Implementation

page i

Implementation

Products, Robotics, and Other Useful Things

Hugh Jack

Copyright, 2006

1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.1 Introduction 1.11.2 Bloom’s Taxonomy 1.11.3 Examples 1.21.4 Summary 1.21.5 References and Bibliography 1.21.6 Problems 1.21.7 Challenge Problems 1.2

2. DRAFTING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.32.1 CONVENTIONAL DRAFTING 2.3

2.1.1 Manual Drafting 2.42.1.2 Turning Three Dimensions Into Two (Multi View Drawings)2.5

The Glass Box 2.52.1.3 Lines 2.82.1.4 Holes 2.102.1.5 Special Cases 2.11

Aligned Features 2.11Incomplete Views 2.14

2.1.6 Section Views 2.16Full Sections 2.16Offset Section 2.17Half Section 2.18Cut Away Sections 2.19Revolved Section 2.20Removed Section 2.20Auxiliary Section 2.22Thin Wall Section 2.23Assembly Section 2.23Special Cases 2.24Fill Patterns 2.26

2.1.7 Auxiliary Views 2.26Secondary Auxiliary Views 2.30Partial Auxiliary Views 2.30

2.1.8 Descriptive Geometry 2.302.1.9 Isometric Views 2.312.1.10 Special Techniques 2.31

2.2 NOTATIONS 2.322.2.1 Basic Dimensions and Tolerances 2.332.2.2 Geometric Dimensioning and Tolerancing (GD & T) 2.33

Feature Control Symbols 2.34Symbols and Meaning 2.35Datums 2.40Modifiers 2.41

2.3 WORKING DRAWINGS 2.42

2.3.1 Drawing Elements 2.42Title Blocks 2.42Drawing Checking 2.43Drawing Revisions 2.43Bill of Materials (BOM) 2.44

2.3.2 Drawing Types 2.44Assembly Drawings 2.44Subassembly Drawings 2.45Exploded Assembly Drawings 2.45Detailed Drawings 2.45

2.4 PRACTICE PROBLEMS 2.462.5 REFERENCES 2.46

3. METROLOGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.473.1 Introduction 3.47

3.1.1 The Role of Metrology 3.473.2 DEFINITIONS 3.483.3 STANDARDS 3.49

3.3.1 Scales 3.493.3.2 Calipers 3.503.3.3 Transfer Gauges 3.50

3.4 Instruments 3.513.4.1 Vernier Scales 3.513.4.2 Micrometer Scales 3.52

The Principle of Magnification 3.53The Principle of Alignment 3.54

3.4.3 Dial Indicators 3.553.4.4 The Tool Makers Microscope 3.573.4.5 Metrology Summary 3.58

3.5 Surfaces 3.593.5.1 Measures of Roughness 3.60

3.6 Measuring Surface Roughness 3.633.6.1 Observation Methods 3.633.6.2 Stylus Equipment 3.633.6.3 Specifications on Drawings 3.683.6.4 Other Systems 3.693.6.5 Roundness Testing 3.72

Intrinsic Roundness Testing 3.73Extrinsic Roundness Testing 3.76

3.7 Gage Blocks 3.783.7.1 Manufacturing Gauge Blocks 3.823.7.2 Compensating for Temperature Variations 3.853.7.3 Testing For Known Dimensions With Standards 3.853.7.4 Odd Topics 3.863.7.5 Limit (GO & NO GO) Gauges 3.87

Basic Concepts 3.87GO & NO GO Gauges Using Gauge Blocks 3.89Taylor’s Theory for Limit Gauge Design 3.90Gauge Maker’s Tolerances 3.91

3.7.6 Sine Bars 3.93Sine Bar Limitations 3.95

3.7.7 Comparators 3.96Mechanical Comparators 3.97Mechanical and Optical Comparators 3.98Optical Comparators 3.99Pneumatic Comparators 3.99

3.8 Measuring Aparatus 3.1013.8.1 Reference Planes 3.101

Granite Surface Plates 3.102Cast Iron Surface Plates 3.102

3.8.2 Squares 3.1033.9 Practice Problems 3.106

4. CUTTING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1194.1 Drilling 4.119

4.1.1 Drill Bits 4.1224.1.2 Reamers 4.1254.1.3 Boring 4.1264.1.4 Taps 4.1274.1.5 Process Parameters 4.1284.1.6 The mrr For Drilling 4.130

4.2 Milling 4.1314.2.1 Types of Milling Operations 4.131

Arbor Milling 4.1334.2.2 Milling Cutters 4.1334.2.3 Milling Cutting Mechanism 4.133

Up-Cut Milling 4.134Down-Cut Milling 4.135

4.3 Feeds and Speeds 4.1364.3.1 The mrr for Milling 4.1394.3.2 Process Planning for Prismatic Parts 4.1394.3.3 Indexing 4.139

4.4 Lathes 4.1424.4.1 Machine tools 4.145

Production Machines 4.1464.4.2 Toolbits 4.1474.4.3 Thread Cutting On A Lathe 4.1504.4.4 Cutting Tapers 4.1524.4.5 Turning Tapers on Lathes 4.1534.4.6 Feeds and Speeds 4.155

4.4.7 The mrr for Turning 4.1564.4.8 Process Planning for Turning 4.157

4.5 Cutting Speeds, Feeds, Tools, and Times 4.1584.6 Cutting Power 4.1594.7 Examples 4.1644.8 Summary 4.1644.9 References and Bibliography 4.1644.10 Problems 4.1644.11 Challenge Problems 4.1644.12 Practice Problems 4.165

5. JOINING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1825.1 Introduction 5.1825.2 ADHESIVE BONDING 5.1835.3 ARC WELDING 5.1845.4 GAS WELDING 5.1865.5 SOLDERING AND BRAZING 5.1875.6 PLASTIC WELDING 5.1885.7 Examples 5.1935.8 Summary 5.1935.9 References and Bibliography 5.1935.10 Problems 5.1935.11 Challenge Problems 5.194

6. ROTATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1956.1 Introduction 6.1956.2 Rotational Masses and Inertia 6.1956.3 Motor Models 6.203

6.3.1 Basic Brushed DC Motors 6.2036.4 Tachometers 6.210

6.4.1 Angular Displacement 6.210Potentiometers 6.210

6.4.2 Encoders 6.211Tachometers 6.215

6.5 Examples 6.2166.6 Summary 6.2166.7 References and Bibliography 6.2166.8 Problems 6.2166.9 Challenge Problems 6.216

7. FEEDBACK CONTROL REVIEW . . . . . . . . . . . . . . . . . . . . . 7.2177.1 Introduction 7.2177.2 OpAmps 7.221

7.3 Examples 7.2257.4 Summary 7.2257.5 References and Bibliography 7.2257.6 Problems 7.2257.7 Challenge Problems 7.225

8. MECHANICAL POWER TRANSMISSION . . . . . . . . . . . . . . 8.2268.1 Mechanisms 8.226

8.1.1 Locking/Engaging 8.2278.1.2 Motion Transmission/Transformation 8.2308.1.3 Four Bar Linkages 8.2308.1.4 Reciprocating 8.2328.1.5 Six Bar Linkages 8.234

8.2 Mechanical Advantage 8.2368.3 Gears 8.238

8.3.1 Spur Gears 8.2388.3.2 Involute Profiles 8.2438.3.3 Design of Gears 8.2458.3.4 Design Issues 8.248

Undercutting and Contact Ratios 8.248Changing the Center Distance 8.250

8.3.5 Helical Gears 8.250Design of Helical Gears 8.251Perpendicular Helical Gears 8.254

8.3.6 Bevel Gears 8.255Design of Bevel Gears 8.257

8.3.7 Other Bevelled Gears 8.2588.3.8 Worm Gears 8.2588.3.9 Harmonic Drives 8.2618.3.10 Design With Gears 8.261

Gear Trains 8.262Examples - Fixed Axis Gears 8.263Examples - Moving Axis Gears 8.267Epicyclic Gear Trains 8.267Differentials 8.270

8.3.11 Gear Forces and Torques 8.2728.4 Cams 8.274

8.4.1 Using Cams in Mechanisms 8.2878.5 Examples 8.2878.6 Summary 8.2878.7 References and Bibliography 8.2878.8 Problems 8.2878.9 Challenge Problems 8.290

9. MECHANICAL ISSUES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.291

9.1 Introduction 9.2919.2 Friction 9.2919.3 Friction 9.2939.4 Contact Points And Joints 9.294

9.4.1 Switching 9.2959.4.2 Deadband 9.2969.4.3 Saturation and Clipping 9.2999.4.4 Hysteresis and Slip 9.2999.4.5 Delays and Lags 9.300

9.5 Wheeled Vehicles 9.3019.6 Examples 9.3019.7 Summary 9.3019.8 References and Bibliography 9.3019.9 Problems 9.3019.10 Challenge Problems 9.301

10. SENSORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.30210.1 INTRODUCTION 10.30210.2 SENSOR WIRING 10.302

10.2.1 Switches 10.30310.2.2 Transistor Transistor Logic (TTL) 10.30310.2.3 Sinking/Sourcing 10.30410.2.4 Solid State Relays 10.311

10.3 PRESENCE DETECTION 10.31210.3.1 Contact Switches 10.31210.3.2 Reed Switches 10.31210.3.3 Optical (Photoelectric) Sensors 10.31310.3.4 Capacitive Sensors 10.32010.3.5 Inductive Sensors 10.32410.3.6 Ultrasonic 10.32610.3.7 Hall Effect 10.32610.3.8 Fluid Flow 10.326

10.4 SUMMARY 10.32710.5 PRACTICE PROBLEMS 10.32710.6 PRACTICE PROBLEM SOLUTIONS 10.33110.7 ASSIGNMENT PROBLEMS 10.337

11. ACTUATORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.33911.1 INTRODUCTION 11.33911.2 SOLENOIDS 11.33911.3 VALVES 11.34011.4 CYLINDERS 11.34211.5 HYDRAULICS 11.34411.6 PNEUMATICS 11.346

11.7 MOTORS 11.34711.8 COMPUTERS 11.34811.9 OTHERS 11.34811.10 SUMMARY 11.34811.11 PRACTICE PROBLEMS 11.34911.12 PRACTICE PROBLEM SOLUTIONS 11.34911.13 ASSIGNMENT PROBLEMS 11.350

12. PROJECT MANAGEMENT . . . . . . . . . . . . . . . . . . . . . . . . . 12.35112.1 Introduction 12.35112.2 An Academic View of Design Revisited 12.35112.3 Project Management 12.355

12.3.1 Timeline - Tentative 12.35512.3.2 Teams 12.35512.3.3 Conceptual Design 12.35612.3.4 Progress Reports 12.35612.3.5 Design Proposal 12.35712.3.6 The Final Report 12.35812.3.7 Gantt Charts 12.35912.3.8 Drawings 12.35912.3.9 Budgets and Bills of Material 12.35912.3.10 Calculations 12.360

12.4 Examples 12.36012.5 Summary 12.36012.6 References and Bibliography 12.36012.7 Problems 12.36112.8 Challenge Problems 12.36112.9 Forms 12.361

13. MOTION CONTROL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.36313.1 INTRODUCTION 1.36313.2 MOTION PROFILES 1.364

13.2.1 Velocity Profiles 1.36413.2.2 Position Profiles 1.373

13.3 MULTI AXIS MOTION 1.37613.3.1 Slew Motion 1.377

Interpolated Motion 1.3781.3.2 Motion Scheduling 1.379

1.4 PATH PLANNING 1.3811.5 CASE STUDIES 1.3831.6 SUMMARY 1.3851.7 PRACTICE PROBLEMS 1.3851.8 PRACTICE PROBLEM SOLUTIONS 1.3861.9 ASSIGNMENT PROBLEMS 1.387

1. INTRODUCTION

1.1 Introduction

1.2 Bloom’s Taxonomy

1. Knowledge: remembering of previously learned material; recall (facts or whole theories); bringing to mind.

Terms: defines, describes, identifies, lists, matches, names.

2. Comprehension: grasping the meaning of material; interpreting (explaining or summarizing); predicting outcome and effects (estimating future trends).

Terms: convert, defend, distinguish, estimate, explain, generalize, rewrite.

3. Application: ability to use learned material in a new situation; apply rules, laws, methods, theo-ries.

Terms: changes, computes, demonstrates, operates, shows, uses, solves.

Topics:

Objectives:•

•

4. Analysis: breaking down into parts; understanding organization, clarifying, concluding.

Identify parts: See Related Order; Relationships; Clarify.

Terms: distinguish, diagrams, outlines, relates, breaks down, discriminates, subdivides.

5. Synthesis: ability to put parts together to form a new whole; unique communication; set of abstract relations.

Terms: combines, complies, composes, creates, designs, rearranges.

6. Evaluation: ability to judge value for purpose; base on criteria; support judgment with reason. (No guessing).

Terms: appraises, criticizes, compares, supports, concludes, discriminates, contrasts, summa-rizes, explains.

1.3 Examples

1.4 Summary

1.5 References and Bibliography

1.6 Problems

1.7 Challenge Problems

2. DRAFTING

• Drafting was previously a set of techniques (using compasses, angles, T-squares, etc.) for creat-ing drawings that could be understood and used in manufacturing.

• More recently drafting is focusing less on techniques and more on conventions, because of CAD systems.

• The conventions of drafting are very important because they allow us to define parts in a way that they will be understood by any engineer, machinist, technologist, etc.

2.1 CONVENTIONAL DRAFTING

• The purpose of drafting is to present technical ideas in precise and concise forms.

• A properly drafted drawing should be understood by any engineer.

• A sample of a drafted drawing is given below.

2.1.1 Manual Drafting

• This is the use of drafting boards, pencils, pens, and a number of specialized tools for drafting. While this method is still very popular, the techniques used in manual drafting are quickly being displaced by CAD (Computer Aided Design) systems.

• I will not cover some of the manual drawing topics list below, but more information on them appears in a large number of drafting books.

- lettering- hand sketching- drawing ellipses- etc

A

A

section A-A

2.500.25

Ø2.

00

Ø1.

62

1.750

Ø0.006 M

Ø1.0005/0.9995

Ø0.25

2Notes:1. Break sharp edges to 0.01 max.

Drill Ø0.985 ream to spec.2

part: bushingdate:etc....

2.1.2 Turning Three Dimensions Into Two (Multi View Drawings)

• The problem with drafting is that the paper is flat, while the object drawn is not.

• To get around this we can develop a number of views to work with.- Front View- Top View (Plan View)- Right Side View- Left Side View

• This method of developing views is known as Orthographic projection

• This method eliminates the perspective distortion in real vision, thus making it easier for techni-cal depiction.

• In this method, object faces that are parallel to the viewing plane are shown as actual size, but objects that are not parallel are foreshortened.

• The number of views used is a function of the geometry. For a simple object such as a washer, only one view is needed. A more complicated object, such as a piston, would require at least two views.

2.1.2.1 - The Glass Box

• The views are developed as if a glass box was placed over the object. The view from each direc-tion was frozen, and when the box is unfolded, the resulting views are seen.

• Imaging the case below of a small tetrahedron (a three pointed triangle),

• The drawings are layed out with certain conventions. The example above is continued below for illustration, In the figure extra construction lines are added to show how the drawings in the different views are related.Note that the top view is related to the side view using a 45° line. These properties are a result of the ‘glass box’ concept. The folding lines are often shown on drawings (they have two dashes and one long). Also note that in the figure shown below, the points in the top view will be the same distance from the folding line as they are in the side view.

the part

cover witha glass box

freeze the viewthrough eachside of the glassbox

Unfold the sides to geta set of views

• The layout of the drawings is done by convention. In this drawing the right side view is to the right of the front view. If this drawing observed european standards, the right side view would be on the left hand side.

• A useful method for keeping the large number of points in a drawing sorted is to number them. For example,

TF

F R-S

• The view that is selected as the front is arbitrary, but it should- be a natural front to the object.- be the most important view- appear stable- chosen to minimize hidden lines in other views- contain most of the detail

2.1.3 Lines

• The number of lines on drawings will become confusing, therefore this calls for some method for differentiating between lines.

• Hidden lines are dashed lines used to show lines that not visible.

TF

F R-S

1

23

4

1

23 4

1

2,34

• Centre lines are used to show the axis of rotation for an object surface. These lines have long/short dashes.

• Construction lines are drawn on to help locate final drawing lines. These lines are so light that they are often not even erased when the drawing is complete.

• Some objects have disproportionate dimensions. As a result, it may be necessary to ‘break’ them to show any reasonable level of detail. There are three types of breaks commonly used,

- S breaks - for round objects- Z breaks - for thin long/wide objects- freehand breaks - for long rectangular objects

hidden line

phantom line

centre line

construction line

drawing line

break line

dimension line

leadercutting plane

2.1.4 Holes

• There are a number of holes commonly depicted in drawings,

S break

Z break

freehand break

2.1.5 Special Cases

2.1.5.1 - Aligned Features

• Aligned features - in some cases, features are revolved, and shown at a consistent radial dis-tance, but not necessarily in the correct position.

• Holes are commonly rotated to simplify views

Through Holes - these are cut all theway through an object

Blind Holes - these holes stop part waythrough an object

Tapered Holes -

Counterbored Holes -

• Ribs and wings are commonly rotated to simplify views

preferred

Preferred

• Large features on parts may be rotated to simplify views. small features, such as slots may also be rotated between views for clarity.

• Sheet metal parts start out flat, but are deformed to new useful shapes. Therefore it is common to draw sheet metal parts in the deformed, and the undeformed state.

Part is imagined in thisposition, but drawncorrectly in the top view

But the part can be drawn, andcorrectly dimensioned in the frontview with the bend artificiallyremoved

2.1.5.2 - Incomplete Views

• Incomplete views - certain details can be omitted to simplify the view. This method produces drawings that are not correct, but they are commonly used in practice.

• Some views will end up having an excessive number of hidden lines. To combat this problem, we may sometimes just leave them out.

• Large radial/cylindrical parts are often cropped to save space. But, enough is shown to make the remainder of the geometry obvious.

use half circles or use a break

2.1.6 Section Views

• when there are complicated internal features, they may be hard to identify in normal views with hidden lines. A view with some of the part “cut away” can make the internal features very easy to see, these are called section views.

• In these views hidden lines are generally not used, except for clarity in some cases.

• The cutting plane for the section is,- shown with thick black dashed lines.- has arrows at the end of the line to indicate the view direction- has letters placed beside the arrow heads. These will identify the section- does not have to be a straight line

• sections can be lined to indicate,- when the section plane slices through material- two methods for representing materials. First, use 45° lines, and refer to material in title

block. If there are multiple materials, lines at 30° and 60° may be used for exam-ple. Second, use a conventional set of fill lines to represent the different types of materials.

2.1.6.1 - Full Sections

• Full sections - generally a straight section line cuts through a part to give a complete view of the inside. This section typically replaces one of the views that is confusing.

2.1.6.2 - Offset Section

• Full sections will experience difficulties when the features do not lie along a single line.

A A

SECTION A-A

A section viewcan clarify a viewappreciably

• We can use a section line that is turned to cut through features. This view can be used to replace one of the principle views.

2.1.6.3 - Half Section

• In some cases it is better to illustrate internal features with both a section, and a full view. In this case we can cut away only part (a quarter) of the object, and draw a view that is half normal, half section.

• this method is well suited to symmetrical parts, with the section starting at the axis of symmetry

• Take note that the section line here only has one arrow head, but the direction must be observed

A A

SECTION A-A

2.1.6.4 - Cut Away Sections

• Instead of doing large scale sections, we can cut away a very specific region of interest.

• In this case a break line is used, and the cutting plane lines used in other cases are not applicable.

A

Section A

2.1.6.5 - Revolved Section

• When we have transition pieces, such as ribs, or airplane wings, we will want to show the shape, but this is not easy with conventional views, In this case we can break out a section.

• The basic procedure is to1. select a characteristic section, and draw cut lines to either side.2. in between the breaks, draw a section that is rotated 90° so that it is obvious on the

drawing.

• This method is useful when space is at a premium

• The cutting plane line is not used with this technique

2.1.6.6 - Removed Section

• this is a more exact alternative to the revolved section method.

• With this method a break is not used, but a cutting plane line is. The sections are then drawn at some other location on the page.

• The only features shown are the features of the section.

• labels such as A-A, B-B, etc are used to avoid ambiguity.

• these views are often placed at a distance and arranged in the same order as the sections.

• These section may also be shown using lines extended from the object

A

A

B

BSECTION B-BSECTION A-A

• modified scales may also be used with appropriate notation

2.1.6.7 - Auxiliary Section

• A section can be done that does not lie in one of the primary planes.

• This done as a normal section

2.1.6.8 - Thin Wall Section

• This method is used for assemblies of thin materials, such as sheet metal.

• This illustrates how the pieces butt up against each other.

• The sections are filled with black, but a small space is left between the piece to indicate the assembled faces (operations such as crimping, spot welding, etc are used for these)

2.1.6.9 - Assembly Section

• When placing parts together we want to verify that they will match, and that they can be assem-bled. We also want to provide assistance to the assembler. To do this a cut away assembly drawing can be used.

• There are a number of elements present in these diagrams,

A

A

SECTION A

-A

two or more parts- a parts list with numbered items- generally section views are used, and oriented along the main assembly axis

2.1.6.10 - Special Cases

• Because sections are to clarify confusing features on diagrams, they are sometimes not theoreti-cally correct.

• A few of the cases that are considered when working with sections are,

1. cutting lines may intersect ribs, but they may be drawn offset somewhat to clarify the rib geometry.

4

3

2

1

ref #

piston

rod

chamber

o-rings

description

M8765

M87101

M8734

P8703

part #

1

1

1

3

qty.

12

3

4

2. sections may be aligned to clarify the views

Preferred

A

A

A

A

3. If a cutting plane cuts through intersecting features, the less important feature may be omitted for clarity, or to save time. For example, two rounds that intersect at an angle other than 90° would have an unusual shape, if one is not drawn, the section becomes much easier to do.

2.1.6.11 - Fill Patterns

• Sections can be filled with a number of patterns to indicate different materials

• This was a common technique in the past. Some examples are given below.

2.1.7 Auxiliary Views

• The glass box can also be folded at odd angles. This technique produces views known as Auxil-iary views.

• These views are useful when we want to draw a view of a surface that is not normal to one of the primary viewing planes.

• common terms used for this method are true size, and true shape. keep in mind that if a feature does not lie parallel to one of the primary viewing planes, it will appear distorted in every view.

• These views can be constructed from any view in a drawing. typical names for these identify the view that they are drawn from,

cast iron and steel and bronze, brasscopper

zinc, lead,alloysmalleable iron wrought iron

aluminum andmagnesium andtheir alloys

front auxiliary view- top auxiliary view- side auxiliary view

• We can also use auxiliary views to project other views for geometric purposes

• hidden lines are typically not used in auxiliary views, unless needed for clarity. Also, a number of surfaces are not included because they are distorted, and of little value.

• typically steps followed to construct an auxiliary view,1. select the face that is to be drawn as i) a true surface, ii) a true length line, iii) an end

view of a line.2. draw construction lines perpendicular to the surface/line/point of interest. This line

should go in a direction, and far enough that leaves enough space for the view.3. draw a folding line at an appropriate distance. This will act as a reference plane.4. transfer distances from another view. This view will typically be the view adjoining the

view that the auxiliary is drawn from.5. Complete the view.

• an example is given below, and all faces are drawn for illustration, but normally only the angled face would be drawn. Because this is the first auxiliary from the drawing, it is called the pri-mary auxiliary view.

Step 1: decide to draw the angledface of the block,using the front view,because an edge viewis available.

Step 2: draw construction linesperpendicular to the face,the view will be drawnin the open space in theupper right opening.

Step 3: draw the fold line in forreference. Just as a visualcheck, each of the constructionlines should be perpendicular

Step 4: Transfer distances to findpoints in the auxiliary view.Here the points are numberedfor the readers benefit. We cantransfer the distances eitherfrom the top or side view.

1,62,7

3,8

4,95,10

6,10 7 8,9

3,421,5

2,1 7,6

8

9,104,5

3

1

2

34

5

67

8

910

d1

d1

d1

d2 d2

d2

Step 5: the view is completed

truesurface

• There are special drafting techniques for rounded, or curved surfaces, these can be found in any drafting textbook.

2.1.7.1 - Secondary Auxiliary Views

• sometimes it is necessary to make an auxiliary view, using an auxiliary view. When this is done, the first auxiliary is constructed as normal. The second auxiliary is made from the first, but the distances can only be transferred from the first auxiliary for the second auxiliary.

• These views can be needed for a number of purposes, but generally they will be needed when the object does not lie perpendicular, or parallel to any of the viewing planes.

2.1.7.2 - Partial Auxiliary Views

• It is not necessary to draw entire auxiliary views, they can be draw in part, and break lines use.

• This technique allows simplified illustrations of features of interest, without full development of an auxiliary view.

2.1.8 Descriptive Geometry

• The use of drafting to determine geometric properties, such as shortest distances between points and lines.

• These methods can also be used to solve statics (vector) problems, etc.

• These methods use extensions to the methods of auxiliary views that allow curved surface to be considered.

• the basic steps in these methods are,1. find the true lengths of a line2. find the end view of a line3. find the edge view of the surface4. find the true shape of the surface

• These steps will allow determination of a number of properties,- points can be projected into other views- lines can be projected into other views

the true length of a line can be determined- a point view of a line can be found- distances between points and lines can be found- distances between lines can be found- distance between a point and plane- angle between two planes- edge view of a plane

2.1.9 Isometric Views

• These views are done as a way of realistically drawing objects. This is not correct, as a perspec-tive drawing would be, but it is very good for engineering problems.

• The viewing directions are skewed so that up is still up, but straight back now goes to the left and back, and right goes to the right and back. Both of the moved axis are drawn at 30° to the horizontal.

• The values measured off these views will be accurate when measured along the axis.

2.1.10 Special Techniques

• There are a number of special techniques of interest when doing manual drafting, but of declin-ing interest in view of modern CAD systems. A list of these techniques are given below, and are described in good detail in most drafting books,

- drawing ellipses- drawing with circles- drawing with revolution- drawing with four centres

- isometric drawing- using 30°/60° angles- using special paper

- Oblique views- cavalier (45°, with full depth size)- cabinet (0-90°, with half depth size)- general (0-90°, with between half and full depth size)

2.2 NOTATIONS

• Typically these are a number of notations added to drawings to describe features, or explain operations.

• Some abbreviated terms are given below,

angledistance

a unit cube is shown forillustration

Abrev.

CBORECSKDIAHDNLLHNCNFPRRcRHTHDTIRTPIUNCUNF

Description

counterborecountersinkdiametercase hardenleadleft handnational coursenational finepitchradiusRockwell C hardnessright handthread(s)total indicated runoutthreads per inchunified national courseunified national fine

2.2.1 Basic Dimensions and Tolerances

• The size of an object, and the required accuracy can have a significant bearing on the cost

• Unilateral Tolerances

• Bilateral Tolerances

• Limits can be used to exactly define the size boundaries of a feature.

• Tolerances use a nominal dimension and differences.

2.2.2 Geometric Dimensioning and Tolerancing (GD & T)

• Specified in standard ANSI Y14.5 (1983).

• Combines rules and independent symbols in addition to the normal tolerancing symbols

• Allows old style tolerances, but adds new methods that cover geometrical forms.

• Allows easy specifications of datums, etc.

1.505”1.495”

+0.005”-0.005”1.500”

• Advantages of this method are,- makes drawings clearer and more ambiguous- allows separated features to be related- uses symbols instead of words to reduce language translation problems- the method helps specify manufacturing and metrology methods

• The main purpose of GD&T is to ensure,size - the overall dimensions are as specifiedform - the shapes specified must have the correct geometrical formfit - two parts must mate as specifiedfunction - the product conforms to performance specification

2.2.2.1 - Feature Control Symbols

• The basic of GD&T is the feature control symbol.

• This indicates what the tolerance is, its value, the reference datums, and any modifiers needed.

• An example of a feature control is given below,

• not all of these symbols/categories will be used on a regular basis, but they provide the designer added flexibility in how they specify tolerances.

0.001 M A B C

datums to be used in this case thepart is placed against A, then B, thenC. This forms a referencecoordinate frame.

The maximum metal modifierthe basic tolerance value

the zone identifier

the type of feature control (parallelism)

2.2.2.2 - Symbols and Meaning

• The basic symbols are shown below,

• Flatness - basically, all the surface elements are constrained to lie within two parallel surface places, separated by the tolerance

characteristic

straightness

flatness

circularity

cylindricity

profile of a line

profile of a surface

angularity

perpendicularity

parallelism

position

concentricity

circular runout

total runout

tolerance type symbol

individualfeatures

for individualor relatedfeatures

relatedfeatures

form

profile

orientation

location

runout

• Straightness - basically, one the surface elements is constrained to lie within two parallel surface places, separated by the tolerance. In effect, this means that if any line across the surface is within two parallel lines, the part is acceptable. This can be tested by running a comparator across the surface (using a reference plane)

• Circularity - all of the points on a cylindrical surface are constrained to lie within two circles. This can be tested with a talyrond.

0.001

<0.001means

tolerancezoneparallel

planes

0.001

<0.001means

tolerancezoneparallel

lines

• Cylindricity - an extension to circularity that specifies the tolerance along the cylinder.

• Concentricity -

• Angularity - requires that all points on a specified feature must form an angle with a datum. This could be measured with a sine bar and a height comparator.

0.01 0.01tolerancezone

means

0.01

0.01tolerancezone

means

0.01tolerancezone

and

• Perpendicularity - this has the same meaning as angularity, but it is specifically applied to 90• angles. This could be measured with squares and reference plates.

• Symmetry -

• Parallelism - all points on a surface are to be parallel to a given datum, within a specified toler-ance

• Line Profile - the amount of deviation that is allowed (typically for irregular lines)

40°

-A-

0.02 A0.02 tolerance

zone

40°

-A-

0.01 A

0.5±0.1

05.

0.01 tolerance zone

• Surface Profile - the amount of deviation that is allowed for a surface

• Circular Runout - when dealing with a surface of revolution, this determines the amount of devi-ation allowed from the central axis. This specifically refers to a specific point

0.01

0.01

0.01

This means over the entiresurface

• Total Runout - similar to circular runout, but this applies to the entire part. In effect, circular runout uses two circles, whereas total runout uses two surface planes.

2.2.2.3 - Datums

• These are reference features, that other features are to be measured against.

• These can be used when setting up parts, for manufacturing or production

• Typical features used are,- axes- cylinders- planes- lines- points

• A datum reference frame can be constructed with,- three perpendicular planes- 3 contact points in the primary plane, 2 in the secondary plane, and 1 in the tertiary

• A datum is specified with a boxed letter with two dashes,

-A-

0.01 A

0.01tolerancezone

this means that at any pointalong the axis, the crosssection of the part willresult in the specifiedtolerance

2.2.2.4 - Modifiers

• to overcome shortcomings in symbols, modifiers can be added to change their meanings.

• in particular,

-A-

-B-

Maximum material condition - the tolerance is at the extreme that would result if too little material was cut off, and the maximum material remains.

M

Least (Minimum) material condition - the tolerance is at the extreme that would result if too much material was cut off, and the minimum mate-rial remains.

L

Regardless of features size (RFS) - this indicates that the tolerance must be maintained, regardless of variations in this size of the object. S

Projected tolerance zone - a tolerance zone can be extended beyond a sur-face. To do this the basic surface must be specified as a datum.P

************ Include an example of tolerances using GD&T

2.3 WORKING DRAWINGS

• The basic skills/topics discussed below lead up to preparing, and understanding a complete set of drawings.

• The purpose of working drawings is to,- describe the exact geometry of parts- indicate other details associated with drawings (for example, material)- show how parts are assembled- indicate manufacturing preferences

• generally, the drawing package will include a number of items,- a drawing (one a separate sheet with a separate title block) for each part- a bill of materials- an assembly drawing

• a typical working drawing package will contain,- a design layout- assembly drawings (and a Bill of Materials)- subassembly drawings- detailed drawing- purchased parts- modified purchased parts

2.3.1 Drawing Elements

2.3.1.1 - Title Blocks

• Most of the important details are put in this block. Each block is individualized to a company, but generally they include,

- company name, and division if applicable- machine or department name- part name- drawing number- part number- the number of parts required- the scale

drafter name/date- drawing checker name/date- material- tolerances- finishing details- units of drawing

• The block is typically located in the bottom right hand corner of the drawing

• The drawing title, and drawing number are commonly printed in large fonts

2.3.1.2 - Drawing Checking

• this is a process whereby a drawing is reviewed for completeness, accuracy, etc.

• modern CAD systems, especially solid modeler should reduce the emphasis on checking the drawings. Some of the main features checked for in manual drawings are,

- appearance - this can be a large issue for hand drawn work- within standards - legal and corporate- clarity - all description, dimensions, etc should be well understood- completeness - sufficient dimensions, etc should be present for production- redundancy - redundant information should be eliminated unless essential- manufacturability - the cost and feasibility of production should be considered. are toler-

ances sufficient/excessive, are other steps sufficient for product life.-

2.3.1.3 - Drawing Revisions

• When a drawing has reached production, it is considered final, but changes are frequently made.

• It is very important that drawing changes are dealt with properly. This means,- all changes are recorded on the drawing, and new drawings made- all old drawing must be collected, or marked void (failure to do this can lead to very

expensive mistakes)- when a drawing has been changed a number of times, it should be redrafted.

• Computer CAD systems still do not sufficiently deal with problems such as these, and often rely on the previous manual drafting systems to process these updates. But, software is available, and is being developed for product information management (PIM) that will deal with these changes in a manner suitable for CAD.

2.3.1.4 - Bill of Materials (BOM)

• An important list on most drawings is a Bill of Materials, this is a list of all required materials/parts required to make to part depicted in the drawing.

• This list contains,- all part numbers- all part names- quantity of parts required- materials required- source

• This is sometimes given on separate sheets, or on the drawing itself

• The typical (but not the only) order for listing parts on a BOM is,1. produced in-house2. specialty purchase (e.g. roller bearings)3. standard purchased hardware (e.g., washers)4. bulk items (e.g. lubricants)

2.3.2 Drawing Types

2.3.2.1 - Assembly Drawings

• These are used to specify an assembly with,- a drawing of the assembled part

• Hidden lines are typically omitted from these drawings. Details may also be omitted if they have no bearing on the product

• assembly instructions may also be included in these drawings to guide workers

• full section assembly drawings are often used

• dimensions not included unless essential

• Small blow-up bubbles are often used to emphasize details

• The parts can be identified using,

numbers with arrows and a block list of parts including,- quantity- part name- part source- part number- reference number- drawing number

- arrows and descriptions- quantity- part name- part source- part number- drawing number

2.3.2.2 - Subassembly Drawings

• these are basically the same as assembly drawings, except that there are components that have already been assembled.

• Modern equipment is complex and is assembled in stages. The final assembly might be some-thing like an automotive body welding shop, whereas a sub-assembly might be the car radio.

2.3.2.3 - Exploded Assembly Drawings

• these are drawings that show each piece separated, and indicates their assembly paths. This can help when determining,

- which part goes where- the orientation of the part- the part of approach- the order of assembly

2.3.2.4 - Detailed Drawings

• These drawings use the techniques discussed earlier in this section to depict, and dimension parts.

************************* INCLUDE A COMPLETE DRAWING PACKAGE TO ILLUS-TRATE

2.4 PRACTICE PROBLEMS

2.5 REFERENCES

Ullman, D.G., The Mechanical Design Process, McGraw-Hill, 1997.

3. METROLOGY

3.1 Introduction

3.1.1 The Role of Metrology

• modern manufacturing can produce features that are more accurate than we can measure by hand, therefore we need tools to assist us.

• These tools allow us to quantitatively evaluate physical properties of objects.

• EVERY industry uses these tools to some extent, for example,- machine shops- tailors- dentists- automotive manufacturers- etc.

Topics:

Objectives:•

•

3.2 DEFINITIONS

Accuracy - The expected ability for a system to discriminate between two settings.

Assembly - the connection of two or more separate parts to make a new single part.

Basic Dimension - The target dimension for a part. This typically has an associated tolerance.

Dimension - A size of a feature, either measured, or specified.

Dimensional Metrology - The use of instruments to determine object sizes shapes, form, etc.

English System - See Imperial.

Error - a discrepency between expected, and actual values.

Imperial System - An older system of measurement, still in use in some places, but generally replaced by the metric system.

Limits - These typically define a dimensional range that a measurement can be expected to fall within.

Machine Tool - Generally use to refer to a machine that performs a manufacturing operation. This is sometimes confused with the actual cutting tools, such as a drill bit, that do the cutting.

Measurement - The determination of an unknown dimension. This requires that known standards be used directly, or indirectly for comparison.

Metric System - A measurement system that has been standardized globally, and is commonly used in all modern engineering projects.

Metrology - The science of measurement. The purpose of this discipline it to establish means of determining physical quantities, such as dimensions, temperature, force, etc.

Precision - Implies a high degree of accuracy.

Repeatability - Imperfections in mechanical systems can mean that during a Mechanical cycle, a process does not stop at the same location, or move through the same spot each time. The vari-ation range is refered to as repeatability.

Standards - a known set of dimensions, or ideals to compare others against.

Standard Sizes - a component, or a dimension that is chosen from a table of standard sizes/forms.

Tolerance - The allowable variation in a basic dimension before a part is considered unacceptable

3.3 STANDARDS

• Standards are the basis for all modern accuracy. As new methods are found to make more accu-rate standards, the level of accuracy possible in copies of the standard increase, and so on.

• A well known metric standard is the metric 1m rod.

• Many standards are available for measuring, and many techniques are available for comparison.

3.3.1 Scales

• The most common tool for crude measurements is the scale (also known as rules, or rulers)

• Although plastic, wood and other materials are used for common scales, precision scales use tempered steel alloys, with graduations scribed onto the surface.

• These are limited by the human eye. Basically they are used to compare two dimensions.

• The metric scales use decimal divisions, and the imperial scales use fractional divisions.

• Some scales only use the fine scale divisions at one end of the scale.

10 20 30 40

8 16 24 32 40 48 561

8 16 24 32 40 48 562

metric(mm)

imperial(inches 1/64)

• It is advised that the end of the scale not be used for measurement. This is because as they become worn with use, the end of the scale will no longer be at a ‘zero’ position. Instead the internal divisions of the scale should be used.

• Parallax error can be a factor when making measurements with a scale.

3.3.2 Calipers

• A tool used to transfer measurements from a part to a scale, or other instrument.

• calipers may be difficult to use, and they require that the operator follow a few basic rules,- do not force them, they will bend easily, and invalidate measurements made- try to get a feel, or personal technique for using these instruments.- if measurements are made using calipers for comparison, one operator should make all of

the measurements (this keeps the feel factor a minimal error source).

• These instruments are very useful when dealing with hard to reach locations that normal measur-ing instruments cannot reach.

• Obviously the added step in the measurement will significantly decrease the accuracy

3.3.3 Transfer Gauges

• Small hole gauges can be inserted into a hole, as an adjustment knob is turned, the head expands

10

20

30

40

If the instrument is not measured directly on,then there may be some error. Note: this wouldnot occur if the scale was perfectly thin.

to the size of the hole. The gauge can be removed and measured to determine the diameter of the hole. The end of this gauge appears as if a sphere with a shaft in it has been split into two halves.

• Telescope gauges have two plungers that are springy, until locked in place. This can be put in holes or hard to reach locations, and used to transfer measurements to other measurement devices.

3.4 Instruments

3.4.1 Vernier Scales

• Vernier scales have normal scale components, but also incorporate a small secondary scale that subdivides major increments.

• This secondary scale is based on a second scale that is one increment shorter than a main scale. If the secondary scale is compared to the main scale, it will indicate relative distance between two offsets.

• The scale pictured above would normally be on an instrument, and the main and vernier scales would slide relative to each other. The ‘0’ on the vernier scale would be used to take the read-ing from the main scale. In this example the main scale would read a value that is between 0.4 and 0.6. (Note: it is not considered good practice to round this to 0.5)

• The vernier scale can then be used to find the internal division, by looking for where the divi-sions in the top and bottom scales align. In this case the second internal division aligns with 1. Using the values on the vernier scale, we can see that the value for this division would be 0.08. The value from the vernier scale is added directly to the main scale value to get the more accu-rate results. 0.4+0.08 = 0.48.

0 1 2

0Vernier scale

Main scale

.2

0.4+0.08=0.48

• On imperial sliding vernier scales the main scale divisions are 0.050” apart, and on the vernier scale they are 0.049”, giving a reading of 0.001” per graduation.

• On metric sliding vernier scales the main scale divisions are 1mm apart, and the vernier scale they are 0.98 mm, giving a reading of 0.02mm per graduation.

• Angular vernier scales are used on protractors, and are identical in use to linear vernier scales. The major protractor scales have divisions of 1 degree, and the vernier scale is divided into 5 minute intervals. One interesting note is that the vernier scale has two halves, one in the posi-tive direction, and one in the negative direction. If reading from the left division, on the main scale, the right vernier scale should be used. And, when measuring from the right hand division on the major scale, the left vernier scale should be used.

3.4.2 Micrometer Scales

• This is a very common method for measuring instruments, and is based on the thread principle.

• In effect, as a thread is turned, a large motion on the outside of the thread will result in a very small advance in the position of the thread.

• The micrometers pictured above have major scales, as well as minor scales. The major scales are read first, and the micrometer scales are read second and the readings added on.

• The metric micrometer above reads 13.5 = 13.5mm on the major scale, and 31 = .31mm on the thimble, for a total of 13.81mm

0 5 10 15

0 1 2 3 4 5

1211

109876

40

35

30

25

Imperial (Inches)

Metric

0.459

13.1

• The Imperial scale above shows a micrometer reading of 4.5 = .45” on the main scale, and 9 = .009” on the thimble, for a total of .459

• On imperial micrometers the divisions are typically .025” on the sleeve, and 0.001” on the thim-ble. The thread used has 40 T.P.I. = a pitch of 0.025”

• Metric micrometers typically have 1 and 0.5 mm divisions on the sleeve, and 0.01mm divisions on the thimble. The thread has a pitch of 0.5mm.

• A vernier micrometer has the scales as pictured above, but also a vernier scale is included to pro-vide another place of accuracy.

• Depth micrometers have an anvil that protrudes, out the end, and as a result the scales are reversed to measure extension, instead of retraction.

3.4.2.1 - The Principle of Magnification

• The operation of micrometers is based on magnification using threads.

• A large movement on the outside of the micrometer thimble will result in a small motion of the anvil.

• There are two factors in this magnification. First, the difference in radius between the thread, and the thimble will give a change in sensitivity relative to the difference in radii. Second, the pitch of the thread will provide a reduction in motion.

• The basic relationship can be seen below,

3.4.2.2 - The Principle of Alignment

• Basically, the line of the physical measurement should be such that it is coincident with the mea-surement axis of the instrument.

• If the measurement is out of line, it may lead to misreadings caused by deflections in the instru-ment.

M CD---- πD

pitch-------------= where,

M = magnification from the moving head to the hand motionC = measuring diameter of the instrumentD = diameter of the threadpitch = the number of threads per unit length

Radial Arm Principle of Magnification

Inclined Plane Principle of Magnification πDpitch-------------=

CD----=

0 5 10 1540353025

CD

pitch

NOTE: magnification can result in greater sensitivity of an insrument to control, and reading by a user.

• micrometers are generally better than sliding vernier calipers when considering this principle.

3.4.3 Dial Indicators

• Converts a linear displacement into a radial movement to measure over a small range of move-ment for the plunger.

0 5 10 1540353025

misalignment isslight, but may stillcause errors.

• The radial arm magnification principle is used here.

• these indicators are prone to errors caused by errors that are magnified through the gear train. Springs can be used to take up any play/backlash in the rack and pinion to reduce these errors.

• The gears are small, but friction can result in sticking, thus reducing accuracy

• A spring is used on the rack to return the plunger after depression.

• The problems mentioned earlier will result in errors in these instruments. If the dial indicator is used to approach a dimension from two different sides, it will experience a form of mechanical

0

10

20

30

4050

60

70

80

90

rack

plunger

pinion

indicator dial

gears

hysteresis that will bias the readings. An example of this effect is given below.

• In the graph shown, as the dial indicator is raised in height (taking care not to change direction), the errors are traced by the top curve. As the height of the dial indicator is decreased, the bot-tom curve is traced. This can be observed using gauge blocks as the known heights to compare the readings against.

• The causes of this hysteresis are bending strain, inertia, friction, and play in the instrument.

• Applications include,- centering workpices to machine tool spindles- offsetting lathe tail stocks- aligning a vise on a milling machine- checking dimensions

• These indicators can be somewhat crude for accurate measurements, comparators have a higher degree of sensitivity.

3.4.4 The Tool Makers Microscope

• Quite basically this is a microscope. But, it has lines added to the optics for visual reference, and micrometer dials, and angular verniers added to the stage to measure distances.

• Parts are put on the stage, and the microscope is focused. The stage can then be rotated, and translated precise distances to allow visually referenced measurements

• Such a microscope might have two micrometer heads for x-y translation of the stage. In addi-tion, the stage can be rotated, and angular positions measures.

+ve errors

-ve errorsmaximum variance

as height is increased

as height is decreased

3.4.5 Metrology Summary

• We can discuss various instruments, and what they are used for.

Table 1: Fill in more later

Feature SizeRange Accuracy Instrument Comments

Angle 90° yes/no square

85°-95° -- cylindrical square

outside dis-tance

depth

3.5 Surfaces

• No surface is perfectly smooth, but the better the surface quality, the longer a product generally lasts, and the better is performs.

• Surface texture can be difficult to analyze quantitatively. Two surfaces may be entirely different, yet still provide the same CLA (Ra) value.

• Recent developments in production technique, and metrology equipment have made it possible to specify and measure surface quality.

• There are standards, such as the CSA B95 1962.

• Surface Quality can be important when dealing with,- lubrication - small indentations can hold lubricant- resistant to wear - smoother surfaces wear less- tool life - rough surfaces will correlate to shorter tool life- fatigue/stress raisers - - corrosion - smoother surfaces easier to clean, less surface area to erode- noise reduction - smooth surfaces make less noise when rubbing, for example meshing

gears.- fit - pressure seals could leak through pits

• Surface geometry can be quantified a few different ways.

• Real surfaces are rarely so flat, or smooth, but most commonly a combination of the two.

Flat and Smooth

Smooth (not flat) - waviness

Rough (flat)

• Some other terms of interest in surface measurement,- Surface texture - all of the details that make up a surface, including roughness, waviness,

scratches, etc.- Lay - the direction of the roughness on a newly manufactured surface. The roughest pro-

file will be perpendicular to the lay.- Flaws - small scratches, cracks, inclusions, etc.- Cutoff - a value selected to be less than the waviness, but greater than the roughness

length. This is controlled using electrical or digital filters. Typical values might be; 0.010”, 0.030”, 0.100”

3.5.1 Measures of Roughness

• A simple measure of roughness is the average area per unit length that is off the centre line (mean). We will call this the Centre Line Average (CLA), or Arithmetic Average (Ra), the units are µinches.

• To calculate the roughness using samples at evenly spaced positions,

• The roughness can also be calculated by area,

• In both cases the mean line is located so the sum of areas above the line is equal to the sum of areas bellow the line.

• As an example we can examine a surface that has a triangular profile,

h1h2 h3 h4

hn

l (and n samples)

CLA Ra

h∑n

----------h1 h2 … hn+ + +

l-----------------------------------------= = =

mean lineArea A1

A2

A3

An

l

CLA Ra

A∑l

----------A1 A2 … An+ + +

l-------------------------------------------= = =

• One of the instruments that we will use is the Surfcom. If we were to have obtained the graph above from this device, we would have to use the following formula to determine the true val-ues,

mean line1 2 1

1 2 1

8

1

CLA Ra

h∑n

---------- 1 2 1 0 1 2 1 0+ + + + + + +8

-------------------------------------------------------------------- 1= = = =

CLA Ra

A∑l

---------- 4 4+8

------------ 1= = = =

We can find the surface roughness using heights,

We can also find the surface areas using areas,

Note the results are the same with both methods. These numbers may varysignificantly if the height method does not take enough samples for a roughersurface texture.

A secondary measure of interest is,

Full Texture Height is 2 - (-2) = 4Full Texture Height/Ra ratio is 4:1

CLA Ra

A 10 6–×∑l vertical magnification×--------------------------------------------------------------µin.= =

measured on trace

3.6 Measuring Surface Roughness

• There are a number of useful techniques for measuring surface roughness,- observation and touch - the human finger is very perceptive to surface roughness- stylus based equipment - very common- interferometry - uses light wave interference patterns (discussed later)

3.6.1 Observation Methods

• Human perception is highly relative. In other words, without something to compare to, you will not be certain about what you are feeling.

• To give the human tester a reference for what they are touching, commercial sets of standards are available.

• Comparison should be made against matched identical processes.

• One method of note is the finger nail assessment of roughness and touch method used for draw dies in the auto industry.

3.6.2 Stylus Equipment

• One example of this is the Brown & Sharpe Surfcom unit.

• Basically this technique uses a stylus that tracks small changes in surface height, and a skid that follows large changes in surface height. The use of the two together reduces the effects of non-flat surfaces on the surface roughness measurement. The relative motion between the skid and the stylus is measured with a magnetic circuit and induction coils.

• The actual apparatus uses the apparatus hooked to other instrumentation. The induction coils drive amplifiers, and other signal conditioning hardware. The then amplified signal is used to drive a recorder that shows stylus position, and a digital readout that displays the CLA/Ra value.

• The paper chart that is recorded is magnified in height by 100000:1, and in length by 82:1 to make the scale suitable to the human eye.

• The datum that the stylus position should be compared to can be one of three,- Skid - can be used for regular frequency roughness- Shoe - can be used for irregular frequency roughness- Independent - can use an optical flat

direction of travel over surface

induction coils

skid/shoestylus

work surface

pivot

magnetic core

Skidskid moves this way

the height of the skid variesslightly, but effectively givesa datum

Skid - used for regular frequencies, and very common.

Flat Shoe: Used for surfaces with irregular frequencies

shoe

• Where the scan is stopped might influence the Ra value. This is especially true if the surface tex-ture varies within a very small section of the surface. For example,

Independent Datum - a separate datum is used for the reference datum.

work pieceoptical flat

This may be a good application for a laboratory.

l1

l2

Measurement of l1, or l2 would yield the same Ra values,or very close.

CASE 1:

• In both cases 2 and 3 above, Ra would be higher over the longer sample (l2) than over the shorter sample (l1).

• The bearing surface that the skid/shoe runs on might also have an effected on the measurement.

The datum changes when the longer sample is taken, thus changing themean line, and the Ra value also.

CASE 2:

mean line for l1

mean line for l2

l1

l2

l1

l2

The surface frequency.amplitude changes over the length of the surfaceCASE 3:

Both of the two surface profiles shown below would result in the same Ravalues

3.6.3 Specifications on Drawings

• The following specification symbol can be used on drawings to specify surface textures desired on a completed part,

Waviness height - the distance from a peak to a valleyWaviness width - the distance between peaks or valleysRoughness width cutoff - a value greater than the maximum roughness width that is the

largest separation of surface irregularities included in the measurements. Typical values are (0.003”, 0.010”, 0.030”, 0.100”, 0.300”)

Lay - the direction the roughness pattern should follow

• The example below shows an upper limit of 40 micro in. roughness

63

32

0.002 0.2

0.030

0.015

Maximum Waviness height

Maximum Ra

Minimum Ra

Cutoff

Maximum waviness width

Maximum roughness width

Lay direction

40

• The symbol below can specify how the roughness is to lay,

• Standards CLA/Ra values used on drawings are: 1, 2, 4, 8, 16, 32, 63, 125, 250, 500 and 1000 µin.

• Stylus travel is perpendicular to the lay specified.

• These symbols can be related to the newer GD&T symbols

3.6.4 Other Systems

• The Root Means Squared (RMS) System (also known as Rq) is not commonly used in Canada,

From the side use this symbol

From this end use this symbol

Other Symbols are,

across both multi(bumpy) radial to centre circular to centre

X M R C

• The Peak to Valley method cuts the peaks off the wave, and then uses the band between to deter-mine the roughness. This method is not commonly used in Canada.

Mean lineh1 h2 h3 h4

h5 h6 hn

l

RMS Rqh1

2 h22 h3

2 … hn2+ + + +

n-----------------------------------------------------= =

**Note: This value is typically 11% higher than CLA or Ra

p1 p2 p3 p4

v1 v2 v3 v4

h

l

L1

L2

The two parallel lines L1 and L2 are positioned such that they cut off thepeaks and valleys, given the mathematical constraints,

h is the measure of peak to valley height

P∑ 0.05l= V∑ 0.10l=

• A simple table that basically outlines the process capabilities of a number of processes is, [ANSI B46.1-1962]

• A table of roughness measurements is given below [Krar],

process

sand castinghot rollingforgingperm. mold castinginvestment castingextrudingcold rolling, drawingdie castingflame cuttingsnaggingsawingplaning. shapingdrillingchemical millingelectrical discharge machiningmillingbroachingreamingboring, turningbarrel finishingelectrolytic grindingroller burnishinggrindinghoningpolishinglappingsuperfinishing

2000

1000

500

250

125

63 32 16 8 4 2 1 0.5

-700

-600

-500

-400

-300

-200

-100

Roughness Height (µin.)

% in

crea

se in

cos

t with

surf

ace

finis

h de

sign

ed b

y th

e cu

rve

Average usage of operation

less common usage

3.6.5 Roundness Testing

• Roundness is of particular importance when designing components for fit and function.

• Most of the methods considered so far are suited to measuring with single points, but a round shape is a collection of points, with each point having significant influence if out of tolerance.

• Precise roundness measurement equipment is expensive

Tool

cutoff saw

shaper

vertical mill

horizontal mill

lathe

Operation

sawing

shaping flat surf.

fly cutting

slab milling

turning

turning

facing

facing

Material

2.5” dia. Al

machine steel

machine steel

cast Al

2.5” dia. Al.

2.5” dia. Al.

2” dia. Al.

2” dia. Al.

speed

320’/min

100’/min

820 rpm

225 rpm

500 rpm

500 rpm

600 rpm

800

feed

0.005”

0.015”

2.5 “/min

0.010”

0.007”

0.010”

0.005”

tool

10 pitch saw

3/64” rad. HSS

1/16” rad.

stellite

4” dia HSS

slab cutter

R3/64” HSS

R5/64” HSS

R1/32” HSS

cutoff

0.030”

0.030”

0.030”

0.030”

0.030”

0.030”

0.030”

0.03

Range

1000

300

300

100

300

100

300

100

100

30

100

surface RMS

300-400

225-250

125-150

40-50

100-200

50-60

200-225

30-

• Two fundamental methods for measuring roundness are,- Intrinsic - uses points on the round surface to measure from- Extrinsic - uses a separate round surface for a reference (e.g. a precision bearing)

3.6.5.1 - Intrinsic Roundness Testing

• Three methods for Intrinsic roundness testing are shown below,

010

20

30

4050

60

70

80

90

Diametrical Intrinsic Method

A dial indicator is positioned over the surfaceto a reference height. The part is then rolledunderneath. The peak height can then becompared to other readings.

Datum Point

Dial Indicator

Rolled this way

dia.

• All three of the intrinsic methods are inexpensive

010

20

30

4050

60

70

80

90

Vee Support Intrinsic Method

A dial indicator is positioned over the surfaceto a reference height. The part is then rolledunderneath. The peak height can then becompared to other readings. The Vee support

Datum Point

Dial Indicator

Rolled this way

reduces the effect of a single datum point.

010

20

30

4050

60

70

80

90 Between Centres

A dial indicator is positioned over the surfaceto a reference height. The part is then rotatedon centres. The variations in the readings arethen used to evaluate the part. Location ofthe centre may lead to problems.

• The Intrinsic methods all have an important limitation. In particular, if the deformation of the round is small, the methods will deal with it reasonably, but if the deformation is large enough to make the shape non-cylindrical, then the results will err significantly.

• When using The Flat Plane, or the Centre to intrinsically measure roundness, the diameters can be directly obtained, but when using the Vee block, some additional calculations are required.

With this test the two readings shown wouldindicate roundness, when in fact this is not true

This test would exaggerate the roundness error suchthat it would be greater than the actual error

• The vee block method has particular disadvantages,- a number of angles are required (the standard angle is 90°)- only suitable for regular odd lobed figures

• The centre support method also has disadvantages,- The part may be bowed, or warped- off centre or degraded centre holes will decrease reading quality- the centres themselves can also affect readings

3.6.5.2 - Extrinsic Roundness Testing

• The features of this methods are,1. the reference datum is not points on the object, but a separate precision bearing2. The axis of the part being measured is aligned with the machine bearing axis

θ

h1

A

indicator reading (IR)

θ

h0B

IR change in centre height change in radii+=

∴ h1 h0–( ) A B–( )+ Aθsin

----------- Bθsin

-----------– A B–( )+= =

∴ A B θcsc( )– A B–( )+=

IR∴ A B 1 θcsc+( )–=

where,θ = 1/2 vee block angle

3. A stylus is moved in to contact the part, and then it moves about in a circular path4. The deflection of the stylus is amplified onto a polar plot to be used in evaluation of the

part

• We can measure the out of roundness value as the minimum distance between two concentric circles that enclose/envelope the trace profile. This distance must obviously be divided by the magnification.

• Only roundness deviations are amplified. This creates distortions in the trace.

• The Talyrond machine also uses a low pass electronic filter to reduce the roughness that is shown on the plot. But this still shows the lobing.

• Eccentricity - the talyrond can also be used to detect concentricity. A simple example is a bear-ing race shown below.

• An example of the part discussed above, is now shown in a trace from the Talyrond

stylus

inside dia.

outside dia.

the stylus measures the profile forboth the inside and outside, and thenthese can be compared to determineconcentricityXXXXXXX

3.7 Gage Blocks

• The purpose of gauge blocks are to provide linear dimensions known to within a given toler-ance.

specimen

talyrond

magn

X10000 BC

filter

Inside circumference

Outside circumference

centres of spheres

X

Y

ECCENTRICITY Y X–2

------------- 1magn--------------× C

magn--------------= =

C

• The requirements of gauge blocks are,- the actual size must be known- the faces must be parallel- the surface must have a smooth finish- the surfaces must be flat

• most gauge blocks are made by normal techniques, but the high accuracy is obtained by a pro-cess called lapping (discussed later)

• The materials gauge blocks are made from are selected for,- hardness- temperature stability- corrosion resistance- high quality finish

• type of gauge blocks- rectangular- hoke (square)

• there are four grades of blocks,- reference (AAA) - high tolerance (± 0.00005mm or 0.000002”)- calibration (AA) (tolerance +0.00010mm to -0.00005mm)- inspection (A) (tolerance +0.00015mm to -0.0005mm)- workshop (B) - low tolerance (tolerance +0.00025mm to -0.00015mm)

• Original gauge block sets had lower tolerances and had a total of 91 pieces with values,0.010” to 0.100” in 0.001” steps

• An 81 piece set of gauge block was developed by Johansson(s??) and is capable of covering wider ranges of dimensions.

0.1001” to 0.1009” in 0.0001” steps0.1010” to 0.1490” in 0.0010” steps0.0500” to 0.9500” in 0.0500” steps1.0000”, 2.0000”, 3.0000”, 4.0000” blocks(2 wear blocks at 0.0500”)

• An 83 piece set has also been developed and it has the values (in inches),

• The metric set has 88 gauge blocks (in mm),

<0.001” divisions

0.001” divisions

0.05” divisions

1” divisions

0.1001

0.1010.1110.1210.1310.141

0.0500.550

1.000

two 0.050” wear blocks

0.1002

0.1020.1120.1220.1320.142

0.1000.600

2.000

0.1003

0.1030.1130.1230.1330.143

0.1500.650

3.000

0.1004

0.1040.1140.1240.1340.144

0.2000.700

4.000

0.1005

0.1050.1150.1250.1350.145

0.2500.750

0.1006

0.1060.1160.1260.1360.146

0.3000.800

0.1007

0.1070.1170.1270.1370.147

0.3500.850

0.1008

0.1080.1180.1280.1380.148

0.4000.900

0.1009

0.1090.1190.1290.1390.149

0.4500.950

0.1100.1200.1300.140

0.500

• Most gauge block sets include thin wear blocks that should be included at the ends of a gauge block stack to protect the other gauge blocks.

• How to select gauge blocks for an application

<0.01mm divisions

0.01mm divisions

0.5mm divisions

1cm divisions

1.001

1.011.111.211.311.41

0.55.5

two 2mm wear blocks

1.002

1.021.121.221.321.42

1.06.0

1.003

1.031.131.231.331.43

1.56.5

1.004

1.041.141.241.341.44

2.07.0

1.005

1.051.151.251.351.45

2.57.5

1.006

1.061.161.261.361.46

3.08.0

1.007

1.071.171.271.371.47

3.58.5

1.008

1.081.181.281.381.48

4.09.0

1.009

1.091.191.291.391.49

4.59.5

1.101.201.301.40

5.0

10 20 30 40 50 60 70 80 90

• To assemble a gauge block stack, 1. remove the gauge blocks required from the protective case2. clean of the oil that they have been coated in using a special cleaner. It is acceptable to

handle the blocks, in fact the oil from your hands will help them stick together.3. one at a time, hold the blocks so that the faces just overlap, push the blocks together,

and slide them until the faces overlap together. This will create a vacuum between the blocks that makes them stick together (this process is known as wringing).

4. Make required measurements with the gauge blocks, being careful not to damage the faces

5. take the blocks apart, and apply the protective coating oil, and return them to their box.

• When using gauge blocks, minimze the number used. Each block will have tolerance errors, and as the stack of blocks becomes larger, so does the error.

• Do not leave gauge blocks wrung together for long periods of time.

3.7.1 Manufacturing Gauge Blocks

• The basic sequence of operations is,1. machine to basic size2. harden blocks and stress relieve3. grind to size4. lap (8 blocks at a time) to obtain tight tolerance

• Johansson’s procedure to make the first set (????)

from the 81 piece set above, build a stack that is 2.5744”

2.5744”-0.1004”

2.4740”-0.1000”

2.3740”-0.1240”

2.2500”-0.2500”

2.0000”-2.0000”

0”

therefore the gauge blocks are,0.1004”2 wear blocks @ 0.0500”0.1240”0.2500”2.0000”

1. make a block with a 100mm length2. Make two 50mm blocks3. Determine the actual size of the 50mm blocks by comparing the difference in height

• Lapping is basically,1. a porous pad is charged with a find grit powder. the excess powder is removed.2. the parts to be lapped are secured to a surface plate magnetically (The positions are as

shown below.3. the lapping plate is placed on the block, and moved about, wearing down the blocks.4. the lapping plate is removed, and the blocks are repositioned on the surface plate (as

shown below) and the process is repeated.5. The blocks are removed from the surface plate, and now are generally the same height.

100mm 50mm

50mmA

B0.0004mm

AB0.0002mm

A + B = 100 - 0.0004 = 99.9996mmA - B = -0.0002mm2A + B - B = 99.9996 - 0.0002 = 99.9994mmA = 49.9947mmB = 49.9949mm

• As each stage of lapping is done, the blocks become more even in size, and the lapping plate become more parallel with the lower plate.

1 3 5 7

2 4 6 8

9 11 13 15

10 12 14 16

In the first lap, there are 8 blocks magneticallyattached to the surface plate. The result is thatthe blocks take on a slight angle as shown belowfor a few of the blocks.

A A

section A-A

misaligned by alphalapping plate

1 3 5 7

lower magnetic plate

1 16 9 8

2 15 10 7

5 12 13 4

6 11 14 3

The blocks are rearranged, and the lappingB B

section B-B

misaligned by θlapping plate

1 16 9 8

lower magnetic plate

process begins again. The figure below showshow rearranging the blocks in the mannershown will wear down the peaks.

• Next, knowing the gauge blocks are all very close in size, the stack of 8 blocks are wrung together into one pile, and compared to the master block using a comparator. The difference in heights, divided by eight, is the error in each block.

3.7.2 Compensating for Temperature Variations

• As gauge blocks change temperature, they also change size. The metals chosen for gauge blocks do resist this dimensional change, but will generally undergo some.

• The gauge block sets will carry dimensional readings, as well as rated temperatures. It is advised that all readings be taken at these temperatures, but if this is not possible, then some estimate of the dimensional change can be done.

• Basically this is done by using the difference between specified measurement temperature, and actual measurement temperature. This difference is multiplied by the coefficient of linear ther-mal expansion to give the change in size. This is obviously for small changes in temperature.

• Typical coefficients of linear thermal expansion is,Steel 9.9 - 13.0 * 10-6 in./(in.°C) (typical is 11.5)Bronze 16.7 * 10-6 in./(in.°C)Aluminum 23.0 * 10-6 in./(in.°C)Chrome carbide 8.4 * Tungsten carbide 4 *Cervit (?) -0.2 *

• Note the units are also ppm/°K

3.7.3 Testing For Known Dimensions With Standards

• When a dimension is well known, it can be measured by comparison to standards, using high precision, but limited range comparison instruments.

• Most gage blocks are steel which has a non-trivial coefficient of thermal expansion. But, consid-ering that many parts are made of steel, these blocks will expand at approximately the same rate as the parts, and therefore no temperature compensation is required.

• If the gage blocks are made of the same material as the parts temperture compensation is less significant.

• For high accuracy measurements we want to allow temperatures of gages and parts to stabilize.

• The ISO 1 and ANSI Y14.5 standards speify a typical dimensional ambient temperature as 20°C.

• Materials may vary widely from the listed coefficient of thermal expansion. As a result it is best to take them to 20±0.1°C for high precision measurements, and 20±0.01°C for critical mea-surements.

3.7.4 Odd Topics

• There are also a number of angular gauge blocks for the measurement of angles. The two com-mon sets are,

• The selection of angular gauge blocks is similar to the selection of linear gauge blocks, except that subtration may also be required. (When the blocks are stacked, then angles are simply reversed.

16 piece setdegreesminutessecond

45°, 30°, 15°, 5°, 3°, 1°30’, 20’, 5’, 3’, 1’30”, 20”, 5”, 3”, 1”

13 piece setdegreesminutes

1°, 3°, 9°, 27°, 41°, 90°1’, 3’, 9’, 27’, 0.1’, 0.3’, 0.5’

tool room accuracy ±1 secondlaboratory accuracy ± 0.25 seconds

3.7.5 Limit (GO & NO GO) Gauges

• These gauges are made for simple pass/fail inspection

• Basically there are two separate, or combined gauges for each feature to be measured.

• One gauge must fit inside the feature, and the second must not. In other words the GO gauge must fit inside/outside the feature, the NO GO gauge must not. If the GO gauge does not fit, the tolerance is above the maximum metal tolerance. If the NO GO gauge goes, the feature is below the minimum metal tolerance.

• This method is best suited to unskilled operators testing many parts, although more modern quality methods suggest this procedure should be replaced with Statistical Process Control (SPC).

• This method can also be used for inspection rooms, and limited runs using gauge blocks.

3.7.5.1 - Basic Concepts

• The GO gauge is made near the maximum metal condition. The GO gauge must be able to slip

For the angle 12°37’13”, find the angular gauge block stack using the 16 piece set.

12°37’13”-3”12°37’10”+30”12°37’40”+20”12°38’-30’12°8’-5’12°3’-3’12°+3°15°-15°0

-3”

+30”+20”-30’

-5’-3’

+3°-15°

inside/over the feature without obstruction.

• The NO GO gauge is made near the minimum metal condition. The NO GO gauge must not be able to slip inside/over the feature.

• The terms minimum metal condition, and maximum metal condition are used to describe the tol-erance state of a workpiece. If we assume (at least for now) that all parts are made by removing metal from larger pieces, then we are trying to remove a certain amount. If we are drilling a hole the maximum metal condition will be when the hole is small, and extra metal is ‘left behind’. The minimum metal condition would be when the hole has been overdrilled and as lit-tle metal as possible is left behind. The tolerances often set the acceptable maximum and mini-mum metal conditions. If features are external, the maximum metal condition is their largest size, and minimum metal condition is their smallest size.

• A basic set of shapes these typically deal with are,- plug- ring- taper- snap- threads

1”±0.5”

3”±0.25”

Maximum Material HoleMinimum Material Bottom

Minimum Material HoleMaximum Material Bottom

0.5”

2.75”

1.5”

3.25”

As Specified

• These are good for work tolerances down to about 0.002” (anything less should use compara-tors)

3.7.5.2 - GO & NO GO Gauges Using Gauge Blocks

• Simple GO & NO GO gauges for internal features can be made from gauge blocks.

• The basic procedure is,1. Determine the dimension and tolerance of the feature to be tested.2. Check the temperature of the measurement environment.3. Determine the upper/lower dimensional limits4. If the gauge blocks are not being used at the rated temperature, adjust the dimensions.5. Determine the gauge block stacks for both the GO and NO GO gauges.6. Test.

3.7.5.3 - Taylor’s Theory for Limit Gauge Design

1. GO gauges should check all features for maximum metal condition at one time2. NO GO gauges should check only one feature at a time for minimum metal condition

• The example below should illustrate the two points,

5.000” +.003”-.001”

If the Part is aluminum the coefficient of linear thermal expansion is

Assume the coefficient for the gauge blocks is C = 0.0000061°F in./in.The temperature in the measurement room is 76°F.The rated temperature for the gauge blocks is 64°F.

Given:

The maximum metal dimension is 5.000-0.001 = 4.999” for the GO gauge.The minimum metal dimension is 5.000+0.003 = 5.003” for the NOGO gauge.

Find the needed change in the gauge block size as a result of the temperature difference.

∆L ∆T( ) ∆C( ) L( )=

C = 0.0000127°F in./in.

∆L∴ 76 64–( ) 0.0000127 0.0000061–( ) 5.000in.( )=∆L∴ 0.0005in.=

The new size for the GO gauge is 4.999”+0.0005” = 4.9995”The new size for the NO GO gauge is 5.003”+0.0005” = 5.0035”

Make up the gauge block stacks. (Note when two stacks are taken from the same set,some planning will be required not to use the same block twice.)

3.7.5.4 - Gauge Maker’s Tolerances

• Because gauges have to be manufactured themselves, they must also have tolerances asigned.

The square hole is to be checkedfor height and width

A GO gauge is designed that must fitinside the hole

If either of the dimensions are toosmall, the gauge will not GO, andthus the part will fail inspection.

Option A: The correct method with two separate gauges each measuringone of the dimensions. If either of the gauges goes into the hole, then thepart will fail inspection.

Option B: This INCORRECT method uses two NO GO gauges joined, this

These gauges could be split into twodifferent gauges without any effect onaccuracy, but they would require moretime for measurement.

results in a gauge as pictured below.

It is possible for one of the gauge dimensions to be stuck (passes inspection),while the other dimension is not stuck (fails inspection), but because one of thedimensions is stuck, the gauge does not go, and the part falsely passes inspection.

• The Unilateral System is very popular,1. A general tolerance is applied to both GO & NO GO gauges of 10% of the work toler-

ances2. If work tolerances are above 0.0035”, a wear allowance of 5% of the work tolerance is

added to the GO gauge only3. All gauge tolerances are made to fall within the work tolerance zones. The effect is that

the gauges will always be between the maximum tolerance limits, and no bad parts should be accepted. The only downside is that some good parts will also be rejected.

• An example of the Unilateral Tolerance System applied to GO & NO GO gauges is given below, as applied to a shaft (here we are measuring external features). The gauge shown is a gap and ring gauge.

• We can also look at an example of a hole that is to be measured with GO & NO GO gauges (an

Shaft (the work)A GO & NO GO gauge combination(Note: a good part will fit inside thefirst hole, but not the second)

D1±T1/2

D2±T2/2D3±T3/2

D1, T1 = The shaft diameter, and tolerance specified by the designerD2, T2 = The GO gauge diameter and toleranceD3, T3 = the NO GO gauge diameter and tolerance

T1

D1+T1/2

D1-T1/2

5% T1wear allowance

10% T1 = T2

10% T1 = T3

GO gauge

NO GO gauge

internal feature). The gauge shown is a Plug Gauge.

3.7.6 Sine Bars

• When a reference for a non-square angle is required, a sine bar can be used.

• Basically a sine bar is a bar of known length. When gauge blocks are placed under one end, the sine bar will tilt to a specific angle.

• The figure below shows a sine bar from the side,

GO & NO GO gauge combination.

A Hole (the work)

D3±T3/2D1±T1/2

D1, T1 = The hole diameter, and tolerance specified by the designerD2, T2 = The GO gauge diameter and toleranceD3, T3 = the NO GO gauge diameter and tolerance

T1

D1+T1/2

D1-T1/2

5% T1wear allowance

10% T1 = T3

10% T1 = T2

NO GO gauge

GO gauge

If the smaller shaft (the GO gauge)fits inside the hole the part is good,if the second NO GO shaft fits in, thepart is rejected.

D2±T2/2

• A simple example is - set up a sine bar with an angle of 24°-57’, if the sine bar has 5” centres.

• The sine bar shown above will only allow a single angle to be set, but in some cases we want to

h

h

surface plate

gauge blocks

hardened and ground bar

hardened and ground cylinders

l

l = distance between centres of ground cylinders (typically 5” or 10”)h = height of the gauge blocksθ = the angle of the plate

θ

θ hl---⎝ ⎠⎛ ⎞asin=

24 5760------+⎝ ⎠

⎛ ⎞sin h5.000-------------=

h∴ 2.1091 inches=

continue on and calculate the gauge blocks required......

set two angles, for this a compound sine plate is used.

3.7.6.1 - Sine Bar Limitations

• When using a sine bar, the height setting is limited by the gauge block divisions available (often 0.0001”). This results in an error that may be negligible, or in some cases quite significant.

• A simple example to illustrate this effect is given below for two extreme cases. In the first case the sine bar is near horizontal, in the second case it is near vertical. Assuming a sine bar with 10” centres, and two angles of 1°-30’ and 88°-00’, and that an 84 piece gauge block set is used.

ASIDE:

SENSITIVITY ∆OUT∆IN

----------------= θ hr---⎝ ⎠⎛ ⎞asin=

∆IN∴ ∆h=

∆OUT∴ ∆θ etc= =

Therefore, as the angle approaches 90°, the error increases

• In any of these cases we can see that at larger angles, the sine bar is susceptible to errors in the length of the sine bar, as well as in the height of the gauge blocks.

3.7.7 Comparators

• Accuracies commonly below 1/10 thousandth of an inch

• These instruments try to reduce the friction that is such a problem for the dial indicators

• There are four common principles used to design these instruments,- mechanical- pneumatic- electrical- optical

• comparators have very limited ranges of motion, but very high sensitivities (and therefore accu-racies). As a result the comparators are often calibrated against standards such as gauge blocks.

First, find the gauge block heights required,

h1 10 1 3060------+⎝ ⎠

⎛ ⎞sin 0.2618in.= = h2 10 88( )sin 9.9939in.= =

Next, find the gauge block heights,

******* DO IN CLASS

Given the actual heights, we can recalculate the actual angle of the sine bar,

θA1h110------⎝ ⎠⎛ ⎞asin= = θA2

h210------⎝ ⎠⎛ ⎞asin= =

This shows the errors of the two angles

***Note: the error for the larger angle is also much larger

θerror1 = θerror2 =

• The basic requirements of these instruments are,- rigidity of the design- linear magnification within the operation range- coarse and fine offset adjustments

3.7.7.1 - Mechanical Comparators

• The Johansson Mikrokator used a twisted strip with a pointer attached. as the plunger is depressed, it causes the strip to stretch. As the twisted strip is stretched, it changes the angle of the pointer, and thus the indicated deflection.

• The Sigma Mechanical Comparator uses a partially wrapped band wrapped about a driving drum to turn a pointer needle.

scale (side view)

pointer (moves in and out of page)

twisted stripbell crank lever

plunger

3.7.7.2 - Mechanical and Optical Comparators

• The Eden-Rolt Reed system uses a pointer attached to the end of two reeds. One reed is pushed by a plunger, while the other is fixed. As one reed moves relative to the other, the pointer that they are commonly attached to will deflect.

plunger

knife edge andsaphire bearingblock (knife edgeposition is adjustable)

arm that is essentiallya pivoting beam

pointer

drum

flexible driving band

3.7.7.3 - Optical Comparators

• These devices use a plunger to rotate a mirror. A light beam is reflected off that mirror, and sim-ply by the virtue of distance, the small rotation of the mirror can be converted to a significant translation with little friction.

XXXXXXXXXXXXXXXX

3.7.7.4 - Pneumatic Comparators

• Flow type- the float height is essentially proportional to the air that escapes from the gauge head- master gauges are used to find calibration points on the scales- the input pressure is regulated to allow magnification adjustment

plunger

pointer

fixed reedmoving reed

a pressure bleed off valve allows changes to the base level for offset- The pressure is similar to that shown in the graph below,

• The Soloflex Back Pressure System uses an orifice with the venturi effect to measure air flow. If the gas is not moving, the pressure on both sides of the orifice will be equal. If the flow is mov-ing quickly, the air pressure on the downstream side of the orifice will be at a lower pressure.

clearance at gaugehead to let air escape

flow throughgauge tube

zero adjust

tapered glass tube

float (with vanes toencourage rotation forballistic stability)

scale

input flow from regulator

output to gauge

• A Differential Back Pressure system uses a split flow channel, one flow goes to the gauge head, the other goes to a zero offset valve. A meter measures the difference in pressures, and thus gives the differences in pressure.

3.8 Measuring Aparatus

3.8.1 Reference Planes

• Very flat surfaces are needed when setting up height or angle measurements. This is because the measuring instruments are moved across the surface, and if the height varies, accuracy will suffer.

• Typical plates are made from cast iron, or granite, and are from a few inches per side, and up. A typical plate might be 2 feet by 2 feet.

air flows in orifice air flows togauge head

manometer tube

water tank

dip tube

height differenceproportional topressure

3.8.1.1 - Granite Surface Plates

• The surfaces are finished by rotary lapping machines.

• When done the flatness of the surfaces are inspected for flatness. This is done with auto-collima-tors or laser alignment equipment followed by geometrical analysis oncomputer.

• The general advantages of these plates over cast iron are,- durability

- closer tolerances- lower cost- lower thermal expansion

- quality- non-rusting- burrs do not occur, but chipping does

- ease of use- non-magnetic- less glare- no oil is required, thus dust does not stick- less wringing- inserts are often provided for clamping

3.8.1.2 - Cast Iron Surface Plates

• Whitworth’s three plate method of manufacture is outlined below. This method is particularly desirable because the flatness is self generating.

3.8.2 Squares

• Squares use known angles as a measurement reference. Generally a square is used to measure 90 degree angles (i.e., square corners)

• The basic types are,

Plate APlate B Plate C

Three plates are shown (with exaggerated curves in the surface). Theseplates will be hand scraped in alternate combinations to reduce the surfacecurvature. As the process continues, the plates will become flatter.

Step 1:plates A and B are scraped.

Step 2:Plate ‘C’ is scraped to match ‘A’

Step 3:The process is repeated by scraping ‘B’ and ‘C’.This reveals errors, and reduces error.

Plate A

Plate B

Plate C

Plate A

Plate B

Plate C

***NOTE: Plate ‘A’ is the master plate

Combination Set - This has a sliding blade and is used for layout.- Standard Square - There are three grades: 1. Reference, 2. Inspection, 3. Workshop

- Toolmakers Square- Cylindrical Square

- Direct Reading Type

• The advantages of the Toolmakers, and cylindrical squares are,1. There is a line of contact between the part and the square.2. More resistant to damage.3. Can be checked by rotation.

• Standard Squares can be checked for errors using a reversal test. In this test an angle plate is placed on a reference plane, and a standard square is placed against the angle plate. A dial indi-cator is run along the square from one end to the other, and the drop/rise is measured. The square is now rotated so that the other side is now measured. The drop/rise in height can be used to calculate the angles of both the square, and the angle plate.

blade

beam

90° Both the object to be measured, andthe square are placed on a referenceplane. The square should provide and90° angle to the reference plane.

test A

test B: With the square reversed

drops 0.0007”

rises 0.0003”

Some values of drops, and distances are given above for illustration. The first stepin calculating the angles is to find the angles in the first, and second tests.

2.00”

4.00”

θA0.0007–4.00

-------------------⎝ ⎠⎛ ⎞asin 0.010°

·–= =

θB0.00032.00

----------------⎝ ⎠⎛ ⎞asin 0.009°= =

Based on these values, the angle of the square is,

θSQUARE 90°θA θB–

2------------------⎝ ⎠⎛ ⎞+ 89.99°= =

θANGLE 90°θA θB+

2------------------⎝ ⎠⎛ ⎞+ 90.00°= =

Likewise, the angle of the angle plate is,

3.9 Practice Problems

1. What are measurement standards?ans. Standards are objects of known size, quantity, roughness, etc. These standards are used to cal-

ibrate and verify measuring instruments. As a result, measured values are more accurate.

2. What effect will temperature variation have on precision measurements?ans. Temperature control during measurement is important because as materials are heated they

expand. Each material expands at a different rate. This leads to distortion of parts and measur-ing devices that results in measurement errors.

3. How can a vernier scale provide higher accuracy?ans. A vernier scale uses a second elongated scale to interpolate values on a major scale.

4. What are dimensional tolerances, and what are their primary uses?ans. Dimensional tolerances specify the amount a dimension may vary about a target value. These

are supplied by a designer to ensure the correct function of a device. If these tolerances are controlled the final product will work as planned.

5. Why is an allowance different from a tolerance?ans. A tolerance is the amount a single dimension can vary. An allowance is an intentional differ-

ence between two dimensions to allow for press fits, running fits, etc.

6. What are fits?ans. There are standard for different types of fits (e.g. press fit, running clearance). These specify

the allowance of two parts, so that they may be made separately and then joined (mated) in an assembly.

7. What is the difference between precision and accuracy?ans. Precision suggests a limit of technology, accuracy is the ability to achieve a value consis-

tently. These are often interchanged because we are usually concerned with the accuracy when producing precision parts.

8. If a steel ruler expands 1% because of a temperature change, and we are measuring a 2” length, what will the measured dimension be?

ans. If we assume that only the steel rule expands, and not the steel part, we can calculate,

9. Draw the scales for a vernier micrometer reading 0.3997”.

lbar100 1+------------------

lmeasures100

--------------------= lmeasures100 2( )

101----------------- 1.98in= =

1. Calculate the CLA/Ra value for the wave form below.

2. What is the difference between surface texture and integrity?

For the 0.3997 value

0 1 2 30

5

20

0

10

5

The vernier scale to the left is shown as flattened out. It would typically be found on the back of the micrometer.

ans.

height

meandistance2

1 4 3

1 3 4 2 1

(um)

ans. CLA Ra2 1 1 4 3 0 1 3 4 2 1+ + + + + + + + + +

11------------------------------------------------------------------------------------------------ 2= = =

ans. Surface integrity refers to all of the properties of the surface of a material, while surface tex-ture on refers to the geometry of the surface.

3. Describe roughness, waviness and lay.ans. Roughness is semi or completely random variation in the surface height, these are typically

smaller in size. Waviness is a period or larger variation in surface height. This can be caused by warping or buckling, ripples, etc. Lay refers to a direction of a roughness pattern. For example when cutting with a lath the roughness will be different in the axial and radial directions.

4. What methods are used for measuring surface roughness?ans. Surface roughness is normally measured with an instrument that drags a stylus across the sur-

face (called a profilometer). The movement up and down is measured and used to calculate a roughness value.

5. Describe cutoff.ans. Cutoff is the length of the surface that the stylus of the profilometer is allowed to move over.

6. Two different surfaces may have the same roughness value. Why?ans. A surface roughness value gives an indication of the rms value, but this can come in many

forms. A regular looking roughness pattern may have the same roughness value as a shallower wave form with an occasional deep pit.

7. What will be the effect of a difference between the stylus path and the surface roughness?ans. If the stylus path does not align with the lay of the roughness, then the roughness reading will

be lower (or higher) than expected.

8. When is waviness a desirable and undesirable design feature?ans. Waviness of a surface can be desirable when the surface is to have a rough appearance. If

there is a moving mechanical contact between two surfaces waviness can lead to premature wearing of the parts.

9. Given the figure below indicating stylus height values for a surface roughness measurement, find the Ra and Rq value.

0 4 3 4 5

0 -2 -4 -3 -3 -5 -3 0

10. How are surface roughness and tolerance of the process related?ans. Surface roughness is a good indication of the ability of a process to control final dimensions.

Therefore if the process cannot control the surface roughness, it will be unlikely that the dimensions can also be controlled.

11. How are tolerances related to the size of a feature?ans. The tolerance/surface roughness graph is based on an important concept in manufacturing.

There is a relationship between the scale of a dimension and the scale of a tolerance. In other words, if we make two parts in the same machine, but one is twice the size of the other, then its tolerance must be twice the size. Here we can see the more precise processes are near the bot-tom with a ratio of tolerance to dimension of 1/10000, the highest is about 1/10. Note: polish-ing and lapping are used to finish the production of gage blocks.

1. Show that the vee block method exaggerates errors using a round that is deformed into a trian-gular shape.

1. Select gauge blocks from an 83 piece set to build up a dimension of 3.2265”

2. Use the Unilateral System for a GO/NO-GO gauge design if the calibrated temperature is 72°F and the actual room temperature is 92°F. The shape to be tested is shown below.

abcdefghijkl

43450-2-4-3-3-5-30

Ra4 3 4 5 0 2– 4– 3– 3– 5– 3– 0–+ + + +

10------------------------------------------------------------------------------------------------------- 0.4–= =

Rq42 32 42 52 0 22 42 32 32 52 32 0+ + + + + + + + + + +

10----------------------------------------------------------------------------------------------------------------------------------- 3.71= =

ans.

3. Find the Running Clearance fit category for the hole and shaft shown below.

4. Set up a sine bar (with 5 inches between cylinder centres) to provide an angle of 15°.

a) What height of gauge blocks is required?b) Suggest an appropriate set of gauge blocks from an 81 piece set.c) What is the actual angle of the sine bar?d) If the room temperature is 95°F and the coefficient of expansion is .000001” per inch per °F,

and the gauge blocks are calibrated to 68°F, what is the actual sine bar angle?e) Suggest a new gauge block stack for the conditions in d).

5. If the scale below reads .48, label the bottom vernier scale.

3.008”2.998”

2.000” ± .005”2.005”

3.000”+.008”-.002”

1.995”

.3004” +.0005”-.0004”

.2992” ±.0003”

6. List four different reasons that a material like cheese would not be good for gauge blocks.

7. When using a dial indicator, is parallax or the principle of alignment more significant? Explain your answer.

8. How can you verify that a standard square is 90°?

9. Design a GO/NO-GO gauge for a 5” by 7” square hole with tolerances of ±.1” on each dimen-sion. Show the tolerances and dimensions for the gauges.

0 1 2

0Vernier scale

10. Write the values displayed on the vernier scales below.

1. If the thimble on a micrometer is made larger, does it affect the ‘radial arm’, or the ‘inclined plane’ principle?

12. When a comparator approaches a workpiece from one direction, it will read a different value than when it approaches from the other way. Explain why.

13. One type of fit is for Interchangeable Assemblies (it uses tolerances to ensure that parts can be made separately, but still fit together). What are the two other types of fits that were described in class? Describe why they are different.

14. A square hole has one dimension that will be checked with a GO-NOGO gauge set. The basic dimension is 2.005” ±0.003”. The gauge and hole are used in a room temperature of 105°F, but they should be accurate when at 60°F. The gauge coefficient of linear thermal expansion is 0.000001”, and the coefficient is 0.000002” for the material of the workpiece with the hole.

a) What sizes should the GO and NOGO gauges be? b) Using the gauge block set shown below, list the gauge block stacks required.

15. A square is set up the two ways shown below, and a comparator is run from one end to the other. The resulting measurements result in the rises, or drops indicated. If the comparator is

0 1 2 3

0 1

Value:

0 1 2 3

0 1

Value:

0 1 2 3

0 1

Value:

0 1 2 3

0 1

Value:

run over a total distance of 5” for both measurements, what is the angle of the squares A and B?

16. The hole shaft pair is assembled with an LN fit.

a) Draw the tolerance diagram.b) Determine what the LN fit number is.

17. A sine bar will be used to give an angle of 82°35’

test A

test B

drops 0.008”

rises 0.002”

3.0000”+.0030”-.0000”

3.0070”+.0000”-.0018”

a) If the sine bar has 5” centres, what height will be needed?b) Calculate the gauge block stack for the height in a).c) What is the actual angle of the sine bar?d) If the temperature in the room is 65°F at calibration, and 85°F at use, what change in angle does

the sine bar have (coefficient of linear thermal expansion 0.000001 “/”°F for the sine bar, and 0.0000005 “/”°F for the gauge blocks)?

e) Could the sine bar be used with other instruments to improve accuracy?

18. Draw the number on the vernier scale below if the reading is 1.12

19. Parallax effects are more important than the principle of alignment for flow type pneumatic comparators - TRUE or FALSE

20. Draw GO/NO-GO gauges for the shaft below.

Select the most significant error that occurs when reading a scale that is properly used.a) parallax errors where the scale is not parallel to the work.

0 1 2

R1.250” ±0.003”

0.250” +.006”-.000”

b) change in the length of the scale due to a temperature change of 1°C.c) reading with a scale that has a damaged end.d) rounding off to the nearest division.

If we wanted to measure the diameter of the inside of a tip of a medical syringe (in the range of 0.005”) what would be the best measuring instrument?

a) transfer gaugeb) tool makers microscopec) GO/NOGO gaugesd) mechanical comparator

Which of the following statements is most correct?a) vernier scales are used for linear measurements only.b) micrometer scales are used for linear measurements only.c) micrometer scales make vernier scales more accurate.d) none of the above.

Which of the statements below is not correct?a) the radial arm principle amplifies the rotation of a screw to a larger surface area and radial

travel.b) the inclined plane principle means that a small axial travel for a thread will be amplified to a

much larger radial travelc) the principle of alignment suggests that the dimension to be measured, and the measuring

instrument should be aligned along the same axis.d) all are correct.

Which of the following physical principles is not used as a basis for comparators.a) air pressure.b) air flow.c) the radial arm principle.d) none of the above.

Surface plates are,a) a surface that can be used to measure flatness without other equipment.b) can be used for measuring small angles without other equipment.c) a surface that can be used for measuring large angles without other equipment.d) all of the above.

Sine bars,a) are more accurate near 90°.b) are more accurate near 0°.c) are used with angular gauge blocks.d) none of the above.

Given the diagram below, what will the average interference/clearance be?a) 0.008”

b) 0.020”c) 0.032”d) none of the above

Given an 83 piece set of gauge blocks, how many different stacks 1.1117” in height can be built from the same set? (do not consider wear blocks)

a) 1b) 2 or 3c) 4 or 5d) more than 5

Select the most appropriate statement.a) dial indicators use the inclined plane principle.b) dial indicators are a crude form of comparator.c) the range of the dial indicator is generally less than standard comparators.d) none of the above.

Briefly describe the relationship between tolerance and accuracy. (2%)

Find a gauge block stack that gives a value of 1.2351°. (3%)

a) given a metric gauge block set that is calibration grade (a tolerance of +0.00010mm to -0.00005mm) find the dimension and tolerance of a stack that is 3.2761cm in height. (4%)

b) If the stack found in a) is increased in temperature from the ambient of 23°C to a higher tem-perature of 41°C, what is the new dimension and tolerance? (assume the coefficient of linear thermal expansion is 10-7K-1. (8%)

Suggest a suitable comparator for measuring the diameter of a threaded nut. (3%)

Two blocks are stacked as shown below. In the first test we measure the drop in height (0.005”) from one side to the other (5.000”). Then the block on top is turned 180° (left to right)and the new drop in height (0.015”) is measured over a distance (4.000”). What are the angles of each of the blocks? (8%)

1. From the same set of gauge blocks build up the dimensions 3.2452” and 3.2462”. You must not use the same gauge blocks twice. Use the 83 piece gauge block set.

3.016”3.000”

2.992”

2.984”

1. Design Plug gauges for holes that are 1.500” +0.0025” - 0.000”. (ans. GO limits are 1.50025”/1.5000” dia., NO GO limits are 1.50250”/1.50225” dia.)

2. Design a gap gauge to inspect shafts that are 0.875” +0.000” -0.008”. (ans. GO limits are 0.8746”/0.8738” dia., NO GO limits are 0.8678”/0.8670” dia.)

3. Design GO and NO GO gauges for the hole shown below.

(ans. the three gauges are pictured below)

4. Design GO/NO GO gauges for an equilateral triangular hole that is to have each side 2.025”±0.002”.

1. Determine what height is required to set up a 5” sine bar for an angle of 11°34’. Specify the gauge block stack required.

2. Why are different grades of gauge blocks used?ans. There are different quality levels for gages blocks. The poorest sets are workshop grade and

are more accurate than most machine tools. The best sets are very accurate, and must be kept in tightly controlled conditions. The bast sets are used for calibrating others.

3. How are a ring gauge and a plug gauge different?ans. A plug gage goes into a hole, a ring gage surrounds a dimension.

1.260”1.254”

1.760”1.754”

RHole

GO gauge1.2549”1.2543”

2.3819”2.3813”

NO GO gaugeN

O G

O

2.3900”2.3891”

1.2600”1.2594”

4. CUTTING

4.1 Drilling

• A very common operation that cuts cylindrical holes.

• General type of drill presses in use are,• Sensitive - typically belt driven, and the bit is fed by hand. There are a limited choice of

speeds. A bench top machine• Vertical or Pillar - has a heavy frame to support a wider range of work. The table height

is adjustable, and power speeds and feeds are available.• Radial Arm - For very large and heavy work. The arm is power driven for the height

location. The drilling head traverses the swinging arm. The workpiece remains sta-tionary on the machine base, or work table. The machine spindle is moved to the location required.

• More specialized drill presses are,• Gang Type - several spindles/or stations are mounted on one long table• Multi Spindle - There are many spindles mounted on one head to allow many holes to be

drilled simultaneously (e.g., up to 24)• Numerical Control Type - The machine can automatically change tooling with a turret or

automatic tool changer. Speeds, feeds and table position are controlled using a computer program.

• Counter Bores - Allows the head of cap screws to be sunk beneath a surface

Topics:

Objectives:•

•

• Spot Face - Allows the head of a bolt to be sunk beneath the surface. This is basically a shallow counter bore.

• Counter Sink - Allows counter sunk head screws to be sunk beneath a surface.

• Center Drilling - Allows parts to be mounted between centers, on lathes typically.

82°

• Tapered Holes - these holes can be cut using reamers.

• Threaded Holes - Taps can be used to add threads to holes

• High tolerance finishes for holes can be made with boring or reaming.

4.1.1 Drill Bits

• The twist drill does most of the cutting with the tip of the bit.

• There are flutes to carry the chips up from the cutting edges to the top of the hole where they are cast off.

• Some of the parts of a drill bit are diagramed below as viewed from the cutting tip of the drill,

• Some other features of the drill bit are shown below for a side view of the drill bit,

• Typical parameters for drill bits are,- Material is High Speed Steel

Tip View of the Drill Web

Body Clearance

Margin

point

flute (space)margin

8° to 12° relief

land

angle

Standard Point Angle is 118°

• Harder materials have higher point angles, soft materials have lower point angles.

• The helix results in a positive cutting rake.

• Drill bits are typically ground (by hand) until they are the desired shape. When done grinding, the lips should be the same length and at the same angle, otherwise and oversized hole may be produced.

• Drill sizes are typically measured across the drill points with a micrometer

• Typical drill sizes are,- FRACTIONAL - 1/64” to 3 1/4” dia. in 1/64” steps- NUMBER - #1 = 0.228” dia. to #80 = 0.0135” dia.- LETTER - A = 0.234” dia. to Z = 0.413” dia.- METRIC - 0.4mm dia. to 50mm dia.

DRILL #

123456789101112131415161718192021222324

dia. (in.)

0.22800.22100.21300.20900.20550.20400.20100.19900.19600.19350.19100.18900.18500.18200.18000.17700.17300.16950.16600.16100.15900.15700.15400.1520

DRILL #

252627282930313233343536373839404142434445464748

dia. (in.)

0.14950.14700.14400.14050.13600.12850.12000.11600.11300.11100.11000.10650.10400.10150.09950.09800.09600.09350.0900.08600.08200.08100.08100.0785

DRILL #

495051525354555657585960616263646566676869707172

dia. (in.)

0.07300.07000.06700.06350.05950.05500.05200.04650.04300.04200.04100.04000.03900.03800.03700.03600.03500.03300.03200.03100.02920.02800.02600.0250

DRILL #

73747576777879808182838485868788899091929394959697

dia. (in.)

0.02400.02250.02100.02000.01800.01600.01450.01350.01300.01250.01200.01150.01100.01050.01000.00950.00910.00870.00830.00790.00750.00710.00670.00630.0059

• Some standard drill types are,- Straight Shank - this type is held in a chuck- Taper shank - this type is held in a sleeve, and a machine spindle. A drift may also be

used.

• Some other types of drills used are,- Core drills - a drill with a small helix, and 3 or 4 flutes. This is used for light drilling,

such as opening holes in castings.- High helix - When drilling a deep hole in a soft material these drills are used to help

remove chips- Straight fluted - Used to drill soft metals and plastics. The straight flutes prevent the bit

from digging in.- Centre drills - A drill with a small entry tip, and a widening profile. The result is a hole

that has a conical shape on the outside, that may be used to mount the part between centres, or to act as a guide for a larger drill.

• Typically an allowance of a third of the drill bit diameter is given for the tip of the drill.

• Center Drill Sizes [Krar],

Letter size

ABCDEFGHIJKLM

dia. (in.)

0.2340.2380.2420.2460.2500.2570.2610.2660.2720.2770.2810.2900.295

Letter size

NOPQRSTUVWXYZ

dia. (in.)

0.3020.3160.3230.3320.3390.3480.3580.3680.3770.3860.3970.4040.413

4.1.2 Reamers

• Reamers are a special class of drill. They are used after a hole has been drilled near to final size. The reamers is then used to remove a small quantity of material, and finish the hole with a good surface texture, roundness, and alignment.

• These are often used to provide holes for bearings, parallel and taper dowels, and various fits with a shaft.

• These are typically made of High Speed Steel, or with carbide tips.

• The main body contains many straight and helical flutes. The tip does not contain any cutting edges.

• Various types are,- Parallel Reamer - Straight fluted reamer held in a drill press spindle with a tapered shank.

Regular Size

12345678

Work Dia.(in.)

3/16-5/163/8-1/25/8-3/41-1.52-33-44-5over 6

Countersinkdia. (in.)

3/329/643/1615/6421/643/815/329/16

Drill pointdia. (in.)

3/645/647/641/83/167/321/45/16

Body Size(in.)

1/83/161/45/167/161/25/83/4

tangtapered holder

side cutting blades no teeth on end

Parallel Hand - Straight flutes, but held in a hand tap wrench.- Taper Reamers - has a taper from one end to the other. These can be used in a spindle

(tapered shank), or by hand (for a taper wrench).- Adjustable Reamer - This uses inserted blades.

• Recommended allowances and speeds for reaming [Krar],

4.1.3 Boring

• Boring is used for high quality finished.

• In boring the tool can be rotated, or the work can be rotated.

Hole Size (in.)

1/41/23/411.251.523

Allowance (in.)

0.0100.0150.0180.0200.0220.0250.0300.045

Material

AluminumBrassBronzeCast IronMachine SteelSteel AlloysStainless SteelMagnesium

Speed (ft/min)

130-200130-18050-10050-8050-7030-4040-50170-270

Workpiece

boring bar

cutting point

4.1.4 Taps

• Taps can use for both internal and external threads.

• A typical set of hand taps consists of- #1 Taper- #2 Plug- #3 Bottoming

• There are flutes in the taps to help remove chips, to provide cutting edges, and channels for lubrication.

• There are a number of sets of threads available,- UNC (Unified National Course)- UNF (Unified National Fine)- ACME- Metric

• To create one of these holes, we must first drill a hole that is slightly smaller. For example,

• Some setups associated with taps are,- alignment of the tap in a drill press- use of taping attachments

• NF/NC Thread Tap Drill Sizes [Krar],

5/8 - 11 - UNCoutside diameter = 5/8 = 0.625”11 threads per inch (T.P.I.)Unified National Coarse is the tooth profile

The tap drill size is Outside Diameter - 1/T.P.I. for UNC, UNF, Metricthreads.

Therefore, the Tap Drill Size (TDS) is,

T.D.S. = 0.625” - 1/11” = 17/32”

4.1.5 Process Parameters

• The parameters for drilling are found in almost the same way as for lathes,

Tap Size

#5#6#8#10#121/45/163/87/161/29/165/83/47/811-1/81-1/41-3/81-1/21-3/42

TPI

4032322424201816141312111098776654-1/2

Tap Drill Size

#38#36#29#25#16#7F5/16U27/6431/6417/3221/3249/647/863/641-7/641-7/321-11/321-9/161-25/32

Tap Size

#5#6#8#10#121/45/163/87/161/29/165/83/47/811-1/81-1/41-3/81-1/2

TPI

44403632282824242020181816141412121212

Tap Drill Size

#37#33#29#21#14#31Q25/6429/6433/6437/6411/1613/1615/161-3/641-11/641-19/641-27/64

National Coarse (NC) National Fine (NF)

• Typical high speed drill speeds are, [Krar]

CS rpm C×=

C πD12------- πD

1000------------= =

rpm 12 CS×π D×

------------------- 1000 CS×π D×

-------------------------= =

metricimperial

where,

CS = cutting speed (fpm or m/s) - can be selected from tablesrpm = revolutions per minute of the machine spindleC = circumference of the drill bit (ft. or m)D = diameter of drill bit (in. or mm)

T LF--- L

rpm F×--------------------= =

where,

C T R×=

L = length of cut (in. or mm)F = feed rate (in./rev. or mm/rev.) - found in tablesR = Machine cost ($/min.)

• Consider also the typical feeds for drilling, [Krar]

4.1.6 The mrr For Drilling

• considering the parameters defined in the discussion of speeds and feeds, etc, the mrr is given below,

Drill dia. (in.)

1/161/3/161/45/163/87/161/25/83/47/81

steel casting40 fpm

24451220815610490405350305245205175155

tool steel60 fpm

366518351220915735610525460365305260230

cast iron80 fpm

4890244516301220980815700610490405350305

machine steel100 fpm

611030552035153012201020875765610510435380

brass/aluminum200 fpm

12225611040753055244520351745153012201020875765

Drill dia. (in.)

1/8 or less1/8 to 1/41/4 to 1/31/2 to 11 to 1.5

Feed per Rev. (in.)

0.001 to 0.0020.002 to 0.0040.004 to 0.0070.007 to 0.0150.015 to 0.025

mrr A F rpm×× πD2

4---------- F rpm××= =

where,A = cutting area of the drill bit (a cross section)

4.2 Milling

• Milling machines typically have a rotating cutting tool mounted in a spindle. The work is mounted on a bed, and then either the spindle, or bed is moved. Cutting is done with different parts of the milling tool, as will be described later.

• Some basic types of milling machines include,- Knee and Column

- vertical- horizontal- horizontal with vertical head attachment- universal (table rotates in plan view) and is used for helical milling

- Ram & Turret - Light weight machine tool with slotter on one end of turret. No power feeds.

- Special Purpose - For production usage. Usually more rigid construction.

4.2.1 Types of Milling Operations

• Typical operations re pictured below

table

column

head

spindle

spindle mill arbor mill

arbor

cutter

bed

spindle

mountedon arbor

Face - cut a face flat

Step

Slots

Pockets/contours

Angles

Gear Teeth

4.2.1.1 - Arbor Milling

• The advantages of arbor milling are,- The cutter is held more rigidly on the spindle nose- There is less variation in the arbor torque- The teeth responsible for surface finish do not encounter the hard mill scale- Lower power requirements- Flatter work surface finish.

• For straddle milling- Two similar side and face cutters are mounted on the same arbor, with spacers to separate

them.- This allows two sides of a part to be cut in a single pass.

• For Gang milling- Many dissimilar cutters are mounted on the same arbor at the same time.- When the work is passed under the cutter, multiple cuts are made in a single pass, reduc-

ing alignment problems, and decreasing operation time.

4.2.2 Milling Cutters

• The family milling cutters include a number of basic operations, but in general they will cut with some combination of the end and/or the sides.

• The basic types include,- End Mills - The face and sides at the bottom end of this tool are used for plunge cutting

(two flutes) and side and end cuts (four flute).- Plain - These mills are used to cut with the sides only. They are generally mounted on an

arbor.- Side or Side & Face - - Face - This cutter is held on a spindle nose.- Shell and adapter - - Form - - T-Slot, Dovetail, Woodruff - - Slitting Saws -

4.2.3 Milling Cutting Mechanism

• In milling each tooth on a tool removes part of the stock in the form of a chip.

• There are two types of cutting actions,Peripheral - The teeth at the periphery do the cuttingFace - The teeth on the face of the cutter remove metal.

• The basic interface between tool and work is pictured below. This shows a peripheral milling tooth.

4.2.3.1 - Up-Cut Milling

• The milling method shown above is called up-cut (or conventional) milling. In this case the table is moving towards the cutter, opposing the cutter direction. The basic steps of chip cutting here are,

1. As the tooth makes contact with the surface, the tooth begins to push down. As the tooth continues to turn, it reaches a point at which the pressure has built up to a high level, and the tooth begin to dig in.

2. As the tooth starts to dig, it cuts down, and the metal chip begins to shear off.3. The tooth continues to cut the chip off, until it reaches the surface of the material. At

this point the chip breaks free, and the cutting forces drop to zero.

• Because the cutter does not start to cut when it makes contact, and because the advance moves

rake(α)

rotation of cutter

centre of cutter

chip being cut off

tooth angle

secondaryangle

clearance orprimary angle

table/work feedat constant rate

high points past the cutter contact, the surface has a natural waviness.

• If a cutter has straight flutes, then a torque profile for it might look like,

• The peak arbor torque can be smoothed out by using helical cutting blades, so that there is always a cutter in contact at any one time.

4.2.3.2 - Down-Cut Milling

• When the cutter rotation is in the same direction as the motion of the work being fed, it is referred to a Down-cut, or climb milling.

cutter rotation

arbor torque chip breaks free

tooth is in contact (time)

• When this cutter makes contact with the work, it must begin cutting at the maximum torque. As a result, a back-lash eliminator must be used to take play out of the system.

• This method has advantages,- The cutter forces are directed into the table, which reduces fixture forces, and allows

thinner workpieces- There is less radial pressure on the arbor- Better surface finishes obtained because there is no “dig-in”

4.3 Feeds and Speeds

• Milling is somewhat different than drilling and turning,

rotation of cutter

centre of cutter

table/work feedat constant rate

• Typical speeds are, [Krar]

CS rpm C×=

C πD12------- πD

1000------------= =

rpm 12 CS×π D×

------------------- 1000 CS×π D×

-------------------------= =

metricimperial

where,

CS = cutting speed (fpm or m/s) - can be selected from tablesrpm = revolutions per minute of the machine spindleC = circumference of the cutter (ft. or m)D = diameter of the cutter (in. or mm)

T LF--- L

rpm F×--------------------= =

where,

C T R×=

L = length of cut (in. or mm)R = Machine cost ($/min.)

F fpt #× t rpm×=

where,F = feed rate (in./min.) - this is independent of the spindle rpmfpt = feed per tooth - found in tables#t = number of teeth on a particular tool

• Typical feed per tooth values for HSS cutters, [Krar]

• Typical feed per tooth values for cemented carbide (tipped) cutters, [Krar]

Work Material

machine steeltool steelcast ironbronzealuminum

HSS tool(fpm)

70-10060-7050-8065-120500-1000

carbide tool(fpm)

150-250125-200125-200200-4001000-2000

Material

aluminumbrass/bronze (medium)cast iron (medium)machine steeltool steel (medium)stainless steel

face mill(in.)

0.0220.0140.0130.0120.0100.006

helical mill (in.)

0.0180.0110.0100.0100.0080.005

slot/side mill (in.)

0.0130.0080.0070.0070.0060.004

end mill (in.)

0.0110.0070.0070.0060.0050.003

form cut (in.)

0.0070.0040.0040.0040.0030.002

circular saws (in.)

0.0050.0030.0030.0030.0030.002

Material

aluminumbrass/bronze (medium)cast iron (medium)machine steeltool steel (medium)stainless steel

face mill(in.)

0.0200.0120.0160.0160.0140.010

helical mill (in.)

0.0160.0100.0130.0130.0110.008

slot/side mill (in.)

0.0120.0070.0100.0090.0080.006

end mill (in.)

0.0100.0060.0080.0080.0070.005

form cut (in.)

0.0060.0040.0050.0050.0040.003

circular saws (in.)

0.0050.0030.0040.0040.0040.003

4.3.1 The mrr for Milling


4.3.2 Process Planning for Prismatic Parts

• The basic steps are,1. Cut off the stock slightly larger than required.2. Cut the basic outside diameter to size using a milling machine.3. Lay out the basic features of the parts (in manual setups, this involves coating the surface with

a blue stain, this is then cut and marked).4. Use a bandsaw to rough cut the work.5. On the mill, cut steps, radii, angles, grooves, etc.6. Lay out the holes to be drilled, and then drill them.7. Ream holes as required8. Grind any surfaces that require it. Ground surfaces should generally have 0.010”

4.3.3 Indexing

• It may sometimes become necessary to rotate parts on a milling machine, beyond the rotation

mrr w d F××=

where,w = width of cut

wd

work fed into cutter

d = depth of cut

offered in some beds (e.g. Universal Milling Machine).

• Some of the applications that require this capability are milling of,- polygons,- splines- gears,- cams- spirals

• This method can be done with a dividing head. This is basically a worm gear unit. As the crank is turned, the cylindrical gear will drive the round gear. This will result in an apparatus that takes large motions in the crank, and results in small rotations of the work. When coupled with a scale of some description this becomes very accurate.

• If a worm wheel has 40 teeth, each rotation of the crank will result in a rotation of 40/360 degrees, or 1/40th of a rotation. This means the rotation is 40:1.

****************************** INCLUDE FIGURES OF INDEXING HEAD

• There are two methods of indexing,- Direct Indexing - A notched plate is located so that the crank shaft can be fixed at set

positions (notches).- Simple Indexing - Work is rotated by turning a crank. The crank is finally positioned

using a plate with holes, and a sector arm. (The sector arm is used to count off the divisions on the plates)

• An example of the calculations involved is,

• Another example of indexing considers a rotation of 50 degrees,

Say that we want to mill a polygon on 11 sides (i.e., 1/11th of a circle).

First, we will assume that we have a worm ratio of 40:1, and that we are using aBrown and Sharp #2 plate.

Next, we will calculate the fraction of the indexed plate to be covered,

INDEX 401------⎝ ⎠⎛ ⎞ 1

11------⎝ ⎠⎛ ⎞ 40

11------ 3 7

11------= = =

So, we must turn the crank handle 3 times, plus a bit more. Next we must determinewhich ring of index holes to use, and how many to count ahead by.

We can do this by looking at the remainder (7/11) and taking the denominator (11).Next we look at the counts available for the Brown and Sharp #2 plate(i.e., 21, 23, 27, 29, 31, 33), and from this we will notice that 33 is a multiple of 11.Therefore we can compute the number of divisions required with,

holes 33 711------⎝ ⎠⎛ ⎞ 21= =

Therefore in total, we must advance the crank 3 full rotations, and 21 holes (in thering of 33) to rotate 1/11th of a circle.

First we will calculate the total indexing required,

INDEX 401------⎝ ⎠⎛ ⎞ 50

360---------⎝ ⎠⎛ ⎞ 2000

360------------ 50

9------ 55

9---= = = =

Therefore there are 5 full rotations of the indexing wheel required. Next welook at the list of indexing plates. Assume we are using the CincinnatiStandards Plates, we should look for the ring that has lowest number ofindex holes and is a multiple of 9. This would be 54 on the other side.Therefore we would advance the sector arm by,

holes 54 59---⎝ ⎠⎛ ⎞ 30= =

• Differential indexing - is sometimes required to move plates both forward and backward part of a turn to obtain correct spacing. i.e., output shaft through gear train drives the index plate. XXXXXXXXXXXXXXXXXX

• Helical milling - the machine table is rotated through a helix angle. The machine lead screw drives the dividing head. Work is rotated while the machine table feeds. XXXXXXXXXXXXXXXX

• CAM Milling - requires a milling machine with a rotating vertical head. The dividing head is driven by the machine lead screw.

4.4 Lathes

• Cutting is performed in lathes by rotating the workpiece, and then holding a relatively stationary tool against it. Where the tool touches, the work is cut down in round patterns.

• A lathe is a large machine that rotates the work, and cutting is done with a non-rotating cutting tool. The shapes cut are generally round, or helical. The tool is typically moved parallel to the axis of rotation during cutting.

• Manual lathes have the following major components,

• General classifications used when describing lathes are,- Swing - the largest diameter of work that can be rotated.- Distance Between Centres - the longest length of workpiece- Length of Bed - Related to the Distance Between Centres- Power - The range of speeds and feeds, and the horsepower available

• The critical parameters on the lathe are speed of rotation (speed in RPM) and how far the tool moves across the work for each rotation (feed in IPR).

• Operations on a lathe include,

head stock

tail stock

ways

carriagelead screwlead rod

bed

head stock - this end of the lathe contains the driving motor and gears. Power to rotate the part is delivered from here. This typically has levers that let the speeds and feeds be set.

ways - these are hardened rails that the carriage rides on.tail stock - this can be used to hold the other end of the part.bed - this is a bottom pan on the lathe that catches chips, cutting fluids, etc.carriage - this part of the lathe carries the cutting tool and moves based on the rotation of

the lead screw or rod.lead screw - a large screw with a few threads per inch used for cutting threadslead rod - a rod with a shaft down the side used for driving normal cutting feeds.

axis of part rotation

Turning - produces a smooth and straight outside radius on a part.

Facing - The end of the part is turned to be square.

Threading - The cutting tool is moved quickly cutting threads.

Tapering - the tool is moves so as to cut a taper (cone shape).

Parting/Slotting/Grooving - A tool is moved in/out of the work. shallow cut will leave a formed cut, a deep cut will cut off the unsupported part.

4.4.1 Machine tools

• There are two tool feed mechanism on most lathes. These cause the cutting tool to move when engaged.

- The larger screw (the lead screw) will cause the lathe cutter to advance quickly. This is used for cutting screws, and for moving the tool quickly. Typical feed rates range from about 0.05” to 0.5” per revolution.

- The small screw (the feed rod) will move the cutter slowly forward. This is largely used when doing rough cuts, or finishing operations. Typical feeds with this screw range from 0.001” to 0.010” per revolution.

• On a lathe the axial distance of the tool on the part is set by the carriage. A compound rest is used on a lathe that allows the radial tool position and orientation or the cutting edges.

• Work is held in the lathe with a number of methods.

Drilling/Boring - a cutter or drill bit is pushed into the end to create an internal feature.

carriage

compound rest

3 jaw self centering chuck- 4 jaw independently adjusted chuck- Between centres- Face Plates- Mandrels- Collets- Soft Jaws

4.4.1.1 - Production Machines

• In production there are a variety of cutting machines used to increase throughput by automati-cally feeding stock (through the headstock).

• Other types of turning centers provide multiple operations on a single machine,- Multispindle - Multiple spindles in a single machine allows parallel operations in a sin-

Collet - Stock is fed through from the back of the machine and clamped by the col-let. The collet is then driven to turn the part and cutting tools cut the exposed stock and then the part is cut off, and the stock is advanced for the next part. This is the most basic process.

collet

bar stock

bar stock

colletbushing Sliding Headstock - In these machines the

collet still grips the part, but it slowly moves forward. The cutting tools only move in a radial direction and are posi-tioned near the bushing (it may have bearings also). Keeping the tools near the bushing reduced bending moments and allows slender parts to be cut.

bar stock

Esco - In this type of machine the bar stock is still held and advanced through a col-let, but the tools rotate on a mounting assembly. The tools on the assembly can be moved in radial distances to change the profile of the part. This machine allows coiled stock to be turned and is suited to simpler parts.

gle lathe. Between each operation the spindles are advanced to the next operations.- Rotary Transfer - Large machines where parts are moved to different stations, typically

over ten stations. These may have other tools such as drills mounted.- CNC machines - These computer controlled machines are typically flexible, but a bit

slower. Flexibility is enhanced by a wider variety of operations and multiple tools in the same machine.

- Cam - For high production rates, cams can be made to drive the cutting heads. Their geometry will move the tool in complex patterns.

4.4.2 Toolbits

• A lathe toolbit is shown in the figure below, with a few terms defined.

• In general, as the rake angle increases (positive), the cutting forces are reduced, the surface fin-ish improves, and tool life increases.

• The side edge cutting angle has two effects outlined below,

back rake

end relief

side relief

side rake

end cuttingedge angle

side cuttingedge angle

noseradius

• The End Relief Angle prevents friction on the flank of the tool. The holder for the bit is often angled, and the end relief angle must be larger than the tool holder angle to prevent rubbing.

Work rotates (the top out ofpage in this example)

Tool is moved slowly

1. The angles edge allows a slow build up of cutting forces

Tool is moved slowly

T1T2

d

T1 < T2 for same area

2. Increase in the side rake angle reduces the chip thickness

• The side relief angle has a function similar to the end relief, This angle must exceed the feed helix angle.

• Increasing the nose radius improves the surface finish. But this reaches a limit.

Effective end relief angle

tool holder

the work rotates this way

work circumference= 3.14159 Dia.

feed/rev.

effective side relief

helix angle

4.4.3 Thread Cutting On A Lathe

• Threads are cut using lathes by advancing the cutting tool quickly so that it cuts in a helical band. This helical band is actually a thread. The procedure calls for correct settings of the machine, and also that the helix be restarted at the same location each time.

• The basic procedure is,1. The tool point must be ground so that it has the same angle as the thread to be cut. Typ-

ical angles are 60° for Vee threads, and 29° for ACME threads. A thread gauge can be used to measure thread angles. (also called Centre Gauge or Fish Tail Gauge).

2. The correct gear ratio is required between the machine spindle to the lead screw. This can be determined with the equation,

3. The compound slide is set at half the thread angle. This is so that as multiple passes are made to cut the thread (most threads require a few passes to cut), the tool will be advanced in by the compound slide in such a way that only one face cuts. If both faces were used for cutting there would be a good chance of vibrations and chatter. For example, if a 60° thread is being cut, the compound rest is often set at 29°.

4. The cutting tool is set in the holder perpendicular to the work, and the fishtail gauge is used to check the angle of the point.

5. The In-feed is set to the surface of the part for the first pass (quite often the first pass just scratches the surface to allow visual checking of the settings). On each subse-quent pass the infeed will be set closer.

6. The cross slide is set at the same location for each cutting pass. i.e., the dial setting is zero.

ratio driverdriven-----------------

TPILEADSCREWTPIWORKPIECE-------------------------------------= =

TPILEADSCREW the threads per inch on the lead screw (typically 4)=

where,

TPIWORKPIECE the TPI to be cut on the workpiece=

For example, to cut 20 TPI we calculate,

ratio 420------ 5 4

20------⎝ ⎠⎛ ⎞ 20

100---------= = =

The increase is made to match thenumber of teeth available in ourlathe (these figures depend on specificmachine tools).

7. The In-feed is adjusted on the compound slide for each pass by moving it in a distance. A simple measure of this distance is,

************************** INCLUDE CHASING DIAL FIG 31-13

8. The chasing dial is used to restart the thread cutting in synchronization with what has been cut before. (If this step is not done properly, the notches in a thread might be cut over existing ridges - effectively cutting the entire thread flat to the bottom). The carriage of the lathe in driven across by a split nut. When the split nut is closed over the lead screw, it begins to move. It must be clamped over the lead screw when it is at the right angle. The method for doing this is with the chasing dial. The chasing dial has 16 different locations to engage at. In some cases you can engage the nut at any time, in other cases there are only a few positions to engage at. The basic rules are,

INFEED∆ 0.75TPI----------=

Calculate the following ratio (the previous example is used for illustration),

RTPIWORKPIECETPILEADSCREW------------------------------------- 20

4------ 5

1---= = =

and reduce the denominator to the smallest integer value.

Other examples could be,

184------ 9

2--- 19

4------

712---

4------, , 15

8------

314---

4------, 13

16------= = =

Then looking at the denominator only, select the positions of the chasingdial that the carriage can be engaged at,

124816

close nut at any positionevery 1/8 of dial (e.g., at any line)every 1/4 of dial (e.g., at any line with number)every 1/2 of dial (e.g., 1 and 3, or 2 and 4)every revolution at the same place (e.g., 1)

DENOMINATOR WHEN TO ENGAGE CARRIAGE

4.4.4 Cutting Tapers

• A taper is a conical shape.

• Tapers can be cut with lathes quite easily.

• The typical measures for tapers are shown below,

• Standard tapers include,- Lathe-Spindle Nose - Used for alignment of hole/shaft pairs

type D-1 (tpf = 3”)type L (tps = 3.5”)

- Self Holding Tapers - Used for stabilityTaper shank drills, reamers, sleeves, etc.Use “Morse Tapers” numbered 1 to 7

D d

TL

where,TL = taper lengthD = the large diameterd = the small diameter

In Imperial:

tpf D d–TL

------------- 12×=

where,

D = large diameter (in.)d = small diameter (in.)TL = the taper length (in.)tpf = taper per foot (in./ft.)

In Metric:Specified as a ratio of mm change in diameter to length in mmFor example, a 20cm long bar that changes in diameter from 3cm to 2.2cmwould result in,

D : TL∆ 30 22–( ) : 200 8 : 200 1 : 25= = =

4.4.5 Turning Tapers on Lathes

• There are some common methods for turning tapers on a lathe,- Off-setting the tail stock- Using the compound slide- using a taper turning attachment- using a form tool

• Off-Set Tail Stock - In this method the normal rotating part of the lathe still drives the workpiece (mounted between centres), but the centre at the tailstock is offset towards/away from the cut-ting tool. Then, as the cutting tool passes over, the part is cut in a conical shape. The method for determining the offset distance is described below.

D

d

OL

TL

where,OL = overall lengthTL = taper lengthD = the large taper diameterd = the small taper diameter

OFFSET OLTL-------- D d–( )

2------------------× tpf OL×

24---------------------= =

tpf = taper per foot (in.)OFFSET = the distance to move the tailstock from the zero setting

• The Compound Slide Method - The compound slide is set to travel at half of the taper angle. The tool is then fed across the work by hand, cutting the taper as it goes.

• Taper Turning Attachment - Additional equipment is attached at the rear of the lathe. The cross slide is disconnected from the cross feed nut. The cross slide is then connected to the attach-ment. As the carriage is engaged, and travels along the bed, the attachment will cause the cutter to move in/out to cut the taper.

• Form Tool - This type of tool is specifically designed for one cut, at a certain taper angle. The tool is plunged at one location, and never moved along the lathe slides.

Metal removed tailstockoffset

It is necessary to measure the tailstock offset when using this method. This can bedone with,

1. A scale2. A dial indicator

This method is limited to small tapers over long lengths.

The misalignment of the centres used in this method can cause damage tothe work, and to the centres.

4.4.6 Feeds and Speeds

• If we consider the speed and feed of a lathe,- Spindle Speed is in revolutions per minute- Feed is in inches per revolution

• The Feed Chart is used to select the speeds and feeds of the lathe, and is often attached to the lathe near the setting levers.

• There are some simple (geometric) equations that can be listed,

• Typical cutting speeds for a high speed steel tool are, [Krar]

CS rpm C×=

C πD12------- πD

1000------------= =

rpm 12 CS×π D×

------------------- 1000 CS×π D×

-------------------------= =

metricimperial

where,

CS = cutting speed (fpm or m/s) - can be selected from tablesrpm = revolutions per minute of the machine spindleC = circumference of the workpiece (ft. or m)D = diameter of workpiece (in. or mm)

T Lrpm F×--------------------=

where,

C T R×=

L = length of cut (in. or mm)F = feed rate (in./rev. or mm/rev.) - found in tablesR = Machine cost ($/min.)

• Typical feeds when using a high speed steel tool are, [Krar]

4.4.7 The mrr for Turning


Material


Rough Cut(fpm)

90706090200

Finish Cut(fpm)

1009080100300

Thread cut(fpm)

3530252560

Material


Rough Cut(in./rev.)

0.010-0.0200.010-0.0200.015-0.0250.015-0.0250.015-0.030

Finish Cut(in./rev.)

0.003-0.0100.003-0.0100.005-0.0120.003-0.0100.005-0.010

4.4.8 Process Planning for Turning

• The general steps when process planning for turning external parts are,1. Rough cuts all diameters to within 1/32” starting with the largest diameters first.2. Rough cut all shoulders and steps to within 1/32”3. Do special operations such as knurling and grooving4. Cool the workpiece to get it close to the final dimension.5. Finish turn the diameters, then the shoulders and steps6. Deburr if necessary

• If the part is to be mounted between centres, plan should precede by,1. cut stock that is 1/8” larger than required.2. Put the work in the lathe, in a chuck, and face and centre drill the end.3. reverse the pice in the chuck and face the piece to size, and centre drill.4. Mount the work between centres

• For work to be mounted in a chuck, (implies internal features),1. cut the stock 1/8” wider in diameter, and 1/2” longer.2. Mount the work in the chuck with 5/16” to 3/8” inside.3. Use a facing operation (lightly) to square the end.4. Rough cut the external diameters, from the largest to the smallest.5. Drill out the centre of the work using a drill chuck mounted in the tailstocks spindle. Start with

a centre drill, and increase drill sizes to increase the hole.6. Mount a boring tool to cut the internal diameter to close to the final diameter.

mrr πD2

4---------- πd2

4---------–⎝ ⎠

⎛ ⎞ F rpm××=

where,D = diameter of workpiece before cutting

D d

d = diameter of workpiece after cutting

7. Cut any special feature now.8. Do finish cuts on outside and inside.9. Reverse the part in the chuck and face off the material to size. Protect the work by placing a

piece of soft metal between it and the chuck.

4.5 Cutting Speeds, Feeds, Tools, and Times

• Cutting is a balance between a number of factors,- cutting slowly will add costly time to manufacturing operations.- cutting faster will lead to decreased tool life, and extra time will be required to repair

tools.

• Some reasonable speeds and feeds for a single cutting point tool are given below [Krar],

MATERIAL

Aluminum

Brass, Bronze

cast iron (medium)

machine steel

tool steel

stainless steel

titanium alloys

DEPTH (in.)

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

0.005-0.0150.020-0.0900.100-0.2000.300-0.700

FEED PER REV. (ipr)

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

0.002-0.0050.005-0.0150.015-0.0300.030-0.090

CUTTING SPEED (fpm)

700-1000450-700300-450100-200

700-800600-700500-600200-400

350-450250-350200-25075-150

700-1000550-700400-550150-300

500-750400-500300-400100-300

375-500300-375250-30075-175

300-400200-300175-20050-125

4.6 Cutting Power

• There are a number of reasons for wanting to calculate the power consumed in cutting. These numbers can tell us how fast we can cut, or how large the motor on a machine must be.

• Having both the forces and velocities found with the Merchant for Circle, we are able to calcu-late the power,

• We can relate the energy used in cutting to the mrr.

PcFcVc33000---------------=

PsFsVs

33000---------------=

PfF Vf×33000---------------=

All have units of Horsepower (i.e., 1/33000)

where,Pc = the total cutting powerPs = the shearing power requiredPf = the friction losses

Energy Consumed

Metal Removal Rate

where,A0 = Area of Cut

Pc Fc Vc×=

Q A0 Vc×=

***Note: both Wc and Q are proportional to Vc

From these basic relationships we can a simple relationship that is the ratio

psPcQ-----

Fc Vc×A0 Vc×------------------

FcA0------= = =

between the energy consumed, and the volume of metal removed,

You will notice that the result is a force over an area, which is a pressure. Asa result Ps will be called the Specific Cutting Pressure.

• The cutting force will vary, thus changing Ps, as the cutting velocities are changed.

• The effects of rake angle on cutting are shown in the graph below, [REF ******]

ps

Vc

This curve turns downward for two reasons,

1. The tool experiences edge forces that are more significant at lower cutting speeds.2. As the velocity increases, the temperature increases, and less energy is required

to shear the metal.

• The horsepower required for cutting can be found using empirical methods,

15°

10°

5°0°-5°

-10°

-15°

500

400

300

200

100

0

Cutting Force (F

c ) (lb.)

Rake Angle

fpm150

200300400500600

Carbide ToolFeed = 0.010”/rev.

The Effect of Rake Angle on Cutting Force

• If we consider the implications these formulas have when cutting on a lathe, we would be able to develop the following equations,

Unit horse power (HPu) is the amount of power to remove a volume of metalin a period of time.

HPu power to cut 1 cubic inch per minute - found in tables=

HPg Q HPu× Gross Horsepower= =

Material

Carbon steels

Leaded steelsCast irons

Stainless steelsAluminum alloysMagnesium alloysCopper

BHN

150-200200-250250-350150-175125-190190-250135-27550-10040-90125-140100-150Copper alloys

HPu (HP/(in3/min.)

1.01.41.60.70.51.61.50.30.20.70.7

Average Unit Horsepower Values of Energy Per Unit Volume [REF]

Q f d V 12×××=

where,f = feedd = depth or cutV = velocity

HPcFc Vc×33000

------------------ HPu Q c××= =

where,c = a feed factor from tables

Feed

(ips or ipr)

0.0020.0050.0080.0120.0200.0300.0400.050

(mm/rev or mm/stroke)

0.050.120.200.300.500.751.001.25

Factor

1.41.21.051.00.90.800.800.75

Horsepower Feed Correction Factors for Turning, Planning and Shaping [RE

4.7 Examples

4.8 Summary


4.10 Problems


We can also consider the efficiency of the machine tool,

HPgHPcem

----------=

where,em the machine tool efficiency factor [0, 1]=

HPM HPI HPg+ HPIHPcem

----------+= =

from this we can determine the minimum machine tool horsepower required,

where,HPM The minimum machine tool horse power required=

HPI The idle horsepower consumed by the machine tool=

4.12 Practice Problems

1. What would happen if a drill without flutes was used?

2. If we want a hole with a 1/2-14-UNC thread, what size of tap drill should be used?

3. What type of drill press is suitable for drilling holes in car engine blocks? Justify your answer.

4. Which of these statements is not correct?a) work is not moved on a radial arm drill press.b) automatic feeds are available on sensitive drill presses.c) multispindle drill presses always drill multiple holes at once.d) all of the above.

5. Which of these statements is correct?a) a margin of a drill bit does most of the cutting.b) the relief angle on the tip of the drill bit makes it not a conical shape.c) a large drill bit point angle is useful for cutting soft materials.d) none of the above.

6. Which of the following statements is not correct?a) core drills have 3 or 4 flutes.b) high helix drills help in chip removal.c) straight fluted drills are used for sheet metal.d) centre drills are for long holes, such as gun barrels.

7. Which of the following is not a typical drill press operation?a) counter boring.b) spot facing.c) counter sinking.d) none of the above.

8. Which of the statements is most correct?a) reamers are used to finish holes with accuracies not possible when a normal drill is used.b) adjustable taps will cut a wide variety of threaded holes.c) taps and reamers can both be used without a machine tool.d) none of the above.

9. Given a hole that is to be drilled then reamed to 3.000”, develop a process plan including speeds and feeds.

10. We want to drill a hole that is 2.369” in diameter. If we know that the accuracy the shop can provide for drilling is +0.030” to -0.010”,a) what is an appropriate fractional drill size to use?b) what operation might follow?

11. Calculate the machine tool spindle speeds for the following,a) drilling with a 19/32” high speed steel bit in mild steel. The CS is 70 ft./min.

12. We are to drill 6 holes in a 2” thick mild steel plate. The plate is held in a jig. We are using a 63/64” high speed steel drill, and the suggested parameters are CS = 80 ft./min. with a feed of 0.004”/rev. After drilling each hole is to be finished with a 1.0” diameter reamer. If the sug-gested parameters for the reamer are CS = 80 ft./min. with a feed of 0.010”/rev.,

a) calculate the time to do all of the operations (and make allowances for drill point travel)b) find the cost to produce 500 parts when each part needs 3 minutes for setup (no opera-

tion), labor rates are $25/hr., and overhead is $25/hr.

13. Which of these statements is not correct?a) work is not moved on a radial arm drill press.b) automatic feeds are available on sensitive drill presses.c) multispindle drill presses must always drill multiple holes at once.d) all of the above.

ans. B

14. Which of these statements is correct?a) a margin of a drill bit does most of the cutting.b) the relief angle on the tip of the drill bit makes it a conical shape.c) a large drill bit point angle is useful for cutting soft materials.d) none of the above.

ans. D

15. Which of the following statements is not correct?a) core drills have a hollow center to remove chips.b) high helix drills help in chip removal.c) straight fluted drills are used for sheet metal.d) centre drills are for long holes, such as gun barrels.

ans. A or D

16. Which of the following is not a typical drill press operation?a) counter boring.b) spot facing.c) counter sinking.d) none of the above.

ans. D

17. Which of the statements is most correct?a) reamers are used to finish holes with accuracies not possible when a normal drill is used.

b) adjustable taps will cut a wide variety of threaded holes.c) taps and reamers can both be used without a machine tool.d) none of the above.

ans. A

18. What are functions of the following parts of a drill bit. a) body, b) web, c) point, d) tang, e) margin, f) flutes, g) body clearance.

19. What are the purposes of the following drill points. a) conventional, b) flat, c) long angle.

20. What applications are the following drill bits well suited to? a) high helix, b) straight flute, c) gun, d) hard steel, e) core, f) oil hole.

21. What will happen if a drill bit has unequal angles on the cutting edges/lips? What if the edges are not of equal length?

22. Why should most holes be started with a center drill?

23. What are the disadvantages of a thick web found on some drills?

24. What is the purpose of pilot holes?

25. What is the main difference between a) threading operations and tapping operations? b) bor-ing and reaming?

26. List 5 ways work can be held in a lathe.

27. Can peripheral and face milling be done with the same cutter? How common is this?

28. Describe the steps in cutting a 3/8-12-UNC taped hole.(ans. center drill, drill 1/4”, drill .292”, starting tap, finishing tap)

29. a) Explain the cutting mechanism of a drill bit, and b) suggest the features of a drill bit for cut-ting a thin piece of sheet metal.

1. A 2” diameter milling cutter with 8 teeth has been selected. What is the table feed if we are milling at 80 ft./min. with a tooth load of 0.004”/tooth?

2. Calculate the machine tool spindle speeds for the following,a) milling with a 3/4” high speed steel cutter in tool steel work. The CS is 60 ft./min.b) milling with a 150mm diameter tungsten carbide tipped face cutter in stainless steel

work. The CS is 65 m/min.

3. You are given a block of aluminum (5” by 5” by 5”) and you must mill off a 1/16” layer. Using the tables for speeds and feeds, and using the other details provided below, determine a cost for the operation.

• Milling cutter- high speed steel- diameter 2”- 10 teeth with a tooth load of 0.004” per tooth- cost for the machine is $20.00 per hours

4. Which of the following statements is true for milling?a) milling cutters can cut with the face and peripheral teeth.b) the cutting edge moves opposite to the direction used in lathes.c) indexing is used to cut rounded surfaces.d) none of the above.

5. What are the advantages of upcut and downcut milling?

ans.

rpm 12 CS×π D×

------------------- 12 750( )π2

-------------------- 1432= = =

T LF--- 21in

57ipm---------------- 0.37min= = =

C T R× 0.3760

----------⎝ ⎠⎛ ⎞ 20( ) 0.12$= = =

F fpt #× t rpm× 0.004( ) 10( ) 1432( ) 57ipm= = =

This cut is more than a finishing cut.

R 20$/hr=

We can assume this is a light rough cut or heavy finishing cut. The cutter type will be assumed to be a face mill. Because the part width is 5” and the cutter is 2” we will need three passes to cut the part.

D 2in= #t 10= fpt 0.004= CS 500 1000+2

--------------------------- 750fpm= =

Lpass 5in D+= L 3Lpass 3 5 2+( ) 21in= = =

Upcut - lower tool impact forcesUpcut - loose work is saferDowncut - pushes work into tableDowncut - better surface finish

ans.

6. Which of the following statements is true for milling?a) milling cutters can cut with the face and not the peripheral teeth.b) the cutting edge moves opposite to the direction used in lathes.c) indexing is only used to cut rounded surfaces.d) none of the above.

ans. D

7. Given a 3” dia. 8 tooth fly cutter, with carbide cutting points, and a steel work piece, recom-mend, a) RPM, b) feed.

8. Given a 6” dia. high speed steel arbor mill with 10 teeth, that will be cutting cast iron work, rec-ommend, a) RPM, feed.

9. Calculate the indexing required when would be cutting a gear with 36 teeth? Use one of the Brown and Sharp indexing plates.

10. Determine the angular indexing required (on Brown and Sharp, and Cincinatti Standard plates) if we want an angle of 23°30’.

1. Given that a tapered piece is to be made with the tailstock offset method, determine the taper per foot, and offset required if, you are starting with a bar of stock that is 8” long, and 1.125” in diameter, and the final taper is to be 6” long and 1” at the small end.

(ans. tpf = 0.25”, offset = 0.0833”)

2. Given the 1/2-12 UNC thread that is to be cut on the lathe,

a) What should the gear ratio between the machine spindle and the lead screw be if the lead screw is 5 t.p.i.?

b) What should the in-feed be for each pass?

3. Which of the following statements about lathe toolbits is correct?a) a small nose radius will result in a smoother surface.b) small relief angles will always increase friction.c) large rake angles will decrease cutting forces.d) none of the above.

(ans. c)

4. Which of the statements about lathes below is most correct?a) jawed chucks hold only standard sizes of pieces.b) mandrels hold work pieces from the outside.c) the chasing dial is used for measuring fine cuts.d) lead screws and feed rods are lathe parts

(ans. d)

5. When turning between centres a dog is required; what is a dog in this context?

(ans. it holds the work piece so that it can be driven with a face plate mounted on the lathe spin-dle)

6. A centre gauge (fish tail gauge) is employed in thread cutting. Suggest two uses for the gauge.

(ans. aligning a cutting tool for threads, )

7. Given an external 9/16-12-UNC thread, determined which tools would be used.

(ans. a turning tool to turn the outside diameter of 9/16”+1/12”, UNC tool to turn thread)

8. If we are rough cutting a 5” diameter bar of bronze on a lathe with a HSS tool,a) what speed and feed should be used?b) if the cut is 12” long, and will be made in two passes, how long will the operation take?c) if the setup time is 5 minutes, and the machine rate is $50/hr., what will the cost of the oper-ation be?

9. List the basic steps for setting up a lathe to cut a thread on a bar of stock, assume the stock is mounted between centres already.

(ans. see thread cutting section)

10. If a taper of 1mm in 10mm is to be cut, what will the offset distance be for a 10cm part?

Using the lookup tables in the notes we pick a surface cutting speed and feed.

CS 90 ftmin---------=

Rough Finish

f 0.015 0.025+2

--------------------------------- inrev-------- 0.02 in

rev--------= =

CS 100 ftmin---------=

f 0.003 0.010+2

--------------------------------- inrev-------- 0.0065 in

rev--------= =

D 5in=

T Lrpm f×------------------ 17.39min= =

C T R× 2 8.82( ) 5+60

----------------------------⎝ ⎠⎛ ⎞ 50 18.71$= = =

rpm 12CSπD

------------- 68= = rpm 12CSπD

------------- 76= =

L 12in= R 50$/hr=

T Lrpm f×------------------ 24.29min= =

C T R× 2 24.29( ) 5+60

-------------------------------⎝ ⎠⎛ ⎞ 50 44.65$= = =

11. Calculate the machine tool spindle speeds for the following,a) turning on a lathe with a high speed steel tool in mild steel work with a diameter of

2.75”. The cutting speeds is 100 ft./min.

12. We have been given a mild steel bar that is to be turned on a lathe. It has a diameter of 14” and a length of 28”. We have been asked to make two rough passes, and one finishing pass. The tool we have selected is Carbide. When doing rough cuts we use a feed of 0.007”/rev., and for finishing cuts we use a feed of 0.004”/rev. How long will this operation take?

13. Which of the following statements about lathe toolbits is correct?a) a small nose radius will result in a smoother surface.b) small relief angles will always increase friction.c) large rake angles will decrease cutting forces.d) none of the above.

ans. C or D

14. Which of the statements about lathes below is most correct?a) jawed chucks hold only standard sizes of pieces.b) collets hold work pieces from the outside.c) the chasing dial is used for measuring fine cuts.d) lead screws and indexers are lathe parts

ans. B

15. Given the non-standard 3/8-19 UNC thread that is to be cut on the lathe,

a) What should the gear ratio between the machine spindle and the lead screw be if the lead screw is 4 t.p.i.?

b) What should the in-feed be for each pass?

ans. a) 4/19, b) 0.039

16. Develop a rough process plan for the part below by clearly listing operation steps in the cor-rect sequence. Feeds, speeds, times and costs are not needed at this time.

ans.

17. The aluminum component below is to be turned on a lathe using a HSS tool. Develop a pro-cess plan, including offset for the taper, speeds, feeds, etc. Put the process plan in a list similar to the format shown. Assume a cost of $45.00/hr. for the lathe, and $25.00/hr. for all other

1.00”1.25”

3.75”

1.25”

0.75” 3” tpf1/8” slot X 1/16” below taper

1/16” slot x 1/16”1.75”

OperationNumber

0010

OperationDescription

Cut off 2” dia. Stock to 4”

0020 Mount in lathe chuck, face and centre drill

OperationNumber

0010

OperationDescription (Note excess details given for beginners)

Cut off 2” dia. Stock to 4”0020003000400050006000700080009001000110

Mount in lathe chuck, face and center drill to 3.75” lengthMount between centersTurn entire length to 1.75” dia.Cut slot with form tool 1” from end to 1/16” depthTurn one end down to 1.25”dia. for 15/16”Reverse part in centers (cover finished end with soft metal)Cut 1/8” by 1/16” slotTurn taper with taper turning attachmentReturn tailstock to normal positionDeburr and inspect

*Note: the implied tolerances +/- 0.005 would not require cooling

pieces of equipment. State all assumption clearly, and justify numbers in the process plan with calculations or references.

18. On a lathe toolbit what are the functions of, a) the side relief angle, b) end relief angle, c) back rake, d) side rake angle, nose radius.

19. What applications are large positive rake angles for? negative rake angles?

20. What is the difference between end and face milling?

21. What RPM should be used to rough cut a cast iron piece with a 3” dia. with a high speed steel tool. What RPM should be used for a similar workpiece of plain carbon steel? What RPM should be used for the two materials if finishing cuts are being made?

22. Calculate the time required to machine a 2” dia. copper rod that is to be turned for a length of 10”.

23. What are rough and finish turning operations used for?

24. What are two methods for cutting stepped shoulders on a lathe?

25. Explain the difference between self holding and steep tapers using the coefficient of friction.

26. Find the tpf and tailstock offset for tapers on the following work.

1.00”1.25”

3.75”

1.25”

0.75” 3” tpf1/8” slot X 1/16” below taper

1/16” slot x 1/16”1.75”

OperationNumber

0010

OperationDescription

Cut off Stock to 4”

Time

6 min.

Cost

$5.00

0020 Mount in lathe chuck, face and centre drill 12 min. $9.00

a) D=1.5”, d=1.25”, TL=4”, OL=8”

27. convert a metric taper of 1:50 to a tpf. Convert a 1”tpf to metric.

28. Define the terms, fit, tolerance, allowance, limits, clearance, press fit, precision.

29. For a 1”-8-NC thread find the minimum and maximum diameters and minimum width of the toolbit point.

30. Describe the differences in speeds, feeds and depths of cuts for roughing and finishing cuts.

31. What types of chips are desirable when setting up automated cutting processes?

32. Compare the time to cut a work piece using a high speed steel tool and a carbide tool. The 4” dia. aluminum work is to be rough turned over a length of 14”.

33. What operations can be performed on a lathe?

34. How are the parameters different for a lathe when turning, as opposed to finishing?

35. A taper is to be cut on the aluminum part below. Indicate how far the tailstock should be offset and the speed and feed settings for the lathe.

(ans. offset=0.4”, feed 0.005-0.010”, speed 760RPM)

1. An orthogonal cut is made with a carbide tool having a 15° positive rake angle. The various parameters were noted,

- the cut width was 0.25”- the feed was set at 0.0125”- the chip thickness was measured to be 0.0375”- the cutting speed was 250 ft./min.- the forces measured were Fc = 375 lb. and Ft = 125 lb.

1.50”

3.00” 5.00”

1.00”

a) Use Merchant’s Circle to scale, and the velocity diagramb) From the Merchant Circle diagram find the shear angle (φ), friction force (F), friction

normal force (N), and shear force (Fs).c) From the Velocity diagram find the friction velocity (Vf).d) Calculate values for the coefficient of friction (mu) and the metal removal rate.e) Calculate values, and compare the results for the results found in a), b) and c).

(ans: F = 218lb., N = 330lb., φ= 19.37°, Fs = 312 lb., µ= 0.948, Vc = 250 ft./min., Vf = 83.5 ft./min. Q = 9.375 in3/min.)

2. The cutting forces for a lathe are listed below,• work RPM = 125• feed/rev = 0.005”• chip thickness = 0.0123”• rake angle of tool = 14°• Ft = 150 lb, Fc = 245 lb• work diameter = 8”

a) Find the horsepower consumed in cutting, shearing and friction.b) Find a maximum lathe horsepower, assuming the machine efficiency is 95% and it

requires 1/8 idle horsepower.c) Based on the cutting horsepower, what material(s) might we be cutting?

3. What roles do rake and relief angles play in cutting tools?

ans. the rake angle will change the basic cutting parameters. A positive rake (sharp tool) will give lower cutting forces, but less edge strength. A negative or neutral rake will give higher cutting forces, but more strength. The relief angle provide a gap behind the cutting edge so that the tool does not rub the work.

4. Which of these statement is the most correct?a) a continuous chip with built up edge may result when we try to cut too much metal.b) a continuous chip will result when cutting very brittle work materials.c) a discontinuous chip will result when we use fine feeds and speeds.d) none of the above.

ans. a

5. One of the assumptions behind orthogonal cutting is,a) that the rake angle is positive.b) that the tool is only cutting with one edge and one point.c) the shear plane is a function of before and after chip thicknesses.d) none of the above.

ans. b

6. Which of these statements is correct?a) the cutting pressure drops as cutting velocity increases.b) power required drops as metal temperature and cutting velocity increase.c) we can use the quantity of metal removed by itself to estimate the required horsepower of a machine tool.d) all of the above.

ans. a

7. A lathe toolbit with a rake angle of 20° is cutting a section of pipe with an inner diameter of 6” and an outer diameter of 6.25”. The cut has a depth of 0.010” and the chip has a thickness of 0.020”. If the lathe is turning at 200 rpm, and the measured cutting forces are Fc = 300 lb, and Ft = 125lb,a) what assumption must you make.b) find the following values using a graphical or numerical solution: (Marks are only awarded for correct answers) Fs, FN, F, N, τ, φ, µ, Vc, Vf, Vs.c) what is the minimum horsepower required for the machine?d) given that the tube is aluminum, use another method to find the required horsepower.

8. Calculate the machine tool spindle speeds for the following:a) Milling with a tungsten carbide tipped face cutter on a stainless steel work piece. C.S. =

65 m/min., cutter dia. = 150mm.b) Drilling with a High Speed Steel drill in Machine Steel work, with C.S. = 70 ft./min.,

and a drill diameter of 19/32”c) Turning on a lathe with a High Speed Steel tool in a mild steel work piece. Surface cut-

ting speed = 100 ft./min., and a workpiece diameter of 2.75”d) Milling with a High Speed Steel cutter in tool steel work with a cutter speed of 60 ft./

min., and a cutter diameter of 3/4”.

9. Short answer,

α 20deg=ans.

Fc 300lbs= Ft 125lbs=

VcRPMπD

12--------------------- 200π 6.125( )

12------------------------------- 321fpm= = =

µ FN---- 0.92= =

F Ft αcos Fc αsin+ 220lbs= =

rct1t2---- 0.5= = φ

rc αcos1 rc αsin–-------------------------⎝ ⎠⎛ ⎞atan 0.56676719( )atan 29.5deg= = =

D 6 6.25+2

-------------------⎝ ⎠⎛ ⎞ 6.125in= =

t1 d 0.010in= =

RPM 200= t2 0.020in=

a) reasonable assumptions are that we are performing orthogonal cutting. This means that we are cutting fully through the wall of the tube. We also want to assume that the effects of the different cutting speeds from the inside to the outside of the tube are negligible.

N Fc αcos Ft αsin– 239lbs= =

Fs Fc φcos Ft φsin– 200lbs= = Fn Ft φcos Fc φsin+ 257lbs= =

τ 0.92( )atan 42.6deg= =

b)

Vs321 20cos

29.5 20–( )cos------------------------------------ 306fpm= = Vf

321 29.5sin29.5 20–( )cos

------------------------------------ 160fpm= =

c)

HPc HPuQ 0.3 9.63( ) 2.9HP= = =

HPcFcVc33000--------------- 300 321( )

33000----------------------- 2.9HP= = =

Q d 6.25 6–( )Vc12 0.010 0.25( )321 12( ) 9.63 in3

min---------= = =

For an efficient machine with no idle horsepower.

d)

a) Why are ceramics normally provided as inserts for tools, and not as entire tools?b) List the important properties of cutting tool materials and explain why each is important.

10. A turning cut was made in a magnesium workpiece with a feed of 0.050ipr. The cutting speed was 300 fpm, and the cutting force was measured as 200lbs. The lathe is 95% efficient and has an idle horsepower of 0.1HP. Using all of the provided information estimate the horsepower required for the cut.

11. Develop an expression that is the ratio friction power over cutting power using the equations for orthogonal cutting power. Simplify the expression to be in terms of measured values (rake angle, Fc, Ft, and chip thicknesses).

12. A new lathe tool is to be used on cast iron work with a 6” diameter to make a 5” long rough cut in 3 passes. The operation conditions listed below were provided by the supplier or

a) Ceramics are brittle materials and cannot provide the structural strength required for a tool.

hardness at high temperatures - this provides longer life of the cutting tool and allows higher cutting speeds.

toughness - to provide the structural strength needed to resist impacts and cutting forces

wear resistance - to prolong usage before replacementdoesn’t chemically react - another wear factorformable/manufacturable - can be manufactured in a useful geometry

b)

ans.

HPcVcFc33000---------------

300 ftmin--------- 200lbs( )

33000 ftlbsminHP------------------

----------------------------------------- 1.82HP= = =

ans.

HPM HPIHPcc

e-------------+ 0.1HP 1.82HP( ) 0.75( )

0.95---------------------------------------+ 1.54HP= = =

ans.R

WfWc-------

FVf33000---------------⎝ ⎠⎛ ⎞

FcVc33000---------------⎝ ⎠⎛ ⎞--------------------

FVfFcVc------------

Ft αcos Fc αsin+( )Vc φsin

φ α–( )cos--------------------------⎝ ⎠⎛ ⎞

FcVc--------------------------------------------------------------------------------= = = =

RFt αcos Fc αsin+( ) φsin( )

Fc φ α–( )cos---------------------------------------------------------------

Ft αcos Fc αsin+( )Fc

----------------------------------------------- φsinφ α–( )cos

--------------------------⎝ ⎠⎛ ⎞= =

RFt αcos Fc αsin+( )

Fc-----------------------------------------------

t1t2----⎝ ⎠⎛ ⎞=

assumed. Calculate the parameters a) to e) as requested.Cutting Speed = 300 fpmFeed Rate = 0.008 iprDepth of Cut = 0.125”Idle Horse Power = 0.25Machine Efficiency = 0.90

a) Spindle RPMb) Time to make the cut (min.)c) Metal Removal Rate Q (in.3/min.)d) Cutting Horse Power (HPc)e) Minimum Machine Tool Motor HP.

13. Which of these statement is most correct?a) a continuous chip with built up edge may result when we try to cut brittle metals.b) a continuous chip will result when cutting very strong work materials.c) a discontinuous chip will result when we use heavy feeds and speeds.d) all of the above.

ans. C

14. One of the assumptions behind calculating orthogonal cutting forces is,a) that the rake angle is positive.b) that the tool is only cutting with one edge and one point.c) the shear plane is a function of before and after chip thicknesses.d) none of the above.

ans. C

15. Which of these statements is most correct?a) the cutting pressure drops as cutting velocity decreases.b) power required to cut each cubic inch drops as cutting velocity increases.c) we can use the quantity of metal removed by itself to estimate the required horsepower of a machine tool.d) all of the above.

ans. B

16. A new lathe tool is to be used on cast iron work with a 6” diameter to make a 36” long rough cut in 4 passes. The operation conditions listed below were provided by the supplier or assumed. Calculate the parameters a) to e) as requested.

Cutting Speed = 200 fpmFeed Rate = 0.010 iprDepth of Cut = 0.100”Idle Horse Power = 0.25Machine Efficiency = 0.90

a) Spindle RPMb) Time to make the cut (min.)c) Metal Removal Rate Q (in.3/min.)d) Cutting Horse Power (HPc)e) Minimum Machine Tool Motor Horse Power.

ans. a) 127rpm, b) 113min., c) 2.4 ipm, d) 1.23 or 3.94HP, e) 1.62 or 4.63HP

17. a) Define machinability. b) What determines the machinability of a metal?

20. What factors will affect surface finish?

21. Sketch a single edge cutting tool and label the a) face, b) flank, c) nose, d) cutting edge, e) relief, f) shank.

22. Why is the cutting speed important? What will happen at different cutting speeds, from very slow to very fast?

23. We have set up a lathe and are doing an orthogonal cut. The feed rate of the lathe is 0.1mm, and the chip thickness after the cut is 0.2mm. The depth of the chip being cut is 5mm. The sur-face cutting speed of the tool is 2m/s. The tool has a rake angle of 10deg. The tangential force is measured as 200N, and the cutting force is 500N. a) Calculate the shear force and velocity. b) Calculate the total energy produced in the cut, c) Calculate the energy used to shear d) Explain the difference between the total and the shear energy. [based on Kalpakjian]

D 6in= CS 200 ftmin---------= d 0.1in= HPI 0.25= e 0.9=

rpm CSπd-------

200 ftmin---------

π 6in( )------------------- 200 12( )in

π 6( )in min--------------------------- 127rpm= = = =a)

T Lf rpm( )----------------- 36in

0.01ipr 127rpm( )------------------------------------------- 28.35min= = =b)

f 0.01ipr=

(for one pass)

Q 12fdCS 12 0.01( ) 0.1( ) 200( ) 2.4 in3

min---------= = =c)

HPc HPuQ 0.5 1.6+2

---------------------⎝ ⎠⎛ ⎞ 2.4 2.5HP= = =d)

HPM HPIHPC

e-----------+ 0.25 2.5

0.9-------+ 3.0HP= = =e)

24. How is machining different than other processes?

25. What is the difference between a roughing and finishing operation? How does this affect the workpiece and the power consumed?

26. What type of chip is expected at higher cutting speeds?

27. Does the friction power in cutting increase more with a feed or speed increase?

28. Why does cost typically increase for finishing operations.

29. Explain the correction factor ‘c’ used with the HPu values.(ans. the HPu values are not linear, and ‘c’ corrects for these non-linear values)

t1 0.1mm=

t2 0.2mm=

Given,α 10°=

depth 5mm=

Vc 2ms----= Fc 500N=

Ft 200N=

Find the total power and shear power.

Wc FcVc 500N( ) 2ms----⎝ ⎠

⎛ ⎞ 1000W 1HP746W--------------⎝ ⎠⎛ ⎞ 1.34HP= = = =

rct1t2---- 0.1

0.2------- 0.5= = =

φrc αcos

1 rc αsin–-------------------------⎝ ⎠⎛ ⎞atan 28.3°= =

FS Fc φcos Ft φsin– 345N= =

VSVc αcos

φ α–( )cos-------------------------- 2.07m

s----= =

WS FSVS 714W 1HP746W--------------⎝ ⎠⎛ ⎞ 0.96HP= = =

Finally the ratio between the cutting power and the shear power

WSWc------- 0.96

1.34---------- 0.71= =

ans.

5. JOINING

5.1 Introduction

• Welding is the process of joining two or more objects together. In general this is done by melting the adjacent surfaces, or by melting a third material that acts as a ‘glue’

• We can categorize welding by processes,

Topics:

Objectives:•

•

5.2 ADHESIVE BONDING

Arc

Brazing

Electron BeamExplosiveFlowForge

Gas

InductionIon BeamLaser BeamResistance

Thermit

Ultrasonic

Carbon Electrode

Metal Electrode

BlockDipFlowFurnaceInductionResistanceTorchTwin Carbon Arc

DieFrictionHammerRollAir AcetyleneOxy AceteleneOxy HydrogenPressure

FlashPercussionProjectionSpotSeamUpsetNonpressurePressure

Shielded

UnShielded

Shielded

Unshielded

ShieldedInert GasCarbon ArcTwin Carbon ArcCO2 Mig

Coated ElectrodeElectro SlagImp. TapeInert Gas

PlasmaStudSubmergedBare ElectrodeStud

CO2FluxArc-Spot

MigTigArc-Spot

5.3 ARC WELDING

• Basically, an electric arc is used to heat base metals and a consumable filler rod.

• This is the most common form of welding and is used in about half of all applications.

• A power supply is used to create a high potential between an electrode (guided by the welder) and a metal work piece. When moved close enough electrodes break down the air and start to flow. The local current of the flow is so high that it heats metals up to 30000C or 54000F.

• Material is added during this welding process.This material can come from a consumable elec-trode, or from a rod of material that is fed separately.

• The electrodes/rods are often coated. This coating serves a number of functions,- it protects the welder from contact- it deoxidizes and provides a gas shield

• Problems that arise in this form of welding is contamination of the metal with elements in the atmosphere (O, H, N, etc.). There can also be problems with surfaces that are not clean. Solu-tions to this include,

Gas shields - an inert gas is blown into the weld zone to drive away other atmospheric gases.

Flux - a material that is added to clean the surface, this may also give off a gas to drive away unwanted gases.

• Common types of processes include,SMAW (Shielded Metal Arc Welding)/Stick Welding - A consumable electrode with a

coating that will act as a flux to clean the metal, and to create a gas shield.

metal welding tablepower

welding stick

work piece

supply

electric arc

current flowsin a loop throughthe metal

MIG (Metal Inert Gas) - A consumable electrode in a gas shield. In addition to simple materials, this can handle aluminum, magnesium, titanium, stainless steel, copper, etc. This torch is normally water or air cooled.

TIG (Tungsten Inert Gas) - A nonconsumable tungsten electrode is used with a filler rods and a gas shield. This can handle aluminum, titanium, stainless steel, copper, etc. This torch is normally water or air cooled.

SAW (Submerged Arc Welding) - A normal wire is used as a consumable electrode, and

filler material

work coating sets up gas shield

coating

coating

gas to shield the arc

electrode and feed rodwork

rod is fed into work

gas to shield the arc

tungsten electrodework


the flux is applied generously around the weld. The weld occurs within the flux, and is protected from the air.

• Process variables include,- electrode current 50-300A is common- voltage- polarity- arc length- speed- materials- flux- workpiece thickness

5.4 GAS WELDING

• Basically, filler and base materials are heated to the point of melting by a burning a gas.

• Two common types are,- oxygen-acetylene- mapp gas

• These are suited to a few applications, but they produce by-products that can contaminate the final weld.

• Typically the flame is adjusted to give a clean burn, and this is applied to the point of the weld.

electrode and feed rodwork


flux covers weld

• A welding rod will be fed in separately to melt and join the weld line.

• Flux can be used to clean the welds.

• Process variables include,- gas and oxygen flow rates- distance from surface- speed- material types- surface preparation of materials

5.5 SOLDERING AND BRAZING

• Basically, soldering and brazing involve melting a filler material that will flow into a narrow gap and solidify. It is distinct because the base materials should not be melted.

• The main difference is,- Soldering is done at a lower temperature, either with a propane torch, or an electric

heater. It is intended for bonds with less required strength, such as electrical and plumbing applications.

- Brazing is done at higher temperatures with oxyacetylene or mapp gas torches. These bonds tend to be higher and can be used for mechanical strength.

• General process considerations include,- Suitable for gaps from 0.001” to 0.01”- Surfaces must be sanded and cleaned before these processes are used.- Flux is often used to deoxidize a surface so that the filler will adhere better. Typical

fluxes include,Brazing flux - fused borax or alcohol and borax pasteSoldering flux - inorganic salts (zinc ammonium chloride), muriatic acid, resin

based- Some fluxes are corrosive and should be removed after use.

• Materials include,

hot point (inside blue flame)

weld tip andgas orifice

Solder is often an alloy combination of two of tin, lead, silver, zinc, antimony or bis-muth.

- Brazing metals are typically alloys such as,brazing brass (60% Cu, 40%Zn)manganese bronzenickel silvercopper siliconsilver alloys (with/without phosphorous)copper phosphorous

5.6 PLASTIC WELDING

• Well suited to joining of thermoplastics.

• Types of plastics used in welding are,

• Plastics to be joined should be compatible. A common method is based on trial and error testing.

• To determine plastic types burning small samples with a low flame gives the following observa-tions,

Material

ABS AcylicsPCPolyamidePolybutylenePolyethylene high densityPolyethylene low densityPolypropylenePolypropylene rubberPolyurethanePVC high densityPVC low density

Welding Temp. (°C)

350350350400350300270300300300-350300400-500

• Welding of thermoplastics involves heating, contact, cooling and bonding.

• Joints may be weakened by incomplete fusion, oxidation or thermal degradation of the plastic.

• Melting may be done by,- gas or electric gun- heated tool- induction heating- friction- spinning

• Sheet welding,- heat and pressure are applied at an overlapped joint between thin sheets.- rollers join the sheets (one roller is often motor driven and heated while the other just

applies pressure).- typical variables are,

- roller temperatures- feed rate- pressure

-underheating leads to a loose seam.- overheating leads to a hole formation.- parametric setting is very sensitive.- welds can be done on tables with a hand roller and a heat gun.- advantages

- simple tools- disadvantages

- hard to set parameters- preparation of welded sheets.

Material

ABSPolyamidePolycarbonatePolyethylenePolypropylenePVC

Observations

smells sweet, black sooty flame, does not extinguishsmells like burnt horn, stringy, does not extinguishblack sooty smoke, may extinguishsmells bad, feels like wax, burns like wax, dripssmells like wax, feels like wax, burns like wax, dripsacrid smell, black smoke, does not extinguish

if welded plastics to repair cracks, drill holes at the ends to stop crack propaga-tion.

• Hot air/gas welding,- Used successfully with molded parts in,

- PVC- polyvinylidene chloride- polyethylene- acrylic- polychlorotriflourethylene

- Operation steps,1. pieces positioned but a gap of 1/16” left2. a suitable welding (often same material) rod is pushed into the gap3. a hot blast often 400-600°F is directed at the tip of the welding rod and surfaces

to be welded. * if a torch to focus distance of 1-2” is used, a drop in tem-perature of 200°F will occur, the resulting temperature should be the melt-ing temperature of the plastic.

-The final strength ranges from 50% maximum for high density materials, to near 100% for low densities.

- The heating guns are similar to common hair dryers with heaters and fans, and vents to control air flow rates.

- The heat calls for safety measures.- Nozzles - a variety of nozzles and tools are available.- Advantages,

- simple tools

Overlap - less than 3 mm

Butt with lap

Single butt

Double butt

Disadvantages,- welding angles hard to set

• Tack welding,- parts are put in position.- the gun temperature is allowed to heat up (a tack welding nozzle is used).- the gun is put at an angle of 30-40°F to the weld and held in place until melting begins.- the gun is slowly drawn along the seam.- since the tack weld is weak (used for positioning) subsequent welding is required.

• General welding,- operation

1. the gun is held 90° to the weld and a rod is inserted.2. once the rod starts to melt, the gun is turned to a 45° angle and moved steadily

along the weld.3. The gun is moved in an elliptical path over the weld with an amplitude of about

1”.4. The rod is forced into the groove with a pressure of about 3-6 lbs. This pressure

prevents air from entering the weld. An angle of 45° to 90° is used for the rod.

5. When ending a weld, the heat is turned off, and after cooling the rod is twisted off, or for continuous welds there should be an overlap of 1/2”.

6. If required a weld can be restarted by cutting the previous weld at an angle, and starting from that point.

• Speed Welding- the rod and gas are fed side by side.- the rod is heated in the gun, and is “wiped” out as it leaves the gun.- when starting pressure is applied to the rod and a sharpened tip is forced into the work.- as the rod starts to melt, the gun is lowered to 45° and drawn along. The welding rod is

pulled in itself.- moving the tip too fast will result in beading and too slow will result in charring.- the weld is stopped by standing the gun at 90° to the surface and pulling the gun off. The

rod is then cut off.

• Tractor Welding (Machine Welding),- a hot air gun and rollers are driven over a surface by motors.- a tape can be dispensed that will join the sheets or the two sheets can be overlapped.- advantages,

- fully automated- easy to set parameters

- disadvantages,- special equipment required

• Ultra Sonic Welding,- basically a high frequency vibration is directed through a plastic joint. The vibration

causes friction, and then heat, often causing a solid bond in less than a second.- frequencies above 20 KHz.- the distance the vibration travels has a great deal to do with determining the classifica-

tion.- very well suited to rigid thermo plastic parts.- good designs make direct application of the vibrations possible.

- a smaller contact area increases the energy concentration. As a result V-notches, tongues, pins, and other special joints are commonly used.

- if remote sealing is necessary, thicker walls should be incorporated into the part design- epoxy molds can be used to reinforce weaker parts when doing this operation.- advantages,

- fast- clean- no extra materials needed

- disadvantages,- tool design required- simple design rules not always available

• Linear Vibration Welding,- similar to Ultrasonic Welding, except that frequencies are about one hundred Hz and

amplitude are mm.

Better

Better

this is best used with high coefficient of friction, low viscosity plastics.

• Spin/Friction Welding,- two parts are spun and the contact area builds up heat through friction and pressure. The

pressure forces a good bond between parts and drives out bobbles.- flashing may occur with this method.- advantages,

- produces a good weld- air does not enter during welding- inexpensive machines, such as drill presses may be used

- disadvantages,- circular weld joints are required

• Testing Plastic Welds,- a handheld gun can be used to generate arcs. The sparks are generated with voltages up

to 55 KV at 200 KHz.- operation,

1. The gun is calibrated to spark at distances just over the weld thickness (to a ground plate).

2. the ground plate is placed behind the weld.3. as the probe is moved over the weld, sparks will jump when a gap in the weld

moves between the probe and the ground plate.

5.7 Examples

5.8 Summary


5.10 Problems

2.What is the purpose of flux in welding?

3. List 20 parts you have seen that are welded. Indicate which welding process is the most appro-priate for each.

4. What types of processes would be best suited for joining the following items? Indicate why.a) two 12” dia. plastic pipes.b) two 12” dia. steel pipes.

c) the sides of a plastic bag for potato chips.d) two aluminum plates along one edge.e) an aluminum and steel plate into a laminated plate.f) steel muffler pipes.

5. What are the primary differences between welding soldering and gluing?


6. ROTATIONS

6.1 Introduction

6.2 Rotational Masses and Inertia

Topics:

Objectives:•

•

θ ω ddt-----⎝ ⎠⎛ ⎞ θ=

α ddt-----⎝ ⎠⎛ ⎞ω d

dt-----⎝ ⎠⎛ ⎞2

θ= =T

θ t( ) ω t( ) td∫ α t( ) td td∫∫= =

ω t( ) α t( ) td∫=

OR

(1)

(2)

(3)

(4)

equations of motion

α t( ) T t( )JM----------= (5)

where,θ ω α, , position, velocity and acceleration=

JM second mass moment of inertia of the body=T torque applied to body=

θ ω α, , T∑ JMα=

JM Ixx Iyy+=

J

T

Ixx y2 Md∫=

Iyy x2 Md∫=

(6)

(7)

(8)

(9)

Note: The ’mass’ moment of inertia will be used when dealing with acceleration of a mass. Later we will use the ’area’ moment of inertia for torsional springs.

Figure 6.1 Parallel axis theorem for shifting a mass moment of inertia

Figure 6.2 Parallel axis theorem for shifting a area moment of inertia

JM JM˜ Mr2+=

where,JM mass moment about the new point=

JM˜ mass moment about the center of mass=

M mass of the object=

r distance from the centroid to the new point=

JA JA˜ Ar2+=

where,JA area moment about the new point=

JA˜ area moment about the centroid=

A mass of the object=

r distance from the centroid to the new point=

Aside: If forces do not pass through the center of an object, it will rotate. If the object is made of a homogeneous material, the area and volume centroids can be used as the center. If the object is made of different materials then the center of mass should be used for the center. If the gravity varies over the length of the (very long) object then the center of gravity should be used.

Ixx y2 Md4–

4

∫ y2ρ2 5m( ) yd4–

4

∫ 1.25Kgm 1– y3

3-----

4–

4

= = =

The rectangular shape to the right is constrained to rotate about point A. The total mass of the object is 10kg. The given dimensions are in meters. Find the mass moment of inertia.

First find the density and calculate the moments of inertia about the centroid.

4

5-5

-4

-2.5

-1

ρ 10Kg2 5m( )2 4m( )-------------------------------- 0.125Kgm 2–= =

∴ 1.25Kgm 1– 4m( )3

3--------------- 4m–( )3

3------------------–⎝ ⎠

⎛ ⎞ 53.33Kgm2= =

Iyy x2 Md5–

5

∫ x2ρ2 4m( ) xd5–

5

∫ 1Kgm 1– x3

3-----

5–

5

= = =

∴ 1Kgm 1– 5m( )3

3--------------- 5m–( )3

3------------------–⎝ ⎠

⎛ ⎞ 83.33Kgm2= =

JM Ixx Iyy+ 53.33Kgm2 83.33Kgm2+ 136.67Kgm2= = =

The centroid can now be shifted to the center of rotation using the parallel axis theorem.

JM JM˜ Mr2+ 136.67Kgm2 10Kg( ) 2.5m–( )2 1m–( )2+( )+ 209.2Kgm2= = =

Figure 6.3 A solid torsional spring

TJAG

L----------⎝ ⎠⎛ ⎞ θ=

L

T

θ

T KS ∆θ( )=

(8)

(9)

Note: Remember to use radians for these calculations. In fact you are advised to use radians for all calculations. Don’t forget to set your calculator to radians also.

Note: This calculation uses the area moment of inertia.

Figure 6.4 A rotational spring example

Model the system above assuming that the center shaft is a torsional spring, and that a torque is applied to the leftmost disk. Leave the results in state variable form.

τJM1

JM2Ks1 Ks2 Ks3

JM1

τ

θ1 Ks2 θ1 θ2–( ) M∑ τ Ks2 θ1 θ2–( )– JM1θ1··= =+

JM2

Ks3θ2

θ2 Ks2 θ2 θ1–( ) M∑ Ks2 θ2 θ1–( )– Ks3θ2– JM2θ2··= =+

JM1θ1·· Ks2θ1– Ks2θ2 τ+ +=

θ1· ω1=

ω1· K– s2

JM1

-----------⎝ ⎠⎛ ⎞ θ1

Ks2JM1

--------⎝ ⎠⎛ ⎞ θ2 τ+ +=

(1)

(2)

θ2·· Ks3– Ks2–

JM2

---------------------------⎝ ⎠⎛ ⎞ θ2

Ks2JM2

--------⎝ ⎠⎛ ⎞ θ1+=

θ2· ω2=

ω2· Ks3– Ks2–

JM2

---------------------------⎝ ⎠⎛ ⎞ θ2

Ks2JM2

--------⎝ ⎠⎛ ⎞ θ1+=

(3)

(4)

ddt-----

θ1

ω1

θ2

ω2

0 1 0 0K– s2

JM1

----------- 0Ks2JM1

-------- 0

0 0 0 1Ks2JM2

-------- 0Ks3– Ks2–

JM2

--------------------------- 0

θ1

ω1

θ2

ω2

0τ00

+=

Figure 6.5 The rotational damping equation

T Kdω=

T Kd ω1 ω2–( )=

Figure 6.6 A System Example

Model the system above assuming that the center shaft is a torsional spring, and that a torque is applied to the leftmost disk. Leave the results in state variable form.

τ

B1 B2

JM1JM2Ks1 Ks2 Ks3

JM1

τ

θ1

B1θ1·

Ks2 θ1 θ2–( ) M∑ τ Ks2 θ1 θ2–( )– B1θ1·– JM1

θ1··= =+

JM2

Ks3θ2

θ2

B2θ2·

Ks2 θ2 θ1–( ) M∑ Ks2 θ2 θ1–( )– B2θ2·– Ks3θ2– JM2

θ2··= =+

JM1θ1·· B1θ1

·– Ks2θ1– Ks2θ2 τ+ +=

θ1· ω1=

ω1· B– 1

JM1

---------⎝ ⎠⎛ ⎞ ω1

K– s2JM1

-----------⎝ ⎠⎛ ⎞ θ1

Ks2JM1

--------⎝ ⎠⎛ ⎞ θ2

τJM1

--------+ + +=

(1)

(2)

θ2·· B– 2

JM2

---------⎝ ⎠⎛ ⎞ θ2

· Ks3– Ks2–JM2

---------------------------⎝ ⎠⎛ ⎞ θ2

Ks2JM2

--------⎝ ⎠⎛ ⎞ θ1+ +=

θ2· ω2=

ω2· B– 2

JM2

---------⎝ ⎠⎛ ⎞ ω2

Ks3– Ks2–JM2

---------------------------⎝ ⎠⎛ ⎞ θ2

Ks2JM2

--------⎝ ⎠⎛ ⎞ θ1+ +=

(3)

(4)

ddt-----

θ1

ω1

θ2

ω2

0 1 0 0K– s2JM1

-----------B– 1

JM1

---------Ks2JM1

-------- 0

0 0 0 1Ks2JM2

-------- 0Ks3– Ks2–

JM2

---------------------------B– 2

JM2

---------

θ1

ω1

θ2

ω2

0τ

JM1

--------

00

+=

6.3 Motor Models

6.3.1 Basic Brushed DC Motors

In a DC motor there is normally a set of coils on the rotor that turn inside a stator populated with permanent magnets. Figure 6.7 shows a simplified model of a motor. The magnetics provide a permanent magnetic field for the rotor to push against. When current is run through the wire loop it creates a magnetic field.

Figure 6.7 A Simplified Rotor

The power is delivered to the rotor using a commutator and brushes, as shown in Figure 6.8. In the figure the power is supplied to the rotor through graphite brushes rub-bing against the commutator. The commutator is split so that every half revolution the polarity of the voltage on the rotor, and the induced magnetic field reverses to push against the permanent magnets.

I

I

magnetic

axis ofrotation ω

field

Figure 6.8 A Split Ring Commutator

The direction of rotation will be determined by the polarity of the applied voltage, and the speed is proportional to the voltage. A feedback controller is used with these motors to provide motor positioning and velocity control.

These motors are losing popularity to brushless motors. The brushes are subject to wear, which increases maintenance costs. In addition, the use of brushes increases resis-tance, and lowers the motors efficiency.

motor

split commutator

brushes

motor

split commutator

brushes

shaft

shaft

Top

Front

V+ V-powersupply

Figure 6.9 Pulse Width Modulation (PWM) For Control

ASIDE: The controller to drive a servo motor normally uses a Pulse Width Modulated (PWM) signal. As shown below the signal produces an effective voltage that is rela-tive to the time that the signal is on. The percentage of time that the signal is on is called the duty cycle. When the voltage is on all the time the effective voltage deliv-ered is the maximum voltage. So, if the voltage is only on half the time, the effective voltage is half the maximum voltage. This method is popular because it can pro-duce a variable effective voltage efficiently. The frequency of these waves is nor-mally above 20KHz, above the range of human hearing.

Vmax

0t

Veff50100---------Vmax=

50% duty cycle

Vmax

0t

Veff20100---------Vmax=

20% duty cycle

Vmax

0t

Veff100100---------Vmax=

100% duty cycle

Vmax

0t

Veff0

100---------Vmax=

0% duty cycle

Figure 6.10 PWM Unidirectional Motor Control Circuit

ASIDE: A PWM signal can be used to drive a motor with the circuit shown below. The PWM signal switches the NPN transistor, thus switching power to the motor. In this case the voltage polarity on the motor will always be the same direction, so the motor may only turn in one direction.

signalsource

V+

com

powersupplyV+

V-

DC motor

Figure 6.11 PWM Bidirectional Motor Control Circuit

ASIDE: When a motor is to be con-trolled with PWM in two directions the H-bridge circuit (shown below) is a popular choice. These can be built with individual components, or purchased as integrated circuits for smaller motors. To turn the motor in one direction the PWM signal is applied to the Va inputs, while the Vb inputs are held low. In this arrangement the positive voltage is at the left side of the motor. To reverse the direction the PWM sig-nal is applied to the Vb inputs, while the Va inputs are held low. This applies the positive voltage to the right side of the motor.

+Vs

-Vs

Va

Va

Vb

Vb

ddt-----⎝ ⎠⎛ ⎞ω ω K2

JR------⎝ ⎠⎛ ⎞+∴ Vs

KJR------⎝ ⎠⎛ ⎞ Tload

JM------------–=

where,

ω the angular velocity of the motor=K the motor speed constant=JM the moment of inertia of the motor and attached loads=R the resistance of the motor coils=

Tload a torque applied to a motor shaft=

Figure 6.12 Model of a permanent magnet DC motor

Figure 6.13 Torque speed curve for a permanent magnet DC motor

ωss

T voltage/current increases

Figure 6.14 Motor speed curve and the derived differential equation

rpm

2400

0.5s

ddt-----⎝ ⎠⎛ ⎞ ωm ωm

K2

JMR----------⎝ ⎠⎜ ⎟⎛ ⎞

+ VsK

JMR----------⎝ ⎠⎛ ⎞ Tload

JM------------–=

The steady-state velocity can be used to find the value of K.

0( ) 2400 rotmin---------⎝ ⎠

⎛ ⎞ K2

JMR----------⎝ ⎠⎜ ⎟⎛ ⎞

+ 15V KJMR----------⎝ ⎠⎛ ⎞ 0( )–=

2400 rotmin---------1min

60s-------------2πrad

1rot----------------⎝ ⎠

⎛ ⎞ K( ) 15V=

K 15V120πrads 1–----------------------------- 39.8 10 3–× Vs

rad---------= =

The time constant can be used to find the remaining parameters.

K2

JMR---------- 1

0.5s---------- 2s 1–= =

ωm' Vs50.3V 1– s 2– rad ωm2s 1–– 50505Kg 1– m 2– Tload–=

(1)θm' ωm=

(2)

ddt-----⎝ ⎠⎛ ⎞ωm ωm2s 1–+ Vs 50.3V 1– s 2– rad( )

Tload

19.8 10 6–× Kgm2------------------------------------------–=

R 40Ω=

J39.8 10 3–× Vs

rad---------⎝ ⎠

⎛ ⎞ 2

40Ω( ) 2s 1–( )--------------------------------------------- 0.198005 -4×10 19.8 10 6–× Kgm2= = =

6.4 Tachometers

6.4.1 Angular Displacement

6.4.1.1 - Potentiometers

Potentiometers measure the angular position of a shaft using a variable resistor. A potentiometer is shown in Figure 6.15. The potentiometer is resistor, normally made with a thin film of resistive material. A wiper can be moved along the surface of the resistive film. As the wiper moves toward one end there will be a change in resistance proportional to the distance moved. If a voltage is applied across the resistor, the voltage at the wiper interpolate the voltages at the ends of the resistor.

Figure 6.15 A Potentiometer

The potentiometer in Figure 6.16 is being used as a voltage divider. As the wiper rotates the output voltage will be proportional to the angle of rotation.

schematic

physical

resistive

wiper

film

V1

V2

Vw

V1

Vw

V2

Figure 6.16 A Potentiometer as a Voltage Divider

Potentiometers are popular because they are inexpensive, and don’t require special signal conditioners. But, they have limited accuracy, normally in the range of 1% and they are subject to mechanical wear.

Potentiometers measure absolute position, and they are calibrated by rotating them in their mounting brackets, and then tightening them in place. The range of rotation is nor-mally limited to less than 360 degrees or multiples of 360 degrees. Some potentiometers can rotate without limits, and the wiper will jump from one end of the resistor to the other.

Faults in potentiometers can be detected by designing the potentiometer to never reach the ends of the range of motion. If an output voltage from the potentiometer ever reaches either end of the range, then a problem has occurred, and the machine can be shut down. Two examples of problems that might cause this are wires that fall off, or the poten-tiometer rotates in its mounting.

6.4.2 Encoders

Encoders use rotating disks with optical windows, as shown in Figure 6.17. The encoder contains an optical disk with fine windows etched into it. Light from emitters passes through the openings in the disk to detectors. As the encoder shaft is rotated, the light beams are broken. The encoder shown here is a quadrature encode, and it will be dis-cussed later.

V2

V1

Vout

Vout V2 V1–( )θwθmax-----------⎝ ⎠⎛ ⎞ V1+=

θmax θw

Figure 6.17 An Encoder Disk

There are two fundamental types of encoders; absolute and incremental. An abso-lute encoder will measure the position of the shaft for a single rotation. The same shaft angle will always produce the same reading. The output is normally a binary or grey code number. An incremental (or relative) encoder will output two pulses that can be used to determine displacement. Logic circuits or software is used to determine the direction of rotation, and count pulses to determine the displacement. The velocity can be determined by measuring the time between pulses.

Encoder disks are shown in Figure 6.18. The absolute encoder has two rings, the outer ring is the most significant digit of the encoder, the inner ring is the least significant digit. The relative encoder has two rings, with one ring rotated a few degrees ahead of the other, but otherwise the same. Both rings detect position to a quarter of the disk. To add accuracy to the absolute encoder more rings must be added to the disk, and more emitters and detectors. To add accuracy to the relative encoder we only need to add more windows to the existing two rings. Typical encoders will have from 2 to thousands of windows per ring.

lightemitters

lightdetectors

Shaft rotates

Note: this type of encoder is commonly used in com-puter mice with a roller ball.

Figure 6.18 Encoder Disks

When using absolute encoders, the position during a single rotation is measured directly. If the encoder rotates multiple times then the total number of rotations must be counted separately.

When using a relative encoder, the distance of rotation is determined by counting the pulses from one of the rings. If the encoder only rotates in one direction then a simple count of pulses from one ring will determine the total distance. If the encoder can rotate both directions a second ring must be used to determine when to subtract pulses. The quadrature scheme, using two rings, is shown in Figure 6.19. The signals are set up so that one is out of phase with the other. Notice that for different directions of rotation, input B either leads or lags A.

relative encoderabsolute encoder(quadrature)

sensors read acrossa single radial line

Figure 6.19 Quadrature Encoders

Interfaces for encoders are commonly available for PLCs and as purchased units. Newer PLCs will also allow two normal inputs to be used to decode encoder inputs.

Quad input A

Quad Input B

total displacement can be determined

Quad input A

Quad Input B

Note the changeas directionis reversed

by adding/subtracting pulse counts(direction determines add/subtract)

Note: To determine direction we can do a simple check. If both are off or on, the first to change state determines direction. Consider a point in the graphs above where both A and B are off. If A is the first input to turn on the encoder is rotating clockwise. If B is the first to turn on the rotation is counterclockwise.

clockwise rotation

counterclockwise rotation

Aside: A circuit (or program) can be built for this circuit using an up/down counter. If the positive edge of input A is used to trigger the clock, and input B is used to drive the up/down count, the counter will keep track of the encoder position.

Normally absolute and relative encoders require a calibration phase when a con-troller is turned on. This normally involves moving an axis until it reaches a logical sensor that marks the end of the range. The end of range is then used as the zero position. Machines using encoders, and other relative sensors, are noticeable in that they normally move to some extreme position before use.

6.4.2.1 - Tachometers

Tachometers measure the velocity of a rotating shaft. A common technique is to mount a magnet to a rotating shaft. When the magnetic moves past a stationary pick-up coil, current is induced. For each rotation of the shaft there is a pulse in the coil, as shown in Figure 6.20. When the time between the pulses is measured the period for one rotation can be found, and the frequency calculated. This technique often requires some signal conditioning circuitry.

Figure 6.20 A Magnetic Tachometer

Another common technique uses a simple permanent magnet DC generator (note: you can also use a small DC motor). The generator is hooked to the rotating shaft. The rotation of a shaft will induce a voltage proportional to the angular velocity. This tech-nique will introduce some drag into the system, and is used where efficiency is not an issue.

Both of these techniques are common, and inexpensive.

rotatingshaft

magnet

pickupcoil

Vout

Vout

t

1/f

6.5 Examples

6.6 Summary


6.8 Problems


7. FEEDBACK CONTROL REVIEW

7.1 Introduction

Topics:

Objectives:•

•

Figure 7.21 An automotive cruise control system

Figure 7.22 Example control rules

INPUT(e.g. θgas) SYSTEM

(e.g. a car)

OUTPUT(e.g. velocity)

Control variable

vdesired verror+

_

control car vactualθgas

Note: The arrows in the diagram indicate directions so that outputs and inputs are unambiguous. Each block in the diagram represents a transfer function.

function

Human rules to control car (also like expert system/fuzzy logic):

1. If verror is not zero, and has been positive/negative for a while, increase/decrease θgas2. If verror is very big/small increase/decrease θgas3. If verror is near zero, keep θgas the same4. If verror suddenly becomes bigger/smaller, then increase/decrease θgas.5. etc.

Figure 7.23 A PID control system

Figure 7.24 A PID controller equation

V V

+

-

amp motor

+

++

proportional

integral

derivative

Ki e∫( )

Kp e( )

Kdddt-----e⎝ ⎠⎛ ⎞

PID function

ue

+V

-V

u Kpe Ki edt∫ Kddedt------⎝ ⎠⎛ ⎞+ +=

Figure 7.25 A motor feedback control system

Figure 7.26 A block diagram for the feedback controller

+5V -5V5K potentiometer

12Vdc motor

shafts are coupled

-+

LM675op-amp

2.2K

1K

PCI-

1200

dat

a ac

quis

ition

car

dfro

m N

atio

nal I

nstr

umen

ts

Computer Running Labview

-+

desired positionvoltage Vd

gain Kp

X

desiredpositionvoltage +

-

potentiometer

gain Kp op-amp motor shaftVd

7.2 OpAmps

The ideal model of an op-amp is shown in Figure 7.27. On the left hand side are the inverting and non-inverting inputs. Both of these inputs are assumed to have infinite impedance, and so no current will flow. Op-amp application circuits are designed so that the inverting and non-inverting inputs are driven to the same voltage level. The output of the op-amp is shown on the right. In circuits op-amps are used with feedback to perform standard operations such as those listed below.

• adders, subtractors, multipliers, and dividers - simple analog math operations• amplifiers - increase the amplitude of a signal• impedance isolators - hide the resistance of a circuit while passing a voltage

Figure 7.27 An ideal op-amp

A simple op-amp example is given in Figure 7.28. As expected both of the op-amp input voltages are the same. This is a function of the circuit design. (Note: most op-amp circuits are designed to force both inputs to have the same voltage, so it is normally rea-sonable to assume they are the same.) The non-inverting input is connected directly to ground, so it will force both of the inputs to 0V. When the currents are summed at the inverting input, an equation including the input and output voltages is obtained. The final equation shows the system is a simple multiplier, or amplifier. The gain of the amplifier is determined by the ratio of the input and feedback resistors.

+

-V-

V+I+

I-Vo I- I+ 0= =

V- V+=

Note: for analysis use,

Figure 7.28 A simple inverting operational amplifier configuration

An op-amp circuit that can subtract signals is shown in Figure 7.29.

-

++Vi-

+Vo-

R1

R2

The voltage at the non-inverting input will be 0V, by design the voltage at the invert-ing input will be the same.

V+ 0V=

V- V+ 0V= =

The currents at the inverting input can be summed.

IV-∑V- Vi–

R1----------------

V- Vo–R2

-----------------+ 0= =

0 Vi–R1

--------------0 Vo–

R2---------------+ 0=

VoR2Vi–R1

--------------=

VoR2–

R1---------⎝ ⎠⎛ ⎞Vi=

Figure 7.29 Op-amp example

For ideal op-amp problems the node voltage method is normally the best choice. The equations for the circuit in Figure 7.29 and derived in Figure 7.30. The general approach to this solution is to sum the currents into the inverting and non-inverting input nodes. Notice that the current into the op-amp is assumed to be zero. Both the inverting and non-inverting input voltages are then set to be equal. After that, algebraic manipula-tion results in a final expression for the op-amp. Notice that if all of the resistor values are the same then the circuit becomes a simple subtractor.

Find the input/output ratio,

+-

+Vi-

+Vo-

+Vref-

R1

R2

R3R4

R5

Figure 7.30 Op-amp example (continued)

An op-amp (operational amplifier) has an extremely high gain, typically 100,000 times. The gain is multiplied by the difference between the inverting and non-inverting terminals to form an output. A typical op-amp will work for signals from DC up to about

VrefR4

R4 R5+------------------⎝ ⎠⎛ ⎞ Vi

R2R1 R2+------------------⎝ ⎠⎛ ⎞ Vo

R1R1 R2+------------------⎝ ⎠⎛ ⎞+=

Now the equations can be combined.

V- V+=

VoR1

R1 R2+------------------⎝ ⎠⎛ ⎞ Vi

R2R1 R2+------------------⎝ ⎠⎛ ⎞ Vref

R4R4 R5+------------------⎝ ⎠⎛ ⎞–=

Vo ViR2R1------⎝ ⎠⎛ ⎞ Vref

R4 R1 R2+( )R1 R4 R5+( )------------------------------⎝ ⎠⎛ ⎞–=

Note: normally node voltage methods work best with op-amp circuits, although others can be used if the non-ideal op-amp model is used.

First sum the currents at the inverting and non-inverting op-amp terminals.

IV+∑V+ Vi–

R1-----------------

V+ Vo–R2

------------------+ 0= =

IV-∑V- Vref–

R5--------------------

V-R4------+ 0= =

V+1

R1------ 1

R2------+⎝ ⎠

⎛ ⎞ Vi1

R1------⎝ ⎠⎛ ⎞ Vo

1R2------⎝ ⎠⎛ ⎞+=

V-1

R4------ 1

R5------+⎝ ⎠

⎛ ⎞ Vref1

R5------⎝ ⎠⎛ ⎞=

V+R1 R2+R1R2

------------------⎝ ⎠⎛ ⎞ Vi

1R1------⎝ ⎠⎛ ⎞ Vo

1R2------⎝ ⎠⎛ ⎞+=

V+ ViR2

R1 R2+------------------⎝ ⎠⎛ ⎞ Vo

R1R1 R2+------------------⎝ ⎠⎛ ⎞+=

V- VrefR4

R4 R5+------------------⎝ ⎠⎛ ⎞=

(1)

(2)

(3)

100KHz. When the op-amp is being used for high frequencies or large gains, the model of the op-amp in Figure 7.31 should be used. This model includes a large resistance between the inverting and non-inverting inputs. The voltage difference drives a dependent voltage source with a large gain. The output resistance will limit the maximum current that the device can produce, normally less than 100mA.

Figure 7.31 A non-ideal op-amp model

7.3 Examples

7.4 Summary


7.6 Problems


+-

V +

V–

rn

roA V+ V-–( )

Vorn 1MΩ>

typically,

A 105>ro 100Ω<

8. MECHANICAL POWER TRANSMISSION

8.1 Mechanisms

• Consider a pair of adjustable vice grips.

•Some definitions,Machine - a collection of components that will do work.Mechanism - a collection of components to transform motionKinematics - consider positions/velocities/accelerations in mechanical systemsStructure - a collection of components to make larger static structuresStatics - estimate forces in mechanisms that are in equilibriumDynamics - determine motion that results when forces are out of balanceLink - rigid body between jointsBinary Link - has two joints onlyTernary Link - has three jointsQuaternary Link - has four jointsPair or Joint - a connection between two linksDriver / Follower - the driver link will be driving the followerKinematic Chain - a sequence of links making up a mechanismOpen Loop - a snake like set of connected linksClosed Loop - a kinematic chain has one or more links that go back in the chain

Topics:

Objectives:•

•

Frame - a grounded or fixed link in a mechanismSpatial - in 3 dimensionsRelative/Absolute - a position, velocity, etc. is measured based on a fixed (absolute) or

moving (relative) point.

• A Degree Of Freedom (DOF) is an independently controllable variable. As an example, a machine that has two degrees of freedom might need two motors to control it.

• Lower Pairs, - constrained position/orientation of both sides of the joints are identicalTurning / Revolute - basically a pin joint (R)Prismatic - a slider (P)Screw/Helix - a nut and screw pair (H)Cylindric - a shaft in a collar (C)Globular/Spherical - a ball joint (S)

• Higher pairs include, - typically other equations are needed to constrain the joints, such as gear ratios (if the joint has more than a single degree of freedom)

- flat/planar - constrained to move over a plane- belt on pulley- meshing gears- sliding wheel on a surface- etc.

• The definition of higher/lower pairs given in Shigley [1995] is, “the lower pairs, such as the pin joint, have surface contact between the pair elements, while higher pairs, such at the connec-tion between a cam and its follower, have line or point contact between the surface elements.” They go on to point out that the definition is not exact, which is somewhat disappointing.

• A better definition of a higher pair is - A higher pair is not a lower pair, where a lower pair per-mits the following relative motions between links; circular, linear, helical, cylindrical, spheri-cal, planar.

• If a link has one joint, it is a unary link. A link with two joints is binary, with three it is ternary, with four it is quaternary, etc.

• Planar linkages use lower order pairs, and are constrained to a single plane of motion.

• Some basic mechanism types are listed below, and split into some suggested categories

8.1.1 Locking/Engaging

Snap-action mechanisms - typically bistable mechanisms, such as electrical breaker

switches, or toggle mechanisms such as XXXX

Clamping Mechanisms - vices, collets, etc.Locational Devices - self alining/centering devicesRatchets and Escapements - A locking mechanism, like a ratchet wrench or winch

Bistable

Toggle

Indexing Mechanisms - e.g. the geneva mechanism

Reversing Mechanisms - A mechanism that can disengage a transmission, and reverse direction of transmission

Ratchet

Escapement

8.1.2 Motion Transmission/Transformation

Linear Actuators - produce a straight line motion. can be done with threads, or hydraulic cylinders

Fine Adjustments - screws, wedges, etc. - these can overcome imperfections during manu-facture

Couplings and Connectors - transmit rotations between rotating shafts. e.g. pulleys and belts

Sliding Connectors - transmit linear motions in different directions

8.1.3 Four Bar Linkages

Swinging or Rocking Mechanism - produce cyclic motions

Stop/Pause/Hesitation - a motion is produced that appears to come to a stop for a short period of time.

Curve Generators - mechanisms set up to follow complex paths - typically four bar link-

Crank

Rockercoupler

frame

In this mechanism the crank is turned, and the follower oscillates. But, the motions forward and then backwards are not at the same rate.

ages.

Straight Line - mechanisms are set up to generate straight line motions

8.1.4 Reciprocating

Reciprocating Mechanisms - converts a rotational motion to a linear motion

Roberts Mechanism - the geometry is such that the mechanism is made of three isosceles triangles (note the dashed lines). When actuated the bottom point ‘A’ follows an approximate straight line.

A

Offset slider crank mechanism - will generate a fast stroke of the slider in one direction. For example, if the crank is turned clockwise then the slider will move fast going right, and slower returning left. The faster stroke will provide less force, if the crank torque is constant.

advance (clockwise rotation only)

return

Q time of advance stroketime of return stroke

------------------------------------------------------ angle of advance strokeangle of return stroke

--------------------------------------------------------= =

where,Q = the advance to return time ratio

8.1.5 Six Bar Linkages

• These allow more complex motion, especially when ternary links are used.

• In Watt linkages there are two ternary links touching,

• In Stephenson linkages the terary links don’t touch,

Scotch Yoke - the crank rotates, and causes a lin-ear motion.

Watt I Watt II

•. Consider the equation, and the four basic kinematic inversions below. Keep in mind that the crank will be the shortest link, with length ‘s, and in all four cases will rotate continuously.

Stephenson III

Stephenson IIStephenson I

Drag link mechanisms

8.2 Mechanical Advantage

• As a mechanism moves over a range of motion its geometry changes. If we are using a mecha-nisms to transmit torque, or force then we must consider the ratio between the input and output force in various positions.

• Transmission angle is the angle between the coupling member and the output member in a mechanism. As this angle approaches ±90°, the mechanical advantage of the mechanism typi-cally increases.

• Toggle positions occur when the input crank has near infinite mechanical advantage. Note: this also applies that the follower has no mechanical advantage on the crank.

• Consider the example below, [prob. 1-3 from Shigley & Uicker],

Crank-Rocker mechanisms

Double-Rocker mechanisms

s = 25mml = 100mm

p = 75mm

q = 90mm Find the maximum and minimum transmission angles. Find the two toggle angles of the crank AB.

100mm

75mm90+25mmFirst, find the minimum toggle angle.

θ1

752 1152 1002 2 115( ) 100( ) θ1cos–+= =115mmθ1∴ 40.1°=

100mm

75mm90-25mmNext, find the maximum toggle angle.

θ2’

752 652 1002 2 65( ) 100( ) θ2cos–+= =65mmθ2′∴ 48.6°°=

θ2 48.6 180+ 228.6°= =

A

B

C

D

Note: these angles are a measure of when the crank torque will create a maximum force.

8.3 Gears

• When forces become large we cannot count on friction for rolling contact (no slip). Gears use metal teeth that are meshed together to transmit motion between moving components.

8.3.1 Spur Gears

• Spur gears are in very wide use throughout engineering.

• These gears are flat, and either circular or straight (a rack).

100-25mm=75mm

75mm90mm

Now, find the minimum transmission angle.γmin

752 752 902 2 75( ) 90( ) γmincos–+=

γmin∴ 53.1°=

100+25mm=125mm

75mm90mm

Finally, find the maximum transmission angle.γmax

1252 752 902 2 75( ) 90( ) γmaxcos–+=

γmax∴ 98.1°=

Note: This occurs as we get the largest angle between the links, or when the bottom two toggle.

Note: These angles show the relationship between tension/compression in the driver and the follower.

• The figure below shows a typical gear with common terms marked,

egr352a0.jpg

• When gears are properly mated their pitch circles will be tangent. And the faces of the teeth will touch along the addendum and dedendum surfaces, down to the clearance circles.

• Some terms of use when discussing gears,backlash - the difference is the gap between gear teeth where they mesh. This leads to

‘play’ in the gears.pinion - a smaller gearwheel - a larger gear

• diametral pitch is defined by,

addendumdedendum

addendum circle

pitch circle

dedendum circle

tooth

width

circular

clearance

thickness

pitch

of space

• module is defined by,

• The following relationships are also applicable,

• The ratio between angular velocities of two gears can be determined with the law of gearing,

• As seen above the law of gearing assumes that the pitch point is found at a constant radius. If this were to move the driven gear would accelerate /decelerate as the teeth mesh and separate.

• To keep the gears meshing constantly an involute profile is typically used for the shape of the teeth.

• To construct an involute profile, we need to construct a line that is tangential to both gears. The teeth on both gear will be constructed to contact only on this line.

P Nd----=

where,N = number of teethd = pitch diameter, in.P = diametral pitch (teeth/in.)

m dN----=

where,m = module (mm)d = pitch diameter (mm)

p πdN------ πm π

P---= = =

where,p = circular pitch

ωiωj-----

rjri---=

where,ωi ωj, the angular velocities of gears i and j=ri rj, the pitch radii of two gears i and j=

• The involute profiles for a single tooth will trace out a line as shown below (later we will develop an equation for the point on the unwrapping string).

• The pressure angle is shown below,

base cylinders

pitchpoint

gear i

gear j

tooth on i

tooth on j

To do this we start with a cord wrapped about the base cylinder of the gear. We pick a point on the cyl-inder where the tooth is to start, and mark the point on the cord. As we then unwrap the cord the point will trace out the involute profile of the tooth.

• Standards geometries for spur gears include, (based on American Gear Manufacturers Associa-tion and ANSI standards)

• Typical diametral pitches and modules include, (based on American Gear Manufacturers Asso-ciation and ANSI standards)

pressureangle

Teeth type

Full depth

Stub

Pressure angle

20 (deg)22.525

20

Addendum ‘a’

1/P1/P1/P

0.8/P

Dedendum ‘b’

1.25/P, 1.35/P1.25/P, 1.35/P1.25/P, 1.35/P

1/P

CoarseP = 2, 2.25, 2.5, 3, 4, 6, 8, 10, 12, 16

FineP = 20, 24, 32, 40, 48, 64, 80, 96, 120, 150, 200

Preferredm = 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25

Less Preferredm = 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36

OR

8.3.2 Involute Profiles

• The figure below shows the triangulated layout for the basic involute function,

• Next the involute curve is applied to the generation of gear teeth.

r

involute curve

x

y

r

α invϕ=ϕ

ρ

We define the involute curve such that,ρ r α ϕ+( )=

l

ρ r ϕtan=Therefore,

ρ r α ϕ+( ) r ϕtan= =

α∴ ϕtan ϕ–=

invϕ∴ ϕtan ϕ–=

ϕinvϕ

b/2

ρ

x

y

d/2

pitch circle

toothprofile

θr

θb

r

base circleθp

φ

• To generate points on these curves we must select values of XXXX and calculate (x, y) posi-tions. These will be correct for one face of the tooth. If these are used to generate a splined curve, or graphed, they will form the tooth profile. The upper and lower bounds are determined by the addendum and dedendum.

8.3.3 Design of Gears

• The basic steps to design a gear are outlines below,

x r θr( )cos=

θb θp invφ+ π2N------- invφ+= =

For the gear we must consider the pitch,

r d2---= ϕ φ=when

where,φ the pressure angle=d diameter of pitch circle=

Therefore,

b diameter of given base circle=

b2--- r φ( )cos=

y r θr( )sin=

p πdN------⎝ ⎠⎛ ⎞ 2 2rθp( ) 2 dθp( )= = = θp∴ π

2N-------=

θrπ2--- θb invϕ–+ π

2--- π

2N------- invφ invϕ–+ + π

2--- 1 1

N----+⎝ ⎠

⎛ ⎞ invφ invϕ–+= = =

ϕ 0deg≥

1. Calculate the desired pitch diameters for both gears, i and j. Note: gear i should be larger than gear j for the formulas given.

diNiP-----= dj

NjP-----=

where,di dj, pitch diameters for meshing gears i and j=P diametral pitch selected=Ni Nj, given number of teeth on gears i and j=

2. Find the base circles of the gears by calculating the shortest perpendicular from the pressure line to the center. This will be the diameters of the base circles.

bi

bj

bi di φcos=bj dj φcos=

where,

bi bj, diameters of base circles=

φ pressure angle=

di

dj

φ

3. Using the base circle calculate the gear tooth profile for both gears using the involute equations described before.

4. Import the geometry into the cad system as a line. convert this to a complete tooth (this may involve a sweep, or mirror and join depending on the CAD software). Trim-ming, fillets, etc will be done later.

5. Calculate the addendum and dedendum radii for both gears - using the standards given earlier. Create dedendum circles for each gear body with the diameters of the pitch diameter minus the dedendum. (In one case this will be below the base, in the other this will be above). In the final application the centers between these gears will be d1+d2 apart.

ai aj1P---= = ci cj

1.25P

----------= =

Bi di 2ci–= Bj dj 2cj–=

where,ai aj, addendums for both gears=

ci cj, dedendums for both gears=

Bi Bj, dedendum diameter - size of circle for the gear bodies=

Ai di 2ai+= Aj dj 2aj+=

Ai Aj, Addendum circles for gears=

6. Calculate the angular spacing of the teeth using the number of teeth. Align the teeth at the right distance from the center of rotation. The addendum, dedendum and pitch circles must cross the tooth at the right location. Then use the cad system to copy the teeth about the center of the gear. If required do a join step and/or cut the base circle for the gear.

∆θi360°

Ni-----------= ∆θj

360°Nj

-----------=

where,

∆θi ∆θj, angles between gear teeth on gear i and j=

• If we are dealing with a rack, it is effectively a circular gear with an infinite radius.

• When we have internal gears (one gear inside another) we need to adjust the methods to reflect that both gears are on the same side of the pitch line.

8.3.4 Design Issues

• During motion the gear teeth undergo a combination of sliding and rolling. The direction of slid-ing reverses at the pitch point, where the motion is pure rolling.

8.3.4.1 - Undercutting and Contact Ratios

• Undercutting occurs on some gears. This is a gouging of teeth that occurs when teeth contact below the base circle of the gear during motion.

• During manufacturing some processes (generation) can remove the excess material that lead to undercutting. But this can reduce the base width of the teeth and weakening of the gear.

• Undercutting problems can be reduced by increasing the radius of the gear, and increasing the number of teeth.

• The gear teeth are in contact along the pressure line between the points where it intersects the addendum lines.

7. Add other features as required, such as holes, keyways, holes, spokes, etc. Note: fil-lets are needed at the base of the teeth to prevent stress concentrations, and small rounds are needed at the top of the teeth to reduce wear. The maximum radii of the fillets can be determined using the clearance.

rifilletci aj–<

rjfilletcj ai–<

where,rjfillet

rifillet, maximum fillet radii at the bottom of the gear teeth=

• The contact ratio is as defined below. A value of 1 means that at any time only one tooth is engaged. A value greater than 1 means that only one tooth is engaged. A value of 2 would mean that at any time 2 teeth are engage. A value less than 1 means that at times the teeth are not in contact (bad).

• Undercutting will not occur during production of the profiles if the following addendum values

pressure line

pitch point ‘P’addendum

circle

dedendum+clearan

motion

addendum circle

pitch circleqa

qr

qa arc of approach=qr arc of recess=

A PB

Note: my artistry is not so good. Here the dis-tances between the circle should be equal.

mcqtp----

ua ur+p φcos----------------= =

where,

mc the contact ratio=ua approach contact path length (from A to P)=

ur recess contact path length (from P to B)=

ua dj aj+( )2 bj2– dj φsin–=

ur di ai+( )2 bi2– di φsin–=

observed,

8.3.4.2 - Changing the Center Distance

• If the center distance between gears is changed, then the pitch circles on both gears will move away from the center.

• The result of the enlarging of pitch circles will be a reduction in the contact ratio. This will lead to a smaller contact ratio.

• This condition also allows some play in the gears (backlash). This play means that a reversing of direction can lead to a small reversing rotation before the other tooth is impacted. This leads to errors and the impact forces can shorten the life of gears significantly.

8.3.5 Helical Gears

• Helical gears are essentially spur gears, but with a bit of a twist in the normally straight profile of the teeth.

egr352a1.jpg

• The helical arrangement means that the teeth engage at one point (as opposed to a line of con-tact), and then slowly mesh along the face of the tooth. This also means that the number of engaged teeth (contact ratio) can be higher.

• If we contrast spur and helical gears. The spur gear teeth contact fully all at once.

• These gears mesh very smoothly, so they find application for,- high speeds- heavy loads- gear noise must be reduced- center lines of shafts do not intersect

• These gears,

aj bj2 di dj+( )2 φsin( )2+ dj–<

ai bi2 di dj+( )2 φsin( )2+ di–<

require more effort during fabrication- need additional bearings to resist axial thrust.

• helical gears can be used to transmit torques between parallel, and non-parallel (often perpendic-ular) shafts.

• To eliminate the need for an axial thrust bearing we can use a herringbone (double helical) gear. This is effectively two helical gears with opposite twists on the helix, and joined down a center line.

8.3.5.1 - Design of Helical Gears

• We can describe the helix with a single angle,

• This helix allows three different ways to measure the pitch and pressure angles.

ψ

ψ helix angle=

• We can now develop an equivalent pitch radius,

pn

px

pt

pn normal diametral pitch=pt tranverse diametral pitch=px axial pitch=

ptpnψcos

------------- px ψcos= =

φn

φt

φn normal pressure angle=φt tranverse pressure angle=

ψcosφntanφttan

-------------=

• Typical design parameters include,- produced for specialized applications, and custom designed- typical pressure angle is 20°- for helix angles from 0 to 30° use the normal diametral pitch to calculate tooth propor-

tions- helix angles greater than 45° are not recommended- in mating parallel gears - one must have a left hand helix, and the other must have a right

hand helix. Use the sign of the helix angle to indicate left or right handed.

• The line of contact in spur gears is straight across the teeth. In helical gears, the line of contact is diagonal.

• To measure the contact ratio we need to use three values to be effective,

dedψcos( )2

--------------------=

where,d the pitch diameter of the helical gear=de the diameter of an equivalent spur gear=

Ne dePnNψcos( )3

--------------------= =

where,N number of teeth on the helical gear=

Pn diametral pitch of gear=Ne equivalent number of teeth on a spur gear=

8.3.5.2 - Perpendicular Helical Gears

• If two shafts intersect at an angle (typically 90°) we can link them using helical gears. The angle between the shafts can be,

• The example below shows one configuration for helical gears, including the location of thrust bearings.

ψb ψtan φcos( )atan=

mnmt

ψbcos( )2----------------------=

mxFpx----- F ψtan

pt----------------= =

where,ψb base helix angle=

mt transverse contact angle=mn normal contact ratio (as found for spur gears)=mx axial contact ratio=F face width=mT total contact ratio=

mT mn mx+=

Aside: Recall, a contact ratio below 1 is unacceptable, and the gears will come out of contact. For helical gears this ratio should be greater than 2.

Σ ψi ψj±=

where,

ψi ψj, helix angle on gears i and j=

Σ angle between shafts=

• When the shaft intersection angle is large (90°) you may use same handed gears to intersect. In the example above the gears are both right handed.

• The pitch diameter for these gears can be found using,

• A minimum contact ratio of 2 is recommended for these gears.

• Typical parameters for cross-axis helical gears are given below,

8.3.6 Bevel Gears

• These gears are like normal spur gears, except that they have a conical form.egr352a2.jpg

Driver Helical Gears

Thrust Bearings

d NPn ψcos-------------------=

Helix Angle

45°60°75°86°

DriverMinimum# teeth

20941

Helix angle

45°30°15°4°

DrivenPressure angle

14.5°17.5°19.5°20°

• Their applications are characterized by,- to couple shafts with intersecting axes-

• Bevel gears are meshed so that the points of their cones are coincident.

• As we move towards the point of the cones, the number of teeth remains the same, but the diam-eter reduces towards zero. This changes the size of the teeth, and the pitch diameter.

• The form of the gears is like that of spur gears, but each has a cone angle, and when added together this gives the angle between the shafts.

• We can apply some of the basic ratios to bevel gears,

γi γj, pitch angles=

γi

γj

Σ γi γj+= Σ shaft angle=

di

dj

P

O

ωiωj-----

djdi----

NjNi-----= =

where,

ωi ωj, input/output angular velocities=di dj, pitch diameters of gears (normally taken at wide base)=Ni Nj, number of teeth on gears=

8.3.6.1 - Design of Bevel Gears

• To determine the pitch angles for the gears we can write the following expressions,

• An approximate methods for creating bevel gears is called ‘Tredgold’s approximation’

• Tredgold’s technique requires that a cone on the bottom of the bear be found. This cone is then flattened out, and normal gear design is done. Finally the cone is mapped back onto the bottom of the gear. The profiles are then projected up to the point of the cone.

OPdi

2 γisin---------------

dj2 γjsin---------------= =

where,

OP the distance from the mesh point to the tips of the cones=

γisindidj---- γjsin

didj---- Σ γi–( )sin

didj---- Σ γisinsin γi Σsinsin–( )= = =

by trigonometry,

γi∴ Σsindjdi---- Σcos+-----------------------atan Σsin

NjNi----- Σcos+------------------------atan= =

γjsindjdi---- γisin

djdi---- Σ γji–( )sin

djdi---- Σ γjsinsin γj Σsinsin–( )= = =

γj∴ ΣsinNiNj----- Σcos+------------------------atan=

• Typical design parameters include,- 20° pressure angle for straight bevel gears- bevel gears are always custom made, and are not interchanged- deflections mean that the wider base tends to take most of the load, so the teeth are

designed with a short length (commonly less tan 1/3 of the total cone length)-

8.3.7 Other Bevelled Gears

• Crown and Face Gear - this gear is much like a rack for spur gears. To get this, one of the gears is given a pitch angle of 90°.

• Spiral Bevel Gears - to reduce noise in beveled gears, a spiral can be added to the teeth.

• Hypoid gears - the centers of the bevelled gears are not coincident - the shaft is offset.

8.3.8 Worm Gears

• Worm gears use a long helical screw that drives a larger helical gear.egr352a3.jpg

• Basically we use a screw like gear (the worm) and a large cylindrical gear (worm gear) that is driven by the worm.

• The worm gear curves to the shape of the worm to increase contact. Also note that worm gear is a helical gear.

dei

diγicos

------------= dej

djγjcos

------------=

Nei

πdei

pi---------= Nej

πdej

pj---------=

where,

deidej, equivalent diameter for spur gear design=

NeiNej, equivalent number of teeth for spur gear design=

• The worm acts very much like a rack, except that it is threaded onto a cylindrical surface.

worm

worm gear

pitch

centerdistance

circle

dg

dg pitch diameter=

dgNgpπ

---------=where,

Ng number of worm gear teeth=

• These gears find their best applications when a large gear ratio is needed in a compact space. The shafts typically intersect at a 90° angle - when this is the case the helix angle on both gears is the same.

• The worm gear can be single enveloping or double enveloping.- The single envelope is the parallel sided gear. This type of gear is more forgiving

for position and alignment tolerance problems.- The double enveloping gear has a curvature that increases the surface on contact

between the gears. This can be very useful for power applications.

• The following values are reasonable for finding the profile dimensions of the teeth,

ψλ

λ lead angle=where,

ψ helix angle=

p

p = axial pitchdw pitch diameter=

dw

dw

dw dg+2

------------------⎝ ⎠⎛ ⎞

0.875

2.2----------------------------------=

This relationship is suggested by the AGMA for a good power capacity.

λ 90° ψ–pNwπdw----------⎝ ⎠⎛ ⎞atan= =

Nw number of teeth (threads) on the worm=

addendum 0.3183p=dedendum 0.3183p=clearance 0.050p=

• Suggested pressure angles for given lead angles are listed below,

8.3.9 Harmonic Drives

• These are actually normal servo motors, but with an integrated harmonic gear. The harmonic gears are very compact, and as a result the overall size is reduced. These gears also allow very high gear ratios (eg, 100:1)

• Harmonic drives are also gaining popularity with smaller manipulators. They use a rotating elliptical core that deforms a flexible section. The flexible section is in contact with an outer section for short periods of time, and as the ellipse rotates, there is a geared down rotation gen-erated. This allows integral gears in motors

8.3.10 Design With Gears

• When we design gears, cams and mechanisms we are free to set and vary parameters. But, above this we often must select these components to start with.

λ φ

0-16°16-25°25-35°35-45°

14.5°20°25°30°

Note that as the inner elliptical spline rotates, the flexible spline counter-rotates. The surface between the wave generator, and flexible spline is smooth, and the surfacer between the flexible spline, and the outer spline is geared.

• The selection of components can be aided by using techniques such as,- schematics- ratios- common approaches

8.3.10.1 - Gear Trains

• When we want to increase/reduce angular displacements/velocities/etc. we can use simple gear trains.

• As we have seen before, gears typically have an input to output ratio. The relationship below is for simple gear trains - only one gear on each axis.

• We can deal with compound gear trains (multiple gears on each axis) by using product of driven and driving teeth.

eωnω2------

NiNi 1+------------

i 2=

n 1–

∏di

di 1+-----------

i 2=

n 1–

∏= = =

where,

ωi The angular velocity from the first gear 2 to the last gear n=

Ni The number of teeth on gear i=di The pitch diameter of gear i=

e The speed ratio of a gear train (can be negative)=

eωnω2------

Ni

i 1=

m

∏

Nj

j 1=

n

∏

---------------= =

where,Ni The number of driving teeth on gear i of m driving gears=Nj The number of driving teeth on gear j of n driving gears=

8.3.10.2 - Examples - Fixed Axis Gears

• A simple gear train has only one gear on each axis.

• A compound gear train has multiple gears on the same axis. Consider the truck transmission example from Shigley and Uicker,

• We may also consider an in-line gear train. These can be used for items such divide by twelve and sixty in clocks,

Motor shaft

clutch

reverse idler

2

3 4

56

7

8

9

1011

stem gear

Speed (gear)

1234reverse

Gear Train

2-3-6-92-3-5-82-3-4-7bypass gear train2-3-6-10-11-9

In this manual transmission the gear shifter will move the gears in and out of con-tact. At this point all of the needed gears will be meshed and turning. The final step is to engage the last gear in the gear train with the clutch (plate) and this couples the gears to the wheels.

• Quite often we will have a particular speed ratio in mind. We can convert this to teeth numbers by finding a suitable fractional value,

• The gear train in the previous example might look something like,

N2 60=

N3 15=

N4 60=

N5 15= eN2N4N3N5------------- 16= =

Input shaft Output shaftIn this case the output shaft

turns 16 times faster than the input shaft. If we reversed directions the output (former input) would now turn 1/16 of the input (former out-put) shaft speed.

e 0.156 1561000------------ 78

500--------- 39

250---------= = = =

assume we are given a value of e=0.156, we can begin by putting this in fraction form with the lowest integer values,

In this case the ratio is very high. We could decide to try to make a gear, or we could split the numerator and denominator into multiples. We can then put the multiples into smaller fractions. In the case below there are no common numerators and denominators so all of the gears will need to be compound.

e 39250--------- 3( ) 13( )

10( ) 25( )---------------------- 3

10------⎝ ⎠⎛ ⎞ 13

25------⎝ ⎠⎛ ⎞= = =

Looking at the numerators and denominators there are a few integers that are small. We should set a minimum number of teeth for practical design purposes. In this case we can set the minimum at 20.

e 310------⎝ ⎠⎛ ⎞ 13

25------⎝ ⎠⎛ ⎞ 21

70------⎝ ⎠⎛ ⎞ 26

50------⎝ ⎠⎛ ⎞ 21

50------⎝ ⎠⎛ ⎞ 26

70------⎝ ⎠⎛ ⎞= = =

• Try the design below,

N2 21=

N3 50=

N4 26=

N5 70=

Output shaftInput shaft

Design a gear train for the value e=-0.2. Next design for e=0.2.

8.3.10.3 - Examples - Moving Axis Gears

• In some cases gears move relative to each other. This can be used to generate some interesting alternatives.

8.3.10.4 - Epicyclic Gear Trains

• Epicyclic gears have many applications, such as automatic transmissions in automobiles.

• In these trains the gears typically orbit each other.

• Consider the basic epicyclic gear train,

• We can represent these gears using a notation developed by Levai. Consider the basic epicyclic gear,

Sun gear

Planet Gear

Planet

2

3

4

CarrierOR KINEMATICEQUIVALENT

• When designing with these gears, we can consider different control modes possible. In the case above we could connect the gear ‘4’ to a ring gear (internal gear) and make a simple multi-speed transmission.

• Some of the other possible gear train types include variations on the number of planets.

• Consider the compact planetary gear shown below,

4

3

2

gear 2 and 3 move relative to ground. This is different that ‘4’ because it is not directly connected to ground.

teeth

collar

43

2

If we ground ‘3’ then we get a speed ratio between ‘2’ (clutch stem gear) and ‘5’ (the output shaft) of,

5

eN2N4------–

N4N5------

N2N5------–= =

If we connect ‘3’ to ‘5’ then we get a speed ratio of,

e 1=

• We can also construct an epicyclic gear using bevel gears. This is called Humpage’s reduction gear.

• Consider the example below,

doublesun gear

planet carrier

fixed gear

planet gear

input shaft

output shaft

double planetbevel gear

planet carrier

fixed gear

output gearand shaft

8.3.10.5 - Differentials

• Differential mechanisms allow us to effectively do subtraction or averaging.

• If we want to determine the difference between two linear motions we could a mechanism like the one shown below,

23

4

5

1 Draw the Levai representation for the epicyclic gear to the left. Assuming that the num-ber of teeth are 20, 30, 15, 50 from gear 2 through to gear 5, find the speed ratio between the input at 2 and the output at 3. What would the speed ratio be if 4 was the output?

• An angular differential is shown below using bevel gears,

************** include image here

• In one type of automotive differential the housing above is driven, and both the output shafts turn. This allows a small difference in wheel speed. Without this simple action like turning cor-ners would exert large forces on the tires and drive train. This is also why one of the drive wheels can spin while the other is fixed.

• Various vehicles can disengage the differential for offroad conditions (where tires can slip),

aa

∆x1

∆x2

∆x12 2∆x1 ∆x2–=

θ2θ4

θ6

θ5

θ5N2N5------θ

1

N4N5------θ

2–=

5

2

3

4

6

θ6 θ2 θ4+=

while others have mechanisms to balance torque to the wheels when an excessive difference in speed is detected.

• Worm gears have also been used in automotive differentials.

8.3.11 Gear Forces and Torques

• The involute profile of the gear means that the force applied at the gear teeth is not tangent to the pitch line, but actually tangent to the base circle. At the pitch point the force between the teeth acts at the pressure angle.

• For spur gears the following values and equations can be used for the applied forces. Clearly there would be a reaction force that is not shown here. (Note: this applies even if multiple teeth are in contact).

• If we are considering helical gears the helical spiral of the teeth adds a second angle to the con-tact force. This means that the contact force requires 3D analysis.

pitchT

F

φ

d

circle

Ft F φcos=

Fn F φsin=

T d2---F φcos=

• Straight bevel gears are not so easy to calculate because the force is applied at a variable dis-tance from the center of the rotational axis.

• In these gears we compromise by assuming the force is applied at the center of the tooth. In actu-ality this force will be further from the center of the gear.

• We can calculate force components using the relationships below,

• For all of these gears we need to use the calculated forces to design bearings and supports. Most notably the axial thrusts require thrust bearings be included in the design.

Given the gear geometry,φt transverse pressure angle=ψ helix angle=

we can calculate,

Ft F ψ φtcoscos=

Fn F ψ φtsincos=

Fa F ψ φtcossin=

T Ftd2--- d

2---F ψ φtcoscos= = F∴ 2T

d ψ φtcoscos------------------------------=

where,

Fa the axial component of the force (need axial bearings)=

Tdavg

2----------Ft

davg2

----------F φcos= =

where,davg the average pitch diameter of the teeth=

Ft F φcos=

γ angle of bevel=

Fr F φ γcossin=

F∴ 2Tdavg φcos----------------------=

Fr the force component normal to the axis of rotation=

Fa F φ γsinsin=

8.4 Cams

• Cams are basically shaped surfaces that are typically not round. The cam is rotated or translated, and a follower (possibly a small wheel) is displaced as it moves over the surface.

• Cams can generate complex motion profiles in a compact area.

• Engine values are a well known application of cams. They can open and close the cylinders with a large force, but will also dwell in open or closed positions.

• Typical cams are pictured below,

Plate (or Disk, or Radial) Cam

Wedge/Translating Cam

• Some types of reciprocating followers include,

End (or Face) Cam

Cylindrical (or Barrel) Cam

Roller Knife Edge flat-face

• Some oscillating followers include,

• We can often describe a cam by drawing the displacement profile on a graph.

• Consider possible displacement curves for,

roller curved shoe

Rise Dwell Return Dwell

lift

displacement

camangle

0 360°

f θ( )

θ

• The curve above can be broken into sections and described with a mathematical function,

• Of these two functions, the parabolic will allow a greater level of control, but harmonic motion will permit smooth transitions between motions.

• Some of the general design rules,1. fulfil the basic motion requirements. (cam profiles are not exact and decisions are

required.)2. The displacement, velocity and acceleration curves must be continuous, but the jerk

- wave machine

- stamping press

- washing machine

y f θ( )=

f θ( ) Aθ2 Bθ C+ += Parabolic motion

f θ( ) A B Cθcos+= Harmonic motion

f θ( ) A= Constant/dwell motion

must not be infinite. This means that the positions and first and second derivatives must be equal at the segment ends.

3. Minimize the velocities and accelerations.

• The first step in developing a cam is to develop a motion profile. Consider example 5-2 from Shigley and Uicker,

We want a cam driving a reciprocating follower. The cam is rotated with an angular veloc-ity of 150 rev/min. The follower should start in a dwell and accelerate to a constant velocity of 25 in./sec. for a rise of 1.25 in. The follower should come to a rest after moving a total height of 3 in. The follower should then drop back, and dwell for 0.10 sec.

First we will draw a function that has the basic components of the motion. In total there is the initial acceleration from a dwell (parabolic segment) to a constant velocity (straight line segment). After the constant velocity there are two transitions back to a final dwell at the starting height (2 parabolic segments). This evens out to a dwell (straight line segment). The displacement magnitudes were calculated and labeled on the graph

θ

f θ( )

accel. 25 in/s decel. reverse dwellβ0 β1 β2 β3 β4

360° β0 β1 β2 β3 β4+ + + +=

β4 0.10s( ) 150 revmin---------⎝ ⎠

⎛ ⎞ 1min60s

-------------⎝ ⎠⎛ ⎞ 360°

1rev-----------⎝ ⎠⎛ ⎞ 90°= =

β11.25in

25ins-----

---------------⎝ ⎠⎜ ⎟⎜ ⎟⎛ ⎞

150 revmin---------⎝ ⎠

⎛ ⎞ 1min60s

-------------⎝ ⎠⎛ ⎞ 360°

1rev-----------⎝ ⎠⎛ ⎞ 45°= =

0.0

3.0

0.875

2.125

225°∴ β0 β2 β3+ +=

Next, determine the obvious angles,

L0

L1

L2 f2 θ( ) f3 θ( )

f4 θ( )f0 θ( )f1 θ( )

C∴ 0=f0 0( ) 0 Aθ2 Bθ C+ += =

f0' 0( ) 0 2Aθ B+= =

f0 β0( ) L0 Aθ2= =

First, consider the first acceleration segment, and the maximums,

B∴ 0=

A∴L0

β02

-----=

second, consider the first decceleration segment (flipped left/right), and the maximums,

f0' β0( ) 2Aθ 2L0

β02

-----β0 2L0β0-----= = =

f0'' β0( ) 2A 2L0

β02

-----= =

C∴ 0=f2 0( ) 0 Aθ2 Bθ C+ += =

f2' 0( ) 0 2Aθ B+= =

f2 β2( ) L2 Aθ2= =

B∴ 0=

A∴L2

β22

-----=

f2' β2( ) 2Aθ 2L2

β22

-----β2 2L2β2-----= = =

f2'' β2( ) 2A 2L2

β22

-----= =

finally, consider the return to the dwell (half the way), and the maximums,

C∴ 0=f3 0( ) 0 Aθ2 Bθ C+ += =

f3' 0( ) 0 2Aθ B+= =

f3β32-----⎝ ⎠⎛ ⎞ L0 L1 L2+ + Aθ2= =

B∴ 0=

A∴4 L0 L1 L2+ +( )

β32

--------------------------------------=

f3'β32-----⎝ ⎠⎛ ⎞ 2Aθ

4 L0 L1 L2+ +( )β3

--------------------------------------= =

f3''β32-----⎝ ⎠⎛ ⎞ 2A

8 L0 L1 L2+ +( )

β32

--------------------------------------= =

• Now,

Now, we need to balance the velocities and accelerations. This can be done by setting all of the maximum velocities and accelerations equal,

Velocities,

2L0β0----- 2

L2β2-----

4 L0 L1 L2+ +( )β3

--------------------------------------= =

Accelerations,

1.75β0

2----------∴ 1.75

β22

---------- 24β3

2------= =

1.75β0

---------- 1.75β2

---------- 12β3------= =∴

2L0

β02

----- 2L2

β22

-----8 L0 L1 L2+ +( )

β32

--------------------------------------= =

At this point we can use either velocities or accelerations to find the times for each segment. I arbitrarily decide to use velocities,

β1 β0= β312

1.75----------β

0=

225° β0 β2 β3+ + β0 β012

1.75----------β

0+ + 8.857β0= = =

β∴ 0 25.4°=

β∴ 2 25.4°=β∴ 3 174.2°=

• You may have recognized that the previous design assumed that the follower must have a point contact with the curve.

• In actual practice we will have surfaces that are in contact, the surfaces can be identified using the equations developed previously.

• Consider the flat-face follower.

Convert the cam profile in the previous problem to a circular cam profile using a knife edge follower.

• We can develop the a modified cam profile based on the flat faced follower. (Note: the proof is done as if a milling cutter is used, but this turns out to be more a matter of convenience)

Note the follower is offset, this will not change the operation, but can be used to reduce moments in the shaft for a clock-wise rotation.

90° dwell45° constantvelocity

We can use the following relationships to plot the cam profile, based on the motion profile. The first relationship will avoid undercutting - this is a case where the cam gets ‘stuck’ on a second peak.

R0 rmin f''min θ( )– f θ( )–>

R0

rmin a minimum allowed curvature for the cam=

The face width must be wider than,

face width f'max θ( ) f'min θ( )–>

• Using the derivation of the basic relationships, we can now develop a method to plot out a com-plete cam profile.

γ

Rc

rmill

φ

L f θ( ) R0+=

a

dL dθ

dθ

φ

R

LR--- φcos=

aL--- φtan= φ∴ a

L---⎝ ⎠⎛ ⎞atan=

θθ

Fixedpositionon cam

R∴ Lφcos

------------=

• Now, develop a cam for example 3-52 from Shigley and Uicker,

1. Pick an angle of 0 for the first iteration, increment this in subsequent calculations.

2. Calculate,

3. Plot the point, pick a subsequent point and then do a new calculation for the new angle.

θi 0°…360°[ ]=

L f φ( ) R0+=

a ddφ------f φ( )=

x R θ φ+( )cos= y R θ φ+( )sin=

φ aL---⎝ ⎠⎛ ⎞atan=

R Lφcos

------------=

• Keep in mind that when designing cam-follower pairs that the radius of the follower is not zero. Therefore it may be necessary to compensate for this during the design.

• Consider the effect of a round follower on a wedge cam.

First, determine the minimum radius of the cam, if the curvature of the cam is to be greater than 0.5” at all points, and the face width.

Develop the equations for the geometry of the cam using the profile calculated earlier.

• Consider the effect of a round follower on a radial cam.

• Other arrangements are possible, and some proofs are provided in the text.

NOTE: The point of contact remains tangential, angle of the cam suggests a different point of contact. In this case the upper position of the cam has a small offset across the surface of the cam.

We can use the following relationships to plot the cam profile, based on the motion profile.

ψ mL----⎝ ⎠⎛ ⎞atan=

x F θ ψ+( )cos rfollower φcos–=

F L2 m2+=

φ α ψ–=

1. Pick a displacement angleθi 0°…360°[ ]=

2. Calculate

3. Plot the point, and select the subsequent point.

mrfollower

L f θ( ) rfollower rmin+ +=

γ

θ

x F θ ψ+( )sin rfollower φsin–=

a ddθ------f θ( )=

α LaF2 ma–-------------------⎝ ⎠⎛ ⎞atan=

8.4.1 Using Cams in Mechanisms

• We can use cams to give complex joint motion,

8.5 Examples

8.6 Summary


Erdman, A.G. and Sandor, G.N., Mechanism Design Analysis and Synthesis, Vol. 1, 3rd Edition, Prentice Hall, 1997.

Shigley, J.E., Uicker, J.J., “Theory of Machines and Mechanisms, Second Edition, McGraw-Hill, 1995.

8.8 Problems

1. Draw the kinematic inversions of the linkage below. Does it satisfy the Grashof Criterion? What is the mobility of the mechanism? Determine the maximum transmission angles, and the toggle angles if CD is the crank. What is the advance-to-return time ratio?

2. Draw the kinematic inversions of the linkage below. Does it satisfy the Grashof Criterion? What is the mobility of the mechanism?

3. The human arm is an open loop kinematic chain. Assuming the shoulder is the ground, what is the mobility of the human arm, including all joints down to the finger tips?

1. We have been asked to design a coarse pitch (P=3) gear pair with stub teeth and a speed ratio of e=0.24. Find the radii and dimensions needed to draw the gears. Find at least one point on a gear tooth profile. Roughly sketch the gears showing the dimensions calculated.

1. You were recently hired as a Fuel Containment and Monitoring specialist for Generous Motors. Your first job is to design a mechanical gauge for an instrument panel. The tank holds up to 10 gallons of fuel. It has been determined that the needle on the gauge should remain steady at the full ‘F’ mark while the tank contains 8 to 10 gallons. When the tank has less than 3 gallons the gauge should read empty ‘E’. The last design was a failure ‘F’, and your boss fired the engineer responsible. It seems that he his design did not follow good cam design rules - the velocity and accelerations were not minimized - and so the gauge would wear out, and jam prematurely. Design a new cam to relate the float in the tank to the gauge on the instrument panel.

10mm

7mm

7mm

5mm

A

B

C

D

10mm

7mm

7mm

5mm

2. The motion profile curve below has 4 segments. Segments A and C are based on polynomials. Segment D is based on a harmonic/cosine function. Segment B is a constant velocity segment.

a) Write the equation for curve segment B.b) What effect does the follower shape have when converting the motion profile to a cam profile. Draw a figure to illustrate this with a round follower.c) Write the coefficients for the curve segment C.

E Fgasoline

float

linear cam

follower

gauge(old design)

tank

3”10”

needle

θ

3 6.242210

1

0.5

0.3

0

y

A

B

C

D


yC ACθ3 BCθ

2 CCθ DC+ + += AC =

BC =CC =

DC =

9. MECHANICAL ISSUES

9.1 Introduction

9.2 Friction

Viscous friction was discussed before, where a lubricant would provide a damping effect between two moving objects. In cases where there is no lubricant, and the touching surfaces are dry, dry coulomb friction may result. In this case the surfaces will stick in place until a maximum force is overcome. After that the object will begin to slide and a constant friction force will result.

Figure 9.32 shows the classic model for (dry Coulomb) friction. The force on the horizontal axis is the force applied to the friction surfaces while the vertical axis is the resulting friction force. Beneath the slip force the object will stay in place. When the slip force is exceeded the object will begin to move, and the resulting kinetic friction force will be relatively constant. (Note: If the object begins to travel much faster then the kinetic friction force will decrease.) It is common to forget that friction forces are bidirectional,

Topics:

Objectives:•

•

but it always opposes the applied force or motion. The friction force is a function of the coefficient of friction and the normal force across the contact surfaces. The coefficient of friction is a function of the materials, surface texture and surface shape.

Figure 9.32 Dry friction

Many systems use kinetic friction to dissipate energy from a system as heat, sound and vibration.

N

Fk µkN=

Fs µsN≤

Fs

Fk Fs,

F

Fk

Fs

Fg

Block begins to slip and the

Note: When solving problems with friction remember that the friction force will always equal the applied force (not the maximum force) until slip occurs. After that the fric-tion is approximately constant. In addition, the friction forces direction opposes applied forces, and motion.

Fresult

Fapplied

applied force exceeds theresultant and acceleration begins.

Figure 9.33 Drill problem: find the accelerations

9.3 Friction

Find the acceleration of the block for both angles indicated.

10 kgθ1 5°=

θ

µs 0.3=

θ2 35°=

µk 0.2=

ans.

Figure 9.34 A friction system example

9.4 Contact Points And Joints

A system is built by connecting components together. These connections can be rigid or moving. In solid connections all forces and moments are transmitted and the two pieces act as a single rigid body. In moving connections there is at least one degree of free-

Model the system and consider the static and kinetic friction forces on the shaft on the right hand side. τ

JM Ts 10Nm≤Ks

Tk 6Nm=

JM

τ

θ

FF

KsθM∑ τ Ksθ– TF– JMθ

··= =+

JMθ·· τ Ksθ– TF–=

θ· ω= (2)

(3)ω·τ TF–

JM--------------⎝ ⎠⎛ ⎞ Ks

JM------⎝ ⎠⎛ ⎞ θ+=

JMω· τ Ksθ– Ttest–=

(1)

Next, the torque force must be calculated, and then used to determine the new torque force.

Ttest τ Ksθ– JMω·–=

Ttest 10Nm≤

cases:

TF Ttest=

Ttest 10Nm> TF 6Nm=

Not slipping previously

Slipping previouslyTtest 6Nm< TF Ttest=

Ttest 6Nm≥ TF 6Nm=

(4)

dom. If we limit this to translation only, there are up to three degrees of freedom, x, y and z. In any direction there is a degree of freedom, a force or moment cannot be transmitted.

When constructing FBDs for a system we must break all of the components into individual rigid bodies. Where the mechanism has been broken the contact forces must be added to both of the separated pieces. Consider the example in Figure 9.35. At joint A the forces are written as two components in the x and y directions. For joint B the force com-ponents with equal magnitudes but opposite directions are added to both FBDs.

Figure 9.35 FBDs for systems with connected members

9.4.1 Switching

- system components turned on/off

- cables in tension/compression

- show an example where input conditions change

A

B

M1

M2

FBxFBy

FAyFAx

M1g

FBx

FBy

M2g

Note: Don’t forget that forces on con-nected FBDs should have equal magnitudes, but opposite direc-tions.

give PWM (Pulse Width Modulation) example with ripple showing equivalent voltage. PWM is used to generate analog voltage equivalents. Show for a system with first order response with tau = 0.1s for a frequency of 1KHz, 10Hz and 1Hz. Point out the rip-ple and effective voltage.

- important to consider when doing system analysis

9.4.2 Deadband

- Friction in all components

- costs money to reduce friction, so it is better to compensate in software

- small actuation signals not large enough to overcome friction

- This effect is normally known as ’stiction’, a combination of the words static and friction.

- Friction is common in less expensive motors, and when a motor is driving a mechanical system.

- In systems there are two type of friction that must be considered.

- The static friction, ’stiction’, will prevent initial motion. If the systems breaks free and starts turning, the kinetic friction will provide a roughly constant friction resis-tance.

- relationship in figure below.

y· 4y+ f t( )= f t 0<( ) 0=

f 0 t 5s<≤( ) 2=

f t 5s≥( ) 0=

the region where the applied voltage has no effect is called the deadband.

-

Figure 9.36 Motor deadband for a bidirectional motor

- deadband compensation as shown in figure below.

Figure 9.37 Deadband approximation for a bidirectional motor

- equations for these are shown in figure

Vapplied

ω

deadbandmotor startsto turn asfriction isovercome

The kineticfriction resultsin a different curvewhile slowing down

Vwanted

Vadjusted

Vstick

-Vstick

Figure 9.38 Deadband approximation for a bidirectional motor

- c-code below

Figure 9.39 Deadband compensation subroutine

Cwanted

Cadjusted

Cstick

-Cstick

Cadjusted CstickCwanted

Cmax------------------⎝ ⎠⎛ ⎞ Cmax Cstick–( )+=Cmin 255=

Cmax 255=

Cmin 255–=Cmax 255=

if(Cwanted > 0)

Cadjusted C– stickCwanted

Cmin------------------⎝ ⎠⎛ ⎞ Cmin Cstick–( )+=

if(Cwanted < 0)

Cadjusted 0=

if(Cwanted = 0)

#define c_stick_pos 100#define c_stick_neg -110 /* make the value positive */#define c_max 255#define c_min -255 /* make the value positive */

int deadband(int c_wanted) /* call this routine when updating */int c_adjusted;

if(c_wanted == 0) /* turn off the output */c_adjusted = 0;

else if(c_wanted > 0) /* a positive output */c_adjusted = c_stick_pos +

(c_max - c_stick_pos) * c_wanted / c_max;if(c_adjusted > c_max) c_adjusted = c_max;

else /* the output must be negative */c_adjusted = -c_stick_neg -

(c_min - c_stick_neg) * c_wanted / c_min;if(c_adjusted < -c_min) c_adjusted = c_min;

return c_adjusted;

9.4.3 Saturation and Clipping

• Some devices have natural maximum values, such as voltage or pressure limita-tions caused by a regulated supply.

•

9.4.4 Hysteresis and Slip

- windup resulting from springiness and friction

- backlash

-

•

- correct by tracking the previous motion direction and taking extra steps when reversing direction

9.4.5 Delays and Lags

• Time delays are common in systems

• In the simplest form this is a period of time between when an event occurs and when the effect occurs.

• If an output delay is larger than the control system step time it may be necessary to predict future states and initiate outputs ahead of those.

• If an input delay is larger than the control system it might be necessary to slow the control action, or build it into the control law.

9.5 Wheeled Vehicles

9.6 Examples

9.7 Summary


9.9 Problems


10. SENSORS

10.1 INTRODUCTION

Sensors allow a PLC to detect the state of a process. Logical sensors can only detect a state that is either true or false. Examples of physical phenomena that are typically detected are listed below.

• inductive proximity - is a metal object nearby?• capacitive proximity - is a dielectric object nearby?• optical presence - is an object breaking a light beam or reflecting light?• mechanical contact - is an object touching a switch?

Recently, the cost of sensors has dropped and they have become commodity items, typically between $50 and $100. They are available in many forms from multiple vendors such as Allen-Bradley, Omron, Hyde Park and Turck. In applications sensors are inter-changeable between PLC vendors, but each sensor will have specific interface require-ments.

This chapter will begin by examining the various electrical wiring techniques for sensors, and conclude with an examination of many popular sensor types.

10.2 SENSOR WIRING

When a sensor detects a logical change it must signal that change to the PLC. This is typically done by switching a voltage or current on or off. In some cases the output of the sensor is used to switch a load directly, completely eliminating the PLC. Typical out-puts from sensors (and inputs to PLCs) are listed below in relative popularity.

Topics:

Objectives:• Understand the different types of sensor outputs.• Know the basic sensor types and understand application issues.

• Sensor wiring; switches, TTL, sourcing, sinking • Proximity detection; contact switches, photo-optics, capacitive, inductive and

ultrasonic

Sinking/Sourcing - Switches current on or off.Plain Switches - Switches voltage on or off.Solid State Relays - These switch AC outputs.TTL (Transistor Transistor Logic) - Uses 0V and 5V to indicate logic levels.

10.2.1 Switches

The simplest example of sensor outputs are switches and relays. A simple example is shown in “An Example of Switched Sensors” on page 303.

Figure 10.40 An Example of Switched Sensors

In the figure a NO contact switch is connected to input 01. A sensor with a relay output is also shown. The sensor must be powered separately, therefore the V+ and V- ter-minals are connected to the power supply. The output of the sensor will become active when a phenomenon has been detected. This means the internal switch (probably a relay) will be closed allowing current to flow and the positive voltage will be applied to input 06.

10.2.2 Transistor Transistor Logic (TTL)

Transistor-Transistor Logic (TTL) is based on two voltage levels, 0V for false and

24 VdcPowerSupply

normally open push-button

PLC Input Card24V DC

00

01

02

03

04

05

06

07

COM

+

-

sensorV+

V-

relayoutput

5V for true. The voltages can actually be slightly larger than 0V, or lower than 5V and still be detected correctly. This method is very susceptible to electrical noise on the factory floor, and should only be used when necessary. TTL outputs are common on electronic devices and computers, and will be necessary sometimes. When connecting to other devices simple circuits can be used to improve the signal, such as the Schmitt trigger in “A Schmitt Trigger” on page 304.

Figure 10.41 A Schmitt Trigger

A Schmitt trigger will receive an input voltage between 0-5V and convert it to 0V or 5V. If the voltage is in an ambiguous range, about 1.5-3.5V it will be ignored.

If a sensor has a TTL output the PLC must use a TTL input card to read the values. If the TTL sensor is being used for other applications it should be noted that the maximum current output is normally about 20mA.

10.2.3 Sinking/Sourcing

Sinking sensors allow current to flow into the sensor to the voltage common, while sourcing sensors allow current to flow out of the sensor from a positive source. For both of these methods the emphasis is on current flow, not voltage. By using current flow, instead of voltage, many of the electrical noise problems are reduced.

When discussing sourcing and sinking we are referring to the output of the sensor that is acting like a switch. In fact the output of the sensor is normally a transistor, that will act like a switch (with some voltage loss). A PNP transistor is used for the sourcing out-put, and an NPN transistor is used for the sinking input. When discussing these sensors the term sourcing is often interchanged with PNP, and sinking with NPN. A simplified exam-ple of a sinking output sensor is shown in “A Simplified NPN/Sinking Sensor” on page 305. The sensor will have some part that deals with detection, this is on the left. The sensor needs a voltage supply to operate, so a voltage supply is needed for the sensor. If the sensor has detected some phenomenon then it will trigger the active line. The active

Vi VoVi

Vo

line is directly connected to an NPN transistor. (Note: for an NPN transistor the arrow always points away from the center.) If the voltage to the transistor on the active line is 0V, then the transistor will not allow current to flow into the sensor. If the voltage on the active line becomes larger (say 12V) then the transistor will switch on and allow current to flow into the sensor to the common.

Figure 10.42 A Simplified NPN/Sinking Sensor

Sourcing sensors are the complement to sinking sensors. The sourcing sensors use a PNP transistor, as shown in “A Simplified Sourcing/PNP Sensor” on page 306. (Note: PNP transistors are always drawn with the arrow pointing to the center.) When the sensor is inactive the active line stays at the V+ value, and the transistor stays switched off. When the sensor becomes active the active line will be made 0V, and the transistor will allow current to flow out of the sensor.

Sensor

V+

V-

Active

physicalphenomenon

Aside: The sensor responds to a physical phenomenon. If the sensor is inactive (nothing detected) then the active line is low and the transistor is off, this is like an open switch. That means the NPN output will have no current in/out. When the sensor is active, it will make the active line high. This will turn on the transistor, and effec-tively close the switch. This will allow current to flow into the sensor to ground (hence sinking). The voltage on the NPN output will be pulled down to V-. Note: the voltage will always be 1-2V higher because of the transistor. When the sensor is off, the NPN output will float, and any digital circuitry needs to contain a pull-up resistor.

V+

NPN

V-

sensoroutput

andDetector

Line

current flows inwhen switched on

Figure 10.43 A Simplified Sourcing/PNP Sensor

Most NPN/PNP sensors are capable of handling currents up to a few amps, and they can be used to switch loads directly. (Note: always check the documentation for rated voltages and currents.) An example using sourcing and sinking sensors to control lights is shown in “Direct Control Using NPN/PNP Sensors” on page 307. (Note: This example could be for a motion detector that turns on lights in dark hallways.)

Sensor

V+

V-

Active

physicalphenomenon

Aside: The sensor responds to the physical phenomenon. If the sensor is inactive (nothing detected) then the active line is high and the transistor is off, this is like an open switch. That means the PNP output will have no current in/out. When the sensor is active, it will make the active line high. This will turn on the transistor, and effectively close the switch. This will allow current to flow from V+ through the sensor to the output (hence sourcing). The voltage on the PNP output will be pulled up to V+. Note: the voltage will always be 1-2V lower because of the transistor. When off, the PNP output will float, if used with digital circuitry a pull-down resistor will be needed.

V+

PNP

V-

sensoroutput

andDetector

Linecurrent flows outwhen switched on

Figure 10.44 Direct Control Using NPN/PNP Sensors

In the sinking system in “Direct Control Using NPN/PNP Sensors” on page 307 the light has V+ applied to one side. The other side is connected to the NPN output of the sensor. When the sensor turns on the current will be able to flow through the light, into the output to V- common. (Note: Yes, the current will be allowed to flow into the output for an NPN sensor.) In the sourcing arrangement the light will turn on when the output becomes active, allowing current to flow from the V+, thought the sensor, the light and to V- (the common).

At this point it is worth stating the obvious - The output of a sensor will be an input for a PLC. And, as we saw with the NPN sensor, this does not necessarily indicate where current is flowing. There are two viable approaches for connecting sensors to PLCs. The first is to always use PNP sensors and normal voltage input cards. The second option is to purchase input cards specifically designed for sourcing or sinking sensors. An example of a PLC card for sinking sensors is shown in “A PLC Input Card for Sinking Sensors” on page 308.

powersupply

sensor V+

V- (common)

V+

V-

NPN

powersupply

sensor V+

V- (common)

V+

V-

PNP

Note: remember to check the current and voltage ratings for the sensors.

sinking

sourcing

Note: When marking power terminals, there will sometimes be two sets of markings. The more standard is V+ and COM, but sometimes you will see devices and power supplies without a COM (common), in this case assume the V- is the common.

Figure 10.45 A PLC Input Card for Sinking Sensors

The dashed line in the figure represents the circuit, or current flow path when the sensor is active. This path enters the PLC input card first at a V+ terminal (Note: there is no common on this card) and flows through an optocoupler. This current will use light to turn on a phototransistor to tell the computer in the PLC the input current is flowing. The current then leaves the card at input 00 and passes through the sensor to V-. When the sen-sor is inactive the current will not flow, and the light in the optocoupler will be off. The optocoupler is used to help protect the PLC from electrical problems outside the PLC.

The input cards for PNP sensors are similar to the NPN cards, as shown in “PLC Input Card for Sourcing Sensors” on page 309.

PLC Input Card for Sinking Sensors

Internal Card Electronics

PLC Data Bus External Electrical

NPNsensor

powersupply

+V

-V

+V

NPN

-V

current flow+V

00

01 Note: When a PLC input card does not have a common but it has a V+ instead, it can be used for NPN sensors. In this case the cur-rent will flow out of the card (sourcing) and we must switch it to ground.

ASIDE: This card is shown with 2 optocouplers (one for each output). Inside these devices the is an LED and a phototransistor, but no electrical connection. These devices are used to isolate two different electrical systems. In this case they pro-tect the 5V digital levels of the PLC computer from the various external voltages and currents.

Figure 10.46 PLC Input Card for Sourcing Sensors

The current flow loop for an active sensor is shown with a dashed line. Following the path of the current we see that it begins at the V+, passes through the sensor, in the input 00, through the optocoupler, out the common and to the V-.

Wiring is a major concern with PLC applications, so to reduce the total number of wires, two wire sensors have become popular. But, by integrating three wires worth of function into two, we now couple the power supply and sensing functions into one. Two wire sensors are shown in “Two Wire Sensors” on page 310.

PLC Input Card for Sourcing Sensors

Internal Card Electronics

PNPsensor

powersupply

+V

-V

+V

PNP

-V

current flow00

01

comNote: When we have a PLC input card that has

a common then we can use PNP sensors. In this case the current will flow into the card and then out the common to the power sup-ply.

Figure 10.47 Two Wire Sensors

A two wire sensor can be used as either a sourcing or sinking input. In both of these arrangements the sensor will require a small amount of current to power the sensor, but when active it will allow more current to flow. This requires input cards that will allow a small amount of current to flow (called the leakage current), but also be able to detect when the current has exceeded a given value.

PLC Input Card two wiresensor

powersupply

+V

-V

+V

-V00

01

comNote: These sensors require a certain leakage

current to power the electronics.

PLC Input Card

two wiresensor

powersupply

+V

-V

+V

-V

00

01

V+

for Sourcing Sensors

for Sinking Sensors

When purchasing sensors and input cards there are some important considerations. Most modern sensors have both PNP and NPN outputs, although if the choice is not avail-able, PNP is the more popular choice. PLC cards can be confusing to buy, as each vendor refers to the cards differently. To avoid problems, look to see if the card is specifically for sinking or sourcing sensors, or look for a V+ (sinking) or COM (sourcing). Some vendors also sell cards that will allow you to have NPN and PNP inputs mixed on the same card.

When drawing wiring diagrams the symbols in “Sourcing and Sinking Schematic Symbols” on page 311 are used for sinking and sourcing proximity sensors. Notice that in the sinking sensor when the switch closes (moves up to the terminal) it contacts the com-mon. Closing the switch in the sourcing sensor connects the output to the V+. On the physical sensor the wires are color coded as indicated in the diagram. The brown wire is positive, the blue wire is negative and the output is white for sinking and black for sourc-ing. The outside shape of the sensor may change for other devices, such as photo sensors which are often shown as round circles.

Figure 10.48 Sourcing and Sinking Schematic Symbols

10.2.4 Solid State Relays

Solid state relays switch AC currents. These are relatively inexpensive and are available for large loads. Some sensors and devices are available with these as outputs.

NPN (sinking)

PNP (sourcing)

V+

V-NPN

V-

V+ PNPblack

brown

white

brown

blue

blue

10.3 PRESENCE DETECTION

There are two basic ways to detect object presence; contact and proximity. Contact implies that there is mechanical contact and a resulting force between the sensor and the object. Proximity indicates that the object is near, but contact is not required. The follow-ing sections examine different types of sensors for detecting object presence. These sen-sors account for a majority of the sensors used in applications.

10.3.1 Contact Switches

Contact switches are available as normally open and normally closed. Their hous-ings are reinforced so that they can take repeated mechanical forces. These often have roll-ers and wear pads for the point of contact. Lightweight contact switches can be purchased for less than a dollar, but heavy duty contact switches will have much higher costs. Exam-ples of applications include motion limit switches and part present detectors.

10.3.2 Reed Switches

Reed switches are very similar to relays, except a permanent magnet is used instead of a wire coil. When the magnet is far away the switch is open, but when the mag-net is brought near the switch is closed as shown in “Reed Switch” on page 312. These are very inexpensive an can be purchased for a few dollars. They are commonly used for safety screens and doors because they are harder to trick than other sensors.

Figure 10.49 Reed Switch

Note: With this device the magnet is moved towards the reed switch. As it gets closer the switch will close. This allows proximity detection without contact, but requires that a separate magnet be attached to a moving part.

10.3.3 Optical (Photoelectric) Sensors

Light sensors have been used for almost a century - originally photocells were used for applications such as reading audio tracks on motion pictures. But modern optical sensors are much more sophisticated.

Optical sensors require both a light source (emitter) and detector. Emitters will produce light beams in the visible and invisible spectrums using LEDs and laser diodes. Detectors are typically built with photodiodes or phototransistors. The emitter and detec-tor are positioned so that an object will block or reflect a beam when present. A basic opti-cal sensor is shown in “A Basic Optical Sensor” on page 313.

Figure 10.50 A Basic Optical Sensor

In the figure the light beam is generated on the left, focused through a lens. At the detector side the beam is focused on the detector with a second lens. If the beam is broken the detector will indicate an object is present. The oscillating light wave is used so that the sensor can filter out normal light in the room. The light from the emitter is turned on and off at a set frequency. When the detector receives the light it checks to make sure that it is at the same frequency. If light is being received at the right frequency then the beam is not broken. The frequency of oscillation is in the KHz range, and too fast to be noticed. A side effect of the frequency method is that the sensors can be used with lower power at longer distances.

An emitter can be set up to point directly at a detector, this is known as opposed mode. When the beam is broken the part will be detected. This sensor needs two separate

oscillator

+V +V

lens lens

square wave

light

LED

phototransistor

amplifierdemodulatordetector andswitching circuits

smaller signal

components, as shown in “Opposed Mode Optical Sensor” on page 314. This arrangement works well with opaque and reflective objects with the emitter and detector separated by distances of up to hundreds of feet.

Figure 10.51 Opposed Mode Optical Sensor

Having the emitter and detector separate increases maintenance problems, and alignment is required. A preferred solution is to house the emitter and detector in one unit. But, this requires that light be reflected back as shown in “Retroreflective Optical Sensor” on page 314. These sensors are well suited to larger objects up to a few feet away.

Figure 10.52 Retroreflective Optical Sensor

emitter object detector

Note: the reflector is constructed with polarizing screens oriented at 90 deg. angles. If the light is reflected back directly the light does not pass through the screen in front of the detector. The reflector is designed to rotate the phase of the light by 90 deg., so it will now pass through the screen in front of the detector.

emitter

detector

object

reflector

emitter

detector

reflector

In the figure, the emitter sends out a beam of light. If the light is returned from the reflector most of the light beam is returned to the detector. When an object interrupts the beam between the emitter and the reflector the beam is no longer reflected back to the detector, and the sensor becomes active. A potential problem with this sensor is that reflective objects could return a good beam. This problem is overcome by polarizing the light at the emitter (with a filter), and then using a polarized filter at the detector. The reflector uses small cubic reflectors and when the light is reflected the polarity is rotated by 90 degrees. If the light is reflected off the object the light will not be rotated by 90 degrees. So the polarizing filters on the emitter and detector are rotated by 90 degrees, as shown in “Polarized Light in Retroreflective Sensors” on page 315. The reflector is very similar to reflectors used on bicycles.

Figure 10.53 Polarized Light in Retroreflective Sensors

For retroreflectors the reflectors are quite easy to align, but this method still requires two mounted components. A diffuse sensors is a single unit that does not use a reflector, but uses focused light as shown in “Diffuse Optical Sensor” on page 316.

object reflectoremitter

detectorhave filters foremitted lightrotated by 90 deg.

reflectoremitter

detector

light reflected withsame polarity

light rotated by 90 deg.

Figure 10.54 Diffuse Optical Sensor

Diffuse sensors use light focused over a given range, and a sensitivity adjustment is used to select a distance. These sensors are the easiest to set up, but they require well controlled conditions. For example if it is to pick up light and dark colored objects prob-lems would result.

When using opposed mode sensors the emitter and detector must be aligned so that the emitter beam and detector window overlap, as shown in “Beam Divergence and Align-ment” on page 317. Emitter beams normally have a cone shape with a small angle of divergence (a few degrees of less). Detectors also have a cone shaped volume of detection. Therefore when aligning opposed mode sensor care is required not just to point the emitter at the detector, but also the detector at the emitter. Another factor that must be considered with this and other sensors is that the light intensity decreases over distance, so the sensors will have a limit to separation distance.

Note: with diffuse reflection the light is scattered. This reduces the quantity of light returned. As a result the light needs to be amplified using lenses.

emitter

detector

object

Figure 10.55 Beam Divergence and Alignment

If an object is smaller than the width of the light beam it will not be able to block the beam entirely when it is in front as shown in “The Relationship Between Beam Width and Object Size” on page 317. This will create difficulties in detection, or possibly stop detection altogether. Solutions to this problem are to use narrower beams, or wider objects. Fiber optic cables may be used with an opposed mode optical sensor to solve this problem, however the maximum effective distance is reduced to a couple feet.

Figure 10.56 The Relationship Between Beam Width and Object Size

Separated sensors can detect reflective parts using reflection as shown in “Detect-ing Reflecting Parts” on page 318. The emitter and detector are positioned so that when a reflective surface is in position the light is returned to the detector. When the surface is not present the light does not return.

effectivebeam angle

emitter

detectoreffectivedetectorangle

effective beam

alignmentis required

intensity 1r2----∝

the smaller beam width is good (but harder to align

emitter detectorobject

Figure 10.57 Detecting Reflecting Parts

Other types of optical sensors can also focus on a single point using beams that converge instead of diverge. The emitter beam is focused at a distance so that the light intensity is greatest at the focal distance. The detector can look at the point from another angle so that the two centerlines of the emitter and detector intersect at the point of inter-est. If an object is present before or after the focal point the detector will not see the reflected light. This technique can also be used to detect multiple points and ranges, as shown in “Multiple Point Detection Using Optics” on page 319 where the net angle of refraction by the lens determines which detector is used. This type of approach, with many more detectors, is used for range sensing systems.

Figure 10.58 Point Detection Using Focused Optics

emitter

detec

tor

reflective surface

emitter

detector

focal point

Figure 10.59 Multiple Point Detection Using Optics

Some applications do not permit full sized photooptic sensors to be used. Fiber optics can be used to separate the emitters and detectors from the application. Some ven-dors also sell photosensors that have the phototransistors and LEDs separated from the electronics.

Light curtains are an array of beams, set up as shown in “A Light Curtain” on page 319. If any of the beams are broken it indicates that somebody has entered a workcell and the machine needs to be shut down. This is an inexpensive replacement for some mechanical cages and barriers.

Figure 10.60 A Light Curtain

The optical reflectivity of objects varies from material to material as shown in

emitter

detector 2

detector 1

distance 1 distance 2lens

lens

“Table of Reflectivity Values for Different Materials [Banner Handbook of Photoelectric Sensing]” on page 320. These values show the percentage of incident light on a surface that is reflected. These values can be used for relative comparisons of materials and esti-mating changes in sensitivity settings for sensors.

Figure 10.61 Table of Reflectivity Values for Different Materials [Banner Handbook of Photoelectric Sensing]

10.3.4 Capacitive Sensors

Capacitive sensors are able to detect most materials at distances up to a few centi-meters. Recall the basic relationship for capacitance.

Kodak white test cardwhite paperkraft paper, cardboardlumber (pine, dry, clean)rough wood palletbeer foamopaque black nylonblack neopreneblack rubber tire wall

clear plastic bottletranslucent brown plastic bottleopaque white plasticunfinished aluminumstraightened aluminumunfinished black anodized aluminumstainless steel microfinishedstainless steel brushed

Reflectivity

90%80%70%75%20%70%14%4%1.5%

40%60%87%140%105%115%400%120%

nonshiny materials

shiny/transparent materials

Note: For shiny and transparent materials the reflectivity can be higher than 100% because of the return of ambient light.

In the sensor the area of the plates and distance between them is fixed. But, the dielectric constant of the space around them will vary as different materials are brought near the sensor. An illustration of a capacitive sensor is shown in “A Capacitive Sensor” on page 321. an oscillating field is used to determine the capacitance of the plates. When this changes beyond a selected sensitivity the sensor output is activated.

Figure 10.62 A Capacitive Sensor

These sensors work well for insulators (such as plastics) that tend to have high dielectric coefficients, thus increasing the capacitance. But, they also work well for metals because the conductive materials in the target appear as larger electrodes, thus increasing the capacitance as shown in “Dielectrics and Metals Increase the Capacitance” on page 322. In total the capacitance changes are normally in the order of pF.

C Akd

------= where, C = capacitance (Farads)k = dielectric constantA = area of platesd = distance between plates (electrodes)

electricfield

objectelectrode

electrode

oscillator

detector

loadswitching

+V

NOTE: For this sensor the proximity of any material near the electrodes will increase the capacitance. This will vary the magnitude of the oscillating signal and the detector will decide when this is great enough to determine proximity.

Figure 10.63 Dielectrics and Metals Increase the Capacitance

The sensors are normally made with rings (not plates) in the configuration shown in “Electrode Arrangement for Capacitive Sensors” on page 322. In the figure the two inner metal rings are the capacitor electrodes, but a third outer ring is added to compensate for variations. Without the compensator ring the sensor would be very sensitive to dirt, oil and other contaminants that might stick to the sensor.

Figure 10.64 Electrode Arrangement for Capacitive Sensors

A table of dielectric properties is given in “Dielectric Constants of Various Materi-als [Turck Proximity Sensors Guide]” on page 324. This table can be used for estimating the relative size and sensitivity of sensors. Also, consider a case where a pipe would carry different fluids. If their dielectric constants are not very close, a second sensor may be desired for the second fluid.

electrode

electrode

electrode

electrode

metal dielectric

electrode

compensatingelectrode

Note: the compensating electrode is used for negative feedback to make the sensor more resistant to variations, such as con-taminations on the face of the sensor.

Material

ABS resin pelletacetoneacetyl bromideacrylic resinairalcohol, industrialalcohol, isopropylammoniaanilineaqueous solutionsash (fly)bakelitebarley powderbenzenebenzyl acetatebutanecable sealing compoundcalcium carbonatecarbon tetrachloridecelluloidcellulosecementcement powdercerealcharcoalchlorine, liquidcokecorneboniteepoxy resinethanolethyl bromideethylene glycolflourFreonTM R22,R502 liq.gasolineglassglass, raw materialglycerine

Constant

1.5-2.519.516.52.7-4.51.016-3118.315-255.5-7.850-801.73.63.0-4.02.351.42.59.12.23.03.2-7.51.5-2.15-103-51.2-1.82.01.1-2.25-102.7-2.92.5-6244.938.72.5-3.06.12.23.1-102.0-2.547

Material

hexanehydrogen cyanidehydrogen peroxideisobutylaminelime, shellmarblemelamine resinmethane liquidmethanolmica, whitemilk, powderednitrobenzeneneoprenenylonoil, for transformeroil, paraffinoil, peanutoil, petroleumoil, soybeanoil, turpentinepaintparaffinpaperpaper, hardpaper, oil saturatedperspexpetroleumphenolphenol resinpolyacetal (Delrin TM)polyamide (nylon)polycarbonatepolyester resinpolyethylenepolypropylenepolystyrenepolyvinyl chloride resinporcelainpress board

Constant

1.995.484.24.51.28.0-8.54.7-10.21.733.64.5-9.63.5-4366-94-52.2-2.42.2-4.83.02.12.9-3.52.25-81.9-2.51.6-2.64.54.03.2-3.52.0-2.29.9-154.93.62.52.92.8-8.12.32.0-2.33.02.8-3.14.4-72-5

Figure 10.65 Dielectric Constants of Various Materials [Turck Proximity Sensors Guide]

The range and accuracy of these sensors are determined mainly by their size. Larger sensors can have diameters of a few centimeters. Smaller ones can be less than a centimeter across, and have smaller ranges, but more accuracy.

10.3.5 Inductive Sensors

Inductive sensors use currents induced by magnetic fields to detect nearby metal objects. The inductive sensor uses a coil (an inductor) to generate a high frequency mag-netic field as shown in “Inductive Proximity Sensor” on page 325. If there is a metal object near the changing magnetic field, current will flow in the object. This resulting cur-rent flow sets up a new magnetic field that opposes the original magnetic field. The net effect is that it changes the inductance of the coil in the inductive sensor. By measuring the inductance the sensor can determine when a metal have been brought nearby.

These sensors will detect any metals, when detecting multiple types of metal mul-tiple sensors are often used.

Material

quartz glassrubbersaltsandshellacsilicon dioxidesilicone rubbersilicone varnishstyrene resinsugarsugar, granulatedsulfursulfuric acid

Constant

3.72.5-356.03-52.0-3.84.53.2-9.82.8-3.32.3-3.43.01.5-2.23.484

Material

Teflon (TM), PCTFETeflon (TM), PTFEtoluenetrichloroethyleneurea resinurethanevaselinewaterwaxwood, drywood, pressed boardwood, wetxylene

Constant

2.3-2.82.02.33.46.2-9.53.22.2-2.948-882.4-6.52-72.0-2.610-302.4

Figure 10.66 Inductive Proximity Sensor

The sensors can detect objects a few centimeters away from the end. But, the direction to the object can be arbitrary as shown in “Shielded and Unshielded Sensors” on page 326. The magnetic field of the unshielded sensor covers a larger volume around the head of the coil. By adding a shield (a metal jacket around the sides of the coil) the mag-netic field becomes smaller, but also more directed. Shields will often be available for inductive sensors to improve their directionality and accuracy.

oscillatorand leveldetector

outputswitching

inductive coilmetal

+V

Note: these work by setting up a high frequency field. If a target nears the field will induce eddy currents. These currents consume power because of resistance, so energy is in the field is lost, and the signal amplitude decreases. The detector exam-ines filed magnitude to determine when it has decreased enough to switch.

Figure 10.67 Shielded and Unshielded Sensors

10.3.6 Ultrasonic

An ultrasonic sensor emits a sound above the normal hearing threshold of 16KHz. The time that is required for the sound to travel to the target and reflect back is propor-tional to the distance to the target. The two common types of sensors are;

electrostatic - uses capacitive effects. It has longer ranges and wider bandwidth, but is more sensitive to factors such as humidity.

piezoelectric - based on charge displacement during strain in crystal lattices. These are rugged and inexpensive.

These sensors can be very effective for applications such as fluid levels in tanks and crude distance measurement.

10.3.7 Hall Effect

Hall effect switches are basically transistors that can be switched by magnetic fields. Their applications are very similar to reed switches, but because they are solid state they tend to be more rugged and resist vibration. Automated machines often use these to do initial calibration and detect end stops.

10.3.8 Fluid Flow

We can also build more complex sensors out of simpler sensors. The example in

shielded unshielded

“Flow Rate Detection With an Inductive Proximity Switch” on page 327 shows a metal float in a tapered channel. As the fluid flow rate increases the pressure forces the float upwards. The tapered shape of the float ensures an equilibrium position proportional to flowrate. An inductive proximity sensor can be positioned so that it will detect when the float has reached a certain height, and the system has reached a given flowrate.

Figure 10.68 Flow Rate Detection With an Inductive Proximity Switch

10.4 SUMMARY• Sourcing sensors allow current to flow out from the V+ supply.• Sinking sensors allow current to flow in to the V- supply.• Photo-optical sensors can use reflected beams (retroreflective), an emitter and

detector (opposed mode) and reflected light (diffuse) to detect a part.• Capacitive sensors can detect metals and other materials.• Inductive sensors can detect metals.• Hall effect and reed switches can detect magnets.• Ultrasonic sensors use sound waves to detect parts up to meters away.


1. Given a clear plastic bottle, list 3 different types of sensors that could be used to detect it.

2. List 3 significant trade-offs between inductive, capacitive and photooptic sensors.

3. Why is a sinking output on a sensor not like a normal switch?

4. a) Sketch the connections needed for the PLC inputs and outputs below. The outputs include a

fluid flow in

fluid flow out

metal inductive proximity sensor

As the fluid flow increases the float is forced higher. A proximity sensor can be used to detect when the float reaches a certain height.

float

24Vdc light and a 120Vac light. The inputs are from 2 NO push buttons, and also from an opti-cal sensor that has both PNP and NPN outputs.

b) State why you used either the NPN or PNP output on the sensor.

5. Select a sensor to pick up a transparent plastic bottle from a manufacturer. Copy or print the specifications, and then draw a wiring diagram that shows how it will be wired to an appropri-ate PLC input card.

6. Sketch the wiring to connect a power supply and PNP sensor to the PLC input card shown

0

1

2

3

4

5

6

7

com

24VDC+

-V+

0

1

2

3

4

5

6

7

24Vdcoutputs

24Vdcinputs

OR

below.

7. Sketch the wiring for inputs that include the following items.3 normally open push buttons1 thermal relay3 sinking sensors1 sourcing sensor

8. A PLC has eight 10-60Vdc inputs, and four relay outputs. It is to be connected to the following devices. Draw the required wiring.

• Two inductive proximity sensors with sourcing and sinking outputs.• A NO run button and NC stop button.• A 120Vac light.• A 24Vdc solenoid.

00

01

02

03

04

05

06

07

COM

24VDC+

-

9. Draw a ladder wiring diagram (as done in the lab) for a system that has two push-buttons and a sourcing/sinking proximity sensors for 10-60Vdc inputs and two 120Vac output lights. Don’t

I:0/0

I:0/1

I:0/2

I:0/3

I:0/4

I:0/5

I:0/6

I:0/7

com

I:0/0

I:0/1

I:0/2

I:0/3

forget to include hard-wired start and stop buttons with an MCR.

10.6 PRACTICE PROBLEM SOLUTIONS

1. capacitive proximity, contact switch, photo-optic retroreflective/diffuse, ultrasonic

2. materials that can be sensed, environmental factors such as dirt, distance to object

3. the sinking output will pass only DC in a single direction, whereas a switch can pass AC and DC.

PLCL1 N

I:0/1

I:0/2

I:0/3

com

I:0/0

O:0/0

O:0/1

O:0/2

O:0/3

Vac

L1 N

4.

0

1

2

3

4

5

6

7

com

24VDC+

-V+

0

1

2

3

4

5

6

7

24Vdcoutputs

24Vdcinputs

hot120Vac

neut.

b) the PNP output was selected. because it will supply current, while the input card requires it. The dashed line indicates the current flow through the sensor and input card.

5.

A transparent bottle can be picked up with a capacitive, ultrasonic, diffuse optical sen-sor. A particular model can be selected at a manufacturers web site (eg., www.ban-ner.com, www.hydepark.com, www.ab.com, etc.) The figure below shows the sensor connected to a sourcing PLC input card - therefore the sensor must be sink-ing, NPN.

V+

0

1

2

3

4

5

6

7

24VDC+

-

6.

00

01

02

03

04

05

06

07

COM

24VDC+

-

7.

00

01

02

03

COM

04

05

06

07

powersupply

+

-24Vdc

V+

00

01

02

03

powersupply

+

-24Vdc

8.

I:0/0

I:0/1

I:0/2

I:0/3

I:0/4

I:0/5

I:0/6

I:0/7

com

O:0/0

O:0/1

O:0/2

O:0/3

powersupply

+

-

V+PNPV-

V+PNPV-

powersupply

+

-

powersupply

120Vac

neut.

9.

10.7 ASSIGNMENT PROBLEMS

1. What type of sensor should be used if it is to detect small cosmetic case mirrors as they pass along a belt. Explain your choice.

2. Summarize the tradeoffs between capacitive, inductive and optical sensors in a table.

3. a) Show the wiring for the following sensor, and circle the output that you are using, NPN or

PLCL1 N

I:0/1

I:0/2

I:0/3

com

I:0/0

O:0/0

O:0/1

O:0/2

O:0/3

Vac

L1 N

MCRstartstop C1

C1

PB1

PB2

PR1

L1

L2

L1 N

V+ V-

C1

PNP. Redraw the sensor using the correct symbol for the sourcing or sinking sensor chosen.

4. A PLC has three NPN and two PNP sensors as inputs, and outputs to control a 24Vdc solenoid and a small 115Vac motor. Develop the required wiring for the inputs and outputs.

V+

0

1

2

3

4

5

6

7

24VDC+

-

24Vdcinputs

11. ACTUATORS

11.1 INTRODUCTION

Actuators Drive motions in mechanical systems. Most often this is by converting electrical energy into some form of mechanical motion.

11.2 SOLENOIDS

Solenoids are the most common actuator components. The basic principle of oper-ation is there is a moving ferrous core (a piston) that will move inside wire coil as shown in Figure 11.69. Normally the piston is held outside the coil by a spring. When a voltage is applied to the coil and current flows, the coil builds up a magnetic field that attracts the piston and pulls it into the center of the coil. The piston can be used to supply a linear force. Well known applications of these include pneumatic values and car door openers.

Figure 11.69 A Solenoid

Topics:

Objectives:• Be aware of various actuators available.

• Solenoids, valves and cylinders• Hydraulics and pneumatics• Other actuators

current off current on

As mentioned before, inductive devices can create voltage spikes and may need snubbers, although most industrial applications have low enough voltage and current rat-ings they can be connected directly to the PLC outputs. Most industrial solenoids will be powered by 24Vdc and draw a few hundred mA.

11.3 VALVES

The flow of fluids and air can be controlled with solenoid controlled valves. An example of a solenoid controlled valve is shown in Figure 11.70. The solenoid is mounted on the side. When actuated it will drive the central spool left. The top of the valve body has two ports that will be connected to a device such as a hydraulic cylinder. The bottom of the valve body has a single pressure line in the center with two exhausts to the side. In the top drawing the power flows in through the center to the right hand cylinder port. The left hand cylinder port is allowed to exit through an exhaust port. In the bottom drawing the solenoid is in a new position and the pressure is now applied to the left hand port on the top, and the right hand port can exhaust. The symbols to the left of the figure show the schematic equivalent of the actual valve positions. Valves are also available that allow the valves to be blocked when unused.

Figure 11.70 A Solenoid Controlled 5 Ported, 4 Way 2 Position Valve

solenoid

solenoid

power inexhaust out

power in exhaust out

The solenoid has two positions and when actuated will change the direction that fluid flows to the device. The symbols shown here are commonly used to represent this type of valve.

Valve types are listed below. In the standard terminology, the ’n-way’ designates the number of connections for inlets and outlets. In some cases there are redundant ports for exhausts. The normally open/closed designation indicates the valve condition when power is off. All of the valves listed are two position valve, but three position valves are also available.

2-way normally closed - these have one inlet, and one outlet. When unenergized, the valve is closed. When energized, the valve will open, allowing flow. These are used to permit flows.

2-way normally open - these have one inlet, and one outlet. When unenergized, the valve is open, allowing flow. When energized, the valve will close. These are used to stop flows. When system power is off, flow will be allowed.

3-way normally closed - these have inlet, outlet, and exhaust ports. When unener-gized, the outlet port is connected to the exhaust port. When energized, the inlet is connected to the outlet port. These are used for single acting cylinders.

3-way normally open - these have inlet, outlet and exhaust ports. When unener-gized, the inlet is connected to the outlet. Energizing the valve connects the out-let to the exhaust. These are used for single acting cylinders

3-way universal - these have three ports. One of the ports acts as an inlet or outlet, and is connected to one of the other two, when energized/unenergized. These can be used to divert flows, or select alternating sources.

4-way - These valves have four ports, two inlets and two outlets. Energizing the valve causes connection between the inlets and outlets to be reversed. These are used for double acting cylinders.

Some of the ISO symbols for valves are shown in Figure 11.71. When using the symbols in drawings the connections are shown for the unenergized state. The arrows show the flow paths in different positions. The small triangles indicate an exhaust port.

Figure 11.71 ISO Valve Symbols

When selecting valves there are a number of details that should be considered, as listed below.

pipe size - inlets and outlets are typically threaded to accept NPT (national pipe thread).

flow rate - the maximum flow rate is often provided to hydraulic valves.operating pressure - a maximum operating pressure will be indicated. Some valves

will also require a minimum pressure to operate.electrical - the solenoid coil will have a fixed supply voltage (AC or DC) and cur-

rent.response time - this is the time for the valve to fully open/close. Typical times for

valves range from 5ms to 150ms.enclosure - the housing for the valve will be rated as,

type 1 or 2 - for indoor use, requires protection against splashestype 3 - for outdoor use, will resists some dirt and weatheringtype 3R or 3S or 4 - water and dirt tighttype 4X - water and dirt tight, corrosion resistant

11.4 CYLINDERS

A cylinder uses pressurized fluid or air to create a linear force/motion as shown in Figure 11.72. In the figure a fluid is pumped into one side of the cylinder under pressure,

Two way, two position

normally closed normally open

normally closed normally open

Three way, two position

Four way, two position

causing that side of the cylinder to expand, and advancing the piston. The fluid on the other side of the piston must be allowed to escape freely - if the incompressible fluid was trapped the cylinder could not advance. The force the cylinder can exert is proportional to the cross sectional area of the cylinder.

Figure 11.72 A Cross Section of a Hydraulic Cylinder

Single acting cylinders apply force when extending and typically use a spring to retract the cylinder. Double acting cylinders apply force in both direction.

For Force:

F PA=P FA---=

Fluid pumped inat pressure P

Fluid flows outat low pressure

F

Fluid pumped inat pressure P

Fluid flows outat low pressure

F

where,P = the pressure of the hydraulic fluidA = the area of the pistonF = the force available from the piston rod

advancing

retracting

Figure 11.73 Schematic Symbols for Cylinders

Magnetic cylinders are often used that have a magnet on the piston head. When it moves to the limits of motion, reed switches will detect it.

11.5 HYDRAULICS

Hydraulics use incompressible fluids to supply very large forces at slower speeds and limited ranges of motion. If the fluid flow rate is kept low enough, many of the effects predicted by Bernoulli’s equation can be avoided. The system uses hydraulic fluid (nor-mally an oil) pressurized by a pump and passed through hoses and valves to drive cylin-ders. At the heart of the system is a pump that will give pressures up to hundreds or thousands of psi. These are delivered to a cylinder that converts it to a linear force and dis-placement.

single acting spring return cylinder

double acting cylinder

Hydraulic systems normally contain the following components;

1. Hydraulic Fluid2. An Oil Reservoir3. A Pump to Move Oil, and Apply Pressure4. Pressure Lines5. Control Valves - to regulate fluid flow6. Piston and Cylinder - to actuate external mechanisms

The hydraulic fluid is often a noncorrosive oil chosen so that it lubricates the com-ponents. This is normally stored in a reservoir as shown in Figure 11.74. Fluid is drawn from the reservoir to a pump where it is pressurized. This is normally a geared pump so that it may deliver fluid at a high pressure at a constant flow rate. A flow regulator is nor-mally placed at the high pressure outlet from the pump. If fluid is not flowing in other parts of the system this will allow fluid to recirculate back to the reservoir to reduce wear on the pump. The high pressure fluid is delivered to solenoid controlled vales that can switch fluid flow on or off. From the vales fluid will be delivered to the hydraulics at high pressure, or exhausted back to the reservoir.

fluid return

air filter

outlet tube

level

refill oil filter

access hatch

gauge

for cleaning

baffle - isolates theoutlet fluid fromturbulence in the inlet

Figure 11.74 A Hydraulic Fluid Reservoir

Hydraulic systems can be very effective for high power applications, but the use of fluids, and high pressures can make this method awkward, messy, and noisy for other applications.

11.6 PNEUMATICS

Pneumatic systems are very common, and have much in common with hydraulic systems with a few key differences. The reservoir is eliminated as there is no need to col-lect and store the air between uses in the system. Also because air is a gas it is compress-ible and regulators are not needed to recirculate flow. But, the compressibility also means that the systems are not as stiff or strong. Pneumatic systems respond very quickly, and are commonly used for low force applications in many locations on the factory floor.

Some basic characteristics of pneumatic systems are,

- stroke from a few millimeters to meters in length (longer strokes have more springiness

- the actuators will give a bit - they are springy- pressures are typically up to 85psi above normal atmosphere- the weight of cylinders can be quite low- additional equipment is required for a pressurized air supply- linear and rotatory

actuators are available.- dampers can be used to cushion impact at ends of cylinder travel.

When designing pneumatic systems care must be taken to verify the operating location. In particular the elevation above sea level will result in a dramatically different air pressure. For example, at sea level the air pressure is about 14.7 psi, but at a height of 7,800 ft (Mexico City) the air pressure is 11.1 psi. Other operating environments, such as in submersibles, the air pressure might be higher than at sea level.

Some symbols for pneumatic systems are shown in Figure 11.75. The flow control valve is used to restrict the flow, typically to slow motions. The shuttle valve allows flow in one direction, but blocks it in the other. The receiver tank allows pressurized air to be accumulated. The dryer and filter help remove dust and moisture from the air, prolonging the life of the valves and cylinders.

Figure 11.75 Pneumatics Components

11.7 MOTORS

Motors are common actuators, but for logical control applications their properties are not that important. Typically logical control of motors consists of switching low cur-rent motors directly with a PLC, or for more powerful motors using a relay or motor starter. Motors will be discussed in greater detail in the chapter on continuous actuators.

Flow control valve

Shuttle valve

Receiver tank

Pump

Dryer

Filter

Pressure regulator

11.8 COMPUTERS

- More complex devices contain computers and digital logic.

- to interface to these we use TTL logic, 0V=false, 5V=true

- TTL outputs cards supply power and don’t need a separeate power supply

- sensitive to electrical noise

11.9 OTHERS

There are many other types of actuators including those on the brief list below.

Heaters - The are often controlled with a relay and turned on and off to maintain a temperature within a range.

Lights - Lights are used on almost all machines to indicate the machine state and provide feedback to the operator. most lights are low current and are connected directly to the PLC.

Sirens/Horns - Sirens or horns can be useful for unattended or dangerous machines to make conditions well known. These can often be connected directly to the PLC.

11.10 SUMMARY

• Solenoids can be used to convert an electric current to a limited linear motion.• Hydraulics and pneumatics use cylinders to convert fluid and gas flows to limited

linear motions.• Solenoid valves can be used to redirect fluid and gas flows.• Pneumatics provides smaller forces at higher speeds, but is not stiff. Hydraulics

provides large forces and is rigid, but at lower speeds.• Many other types of actuators can be used.


1. A piston is to be designed to exert an actuation force of 120 lbs on its extension stroke. The inside diameter of the cylinder is 2.0” and the ram diameter is 0.375”. What shop air pressure will be required to provide this actuation force? Use a safety factor of 1.3.

2. Draw a simple hydraulic system that will advance and retract a cylinder using PLC outputs. Sketches should include details from the PLC output card to the hydraulic cylinder.

3. Develop an electrical ladder diagram and pneumatic diagram for a PLC controlled system. The system includes the components listed below. The system should include all required safety and wiring considerations. a 3 phase 50 HP motor1 NPN sensor1 NO push button1 NC limit switch1 indicator lighta doubly acting pneumatic cylinder


1. A = pi*r^2 = 3.14159in^2, P=FS*(F/A)=1.3(120/3.14159)=49.7psi. Note, if the cylinder were retracting we would need to subtract the rod area from the piston area. Note: this air pressure is much higher than normally found in a shop, so it would not be practical, and a redesign would be needed.

2.

+24Vdc

-

V

00

01

02

03

sump pump

cylinder

regulator

pressure

release

S1

S1

3.


1. Draw a schematic symbol for a solenoid controlled pneumatic valve and explain how the valve operates.

3. A PLC based system has 3 proximity sensors, a start button, and an E-stop as inputs. The sys-tem controls a pneumatic system with a solenoid controlled valve. It also controls a robot with a TTL output. Develop a complete wiring diagram including all safety elelemnts.

4. A system contains a pneumatic cylinder with two inductive proximity sensors that will detect when the cylinder is fully advanced or retracted. The cylinder is controlled by a solenoid con-trolled valve. Draw electrical and pneumatic schematics for a system.

5. Draw an electrical ladder wiring diagram for a PLC controlled system that contains 2 PNP sen-sors, a NO pushbutton, a NC limit switch, a contactor controlled AC motor and an indicator light. Include all safety circuitry.

2. We are to connect a PLC to detect boxes moving down an assembly line and divert larger boxes. The line is 12 inches wide and slanted so the boxes fall to one side as they travel by. One sensor will be mounted on the lower side of the conveyor to detect when a box is present. A second sensor will be mounted on the upper side of the conveyor to determine when a larger box is present. If the box is present, an output to a pneumatic solenoid will be actuated to divert the box. Your job is to select a specific PLC, sensors, and solenoid valve. Details (the absolute minimum being model numbers) are expected with a ladder wiring diagram. (Note: take advantage of manufacturers web sites.)

ADD SOLUTION

12. PROJECT MANAGEMENT

12.1 Introduction

12.2 An Academic View of Design Revisited

• Before we begin,1. Design is the satisfaction of need.2. Design is never an exact process, and each design will differ.3. Try to do it right the first time.4. Most design methods try to cut the problems into smaller problems.

• One of the common problems encountered by designers is the overwhelming number of details. Most design methods focus on dealing with detail overload. The challenges a designer faces are,

- multiple technologies require arbitrary decisions- a design will have many components that interact, and the effects of changes can be

widespread

Topics:

Objectives:•

•

economics- other competitive designs

• Design is typically referred to as having certain stages,- Conceptual- Synthesis- Detailed- Analysis

• The typical stages of design include,

• Design factors commonly considered are,- functional requirements- physical constraints- specifications- aesthetics- usability/user interface

Need

Specifications

feasibility

conceptual

detailed

analysis/testing

review

cost- manufacturing- evaluation/testing/analysis- maintenance- retirement

• A more detailed design sequence is shown below,

• Conceptual - The selection of general components to go into a system. At this point the exact form of final point is inexact. At this point we might be deciding to put wheels on a car.

• Synthesis - The selection of components or devices for the system. At this point the general geometry, and components for the system are selected.

• Detailed - Exact dimensions are finally assigned to parts in the system.

• Analysis - The review of design details to determine suitability. This is done after the exact design is complete. It may lead to redesign.

START: select aneeded design

determinefunctional elements

pick membersgeometry, etc.

select valuesand properties

analyze forsuitability

decide on

DONE: Approvedesign

deficiencies

major changes e.g.replace cable with beam

small changese.g. diameter

no deficiencies

e.g. a support cable

e.g. maximum tension

e.g 1/2” steel cable

e.g. find stress and compareto ultimate strength

e.g. use factorof safety

• The activity of design creates a dilemma for management in that it adds to the overall cost of a product, but it can also reduce the final cost of a product.

• We can draw graphs that illustrate the total amount committed in the final cost from the first concept, to the final product. Most of the final cost is determined by decisions early in the design phase.

• By planning for design, and then committing fully, we can obtain a better product.

• Over-the-wall is an engineering approach that has developed because of management pressures. It helps split designs into clean stages and responsibilities. This approach does simplify man-agement up front, but requires fire fighting as problems arise.

• A product life cycle has four phases,1. Identify needs, plan and design

costs committed

money spenttime

product costs

design only preparation for manufacturing

product starts shipping

design

timeprojectproduction

quality design process(do it right the first time)

rush to design(just do it and fix it later)

changes

releasestart

2. Manufacture and deliver3. Use, maintain, repair4. Retire

12.3 Project ManagementThe management details for the project are outlined below. These may evolve as

the semester progresses.

12.3.1 Timeline - TentativeSept 3 - Teams assignedSept 19 - Preliminary design concept submitted with specifications

materials list estimatebudget estimateGantt charts

Sept 21 - Design concept approvedOct 17 - Proposal submitted

detailed drawings (CAD)materials listbudgetcalculations/simulations

Oct 19 - Proposals approved, mechanical building beginsOct 26 - First mechanical test Nov 2 - Second mechanical testNov 7 - First controlled testNov 14 - Second controlled testNov 16 - First draft of report posted to the web used to determine the score for

competitionNov 21 - EGR 345 competition and judgingNov 30 - Final report draft posted to the webDec 10 - EGR 101 final project morning (EGR 345 students attend)

12.3.2 TeamsThe teams are typically composed of up to 5 students from EGR 345. The EGR

345 teams are grouped using the self evaluation attached in the appendices.

Teams are expected to divided tasks for members to work in parallel. It is also expected that team members will review the work of others to ensure accuracy and completeness. This is particularly true of calculations, materials lists draw-ings and budgets.

All team members are expected to work in a professional manner. The general

rules of conduct in a team are,- treat others as you want to be treated- communicate expectations and problems clearly- be polite and accommodating- when problems arise, help to solve them, even if they are not your fault.

Don’t lay the blame for problems on others.

Personal conflicts must be resolved by team members in a professional manner. The performance of the team will be assessed using peer evaluations on a regu-lar basis and may impact individual grades. In the case of non-participation the penalty may be up to 100% of the project grade.

In the event that team members cannot resolve differences with a team member, the team may ’fire’ the team member by a vote (it must be unanimous, except for the member in question). In this case the ’fired’ student is responsible for finding another team that will accept (or ’hire’) them. In the event they cannot find another team to join, they will be expected to perform all of the work them-self. The firing mechanism is intended to deal with individuals who ’harm’ the progress of the team, not for non-participants.

12.3.3 Conceptual DesignConcepts are normally communicated with sketches that make the overall design

clear. Components that would be expected for this type of design are,- There should be a minimum of 3 conceptual designs, however more are

recommended.- A decision matrix should be used to justify the design selected.- Sketches should be done using normal drafting practices. A good set of

sketches will include 2-D, isometric and pictorial views.- Calculations should be provided that support the design concept.- Electrical schematics to describe the control system.- Block diagram to describe the control system.- Lists of components and budgets to indicate the major parts of the system.- Other items, such as flowcharts, are often required to clarify the design

concept.

12.3.4 Progress ReportsTeams are expected to submit progress reports on a weekly basis. These reports

will include the following elements divided into sections with a heading for each. Point form is preferred, but complete sentences must be used. (Note: ’Parts purchased’ should be ’Parts were purchased for the cart assembly.’) Each section should include items completed since the last report, and current action items. If there is nothing to be said about a category use ’no changes’, ’nothing done’, or ’complete’ as appropriate.

Cover Page - a cover sheet indicating the course, project and team num-bers. The names of all team members should be listed on the cover.

Gantt chart - updated on a weekly basis and included each time. Budget - when changes are made, include an updated budget. The budget

table should include descriptions, suppliers, quantity, price, and status.Mass Table - when changes are made, include an updated mass table. Note:

this can be combined with the budget.Design - Design changes should be indicated. Appropriate drawings, sche-

matics, or equivalent should be included. When appropriate, these should normally be accompanied by a new set of models, calculations and/or simulations to verify the new design.

Software - The current status of software development should be indicated, including major accomplishments and issues.

Fabrication - The status of items being built/assembled should be indicated.Purchasing - The status of ordered items should be indicated.Testing - The testing progress should be indicated, including any numerical

results when available.Other Issues - Items that may impact the success of the team should be

indicated.Performance - A prediction of performance, including the overall perfor-

mance equation.

Early in the semester, other items will be requested, such as a combined timetable for all team members, and a skills inventory. These should only be included in the reports the weeks they are requested.

12.3.5 Design Proposal

The design proposal is used to present all of the design details in a single docu-ment. The focus of this document is a MINIMAL AMOUNT OF TEXT, but a thorough presentation of the design details. Typical elements are listed below in a typical sequence;

- a cover page indicating all of the team members and all other pertinent informa-tion.

- a table of contents- three view drawings of each significant part- block diagram of the control system (s)- block diagrams showing the system architecture- circuit schematic- if a motion profile will be used, it should be documented- an assembly drawing of the mechanism, including a BOM- a budget listing each of the parts that must be purchased/acquired. Catalog pages

and quotes can be used to validate the budget. In the final report, copies of receipts, or catalog pages will be required.

a weight inventory, itemized by each part of the design- additional calculations for mechanical design issues, such as stress that may

result in failure. Normally these result in a factor of safety.- the equations of motion for the system- a Scilab program that verifies the operation of the system using the equations of

motion.- a prototype C program that implements the controller as designed.

12.3.6 The Final Report

The final design will follow the same structure as the Design Proposal, with the addition of the following elements.

- test results- the drawings, calculations, etc. should be based upon the final design. It is rea-

sonable to write a page or less about the modifications that were required, but it is a minimal/optional part of the report.

The report should concisely and clearly describe the design, as shown in the dia-grams, drawings, calculations, etc. In general the format of the report is as outlined below. Sections and subsections should be numbered.

Cover pageExecutive summary - one page or less that summarizes the design and results.Table of contentsDesign description - this section should describe the mechanical, electrical and

software design aspects. Subsections should focus on the following elements;Drawing summary - selected isometric and assembly drawingsSystem block diagramsDescription of control scheme, such as the motion profileSchematicsCalculations - FBDs and differential equationsProject budget and BOMWeight inventory

Test results - this section should describeSimulation resultsThe tests that were done to describe the overall performance. There should

be a comparison of the cart with and without sway compensation.The results of formal tests should also be described.A comparison of overall score estimates.

Conclusions - A brief description of the overall results indicating what the strengths and weaknesses of the design.

Recommendations - Suggestions for improvement.

Appendix - DrawingsAppendix - Stress and other similar calculationsAppendix - Controller C programAppendix - Simulation programAppendix - Receipts and cost evidence

Final reports will be evaluated on numerous factors including the clarity for the design documentation (i.e., how clear is what has been done?), theory to backup the design (does the theory match the actual design?), did the theory and actual match?

12.3.7 Gantt ChartsGantt charts are used to track major project tasks, including sequencing. a brief

tutorial on project management with Microsoft Project is available at http://www.stylusinc.net/ms_project_tutorial/project_management.shtml. You may also download (free for 30 days) project management software from http://www.smartworks.us/htm/downloads.htm. In general a Gantt chart should include,

- Most project parts should have a development, and then a review stage by another team member. For example, if one task is preparing mechanical drawings, this should be followed by a review task - done by another team member.

- Most tasks that stretch more than a week should be broken into smaller sub tasks.

- A Gantt chart should be presented on one single sheet. If it is small enough this can be one page, otherwise a larger sheet can be used.

- On a weekly basis a Gantt chart should be updated to include the comple-tion of tasks.

12.3.8 DrawingsAll drawings will observe the standards used in EGR 101 (see the EGR 101 or 345

course pages). This includes dimensions and tolerances that can be produced using the available equipment and materials. Please note that ’sketch’ means that it is done by hand, approximately, while drawing means it is done formally in a CAD package. Normally you should create solid models, and then generate multiview drawings. Note: All drawings must have a title block.

Shaded views have very little value and should be avoided, wireframe drawings are much more useful. If there is a definite need to include a shaded drawing, change the background to change the quantity of toner used.

12.3.9 Budgets and Bills of MaterialBudgets should list all substantial components. Consumables, such as bolts are

normally listed under a ’miscellaneous’ heading. However, all other compo-

nents should be listed, and prices provided. If the components have been drawn from the engineering stores, similar devices can be identified from catalogs and those prices may be used. If your design calls for parts not commonly issued to engineering students, you may be required to purchase these yourself. An excel-lent local source of small parts is hobby stores, such as Ryder’s Hobbies on 28th St. The budget should also list the quantity of parts/material, price, source/sup-plier and status (eg., not ordered, received, due 2 weeks, late 1 week).

Don’t forget to include cost of the controller and other components used. Assume the power supply is provided as part of the crane system and therefore has no cost is associated. Some of the commonly available components are listed below. List simple commodity items such as wires, bolts, etc., under a miscella-neous category with a general cost estimate.

ATMega32 circuit board $30Breakout board $20motors assume $10CT3001-ND 16 position, non-detent mechanical encoder $2.58Potentiometer 10K (Digikey #296xd103b1n) $3.61 (???)

A Bill of Materials (BOM) lists all of the parts required to produce or assemble some other device. This is different from parts listed in a budget in that some of the parts will be work in process. In other words, the original material has been worked on to produce new parts. A BOM is normally found on assembly draw-ings.

Notes;- there are different type of plastic, some are more brittle, others are

tougher.

12.3.10 CalculationsCalculations are required to justify the design work. These should follow the con-

ventions used in EGR 345. When computer programs are written, they should be commented and included.

12.4 Examples

12.5 Summary


12.7 Problems


12.9 Forms

Skills Self Evaluation

Your Name:

Hands-on Mechanical: 1 2 3 4 5

none proficient

Personal/Technical Strengths:

Hands-on Electrical: 1 2 3 4 5

Hands-on Computer Usage: 1 2 3 4 5

Mathematical Problem Solving: 1 2 3 4 5

Writing: 1 2 3 4 5

The ability to build components with wood, plastic, metal or other materials.

CAD, Spreadsheets, creating web pages, etc.

Basic wiring skills, soldering, etc.

Ability to formulate and solve complex problems

Layout and write complex documents

Teamwork Skills: 1 2 3 4 5The ability to work with others in a team environment.

Personal/Technical Weaknesses:

Hands-on Computer Application: 1 2 3 4 5Programming and computer interfacing

Leadership Skills: 1 2 3 4 5The ability to act as a role model that teammates will follow.

People you would NOT like to work with:People you would like to work with:

Design Skills: 1 2 3 4 5Work in unstructured/semistructured problem solving.

Other Items of Interest:

Other Commitments (courses, work, etc. - give hours for each)

13. MOTION CONTROL

13.1 INTRODUCTION

A system with a feedback controller will attempt to drive the system to a state described by the desired input, such as a velocity. In earlier chapters we simply chose step inputs, ramp inputs and other simple inputs to determine the system response. In practical applications this setpoint needs to be generated automatically. A simple motion control system is used to generate setpoints over time.

An example of a motion control system is shown in Figure 1.76. The motion con-troller will accept commands or other inputs to generate a motion profile using parameters such as distance to move, maximum acceleration and maximum velocity. The motion pro-file is then used to generate a set of setpoints, and times they should be output. The set-point scheduler will then use a real-time clock to output these setpoints to the motor drive.

Topics:

Objectives:• To understand single and multi axis motion control systems.

• Motion controllers• Motion profiles, trapezoidal and smooth• Gain schedulers

Figure 1.76 A motion controller

The combination of a motion controller, drive and actuator is called an axis. When there is more than one drive and actuator the system is said to have multiple axes. Com-plex motion control systems such as computer controlled milling machines (CNC) and robots have 3 to 6 axes which must be moved in coordination.

13.2 MOTION PROFILES

13.2.1 Velocity Profiles

A simple example of a velocity profile for a point-to-point motion is shown in Fig-ure 1.77. In this example the motion starts at 20 deg and ends at 100 deg. (Note: in motion controllers it is more common to used encoder pulses, instead of degrees, for positions velocities, etc.) For position control we want a motion that has a velocity of zero at the start and end of the motion, and accelerates and decelerates smoothly.

ServoDrive

SetpointScheduler

SetpointGenerator

Motormotion controller

motioncommands

t

ωt (s)

00.10.20.30.4.....

setpoint

0.00.20.40.40.4....

0 0.2 0.6 0.8

0.4

motion profile

setpoint schedule

Figure 1.77 An example of a desired motion (position)

A trapezoidal velocity profile is shown in Figure 1.78. The area under the curve is the total distance moved. The slope of the initial and final ramp are the maximum acceler-ation and deceleration. The top level of the trapezoid is the maximum velocity. Some con-trollers allow the user to use the acceleration and deceleration times instead of the maximum acceleration and deceleration. This profile gives a continuous acceleration, but there will be a jerk (third order derivative) at the four sharp corners.

Figure 1.78 An example of a velocity profile

θ deg( )

20

100

t (s)

ω deg( )

0t (s)

ωmax

αmax

αmax–

tacc tdec ttotal

where,ωmax the maximum velocity=

αmax the maximum acceleration=

tacc tdec, the acceleration and deceleration times=

tmax the times at the maximum velocity=

0

ttotal the total motion time=

tmax

The basic relationships for these variables are shown in Figure 1.79. The equations can be used to find the acceleration and deceleration times. These equations can also be used to find the time at the maximum velocity. If this time is negative it indicates that the axis will not reach the maximum velocity, and the acceleration and deceleration times must be decreased. The resulting velocity profile will be a triangle.

Figure 1.79 Velocity profile basic relationships

For the example in Figure 1.80 the move starts at 100deg and ends at 20 deg. The acceleration and decelerations are completed in half a second. The system moves for 7.5 seconds at the maximum velocity.

tacc tdecωmaxαmax------------= =

θ 12---taccωmax tmaxωmax

12---tdecωmax+ + ωmax

tacc2

-------- tmaxtdec2

--------+ +⎝ ⎠⎛ ⎞= =

ttotal tacc tmax tdec+ +=

(1)

(2)

(3)

tmaxθ

ωmax------------

tacc2

-----------–tdec2

-----------–=

Note: if the time calculated in equation 4 is negative then the axis never reaches maximum velocity, and the velocity profile becomes a triangle.

(4)

Figure 1.80 Velocity profile example

The motion example in Figure 1.81 is so short the axis never reaches the maximum velocity. This is made obvious by the negative time at maximum velocity. In place of this the acceleration and deceleration times can be calculated by using the basic acceleration position relationship. The result in this example is a motion that accelerates for 0.316s and then decelerates for the same time.

tacc tdecωmaxαmax------------

10degs

---------

20degs2

------------------------ 0.5s= = = =

ttotal tacc tmax tdec+ + 0.5s 7.5s 0.5s+ + 8.5s= = =

tmaxθ

ωmax------------

tacc2

-----------–tdec2

-----------– 80deg

10degs

------------------------ 0.5s

2----------– 0.5s

2----------– 7.5s= = =

Given, θstart 100deg= θend 20deg=

ωmax 10degs

---------= αmax 20degs2

---------=

The times can be calculated as,

θ θend θstart– 20deg 100deg– 80deg–= = =

Figure 1.81 Velocity profile example without reaching maximum velocity

Given the parameters calculated for the motion, the setpoints for motion can be calculated with the equations in Figure 1.82.


10degs

---------

20degs2

------------------------ 0.5s= = = =

tmaxθ

ωmax------------

tacc2

-----------–tdec2

-----------– 2deg

10degs

------------------------ 0.5s

2----------– 0.5s

2----------– 0.3– s= = =

Given, θstart 20deg= θend 22deg=

ωmax 10degs

---------= αmax 20degs2

---------=

The times can be calculated as,

θ θend θstart– 22deg 20deg– 2deg= = =

The time was negative so the acceleration and deceleration times become,

θ2--- 1

2---αmaxtacc

2=

taccθ

αmax------------ 2deg

20degs2

------------------------ 0.1s2 0.316s= = = =

tmax 0s=

Figure 1.82 Generating points given motion parameters

A subroutine that implements these is shown in Figure 1.83. In this subroutine the time is looped with fixed time steps. The position setpoint values are put into the setpoint array, which will then be used elsewhere to guide the mechanism.

0s t tacc<≤

Assuming the motion starts at 0s,

θ t( ) 12---αmaxt2 θstart+=

tacc t tacc tmax+<≤

θ t( ) 12---αmaxtacc

2 ωmax t tacc–( ) θstart+ +=

tacc tmax+ t tacc tmax tdec+ +<≤

θ t( ) 12---αmaxtacc

2 ωmaxtmax12---αmax t tmax tacc––( )2 θstart+ + +=

tacc tmax tdec+ + t≤

θ t( ) θend=

Figure 1.83 Subroutine for calculating motion setpoints

In some cases the jerk should be minimized. This can be achieved by replacing the acceleration ramps with a smooth polynomial, as shown in Figure 1.84. In this case two quadratic polynomials will be used for the acceleration, and another two for the decelera-tion.

void generate_setpoint_table( double t_acc, double t_max, double t_step, double vel_max, double acc_max, double theta_start, double theta_end, double setpoint[], int *count)

double t, t_1, t_2, t_total;t_1 = t_acc;t_2 = t_acc + t_max;t_total = t_acc + t_max + t_acc;*count = 0;for(t = 0.0; t <= t_total; t += t_step)

if( t < t_1) setpoint[*count] = 0.5*acc_max*t*t + theta_start; else if ( (t >= t_1) && (t < t_2)) setpoint[*count] = 0.5*acc_max*t_acc*t_acc

+ vel_max*(t - t_1) + theta_start; else if ( (t >= t_2) && (t < t_total))

setpoint[*count] = 0.5*acc_max*t_acc*t_acc + vel_max*(t_max) + 0.5*acc_max*(t-t_2)*(t-t_2) + theta_start;

else setpoint[*count] = theta_end;

*count++;

setpoint[*count] = theta_end;*count++;

Figure 1.84 A smooth velocity profile

An example of calculating the polynomial coefficients is given in Figure 1.85. The curve found is for the first half of the acceleration. It can then be used for the three other required curves.

ω deg( )

0t (s)

ωmax

αmax αmax–

tacc tdec ttotal

where,ωmax the maximum velocity=

αmax the maximum acceleration=

tacc tdec, the acceleration and deceleration times=

tmax the times at the maximum velocity=

0

ttotal the total motion time=

tmax

ω t( ) At2 Bt C+ +=

Figure 1.85 A smooth velocity profile example

Given, θstart θend ωmaxαmax

The constraints for the polynomial are,

ω 0( ) 0= ωtacc2

--------⎝ ⎠⎛ ⎞ ωmax

2------------=

ddt-----ω 0( ) 0= d

dt-----ω

tacc2

--------⎝ ⎠⎛ ⎞ αmax=

These can be used to calculate the polynomial coefficients,

0 A02 B0 C+ += C∴ 0=

ωmax Atacc2=

0 2A0 B+=

The equation for the first segment is,

ω t( )αmax

2

4ωmax---------------t2=

B∴ 0=

αmax 2Atacc=

Aωmax

tacc2

------------=

Aαmax2tacc------------=

Aωmax

tacc2

------------αmax2tacc------------= = tacc

2ωmaxαmax

---------------=

Aαmax2tacc------------

αmax

22ωmaxαmax

---------------⎝ ⎠⎛ ⎞

------------------------αmax

2

4ωmax---------------= = =

The equation for the second segment can be found using the first segment,

ω t( ) ωmaxαmax

2

4ωmax--------------- tacc t–( )2–=

0 ttacc2

--------<≤

tacc2

-------- t tacc<≤ω t( ) ωmax

αmax2

4ωmax--------------- t2 2tacct– tacc

2+( )–=

Figure 1.86 A smooth velocity profile example (cont’d)

13.2.2 Position Profiles

A motion can be described using position points along a path. These methods are normally used when a controller does not have any velocity or acceleration limits. The method shown in Figure 1.87 controls motion using a parametric function ’p(u)’. The function value varies from 0 to 1 as the parameter ’u’ varies from 0 to 1. However, the parameters of the function are selected so that the motion starts and stops with a velocity of zero. In this case the final polynomial equation, (3), is fairly simple. This equation can then be used in equation (1) to generate a smooth motion path between any arbitrary start

The distance covered during acceleration, the area under the curves, is,

so the time required at the maximum velocity is,

tmaxθ 2θacc–( )ωmax

---------------------------=

θaccαmax

2

4ωmax---------------t2 td

0

tacc

2--------

∫ ωmaxαmax

2

4ωmax--------------- t2 2tacct– tacc

2+( )–⎝ ⎠⎜ ⎟⎛ ⎞

tdtacc

2--------

tacc

∫+=

θaccαmax

2

12ωmax------------------t3

0

tacc

2--------

ωmaxtαmax

2

4ωmax--------------- t3

3---- tacct2– tacc

2 t+⎝ ⎠⎛ ⎞–

⎝ ⎠⎜ ⎟⎛ ⎞

tacc

2--------

tacc

+=

accαmax

2

12ωmax------------------

tacc3

8-------- ωmaxtacc

αmax2

4ωmax---------------

tacc3

3-------- tacc

3– tacc3+

⎝ ⎠⎜ ⎟⎛ ⎞

– ωmaxtacc2

--------αmax

2

4ωmax---------------

tacc3

24--------

tacc3

4--------–

tacc3

2--------+

⎝ ⎠⎜ ⎟⎛ ⎞

+–+=

θaccαmax

2

96ωmax------------------tacc

3 ωmaxtacc2

---------------------αmax

2

12ωmax------------------tacc

3–7αmax

2

96ωmax------------------tacc

3–+=

θacc14α– max

2

96ωmax--------------------- tacc

3 ωmaxtacc2

---------------------+=

and end point, with arbitrary start and end times.

Figure 1.87 Generating smooth motion paths

The example in Figure 1.88 shows the use of a trigonometric function, instead of a polynomial. This function was used to generate the points in the following sample pro-gram in Figure 1.89.

where,θstart θend, start and end positions of motion=

The constraints for the polynomial are,

p 0( ) 0= p 1( ) 1=

ddt-----p 0( ) 0=

ddt-----p 1( ) 0=

These can be used to calculate the polynomial coefficients,

0 A03 B02 C0 D+ + +=

C∴ 0=

B∴ 32---–⎝ ⎠

⎛ ⎞A=

θ t( ) θstart θend θstart–( )pt tstart–

tend tstart–--------------------------⎝ ⎠⎛ ⎞+=

tstart tend, start and end times for the motion=

p u( ) Au3 Bu2 Cu D+ + +=

D∴ 0=

0 3A02 2B0 C+ +=

0 3A12 2B1+=

1 A13 B12 0( )0 0+ + += A∴ 2–=

B∴ 3=

p u( ) 2– u3 3u2+=

(1)

(2)

(3)

Figure 1.88 Generating smooth motion paths

The program in Figure 1.89 generates a motion table that can then be used to update setpoints. The function ’table_init()’ must be called once when the program starts to set up global time and table values. When a new target position has been specified the ’table_generate()’ function is called to generate the setpoint table. The ’table_update()’ function is called once every interrupt scan to check the setpoint table, and update the glo-bal setpoint variable, ’point_current’ at scheduled times. This function also includes a simple clock to keep track of the system time.

where,

The coefficients can be calculated using the conditions used previously,

p 0( ) A B 0( ) π2---–⎝ ⎠

⎛ ⎞sin D+ 0= =

p u( ) A Bt C+( )sin D+=

ddt-----p 0( ) AB B 0( ) C+( )cos 0= =

p u( ) 12--- πt π

2---–⎝ ⎠

⎛ ⎞sin 12---+=

C∴ π2---–=

ddt-----p 1( ) AB B 1( ) C+( )cos 0= =

p 1( ) A π 1( ) π2---–⎝ ⎠

⎛ ⎞sin D+ 1= =

A∴ D=

C( )cos∴ 0=

B C+( )cos∴ 0=

B C+∴ π2---= B∴ π=

A∴ 1( ) A+ 1= A∴ 12---=

The final relationship is,

Figure 1.89 Subroutines for motion profile generation and use

13.3 MULTI AXIS MOTION

In a machine with multiple axes the motions of individual axes must often be coor-dinated. A simple example would be a robot that needs to move two joints to reach a new position. We could extend the motion of the slower joints so that the motion of each joint would begin and end together.

#define TABLE_SIZE 11int point_master[TABLE_SIZE] = 0, 24, 95, 206, 345, 500, 655, 794, 905, 976, 1000;int point_position[TABLE_SIZE];int point_time[TABLE_SIZE];int point_start_time;int point_index;

int ticks; /* variables to keep a system clock count */int point_current; /* a global variable to track position */

int table_init() /* initialize the setpoint table */ticks = 0; /* set the clock to zero */point_current = 0; /* start the system at zero */point_index = TABLE_SIZE; /* mark the table as empty */

void table_generate(int start, int end, int duration_sec) unsigned i;

point_time[0] = ticks + 10; /* delay the start slightly */point_position[0] = start;

for(i = 1; i < TABLE_SIZE; i++)point_time[i] = point_time[0] +

(unsigned long)i * duration_sec * 250 / (TABLE_SIZE - 1);point_position[i] = start + (long int)(end - start) * point_master[i] / 1000;

point_index = 0;

int table_update()/* interrupt driven encoder update */ticks++; /* update the clock */

if(point_index < TABLE_SIZE)if(point_time[point_index] == ticks)

point_current = point_position[point_index++];outint16(point_current);putch("\n");

return point_current;

13.3.1 Slew Motion

When the individual axis of a machine are not coordinated this is known as slew motion. Each of the axes will start moving at the same time, but finish at separate times. Consider the example in Figure 1.90. A three axis motion is required from the starting angles of (40, 80, -40)deg, and must end at (120, 0, 0)deg. The maximum absolute acceler-ations and decelerations are (50, 100, 150) degrees/sec/sec, and the maximum velocities are (20, 40, 50) degrees/sec.

Figure 1.90 Multi-axis slew motion

The calculations for the motion parameters are shown in Figure 1.91. These are done in vector format for simplicity. All of the joints reach the maximum acceleration. The fastest motion is complete in 1.13s, while the longest motion takes 4.4s.

-90

90

180

time(sec)

Joint angle (degrees)

Joint velocity (degrees/sec)

θ2 θ1

θ3

αmax–ωmax

tacc tmax tdec

Figure 1.91 Calculated times for the slew motion

1.3.1.1 - Interpolated Motion

In interpolated motion the faster joints are slowed so that they finish in coordina-tion with the slowest. This is essential in devices such as CNC milling machines. If this did not occur a straight line cut in the x-y plane would actually be two straight lines. The slew motion example can be extended to be slew motion where all joints finish their motion at 4.4s. This can be done by accelerating at the maximum acceleration, but setting a new maximum velocity. This is shown in the example in Figure 1.92 using the results from the example in Figure 1.91.

tacc tdecωmaxαmax------------ 20

50------ 40

100--------- 50

150---------, ,⎝ ⎠

⎛ ⎞ 0.4 0.4 0.333, ,( )sec.= = = =

θacc. θdec.taccωmax.vel.

2---------------------------- 0.4 20( )

2------------------ 0.4 40( )

2------------------ 0.333 50( )

2------------------------, ,⎝ ⎠

⎛ ⎞ 4 8 8.33, ,( )deg.= = = =

The next step is to examine the moves specified,

θmove θend θstart– 120 40– 0 80– 0 40–( )–, ,( ) 80 80– 40, ,( )deg.= = =

Remove the angles covered during accel./deccel., and find the travel time at maximumvelocity.

tmaxθmove 2θacc–

ωmax----------------------------------- 80 2 4( )–

20----------------------- 80 2 8( )–

40----------------------- 40 2 8.333( )–

50---------------------------------, ,⎝ ⎠

⎛ ⎞= =

tmax 3.6 1.6 0.46668, ,( )sec.=

ttotal tacc tmax tdec+ + 4.4 2.4 1.13, ,( )s= =

The area under the velocity curve is the distance (angle in this case) travelled. First we can determine the distance covered during acceleration, and deceleration and the time during acceleration, and deceleration.

Figure 1.92 Interpolated motion based upon Figure 1.91

1.3.2 Motion Scheduling

After the setpoint schedule has been developed, it is executed by the setpoint scheduler. The setpoint scheduler will use a clock to determine when an output setpoint should be updated. A diagram of a scheduler is shown in Figure 1.93. In this system the setpoint scheduler is an interrupt driven subroutine that compares the system clock to the

The longest motion time is 4.4s for joint 1, and this can be used to prolong the other motions. The calculation begins by rewriting the velocity/position relationship using a new maximum velocity.

∆θ 12---ωmax'αmax-------------ωmax' ttotal 2

ωmax'αmax-------------–⎝ ⎠

⎛ ⎞ωmax'12---ωmax'αmax-------------ωmax'+ +=

A new maximum velocity can be calculated for joint 2 using this equation.

∆∴ θ ωmax' ttotalωmax'αmax-------------–⎝ ⎠

⎛ ⎞=

ωmax'( )2 ωmax' ttotalαmax–( ) ∆θαmax+ +∴ 0=

ωmax'∴ttotalαmax ttotal

2 αmax2 4∆θαmax–±

2--------------------------------------------------------------------------------------=

ωmax'∴ 4.4 100( ) 4.4( )2 100( )2 4 80( ) 100( )–±2

--------------------------------------------------------------------------------------------------=

ωmax'∴ 440 401.99502±2

---------------------------------------- 421 19.0,= =

tacc' tacc'ωmax'αmax-------------= =

tacc' tacc'19.0100---------- 0.19s= = =

A new maximum velocity can be calculated for joint 3 using this equation.

ωmax'∴ 4.4 150( ) 4.4( )2 150( )2 4 40( ) 150( )–±2

--------------------------------------------------------------------------------------------------=

ωmax'∴ 651 9.22,=

tacc' tacc'9.22100---------- 0.092s= = =

total motion time. When enough time has elapsed the routine will move to the next value in the setpoint table. The frequency of the interrupt clock should be smaller than or equal to the time steps used to calculate the setpoints. The servo drive is implemented with an algorithm such a PID control.

Figure 1.93 A setpoint scheduler

The output from the scheduler updates every time step. This then leads to a situa-tion where the axis is always chasing the target value. This leads to small errors, as shown in Figure 1.94.

InterruptClock

Setpointtable

Choose newpoint fromtrajectory table

Return

θdesired

Time based interruptroutine

Servo motor routine runsfor each axis

Computeerror

Output actuatorsignal

Read θdesired

Figure 1.94 Errors in path following

1.4 PATH PLANNING• When we have simple features, paths are easy to generate. These features include,

- steps- pockets- holes- etc.

• Typically paths for these will repeat as shown below,

speed

trajectory table

actual position

timetrajectory table time step

required

actual

position

time

• For complex surfaces we want to contour appropriately. These surfaces will almost always

be represented with spline patches.

• Recall that a spline patch can be represented parametrically

• A simple algorithm to cut the surface is shown below.

1.5 CASE STUDIES

- the controller described in the block diagram below uses a model for a DC per-manent magnet DC motor to estimate a voltage based upon a predicted velocity and posi-tion.

- the desired position and velocity are given in figure xxx based upon the motion position control derived in the earlier section

(u=0,v=0)

(u=0,v=1)

(u=1,v=0)

(u=1,v=1)

p u v,( )xp

yp

zp

=

dirn_flag = 1; a direction flagn=10 ; number of passes to cut the surfacestep=1.0/n ; step sizes for u and v directionsstart=step/2 ; the start offset in the u and v directions[xp,yp,zp] = p(start,start) ; calculate the start positionprint(“G00 X”,xp,” Y”,yp,” Z”,zp+0.2) ; move the tool to above the start positionfor i=0 to (n-1) ; will increment in the u direction

for j=0 to (n-1) ; will increment in the v direction; calculate next point

if dirn_flag=-1 then [xp,yp,zp]=p(start+i*step,start+j*step)if dirn_flag=1 then [xp,yp,zp]=p(start+i*step,start+(n-j)*step)print(“G01 X”,xp,” Y”,yp,” Z”,zp) ; instruction to cut to next point

next j ; make next step in v direction until donedirn_flag = -dirn_flag ; reverse direction to cut in opposite direction

next i ; move to next cut line in the u directionprint(“G00 Z”,zp+0.2) ; move the tool to above the end position

Figure 1.95 Motion position control equation

- the controller that uses the desired position and velocity is shown in figure xxxx

Figure 1.96 Feedforward trajectory controller

θ t( ) θstart θend θstart–( )pt tstart–

tend tstart–--------------------------⎝ ⎠⎛ ⎞+=

p u( ) 2– u3 3u2+=

ω t( ) θend θstart–( )p·t tstart–

tend tstart–--------------------------⎝ ⎠⎛ ⎞=

p· u( ) 6– u2 6u+=

KJRD K2+------------------------

JRD K2+K

------------------------

+

++

-

1Kenc-----------

P

Kenc

ωd

θd θa

real-time control loop

physical system

1.6 SUMMARY• Axis limits can be used to calculate motion profiles.• Trapezoidal and smooth motion profiles were presented.• Motion profiles can be used to generate setpoint tables.• Values from the setpoints can then be output by a scheduler to drive an axis.


1. a) Develop a motion profile for a joint that moves from -100 degrees to 100 degrees with a maximum velocity of 20 deg/s and a maximum acceleration of 100deg/s/s. b) Develop a set-point table that has values for positions every 0.5 seconds for the entire motion.

2. Consider a basic servo controller with encoder feedback. The motor will start at a position count of 100, and end the motion at a count of 3000. The motion is to have a maximum accel-eration of 300 counts/s/s, and a maximum velocity of 100 counts/s. Find a motion profile that satisfies these constraints. Generate a table of setpoints for the desired position every 2 sec-onds.


1.

ωmax 20degs

---------=

Given,


20degs

---------

100degs2

--------------------------- 0.2s= = = =

The motion times can be calculated.

αmax 100degs2

---------=

tmax∆θ ωmaxtacc–

ωmax----------------------------------

200deg 20degs

---------0.2s–

20degs

------------------------------------------------------------ 9.8s= = =

∆θ 100 100–( )– 200°= =

ttotal tacc tmax tdec+ + 0.2s 9.8s 0.2s+ + 10.2s= = =

t (s)

0.00.51.01.52.02.53.03.54.04.55.05.56.06.57.07.58.08.59.09.510.010.511.0

angle(deg)-100-92-82-72-62-52-42-32-22-12-28182838485868788898100100

θ0.5s12---100deg

s2--------- 0.2s( )2 20deg

s--------- 0.5s 0.2s–( ) 100deg–+=

θ0.5s 92deg–=

θ1.0s θ0.5s 20degs

--------- 0.5s( )+ 92deg– 10deg+= =

2.


1. Find a smooth path for a robot joint that will turn from θ= 75° to θ = -35° in 10 seconds. Do this by developing an equation then calculating points every 1.0 seconds along the path for a total motion time of 10 seconds. Do not assume a maximum velocity.

2. Paths are to be planned for a three axis motion controller. Each of the joints has a maximum velocity of 20 deg/s, and a maximum acceleration of 30 deg/s/s. Assuming all of the joints start at an angle of 0 degrees. Joints 1, 2 and 3 move to 40 deg, 100deg and -50deg respectively. Develop the motion profiles assuming,

a) slew motionb) interpolated motion

3. Develop a smooth velocity profile for moving a cutting tool that starts at 1000 inches and moves to -1000 inches. The maximum velocity is 100 in/s and the maximum acceleration is 50in/s/s.

4. An axis has a maximum velocity of 2.0 rad/s and a maximum acceleration/deceleration of 0.5 rad/s^2. Sketch a trapezoidal motion profile and find the motion time for a -4.0 rad motion.

5. An axis has a maximum velocity of 5m/s and an acceleration/deceleration time of 1s. Sketch a trapezoidal motion profile and find the motion time for a 25m motion.

6. Develop a smooth velocity profile for moving a cutting tool that starts at 1000 inches and moves to -1000 inches. The maximum velocity is 100 in/s and the maximum acceleration is

t(s)024681012141618202224262830

counts

10028348368388310831283148316831883208322832483268328833000

50in/s/s.

Implementation

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