Pascal's Law Simulator

What Is the Pascal's Law Simulator?

How to use the Pascal's law simulator

1. Set the push. Drag Input force F1 anywhere from 20 to 500 N. This is the force your hand puts on the narrow piston; the figure beside the slider echoes it back.
2. Size the small piston. Drag Small piston Ø d1 between 2 and 10 cm. This slider alone decides the pressure, because P = F1/A1 — watch Fluid pressure P jump as you shrink it, and the fluid in the drawing change colour from blue through gold to deep red as the gauge climbs.
3. Size the large piston. Drag Large piston Ø d2 between 10 and 40 cm. Now watch what does not happen: the pressure stays exactly where it was while Output force F2 and Mech. advantage climb.
4. Read the consequences. Small piston A1 and Large piston A2 show the areas your diameters produced, and Distance trade-off shows how far the input piston must travel for each centimetre the load rises. The line under the big number converts the output force into the mass it would hold, using m = F2/g.
5. Start over. The Reset button returns the lab to 100 N, 4 cm and 20 cm and replays the stroke, so you can watch the pistons move again. The four preset machines above are quicker if you want a specific real-world configuration.

Worked example: change one thing at a time

Start from the Reset state and move a single slider per step. Every figure in the table is the string the simulator itself printed for those slider positions — the values were read back out of the running lab, not worked out on paper, so if the table and the tool ever disagree, the tool is right.

Read down the rows and the machine explains itself. Row 1 to row 2: doubling the push doubles the pressure and doubles the output, but the mechanical advantage does not budge. Geometry sets the multiplier; your hand only sets the scale. Row 2 to row 3: halving d1 from 4 cm to 2 cm cuts the input area from 12.6 cm² to 3.14 cm², so the pressure quadruples to 637 kPa and the output goes with it. Row 3 to row 4: doubling d2 from 20 cm to 40 cm quadruples the output area from 314 cm² to 1260 cm² and quadruples the force again — while the pressure does not move at all, because the output piston has no vote in what the pressure is. Row 5 is the control: with both bores at 10 cm the advantage is 1× and 200 N in gives 200 N out. Matched cylinders make a rigid rod that happens to bend round corners, which is exactly what a brake line is when the pistons are the same size.

There are also two different ways to buy the same multiplier, and the simulator prices them differently. Hold F1 at 100 N and set d1 = 2 cm with d2 = 20 cm: you get 10 kN out at 318 kPa. Now get to the same 100× the other way, d1 = 4 cm with d2 = 40 cm: still 10 kN out, but now at 79.6 kPa. Identical force, a quarter of the pressure. That second number is the one an engineer actually designs around, because pressure is what bursts hoses and blows seals — force is only what the machine does for a living.

Formula and symbol reference

Everything the lab computes comes from three relationships: A = pi·(d/2)², P = F1/A1 and F2 = P·A2. The ranges marked “in this lab” are the simulator's own displayed values at the ends of its slider travel.

The physics: why pressure spreads and area amplifies

A liquid at rest cannot hold a shear stress. Push it sideways and it simply flows until the sideways stress has gone, which leaves only stress perpendicular to any surface you imagine inside it — and, at any given point, that normal stress turns out to be the same in every direction. That single number is the pressure. Now squeeze the fluid at one boundary. The molecules there are crowded slightly closer, they crowd their neighbours, and the excess propagates through the confined volume at the speed of sound in the liquid — about 1,500 m/s in water, roughly 1,300 m/s in hydraulic oil — until every part of the fluid carries the same increase. That is Pascal's principle stated properly: it is the change in pressure that is transmitted undiminished, added on top of whatever pressure the fluid already had.

Nothing in that story amplifies anything. The amplification is entirely a matter of bookkeeping at the boundaries, because force is pressure collected over an area: F = P·A. Set d1 = 2 cm and d2 = 40 cm in the lab and it reports the same 1590 kPa pressing on both piston faces — but one face is 3.14 cm² and the other is 1260 cm², four hundred times larger, so it gathers four hundred times the force. Area grows as the square of diameter, which is why the multiplier is (d2/d1)² rather than d2/d1, and why the small-piston slider feels so much more violent than the large one: shrinking d1 raises the pressure and widens the ratio, so it hits the output twice.

The energy ledger closes exactly. Incompressible means volume in equals volume out, so A1·s1 = A2·s2, and dividing through gives s2/s1 = A1/A2 = 1/MA. The factor by which the force grows is precisely the factor by which the travel shrinks, so the work done is F1·s1 = P·A1·s1 = P·A2·s2 = F2·s2 — the same joules on both sides. A hydraulic press is a lever made of liquid: the fluid is the rigid beam, the piston areas are the arms, and like every lever it repackages energy rather than creating it. The one thing the fluid adds over a metal beam is that its “beam” can be routed round corners, up a wall and through a flexible hose, which is the whole reason hydraulics exists as an engineering discipline.

The lab does leave out the weight of the fluid itself. In a real machine the two pistons are rarely at the same height, and the pressure at the lower one is greater by rho·g·h — about 8.6 kPa for every metre of head in a typical mineral oil at 875 kg/m³. Against the 1590 kPa this simulator can reach that is barely half a percent, which is exactly why the idealisation survives in practice. In a low-pressure system, or one with a tall vertical run, it does not.

Where the ideal press breaks down

The formula on this page describes a rigid container full of a perfectly incompressible, frictionless liquid that never moves. Every one of those assumptions fails somewhere in a real machine, and each failure has a recognisable symptom.

Entrained air

Hydraulic oil has a bulk modulus around 1.5 to 2 GPa: press it to 200 bar and it yields about one per cent of its volume. Air is roughly four orders of magnitude softer, so even one per cent of air by volume can cut the effective stiffness of the column by a factor of a hundred at low pressure. The swept volume then goes into squashing bubbles instead of moving the output piston, and A1·s1 = A2·s2 stops holding.

Symptom: a spongy, long-travel brake pedal, or a jack whose handle strokes freely while the load barely rises. The cure is bleeding, and it is the single most common hydraulic fault.

Seal leakage and internal bypass

A worn master-cylinder cup or a scored bore lets fluid slip past the piston instead of being pushed ahead of it. Pressure still equalises across whatever fluid is still connected, so the readouts look right for a moment and then sag while you hold the load. The idealised law says what the pressure will be; it says nothing about how long it lasts.

Symptom: a pedal that sinks slowly to the floor under steady foot pressure, or a jack that has settled by morning.

Line and hose friction

Pascal's law is a statics law: it is exact when nothing is flowing. Move the fluid and you pay a viscous pressure drop down the line. For laminar flow that drop scales with flow rate, viscosity and length, and inversely with the fourth power of the bore — so halving a hose's internal diameter multiplies the loss sixteenfold. While the machine is moving, the pressure at the output cylinder is genuinely lower than at the pump; only when the flow stops do the two agree again.

Symptom: full force at rest, noticeably less while the ram is travelling, and an undersized hose that gets warm.

Hose compliance

A steel line is effectively rigid; a rubber hose is not. Pressurise it and it swells, and every cubic centimetre that goes into that swelling never reaches the output piston. The distance trade-off the lab prints is therefore a floor, not a promise — a real machine always needs more input stroke than the ideal ratio predicts.

Symptom: more pumping than the geometry says it should take. It is also why high-pressure brake hoses are steel-braided rather than plain rubber.

Viscosity against temperature

The multiplication ratio is pure geometry and does not care what temperature the oil is. Everything about how the machine behaves does. An ISO VG 46 oil is 46 mm²/s at 40 °C by definition and falls to roughly 7 mm²/s at 100 °C — a factor of six across an ordinary working day.

Symptom: cold oil gives sluggish response, large line losses and a pump that struggles to draw; hot thin oil leaks past clearances and loses film strength where metal meets metal.

Cavitation

The law assumes an unbroken column of liquid. Drop the absolute pressure anywhere below the fluid's vapour pressure — classically at a pump inlet drawing through a clogged filter — and the oil boils into vapour cavities. Those cavities collapse violently once they reach the high-pressure side, pitting metal as they go. With vapour in the line the fluid is no longer incompressible and the volume bookkeeping fails outright.

Symptom: a distinctive gravel-in-a-tin rattle from the pump, erratic output, and pitting on pump and valve faces.

The lab's own pressure ceiling

With F1 capped at 500 N and d1 no smaller than 2 cm, the highest pressure this simulator can produce is 1590 kPa, or about 16 bar. Production hydraulics run at 150 to 350 bar. Real machines close that gap not with a stronger hand but with a lever ahead of the plunger and many strokes, or with an engine-driven pump — so treat the pressure readout here as a demonstration of the relationship, not as the pressure a comparable machine would run at.

Where hydraulic force multiplication is actually used

Car brake systems

A passenger-car master cylinder has a bore of roughly 19 to 25.4 mm, and the pedal itself is a lever with a ratio between about 4:1 and 6:1 before any servo assistance. Across the front axle, four 50 mm caliper pistons add up to the same area as one 100 mm piston — which is exactly the Car brake circuit preset above. Load it and the lab reports 884 kPa and 6.94 kN of clamping force from 400 N at the pushrod. Add the vacuum servo a real car has and you multiply that pushrod force again; hard braking takes the circuit to 80 to 100 bar.

Why the master bore is small: the same modest pedal travel has to fill four calipers, and a narrow bore buys pressure at the cost of stroke — the trade-off readout, in reverse.

Excavator and backhoe arms

The boom cylinder on a 20-tonne excavator has a bore around 120 to 140 mm and the machine runs its circuit at 300 to 350 bar. The Excavator boom cylinder preset uses that real 120 mm bore, driven by a 20 mm hand-pump plunger, and the lab reports 955 kPa and 10.8 kN. The geometry is honest; the pressure is not, and the size of that gap is precisely why an excavator carries an engine-driven pump rather than a handle. At 350 bar the same cylinder would push about thirty-seven times harder than the lab shows.

Workshop jacks and vehicle lifts

A bottle jack pairs a 12 to 20 mm plunger with a 30 to 50 mm ram and puts a long hand lever in front of the plunger; a single-post in-ground vehicle lift uses a ram 200 to 250 mm across fed by a small motor pump. The Vehicle lift preset takes that 250 mm ram and a 25 mm pump piston: 500 N in gives 1020 kPa and 50 kN out, which is 5100 kg of holding force — the class of a five-tonne lift.

Industrial presses

Press rams run from about 100 mm to 400 mm in bore at 250 to 700 bar, giving anything from 100 kN on a benchtop arbor press to several meganewtons on a forging press. The Industrial press preset puts the lab at its ceiling: a 20 mm plunger into a 400 mm ram, 400× advantage, 200 kN of output. Deep-drawing a car body panel or pressing a bearing into a housing needs force far more than speed, which is exactly the side of the trade a large area ratio buys.

Aircraft landing gear and brakes

Civil aircraft hydraulics standardised on 3,000 psi (207 bar) for decades, and the A380 and 787 moved to 5,000 psi (345 bar). The reason is this page's equation read backwards: since F = P·A, raising the pressure lets every cylinder, line and fitting shrink for the same output force, and on an aircraft that saved weight is worth a great deal. It is why the actuator that swings a multi-tonne gear leg into the wheel well is not itself an enormous object.

Where to go next

Put numbers to a specific pair of pistons with the Pascal's law calculator, size a real cylinder including its rod side with the cylinder force calculator, or see the same force-for-distance bargain without the fluid in the mechanical advantage simulator. Everything else we build is in the library of physics simulations.