Exercises — Wave drag — transonic and supersonic

WHAT we're asked: identify the drag type and its microscopic cause.

• The drag born from the shock is wave drag.
• The quantity that rises across any shock is entropy ($s_2>s_1$). Rising entropy means falling stagnation pressure ($p_{0,2}<p_{0,1}$), which shows up as a rearward force. See Entropy and stagnation pressure loss. Answer: wave drag, caused by an entropy increase.

WHY this formula: the Mach cone is pure geometry — a weak pulse spreads to radius $at$ while the body moves $ut$, and $\sin \mu =at/ut=1/M$. $\sin \mu =\frac{1}{M}=\frac{1}{2}=0.5\Rightarrow \mu =\arcsin (0.5)=30^{\circ }.$ Answer: $\mu =30^{\circ }$.

False. The Mach angle $\mu$ is the envelope of infinitesimally weak signals. A real shock turns the flow by a finite angle, which needs finite compression, so it always leans steeper: $\beta >\mu$. Equality only happens in the limit of zero turning ($\theta \to 0$). See Normal and oblique shock waves.

$\mu (1.4)=\arcsin \left(\frac{1}{1.4}\right)=\arcsin (0.714)=45.6^{\circ },$ $\mu (4)=\arcsin \left(\frac{1}{4}\right)=\arcsin (0.25)=14.5^{\circ }.$ WHAT IT LOOKS LIKE (see figure): faster flight ⇒ the cone wraps tighter to the body ($\mu$ shrinks). As $M\to 1^{+}$, $\mu \to 90^{\circ }$ (cone opens flat, almost a wall). As $M\to \infty$, $\mu \to 0$.

WHY these formulas: for a thin section in supersonic flow, linearized (Ackeret) theory gives $C_p=2\theta /\sqrt{M^2-1}$; integrating a flat plate's surfaces gives the boxed results. Here $\beta \equiv \sqrt{M^2-1}$ is the compressibility factor (NOT a shock angle). $\beta =\sqrt{M^2-1}=\sqrt{4-1}=\sqrt{3}=\mathrm{1.732.}$ $c_l=\frac{4\alpha }{\beta }=\frac{4(0.0524)}{1.732}=\mathrm{0.121.}$ $c_{d,\text{wave}}=\frac{4{\alpha }^2}{\beta }=\frac{4(0.0524{)}^2}{1.732}=\mathrm{0.00634.}$ Answer: $c_l=0.121$, $c_{d,\text{wave}}=0.00634$.

$\frac{L}{D_{\text{wave}}}=\frac{c_l}{c_{d,\text{wave}}}=\frac{4\alpha /\beta }{4{\alpha }^2/\beta }=\frac{1}{\alpha }=\frac{1}{0.0524}=\mathrm{19.1.}$ WHY so clean: $\beta$ cancels and one power of $\alpha$ cancels, leaving $L/D_{\text{wave}}=\cot \alpha \approx 1/\alpha$. Small $\alpha$ is efficient — but note the lift itself is small too, so you can't just make $\alpha$ tiny for free. Answer: $L/D_{\text{wave}}\approx 19.1$.

Mach angle: $\mu =\arcsin (1/2)=30^{\circ }$. Real shock: $\beta \approx 39.3^{\circ }$ from the θ–β–M relation. WHY $\beta >\mu$: the Mach cone is the zero-strength limit. Turning the flow by a finite $10^{\circ }$ needs finite compression, so the wave must lean more steeply into the flow. The gap $\beta -\mu =9.3^{\circ }$ is a direct measure of shock strength. WHAT IT LOOKS LIKE (see figure): two lines from the wedge tip — the shallow dashed Mach cone ($30^{\circ }$) and the steeper solid shock ($39.3^{\circ }$).

WHY this formula: the flow across a shock is adiabatic but irreversible, so Gibbs gives $p_{0,2}/p_{0,1}=e^{-\Delta s/R}$. $\frac{p_{0,2}}{p_{0,1}}=e^{-0.06}=\mathrm{0.942.}$ So stagnation pressure falls to $94.2\%$ — a $5.8\%$ loss. WHY it is drag: a momentum balance on a control volume around the body says a stagnation-pressure deficit in the wake equals a net rearward force. That force is wave drag — no viscosity needed. See Entropy and stagnation pressure loss. Answer: ratio $\approx 0.942$ (a $5.8\%$ stagnation-pressure loss).

WHY: only the velocity component normal to the leading edge drives the compressibility on the section: $M_n=M_{\infty }\cos \Lambda$. Solve for $M_{\infty }$. $M_{\infty }=\frac{M_n}{\cos \Lambda }=\frac{0.7}{\cos 30^{\circ }}=\frac{0.7}{0.8660}=\mathrm{0.808.}$ Meaning: sweep lets you fly faster ($0.808$ vs $0.7$) before the section "feels" sonic flow — this is why fast wings are swept. See Critical and drag-divergence Mach number. Answer: $M_{\infty }\approx 0.808$.

WHY: at zero lift and zero camber only the thickness term survives: $c_{d,\text{wave}}=\frac{4}{\sqrt{M^2-1}}\overline{{\left(\frac{d{y}_t}{dx}\right)}^2}.$ Since the slope magnitude is $t$ everywhere, its mean square is $\overline{(d{y}_t/dx{)}^2}=t^2=0.05^2=0.0025$. $c_{d,\text{wave}}=\frac{4}{\sqrt{3}}(0.0025)=\frac{4(0.0025)}{1.732}=\mathrm{0.00577.}$ Insight: wave drag exists even with no lift — it's the price of having volume in supersonic flow. Halving thickness quarters this drag (it's a square). Answer: $c_{d,\text{wave}}\approx 0.00577$.

$c_{d,\text{wave}}=\frac{4}{\sqrt{M^2-1}}({\alpha }^2+\overline{(d{y}_c/dx{)}^2}+\overline{(d{y}_t/dx{)}^2}),\sqrt{M^2-1}=\mathrm{1.732.}$

• Lift term: $\frac{4}{1.732}{\alpha }^2=2.309(0.0349^2)=2.309(0.001218)=\mathrm{0.00281.}$
• Camber term: $2.309(0.0009)=\mathrm{0.00208.}$
• Thickness term: $2.309(0.0025)=\mathrm{0.00577.}$
• Total: $c_{d,\text{wave}}=0.00281+0.00208+0.00577=\mathrm{0.01066.}$ Reading it: thickness dominates here — that's the volume drag the Area Rule targets. Every term is a square, so wave drag can never be negative. Answer: total $\approx 0.01066$.

$c_d(M)=\frac{4t^2}{\sqrt{M^2-1}},t^2=\mathrm{0.0025.}$ $c_d(2)=\frac{0.01}{\sqrt{3}}=0.00577,c_d(3)=\frac{0.01}{\sqrt{8}}=\frac{0.01}{2.828}=\mathrm{0.00354.}$ $\frac{c_d(3)}{c_d(2)}=\frac{\sqrt{3}}{\sqrt{8}}=\mathrm{0.612.}$ Meaning: the coefficient drops as $1/\sqrt{M^2-1}$ because oblique shocks become weaker per unit length as the flow leans more supersonic. (This is coefficient only — actual force still depends on dynamic pressure, which grows with speed.) Answer: $0.00354$ at $M=3$; ratio $\approx 0.612$.

Step 1 (WHY normal Mach): criticality is governed by the section's normal Mach. For the swept wing, the free-stream Mach at which the section sees $M_n=0.72$ is $M_{\infty }^{crit}=\frac{M_n}{\cos \Lambda }=\frac{0.72}{\cos 35^{\circ }}=\frac{0.72}{0.8192}=\mathrm{0.879.}$ Step 2 (compare to cruise):

• Wing A goes critical at $M_{\infty }=0.72$ — well below the $0.82$ cruise ⇒ it is already past $M_{cr}$, deep into drag rise. Bad.
• Wing B goes critical at $M_{\infty }=0.879>0.82$ ⇒ at cruise it hasn't reached sonic flow yet. Good — margin of $0.879-0.82=0.059$. Decision: choose the swept wing B; sweep pushes $M_{cr}$ above the cruise Mach, delaying wave drag. See Critical and drag-divergence Mach number and Prandtl-Glauert compressibility correction. Answer: B (critical at $M_{\infty }\approx 0.879$, comfortably above the $0.82$ cruise).

WHY ${\theta }_{max}$ matters: the $\theta$–$\beta$–$M$ relation has a solution only for $\theta \le {\theta }_{max}$. The nose demands a turn of $25^{\circ }$. Since $25^{\circ }>22.97^{\circ }$, no attached shock can turn the flow that sharply ⇒ the shock detaches into a curved bow shock standing ahead of the nose. Drag consequence: near the centreline the bow shock is essentially a normal shock — the strongest possible ⇒ largest entropy jump ⇒ largest stagnation-pressure loss ⇒ high wave drag. This is why supersonic noses are kept slender ($\theta <{\theta }_{max}$). See Theta-beta-M relation and detached bow shocks. Answer: shock detaches (bow shock); wave drag is high because the near-centreline shock is normal.

Step 1 (WHY $\mu$): the boom front is the Mach cone, whose half-angle obeys $\sin \mu =1/M$. $\mu =\arcsin \left(\frac{1}{1.6}\right)=\arcsin (0.625)=38.68^{\circ }.$ Step 2 (geometry, see figure): the cone surface makes angle $\mu$ with the flight path. The ground is a vertical drop $h$ below the aircraft. The horizontal lag from the aircraft to where the cone meets the ground is $x=\frac{h}{\tan \mu }=\frac{12000}{\tan 38.68^{\circ }}=\frac{12000}{0.8005}=14,991\text{m}\approx 15.0\text{km}.$ Meaning: you hear the boom only after the jet has already passed roughly $15$ km ahead of you along its track. See Sonic boom and the Mach cone. Answer: $x\approx 15.0$ km behind the aircraft's ground position.

The three faces of "$\beta$" in this chapter ::: shock angle (oblique shock tilt), the letter overloaded as $\sqrt{M^2-1}$ in Ackeret, and $\Lambda$ is sweep — keep them separate. Why wave drag is always $\ge 0$ ::: every Ackeret term is a square (${\alpha }^2$, camber${}^2$, thickness${}^2$). Why sweep raises $M_{cr}$ ::: only $M_n=M_{\infty }\cos \Lambda$ drives criticality, so the free-stream Mach can be higher. Why blunt noses have high wave drag ::: their shock detaches into a near-normal bow shock — maximum entropy jump, maximum stagnation-pressure loss.