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Product category: Linear Drives and Motors
News Release from: Aerotech | Subject: Linear motors
Edited by the Engineeringtalk Editorial Team on 06 December 2001

An introduction to linear motors

Applications for linear motors are growing rapidly, but the one thing impeding even faster growth is engineers' lack of understanding of the devices. Simon Smith of Aerotech makes an introduction

Once the subject of futuristic speculation about maglev trains and super guns, linear motors have come of age as advanced and reliable alternatives for ball and roller screws for linear motion Not only are the motors now affordable and practical, but the they are easy to control

Applications for linear motors are growing at a rapid rate, but the one thing impeding even faster growth is engineers' lack of understanding of the devices.

Without going into the basic physics, which are quite simple anyway, a linear motor is essentially a rotary motor that has been cut and made flat.

The forcer (rotor) is made up of coils of wires encapsulated in epoxy and the track is constructed by placing magnets on steel.

The forcer of motor contains the windings, Hall Effect board and the electrical connections.

The control for linear motors is identical as with rotary motors.

Like a brushless rotary motor, the forcer and track have no mechanical connection, i.e, no brushes.

Unlike rotary motors, where the rotor spins and the stator is held fixed, a linear motor system can have either the forcer or the magnet track move.

Most applications for linear motors, at least in positioning systems use a moving forcer and static track, but linear motors can also be used with a moving track and static forcer.

With a moving forcer motor the forcer weight is small compared to load, but there is the need for a cable management system with high flex cable, since the cable has to follow the moving forcer.

With a moving track arrangement, the motor must move the load plus the mass of the track.

However, there is the advantage that no cable management system is required.

Why use linear motors? Linear motors overcome most of the disadvantages of the most commonly used ballscrews.

Ball screw systems are subject to screw wear and backlash and cannot tolerate high speeds or acceleration rates.

There is a temperature effect on the screw which reduces accuracy.

Ball bearings also reduce the smoothness in velocity and there is windup/compliance.

Because of wear, the characteristics change over time.

A linear motor directly converts electrical energy to linear mechanical force and is directly coupled to the load.

There is no compliance or windup, higher accuracy, and unlimited travel.

Today, linear motors typically reach high speeds, for example, 5msec, with high accelerations of 5g but can now reach up to 10g with 10msec velocity.

There is no wear, no lubrication and therefore minimal or no maintenance cost.

Finally, there is higher system bandwidth and stiffness.

A linear motor can be flat, U-channel, or tubular in shape.

The moving part of a flat motor, which is typically an iron core design, rides close to the surface of the secondary.

A U-channel motor's moving part rides within two rows of permanent magnets.

In all cases, the moving copper coil assembly is ironless, and thus has no attractive force, resulting in smooth velocities without mechanical disturbances.

The configuration that is most appropriate for a particular application depends on the specifications and operating environment.

Flat motors are often cheaper because they require fewer machined parts and magnets.

However, since the surface of the magnets is exposed, one limitation is that they cannot be used in environments that will be affected by magnetic flux that will "leak" out of the system.

In U-channel designs, this problem is significantly reduced since the magnets are contained within the motor casing.

As with any technology, there are always limitations and caution must be used to employ the correct solution in any application.

While cost was once a limitation over ballscrews, improved manufacturing methods and increasing volume, coupled with higher performance requirements of today's machine manufacturers, costs have reduced to be comparable with a typical ballscrew and motor alternative.

Indeed when cost of ownership is taken into account, in many cases a linear motor system may over time prove to be a considerably less expensive solution than the traditional screw alternative.

One area where a mechanical system will always be preferred is where a high load and hence high inertia is encountered.

Whereas a high torque to inertia ratio is beneficial in many instances, it must be considered that a linear motor is a servo system and therefore the usual inertia matching rules relating to motor and load must be applied.

There is none of the mechanical advantage inherent in a ballscrew and an unstable system running at high accelerations and velocity is a recipe for disaster! However, any reputable manufacturer or supplier would be pleased to aid in the sizing of a suitable motor, and should have a wide experience in applying the correct solution to each application.

Another disadvantage with linear motors is they are not inherently suitable for use in a vertical axis.

Due to its non contact operation, if the motor is shut down, any load that is been held vertically would be allowed to fall.

There are also no failsafe mechanical brakes for linear motors at present.

The only solution that some manufacturers have achieved is by using an air counterbalance.

Environmental conditions must also be considered.

Although the motor itself is quite robust, it cannot be readily sealed to the same degree that a rotary motor could be.

In addition, linear encoders are often employed as feedback devices and therefore care must be taken to ensure that the encoder is suited to the environment too.

That said, linear motors have been successfully employed in wafer dicing, an environment where highly abrasive ceramic dust has lead to the downfall of many supposedly more robust solutions.

Again, the motor supplier should be familiar with all the options, and would be pleased to offer advice in each case.

So how does one take advantage of the linear motor's superior performance? What are the correct procedures when sizing and applying a linear motor? We have considered an example of a 50kg load (mass) that is required to move 500mm in 250ms, dwell for 275ms, and then repeat.

What are the required forces and what size linear motor is needed? The fundamental equations for calculating forces during a trapezoidal move are: Fa = ma + Ff, Ft = Ff, Fd = ma - Ff and Fw = 0 (when horizontal) where m = mass, kg; a = acceleration, m/sec/sec; Fa = force required to accelerate the load, N; Ft = force required during traverse motion of the load, N; Fd = force required to decelerate the load, N; Ff = frictional force, N; and Fw = force at dwell, N.

First, determine the average speed required to make the move; Sa = 500mm/250msec = 2,000 mm/sec.

The load cannot accelerate instantaneously from 0 to 2m/sec, so apply the equation for the most efficient move - a trapezoidal move with constant acceleration.

Divide 250 msec into three equal parts: one-third acceleration, one- third traverse, and one-third deceleration, or 250 msec/3 = 83.3 msec.

The average speed must be multiplied by a factor of 1.5 to ensure the load will make the move in 250 m/sec over the three symmetrical segments.

With this technique, the peak speed of the move is Sp = 1.5 x 2000 mm/s = 3000 mm/s.

This means the load must accelerate from rest to 3,000 mm/sec (3m/sec).

Newton's equation finds the force Fa = ma = (50kg)(3m/sec)/83.3 msec = (50kg)(36 m/sec/sec) = 1,800 N.

This is the peak rating needed from the prospective motor, derived only from acceleration force.

It does not account for friction or other opposing forces.

For example, a quality cross-roller bearing used to carry the load has a coefficient of friction of about 0.0005 to 0.003.

When the 50kg rides on these bearings, the frictional force is Ff = (0.003)(50kg) = 0.15kg = 1.47N.

The frictional force is low because the linear bearings are highly efficient.

And because friction always opposes motion, it adds to the driving force required 1800N + 1.47N = 1801.47N.

Next, with a known total acceleration force, determine the rms or continuous-force requirement.

The rms force is the major contributor to the temperature rise of the linear motor's forcer coil, which ultimately limits the power output.

This is where the duty cycle enters the equation.

For the example, the duty cycle tdc = ton/(ton + toff) = 250 msec/(250 msec + 275 msec) = 0.476 or 47.6 %.

Apply the rms force equation to calculate the system needs on a continuous basis Frms = sq.rt([(Fa)(Fa)(t1) + (Ff)(Ff)(t2) + (Fd)(Fd)(t3)]/(t1 + t2 + t3 + toff)) = 1012.16N.

Use this rms force, 1012N, to choose a specific model from a linear motor catalogue that can apply this force continuously.

Adding air-cooling can significantly increase the rms output force of a particular motor, which allows a smaller forcer coil to maximise stroke length.

If there is any doubt about the application characteristics or any other parameters, it is always best to consult the motor supplier. Request a free brochure from Aerotech ...

(This was Engineeringtalk's Top Story on 5 December 2001).

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