A customer called me last week. “My cylinder won’t push the slide. I turned the air up to 8 bar and it still stalls.” I asked him what bore it was. “Ø32.” I asked what the load was. “50 kg on linear bearings.” I told him to put in a Ø63. He pushed back — “Ø32 should be enough, it’s just 50 kg.” This is the calculation I walk through on the phone.
The force formula
A cylinder’s push force is:
F = P × A
Where P is gauge pressure in N/mm² (1 bar = 0.1 N/mm²) and A is the piston area in mm². For a double-acting cylinder, the push (extend) uses the full bore area. The pull (retract) uses the bore area minus the rod area. Always size for the extend stroke unless the load is pulling on retract.
For Ø32 at 6 bar (0.6 N/mm²): A = π × 16² = 804 mm². F = 0.6 × 804 = 483 N.
The load is 50 kg = 490 N. The cylinder’s rated force (483 N) is barely more than the weight (490 N). That’s not a safety factor — that’s a coin flip. It moves when the air is at 6 bar, but the plant air is actually 5.5 bar at the machine. F = 0.55 × 804 = 442 N. The load is 490 N. It stalls. That’s exactly what happened.
The safety factor that matters
For a horizontal slide on linear bearings, friction is small (μ ≈ 0.005). The force is mostly acceleration. But I don’t use the theoretical minimum. I use a safety factor of 2 to 3 for pushing loads. Why? Because:
- Plant air pressure varies (6 bar at the compressor, 5 bar at the machine after filters and drops)
- The slide gets dirty, increasing friction
- The customer might add a heavier part later
- Cylinder seal wear reduces output over time
For a 50 kg horizontal load, the needed force is F_load × SF = 490 × 2.5 = 1,225 N. At 6 bar, I need A = 1,225 / 0.6 = 2,042 mm². That’s a bore of √(2,042 × 4 / π) = 51 mm. So Ø50 is the minimum. Ø63 gives margin.
The quick-reference table
Push force in Newtons at 6 bar, with 2.5x safety factor for horizontal loads:
| Bore | Area (mm²) | Theoretical F (N) | With 2.5x SF (N) | Max horizontal load (kg) |
|---|---|---|---|---|
| Ø20 | 314 | 188 | 75 | 7.7 |
| Ø25 | 491 | 295 | 118 | 12.0 |
| Ø32 | 804 | 483 | 193 | 19.7 |
| Ø40 | 1,257 | 754 | 302 | 30.8 |
| Ø50 | 1,963 | 1,178 | 471 | 48.1 |
| Ø63 | 3,117 | 1,870 | 748 | 76.3 |
| Ø80 | 5,027 | 3,016 | 1,206 | 123 |
| Ø100 | 7,854 | 4,712 | 1,885 | 192 |
For 50 kg, I need Ø50 minimum. The customer’s Ø32 was rated for 20 kg max. He was overloading it by 2.5x.
The vertical load exception
If the cylinder is lifting a vertical load (not pushing horizontally), I use a safety factor of 4. If the air fails, the load drops. A counterbalance or check valve is also needed. For a vertical lift of 20 kg, the force is 20 × 9.8 × 4 = 784 N. That needs Ø50 (754 N is close, Ø63 is safer). I never use a 2x factor for vertical loads. If the air bleeds off, the part crashes.
Don’t turn up the pressure
The customer turned the regulator to 8 bar to compensate. At 8 bar, Ø32 gives F = 0.8 × 804 = 643 N. That pushes the 50 kg load. But: higher pressure costs more air (50% more), the cylinder seals wear faster, and the impact at end-of-stroke is harder. The cylinder was the wrong size. Turning up the pressure is a band-aid that buys a few months before the seals blow out. I told him to keep the regulator at 6 bar and change the bore.
Retract force is less than extend
On the retract stroke, the rod takes up some of the piston area. For a Ø32 cylinder with Ø12 rod: retract area = 804 – 113 = 691 mm². F = 0.6 × 691 = 415 N (vs 483 N extend). If the load is pulling on retract (like a spring return or a lifting motion on pull), I size for the retract area. People forget this and wonder why the cylinder pulls slower than it pushes.
The number I check: rated force at actual plant pressure (not 6 bar nameplate, but the real 5-5.5 bar at the machine), divided by the load, must be at least 2.5 for horizontal, 4 for vertical. If it’s under, I go up a bore size. Turning up the regulator isn’t a fix — it’s a delay. The cylinder that stalled wasn’t weak; it was half the size it should have been.
