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

WHAT we do: plug numbers into the definition — no physics beyond arithmetic. $q_{\infty }=\frac{1}{2}(1.225)(50{)}^2=\frac{1}{2}(1.225)(2500)=1531.25\text{Pa}.$ WHY: $q_{\infty }$ is the "pressure scale" of the flow; every force coefficient divides by it so that size and speed drop out.

(a) $M_{\infty }=0.5$ is subsonic → Prandtl–Glauert: divide the incompressible $C_L$ by $\sqrt{1-M_{\infty }^2}=\sqrt{1-0.25}=\sqrt{0.75}\approx 0.866$. (b) $M_{\infty }=2.0$ is supersonic → Ackeret: divide by $\sqrt{M_{\infty }^2-1}=\sqrt{4-1}=\sqrt{3}\approx 1.732$. WHY the flip: the same square root appears, but the sign under it flips as you cross $M=1$. Below $M=1$ the factor amplifies lift ($<1$, so dividing makes it bigger); above $M=1$ the factor $>1$, so lift shrinks with increasing Mach.

$q_{\infty }S$ has units $(\text{Pa})({\text{m}}^2)=({\text{N/m}}^2)({\text{m}}^2)=\text{N}$ — exactly a force. A force $L$ divided by a force is dimensionless → $C_L$ works. A moment $M$ has units $\text{N}\cdot \text{m}$. Dividing by $q_{\infty }S$ (units N) leaves metres, not a pure number. We need one more length, $c$, so $q_{\infty }Sc$ has units $\text{N}\cdot \text{m}$, matching the moment. WHY it matters: forgetting the $c$ gives a "coefficient" that still has units and changes with model scale — defeating the whole purpose.

WHAT: rotate the force components from body axes to wind axes (see figure — same vector, two rulers rotated by $\alpha$).

Convert to radians first (the slope $2\pi$ is per radian): $\alpha =6^{\circ }\times \frac{\pi }{180}=0.10472\text{rad}$. $C_L=2\pi (\alpha -{\alpha }_{L=0})=2\pi (0.10472)=0.658.$ WHY radians: the $2\pi$ slope comes from an integral over the chord in Thin Airfoil Theory that is naturally in radians. Plugging degrees would inflate the answer ~57×.

$\beta =\sqrt{1-M_{\infty }^2}=\sqrt{1-0.49}=\sqrt{0.51}=0.7141$. $C_L=\frac{C_{L,\text{incomp}}}{\beta }=\frac{0.658}{0.7141}=0.921.$ WHY dividing amplifies: as the flow speeds toward $M=1$, density variations let the same shape deflect more air, so lift per degree rises. See Prandtl-Glauert Compressibility Correction.

$\alpha =3^{\circ }=0.05236\text{rad}$, $\sqrt{M_{\infty }^2-1}=\sqrt{6.25-1}=\sqrt{5.25}=2.2913$. $C_L=\frac{4\alpha }{\sqrt{M_{\infty }^2-1}}=\frac{4(0.05236)}{2.2913}=\frac{0.20944}{2.2913}=0.0914.$ $C_{D,\text{wave}}=\frac{4{\alpha }^2}{\sqrt{M_{\infty }^2-1}}=\frac{4(0.05236{)}^2}{2.2913}=\frac{0.010966}{2.2913}=0.00479.$ WHY wave drag exists: supersonically, the plate makes shock/expansion waves that carry energy away — a drag with no subsonic counterpart, and it scales like ${\alpha }^2$. See Supersonic Linearized (Ackeret) Theory.

Induced-drag term: $\frac{C_L^2}{\pi eAR}=\frac{0.9^2}{\pi (0.85)(8)}=\frac{0.81}{21.362}=0.03792$. $C_D=C_{D0}+\frac{C_L^2}{\pi eAR}=0.020+0.03792=0.0579.$ Induced fraction $=\frac{0.03792}{0.0579}=0.655$ i.e. about $65.5\%$. WHY it's parabolic: since $C_L\propto \alpha$, the induced term $\propto C_L^2\propto {\alpha }^2$ — the drag-vs-$\alpha$ curve is a parabola opening upward. See Induced Drag and Wingtip Vortices.

Use the inverse rotation (rotate the other way by $\alpha$): $\cos 12^{\circ }=0.9781$, $\sin 12^{\circ }=0.2079$. $C_N=C_L\cos \alpha +C_D\sin \alpha =0.75(0.9781)+0.045(0.2079)=0.73359+0.00936=0.743.$ $C_A=-C_L\sin \alpha +C_D\cos \alpha =-0.75(0.2079)+0.045(0.9781)=-0.15593+0.04401=-0.112.$ WHAT the negative $C_A$ means: a negative axial coefficient points forward along the chord — here the strong forward-tilted suction of the lifting force overwhelms the small backward friction drag. That is physically normal for a lifting airfoil at moderate $\alpha$ (leading-edge suction).

$C_N\sin \alpha =0.743(0.2079)=0.15447$. $C_A\cos \alpha =-0.112(0.9781)=-0.10955$. Sum $=0.15447-0.10955=0.0449\approx 0.045✓$ (matches given $C_D$). Interpretation: the normal-force tilt contributes a positive streamwise push (like induced drag), while the forward-pointing axial term subtracts (leading-edge suction recovers thrust). Net drag is the small residual — this is exactly why inviscid lifting airfoils can have very low drag.

(a) Effective angle $=\alpha -{\alpha }_{L=0}=5^{\circ }-(-2^{\circ })=7^{\circ }=0.12217\text{rad}$. $C_{L,\text{incomp}}=2\pi (0.12217)=0.7676$. (b) $\beta =\sqrt{1-0.36}=\sqrt{0.64}=0.8$. $C_L=0.7676/0.8=0.9595$. (c) With $C_N\approx C_L=0.9595$, $C_A=0.010$, $\alpha =5^{\circ }$ ($\sin =0.08716$, $\cos =0.99619$): $C_D=C_N\sin \alpha +C_A\cos \alpha =0.9595(0.08716)+0.010(0.99619)=0.08363+0.00996=0.0936.$ WHY the approximation is fair: at $5^{\circ }$, $\cos \alpha \approx 1$ so $C_L\approx C_N$ to under $0.4\%$ error — the small-angle sanity check from the parent note.

(a) $\beta =\sqrt{1-0.9025}=\sqrt{0.0975}=0.31225$. $C_L=0.5/0.31225=1.601$. (b) The formula amplifies lift by $1/\beta \approx 3.2\times$ — a huge jump. But near $M_{\infty }=1$, local flow over the airfoil is already supersonic, shocks form, and the linearized subsonic potential-flow assumption is broken. As $M_{\infty }\to 1$, $\beta \to 0$ and the formula predicts infinite lift, which is physically impossible. Valid only for $M_{\infty }\lesssim 0.7$. The lesson: a formula can be arithmetically fine and physically nonsense outside its domain.

(a) Stability needs $\frac{d{C}_m}{d\alpha }<0$. Here $\frac{d{C}_m}{d\alpha }=-0.9<0$ → statically stable ✔. WHY: if a gust raises $\alpha$, the moment becomes negative (nose-down), pushing $\alpha$ back down — a restoring response. See Static Longitudinal Stability. (b) Trim where $C_m=0$: $0.06-0.9\alpha =0\Rightarrow {\alpha }_{\text{trim}}=0.06/0.9=0.06\overline{6}\text{rad}=3.82^{\circ }$. WHY it matters: $C_{m0}>0$ (positive intercept) plus negative slope guarantees a positive, flyable trim angle — the aircraft naturally settles at $3.82^{\circ }$.

WHAT: the AC is the point where $C_m$ does not change with $C_L$ (equivalently with $\alpha$). Subtract the two equations to kill $C_{m,\text{ac}}$: $C_{m,\text{LE}}(0.8)-C_{m,\text{LE}}(0.4)=-\frac{x_{ac}}{c}(0.8-0.4).$ $-0.240-(-0.140)=-0.100=-\frac{x_{ac}}{c}(0.4)\Rightarrow \frac{x_{ac}}{c}=\frac{0.100}{0.4}=0.25.$ Back-substitute at $C_L=0.4$: $-0.140=C_{m,\text{ac}}-0.25(0.4)=C_{m,\text{ac}}-0.100$ $\Rightarrow C_{m,\text{ac}}=-0.040$. WHY $0.25$ is famous: thin-airfoil theory predicts the AC at the quarter-chord — this data confirms it. See Aerodynamic Center vs Center of Pressure.

WHY a derivative here: $L/D$ is a ratio that rises then falls with $C_L$; its peak is where the slope is zero. Maximize $f(C_L)=\frac{C_L}{C_{D0}+k{C}_L^2}$. Set $f^{\prime }=0$:

(a) Thin-airfoil: $\frac{d{C}_L}{d\alpha }=2\pi =6.283$ /rad. (b) Prandtl–Glauert scales the slope by $1/\beta$: $\beta =\sqrt{1-0.64}=0.6$, $\frac{d{C}_L}{d\alpha }=2\pi /0.6=10.47$ /rad. (c) Ackeret flat plate: $C_L=4\alpha /\sqrt{M^2-1}$, so $\frac{d{C}_L}{d\alpha }=\frac{4}{\sqrt{3}}=2.309$ /rad.

$q_{\infty }=\frac{1}{2}\rho V^2$ · $C_L=C_N\cos \alpha -C_A\sin \alpha$ · $C_D=C_N\sin \alpha +C_A\cos \alpha$ · $C_L=2\pi (\alpha -{\alpha }_{L=0})$ · PG: $/\sqrt{1-M^2}$ · Ackeret: $C_L=4\alpha /\sqrt{M^2-1}$, $C_{D,w}=4{\alpha }^2/\sqrt{M^2-1}$ · polar $C_D=C_{D0}+k{C}_L^2$ · AC where $d{C}_m/d{C}_L=0$ · best $L/D$ where $k{C}_L^2=C_{D0}$.