Servo Driven Circular Conveyor

Servo Driven Circular Conveyor: A Six-Carriage Oval Sized Around the Customer’s Own Servo

Author: ShanGong, Director of Engineering Department | Sept 2026| Category: Circular Motion

On this project the customer brought the motor. They asked for a servo driven circular conveyor they could fit with their own 750 W Omron servo and its 50:1 gear reduction, while we built everything from the motor bracket outward. That split changes what a supplier owes the buyer. A loop is not finished when it is assembled; the buyer’s engineers also need the numbers to size a drive for it.

This page sets out those numbers for a six-carriage horizontal oval: how the belt loop closes, how fast the carriages move through the arcs, what ramp time to use, what gearbox ratio the load calls for, and where the ±0.05 mm at the working station comes from.

Confirmed 3D model. The 80-tooth pulleys sit inside the two 400 mm arcs; the swing-arm locator is on the near straight, and the buyer’s servo and gearbox mount beneath the right-hand pulley.

The Signed Specification of this Servo Driven Circular Conveyor

The customer confirmed the following on our technical specification confirmation form before production:

Two lines on that form shaped everything else. A 1.2 s stepping time for a 540 mm move fixes the motion profile, and the remark that the motor and gearbox are supplied and mounted by the buyer made the drive sizing a shared job.

How the Belt Loop Closes in this Servo Driven Circular Conveyor

Every carriage is clamped to an AT10 timing belt, so the belt’s pitch line is the length that has to close. It runs straight between two 80-tooth pulleys 1,220 mm apart and wraps half of each:

6 × 540  =  2 × 1,220  +  80 × 10  =  3,240 mm

The fact that 540 mm is exactly 54 belt teeth matters for the carriage clamps. Each 25 × 16 × 15 mm belt buckle locks into the teeth and connects to its carriage through a steel link running in a bronze bushing, so the carriage can pivot as it follows the arc, and a whole-number tooth count means all six buckles sit on the same tooth phase. The pitch therefore repeats around the whole loop instead of drifting by a fraction of a tooth from one carriage to the next.

The carriages themselves ride the guideway on a 400 mm arc, while the belt turns on a 254.6 mm pitch circle. Measured along the guideway, the loop is 2 × 1,220 + π × 400 = 3,697 mm, which is 457 mm longer than the belt. Carriage spacing opens up by that amount as carriages pass through the arcs, and closes again on the straights.

Motion Profile: Choosing the Ramp Time

Because the arc radius (200 mm) is larger than the pulley’s pitch radius (127.3 mm), a carriage in an arc moves 1.57 times faster than the belt. That makes the ramp time a real engineering choice, not a comfort setting. For a 540 mm move in 1.2 s on a trapezoidal profile:

peak speed  =  540 ÷ (1.2 − t(ramp))        acceleration  =  peak speed ÷ t(ramp)

The table shows what a 10 kg part experiences on a carriage passing through an arc. The resultant combines the centripetal load at full speed with the tangential ramp acceleration, which is also multiplied by 1.57 in the arc.

A short ramp keeps the peak speed down but loads the part hard along the track. A long ramp does the opposite, and past about 0.4 s the centripetal term takes over quickly. The minimum falls between 0.25 and 0.3 s, where a 10 kg part pushes on its fixture with about 54 N. The last row also exceeds what the drive can deliver, as the next section shows. The form’s instruction to set acceleration and deceleration no shorter than 0.2 s sits just below that band, and 0.25 to 0.3 s is where we recommend starting.

The general method for building a cycle from a move profile is on conveyor cycle time calculation. On this line, the full cycle is the 1.2 s index plus the locator stroke plus the process dwell at the station, and every carriage passes the located station once per six indexes.

Sizing the Buyer’s Servo and Gearbox for this Servo Driven Circular Conveyor

The customer’s motor is an Omron R88M-1L75030C-S2 from the 1S series: 750 W, 3,000 rpm and 2.39 N·m rated, with a 23-bit encoder. Omron publishes its data on the R88M-1L75030C-S2 product page . The motor arrived as one unit with a built-in 50:1 planetary reduction, so no separate gearbox was needed; we made a custom motor bracket for it and coupled its output to the drive pulley shaft through a rigid steel coupling with 22 mm bores on both sides.

Speed and torque are not the hard part of this sizing. The 80-tooth pulley moves 800 mm of belt per revolution, so 600 mm/s needs only 45 rpm at the pulley. Even with little reduction the motor would run slowly. The number that decides the ratio is inertia.

Seen from the pulley shaft, the loop carries a large moment of inertia. Four carriages on the straights contribute their mass times the pulley radius squared. The two sitting at the arc apexes, where the drawing places them at rest, contribute their mass times the 200 mm arc radius squared, which is 2.5 times as much each. Taking 10 kg of payload plus about 0.8 kg of carriage per station, and two aluminium pulleys, the total is about 1.63 kg·m². Divided by the square of the gearbox ratio, that is what the motor sees:

The torque column assumes the 2.0 m/s² ramp of a 0.3 s profile and 95% efficiency for the reduction stage. From 10:1 upward, torque is never the limit; the motor’s 7.16 N·m peak covers the ramp easily. The deciding column is the load-to-rotor inertia ratio. With a rotor inertia of about 1.3 kg·cm², a single-stage 10:1 would leave the motor controlling a load about 126 times its own inertia, and the position loop would settle slowly at each stop.

The customer’s built-in 50:1 brings that ratio down to about 5:1, which lets the servo stop six loaded carriages crisply. The ramp needs only about 0.6 N·m, a quarter of rated torque, so the motor runs cool with plenty in reserve. The cost of the high ratio is speed. At 3,000 rpm rated, 50:1 turns the pulley at 60 rpm, which caps the belt at 800 mm/s.

That ceiling sets the usable ramp window. A 1.2 s index needs a peak of 540 ÷ (1.2 − t(ramp)), which stays at or below 800 mm/s only for ramps up to about 0.52 s. The recommended 0.25 to 0.3 s ramp needs 568 to 600 mm/s, or 2,130 to 2,250 rpm at the motor. That uses about three quarters of rated speed and leaves the rest for tuning. The 0.6 s row in the ramp table is out of reach with this reduction.

The pulley size cuts both ways. An 80-tooth pulley closes the loop neatly with a 400 mm arc, but it has a large radius, and every quantity at the pulley scales with that radius. The next section shows the same effect on accuracy.

Where the ±0.05 mm Comes From in this Servo Driven Circular Conveyor

The single working station repeats to ±0.05 mm. Neither the servo nor its gear reduction delivers that figure. The table lists what lies between the motor and the carriage:

On a 127.3 mm pulley radius, each arc-minute of gearhead backlash moves the belt 0.037 mm. Because the large pulley multiplies every arc-minute at the gearhead output, even a few arc-minutes of backlash add up to more than the ±0.05 mm the station has to hold, before the pitch tolerance and belt stretch are counted. The 23-bit encoder is thousands of times finer than any of these.

So the station uses a mechanical locator. When the index stops, a 32 mm bore, 15 mm stroke compact cylinder swings an arm carrying a CF8 cam follower. The follower drops into a 70 × 28 × 9 mm locating slot fitted to each carriage and pulls the carriage onto the station datum. From that point the carriage position depends on the slot and the arm, not on the belt, the gearbox or the pitch. An M8 inductive sensor confirms the carriage is in position.

A timing belt with a positioning mechanism repeats to ±0.05 mm, which is what this station delivers. Chain drive with a positioning mechanism reaches ±0.03 mm, and a belt with no locator holds ±0.5 mm. Where a station needs tighter than ±0.05 mm, the chain oval conveyor system  is the route. The principle is covered in more depth in circular conveyor positioning accuracy

Carriage Plate Interface

Each carriage is an HST400 slide, 108 × 95 × 18 mm overall, with a 92 × 95 mm fixture face rated for 10 kg. The customer’s fixtures bolt down through eight M6-6H tapped holes on a 76 × 80 mm pattern, drilled through. Two Ø5 mm dowel holes, reamed to +0.02/0 and 7 mm deep on a 30 × 50 mm pattern, register the fixture so it goes back in the same place after removal. Because the arcs put about 54 N of side load on a 10 kg part at the recommended ramp, the fixture has to hold the part sideways as well as down.

Confirmed Specification of this Servo Driven Circular Conveyor

Other Servo Driven Loops in the TallMan Range

• Belt driven circular conveyor for the belt drive family this oval belongs to.
• Horizontal circular conveyor systems for other layouts in the horizontal plane.
• Special shaped closed loop conveyor for a 24-carriage rectangle with tight corners.
• Multi station circular conveyor for loops that feed robot cells.
• Precision circular conveyor system  for the wider precision range.

From Specification Form to Shipment

The route was the same as for any loop we build. The customer returned our specification confirmation form with the stepping time, payload and station count, and we returned a basic drawing and 3D model for signature. For buyer-supplied drives we add one more step: we send the loop data the drive has to be sized against, namely pulley radius, arc radius, carriage and payload mass, and the recommended ramp, so the buyer’s engineers can check their motor and gearbox before the machine arrives. The confirmation form also lists the wear parts that can be ordered as spares with the machine.

Servo Driven Circular Conveyor FAQ

Can I fit my own servo motor and gearbox?

Yes, as this customer did. Send us the motor or gearhead flange and output shaft drawing, and we supply the motor bracket, coupling and pulley shaft to suit. On this line the motor arrived with its 50:1 reduction built in, so we made a custom bracket for it. In return we give you the loop’s pulley radius, reflected inertia and recommended ramp so you can size the drive. The two parties agree in writing where the supply boundary sits, which here was the motor bracket.

Why does a servo driven circular conveyor need a gearbox at all when the pulley turns so slowly?

Mainly for inertia matching, not speed. A large pulley carrying heavy carriages presents far more inertia than a servo rotor. The gearbox divides it by the square of the ratio, so on this line the motor’s built-in 50:1 reduction cuts the inertia the motor sees by a factor of 2,500. The trade-off is a lower top speed, which is why the ratio and the ramp time are chosen together.

Which parts should I order as spares?

On this line the confirmation form lists four wear items: the CF8 cam follower on the locator arm, the 70 × 28 × 9 mm locating slots on the carriages, the 25 × 16 × 15 mm belt buckles, and the AT10-3240 timing belt. The slots, buckles and belt are made for this loop, so ordering them with the machine avoids a lead time later.

Can one located station serve all six carriages?

Yes. Each carriage carries its own locating slot, so the arm locates whichever carriage stops in front of it, and every carriage passes the station once per six indexes. Adding a second located station means a second arm, cylinder and sensor, not changes to the carriages.

What happens if the ramp time is set too short?

Below 0.2 s the tangential acceleration in the arcs climbs quickly, the belt stretches more at each start and stop, and parts on the carriages see harder jolts. Keep acceleration and deceleration at 0.2 s or longer; on this loop 0.25 to 0.3 s gives the lowest combined load on the part.

Bringing your own drive? Request a Quote — Circular Conveyor System [link: quotation and technical requirement form page]