Visual walkthrough — Area rule — Whitcomb's rule for transonic drag reduction

We lean on the parent note, on Slender-body theory, on Shock waves and wave drag, and on the Sears–Haack body as the finish line.

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Step 0 — The speed measure we need: Mach number and Mach angle

WHAT. Every claim below depends on how fast we fly compared to the speed of sound. Define the Mach number

$$M_{\infty }=\frac{U_{\infty }}{a_{\infty }},$$

• $U_{\infty }$ ::: the free-stream speed of the aircraft through the air.
• $a_{\infty }$ ::: the speed of sound in that undisturbed air far away.
• $M_{\infty }$ ::: their ratio — how many "sound-speeds" fast we are going. $M_{\infty }=1$ is exactly the speed of sound; $M_{\infty }>1$ is supersonic; $M_{\infty }\approx 1$ is transonic (see Transonic flow).

WHY. When $M_{\infty }>1$, no disturbance can outrun the aircraft, so every little source's influence is trapped inside a backward-swept cone — the Mach cone. Its half-angle, the Mach angle $\mu$, is set by simple geometry (see Mach angle and Mach cone):

$$\mu =\arcsin \left(\frac{1}{M_{\infty }}\right),\tan \mu =\frac{1}{\sqrt{M_{\infty }^2-1}}.$$

• $\mu$ ::: the tilt of the Mach cone — steep (near $90^{\circ }$) just above $M_{\infty }=1$, flattening as $M_{\infty }$ grows.
• $\sqrt{M_{\infty }^2-1}$ ::: the crucial combination that measures "how supersonic" we are; it appears everywhere below. Note it vanishes as $M_{\infty }\to 1$ — a warning flag we return to in Step 6.

PICTURE. The Mach cone trailing a source, with half-angle $\mu$, and how it flattens as $M_{\infty }$ increases.

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Step 1 — What a "cross-section" even is

WHAT. Take the aircraft (or any long thin body). Point it along an arrow we call the flight axis. Now slide a flat sheet of glass down that axis, always kept perpendicular to it. Where the glass slices the body, it cuts out a flat 2-D shape — a disc, an oval, a wing-plus-fuselage blob. The area of that slice is one number. Call the position of the glass $x$ (distance from the nose), and call the slice area $A(x)$.

WHY. Everything in the area rule is a statement about this one function $A(x)$. Before we can say "drag depends on how $A$ changes," we must be crystal clear that $A(x)$ is just: pick a distance $x$, slice, measure the area you cut.

PICTURE. The red slice below is at one value of $x$; the number $A(x)$ is the red area. Slide the slice and you trace the whole body.

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Step 2 — The air only "sees" $A(x)$ (the slender-body idea)

WHAT. Imagine the body is thin — long compared to how fat it ever gets. As air flows past, each thin ring of air is pushed outward just enough to slide around the body. How much outward push? Exactly enough to make room for the body's growing thickness.

WHY. Here is the key claim we will make precise: a ring of air does not know whether the width it must flow around came from a wing or from a fuselage. It only feels the total area it must dodge at that station. So two aircraft with the same $A(x)$ shove the air identically — and therefore make the same waves. This is Slender-body theory in one sentence.

PICTURE. Below: two very different bodies (a fat tube, and a thin tube with wings) sliced at the same $x$. If their red slice areas are equal, the air is pushed out by the same amount at that station.

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Step 3 — Turning "make room" into a source (defining $f(x)$)

WHAT. To model "the body makes room for itself," we replace the solid body with a line of tiny air-emitters running down the axis. Each emitter, called a source, blows air outward. A source of strength $f(x)$ blows out a volume of air per second, per unit length of axis.

WHY this tool. We cannot easily solve flow around a weird solid shape. But flow away from a point source is one of the simplest flows there is (air spreading outward, symmetric). By stacking sources along the axis and choosing their strengths, we can reproduce the exact outward-push the body demands — without ever solving the hard solid-boundary problem. This substitution is the whole trick of Slender-body theory.

How strong must the source be? Over a slice of length $dx$, the body's area grows from $A$ to $A+dA$. The oncoming air arrives at speed $U_{\infty }$ (free-stream speed). In one second, a column of air of length $U_{\infty }$ passes; the extra volume it must be shoved aside to accommodate is $U_{\infty }dA$. Per unit length that is:

$$f(x)=U_{\infty }\frac{dA}{dx}.$$

Term by term:

• $f(x)$ ::: source strength at station $x$ — volume of air emitted per second per unit axis-length.
• $U_{\infty }$ ::: free-stream speed of the oncoming air (subscript $\infty$ = "far away, undisturbed").
• $\frac{dA}{dx}$ ::: the rate of change of area with distance — "how fast the body is fattening (or thinning) right here."

PICTURE. Where the body fattens, $dA/dx>0$ → sources blow out (red, positive). Where it slims, $dA/dx<0$ → sources suck in (a negative source, a "sink"). Where area is constant, $f=0$: no disturbance at all.

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Step 4 — Waves carry away momentum, and momentum lost = drag

WHAT. At supersonic speed each source doesn't push air smoothly outward — its influence is trapped along the Mach cones of Step 0 and radiates as pressure waves / shocks (see Shock waves and wave drag). These waves stream backward off the body and never come back. They carry momentum away.

WHY. Drag is, by Newton, the rate at which the body loses forward momentum to the air. If waves leave carrying backward-momentum "debt," the body must supply it — that supply is the wave drag $D_w$. So: stronger, more abrupt sources → stronger waves → more momentum bled off → more drag.

PICTURE. A smoothly changing body (gentle sources) sheds weak, spread-out waves. A body with a sudden fattening (a violent source spike) sheds a strong shock. Same air, very different momentum loss.

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Step 5 — Abruptness is curvature: enter $A^{\prime \prime }(x)$

WHAT. "How abruptly the source strength changes" is the rate of change of $f(x)$. Since $f=U_{\infty }dA/dx$, its rate of change is $U_{\infty }\frac{d}{dx}\left(\frac{dA}{dx}\right)$. The thing in the middle, "the slope of the slope," is the second derivative, written

$$A^{\prime \prime }(x)=\frac{d}{dx}\left(\frac{dA}{dx}\right)=\frac{d^{2}A}{d{x}^2}.$$

• $A^{\prime \prime }(x)$ ::: the curvature of the area curve — how fast the slope $dA/dx$ is itself changing.

WHY this tool. The waves are launched by changes in the source strength, not by the source strength itself. A body that fattens at a constant rate ($dA/dx$ constant, so $A^{\prime \prime }=0$) launches no new disturbance — the sources are all equal, nothing changes downstream. It is only where the fattening-rate changes — where $A^{\prime \prime }\ne 0$ — that fresh waves are thrown off. That is why the second derivative, not the first, is the drag driver.

PICTURE. Three stacked curves for the same body: $A(x)$ on top, its slope $A^{\prime }(x)$ in the middle, its curvature $A^{\prime \prime }(x)$ on the bottom (red). Notice a kink in $A(x)$ (from a wing bolted on) becomes a step in $A^{\prime }$ and an enormous spike in $A^{\prime \prime }$.

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Step 6 — Adding up the waves: the von Kármán–Sears integral

WHAT. Every little piece of the axis at position $x_1$ launches a wave of strength $\propto A^{\prime \prime }(x_1)$. That wave sweeps back along its Mach cone and interacts with the wave from another piece at $x_2$ (strength $\propto A^{\prime \prime }(x_2)$). Their combined momentum loss depends on both strengths and on how far apart the two pieces are, $|x_1-x_2|$. Summing every pair of pieces gives the supersonic slender-body wave drag:

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

Reading it symbol by symbol:

• $D_w$ ::: the wave drag we are computing (a force).
• ${\rho }_{\infty }$ ::: free-stream air density; $U_{\infty }$ ::: free-stream speed. Together ${\rho }_{\infty }{U}_{\infty }^2$ sets the scale of dynamic pressure — how "punchy" the flow is.
• $\frac{1}{4\pi }$ ::: a geometric constant from spreading in the flow (don't fear it — it's just bookkeeping).
• ${\int }_0^{\ell }{\int }_0^{\ell }\cdots d{x}_{1}d{x}_2$ ::: the double sum over every pair of stations $(x_1,x_2)$ along the length — the first integral picks $x_1$, the second picks $x_2$, and we add up all combinations.
• $A^{\prime \prime }(x_1)A^{\prime \prime }(x_2)$ ::: strength of the wave from station $x_1$ times strength from station $x_2$ — a self-correlation of the curvature.
• $\ln \frac{|x_1-x_2|}{\ell }$ ::: the natural logarithm of the separation divided by the body length $\ell$. Dividing by $\ell$ makes the argument a pure number (dimensionless) — you cannot take the log of "so many metres." The choice of reference length only shifts $D_w$ by a term proportional to $(\int A^{\prime \prime }dx{)}^2=(A^{\prime }(\ell )-A^{\prime }(0){)}^2$, which vanishes for a closed body (pointed nose and tail), so the answer is well-defined.

WHY this exact form. The two sources interact through the linearized supersonic influence (a Green's function). The Mach cone geometry enters through the factor $\sqrt{M_{\infty }^2-1}$: after aligning coordinates along the cone this factor is absorbed into the effective streamwise length, and for the leading term it drops out of the shape of the integral — leaving the logarithm of the separation. This is why the classic form looks $M_{\infty }$-free even though the geometry is Mach-cone geometry.

PICTURE. The double integral drawn as a grid over the square $[0,\ell ]\times [0,\ell ]$. Each cell is one pair $(x_1,x_2)$; we colour it by $A^{\prime \prime }(x_1)A^{\prime \prime }(x_2)$. A body with a kink lights up bright hot cells (red) → big total → big drag. A smooth body stays cool everywhere.

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Step 7 — Edge cases: constant, linear, and kinked bodies

WHAT. Let's test the formula on the extreme cases so no scenario surprises the reader.

Case (a) — constant area (a plain cylinder, $A(x)=\text{const}$). Then $dA/dx=0$ and $A^{\prime \prime }=0$ everywhere. The integrand is $0\cdot 0$ → $D_w=0$. A parallel tube (in this idealization) makes no wave drag: nothing is changing.

Case (b) — linearly growing area ($A(x)$ a straight ramp). Then $dA/dx=$ const, so $A^{\prime \prime }=0$ in the interior. But at the two ends the slope suddenly turns on and off — kinks — producing spikes in $A^{\prime \prime }$ there. Drag comes entirely from the nose and tail corners. Lesson: round the corners.

Case (c) — bolted-on wing (a kink in $A$). As in Step 5, a kink → a delta-spike in $A^{\prime \prime }$. In the grid picture (Step 6) that spike lights up an entire hot row and column → large drag. This is the villain the Coke-bottle fuselage exists to erase.

WHY show all three. They map the whole landscape: $A^{\prime \prime }=0$ is free, corners cost a little, kinks cost a lot. The design goal writes itself: kill $A^{\prime \prime }$ spikes; spread curvature out gently.

PICTURE. The three area curves side by side with their $A^{\prime \prime }$ underneath (red). Watch the $A^{\prime \prime }$ go from flat (free) → two end-spikes (cheap) → one tall interior spike (expensive).

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Step 8 — The winner: smoothest possible curve = Sears–Haack

WHAT. Ask the optimization question: over all area curves $A(x)$ with a fixed length $\ell$ and fixed volume $V={\int }_0^{\ell }A(x)dx$ (from Step 1), which one makes the double integral smallest? The answer (a calculus-of-variations result) is the Sears–Haack body:

$$A(x)=A_{\max }[4\xi (1-\xi ){]}^{3/2},\xi =\frac{x}{\ell }\in [0,1].$$

• $\xi$ ::: fractional position, $0$ at nose, $1$ at tail.
• $A_{\max }$ ::: the largest cross-sectional area anywhere on the body, $A_{\max }={\max }_{x}A(x)$. It occurs at the middle ($\xi =\frac{1}{2}$), where the bracket equals $1$.
• $4\xi (1-\xi )$ ::: a smooth hump, zero at both ends, peaking at the middle.
• power $3/2$ ::: sharpens the nose/tail to zero gently, so $A^{\prime \prime }$ never spikes.

Its minimum wave drag, for fixed volume, is

$$D_w=\frac{128q_{\infty }{V}^2}{\pi {\ell }^4},q_{\infty }=\frac{1}{2}{\rho }_{\infty }{U}_{\infty }^2.$$

• $q_{\infty }$ ::: dynamic pressure, the flow's "push."
• $V={\int }_0^{\ell }A(x)dx$ ::: the fixed total volume being carried.
• $V^2/{\ell }^4$ ::: the scaling — more volume hurts (squared); more length helps enormously (fourth power).

WHY. This is the concrete target Whitcomb tells you to aim your total $A(x)$ at. The fuselage is pinched near the wings precisely so the sum fuselage+wing+tail traces this gentle hump instead of a lumpy one.

PICTURE. The Sears–Haack area curve (red) laid over a lumpy "components-added" curve. Same volume, same length — but the smooth one has small $A^{\prime \prime }$ everywhere.

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The one-picture summary

Everything on this page compressed: choose a speed (Mach $M_{\infty }$) → slice the body → get $A(x)$ (its volume $V=\int Adx$) → its slope $A^{\prime }$ → its curvature $A^{\prime \prime }$ → feed pairs of $A^{\prime \prime }$ into the double integral → out comes $D_w$. Smooth $A(x)$ ⇒ tiny $A^{\prime \prime }$ ⇒ tiny drag; a kink ⇒ spike ⇒ big drag.

Recall Feynman retelling — say it back in plain words

First fix how fast you fly: the Mach number $M_{\infty }$, your speed over the speed of sound. Above $M_{\infty }=1$ nothing you disturb can get ahead of you, so every little push is trapped inside a backward cone — the Mach cone, tilted at the Mach angle $\mu$. Now point the plane along an arrow. Slide a sheet of glass down the arrow; where it cuts the plane, measure the area. That gives one curve: area versus distance, $A(x)$; add up all the slices and you get the volume $V=\int Adx$. The air rushing by has to be shoved outward just enough to make room for the body getting fatter — so I replace the body with a row of little blowers whose strength is how fast the area is growing. Blowers all blowing the same amount make no waves; it's when the blowing rate changes that waves get thrown off. "How fast the blowing rate changes" is the curvature of my area curve, $A^{\prime \prime }$. Those waves race backward along their Mach cones and never return, carrying momentum away — and lost momentum is exactly drag. To total it up I let every point's wave interact with every other point's wave: that's the double integral, weighted by the log of their scaled separation (scaled by the length $\ell$ so the log has a plain number inside). It depends on nothing but $A^{\prime \prime }$, which comes from nothing but $A(x)$ — so two totally different-looking aircraft with the same area curve have the same wave drag. This clean formula is the supersonic slender-body limit; right at $M_{\infty }\approx 1$ the linear theory strains, but the design lesson survives: want low drag, want a smooth area curve, and the smoothest one is the Sears–Haack shape. That's why designers pinch the fuselage where the wings add area. And since drag scales as $1/{\ell }^4$, making the whole thing longer is the biggest win of all.

Recall Quick self-test

Why does a constant-area cylinder have zero wave drag here? ::: Because $A^{\prime \prime }=0$ everywhere, so every term in the integral is zero — nothing about the flow is changing. Why $A^{\prime \prime }$ and not $A$ or $A^{\prime }$? ::: Waves are launched by changes in source strength; source strength $\propto A^{\prime }$, so its change $\propto A^{\prime \prime }$ — the curvature. What does the double integral physically represent? ::: Every station's wave interacting with every other station's wave, weighted by $\ln$ of their (length-scaled) separation — a self-correlation of the curvature. What is $V$ in the Sears–Haack drag formula? ::: The body's total volume, $V={\int }_0^{\ell }A(x)dx$. Why divide inside the $\ln$ by $\ell$? ::: A logarithm needs a dimensionless argument; $|x_1-x_2|/\ell$ is a pure number, and the choice of reference length cancels for a closed body. Is the boxed integral exact at $M_{\infty }=1$? ::: No — it is the linearized supersonic ($M_{\infty }>1$) slender-body result; at transonic $M_{\infty }\approx 1$ linear theory breaks down, though the smooth-$A(x)$ design rule still holds.

Related: Transonic flow, Prandtl–Glauert and compressibility corrections, Mach angle and Mach cone.