Angle of attack, lift coefficient, drag coefficient: Exercises

Quick symbol reminder (everything below leans on these):

Level 1 — Recognition

L1.1 — Read off the dynamic pressure

Air of density $ρ = 1.225 kg/m^{3}$ flows at $V = 50 m/s$ . Compute the dynamic pressure $q$ .

Recall Solution

WHAT: we just plug into the definition $q = \frac{1}{2} ρ V^{2}$ . WHY: $q$ is the single quantity that bundles density and speed into the "available pressure" that all aerodynamic force scales with. $q = \frac{1}{2} (1.225) (50)^{2} = \frac{1}{2} (1.225) (2500) = 1531.25 Pa .$ So $q \approx 1531 Pa$ .

L1.2 — Identify the coefficient

A wind-tunnel model produces $L = 240 N$ in a stream with $q = 1000 Pa$ over a reference area $S = 0.3 m^{2}$ . What is $C_{L}$ ?

Recall Solution

WHAT: rearrange $L = C_{L} q S$ to solve for the dimensionless number. WHY: $C_{L}$ strips out the airstream ( $q$ ) and the size ( $S$ ), leaving pure shape-and-angle physics. $C_{L} = \frac{L}{q S} = \frac{240}{1000 \times 0.3} = \frac{240}{300} = 0.80.$ Dimensionless, as required (newtons ÷ [Pa · m²] = N ÷ N = 1).

Level 2 — Application

L2.1 — Lift on the linear part of the curve

An airfoil has zero-lift angle $α_{L = 0} = - 3^{\circ}$ and lift-curve slope $a_{0} = 0.10 per degree$ . Find $C_{L}$ at $α = 6^{\circ}$ .

Recall Solution

WHAT: apply $C_{L} = a_{0} (α - α_{L = 0})$ . WHY the subtraction: the slope multiplies the angle measured from the zero-lift line, not from the chord. The airfoil is already making zero lift at $- 3^{\circ}$ , so the effective angle is $6 - (- 3) = 9^{\circ}$ . $C_{L} = 0.10 (6 - (- 3)) = 0.10 \times 9 = 0.90.$

L2.2 — Coefficient to actual force

Aircraft data: $S = 16 m^{2}$ , $V = 80 m/s$ , $ρ = 1.2 kg/m^{3}$ , $C_{L} = 0.6$ . Find the lift $L$ in newtons.

Recall Solution

Step 1 — dynamic pressure. WHY: build $q$ first, then multiply. $q = \frac{1}{2} (1.2) (80)^{2} = \frac{1}{2} (1.2) (6400) = 3840 Pa .$ Step 2 — lift. WHY: $L = C_{L} q S$ is coefficient × airstream × area. $L = 0.6 \times 3840 \times 16 = 36, 864 N .$ Sanity check: that supports $36864/9.81 \approx 3758 kg$ — plausible for a light aircraft.

L2.3 — Induced drag term

A wing operates at $C_{L} = 0.9$ , aspect ratio $AR = 7$ , Oswald efficiency $e = 0.85$ . Compute the induced-drag coefficient.

Recall Solution

WHAT: plug into $C_{D, i} = \frac{C _{L}^{2}}{π e AR}$ . WHY the $C_{L}^{2}$ : making lift bends the flow downward (downwash, see Induced Drag & Wingtip Vortices), which tilts the lift vector back — the drag this creates grows with the square of the lift. $C_{D, i} = \frac{0. 9 ^{2}}{π ( 0.85 ) ( 7 )} = \frac{0.81}{18.692} = 0.04333.$ So $C_{D, i} \approx 0.0433$ .

Level 3 — Analysis

L3.1 — Full drag polar and $L / D$

A wing has $C_{D, 0} = 0.018$ , $AR = 8$ , $e = 0.9$ , operating at $C_{L} = 0.75$ . Find (a) total $C_{D}$ and (b) the lift-to-drag ratio.

Recall Solution

Step 1 — induced drag. $C_{D, i} = \frac{0.7 5 ^{2}}{π ( 0.9 ) ( 8 )} = \frac{0.5625}{22.619} = 0.02487.$ Step 2 — total drag. WHY: profile drag (friction + form) and induced drag simply add. $C_{D} = C_{D, 0} + C_{D, i} = 0.018 + 0.02487 = 0.04287.$ Step 3 — glide ratio. WHY: $L / D = C_{L} / C_{D}$ (the $q S$ factor cancels top and bottom). $\frac{L}{D} = \frac{0.75}{0.04287} = 17.49.$ So $C_{D} \approx 0.0429$ and $L / D \approx 17.5$ .

L3.2 — Same coefficient, doubled speed

At fixed angle of attack a wing has $C_{L} = 0.5$ and produces $L_{1} = 20, 000 N$ at speed $V_{1}$ . If the speed is doubled to $2 V_{1}$ (same $α$ , same altitude), what is the new lift $L_{2}$ ?

Recall Solution

WHAT/WHY: at fixed $α$ the coefficient $C_{L}$ is unchanged — it depends on $α$ , Mach, Reynolds, not directly on $V$ . Only $q \propto V^{2}$ changes. Since $L = C_{L} q S$ and $q \propto V^{2}$ , $\frac{L _{2}}{L _{1}} = \frac{V _{2}^{2}}{V _{1}^{2}} = (2)^{2} = 4.$ $L_{2} = 4 \times 20, 000 = 80, 000 N .$ The lift quadruples — not because $C_{L}$ grew, but because the dynamic pressure did.

Level 4 — Synthesis

L4.1 — Best glide angle of attack

For a wing $C_{D, 0} = 0.020$ , $AR = 10$ , $e = 0.8$ , the drag polar is $C_{D} = C_{D, 0} + \frac{C _{L}^{2}}{π e AR}$ . Find the $C_{L}$ that maximises $L / D$ , then the resulting maximum $L / D$ .

Figure — Angle of attack, lift coefficient, drag coefficient

Recall Solution

WHAT we want: maximise $\frac{L}{D} = \frac{C _{L}}{C _{D, 0} + k C _{L}^{2}}$ where $k = \frac{1}{π e AR}$ . WHY a derivative: $L / D$ first rises (lift grows faster than the small drag) then falls (induced drag $\propto C_{L}^{2}$ eventually dominates). The peak is where the slope is zero — exactly the question a derivative answers. Look at the curve in the figure: it climbs, peaks, then droops.

Step 1 — set up $k$ . $k = \frac{1}{π ( 0.8 ) ( 10 )} = \frac{1}{25.133} = 0.039789.$ Step 2 — differentiate and set to zero. Let $f (C_{L}) = \frac{C _{L}}{C _{D, 0} + k C _{L}^{2}}$ . Using the quotient rule, the numerator of $f^{'}$ is $(C_{D, 0} + k C_{L}^{2}) - C_{L} (2 k C_{L}) = C_{D, 0} - k C_{L}^{2} .$ Setting this to zero: $C_{D, 0} = k C_{L}^{2}$ , i.e. induced drag equals profile drag at best glide. $C_{L}^{⋆} = \frac{C _{D, 0}}{k} = \frac{0.020}{0.039789} = 0.50264 = 0.70897.$ Step 3 — the drag there. Since $C_{D, i} = C_{D, 0}$ at the optimum, $C_{D} = 2 C_{D, 0} = 0.040$ . Step 4 — maximum $L / D$ . $(\frac{L}{D})_{m a x} = \frac{C _{L}^{⋆}}{2 C _{D, 0}} = \frac{0.70897}{0.040} = 17.72.$ So best glide sits at $C_{L}^{⋆} \approx 0.709$ with $L / D_{m a x} \approx 17.7$ .

L4.2 — Wind-tunnel model scaling

A $1 : 8$ scale model ( $S_{m} = 0.25 m^{2}$ ) is tested at $q_{m} = 2000 Pa$ and gives $L_{m} = 350 N$ at $α = 4^{\circ}$ . The full aircraft ( $S = 16 m^{2}$ ) flies at $ρ = 1.0 kg/m^{3}$ , $V = 120 m/s$ , same $α$ (assume matched Reynolds & Mach so $C_{L}$ transfers). Find the full-scale lift.

Recall Solution

WHY the coefficient transfers: $C_{L}$ is the "DNA of the shape" — dimensionless, the same for model and full scale at the same $α$ (and matched Reynolds Number, Mach Number). This is the whole reason coefficients exist. Step 1 — model coefficient. $C_{L} = \frac{L _{m}}{q _{m} S _{m}} = \frac{350}{2000 \times 0.25} = \frac{350}{500} = 0.70.$ Step 2 — full-scale dynamic pressure. $q = \frac{1}{2} (1.0) (120)^{2} = \frac{1}{2} (14400) = 7200 Pa .$ Step 3 — full-scale lift. $L = C_{L} q S = 0.70 \times 7200 \times 16 = 80, 640 N .$

Level 5 — Mastery

L5.1 — Stall speed and the $C_{L, m a x}$ ceiling

An aircraft of mass $m = 1200 kg$ has wing area $S = 18 m^{2}$ and maximum lift coefficient $C_{L, m a x} = 1.4$ (just before stall). At sea level $ρ = 1.225 kg/m^{3}$ , take $g = 9.81 m/s^{2}$ . Find the stall speed $V_{s}$ (the slowest speed at which lift can still equal weight in level flight).

Recall Solution

WHAT/WHY: In level flight lift = weight: $C_{L} \frac{1}{2} ρ V^{2} S = m g$ . As the plane flies slower, $q$ drops, so to keep lift = weight the coefficient $C_{L}$ must rise. But $C_{L}$ cannot exceed $C_{L, m a x}$ (beyond it the flow separates — Boundary Layer Separation & Stall). The slowest possible level-flight speed is therefore the one where $C_{L}$ is maxed out. Set up at $C_{L} = C_{L, m a x}$ : $C_{L, m a x} \frac{1}{2} ρ V_{s}^{2} S = m g \Rightarrow V_{s} = \frac{2 m g}{ρS C _{L, m a x}} .$ Plug in. Numerator: $2 m g = 2 (1200) (9.81) = 23, 544$ . Denominator: $ρS C_{L, m a x} = 1.225 \times 18 \times 1.4 = 30.87$ . $V_{s} = \frac{23544}{30.87} = 762.68 = 27.62 m/s .$ So $V_{s} \approx 27.6 m/s$ ( $\approx 99 km/h$ ). Below this, no angle of attack can hold the plane up.

L5.2 — Trading altitude: thin air raises stall speed

The same aircraft climbs to where $ρ = 0.90 kg/m^{3}$ . By what factor does the stall speed change, and what is the new $V_{s}$ ?

Recall Solution

WHY: from $V_{s} = 2 m g / (ρS C_{L, m a x})$ , everything except $ρ$ is fixed, so $V_{s} \propto 1/ ρ$ . Thinner air → higher stall speed. Factor: $\frac{V _{s, 2}}{V _{s, 1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{1.225}{0.90} = 1.3611 = 1.1667.$ New stall speed: $V_{s, 2} = 1.1667 \times 27.62 = 32.23 m/s .$ So stall speed rises about $17%$ — the wing must fly faster in thin air to make the same lift. This is why high-altitude airfields demand longer take-off runs.

Connections

Bernoulli's Equation — the origin of $q = \frac{1}{2} ρ V^{2}$ used in every problem.
Kutta-Joukowski Theorem — behind the linear lift curve of L2.1.
Boundary Layer Separation & Stall — the ceiling $C_{L, m a x}$ in L5.
Induced Drag & Wingtip Vortices — the $C_{L}^{2}$ term in L2.3, L3, L4.
Reynolds Number & Mach Number — the conditions that let L4.2's coefficient transfer.
Compressible Flow — where high-speed corrections to these coefficients arise.

Recall One-line self-test

Why does doubling speed at fixed $α$ quadruple lift but leave $C_{L}$ unchanged? ::: Because $L = C_{L} q S$ and $q = \frac{1}{2} ρ V^{2} \propto V^{2}$ ; the coefficient depends on $α$ (and Mach/Reynolds), not on $V$ .