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Cam Designing

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Page 1: Cam Designing

Chapter 8: Cam Design

Chapter Outcomes

• At the end of this chapter, you should be able to:– Derive SVAJ functions to fulfill cam design

specifications

8.0: Introduction

• A cam-follower system:

8.0: Introduction

• This chapter will focus on designing the cam, and not the follower.

• Advantages of cam-follower system: FLEXIBILITY– Can specify any desired output function– Create a curved surface on the cam to generate the function– The follower will trace the cam profile and hence generate the

desired function• Disadvantage of cam-follower system:

– Follower output rocking motion

Page 2: Cam Designing

8.0: Introduction

• Example of real world application of cam-follower system:

Engine Valve Timing Control

8.1: Cam Terminology

Follower Motion

Rotating follower (analogous to crank-rocker linkage)

Translating follower (analogous to slider-crank linkage)

Follower Type

Flat-faced follower Mushroom follower Roller follower

Page 3: Cam Designing

Cam Type

Radial cam (follower motion in radial direction of cam)

Axial cam (follower motion in axial direction of cam)

Type of Joint Closure

Force closure: Use force, usually through a spring

Form closure: Use geometry, usually through a track/groove

How to keep cam and follower always in contact?

Type of Motion Constraints

• Critical Extreme Position (CEP):– Design specifications define only start and

finish positions of the follower– Designer has freedom to determine motion

between start and finish position• Critical Path Motion (CPM):

– Design specifications define motion path / velocity / acceleration over entire interval of motion

• Whiteboard example

Type of Motion Program

• Rise Fall (RF)• Rise Fall Dwell (RFD)• Rise Dwell Fall Dwell (RDFD)• Dwell: Period when follower has no motion

(not doing anything)• Computer examples

Page 4: Cam Designing

8.2: SVAJ Diagrams

• S: Position• V: Velocity• A: Acceleration• J: Jerk

Obtaining the S Diagram

• Base Circle: Largest circle centered at axis of rotation that could fit into the cam

Cam Profile

Axis of Rotation

Base Circle

NOT Base Circle

0°, 360°

330°

300°

60°

30°

240°

210°

150°

120°90°

270°

180°

Unwrapping/Linearizing a Cam

Obtain distance between base circle and cam profile for every angle from 0° to 360°

0

2

4

6

8

10

12

14

16

0 30 60 90 120 150 180 210 240 270 300 330 360

Cam Angle (°)

s (m

m)

Unwrapping/Linearizing a Cam

• Transfer distances between cam profile and base circle for every angle (0° – 360°) to a set of linear axes

0°, 360°

330°

300°

60°

30°

240°

210°

150°

120°90°

270°

180°

0

2

4

6

8

10

12

14

16

0 30 60 90 120 150 180 210 240 270 300 330 360

Cam Angle (°)

s (m

m)

0

2

4

6

8

10

12

14

16

0 30 60 90 120 150 180 210 240 270 300 330 360

Cam Angle (°)

s (m

m)

Page 5: Cam Designing

The S Diagram

• A curve that describes cam profile displacement from base circle for every angle (0° – 360°)

• Produced by unwrapping/linearizing the cam• Angles 0° and 360° are the same, hence they must have the same s

value• Minimum s value will always be 0• Can never have negative s values

0

2

4

6

8

10

12

14

16

0 30 60 90 120 150 180 210 240 270 300 330 360

Cam Angle (°)

s (m

m)

Why S Diagram Cannot Have Negative Values

• To get negative S values, cam profile needs to go ‘below’ the base circle

• This violates the definition of a base circle

• Need to define new base circle, which will NOT produce negative s values

Cam Profile

Base Circle

New Base Circle

Low Dwell, High Dwell, Rise and Fall Segments in an S Diagram

360°180° 270°90° 330°240°60° 150°0°

Low Dwell

High DwellRise Fall

8.2: SVAJ Diagrams

θ

θ

θ

θ90

18090

90 180

180 360

360

360270

270

270θddsv =

θddva =

θddaj =

Page 6: Cam Designing

8.2: SVAJ Diagrams• Angular velocity is assumed to be constant

throughout this chapter: θ = ωt• Conversion between v and a to actual

velocity and acceleration:• Actual velocity:

• Actual acceleration:

cam s a jvunwrap θd

dsθddv

θdda

ωθdds

dtds

=

2ωθddv

dtdv

=

8.3: Double Dwell Cam Design –Choosing SVAJ Functions

• Double dwell Rise Dwell Fall Dwell (RDFD) motion• CEP specifications:

– Dwell at 0 for 90°– Rise h mm in 90°– Dwell at h in 90°– Fall h mm in 90°– Cam ω = 2π rad/s

• Specifications easier to comprehend by conveying them in the form of a timing diagram

• Critical Extreme Position (CEP): nothing is specified about the functions needed to be used to get from low dwell (one extreme) to high dwell (another extreme)

• Cam designer’s task: Find suitable function to ‘connect’ (define motion between) low dwell and high dwell

h

Timing Diagram

How NOT to Meet Cam Design Specifications (Linear Function)

• Timing diagram (obtained by translating CEP specs.):

• Cam designer’s task: Find suitable function to ‘connect’ (define motion between) low dwell and high dwell

• Easy function to connect dwell intervals - Linear function: y = mx + b

90 180 270 3600

SVAJ Diagram

Discontinuities

Infinite spikes

θddsv =

θddva =

θddaj =

Page 7: Cam Designing

SVAJ Diagram

• Discontinuities exist at boundaries of each interval in V diagram

• Leads to infinite spikes of acceleration in Adiagram

• Force is proportional to acceleration• Dynamic forces will be very large due to

the spikes lead to high stresses• Cam will wear out quickly unacceptable

Fundamental Law of Cam Design• The cam function must be continuous through the first

and second derivatives across the entire interval (360°)• In other words, s,v and a functions must have no

discontinuities

90 180 270 360

Discontinuities

Fundamental Law of Cam Design

• Cam functions discussed in text:– Linear– Simple Harmonic Motion– Cycloidal Displacement– Combined– Polynomial

(will not work)

Cycloidal Displacement Function• CEP specifications:

– Dwell at 0 when: 0° ≤ θ ≤ 90°– Rise h mm when: 90° ≤ θ ≤ 180°– Dwell at h when: 180° ≤ θ ≤ 270°– Fall h mm when: 270° ≤ θ ≤ 360°– Cam ω = 2π rad/s

• Cam designer’s task: Find a suitable function to define motion between dwell intervals

• Fundamental Law of Cam Design: s,v and a functions must have no discontinuities

90 180 270 3600

Page 8: Cam Designing

SVAJ Diagrams (focus on rise interval)

θ (°)

θ (°)

θ (°)

θ (°)

As F α a, we start with a diagram, and ensure that it is continuous

θddsv =

θddva =

θddaj =

a Diagram• Apply a full period sinusoid for acceleration

• Equation: – C: amplitude (actual value will be determined later)– β: interval length = π/2 (always express in radians)

• Why argument for sine function is and not just θ?

⎟⎟⎠

⎞⎜⎜⎝

⎛=

βθπ2sinCa

βθπ2

θ (°)90 180 2700 β

β

C

Low Dwell High DwellRise

v Diagram

• For v to be continuous:– At θ = 0, v = 0– At θ = β, v = 0(boundary conditions)

∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛== θ

βθπ2sinθ dCadv

π2β

1 Ck =π2β

βθπ2cos

π2β CCv +⎟⎟

⎞⎜⎜⎝

⎛−=

1βθπ2cos

π2β kCv +⎟⎟

⎞⎜⎜⎝

⎛−=

θ (°)

0 β

Low Dwell High DwellRise

s Diagram

• For s to be continuous:– At θ = 0, s = 0– At θ = β, s = h

θπ2β

βθπ2sin

π2βθ dCCvds ∫∫ ⎥

⎤⎢⎣

⎡+⎟⎟

⎞⎜⎜⎝

⎛−==

22

2

θπ2β

βθπ2sin

π4β kCCs ++⎟⎟

⎞⎜⎜⎝

⎛−=

02 =k 2βπ2 hC = θ

ββθπ2sin

π2hhs +⎟⎟

⎞⎜⎜⎝

⎛−=

θ (°)0 β

Page 9: Cam Designing

Cycloidal Displacement Function

θββ

θπ2sinπ2

hhs +⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ββθπ2cos

βhhv +⎟⎟

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

βθπ2sin

βπ2 2ha

⎟⎟⎠

⎞⎜⎜⎝

⎛=

βθπ2cos

βπ4 32 hj

θ (°)

θ (°)

θ (°)

θ (°)

0

0 β

0 β

0 β

90 180 270

θddsv =

θddva =

θddaj =

Polynomial Function• CEP specifications:

– Dwell at 0 when: 0° ≤ θ ≤ 90°– Rise h mm when: 90° ≤ θ ≤ 180°– Dwell at h when: 180° ≤ θ ≤ 270°– Fall h mm when: 270° ≤ θ ≤ 360°– Cam ω = 2π rad/s

• Cam designer’s task: Find a suitable function to define motion between dwell intervals

• Fundamental Law of Cam Design: s,v and a functions must have no discontinuities

90 180 270 3600

Polynomial Function

• General form:– s: displacement– x: independent variable (θ/β in our case)

• Why is θ/β used instead of just θ?– Cn: constant coefficients (unknowns to be

solved for)• Degree of polynomial: Highest power of

the function

nn xCxCxCxCxCCs ++++++= ...4

43

32

210

Polynomial Function

• Using polynomial functions for cam design:

1. Decide on number of boundary conditions, k(BCs)- Boundary Conditions: Conditions that must be

met to ensure that s, v and a are continuous2. Number of BCs determine the degree of

polynomial needed- Degree of polynomial needed, n = k – 1

3. Use BCs to solve for all unknown polynomial coefficients

Page 10: Cam Designing

Polynomial Function• CEP specifications:

– Dwell at 0 when: 0° ≤ θ ≤ 90°– Rise h mm when: 90° ≤ θ ≤ 180°– Dwell at h when: 180° ≤ θ ≤ 270°– Fall h mm when: 270° ≤ θ ≤ 360°– Cam ω = 2π rad/s

• Cam designer’s task: Find a suitable function to define motion between dwell intervals

• Fundamental Law of Cam Design: s,v and a functions must have no discontinuities

90 180 270 3600

SVAJ Diagrams (focus on rise interval)

θ (°)

θ (°)

θ (°)

θ (°)

To ensure continuity in s, v and a:

At θ = 0, s = 0At θ = β, s = h

At θ = 0, v = 0At θ = β, v = 0

At θ = 0, a = 0At θ = β, a = 0

These are theBCs

0 β

0 β

0 βTotal BCs = 6

0 90 180 270

What is β in this example?

θddsv =

θddva =

θddaj =

Polynomial Function

• Number of Boundary Conditions, k = 6• Degree of polynomial needed,

n = k – 1 = 5• General form of 5 degree polynomial:

• Replace independent variable x with θ/β

55

44

33

2210 xCxCxCxCxCCs +++++=

5

5

4

4

3

3

2

210 βθ

βθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+= CCCCCCs

Polynomial Function

• 6 unknowns (C0, C1, C2, C3, C4, C5), need 6 equations

• Boundary conditions:– At θ = 0, s = 0– At θ = β, s = h– At θ = 0, v = 0

– At θ = β, v = 0 – At θ = 0, a = 0– At θ = β, a = 0

5

5

4

4

3

3

2

210 βθ

βθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+= CCCCCCs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

4

5

3

4

2

321 βθ5

βθ4

βθ3

βθ2

β1

θCCCCC

ddsv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

3

5

2

4322 βθ20

βθ12

βθ62

β1

θCCCC

ddva

Page 11: Cam Designing

Polynomial Function• Use 6 BCs to solve for 6 unknown coefficients• C0 = 0, C1 = 0, C2 = 0• C3 = 10h, C4 = -15h, C5 = 6h• Polynomial is called the 3-4-5 polynomial

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

543

βθ6

βθ15

βθ10hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

432

βθ

βθ2

βθ

β30 hv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

32

2 βθ2

βθ3

βθ

β60 ha

Polynomial Function

θ (°)

θ (°)

θ (°)

θ (°)

0 β

0 β

0 β

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

543

βθ6

βθ15

βθ10hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

432

βθ

βθ2

βθ

β30 hv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

32

2 βθ2

βθ3

βθ

β60 ha

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

3 βθ6

βθ61

β60 hj

0 90 180 270

Double Dwell Cam Design

• Cycloidal displacement function• Polynomial function• Both are able to produce cams that obey

the fundamental law of cam design• So what’s the difference between these 2

functions?• Which is better?

S Diagram of Rise Interval

• Design a RDFD cam that has a rise and fall distance of h = 20 mm with a rise and fall interval of β = 90°using both cycloidaldisplacement and polynomial functions, and compare their characteristics

• s diagram:– Both functions produce a

smooth 20 mm rise in a 90° interval

θ (rad)

s(m

m)

Page 12: Cam Designing

V Diagram of Rise Interval

• v diagram• Both functions able to

produce continuous v• Peak velocity of

cycloidal displacement (red) is larger than that of polynomial (blue)

θ (rad)

v(m

m/ra

d)

A Diagram of Rise Interval

• a diagram• Both functions able to

produce continuous a• Peak acceleration of

cycloidal displacement (red) is once again larger than that of polynomial (blue)

θ (rad)

a(m

m/ra

d2)

Double Dwell Cam Design Case Study

• Kinetic energy is proportional to velocity squared• Dynamic force is propotional to acceleration• Cycloidal displacement function produced both larger

peak velocity and peak acceleration than polynomial function

• Hence follower for cycloidal displacement will experience larger kinetic energies and dynamic forces

• If this is a main concern for the cam design, the designer would be better off opting for the polynomial function, as it produces lower peak velocities and peak accelerations

8.4: Single Dwell Cam Design• Single Dwell – Rise Fall Dwell (RFD) motion• CEP specifications:

– Rise-Fall: Rise h mm in 90° and fall h mm in the next 90° (symmetric)

– Dwell at 0 mm for the remaining 180°– Cam ω = 1 rad/s

• Will only consider polynomial functions to fulfill these specifications

• First task: Convert CEP specifications into timing diagram

Page 13: Cam Designing

SVAJ Diagrams

0 β

0 β

0 β

0 β

θ = 0, s = 0θ = β, s = 0

θ = 0, v = 0θ = β, v = 0

θ = 0, a = 0θ = β, a = 0

h

θ = β/2, s = hv

s

a

j

β/2

Total of 7 BCs

Rise Fall Dwell

What is β?

Single Dwell Cam Design (Polynomial Function)

• Number of Boundary Conditions, k = 7• Degree of polynomial needed, n = k – 1 = 6• Degree 6 polynomial with θ/β as independent

variable:6

6

5

5

4

4

3

3

2

210 βθ

βθ

βθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCCCs

Single Dwell Cam Design (Polynomial Function)

• 7 Boundary conditions:– At θ = 0, s = 0– At θ = β, s = 0– At θ = β/2, s = h

– At θ = 0, v = 0– At θ = β, v = 0 – At θ = 0, a = 0– At θ = β, a = 0

6

6

5

5

4

4

3

3

2

210 βθ

βθ

βθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+= CCCCCCCs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+==

5

6

4

5

3

4

2

321 βθ6

βθ5

βθ4

βθ3

βθ2

β1

θCCCCCC

ddsv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+==

4

6

3

5

2

4322 βθ30

βθ20

βθ12

βθ62

β1

θCCCCC

ddva

7 unknown coefficients (C0, C1, C2, C3, C4, C5, C6), need 7 equations

Single Dwell Cam Design (Polynomial Function)

• Use 7 BCs to solve for 7 unknown coefficientsC0 = 0, C1 = 0, C2 = 0C3 = 64h, C4 = -192h, C5 = 192h, C6 = -64h

• Hence polynomial is called the 3-4-5-6 polynomial

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

6543

βθ64

βθ192

βθ192

βθ64hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

5432

βθ384

βθ960

βθ768

βθ192

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

432

2 βθ1920

βθ3840

βθ2304

βθ384

βha

Page 14: Cam Designing

SVAJ Diagrams

0 β

0 β

0 β

0 β

h

β/2

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

6543

βθ64

βθ192

βθ192

βθ64hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

5432

βθ384

βθ960

βθ768

βθ192

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

432

2 βθ1920

βθ3840

βθ2304

βθ384

βha

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−=

32

3 βθ7680

βθ11520

βθ4608384

βhj

Rise FallAsymmetrical Single Dwell Cam Design• Previous example: Rise in 90°, fall in 90° symmetrical• Asymmetrical: Rise interval ≠ Fall interval• Asymmetrical CEP specifications:

– Rise h mm in 45°– Fall h mm in next 135 °– Dwell at 0 mm for the remaining 180°– Cam ω = 1 rad/s

• Will only consider polynomial functions to fulfill these specifications

• Need to use 2 different functions for each interval (1 function for rise interval, 1 function for fall interval)

• Start with larger interval first (Fall interval = 135° > Rise interval = 45°)

Start with Larger Interval (Fall)

S

V

A

J

0 45 180 360

h

0 β1

0 β1

0 β1

At θ = β1, s = 0

At θ = β1, v = 0

At θ = β1, a = 0

At θ = 0, s = h

At θ = 0, v = 0

Total of 5 BCs

Rise Fall Dwell

What is β1 in this example?

Slope at this location must be zero

Polynomial Function for Fall Interval

• Number of Boundary Conditions, k = 5• Degree of polynomial needed, n = k – 1 = 4• Degree 4 polynomial with θ/β1 as independent

variable:4

14

3

13

2

12

110 β

θβθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCs

Page 15: Cam Designing

Polynomial Function for Fall Interval

• 5 Boundary conditions:– At θ = 0, s = h– At θ = β1, s = 0

– At θ = 0, v = 0– At θ = β1, v = 0 – At θ = β1, a = 0

4

14

3

13

2

12

110 β

θβθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

3

14

2

13

121

1 βθ4

βθ3

βθ2

β1

θCCCC

ddsv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

2

14

1322

1 βθ12

βθ62

β1

θCCC

ddva

5 unknown coefficients (C0, C1, C2, C3, C4), need 5 equations

Polynomial Function for Fall Interval

• Use 5 BCs to solve for 5 unknown coefficientsC0 = h, C1 = 0C2 = -6h, C3 = 8h, C4 = -3h

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−=

4

1

3

1

2

1 βθ3

βθ8

βθ61hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

3

1

2

111 βθ12

βθ24

βθ12

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

2

1121 β

θ36βθ4812

βha

Fall Interval SVAJ

S

V

A

J

0 45 180 360

h

0 β1

0 β1

0 β1

Rise Fall

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−=

4

1

3

1

2

1 βθ3

βθ8

βθ61hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

3

1

2

111 βθ12

βθ24

βθ12

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

2

1121 β

θ36βθ4812

βha

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

131 β

θ7248βhj

Next Work on Smaller Interval (Rise)

S

V

A

J

0 45 180 360

h

0 β2

0 β2

0 β2

At θ = 0, s = 0

At θ = 0, v = 0

At θ = β2, s = h

At θ = β2, v = 0

At θ = 0, a = 0

Rise Fall

a =???

Page 16: Cam Designing

Fall Interval SVAJ

S

V

A

J

0 45 180 360

h

0 β1

0 β1

0 β1

Rise Fall

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−=

4

1

3

1

2

1 βθ3

βθ8

βθ61hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

3

1

2

111 βθ12

βθ24

βθ12

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

2

1121 β

θ36βθ4812

βha

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

131 β

θ7248βhj

Next Work on Smaller Interval (Rise)

S

V

A

J

0 45 180 360

h

0 β2

0 β2

0 β2

At θ = 0, s = 0

At θ = 0, v = 0

At θ = β2, s = h

At θ = β2, v = 0

At θ = 0, a = 0

At θ = β2, a = -12h/β12

Total of 6 BCs

Rise Fall

What is β2 in this example?

Polynomial Function for Rise Interval

• Number of Boundary Conditions, k = 6• Degree of polynomial needed, n = k – 1 = 5• Degree 5 polynomial with θ/β2 as independent

variable:5

25

4

24

3

23

2

22

210 β

θβθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCCs

Polynomial Function for Rise Interval

• 6 Boundary conditions:– At θ = 0, s = 0– At θ = β2, s = h– At θ = 0, v = 0

– At θ = β2, v = 0 – At θ = 0, a = 0– At θ = β2, a = -12h/β1

2

5

25

4

24

3

23

2

22

210 β

θβθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCCs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

4

25

3

24

2

23

221

2 βθ5

βθ4

βθ3

βθ2

β1

θCCCCC

ddsv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

3

25

2

24

2322

2 βθ20

βθ12

βθ62

β1

θCCCC

ddva

6 unknown coefficients (C0, C1, C2, C3, C4, C5), need 6 equations

Page 17: Cam Designing

Polynomial Function for Rise Interval

• Use 6 BCs to solve for 6 unknown coefficientsC0 = 0, C1 = 0, C2 = 0C3 = 9.333h, C4 = -13.667h, C5 = 5.333h

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

5

2

4

2

3

2 βθ333.5

βθ667.13

βθ333.9hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

4

2

3

2

2

22 βθ667.26

βθ667.54

βθ28

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

3

2

2

2222 β

θ667.106βθ164

βθ56

βha

Rise Interval SVAJ

S

V

A

J

0 45 180 360

h

0 β2

0 β2

0 β2

Rise Fall

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

5

2

4

2

3

2 βθ333.5

βθ667.13

βθ333.9hs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛=

4

2

3

2

2

22 βθ667.26

βθ667.54

βθ28

βhv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

3

2

2

2222 β

θ667.106βθ164

βθ56

βha

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

2232 β

θ320βθ32856

βhj

Asymmetrical Single Dwell Cam Design

• What happens if we start with the smaller interval (rise) first???

• What happens if only 4 boundary conditions were used for the first (fall) interval??

What happens if we start with smaller interval (rise) first…

S

V

A

J

0 45 180 360

Fallh

Rise

Negative Displacement !?!?Negative Displacement !?!?

Unacceptable!

Page 18: Cam Designing

What happens if only 4 BCs were used for the fall interval…

S

V

A

J

0 45 180 360

h

0 β1

0 β1

0 β1

At θ = β1, s = 0

At θ = β1, v = 0

At θ = β1, a = 0

At θ = 0, s = h

At θ = 0, v = 0

Total of 5 BCs

Rise Fall Dwell

What happens if this BC were removed?

So remember: Start with larger interval first, and use 5 BCs for it

S

V

A

J

0 45 180 360

h

0 β2

0 β2

0 β2

Rise FallAt θ = β1, s = 0

At θ = β1, v = 0

At θ = 0, s = h

At θ = 0, v = 0

Important!!!

At θ = β1, a = 0

Total of 5 BCs

Conclusion For Single Dwell Cam Design

• Symmetrical (Rise = Fall)– Just need one polynomial function for both

rise & fall intervals• Asymmetrical (Rise ≠ Fall)

– Need two different polynomial functions for each interval. 1 function for rise interval, 1 function for fall interval.

– Need to start with the larger interval first, and then move to the smaller interval

– Use 5 BCs for the larger interval

8.5: CPM Cam Design - Constant Velocity

• Critical Path Motion (CPM): Design specifications define motion path / velocity / acceleration over specific interval of motion

• Will focus on constant velocity motion as it is the most common application in machinery design

• Will only consider polynomial functions to fulfill these specifications

Page 19: Cam Designing

CPM Cam Design – Constant Velocity

• CPM specifications:– Maintain constant v of 10 mm/s for first 180°– Return follower to initial position in remaining

180°– Cam ω = 2π rad/s

• Convert specifications into timing diagram

SVAJ Timing Diagram for First Interval

θ (rad)

θ (rad)

θ (rad)

θ (rad)

S

V

A

J

0 β

0 β

0 β

2π0 π

Constant v = 10 mm/s5/π mm/rad

0=a

0=j

θπ5

=s

CPM Cam Design – Constant Velocity

• CPM specifications:– Maintain constant v of 10 mm/s for first 180°– Return follower to initial position in remaining

180°– Cam ω = 2π rad/s

• Convert specifications into timing diagram

Boundary Conditions for 2nd Interval

S

V

A

J

0 β

0 β

0 β

θ (rad)

θ (rad)

θ (rad)

θ (rad)

2π0 π

θ = 0, s = 5 θ = β, s = 0

θ = 0, v = 5/π θ = β, v = 5/π

θ = 0, a = 0 θ = β, a = 0

Total of 6 BCs

What is β in this example?

0=a

0=j

θπ5

=s

π5

=v

Page 20: Cam Designing

Polynomial Function for 2nd Interval

• Number of Boundary Conditions, k = 6• Degree of polynomial needed, n = k – 1 = 5• Degree 5 polynomial with θ/β as independent

variable:5

5

4

4

3

3

2

210 βθ

βθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCCs

Polynomial Function for 2nd Interval

• 6 Boundary conditions:– At θ = 0, s = 5– At θ = β, s = 0– At θ = 0, v = 5/π

– At θ = β, v = 5/π– At θ = 0, a = 0– At θ = β, a = 0

5

5

4

4

3

3

2

210 βθ

βθ

βθ

βθ

βθ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+= CCCCCCs

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

4

5

3

4

2

321 βθ5

βθ4

βθ3

βθ2

β1

θCCCCC

ddsv

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+==

3

5

2

4322 βθ20

βθ12

βθ62

β1

θCCCC

ddva

6 unknown coefficients (C0, C1, C2, C3, C4, C5), need 6 equations

Polynomial Function for 2nd Interval

• Use 6 BCs to solve for 6 unknown coefficientsC0 = 5, C1 = 5β/π, C2 = 0C3 = -50(π+β)/π, C4 = 75(π+β)/π, C5 = -30(π+β)/π

543

βθ60

βθ150

βθ100

βθ55 ⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+=s

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−=

432

βθ300

βθ600

βθ3005

β1v

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

32

2 βθ1200

βθ1800

βθ600

β1a

β = π

C0 = 5

C1 = 5

C2 = 0

C3 = -100

C4 = 150

C5 = -60

SVAJ Diagram for 2nd Interval

S

V

A

J

0 β

0 β

0 β

θ (rad)

θ (rad)

θ (rad)

θ (rad)

2π0 π

543

βθ60

βθ150

βθ100

βθ55 ⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+=s

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−=

432

βθ300

βθ600

βθ3005

β1v

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−=

32

2 βθ1200

βθ1800

βθ600

β1a

⎥⎥⎦

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

2

3 βθ3600

βθ3600600

β1j

Page 21: Cam Designing

SVAJ Diagram for 2nd Interval0θ

π5

+=s 0.484θπ5

+=s

S

V

A

J

0 β

0 β

0 β

θ (rad)

θ (rad)

θ (rad)

θ (rad)

2π0 π

V still constant at 5/π mm/rad s = -0.484 mm

543

βθ60

βθ150

βθ100

βθ55 ⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+=s 484.0

βθ60

βθ150

βθ100

βθ55

543

+⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+⎟⎟

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛+=s

Cam Functions Summary• Fundamental Law of Cam Design: S, V and A must be

continuous• Double Dwell Cam Design (CEP)

– Cycloidal Displacement Function– Polynomial Function

• Single Dwell Cam Design (CEP)– Only Polynomial Function– Symmetrical (Rise = Fall)– Asymmetrical (Rise ≠ Fall)

• CPM Cam Design: Constant velocity– Only Polynomial Function

Cam Design Process

Design Specifications

s, v, a and j Functions

SVAJ Diagrams

Cam Sizing

Cam Profile

8.7: Sizing the Cam

• Major factors influencing cam size:– Pressure Angle– Radius of Curvature (beyond coverage of this

course)

Page 22: Cam Designing

Terminologies

Base Circle: Largest circle centered at axis rotation that could fit into cam (radius = Rb)

Cam Profile

Axis of rotation

Base Circle

TerminologiesOnly applicable to cams with roller / mushroom followers:

Pitch Curve: Locus / Path traced by of centerline of the follower

Prime Circle: Largest circle centered at axis rotation that could fit into the pitch curve(radius = Rp)

Rp = Rb + Rf

Cam Profile

Axis of rotation

Base CirclePitch Curve

Prime Circle

Pressure Angle, Φ

• Angle between the direction of motion (velocity) of the follower and the axis of transmission

• Measure of quality of force transmission between cam and follower (analogous to transmission angle for a linkage)

ΦCommon tangent (axis of slip)

Common normal (axis of transmission)

Pressure Angle, Φ

Vfollower

Axis of Transmission VfollowerΦ

Axis of Transmission

• Φ = 0°

• All transmitted force goes into motion of the follower

• Good force transmission

• Φ = 90°

• None of transmitted force goes into motion of the follower (follower does not move)

• Very poor force transmission

Page 23: Cam Designing

Pressure Angle, Φ

• Rule of thumb:– Translating follower:

-30° ≤ Φ ≤ 30°– Rotating follower:

-35° ≤ Φ ≤ 35°• Eccentricity, ε:

Perpendicular distance between follower’s axis of motion and center of rotation

Φ

Axis of transmission

ε

• Where:– Rp: Radius of prime circle– ε: Eccentricity– s: Cam displacement at

that instant– v: Cam velocity at that

instant (in units of length/rad)

• Must have s and v diagrams available to use formula

Formula for Pressure Angle, Φ

Φ

Axis of transmission

ε

⎟⎟⎟

⎜⎜⎜

−+

−=

22 εεarctanpRs

Rp

Numerical Example

• Determine the pressure angle, Φ, when the following cam is at an angle of θ = 60°

0

4

8

12

16

20

24

0 20 40 60 80 100 120 140 160 180

Cam angle, θ (°)

Cam

dis

plac

emen

t, s

(mm

)

-30

-20

-10

0

10

20

30

0 20 40 60 80 100 120 140 160 180

Cam angle, θ (°)

Cam

vel

ocity

, v (m

m/ra

d)

Rf = 5 mm

Rb = 45mm

3 mm

Numerical Example

Rf = 5 mm

Rb = 45mm

3 mm

ε = 3 mm

Rf = 5 mm

Rb = 45 mm

Rp = Rf + Rb = 50 mm

⎟⎟

⎜⎜

−+

−=

22 εεarctanpRs

Page 24: Cam Designing

• Determine the pressure angle, Φ, when the following cam is at an angle of θ = 60°

0

4

8

12

16

20

24

0 20 40 60 80 100 120 140 160 180

Cam angle, θ (°) C

am d

ispl

acem

ent,

s (m

m)

-30

-20

-10

0

10

20

30

0 20 40 60 80 100 120 140 160 180

Cam angle, θ (°)

Cam

vel

ocity

, v (m

m/ra

d)

s = 14.05 mm

v = 20.12 mm/rad

Pressure Angle Plot

• In the previous numerical example, the pressure angle Φ was calculated only for cam angle θ = 60°

• The same process can be repeated for all cam angles 0° ≤ θ ≤ 360°

• Doing so will produce the pressure angle plot

Pressure Angle Plot

0

5

10

15

20

25

0 40 80 120 160 200 240 280 320 360

Cam angle, θ (°)

Cam

dis

plac

emen

t, s

(mm

)

-30

-20

-10

0

10

20

30

0 40 80 120 160 200 240 280 320 360

Cam angle, θ (°)

Cam

vel

ocity

, v

(mm

/rad

)

S Diagram

V Diagram

Φ Diagram(shape will almost be similar to V diagram)

-30

-20

-10

0

10

20

30

0 40 80 120 160 200 240 280 320 360

Cam angle, θ (°)

Pres

sure

ang

le, Φ

(°)

Rp Selection ProcessAssumption: You already produced the SVAJ diagrams based on cam specifications1. Pick any value for Rp

2. Assume eccentricity, ε = 03. Calculate pressure angle Φ for 0° ≤ θ ≤ 360° (produce

plot)4. Check whether values of Φ is in acceptable range

• Translating follower: -30° ≤ Φ ≤ 30°• Rotating follower: -35° ≤ Φ ≤ 35°

5. If not within range, repeat steps 1 to 4 with slightly larger Rp

Iterative process usually done using a computer program

Page 25: Cam Designing

Rp Selection Process Example

• Cam specifications:– Rise-Fall: Rise 20 mm in 90° and fall 20 mm

in the next 90° (symmetric)– Dwell at 0 mm for the remaining 180°

• Cam will have translating follower, hence -30° ≤ Φ ≤ 30°

Rp Selection Process

• Rp is inversely proportional to magnitude of Φ• Larger Rp will result in smaller Φ• However larger Rp will also result in larger cam

size, hence more material usage• Hence there is a trade-off in choosing a suitable

value for Rp

• Designer needs to find right balance of Rp and Φ

Effect of Eccentricity, ε• Positive eccentricity will essentially shift the pressure

angle curve downwards• For a symmetrical cam, it will reduce maximum Φ, but it

also will increase minimum Φ (make it more negative)• Most value gained when introduced for an asymmetrical

cam where a significant difference exist between maximum and minimum pressure angles

• Eccentricity helps balance the pressure angles for asymmetrical cams

Creation of Physical Cam from S Diagram

• Process is the reverse of cam unwrapping/linearizing• Derive S diagram based on cam specifications• Select a suitable radius for the prime circle, Rp

• Determine radius of base circle from radius of prime circle: Rb = Rp – Rf

Page 26: Cam Designing

02468

10121416

0 30 60 90 120 150 180 210 240 270 300 330 360

Cam Angle (°)

s (m

m)

Rb

120°90°

60°

30°

0°, 360°

330°

300°270°

240°

210°

180°

150°• Divide base circle into

angular intervals

• Transfer displacement from S diagram to distances on base circle for every angle (0° - 360°)

• Connect displacement points on base circle to form cam profile

Course OutcomesIdentify mechanisms and predict their motionCalculate the degrees of freedom of mechanismsDesign mechanisms to fulfill motion generation and quick return requirementsDetermine the positions, velocities and accelerations of links and points on mechanismsDerive SVAJ functions to fulfill cam design specificationsCalculate dynamic joint forces of mechanismsBalance simple rotating objects and pin-jointed fourbarlinkagesWork in a team to analyze and modify existing mechanismsPresent completed work in oral and written formUse related computer programs to design, model and analyze mechanisms

Stuff You Should Know…

• Students should know how to:– Derive SVAJ functions for a cam based

on CEP/CPM specifications– Calculate pressure angle and explain

how it influences the size of a cam

End of Chapter 8

Thank You