Enter nonnegative pounds-force and a positive area in square inches or square feet. The default 100 lb-force ÷ 10 in² displays 10 psi, 1,440 psf and 68.948 kPa. This is supplied force divided by supplied area, not pounds-mass converted into pressure.
The form is useful for contact pressure checks, simple piston loads, flat-pad loading, gasket estimates, fixture clamping, and classroom force-over-area problems. It is not a substitute for a structural or pressure-vessel design calculation, and it does not infer the contact area from shape dimensions. If you need to convert pressure units after finding psi, use the pressure calculator, psi conversion, or kPa to psi conversion. If you need force units before pressure, use the force calculator, and if your area is in another unit, use the area converter.
What the calculator does
Enter three values:
- Force, labeled in lb-force
- Contact area
- Area unit, either square inches or square feet
If the area unit is square feet, the form multiplies the area by 144 because one square foot contains 144 square inches. It then divides the force by the area in square inches. The result panel shows the primary pressure in psi, the force used, the area used in square inches, pounds per square foot, and kilopascals.
This means the page has no hidden “one square inch” assumption once you enter your own area. The default 10 square inches is only a starting example. If you leave it unchanged, then yes, your result is based on a 10 in² area.
Formula used by the form
For area entered in square inches:
For area entered in square feet:
The supporting pressure units are:
The form rejects negative force values and rejects zero or negative area values. A zero area would require division by zero, and a negative contact area has no physical meaning in this context.
Conversion example matching the default
Use the default values:
- Force: 100 lb-force
- Contact area: 10
- Area unit: Square inches
The area is already in square inches, so the calculator does not multiply by 144:
Then it divides force by area:
The primary result is 10 psi. The support rows show 100 lb-force, 10 in², 1,440 psf, and 68.948 kPa after display rounding. The note states that 100 lb-force spread over 10 square inches equals 10 psi, matching the calculation exactly.
If you change only the area unit to square feet and leave the area value as 10, the calculation changes dramatically:
That example shows why area units matter as much as the force value.
Reference table
| Force | Area entered | Area used | Calculator PSI | kPa |
|---|---|---|---|---|
| 100 lb-force | 10 in² | 10 in² | 10 | 68.948 |
| 250 lb-force | 10 in² | 10 in² | 25 | 172.369 |
| 500 lb-force | 25 in² | 25 in² | 20 | 137.895 |
| 1000 lb-force | 1 ft² | 144 in² | 6.9444 | 47.880 |
| 50 lb-force | 0.5 in² | 0.5 in² | 100 | 689.476 |
Notice how a smaller contact patch can create a much larger psi from a modest force. That is why a narrow punch, high heel, cutting edge, or small piston face can produce high pressure even when the total force is not large.
Pressure, stress, and practical use
The output is a pressure or average contact stress over the entered area. Real contact pressure may not be perfectly uniform. A soft gasket, uneven floor pad, curved tire patch, or rough surface can concentrate load in some regions and reduce it in others. The calculator gives the simple average, which is often the first step before a more detailed design check.
For hydraulics, a piston calculation often starts with pressure and area to find force. This page runs the opposite direction: force and area produce pressure. If your problem asks for cylinder force from psi, use a dedicated piston or force calculation rather than reversing this page mentally without checking units.
For structural supports and pads, the area should be the bearing area that transfers load. Do not use the full plan dimensions of a machine if only small feet touch the slab. For soil, floor, or material limits, compare the resulting pressure with the allowable bearing value in the same units.
Common mistakes
- Entering pounds of mass without considering whether it is a force. Static weight on Earth often has the same number in pounds-force, but dynamic loads do not.
- Forgetting to replace the default 10 in² area.
- Using square feet as if one foot equals 12 inches; area requires 144 square inches per square foot.
- Dividing by the total object area instead of the actual contact area.
- Treating the result as a pressure unit conversion rather than a force-over-area calculation.
Match force and contact area
Keep the entered force and the area over which it acts together, including their units. Compare another contact area while holding force fixed, then explore a different force separately. This makes the pressure scenario reproducible instead of treating a pound-valued quantity alone as a pressure specification.
Sources
- NIST, Definitions of the SI base units and SI prefixes — SI unit and prefix definitions for pressure comparisons.
- BIPM, The International System of Units, 9th edition — pascal definition as a coherent derived unit.
- NIST, Handbook 44, Appendix C — exact international inch, foot, and pound definitions used for psi and square-foot area.
- NIST, standard acceleration of gravity — exact standard-gravity value used to define pound-force.