Worked examples — Area rule — Whitcomb's rule for transonic drag reduction

This page is the stress-test for the area rule. The parent note Area Rule gave you the machinery. Here we push it into every corner: smooth bodies, kinked bodies, degenerate (zero-length or zero-area) bodies, the perpendicular-cut $M\approx 1$ case and the tilted Mach-plane case, and a real aircraft. If a scenario can happen, it is worked below.

Before we start, one reminder in plain words, because every example leans on it:

We will also need two quantities the parent note used but this page should recall in plain words, because Examples 4, 5 and 7 lean on them:

Recall Two quantities from the parent:

$q_{\infty }$ and $V$ Dynamic pressure ::: $q_{\infty }=\frac{1}{2}{\rho }_{\infty }{U}_{\infty }^2$ — the "push per unit area" the moving air carries, in $\text{Pa}={\text{N/m}}^2$. Here ${\rho }_{\infty }$ is the free-stream air density and $U_{\infty }$ the flight speed. Body volume ::: $V={\int }_0^{\ell }A(x)dx$ — literally the sum of all the slice-areas along the length, in ${\text{m}}^3$. It is what the body must "make room for" in the air.

Everything below is one of these, or the master integral from the parent:

$$D_w=-\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }{\int }_0^{\ell }{\int }_0^{\ell }{A}^{\prime \prime }(x_1)A^{\prime \prime }(x_2)\ln |x_1-x_2|d{x}_{1}d{x}_2.$$

Written with dynamic pressure, the leading $\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }=\frac{q_{\infty }}{2\pi }$, since $q_{\infty }=\frac{1}{2}{\rho }_{\infty }{U}_{\infty }^2$.

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The scenario matrix

Think of "every scenario" this topic can throw at you as filling this grid. Each later example is tagged with the cell(s) it covers.

Cell	What makes it distinct	Covered by
C1 Smooth body, positive curvature everywhere	textbook Sears–Haack, $A^{\prime \prime }$ finite & smooth	Ex 1
C2 Kink in $A(x)$ (slope jump)	$A^{\prime \prime }$ becomes a spike (impulse) → drag blows up	Ex 2
C3 Superposition: fuselage + wing bump	the classic "Coke-bottle" fix, $A_{\text{tot}}=A_{\text{fus}}+A_{\text{wing}}$	Ex 3
C4 Degenerate: zero volume / zero length	limiting inputs — does the formula stay sane?	Ex 4
C5 Scaling limit: $\ell \to$ large at fixed $V$	limiting behaviour of $D_w\propto V^2/{\ell }^4$	Ex 5
C6 Speed regime split: $M\approx 1$ vs $M>1$	perpendicular cut vs tilted Mach plane $\mu =\arcsin (1/M)$	Ex 6
C7 Real-world word problem	YF-102 → F-102A, percentage drag drop	Ex 7
C8 Exam twist: sign/units trap	why $D_w$ from the integral comes out positive despite the minus	Ex 8

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Example 1 — C1: the smooth Sears–Haack slice (positive-curvature baseline)

Forecast: Guess before computing — is the maximum in the middle? Is $A(0.25)$ closer to half of $A_{\max }$ or much less?

Steps.

1. Find the maximum. The bracket $4\xi (1-\xi )$ is a downward parabola in $\xi$, peaking where $\xi (1-\xi )$ is largest, i.e. $\xi =\frac{1}{2}$. There $4\cdot \frac{1}{2}\cdot \frac{1}{2}=1$, so $A=A_{\max }\cdot 1^{3/2}=1{\text{m}}^2$. Why this step? The area rule wants a smooth, single-humped curve; the peak location tells you where to expect the widest fuselage, and it must sit mid-body for the Sears–Haack optimum.

2. Evaluate $\xi =0.25$. $4(0.25)(0.75)=0.75$, and $0.75^{3/2}=0.6495$. So $A=0.6495{\text{m}}^2$. Why this step? Shows the curve is not a triangle — at a quarter of the way in you're already at $65\%$ of peak area. Gentle, no kinks.

3. Check the ends. At $\xi =0$: $4\cdot 0\cdot 1=0$, $0^{3/2}=0$ → $A=0$. Same at $\xi =1$. Nose and tail taper to a point. Why this step? This is our first taste of a degenerate input — the ends are legitimate cross-sections of zero area, and the formula handles them cleanly.

Verify: Units: $A_{\max }$ in ${\text{m}}^2$, the bracket is dimensionless, so $A$ is ${\text{m}}^2$. Peak value $1{\text{m}}^2$ matches $A_{\max }$ exactly ✓. Symmetry: $A(\xi )=A(1-\xi )$, so $A(0.75)=A(0.25)=0.6495{\text{m}}^2$ — a fore-aft symmetric body, as Sears–Haack should be ✓.

The figure below plots this area law. Look at the blue curve: it is the smooth, single hump the area rule prizes — no corners anywhere. The yellow dot marks the mid-body peak from step 1; the green dot is $A(0.25)=0.6495{\text{m}}^2$ from step 2 (already two-thirds of the way up, not a straight-line ramp); the two red dots are the degenerate zero-area nose and tail from step 3. Read left to right and you are literally walking the ruler down the aircraft.

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Example 2 — C2: a kink turns $A^{\prime \prime }$ into a spike

Forecast: Guess — is $A^{\prime \prime }$ merely "large" at the join, or is it infinite?

Steps.

1. First derivative $A^{\prime }$. For $x<a$, $A^{\prime }=k$ (constant slope). For $x>a$, $A^{\prime }=0$. So $A^{\prime }$ jumps from $k$ down to $0$ at $x=a$ — a step. Why this step? $A^{\prime }=dA/dx$ is the rate the slice grows. A cone grows at constant rate; a cylinder not at all. The abrupt handover is the "kink."

2. Second derivative $A^{\prime \prime }$. The derivative of a step is a Dirac impulse (a delta spike $\delta$). So $A^{\prime \prime }(x)=-k\delta (x-a)$: zero everywhere except an infinitely tall spike at the join. Why this step? The parent's chain says strong $A^{\prime \prime }$ ⇒ strong shock ⇒ drag. An impulse is the most violent possible $A^{\prime \prime }$.

3. Feed it to the drag integral. With $A^{\prime \prime }\propto \delta (x-a)$, the double integral hits the $\ln |x_1-x_2|$ kernel right at $x_1=x_2=a$ where $\ln 0=-\infty$. The self-correlation diverges — the linearized model literally predicts unbounded drag from a slope discontinuity. Why this step? This is the mathematical scream that says "never kink your area curve." Real air can't produce infinite drag, but it produces a strong shock, which is exactly what we want to avoid.

Verify: Sanity via smoothing. Round the corner over a small length $\epsilon$: then $A^{\prime }$ ramps down over $\epsilon$ instead of jumping, so $A^{\prime \prime }\sim k/\epsilon$ (large but finite). As $\epsilon \to 0$, $A^{\prime \prime }\to \infty$ — confirming the delta limit and confirming drag grows without bound as the corner sharpens ✓. Units check: $[A^{\prime }]={\text{m}}^2/\text{m}=\text{m}$, $[A^{\prime \prime }]={\text{m}}^2/{\text{m}}^2=$ dimensionless density, consistent with $k$ having units $\text{m}$ ✓.

The three stacked panels below tell the story top-to-bottom. Top (blue): $A(x)$ ramps up then flattens — the yellow arrow points at the corner. Middle (green): its slope $A^{\prime }$ jumps abruptly from $k$ to $0$ (step 1). Bottom (red): the second derivative $A^{\prime \prime }$ is zero everywhere except the tall red arrow — the delta spike of step 2. Trace one vertical line through all three panels at $x=a$ and you see how a barely-visible corner in $A$ becomes an infinite spike in $A^{\prime \prime }$.

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Example 3 — C3: superposition, the Coke-bottle cancellation

Forecast: Guess — must the fuselage dip by exactly the wing bump, more, or less?

Steps.

1. Write the total. $A_{\text{tot}}(x)=A_{\text{fus}}(x)+A_{\text{wing}}(x)$. The area rule cares only about this sum, not the split. Why this step? This is the whole theorem in one line — two shapes with identical $A_{\text{tot}}$ have identical slender-body wave drag.

2. Demand smoothness. For no bump at the wing station, require $A_{\text{tot}}$ there to equal the surrounding tube value $1.00{\text{m}}^2$: $A_{\text{fus}}^{\text{pinched}}+0.30=1.00\Rightarrow A_{\text{fus}}^{\text{pinched}}=0.70{\text{m}}^2.$ Why this step? The fuselage must shed exactly the wing's added area — a $0.30{\text{m}}^2$ dip. Cancel the hump, cancel its curvature $A^{\prime \prime }$, cancel the drag it caused.

3. Interpret the pinch. The fuselage cross-section drops from $1.00$ to $0.70{\text{m}}^2$, i.e. a $30\%$ area reduction at that station — the visible "wasp waist." Why this step? Ties the abstract cancellation to the physical Coke-bottle shape the parent note describes.

Verify: $A_{\text{tot}}=0.70+0.30=1.00{\text{m}}^2$ — flat across the wing region, so $A_{\text{tot}}^{\prime }=0$ and $A_{\text{tot}}^{\prime \prime }=0$ there. No curvature spike, no wing-induced shock ✓. The fractional pinch $0.30/1.00=30\%$ matches the "~25–30% drag reduction" era numbers as an order of magnitude ✓.

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Example 4 — C4: degenerate inputs (zero volume, zero length)

Forecast: Guess — which degenerate case gives zero drag and which blows up?

Steps.

0. Where does the $128/\pi$ come from? It is not magic — it is what you get after doing the calculus-of-variations minimization the parent flagged. Feed the Sears–Haack area law $A(x)=A_{\max }[4\xi (1-\xi ){]}^{3/2}$ into the von Kármán–Sears double integral and evaluate it. Two pieces combine: (i) the volume of that specific shape works out to $V=\frac{3\pi }{16}{A}_{\max }\ell$, so $A_{\max }=\frac{16V}{3\pi \ell }$; (ii) the $\iint A^{\prime \prime }{A}^{\prime \prime }\ln |\cdot |$ integral for this smooth curve evaluates to a pure number times $A_{\max }^2/{\ell }^2$. Substituting $A_{\max }$ in terms of $V$ and collecting the leading $\frac{\rho U^2}{4\pi }=\frac{q_{\infty }}{2\pi }$ leaves exactly $\frac{128q_{\infty }{V}^2}{\pi {\ell }^4}$. So $128/\pi$ is the fingerprint of this optimal shape, nothing more. Why this step? Learners should see the constant is earned by the minimization, not handed down; every other body gives a bigger number in front of $q_{\infty }{V}^2/{\ell }^4$.

1. Zero-volume body ($V=0$). Then $D_w=\frac{128q_{\infty }\cdot 0}{\pi {\ell }^4}=0$. Why this step? A body of zero volume is a needle/streamline with no cross-section to shove air around. No displaced volume ⇒ no source strength $f=U_{\infty }{A}^{\prime }=0$ ⇒ no waves ⇒ no wave drag. The formula agrees.

2. Vanishing length ($\ell \to 0$, $V$ fixed). $D_w\propto \frac{1}{{\ell }^4}\to \infty$. Why this step? Cramming a fixed volume into zero length forces the area to spike over a tiny distance — enormous $A^{\prime \prime }$. The formula correctly predicts unbounded wave drag: you cannot make a fat body infinitely short and stay slender.

3. Boundary of validity. Case (b) is exactly where slender-body theory dies: it assumed thickness $\ll$ length. The $1/{\ell }^4$ growth is the theory politely blowing up at its own edge. Why this step? Every model has a domain; naming it is part of "covering every scenario."

Verify: Dimensional check on $D_w=128q_{\infty }{V}^2/(\pi {\ell }^4)$: $[q_{\infty }]=\text{Pa}={\text{N/m}}^2$, $[V^2]={\text{m}}^6$, $[{\ell }^4]={\text{m}}^4$, so $[D_w]=\frac{\text{N}}{{\text{m}}^2}\cdot \frac{{\text{m}}^6}{{\text{m}}^4}=\text{N}$ — a force ✓. Monotonic: $D_w$ increases as $V$ up or $\ell$ down, both physically correct ✓.

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Example 5 — C5: the length-scaling limit

Forecast: Guess the drag ratio for the $2\times$ length case, and whether the $1\%$ target needs roughly a $2\times$, $3\times$, or $4\times$ longer body.

Steps.

1. Double-length drag ratio. $\frac{D_{w,2}}{D_{w,1}}=\frac{{\ell }_1^4}{{\ell }_2^4}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}=0.0625$. Why this step? $D_w\propto V^2/{\ell }^4$ with $V,q_{\infty }$ fixed leaves only the ${\ell }^{-4}$ factor. Doubling length ⇒ $2^4=16$-fold drop.

2. Solve for the $1\%$ target. Want $\frac{{\ell }_1^4}{{\ell }_3^4}=0.01$, so ${\left(\frac{{\ell }_3}{{\ell }_1}\right)}^4=100$, giving $\frac{{\ell }_3}{{\ell }_1}=100^{1/4}=\sqrt{10}\approx 3.162$. Why this step? Shows the steep payoff and its limit: slenderness buys huge drag cuts, but the fourth-power means you need only ~$3.16\times$ length for a $100\times$ drag cut — geometry rewards length dramatically.

3. Reading the limit. As $\ell \to \infty$ at fixed $V$, $D_w\to 0$: the ideal slender needle. Real limits (weight, structure, friction) stop you, but wave drag alone vanishes. Why this step? Names the limiting behaviour cell explicitly.

Verify: $2^4=16$ ⇒ ratio $1/16=0.0625$ ✓. $(\sqrt{10}{)}^4=10^2=100$ ⇒ drag ratio $1/100=0.01$ ✓. Both consistent with $D_w\propto {\ell }^{-4}$.

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Example 6 — C6: perpendicular cut vs tilted Mach-plane cut

Forecast: Guess $\mu$ at $M=2$ — bigger or smaller than $45^{\circ }$? And does the tilt make the equivalent body look longer or shorter in $x$?

Steps.

1. Why an angle at all? In Transonic flow at $M\approx 1$, disturbances spread nearly straight sideways, so the "footprint" a wave feels is the ordinary perpendicular cross-section. Above $M=1$, every disturbance is swept back into a Mach cone; the plane the wave actually "sees" is tilted. Why this step? Justifies switching tools — we don't change the physics, we change the cutting plane to match how signals propagate.

2. Compute $\mu$ at $M=2$. $\mu =\arcsin (1/M)=\arcsin (0.5)=30^{\circ }$. Why this step? $\arcsin$ answers "which angle has this sine?" Here $\sin \mu =1/M=1/2$, and the angle with sine $\frac{1}{2}$ is $30^{\circ }$. Smaller than $45^{\circ }$, so at higher $M$ the cone is narrower and the cut is more tilted toward the flow direction.

3. Effect on the area law. Tilting the cut and averaging over roll angle stretches and smears the area contributions along $x$ (a station's area gets projected forward/back). The equivalent body is longer and its area curve smoother — which is why supersonic designs are area-ruled for a chosen Mach number, not just $M=1$. Why this step? Connects the geometric tilt to a different $A(x)$, hence different $D_w$ — the whole reason for a separate supersonic rule.

Verify: $M\to 1^{+}$ gives $\mu =\arcsin (1)=90^{\circ }$ — the Mach plane becomes perpendicular, smoothly recovering case (a) ✓. $M\to \infty$ gives $\mu =\arcsin (0)=0^{\circ }$ — the cone flattens onto the axis ✓. At $M=2$, $\sin (30^{\circ })=0.5=1/M$ ✓.

The curve below plots the Mach angle $\mu$ against Mach number $M$. Follow the blue curve from the left: near $M=1$ it starts up at $90^{\circ }$ (the green arrow) — that is the perpendicular cut of case (a), so the transonic rule is just the supersonic rule's endpoint. The yellow dot marks our answer $\mu =30^{\circ }$ at $M=2$, sitting below the red dashed $45^{\circ }$ line, confirming the cone narrows as you fly faster. Reading further right, $\mu$ keeps dropping toward $0^{\circ }$, the hypersonic limit where disturbances lie almost along the axis.

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Example 7 — C7: real-world word problem (YF-102 → F-102A)

Forecast: Guess — is the total drag drop bigger or smaller than $28\%$, and why?

Steps.

1. New wave drag. Remove $28\%$: $C_{D,w}^{\text{new}}=0.0200\times (1-0.28)=0.0200\times 0.72=0.0144$. Why this step? Area ruling attacks only the wave-drag piece (parent mistake #3), so we scale just that term and leave the rest alone.

2. New total. Friction+induced is untouched at $0.0300$. So $C_D^{\text{new}}=0.0300+0.0144=0.0444$. Why this step? Drag breakdown is additive; you only edited the one component that area ruling touches, then re-add the parts it cannot touch.

3. Percentage total drop. $\frac{0.0500-0.0444}{0.0500}=\frac{0.0056}{0.0500}=0.112=11.2\%$. Why this step? Shows the real-world lesson: a $28\%$ wave-drag cut becomes a smaller total cut because friction and induced drag ride along unchanged — yet $11.2\%$ was still enough to push the F-102A past Mach 1 when the un-ruled YF-102 could not.

Verify: $0.02\times 0.72=0.0144$ ✓; $0.03+0.0144=0.0444$ ✓; $(0.05-0.0444)/0.05=0.112$ ✓. The total drop ($11.2\%$) is less than the wave-drag drop ($28\%$) exactly because wave drag is only $0.02/0.05=40\%$ of the total, and $0.28\times 0.40=0.112$ ✓.

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Example 8 — C8: exam twist — why the minus sign gives positive drag

Forecast: Guess — do stations closer together or farther apart contribute more drag through the $\ln$ kernel?

Steps.

1. Handle the sign of $\ln$. For $|x_1-x_2|<1$ (measuring separation in the body's own length units), $\ln |x_1-x_2|<0$. Multiply by the leading minus $-\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }$ and you get a positive contribution from those close-station pairs. Why this step? The scary minus sign is doing bookkeeping, not producing thrust; combined with the negative log it yields $D_w>0$ as physics demands.

2. Probe with two curvature impulses. Model $A^{\prime \prime }$ as two spikes; the kernel value is $\propto -\ln (\text{separation})$. At separation $s$: weight $-\ln s$. At $2s$: weight $-\ln (2s)=-\ln s-\ln 2$. Since $-\ln 2\approx -0.6931$, the $2s$ case has a smaller $-\ln$ weight. Why this step? It shows spreading the curvature out (bigger separation) lowers drag — the mathematical reason smooth, gently varying $A(x)$ wins, and abrupt back-to-back area changes lose.

3. Tie to design. Closely bunched area changes (a kink, or wing area piled where the fuselage is already fat) correlate strongly through the log kernel ⇒ high drag; a body that spreads its area changes over its whole length minimizes the self-correlation. Why this step? Recovers the area-rule design principle straight from the integral's sign structure.

Verify: $-\ln 2=-0.6931$; so relative to separation $s$, doubling to $2s$ subtracts $0.6931$ from the (positive-drag) weight — less drag at wider spacing ✓. Sign check: with ${\rho }_{\infty },U_{\infty }^2>0$ and typical negative $\ln$ over the interacting region, $-(\text{neg})=\text{positive}$, so $D_w>0$ ✓.

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Recall Did every cell get covered?

C1 Ex1 ::: smooth Sears–Haack baseline C2 Ex2 ::: kink → delta-spike $A^{\prime \prime }$ C3 Ex3 ::: fuselage+wing superposition (Coke-bottle) C4 Ex4 ::: zero-volume and zero-length degenerates, plus origin of the $128/\pi$ prefactor C5 Ex5 ::: ${\ell }^{-4}$ scaling limit C6 Ex6 ::: perpendicular vs Mach-plane cut, $\mu =\arcsin (1/M)$ C7 Ex7 ::: real F-102A drag-count word problem C8 Ex8 ::: sign/units exam twist