A hydraulic system multiplies force by the ratio of piston areas and divides travel by the same ratio, so work in equals work out: a 20 mm pump piston driving a 100 mm ram multiplies force 25 times, and 196 N of hand force through a 5:1 lever becomes 24.5 kN at the ram at a pressure of 3.1 MPa. Lifting the ram 150 mm then takes 3,750 mm of pump-piston travel.
Part of Hydraulic Systems
Two pistons of different areas connected by oil trade distance for force. The ratio of the areas is the ratio of the forces, and the inverse of the ratio of the distances.
The Area Ratio Is the Force Ratio
By Pascal’s law the pressure is the same on both pistons (Pascal’s Law). Force = pressure x area, so
F2 = F1 x (A2 / A1)
Areas go as the square of the diameter, so the ratio is (D2 / D1)².
Worked example, following the carrier but in newtons. A pump piston of 20 mm diameter drives a cylinder piston of 100 mm diameter.
- Pump piston area = π / 4 x 0.02² = 3.142 x 10⁻⁴ m²; ram area = π / 4 x 0.1² = 7.854 x 10⁻³ m²
- Area ratio = (100 / 20)² = 25
- A hand force of 20 kgf = 196 N pushes the pump piston: pressure = 196 / 3.142 x 10⁻⁴ = 624,000 Pa (0.62 MPa)
- Force at the ram = 624,000 x 7.854 x 10⁻³ = 4,905 N, which is 25 x 196 N, about 500 kgf
Add a pump handle with a 5:1 lever:
- Force on the pump piston = 5 x 196 = 981 N, pressure = 981 / 3.142 x 10⁻⁴ = 3.12 MPa, about 3.1 MPa
- Force at the ram = 3.12 x 10⁶ x 7.854 x 10⁻³ = 24,525 N (24.5 kN), about 2,500 kgf
The units are newtons throughout; the carrier’s “kg” figures are kilograms-force, 9.81 N each.
Distance Is Divided by the Same Ratio
The oil is incompressible, so the volume pushed out by the pump piston equals the volume that enters the ram. A 150 mm lift of the ram needs a volume of 7.854 x 10⁻³ x 0.15 = 1.18 x 10⁻³ m³ (1.18 L).
- Pump piston travel for that volume = 1.18 x 10⁻³ / 3.142 x 10⁻⁴ = 3.75 m = 3,750 mm, which is 25 x 150 mm
- With a 20 mm pump stroke that is 187.5, so about 188 strokes
Work in equals work out. The hand does 196 N x 3.75 m = 735 J (ignoring the lever, which trades handle travel for force in the same way), and the ram delivers 4,905 N x 0.15 m = 736 J. Friction and seal drag make real systems return less.
The Pressure Is Set by the Pump Piston, Not by the Ram
Hand force is fixed, so a smaller pump piston means a higher pressure: p = F1 / A1. Halve the pump diameter and the same hand force makes four times the pressure, a force ratio of 100 to a 100 mm ram. The gain is not free: pressure rises until some part reaches its rating, so the multiplication a design can use is capped by the weakest component’s marked rating and by the relief valve (Pressure Relief Valves). It is also paid for in strokes: 4 times the force costs 4 times the pump travel.
Common Mistakes
- Mixing kilograms and newtons. Force is in newtons; a “kg” of force is 9.81 N.
- Forgetting the lever. The total gain is lever ratio x area ratio.
- Ignoring friction. Seal drag and valve losses take a share; the corpus gives no efficiency figure, so measure the real force with a known test weight.
- Raising pressure by shrinking the pump piston without checking the parts’ ratings.
Safety
The force is large and the pressure is higher than it feels
A hand on a lever can make pressures that burst weak fittings and drive a pinhole jet through skin. Keep every part rated above the pressure the arithmetic gives, fit a relief valve, and never loosen a fitting under load. Never search for a leak with a hand; use card or wood. Stay out of the line of the ram and support what it lifts.
What’s Next
- Learn next: Single-Acting Pump — the small piston that makes the pressure
- Combine with: Simple Machines — the lever half of the gain