What is the drag force?
Choose what to find, pick the shape and the fluid, then type the other values. The drag force calculator uses F = ½CρAv² and also gives the dynamic pressure and the power needed to push through the air or water.
- Answer
- F = 741.189 N (166.626 lbf)
F = 741.189 N (166.626 lbf), with C = 0.7, ρ = 1.225 kg/m³, A = 0.18 m² and v = 98 m/s.
- Drag force (N)
- 741.189
- Drag force (lbf)
- 166.626
- Drag coefficient C
- 0.7
- Frontal area (m²)
- 0.18
- Frontal area (ft²)
- 1.9375
- Speed (m/s)
- 98
- Speed (mph)
- 219.22
- Speed (km/h)
- 352.8
- Dynamic pressure (Pa)
- 5,882.45
- Power to overcome drag (W)
- 72,636.5
- Power to overcome drag (hp)
- 97.4071
- Density used (kg/m³)
- 1.225
Answer: F = 741.189 N (166.626 lbf). F = 741.189 N (166.626 lbf), with C = 0.7, ρ = 1.225 kg/m³, A = 0.18 m² and v = 98 m/s.
How to calculate
Drag force calculator: F = ½CρAv² from the drag coefficient, fluid density, frontal area and speed, or the drag coefficient, area or speed from a known force, with the power to overcome drag.
Example with the default inputs (Find Drag force, Shape Skydiver, feet first, 0.70, Frontal area (A) 1.93750387500775 ft², Speed (v) 219.219756621331 mph, Fluid Air at sea level, 15 °C (1.225 kg/m³)): F = 741.189 N (166.626 lbf), with C = 0.7, ρ = 1.225 kg/m³, A = 0.18 m² and v = 98 m/s.
Method: F = ½ × C × ρ × A × v²; C = 2F ÷ (ρAv²); A = 2F ÷ (Cρv²); v = √(2F ÷ (CρA)); dynamic pressure q = ½ρv²; power P = F × v.
- Drag grows with the square of the speed: large objects at moderate to high speed in air or water (OpenStax §6.4). Very small or slow objects follow Stokes’ law instead.
- C and A are for the same reference area (the frontal area); the fluid is still, and v is the relative speed.
Worked examples
Each example is checked against the calculator on every build.
- Find Drag force, Shape Skydiver, feet first, 0.70, Frontal area (A) 0.18, Speed (v) 98, Fluid Other: type the density, Fluid density (ρ) 0.07554 lb/ft³ gives Drag force (F) 732.11292, Drag force (lbf) 164.585532, Answer F = 732.113 N (164.586 lbf), Dynamic pressure (Pa) 5,810.42.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½CρAv²; Table 6.2; skydiver: C = 0.70, A = 0.18 m², ρ = 1.21 kg/m³, 75 kg, v_T = 98 m/s), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (retrieved 2026-10-18)
- Find Speed, Shape Skydiver, feet first, 0.70, Frontal area (A) 0.18, Drag force (F) 735, Fluid Other: type the density, Fluid density (ρ) 0.07554 lb/ft³ gives Speed (v) 98.193041.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½CρAv²; Table 6.2; skydiver: C = 0.70, A = 0.18 m², ρ = 1.21 kg/m³, 75 kg, v_T = 98 m/s), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (retrieved 2026-10-18)
- Find Drag coefficient, Frontal area (A) 0.7, Speed (v) 44, Drag force (F) 819.896, Fluid Other: type the density, Fluid density (ρ) 0.07554 lb/ft³ gives Drag coefficient (C) 1, Answer C = 1.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½CρAv²; Table 6.2; skydiver: C = 0.70, A = 0.18 m², ρ = 1.21 kg/m³, 75 kg, v_T = 98 m/s), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (retrieved 2026-10-18), Example 6.17 (A = 0.70 m², C = 1.0, v_T = 44 m/s)
- Find Area, Shape Sphere, 0.45, Speed (v) 10, Drag force (F) 2.75625, Fluid Air at sea level, 15 °C (1.225 kg/m³) gives Frontal area (A) 0.1, Power to overcome drag (W) 27.5625.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½CρAv²; Table 6.2; skydiver: C = 0.70, A = 0.18 m², ρ = 1.21 kg/m³, 75 kg, v_T = 98 m/s), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (retrieved 2026-10-18), Table 6.2 (sphere 0.45)
How it works
- Drag force: F = ½ × C × ρ × A × v², with C the drag coefficient, ρ the fluid density, A the frontal area and v the speed.
- Drag coefficient: C = 2F ÷ (ρ × A × v²).
- Frontal area: A = 2F ÷ (C × ρ × v²).
- Speed: v = √(2F ÷ (C × ρ × A)). With the weight mg as F, this is the terminal speed.
- Dynamic pressure q = ½ × ρ × v², and the power to keep moving against the drag is P = F × v.
Shapes (OpenStax Table 6.2): airfoil 0.05; sphere 0.45; skydiver feet first 0.70; bicycle 0.90; skydiver horizontal 1.0; circular flat plate 1.12. Or type your own C.
Fluids: air at sea level, 15 °C, 1.225 kg/m³ (U.S. Standard Atmosphere, 1976); air at 0 °C and 1 atm, 1.29 kg/m³; water at 20 °C, 998.2 kg/m³; sea water at 0 °C, 1,030 kg/m³ (OpenStax Tables 14.1 and 14.2). Or type your own density.
Units (exact): 1 km/h = 5/18 m/s; 1 mph = 0.44704 m/s; 1 ft/s = 0.3048 m/s; 1 knot = 1,852/3,600 m/s; 1 lbf = 4.4482216152605 N; 1 ft² = 0.09290304 m²; 1 lb/ft³ = 0.45359237 ÷ 0.3048³ kg/m³ (16.0185); 1 hp = 745.69987158227022 W.
Rules
- Force from 10⁻¹² N to 10¹² N; C from 0.001 to 100; area from 10⁻¹² m² to 10⁶ m²; speed from 10⁻⁹ m/s to 10⁵ m/s; a typed density from 10⁻⁶ to 30,000 kg/m³.
- If the value found (F, C, A or v) or a typed one is 0, not finite, or over 10¹⁵, there is no answer. The other results (power, dynamic pressure, unit conversions) show as they come, in a × 10ⁿ form when large.
Output format
The answer line reads F = a N (b lbf), C = a, A = a m² (b ft²) or v = a m/s (b mph) for the value found; it and every number show to 6 significant figures, rounded half up. Numbers in text are written with thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ. Typed values are read back as exact decimals and every unit size is exact, so F, C and A are exact fractions rounded once; the speed is a square root, exact when the root is exact and a 64-bit float otherwise.
Worked examples by hand
A skydiver. C = 0.70, A = 0.18 m², ρ = 1.21 kg/m³, v = 98 m/s: F = ½ × 0.70 × 1.21 × 0.18 × 98² = 0.07623 × 9,604 = 732.113 N = 164.586 lbf. The dynamic pressure is ½ × 1.21 × 98² = 5,810.42 Pa.
Terminal speed. The same skydiver weighs 75 kg × 9.80 m/s² = 735 N. With F = 735 N: v = √(2 × 735 ÷ (0.70 × 1.21 × 0.18)) = √9,641.87 = 98.193 m/s.
A drag coefficient. A = 0.70 m², v = 44 m/s, ρ = 1.21 kg/m³ and F = 819.896 N: C = 2 × 819.896 ÷ (1.21 × 0.70 × 44²) = 1,639.792 ÷ 1,639.792 = 1.
An area. A sphere (C = 0.45) in sea-level air (1.225 kg/m³) at 10 m/s with 2.75625 N of drag: A = 2 × 2.75625 ÷ (0.45 × 1.225 × 100) = 0.1 m². Power P = 2.75625 × 10 = 27.5625 W.
Other questions people ask
How do you calculate drag force?
Multiply half the fluid density by the drag coefficient, the frontal area and the square of the speed: F = ½ × C × ρ × A × v². A skydiver falling feet first (C = 0.70, A = 0.18 m²) at 98 m/s through air of 1.21 kg/m³ feels ½ × 0.70 × 1.21 × 0.18 × 98² = 732 N.
How do you calculate the drag coefficient?
Rearrange the drag equation: C = 2F ÷ (ρ × A × v²). With 819.9 N of drag on an area of 0.70 m² at 44 m/s in air of 1.21 kg/m³, C = 2 × 819.9 ÷ (1.21 × 0.70 × 44²) = 1.0.
Why does drag grow with the square of the speed?
At the speeds of cars, cyclists and falling people, the force comes from pushing air out of the way. Twice the speed moves twice as much air each second, each with twice the speed, so the drag is 4 times larger.
What are typical drag coefficients?
From OpenStax Table 6.2: an airfoil 0.05, a sphere 0.45, a skydiver feet first 0.70, a bicycle 0.90, a skydiver lying flat 1.0 and a circular flat plate 1.12. The cars in the table run from 0.28 to 0.37, a pickup truck is 0.43 and an SUV 0.64.
How is drag force related to terminal velocity?
An object stops speeding up when drag equals its weight: mg = ½CρAv². Solving for v gives the terminal speed √(2mg ÷ (CρA)). Use Find: Speed with the weight as the force.
How much power does it take to overcome drag?
Power is force times speed, P = F × v, so it grows with the cube of the speed. A sphere with 2.76 N of drag at 10 m/s needs 27.6 W.