, isliye Barrowman kuch useful nahi deta.

False. Barrowman deliberately slope

C_{N α} = \partial C_{N} / \partial α

compute karta hai (force ka rate per unit tilt), jo exactly woh cheez hai jo ek chote disturbance ke response ko govern karta hai, chahe

α = 0

par force zero ho.

CP location

\overset{ˉ}{X}_{c p}

badal jaata hai agar tum faster fly karo (abhi bhi subsonic, abhi bhi chota

α

Reality mein True lekin Barrowman ise False treat karta hai. Linear equations ek angle- aur speed-independent CP dete hain; yeh ek approximation hai jo sirf subsonic, small-

α

flow ke liye valid hai — Mach 1 ke paas real CP aft shift ho jaata hai.

Ek two-fin rocket mein teesra fin add karna fin normal-force slope ko triple kar deta hai.

False. Factor

n

ise scale karta hai, isliye

n = 2 \to 3

jaana

C_{N α, fins}

3/2

se multiply karta hai, 3 se nahi — aur yeh tail mass bhi add karta hai jo CG shift karta hai, isliye stability expect ke according improve nahi ho sakti.

Body-interference factor

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

1 se kam ho sakta hai.

False. Kyunki radius

r > 0

aur semi-span

s > 0

hai, fraction hamesha positive hai, isliye

K_{f b} > 1

hamesha — body hamesha fin ki effective force badhati hai, kabhi ghatati nahi.

Spot the error

"Main nose ke liye

C_{N α} = 2

per degree use karunga kyunki slope ek simple 2 hai."

Error: units. Saare Barrowman slopes per radian hain. Per-degree use karna secretly har force ko 57.3 se divide kar deta hai, components ke beech koi bhi comparison bigad jaata hai (dekho degrees/radians angle convention).

"Mera fin term

K_{f b}

ignore karta hai — pure-fin formula already complete hai, isliye mera CP conservative hai."

Error.

K_{f b}

drop karna

C_{N α, fins}

ko shrink karta hai, jo

\overset{ˉ}{X}_{c p} = \frac{\sum C _{N α, i} X ˉ _{i}}{\sum C _{N α, i}}

mein sabse bada, sabse aft weight hai. Woh bada rear weight kam karna weighted mean ko forward (chota

\overset{ˉ}{X}_{c p}

) kheenchta hai — isliye tum stability ko underestimate karte ho, jo automatically "conservative" nahi hai.

"Bade fins → zyada restoring force → hamesha zyada stable."

Error. Bade fins

C_{N α, fins}

badhate hain aur CP ko aft move karte hain (accha), lekin saath mein tail par mass bhi add karte hain, CG ko aft drag karte hain (bura). Net static margin shrink ho sakta hai; CP aur CG dono recompute karo.

"CP saari aerodynamic surfaces ka geometric middle hai."

Error. CP normal-force-weighted mean

\overset{ˉ}{X}_{c p} = \frac{\sum C _{N α, i} X ˉ _{i}}{\sum C _{N α, i}}

hai; ek component jiska tiny

C_{N α}

hai woh barely count karta hai chahe woh kitna bhi door ho.

"Maine Barrowman se ek bada

α = 2 5^{\circ}

manoeuvre compute kiya aur ek clean CP number mila."

Error. Numbers compute hote hain lekin theory invalid hai: yeh linear, small-

α

(roughly

α ≲ 1 0^{\circ}

) slender-body flow assume karta hai. Usse aage real CP move karta hai aur answer meaningless hai.

"Mere rocket ka CP exactly CG par hai, isliye static margin zero hai — woh neutrally theek hai."

Error. Zero margin ka matlab koi restoring torque nahi; koi bhi gust rocket ko tumbling chhod deta hai aur kuch correct nahi karta. Tumhe roughly 1–2 calibres ka positive margin chahiye.

"Fin CP

\overset{ˉ}{X}_{fins}

forward move karta hai jab main tips ko peeche sweep karta hoon."

Error. Sweep load ko aft push karta hai:

X_{r}

(sweep distance) term

\overset{ˉ}{X}_{fins}

mein add hota hai, fin CP ko rearward (bada

X

) move karta hai — jo typically stability help karta hai.

Why questions

Barrowman force slopes

C_{N α}

kyun compute karta hai, absolute forces kyun nahi?

Kyunki ek symmetric rocket ka

α = 0

par zero side force hota hai; stability ek chote tilt ke response ke baare mein hai, isliye meaningful quantity woh rate

\partial C_{N} / \partial α

hai jis par restoring force har radian tilt par appear hoti hai.

Nose CP uski length ka

k

factor kyun hai, cone ke liye

k = 2/3

Local lift

d S / d x

ke proportional hai (cross-section area

S

kaise tezi se

x

ke along badhti hai); ek cone ke liye radius linearly badhta hai isliye

S \propto x^{2}

aur

d S / d x \propto x

hai, aur CP area-weighted centroid

\frac{\int _{0}^{L} x \cdot x d x}{\int _{0}^{L} x d x} = \frac{L ^{3} /3}{L ^{2} /2} = \frac{2}{3} L

hai. Neeche ki figure dekho.

Total CP formula

\sum C_{N α, i}

se kyun divide karta hai?

Kyunki CP is se define hota hai ki "net moment = net force × lever arm":

\sum N_{i} \overset{ˉ}{X}_{i} = \overset{ˉ}{X}_{c p} \sum N_{i}

, aur dono sides ko

\frac{1}{2} ρ v^{2} A_{r e f}

se divide karna har

N_{i}

C_{N α, i}

mein turn karta hai, ek weighted average chhod ke.

Fins usually CP ko kyun dominate karte hain chahe nose bhi force banata ho?

Fin slope

C_{N α, fins}

typically nose ke 2 se kai guna bada hota hai, isliye weighted mean mein fins CP ko strongly tail ki taraf kheenchte hain — yahi exactly woh cheez hai jo rocket ko stable banati hai.

Conical transition ka slope squared diameter ratios ke difference ki tarah kyun likha jaata hai?

Lift sirf wahan generate hoti hai jahan area change hoti hai; term

2 [(d_{2} / d)^{2} - (d_{1} / d)^{2}]

fore diameter

d_{1}

aur aft diameter

d_{2}

ke beech add (ya remove) ki gayi net cross-section area measure karta hai.

Body-interference factor 1 se zyada kyun hota hai?

Ek mote body ke paas se flow bahar ki taraf deflect hoti hai, isliye us body par mounted fins "extra sideways flow" dekhte hain aur zyada force produce karte hain jo free air mein identical fin produce karta — isliye

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

. Neeche deflected streamlines dekho.

CP location ko safe kyun compute kiya ja sakta hai chahe tumne accidentally degrees aur radians mix kar diye ho?

Kyunki CP do force-weighted terms ka ratio hai; ek common per-degree/per-radian scale factor upar aur neeche cancel ho jaata hai. Slope magnitudes (aur isliye margin comparisons) corrupt hote hain, CP position nahi.

Edge cases

Transition slope kya hoga agar

d_{1} = d_{2}

ho (ek cylinder, koi taper nahi)?

Zero.

(d_{2} / d)^{2} - (d_{1} / d)^{2} = 0

: ek plain cylinder koi cross-section area change nahi karta, isliye slender-body theory predict karta hai ki usse koi normal force nahi aati.

Fin term ka kya hoga jab semi-span

s \to 0

ho jaaye (fins shrink ho kar kuch nahi bache)?

Factor

(s / d)^{2} \to 0

, isliye

C_{N α, fins} \to 0

— vanishing fins koi restoring force nahi dete, aur CP sirf nose aur body se set hota hai.

Zero tip chord (

C_{t} = 0

, ek triangular fin) wale fin ke liye, kya CP formula abhi bhi kaam karta hai?

Haan.

C_{t} = 0

substitute karne par: taper term ban jaata hai

\frac{X _{r}}{3} \cdot \frac{C _{r} + 0}{C _{r} + 0} = \frac{X _{r}}{3}

, aur bracket ban jaata hai

\frac{1}{6} [C_{r} - 0] = \frac{1}{6} C_{r}

— har term finite aur well-defined rehta hai.

Ek negative total static margin physically kya matlab rakhta hai?

CP CG ke aage baitha hai (

\overset{ˉ}{X}_{c p} < X_{c g}

), isliye air-force torque kisi bhi tilt ko amplify karta hai — rocket diverge karta hai aur badly thrown dart ki tarah tumble karta hai, self-correct nahi karta.

Bahut mote body ki limit mein jahan

r ≫ s

ho,

K_{f b}

kiske paas approach karta hai?

K_{f b} = 1 + \frac{r}{s + r} \to 1 + 1 = 2

— jab body radius fin span se bahut bada ho to interference fins ki effective force ko lagbhag double kar deta hai.

Agar CG, CP se bahut aage ho, jaise 10 calibres ka margin?

Rocket over-stable ho jaata hai: yeh kisi bhi crosswind mein aggressively weathercock karta hai, hawa mein tilt karta hai aur off-course fly karta hai — bahut zyada margin utna hi unwanted hai jitna bahut kam (ideal ≈ 1–2 calibres).

Kya boat-tail ka negative

C_{N α}

kabhi rocket ko destabilise kar sakta hai?

Ho sakta hai. Uska negative force tail ke paas act karta hai, aur weighted mean mein bade

\overset{ˉ}{X}

par ek negative term total CP ko forward (chota

X

) drag kar sakta hai, static margin shrink karta hai — hamesha combined result check karo.

Recall Ek-line self-test

Answers cover karo: kya tum kyun bol sakte ho CP ek force-weighted mean hai, kyun $K_{f b} > 1$ hai, aur kab Barrowman valid nahi rehta? Force-weighted kyunki moment = force × arm ::: total force se divide karna balance point deta hai. $K_{f b} > 1$ kyunki ::: body flow ko outward deflect karti hai, isliye fins extra deflected air feel karte hain. Barrowman fail karta hai jab ::: $α$ bada ho ( $≳ 1 0^{\circ}$ ) ya speed transonic/supersonic ho.