Interactive Lecture
External Flow Forces, Drag Coefficients, and Aerodynamic Lift
by Dr Arnaldo Delli Carri
When a body moves through a fluid — or a fluid flows past a stationary body — the fluid exerts forces on it. These forces arise from two physical mechanisms: pressure acting normal to the surface, and viscous (friction) stress acting tangentially. The vector sum of all these distributed surface forces can be decomposed into two components: drag, acting parallel to the flow direction, and lift, acting perpendicular to it.
Drag and lift are not separate phenomena — they are orthogonal components of the same resultant aerodynamic force. Drag opposes motion and costs energy; lift supports or deflects the body. Minimising one and maximising the other sits at the heart of aircraft, turbine, and vehicle design.
The total drag on a body has two contributions: pressure drag (form drag), caused by the asymmetric pressure distribution front-to-back, and friction drag, caused by viscous shear stress along the surface. Their relative magnitudes depend strongly on body shape and flow regime.
Drag Force:
\[F_D = \frac{1}{2}\,\rho\,U^2\,A\,C_D\]
Where:
The term \(\frac{1}{2}\rho U^2\) is the dynamic pressure, \(q\), the kinetic energy per unit volume of the free-stream. The drag coefficient \(C_D\) encapsulates all the complexity of the flow into a single non-dimensional number that depends on body shape, Reynolds number, Mach number, and surface roughness.
Lift arises from an asymmetric pressure distribution around a body. For an aerofoil at angle of attack, the curved upper surface accelerates the flow (lowering pressure via Bernoulli's equation), while the lower surface decelerates it (raising pressure). This pressure difference produces a net upward force.
Lift Force:
\[F_L = \frac{1}{2}\,\rho\,U^2\,A\,C_L\]
Where:
For simple geometries, \(C_D\) varies with Reynolds number \(Re = \rho U D / \mu\). At very low Re (Stokes flow), \(C_D \propto 1/Re\) and drag is dominated by viscosity. At high Re, \(C_D\) becomes roughly constant (dominated by form drag). The drag crisis is a dramatic drop in \(C_D\) at a critical Re when the boundary layer transitions to turbulence, delaying separation and reducing the wake.
Stokes Drag (very low Re, Re < 1):
\[F_D = 3\pi\,\mu\,D\,U \qquad \Rightarrow \qquad C_D = \frac{24}{Re}\]
Empirical fit for a sphere (Re < 2×105):
\[C_D \approx \frac{24}{Re} + \frac{6}{1+\sqrt{Re}} + 0.4\]
For a thin aerofoil at a small angle of attack \(\alpha\) in inviscid, incompressible flow, thin aerofoil theory gives a beautifully simple result: lift is proportional to angle of attack, and the lift slope is exactly \(2\pi\) per radian.
Thin Aerofoil Theory — Lift Slope:
\[\frac{dC_L}{d\alpha} = 2\pi \quad \text{(per radian)}\]
Lift coefficient for symmetric aerofoil:
\[C_L = 2\pi\,\alpha\]
Where \(\alpha\) is the angle of attack in radians. For cambered aerofoils, lift is produced even at \(\alpha = 0\); the zero-lift angle \(\alpha_{L=0}\) shifts the curve.
Stall: As the angle of attack increases, \(C_L\) rises linearly until the boundary layer separates from the upper surface. Beyond this critical angle (typically 12–18°), \(C_L\) drops sharply and \(C_D\) rises dramatically — the aerofoil has stalled. Recovery requires reducing the angle of attack.
The aerodynamic efficiency of a body is characterised by its lift-to-drag ratio, \(L/D\). A high \(L/D\) means the body generates large lift for little drag penalty, so less thrust — and less fuel — is needed to sustain flight. Modern commercial aircraft achieve \(L/D \approx 15\text{--}20\); gliders can exceed 60.
Glide Angle:
\[\tan\theta = \frac{D}{L} = \frac{C_D}{C_L} = \frac{1}{L/D}\]
A higher \(L/D\) means a shallower glide angle and longer range for a given altitude loss.
The drag coefficient varies enormously with geometry. Streamlining a body (making it elongated in the flow direction) can reduce \(C_D\) by more than an order of magnitude compared to a bluff body of the same frontal area:
Approximate \(C_D\) values (high Re):
Three complementary views reveal the physics of lift and drag. Use the tabs to switch between them — each rewards careful exploration.
| Concept | Equation | Description |
|---|---|---|
| Drag Force | \(F_D = \frac{1}{2}\rho U^2 A C_D\) | Total drag on a body |
| Lift Force | \(F_L = \frac{1}{2}\rho U^2 A C_L\) | Lift perpendicular to flow |
| Dynamic Pressure | \(q = \frac{1}{2}\rho U^2\) | Kinetic energy per unit volume |
| Reynolds Number | \(Re = \frac{\rho U D}{\mu} = \frac{U D}{\nu}\) | Inertial-to-viscous force ratio |
| Stokes Drag (Re < 1) | \(C_D = 24/Re\) | Viscous-dominated regime |
| Thin Aerofoil Lift Slope | \(dC_L/d\alpha = 2\pi\) rad−1 | Incompressible, inviscid theory |
| Lift Coefficient | \(C_L = 2\pi\,\alpha\) | Symmetric aerofoil, small \(\alpha\) |
| Lift-to-Drag Ratio | \(L/D = C_L / C_D\) | Aerodynamic efficiency |
| Glide Angle | \(\tan\theta = C_D/C_L\) | Unpowered descent angle |
| Induced Drag Coefficient | \(C_{D,i} = \frac{C_L^2}{\pi\,e\,AR}\) | Drag due to lift; \(e\) = Oswald efficiency, \(AR\) = aspect ratio |
A football (diameter D = 0.22 m, \(C_D = 0.47\)) travels through air at ρ = 1.2 kg/m³ at a speed of 25 m/s. Calculate:
Show Solution
Solution:
Given:
Step 1: Reynolds number
\[Re = \frac{\rho U D}{\mu} = \frac{1.2 \times 25 \times 0.22}{1.8 \times 10^{-5}} = \frac{6.6}{1.8 \times 10^{-5}} = 3.67 \times 10^5\]
This is in the subcritical range, so \(C_D \approx 0.47\) is appropriate.
Step 2: Frontal area
\[A = \frac{\pi D^2}{4} = \frac{\pi \times 0.22^2}{4} = 0.03801\,\text{m}^2\]
Step 3: Drag force
\[F_D = \frac{1}{2}\rho U^2 A\,C_D = \frac{1}{2} \times 1.2 \times 25^2 \times 0.03801 \times 0.47\]
\[F_D = 0.5 \times 1.2 \times 625 \times 0.03801 \times 0.47 = 6.70\,\text{N}\]
Step 4: Deceleration
\[a = \frac{F_D}{m} = \frac{6.70}{0.43} = 15.6\,\text{m/s}^2\]
This is approximately 1.6g — drag has a significant effect even at moderate football speeds.
Answer: \(Re = 3.67 \times 10^5\); drag force \(F_D = 6.70\,\text{N}\); deceleration \(a = 15.6\,\text{m/s}^2\) (about 1.6g).
A light aircraft has a total mass of 1200 kg and cruises at 60 m/s at sea level (ρ = 1.225 kg/m³). The wing aerofoil has \(C_L = 0.8\) at the cruise angle of attack. Calculate the required wing planform area for level flight, and then find the drag force if \(C_D = 0.05\).
Show Solution
Solution:
Given:
Step 1: Lift must equal weight in level flight
\[F_L = W \qquad \Rightarrow \qquad \frac{1}{2}\rho U^2 A\,C_L = mg\]
Step 2: Solve for wing area \(A\)
\[A = \frac{mg}{\frac{1}{2}\rho U^2 C_L} = \frac{1200 \times 9.81}{\frac{1}{2} \times 1.225 \times 60^2 \times 0.8}\]
\[A = \frac{11772}{1764} = 6.67\,\text{m}^2\]
Step 3: Drag force
\[F_D = \frac{1}{2}\rho U^2 A\,C_D = \frac{1}{2} \times 1.225 \times 3600 \times 6.673 \times 0.05\]
\[F_D = 736\,\text{N}\]
Lift-to-Drag ratio:
\[L/D = \frac{C_L}{C_D} = \frac{0.8}{0.05} = 16\]
This confirms: \(F_D = W/(L/D) = 11772/16 = 736\,\text{N}\) — consistent.
Answer: Wing planform area \(A = 6.67\,\text{m}^2\); drag force \(F_D = 736\,\text{N}\); lift-to-drag ratio \(L/D = 16\).
Two aircraft wings have the same area (\(A = 20\) m²) but different aspect ratios: Wing A has \(AR = 8\), Wing B has \(AR = 12\). Both have Oswald efficiency \(e = 0.9\) and are generating \(C_L = 1.2\) at low speed. Calculate the induced drag coefficient for each, and comment on the trade-off involved in choosing a high aspect ratio.
Show Solution
Solution:
Given:
Step 1: Induced-drag coefficient
\[C_{D,i} = \frac{C_L^2}{\pi\,e\,AR}\]
Step 2: Wing A (\(AR = 8\))
\[C_{D,i,A} = \frac{1.2^2}{\pi \times 0.9 \times 8} = \frac{1.44}{22.62} = 0.0637\]
Step 3: Wing B (\(AR = 12\))
\[C_{D,i,B} = \frac{1.2^2}{\pi \times 0.9 \times 12} = \frac{1.44}{33.93} = 0.0424\]
Step 4: Compare
Wing B has \(33\%\) less induced drag than Wing A at this \(C_L\).
Answer: \(C_{D,i} = 0.0637\) for Wing A (\(AR = 8\)) and \(0.0424\) for Wing B (\(AR = 12\)) — a 33% reduction.
Trade-offs of high \(AR\):
A lorry (truck) has a frontal area of 6.5 m² and a drag coefficient of \(C_D = 0.6\). Estimate the drag force and the engine power required to overcome aerodynamic drag alone at highway speed (90 km/h) in air at ρ = 1.2 kg/m³. How does the drag power change if speed increases to 110 km/h?
Show Solution
Solution:
Given:
Step 1: Convert speeds
Step 2: Drag and power at 90 km/h
\[F_{D1} = \frac{1}{2} \times 1.2 \times 25^2 \times 6.5 \times 0.6 = \frac{1}{2} \times 1.2 \times 625 \times 6.5 \times 0.6 = 1463\,\text{N}\]
\[\dot{W}_1 = F_{D1} \times U_1 = 1463 \times 25.0 = 36.6\,\text{kW}\]
Step 3: Drag and power at 110 km/h
\[F_{D2} = \frac{1}{2} \times 1.2 \times 30.56^2 \times 6.5 \times 0.6 = 2185\,\text{N}\]
\[\dot{W}_2 = 2185 \times 30.56 = 66.8\,\text{kW}\]
Step 4: Power scales as \(U^3\)
\[\frac{\dot{W}_2}{\dot{W}_1} = \left(\frac{U_2}{U_1}\right)^3 = \left(\frac{30.56}{25.0}\right)^3 = 1.222^3 = 1.82\]
A 22% increase in speed requires 82% more power to overcome drag. This is why aerodynamic drag dominates fuel consumption at highway speeds.
Answer: At 90 km/h, \(F_D = 1463\,\text{N}\) and \(\dot{W} = 36.6\,\text{kW}\); at 110 km/h, \(F_D = 2185\,\text{N}\) and \(\dot{W} = 66.8\,\text{kW}\) — a 1.82× rise in power for a 22% speed increase.