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Square Inches to Square Centimeters

Square Inches to Square Centimeters

Restate a hydraulic cylinder's effective area from in² to cm² so a psi rating and a bar gauge agree on the same force in pounds, newtons or kilograms-force.

A Bore Quoted in Inches, a Commissioning Sheet in Metric

The press cylinder on the order confirmation reads "4 in bore, 12.57 in² effective area, 3,000 psi rated". The commissioning paperwork sitting next to it wants the same cylinder described in square centimetres and bar, with the clamping force in kilonewtons so it can be checked against the tooling limit.

American fluid-power catalogues have never left the inch. Bore and rod diameters, effective areas, seal kits and port threads are all listed that way, while the machine build, the risk assessment and the operator's gauge are metric. Converting the area is the single step that lets the two halves of the job agree on a force.

Conversion factor: 1 in² = 6.4516 cm² exactly, because an inch is exactly 2.54 cm and 2.54² = 6.4516. So that 12.566 in² piston face is 81.07 cm², and 3,000 psi is 206.8 bar.

Where Inch Areas Show Up in a Fluid-Power System

Effective areas printed on the catalogue page

Cylinder tables list a cap-end area and a rod-end area in square inches for every bore and rod combination. Those two figures, not the bore letter, are what any force calculation actually uses.

Gauges in psi, paperwork in bar

A relief valve set at 3,000 psi is 206.8 bar; 2,000 psi is 137.9 bar. Pair a psi setting with an inch area or a bar setting with a centimetre area, never one of each.

Force is nothing but pressure times area

Multiply and you are done, provided the units match. The same cylinder gives 37,699 lbf or 167.7 kN — one number, two vocabularies, and a rejected sign-off if they disagree.

Seals and rods carry inch part numbers

A 2 in rod takes an inch-series wiper and rod seal even when the machine around it is metric, so the annulus area you subtract comes off an imperial table as well.

Working a Bore Figure Through the Page

Have the cylinder data sheet and the machine spec side by side. You are usually doing one of two things: restating a supplied area in metric, or starting from a force the machine has to produce and finding out which bore delivers it.

1

Put the catalogue area in the left box

Enter the effective area straight from the table — 12.566 for a 4 in bore cap end. A comma works as the decimal mark just as well as a dot, and stray spaces are ignored, so a figure pasted out of a PDF still parses.

2

Read the metric area and multiply

81.07 cm² appears as you type. Since one bar acting on one square centimetre is ten newtons, 81.07 × 206.8 × 10 works out at about 167,700 N — the same 167.7 kN the imperial route produces.

3

Reverse it when sizing from a force

If the duty says 200 kN at 207 bar, the area you need is 96.7 cm². Type that into the right-hand box, or hit the swap arrows, and 14.99 in² comes back — which sends you to the 5 in bore, not the 4 in.

4

Copy it into the build documents

Each field has a copy button that hands over the digits alone, without the unit or the thousands spacing, so the value drops into a commissioning form or a force calculation sheet ready to use. Ctrl + C over a selection does the same.

Retracting is not the same stroke: the rod eats into the piston face, so a 4 in cylinder with a 2 in rod pulls on 9.425 in² (60.81 cm²), a quarter less force than it pushes with.

Bore Size, Effective Area and the Force It Delivers

Cap-end areas for the standard bore sizes, each shown in both unit systems with the push force the cylinder develops at a 3,000 psi (206.8 bar) setting. Rod-end figures are always lower; subtract the rod area before you quote a pulling force.

Bore Cap-end area (in²) Cap-end area (cm²) Push at 3,000 psi
1.5 in 1.767 11.40 5,301 lbf · 23.6 kN
2 in 3.142 20.27 9,425 lbf · 41.9 kN
2.5 in 4.909 31.67 14,726 lbf · 65.5 kN
3 in 7.069 45.60 21,206 lbf · 94.3 kN
4 in 12.566 81.07 37,699 lbf · 167.7 kN
5 in 19.635 126.68 58,905 lbf · 262.0 kN
6 in 28.274 182.41 84,823 lbf · 377.3 kN
8 in 50.265 324.29 150,796 lbf · 670.8 kN

Notice how fast the column climbs. Stepping from 4 in to 5 in is only a quarter more diameter but 56 % more piston face, which is why an underspecified press is nearly always fixed by dropping in the next bore up rather than by winding the relief valve higher. Pumps have to keep up too: the larger face swallows proportionally more oil per millimetre of travel, so the same flow gives a slower stroke.

Older gauges: a dial marked kgf/cm² reads almost the same as bar — 1 kgf/cm² = 0.980665 bar — and multiplied by an area in cm² it gives force directly in kilograms-force.

What Earns Its Keep During a Cylinder Swap

Sizing and checking in one place

Either field accepts input, so you can restate a supplied area in metric or start from the area a required force demands and see which imperial bore covers it.

Unit list you can search

Both dropdowns hold all thirteen area units and filter as you type, so switching the output to mm² for a seal groove drawing takes one keystroke.

Values ready to paste

The copy control strips the unit and the digit grouping, which keeps a converted area usable in a spreadsheet formula instead of arriving as text.

Runs entirely on your machine

Every figure is worked out inside the page, so customer equipment data typed in on a service laptop is never transmitted to a server.

Questions from the Fluid-Power Bench

Why does the rod end of a cylinder pull less than the cap end pushes?

On the retract stroke the oil only acts on the ring of piston left around the rod, so the working surface is an annulus rather than a full disc. Take the 4 in bore: the cap end is 12.566 in² (81.07 cm²), while a 2 in rod removes 3.142 in² and leaves 9.425 in², or 60.81 cm². At the same 206.8 bar that is 125.8 kN pulling against 167.7 kN pushing. Fit a fatter 2.5 in rod and the annulus drops to 7.658 in² (49.4 cm²) and the pull to 102.2 kN. Machines that have to grip or draw hard on the way back need that annulus checked first, because catalogues put the flattering cap-end number in the headline column.

How do I turn psi times square inches into bar times square centimeters?

Convert both quantities and the two factors settle into a single tidy constant. Dividing psi by 14.5038 gives bar, and multiplying in² by 6.4516 gives cm²; 6.4516 ÷ 14.5038 comes to 0.44482, and since a bar on a square centimetre is ten newtons, the product is 4.4482 N per pound-force — exactly the pound-force to newton factor. So the metric route can never disagree with the imperial one. Test it on the 4 in bore: 12.566 × 3,000 = 37,699 lbf, and 81.07 × 206.8 × 10 = 167,700 N, which is 37,699 × 4.4482. Use whichever pair the paperwork asks for and treat the other as a check.

What bore do I need to produce 200 kN at 207 bar?

Divide the force by the pressure in newtons per square centimetre: 200,000 N ÷ (207 × 10) = 96.6 cm², which this page returns as 14.98 in². Working back to a diameter, that is a hair over 4.36 in — above the 4 in bore's 12.566 in² and comfortably inside the 5 in bore's 19.635 in². Take the 5 in and you have real margin: it reaches 200 kN at about 158 bar, so the relief sits well below its ceiling, the pump runs cooler and the seals see an easier life. Squeezing the same duty out of the 4 in bore would mean winding the setting up to nearly 247 bar, past the rated 207, with nothing left in reserve for a sticky workpiece.

Does it matter whether the pressure is gauge or absolute?

Not for the area arithmetic, which is why the distinction rarely trips anyone up on a hydraulic bench. The piston has atmosphere on its far side pressing back with the same one bar or so, and the two cancel — a gauge reading is already the differential that does the work. Every catalogue rating and every figure in the table above is gauge pressure, written psig or barg when someone is being careful. It starts to matter when a chamber is not vented to atmosphere: a cylinder working inside a pressurised vessel, or one whose return line sits at 8 bar of back pressure, loses that back pressure multiplied by the opposing area. On the 4 in bore, 8 bar on the 60.81 cm² annulus is 4.9 kN removed from the push.

My gauge reads kgf/cm² — how does that fit with an inch area?

Convert the area first and the rest is a single multiplication. One kilogram-force per square centimetre is 0.980665 bar, near enough to bar that older Japanese and European machines were often specified interchangeably in the two. The convenience is that kgf/cm² multiplied by an area in cm² yields force straight in kilograms-force, no constant needed. A gauge showing 210 kgf/cm² on the 81.07 cm² face is 17,025 kgf, or 167 kN once multiplied by 9.80665. Convert 210 kgf/cm² to 205.9 bar and the newton route lands within a fraction of a percent, so either path is fine as long as the area has already been moved out of square inches.

in²
cm²

Cylinder Bore Areas in Square Inches and Square Centimeters

1.767 in² (1.5 in bore)=11.40 cm²
3.142 in² (2 in bore)=20.27 cm²
7.069 in² (3 in bore)=45.60 cm²
12.566 in² (4 in bore)=81.07 cm²
19.635 in² (5 in bore)=126.68 cm²
28.274 in² (6 in bore)=182.41 cm²

Square Inch (in²)

The unit American fluid-power catalogues quote effective piston areas in. Paired with psi it gives force in pounds directly, which is why bore tables, rod sizes and seal part numbers all stayed imperial.

Square Centimeter (cm²)

Exactly 0.15500031 in², or 6.4516 cm² to the square inch. Metric fluid power leans on it because a bar acting on a square centimetre is ten newtons flat, turning any force check into one multiplication.

Multiply by 6.4516 — the inch is exactly 2.54 cm, so the area factor is 2.54²
One bar on one cm² is 10 N, so cm² × bar × 10 gives force straight in newtons
For the retract stroke, subtract the rod area before converting
Sizing from a duty? Type the cm² you need on the right and read the bore area in in²
Want to learn more? Read documentation →
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