Hydraulic Cylinder Calculation Formula
F_push = P × π × D² / 4F_pull = P × π × (D² − d²) / 41 bar = 0.1 N/mm² = 14.504 psi
Use the psi / inch helper fields above for US units; results are shown in metric tons, kN, lbf and US tons.
How to Calculate Hydraulic Cylinder Force: Worked Examples
Metric: 160 bar, 80 mm bore, 45 mm rod. Area = 5,027 mm²; push = 16 N/mm² × 5,027 = 80.4 kN (8.2 tf); pull = 55.0 kN (5.6 tf).
US: 2,500 psi, 3 in bore, 1.5 in rod. Area = 7.07 in²; push = 17,671 lbf (8.8 US tons); annulus = 5.30 in², pull = 13,254 lbf.
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Sizing: Safety Factor and Rod Buckling
Size the cylinder for 1.2–1.3 × the required force to cover seal friction, pressure losses and dynamic loads; real output is typically 90–95% of the theoretical value.
For push loads with a stroke longer than about 10 × rod diameter, check rod buckling (Euler) for your mounting style. Pivot (clevis/trunnion) mounts have a longer free buckling length than rigid flange mounts, so they need a thicker rod or a stop tube. For a custom cylinder, send us the stroke and mounting style and we will check buckling for your case.
Required Bore for a Given Force
D = √(4 × F / (π × p))p in N/mm² = bar × 0.1
Metric: a press needs 25 t at 160 bar. F = 25 × 9,807 = 245,166 N; D = √(4 × 245,166 / (π × 16)) = 139.7 mm. The next standard bore is Ø160 mm, which gives 32.8 t at 160 bar.
US: 20 US tons (40,000 lbf) at 2,500 psi needs A = 40,000 / 2,500 = 16.0 in², so D = 4.51 in (114.6 mm). The next metric bore is Ø125 mm (4.92 in), giving about 47,553 lbf (23.8 US tons) at 2,500 psi.
| Load | Calculated bore | Standard bore | Force of the standard bore |
|---|---|---|---|
| 5 t | 62.5 mm | Ø63 mm | 5.1 t |
| 10 t | 88.3 mm | Ø100 mm | 12.8 t |
| 20 t | 124.9 mm | Ø125 mm | 20.0 t |
| 30 t | 153.0 mm | Ø160 mm | 32.8 t |
| 50 t | 197.5 mm | Ø200 mm | 51.3 t |
Round up to the next standard bore and keep 20–30 % in hand for friction and pressure losses — 20 t at 160 bar needs exactly Ø125 mm, so choose Ø160 mm or a higher pressure if you need a reserve.
Force Table at 160 bar (kN, Metric Tons and US Tons)
| Bore | kN | metric t | US tons |
|---|---|---|---|
| Ø40 mm (1.57 in) | 20.1 | 2.1 | 2.3 |
| Ø50 mm (1.97 in) | 31.4 | 3.2 | 3.5 |
| Ø63 mm (2.48 in) | 49.9 | 5.1 | 5.6 |
| Ø80 mm (3.15 in) | 80.4 | 8.2 | 9.0 |
| Ø100 mm (3.94 in) | 125.7 | 12.8 | 14.1 |
| Ø125 mm (4.92 in) | 196.3 | 20.0 | 22.1 |
| Ø160 mm (6.30 in) | 321.7 | 32.8 | 36.2 |
| Ø200 mm (7.87 in) | 502.7 | 51.3 | 56.5 |
160 bar is the working pressure of the Fenitsa FDH series (240 bar test, 250 bar option). For every bore at 100, 160, 210 and 250 bar, with rods and strokes, see the hydraulic cylinder tonnage chart.
Force Chart – Metric (kN)
| Bore | 100 bar | 160 bar | 250 bar |
|---|---|---|---|
| 32 mm | 8.0 kN | 12.9 kN | 20.1 kN |
| 40 mm | 12.6 kN | 20.1 kN | 31.4 kN |
| 50 mm | 19.6 kN | 31.4 kN | 49.1 kN |
| 63 mm | 31.2 kN | 49.9 kN | 77.9 kN |
| 80 mm | 50.3 kN | 80.4 kN | 125.7 kN |
| 100 mm | 78.5 kN | 125.7 kN | 196.3 kN |
| 125 mm | 122.7 kN | 196.3 kN | 306.8 kN |
| 160 mm | 201.1 kN | 321.7 kN | 502.7 kN |
| 200 mm | 314.2 kN | 502.7 kN | 785.4 kN |
Force Chart – US Units (lbf)
| Bore | Area (in²) | 1,000 psi | 2,000 psi | 3,000 psi |
|---|---|---|---|---|
| 1.5 in | 1.77 | 1,767 | 3,534 | 5,301 |
| 2 in | 3.14 | 3,142 | 6,283 | 9,425 |
| 2.5 in | 4.91 | 4,909 | 9,817 | 14,726 |
| 3 in | 7.07 | 7,069 | 14,137 | 21,206 |
| 3.5 in | 9.62 | 9,621 | 19,242 | 28,863 |
| 4 in | 12.57 | 12,566 | 25,133 | 37,699 |
| 5 in | 19.63 | 19,635 | 39,270 | 58,905 |
| 6 in | 28.27 | 28,274 | 56,549 | 84,823 |
| 8 in | 50.27 | 50,265 | 100,531 | 150,796 |
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Hydraulic Cylinder Force – FAQ
How do you calculate hydraulic cylinder force?
F = P × A. In metric units use bar × 0.1 (N/mm²) × area in mm²; in US units use psi × area in in². An 80 mm bore at 160 bar gives 80.4 kN (8.2 tf); a 3 in bore at 2,000 psi gives 14,137 lbf.
How do I calculate pull (retract) force?
Subtract the rod area: F = P × π × (D² − d²) / 4.
What is the difference between tons and US tons?
One metric ton-force is 9.807 kN (2,204.6 lbf); one US (short) ton is 2,000 lbf (8.9 kN).
Should I add a safety factor?
Yes, typically 20–30% over the required force, and check rod buckling for long push strokes.
How do I find the bore for a required force?
Use D = √(4F / (π × p)) with F in N and p in N/mm² (bar × 0.1), then round up to the next standard bore. Example: 25 t at 160 bar needs 139.7 mm, so a Ø160 mm cylinder.
Where can I find a hydraulic cylinder tonnage chart?
The hydraulic cylinder sizes and tonnage chart lists every standard bore from Ø40 to Ø200 mm with the force in metric tons and US tons at 100 to 250 bar.