Linear Actuator Force Calculator

Use F = P × A, where A = π × d² ÷ 4 is the piston face area. Worked example: 100 bar (10 MPa) on a 50 mm bore. A = π × 0.05² ÷ 4 = 0.0019635 m², so F = 10,000,000 × 0.0019635 = 19,634.95 N, about 19.6 kN or 4,414 lbf. Keep units consistent — convert bar to pascals and millimeters to meters before multiplying.

Push (extend) uses the full bore area; pull (retract) uses the annulus — bore area minus rod area — because the rod occupies part of the piston face. Worked example: 100 bar, 50 mm bore, 25 mm rod. Push area = π × 50² ÷ 4 = 1,963.5 mm² → 19,634.95 N. Pull area = π × (50² − 25²) ÷ 4 = 1,472.6 mm² → 14,726.22 N, about 25% less. Always size for the weaker direction if the load reverses.

Multiply bar by 14.5038 to get psi (1 bar = 100,000 Pa, 1 psi = 6,894.76 Pa, so the ratio is exact to 4 decimals). Worked example: a 3-inch bore at 1,500 psi. A = π × (3 × 0.0254)² ÷ 4 = 0.0045604 m², F = 10,342,136 Pa × 0.0045604 = 47,163.94 N, which is 10,602.88 lbf (divide newtons by 4.44822). Enter psi directly in the calculator — the conversion is handled for you.

Use F = 2π × T × η ÷ lead, where T is motor torque, η is screw efficiency, and lead is travel per revolution. Worked example: T = 2 N·m, η = 0.9 (ball screw), lead = 5 mm. F = 2π × 2 × 0.9 ÷ 0.005 = 2,261.95 N, about 2.26 kN or 509 lbf. Smaller lead or higher torque raises force; efficiency is the dominant term, so pick it honestly (see below).

Ball screw: η ≈ 0.85–0.95, use 0.9 as a typical value — rolling balls waste little energy. Lead (sliding) screw: η ≈ 0.3–0.5, use 0.4 as typical — thread friction eats most of the torque. Low efficiency also means the screw is usually self-locking (it will not back-drive under load), which is a feature for holding positions without power but a penalty for speed and heat. Acme screws sit in the lead-screw range; roller screws behave like ball screws.

Rearrange to lead = 2π × T × η ÷ F or T = F × lead ÷ (2π × η) and use the 3-way solver. Worked example: you need 2,261.95 N and have a 2 N·m motor with a 0.9-efficiency ball screw. lead = 2π × 2 × 0.9 ÷ 2,261.95 = 0.005 m = 5 mm. Rule of thumb: halve the lead to double the force at the same torque — but speed halves too, so check your required travel speed before committing.

Solve d = √(4 × A ÷ π), with A = F ÷ P (add rod² under the root for the pull direction). Worked example: need 19,634.95 N at 100 bar. A = 19,634.95 ÷ 10,000,000 = 0.0019635 m², d = √(4 × 0.0019635 ÷ π) = 0.05 m = 50 mm — then round UP to the next standard bore and re-check force. Oversizing slightly also buys margin against seal friction, which typically steals 3–10% of theoretical force.

Yes — F = P × A is identical; only the pressure numbers are smaller. Typical shop air is 6 bar (87 psi). Worked example: 63 mm bore at 6 bar. A = π × 0.063² ÷ 4 = 0.0031172 m², F = 600,000 × 0.0031172 = 1,870.35 N, about 191 kgf. Because air compresses, pneumatic force sags under load and mid-stroke stiffness is poor — size generously and never use pneumatics where a stalled, rigid hold matters.

Yes, in the same proportion as force: for a fixed pump flow, retract is faster than extend because the annulus holds less oil. Speed ratio = bore area ÷ annulus area = d² ÷ (d² − d_rod²). Worked example: 50 mm bore, 25 mm rod → 2,500 ÷ (2,500 − 625) = 1.333, so retract is 33% faster (and 25% weaker, matching the force FAQ above). Regenerative circuits exploit this by routing rod-side oil back to the cap end for high-speed, low-force approach strokes.

Derate theoretical force by 10–20% for fluid-power cylinders: seal and wiper friction typically consumes 3–10%, pressure drops and temperature drift take the rest. For electric actuators, also respect the duty cycle — a screw rated 2,261 N intermittently may only sustain half that continuously before the motor overheats. For long strokes, check column buckling of the rod/screw, which can fail far below the force rating. When in doubt, one standard bore size up is the cheapest insurance.