Hydraulic, Pneumatic & Fluid Power Systems worked example

Cylinder Force with cylinder force input load of 250 units: a worked example

What does the result look like when cylinder force input load reaches 250 units? The full calculation is worked below with real intermediate numbers. Use it when cylinder force in hydraulic, pneumatic and fluid power systems is being sized against an asset rating.

The inputs for this scenario

  • Cylinder Force input load: 250 units (raised for this scenario; the documented default is 100)
  • Cylinder Force load factor: 1.2 x (unchanged)
  • Cylinder Force operating time: 8 hr (unchanged)

Working through the calculation

  • Applying the documented formula (Cylinder Force load = input load × load factor) to the inputs above produces each figure below.
  • At this operating point the engine returns 300 units for total load, the number this scenario is built around.
  • At this operating point the engine returns 37.5 units / hr for hourly equivalent.
  • At this operating point the engine returns 250 units for input load.
  • At this operating point the engine returns 1.2 x for load factor.

How this compares with the baseline

  • Against the tool's baseline example, where cylinder force input load sits at 100 units and the headline result is 120 units, this scenario comes in 150% above the baseline at 300 units.
  • A figure at this level is achievable when cylinder force input load is genuinely sustained, not just peaked for a shift. It is a simplified load-factor model — it does not account for bore area, system pressure, seal friction, or back-pressure, so use it for first-pass sizing, not final pressure calculations.

Results at a glance

  • Total load: 300 units (headline result)
  • Hourly equivalent: 37.5 units / hr
  • Input load: 250 units
  • Load factor: 1.2 x

Run it with your numbers

  • Every input above is editable in the live Cylinder Force calculator, which recalculates instantly and can be shared with the inputs intact.

Last reviewed 2026-05-12.