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

Step 1 — WHAT is the force, and what are we allowed to draw?

WHAT. Air flowing past a wing pushes on it. Add up every tiny push (pressure) and every tiny drag-along (shear) over the whole surface. The grand total is one single arrow: the resultant aerodynamic force. Call it $R$ . That is the only physics on this page — the rest is geometry.

WHY start here. Before we split anything, we must agree there is one thing to split. $L$ , $D$ , $N$ , and $A$ are not four separate forces — they are four shadows of the same arrow $R$ cast onto two different pairs of directions. Miss this and every sign looks arbitrary.

PICTURE. The chord line (the straight line from nose to tail of the airfoil) and the freestream velocity $V_{\infty}$ (the wind the wing flies into). The angle between them is the angle of attack $α$ . The single black arrow is $R$ .

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

Step 2 — WHY two frames at all? Because two people ask two questions

WHAT. We lay down two sets of perpendicular axes on top of the same arrow $R$ :

the body frame: one axis along the chord, one perpendicular to it → gives $(A, N)$ ;
the wind frame: one axis along $V_{\infty}$ , one perpendicular to it → gives $(D, L)$ .

WHY. The structural engineer cares about the body frame — bending and twisting happen along the wing's own axes. The performance engineer cares about the wind frame — lift holds you up, drag slows you down, both defined relative to the oncoming air. Same arrow, two honest questions.

PICTURE. Both axis-crosses drawn on the same $R$ . Notice the wind cross is just the body cross spun by $α$ . That is the whole secret: the two frames differ by a single rotation of angle $α$ .

Step 3 — WHY sine and cosine? Earning the tools on a triangle

WHAT. We need a tool that, given an angle, tells us how much of one direction points along another direction. That tool is cosine and sine, read straight off a right triangle.

WHY these tools and not others? We are decomposing an arrow onto tilted axes. Projecting an arrow of length $r$ onto a direction tilted by angle $ϕ$ from it gives $r cos ϕ$ (the "aligned" part) and $r sin ϕ$ (the "sideways" part). Cosine answers "how much survives along this axis?"; sine answers "how much leaks into the perpendicular axis?" No other function measures projection.

PICTURE. A clean right triangle. The hypotenuse is $R$ . Drop it onto a tilted axis: the piece lying along the axis is the adjacent side $= R cos ϕ$ ; the piece across it is the opposite side $= R sin ϕ$ .

Step 4 — Build $L$ : project $N$ and $A$ onto the lift direction

WHAT. The lift direction is perpendicular to $V_{\infty}$ . We ask each body-frame piece, "how much of you points along the lift direction?" and add the answers.

WHY the two contributions. $N$ (perpendicular to the chord) is tilted by exactly $α$ from the lift axis, so its aligned share is $N cos α$ . The axial force $A$ points toward the tail along the chord; when you resolve it onto the lift axis it points slightly downward (opposite to lift), so it subtracts — its share is $- A sin α$ .

PICTURE. Two dashed projection lines from the tips of $N$ and $A$ onto the lift axis. The $N cos α$ arrow points up along lift; the $A sin α$ arrow points down — hence the minus.

L = perp-to-chord force N fraction along lift cos α - aft chord force A fraction leaking down into lift axis sin α

The minus sign is geometry, not a rule to memorize: $A$ tilts backward, so on the lift axis it opposes $N$ .

Step 5 — Build $D$ : project $N$ and $A$ onto the drag direction

WHAT. The drag direction is along $V_{\infty}$ . Ask the same two forces how much of each points downstream, and add.

WHY both add this time. Rotate your eye 90°. Now $N$ leans downstream (a $sin α$ share) and $A$ points mostly downstream already (a $cos α$ share). Both contribute with the same sign — nothing opposes the flow direction here, so drag is a pure sum.

PICTURE. Projection lines onto the drag axis: $N sin α$ and $A cos α$ both point downstream, laid head-to-tail to form $D$ .

D = perp-to-chord force N fraction leaking downstream sin α + aft chord force A fraction along flow cos α

Divide both equations by the same $q_{\infty} S$ (dynamic pressure × reference area — see Dynamic Pressure and Non-dimensionalization). Every force becomes its coefficient and the trigonometry is untouched:

Step 6 — Edge case: $α = 0$ (flying dead-on)

WHAT. Set $α = 0$ : the chord is exactly aligned with the wind, so the two frames coincide.

WHY check this. A correct formula must collapse to the "obvious" answer when the rotation vanishes. If it doesn't, we made a sign error somewhere.

PICTURE. Both axis-crosses on top of each other; $L$ sits exactly on $N$ , $D$ exactly on $A$ .

Plug in $cos 0 = 1$ , $sin 0 = 0$ : $C_{L} = C_{N} (1) - C_{A} (0) = C_{N}, C_{D} = C_{N} (0) + C_{A} (1) = C_{A} .$

So at zero angle of attack, lift is the normal force and drag is the axial force — exactly what "same frame" should give. This is why the small-angle habit " $L \approx N$ " feels true.

Step 7 — Edge case: large and negative $α$ (where the habit breaks)

WHAT. Now let $α$ grow, and also let it go negative (nose below the wind).

WHY. Two dangers hide at the extremes:

Large $α$ — $sin α$ is no longer small, so the $- C_{A} sin α$ term bites; lift and normal force visibly diverge. The " $L = N$ always" myth dies here.
Negative $α$ — $sin (- α) = - sin α$ flips the sign of the sine terms automatically. The same formula handles nose-down flight with no special casing, because sine is an odd function and cosine is even ( $cos (- α) = cos α$ ).

PICTURE. Left: a big- $α$ triangle where $L$ and $N$ point in clearly different directions. Right: a negative- $α$ triangle where the wind axes rotate the other way and the sine terms change sign on their own.

The one-picture summary

Everything on this page is a single arrow $R$ read on two rotated crosses. The final figure stacks the whole story: the body cross $(N, A)$ , the wind cross $(L, D)$ rotated by $α$ , and the four projection segments $N cos α, A sin α, N sin α, A cos α$ colour-matched to the two boxed equations.

Recall Feynman retelling — say it back in plain words

There is only one push from the air, one arrow. I want to describe it two ways. The wing's body has its own natural directions — along the chord and across it — and there the push reads as axial and normal. But what I feel as a pilot is up-force and slow-down-force, measured against the wind, and the wind points at an angle $α$ to the chord. So I take the same arrow and read it on a cross that's been spun by $α$ .

Reading a component on a tilted axis is just projection: the part that stays lined up is a cosine, the part that spills sideways is a sine. Lift picks up $N cos α$ (most of the normal force, since it nearly points up) minus $A sin α$ (the aft-pointing axial force leans a little down on the lift axis). Drag adds them the other way, $N sin α + A cos α$ , because downstream nothing fights back. Divide by $q_{\infty} S$ to swap forces for coefficients — the trig is unchanged.

Test it: at $α = 0$ the two crosses land on each other, so lift equals normal, drag equals axial — the obvious answer. Crank $α$ up and lift and normal split apart. Make $α$ negative and the sines flip on their own. One rotation, no new physics, every case covered.

Where this leads: the values of $C_{N}$ and $C_{L}$ vs $α$ come from Thin Airfoil Theory; their growth with Mach from Prandtl-Glauert Compressibility Correction and Supersonic Linearized (Ackeret) Theory; the drag half connects to Induced Drag and Wingtip Vortices; and the moment's sign to Static Longitudinal Stability and Aerodynamic Center vs Center of Pressure.