Wave drag — transonic and supersonic: Question bank

Before you attempt the bank, lock down the notation and geometry it leans on. Every symbol used below is defined here first, and the four core geometric pictures are drawn so you never have to imagine them from text alone.

Definition Notation used everywhere on this page

== $M_{\infty}$ == — the free-stream Mach number, the Mach of the undisturbed air far ahead of the whole body. This is the aircraft's flight Mach. We use $M_{\infty}$ throughout for the free stream.
== $M_{1}$ == — the local Mach number entering one specific shock. It is $M_{\infty}$ only when that shock sits in undisturbed flow (e.g. the leading edge of an isolated wedge). We write $M_{1}$ whenever we mean "the Mach just upstream of the shock we are looking at," to keep it distinct from the flight Mach.
== $C_{D}$ == (capital, subscript $D$ ) — the total drag coefficient of the whole body: total drag force $\div (\frac{1}{2} ρ V^{2} \times reference area)$ . It bundles all drag sources (friction + wave + induced).
== $c_{d, w a v e}$ == (lower-case, per-unit-span) — the wave-drag coefficient of a 2-D airfoil section, drag per unit span $\div (\frac{1}{2} ρ V^{2} \times chord)$ . Lower-case = "section" (2-D strip); capital = "whole 3-D body." $c_{d, w a v e}$ is one slice of $C_{D}$ .
== $θ$ == — a flow-turning angle: how many degrees the streamline is deflected. For a wedge it equals the wedge's half-angle.
== $θ_{ma x} (M_{1})$ == — the largest turn a shock can make while staying attached to the body at that $M_{1}$ . Ask for more turning than $θ_{ma x}$ and the shock detaches (becomes a bow shock). It is a fixed ceiling set by $M_{1}$ .
== $d y_{t} / d x$ and $d y_{c} / d x$ == — local surface slopes of the airfoil: $y_{t} (x)$ is the thickness distribution (half-thickness above the chord line) and $y_{c} (x)$ is the camber (mean-line) shape. Their slopes tell you how steeply the surface tilts into the flow at each station $x$ ; the overbar $\overline{(\cdot)}$ means "averaged along the chord."

The four pictures below are the ones this bank keeps referring to.

Figure — Wave drag — transonic and supersonic

Recall The three "angles" that all live near

$M > 1$ (see figure s01)

Mach angle $μ = arcsin (1/ M_{\infty})$ — the envelope of infinitesimally weak pressure pulses (the teal cone in s01).
Shock angle $β$ — the lean of a finite-strength oblique shock (the orange line in s01), from the $\theta$–$\beta$–$M$ relation. Always $β > μ$ .
Prandtl–Glauert factor $β_{P G} = M_{\infty}^{2} - 1$ — a number, not an angle, appearing in Ackeret theory. It equals $M_{\infty} cos μ$ : look at s01 and read $cos μ = M_{\infty}^{2} - 1 / M_{\infty}$ straight off the little Mach-angle triangle. Same letter as the shock angle by cruel tradition.

Recall What the linked pages contribute (standalone summary)

Speed of sound and Mach number — pressure signals travel at $a$ ; $M = V / a$ compares flight speed to signal speed.
Normal and oblique shock waves — a shock is a thin, abrupt, irreversible compression; entropy jumps up across it.
Theta-beta-M relation and detached bow shocks — links turn angle $θ$ , shock angle $β$ , upstream $M_{1}$ ; sets $θ_{ma x}$ and detachment.
Entropy and stagnation pressure loss — $p_{0, 2} / p_{0, 1} = e^{- Δ s / R} < 1$ : every shock loses stagnation pressure.
Critical and drag-divergence Mach number — $M_{cr}$ (first sonic point) and $M_{dd}$ (steep drag rise) delimit the transonic regime.
Whitcomb Area Rule and supercritical airfoils — smooth total cross-sectional area vs length minimizes transonic wave drag (figure s04).

True or false — justify

State true/false and give the physical reason before revealing.

Wave drag requires viscosity to exist.

False. Wave drag is born from entropy rise across a shock (an inviscid, adiabatic-but-irreversible process); it appears even in a perfectly frictionless fluid. Skin-friction drag is the separate viscous effect.

A body flying at

M_{\infty} < 1

can never experience wave drag.

False. Flow accelerating over a curved surface can locally exceed

M = 1

even when the free stream is subsonic; that local supersonic pocket ends in a shock. This is exactly the transonic drag-divergence mechanism.

Wave drag can become negative if the airfoil is shaped cleverly.

False. In Ackeret theory every wave-drag term is a square (

α^{2}

\overline{(d y_{c} / d x)^{2}}

\overline{(d y_{t} / d x)^{2}}

), so

c_{d, w a v e} \geq 0

always. Squares encode "any deflection, either sign, costs energy."

For a slender wedge at

M_{\infty} > 1

, the shock lies along the Mach cone.

False. The Mach cone is the weak-signal envelope; a real attached oblique shock has

β > μ

because finite turning demands finite compression (compare orange vs teal in s01). They coincide only in the zero-strength limit

θ \to 0

M_{\infty}

increases through the pure supersonic range, wave drag keeps rising.

False. Because

c_{d, w a v e} \propto 1/ M_{\infty}^{2} - 1

, wave drag actually falls with increasing supersonic

M_{\infty}

(shocks become more oblique and weaker per unit length). The big rise happens only in the transonic bump.

Drag-divergence Mach number equals the critical Mach number.

False.

M_{cr}

is where the flow first reaches

M = 1

somewhere (no real drag yet);

M_{dd} > M_{cr}

is where the shock has grown enough that the total drag coefficient

C_{D}

climbs steeply.

A thicker airfoil has a higher critical Mach number.

False. Thicker sections force the flow to accelerate more over the surface, so local

M = 1

is reached at a lower free-stream Mach —

M_{cr}

is lower, which is why fast wings are thin.

Sweeping a wing helps because it makes the wing longer.

False. Sweep helps because only the component of Mach normal to the leading edge drives compressibility:

M_{n} = M_{\infty} cos Λ < M_{\infty}

(see s03). The reduced effective Mach delays the shock. See Critical and drag-divergence Mach number.

Across a shock the stagnation pressure stays constant because the flow is adiabatic.

False. Adiabatic conserves total enthalpy

h_{0}

, not total pressure. Entropy rises, so

p_{0, 2} / p_{0, 1} = e^{- Δ s / R} < 1

— the stagnation pressure always drops.

Spot the error

Each statement contains one flaw. Name it.

"At

M_{\infty} = 2

the Mach angle is

30°

, so the shock off a

10°

wedge sits at

30°

The error is equating the Mach cone with the shock (s01 shows them apart). The wedge's attached shock is at

β \approx 39.3°

— steeper — because it must turn the flow a finite

10°

;

μ = 30°

is only the weak-disturbance envelope.

"Blunt noses are best for minimizing wave drag because they spread the shock out."

Backwards. A blunt body forces turning beyond

θ_{ma x}

, so the shock detaches into a strong near-normal bow shock (s02, right) with large entropy jump — maximal wave drag. Slender bodies keep shocks attached and weak (s02, left).

c_{d, w a v e} = \frac{4}{M _{\infty}^{2} - 1} (α + d y_{t} / d x + d y_{c} / d x)

The slopes and

α

must be squared, not summed linearly. Linear terms could go negative and would wrongly predict thrust; the squares guarantee

c_{d, w a v e} \geq 0

"Using

sin μ = 1/ M_{\infty}

, at

M_{\infty} = 1

the Mach angle is

0°

, so the cone collapses onto the body."

Sign/limit error. At

M_{\infty} = 1

sin μ = 1

, so

μ = 90°

— the "cone" is a flat plane perpendicular to the flow. The angle shrinks toward

0°

only as

M_{\infty} \to \infty

"The Prandtl–Glauert

β = M_{\infty}^{2} - 1

is the shock angle in the Ackeret formula."

It is a compressibility factor (a pure number equal to

M_{\infty} cos μ

), not the geometric shock angle

β

. They share a letter but are unrelated quantities — a classic notation trap.

"Since Ackeret's

C_{p} = 2 θ / M_{\infty}^{2} - 1

is linear in

θ

, a downward-sloping surface gives negative

C_{p}

that cancels the drag."

The negative

C_{p}

on a rearward-facing surface is exactly what a Prandtl–Meyer expansion gives, but it acts on a surface facing the other way, so its contribution to drag is still positive once you resolve the force along the flow. Drag terms end up as squares.

Why questions

Answer the "why" in one or two sentences of mechanism.

Why does wave drag appear only above the critical Mach number?

Below

M_{cr}

the entire flow is subsonic, so compressions are smooth and reversible with no shock. Only once some point reaches

M = 1

can a shock form, and a shock is the sole source of the entropy loss that becomes wave drag.

Why does a stagnation-pressure loss in the wake mean a force on the body?

A control-volume momentum balance ties the momentum deficit (from the reduced

p_{0}

) directly to a net rearward pressure force on the body — that force is the wave drag. Lost stagnation pressure = lost push-back on the surface.

Why does the transonic drag rise so steeply compared with the gentle subsonic drag?

Near

M_{\infty} = 1

a supersonic pocket suddenly forms and its terminating shock both loses stagnation pressure and thickens/separates the boundary layer behind it, so two drag sources switch on at once over a narrow Mach band.

Why is

L / D_{w a v e} = cot α

for a flat plate a warning, not just a win?

Small

α

gives huge

L / D_{w a v e}

(efficient), but lift itself scales as

α

, so you generate almost no lift — you cannot fly on efficiency alone.

Why does the Area Rule focus on the total cross-sectional area distribution rather than the wing alone?

Transonic wave drag depends chiefly on how smoothly the aircraft's combined cross-section grows and shrinks along its length (s04), so a bulge from the wing must be offset (e.g. by pinching the fuselage) to keep the area curve smooth.

Why does the linearized Ackeret pressure use the same

M_{\infty}^{2} - 1

that appears in the Mach angle?

In the small-deflection limit, real oblique waves approach the Mach angle

μ = arcsin (1/ M_{\infty})

, and

cos μ = M_{\infty}^{2} - 1 / M_{\infty}

sets the geometry (read it off s01) — so weak-wave theory inherits the same

M_{\infty}^{2} - 1

factor.

Why can a bow shock never be described by

sin μ = 1/ M_{\infty}

A bow shock is a finite, curved, near-normal compression whose strength and angle come from the full shock relations, whereas

sin μ = 1/ M_{\infty}

describes only the envelope of vanishingly weak Mach waves.

Edge cases

Push each formula to its boundary and say what happens.

What is the Mach angle exactly at

M_{\infty} = 1

, and what does the "cone" become?

sin μ = 1/1 = 1

μ = 90°

; the Mach cone flattens into a plane standing perpendicular to the flow — signals just barely fail to outrun the body. (This is the value the mnemonic below points to — not

0°

What happens to

c_{d, w a v e} = 4 (\dots) / M_{\infty}^{2} - 1

M_{\infty} \to 1^{+}

The denominator

M_{\infty}^{2} - 1 \to 0

, so linearized theory predicts

c_{d, w a v e} \to \infty

. This is a breakdown of the thin, small-perturbation assumption near sonic; real transonic drag is large but finite and needs nonlinear theory.

What happens to the Mach cone as

M_{\infty} \to \infty

sin μ = 1/ M_{\infty} \to 0

, so

μ \to 0°

— the cone wraps tightly around the body and the disturbance region shrinks to a thin sliver along the surface.

For a wedge whose half-angle exceeds

θ_{ma x} (M_{1})

, what does the shock do?

No attached oblique-shock solution exists at that upstream Mach

M_{1}

, so the shock detaches and stands off as a curved bow shock (s02, right) — near the centreline it is locally a normal shock with maximal entropy jump and drag.

At exactly zero lift (

α = 0

) and zero camber, is wave drag zero?

No. The thickness term

\overline{(d y_{t} / d x)^{2}}

survives, giving nonzero "volume" wave drag — a body of finite thickness must still push air aside, forming waves.

In the limit

θ \to 0^{+}

, what does the weak-shock root of the

θ

–

β

–

M

relation give for

β

It tends to the Mach angle:

β \to μ = arcsin (1/ M_{1})

. This zero-strength limit is the only case where a real shock angle equals the Mach angle.

What is the sweep component

M_{n}

when a wing is swept a full

90°

(theoretical limit)?

M_{n} = M_{\infty} cos 90° = 0

, so the normal Mach vanishes and compressibility effects along the leading edge disappear — an idealized extreme showing why sweep delays the shock (s03). See Speed of sound and Mach number.