Cylinder force calculator
Force = pressure × area. A Ø63 mm bore at 160 bar pushes 49.9 kN; with a Ø36 mm rod it pulls 33.6 kN, because the pressure on the rod side acts only on the ring round the rod. On 20 L/min it goes out at 0.107 m/s and comes back faster, at 0.159 m/s.
Enter the bore, rod, pressure and stroke, with the pump’s flow for a hydraulic cylinder or the cycles a minute for a pneumatic one. The cylinder is drawn in section, pushing or pulling, to scale.
Theoretical force: seal friction and back pressure in the other side take some of it.
- Push F1
- Pull F2
- Areas A1 / A2
- 31.17 / 20.99cm²
- Speed out / in
- 0.107 / 0.159m/s
- Stroke time out / in
- 2.8 / 1.9s
How it’s calculated
F1 = p·π/4·D² pushing, F2 = p·π/4·(D² − d²) pulling (1 bar = 0.1 N/mm²). Speed v = Q / A: v [m/s] = Q [L/min] / (6 × A [cm²]), and the stroke time is L / v. Free air a stroke (SMC) = A × L × (P + 0.1) / 0.1, with P the gauge pressure in MPa; a cycle is the stroke out on A1 and back on A2.
Mill-type cylinders: bore and rod (Bosch Rexroth CDH1), forces at 160.0 bar
| Bore D | Rod d | A1, cm² | A2, cm² | Push F1, kN | Pull F2, kN |
|---|---|---|---|---|---|
| 40 | 22 | 12.57 | 8.77 | 20.11 | 14.02 |
| 40 | 28 | 12.57 | 6.41 | 20.11 | 10.25 |
| 50 | 28 | 19.63 | 13.48 | 31.42 | 21.56 |
| 50 | 36 | 19.63 | 9.46 | 31.42 | 15.13 |
| 63 | 36 | 31.17 | 20.99 | 49.88 | 33.59 |
| 63 | 45 | 31.17 | 15.27 | 49.88 | 24.43 |
| 80 | 45 | 50.27 | 34.36 | 80.42 | 54.98 |
| 80 | 56 | 50.27 | 25.64 | 80.42 | 41.02 |
| 100 | 56 | 78.54 | 53.91 | 125.66 | 86.26 |
| 100 | 70 | 78.54 | 40.06 | 125.66 | 64.09 |
| 125 | 70 | 122.72 | 84.23 | 196.35 | 134.77 |
| 125 | 90 | 122.72 | 59.10 | 196.35 | 94.56 |
| 140 | 90 | 153.94 | 90.32 | 246.30 | 144.51 |
| 140 | 100 | 153.94 | 75.40 | 246.30 | 120.64 |
| 160 | 100 | 201.06 | 122.52 | 321.70 | 196.04 |
| 160 | 110 | 201.06 | 106.03 | 321.70 | 169.65 |
| 180 | 110 | 254.47 | 159.44 | 407.15 | 255.10 |
| 180 | 125 | 254.47 | 131.75 | 407.15 | 210.80 |
| 200 | 125 | 314.16 | 191.44 | 502.65 | 306.31 |
| 200 | 140 | 314.16 | 160.22 | 502.65 | 256.35 |
| 220 | 140 | 380.13 | 226.19 | 608.21 | 361.91 |
| 220 | 160 | 380.13 | 179.07 | 608.21 | 286.51 |
| 250 | 160 | 490.87 | 289.81 | 785.40 | 463.70 |
| 250 | 180 | 490.87 | 236.40 | 785.40 | 378.25 |
| 280 | 180 | 615.75 | 361.28 | 985.20 | 578.05 |
| 280 | 200 | 615.75 | 301.59 | 985.20 | 482.55 |
| 320 | 200 | 804.25 | 490.09 | 1286.80 | 784.14 |
| 320 | 220 | 804.25 | 424.12 | 1286.80 | 678.58 |
Why does a cylinder pull less than it pushes?
Pushing, the oil or air presses on the whole piston, π/4·D². Pulling, it presses on the rod side, where the rod takes up the middle: only the ring π/4·(D² − d²) is left. The same flow fills that smaller side sooner, so the cylinder comes in faster than it goes out, by the ratio of the two areas.
How much air does a pneumatic cylinder use?
Each stroke fills the swept volume with air at the working pressure; as free air that is the volume times (gauge pressure + atmosphere) / atmosphere. SMC’s example, a Ø50 mm cylinder with a 600 mm stroke at 0.5 MPa, takes about 13.0 litres of free air a cycle with a Ø20 mm rod, before the air in its tubing. Multiply by the cycles a minute and the cylinders for the compressor’s load.
Will the cylinder really give this force?
No: this is the theoretical force. Seals rub and the other side has to exhaust, so the rod gives less. Hydraulic catalogues print the theoretical figure; for air, SMC sizes a cylinder so the load is at most 70 % of it standing still and 50 % moving, and Festo allows about 10 % for friction.