Critical Mach number — onset of local supersonic flow: Exercises

Two reference tools you will reuse constantly:

Intuition Where the universal curve comes from (the "why" in three moves)

You already have the $C_{p}$ -in-Mach form. Build the bracket in three honest steps:

Take the $C_{p}$ formula at the surface point: $C_{p} = \frac{2}{γ M _{\infty}^{2}} (\frac{p}{p _{\infty}} - 1)$ . We just need the pressure ratio $p / p_{\infty}$ at that point.
Route through the shared $p_{0}$ . Since $p_{0}$ is common on the streamline, $\frac{p}{p _{\infty}} = \frac{p / p _{0}}{p _{\infty} / p _{0}}$ , and each of $p / p_{0}$ , $p_{\infty} / p_{0}$ is given by the isentropic relation. This yields $\frac{p}{p _{\infty}} = (\frac{1 + \frac{γ - 1}{2} M _{\infty}^{2}}{1 + \frac{γ - 1}{2} M _{l oc a l}^{2}})^{\frac{γ}{γ - 1}}$ .
Demand the surface point be exactly sonic: set $M_{l oc a l} = 1$ (so $1 + \frac{γ - 1}{2} = 1.2$ ) and rename the free stream $M_{\infty} \to M_{cr}$ . Substituting that pressure ratio back into step 1 is the boxed universal formula. So the curve is nothing more than " $C_{p}$ written out for a point that has just reached Mach 1." It carries only $γ$ — no airfoil geometry — which is exactly why it is universal.

Figure — Critical Mach number — onset of local supersonic flow

Level 1 — Recognition

L1.1

State whether each is true or false, and why. (a) $M_{cr} = 1$ . (b) At $M_{\infty} = M_{cr}$ , exactly one surface point has $M_{l oc a l} = 1$ . (c) $C_{p, cr}$ depends on the airfoil shape.

Recall Solution L1.1

(a) False. The whole aircraft is still subsonic; only a local patch over the curved surface has reached sonic. Air accelerates over curvature (thumb-on-hose picture), so $M_{l oc a l} > M_{\infty}$ . Hence $M_{cr} < 1$ always — typically $0.7$ – $0.85$ . (b) True. At exactly $M_{\infty} = M_{cr}$ , the single lowest-pressure point (the suction peak) just touches $M_{l oc a l} = 1$ . Below it: fully subsonic. Above it: a growing supersonic pocket. (c) False. The $C_{p, cr}$ curve comes purely from isentropic gas dynamics — it contains only $γ$ and $M_{cr}$ . Every airfoil in air shares the same $C_{p, cr}$ curve. What differs between airfoils is the suction curve $C_{p, m i n}$ .

L1.2

The two curves in the figure both slope differently as $M_{\infty}$ rises. Which curve is universal, and which is airfoil-specific?

Recall Solution L1.2

Blue, $C_{p, cr}$ : universal. Its value at each $M_{\infty}$ is the pressure coefficient a point must reach to be sonic. It gets less negative as $M_{\infty}$ grows (a faster free stream needs a smaller pressure dip to hit Mach 1 locally).
Yellow, $C_{p, m i n}$ : airfoil-specific. It starts at the airfoil's incompressible suction value $C_{p, 0}$ and grows more negative through the $1/ 1 - M_{\infty}^{2}$ factor as compressibility sharpens the suction peak. The crossing is $M_{cr}$ .

Level 2 — Application

L2.1

Compute $C_{p, cr}$ for $M_{cr} = 0.75$ .

Recall Solution L2.1

Step 1 — inner ratio. $\frac{1 + 0.2 ( 0.7 5 ^{2} )}{1.2} = \frac{1 + 0.2 ( 0.5625 )}{1.2} = \frac{1.1125}{1.2} = 0.92708$ . Why: this is $p_{\infty} / p^{*}$ — the free-stream pressure measured against the sonic pressure $p^{*}$ defined above (matched stagnation pressure on the streamline). Step 2 — isentropic power. $0.9270 8^{3.5} = 0.76505$ . Step 3 — assemble. $C_{p, cr} = \frac{2}{1.4 ( 0.5625 )} (0.76505 - 1) = \frac{2}{0.7875} (- 0.23495) = - 0.5967$ . Answer: $C_{p, cr} \approx - 0.597$ .

L2.2

Compute $C_{p, cr}$ for $M_{cr} = 0.60$ , and confirm it is more negative than the L2.1 answer.

Recall Solution L2.2

Step 1. $\frac{1 + 0.2 ( 0.36 )}{1.2} = \frac{1.072}{1.2} = 0.89333$ . Step 2. $0.8933 3^{3.5} = 0.67243$ . Step 3. $C_{p, cr} = \frac{2}{1.4 ( 0.36 )} (0.67243 - 1) = \frac{2}{0.504} (- 0.32757) = - 1.3000$ . Answer: $C_{p, cr} \approx - 1.300$ . Indeed $- 1.300 < - 0.597$ : at the lower free-stream Mach a deeper pressure dip is required to reach local sonic. This confirms the blue curve's downward-left rise.

L2.3

An airfoil has incompressible minimum $C_{p, 0} = - 0.43$ . What is its compressible suction $C_{p, m i n}$ at $M_{\infty} = 0.65$ ?

Recall Solution L2.3

$C_{p, m i n} = \frac{- 0.43}{1 - 0.6 5 ^{2}} = \frac{- 0.43}{1 - 0.4225} = \frac{- 0.43}{0.5775} = \frac{- 0.43}{0.75993} = - 0.5658$ . Answer: $C_{p, m i n} \approx - 0.566$ . The suction peak has deepened from $- 0.43$ to $- 0.566$ purely because of compressibility. See Prandtl–Glauert Compressibility Correction.

Level 3 — Analysis

L3.1

Airfoil A has $C_{p, 0} = - 0.55$ . Find $M_{cr}$ by locating the crossing of its suction curve with the universal curve (work to $\pm 0.005$ ).

Recall Solution L3.1

We need $f (M) = \frac{- 0.55}{1 - M ^{2}} - C_{p, cr} (M) = 0$ .

$M = 0.70$ : suction $= - 0.55/ 0.51 = - 0.7702$ ; required $C_{p, cr} (0.70) = - 0.7790$ . Then $f = - 0.7702 - (- 0.7790) = + 0.0088 > 0$ → suction not yet deep enough, still subsonic.
$M = 0.71$ : suction $= - 0.55/ 0.4959 = - 0.7810$ ; $C_{p, cr} (0.71) = - 0.7376$ . Then $f = - 0.7810 - (- 0.7376) = - 0.0434 < 0$ → past sonic. $f$ flips from $+$ to $-$ between $0.70$ and $0.71$ , so the root (crossing) lies there, close to $0.70$ . Answer: $M_{cr} \approx 0.702$ .

L3.2

Airfoil B is thinner: $C_{p, 0} = - 0.35$ . Predict qualitatively whether $M_{cr}$ rises or falls versus airfoil A, then find it numerically.

Recall Solution L3.2

Prediction: thinner ⇒ gentler curvature ⇒ smaller suction ⇒ its yellow curve sits higher (less negative) ⇒ it must climb further right before meeting the universal curve ⇒ $M_{cr}$ rises. Numerics. Solve $f (M) = - 0.35/ 1 - M^{2} - C_{p, cr} (M) = 0$ .

$M = 0.76$ : suction $= - 0.35/ 0.4224 = - 0.5385$ ; $C_{p, cr} (0.76) = - 0.5583$ . Then $f = - 0.5385 - (- 0.5583) = + 0.0198 > 0$ → not yet sonic.
$M = 0.78$ : suction $= - 0.35/ 0.3916 = - 0.5593$ ; $C_{p, cr} (0.78) = - 0.4952$ . Then $f = - 0.5593 - (- 0.4952) = - 0.0641 < 0$ → past sonic. $f$ flips sign between $0.76$ and $0.78$ , so the crossing sits there, near $M_{cr} \approx 0.765$ . Answer: $M_{cr} \approx 0.77$ — higher than A's $0.702$ , as predicted. This is exactly why fast jets use thin wings.

L3.3

Airfoil C is extremely thin with $C_{p, 0} = - 0.05$ . Try to find its $M_{cr}$ . What happens, and what does it mean physically?

Recall Solution L3.3

Form $f (M) = - 0.05/ 1 - M^{2} - C_{p, cr} (M)$ .

$M = 0.90$ : suction $= - 0.05/ 0.19 = - 0.1147$ ; $C_{p, cr} (0.90) = - 0.1876$ . $f = + 0.0729 > 0$ .
$M = 0.95$ : suction $= - 0.05/ 0.0975 = - 0.1601$ ; $C_{p, cr} (0.95) = - 0.0863$ . $f = - 0.0738 < 0$ . A sign flip does occur, but only very close to $M_{\infty} = 1$ : solving gives $M_{cr} \approx 0.93$ . Edge-case lesson: for a section this thin the suction is so weak that the free stream must be pushed almost to sonic before the tiny suction peak can reach Mach 1. In the limit $C_{p, 0} \to 0^{-}$ (a flat plate at zero incidence — no thickness, no suction), the yellow curve hugs the axis, and the crossing pushes all the way to $M_{cr} \to 1$ : such a body never develops a local supersonic pocket before the whole flow is sonic. So $M_{cr}$ is bounded above by $1$ , approached only as the suction vanishes.

L3.4

Airfoil D is deeply cupped with a very strong suction peak, $C_{p, 0} = - 1.60$ . Find $M_{cr}$ and note the low-speed edge case.

Recall Solution L3.4

$M = 0.40$ : suction $= - 1.60/ 0.84 = - 1.7457$ ; $C_{p, cr} (0.40) = - 3.6607$ . $f = + 1.915 > 0$ → not sonic yet.
$M = 0.55$ : suction $= - 1.60/ 0.6975 = - 1.9158$ ; $C_{p, cr} (0.55) = - 1.6606$ . $f = - 1.9158 - (- 1.6606) = - 0.2552 < 0$ → past sonic. Sign flip between $0.40$ and $0.55$ → $M_{cr} \approx 0.52$ . Edge-case lesson: a very deep suction ( $∣ C_{p, 0} ∣$ large) drags the crossing to a low $M_{\infty}$ . In the extreme, if the incompressible suction already exceeds the $C_{p, cr}$ value at very low $M_{\infty}$ , the section is critical almost as soon as it flies — such shapes are useless for high-speed flight. So the two ways the crossing can escape the useful $0.5$ – $0.9$ band are: (i) suction so weak the crossing races to $M_{\infty} \to 1$ (L3.3), and (ii) suction so strong the crossing collapses toward low $M_{\infty}$ (here).

Level 4 — Synthesis

L4.1

A straight wing has $M_{cr} = 0.72$ . It is swept back by $Λ = 3 0^{\circ}$ . Estimate the new $M_{cr}$ using the simple cosine (component) rule.

Recall Solution L4.1

The idea. Only the velocity component normal to the leading edge "sees" the airfoil. Sweep hides the streamwise speed: the effective normal Mach is $M_{\infty} cos Λ$ . The airfoil goes critical when this normal component equals the straight-wing critical value: $M_{cr, straight} = M_{cr, swept} cos Λ \Rightarrow M_{cr, swept} = \frac{M _{cr, straight}}{c o s Λ} .$ Compute. $cos 3 0^{\circ} = 0.86603$ , so $M_{cr, swept} = \frac{0.72}{0.86603} = 0.8314$ . Answer: $M_{cr} \approx 0.831$ . Sweep bought roughly $+ 0.11$ in critical Mach — see Swept Wings & Transonic Design.

L4.2

Two designers argue. Designer 1 wants a thinner wing; designer 2 wants more sweep. Starting from airfoil A ( $M_{cr} = 0.702$ , straight), which single change reaches $M_{cr} = 0.80$ first: thinning to airfoil B (from L3.2, $M_{cr} = 0.77$ , straight), or sweeping airfoil A to $Λ = 2 5^{\circ}$ ?

Recall Solution L4.2

Thinning alone: airfoil B gives $M_{cr} = 0.77 < 0.80$ — falls short. Sweeping A by $2 5^{\circ}$ : $cos 2 5^{\circ} = 0.90631$ , so $M_{cr} = 0.702/0.90631 = 0.7746 < 0.80$ — also falls short. Neither single change reaches $0.80$ . Combine them: sweep airfoil B by $2 5^{\circ}$ : $M_{cr} = 0.77/0.90631 = 0.8496$ . Answer: neither change alone suffices; combining thin + swept clears $0.80$ comfortably ( $\approx 0.85$ ). Real transonic wings do exactly this — plus Area Rule (Transonic) on the fuselage.

Level 5 — Mastery

L5.1

Prove analytically that as $M_{cr} \to 1^{-}$ , the universal critical pressure coefficient $C_{p, cr} \to 0^{-}$ . Interpret physically.

Recall Solution L5.1

Set $M_{cr} = 1$ in the bracket first. The inner ratio becomes $\frac{1 + 0.2 ( 1 )}{1.2} = \frac{1.2}{1.2} = 1$ . Raise to $3.5$ : still $1$ . So the bracket is $1 - 1 = 0$ . Prefactor. $\frac{2}{γ M _{cr}^{2}} = \frac{2}{1.4}$ stays finite. Finite $\times$ zero $= 0$ . $lim_{M_{cr} \to 1} C_{p, cr} = 0.$ Approach from below ( $M_{cr} < 1$ ): the inner ratio is $< 1$ , its $3.5$ power is $< 1$ , so the bracket is negative → $C_{p, cr} \to 0^{-}$ . Physical meaning: when the free stream itself is nearly sonic, the surface point barely has to accelerate to reach Mach 1 — so almost no pressure dip is required, $C_{p, cr} \to 0$ . The blue curve meets the horizontal axis at $M_{\infty} = 1$ , which the figure shows.

L5.2

An airfoil designer wants $M_{cr} = 0.68$ exactly on a straight wing. What incompressible minimum pressure coefficient $C_{p, 0}$ must the airfoil be designed to?

Recall Solution L5.2

At the crossing, the two curves are equal: $\frac{C _{p, 0}}{1 - M _{cr}^{2}} = C_{p, cr} (M_{cr})$ , so $C_{p, 0} = C_{p, cr} (M_{cr}) 1 - M_{cr}^{2} .$ Compute $C_{p, cr} (0.68)$ . Inner ratio $= \frac{1 + 0.2 ( 0.4624 )}{1.2} = \frac{1.09248}{1.2} = 0.9104$ . Power: $0.910 4^{3.5} = 0.71611$ . Then $C_{p, cr} = \frac{2}{1.4 ( 0.4624 )} (0.71611 - 1) = \frac{2}{0.64736} (- 0.28389) = - 0.8771$ . Multiply by $1 - 0.6 8^{2}$ . $1 - 0.4624 = 0.5376 = 0.73321$ . So $C_{p, 0} = - 0.8771 \times 0.73321 = - 0.6431$ . Answer: $C_{p, 0} \approx - 0.643$ . The airfoil's incompressible suction peak must be designed no deeper than about $- 0.643$ to hold $M_{cr} = 0.68$ .

L5.3

For airfoil A ( $C_{p, 0} = - 0.55$ , $M_{cr} \approx 0.702$ ), sweeping to $Λ$ pushes $M_{cr}$ to $0.80$ . What sweep angle $Λ$ is required?

Recall Solution L5.3

Rearrange the sweep rule. $M_{cr, swept} = \frac{M _{cr, straight}}{cos Λ} \Rightarrow cos Λ = \frac{M _{cr, straight}}{M _{cr, swept}} = \frac{0.702}{0.80} = 0.8775$ . Invert. $Λ = arccos (0.8775) = 28.7 1^{\circ}$ . Answer: $Λ \approx 28. 7^{\circ}$ . A sweep near $2 9^{\circ}$ lifts airfoil A's critical Mach from $0.702$ to $0.80$ — consistent with real transonic transports and the Drag Divergence Mach Number margin they need.

Recall One-screen summary

L1: $M_{cr} < 1$ ; blue curve universal, yellow curve airfoil-specific; answer = intersection.
L2: plug into $C_{p, cr}$ ; square $M$ before the $0.2$ factor; $p^{*}$ = sonic-point pressure.
L3: bracket the root by the sign flip of $f =$ (suction − requirement) — the intermediate-value guarantee; the crossing can escape toward $1$ (too thin) or toward $0$ (too cupped).
L4: thin lifts $M_{cr}$ ; sweep multiplies by $1/ cos Λ$ ; combine for transonic wings.
L5: limits ( $C_{p, cr} \to 0^{-}$ at $M_{\infty} \to 1$ ), inverse design ( $C_{p, 0} = C_{p, cr} 1 - M_{cr}^{2}$ ), sweep sizing.

Reveal-drill:

What is

C_{p}

The dimensionless local pressure deficit

(p - p_{\infty}) / (\frac{1}{2} ρ_{\infty} V_{\infty}^{2})

; negative means suction.

What is

p^{*}

The static pressure at the point where local flow is exactly sonic (

M_{l oc a l} = 1

What must you multiply

C_{p, cr} (M_{cr})

by to recover the required

C_{p, 0}

1 - M_{cr}^{2}

Sweep rule for critical Mach?

M_{cr, swept} = M_{cr, straight} / cos Λ

M_{\infty} \to 1

, where does

C_{p, cr}

go?

0^{-}

(the curve meets the axis).

Why does a sign flip of (suction − requirement) guarantee a crossing?

The difference is continuous, so it cannot pass from

+

-

without hitting

0

(intermediate-value theorem).