Hydraulic Motor Torque & Speed Calculator
Estimate the running speed and torque of a hydraulic motor from its displacement, inlet oil flow, and pressure drop. Separate leakage-related speed loss from mechanical torque loss.
HYDRAULIC DRIVE MATCHING
Match oil delivery to the torque and speed needed at the shaft.
TWO SIDES OF MOTOR OUTPUT
Flow and pressure answer different questions.
Oil flow sets the speed estimate
For a fixed displacement, more inlet flow produces a higher calculated running speed. Volumetric efficiency reduces the useful displacement flow to account for leakage. The inlet flow should describe what reaches the motor, not simply the hydraulic power unit’s maximum catalog delivery.
Pressure drop sets the torque estimate
Motor torque depends on the pressure difference across the motor and its displacement. The relevant pressure is inlet minus outlet pressure in the motoring direction. Return-line pressure therefore affects the available differential even if the supply gauge reading stays unchanged.
MOTOR POWER BALANCE
Displacement connects hydraulic delivery to shaft motion.
Displacement is the volume associated with one motor revolution. A larger displacement produces more theoretical torque at a given pressure drop, while requiring more flow for a given speed. This gives a practical way to compare candidate drive conditions without assuming that more pressure also means more RPM.
The calculator applies separate volumetric and mechanical efficiencies. The first reduces estimated speed; the second reduces estimated torque. Their product sets the overall conversion of hydraulic input power to shaft output power. Values must come from information appropriate to the motor and operating point.
RUNNING-CONDITION EQUATIONS
Calculate speed, torque, and power on one consistent basis.
Running speed
n = 60Qηv/D
In this SI expression, Q is m³/s and D is m³/revolution. The factor 60 converts revolutions per second to RPM.
Running shaft torque
T = ΔpDηm/(2π)
With pressure difference in pascals and displacement in m³/revolution, torque is in N·m. Mechanical efficiency is entered as a fraction in the equation.
Hydraulic and shaft power
Pin = ΔpQ; Pout = Pinηvηm
The difference is estimated power lost within the motor under the entered assumptions. The upstream hydraulic power unit needs its own loss assessment.
DRIVE-MATCHING EXAMPLE
An 80 L/min supply to a 160 cm³/rev motor.
Enter 80 L/min at the motor, a 200 bar pressure drop, and 160 cm³/revolution displacement. With volumetric efficiency of 90%, estimated speed is 450 RPM rather than the ideal 500 RPM. With mechanical efficiency of 85%, calculated running torque is approximately 432.9014 N·m.
Hydraulic input at the motor is 26.6667 kW. The combined efficiency is 76.5%, producing 20.4 kW at the shaft and an estimated 6.2667 kW motor loss. As a cross-check, multiplying torque by angular velocity gives the same 20.4 kW mechanical output.
If outlet backpressure rises while inlet pressure stays fixed, the pressure drop falls and so does calculated torque. If motor inlet flow falls instead, speed changes. The two adjustments should be examined separately when diagnosing a drive that runs slowly or struggles to develop torque.
EXAMPLE OUTPUT
432.90
N·m estimated running torque
Change one input and watch which output moves
Each row changes one condition from the worked example. Notice that flow and efficiency move speed, while pressure drop and displacement move torque.
Baseline: 80 L/min, 200 bar pressure drop, 160 cm³/rev, volumetric efficiency 90%, mechanical efficiency 85%.
| Change from baseline | Speed (RPM) | Torque (N·m) | Shaft power (kW) |
|---|---|---|---|
| Baseline | 450 | 432.9 | 20.40 |
| Flow reduced to 60 L/min | 337.5 | 432.9 | 15.30 |
| Pressure drop reduced to 150 bar | 450 | 324.7 | 15.30 |
| Return backpressure raises the drop to 180 bar | 450 | 389.6 | 18.36 |
| Displacement increased to 200 cm³/rev | 360 | 541.1 | 20.40 |
| Volumetric efficiency falls to 80% | 400 | 432.9 | 18.13 |
Flow sets speed
Less oil to the motor means fewer revolutions.
Pressure drop sets torque
Lower differential pressure means less torque, even if speed is unchanged.
Steady running estimates from the same equations as the calculator. They do not cover starting torque or the rest of the circuit.
FIELD DATA TO BRING TO A REVIEW
Collect measurements at the motor, not only at the power unit.
Record inlet flow, inlet pressure, outlet pressure, and shaft speed at the operating condition of interest. Identify the motor model and its current displacement setting. A variable-displacement motor needs its actual setting for this calculation; a maximum displacement taken from a catalog may not describe its present operation.
Keep running and starting requirements separate. The equations describe a steady operating balance with chosen efficiencies. They do not predict breakaway torque, acceleration of a loaded shaft, a jammed attachment, or pressure spikes. Check manufacturer starting-torque data and the complete control circuit for those conditions.
Also record the shaft connection and any downstream transmission. The displayed torque is at the hydraulic motor shaft. A gearbox, belt drive, or coupling arrangement can change the torque and speed delivered to the driven equipment. This page does not silently apply a gear ratio.
The estimated motor loss is useful for understanding the energy balance, but it is not a complete cooler selection. An actual hydraulic circuit may have losses in its pump, valves, hoses, and other components. Pair this motor calculation with the circuit’s duty cycle and operating-temperature review.
Start from what the pump needs, then find the motor
If a hydraulic motor drives a pump, begin with the pump's shaft power and speed. This example uses the 37.72 kW shaft demand from the Pump Power Calculator example and assumes 1,500 RPM, a 200 bar pressure drop, 90% volumetric efficiency and 85% mechanical efficiency.
01
Find the torque
T = P ÷ ω
At 1,500 RPM (157.08 rad/s), 37,716 W needs about 240.1 N·m.
02
Find the displacement
D = 2πT ÷ (Δp × ηm)
With a 200 bar drop and 85% mechanical efficiency, D is about 88.7 cm³/rev.
03
Find the flow
Q = n × D ÷ ηv
At 1,500 RPM and 90% volumetric efficiency, Q is about 147.9 L/min.
04
Cross-check the power
Δp × Q × ηv × ηm
Hydraulic input is about 49.3 kW. Multiplying by 0.90 × 0.85 gives 37.7 kW at the shaft, matching the pump demand.
≈ 89 cm³/rev
at about 148 L/min
Then pick the nearest available motor size and confirm its speed, pressure and starting-torque limits with the manufacturer.
PRACTICAL QUESTIONS
Hydraulic motor selection and troubleshooting questions
Should I enter pump outlet pressure or motor pressure drop?
Enter the positive pressure difference across the motor for the motoring direction. A supply reading at the hydraulic pump can include upstream line and valve losses. Subtracting motor outlet pressure from motor inlet pressure defines the differential used by this calculation.
Why are there two efficiency inputs?
Volumetric efficiency affects the speed obtained from the supplied flow. Mechanical efficiency affects the torque obtained from the pressure drop. Combining them into one input would hide which output each assumption changes. Their product gives the overall motor efficiency used for power.
Does higher pressure make a fixed-displacement motor spin faster?
In these steady equations, pressure drop changes torque while flow changes speed. Real leakage and load interactions can influence operation, but those effects are represented only through the entered efficiencies. Raising pressure is not treated as a direct RPM multiplier.
Can I compare different motor displacements?
Yes. Keep the same flow, pressure drop, and efficiency assumptions while changing displacement to see the torque-speed tradeoff. Then replace those assumed efficiencies with candidate-specific performance data before making a selection.
Is the reported torque available for starting a loaded pump?
Not necessarily. It is a running-torque estimate. Breakaway and starting performance can differ from steady operation. Use the motor manufacturer’s starting data and the driven equipment’s starting requirement for that check.
What does the power-loss result include?
It is hydraulic input at the motor minus calculated mechanical shaft output. It excludes losses upstream in the power unit, hoses, or valves and downstream in a gearbox or other transmission. It should not be used alone to specify the entire circuit’s cooling system.
Can I use US gallons and cubic inches per revolution?
Yes. The calculator converts supported inputs to SI internally, then displays torque and power in the selected units. US gallons are used, not Imperial gallons. Both lbf·ft and lbf·in torque outputs are available.
Does a calculated point confirm the motor can run continuously?
No. Check the particular motor’s continuous and intermittent speed, pressure, temperature, case-drain, and shaft-load limits. A numerical torque-speed balance does not establish an approved operating envelope for a specific model.