Flying Shear Servo Motor Sizing: 6 m Cuts on a 120 m/min Line, Step by Step

The carriage is driven by a timing belt from a pulley on the gearbox output; the belt runs 1:1 between two equal pulleys. With a 500 mm pulley circumference, each turn of the gearbox output moves the carriage 500 mm. Rows marked "Data" come from the machine, rows marked "Assumption" were chosen for this example; replace them with measured values for your machine.

Because the line never stops, the time between two cuts is the cut length divided by the line speed. In this time the carriage must go forward, make the cut and return to its start position.

The carriage accelerates from standstill to line speed, travels with the material during the cut and then stops. During the 1 s cut the carriage covers 2.0 m; this is the largest part of the stroke.

When synchronisation starts, the blade must be exactly above the cut point. While accelerating the carriage covers 0.31 m and the material 0.62 m, so the carriage starts when the cut point is 0.31 m behind the blade. In CODESYS SoftMotion, MC_GearInPos or MC_CamIn calculates this start point itself.

A longer stroke means gentler acceleration, but the longer the forward move, the less time is left for the return. Using all 4 m brings the forward acceleration down to 2.0 m/s², and the forward move alone then takes 3.0 s: no time is left to return.

The table shows how stroke, return speed and motor torque change with the forward acceleration (motor: HPB PM-109-20). Peak torque is lowest when the forward and return accelerations are close to each other; 6.5 m/s² was chosen for this example. The stroke stays at 2.62 m, leaving 1.38 m of the 4 m travel as margin at both ends.

In the remaining time the carriage travels 2.62 m back. A 1/3 profile (one third acceleration, one third constant speed, one third deceleration) gives a good balance between peak speed and acceleration for a given time.

Each pulley turn is 0.5 m and the gearbox is 1:5, so each motor turn moves the carriage 0.1 m. The highest speed occurs on the return, not during the cut.

A linearly moving mass becomes rotary inertia with the square of the pulley radius, and is reflected to the motor divided by the square of the reduction ratio. Almost all of the load is the carriage.

Motor torque has three parts: the force that accelerates the carriage, friction, and the torque needed to accelerate the motor's own inertia and the gearbox. When the motor drives, efficiency increases the torque demand (÷η); when it brakes, it reduces it (×η). The calculation below is for the forward acceleration with an HPB PM-109-20 motor (rotor inertia 0.0165 kg·m²).

What heats the motor is the root mean square of the torque over the cycle (RMS). The peak torque must stay below the motor's short-time capacity, the RMS torque below its continuous torque, and the maximum speed within its speed limit. Three candidate motors were calculated with the same cycle, each with its own inertia.

The carriage stops twice per cycle; its kinetic energy flows back into the drive's DC bus. It is either turned into heat in a braking resistor or passed to other axes on a shared DC bus.

ONX Motion Sizing calculates a point-to-point move. For a flying shear, enter the forward and return moves separately: the peak torque and speed the tool reports can be compared with this guide, and the RMS torque is found by combining both moves over the cycle as in section 8. The tool's default inertia ratio limit is 10:1, so expect a warning with the 1:5 gearbox.