Worked examples — Barrowman equations — centre of pressure calculation for finned rockets

3.4.8 · D3 · Physics › Rocket Flight Mechanics › Barrowman equations — centre of pressure calculation for fin

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Scenario matrix

Kuch bhi solve karne se pehle, chalte hain saare case-classes list karte hain jo is topic mein aa sakte hain. Neeche har worked example us cell ko fill karta hai jiska usse tag mila hai.

Cell	Case class	Kya khaas hai	Example
A	Baseline: nose + fins	"Rozana" wala rocket	Ex 1
B	Positive transition (flare)	Area badhti hai, term $>0$	Ex 2
C	Negative transition (boat-tail)	Area ghatati hai, term $<0$	Ex 3
D	Degenerate fin: $C_t\to 0$	Triangular fin, dekho $0/0$	Ex 4
E	Limiting body: $r\to 0$ (skinny)	Interference $K_{fb}\to 1$	Ex 5
F	Limiting body: $r\gg s$ (fat)	Interference $K_{fb}\to 2$	Ex 5
G	Real-world word problem	Prose → numbers mein translate karo	Ex 6
H	Exam twist: CG target ke liye solve karo	Static-margin formula ko invert karo	Ex 7
I	Sanity: symmetric = $\alpha =0$ par zero force	Sirf slopes kyun matter karte hain	Ex 8

Har jagah use hone wale symbols:

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Example 1 — Cell A: baseline (nose + fins)

1. Nose. $C_{N\alpha ,n}=2$, ${\bar{X}}_n=\frac{2}{3}(15)=10$ cm. Yeh step kyun? Slender-body theory nose slope ko 2 fix karti hai chahe shape koi bhi ho; cone ka load $x$ ke saath badhta hai toh uska centroid length ke $2/3$ par baithta hai (figure mein amber load triangle dekho).
2. Mid-chord length ${\ell }_f=\sqrt{s^2+{\left(X_r+\frac{C_t-C_r}{2}\right)}^2}=\sqrt{25+(4-1.5{)}^2}=\sqrt{31.25}=5.590$ cm. Kyun? Ek swept fin par lift uski tirchi mid-chord line ke along kaam karti hai, axis ke along nahi, isliye hum woh sahi length measure karte hain (Tool 3 ka ${\ell }_f$).
3. Interference $K_{fb}=1+\frac{2}{5+2}=1.2857$. Kyun? Body ke around deflect hui hawa fins ko extra zyada hit karti hai; yeh factor fin force ko scale up karta hai.
4. Fin slope $\frac{2{\ell }_f}{C_r+C_t}=\frac{11.18}{9}=1.2423$; $\sqrt{1+1.2423^2}=1.5981$. $C_{N\alpha ,f}=1.2857\cdot \frac{4\cdot 3\cdot (5/4{)}^2}{1+1.5981}=1.2857\cdot \frac{18.75}{2.5981}=\mathrm{9.279.}$ Kyun? Seedha Tool 3 mein substitution; note karo $9.28\gg 2$, toh fins dominate karte hain.
5. Fin CP (Tool 3b) ${\bar{X}}_f=50+\frac{4}{3}\cdot \frac{6+2\cdot 3}{9}+\frac{1}{6}\left[9-\frac{18}{9}\right]=50+1.778+1.167=52.944$ cm. Kyun? Pehla term taper-weighted centroid hai; bracket chords ko average karta hai.
6. Combine (Tool 4, weighted mean): ${\bar{X}}_{cp}=\frac{2(10)+9.279(52.944)}{2+9.279}=\frac{20+491.3}{11.279}=45.33\text{ cm}.$

Recall Verify karo

Weighted mean 10 aur 52.9 ke beech hona chahiye — hai (45.3). Kyunki fin weight (9.28) nose weight (2) se ~4.6× zyada hai, CP fins ki taraf strongly khicha jaata hai, forecast se match karta hai. ✓

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Example 2 — Cell B: ek flare (positive transition)

Figure mein yeh flare cyan mein draw ki gayi hai: body walls (white) $d_1=4$ se $d_2=6$ tak shaded cone ke upar bahar ki taraf step karti hain. Amber dashed line woh jagah mark karti hai jahan transition CP land karta hai — neeche ke numbers ke against isko read karo.

1. Slope (Tool 2). $C_{N\alpha ,t}=2\left[{\left(\frac{6}{4}\right)}^2-{\left(\frac{4}{4}\right)}^2\right]=2[2.25-1]=2.5$. Kyun? Lift sirf wahin generate hoti hai jahan cross-section area change hoti hai; figure mein walls bahar ki taraf splay karti hain, toh bracket positive hai.
2. CP (Tool 2b) $d_1/d_2=4/6=0.6667$ ke saath: ${\bar{X}}_t=X_t+\frac{L_t}{3}\left(1+\frac{1-0.6667}{1-0.6667^2}\right)=40+\frac{8}{3}\left(1+\frac{0.3333}{0.5556}\right).$ $\frac{0.3333}{0.5556}=0.6$, toh ${\bar{X}}_t=40+2.667(1.6)=40+4.267=44.27$ cm — figure mein bilkul amber dashed line par. Kyun? Yeh truncated cone ke along linearly-growing load ka centroid hai.

Recall Verify karo

$C_{N\alpha ,t}=2.5>0$ ✓ (flare force add karta hai). CP $44.27$ flare $[40,48]$ ke andar hai ✓ aur amber marker se match karta hai. Yeh flare, nose ke peeche baithke, ${\bar{X}}_{cp}$ ko aft ki taraf kheenchega — stability ke liye acha.

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Example 3 — Cell C: ek boat-tail (NEGATIVE transition)

1. Slope (Tool 2). $C_{N\alpha ,t}=2\left[{\left(\frac{2.5}{4}\right)}^2-{\left(\frac{4}{4}\right)}^2\right]=2[0.3906-1]=-1.2188$. Kyun? Aft diameter fore se chota ⇒ bracket negative ⇒ boat-tail wahaan normal force hatata hai jahan woh baithta hai (figure mein amber walls andar ki taraf pinch karti hain).
2. CP (Tool 2b) $d_1/d_2=4/2.5=1.6$ ke saath: ${\bar{X}}_t=44+\frac{6}{3}\left(1+\frac{1-1.6}{1-1.6^2}\right)=44+2\left(1+\frac{-0.6}{-1.56}\right).$ $\frac{-0.6}{-1.56}=0.3846$, toh ${\bar{X}}_t=44+2(1.3846)=44+2.769=46.77$ cm. Kyun? Wohi centroid formula; note karo ratios valid rehte hain chahe $d_1>d_2$ ho (numerator aur denominator dono sign flip karte hain, ek legitimate positive fraction dete hain).

Recall Verify karo

$C_{N\alpha ,t}=-1.219<0$ ✓. CP $46.77$ $[44,50]$ ke andar ✓. Average mein ek negative weight legal hai; agar $|\sum C_{N\alpha }|$ overall positive rehta hai toh mean abhi bhi meaningful hai.

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Example 4 — Cell D: degenerate fin ($C_t\to 0$, triangular)

1. Mid-chord ${\ell }_f=\sqrt{6^2+{\left(8+\frac{0-8}{2}\right)}^2}=\sqrt{36+16}=\sqrt{52}=7.211$ cm. Kyun? $\frac{C_t-C_r}{2}=-4$; $X_r=8$ mein add karo toh $4$ milta hai; slant length delta ke liye bhi honest hai.
2. Interference $K_{fb}=1+\frac{2.5}{6+2.5}=1.2941$.
3. Slope (Tool 3) $\frac{2{\ell }_f}{C_r+C_t}=\frac{14.42}{8}=1.8028$; $\sqrt{1+1.8028^2}=2.0592$. $C_{N\alpha ,f}=1.2941\cdot \frac{4\cdot 4\cdot (6/5{)}^2}{1+2.0592}=1.2941\cdot \frac{23.04}{3.0592}=\mathrm{9.746.}$ Kyun? $C_r+C_t=8$ finite hai chahe $C_t=0$ ho; yahaan koi division problem nahi.
4. Fin CP (Tool 3b) — danger term check karo. ${\bar{X}}_f=X_f+\frac{X_r}{3}\cdot \frac{C_r+2C_t}{C_r+C_t}+\frac{1}{6}\left[(C_r+C_t)-\frac{C_r{C}_t}{C_r+C_t}\right].$ $C_t=0$ ke saath: $\frac{C_r+0}{C_r}=1$ aur $\frac{C_r{C}_t}{C_r+C_t}=\frac{0}{8}=0$ — $0/0$ nahi, kyunki denominator $C_r+C_t=8\ne 0$. ${\bar{X}}_f=60+\frac{8}{3}(1)+\frac{1}{6}[8-0]=60+2.667+1.333=64.00\text{ cm}.$ Yeh step kyun? Apparent singularity dissolve ho jaati hai: product $C_r{C}_t$ zero hit karta hai jabki uska denominator positive rehta hai.

Recall Verify karo

$\frac{C_r{C}_t}{C_r+C_t}=0$ (finite), toh koi blow-up nahi ✓. CP $64.0$ cm fin ke chord $[60,68]$ ke andar hai ✓. Ek delta fin ka CP yahaan bilkul $C_r/2$ peeche plus sweep-share par land karta hai.

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Example 5 — Cells E & F: skinny vs fat body ($K_{fb}$ limits)

1. Skinny (E): $K_{fb}=1+\frac{0.1}{5.1}=1.0196\approx 1$. Kyun? Almost koi body nahi, toh flow ko around deflect karne ke liye kuch nahi, isliye fins free air mein act karte hain jaison ($K_{fb}\to 1$).
2. Fat (F): $K_{fb}=1+\frac{100}{105}=1.9524\approx 2$. Kyun? Ek huge body fins jo effective flow dekhte hain use double kar deti hai; ratio $\frac{r}{s+r}\to 1$ jab $r\to \infty$, toh $K_{fb}\to 2$.
3. Consequence. Fin slope $K_{fb}$ ke saath linearly scale hoti hai, toh ek fat body fin ke $C_{N\alpha }$ contribution ko skinny wale ke versus almost double kar deti hai — CP noticeably aft shift hota hai.

Recall Verify karo

${\lim }_{r\to 0}{K}_{fb}=1$ ✓, ${\lim }_{r\to \infty }{K}_{fb}=2$ ✓. Dono computed values bilkul in limits ke beech hain: $1.0196$ aur $1.9524$.

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Example 6 — Cell G: real-world word problem

1. Prose → symbols translate karo. $d=5.4$, $r=2.7$; $n=4$, $C_r=7$, $C_t=2$, $s=5.5$, $X_r=5$, $X_f=68$; $L_n=16$, $k=0.466$. Sab cm mein. Kyun? Aadha kaam naming hai: "chauda" = diameter, "bahar khada" = semi-span, "root ke aaglay edge se peeche" = sweep.
2. Nose (Tool 1): $C_{N\alpha ,n}=2$, ${\bar{X}}_n=0.466\cdot 16=7.456$ cm. Kyun? Ogive apna $k$ use karta hai; slope abhi bhi 2 hai (shape-independent).
3. Mid-chord ${\ell }_f=\sqrt{5.5^2+{\left(5+\frac{2-7}{2}\right)}^2}=\sqrt{30.25+(2.5{)}^2}=\sqrt{36.5}=6.042$ cm.
4. Interference $K_{fb}=1+\frac{2.7}{5.5+2.7}=1.3293$.
5. Fin slope (Tool 3) $\frac{2{\ell }_f}{C_r+C_t}=\frac{12.083}{9}=1.3426$; $\sqrt{1+1.3426^2}=1.6739$. $C_{N\alpha ,f}=1.3293\cdot \frac{4\cdot 4\cdot (5.5/5.4{)}^2}{1+1.6739}=1.3293\cdot \frac{16.596}{2.6739}=\mathrm{8.251.}$
6. Fin CP (Tool 3b) ${\bar{X}}_f=68+\frac{5}{3}\cdot \frac{7+4}{9}+\frac{1}{6}\left[9-\frac{14}{9}\right]=68+2.037+1.241=71.278$ cm.
7. Combine (Tool 4): ${\bar{X}}_{cp}=\frac{2(7.456)+8.251(71.278)}{2+8.251}=\frac{14.912+588.12}{10.251}=58.83$ cm.

Recall Verify karo

CP 7.46 aur 71.3 ke beech ✓. Forecast ke mutabiq, longer body CP ko aft push karta hai (58.8 vs Ex 1 ka 45.3) ✓. Ogive ka chota $k$ barely matter karta hai kyunki fins zyaadatar weight hold karte hain.

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Example 7 — Cell H: exam twist (CG target ke liye solve karo)

1. Formula invert karo. $\text{SM}=\frac{{\bar{X}}_{cp}-X_{cg}}{d}\Rightarrow X_{cg}={\bar{X}}_{cp}-\text{SM}\cdot d$. Kyun? Static margin ek definition hai jise hum algebraically rearrange kar sakte hain — CP aur $d$ known hain, $X_{cg}$ ke liye solve karo.
2. 1.5 calibres: $X_{cg}=58.83-1.5(5.4)=58.83-8.1=50.73$ cm.
3. 1.0 calibre (minimum): $X_{cg}=58.83-1.0(5.4)=58.83-5.4=53.43$ cm. Kyun? Chota margin matlab CG ko CP ke zyada paas allow kiya gaya, yaani zyada aft (53.4 > 50.7) — forecast se match karta hai.
4. Isko padhna: margin $\ge 1.0$ maintain karne ke liye, CG 53.43 cm par ya usse forward hona chahiye; target 1.5 ke liye, 50.73 cm aim karo.

Recall Verify karo

Plug back karo: $(58.83-50.73)/5.4=8.1/5.4=1.5$ ✓; $(58.83-53.43)/5.4=5.4/5.4=1.0$ ✓. Higher margin ⇒ CG forward, confirmed. CG side ke liye Static stability and static margin aur Centre of gravity determination dekho.

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Example 8 — Cell I: sanity case ($\alpha =0$ par zero force)

1. Zero tilt par force. Har component par normal force $N_i=\frac{1}{2}\rho v^2{A}_{ref}{C}_{N\alpha ,i}\alpha$ hai. $\alpha =0$ par, har $N_i=0$. Kyun? Ek symmetric body seedha hawa mein point karti hui dono sides par equal push feel karti hai — woh cancel ho jaate hain, toh net side force bilkul zero hai.
2. Toh absolute forces CP dhundhne ke liye bekar hain: moment balance padh kar $0={\bar{X}}_{cp}\cdot 0$, yaani $0/0$ — undefined. Kyun? Tum un forces ka balance point locate nahi kar sakte jo sab zero hain.
3. Slopes limit survive karte hain. Moment balance $\sum N_i{\bar{X}}_i={\bar{X}}_{cp}\sum N_i$ ko $\frac{1}{2}\rho v^2{A}_{ref}\alpha$ se divide karo — $\alpha$ ka har factor cancel ho jaata hai, leaving ${\bar{X}}_{cp}=\frac{\sum C_{N\alpha ,i}{\bar{X}}_i}{\sum C_{N\alpha ,i}},$ jo finite aur angle-independent hai. Yeh step kyun? Isliye parent ne kaha tha "har Barrowman term ek ${}_{\alpha }$ quantity hai" — yeh derivative hi hai jo $\alpha \to 0$ par zinda rehti hai.
4. Numeric demonstration. Ex 1 ke numbers lo ($C_{N\alpha ,n}=2$ at ${\bar{X}}_n=10$; $C_{N\alpha ,f}=9.279$ at ${\bar{X}}_f=52.944$). Koi bhi tilt lo, jaise $\alpha =2^{\circ }=0.0349$ rad, aur $q=\frac{1}{2}\rho v^2{A}_{ref}$ lo. Tab $N_n=q(2)(0.0349),N_f=q(9.279)(0.0349).$ Weighted mean: ${\bar{X}}_{cp}=\frac{N_n(10)+N_f(52.944)}{N_n+N_f}$. $q$ aur $0.0349$ dono har term mein appear karte hain aur cancel ho jaate hain, $\frac{2(10)+9.279(52.944)}{2+9.279}=45.33$ cm milta hai — bilkul Ex 1 jaisa, kisi bhi $\alpha$ ke liye jo tune choose ki ho. Kyun? Yeh prove karta hai ki CP purely ek geometric property hai; tilt sirf force ki size set karta hai, kabhi nahi ki woh kahaan kaam karta hai.

Recall Verify karo

$\alpha =0$ set karo: numerator aur denominator forces dono vanish ho jaate hain ($0/0$), lekin unki slopes nahi — ratio well-defined hai aur 45.33 cm equal hai. $\alpha =2^{\circ }$ choose karna bhi wohi 45.33 cm deta hai kyunki $q\alpha$ cancel ho jaata hai. ✓ (Yeh Angle of attack and restoring moment se link karta hai.)

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Recall Matrix ka ek-line recap

Baseline weighted mean :: Ex 1 — CP 45.3 cm Flare = positive term :: Ex 2 — $C_{N\alpha }=+2.5$ Boat-tail = negative term :: Ex 3 — $C_{N\alpha }=-1.22$ Delta fin $C_t\to 0$ safe hai :: Ex 4 — CP 64.0 cm, koi $0/0$ nahi $K_{fb}$ $1\to 2$ run karta hai :: Ex 5 — skinny 1.02, fat 1.95 Word problem :: Ex 6 — CP 58.8 cm CG ke liye solve karo :: Ex 7 — 50.73 cm at 1.5 cal Sirf slopes matter karte hain $\alpha =0$ par :: Ex 8 — CP 45.3 cm kisi bhi $\alpha$ ke liye

Yeh bhi dekho: Slender-body aerodynamic theory · Nose cone shapes and drag · Rocket flight simulation (6-DOF).