Wave drag — transonic and supersonic: Worked examples

Everything we use is restated in plain words below before it appears in a calculation, so this page stands on its own.

Formula The

$θ$ – $β$ – $M$ relation (stated here, used in Ex 4 & 5) For a supersonic flow at Mach $M_{1}$ meeting a surface that turns it by angle $θ$ , an attached oblique shock forms at tilt $β_{shock}$ . Conservation of mass, momentum and energy across the shock give $tan θ = 2 cot β_{shock} \frac{M _{1}^{2} s i n ^{2} β _{shock} - 1}{M _{1}^{2} ( γ + c o s 2 β _{shock} ) + 2} .$ How to read it: you know the turn $θ$ and the incoming $M_{1}$ ; you solve for the shock tilt $β_{shock}$ . Two things to remember:

As $θ \to 0$ (vanishingly weak turn) the weak-shock root $β_{shock} \to μ = arcsin (1/ M_{1})$ — the Mach angle. That is the only case where a real shock equals the Mach cone.
For each $M_{1}$ there is a maximum $θ_{ma x}$ . Ask for more turning than that and no attached shock exists — it detaches (Ex 5). We will never re-derive this on the page; we just use it (numerical roots come from the standard tables/solver for air, $γ = 1.4$ ).

Definition Two more symbols we will need

Mach angle $μ$ : the half-angle of the Mach cone, the cone swept out by the weakest possible pressure signals when a body moves faster than sound. From the parent's pure geometry (pulse radius $a t$ vs distance travelled $u t$ ), $sin μ = 1/ M$ , so $μ = arcsin (1/ M)$ . It is the tilt of a zero-strength wave — always the shallowest line in the flow.
Ratio of specific heats $γ$ : a property of the gas — the ratio of its heat capacity at constant pressure to that at constant volume. For ordinary air (a diatomic gas: mostly $N_{2}$ and $O_{2}$ ) it is $γ = 1.4$ . Every shock relation carries $γ$ because compressing a gas heats it, and $γ$ measures how strongly. We use $γ = 1.4$ throughout because the working fluid is air.
Mean-square slope $\overline{(d y / d x)^{2}}$ : the overline means "average along the chord." You take the surface slope $d y / d x$ at every point from leading to trailing edge, square it, then average all those squares over the chord length. It is a single number summarising how steep, on average, the surface is — and squaring first means both up-slopes and down-slopes add drag, never cancel.

The scenario matrix

Every wave-drag problem lands in one of these cells. The worked examples below are labelled with the cell(s) they cover, so together they fill the whole grid.

Cell	Case class	What decides the physics	Example(s)
A	$M_{\infty} < 1$ , subsonic everywhere	no shock ⇒ no wave drag	Ex 1
B	$M_{\infty} < 1$ but local pocket hits $M = 1$	onset — defines $M_{cr}$	Ex 2
C	$M_{\infty} = 1$ exactly (degenerate)	$β_{P G} = 0$ ⇒ formulas blow up	Ex 3
D	$M_{\infty} > 1$ , slender body, attached shock	$θ < θ_{ma x}$ , real $β_{shock} > μ$	Ex 4
E	$M_{\infty} > 1$ , blunt/over-turned, detached bow shock	$θ > θ_{ma x}$	Ex 5
F	Supersonic thin airfoil, lifting	Ackeret $c_{l}, c_{d}$ with $α \neq = 0$	Ex 6
G	Supersonic thin airfoil, zero lift	thickness/camber term only	Ex 7
H	Sign / limiting behaviour	$c_{d} \geq 0$ always; $M \to \infty$ trend	Ex 7, Ex 8
I	Real-world word problem	sweep, $M_{n} = M_{\infty} cos Λ$	Ex 8
J	Exam twist — the two $β$ 's	catch the symbol clash	Ex 9

Example 1 — Cell A: subsonic, no wave drag

Worked example Statement

A wing flies at $M_{\infty} = 0.5$ . Its fastest local point reaches $M_{l oc a l} = 0.82$ . How much wave drag does it make?

Forecast: Is any part of the flow supersonic? If not, what does that force wave drag to be?

Find the largest local Mach number anywhere. It is given: $M_{l oc a l} = 0.82$ . Why this step? Wave drag exists only if the flow reaches $M \geq 1$ somewhere. So the whole question reduces to "does any point hit sonic?"
Compare to 1. $0.82 < 1$ , so the flow is subsonic everywhere. No supersonic pocket ⇒ no shock can form. Why this step? No shock means no entropy jump, no stagnation-pressure loss, no wave drag — by the definition in the parent.
Conclude: $c_{d, wave} = 0$ .

Verify: The largest Mach number anywhere is $M_{l oc a l} = 0.82 < 1$ . So even the fastest point never reaches sonic, the supersonic factor $M_{l oc a l}^{2} - 1$ would be imaginary, and the supersonic machinery simply does not apply. Answer $0$ is consistent.

Example 2 — Cell B: onset, finding $M_{cr}$ conceptually

Worked example Statement

By what rule do we locate the critical Mach number $M_{cr}$ , and why does a thicker airfoil have a lower $M_{cr}$ ?

Forecast: At $M_{cr}$ , what is the local Mach number at the fastest point — guess the exact value.

State the defining condition: set $M_{l oc a l} = 1$ at the airfoil's minimum-pressure (maximum-velocity) point. Why this step? $M_{cr}$ is defined as the free-stream $M_{\infty}$ at which sonic flow first appears anywhere. The first place to go sonic is the fastest point.
Link local to free-stream: the local speed-up factor $k = M_{l oc a l} / M_{\infty} > 1$ depends on geometry. Thicker section ⇒ air must accelerate more to get over the bulge ⇒ larger $k$ . Why this step? A bigger $k$ means the local flow reaches $M = 1$ at a smaller $M_{\infty}$ .
Conclude: larger $k$ ⇒ $M_{cr} = 1/ k$ (schematically) is smaller. Thick wings hit trouble earlier.

Verify: Consistent with the parent's design rule — high-speed wings are thin and swept precisely to keep $k$ small and $M_{cr}$ high. The logic direction (thicker → lower $M_{cr}$ ) matches.

Example 3 — Cell C: the degenerate case $M_{\infty} = 1$

Worked example Statement

Try to use the Ackeret lift formula $c_{l} = \frac{4 α}{M _{\infty}^{2} - 1}$ at exactly $M_{\infty} = 1$ . What happens, and what does it mean physically?

Forecast: Look at the denominator. What does it become at $M = 1$ ?

Substitute $M_{\infty} = 1$ : $1^{2} - 1 = 0 = 0$ . Why this step? The whole formula is built on the Prandtl–Glauert factor $β_{P G} = M^{2} - 1$ ; testing its edge tells us where the theory dies.
Divide by it: $c_{l} = 4 α /0 \to \infty$ . Why this step? The prediction is infinite — a clear signal that linearized theory is invalid at $M = 1$ .
Interpret physically: at $M = 1$ the Mach angle $μ = arcsin (1/1) = 90°$ ; disturbances pile up perpendicular to the flow, the small-perturbation assumption collapses, and the true drag is finite but the formula is not. Why this step? The math singularity mirrors the real transonic drag spike — the regime where you must use full nonlinear (or experimental) methods, not Ackeret.

Verify: $μ = 90°$ at $M = 1$ and $μ \to 0$ as $M \to \infty$ (from $sin μ = 1/ M$ ), so the singularity sits exactly where the physics is hardest — internally consistent.

Example 4 — Cell D: attached oblique shock, real $β > μ$

The figure below shows an $M_{1} = 2$ flow hitting a $10°$ half-angle wedge (grey), drawn head-on. Three lines leave the wedge tip: the black dashed Mach line at $μ = 30°$ (weakest signal), and the red real shock at $β_{shock} = 39.3°$ (steeper). The incoming flow arrow enters from the left. The point of the figure: the red shock always leans more than the dashed Mach line.

Figure — Wave drag — transonic and supersonic

Worked example Statement

A $10°$ half-angle wedge flies at $M_{1} = 2.0$ . Compare the Mach angle $μ$ (envelope of weak signals) with the real attached shock angle $β_{shock}$ .

Forecast: Which is bigger — the weak-signal cone, or the actual shock that has to turn the flow by $10°$ ?

Mach angle: $sin μ = 1/ M_{1} = 1/2 \Rightarrow μ = arcsin (0.5) = 30°$ . Why this step? $μ$ is the pure geometry of "pulse radius $a t$ vs distance $u t$ " — the weakest possible disturbance. It is our lower bound.
Real shock angle: solve the $\theta$–$\beta$–$M$ relation (stated in the box above) with $θ = 10°$ , $M_{1} = 2$ , $γ = 1.4$ (air). The weak-shock root is $β_{shock} \approx 39.3°$ . Why this step? A finite $10°$ turn needs finite compression, and finite compression means the shock leans more than the zero-strength Mach cone.
Compare: $39.3° > 30°$ . The real shock is steeper — in the figure the red shock line sits above the black dashed Mach line. Why this step? This is the parent's key caution made concrete — never set a real shock angle equal to $μ$ .

Verify: As $θ \to 0$ , the $θ$ – $β$ – $M$ weak root must relax back to $β = μ = 30°$ . Plugging $β = 30°$ , $M_{1} = 2$ into the numerator $M_{1}^{2} sin^{2} β - 1 = 4 (0.25) - 1 = 0$ , so $θ = 0$ . ✓ Consistent — the Mach angle is exactly the zero-turn shock.

Example 5 — Cell E: over-turning ⇒ detached bow shock

The figure below shows the same $M_{1} = 2$ flow hitting a much blunter wedge (grey, half-angle $40°$ ). Because the wedge demands more turning than an attached shock can give, the shock (red curve) cannot touch the tip — it bows out and stands ahead of the body, with a visible stand-off gap marked between the red curve and the wedge tip. The incoming flow arrow enters from the left.

Worked example Statement

A wedge with a very large half-angle $θ = 40°$ is placed in an $M_{1} = 2.0$ flow. Can the shock stay attached?

Forecast: Is $40°$ a gentle turn or a violent one for Mach 2? Guess whether the shock hugs the tip or stands off in front.

Find the maximum turning angle for $M_{1} = 2$ . From the $θ$ – $β$ – $M$ relation (with $γ = 1.4$ for air), $θ_{ma x} \approx 22.97°$ at $M_{1} = 2$ . Why this step? For every Mach number there's a hard ceiling on how much an attached shock can turn the flow. Beyond it, no attached solution exists.
Compare: requested turn $40° > θ_{ma x} \approx 23°$ . Why this step? The body is asking for more turning than any attached oblique shock can provide.
Conclude: the shock detaches into a curved bow shock standing ahead of the wedge (the red curve in the figure). Near the centreline it is locally a normal shock — the strongest, most entropy-generating kind. Why this step? Stronger shock ⇒ larger $Δ s$ ⇒ larger stagnation-pressure loss ⇒ maximum wave drag. This is exactly why blunt/over-angled bodies are draggy.

Verify: $θ_{ma x} (M = 2) \approx 23°$ and our turn is $40°$ ; since $40 > 23$ the "no attached solution" verdict is forced. A normal shock at $M_{1} = 2$ gives a stagnation-pressure ratio $p_{0, 2} / p_{0, 1} \approx 0.721$ — a big loss, confirming high drag. ✓

Example 6 — Cell F: lifting supersonic flat plate

Worked example Statement

A flat plate at $M_{\infty} = 2.0$ , angle of attack $α = 3° = 0.0524$ rad. Find $c_{l}$ , $c_{d, wave}$ , and $L / D_{wave}$ .

Forecast: Will $L / D$ be around 2, 20, or 200? (Hint: it's roughly $1/ α$ .)

Compute the Prandtl–Glauert factor: $β_{P G} = M_{\infty}^{2} - 1 = 4 - 1 = 3 = 1.732$ . Why this step? Every Ackeret result is divided by $β_{P G}$ ; it's the compressibility strength for the supersonic (not shock-angle) sense.
Lift coefficient: $c_{l} = \frac{4 α}{β _{P G}} = \frac{4 ( 0.0524 )}{1.732} = 0.1210$ . Why this step? Ackeret's linear theory says lift grows linearly with tilt $α$ — the small- $θ$ limit of the oblique-wave pressure.
Wave drag (flat plate, no thickness/camber): $c_{d, wave} = \frac{4 α ^{2}}{β _{P G}} = \frac{4 ( 0.0524 ) ^{2}}{1.732} = 0.006340$ . Why this step? For a flat plate only the lift-induced wave-drag term $α^{2}$ survives (no thickness, no camber).
Efficiency: $L / D_{wave} = c_{l} / c_{d, wave} = \frac{4 α}{4 α ^{2}} = \frac{1}{α} = \frac{1}{0.0524} = 19.1$ . Why this step? The $β_{P G}$ and the $4$ cancel, leaving $L / D_{wave} = 1/ α$ in this linearized theory. The exact geometric result for a flat plate is $L / D = cot α$ ; because $cot α = cos α / sin α \approx 1/ α$ only when $α$ is small (so $sin α \approx α$ , $cos α \approx 1$ ), the $1/ α$ form is an approximation valid for small $α$ . At $α = 3°$ they agree to better than $0.2%$ .

Verify: $c_{l} / c_{d} = 0.1210/0.006340 = 19.1$ and $1/0.0524 = 19.1$ . The exact $cot (3°) = 19.08$ — within $0.1%$ of $1/ α = 19.1$ , confirming the small-angle approximation. ✓ Units: all coefficients dimensionless, ratio dimensionless. Small $α$ is efficient — but the absolute lift $0.1210$ is then small too, the eternal supersonic trade-off.

Example 7 — Cells G & H: zero-lift thickness drag, and the "never negative" fact

Worked example Statement

A symmetric diamond (double-wedge) airfoil flies at $M_{\infty} = 2$ at zero lift ( $α = 0$ , no camber). Its thickness slope has mean-square value $\overline{(d y_{t} / d x)^{2}} = 0.01$ . Find $c_{d, wave}$ and confirm it cannot be negative.

Forecast: With no lift and no camber, is the wave drag zero? Or does just having volume still cost drag?

Drop the lift and camber terms. $α = 0$ and $\overline{(d y_{c} / d x)^{2}} = 0$ , so only the thickness term remains: $c_{d, wave} = \frac{4}{β _{P G}} \overline{(d y_{t} / d x)^{2}} .$ Why this step? The parent's Ackeret drag has three additive squares; zeroing lift and camber isolates the volume (thickness) contribution.
Insert numbers: $β_{P G} = 3 = 1.732$ , so $c_{d, wave} = \frac{4 ( 0.01 )}{1.732} = 0.0231$ . Why this step? Shows wave drag is nonzero even at zero lift — pure volume makes shocks.
Sign argument (Cell H). Every term is a square ( $α^{2}, \overline{(\cdot)^{2}}$ ) and $β_{P G} > 0$ , so $c_{d, wave} \geq 0$ always — it can never be negative, and is minimized (→0) only by an infinitely thin, uncambered plate at zero lift. Why this step? Wave drag is a loss (entropy rises across every shock); a negative drag would mean the shock returns energy to the body, which the second law forbids.

Verify: $4 (0.01) /1.732 = 0.0231$ . ✓ Squares guarantee non-negativity: even flipping the sign of any slope leaves the square positive. Consistent with entropy-only-rises.

Example 8 — Cells H & I: the $M \to \infty$ trend, and sweep to escape drag divergence

Worked example Statement

Part (a) — limiting behaviour (Cell H): Take the flat-plate wave drag $c_{d, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1}$ at fixed $α$ . What happens to it as $M_{\infty} \to \infty$ , and as $M_{\infty} \to 1^{+}$ ? Part (b) — word problem (Cell I): An airliner cruises at $M_{\infty} = 0.85$ . Its unswept wing section would have $M_{cr} = 0.72$ (drag already diverging). The designer sweeps the wing by $Λ = 35°$ . What normal Mach number does the wing section actually "feel", and does that fix the drag rise?

Forecast: (a) Does going faster supersonically make wave drag rise or fall? (b) Will $M_{n} = M_{\infty} cos Λ$ drop below $0.72$ ?

(a) High-Mach limit. As $M_{\infty} \to \infty$ , $M_{\infty}^{2} - 1 \to \infty$ , so $c_{d, wave} = \frac{4 α ^{2}}{M _{\infty}^{2} - 1} \to 0$ . Why this step? The lone $M$ -dependence sits in the denominator; a growing denominator drives the coefficient down. Physically, oblique shocks lean tighter and get weaker per unit length, so the wave-drag coefficient falls with increasing supersonic speed.
(a) Low-Mach (transonic) limit. As $M_{\infty} \to 1^{+}$ , $M_{\infty}^{2} - 1 \to 0^{+}$ , so $c_{d, wave} \to + \infty$ . Why this step? This is the same singularity as Ex 3 — linearized theory over-predicts near $M = 1$ , mirroring the real transonic drag spike. Between the two limits, $c_{d, wave}$ falls monotonically for $M_{\infty} > 1$ . Sample numbers at $α = 0.0524$ : $M = 1.4 \Rightarrow \frac{4 α ^{2}}{0.96} = 0.01121$ ; $M = 2 \Rightarrow 0.006340$ ; $M = 5 \Rightarrow \frac{4 α ^{2}}{24} = 0.002242$ — steadily shrinking. ✓ trend.
(b) Compute the normal Mach number: $M_{n} = M_{\infty} cos Λ = 0.85 \times cos 35° = 0.85 \times 0.8192 = 0.696$ . Why this step? Only the velocity component perpendicular to the wing's leading edge drives the pressure field over the section, so that component sets the local sonic onset — not the full flight speed.
(b) Compare to the section $M_{cr} = 0.72$ : $0.696 < 0.72$ . Why this step? If the felt Mach number stays below critical, no supersonic pocket forms on the section ⇒ wave drag is suppressed even though the plane flies at $0.85$ .
(b) Limiting check. More sweep ⇒ smaller $cos Λ$ ⇒ smaller $M_{n}$ ⇒ higher effective cruise Mach before divergence. As $Λ \to 90°$ , $M_{n} \to 0$ (but the wing becomes useless for lift — the practical trade-off).

Verify: (a) $M = 5$ : $4 (0.0524)^{2} / 24 = 0.002242 < 0.006340$ (the $M = 2$ value) $< 0.01121$ (the $M = 1.4$ value) — monotone decrease confirmed. (b) $0.85 cos 35° = 0.696 < 0.72$ ✓ — the sweep buys back margin so the section no longer goes supersonic at cruise. All Mach numbers dimensionless.

Example 9 — Cell J: the exam twist — two $β$ 's in one problem

Worked example Statement

At $M_{\infty} = 2.0$ : a student writes "the shock angle is $β = M^{2} - 1 = 1.732$ ." What's wrong, and what are the two correct quantities that this one symbol is being confused between?

Forecast: Can a physical angle equal the bare number $1.732$ ? (Think about the units of an angle versus a pure number.)

Spot the category error. $M^{2} - 1 = 4 - 1 = 1.732$ is a dimensionless number, the Prandtl–Glauert factor $β_{P G}$ — it is not an angle. Read as an angle it would be either $1.732$ rad $= 99.2°$ (absurd for a Mach-2 shock) or a naked number with no degrees at all. Why this step? The student collided the two meanings of $β$ from the opening definition box — the exact trap this cell exists to catch.
Give the real shock angle. For a concrete body — say a $10°$ wedge — the physical shock tilt from the $θ$ – $β$ – $M$ relation is $β_{shock} \approx 39.3°$ (Example 4). Why this step? The real shock angle comes from the shock relations and depends on the turning angle $θ$ ; it is never $M^{2} - 1$ .
Give the Mach angle for contrast. $μ = arcsin (1/ M) = arcsin (0.5) = 30°$ . Why this step? This nails down that at $M = 2$ there are three distinct numbers: $β_{P G} = 1.732$ (a pure number in the Ackeret denominator), $μ = 30°$ (weak-signal cone angle), and $β_{shock} = 39.3°$ (real shock tilt for a $10°$ wedge). Keep them in separate mental boxes.

Verify: $4 - 1 = 1.732$ (a number); $arcsin (0.5) = 30°$ ; the $10°$ wedge shock $\approx 39.3°$ . Three different values with three different meanings — and $1.732 rad = 99.2° \neq = 39.3°$ , proving $β_{P G}$ is not the shock angle. ✓

Recall Self-test

Define the Mach number in one line. ::: $M = v / a$ , the ratio of flow speed to the local speed of sound (a pure number). No shock anywhere means wave drag equals what? ::: Exactly zero (Cell A). At exactly $M = 1$ , why does $c_{l} = 4 α / M^{2} - 1$ fail? ::: Denominator $\to 0$ , prediction $\to \infty$ ; linearized theory is invalid at $M = 1$ (Cell C). For a $10°$ wedge at $M = 2$ , order $μ$ , $β_{shock}$ , and does $β_{P G}$ belong? ::: $μ = 30° < β_{shock} = 39.3°$ ; $β_{P G} = 1.732$ is a pure number, not an angle (Cells D, J). When does an oblique shock detach into a bow shock? ::: When the required turn $θ$ exceeds $θ_{ma x}$ (≈ $23°$ at $M = 2$ ) — Cell E. Why can $c_{d, wave}$ never be negative? ::: Every term is a square and $M^{2} - 1 > 0$ ; a negative wave drag would violate entropy-rise (Cell H). As $M_{\infty} \to \infty$ at fixed $α$ , what does $c_{d, wave}$ do? ::: It falls to zero like $1/ M^{2} - 1$ (Cell H). What Mach number does a swept wing section "feel"? ::: The normal component $M_{n} = M_{\infty} cos Λ$ (Cell I).