One number splits into two
Displacement is cubic inches per revolution, and it is doing double duty. Divide 231 by it and you get revolutions per gallon: a 15 cu in/rev motor turns 15.4 times for every gallon that goes through it. Multiply it by pressure and divide by 2π and you get torque per psi.
So a bigger motor is slower and stronger, in exact proportion, at any fixed flow and pressure. Double the displacement and the speed halves and the torque doubles. There is no free choice there — the only thing that gets you both is more flow and more pressure, which is a pump and a drive question rather than a motor one.
The default numbers
Fifteen cubic inches per revolution, 20 GPM in, 2,500 psi across it. Swept speed is 231 × 20 ÷ 15 = 308 rpm, and at 93 percent volumetric the shaft actually turns 286 rpm. Work per revolution is 15 × 2,500 = 37,500 lbf-in, which over 2π is 5,968 lbf-in of ideal torque; at 92 percent mechanical the shaft delivers 5,491 lbf-in, or 458 lbf-ft.
Power in is 20 × 2,500 ÷ 1,714 = 29.17 hp. Power out at the shaft is 457.6 lbf-ft × 286.4 rpm ÷ 5,252 = 24.96 hp. The ratio is 85.6 percent, which is 0.93 × 0.92 exactly — the two efficiencies multiply, because one of them is losing oil and the other is losing torque and they are independent losses.
Torque does not have a curve
This is the thing that makes hydraulic drives feel different. An engine makes different torque at different speeds and you shift gears to stay near its peak. A hydraulic motor makes torque strictly in proportion to the pressure drop across it, and the pressure drop is whatever the load is demanding at that instant. Flow controls speed, pressure controls torque, and the two are set by different parts of the circuit.
The flow table on the page shows it plainly: change the flow from 5 GPM to 40 and the speed moves eightfold while the torque column would not move at all, because nothing there touched the pressure. Full torque is available from a standstill, which is why a hydraulic winch holds without stalling and why a hydrostatic drive has no gears in it.
Through a reduction and onto a wheel
Put a planetary hub on the motor and the arithmetic is ordinary gearing: divide the speed by the ratio, multiply the torque by it, and take the gearbox efficiency off the torque. Then a radius converts torque into a straight pull — torque in lbf-ft times twelve, divided by the radius in inches.
The default with no reduction and a 12 in radius gives 458 lbf of rim pull at 1,800 ft/min, which is 20.5 mph. Put a 20 to 1 hub on it at 97 percent and it becomes 8,877 lbf at 1.02 mph. Same motor, same oil, same pressure — the hub is doing all of it, and the horsepower crossing the rim is identical either way apart from the three percent the gearbox takes.
Where the model gives up
Low speed is the honest gap. Mechanical efficiency on most motor types falls away sharply below some speed on the maker curve, and a drive that runs fine at 200 rpm can refuse to start turning at 15. This page takes one flat mechanical efficiency figure because it has no curve to read, so it will be optimistic down there. A winch drum also changes radius as it fills, so line pull and line speed both drift through a lift, and one radius cannot describe that. And nothing here speaks to what a motor, a shaft, a hub or a rope is rated for.
Questions people ask
How do I calculate hydraulic motor rpm from GPM?
Multiply the flow by 231 and divide by the displacement in cubic inches per revolution, then multiply by the volumetric efficiency. Twenty GPM into a 15 cu in/rev motor sweeps 308 rpm, and at 93 percent volumetric the shaft turns 286. The efficiency figure comes from the motor curve at your pressure and speed, not from a rule of thumb, because it drops with wear and with pressure.
What is the torque formula for a hydraulic motor?
Displacement times pressure drop, divided by 2π, times mechanical efficiency. That gives lbf-in, and dividing by twelve gives lbf-ft. A 15 cu in/rev motor at 2,500 psi has 5,968 lbf-in of ideal torque and 5,491 at 92 percent mechanical, which is 458 lbf-ft. The pressure drop is inlet minus outlet, measured on gauges, not the relief setting.
Why does a hydraulic motor have flat torque?
Because torque depends only on displacement and pressure drop, and neither of those has anything to do with speed. Full torque is available at zero rpm, which is why hydraulic winches hold a load without a brake stalling and why hydrostatic drives have no gearbox. What changes with speed is horsepower, since horsepower is torque times speed.
Do volumetric and mechanical efficiency multiply?
Yes, and they cost you different things. Volumetric efficiency is oil slipping past without turning the shaft, so it costs speed. Mechanical efficiency is friction inside the motor, so it costs torque. Ninety-three percent volumetric times 92 percent mechanical gives 85.6 percent overall, and the power figures on the page confirm it: 29.17 hp in, 24.96 out.
How do I turn motor torque into wheel pull?
Multiply the torque in lbf-ft by twelve and divide by the loaded radius in inches. The default, with no reduction and a 12 in radius, gives 458 lbf at the rim. Add a 20 to 1 hub at 97 percent and the same motor gives 8,877 lbf at a twentieth of the speed. Whether the tyre, hub, axle or rope carries that pull is a question for their own ratings.
Why will my hydraulic motor not start turning under load?
Most motor types lose mechanical efficiency steeply at very low shaft speed, so the torque that reaches the shaft near standstill is well below what the flat percentage on this form implies. The maker publishes a curve of efficiency against speed and pressure for exactly this reason. This page takes one flat figure and will therefore be optimistic at the bottom of the speed range.