July 11, 2002

There are many factors to consider when selecting a drive and control system for a cut-to-length line. After choosing the line, you need to choose the drive, calculate the load inertia, calculate the feeder speed, and choose a motion controller.

Many factors should be considered when selecting a drive and control system for a cut-to-length (CTL) line.

First, decide the type of CTL line best-suited for your application, specifications, and budget. Let’s assume you’ve chosen a loop roll feed. Where do you go from there?

**Figure 1**shows some typical frequency response ranges for all types of drives available today. Performance varies widely among manufacturers.

Figure 1:Drive frequency response varies widely among manufacturers, but ranges can be identified for all types of drives available today. |

In general, AC drives are preferred when power requirements are lower than 75 HP, when significant numbers of driven sections are likely to be braking continuously, for high-performance applications, and for large process lines using many driven sections.

DC drives are most suitable when power requirements are higher than 75 HP; when power requirements are positive for all sections; when there is a large DC installed base; and when the application requires a wide, constant-horsepower speed range.

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For the roll feed CTL line, an AC vector drive probably would be the most suitable choice.

Now that you've settled on an AC drive for your roll feed, you need to size the motor for the application by determining how much torque is required at what speed and what kind of performance you can expect. Because the line most likely will run materials of various lengths, you'll need to calculate for the entire product range.

This application calls for you to move a strip of material a known distance. You can record and graph the velocity over a period of time, resulting in a motion profile chart from which you can determine distance, maximum speed, and acceleration rates.

One of the most commonly used motion profiles is the 1/3-1/3-1/3 profile. The time allocated for strip movement is divided into three equal time segments: one-third for acceleration, one-third for traverse, and one-third for deceleration (see**Figure 2**). The distance and time requirements are known, so you can solve for maximum speed N(max):

N(max) = 1.5 ¥ Total distance/ Total time (inches per second)

This value for N(max) can be converted to RPM:

Inches/second ¥ 60 seconds/minute ¥ 1 rotation/P inches

where: P = inches per revolution

Figure 2:In the 1/3-1/3-1/3 motion profile, the time allocated for strip movement is divided into three equal time segments: one-third for acceleration, one-third for traverse, and one-third for deceleration. |

Using the required application data information gathered for the machine, you can calculate the total connected load inertia. Two of the most common geometric shapes encountered in these applications are solid and hollow cylinders. To calculate the inertia of each, use the following formulas:

Solid cylinder for known weight and radius:

J = (1/2) ¥ (W/g) ¥ r^{2}

Solid cylinder for known density, radius, and length:

J = (1/2) ¥ [(Pi ¥ 1 ¥ p)/g] ¥ r^{4}

Hollow cylinder for known weight and radius:

J = (1/2) ¥ (W/g) ¥ (or^{2}+ ir^{2})

Hollow cylinder for known density, radius, and length:

J = (1/2) ¥ [(Pi ¥ 1 ¥ p)/g]

¥ (or4 - ir4)

where: J = inertia (lb. in. sec.^{2})

W = weight (lbs.)

g = gravitational constant (386 in. per sec.^{2})

l = length (in.)

p = density (lbs./in.^{3})

r = radius (in.)

or^{2}= outside radius (in.)

ir^{2}= inside radius (in.)

To find the total connected load inertia, add together the inertia of each driven element.

Now that you know the load inertia and motion profile, you can select the correct motor and gearbox combination for load acceleration.

First, look at the maximum required feeder load speed you calculated using the motion profile. If you know the base speed of the motor, you can calculate a reducer gear ratio. You now can calculate J(Re), the load inertia reflected through the gearbox back to the motor shaft:

J(Re) = J(1) / R2 (lb. in. sec.^{2})

where: J(Re) = reflected inertia

J(1) = load inertia

R = gearbox ratio

If there is any load torque, the gearbox will reduce the amount at the motor shaft. You can calculate the load torque reflected back to the motor shaft as follows:

T(Re) = T(1)/R

where: T(Re) = reflected torque

T(1) = load torque

R = gearbox ratio

Once the reflected inertia and reflected torque are known, you can solve for the required motor acceleration torque T(a) using the following formula:

T(a) = {[(J(m) + J(Re)) ¥ N(max)]/(9.55 ¥ t)} + T(Re)

where: T(a) = motor acceleration torque (in. lb.)

J(Re) = reflected inertia (lb. in. sec.^{2})

J(m) = motor inertia (lb. in. sec.^{2})

N(max) = motor maximum speed (RPM)

t = acceleration time (sec.)

T(Re) = reflected torque (lb. in.)

The motor you select should be able to produce the amount of acceleration torque required.

Because the load in this example is cyclical, you must check the average amount of torque the application requires to be certain that your motor can deliver that torque without overheating.

Calculate the average required torque using the following equation:

Trms = [(T_{1}^{2}t_{1}+ T_{2}^{2}t_{2}+ T_{3}^{2}t_{3}+ ***)/(t_{1}+ t_{2}+ t_{3}+ ***)]^{-2}

where:

T_{1}= acceleration torque (lb. in.)

T_{2}= traverse torque (lb. in.)

T_{3}= deceleration torque (lb. in.)

t_{1}= acceleration time (sec.)

t_{2}= traverse time (sec.)

t_{3}= deceleration time (sec.)

The continuous torque rating of the motor must be greater than the calculated average required torque.

Once you've sized the motor for a particular cut length, you need to check the expected performance at all of the other possible cut lengths by rerunning all of the calculations for each cut length. Usually, a motor and gearbox sized to perform adequately at short lengths may have limited performance at other lengths. You might have to give up performance in one area to gain it back in another.Constraints

One of the constraints you will have to deal with is the type of shear and the maximum number of strokes per minute it can handle. This usually sets the maximum number of pieces per minute (PPM) that the line can handle.

Another short-length constraint is the ability of the motor to accelerate fast enough to make the move. The motor must have enough torque to bring the connected inertia up to the maximum speed in the time allowed. Sometimes the acceleration constraint is not with the motor but with the machine itself. For example, the motor may have enough torque to accelerate the load, but the feed rolls may slip on the material at that particular acceleration rate, causing inaccurate cut lengths. Or the acceleration rate may be too high for the mechanical portion of the machine to handle.

Gearboxes, belts, and other driven equipment could become worn out quickly, chatter, or exhibit other unstable characteristics. Also, consider the control of the free loop. If the acceleration rate approaches or even exceeds the force of gravity, you will not have control over the loop, leading to an unstable condition.

On longer lengths, a constraint that may be encountered is that the maximum feeder speed is too slow. You could change the selected gear ratio to allow for a faster maximum motor speed, but you would be giving up torque and losing some of the advantage of smaller reflected inertia.

Another long-sheet constraint is the leveler gear in speed. The maximum line speed at the leveler divided by the cut length gives you the maximum PPM for longer lengths. Also, the amount of material stored in the loop must be considered for longer lengths.

These constraints must be observed for all cut lengths. You can do some things to overcome them. Sometimes it is a trade-off with other constraints, and sometimes a larger, faster motor will do the trick. But oftentimes you will find a situation of diminishing returns as you try to apply bigger motors. That is, the larger torques require larger frames that have larger inertias, and as you apply the formulas, you will find that the expected results begin to drop off even though you have a larger motor.

A motion controller handles all of the computations to generate the motion profile used by the feeder drive. It usually is microprocessor-based and can be a stand-alone box or part of the programmable logic control system. It can reside in a PC or industrial computer, or it can be part of the AC, DC, or servo drives.

The motion controller calculates distances, velocities, accelerations, torque, and speed. It may have to accommodate several encoders and be capable of both analog and digital input and output. It has to be able to communicate to the real world to accept set points and other commands. And last, the motion controller must be fast enough to send its calculations to the drive in time for the drive to react.

A few final checks remain:

• Is the inertia of the motor equal to or bigger than the total reflected load inertia divided by 3? This is necessary to ensure system stability.

• Is the continuous motor torque still greater than the required average torque? If you’ve made any changes, especially the gear ratio, you may have to recalculate.

• Is the peak torque of the motor high enough to accelerate the load fast enough? Again, changes can affect peak torque requirements.

• Can the motor run at the top speed you've designed? In your efforts to reduce load torque and inertia, you might have selected a gear ratio that requires the motor's top speed to be higher than the manufacturer recommends.

• Is the acceleration rate acceptable? Is it high enough for good production but below a threshold for good control of the strip and acceptable wear on the machine?

• Have the performance specifications been met? If not, you need to modify your design choices or modify your performance specifications.

*Martin Marincic is president of New Era Controls, P.O. Box 25630, Garfield Heights, OH 44125, phone 216-901-1300, fax 216-901-1305, e-mail neweracontrols@worldnet.att.net, Web site neweracontrols.com. New Era Controls is a systems integrator that specializes in providing drives and controls for cut-to-length lines and other coil processing applications.*

*This article is adapted from Martin Marincic's conference presented at Coil Cut-to-length Workshop, Aug. 2 and Oct. 18, 2001, ©2001 by the Fabricators & Manufacturers Association, Intl.*

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