Indexing Packaging Machines: Calculating Acceleration Torque From a Motion Profile

An indexing table on a packaging machine moves a fixture through 180 degrees, dwells while the seal forms, then moves again. The first attempt at sizing often comes from the average speed, total angle over total cycle time. That number is real, but insufficient for sizing the drive on its own, because the table moves for only part of the cycle and the motor has to accelerate the load in that fraction.

Getting acceleration torque right comes down to reading the motion profile, not the average. Here is the sequence, with a worked example that is purely illustrative.

Why Average RPM Is Not Enough

Average speed spreads the motion evenly over the whole cycle; an indexing machine does the opposite, moving fast then stopping. The motor sees the whole motion compressed into the move time, so its peak speed exceeds the cycle average, while the required acceleration depends on how quickly that speed is reached.

The average tells you how productive the machine is. The move time tells you what the drive has to do, and everything downstream, the acceleration, peak speed, and torque, starts from the move time and the displacement.

Read the Trapezoidal Profile: Accel, Cruise, Decel

A common starting profile for an index is a trapezoidal velocity curve: constant acceleration from rest, constant-velocity cruise, and constant deceleration back to rest. Angular displacement is the area under the angular-velocity curve. For an S-curve or another programmed profile, use its actual velocity and acceleration rather than this trapezoidal approximation.

Two unit rules keep the math clean. Work in radians per second for angular speed and radians per second squared for acceleration, so displacement is in radians, and keep time in seconds. If the displacement is given in degrees or revolutions, convert it to radians first.

A useful check before any torque work: the area under the velocity curve must equal the displacement. For a symmetric trapezoid with equal acceleration and deceleration times t_a, cruising at angular speed v for t_c:

θ = ½ v t_a + v t_c + ½ v t_a

If the area does not come out equal to the required index angle, the profile is not yet consistent, and every torque built on it will be wrong.

Reflect Inertia to One Side, and Count It Once

Torque during acceleration depends on the total inertia the motor accelerates, worked out on one side of the gearbox, not both.

A servo or gear motor drives a load through a reduction ratio i = ω_motor / ω_output. For the following ideal inertia comparison, neglect losses and gearbox and coupling inertia; include those contributions when sizing the real assembly. The load has inertia at the output and the motor has inertia at the input, and to combine them, reflect one to the other. Reflected to the motor shaft, the load inertia becomes:

J_reflected = J_load / i²

The motor-side equivalent inertia is then J_motor + J_load/i². Reflected the other way, to the output, it is J_load + J_motor × i². Both are valid equivalent-inertia descriptions, provided the acceleration and the resulting torque are referred to that same shaft. With α_motor = i × α_output, the output-equivalent torque is i times the motor torque in the lossless model. Mixing motor-side inertia with output-side acceleration misses that speed ratio; it is not generally an error of i².

The output-equivalent total includes motor rotor inertia. It expresses the motor torque at the output speed; it is not the torque actually transmitted to the downstream load through the gearbox output shaft. Keep that distinction when comparing ratings.

The i² is why a servo drive sees the reflected load inertia, not the raw output inertia: a load that looks heavy at the output looks far lighter at the motor after a large reduction. Count the reflection once, on one side, and use the acceleration that belongs to that same side.

Build the Staged Torque, Mind the Sign, and Watch the Power Flow

Acceleration torque is T = J × α, where J is the equivalent inertia on the side you are working and α is the angular acceleration on that same side. Add the external load torque to get the total the drive must deliver at each stage.

Break the cycle into stages and write the torque for each. For positive rotation against a resisting load, acceleration requires inertia torque plus load torque; cruise requires load torque; deceleration reverses the inertia term. Define the load sign explicitly: gravity or a process load can assist motion instead of resisting it. Include any holding or process torque during dwell.

Power flow follows torque times angular velocity. When the motor torque opposes its rotation, the motor is in a generating condition; check the energy that reaches the drive after losses, and the drive or braking resistor capacity. This does not mean the system can return energy to the supply.

Efficiency must follow the direction of power flow. Let T_out = J_load × α_output + T_resist be the torque the gearbox supplies to the downstream load, with T_resist positive when it resists positive motion. In a simple model that neglects gearbox inertia, the motor torque is:

Forward power flow: T_motor = J_motor × α_motor + T_out / (i × η_forward)

Backdriven power flow: T_motor = J_motor × α_motor + η_back × T_out / i

Use the first branch when T_out × ω_output is positive, and the second when it is negative, with confirmed efficiencies for those directions and operating points. Do not divide by forward efficiency for every deceleration segment. Gearbox and coupling inertia, motor losses and any separately modeled friction still need to be included without double-counting losses. At rest, use the actual holding-load model.

Check Peak Torque, Repeat Rate, and RMS Thermal Rating Separately

A profile that works on peak torque alone is only half checked. Three ratings are independent.

Peak motor torque is the largest absolute motor torque in the cycle. Check it at the actual speed, for the permitted duration and repetition rate, against both the motor capability and the drive's current limit. A peak that is permitted occasionally may not be permitted several times a second.

RMS (root-mean-square) motor torque is a useful motor-heating check when torque is proportional to current. It does not capture every loss or cooling condition. Check the motor's continuous capability and the manufacturer's duty model, including speed, ambient temperature and low-speed or standstill cooling. A motor can pass the peak check and still overheat over repeated cycles.

T_rms = √[ Σ (T_i² × t_i) / Σ t_i ]

Use motor-referred torques for this motor RMS calculation, including every stage of the cycle. Check the gearbox's actual transmitted torque, permissible peak duration, load spectrum and thermal duty separately. A motor RMS result is not a gearbox fatigue or life rating.

Worked Example (Illustrative)

This is a hypothetical horizontal, balanced index, not a machine or product rating. The motion data, inertias, ratio and load torque are chosen assumptions. The example uses an ideal lossless gearbox (η = 1), omits gearbox and coupling inertia, and assumes no holding or process torque during dwell. Actual selection must restore those checks.

Given: index angle 180° = π rad, total cycle 1.0 s, dwell 0.4 s, so move time is 0.6 s. Split the move into 0.2 s accelerate, 0.2 s cruise, 0.2 s decelerate.

1. Area check: θ = ½ v (0.2) + v (0.2) + ½ v (0.2) = 0.4 v. Setting 0.4 v = π gives v ≈ 7.854 rad/s. The average output speed over the whole cycle is π/1.0 ≈ 3.142 rad/s, or 30 rpm, while the cruise speed is 75 rpm. Acceleration is α_output = v/0.2 ≈ 39.270 rad/s²; deceleration has the same magnitude and the opposite sign.
2. Inertia, motor side: take J_motor = 5 × 10⁻⁵ kg·m², J_load = 0.5 kg·m² and i = 20. Reflected load inertia = 0.5/400 = 1.25 × 10⁻³ kg·m², so J_eq = 1.30 × 10⁻³ kg·m². Motor acceleration is 20 × α_output ≈ 785.398 rad/s² and motor cruise speed is 1500 rpm. Inertia torque at the motor is J_eq × α_motor ≈ 1.021 N·m.
3. Cross-check, output-equivalent side: J_eq,out = 0.5 + 5 × 10⁻⁵ × 400 = 0.52 kg·m². Its equivalent inertia torque is 0.52 × α_output ≈ 20.420 N·m. Divide by i = 20 to recover 1.021 N·m at the motor. This includes the motor rotor inertia; it is an equivalence check, not the gearbox output-shaft load.
4. Add a constant resisting load of 5 N·m at the output during motion, equivalent to 5/20 = 0.250 N·m at the motor for η = 1. The motor stage torques are approximately 1.271 N·m during acceleration, 0.250 N·m during cruise, −0.771 N·m during deceleration, and zero during the assumed unloaded dwell. During deceleration the motor is still rotating positively, so the negative torque indicates generating operation.
5. Calculate motor RMS over the complete 1.0 s cycle, using the 0.2 s acceleration, 0.2 s cruise, 0.2 s deceleration and 0.4 s dwell. Keep unrounded values in the calculation:

T_motor,rms ≈ √[(1.271018² × 0.2 + 0.250² × 0.2 + (−0.771018)² × 0.2 + 0² × 0.4) / 1.0] ≈ 0.674 N·m.

The motor therefore needs a peak check at about 1.271 N·m for 0.2 s, repeated once per second, and a separate thermal check using about 0.674 N·m RMS, with its speed and cooling conditions. These calculated requirements do not establish that any particular motor or drive is suitable.

For the gearbox output shaft, use only the downstream demand: T_out = 0.5 × α_output + 5. That gives approximately 24.635 N·m in acceleration, 5 N·m in cruise and −14.635 N·m in deceleration. Check those transmitted torques and their durations against the gearbox data. Do not compare the motor's 1.271 N·m peak or the output-equivalent total of 25.420 N·m directly with the gearbox output rating.

Motion Profile and Sizing Record

FAQ

Can I size the drive from the average output speed? No. Average speed spreads the motion over the whole cycle, but an indexing machine moves in a fraction of the cycle, so the average hides the peak speed and the acceleration required by the move. Size from the motion profile, then check peak and RMS torque separately.

Do I add the load inertia at the motor or the output? Either, for a consistent equivalent-inertia calculation. Reflected to the motor, the load inertia is J_load/i²; reflected to the output, the motor inertia is J_motor × i². Use α_motor = i × α_output and refer the resulting torque to the same shaft before comparing it. An output-equivalent total that includes the motor rotor is not the actual gearbox output-shaft load.

Technical References

Oriental Motor’s selection-calculation reference, pages H-4 to H-7, covers inertia, acceleration torque, effective torque and motor selection checks. SEW-EURODRIVE’s drive-planning examples, page 53 (section 3.6.1, “Motor torques”), illustrate the different treatment of gearbox efficiency in motoring and generator operation. The calculations above are an independently worked ideal example; neither source establishes a Mingye product rating.

Next Step

If you are sizing a drive for an indexing packaging machine and want the motion profile turned into real torque figures, send us the index angle, the cycle and dwell times, the load and fixture inertia, and the proposed ratio. Include the stage loads, motor and gearbox data, holding requirements and planned control system so the motor, drive and gearbox can be checked on the correct basis. Explore our packaging machinery solutions, or contact us to start a configuration review.