# Kinematics Project (CAM Design)

# the question of the project is in the attachment 6.7The due date is in a day and a half.Following information should be included in the project reportMotion Requirement description (problem description)Sketches of displacement diagram, velocity, and acceleration together with answers for following questions:The number of segments neededThe reasons for the existence of each segment.Detailed calculation of each segment:The given parameters and boundary conditionunknown parameters of the segmentThe function chosen for the segment and reason to chooseparameter calculationFinal result:The parameter value of each segment (ßi and Li)The function of each segment# I’ve attached a PDF file has an example of the project. So, the project youre going to do should be similar to it. It’s example 6.2 in the pdf file at the bottom. Also, it should be handwritten or typed. thanks
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Chapter 6
Cam Design
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
What is a CAM
 What Is A Cam?
Components in CAM
 Cam: The cam is the teardrop shaped device.
 Shaft (Rotating Wheel): The cam is typically
mounted in a fixed position on a shaft or rotating
wheel. As the shaft rotates the pointy end of the
cam comes into contact with the cam follower
once every 360 degrees of angular motion.
 Cam Follower (Lever): The cam follower (or lever)
is a long, linear part that comes into contact with
and is linearly displaced by the pointy end of the
cam once every 360 degrees of cam motion.
Classification of CAMs and Follower
 CAM-follower mechanisms are:
 Simple
 In-expensive
 Fewer moving parts
 Occupy little space
Types of CAM:
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Types of Follower:
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Displacement Diagram



Displacement Diagram: shows the movement of follower during one
revolution of CAM.
During on cycle of CAM, the follower execute following event: Rise,
Dwell, Return.
The total rise is called lift.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley




Section ß1: lift=L1, parabolic motion
Section ß2: lift=L2, linear motion
Section ß3: lift=L3, parabolic motion
Boundary condition must be met.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Graphic Layout of CAM Profile
 Given: The followers motion represented by
the displacement diagram.
 Design Goal: Design the exact shape of the
cam profile to deliver the specified motion of
follower.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Design of CAM
 Trace point: center of roller. Used to
represent the motion of follower.
 Pitch curve: locus on the cam generated by
the trace point of follower moves with respect
to cam
 Prime circle: the smallest circle that is tangent
to pitch curve with center at cam rotation axis.
 Basic circle: smallest circle that is tangent to
cam profile.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Kinematic Coefficient of Follower
 Displacement diagram:
 abscissa: the angle of cam input ?
 ordinate: the output displacement of follower Y-Y0
 The diagram has several segments
Kinematic Coefficient of Follower
 The displacement diagram represents a
mathematical function relating to the input.
 y=y(?)
??
??
 y(?) =
(velocity)
first order kinematic coefficient
?2 ?
??2
 y(?) =
second order kinematic coefficient
(acceleration)
 y(?) =
?3 ?
??3
third order kinematic coefficient
Design Example 6.1
 Requirement:
1. Design a segment of a displacement diagram of
cam.
2. Rise segment from one dwell to another dwell
needs to be parabolic motion
3. The total lift is L
4. Total cam angle is ß
Solution for Example 6.1
 Draw the diagram
 The function needs to be in two segments:
 First segment moves from dwell (steady state) to
mid point, i.e. is acceleration motion.
 Second segment moves from mid point to dwell
(steady state), i.e. is deceleration motion.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley





First Segment Function Design
y = A ?2 +B ? +C
Derivative:
 y = 2A ? +B
 y = 2A
 y=0
Boundary condition: start from dwell
 y(0) = 0 ? C=0
 y(0) = 0 ? B=0
In order for cam to go back to dwell at end, the cam needs to decelerate in
second half.
 When ? = ß/2, y=L/2 ?A = 2?/ ß2
First segment function Equation:
?
 y= 2L(ß )2
4? ?
(
ß ß
4?
= ß2
 y =
 y
 y=0
),
ymax =
2?
ß
Second Segment Function Design



y = A ?2 +B ? +C
Boundary condition: ends at dwell
 y(ß) = L
 y(ß) = 0
At mid point, speed must match that of section 1s
 When ? = ß/2, y=
2?
 A = – ß2 , B = 
4?
ß
2?
ß
, C=-L,
Second segment function Equations:
?
 y= L[ 1- 2(1- ß )2 ]
 y =
4?
ß
 y =  y=0
?
(1- ß ),
4?
ß2
Requirement for High-Speed CAMS
 A follower has mass m.
 Force on the follower and cam F=m*a, here a is the
acceleration from the displacement curve
 An acceleration curve with abrupt change (jerk becomes
infinity) exert an abrupt force on the contact surfaces that
leads to noise and surface wearing.
 Requirement: First order and second order kinematic
coefficient (velocity and acceleration) must be continuous.
Iteration Design Approach for low speed
cam:
 Four circle-arcs are used to meet
follower displace requirements.
 Tangent lines are used to blend in the
arcs.
 Acceleration changes abruptly at each
blending point because of instantaneous
change in the radius of curvature of cam
profile.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Kinematic Coefficients w.r.t. Input Variable ?
 First segment function Equation:
?
ß
4? ?
y = ( ),
ß ß
4?
y = 2
ß
 y= 2L( )2


ymax =
2?
ß
 y=0
 Second segment function Equations:
?
ß
 y= L[ 1- 2(1- )2 ]
 y =
4?
ß
 y = –
?
ß
(1- ),
4?
ß2
 y=0
??
 Here y=
??
?2?
y= 2
??
?3?
y= 3
??
Kinematic Coefficients w.r.t. Angular Velocity
and Time

Input angle ? is a function of time t: ?(t)

Angular velocity: ? = ??
??
?2?
??2

Angular acceleration: a =

Usually, plate cam is driven by a constant angular velocity input shaft.
 ? = constant
 ? = ?t

?2?
a= 2
??
?? ?? ??
=
= y?
?? ?? ??
?2?
= y?2 + ya
??2
?3?
3
3 = y? + 3ya
??

?? =

?? =

?? =




When a = 0
?=
? y?
?? = y?2
?? = y?3
? + ya?
Motion Type and Comparison





Parabolic
Harmonic
Cycloidal
Polynomial
Rise vs. Return motion
Standard CAM Motion Parabolic, Full-rise
?
ß
y= 2L( )2
y =
4? ?
( )
ß ß
y =
4?
ß2
y=0
 Advantage: velocity, y is continuous
 Disadvantage: acceleration is not continuous, jerk in infinity
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Standard CAM Motion Harmonic, Full-rise
p?
ß
 y= 2L(1 – cos
 y =
p?
ß
 y =
sin(
p2?
2ß2
 y= –
p?
ß
)
cos (
p3?
2ß3
)
p?
ß
sin (
)
p?
ß
)
 Advantage: no discontinuity for y and y
 Disadvantage: acceleration is not zero at beginning and end
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Standard CAM Motion Cycloidal, Full-rise
?
ß
 y= L( –
1
2p?
sin
2p
ß
)
2p?
ß
)
?
ß
 y = (1 – cos
 y =

2p?
ß2
4p4?
y= 3
ß
sin (
2p?
ß
cos (
2p?
ß
)
)
 Advantage: y and y are all zero at boundary
 Disadvantage: peak value of acceleration and jerk are too high.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Standard CAM Motion 8th order
Polynomial, Full-rise
?
?
?

y = L[6.0955( )3 – 20.78040( )5 +26.73155( )6

y = [18.29265( )2 – 103.90200( )4 +160.38930( )5
ß
?
– 13.60965 ( )7
ß
?
ß
?
+2.56095( )8 ]
ß
ß
?
ß
ß
?
?
?
?
ß
ß
– 95.26755 ( )6 +20.48760( )7 ]
ß

?
y=
ß
?
2
ß
?
?
ß
ß
[36.58530( ) – 415.60800( )3 + 801.94650( )4
?
ß
?
– 571.60530 ( )5 + 143.41320( )6 ]
ß

y=
?
3
ß
[36.58530-
?
ß
? 2
1246.82400( )
ß
? 5
?
+ 3207.78600( )3 –
ß
2858.02650 ( )4 + 860.47920( ) ]
ß
ß
 Advantage: y is non-symmetric(positive and negative accelerations
are possible
 peak value of acceleration is small.
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
Standard CAM Motion Harmonic, Full-return
p?
ß
 y= 2L(1 + cos
 y = –
p?
ß
 y = 
sin(
p2?
2ß2
p3?
y= 3

p?
ß
)
cos (
sin (
)
p?
ß
p?
ß
Theory of Machines and Mechanisms, 5e
)
)
Uicker, Pennock, Shigley
Standard CAM Motion Cycloidal, Fullreturn
 y= L(1 –
?
1
2p?
+ sin
ß 2p
ß
?
ß
2p?
ß
 y = – (1 – cos
 y = –
2p?
ß2
 y= –
4p4?
ß3
sin (
)
2p?
ß
cos (
)
)
2p?
ß
Theory of Machines and Mechanisms, 5e
)
Uicker, Pennock, Shigley
Standard CAM Motion 8th order
Polynomial, Full-return

?
ß
?
ß
y = L[1  2.63415 ( )2 + 2.78055( )5
?
ß
?
ß
+ 3.1706( )6 -6.87795( )7
?
+ 2.56095(ß )8 ]



?
?
?
[5.62830( ) – 13.90275( )4
ß
ß
ß
? 5
? 6
– 19.02360(ß ) +48.14565(ß )
?
– 20.48760( )7 ]
ß
?
?
y= – ß2 [5.2683 – 55.611 (ß )3 –
?
?
95.118(ß )4 + 288.8739 (ß )5 –
?
143.41320(ß )6 ]
?
?
y= ß3 [166.833(ß )2 +
?
?
380.472(ß )3 – 1444.3695 (ß )4 +
?
860.47920( )5 ]
ß
y = –
Theory of Machines and Mechanisms, 5e
Uicker, Pennock, Shigley
p?

 y= L(1 – cos
 y =
p?

 y =
sin(
p2?
4ß2
 y= –
p?

)
 y= L(sin
)
 y =
p?
cos (

p3?
4ß3
Theory of Machines and Mechanisms, 5e
)
p?
sin (

)
Uicker, Pennock, Shigley
p?

p?

)
p?

cos(
)
 y = –
p2?
4ß2
???(
 y= –
p3?
8ß3
cos (
p?

p?

)
)
p?

 = L(cos
 y = –
p?

 y = –
 y=
sin(
p2?
4ß2
p3?
8ß3
 y= L(1- sin
)
p?

)
p?
cos (

p?
sin (

Theory of Machines and Mechanisms, 5e
)
)
Uicker, Pennock, Shigley
p?

)
 y = –
p?

 y =
p2?
4ß2
???(
 y=
p3?
8ß3
cos (
cos(
p?

)
p?

p?

)
)
 y= L( 1  y =
?
(1
ß
 y =
p?
ß2
 y= –
?
ß
sin (
p?
ß
)
 y= L( + sin
?
ß
1
p
p?
ß
)
?
ß
p?
ß
)
p?
ß
)
 y = (1 + cos
p?
ß
)
 y = –
p?
ß2
 y= –
p4?
ß3
– cos
p2?
ß3
Theory of Machines and Mechanisms, 5e
1
p
– sin
p?
cos (
ß
)
Uicker, Pennock, Shigley
sin (
p?
ß
cos (
)
p?
ß
)
 y= L( 1  y = –
?
ß
 y =
p?
ß2
 y= –
?
ß
+
(1 –
1
p?
sin
p
ß
p?
cos
ß
p?
sin (
ß
p2?
ß3
Theory of Machines and Mechanisms, 5e
)
)
p?
cos (
ß
)
)
?
ß
1
p
 y= L( 1 – – sin
?
ß
 y = – (1 + cos
 y = –
p?
ß2
 y= –
p2?
ß3
Uicker, Pennock, Shigley
sin (
p?
ß
p?
ß
p?
ß
cos (
)
)
)
p?
ß
)
Match Derivatives of Displacement
Diagrams
 Motion requirements must be met
 Kinematic coefficients are continuous
 Maximum magnitude of velocity, acceleration
peak need to be as small as possible
Example 6.2 Cam Design example
A plate cam with a reciprocal follower is to be driven by a constant
speed motor at 150rpm. The follower is to start from a dwell, accelerate
to a uniform velocity of 25in/s, maintain this velocity for a 1.25 in of rise,
decelerate to the top of the lift, return, and then dwell for 0.1 s. The
total lift is 3.00 in. Determine the complete specification of the
displacement diagram.
Problem analysis:
1. How many segments are needed to implement the above
requirements?
2. The velocity and acceleration curve need to be continuous. This
means the connection between each segment must be continuous.
Example 6.2 Problem Analysis
1.
2.
3.
4.
5.
First segment AB:


Continuity: yAB(B)=yBC(B)

Accelerate: y>0
Second segment BC:

Constant velocity:
y=constant, y=0

Continuity: yBC(C)=yCD(C)
Third segment CD:


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