Rule of thumb · MechanicalNº 34 / 149

On retract, the return line carries more oil than the pump sends

Retracting, the small annulus end takes the pump flow while the big bore end empties. A 100 mm bore with a 50 mm rod expels about 53 L/min while the pump delivers 40 — so the return line is the one that has to be bigger.

Why it works

The piston moves at flow divided by the annulus area, and the whole bore area empties at that same speed. The ratio out to in is exactly the ratio of the two areas, so the more slender the rod, the worse it gets. Size the return for the supply flow and the restriction builds back-pressure on the annulus face, which subtracts directly from the force you calculated — the cylinder quietly gets weaker rather than obviously failing.

When it fails

Only on retract, and only for a single-rod cylinder. Extending, the flows reverse and the return is the smaller one. Regenerative circuits deliberately send the bore-end oil back to the pump side instead, which is the whole trick.

Do it exactly

Estimate with the rule, then check it against the calculator that models it properly.

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How do I size a hydraulic return line?

Retracting, the small annulus end takes the pump flow while the big bore end empties. A 100 mm bore with a 50 mm rod expels about 53 L/min while the pump delivers 40 — so the return line is the one that has to be bigger. The piston moves at flow divided by the annulus area, and the whole bore area empties at that same speed.

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