FIGURE 40.6 Types
of
follower
motion.
In all the
cases, increasing
k
e
and
reducing
m
e
are
recommended, because
it
reduces
ji
rf
.
There
are two
basic phases
of the
follower motion,
rise
and
return. They
can be
combined
in
different
ways, giving types
of
cams classifiable
in
terms
of the
type
of
follower
motion,
as in
Fig. 40.6.
For
positive drives,
the
symmetric acceleration curves
are to be
recommended.
For cam
systems with spring restraint,
it is
advisable
to use
unsymmetric curves
because they allow smaller springs. Acceleration curves
of
both
the
symmetric
and
unsymmetric
types
are
depicted
in
Fig. 40.7.
FIGURE 40.7
Acceleration
diagrams: (a),
(b)
spring loading;
(c),
(d)
positive
drive.
40.2
BASICCAMMOTIONS
Basic
cam
motions consist
of two
families:
the
trigonometric
and the
polynomial.
40.2.1
Trigonometric
Family
This
family
is of the
form
s
"=
C
0
+
C
1
sin
00
+
C
2
cos
bQ
(40.9)
where
C
0
,
C
1
,
a,
and b are
constants.
For the
low-speed systems where
\i
d
<
10"
4
,
we can
construct
all the
necessary dia-
grams,
symmetric
and
unsymmetric,
from
just
two
curves:
a
sine curve
and a
cosine
curve.
Assuming
that
the
total rise
or
return motion
S
0
occurs
for an
angular displace-
ment
of the cam 0 =
p
0
,
we can
partition acceleration curves into
i
separate segments,
where
/
=
1,2,3,
with subtended angles
P
1
,
p
2
,
P
3
,
so
that
P
1
+
P
2
+
P
3
+ - =
Po-
The sum of
partial
lifts
S
1
,
S
2
,
S
3
,
in the
separate segments should
be
equal
to the
total rise
or
return
S
0
:
^i
+
S
2
+
S
3
+
—
=
SQ.
If a
dimensionless description
0/p of cam
rotation
is
introduced into
a
segment,
we
will
have
the
value
of
ratio
0/p
equal
to
zero
at the
beginning
of
each segment
and
equal
to
unity
at the end of
each segment.
All the
separate segments
of the
acceleration curves
can be
described
by
equa-
tions
of the
kind
s"=Asin^
/1
=
^,1,2
(40.10)
P
2
or
s"=
A
cos^
(40.11)
where
A is the
maximum
or
minimum value
of the
acceleration
in the
individual
segment.
The
simplest case
is
when
we
have
a
positive drive with
a
symmetric acceleration
curve
(Fig.
40.7d).
The
complete rise motion
can be
described
by a set of
equations
/0 1
2710
\
„
27W
0
27C0
J=ffo
fe
-
&
sm
"T
J
5=
~F
sm
~T
(40.12)
,
S
0
(
-
2710
\
,„
4n
2
s
0
2710
^Tl
1
-
008
!-)
s
=
-p-
cos
T
The
last term
is
called geometric jerk
(s'
=
coY").
Traditionally, this motion
is
called
cycloidal
The
same equations
can be
used
for the
return motion
of the
follower.
It is
easy
to
prove that
^return
~
^O
~~
^rise
$
return
~
~$
rise
(40.13)
v'
—
-v'
?'"——?'"
1
^
return
3
rise
J
return
1
^
rise
FIGURE 40.8 Trigonometric standard
follower
motions (according
to the
equation
of
Table 40.1,
for c = d =
O).
All the
other acceleration curves, symmetric
and
unsymmetric,
can be
constructed
from
just
four
trigonometric standard
follower
motions. They
are
denoted
further
by
the
numbers
1
through
4
(Fig. 40.8).
These
are
displayed
in
Table 40.1.
Equations
in
Table 40.1
can be
used
to
represent
the
different
segments
of a
fol-
lower's displacement diagram. Derivatives
of
displacement diagrams
for the
adja-
cent segments should match each other; thus several requirements must
be met in
order
to
splice them together
to
form
the
motion specification
for a
complete cam.
Motions
1
through
4
have
the
following
applications:
Motion
1 is for the
initial part
of a
rise motion.
Motion
2 is for the end
and/or
the
middle part
of a
rise motion
and the
initial part
of
a
return motion.
The
value
c is a
constant, equal
to
zero only
in
application
to
the end
part
of a
rise
motion.
Motion
3 is for the end
part
of a
rise motion and/or
the
initial
or
middle part
of a
return motion.
The
value
d is a
constant, equal
to
zero only
in
application
to the
initial
part
of a
return motion.
Motion
4 is for the end
part
of a
return motion.
The
procedure
of
matching
the
adjacent
segments
is
best understood through
examples.
Example
1.
This
is an
extended version
of
Example
5-2
from
Shigley
and
Uicker
[40.8],
p.
229. Determine
the
motion specifications
of a
plate
cam
with
reciprocating
TABLE
40.1
Standard Trigonometric Follower Motions
Parameter
Motion
1
Motion
2
Motion
3
Motion
4
5
j,/V0
.
7r0\
.
vo
e
re
(e
i\
f e
i
*e\
T~
sm
T
S2
*
m
™"*
c
~n
53008
TF
+
^U"?
M
1
T-^
11
T
If
\P\
PlJ
2/3
2
P2
2^
3
\0
3
2/ V
0
4
TT
#4/
5'
Sj/.
7T^\
5
2
1T
*0
C
SjT
.
IfB
d
S
4
L
*6\
A
I
1
-""ft)
2fe
C
°
S
2ft
+
^
-2^
Wn
2ft
+
^
-id
1+C
°
S
£J
5*
tr5,
.
IT^
S
2
ir
2
.
*0
5
3
ir
2
TT^
TTS
4
.
ir0
W-
11
A'
~^
sm
%
-^'^
^r
sin
^
J^"
TT
2
^
1
TT^
J
2
*
3
TO
Si**
•
*0
T
2
S*
V0
1T
COS
^
-8^
C
°
S
%
8l
Sm
2^
7T
C
°
S
£
S
'
/1 -
O^
&•
- -
^
4
"Ur
}
2^
A
ft
^.
(L-t\
25,
£
_M
+
^
*"*U-
/
/J
1
ft
2ft
+
ft
**«.;**.
.»
-2*1
__a!
.»
_af
,"-2^
^
max
^j
S
min
-
^
*
™"
4j
g2
m
"
~
/Sj
FIGURE
40.9 Example
1: (a)
displacement diagram,
in; (b)
geometric velocity diagram, in/rad;
(c)
geomet-
ric
acceleration diagram,
in/rad
2
.
follower
and
return spring
for the
following
requirements:
The
speed
of the cam is
con-
stant
and
equal
to 150
r/min.
Motion
of the
follower consists
of six
segments (Fig.
40.9):
1.
Accelerated motion
to
s^
end
= 25
in/s (0.635 m/s)
2.
Motion with constant velocity
25
in/s, lasting
for
1.25
in
(0.03175
m) of
rise
3.
Decelerated
motion (segments
1 to 3
describe
rise of the
follower)
4.
Return motion
5.
Return motion
6.
Dwell, lasting
for t
>
0.085
s
The
total
lift
of the
follower
is 3 in
(0.0762
m).
Solution.
Angular velocity
CG
=
15071730
=
15.708 radians
per
second (rad/s).
The
cam
rotation
for
1.25
in of
rise
is
equal
to
p
2
=
1.25
mlS
2
=
1.25 in/1.592 in/rad
=
0.785
rad
-
45°, where
si =
25/15.708
-
1.592 in/rad.
The
following decisions
are
quite arbitrary
and
depend
on the
designer:
1. Use
motion
1;
then
S
1
= 0.5 in,
<
ax
-
0.057C/P
2
!
-
0.5jc/(0.628)
2
-
4
in/rad
2
(0.1016
m/rad
2
).
s"^
d
=
2(0.5)/pi;
so
P
1
-
1/1.592
=
0.628 rad,
or
36°.
2. For the
motion with constant velocity,
S
2
-1.592
in/rad (0.4044
m/rad);
S
2
=
1.25
in.
3.
Motion type
2:
S
3
=
S
2
=
1.25
in,
$3'^
=
s
3
7i/(2p
3
)
=
1.592 in/rad; therefore
p
3
=
1.257C/[2(1.592)]
-1.233
rad
=
71°,
^n
=
-(1.257i
2
)/[4(1.233)
2
]
= -2
in/rad
2
.
(Points
1
through
3
describe
the
rise motion
of the
follower.)
4.
Motion type
3:s4'
init
=s
4
7r
2
/(4p
2
)
= -2
in/rad
2
(the same value
as
that
of
s^),
s£
end
=
-7tt4/(2p
4
),
S
4
+
S
5
= 3 in.
5.
Motion type
4:
s
5
"
max
=
7W
5
/P
2
,
s
5
'i
nit
= -
s(
end
=
-2s
5
/fi
5
.
We
have here
the
four
unknowns
p
4
,
S
4
,
p
5
,
and
S
5
.
Assuming time
I
6
=
0.85
s for the
sixth segment
(a
dwell),
we can
find
(3
6
=
COf
6
=
15.708(0.08)
=
1.2566 rad,
or
72°.
Therefore
P
4
+
P
5
=
136°,
or
2.374
rad
(Fig. 40.9). Three other equations
are
S
4
+
S
5
=
3,s
4
n
2
/(4$)
= 2,
and
7cs
4
/(2p
4
)
=
2s
5
/p
5
.
From these
we can
derive
the
quadratic equation
in
p
4
.
0.696Pi
+
6.044p
4
-12
=
Q
Solving
it, we
find
p
4
=
1.665
848 rad
=
95.5°
and
p
5
=
40.5°. Since
S
4
Is
5
=
4p
4
/(7ip
5
)
=
3.000
76, it is
easy
to
find
that
S
5
=
0.75
in
(0.019
05 m) and
S
4
=
2.25
in
(0.057
15
m).
Maximum geometric acceleration
for the
fifth
segment
s
5
'
max
=
4.7
in/rad
2
(0.0254
m/rad
2
),
and the
border (matching) geometric velocity
s
4>end
=
s^
=
2.12
in/rad
(0.253 m/rad).
Example
2. Now let us
consider
a cam
mechanism with spring loading
of the
type
D-R-D-R (Fig.
40.70).
The
rise part
of the
follower motion might
be
constructed
of
three segments
(1,2,
and 3)
described
by
standard follower motions
1,2,
and 3
(Fig.
40.8).
The
values
of
constants
c and d in
Table 40.1
are no
longer zero
and
should
be
found
from
the
boundary conditions. (They
are
zero only
in the
motion case R-R-D,
shown
in
Fig.
40.Jb,
where there
is no
dwell between
the
rise
and
return motions.)
For a
given motion specification
for the
rise motion,
the
total follower stroke
S
0
,
and the
total angular displacement
of the cam
p
0
,
we
have eight unknowns:
P
1
,
Si,
P
2
,
$2,
Ps>
S
3
,
and
constants
c and d. The
requirements
of
matching
the
displacement
derivatives
will
give
us
only
six
equations; thus
two
more must
be
added
to get a
unique
solution.
Two
additional equations
can be
written
on the
basis
of two
arbi-
trary
decisions:
1. The
maximum value
of the
acceleration
in
segment
l,s"
tmsa
should
be
greater than
that
in
segment
2
because
of
spring loading.
So
s"
max
=
-as"
min
where
s^'mm
is
the
minimum
value
of the
second-segment acceleration
and a is any
assumed num-
ber, usually greater than
2.
2. The end
part
of the
rise (segment
3), the
purpose
of
which
is to
avoid
a
sudden
drop
in a
negative accelerative curve, should have
a
smaller duration than
the
basic
negative part (segment
2).
Therefore
we can
assume
any
number
b
(greater
than
5) and
write
p
2
=
&p
3
.The
following
formulas were
found
after
all
eight equa-
tions
for the
eight unknowns were solved simultaneously:
R
Po Q
_«
a
^~l
+
a +
alb
^'
1
S(I
+
*)
+
*
TC^
,_
4a
Sl
~
S
°
b
2
(n
+
4a)
+
4a(2a
+
l)
"
2
~
Sl
K
4a
Sa
2
53
"
"
1
TtZ)
2
C
~
Sl
nb
2
d =
2s
3
SQ
=
Si+
52
+c
+
S
3
We
can
assume practical values
for a and b
(say
a =
2,
b = 10) and
find
from
the
above
equations
the set of all the
parameters
(as
functions
of
S
0
and
p
0
)
necessary
to
form
the
motion specification
for the
rise motion
of the
follower
and the
shape
of the cam
profile.
The
whole
set of
parameters
is as
follows:
5j
=
0.272
198so
Pi
=
0.312
5p
0
s'
2
=
0.693
147*0
c =
0.027
726s
0
P
2
=
0.625p
0
*3
=
0.006931*o
d
=
2s
3
(always!)
p
3
=
0.0625po
These
can be
used
for
calculations
of the
table
s =
5(6),
which
is
necessary
for
manu-
facturing
a cam
profile.
For
such
a
table,
we use as a
rule increments
of 0
equal
to
about
1°
and
accuracy
of s up to 4 x
10~
5
in 1
micrometer
(um).
The
data
of
such
a
table
can be
easily used
for the
description
of
both
the
return motion
of the
follower
and a cam
profile,
providing
p
0
(return)
=
J3
0
(rise),
and the
acceleration diagram
for
the
return motion
is a
mirror image
of the
acceleration diagram
for the
rise motion.
Table 40.2
can be of
assistance
in
calculating
the
return portion
of the cam
profile.
The
column
s(return)
is the
same
as the
column
s(rise).
TABLE
40.2
Data
of
Rise
Motion
Used
for
Calculations
of
Return
Portion
of Cam
Profile
Rise Return
0(rise)
s(rise)
^(return)
^return)
O
O
20Q
+
&/-0
O
Bi
s,
2A>
+
&/-0/
J/
0o
^
0
20o
+
0</
— 0o
S
0
The
trigonometric acceleration diagram
for the
positive drive
was
described
at
the
beginning
of
this section
by Eq.
(40.12).
The
improved diagram (smaller maxi-
mum
values
of
acceleration
for the
same values
of
S
0
and
p
0
f
)
can be
obtained
if we
combine sine segments with segments
of
constant acceleration. Such
a
diagram,
called
a
modified
trapezoidal
acceleration
curve,
is
shown
in
Fig. 40.10. Segments
1,
3,
4, and 6 are the
sinusoidal type. Sections
2 and 5 are
with
s" =
constant.
It was
assumed
for
that diagram that
all the
sine segments take one-eighth
of the
total
angular displacement
p
0
of the cam
during
its
rise
motion.
The
first
half
of the
motion
has
three
segments.
The
equations
for the
first
segment
are O
<
0/p
0
<
1
^,
and so
f
The
maximum
acceleration
ratio
is
4.9/6.28.
FIGURE 40.10
A
modified
trapezoidal acceleration diagram.
S
0
-
/47C0
.
4710
\
,
2s'
Q
/
47c9\
S
= —
-;r~
-
Sin
—:—
S
=-r—
1-COS
——
271
V Po Po
/
Po V Po
/
(40.14)
„
0
SQ .
47T0
„,
^
4rc9
5
=871-rj
sm-r—
s
=
327T
-TJ
cos
——
Po
Po Po Po
For the
second segment,
we
have
1
A
<
0/p
0
<
%,
and so
,[
1
29
/9
IVl
5:=5o
~^;
+
ir
+47C
U~~^
L
2n
Po VPo
8/J
/
=
^[
2
+
8icg i)l
(40.15)
Po
L
\Po
o
/J
„
8ns'
0
,„
s=
~w
s
=0
The
relations
for the
third segment
are
3
A
<
9/P
0
<
%
([40.7]);
therefore,
J
TC
,9
1 .
I"
/0
2\11
J
=
^o
+
2(1+7C)
—
sin
UTC
—
[2
Po
27C
L VPo
8
/JJ
'-f{—Nt-I)Il
(40.16)
„
87K
0
'
.
r
/e
2\]
s
=
ir
sm
rte-8JJ
^"=-i
cos
h(M)]
where
J
0
'
=
s<J(2
+
n)
=
0.194
492^
0
.
Using
Eqs. (40.14) through (40.16)
for all
three segments,
we can
calculate
the s
values
for the
first
half
of the
rise motion, where
6/p
0
-
%
and s =
s
0
/2.
Since
the
neg-
ative
part
of the
acceleration diagram
is a
mirror image
of the
positive part,
it is
easy
to
calculate
the s
values
for the
second
half
of the
rise motion
from
the
data obtained
for
the
first
half.
The
necessary procedure
for
that
is
shown
in
Table 40.3.
The
proce-
dure concerns
the
case with
the
modified trapezoidal acceleration diagram,
but it
could
be
used
as
well
for all the
cases with symmetric acceleration diagrams
for the
rise motion.
For the
return motion
of the
follower, when
its
acceleration diagram
is
a
mirror image
of the
rise diagram,
we can use
again
the
technique shown
in
Table
40.2.
All the
calculations
can be
done simultaneously
by the
computer
after
a
simple
program
is
written.
TABLE
40.3
Data
of
First
Half
of
Rise
Motion
Used
for
Calculations
of
Second
Half
7
-
W
0
s y s
0
O 1
S
0
1
T/
Sj
1 -
7/
SQ
-
Si
6
1
S
1
=
s'
0
(*/2
-
l)/27r
i
S
0
-S
1
t
*
f
'
0
T'
1
2
]>
sj
!
T
Tj
S
°~
Sj
5
1
S
2
-
S
0
(^
-'
1/2»
+
x/4)
J
5
°~
52
i
S
2
I
S
0
-S
2
y
Sk 1 -
7*
So-Sk
4
k
'
:
'
:
:
:
S
3
=
S
0
/2
I
S
0
-
S
3
=
S
0
/2
I
All the
trigonometric curves
of
this section were calculated with finite values
of
jerk,
which
is of
great importance
for the
dynamic behavior
of the cam
mechanism.
An
example
of the
jerk diagram
is
given
in
Fig.
40.10.
The
jerk curve
;
was
plotted
by
using
the
dimensionless expression
/=
^
(4ai7)
This
form
of the
jerk description
can
also
be
used
to
compare properties
of
different
acceleration diagrams.
Segments
40.2.2
Polynomial Family
The
basic polynomial equation
is
Q
/
Q
\2
/ 0
\
3
5-C
0
+
C
1
-
+C
2
-
+C
3
-
+».
(40.18)
Po
\Po/
\Po/
with
constants
C/
depending
on
assumed initial
and
final
conditions.
This
family
is
especially
useful
in the
design
of
flexible
cam
systems, where values
of
the
dynamic
factor
are
U^
>
10~
2
.
Dudley (1947)
first
used polynomials
for the
syn-
thesis
of
flexible
systems,
and his
ideal later
was
improved
by
Stoddart [40.9]
in
application
to
automotive
cam
gears.
The
shape
factor
s of the cam
profile
can be
found
by
this method
after
a
priori
decisions
are
made about
the
motion
y of the
last link
in the
kinematic chain. Cams
of
that kind
are
called
poly
dyne
cams.
When
flexibility
of the
system
can be
neglected,
the
initial
and
final
conditions
([40.3],
[40.4],
and
[40.8])
might
be as
follows
(positive drive):
1.
Initial conditions
for
full-rise
motion
are
-|-
= 0 s =
0
s'
=
0
s"
=
0
Po
2.
Final (end) conditions
are
-|-
= 1 s =
s
0
s'
= 0
s"
=
Q
Po
The
first
and
second derivatives
of Eq.
(40.18)
are
/
=
C
1
+
2C
2
|
+
3C
3
(|)
2
+
4C
4
(|)
3
+
Po
\
Po
/
\
Po
/
(40.19)
9 / 9 V
^
=
2C
2
+
6C
3
-T-+
12C
4
-
+•••
Po
\Po/
Substituting
six
initial
and
final
conditions into Eqs.
(40.18)
and
(40.19)
and
solving
them simultaneously
for
unknowns
C
0
,
Ci,
C
2
,
C
3
,
C
4
,
and
C
5
,
we
have
I79\
3
/9V
/9
Vl
"My-
L
4r
a6
(id]
'-"SKiMiHi)I
'-S[HiMi)I
and for a
jerk
s'"
=
ds"ld§,
or
—SHHifl
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