The Cylinder That Couldn’t Push (Undersized Bore)
We sized a clamp cylinder (article 85). The load was 500 N (the part to clamp). We picked a Ø32 cylinder (at 6 bar, it gives ~480 N). On the floor, the cylinder couldn’t clamp (it stalled at 450 N). The problem: the bore was undersized (Ø32 gives 480 N, but the actual force needed was 500 N, and the pressure dropped to 5.5 bar on the floor — the force dropped to 440 N). We upsized to Ø40 (at 5.5 bar, it gives ~690 N). It clamped. The mistake was calculating the bore at 6 bar (nominal) without the pressure drop (article 149) and the margin.
Cylinder bore sizing (deep) accounts for the pressure drop and the margin. This article revisits article 85 (more detail).
The Force Formula (Revisit)
The cylinder force (push) is:
F = P × A = P × π × (D/2)²
Where P is the pressure (bar, or N/mm²), A is the piston area (mm²), D is the bore (mm). For Ø32: A = π × 16² = 804 mm². At 6 bar (0.6 N/mm²): F = 482 N. At 5.5 bar (0.55 N/mm²): F = 442 N. The pressure drop (article 149) reduces the force.
Step 1: The Actual Pressure (Article 149)
Don’t use 6 bar (nominal) — use the actual pressure at the cylinder (on the floor). The FRL (article 119) drops 0.5 bar. The tubing (article 117) drops more. The actual pressure at the cylinder is 5.5 bar (not 6 bar). Calculate the force at 5.5 bar (the real pressure).
The bore sizing rule: Calculate the force at the actual pressure (5.5 bar), not the nominal (6 bar). The cylinder that stalled was Ø32 at nominal 6 bar (480 N), but the actual was 5.5 bar (440 N) on a 500 N load. Size at the actual pressure, with a 1.5× margin.
Step 2: The Return Stroke (Annular Area)
The return stroke (pull) has less area (the rod takes space). The annular area (minus the rod) is smaller. For Ø32, rod Ø12: the return area is 804 – 113 = 691 mm². The return force (at 5.5 bar) is 380 N (vs 442 N push). For a pulling load, size for the return (annular area). Don’t size the push (and forget the return is weaker).
Step 3: The Margin (1.5×)
Add a 1.5× margin (the friction, the part variation). For a 500 N load, the cylinder must give 750 N. For Ø40 at 5.5 bar: A = 1257 mm², F = 691 N (close). For Ø50: A = 1963 mm², F = 1080 N (more margin). Pick the next standard bore (Ø40, or Ø50).
Step 4: The Rod Buckling (Article 161)
For a long stroke (article 161), the rod buckles (Euler). Even if the bore gives the force, the rod buckles. Check the rod (article 161). For a 500 mm stroke, the rod needs to be bigger (or guided, article 161). Don’t pick a bore that buckles.
| Bore | Push (5.5 bar) | Return (rod Ø12) |
|---|---|---|
| Ø32 | 442 N | 380 N |
| Ø40 | 691 N | 615 N |
| Ø50 | 1080 N | 980 N |
| Ø63 | 1720 N | 1580 N |
A Bore Sizing Checklist
- What is the load? (N?)
- Is it push or pull? (Return area?)
- What is the actual pressure? (Article 149? 5.5 bar?)
- Calculate F = P × A at 5.5 bar?
- Is the margin 1.5×? (Load × 1.5?)
- Does the rod buckle? (Article 161?)
- Is the next standard bore picked?
- Does the cylinder push on the floor? (Test?)
- Is the pressure OK? (Gauge? Article 119?)
- Is the speed OK? (Article 171?)
The Bottom Line
Cylinder bore sizing (deep) uses the actual pressure (5.5 bar), the return area (annular), and a 1.5× margin. The cylinder that stalled was undersized (Ø32 at nominal 6 bar). Size at the actual pressure (5.5 bar), with margin. The cylinder that clamped wasn’t the smallest bore — it had margin.
