Worked examples — Angle of attack, lift coefficient, drag coefficient

This deep dive drills the parent topic note by walking through every kind of case the lift/drag machinery can throw at you. We start by mapping the whole territory, then hit each cell with a full worked example.

Everything here rests on three formulas from the parent, which we restate so no symbol is unearned:

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The scenario matrix

Think of every problem as living in one cell of this grid. The examples below tag which cell they hit, and together they fill the whole table.

Cell	Case class	What makes it tricky	Example
A	Positive $\alpha$, linear region	the "textbook" case	Ex 1
B	$\alpha$ below ${\alpha }_{L=0}$ → negative $C_L$	sign flip, downward lift	Ex 2
C	$\alpha ={\alpha }_{L=0}$ (zero-lift)	degenerate: $C_L=0$	Ex 3
D	Stall / beyond linear region	formula breaks — must NOT use it	Ex 4
E	Coefficient → actual newtons (full force)	units, factor of $\frac{1}{2}$	Ex 5
F	Drag split + best glide ($L/D$)	two-term $C_D$, optimum	Ex 6
G	Limiting input: $V\to 0$, and $\mathrm{AR}\to \infty$	degenerate/boundary behaviour	Ex 7
H	Real-world word problem (steady climb)	translate words → equation	Ex 8
I	Exam twist: solve backwards for $\alpha$	invert the linear law	Ex 9
J	Drag coefficient → force ($D=C_{D}qS$)	practise the drag side	Ex 10

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Ex 1 — Cell A: the textbook positive angle

Forecast: guess before computing — will $C_L$ be near $0.5$, $0.9$, or above $1$?

1. Find the effective angle $\alpha -{\alpha }_{L=0}=6-(-3)=9^{\circ }$. Why this step? Lift is generated by tilt measured from the zero-lift line, not from the chord — the airfoil's camber already makes lift at $\alpha =0$.
2. Multiply by the slope: $C_L=0.10\times 9=0.90$. Why this step? In the linear region each extra degree adds a fixed $a_0$ of lift coefficient.

Verify: at $\alpha ={\alpha }_{L=0}=-3^{\circ }$ the formula gives $0.10\times 0=0$ ✓ (consistent with the meaning of zero-lift angle). $0.90$ is comfortably below the typical stall value $C_{L,\max }\approx 1.3$–$1.6$, so we are still in the valid linear zone. ✓

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Ex 2 — Cell B: below zero-lift → negative $C_L$

Forecast: will the lift point up or down?

1. Effective angle: $\alpha -{\alpha }_{L=0}=-8-(-3)=-5^{\circ }$. Why this step? We are now more nose-down than the zero-lift line, so the effective angle is negative.
2. Coefficient: $C_L=0.10\times (-5)=-0.50$. Why this step? The linear law is symmetric; a negative effective angle gives a negative coefficient.
3. Interpret the sign: negative $C_L$ means the aerodynamic "lift" points downward (toward the belly of the wing). This is exactly what an inverted race-car wing or a diving stunt uses.

Verify: the picture shows the sign convention — above the zero-lift line the arrow is up, below it the arrow flips down through the $C_L=0$ crossing. Magnitude check: same $5^{\circ }$ distance from the crossing as $\alpha =+2^{\circ }$ would give $+0.5$, mirror image ✓. Units: $C_L$ is dimensionless, so a bare number is correct. ✓

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Ex 3 — Cell C: the degenerate zero-lift point

Forecast: the airfoil is tilted nose-down. Does it still make lift?

1. Effective angle: $\alpha -{\alpha }_{L=0}=-4-(-4)=0^{\circ }$. Why this step? We are sitting on the zero-lift line — the one angle where camber's upward bias and the nose-down tilt exactly cancel.
2. Coefficient: $C_L=a_0\times 0=0$, for any slope $a_0$. Why this step? Zero effective angle → zero coefficient regardless of the slope's value.
3. Lift force: $L=C_{L}qS=0\times 5000\times 15=0\text{N}$. Why this step? Lift force is coefficient × dynamic pressure × area; with $C_L=0$ the whole product must vanish no matter how big $q$ and $S$ are.

Verify: the key insight — a cambered wing at $\alpha =0$ (chord level) is not at zero lift; it must be pitched down to $-4^{\circ }$ to kill all lift. That is why ${\alpha }_{L=0}$ is negative for cambered airfoils. $0\times \text{anything}=0$ ✓, units N·(dimensionless) = N ✓.

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Ex 4 — Cell D: past stall (where the formula is illegal)

Forecast: does lift keep rising to $\alpha =20^{\circ }$?

1. The naive plug-in: $C_L=0.10(20-(-3))=0.10\times 23=2.3$. Why this step? We do it deliberately to expose the trap — the linear law has no knowledge of stall.
2. Reality check against $C_{L,\max }$: $2.3>1.4$, which is impossible. Beyond $15^{\circ }$ the boundary layer separates (see Boundary Layer Separation & Stall) and $C_L$ drops — a real airfoil at $20^{\circ }$ might give only $C_L\approx 1.0$. Why this step? The linear formula is valid only in $\alpha <{\alpha }_{\text{stall}}$; outside it we must use experimental data, not the line.

Verify: the figure shows the curve rising linearly, peaking at $(15^{\circ },1.4)$, then collapsing — the amber "forbidden extrapolation" dashed line shoots to $2.3$ while the true curve turns down. Lift is maximum at stall, not monotonic. ✓

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Ex 5 — Cell E: coefficient → newtons (mind the $\frac{1}{2}$)

Forecast: more or less than the weight of a 10-tonne aircraft ($\approx 98000$ N)?

1. Dynamic pressure: $q=\frac{1}{2}\rho V^2=\frac{1}{2}(1.15)(80{)}^2=0.5\times 1.15\times 6400=3680\text{Pa}$. Why this step? $q$ is the kinetic-energy density of the airstream (see Bernoulli's Equation); the $\frac{1}{2}$ is not optional — dropping it doubles the answer.
2. Lift: $L=C_{L}qS=0.6\times 3680\times 25=55200\text{N}$. Why this step? Force = coefficient × dynamic pressure × area — the coefficient carries all the shape/angle physics.

Verify: units — $\text{Pa}\cdot {\text{m}}^2=\frac{\text{N}}{{\text{m}}^2}\cdot {\text{m}}^2=\text{N}$ ✓. Mass supported $=55200/9.81\approx 5627\text{kg}$, a plausible small aircraft — and less than 10 tonnes as expected. ✓

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Ex 6 — Cell F: drag split and lift-to-drag ratio

Forecast: will $L/D$ be closer to $5$, $18$, or $40$?

1. Induced drag: $C_{D,i}=\frac{C_L^2}{\pi e\mathrm{AR}}=\frac{0.8^2}{\pi (0.85)(10)}=\frac{0.64}{26.70}=0.02397$. Why this step? Making lift sheds tip vortices (Induced Drag & Wingtip Vortices); their cost grows as $C_L^2$, cheaper for long slender wings (big $\mathrm{AR}$).
2. Total drag coefficient: $C_D=0.018+0.02397=0.04197$. Why this step? Profile drag (friction+form) and induced drag simply add.
3. Glide ratio: $L/D=C_L/C_D=0.8/0.04197=19.06$. Why this step? At the same $q$ and $S$ the areas and pressures cancel, so $L/D=C_L/C_D$ — pure aerodynamic efficiency.

Verify: $L/D\approx 19$ sits in the expected $5$–$40$ band for a clean subsonic wing ✓. Sanity: if $\mathrm{AR}$ were halved to $5$, $C_{D,i}$ would double, dropping $L/D$ — long wings glide farther, matching gliders' huge spans. ✓

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Ex 7 — Cell G: limiting / degenerate inputs

Forecast: does a wing at zero speed make any lift? Does an infinitely long wing make induced drag?

1. Part (a): $L=C_L\frac{1}{2}\rho V^{2}S$. As $V\to 0$, $V^2\to 0$, so $L\to 0$. Why this step? $C_L$ is fixed by $\alpha$, but there is no airstream to push on — lift vanishes with the square of speed. This is why aircraft need a takeoff run.
2. Part (b): $C_{D,i}=\frac{C_L^2}{\pi e\mathrm{AR}}\to 0$ as $\mathrm{AR}\to \infty$. Why this step? An infinitely long wing has "no tips," so no tip vortices, so no induced drag — this is the ideal 2-D airfoil of thin-airfoil theory.

Verify: both limits are physically sensible boundary cases. (a) confirms the parent's "$C_L$ does not depend on $V$, but $L$ does" — at $V=0$, $C_L$ is still whatever $\alpha$ gives, yet $L=0$. (b) explains why gliders chase huge aspect ratios. ✓

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Ex 8 — Cell H: real-world word problem (steady level flight)

Forecast: in level flight, what must lift equal?

1. Translate the words: "straight and level" ⇒ lift balances weight, so $L=mg$. Why this step? No vertical acceleration means the vertical forces cancel — this is the physics hidden in the phrase.
2. Compute required lift: $L=mg=1200\times 9.81=11772\text{N}$. Why this step? Weight is the target lift the wing must produce.
3. Dynamic pressure: $q=\frac{1}{2}(1.10)(55{)}^2=0.5\times 1.10\times 3025=1663.75\text{Pa}$. Why this step? Every lift/drag formula is built on $q$, so we must compute the airstream's kinetic-energy density before we can extract the coefficient.
4. Solve for the coefficient: $C_L=\frac{L}{qS}=\frac{11772}{1663.75\times 16}=\frac{11772}{26620}=0.4422$. Why this step? Invert $L=C_{L}qS$ to find the coefficient the pilot must "dial in" via $\alpha$.

Verify: $C_L\approx 0.44$ is a typical cruise value (well below stall) ✓. Cross-check: $L=0.4422\times 1663.75\times 16=11772\text{N}=mg$ ✓. Units: $\frac{\text{N}}{\text{Pa}\cdot {\text{m}}^2}$ dimensionless ✓.

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Ex 9 — Cell I: exam twist — solve backwards for $\alpha$

Forecast: a small nose-up angle, or something dramatic near stall?

1. Start from the linear law $C_L=a_0(\alpha -{\alpha }_{L=0})$ and invert it: $\alpha ={\alpha }_{L=0}+\frac{C_L}{a_0}.$ Why this step? We know the coefficient and want the cause — so we run the formula backwards.
2. Effective angle: $C_L/a_0=0.4422/0.11=4.02^{\circ }$. Why this step? This is the tilt above the zero-lift line.
3. Add the offset: $\alpha =-2+4.02=2.02^{\circ }$. Why this step? The zero-lift line sits at $-2^{\circ }$, so the true angle is measured from there.

Verify: forward check — $C_L=0.11(2.02-(-2))=0.11\times 4.02=0.4422$ ✓. And $2.02^{\circ }$ is far below the $\sim 15^{\circ }$ stall angle, so the linear formula is legal here (unlike Ex 4). ✓

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Ex 10 — Cell J: drag coefficient → drag force

Forecast: with $C_D$ about $23\times$ smaller than a typical $C_L$, will $D$ be a few hundred or a few thousand newtons?

1. Dynamic pressure: $q=\frac{1}{2}\rho V^2=0.5\times 1.10\times 60^2=0.5\times 1.10\times 3600=1980\text{Pa}$. Why this step? Drag, like lift, is $C\cdot q\cdot S$; we must build the airstream's kinetic-energy density first.
2. Drag force: $D=C_{D}qS=0.04197\times 1980\times 18=1495.8\text{N}$. Why this step? The drag equation is the exact twin of the lift equation — swap $C_L\to C_D$ in $D=C_{D}qS$; this practises the drag side of the matrix.

Verify: units $\text{Pa}\cdot {\text{m}}^2=\text{N}$ ✓. Cross-check via glide ratio: the same wing at $C_L=0.8$ would give lift $L=0.8\times 1980\times 18=28512\text{N}$, and $L/D=28512/1495.8=19.06$ — exactly the $L/D$ from Ex 6, confirming both forces are consistent. ✓

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Recall Quick self-test

Which cell forbids using the linear $C_L$ formula? ::: Cell D — beyond the stall angle the flow separates and $C_L$ must come from data, not the line. In level flight what does $C_L$ satisfy? ::: $C_L=mg/(qS)$, because lift balances weight. As $V\to 0$ at fixed $\alpha$, what happens to $C_L$ and to $L$? ::: $C_L$ is unchanged (it depends on $\alpha$), but $L\to 0$ because $L\propto V^2$. Why is ${\alpha }_{L=0}$ negative for a cambered airfoil? ::: Camber makes lift even at $\alpha =0$, so you must pitch nose-down to reach zero lift. How do you turn a drag coefficient into a drag force? ::: $D=C_{D}qS$ — the exact twin of $L=C_{L}qS$.

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Connections

• Angle of attack, lift coefficient, drag coefficient — the parent topic these examples drill.
• Bernoulli's Equation — source of $q=\frac{1}{2}\rho V^2$ used in every force example.
• Boundary Layer Separation & Stall — why Ex 4's formula fails past stall.
• Induced Drag & Wingtip Vortices — the $C_L^2$ term in Ex 6 and the limit in Ex 7.
• Kutta-Joukowski Theorem — origin of the lift-curve slope used throughout.
• Reynolds Number & Mach Number — secondary parameters shifting these curves.
• Compressible Flow — high-speed corrections to $a_0$.