Hydraulic Motor Acceleration Torque: Inertia and Ramp Time

Hydraulic motor acceleration torque follows Ta = Jα: shaft-referred inertia multiplied by angular acceleration. Add the resisting process torque for the accelerating shaft requirement. The example below starts with an assumed 8 N·m process load and calculates what a one-second start adds to it, before reviewing the supply and the stop.

The calculations here build one hypothetical direct-drive example from shaft motion to hydraulic demand. They are screening calculations, with assumed efficiencies and an ideal speed ramp. They do not represent a BLINCE test, a stock motor specification or a setting for a relief valve. The final selection needs the offered motor's data and the machine's actual motion requirements.

Put the required motion on the specification

Sketch acceleration, steady work, deceleration and dwell, with the process load underneath. On a mixer that may include settled material at the start; on an indexing table, the loaded fixture. A fan needs its resistance as speed changes. The calculation here deliberately uses a constant resistance so that the inertial part can be seen separately.

Mark where the clock starts and stops. A valve-command timestamp includes a response delay that first-motion timing leaves out. The end can be first arrival at target rpm or settling within the required tolerance. Apply the chosen definition to the requirement, the offer and the test record.

The inertia equation uses the shaft's actual speed change. The commanded profile still has to pass through the valve, oil circuit and load before it becomes motion. Pressure buildup, compliance and flow delivery can reshape it. Preserve the speed trace when checking an existing machine; replacing it with the controller setting discards part of the evidence.

If an existing drive runs at the correct speed but starts too slowly, preserve its startup trace before choosing a replacement. BLINCE's motor torque calculation guide covers the pressure-to-torque relationship. The extra task here is to establish the torque demanded by the change in speed.

Establish inertia at the motor shaft

An inertia figure needs a rotation axis and the unit kg·m². A note saying “30 kg rotor” is incomplete: moving that mass farther from the axis changes the torque needed to accelerate it. The shaft torque is a separate quantity, expressed in N·m.

For a geometry comparison, take two hypothetical 30 kg parts with a 0.20 m outer radius:

The ring in this comparison has twice the acceleration demand, although the two parts have the same mass. Choose a geometry that represents the actual part before using either relationship.

Retain the CAD parts list, densities and axis with an assembly inertia result. Check whether it includes the hub, rim, blades, tooling and carried material. A figure for the bare rotor leaves any added coupling or fixture to be accounted for.

The axis-specific cylinder illustrations in Oriental Motor's sizing reference are useful for this mechanical step. The reference also separates load torque from acceleration torque. Its electrical motor ratings stay outside the hydraulic selection.

Uncertainty in material loading belongs in the calculation too. A rotor with a partly filled vessel or interchangeable tool can have several credible inertia cases. List the lightest and heaviest intended configurations with their process loads. A motor selected from the empty fixture may meet commissioning tests and then fail to accelerate the normal production load.

Work through a one-second acceleration

For the main hypothetical example, assume a direct drive with:

This 0.60 kg·m² is an assumed total for the main drive example. The earlier disc was a separate geometry illustration and is not added again.

The calculation uses radians for angular motion. Each row feeds the next:

The relationships are given in SI form in Tokyo Keiki's hydraulic formulas, Appendix 1-5. This calculation uses mass inertia in kg·m² throughout. Converting a flywheel-effect value from another convention would require its definition and units first.

The calculation has no added design margin. Its open items are breakaway resistance, load uncertainty, the actual ramp shape and component limits. Identify those beside the result. An unexplained percentage cannot tell a reviewer which missing condition it was intended to cover.

See what changing the ramp time buys

Use the same inertia, target rpm and constant resistance to compare four possible linear starts:

A longer ramp can make the existing displacement a more plausible candidate. The cost is time in the machine cycle, and possibly more process work while the shaft is accelerating. The production requirement decides whether that trade is acceptable. Extending a ramp is not available when another machine function depends on reaching speed within a fixed interval.

Keep the table's straight speed ramp attached to its results. An S-shaped profile redistributes the acceleration within the available time, and its peak may be greater than the endpoint-based average. For that profile, use its peak acceleration in J × acceleration when checking peak inertial torque.

There is also a timing trap in a one-second command. In a separate hypothetical trace, the first 0.3 s contains no shaft movement; acceleration occupies the remaining 0.7 s. The inertial calculation becomes 0.60 × 157.08/0.7, giving 134.64 N·m before process resistance. Counting the stationary delay as moving time would have produced the lower 94.25 N·m result.

A recorder that shows only a few points through the ramp may locate the final rpm but leave the steepest part unresolved. State the sensor and recording response with the trace. The number of digits on a speed display supplies no missing time history. Where no suitable trace exists, retain the straight-line assumption as an estimate.

Convert the shaft demand into useful pressure difference

For a preliminary motoring calculation in metric units:

Tshaft ≈ D × Δp × ηhm ÷ 62.83

The units for this version are N·m for torque, cm³/rev for displacement and bar for the measured inlet-minus-outlet pressure difference. Enter ηhm as a fraction. To calculate pressure, use Δp ≈ 62.83 × Tshaft/(D × ηhm). The conversion to bar is included in 62.83; the Tokyo Keiki reference presents the corresponding relationship using SI pressure units.

For an assumed 40 cm³/rev motor at ηhm = 0.90, the pressure screen is below. These assumptions are not a BLINCE model specification.

Suppose a circuit has only 200 bar of usable pressure difference available across the motor. In this simplified model, the half-second target needs more than that supply can provide. A 400 bar pressure rating on a replacement would not create the missing circuit pressure. The one-second case clears this particular arithmetic screen, but still needs the actual torque curve, ratings and other exclusions reviewed.

Usable motor Δp is measured between its work ports during the same event. Pump pressure can include losses before the inlet, and pressure at the outlet consumes part of the available difference. Keep case pressure as its own constraint under the selected motor's documentation. Do not subtract it a second time from a measured inlet-minus-outlet value as a generic correction.

An efficiency assumption is especially uncertain near the beginning of motion. Danfoss's general orbital-motor reference distinguishes minimum starting torque from running torque and explains the change as the lubricating film becomes established. Use a starting-torque value or curve for the candidate where breakaway is decisive, instead of carrying a running-efficiency assumption down to zero speed. Danfoss General Orbital Motors, December 2021, pages 14–15.

Check flow and peak power at the same operating point

The flow endpoint in this example is about 65.22 L/min. It comes from Q ≈ D × n ÷ (1,000 × ηv) with an assumed 40 cm³/rev, 1,500 rpm and volumetric efficiency ηv = 0.92. Q is in L/min. Keep that final flow alongside each ramp's pressure requirement on the supply worksheet.

Both ramps end at that flow under the assumption. Compare the pump's delivered flow at the accelerating pressure, together with the prime mover and control response. A nominal 65 L/min figure without its pressure condition leaves that comparison unfinished.

For the same ideal model, integrating flow over the start gives another useful supply quantity. Assume ηv stays constant as speed rises linearly from zero. Average flow is then 65.22/2 = 32.61 L/min: approximately 0.543 L through the motor in one second, or 2.174 L in four seconds. Transient filling is excluded. The extra revolutions in the longer start account for its larger volume. Send the flow endpoint and duration together when reviewing what the supply must deliver.

Just before the ideal one-second ramp ends, the example still needs 102.25 N·m at 1,500 rpm. The power check at that point is:

The prime mover also supplies pump and upstream losses, which are outside these work-port calculations.

For the four-second ramp, the corresponding shaft-power peak falls to about 4.96 kW. After either ramp, the assumed 8 N·m at 1,500 rpm is only about 1.26 kW of process output. Selecting a prime mover solely from that steady output would miss the example's acceleration peak. A rated steady output also says little about the supply's permitted short-time overload response.

The flow expression describes the revolutions and leakage included in its assumed efficiency. A real pressure transient also involves fluid compression, hose expansion, valve response and supply dynamics. Do not use the simple flow ramp as a complete transient simulation. The calculation identifies a demand for the circuit review; the measured response establishes how the assembled machine meets it.

A longer ramp can reduce peak demand without reducing total work

At the example's target speed, ½Jω² = 0.5 × 0.60 × 157.08² gives about 7,402 J (7.40 kJ) stored in rotation. The choice of ramp time has not changed either J or the final ω, so this endpoint energy is unchanged.

The assumed 8 N·m resistance also consumes work as the shaft turns. A linear ramp travels through ½ωfinal × t: approximately 78.54 rad in one second or 314.16 rad in four seconds. Multiplying by 8 N·m gives process work of about 628 J and 2,513 J respectively.

The four-second option therefore buys a lower peak while the shaft makes more turns against the assumed resistance. An energy-saving claim would need a full cycle comparison, including production output, dwell, pump control and losses. Lower peak pressure alone does not establish that saving.

For a speed-dependent load, such as a fan, the work calculation needs the applicable torque-versus-speed relationship through the ramp. A steady reading at target rpm gives only one point. Retain an identified estimate for the rest until suitable process data is available.

The production comparison needs the output delivered during the ramp too. A longer start can leave fewer seconds at full speed in a fixed cycle. Raising final rpm to recover that output would change the flow endpoint and stored energy.

Reflect inertia correctly when adding a gearbox

Use i = motor speed ÷ load speed. Then the ideal rigid-drive relationship Jreflected = Jload ÷ i² gives the output load's contribution at the motor shaft. Add the relevant input-side rotating parts at that reference shaft.

In a separate ideal case, 0.60 kg·m² at a 4:1 reduction's output contributes 0.60/16 = 0.0375 kg·m² at the motor. Motor, input-coupling and gearbox entries remain to be added; losses are excluded.

The motor also runs four times faster than the output in that ratio. A 1,500 rpm output would require 6,000 motor rpm. The reduction may lower the load-related torque at the motor while creating a speed requirement outside a candidate's range. BLINCE's high-speed hydraulic motor selection guide addresses the resulting motor, flow and installation checks.

Include efficiency and transient-duty data in the gearbox enquiry. Reversals involve its backlash, and repeated starts involve more than steady transmission. The mounting drawing and cycle-duty review are both needed before the reduction becomes a selection result.

Relative oscillation between motor and load calls for a mechanical response review of the flexible shaft or coupling. Attach any observed overshoot to the problem description. A larger torque number does not resolve the limitations of a rigid-drive model in that situation.

Account for the stop as well as the start

The stop begins with the example's 7.40 kJ still in the rotating assembly. Identify where it will go: process resistance, a brake, the hydraulic circuit or an engineered recovery arrangement. The valve command alone does not assign those shares or define how quickly the shaft stops.

A separate ideal two-second linear stop gives 0.60 × 157.08 ÷ 2 = 47.12 N·m of inertial deceleration torque. Resistance from the process can help provide it. A load driving the shaft can add energy instead. Establish that load direction before assigning the required torque to a brake or to hydraulic braking.

If one path absorbs the assumed 7.40 kJ in two seconds, its event average is 7.40/2 = 3.70 kW. With the ideal constant braking torque, absorption falls as speed falls, starting above that average. Give the component supplier the shape of the event and its repetitions, plus the temperature conditions for the review.

For the hydraulic part, Danfoss's General Orbital Motors reference, pages 20–22, treats the braking motor as a pump and discusses replenishment for high-inertia loads. Its motoring and pumping efficiency relationships differ. The earlier pressure screen cannot be reused unchanged as a braking-valve setting calculation.

Holding after the stop is another requirement. Internal leakage can permit movement, and a holding brake has its own purpose and rating. Obtain the machine's approved stopping and holding arrangement, including the consequences of supply loss. The calculations in this article provide mechanical inputs to that review; they do not authorize closing ports to stop a spinning load or adjusting a brake-control valve.

Put repeated events into the duty review

A short ramp can occur many times per hour. Give the supplier the event duration, number of events and rest of the cycle, rather than labelling the whole duty “intermittent.”

For an illustrative hourly budget, assume 360 identical full stops with no recovery, and allocate all the stored energy to dissipation within the drive:

Keep this allocation separate from pump, motor, valve and other running losses. The part reaching the oil must also be established before the hydraulic heat balance is complete. Carry the two-second 3.70 kW event average to the component-duty review, including its peak and repetitions; carry the hourly 0.74 kW to the energy balance.

With variable displacement, attach its schedule to the speed profile. Each displacement changes both the torque obtained per bar and the flow required per revolution. Include the control option, displacement limits and transition conditions in the proposal. The fixed 40 cm³/rev example contains no shift sequence to assess.

For a shared supply, identify the other function commanded during startup and the flow it receives. A motor-only test may leave that simultaneous demand absent. Send the pump and control description with the event record so that a load-sensing, pressure-limited or fixed-flow arrangement is assessed as fitted.

Compare the candidate with the whole ramp

A candidate review should follow the operating points from rest through maximum speed. At the beginning, check breakaway torque and any brake release. During movement, compare available motor torque with process and acceleration demand. Near target speed, check the flow and power endpoint. Then examine deceleration, possible overrun and holding.

BLINCE lists a CMF series gear motor and an axial-piston motor category, among other motor families. These pages establish product routes for enquiry. They do not approve the assumed 40 cm³/rev example or publish a universal allowable acceleration time. Request the data for the specific offered build before treating a family name as a selection result.

Compare a proposed larger displacement using its revised flow requirement at target rpm. Compare a new reduction using its revised motor speed. A changed ramp needs a production-cycle check. For a proposed pressure increase, review the affected circuit as well as the offered motor rating. Keep those revisions in the offer so that its scope can be priced.

Include the shaft and mounting arrangement when requesting the peak-duty review. A sprocket or pulley adds bearing forces as well as transmitted torque. An unsupported load can add more. These belong on the mechanical drawing and load record; they are absent from J × acceleration.

For an existing machine, arrange an approved test with a qualified person. Record speed and both work-port pressures on the same time base, with the load configuration, oil conditions and relevant flow information. Choose instrument range, accuracy and response for the ramp. Keep loads secured and people clear of rotating parts; use the approved test connections and procedure, with no improvised stall or opening of a pressurized fitting.

Send a motion profile with the motor enquiry

Start the enquiry with the motion requirement: speeds, start and stop times, inertia and its reference shaft, process torque and repetitions. Attach the supply conditions and interface drawing, including a geared drive's ratio convention. The reviewer should be able to reproduce the load calculation from those inputs.

Label how each input was obtained. A CAD inertia can exclude tooling; a torque inferred from pressure depends on the motor data used; a speed trace has its measurement limits. State the missing part of each estimate alongside it. That makes the next request for data specific.

The technical reply should name the motor build and the performance data used to assess the start, ramp and stop. It should identify its speed/pressure duty and inlet/drain conditions, together with unresolved overrun or holding questions. Have any proposed change to displacement, timing or supply recorded in that reply, so the commercial offers describe comparable duties.

Send the profile and the relevant drawing through the BLINCE contact page. For the hypothetical example, a useful enquiry states “0.60 kg·m² total motor-shaft inertia, 0–1,500 rpm, proposed one-second linear ramp, estimated 8 N·m resistance,” together with the actual circuit limits. Those inputs give the supplier a defined acceleration duty to assess.

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Disclaimer

This article is a general engineering guide. Final component selection should be based on machine drawings, measured hydraulic data, working conditions, safety requirements, and confirmation from a qualified hydraulic engineer or supplier.

Blince Hydraulic Team

Blince Hydraulic is an industry-leading company dedicated to precision-engineered fluid power manufacturing and custom hydraulic solutions. Backed by decades of deep field expertise in industrial machinery and thousands of successful global deployments, our engineering team focuses entirely on high-performance hydraulic component manufacturing, including specialized orbital motors, high-pressure travel drives motor, and robust directional control valves. Our production infrastructure utilizes state-of-the-art multi-axis CNC machining systems and is fully ISO 9001 certified to guarantee repeatable volumetric accuracy across every single manufacturing run.

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