Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions of angle of attack, Mach: Worked examples

This page is the drill-ground for the parent topic. We will not introduce new theory — instead we hit every case class the coefficient relations can throw at you, one worked example per cell, so that no exam scenario is a surprise.

Everything here rests on four building blocks from the parent note. Before we compute anything, let us re-earn each symbol in plain words so a first-time reader is never lost.

Definition The symbols we will use (all defined in plain words)

$C_{N}$ = normal-force coefficient: the push perpendicular to the chord line (the straight line from nose to tail of the wing section), turned into a pure number.
$C_{A}$ = axial-force coefficient: the push along the chord, pointing aft (toward the tail), as a pure number.
$V_{\infty}$ = freestream velocity: the speed and direction of the oncoming air far ahead of the wing, before the wing disturbs it. It is the arrow the "wind frame" is built around.
$C_{L}, C_{D}$ = lift and drag coefficients: the same total push, but measured perpendicular to and along that oncoming wind $V_{\infty}$ .
$α$ = angle of attack: the angle between the chord line and the wind $V_{\infty}$ . Small $α$ = wing nearly aligned with wind; large $α$ = wing tilted up.
$M_{\infty}$ = freestream Mach number: the flight speed $V_{\infty}$ divided by the local speed of sound in the undisturbed air far ahead. $M_{\infty} < 1$ = subsonic, $M_{\infty} > 1$ = supersonic. Throughout this page we always write the Mach number with its subscript as $M_{\infty}$ so it is never confused with a moment.
$C_{m}$ = pitching-moment coefficient: the twisting (nose-up / nose-down) moment about a reference point, non-dimensionalized with an extra chord length.

The bridge between the two frames (proved in the parent note by rotating one vector): $C_{L} = C_{N} cos α - C_{A} sin α, C_{D} = C_{N} sin α + C_{A} cos α$

The remaining tools — thin-airfoil lift, the drag polar, Prandtl–Glauert, Ackeret — are each restated inside the example that first needs them, so you never meet a formula cold.

The scenario matrix

Every problem this topic can pose falls into one of these cells. The rightmost column names the example that covers it.

Cell	Case class	What is tricky about it	Covered by
A	Small $α$ , ordinary signs	baseline conversion, lift ≈ normal	Ex 1
B	$α = 0$ (degenerate)	$sin α = 0$ collapses the rotation	Ex 2
C	Large $α$ (near/at stall)	lift and normal force differ a lot	Ex 3
D	Negative $α$ (sign flip)	which coefficients change sign?	Ex 4
E	Inverse direction (wind → body)	undo the rotation	Ex 5
F	Subsonic compressibility limit	Prandtl–Glauert as $M_{\infty} \to 1$	Ex 6
G	Supersonic (new physics: wave drag)	different square root, extra drag	Ex 7
H	Real-world word problem	strip the words, pick the cell	Ex 8
I	Exam twist (stability sign, $C_{m}$ )	$d C_{m} / d α$ meaning, not just plug-in	Ex 9

The figure below is the master picture for Cells A–E: one resultant force $R$ (the red arrow) seen in two axis systems rotated by $α$ . Every rotation example is just this diagram with different numbers.

Figure — Aerodynamic coefficients — CN, CA, CL, CD, Cm as functions of angle of attack, Mach

Example 1 — Cell A: small angle, ordinary signs

Worked example Body → wind at

$α = 6^{\circ}$ Given $C_{N} = 0.80$ , $C_{A} = 0.04$ , $α = 6^{\circ}$ . Find $C_{L}$ and $C_{D}$ .

Forecast: guess before computing — will $C_{L}$ be bigger or smaller than $C_{N}$ ? Will $C_{D}$ be tiny or large?

Convert the angle: $cos 6^{\circ} = 0.9945$ , $sin 6^{\circ} = 0.1045$ . Why this step? The rotation formulas need cosine and sine of $α$ , not the degree value — trig is what turns an angle into a projection ratio. On the master figure above, $α$ is small, so the two frames nearly overlap.
Lift: $C_{L} = 0.80 (0.9945) - 0.04 (0.1045) = 0.7956 - 0.0042 = 0.7914$ . Why this step? $N$ projects onto the lift direction as $N cos α$ ; the aft-pointing $A$ tilts slightly against lift, hence the minus.
Drag: $C_{D} = 0.80 (0.1045) + 0.04 (0.9945) = 0.0836 + 0.0398 = 0.1234$ . Why this step? Both forces have a piece pointing downstream; they add.

Verify: $C_{L} \approx C_{N} = 0.80$ as the small-angle sanity check predicts (lift ≈ normal at low $α$ ). ✓ Units: all four are dimensionless, consistent. ✓

Example 2 — Cell B: the degenerate case $α = 0$

Worked example Wing perfectly aligned with the wind

Given $C_{N} = 0.30$ , $C_{A} = 0.02$ , $α = 0^{\circ}$ . Find $C_{L}$ and $C_{D}$ .

Forecast: with the chord parallel to the wind, which body force becomes lift, and which becomes drag?

At $α = 0$ : $cos 0 = 1$ , $sin 0 = 0$ . Why this step? This is the degenerate input — the whole rotation switch turns off. On the master figure, imagine the wind axes rotated back until they sit exactly on the body axes.
$C_{L} = 0.30 (1) - 0.02 (0) = 0.30$ . Why this step? Plugging the trig values in: the $sin$ -term is killed by the zero, so only the normal force survives into lift. This makes the "frames coincide" idea concrete rather than abstract.
$C_{D} = 0.30 (0) + 0.02 (1) = 0.02$ . Why: with chord along the wind, "perpendicular to chord" is "perpendicular to wind," so $C_{L} = C_{N}$ exactly; "along chord" is "along wind," so $C_{D} = C_{A}$ exactly.

Verify: at $α = 0$ the two frames are the same frame, so lift must equal normal and drag must equal axial. Both hold exactly. ✓ This is the anchor that tells you a nonzero $α$ only rotates these numbers.

Example 3 — Cell C: large angle (near stall)

Worked example High angle of attack,

$α = 3 0^{\circ}$ Given $C_{N} = 1.20$ , $C_{A} = 0.25$ , $α = 3 0^{\circ}$ . Find $C_{L}$ , $C_{D}$ , and the ratio $C_{L} / C_{N}$ .

Forecast: at $3 0^{\circ}$ , is lift still ≈ normal force, or has the gap opened up?

$cos 3 0^{\circ} = 0.8660$ , $sin 3 0^{\circ} = 0.5000$ . Why this step? At a large angle the sine is no longer tiny (it is a full $0.5$ ), so we need its exact value — this is precisely the regime where dropping $sin α$ would give a wrong answer. On the master figure, picture the wind axes swung a full $3 0^{\circ}$ off the body axes.
$C_{L} = 1.20 (0.8660) - 0.25 (0.5000) = 1.0392 - 0.1250 = 0.9142$ . Why this step? Now $sin α$ is large, so the axial term subtracts a real chunk — the "lift ≈ normal" shortcut has died.
$C_{D} = 1.20 (0.5000) + 0.25 (0.8660) = 0.6000 + 0.2165 = 0.8165$ . Why: a big slice of the normal force ( $N sin α = 0.6$ ) now points downstream — this is the dominant drag at high $α$ .
Ratio $C_{L} / C_{N} = 0.9142/1.20 = 0.762$ . Why this step? The ratio is the single number that exposes how far lift has drifted from normal force — the headline lesson of this cell.

Verify: the ratio dropped from ~0.99 (Ex 1) to 0.76 — exactly the parent note's warning that "lift = normal" fails at high $α$ . ✓ Also $C_{D}$ is now comparable to $C_{L}$ : high- $α$ flight is very draggy, as expected near stall. ✓

Example 4 — Cell D: negative angle of attack

The figure below shows exactly what "negative $α$ " does geometrically. The black arrow is the resultant force $R$ at a positive angle; the red arrow is the same force at the negative angle. Notice they have the same length and the same horizontal reach (cosine is unchanged), but their vertical parts point opposite ways (sine flips). This picture is the reason cosine terms survive a sign change while sine terms reverse — keep it in mind as you read step 1.

Worked example Nose-down attitude,

$α = - 1 0^{\circ}$ Given $C_{N} = 0.55$ , $C_{A} = 0.03$ , $α = - 1 0^{\circ}$ . Find $C_{L}$ and $C_{D}$ .

Forecast: which flips sign — the drag, the lift, or both?

Negative angle: $cos (- 1 0^{\circ}) = cos 1 0^{\circ} = 0.9848$ (cosine is even — unchanged by sign), and $sin (- 1 0^{\circ}) = - sin 1 0^{\circ} = - 0.1736$ (sine is odd — it flips). Why this step? This is the whole point of the sign case: cosine ignores the sign of $α$ , sine tracks it. Look at the figure above — the red vector $R$ is the same length as the black one, only tilted the other way.
$C_{L} = 0.55 (0.9848) - 0.03 (- 0.1736) = 0.5416 + 0.0052 = 0.5468$ . Why: the axial term now adds to lift (double negative), the opposite of Example 1.
$C_{D} = 0.55 (- 0.1736) + 0.03 (0.9848) = - 0.0955 + 0.0295 = - 0.0660$ . Why this step? A negative $C_{D}$ ? That is not real drag — it is telling you that with this contrived (positive) $C_{N}$ at negative $α$ , the streamwise projection of the normal force points upstream. Physically, at negative $α$ a real wing would have negative $C_{N}$ too, and drag would come out positive. The math is faithfully warning us the input combination is unphysical.

Verify: compare with Example 1 ( $+ 6^{\circ}$ -style): flipping $α$ 's sign keeps $cos$ terms and flips $sin$ terms — exactly what we see (lift barely changed, the $sin$ -driven parts reversed). ✓

Example 5 — Cell E: inverse rotation (wind → body)

Worked example Measured

$C_{L}, C_{D}$ — recover $C_{N}, C_{A}$ A wind-tunnel run gives $C_{L} = 0.90$ , $C_{D} = 0.12$ at $α = 1 0^{\circ}$ . Find $C_{N}$ and $C_{A}$ .

Forecast: we are un-rotating. Do we swap the signs on the $sin$ terms?

Use the inverse relations from the parent note: $C_{N} = C_{L} cos α + C_{D} sin α, C_{A} = - C_{L} sin α + C_{D} cos α .$ Why this step? Rotating back by $- α$ flips the sign inside every $sin$ ; this is the same rotation matrix transposed. On the master figure, we now know the wind-frame arrow and want its body-frame shadow.
$cos 1 0^{\circ} = 0.9848$ , $sin 1 0^{\circ} = 0.1736$ . Why this step? The inverse relations still need the same two trig values — un-rotating uses the same angle $α$ , just wired into the formula with swapped signs, so we compute its cosine and sine once here.
$C_{N} = 0.90 (0.9848) + 0.12 (0.1736) = 0.8863 + 0.0208 = 0.9071$ . Why this step? Both wind-frame forces have a piece along the body-normal direction; here they add (note the plus, the mirror image of the forward formula's minus).
$C_{A} = - 0.90 (0.1736) + 0.12 (0.9848) = - 0.1562 + 0.1182 = - 0.0380$ . Why: a negative $C_{A}$ here means the chordwise force actually points forward — plausible when a suction peak near the leading edge pulls the section along the chord (leading-edge thrust).

Verify (round-trip): plug $C_{N} = 0.9071, C_{A} = - 0.0380$ back into the forward relations: $C_{L} = 0.9071 (0.9848) - (- 0.0380) (0.1736) = 0.8933 + 0.0066 = 0.90$ ✓ and $C_{D} = 0.9071 (0.1736) + (- 0.0380) (0.9848) = 0.1575 - 0.0374 = 0.120$ ✓. The forward and inverse maps undo each other.

Example 6 — Cell F: subsonic compressibility, pushing the limit

The figure below plots the Prandtl–Glauert factor: as $M_{\infty}$ climbs toward 1, the red curve for $C_{L}$ rockets upward. That runaway is what this example makes numeric.

Worked example Prandtl–Glauert from

$M_{\infty} = 0.3$ to $M_{\infty} = 0.85$ A thin airfoil has incompressible $C_{L, incomp} = 0.50$ . Restated tool: $C_{L} = \frac{C _{L, incomp}}{1 - M _{\infty}^{2}}, β = 1 - M_{\infty}^{2}$ ( $β$ is the compressibility factor: how much subsonic air "stiffens" the flow.) Find $C_{L}$ at $M_{\infty} = 0.3$ , $0.6$ , and $0.85$ .

Forecast: as $M_{\infty}$ climbs toward 1, does lift grow slowly or run away?

$M_{\infty} = 0.3$ : $β = 1 - 0.09 = 0.91 = 0.9539$ , so $C_{L} = 0.50/0.9539 = 0.5242$ . Why this step? This is why we call low-speed flow "incompressible enough" — a 5% correction only.
$M_{\infty} = 0.6$ : $β = 1 - 0.36 = 0.64 = 0.80$ , so $C_{L} = 0.50/0.80 = 0.625$ . Why this step? We recompute $β$ at a higher Mach to see the correction grow — the factor $1/ β$ is now $1.25$ , a solid 25% boost, no longer negligible.
$M_{\infty} = 0.85$ : $β = 1 - 0.7225 = 0.2775 = 0.5268$ , so $C_{L} = 0.50/0.5268 = 0.9492$ . Why this step? Pushing $M_{\infty}$ almost to the edge of validity shows the denominator collapsing toward zero, so $C_{L}$ nearly doubled — this is the runaway growth as $M_{\infty} \to 1$ that we needed to witness.

Verify (limiting behaviour): the three answers increase monotonically with $M_{\infty}$ , and the jump accelerates ( $+ 0.024$ , then $+ 0.10$ , then $+ 0.32$ ). ✓ Caution flag: at $M_{\infty} = 0.85$ we are past the parent note's validity band ( $M_{\infty} ≲ 0.7$ ); the formula still computes but real shocks would have appeared. Faithful math, physically stretched. See Prandtl-Glauert Compressibility Correction.

Example 7 — Cell G: supersonic, new physics (wave drag)

The figure below contrasts the two square-root factors on one axis: below Mach 1 the factor $1/ 1 - M_{\infty}^{2}$ amplifies lift, above Mach 1 the factor $1/ M_{\infty}^{2} - 1$ shrinks it, and a brand-new red wave-drag term appears.

Worked example Ackeret flat plate at

$M_{\infty} = 2$ Restated tool (Supersonic Linearized (Ackeret) Theory, 2D flat plate): $C_{L} = \frac{4 α}{M _{\infty}^{2} - 1}, C_{D, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1}$ with $α$ in radians. Take $M_{\infty} = 2$ , $α = 3^{\circ}$ . Find $C_{L}$ and $C_{D, wave}$ , and the ratio $C_{D} / C_{L}$ .

Forecast: notice the square root flipped to $M_{\infty}^{2} - 1$ . Above Mach 1, does more $α$ cost drag linearly or quadratically?

Convert to radians: $α = 3^{\circ} \times π /180 = 0.05236$ rad. Why this step? The Ackeret formulas are derived with $α$ small in radians; degrees would give nonsense.
Denominator: $2^{2} - 1 = 3 = 1.7321$ . Why this step? Supersonically the compressibility factor becomes $M_{\infty}^{2} - 1$ — the same square root as subsonic but with the sign under the root flipped.
$C_{L} = 4 (0.05236) /1.7321 = 0.20944/1.7321 = 0.1209$ . Why this step? Plugging the radian angle and the denominator into the lift formula produces the actual supersonic lift — smaller than a subsonic wing would give at the same $α$ , because the $M_{\infty}^{2} - 1$ factor now divides down rather than up.
$C_{D, wave} = 4 (0.05236)^{2} /1.7321 = 4 (0.002742) /1.7321 = 0.010967/1.7321 = 0.006332$ . Why this step? This drag exists only supersonically — it comes from shock losses, not friction. Subsonic inviscid flow has zero wave drag.
Ratio $C_{D} / C_{L} = 0.006332/0.1209 = 0.05236 = α$ (in radians!). Why: dividing the two formulas cancels everything except one factor of $α$ — an elegant check.

Verify: the ratio $C_{D, wave} / C_{L}$ came out exactly $α$ in radians (0.05236). That algebraic identity confirms both numbers are consistent. ✓ And wave drag is quadratic in $α$ (the squared term), matching the parent note's statement that wave drag $\propto α^{2}$ . ✓

Example 8 — Cell H: real-world word problem

Worked example Reading a flight-test report

"A UAV cruises at $M_{\infty} = 0.6$ . Its 2-D airfoil section produces an incompressible lift coefficient of $0.45$ at cruise angle. At the same instant, sensors resolve the section force as $C_{N} = 0.60$ along the body normal and $C_{A} = 0.05$ axial, with $α = 4^{\circ}$ . (a) What lift coefficient does compressibility give? (b) Cross-check by converting the body-frame numbers to wind-frame lift."

Forecast: two independent routes should land on roughly the same lift — spot the cell for each sentence.

Part (a) is Cell F. $β = 1 - 0.36 = 0.80$ , so $C_{L} = 0.45/0.80 = 0.5625$ . Why this step? The phrase "incompressible ... at $M_{\infty} = 0.6$ " is Prandtl–Glauert's exact trigger.
Part (b) is Cell A. $cos 4^{\circ} = 0.99756$ , $sin 4^{\circ} = 0.06976$ . Why this step? Part (b) is a straight body-to-wind rotation, so we first need the trig of the given $α = 4^{\circ}$ before we can project the forces.
$C_{L} = 0.60 (0.99756) - 0.05 (0.06976) = 0.59854 - 0.00349 = 0.5951$ . Why this step? The body-frame reading is the measured, already-compressible force, so no PG factor is applied here — it is a direct rotation.

Verify: the two routes give $0.5625$ and $0.5951$ — same ballpark (within ~6%), the small gap being real 3-D and measurement effects. The word problem is decoded: sentence 1 → compressibility cell, sentence 2 → rotation cell. ✓

Example 9 — Cell I: exam twist (stability sign, $C_{m}$ )

The figure below plots $C_{m}$ versus $α$ for the two wings. The red line slopes down (stable: any nudge is pushed back); the black line slopes up (unstable: any nudge grows). Read the sign of the slope, not the height of the line.

Worked example Which wing is statically stable?

Two configurations are tested about their center of gravity ( $α$ in radians, $C_{m}$ = pitching-moment coefficient):

Wing P: $C_{m} = 0.04 - 0.90 α$
Wing Q: $C_{m} = - 0.02 + 0.30 α$

Which is statically stable, and at what $α$ does each trim (i.e. $C_{m} = 0$ )?

Forecast: stability is about the slope's sign, not the value of $C_{m}$ . Guess which wing before computing.

Read the slope $d C_{m} / d α$ off each line: Wing P $\to - 0.90$ , Wing Q $\to + 0.30$ . Why this step? $d C_{m} / d α < 0$ is the stability condition (parent note): a small pitch-up must generate a nose-down restoring moment. The coefficient multiplying $α$ is that slope, so we read it directly off the figure's line steepness.
Wing P: slope $- 0.90 < 0$ → statically stable. Wing Q: slope $+ 0.30 > 0$ → unstable (pitches further away from equilibrium). See Static Longitudinal Stability. Why this step? We compare each slope against the stability threshold of zero — this is the whole "twist": the answer hinges on a sign, not a magnitude.
Trim angle for Wing P (set $C_{m} = 0$ ): $0 = 0.04 - 0.90 α \Rightarrow α = 0.04/0.90 = 0.04444$ rad $= 2.54 6^{\circ}$ . Why this step? Trim is where the net pitching moment vanishes — the equilibrium the aircraft would actually fly at. We solve the linear equation for the $α$ that zeroes $C_{m}$ ; on the figure it is where the red line crosses zero.
Trim angle for Wing Q (set $C_{m} = 0$ ): $0 = - 0.02 + 0.30 α \Rightarrow α = 0.02/0.30 = 0.06667$ rad $= 3.82 0^{\circ}$ . Why this step? Same algebra for Wing Q — it has a trim point too (the black line's zero-crossing), but as step 2 showed it cannot hold it, so the trim angle is a paper equilibrium the aircraft would diverge from.

Answer: Wing P is statically stable (negative slope) and trims at $α = 0.0444$ rad ( $2.5 5^{\circ}$ ); Wing Q is unstable (positive slope) and, although it trims at $α = 0.0667$ rad ( $3.8 2^{\circ}$ ), it cannot maintain that equilibrium.

Verify: nudge Wing P to $α = 0.05$ (just above its trim of $0.0444$ ): $C_{m} = 0.04 - 0.90 (0.05) = - 0.005 < 0$ → nose-down, pushes back toward trim. ✓ Now nudge Wing Q above its trim, to $α = 0.08 > 0.0667$ : $C_{m} = - 0.02 + 0.30 (0.08) = + 0.004 > 0$ → nose-up, pushes $α$ further up, away from trim → confirms instability. ✓ The twist: identical-looking algebra, opposite physics, decided entirely by one sign. See also Aerodynamic Center vs Center of Pressure.

Recall Which cell is this? (self-test)

"Given $C_{L} = 0.7$ , $C_{D} = 0.1$ , $α = 1 2^{\circ}$ , find $C_{N}$ ." ::: Cell E (inverse rotation, wind → body). " $C_{L} = 0.6$ incompressible, $M_{\infty} = 0.7$ , find compressible $C_{L}$ ." ::: Cell F (Prandtl–Glauert). " $M_{\infty} = 2.5$ flat plate, find wave drag." ::: Cell G (Ackeret, new physics). " $d C_{m} / d α = + 0.2$ — stable?" ::: Cell I: no, positive slope is unstable.