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

Before anything, one promise: every symbol below was earned in the parent note. To be safe, here they are in one place.

Definition Every symbol on this page (tap to expand)

$q = \frac{1}{2} ρ V^{2}$ — dynamic pressure, the "push per unit area" the air carries. Units: pascals ( $Pa = N/ m^{2}$ ). See Dynamic pressure and Bernoulli.
$ρ$ — air density (kg/m³). $V$ — flight speed (m/s).
$S$ — reference area (m²), a fixed number chosen for the rocket (often the cross-section of the body tube).
$d$ — reference length (m), usually the body diameter — needed only for moments.
$A$ — axial force: air pushing straight back along the nose–tail line.
$N$ — normal force: air pushing sideways, perpendicular to the body.
$M$ — pitching moment: air twisting the rocket nose-up (+) or nose-down (−).
$α$ — angle of attack: how far the nose points away from where the rocket is actually going.
$D$ — drag: force along the flight path, opposing the velocity. $L$ — lift: force perpendicular to the flight path. These live in the wind frame (see Drag and Lift in wind axes), whereas $A$ and $N$ live in the body frame.
$C_{A} = \frac{A}{q S}$ , $C_{N} = \frac{N}{q S}$ , $C_{m} = \frac{M}{q S d}$ , $C_{D} = \frac{D}{q S}$ , $C_{L} = \frac{L}{q S}$ — the pure-number "shape factors".
$C_{N α} = \frac{\partial C _{N}}{\partial α}_{0}$ — normal-force slope, per radian. See Fin design and normal-force slope $C_{N\alpha}$.
$C_{m α} = \frac{\partial C _{m}}{\partial α}_{0}$ — moment slope (per radian): how fast the twisting coefficient grows as the nose is tilted. Its sign decides stability (negative = restoring = stable). See Static and dynamic stability of rockets.
static margin $= \frac{x _{c p} - x _{c g}}{d}$ — gap between centre of pressure and centre of gravity in diameters.
$x_{c p}$ , $x_{c g}$ — distances of the centre of pressure and centre of gravity from the nose (m).
Mach number $Ma = \frac{V}{a}$ — flight speed divided by the local speed of sound $a$ . "Mach 2" means $V = 2 a$ . It is the number that must match between a model and the full-size rocket for the coefficients to transfer. See Reynolds and Mach scaling.

From body forces to wind forces (the rotation we will keep reusing)

Two of the examples below (Ex 1 and Ex 6) ask for drag $C_{D}$ and lift $C_{L}$ , not the body forces $C_{A}$ and $C_{N}$ . So we must first derive the bridge between the two frames — never use it before building it.

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

Intuition WHY the two frames differ, and WHY a rotation connects them

Look at the figure. The body axis (chalk-white) points along the nose–tail line. The wind axis (blue) points along the flight velocity $V$ . When the rocket flies at angle of attack $α$ , these two directions are misaligned by exactly $α$ .

The axial force $A$ points backward along the body. Its shadow (projection) onto the wind direction is $A cos α$ — that part opposes $V$ , so it is drag.
The normal force $N$ points perpendicular to the body. Tilt the body up by $α$ and part of $N$ , namely $N sin α$ , now also points backward along the wind (the pink dashed drop in the figure) — extra drag. Adding the two drag contributions, and doing the same bookkeeping perpendicular to the wind for lift:

$D = A cos α + N sin α, L = N cos α - A sin α .$

The lift gets $+ N cos α$ (most of $N$ ) but $- A sin α$ (the axial force leans slightly against the lift direction). Dividing every term by $q S$ turns forces into coefficients, giving the two formulas we will reuse:

The pitching-moment law (the second tool we will reuse)

Several examples below need the moment coefficient $C_{m}$ , and just as we derived the drag/lift rotation before using it, we build the moment law here — never inside an example.

The scenario matrix

Every problem this topic throws is one of these cells. The examples below are tagged [Cell n] so you can see the whole grid is covered.

#	Cell class	What is tricky about it	Covered by
1	Force from coefficient, $α = 0$	pure plug-in; $C_{D} = C_{A}$ here	Ex 1
2	Positive $α$ ( $α > 0$ )	$C_{N} > 0$ , nose-up angle	Ex 2
3	Negative $α$ ( $α < 0$ )	every sign flips — must track them	Ex 3
4	Zero / degenerate: symmetric rocket at $α = 0$ , or CP = CG	$C_{N} = 0$ , $C_{m} = 0$ , neutral stability	Ex 4
5	Unstable rocket ( $C_{m α} > 0$ )	wrong-sign moment, diverges	Ex 5
6	Large $α$ (body vs wind axes)	$cos α, sin α$ no longer $\approx 1, α$	Ex 6
7	Real-world word problem	you must pick out $q$ , $S$ , $α$ yourself	Ex 7
8	Exam twist: solve backwards for static margin	rearrange the moment equation	Ex 8

We will reuse one "base rocket" so numbers stay comparable: $ρ = 0.9 kg/ m^{3}, V = 300 m/s, S = 0.05 m^{2}, d = 0.25 m .$ So its dynamic pressure is always $q = \frac{1}{2} (0.9) (300)^{2} = 40, 500 Pa .$

Example 1 — Plug-in at zero angle · [Cell 1]

Worked example Axial force flying dead straight

The base rocket has $C_{A} = 0.35$ at $α = 0$ . Find the axial force $A$ , and state its drag coefficient $C_{D}$ .

Forecast: guess whether $A$ is closer to 100 N, 700 N, or 7000 N — and whether $C_{D}$ equals $C_{A}$ here.

Step 1. $q = 40, 500 Pa$ (from the base numbers). Why this step? The coefficient alone tells you nothing in Newtons; $q$ carries all the speed-and-density scaling. Compute it first, always.

Step 2. $A = C_{A} q S = 0.35 \times 40, 500 \times 0.05 = 708.75 N$ . Why this step? Just $C_{A} = A / (q S)$ rearranged. Multiply the shape number back onto the physical scale $q S$ .

Step 3. At $α = 0$ : $C_{D} = C_{A} cos 0 + C_{N} sin 0 = C_{A} \cdot 1 + 0 = C_{A} = 0.35$ . Why this step? Use the rotation just derived. Body axis and wind axis coincide when the nose points exactly where the rocket flies, so "along the body" is "along the wind". This is the only case where $C_{D} = C_{A}$ .

Verify: units — $[Pa] \cdot [m^{2}] = N/ m^{2} \cdot m^{2} = N$ . ✔ Magnitude is hundreds of newtons, matching the "closer to 700 N" guess.

Example 2 — Positive angle of attack · [Cell 2]

Worked example Normal force from a nose-up angle

Base rocket, $C_{N α} = 2.0 ra d^{- 1}$ , angle of attack $α = + 4^{\circ}$ . Find $C_{N}$ and the normal force $N$ .

Forecast: will $N$ be bigger or smaller than the axial force from Ex 1?

Step 1. Convert: $α = 4^{\circ} \times \frac{π}{180} = 0.06981 rad$ . Why this step? The slope $C_{N α}$ is defined per radian. Feeding degrees in gives a factor-of-57 blunder.

Step 2. $C_{N} = C_{N α} α = 2.0 \times 0.06981 = 0.13963$ . Why this step? For small $α$ , $C_{N}$ grows linearly — this is the linearized rule $C_{N} \approx C_{N α} α$ .

Step 3. $N = C_{N} q S = 0.13963 \times 40, 500 \times 0.05 = 282.72 N$ . Why this step? Same $q S$ as Ex 1 — that's the whole beauty of coefficients, one scale factor for every force.

Verify: $N = 282.72 N$ is a bit smaller than $A = 708.75 N$ . Sensible — at a gentle $4^{\circ}$ tilt the sideways shove is real but smaller than the head-on push. Sign check: $α > 0 \Rightarrow C_{N} > 0$ (nose-up gives an "upward" normal force). ✔

Example 3 — Negative angle of attack · [Cell 3]

Worked example Every sign flips

Same rocket and slope, but now $α = - 4^{\circ}$ (nose points below the flight path). Find $C_{N}$ , $N$ , and the pitching-moment sign, with static margin $= 1.5$ .

Forecast: predict the sign of $C_{N}$ and the sign of $C_{m}$ before computing.

Step 1. $α = - 0.06981 rad$ . Why this step? Convert, keeping the minus sign — the direction of the tilt lives entirely in that sign.

Step 2. $C_{N} = 2.0 \times (- 0.06981) = - 0.13963$ , so $N = - 282.72 N$ . Why this step? Linearity carries the sign straight through: nose-down ⇒ negative normal force (air now pushes the nose region the other way).

Step 3. $C_{m} = - C_{N} \times margin = - (- 0.13963) (1.5) = + 0.20944$ . Why this step? Use the pitching-moment law $C_{m} = - C_{N} \cdot (static margin)$ built above. A negative $C_{N}$ makes $C_{m}$ positive — a nose-up twist.

Verify: does this restore? Nose was pushed down ( $α < 0$ ); the moment came out positive = nose-up, pushing it back toward zero. ✔ That is exactly what a stable rocket must do — the restoring twist always opposes the disturbance. Compare with Ex 3 of the parent (positive $α$ gave negative $C_{m}$ ): the mirror image, as it should be.

Example 4 — Zero and degenerate cases · [Cell 4]

Worked example The symmetric rocket, and the neutral rocket

(a) A fin-symmetric rocket flies at exactly $α = 0$ . What are $C_{N}$ and $C_{m}$ ? (b) A rocket has its centre of pressure exactly on its centre of gravity, $x_{c p} = x_{c g}$ , at $α = + 4^{\circ}$ . What is $C_{m}$ ? Is it stable?

Forecast: guess both moments before reading on.

Step 1 (a). By symmetry, no sideways preference at $α = 0$ : $C_{N} = C_{N α} \cdot 0 = 0$ , hence $N = 0$ and $C_{m} = - C_{N} (\dots) = 0$ . Why this step? A symmetric body flying straight has no reason to push or twist in any particular sideways direction — the two sides cancel. Only $C_{A}$ survives.

Step 2 (b). static margin $= \frac{x _{c p} - x _{c g}}{d} = \frac{0}{d} = 0$ . Then $C_{m} = - C_{N} \times 0 = 0$ even though $C_{N} = 0.13963 \neq = 0$ . Why this step? The lever arm is the distance from CG to where the force acts. Zero distance ⇒ zero torque, no matter how big the force.

Verify: In (a) both are zero because the cause ( $α$ ) is zero. In (b) $C_{m} = 0$ because the lever arm is zero. Two different roads to "no twist" — and case (b) is neutrally stable: nudge it and it neither returns nor diverges, since $C_{m α} = C_{m} / α = 0$ . This is the exact borderline between the stable Ex 3 and the unstable Ex 5. ✔

Example 5 — The unstable rocket · [Cell 5]

Worked example CG behind the CP — a tumbler

A rocket has its CG placed behind its CP: static margin $= - 0.8$ (negative!). At $α = + 4^{\circ}$ , $C_{N} = 0.13963$ . Find $C_{m}$ and $C_{m α}$ , and decide stability.

Forecast: for a nose-up disturbance, will the moment push the nose further up (bad) or back down (good)?

Step 1. $C_{m} = - C_{N} \times margin = - (0.13963) (- 0.8) = + 0.11170$ . Why this step? Same moment law; a negative margin flips the usual sign of $C_{m}$ .

Step 2. $C_{m α} = \frac{C _{m}}{α} = \frac{0.11170}{0.06981} = + 1.6 ra d^{- 1}$ . Why this step? The stability verdict rides entirely on the sign of the moment slope, not its size.

Verify: $α > 0$ (nose up) gave $C_{m} > 0$ (nose up twist) — the disturbance is amplified. Since $C_{m α} = + 1.6 > 0$ , the rocket is statically unstable and tumbles. This is the mirror-warning of Ex 3: positive margin restores, negative margin diverges, zero margin (Ex 4b) sits on the fence. See Static and dynamic stability of rockets. ✔

Example 6 — Large angle: body axes vs wind axes matter · [Cell 6]

Step 1. $cos 3 0^{\circ} = 0.86603$ , $sin 3 0^{\circ} = 0.5$ . Why this step? At $3 0^{\circ}$ the small-angle shortcuts ( $cos \approx 1$ , $sin \approx α$ ) are wrong by tens of percent — we need the real trig. The figure shows the body axis tilted $3 0^{\circ}$ off the wind axis; the projections onto the wind axis are exactly these cosines and sines.

Step 2. $C_{D} = C_{A} cos α + C_{N} sin α = 0.35 (0.86603) + 0.90 (0.5) = 0.30311 + 0.45 = 0.75311$ . Why this step? Drag is measured along the wind, so we apply the body-to-wind rotation derived above. The axial force projects onto the wind by $cos α$ ; the normal force, once you tilt the body, spills a component $sin α$ into the drag direction too (look at the pink arrow in the figure).

Step 3. $C_{L} = C_{N} cos α - C_{A} sin α = 0.90 (0.86603) - 0.35 (0.5) = 0.77942 - 0.175 = 0.60442$ . Why this step? Lift is perpendicular to the wind. Here the normal force projects with $cos α$ , but the axial force subtracts $A sin α$ (it leans partly against the lift direction).

Verify: $C_{D} = 0.753$ is more than double $C_{A} = 0.35$ — so " $C_{D} = C_{A}$ " is badly wrong at $3 0^{\circ}$ ; it holds only at $α = 0$ (Ex 1). Consistency check: the total force magnitude must be frame-independent. $C_{A}^{2} + C_{N}^{2} = 0.3 5^{2} + 0.9 0^{2} = 0.9656$ and $C_{D}^{2} + C_{L}^{2} = 0.75 3^{2} + 0.60 4^{2} = 0.9656$ — identical, because rotating axes never changes a vector's length. ✔ See Drag and Lift in wind axes.

Example 7 — Real-world word problem · [Cell 7]

Worked example A wind-tunnel model transferred to full scale

A wind-tunnel test of a model rocket at Mach 2 measures $C_{A} = 0.40$ . The full-size rocket flies the same Mach 2 at altitude where $ρ = 0.30 kg/ m^{3}$ and the local speed of sound is $a = 340 m/s$ ; its reference area is $S = 0.20 m^{2}$ . What axial force does the full-size rocket feel?

Forecast: does the model's $C_{A}$ apply to the big rocket unchanged? (Yes — that's why coefficients exist. Guess the force in kN.)

Step 1. Recover the speed from the Mach number: $Ma = V / a$ , so $V = Ma \times a = 2 \times 340 = 680 m/s$ . Why this step? The problem gives speed only as "Mach 2", which by definition means $V = 2 a$ . We must turn it into m/s before we can build $q$ . Matching the Mach number between model and full scale is exactly what lets us borrow the model's $C_{A}$ (see Reynolds and Mach scaling).

Step 2. $q = \frac{1}{2} ρ V^{2} = 0.5 (0.30) (680)^{2} = 0.5 \times 0.30 \times 462, 400 = 69, 360 Pa$ . Why this step? Same recipe for $q$ as always; only $ρ$ and $V$ are the local flight values, not the wind tunnel's. $q$ carries all the scaling that differs between model and full scale.

Step 3. $A = C_{A} q S = 0.40 \times 69, 360 \times 0.20 = 5548.8 N \approx 5.5 kN$ . Why this step? We borrowed the dimensionless $C_{A}$ from the small model but applied the full-size $q$ and $S$ . That is exactly the point of coefficients — the shape number transfers, the scale factor is local.

Verify: units give newtons ✔; magnitude ~5.5 kN is reasonable for a launch vehicle at high dynamic pressure. Had we (wrongly) reused the tunnel's tiny $q$ or the model's tiny $S$ , we'd get a laughably small force — the transfer only works through the coefficient, never the raw newtons. ✔

Example 8 — Exam twist: solve backwards for static margin · [Cell 8]

Worked example Given the measured moment, find where the CP is

Flight telemetry on the base rocket ( $q = 40, 500$ , $S = 0.05$ , $d = 0.25$ ) at $α = + 4^{\circ}$ records a pitching moment $M = - 70 N m$ with $C_{N α} = 2.0 ra d^{- 1}$ . Find the static margin and how many diameters the CP sits behind the CG.

Forecast: the moment is negative at nose-up $α$ — before computing, is this rocket stable?

Step 1. $C_{m} = \frac{M}{q S d} = \frac{- 70}{40 , 500 \times 0.05 \times 0.25} = \frac{- 70}{506.25} = - 0.13827$ . Why this step? Undo the moment definition $C_{m} = M / (q S d)$ to get the dimensionless $C_{m}$ — we must compare coefficients, not raw newton-metres.

Step 2. $C_{N} = C_{N α} α = 2.0 \times 0.06981 = 0.13963$ (as in Ex 2). Why this step? The margin formula $C_{m} = - C_{N} \cdot margin$ needs $C_{N}$ ; we build it from the angle.

Step 3. Rearrange the moment law for the geometry: $margin = - \frac{C _{m}}{C _{N}} = - \frac{- 0.13827}{0.13963} = 0.99026 \approx 0.99$ diameters. Why this step? Solving $C_{m} = - C_{N} \cdot (static margin)$ for the margin instead of the moment — the exam's favourite inversion.

Verify: margin $\approx + 0.99 > 0$ ⇒ CP is about one diameter behind the CG ⇒ statically stable, matching the negative moment at nose-up $α$ . ✔ Sanity: plug back, $C_{m} = - 0.13963 \times 0.99026 = - 0.13827$ ✔, and $M = C_{m} q S d = - 0.13827 \times 506.25 = - 70.0 N m$ ✔. So a stable rocket at nose-up $α$ must show a negative moment, exactly as recorded.

Recall

Recall Which cell needs the full

$cos α, sin α$ and why? Large-angle problems (Cell 6). Below ~ $5^{\circ}$ you may use $cos \approx 1, sin \approx α$ , but at $3 0^{\circ}$ the body-to-wind rotation changes $C_{D}$ from $0.35$ to $0.75$ — you must project with real trig. ::: Cell 6.

The sign of $C_{m α}$ for a stable rocket is ::: negative ( $C_{m α} < 0$ ): a nose-up disturbance must produce a nose-down (negative) restoring moment.

Zero moment can arise two ways — name them ::: either the cause is zero ( $α = 0 \Rightarrow C_{N} = 0$ ) or the lever arm is zero (CP = CG ⇒ margin $= 0$ ).

Why does the model's $C_{A}$ transfer to the full-size rocket in Ex 7? ::: Because $C_{A}$ is dimensionless and both fly at the same Mach number; only $q$ and $S$ (the local scale factors) differ.