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

We will lean on these vault topics: 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

(Can you read a definition and plug numbers in?)

Recall Solution L1.1

What we do: rearrange $C_{N} = N / (q S)$ to $N = C_{N} q S$ . Why: the coefficient is the pure shape/attitude number; $q S$ carries all the dimensions (Newtons). $N = 0.18 \times 20000 \times 0.04 = 144 N .$ Units of $C_{N}$ : $\frac{[ N ]}{[ Pa ] [ m ^{2} ]} = \frac{N}{( N/ m ^{2} ) ( m ^{2} )} = 1$ — it is dimensionless. ✔

Recall Solution L1.2

What / Why: $q$ is the kinetic energy per unit volume of oncoming air — the thing that "does the pushing" (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. A nose-up disturbance ( $α > 0$ ) must produce a nose-down (negative) restoring moment. Negative slope $C_{m α} < 0$ does exactly that (Static and dynamic stability of rockets). ✔

Level 2 — Application

(Chain two or three steps together.)

Recall Solution L2.1

Step 1 (why): get $q$ — it holds all the speed/density scaling. $q = 0.5 \times 1.1 \times 18 0^{2} = 17, 820 Pa .$ 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 (why radians): slopes are per radian. Convert first (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: $C_{m} = - C_{N} \times margin = - 0.1920 \times 1.8 = - 0.3456$ . The minus: nose-up produces nose-down twist — 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

(Pull relationships apart; reason about frames and geometry.)

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

Recall Solution L3.1

What we use (from the parent, dividing the force relations by $q S$ ): $C_{D} = C_{A} cos α + C_{N} sin α, C_{L} = C_{N} cos α - C_{A} sin α .$ Why these signs: look at figure s01. Tilt the body up by $α$ . The axial force $A$ (along the body) projects onto the wind's drag direction as $A cos α$ ; part of the normal force, $N sin α$ , also points backward and adds to drag. For lift (perpendicular to wind), $N cos α$ helps but $A sin α$ subtracts (Drag and Lift in wind axes).

With $α = 8^{\circ}$ : $cos 8^{\circ} = 0.99027$ , $sin 8^{\circ} = 0.13917$ . $C_{D} = 0.40 (0.99027) + 0.30 (0.13917) = 0.3961 + 0.0418 = 0.4379.$ $C_{L} = 0.30 (0.99027) - 0.40 (0.13917) = 0.2971 - 0.0557 = 0.2414.$ Interpretation: $C_{D} > C_{L}$ here because $C_{A}$ (which dominates drag but subtracts from lift) is large. At small $α$ the body-axis drag $C_{A}$ still rules the wind-frame drag. ✔

Recall Solution L3.2

What we use: $C_{m} = - C_{N} \frac{x _{c p} - x _{c g}}{d}$ . Solve for the margin. $\frac{x _{c p} - x _{c g}}{d} = - \frac{C _{m}}{C _{N}} = - \frac{- 0.45}{0.25} = 1.80.$ So $x_{c p} - x_{c g} = 1.80 \times 0.18 = 0.324 m$ , giving $x_{c p} = 1.20 + 0.324 = 1.524 m from the nose .$ CP is behind CG (positive margin) → this rocket is stable. ✔

Level 4 — Synthesis

(Combine multiple ideas into one design decision.)

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 — the slope (needed for the stability verdict, independent of $α$ in the linear regime): $C_{m α} = - C_{N α} \times margin = - 2.5 \times 1.5 = - 3.75 ra d^{- 1} < 0.$ Negative slope → statically stable; the $- 26 N m$ twist pushes the nose back down. ✔

Recall Solution L4.2

Idea (why this works): normal forces add, and the combined CP is the normal-force-weighted average of the parts' CPs — because each part's contribution to the total moment is (its force) × (its arm). $x_{c p} = \frac{C _{N α, body} x _{c p, body} + C _{N α, fin} x _{c p, fin}}{C _{N α, body} + C _{N α, fin}} .$ $x_{c p} = \frac{1.2 ( 0.9 ) + 1.8 ( 2.4 )}{1.2 + 1.8} = \frac{1.08 + 4.32}{3.0} = \frac{5.40}{3.0} = 1.80 m .$ Static margin: $\frac{x _{c p} - x _{c g}}{d} = \frac{1.80 - 1.50}{0.15} = \frac{0.30}{0.15} = 2.0 (calibers) .$ The fins pulled the CP from $0.9 m$ (in front of the CG — unstable!) to $1.8 m$ (behind CG), giving a healthy 2-caliber margin (Fin design and normal-force slope $C_{N\alpha}$). ✔

Level 5 — Mastery

(Edge cases, limits, and full transfer between models.)

Recall Solution L5.1

What neutral means: $C_{m α} = - C_{N α} \cdot \frac{x _{c p} - x _{c g}}{d} = 0$ requires $x_{c g} = x_{c p}$ . The combined CP is (from L4.2) $x_{c p} = 1.80 m$ . So: $x_{c g, neutral} = 1.80 m .$ Limits: if $x_{c g} < 1.80 m$ (CG ahead of CP), margin $> 0$ → stable. If $x_{c g} > 1.80 m$ (CG behind CP), margin $< 0$ → $C_{m α} > 0$ → divergent tumble. Right at $1.80 m$ a disturbance neither grows nor decays. See figure s02. ✔

Recall Solution L5.2

Key idea (why this is legal): coefficients are pure numbers that depend only on shape, $α$ , Mach, Reynolds — not on physical size. Matching Mach and Reynolds (Reynolds and Mach scaling) means the model's coefficients equal the full rocket's. So we just re-dimensionalize with full-scale $q, S, d$ . $A = C_{A} q S = 0.38 \times 40000 \times 0.20 = 3040 N .$ $N = C_{N} q S = 0.22 \times 40000 \times 0.20 = 1760 N .$ $M = C_{m} q S d = - 0.33 \times 40000 \times 0.20 \times 0.30 = - 792 N m .$ The model's tiny forces scale up cleanly — this is the entire reason coefficients exist. ✔

Recall Solution L5.3

At $α = 0$ : $cos 0 = 1$ , $sin 0 = 0$ , and by symmetry $C_{N} (0) = 0$ , $C_{m} (0) = 0$ . $C_{D} = C_{A} (1) + C_{N} (0) = C_{A}, C_{L} = C_{N} (1) - C_{A} (0) = 0.$ So drag equals axial force and lift vanishes — the body and wind axes are aligned, exactly as the " $C_{A}$ is not $C_{D}$ " caveat predicted: they coincide only here. And $C_{m} = 0$ means no twist: a symmetric rocket flying dead straight feels pure drag and nothing else. ✔

Recall One-line self-check before you close this page

Why is a coefficient more useful than the raw force in Newtons? ::: Because it strips out speed, density, and size ( $q S$ ) and keeps only the shape-and-attitude number — so it transfers unchanged from a wind-tunnel model to the full rocket at matched Mach/Reynolds.