Aerodynamic coefficients — CA (axial force), CN (normal force), Cm (pitching moment): Exercises

3.4.7 · D4 · Physics › Rocket Flight Mechanics › Aerodynamic coefficients — CA (axial force), CN (normal forc

Hum in vault topics ka sahara lenge: Dynamic pressure and Bernoulli, Angle of attack $\alpha$, Center of pressure and center of gravity, Static and dynamic stability of rockets, Fin design and normal-force slope $C_{N\alpha}$, Drag and Lift in wind axes, Reynolds and Mach scaling.

Level 1 — Recognition

(Kya aap ek definition padh ke numbers plug in kar sakte ho?)

Recall Solution L1.1

Hum kya karte hain: $C_{N} = N / (q S)$ ko rearrange karke $N = C_{N} q S$ banate hain. Kyun: coefficient ek pure shape/attitude number hai; $q S$ saari dimensions carry karta hai (Newtons). $N = 0.18 \times 20000 \times 0.04 = 144 N .$ $C_{N}$ ke units: $\frac{[ N ]}{[ Pa ] [ m ^{2} ]} = \frac{N}{( N/ m ^{2} ) ( m ^{2} )} = 1$ — yeh dimensionless hai. ✔

Recall Solution L1.2

Kya / Kyun: $q$ aanewali hawa ki kinetic energy per unit volume hai — woh cheez jo "pushing" karti hai (Dynamic pressure and Bernoulli). $q = \frac{1}{2} ρ V^{2} = 0.5 \times 1.0 \times 25 0^{2} = 31, 250 Pa .$

Recall Solution L1.3

Stable. Nose-up disturbance ( $α > 0$ ) ko ek nose-down (negative) restoring moment produce karna chahiye. Negative slope $C_{m α} < 0$ exactly wahi karta hai (Static and dynamic stability of rockets). ✔

Level 2 — Application

(Do ya teen steps ek saath chain karein.)

Recall Solution L2.1

Step 1 (kyun): $q$ nikaalein — isme saari speed/density scaling hoti hai. $q = 0.5 \times 1.1 \times 18 0^{2} = 17, 820 Pa .$ Step 2: $A = C_{A} q S = 0.42 \times 17820 \times 0.03 = 224.5 N$ (≈ $225 N$ ). ✔

Recall Solution L2.2

Step 1 (kyun radians): slopes per radian hote hain. Pehle convert karein (Angle of attack $\alpha$). $α = 5^{\circ} = 5 \times \frac{π}{180} = 0.08727 rad .$ Step 2: $C_{N} = C_{N α} α = 2.2 \times 0.08727 = 0.1920$ . Step 3: $N = C_{N} q S = 0.1920 \times 17820 \times 0.03 = 102.6 N$ (≈ $103 N$ ). ✔

Recall Solution L2.3

Step 1: $C_{m} = - C_{N} \times margin = - 0.1920 \times 1.8 = - 0.3456$ . Minus kyun: nose-up hone par nose-down twist hoti hai — restoring (Center of pressure and center of gravity). Step 2: $M = C_{m} q S d = - 0.3456 \times 17820 \times 0.03 \times 0.20 = - 36.95 N m$ (≈ $- 37 N m$ ). ✔

Level 3 — Analysis

(Relationships ko alag karein; frames aur geometry ke baare mein reason karein.)

Figure — Aerodynamic coefficients — CA (axial force), CN (normal force), Cm (pitching moment)

Recall Solution L3.1

Hum kya use karte hain (parent se, force relations ko $q S$ se divide karke): $C_{D} = C_{A} cos α + C_{N} sin α, C_{L} = C_{N} cos α - C_{A} sin α .$ Yeh signs kyun: figure s01 dekho. Body ko $α$ se upar tilt karein. Axial force $A$ (body ke saath) drag direction par $A cos α$ project karti hai; normal force ka kuch hissa, $N sin α$ , bhi backward point karta hai aur drag mein add hota hai. Lift ke liye (wind ke perpendicular), $N cos α$ help karta hai lekin $A sin α$ subtract karta hai (Drag and Lift in wind axes).

$α = 8^{\circ}$ ke saath: $cos 8^{\circ} = 0.99027$ , $sin 8^{\circ} = 0.13917$ . $C_{D} = 0.40 (0.99027) + 0.30 (0.13917) = 0.3961 + 0.0418 = 0.4379.$ $C_{L} = 0.30 (0.99027) - 0.40 (0.13917) = 0.2971 - 0.0557 = 0.2414.$ Interpretation: yahan $C_{D} > C_{L}$ hai kyunki $C_{A}$ (jo drag mein dominate karta hai lekin lift se subtract hota hai) bada hai. Chhote $α$ par body-axis drag $C_{A}$ abhi bhi wind-frame drag mein rule karta hai. ✔

Recall Solution L3.2

Hum kya use karte hain: $C_{m} = - C_{N} \frac{x _{c p} - x _{c g}}{d}$ . Margin ke liye solve karein. $\frac{x _{c p} - x _{c g}}{d} = - \frac{C _{m}}{C _{N}} = - \frac{- 0.45}{0.25} = 1.80.$ Toh $x_{c p} - x_{c g} = 1.80 \times 0.18 = 0.324 m$ , jisse $x_{c p} = 1.20 + 0.324 = 1.524 m nose se .$ CP, CG ke peeche hai (positive margin) → yeh rocket stable hai. ✔

Level 4 — Synthesis

(Multiple ideas ko ek design decision mein combine karein.)

Recall Solution L4.1

Step 1 — radians: $α = 6^{\circ} = 0.10472 rad$ . Step 2 — normal-force coeff: $C_{N} = C_{N α} α = 2.5 \times 0.10472 = 0.2618$ . Step 3 — moment coeff: $C_{m} = - C_{N} \times margin = - 0.2618 \times 1.5 = - 0.3927$ . Step 4 — dimensional moment: $M = C_{m} q S d = - 0.3927 \times 25000 \times 0.0177 \times 0.15 = - 26.06 N m .$ Step 5 — slope (stability verdict ke liye zaroori, linear regime mein $α$ se independent): $C_{m α} = - C_{N α} \times margin = - 2.5 \times 1.5 = - 3.75 ra d^{- 1} < 0.$ Negative slope → statically stable; $- 26 N m$ ka twist nose ko wapas neeche push karta hai. ✔

Recall Solution L4.2

Idea (kyun kaam karta hai): normal forces add hoti hain, aur combined CP parts ke CPs ka normal-force-weighted average hota hai — kyunki total moment mein har part ka contribution (uski force) × (uska arm) hota hai. $x_{c p} = \frac{C _{N α, body} x _{c p, body} + C _{N α, fin} x _{c p, fin}}{C _{N α, body} + C _{N α, fin}} .$ $x_{c p} = \frac{1.2 ( 0.9 ) + 1.8 ( 2.4 )}{1.2 + 1.8} = \frac{1.08 + 4.32}{3.0} = \frac{5.40}{3.0} = 1.80 m .$ Static margin: $\frac{x _{c p} - x _{c g}}{d} = \frac{1.80 - 1.50}{0.15} = \frac{0.30}{0.15} = 2.0 (calibers) .$ Fins ne CP ko $0.9 m$ (CG ke aage — unstable!) se $1.8 m$ (CG ke peeche) par kheench liya, jisse ek healthy 2-caliber margin mila (Fin design and normal-force slope $C_{N\alpha}$). ✔

Level 5 — Mastery

(Edge cases, limits, aur models ke beech poora transfer.)

Recall Solution L5.1

Neutral ka matlab: $C_{m α} = - C_{N α} \cdot \frac{x _{c p} - x _{c g}}{d} = 0$ ke liye $x_{c g} = x_{c p}$ chahiye. Combined CP (L4.2 se) $x_{c p} = 1.80 m$ hai. Toh: $x_{c g, neutral} = 1.80 m .$ Limits: agar $x_{c g} < 1.80 m$ (CG, CP ke aage), margin $> 0$ → stable. Agar $x_{c g} > 1.80 m$ (CG, CP ke peeche), margin $< 0$ → $C_{m α} > 0$ → divergent tumble. Exactly $1.80 m$ par ek disturbance na badhti hai na khatam hoti hai. Figure s02 dekho. ✔

Recall Solution L5.2

Key idea (kyun yeh legal hai): coefficients pure numbers hote hain jo sirf shape, $α$ , Mach, Reynolds par depend karte hain — physical size par nahi. Mach aur Reynolds match karna (Reynolds and Mach scaling) matlab model ke coefficients full rocket ke coefficients ke barabar hain. Toh hum sirf full-scale $q, S, d$ se re-dimensionalize karte hain. $A = C_{A} q S = 0.38 \times 40000 \times 0.20 = 3040 N .$ $N = C_{N} q S = 0.22 \times 40000 \times 0.20 = 1760 N .$ $M = C_{m} q S d = - 0.33 \times 40000 \times 0.20 \times 0.30 = - 792 N m .$ Model ki chhoti si forces cleanly scale up hoti hain — yahi poora reason hai ki coefficients exist karte hain. ✔

Recall Solution L5.3

$α = 0$ par: $cos 0 = 1$ , $sin 0 = 0$ , aur symmetry se $C_{N} (0) = 0$ , $C_{m} (0) = 0$ . $C_{D} = C_{A} (1) + C_{N} (0) = C_{A}, C_{L} = C_{N} (1) - C_{A} (0) = 0.$ Toh drag, axial force ke barabar ho jaata hai aur lift khatam ho jaati hai — body aur wind axes align ho jaate hain, exactly waise jaisa " $C_{A}$ is not $C_{D}$ " caveat ne predict kiya tha: yeh sirf yahan coincide karte hain. Aur $C_{m} = 0$ matlab koi twist nahi: ek symmetric rocket bilkul seedha fly karta hua pure drag feel karta hai aur kuch nahi. ✔

Recall Yeh page band karne se pehle ek-line self-check

Ek coefficient, raw force in Newtons se zyada useful kyun hai? ::: Kyunki yeh speed, density, aur size ( $q S$ ) strip kar deta hai aur sirf shape-and-attitude number rakhta hai — toh yeh matched Mach/Reynolds par wind-tunnel model se full rocket par unchanged transfer ho jaata hai.