Worked examples — Airfoil aerodynamics — camber, chord, thickness

Before anything, three tiny plain-language reminders so no symbol is unearned:

• $x$ is the distance from the nose, measured along the chord. At the nose $x=0$, at the tail $x=c$.
• $c$ is the chord length — the straight nose-to-tail ruler. When we write "$0.4c$" we mean "four tenths of the way back."
• A percent like "$2\%$ camber" just means the fraction $0.02$. So $2\%$ of a $1$-metre chord is $0.02\text{m}=2\text{cm}$.

---

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below are labelled with the cell(s) they cover.

Cell	Case class	What makes it tricky
A	Read a NACA 4-digit name → geometry	decode digits, convert $\%$ to real length
B	Zero camber (symmetric, e.g. 0012)	degenerate: $y_c=0$, $\theta =0$ everywhere
C	Nonzero camber, build a surface point	full $\sin \theta /\cos \theta$ offset, both surfaces
D	Thin-airfoil limit ($\theta \to 0$)	check the "$\approx$" collapse is valid
E	Zero-lift angle ${\alpha }_{L0}$ from camber	sign of ${\alpha }_{L0}$, the integral shortcut
F	Lift line: solve for $c_{\ell }$ or invert for $\alpha$	radians vs degrees, negative $\alpha$
G	Real-world word problem	pick chord as reference, Reynolds link
H	Exam twist / trap	flat plate, inverted flight, "which is bigger"

---

Example 1 — Reading the name (Cell A)

Forecast: guess now — is $4415$ thicker or thinner than the $2412$ from the parent note? Is it more or less cambered?

1. Split the digits: $4$, $4$, $15$. Why this step? The 4-digit code is literally (camber $\%$, position in tenths of chord, thickness $\%$). Splitting is decoding the sentence.
2. Max camber $=4\%=0.04$ of chord $=0.04\times 2=0.08\text{m}=8\text{cm}$. Why this step? The first digit is a percent of $c$; convert the fraction to a real length using the given $c$.
3. Position of max camber $=$ second digit in tenths $=4\times 0.1=0.4c=0.8\text{m}$ from the nose. Why this step? One digit means "tenths of chord," so $4\to 0.4c$.
4. Max thickness $=15\%=0.15\times 2=0.30\text{m}=30\text{cm}$. Why this step? The last two digits are one number, the thickness percent.

Verify: Compared with $2412$ ($2\%$ camber, $12\%$ thick): $4415$ has double the camber and more thickness — matches the forecast that it's a fatter, more strongly curved section. Units: all lengths in metres, all consistent with $c=2\text{m}$. ✓

---

Example 2 — The symmetric / degenerate case (Cells B, D)

Forecast: with zero camber, what is the slope angle $\theta$ of the mean line? What does that do to $\sin \theta$ and $\cos \theta$?

1. Zero camber ⇒ mean camber line $y_c(x)=0$ everywhere. Why this step? Digits $00$ mean max camber $=0$, so the skeleton is the chord line.
2. Slope of the mean line $\frac{d{y}_c}{dx}=0$, so $\theta =\arctan (0)=0$. Why this step? $\theta$ is defined as $\arctan (d{y}_c/dx)$; a flat line has zero slope, and $\arctan (0)=0$ rad.
3. Therefore $\sin \theta =\sin 0=0$ and $\cos \theta =\cos 0=1$. Why this step? These are the exact factors in the surface formula; knowing them lets the formula simplify with no approximation.
4. Plug into the surface formula: $x_U=x-y_t\sin \theta =x,y_U=y_c+y_t\cos \theta =+y_t,$ $x_L=x+y_t\sin \theta =x,y_L=y_c-y_t\cos \theta =-y_t.$ Why this step? This shows the offset happens purely vertically — the surfaces sit at $\pm y_t$ about the chord.

Look at the figure: both surface points share the same $x$, mirror images across the flat chord.

Verify: Upper and lower $y$ are $+y_t$ and $-y_t$ — equal magnitude, opposite sign ⇒ perfectly symmetric ⇒ camber $=0$. Since $\theta =0$ exactly here, this is also the exact case of the thin-airfoil collapse (Cell D): no approximation was needed. ✓

---

Example 3 — Building a real surface point with camber (Cell C)

Forecast: the upper point — will its $x$ be a little less than $0.25$ or a little more? (Hint: the offset is perpendicular to a line that tilts up.)

1. Find the slope angle: $\theta =\arctan (0.20)=0.1974\text{rad}(\approx 11.3^{\circ })$. Why this step? We need $\theta$ so we can move along the normal to the tilted mean line, not straight up. We use $\arctan$ because it answers "which angle has this slope?"
2. Compute the trig factors: $\sin \theta =0.1961$, $\cos \theta =0.9806$. Why this step? The unit normal to the mean line is $(-\sin \theta ,\cos \theta )$; these numbers are the normal direction.
3. Upper surface: $x_U=0.25-(0.05)(0.1961)=0.2402,$ $y_U=0.03+(0.05)(0.9806)=\mathrm{0.0790.}$ Why this step? Move a distance $y_t$ from the skeleton point $(0.25,0.03)$ along $(-\sin \theta ,\cos \theta )$.
4. Lower surface (opposite normal): $x_L=0.25+(0.05)(0.1961)=0.2598,$ $y_L=0.03-(0.05)(0.9806)=-\mathrm{0.0190.}$ Why this step? The lower point uses the negative normal $(+\sin \theta ,-\cos \theta )$.

Verify: The straight-line distance between the two surface points should equal the full thickness $2y_t=0.10$: $\sqrt{(x_U-x_L{)}^2+(y_U-y_L{)}^2}=\sqrt{(-0.0196{)}^2+(0.0980{)}^2}=\mathrm{0.10.}✓$ Also $x_U<0.25<x_L$ — the upper point shifted forward, exactly the forecast, because tilting up rotates "straight up" toward the nose. ✓

---

Example 4 — The thin-airfoil approximation error (Cell D)

Forecast: will the approximate $y_U=0.08$ be too big or too small compared with the exact $0.0790$?

1. Exact value (Example 3): $y_U=y_c+y_t\cos \theta =0.03+0.05(0.9806)=0.0790$. Why this step? This is the truth we compare against.
2. Approximate value: $y_U^{approx}=y_c+y_t=0.03+0.05=0.0800$. Why this step? The shortcut replaces $\cos \theta$ by $1$ — valid only if $\theta$ is small.
3. Absolute error $=0.0800-0.0790=0.0010$; relative error $=0.0010/0.0790=1.27\%$. Why this step? Quantifies how much the shortcut costs. The error is $y_t(1-\cos \theta )$.
4. Why it's small: $1-\cos \theta$ grows like ${\theta }^2/2$. At $\theta =0.1974$, ${\theta }^2/2=0.0195$, and $y_t\times 0.0195=0.00097\approx 0.0010$. Why this step? Shows the error is second order in slope — halve the slope and the error drops by four. That's why thin, gently-cambered airfoils tolerate the shortcut.

Verify: Approx $0.08>$ exact $0.079$ — the shortcut overestimates, matching the forecast (dropping $\cos \theta <1$ removes a shrink). Error $1.3\%$: acceptable for a rough plot, not for a CNC toolpath. ✓

---

Example 5 — Zero-lift angle and its sign (Cell E)

Forecast: positive camber bows the mean line upward. Do you expect ${\alpha }_{L0}$ to be negative or positive?

1. Convert: ${\alpha }_{L0}=-0.035\text{rad}\times \frac{180}{\pi }=-2.005^{\circ }$. Why this step? Named airfoils quote the zero-lift angle in degrees; the factor $180/\pi$ converts radians.
2. Plug into the lift line $c_{\ell }=2\pi (\alpha -{\alpha }_{L0})$ with $\alpha =0$: $c_{\ell }=2\pi (0-(-0.035))=2\pi (0.035)=\mathrm{0.2199.}$ Why this step? At level chord the only thing making lift is the camber, encoded entirely in ${\alpha }_{L0}$.
3. Interpret the sign: $c_{\ell }=+0.22>0$ ⇒ positive (upward) lift. Why this step? A negative ${\alpha }_{L0}$ shifts the lift line left, so at $\alpha =0$ we're already above the zero crossing.

Verify: $-2.005^{\circ }\approx -2^{\circ }$, matching the parent note's NACA 2412 estimate (${\alpha }_{L0}\approx -2^{\circ }$, $c_{\ell }\approx 0.23$). Sign of ${\alpha }_{L0}$ is negative, as forecast for upward camber. ✓

---

Example 6 — Inverting the lift line for angle of attack (Cell F, negative sign)

Forecast: to cancel the camber's built-in lift, do you push the nose up (positive $\alpha$) or down (negative $\alpha$)?

1. Set $c_{\ell }=0$ in $c_{\ell }=2\pi (\alpha -{\alpha }_{L0})$: $0=2\pi (\alpha -{\alpha }_{L0})\Rightarrow \alpha ={\alpha }_{L0}$. Why this step? Zero lift happens exactly at the zero-lift angle — that's its definition.
2. So $\alpha ={\alpha }_{L0}=-0.035\text{rad}=-2.005^{\circ }$. Why this step? The value we already computed is the answer; the algebra confirms it directly.
3. Interpret: the chord must be pitched nose-down by $2^{\circ }$. Why this step? A negative angle of attack tilts the flow to undo the camber's downward deflection.

Verify: Plug back: $c_{\ell }=2\pi (-0.035-(-0.035))=0$. ✓ Sign is negative, matching the forecast that you fight upward camber by pitching down. ✓

---

Example 7 — Real-world word problem (Cell G)

Forecast: will the Reynolds number be in the hundreds, thousands, or hundred-thousands?

1. Max thickness: $12\%$ of $c$ $=0.12\times 0.20=0.024\text{m}=24\text{mm}$. Why this step? The parent note says chord is the denominator for every geometric quote; $12\%$ decodes to a real length here.
2. Reynolds number uses chord as the length scale: $\text{Re}=\frac{Vc}{\nu }$. Why this step? Reynolds number compares inertia to viscosity; the airfoil's natural length is its chord (see Reynolds Number).
3. Compute: $\text{Re}=\frac{12\times 0.20}{1.5\times 10^{-5}}=\frac{2.4}{1.5\times 10^{-5}}=1.6\times 10^5$. Why this step? Plug the numbers; units cancel: $(\text{m/s})(\text{m})/({\text{m}}^2/\text{s})$ is dimensionless. ✓

Verify: $1.6\times 10^5$ — hundred-thousands, as forecast, and typical for a small model glider. Thickness $24\text{mm}$ is a sensible spar depth for a $200\text{mm}$ chord. ✓

---

Example 8 — Exam twist: flat plate & inverted flight (Cell H)

Forecast: does a flat plate make lift at all? And does flipping the airfoil flip the sign of its camber contribution?

1. (a) Flat plate: $y_c=0$ ⇒ ${\alpha }_{L0}=0$. Convert $\alpha$: $5^{\circ }=0.0873$ rad. Why this step? No camber means the lift line passes through the origin; only $\alpha$ makes lift — this kills the "top is longer" myth.
2. $c_{\ell }=2\pi (0.0873-0)=0.5483$. Why this step? Direct plug — proves a plate with no shape still lifts, purely by tilting the flow.
3. (b) Inverted: flipping the airfoil reverses the effective camber, so its zero-lift angle becomes ${\alpha }_{L0}^{inv}=+2^{\circ }=+0.0349$ rad. Why this step? Upside-down camber now bends flow up when the chord is level, so you must over-rotate to recover lift.
4. Solve $0.5=2\pi (\alpha -0.0349)\Rightarrow \alpha =0.0349+\frac{0.5}{2\pi }=0.0349+0.0796=0.1145\text{rad}=6.56^{\circ }$. Why this step? Same lift line, shifted the other way; we invert it for the required chord angle.

Verify: (a) $0.548$ — a flat plate does lift, killing the equal-transit myth (as the parent's mistake box warns). (b) The inverted airfoil needs $6.56^{\circ }$ vs the upright $\alpha =0.0349+0.0796=\dots$; note it needs a larger angle than upright would, confirming inverted flight is less efficient — matching real aerobatic experience. ✓

---

Recall Quick self-test (click to open)

Digits of NACA 4415 mean camber, position, thickness ::: $4\%$ camber, at $40\%$ chord, $15\%$ thick Why does the symmetric 0012 have $\theta =0$ everywhere ::: its mean camber line is flat ($d{y}_c/dx=0$), and $\arctan 0=0$ Sign of ${\alpha }_{L0}$ for upward (positive) camber ::: negative — the lift line shifts left A flat plate at $\alpha >0$ makes lift because ::: it still deflects air downward; path-length equality is irrelevant Reference length for an airfoil's Reynolds number ::: the chord $c$

Connections

• Thin-Airfoil Theory — source of $c_{\ell }=2\pi (\alpha -{\alpha }_{L0})$ used in Examples 5–8.
• Kutta–Joukowski Theorem — why lift is linear in circulation and hence in $\alpha$.
• Lift and Drag Coefficients — $c_{\ell }$ non-dimensionalisation used throughout.
• Reynolds Number — chord as reference length (Example 7).
• NACA Airfoil Series — the naming rules decoded in Examples 1–2.
• Boundary Layer & Flow Separation — why too much camber/thickness stalls early.