Drag Force (Quadratic)

Equivalent forms

F_d = \frac{1}{2} \rho v^2 C_d A

P = \frac{1}{2} C_d \rho A v^3

Half-rho-v-squared times a shape factor — universal across cars, planes, and skydivers.

Symbol	Name	SI unit	Dimension	Range
$C_d$	Drag Coefficient Dimensionless shape factor	—	1	0.04 – 2
$\rho$	Fluid Density Density of surrounding fluid	kg/m³	M·L⁻³	0.1 – 14000
$A$	Reference Area Frontal cross-sectional area	m²	L²	0.001 – 100
$v$	Velocity Speed relative to fluid	m/s	L·T⁻¹	0 – 300
$F_d$	Drag Forceoutput Aerodynamic drag force	N	M·L·T⁻²	0 – 100000

Unit systems

Symbol	SI	CGS	Imperial
$C_d$	—	—	—
$\rho$	kg/m³	g/cm³	lb/ft³
$A$	m²	cm²	ft²
$v$	m/s	cm/s	ft/s
$F_d$	N	dyn	lbf

Where it holds

High Reynolds-number flow (Re >

\sim 1000)

, incompressible (Mach < 0.3), steady. For compressible flow (Mach > 0.3) the wave drag adds. For very low Re use Stokes' law instead. C_d itself depends weakly on Re and surface roughness.

Dimensional analysis

$[C_d]\cdot [\rho ]\cdot [$ A $]\cdot [v^{2}] = (1)(M\cdot L^{-3})(L^{2})(L^{2}\cdot T^{-2}) = M\cdot L\cdot T^{-2} \checkmark$ (force)

Discovery

Lord Rayleigh · 1876

While Newton (1726) proposed a v² drag from momentum considerations, Rayleigh formalized the quadratic drag equation including the dimensionless coefficient C_d, which absorbed all shape and Reynolds-dependent details.

Try this

Why does doubling your car's speed make air resistance four times worse?

A car (drag coefficient 0.3, frontal area 2.2 m²) drives at 30 m/s through air (density 1.225 kg/m³). What is the aerodynamic drag force?

Research status: stable

Real-world applications

Vehicle aerodynamic design and fuel economy
Skydiving and parachute sizing
Wind loading on buildings
Sports physics (golf balls, cycling, swimming)

Common misconceptions

C_d is NOT constant — it varies with Reynolds number (e.g., the 'drag crisis' on a sphere at $Re \approx 3\times 10^{5}$ where C_d drops sharply)
Drag goes as $v^{2}$ — so doubling speed quadruples drag and OCTUPLES power consumption
A is the FRONTAL area (projected), not total surface area

Experimental verification

Wind tunnel testing across a century has tabulated C_d for thousands of shapes. SAE coast-down tests on production cars confirm the

v^{2}

scaling within 5%. Skydiver terminal velocity

(\sim 55\,\mathrm{m/s})

is correctly predicted to within 10%.

Derivation

From dimensional analysis, drag force must depend

on \rho

, v, A, and a shape factor C_d as

F_d \propto \rho \cdot v^{2}\cdot A

.

The factor of 1/2 is conventional, tying F_d to dynamic pressure

(1/2)\rho v^{2}

.

The drag coefficient C_d packages all viscous and shape effects, and is determined empirically or by CFD.

Typical values: smooth sphere

\approx 0.47

, sleek

car \approx 0.25

, parachute

\approx 1.3

, flat plate

\approx 2.0

.

Limiting cases

v \to 0

⟶

F_d \to 0

No motion means no drag.

v doubles⟶

F_d \times 4

Quadratic scaling — speed kills fuel economy.

Low Re (creeping flow)⟶ Formula failsFor Re < 1, drag is linear in v (Stokes' law) rather than quadratic.

What if…

What if you halve the drag coefficient?

Drag halves at the same speed; equivalently, you can drive 41% faster for the same drag (since drag $\propto Cd\cdot v^{2})$ .

What if you double your speed?

Drag quadruples, fuel consumption octuples (power $= F\cdot v$ scales as $v^{3})$ . Why highway mileage tanks at 130 km/h.

What if you drive at altitude?

Air density drops, drag falls. Hypersonic aircraft and Indycars at Denver benefit substantially.

1

Aerodynamic drag on a car

Given ·

C d:: 0.3
\rho:: 1.225
A:: 2.2
v:: 30

Find · F_d

Steps

$F_d = 0.5 \times C_d \times \rho \times A \times v^{2}$
$F_d = 0.5 \times 0.3 \times 1.225 \times 2.2 \times 900$
$F_d \approx 364\,\mathrm{N} (\approx 10.9\,\mathrm{kW}$ at 30 m/s)

Result ·

F_d = 0.5 \times 0.3 \times 1.225 \times 2.2 \times 900 \approx 364\,\mathrm{N}

2

Skydiver terminal velocity

Given ·

C d:: 1
\rho:: 1.225
A:: 0.7
v:: 55

Find · F_d

Steps

$F_d \approx 1297\,\mathrm{N}$ at $v = 55\,\mathrm{m/s}$
At terminal velocity $F_d = mg$ , so $m \approx 132\,\mathrm{kg}$ (close to body+gear)
Skydiver-belly position confirmed

Result ·

F_d = 0.5 \times 1.0 \times 1.225 \times 0.7 \times 3025 \approx 1297\,\mathrm{N}

Stokes' Law→Reynolds Number→Bernoulli's Equation→