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Two-Piston Hydraulic Force, Pressure, Travel and Work

Relate two piston areas, forces, common applied pressure, travel distances and ideal mechanical work using selectable units.

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

A 100 N input on 10 cm² produces 100 kPa applied pressure. A 100 cm² output piston produces 1,000 N; input travel 0.2 m gives output travel 0.02 m and ideal work 20 J.

p = F₁/A₁ = F₂/A₂; A₁x₁ = A₂x₂; W = F₁x₁ = F₂x₂

Worked example

Enter these known values and leave the other values blank.

Input piston area A₁
10 cm²
Input force F₁
100 N
Output piston area A₂
100 cm²
Input piston travel magnitude x₁
0.2 m
Output force F₂
1000 N
Common applied pressure
100 kPa
Output piston travel magnitude x₂
0.02 m
Ideal work W
20 J

Assumptions and limitations

  • These are positive force and area magnitudes, common applied pressure and nonnegative travel/work. Applied pressure means the added piston pressure under a common reference; it is not a calculation of total atmospheric plus hydrostatic pressure.
  • The pistons have the same hydraulic elevation. Incompressible fluid, no leakage, negligible friction, quasi-static force balance and constant piston areas give equal applied pressure, conserved swept volume and equal input/output work. Opposite piston motion is represented by nonnegative travel magnitudes. The inches-of-mercury selector uses the conventional 0°C factor, based on 13,595.1 kg/m³ mercury and standard gravity.
  • Positive pressure and piston areas determine the force magnitudes even when work is zero. Known positive forces with zero work determine zero travel; areas with one zero travel determine the other zero travel. Reviewed inverse guards skip divisions by zero, so independently determinable forces do not disappear. Two zero travels alone do not identify an area ratio, and zero work with zero travel alone does not identify force.
  • A larger output area multiplies force while reducing its travel by the same factor. Real equipment needs friction, leakage, strength, dynamic and elevation effects evaluated separately; this ideal model does not supply equipment ratings.

Common questions

How can force increase without creating energy?

The larger piston travels less. In the ideal model, its force multiplied by travel equals the input work, and each piston sweeps the same fluid volume.

Should I include atmospheric pressure?

The common pressure here is the piston-applied increment relative to a shared reference. The model does not add atmospheric pressure or a vertical fluid head.

What does zero travel mean?

Zero travel with positive geometry and forces produces zero work and zero travel at the other piston. It cannot by itself reveal a missing area ratio or force.

What if work and travel are both zero?

A force can still be determined from positive pressure and piston area. The work/travel ratio alone is indeterminate, so that inverse is skipped while other applicable equations remain available.

References

A positive-work ideal piston model requires positive travel, and positive finite piston areas require both travel magnitudes to be zero together or positive together. Skip force and piston-area inverses whose travel denominator is zero or whose zero numerator cannot produce a positive force/area, preserving independently determinable pressure/area forces and valid zero-work travel results. Other applicable captured equations remain available; two zero travels or zero work/travel alone do not determine a unique area or force. Mercury pressure columns use conventional density and standard gravity, with the exact international inch. Pounds-force per square inch use the exact international pound and inch with standard gravity, replacing the rounded captured factor. One technical atmosphere is exactly one kilogram-force per square centimeter under standard gravity.

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