This calculator shows the ideal static force ratio for two hydraulically connected piston areas. It does not rate a jack, calculate pressure, or establish a safe lifted load.
Hydraulic Transfer Jack Interactive Calculator
Explore the ideal Pascal-law relation F2/F1=A2/A1 and the reciprocal displacement tradeoff.
Equation Used
- Two effective piston areas share the same ideal static pressure.
- Entered input is pound-force and the area ratio is A2/A1.
- Rigid incompressible-fluid, quasistatic, lossless comparison for the ratio statement.
- No actual area, pressure, stroke, volume, friction, leakage, backpressure, cylinder/structure strength, stability, load holding, pump, valve, relief or safety model.
The force gain is the arithmetic difference F2−F1, not created energy. In an ideal lossless system, the larger output force moves through proportionally less distance: s2/s1=A1/A2.
Ideal equal-pressure area ratio
With the same static pressure on two pistons, F1/A1=F2/A2, so F2=F1(A2/A1). Conservation of ideal work gives the reciprocal displacement relation s2/s1=A1/A2.
The animation illustrates reciprocal piston motion. Piston widths are not drawn to the entered area ratio, so use the numerical ratio for comparisons; the drawing is not a jack design or lift cycle.
Use as a ratio check
The identities can illustrate force/displacement reciprocity and verify unit conversion. They cannot determine actual pressure, usable stroke, speed, capacity or safe load.
Force multiplication trades displacement
The ideal force ratio equals the area ratio. The displayed force difference is F2−F1, but energy is not amplified: the larger piston moves a smaller distance for the same displaced volume. Real losses reduce delivered force.
Worked force-ratio example
For an entered 25 lbf on an ideal 50:1 area ratio, ideal output force is 1,250 lbf or about 5,560 N. The arithmetic force difference is 1,225 lbf, while ideal output displacement is 1/50 of input displacement.
Scope and limitations
A real hydraulic jack requires actual piston areas and strokes, pressure and pump force, fluid volume, seals/friction/leakage, opposing pressure, cylinder and frame strength, buckling and side load, base and load stability, mechanical load holding, overload relief, hose/fittings, controls, guarding, inspection and applicable lifting rules. Never rely on hydraulic pressure alone to support a load.
Questions
Does the force difference represent free energy?
No. Output displacement falls reciprocally in the ideal model.
Is the output a safe lifting capacity?
No. No component, structure or stability rating is included.
What pressure is required?
It cannot be calculated without an actual input or output piston area.
Calculation basis
Equal static pressure gives F₁/A₁ = F₂/A₂. Multiplying the entered input force by the area ratio therefore gives ideal output force. With incompressible fluid and no leakage, equal displaced volumes give A₁s₁ = A₂s₂, so greater output force trades against output travel.
The force-difference tile is simply F₂ − F₁. The newton result uses 1 lbf = 4.4482216153 N. None of these ratios establishes a real jack’s load capacity, which also depends on component ratings, structure and stability.
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