551.25
Pa
569.9925
N
0.569992
kN
17,099.775
W
17.0998
kW
259.0875
N/m²
551.25
Pa
569.9925
N
0.569992
kN
17,099.775
W
17.0998
kW
259.0875
N/m²
The Drag Equation Calculator computes the aerodynamic or hydrodynamic drag force on an object moving through a fluid. Drag is the resistive force that opposes motion through a fluid — it determines fuel economy in vehicles, terminal velocity of falling objects, wind loads on structures, and performance of aircraft and ships.
Understanding drag is essential for optimizing shapes in automotive, aerospace, civil, and marine engineering.
The drag force is given by the drag equation:
$$F_d = \frac{1}{2} C_d \rho A v^2$$
where:
The term ½ρv² is the dynamic pressure (q), representing the kinetic energy per unit volume of the flow:
$$q = \frac{1}{2}\rho v^2$$
The power required to overcome drag at constant velocity:
$$P = F_d \cdot v = \frac{1}{2} C_d \rho A v^3$$
Note the cubic dependence on velocity for power — doubling speed requires 8× the power to overcome drag alone.
Typical drag coefficients:
The drag coefficient depends on the body shape, surface roughness, and Reynolds number. At very low Re (Stokes flow), C_d = 24/Re for a sphere. At high Re, C_d becomes approximately constant — the regime where this equation is most useful.
The drag force increases with the square of velocity, meaning aerodynamic drag dominates at high speeds. The power output shows the energy rate needed just to overcome drag — at highway speeds, aerodynamic drag typically accounts for 50–70% of a car's total resistance. The dynamic pressure provides a useful reference for comparing drag across different conditions.
Inputs
Results
A sedan (Cd = 0.30, A = 2.2 m²) at 108 km/h experiences ~364 N drag, requiring ~11 kW (15 hp) just for aerodynamic resistance.
Inputs
Results
A skydiver (spread-eagle, Cd ≈ 1.0, A ≈ 0.7 m²) at terminal velocity ~55 m/s: drag ≈ 1296 N ≈ 132 kg — equals body weight.
Drag comes from the momentum transferred from the object to the fluid. At higher velocity, the object encounters more fluid per unit time (proportional to v) and each interaction transfers more momentum (also proportional to v). The combined effect gives v² dependence.
For bluff bodies (cars, buildings), use the frontal projected area — the shadow the object casts when light shines along the flow direction. For airfoils and wings, the planform area (top-view area) is used. Be consistent with the Cd value's reference area definition.
Power = Force × Velocity. Since drag force ∝ v², power = F_d × v ∝ v³. This means going from 100 km/h to 200 km/h requires 8× the power for aerodynamic drag alone. This cubic relationship is why fuel consumption increases dramatically at high speeds.
Terminal velocity is reached when drag force equals gravitational force: ½CdρAv² = mg. Solving: v_terminal = √(2mg/(CdρA)). For a human skydiver in spread-eagle position, this is approximately 55 m/s (200 km/h).
Reduce Cd by streamlining the shape (smooth contours, tapered rear), minimize frontal area, smooth surfaces, cover wheel wells, and manage airflow separation. Each 10% reduction in CdA reduces fuel consumption by approximately 2–3% at highway speeds.
Yes, the same equation applies. Use the density of water (ρ ≈ 1000 kg/m³) instead of air. Since water is ~800× denser than air, drag forces are dramatically higher — a swimmer at 2 m/s experiences as much drag as a car at about 55 m/s.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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