Exercises — Area rule — Whitcomb's rule for transonic drag reduction

Physical quantity: the longitudinal distribution of total cross-sectional area, $A(x)$ — not how that area is split between fuselage, wings, and tail. Mathematical driver: the second derivative $A^{\prime \prime }(x)$ (the curvature of the area curve). Abrupt changes / kinks in $A(x)$ make $A^{\prime \prime }$ spike, which makes the double integral large.

For $M>1$ you cut the body with Mach planes, tilted at the Mach angle $\mu$: $\mu =\arcsin \left(\frac{1}{M}\right),$ and you average the resulting area over roll angle (the supersonic / Jones area rule). See Mach angle and Mach cone. At exactly $M=1$, $\arcsin (1)=90^{\circ }$, so the Mach plane stands perpendicular to the flow — the two rules agree there.

False. It reduces wave drag (the transonic/supersonic drag from shock waves) only. Skin-friction drag and induced drag are untouched, and the wasp-waist can even add a little friction/structural penalty at low speed. See Drag breakdown: friction, induced, wave.

$\xi =x/\ell$ is just the fraction of the way along the body (0 at nose, 1 at tail). Mid-station $x=5$: $\xi =0.5$. Then $4\xi (1-\xi )=4(0.5)(0.5)=1$, and $1^{3/2}=1$, so $A=1.2\times 1=1.20{\text{m}}^2.$ As expected the area peaks at the middle — that is where $A_{\max }$ sits. Quarter-station $x=2.5$: $\xi =0.25$. Then $4(0.25)(0.75)=0.75$, and $0.75^{3/2}=0.6495$, so $A=1.2\times 0.6495=0.779{\text{m}}^2.$

With $V$, $q_{\infty }$ fixed, $D_w\propto 1/{\ell }^4$. So $\frac{D_{w,2}}{D_{w,1}}={\left(\frac{{\ell }_1}{{\ell }_2}\right)}^4={\left(\frac{3}{4.5}\right)}^4={\left(\frac{2}{3}\right)}^4=\frac{16}{81}\approx \mathrm{0.198.}$ Wave drag drops to about 20% of the original — a factor-5 reduction just by making the body 50% longer. This is why supersonic bodies are long and slender.

$D_w=\frac{128q_{\infty }{V}^2}{\pi {\ell }^4}=\frac{128\times 25000\times 3^2}{\pi \times 12^4}.$ Numerator: $128\times 25000=3.2\times 10^6$; times $9=2.88\times 10^7$. Denominator: $\pi \times 20736=6.5145\times 10^4$. $D_w=\frac{2.88\times 10^7}{6.5145\times 10^4}\approx 442\text{N}.$

First, a quick tool. A Dirac delta $\delta (x-a)$ is an idealized "infinitely tall, infinitely thin spike" sitting at $x=a$, so narrow that its total area (integral) is exactly $1$. It is what the derivative of a sudden step looks like: if a quantity jumps instantly from one value to another, its rate of change is zero everywhere except an instant, where it is a spike — a delta. (See Slender-body theory for where these source distributions come from.)

Now take derivatives of Body A step by step.

• $A(x)$ has a kink ⇒ its slope $A^{\prime }(x)$ jumps (steps from one value to another) at that station.
• The derivative of a step is a delta: so $A^{\prime \prime }(x)=(\text{jump size})\times \delta (x-x_k)$ at the kink station $x_k$. The curvature is a single delta spike.
• Feed this into the integral. The two deltas, one at $x_1=x_k$ and one at $x_2=x_k$, collapse both integrals onto the single station $x_k$, and there the kernel $\ln |x_1-x_2|$ is evaluated at $x_1=x_2=x_k$, i.e. $\ln |0|=\ln 0=-\infty$.
• So the integral diverges: a true kink gives infinite wave drag. This is the honest, correct conclusion — a genuine slope-discontinuity in $A(x)$ is not physically realizable precisely because it would demand infinite drag. Real "kinks" are slightly rounded, giving a large-but-finite $A^{\prime \prime }$ and a large-but-finite drag; the sharper the corner, the closer to infinity the drag climbs.
• Body B's $A^{\prime \prime }$ is smooth and bounded ⇒ small, spread-out integrand ⇒ small, finite drag. Sears–Haack is in fact the exact minimizer. Conclusion: Body A (the kink) has enormously more wave drag — diverging to infinity in the idealized sharp-corner limit — despite identical volume and length. It is not size but smoothness of $A(x)$ that sets wave drag.

A note on the "off-diagonal" contributions. Away from the coincident point ($x_1\ne x_2$) the kernel $\ln |x_1-x_2|$ is perfectly finite, and even near it the log-singularity is mild: it is integrable. Why "mild"? Recall the elementary integral ${\int }_0^{\epsilon }\ln xdx=[x\ln x-x{]}_0^{\epsilon }=\epsilon \ln \epsilon -\epsilon ,$ which is a finite number (as $x\to 0^{+}$, $x\ln x\to 0$). So $\ln$ integrated against a bounded, spread-out $A^{\prime \prime }$ (Body B) gives a finite drag. The trouble in Body A comes not from the log itself but from squeezing all the curvature into a point (the delta), which forces the kernel to be sampled exactly at its singular point $\ln 0$.

The figure shows both area curves (left) and the contrast between Body A's delta-spike $A^{\prime \prime }$ and Body B's smooth, bounded $A^{\prime \prime }$ (right).

Write the shape function $g(\xi )=[4\xi (1-\xi ){]}^{3/2}=[4\xi -4{\xi }^2{]}^{3/2}$, so $A=A_{\max }g(\xi )$. Where $f=0$: $f=U_{\infty }{A}^{\prime }(x)$, and $A^{\prime }=0$ where $g^{\prime }(\xi )=0$. $g^{\prime }(\xi )=\frac{3}{2}[4\xi -4{\xi }^2{]}^{1/2}(4-8\xi ).$ The bracket $(4-8\xi )=0$ gives $\xi =0.5$; the square-root factor gives the endpoints $\xi =0,1$. So the maximum sits at mid-body $\xi =0.5$, where $A^{\prime }=0\Rightarrow f=0$. Physical meaning: at the widest station the body is neither growing nor shrinking, so it neither injects nor removes streamtube volume — no source, no sink. Where $A^{\prime \prime }=0$ (inflection points): differentiate $g^{\prime }$ again. Writing $u=4\xi -4{\xi }^2$ so $u^{\prime }=4-8\xi$, $g^{\prime \prime }(\xi )=\frac{3}{2}\left[\frac{1}{2}{u}^{-1/2}(u^{\prime }{)}^2+u^{1/2}(-8)\right]=\frac{3}{4}{u}^{-1/2}[(4-8\xi {)}^2-16u].$ Set the bracket to zero (the $u^{-1/2}$ factor blows up only at the pointed ends): $(4-8\xi {)}^2-16(4\xi -4{\xi }^2)=0.$ Expand: $(4-8\xi {)}^2=16-64\xi +64{\xi }^2$ and $16(4\xi -4{\xi }^2)=64\xi -64{\xi }^2$, so $16-64\xi +64{\xi }^2-64\xi +64{\xi }^2=128{\xi }^2-128\xi +16=0\Rightarrow 8{\xi }^2-8\xi +1=0.$ Quadratic formula: $\xi =\frac{8\pm \sqrt{64-32}}{16}=\frac{8\pm \sqrt{32}}{16}=\frac{1}{2}(1\pm \frac{1}{\sqrt{2}})$, i.e. $\xi \approx 0.1464\text{and}\xi \approx \mathrm{0.8536.}$ These are the inflection points, where the area curve switches from concave-up (near the ends) to concave-down (near the middle). They are where the curve is "straightest," contributing least local curvature to $D_w$.

(a) Adding a Gaussian hump to a flat line gives a bump in $A_{\text{tot}}$: it starts flat at $0.8{\text{m}}^2$, climbs to a peak of $0.8+0.5=1.3{\text{m}}^2$ at $x_0=5$ m, then falls back to $0.8{\text{m}}^2$. That single rise-and-fall creates a region of large positive-then-negative curvature $A^{\prime \prime }$ — exactly what the von Kármán–Sears integral punishes. Your sketch should be a horizontal line with one smooth bump in the middle (the teal-plus-plum → orange "hump" panel of the figure below). (b) Pinch the fuselage by subtracting the same hump: $\Delta A_{\text{fus}}(x)=-0.5\exp \left[-{\left(\frac{x-x_0}{w}\right)}^2\right].$ Then $A_{\text{tot}}=A_{\text{fus}}+\Delta A_{\text{fus}}+A_{\text{wing}}=0.8{\text{m}}^2,$ flat again — $A^{\prime \prime }=0$ everywhere (figure, right panel). (In practice you cannot subtract all of it — the fuselage must stay big enough for payload — but you cancel as much curvature as structurally possible.) (c) Because $D_w$ depends only on $A_{\text{tot}}(x)$, cancelling the wing's hump against a fuselage dip flattens $A_{\text{tot}}^{\prime \prime }\to 0$ and drives wave drag toward the smooth-body minimum. This is Whitcomb's insight and the origin of the "wasp-waist" (see Sears–Haack body).

The three panels below walk left→middle→right: the separate parts, the bad naive total, and the good area-ruled total.

(a) The Mach angle is the half-angle of the cone of disturbances a supersonic body makes: $\mu =\arcsin \left(\frac{1}{M}\right)=\arcsin \left(\frac{1}{1.4}\right)=\arcsin (0.7143)=45.58^{\circ }.$ (b) Instead of cutting the body with planes perpendicular to the flow, you cut with Mach planes tilted at $\mu =45.6^{\circ }$ from the axis. Each such plane is tangent to the Mach cone, so it captures how disturbances actually propagate at $M>1$ (see Mach angle and Mach cone). Because a Mach plane can be tilted around the flight axis in any roll direction, you compute the "cut area" for many roll angles and average them — the flow's wave drag responds to that roll-averaged area distribution, which you then smooth as before.

Write the area law in nondimensional form. Let $\xi =x/\ell \in [0,1]$ and $A(x)=Sg(\xi )$, where $g$ is the fixed shape function (dimensionless, same for all members of the family) and $S$ is the peak-area scale (units ${\text{m}}^2$). Step 1 — rewrite $A^{\prime \prime }$. Since $x=\ell \xi$, each $d/dx=\frac{1}{\ell }d/d\xi$. Thus $A^{\prime \prime }(x)=\frac{S}{{\ell }^2}{g}^{\prime \prime }(\xi ).$ Why: two derivatives in $x$ pull out two factors of $1/\ell$. Step 2 — substitute into the integral. Change variables $x_i=\ell {\xi }_i$, so $d{x}_{1}d{x}_2={\ell }^{2}d{\xi }_{1}d{\xi }_2$, and $\ln |x_1-x_2|=\ln \ell +\ln |{\xi }_1-{\xi }_2|$. The $\ln \ell$ term integrates against $\iint g^{\prime \prime }{g}^{\prime \prime }d{\xi }_{1}d{\xi }_2$; but ${\int }_0^1{g}^{\prime \prime }(\xi )d\xi =g^{\prime }(1)-g^{\prime }(0)=0$ for a closed body (pointed ends, zero slope), so the $\ln \ell$ piece vanishes. What survives: $D_w=-\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }\cdot \underset{(A^{\prime \prime }{)}^2}{\underbrace{\frac{S^2}{{\ell }^4}}}\cdot \underset{d{x}_{1}d{x}_2}{\underbrace{{\ell }^2}}\iint g^{\prime \prime }({\xi }_1)g^{\prime \prime }({\xi }_2)\ln |{\xi }_1-{\xi }_2|d{\xi }_{1}d{\xi }_2.$ The last integral is a pure number $I$ fixed by the shape. So $D_w=-\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }I\frac{S^2}{{\ell }^2}=\left(\text{const}\right)q_{\infty }\frac{S^2}{{\ell }^2}.$ Step 3 — bring in volume. Volume $V={\int }_0^{\ell }Adx=S\ell {\int }_0^{1}gd\xi =cS\ell$ with $c$ a shape constant. Hence $S=V/(c\ell )$ and $S^2=V^2/(c^2{\ell }^2)$. Substitute: $D_w\propto q_{\infty }\frac{S^2}{{\ell }^2}=q_{\infty }\frac{V^2}{c^2{\ell }^2\cdot {\ell }^2}\propto \frac{q_{\infty }{V}^2}{{\ell }^4}.■$ The precise Sears–Haack constant is $128/\pi$, giving $D_w=128q_{\infty }{V}^2/(\pi {\ell }^4)$, consistent with the parent note.

Causal chain:

1. Wing added at mid-body ⇒ a hump in total $A(x)$.
2. Hump ⇒ large $A^{\prime \prime }$ (sharp curvature) at that station.
3. Large $A^{\prime \prime }$ ⇒ strong local flow acceleration/deceleration ⇒ strong shock waves (see Shock waves and wave drag).
4. Strong shocks ⇒ large wave drag near $M=1$ (the "transonic drag rise," see Transonic flow).
5. That drag wall ⇒ available thrust could not push the YF-102 through $M=1$. Area-ruling flattens $A(x)$ ⇒ small $A^{\prime \prime }$ ⇒ weak shocks ⇒ drag wall lowered ⇒ F-102A punches through.

Estimate: if $D_w\sim (A_{\text{peak}}^{\prime \prime }{)}^2$ and $A_{\text{peak}}^{\prime \prime }$ is halved, then $\frac{D_{w,\text{new}}}{D_{w,\text{old}}}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}.$ So wave drag drops to about 25% of its former value — consistent with the ~25–30% transonic drag reductions Whitcomb measured in the wind tunnel.