cam design
DESCRIPTION
desTRANSCRIPT
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
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
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
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)
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 =
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 =
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
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
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
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
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)
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
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
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
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 =???
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
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!
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
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
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
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)
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
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
vφ
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
vφ
• 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
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
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