In short

Pascal's law says that when you apply an extra pressure to any part of an enclosed, incompressible fluid, that extra pressure is transmitted undiminished and equally to every other part of the fluid, and acts on the walls of the container perpendicular to the surface. The payoff is the hydraulic press: push on a small piston of area A_1 with force F_1, and the pressure P = F_1/A_1 reappears at every other piston in the circuit. On a large piston of area A_2 it produces a force F_2 = P A_2 = F_1 \cdot A_2/A_1. The mechanical advantage is the area ratio A_2/A_1. Energy is not amplified — the small piston moves through a large distance d_1, the large piston through a small distance d_2, and F_1 d_1 = F_2 d_2 — but force is amplified as much as you like. Hydraulic lifts in service stations, hydraulic brakes in buses and trucks, power-steering in cars, and bulldozers all run on this single line of physics.

Walk into any service station near your locality and look at the bay floor. There, in the middle, is a small hand lever or a foot pedal, connected by a thin steel pipe to a wide steel cylinder buried in the ground. The mechanic pumps the lever a few times — not hard, just with the weight of his forearm — and a two-tonne auto-rickshaw rises three feet into the air. If you did the same lift with your bare hands you would need a crane. What just happened between the pedal and the piston?

The answer is a single idea discovered by Blaise Pascal in the 1650s and stated in one breath: pressure applied to an enclosed fluid is transmitted unchanged to every point of the fluid, and acts on every surface perpendicular to it. That one sentence is why the mechanic's forearm is enough. It is also why your father's Tata bus stops safely at a red light, why the power steering in your cousin's Maruti feels like it does, why an airliner's landing gear retracts the way it does, and why the JCB excavator digging the metro trench swings its bucket with the casual force of something twenty times its operator's weight.

This article derives Pascal's law from the pressure physics you already know, shows why a small-piston-to-large-piston setup multiplies force by exactly the area ratio, carefully traces what happens to energy (nothing is created, you just trade distance for force), walks through Indian examples of hydraulic machines in detail, and then addresses the real-world limits: fluids are not perfectly incompressible, and pistons have friction.

The setup — pressure inside an enclosed fluid

You have already met the pressure of a static fluid in Pressure in Fluids: at depth h below the top surface of an open container, the pressure is

P(h) = P_0 + \rho g h,

where P_0 is the pressure at the top surface (atmospheric, typically). The \rho g h part comes from the weight of the fluid column sitting above the point.

Pascal's law is about what happens when you change P_0. Suppose you seal the top of the container with a piston and press down on the piston with an extra force \Delta F spread over its area A. You have just added an extra pressure

\Delta P = \frac{\Delta F}{A}

to the fluid right below the piston. The question is: what does this extra pressure do to the fluid elsewhere?

Assumptions. The fluid is incompressible (its density \rho does not change when you squeeze it) and non-viscous (it moves smoothly without internal friction). The container is rigid, the fluid fills it completely, and the process is slow enough that the fluid is in equilibrium at every instant.

Under these assumptions, the pressure at every point of the fluid is still given by

P(h) = P_0 + \rho g h,

because the derivation from the force balance on a fluid column only used the weight of the fluid — it did not care what P_0 was. Now increase P_0 by \Delta P. Every point in the fluid feels the new pressure

P'(h) = (P_0 + \Delta P) + \rho g h = P(h) + \Delta P.

Every point went up by the same \Delta P. That is Pascal's law.

Pascal's law

In an enclosed, incompressible fluid in equilibrium, any change in pressure applied at one part of the fluid is transmitted undiminished to every other part of the fluid and to the walls of the container. The transmitted pressure acts perpendicular to any surface it touches.

Three things in that statement deserve unpacking.

  1. "Undiminished." The increment \Delta P you apply at the piston shows up as exactly the same \Delta P at a point one metre away, or ten metres away, or at the far end of a branching network of pipes — as long as the fluid is continuous and enclosed. Nothing is lost in transit.
  2. "Equally in all directions." Pressure in a static fluid is a scalar. At a given point, the pressure is the same whether you measure it by the force per unit area on a horizontal membrane, a vertical membrane, or a membrane tilted at 37°. This is a separate physical fact about fluid statics — see Pressure in Fluids for the wedge-of-fluid argument — but Pascal's law inherits it. When you apply \Delta P at the top piston, that \Delta P pushes sideways on the cylinder walls, downward on the bottom, upward on any piston sitting lower in the system, and so on.
  3. "Perpendicular to the surface." Fluid at rest cannot exert a shear force on a surface — if it did, the fluid would flow. So the only force the fluid exerts on a wall is along the normal to that wall.

The wedge-of-fluid picture — why pressure is isotropic

If you did not read the wedge-of-fluid argument in Pressure in Fluids, here is the compact version. Carve a tiny wedge-shaped volume out of the fluid with three flat faces. Let the wedge be small enough that you can neglect its own weight compared to the pressure forces on its faces. Newton's first law demands the three pressure forces balance. Each force equals pressure times area of its face, and it points inward, perpendicular to the face. Setting the vector sum to zero — after projecting on two perpendicular directions — forces the pressure to be the same on all three faces regardless of their orientation.

A small wedge of fluid showing pressure forces perpendicular to each face A right-angled triangular wedge of fluid with pressure forces drawn perpendicular to each of its three faces, all pointing inward. The three forces balance because the wedge is in equilibrium. This shows that pressure at a point is the same in every direction. horizontal face, area $A_h$ vertical face, area $A_v$ tilted face, area $A_t$ $P A_h$ $P A_v$ $P A_t$ θ
A small wedge of fluid in equilibrium. The pressure forces on the three faces (each perpendicular to the face, each proportional to its area) balance independently of the wedge's orientation. The only way this can happen is if the pressure $P$ at the point is the same on all three faces — pressure is a scalar, not a vector.

This is the fact that lets Pascal's law apply "equally in all directions." When you push on the top piston, the pressure increment \Delta P that shows up at a point deep inside the fluid pushes on every surface at that point — every direction, every orientation — with the same \Delta P. No direction is privileged.

The hydraulic lift — deriving the force multiplier

Now build the canonical machine: two cylinders of different cross-section, connected at the bottom by a pipe, filled with a single body of fluid (oil, usually — it lubricates the seals and doesn't boil). A piston sits in each cylinder.

Schematic of a hydraulic lift with two connected cylinders Two vertical cylinders connected by a horizontal pipe at their bottoms and filled with oil. The left cylinder is narrow (small area A1) with a small piston pushed down by a small force F1. The right cylinder is wide (large area A2) with a large piston supporting a heavy load F2. The oil transmits the pressure between them. piston 1 piston 2 $F_1$ (small) $F_2$ (load) load $A_1$ (small) $A_2$ (large) oil same pressure $P$ everywhere
A hydraulic lift. Push down on the small piston of area $A_1$ with force $F_1$. The pressure $P = F_1/A_1$ that you create is transmitted through the oil to the large piston of area $A_2$, where it produces an upward force $F_2 = P A_2 = F_1 (A_2/A_1)$.

Step 1. Compute the pressure created under piston 1.

Pressing down on piston 1 with force F_1 over area A_1 adds a pressure

\Delta P = \frac{F_1}{A_1}

to the oil just below the piston.

Why: pressure is force per unit area applied perpendicular to the surface — this is the definition of pressure. The piston pushes the oil, and the oil pushes back on the piston with the same pressure (Newton's third law).

Step 2. Use Pascal's law to transport the pressure.

By Pascal's law, the increment \Delta P is transmitted undiminished to every other point in the connected oil. In particular, it reaches the underside of piston 2. Since the two pistons are at the same height (the lift has levelled out and is in equilibrium), the \rho g h contribution is the same under both pistons and cancels when you compare them. Only the applied \Delta P makes a difference.

Why: the P_0 + \rho g h part was already there before you pushed on piston 1. Your push adds exactly \Delta P to every point. If piston 1 and piston 2 are at the same height, the pressures underneath them differ by zero before the push and by zero after the push — the only thing that changed is the applied pressure, which is the same under both.

Step 3. Compute the force that \Delta P exerts on piston 2.

The oil pushes up on piston 2 with a pressure \Delta P spread over the area A_2:

F_2 = \Delta P \cdot A_2 = \frac{F_1}{A_1} \cdot A_2.

Rearranging:

\boxed{\; F_2 = F_1 \cdot \frac{A_2}{A_1} \;}

Why: force equals pressure times area. The pressure is the same everywhere, so whichever piston has the larger area delivers the larger force. The ratio of forces is the ratio of areas.

The ratio A_2/A_1 is called the mechanical advantage of the hydraulic press. If A_2 is 100 times A_1, then F_2 is 100 times F_1. A 50 N push on a small piston becomes a 5000 N push on a large one — enough to lift a 500 kg load.

Number check — the auto-rickshaw lift

A typical small auto-service-station lift has a small piston of radius r_1 = 1 cm and a large piston of radius r_2 = 12 cm. The area ratio is

\frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} = \left(\frac{r_2}{r_1}\right)^2 = 12^2 = 144.

An auto-rickshaw weighing about 400 kg puts a load F_2 = 400 \times 9.8 \approx 3920 N on the large piston. The force required on the small piston is

F_1 = F_2 \cdot \frac{A_1}{A_2} = \frac{3920}{144} \approx 27 \text{ N},

about the weight of a 2.7-kilogram textbook. That is why the mechanic can do it with his forearm.

Energy is conserved — you pay in distance

A machine that amplifies force for free would violate conservation of energy. Hydraulic lifts do not violate conservation of energy. The force amplification is exactly matched by a distance de-amplification — whatever you gain in force at piston 2 you lose in how far piston 2 moves compared to piston 1.

Here is why, in one clean derivation.

Step 1. Volume of oil conserved.

The oil is incompressible. When piston 1 goes down by a distance d_1, it pushes a volume V = A_1 d_1 of oil into the connecting pipe. That same volume must emerge under piston 2 and push it up by a distance d_2, where V = A_2 d_2. Equating the two expressions for V:

A_1 d_1 = A_2 d_2
\boxed{\; d_2 = d_1 \cdot \frac{A_1}{A_2} \;}

Why: incompressibility says the oil cannot be squeezed into a smaller volume — so every cubic centimetre pushed down by piston 1 must come out somewhere else. In a closed loop, it comes out at piston 2.

Step 2. Multiply force and distance for each piston.

Work done by you on piston 1: W_1 = F_1 d_1.

Work done by the oil on piston 2 (and hence by piston 2 on the load): W_2 = F_2 d_2.

Substitute F_2 = F_1 (A_2/A_1) and d_2 = d_1 (A_1/A_2):

W_2 = \left[F_1 \cdot \frac{A_2}{A_1}\right] \cdot \left[d_1 \cdot \frac{A_1}{A_2}\right] = F_1 d_1 = W_1.
\boxed{\; W_2 = W_1 \;}

Why: the area ratios A_2/A_1 and A_1/A_2 are reciprocals — they cancel in the product F_2 d_2. Whatever you gain in force, you lose proportionally in distance. Work in equals work out. Conservation of energy is safe.

This is the honest price of hydraulic force amplification. The mechanic's pedal has to travel 144 times as far, in total, as the auto-rickshaw's platform rises. That is why hydraulic lifts are slow, and why the pedal has a ratchet — a few strokes, each short, each light, slowly accumulate into a large lift.

Explore the area ratio yourself

Drag the area ratio A_2/A_1 below and watch how the force multiplier and the distance ratio change in opposite directions. The input force is fixed at F_1 = 100 N and the input stroke at d_1 = 0.1 m.

Interactive: force output and stroke distance as area ratio varies Two curves showing output force F2 and input distance to lift by 1 cm as functions of the area ratio A2 over A1, with a draggable vertical line at the current ratio. area ratio $A_2/A_1$ newtons or millimetres × 10 0 2000 4000 6000 10 20 30 40 50 $F_2$ (N) $d_1/d_2$ scaled (mm × 10) drag the red point along the axis
Drag the red dot. As the area ratio grows, the output force $F_2$ (red, solid) rises linearly, while the output stroke $d_2$ (dark, dashed; plotted as $d_1/d_2$ scaled) falls as the reciprocal. Their product — the work done — stays constant at $F_1 d_1 = 10$ J.

Hydraulic machines in India — a tour

The hydraulic idea shows up in almost every heavy machine you see.

Auto-service-station lifts

The two-post and single-post lifts that raise your father's Honda City to the mechanic's waist height are the most literal hydraulic press: a small electric pump creates oil pressure, a large piston lifts the car. Area ratios of 200:1 are common; a 5 kW pump lifts a 1500 kg car easily. The pump is slow on purpose — not because hydraulic amplification is inefficient, but because moving 200 times the fluid takes 200 times the time.

Hydraulic brakes — the reason Indian buses don't kill everyone

A fully loaded Tata Marcopolo bus weighing 16 tonnes coming down a Himalayan ghat at 40 km/h has a kinetic energy of \tfrac{1}{2} \times 16000 \times 11.1^2 \approx 990{,}000 J — a million joules. The driver's foot cannot supply the force to dissipate that much energy as heat in the brake pads through a direct mechanical linkage. Hydraulic brakes amplify the driver's push by area ratios of typically 10–15 at the wheel cylinder and another factor from lever geometry, bringing total mechanical advantage to around 40–60. One sentence of Pascal's law is what makes the bus stoppable.

Schematic of a hydraulic brake system A brake pedal on the left presses a small master-cylinder piston, sending brake fluid through pipes to four wheel cylinders, each pressing a brake pad against a disc or drum at each wheel. master cyl pedal FL wheel FR wheel RL wheel RR wheel brake fluid (DOT 4)
A hydraulic brake system. The driver's foot presses the master-cylinder piston. The pressure is transmitted through a sealed fluid line to all four wheel cylinders simultaneously — Pascal's law guarantees that every wheel gets the same pressure, so every wheel brakes equally (a safety-critical requirement).

There is a deeper point. Because Pascal's law says the pressure is transmitted equally to every branch of the fluid, all four wheel cylinders feel the same brake pressure. That means all four wheels decelerate together. A purely mechanical linkage would have to send separate wires or rods to each wheel and balance their forces manually. Hydraulics does it automatically.

Power steering

When your older cousin parks his Maruti Swift, the steering wheel feels nearly weightless even though the front wheels are carrying half the car's mass against carpet friction. A small engine-driven pump maintains a reservoir of hydraulic fluid at about 80 bar (8 MPa). When you turn the wheel, a valve opens, letting fluid into a cylinder on whichever side assists the turn. Area ratios of 20:1 translate the light turn of the steering wheel into the heavy turn of the front axle.

The JCB excavator

The iconic JCB digs with three hydraulic cylinders per arm — one for the boom, one for the dipper, one for the bucket. Each cylinder has a piston area of tens of square centimetres, fed by a pump delivering oil at up to 300 bar (30 MPa). A 20 cm² piston at 30 MPa produces

F = P A = 30 \times 10^6 \times 20 \times 10^{-4} = 60{,}000 \text{ N},

or about 6 tonnes of force from a single cylinder. That is why one operator with a small hand-lever panel can dig a metro trench.

Worked examples

Example 1: A garage lift for a hatchback

A service-station hydraulic lift has a small piston of diameter d_1 = 2.0 cm and a large piston of diameter d_2 = 30 cm. The oil is incompressible and the pistons are at the same height. A Maruti Alto of mass 750 kg sits on the large piston. What minimum force on the small piston will start to lift the car? If the car must rise by 1.2 m, through what total distance must the small piston be pushed?

Maruti Alto on a hydraulic lift with small and large pistons A hydraulic lift showing the small input piston on the left with a force F1 pushing down, connected by oil to the large output piston on the right carrying the weight W of a 750 kg car. $F_1 = ?$ Alto 750 kg $W = 7350$ N $d_1 = 2$ cm $d_2 = 30$ cm
A Maruti Alto weighing 7350 N sits on the large piston. Pascal's law turns a small push on the left into the required 7350 N of lift on the right.

Step 1. Identify the known quantities and the area ratio.

d_1 = 2.0 cm, d_2 = 30 cm, m = 750 kg.

\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{30}{2.0}\right)^2 = 15^2 = 225.

Why: for circular pistons, area goes as diameter squared. The mechanical advantage depends on diameter squared, not diameter — doubling the large piston's diameter quadruples the force you can lift.

Step 2. Compute the load on the large piston.

F_2 = mg = 750 \times 9.8 = 7350 \text{ N}.

Step 3. Use the mechanical-advantage formula to find F_1.

F_1 = F_2 \cdot \frac{A_1}{A_2} = \frac{7350}{225} = 32.7 \text{ N}.

Why: the force multiplier works both ways. Dividing F_2 by the area ratio gives the input force.

Step 4. Use volume conservation to find the input distance.

A_1 d_1 = A_2 d_2 \quad\Longrightarrow\quad d_1 = d_2 \cdot \frac{A_2}{A_1} = 1.2 \times 225 = 270 \text{ m}.

Why: the oil displaced by the small piston must equal the oil that emerges under the large piston. Since the area ratio is 225, the small piston moves 225 times as far as the large piston does.

The 270 m figure sounds absurd — no piston in a garage travels 270 metres. The catch is that the small piston is cycled: a ratchet mechanism lets the mechanic push it down, release it, let it refill from a reservoir, and push again. Each stroke might be 30 cm, so the mechanic pumps the handle about 900 times. In practice, a motor-driven pump does this far faster.

Step 5. Energy check.

Input work: W_1 = F_1 d_1 = 32.7 \times 270 = 8830 J.

Output work: W_2 = F_2 d_2 = 7350 \times 1.2 = 8820 J.

Why: the small rounding gap (about 10 J out of 8820) is because 7350/225 is not exactly 32.67. Exact arithmetic gives W_1 = W_2 identically.

Result: F_1 \approx 33 N (lighter than a full water bottle); total stroke of the small piston = 270 m (cycled in many short strokes); work done in equals work done out (energy is not amplified, only force is).

What this shows: Force is multiplied by the area ratio of 225. Distance is divided by the same factor. The mechanic trades a small force over a long accumulated distance for a large force over a short distance. Conservation of energy is respected precisely because the distance ratio is the exact reciprocal of the force ratio.

Example 2: Hydraulic brake — estimating wheel-cylinder force

In a Tata Nano, the driver's foot presses the master-cylinder piston with a force of F_1 = 400 N. The master cylinder has a diameter of d_1 = 1.9 cm. Each of the four wheel cylinders has a diameter of d_2 = 2.8 cm. Compute the pressure in the brake line and the force pressing each brake pad against its disc.

Pedal, master cylinder, and wheel cylinder force diagram A force of 400 N on the brake pedal creates pressure in the master cylinder, which is transmitted through brake fluid to the wheel cylinder, where the pressure pushes the brake pad outward against the disc. master $F_1 = 400$ N brake fluid at pressure $P$ wheel cyl $F_2 = ?$ disc $d_1 = 1.9$ cm $d_2 = 2.8$ cm
A 400 N push on the master cylinder creates a pressure $P = F_1/A_1$ in the brake fluid, transmitted undiminished to each wheel cylinder. The wheel cylinder converts the pressure back into a force $F_2$ pressing the brake pad against the disc.

Step 1. Compute the pressure in the brake line.

Master-cylinder area: A_1 = \pi (d_1/2)^2 = \pi (0.0095)^2 = 2.835 \times 10^{-4} \text{ m}^2.

P = \frac{F_1}{A_1} = \frac{400}{2.835 \times 10^{-4}} = 1.41 \times 10^{6} \text{ Pa} = 14.1 \text{ bar}.

Why: pressure is force per unit area. A real car brake line sees pressures of about 50–100 bar during hard braking, which corresponds to much higher driver forces (or servo-amplified effective forces); 400 N is a gentle pedal press, and gives a correspondingly modest brake-line pressure.

Step 2. Compute each wheel-cylinder area.

A_2 = \pi (d_2/2)^2 = \pi (0.014)^2 = 6.158 \times 10^{-4} \text{ m}^2.

Step 3. Apply Pascal's law — the same pressure reaches each wheel.

F_2 = P A_2 = 1.41 \times 10^{6} \times 6.158 \times 10^{-4} = 868 \text{ N}.

Step 4. Check the mechanical advantage directly.

\frac{F_2}{F_1} = \frac{A_2}{A_1} = \frac{(2.8)^2}{(1.9)^2} = \frac{7.84}{3.61} = 2.17.

And 400 \times 2.17 = 868 N. The numbers agree.

Why: computing through the pressure P and computing through the area-ratio shortcut must give the same answer — they are the same physics. In a real car, additional mechanical advantage comes from the brake-pedal lever (typically 4:1) and from hydraulic servo-assist (vacuum booster), bringing the overall multiplier to 30:1 or more.

Result: Brake-line pressure P = 14.1 bar; force on each wheel cylinder = 868 N; total force across four wheels = 3472 N.

What this shows: A 400 N pedal push produces 868 N at every wheel — a 2.17× amplification per wheel, 8.7× total across all four wheels. Pascal's law distributes the driver's one action to all four wheels simultaneously and equally, which is exactly what a safety-critical system requires.

Common confusions

If you have understood Pascal's law, the hydraulic press, and the energy balance, you have the full working content of this topic. What follows is for readers who want to see the limits of the idealisation, the connection to thermodynamics, and how real systems deal with real fluids.

What "incompressible" really means

No real fluid is exactly incompressible. The compressibility \kappa of a fluid is defined by

\kappa = -\frac{1}{V}\frac{\partial V}{\partial P},

measured at constant temperature. For water, \kappa \approx 4.6 \times 10^{-10} \text{ Pa}^{-1} — applying a pressure of 10^8 Pa (1000 atmospheres) compresses water by about 4.6%. For a hydraulic press operating at 300 bar (3 \times 10^7 Pa), water compresses by about 1.4% — small, but not zero.

What does this mean for Pascal's law? Each increment of applied pressure still reaches every point of the fluid, but the fluid's slight compression at the input end means some of the applied work goes into storing elastic energy in the compressed fluid instead of pushing the output piston. In steady state (after the compression has saturated), the transmitted pressure is still equal to the applied pressure minus a tiny correction. Real hydraulic presses achieve 90–98% efficiency, with the losses divided between fluid compressibility, piston seal friction, and pipe flow resistance.

Hydraulic brakes and the boiling of brake fluid

Brake fluid specifications like DOT 3, DOT 4, and DOT 5.1 are based on boiling points, because in heavy braking — say, a truck descending the Konkan Railway ghat section — the brake pads dissipate kinetic energy as heat that partly travels through the wheel cylinder into the fluid. If the fluid boils, a bubble of vapour forms in the line. Vapour is compressible. Suddenly the Pascal's-law chain breaks: the pedal compresses the vapour instead of pushing the pads. This is brake fade, and it is the reason long mountain descents require gear-braking in addition to pedal braking. DOT 4 fluid boils at about 230°C dry; after absorbing moisture from the air for a year, its "wet boiling point" falls to about 155°C. That is why brake fluid is changed every two years, not when it looks dirty.

The hydraulic amplifier as an impedance transformer

A deeper way to understand the hydraulic press is by analogy with electrical transformers. In electricity, a transformer with turn ratio n_2/n_1 maps an input voltage V_1 to an output voltage V_2 = V_1 (n_2/n_1) while transforming current in the opposite sense: I_2 = I_1 (n_1/n_2). Power is conserved: V_2 I_2 = V_1 I_1.

The hydraulic press is the mechanical analogue. The area ratio A_2/A_1 plays the role of the turn ratio. Force is the analogue of voltage, velocity (or displacement rate) is the analogue of current. The "power" — force times velocity — is conserved. The hydraulic system performs mechanical impedance transformation: it converts a high-velocity, low-force input into a low-velocity, high-force output.

This analogy is not a coincidence. Both systems are governed by the same mathematics of linear conservation laws — charge for electricity, volume for incompressible fluids.

Piston friction and the real force equation

A real piston does not slide freely. The seal between piston and cylinder wall exerts a friction force F_\text{fric} that opposes motion. The force balance on the small piston during downstroke is

F_1 - P A_1 - F_\text{fric,1} = 0,

so the pressure actually generated in the fluid is

P = \frac{F_1 - F_\text{fric,1}}{A_1}.

Similarly, at the large piston, the delivered force is

F_2 = P A_2 - F_\text{fric,2}.

Combining,

F_2 = (F_1 - F_\text{fric,1}) \cdot \frac{A_2}{A_1} - F_\text{fric,2}.

Two consequences. First, the measured mechanical advantage is always a little less than the area ratio. Second — and more subtly — friction scales roughly with piston circumference (which goes as diameter), while load capacity scales with piston area (diameter squared), so friction becomes less important as piston diameter grows. That is one reason industrial presses use large pistons: the friction-to-load ratio is favourable.

Historical note on Blaise Pascal

Blaise Pascal (1623–1662) was a French mathematician and physicist who published his treatise on fluid pressure in 1653. He designed an early barometer experiment with his brother-in-law Florin Périer — carried out on the summit of the Puy de Dôme — to show that atmospheric pressure decreases with altitude. The law bearing his name was stated in his Traité de l'équilibre des liqueurs. Today the SI unit of pressure — the pascal (1 Pa = 1 N/m²) — carries his name. The word hydraulics itself comes from the Greek hydor (water) and aulos (pipe), but the ideas that turned it into a modern engineering discipline are Pascal's.

Where this leads next