Critical Mach number — onset of local supersonic flow: Worked examples

This page drills the critical Mach number from every angle it can appear on a problem set: forward direction ( $M_{cr} \to C_{p, cr}$ ), backward direction ( $C_{p, 0} \to M_{cr}$ ), the two limiting ends of the curve, degenerate inputs, a real-world word problem, and an exam-style twist. Before the examples, we lay out a scenario matrix so you can see exactly which "cell" each problem fills — and confirm no case is left unshown.

Everything rests on two objects you already met in the parent note. We restate them here so no symbol is used before it is anchored.

Definition The symbols on this page, in plain words

Read these once — every later line reuses them and nothing else.

$M_{\infty}$ (say "M-infinity") — the free-stream Mach number: the whole aircraft's flight speed divided by the local speed of sound. It describes how fast you are flying.
$M_{cr}$ — the critical value of $M_{\infty}$ : the one particular flight speed at which the fastest point on the wing first reaches sound. So $M_{cr}$ is a special reading of $M_{\infty}$ , never a different quantity. Numerically $M_{cr} < 1$ always.
$M_{l oc a l}$ — the Mach number at a single point on the wing's skin; it can exceed $M_{\infty}$ because air speeds up over curvature.
$p$ — static pressure at a point; $p_{\infty}$ — static pressure far upstream (undisturbed air); $p_{0}$ — stagnation pressure, the pressure air would reach if brought smoothly to rest (constant along a streamline in isentropic flow, per Isentropic Flow Relations); $p^{*}$ — the static pressure at the point where $M_{l oc a l} = 1$ (the "sonic" or "starred" state).
$γ$ — ratio of specific heats ( $1.4$ for air, $5/3$ for helium).
$Λ$ (capital "lambda") — the sweep angle of the wing's leading edge, measured back from straight across the flow. $Λ = 0$ is an unswept wing; a bigger $Λ$ means the wing is angled further back. It appears only in Example 7.

Definition The two curves, in words

$C_{p, cr} (M_{cr})$ — the gas-dynamics curve. "How low must the surface pressure dip so that the fastest point on the wing just reaches sound?" It depends only on $γ$ . $C_{p, cr} = \frac{2}{γ M _{cr}^{2}} [(\frac{1 + \frac{γ - 1}{2} M _{cr}^{2}}{1 + \frac{γ - 1}{2}})^{\frac{γ}{γ - 1}} - 1]$
$C_{p, m i n} (M_{\infty}) = \frac{C _{p, 0}}{1 - M _{\infty}^{2}}$ — the airfoil curve (a Prandtl–Glauert estimate). "What is the actual lowest pressure coefficient on THIS wing at speed $M_{\infty}$ ?" It depends on the shape through the single fixed number $C_{p, 0} < 0$ .

Notation note: $C_{p, m i n}$ (the airfoil's lowest $C_{p}$ at a given speed) and $C_{p, cr}$ (the gas-dynamics requirement) are two different curves. They coincide only at the crossing point, and that crossing speed is $M_{cr}$ . We never abbreviate either one differently anywhere below.

The critical Mach number is where these two curves cross. Gas-dynamics requirement meets airfoil reality.

The scenario matrix

Every problem this topic throws is one of these cells. Each example below is tagged with the cell(s) it fills.

Cell	What varies	The trap / lesson	Example
A Forward	Given $M_{cr}$ → find $C_{p, cr}$	Just plug in — but keep all digits	Ex 1
B Backward	Given $C_{p, 0}$ → find $M_{cr}$	Must intersect two curves numerically	Ex 2
C Low-Mach limit	$M_{cr} \to 0$	$C_{p, cr} \to - \infty$ (need infinite suction)	Ex 3
D High-Mach limit	$M_{cr} \to 1$	$C_{p, cr} \to 0$ (no suction needed)	Ex 3
E Degenerate airfoil	$C_{p, 0} = 0$ (flat plate at $0°$ )	No suction ⇒ $M_{cr} = 1$	Ex 4
F Sign check	$C_{p, cr}$ always $< 0$ ?	Yes for $M_{cr} < 1$ ; verify the sign machinery	Ex 5
G Real-world word	Sea-level jet, altitude, speed	Convert speed → $M_{\infty}$ before comparing	Ex 6
H Design twist	Thinner / swept wing	Which way does $M_{cr}$ move?	Ex 7
I Non-air gas	$γ \neq = 1.4$ (e.g. helium $γ = 5/3$ )	The "universal" curve shifts with $γ$	Ex 8

The figure below is the whole game board — the two curves and their crossing. Every example is a move on it; Example 2 walks it in detail.

Figure — Critical Mach number — onset of local supersonic flow

Intuition What the figure shows (read this even if you cannot see it)

The horizontal axis is the flight speed $M_{\infty}$ (from $0.30$ to nearly $1$ ); the vertical axis is pressure coefficient $C_{p}$ , running downward into negative (suction) values. Two curves are drawn:

The orange curve is $C_{p, cr} (M_{\infty})$ , the gas-dynamics requirement. It starts far down at the left ( $M_{\infty} \to 0$ plunges toward $- \infty$ ) and rises smoothly to touch $0$ at the right end ( $M_{\infty} = 1$ ).
The teal curve is the airfoil's own minimum pressure $C_{p, 0} / 1 - M_{\infty}^{2}$ for $C_{p, 0} = - 0.43$ (Example 2's wing). It starts near $- 0.43$ on the left and sinks steadily as $M_{\infty}$ grows.
A single plum dot marks where orange and teal cross, at $M_{\infty} \approx 0.727$ . That crossing speed is $M_{cr}$ . To the right of the dot the teal (airfoil) line dips below the orange (requirement) line — meaning the wing has already gone supersonic somewhere. To the left, the airfoil stays above the requirement — still fully subsonic on the skin. Labels on the figure call out both limits ( $- \infty$ at the left, $0$ at $M_{\infty} = 1$ ).

Worked examples

Cell A — forward direction

Worked example Example 1 — Compute

$C_{p, cr}$ from a given $M_{cr}$ Given $M_{cr} = 0.75$ , $γ = 1.4$ . Find $C_{p, cr}$ .

Forecast: In Example 1 of the parent, $M_{cr} = 0.70$ gave $C_{p, cr} \approx - 0.79$ . We raised $M_{cr}$ a little. Do you expect $C_{p, cr}$ to become more negative or less negative? (Guess before reading.)

Step 1. Inner ratio (numerator = free-stream side, denominator = sonic side): $\frac{1 + 0.2 ( 0.7 5 ^{2} )}{1 + 0.2} = \frac{1 + 0.2 ( 0.5625 )}{1.2} = \frac{1.1125}{1.2} = 0.927083$ Why this step? Both the free stream and the sonic point sit on one streamline, so they share the same stagnation pressure $p_{0}$ (defined above). Writing $p_{\infty} / p_{0}$ over $p^{*} / p_{0}$ cancels $p_{0}$ and leaves exactly this ratio. The sonic point ( $M_{l oc a l} = 1$ ) fixes the denominator once and for all. Step 2. Raise to $\frac{γ}{γ - 1} = 3.5$ (keeping all digits): $0.92708 3^{3.5} = 0.76749$ Why? Isentropy turns that temperature/density-flavoured ratio into the pressure ratio $p / p_{\infty}$ . Step 3. Front factor and subtract 1: $C_{p, cr} = \frac{2}{1.4 ( 0.5625 )} (0.76749 - 1) = \frac{2}{0.7875} (- 0.23251) = - 0.59050$

Answer: $C_{p, cr} \approx - 0.590$ .

Verify: Raising $M_{cr}$ from $0.70$ ( $- 0.79$ ) to $0.75$ made $C_{p, cr}$ less negative ( $- 0.59$ ). ✅ That matches the curve's rise-toward-zero trend in the figure. It also debunks the parent's second [!mistake]: higher $M_{cr}$ ⇒ easier suction requirement.

Cell B — backward direction (the design question)

Worked example Example 2 — Find

$M_{cr}$ from the airfoil's $C_{p, 0}$ An airfoil has incompressible minimum pressure coefficient $C_{p, 0} = - 0.43$ . Estimate $M_{cr}$ ( $γ = 1.4$ ). This is the case drawn in the figure above — follow it there as you read.

Forecast: This suction peak is milder than the parent's $- 0.55$ (which gave $M_{cr} \approx 0.70$ ). Milder suction is easier to keep subsonic — so do you expect $M_{cr}$ higher or lower than $0.70$ ?

Step 1. Write both curves as functions of $M_{\infty}$ — in the figure these are the orange (gas) and teal (airfoil) lines: $airfoil: C_{p, m i n} = \frac{- 0.43}{1 - M _{\infty}^{2}}, gas: C_{p, cr} (M_{\infty}) .$ Why? The crossing of these — the plum dot in the figure — is the definition of $M_{cr}$ . Two unknowns collapse to one equation. Step 2. Probe $M = 0.75$ : airfoil $= - 0.43/ 1 - 0.5625 = - 0.43/ 0.4375 = - 0.65017$ ; gas $C_{p, cr} (0.75) = - 0.59050$ . In the figure the teal line sits below orange here → suction already too strong → past sonic. Step 3. Probe $M = 0.74$ : airfoil $= - 0.43/ 1 - 0.5476 = - 0.43/ 0.4524 = - 0.63930$ ; gas $C_{p, cr} (0.74) \approx - 0.60747$ . Still just past. Probe $M = 0.735$ : airfoil $= - 0.43/ 0.45978 \approx - 0.63417$ ; gas $C_{p, cr} (0.735) \approx - 0.61629$ . The two lines now nearly meet — this is the plum dot.

Answer: $M_{cr} \approx 0.734$ — read straight off the crossing in the figure.

Verify: $0.734 > 0.70$ . ✅ Milder suction → higher $M_{cr}$ , exactly as forecast, and consistent with "thinner wing ⇒ higher $M_{cr}$ ."

Cells C & D — the two limits of the universal curve

Worked example Example 3 — Behaviour at

$M_{cr} \to 0$ and $M_{cr} \to 1$ Without a calculator, describe the two ends of the $C_{p, cr}$ curve.

Forecast: At one end you'd need an infinitely deep pressure well; at the other, essentially none. Which is which?

Step 1 (Cell D, $M_{cr} \to 1$ ). Plug $M_{cr} = 1$ : inner ratio $= \frac{1 + 0.2}{1 + 0.2} = 1$ , so $1^{3.5} - 1 = 0$ , giving $C_{p, cr} = 0$ . In the figure this is the right end of the orange curve rising to touch zero. Why? If the free stream is itself sonic, the surface pressure needn't dip at all to reach $M_{l oc a l} = 1$ . Zero suction required. Step 2 (Cell C, $M_{cr} \to 0$ ). As $M_{cr} \to 0$ the bracket $[\dots - 1]$ tends to a finite negative number, but the front factor $\frac{2}{γ M _{cr}^{2}}$ blows up like $1/ M_{cr}^{2}$ . So $C_{p, cr} \to - \infty$ (left end of the orange curve plunging). Why? At a crawl, air is nearly incompressible; to accelerate a nearly-still stream all the way to sound you must remove almost all the pressure — an unbounded suction. Numerically at $M_{cr} = 0.1$ : inner $= \frac{1.002}{1.2} = 0.835$ ; $0.83 5^{3.5} = 0.531$ ; $C_{p, cr} = \frac{2}{1.4 ( 0.01 )} (- 0.469) = - 67.0$ .

Answer: $C_{p, cr} \to - \infty$ as $M_{cr} \to 0$ ; $C_{p, cr} \to 0^{-}$ as $M_{cr} \to 1$ . A monotone rise across the whole band.

Verify: At $M = 0.1$ the value $- 67.0$ is huge and negative (matches Cell C); at $M = 1$ it is exactly $0$ (Cell D). ✅ Both endpoints reproduced. See the two ends of the orange curve in the figure.

Cell E — degenerate airfoil (no suction)

Worked example Example 4 — A shape with

$C_{p, 0} = 0$ A vanishingly thin flat plate at zero angle of attack has essentially no pressure dip: $C_{p, 0} = 0$ . What is $M_{cr}$ ?

Forecast: If the flow never speeds up anywhere, when does a local point reach sound? Guess before the algebra.

Step 1. Airfoil curve: $C_{p, m i n} = \frac{0}{1 - M _{\infty}^{2}} = 0$ for all $M_{\infty} < 1$ . Why? Prandtl–Glauert scales the incompressible peak; scaling zero gives zero. Step 2. Set equal to gas curve: $0 = C_{p, cr} (M_{\infty})$ . From Example 3, $C_{p, cr} = 0$ only at $M_{\infty} = 1$ . Why? The crossing is the answer, and the gas curve touches zero solely at unity.

Answer: $M_{cr} = 1$ .

Verify: A plate with no suction cannot make any patch supersonic until the whole stream is sonic — so $M_{cr} = 1$ , the theoretical ceiling. ✅ This is the boundary case that proves "less suction ⇒ higher $M_{cr}$ " pushed to its limit.

Cell F — the sign machinery

Worked example Example 5 — Prove

$C_{p, cr} < 0$ for every $M_{cr} \in (0, 1)$ Show the critical pressure coefficient is always negative below $M_{\infty} = 1$ .

Forecast: Suction (pressure below free-stream) means $C_{p} < 0$ . Do you expect the gas curve to ever cross into positive territory before $M = 1$ ?

Step 1. Look at the inner ratio $R = \frac{1 + \frac{γ - 1}{2} M _{cr}^{2}}{1 + \frac{γ - 1}{2}}$ . For $M_{cr} < 1$ the numerator is smaller than the denominator, so $R < 1$ . Why? Both share the same additive $1$ ; only the $M^{2}$ term differs, and $M_{cr}^{2} < 1$ . Step 2. Then $R^{γ / (γ - 1)} < 1$ , so the bracket $[R^{3.5} - 1] < 0$ . Why? Raising a number in $(0, 1)$ to a positive power keeps it in $(0, 1)$ . Step 3. The front factor $\frac{2}{γ M _{cr}^{2}} > 0$ . Positive $\times$ negative $=$ negative. Why? $γ > 0$ and $M_{cr}^{2} > 0$ , so the prefactor never flips the sign.

Answer: $C_{p, cr} < 0$ strictly for all $0 < M_{cr} < 1$ , hitting $0$ only at $M_{cr} = 1$ .

Verify: Check a middle point $M_{cr} = 0.5$ : inner $= \frac{1.05}{1.2} = 0.875$ ; $0.87 5^{3.5} = 0.62666$ ; $C_{p, cr} = \frac{2}{1.4 ( 0.25 )} (- 0.37334) = - 2.1334 < 0$ . ✅ Negative, as proven.

Cell G — real-world word problem

Worked example Example 6 — Cruise jet at altitude

A jet cruises at $V_{\infty} = 250 m/s$ where the local speed of sound is $a = 300 m/s$ . Its wing has $M_{cr} = 0.80$ . Is a supersonic pocket present on the wing? By how much margin?

Forecast: Compare the flight Mach number to $M_{cr}$ . Guess whether $M_{\infty}$ beats $0.80$ .

Step 1. Flight Mach number: $M_{\infty} = \frac{V _{\infty}}{a} = \frac{250}{300} = 0.8333.$ Why? Mach number is speed relative to the local sound speed — always convert before comparing to $M_{cr}$ . (Units cancel: m/s ÷ m/s = dimensionless. ✅) Step 2. Compare: $M_{\infty} = 0.833 > M_{cr} = 0.80$ . Why? Above $M_{cr}$ , the fastest surface point has already passed $M_{l oc a l} = 1$ → a supersonic pocket exists, likely closing in a shock, driving drag divergence. Step 3. Margin: $M_{\infty} - M_{cr} = 0.833 - 0.80 = 0.033$ .

Answer: Yes — the wing is $0.033$ Mach over critical; a shock-terminated supersonic pocket is present.

Verify: If the pilot slowed to $V = 0.80 \times 300 = 240 m/s$ , then $M_{\infty} = 0.80 = M_{cr}$ exactly — the pocket just vanishes. ✅ Consistent.

Cell H — design twist (sweep)

Worked example Example 7 — Sweep the wing back

$30°$ The wing of Example 6 (straight $M_{cr} = 0.80$ ) is swept back by $Λ = 30°$ (sweep angle, defined in the glossary). What effective $M_{cr}$ does the aircraft now enjoy, and does the pocket still form at $M_{\infty} = 0.833$ ?

Forecast: Sweep hides speed from the airfoil. Does the whole-aircraft $M_{cr}$ go up or down?

Step 1. Only the velocity component normal to the leading edge drives the pressure dip: $M_{eff} = M_{\infty} cos Λ$ . Why? The spanwise component slides along the wing and does no compressing; simple-sweep theory keeps just the normal part. Step 2. The airfoil section still goes critical when its normal Mach equals $0.80$ . So the aircraft's true critical Mach is: $M_{cr}^{swept} = \frac{0.80}{c o s 30°} = \frac{0.80}{0.86603} = 0.92376.$ Why? Rearranging $M_{\infty} cos Λ = 0.80$ for $M_{\infty}$ . Step 3. Compare to the cruise $M_{\infty} = 0.833$ : now $0.833 < 0.924$ , so no pocket.

Answer: Swept $M_{cr}$ rises to $\approx 0.924$ ; the cruise of Ex 6 is now comfortably subcritical — the shock is gone.

Verify: Sweep raised $M_{cr}$ ( $0.80 \to 0.924$ ), matching the parent's Example 3 forecast and the mnemonic "skinny swept wings sneak past sonic." ✅ (Note: real sweep gains are smaller than $1/ cos Λ$ — a fully swept ideal is an upper bound. Compare also the Area Rule (Transonic) as a complementary transonic fix.)

Cell I — a different gas

Worked example Example 8 — Helium,

$γ = 5/3$ Recompute $C_{p, cr}$ at $M_{cr} = 0.70$ but in helium ( $γ = 5/3 \approx 1.6667$ ) instead of air. Does the "universal" curve move?

Forecast: The parent got $- 0.79$ in air. Will a higher $γ$ make the required suction deeper or shallower?

Step 1. Compute the coefficient $\frac{γ - 1}{2} = \frac{0.6667}{2} = 0.33333$ and the inner ratio: $\frac{1 + 0.33333 ( 0.7 0 ^{2} )}{1 + 0.33333} = \frac{1 + 0.33333 ( 0.49 )}{1.33333} = \frac{1.16333}{1.33333} = 0.87250.$ Why this step? Same isentropic structure as air, but the $\frac{γ - 1}{2}$ coefficient is larger for helium — that is the only thing that changes when we swap gases. Step 2. Exponent $\frac{γ}{γ - 1} = \frac{1.66667}{0.66667} = 2.5$ . Raise: $0.8725 0^{2.5} = 0.71124.$ Why? Isentropy again turns the ratio into a pressure ratio — but with helium's exponent $2.5$ , not air's $3.5$ . Step 3. Front factor $\frac{2}{γ M _{cr}^{2}} = \frac{2}{1.66667 ( 0.49 )} = \frac{2}{0.81667} = 2.44898$ , then subtract 1 and multiply: $C_{p, cr} = 2.44898 (0.71124 - 1) = 2.44898 (- 0.28876) = - 0.70717.$

Answer: $C_{p, cr} \approx - 0.707$ in helium, versus $- 0.793$ in air, at the same $M_{cr} = 0.70$ .

Verify: The curve shifts with $γ$ — it is "universal" only for a fixed gas. Helium ( $- 0.707$ ) is less negative than air ( $- 0.793$ ) at $M_{cr} = 0.70$ . ✅ The dependence on $γ$ (and nothing about the airfoil) is exactly what the parent's [!formula] callout claimed.

Recall Which cell was which?

Cell A forward — Ex1 answer ::: $C_{p, cr} \approx - 0.590$ at $M_{cr} = 0.75$ . Cell B backward — Ex2 answer ::: $M_{cr} \approx 0.734$ for $C_{p, 0} = - 0.43$ . Cell C limit — trend as $M_{cr} \to 0$ ::: $C_{p, cr} \to - \infty$ . Cell D limit — value at $M_{cr} = 1$ ::: $C_{p, cr} = 0$ (no suction needed). Cell E degenerate — flat plate $C_{p, 0} = 0$ ::: $M_{cr} = 1$ . Cell F sign check — sign of $C_{p, cr}$ for $M_{cr} \in (0, 1)$ , e.g. at $0.5$ ::: strictly negative; $C_{p, cr} (0.5) \approx - 2.133$ . Cell G word problem — is $M_{\infty} = 0.833$ supercritical if $M_{cr} = 0.80$ ? ::: Yes, by $0.033$ Mach. Cell H sweep $30°$ — new $M_{cr}$ ::: $\approx 0.924$ . Cell I helium — $C_{p, cr}$ at $M_{cr} = 0.70$ ::: $\approx - 0.707$ .

Connections

Universal curve rests on Isentropic Flow Relations and Pressure Coefficient Cp.
Airfoil curve is the Prandtl–Glauert Compressibility Correction.
Crossing consequences: Drag Divergence Mach Number, Shock Waves & Normal Shocks.
Design fixes: Swept Wings & Transonic Design, Area Rule (Transonic).