Hydrostatic pressure calculator

The formula

$$P = P_0 + \rho g h$$

$P$ is the absolute pressure at depth $h$, $P_0$ the pressure at the surface (1 atm for open water at sea level), $\rho$ the fluid’s density and $g$ the acceleration due to gravity. The term $\rho g h$ alone is the gauge pressure, which is what a tire gauge or a depth gauge reads.

Where it comes from

Picture a column of fluid above a small patch of area $A$ at depth $h$. The patch holds up everything above it.

The column’s volume is $A h$, so its mass is $\rho A h$ and its weight is:

$$F = m g = \rho A h\,g$$

Pressure is force per unit area. Dividing by $A$, the area cancels:

$$\frac{F}{A} = \rho g h$$

So pressure depends only on depth, not on the shape or width of the container. A narrow tube and a lake have the same pressure at the same depth. Add whatever pushes down on the surface, usually the atmosphere, and you have the absolute pressure.

Worked example

At 10 m in a freshwater lake:

$$\rho g h = 1000 \times 9.81 \times 10 = 98{,}100\ \text{Pa}$$

That’s 98.1 kPa, almost exactly one atmosphere. Adding the 101.3 kPa of air above gives an absolute pressure of 199.4 kPa, about 2 atm. This is why divers say pressure goes up one atmosphere for every 10 m.

Watch out for

• Gauge or absolute. Gas law calculations need absolute pressure. Tire and tank gauges read gauge pressure. Mixing them up is the most common mistake.
• Seawater is denser. At about 1025 kg/m³, seawater gives 2.5% more pressure than fresh water at the same depth.
• Liquids only, roughly. The formula assumes density doesn’t change with depth. That’s true for water to a very good approximation, but not for the atmosphere, which thins out as you go up.

Sources

• OpenStax, University Physics Volume 1, section 14.1 "Fluids, Density, and Pressure"

Last checked November 11, 2026