Hydraulic cylinder thrust
Cylinder thrust from pressure × piston area using 1 MPa = 1 N/mm², reported in newtons and tonnes-force with piston area plus bar and kgf/cm² conversions; add a rod diameter for the retract force and pull/push ratio, with efficiency down to 85–95%.
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Thrust result
Theoretical thrust — efficiency and back pressure reduce it.
Enter the bore to calculate the thrust
Thrust: 0 · Tonne-force: 0 · Piston area: 0
What this tool does
- Size a cylinder before buying: check whether the bore and rated pressure can hold the die, shear or punch load.
- Verify the retract stroke: the rod-side area is smaller, so pulling and retracting may need a bigger bore.
- Turn a pressure-gauge reading into tonnage: recompute the real thrust from MPa on site.
Example
Input
63
Output
缸径 63 mm、压力 10 MPa 的油缸推力：31172.453105 N（3.178706 tf） 活塞面积 πd²/4：3117.245311 mm²（31.172453 cm²） 压力换算：10 MPa = 100 bar = 101.97162 kgf/cm² 公式：10 MPa × 3117.245311 mm² × 100% = 31172.453105 N 活塞杆直径 35 mm，杆截面积：962.11275 mm² 回程拉力（有效面积 2155.13256 mm²）：21551.325604 N（2.197624 tf） 回程推力比：69.135802%
A 100 mm bore at 16 MPa pushes about 125.66 kN (12.81 tf) and retracts about 94.25 kN with a 50 mm rod.
Frequently asked questions
Why does pressure times area give newtons?
Because 1 MPa = 1 N/mm², so the units cancel: 1 MPa × 1000 mm² = 1000 N. Use MPa and mm² and the answer comes out in newtons with no extra factor.
Why is the retract force smaller?
The rod-side effective area is the piston area minus the rod section, so a thicker rod pulls less. At a 100 mm bore with a 50 mm rod the retract area is only 75% of the piston area, giving roughly 75% of the push force.
Should I account for efficiency?
Yes. Mechanical and volumetric losses typically total 5%–15%; use 90% for a realistic estimate. The 100% default is the theoretical upper bound.
How is the tonne-force figure derived?
One tonne-force is 9806.65 N (g = 9.80665 m/s²), so 125663.7 N ≈ 12.81 tf. Note that Chinese practice sometimes folds a tonne of mass at 9.8 or 10 — state your convention when quoting.
What happens if I leave the rod diameter at 0?
Only the push force is reported; the retract force and ratio are omitted. A rod diameter at or above the bore is rejected as invalid.
Can I enter pressure in bar?
No, the field is read as MPa. Use 1 MPa = 10 bar = 10.197162 kgf/cm², so divide bar by 10 first.
Keywords：油缸推力液压缸推力cylinder force挤压力计算活塞面积吨力换算hydraulic thrust回程拉力