Torque and Inertia Calculations

7/29/2019 Torque and Inertia Calculations

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Torque and inertia calculationsThe first step in sizing a positioning system is to calculatethe torque and speed requirements of the load. Speed is

normally determined by the required throughput rate,although it is worth reading the notes later in this sectionabout the relationship between shaft power and move time- even small concessions in move times can have asignificant effect on the power required from the motor.

To establish the torque requirements normally demands acertain amount of calculation, depending on the nature ofthe load and the motion required. Parker has produced asizing program for this purpose which greatly simplifies thisprocess and recommends suitable motor-drive packages forthe application. However we will cover the basics ofmanual calculation here since these can be applied invirtually any situation.

Components of the load torque

In a typical positioning system, the motor must perform twodistinct tasks - it must accelerate and decelerate themechanical components, and it must overcome any frictionin the system. Both require torque from the motor, but eachdepends on totally different mechanical properties.

Calculating moment of inertia

The acceleration/deceleration torque depends only on twoparameters - the moment of inertia of the entire system,including the motor, and the required acceleration rate.Whereas friction is resistance to movement, moment ofinertia can be described as resistance to change in speed

(i.e. acceleration or deceleration). Its a property of anymechanical component, and it depends mainly on weight;but in the case of rotating components it depends on shapeas well.

With circular or cylindrical parts like shafts and pulleys,material further from the axis of rotation has a much greatereffect on inertia than material near the centre. A flywheel,for example, which is deliberately designed to have a highinertia to minimise speed changes caused by fluctuations inengine torque, has its mass concentrated near the edge. Ifit had uniform thickness, it would have lower inertia for thesame weight.

In high-speed positioning systems its essential to keep the

total inertia to a minimum. Perhaps it goes without sayingthat you should concentrate on those components whichmake the major contribution to the total inertia, but its notalways obvious which they are. In many leadscrewsystems, for example, the screw itself is the dominantinertia rather than the load actually being moved (seeexample later).

Rotating components

To calculate the moment of inertia of the rotatingcomponents in a mechanical system, we need to know boththeir shape and their weight. Fortunately most of thesecomponents like shafts, leadscrews, couplings andgearwheels approximate to simple cylindrical shapes. And

since they are usually made either of steel or aluminium, wecan come up with a simple formula based only ondimensions. Heres how to calculate the inertia of thesecomponents:

Fig 7.1 Rotating cylindrical components

When the component is made of steel:

J = 761 x D4 x L (J in Kg.m2, D & L in metres)

For aluminium:

J = 261 x D4 x L (J in Kg.m2, D & L in metres)

When you know (or can calculate) the weight:

J = WD2 8 (J in Kg.m2, W in Kg, D in metres)

(Note that if you double the length, you double the inertia;but if you double the diameter, the inertia goes up by 16times).

Components which move in a straight line

Moving tables and carriages are normally driven either by abelt and pulley system or by a screw. In either case thereis an equivalent inertia seen by the motor which dependson the moving weight and also on the pulley diameter or thepitch of the screw.

Fig 7.2 Belt-driven system

For a belt or chain-driven system:

Equivalent inertia of weight W:

Jw

= WD2 4 (J in Kg.m2, W in Kg, D in metres)If the belt or chain has significant weight, add it to that ofthe moving carriage before doing the calculation.Remember to add the inertia of both pulleys (calculate as

for a cylinder).

Fig 7.3 Screw-driven system

For a screw-driven system:

Equivalent inertia of weight W:

Jw

= Wp2 (4x107) (J in Kg.m2, W in Kg,p = linear travel/turn in mm)

Remember to add the inertia of the leadscrew, againcalculated as for a cylinder.

System calculations

W

p

L

D

W

D


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Dealing with friction torque

Unlike accelerating torque, friction torque is seldom easy tocalculate. For a start it can be extremely variable,

depending on factors such as lubrication and the degree ofbearing preload. Proprietary positioning tables usually havea specified no-load or breakaway torque which is a usefulstarting point. With larger systems or custom-builtassemblies its a good idea to measure the friction torque,making an allowance for variations during service andbetween one system and another. Use a torque spanner ifpossible, alternatively wind thread round a pulley or shaftand use a spring balance to measure the force needed tomake it turn. The friction torque is then:

T = WD x 4.9 (T in Nm, W is balance reading in Kg,D is pulley dia. in metres)

For a screw-driven system, you can calculate the torqueneeded to overcome a linear force acting on the movingcarriage (this may be friction or some other kind of thrustload).

Fig 7.4 Screw driven system with friction

Torque due to linear force F:

T = Fp 6284e(T in Nm, F in Newtons, p is the linear travel per turn in mm,e is the screw efficiency, e.g. 0.8)

For a belt or chain drive system, the equivalent formula is:

Fig 7.5 Belt driven system with friction

Torque due to linear force F:

T = FD 2 (T in Nm, F in Newtons, Din metres)

If the load moves vertically, the linear force F due to gravityis:

F = 9.8W (F in Newtons, W in Kg)

The effect of a reduction ratio

Introducing a gearbox or toothed belt reduction systembetween motor and load can be a very useful techniquebecause the reflected load inertia is reduced by the squareof the gear ratio. This can help to match the load inertia tothe motor inertia, and in the case of screw driven systemscan sometimes be used to improve the dynamicperformance or reduce the torque requirement (seeexample later).

For a reduction system with a ratio N and efficiency e:Torque out = torque in x Ne

Inertia at input shaft = load inertia N2

Calculation of accelerating torque

Once the total system inertia has been worked out, therequired accelerating torque is the product of inertia and

acceleration rate:T = J x revs/sec2 x 2 (T in Nm, J in Kg.m2)

Torque needed to overcome friction or other static loadsmust be added to this to get the total torque required.

Calculating shaft power

When you have selected a suitable motor, its worth whilecalculating the peak power required by the load andcomparing this with the peak power available from themotor. This will confirm that the sizing is realistic and mayhelp to avoid the inefficient solution in which the motor isonly producing a fraction of its potential power output,usually by running at low speed.

Power is the product of torque and speed:Shaft power:

W = T x revs/sec x 2 (W in watts, T in Nm)Assuming a typical trapezoidal move profile, the peakpower required by the load is the torque-speed product atthe end of acceleration or beginning of deceleration. Thepeak power available from a stepper motor isapproximately where the torque is down to 50% of its low-speed value. Peak power from a servo occurs at the kneepoint of the peak torque curve.

System calculations

Fig 7.6 Peak power points

W F

D

WF

p

Time

Speed

Speed

50%

100%

A

A

B

B

Speed

Torque

C

C

Peak power points:

stepper motor

servo motor

load, trapezoidial move

Torque


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The peak power in the load must always be less than thepeak power available from the motor (no more than about70% in the case of a stepper). Torque and speed may betraded by changing the drive ratio, but power can never be

increased.

Motion profiles

Motor sizing for high-speed positioning applications isfrequently based on the equal thirds trapezoidal profile, inwhich acceleration, constant-speed motion and decelerationeach occupy one third of the move time. This profilerequires the lowest peak power during the move; atriangular profile, or one having a longer constant-speedsegment, will both require greater peak shaft power for thesame move time and distance.

Example application

System calculations

Fig 7.7 Triangular & trapezoidal profiles

It is obviously a simple matter to calculate this optimumprofile where the move distance is fixed, such as repetitiveindexing in a packaging machine for example. Theexception is the case where this would result in anexcessive maximum speed, in which case a longerconstant-speed segment will be needed.

However in many positioning applications the movedistance is variable, so the profile can only be optimised forone specific distance (it is not normal to changeacceleration and velocity values as the move distancevaries). Moves shorter than the specified distance willresult in a shorter constant-speed segment or a triangularprofile; longer moves will simply extend the constant-speedsegment. In either case there will be no increase in thepeak power if the acceleration and velocity values areunchanged.

The triangular profile may be useful where torque is alimitation rather than speed - the triangle represents theminimum-torque profile for a given move, albeit at the

expense of a higher maximum speed.The following comparison shows relative maximum velocity,acceleration rate and peak power for triangular and equal-thirds trapezoidal profiles for a move of D revolutions in timet seconds.

Fig 7.8 Example leadscrew system

A table weighing 100Kg is driven by a steel leadscrew40mm diameter and one metre long with a 10mm pitch.The load needs to move 5mm in 120mS. The motor has arotor inertia of 5 Kg-cm2,and friction is negligible.

Using an equal-thirds move profile, how much torque will be

needed?Reflected inertia of the table

Jw

= 100 x 102 4 x 107

= 250 x 10-6 Kg-m2

Inertia of the screw (a steel cylinder)J

s= 761 x (0.04)4 x 1= 1900 x 10-6 Kg-m2

Inertia of the motor (5 Kg-cm2)(remember 1 Kg-m2 = 104 Kg-cm2)

Jm

= 500 x 10-6 Kg-m2

Total inertia

Jt

= 2650 x 10-6 Kg-m2

Acceleration rate (distance=0.5 rev)A = 4.5 x 0.5 0.122

= 156 revs/sec2

Accelerating torque

T = 2650 x 10-6 x 156 x 2= 2.6Nm

How would the torque be affected if the screw is only 70%efficient?

The reduced efficiency will only affect the torque needed toaccelerate the mass of the table, since the leadscrew itselfis directly connected. However, reduced screw efficiencynormally implies friction torque in the nut as well.

Proportion of inertia from the table

= 250 2650= 0.094

Torque to accelerate table mass

= 2.6 x 0.094= 0.24 Nm

New value with 70% efficiency

= 0.24 0.7= 0.34 Nm

Increase in torque

= 0.1 Nm

New total torque demand

T = 2.7Nm

Triangle Trapezoidal Change,

trap/tri.

Maximum velocity 2D/t 5D/t 75%

Acceleration rate 4D/t2 4.5D/t2 112.5%

Peak power Jx8D2/t3 Jx6.75D2/t3 84%

V

tTime

Speed

t/3 2t/3

V

t

Speed

Time

100Kg Motor

10mm

5mm in 120mS

Jm = 5Kg-cm2


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What happens if the slideways carrying the table have afriction coefficient of 0.2 (using the screw with 70%efficiency)?

Gravitational force due to table mass= 100 x 9.81= 981 Newtons

Friction force ( = 0.1)

F = 981 x 0.2= 196 Newtons

Torque to overcome this force

T = 196 x 10 6284 x 0.7= 0.44 Nm

New torque demand

T = 3.14Nm

What acceleration & velocity values would you need toprogram?

Acceleration rate (from above)

A = 156 revs/sec2

Velocity

V = 1.5 x 0.5 0.12= 6.25 revs/sec

Going back to the simple case with a perfect leadscrew &no friction, what is the peak power demand?

Peak power

W = 2.6 x 6.25 x 2

= 102 watts

What is the load/rotor inertia ratio?

Table + screw inertia

Jw+ J

s= 2150 x 10-6 Kg-m2

Motor inertiaJ

m= 500 x 10-6 Kg-m2

Inertia ratio

= 2150 500= 4.3:1

If we introduce a 2:1 reduction and double the screw pitchto 20mm, what happens to the torque requirement? (Ignorethe inertia of the reduction system).

Reflected inertia of the table

Jw

= 100 x 202 4 x 107

= 1000 x 10-6 Kg-m2

The screw inertia will be the same as before. The inertia ofthe table plus the screw will now be reduced by the squareof the reduction ratio.

Table + screw inertia

Jw+ J

s= 2900 x 10-6 Kg-m2

Reflected inertia at motor

= 2900 x 10-6 4 Kg-m2

= 725 x 10-6 Kg-m2

New total inertia (adding the motor)J

t= 1225 x 10-6 Kg-m2

The maximum speed, acceleration rate and distancetravelled by the motor remain the same.

New torque demand

T = 1225 x 10-6

x 156 x 2= 1.23Nm

What is the new load/rotor inertia ratio?

New inertia ratio

= 725 500= 1.45:1

How is the peak power affected?

New peak power

W = 1.23 x 6.25 x 2= 48 watts

Here is a situation where increasing the screw pitch and

introducing a reduction ratio has offered a significantimprovement - the screw only rotates at half its previousspeed due to the increased pitch, and the effect of itsinertia is reduced by the reduction ratio. The torque andpower required from the motor have both been reduced by50%, and the load/motor inertia ratio is more favourable.

Since the torque demand is now around half what it wasbefore, using the same motor we could double theacceleration rate and reduce the overall move time by30%. In practice of course the reduction components willalso introduce some additional inertia and friction.

Relationship between shaft power, move

distance and move timeThe expressions in the last section showed that maximumvelocity is proportional to D/t, and acceleration rate(therefore torque) to D/t2. Power is the product of torqueand speed, and is therefore proportional to D2 and to 1/t3.From this we can conclude that to move twice the distancein the same time requires 4 times as much shaft power,and to move the same distance in half the time requires 8times as much shaft power.

The maximum power output of a motor is related to its size,and a motor capable of delivering 8 times the power will bevery much larger. It will also have significantly more inertiawhich will increase the torque demand still further. So in

practice the additional power required to halve the movetime can be as much as 50 times.

There are two important conclusions to be drawn here -

1. Move time is a critical factor in system powerrequirements. Significant economies can beobtained by relaxing move time requirements wherepossible, and its always worth looking at otheraspects of a machine cycle to see if savings can bemade elsewhere.

2. A well-designed positioning system cannot be made torun significantly faster than its design speed. Forexample, with a system having a power margin of 20%,you would only expect to be able to trim around 5% off

the move time by pushing everything to the limit.

System calculations


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RMS torque calculation

When using a servo motor in intermittent duty applications,its possible to overdrive the motor for part of the operating

cycle provided its given time to cool down afterwards. Atypical intermittent duty rating is anything up to four timesthe continuous torque, and is usually limited only by theamount of current available from the drive. As long as thetotal heat produced during an operating cycle is no morethan when the motor runs continuously at its rated torque,its safe to assume that the motor wont overheat.

The main source of heat is resistive loss in the motorwindings, which is proportional to current squared(remember W = RI2?) Since current and torque are directlyrelated (from the Torque Constant), the heat generated isalso proportional to torque squared. For example, at doublethe torque there will be four times as much heat produced.

We therefore cant take a simple arithmetic average of thetorque to calculate the total heat input during the cycle. Weneed to use a root-mean-square (RMS) average. Theformula looks like this:

T1

2t1+T

22t

2+T

32t

3+......

Trms

=t1

+ t2

+ t3+ t

4+.....

slower the temperature rises. The thermal time constant isthe time taken to reach 63% of the final temperature; ittakes 3 time constants to reach approximately 95% of thefinal temperature, so after that things obviously change

quite slowly.

For our RMS heating calculations to be valid, its importantthat the complete operating cycle is short compared withthe motors thermal time constant - 10% is a good guide.You can usually find the thermal time constant in the motordata, and it typically varies from 30 minutes for a smallmotor to 90 minutes for a large one. An operating cycle upto 3 minutes long should therefore be acceptable in mostcases. In practice this would be quite a long cycle, so thisis not a problem in the majority of applications.

Sizing mains transformers forpositioning systemsTransformer-fed stepper and servo drives are normallyspecified with a standard transformer which will beadequate in all practical applications. In reality this meansthat the transformer is over-specified in the majority ofcases. Nevertheless if the quantities involved are small, itis better to have a guaranteed working solution rather thanspend a great deal of engineering effort optimising thetransformer design in order to save a comparatively smallamount of money.

Where larger quantities are involved or where space is at apremium then an optimised design is more likely to bejustified. However most positioning systems involvingpoint-to-point moves can be tricky to size because the

power changes significantly throughout the operating cycle.

The basic requirements of the transformer are:

1. To supply the long-term average power withoutoverheating.

2. To supply the short-term peak demand withoutexcessive voltage drop.

The long-term average power requirement can beestimated once the RMS torque calculations have been

done.

When the transformer is delivering peak powers greaterthan its continuous rating for short periods (e.g. 2 secondsor less), a good guide to the required VA rating will be the

greater of the following:1 Peak load x peak load duration/total cycle time.

2 70% of the peak load.

For example, a peak load of 1000VA for a total of 400mSevery 1.2 seconds would require a transformer rating of1000 x 0.4/1.2 = 577VA. However, in this case thesuggested rating is 700VA (i.e. 70% of the peak load).

A drop in DC bus voltage caused by transformer regulationwill affect the high-speed performance. The greatest dropwill occur when the system is delivering maximum power,which coincides with the knee of the peak torque curve.This has the effect of rounding off the torque-speed curveso that it becomes impossible to operate right up at the

knee point.

Most proprietary transformers have a secondary voltagerating which is specified at full load current. This can result

Fig 7.9 Simple machine cycle

All were doing here is adding up the heat produced duringeach part of the cycle, dividing it by the total time and then

taking the square root to get back to an equivalentcontinuous torque.

What happens with very long operating cycles?

If the torque exceeds the continuous rating for a longperiod, its obviously possible for the motor to overheatduring this time even though the RMS torque is acceptable.For example, at double the continuous torque we arerestricted to a 25% duty cycle (i.e. four times as much heatfor a quarter of the time). If the torque is required for onesecond in a four-second cycle that sounds fine - but onehour out of every four hours seems like bad news, eventhough this is still a 25% duty cycle. So where is thedividing line?

It comes down to the Thermal Time Constant of the motor.When current passes through a motor, the temperaturefollows an exponential rise. The larger the motor, the

System calculations

Torque (T)

Velocity

T1

T2

T3

(dwell)

t1 t2 t3 t4


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in an unacceptably high no-load voltage on a transformerwith poor regulation. On Parker-supplied transformers thesecondary voltage is specified as an open-circuit value,ensuring that the no-load DC bus voltage does not rise to a

level which could damage the drive. A reduction in busvoltage at full load will degrade the high-speed performancebut will not represent a threat to the drive.

Fig 7.10 Effect of transformer regulation on servoperformance

Typical transformer regulation figures depend on VA rating,but the following will be a useful guide:

VA rating Typical regulation

100 - 500 8%

500 - 1000 6%

1000 - 1500 4 - 5%

1500 - 2000 3 - 4%

>2000 2 - 2.5%

Dynamic braking of servomotorsPart of the information contained in this section has beenkindly supplied by SEM Limited.

In an emergency situation, when the motor needs to bestopped following a drive fault or loss of electrical power,

dynamic braking can be used to minimise the decelerationtime. This is achieved by connecting a resistive load acrossthe motor terminals, forcing the motor to act as a generatorand produce a braking torque. This torque will vary withmotor speed, and will inevitably fall off as the motor reacheslow speeds. The way in which the torque varies with speedis shown in Fig 7.11 - it reaches a maximum at a speedreferred to as the centre speed.

The maximum braking torque depends only on the motor,whereas the speed at which the maximum torque isproduced depends on the value of the braking resistor. Todecelerate the motor as quickly as possible, the resistorvalue must be chosen to create the greatest torque over thewidest possible speed range. This requires a knowledge of

the maximum operating speed as well as certain motorparameters.

Fig 7.11 Variation of dynamic braking torque withspeed

In practice, the value of the load resistor is normally chosento give maximum torque at half the maximum operatingspeed. This results in a braking torque of more than 80%of maximum over the top 75% of the speed range.

Resistance Value Calculation

The required load resistance per phase to produce theoptimum braking torque can be calculated using thefollowing formula:

Load resistance= 0.013 x Inductance x No of motor poles x Max

speed in rpm

The inductance is the line-to-line value measured inHenrys, and the calculated load resistance will be the totalcircuit resistance in ohms including the motor resistance.

Example: SEM HDX142C6-44S motor running at amaximum speed of 3,000rpm.

From the motor data sheet:

Line-to-line inductance= 2.3mH (phase inductance = 1.15mH)

Line-to-line resistance

= 0.2 (phase resistance = 0.1)

No of poles = 6

The calculation gives a total load resistance per phase of0.54. This includes the motor phase resistance of 0.1,so the external resistance required will be 0.44.

Resistor Power Rating

To a first approximation, the required power rating for theload resistors can be estimated from the kinetic energystored in the system. For example, taking a relativelysevere condition where the load has an inertia of ten timesthe motor inertia:

For the HDX142C6-44S the motor inertia is 0.00115 kgm2

Assuming a load inertia of 0.0115 kgm2

Total inertia= 0.0126 kgm2

System calculations

ContinuousTorque

PeakDC bus dropreduces available

torque & speed

Speed

100

80

60

40

20

00 10 20 30 40 50 60 70 80 90 100

% of

maximumtorque

% of maximum speed

CENTRE

SPEED


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Total energy stored in the system at 3,000rpm

= 0.5Jw2

= 622 watt-seconds

The time taken for the motor to slow down will depend on itsmaximum speed, the system inertia and the availablebraking torque.

The maximum torque at the centre speed is given by:

T = 3 Kt2 (L x No of poles)

where Kt= torque constant per phase (rms)

L = phase inductance

The approximate time taken to slow down to 5% ofmaximum speed is given by:

t = 0.27 x Maximum speed in rpm x system inertiamaximum braking torque

For the HDX142C6-44S, Kt= 0.17 and the maximum

torque (from the equation above) is 13 Nm (this means thatthe decelerating torque is in excess of 10Nm between3000 rpm and 750 rpm). The calculation shows that themotor should come to rest in approximately 0.8s.

Generally, power resistors have an overload rating of fivetimes the normal wattage rating over a period of 5seconds. As the time taken for the motor to come to rest isless than 1 second and the power to be dissipated over 5seconds is 622/5 = 125 watts, 25-watt resistors should beadequate.

System calculations

Fig 7.12 Typical dynamic braking circuit

0V

CONTACTOR

A1 output transistorin the OFF condition.

DRIVE

U

V

W

E

DYNAMIC

BRAKE

LOAD

MOTOR

screenGND 24V

10nf 63V

X11/7X10/10

X10/16

X10/15

X10/14

+24V

Contactor

supply

Relay

1F

SV DRIVE

A1 OUTPUT


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An introduction to X-Codeprogramming

X-Code is a command language devised by Parkerspecifically for motion control. It is used both in standalonecontrollers and in combined controller-drive products.X-Code is particularly simple and straightforward to learn;most of the commands use the initial letter or letters of theirfunction name, for example A for acceleration and V forvelocity. Although many X-Code products offer in excess of150 commands, the majority of applications only make useof a handful of these. Most users therefore start by learningonly the basic motion commands, together with the

additional commands specific to their particular application.

A typical installation

Apart from the AC supply and motor cables, the only

additional connection required for a basic indexer-driveinstallation is a 3-wire RS232 cable from the drive to a PC.

X-Code programming

Fig 8.1 Single-axis RS232 connection

The PC can be set up as a simple terminal by making useof X-Ware, a software package supplied with X-Codeproducts which also incorporates program storage andretrieval facilities as well as a program editor. The X-Codecommands are simply typed on the PC keyboard and are

transmitted to the controller as ASCII characters.

Some notes on setting up and troubleshooting RS232communications will be found at the end of this section.

Multi-axis systems

X-code can be used where there are two or morecontrollers connected back to the PC. The RS232connection is made using a daisy chain so that the signal

is passed from one controller to the next and finally back tothe PC. In this arrangement, each controller is given adifferent address to distinguish it from the others. Theaddress is simply a number (usually starting from 1) which

is set up on bit switches, jumper links or in software. In thisway, commands which incorporate a specific address willonly be accepted by the corresponding controller.

All X-Code products have the ability to store completemotion programs within non-volatile memory, either as abuilt-in or optional facility. Once programmed it is thereforefrequently possible to remove the RS232 connection. Astored program may then be selected and started byexternal signals or switch inputs, or it may be arranged torun automatically on power-up. Alternatively the indexer-drive may accept real-time X-Code commands via RS232from a host controller such as a PLC or industrial PC.

Basic motion commands

The following commands will be common to everyapplication. They define the basic parameters associatedwith a move; in many positioning applications, theacceleration and velocity values will apply to every moveand are defined only at the start of the program. Thecontroller always uses the last-specified value for anyparameter until such time as it is overwritten.

V - Velocity, revs/sec

A - Acceleration rate, revs/sec2

D - Distance, motor/user steps

G - Go

Example:

V10 - set velocity to 10 revs/sec

A150 - set acceleration rate to 150 revs/sec2

D4000 - set distance to 4000 motor steps

G - go (i.e. make the move)

Before going on to look at more commands, well look atthe command format and review the different types ofcommand and alternative operating modes for thecontroller.

Command format

All X-Code commands consist of upper case charactersarranged as follows:

[device address] [command] [numerical value] [delimiter]

The device address is not always required (see below).The delimiter marks the end of the command and can beeither a space (i.e. space bar) or a carriage return.

Example:

2V10

Command types

Commands can be grouped into different types accordingto whether they join a queue or take immediate effect, alsowhether they apply just to one controller or to all controllersin the system.

IMMEDIATE commands are executed as soon as they arereceived, regardless of what is going on at the time. Anexample is the command S which is to Stop. Not manycommands come into this category.Fig 8.2 RS232 daisy chain

RS232

cable

PC

Indexer/

drive

Motor

PC drive 1 drive 2 drive 3


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BUFFERED commands go into a first-in, first-out store orbuffer and are executed in the order they are received.Each command will be completed before the next onestarts. This means that a string of commands can be

downloaded from the host PC without waiting for each oneto finish. Theres a limit to the number of commands whichcan be stored, typically around 2000 characters. Examplesof buffered commands are V for Velocity and G for Go.Most commands are buffered, and only buffered commandscan be stored in memory for execution as a sequence.

DEVICE SPECIFIC commands are executed by onecontroller or device only, as specified by the deviceaddress. Commands in this category must include anaddress; they are mostly commands which requestinformation back, such as whether the drive is ready for anew command or busy, so its essential that only the correctaxis replies. Commands which are defined in the manual as

device specific wont be actioned without an address. Anexample is R for Report Status - this would have to be sentfor example as 1R for axis 1.

UNIVERSAL commands are executed by all devices in thechain. In this case there is no address included, so thecommand will be accepted by all axes. An example wouldbe V10 which would set the velocity on all axes to 10 revs/sec. However, most universal commands may be madedevice specific simply by including an address, for example3V10 would set the speed on axis 3 only to 10 revs/sec.

Operating modes

All X-Code controllers can operate in either of two basic

modes - preset (normal) or continuous.MODE NORMAL (MN) - the move distance is pre-determined using the D (Distance) command. This mode isused in all normal point-to-point positioning operations.

MODE CONTINUOUS (MC) - in this mode the motor runscontinuously at the specified velocity until it is stopped or anew velocity command is entered. Distance data set by theD command is ignored.

Within the normal mode (MN) there are two further options:

MODE PRESET INCREMENTAL (MPI) - here each moveis performed as an increment, with the distance being set bythe D command. In other words, the distance travelled isrelative to the current position.

MODE PRESET ABSOLUTE (MPA) - in this mode the Dor distance data is interpreted as an absolute positionrelative to position zero. Therefore each move will be to adefined absolute position regardless of the current position.The controller calculates the required distance and direction

of travel to achieve the new position.

You can switch freely between the absolute and incrementalpositioning modes since the controller always stores itscurrent absolute position.

Additional basic commands

Direction control

In the normal incremental mode you determine the directionof shaft rotation by the sign of the distance value. So D4000will set a move of 4000 steps in the clockwise (CW)direction, while D-6000 will give a 6000-step CCW move.

The direction is assumed to be CW unless a minus sign isincluded, although you can include a plus sign for CWmoves if you wish.

In addition you can control direction by using H commands.The H command allows you to set or change directionwithout re-specifying the distance, and is also used in theContinuous Mode (where D values are ignored).

H+ sets the direction CW

H- sets the direction CCW

H (with no sign) reverses direction from whatever itcurrently is

Example: D25000 G H- G

Motor runs 25000 steps CW, then 25000 steps CCW.

Creating loops

Where a group of commands must be repeated a set

number of times, or even repeated continuously, this isachieved simply by enclosing the commands within a loop.The Loop command L marks the beginning of the loop,followed by the number of times you want to go round it,e.g. L6. The finish of the loop is marked by the Endcommand N.

Example: L20 D500 G N

The motor will make 20 moves of 500 steps each.

If there is no number included after the L command, theloop will run continuously. It can be stopped either with theS (Stop) command, which will abort any move in progress,or with the Y command which stops at the end of thecurrent loop.

With most controllers its possible to nest loops together, inother words you can have one or more loops inside anouter loop. However you must make sure that the numberof N commands matches the number of L commands, evenif they all come together.

Example: L5 D1000 G L10 D200 G N N

Here each of the five 1000-step moves will be followed bythe loop of ten 200-step moves.

Adding time delays

Its common to want to include a simple time delay betweenmoves, for instance to allow for an external operation to becompleted. This is achieved simply with the T (Time)

command followed by the delay time in seconds.

Examples: T2 gives a 2 second delay

T0.05 gives a 50mS delay

L20 D500 G T0.5 N

In the last example the T command adds a half-seconddelay between each move in the loop.

Using trigger inputs

The Trigger command TR allows you to specify a patternon the controller inputs which will pass on to or trigger thenext command. The input signal could come for instancefrom an operator pushbutton or a safety interlock switch.

The options on the inputs are:1 = input high0 = input lowX = dont care (input can be high or low)

X-Code programming


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The input pattern is specified in the numerical order of theinputs. For example: TR01X - continue to the nextcommand when input 1 is low and input 2 is high. Ignorewhats happening on input 3. In fact you can ignore any

trailing X values, so in this case TR01 would work just aswell.

Some controllers offer the option of alternative versions ofthe trigger command, i.e.

TRE- trigger when inputs equal to specified pattern(equivalent to TR)

TRN- trigger when inputs not equal to specified pattern

Using programmable outputs

The O (Output) command is used to turn programmableoutputs on and off. It operates in a similar way to the inputtrigger command in that a required pattern is specified forthe outputs. The options are:

1 = output on

0 = output off

X = leave unchanged

Example: 2O1X1 - for axis address 2, turn on outputs 1 &3 but leave output 2 unchanged. Again any trailing Xs canbe ignored.

Going to the Home position

Most positioning systems not fitted with some form ofabsolute positioning device must move to a mechanicalreference or home position on power-up. This is toestablish where the mechanics are located and all

subsequent moves will be relative to the home position.This position is normally determined by a switch or proximitydetector.

The Go Home command (GH) initiates an automatic returnto the home position, beginning with a fast approach andusually followed by a slow search for the switch operatingpoint. The number following the command represents thefast approach speed.

Example: GH5 - go home at 5 revs/sec. (a direction signcan be included)

The approach pattern will resemble Fig. 8.3. During thefinal slow approach, the absolute position counter is reset tozero as the switch edge is detected. Depending on the

controller type, the system does not necessarily stop at thehome position - time may still be required for deceleration.

Fig 8.3 Typical home approach pattern

However the position has been captured as the switchedge is crossed and an accurate current positionestablished.

There are a number of options available with the homingroutine which vary according to the type of controller. Theyinclude a choice of final approach speed and the ability toselect which edge of the home switch is taken as thestopping position.

Programming complete motionsequences

A sequence is a series of commands that will be performedin the programmed order. The sequence can be initiatedby a single command, making it easier to programrepeated operations. Sequences can be stored in non-volatile memory and then selected and initiated by triggerinputs without the need for an RS232 connection. This

style of operation is particularly useful when the overallmachine control is via a PLC.

Note that only buffered commands can be used and storedin a sequence - by definition, immediate commands willbe executed as soon as they are received and not stored inthe command buffer.

All sequence commands begin with X.

XDn starts the definition of a sequence. For example, XD1marks the beginning of sequence 1.

XT terminates a sequence. Here is an example of a singlesequence:

XD1 A2 V10 D2000 G H- G XT

XRn runs the sequence, so XR1 runs sequence 1. Youcan use the XR command to run a sequence from withinanother sequence, rather like a subroutine, for example:

XD2 A2 V10 D2000 G XR1 XT

In this case, sequence 2 incorporates sequence 1. (Withcertain controller types, XR acts like a GoTo rather than aGoSub, in which case it wont return to the primarysequence. Check with the User Guide when embeddingXRs within a sequence).

To change a sequence, the old one must first be erased.This is a safety feature that minimises the chance ofoverwriting a sequence by accident.

XEn erases a sequence, i.e. XE2 erases sequence 2.

To check what is programmed in a sequence, it can bereturned to the terminal.

aXUn uploads a sequence (this is a device-specificcommand so must always include the address). Forexample, 1XU3 returns sequence 3 to the controller fromdevice number 1.

For further information

This is only a brief introduction to X-Code, but it illustratesthe simple nature of the language and the ease with whichroutine motion control functions can be programmed.Individual product User Guides provide comprehensiveinformation on all the available commands and ways in

which they can be used. If you are considering the use ofany X-Code product, a copy of the User Guide can bemade available on request.

X-Code programming

Home Switch

Fast return (n)

Final slowapproach


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X-Code programming

Alternative programminglanguages

6000 language

Parkers 6000 language has been developed from X-Codein order to provide convenient programming for multi-axiscontrollers. Whereas X-Code is ideal for standalone single-axis systems, and has the capability of addressing multipleunits via a daisy chain, 6000 Language simplifies theprogramming process within controllers with two or moreaxis outputs.

6000 Language generally uses longer mnemonics than X-Code which allows them to be more descriptive. Inaddition, 6000 offers more advanced features than areavailable within X-Code, many of which relate to multi-axiscontrol systems. These include contouring, cam functions,joystick control, teach operations and real-time datacapture. The high-functionality error programming availablewith 6000 code allows for more efficient recovery from errorconditions.

6000 is a very versatile, universal language which can beapplied in an extremely wide range of applications. Theextra capabilities of this language naturally demandincreased processing power, therefore X-Code is moreappropriate in low-cost systems with limited functionalityrequirements.

COMPAX control language

The Compax is a fully-digital servo controller including the

power stage. It uses its own unique programming languagewhich has very BASIC-like commands. Unlike the othercontrollers in Parkers range, the Compax may be equippedwith application-specific firmware which simplifies theprogramming of a range of advanced machine functions.These include time-based velocity profiling, cam generation,cutting on the fly, following, electronic gearbox andregistration applications. By using dedicated firmware,programming these specific functions involves only theentry of a few parameters. The cam generation facility isparticularly powerful, with a dedicated PC software packageto simplify profile generation. There is an option to enter orleave the cam profile at any point for complex machine

operations.

Although rather less flexible than products using 6000Language, Compax-based systems are much simpler toimplement when a firmware option is available to suit theapplication. The event-driven programming not onlysimplifies and shortens programming time, it also minimisesthe amount of data that needs to be transferred. Thismeans that Compax is particularly suited to systemsemploying Fieldbus communication. In addition to standardRS485 Fieldbus, it offers the option of Interbus-S, Profibusand CANbus. It is also equipped with Parkers HEDA(High-Efficiency Data Access) bus which provides veryaccurate time synchronisation (within 2.5S) between allaxes in a system. This is combined with a sophisticated

servo algorithm to give extremely high dynamicperformance.

RS232 troubleshooting

Checking the terminal

Disconnect the RS232 cable from the terminal or computer.Identify pins 2 and 3 on the serial port - Fig. 8.4 shows thepin layout for 9-way and 25-way connectors. Then shortpins 2 and 3 together. If the serial connector is female, usea paper clip to short out the pins. If its male, the blade of asmall screwdriver can be used to connect the pins together.Now type some characters on the keyboard. If nothingappears on the screen, you may not be communicating withthe correct port. Use your communications software tochange the COM port and try again. If characters nowappear or if double characters appear, remove the shortbetween pins 2 & 3 and type some more. If characters stillappear, either you are communicating with another deviceon COM port (such as a mouse or network card), or the

local echo is switched on. Use your communicationssoftware to check that the local echo is turned off, and if sotry changing the COM ports. When the terminal is operatingcorrectly, single characters should appear with pins 2 & 3shorted, and should not appear with the short removed.

Fig 8.4 25-way & 9-way RS232 connectors

Checking the RS232 cable

Reconnect the RS232 cable to the terminal and leave theother end free. Repeat the last exercise by shorting thepins on the remote end of the cable. If no charactersappear, there is obviously a problem with the cable. Ifcharacters do appear, reconnect the cable to the devicebeing controlled and try again.

If there is no response, check that the echoback isswitched on in the device. If there is still no response, tryswapping the Rx and Tx connections and try again. Alsocheck that there is good ground connection between the

terminal and the device (pin 5 on a 9-way connector, andpin 7 on a 25-way). If you have reached this stage withoutsuccess, check the manufacturers documentation onRS232 communication to see if it gives any further clues.As a last resort, contact the manufacturer for assistance.

Verifying correct communication

Once you have confirmed that characters are being echoedback from the device, try sending a command that willalways produce a response. Typical examples would be1R for X-Code products and !TAS for 6000-seriesproducts. If there is no response, check that the correctaddress is being used (e.g. try 2R, 3R etc.). Note that in

certain products the address can be selected by software.It is also worth checking that communication has not beenturned off at the device - send the command E to turn itback on.

Pin: 2 3 7Tx Rx G

25 Pin Connector(Male)

Pin: 5 3 2G Tx Rx

9 Pin Connector(Female)

1

25

1

9


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EMC installation principlesEMC, or electromagnetic compatibility, has been given anumber of definitions. In simple terms it means that a piece

of equipment neither generates excessive electrical noisenor is unduly sensitive to external noise. The basic intentionof the European EMC Directive is to ensure that items ofelectrical or electronic equipment will operate in the sameenvironment without disturbing each other. It also helps torestrict the rise in background interference created by theever-increasing use of electronic controls.

The installation requirements for EMC compatibility will varywith the particular product. Equipment which is CE-markedand certified as having inherent EMC compliance may beinstalled in any location provided that the installationinstructions are followed. Other equipment intended only foruse by qualified system builders requires additional

measures to ensure EMC compliance and the generalprinciples are outlined below.

External enclosure

In order to control radiated emissions, all drive and controlequipment must be installed in a steel equipment cabinetwhich will provide adequate screening.

Filtering

Install an AC supply filter on the power input cable of theunit to suppress any power line interference. Mount the filterwithin 50mm of the drive or transformer, and run the inputcable and any earth cable close to the panel. Try to arrangethe layout of the drive and filter so that the AC input cable is

kept away from the the filter output leads.

Suppression

External Electromagnetic Interference suppression devices,such as ferrite absorbers should be installed on Signal andControl cables as close to the unit as practically possible.

Twisted Pairs

In the case of drives having differential inputs, it ispreferable to use cable with twisted pairs in order tominimise magnetic coupling. This applies to both analogueand digital signals.

Screening

Use high quality shielded (screened) cables for all Signaland Control inputs. The shield (screen) pigtail connectionshould be made as short as possible. The connection pointfor the shield depends upon the individual application.Some of the recommended methods of connecting theshield, in order of effectiveness are;

a) Connect the shield only at the panel where the unitis mounted to earth ground (protective earth). Usingthis method will reduce the overall loop area, andhence provide improved protection.

b) Connect the shield to earth ground at both ends ofthe cable, usually when the noise source frequencyis above 1 MHz.

c) Connect the shield to common of the unit and leavethe other end of the shield unconnected andinsulated from earth ground.

Equipment installation

Segregation

Never run Signal or Control cables in the same conduit withAC power lines, conductors feeding motors, solenoids etc.

The cables should be run in metal conduit that is properlygrounded. Also Signal and Control cables within anenclosure should be routed as far away as possible fromcontactors, control relays, transformers, and any othernoisy components.

P-Clip Installation

The function of a P-Clip is to provide 360 degree metalliccontact and thus a convenient means of ensuring a properR.F ground. Install as close to the cable end as possibleprovided a suitable ground, backplane or earth stud isaccessible. The use of brass or other inert conductive metalP-Clip is recommended.

Surge SuppressionPlace surge suppression components, such as resistor/capacitor filter or zener and clamping diodes, on allelectrical coils.eg contactors.

Opto Isolation

Isolation of remote signals with the use of solid state relaysor opto isolators is recommended.

Motors, electrical installationAlthough there are some well-accepted general guidelines,the electrical installation requirements tend to be specific toa particular motor type. The following comments therefore

apply mainly to Parker servo motors, although many pointswill be relevant to other motor types as well.

Fitting the cables into conduits

The most common reason for encoder failure is incorrectrewiring after a cable has been removed, usually to feed itthrough a conduit.

Beforeyou disconnect any individual wires, make anaccurate sketch of the connections to any terminal strips ormulti-way connectors. If twisted-pair cable is used, payspecial attention to how the wires are paired. Since manyof the pairs have one black wire, simply labelling the wireon a sketch as black is not sufficient to identify it. Its agood idea to slip a short piece of sleeving over each pair of

wires as you disconnect them - this keeps the two wirestogether.

When preparing and stripping twisted-pair cable, alwaysstrip back sufficient of the outer sheath (at least 100mm) tobe certain that you have the correct wires twisted together.Then fit a short piece of sleeving over the pair beforecutting to the required length.

When feeding the cable from an ML motor through conduit,its easier to disconnect the cable at the motor end. Firstremove the terminal box and sketch the connections asdescribed above. You can then unscrew the terminals andpull the cable through the gland nuts.

Wiring stepper motorsMost stepper motors are supplied with short flying leads or,in the case of larger motors, a terminal box. How these are


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connected back to the drive depends very much on theapplication and the distance involved. If the motor is within3m of the drive, the type of cable used is usually not toocritical provided of course that it can carry the necessary

current without overheating.

If the distance is greater than this or the environment iselectrically noisy, it may be better to use screened cable.Noise picked up by the motor cable is unlikely to affect themotor directly, but it will get conducted back into the driveand may cause problems at the signal inputs. Screenedcable will also be necessary when EMC compliance isrequired.

Motors, mechanical installation

Mounting the motor - the pilot register

Many people think that a pilot register on the front of a motor

is a nuisance. It means that if you try and mount the motoron a flat plate, you need to put washers between the motorflange and the plate to stop the flange from bending.

Fig 9.1 Motor pilot register

The pilot register is provided so that you can locate themotor concentrically with the load. This is particularlyimportant when the motor is attached to something like agearbox or an X-Y table. The register is intended to fit into amating recess on the mounting plate - this is what locatesthe motor, not the mounting bolts. The pilot register ismachined to a tight tolerance, typically 0.05mm. When youspecify the mating recess, make the lower tolerance of therecess the same as the upper tolerance of the pilot. In other

words, the largest pilot should just fit into the smallestrecess. This gives the best location whilst guaranteeing thatit will always fit. The depth of the recess must always begreater than the thickness of the pilot register so that theregister cant bottom in the recess.

If you mustmount the motor to a flat plate and concentricityis not important, the best way is to use a machined spacerwhich accommodates the pilot. Since in this case you wontbe using the pilot to locate the motor, the tolerance on therecess machined in the spacer is not critical.

Coupling the motor to the load

Why should you bother to use an expensive coupling when

you could get away with a simple steel sleeve? No reason,provided you can guarantee that the motor shaft and theload shaft are perfectly aligned and that your steel sleeve isa perfect fit. In practice we seldom achieve perfection, so

the coupling must have sufficient compliance or flexibilityto accommodate any misalignment without stressing themotor bearing.

There are two types of misalignment that every coupling iscertain to encounter to some extent. One is angularmisalignment, which occurs when the two shafts are at aslight angle to each other. The other is radial misalignment,in which the shafts are parallel but not concentric - theiraxes are displaced. The coupling may also have to copewith axial movement due, for example, to thermalexpansion.

Fig 9.2 Angular and radial misalignment

A coupling designed for servo and stepper applicationsshould accommodate any misalignment without introducingbacklash (lost motion when changing direction) or torsional

compliance (meaning the coupling can be twisted tooeasily). Backlash and excessive torsional compliance canlead to instability, making servo tuning extremely difficultand imposing a severe limit on performance.

The number of coupling designs available today seemsalmost endless, so the next problem is deciding which typeto use. Well concentrate on two designs which are suitablefor a wide range of applications, the membrane couplingand the Oldham coupling. These couplings willaccommodate both types of misalignment and have verylittle torsional compliance or backlash.

Diagram by courtesy of Huco Engineering Ltd.

Fig 9.3 A typical membrane coupling

pilotregister

Angularerror

Radial error

Axial movement


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Membrane couplings usually have two discs as shown. Asingle-disc version is also available but this wont tolerateany radial misalignment. Single-disc couplings are usuallyused in pairs with a floating shaft in between. A complete

assembly consisting of two single-disc couplings plus a linkshaft is known as a Cardan; it operates in the same way asthe two-disc coupling except that the longer the link shaft,the greater the radial misaligment that can beaccommodated. The three elements of the Cardan canusually be assembled in situ.

Other coupling types which are sometimes used in low-power applications are the bellows and helical beamcouplings. Bellows couplings are particularly good ataccommodating axial movement and offer extremely high

torsional stiffness. The helical beam coupler has thebenefit of low cost but has relatively poor torsional stiffnessand generates high bearing loads when misalignment ispresent. Both coupling types are more prone to eventualfailure unless used well within their rating.

Attaching the coupling - clamping methods

Once the type of coupling is decided, the next question ishow to attach it to the motor and load shafts. A bi-conicalor Taper lock bush, used in conjunction with a key, isprobably the favourite method. It maintains concentricityand grips the shaft over a large area, which is highlydesirable where high torques and rapid reversals areinvolved. In this situation the key is used primarily as asafety device to prevent slippage in the event of anoverload - its not being used to transmit the torque.Unfortunately, many couplings are too small toaccommodate this type of bush.

Clamps are probably the most common fixing for smallercouplings and are generally reliable if you use a thread-locking compound on the screws. This will act as alubricant before it sets, reducing frictional losses andmaximising the clamping force - but don't take too longtightening the screws! Use washers under the screw headswherever possible.

Fig 9.4 An unsupported (floating) shaft

Membrane couplings can operate at high speeds, typicallyup to 25,000 rpm, and have a high life expectancy sincethere are no sliding parts. However they are moreexpensive than most other coupling types.

Diagram by courtesy of Huco Engineering Ltd.

Fig 9.5 A typical Oldham coupling

As the Oldham coupling has three separate elements, thetwo hubs can be fitted loosely to the shafts before assemblyand the central disc dropped in once the motor and loadshaft are in place. This can be very useful in situations withpoor access. Similarly, the shafts can be uncoupled simplyby loosening one of the hubs. The central disc is availablein different materials to give the desired properties. For

example, acetal gives a long life and good torsionalstiffness; nylon gives good vibration absorption and quietrunning. Since the disc is continually sliding, it graduallywears and introduces backlash so eventually it must bereplaced. Oldham couplings cant be used on unsupportedshafts. They are one of the least expensive types ofcoupling, but are intended only for operation at speeds up toabout 3,000 rpm.

Here are some typical figures showing how well each of thecouplings can accommodate misalignment:

Membrane Oldham

Max. Angular Misalignment 4 degrees 1 degree

Max. Radial Misalignment 0.4mm 11% ofcoupling diam.

Max. Axial Movement 0.2mm 0.3mm

Fig 9.6 A leaf clamp and a pinch clamp

Set screws

Set screw (or grub screw) fixings are another commonmethod, but you should only use these for low torqueapplications. Dont tighten the set screw into the keyway-you can machine a dimple or a flat on the shaft, providedyou take the same precautions as when shortening theshaft, and these are described later. One drawback withthe setscrew is that, unlike a clamp, it makes a small recessin the shaft so that minor adjustment becomes almostimpossible.

It is essential that the coupling bore and shaft diameter areclose-fitting when a set screw is used - the clearanceshould not be more than 0.025mm. This is because the setscrew always forces the shaft to one side of the coupling

bore, and a larger clearance can allow the coupling topivot around the set screw, leading to rapid wear. Two setscrews, 90 apart, are a significant improvement on justone and also double the torque which can be transmitted



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(see table following). Forcing the shaft to one side of thebore also creates eccentricity, so you should use a couplingwhich will accommodate this. Always try to locate thesetscrew as centrally as possible along the coupling or

pulley length.

The following table is a guide to the maximumrecommended torque using couplings or pulleys attached bysetscrews. The figures apply to a single setscrew and canbe doubled if you use two screws 90 apart.

Shaft dia. Screw Max. torque (mm) size (Nm)

6 M2 0.1

6 M3 0.2

11 M3 0.4

11 M4 0.8

15 M3 0.5

15 M4 1

15 M5 2

Pinning

Drilling and pinning the shaft is a method which isoccasionally used, but again its only suitable for low-torqueapplications. The hole considerably reduces the cross-sectional area of the shaft and therefore the torque it cantransmit. If the shaft already has a keyway, the situationbecomes even worse. You should remember that the shearstrength of a 3mm pin is only about 20-30% of that of a3mm key. If you do opt for this fixing, you should take thesame precautions when drilling the shaft as whenshortening it, and these are described later. The comments

on coupling/shaft tolerances when using set screws alsoapply to pinning.

Using adhesives

Its often beneficial to use an anaerobic adhesive such asLoctite between the shaft and the coupling. This type ofadhesive cures in the absence of air. You should only usean adhesive in conjunction with one of the fixing methodslisted above. If you use a thick high-strength adhesive itmust be spread onto the shaft before assembly, whereas athinner adhesive may be run into the gap after the couplinghas been fitted. If you use an adhesive, youll find it almostimpossible to remove the coupling without damaging themotor unless youve left sufficient room between the

coupling and the motor to insert the jaws of a bearing puller.Another point to consider is that in the event of a failure, itmay not be possible to repair the motor without undertakingan expensive rebuild.

Keys and keyways

Youve sent us the wrong key - it doesnt fit into thekeyway. This is a common cry among those new to thebusiness of fitting motors.

A standard 3mm key, for example, is about 0.04mm widerthan the shaft keyway and is intended to be filed to be atight fit. So why doesnt the motor manufacturer make thekeyway the right width in the first place? Well its all a

matter of tolerances - the key should be a very tight fit inthe shaft keyway because the coupling keyway is about0.025mm larger than the one in the shaft. This is to makesure that once the key has been filed to fit the shaft keyway,

the resulting clearance allows the coupling to slide easilyonto the shaft. However this small clearance will inevitablylead to some backlash (the keyways in both the shaft andthe pulley should be deep enough to ensure that there is

adequate clearance for the key).

The best way to overcome this problem is to have a setscrew in the coupling which clamps down into a dimple inthe top face of the key. Another is to machine the key witha step so that its a tight fit into both keyways, but since thisprevents the coupling from sliding easily onto the shaft, youmust take great care not to damage the motor bearings.Applying a large axial load (for instance with a fly press)and hammering the coupling onto the shaft are among thecommonest causes of bearing damage leading topremature failure. If a Taper lock bush is used, theseproblems dont arise because the bush slides onto the shaftvery easily and theres no backlash because the bush

clamps tightly round the shaft.Bear in mind that a 5mm key in a 15mm shaft transmitting atorque of 10Nm is subjected to a shearing force of about 30MegaNewtons/m2. A badly fitted key will rock as the motorchanges direction and, under this sort of force, it is notunheard of for the key to wear into a perfect cylinder whichthen reduces in diameter until the system fails.

Shaft tolerances

There is a popular theory that the coupling should be a verytight fit on the shaft and should be carefully installed usingthe largest available hammer. Its true that if the coupling istoo loose, slight movement between the shaft and couplingwill cause both components to wear and eventually fail. Butbearings are relatively fragile components and ideally, theforce used to fit the coupling should not exceed themaximum axial force, as taken from Fig. 9.10. Usually theupper tolerance of the shaft will equal the lower tolerance ofthe coupling bore, so there should always be a smallclearance between the two.

If youre unlucky enough to have a shaft on its maximumtolerance and a coupling on its minimum tolerance, youwould have a size fit which would be very difficult to slideon. The easiest solution in this situation is to push a ring ofcardboard or thin plastic over the shaft, to prevent dust fromentering the bearing, and then run the motor while you holda piece of carborundum paper round the shaft. Keep trying

the shaft for size to make sure you dont take too much off.

Shortening the shaft

Its surprising how frequently users find it necessary toshorten a motor shaft. The problem can often be avoided ifa little more thought is given at the design stage. Ifamputation is unavoidable, its essential that the shaft issupported in such a way that the motor bearingexperiences no shock or strain. There must be no risk ofswarf or coolant entering the bearing. Dont let the shaftget too hot - if it gets above 120C you run the risk ofpartial demagnetisation of the motor.

Support the shaft on the motor side of the cut, as close tothe cut as possible, using a vee block or clamp. The motoritself should be loosely supported to prevent strain on thebearing. Use a ring of adhesive tape or mouldableadhesive compound (such as Blue-Tack) to prevent coolant



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or swarf from entering the bearing. Please remember thatany form of shaft modification made after despatch fromParker invalidates the warranty in respect of the shaft,bearings, encoder or resolver.

Fig 9.7 A motor shaft supported for cutting

Radial bearing loadIts not unknown for a motor to be returned with its shaftsheared either immediately before or behind the frontbearing. Motors are also returned with bearings that havefailed after only a few months. In almost every case of shaftfailure, and most bearing failures, the cause is the same -too high a radial load acting too near the end of the shaft.By far the most common reason for this is an over-tensioned drive belt, often on a pulley which is overhangingthe end of the shaft because the motor mounting plate istoo thick.

Fig 9.8 Radial and Axial Load

High radial loads applied to the motor shaft will affect bothbearing life and the likelihood of shaft failure.

Bearing life

If there is a large radial force on the inner race of a bearing,this tends to crush the balls between the two races. This inturn results in scoring of the races, leading to unevenrunning and excessive wear. Figs. 10 is a graph of bearinglife against radial load for a typical servo motor, assumingthat the load acts half way along the shaft extension.

Lifetimes are given for specific constant operating speeds,and you can use the curves as a guide to estimate bearinglife for intermediate speeds. If the motor is accelerating ordecelerating for a significant proportion of the time, work outthe average speed over the whole operating cycle and usethis figure. (A simple average is quite good enough here -bearing life depends not so much on absolute speed as onthe total number of revolutions). If the load is acting at the

end of the shaft, you can expect the lifetime to be reducedby about 15%. These graphs assume that the axial load isless than 30% of the radial load.

Shaft fatigue

Its clear that any radial load will compress one side of theshaft and create tension on the other. As the shaft rotates,

each individual element of the shaft experiences alternatingtension and compression which can lead to metal fatigue.The vertical line on Fig. 9.9 represents the maximum radialload that can be applied half way along the shaft withoutrisk of shaft failure through metal fatigue. You will need tohalve this figure if the load is applied at the end of the shaft

Axial bearing load

Motors are normally fitted with spring washers which applya small axial load to the bearings all the time. This takes upany clearance between the balls and the races to preventthe balls from rattling, which would cause wear. A smallamount of additional axial load does no harm, but once theload reaches a level where the balls are being forced into

the races, the bearing will wear rapidly.

Fig. 9.10 shows how the radial load must be reduced as theaxial load increases. Both radial and axial loads areexpressed as a percentage of the radial load Fr, which isthe value read from Fig. 9.9 for the appropriate mean speedand required bearing life.

Fig 9.9 Typical bearing life curves (Parker ML34servo motors)

Fig 9.10 Permitted radial load vs axial load (ParkerML34 servo motors)


clamp here

sawcut

vee blocks

rubber block

Motor

radial load

axial loadMotor

50,000

40,000

30,000

20,000

10,000

00 400 600 800200

Bearinglife in

hours

Radial load (Newtons) half way along shaft

100 300 500 700

4000

rpm

3000

rpm

2000

rpm

1000

rpm

*Shaft fatigue limit for ML3450A & ML3475A is 1000 Newtons

shaftfatiguelimitfor

ML3450B&ML3475B*

125% Fr

100% Fr

75% Fr

50% Fr

25% Fr

00 50% Fr 75% Fr 100% Fr25% Fr

30% Fr

Axial load

Fr is the radial load from bearing life curve

Deratedradialload


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51

Automation

Drive belts

Tensioning a drive belt is frequently regarded as an artrather than a science, partly because there are no simple

guidelines that can be applied. As a result, the tendency isto tension the belt as tightly as possible on the basis thatthis is the only way to get rid of the backlash. Theconsequence is rapid belt wear, reduced bearing life or evenshaft failure. Nevertheless there is a more scientificapproach which is strongly recommended.

To estimate the correct belt tension, you need to know theperipheral force on the motor pulley (Fp) when the motor isproducing maximum torque. You can calculate this force bydividing the maximum torque in Nm by the radius of themotor pulley in metres (remember to divide by the radius,not the diameter). A belt tension between 30% and 50% ofFp is normally used for short or stiff (inelastic) belts,whereas higher tensions (50-65% of Fp) are used for long or

elastic belts. You should use the lowest tension that willensure that the belt never becomes completely slack. If thishappens, the belt teeth can start to climb up the pulley teethand this will accelerate wear.

The simplest way of tensioning a belt, assuming that theaxis of one shaft can be moved freely, is to use a springbalance to apply a force of twice the required belt tension tothat shaft.


Fig 9.11 Tensioning a drive belt

Another method is to measure the force needed to producea known deflection in the centre of the belt span. Nogeneral guidelines can be given here because the deflectiondepends on an elongation factor determined by belt materialand construction, and this factor can vary by 10:1 or more.However the belt manufacturer will usually provide aformula for this deflection. When installing a belt drivesystem, always mount the pulley as close to the motor aspossible to avoid an excessive bending moment on the

shaft.

Vibration

Most of the problems caused by excessive vibration will befairly obvious, such as screws working loose. One that isntquite so obvious is the damage that vibration can cause to astationary bearing. If a motor is subject to prolongedvibration, either in storage or even mounted on a machinebut rarely used, the balls will gradually make very smalldents in the race at the points of contact. This will causethe bearing to become noisy and lead to increased wear.

IP ratings

IP ratings are listed in the following table. The first IPnumber relates to protection against solid objects, and thesecond number protection against liquids.

Most stepper and brushless servo motors are rated at IP54,though to be strictly accurate this only applies from theflange backwards. In other words, the body of the motor isreasonably well sealed but there is no positive seal on the

front bearing. In the majority of applications, the mountingsurface gives a degree of protection to the bearing and forpractical purposes an overall IP54 rating applies. However,if you mount the motor vertically with the shaft upwards,excessive condensation can form a pool in the bearinghousing and enter the motor. In these situations you willneed the additonal protection of a shaft seal.

To improve the rating to IP65, a shaft seal is added and allmetal-to-metal joints are fitted with gaskets. All shaft sealsexert a frictional force on the shaft in some way so they willeventually wear out. However, most IP65 motors still havemild steel shafts (except for stepper motors which alwayshave stainless steel shafts), and these will corrode in moist

conditions. This accelerates the wear rate considerably.Therefore any steps you can take to keep moisture,coolant, fine dust, corrosive fumes etc. away from the motorare usually well worth the effort.

Table of ratings for 1st and 2nd IP numbers

1st: Protection against:0 No protection1 Objects over 50mm e.g. accidental touch by hand2 Objects over 12mm, e.g. fingers3 Objects over 2.5mm, e.g. tools & wires4 Objects over 1mm, e.g. small wires & tools5 Dust, limited ingress permitted (no harmful deposit)6 Total protection against dust

2nd: Protection against:1 Vertically-falling water drops e.g. condensation2 Direct sprays of water up to 15 from the vertical3 Direct sprays of water up to 60 from the vertical4 Water sprayed from all directions, limited ingress

permitted5 Low-pressure water jets from all directions, limited

ingress permitted6 Strong jets of water, limited ingress (e.g. on

shipdecks)7 Immersion between 15cm & 1m8 Long periods of immersion under pressure

Thermal considerationsContinuous torque (or stall torque) is the constant torquethat the motor can deliver without overheating. This torquemay be quoted with the motor mounted in three ways - onan infinite heatsink, mounted on a standard heatsink,usually with a thermal resistance of around 0.5C per watt,and not mounted on any form of heat sink. By far the mostuseful value is that quoted on a standard heatsink, as thisis a good approximation to a typical mounting bracket. Avery thick mounting bracket may allow you to increase thecontinuous torque by up to 20%; mounting on a thin platemay mean derating by 20%.

Torques quoted on the other bases are of limited practical

value. Good unobstructed vertical air flow, or better stillforced cooling, will help to increase the continuous torquerating.

2Fp


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52

Automation

Stopping in an emergencyFor safety reasons, its usually necessary to incorporatesome form of emergency stop system into machinery fitted

with stepper or servo motors. There may be severalreasons for needing to stop quickly, the more obvious onesbeing:

s To prevent injury to the operator if he makes amistake or operates the machine incorrectly

s To prevent damage to the machine or to the productas a result of a jam

s To guard against the consequences of machine faults

You should think about all the possible reasons for stoppingto make sure that they are adequately covered.

Standards which support the European Machinery Directivestate that Emergency Stops must be hard-wired and must

not depend on software or electronic logic.

Methods of stopping the motor

There are several ways to bring a motor to a rapid stop.The choice depends partly on whether its more important tostop in the shortest possible time or to guarantee a stopunder all circumstances. For instance, to stop as quickly aspossible normally means using the decelerating power ofthe servo system. However, if the servo has failed orcontrol has been lost for any reason, this is not an optionopen to you. In this case removing the power will guaranteethat the motor stops, but if the load has a high inertia it maytake an unacceptably long time to come to rest. If the loadis moving vertically and can back-drive the motor, thisintroduces additional complications. In extreme caseswhere personal safety is at risk, it may be necessary tomechanically lock the system even at the expense ofpossible damage to the machine.

The European standards describe two categories ofEmergency Stop:

Category 0 involves stopping by immediate removal ofpower using only hard wired electromechanicalcomponents. This is intended to be used where removal ofpower alone will ensure a fast and safe stop.

Category 1 refers to a controlled stop with powermaintained to the motor to achieve a rapid stop followed byremoval of power once the stop is achieved.

In Category 0 applications the situation is quitestraightforward - the Emergency Stop circuit simplyremoves all AC power from the drives. On drives thatincorporate a power dump circuit, a degree of dynamicbraking may still be provided after you remove the power.However the power supply capacitors may take some timeto decay and this can extend the stopping distance. Apossible alternative is to disconnect the motor as well asremoving the AC power, but this is not normallyrecommended. Not only is the time to stop dependentpurely on load inertia and friction, but you can actuallydamage certain types of drive by disconnecting the motorunder power.

Emergency stopping by removal of AC power should onlybe considered where this alone will guarantee a rapid stop.

For Category 1 applications we need to consider the bestmethod of achieving a controlled stop prior to removal of

power.

Using a full-torque controlled stopGrounding the input to a servo amplifier operating invelocity mode will cause it to decelerate hard in currentlimit, in other words using the maximum available torque.This will create the fastest possible deceleration to rest, butcan only be used with a velocity servo - the sametechnique cannot be used with a torque amplifier, sincegrounding the input will simply produce zero torque. In thelatter case you will normally have to rely on a facility withinthe controller to achieve a rapid stop. If you are using adigital servo with step and direction inputs, cutting off thestep pulses will also produce a rapid deceleration to rest -but please see the warning at the end of this section.

The situation is different for an open-loop stepper drive.You need to ramp the step pulse frequency down to zero inorder to utilise the available torque. Simply cutting off thestep pulses at speeds above the start-stop rate will de-synchronise the motor and the decelerating torque will nolonger be available.

Many stepper and servo controllers are able to generate arapid deceleration rate which is independent of the normalprogrammed rate, to be used only for overtravel limit andemergency stop functions. This deceleration should be setto the highest rate which the system can safely handle.

Using dynamic braking

Conventional servo motors will act as generators whendriven mechanically. By applying a resistive load to themotor, a braking effect is produced that is speed-dependent. Deceleration therefore tends to be rapid at highspeeds, but falls off as the motor slows down. By carefulchoice of load resistor in relation to maximum operatingspeed, the motor can be made to produce substantialbraking torque over a wide part of its speed range. Achangeover contactor can be arranged to switch the motorconnections from the drive to the resistive load, and it canbe made failsafe by ensuring that braking occurs if thepower supply fails.

You can find more information on dynamic braking,including how to calculate the optimum load resistance, in

an earlier section of this handbook.

Using a mechanical brake

Its very often possible to fit a mechanical brake eitherdirectly on the motor or on some other part of themechanism. However, such brakes are usually intended toprevent movement at power-down and are seldomadequate to bring the system to a rapid halt, particularly ifthe drive is delivering full current at the time. Brakes canintroduce friction even when released, and also add inertiato the system both effects will increase the drive powerrequirements. Nevertheless if a mechanical brake isrequired for other reasons, for example to prevent a verticalaxis from falling or to lock the system while stationary, then

it makes sense to utilise it as part of the emergency stoproutine.

Safety considerations


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53

Automation

Emergency stop control

Conventional electromechanical relays are generally theweak point in safety circuits. Their failure mode is

indeterminate but is frequently with contacts weldedtogether. Relays designed for use in emergency stopcircuits have comprehensive protection against failure bymeans of redundancy (a duplicate relay which will operate ifthe primary relay fails), and cross-monitoring (additionalcontacts which prevent resetting if one relay has failed).They also incorporate positive guidance, which ensuresthat normally-open contacts cannot close before normally-closed contacts have opened.

To satisfy Category 1 requirements, a two-stage operation isneeded in which a controlled stop is followed by removal ofAC power. The first controlled-stop stage can rely ultimatelyon software and the use of electronic logic. The secondstage must then guarantee the removal of AC power by

means of hard wired electromechanical components.Dynamic braking may also be introduced at this point toassist a rapid stop in the event of controller failure.

Emergency stop relays designed for Category 1 incorporatedelay on de-energisation contacts which are inherently fail-safe (see Fig. 10.1). Conventional delay-off timers cannotbe used since they do not have the required fail-saferedundancy.

Safety earth wiringAll equipment housed in a metal case and which carries ACline voltage must have a protective earth connection.

Under fault conditions, this connection must be capable ofhandling any resulting fault current until such time as theovercurrent protection disconnects the supply. In practicethis means that the cable used for the protective earthconnection must be at least equivalent in current-carryingcapacity to the mains supply cables.

The protective earth connection can be made either bymeans of discrete copper conductors or by structural partswhich are electrically connected together. If the earthconnection is formed by structural components, make surethat they have a cross-sectional area at least equivalent tothe copper conductor required to do the same job.

It is not necessary for the protective earth conductors inside

a cabinet to be insulated; if they are, the insulation shouldbe green/yellow. Exempted from this are internal protectiveconductors in assemblies such as ribbon cables andflexible printed wiring.

The terminal for the external protective earth connection tothe machine should be identified with the letters PE (seeFig 10.2). Dont use the PE identification for any otherterminals in the system - protective earth connections fromother components such as motors should be identified withthe symbol as shown in the diagram or the colour green/yellow. Dont use the PE terminal for any other purpose,such as a 0v connection.

Fig 10.1 Two stage emergencystop circuit

WARNING you should ensure that the deceleration rateused under emergency stop conditions can be handledsafely by the system mechanics. If the peak torqueavailable from your servo is well in excess of that normallyused, there may be a risk of mechanical damage whendecelerating in full current limit. This is particularly true ifyou are using a high-ratio gearbox.

Further information

A booklet giving a great deal of useful information on manyaspects of machine safety has been published by Pilz UK.It includes sections on European standards, riskassessment, components and wiring diagrams for manytypes of safety circuit. We gratefully acknowledge the

permission of Pilz UK to make use of this information in thecompilation of this handbook.

Fig 10.2 Safety earth connections and symbols

Take particular care if there are any connectors or plug-socket combinations which can interrupt the protectivebonding. The protective conductor circuit must only beinterrupted after the live conductor circuit is broken, andmust be re-established before the live circuit is restored.

Further information on safety earth connections can be

found in the Machinery Safety Standard BS EN60204-1Part 1.

Safety considerations

instant

'stop'

signal

E-stop relay

delayed

final

isolation

Servo

drive/

controller

E-stop

button

motor

Internalprotectiveconductor

Protective earthfor motor case

PEterminal

External protectiveearth connection

Internalequipment


20/20

Limit switches in safety-criticalapplications

In many low-power applications, the provision of overtravellimit switches is either a matter of operational convenienceor a means of avoiding mechanical damage. In either casethere is usually a choice of suitable switch types, dependingon factors such as the available space and ease ofincorporating an appropriate actuator. But in situationswhere safety is paramount, such as when failure to stop ona limit may cause personal injury, it is essential to use thecorrect type of switch and to wire it in an intrinsically-safemanner.

We can divide machine limit switches into two broadcategories - contacting and non-contacting. Contactingswitches are normally conventional microswitches; non-

contacting types include proximity detectors (capacitive,inductive or optical) as well as reed switches. All types canbe used in non-critical applications, but mechanically-operated roller drive microswitches are inherently muchsafer than non-contacting types.

Safety-critical applications require that limit switches arepositive drive devices. This means that the switch contactsare directly coupled to the actuator via a non-resilientcomponent, ensuring that the contacts are forced open evenin the event of welding - there is no reliance on an internalspring to open the contacts. Dont use proximity switcheswhich are more easily defeated.

The switches must be arranged in such a way that theycant be overriden; use double switches if necessary. They

should also be wired into the Emergency Stop circuit (in thiscase the Estop must be resettable even when the limit is stillpresent to allow the system to be driven off).

Limit switch input circuits are arranged so that a closedcircuit is required for normal oper

Torque and Inertia Calculations

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