# Hydrostatic pressure

This online calculator can solve hydrostatic pressure problems by finding unknown values in the hydrostatic equation.

This online calculator can solve hydrostatic pressure problems by finding unknown values in the hydrostatic equation.
The equation is as follows:
$P=\rho gh$
It states that the pressure difference between two elevations in a fluid is the product of elevation change, gravity, and density.
The calculator can solve this equation

• for pressure using known density, height, and gravity
• for density using known pressure, height, and gravity
• for height using known pressure, density, and gravity
• for gravity using known pressure, height, and density

All formulas are trivial. The input's default values are as follows:

• density is the density of water
• gravity is gravity on Earth
• pressure is 1 atm.

You can also find some theory below the calculator.

#### Hydrostatic pressure

Digits after the decimal point: 2
Pressure, Pa

Height, m

Density, kg/m3

Gravity, m/s2

Hydrostatic pressure — pressure difference between two elevations.

The simplified formula, which does not consider, for example, fluid's compression, yet gives good estimations, can be obtained as follows:
$P=\frac{F}{S} \longrightarrow P=\frac{mg}{S} \longrightarrow P=\frac{\rho Vg}{S} \longrightarrow P=\frac{\rho Shg}{S} \longrightarrow P=\rho gh$

The formula depends only on the fluid chamber's height, not on its width or length. Given a large enough height, any pressure may be attained. It is believed that this was first demonstrated by Pascal in his barrel experiment.
Pascal's barrel is the name of a hydrostatics experiment allegedly performed by Blaise Pascal in 1646. Pascal inserted a 10-m long (32.8 ft) vertical tube into a barrel filled with water in the experiment. When water was poured into the vertical tube, Pascal found that the increase in hydrostatic pressure caused the barrel to burst.

This feature of hydrostatics has been called the hydrostatic paradox. As expressed by W. H. Besant,

• Any quantity of liquid, however small, may be made to support any weight, however large.

The picture above shows that the static fluid pressure at a given depth does not depend upon the total mass, surface area, or container geometry. It depends only on the height of the fluid. This is because the pressure to the inclined walls has a vertical part, and in the left vessel, it points up, while in the middle vessel, it points down, thus, adding or removing additional pressure to the bottom pressure, so it is proportional only to the height of fluid.

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