Mechatronics - Higher - Robotic and Automated Systems

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Mechatronics

Robotic and Automated
Systems

Teacher/Lecturer Notes

(Higher)


6777









Spring

2000





Mechatronics

Robotic and Automated
Systems

Teacher/Lecturer Notes

and Solutions to Assignments

Higher








Support Materials



HIGHER STILL





CONTENTS





Introductio
n


Outline of Unit outcomes, Assignments,
Practical work and Equipment suppliers






Solutions to Assignments






Using Alternate Robot Arm in Outcome 4






























Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


1





MECHATRONICS


HIGHER


ROBOTIC AND AUTOMATE
D SYSTEMS


TEACHER/LECT
URER NOTES AND

SOLUTIONS TO ASSIGNM
ENTS





INTRODUCTION






Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


2




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


3

UNIT OUTCOMES



Outcome 1


Robot Anatomy

This outcome introduces the various motion capabilities of industrial robots, the
different methods of holding objects, the drive systems available to po
wer them and
the safety aspects of their use and application.


When students have completed this outcome they should be able to:



describe the axes of motion and degrees of freedom of typical robots



describe the operation of various end effector grippers



co
mpare pneumatic, hydraulic and electric drive systems for robot application.


Outcome 2


Robot sensory systems

This outcome examines some sensor systems and discovers how they function and
how they are applied to robotic systems.


When students have comp
leted this outcome they should be able to:



describe incremental and absolute encoders



compare Binary, BCD and Gray codes as used with encoders



calculate positional accuracy using encoders



explain tactile sensing.


Outcome 3


Control strategies in Automate
d systems

The purpose of this outcome is to examine basic control strategies, how they operate,
how they are applied and to understand important elements of their operation.


When students have completed this outcome they should be able to:



describe sequen
tial control strategy



compare open loop and closed loop control



describe the output response of proportional control



describe the constituent elements of PID control.


Outcome 4


Programming a robot

The purpose of this outcome is to know about programming

methods and how to
program a typical robot


When students have completed this outcome they should be able to:



clearly describe robotic programming methods



correctly analyse a sequential programming task



correctly translate the task analysis into functiona
l software



verify the correct operation of the software on the given hardware.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


4

ASSIGNMENTS


Outcome 1

The assignments comprise two categories,


Assignments 1


5

Involve the student in giving simple answers to the questions.


Assignments 6


10

Assignment
s 6
-
8 require the student to evaluate the data given and make a selection
supported by reasons for the selection
, 9
-
10 involve power calculations.


Outcome 2

Assignments 1


5

Require the student to (a) identify absolute and incremental encoder character
istics,
(b) evaluate Binary, Gray and BCD number systems to identify a pattern, perform a
graphical exercise on encoder discs and pulse patterns and do some simple
calculations on resolution.


Outcome 3

Assignments 1


7

Require the student to (a) explain
sequence control and identify two different forms
from examples given, (b) describe the basic attributes of open loop and closed loop
control in reference to a specified application, (c) complete graphical and written
exercises to illustrate the operation
of a proportional controller with different levels of
gain, (d) complete written and graphical exercise to illustrate the effect of the
individual elements of a PID controller.


Outcome 4

A set assignment is not provided but a practical class exercise with

a specified robotic
arm is described. The teacher/lecturer will take the students through the practicalities
of using the robot and the software.


Warning! Most educational robot arms are driven by model servomotors and
have limited force capability. Th
e axes can be damaged if students force them to
move physically by hand. When robots are in use in a class, ensure a responsible
person is supervising.


Schools and colleges may use their own robots but the basic approach to designing a
program as describ
ed in the notes should be followed, at least in outline.





Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


5

The student should be guided through a logical approach. The initial specification
needs to be fully understood. The problem should be analysed into a series of tasks.
Next key positions of the ro
bot arm should be identified with particular attention to
avoiding any collisions in moving from one position to another. The final planning
stage will depend on the robot and programming arrangements available but a more
detailed plan of the movements is
useful before the final stage of moving the arm to
each required position and then inserting the details in the program.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


6

PRACTICAL


Outcome 4

The only specified practical element appears in outcome four. The student has to
operate and program a robot arm.


There are various suppliers of robot equipment specifically designed for educational
and training purposes. The prices range from cheap to costly but all will meet the
needs of the unit. In many instances schools and colleges will have equipment that
wi
ll be adequate. Lego kits with robot capabilities spring to mind.


Suppliers of Educational Robot Kits


Milford instruments

Milford House

120 High Street

South Milford

Leeds

LS25 5AQ

Tel: 01977 683665

Fax: 01977 681465

e
-
mail:
info@milinst.demon.co.uk

Web address:
www.milinst.demon.co.uk


Supplies a six
-
servo self
-
assembly robotic arm with associated Windows or DOS
software. Limited to light loads.


Robotica L
imited

17
-
23 Park Terrace Lane

Glasgow

G3 6BQ

Tel: 0141 353 2261

Fax: 0141 353 2614

e
-
mail:
info@robotica.co.uk

Web address:
www.robotica.co.uk

Manufactures robots and

motion control products for education. The robot
manipulator has six axes controlled by Windows software. Limited to light loads.


Bytronic International Ltd

The Courtyard

Reddicap Trading Estate

Sutton Coldfield

West Midlands

B75 7BU

Tel: 0121 378 0613

F
ax: 0121 311 1774

e
-
mail:
sales@bytronic.co.uk

Web address:
www.bytronic.co.uk



A choice of two robots, one open
-
loop driven by stepper motors, the second closed
-
loop d
riven by servomotors. These are costly but substantial machines with lifting
capability of half a kilogram plus.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


7

PRACTICAL SUGGESTION
S.

There is no other requirement in the unit for practical assessment but the learning
experience can be enhanced where pr
actical exercises, demonstrations or simulations
are employed.


Suggestions


Control in Outcome 3

Bytronic supply software simulation for PID control. One package employs the
concept of a liquid level control system with associated graphical displays to
i
ntroduce students to PID control. Gain, integral and derivative values can be entered
to illustrate the effect on the process under control. This package is purely simulation
with no associated hardware elements.


A second training package uses a small D.C

motor and associated software. The
package allows the student to input various values to see the effect on the control of
an electric motor when employing PID. The software produces a graph to show the
reaction of the motor to input changes.


Note: both
packages are DOS based only at the moment but the liquid level
control package will be available in a Windows format in the year 2000.


Suggestions


Encoders in Outcome 2

Encoders are ready available from many suppliers such as RS components and
Farnell.
They can be connected up to lead screws driven by electric motors (or simply
turned by hand) to demonstrate their use in position control. Simple counters are
available from the same suppliers to memorise counts so that students can associate
counts with d
istance or angle moved.


In relation to lead screw use, a marker, which moves along the lead screw as it is
turned, gives an indication of distance moved. This can be related to the number of
pulses generated by an incremental encoder and stored in a count
er memory.







Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


8

SOLUTIONS TO ASSIGNM
ENTS (KEY POINTS AR
E HIGHLIGHTED)


Outcome 1


(1)

Name the following robot configurations from the description given


(i)

Most resembles the human arm in operation.

(ii)

Has only translational axes of movement.

(iii)

Has two horizontal rot
ational axes and one linear axis.

(iv)

Is good at reaching confined spaces.

(v)

Rotates about its base axis and can move vertically up and down and
horizontally in and out.


Solutions

(i) Revolute (ii) Cartesian (iii) Scara (iv) Polar (v) Cylindrical



(2
)

Explain the term
‘degrees of freedom’
.


Solution


Degrees of freedom refers to the number of axes or pivot points on a
robot. These degrees of freedom can supply rotational or linear motion.



(3)

Explain the term ‘
work envelope’

with reference to t
he manufacturers

and users’ definition.


Solution


‘Work Envelope’ is the volume of space around a robot that can be
accessed by the mounting plate for the end effector (Manufacturers). For a user
the ‘Work envelope’ will be extended with the fitting of

the end effector to the
mounting plate.




(4)

Explain how the lifting force of a vacuum gripper is determined.


Solution


the magnitude of the lifting force is determined from the pressure
difference between the inside and outside of the suction cup and

by the cross
-
sectional area of the cup.



(5)

A mechanical gripper employs ‘
friction force’ to
hold an object. Explain how

this is applied. Explain the forces involved in a worst case situation for lifting
an object.


Solution


The motive power (electrical,

hydraulic or pneumatic) is applied to the
two fingers to move them against the object. Sufficient ‘friction force’ must be
applied to prevent the object slipping against the ‘force of gravity’ and any
‘acceleration forces’ (inertial) as the object is move
d.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


9

(6)

Three actuators are to be used on a Cartesian
-
style robot for pick and place
operations related to picking up items from one conveyor and placing them on
another. Each actuator will work to mechanical limits
-

up and down, left to right, and
side t
o side. Full range of movement on each axis is 0 to 0.4 m The lifting and
moving forces required are fairly low but speed requirements are fairly fast. The
environment is non
-
hazardous and power supplies of hydraulic, pneumatic and
electric are readily av
ailable. Careful costing should be observed as there is a limited
budget. Using the table of data given, select suitable actuators and give your reasons
for the final selection



PNEUMATIC

HYDRAULIC

ELECTRICAL

Criteria

Linear

Rotary

Linear

Rotary

Linear

Rotary

Range of movement

10mm to
1m

0 to360
0


1mm to
2m

0 to360
0


1mm to
1m

0 to360
0


Positional Accuracy

Pick &
place

Pick &
place



0.1%



0.1%



0.05%



0.05%

Safety

See Note 1

See Note 2

See Note 3

Force/Torque

Medium

Medium

High

High

Medium

Mediu
m

Cost

Low

Low

High

High

Medium

Low

Speed

Medium

Medium

High

High

Low

High


Solution
-

A Cartesian robot only requires linear or translational actuation.

Any system will meet this requirement.



The limit of movement on each axis is 0.4 m to end stops

Any system will meet the movement limits. No close position control
required (driven to end stops) so again any system will meet the
requirements.


Force requirements are low.

Any system will meet the low force requirements.


Speed of operation is fairly
fast.

The only doubt here might be the electrical linear. It may be possible
to use a lead screw driven by an electric rotational drive but it would
probably be too slow.



Cost is a major concern as the money available to spend on the project is
limited.

The cost limitation must rule out hydraulic. Power supplies are
available for pneumatic.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


10



No environmental constraints
.


There are no safety or clean room requirements so there is no
restrictions on any chosen system.



RECOMMENDATION :
All the system
s could meet most of the
specification but there is a limitation of speed on the electric linear drive.
The major factor in eliminating the hydraulic system is cost but it is too
powerful anyway. The pneumatic system is the ideal choice for this
applicatio
n.



(7)

An actuator is required to operate one axis of a robot. The actuator is required
to move very heavy loads to precise positions within a range of 0 to 1metre.
All power sources are available and the environment is non
-
hazardous. If cost
is no re
striction, choose a suitable actuator and give your reasons for the
selection. Use the table of data previously supplied.


Solution

=
i楮ia爠潰r牡瑩潮⁩猠aga楮i牥煵楲q搮
=
Any system specified in the table will meet the linear requirement.


Linear movement r
equired is up to 1metre.

Any system specified in the table will meet the required distance.


There is no speed specified.

Any system in the table will be satisfactory in terms of speed.


Force requirement is very high.

It is unlikely that electrical or pn
eumatic would meet the force
requirement. Hydraulic will meet the requirement.


There are no cost or environmental restrictions

Any system would be suitable under these two terms.


RECOMMENDATION :
It is clear from the analysis that only a
hydraulic syst
em would meet the force requirement. Electric and
pneumatic Linear devices do not have the power.












Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


11

(8)

Two actuators are required for the positioning of electronic components on a
printed circuit board, one to drive the X axis and the other the Y

axis of a chip
placement device. Precise and repeatable position control is vital with a fairly
fast response to place a specified number of components in a given time.
Moderate force or torque will be required from the drives. Pneumatic and
electrical s
upplies are available but no hydraulic. Cost and minimum
maintenance downtime have to be considered. Choose suitable actuators and
give your reasons for the selection.


Solution
-


The key feature required in this application is high precision
positioning
.

This immediately rules out a pneumatic system. Precise positioning
is difficult in a pneumatic system because of the compressibility of
air.


A moderate speed of operation is needed with moderate force to be
applied.

Electrical rotational and hydraulic (
both types) could meet the
force/torque and speed requirements. This does not resolve a
choice although it is likely that if electric is chosen it will be
rotational high
-
speed servomotors driving lead screws or ball and
screw.


Cost is an important factor

in the choice.

This probably rules out hydraulic as there is no hydraulic supply
pump installed.


Minimum interruption to production is an important part of the


requirement.

This should militate against the hydraulic system, as it requires
regular mainte
nance.


RECOMMENDATION


Electric servomotors appear to be the best
choice for accurate positioning. Electric stepper motors are not as
good as servomotors for precise positioning. Pneumatics are ruled out
because of the poor positioning performance. Hydr
aulics would have
been an option but ruled out in this case because of cost and
maintenance limitations.



(9)

Calculate the power in watts developed by an linear actuator which needs to
generate a force of 150 N over a forward stroke distance of 350mm in
a time
of 450msecs.


Solution
-

Power in watts = (Force x Distance) / Time


= (150 x 0.35) / 0.45 = 117 watts






Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


12


(10)

Calculate the output power in watts developed by a hydraulic motor with an
overall efficiency of 0.85. The oil flow rate through

the motor is determined
to be 0.55 x 10
-
3
m
3

/ sec and the pressure drop across the motor is measured
at 50 bar. (1 bar = 10
5
NM / m
2

).


Solution
-

Power

= Q
m

x

P
m

x

o



= 0.55 x 10
-
3
x 50 x 10
5
x 0.85


= 2337.5 watts = 2.34 KW




SOLUTIONS TO

ASSIGNMENTS


Outcome 2


(1)

Each statement below relates to either an absolute or an incremental optical
encoder. Write down the line references pertaining to each encoder type in the
form of Absolute = ( ), ()…. And Incremental = ( ), ( )…..


(a)

Positional dat
a lost if power fails.

(b)

Produces parallel output data.

(c)

Angular Resolution is calculated from 360
0
/ number of bits.

(d)

Requires an external counter to track speed and position.

(e)

Angular resolution is determined from 360
0
/ ppr

(f)

Can employ anti
-
ambiguity techni
ques.

(g)

Direction of rotation can be determined from pulse phase relationship.

(h)

Low noise immunity.



(i)

Provides absolute position.

(j)

Position is relative to an initial count.

(k)

Serial data output.

(l)

Has high noise immunity.

(m)

If power fails positional data is retained

for restart.

(n)

Can use Pure Binary, Gray or BCD code.



Solution
-

Absolute

Incremental


(b)

(a)


(c)

(e)


(f)

(d)


(i)

(h)


(l)

(g)


(m)

(k)


(n)

(j)






Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


13


(2)

Examine the completed Pure Binary and Gray codes in the table below.
Identify a possible patt
ern of ones and zeros in each case to make it easier to
remember each code.


DECIMAL

Gray Code

Pure Binary Code

BCD (2 decades)

00

0000

0000

0000 0000

01

0001

0001

0000 0001

02

0011

0010

0000 0010

03

0010

0011

0000 0011

04

0110

0100

0000 0100

05

011
1

0101

0000 0101

06

0101

0110

0000 0110

07

0100

0111

0000 0111

08

1100

1000

0000 1000

09

1101

1001

0000 1001

10

1111

1010

0001 0000

11

1110

1011

0001 0001

12

1010

1100

0001 0010

13

1011

1101

0001 0011

14

1001

1110

0001 0100

15

1000

1111

0001 0101


The object of this exercise is to encourage the student to examine the binary patterns.
Instead of relying on pure rote memorising they should look for patterns or clues to
the structure of each binary code. We concentrate on a four bit binary number.


Solution


the binary pattern can be constructed from the simple rule of the column

Values. Add the values only if a one appears in the column.


8

4

2

1

0

0

0

0 (zero)

1

1

0

1 (thirteen)


Thus the number zero is 0000 in four bit binary number forma
t.

The number thirteen is 1101 in four bit binary number format.


Pattern recognition.

Binary numbers follow a definite pattern.

LSB pattern is 01 01 01 01 01 01 01 01

LSB + 1 is


00 11 00 11 00 11 00 11

LSB + 2 is


0000 1111 0000 1111

MSB is


00000000 1
1111111




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


14


Binary Coded Decimal


BCD follows the same pattern as standard decimal but only up to the number 9.

Thus the maximum binary pattern in any decade of a BCD number is 1001


Gray Code

More difficult to construct a pattern. The basic rule must be fol
lowed in that from one
number to the next only one bit must change.


One guide is to change the least significant bit possible on each successive number to
abide by the rule, and without repeating a pattern of bits.



0 0 0 0

0 0 0 1

0 0 1
1

0 0 1 0

0 1 1 0

0 1 1 1

0 1 0 1

0 1 0 0

1 1 0 0 This is the LSB change at this point without repeating a previous pattern.

1 1 0 1

1 1 1 1

1 1 1 0

1 0 1 0

1 0 1 1

1 0 0 1

1 0 0 0



A r
ecognisable pattern can be discerned for the four bit number. The students will
come up with different ways of grouping or describing the patterns.


LSB pattern is 0 11 00 11 00 11 00 11 0

LSB + 1 is 00 1111 0000 1111 00

LSB + 2 i
s 0000 11111111 0000

MSB 00000000 11111111














Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


15


(3)

(i)

Figure 11 shows two incomplete discs for absolute encoders. A blank area will
transmit a light beam to a photo transistor and a shaded area will block
the light. Shad
e in the correct sections in each disc to produce a pure
binary code on one and a gray code on the other.



(ii)

Calculate the angular resolution of the completed discs.



(iii)

If the smallest angle to be resolved is to be half of that achieved with
the

above discs, how many concentric tracks would be needed on the
disc.



Solutions
:(
i )



1`
0`
2
3
4
5
6
7
HUB
HUB
1`
0`
2
3
4
5
6
7
BINARY
GRAY




Figure 11



( ii ) Resolution of each disc

= 360/ 2
n

where n = number of bits



= 360 / 2
3

= 360 / 8 = 45
0



(iii) Angle of
resolution = 22.5
0


2
n
= 360 / angle = 360 / 22.5 = 16


n = 4 because 2
4
= 16


Therefore disc requires four concentric tracks.





Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


16


(4 )

Figure 12 shows a set of pulses for bit 0 on a four bit encoder. Draw the pulses
with solid lines for bit 1, 2

and 3 to produce a single decade binary coded
decimal unit.


Solution:


Figure 12















0
1
2
3
4
5
6
7
8
9
BIT 0
BIT 1
BIT 2
BIT 3
SINGLE DECADE
-
BCD



Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


17


(5)

Figure 14 shows an incremental linear encoder. The scanning head,
comprising the phototransistor and light emitter, can travel the full distance of
the

linear tape, which comprises 5500 holes over a distance of 1.2 metres.


Assume that the scanning head moves in synchronism with a guillotine blade.
The blade can be moved to different positions over plastic sheets to cut them
to specified lengths. Calcul
ate,


(a)

The size of counter to store the maximum count for 1.2m distance.



(b)

The resolution in terms of the smallest distance that can be resolved.


(c)

The distance moved from a reference or zero position if the counter
reads 1600.


Figure 14


Solution


⡡)

Counter size, test 2
12
= 4096, too small; try 2
13
= 8192


We need to use a counter with thirteen bits to hold a count of 5500




(b)

Resolution = 1200 / 5500 mm = 0.218 mm



(c)

Distance moved from reference when hole count is 1600


= 1600 x 0.218 = 348.8 mm




Linear Tape with holes

Light Sensor

Light emitter

Linear Incremental Optical Encoder

Digital Counter and Conditioning Electronics




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


18

SOLUTIONS TO ASSIGNM
ENTS

Outcome 3


(1)

Explain what is meant by ‘sequence control’.


Solution


‘Sequence control’ means that a series of tasks are carried out one
after the other and in a specified order.



(2)

(i)

A robot
arm picks up a metal part from a conveyor. When grasped the
arm is raised to clear obstacles. When raised the arm moves the part
over a container. When over the container the arm lowers into the
container. When in the container the arm releases the part. W
hen
released the arm is raised to clear the box. When clear the arm moves
back to the conveyor.


Describe two ways in which these tasks could be executed. Select the
best method for the task, giving a reason for your choice and why the
alternative method m
ay not be as suitable.


(ii)


Suggest a suitable application for the alternative control method.



Solution
-

One method of carrying out the above tasks is by the sequential
method called ‘event based’ sequential control. Each task in the sequence can
only

be initiated when a sensor identifies that the previous task has been
completed.


A second technique is the ‘time based’ method. When a task is initiated a time
out period starts. When the time out period expires, the next task and the next
time out perio
d are initiated


The ‘event based’ method is the best method for this sequence. The system will
know that it has picked up an item before proceeding to the next step. With
‘time based’ a certain time is allocated to the task of picking up the item. If the
re
is no item present the system will continue to the next step anyway.


(ii)

A time based sequence would be best in sequences such as mixing
ingredients. Each stage in the mixing process requires a set time. The cost
of sensors is eliminated in such an ar
rangement.









Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


19

(3)



A D.C. electric motor is chosen to drive a section of a production line.
Whether to control the motor speed by open or closed loop has not yet been
decided.



(a)

Describe how each methods works and comment on the suitability of
each
in terms of,



(b)

How easily they could be applied



(c)

The cost of each method



(d)

The accuracy of each technique and finally



(e)

The stability of each during step changes.



Solution
-

The D.C. motor speed is controlled by varying the applie
d voltage.


Open Loop Control


Would use a controller that provides a variable output
voltage for various input signals. The controller also has to provide sufficient
power to run the motor. Open loop does not monitor itself to see if the selected
speed i
s correct (no sensor). The various running speeds would be marked on
the input signal control during commissioning.


Closed Loop Control


Uses a voltage controller as before but additionally
employs a sensor to monitor motor speed. The required speed an
d the actual
speed are compared. If there is any difference the controller output voltage is
modified to correct the speed error.


Open loop is a much cheaper method than closed loop because it doesn’t employ
sensors or comparators. Consequently open loop
is easier to apply and set up.
Closed loop requires adjustments and fine tuning to get the best response.


Open loop is not as accurate as closed loop because it has no self
-
monitoring. It
will not detect disturbances which can easily change the output wit
hout being
noticed


When a set point or desired value change is made to open loop it will, in most
applications, respond smoothly and in a stable manner. Closed loop has to be set
up very carefully but even with optimum adjustment, there will be some
overs
hoot before it settles down to a steady value.








Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


20



(4)

A proportional controller is applied to the temperature control of a process
liquid in a tank. Draw three graphs of actual value against time to show the
following control effects after a step change,


(i)

An optimum acceptable response

(ii)

A slow sluggish response

(iii)

An excessively fast response.




Solution
-



Time

40

Temp

0

C

30

0

Original set point

New set point (step
change)

Curve (ii)

Curve (i)

Curve (iii)

offset



Curve (i)

An optimum response after a step change. For a slow process this
would be a 4:1 decay ratio.

Curve (ii)

A sluggis
h response.

Curve (iii)

An excessively fast response.







Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


21



(5)

From the previous exercise;

(i)

Comment on the effect of increasing the gain.

(ii)

Comment on the steady state error.

(iii)

Comment on the stability.


Solution

=
(i)

The larger the gain, the faster the res
ponse.


(ii)

A proportional only controller will always have offset or steady
state error. As the gain increases, the steady state error reduces
.


(iii
) Increases the gain improves the response time and reduces the
steady state error but does not eliminate

it. Increasing gain
eventually introduces instability or overshoot. A four to one decay
ratio is acceptable in slow processes. Faster response systems such as
motor drives must be set up with less overshoot
.



(6)

Many modern process controllers are classi
fied as PID controllers. Describe
each element of a PID controller in the following way;

(i)

Give an accepted standard definition for each function.

(ii)

Explain each of these definitions in terms of the practical effect.


Solution

( i )

Proportional control


Ou
tput is proportional to the magnitude of the error.

Integral Action


The output is proportional to the integral of the error over time
.

Derivative Action


The output is proportional to the rate of change of the error.


(ii)

Proportional
action will respo
nd immediately to any error or increased error that
develops.

Integral
action removes any steady state error that develops
.

Derivative action
improves the stability and reduces the overshoot
.














Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


22



(7)

Illustrate the individual effects of the elements o
f PID control after a step
change as follows;


(i)

A response curve for proportional control only.

(ii)

A second response curve showing the effect of adding integral action
to the proportional action.

(iii)

A third response curve showing the effect of adding derivative a
ction
to the proportional action.

(iv)

Comment on the effects of each on the control action.



Solution:

(i
-
iii)

Time
0
100
Curve ( i )
Offset
Curve (iii)
Curve (ii)
Original set point
New set point

(iv)


Curve (i)
is a pure proportional controller response with offset or steady state
error.


Curve (ii)
is a proportional + integral controller
which eliminates the steady state
error. Set point and actual value are the same after the transient step change has
passed.


Curve (iii
) is a proportional + derivative controller. Compare the transient step
change effect of proportional only with proport
ional + derivative. The derivative
element has reduced the magnitude of the overshoots and allowed the system to
stabilise sooner.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


23

Outcome 4

Alternative Robotic Arm applied to the Programming Exercise.

The following notes refer to the same practical exerc
ise in Outcome 4 of the student
notes. A Robotica robot arm was used as an example in the student notes. The
lecturer/teacher notes here suggest an alternative robot arm for the exercise.
Remember that any suitable manipulator with programming capability c
an be used for
the exercise.


The Task Analyses are much the same but the exercise is reproduced here for
convenience.


Exercise

Figure 28 shows two blocks, ‘A’ and ‘B’, and three positions, 1,2 and 3. A robot arm
is positioned behind and in line with the

three positions. A robot arm is to be used to
move block ‘B’ to position 2 and block ‘A’ to position 3. Design a robot program to
execute the task. Assume that due to situation restrictions the arm can only move in
straight line back and forward or up and

down, but not sideways.







Position 1

Position 2

Position 3

Robot arm

Block A

Block B

Plan view

Direction
of Travel


Figure 28





Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


24

Task Analysis

After a preliminary look at the task, ask the initiator if the task has to be carried out in
a particular manner. If the answer is no then the task will be successfully

completed if
each block is placed in the correct position. If yes, then the desired sequence must be
given by the user.


Assume the answer is ‘no’ so we can use any sequence.


We can now draw a preliminary flowchart (figure 29) for the steps or tasks invo
lved.


Flowchart


START
TRANSFER
BLOCK ‘A’
TO POSITION 1
TRANSFER
BLOCK ‘A’
TO POSITION 3
TRANSFER
BLOCK ‘B’
TO POSITION 2
ARM TO PARK
POSITION
END


Figure 29





Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


25

Key Positions

Now we can identify key positions included in the complete task.


1.

Initial Home position (jaws open and high to clear blocks)

2.

Grab block A

3.

Lift block A

4.

Transfer to position 1

5.

Clear block A

6.

Above position

3

7.

Grab block B

8.

Lift block B

9.

Transfer to position 2

10.

Clear block B

11.

Above position 1

12.

Grab block A

13.

Lift block A to clear block B

14.

Transfer to position 3.

15.

Park position.


These are key positions for the arm during its movements. We start from a parked
posit
ion with jaws open and high to clear any blocks when the first move is made. It is
assumed we can go straight to ‘grab block A’. At position 5 the jaws have to clear
block A before moving above block B. In the final stage at position 13 we lift block
‘A’ c
lear of block ‘B’, before we travel over it to deposit block ‘A’ at position 3.


Home position.

The waist should be positioned to allow the arm to go straight down the line of
movement shown in Figure 28. The jaws should be fully open and horizontal. The
s
houlder, elbow and wrist should set the arm back from and high enough to clear both
blocks, but ready to make the first move to pick up block A. The height of the blocks
must be decided early on in the preparations for the exercise.






Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


26

The Lynxarm Robot fr
om Milford Instruments.

The following instructions relate to using the DOS version of the programming
software. A windows version is now available which uses an onscreen method of
moving and saving the servo positions.


The DOS version can apply a simulate
d Teach Pendant method. Selected computer
keys are used to move the axes as follows,


A serial cable needs to be connected between the robot and a computer serial port.


Axis

Keys



Base (left or right)

S

=
汥lt
=


=
物r桴
=
=
p桯畬摥h
=潲oa牤r⁢ac欩
=


=
景f睡牤
=


=
扡ck
=
=
䕬扯眠b異u⁤潷温
=


=

=


=
摯睮
=
=
t物獴
異r⁤潷温
=


=

=


=
摯睮
=
=
䝲楰灥i
潰e渠潲⁣汯le搩
=


=
潰on
=


=
c汯獥
=
=
o畮⁴桥⁓䕔⁕倠灲潧ra洠晩牳琠m漠o獴慢汩s栠瑨攠獥物r氠捯浭畮lca瑩潮⁰潲琠慮搠瑨t=
牡nge⁡湤⁨潭e⁰潳=瑩潮o⁥ac栠ax楳i
=
=
o畮⁴桥ui奎堳⁰uogra活⁥湴敲⁴桥⁲潢潴畭扥爠r湤⁳灥e搠⡳灥e搠da汵攠㌠楳⁢e獴s
=
=
m牥獳⁆㜠瑯⁴a步⁴桥=a牭⁴漠瑨攠桯浥⁰潳=瑩潮o
=
=

Programming the Robot.


The specification for the task states that all operations are carried out in a straight line.
Once t
he waist axis is set up for straight
-
line operation in the home position it will not
need to be altered.



This leaves the shoulder, elbow, wrist and jaws servos to be moved as required.


A more detailed sequence of movements can now be planned. Each move
ment will
be initiated using the listed keys.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


27

Programming Steps


1.

Move to grab position over block A.

2.

Close jaws around block A.

3.

Lift block to clear ground.

4.

Move block A to position 1.

5.

Open jaws.

6.

Lift jaws up clear of block A.

7.

Move jaws over block B.

8.

Clos
e jaws around block B.

9.

Lift block clear of ground.

10.

Move block b to position 2.

11.

Open jaws.

12.

Clear block B.

13.

Move jaws over block A.

14.

Grab block A.

15.

Lift bottom of block A clear of block B.

16.

Move block A to position 3.

17.

Open jaws.

18.

Home


Recording the Steps

Use the

appropriate keys to carry out each movement. After each movement is
complete press

F4
to record the position in memory.


Saving the Program

When all the movements have been completed, press
F2
and save the program

under
a file name.


Loading the Program

Load the program by pressing
F1
and typing in the program name.


Single Stepping the Program

For safety the program should be tested for the first time using single step mode.

Press
F5
and the
+
key for each step through the program.


Full Test.

When sat
isfied that the program is satisfactory, run a complete sequence test by
pressing
F6.




Mechatronics:
Robotic and Automated Systems, Teacher/Lecturer Notes and Solutions


28