Hydraulic Cylinder Calculator - Force, Pressure, Flow & Speed

Calculate hydraulic cylinder push and pull force, piston and rod annular area, extension speed, cycle time, and required hydraulic pump horsepower (HP).

Advertisement
Responsive Leaderboard Ad Unit
⚙️ Configuration & Inputs
Standard mobile hydraulics: 2000 - 3000 PSI
Inside diameter of cylinder tube
Piston rod outside diameter
Gallons per minute from hydraulic pump
Total linear travel distance
📊 Results & Analysis
Push Force (Extension) 31,416 lbs 15.71 US Tons (139.7 kN)
Pull Force (Retraction) 23,562 lbs 11.78 US Tons (104.8 kN)
Extension Speed 3.06 in/sec Stroke Time: 7.84 sec
Retraction Speed 4.08 in/sec Stroke Time: 5.88 sec
Hydraulic Pump Power 14.59 HP 10.88 kW (at 100% efficiency)
Piston Area (Cap vs Annular) 12.57 sq in Annular (Rod End): 9.42 sq in
Advertisement
Responsive In-Content Ad

Precision Hydraulic Cylinder Engineering Formulas

Hydraulic cylinders convert pressurized fluid energy into linear mechanical force. Pascal's Law states that pressure applied to an enclosed fluid is transmitted undiminished in all directions.

1. Hydraulic Force Equations

During extension (push), the pressurized oil acts against the entire circular piston cap area. During retraction (pull), the effective area is reduced by the presence of the piston rod:

  • Piston Cap Area: A_cap = (π × D_bore²) / 4
  • Rod End Annular Area: A_annular = π × (D_bore² - D_rod²) / 4
  • Push Force (lbs): F_push = Pressure (PSI) × A_cap (in²)
  • Pull Force (lbs): F_pull = Pressure (PSI) × A_annular (in²)

2. Cylinder Velocity & Cycle Time

The speed of cylinder actuation depends directly on fluid flow rate (Gallons Per Minute) delivered by the hydraulic pump:

  • Velocity (inches/second) = (231 × GPM) / (60 × Area)
  • Stroke Time (seconds) = Stroke (inches) / Velocity (in/sec)

3. Theoretical Pump Power

The mechanical input horsepower required to drive a pump at a given pressure and flow is calculated as:

Horsepower (HP) = (Pressure in PSI × Flow in GPM) / 1714

Tip: For real-world systems, assume 85% to 90% hydraulic efficiency (multiply required engine/motor HP by 1.15 to account for friction and piping heat losses).

Frequently Asked Questions

In a single-rod double-acting cylinder, the steel rod occupies space inside the cylinder during retraction. The pressurized oil only acts against the annular ring between the tube bore and the rod, resulting in less surface area and lower force at equal pressure.
One horsepower equals 33,000 ft-lbs/min. Converting gallons (231 cubic inches) and square inches into feet and pounds yields the exact engineering constant: (33,000 × 12) / 231 = 1,714.28.
Multiply target velocity (in/sec) by the piston area (in²) and divide by 3.85: GPM = (Velocity × Area) / 3.85.