Appendix B Appxb

User Manual: Appendix B SureServo™ AC Servo Systems User Manual - AutomationDirect

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B

B

B

APPENDIX

APPENDIX

APPENDIX

SELECTING THE

SureServo™

SERVO SYSTEM

In This Appendix ...

Selecting the SureServo™ Servo System . . . . . . . . . . . .B–2

The Selection Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .B–2

How many pulses from the PLC to make the move? . . . . . . . . . . . . . . . .B–2

What is the positioning resolution of the load? . . . . . . . . . . . . . . . . . . . .B–3

What is the indexing speed to accomplish the move time? . . . . . . . . . . .B–3

Calculating the Required Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .B–4

Leadscrew - Example Calculations . . . . . . . . . . . . . . . . .B–8

Step 1 - Define the Actuator and Motion Requirements . . . . . . . . . . . . . .B–8

Step 2 - Determine the Positioning Resolution of the Load . . . . . . . . . . . .B–8

Step 3 - Determine the Motion Profile . . . . . . . . . . . . . . . . . . . . . . . . . . .B–9

Step 4 - Determine the Required Motor Torque . . . . . . . . . . . . . . . . . . . .B–9

Step 5 - Select and Confirm the Servo Motor and Driver System . . . . . .B–10

Belt Drive - Example Calculations . . . . . . . . . . . . . . . .B–11

Step 1 - Define the Actuator and Motion Requirements . . . . . . . . . . . . .B–11

Step 2 - Determine the Positioning Resolution of the Load . . . . . . . . . . .B–11

Step 3 - Determine the Motion Profile . . . . . . . . . . . . . . . . . . . . . . . . . .B–12

Step 4 - Determine the Required Motor Torque . . . . . . . . . . . . . . . . . . .B–12

Step 5 - Select and Confirm the Servo Motor and Driver System . . . . . .B–13

Index Table - Example Calculations . . . . . . . . . . . . . . .B–14

Step 1 - Define the Actuator and Motion Requirements . . . . . . . . . . . . .B–14

Step 2 - Determine the Positioning Resolution of the Load . . . . . . . . . . .B–14

Step 3 - Determine the Motion Profile . . . . . . . . . . . . . . . . . . . . . . . . . .B–15

Step 4 - Determine the Required Motor Torque . . . . . . . . . . . . . . . . . . .B–15

Step 5 - Select and Confirm the Servo Motor and Driver System . . . . . .B–16

Engineering Unit Conversion Tables, Formulas, &

Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .B–17

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–2 2nd Ed, Rev B 08/2011

Selecting the SureServo™ Servo System

The selection of your SureServo™ servo system follows a defined process. Let's go

through the process and define some useful relationships and equations. We will

use this information to work some typical examples along the way.

The Selection Procedure

The motor provides for the

required motion of the load

through the actuator (mechanics

that are between the motor shaft

and the load or workpiece). Key

information to accomplish the

required motion is:

• total number of pulses from

the PLC

• positioning resolution of the

load

• indexing speed (or PLC pulse

frequency) to achieve the

move time

• required motor torque

(including the 25% safety

factor)

• load to motor inertia ratio

In the final analysis, we need to achieve the required motion with acceptable

positioning accuracy.

How many pulses from the PLC to make the move?

The total number of pulses to make the entire move is expressed with the equation:

Equation 햲: Ptotal = total pulses = (Dtotal ÷ (dload ÷ i)) x θcount

Dtotal = total move distance

dload = lead or distance the load moves per revolution of the actuator's drive shaft

(P= pitch = 1/dload)

θcount = servo resolution (counts/revmotor) (default = 10,000)

i= gear reduction ratio (revmotor/revgearshaft)

Example 1: The motor is directly attached to a disk and we need to move the disk

5.5 revolutions. How many pulses does the PLC need to send to the driver?

Ptotal = (5.5 revdisk ÷ (1 revdisk/revdriveshaft ÷ 1 revmotor/revdriveshaft))

5.5 ÷ (1.0 ÷ 10) x 10,000 =550,000

x 10,000 counts/revmotor

= 55,000 pulses

Acceleration Deceleration

Indexing

Speed

Move Time

SureServo™ Servo Systems User Manual B–3

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Example 2: The motor is directly attached to a ballscrew where one turn of the

ballscrew results in 20 mm of linear motion and we need to move 45 mm. How

many pulses do we need to send the driver?

Ptotal = (45 mm ÷ (20 mm/revscrew ÷ 1 revmotor/revscrew)) x 10,000 counts/revmotor

=22,500 pulses

Example 3: Let's add a 2:1 belt reduction between the motor and ballscrew in

example 2. Now how many pulses do we need to make the 45 mm move?

Ptotal = (45 mm ÷ (20mm/revscrew ÷ 2 revmotor/revscrew)) x 10,000 counts/revmotor

= 45,000 pulses

What is the positioning resolution of the load?

We want to know how far the load will move for one command pulse. The equation

to determine the positioning resolution is:

Equation 햳: L= load positioning resolution = (dload ÷ i) ÷ count

Example 4: What is the positioning resolution for the system in example 3?

L= (dload ÷ i) ÷ count

= (20 mm/revscrew ÷ 2 revmotor/revscrew) ÷ 10,000 counts/revmotor

= 0.001mm/count

0.00004"/count

What is the indexing speed to accomplish the move time?

The most basic type of motion profile is a

"start-stop" profile where there is no

acceleration or deceleration period. This

type of motion profile is only used for

low speed applications because the load

is "jerked" from one speed to another and

the servo system may experience a

position deviation error if excessive

speed changes are attempted. The

equation to find indexing speed for "start-

stop" motion is:

Equation 햴: fSS = indexing speed for start-stop profiles = Ptotal ÷ ttotal

ttotal = move time

Indexing

Speed

Move Time

Start - Stop Profile

45 mm 1 revscrew 1 revmotor 10,000 pulses

move 20 mm 1 revscrew 1 revmotor

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–4 2nd Ed, Rev B 08/2011

Example 5: What is the indexing speed to make a "start-stop" move with 10,000

pulses in 800 ms?

fSS = indexing speed = Ptotal ÷ ttotal = 10,000 pulses ÷ 0.8 seconds

= 12,500 Hz.

For higher speed operation, the "trapezoidal" motion profile includes controlled

acceleration & deceleration and, in

some cases, an initial non-zero

starting speed. With the acceleration

and deceleration periods equally set,

the indexing speed can be found

using the equation:

Equation 햵: fTRAP = (Ptotal - (fstart

x tramp)) ÷ (ttotal -

tramp)

for trapezoidal

motion profiles

fstart = starting speed

tramp = acceleration or deceleration time

Example 6: What is the required indexing speed to make a "trapezoidal" move in

1.8s, accel/decel time of 200 ms each, 100,000 total pulses, and a starting speed of

40 Hz?

fTRAP = (100,000 pulses - (40 pulses/sec x 0.2 sec)) ÷

(1.8 sec - 0.2 sec)

62,375 Hz.

Calculating the Required Torque

The required torque is the sum of acceleration (or

deceleration) torque and the running torque. The

equation for required motor torque is:

Equation 햶: Tmotor = Taccel (or decel) + Trun

Taccel = motor torque required to accelerate the total

system inertia (including motor inertia).

Tdecel = motor torque required to decelerate; not

always the same as acceleration.

Trun = constant motor torque requirement to run the

mechanism due to friction, external load

forces, etc.

Continuous Duty Zone means the system can provide the torque under the curve

100% of the time.

Intermittent Duty Zone means the system can provide the torque under the curve

LESS THAN 100% of the time.

The amount of time the system can operate in this region depends on the amount

of torque. In general, the higher the torque, the shorter period of time is allowed.

See overload curves information in Chapter 1. If a system requires more than

rated torque occasionally, but only for a short time, the system can do it. Running

in this zone continuously will result in an overload fault.

In Table 1 we show how to calculate torque required to accelerate or decelerate an

inertia from one speed to another and the calculation of running torque for common

mechanical actuators.

Acceleration Deceleration

Indexing

Speed

Move Time

Start

Speed

Trapezoidal Profile

Torque

(N•m)

400W Low Inertia

SVL-204

Speed

(rpm)

Intermittent

Duty Zone

Continuous

Duty Zone

SureServo™ Servo Systems User Manual B–5

Table 1 - Calculate the Torque for "Acceleration" and "Running"

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

The torque required to accelerate or decelerate a constant inertia with a linear

change in velocity is:

Equation 햷: Taccel = Jtotal x (speed ÷ time) x (2÷ 60)

Jtotal is the motor inertia plus load

inertia ("reflected" to the motor

shaft). The (2÷ 60) is a factor used

to convert "change in speed"

expressed in rpm into angular

speed (radians/second). Refer to

information in this table to

calculate "reflected" load inertia for

several common shapes and

mechanical mechanisms.

Example 7: What is the required

torque to accelerate an inertia of

0.002 lb-in-sec2(motor inertia is 0.0004 lb-in-sec2and "reflected" load inertia is

0.0016 lb-in-sec2) from zero to 600 rpm in 50 ms?

Taccel = 0.002 lb-in-sec2x (600 rpm ÷ 0.05 seconds) x (2÷ 60)

2.5 lb-in

Accel

Period

Velocity Indexing Velocity

Torque

Decel

Period

time

time

T1

T2

T3

Leadscrew Equations

J

W

motor

Jgear

Jcoupling

Fext

Jscrew

Fgravity



JW

Description: Equations:

Motor rpm nmotor = (vload x P) x i, nmotor (rpm), vload (in/min)

Torque required to accelerate

and decelerate the load Taccel Jtotal x (speed ÷ time) x 0.1

Motor total inertia Jtotal = Jmotor + Jgear + ((Jcoupling + Jscrew + JW) ÷ i2)

Inertia of the load JW= (W ÷ (g x e)) x (1 ÷ 2  P)2

Pitch and Efficiency P = pitch = revs/inch of travel, e = efficiency

Running torque Trun = ((Ftotal ÷ (2 P)) + Tpreload) ÷ i

Torque due to preload

on the ballscrew Tpreload = ballscrew nut preload to minimize backlash

Force total Ftotal = Fext + Ffriction + Fgravity

Force of gravity and

Force of friction Fgravity = Wsinθ, Ffriction = µWcosθ

Incline angle and

Coefficient of friction θ= incline angle, µ = coefficient of friction

i = gear ratio

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–6 2nd Ed, Rev B 08/2011

Table 1 (cont’d)

Belt Drive (or Rack & Pinion) Equations

J

W

motor

Jgear



Fext

Fgravity

JW

Jpinion



Jmotor Jgear

JW

W

Fext

Fgravity

1

W2

Jpinion

Description: Equations:

Motor rpm nmotor = (vload x 2  r) x i

Torque required to accelerate

and decelerate the load Taccel Jtotal x (speed ÷ time) x 0.1

Inertia of the load Jtotal = Jmotor + Jgear + ((Jpinion + JW) ÷ i2)

Inertia of the load JW= (W ÷ (g x e)) x r2; JW= ((W1+ W2) ÷ (g x e)) x r2

Radius of pulleys r = radius of pinion or pulleys (inch)

Running torque Trun = (Ftotal x r) ÷ i

Force total Ftotal = Fext + Ffriction + Fgravity

Force of gravity and

Force of friction Fgravity = Wsinθ; Ffriction = µWcosθ

Belt (or Gear) Reducer Equations

Jmotor Jmotor

Jmotorpulley Jmotorpulley

Jloadpulley

Jloadpulley

JLoad JLoad

Description: Equations:

Motor rpm nmotor = nload x i

Torque required to accelerate

and decelerate the load Taccel Jtotal x (speed÷time) x 0.1

Inertia of the load Jtotal = Jmotor + Jmotorpulley + ((Jloadpulley + JLoad) ÷ i2)

Motor torque Tmotor x i = TLoad

Typical Leadscrew Data

Material: e=

efficiency Material: µ =

coef. of friction

ball nut 0.90 steel on steel 0.580

acme with plastic nut 0.65 steel on steel (lubricated) 0.150

acme with metal nut 0.40 teflon on steel 0.040

ball bushing 0.003

i = gear ratio

SureServo™ Servo Systems User Manual B–7

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Table 1 (cont’d)

Inertia of Hollow Cylinder Equations

LDo = 2ro

Di = 2ri

Description: Equations:

Inertia (known weight) J = (W x (ro2+ ri2)) ÷ (2g)

Inertia (known density) J = (x L x x (ro4– ri4)) ÷ (2g)

Volume volume = 4 x (Do2- Di2) x L

Inertia of Solid Cylinder Equations

LD = 2r

Description: Equations:

Inertia (known weight) J = (W x r2) ÷ (2g)

Inertia (known density) J = (x L x x r4) ÷ (2g)

Volume volume = x r2x L

Inertia of Rectangular Block Equations

hw

l

Description: Equations:

Inertia (known weight) J = (W ÷ 12g) x (h2+ w2)

Volume volume = l x h x w

Symbol Definitions

J = inertia lb-in-s2 (Kg-m-s2)= density

L = Length, inches (m) = 0.098 lb/in3(aluminum)

h = height, inches (m) = 0.28 lb/in3(steel)

w = width, inches (m) = 0.04 lb/in3(plastic)

W = weight, lbs. (Kg) = 0.31 lb/in3(brass)

D = diameter, inches (m) = 0.322 lb/in3 (copper)

r = radius, inches (m)

g = gravity = 386 in/sec2(9.8 m/s2)3.14

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–8 2nd Ed, Rev B 08/2011

Leadscrew - Example Calculations

Step 1 - Define the Actuator and Motion Requirements

Weight of table and workpiece = 150 lb

Angle of inclination = 0°

Friction coefficient of sliding surfaces = 0.05

External load force = 0

Ball screw shaft diameter = 0.8 inch

Ball screw length = 96 inch

Ball screw material = steel

Ball screw lead = 8.0 inch/rev (P 0.125 rev/in)

Desired Resolution = 0.0005 inches/count

Gear reducer = 2:1

Stroke = 78 inches

Move time = 12 seconds

Step 2 - Determine the Positioning Resolution of the Load

The resolution of the load can be determined using Equation 햳. If the servo motor

is connected directly to the ballscrew, then the best resolution possible would be:

Lθ= (dload ÷ i) ÷ θcount

= (8 ÷ 1) ÷ 10,000

= 0.0008

This does not meet the system requirements; however, if we add a 2:1 transmission

with no lost motion (backlash, etc.) to the output of the motor, the resolution gets

better by a factor of 2, so the minimum requirements would be met.

Lθ= (8 ÷ 2) ÷ 10,000

= 0.0004

J

W

motor

Jgear

Jcoupling

Fext

Jscrew

Fgravity



JW

Definitions

dload = lead or distance the load moves per revolution of the actuator’s drive shaft (P = pitch = 1/dload)

Dtotal = total move distance

θcount = servo resolution (counts/revmotor)

i = gear reduction ratio (revmotor/revgearshaft)

Taccel = motor torque required to accelerate and decelerate the total system inertia (including motor inertia)

Trun = constant motor torque requirement to run the mechanism due to friction, external load forces, etc.

ttotal = move time

SureServo™ Servo Systems User Manual B–9

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Step 3 - Determine the Motion Profile

From Equation 햲, the total pulses to make the required move is:

Ptotal = (Dtotal ÷ (dload ÷ i)) x count

= (78 ÷ (8 ÷ 2)) x 10,000 = 195,000 pulses

From Equation 햵, the indexing frequency for a trapezoidal move is:

fTRAP = (Ptotal - (fstart x tramp)) ÷ (ttotal - tramp)

= (195,000 - (100 x 0.6)) ÷ (12 - 0.6) 17.1 KHz

where accel time is 5% of total move time and starting speed is 100 Hz.

=17.1 KHz x (60 sec/1 min) ÷ 10,000 counts/rev

103 rpm

Step 4 - Determine the Required Motor Torque

Using the equations in Table 1:

Jtotal = Jmotor + Jgear + ((Jcoupling + Jscrew + JW) ÷ i2)

For this example, let's assume the gearbox and coupling inertia are zero.

JW= (W÷ (g x e)) x (1 ÷ 2P)2

= (150 ÷ (386 x 0.9)) x (1 ÷ 2 x 3.14 x 0.125)2

0.700 lb-in-sec2

Jscrew (x L x x r4) ÷ (2g)

(3.14 x 96 x 0.28 x 0.0256) ÷ (2 x 386)

0.0028 lb-in-sec2

The inertia of the load and screw reflected to the motor is:

J(screw + load) to motor = ((Jscrew + JW) ÷ i2)

((0.0028 + 0.700) ÷ 22) = 0.176 lb-in-sec2

The torque required to accelerate the inertia is:

Taccel Jtotal x (speed ÷ time) x 0.1

= 0.176 x (103 ÷ 0.6) x 0.1 1.08 lb-in

Next, we need to determine running torque. If the machine already exists then it is

sometimes possible to actually measure running torque by turning the actuator

driveshaft with a torque wrench.

Trun = ((Ftotal ÷ (2 P)) + Tpreload) ÷ i

Ftotal = Fext + Ffriction + Fgravity

= 0 + µWcos+ 0 = 0.05 x 150 = 7.5 lb

Trun = (7.5 ÷ (2 x 3.14 x 0.125)) ÷ 2

4.77 lb-in

where we have assumed preload torque to be zero.

From Equation 햶, the required motor torque is:

Tmotor = Taccel + Trun = 1.08 + 4.77 5.85 lb-in 0.66 N•m

However, this is the required motor torque before we have picked a motor and

included the motor inertia.

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–10 2nd Ed, Rev B 08/2011

Step 5 - Select and Confirm the Servo Motor and Driver System

It looks like a reasonable choice for a motor would be the SVL-207. This motor has

an inertia of:

Jmotor = 0.00096 lb-in-sec2

The actual motor torque would be modified:

Taccel = Jtotal x (speed ÷ time) x 0.1

= (0.176 + 0.00096) x (103 ÷ 0.6) x 0.1

1.09 lb-in

so that:

Tmotor = Taccel + Trun

= 1.09 + 4.77 5.86 lb-in  0.66 N•m

It looks like the 750W system will work. However, we still need to check the load

to motor inertia ratio:

Ratio = J(screw + load) to motor ÷ Jmotor

= 0.176 ÷ 0.00096 = 183.3

It is best to keep the load to motor inertia ratio below 10, so 183 is well outside this

guideline. Although the servo has enough power to control the system, the large

mismatch ratio may prevent proper tuning and faster acceleration settings in the

future. Since the motor speed required to move the system is well within the motor

specs, we can change the gear ratio to use a 750W motor or select a much larger

motor such as the SVM-220. Because the reflected inertia is decreased by the square

of the ratio, we will change the gear ratio to 10:1. By doing this, the mismatch ratio

is now 7.3 (before we consider any added inertia due to the reducer).

Reflected J = Jscrew+Jload = .176, so

22

New Reflected J = Jscrew+Jload = .00704

102

New J Ratios = .00704 = 7.33

.00096

Torque

(N•m)

Intermittent

Duty Zone

Continuous

Duty Zone

750W Low Inertia

SVL-207

Speed

(rpm)

SureServo™ Servo Systems User Manual B–11

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Belt Drive - Example Calculations

Step 1 - Define the Actuator and Motion Requirements

Weight of table and workpiece = 90 lb

External force = 0 lb

Friction coefficient of sliding surfaces = 0.05

Angle of table = 0º

Belt and pulley efficiency = 0.8

Pulley diameter = 2.0 inch

Pulley thickness = 0.75 inch

Pulley material = aluminum

Desired Resolution = 0.0005 inch/step

Gear Reducer = 10:1

Stroke = 50 inch

Move time = 4.0 seconds

Accel and decel time = 1.0 seconds

Step 2 - Determine the Positioning Resolution of the Load

The resolution of the load can be determined using Equation 햳. If the servo

motor is connected directly to the pulley, then the best resolution possible would

be:

L= (dload ÷ i) ÷ count

= ((x 2.0) ÷ 1) ÷10,000

= =0.00063

where dload = x Pulley Diameter.

This does not meet the system requirements. However, if we add a 10:1

transmission to the output of the motor, the resolution improves by a factor of 10,

meeting the minimum system requirements.

Lu = ((p x 2.0) ÷ 10) ÷10,000

= 0.000063

Definitions

dload = lead or distance the load moves per revolution of the actuator’s drive shaft (P = pitch = 1/dload)

Dtotal = total move distance

count = servo resolution (counts/revmotor)

i = gear reduction ratio (revmotor/revgearshaft)

Taccel = motor torque required to accelerate and decelerate the total system inertia (including motor inertia)

Trun = constant motor torque requirement to run the mechanism due to friction, external load forces, etc.

ttotal = move time

J

W

motor

Jgear



Fext

Fgravity

JW

Jpinion

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–12 2nd Ed, Rev B 08/2011

Step 3 - Determine the Motion Profile

From Equation 햲, the total pulses to make the required move is:

Ptotal = (Dtotal ÷ (dload ÷ i)) x count

= 50 ÷ ((3.14 x 2.0) ÷ 10 x 10,000

795,775 pulses

From Equation 햵, the running frequency for a trapezoidal move is:

fTRAP = (Ptotal - (fstart x tramp)) ÷ (ttotal - tramp)

= 795,775 ÷ (4 - 1)

265,258 Hz or 265.3 KHz

where accel time is 25% of total move time and starting speed is zero.

= 265.3 KHz x (60 sec/1 min) ÷ 10,000 counts/rev

1,592 rpm motor speed

Step 4 - Determine the Required Motor Torque

Using the equations in Table 1:

Jtotal = Jmotor + Jgear + ((Jpulleys + JW) ÷ i2)

For this example, let's assume the gearbox inertia is zero.

JW= (W÷ (g x e)) x r2

= (90 ÷ (386 x 0.8)) x 1

0.291 lb-in-sec2

Pulley inertia (remember, there are two pulleys) can be calculated as:

Jpulleys ((x L x x r4) ÷ (2g)) x 2

((3.14 x 0.75 x 0.098 x 1) ÷ (2 x 386)) x 2

0.0006 lb-in-sec2

The inertia of the load and pulleys reflected to the motor is:

J(pulleys + load) to motor = ((Jpulleys + JW) ÷ i2)

((0.291 + 0.0006) ÷ 100) 0.0029 lb-in-sec2

The torque required to accelerate the inertia is:

Tacc Jtotal x (speed ÷ time) x 0.1

= 0.0029 x (1592 ÷ 1) x 0.1

= 0.46 lb-in

Trun = (Ftotal x r) ÷ i

Ftotal = Fext + Ffriction + Fgravity

= 0 + µWcos+ 0 = 0.05 x 100 = 5.0 lb

Trun = (5.0 x 1) ÷ 10

0.50 lb-in

From Equation 햶, the required motor torque is:

Tmotor = Taccel + Trun = 0.46 + 0.50 0.96 lb-in  0.11 N•m

However, this is the required motor torque before we have picked a motor and

included the motor inertia.

SureServo™ Servo Systems User Manual B–13

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Step 5 - Select and Confirm the Servo Motor and Driver System

It looks like a reasonable choice for a motor would be the SVL-2040. This motor has

an inertia of:

Jmotor = 0.0003 lb-in-sec2

The actual motor torque would be modified:

Taccel = Jtotal x (speed ÷ time) x 0.1

= (0.0029 + 0.0003) x (1592 ÷ 1) x 0.1 0.51 lb-in

so that:

Tmotor = Taccel + Trun

= 0.51 + 0.5 1.01 lb-in 0.12 N•m

It looks like the 400W system will work. However, we still need to check the load

to motor inertia ratio:

Ratio = J(pulleys + load) to motor ÷ Jmotor

= 0.0029 ÷ 0.0003 = 9.6

It is best to keep the load to motor inertia ratio at or below 10, so 9.6 is within an

acceptable range.

Torque

(N•m)

400W Low Inertia

SVL-204

Speed

(rpm)

Intermittent

Duty Zone

Continuous

Duty Zone

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–14 2nd Ed, Rev B 08/2011

Index Table - Example Calculations

Step 1 - Define the Actuator and Motion Requirements

Diameter of index table = 12 inch

Thickness of index table = 3.25 inch

Table material = steel

Number of workpieces = 8

Desired Resolution = 0.006º

Gear Reducer = 6:1

Index angle = 45º

Index time = 0.5 seconds

Step 2 - Determine the Positioning Resolution of the Load

The resolution of the load can be determined using Equation 햵If the servo

motor is connected directly to the table, then the best resolution possible would

be:

L= (dload ÷ i) ÷ count

= (360º ÷ 1) ÷ 10,000

= 0.036°

This does not meet the system requirements. However, if we add a 6:1 transmission

to the output of the motor, the resolution gets better by a factor of 6, meeting the

minimum system requirements.

= (360º ÷ 6) ÷ 10,000

= 0.006°

Jmotor

Jgear

Definitions

dload = lead or distance the load moves per revolution of the actuator’s drive shaft (P = pitch = 1/dload)

Dtotal = total move distance

count = servo resolution (counts/revmotor)

i = gear reduction ratio (revmotor/revgearshaft)

Taccel = motor torque required to accelerate and decelerate the total system inertia (including motor inertia)

Trun = constant motor torque requirement to run the mechanism due to friction, external load forces, etc.

ttotal = move time

SureServo™ Servo Systems User Manual B–15

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Step 3 - Determine the Motion Profile

From Equation 햲, the total pulses to make the required move is:

Ptotal = (Dtotal ÷ (dload ÷ i)) x count

= (45º ÷ (360º ÷ 6) x 10,000

= 7,500 pulses

From Equation 햵, the running frequency for a trapezoidal move is:

fTRAP = (Ptotal - (fstart x tramp)) ÷ (ttotal - tramp)

= 7,500 ÷ (0.5 - 0.13) 20.27 kHz

where accel time is 25% of total move time and starting speed is zero.

= 20.27 kHz x (60 sec/1 min) ÷ 10,000 counts/rev

121 rpm

Step 4 - Determine the Required Motor Torque

Using the equations in Table 1:

Jtotal = Jmotor + Jgear + (Jtable ÷ i2)

For this example, let's assume the gearbox inertia is zero.

Jtable (x L x x r4) ÷ (2g)

(3.14 x 3.25 x 0.28 x 1296) ÷ (2 x 386)

4.80 lb-in-sec2

The inertia of the indexing table reflected to the motor is:

Jtable to motor = Jtable ÷ i2

0.133 lb-in-sec2

The torque required to accelerate the inertia is:

Taccel Jtotal x (speed ÷ time) x 0.1

= 0.133 x (121 ÷ 0.13) x 0.1

12.38 lb-in

From Equation 햶, the required motor torque is:

Tmotor = Taccel + Trun

=12.38 + 0 = 12.38 lb-in 1.40 N•m

However, this is the required motor torque before we have picked a motor and

included the motor inertia.

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–16 2nd Ed, Rev B 08/2011

Step 5 - Select and Confirm the Servo Motor and Driver System

It looks like a reasonable choice for a motor would be the SVM-220. This motor has

an inertia of:

Jmotor = 0.014 lb-in-sec2

The actual motor torque would be modified:

Taccel = Jtotal x (speed ÷ time) x 0.1

= (0.133 + 0.014) x (121 ÷ 0.13) x 0.1

13.68 lb-in

so that:

Tmotor = Taccel + Trun

= 13.68 + 0

= 13.68 lb-in  1.55 N•m

It looks like the 2 kW medium inertia system will work. However, we still need to

check the load to motor inertia ratio:

Ratio = Jtable to motor ÷ Jmotor

= 0.133 ÷ 0.014 = 9.5

It is best to keep the load to motor inertia ratio at or below 10, so 9.5 is within an

acceptable range.

Speed

(rpm)

2kW Medium Inertia

SVM-220

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Torque

(N•m)

25

20

15

10

5

0

0 1000 2000 3000 4000 5000

IntermittentIntermittent

Duty ZoneDuty Zone

Continuous

Duty Zone

SureServo™ Servo Systems User Manual B–17

Appendix B: Selecting the SureServo™ Servo System

2nd Ed, Rev B 08/2011

Engineering Unit Conversion Tables, Formulas, &

Definitions

Conversion of Length

To convert A to B,

multiply A by the

entry in the table.

B

µm mm mmil in ft

A

µm 11.000E–03 1.000E–06 3.937E–02 3.937E–05 3.281E–06

mm 1.000E+03 11.000E–03 3.937E+01 3.937E–02 3.281E–03

m1.000E+06 1.000E+03 13.937E+04 3.937E+01 3.281E+00

mil 2.540E+01 2.540E–02 2.540E–05 11.000E–03 8.330E–05

in 2.540E+04 2.540E+01 2.540E–02 1.000E+03 18.330E–02

ft 3.048E+05 3.048E+02 3.048E–01 1.200E+04 1.200E+01 1

Conversion of Torque

To convert A to B,

multiply A by the

entry in the table.

B

Nm kpm(kg-m) kg-cm oz-in lb-in lb-ft

A

Nm 11.020E–01 1.020E+01 1.416E+02 8.850E+00 7.380E-01

kpm(kg-m) 9.810E+00 11.000E+02 1.390E+03 8.680E+01 7.230E+00

kg-cm 9.810E–02 1.000E–02 11.390E+01 8.680E–01 7.230E–02

oz-in 7.060E–03 7.200E–04 7.200E–02 16.250E–02 5.200E–03

lb-in 1.130E–01 1.150E–02 1.150E+00 1.600E+01 18.330E–02

lb-ft 1.356E+00 1.380E–01 1.383E+01 1.920E+02 1.200E+01 1

Conversion of Moment of Inertia

To convert A to B,

multiply A by the

entry in the table.

B

kg-m2kg-cm-s2oz-in-s2lb-in-s2oz-in2lb-in2 lb-ft2

A

kg-m211.020E+01 1.416E+02 8.850E+00 5.470E+04 3.420E+03 2.373E+01

kg-cm-s29.800E–02 11.388E+01 8.680E–01 5.360E+03 3.350+02 2.320E+00

oz-in-s27.060E–03 7.190E–02 16.250E–02 3.861E+02 2.413E+01 1.676E–01

lb-in-s21.130E–01 1.152E+00 1.600E+01 16.180E+03 3.861E+02 2.681E+00

oz-in21.830E–05 1.870E–04 2.590E–03 1.620E–04 16.250E–02 4.340E–04

lb-in22.930E–04 2.985E–03 4.140E–02 2.590E–03 1.600E+01 16.940E–03

lb-ft24.210E–02 4.290E–01 5.968E+00 3.730E–01 2.304E+03 1.440E+02 1

Appendix B: Selecting the SureServo™ Servo System

SureServo™ Servo Systems User Manual

B–18 2nd Ed, Rev B 08/2011

Engineering Unit Conversion Tables, Formulas, & Definitions (continued)

General Formulae & Definitions

Description: Equations:

Gravity gravity = 9.8 m/s2= 386 in/s2

Torque T = J , = rad/s2

Power (Watts) P(W) = T(N·m) · (rad/s)

Power (Horsepower) P(hp) = T(lb·in) · ν(rpm) / 63,024

Horsepower 1 hp = 746 W

Revolutions 1 rev = 1,296,000 arc·sec = 21,600 arc·min = 360 degrees

Equations for Straight-Line Velocity & Constant Acceleration

Description: Equations:

Final velocity vf= vi+ at

final velocity = initial velocity + (acceleration · time)

Final position xf= xi+ ½(vi+vf)t

final position = initial position + [1/2 · (initial velocity + final velocity) · time]

Final position xf= xi+ vit + ½at2

final position = initial position + (initial velocity · time) + (1/2 · acceleration · time squared)

Final velocity

squared vf2= vi2+ 2a(xf– xi)

final velocity squared = initial velocity squared + [2 · acceleration · (final position – initial

position)]