Hydrostatic Pressure Calculator
Solve P = ρgh for any variable — pressure, fluid density, gravity, or depth — with fluid presets, full unit conversion, and a bonus force-on-surface mode (F = Pavg × A).
Densities in kg/m³, typical values.
Hydrostatic pressure P
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Force on a flat submerged surface: F = ρ · g · hcentroid · A. Enter the depth of the surface's centroid.
Total hydrostatic force F
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Worked example
How much pressure is 10 m underwater? P = ρgh = 1000 × 9.80665 × 10 = 98,066.5 Pa = 98.0665 kPa ≈ 14.2233 psi. In seawater (ρ = 1025 kg/m³) the same depth gives 100.518 kPa — about 2.5% more. One atmosphere (101,325 Pa) corresponds to a 10.3323 m column of fresh water.
Frequently asked questions
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above a point. It is calculated with the formula P = ρgh, where ρ (rho) is the fluid density in kg/m³, g is gravitational acceleration in m/s², and h is the depth below the surface in meters. For example, 10 m below the surface of fresh water (ρ = 1000 kg/m³, g = 9.80665 m/s²) the pressure is 1000 × 9.80665 × 10 = 98,066.5 Pa, or 98.0665 kPa.
Because pressure at a depth depends only on the vertical height of the fluid column above that point, not on the total volume or the shape of the vessel. Two containers holding water at the same depth — one narrow, one wide — produce exactly the same pressure at the bottom: at 10 m depth, both give 98.0665 kPa. The extra fluid in the wide container is supported by the container walls, not by the fluid at the bottom.
The hydrostatic paradox is the surprising observation that the force on the bottom of a vessel depends only on the depth and the base area — not on the weight of the fluid it contains. A tall, narrow funnel and a wide beaker filled to the same depth exert the same pressure on their bases, even though one holds far more water. It is not truly a paradox: pressure is P = ρgh, and the slanted walls of the funnel carry the rest of the fluid's weight.
Use P = ρgh: multiply fluid density (ρ, kg/m³) by gravity (g, m/s²) by depth (h, m) to get pressure in pascals (Pa). Worked example: at 10 m depth in fresh water, P = 1000 × 9.80665 × 10 = 98,066.5 Pa = 98.0665 kPa ≈ 14.2233 psi. In seawater (ρ = 1025 kg/m³) the same 10 m gives 100,518.1625 Pa ≈ 100.518 kPa — about 2.5% more.
The SI unit is the pascal (Pa); common alternatives are kilopascal (kPa), megapascal (MPa), bar, psi, atmosphere (atm), and millimetres of mercury (mmHg). Conversions: 1 bar = 100,000 Pa, 1 atm = 101,325 Pa, 1 psi = 6,894.757 Pa, 1 mmHg = 133.322387415 Pa. So 98,066.5 Pa (10 m of water) equals 98.0665 kPa, 0.980665 bar, 14.2233 psi, 0.9678 atm, or 735.56 mmHg. This calculator converts between all of them.
The formula P = ρgh gives gauge pressure — the pressure above the ambient atmosphere at the surface. Absolute pressure at depth is P_abs = P_atm + ρgh. Divers and engineers usually work with gauge pressure (their gauges read zero at the surface), but for physics problems involving gas compression at depth you must add atmospheric pressure: at 10 m in water, absolute pressure is about 101,325 + 98,066.5 = 199,391.5 Pa ≈ 1.968 atm.
Set P = ρgh and solve for h. For 1 atm (101,325 Pa) in fresh water: h = 101,325 / (1000 × 9.80665) = 10.3323 m. For 1 bar (100,000 Pa): h = 100,000 / (1000 × 9.80665) = 10.1972 m. This is why divers say every 10 m of depth adds roughly one atmosphere of pressure.
A mercury column balances atmospheric pressure: h = P / (ρg). With mercury density ρ = 13,534 kg/m³ and P = 101,325 Pa, h = 101,325 / (13,534 × 9.80665) = 0.7634 m ≈ 763.4 mm. Conversely, a 760 mm column of this mercury produces 13,534 × 9.80665 × 0.760 = 100,869.63 Pa ≈ 100.870 kPa — very close to one standard atmosphere. Mercury is used because it is about 13.5 times denser than water, so the column is a manageable height instead of 10.3 m.
Force equals average pressure times area: F = P_avg × A. For a flat surface, the average pressure acts at the centroid, so F = ρ × g × h_centroid × A. Example: a vertical plate whose centroid is 2 m deep with area 1.5 m² in fresh water feels F = 1000 × 9.80665 × 2 × 1.5 = 29,419.95 N ≈ 29.420 kN. The Force tab of this calculator does this computation, including unit conversions for area.
At the Challenger Deep (~10,935 m) in seawater (ρ = 1025 kg/m³, g = 9.80665 m/s²), P = 1025 × 9.80665 × 10,935 = 109,916,610.69 Pa ≈ 109.917 MPa — about 1,085 times atmospheric pressure. That is why deep-submergence vehicles need hulls rated for over a thousand bar.
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