Two areas, not one
The cap side of a piston is the full bore. The rod side is the bore minus whatever the rod covers. For the 3 in bore and 1.5 in rod sitting in the form, that is 7.069 square inches one way and 5.301 the other — the rod, at half the bore diameter, takes a quarter of the area, because area goes as the square. At 2,000 psi with nothing behind the piston, that is 14,137 lbf pushing and 10,603 lbf pulling.
The default form knocks that down: 75 psi of back pressure on the return side and 95 percent for seal drag give 13,053 lbf extending and 9,569 lbf retracting. Notice which way the back pressure hurts more. Extending, it works on the annulus; retracting, it works on the full bore, so the same 75 psi costs 398 lbf one way and 530 lbf the other.
The return line carries more than the pump sends
This is the one that bites. Feed 10 GPM into the rod side to retract and the cap side has to empty at the same rate the rod side fills — but the cap side is 1.33 times the volume, so 13.3 GPM comes out of it. The return filter, the cooler and the valve spool on the way back see 13.3, not 10.
Widen the ratio and it gets worse fast. A 4 in bore with a 2.5 in rod has an annulus of 7.66 square inches against a bore of 12.57, so retracting at 10 GPM pushes 16.4 GPM back through the return line — 64 percent above pump flow. A big-rod cylinder chosen for buckling can quietly double the flow through components sized off the pump plate.
Speed is flow over area, and nothing else
One GPM is 231 cubic inches a minute. Divide by the area and you have inches a minute. At 10 GPM the 3 in bore extends at 326.8 in/min, or 5.45 in/s, and retracts at 435.7 in/min. A 24 in stroke takes 4.41 seconds out and 3.30 back.
Those are floor times. Nothing in the arithmetic accounts for the ramp at each end, the time the valve spool takes to shift, or a cushion slowing the last inch. On a machine with a hand lever the operator is usually the slowest part of the cycle. Time it and take the difference as your own correction.
Regeneration trades force for speed at a fixed exchange rate
Plumb the rod port back into the cap port and the annulus oil rejoins the supply instead of going to tank. The cylinder then only has to fill the rod volume, so it extends at 231 × GPM divided by the rod area — 1,307 in/min for the default, four times the normal extend speed, because the rod area is a quarter of the bore. Force falls to 2,000 psi on 1.767 square inches, which is 3,534 lbf before the friction deduction and 3,358 after it: a quarter of the normal push.
The exchange is exact and it runs both ways. The speed multiplier and the force divisor are the same number, the bore-to-rod area ratio, so a regen circuit never buys anything on net — it just moves the cylinder fast through the part of the stroke where there is no load, which is what it is for.
What the page will not tell you
It will not say whether a cylinder suits a load. Rod buckling at a given extension, side load on the rod bearing, mounting style, cushion capacity and the pressure the tube is built for are all cylinder-specific numbers that come from the maker, and none of them are here. Enter a load and the page prints the pressure that load needs at your bore — an arithmetic statement, not an approval.
Questions people ask
Why does a hydraulic cylinder push harder than it pulls?
Because the rod is in the way on one side. The cap side is the full bore area; the rod side is the bore minus the rod cross-section. A 3 in bore is 7.069 sq in, and a 1.5 in rod takes 1.767 of that away, leaving 5.301 sq in on the annulus. Same pressure, three quarters the area, three quarters the force — 14,137 lbf against 10,603 lbf at 2,000 psi.
How do I work out cylinder speed from GPM?
A gallon is 231 cubic inches, so one GPM is 231 cubic inches a minute. Divide that by the area being filled and you have inches a minute. At 10 GPM into a 3 in bore that is 2,310 divided by 7.069, or 326.8 in/min extending. Retracting fills the 5.301 sq in annulus instead, so it goes 435.7 in/min. The same flow always retracts faster than it extends.
Why is the return flow higher than the pump flow when retracting?
The cap side empties as fast as the rod side fills, but it holds more oil per inch of travel. The extra is exactly the area ratio: 10 GPM into a 5.301 sq in annulus pushes 13.3 GPM out of a 7.069 sq in bore. The return filter and cooler see that larger number, which is why a large-rod cylinder can overrun return components picked off the pump nameplate.
What does back pressure do to cylinder force?
It subtracts, and it subtracts unequally. Back pressure acts on whichever area is discharging, so extending it presses on the annulus and retracting it presses on the full bore. Seventy-five psi on this cylinder costs 398 lbf while extending and 530 lbf while retracting. Counterbalance valves, return filters and coolers all put pressure there, and a gauge in the return line is the only way to know your figure.
How much faster is a regeneration circuit?
By the bore-to-rod area ratio, and it loses force by the same factor. Joining both ports means the cylinder only fills the rod volume, so a 3 in bore with a 1.5 in rod extends four times faster and pushes a quarter as hard. The trade is fixed by the two diameters and nothing about the circuit changes it.
Does this account for friction inside the cylinder?
Only through the efficiency percentage you enter, which is a single flat deduction. Real seal drag varies with pressure, temperature, rod speed and how worn the seals are, and breakaway force at the start of a stroke runs well above running friction. The field is there so the number is yours rather than a constant somebody baked in.